Split (2)

train · ~239k rows (showing the first 191k)

id stringlengths 1 6	url stringlengths 16 1.82k	content stringlengths 37 9.64M
13300	https://www.dcode.fr/chinese-remainder	Chinese Remainder Tool to compute congruences with the chinese remainder theorem. The Chinese Remainder Theorem helps to solve congruence equation systems in modular arithmetic. Results Summary Chinese Remainder Calculator \| \| \| --- \| \| Display \| The smallest (positive) solution All solution in general form (if possible) \| Answers to Questions (FAQ) What is the Chinese Remainder Theorem? (Definition) The Chinese remainder theorem is the name given to a system of congruences (multiple simultaneous modular equations). The original problem is to calculate a number of elements which remainders (of their Euclidean division) are known. Example: If they are arranged by 3 there remains 2. If they are arranged by 5, there remain 3 and if they are arranged by 7, there remain 2. How many objects are there? This exercise implies to calculate $ x $ such that $ x \equiv 2 \mod 3 $ and $ x \equiv 3 \mod 5 $ and $ x \equiv 2 \mod 7 $ Take a list of $ k $ coprimes integers $ n_1, ..., n_k $ and their product $ n = \prod_{i=1}^k n_i $. For all integers $ a_1, ... , a_k $, it exists another integer $ x $ which is unique modulo $ n $, such as: $$ \begin{array}{c} x \equiv a_1\pmod{n_1} \ \ldots \ x \equiv a_k\pmod{n_k} \end{array} $$ How to calculate Chinese remainder? To find a solution of the congruence system, take the numbers $ \hat{n}_i = \frac n{n_i} = n_1 \ldots n_{i-1}n_{i+1}\ldots n_k $ which are also coprimes. To find the modular inverses, use the Bezout theorem to find integers $ u_i $ and $ v_i $ such as $ u_i n_i + v_i \hat{n}_i = 1 $. Here, $ v_i $ is the modular inverse of $ \hat{n}_i $ modulo $ n_i $. Take then the numbers $ e_i = v_i \hat{n}_i \equiv 1 \mod{n_i} $. A particular solution of the Chinese remainders theorem is $$ x = \sum_{i=1}^k a_i e_i $$ dCode accepts numbers as pairs (remainder A, modulo B) in equations of the form x = A mod B Example: $ (2,3),(3,5),(2,7) \iff \left{ \begin{array}{ll} x = 2 \mod 3 \ x = 3 \mod 5 \ x = 2 \mod 7 \end{array} \right. \Rightarrow x = 23 $ When does the Chinese Remainder Theorem have no solution? The system of equations with remainders $ r_i $ and modulos $ m_i $ has solutions only if the following modular equation is true: $$ r_1 \mod d = r_2 \mod d = \cdots r_n \mod d $$ with $ d $ the GCD of all modulos $ m_i $. Summary Similar pages Share Support Forum/Help Keywords Source code dCode retains ownership of the "Chinese Remainder" source code. Any algorithm for the "Chinese Remainder" algorithm, applet or snippet or script (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, breaker, translator), or any "Chinese Remainder" functions (calculate, convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (Python, Java, PHP, C#, Javascript, Matlab, etc.) or any database download or API access for "Chinese Remainder" or any other element are not public (except explicit open source licence). Same with the download for offline use on PC, mobile, tablet, iPhone or Android app. Reminder: dCode is an educational and teaching resource, accessible online for free and for everyone. Cite dCode The content of the page "Chinese Remainder" and its results may be freely copied and reused, including for commercial purposes, provided that dCode.fr is cited as the source (Creative Commons CC-BY free distribution license). Exporting the results is free and can be done simply by clicking on the export icons ⤓ (.csv or .txt format) or ⧉ (copy and paste). To cite dCode.fr on another website, use the link: In a scientific article or book, the recommended bibliographic citation is: Chinese Remainder on dCode.fr [online website], retrieved on 2025-09-29, Need Help ? Please, check our dCode Discord community for help requests! NB: for encrypted messages, test our automatic cipher identifier! Questions / Comments Feedback and suggestions are welcome so that dCode offers the best 'Chinese Remainder' tool for free! Thank you! dCode and more dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day! A suggestion ? a feedback ? a bug ? an idea ? Write to dCode! Please, check our dCode Discord community for help requests! NB: for encrypted messages, test our automatic cipher identifier! Feedback and suggestions are welcome so that dCode offers the best 'Chinese Remainder' tool for free! Thank you!
13301	https://www.sciencedirect.com/topics/mathematics/congruence-modulo	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. The Theory of Divisibility 6.1. Congruences Modulo Divisors We consider a ring with quotient field K for which there exists a theory of divisors . Definition. We say that the elements α and β of the ring are congruent modulo the divisor ∈ , and write In the case of a principal divisor (μ) the congruence α ≡ β (mod (μ)) is clearly equivalent to the congruence α ≡ β (mod μ) in the sense of the definition of Section 4.1 of the Supplement. We indicate some elementary properties of congruences which easily follow from the definition. (1) : Congruences modulo can be added and multiplied termwise. (2) : If a congruence holds modulo , then it also holds modulo for any divisor divisible by . (3) : If a congruence holds modulo and modulo , then it also holds modulo their least common multiple. (4) : If an element α ∈ is relatively prime to [that is, if the divisors (α) and are relatively prime], then from the congruence αβ ≡ 0 (mod ) it follows that β ≡ 0 (mod ). (5) : If α divides both sides of a congruence modulo and α is relatively prime to , then we may cancel α from the congruence. (6) : If is a prime divisor and αβ ≡ 0 (mod ) then either α ≡ 0 (mod ) or β ≡ 0 (mod ). It follows from property (1) that the residue classes of the ring modulo a given divisor can be added and multiplied. It is easily verified that under these operations the set of residue classes becomes a ring. It is called the ring of residue classes modulo the divisor a and is denoted by / . Property (6) can then be interpreted as saying that for a prime divisor the ring / has no divisors of zero. Assume now that is the maximal order of an algebraic number field K. The divisors of the ring we call in this case the divisors of the field K. Since every divisor of the field K divides some nonzero number α ∈ , and the number α in its turn divides some natural number a [for example, \|N(α)\| is divisible by α], then for each divisor there is a natural number a which is divisible by . By property (2) numbers in distinct residue classes modulo remain in distinct classes modulo a. Recalling now that in the order the number of residue classes modulo a is finite (actually equal to an, where n is the degree of the field K; see the proof of Theorem 5 of Section 2, Chapter 2), we obtain the following theorem. Theorem 1. For any divisor of the algebraic number field K, the residue class ring / is finite. Let be any prime divisor of the field K. The corresponding valuation induces on R the p-adic valuation for some prime p. Since (p) = 1, then (p) > 0; that is, p ≡ (mod ). If the prime number q is different from p, then (q) = 0, and therefore (q) = 0; that is, q ≢ 0 (mod p). The residue class ring / , being finite and without divisors of zero, is a finite field (Supplement, Section 3). Since for any α ∈ we have pα ≡ 0 (mod ), then the characteristic of this field is p. Hence we have Theorem 2. Any prime divisor of an algebraic number field divides one and only one rational prime p. The residue class ring / is a finite field of characteristic p. A theory of divisors for an algebraic number field hence has the property that the residue class ring modulo a prime divisor is a field. In general, this is not the case. For example, in the ring of polynomials k[x, y] in two variables over a field k the residue class ring of the prime divisor (x) is isomorphic to the ring of polynomials k[y] and hence is not a field. The residue class ring / is a field if and only if the congruence αξ ≡ 1 (mod ) is always solvable when α ≢ 0 (mod ). Hence only under this assumption can we expect to construct a completely adequate theory of congruences in the ring . View chapterExplore book Read full chapter URL: Chapter The Theory of Divisibility 1966, Pure and Applied Mathematics 6. Dedekind Rings 6.1. Congruences Modulo Divisors We consider a ring with quotient field K for which there exists a theory of divisors . Definition. We say that the elements α and β of the ring are congruent modulo the divisor ∈ , and write In the case of a principal divisor (μ) the congruence α ≡ β (mod (μ)) is clearly equivalent to the congruence α ≡ β (mod μ) in the sense of the definition of Section 4.1 of the Supplement. We indicate some elementary properties of congruences which easily follow from the definition. (1) : Congruences modulo can be added and multiplied termwise. (2) : If a congruence holds modulo , then it also holds modulo for any divisor divisible by . (3) : If a congruence holds modulo and modulo , then it also holds modulo their least common multiple. (4) : If an element α ∈ is relatively prime to [that is, if the divisors (α) and are relatively prime], then from the congruence αβ ≡ 0 (mod ) it follows that β ≡ 0 (mod ). (5) : If α divides both sides of a congruence modulo and α is relatively prime to , then we may cancel α from the congruence. (6) : If is a prime divisor and αβ ≡ 0 (mod ) then either α ≡ 0 (mod ) or β ≡ 0 (mod ). It follows from property (1) that the residue classes of the ring modulo a given divisor can be added and multiplied. It is easily verified that under these operations the set of residue classes becomes a ring. It is called the ring of residue classes modulo the divisor a and is denoted by / . Property (6) can then be interpreted as saying that for a prime divisor the ring / has no divisors of zero. Assume now that is the maximal order of an algebraic number field K. The divisors of the ring we call in this case the divisors of the field K. Since every divisor of the field K divides some nonzero number α ∈ , and the number α in its turn divides some natural number a [for example, \|N(α)\| is divisible by α], then for each divisor there is a natural number a which is divisible by . By property (2) numbers in distinct residue classes modulo remain in distinct classes modulo a. Recalling now that in the order the number of residue classes modulo a is finite (actually equal to an, where n is the degree of the field K; see the proof of Theorem 5 of Section 2, Chapter 2), we obtain the following theorem. Theorem 1. For any divisor of the algebraic number field K, the residue class ring / is finite. Let be any prime divisor of the field K. The corresponding valuation induces on R the p-adic valuation for some prime p. Since (p) = 1, then (p) > 0; that is, p ≡ (mod ). If the prime number q is different from p, then (q) = 0, and therefore (q) = 0; that is, q ≢ 0 (mod p). The residue class ring / , being finite and without divisors of zero, is a finite field (Supplement, Section 3). Since for any α ∈ we have pα ≡ 0 (mod ), then the characteristic of this field is p. Hence we have Theorem 2. Any prime divisor of an algebraic number field divides one and only one rational prime p. The residue class ring / is a finite field of characteristic p. A theory of divisors for an algebraic number field hence has the property that the residue class ring modulo a prime divisor is a field. In general, this is not the case. For example, in the ring of polynomials k[x, y] in two variables over a field k the residue class ring of the prime divisor (x) is isomorphic to the ring of polynomials k[y] and hence is not a field. The residue class ring / is a field if and only if the congruence αξ ≡ 1 (mod ) is always solvable when α ≢ 0 (mod ). Hence only under this assumption can we expect to construct a completely adequate theory of congruences in the ring . 6.2. Congruences in Dedekind Rings Definition. A ring is called a Dedekind ring if it has a theory of divisors → and for every prime divisor ∈ the residue class ring / is a field. Examples of Dedekind rings, other than the maximal orders of algebraic number fields, can be obtained by taking the integral closure of the polynomial ring k[x] in a single variable in a finite extension of the field of rational functions f(x) (Problems 1 and 2). The valuation ring v of any valuation v is also a Dedekind ring (see Section 4.1), as is any ring which has a theory of divisors with only a finite number of prime divisors (Problem 3). Lemma 1. If is a Dedekind ring and α ∈ is not divisible by the prime divisor , then the congruence αξ ≡ 1 (mod m) is solvable in for any natural number m. Proof. For m = 1 the congruence is solvable by the definition of a Dedekind ring. The lemma will be proved by induction on m. Suppose that for some ξo ∈ we have αξ0 ≡ 1 (mod m). Choose an element ω in the ring for which (ω) = m. The principal divisor (ω) has the form (ω) = m , where is not divisible by . Choose an element γ ∈ for which (γ) = 0 and γ ≡ 0 (mod ). The product γ(αξ0 −1) will be divisible by m = (ω), and hence γ(αξ0 −1) = ωμ with μ ∈ . We now try to solve the congruence αxi; ≡ 1 (mod m + 1), taking as ξ an element in the form ξ = ξ0 + ωλ, where λ is to be chosen suitably in . Since and ω ≡ 0 (mod m), then we shall achieve our goal if λ satisfies the congruence λαγ ≡ – μ (mod ). But since αγ is not divisible by , this congruence is solvable. Hence there is an element ξ ∈ for which γ(αξ −1) ≡ 0 (mod m + 1) and since (γ) = 0, dividing by γ, we obtain αξ −1 ≡ 0 (mod m + 1). Lemma 1 is proved. Theorem 3. If 1,…, m, are distinct prime divisors of the Dedekind ring , and β1,…, βm are any elements of , then there is an element ξ in which satisfies (k1,…, km are any natural numbers). Proof. For each divisor we can find an element αi ∈ which is divisible by i but not divisible by i. Lemma 1 guarantees that we can solve the congruence αiξi ≡ βi (mod ) in ξi ∈ . It is easily seen that the element satisfies the requirements of the theorem. Theorem 4. If α ≠ 0 and β are elements of a Dedekind ring , then the congruence (6.1) is solvable if and only if β is divisible by the greatest common divisor of the divisors (α) and . Proof. We first assume that the divisors (α) and are relatively prime, and will show that in this case the congruence (6.1) is solvable for any β. Let , where 1,…, m are distinct prime divisors. By Lemma 1 for each i = 1,…, m in the ring there is an element ξi′ such that at αξi′ ≡ β (mod iki). By Theorem 3 we can find for each i an element ξi for which ξi ≡ ξi′ (mod ) and ξi ≡ 0 (mod i). It is now clear that the sum ξl + · · · + ξm = ξ will satisfy the congruences αξ ≡ β (mod ) for i = 1,…, m, and hence will also satisfy (6.1). We now prove the theorem in general. Let be the greatest common divisor of the divisors (α) and . If (6.1) holds modulo , then it also holds modulo δ, and since α ≡ 0 (mod δ), then we must also have β = 0 (mod δ). This proves the necessity of this condition. Assume now that β is divisible by δ. By Theorem 3 of Section 4, there is an element μ ∈ K for which (6.2) We shall show that we can choose μ so that also (6.3) for all prime divisors , distinct from 1,…, m. Suppose that μ does not satisfy condition (6.3), and let, 1,…, s be the prime divisors, different from 1,…, m, for which . Choose in an element γ such that and . It is clear that the element μ′ = μγ satisfies both conditions (6.2) and (6.3), and our assertion is proved. Let the divisor be determined by = d . If μ satisfies conditions (6.2) and (6.3), then the element αμ belongs to and is relatively prime to . Since we have assumed β divisible by d, then βμ also belongs to . We have already proved that then there exists an element ξ in the ring such that αμξ ≡ βμ (mod ). For i = 1,…, m we have and this means that ξ satified (6.1). 6.3. Divisors and Ideals In this section we show that there is a one-to-one correspondence between divisors and nonzero ideals in a Dedekind ring. For each divisor denote by the set of all elements of the ring which are divisible by . It is clear that is a nonzero ideal of the ring . Theorem 5. In a Dedekind ring the mapping → ( ∈ ) is an isomorphism of the semigroup of divisors onto the semigroup of all nonzero ideals of the ring . We first verify the following lemma. Lemma 2. If α1,…, αs are any nonzero elements of the Dedekind ring and δ is the greatest common divisor of the principal divisors (α1),…, (αs), then any element α ∈ which is divisible by δ can be written in the form The proof of the lemma is by induction on s. For s = 1 the lemma is obvious. Let s ≥. Let δ1 denote the greatest common divisor of the divisors (α1),…, (alpha;s−1). Then δ is the greatest common divisor of the divisors δ1 and (αs). Let α be divisible by δ. By Theorem 4 the congruence αsξ ≡ α (mod δ1) has a solution ξ ∈ . By the induction hypothesis there are elements ξs−1 in the ring such that . Lemma 2 is proved. Proof of Theorem 5. By condition (3) of the definition of a theory of divisors the mapping takes distinct divisors to distinct ideals. Let A be any nonzero ideal of the ring . For each prime divisor set It is clear that a( ) will be nonzero for only a finite number of prime divisors . Hence the product , in which runs through all prime divisors for which a( ) ≠ 0, is a divisor. We shall show that . Let α be any element of . It is clear that we can find a finite set of elements αs in A such that a( ) = min ( (α1),…, (αs)) for all . This means that the divisor is the greatest common divisor of the principal divisors (α1),…, (αs). By Lemma 2 the element α can be represented in the form α ξ1α1 + · · · + ξsαs with coefficients ξi ∈ . It follows that α ∈ A, and hence that . Since it is clear that A ⊂ , we obtain A = . We have thus proved that is a one-to-one mapping of the set of all divisors of the ring onto the set of all nonzero ideals of . We shall now show that this mapping is an isomorphism, that is, that for any two divisors and we have (6.4) Denote the product by C. Since C is a nonzero ideal of , there is a divisor such that . We must prove that = . Let the prime divisor enter into the divisors and with exponents a and b. Then Since this is true for all prime divisors , then = and (6.4) is proved. From the fact that the mapping → is an isomorphism, it follows, in particular, that the set of all nonzero ideals of the Dedekind ring form a semigroup with unique factorization under the operation of multiplication. To construct a theory of divisors in Dedekind rings (in particular, in the maximal order of an algebraic number field), we could take the semigroup of nonzero ideals for the semigroup . The image of the element α under the homomorphism → would then be the principal ideal (α) generated by this element. This construction of a theory of divisors is due to Dedekind. 6.4. Fractional Divisors If we construct a theory of divisors → for the ring , then we obtain some information on the structure of the semigroup . It is natural to try an analogous procedure with the multiplicative group K of the quotient field K. To do this we need to extend the concept of a divisor. Following an established tradition, we shall reserve the term “divisor” for this broader concept, and will call divisors in the earlier sense “integral divisors.” Definition. Let be a ring with a theory of divisors, with quotient field K, and let 1,…, m be a finite system of prime divisors. An expression (6.5) with integer exponents k1,…, km (not necessarily positive) is called a divisor of the field K. If all the exponents ki are nonnegative, then the divisor is called integral (or a divisor of the ring ). Otherwise it is called fractional. It is sometimes convenient to write a divisor (6.5) as a formal infinite product (6.6) over all prime divisors , in which almost all exponents a( ) are zero. Multiplication of divisors is determined by the formula For integral divisors this definition coincides with the definition of multiplication in the semigroup . It is easily seen that under this operation the set of all divisors of the field K is an Abelian group, which we shall denote by . The unit element of this group is the divisor for which all exponents a( ) in (6.6) are zero. Since every nonzero element ξ ∈ K is the quotient of two elements of , it follows from condition (1) of Theorem 4 of Section 3 that for all but a finite number of the valuations , which correspond to the prime divisors , we have . We denote this finite set by . The divisor is called the principal divisor corresponding to the element ξ ∈ K, and is denoted by (ξ). When applied to elements of the ring , the new concept of a principal divisor coincides with the previous one (Section 3.4). By condition (2) of Theorem 4 of Section 3 the principal divisor (ξ) will be integral if and only if ξ belongs to . From the definition of a valuation (Section 3.4) it easily follows that the mapping ξ → (ξ), ξ ∈ K, is a homomorphism of the multiplicative group of the field K to the group of divisors . By Theorem 2 of Section 3 this homomorphism maps onto the entire group (that is, is an epimorphism) if and only if has unique factorization. The kernel of this map clearly is the group of units of the ring , and this means that for elements ξ, η of K we have (ξ) = (η) if and only if ξ = ηε, where ε is a unit of the ring . We now define a concept of divisibility for arbitrary divisors. Let and be two divisors (not necessarily integral). We say that is divisible by if there is an integral divisor such that = . In other words, is divisible by if and only if a( ) ≥ b( ) for all . For any and set d( ) = min (a( ), b( )). Since the rational integer d( ) is equal to zero for almost all , then the product is a divisor. The divisor is called the greatest common divisor of the divisors and ( and are both divisible by and is divisible by every common divisor of and ). The least common multiple of the divisors and is defined analogously. The element α ∈ K is called divisible by the divisor , if α = 0 or the principal divisor (α) is divisible by . In terms of valuations this is characterized by for all . The correspondence of the preceding section between integral divisors and ideals of a Dedekind ring can be extended to fractional divisors, providing the proper generalization of the concept of ideal is used. As in Section 6.3, we denote the set of all elements of the field K which are divisible by the divisor by (these elements may now be nonintegral). From condition (3) in the definition of a valuation (Section 3.4) it follows that if α and β are divisible by , then α ± β are also divisible by . This means that the set is a group under addition. Further, if α ∈ and ξ ∈ , then the product ξα also belongs to . We now verify the following formula: (6.7) For the element ξ is divisible by (γ) if and only if any of the following hold: for all ; vp(ξ/γ) ≥ a( ) for all ; ξ/γ ∈ ; ξ ∈ γ [here a( ) denotes the power to which appears in the divisor ]. It is clear that for any divisor we can find an element γ ∈ such that the divisor (γ) is integral. Formula (6.7) shows that for such a γ we will have . Definition. Let be a Dedekind ring with quotient field K. A subset A ⊂ K, containing at least one nonzero element, is called an ideal of the field K (relative to ), if it satisfies: (1) : A is a group under the operation of addition. (2) : For any α ∈ A and any ξ ∈ , the product ξα lies in A. (3) : There is a nonzero element γ of the field K such that γA ⊆ . The ideal A is called integral if it is contained in and otherwise is called fractional. An integral ideal in K is clearly just a nonzero ideal in . If A and B are two ideals of the field K, then by their product AB we mean the set of all elements γ ∈ K which can be represented in the form It is clear that the product of two ideals of a field K is again an ideal of the field K. (When the ideals are integral, the definition of product coincides with the usual notion of the product of two ideals in a commutative ring.) We have already verified that for any divisor of K, the set is an ideal in K. Assume that for two divisors and we have . Choose a nonzero element γ so that the divisors (γ) and (γ) are both integral. From formula (6.7) we have , so that (γ) = (γ) and hence = . Hence the mapping is a monomorphism. Now let A be any ideal of the field K. If the element γ ≠ 0 is chosen so that γA ⊂ , then γA will be a nonzero ideal of the ring , and hence by Theorem 5 there exists an integral divisor such that . Set = (γ)−1. Then , so that A = . Thus each ideal of the field K is the image of some divisor under the mapping → . If and are two divisors, then, taking elements γ ≠ 0 and γ′ ≠ 0 so that (γ) and (γ′) are integral divisors, we have [from Theorem 5 and formula (6.7)] so that . The mapping → hence is an isomorphism. It follows that the set of all ideals of the field K is a group under multiplication. The unit element in this group is the ring = . The inverse of the ideal will be the ideal . We formulate this generalization of Theorem 5. Theorem 6. Let be a Dedekind ring with quotient field K. For every divisor , denote by the set of all elements of K which are divisible by . The mapping → is an isomorphism of the group of all divisors of the field K onto the group of all ideals of the field K. This mapping takes integral divisors to integral ideals and conversely. View chapterExplore book Read full chapter URL: Book series1966, Pure and Applied Mathematics Chapter The Theory of Divisibility 1966, Pure and Applied Mathematics Remark 1. The main part of the proof of the theorem is the verification of the impossibility of certain congruences modulo l. Of course it does not follow from this that the congruence xl + yl ≡ zl (mod l) is impossible, since this congruence is equivalent to x + y ≡ z (mod l), which always has solutions in integers not divisible by l. Moreover, it can be shown that, for example, when l = 7, the equation xl + yl = zl, when considered as a congruence, has, for any modulus, solutions not divisible by 7. Thus the proof of the unsolvability of (1.1) is achieved first by using unique factorization in the ring Z[ζ] to obtain Equations (1.3) and (1.3′), and then by applying the theory of congruences to these latter equations. View chapterExplore book Read full chapter URL: Book series1966, Pure and Applied Mathematics Chapter Congruences 1966, Pure and Applied Mathematics 3. p-Adic Numbers 3.1. p-Adic Integers We now turn to congruences modulo a power of a prime. We start with an example. Consider the congruence modulo a power of the prime 7. For n = 1 the congruence has two solutions, (3.1) Now set n = 2. From (3.2) it follows that x2 ≡ 2 (mod 7), and hence any solution of (3.2) must be of the form x0 + 7t1, where x0 is a number satisfying the congruence (3.1). We now look for a solution in the form x1 = 3 + 7t1. (Solutions of the type −3 + 7t1 are found in precisely the same way.) Substituting this expression for x1 in (3.2), we obtain We thus have the solution x1 ≡ 3 + 7·1 (mod 72). Similarly, when n = 3 we have x2 = x1 + 72t2 and from the congruence we find that t2 ≡ 2 (mod 7); that is, It is easily seen that this process can be continued indefinitely. We obtain a sequence (3.3) satisfying the conditions The construction of the sequence (3.3) is reminiscent of the process for finding the square root of 2. Indeed, the computation of consists of finding a sequence of rational numbers r0, r1,…, rn,…, the squares of which converge to 2, for example: In our case we construct a sequence of integers x0, x1,…, xn, …, for which xn2 −2 is divisible by 7n+1. This analogy becomes more precise if we say that two integers are close (more precisely, p-close, where p is some prime), when their difference is divisible by a sufficiently large power of p. With this concept of closeness we can say that the squares of the numbers in the sequence (3.3) become arbitrarily 7-close to 2 as n increases. By giving the sequence {rn} we determine the real number . One might suppose that the sequence (3) also determines a number α, of a different type, such that α2 = 2. We now note the following fact. If the sequence {rn'} of rational numbers satisfies \|rn – rn'\| < 1/10n for all n, then its limit is also . One would naturally assume that a sequence {xn'}, for which xn ≡ xn' (mod 7n+1), would determine the same new number α [the new sequence {xn'} clearly, also satisfies xn'2 ≡ 2 (mod 7n+1) and xn' ≡ x′n−1 (mod 7n)]. These remarks lead to the following definition. Definition. Let p be some prime number. A sequence of integers satisfying (3.4) for all n ≥ 1, determines an object called a p-adic integer. Two sequences {xn} and {xn'} determine the same p-adic integer if and only if for all n ≥ 0. If the sequence {xn} determines the p-adic integer α, we shall write The set of all p-adic integers will be denoted by Op. To distinguish them from p-adic integers, ordinary integers will be called rational integers. Each rational integer x is associated with a p-adic integer, determined by the sequence {x, x,…, x,… }. The p-adic integer corresponding to the rational integer x will also be denoted by x. Two distinct rational integers x and y correspond to distinct p-adic integers. Indeed, if they are equal as p-adic integers, then x ≡ y (mod pn) for all n, which is possible only if x = y. Hence we may assume that the set Z of all rational integers is a subset of the set Op of all p-adic integers. To clarify the nature of the set Op, we shall describe a method for choosing, from the set of all possible sequences which determine a given p-adic integer, one standard sequence. Let a p-adic integer be given by the sequence {xn}. Denote the smallest nonnegative integer, congruent to xn modulo pn+1 by (3.5) (3.6) The congruence (3.6) shows that so that the sequence determines some p-adic integer, which by (3.5) is the same as that determined by the sequence {xn}. A sequence, each term of which satisfies conditions (3.4) and (3.6), will be called canonical. Hence we have shown that every p-adic integer is determined by some canonical sequence. It is easy to see that two distinct canonical sequences determine distinct p-adic integers. If the canonical sequences and determine the same p-adic integer, then from the congruence and the conditions , , we obtain for all n ≥ 0. Thus the p-adic integers are in one-to-one correspondence with the canonical sequences. From (3.4) it follows that , and since and , we have 0 ≤ an+1 < p. Hence every canonical sequence has the form where 0 ≤ ai < p. On the other hand, every sequence of this type is a canonical sequence, which determines some p-adic integer. From this it follows that the set of all canonical sequences, and also the set of all p-adic integers, have the cardinality of the continuum. 3.2. The Ring of p-Adic Integers Definition. Let the p-adic integers α and β be determined by the sequences {xn} and {yn}. Then the sum (respectively, product) of α and β is the p-adic integer determined by the sequence {xn + yn} (respectively, {xnyn}). To verify that this definition makes sense, we must show that the sequences {xn + yn} and {xnyn} do indeed determine some p-adic integer, and that this integer depends only on α and β and not on the choice of the sequences which determine them. Both of these assertions are easily verified, and we shall omit the details. It is now obvious that under these operations the set of p-adic integers becomes a commutative ring, which contains the ring of rational integers as a subring. Divisibility of p-adic integers is defined as in any commutative ring (see the Supplement, Section 4.1); α divides β if there is a p-adic integer γ such that β = αγ. To investigate the divisibility properties of p-adic integers we must know for which p-adic integers there exists a multiplicative inverse. Such numbers, by Section 4.1 of the Supplement, are called divisors of unity or units. We shall call them p-adic units. Theorem 1. A p-adic integer α, which is determined by a sequence {x0, x1,…, xn,… }, is a unit if and only if x0 ≢ 0 (mod p). Proof. Let α be a unit. Then there is a p-adic integer β such that αβ = 1. If β is determined by the-sequence {yn}, then the fact that αβ = 1 implies that (3.7) In particular, x0y0 ≡ 1 (mod p) and hence x0 ≡ 0 (mod p). Conversely, let x0 ≢ 0 (mod p). From (3.4) it easily follows that so that xn ≢ 0 (mod p). Consequently, for any n, we may find a yn such that (3.7) holds. Since xn ≡ xn-1 (mod pn) and xn-n ≡ xn-1 yn-1 (mod pn), then also yn ≡ yn-1 (mod pn.) This means that the sequence {yn} determines some p-adic integer β. Equation (3.7) implies that αβ = 1, which means that α is a unit. From this theorem it follows that a rational integer a, considered as an element of Op, is a unit if and only if a ≢ 0 (mod p). If this condition holds, then a-1 belongs to Op. Hence any rational integer b is divisible by such an a in Op, that is, any rational number of the form b/a, where a and b are integers and a ≢ 0 (mod p), belongs to Op. Rational numbers of this type are called p-integers. They clearly form a ring. We can now formulate the above result as follows: Corollary. The ring Op of p-adic integers contains a subring isomorphic to the ring of p-integral rational numbers. Theorem 2. Every p-adic integer, distinct from zero, has a unique representation in the form (3.8) where ε is a unit of the ring O. Proof. If α is a unit, then (3.8) holds with m = 0. Let {xn} → α, where α is not a unit, so that by Theorem 1, x0 ≡ 0 (mod p). Since α ≠ 0, the congruence xn ≡ 0 (mod pn+1) does not hold for all n. Let m be the smallest index for which (3.9) For any s ≥ 0, and therefore the number ys = xm+s/pm is an integer. From the congruences it follows that for all s ≥ 0. Thus the sequence {ys} determines some ε ∈ Op. Since y0 = xm/pm ≢ 0 (mod p), ε is a unit by Theorem 1. Finally, from it follows that pmε = α, which is the desired representation. We assume now that α has another representation α = pkη, where k ≥ 0 and η is a unit. If {zs} → η then (3.10) for all s ≥ 0, and, by Theorem 1, p never divides ys or zs, since ε and η are units. Setting s = m in (3.10), we obtain from which we deduce that k ≤ m. By symmetry we also have m ≤ k, i.e., k = m. Replacing s by s + m in (3.10) and dividing by pm we find that Since by condition (3.4) ym+s ≡ ys (mod ps+1) and zm+s ≡ zs, (mod ps + 1), we obtain Since this congruence holds for all s ≥ 0, ε = η, and Theorem 2 is proved. Corollary 1. The p-adic integer α, determined by the sequence {xn}, is divisible by pk if and only if xn ≡ 0 (mod pn+1) for all n = 0, 1,…, k −1. Indeed, we find the exponent m of expression (3.8) as the smallest index m for which (3.9) holds. Corollary 2. The ring Op does not have any zero divisors. If α ≠ 0 and β ≠ 0, then we have the representations in which ε and η are units. (Thus ε and η have inverses ε-1 and η-1 in the ring Op.) If we had αβ = 0, then, multiplying the equation pm+kεη = 0 by ε-1η-1, we would obtain pm+k = 0, which is impossible. Definition. The number m in the representation (3.8) of a nonzero p-adic integer α is called the p-adic value, or simply the p-value, of α and is denoted by vp(α). In case it is clear which prime p is intended, we shall speak simply of the value, and write v(α). In order that the function v(α) be defined for all p-adic integers, we set v(0) = ∞. (This convention is appropriate since 0 is divisible by arbitrarily high powers of p.) It is easy to verify the following properties of the value function: (3.11) (3.12) (3.13) The divisibility properties of p-adic integers are concisely expressed in terms of the value function. From Theorem 2 we immediately deduce Corollary 3. The p-adic integer α is divisible by β if and only if v(α) ≥ v(β). Thus the arithmetic of the ring Op is very simple. There is a unique (up to associates) prime element, namely, p. Every nonzero element of Op is a product of a power of p and a unit. Finally, we turn to congruences in the ring Op. Congruence of elements is defined here exactly as it is for rational integers, or, more generally, for elements of any ring (see the Supplement, Section 4.1): α ≡ β (mod γ) means that α – β is divisible by γ. If γ = pn — ε, where ε is a unit, then any congruence modulo γ is equivalent to a congruence modulo pn. We thus confine our attention to congruences modulo pn. Theorem 3. Any p-adic integer is congruent to a rational integer modulo pn. Two rational integers are congruent modulo pn in the ring Op if and only if they are congruent modulo pn in the ring Z. Proof. To prove the first assertion we shall show that if α is a p-adic integer and {xn} is a sequence of rational numbers determining α, then (3.14) Since xn-1 is determined by the sequence {xn-1, xn-1,… }, the sequence {x0 – xn-1, x1 – x1 – xn-1,… } determines the number α – xn−1. We apply Corollary 1 of Theorem 2 to the p-adic integer α – xn-1. We see that the congruence (3.14) is equivalent to the congruence which is in turn implied by condition (3.4) in the definition of p-adic integers. We now show that for two rational integers x and y, congruence modulo p in the ring Op is equivalent to congruence modulo p in the ring Z. Set (3.15) (we assume that x ≠ y). The congruence (3.16) in the ring Z is equivalent to the condition n ≤ m. On the other hand, (3.15) is a representation of the type (3.8) for the number x – y, since a is a p-adic unit. Consequently, vp(x – y) = m, and the condition n ≤ m can be written in the form vp(x — y) ≥ n. But this is equivalent to the congruence (3.16) in Op since v(pn) = n (see Corollary 3 of Theorem 2). Corollary. There are pn residue classes in Op, modulo pn. 3.3. Fractional p-adic Numbers Since the ring Op has no zero divisors (Corollary 2 of Theorem 2), it can be embedded in a field, using the standard construction of a field from an integral domain. Application of this construction to our situation leads to consideration of fractions of the form α/pk, where α is some p-adic integer, and k ≥ 0. The fractions considered here could more suitably be written as pairs (α, pk). Definition. A fraction of the form α/pk, α ∈ Op, k ≥ 0, determines a fractional p-adic number, or, more simply, a p-adic number. Two fractions, α/pk and β/pm, determine the same p-adic number if and only if αpm = βpk in Op. The set of all p-adic numbers will be denoted by Rp. A p-adic integer determines an element α/1 = α/p0 in Rp. It is clear that distinct p-adic integers determine distinct elements of Rp. Hence we shall assume that Op is a subset of the set Rp. Addition and multiplication are defined in Rp by the rules It is a simple exercise to verify that the result of these operations does not depend on the choice of fractions to represent the elements of Rp, and that under these operations Rp is turned into a field—the field of all p-adic numbers. It is clear that the field Rp has characteristic zero and thus contains the field of rational numbers. Theorem 4. Any nonzero p-adic number α is uniquely representable in the form (3.17) where m is an integer and ε is a unit of Op. Proof. Let ξ = α/pk, α ∈ Op. By Theorem 2, α can be represented in the form α = p1ε, l ≥ 0, where ε is a unit of the ring Op. Thus ξ = pmε, where m = l — k. The uniqueness of the representation (3.17) follows from the corresponding assertion for p-adic integers, proved in Theorem 2. The concept of the value of an element, introduced in Section 2, easily generalizes to any p-adic number. We set where m is the exponent in (3.17). It is easily seen that properties (3.11), (3.12), and (3.13) of the value automatically carry over to the field Rp. The p-adic number ζ is a p-adic integer if and only if vp(ξ) ≥ 0. 3.4. Convergence in the Field of p-Adic Numbers In Section 3.1 we noted the analogy between p-adic integers and real numbers, in that both are determined by sequences of rational numbers. Just as every real number is the limit of any sequence of rational numbers which determines it, it would be natural to conjecture that the same fact should hold for p-adic numbers, if the correct definition of the concept of convergence is given. The definition of limit for real or rational numbers can be based, for example, on the notion of nearness; two real or rational numbers being near if the absolute value of their difference is small. For the definition of convergence for p-adic numbers we thus must decide under what conditions two p-adic numbers are to be considered close to one another. In the example of the first section, we spoke of the p-nearness of two p-adic integers x and y, meaning by this that the difference of x and y should be divisible by a high power of p. It was under this definition of nearness that the analogy between the definitions of real numbers and of p-adic integers became apparent. If we use the concept of the p-value vp, then the p-nearness of x and x will be characterized by the value of vp(x – y). Thus we may speak of two p-adic numbers ξ and η (not necessarily integers) as being near when the value of vp(ξ – η) is sufficiently large. Thus “small” p-adic numbers are characterized by the large value of their p-value. After these remarks we turn to precise definitions. Definition. The sequence of p-adic numbers converges to the p-adic number ξ (we denote this by limn → ∞ ξn = ξ or {ξn} → ξ) if A singular feature of this definition (which distinguishes it from the usual definition of convergence for real numbers) is that the convergence of {ξn} to ξ is determined by the sequence of rational integers vp(ξn – ξ), which must converge to infinity. We can put the definition in a more familiar form if, instead of vp, we consider another nonnegative real-valued function on the field Rp, which will converge to zero as vp goes to infinity. Namely, choose some real number ρ, satisfying 0 < ρ < 1, and set (3.18) Definition. The function φp(ξ), ξ ∈ Rp, defined by (3.18), is called a p-adic metric. The number φp(ξ) is called the p-adic size of ξ. As in the case of the value function, we shall sometimes simply call φp a value and denote it by ϕ. Properties (3.11) and (3.12) of the value clearly imply the following properties of the metric: (3.19) (3.20) From the last inequality we also obtain (3.21) Properties (3.19) and (3.21) [and also the fact that ϕ(ξ) > 0 for ξ ≠ 0] show that the concept of metric for p-adic numbers is analogous to the concept of absolute value in the field of real (or complex) numbers. In terms of the valuation φp the definition of convergence in the field Rp takes the following form: The sequence {ξn}, ξn ∈ Rp, converges to the p-adic number ξ if We may formulate and prove, for the field Rp, general theorems on the limits of sequences, well known in analysis. As an example we shall show that if {ξn} → ξ and ξ ≠ 0, then {1/ξn} → 1/ξ. First, from some point on, that is, for all n ≥ n0, we have v(ξn – ξ) > v(ξ), from which, by (3.13), v(ξn) = min (v(ξn – ξ), v(ξ)) = v(ξ). In particular, v(ξn) ≠ ∞, that is, ξn ≠ 0, so that 1/ξn makes sense for all n ≥ n0. Further, as n → ∞, and our assertion is proved. Theorem 5. If the p-adic integer α is determined by the sequence {xn} of rational integers, then this sequence converges to α. An arbitrary p-adic number ξ is a limit of a sequence of rational numbers. Proof. From the congruence (3.14) it follows that vp(xn – α) ≥ n + 1. Consequently, v(xn – α) → ∞ as n → ∞, and this means that {xn} converges to α. Consider now the fractional p-adic number ξ = α/pk. Since as n → ∞, then ξ is the limit of the rational sequence {xn/pk}. The theorem is proved. From any bounded sequence of real numbers it is possible to choose a convergent subsequence. An analogous property also holds for p-adic numbers. Definition. The sequence {ξn} of p-adic numbers is called bounded if the numbers φp(ξn) are bounded from above, or equivalently, if the numbers vp(ξn) are bounded from below. Theorem 6. From any bounded sequence of p-adic numbers (in particular, from any sequence of p-adic integers) it is possible to choose a convergent subsequence. Proof. We first prove the theorem for a sequence {αn} of p-adic integers. Since in the ring Op the number of residue classes modulo p is finite (corollary of Theorem 3), there are an infinite number of terms in the sequence {αn} which are congruent modulo p to some rational integer x0. Choosing all such terms, we obtain a subsequence {αn(1)}, all terms of which satisfy the congruence Analogously, applying the corollary of Theorem 3 to the case n = 2, we choose from the sequence {αn(1)} a subsequence with the condition where x1 is some rational integer. Here, clearly, x1 ≡ x0 (mod p). Continuing this process indefinitely, we obtain for each k a sequence {αn(k)}, which is a subsequence of the preceding sequence {αn(k −1)} and for all terms of which the congruence holds for some rational integer xk−1. Since (mod pk+1) and all belong among the , for all k ≥ 1. Thus the sequence {xn} determines some p-adic integer α. We now take the “diagonal” sequence . It clearly is a subsequence of the initial sequence {αn}. We claim that . Indeed, by (3.14) we have α ≡ xn-1 (mod pn). But, on the other hand, (mod pn) and hence (mod pn); that is, . From this it follows that as n → ∞, and thus converges to α. We now turn to the proof of the theorem in the general case. If the sequence {ξn} of p-adic numbers satisfies v(ξn) = −k (k some rational integer), then for αn = ξnpk we have v(αn) ≥ 0. By the above we may extract a convergent subsequence from the sequence {αn} of p-adic integers. But then the sequence is a convergent subsequence of the sequence {ξn}. Theorem 6 is completely proved. The Cauchy convergence criterion also holds for p-adic numbers: The sequence (3.22) converges if and only if (3.23) The necessity of the condition is clear. For the proof of sufficiency we first note that (3.23) implies that the sequence (3.22) is bounded. Indeed, from (3.23) it follows that there is an n0, such that for all m ≥ n0. But then by (3.12) for all m ≥ n0, and from this it follows that (3.22) is bounded. By Theorem 6 we may extract from (3.22) a convergent subsequence with limit, say ξ. We now show that the sequence (3.22) converges to ξ. Let M be an arbitrarily large number. From (3.23) and the definition of convergence we can find a natural number N so that, first, v(ξm — ξn) ≥ M for m, n ≥ N, and, second, for ni ≥ N. Then for all m ≥ N. Thus limm→∞ v(ξm — ξ) = ∞, that is, the sequence (3.22) converges. The principle of convergence proved above can be put in a stronger form. If the sequence (3.22) satisfies (3.23), then it clearly also satisfies (3.24) We shall show that, conversely, (3.24) implies (3.23). For if v(ξn+1 — ξn) ≥ M for all n > N, then by (12) from the equation it follows that that is, v(ξm — ξn) → ∞ as m, n → ∞. Thus we have Theorem 7. For the convergence of the sequence {ξn} of p-adic numbers, it is necessary and sufficient that limn→∞ v(ξn+1 — ξn) = ∞. Having a concept of convergence in the field Rp, we may speak of continuous p-adic functions of p-adic variables. Their definition does not differ at all from the usual one. That is, the function F(ξ) is called continuous at ξ = ξ0 if for any sequence {ξn} which converges to ξ0, the sequence of values {F(ξn)} converges to F(ξ0). A similar definition holds for functions of several variables. Just as in real analysis it is easy to prove the usual theorems on arithmetic operations with continuous p-adic functions. In particular, it is easily verified that polynomials in any number of variables with p-adic coefficients are continuous p-adic functions. This simple fact will be used (Section 5.1) in the future. To conclude this section we make some remarks on series with p-adic terms. Definition. If the sequence of partial sums , of the series (3.25) with p-adic terms converges to the p-adic number α, then we shall say that the series converges and that its sum is α. From Theorem 7 we immediately deduce the following convergence criterion for series. Theorem 8. In order that the series (3.25) converge, it is necessary and sufficient that the general term converge to zero, that is, that v(αn) → ∞ as n → ∞. Convergent p-adic series can clearly be termwise added and subtracted and multiplied by a constant p-adic number. The associativity property of series also holds for them. Theorem 9. If the terms of a convergent p-adic series are rearranged, its convergence is not affected and its sum does not change. The simple proof of this theorem is left to the reader. In analysis it is proved that the property described in Theorem 9, when applied to real numbers, characterizes absolutely convergent series. Thus every convergent p-adic series is “absolutely convergent.” From this it follows that convergent p-adic series can be multiplied in the usual manner. If the p-adic integer α is defined by the canonical sequence {a0, a0 + a1p, a0 + a1p + a2p2,… } (Section 3.1), then, by the first assertion of Theorem 5, it will equal the sum of the convergent series (3.26) Since distinct canonical sequences determine distinct p-adic integers, the representation of α in the form of the series (3.26) is unique. Conversely, any series of the form (3.26) converges to some p-adic integer. The representation of p-adic integers in series (3.26) is reminiscent of the expansion of real numbers as infinite decimals. If we consider the series (3.27) in which the coefficients are arbitrary rational integers, then it clearly converges [since v(bnpn) ≥ n], and its sum will equal some p-adic integer α. To obtain the representation (3.26) for this number α, we must successively replace each coefficient in (3.27) by its remainder after division by p, and carry over the quotient at each step to the coefficient of the next term. This observation can be used for computations in the ring Op. That is, after addition, subtraction, or multiplication of series of the form (3.26) according to the rules for operating with series, we shall obtain a series in the form (3.27), in which the coefficients, in general, will not be the smallest nonnegative residues modulo p. To transform this series into the form (3.26) we need only apply the rule just mentioned. This method of carrying out operations with p-adic integers is easily seen to be analogous to the usual method for operating with real numbers which are expressed as infinite decimals. From Theorem 1 it easily follows that a p-adic integer, represented in the form of a series (3.26), is a unit in the ring Op if and only if a0 ≠ 0. Along with Theorem 4 this gives us the following result. Theorem 10. Every nonzero p-adic number ξ is uniquely representable in the form (3.28) where m = vp(ξ), 1 ≤ a0 ≤ p −1, 0 ≤ an ≤ p −1 (n = 1, 2,…). View chapterExplore book Read full chapter URL: Book series1966, Pure and Applied Mathematics Chapter Congruences 1966, Pure and Applied Mathematics Corollary 3. The p-adic integer α is divisible by β if and only if v(α) ≥ v(β). Thus the arithmetic of the ring Op is very simple. There is a unique (up to associates) prime element, namely, p. Every nonzero element of Op is a product of a power of p and a unit. Finally, we turn to congruences in the ring Op. Congruence of elements is defined here exactly as it is for rational integers, or, more generally, for elements of any ring (see the Supplement, Section 4.1): α ≡ β (mod γ) means that α – β is divisible by γ. If γ = pn — ε, where ε is a unit, then any congruence modulo γ is equivalent to a congruence modulo pn. We thus confine our attention to congruences modulo pn. View chapterExplore book Read full chapter URL: Book series1966, Pure and Applied Mathematics Chapter General Lattice Theory 1978, Pure and Applied Mathematics 3 Congruence Relations Let L be a lattice and let H ⊆ L2. We denote by Θ(H) the smallest congruence relation such that a ≡ b for all 〈a, b〉 ∈ H, and call it the congruence relation generated by H. Lemma 1 For any H ⊆ L2, Θ(H) exists. Proof. Let Φ = ∧ (Θ \| a ≡ b (Θ) for all 〈a, b〉 ∈ H). Since in the lattice C(L) the meet is intersection, it is obvious that a ≡ b (Φ) for all 〈a, b〉 ∈ H; thus Φ = Θ(H). We will use special notations in two cases: If H = {〈a, b〉}, we write Θ(a, b) for Θ(H). If H = I2, where I is an ideal, we write Θ[I] for Θ(H). The congruence relation Θ(a, b) is called principal; its importance is revealed by the following formula. Lemma 2. Proof. The proof is obvious. Note that Θ(a, b) is the smallest congruence relation under which a ≡ b, whereas Θ[I] is the smallest congruence relation under which I is contained in a single class. In general lattices, not much can be said about Θ(H). In distributive lattices, the following description of Θ(a, b) is important (G. Grätzer d E. T. Schmidt [1958d]): Theorem 3 Let L be a distributive lattice, a, b, x, y ∈ L, and let a ≤ b. Then x ≡ y(Θ(a, b)) iff x∧a = y∧a and x∨b = y∨b. Remark. This situation is illustrated in Figure 1. Proof. Let Φ denote the binary relation under which x ≡ y (Φ) iff x∧a = y∧a and x∨b = y∨b. Φ is obviously an equivalence relation. If x ≡ y (Φ) and z ∈ L, then , and ; thus x∨z ≡ y∨z (Φ). Similarly, x∧z ≡ y∧z (Φ), and so Φ is a congruence relation. That a ≡ b (Φ) is obvious. Finally, let Θ be any congruence relation such that a ≡ b (Θ) and let x ≡ y (Φ). Therefore, , and x∧b x∧a (Θ). Then, computing modulo Θ, we obtain that is, x ≡ y (Θ), proving that Φ ≤ Θ. Explanation. Since a ≡ b implies that (a∨p)∧q ≡ (b∨p)∧q, we must have x ≡ y (Θ(a, b)) if . It is easy to check that the x, y satisfying the conditions of Theorem 3 are exactly the same as those for which such p, q exist. Thus Theorem 3 can be interpreted as follows: We get all pairs x, y with x ≡ y (Θ(a, b)) and x ≤ y by applying the Substitution Property “twice”. No further application of the Substitution Property or transitivity is needed. An equivalent form of Theorem 3 is to require x∨y ≤ b∨(x∧y) and (a∨(x∧y))∧(x∨y) = x∧y. Some applications of Theorem 3 follow. Corollary 4 Let I be an ideal of the distributive lattice L. Then x ≡ y (Θ[I]) iff x∨y = (x∧y)∨i for some i ∈ I. Therefore, I is a congruence class modulo Θ[I]. Remark. This situation is illustrated in Figure 2, in which the wavy line indicates congruence modulo Θ[I]. Proof. If x∨y = (x∧y)∨i, then , and so x ≡ y(Θ[I]). Conversely, by Lemma 2. However, ; therefore . If , u, v ∈ I, u ≤ v, then x∨v = y∨v, and so (x∧y)∨(v∧(x∨y)) = x∨y; thus the condition of Corollary 4 is satisfied with i = v∧(x∨y) ∈ I. Finally, if a ∈ I and a ≡ b(Θ[I]), then a∨b = (a∧b)∨i, i ∈ I, and so a∨b ∈ I and b ∈ I, showing that I is a full congruence class. Corollary 5 Let L be a distributive lattice, x, y, a, b ∈ L, and let x ≤ y ≤ a ≤ b or a ≤ b ≤ x ≤ y. Then x ≡ y (Θ(a, b)) implies that x = y. A very important congruence relation has already been used in the proof of Lemma I.3.5(ii): Given a prime ideal P of the lattice L, we can construct a congruence relation that has exactly two congruence classes, P and L – P. This statement can be generalized as follows: Let A be a set of prime ideals of a lattice L and let us call two elements x and y congruent modulo A iff, for every P ∈ A, either x, y ∈ P or x, y ∈ L – P; this describes a congruence relation on L. For instance, if A = {P, Q, R}, Q ⊂ P, R ⊂ P, we get five congruence classes as shown in Figure 3; the quotient lattice is shown in Figure I.5.8. This principle will be used often. An interesting application is: Theorem 6 Let K be a sublattice of a distributive lattice L and let Θ be a congruence relation of K. Then Θ can be extended to L; that is, there exists a congruence relation Φ on L such that, for x, y ∈ K, x ≡ y (Φ) iff x ≡ y (Θ). Proof. Let φ be the natural homomorphism of K onto K/Θ, that is, φ: x → [x]Θ; then, for every prime ideal P of K/Θ, Pφ−1 is a prime ideal of K. Therefore, (Pφ−1] is an ideal of L and [K – Pφ−1) is a dual ideal of L; thus by Theorem 1.15 we can choose a prime ideal P1 of L such that P1 ⊇ Pφ−1 and P1 ∩ (K – Pφ−1) = Ø. For every prime ideal P of K we choose such a prime ideal P1 of L; let A denote the collection of all such prime ideals. Let Φ be the congruence relation associated with A as previously described. Now for x, y ∈ K, the condition x ≡ y (Θ) is equivalent to xφ = yφ, and so for every P1 ∈ A either x, y ∈ P1 or x, y ∉ P1; thus x ≡ y (Φ). Conversely, if x ≡ y (Φ), then for every P1 ∈ A either x, y ∈ P1 or x, y ∉ P1, and so either xφ, yφ ∈ P or xφ, yφ ∉ P. Since every pair of distinct elements of K/Θ is separated by a prime ideal (Corollary 1.17), we conclude that xφ = yφ and thus x ≡ y (Θ). It is well known that in rings, ideals are in a one-to-one correspondence with congruence relations. In one class of lattices the situation is exactly the same as in the class of rings. Theorem 7 Let L be a Boolean lattice. Then is a one-to-one correspondence between congruence relations and ideals of L. Proof. By Corollary 4 the map is onto; therefore, we have only to prove that it is one-to-one, that is, that Θ determines Θ. This fact, however, is obvious, since a ≡ b (Θ) iff a∧b ≡ a∨b (Θ), which in turn is equivalent to c ≡ 0 (Θ), where c is the relative complement of a∧b in [0, a∨b] (see Figure 4). Thus a ≡ b (Θ) iff c ∈ Θ. This proof does not make full use of the hypothesis that L is a complemented distributive lattice. In fact, all we need to make the proof work is that L has a zero and is relatively complemented. Such a distributive lattice is called a generalized Boolean lattice. Theorem 8 (J. Hashimoto ) Let L be a lattice. There is a one-to-one correspondence between ideals and congruence relations of L under which the ideal corresponding to a congruence relation Θ is a whole congruence class under Θ iff L is a generalized Boolean algebra. Proof (G. Grätzer and E. T. Schmidt [1958d]). The proof of the “if” part is the proof of Theorem 7. We proceed with the “only if” part. The ideal corresponding to ω has to be (0], and thus L has a 0. If L contains a diamond {o, a, b, c, i}, then (a] cannot be a congruence class, because a ≡ o implies that i = a∨c ≡ o∨c = c, and b = b∧i ≡ b∧c = o. But o ∈ (a], and thus any congruence class containing (a] contains b, and b ∉ (a]. Similarly, if L contains a pentagon {o, a, b, c, i} and a congruence class contains (b], then b ≡ o; thus i = b∨c ≡ o∨c = c, and so a = a∧i ≡ a∧c = o. Therefore, this congruence class has to contain a, and a ∉ (b]. Thus, by Theorem 1.1, L is distributive. Let a < b and I = Θ(a, b). By Corollary 4, Θ[I] is also a congruence relation of L having I as a whole congruence class; consequently, Θ[I] = Θ(a, b), and so a ≡ b (Θ[I]). Thus, again by Corollary 4, b = a∨i for some i ∈ I, and i ≡ 0 (Θ(a, b)). The latter is equivalent to i∨b = 0∨b and i∧a = 0∧a. Thus a∨i = b and a∧i = 0, and so i is the relative complement of a in [0, b]. It is no coincidence that, in the class of generalized Boolean lattices, congruences and ideals behave as they do in rings. Indeed, generalized Boolean lattices are rings in disguise. Theorem 9 (M. H. Stone ) (i) : Let B = 〈B; ∧, ∨〉 be a generalized Boolean lattice. Define the binary operations · and + on B by setting and by defining x + y as the relative complement of x∧y in [0, x∨y] (see Figure 5). Then BR = 〈B; +, ·〉 is a Boolean ring–that is, an (associative) ring satisfying x2 = x, for all x ∈ B (and, consequently, satisfying xy = yx and x + x = 0, for x, y ∈ B). (ii) : Let B = 〈B; +, ·〉 be a Boolean ring. Define the binary operations ∧ and ∨ on B by andThen BL = 〈B; ∧, ∨〉 is a generalized Boolean lattice. (iii) : Let B be a generalized Boolean lattice. Then (BR)L = B. (v) : Let B be a Boolean ring. Then (BL)R = B. The proof of this theorem is purely computational. Some steps will be given in the exercises. The correspondence between Boolean rings and generalized Boolean lattices preserves many algebraic properties. Theorem 10 Let B0 and B1 be generalized Boolean lattices. (i) : Let I ⊆ B0. Then I is an ideal of B0 iff I is an ideal of . (ii) : Let φ: B0 → B1. Then φ is a {0}-homomorphism of B0 into B1 iff φ is a homomorphism of into . (iii) : B0 is a {0}-sublattice of B1 iff is a subring of . The proof is again left to the reader. Congruence relations on an arbitrary lattice have an interesting connection with distributive lattices: Theorem 11 (N. Funayama and T. Nakayama ) Let L be an arbitrary lattice. Then C(L), the lattice of all congruence relations of L, is distributive. Proof. Let . Since , it suffices to prove that implies that . So let a ≡ b (Θ∧(Φ∨Ψ)); that is, a ≡ b (Θ) and a ≡ b (Φ∨Ψ). By Theorem I.3.9, there exists a sequence such that zi ≡ zi+1 (Φ) or zi ≡ zi+1 (Ψ) for every 0 ≤ i < n. Since a ≡ b (Θ), we also have a∧b ≡ a∨b (Θ), and so zi ≡ zi+1 (Θ) for every 0 ≤ i < n. Thus for every 0 ≤ i < n, zi ≡ zi+1 (Θ∧Φ) or zi ≡ zi+1 (Θ∧Ψ), implying that . Another property of congruence lattices is given in the following definition. Definition 12 (i) : Let L be a complete lattice and let a be an element of L. Then a is called compact iff a ≤ ∨ X for some X ⊆ L implies that a ≤ ∨ X1 for some finite X1 ⊆ X. (ii) : A complete lattice is called algebraic iff every element is the join of compact elements. The name, algebraic lattice, is due to G. Birkhoff , however Birkhoff does not assume completeness. In the literature, algebraic lattices are also called compactly generated lattices. Just as for lattices, a nonvoid subset I of a join-semilattice F is an ideal iff, for a,b ∈ F, we have a∨b ∈ I exactly if a and b ∈ I. Again, I(F) is the poset (not necessarily a lattice) of all ideals of F partially ordered under set inclusion. If F has a zero, then I(F) is a lattice. Using I(F), we give a useful characterization of algebraic lattices: Theorem 13 A lattice L is algebraic iff it is isomorphic to the lattice of all ideals of a join-semilattice with 0. Proof. Let F be a join-semilattice with 0; we prove that I(F) is algebraic. We know that I(F) is complete. We claim that for a ∈ F, (a] is a compact element of I(F). Let X ⊆ I(F) and let Just as in the proof of Corollary I.3.2, Therefore, . Thus with X1 ={I0, …,In-1}, Since for any I ∈ I(F) we have we see that I(F) is algebraic. Now let L be an algebraic lattice and let F be the set of compact elements of L. Obviously, 0 ∈ F. Let a, b ∈ F and a∨b ≤ ∨X for some X⊆L. Then a ≤ a∨b ≤ ∨ X, and so a ≤ ∨ X0, for some finite X1 ⊆ X. Similarly, b ≤ ∨ X1, for some finite X0 ⊆ X. Thus a∨b≤(X0 ∪ X1), and X0 ∪ X1 is a finite subset of X. So a∨b ∈ F. Therefore, 〈F; ∨〉 is a join-semilattice with 0. Consider the map: Obviously, φ maps L into I(F). By the definition of an algebraic lattice, a = ∨aφ, and thus φ is one-to-one. To prove that φ is onto, let I ∈ I(F), a = ∨I. Then aφ ⊇ I. Let x ∈ aφ. Then x ≤ ∨I, so that by the compactness of x, x ≤ ∨ I1 for some finite I1 ⊆ I. Therefore x ∈ I, proving that aφ ⊆ I. Consequently, aφ = I, and so φ is onto. Thus φ is an isomorphism. Now we connect the foregoing with congruence lattices. Lemma 14 Every principal congruence relation is compact. Proof. Let L be a lattice, a, b ∈ L, X ⊆ C(L), and Then a ≡ b (∨X), and thus (just as in Theorem I.3.9) there exists a sequence a = x0, x1, …,xn-1 = b, in which xi ≡ xi+1 (Θi) for some Θi ∈ X. Therefore, a ≡ b (∨X0), where X0 = {Θ0, …,Θn-1}, and so Θ(a, b) ≤ (∨X0), where X0 is a finite subset of X. Theorem 15 Let L be an arbitrary lattice. Then C(L) is an algebraic lattice. Proof. For every Θ ∈ C(L), Consequently, this theorem follows from Lemma 14 and Corollary I.3.15. Combining Theorems 11 and 15 we get: Corollary 16 Let L be an arbitrary lattice. Then C(L) is a distributive algebraic lattice. The converse of Corollary 16 is a long-standing conjecture of lattice theory. We shall verify the conjecture in the finite case. This was first established by R. P. Dilworth. The present proof combines a construction of G. Grätzer and E. T. Schmidt (Lemma 18) with a result of G. Grätzer and H. Lakser (Lemma 19). The reader is advised to page over the proof of Theorem 17 at the first reading of this book. Theorem 17 Let K be a finite distributive lattice. Then there exists a finite lattice L such that K is isomorphic to C(L). Proof. The proof of Theorem 17 is immediate from Lemmas 18 and 19. We take L=I(M) where M is given in Lemma 18. By Lemma 19, C(L) ≅ C(M); by Lemma 18, C(M) ≅ K. Since M is finite, so is L. Let M be a finite poset such that inf {a, b} exists in M for any a, b ∈ M. We define in M: a∧b = inf {a, b} and a∨b = sup {a, b} whenever it exists. This makes M into a partial lattice. An equivalence relation Θ on M is a congruence relation iff a0 ≡ b0 (Θ) and a1 ≡ b1 (Θ) imply that a0∧a1 ≡ b0∧b1 (Θ) and that a0∧a1 ≡ b0∨b1 (Θ) whenever a0∨a1 and b0∨b1 exist. Then the set C(M) of all congruence relations is again a lattice. Lemma 18 Let K be a finite distributive lattice. Then there exists a finite poset M such that inf {a, b} exists for all a, b ∈ M and C(M) is isomorphic to K. Proof. Take the set M0 = J(K) ∪ {0} and make it a meet-semilattice by defining inf {a, b} = 0 if a ≠ b (J(K) is the set of nonzero join-irreducible elements of K; see Section 1), as illustrated in Figure 6. Note that a ≡ b (Θ) and a ≠ b imply in M0 that a≡0 (Θ) and b≡0(Θ) therefore, congruence relations of M0 are in one-to-one correspondence with subsets of J(K). Thus C(M0) is a Boolean lattice whose atoms are associated with elements of J(K); the congruence Φa associated with a ∈ J(K) is: x ≡ y (Φ)a if {x, y} ≠ {a, 0}, then x≡y(Φa) implies that x=y. If J(K) is unordered in K, then we are ready. However, if, say, a, b ∈ J(K), a > b in K, then we must Φa > Φb. The simplest way to make this happen is to use the lattice M(a, b) of Figure 7. Note that M(a, b) has three congruence relations, namely, ω, ι, and Θ, where Θ is the congruence relation with congruence classes {0, b1; b2, b}, {a1, a(b)}. Thus Θ(a1, 0) = ι. In other words, a1≡0 “implies” that b1≡0, but b1≡0 does not “imply” that a1 ≡ 0. We construct M by “inserting” M(a, b) in M0 whenever a > b in J(K). Figure 8 gives M for the three-element chain. More precisely, M consists of four kinds of elements: (i) 0; (ii) all maximal join-irreducible elements of K (that is, all a ∈ J(K) such that there is no x ∈ J(K) with a < x in K); (iii) for any nonmaximal join-irreducible element a of K, three elements: a, a1, a2; (iv) for each pair a, b ∈ J(K) with a > b, a new element, a(b). To simplify the notation, for each maximal join-irreducible element a, we write a = a1 = a2. For a, b ∈ J(K) with a > b, we set M(a, b) = {0, a1, b, b1, b2, a(b)}. Observe that if a=c and if a ≠ c and if a=c and otherwise, . For x, y ∈ M, let us define x ≤ y to mean that for some a, b ∈ J(K) with a > b, we have x, y ∈ M(a, b) and x ≤ y in the lattice M(a, b) as illustrated in Figure 7. It is easily seen that x ≤ y does not depend on the choice of a and b, and that ≤ is a partial ordering relation. Since, under this partial ordering, all M(a, b) and M(a, b) ∩ M(c, d) are lattices and x, y ∈ M, x ∈ M(a, b), and y ≤ x imply that y ∈ M(a, b), we conclude that inf {u, v} exists for all u, v ∈ M. Now we describe C(M). Let H ∈ H(J(K)) (notation of Definition 1.8). We define a binary relation ΘH on M: iffeither , or , where a > b, a > c, b, c ∈ H, or x = y. In other words, ΘH contains all a1, a2, a with a ∈ H; and if a > b, a,b ∈ H, then it also contains a(b). Outside this class the only nontrivial congruences are a(b) ≡ a1 ≡ a(c), where a∉H, and b, c ∈ H, a > b, a > c. ΘH is obviously an equivalence relation. The fact that ΘH restricted to any M(a, b) is a congruence relation easily implies that ΘH is a congruence relation. Given a ΘH we get thus the mapis a one-to-one order-preserving map of H(J(K)) into C(M). To show that φ is an isomorphism, we have to show that φ is onto. So let Θ be a congruence relation of M, and Since in M(a, b) every congruence Θ is determined by the atoms in Θ, the same holds in M. Therefore, Θ = ΘH. Thus H(J(K)) ≅ C(M). By Theorem 1.9, K ≅ H(J(K)), and so K ≅ C(M). Lemma 19 Let M be a finite poset with the property that inf (a, b} exists for all a, b ∈ M. Then for every congruence relation Θ there exists exactly one congruence relation Θ of I(M) such that for . Proof. Since arbitrary meets exist in M, for every element m ∈ M, (m] is a (finite) lattice, and so if {x, y} has an upper bound, then x∨y exists. Let Θ be a congruence relation of M. For X ⊆ M, set that is, for some x ∈ X}. If I J ∈ I(M), define . Obviously, is an equivalence relation. Let , and x ∈ I ∩ N. Then x ≡ y (Θ) for some y ∈ J, and so x ≡ x∧y (Θ) and . This shows that . Similarly, , so . To show that , recall the description of I ∨ N given in Exercise I.5.22: Set A0 = I ∪ N and, for , for some then . We prove by induction on n that . For . Suppose that and let x ∈ An. Then x ≤ t0∨t1 for some . Thus t0 ≡ u0 (Θ) and t1 ≡ u1 (Θ) for some u0, u1 ∈ J∨N, and so t0≡t0∧u0 (Θ) and t1 ≡ t1∧u1 (Θ). Observe that t0∨t1 is an upper bound for {t0∧u0, t1∧u1}; consequently, (t0∧u0)∨(t1∧u1) exists. Therefore, . Finally, and . Thus x ∈ [J ∨ N]Θ. Since , we conclude that . Similarly, J ∨ N ⊆ [I ∨ N]Θ, proving that . This completes the verification that is a congruence relation of I(M). If a ≡ b (Θ) and x ∈ (a], then x ≡ x∧b (Θ). Thus (a] ⊆ [(b]]Θ. Similarly, (b] ⊆ [(a]]Θ, and so (a] ≡ (b] ( ). Conversely, if (a] ≡ (b] ( ), then a ≡ b1 (Θ) and a1≡b (Θ) for some a1 ≤ a and b1 ≤ b. Forming the join of the two congruences, we get a ≡ b (Θ). Thus has all the properties required by Lemma 19. To show the uniqueness, let Φ be a congruence relation of I(M) satisfying (a] ≡ (b](Φ) iff a ≡ b(Θ). Let I, J ∈ I(M), I ≡ J (Φ), and x ∈ I. Then , and for some y ∈ J. Thus (x] ≡ (y] (Φ), and so x ≡ y (Θ), proving that . Similarly, , and so . Conversely, if , then take all congruences of the form . By our assumption regarding , and by our definition of , the join of all these congruences yields I ≡ J (Φ). Thus Φ = . More general results along these lines can be found in G. Grätzer and E. T. Schmidt , G. Grätzer and H. Lakser , and E. T. Schmidt , , and . View chapterExplore book Read full chapter URL: Book series1978, Pure and Applied Mathematics Chapter Congruences 1966, Pure and Applied Mathematics Theorem 3. Any p-adic integer is congruent to a rational integer modulo pn. Two rational integers are congruent modulo pn in the ring Op if and only if they are congruent modulo pn in the ring Z. Proof. To prove the first assertion we shall show that if α is a p-adic integer and {xn} is a sequence of rational numbers determining α, then (3.14) Since xn-1 is determined by the sequence {xn-1, xn-1,… }, the sequence {x0 – xn-1, x1 – x1 – xn-1,… } determines the number α – xn−1. We apply Corollary 1 of Theorem 2 to the p-adic integer α – xn-1. We see that the congruence (3.14) is equivalent to the congruence which is in turn implied by condition (3.4) in the definition of p-adic integers. We now show that for two rational integers x and y, congruence modulo p in the ring Op is equivalent to congruence modulo p in the ring Z. Set (3.15) (we assume that x ≠ y). The congruence (3.16) in the ring Z is equivalent to the condition n ≤ m. On the other hand, (3.15) is a representation of the type (3.8) for the number x – y, since a is a p-adic unit. Consequently, vp(x – y) = m, and the condition n ≤ m can be written in the form vp(x — y) ≥ n. But this is equivalent to the congruence (3.16) in Op since v(pn) = n (see Corollary 3 of Theorem 2). View chapterExplore book Read full chapter URL: Book series1966, Pure and Applied Mathematics Chapter The Theory of Divisibility 1966, Pure and Applied Mathematics Theorem 2. The ring v has (up to associates) a single prime element π, and any element α ≠ 0 of v has a unique (for fixed π) representation in the form α = επm where ε is a unit in v (m ≥ 0). The prime element π is clearly characterized by v(n) = 1. In the ring v as in any ring, we can consider congruences with respect to the elements of v (see the Supplement, Section 4.1). Since congruences modulo associate elements are equivalent, the ring of residue classes modulo the prime element π does not depend on the choice of π but is completely determined by the ring v. We denote this ring of residue classes by Σv and will now show that it is a field. For if α ∈ v and α ≢ 0 (mod π), then v(α) = 0 and this means that α is a unit in v. Then α has an inverse ξ and αξ ≡ 1 (mod π), since αξ = 1. The field Σv is called the residue class field of the valuation v. View chapterExplore book Read full chapter URL: Book series1966, Pure and Applied Mathematics Chapter Analytic Methods 1966, Pure and Applied Mathematics Remark. To determine whether h and l are relatively prime, it is not necessary to find the precise value of the Bernoulli numbers. It suffices to consider the recurrence relation (8.2) as a congruence modulo l and to use these congruences to compute the sequence B2, B4,…, Bi-3. The number h will be relatively prime to l if and only if none of these numbers is divisible by l. View chapterExplore book Read full chapter URL: Book series1966, Pure and Applied Mathematics Chapter Classical Error-Correcting Codes 2012, Classical and Quantum InformationRichard Hamming 4.6 SYNDROME AND STANDARD ARRAY DECODING OF LINEAR CODES Now we discuss the partitioning of a vector space, Vn, over the finite field, F, into equivalence classes such that all n-tuples in the same class have the same syndrome; then we present a decoding procedure that exploits this partitioning. We start with a few definitions of algebraic structures. Let G be an additive Abelian group and S be a subgroup of G, S ⊆ G. Two elements, g1, g2 e G, are congruent modulo the subgroup, S, if g1 – g2 e S,1 or Let G be an additive Abelian group and S be a subgroup of G, S ⊆ G; the equivalence classes for congruence modulo S are called the cosets of the subgroup S.2 Proposition. The order, m, of a subgroup, S ⊆ G, divides, n, the order of the group (Lagrange theorem). Proof. The elements of the subgroup are S = {s1, s2,…, sm}. Coset i consists of the elements congruent with si and is denoted as Ssi, and coset j consists of the elements congruent with sj and is denoted as Ssj. Let f be the mapping The function, f, maps distinct elements of the coset, Ssi, into distinct elements of the coset, Ssj. All elements of Ssj are mapped onto. Thus, f is a one-to-one map. This implies that the two cosets, and then obviously all cosets, have the same cardinality, m. But congruence does partition the set, G, of n elements into a number of disjoint cosets and n = (number of cosets of S) × m. We now present an alternative definition of the cosets useful for syndrome calculation and decoding of linear codes. We discuss only binary codes, codes over GF(2n), but similar definitions and properties apply to codes over GF(qn). Throughout this section, Vn is GF(2n), and we consider binary n-tuples; the code, c, is a subset of GF(2n). Let c be an [n, k] code, c c GF(2n). The coset of the code, C, containing v GF(2n) is the set v + c = {v + c, c c}. The binary n-tuple of minimum weight in Ci is called the leader of the coset Ci. The decoding procedure we present later is based on the following two propositions. Proposition. Two binary n-tuples are in the same coset if and only if they have the same syndrome. Proof. Recall that if H is the parity-check matrix of the code, c, and c c, then HcT = 0. The syndrome corresponding to an n-tuple, v GF(2n), is σv = HvT. If the two binary n-tuples are in the same coset, they can be written as v1 = c1 + e and v2 = c2 + e, with c1, c2 c, e GF(2n). The two syndromes are equal: If the two n-binary tuples, v1 and v2, have the same syndrome, then Because (v1 – v2) is a codeword, v1 and v2 must be in the same coset. It is easy to see that there is a one-to-one correspondence between the coset leaders and the syndromes. Proposition. Let C be an [n, k] binary linear code. C can correct t errors if and only if all binary n-tuples of weight, t, or less, are coset leaders. Proof. To prove that this condition is necessary, assume that c has the distance, d = 2t + 1; thus, it can correct, t, errors. Let v1, v2 e GF(2n); if v1 and v2 belong to the same coset, then the binary n-tuple, (v1 – v2), is a codeword, (v1 – v2) e c, according to the definition of the coset and to the previous proposition. If their weight is at most, t, w(v1), w(v2) ≤ t, then the weight of the codeword, (v1 – v2), is w(v1 – v2) ≤ 2t, which contradicts the fact that the distance of the code is d = 2t + 1. Thus, if the code can correct t errors, all binary n-tuples of weight less than or equal to t must be in distinct cosets and each one can be elected as coset leader. To prove that this condition is sufficient, assume that all binary n-tuples of weight less than or equal to t are coset leaders but there are two codewords, c1 and c2, such that d(c1, c2) ≤ 2t. There exist two coset leaders, v1 and v2, such that w(v1), w(v2) ≤ t, and c1 + v1 = c2 + v2. Indeed, we could choose them such that c1 – c2 = v2 – v1. This is possible because d(c1, c2) ≤ 2t and also d(v1, v2) ≤ 2t as each coset leader has a weight of at most, t. But c1 – c2 = v2 – v1 implies that the two syndromes are equal, . This is a contradiction because we assumed that two coset leaders, v1 and v2, are in different cosets. Thus, the assumption that d(c1, c2) ≤ 2t is incorrect and the code is capable of correcting t errors. Example. Let C be a [7, 4, 3] code, with the parity-check matrix Table 4.3 shows the correspondence between the coset leaders and the syndromes for code c. Next, we discuss an efficient decoding procedure for a linear code c. Table 4.3. The Correspondence Between the Coset Leaders and the Syndromes \| Coset Leader \| Syndrome \| --- \| \| 0000000 \| (000)T \| \| 0000001 \| (111)T \| \| 0000010 \| (011)T \| \| 0000100 \| (110)T \| \| 0001000 \| (101)T \| \| 0010000 \| (001)T \| \| 0100000 \| (010)T \| \| 1000000 \| (100)T \| Standard Array Decoding Consider an [n, k] code, c = {c1, c2,…,c2k}, over GF(2). Let V2n be the vector space of all n-tuples over GF(2). c is a subgroup of order 2k of V2n, with 2n elements. According to the Lagrange theorem, there are 2n/2k = 2n–k cosets of c denoted as {c0,c1,…,Ci,…,c2n-k-1}. We start with the coset leader, £0 = 00 … 0, the all-zero n-tuple, and then construct the coset c0 = {£0, c1, c2,…,c2k}. Then we select as coset leaders n-tuples of weight 1. After we select all n-tuples of weight 1 as coset leaders, we move to select those n-tuples of weight 2 that do not appear in any previously constructed coset and continue the process until all 2n–k cosets have been constructed. Once we select the leader of a coset j, £j, the i-th element of the coset is £j + ci. The cosets could be arranged in order as rows of a table, with 2n-k rows and with 2k columns. This table is called a standard array for code c. Example. Consider the binary [6, 3] code, with distance 3 and the generator matrix The standard array has in the first row the codewords, c1,c2,c3,c1 + c2,c1 + c3,c2 + c3,c1 + c2 + c3, and in the first column, the coset leaders, \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| 000000 \| 001011 \| 010110 \| 100101 \| 011101 \| 101100 \| 110011 \| 111000 \| \| 000001 \| 001010 \| 010111 \| 100100 \| 011100 \| 101101 \| 110010 \| 111001 \| \| 000010 \| 001001 \| 010100 \| 100111 \| 011111 \| 101110 \| 110001 \| 111010 \| \| 000100 \| 001111 \| 010010 \| 100001 \| 011001 \| 101000 \| 110111 \| 111100 \| \| 001000 \| 000011 \| 011110 \| 101101 \| 010101 \| 100100 \| 111011 \| 110000 \| \| 010000 \| 011011 \| 000110 \| 110101 \| 001101 \| 111100 \| 100011 \| 101000 \| \| 100000 \| 101011 \| 110110 \| 000101 \| 111101 \| 001100 \| 010011 \| 011000. \| Proposition. Consider an [n, k] code, C = {c1, c2,…, cqk–1}, over GF(2). Let SA be a standard array, with cosets Ci : i e {1, 2n-k} and coset leaders £i: i ∈ {1, 2n-k}. The (i + 1, j + 1) entry is £i + cj, as described earlier. Then ∀h ∈ {0, 2k}: In other words, the j + 1 entry in column, i, is closer to the codeword at the head of its column, cj, than to any other codeword of c. Proof. Recall that(¿) is the weight of the codeword, c¿. Then from the definition of the entries of the standard table, Coset Ci is obtained by adding the coset leader, £i, to the codewords of C, ci ∈ C. Clearly, the n-tuple, (£i + cj – ch), is a member of the same coset, Ci because cj – ch c. Yet, by construction, the coset leader, £i, is the n-tuple with the minimum weight in Ci: Proposition. The following procedure based on a standard table can be used to decode a received n-tuple, v, of a linear code: 1. : Locate v in the standard table. 2. : Correct v as the codeword at the top of its column. Example. If we receive v = 111101 we identify it at the fifth element of the last row of the standard table constructed for the previous example and decode it as the codeword at the top of the fifth column, c = 011101. In the next section, we discuss the limitations imposed on our ability to construct “good” linear codes. View chapterExplore book Read full chapter URL:
13302	https://runestone.academy/ns/books/published/orcca/section-special-solution-sets.html	Skip to main content Section 2.4 Special Solution Sets Objectives: PCC Course Content and Outcome Guide MTH 60 CCOG 2.3 Most of the time after solving a linear equation in one variable, you end with a direct statement of what the variable must equal. For example, concludes with . There is only one solution, in this case. Similarly, after solving a linear inequality, you typically end with a statement like , and the solution set is represented with either interval notation like , or with set-builder notation like . Visually on a number line, it is either the left half or the right half of the number line. Occasionally you will have a linear equation or inequality that doesn’t work this way. A linear equation might have many solutions, not just one. Or it might have none at all. An inequality might also have no solutions. Or maybe every number on the entire number line will be a solution. In this section we explore these kinds of equations and inequalities, where the solution set is “special”. Subsection 2.4.1 Special Solution Sets Recall that for the equation , there is only one number which will make the equation true: . This means that the solution is , and we write the solution set as . We say the equation’s solution set has one element, . We’ll now explore equations where all real numbers are solutions, and other equations where no real number is a solution. Example 2.4.2. Solve for in . To solve this equation, we need to move all terms containing to one side of the equal sign. We set out to remove the from the right side of the equation, by subtracting from each side: But now is no longer present in the equation. Our effort to remove from the right side coincidentally removed it from the left side too. What value can we substitute in for to make become true? That’s a trick question; you could never make become true. We say this equation has “no solution”. Or we say that the equation has an “empty solution set”. We can write an empty solution set as , or . (Be careful notto confuse the empty set symbol with the number zero, .) The equation isunambiguously false no matter what might be. This shows us there is no solution to the original equation. Example 2.4.3. Solve for in . We will move all terms containing to one side of the equal sign: Once again, is no longer present in the equation. What value can we substitute in for to make be true? This is another trick question; any value for would work. This means that all real numbers are solutions to the equation . We say this equation’s solution set contains all real numbers. We can write this set using interval notation as , or use as an abbreviation for the set of all real numbers. The equation isunambiguously true no matter what might be. This shows us that all real numbers are solutions to the original linear equation. Remark 2.4.4. What would have happened if we had continued the solving process after we obtained in Example 3? All we found was another unambiguously true equation, and we still conclude that all real numbers are solutions. Warning 2.4.5. Note that there is a very important difference when our process ends with , compared to when it ends with . The first equation is true for all values that might have, and the solution set is . The second situation has only one solution, , and the solution set is . Example 2.4.6. Solve for in the inequality . To solve for , we aim to eliminate that on the right side by subtracting from each side: We again find ourselves with the variable completely eliminated from both sides. What values of would make the inequality true? The answer is that all values of make , which we know is a strange sounding sentence. So our solution set is all real numbers, which we can write as or . Example 2.4.7. Solve for in the inequality . To solve for , we aim to eliminate the from the right side by adding to each side: And yet again, the variable has gone missing. We can ask ourselves, “For which values of is true?” The answer is that this is impossible. There is no solution to this inequality. We can write the solution set using or just say that there are “no solutions”. Let’s summarize these two special cases that sometimes arise when solving linear equations and inequalities. Subsection 2.4.2 Further Examples These examples may have no solutions or all real numbers as the solution set. Each example calls back to the fundamental steps for solving linear equations and inequalities that we learned in Section 1, Section 2, and Section 3. Checkpoint 2.4.9. Solve for in . Explanation. We recall the technique from Section 3 where we clear denominators by multiplying each side of the equation by the least common denominator. Here, we will multiply each side by . After that, we’ll be able to simplify each side of the equation and continue: The statement is unambiguously false, so the equation has no solution. Checkpoint 2.4.10. Solve for in the equation . Explanation. A lot could be simplified on each side before continuing, by distributing and combining like terms. From here, we subtract from each side: As the equation is outright false, there is no solution to this equation. Checkpoint 2.4.11. Solve for in the inequality . Explanation. We start by multiplying each side of the inequality by the LCD, which is . Then we can continue: As the equation is true for all values of , all real numbers are solutions to the original inequality. So the solution set is , or just . Reading Questions 2.4.3 Reading Questions 1. shortanswer (orcca_3_how-many-solutions-including-special) 2. shortanswer (orcca_3_how-do-you-know-no-solution) 3. shortanswer (orcca_3_how-do-you-know-all-are-solutions) Exercises 2.4.4 Exercises Notation 1. What is one valid way to communicate that there are no solutions to an equation? (There are several correct answers.) 2. What is one valid way to communicate that all real numbers are solutions to an equation? (There are several correct answers.) Skills Practice Equations with Special Solution Sets. Solve the equation for its variable. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. Inequalities with Special Solution Sets. Solve the inequality for its variable. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. Challenge 41. Fill in the right side of the equation to create a linear equation that meets the description. Create a linear equation with solution set . Create a linear equation with infinitely many solutions. You have attempted 1 of 48 activities on this page. PrevTopNext
13303	https://www.youtube.com/watch?v=X_07h29BSlQ	MIT 2022 Integration Bee, Quarterfinal (Floor/Ceiling Functions)-Part 2 Everyday Mathematics 2770 subscribers 32 likes Description 1248 views Posted: 23 Nov 2023 We move to part 2 of the ceiling/floor functions of the MIT 2022 integration bee quarterfinal. This is a problem involving nested ceiling functions of x. We hope you enjoy following along. #mit #integration #calculus #ceiling #floor #mathematics #maths 2 comments Transcript: welcome welcome to another session of Everyday Mathematics here at Everyday Mathematics as we always say we do enjoy solving the harder problems uh but above all we also see and appreciate the beauty in the simpler problems so today uh we move from uh part one to part two on the fla caling function uh series of the MIT 2022 integration B quar ay and today's problem is really intricate uh even from just uh the visuals so looking at our today's problem it is the integral of uh the ceiling function of one over the ceiling function of x - x uh to the power negative ciling function of x from X = 0 to X = to Infinity so uh before I jump on to the solution I would like to uh really thank our subscribers we thank you for uh the support we thank you for the continued interest uh we are seeing uh just uh incredible numbers in terms of the views and also the comments that you leave behind uh for our firsttime visitors we really encourage you to subscribe to our Channel um we will promise you that we'll keep uh feeding you with uh very interesting solutions to very interesting problems uh for visitors who are here for the second third or fourth or fifth time but I haven't gotten uh that uh extra extra motivation to subscribe to our Channel we can only say we will use your support and uh you won't do great um thank you so much so let's uh quickly jump into the solution um so this problem is obviously a nested problem of seiling functions and what is a seiling function a sing function uh if taken in between when X is between K and K + one that means that the ceiling function is going to be a constant uh sing function of X going to be just k + 1 um and so um what that means for problem is that uh this sing function of one over uh x minus X to the^ negative sing function of X is just a summation uh of uh strips so we have the strip when X is between K and K + one when X is between k + 1 and k + 2 or k minus1 to 2 K so we have those strips and we see here that we have infinite number of uh strips from K = to Z is integer intervals up to Infinity uh such that for every strip we are integrating obviously from K to k + one one over uh in place of uh sing function of X we have k + 1 and since this there's a negative here so that's a matter of just getting the reciprocal of this number here which is now one over um seing function of 1 / k + 1 - x^ k + one since we still have an X component here and still a sing function here we have to keep working uh this um content the content of the function here is also discret and so it it's when we consider this content to be between that is 1 / k + 1 - x between another integer interval L and L + one um what happens there is that this sing function equals to L + one that is uh the integer above um the number between the real number between this interval now um what that means is that um we have to figure out what interval X must lie in between so that uh this expression is between this integer interval so we have to take the the reciprocal uh of this bounce so that means that k + 1 - x must be between one over L + 1 and 1 / L and since we are taking the inverse so we have to swap uh this uh inequality sign which means now if we apply negative sign again we swap them since we know the negative values the absolute higher value is less than the absolute lower value so just swap around and then now we add k + 1 to both sides now k + 1 - 1 / L less than X or and X less than k + 1 - 1 / L +1 in this interval is when uh this expression holds so what that means is that our integration since we had created a discretization uh interval here uh under this integral sign we also still have um the indexing of L taken into consideration so we have a nested sum summation here uh that is for L = to 1 to Infinity of the integral in that of k + 1 - 1 / L to k + 1 1 - l/ 1 one / L + one and so uh whatever is inside this uh ceiling function now becomes L + one as we had indicated here and that's now L + one the denominator the power k + 1 and since um there's no X here so this is just taken to be a constant in this interval and so the difference let's continue with the integration uh the integration now is results into obviously x with the limits being k + 1 - 1 / l k + 1 - 1 / L + one so the difference of those limits since we're looking at just a an X there um it's taken into consideration and so uh we have a summation from K = 0 to Infinity inside that we have summation of L is = 1 to infinity or 1 over L + 1^ k + 1 / 1/ L -1 over L + 1 so that is our uh sum we multiply that through we get that our nested summation here is of one/ L L + 1^ k + 1us 1 L + 1^ k + 2 so this is the same as when L is equal to 1 this is obviously one here and then this is 2^ k + 1 minus 2 the^ k + 2 plus when L is = 2 is 2 3 ^ k + 1 - 1 / 3 + k + + 2 K to the^ k + 2 plus when L is = 3 3 4 the^ k + 1 minus 1 over 4 + k + 2 okay then um considering the K summation so for every um bracket sign here we have one/ 2 when K is = to 0 1 / 2 - 1 / 2^ 2 + 1 / 2^ 2 - 1 / 2^ 3 this is when K = to to 1 and and so on and now we realize that here is two one to the power 2 SAR and the next time this cancels out and this process continues so at the end of the day you're only left with a half for when um L is equals to two now here we have 2 to the power 1/ 2 3 when K is = to 0 minus 3^ 2 + 1 / 2 3^ 2 - 1 / 3^ 3 this is when K is equals to one and so on and so on and as we can see this interacts with this uh so we have positive 1 / 2 3^ 2 - 1 / 3² this just becomes half of 1/ 1 over 3^ 2 and that continues we'll look at how that will be treed later so the same thing here when L is = to 2 so 1/ 3 4 this is when K is = 0 - 1 4^ 2 + 1 3^ 3 4^ 2 - 1 um over 4^ 3 and here now we see this interacts with this so 1 / positive 1 over 3 4^ 2 - 1/ 4 2 that means that the difference here will be-2 over3 okay and so we continue on and on and on now the result here is interesting so we have for for the first time we have only half left for the second time we have 1/ 2 3 - 1 / 2 3^ 2 + 1 / 2 3^ 3 and so on and so on here for the third term 1 3 4- 2 3 4 2 + 2 3 4 Cub Plus on and on and on and then + 1 4 5 - 3 4 5 2 + 3 4^ 5^ 3+ On and On To Infinity so uh here we see that a half can be pulled out and now we have a summation of a geometrical progression of initial term 1 / 3^ s but the common ratio is 1 over 3 so we have 1 / 3^ 2 minus divided 1 1 - A3 the second one we have 1 / 3 4 minus we pull 2 over three out and then we have a geometrical Series where 1 over 4 S is the initial term and 1 over 4 is the common ratio and this series goes to Infinity so we have 1 over 4^ 2 over 1 minus uh quar inside that this bracket and so on and so on and so on so the next obviously term 1 over five squar is the initial term and one over five is the common ratio and the next one 1/ 6^ squ is the initial term then 1/ 6 is the common ratio um so taking that into consideration what we left here is 1/ 2 3 - 1 / 2 here you get 1 / 6 + 1 over 3 4 minus we pull 2 over 3 out we have 1 over 12 in the bracket 1 over 4 5 we pull minus 34s out and 1 over 20 is in bracket 1 over 5 6 - 4 5 into bracket 1 over 30 now you compute that we get a half if you pull a half out we have a third minus a 6 we have a third pulled out we have A4 minus a 6 we have a quarter pulled up we have 1 over 5 minus 3 over 20 you have a fifth pull up we 1 over five 1 6 - 4 30 and then um we have now half plus a half uh the difference here is 1/ 6 and then a third the difference here is 1 over 12 the difference here here is 1 over 20 difference here is one /3 here and you see the constant a half a third a fourth a fifth so now now we can also express it differently this is the same as 1 / 1 2 a half this is when we pull a half out we left with a third so we have a half 1 over 2^ 2 a thir when we pull another third out here so that we have 1 over 3 S we left with a quarter so 1 over 3^ 2 A4 and then the same thing 1 over 4 2 um 5th plus 1 5 squ time 6 so we can express that in a generalized series format this is the same as uh the summation of K isal to 1 to Infinity of the squared part is K K2 times k + 1 for example for 1 s and two this is K squar and then 1 + 1 is 2 and this is how we've created that you can simplify that further by uh pulling one/ k out and then having one over K k + 1 such that when you use partial fractions you have 1 / K 1 / K - 1 / k + 1 1 / K of course we have 1 / k^ 2 - 1 / k + 1 and so we have 1 / k^ 2 minus again in this bracket here we have 1 / Kus 1/ k + 1 so we're going to since we have this expression we can use what we call the baso problem the Basel problem tells us that if you have a summation of um 1 / k 2 from K = 1 to Infinity that summation is equal to pi^ 2/ 6 this is a well-known problem you can go online and and check uh the different derivations that are out there uh but this is what somebody um who has some indepth knowledge of algebra or um series should should know this is one of the really popular series so we see that this fast time here can decompose or can be approx more abbreviated to pi over squ pi^ squ over 6 minus and then the second time we have 1/ 1 when K is = to 1 1 / 2us 1 / 2 Then when K is = to 2 is 1 / 2 - 1 over 3 when K is = to 3 we have 1 over 3 minus a quar and so on and so on and but we see that this is a telescoping series because this part here can with this and this part cancels with this and that continues onwards and we left the initial term which is one so the final um answer here is pi^ 2 / 6 minus one is the answer for the very complicated integral of um sing function one/ selling function of x - x the power negative selling function of x from x0 to Infinity so this is actually a very beautiful problem I hope you are able to follow if you have any questions uh please please uh feel free to leave uh those questions in the comment below uh but it was an absolute pleasure um solving this problem um and really if you haven't subscribed to our Channel we do encourage you to subscribe to our Channel thank you so much until next time
13304	https://fctemis.org/notes/11081_Trigo%201.pdf	GE LESSON PLAN (MODERN TEACHING APPROACH) A. LESSON INFORMATION DATE: CLASS: TIME: DURATION: PERIOD: SUBJECT: Mathematics THEME: Geometry TOPIC: Trigonometrical Ratios SUB-TOPIC: Problem Solving SPECIFIC OBJECTIVES: At the end of the lesson, the student should to: (i) Give sine, cosine and tangent ratios. (ii) Solve problems involving right angle triangle. INSTRUCTIONAL RESOURCES: Right angle triangle. B. PRESENTATION: The teacher presents the lesson with the steps below: STEP I: Identification of prior idea Mode: Entire class Teacher’s Activities: Instructs students to identify the right-angled triangle in their mathematical set. Students’ Activities: Students identifies the right-angled triangle in their mathematical set. The trig ratios can be used to find lots of information, and one of their main purposes is to help solve triangles. To solve a triangle means to find the length of all the sides and the measure of all the angles. This lesson will cover how to use trig ratios to find the side lengths of a triangle. STEP II: Exploration Mode: Entire class Teacher’s Activities:The teacher leads students to give the trigonometrical ratios. Students’ Activities: Students gives the trigonometrical ratios. , hyp opp  sin STEP III: Discussion Mode: Entire class Teacher’s Activities: Leads students to solve problems involving right angles Students’ Activities: Students solve problems involving right angles. There are three steps: 1. Choose which trig ratio to use. - Choose either sin, cos, or tan by determining which side you know and which side you are looking for. 2. Substitute - Substitute your information into the trig ratio. 3. Solve - Solve the resulting equation to find the length of the side. Example: 1. Find b. 1: Choose which trig ratio to use. First, we know we must look at angle B because that is the angle we know the measure of.(Now, you could find the measure of angle A and then use that but that's less reliable since you could make a mistake.) So, looking at angle B, we want to identify which sides are involved. We know one side is 8m, and that side is adjacent to angle B. The side we're looking for is opposite angle B. So we need to choose the trig ratio that has opposite and adjacent. This of course is the tangent. 2: Substitute Next, we write our trig ratio: Then, we substitute in the angle and the side we know: 3: Solve Now move the 8 to the other side by multiplying both sides by 8: And use a calculator to find the answer. Well round to the nearest tenth: 3.7 m. STEP IV: Discussion Mode: Entire class Teacher’s Activities:The teacher leads students to solve problem involving exterior angles of polygon Students’ Activities: Students solves problem involving interior angles of polygon. Find c in the question above. Now that we know two sides, you could use the Pythagorean Theorem to find the third. But that's less reliable because if you made a mistake on side b, then side c will also be wrong. So we are going to repeat the same process for side c. 1: Choose the trig ratio to use. We're still using angle B. 8m is the adjacent and c is the hypotenuse. The trig ratio that uses the adjacent and hypotenuse is the cosine. 2: Substitute Write our trig ratio: Then, we substitute in the angle and the side we know: 3: Solve Since our variable is on the bottom, we can start by cross multiplying: Then we'll divide both sides by cos 25° .: And use a calculator to find the answer. Well round to the nearest tenth: 8.8 m. EVELUATION: The teacher assesses the lesson through the following questions: Find the following. (1) a 2) c 3) x 4) y C. CONCLUSION: The teacher goes round to assess the student’s work and gives correction on the board for student to copy. Answers: 1) 7.4 in 2) 9.5 in. 3) 7.0 ft 4) 5.7 ft. D. ASSIGNMENT: Go through today’s work at home. E. REFERENCES: a. H. N. Odogwu etal (2012), New Concept Mathematics for Senior Secondary Schools 1, Pages 164 – 167. b. J. Olowofeso (2012), Multipurpose Mathematics for SSCE, pages 136 – 139. c.
13305	https://www.math.cmu.edu/~mlavrov/arml/16-17/number-theory-09-11-16.pdf	Modular Arithmetic and Divisibility Number Theory Annie Xu and Emily Zhu September 11, 2016 1 Introduction Deﬁnition 1 (Divisibility). An integer a is said to be divisible by some nonzero integer b if there exists an integer c such that a = bc. Alternatively, for b ̸= 0, a b is an integer. Deﬁnition 2 (Euclidean Algorithm). Given integers a, b, the series of divisors q1, q2, . . . such that a = bq1 + q2, b = q2q3 + q4, q2 = q4q5 + q6, . . . (see example). The ﬁnal value (when the other is 0) gives gcd(a, b), i.e. the greatest common divisor of a and b. Example 3. Find gcd(126, 224). Solution. gcd(126, 224) = gcd(126, 224 −126) 224 = 1 × 126 + 98 gcd(126, 98) = gcd(98, 126 −98) 126 = 1 × 98 + 28 gcd(98, 28) = gcd(28, 98 −3 · 28) 98 = 3 × 28 + 14 gcd(28, 14) = gcd(14, 28 −2 · 14) 28 = 2 × 14 + 0 = gcd(14, 0) Thus, gcd(126, 224) = 14 . Deﬁnition 4 (Relatively Prime). Given integers a, b, they are called relatively prime or coprime if gcd(a, b) = 1. Deﬁnition 5 (Prime). An integer p is called prime if when p divides a product ab (where a, b are integers), then p divides a or p divides b. Equivalently p is prime if p = ab (where a, b are integers), then either a = 1 or b = 1. Theorem 6 (Fundamental Theorem of Arithmetic). Every nonzero integer can be written uniquely (up to order) as a product of primes. Deﬁnition 7 (Modular Arithmetic). Given integers a, b, c, b ̸= 0, a ≡c (mod b) if b divides (a−c). Example 8. 5 ≡2 (mod 3) 17 ≡12068357 (mod 10) 54 ≡42 ≡0 (mod 6) 2 ̸≡1 (mod 3) 4 + 1 ≡29 + 6 (mod 5) 3 × −1 ≡19 × 15 (mod 8) 2 Problems 1. Using modular arithmetic, show that 3 divides n if and only if 3 divides the sum of the digits of n. Do the same for 9. Can you ﬁnd something similar for 11? 1 2. Find gcd(221, 299) and gcd(2520, 399). 3. Compute the remainder when 333 + 999 and 3333 × 7777 are divided by 5. 4. How many steps does it take the Euclidean Algorithm to reach (1, 0) when the input is (n + 1, n)?1 5. Let n be a positive integer. Construct a set of n consecutive positive integers that are not prime.1 6. Find all positive integers n such that (n + 1) divides (n2 + 1).2 7. Find all primes in the form n3 −1.2 8. What is the largest positive integer n for which (n + 10) divides n3 + 100?3 9. Show that 1 . . . 1 \| {z } 91 ones is composite.2 10. A year is a leap year if and only if the year number is divisible by 400 (such as 2000) or is divisible by 4 but not 100 (such as 2012). The 200th anniversary of the birth of novelist Charles Dickens was celebrated on February 7, 2012, a Tuesday. On what day of the week was Dickens born?4 11. What is the largest prime factor of 7999488?5 12. An n-digit number is cute if its n digits are an arrangement of the set {1, 2, . . . , n} and its ﬁrst k digits form an integer that is divisible by k, for k = 1, 2, . . . , n. For example, 321 is a cute 3-digit integer because 1 divides 3, 2 divides 32 and 3 divides 321. How many cute 6-digit numbers are there?6 13. An old receipt has faded. It reads 88 chickens at the total of $x4.2y, where x and y are unreadable digits. How much did each chicken cost?2 14. Find the smallest positive integer such that n 2 is a square and n 3 is a cube.2 15. How many primes have alternating 1s and 0s in base 10 (like 101)?7 16. If a, b ∈N such that a b = 1 −1 2 + 1 3 −1 4 + · · · − 1 1318 + 1 1319, show that 1979 divides a.8 1From Mathematical Thinking by John P. D’Angelo and Douglas B. West 2From Number Theory for Mathematical Contests by David A. Santos 3From AIME 1986 4AMC 10/12A 2012 5PUMaC 2011 6AHSME 1991 7Putnam 1989 8IMO 1979 2
13306	http://www.kentchemistry.com/links/GasLaws/densitygasesnotatSTP.htm	KentChemistry HOME \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| \| --- \| \| \| \| Google Custom Search More Gas Law links \| \| \| \| --- \| \| Gas Density, Molar Mass not at STP \| \| \| \| \| Another really handy rearrangement of the ideal gas equation can be used to find the molecular weight of an unknown gas . You’ll get a chance to practice using these in the problems at the end of the chapter. However, there is no need to memorize these last equations since they are all rearrangements of the ideal gas law. Okay, two more important laws and then we’re finished with our discussion of gases, and we move on to solutions. So, for gases, we speak of "standard gas density." This is the density of the gas (expressed in grams per liter) at STP. If you discuss gas density at any other set of conditions, you drop the word standard and specify the pressure and temperature. Also, when you say "standard gas density," you do not need to add "at STP." STP is part of the definition of the term. It does no harm to say "standard gas density at STP," it just looks a bit silly. \| Chemical Demonstration Videos \| \| \| \| \| \| --- \| \| \| \| More Gas Law links \| \| \| \| --- \| \| Gas Density, Molar Mass not at STP \| \| Gas Density, Molar Mass not at STP \| \| \| Another really handy rearrangement of the ideal gas equation can be used to find the molecular weight of an unknown gas . You’ll get a chance to practice using these in the problems at the end of the chapter. However, there is no need to memorize these last equations since they are all rearrangements of the ideal gas law. Okay, two more important laws and then we’re finished with our discussion of gases, and we move on to solutions. So, for gases, we speak of "standard gas density." This is the density of the gas (expressed in grams per liter) at STP. If you discuss gas density at any other set of conditions, you drop the word standard and specify the pressure and temperature. Also, when you say "standard gas density," you do not need to add "at STP." STP is part of the definition of the term. It does no harm to say "standard gas density at STP," it just looks a bit silly. \| Another really handy rearrangement of the ideal gas equation can be used to find the molecular weight of an unknown gas . You’ll get a chance to practice using these in the problems at the end of the chapter. However, there is no need to memorize these last equations since they are all rearrangements of the ideal gas law. Okay, two more important laws and then we’re finished with our discussion of gases, and we move on to solutions. So, for gases, we speak of "standard gas density." This is the density of the gas (expressed in grams per liter) at STP. If you discuss gas density at any other set of conditions, you drop the word standard and specify the pressure and temperature. Also, when you say "standard gas density," you do not need to add "at STP." STP is part of the definition of the term. It does no harm to say "standard gas density at STP," it just looks a bit silly. Chemical Demonstration Videos
13307	https://www.epa.gov/sites/default/files/2014-03/documents/human_health_effects_from_chronic_arsenic_poisoning_3v.pdf	Journal of Environmental Science and Health Part A, 41:2399–2428, 2006 Copyright C ⃝Taylor & Francis Group, LLC ISSN: 1093-4529 (Print); 1532-4117 (Online) DOI: 10.1080/10934520600873571 Human Health Effects From Chronic Arsenic Poisoning– A Review Simon Kapaj,1 Hans Peterson,1 Karsten Liber,2 and Prosun Bhattacharya3 1The Safe Drinking Water Foundation, Saskatoon, Saskatchewan, Canada 2Toxicology Centre, University of Saskatchewan, Saskatoon, Saskatchewan, Canada 3KTH-International Groundwater Arsenic Research Group, Department of Land and Water Resources Engineering, Royal Institute of Technology (KTH), Stockholm, Sweden The ill effects of human exposure to arsenic (As) have recently been reevaluated by government agencies around the world. This has lead to a lowering of As guidelines in drinking water, with Canada decreasing the maximum allowable level from 50 to 25 µg/L and the U.S. from 50 to 10 µg/L. Canada is currently contemplating a further decrease to 5 µg/L. The reason for these regulatory changes is the realization that As can cause deleterious effects at lower concentrations than was previously thought. There is a strong relationship between chronic ingestion of As and deleterious human health effects and here we provide an overview of some of the major effects documented in the scientiﬁc literature. As regulatory levels of As have been decreased, an increasing number of water supplies will now require removal of As before the water can be used for human consumption. While As exposure can occur from food, air and water, all major chronic As poisonings have stemmed from water and this is usually the predominant exposure route. Exposure to As leads to an accumulation of As in tissues such as skin, hair and nails, resulting in various clinical symptoms such as hyperpigmentation and keratosis. There is also an increased risk of skin, internal organ, and lung cancers. Cardiovascular disease and neuropathy have also been linked to As consumption. Verbal IQ and long term memory can also be affected, and As can suppress hormone regulation and hormone mediated gene transcription. Increases in fetal loss and premature delivery, and decreased birth weights of infants, can occur even at low (<10 µg/L) exposure levels. Malnourished people have been shown to be more predisposed to As-related skin lesions. A large percentage of the population (30–40%) that is using As-contaminated drinking water can have elevated As levels in urine, hair and nails, while showing no noticeable clinical symptoms, such as skin lesions. It is therefore important to carry out clinical tests of As exposure. Received January 3, 2006. Address correspondence to Simon Kapaj, The Safe Drinking Water Foundation, 912 Idylwyld Drive, Saskatoon, SK; S7L OZ6, Canada; E-mail: skapaj@sasktel.net 2399 2400 Kapaj et al. Factors combining to increase/decrease the ill effects of As include duration and magnitude of As exposure, source of As exposure, nutrition, age and general health status. Analytical determinations of As poisoning can be made by examining As levels in urine, hair and toenails. Communities and individuals relying on groundwater sources for drinking water need to measure As levels to ensure that their supplies are safe. Communities with water As levels greater than 5 µg/L should consider a program to document As levels in the population. Key Words: Arsenic; Drinking water; Chronic toxicity; Cancer; Hyperpigmentation; Hair and toenail arsenic. INTRODUCTION Human health effects caused by exposure to arsenic (As) have been highlighted by recent regulatory initiatives in the U.S. This includes three panel reviews: The National Academy of Sciences, The National Drinking Water Advisory Council, and the U.S. Environmental Protection Agency’s (EPA’s) Science Advisory Board. This work led the US EPA to lower the maximum As contaminant level in drinking water to 10 µg/L. All public water systems in the U.S. must comply with the 10 µg/L standard beginning January 23, 2006. A brief review of the panels ﬁndings are highlighted below.[1,2] Human health effects caused by As exposure were key in panel assess-ments that resulted in the lowered U.S. As guideline. Health effects were dependent on the duration and dose of exposure. The National Research Council (NRC, the operating arm of the National Academy of Sciences) conﬁrmed that the chronic effects of inorganic As exposure via drinking water include skin lesions, such as hyperpigmentation, and respiratory symptoms, such as cough and bronchitis. The cardiovascular, gastrointestinal and urinary systems were some of the other systems most affected in humans. This review also concluded that there was sufﬁcient evidence to link bladder and lung cancers with ingestion of inorganic As. In addition, ailments linked to As included increased risk of high blood pressure and diabetes. The NRC accepted the fact that there is a need for more data to conﬁrm the link between the As ingestion and negative effects on reproductive outcomes. The NRC underlined that there are differences in outcomes due to factors contributing to risk, such as exposure in different population groups. Evidence adds to the fact that As exposure may cause adverse effects, but this evidence is not conclusive because studies lack information on lifestyle, and other exposures that could affect health outcomes. The NRC also concluded that infants and children may be at greater risk for both cancer and non-cancer effects because of greater consumption via drinking water on a body-weight basis. The US EPA indicated that lowering of the As guideline from 50 to 10 µg/L could prevent deaths from bladder, lung and skin cancers, and from heart disease. Review of Chronic Arsenic Poisoning 2401 Similarly, Canada is also reevaluating As contamination of drinking water. The proposed maximum acceptable concentration for As in drinking water is 5 µg/L. However, Health Canada has set the current guideline at 25 µg/L. This guideline has been put in place temporarily until improved treatment technologies have been developed to further reduce As levels in drinking water to 5 µg/L. As is of greatest concern in groundwater supplies where it is a naturally occurring mineral. Reducing the current level of As in drinking water will have a signiﬁcant impact on communities across Canada because groundwater is a major water source in many rural, small communities. The Word Health Organization (WHO) has also reviewed As guidelines in drinking water and established a provisional guideline of 10 µg/L after concluding that inorganic As is a human carcinogen and that the main route of exposure is through drinking water and food. The interest in As has also resulted in the publication of several re-cent books, including Aquatic Arsenic Toxicity and Treatment and Natural Arsenic in Groundwater: Occurrence, Remediation and Management. Several review articles have also been published, including Yoshida et al., Luster and Simeonova, Watanabe, Tchounwou et al., Rossman et al., Mahata et al., Kitchin, and Ratnaike. Throughout this review it is important to mention that, whenever not stated otherwise, “arsenic” means total inor-ganic As. PATHWAYS OF ARSENIC EXPOSURE Arsenic Exposure Through Drinking Water Groundwater with elevated concentrations of As has been recognized as a problem of global concern.[16-18] As contamination of groundwater is one of the principal pathways of human exposure to inorganic As. Elevated concentrations of As have been reported from several regions of the world (Table 1) resulting primarily from natural sources, such as erosion and leaching from geological formations, although sometimes from anthropogenic sources, such as uses of As for industrial purposes, mining activities and metal processing, and application of pesticides and fertilizers containing As. The risk of As contamination is generally much higher in groundwater compared to surface water. Natural occurrence of As in groundwater (>10 µg/L) is reported from many parts of the United States, such as California, Alaska, Arizona, Indiana, Idaho, Nevada, Washington, Missouri, Ohio, Wisconsin, and New Hampshire.[20–25] Higher concentrations of As are also found close to areas of geothermal ﬁelds, uranium and gold mining. Natural occurrences of As have also been found in Canada, Argentina,[28–31] Mexico,[32–34] Chile,[35,36] Taiwan,[37,38] Table 1: Comparison of arsenic occurrences in groundwater from selected parts of the world (Courtesy of Naidu and Bhattacharya). Country/Region Area affected Depth of well Arsenic conc. (µg/L) Mechanism of contamination Bangladesh, BDP (52 districts) 118, 012 km2 8–260 m <2→900 Reduction of Fe-oxyhydroxides/Sulﬁde oxidation(?) in alluvial sediments West Bengal, India, BDP (8 districts) 34 000 km2 14–132 m <1–1300 Reduction of Fe-oxyhydroxides/Sulﬁde oxidation (?) in alluvial sediments China, Xinjiang Inner Mongolia (HAB) 4800 km2 Shallow/Deep <50–1860 Reducing environment in alluvial sediments Taiwan — Deep Up to 1820 Oxidation of pyrite in mine tailings Thailand (10 districts) 10 districts Shallow 120–6700 Oxidation of mine wastes and tailings Ghana 1600 km2 70–100 m Oxidation of arsenopyrite in mine tailings Argentina (Chaco-Pampean Plains) 10 million km2 Shallow aquifers 100–4800 Volcanic ash with 90% rhyolitic glass Chile — Shallow and deep wells 100–1000 Volcanic ash Mexico, Zimapan, Lagunera — Shallow and deep wells 300–1100 Oxidation of sulﬁde from mine wastes Hungary (Great Hungarian Plain) 4263 km2 80–560 m 25→50 Complexation of arsenic with humic substances USA Large areas 53–56 m 100→500 Desorption of arsenic from Fe-oxyhydroxides/Sulﬁde oxidation Canada (Nova Scotia) — 8–53 m 18–146 Oxidation of sulﬁdes United Kingdom (Cornwall) — Shallow wells >10 Oxidation of sulﬁdes from mine wastes 2402 Review of Chronic Arsenic Poisoning 2403 China,[39,40] Japan, southern Thailand, Ghana, Hungary, and Finland. Occurrence of As in groundwater of the Bengal Delta Plain in West Bengal, India and Bangladesh, is the region’s single largest emerging societal and environmental problem of the present century.[46–56] Similar As problems also exist in the Flood Plain aquifers of the Mekong Delta in Cambodia and the Red River Delta in Vietnam, where drinking water supplies are primarily based on groundwater resources.[58-62] Here, a population of over 20 million has resorted to groundwater use to meet agricultural productivity and increased drinking water demand.[57,60] Drinking Water Criteria for Arsenic Arsenic in drinking water can affect human health and is considered as one of the most signiﬁcant environmental causes of cancer in the world. Therefore, it is necessary to document the levels of As in drinking water, and its chemical speciation, and for establishing regulatory standards and guidelines. The FAO health limit for As in groundwater was until recently 50 µg/L, but in view of recent incidences of As poisoning in the Indian subcontinent, a decrease to 5–10 µg/L is being considered by a number of regulatory bodies throughout the world. The temporary WHO guideline for As in drinking water is 10 µg/L. This is based on a 6×10−4 excess skin cancer risk, which is 60 times higher than the factor normally used to protect human health. However, the WHO states that the health-based drinking water guideline for As should in reality be 0.17 µg/L. Previously, such low levels were not feasible to determine as many analytical techniques had detection limits of 10 µg/L, which is why the less protective guideline was adopted.[64–66] The US EPA drinking water standard for As was set at 50 µg/L in 1975, based on a Public Health Service standard originally established in 1942. On the basis of investigations initiated by the National Academy of Sciences, it was concluded that this standard did not eliminate the risks of skin, lung, and prostate cancer from long-term exposure to low As concentrations in drinking water. In addition, there are several non-cancer effects related to ingestion of As at low levels, which include cardiovascular disease, diabetes, and anemia, as well as reproductive, developmental, immunological, and neurological disorders. In order to achieve the EPA’s goal of protecting public health, recommendations were made to lower the safe drinking water limit to 5 µg/L, slightly higher than what is considered the technically feasible measurable level (3 µg/L). Recently, the US EPA has established a health-based, non-enforceable Maximum Contaminant Level Goal (MCLG) of zero As and an enforceable Maximum Contaminant Level (MCL) of 10 µg As/L in drinking water. This would apply to both non-transient, non-community water systems, as well as to the community water systems, as opposed to the previous MCL of 50 µg As/L set by the US EPA in 1975. However, the current 2404 Kapaj et al. drinking water guideline for As adopted by both the WHO and the US EPA is 10 µg/L. This is higher than the proposed Canadian and Australian maximum permissible concentrations of 5 and 7 µg As/L, respectively. Arsenic Exposure Through Coal Combustion and Incineration of Preserved Wood Products Combustion of high As bearing coals is known to be a principal pathway of As emission in Guizhou province of southwestern China.[70,71] Open coal-burning stoves used for drying chili peppers have been the principal cause for chronic As poisoning to a population of nearly 3,000. Fresh chili peppers have less than 1 mg/kg As, while the chili peppers dried over high-As coal ﬁres were reported to contain more than 500 mg/kg As. Consumption of other tainted foods, ingestion of kitchen dust containing as high as 3000 mg/kg As, and inhalation of indoor air polluted by As from coal combustion are the other causes of chronic As poisoning. A possible pathway for exposure through air-particulates is the incidental use of preserved wood in open ﬁres, indoors or outdoors. Incineration of preserved wood products, pressure treated with chro-mated copper arsenate was found to be a source of As contamination to the environment. The content of As in air-particulates from open ﬁres was found to exceed the German air quality standards by a 100-fold. The ashes, spread on lawns and vegetable cultivations, pose further risk to human health. In addition, tobacco smoke is another source of As emission in indoor environment. It is interesting to note that mainstream cigarette smoke contains 0.04 to 0.12 µg As per cigarette. HEALTH EFFECTS OF ARSENIC EXPOSURE Terminology Arsenicosis is a chronic illness resulting from drinking water with high levels of As over a long period of time. It is commonly known as As poisoning. Arseniasis means chronic arsenical poisoning, also called arsenicalism; the term arsenicism refers to a disease condition caused by slow poisoning with As. Despite the existence of recent reviews, there does not appear to be a concise overview of the human health issues caused by As. This review is an attempt to address that gap. In addition, we address analytical approaches that can be used to determine human As exposures even after the As has been removed from the drinking water. The review is aimed to help health workers, practicing rural physicians, water treatment plant operators, government agencies, and community groups dealing with issues of human As exposure Review of Chronic Arsenic Poisoning 2405 and effects. Together, we need to put the safeguards in place to avoid adverse human health effects from chronic As exposure. Arsenic and Cancer The International Agency for Research on Cancer (IARC) has listed As as a human carcinogen since 1980. Many researchers have underlined the potential risk that As in drinking water plays in human health. The positive association between As exposure and cancer has been evaluated by many researchers in different countries including the USA, Taiwan, Bangladesh, India, Argentina, and Chile, to name a few. This section highlights work performed during the last 4 to 5 years. In a recent publication, Centeno et al. report that As is a unique carcinogen. It is the only known human carcinogen for which there is adequate evidence of carcinogenic risk by both inhalation and ingestion. In a very detailed study spanning a 7-year period, Rahman et al. indicated that As-affected patients in West Bengal had severe skin lesions. It was not clear what number of patients suffered from cancers, because they were too poor to afford the investigations. However, patients that had premature death due to cancer had serious arsenical skin lesions prior to that. Also, in follow-up visits, people that were exposed to high levels of As from drinking water and/or food for many years were frequently developing cancer. These small communities in West Bengal use groundwater sources for drinking, and this study showed that intervention of water management is critical. Taiwanese studies investigated the risk association at 50 µg/L As in drinking water, the standard that was being reevaluated by the US EPA at that time. Data from Taiwan indicated that there is increased risk of internal cancers from As exposure through drinking water. In a follow-up study of 8102 residents from an arseniasis-endemic area in Northeastern Taiwan, the association between ingested As and risk of cancers of urinary organs was investigated. It indicated that residents being exposed to well water As for 40 years or more had greater chances of getting urinary tract cancer than residents that had less than 40 years of exposure. Conclusions from these studies suggested that the US EPA needed to revise the 50 µg/L As standard, which has now been done. It is believed that there is a long latent stage between the time that humans are exposed to As and ﬁnal cancer diagnosis.[79–81] In addition, Ferrecio et al. presented a positive correlation between ingestion of inorganic As and lung cancer in humans in Chile. It is already known that cigarette smoking is a main risk factor for lung cancer, but the authors found that cigarette smoking plus ingestion of As from drinking water had a synergistic effect. 2406 Kapaj et al. Skin Cancer A signiﬁcant relationship between As exposure and skin cancer has been observed. In a review, Rossman et al. pointed out that arsenite can play a role in the enhancement of UV-induced skin cancers. The mechanism of action may involve effects on DNA methylation and DNA repair. In addition, Luster and Simeonova reported epidemiological evidence indicating that As is associated with cancers of skin and internal organs, as well as with vascular disease. Bladder Cancer In a major U.S. study conducted on a population with chronic As exposure through drinking water, Steinmaus et al. did not ﬁnd a clear association between bladder cancer risk and exposure. The risks were lower than those in Taiwan with high As exposure. However, in the U.S. study there was an elevated risk of bladder cancer in smokers that were exposed to As in drinking water near 200 µg/L, compared with smokers consuming lower As levels. These data suggest that As is synergistic with smoking at relatively high As levels (200 µg/L). Steinmaus et al. highlighted that latency of As exposure causing bladder cancer can be very long (more than 40 years). Lung Cancer Hopenhayn-Rich et al. found that mortality from lung cancer was signiﬁcantly increased with increasing As ingestion. In addition, As and cigarette smoke are synergistic, thus increasing the risk of lung cancer. In a recent Taiwanese study, residents in arseniasis-endemic areas were followed during an 8-year period. An increased risk of lung cancer was associated with high levels of As exposure via drinking water. The authors suggested that reduction in As exposure should reduce the lung cancer risk in cigarette smokers. Southwest Taiwan has been a region that used wells with high As levels for the past 5 decades. Researchers looked at lung cancer mortality versus standard mortality ratio (SMR). Their study further indicated that the mortality from lung cancer declined after the levels of As in the well water were reduced. China is another country where millions of people are exposed to elevated levels of As. In the review of Xia and Liu, it was stressed that chronic arsenism in China is a serious health issue, which the authorities are now trying to tackle. Measures are being implemented to improve drinking water sources, patient treatment, and health education. However, in As-endemic areas it is predicted that cancer incidence may increase over the next 10– 20 years mainly due to previous exposures. This shows that urgent effective prevention is needed. Often in China, areas that have chronic arsenism also Review of Chronic Arsenic Poisoning 2407 have increased levels of ﬂuoride in the drinking water. There are suggestions that the combination of the two could increase the risk to human health due to potential synergism. This should be further evaluated. In a study with mice, Wu et al. found that chronic low-level As exposure may affect heme metabolism, causing porphyrin changes. These changes may appear in the beginning stages of arsenicosis, before the carcinogenesis and can be a clinical indicator to diagnosis. NON-CARCINOGENIC EFFECTS OF CHRONIC ARSENIC EXPOSURE Neurobehavioral and Neuropathic Effects In a cross-sectional study in Taiwan, Tsai et al. suggested that long-term accumulated As may cause neurobehavioral effects in adolescence; therefore consumption of As in childhood may affect behavior later in life. In addition, these effects will be more severe if lead is present, because of synergistic effects. This facet of As toxicity needs to be addressed further. Arsenic neuropathy is a recognized complication of As toxicity. Peripheral neuropathy (an abnormal and usually degenerative state of the peripheral nerves) due to chronic As exposure is one of the most common complications of the nervous system. The neuropathy is usually sensor (affects sensation), and the course of development is chronic. Patients can suffer from constant pain, hypersensitivity to stimuli, muscle weakness, or atrophy.[89,90] Sensory and sensorimotor (sensation and muscles are affected) neuropathy have also been observed. The authors suggest that neurological symptoms are more frequently associated with people that have chronic As exposure, so duration, amount of As exposure, and nutritional factors together may affect As toxicity. Effects on Memory and Intellectual Function A study of children in Mexico found that urinary As concentration was inversely associated with verbal IQ and long-term memory. In addition, it was found that long-term memory, attention and the ability to understand speech may be affected by exposure to As in people with chronic malnutrition. Wasserman et al. have also shown that children’s intellectual function can be decreased by increased As exposure. This correlation was proportional to the dose, which means children that had more than 50 µg/L As exposure had lower performance scores than children with less than 5.5 µg/L exposure. However, this study was limited to a certain period of time for a certain group of the population and some questions remained unanswered, like the role of exposure to As on the intellectual functions, and developing a better understanding of exposure-outcome by follow-up at an earlier age. 2408 Kapaj et al. In addition, Watanabe et al., evaluating the effects of As at different ages, found that age is a very important factor when evaluating effects. In younger generations, clinical manifestations are not always obvious and, as a result, can be missed or underestimated, producing complications later. Effects of early-life exposure are not well understood compared with the effects of adult exposure. Reproductive Effects In a study by Chakraborti et al., pregnancy complications were found to be due to chronic exposure from groundwater As. They found a positive trend in women, with increased As exposure leading to increased fetal loss and premature delivery. Furthermore, research on the effects of As exposure in rats has shown that As causes necrosis (death of living tissue), apoptosis (programmed cell death), loss of conception in the uterus, and death of the newborn. Toxic effects on the fetus were also suggested by Hopenhayn et al., who reported that women with chronic exposure to As (less than 50 µg/L) in drinking water were predisposed to decreased birth weight of infants, suggesting that As may reduce the development of the fetus in utero. Reproductive effects should be further studied to conﬁrm the risks to humans. A separate study by Hopenhayn et al. found that women exposed to As in drinking water during pregnancy have changes in urinary excretion and metabolite distribution that can cause toxic effects on the developing fetus. As metabolism changes during pregnancy, so the impact on the fetus may be different at different stages of pregnancy. It is suggested that this may affect the health in premature and full-term babies. The effects of As exposure through drinking water on pregnancy outcomes were also assessed in a recent study by Milton et al. This study indicated a strong link between chronic As exposure and spontaneous abortion and stillbirth. However, further studies are needed to conﬁrm the association between As and negative pregnancy outcomes. Steatosis (Fatty Liver) Chen et al. studied the effects of As in mouse liver and concluded that chronic oral inorganic As exposure caused cellular hypertrophy (enlargement of the cell) and steatosis. It was suggested that this may cause DNA methyla-tion, which is thought to play a key role in the control of gene expression in mammalian cells, which is important in oncogenesis in mammals. Review of Chronic Arsenic Poisoning 2409 Cardiovascular Disease Lee et al. reported that As ingestion affects the platelets. Platelets are key players in cardiovascular disease. In the presence of thrombin, trivalent As (arsenite) was observed to increase platelet aggregation. In vivo, As in drinking water increased arterial thrombus formation in rats. The authors indicated that platelet aggregation increased with long-term exposure to As in drinking water, being one of the factors causing cardiovascular disease. The authors proposed that their results may be used for estimation of risks from thrombosis and cardiovascular disease in humans, but further evidence is necessary to support their ﬁndings. Ischemic Heart Diseases (IHD) Ischemia is localized tissue anemia due to obstruction of the inﬂow of arterial blood. In a study in arseniasis hyper-endemic villages in southwest Taiwan, researchers evaluated a possible relationship between long-term As exposure and IHD. This study included 462 individuals living in a blackfoot-disease (BFD) area that were drinking well-water for many years. The study indicated that 78 subjects (16.9%) had IHD. Looking at age groups, the highest rate of IHD was for individuals ≥60 years old (about 31%). This suggests that the prevalence of IHD increased with increasing duration of consuming artesian well-water. Carotid Atherosclerosis The carotid arteries are a chief pair of arteries that pass up the neck and supply the head including the brain. Wang et al. highlighted that long-term exposure to As is an independent risk factor for atherosclerosis. Long-term exposure to As is associated with increased risk of carotid atherosclerosis and they suggested that carotid atherosclerosis is an excellent biomarker for arseniasis. Respiratory System Diseases Based on separate studies in Bangladesh and West Bengal (India), it was concluded that, in addition to skin lesions, chronic exposure to As can cause respiratory system effects such as chronic cough and chronic bronchitis.[90,102] In another study, Milton et al. underlined the fact that patients with chronic As exposure have skin manifestations associated with weakness, con-juctival congestion, redness of the eyes, chronic cough, and chronic bronchitis (inﬂammation of the respiratory tract). This work strengthens the evidence that long-term ingestion of As can cause adverse effects on the respiratory system.[102,103] 2410 Kapaj et al. Effects on Hormonal System Arsenic is thought to be an endocrine disruptor, able to alter hormone gene transcription at doses as low as 0.4 µg/L arsenite. Different doses of As can affect hormone regulation in cells at different levels. It is suggested that As effects on gene expression may depend on internal conditions in the human body. Different organs in the body will respond differently to As exposure. Diabetes Mellitus-Type Two Diabetes Type-two diabetes mellitus is non-insulin dependent diabetes, which gen-erally occurs after 40 years of age, with the highest risk in obese people and people that have a family history of diabetes. Tseng et al.[105,106] suggest that inorganic As is diabetogenic in humans, but little is known about patho-physiological mechanisms. They underline the fact that people exposed to As suffer from type two-diabetes. However, there are some limitations in the study design that weakens the evidence reported. Other Effects Guha Mazumder conﬁrms the ﬁndings of previous studies in that chronic exposure to As is associated with pigmentation, keratosis, skin cancer, weakness, anemia, dyspepsia, enlargement of the liver, spleen, and ascites (ﬂuid in abdomen). Other symptoms included chest problems like cough, restrictive lung disease, polyneuropathy, altered nerve conduction velocity, and hearing loss. In West Bengal, India, people are endemically exposed to more than 50 µg/L As in drinking water. Patients reported having irritability, lack of concentration, depression, sleep disorders, headaches, fatigue, skin itching, burning of eyes, weight loss, anemia, chronic abdominal pain, diarrhea, edema of feet, liver enlargement, spleen enlargement, cough, joint pain, decreased hearing, decreased vision, loss of appetite, and weakness. Liver enzymes were increased and liver histology showed ﬁbrosis (ﬁbrotic tissue in liver). Other symptoms included cirrhosis (end stage of hepatic reaction to liver parencymal cell injury), hematemesis (vomiting with blood), and melena (the passage of dark, pitchy and grumous stools stained with blood pigments or with altered blood). It was found that the longer the time of exposure, the more severe the signs and symptoms of As toxicity.[89,90] Table 2 shows some of the most common toxic effects that can result from chronic As exposure. Subclinical Effects Clinical As symptoms depend on the duration of exposure, with signs and symptoms appearing at later stages and with diseases progressing in silent conditions at earlier stages. Rahman et al.’s study in West Bengal Table 2: Studies documenting toxic effects of chronic As exposure. Study Study type Toxic effects Country Mukherjee et al. Cohort Peripheral neuropathy India Milton et al. Cross-sectional Fetal and infant death, spontaneous abortion Bangladesh Hopenhayn et al. Cohort Reduction in birth weight Chile Steinmaus et al. Case-control Bladder cancer USA Tseng et al.[100,101] Cohort Diabetes Taiwan Hopenhayn-Rich et al. Ecological Lung, kidney cancers Argentina Smith et al. Cross-sectional Skin lesions Chile Kurttio et al. Cohort Bladder cancer Finland Lee et al. in vivo/in vitro rats Platelet aggregation, thrombus formation Korea Milton and Rahman Case-control Cough, bronchitis Bangladesh Morales et al. Case-control Lung, bladder cancer Taiwan Rahman et al. Cross-sectional Hypertension Bangladesh Rahman et al. Cohort Skin cancer, gangrene, neuropathy West Bengal (India) Tsai et al. Cross-sectional Neurobehavioral function Taiwan Tseng et al. Cross-sectional Ischemic heart disease Taiwan 2411 2412 Kapaj et al. included one that followed a large population during a seven year period. About 0.1 million people out of 7.3 million in the area evaluated had As-associated skin lesions. In addition, in small villages affected by As exposure, 30–40% of the population drinking the same As-contaminated drinking water had high As levels in urine, hair, and nails, but they did not have As associated skin lesions, indicating that sub-clinical effects may be more widespread than clinical effects. The authors found that families that had safe water for drinking and cooking during a 2-year period, but that had been previously exposed to As, still had high levels of As because of intake from food grown in contaminated areas and washing of food with contaminated water. Thus, if you minimize As contamination in drinking water, concentration of As in tissue still remains above normal, mainly due to consumption of food grown in contaminated areas. Skin Lesions, Drinking Water and Urinary Arsenic In a cross-sectional study in Bangladesh, Ahsan et al. reported that 21.6% of participants in the study had skin lesions such as melanosis and/or keratosis. Of these subjects, 13.9% were currently drinking water with As levels less than 10 µg/L. This either points to previous higher-level exposures, or suggests that even levels below current guidelines are not safe. In a West Bengal study patients that had As-related skin lesions were using water with As levels of 800 µg/L, as a result, many patients with skin lesions also suffered from cancer. In addition, Ahsan et al. underlined the fact that skin lesions were three times more likely in subjects with the highest levels of urinary As. This may be because urinary As is a cumulative exposure indicator, suggesting that urinary As concentration may be a good indicator for predicting negative health effects in humans. Dose-Response Relationship between Arsenic Exposure and Chronic Health Effects Smith et al. reported that chronic health effects of inorganic As exposure in Northern Chile included As-induced skin lesions. Skin lesions were evident despite good nutritional status. Although previous generations have potentially been exposed to As in the Andes mountains, the dose-response link in the current generation was not inﬂuenced by As exposure of previous generations. Guo et al. indicated that the prevalence of As dermatosis was highest in the regions that drank water from wells with higher concentrations of inorganic As. The prevalence of skin lesions was greatest in people over 40 years of age. Kurttio et al. reported that a signiﬁcant increase in the risk of bladder cancer was seen at levels of As >0.5 µg/L in people from Finland. Review of Chronic Arsenic Poisoning 2413 This correlation was seen at exposure concentrations many times lower than any jurisdiction’s current drinking water quality guideline. However, further research should be conducted to conﬁrm this link. Rahman et al. indicated that there was a dose-response relationship between risk of hypertension and drinking water contaminated with inorganic As. The prevalence of hypertension increased in middle-aged men and in women over the age of 60 years. A clear dose-response link was shown as increased exposures were associated with increases in the prevalence ratio. Guha Mazumder reported that chronic respiratory diseases increased signiﬁcantly with increasing As concentrations in drinking water. Among clin-ical manifestations described were cough, crepitations (to make small sharp sudden repeated noises), and shortness of breath. In males, the prevalence of cough adjusted for age was twice as high as for females. With increasing As concentration in water, the prevalence of keratosis and pigmentation also increased. The association between exposure and response, and the prevalence of skin effects, were evident. In addition, in people already identiﬁed with skin lesions, the strongest correlation was with weakness, which increased with increased As exposure. The same link was conﬁrmed by Milton and Rahman, who showed that the prevalence ratios for chronic bronchitis increased with increasing As exposure. It appears that long-term ingestion of As may be a cause for chronic respiratory diseases and skin lesions. ARSENIC IN HUMAN TISSUE Drugs with Arsenic as an Ingredient Homeopathic medicine is frequently used in countries such as India. In some cases, patients use non-doctor prescriptions containing As compounds to treat their disorder. As has caused health problems when used inappropri-ately and patients have represented with hyperpigmentation, keratosis, and increased As in tissues such as skin, hair and nails. This shows that people using As in homeopathic medicine may be at risk of toxicity and discouraging their use to be appropriate. Arsenic Accumulation in Tissues Lin et al. studied biomarkers in BFD patients in southwest Taiwan. Patients having BFD were linked with the presence of high concentrations of inorganic As (the most common As form) in well drinking water. A signiﬁcant increase in inorganic As in urinary excretion, hair, and ﬁngernails of BFD patients was observed, underlining that As in urine, hair, and ﬁngernails are biomarkers of similar value when evaluating As exposure in humans. 2414 Kapaj et al. In an Australian study by Hinhood et al., analysis of As in hair and toenails showed that there was a clear association with As in drinking water and residential soil. Their results also indicated that hair As concentrations were higher in people consuming greater amounts of As in drinking water than people exposed to As from other sources. Children had higher As concentra-tions in both hair and toenails compared with other age groups tested, probably because of more environmental As exposure from their daily life activities. The study also indicated that toenail As concentrations were more strongly linked with external As exposures than was hair As concentration. In a separate study, Chakraborti et al. showed that As levels were high in hair, nail, and skin tissue of individuals with As-associated skin lesions. However, As levels were also high in people that had no skin lesions, but who lived in the same villages in the Ganga-Meghna-Brahmaputra plain of India and Bangladesh. These individuals may not suffer from physical symptoms at the present time, but they may be sub-clinically affected. Skin lesions and use of biological markers like As concentration in hair and nails may help in early diagnosis of chronic As poisoning. Urinary Arsenic Species In studying BFD patients in Taiwan, Lin et al. found that individuals using well-water contaminated by As excreted higher total urinary As. In studying the health effects to Mexican populations from chronic As exposure, Meza et al. found a weak link between total As in water and total As eliminated in urine. Among the urinary As species, dimethyl arsenic (DMA), inorganic As (in the form of trivalent As) and monomethyl As (MMA) were most common. In this study, the methylated As metabolites, such as DMA, were excreted at a level of about 50%. This was considered a very low percentage of methylated As metabolites. Different communities that had experienced chronic As exposure did not have the same level of As metabolism, suggesting that the main reason may be individual ability to metabolize and excrete As due to ethnic differences such as the presence of native Indian, Mexican and Spanish mixture (genetic polymorphism). Urinary Porphyrins The impact of As among people who use As-rich coal for heating, cooking, and drying of food in poorly ventilated dwellings in Guizhou province, China was studied by Ng et al. It was found that burning As-contaminated coal causes effects on porphyrin metabolism. The study indicated signiﬁcant positive association between urinary As concentration and porphyrin concen-tration. Porphyrin levels were higher in the young, women, and old age groups compare to controls (<20, and >40), suggesting that people spending more Review of Chronic Arsenic Poisoning 2415 time indoors are at greater risk of increased As exposure, resulting in higher porphyrin levels. However, the most interesting ﬁnding was that younger age groups had higher levels of uroporphyrin and coproporphyrin III, which can be used as early biomarkers of chronic As exposure. Since As affects porphyrin excretion and the heme biosynthetic pathway, there is a need for further investigation into possible associations between urinary porphyrins and both As-induced cancer and non-cancer clinical manifestations. ARSENIC EXPOSURE FROM FOOD The study by Rahman et al. in West Bengal, India, looked at other As exposure sources. The study revealed that consumption of food from contam-inated areas was another source of chronic As poisoning, since food products like vegetables and rice were cultivated using As-contaminated groundwater. The level of As in groundwater used to cultivate rice and vegetables ranged from 103 µg/L to 827 µg/L. The average As levels in rice and vegetables were 0.323 µg/g and 0.027 µg/g. It was estimated that in villages where people consume such agricultural products, the mean daily individual exposure was about 100 µg. In addition, Chakraborti et al. conﬁrmed that contaminated groundwater used to cultivate vegetables and rice consumed by people may be an important pathway of ingesting As. Urinary As concentration in control subjects drinking “safe” water was higher than the norm, most likely as a direct result of contamination of food products. Huq and Naidu also suggest from their study in Bangladesh that food is another pathway of As exposure. Different foods have different As concentrations. However, there is uncertainty about the bioavailability and associated toxicity of As from different foods. Data from As concentrations in certain vegetables from where As poisoning is documented show that people using the same water source are not affected the same way. This raises the need for more investigation related to speciation and bioaccumulation of As.[118,119] Rmalli et al. investigated As levels in food imported from Bangladesh to the United Kingdom. Results showed that imported vegetables from Bangladesh have from 2-fold to 100-fold higher concentrations of As than vegetables cultivated in the United Kingdom, European Union, and North America. Average As concentrations found were for the skin of arum tuber, 540 µg/kg, arum stem, 168 µg/kg, and amaranthus, 160 µg/kg. The study did not determine the As species found in the foods, which is necessary to asses the risk to humans. This does, however, further support the fact that food may be an important route of As exposure in some regions and that such exposure could have long-term health effects in people. 2416 Kapaj et al. Social Impact of Chronic Arsenic Exposure and Safety Caution In their review of chronic As toxicity, Ratnaike et al. stressed the impact of As contamination, not only on people’s health, but also on the economy, personal incomes and crop productivity. Furthermore, Moyad explained the importance of public awareness of As toxicity. As has not been routinely tested in Canada in the past, but the data in this review clearly show the need to test all groundwater supplies for As compounds, especially inorganic As. Populations most at risk are those using private well-water as a drinking water source. Frequently, such sources are only tested for coliforms and nitrates. Therefore, public education and groundwater monitoring are the most effective ways to provide people with needed information related to As and its negative impact on human health. Mechanism of Arsenic Toxicity and Carcinogenicity The detailed mechanisms of As toxicity and carcinogenicity are not well understood. Experiments in animals and in vitro indicate that As acts at the cellular level at low doses of exposure. In reviewing the literature on As carcinogenicity, Kitchin found that mechanisms of action included chromosomal abnormalities, oxidative stress, altered DNA repair, altered DNA methylation, altered growth factors, cell proliferation, promotion/progression, gene ampliﬁcation, and p53 gene suppression. In addition, the author sug-gested that the methylation of inorganic As can actually be a toxiﬁcation, not a detoxiﬁcation pathway as methylated trivalent As metabolites play an important role in carcinogenesis. In a study by Nesnow et al., it was stressed that DNA damage induced by methylated trivalent arsenicals is mediated by reactive oxygen species. Mass et al. indicated that exposure of human lymphocytes to methylated trivalent As causes direct DNA damage. Their study suggested that As can be carcinogenic and/or genotoxic by direct and/or indirect changes in the structure of DNA and chromosomes. Mahata et al. compared the As-induced cytogenetic damage between symptomatic (having skin mani-festations) and asymptomatic individuals who drank As-contaminated wa-ter. Results indicated that the frequency of genetic damage was higher in symptomatic than asymptomatic individuals. In addition, using a sodium arsenite treatment in vitro, it appeared that lymphocytes of the control group were more sensitive than those of symptomatic or asymptomatic groups. The authors suggested that lymphocytes of people exposed to As for long periods of time reply weaker than the unexposed control group, which may be due to acquired susceptibility. Mahata et al. suggested that the higher numbers of chromosomal aberrations occurring in lymphocytes was related to chronic As exposure, and cancer risks may be predicted by knowing in advance the predisposition to genetic alterations of this population group. Also, Mahata Review of Chronic Arsenic Poisoning 2417 et al. and Yamauchi et al. reported that after stopping As exposure, patients may have less frequent chromosomal aberrations, or DNA damage was reversible and returned to the previous state. Further research is needed to fully understand the biochemical and cytotoxic mechanisms of As toxicity. Treatment of Chronic Arsenic Toxicity There is no clear treatment for chronic As toxicity. The effectiveness of drugs was studied in clinical trials which showed that one form of chelation therapy can stop deterioration of chronic toxicity symptoms, while at the same time preventing outcomes such as cancer. Two main treatments were dimercapto succinic acid (DMSA) and 2, 3-dimercapapto-1-propanesulfonate (DMPS). The study indicated that DMSA did not improve the skin lesions in chronic arsenicosis patients. In contrast, DMPS improved signiﬁcantly chronic arsenicosis. In a separate study, Guha Mazumder et al. found that DMSA did not improve patient health status, nor did it beneﬁt patients with skin lesions from chronic arsenicosis. However, DMPS increased excretion of As in the urine several-fold. Patients under this chelation therapy had signiﬁcant improvements in symptoms. The increase in urinary As excretion during chelation therapy may be the key factor in DMPS therapy. Further research is needed to conﬁrm the efﬁcacy of this drug. Guha Mazumder indicated that proteins in food may increase the elimination of inorganic As by increasing methylation. Hence, people exposed to As are advised to increase protein consumption from both animal and plant origins. In addition, retinoids and antioxidants have anti-keratinizing effects and may prevent cancer. The study underlines that clinical presentations, like chronic bronchitis, interstitial lung disease, portal hypertension, peripheral vascular disease, and peripheral neuropathy, must be treated regularly, so that the patient’s health will not deteriorate. Early detection of cancers due to chronic arsenicosis, especially skin, urinary bladder and lung, can improve the intervention and slow the progress of disease. METHODS FOR ESTIMATING ARSENIC EXPOSURE IN HUMANS A detailed review of the methods for evaluation of individual As exposure was recently compiled by Yoshida et al. and is summarized below. Evaluations of As exposure among individuals are classiﬁed on the basis of: (i) monitoring As concentration in drinking water, and (ii) biological monitoring for As exposure. Arsenic Concentration in Drinking Water Four methods have been described for evaluating As exposure in humans based on As concentration in drinking water. The ﬁrst method uses the 2418 Kapaj et al. concentration of As in drinking water as an index of exposure, but it does not consider individual consumption volume. This reﬂects only current exposure that correlates with short-term effects, but provides less information about long-term effects. The second method seeks to establish the daily body burden of As from the amount of drinking water consumed. Air temperature and humidity may effect the daily individual consumption. The third method is based on average As exposure. This is an advanced index because it can assess the link between exposure and chronic health effects, such as cancer, occurring after long-term exposure. The last method is a cumulative As expo-sure index, which is more appropriate for cases where As levels in drinking water have changed, or where there has been a long period of low level As exposure. Biological Monitoring of Arsenic Exposure Drinking water from wells can contain inorganic As in both the trivalent or pentavalent oxidation states. Inorganic As is metabolized by two-step methylation and the total amount of inorganic As, monomethylarsonic acid (MMA) and dimethylarsinic acid (DMA) can be used as biomarkers of As expo-sure. In one study, Styblo et al. compared in vitro methylation of trivalent and pentavalent As. They concluded that trivalent arsenicals methylated more rapidly than pentavalent arsenicals. In general, there are four main methods of As biomonitoring. The ﬁrst method determines the concentration of As in voided urine. Calderon et al. indicated that urinary As is a good index for estimating As exposure. As concentrations in urine were evaluated in a U.S. population that was exposed to inorganic As in drinking water in the range of 8 to 620 µg/L. The authors found a strong link between the concentration of urinary As and the concentration of inorganic As in drinking water. It was suggested that a few urine samples are able to evaluate the inorganic As burden an individual has received from drinking water. The second method measures the amount of As in blood. It is preferred to use peripheral blood samples for the evaluation of As exposure. Blood and urine samples reﬂect individual As intake and are not contaminated from external factors (dust, hands, contaminated water). The third method determines the amount of As in hair. Hair samples are used as a biomarker for As exposure because inorganic As and DMA are stored in the hair root and thus reﬂect past exposure. The last method is to estimate the amount of As in nails. Nails of ﬁngers or toes are used as they reﬂect As storage 3 months ago in ﬁngers and 6 months ago in toes. Hair and nails are used as biomarkers to estimate average As exposure. Review of Chronic Arsenic Poisoning 2419 CONCLUSIONS AND RECOMMENDATIONS The ingestion of As by humans can cause a variety of disorders, including skin lesions (e.g., hyperpigmentation, melanosis, keratosis), respiratory sys-tem problems (e.g., chronic cough, shortness of breath, bronchitis), nervous system effects (e.g., neuropathy, neurobehavioral, weakened memory, lower IQ, decreased attention), cancers of different organs (e.g., skin, lung, bladder), and reproductive effects (e.g., pregnancy complications, fetus abnormalities, premature deliveries, reduced birth weight). There are, in addition, potential links to heart disease and diabetes, but further evidence is needed to support these relationships. Approaches available to document chronic As exposure include analysis of As levels in drinking water, and measurement of urinary, nail, hair and blood As levels (biological monitoring). It has been shown that even low level As-exposures may affect human health, with greater effects in malnourished people. Recent evidence also implicates ethnic origin as a potential variable when determining As effects. It is becoming clear that a drinking water quality guideline of 50 µg/L As is not protective, and while guidelines have decreased (to 25 µg/L in Canada and 10 µg/L USA and WHO), attempts to lower them to ≤5 µg/L (Canada) must be encouraged. Because groundwater can contain high levels of As, most groundwater sources used for drinking water should be tested for As. If total As concentrations are above 5 µg/L, then it is suggested that biological monitoring should be carried out. This includes measuring As levels in urine, blood, toenails and hair. There are a few promising treatment methods currently in use, including chelation therapy, that may reduce or at least arrest deterioration of chronic As-poisoned individuals. At any one time, it is only a small percentage of a population that shows clinical symptoms, making it a prerequisite to test drinking water, and potentially humans, in order to prevent this element from causing systemic illness. In addition, education of the public about the consequences of drinking water contaminated with As is a necessity. Research needs include the improvement of As quantiﬁcation in both water and human samples, as well as improving our understanding of the environmental occur-rence and cycling of As. A key aspect is to develop a much better understanding of the relationship between chronic As exposure and various adverse effects (i.e., quantitative) in humans, and a better understanding of the underlying mechanisms of action (chronic toxicity). Unfortunately, rural water users have received little research and monitoring support from government agencies, corporations, and Non-Government Organizations. The on-going challenges to produce safe drinking water from typically poor quality rural water sources therefore continue to provide extreme challenges for communities and individ-uals that commonly have few resources available to them, even in developed countries such as Canada. 2420 Kapaj et al. ACKNOWLEDGMENTS The ﬁnancial support of the Safe Drinking Water Foundation’s Dr. Simon Kapaj by the W. Garﬁeld Weston Foundation and George Gordon Groundwater Research Centre is gratefully acknowledged. This study was carried out for the beneﬁt of rural water users including physicians, water treatment plant operators and the general public. Through improved knowledge it is possible for rural water users to protect themselves from the ill effects of unsafe drinking water REFERENCES 1. USEPA. 2001. Arsenic in Drinking Water, ⟨ history.html-expertpanel⟩(accessed May 2005). 2. USEPA. 2002. National primary drinking water regulations. Current drink-ing water regulations. U.S. Environmental Protection Agency. Ofﬁce of Water. ⟨www.epa.gov/safewater/mcl.html mcls⟩(accessed May 2005). 3. NRC. 2001. Arsenic in Drinking Water. National Research Council, Washington, DC, ⟨ (accessed June 2005). 4. Health Canada. 2001. Arsenic in Drinking water. It’s your health. ⟨ July 2005). 5. WHO. Guidelines for Drinking Water Quality, Third edition. World Health Or-ganization, Geneva, Switzerland, 2003. ⟨ sanitation health/ dwq/en/gdwq3 12.pdf⟩(accessed June 2005). 6. Murphy, T.; Guo, J.; Eds. Aquatic Arsenic Toxicity and Treatment; Backhuys Publishers, Leiden, The Netherlands, 2003. 7. Bundschuh, J.; Bhattacharya, P.; Chandrasekharam, D.; Eds. A.A. Balkema: Natural Arsenic in Groundwater: Occurrence, Remediation and Management; Leiden, The Netherlands, 2005; 339 p. 8. Yoshida, T.; Yamauchi, H.; Fan, S.G. Chronic health effects in people exposed to arsenic via the drinking water: Dose-response relationships in review. Toxicol. Appl. Pharmacol. 2004, 198, 243–252. 9. Luster, M.I.; Simeonova, P.P. Arsenic and urinary bladder cell proliferation. Toxicol. Appl. Pharmacol. 2004, 198, 419–423. 10. Watanabe, C.; Inaoka, T.; Matsui, T.; Ishigaki, K.; Murayama, N.; Ohtsuka, R. Effects of arsenic on younger generations. J. Environ. Sci. Health 2003, A38, 129–139. 11. Tchounwou, B.P.; Patlolla, A.K.; Centeno, J.A. Carcinogenic and systemic health effects associated with arsenic exposure—a critical review. Toxicol. Pathol. 2003, 31, 575–588. 12. Rossman, T.G.; Uddin, A.N.; Burns, F.J. Evidence that arsenite acts as a cocar-cinogen in skin cancer. Toxicol. Appl. Pharmacol. 2004, 198, 394–404. 13. Mahata, J.; Chaki, M.; Ghosh, P.; Das, L.K.; Baidya, K.; Ray, K.; Natarajan, A.T.; Giri, A.K. Chromosomal aberrations in arsenic-exposed human populations: A review with special reference to a comprehensive study in West Bengal, India. Cytogenet. Genome Res. 2004, 104, 359–364. Review of Chronic Arsenic Poisoning 2421 14. Kitchin, K.T. Recent advances in arsenic carcinogenesis: Modes of action, animal model systems, and methylated arsenic metabolites. Toxicol. Appl. Pharm. 2001, 172, 249–61. 15. Ratnaike, R.N. Acute and chronic arsenic toxicity. Postgrad. Med. J. 2003, 79, 391– 396. 16. Bhattacharya, P.; Chatterjee, D.; Jacks, G. Occurrence of As-contaminated groundwater in alluvial aquifers from the Delta Plains, eastern India: Option for safe drinking water supply. Int. J. Water Res. Dev. 1997, 13, 79–92. 17. Bhattacharya, P.; Frisbie, S.H.; Smith, E.; Naidu, R.; Jacks, G.; Sarkar, B. Arsenic in the Environment: A Global Perspective. In Handbook of Heavy Metals in the Environment; Sarkar, B., Ed.; Marcel Dekker: New York, 2002; 145–215. 18. Bhattacharya P.; Jacks, G.M.; Khan, A.A. Arsenic in groundwater of the Bengal delta plain aquifers in Bangladesh. Bull. Environ. Cont. Toxicol. 2002, 69, 538–545. 19. Naidu, R.; Bhattacharya, P. Management and remediation of arsenic from con-taminated water. In. Managing Arsenic in the Environment: From Soil to Human Health, Naidu, R.; Smith, E.; Owens, G.; Bhattacharya, P.; Nadebaum, P. Eds.; CSIRO Publishing Melbourne, Australia, 2006; 331–354. 20. Welch, A.H.; Lico, M.S. Factors controlling As and U in shallow groundwater, southern Carson Desert, Nevada. Appl. Geochem 1998, 13 (4), 521–539. 21. Welch, A.H.; Westjohn, D.B.; Helsel, D.R.; Wanty, R.B. Arsenic in ground water of the United States-occurrence and geochemistry. Ground Water 2000, 38 (4), 589–604. 22. Moncure, G.; Jankowski, P.A.; Drever, J.I. The hydrochemistry of arsenic in reser-voir sediments, Milltown, Montana, USA. In Water-Rock Interaction, Low Temperature Environments; Kharaka, Y.K.; Maest, A.S., Eds.; A.A. Balkema: Rotterdam, 1992; Vol. 1, 513–516. 23. Schlottmann, J.L.; Breit, G.N.; Mobilizagtion of As and U in the General Oklahoma aquifer, USA. In: Water-Rock Interaction, Low Temperature Environments, Kharaka, Y.K.; Maest, A.S.; Eds.; A.A. Balkema, Rotterdam, 1992; Vol. 1, 835–838;. 24. Frost, F.; Frank, D.; Pierson, K.; Woodruff, L.; Raasina, B.; Davis, R.; Davies, J.; A seasonal study of arsenic in groundwater, Snohomish County, Washington. DC. Environ. Geochem. Health. 1993, 15, 209–214;. 25. McArthur, J.M.; Ravenscroft,; Saﬁulla, S.; Thirlwall, M.F. Arsenic in groundwater: Testing pollution mechanisms for sedimentary aquifers in Bangladesh. Water Resour. Res. 2001, 37, 109–117. 26. Nordstrom, D.K.; An overview of arsenic mass-poisoning in Bangladesh and West Bengal, Inida. In: Minor Elements. Arsenic, Antimony, Selenium, Tellurium and Bismuth, Young, C., Eds.; Soc. Mining Metallurgy and Exploration. 2000; 21–30. 27. Grantham, D.A.; Jones, F.J. Arsenic contamination of water wells in Nova Scotia. J. Amer. Water Works Assoc. 1977, 69, 653–657. 28. Smedley, P.L.; Nicolli, H.B.; Macdonald, D.M.J.; Barros, A.J.; Tullio, J.O. Hydro-chemistry of arsenic and other inorganic constituents in groundwater from La Pampa, Argentina. Appl. Geochem. 2002, 17, 259–284. 29. Smedley, P.L.; Kinniburgh, D.G.; Macdonald, D.M.J.; Nicolli, H.B.; Barros, A.J.; Tullio, J.O.; Pearce, J.M.; Alonso, M.S. Arsenic associations in sediments from the loess aquifer of La Pampa, Argentina. Appl. Geochem. 2005, 20, 989–1016. 30. Bundschuh, J.; Farias, B.; Martin, R.; Storniolo, A.; Bhattacharya,; Cortes, J.; Bonorino, G.; Albouy, R. Groundwater arsenic in the Chaco-Pampean Plain, Argentina: 2422 Kapaj et al. Case study from Robles County, Santiago del Estero Province. Appl. Geochem. 2004, 19 (2), 231–243. 31. Bhattacharya, P., Claesson, M., Bundschuh, J., Sracek, O., Fagerberg, J., Jacks, G., Martin, R.A., Storniolo, A.R., Thir, J.M. Distribution and mobility of arsenic in the R´ ıo Dulce Alluvial aquifers in Santiago del Estero Province, Argentina. Sci. Total Environ. 2005, 358, 97–120, DOI information 10.1016/j.scitotenv.2005.04.048. 32. Cebrian, M.E.; Albores, A.; Aquilar, M.; Blakely, E. Chronic arsenic poisoning in the north of Mexico. Human Toxicol. 1983, 2, 121–133. 33. Armienta, M.A.; Rodriguez, R.; Aguayao, A.; Ceniceros, N.; Juraez, F.; Cruz, O.; Villasenor, G. Point and regional sources of arsenic in the groundwater of Zimapan, Mexico. Acta. Universitatis Caroline Geologica 1995, 39, 285–290. 34. Rodr´ ıguez, R.; Ramos, J.A.; Armienta, A. Groundwater arsenic variations: The role of local geology and rainfall. Appl. Geochem. 2004, 19 (2), 245–250. 35. Borogo˜ no, J.M.; Greiber, R. Epidemiological study of arsenicism in the city of Antofagasta. In: Trace Substances in Environmental Health-V.A. Symposium, Hemphill, D.D., Univ. of Missouri Press, Columbia, MO, 1972; 13–24;. 36. Borogo˜ no, J.M.; Vicent,; Venturino, H.; Infante, H. Arsenic in the drinking water of the city of Antofagasta: Epidemiological and clinical study before and after the installation of a treatment plant. Environ. Health Perspect. 1977, 19, 103–105. 37. Lu, F.J. Blackfoot disease: Arsenic or humic acid ?. Lancet 1990, 336, 115. 38. Chen, S.L.; Dzeng, S.R.; Yang, M.; Chiu,; Shieh, K.; Wai, C.M. Arsenic species in groundwaters of the Blackfoot disease area, Taiwan. Environ. Sci. Technol. 1994, 28, 877–881. 39. Lianfang, W.; Jianzhong, H. Chronic arsenicism from drinking water in some areas of Xinjiang, China. In Arsenic in the Environment, Part II: Human Health and Ecosystem Effects; Nriagu, J., Ed.; John Wiley Inc.: New York, 1994; 159–172. 40. Luo, Z.D.; Zhang, Y.M.; Ma, L.; Zhang, G.Y.; He, X.; Wilson, R.; Byrd, D.M.; Grifﬁths, J.G.; Lai, S.; He, L.; Grumski, K.; Lamm, S.H.; Chronic arsenism and cancer in Inner Mongolia-consequence of well-water arsenic levels greater than 50 µg/L. In: Arsenic: Exposure and Health Effects, Abernathy, C.O.; Calderon, R.L.; Chappell, W.R.; Eds. Chapman and Hall, New York, 1997 55–68;. 41. Terade, H.; Katsuta, K.; Sasagawa, T.; Saito, H.; Shirata, H.; Fukuchi, H.; Sekiya, K.; Yokoyama, Y.; Hirokawa, S.; Watanabe, Y.; Hasegawa, K.; Oshina, T.; Sekiguchi, T. Clinical observations of chronic toxicosis by arsenic. Nihon Rinsho 1960, 118, 2394– 2403. 42. Choprapawon, C.; Rodcline, A.; Chronic arsenic poisoning in Ronphibool Nakhon Sri Thammarat, the southern province of Thailand. In: Arsenic: Exposure and Health Effects, Abernathy, C.O.; Calderon, R.L.; Chappell, W.R.; Eds.; Chapman and Hall, New York, 1997; 69–77. 43. Smedley, P. Arsenic in rural groundwater in Ghana. J. Afr. Earth. Sci. 1996, 22 (4), 459–470. 44. Varsan´ ayi, I.; Zsoﬁa, F.; Bartha, A. Arsenic in drinking water and mortality in Southern Great Plain, Hungary. Environ. Geochem. Health 1991, 13, 14–22. 45. Minkkinen, P.; Yliruokanen, I. The arsenic distribution in Finnish peat bogs. Kemia-Kemi 1978, 7–8, 331–335. 46. Public Health Engineering Department. National Drinking Water Mission: Sub-mission Project on Arsenic Pollution in Groundwater in West Bengal. Final Report, Review of Chronic Arsenic Poisoning 2423 Steering Committee, Arsenic Investigation Project, Government of West Bengal 1991, 57 p. 47. Public Health Engineering Department. First phase Action Plan Report on Arsenic Pollution in Groundwater in West Bengal. Arsenic Investigation Project, Government of West Bengal 1993, 46 p. 48. Bagla,; Kaiser, J. India’s spreading health crisis draws global arsenic experts. Science 1996, 274, 174–175. 49. Bhattacharya, P.; Chatterjee, D.; Jacks, G.; Options to safeguard groundwater from arseniferous aquifers in West Bengal, India, In: Reaching the Unreached: Chal-lenges for the 21st Century, Proceedings of 22nd WEDC Conference, Pickford, J.; House, S.; Miles, D.; Ockelford, J.; Parr, J.; Saywell, D.; Shaw, R.; Skinner, B.; Smout, I.; Stear, R.; Eds.; New Delhi, India, University of Loughborough, UK, 1996; 258– 261. 50. Bhattacharya, P.; Welch, A.H.; Ahmed, K.M.; Jacks, G.; Naidu, R. Arsenic in groundwater of sedimentary aquifers. Appl. Geochem. 2004, 19 (2), 163–167. 51. Dhar, R.K.; Biswas, B.K.; Samanta, G.; Mandal, B.K.; Chakraborti, D.; Roy, S.; Jafar, A.; Islam, A.; Ara, G.; Kabir, S.; Khan, A.W.; Ahmed, S.A.; Hadi, S.A. Groundwater arsenic calamity in Bangladesh. Curr. Sci. 1997, 73 (1), 48–59. 52. von Br¨ omssen, M.; Genesis of high arsenic groundwater in the Bengal Delta Plains, West Bengal and Bangladesh. M.Sc. Thesis, Division of Land and Water Resources, Dept. of Civil and Environ. Engineering, Royal Institute of Technology, Stockholm, Sweden, 1999; 49 p. 53. ACIC. Arsenic Crisis Information Centre, ⟨ October 2005). 54. Karim, M.M. Arsenic in groundwater and health problem in Bangladesh. Water Res. 2000, 34 (1), 304–310. 55. Mukherjee, A.B.; Bhattacharya, P. Arsenic in groundwater in the Belgal Delta Plain: slow poisoning in Bangladesh. Environ. Rev. 2001, 9, 189–220. 56. Mukherjee, A.B.; Bhattacharya, P.; Jacks, G.; Banerjee, D.M.; Ramanathan, A.L.; Mahanta, C.; Chandrashekharam, D.; Chatterjee, D.; Naidu, R.; Groundwater arsenic contamination in India: Extent and severity. In: Managing Arsenic in the Environment: From soil to human health. Naidu, R.; Smith, E.; Owens, G.; Bhattacharya, P.; Nadebaum, P.; Eds.; CSIRO Publishing: Melbourne, Australia, 2006; 553–593; (ISBN: 0643068686). 57. Berg, M.; Tran, H.C.; Nguyen, T.C.; Pham, H.V.; Schertenleib, R.; Giger, W. Arsenic contamination of groundwater and drinking waiter in Vietnam: A human health threat. Environ. Sci. Technol. 2001, 35, 2621–2626. 58. Polizzoto, M.L.; Kocar, B.D.; Sampson, M.; Benner, S.G.; Ouch, K.; Fendorf, S.; Ung, M.; Oum, R. Arsenic contamination in the ﬂood plain aquifer of the Mekong Delta, Cambodia. Geol. Soc. America. Abstracts with Program 2005, 37 (7), 376 p. 59. Yount, J.; Breit, G.; Sopheap, S.; Moniphea, L.; Mengieng, U.; Kagna, O.; Sampson, M.; Rosemboom, J.W. Geologic controls on groundwater arsenic distribution in the Up-per Mekong Floodplain, Kien Svay region of Cambodia. Geol. Soc. America. Abstracts with Program 2005, 37 (7), 376. 60. Berg, M.; Stengel, C.; Trang, P.T.; Viet Pham, K.H.; Sampson, M.L.; Leng, M.; Samreth, S.; Fredericks, D. Magnitude of arsenic pollution in the Mekong and Red River Deltas—Cambodia and Vietnam. Sci. Tot. Environ. 2006, Special Issue: The Science of the Total Environment: Biology and Chemistry. 2424 Kapaj et al. 61. Fredericks, D.; Situation Analysis: Arsenic Contamination of Groundwater in Cambodia. Report, January 2004, UNICEF Phonom Penh, Cambodia 2004. 62. Fieldman, P.R., Rosenboom, J.W. Cambodia Drinking Water Quality Assessment. World Health Organisation of the UN [WHO] in cooperation with Cambodian Ministry of Rural Development and the Ministry of Industry, Mines and Energy, Phnom Penh, Cambodia, 2001. 63. Smith, A.H.; Hopenhayn-Rich, C.; Bates, M.N.; Goeden, H.M.; Hertz-Picciotto, I.; Duggan, H.M.; Wood, R.; Kosnett, M.J.; Smith, M.T. Env. Health Perspect. 1992, 97, 259–267. 64. WHO Guidelines for Drinking-Water Quality, Volume 1 Recommendations. 2nd edition Geneva 1993. 65. WHO. Guidelines for Drinking-Water Quality Addendum to Volume 1: Recommen-dations, 2nd edition. Geneva, 1998. 66. WHO. Fact Sheet #210 Arsenic contamination of drinking water in Bangladesh, 1999. ⟨www.who.int/inf-fs/en/fact210.html⟩. 67. US EPA. US EPA Drinking Water Regulations and Health Advisories; EPA 822-B-96-002: Washington, DC, 1996. 68. NRC. Arsenic in Drinking Water; National Academy Press: Washington, DC, 1999; 310p. 69. US EPA. National Primary Drinking Water Regulations; Arsenic and Clariﬁca-tions to Compliance and New Source Contaminants Monitoring. Final Rule; Federal Register, Part VIII. Environmental Protection Agency: Washington, DC, 2001; Vol. 66(14), 6976–7066. 70. Zheng, B.; Yu, X.; Zhand, J.; Zhou, D.; Environmental geochemistry of coal and endemic arsenism in southwest, Guizhou, P.R. China. 30th Int. Geol. Cong., Abstracts, 1996, Vol. 3, 410 p;. 71. Finkelman, R.B.; Belkin, H.E.; Zheng, B.; Centeno, J.A.; Arsenic poisoning caused by residential coal combuation in Guizhou Province. In: Arsenic in Groundwater of Sedimentary Aquifer, Bhattacharya, P. Welch, A.; Eds.; Pre-Congress Workshop, 31st Int. Geol. Cong., Rio de Janeiro, 2000, 41–42;. 72. Bhattacharya, P.; Jacks, G.; Nordqvist, S.; Mukherjee, A.B. Metal contamination at a wood preservation site: characterization and experimental studies on remediation. Science of the Total Environment 2002, 290 (1–3), 165–180. 73. Bringezu, S. Arsen im Holzschutz—m¨ ogliche Gesundheits-und Umweltgef¨ ahrdun-gen. Holz als Roh-und Werkstoff 1990, 48, 237–243. 74. US EPA. Review Draft of the Health Effects Notebook for Hazardous Air Pollutants (US EPA Contract # 68-D2-0065).;. Air Risk Information Support Center Research Triangle Park, NC, 1994. 75. International Agency for Research on Cancer. Arsenic and arsenic compounds. 1980 Vol. 23, 39p. (ac-cessed July 2005). 76. Centeno, J.A.; Tchounwou, P.B.; Patlolla, A.K.; Mullick, F.G.; Murakata, L.; Meza, E.; Gibb, H.; Longfellow, D.; Yedjou, C.G. Environmental Pathology and Health Effects of Arsenic Poisoning: A Critical Review. In Managing Arsenic in the Environment: From Soil to Human Health; Naidu, R.; Smith, E.; Owens, G.; Bhattacharya, P.; Nadebaum,, Eds.; CSIRO Publishing: Melbourne, Australia, 2006; 311–327. (ISBN: 0643068686). Review of Chronic Arsenic Poisoning 2425 77. Rahman, M.M.; Mandal, B.K.; Chowdhury, T.R.; Sengupta, M.K.; Chowdhury, U.K.; Lodh, D.; Chanda, C.R.; Basu, G.K.; Mukherjee, S.C.; Saha, K.C.; Chakraborti, D. Arsenic groundwater contamination and sufferings of people in North 24-Parganas, one of the nine arsenic affected districts of West Bengal, India. J. Environ. Sci. Health, 2003, A38, 25–59. 78. Morales, K.H.; Ryan, L.; Kuo, T.L.; Wu, M.M.; Chen, C.J. Risk of internal cancers from arsenic in drinking water. Environ. Health. Perspect. 2000, 108, 655–661. 79. Chiou, H.Y.; Chiou, S.T.; Hsu, Y.H.; Chou, Y.L.; Tseng, C.H.; Wei, M.L.; Chen, C.J. Incidence of transitional cell carcinoma and arsenic in drinking water: a follow-up study of 8,102 residents in an arseniasis-endemic area in northeastern Taiwan. Am. J. Epidemiol. 2001, 153 (5), 411–418. 80. Steinmaus, C.; Yuan, Y.; Bates, M.N.; Smith, A.H. Case-control study of bladder cancer and drinking water arsenic in the western United States. Am. J. Epidemiol. 2003, 158, 1193–1201. 81. Bates, M.N. Case-control study of bladder cancer and exposure to arsenic in Argentina. Amer. J. Epidemiol. 2004, 159, 381–389. 82. Ferreccio, C.; Gonzalez, C.; Milosavjlevic, V.; Marshall, G.; Sancha, A.M.; Smith, A.H. Lung cancer and arsenic concentrations in drinking water in Chile. Epidemiology 2000, 11 (6), 673–679. 83. Hopenhayn-Rich, C.; Biggs, M.L.; Smith, A.H. Lung and kidney cancer mortality associated with arsenic in drinking water in Cordoba, Argentina. Int. J. Epidemiol. 1998, 27 (4), 561–569. 84. Chen, C.L.; Hsu, L.I.; Chiou, H.Y.; Hsueh, Y.M.; Chen, S.Y.; Wu, M.M.; Chen, C. Ingested arsenic, cigarette smoking, and lung cancer risk: a follow-up study in arseniasis-endemic areas in Taiwan. JAMA 2004, 292 (24), 2984–2990. 85. Chiu, H.F.; Ho, S.C.; Yang, C.Y. Lung cancer mortality reduction after installation of tap-water supply system in an arseniasis-endemic area in Southwestern Taiwan. Lung Cancer 2004, 46 (3), 265–270. 86. Xia, Y.; Liu, J. An overview on chronic arsenism via drinking water in PR China. Toxicology 2004, 198, 25–29. 87. Wu, H. Urinary arsenic speciation and porphyrins in C57Bl/6J mice chronically exposed to low doses of sodium arsenate. Toxicol. Lett. 2004, 154 (1–2), 149–157. 88. Tsai, S.Y.; Chou, H.Y.; The, H.W.; Chen, C.M.; Chen, C.J. The effects of chronic arsenic exposure from drinking water on the neurobehavioral development in adoles-cence. Neurotoxicology 2003, 24, 747–753. 89. Mukherjee, S.C.; Rahman, M.M.; Chowdhury, U.K.; Sengupta, M.K.; Lodh, D.; Chanda, C.R.; Saha, K.C.; Chakraborti, D. Neuropathy in arsenic toxicity from ground-water arsenic contamination in West Bengal, India. J. Environ. Sci. Health 2003, A38, 165–183. 90. Guha Mazumder, D.N. Chronic arsenic toxicity: clinical features, epidemiology, and treatment: experience in West Bengal. J. Environ Sci Health 2003, A38, 141– 163. 91. Calderon, J.; Navarro, M.E.; Jimenez-Capdeville, M.E.; Santos-Diaz, M.A.; Golden, A.; Rodriguez-Leyva, I.; Borja-Aburto, V.; Diaz-Barriga, F. Exposure to arsenic and lead and neuropsychological development in Mexican children. Environ. Res. 2001, 85, 69–76. 92. Wasserman, G.A.; Liu, X.; Parvez, F.; Ahsan, H.; Factor-Litvak, P.; van Geen, A.; Slavkovich, V.; LoIacono, N.J.; Cheng, Z.; Hussain, I.; Momotaj, H.; Graziano, J.H. 2426 Kapaj et al. Water arsenic exposure and children’s intellectual function in Araihazar, Bangladesh. Environ. Health Perspect. 2004, 112 (13), 1329–1333. 93. Chakraborti, D.; Mukherjee, S.C.; Pati, S.; Sengupta, M.K.; Rahman, M.M.; Chowdhury, U.K.; Lodh, D.; Chanda, C.R.; Chakraborti, A.K.; Basu, G.K. Arsenic groundwater contamination in Middle Ganga Plain, Bihar, India: A future danger?. Environ. Health Perspect. 2003, 111, 1194–2201. 94. Chattopadhyay, S.; Bhaumik, S.; Purkayastha, M.; Basu, S.; Nag Chaudhuri, A.; Das Gupta, S. Apoptosis and necrosis in developing brain cells due to arsenic toxicity and protection with antioxidants. Toxicol. Lett. 2002, 136, 65–76. 95. Hopenhayn, C.; Ferreccio, C.; Browning, S.R.; Huang, B.; Peralta, C.; Gibb, H.; Hertz-Picciotto, I. Arsenic exposure from drinking water and birth weight. Epidemiol-ogy 2003, 14, 593–602. 96. Hopenhayn, C.; Huang, B.; Christian, J.; Peralta, C.; Ferreccio, C.; Atallah, R.; Kalman, D. Proﬁle of urinary arsenic metabolites during pregnancy. Environ. Health Perspect. 2003, 111, 1888–1891. 97. Milton, A.H.; Smith, W.; Rahman, B.; Hasan, Z.; Kulsum, U.; Dear, K.; Rakibuddin, M.; Ali, A. Chronic arsenic exposure and adverse pregnancy outcomes in Bangladesh. Epidemiology 2005, 16, 82–86. 98. Chen, H.; Li, S.; Liu, J.; Diwan, B.A.; Barrett, J.C.; Waalkes, M. Chronic inorganic arsenic exposure induces hepatic global and individual gene hypomethylation: Implica-tions for arsenic hepatocarcinogenesis. Carcinogenesis 2004, 25, 1779–1786. 99. Lee, M.Y.; Bae, O.N.; Chung, S.M.; Kang, K.T.; Lee, J.Y.; Chung, J.H. Enhancement of platelet aggregation and thrombus formation by arsenic in drinking water: A contributing factor to cardiovascular disease. Toxicol. Appl. Pharmacol. 2002, 179, 83–88. 100. Tseng, C.H.; Chong, C.K.; Tseng, C.; Hsueh, Y.M.; Chiou, H.Y.; Tseng, C.C.; Chen, C.J. Long-term arsenic exposure and ischemic heart disease in arseniasis-hyperendemic villages in Taiwan. Toxicol. Lett. 2003, 137, 15–21. 101. Wang, C.H.; Jeng, J.S.; Yip,K.; Chen, C.L.; Hsu, L.I.; Hsueh, Y.M.; Chiou, H.Y.; Wu, M.M.; Chen, C.J. Biological gradient between long-term arsenic exposure and carotid atherosclerosis. Circulation 2002, 105, 1804–1809. 102. Milton, A.H.; Rahman, M. Respiratory effects and arsenic contaminated well water in Bangladesh. Int. J. Environ. Health. Res. 2002, 12, 175–179. 103. Milton, A.H.; Hasan, Z.; Rahman, A.; Rahman, M. Non-cancer effects of chronic arsenicosis in Bangladesh: preliminary results. J. Environ. Sci. Health 2003, A38, 301– 305. 104. Bodwell, J.E.; Kingsley, L.A.; Hamilton, J.W. Arsenic at very low concentrations alters glucocorticoid receptor (GR)-mediated gene activation but not GR-mediated gene repression: complex dose-response effects are closely correlated with levels of activated GR and require a functional GR DNA binding domain. Chem. Res. Toxicol. 2004, 17, 1064–1076. 105. Tseng, C.H.; Tai, T.Y.; Chong, C.K.; Tseng, C.; Lai, M.S.; Lin, B.J.; Chiou, H.Y.; Hsueh, Y.M.; Hsu, K.H.; Chen, C.J. Long-term arsenic exposure and incidence of non-insulin-dependent diabetes mellitus: a cohort study in arseniasis-hyperendemic villages in Taiwan. Environ. Health Perspect. 2000, 108, 847–851. 106. Tseng, C.H.; Tseng, C.; Chiou, H.Y.; Hsueh, Y.M.; Chong, C.K.; Chen, C.J. Epidemiologic evidence of diabetogenic effect of arsenic. Toxicol. Lett. 2002, 133, 69– 76. Review of Chronic Arsenic Poisoning 2427 107. Ahsan, H.; Perrin, M.; Rahman, A.; Parvez, F.; Stute, M.; Zheng, Y.; Milton, A.H.; Brandt-Rauf,; Geen, A.V.; Graziano, J. Associations between drinking water and urinary arsenic levels and skin lesions in Bangladesh. J. Occup. Environ. Med. 2000, 42, 1195–1201. 108. Smith, A.H.; Arroyo, A.; Guha Mazumder, D.N.; Kosnett, M.J.; Hernandez, A.L.; Beeris, M.; Smith, M.M.; Moore, L.E. Arsenic-induced skin lesions among Atacameno people in Northern Chile despite good nutrition and centuries of exposure. Environ. Health Perspect 2000, 108, 617–620. 109. Guo, X.; Fujino, Y.; Kaneko, S.; Wu, K.; Xia, Y.; Yoshimura, T. Arsenic contami-nation of groundwater and prevalence of arsenical dermatosis in the Hetao plain area, Inner Mongolia, China. Mol. Cell. Biochem. 2001, 222, 137–140. 110. Kurttio, P.; Pukkala, E.; Kahelin, H.; Auvinen, A.; Pekkanen, J. Arsenic con-centrations in well water and risk of bladder and kidney cancer in Finland. Environ. Health Perspect 1999, 107, 705–710. 111. Rahman, M.; Tondel, M.; Ahmad, S.A.; Chowdhury, I.A.; Faruquee, M.H.; Ax-elson, O. Hypertension and arsenic exposure in Bangladesh. Hypertension. 1999, 33, 74–78. 112. Chakraborti, D.; Mukherjee, S.C.; Saha, K.C.; Chowdhury, U.K.; Rahman, M.M.; Sengupta, M.K. Arsenic toxicity from homeopathic treatment. J. Toxicol-Clin Toxic 2003, 41, 963–967. 113. Lin, T.H.; Huang, Y.L.; Wang, M.Y. Arsenic species in drinking water, hair, ﬁngernails, and urine of patients with blackfoot disease. J. Toxicol. Environ. Health 1998, A53, 85–93. 114. Hinwood, A.L.; Sim, M.R.; Jolley, D.; de Klerk, N.; Bastone, E.B.; Gerostamoulos, J.; Drummer, O.H. Hair and toenail arsenic concentrations of residents living in areas with high environmental arsenic concentrations. Environ. Health Perspect 2003, 111, 187–193. 115. Chakraborti, D.; Sengupta, M.K.; Rahman, M.M.; Ahamed, S.; Chowdhury, U.K.; Hossain, M.A.; Mukherjee, S.C.; Pati, S.; Saha, K.C.; Dutta, R.N.; Quamruzzaman, Q. Groundwater arsenic contamination and its health effects in the Ganga-Meghna-Brahmaputra plain. J. Environ. Monit. 2004, 6 (6), 74–83. 116. Meza, M.M.; Kopplin, M.J.; Burgess, J.L.; Gandolﬁ, A.J. Arsenic drinking water exposure and urinary excretion among adults in the Yaqui Valley, Sonora, Mexico. Environ. Res. 2004, 96, 119–126. 117. Ng, J.C.; Wang, J.; Zheng, B.; Zhai, C.; Maddalena, R.; Liu, F.; Moore, M.R. Urinary porphyrins as biomarkers for arsenic exposure among susceptible populations in Guizhou province, China. Toxicol. Appl. Pharmacol. 2005, 206 (2), 176–184. 118. Huq, S.M.I.; Naidu, R. Arsenic in groundwater and contamination of the food chain: Bangladesh scenario. In Natural Arsenic in Groundwater, Remediation and Man-agement; Bundschuh, J.; Bhattacharya,; Chandrashekharam, D., Eds.; A.A. Balkema,: Leiden, The Netherlands, 2005; 95–101. (ISBN 04 1536 700 X). 119. Huq, S.M.I.; Ara, Q.A.J.; Khaleda, I.; Zaher, A.; Naidu, R. The possible contami-nation from arsenic through food chain In. Groundwater Arsenic Contamination in the Bengal Delta Plain of Bangladesh;. Proceedings of the KTH-Dhaka University Seminar, University of Dhaka, Bangladesh. Jacks G., Bhattacharya P., Khan A.A., Eds.; 2001 KTH Special Publication, TRITA-AMI REPORT 3084, p.91–96. 120. Al Rmalli, S.W. A survey of arsenic in foodstuffs on sale in the United Kingdom and imported from Bangladesh. The Sci. of the Total Environment 2005, 337 (1–3), 23–30. 2428 Kapaj et al. 121. Moyad, M.A. What do I tell my patients about drinking water and the risk of bladder cancer?. Urol. Nurs. 2003, 23, 371–377. 122. Nesnow, S.; Roop, B.C.; Lambert, G.; Kadiiska, M.; Mason, R.; Cullen, W.R.; Mass, M.J. DNA damage induced by methylated trivalent arsenicals is mediated by reactive oxygen species. Chem. Res. Toxicol. 2002, 15, 1627–1634. 123. Mass, M.J.; Tennant, A.; Roop, B.C.; Cullen, W.R.; Styblo, M.; Thomas, D.J.; Kligerman, A.D. Methylated trivalent arsenic species are genotoxic. Chem. Res. Toxicol. 2001, 14, 355–361. 124. Mahata, J.; Ghosh,; Sarkar, J.N.; Ray, K.; Natarajan, A.T.; Giri, A.K. Effect of sodium arsenite on peripheral lymphocytes in vitro: Individual susceptibility among a population exposed to arsenic through the drinking water. Mutagenesis 2004, 19, 223–229. 125. Yamauchi, H.; Aminaka, Y.; Yoshida, K.; Sun, G.; Pi, J.; Waalkes, M.P. Evaluation of DNA damage in patients with arsenic poisoning: urinary 8-Hydroxydeoxyguanine. Toxicol. Appl. Pharmacol 2004, 198, 291–296. 126. Guha Mazumder, D.N.; Ghoshal, U.C.; Saha, J.; Santra, A.; De, B.K.; Chatterjee, A.; Dutta, S.; Angle, C.R.; Centeno, J.A. Randomized placebo-controlled trial of 2,3-dimercaptosuccinic acid in therapy of chronic arsenicosis due to drinking arsenic-contaminated subsoil water. J. Toxicol. Clin. Toxicol. 1998, 36, 683–90. 127. Guha Mazumder, D.N.; De, B.K.; Santra, A.; Ghosh, N.; Das, S.; Lahiri, S.; Das, T. Randomized placebo-controlled trial of 2,3-dimercapto-1-propanesulfonate (DMPS) in therapy of chronic arsenicosis due to drinking arsenic-contaminated water. J. Toxicol. Clin. Toxic. 2001, 39, 665–674. 128. Styblo, M.; Yamauchi, H.; Thomas, J.D. Comparative in vitro methylation of trivalent and pentavalent arsenicals. Toxicol. Appl. Pharmacol 1995, 135 (2), 172–178. 129. Calderon, R.L.; Hudgens, E.; Le, X.C.; Schreinemachers, D.; Thomas, D.J. Excre-tion of arsenic in urine as a function of exposure to arsenic in drinking water. Environ. Health Perspect. 1999, 107, 663–667.
13308	https://www.chegg.com/homework-help/questions-and-answers/9-7-pts-use-washer-method-find-volume-solid-obtained-rotating-region-bounded-graphs-y-4-x--q108970832	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: 9. (7 pts) Use the washer method to find the volume of the solid obtained by rotating the region bounded by the graphs of y=4x and y=x2 about the line y=−2. Sketeh the resion and the axis of resolution. outer radius: inner radius = the required solution Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
13309	https://www.leybold-shop.com/physics/physics-experiments/heat/thermal-expansion/thermal-expansion-of-liquids/determination-of-volumetric-expansion-coefficient-of-liquids/vp2-1-2-1.html	Determination of volumetric expansion coefficient of liquids - Determination of volumetric expansion coefficient of liquids - Thermal expansion of liquids - Thermal expansion - Heat - Physics Experiments - Physics Forgot password? Register Deutsch English Français Physics Physics Equipment Systems CASSY Students experiments Mechanics Stand material Measuring devices Simple machines Mechanics on the adhesive magnetic board Dynamics and Kinematics Rotational motion Mechanical oscillations Mechanical waves Acoustics Mechanics of liquids Mechanics of gases Vacuum Aerodynamics Manuals Heat Heaters Temperature measurement Themal expansion of bodies and liquids Calorimetry Thermal behavior of gases Aggregat states and phase transitions Heat transfer Heat and work Meteorology Problems of climate Manuals Electricity/Electronics Electrical power supply units Electrical measuring instruments Electricity and electronics accessories Electrostatics Electrical conductivity Magnetostatics Electromagnetism and induction Electrical machine teaching models (ELM) Electricity/electronics (BST) Electricity/electronics (STE) Digital technology Electromagnetic oscillations and waves Electroacoustics Thermoelectricity Manuals Optics Geometrical optics Optical benches and accessories Light sources and accessories Optical components Wave optics Interferometry and holography Spectroscopy Velocity of light Electro-optic and magneto-optic effects Laser optics Manuals Atomic and nuclear physics Fundamentals Physics of the electron Franck-Hertz experiments Electron spin resonance Nuclear magnetic resonance - NMR Zeeman effect X-ray apparatus Sensors for α-, β-, γ- and X-rays Radioactivity Energy analysis Manuals Solid-state physics Structure of crystals Conduction phenomena Manuals Auxiliary equipment Erlenmeyer flasks Beakers Volumetric flasks, Pycnometers Funnels Self-supporting cylinders Graduated pipettes Graduated cylinders/measuring beakers Glass tubes, glass rods Adapter Test tubes and racks Ruber stoppers and accessories Tubing and accessories Screw caps and accessories Stopcock, glass Brushes Spatulas and tweezers Glass working Tools Laboratory gases and accessories Storage Projectors Physics Experiments Mechanics Measuring methods Forces Translational motions of a mass point Rotational motions of a rigid body Oscillations Wave mechanics Acoustics Aero- and hydrodynamics Heat Thermal expansion Heat transfer Heat as a form of energy Phase transitions Kinetic theory of gases Thermodynamic cycle Electricity Electrostatics Fundamentals of electricity Magnetostatics Electromagnetic induction Electrical machines DC and AC circuits Electromagnetic oscillations and waves Free charge carriers in a vacuum Electrical conduction in gases Electronics Components and basic circuits Operational amplifier Digital electronics Optics Geometrical optics Dispersion and chromatics Wave optics Polarization Light intensity Velocity of light Spectrometer Photonics Atomic and nuclear physics Introductory experiments Atomic shell X-ray physics Radioactivity Nuclear physics Quantum physics Solid-state physics Properties of crystals Conduction phenomena Magnetism Applied solid-state physics Demonstration Experiments Mechanics Common properties of bodies Forces and Work Motion Mechanical oscillations and waves Acoustics Structure of substances and particles Mechanics of liquids and gases Heat Thermal behaviour of bodies Thermal energy Heat transfer Changes of the state of aggregation Conservation of energy Electricity Electrostatics Basic electric circuits Electromagnetism and induction Motors and generators Electrical safety in the household Phenomena of electrical conduction Electronics Basic electronic circuits Optics Sources and propagation of light Formation of images by lenses and mirrors Optical instruments Light and colour Radioactivity Detection and properties of radioactive radiation Basic Science Kits Basic Science Kits Natural Sciences Perception with all senses Journey to the world of the small Sun, earth, moon and heat My body, my health Everyday substances and devices Soil, water and air Mechanics Hydrostatic / Heat Electricity Optics Advanced Student Experiments Mechanics Measuring methods, properties of matter & liquid Forces, simple machines and oscillations Linear motion, free fall and collision experiments Acoustics Energy Heat Renewable energies Fuel cells Electrics Electrostatics Magnetism Basic electrical circuits Electromagnetism and induction Motors and generators Electronics Basic electronic circuits Optics Ray optics and geometrical optics Chromatics Wave optics Polarisation Atomic and nuclear physics Environmental radioactivity Projects Programming with Arduino Advanced experiments Physics Standard Lab Proposals Chemistry Product catalogue chemistry/biology Natural sciences - student experiments Chemistry - student experiments Student experiment systems Measuring in student experiments Molecule building kits for students Student experiment literature Chemistry - demonstration experiments Demonstration experiment systems Measuring in demonstration experiments Inorganic Chemistry Organic Chemistry Analytical Chemistry Physical Chemistry Chemical engineering Illustrative material for chemistry Literature overview Measuring technology Measuring and evaluating with CASSY Electrical measuring instruments Laboratory equipment Glass, plastic, porcelain instruments Stand material Electrical power supply Cables and lamps Heaters, stirrers, pumps, laboratory accessories Tools/auxiliary equipment Laboratory auxiliary equipment Safety Chemicals Chemical kits Individual chemicals Laboratory gases Chemicals by application Catalogue of experiments - Chemistry General and inorganic chemistry Material properties Chemical reactions and stochiometry The compound water Air and other gases Metals and their properties Non-metals and their properties Acids, bases and salts Organic chemistry Organic compounds Reactions in organic chemistry Petrochemistry Synthesis and purification of organic compounds Analytical chemistry Determination of physical properties Chromatography Optical analysis methods Structural analysis Quantitative analysis Applied analytics Physical chemistry Reaction kinetics Chemical equilibrium Thermochemistry Electrochemistry Transport processes Chemical engineering Industrial-scale processes Products of the chemical industry Chemistry and the environment Products made from renewable raw materials Biochemistry The building blocks of life Biotechnology Basic Science Kits Chemistry Student experiments secondary schools Inorganic Chemistry Organic Chemistry Analytical Chemistry Physical Chemistry Advanced Student Experiments Inorganic and general chemistry General methods & separation methods Water Air, gases and their properties Acids and bases Salts Metals Redox reactions Chemical reaction New fields in chemistry Organic Chemistry Organic substances Hydrocarbons Alcohols, aldehydes and ketones Carboxylic acids and esters Reactions in organic chemistry Methods of organic chemistry Physical Chemistry Electrochemistry Physical processes Energy in chemical reactions Rate of reaction The Chemical equilibrium Technical Chemistry Building material Glass Metals Chemical applications Products of the organic industry Biochemistry Fats Carbohydrates Amino acids and proteins Food Projects Fuel Cell Detailed experiments Biology Devices Natural sciences - student experiments Biology - student experiments Student experiment systems Measuring in student experiments Student microscopes Student experiment literature Biology demonstration experiments Demonstration systems Microscopy Laboratory gases Topics Illustrative material for Biology Literature overview Measuring technology Measuring and evaluating with CASSY Electrical measuring instruments Laboratory equipment Glass, plastic, porcelain instruments Stand material Electrical power supply Cables and lamps Heaters, stirrers, pumps, laboratory accessories Tools/auxiliary equipment Laboratory auxiliary equipment Safety Chemicals Chemical kits Individual chemicals Laboratory gases Chemicals by application Versuchekatalog Biologie Human Biology The Nervous System Plant biology Plant physology Ecology & Evolution Basic Science Kits Biology Botany Ecology Microbiology Sense Physiology Microscopy Genetics Advanced Student Experiments Human Biology Our senses Our Body Maintaining a healthy body Botany Introduction to methods The shape of plants Function of plants Ecology Introduction to methods Ecosystems Investigation of ecosystems Humans and the environment Evolution Cellular Biology Introduction to methods Structure of the cell Processes in the cell Genetics Genetics - Classic Projects Hygiene Technology Automotive Fundamentals of car electr(on)ics Fundamentals electr(on)ics and Multimedia Automotive Technology Vehicle Electrics Combustion Engine Driver assistance and comfort systems Chassis and Drive Train Automotive networking and diagnosis systems Electromobility Electrical Engineering Electric circuits and components Fundamentals Basics of electricity Basics of electronics Electrical drives Fundamentals of electrical machines and networks Industrial machines, 300 W Industrial machines, 1 kW Power electronics Drive technology Servo technology Electrical power engineering Electrical power generation Electrical power transmission and distribution Energy consumption Protective systems Plug-in models for power engineering Smart Grid Building technology House installation technology Photovoltaic systems Protection circuits Smart Building Communications technology Transmission technology High frequency technology Control Engineering and Automation Measurement Technology and Sensors Didactic Control Engineering Applied Control Engineering Industrial Control Technology Automation Technology Open Loop Control Technology Logik Controllers and Bus Systems PLC Programming Process Automation Hydraulics and Pneumatics COM4LAB COM4LAB Automotive Technology Digital Technology COM4LAB Electrical Engineering Circuit Technology Electrical Drives Electrical Power Engineering Telecommunications Engineering Control Engineering and Automation COM4LAB product catalogue Master Unit and Accessories Boards and Accessories Courses Visit LD DIDACTIC Cart 0 Search: Search Services Success Stories Contact LEYBOLD Vacuum Home/ Physics/ Physics Experiments/ Heat/ Thermal expansion/ Thermal expansion of liquids/ Determination of volumetric expansion coefficient of liquids/ Determination of volumetric expansion coefficient of liquids Determination of volumetric expansion coefficient of liquids 1. Thermal expansion Thermal expansion of solids Thermal expansion of liquids Determination of volumetric expansion coefficient of liquids Determination of volumetric expansion coefficient of liquids - Measuring with Mobile-CASSY Thermal anomaly of water Heat transfer Heat as a form of energy Phase transitions Kinetic theory of gases Thermodynamic cycle Determination of volumetric expansion coefficient of liquids P2.1.2.1 Determination of volumetric expansion coefficient of liquids Add to product list This product is classified as a dangerous good and is not available for online purchase. For ordering the dangerous good or other information please contact our customer service center. This experiment/set up contains dangerous goods. These are not available for online purchase. You may order the experiment/set up without the dangerous goods. For ordering the dangerous good or other information please contact our customer service center. Email: webshopuk@ld-didactic.de Telephone:+49 (0) 22 33 / 604 – 319 Go to contact form Order without dangerous goods ˟ Request an offer Name Email Submit Thank you for your inquiry. We will get back to you as soon as possible. ˟ This product contains dangerous substances! Description Components of equipment sets Experiment Instructions Description In the experiment P2.1.2.1, the volumetric expansion coefficients of water and ethanol are determined using a volume dilatometer made of glass. An attached riser tube with a known cross-section is used to measure the change in volume. i.e. the change in volume is determined from the rise height of the liquid. Components of equipment sets 1382 154 Dilatometer 1382 348 Thermometer, -10...+110 °C/0.2 K 1666 7678 Hotplate, 1500 W, 185 mm Ø 1664 1044 Beaker, DURAN, 400 ml, squat 1315 058 Single-pan suspension balance 311 1300 024 Stand base, V-shaped, small 1300 424 Stand rod, 47 cm, 12 mm diam. 2301 014 Leybold multiclamp 2666 5554 Universal clamp 0...80 mm 1671 97204 Ethanol, denaturated, 1 l Danger H225 H319 Experiment Instructions PDF (Experiment description)P2.1.2.1 Determination of volumetric expansion coefficient of liquids Print page Physics Physics Equipment Physics Experiments Demonstration Experiments Basic Science Kits Advanced Student Experiments Standard Lab Proposals Chemistry Product catalogue chemistry/biology Catalogue of experiments - Chemistry Basic Science Kits Chemistry Advanced Student Experiments Biology Devices Versuchekatalog Biologie Basic Science Kits Biology Advanced Student Experiments Technology Automotive Electrical Engineering COM4LAB DON'T MISS A BIT! Subscribe to Newsletter Contact Us Technical Service & Support: Online Service Portal Do you have any questions or suggestions regarding our devices, products, experiments, equipment sets or about our software? You need spare parts?FORM Repair Service You would like to register a repair? FORM Europe For European distributors, please visit: European distributors Worldwide requests Please contact us per email: sales@ld-didactic.de Click here for our contact form. Contact Us LD DidacticLeybold ShopFeedback InstrumentsELWE Technik LD Didactic Group General Business Terms Privacy Policy Sitemap Imprint
13310	https://mechfamily-ju.com/storage/images/files/file_17314458349Odfj.pdf	ELECTRIC MACHINERY FUNDAMENTALS ELECTRIC MACHINERY FUNDAMENTALS FOURTH EDITION Stephen J. Chapman BAE SYSTEMS Australia Higher Education Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco SI. l ouis Bangkok Bogota Caracas Kuala l umpur Lisbon London Madrid Mexico City Milan Montreal New Delhi Santiago Seoul Singapore Sydney T aipei Toronto • Higher Education ELECTRIC MACHINERY RJNDAMENTALS. FOURTH EDITION Published by McGraw-Hill. a business unit of The McGraw-Hill Companies. Inc., 1221 Avenue of the Americas, New Yort. NY 10020. Copyright 0 2005, 1999. 1991. 1985 by The McGraw,Hill Companies. Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means. or stored in a database or retrieval system. without the prior written con' sent of The McGraw-Hill Companies. Inc., including. but not limited to, in any network or other electronic storage or transmission. or broadcast for distance learning. Some ancillaries. including electronic and prim components. may not be available to customers out, side the United States. This book is printed on acid'free paper. 1234567890DOaDOC09876543 ISBN 0--07- 246523--9 Publisher: Elizabeth A. Jones Senior sponsoring editor: Carlise Paulson Managing developmental editor: EmilyJ. Lupash Marketing manager: Val''"" R. Bercier Senior project manager: Sheila M. Frank Senior production supervisor: Laura Fuller Senior media project manager: Tammy Juran Senior designer: Da\·id W. Hash Lead photo research coordinator: Carrie K. Burger Compositor: GAC- Indianapolis Typeface: /0//2 Times Rotnlln Printer: R. R. Donnelley Crawfordsville. IN Libmry of Co n ~ress Gltalo~in~-in-l'ublic:ltion Data Chapman. Stephen J. Electric machinery fundamentals / Stephen Chapman. -4th ed. p. em. Includes index. ISBN 0-07- 246523--9 I. Electric machinery. I. Title. TK2000.C46 2005 621.31 ·042---dc22 www.mhhe.oom 2003065174 CIP THIS WORK IS DEDICATED WITH LOVE TO MY MOTHER, LOUISE G. CHAPMAN, ON THE OCCASION OF HER EIGHTY-RFfH BIRTHDAY. ABOUT THE AUTHOR Stephen J. Chapman received a B.S. in Electrical Engineering from Louisiana State University (1975) and an M.S.E. in Electrical Engineering from the Univer-sity of Central Florida ( 1979), and pursued further graduate studies at Rice University. From 1975 to 1980, he served as an officer in the U.S. Navy, assigned to teach electrical engineering at the U.S. Naval Nuclear Power School in Orlando, Florida. From 1980 to 1982, he was affiliated with the University of Houston, where he ran the power systems program in the College of Technology. From 1982 to 1988 and from 1991 to 1995, he served as a member of the technical stafT of tile Massachusetts Institute of Technology's Lincoln Laboratory, both at the main facility in Lexington, Massachusetts, and at the field site on Kwa-jalein Atoll in the Republic of the Marshall Islands. While there, he did research in radar signal processing systems. He ultimately became the leader of four large operational range instrumentation radars at the Kwajalein field site (TRADEX, ALTAIR, ALCOR, and MMW). From 1988 to 1991 , Mr. Chapman was a research engineer in Shell Devel-opment Company in Houston, Texas, where he did seismic signal processing re-search. He was also affiliated with the University of Houston, where he continued to teach on a part-time basis. Mr. Chapman is currently manager of systems modeling and operational analysis for BAE SYSTEMS Australia, in Melbourne. Mr. Chapman is a senior member of the Institute of Electrical and Elec-tronic Engineers (and several of its component societies). He is also a member of the Association for Computing Machinery and the Institution of Engineers (Australia). vu BRIEF CONTENTS Chapter 1 Introduction to Machinery Principles Chapter 2 Transformers 65 Chapter 3 Introduction to Power Electronics 152 Chapter 4 AC Machinery Fundamentals 230 Chapter 5 Synchronolls Generators 267 Chapter 6 Synchronolls Motors 346 Chapter 7 Induction Motors 380 Chapter 8 DC Machinery Fundamentals 473 Chapter 9 DC Motors and Generators 533 Chapter 10 Single-Phase and Special-Purpose Motors 633 Appendix A Three-Phase Circuits 681 Appendix B Coil Pitch and Distributed Windings 707 Appendix C Salient-Pole Theory ofSynchronolls Machines 727 Appendix D Tables of Constants and Conversion Factors 737 " TABLE OF CONTENTS Chapter 1 Introduction to Machinery Principles 1.1 Electrical Machines, Transformers, and Daily Life 1.2 A Note on Units and Notation Notation 2 1.3 Rotational Motion, Newton's Law, and Power Relationships 3 Angular Position (J I Angular Velocity w / Angular Acceleration a / Torque T / Newton 's Law o/Rotation I W ork W Power P I.. The Magnetic Field 8 Production of a Magnetic Field / Magnetic Circuits / Magnetic Behavior 01 Ferromagnetic Materials I Energy Losses in a Ferromagnetic Core 1.5 Faraday's Law-Induced Voltage from a Time-Changing Magnetic Field 28 1.6 Production of Induced Force on a Wire 32 1.7 Induced V oltage on a Conductor Moving in a Magnetic Field 34 I." The Linear OC Machine- A Simple Example 36 Starting the Linear DC Machine / The linear DC Machine as a Motor I The Linear DC Machine as a Generator I Starting Problems with the Linear Machine I.. Real, Reactive, and Apparent Power in AC Circuits 47 Alternative Fonns of the Power Equations I Complex Power I The Relationships beflt'een Impedance Angle, Current Angle, and Power I The Power Triangle 1.10 Summary 53 Questions 54 Problems 55 References 64 " XII TABLE OF CONTENTS Chapter 2 Transformers 65 2.1 Why Transfonners Are Important to Modern Life 66 2.2 Types and Construction of Transformers 66 2.3 The Ideal Transfonner 68 Power in an Ideal Transformer I Impedance TransfornUltion through a Transfornler I Analysis of Circuits Containing Ideal Transformers 2.4 Theory of Operation of Real Single-Phase Transformers 76 The Voltage Ratio across a Transformer I The Magnetization Current in a Real Transformer I The Current Ratio on a Transformer and the Dot Conrention 2.5 The Equivalent Circuit of a Transformer 86 The Exact Equivalent Circuit of a Real Transformer I ApproxinUlte Equivalent Circuits of a Transformer I Determining the Values of Components in the Transfonner Model 2.6 The Per-Unit System of Measurements 94 2.7 Transfonner Voltage Regulation and Efficiency 100 The Transformer Phasor Diagram I Transfonner E fficiency 2.8 Transfonner Taps and V oltage Regulation 108 2.9 The Autotransfonner 109 Voltage and Current Relationships in an Autotransformer I The Apparent Power Rating Advantage of Autotransfornlers I The Internal Impedance of an Autotransformer 2.10 Three-Phase Transfonners 11 6 Three-Phase Transformer Connections I The Per-Unit System for Three-Phase Transformers 2.11 Three-Phase Transfonnation Using Two Transformers 126 The Open-il (or V-V) Connection I The Open-"3'e-Open-Delta Connection I The Scott-T Connection I The Three-Phase T Connection 2.12 Transfonner Ratings and Related Problems 134 The Voltage and Frequency Ratings of a Transformer I The Apparent Power Rating of a Transfornler I The Problem of Current Inrnsh I The Transformer Nameplate 2.13 Instnunent Transformers 140 2.14 Swnmary 142 Questions 143 Problems 144 References 15 1 Chapter 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.' 3.' Chapter 4 4.1 TABLE OF CONTENTS XlU Introduction to Power Electronics Power Electronic Components The Diode / The Two- Wire Thyristor or PNPN Diode / The Three-Wire Thyristor of SCR / The Gate Turnoff Thyristor / The DlAC / The TRIA C / The Power Transistor / The Insulated-Gate Bipolar Transistor / Power atui Speed Comparison of Power Electronic Components Basic Rectifier Circuits The Half-Wave Rectifier / The Full-Wave Rectifier / The Three-Phase Half-Wave Rectifier / The Three-Phase Full-Wave Rectifier / Filtering Rectifier Output Pulse Circuits A Relaxation Oscillator Using a PNPN Diode / Pulse Synchronization Voltage Variation by AC Phase Control AC Phase Controlfora DC Load Drivenfrom an AC Source / AC Phase Angle Control for an AC Load / The Effect of Inductive Loads on Phase Angle Control DC-to-DC Power Control-Choppers Forced Commutation in Chopper Circuits / Series-Capacitor Commutation Circuits / Parallel-Capacitor Commutation Circuits Inverters The Rectifier / External Commutation lnverters / Self-Commutation Inverters / A Single-Phase Current Source Inverter / A Three-Phase Current Source lnverter / A Three-Phase Voltage Source Inverter / Pulse-Width Modulation lnverters Cycloconverters Basic Concepts / Noncirculating Current Cycloconverters / Circulating Current Cycloconverters Hannonic Problems Summary Questions Problems References AC Machinery Fundamentals A Simple Loop in a Uniform Magnetic Field The Voltage Induced in a Simple Rotating Loop / The Torque lnduced in a Current-Cart}'ing Loop 152 152 163 171 177 186 193 209 218 221 223 223 229 230 230 XIV TABLEOF CONTENTS 4.2 The Rotating Magnetic Field 238 Proof of the Rotating Magnetic Field Concept I The Relationship between Electrical Frequency and the Speed of Magnetic Field Rotation I Reversing the Direction of Magnetic Field Rotation 4.3 Magnetomotive Force and Flux Distribution on AC Machines 246 4.4 Induced Voltage in AC Machines 250 The Induced Voltage in a Coil on a Two-Pole Stator I The Induced Voltage in a Three-Phase Set of Coils I The RMS Voltage in a Three-Phase Stator 4.5 Induced Torque in an AC Machine 255 4 .• Wmding Insulation in an AC Machine 258 4.7 AC Machine Power Flows and Losses 26 1 The Losses in AC Machines I The Power-Flow Diagram 4.S Voltage Regulation and Speed Regulation 262 4.9 Swnmary 264 Questions 265 Problems 265 References 266 Chapter 5 Synchronous Generators 267 5. 1 Synchronous Generator Construction 267 5.2 The Speed of Rotation of a Synchronous Generator 272 5.3 The Internal Generated Voltage of a Synchronous Generator 273 5.4 The Equivalent Circuit of a Synchronous Generator 274 5.5 The Phasor Diagram of a Synchronous Generator 279 5 .• Power and Torque in Synchronous Generators 280 5.7 Measuring Synchronous Generator Model Parameters 283 The Short-Circuit Ratio 5.8 The Synchronous Generator Operating Alone 288 The E ffect of Load Changes on Synchronous Generator Operating Alone I Example Problems 5.9 Parallel Operation of AC Generators 299 The Conditions Requiredfor Paralleling I The General Procedure for Paralleling Generators I Frequency-Power and Voltage-Reactive Power Characteristics of a Synchronous Generator I Operation of Generators in Parallel with Large Power Systems I Operation of Generators in Parallel with Other Generators of the Same Size 5.10 Synchronous Generator Transients 319 Transient Stability of Synchronous Generators I Short-Circuit Transients in Synchronous Generators 5.11 5. 12 Chapter 6 6.1 6.2 6.3 6.4 6.5 6.6 Chapter 7 7.1 7.2 7.3 TABLE OF CONTENTS XV Synchronous Generator Ratings The Voltage, Speed, and Frequency Ratings / Apparent Power atui Power-Factor Ratings / Synchronous Generator Capability CUf1Jes / Short-Time Operation and Sef1Jice Factor Summary Questions Problems References Synchronous Motors Basic Principles of Motor Operation The Equiralent Circuit of a Synchronous Motor / The Synchronous Motor from a Magnetic Field Perspective Steady-State Synchronous Motor Operation The Synchronous Motor Torque-Speed Characteristic CUf1Je / The Effect of Load Changes on a Synchronous Motor / The Effect of Field Changes on a Synchronous Motor / The Synchronous Motor atui Power, Factor Correction / The Synchronous Capacitor or Synchronous Condenser Starting Synchronous Motors Motor Starting by Reduced Electrical Frequency / Motor Starting with an utemal Prime Mover / Motor Starting by Using Amortisseur Windings / The Effect of Amortisseur Windings on Motor Stability Synchronous Generators and Synchronous Motors Synchronous Motor Ratings Summary Questions Problems References Induction Motors Induction Motor Construction Basic Induction Motor Concepts The Development of Induced Torque in an ltuiuction Motor / The Concept of Rotor Slip / The Electrical Frequency on the Rotor The Equivalent Circuit of an Induction Motor The Transformer Model of an Induction Motor / The Rotor Circuit Model/The Final Equiralent Circuit 326 336 337 338 345 346 346 350 364 37 1 372 373 374 374 379 380 380 384 388 XVI TABLE OF CONTENTS 7.4 Power and Torque in Induction Motors 394 Losses and the Pml'er-Flow Diagram I Power atui Torque in an Induction Motor I Separating the Rotor Copper Losses and the Pmwr Converted in an lnduction Motor S Equivalent Cirr:uit 7.5 Induction Motor Torque-Speed Characteristics 401 lnduced Torque from a Physical Statuipoint IThe Derivation of the lnduction Motor ltuiuced-Torque Equation I Comments on the Induction Motor Torque-Speed Cun'e I Maximum (Pullout) Torque in an ltuiuction Motor 7 .• Variations in Induction Motor Torque-Speed Characteristics 416 Control of Motor Characteristics by Cage Rotor Design I Deep-Bar and Double-Cage Rotor Designs I lnduction Motor Design Classes 7.7 Trends in Induction Motor Design 426 7.8 Starting Induction Motors 430 lnduction Motor Starting Circuits 7.9 Speed Control of Induction Motors 434 lnduction Motor Speed Control by Pole Changing I Speed Control by Changing the Line Frequency I Speed Control by Changing the Line Voltage I Speed Control by Changing the Rotor Resistance 7.10 Solid-State Induction Motor Drives 444 Frequency (Speed) Adjustment I A Choice of Voltage and Frequency Patterns I Independently Adjustable Acceleration atui Deceleration Ramps I Motor Protection 7.11 Detennining Circuit Model Parameters 452 The No-Load Test I The DC Test for Stator Resistance I The Locked-Rotor Test 7.12 The Induction Generator 460 The lnduction Generator Operating Alone I lnduction Generator Applications 7.13 Induction Motor Ratings 464 7.14 Swnmary 466 Questions 467 Problems 468 Rererences 472 Chapter 8 DC Machinery Fundamentals 473 8.1 A Simple Rotating Loop between Curved Pole Faces 473 TABLE OF CONTENTS XVU The lliltage lnduced in a Rotating Loop / Getting DC V oltage out of the Rotating Loop / The Induced T orque in the Rotating Loop 8.2 Commutation in a Simple Four-Loop IX Machine 485 8.3 Commutation and Armature Construction in Real DC Machines 490 The Rotor Coils / Connections to the Commutator Segments / The Lap Winding / The Wave Winding / The Frog-Leg Winding 8.4 Problems with Conunutation in Real Machines 502 Armature Reaction / L dildt Voltages / Solutions to the Problems with Commutation 8.5 The Internal Generated Voltage and Induced Torque Equations of Real DC Machines 514 8.6 The Construction of DC Machines 518 Pole and Frame Construction / Rotor or Armature Constrnction / Commutator and Brushes / Winding Insulation 8.7 Power Flow and Losses in DC Machines 524 The Losses in DC Machines / The Power-Flow Diagram 8.8 Summary 527 Questions 527 Problems 527 References 530 Chapter 9 DC Motors and Generators 533 9.1 Introduction to DC Motors 533 9.2 The Equivalent Circuit of a IX Motor 535 9.3 The Magnetization Curve of a DC Machine 536 9.4 Separately Excited and Shunt IX Motors 538 The Tenninal Characteristic of a Shunt DC Motor / Nonlinear Analysis of a Shunt DC Motor / Speed Control of Shunt DC Motors / The Effect of an Open Field Circuit 9.5 The Pennanent-Magnet DC Motor 559 9.6 The Series IX Motor 562 Induced Torque in a Series DC Motor / The Terminal Characteristic of a Series DC Motor / Speed Control of Series DC Motors 9.7 The Compounded DC Motor 568 The Torque-Speed Characteristic of a Cumulatively Compounded DC Motor / The Torque- Speed XVIII TABLE OF CONTENTS Characteristic of a Differentially Compoutuied DC Motor / The Nonlinear Analysis of Compounded DC Motors / Speed Control in the Cumulatively Compoutuied DC Motor 9.8 DC Motor Starters 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 9.17 Chapter 10 10.1 DC Motor Problems on Starting / DC Motor Starting Circuits The Ward-Leonard System and Solid-State Speed Controllers Protection Circuit Section / StartlStop Circuit Section / High.Power Electronics Section / Low-Power Electronics Section DC Motor Efficiency Calculations Introduction to IX Generators The Separately Excited Generator The Terminal Characteristic of a Separately Excited DC Generator / Control of Terminal Voltage / Nonlinear Analysis of a Separately Excited DC Generator The Shunt DC Generator Voltage Buildup in a Shunt Generator / The Tenninal Characteristic of a Shunt DC Generator / V oltage Control fo r a Shunt DC Generator / The Analysis of Shunt DC Generators The Series DC Generator The Terminal Characteristic of a Series Generator The Crunul atively Compounded DC Generator The Terminal Characteristic of a Cumulatively Compounded DC Generator / V oltage Control of Cumulatively Compounded DC Generators / Analysis of Cumulatively Compounded DC Generators The Differentially CompolUlded DC Generator The Terminal Characteristic of a Differentially Compounded DC Generator / Voltage Control of Differentially Compounded DC Generators / Graphical Analysis of a Differentially Compounded DC Generator Srunmary Questions Problems References Single-Phase and Special-Purpose Motors The Universal Motor Applications of Universal Motors / Speed Control of Universal Motors 573 582 592 594 596 602 608 611 615 619 620 621 631 633 634 TA BLE OF CONTENTS XIX 10.2 Introduction to Single-Phase Induction Motors 637 The Double.Rerolving-Field Theory of Single.Phase Induction Motors / The Cross· Field Theory of Single. Phase Induction Motors 10.3 Starting Single-Phase Induction Motors 646 Split-Phase Windings / Capacitor.Start Motors / Pennanent Split-Capacitor and Capacitor.Start, Capacitor.Run Motors / Shaded-Pole Motors / Comparison of Single.Phase Induction Motors 10.4 Speed Control of Single-Phase Induction Motors 656 10.5 The Circuit Model of a Single-Phase Induction Motor 658 Circuit Analysis with the Single-Phase Induction Motor Equiralent Circuit 10.6 Other Types of Motors 665 Reluctance Motors / Hysteresis Motors / Stepper Motors / Brushless DC Motors 10.7 Summary 677 Questions 678 Problems 679 References 680 Appendix A Three-Phase Circuits 68 1 A.I Generation of Three-Phase Voltages and Currents 68 1 A.2 Voltages and Currents in a Three-Phase Circuit 685 Voltages and Currents in the ~~'e (Y) Connection / Voltages and Currents in the Delta (8) Connection A.3 Power Relationships in lbree-Phase Circuits 690 Three-Phase Po .... er Equations Involving Phase Quantities / Three-Phase Po .... er Equations Involving Line Quantities A.4 Analysis of Balanced Three-Phase Systems 693 A.5 One-Line Diagrams 700 A.6 Using the Power Triangle 700 Qnestions 703 Problems 704 References 706 Appendix B Coil Pitch and Distributed Windings 707 8.1 The Effect of Coil Pitch on AC Machines 707 The Pitch of a Coil / The Induced Voltage of a Fractional· Pitch Coil / Harmonic Problems and Fractional-Pitch Windings XX TABLE OF CONTENTS 8.2 Distributed Windings in AC Machines 71 6 The Breadth or Distribution Factor I The Generated Voltage Including Distribution E ffects / Tooth or Slot Harmonics 8.3 Swnmary 724 Questions 725 Problems 725 References 726 Appendix C Salient-Pole Theory of Synchronous Machines 727 C. I Development of the Equivalent Circuit of a Salient-Pole Synchronous Generator 728 C.2 Torque and Power Equations of Salient-Pole Machine 734 Problems 735 Appendix D Tables of Constants and Conversion Factors 737 PREFACE I n the years since the first edition of Electric Machinery Fundamentals was published, there has been rapid advance in the development of larger and more sophisticated solid-state motor drive packages. The first edition of this book stated that de motors were the method of choice for demanding variable-speed applica-tions. 11131 statement is no longer true today. Now, the system of choice for speed control applications is most often an ac induction motor with a solid-state motor drive. DC motors have been largely relegated to special-purpose applications where a de power source is readily available, such as in automotive electrical systems. The third edition orthe book was extensively restructured to reflect these changes. 1lle material on ac motors and generators is now covered in Chapters 4 through 7, before the material on dc machines. In addition, the dc machinery cov-erage was reduced compared to earlier editions. 1lle fourth edition continues with this same basic structure. Chapter I provides an introduction to basic machinery concepts, and con-cludes by applying those concepts to a linear dc machine, which is the simplest possible example of a machine. Glapter 2 covers transformers, and Chapter 3 is an introduction to solid-state power electronic circuits. The material in Chapter 3 is optional, but it supports ac and dc motor control discussions in Chapters 7, 9, and 10. After Chapter 3, an instructor may choose to teach either dc or ac machin-ery first. Chapters 4 through 9 cover ac machinery, and Chapters 8 and 9 cover dc machinery. 1llese chapter sequences have been made completely independent of each other, so that instructors can cover the material in the order that best suits their needs. For example, a one-semester course with a primary concentration in ac machinery might consist of parts of Chapters I to 7, with any remaining time devoted to dc machinery. A one-semester course with a primary concentration in dc machinery might consist of parts of Chapters I, 3, 8, and 9, with any remain-ing time devoted to ac machinery. Chapter \0 is devoted to single-phase and special-purpose motors, such as universal motors, stepper motors, brushless dc motors, and shaded-pole motors. XXI XXII PREFACE TIle homework problems and the ends of chapters have been revised and corrected, and more than 70 percent of the problems are either new or modified since the last edition. In recent years, there have been major changes in the methods used to teach machinery to electrical engineering and electrical technology students. Excellent analytical tools such as MATLAB have become widely available in university en-gineering curricula. TIlese tools make very complex calculations simple to per-form, and allow students to explore the behavior of problems interactively. This edition of Electric Machinery Fundamentals makes sclected use of MATLAB to enhance a student's learning experience where appropriate. For example, students use MATLAB in Chapter 7 to calculate the torque-speed characteristics of induc-tion motors and to explore the properties of double-cage induction motors. TIlis text does not teach MATLAB; it assumes that the student is familiar with it through previous work. Also, the book does not depend on a student hav-ing MATLAB. MATLAB provides an enhancement to the learning experience if it is available, but if it is not, the examples involving MATLAB can simply be skipped, and the remainder of the text still makes sensc. Supplemental materials supporting the book are available from the book's website, at www.mhhe.com/engcslelectricallchapman. The materials available at that address include MATLAB source code, pointers to sites of interest to ma-chinery students, a list of errata in the text, some supplemental topics that are not covered in the main text, and supplemental MATLAB tools. TIlis book would never have been possible without the help of dozens of people over the past 18 years. I am not able to acknowledge them al l here, but I would especially like to thank Charles P. LeMone, Teruo Nakawaga, and Tadeo Mose of Toshiba International Corporation for their invaluable help with the solid-state machinery control material in Chapter 3. I would also like to thank Jeffrey Kostecki, Jim Wright, and others at Marathon Electric Company for sup-pi ying measured data from some of the real generators that the company builds. TIleir material has enhanced this revision. Finally, I would like to thank my wife Rosa and our children Avi, David, Rachel, Aaron, Sarah, Naomi, Shira, and Devorah for their forbearance during the revision process. I couldn't imagine a better incentive to write! Stepllell J. Chapman Metboume, Victoria, Australia 1.1 ELECTRICAL MACHINES, TRANSFORMERS, AND DAILY LIFE CHAPTER 1 INTRODUCTION TO MACHINERY PRINCIPLES An electrical machine is a device that can convert either mechanical energy to electrical energy or electrical energy to mechanical energy. When such a device is used to convert mechanical energy to e lectrical energy, it is called a generator. When it converts electrical energy to mechanical energy, it is called a motor. Since any given electrical machine can convert power in either direction, any machine can be used as either a generator or a motor. Almost all practical motors and gen-erators convert energy from one form to another through the action of a magnetic fie ld, and only machines using magnetic fields to perform such conversions are considered in this book. The transformer is an electrical device that is closely related to electrical machines. It converts ac electrical energy at one voltage level to ac electrical en-ergy at another voltage level. Since transfonners operate on the same principles as generators and motors, depending on the action ofa magnetic field to accomplish the change in voltage level, they are usually studied together with generators and motors. These three types of electric devices are ubiquitous in modern daily life. Electric motors in the home run refrigerators, freezers, vacuum cleaners, blenders, air conditioners, fans, and many similar appliances. In the workplace, motors pro-vide the motive power for almost all tools. Of course, generators are necessary to supply the power used by alJ these motors. I 2 ELECTRIC MACHINERY FUNDAMENTALS Why are electric motors and generators so common? The answer is very simple: Electric power is a clean and efficient energy source that is easy to trans-mit over long distances, and easy to control. An electric motor does not require constant ventilation and fuel the way that an internal-combustion engine does, so the motor is very well suited for use in environments where the pollutants associ-ated with combustion are not desirable. Instead, heat or mechanical energy can be converted to electrical fonn at a distant location, the energy can be transmitted over long distances to the place where it is to be used, and it can be used cleanly in any horne, office, or factory. Transfonners aid this process by reducing the en-ergy loss between the point of electric power generation and the point of its use. 1.2 A NOTE ON UNITS AND NOTATION TIle design and study of electric machines and power systems are among the old-est areas of electrical engineering. Study began in the latter part of the nineteenth century. At that time, electrical units were being standardized internationally, and these units came to be universally used by engineers. Volts, amperes, ohms, watts, and similar units, which are part of the metric system of units, have long been used to describe electrical quantities in machines. In English-speaking countries, though, mechanical quantities had long been measured with the English system of units (inches, feet, pounds, etc.). This prac-tice was followed in the study of machines. TIlerefore, for many years the electri-cal and mechanical quantities of machines have been measured with different sys-tems of units. In 1954, a comprehensive system of units based on the metric system was adopted as an international standard. This system of units became known as the Systeme International (SI) and has been adopted throughout most of the world. The United States is practically the sole holdout--even Britain and Canada have switched over to Sl. TIle SI units will inevitably become standard in the United States as time goes by, and professional societies such as the Institute of Electrical and Elec-tronics Engineers (IEEE) have standardized on metric units for all work. How-ever, many people have grown up using English units, and this system will remain in daily use for a long time. Engineering students and working engineers in the United States today must be familiar with both sets of units, since they will en-counter both throughout their professional lives. Therefore, this book includes problems and examples using both SI and English units. TIle emphasis in the ex-amples is on SI units, but the older system is not entirely neglected. Notation In this book, vectors, electrical phasors, and other complex values are shown in bold face (e.g., F), while scalars are shown in italic face (e.g., R). In addition, a special font is used to represent magnetic quantities such as magnetomotive force (e.g., 'iJ). INTRODUCTION TO MACHINERY PRINCIPLES 3 1.3 ROTATIONAL MOTION, NEWTON'S LAW, AND POWER RELATIONSHIPS Almost all electric machines rotate about an axis, called the shaft of the machine. Because of the rotational nature of machinery, it is important to have a basic un-derstanding of rotational motion. This section contains a brief review of the con-cepts of distance, velocity, acceleration, Newton's law, and power as they apply to rotating machinery. For a more detailed discussion of the concepts of rotational dynamics, see References 2, 4, and 5. In general, a three-dimensional vector is required to completely describe the rotation of an object in space. However, machines nonnally turn on a fixed shaft, so their rotation is restricted to one angu lar dimension. Relative to a given end of the machine's shaft, the direction of rotation can be described as either clockwise (CW) or counterclockwise (CCW). For the purpose of this volume, a counter-clockwise angle of rotation is assumed to be positive, and a clockwise one is as-sumed to be negative. For rotation about a fixed shaft, all the concepts in this sec-tion reduce to scalars. Each major concept of rotational motion is defined below and is related to the corresponding idea from linear motion. Angular Position 0 The angular position () of an object is the angle at which it is oriented, measured from some arbitrary reference point. Angular position is usually measured in radians or degrees. It corresponds to the linear concept of distance along a line. Angular Velocity w Angular velocity (or speed) is the rate of change in angular position with respect to time. It is assumed positive if the rotation is in a counterclockwise direction. Angular velocity is the rotational analog of the concept of velocity on a line. One-dimensional linear velocity along a line is defmed as the rate of change of the dis-placement along the line (r) with respect to time d, v~ -dt (I-I ) Similarly, angular velocity w is defined as the rate of change of the angular dis-placement () with respect to time. w = de dt (1- 2) If the units of angular position are radians, then angular velocity is measured in ra-dians per second. In dealing with ordinary electric machines, engineers often use units other than radians per second to describe shaft speed. Frequently, the speed is given in 4 ELECTRIC MACHINERY FUNDAMENTALS revolutions per second or revolutions per minute. Because speed is such an im-portant quantity in the study of machines, it is customary to use different symbols for speed when it is expressed in different units. By using these different symbols, any possible confusion as to the units intended is minimized. TIle following sym-bols are used in this book to describe angular velocity: Wm angular velocity expressed in radians per second f.. angular velocity expressed in revolutions per second nm angular velocity expressed in revolutions per minute TIle subscript m on these symbols indicates a mechanical quantity, as opposed to an electrical quantity. If there is no possibility of confusion between mechanical and electrical quantities, the subscript is often left out. TIlese measures of shaft speed are related to each other by the following equations: (l- 3a) ( 1 - 3b) Angular Acceleration a Angular acceleration is the rate of change in angular velocity with respect to time. It is assumed positive if the angular velocity is increasing in an algebraic sense. Angular acceleration is the rotational analog of the concept of acceleration on a line. Just as one-dimensional linear acceleration is defined by the equation angular acceleration is defined by d, a~ -dt a = dw dt (1-4) (1- 5) I f the units of angular velocity are radians per second, then angular acceleration is measured in radians per second squared. Torque "T In linear motion, aforce applied to an object causes its velocity to change. In the absence of a net force on the object, its velocity is constant. TIle greater the force applied to the object, the more rapidly its velocity changes. TIlere exists a similar concept for rotation. When an object is rotating, its angular velocity is constant unless a torque is present on it. The greater the torque on the object, the more rapidly the angular velocity of the object changes. What is torque? It can loosely be called the "twisting force" on an object. Intuitively, torque is fairly easy to understand. Imagine a cylinder that is free to FIGURE I- I , , , , • (a) r=O Torque is zero INTRODUCTION TO MACHINERY PRINCIPLES 5 • F , Torque is counterclockwise 'bJ (a) A force applied to a cylinder so that it passes through the axis of rotation. T = O. (b) A force applied to a cylinder so that its line of action misses the axis of rotation. Here T is counterclockwise. rotate aboul its axis. If a force is applied to Ihe cylinder in such a way thai its line of action passes through the axis (Figure I-Ia), then the cylinder will not rotate. However, if the same force is placed so that its line of action passes 1 0 Ihe righl of Ihe axis (Figure I-I b), then Ihe cylinder will lend 1 0 rotate in a counterclockwise direction. The torque or twisting action on the cylinder depends on ( I) the magni-tude of the applied force and (2) the distance between the axis of rotation and the line of action of the force. The torque on an object is defined as the product of the force applied 1 0 the object and the smallest distance between the line of action of the force and the ob-ject's axis of rotation. If r is a vector pointing from the axis of rotation to the poinl of applicalion of the force, and if F is the applied force, then the torque can be de-scribed as 7" = (force applied)(perpendicular distance) = (F) (r sin ()) = rF sin () (1-6) where () is the angle between the vector r and the vector F. The direction of the lorque is clockwise if it would tend 1 0 cause a clockwise rotation and counler-clockwise if it wou Id tend to cause a counterclockwise rotalion (Figure 1-2). The units of torque are newton-meters in SI units and pound-feel in lhe Eng-lish system. 6 ELECTRIC MACHINERY FUNDAMENTALS rsin(1800 - 1I)=rsinll ~ , __ _ J , , , , , , I 1800 _ II T = (perpendicular distance) (force) T = (r sin 9)F. cou nterclockwise Newton's Law of Rotation , , , , F ' FIGURE 1 -1 Derivation of the equation for the torque on an object. Newton's law for objects moving along a straight line describes the relationship between the force applied to an object and its resulting acceleration. This rela-tionship is given by the equation where F = nuJ F = net force applied to an object m = mass of the object a = resulting acceleration (1- 7) In SI units, force is measured in newtons, mass in kilograms, and acceleration in meters per second squared. In the English system. force is measured in pounds, mass in slugs, and acceleration in feet per second squared. A similar equation describes the relationship between the torque applied to an object and its resulting angular acceleration. This relationship, cal led Newton S law ofrotation, is given by the equation 7" = Ja (1- 8) where 7" is the net applied torque in newton-meters or pound-feet and a is the re-sulting angular acceleration in radians per second squared. 1lle tenn J serves the same purpose as an object's mass in linear motion. It is called the moment of ineT1ia of the object and is measured in kilogram-meters squared or slug-feet squared. Calculation of the moment of inertia of an object is beyond the scope of this book. For infonnation about it see Re f. 2. INTRODUCTION TO MACHINERY PRINCIPLES 7 Work W For linear motion, work is defined as the application of a force through a distance. In equation fonn, W = f Fdr (1- 9) where it is assumed that the force is coil inear with the direction of motion. For the special case of a constant force applied collinearly with the direction of motion, this equation becomes just W = Fr (1 -1 0) The units of work are joules in SI and foot-pounds in the English system. For rotational motion, work is the application of a torque through an angle. Here the equation for work is w ~ f ,dO (I -II ) and if the torque is constant, W = TO ( 1-12) Power P Power is the rate of doing work, or the increase in work per unit time. The equa-tion for power is p = dW dt ( 1-13) It is usually measured in joules per second (watts), but also can be measured in foot-pounds per second or in horsepower. By this defmition, and assuming that force is constant and collinear with the direction of motion, power is given by p = dd~ = :r (Fr) = F (~;) = Fv ( 1-1 4) Similarly, assuming constant torque, power in rotational motion is given by p = dd~ = :r (TO) = T(~~) = TW p = TW ( 1-15) Equation (1-1 5) is very important in the study of electric machinery, because it can describe the mechanical power on the shaft of a motor or generator. Equation (1-I 5) is the correct relationship runong power, torque, and speed if power is measured in watts, torque in newton-meters, and speed in radians per sec-ond. If other units are used to measure any of the above quantities, then a constant 8 ELECTRIC MACHINERY FUNDAMENTALS must be intnx:luced into the equation for unit conversion factors. It is still common in U.S. engineering practice to measure torque in pound-feet, speed in revolutions per minute, and power in either watts or horsepower. If the appropriate conversion factors are included in each tenn, then Equation (I-IS) becomes T (lb-ft) n (r/min) P (watts) = 7.04 ( 1-16) P (h ) _ T (lb-ft) n (r/min) orsepower -5252 ( 1-17) where torque is measured in pound-feet and speed is measured in revolutions per minute. 1.4 THE MAGNETIC FIELD As previously stated, magnetic fields are the fundrunental mechanism by which en-ergy is converted from one fonn to another in motors, generators, and transfonn-ers. Four basic principles describe how magnetic fields are used in these devices: I. A current-carrying wire produces a magnetic field in the area around it. 2. A time-changing magnetic field induces a voltage in a coil of wire ifit passes through that coil. (This is the basis of transfonner action.) 3. A current-carrying wire in the presence of a magnetic field has a force in-duced on it. (This is the basis of motor action.) 4. A moving wire in the presence of a magnetic field has a voltage induced in it. (This is the basis of generator action.) TIlis section describes and elaborates on the production of a magnetic field by a current-carrying wire, while later sections of this chapter explain the remaining three principles. Production of a Magnetic Field TIle basic law governing the production of a magnetic field by a current is Ampere's law: ( 1-18) where H is the magnetic field intensity produced by the current I""" and dl is a dif-ferential element of length along the path of integration. In SI units, I is measured in amperes and H is measured in ampere-turns per meter. To better understand the meaning of this equation, it is helpful to apply it to the simple example in Figure -3. Figure 1-3 shows a rectangular core with a winding of N turns of wire wrapped about one leg of the core. If the core is composed of iron or certain other similar metals (collectively calledferromagnefic mnterials), essentially all the magnetic field produced by the current will remain inside the core, so the path of integration in Ampere's law is the mean path length of the core (. TIle current INTRODUCTION TO MACHINERY PRINCIPLES 9 Mean path length Ie FIGURE 1 -3 A simple magnetic core. Nturns It~'1I--- Cross-sectional =.A passing within the path of integration I""" is then Ni, since the coil of wire cuts the path of integration Ntimes while carrying current i. Ampere's law thus becomes H( = Ni ( 1-19) Here H is the magnitude of the magnetic field intensity vector H. Therefore, the magnitude or the magnetic field intensity in the core due to the applied current is H =Ni Ie ( 1-20) The magnetic field intensity H is in a sense a measure of the "effort" that a current is putting into the establishment of a magnetic field. The strength of the magnetic field nux prOOuced in the core also depends on the material of the core. The relationship between the magnetic field intensity H and the resulting mag-netic flux density B produced within a material is given by ( 1-21) where H = magnetic field intensity /L = magnetic penneabi/ity of material B = resulting magnetic flux density prOOuced TIle actual magnetic flux density produced in a piece of material is thus given by a product of two tenns: H, representing the effort exerted by the current to establish a magnetic field /L, representing the relative ease of establishing a magnetic field in a given material 10 ELECIRIC MACHINERY FUNDAMENTALS The units of magnetic field intensity are ampere-turns per meter, the units of per-meability are henrys per meter, and the units of the resulting flux density are webers per square meter, known as teslas (T). TIle penneability of free space is called J.Lo, and its value is /J.o = 47T X 10- 7 Him ( 1-22) TIle penneability of any other material compared to the penneability of free space is called its relative permeability: - " /L, -J.Lo (1-23) Relative penneability is a convenient way to compare the magnetizability of materials. For example, the steels used in modern machines have relative penne-abilities of 2000 to 6000 or even more. This means that, for a given amount of current, 2000 to 6000 times more flux is established in a piece of steel than in a corresponding area of air. (The penneability of air is essentially the same as the penneability of free space.) Obviously, the metals in a transformer or motor core play an extremely important part in increasing and concentrating the magnetic flux in the device. Also, because the permeability of iron is so much higher than that of air, the great majority of the flux in an iron core like that in Figure 1-3 remains inside the core instead of traveling through the surrounding air, which has much lower per-meability. The small leakage flux that does leave the iron core is very important in detennining the flux linkages between coils and the self-inductances of coils in transformers and motors. In a core such as the one shown in Figure 1-3, the magnitude of the flux density is given by ( 1-24) Now the total flux in a given area is given by (1-25a) where dA is the differential unit of area. If the flux density vector is perpendicu-lar to a plane of areaA, and if the fl ux density is constant throughout the area, then this equation reduces to (I - 25b) TIlUS, the total flux in the core in Figure 1-3 due to the current i in the wind-ing is ( 1-26) where A is the cross-sectional area of the core. INTRODUCTION TO MACHINERY PRINCIPLES II --I " v + ) -R + g= Ni -I v = -R ", ,b, FIGURE 1-4 (3) A simple electric cin:uit. (b) The magnetic circuit analog to a transformer COTe. Magnetic Circuits In Equation ( 1 - 26) we see that the current in a coil of wire wrapped around a core produces a magnetic nux in the core. This is in some sense analogous to a voltage in an electric circuit producing a current now. It is possible to define a "magnetic circuit" whose behavior is governed by equations analogous to those for an elec-tric circuit. The magnetic circuit model of magnetic behavior is often used in the design of electric machines and transfonners to simplify the otherwise quite com-plex design process. In a simple electric circuit such as the one shown in Figure 1-4a, the volt-age source V drives a current J around the circuit through a resistance R. The rela-tionship between these quantities is given by Ohm's law: V = JR In the electric circuit, it is the voltage or electromotive force that drives the cur-rent now. By analogy, the corresponding quantity in the magnetic circuit is called the magnetomotive force (mmI). The rnagnetomotive force of the magnetic circuit is equal to the effective current now applied to the core, or g=Ni ( 1 - 27) where gis the symbol for magnetomotive force, measured in ampere-turns. Like the voltage source in the electric circuit, the magnetomotive force in the magnetic circuit has a polarity associated with it. The positive end of the mmf source is the end from which the nux exits, and the negative end of the mmf source is the end at which the nux reenters. The polarity of the mmf from a coil of wire can be detennined from a modification of the right-hand rule: If the fillgers of the right hand curl in the direction of the current now in a coil of wire, then the thumb wi ll point in the direction of the positive rnrnf (see Figure 1 - 5). In an electric circuit, the applied voltage causes a current J to flow. Simi-larly, in a magnetic circuit, the applied magnetomotive force causes nux <p to be produced. The relationship between voltage and current in an electric circuit is 12 ELECIRIC MACHINERY FUNDAMENTALS / / ; rr='-N I I ""GURE l-S Determining the polarity of a magnetomotive force source in a magnetic cirwit. Ohm's law (V = IR); similarly, the relationship between magnetomotive force and flux is where g = magnetomotive force of circuit <p = flux of circuit CR = reluctance of circuit ( 1 - 28) 1lle reluctance of a magnetic circuit is the counterpart of electrical resistance, and its units are ampere-turns per weber. 1llere is also a magnetic analog of conductance. Just as the conductance of an electric circuit is the reciprocal of its resistance, the permeance cP of a magnetic circuit is the reciprocal of its rei uctance: ( 1 - 29) The relationship belween rnagnetomotive force and flux can lhus be expressed as ( 1 - 30) Under some circumstances, it is easier to work with the penneance of a magnetic circuit than with its reluctance. INTRODUcnONTO MACHINERY PRINCIPLES 13 What is the reluctance of the core in Figure 1-3? The resulting flux in this core is given by Equation (1-26): (1-26) ( 1-31) By comparing Equation ( 1-31) with Equation ( 1-28), we see that the reluctance of the core is (1-32) Reluctances in a magnetic circuit obey the same rules as resistances in an electric circuit. TIle equivalent reluctance of a number of reluctances in series is just the sum of the individual reluctances: (1-33) Similarly, reluctances in parallel combine according to the equation (1-34) Penneances in series and parallel obey the same rules as electrical conductances. Calculations of the flux in a core performed by using the magnetic circuit concepts are always approximations-at best, they are accurate to within about 5 percent of the real answer. lllere are a number of reasons for this inherent inaccuracy: I. TIle magnetic circuit concept assumes that all flux is confilled within a mag-netic core. Unfortunately, this is not quite true. The permeability of a ferro-magnetic core is 2(x)() to 6()(x) times that of air, but a small fraction of the flux escapes from the core into the surrounding low-permeability air. This flux outside the core is called leakage flux, and it plays a very important role in electric machine design. 2. TIle calculation of reluctance assumes a certain mean path length and cross-sectional area for the core. These assumptions are not really very good, espe-cially at corners. 3. In ferromagnetic materials, the permeabi lity varies with the amount of flux already in the material. This nonJinear effect is described in detail. It adds yet another source of error to magnetic circuit analysis, since the reluctances used in magnetic circuit calculations depend on the penneability of the material. 14 ELECIRIC MACHINERY FUNDAMENTALS N s ""GURE 1-6 The fringing effect of a magnetic field at an air gap. Note the increased cross-sectional area of the air gap compared with the cross-sectional area of the metal. 4. If there are air gaps in the flux path in a core, the effective cross-sectional area of the air gap will be larger than the cross-sectional area of the iron core on either side. The extra effective area is caused by the "fringing effect" of the magnetic field at the air gap (Figure 1-6). It is possible to partially offset these inherent sources of error by using a "cor-rected" or "effective" mean path length and the cross-sectional area instead of the actual physical length and area in the calculations. TIlere are many inherent limitations to the concept of a magnetic circuit, but it is still the easiest design tool avai lable for calculating fluxes in practical ma-chinery design. Exact calculations using Maxwell's equations are just too diffi-cult, and they are not needed anyway, since satisfactory results may be achieved with this approximate method. TIle following examples illustrate basic magnetic circuit calculations. Note that in these examples the answers are given to three significant digits. Example I-t. A ferromagnetic core is shown in Figure 1-7a. Three sides of this core are of unifonn width. while the fourth side is somewhat thinner. The depth of the core (into the page) is 10 cm. and the other dimensions are shown in the figure. There is a 2()()'" turn coil wrapped around the left side of the core. Assruning relative penneability I-Lr of 2500. how much flux will be produced by a I-A input current? Solutio" We will solve this problem twice. once by hand and once by a MATLAB program. and show that both approaches yield the same answer. Three sides of the core have the same cross-sectional areas. while the fourth side has a different area. Thus. the core can be divided into two regions: (I) the single thiImer side and (2) the other three sides taken together. The magnetic circuit corresponding to this core is shown in Figure 1-7b. INTRODUcnONTO MACHINERY PRINCIPLES 1 5 , , f--- 15 cm-!-____ 30 em -----.,_10 cm , , , , , , -----t---------t-------;------t---T-15 cm ----- ---- -----i------i--- -- ---- ---; --- -N", 200 turns 30 em ---- --------- --------+-------~-- -- ---- ---'-----i---- I, -----i----' _ +-----+ ___ -+----+ L 15 cm , , , , , , f--- 15 cm-i-_ 30 = ------i-1O cm_ , , , (.j Depth = IOcm ; -+ 'iJ( '" NO (b j FIGURE 1-7 (a) The ferromagnetic core of Example I- I. (b) The magnetic circuit corresponding to (a). 16 ELECIRIC MACHINERY FUNDAMENTALS The mean path length of region I is 45 cm, and the cross-sectional area is 10 X 10 cm = 100 cm2. Therefore, the reluctance in the first region is cc~~-, O~ .4~ 5~ m~~c-cc = (25DOX47T X 10 7 )(0.01 m2) = 14,300 A ° turns/Wb (1- 32) The mean path length of region 2 is 130 cm, and the cross-sectional area is 15 X 10 cm = 150 cm2. Therefore, the reluctance in the second region is l2 l2 "" = _ = c:-'-, J.Ul.2 14 ~2 ccccccc-l c ·3 ~m~cccc-c~ = (25DOX47T X 10 7 )(0.015 m2) = 27,600 A ° turns/Wb Therefore, the total reluctance in the core is ~ = 'l:!l + 'l:!2 = 14,3DO A ° tumslWb + 27,600 A ° tlUlls/Wb = 41,900 A ° tumslWb The total magnetomotive force is g = Ni = (2DO turnsXI.O A) = 200 A ° turns The total flux in the core is given by g 200 A ° tlUllS cp = CR = 4 1,900 A otlUllsl Wb = 0.0048 \Vb (1- 32) This calculation can be perfonned by using a MATLAB script file, if desired. A sim-ple script to calculate the flux in the core is shown below. !l; M -file : exl _ 1.m !l; M -file t o ca l c ula t e the fl ux 11 0 .4 5; i n Exa mp l e 1-1 . 12 1. 3; "' 0 . 01 ; "' 0 . 01 5 ; or 2500; 0 0 4 p i l E-7; 0 200; i ~ " !l; Ca l c ula t e the fi r s t r e l u c t a n ce rl = 11 I (ur u O a l ) ; d i sp ( ['rl = ' num2str (rl ) I ) ; !l; Ca l c ula t e the second r e l u c t a n ce r 2 = 12 I (ur u O a2 ) ; d i sp ( ['r 2 = ' num2str (r 2 ) I ) ; % Le ng t h of r egi on 1 • • • • • • • Le ng t h of r egi on 2 Ar ea of r egi on 1 Ar ea of r egi on 2 Re l a t i ve permeabili t y Pe rmeabili t y of f r ee space Number of turns on cor e CUrre nt in amps INTRODUCTION TO MACHINERY PRINCIPLES 17 % Ca l c u l a t e the t ot a l r e l uc t a nce rt ot = rl + r 2; % Ca l c u l a t e the rumf rumf= n i ; % Fina lly, get the fl ux i n the co r e f l ux = rumf I rt ot ; % Di sp l ay r esult d i sp ( 'Flux = ' num2str (fl ux ) I ) ; When this program is executed, the results are: ,. e1:1_ 1 rl = 14323.9 44 9 r 2 = 27586 , 8568 Flux = 0 , 00 4772 This program produces the same answer as our hand calculations to the number of signifi-cant digits in the problem. Example 1-2. Figure 1- 8a shows a ferromagnetic core whose mean path length is 40 cm. There is a small gap of 0.05 cm in the structure of the otherwise whole core. The cross-sectional area of the core is 12 cm2, the relative permeability of the core is 4(x)(), and the coil of wire on the core has 400 turns. Assmne that fringing in the air gap increases the effective cross-sectional area of the air gap by 5 percent. Given this infonnation, find (a) the total reluctance of the flux path (iron plus air gap) and (b) the current required to produce a flux density of 0.5 T in the air gap. Solutioll The magnetic circuit corresponding to this core is shown in Figure 1-8b. (a) The reluctance of the core is I, 1 < "" = -= J.Ul. < /.t r IJ.ty'I. < (1- 32) cccccc--c O c·4 ~m~=ccc-cc = (4000X47T X 10 7XO.OO2 m2 ) = 66,300 A • turns/Wb The effecti ve area of the air gap is 1.05 X 12 cm2 = 12.6 cm2, so the reluctance of the air gap is I. "'" = -wi, 0.0005 m = C (4 C~--C X~1 ~Oa'X~O~ .OO ~1~ 276~ m "') = 316,OOO A · turns/Wb (1- 32) 18 ELECIRIC MACHINERY FUNDAMENTALS N=400 turns g(=Ni) fo'IGURE 1-8 + ,b, " J 1- 0.05 em T A=12cm2 CJ;>.. (Reluctance of core) CJ:.>., (Reluctance of air gap) (a) The ferromagnetic core of Example 1- 2. (b) The magnetic circuit corresponding to (a). Therefore, the total reluctance of the flux path is 1l! ... = CJ:l" + CQ" = 66,300 A· turns/Wb + 3 16,OOO A· turnsIWb = 382,300 A • tumslWb Note that the air gap contributes most of the reluctance even though it is 800 times shorter than the core. (b) Equation (1 - 28) states that Since the flux cp = BA and 'if = Ni, this equation becomes Ni = BACl! (1 - 28) INTRODUcnONTO MACHINERY PRINCIPLES 19 . BAct! ,=--N (0.5 D(0.00126 m2)(383,200 A ° tlU1lsi Wb) = 400 turns = 0.602 A Notice that, since the air-gap flux was required, the effective air-gap area was used in the above equation. EXllmple 1-3. Figure 1-9a shows a simplified rotor and stator for a dc motor. The mean path length of the stator is 50 cm, and its cross-sectional area is 12 cm2. The mean path length of the rotor is 5 em, and its cross-sectional area also may be assruned to be 12 cm2. Each air gap between the rotor and the stator is 0.05 cm wide, and the cross-sectional area of each air gap (including fringing) is 14 cm2. The iron of the core has a rel-ative penneability of 2()(x), and there are 200 turns of wire on the core. If the current in the wire is adjusted to be I A, what will the resulting flux density in the air gaps be? Solutioll To detennine the flux density in the air gap, it is necessary to first calculate the magneto-motive force applied to the core and the total reluctance of the flux path. With this infor-mation, the total flux in the core can be found. Finally, knowing the cross-sectional area of the air gaps enables the flux density to be calculated. The reluctance of the stator is ---""c-'ll,= /-t r IJ.QI\ I ccccccc-~OC·50m",ccc~o-" = (2000X47T X 10 7)(0.0012 m2) = 166,oooA o tlUllsIWb The reluctance of the rotor is ---""~ 'll,= /-tr IJ.QI\r O.05m = C(2;;;()()()=X:C4C-~-;X""'lo;O'i'C;;)(;;CO.;;:OO;;;1;;:2C:m:h') = 16,600A o tumslWb The reluctance of the air gaps is '. "'" = ---'"c-/-t r IJ.QI. 0.0005 m = C(1~ )( ~4-~~X~10~ 'X~O ~.OO~1~4-m" ) = 284,000 A ° tlUllsIWb The magnetic circuit corresponding to this machine is shown in Figure 1-9b. The total re-luctance of the flux path is thus 20 ELECIRIC MACHINERY FUNDAMENTALS ~I~ I,="m Ic=50cm ,., Stator reluctance ,b, ""GURE 1-9 (a) A simplified diagram of a rotor and stator for a de motor. (b) The magnetic circuit corresponding to (a). CJ:l..q = Ci:!, + CJ:!"t + Ci:!, + Ci:!~2 = 166,000 + 284,000 + 16,600 + 284,000 A • tumslWb = 75 1,ooo A · turns/Wb The net magnetomotive force applied to the core is g = Ni = (200 turnsXI.OA) = 200 A • turns Therefore, the total fl ux in the core is INTRODUCTION TO MACHINERY PRINCIPLES 21 g 200 A • turns cp = CR = 751.0CXl A • turnsJ \Vb = 0.cXl266 Wb Finally, the magnetic flux density in the motor's air gap is B = cp = 0.000266 Wb A 0.0014 m2 0.19T Magnetic Behavior of Ferromagnetic Materials Earlier in this section, magnetic permeability was defined by the equation ( 1-21) It was explained that the penneability of ferromagnetic materials is very high, up to 6(X)Q times the penneability of free space. In that discussion and in the examples that followed, the penneability was assumed to be constant regardless of the mag-netomotive force applied to the material. Although permeability is constant in free space, this most certainly is not true for iron and other ferromagnetic materials. To illustrate the behavior of magnetic penneability in a ferromagnetic ma-terial, apply a direct current to the core shown in Figure 1-3, starting with 0 A and slowly working up to the maximum permissible current. When the flux prOOuced in the core is plotted versus the magnetomotive force producing it, the resulting plot looks like Figure I-lOa. lllis type of plot is called a saturation curoe or a magnetization culVe. At first , a small increase in the magnetomotive force pro-duces a huge increase in the resulting flux. After a certain point, though, further increases in the magnetomotive force produce relatively smaller increases in the flux. Finally, an increase in the magnetomotive force produces almost no change at all. The region of this figure in which the curve flattens out is called the satu-ration region, and the core is said to be saturated. In contrast, the region where the flux changes very rapidly is calJed the unsaturated region of the curve, and the core is said to be unsaturated. The transition region between the unsaturated re-gion and the saturated region is sometimes called the knee of the curve. Note that the flux produced in the core is linearly related to the applied magnetomotive force in the unsaturated region, and approaches a constant value regardless of magnetomotive force in the saturated region. Another closely related plot is shown in Figure I-lOb. Figure I-lOb is a plot of magnetic flux density B versus magnetizing intensity H. From Equations (1- 20) and (I- 25b), Ni g H ~ - ~ -( Ie '" ~ BA ( 1-20) (I- 25b) it is easy to see that magnetizing intensity is directly proP011io1UJi to magnetomotive force and magnetic flux density is directly propoT1ional to flux for any given core. Therefore, the relationship between B and H has the same shape as the relationship 22 ELECIRIC MACHINERY FUNDAMENTALS ..p.Wb B. T 2.8 2.6 2.4 2.2 2.0 E 1.8 • . ~ 1.6 ~ 1.4 1.0 0.8 0.6 0.4 0.2 o \0 (a) 20 30 40 50 F. A · turns v / 100 200 300 500 1000 Magnetizing iotensity H. A · turnslm "I ""GURE - 10 H. A · turnslm ,b, 2000 5000 (a) Sketch of a dc magnetization curve for a ferromagnetic core. (b) The magnetization curve expressed in terms of flux density and magnetizing intensity. (c) A detailed magnetization curve for a typical piece of steel. (d) A plot of relative permeability /J., as a function of magnetizing intensity H for a typical piece of steel. 7(XX) 2(xx) o 10 FIGURE - \0 (continued) INTRODUCTION TO MACHINERY PRINCIPLES 23 / ~ '\ '\ '\ 20 30 40 50 ]00 200 Magnetizing intensity H (A ' turnslm) I" "-300 '" i'-500 between flux and magnetomotive force. The slope of the curve of flux density ver-sus magnetizing intensity at any value of H in Figure I- I Db is by definition the per-meability of the core at that magnetizing intensity. The curve shows that the penne-ability is large and relatively constant in the unsaturated region and then gradually drops to a very low value as the core becomes heavily saturated. 1000 Figure I- IDe is a magnetization curve for a typical piece of steel shown in more detail and with the magnetizing intensity on a logarithmic scale. Only with the magnetizing intensity shown logarithmically can the huge saturation region of the curve fit onto the graph. The advantage of using a ferromagnetic material for cores in electric ma-chines and transfonners is that one gets many times more flux for a given magne-tomotive force with iron than with air. However, if the resulting flux has to be pro-portional, or nearly so, to the applied magnetomotive force, then the core must be operated in the unsaturated region of the magnetization curve. Since real generators and motors depend on magnetic flux to produce volt-age and torque, they are designed to produce as much flux as possible. As a result, most real machines operate near the knee of the magnetization curve, and the flux in their cores is not linearly related to the magnetomotive force producing it. This 24 ELECIRIC MACHINERY FUNDAMENTALS nonlinearity accounts for many of the peculiar behaviors of machines that will be explained in future chapters. We will use MATLAB to calculate solutions to prob-lems involving the nonlinear behavior of real machines. Example 1-4. Find the relative penneability of the typical ferromagnetic material whose magnetization curve is shown in Figure l-lOc at (a) H = 50. (b) H = 100. (c) H = 500. and (d) H = 1000 A ° turns/m. Solutio" The penneability of a material is given by and the relative permeability is given by IJ- = B H (1- 23) Thus. it is easy to detennine the penneability at any given magnetizing intensity. (a) AtH = 50A otums/m.B = 0.25T. so B 0.25 T IJ- = H = 50 A ° turns/m = O.OO5Q Him = ~ = 0.0050 HIm = 3980 IL, f.1.O 47T X 10 7 Hhn (b) At H = lOOA ° turns/m. B = 0.72 T. so B 0.72 T IJ- = H = 100 A ° turns/m = 0.0072 Him = ~ = 0.0072 HIm = 5730 IJ-, f.1.O 47T X 10 7 Hhn (c) AtH = 500 A ° turns/m.B = I.40 T, so B 1.40 T IJ- = H = 500 A ° turns/m = 0.0028 Him = ~ = 0.0028 HIm = 2230 IJ-, f.1.O 47T X 10 7 Hhn (d) AtH = lOOOA oturns/m,B = 1.51 T, so B 1.51 T IJ- = H = 1000 A ° turns/m = 0.00151 Him = ~ = 0.00151 HIm = 1200 IJ-, f.1.O 47T X 10 7 Hhn INTRODUCTION TO MACHINERY PRINCIPLES 25 Notice that as the magnetizing intensity is increased, the relative penne-ability first increases and then starts to drop off. The relative permeability of a typ-ical ferromagnetic material as a function of the magnetizing intensity is shown in Figure 1-lOd. This shape is fairly typical of all ferromagnetic materials. It can easi ly be seen from the curve for /L. versus H that the assumption of constant rel-ative penneability made in Examples 1-1 to -3 is valid only over a relatively narrow range of magnetizing intensities (or magnetomoti ve forces). In the fo llowing example, the relative penneability is not assumed constant. Instead, the relationship between Band H is given by a graph. Example 1-5. A square magnetic core has a mean path length of 55 cm and a cross-sectional area of 150 cm2. A 2()()...tum coil of wire is wrapped arOlUld one leg of the core. The core is made of a material having the magnetization curve shown in Figure l-lOc. (a) How much current is required to produce 0.012 Wb of flux in the core? (b) What is the core's relative permeability at that current level? (c) What is its reluctance? Solutioll (a) The required flux density in the core is B = q, = 1.012 Wb = 0.8T A 0.015 m2 From Figure l-lOc, the required magnetizing intensity is H = 115 A" turns/m From Equation (1-20), the magnetomotive force needed to produce this magnetizing in-tensity is q= Ni = Hlc = (115 A" turns/mXD.55 m) = 63.25 A" turns so the required current is i = q = 63.25 A" turns = 0.316A N 200 turns (b) The core's permeability at this current is B 0.8T I.L = H = liS A "turnshn = 0.00696 Him Therefore, the relative permeability is I.L 0.00696 Him I.Lr = 1.1.0 = 47T X 10 7 Him 5540 (c) The reluctance of the core is tT'I = q = 63.25 A" turns = 5270 A. _ .. ~ -'" q, 0.012Wb turn", .. u 26 ELECIRIC MACHINERY FUNDAMENTALS i (t) Residual flux -------= ,., ,,' ¢ (or 8) b -----------.,f-J~ " 'I_-------- 'J(or H) ,b, ""GURE I- II The hysteresis loop traced out by the flux in a core when the current i{l) is applied to it. Energy Losses in a Ferromagnetic Core Instead of applying a direct current to the windings on Ihe core, let us now apply an alternating currenl and observe what happens. TIle currenl to be applied is shown in Figure I- ila. Assume that the flux in the core is initially zero. As the current increases for the first time, the flux in the core traces out path ab in Figure I- lib. lllis is basically the saturation curve shown in Figure 1- 10. However, when Ihe current falls again, thef1ux traces out a different path from the one itfol-lowed when the current increased. As the current decreases, the flux in the core traces out path bcd, and later when the current increases again, the flux traces out path deb. Notice that the amount of flux present in Ihe core depends nol only on Ihe amount of current applied to the windings of the core, but also on the previous history of the flux in Ihe core.lllis dependence on the preceding flux history and the resulting failure to retrace flux paths is cal led hysteresis. Path bcdeb traced out in Figure I- II b as the applied current changes is called a hysteresis loop. INTRODUCTION TO MACHINERY PRINCIPLES 27 -/' '- I \ - -- - - - - -I X I -X -I - -/' - --- --I -/' \ "-X /' - - - - -'\ -\ - "- /' -" - ---- " -/ X / I • • ---.. -/' --- -1 I \ I " t / - - --.. - " ,b, FIGURE 1-12 (a) Magnetic domains oriented randomly. (b) Magnetic domains lined up in the presence of an external magnetic field. --/' -----.. Notice that if a large magnetomolive force is first applied to the core and then removed, the flux path in the core will be abc, When the magnetomotive force is removed, the flux in the core does not go to zero. Instead, a magnetic field is left in the core. This magnetic field is called the residual flux in the core. It is in precisely this manner that pennanent magnets are produced. To force the flux to zero, an amount of magnetomotive force known as the coercive magnetomotive force '(tc must be applied to the core in the opposite direction. Why does hysteresis occur? To understand the behavior of ferromagnetic materials, it is necessary to know something about their structure. TIle atoms of iron and similar metals (cobalt, nickel, and some of their alloys) tend to have their magnetic fields closely aligned with each other. Within the metal, there are many small regions called domnins, In each domain, all the atoms are aligned with their magnetic fields pointing in the same direction, so each domain within the material acts as a small permanent magnet. The reason that a whole block of iron can ap-pear to have no flux is that these numerous tiny domains are oriented randomly within the material. An example of the domain structure within a piece of iron is shown in Figure 1- 12. When an external magnetic field is applied to this block of iron, it causes do-mains that happen to point in the direction of the field to grow at the expense of domains pointed in other directions. Domains pointing in the direction of the mag-netic field grow because the atoms at their boundaries physically switch orientation to align themselves with the applied magnetic field. The extra atoms aligned with the field increase the magnetic nux in the iron, which in turn causes more atoms to switch orientation, further increasing the strength of the magnetic field. It is this pos-itive feedback effect that causes iron to have a penneabiJity much higher than air. As the strength of the external magnetic field continues to increase, whole domains that are aligned in the wrong direction eventually reorient themselves as 28 ELECIRIC MACHINERY FUNDAMENTALS a unit to line up with the field. Finally, when nearly all the atoms and domains in the iron are lined up with the external field, any further increase in the magneto-motive force can cause only the same flux increase that it would in free space. (Once everything is aligned, there can be no more feedback effect to strengthen the field.) At this point, the iron is saturated with flux. This is the situation in the saturated region of the magnetization curve in Figure 1-10. TIle key to hysteresis is that when the external magnetic field is removed, the domains do not completely randomize again. Why do the domains remain lined up? Because turning the atoms in them requires energy. Originally, energy was provided by the external magnetic field to accomplish the alignment; when the field is removed, there is no source of energy to cause all the domains to rotate back. The piece of iron is now a penn anent magnet. Once the domains are aligned, some of them wi ll remain aligned until a source of external energy is supplied to change them. Examples of sources of ex-ternal energy that can change the boundaries between domains and/or the align-ment of domains are magnetomotive force applied in another direction, a large mechanical shock, and heating. Any of these events can impart energy to the do-mains and enable them to change alignment. (It is for this reason that a permanent magnet can lose its magnetism if it is dropped, hit with a hammer, or heated.) TIle fact that turning domains in the iron requires energy leads to a common type of energy loss in all machines and transfonners. The hysteresis loss in an iron core is the energy required to accomplish the reorientation of domains during each cycle of the alternating current applied to the core. It can be shown that the area enclosed in the hysteresis loop formed by applying an alternating current to the core is directly proportional to the energy lost in a given ac cycle. The smaller the applied magnetomotive force excursions on the core, the smaller the area of the resulting hysteresis loop and so the smaller the resulting losses. Figure 1-1 3 illustrates this point. Another type of loss should be mentioned at this point, since it is also caused by varying magnetic fields in an iron core. This loss is the eddy current loss. The mechanism of eddy current losses is explained later after Faraday's law has been introduced. Both hysteresis and eddy current losses cause heating in the core material, and both losses must be considered in the design of any machine or transformer. Since both losses occur within the metal of the core, they are usually lumped together and called core losses. 1.5 FARADAY'S LAW-INDUCED VOLTAGE FROM A TIME-CHANGING MAGNETIC FIELD So far, attention has been focused on the pnxluction of a magnetic field and on its properties. It is now time to examine the various ways in which an existing mag-netic field can affect its surroundings. TIle first major effect to be considered is cal led Faraday s law. It is the ba-sis of transfonner operation. Faraday's law states that if a flux passes through a INTRODUCTION TO MACHINERY PRINCIPLES 29 ¢ (or 8) , , -' " -----,Hf,',fr-!,f-f------- 'J(or H) , FIGURE -13 , , Area c< hysteresis loss The elTect of the size of magoetomotive force excursions on the magnitude of the hysteresis loss. turn of a coil of wire, a voltage will be induced in the turn of wire that is directly proportional to the rate of change in the flux with respect to time. In equation fonn, (1-35) where eind is the voltage induced in the turn of the coil and <P is the flux passing through the turn. If a coil has N turns and if the same flux passes through all of them, then the voltage induced across the whole coil is given by where eioo = voltage induced in the coil N = number of turns of wire in coil <p = nux passing through coil (1-36) The minus sign in the equations is an expression of Lenz slaw. Lenz's law states that the direction of the voltage buildup in the coil is such that if the coil ends were short circuited, it would produce current that would cause a flux opposing the original nux change. Since the induced voltage opposes the change that causes it, a minus sign is included in Equation ( 1-36). To understand this concept clearly, 30 ELECIRIC MACHINERY FUNDAMENTALS Direction of j required ; + '00 ( N turns I. (. ) ""GURE 1- 14 , , • I· (b) Direction of opposing flux ¢ increasing The m eaning of Lenz's law: (a) A coil enclosing an in creasing magnetic flux: (b) determining the resulting voltage polarity. examine Figure 1- 14. If the nux shown in the figure is increasing in strength, then the voltage built up in the coil will tend to establish a flux that will oppose the in-crease. A current flowing as shown in Figure 1-1 4b would produce a nux oppos-ing the iflcrease, so the voltage O fl the coil must be built up with the polarity re-quired to drive that current through the external circuit. 1l1erefore, the voltage must be buill up with Ihe polarity shown in the figure. Since the polarity of the re-sulting voltage can be detennined from physical considerations, Ihe minus sign in Equalions (1- 35) and (1- 36) is often left out. It is left out of Faraday's law in the remainder of this book. 1l1ere is one major difficu lIy involved in using Equation (1- 36) in practical problems. That equation assumes that exactly Ihe same flux is present in each tum of the coil. Unfortunately, the flux leaking oUI of the core inlo the surrounding air prevents this from being lrue. If the windings are tightly coupled, so that the vast majority of the flux passing through one turn of the coil does indeed pass through all of them, then Equation ( 1- 36) will give valid answers. Bul if leakage is quite high or if extreme accuracy is required, a different expression that does not make that assumption will be needed. The magnitude of the voliage in the ith tum of the coil is always given by I f there are N turns in the coi I of wire, the total vollage on the coil is N eind = ~ ei i - I ( 1- 37) ( 1- 38) INTRODUcnONTO MACHINERY PRINCIPLES 31 (1-39) ( 1-40) The term in parentheses in Equation (1--40) is cal led the flux linkage A of the coil, and Faraday's law can be rewritten in terms of flux linkage as ( 1-41) where ( 1-42) The units of flux linkage are weber-turns. Faraday's law is the fundamental property of magnetic fields involved in transformer operation. The effect of Lenz's law in transformers is to predict the polarity of the voltages induced in transformer windings. Faraday's law also explains the eddy current losses mentioned previously. A time-changing flux induces voltage within a ferromagnetic core in just the same manner as it would in a wire wrapped around that core. TIlese voltages cause swirls of current to fl ow within the core, much like the eddies seen at the edges of a river. It is the shape of these currents that gives rise to the name eddy currents. These eddy currents are fl owing in a resistive material (the iron of the core), so energy is dissipated by them. The lost energy goes into heating the iron core. The amount of energy lost to eddy currents is proportional to the size of the paths they follow within the core. For this reason, it is customary to break up any ferromagnetic core that may be subject to alternating fluxes into many small strips, or laminntions, and to build the core up out of these strips. An insulating oxide or resin is used between the strips. so that the current paths for eddy currents are limited to very small areas. Because the insulating layers are extremely thin, this action reduces eddy current losses with very little effect on the core's mag-netic properties. Actual eddy current losses are proportional to the square of the lamination thickness, so there is a strong incentive to make the laminations as thin as economically possible. EXllmple 1-6. Figure 1-15 shows a coil of wire wrapped around an iron core. If the flux in the core is given by the equation cp = 0.05 sin 377t Wb If there are 100 turns on the core. what voltage is produced at the terminals of the coil? Of what polarity is the voltage during the time when flux is increasing in the reference 32 ELECIRIC MACHINERY FUNDAMENTALS /' / Require d direction of i ; -+ N= 100 turns Opposing ~ .-H , t I / ~ _ 0.05 Sin 3771 Wb ""GURE 1-15 The core of Example 1--6. Determination of the voltage polarity at the terminals is shown. direction shown in the figure? Asswne that all the magnetic flux stays within the core (i.e., assume that the flux leakage is zero). Solutioll By the same reasoning as in the discussion on pages 29- 30, the direction of the voltage while the flux is increasing in the reference direction must be positive to negative, as shown in Figure 1- 15. The magnitude of the voltage is given by or alternatively, d~ e;"" = NYt = (100 turns) :, (0.05 sin 377t) = 1885 cos 377t eiod = 1885 sin(377t + goO) V 1.6 PRODUCTION OF INDUCED FORCE ONAWIRE A second major effect of a magnetic field on its surroundings is that it induces a force on a current-carrying wire within the field. The basic concept involved is il-lustrated in Figure 1- 16. The figure shows a conductor present in a unifonn mag-netic field of flux density D, pointing into the page. 1lle conductor itself is I me-ters long and contains a current of i amperes. The force induced on the conductor is given by F = i(I X D) ( 1-43) INTRODUCTION TO MACHINERY PRINCIPLES 33 , , -, , " J , , , , , , , , , , I , , , , " , I , , J , , FlGURE 1- 16 h A current-carrying wire in the presence of 3. , , -, , magnetic field. where i = magnitude of current in wire I = length of wire, with direction of I defined to be in the direction of current flow B = magnetic flux density vector The direction of the force is given by the right-hand rule: If the index finger of the right hand points in the direction of the vector I and the middle finger points in the direction of the flux density vector B, then the thumb points in the direction of the resultant force on the wire. TIle magnitude of the force is given by the equation F = UR sin () ( 1-44) where () is the angle between the wire and the flux density vector. Example 1-7. Figure 1- 16 shows a wire carrying a current in the presence ofa magnetic field. The magnetic flux density is 0.25 T. directed into the page. If the wire is 1.0 m long and carries 0.5 A of current in the direction from the top of the page to the bot-tom of the page. what are the magnitude and direction of the force induced on the wire? Solutioll The direction of the force is given by the right-hand rule as being to the right. The magni-tude is given by F = ilB sin (J (1-44) = (0.5 AXI.O m)(0.25 T) sin 90° = 0.125 N Therefore. F = 0.125 N. directed to the right The induction of a force in a wire by a current in the presence of a magnetic fie ld is the basis of motor action. Almost every type of motor depends on this basic principle for the forces and torques which make it move. 34 ELECIRIC MACHINERY FUNDAMENTALS 1.7 INDUCED VOLTAGE ON A CONDUCTOR MOVING IN A MAGNETIC FIELD There is a third major way in which a magnetic field interacts with its surround-ings. If a wire with the proper orientation moves through a magnetic field, a volt-age is induced in it. TIlis idea is shown in Figure -\7. TIle voltage induced in the wire is given by eiD<! = (v X B) • I ( 1-45) where v = velocity of the wire B = magnetic nux density vector I = length of conductor in the magnetic field Vector I points along the direction of the wire toward the end making the smallest angle with respect to the vector v X B. The voltage in the wire will be built up so that the positive end is in the direction of the vector v X B. TIle following exam-ples illustrate this concept. Example 1-8. Figure 1-17 shows a conductor moving with a velocity of 5.0 m1s to the right in the presence of a magnetic field. The flux density is 0.5 T into the page, and the wire is 1.0 m in length, oriented as shown. What are the magnitude and polarity of the resulting induced voltage? Solutio" The direction of the quantity v X B in this example is up. Therefore, the voltage on the con-ductor will be built up positive at the top with respect to the bottom of the wire. The direc-tion of vector I is up, so that it makes the smallest angle with respect to the vector \' X B. Since \' is perpendicular to B and since v X B is parallel to I, the magnitude of the induced voltage reduces to ejmd = (\' X B) ·I = (vB sin 90°) I cos 0° = vBI = (5.0 mls)(0.5 TXI.O m) = 2.5 V Thus the induced voltage is 2.5 V, positive at the top of the wire. (1---45) Example 1-9. Figure 1-18 shows a conductor moving with a velocity of 10 m1s to the right in a magnetic field. The flux density is 0.5 T, out of the page, and the wire is 1.0 m in length, oriented as shown. What are the magnitude and polarity of the resulting induced voltage? Solutioll The direction of the quantity v X B is down. The wire is not oriented on an up-down line, so choose the direction of I as shown to make the smallest possible angle with the direction INTRODUCTION TO MACHINERY PRINCIPLES 35 x x + ~ ,:X B x • ++ x x x x ,~ x X I X X , X X X X Jo'IGURE 1- 17 X X - '" X X A conductor moving in the presence of a magnetic field. • • • • II • • • • 3<1' • I • • vX II • • • Jo'IGURE 1- 18 The conductor of Example 1--9. of \' X B. The voltage is positive at the bottom of the wire with respect to the top of the wire. The magnitude of the voltage is eiod = (\' X B) ' I = (vB sin 90°) l cos 30° = (10.0 m1sX0.51)(I.O m) cos 30° = 4.33 V (1--45) The induction of voltages in a wire moving in a magnetic field is funda-mental to the operation of all types of generators. For this reason, it is called gen-erator action. 36 ELECIRIC MACHINERY FUNDAMENTALS Switch Magnetic field into page R 1 X X X + -=- VB e ;00 X X X ""GURE 1-19 A linear dc machine. The magnetic field points into the page. 1.8 THE LINEAR DC MACHINE- A SIMPLE EXAMPLE • A linear dc machine is about the simplest and easiest-to-understand version of a dc machine, yet it operates according to the same principles and exhibits the same behavior as real generators and motors. It thus serves as a good starting point in the study of machines. A linear dc machine is shown in Figure - \9. It consists of a battery and a resistance connected through a switch to a pair of smooth, frictionless rails. Along the bed of this "rai lroad track" is a constant, uni form-density magnetic field di-rected into the page. A bar of conducting metal is lying across the tracks. How does such a strange device behave? Its behavior can be determined from an application of four basic equations to the machine. These equations are I. The equation for the force on a wire in the presence ofa magnetic field: F i(I X B) I ( 1 - 43) where F = force on wire i = magnitude of currenl in wire I = length of wire, with direction of! defined to be in the direction of current now B = magnetic nux density vector 2. The equation for the voltage induced on a wire moving in a magnelic field: l e;Dd (vXB) -1 where e;Dd = voltage induced in wire v = velocity of the wire B = magnetic nux density vector I = length of conductor in the magnelic field ( 1 - 45) INTRODUCTION TO MACHINERY PRINCIPLES 37 ~o R • , x x X i (/) + -Find -=- VB e;!!d ,. X X X FIGURE 1-10 Starting a linear dc machine. 3. Kirchhoff's voltage law for Ihis machine. From Figure J- J9lhis law gives VB - iR -ei!!d = 0 ( 1 - 46) 4. Newton's law for the bar across the tracks: (1- 7) We will now explore the fundamental behavior of this simple dc machine using these four equations as lools. Starting the Linear DC Machine Figure 1- 20 shows the linear dc machine under starting conditions. To start this machine, simply close the switch. Now a currenl fl ows in the bar, which is given by Kirchhoff's voltage law: ( 1 - 47) Since the bar is initially at rest, e;Dd = 0, so i = VBIR. The current flows down through the bar across the tracks. But from Equation (1- 43), a currenl flowing Ihrough a wire in the presence ofa magnetic field induces a force on Ihe wire. Be-cause of the geometry of the machine, this force is Find = ilB to the right ( 1 - 48) Therefore, the bar will accelerale to the right (by Newton's law). However, when Ihe velocity of the bar begins 1 0 increase, a voltage appears across the bar. The voltage is given by Equation (1 - 45), which reduces for this geometry to e;Dd = vBI positive upward ( 1 - 49) The voliage now reduces the current fl owing in the bar, since by Kirch-hoff's voliage law 38 ELECIRIC MACHINERY FUNDAMENTALS v (t) V, BI o e;oo (t) V, o i (t) V, R o F;oo (t) VBIB R o (a) 'hI "I ,dl ""GURE 1-21 The linear de machine on starting. (a) Velocity v(t) as a function of time; (b) induced voltage e .... (/); (c) currem i(t); (d) induced force Fu.il). ( 1-47) As eioo increases, the current i decreases. TIle result of this action is that eventually the bar wi ll reach a constant steady-state speed where the net force on the bar is zero. TIlis will occur when eioo has risen all the way up to equal the voltage VB. At that time, the bar will be mov-ing at a speed given by VB = e;Dd = v" BI V, v'" = BI ( I-50) TIle bar will continue to coast along at this no-load speed forever unless some ex-ternal force disturbs it. When the motor is started, the velocity v, induced voltage eiDd , current i, and induced force F ind are as sketched in Figure 1-21. To summarize, at starting, the linear dc machine behaves as follows: I, Closing the switch produces a current flow i = VB /R. 2, The current flow produces a force on the bar given by F = ilB. INTRODUCTION TO MACHINERY PRINCIPLES 39 R " i (/) X X X + -F ;md _ fo'_ e;md I , X X X FIGURE 1-22 The linear dc machine as a motor. 3. The bar accelerales to the right, producing an induced voltage e;md as it speeds up. 4. lllis induced voltage reduces the current flow i = (VB -e;Dd t)! R. 5. The induced force is thus decreased (F = i J.IB) until eventually F = o. At that point, e;Dd = VB, i = 0, and the bar moves at a constant no-load speed v" = VB ! BI. This is precisely the behavior observed in real motors on starting. The Linear DC Machine as a Molor Assume that the linear machine is initial ly running at the no-load steady-state con-ditions described above. What wi ll happen to this machine if an external load is applied to it? To find out, let's examine Figure 1- 22. Here, a force Fto.d is applied to the bar opposite the direction of motion. Since the bar was initially at steady state, application of the force Ftoad will result in a net force on the bar in the direc-tion opposite the direction of motion (F Del = Fto.d -Find). The effect of this force will be to slow the bar. But just as soon as the bar begins to slow down, the in-duced voltage on the bar drops (e;Dd = vJ.BI). As the induced voltage decreases, the current flow in the bar rises: ( 1 - 47) 1l1erefore, the induced force rises too (Find = itIB). The overall result of this chain of events is that the induced force rises until it is equal and opposite to the load force, and the bar again travels in steady state, but at a lower speed. When a load is attached to the bar, the velocity v, induced voltage eind, current i, and in-duced force Find are as sketched in Figure 1- 23. 1l1ere is now an induced force in the direction of motion of the bar, and power is being convenedJrom electricalfonn to mechanical Jonn to keep the bar moving. The power being converted is 40 ELECIRIC MACHINERY FUNDAMENTALS v (I) V, BI o eiOO (I) V, o j (I) F BI o FiOO (I) F~ o '" 'hI ' VB, the currenl reverses direction and is now given by the equation . eiDd -VB ,~ R ( I- 53) Since this current now flows up through the bar, it induces a force in the bar given by Find = ilB to Ihe left ( I- 54) TIle direction of the induced force is given by the right-hand rule. TIlis induced force opposes the applied force on the bar. Finally, the induced force will be equal and opposite to the applied force, and the bar wi ll be moving at a higher speed than before. Notice Ihat now the bat-tery is charging. The linear machine is now serving as a generator, converting me-chanical power Findv into electric power ejDdi . To summarize this behavior: 42 ELECIRIC MACHINERY FUNDAMENTALS I. A force F app is applied in the direction of motion; F oet is in the direction of motion. 2. Acceleration a = F"",/m is positive, so the bar speeds up (vt). 3. The voltage eiod = vtBl increases, and so i = (eiod t-VBYR increases. 4. The induced force F ;od = itlB increases until I Find I = I Fload I at a higher speed v. 5. An amount of mechanical power equal to F ;odv is now being converted to electric power e;odi, and the machine is acting as a generator. Again, a real dc generator behaves in precisely this manner: A torque is ap-plied to the shaft in the direction of motion, the speed of the shaft increases, the in-ternal voltage increases, and current fl ows out of the generator to the loads. The amount of mechanical power converted to electrical form in the real rotating gen-erator is again given by Equation (I-52): (I-52) It is interesting that the same machine acts as both motor and generator. The only difference between the two is whether the externally applied forces are in the direction of motion (generator) or opposite to the direction of motion (motor). Electrically, when eind > VB, the machine acts as a generator, and when e iod < VB, the machine acts as a motor. Whether the machine is a motor or a generator, both induced force (motor action) and induced voltage (generator action) are present at all times. nlis is generally true of all machines- both actions are present, and it is only the relative directions of the external forces with respect to the direction of motion that determine whether the overall machine behaves as a motor or as a generator. Another very interesting fact should be noted: This machine was a genera-tor when it moved rapidly and a motor when it moved more slowly, but whether it was a motor or a generator, it always moved in the same direction. Many begin-ning machinery students expect a machine to turn one way as a generator and the other way as a motor. This does not occur. Instead, there is merely a small change in operating speed and a reversal of current fl ow. Starting Problems with the Linear Machine A linear machine is shown in Figure 1-25. This machine is supplied by a 250-V dc source, and its internal resistance R is given as about 0. 10 n. (1lle resistor R models the internal resistance of a real dc machine, and this is a fairly reasonable internal resistance for a medium-size dc motor.) Providing actual numbers in this figure highlights a major problem with ma-chines (and their simple linear model). At starting conditions, the speed of the bar is zero, so eind = O. TIle current flow at starting is . VB 250 V I su.n= R" = 0.1 n = 2500 A FIGURE 1- 15 o.lOn " -i (I) INTRODUCTION TO MACHINERY PRINCIPLES 43 U =0.5 T . directed into the page x x X 0.5 m X X X The linear dc machine with componem values illustrating the problem of excessive starting currem. fa o.lOn " i (t) -=- VB =250V FIGURE 1- 16 R",., " X X X 0.5 m X X X A linear dc machine with an extra series resistor inserted to control the starting currem. This current is very high, often in excess of 10 times the rated current of the ma-chine. Such currents can cause severe damage to a motor. Both real ac and real dc machines suffer from similar high-currenl problems on starting. How can such damage be prevented? TIle easiest method for this simple lin-ear machine is 1 0 insert an extra resistance into Ihe circuit during starting 1 0 limit the current fl ow until ej!>d builds up enough to limit it. Figure -26 shows a start-ing resistance inserted into the machine circuitry. The same problem exists in real dc machines, and it is handled in precisely Ihe same fashion-a resistor is inserted into Ihe motor annature circuit during starting. TIle control of high starting current in real ac machines is handled in a different fashion, which will be described in Chapter 8. EXllmple 1-10. The linear dc machine shown in Figure 1-27a has a battery volt-age of 1 20 V. an internal resistance of 0.3 n. and a magnetic flux density of 0.1 T. (a) What is this machine's maximum starting current? What is its steady-state velocity at no load? (b) Suppose that a 30-N force pointing to the right were applied to the bar. What would the steady-state speed be? How much power would the bar be producing or consruning? How much power would the battery be producing or consuming? 44 ELECIRIC MACHINERY FUNDAMENTALS -=- 120V -::~ 120V -::~ 120V ""GURE -27 0.30 " 0.30 0.30 -x X "J X F oo 30 N X 'bJ X F toad =30 N X ,< J U =O.1 T . directed into the JX1ge X X 10m X X U =O.1 T . directed into the JX1ge X X + -F 'PP- 3O N e;nd -, X X U =O.1 T . directed into the JX1ge X X + -F;nd- 3O N e;nd -, X X The linear de machine of Example 1 - 10. (a) Starting conditions; (b) operating as a generator; (e) operating as a motor. Explain the difference between these two figures. Is this machine acting as a motor or as a generator? (c) Now suppose a 30-N force pointing to the left were applied to the bar. What would the new steady-state speed be? Is this machine a motor or a generntor now? (d) Assrune that a force pointing to the left is applied to the bar. Calculate speed of the bar as a flUlction of the force for values from 0 N to 50 N in IO-N steps. Plot the velocity of the bar versus the applied force. (e) Assume that the bar is lUlloaded and that it suddenly nms into a region where the magnetic field is weakened to 0.08 T. How fast will the bar go now? Solutioll (a) At starting conditions, the velocity of the bar is 0, so em = O . Therefore, i = VB -eiAd = l20Y - OV = 400 A R 0.3 0 INTRODUCTION TO MACHINERY PRINCIPLES 45 When the machine reaches steady state, Find = 0 and i = O . Therefore, VB = eind = v,J31 V , v ... = BI 120 V = (0.1 TXlOm) = 120mls (b) Refer to Figure 1-27b. If a 30-N force to the right is applied to the bar, the final steady state will occur when the induced force Find is equal and opposite to the applied force F OPP' so that the net force on the bar is zero: F.pp = Find = ilB Therefore, . Find 30 N 1 = IB = (IOmXo. l T) = 30 A flowing up through the bar The induced voltage eind on the bar must be eind = VB+iR = 120 V + (30AX0.3 0 ) = 129 V and the final steady-state speed must be ,~ v" =m 129 V = (0.1 TXlOm) = 129m1s The bar is producing P = (129 VX30 A) = 3870 W of power, and the battery is consuming P = (120 VX30 A) = 3600 W. The difference between these two num-bers is the 270 W of losses in the resistor. This machine is acting as a generator. (c) Refer to Figure 1-25c. This time, the force is applied to the left, and the induced force is to the right. At steady state, Fopp = Find = ilB . Find 30 N 1 = IB = (IO mXO.IT) = 30 A flowing down through the bar The induced voltage eind on the bar must be eind= VB - iR = 120 V - (30 AX0.3 0 ) = III V and the final speed must be ,~ v .. = m 11I V = (0.1 TXIO m) = 111 mls This machine is now acting as a motor, converting electric energy from the bat-tery into mechanical energy of motion on the bar. 46 ELECIRIC MACHINERY FUNDAMENTALS (d) This task is ideally suited for MATLAB. We can take advantage ofMATLAB's vectoriled calculations to detennine the velocity of the bar for each value of force. The MATLAB code to perform this calculation is just a version of the steps that were performed by hand in part c. The program shown below calcu-lates the current, induced voltage, and velocity in that order, and then plots the velocity versus the force on the bar. % M - f ile, exl _ 10 .m % M - f il e to ca l cu l ate and p l ot the vel oci t y of f unc t i on of l oad . % a linear motor as a VB = 120; r = 0 . 3; % Battery vol tage (V) % Res i s tance (ohms ) 1 = 1 ; B = 0 . 6; % Sel ect the F = 0, 10,50; • Ca l cu l ate 1 0 , . / (1 • Ca l cu l ate e i nd = VB -• Ca l cu l ate v_ba r 0 e i nd % Bar l ength (m) % Flux density (T) forces to appl y t o • the bar Force (N) 'he current s f l owi ng 1 0 'he mot or. • B) ; • CUrrent (A ) 'ho i nduced vol tages 0 0 'ho bar. i • c, • I nduced vol tage 'ho vel oci t i es of the bar. . / (1 • B) ; % Vel oci t y (m/ s) % Pl ot the vel oci t y of the bar ver s u s f or ce . p l ot (F,v_ba r ) ; t i t l e ('Pl ot of Vel oci t y ve r s u s Applied For ce'); x l abel (' Force (N) ' ) ; y l abel ('Vel oci t y (m/ s)'); axi s ( [0 500 200 ] ) ; (V) The resulting plot is shown in Figure 1- 28. Note that the bar slows down more and more as load increases. (e) If the bar is initially unloaded, then eind = VB. If the bar suddenly hits a region of weaker magnetic field, a transient will occur. Once the transient is over, though, eind will again equal VB. This fact can be used to determine the final speed of the bar. The initial speed was 120 rnls. The final speed is VB = eind = vllBl V, v .. = Bl 120 V = (0.08 TXIO m) = 150 mls Thus, when the flux in the linear motor weakens, the bar speeds up. The same behavior oc-curs in real dc motors: When the field flux of a dc motor weakens, it turns faster. Here, again, the linear machine behaves in much the same way as a real dc motor. INTRODUCTION TO MACHINERY PRINCIPLES 47 200 ISO 160 140 ~ 120 60 40 20 0 o FIGURE 1-28 , \0 15 20 25 Force (N) Plot of velocity versus force for a linear de machine. ~ , 30 1.9 REAL, REACTlVE,AND APPARENT POWER IN AC CIRCUITS , 40 45 50 In a dc circuit such as Ihe one shown in Figure l- 29a, the power supplied to the dc load is simply Ihe product of the voltage across the load and the current flow-ing through it. p = VI ( I- 55) Unfortunately, the situation in sinusoidal ac circuits is more complex, be-cause there can be a phase difference between the ac voltage and the ac currenl supplied to Ihe load. TIle instantaneous power supplied to an ac load will still be Ihe product of the instantaneous voltage and the instantaneous currenl, but the av-erage power supplied 1 0 the load wi ll be affected by the phase angle between the voltage and the current. W e will now explore the effects of this phase difference on the average power supplied to an ac load. Figure l- 29b shows a single-phase voltage source supplying power 1 0 a single-phase load with impedance Z = ZL O O. If we assume that the load is in-ductive, then the impedance angle 0 of the load will be positive, and the currenl will lag the voltage by 0 degrees. The voltage applied to this load is vet) = yI1V cos wi ( I- 56) 48 ELECIRIC MACHINERY FUNDAMENTALS 1 v + ) -1 1 + v(t) '" -1 I (a) 1- / L 0° --V = VLO" 'hI I R I I Z I z = Z L 0 ""GURE 1-29 (a) A de voltage source supplying a load with resistance R. (b) An ac voltage source supplying a load with impedance Z = Z L (J fl. where V is the nns value of the voltage applied to the load, and the resulting cur-rent flow is i(t) = V2I COS(wl -()) where I is the rms value of the current fl owing through the load. 1lle instantaneous power supplied to this load at any time t is pet) = v(t)i(t) = 2VI cos wt COS(wl -()) ( I-57) ( I-58) 1lle angle () in this equation is the impedance angle of the load. For inductive loads, the impedance angle is positive, and the current waveform lags the voltage waveform by () degrees. Ifwe apply trigonometric identities to Equation (1-58), it can be manipu-lated into an expression of the form pet) = VI cos () (1 + cos 2wt) + VI sin () sin 2wt ( I-59) 1lle first tenn of this equation represents the power supplied to the load by the component of current that is in phase with the voltage, while the second tenn rep-resents the power supplied to the load by the component of current that is 90° out of phase with the voltage. The components of this equation are plotted in Figure 1-30. Note that the first term of the instantaneous power expression is always pos-itive, but it produces pulses of power instead of a constant value. The average va lue of this term is p = Vl cos () ( 1-60) which is the average or real power (P) supplied to the load by term 1 of the Equa-tion (I- 59). The units of real power are watts (W), where 1 W = 1 V X 1 A. INTRODUCTION TO MACHINERY PRINCIPLES 49 p(' 1 Component I /\/ f\ f\ f , , " , , , , , , , , , , , , , , , , , , , , , , , , , C omponent 2 , , , 0 , , , , , , , \,~2 I 0 ,2 , 4 , 6' • , 10 w o. , , , , , , " ""GURE 1-30 The components of power supplied to a single-phase load versus time. The first component represents the power supplied by the component of current j" phase with the voltage. while the second term represents the power supplied by the component of current 90° OUI Of phase with the voltage. Nole that Ihe second tenn of the instantaneous power expression is positive half of the time and negative half of the time, so Ihal the average power supplied by this term is zero. This tenn represents power that is first transferred from the source 1 0 Ihe load, and then returned from Ihe load to the source. The power that continually bounces back and forth between the source and the load is known as re-active power (Q). Reactive power represents the energy thai is first stored and then released in the magnelic field of an inductor, or in the electric field of a capacitor. The reactive power of a load is given by Q = v/ sin() ( 1-61) where () is the impedance angle of the load. By convention, Q is positive for in-ductive loads and negative for capacitive loads, because Ihe impedance angle () is positive for inductive loads and negative for capacitive loads. TIle units of reac-tive power are voH-amperes reactive (var), where I var = 1 V X 1 A. Even though the dimensional units are the same as for watts, reactive power is traditionally given a unique name to distinguish it from power actually supplied 1 0 a load. TIle apparent power (S) supplied to a load is defined as the product of the voHage across the load and the current Ihrough the load. TIlis is the power thai "appears" to be supplied to the load if the phase angle differences between volt-age and current are ignored. Therefore, the apparenl power of a load is given by 50 ELECIRIC MACHINERY FUNDAMENTALS S = V[ ( 1-62) 1lle units of apparent power are volt-amperes (VA), where I VA = I V X 1 A. As with reactive power, apparent power is given a distinctive set of units to avoid confusing it with real and reactive power. Alternative Forms of the Power Equations If a load has a constant impedance, then Ohm's law can be used to derive alterna-tive expressions for the real, reactive, and apparent powers supplied to the load. Since the magnitude of the voltage across the load is given by V = JZ ( 1-63) substituting Equation ( 1--63) into Equations (1--60) to (1--62) produces equations for real, reactive, and apparent power expressed in tenns of current and impedance: P = [lZcos () Q = [ lZ sin () S = [ lZ where Izi is the magnitude of the load impedance Z. Since the impedance of the load Z can be expressed as Z ~ R + jX ~ Izl co, 0 + j Izl ' in 0 ( 1-64) ( 1-65) ( 1-66) we see from this equation that R = Izi cos () and X = Izi sin (), so the real and reacti ve powers of a load can also be expressed as where R is the resistance and X is the reactance of load Z. Complex Power ( 1-67) ( 1-68) For simplicity in computer calculations, real and reactive power are sometimes represented together as a complex power S, where s ~ P + jQ ( 1-69) 1lle complex power S supplied to a load can be calculated from the equation S = VI ( 1-70) where the asterisk represents the complex conjugate operator. To understand this equation, let's suppose that the voltage applied to a load is V = V L a and the current through the load is I = [ L {3. 1llen the complex power supplied to the load is INTRODUcnONTO MACHINERY PRINCIPLES 51 p --1 -+ 1 Q (~ v z Z = Izi Lon T -I FIGURE 1-31 An inductive load has a posilil'e impedance angle (J. This load produces a lngging current. and it consumes both real power P and reactive power Q from the source. S = VI = (VL a )(JL-f3) = VI L(a -(3) = VI cos(a -(3) + jVI sin(a -(3) The impedance angle () is the difference between the angle of the voltage and the angle of the current (() = a - /3), so this equation reduces to S = VI cos () + jVI sin () ~ P +jQ The Relationships between Impedance Angle, Current Angle, and Power As we know from basic circuit theory, an inductive load (Figure 1- 31) has a pos-itive impedance angle (), since the reactance of an inductor is positive. If the im-pedance angle () of a load is positive. the phase angle of the current flowing through the load will lag the phase angle of the voltage across the load by (). I = V = VLoo= ~ L_ () Z IzlL6 Izl Also, if the impedance angle () of a load is positive, the reactive power consumed by the load wi ll be positive (Equation 1-65), and the load is said to be consuming both real and reactive power from the source. In contrast, a capacitive load (Figure 1- 32) has a negative impedance angle (), since the reactance of a capacitor is negative. If the impedance angle () of a load is negative, the phase angle of the current flowing through the load wi ll lead the phase angle of the voltage across the load by (). Also, if the impedance an-gie () of a load is negative, the reactive power Q consumed by the load will be negative (Equation 1-65). In this case, we say that the load is consuming real power from the source and supplying reactive power to the source. The Power Triangle The real, reactive, and apparent powers supplied to a load are related by the power triangle. A power triangle is shown in Figure 1- 33. The angle in the lower left 52 ELECIRIC MACHINERY FUNDAMENTALS p --1 -+ I Q + rv v z Z = 121 Lon -1 -I ""GURE 1-32 A capacitive loo.d has a nega/il'e impedance angle (j, This load produces a leading current, and it consumes real pO\'er P from the source and while supplying reactive power Q to the source, s Q = SsinO o P = ScosO p cosO =-S sinO = SJ S tanO = ~ FI GURE 1-33 The power triangle, corner is the impedance angle (), The adjacent side of Ihis triangle is Ihe real power P supplied to the load, the opposite side of the triangle is the reactive power Q supplied to the load, and the hypotenuse of the triangle is the apparent power S of the load, 1lle quantity cos () is usually known as the power factor of a load, The power factor is defined as the fraction of the apparent power S that is actually sup-plying real power to a load, TIlUS, PF = cos () ( 1 - 71) where () is the impedance angle of the load, Note that cos () = cos (- ()), so the power factor produced by an impedance angle of +30° is exactly the same as the power factor produced by an impedance angle of -30°, Because we can't tell whether a load is inductive or capacitive from the power factor alone, it is customary to state whether the current is leading or lagging the voltage whenever a power factor is quoted, TIle power triangle makes the relationships among real power, reactive power, apparent power, and the power factor clear, and provides a convenient way to calculate various power-related quantities if some of them are known, Example I- II. Figure 1- 34 shows an ac voltage source supplying power to a load with impedance Z = 20L - 30° n. Calculate the current I supplied to the load, the power factor of the load, and the real, reactive, apparent, and complex power supplied to the load, INTRODUCTION TO MACHINERY PRINCIPLES 53 -1 I + '" \' = 120LO"V Z Z = 20L - 30"n -T I FIGURE 1-34 The circuit of Example I- II. Solutioll The current supplied to this load is I = V = 120LO° V = 6L300 A Z 20L 300 n The power factor of the load is PF = cos (J = cos (-30°) = 0.866 leading (1 - 71) (Note that this is a capacitive load, so the impedance angle (J is negative, and the current leads the voltage.) The real power supplied to the load is P = Vlcos (J P = (120 VX6A) cos (-30°) = 623.5 W The reactive power supplied to the load is Q=Vlsin(J Q = (120 V)(6A) sin (-30°) = -360 VAR The apparent power supplied to the load is S = VI Q = (120 V)(6A) = 720 VA The complex power supplied to the load is S = VI 1.10 SUMMARY = (l20LOOV)(6L-30° A) = (l20LO° V)(6L30° A) = 720L30° VA = 623.5 - j360 VA (1- 60) (1- 61) (1- 62) (1- 70) This chapter has reviewed briefly the mechanics of systems rotating about a sin-gle axis and introduced the sources and effects of magnetic fields important in the understanding of transformers, motors, and generators. Historically, the English system of units has been used to measure the mechanical quantities associated with machines in English-speaking countries. 54 ELECIRIC MACHINERY FUNDAMENTALS Recently, the 51 units have superseded the English system almost everywhere in the world except in the United States, but rapid progress is being made even there. Since 51 is becoming almost universal, most (but not all) of the examples in this book use this system of units for mechanical measurements. Electrical quantities are always measured in 51 units. I n the section on mechanics, the concepts of angu lar position, angular veloc-ity, angular acceleration, torque, Newton's law, work, and power were explained for the special case of rotation about a single axis. Some fundamental relationships (such as the power and speed equations) were given in both 51 and English units. TIle prOOuction of a magnetic field by a current was explained, and the spe-cial properties of ferromagnetic materials were explored in detail. The shape of the magnetization curve and the concept of hysteresis were explained in terms of the domain theory of ferromagnetic materials, and eddy current losses were discussed. Faraday's law states that a voltage will be generated in a coil of wire that is proportional to the time rate of change in the flux passing through it. Faraday's law is the basis oftransfonner action, which is explored in detail in Chapter 3. A current-carrying wire present in a magnetic field, if it is oriented properly, wi ll have a force induced on it. This behavior is the basis of motor action in all real machines. A wire moving through a magnetic field with the proper orientation will have a voltage induced in it. TIlis behavior is the basis of generator action in all real machines. A simple linear dc machine consisting of a bar moving in a magnetic field illustrates many of the features of real motors and generators. When a load is at-tached to it, it slows down and operates as a motor, converting electric energy into mechanical energy. When a force pulls the bar faster than its no-load steady-state speed, it acts as a generator, converting mechanical energy into electric energy. In ac circuits, the real power P is the average power supplied by a source to a load. TIle reactive power Q is the component of power that is exchanged back and forth between a source and a load. By convention, positive reactive power is consumed by inductive loads (+ 0) and negative reactive power is consumed (or positive reactive power is supplied) by capacitive loads (- 0). TIle apparent power S is the power that "appears" to be supplied to the load if only the magnitudes of the voltages and currents are considered. QUESTIONS I-I. What is torque? What role does torque play in the rotational motion of machines? 1-2. What is Ampere's law? 1-3. What is magnetizing intensity? What is magnetic flux density? How are they related? 1-4. How does the magnetic circuit concept aid in the design of transformer and machine cores? 1-5. What is reluctance? 1-6. What is a ferromagnetic material? Why is the permeability of ferromagnetic mate-rials so high? INTRODUCTION TO MACHINERY PRINCIPLES 55 1-7. How does the relative penneability of a ferromagnetic material vary with magneto-motive force? 1-8. What is hysteresis? Explain hysteresis in tenns of magnetic domain theory. 1-9. What are eddy current losses? What can be done to minimize eddy current losses in a core? 1- 10. Why are all cores exposed to ac flux variations laminated? I- II. What is Faraday's law? 1 - 12. What conditions are necessary for a magnetic field to produce a force on a wire? 1 - 13. What conditions are necessary for a magnetic field to produce a voltage in a wire? 1 - 14. Why is the linear machine a good example of the behavior observed in real dc machines? 1 - 15, The linear machine in Figure 1 - 19 is running at steady state. What would ha~n to the bar if the voltage in the battery were increased? Explain in detail. 1- 16. Just how does a decrease in flux produce an increase in speed in a linear machine? 1 - 17, Will current be leading or lagging voltage in an inductive load? Will the reactive power of the load be positive or negative? 1 - 18. What are real, reactive, and apparent power? What lUlits are they measured in? How are they related? 1 - 19. What is power factor? PROBLEMS 1- 1. A motor's shaft is spinning at a speed of 3000 r/min. What is the shaft speed in radians per second? 1- 2. A flywheel with a moment of inertia of 2 kg 0 m2 is initially at rest. If a torque of 5 N o m (cOlUlterc1ockwise) is suddenly applied to the flywheel, what will be the speed of the flywheel after 5 s? Express that speed in both radians per second and revolutions per minute. 1-3. A force of 10 N is applied to a cylinder, as shown in Figure PI- I. What are the mag-nitude and direction of the torque produced on the cylinder? What is the angular ac-celeration a of the cylinder? 3D' , r= 0.25 m J=5k:s o m2 F = ION fo'IGURE 1 '1-1 The cylinder of Problem - 3. 56 ELECIRIC MACHINERY FUNDAMENTALS 1-4. A motor is supplying 60 N · m of torque to its load. If the motor's shaft is turning at 1800 r/min. what is the mechanical power supplied to the load in watts? In horse-power? 1-5. A ferromagnetic core is shown in Figure PI- 2. The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb. With this current. what is the flux density at the top of the core? What is the flux density at the right side of the core? Assrune that the relative permeability of the core is I()(x). I. I' ,m 1 r lOem - i--- 20 cm---+ - =~-T ", m ; [. + -I-e--- - ---- - --- - -400 turns ", m - - --- - -I-e-[. ", m 1 Core depth - 5 em fo'IGURE PI- 2 The core of Problems 1 - 5 and 1- 16. 1-6. A ferromagnetic core with a relative permeability of 1500 is shown in Figure PI- 3. The dimensions are as shown in the diagram. and the depth of the core is 7 cm. The air gaps on the left and right sides of the core are 0.070 and 0.050 cm. respectively. Because of fringing effects. the effective area of the air gaps is 5 percent larger than their physical size. If there are 400 IlU1lS in the coil wrapped arOlUld the center leg of the core and if the current in the coil is 1.0 A. what is the flux in each of the left. center. and right legs of the core? What is the flux density in each air gap? 1-7. A two-legged core is shown in Figure PI-4. The winding on the left leg of the core (Nt) has 400 turns. and the winding on the right (N2) has 300 turns. The coils are wound in the directions shown in the figure. If the dimensions are as shown. then what flux would be produced by currents il = 0.5 A and i2 = 0.75 A? Assume I-L, = J(XXl and constant. 1-8. A core with three legs is shown in Figure PI- 5. Its depth is 5 cm. and there are 200 IlU1lS on the leftmost leg. The relative penneability of the core can be assruned to be 1500 and constant. What flux exists in each of the three legs of the core? What is the flux density in each of the legs? Assume a 4 percent increase in the effective area of the air gap due to fringing effects. ---r 'om - f-; -----400 turns JOcm -O.07cm 0.05 cm --- --- f-'om ---.L Core depth = 7 em H GURE )'1-3 The core of Problem 1-6. r-- 15 cm-+------- 50'm-------+-15 Cffi---1 T 15 em ; , ;, -- -- --- -- -- --- -50 om - --400 turns 300 turns ----- ---- -N, N, -----------15 em ~ Core depth'" 15 em FIGURE PI- 4 The core of Problems - 7 and - 12. 57 58 ELECIRIC MACHINERY FUNDAMENTALS ~r- 25cm----t- 15cm-+--- 25'm---r l C 9 C,"m'-ll T ',m ; --2A -- -200 turns -O.04cm -- ---t 25cm + ',m ~ Core depth = 5 cm ""GURE PI-S The core of Problem 1--8. 1-9. The wire shown in Figure PI--6 is carrying 5.0 A in the presence of a magnetic field. Calculate the magnitude and direction of the force induced on the wire. ----1= 1 m I -b n =O.25 T. ----- to the right i=5.0A _ ""GURE 1'1-6 A current-carrying wire in a magnetic field (Problem 1 - 9). 1- 10. The wire shown in Figure PI- 7 is moving in the presence of a magnetic field. With the information given in the figure. detennine the magnitude and direction of the in-duced voltage in the wire. I- II. Repeat Problem 1 - 10 for the wire in Figure PI-8. 1- 12. The core shown in Figure PI-4 is made of a steel whose magnetization curve is shown in Figure PI-9. Repeat Problem 1 - 7. but this time do not asswne a constant value of Pro How much flux is produced in the core by the currents specified? What is the relative permeability of this core under these conditions? Was the asswnption x x x X X X X X X X X \'5m1s X X X X , {:O.Sm INTRODUCTION TO MACHINERY PRINCIPLES 59 x x x x ~ 45' ~ X ~ X ~ ~ ~ X ~ X X , ~ ~ • / ' X x X I=O.SOm X X X X X X ') X X HGURE 1 '1-7 A wire moving in a magnetic field (Problem 0 ", 0.25 T. into the page 1- 10). lV=,mlS U =O.5 T HGURE " 1-8 A wire moving in a magnetic field (Problem I- II). in Problem 1- 7 that the relative penneability was equal to 1()(x) a good assumption for these conditions? Is it a good assumption in general? 1-13. A core with three legs is shown in Fig ure Pi- IO. Its depth is 8 em, and there are 400 turns on the center leg. The remaining dimensions are shown in the figure. The core is composed of a steel having the magnetization curve shown in Figure I- Uk . An-swer the following questions about this core: (a) What current is required to produce a flux density of 0.5 T in the central1eg of the core? (b) What current is required to produce a flux density of 1.0 T in the central leg of the core? Is it twice the current in part (a)? (c) What are the reluctances of the central and right legs of the core under the con-ditions in part (a)? (d) What are the reluctances of the central and right legs of the core under the con-ditions in part (b)? (e) What conclusion can you make about reluctances in real magnetic cores? 1- 14. A two-legged magnetic core with an air gap is shown in Figure PI - II. The depth of the core is 5 cm. the length of the air gap in the core is 0.06 cm. and the munber of turns on the coil is I(x)(). The magnetization curve of the core material is shown in 60 ELECIRIC MACHINERY FUNDAMENTALS 1.00 E • ~ 0.75 ." , ~ " B 0.50 " 0.25 0.0 100 1000 Magnetizing intensity H (A ' turns/m) ""GURE 1 '1-9 The magnetization curve for the core material of Problems 1- 12 and 1- 14. ; -N 400 turns ~gCm1- 16cm -----1-8cm+- 16C1l1-+8cm -1 Depth '" 8 em ""GURE Pi- IO The core of Problem 1- 13. T 8cm t 16 em 1 I 80m -.L Figure PI-9. Assume a 5 percent increase in effective air-gap area to account for fringing. How much current is required to produce an air-gap flux density of 0.5 T? What are the flux densities of the four sides of the core at that current ? What is the total flux present in the air gap? INTRODUcnONTO MACHINERY PRINCIPLES 61 T \Oem ; 1)-- ---------- -------N: \,O(X) turns O06om1= JOem ----------- -- - -I-f-\Oem ~ L \0 em ~--- 30 em------ll -;-:--ll r l 'om Depth : 5 em ""GURE Pl-ll The core ofProbJem - \4. 1 - 15. A transformer core with an effective mean path length of JO in has a 300-tW1l coil wrapped arOlUld one leg. Its cross-sectional area is 0.25 inl. and its magnetization curve is shown in Figure 1- IOc. If current of 0.25 A is flowing in the coil. what is the total flux in the core? What is the flux density? 1 - 16. The core shown in Figure PI- 2 has the flux cp shown in Figure PI- l2. Sketch the voltage present at the terminals of the coil. 1 - 17. Figure PI- 13 shows the core ofa simple dc motor. The magnetization curve for the metal in this core is given by Figure I- JOc and d. Assume that the cross-sectional area of each air gap is 18 cm2 and that the width of each air gap is 0.05 em. The ef-fective diameter of the rotor core is 4 em. (a) It is desired to build a machine with as great a flux density as possible while avoiding excessive saturation in the core. What would be a reasonable maxi-mum flux density for this core? (b) What would be the total flux in the core at the flux density of part (a)? (c) The maximum possible field current for this machine is I A. Select a reasonable nwnber of turns of wire to provide the desired flux density while not exceeding the maximum available current. 1 - 18. Asswne that the voltage applied to a load is V = 208L -30° V and the current flow-ing through the load is I = 5L 15° A. (a) Calculate the complex power S consruned by this load. (b) Is this load inductive or capacitive? (c) Calculate the power factor of this load. 62 ELECIRIC MACHINERY FUNDAMENTALS omo -------- -------------------------0.005 O~---}----' 2 ----'3-'~"4----~ 5----" 6C--->C---. 8 C---- f(ms) - 0.005 - 0.010 --------------------- -------------FIGURE 1'1 - 12 Plot of flux 4> as a function of time for Problem 1- 16. 4,m N=? Ntums 4,m Depth = 4cm FIGURE 1 '1- 13 The core of Problem 1- 17. (d) Calculate the reactive power consmned or supplied by this load. Does the load consume reactive power from the source or supply it to the source? 1- 19. Figure PI - 14 shows a simple single-phase ac power system with three loads. The voltage source is V = l20LO° V. and the impedances of the three loads are 2:J = 5L _90° n Answer the following questions about this power system. (a) Assrune that the switch shown in the figure is open. and calculate the current I. the power factor. and the real. reactive. and apparent power being supplied by the load. INTRODUCTION TO MACHINERY PRINCIPLES 63 (b) Assume that the switch shown in the figure is closed, and calculate the current I, the power factor, and the real, reactive, and apparent power being supplied by the load. (c) What happened to the current flowing from the source when the switch closed? Why? -1 +1 +1 +1 + '"., V z, Z, Z, -T 1 1 1 FIGURE PI- 14 The circuit of Problem 1- \9. 1-20. Demonstrate that Equation (I- 59) can be derived from Equation (I- 58) using the simple trigonometric identities: pet) = v(t)i(t) = 2VI cos wt cos(wt -(J) pet) = VI cos (J (I + cos 2wt) + VI sin e sin 2wt (I- 58) (I- 59) 1-2 1. The linear machine shown in Figure PI- IS has a magnetic flux density of 0.5 T directed into the page, a resistance of 0.25 n, a bar length I = 1.0 m, and a battery voltage of 100 V. (a) What is the initial force on the bar at starting? What is the initial current flow? (b) What is the no-load steady-state speed of the bar? (c) If the bar is loaded with a force of 25 N opposite to the direction of motion, what is the new steady-state speed? What is the efficiency of the machine under these circrunstances? 1=0 1 0.25 n i " -X VB = \00 V -=-X FIGURE PI- IS The linear machine in Problem - 21. 1-22. A linear machine has the following characteristics: B = 0.33 T into page 1= 0.5 m R = 0.50 n VB = 120V H =0.5 T x X X 1m X X X 64 ELECIRIC MACHINERY FUNDAMENTALS (a) If this bar has a load of 10 N attached to it opposite to the direction of motion, what is the steady-state speed of the bar? (b) If the bar nms off into a region where the flux density falls to 0.30 T, what hap-pens to the bar? What is its fmal steady-state speed? (c) Suppose VB is now decreased to 80 V with everything else remaining as in part b. What is the new steady-state speed of the bar? (d) From the results for parts band c, what are two methods of controlling the speed of a linear machine (or a real dc motor)? REFERENCES I. Alexander. Charles K., and Matthew N. O. Sadiku: Fundamentals of Electric Circuits, McGraw-Hill. 2CXXl. 2. Beer. E , and E. Johnston. Jr.: Vector Mechanicsfor Engineers: Dynamics, 6th ed., McGraw-Hill. New Y ort.I 997. 3. Hayt. William H.: Engineering Electromllgnetics, 5th ed .. McGraw-Hill. New Yort. 1989. 4. Mulligan. J. E : Introductory College Physics, 2nd ed .. McGraw-HilL New Y ork. 1991. 5. Sears. Francis W., Mark W. Zemansky. and Hugh D. Young: University Physics, Addison-Wesley. Reading. Mass., 1982. CHAPTER 2 TRANSFORMERS A transformer is a device that changes ac electric power at one voltage level to ac electric power at another voltage level through the action of a magnetic field. It consists of two or more coils of wire wrapped around a common ferromagnetic core. These coils are (usually) not directly connected. The only connection be-twecn the coils is the common magnetic nux present within the core. FIGURE 2-1 The fi rst practical modern transformer. built by William Stanley in 1885. Note that the core is made up of individual sheets of metal (laminations). (Courtesy ofGeneml Electric Company.) 65 66 ELECIRIC MACHINERY FUNDAMENTALS One of the transfonner windings is connected to a source of ac electric power, and the second (and perhaps third) transformer winding supplies electric power to loads. 1lle transfonner winding connected to the power source is called the primary winding or input winding, and the winding connected to the loads is called the secondnry winding or output winding. I f there is a third winding on the transformer, it is called the tertiary winding. 2.1 WHY TRANSFORMERS ARE IMPORTANT TO MODERN LIFE TIle first power distribution system in the United States was a 120-V dc system in-vented by Thomas A. Edison to supply power for incandescent light bulbs. Edi-son's first central power station went into operation in New York City in Septem-ber 1882. Unfortunately, his power system generated and transmitted power at such low voltages that very large currents were necessary to supply significant amounts of power. These high currents caused huge voltage drops and power losses in the transmission lines, severely restricting the service area of a generat-ing station. In the 1880s, central power stations were located every few city blocks to overcome this problem. The fact that power could not be transmitted far with low-voltage dc power systems meant that generating stations had to be small and localized and so were relatively inefficient. TIle invention of the transfonner and the concurrent development of ac power sources eliminated forever these restrictions on the range and power level of power systems. A transfonner ideally changes one ac voltage level to another voltage level without affecting the actual power supplied. If a transfonner steps up the voltage level of a circuit, it must decrease the current to keep the power into the device equal to the power out of it. 1llcrefore, ac electric power can be gener-ated at one central location, its voltage stepped up for transmission over long dis-tances at very low losses, and its voltage stepped down again for fmal use. Since the transmission losses in the lines of a power system are proportional to the square of the current in the lines, raising the transmission voltage and reducing the resulting transmission currents by a factor of 10 with transformers reduces power transmission losses by a factor of lOll Without the transfonner, it would simply not be possible to use electric power in many of the ways it is used today. In a rmx:lern power system, electric power is generated at voltages of 12 to 25 kV. Transfonners step up the voltage to between 110 kV and nearly 1000 kV for transmission over long distances at very low losses. Transfonners then step down the voltage to the 12- to 34.5-kV range for local distribution and fmally pennit the power to be used safely in homes, offices, and factories at voltages as low as 120 V. 2.2 TYPES AND CONSTRUCTION OF TRANSFORMERS The principal purpose of a transformer is to convert ac power at one voltage level to ac power of the same frequency at another voltage level. Transfonners are also TRANSFORMERS 67 i, (I) --+ / \ + ") N, N, ) v, (t \ . HGURE 2-2 Core-foml transfomler construction. used for a variety of other purposes (e.g., voltage sampling, current sampling, and impedance transformation), but this chapter is primarily devoted to the power transformer. Power transfonners are constructed on one of two types of cores. One type of construction consists of a simple rectangular laminated piece of steel with the transformer windings wrapped around two sides of the rectangle. This type of construction is known as corefonn and is illustrated in Figure 2- 2. The other type consists of a three-legged laminated core with the windings wrapped around the center leg. nlis type of construction is known as shell form and is illustrated in Figure 2- 3. In either case, the core is constructed of thin laminations electrically isolated from each other in order to minimize eddy currents. The primary and secondary windings in a physical transformer are wrapped one on top of the other with the low-voltage winding innermost. Such an arrange-ment serves two purposes: I . It simplifies the problem of insu lating the high-voltage winding from the core. 2. It results in much less leakage nux than would be the case if the two windings were separated by a distance on the core. Power transformers are given a variety of different names, depending on their use in power systems. A transformer connected to the output of a generator and used to step its voltage up to transmission levels (110+ kV) is sometimes called a unit transformer. The transfonner at the other end of the transmission line, which steps the voltage down from transmission levels to distribution levels (from 2.3 to 34.5 kV), is called a substation transfonner. Finally, the transformer that takes the distribution voltage and steps it down to the final voltage at which the power is actually used (110, 208, 220 V, etc.) is called a distribution transformer. All these devices are essentially the same- the only difference among them is their intended use. 68 ELECIRIC MACHINERY FUNDAMENTALS (a) ""GURE 2-3 (a) Shell-form transformer construction. (b) A typical shell-form transformer. (Courtesy ofGeneml Electric Company.) In addition to the various power transfonners, two special-purpose trans-fonners are used with electric machinery and power systems. TIle first of these special transformers is a device specially designed to sample a high voltage and produce a low secondary voltage directly proportional to it. Such a transfonner is called a potential transfonner. A power transformer also produces a secondary voltage directly proportional to its primary voltage; the difference between a po-tential transfonner and a power transfonner is that the potential transformer is de-signed to handle only a very small current. The second type of special transfonner is a device designed to provide a secondary current much smaller than but directly proportional to its primary current. This device is called a current transformer. Both special-purpose transformers are discussed in a later section of this chapter. 2.3 THE IDEAL TRANSFORMER An ideal transformer is a lossless device with an input winding and an output winding. The relationships between the input voltage and the output voltage, and betwccn the input current and the output current, are given by two simple equa-tions. Figure 2- 4 shows an ideal transfonner. TIle transformer shown in Figure 2- 4 has N p turns of wire on its primary side and Ns turns of wire on its secondary side. 1lle relationship betwccn the volt-TRANSFORMERS 69 ip (t) i, (t) --• • + + ( \ N, N, "'p(1) ",,(t) \ J ,., --,b, H GURE 2-4 (a) Sketch of an ideal transformer. (b) Schematic symbols of a transformer. age vp(t) applied to the primary side of the transformer and the voltage vsCt) pro-duced on the secondary side is ~ 'p"(t") -!iJ'~-, vsCt) = Ns = a where a is defined to be the turns ratio of the transformer: Np a~ N, (2- 1) (2- 2) llle relationship between the current il...t) flowing into the primary side of the trans-fonner and the current isCt) flowing out of the secondary side of the transfonner is ;"I,IL 1 isCt) - a (2- 3a) (2- 3b) 70 ELECIRIC MACHINERY FUNDAMENTALS In tenns of phasor quantities, these equations are ,nd ~ ~ I" ~ 11 I, a (2- 4) (2- 5) Notice that the phase angle of Vp is the same as the angle of Vs and the phase an-gie of Ip is the same as the phase angle of Is. TIle turns ratio of the ideal trans-fonner affects the magnitudes of the voltages and currents, but not their angles. Equations (2-1 ) to (2- 5) describe the relationships between the magnitudes and angles of the voltages and currents on the primary and secondary sides of the transformer, but they leave one question unanswered: Given that the primary cir-cuit's voltage is positive at a specific end of the coil, what would the polarity of the secondary circuit's voltage be? In real transformers, it would be possible to tell the secondary's polarity only if the transformer were opened and its windings ex-amined. To avoid this necessity, transfonners utilize the dot convention. The dots appearing at one end of each winding in Figure 2-4 tell the polarity of the voltage and current on the secondary side of the transformer. TIle relationship is as follows: I. If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the dots on each side of the core. 2. If the primary current of the transformer fl ows into the dotted end of the pri-mary winding, the secondary current wi ll flow out of the dotted end of the secondary winding. TIle physical meaning of the dot convention and the reason polarities work out this way wi ll be explained in Section 2.4, which deals with the real transfonner. Power in an Ideal Transformer TIle power supplied to the transformer by the primary circuit is given by the equation (2-6) where ()p is the angle between the primary voltage and the primary current. The power supplied by the transformer secondary circuit to its loads is given by the equation (2- 7) TRANSFORMERS 71 where ()s is the angle between the secondary voltage and the secondary current. Since voltage and current angles are unaffected by an ideal transfonner, ()p - ()s = (). The primary and secondary windings of an ideal transfonner have the same power factor. How does the power going into the primary circuit of the ideal transformer compare to the power coming out of the other side? lt is possible to find out through a simple application of the voltage and current equations [Equations (2-4) and (2- 5)]. 1lle power out of a transformer is P oot = Vsls cos () (2-8) Applying the turns-ratio equations gives Vs = Vp /a and Is = alp, so -"" P out -a (alp) cos () I Poot -VpIp cos () -P;n (2- 9) Thus, the output power of an ideal transfonner is equal to its input power. The same relationship applies to reactive power Q and apparent power S: IQ ;" VpIp sin () VsIs sin () Qoo, I (2- 10) and Is;, ~ Vplp = VsIs = Soot I (2- 11 ) Impedance Transformation through a Transformer The impedance of a device or an element is defined as the ratio of the phasor volt-age across it to the phasor current flowing through it: V, ZL = 1; (2- 12) One of the interesting properties of a transfonner is that, since it changes voltage and current levels, it changes the ratio between voltage and current and hence the apparent impedance of an element. To understand this idea, refer to Figure 2- 5. If the secondary current is calJed Is and the secondary voltage Vs, then the imped-ance of the load is given by V , ZL = 1; The apparent impedance of the primary circuit of the transfonner is V Z' = --.f.. , Ip Since the primary voltage can be expressed as Vp = aVs (2- 13) (2- 14) 72 ELECIRIC MACHINERY FUNDAMENTALS -+ v, Z, v, z,= -I, ,,' -+ + • • V, -,b, ""GURE 2--.5 (a) Definition of impedance. (b) Impedance scaling through a transformer. and the primary current can be expressed as " Ip = -a the apparent impedance of the primary is V aV V Z' -..:...t!. - ~ -a2.:...s. L -Ip -Isla -Is I Z~ =a2 ZL I Z, (2- 15) With a transfonner, it is possible to match the magnitude of a load imped-ance to a source impedance simply by picking the proper turns ratio. Analysis of Circuits Containing Ideal Transformers If a circuit contains an ideal transformer, then the easiest way to analyze the cir-cuit for its voltages and currents is to replace the portion of the circuit on one side of the transfonner by an equivalent circuit with the same tenninal characteristics. After the equivalent circuit has been substituted for one side, then the new circuit (without a transformer present) can be solved for its voltages and currents. In the portion of the circuit that was not replaced, the solutions obtained will be the COf--,I Ztime + V =4S0LO"V -,., T, I tiDe O.ISo. 1 ,I • :10 -• Zlime + -V =4S0LO"V ,b, FIGURE 2-6 j 0.24 0. j 0.24 0. TRANSFORMERS 73 + V~~ -T, 10: 1 • • I' z~ 4+j3o. 1 ~ -Z + 4+j ~ -V ~ The power system of Example 2- 1 (a) without and (b) with transformers at the ends of the transmission line. rect values of voltage and current for the original circuit. 1llen the turns ratio of the transfonner can be used to detennine the voltages and currents on the other side of the transfonner. TIle process of replacing one side of a transformer by its equivalent at the other side's voltage level is known as referring the first side of the transfonner to the second side. How is the equivalent circuit fonned? Its shape is exactly the same as the shape of the original circuit. TIle values of voltages on the side being replaced are scaled by Equation (2-4), and the values of the impedances are scaled by Equa-tion (2- 15). TIle polarities of voltage sources in the equivalent circuit will be re-versed from their direction in the original circuit if the dots on one side of the transformer windings are reversed compared to the dots on the other side of the transformer windings. The solution for circuits containing ideal transformers is illustrated in the following example. Example 2-1. A single-phase power system consists of a 4SO-V 60-Hz gen-erator supplying a load Z = 4 + )3 0 through a transmission line of impedance Ztm. = O.IS + jO.24 O. Answer the following questions about this system. (a) If the power system is exactly as described above (Figure 2-6a), what will the voltage at the load be? What will the transmission line losses be? 74 ELECIRIC MACHINERY FUNDAMENTALS (b) Suppose a I: 10 step-up transformer is placed at the generator end of the trans-mission line and a 10: I step-down transfonner is placed at the load end of the line (Figure 2- 6b). What will the load voltage be now? What will the transmis-sion line losses be now? Solutioll (a) Figure 2-6a shows the power system without transfonners. Here IG = IIu.. = Ilood' The line current in this system is given by IIu.. = ,----'V-,- ZIi". + Zioad 480 L O ° V = " (O".1"' 8'"~ +' jio.'!! 24'f!i"~ ) "; + -' ( "4'"~ +' j" 3 "'ll ) 480 LO° = = 4.18 + j3.24 = 9O.8L-37.8° A Therefore the load voltage is V10ad = I line~ 480 LO° 5.29L37.8° = (90.8 L -37.8° A)(4 n + j3 n ) = (90.8 L -37.8° A)(5 L36.9° fl) = 454 L - 0.9° V and the line losses are Pim• = (tUDe)' R line = (90.8 A)' (0.1 8 n) = 1484 W (b) Figure 2-6b shows the power system with the transfonners. To analyze this sys-tem, it is necessary to convert it to a common voltage level. This is done in two steps: I. Eliminate transfonner T2 by referring the load over to the transmission line's voltage level. 2. Eliminate transformer TI by referring the transmission line's elements and the equivalent load at the transmission line's voltage over to the source side. The value of the load's impedance when reflected to the transmission system's voltage is Z · - , ' Z load -load = (1f)\4n + j3 n) = 4000 + j300n The total impedance at the transmission line level is now Zeq = ~ine + Z io.t = 400.18 + j300.24 n = 500.3 L36.88° n TRANSFORMERS 75 V =480LO"V 0.18 n jO.24 n -:10 " , , • • , ZHDO , Z'= : , + 4OO+j300n , , , , ,~ , ,' : Equivalent cin:uit 0.0018 fi jO.0024fi I ,I , " + Z'line ( V =480LO"V Z ' =4+j3fi -. Equivalent cin:uit ,b , FIGURE 2-7 (a) System with the load referred to the transmission system voltage level. (b) System with the load and transmission line referred to the generator's voltage level. This equi valent circuit is shown in Figure 2- 7a. The total impedance at the transmission line level (4". + Zl~ is now reflected across Tl to the source's voltage level: Z ' = a2 Z ~ ~ = a 2(l1ine + Z k...i) = ( lb)\0.18 0 + jO.24 0 + 400 0 + j300f.l) = (0.0018 0 + jO.0024 0 + 4 n + j3 f.!) = 5.003 L36.88° n Notice that Z'k-d = 4 + j3 0 and:!.time = 0.0018 + jllOO24 n. The resulting equi valent cir-cuit is shown in Figure 2- 7b. The generator's current is _ 480LOo V _ ° 1 0 - 5.003 L36.88° n - 95 .94L-36.88 A Knowing the current Ie. we can now work back and find II;'" and 1 . Working back through Tl , we get 76 ELECIRIC MACHINERY FUNDAMENTALS = 1~(95.94 L-36.88° A) = 9.594L-36.88° A Working back through T2 gives NnIline = NSlIload "n Ilood = N IUDe n = \0 (9.594 L-36.880 A) = 95.94L-36.88° A It is now possible to answer the questions originally asked. The load voltage is given by V10ad = Ibdl10ad = (95.94 L-36.88° A)(5 L36.87° 0 ) = 479.7 L-O.Olo V and the line losses are given by PIo!;. = (/n ... )lRnne = (9.594 A)l (0.18 n) = 16.7 W Notice that raising the transmission voltage of the power system reduced transmission losses by a factor of nearly 90! Also, the voltage at the load dropped much less in the system with transformers compared to the system without trans-fonners. This simple example dramatically illustrates the advantages of using higher-voltage transmission lines as well as the extreme importance of transform-ers in modern power systems. 2.4 THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS TIle ideal transformers described in Section 2.3 can of course never actually be made. What can be produced are real transformers-two or more coils of wire physically wrapped around a ferromagnetic core. The characteristics of a real transformer approximate the characteristics of an ideal transfonner, but only to a degree. This section deals with the behavior of real transformers. To understand the operation of a real transfonner, refer to Figure 2--8. Fig-ure 2- 8 shows a transfonner consisting of two coils of wire wrapped around a transfonner core. 1lle primary of the transfonner is connected to an ac power source, and the secondary winding is open-circuited. TIle hysteresis curve of the transformer is shown in Figure 2- 9. TRANSFORMERS 77 iP(t) -+ + l' 0-N, vP(t) FIGURE l-8 Sketch of a real transformer with no load attached to its secondary. q, Rux --------ttt--------- Magnetomotive force FIGURE 2-9 The hysteresis curve of the transformer. The basis of transfonner operation can be derived from Faraday's law: dA e iod = dt ( 1-41 ) where A is the flux linkage in the coil across which the voltage is being induced. The flux linkage A is the sum of the flux passing through each turn in the coil added over all the turns of the coil: N A ~ '2:; ( 1-42) ; = 1 78 ELECIRIC MACHINERY FUNDAMENTALS The total flux linkage through a coil is not just N<p, where N is the number of turns in the coil, because the flux passing through each turn of a coil is slightly differ-ent from the flux in the other turns, depending on the position of the turn within the coil. However, it is possible to define an average flux per turn in a coil. If the total flux linkage in all the turns of the coils is A and if there are N turns, then the average flux per turn is given by and Faraday's law can be written as -A q,~ N -<& eind -N dt The Voltage Ratio across a Transformer (2-1 6) (2-1 7) Ifthe voltage of the source in Figure 2--8 is vp(t), then that voltage is placed di-rectly across the coils of the primary winding of the transformer. How wi ll the transformer react to this applied voltage? Faraday's law explains what wi ll hap-pen. When Equation (2- 17) is solved for the average flux present in the primary winding of the transfonner, the result is -1 I <P = N Vp(t) dt p (2- 18) TIlis equation states that the average flux in the winding is proportional to the in-tegral of the voltage applied to the winding, and the constant of proportionality is the reciprocal of the number of turns in the primary winding IINp. TIlis flux is present in the primary coil of the transformer. What effect does it have on the secondary coil of the transfonner? TIle effect depends on how much of the flux reaches the secondary coil. Not all the flux produced in the primary coil also passes through the secondary coil-some of the flux lines leave the iron core and pass through the air instead (see Figure 2- 10). TIle portion of the flux that goes through one of the transfonner coils but not the other one is called leak-age flux. The flux in the primary coil of the transformer can thus be divided into two components: a mutual flux, which remains in the core and links both wind-ings, and a small leakage flux, which passes through the primary winding but re-turns through the air, bypassing the secondary winding: I s = 4>M + 4>LS 4>s = total average secondary flux (2- 20) 4>M = flux component linking both primary and secondary coils fu = secondary leakage flux With the division of the average primary flux into mutual and leakage com-ponents, Faraday's law for the primary circuit can be reexpressed as (2- 21) The first term of this expression can be called ep(l), and the second term can be called eLP(l). If this is done, then Equation (2- 21) can be rewritten as (2- 22) 80 ELECIRIC MACHINERY FUNDAMENTALS TIle voltage on the secondary coil of the transfonner can also be expressed in terms of Faraday's law as ds vs<.t) = NSdt _ dM dfu - Ns dt + Ns dt = esCt) + eL'>(t) TIle primary voltage due to the mutualflux is given by _ d4>M ep(t) - Np dt and the secondary voltage due to the mutualflux is given by _ d4>M es(t) - Ns dt Notice from these two relationships that TIlerefore, ep(t) _ d4>M _ edt) Np -dt -Ns (2- 23) (2- 24) (2- 25) (2- 26) (2- 27) TIlis equation means that the ratio of the primary voltage caused by the mutual flux to the secondary voltage caused by the mutualflux is equal to the turns ratio of the transformer. Since in a well-designed transfonner 4>M » M » 4>u" the ratio of the total voltage on the primary of a transformer to the to-tal voltage on the secondary of a transfonner is approximately vp(t) !i.E. vs<.t) = Ns = a (2- 28) TIle smaller the leakage fluxes of the transfonner are, the closer the total trans-fonner voltage ratio approximates that of the ideal transfonner discussed in Sec-tion 2.3. The Magnetization Current in a Real Transformer When an ac power source is connected to a transformer as shown in Figure 2--8, a current flows in its primary circuit, even when the secondary circuit is open-circuited. TIlis current is the current required to produce flux in a real ferromag-netic core, as explained in Chapter I. It consists of two components: TRANSFORMERS 81 I. The magnetization current iM , which is the current required to produce the flux in the transformer core 2. llle core-loss current iH " which is the current required to make up for hys-teresis and eddy current losses Figure 2-11 shows the magnetization curve of a typical transformer core. If the fl ux in the transformer core is known, then the magnitude of the magnetization current can be found directly from Figure 2-11. Ignoring for the moment the effects of leakage flux, we see that the average flux in the core is given by -1 f cf> = Np vp(t)dt (2-1 8) If the primary voltage is given by the expression vp(t) = VM cos wt V, then the re-sulting flux must be cf> = ~ f VM cos wtdt p VM . ~ -- smwt Wb wNp (2- 29) If the values of current required to produce a given flux (Figure 2-11 a) are com-pared to the flux in the core at different times, it is possible to construct a sketch of the magnetization current in the winding on the core. Such a sketch is shown in Figure 2-11 b. Notice the following points about the magnetization current: I. The magnetization current in the transfonner is not sinusoidal. The higher-frequency components in the magnetization current are due to magnetic sat-uration in the transfonner core. 2. Once the peak flux reaches the saturation point in the core, a small increase in peak flux requires a very large increase in the peak magnetization current. 3. The fundamental component of the magnetization current lags the voltage ap-plied to the core by 90°. 4. llle higher-frequency components in the magnetization current can be quite large compared to the fundamental component. In general, the further a trans-fonner core is driven into saturation, the larger the hannonic components wil I become. The other component of the no-load current in the transformer is the current required to supply power to make up the hysteresis and eddy current losses in the core. lllis is the core-loss current. Assume that the flux in the core is sinusoidal. Since the eddy currents in the core are proportional to d<PIdt, the eddy currents are largest when the flux in the core is passing through 0 Wb. lllerefore, the core-loss current is greatest as the flux passes through zero. The total current required to make up for core losses is shown in Figure 2-1 2. ,.Wb --------,/---------- ~,A · turns (a) " ---7'C---------------------t---~ , , , , , , " , " , , , , , , ~~ 'c----r'--'"--",C-'--- , , , , , , , " , " , " , , , ------~I---+---- ~, '. ... ./ '-'---------' , ------------c:01' v. ifi(l) '" - N sin WI w, (bj -------t----"'--+----- 'm , H GURE 2-11 (a) The magnetization curve of the transformer core. (b) The magnetization current caused by the flux in the transformer core. 82 FIGURE 2-12 The core-loss current in a transformer. f\ 1 ;\ FIGURE 2-13 f\ -, " , , , , , , , , , v , , , , , The total excitation current in a transformer. , , '-, , , , v TRANSFORMERS 83 Notice the following points about. the core-loss current: I. The core-loss current is nonlinear because of the nonline.1.r effects of hysteresis. 2. 1lle fundamental component of the core-loss current is in phase with the volt-age applied to the core. The total no-load current in the core is called the excitation current of the transfonner. It is just the sum of the magnetization current and the core-loss cur-rent in the core: (2- 30) The total excitation current in a typical transfonner core is shown in Figure 2-1 3. 84 ELECIRIC MACHINERY FUNDAMENTALS I, -• + V , N, ""GURE 2-14 A real transformer with a load connected to its secondary. The Current Ratio on a Transformer and the Dot Convention I, • -+ V, N, '"''' Now suppose that a load is connected to the secondary of the transformer. The re-sulting circuit is shown in Figure 2- 14. Notice the dots on the windings of the transformer. As in the ideal transfonner previously described, the dots help deter-mine the polarity of the voltages and currents in the core without having physi-cally to examine its windings. The physical significance of the dot convention is that a current flowing into the doffed end of a winding produces a positive mng-netomotive force '?J', while a current flowing into the undotted end of a winding produces a negative rnagnetomotive force. Therefore, two currents fl owing into the dotted ends of their respective windings produce rnagnetomotive forces that add. If one current flows into a dotted end of a winding and one flows out ofa dot-ted end, then the magnetornotive forces will subtract from each other. In the situation shown in Figure 2- 14, the primary current produces a posi-tive magnetornotive force '?J'p = Npip, and the secondary current produces a neg-ative rnagnetomotive force:lis = - Nsis. Therefore, the net rnagnetomotive force on the core rnust be (2- 31) lllis net magnetornotive force must produce the net flux in the core, so the net magnetornotive force must be equal to (2- 32) TRANSFORMERS 85 ;.Wb -------I-------- ~.A .turns FIGURE 2- 15 The magnetization curve of an ideal transformer. where m is the reluctance of the transfonner core. Because the reluctance of a well-designed transfonner core is very small (nearly zero) until the core is saturated, the relationship between the primary and secondary currents is approximately 2i'Det = Npip - Nsis "'" 0 as long as the core is unsaturated. TIlerefore, I Npip "'" Nsis I (2- 33) (2- 34) (2- 35) It is the fact that the magnetomotive force in the core is nearly zero which gives the dot convention the meaning in Section 2.3. In order for the magnetomotive force to be nearly zero, current must flow into one dotted end and out of the other dotted end. The voltages must be built up in the same way with respect to the dots on each winding in order to drive the currents in the direction required. (TIle po-larity of the voltages can also be determined by Lenz' law if the construction of the transfonner coils is visible.) What assumptions are required to convert a real transformer into the ideal transfonner described previously? 1lley are as follows: I. 1lle core must have no hysteresis or eddy currents. 2. 1lle magnetization curve must have the shape shown in Figure 2- 15. Notice that for an unsaturated core the net magnetomotive force 2i'nel = 0, implying that Npip = Nsis. 3. The leakage flux in the core must be zero, implying that all the flux in the core couples both windings. 4. 1lle resistance of the transfonner windings must be zero. 86 ELECIRIC MACHINERY FUNDAMENTALS While these conditions are never exactly met, well-designed power transformers can come quite close. 2.5 THE EQUIVALENT CIRCUIT OF A TRANSFORMER TIle losses that occur in real transformers have to be accounted for in any accurate model oftransforrner behavior. The major items to be considered in the construc-tion of such a model are I. Copper cPR) losses. Copper losses are the resistive heating losses in the pri-mary and secondary windings of the transformer. They are proportional to the square of the current in the windings. 2. Eddy current losses. Eddy current losses are resistive heating losses in the core of the transformer. They are proportional to the square of the voltage ap-plied to the transformer. 3. Hysteresis losses. Hysteresis losses are associated with the rearrangement of the magnetic domains in the core during each half-cycle, as explained in Chapter 1. They are a complex, nonlinear function of the voltage applied to the transformer. 4. Leakagef1ux. TIle fluxes u. which escape the core and pass through only one of the transformer windings are leakage fluxes. These escaped fluxes produce a self-inductance in the primary and secondary coils, and the effects of this inductance must be accounted for. The Exact Equivalent Circuit of a Real Transformer lt is possible to construct an equivalent circuit that takes into account all the ma-jor imperfections in real transformers. E:1.ch major imperfection is considered in turn, and its effect is included in the transformer model. TIle easiest effect to model is the copper losses. Copper losses are resistive losses in the primary and secondary windings of the transformer core. They are modeled by placing a resistor Rp in the primary circuit of the transformer and a re-sistor Rs in the secondary circuit. As explained in Section 2.4, the leakage flux in the primary windings <PLP produces a voltage u. produces a voltage eLS given by (2- 36b) TRANSFORMERS 87 Since much of the leakage flux path is through air, and since air has a constant re-luctance much higher than the core reluctance, the flux fu is directly proportional to the primary circuit current ip and the flux dt The constants in these equations can be lumped together. Then _ dip eLP(t) -Lp dt (2- 37a) (2- 37b) (2- 38a) (2- 38b) (2- 39a) (2-39b) where Lp = N}1lP is the self-inductance of the primary coil and Ls = NIIlP is the self-inductance of the secondary coil. Therefore, the leakage flux will be modeled by primary and secondary inductors. How can the core excitation effects be modeled? TIle magnetization current im is a current proportional (in the unsaturated region) to the voltage applied to the core and lagging the applied voltage by 900, so it can be modeled by a reactance XM connected across the primary voltage source. TIle core-loss current i H< is a current proportional to the voltage applied to the core that is in phase with the ap-plied voltage, so it can be modeled by a resistance Re connected across the pri-mary voltage source. (Remember that both these currents are really nonlinear, so the inductance XM and the resistance Re are, at best, approximations of the real ex-citation effects.) The resulting equivalent circuit is shown in Figure 2- 16. Notice that the ele-IllCnts forming the excitation branch are placed inside the primary resistance Rp and the primary inductance Lp. lllis is because the voltage actually applied to the core is really equal to the input voltage less the internal voltage drops of the winding. Although Figure 2- 16 is an accurate model ofa transfonner, it is not a very useful one. To analyze practical circuits containing transformers, it is normally necessary to convert the entire circuit to an equivalent circuit at a single voltage level. (Such a conversion was done in Example 2- 1.) Therefore, the equivalent 88 ELECIRIC MACHINERY FUNDAMENTALS I, -+ v, ~ R,~ -""GURE 2-16 The model of a real transformer. I, R, j Xp + V, R, ~ R, . X, " I, ? J ? -+ R, ~ ? -""GURE 2-17 [ • • JX. N, N, d'R, [ " j XM j ,., R, I .X. J e> " [ ,b, Ideal transformer ja2 X, JX, I, -j Xs I, -+ aV, + V, (a) The transformer model referred to its primary voltage level. (b) The transfonner model referred to its secondary voltage level. + v, -circuit must be referred either to its primary side or to its secondary side in problem solutions. Figure 2- 17a is the equivalent circuit of the transfonner re-ferred to its primary side, and Figure 2-17b is the equivalent circuit referred to its secondary side. TRANSFORMERS 89 Approximate Equivalent Circuits of a Transformer The transfonner models shown before are often more complex than necessary in order to get good results in practical engineering applications. One of the princi-pal complaints about them is that the excitation branch of the model adds another node to the circuit being analyzed, making the circuit solution more complex than necessary. The excitation branch has a very small current compared to the load current of the transfonners. In fact, it is so small that under nonnal circumstances it causes a completely negligible voltage drop in Rp and Xp. Because this is true, a simplified equivalent circuit can be produced that works almost as well as the original model. The excitation branch is simply moved to the front of the trans-fonner, and the primary and secondary impedances are left in series with each other. TIlese impedances are just added, creating the approximate equivalent cir-cuits in Figure 2- 18a and b. In some applications, the excitation branch may be neglected entirely with-out causing serious error. In these cases, the equivalent circuit of the transformer reduces to the simple circuits in Figure 2- 18c and d. I, I, " ,I, I, -)X"lP --jXeq. -+ j v, < R, J -,,' I, -R.~ v, jXM + aV, Reqp '" Rp + rrR, xeqp '" xp + rrx, .. , -jXeqp -- ,~---------~, -,,' FIGURE 2-18 + V , , -,b, V , , ,I, -: j R , .x. if )-;;> J I, -V, - ,~---------~, -,d, Approximate transformer models. (a) Referred to the primary side; (b) referred to the secondary side; (c) with no excitation branch. referred to the primary side; (d) with no excitation branch. referred to the secondary side. + V, -90 ELECIRIC MACHINERY FUNDAMENTALS r:0 at me er ip (t) -\J + • • w, + v (t) ""' V vp (t) --Transformer -0-Ammeter --0- Voltmeter ""GURE 2-19 Connection for transformer open-cirwit test. Detennining the Values of Components in the Transformer Model It is possible to experimentally detennine the values of the inductances and resis-tances in the transfonner model. An adequate approximation of these values can be obtained with only two tests, the open-circuit test and the short-circuit test. In the open-circuit test, a transfonner's secondary winding is open-circuited, and its primary winding is connected to a full-rated line voltage. Look at the equivalent circuit in Figure 2- 17. Under the conditions described, all the in-put current must be fl owing through the excitation branch of the transfonner. The series elements Rp and Xp are too small in comparison to Rcand XM to cause a sig-nificant voltage drop, so essentially all the input voltage is dropped across the ex-citation branch. TIle open-circuit test connections are shown in Figure 2- 19. Full line volt-age is applied to the primary of the transfonner, and the input voltage, input cur-rent, and input power to the transfonner are measured. From this information, it is possible to detennine the power factor of the input current and therefore both the mngnitude and the angle of the excitation impedance. TIle easiest way to calculate the values of Rc and XM is to look first at the admittance of the excitation branch. TIle conductance of the core-loss resistor is given by 1 GC=R c and the susceptance of the magnetizing inductor is given by 1 BM = -XM (2- 40) (2- 41) Since these two elements are in parallel, their admittances add, and the total exci-tation admittance is r:0 a meter ip (t) -\J + • • w" + v (t) '" V vp (t) --Transformer FIGURE 2-10 Connection for transformer shon-circuit test. Y E = Gc - JBM ~ I _j ' Rc XM TRANSFORMERS 91 i, (t) -(2- 42) (2- 43) The magnitude of the excitation admittance (referred to the primary circuit) can be found from the open-circuit test voltage and current: IYEI ~ ~oe DC (2- 44) The angle of the admittance can be found from a knowledge of the circuit power factor. 1lle open-circuit power factor (PF) is given by and the power-factor angle () is given by Poe () = cos- 1 ,,-'7~ YclC10c (2- 45) (2- 46) The power factor is always lagging for a real transfonner, so the angle of the current always lags the angle of the voltage by () degrees. 1llCrefore, the admittance YE is I YE = V OC L-() oe loe ~ -- L-cos- 1 PF Voe (2- 47) By comparing Equations (2-43) and (2-47), it is possible to determine the values of Rc and XM directly from the open-circuit test data. In the shott-circuit test, the secondary tenninals of the transformer are short-circuited, and the primary tenninals are connected to a fairly low-voltage source, as shown in Figure 2- 20. The input voltage is adjusted until the current in the short-circuited windings is equal to its rated value. (Be sure to keep the primary voltage at a saJe level. It would not be a good idea to burn out the transformer's windings while trying to test it.) 1lle input voltage, current, and power are again measured. 92 ELECIRIC MACHINERY FUNDAMENTALS Since the input voltage is so low during the short-circuit test, negligible cur-rent flows through the excitation branch. If the excitation current is ignored, then all the voltage drop in the transformer can be attributed to the series elements in the circuit. The magnitude of the series impedances referred to the primary side of the transformer is TIle power factor of the current is given by P,c PF = cos (J = u-''i-VscIsc (2- 48) (2- 49) and is lagging. The current angle is thus negative, and the overall impedance an-gle (J is positive: (2- 50) TIlerefore, VscLO° = Vsc L (J 0 ZSE = lsc L (J 0 lsc (2- 51) TIle series impedance ZSE is equal to ZSE = Req + jXeq = (Rp + a2RS) + j(Xp + a2Xs) (2- 52) It is possible to detennine the total series impedance referred to the primary side by using this technique, but there is no easy way to split the series impedance into primary and secondary components. Fortunately, such separation is not nec-essary to solve nonnal problems. TIlese same tests may also be perfonned on the secondary side of the trans-fonner if it is more convenient to do so because of voltage levels or other reasons. If the tests are performed on the secondary side, the results will naturally yield the equivalent circuit impedances referred to the secondary side of the transfonner in-stead of to the primary side. EXllmple 2-2. The equivalent circuit impedances of a 20-kVA, 800CV240-V, 6O-Hz transformer are to be determined. The open-circuit test and the short-circuit test were perfonned on the primary side of the transfonner, and the following data were taken: Open-circuit tcst (on prinmry) Voc = 8000 V loc = O.214A Voc = 400W Short-circuit tcst (on prinmry) Vsc = 489V Isc = 2.5 A Psc = 240 W TRANSFORMERS 93 Find the impedances of the approximate equivalent circuit referred to the primary side, and sketch that circuit. Solution The power factor during the open,circuit test is Poc PF = cos (J = oi-~ V oc loc 400 W = cos (J = (8000 V)(0.214 A) = 0.234 lagging The excitation admittance is given by = 0.214 A L-- , 0 23' 8(x)() V cos . = 0.cX)OO268 L -76.5° n = 0.0000063 - j O.OOOO261 = i -j i C M Therefore, 1 Rc = 0.0000Cl63 = 159 kO 1 XM = 0.000026 1 = 38.4 k!l The power factor during the short-circuit test is P ", PF = cos (J = oi--~ Vsclsc = cos (J = (489~~(~5 A ) = 0.196 lagging The series impedance is given by V ZsE = .....K L -cos- l PF I", = i~; X L78.7° = 195.6 L78.7° = 38.4 + j l92 0 Therefore, the equivalent resistance and reactance are Req = 38.4 0 Xeq= 192 0 The resulting simplified equivalent circuit is shown in Figure 2- 21. (2- 45) (2- 47) (2- 49) 94 ELECIRIC MACHINERY FUNDAMENTALS jXeq --+ I ;8.40 + j1920 ' ... I I" v ~ R, jX .. , j38.4 kD. 159kD. ,v, \ I -""GURE 2-21 The equivalent cin:uit of Example 2- 2. 2.6 THE PER-UNIT SYSTEM OF MEASUREMENTS As the relatively simple Example 2- 1 showed, solving circuits containing trans-fonners can be quite a tedious operation because of the need 1 0 refer all the dif-ferent voltage levels on differenl sides of the transfonners in the system to a com-mon level. Only after this step has been taken can the system be solved for its voltages and currents. TIlere is another approach 1 0 solving circuits containing transfonners which eliminates the need for explicit voltage-level conversions at every transformer in the system. Instead, the required conversions are handled automatically by the method itself, without ever requiring the user to worry about impedance transfor-mations. Because such impedance transfonnations can be avoided, circuits con-taining many transfonners can be solved easily with less chance of error. This method of calculation is known as the per-unit (pu) system of measurements. There is yet another advantage to the per-unit system that is quite significant for electric machinery and transfonners. As the size of a machine or transfonner varies, its internal impedances vary widely. Thus, a primary circuit reactance of O. I n might be an atrociously high number for one transfonner and a ridiculously low number for another-it all depends on the device's voltage and power ratings. However, it turns out that in a per-unit system related to the device's ratings, ma-chine and transformer impednnces fall within fairly nanvw ranges for each type and construction of device. This fact can serve as a usefu I check in problem solutions. In the per-unit system, the voltages, currents, powers, impedances, and other electrical quantities are not measured in their usual SI units (volts, amperes, watts, ohms, etc.). Instead, each electrical quantity is measured as a decimal fraction of some base level. Any quantity can be expressed on a per-unit basis by the equation Quantit r unit = Actual value. y pe base value of quantity (2- 53) where "actual value" is a value in volts, amperes, ohms, etc. TRANSFORMERS 95 It is customary to select two base quantities to define a given per-unit sys-tem. The ones usually selected are voltage and power (or apparent power). Once these base quantities have been selected, all the other base values are related to them by the usual electrical laws. In a single-phase system, these relationships are (2- 54) (2- 55) (2- 56) and (2- 57) Once the base values of S (or P) and V have been selected, all other base values can be computed easily from Equations (2- 54) to (2- 57). In a power system, a base apparent power and voltage are selected at a spe-cific point in the system. A transfonner has no effect on the base apparent power of the system, since the apparent power into a transfonner equals the apparent power out of the transfonner [Equation (2-11 )] . On the other hand, voltage changes when it goes through a transformer, so the value of VI>a .. changes at every transformer in the system according to its turns ratio. Because the base quantities change in passing through a transfonner, the process of referring quantities to a common voltage level is automatically taken care of during per-unit conversion. EXllmple 2-3. A simple power system is shown in Figure 2- 22. This system con-tains a 480-V generator connected to an ideal I: 10 step-up transfonner, a transmission line, an ideal 20: I step-down transformer, and a load. The impedance of the transmission line is 20 + j60 n, and the impedance of the load is IOL30on. The base values for this system are chosen to be 480 V and 10 kVA at the generator. (a) Find the base voltage, current, impedance, and apparent power at every point in the power system. (b) Convert this system to its per-unit equivalent circuit. (c) Find the power supplied to the load in this system. (d) Find the power lost in the transmission line. YG 480LOo y '-''-' Region I RegIOn 2 FIGURE 2-22 The power system of Example 2- 3. RegIOn 3 96 ELECIRIC MACHINERY FUNDAMENTALS Solutio" (a) In the generator region. Vbo .. = 480 V and 5_ = 10 kVA, so s~. lbase I = -~ -- = ~, 10,000 VA = 2083A 480 V . V base I 480 V II z."... I = l ba .. I = 20.83 A = 23.04 The turns ratio oftransfonner Tl is a = 1110 = 0.1, so the base voltage in the transmission line region is v. = V base1 = 480V = 4800 V bo. •• 2 a 0.1 The other base quantities are Sbasel = 10 kVA k m~ = 10,000 VA = 2083A .,....~ 4800 V . 4800 V Z basel = 2.083 A = 2304 n The turns ratio of transfonner Tl is a = 2011 = 20, so the base voltage in the load region is _ ~ = 4800 V = 240 V V base ) -a 20 The other base quantities are 5110..,)= IOkVA lbo.se) = lOi~\; A = 41.67 A 240 V Z~ 1 = 41.67 A = 5.76 n (b) To convert a power system to a per-lUlit system, each component must be di-vided by its base value in its region of the system. The generator s per-lUlit volt-age is its actual value divided by its base value: _ 480LOoV _ ° Vo."" -480V -1.0LO pu The transmission line s per-unit impedance is its actual value divided by its base value: 20+j60n . ~iDe."" = 2304 n = 0.0087 + )0.0260 pu The loads per-lUlit impedance is also given by actual value divided by base value: The per-unit equivalent circuit of the power system is shown in Figure 2- 23. TRANSFORMERS 97 ,~ IHme 0.0087 pu jO.0260 pu I ,~ --I -1 I I I + I "G ",lLOO ~ '" 1.736 L 30° per unit -I IG . I""' '" llino• "" '" IJo.d. I""' '" I"" Jo'IGURE 2-23 The per-unit equivalent circuit for Example 2- 3. (c) The current flowing in this per-unit power system is V I = ~ I""' z.."."" I LO° = " CO '".OOmoS' 7 " + C-J "'·OO.O" 26 fi O;C ) '; + CCC"".7'36'L 7i0 30 WO ") I I I I 1 LO° = " CO ,".OOmoS' 7 ~+ -J "'·OO.O~ 2 6f,O~ ) ~+~ C "'<.5"m,"+' j"O."S6as" ) 1.512 + jO.894 1.757 L30.6° = 0.569 L -30.6° pu Therefore, the per-unit power of the load is p load.1""' = PI""'Rpu = (0.569)2(1.503) = 0.487 and the actual power supplied to the load is PIo.! = flo.l.I""'Sbo>e = (0.487)( 10,000 VA) = 4870W (d) The per-unit power lost in the transmission line is pu .... 1""' = PI""'R1ine.pu = (0.569)2(0'(XJ87) = 0'(xl282 and the actual power lost in the transmission line is fline = fli .... ""St-. = (0.00282)(10,000 VA) = 28.2 W When only one device (transfonner or motor) is being analyzed, its own rat-ings are usually used as the base for the per-unit system. If a per-unit system based on the transfonner's own ratings is used, a power or distribution transformer's characteristics will not vary much over a wide range of voltage and power ratings. For example, the series resistance of a transfonner is usually about 0.01 per unit, 98 ELECIRIC MACHINERY FUNDAMENTALS .. 2 1' ,.J (a) ,b, ""GURE 2-14 (a) A typical 13.2---kY to 1201240-Y distribution transformer. (Courtesy ofGeneml Electric Company.) (b) A cutaway view of the distribution transformer showing the shell-form transfonner inside it. (Courtesy ofGeneml Electric Company. ) and the series reactance is usually between 0.02 and 0.10 per unit. In general, the larger the transformer, the smaller the series impedances. 1lle magnetizing reac-tance is usually between about 10 and 40 per unit, while the core-loss resistance is usually between about 50 and 200 per unit. Because per-unit values provide a con-venient and meaningful way to compare transformer characteristics when they are of different sizes, transformer impedances are normally given in per-unit or as a percentage on the transformer's nameplate (see Figure 2- 46, later in this chapter). 1lle same idea applies to synchronous and induction machines as well: Their per-unit impedances fall within relatively narrow ranges over quite large size ranges. If more than one machine and one transformer are included in a single power system, the system base voltage and power may be chosen arbitrarily, but the entire system must have the same base. One common procedure is to choose the system base quantities to be equal to the base of the largest component in the system. Per-unit values given to another base can be converted to the new base by converting them to their actual values (volts, amperes, ohms, etc.) as an in-between step. Alternati vely, they can be converted directly by the equations TRANSFORMERS 99 Ip,po R., jXoq I" po --+ j + , 0.012 fJ·06 I~H 1 I'm V R, JXm V"po ,,~ 49.7 jl2 FIGURE 2-15 The per-unit equivalent circuit of Example 2-4. Sba.se t (P, Q, S)poon base 2 = (P, Q, S)poon base ]-S--"=, v.: v.: V base ] P" on base 2 = po on base ] Vbase2 (R, X, Z)P" 00 base 2 = (Vbase t?(Sbase 2) (R, X, Z)pu on base t(l< )'(S ) base 2 base ] (2- 58) (2- 59) (2-60) Example 2-4. Sketch the approximate per-unit equivalent circuit for the trans-fonner in Example 2- 2. Use the transformer's ratings as the system base. Solutioll The transfonner in Example 2- 2 is rated at 20 kVA, 8()(x)/240 V. The approximate equiva-lent circuit (Figure 2- 21) developed in the example was referred to the high-voltage side of the transfonner, so to convert it to per-unit, the primary circuit base impedance must be fOlUld. On the primary, Therefore, V!>Me I = 80CXl V Sbo>e I = 20,(XXl VA (Vb ... t)l (8()(x) V)2 z.... •• t= S = 20 00Cl VA =3200 0 ~, ' _ 38.4 + jl92 0 _ . ZsE,po -3200 0 - 0.012 + jO.06 pu 159 ill Rc.pu = 3200 0 = 49.7 pu 38.4 kO ZM.pu = 3200 0 = 12 pu The per-unit approximate equivalent circuit, expressed to the transfonner's own base, is shown in Figure 2- 25. 100 ELECTRIC MACHINERY RJNDAMENTALS 2.7 TRANSFORMER VOLTAGE REGULATION AND EFFICIENCY Because a real transformer has series impedances within it, the output voltage of a transfonner varies with the load even if the input voltage remains constant. To conveniently compare transfonners in this respect, it is customary to define a quantity called voltage regulation (VR). Full-load voltage regulation is a quantity that compares the output voltage of the transformer at no load with the output voltage at full load. lt is defined by the equation I VR = VS.nlV: n VS. fl x 100% I (2-6 1) Since at no load, Vs = Vp /a, the voltage regulation can also be expressed as (2-62) I f the transformer equivalent circuit is in the per-unit system, then voltage regula-tion can be expressed as VR = ~ -~ p.P" S.fl.p" ~ X S.fl.pu 100% (2-63) Usually it is a good practice to have as small a voltage regulation as possible. For an ideal transfonner, VR = 0 percent. lt is not always a good idea to have a low-voltage regulation, though-sometimes high-impedance and high-voltage reg-ulation transfonners are deliberately used to reduce the fault currents in a circuit. How can the voltage regulation of a transfonner be detennined? The Transformer Phasor Diagram To detennine the voltage regu lation of a transfonner, it is necessary to understand the voltage drops within it. Consider the simplified transfonner equivalent circuit in Figure 2-1 Sb. TIle effects of the excitation branch on transformer voltage reg-ulation can be ignored, so only the series impedances need be considered. The voltage regulation of a transfonner depends both on the magnitude of these series impedances and on the phase angle of the current fl owing through the transformer. TIle easiest way to detennine the effect of the impedances and the current phase angles on the transformer voltage regulation is to examine a phasor diagram, a sketch of the phasor voltages and currents in the transformer. In all the following phasor diagrams, the phasor voltage V s is assumed to be at an angle of 0°, and all other voltages and currents are compared to that refer-ence. By applying Kirchhoff's voltage law to the equivalent circuit in Figure 2-I Sb, the primary voltage can be found as TRANSFORMERS 101 (2- 64) A transfonner phasor diagram is just a visual representation of this equation. Figure 2- 26 shows a phasor diagram of a transformer operating at a lagging power factor. It is easy to see that Vp la > ~ for lagging loads, so the voltage reg-ulation of a transformer with lagging loads must be greater than zero. A phasor diagram at unity power factor is shown in Figure 2- 27a. Here again, the voltage at the secondary is lower than the voltage at the primary, so VR > O. However, this time the voltage regulation is a smaller number than it was with a lag-ging current. If the secondary current is leading, the secondary voltage can actually be higher than the referred primary voltage. If this happens, the transformer actually has a negative voltage regulation (see Figure 2- 27b). FIGURE 2-26 Phasor diagram of a traruformer operating at a lagging power factor. (a) I, ,b, FIGURE 2-27 v, , v , v , , Phasor diagram of a transformer operating at (a) unity and (b) teading power factor. 102 ELECTRIC MACHINERY RJNDAMENTALS , v, I, , v -" I , I jX"jI, I I -----t--Vp ... V,+Roq l,cos0 +Xoql.Si~O I " ""GURE 2-28 Derivation of the approximate equation for Vpla. Transformer Efficiency Transformers are also compared and judged on their efficiencies. The efficiency of a device is defined by the equation Pout " ~ - X 100% flo Pout 1/ = x 100% ~ut+ ~oss (2-65) (2-66) TIlese equations apply to motors and generators as well as to transfonners. TIle transformer equivalent circuits make efficiency calculations easy. There are three types of losses present in transfonners: I. Copper (PR) losses. These losses are accounted for by the series resistance in the equivalent circuit. 2. Hysteresis losses. These losses were explained in Chapter I and are ac-counted for by resistor Re. 3. Eddy current losses. lllese losses were explained in Chapter I and are ac-counted for by resistor Re. To calculate the efficiency of a transfonner at a given load, just add the losses from each resistor and apply Equation (2-67). Since the output power is given by (2- 7) the efficiency of the transfonner can be expressed by (2-67) TRANSFORMERS 103 Example 2-5. A 15-kVA, 23001230-V transformer is to be tested to detennine its excitation branch components, its series impedances, and its voltage regulation. The fol-lowing test data have been taken from the primary side of the transformer: Open-ciITuit tcsl Voc = 2300 V loc = 0.21 A Poc = SOW Short-circuillesl Vsc = 47V Isc = 6.0A Psc = I60W The data have been taken by using the connections shown in Figures 2- 19 and 2- 20. (a) Find the equivalent circuit of this transformer referred to the high-voltage side. (b) Find the equivalent circuit of this transformer referred to the low-voltage side. (c) Calculate the full-load voltage regulation at O.S lagging power factor, 1.0 power factor, and at O.Sleading power factor. (d) Plot the voltage regulation as load is increased from no load to full load at power factors of O.S lagging, 1.0, and O.S leading. (e) What is the efficiency of the transformer at full load with a power factor of O.S lagging? Solutioll (a) The excitation branch values of the transformer equivalent circuit can be calcu-lated from the open-circuit test data, and the series elements can be calculated from the short-circuit test data. From the open-circuit test data, the open-circuit impedance angle is _ - t Poe 60c -cos V oc:ioc _ - t SOW _ 84" - cos (2300 VXO.21 A) -The excitation admittance is thus = 0.21 A L -S40 2300 V = 9.13 x 1O - ~ L-84°0 = 0.0000095 - jO.OOOO9OS0 The elements of the excitation branch referred to the primary are 1 Rc = 0.0000095 = 105 kO 1 XM = O.()()(X)9()S = II kf! From the short-circuit test data, the short-circuit impedance angle is 1 04 ELECTRIC MACHINERY RJNDA MENTALS V, " , - I Psc sc=cos V. J sc sc _ - I l60 W _ 554" - cos (47 V)(6 A) -. The equi valent series impedance is thus V " ZsE = -[- L ' oc " = ~: L55.4° n = 7.833L55.4° = 4.45 + j6.45 The series elements referred to the primary are Xeq = 6.45 n This equivalent circuit is shown in Figure 2- 29a. (b) To find the equivalent circuit referred to the low-voltage side, it is simply neces-sary to divide the impedance by il. Since a = NpiNs = 10, the resulting values are " R.., jXoq I, " , -" -+ + I 4.450 j6.45 0 IhH j j'm V Rc ~ jXm aV, PJ05 k fl +jll k fl I , ,' "', R"" jXoq, , , --+ j + 0.04450 .fl.0645 fl a l~ + ~ I "'m ~= J0500 V, ~=jIJOO " I ,b , fo'IGURE 2- 29 The lransfer equivatent circuit for Example 2- 5 referred 10 (a) its primary side and (b) its secondary side. TRANSFORMERS 105 Rc = 1050 n XM = lIOn Roq = 0.0445 n Xoq = 0.0645 n The resulting equivalent circuit is shown in Figure 2- 29b. (c) The full-load current on the secondary side of this transfonner is _ ~ _ 15,OCXl VA _ IS,med -V. -230 V - 65.2 A S,med To calculate Vpla, use Equation (2-64): V , . a = V s + RoqIs + JXoqIs At PF = 0.8 lagging, current Is = 65.2 L - 36.9° A. Therefore, (2-64) ~ = 230L O ° V + (0.0445 fl)(65.2L-36.9° A) + j(0.0645 OX65.2L -36.9° A) = 230 LO° V + 2.90L -36.9° V + 4.21 L53 .10 V = 230 + 2.32 - j l.74 + 2.52 + j3.36 = 234.84 + j l.62 = 234.85 L0.40° V The resulting voltage regulation is Vp/a - VS() VR = x 100% V s.() = 234. 85; 0~ 230 V x 100% = 2.1% At PF = 1.0, current Is = 65.2 L 0° A. Therefore, (2-62) ':; = 230 LO° V + (0.0445 OX65.2 LO° A) + j(0.0645 ll)(65.2 LO° A) = 230 LOo V + 2.90L O oV + 4.2 I L90o V = 230 + 2.90 + j4.2 1 = 232.9 + j4.2 l = 232.94 L 1.04° V The resulting voltage regulation is VR = 232.9ijo ~ 230 V x 100% = 1.28% At PF = 0.8 leading, current Is = 65.2 L36.9° A. Therefore, V : = 230 LO° V + (0.0445 n X65.2 ":::::36.9° A) + j(0.0645 0 )(65.2 L36.9° A) = 230 LO° V + 2.90 L36.9° V + 4.21 L 126.9° V = 230 + 2.32 + j l.74 - 2.52 + j3.36 = 229.80 + j5.10 = 229.85 L 1.27° V The resulting voltage regulation is 106 ELECTRIC MACHINERY RJNDA MENTALS v -!- '" 234.9 L 0.4° Y V,,,,230LOoy jXoql, '" 4.21 L 53.1° Y Roql, '" 2.9 L - 36.9° Y I,'" 65.2 L - 36.9° A V -.l'",2329L 104°Y , . . 1.1 '~ 6~5~ .2~L ~ O' ~A ~==:::::::==:::=~ 2:30 ~ L~o : ,~ v~::JJ ) 4.21 L 90' V 2.9LOoy ,hI I, '" 65.2 L 36.9° A V -.l'",2298L 127°Y " . . L /==:::=========:}I L 1269' V ~ a.:L36.90 Y 230LOoy "I fo'IGURE 2- 30 Transformer phasor diagrams for Example 2- 5. VR = 229 .852~0 ~ 230 V x 100% = -0.062% Each of these three phasor diagrams is shown in Figure 2- 30. (d) The best way to plot the voltage regulation as a function of load is to repeat the calculations in part c for many different loads using MATLAB. A program to do this is shown below. % M-fil e : tra n s_vr.m % M-fil e t o ca l cula t e a nd p l ot the voltage r egul ation % o f a tra n s f o rme r as a fun cti on of l oad f or power % f act o r s o f 0 . 8 l agging, 1. 0, a nd 0 . 8 l eading . VS = 230; % Secondary voltage (V) a mps = 0: 6.52: 65 . 2; % CUrrent values (A) TRANSFORMERS 107 'oq xoq 0.0445; 0.0645; !l; Equ i va l ent R (ohms) !l; Equ i va l ent X (ohms) !l; Ca l c u l ate the c urrent va l u es for the three !l; power f actors. The fi r s t row of I contains !l; the l agg i ng c urrent s, the second row contains !l; the uni t y c urrent s, and the third row contains , "0 l eadi ng c urrent s. I (1 , : ) ~P' • ( 0.8 j0.6) ; I (2, : ) ~P' • ( 1. 0 I , I (3, : ) amps • ( 0.8 + j 0.6) ; !l; Ca l c u l ate VP/ a. VPa = VS + Req. I + j. Xeq. I ; !l; Ca l c u l ate vo l tage regul at i on VR = (abs (VPa ) -VS) . / VS . 100; !l; Pl ot the vo l tage regu l at i on p l ot (amps, VR( l ,:), 'b- ' ) ; h o l d on; p l ot (amps, VR(2,:), 'k- ' ) ; p l ot (amps, VR (3, : ), 'r- .' ) ; • • • t i t l e ('Vo l tage Regu l at i on Ver s u s Load' ) ; x l abe l ('Load (A) ' ) ; y l abe l ('Voltage Regu l at i o n (%) ' ) ; Lagg i ng Unity Leadi ng l egend (' O.8 PF l agg i ng' , 'l .O PF' ,'0 . 8 PF l eadi ng' ) ; h o l d o ff ; The plot produced by this program is shown in Figure 2- 31. (e) To find the efficiency of the transformer. first calculate its losses. The copper losses are Peu = (ls)2Req = (65.2 A)2(0.0445ll) = 189 W The core losses are given by (Vp/a)2 (234.85 V)l P core = Rc = 1050 n = 52.5 W The output power of the transformer at this power factor is = (230 VX65.2 A) cos 36.9° = 12.()(X) W Therefore. the efficiency of the transformer at this condition is Vslscos () 7f = x 100% Peu + P.: .... + Vslscos () = 189W + 52.5 W + 12.()(X) W x = 98.03% 100% (2- 68) 108 ELECTRIC MACHINERY RJNDAMENTALS Voltage regulation versus load 2.5~~Fln I I 0.8 PF lagging ---- 1.0 PF ..- 0.8 PF leading 2 / ......... :---o .-.. . .••. _ . . __ . .. __ .. _ •. . .-'-. . --- .- . --O.50 !--..J IOc---2 eO~-C 3 ± O--C 4"O---!50~--60 ±---! 70 Load (A) fo'IGURE 2- 31 Plot of voltage regulation versus load for the transformer of Example 2--5. 2.8 TRANSFORMER TAPS AND VOLTAGE REGULATION In previous sections of this chapter, transformers were described by their turns ra-tios or by their primary-to-secondary-voltage ratios. Throughout those sections, the turns ratio of a given transformer was treated as though it were completely fixed. In almost all real distribution transfonners. this is not quite true. Distribution trans-fonners have a series of taps in the windings to pennit small changes in the turns ratio ortile transfonner after it has left the factory. A typical installation might have four taps in addition to the nominal setting with spacings of 2.5 percent of full-load voltage between them. Such an arrangement provides for adjustments up to 5 per-cent above or below the nominal voltage rating of the transfonner. Example 2-6. A 500-kVA, 13,200/480-V distribution transfonner has four 2.5 percent taps on its primary winding. What are the voltage ratios of this transfonner at each tap setting? Solutioll The five possible voltage ratings of this transfonner are +5.0% tap +2.5% tap Nominal rating -2.5% tap -5.0% tap 13,8601480 V 13,5301480 V 13,2001480 V 12,8701480 V 12,540/480 V TRANSFORMERS 109 The taps on a transfonner permit the transfonner to be adjusted in the field to accommodate variations in local voltages. However, these taps nonnally can-not be changed while power is being applied to the transformer. They must be set once and left alone. Sometimes a transfonner is used on a power line whose voltage varies widely with the load. Such voltage variations might be due to a high line imped-ance between the generators on the power system and that particular load (perhaps it is located far out in the country). Nonnal loads need to be supplied an essen-tially constant voltage. How can a power company supply a controlled voltage through high-impedance lines to loads which are constantly changing? One solution to this problem is to use a special transformer called a tap changing under load (TCUL) transformer or voltage regulator. Basically, a TCUL transformer is a transformer with the ability to change taps while power is con-nected to it. A voltage regulator is a TCUL transfonner with built-in voltage sens-ing circuitry that automatically changes taps to keep the system voltage constant. Such special transformers are very common in modem power systems. 2.9 THE AUTOTRANSFORMER On some occasions it is desirable to change voltage levels by only a small runount. For example, it may be necessary to increase a voltage from 110 to 120 V or from 13.2 to 13.8 kV These small rises may be made necessary by voltage drops that occur in power systems a long way from the generators. In such circumstances, it is wasteful and excessively expensive to wind a transfonner with two full wind-ings, each rated at about the same voltage. A special-purpose transformer, called an autotransformer. is used instead. A diagram of a step-up autotransfonner is shown in Figure 2- 32. In Figure 2- 32a, the two coils of the transformer are shown in the conventional manner. In Figure 2- 32b, the first winding is shown connected in an additive manner to the second winding. Now, the relationship between the voltage on the first winding and the voltage on the second winding is given by the turns ratio of the trans-former. However, the voltage at the output of the whole transformer is the sum of the voltage on the first winding and the voltage on the second winding. TIle first winding here is called the common winding, because its voltage appears on both sides of the transfonner. The smaller winding is called the series winding, because it is connected in series with the common winding. A diagram of a step-down autotransformer is shown in Figure 2- 33. Here the voltage at the input is the sum of the voltages on the series winding and the common winding, while the voltage at the output is just the voltage on the com-mon winding. Because the transformer coils are physically connected, a different tenni-nology is used for the autotransformer than for other types of transformers. The voltage on the common coil is called the common voltage Vc, and the current in that coil is called the common current [c. The voltage on the series coil is called the series voltage VSE, and the current in that coil is called the series current IsE. 110 ELECTRIC MACHINERY RJNDAMENTALS (a) 'bJ ""GURE 2-32 A transfomler with its windings (a) connected in the conventional manner and (b) reconnected as an autotransformer. 'H -IH", ISE • IL"' ISE+ Ic I~ I N~ I, VH -• ) V, Ic\ N, ""GURE 2-.33 A step-down autotransformer connection. TIle voltage and current on the low-voltage side of the transfonner are called VL and IL , respectively, while the corresponding quantities on the high-voltage side of the transformer are called VH and IH . The primary side of the autotransfonner (the side with power into it) can be either the high-voltage side or the low-voltage side, depending on whether the autotransfonner is acting as a step-down or a step-up transfonner. From Figure 2- 32b the voltages and currents in the coils are re-lated by the equations Vc _ Nc VSE -NSE Nc Ic = NSE IsE (2-68) (2-69) TRANSFORMERS III The voltages in the coils are related to the voltages at the tenninals by the equations YL = Ye YH = Ye + VSE (2- 70) (2- 71) and the currents in the coils are related to the currents at the terminals by the equations IL= l e+ lsE IH = I SE Voltage and Current Relationships in an Autotransformer (2- 72) (2- 73) What is the voltage relationship between the two sides of an autotransfonner? It is quite easy to determine the relationship between Y Hand Vv The voltage on the high side of the autotransfonner is given by (2- 71 ) (2- 74) Finally, noting that Y L = Ve, we get (2- 75) (2- 76) The current relationship between the two sides of the transformer can be found by noting that IL = Ie + ISE From Equation (2--69), Ie = (NSEINc) lsE' so N>E IL = N ISE + ISE C Finally, noting that Iy = ISE' we find (2- 72) (2- 77) 11 2 ELECTRIC MACHINERY RJNDAMENTALS NSE + Nc -N IH C "' I, NSE + Nc IH -Nc The Apparent Power Rating Advantage of Autotransformers (2- 78) (2- 79) It is interesting to note that not all the power traveling from the primary to the sec-ondary in the autotransformer goes through the windings. As a result, if a con-ventional transformer is reconnected as an autotransfonner, it can handle much more power than it was originally rated for. To understand this idea, refer again to Figure 2- 32b. Notice that the input apparent power to the autotransformer is given by Sin = VLIL and the output apparent power is given by (2-80) (2-81) It is easy to show, by using the voltage and current equations [Equations (2- 76) and (2- 79)], that the input apparent power is again equal to the output apparent power: (2-82) where SIO is defined to be the input and output apparent powers of the transformer. However, the apparent power in the transfonner windings is (2-83) 1lle relationship between the power going into the primary (and out the sec-ondary) of the transformer and the power in the transfonner's actual windings can be found as follows: Sw = Vcl c Using Equation (2- 79), we get = VL(JL -IH) = VLIL -~IH -s -10NsE + Nc (2-84) (2-85) TRANSFORMERS 113 Therefore, the ratio of the apparent power in the primary and secondary of the autotransfonner to the app.:1.rent power actually traveling through its windings is (2-86) Equation (2--86) describes the apparent power rating advantage of an auto-transformer over a conventional transfonner. Here 5[0 is the apparent power enter-ing the primary and leaving the secondary of the transformer, while Sw is the ap-parent power actually traveling through the transfonner's windings (the rest passes from primary to secondary without being coupled through the transfonner 's wind-ings). Note that the smaller the series winding, the greater the advantage. For example, a SOOO-kVA autotransfonner connecting a 11 O-kV system to a 138-kV system would have an NelNsE turns ratio of 110:28. Such an autotrans-fonner would actually have windings rated at N" Sw= SION + N. " c 28 (2-85) The autotransfonner would have windings rated at on ly about lOIS kVA, while a conventional transformer doing the same job would need windings rated at S(x)() kVA. The autotransfonner could be S times smaller than the conventional trans-fonner and also would be much less expensive. For this reason, it is very advanta-geous to build transfonners between two nearly equal voltages as autotransfonners. The following example illustrates autotransformer analysis and the rating advantage of autotransformers. Example 2-7. A 100-VA 120/12-V transformer is to be connected so as to form a step-up autotransfonner (see Figure 2- 34). A primary voltage of 120 V is applied to the transformer. (a) What is the secondary voltage of the transfonner? (b) What is its maximum voltampere rating in this mode of operation? (e) Calculate the rating advantage of this autotransfonner connection over the trans-fonner 's rating in conventional 120112- V operation. Solutioll To accomplish a step-up transfonnation with a 120-V primary, the ratio of the HUllS on the common winding Ne to the turns on the series winding NSE in this transfonner must be 120:12 (or 10:1). (a) This transfonner is being used as a step-up transformer. The secondary voltage is VH, and from Equation (2- 75), NSE + Ne VH = Ne VL = 12 + 120 120 V = 132 V 120 (2- 75) 114 ELECTRIC MACHINERY RJNDAMENTALS -,-------~+ • -+~------+ Nd - 120) V'= 1 20LO OV( • ""GURE 2-34 The autotransformer of Example 2--7. (b) The maximwn voltampere rating in either winding of this transfonner is 100 VA. How much input or output apparent power can this provide? To fmd out, examine the series winding. The voltage VSE on the winding is 12 V, and the voltampere rat-ing of the winding is 100 VA. Therefore, the maximum series winding current is 100 VA = 8.33A 12V Since ISE is equal to the secondary current Is (or IH ) and since the secondary voltage Vs = V H = 132 V, the secondary apparent power is SOUl = V s Is = VHIH = (132 V)(S.33 A) = 1100 VA = Sin (e) The rating advantage can be calculated from part (b) or separately from Equa-tion (2-86). From part b, From Equation (2-86), 1100 VA = II 100 VA S[o N SE + Ne = Sw N SE = 12+120 = 132 = 11 12 12 By either equation, the apparent power rating is increased by a factor of II. (2-86) It is not nonnaJly possible to just reconnect an ordinary transformer as an autotransfonner and use it in the manner of Example 2- 7, because the insulation on the low-voltage side of the ordinary transfonner may not be strong enough to withstand the full output voltage of the autotransfonner connection. In transform-TRANSFORMERS 115 FIGURE 2-35 (a) A variable-voltage autotransformer. (b) Cutaway view of the autotransformer. (Courtesy of Superior Electric Company.) ers built specifically as autotransfonners, the insulation on the smaller coil (the se-ries winding) is made just as strong as the insulation on the larger coil. It is common practice in power systems to use autotransfonners whenever two voltages fairly close to each other in level need to be transfonned, because the closer the two voltages are, the greater the autotransformer power advantage be-comes. 1lley are also used as variable transfonners, where the low-voltage tap moves up and down the winding. This is a very convenient way to get a variable ac voltage. Such a variable autotransfonner is shown in Figure 2- 35. The principal disadvantage of autotransformers is that, unlike ordinary transformers, there is a direct physical connection between the primary and the secondary circuits, so the electrical isolation of the two sides is lost. If a particu-lar application does not require electrical isolation, then the autotransfonner is a convenient and inexpensive way to tie nearly equal voltages together. The Internal Impedance of an Autotransformer Autotransformers have one additional disadvantage compared to conventional transformers. It turns out that, compared to a given transformer connected in the conventional manner, the effective per-unit impedance of an autotransformer is smaller by a factor equal to the reciprocal of the power advantage of the auto-transfonner connection. The proof of this statement is left as a problem at the end of the chapter. The reduced internal impedance of an autotransfonner compared to a con-ventional two-winding transformer can be a serious problem in some applications where the series impedance is needed to limit current flows during power system faults (short circuits). The effect of the smaller internal impedance provided by an autotransformer must be taken into account in practical applications before auto-transfonners are selected. 116 ELECTRIC MACHINERY RJNDAMENTALS Example 2-8. A transfonner is rated at 1000 kVA, 1 211.2 kY, 60 Hz when it is op-erated as a conventional two-winding transformer. Under these conditions, its series resis-tance and reactance are given as I and 8 percent per unit, respectively. This transfonner is to be used as a 13.2I12-kV step-down autotransformer in a power distribution system. In the autotransformer connection, (a) what is the transformer's rating when used in this man-ner and (b) what is the transformer's series impedance in per-unit? Solutio" (a) The NclNsE turns ratio must be 12:1.2 or 10:1. The voltage rating of this trans-former will be 13.2112 kV, and the apparent power (voltampere) rating will be NSF. + Nc 5[0 = NSF. Sw = 1+ IO I()(X)kVA = IIOOOkVA 1 ' (b) The transfonner's impedance in a per-unit system when cOIUlected in the con-ventional manner is Zoq = 0.01 + jO.08 pu separate windings The apparent power advantage of this autotransfonner is II, so the per-unit im-pedance of the autotransfonner connected as described is 0.01 + jO.08 Zoq= II = 0.00091 + jOJ'lJ727 pu autotransformer 2,10 THREE-PHASE TRANSFORMERS Almost all the major power generation and distribution systems in the world today are three-phase ac systems. Since three-phase systems play such an important role in modern life, it is necessary to understand how transformers are used in them. Transformers for three-phase circuits can be constructed in one of two ways. One approach is simply to take three single-phase transformers and connect them in a three-phase bank. An alternative approach is to make a three-phase transfonner consisting of three sets of windings wrapped on a common core. TIlese two possible types of transfonner construction are shown in Figures 2- 36 and 2- 37. The construction of a single three-phase transformer is the preferred practice today, since it is lighter, smaller, cheaper, and slightly more efficient. The older construction approach was to use three separate transfonners. That approach had the advantage that each unit in the bank could be replaced individually in the event of trouble, but that does not outweigh the ad vantages of a combined three-phase unit for most applications. However, there are still a great many installa-tions consisting of three single-phase units in service. A discussion of three-phase circuits is included in Appendix A. Sorne read-ers may wish to refer to it before studying the following material. TRANSFORMERS 117 N" N" N~ ~ f---< ~ f No; N ~ f---< " FIGURE 2-36 A three-phase transformer bank composed of independent transformers. N " N~ No; ~ ~ ~ N " N" N" ~ ~ ~ FIGURE 2-37 A three-phase transformer wound on a single three-legged COTe. 118 ELECTRIC MACHINERY RJNDAMENTALS Three-Phase Transformer Connections A three-phase transfonner consists of three transformers, either separate or com-bined on one core. The primaries and secondaries of any three-phase transfonner can be independently connected in either a wye (Y) or a delta (d ). This gives a to-tal of four possible connections for a three-phase transfonner bank: I. Wye-wye (Y-Y) 2. Wye-delta (Y -d) 3. Oelta-wye (d-Y) 4. Oelta-delta (d--&) 1llese connections are shown in Figure 2- 38. 1lle key to analyzing any three-phase transformer bank is to look at a single transfonner in the bank. Any single transfonner in the bank behaves exactly like the single-phase transformers already studied. 1lle impedance, voltage regula-tion, efficiency, and similar caJcu lations for three-phase transfonners are done on a per-phase basis, using exactly the same techniques already developed for single-phase transfonners. 1lle advantages and disadvantages of each type of three-phase transfonner connection are discussed below. WYE-WYE CONNECTION. TIle Y-Y connection of three-phase transformers is shown in Figure 2- 38a. In a Y-Y connection, the primary voltage on each phase of the transformer is given by V4>P = V LP / \G. The primary-phase voltage is re-lated to the secondary-phase voltage by the turns ratio of the transformer. The phase voltage on the secondary is then related to the line voltage on the secondary by Vu; = \GV4>S.1llerefore, overall the voltage ratio on the transformer is y - y (2-87) 1lle Y-Y connection has two very serious problems: I. If loads on the transfonner circuit are unbalanced, then the voltages on the phases of the transfonner can become severely unbalanced. 2. Third-harmonic voltages can be large. If a three-phase set of voltages is applied to a Y - Y transfonner, the voltages in any phase wi ll be 120 0 apart from the voltages in any other phase. However, the third-hannonic components of each of the three phases will be in phase with each other, since there are three cycles in the third hannonic for each cycle of the fun-damental frequency. There are always some third-harmonic components in a transfonner because of the nonlinearity of the core, and these components add up. TRANSFORMERS 119 " " • • N" N" b + + b' I v .:( • Nn • )+ v " N" --, -" • • Nn N" " " (.j FIGURE 2-38 Three-phase transfonner connections and wiring diagrams: (a) Y - V: (b) y-~: (e) ~ Y; (d) 6.~. The result is a very large third-harmonic component of voltage on top of the 50-ar 6O-Hz fundamental voltage. This third-harmonic voltage can be larger than the fundamental voltage itself. Both the unbalance problem and the third-harmonic problem can be solved using one of two techniques: I. Solidly ground the neutrals of the transfonners, especially the primary wind-ing's neutral. nlis connection permits the additive third-hannonic components to cause a current flow in the neutral instead of building up large voltages. The neutral also provides a return path for any current imbalances in the load. 2. Add a third (tel1iary) winding connected in 11 to the transfonner bank. Ifa third l1-connected winding is added to the transfonner, then the third-hannonic 120 ELECTRIC MACHINERY RJNDAMENTALS components of voltage in the.1. will add up, causing a circulating current flow within the winding. nlis suppresses the third-hannonic components of voltage in the same manner as grounding the transfonner neutrals. The .1.-connected tertiary windings need not even be brought out of the transformer case, but they often are used to supply lights and auxiliary power within the substation where it is located. The tertiary windings must be large enough to handle the circulating currents, so they are usually made about one-third the power rating of the two main windings. One or the other of these correction techniques must be used any time a Y-Y transfonner is installed. In practice, very few Y-Y transfonners are used, since the same jobs can be done by one of the other types of three-phase transformers. WYE-DELTA CONNECTION. TIle Y --d connection of three-phase transformers is shown in Figure 2- 38b. In this connection, the primary line voltage is related to the primary phase voltage by VLP = V3V4>p, while the secondary line voltage is equal to the secondary phase voltage VLS = V <!>S' The voltage ratio of each phase is V ~ =a V., so the overall relationship betwccn the line voltage on the primary side of the bank and the line voltage on the secondary side of the bank is V LP _ V3Vp.p VL'> -V4>S I~~ = V3a (2-88) TIle Y -.6. connection has no problem with third-hannonic components in its voltages, since they are consumed in a circulating current on the.1. side. nlis con-nection is also more stable with respect to unbalanced loads, since the .1. partially redistributes any imbalance that occurs. TIlis arrangement does have one problem, though. Because of the connec-tion, the secondary voltage is shifted 30" relative to the primary voltage of the transformer.llle fact that a phase shift has occurred can cause problems in paral-leling the secondaries of two transformer banks together. TIle phase angles of transformer secondaries must be equal if they are to be paralleled, which means that attention must be paid to the direction of the 3~ '' phase shift occurring in each transformer bank to be paralleled together. In the United States, it is customary to make the secondary voltage lag the primary voltage by 30°. Although this is the standard, it has not always been ob-served, and older installations must be checked very carefully before a new trans-fonner is paralleled with them, to make sure that their phase angles match. TRANSFORMERS 121 V[J> r " b • • N" ::{ • N n • N n b' N" N" -:A.S V~ • , " " v ., ( N" • "LP • N" ) V ., ,----' b j b' • • N" N n , ~ " • • N" N n ,b , FIGURE 2-38 (b) Y -b. (continued) The connection shown in Figure 2- 38b will cause the secondary voltage to be lagging if the system phase sequence is abc. Ifthe system phase sequence is acb, then the connection shown in Figure 2- 38b will cause the secondary voltage to be leading the primary voltage by 30°. DELTA-WYE CONNECTION. A !:J..- Y connection of three-phase transformers is shown in Figure 2- 38c. In a !:J..- Y connection, the primary line voltage is equal to the primary-phase voltage V LP = Vo/>p, while the secondary voltages are related by VLS = V3V ¢S' TIlerefore, the line-to-line voltage ratio of this transformer con-nection is 122 ELECTRIC MACHINERY RJNDAMENTALS lj: .' + • " V LP [ V" "..-::;0+ N~ N" N" V~ b • Nn N" Nn • • 0 b' " " + + • V ,,( N" --• N" } " + b ~ I • • Nn N~ , -b' • • N" N" (0) ""GURE 2-38 (e) d.- Y (continued) VLP _ V4>P VLS -V3"V¢>S (2-89) TIlis connection has the same advantages and the same phase shift as the Y- .d transformer. TIle connection shown in Figure 2- 38c makes the secondary voltage lag the primary voltage by 30°, as before. TRANSFORMERS 123 N n • +d " + + • )"LS v~[ V ii Nn • N" N n , - b' N" • v ., • b + • c " + + • v ,,[ N" • 1 v " N" d -b _ I I b' • • N" N n , ~ • • Nn N n ,d , FIGURE 2-38 (d) ~6. (concluded) DELTA-DELTA CONNECTION. 1lle .6.---d connection is shown in Figure 2- 38d. In a 11- 11 connection, V LP = Vq.p and V LS = V.jtS, so the relationship between pri-mary and secondary line voltages is (2- 90) This transformer has no phase shift associated with it and no problems with unbalanced loads or hannonics. The Per-Unit System for Three-Phase Transformers The per-unit system of measurements applies just as well to three-phase trans-fonners as to single-phase transformers. TIle single-phase base equations (2- 53) 124 ELECTRIC MACHINERY RJNDAMENTALS to (2- 56) apply to three-phase systems on a per-phase basis. If the total base voltampere value of the transfonner bank is called Sb ... , then the base voltampere value of one of the transfonners SI4>.I>o .. is S"'. SI4>.hase = - 3 -(2- 9) and the base phase current and impedance of the transfonner are (2- 92a) (2- 92b) (2- 93a) (2- 93b) Line quantities on three-phase transformer banks can also be represented in the per-unit system. 1lle relationship between the base line voltage and the base phase voltage of the transformer depends on the connection of windings. If the windings are connected in delta, VL.l>ose = V •• b . .. , while if the windings are con-nected in wye, VL hase = V3"V4>.ba ... 1lle base line current in a three-phase trans-fonner bank is given by (2- 94) 1lle application of the per-unit system to three-phase transformer problems is similar to its application in the single-phase examples already given. Example 2-9. A 50-kVA 13.S0CV20S-V 6.-Y distribution transformer has a resis-tance of I percent and a reactance of 7 percent per lUlit. (a) What is the transfonner's phase impedance referred to the high-voltage side? (b) Calculate this transfonner's voltage regulation at full load and O.S PF lagging, using the calculated high-side impedance. (c) Calculate this transformer's voltage regulation under the same conditions, using the per-unit system. Solutioll (a) The high-voltage side of this transfonner has a base line voltage of 13,800 V and a base apparent power of 50 kVA. Since the primary is 6.-connected, its phase voltage is equal to its line voltage. Therefore, its base impedance is (2-93b) TRANSFORMERS 125 = 3(13,SOOV)2 = II 426ll 50,DOOVA ' The per-unit impedance of the transfonner is Zoq = 0.0 I + jJ.07 pu so the high-side impedance in ohms is Zoq = Zoq.~ = (0.01 + jJ.07 pu)(11,426 ll) = 114.2 + jSOOll (b) To calculate the voltage regulation of a three-phase transfonner bank, determine the voltage regulation of any single transformer in the bank. The voltages on a single transfonner are phase voltages, so V - aV VR = #' V # x 100% " " The rated transformer phase voltage on the primary is 13,SOO V, so the rated phase current on the primary is given by S 14> = 3V • The rated apparent power S = 50 kVA, so _ 50,OOOVA _ 1 4> - 3(13,SOO V) -1.20S A The rated ph~ voltage on the secondary of the transfonner is 20S VI v'1 = 120 V. When referred to the high-voltage side of the transformer, this voltage becomes V~ = aV otS = 13,SOO V. Assume that the transfonner secondary is operating at the rated voltage and current, and find the resulting primary phase voltage: V4>P = aV# + Roq l 4> + jXoql 4> = 13,SOOLO° V + (114.2 llX1.20SL -36.S7° A) + (iSOO llX1.20SL-36.S7° A) = 13,SOO + 13SL -36.S7° + 966.4L53.13° = 13,SOO + 110.4 - jS2.S + 579.S + )773.1 = 14,490 + j690.3 = 14,506 L2.73° V Therefore, VR = V#, ~ aV# x 100% "" = 14,5~3 ~~3,SOO x 100% = 5.1% (c) In the per-unit system, the output voltage is I L 0", and the current is I L - 36.S7°. Therefore, the input voltage is V p = I LO° + (O.OIXI L -36.S7°) + (i0.07XI L-36.S7°) = I + O.OOS - jJ.OO6 + 0.042 + jO.056 = 1.05 + jO.05 = 1.051 L2.73° 126 ELECTRIC MACHINERY RJNDAMENTALS The voltage regulation is VR = 1.05~.~ 1.0 x 100% = 5.1% Of course, the voltage regulation of the transfonner bank is the same whether the calculations are done in actual ohms or in the per-unit system. 2.11 THREE-PHASE TRANSFORMATION USING TWO TRANSFORMERS In addition to the standard three-phase transfonner connections, there are ways to perform three-phase transformation with only two transformers. All techniques that do so involve a reduction in the power-handling capability of the transform-ers, but they may be justified by certain economic situations. Some of the more important two-transfonner connections are I. The open-.6. (or V- V) connection 2. The open-Y -open-.6. connection 3. The Scott-T connection 4. The three-phase T connection E:1.ch of these transfonner connections is described below. The Open-.6. (or V-V) Connection In some situations a full transformer bank may not be used to accomplish three-phase transformation. For example, suppose that a .6.-.6. transformer bank com-posed of separate transformers has a damaged phase that must be removed for re-pair. 1lle resulting situation is shown in Figure 2- 39. If the two remaining secondary voltages are VA = V L 0° and VA = V L 120° V, then the voltage across the gap where the third transfonner used 1.0 be is given by Vc= - VA -VB -- V LO° -V L-120° -- V - (-0.5V - jO.866V) - -0.5 V + jO.866V = V L 120° V 1llis is exactly the same voltage that wou ld be present if the third transformer were still there. Phase C is sometimes called a ghost phase. Thus, the open-delta connection lets a transfonner bank get by with only two transfonners, allowing some power flow to continue even with a damaged phase removed. How much apparent power can the bank supply with one of its three trans-fonners removed? At first, it seems that it could supply two-thirds of its rated TRANSFORMERS 127 , ~----------------~ HGURE 2-39 The open-a or v- v transformer connection. v, ------:--+ v, VA ",VLOO V VB '" V L 120" V '---~ b' apparent power, since two-thirds of the transfonners are still present. Things are not quite that simple, though. To understand what happens when a transfonner is removed, see Figure 2-40. Figure 2-40a shows the transformer bank in nonnal operation connected to a resistive load. I f the rated voltage of one transformer in the bank is Vol> and the rated current is It/» then the maximum power that can be supplied to the load is P = 3V4 ,!4> cos (J The angle between the voltage Vol> and the current 101> in each phase is 0° so the to-tal power supplied by the transformer is P = 3V4>I4> cos (J = 3 V4>I4> (2- 95) The open-delta transfonner is shown in Figure 2-40b. It is important to note the angles on the voltages and currents in this transfonner bank. Because one of the transformer phases is missing, the transmission line current is now equal to the phase current in each transformer, and the currents and voltages in the transfonner bank differ in angle by 30°. Since the current and voltage angles differ in each of the two transfonners, it is necessary to examine each transfonner individually to determine the maximum power it can supply. For transfonner I, the voltage is at an angle of 150° and the current is at an angle of 120°, so the expression for the maximum power in transformer I is PI = 3V4> 14> cos (150° -120°) = 3V4>I4> cos 30° V1 = 2 V4>I4> (2- 96) 128 ELECTRIC MACHINERY RJNDAMENTALS ..[f /. LOOA Nn I. L300A • -+ • \ V.L300y N" + N" • Nn V. L - 90oy V. LI50o y • I. L --SXf' A ..[3 /. L 120° A • -Nn ..[f l. L _ 120° A I. L 150° A -(., I. LOo A I. L60oA -Nn • + V.L300y N" V. LI50o y • ~ L 120° A I. L 120° A • Nn • I. L _ 120° A -(b' ""GURE 2-40 (a) Voltages and currents in a ~--a transformer banlc. (b) Voltages and currents in an open-~ transformer banlc. R , • • , , 0 • d R , • • , , 0 • d For transfonner 2, the voltage is at an angle of 30° and the current is at an angle of 60°, so its maximum power is P2 = 3V 4 ,!4> cos (30° - 60°) = 3 V4> J4> cos (-30°) v:l =""2 ~J4> TIlerefore, the total maximum power of tile open-delta bank is given by P = V3V4>J4> (2- 97) (2- 98) TIle rated current is the same in each transfonner whether there are two or three of them, and the voltage is the same on each transfonner; so the ratio of the output power available from the open-delta bank. to the output power available from the normal three-phase bank is TRANSFORMERS 129 Popen<1 _ V1"V,f,!1> _ _ , _ _ P -3\1:[ -V3 -0.577 3 phase 1> 1> (2- 99) The available power out of the open-delta bank is only 57.7 percent of the origi-nal bank's rating. A good question that could be asked is: What happens to the rest of the open-delta bank's rating? After all, the total power that the two transformers to-gether can produce is two-thirds that of the original bank's rating. To find out, ex-amine the reactive power of the open-delta bank. The reactive power of trans-fonner I is Ql = 3V1>J1> sin (150° -120°) = 3V1>J1> sin 30° = Yl V1>[1> The reactive power of transfonner 2 is Q2 = 3 V1> [1> sin (30° -60°) = 3 ~J1>sin (-300) = --1; V1>J1> Thus one transfonner is producing reactive power which the other one is con-suming. It is this exchange of energy between the two transformers that limits the power output to 57.7 percent of the orig inal bank s rating instead of the otherwise expected 66.7 percent. An alternative way to look at the rating of the open-delta connection is that 86.6 percent of the rating of the two remaining transformers can be used. Open-delta connections are used occasionally when it is desired to supply a small amount of three-phase power to an otherwise single-phase load. In such a case, the connection in Figure 2-4 1 can be used, where transformer Tl is much larger than transfonner Tt . c -----------------, "--, • Th_· T, T, T, ) Single-phase pJu~ power • power • T, • b----y FIGURE 2-41 Using an open-<1 lransformer connection to supply a small amount of lhree-phase power along with a lot of single-phase power. Transformer Tl is much larger than transformer h 130 ELECTRIC MACHINERY RJNDAMENTALS V~C , 0 • Missing ph= ""GURE 2-42 • N" :;." • N" . ~ '" • • ,-----,-~o ' b ~--+, r-'--i-~ b' • • co- ----I--+-~,' " b' V~ 0' The open-Y -open-~ transfonner connection and wiring diagram. Note that this connection is identical to the y....a connection in Figure 3- 38b. except for the absence of the thinl transformer and the presence of the neutral lead. The Open-Wye-Open-Delta Connection TIle open-wye-open-delta connection is very similar to the open-delta connection except that the primary voltages are derived from two phases and the neutral. This type of connection is shown in Figure 2-42. It is used to serve small commercial customers needing three-phase service in rural areas where all three phases are not yet present on the power poles. With this connection, a customer can get three-phase service in a makeshift fashion until demand requires installation of the third phase on the power poles. TRANSFORMERS 131 A major disadvantage of this connection is that a very large return current must flow in the neutral of the primary circuit. The Scott-T Connection The Scott-T connection is a way to derive two phases 90° apart from a three-phase power supply. In the early history of ac power transmission, two-phase and three-phase power systems were quite common. In those days, it was routinely neces-sary to interconnect two- and three-phase power systems, and the Scott-T trans-fonner connection was developed for that purpose. Today, two-phase power is primarily limited to certain control applications, but the Scott T is still used to produce the power needed to operate them. The Scott T consists of two single-phase transformers with identical ratings. One has a tap on its primary winding at 86.6 percent of full-load voltage. 1lley are connected as shown in Figure 2-43a. TIle 86.6 percent tap of transfonner T2 is connected to the center tap of transformer Tl . The voltages applied to the primary winding are shown in Figure 2-43b, and the resulting voltages applied to the pri-maries of the two transformers are shown in Figure 2- 43c. Since these voltages are 90° apart, they result in a two-phase output. It is also possible to convert two-phase power into three-phase power with this connection, but since there are very few two-phase generators in use, this is rarely done. The Three-Phase T Connection The Scott-T connection uses two transformers to convert three-phase power to two-phase power at a different voltage level. By a simple modification of that connection, the same two transfonners can also convert three-phase power to three-phase power at a different voltage level. Such a connection is shown in Fig-ure 2- 44. Here both the primary and the secondary windings of transformer Tl are tapped at the 86.6 percent point, and the taps are connected to the center taps of the corresponding windings on transformer Tl . In this connection T] is called the main transfonner and Tl is called the teaser transfonner. As in the Scott T, the three-phase input voltage produces two voltages 90° apart on the primary windings of the transformers. TIlese primary voltages pro-duce secondary voltages which are also 90° apart. Unlike the Scott T, though, the secondary voltages are recombined into a three-phase output. One major advantage of the three-phase T connection over the other three-phase two-transfonner connections (the open-delta and open-wye-open-delta) is that a neutral can be connected to both the primary side and the secondary side of the transfonner bank. This connection is sometimes used in self-contained three-phase distribution transformers, since its construction costs are lower than those of a full three-phase transformer bank. Since the bottom parts of the teaser transfonner windings are not used on ei-ther the primary or the secondary sides, they could be left off with no change in perfonnance. 1llis is, in fact, typically done in distribution transformers. 132 ELECTRIC MACHINERY RJNDAMENTALS T, + b 86.6% / tap .<-----V" Center mp V :( , + -O-----~--~ T, " ab'" V L 120" " /r<-",VLO" "c~"'VL - 12O" (a) T, ~-----~. + • V " T, " p2 '" 0.866 V L 90" )---- V . (b' V " S2"' -L'Xf' , (d' ""GURE 2-43 " St '" -...t. L 0" , (,' The Scolt-T transformer conneeliOll. (a) Wiring diagram; (b) the three-phase input voltages; (e) the vollages on the transformer primary windings; (d) the two-phase secondary voltages. + V oo V ro V :( " T, ( N, 86.6% ''P V " Center b N, "p " + '-----'" e Vah ", V L 1200 Vbc",VLOo V ...,'" V L _ 1200 >-- V~ ,b , N, a"' -N, \~t T, '-----'-__ .>V" Vct ",.!':L - 1200 " Note: FIGURE 2-44 V VSt '" VBC '" -Lr:f' " Va ", - VSt - VS2 ,d , , ,' r---------------- -Q n , , T, A , + , 57.7% N, , "p +" , , , l ______ ) V n 86.6% V~ "p N, + B " J,e -------V " T, C + ,e, }-----V BC'" L 0 0 Va ",.!':L - 1200 " ,., The three-phase T tr:I.nsformer connection. (a) Wiring diagram; (b) the three-phase input voltages; (e) the voltages on the transfonner primary windings; (d) the voltages on the transformer secondary windings; (e) the resulting three-phase secondary voltages. 133 V a 134 ELECTRIC MACHINERY RJNDAMENTALS 2.12 TRANSFORMER RATINGS AND RELATED PROBLEMS Transfonners have four major ratings: apparent power, voltage, current, and fre-quency. This section examines the ratings of a transfonner and explains why they are chosen the way they are. It also considers the related question of the current inrush that occurs when a transformer is first connected to the line. The Voltage and Frequency Ratings of a Transformer TIle voltage rating of a transformer serves two functions. One is to protect the winding insulation from breakdown due to an excessive voltage applied to it. TIlis is not the most serious limitation in practical transfonners. TIle second function is related to the magnetization curve and magnetization current of the transformer. Figure 2-11 shows a magnetization curve for a transformer. If a steady-state voltage vet) = V M sin wi V is applied to a transfonner's primary winding, the nux of the transfonner is given by 4>(t) = ~p f v(t) dt = ~ f VM sinwtdt p VM - -- cos wt wN p ('-1 00) If the applied voltage v(t) is increased by 10 percent, the resulting maximum nux in the core also increases by 10 percent. Above a certain point on the magne-tization curve, though, a 10 percent increase in nux requires an increase in mag-netization current much larger than 10 percent. TIlis concept is illustrated in Fig-ure 2- 45. As the voltage increases, the high-magnetization currents soon become unacceptable. TIle maximum applied voltage (and therefore the rated voltage) is set by the maximum acceptable magnetization current in the core. Notice that voltage and frequency are related in a reciprocal fashion if the maximum nux is to be held constant: V _ q,max = wN p (2- 101) TIlUS, if a 60-Hz transfonner is to be operated on 50 Hz, its applied voltage must also be reduced by one-sixth or the peak nux in the core will be too high. TIlis re-duction in applied voltage with frequency is called derating. Similarly, a 50-Hz transfonner may be operated at a 20 percent higher voltage on 60 Hz if this action does not cause insulation problems. TRANSFORMERS 135 ------+++-----1- 3' ('" N/). A • turns I 2 3 FIGURE 2-45 The elTect of the peak flux in a tnlnsformer core upon the required magnetization current. Example 2-10. A l-kVA. 230/115-V. 60-Hz single-phase transfonner has 850 turns on the primary winding and 425 llU1lS on the secondary winding. The magnetization curve for this transfonner is shown in Figure 2-46. (a) Calculate and plot the magnetization current of this transfonner when it is run at 230 V on a 60-Hz power source. What is the rms value of the magnetization clUTent ? (b) Calculate and plot the magnetization C lUTent of this transfonner when it is nUl at 230 V on a 50-Hz power source. What is the rms value of the magnetization cur-rent ? How does this C lUTent compare to the magnetization current at 60 Hz? Solutioll The best way to solve this problem is to calculate the flux as a function of time for this core. and then use the magnetization curve to transform each flux value to a corresponding magnetomotive force. The magnetizing current can then be determined from the equation 136 ELECTRIC MACHINERY RJNDA MENTALS Magnetization curve for 2301115· V transfonner 1.4 ~~=~=~:CC:;==;:=-';-'=;==-,-~ 1.2 ~ 0.8 " , ti: 0.6 0.4 0.2 °0 ~-C 2~ OO ~-~ ~c-0600 ~-C 8 ~ OO ~-I " OOO ~-c 12 0 00 OCC1C~ ~-C I60000-c l~ 800 MMF. A - turns HGURE 2-46 Magnetization curve for the 2301115· V transfonner of Example 2- 10. i=~ N, (2- 102) Assruning that the voltage applied to the core is v(t) = VM sin w t volts, the flux in the core as a function of time is given by Equation (2-101 ): ~(t) = VM - -- coswt wN, (2-1 00) The magnetization curve for this transfonner is available electronically in a file called mag_curve_ l . dat . This file can be used by MATLAB to translate these flux values into corresponding mmf values, and Equation (2- 102) can be used to fmd the required magnetization current values. Finally, the nns value of the magnetization current can be calculated from the equation I = l IT Pdt -T O A MATLAB program to perform these calculations is shown below: ~ M-file, mag_current.m ~ M-file t o ca l c ula t e a nd p l o t the magn e tiza tio n ~ c urre nt o f a 230 / 115 tra n s f o rme r oper a ting a t ~ 230 volt s a nd 50 /60 Hz. Thi s p r ogr a m a l so ~ ca l c ula t es the rms va lu e o f the mag . current . ~ Load the magne tizatio n cu rve. It i s in t wo ~ co lumns, with the first column being mmf and ~ the second co lumn being flux . l oad mag_curve_l. dat ; mmf_dat a = mag_curve_l ( : , l ) ; (2-103) TRANSFORMERS 137 % Initia liz e va lues VM 325; NP = 850; % M ax imum voltage (V) % Prima r y turn s % Ca l c ula t e a ngula r ve l ocity f or 60 Hz freq = 60; % Freq (Hz ) w = 2 p i .. freq; % Ca l c ula t e flu x ver s u s time time 0, 1 /300 0, 1 /3 0; % 0 t o 1 /3 0 sec flu x = - VM /(wNP ) cos(w . time ) ; % Ca l c ula t e the mmf correspo nd ing t o a g i ven flu x % u s ing the flu x ' s int e r po l ation function. mmf = int e r p l (flu x_dat a, mmf_dat a, flu x ) ; % Ca l c ula t e the magne tiz atio n c urre nt im = mmf 1 NP ; % Ca l c ula t e the rms va lue o f the c urre nt irms = sqrt (sum ( im. A 2 )/ l e ngth ( im )) ; d i sp( ['The rms c urre nt a t 60 Hz i s " num2str (inns) ] ) ; % Plot the magne tiz ation c urre nt. fi gure ( l ) s ubp l ot (2, 1 , 1 ) ; p l ot (time, im ) ; title ('\b fMagne tiz ation Curre nt a t 60 Hz' ) ; x l abe l ('\b fTime (s) ' ) ; y l abe l ('\b f \ itI_( m) \ rm (A) ' ) ; ax i s( [ O 0.04 - 2 2 ] ) ; g rid on ; % Ca l c ula t e a ngula r ve l ocity f or 50 Hz freq = 50; % Freq (Hz ) w = 2 p i .. freq; % Ca l c ula t e flu x ver s u s time time 0, 1 / 2500, 1 / 25; % 0 t o 1 / 25 sec flu x = - VM /(wNP ) cos(w . time ) ; % Ca l c ula t e the mmf correspo nd ing t o a g i ven flu x % u s ing the flu x ' s int e r po l ation function. mmf = int e r p l (flu x_dat a, mmf_dat a, flu x ) ; % Ca l c ula t e the magne tiz atio n c urre nt im = mmf 1 NP ; % Ca l c ula t e the rms va lue o f the c urre nt irms = sqrt (sum ( im. A 2 )/ l e ngth ( im )) ; d i sp( ['The rms c urre nt a t 50 Hz i s " num2str (inns) ] ) ; 138 ELECTRIC MACHINERY RJNDAMENTALS ~ '" -' 1.414 0.707 0 -0.7(17 - 1.414 0 0.005 om om5 om 0.025 0.Q3 0.Q35 0.04 Time (s) ,,' 1.414,-,--"'C---,--,--,--"' __ ---,----, 0.707 50Hz o -0.7(17 - 1.414:'---o~~~~cc!~--c'~--o+.co-~o-cc!=~ o 0.005 om om5 om 0.025 0.Q3 0.Q35 0.04 Time (s) ,b, ""GURE 2-47 (a) Magnetization current for the transformer operating at 60 Hz. (b) Magnetization current for the transformer operating at 50 Hz. % Pl ot the magnet i zat i on curre nt. s ubpl ot (2, 1 ,2); p l ot (t i me , i m) ; tit l e ('\ b f Magnet i zat i on Curren t at 50 Hz' ) ; x l abe l (' \ bfTime (s) ' ) ; y l abe l (' \ b f \ it I {m) \ nn (A) ' ) ; axi s( [O 0 . 04 - 2 2 ] ) ; gri d on; When this program executes, the results are ,.. mag_current The nns c urrent at 60 Hz i s 0 .4 894 The nns c urrent at 50 Hz i s 0 . 79252 The resulting magnetization currents are shown in Figure 2-47. Note that the nns magne-tization current increases by more than 60 percent when the frequency changes from 60 Hz to 50 Hz. The Apparent Power Rating of a Transformer TIle principal purpose of the apparent power rating of a transfonner is that, to-gether with the voltage rating, it sets the current flow through the transformer windings. TIle current now is important because it controls the jlR losses in trans-fonner, which in turn control the heating of the transformer coils. It is the heating TRANSFORMERS 139 that is critical, since overheating the coi Is of a transformer drastically shortens the life of its insulation. The actual voltampere rating of a transfonner may be more than a single value. In real transfonners, there may be a voltampere rating for the transformer by itself, and another (higher) rating for the transformer with forced cooling. The key idea behind the power rating is that the hot-spot temperature in the trans-fonner windings must be limited to protect the life of the transfonner. If a transfonner's voltage is reduced for any reason (e.g., if it is operated at a lower frequency than normal), then the transfonner's voltrunpere rating must be re-duced by an equal amount. If this is not done, then the current in the transfonner's windings wi ll exceed the maximum pennissible level and cause overheating. The Problem of Current Inrush A problem related to the voltage level in the transfonner is the problem of current inrush at starting. Suppose that the voltage v(t) = VM sin (wi + 6) v (2-104) is applied at the moment the transfonner is first connected to the power line. The maximum flux height reached on the first half-cycle of the applied voltage depends on the phase of the voltage at the time the voltage is applied. If the initial voltage is vet) = V M sin (wi + 90°) = V M cos wt v (2-1 05) and if the initial flux in the core is zero, then the maximum flux during the first half-cycle wi ll just equal the maximum flux at steady state: V=, q,max = wN p (2-1 01) This flux level is just the steady-state flux, so it causes no special problems. But if the applied voltage happens to be vet) = VM sin wi V the maximum flux during the first half-cycle is given by I f"'· q,(t)= N p 0 VM sinwidt VM I " '· ~ - -- cos wt wNp 0 V M ~ - -[(- 1) -(1) wNp (2-106) This maximum flux is twice as high as the normal steady-state flux. If the magnetization curve in Figure 2-11 is examined, it is easy to see that doubling the 140 ELECTRIC MACHINERY RJNDAMENTALS Rated current ""GURE 2-48 v(I)=Vm sinrot f\ The current inrush due to a transformer's magnetization current on starting. maximum flux in the core results in an enonnous magnetization current. In fact, for part of the cycle, the transformer looks like a short circuit, and a very large current fl ows (see Figure 2-48). For any other phase angle of the applied voltage between 90°, which is no problem, and 0°, which is the worst case, there is some excess current flow. The applied phase angle of the voltage is not Ilonnally controlled on starting, so there can be huge inrush currents during the first several cycles after the transfonner is connected to the line. The transfonner and the power system to which it is con-nected must be able to withstand these currents. The Transformer Nameplate A typical nameplate from a distribution transformer is shown in Figure 2-49. TIle infonnation on such a nameplate includes rated voltage, rated kilovoltamperes, rated frequency, and the transfonner per-unit series impedance. lt also shows the voltage ratings for each tap on the transfonner and the wiring schematic of the transfonner. Nameplates such as the one shown also typically include the transformer type designation and references to its operating instructions. 2.13 INSTRUMENT TRANSFORMERS Two special-purpose transfonners are used with power systems for taking mea-surements. One is the potential transfonner, and the other is the current transfonner. TRANSFORMERS 141 I ~ ! , ." j ~~ 3 PHASI: ClASS 0 A iASIC 1Mf'UlSl UYEl II'IWHtDHtG l V I'll NIlItIG KV Wl:IGHTSI~ POINUS 1m:IIIOR .. NPlDOIC(WC U.UDVOlTS ~ xI "0 "l ~ •••• ,r;::;::;::;::;: ~ "0 x , "i "i • , • ~ :., .1:;-""1 I I I ! HI~I HO~ x, xl "l ~ 00 GROUND STlW' 11! = I DISTRIBUTION TRANSFORMER COIITAiNSHON-PCaATnr.(J' a ~ ,-cccccccccc~cccccccc~ ______ -''''''' __ .. ~·"~ ··"'""" .. '·"·"·····"' .. " ,-~ • f-a ~ o ~ FIGURE 2-49 , , ! , j A sample distribution transfonner nameplate. Note the ratings li5ted: voltage, frequen>;y, apparem power. and tap settings. (Courtesy o/General Electric Company.) A potential transformer is a specially wound transformer with a high-voltage primary and a low-voltage secondary. It has a very low power rating, and its sole purpose is to provide a sample of the power system's voltage to the in-struments monitoring it. Since the principal purpose of the transfonner is voltage sampling, it must be very accurate so as not to distort the true voltage values too badly. Potential transfonners of several accuracy classes may be purchased de-pending on how accurate the readings must be for a given application. Current transformers sample the current in a line and reduce it to a safe and measurable level. A diagram of a typical current transformer is shown in Figure 2- 50. The current transfonner consists of a secondary winding wrapped around a ferromagnetic ring, with the single primary line running through the center of the ring. 1lle ferromagnetic ring holds and concentrates a small sample of the flux from the primary line. That flux then induces a voltage and current in the sec-ondary winding. A current transformer differs from the other transformers described in this chapter in that its windings are loosely coupled. Unlike all the other transfonners, the mutual flux <PM in the current transfonner is smaller than the leakage flux d from primary side) Vsc = 1130 V Isc = UX) A Psc =260W (a) Find the per-lUlit equivalent circuit for this transformer at 60 Hz. (b) What would the rating of this transfonner be if it were operated on a 50-Hz power system? (c) Sketch the per-lUlit equivalent circuit of this transfonner referred to the primary side if it is operating at 50 Hz. 2-20. Prove that the three-phase system of voltages on the secondary of the Y -/1 trans-fonner shown in Figure 2- 38b lags the three-phase system of voltages on the pri-mary of the transformer by 30°. 2-21. Prove that the three-phase system of voltages on the secondary of the /1-Y trans-fonner shown in Figure 2- 38c lags the three-phase system of voltages on the pri-mary of the transformer by 30°. 2-22. A single-phase lO-kVA, 4801120-V transformer is to be used as an autotransfonner tying a 600-V distribution line to a 480-V load. When it is tested as a conventional 150 ELECTRIC MACHINERY RJNDAMENTALS transfonner, the following values are measured on the primary (480-V) side of the transformer: Open,circuit tcst Voe = 480 V loe = 0.41 A P oe = 38W Short,circuit test Vsc = 10.0 V Isc = 1O.6A Psc = 26W (a) Find the per-unit equivalent circuit of this transfonner when it is connected in the conventional ma/Uler. What is the efficiency of the transfonner at rated condi-tions and lUlity power factor? What is the voltage regulation at those conditions? (b) Sketch the transfonner connections when it is used as a 600/480-V step-down autotransfonner. (c) What is the kilovoltampere rating of this transformer when it is used in the au-totransformer connection? (d) Answer the questions in a for the autotransformer connection. 2-23. Figure P2-4 shows a power system consisting of a three-phase 480-V 60-Hz gener-ator supplying two loads through a transmission line with a pair of transfonners at either end. Generator 41lO V T, ~~~~~ _ _ _ T, 480114.400 V ; x l abel ( ' Fir i ng a ng l e (deg ) ' ) ; y l abel ( 'RMS voltage (V) ' ) ; g r i d on ; Two examples of the waveform generated by this function are shown in Figure 3- 35. 0.' 0.' Load ,·ol!aJ!e for a 45° firi nJ! anJ!le V / 1.0 1.0 1.5 2.0 w i (radians) (a) -----1.5 2.0 WI (radians) ,b , \ \ \ 1 \ 2.5 3.0 \ \ \ 1 \ 2.5 3.0 3.5 3.5 FIGURE 3-35 Waveform produced by vo l t s _ vs...Jlh ase_ a n g 1 e for a firing angle of (a) 45°; (b) 90°. 184 ELECTRIC MACHINERY RJNDAMENTALS 140 120 ~ 100 ~ 80 ~ 60 ~ • " 40 20 ------'" "-"-" w ~ ro 80 100 120 140 lro IW Firing angle (deg) ""GURE 3-36 Plot of rms load voltage versus TRIAC firing angle. When this m-fIle is executed, the plot shown in Figure 3- 36 results. Note that the earlier the firing angle, the greater the rms voltage supplied to the load. However, the relationship between firing angle and the resulting voltage is not linear, so it is not easy to predict the required firing angle to achieve a given load voltage. (b) The firing angle required to supply 75 V to the load can be fOlUld from Figure 3- 36. It is about 99°. The Effect of Inductive Loads on Phase Angle Control I f the load attached to a phase angle controller is inductive (as real machines are), then new complications are introduced to the operation of the controller. By the nature of inductance, the current in an inductive load cannot change instanta-neously. nlis means that the current to the load will not rise immediately on firing the SCR (or TRIAC) and that the current will not stop flowing at exactly the end of the half-cycle. At the end of the half-cycle, the inductive voltage on the load will keep the device turned on for some time into the next half-cycle, until the cur-re nt flowing through the load and the SCR finally fall s below ly. Fig ure 3- 37 shows the effect of this delay in the voltage and current wavefonns for the circuit in Fig ure 3- 32. A large inductance in the load can cause two potentially serio us proble ms with a phase controller: I. The inductance can cause the current buildup to be so slow when the SCR is switched on that it does not exceed the holding current before the gate current disappears. If this happens, the SCR will not remain on, because its current is less than [y. INTRODUCTION TO POWER ELECTRONICS 185 , , , , , , / / / / / / / / / FIGURE 3-37 \ \ , , , \ \ \ , \ , , , , / / The elTect of an inductive load on the current and voltage waveforms of the cin:uit shown in Figure 3- 32. 2. If the current continues long enough before decaying to IH after the end of a given cycle, the applied voltage could build up high enough in the next cycle to keep the current going, and the SCR wi II never switch off. The nonnal solution to the first problem is to use a special circuit to provide a longer gating current pulse to the SCR. nlis longer pulse allows plenty of tirne 186 ELECTRIC MACHINERY RJNDAMENTALS Freewheeling diod ~---r--;='I "d"t:::;"i" I) R2 load SCR c R, ""GURE 3-38 A phase angle controller illustrating the use of a free-wheeling diode with an inductive load. for the current through the SCR to rise above IH, pennitting the device to remain on for the rest of the half-cycle. A solution to the second problem is to add afree-wheeling diode. A free-wheeling diode is a diode placed across a load and oriented so that it does not con-duct during nonnal current flow. Such a diode is shown in Figure 3- 38. At the end ofa half-cycle, the current in the inductive load will attempt to kccp fl owing in the same direction as it was going. A voltage will be built up on the load with the po-larity required to keep the current flowing. This voltage will forward-bias the free-wheeling diode, and it will supply a path for the discharge current from the load. In that manner, the SCR can turn off without requiring the current of the inductor to instantly drop to zero. 3.5 DC-TO-DC POWER CONTROL-CHOPPERS Sometimes it is desirable to vary the voltage available from a dc source before ap-plying it to a load. TIle circuits which vary the voltage of a dc source are called de-to-de conveners or choppers. In a chopper circuit, the input voltage is a constant dc voltage source, and the output voltage is varied by varying thefmerion of the time that the dc source is connected to its load. Figure 3- 39 shows the basic prin-ciple of a chopper circuit. When the SCR is triggered, it turns on and power is sup-plied to the load. When it turns off, the dc source is disconnected from the load. In the circuit shown in Figure 3- 39, the load is a resistor, and the voltage on the load is either Voc or O. Similarly, the current in the load is either VoclR or O. lt is possible to smooth out the load voltage and current by adding a series inductor to filter out some of the ac components in the waveform. Figure 3-40 shows a chopper circuit with an inductive filter. The current through the inductor increases exponentially when the SCR is on and decreases exponentially when the SCR is off. If the inductor is large, the time constant of the current changes (T = LIR) will INTRODUCTION TO POWER ELECTRONICS 1 87 SCR + L~ + Voc ,~ ( R~ ,,' Voc ,b, ,,' FIGURE 3-39 (a) The basic principte of a chopper circuit. (b) The input voltage to the circuit. (c) The resulting voltage on the load. be long relati ve to the on/off cycle of the SCR and the load voltage and current will be almost constant at some average value. In the case of ac phase controllers, the SCRs automatically turn off at the end of each half-cycle when their currents go to 7..ero. For dc circuits, there is no point at which the current naturally falls below IH, so once an SCR is turned on, it never turns off. To turn the SCR off again at the end of a pulse, it is necessary to apply a reverse voltage to it for a short time. TIlis reverse voltage stops the current flow and turns off the SCR. Once it is off, it will not turn on again until another pulse enters the gate of the SCR. TIle process of forcing an SCR to turn off at a desired time is known as forced commutation. GTO thyristors are ideally suited for use in chopper circuits, since they are self-commutating. In contrast to SCRs, GTOs can be turned off by a negative cur-rent pulse applied to their gates. Therefore, the extra circuitry needed in an SCR 188 ELECTRIC MACHINERY RJNDAMENTALS SCR + "j I ;~ + L + Voc vI(t) ,~ ( D, R~ , ,' Voc e------------------------------"------------------------------ , ,b, , --- vI(t) , ------ vlood(l) ------\ -\ -\ , / , / , / / / / , , , " , ""GURE 3-40 A chopper circuit with an inductive filter 10 smooth oUllhe load voltage and current. chopper circuit to turn off the SCR can be eliminated from a GTO thyristor chop-per circuit (Figure 3---4 1 a). Power transistors are also self-commutating and are used in chopper circuits that fall within their power limits (Figure 3---41 b). Chopper circuits are used with dc power systems to vary the speed of dc motors. TIleir greatest advantage for dc speed control compared to conventional methods is that they are more efficient than the systems (such as the Ward-Leonard system described in Chapter 6) that they replace. Forced Commutation in Chopper Circuits When SCRs are used in choppers, a forced-commutation circuit must be included to turn off the SCRs at the desired time. Most such forced-commutation circuits + VOC INTRODUCTION TO POWER ELECTRONICS 189 L Lo,d ;~ Voc -I, I + }~ ; J Lood I ;, f/ - " (a) (b' FIGURE 3-41 (a) A chopper cin:uit made with a GTO thyristor. (b) A chopper cin:uit made with a transistor. + c ~----------------------, I D \ FIGURE 3-42 I SCR ~-T +,---f--, R L - '--+---' A series-capacitor forced-commutation chopper cin:uit. Lo,d + depend for their turnoff voltage on a charged capacitor. Two basic versions of ca-pacitor commutation are examined in this brief overview: I . Series-capacitor commutation circuits 2. Parallel-capacitor commulalion circuits Series-Capacitor Commutation Circuits Figure 3-42 shows a simple dc chopper circuit with series-capacitor commuta-lion. Ii consists of an SCR, a capacitor, and a load, all in series with each other. 190 ELECTRIC MACHINERY RJNDAMENTALS Voc --------Discharge 1" '" RC L--1 __ ~~ _ ~ _ , Voc --- --------- --------- --------""GURE 3-43 The capacitor and load voltages in the series chopper circuit. TIle capacitor has a shunt discharging resistor across it, and the load has a free-wheeling diode across it. TIle SCR is initially turned on by a pulse applied to its gate. When the SCR turns on, a voltage is applied to the load and a current starts flowing through it. But this current flows through the series capacitor on the way to the load, and the capacitor gradually charges up. When the capacitor's voltage nearly reaches Voc. the current through the SCR drops below iH and the SCR turns off. Once the capacitor has turned off the SCR, it gradually discharges through resistor R. When it is totally discharged, the SCR is ready to be fired by another pulse at its gate. The voltage and current wavefonns for this circuit are shown in Figure 3-43. Unfortunately, this type of circuit is limited in tenns of duty cycle, since the SCR cannot be fired again until the capacitor has discharged. The discharge time depends on the time constant 1" = RC, and C must be made large in order to let a lot of current flow to the load before it turns off the SCR. But R must be large, since the current leaking through the resistor has to be less than the holding cur-rent of the SCR. These two facts taken together mean that the SCR cannot be re-fired quickly after it turns off. It has a long recovery time. An improved series-capacitor commutation circuit with a shortened recov-ery time is shown in Figure 3-44. TIlis circuit is similar to the previous one except that the resistor has been replaced by an inductor and SCR in series. When SCR is fired, current will flow to the load and the capacitor will charge up, cutting off SCRI. Once it is cut off, SCR2 can be fired, discharging the capacitor much more INTRODUCTION TO POWER ELECTRONICS 191 ;" [ n n n , ;" [ n n n , i i!l SCR I vc(l) i!2 L + co ~ vc(l) -SCR, ~ -+ Inductive D load I ~~-~ -~ .= -!--~~--~- , Ready '" -fire (a) (bj FIGURE 3-44 (a) A series-capacitor forced-commutation chopper cin:uit with improved capacitor recovery time. (b) The resulting capacitor and load voltage waveforms. Note that the capacitor discharges much more rapidly, so SCR[ could be refired sooner than before. quickly than the resistor would. TIle inductor in series with SCR1 protects SCR1 from instantaneous current surges that exceed its ratings. Once the capacitor dis-charges, SCR1 turns off and SCR] is ready to fi re again. Parallel-Capacitor Commutation Circuits The other common way to achieve forced commutation is via the parallel-capacitor commutation scheme. A simple example of the parallel-capacitor scheme is shown in Figure 3-45. In this scheme, SCR] is the main SCR, supply-ing power to the load, and SCR2 controls the operation of the commutating capac-itor. To apply power to the load, SCR I is fired. When this occurs, a current flows through the SCR to the load, supplying power to it. Also, capacitor C charges up through resistor R to a voltage equal to the supply voltage Voc. When the time comes to turn off the power to the load, SCR2 is fired. When SCR1 is fired, the voltage across it drops to zero. Since the voltage across a 192 ELECTRIC MACHINERY RJNDAMENTALS • R \ D ,~ v OC L " -- C. SCR] ~ ~ ""GURE 3-45 A parallel-<:apacitor forced-commuta.tion chopper circuit . • R \ L D ,~ L oc I o---J v _ Vc + " SCR] ~ ~ ""GURE 3-46 A parallel-<:apacitor forced-commuta.tion chopper circuit with improved capacitor charging time. SCRJ permits the load power to be turned off more quickly thaD it could be with the basic parallel-capacitor circu it. capacitor cannot change instantaneously, the voltage on the left side of the capaci-tor must instantly drop to -Voc volts. This turns off SCRb and the capacitor charges through the load and SCR2 to a vol tage of Voc volts positive on its left side. Once capacitor C is charged, SCRl turns off, and the cycle is ready to begin again. Again, resistor R] must be large in order for the current through it to be less than the holding current of SCR2. But a large resistor Rt means that the capacitor will charge only slowly after SCR] fi res. This limits how soon SCR] can be turned off after it fires, setting a lower limit on the on time of the chopped waveform. A circuit with a reduced capacitor charging time is shown in Figure 3-46. In this circuit SCR1 is triggered at the same time as SCR] is, and the capacitor can INTRODUCTION TO POWER ELECTRONICS 193 charge much more rapidly. This allows the current to be turned off much more rapidly if it is desired to do so. In any circuit of this sort, the free-wheeling diode is extremely important. When SCR[ is forced off, the current through the inductive load must have an-other path available to it, or it could possibly damage the SCR. 3.6 INVERTERS Perhaps the most rapidly growing area in modern power electronics is static fre-quency conversion, the conversion of ac power at one frequency to ac power at another frequency by means of solid-state electronics. Traditionally there have been two approaches to static ac frequency conversion: the cycloconverter and the rectifier-inverter. The cycJoconverter is a device for directly converting ac power at one frequency to ac power at another frequency, while the rectifier-inverter first converts ac power to dc power and then converts the dc power to ac power again at a different frequency. lllis section deals with the operation of rectifier-inverter circuits, and Section 3.7 deals with the cycJoconverter. A rectifier-inverter is divided into two parts: I. A rectifier to produce dc power 2. An inveT1er to produce ac power from the dc power. Each part is treated separately. The Rectifier The basic rectifier circuits for converting ac power to dc power are described in Section 3.2. These circuits have one problem from a motor-control point of view-their output voltage is fixed for a given input voltage. This problem can be overcome by replacing the diodes in these circuits with SCRs. Figure 3-47 shows a three-phase full-wave rectifier circuit with the diodes in the circuits replaced by SCRs. The average dc output voltage from this circuit depends on when the SCRs are triggered during their positive half-cycles. If they are triggered at the beginning of the half-cycle, this circuit will be the same as that of a three-phase full-wave rectifier with diodes. Ifthe SCRs are never triggered, the output voltage wi ll be 0 V. For any other firing angle between 0° and 180 0 on the wavefonn, the dc output voltage will be somewhere between the maximum value and 0 V. When SCRs are used instead of diodes in the rectifier circuit to get control of the dc voltage output, this output voltage wi ll have more harmonic content than a simple rectifier would, and some fonn of filter on its output is important. Figure 3-47 shows an inductor and capacitor filter placed at the output of the rectifier to help smooth the dc output. 1 94 ELECTRIC MACHINERY RJNDAMENTALS • L , , -, vB<t) vc<t) c , , -, ""GURE 3-47 A three-phase rectifier circuit using SCRs to provide control of the dc output voltage level. L, I, J -"-' S~ SCR2 SCR1 ;C C·l r{ '-'I Rectifier Synchronous motor SCR4 <>-'-' SCR~ o-C-J SCR6 o-C-J ""GURE 3-48 An external commutation inverter. External Commutation Inverters Inverters are classified into two basic types by the commutation technique used: external commutation and self-commutation. Extenwl commutation inverters are inverters in which the energy required to turn off the SCRs is provided by an ex-ternal motor or power supply. An example of an external commutation inverter is shown in Figure 3-48. The inverter is connected to a three-phase synchronous motor, which provides the countervoltage necessary to turn off one SCR when its companion is fired. llle SCRs in this circuit are triggered in the following order: SCRj, SC~, SCR2, SCR4, SCR1, SCR5. When SCR t fi res, the internal generated voltage in the synchronous motor provides the voltage necessary to turn off SCRJ . Note that if the load were not connected to the inverter, the SCRs would never be turned off and after ~ cycle a short circuit would develop through SCR t and SC~. lllis inverter is also called a load-commutated inverter. INTRODUCTION TO POWER ELECTRONICS 195 Self-Commutation Inverters Ifit is not possible to guarantee that a load will always provide the proper coun-tervoltage for commutation, then a self-commutation inverter must be used. A self-commutation inverter is an inverter in which the active SCRs are turned off by energy stored in a capacitor when another SCR is switched on. It is also possi-ble to design self-commutation inverters using GTOs or power transistors, in which case commutation capacitors are not required. There are three m~ o r types of self-commutation inverters: current source in-verters (CSls), voltage source inverters (VSls), and pulse-width modulation (PWM) inverters. Current source inverters and voltage source inverters are simpler than PWM inverters and have been used for a longer time. PWM inverters require more complex control circuitry and faster switching components than CSls and VSls. CSls and VSls are discussed first. Current source inverters and voltage source inverters are compared in Figure 3-49. In the current source inverter, a rectifier is connected to an inverter through a large series inductor Ls. The inductance of Ls is sufficiently large that the direct current is constrained to be almost constant. TIle SCR current output waveform will be roughly a square wave, since the current flow Is is constrained to be nearly constant. The line-to-Iine voltage will be approximately triangular. It is easy to limit overcurrent conditions in this design, but the output voltage can swing widely in response to changes in load. In the voltage source inverter, a rectifier is connected to an inverter through a series inductor Ls and a parallel capacitor C. The capacitance of C is sufficiently large that the voltage is constrained to be almost constant. The SCR line-to-line voltage output wavefonn wi ll be roughly a square wave, since the voltage Vc is constrained to be nearly constant. TIle output current now wi ll be approximately triangular. Voltage variations are small in this circuit, but currents can vary wildly with variations in load, and overcurrent protection is difficult to implement. TIle frequency of both current and voltage source inverters can be easily changed by changing the firing pulses on the gates of the SCRs, so both inverters can be used to drive ac motors at variable speeds (see Chapter 10). A Single-Phase Current Source Inverter A single-phase current source inverter circuit with capacitor commutation is shown in Figure 3- 50. It contains two SCRs, a capacitor, and an output trans-former. To understand the operation of this circuit, assume initially that both SCRs are off. If SCR[ is now turned on by a gate current, voltage Voc will be applied to the upper half of the transfonner in the circuit. This voltage induces a voltage Voc in the lower half of the transfonner as well, causing a voltage of 2 Voc to be built up across the capacitor. The voltages and currents in the circuit at this time are shown in Figure 3- 50b. Now SCRl is turned on. When SCR2 is turned on, the voltage at the cathode of the SCR will be Voc. Since the voltage across a capacitor cannot change 196 ELECTRIC MACHINERY RJNDAMENTALS Current source invener Voltage source invener L, I, L, -I I r + I Main circuit ~ ~ ~ y,( f--o configuration ~ ~ ~ c f--o ~ ~ ~ f--o I I I -I Rectifier Inverter Rectifier Inverter Type of source Current source - Is almost constant Voltage source - Vs almost constam Output impedance High Low Line /"""-.. . , ', ' Vi r-Line voltage / voltage lO'\J"ilf a If 21f (1800 conduction) Output waveform =t1 , CUrrent OJ /tv Currem (1200 conduction) I. Easy to control overcurrent I. Difficult to limit current Characteristics conditions with this design because of capacitor 2. Output voltage varies widely 2. Output voltage variations with changes in load small because of capacitor ""GURE 3-49 Comparison of current source inveners and voltage source inverters. instantaneously, this forces the voltage at the top of the capacitor to instantly be-come 3 Voc, turning off SCR t . At this point, the voltage on the bottom half of the transfonner is built up positive at the bottom to negative at the top of the winding, and its magnitude is V oc. The voltage in the bottom half induces a voltage Voc in the upper half of the transformer, charging the capacitor C up to a voltage of 2Voc, oriented positive at the bottom with respect to the top of the capacitor. The condi-tion of the circuit at this time is shown in Figure 3- 5Oc. When SCR] is fired again, the capacitor voltage cuts off SCR2, and this process repeats indefinitely. The resulting voltage and current wavefonns are shown in Figure 3- 5 1. + (a) + (b' + «, oc L ' ",-~ -FIGURE 3-50 INTRODUCTION TO POWER ELECTRONICS 197 + J i...(t) C ", -SCR 2 4----' SCRl -c-' := ;, Voc + + ;, if. ;~ Lo,' • -• ;~ (a) A simple single-phase inverter circuit. (b) The voltages and currems in the circuil when SCR[ is triggered. (c) The voltages and currents in the circuit when SCR1 is lriggered. A Three-Phase Current Source Inverter Figure 3- 52 shows a three-phase current source inverter. In this circ uit, the six SCRs fire in the order SCR], SC~, SCR2, SC~ , SCR1, SCR5. Capacitors C l through C6 provide the commutation required by the SCRs. 198 ELECTRIC MACHINERY RJNDA MENTALS 3Voc SCR2 turned SCR] turned ~ off ~] I~ " 2Voc " I ----I I I I I ; I I o -, ----" vif), ;,.(1) 2Voc Voc o I -H GURE 3-51 I / ~ , SCR2 turned SCR] tu ~ off off I~ " I I I -I I I I I ; I I , ----" ~ I, I, I No< t ~d --- "'SCRI cathode ---"'SCR2 cathode __ v)/) --- jefl) Plots of the voltages and current in the inverter circuit: VI is the voltage al the cathode of seRlo and V 1 is the voltage at the cathode of SeRlo Since the voltage at their anodes is Voc. any time VI or V 1 exceeds Voc. that SCR is turned off. il.ood is the current supplied to the invener's load. INTRODUCTION TO POWER ELECTRONICS 199 Three-ph~ input ~ ~ ~ I, I L, Rectifier FIGURE 3-52 SCR[ ~ D, D, SCI<, ~ A three-phase current source inverter. SCR2 ~ C, " " D, D, C, " " SCR~ ~ SCRJ ~ C, C, D, 0 b f '\ Motor , D, C, C, SCR,; ~ To understand the operation of this circuit, examine Figure 3- 53. Assume that initially SCR[ and SCR~ are conducting, as shown in Figure 3- 53a. Then a voltage will build up across capacitors Ct , C3, C4, and Cs as shown on the dia-gram. Now assume that SCR6 is gated on. When SC~ is turned on, the voltage at point 6 drops to zero (see Figure 3- 53b). Since the voltage across capacitor Cs cannot change instantaneously, the anode of SCRsis biased negati ve, and SCRs is turned off. Once SC~ is on, all the capacitors charge up as shown in Figure 3- 53c, and the circuit is ready to turn off SCR6 whenever SCR4 is turned on. This same commutation process applies to the upper SCR bank as well. The output phase and line current from this circuit are shown in Figure 3- 53d. A Three-Phase Voltage Source Inverter Figure 3- 54 shows a three-phase voltage source inverter using power transistors as the active elements. Since power transistors are self-commutating, no special commutation components are included in this circuit. N Q Q Three-p,,", input -I, I L, Rectifier FlG URE J-..5 3 ---. SCR] J D, L -D, SCR4 J SCR2 SCR3 C, J + ,;-C, J " + -D, C] D3 " -b Motor I -./ c "-D, D. C, C, " ,, ~ +"--" + " C. " SCR, ~ SCRo -' , ,' T ~ p"'~ input I, I L, Rectifier SCR ] J D, D, SCR., J SCR2 SCR1 C, J + ,,'-C, ~ " + -D, " C1 D3 D, D. C, ~,' + -v~ 11 + ~,' " SCR, SCRo -' ~ ,b, " b Motor c ./ '. WhenSC~fires. "'6 ---+ O. Therefore the anode voltage of SCR~ (\'~ ) becomes negative. and SC R~ turns off. 1be operation of the three-phase CSI. (3) Initially. SCRI and SCRI are conducting. Note how the commutating capacitocs have charged up. (b) 1be situation when SC~ fires. 1be voltage at the aoode of SCR6 falls almost instantaneously to zero. Since the voltage across capacitor Cl cannot change instantaneously. the voltage at the anode ofSeRl will become negative. arxI SCRI will tum off. Three-p,,", input I, I L, Rectifier ---. SCR] J D, I D, SCR4 J N FlGURE 3-S3 (concluded) -SCR2 SCR3 C, J C, -' + ,,-" + -D, ~3 D3 D, D. C, C, " + " " C. + " SCI<, ~J SCR, ..i , " " -b MOlOr , I -V Gale pulses SCR 1 1 6 2 4 3 S 6 conducting I SCR t SCR2 SCRJ SCR t nlerva/s SCI<, SCR, SCR, SCI<, SCR, , - , '.(1'1 I I , I, I I I - I, ]1----,--------'---'----,-------,----"(:~II------,-----,-------'----'--------,------ I, ,d, 8: (c) Now SCR[ and SC~ are conducting, and the commutating capacitors charge up as shown, (d) The gating pulses, SCR conducting intervals, and the output current from this invener, 202 ELECTRIC MACHINERY RJNDAMENTALS +~~ Th~ ph~ invene ~ ~ ~ Rectifier -""GURE 3-54 -~ + /V, T + T, D, V, = = C T, D, -, ,' T, T, (a) A three-phase voltage source inverter using power transistors. T, D, D, T. D, D. I n this circuit, the transistors are made to conduct in the order Tb T6 , T2, T4, T1, T5. The output phase and line voltage from this circuit are shown in Fig-ure 3- 54b. Pulse-Width Modulation Inverters Pulse-width modulation is the process of modifying the width of the pulses in a pulse train in direct proportion to a small control signal; the greater the control voltage, the wider the resulting pulses become. By using a sinusoid of the desired frequency as the control voltage for a PWM circuit, it is possible to produce a high-power wavefonn whose average voltage varies sinusoidally in a manner suitable for driving ac motors. 1lle basic concepts of pulse-width modulation are illustrated in Figure 3- 55. Figure 3- 55a shows a single-phase PWM inverter circuit using IGBTs. The states of IGSTt through IGBT4 in this circuit are controlled by the two comparators shown in Figure 3- 55b. A comparator is a device that compares the input voltage Vinet) to a refer-ence signal and turns transistors on or off depending on the results of the test. Comparator A compares VinCt) to the reference voltage v..(t) and controls IGBTs Tt and Tl based on the results of the comparison. Comparator B compares Vinet) to the reference voltage v,(t) and controls IGBTs Tl and T4 based on the results of the comparison. If VinCt) is greater than v..(t) at any given time t, then comparator A will turn on Tt and turn off Tl . Otherwise, it will turn off Tt and turn on T2. Simi-larly, if Vinet) is greater than vy(t) at any gi ven time t, then comparator B will turn b , T, T, T, , , -v, '-------' "'c(l) Vs f-----, -v, "'w,(/) 2V , v , o -v , - 2V , "'bc(1) 2V , v , o -v , - 2V , "'c,.(1) 2V , v , o -v , - 2V , FIGURE 3-54 (concluded) INTRODUCTION TO POWER ELECTRON ICS 203 T, T, T, T, T, T. T, T. L ,b , (b) The output phase and line voltages from the inverter. 204 ELECTRIC MACHINERY RJNDAMENTALS + VBI~ T, VB3~ T v(I) , , + -Lo,d + + VB2~ T, ) 'J" ('in VB4~ T , --(., ""GURE J-S5 The basic concepts of pulse-width modulation. (a) A single-phase PWM cin:uil using IGBTs. off TJ and tum on T4. Otherwise, it will turn on T) and turn off T4 . The reference voltages vit) and vy(!) are shown in Figure 3-55c. To understand the overall operation of this PWM inverter circuit, see what happens when different control voltages are applied to it. First, assume that the control voltage is a v. 1llen voltages vit) and v.(t) are identical, and the load volt-age out of the circuit V1oad(t) is zero (see Figure 3- 56). Next, assume that a constant positive control voltage equal to one-half of the peak reference voltage is applied to the circuit. 1lle resulting output voltage is a train of pulses with a 50 percent duty cycle, as shown in Figure 3- 57. Finally, assume that a sinusoidal control voltage is applied to the circuit as shown in Figure 3- 58. 1lle width of the resulting pulse train varies sinusoidally with the control voltage. 1lle result is a high-power output wavefonn whose aver-age voltage over any small region is directly proportional to the average voltage of the control signal in that region. 1lle fundamental frequency of the output waveform is the same as the frequency of the input control voltage. Of course, there are hannonic components in the output voltage, but they are not usually a concern in motor-control applications. 1lle hannonic components may cause ad-ditional heating in the motor being driven by the inverter, but the extra heating can be compensated for either by buying a specially designed motor or by derating an ordinary motor (running it at less than its full rated power). A complete three-phase PWM inverter would consist of three of the single-phase inverters described above with control voltages consisting of sinusoids INTRODUCTION TO POWER ELECTRONICS 205 Comparator A '" '" I'in > I'x 0" Off viI) l'in < I'x Off 0" Comparator B '~ '" I'in > I'y Off 0" vft) l'in < I'y 0" Off ,b, ,,' FIGURE 3-55 (col/eluded) (b) The comparators used to control the on and off states ofttle transistors. (c) The reference voltages used in the comparators. shifted by 120 0 between phases. Frequency control in a PWM inverter of this sort is accomplished by changing the frequency of the input control voltage. A PWM inverter switches states many times during a single cycle of the re-sulting output voltage. At the time ofthis writing, reference voltages with frequen-cies as high as 12 kHz are used in PWM inverter designs, so the components in a PWM inverter must change states up to 24,(X)Q times per second. This rapid switch-ing means that PWM inverters require faster components than CSls or YSls. PWM inverters need high-power high-frequency components such as GTO thyristors, 206 ELECTRIC MACHINERY RJNDAMENTALS • " , , , , , , • " , , , , , , V.jt) "'. " , , , , , , • " , , , , , , • " , , , , , , Vio '" 0 f---~'----'--~-~I-----il----lf---- , , , , , r-r-, , , , , , " • , , , , , , " • , , , , , , " • , , , , , , " • , , , , , ,--,--, , Vmd(l) ", 0 r-------------------- , ""GURE 3-56 The output of the PWM circuit with an input voltage of 0 V. Note that v,(t) '" v,(t). so v~t) '" o. ) , , , , , " , , FIGURE 3-57 , , , , , , " • ,-, , , , , , " • ,-INTRODUCTION TO POWER ELECTRONICS 207 , , , , , , " • ,-, , , , , , " • , , , , , , " • r , The output of the PWM circuit with an input voltage equal to one-half of the peak comparator voltage. ------<-------<--< -----208 --L -- -. -;. -----;. ------. ------. ---~ i INTRODUCTION TO POWER ELECTRONICS 209 IGBTs, and/or power transistors for proper operation. (At the time of this writing, IGBTs have the advantage for high-speed, high-power switching, so they are the preferred component for building PWM inverters.) The control voltage fed to the comparator circuits is usually implemented digitally by means of a microcomputer mounted on a circuit board within the PWM motor controller. The control voltage (and therefore the output pulse width) can be controlled by the microcomputer in a manner much more sophisticated than that described here. It is possible for the microcomputer to vary the control voltage to achieve different frequencies and voltage levels in any desired manner. For example, the microcomputer could im-plement various acceleration and deceleration ramps, current limits, and voltage-versus-frequency curves by simply changing options in software. A real PWM-bascd induction motor drive circuit is described in Section 7.10. 3.7 CYCLOCONVERTERS The cyc1oconverter is a device for directly converting ac power at one frequency to ac power at another frequency. Compared to rectifier-inverter schemes, cyc1o-converters have many more SCRs and much more complex gating circuitry. De-spite these disadvantages, cyc1oconverters can be less expensive than rectifier-inverters at higher power ratings. Cyc1oconverters are now available in constant-frequency and variable-frequency versions. A constant-frequency cyc1oconverter is used to supply power at one frequency from a source at another frequency (e.g., to supply 50-Hz loads from a 6O-Hz source). Variable-frequency cyc1oconverters are used to provide a variable output voltage and frequency from a constant-voltage and constant-frequency source. They are often used as ac induction motor drives. Although the details of a cyc1oconverter can become very complex, the ba-sic idea behind the device is simple. TIle input to a cyc1oconverter is a three-phase source which consists of three voltages equal in magnitude and phase-shifted from each other by 120°. TIle desired output voltage is some specified wavefonn, usu-ally a sinusoid at a different frequency. The cycloconverter generates its desired output waveform by selecting the combination of the three input phases which most closely approximates the desired output voltage at each instant of time. There are two major categories of cycloconverters, noncirculating current cycloconverters and circulating current cycloconverters. These types are distin-guished by whether or not a current circulates internally within the cyclocon-verter; they have different characteristics. The two types of cycloconverters are described following an introduction to basic cycloconverter concepts. Basic Concepts A good way to begin the study of cycloconverters is to take a closer look at the three-phase fu ll-wave bridge rectifier ci rcuit described in Section 3.2. TIlis circuit is shown in Figure 3- 59 attached to a resistive load. In that figure, the diodes are divided into two halves, a positive half and a negative half. In the positive half, the 210 ELECTRIC MACHINERY RJNDAMENTALS D, D, VA(I) '" V Msin WI V VB(I) '" VMsin (ro/- 120") V Vc(l) '" VMsin (&1- 240°) V fo'IGURE 3-59 D, D, Lo.d D, D, + }~ (') -A three-phase full-wave diode bridge ci['(;uit connected to a resistive load. diode with the highest voltage applied to it at any given time will conduct, and it will reverse-bias the other two diodes in the section. In the negative half, the diode with the lowest voltage applied to it at any given time will conduct, and it will reverse-bias the other two diodes in the section. TIle resulting output voltage is shown in Figure 3--60. Now suppose that the six diodes in the bridge circuit are replaced by six SCRs as shown in Figure 3--61. Assume that initially SCR] is conducting as shown in Figure 3--61 b. nlis SCR will continue to conduct until the current through it falls below IH. Ifno other SCR in the positive halfis triggered, then SCR[ will be turned ofT when voltage VA goes to 7..ero and reverses polarity at point 2. However, if SCRl is triggered at any time after point I, then SCR[ will be instantly reverse-biased and turned off. TIle process in which SCR2 forces SCR[ to turn off is calJed/orced com-mutation; it can be seen that forced commutation is possible only for the phase an-gles between points I and 2. 1lle SCRs in the negative half behave in a similar manner, as shown in Figure 3--61 c. Note that if each of the SCRs is fired as soon as commutation is possible, then the output of this bridge circuit will be the same as the output of the full-wave diode bridge rectifier shown in Figure 3- 59. Now suppose that it is desired to produce a linearly decreasing output volt-age with this circuit, as shown in Figure 3--62. To produce such an output, the con-ducting SCR in the positive half of the bridge circuit must be turned off whenever its voltage falls too far below the desired value. 1llis is done by triggering another SCR voltage above the desired value. Similarly, the conducting SCR in the nega-tive half of the bridge circuit must be turned ofT whenever its voltage rises too far above the desired value. By triggering the SCRs in the positive and negative halves at the right time, it is possible to produce an output voltage which de-creases in a manner roughly corresponding to the desired wavefonn. It is obvious from examining Figure 3--62 that many harmonic components are present in the resulting output voltage. (,) (q) (.) 'Pro1 ;KJ] 0] P"!1ddil "iil'HOA 1C10] ;KU (:l) 's1.'1 ,,",' ,,- - ~.... ,,'--..... ,--......... ,,-- ~ .... /' " /" .... , /" "v" .... , / ',// .... , /" ~ )-/' X /, < , /~ /' /' /\ /~ I, ~ /, /' /' I~ I, /~ '/,1' ,/~/,I, '1,1,1 1 \ 1\ \ 1 ,I ,I \1 '/ \1 \ , I ,I ' I .1 'I vI , Y A v . ~ f ,\ /\ " \ 1/\ / \ / \ / \ / 1\ I, \ 1'1\ 1\ I 1 \ , 1 \ 1 \1 I " ",,"\', 1'1 \1 ,I 1 'I , 'I ,I " \1 I , 1 '/ 'I 'I ,I ,I I '( ) I~ I )1 ~ , I 1\ I 1\ 1\ \ 1\ " /\ II /\ II \// '\// \,/ '\x'/ '\,,/ '\;1// \1\ o , , I' /' 1 \ I' I, 1\ 1\ , I, I' 1\ I' f\ I, I~/\ 1,1' , , , , I\I\I'I\/~/\ I \/1 ./ ',/ \/ ~./ \, , , iiu!I:lnpuOJ ta iiunJnpUOJ 9 a l!unJnpUO:l 'a /.... " /'. /' /, /, /' " ~ /, '" / ~ I , I / '. / • / , ~ / 1 t , / r t , / 1 ~ , / / .... 1 ' / ~/ / ,/ '/ ,/ / ,I / '/ / , " , / ," , .( A /, .... /, /' /' , / , / , / , , /" .... / .... ',,-----'\ ',,-----'''\ ',------'''\ ' (/j-J~ (lfI~ (I)V~ ..... ..... --'-, /"....... ",,--......... ,,- --- ~ ..... , / /..." '" / .... / ....". , / ' / ,,/ ,/ ' " " / /<. X ',,~ / ~, / , / .... / / , / , / ' / ~ /, I / '. / , /, / 1 t ( . 1 , ~ , / t / , / , /1 ~ / ,/ ' / , / ' / ; ; / l!unJnpUO:l 'a \ (lj-J~ , / ' / , ; ~\ (1)11 ... l!unJnpUO:l Za l!ufpnpUOJ la II Z S:JINOlli.:J3"l3 M3MOd 0.1 NOIl.JOamUNI , , ; , ; , ; " (!)V~ " A-" A IVA £L'I (!)JlOO'I~ " A (11"",\ " A (!)rod~ 212 ELECTRIC MACHINERY RJNDAMENTALS Posit holf Negal half ,,, ! lrSCRI ,,, ! lr SCR! SCR I conducting , " , ~ , ~ , ~ , .-,-'/ '" V-SCR2 lrSCR~ , , , , , ----~ -- - -~ , ""GURE 3-61 , , , , lr SCR3 lrSC~ (a) ,b, " , + Lo'" }~(" -(a) A three-phase full-wave SCR bridge cirwit connected to a resistive load. (b) The operation of the positive half of the SCRs. Assume that initially SCRI is conducting. If SCR1 is triggered at any time after point I. then SCRI will be reverse-biased and shut off. (c) The operation of the negative half of the SCRs. Assume that initially SC~ is conducting. If SCR! is triggered at any time after point I. then SC~ will be reverse-biased and shut off. , , , , , , , , v " , , , , , , , , , , " , , , , , , , , , , , , INTRODUCTION TO POWER ELECTRONICS , , , , , , , , , , , -, , , , , , , , , , , , , , , , , ' , 'i " , " ," \ ' , , , \ " " '\ " / ' , " ,\ " f', ' " \ , \ 1 " I \'\1'/\"", \/ 'f \/ ,I \/ \' " \' " 't' ,I '/ 213 , , , , " I, /' \ I, I' , I' '\ " '\ " " f ,\ '\ " "1, ' I "'\1' I, I," I ", \' I", ,'\' ' I \/ 1/ \/ \/ 1/ \/ \' I ,/ ! ' , " \ ," l ,\ " 1\ " , " ,\ '\ I' " ,\ \ " I", , II ,I \1\'1 \'\/11\/\"1 , " 'I ,I ,I \1 ' I , ) / \/ ' I ,I " / , I , , /' X /, >( >, / ,/" ", ,/ ~" /" // '" /" / " ~- ," "~ - / ,,- - /" "~ - -" ,,~" '~ - -~ FIGURE 3-62 Approximating a linearly decreasing voltage with the three-phase full-wave SCR bridge circuit. 214 ELECTRIC MACHINERY RJNDAMENTALS Positive group Negative group j ;;<) , SCRI r SCR2 r SCR1 r SCR7 SCR, SCR, -' -' -' + ,,(') ( Lood -SCR4 r SCR ~ r SC~ r SCRIO SCRll SCRl2 J J J ""GURE 3-63 One phase of a noncin:ulating current cycloconvener cin:ui1. I f two of these SCR bridge circuits are connected in parallel with opposite polarities, the result is a noncirculating current cycloconverter. Noncirculating Current Cycloconverters One phase of a typical noncirculating current cycJoconverter is shown in Figure 3--63. A fuJi three-phase cycloconverter consists of three identical units of this type. Each unit consists of two three-phase full-wave SCR bridge circuits, one conducting current in the positive direction (the positive group) and one conduct-ing current in the negative direction (the negative group).llle SCRs in these cir-cuits are triggered so as to approximate a sinusoidal output voltage, with the SCRs in the positive group being triggered when the current fl ow is in the positive di-rection and the SCRs in the negative group being triggered when the current flow is in the negative direction. The resulting output voltage is shown in Figure 3--64. As can be seen from Figure 3--64, noncirculating current cycJoconverters produce an output voltage with a fairly large harmonic component. TIlese high harmonics limit the output frequency of the cycloconverter to a value less than about one-third of the input frequency. In addition, note that current flow must switch from the positive group to the negative group or vice versa as the load current reverses direction. The cycJo-converter pulse-control circuits must detect this current transition with a current polarity detector and switch from triggering one group of SCRs to triggering the other group. There is generally a brief period during the transition in which nei-ther the positive nor the negative group is conducting. This current pause causes additional glitches in the output waveform. TIle high harmonic content, low maximum frequency, and current glitches associated with noncirculating current cycloconverters combine to limit their use. INTRODUCTION TO POWER ELECTRONICS 21 5 c--- Negative + Positive -------------------group group FIGURE 3-64 The output voltage and current from a noncin:ulating current cycJoconvener connected to an inductive load. Note the switch from the operation of the negative group to the operation of the positive group at the time the current changes direction. In any practical noncircu lating current cyc1oconverter, a fi Iter (usually a series in-ductor or a transformer) is placed between the output of the cyc1oconverter and the load, to suppress some of the output hannonics. Circulating Current Cyclocollver ters One phase of a typical circulating current cyc1oconverter is shown in Figure 3- 65. It differs from the noncirculating current cyc1oconverter in that the positive and negative groups are connected through two large inductors, and the load is sup-plied from center taps on the two inductors. Unlike the noncirculating current cy-c1oconverter, both the positive and the negative groups are conducting at the same time, and a circulating current fl ows around the loop fonned by the two groups and the series inductors. The series inductors must be quite large in a circuit ofthis sort to limit the circulating current to a safe value. The output voltage from the circulating current cyc1oconverter has a smaller hannonic content than the output voltage from the noncirculating current cyc1o-converter, and its maximum frequency can be much higher. It has a low power factor due to the large series inductors, so a capacitor is often used for power-factor compensation. The reason that the circulating current cyc1oconverter has a lower hannonic content is shown in Figure 3--66. Figure 3--66a shows the output voltage of the positive group, and Figure 3--66b shows the output voltage of the negative group. The output voltage V1oad(t) across the center taps of the inductors is _ vpo,(t) -vrw:s(t) Vtoait) -2 (3- 9) N -~ v, v, Vc ~ ~ ~ isolation transformer HGURE 3-65 SCR[ V,\ V C/ SCR4 + "~ lr SCR2 lr SCR3 lr lr SC R ~ lr SCR6 lr "'(1) '" Vp:>«t) -'>' .. g(t) o -One phase of a six-pulse type of cin:ulaling current cycloconverter. L, "., + J SCR7 J SCR8 J SCI<, v, \ + V/ Lo., J seRlO J seRll .J SCRl2 > .. --L, INTRODUCTION TO POWER ELECTRONICS 217 Many of the high-frequency harmonic components which appear when the posi-ti ve and negative groups are examined separately are common to both groups. As such, they cancel during the subtraction and do not appear at the tenninals of the cycloconverter. Some recirculating current cycloconverters are more complex than the one shown in Figure 3- 65. With more sophisticated designs, it is possible to make cycloconverters whose maximum output frequency can be even higher than their input frequency. These more complex devices are beyond the scope of this book. , , , , , , , , , , , , , , / , , , , , , , , , , , ,\ 1\ ,\ 1\ 1 \ , \ "",,\ \1\'\1\ \1 \' " \ " \' \, \, , > ' \ 1 \ '\ ,\ I' \ 1\ '\ ,\ 1\ " \1\ 1,1'" I \ I \ I \, \ \/ ~/ ~I \I ,/ ) ' i< >( / /' .... /" .... /', /' / , " .... -" '" ,. ,"--' ,"--,,/ .... , , , , , , , , v " , , , , , , , , , , , , " , " .... -" FIGURE 3-66 , " , , , , , , , , , (a) / -' ,b, , , " , , , , , , '-' Voltages in the six-pulse circulating current cycloconverter. (a) The voltage out of the positive group; (b) the voltage out of the negative group. 218 ELECTRIC MACHINERY RJNDAMENTALS , ,-~, ,--, ,--,'-' , ,' - ' ,.- , ',,' ' ,/' '/' '" , , " ' "-A P;:-'/ /' /\ /\ \/ I\/\/~/ /\/\ \'~/\/'/ , " 1 \ 'I '/ \1 ~ 'I \ '( \ I' , ' /' , , " I / ' , I' 1"'1' \ I' \/\/\/\/1/1/1 , 1/ ' 1 ' / 1 "1 ' \ II 'I 'I \1 " ' \ I \1 \1 \1 \ " 1\ " II I, I' I ,II II II " I , /1/ /\/\/\,'\/\'\1 / 'I ,I 'I \, \' \' " \' " " '-/ ,I I'-, > 'I I' , /\ I, I' ,I /' I, , " /, /' /' /' /' ,/,1,1'/','", / 'I" 'I , ,/ ~I ,I ~I V ,/ )/ )l -( ( /~, ,A, ,', /' /, /, " '--" '--" ,-,-,~.," '- - ," '-.'/ HGURE 3--66 (concluded) (e) the resulting load voltage. 3.8 HARMONIC PROBLEMS " ) Power electronic components and circuits are so flexible and useful that equip-menl control led by them now makes up 50 to 60 percent of the total load on most power systems in the developed world. As a result, the behavior of these power electronic circuits strongly inn uences the overall operation of the power systems that they are connected to. TIle principal problem associated with power electronics is the harmonic components of voltage and current induced in the power system by the switching transients in power electronic controllers. TIlese hannonics increase the total cur-rent flows in the lines (especially in the neutral of a three-phase power system). The extra currents cause increased losses and increased heating in power system components, requiring larger components to supply the same total load. In addi-tion, the high neutral currents can trip protective relays, shutting down portions of a power system. As an example of this problem, consider a balanced three-phase motor with a wye connection that draws 10 A at full load. When this motor is connected to a power system, the currents flowing in each phase will be equal in magnitude and 120 0 out of phase with each other, and the return current in the neutral will be 0 (see Figure 3--67). Now consider the same motor supplied with the same total power through a rectifier-inverter that pn:x:luces pulses of current. TIle currents in the power line now are shown in Figure 3--68. Note that the nns current of each line is still lOA, but the neutral also has an rms current of 15 A! The current in the neutral consists entirely of hannonic components. Ie. /" '--o o I u<t.un:J r-'--o o IU o "1 o , v ) o "1 o , o o Ie--/' 1 "---o o o , ~ ') o , 219 N N o 2' ) , ) 5 , ~ 0 5 - 10 ) - I , - 2' ) 20 , ) , • ~ ) 5 - 10 ) - I 5 - 20 Time ,,' Time ,b, 20 1 }, 1,,1 ' I 'I' I' 'I I \ I \ 10 1 I III I III ! :11 I r I II 1 II II 1 I u -51 I II I I 1 I I - 10 I II I I \ I I I - 151 I" " I \ i \ I -20 1 \I \/1 Time 'e' 20 1 1\1 It h 11\ 1/\ 11\1 ' I I I III II I I I II I II II 10 1 I I III II I I I II I II II ! :1 11\ II \ II \ II \ I \ III u -51 I II I II II I II I I I II I - 10 II I II I I I I I I II I II I I I - 15 II I II I I I I I I II I II I I I - 20 I \I \I \I \I \I I " I Time ,d , Jo'IG URE 3-68 Current flow for a balanced three-phase. wye<onnected mOlOr connected to the power line through a power electronic controller that produces current pulses: (a) phase a; (b) phaseb; (e) phasec; (d) neutral The nlls current flow in phases a. b, and c is \0 A. wh.ile the nns current flow in the neutral is 15 A. INTRODUCTION TO POWER ELECTRONICS 221 The spectra of the currents in the three phases and in the neutral are shown in Figure 3--69. For the motor connected directly to the line, only the fundamen-tal frequency is present in the phases, and nothing at all is present in the neutral. For the motor connected through the power controller, the current in the phases in-cludes both the fundamental frequency and all of the odd hannonics. TIle current in the neutral consists principally of the third, ninth, and fifteenth harmonics. Since power electronic circuits are such a large fraction of the total load on a modern power system, their high hannonic content causes significant problems for the power system as a whole. New standards have been created to limit the amount of harmonics produced by power electronic circuits, and new controllers are designed to minimize the hannonics that they produce. 3.9 SUMMARY Power electronic components and circui ts have produced a m~ o r revolution in the area of motor controls during the last 35 years or so. Power electronics provide a convenient way to convert ac power to dc power, to change the average voltage level of a dc power system, to convert dc power to ac power, and to change the frequency of an ac power system. The conversion of ac to dc power is accomplished by rectifier circuits, and the resulting dc output voltage level can be controlled by changing the firing times of the devices (SCRs, TRIACs, GTO thyristors, etc.) in the rectifier circuit. Adjustment of the average dc voltage level on a load is accomplished by chopper circuits, which control the fraction of time for which a fixed dc voltage is applied to a load. Static frequency conversion is accomplished by either rectifier-inverters or cycloconverters. Inverters are of two basic types: externally cornrnutated and self-commutated. Externally commutated inverters rely on the attached load for com-mutation voltages; self-commutated inverters either use capacitors to produce the required commutation voltages or use self-commutating devices such as GTO thyristors. Self-commutated inverters include current source inverters, voltage source inverters, and pulse-width modu lation inverters. Cycloconverters are used to directly convert ac power at one frequency to ac power at another frequency. There are two basic types of cycloconverters: non-circulating current and circulating current. Noncirculating current cycloconverters have large harmonic components and are restricted to relatively low frequencies. In addition, they can suffer from glitches during current direction changes. Circu-lating current cycloconverters have lower hannonic components and are capable of operating at higher frequencies. TIley require large series inductors to limit the circulating current to a safe value, and so they are bulkier than noncirculating cur-rent cycloconverters of the same rating. See IEC l00Q.3-2. EMC: Part 3. Section 2. "Limits for harmonic current emission (equipment input current s 16 A per phase)," and ANSI/IEEE Standard 519-1992, "IEEE recommended practices and requiremems for harmonic control in power systems." N N N 30.000 J 25.000 J 20.000 i J .• '0.. 15.000 ~ • J lOilOO 0 '.000 " 45.000 40.000 35.000 30.000 .a 25.000 o ~ ~ 20.000 15ilOO lOilOO '.000 J o 2 4 o -2 4 6 8 10 12 14 Harrnonicnumber ,,' I -6 -8 \0 12 14 Harmonic number 'e' 16 16 25.000 J 20.000 • " 15.000 , .• f ) 10000 ) 5.000 J o FIGURE 3-69 -2 • 4 I • 6 8 10 12 14 16 Harmonic number ,b, (a) The spectrum of the phase current in the balanced three-phase. wye-connected motor connected directly to the power line. Only the fundamental frequency is present. (b) The spectrum of the phase current in the balanced three-phase. wye-<:onnected motor connected through a power electronic controller that produces current pulses. T he fundamental frequency and all odd harmonics are pre.-;ent. (c) The neutral current for the motor connected through a electronic power controller. The third. ninth. and fifteenth harmonics are present in the current INTRODUCTION TO POWER ELECTRONICS 223 QUESTIONS 3-1. Explain the operation and sketch the output characteristic of a diode. 3-2. Explain the operation and sketch the output characteristic of a PNPN diode. 3-3. How does an SCR differ from a PNPN diode? When does an SCR conduct? 3-4. What is a GTO thyristor? How does it differ from an ordinary three-wire thyristor (SCR)? 3-5. What is an IG8T? What are its advantages compared to other power electronic devices? 3-6. What is a DIAC? A TRIAC? 3-7. Does a single-phase full-wave rectifier produce a better or worse dc output than a three-phase half-wave rectifier? Why? 3-8. Why are pulse-generating circuits needed in motor controllers? 3-9. What are the advantages of digital pulse-generating circuits compared to analog pUlse-generating circuits? 3-10. What is the effect of changing resistor R in Figure 3- 32? Explain why this effect occurs. 3-11. What is forced conunutation? Why is it necessary in dc-to-dc power-control circuits? 3-12. What device(s) could be used to build dc-to-dc power-control circuits without forced conunutation? 3-13. What is the purpose of a free-wheeling diode in a control circuit with an inductive load? 3-14. What is the effect of an inductive load on the operation of a phase angle controller? 3-15. Can the on time of a chopper with series-capacitor commutation be made arbitrarily long? Why or why not? 3-16. Can the on time of a chopper with parallel-capacitor conunutation be made arbitrar-ily long? Why or why not? 3-17. What is a rectifier-inverter? What is it used for? 3-18. What is a current-source inverter? 3-19. What is a voltage-source inverter? Contrast the characteristics of a VSI with those of a CSI. 3-20. What is pulse-width modulation? How do PWM inverters compare to CSI and VSI inverters? 3-21. Are power transistors more likely to be used in PWM inverters or in CSI inverters? Why? PROBLEMS 3-1. Calculate the ripple factor of a three-phase half-wave rectifier circuit. both analyti-cally and using MATLAB. 3-2. Calculate the ripple factor of a three-phase full-wave rectifier circuit. both analyti-cally and using MATLAB. 3-3. Explain the operation of the circuit shown in Figure P3-1. What would happen in this circuit if switch S, were closed? 224 ELECTRIC MACHINERY RJNDA MENTALS + 21 D, • • I + ) '~(') ") Y. Lood S, /-(-~I SCR 0, v,...(l) '" 339 sin 3771 V C2 C] FlGURE P3-1 The cin:uit of Problems 3- 3 through 3--6. 3-4. What would the nns voltage on the load in the circuit in Figure P3-1 be if the firing angle of the SCR were (a) 0°, (b) 30°, (c) 90°? '3-5. For the circuit in Figure P3-1, assume that V BO for the DlAC is 30 V, Ct is I p.F, R is adjustable in the range I to 20 ill, and switch SI is open. What is the firing angle of the circuit when R is 10 kO? What is the rillS voltage on the load lUlder these con-ditions? (Caution: This problem is hard to solve analytically because the voltage charging the capacitor varies as a function of time.) 3-6. One problem with the circuit shown in Figure P3-1 is that it is very sensitive to vari-ations in the input voltage v.At). For example, suppose the peak. value of the input voltage were to decrease. Then the time that it takes capacitor C] to charge up to the breakover voltage of the DIAC will increase, and the SCR will be triggered later in each half-cycle. Therefore, the rillS voltage supplied to the load will be reduced both by the lower peak voltage and by the later firing. This same effect happens in the opposite direction if voc(t) increases. How could this circuit be modi fied to reduce its sensitivity to variations in input voltage? 3-7. Explain the operation of the circuit shown in Figure P3- 2, and sketch the output voltage from the circuit. 3-8. Figure P3- 3 shows a relaxation oscillator with the following parameters: R] = variable C= IJ.tF VBO = 30V R2 = 1500 0 Voc = 100 V lH = 0.5 mA (a) Sketch the voltages vc(t), vo(t), and rrJt) for this circuit. (b) If R, is currently set to 500 kO, calculate the period of this relaxation oscillator. 3-9. In the circuit in Figure P3-4, T] is an autotransformer with the tap exactly in the center of its winding. Explain the operation of this circuit. Assruning that the load is inductive, sketch the voltage and current applied to the load. What is the purpose of SCR2? What is the purpose of D2? (This chopper circuit arrangement is known as a Jones circuit.) The asterisk in front of a problem number indicates that it is a more difficult problem. INTRODUCTION TO POWER ELECTRONICS 225 • "~---T ~ • • c, Lood --.,'----.L T, T, FIGURE 1'3-2 The inverter circuit of Problem 3- 7. +~-----; Voc'" lOOV " J + + Vc (I) ( '>'0 (/) R2", 1500 n C'" 1.0~ FIGURE 1'3-3 The relaxation oscillator circuit of Problem 3--8. + J SCR] , icC ~CR2 D, • T] (Autotnlnsformer) D, Lo,d FlGURE P3-4 The chopper ci["(;uit of Problem 3- 9. 226 ELECTRIC MACHINERY RJNDAMENTALS 3-10. A series-capacitor forced commutation chopper circuit supplying a purely resistive load is shown in Figure P3- 5. Voc = 120 V lH = 8 rnA VBO = 200 V R l =20 kll Rlood = 250 0 C = 150 /lF (a) When SCRl is turned on. how long will it remain on? What causes it to tum off? (b) When SCRl turns off. how long will it be until the SCR can be turned on again? (Assume that 3 time constants must pass before the capacitor is discharged.) (c) What problem or problems do these calculations reveal about this simple series-capacitor forced-commutation chopper circuit ? (d) How can the problem(s) described in part c be eliminated? + SCR 0---/ + R\ C ) " -v oc / D ,~ ~ Rwm Lo,' \ ""GURE 1'3-5 The simple series-capacitor forced-commulation cin:uit of Problem 3-10. 3-11. A parallel-capacitor forced-conunutation chopper circuit supplying a purely resis-tive load is shown in Figure P3-6. Voc = 120 V lH = 5 rnA V BO = 250V R] =20kfi Rlood = 250 0 C= 15 /lF (a) When SCR] is turned on, how long will it remain on? What causes it to IlUll off? (b) What is the earliest time that SCR] can be turned off after it is turned on? (Assume that 3 time constants must pass before the capacitor is charged.) (c) When SCR] turns off, how long will it be until the SCR can be tlUlled on again? (d) What problem or problems do these calculations reveal about this simple parallel-capacitor forced-commutation chopper circuit? (e) How can the problem(s) described in part d be eliminated? 3-12. Figure P3- 7 shows a single-phase rectifier-inverter circuit. Explain how this circuit ftmctions. What are the purposes of C] and C2? What controls the output frequency of the inverter? INTRODUCTION TO POWER ELECTRONICS 227 + + , D RLOAD " ~ R, " -C -+ 0--/ SCRl 0--/ FIGURE P3-6 The simple parallel-capacitor forced commutation circuit of Problem 3- 11. FIGURE P3-7 The single-phase rectifier-inverter circuil of Problem 3- 12. '3- 13. A simple full-wave ac phase angle voltage controller is shown in Figure P3-8. The component values in this circuit are R = 20 to 300 kf.!. currently set to 80 kf.! C = 0.1 5 p.,F V BO = 40 V (for PNPN diode DJ) V BO = 250 V (for SCR l ) rs(t) = VM sin wt V where VM = 169.7 V and w = 377 radls (a) At what phase angle do the PNPN diode and the SCR tum on? (b) What is the rms voltage supplied to the load under these circwnstances? '3- 14. Figure P3-9 shows a three-phase full -wave rectifier circuit supplying power to a dc load. The circuit uses SCRs instead of diodes as the rectifying elements. 228 ELECTRIC MACHINERY RJNDAMENTALS + R SCR I ) vD (t) v,m C D, ""GURE I'J-S The full-wave phase angle voltage controller of Problem 3- t3. lr r lr SCRI SCR2 SCR3 + Lo., ir' r I r SCR, SCR3 SC", ""GURE PJ-9 The lhree-phase full-wave reclifier circuit of Problem 3- t4. (a) What will the nns load voltage and ripple be if each SCR is triggered as soon as it becomes forward-biased? At what phase angle should the SCRs be triggered in order to operate this way? Sketch or plot the output voltage for this case. (b) What will the rms load voltage and ripple be if each SCR is triggered at a phase angle of 90° (that is, halfway through the half-cycle in which it is forward bi-ased)? Sketch or plot the output voltage for this case. '3-15. Write a MATLAB program that imitates the operation of the pulse-width modula-tion circuit shown in Figure 3-55, and answer the following questions. (a) Assume that the comparison voltages vjt) and vI') have peak amplitudes of 10 V and a frequency of 500 Hz. Plot the output voltage when the input voltage is Vinet) = 10 sin 2'lT ft V, andf = 60 Hz . (b) What does the spectrwn of the output voltage look like? What could be done to reduce the hannonic content of the output voltage? (c) Now assume that the frequency of the comparison voltages is increased to 1000 Hz. Plot the output voltage when the input voltage is Vinet) = 10 sin 2'lTft V and/ = 60 Hz. (d) What does the spectrum of the output voltage in c look like? (e) What is the advantage of using a higher comparison frequency and more rapid switching in a PWM modulator? INTRODUCTION TO POWER ELECTRONICS 229 REFERENCES I. Dewan. S. 8.. G. R. Siemon. and A. Straughen. Power Semiconductor Drives. New York: Wiley-Interscience.1984. 2. IEEE. Graphic Symbols for Electrical and Electronics Diagrams. IEEE Standard 315-19751ANSI Standard Y32.2-1975. 3. Millman. Jacob. and Christos C. Halkias. Integrated Electronics: Analog and Digital Circuits and Systems. New York: McGraw-Hill. 1972. 4. Vithayathil. Joseph. Pov.·er Electronic:;: Principles and Applications. New York: McGraw-Hill. 1995. 5. Werninck. E. H. (ed.). Electric Motor Handbook. London: McGraw-Hill. 1978. CHAPTER 4 AC MACHINERY FUNDAMENTALS AC machines are generators that convert mechanical energy to ac electrical energy and motors that convert ac electrical energy to mechanical en-ergy. The fundamental principles of ac machines are very simple, but unfortu-nately, they are somewhat obscured by the complicated construction of real ma-chines. This chapter will first explain the principles of ac machine operation using simple examples, and then consider some of the complications that occur in real ac machines. TIlcre are two major classes of ac rnachines-synchronous machines and in-duction machines. Synchronous machines are motors and generators whose mag-netic field current is supplied by a separate de power source, while induction ma-chines are motors and generators whose field current is supplied by magnetic induction (transformer action) into their fi eld windings. The field circuits of most synchronous and induction machi nes are located on their rotors. nlis chapter cov-ers some of the fundamentals common to both types of three-phase ac machines. Synchronous machines will be covered in detail in Chapters 5 and 6, and induc-tion machines will be covered in Chapter 7. 4.1 A SIMPLE LOOP IN A UNIFORM MAGNETIC FIELD We wil l start our study of ac machines with a simple loop of wire rotating within a uniform magnetic field. A loop of wire in a uniform magnetic field is the sim-plest possible machine that produces a sinusoidal ac voltage. nlis case is not rep-resentative of real ac machines, since the flux in real ac machines is not constant in either magnitude or direction. However, the factors that control the voltage and torque on the loop wi ll be the same as the factors that control the voltage and torque in real ac machines. 230 ACMACHINERYFUNDAMENTALS 231 rom -------= ----------~---~----~~~ VcJ \ a/ -----L -----------T --- ' .... - .... N s -~ ---------------------ii- "----" is a uniform magnetic field, aligned as shown. , ., FIGURE 4- 1 , + d I , I , I , I , I , I 'h' b , + r- " A simple rotating loop in a uniform magnetic field. (a) Front view; (b) view of coil. Figure 4- 1 shows a simple machi ne consisting of a large stationary magnet producing an essentially constant and uniform magnetic field and a rotating loop of wire within that field. The rotating part of the machine is called the rotor, and the stationary part of the machine is called the stator. W e will now deterrnine the voltages present in the rotor as it rotates within the magnetic field. The Voltage Induced in a Simple Rotating Loop I f the rotor of this machine is rotated, a voltage will be induced in the wire loop. To detennine the magnitude and shape of the voltage, examine Figure 4- 2. 1lle loop of wire shown is rectangular, with sides ab and cd perpendicular to the plane of the page and with sides be and da parallel to the plane of the page. The mag-netic field is constant and unifonn, pointing from left to right across the page. To determine the total voltage e,OI on the loop, we will examine each seg-ment of the loop separately and sum all the resulting voltages. The voltage on each segment is given by Equation (1-45): eind = (v x H) -' ( 1-45) I. Segment abo In this segment, the velocity of the wire is tangential to the path of rotation, while the magnetic field B points to the right, as shown in Figure 4- 2b. The quantity v x B points into the page, which is the same direction as segment abo Therefore, the induced voltage on this segment of the wire is eoo = (v x H) ·' = vBI sin (Jab into the page (4-1) 2. Segment be. In the first half of this segment, the quantity v x B points into the page, and in the second half of this segment, the quantity v x B points out of 232 ELECTRIC MACHINERY RJNDAMENTALS 8 (a) (b) ( ,) ""GURE 4-2 (a) Velocities and oriemations of the sides of the loop with respect to the magnetic field. (b) The direction of motion with respect to the magnetic field for side abo (c) The direction of motion with respect to the magnetic field for side cd. the page. Since the length I is in the plane of the page, v x B is perpendicular to I for both portions of the segment. lllerefore the voltage in segment be will be zero: (4-2) ), Segment ed, In this segment, the velocity of the wire is tangential to the path of rotation, while the magnetic field B points to the right, as shown in Figure 4- 2c. The quantity v x B points into the page, which is the same direction as segment ed. TIlerefore, the induced voltage on this segment of the wire is edc = (v x B) ·1 = vBI sin (Jed out of the page (4-3) 4. Segment da. Just as in segment be, v x B is perpendicular to I. TIlerefore the voltage in this segment will be zero too: ead = 0 (4-4) 1lle total induced voltage on the loop ei!>d is the sum of the voltages on each of its sides: = vBI sin (Jab + vBI sin (Jed (4-5) Note that (Jab = 180 0 -(Jed, and recall the trigonometric identity sin (J = sin (180 0 -(J). Therefore, the induced voltage becomes eind = 2vBL sin (J (4-6) 1lle resulting voltage eind is shown as a function of time in Figure 4- 3. TIlere is an alternative way to express Equation (4-6), which clearly relates the behavior of the single loop to the behavior of larger, real ac machines. To de-rive this alternative expression, examine Figure 4- 2 again. If the loop is rotating at a constant angular velocity w, then angle (J of the loop will increase linearly with time. In other words, ACMACHINERYFUNDAMENT ALS 233 • 2 FIGURE 4-3 Plot of e ... versus a. 3 • -,-(J = wt e. radians Also, the tangential velocity v of the edges of the loop can be expressed as v = rw (4-7) where r is the radius from axis of rotation out to the edge of the loop and w is the an-gular velocity of the loop. Substituting these expressions into Equation (4-6) gives e;nd = 2rwBI sin wi ( 4-!l) Notice also from Figure 4-1 b that the area A of the loop is just equal to 2rl. Therefore, e;nd = ABw sin wt (4-9) Finally, note that the maximum flux through the loop occurs when the loop is per-pendicular to the magnetic flux density lines. This flux isjust the product of the loop's surface area and the flux density through the loop. q,max = AB Therefore, the final fonn of the voltage equation is I e;nd q,rmu.w sin wt I (4-1 0) (4-11 ) Thus, the voltage generated in the loop is a sinusoid whose magnitude is equal to the product oftheJ1ux inside the machine and the speed of rotation of the machine. This is also true of real ac machines. In general, the voltage in any real machine will depend on three factors: I. TIle flux in the machine 2. TIle speed of rotation 3. A constant representing the construction of the machine (the number of loops, etc.) 234 ELECTRIC MACHINERY RJNDAMENTALS ------------------------,,---~ ---~R---~ ... \ ---J~: \ ""'""--(j I ----~--------!"-~---, --~---------------------------II n is a uniform magnetic field. aligned as shown. The x in a wire indicates current flowing into the page. and the • in a wire indicates current flowing out of the page. (a) HGURE 4-4 , , I , I I , I , d -I , I ,b, A current-carrying loop in a unifonn magnetic field. (a) Front view; (b) view of coil. I into page ~ ' .~ , F ,., l out of page ,,' ,"'IGURE 4-5 ~ r. ,"' into page n roc'" 0 ,b, r. ,"' out of page • ,d, b , I, " (a) Derivation of force and torque on segment ab. (b) Derivation of force and torque on segment bc. (c) Derivation of force and torque on segment cd. (d) Derivation of force and torque on segment da. The Torque Induced in a Current-Carrying Loop Now assume that the rotor loop is at some arbitrary angle () with respect to the magnetic field, and that a current i is fl owing in the loop, as shown in Figure 4--4. If a current flows in the loop, then a torque will be induced on the wire loop. To detennine the magnitude and direction of the torque, examine Figure 4- 5. The force on each segment of the loop will be given by Equation (1--43), F = i(l x B) ( 1-43) where ACMACHINERYFUNDAMENTALS 235 i = magnitude of current in the segment I = length of the segment, with direction of I defined to be in the direction of current flow B = magnetic flux density vector The torque on that segment will then be given by 7" = (force applied)(perpendicular distance) = (F) (r sin (j) = rF sin (j (1-6) where (J is the angle between the vector r and the vector F. The direction of the torque is clockwise if it would tend to cause a clockwise rotation and counter-clockwise if it wou Id tend to cause a counterclockwise rotation. I. Segment abo In this segment, the direction of the current is into the page, while the magnetic field B points to the right, as shown in Figure 4- 5a. The quantity I x B points down. Therefore, the induced force on this segment of the wire is TIle resulting torque is F =i(lxB) = ilB down 7"ab = (F) (r sin (jab) = rilB sin (jab clockwise (4-1 2) 2. Segment be. In this segment, the direction of the current is in the plane of the page, while the magnetic field B points to the right, as shown in Figure 4- 5b. TIle quantity I x B points into the page. Therefore, the induced force on this segment of the wire is F =i(lxB) = ilB into the page For this segment, the resulting torque is 0, since vectors r and I are parallel (both point into the page), and the angle (jbc is O. 7"bc = (F) (r sin (jab) ~O (4-1 3) 3. Segment ed. In this segment, the direction of the current is out of the page, while the magnetic field B points to the right, as shown in Figure 4- 5c. The quantity I x B points up. Therefore, the induced force on this segment of the wire is F =i(lxB) = ilB up 236 ELECTRIC MACHINERY RJNDAMENTALS The resulting torque is Ted = (F) (r sin Oed) = rilB sin Oed clockwise (4-1 4) 4. Segmentda. In this segment. the direction of the current is in the plane of the page, while the magnetic field B points to the right, as shown in Figure 4- 5d. The quantity I x B points out of the page. 1llerefore, the induced force on this segment of the wire is F = i(l x B) = ilB out of the page For this segment, the resulting torque is 0, since vectors r and I are parallel (both point out of the page), and the angle 000 is O. Too = (F) (r sin O "J ~O (4- 15) 1lle total induced torque on the loop Tind is the sum of the torques on each of its sides: = rilB sin O ab + rilB sin O ed Note that O ab = O c.t, so the induced torque becomes TiDd = 2rilB sin 0 (4-1 6) (4-1 7) TIle resulting torque TiDd is shown as a function of angle in Figure 4-6. Note that the torque is maximum when the plane of the loop is parallel to the magnetic field, and the torque is zero when the plane of the loop is perpendicular to the mag-netic field. TIlere is an alternative way to express Equation (4-17), which clearly re-lates the behavior of the single loop to the behavior of larger, real ac machines. To derive this alternative expression, examine Figure 4-7. I f the current in the loop is as shown in the figure, that current will generate a magnetic flux density Bloop with the direction shown. The magnitude of Bloop will be _ 1!i.. Bloop -G where G is a factor that depends on the geometry of the loop. Also, note that the area of the loop A is just equal to 2rl. Substituting these two equations into Equa-tion (4-17) yields the result (4- 18) If the loop were a cirde. then G", 2r. where r is the radius of the circle. so B""" '" lJ.inr. For a rec-tangular loop. the value of G will vary depending on the exact length-to-width ratio of the loop. FIGURE 4-6 Plot of 1' ... versus (J . .... (.) (b) ACMACHINERYFUNDAMENTALS 237 e. radians Jo'IGURE4-7 Derivation of the induced torque equation. (a) The current in the loop produces a magnetic flul( density "loop perpendicular to the plane of the loop; (b) geometric relationship between Dioop and Os. (4-1 9) where k = AGIJ1 is a factor depending on the construction of the machine, Bs is used for the stator magnetic field to distinguish it from the magnetic field gener-ated by the rotor, and () is the angle between B loop and Bs. The angle between B loop and Bs can be seen by trigonometric identities to be the same as the angle () in Equation (4-1 7). Both the magnitude and the direction of the induced torque can be deter-mined by expressing Equation (4--19) as a cross product: (4- 20) Applying this equation to the loop in Fig ure 4- 7 produces a torque vector into the page, indicating that the torque is clockwise, with the magnitude given by Equa-tion (4-1 9). Thus, the torque induced in the loop is proportional to the strength of the loop's magnetic field, the strength of the external magnetic field, and the sine of the angle between them. This is also true of real ac machines. In general, the torque in any real machine will depend on four factors: 238 ELECTRIC MACHINERY RJNDAMENTALS I. The strength of the rotor magnetic field 2. The strength of the external magnetic field 3. The sine of the angle between them 4. A constant representing the construction of the machine (geometry. etc.) 4.2 THE ROTATING MAGNETIC FIELD In Section 4.1, we showed that if two magnetic fields are present in a machine, then a torque will be created which will tend to line up the two magnetic fields. If one magnetic field is produced by the stator of an ac machine and the other one is produced by the rotor of the machine, the n a torque will be induced in the rotor which will cause the rotor to turn and align itself with the stator magnetic field. I f there were some way to make the stator magnetic field rotate, then the in-duced torque in the rotor would cause it to constantly "chase" the stator magnetic field around in a circle. lllis, in a nutshell, is the basic principle of all ac motor operation. How can the stator magnetic field be made to rotate? llle fundamental prin-ciple of ac machine operation is that if a three-phase set of currents, each of equal mngnitude and differing in phase by 120°,flows in a three-phase winding, then it will produce a rotating mngnetic field of constant mngnitude. The three-phase winding consists of three separate wi ndi ngs spaced 120 electrical degrees apart around the surface of the machine. llle rotating magnetic field concept is illustrated in the simplest case by an empty stator contai ning just three cai Is, each 120 0 apart (see Figure 4-8a). Since such a winding produces only one north and one south magnetic pole, it is a two-pole winding. To understand the concept of the rotating magnetic field, we will apply a set of currents to the stator of Figure 4--8 and see what happens at specific instants of time. Assume that the currents in the three coils are given by the equations iaa' (1) = 1M sin wt A ibb' (1) = 1M sin (wt - 120°) icc' (1) = 1M sin (wt - 240°) A A (4- 2 I a) (4- 21 b) (4- 2Ic) llle current in coil aa' flows into the a end of the coil and out the a' end of the coil. It produces the magnetic field inte nsity A ·turns/ m (4- 22a) where 0° is the spatial angle of the magnetic field intensity vector, as shown in Figure 4-8b. llle direction of the magnetic field intensity vector Had(t) is given by the right-hand rule: If the fingers of the right hand curl in the direction of the current flow in the coil, then the resulting magnetic field is in the direction that the thumb points. Notice that the magnitude of the magnetic field intensity vector H"..,(l) varies sinusoidally in time, but the direction of Had(l) is always constant. Similarly, the magnetic field intensity vectors Hb/;o,(l) and H«(l) are ACMACHINERYFUNDAMENT ALS 239 ° , ° b' FIGURE 4- 8 11",, -(/) 11",-(/) " ",,(0 0" "I o a' (a) A simple three-phase stator. Currents in this stator are assumed positive if they flow into the unprimed end and out the primed end of the coils. The magnetizing intensities produced by each coil are also shown. (b) The magnetizing intensity vector H .... (/) produced by a current flowing in coil 00'. A · turns/ m (4-22b) A·turns/ m (4- 22c) The flux densities resulting from these magnetic field intensities are give n by Equation ( 1-2 1): They are Baa' (1 ) = BM sin wl L 0° T Bbb' (1 ) = BM sin (wl- 120°) L 120° BC<", (1 ) = BM sin (wl- 240°) L 240° T T ( 1-21) (4- 23a) (4-23b) (4- 23c) where BM = J1HM . 1lle currents and their corresponding flux densities can be ex-amined at specific times to detennine the resulting net magnetic field in the stator. For example, at time wt = 0°, the magnetic field from coil ad will be B"",,= 0 The magnetic field from coil bb' will be BbI>' = BM sin (_1 20°) L 120 0 and the magnetic field from coil ee' wi] I be BC<", = BM sin (_240°) L 240° (4- 24a) (4-24b) (4- 24c) 240 ELECTRIC MACHINERY RJNDAMENTALS , '" b' ""GURE 4-9 0, WI = 0" (a) o b o , o '" b' n cr . ~ R~ " w '" , c '" , WI =90° (b) (a) The vector magnetic field in a stator at time WI = 0°. (b) The vector magnetic field in a sta.tor a.t time WI = 90°. TIle total magnetic field from all three coils added together wi ll be Bne! = Baa' + Bw + B ee' = 0 + (-f BM ) L 120° + (f BM) L240° = 1.5BML-9Q0 TIle resulting net magnetic field is shown in Figure 4- 9a. b As another example, look at the magnetic field at time wt = 90°. At that time, the currents are i",,' = 1M sin 90° A i"",= IM sin (- 300) A iec,= IM sin (-1 500) A and the magnetic fields are B"",= BM LO° B"", = -0.5 BM L 120 0 Bec' = -0.5 BM L 240 0 The resulting net magnetic field is Bn .. = Baa' + B"". + B ..... = BM L 0° + (-O.5BM) L 120° + (-O.5BM ) L 240° = 1.5 BM LO° ACMACHINERYFUNDAMENTALS 241 The resulting magnetic field is shown in Figure 4- 9b. Notice that although the di-rection of the magnetic field has changed, the magnitude is constant. TIle mag-netic field is maintaining a constant magnitude while rotating in a counterclock-wise direction. Proof of the Rotating Magnetic Field Concept At any time t, the magnetic field wi ll have the same magnitude I.5BM , and it wi ll continue to rotate at angu lar velocity w. A proof of this statement for all time t is now given. Refer again to the stator shown in Figure 4-8. In the coordinate system shown in the fi gure, the x direction is to the right and the y direction is upward. TIle vector:l1 is the unit vector in the horizontal direction, and the vector S' is the unit vector in the vertical direction. To find the total magnetic flux density in the stator, simply add vectorially the three component magnetic fields and detennine their sum. The net magnetic nux density in the stator is given by B •• (I) ~ B, (I) + B~, (I) + B~ , (I) = BM sin wt LO° + BMsin (wt - 120°) L 120° + BM sin (wi_ 240°) L 2400T Each of the three component magnetic fields can now be broken down into its x and y components. Bnet(t) = BM sin wt x -[O.5BM sin (wt -1200)]x + ['] BM sin (wt -1200) ]y -[O.5BM sin (wt - 2400)]x -['] BM sin (wt - 2400) ]y Combining x and y components yields Dnet(t) = [BM sin wi -O.SBM sin (wt -120°) - O.5BM sin (wt - 240°)] x + [ '7 BMsin(wt -120°) - '7 BMsin (wt - 2400) ]y By the angle-addition trigonometric ide ntities, BnetCt) = [BM sin wi + iBM sin wt + 1 BM cos wi + i BM sin wt -1BM cos wt]x + [-1BMsinwt - ~BMCOSWt + ~BM sinwt - ~BMCOSWt] S' I Bneln = (1.5BM sin wt):I1 - (1.5BM cos wI)y I (4- 25) Equation (4- 25) is the final expression for the net magnetic flux density. Notice that the magnitude of the field is a constant I.SBM and that the angle changes con-tinually in a counterclockwise direction at angular velocity w. Notice also that at 242 ELECTRIC MACHINERY RJNDAMENTALS N -\ 0 .\ b ""GURE 4-10 The rotating magnetic field in a stator represented as moving north and south stator poles. wi = 0°, BDeI = I .SBM L _90° and that at wt = 90°, Bne, = 1.58M L 0°. 1l1ese re-sults agree with the specific examples examined previously. The Relationship between Electrical Frequency and the Speed of Magnetic Field Rotation Figure 4-1 0 shows that the rotating magnetic field in this stator can be represented as a north pole (where the flux leaves the stator) and a south pole (where the flux enters the stator). These magnetic poles complete one mechanical rotation around the stator surface for each electrical cycle of the applied current. 1l1erefore, the mechanical speed of rotation of the magnetic field in revolutions per second is equal to the electric frequency in hertz: two poles two poles (4- 26) (4- 27) Here 1m and w,., are the mechanical speed in revol utions per second and radians per second, while!. and W e are the electrical speed in hertz and radians per second. Notice that the windings on the two-pole stator in Figure 4- 10 occur in the order (taken counterclockwise) a-c'-b-a '-c-b' What would happen in a stator if this pattern were repeated twice within it? Fig-ure 4-ll a shows such a stator. There, the pattern of windings (taken counter-clockwise) is a-c '-b-a' -c-b '-a-c '-b-a '-c-b' which is just the pattern of the previous stator repeated twice. When a three-phase set of currents is applied to this stator, two north poles and two south poles are pro-duced in the stator winding, as shown in Figure 4-11 b. In this winding, a pole ACMACHINERYFUNDAMENTALS 243 b, b,~ ~ @ ~ s @ " ai / II " "; • 0 0 0 w. w. 0 "i / oi " i!1 @ ~i!1 b, b, b' '\ll (a) (b) , " b , " b Back: 1 '"' of stator coils • • 8 • X • X • II ,s, II ,N , I-I ,s, II ,N , , , , , , , , , ! ! ! I I I ", '; b, "i " bi ", 'i b, "; C2 bi ) j j I , b , Counterclock:wise b' f') FIGURE 4- 11 (a) A simple four-pole stator winding. (b) The resulting stator magnetic poles. Notice that there are moving poles of alternating polarity every 90° around the stator surface. (c) A winding diagram of the stator as seen from its inner surface, showing how the stator currents produce north and south magnetic poles. moves only halfway around the stator surface in one electrical cycle. Since one electrical cycle is 360 electrical degrees, and since the mechanical motion is 180 mechanical degrees, the relationship between the electrical angle O e and the me-chanical angle 0", in this stator is (4- 28) Thus for the four-pole winding, the electrical frequency of the current is twice the mechanical frequency of rotation: 244 ELECTRIC MACHINERY RJNDAMENTALS fe = 2fm four poles W e = 2wm four poles (4- 29) (4- 30) I n general, if the number of magnetic poles on an ac machine stator is P, then there are PI2 repetitions of the winding sequence a-c '-b-a '-e-b' around its inner surface, and the electrical and mechanical quantities on the stator are related by le, ~ iem l (4- 3 1) (4- 32) (4- 33) Also, noting that fm = n,,/60, it is possible to relate the electrical frequency in hertz to the resulting mechanical speed of the magnetic fields in revol utions per minute. nlis relationship is Reversing the Direction of Magnetic Field Rotation (4- 34) Another interesting fact can be observed about the resulting magnetic field.lfthe current in any two of the three coils is swapped, the direction of the mngnetie field's rotation will be reversed. This means that it is possible to reverse the direc-tion of rotation of an ac motor just by switching the connections on any two of the three coils. lllis result is verified below. To prove that the direction of rotation is reversed, phases bb' and ee' in Fig-ure 4-8 are switched and the resulting flux density Bn .. is calculated. llle net magnetic flux density in the stator is given by B ... ,(t) = B"".(t) + Bw(t) + BeAt) = BM sin wi L 0° + BM sin (wi- 240°) L 120° + BM sin (wI- 120°) L 240° T Each of the three component magnetic fie lds can now be broken down into its x and y components: BDeI(t) = BM sin wt5i. -[0.5BM sin (wt -2400)] 5i. + [ 1" BM sin (wt -240 0) ]y -[0.5BM sin (wt -I 200)] 5i. - [ 1" BM sin (wt -120 0) ]y Combining x and y components yields ACMACHINERYFUNDAMENTALS 245 Hnem = [BM sin wt - O.5BM sin (wt -240°) - 0.5BM sin(WI' -1200jx + [ '7 BM sin (WI' -240°) - '7 BM sin (wt -l200)]y By the angle-addition trigonometric identities, S nelt) = [BM sin WI' + iBM sin wt -1 BM cos WI' + iBM sin wt + IBM cos wt]x + [-1BMsinwt + ~BM COS Wt + ~BM sinwt + ~BMCOSWt]S I S nell) = (1.5BM sin wt) J1 + (1.5BM cos WI')Y I (4- 35) This time the magnetic fie ld has the same magnitude but rotates in a clock-wise direction. 1l1erefore, switching the currents in two stator phases reverses the direction of magnetic field rotation in an ac machine. EXllmple 4-1. Create a MATLAB program that models the behavior of a rotating magnetic field in the three-phase stator shown in Figure 4-9. Solutioll The geometry of the loops in this stator is fIXed as shown in Figure 4-9. The currents in the loops are i"",(t) = 1M sin wt A iw(t) = 1M sin (wt - 120°) iec,(t) = 1M sin (wt - 240°) and the resulting magnetic flux densities are B"",(t)= BM sinwt LO° T A A B"",(t) = BM sin (wt - 120°) L 120° Bec,(t) = BM sin (wt - 240°) L 240° d = vB&I cos wmt + vB&I cos wmt = 2vB&I cos wmt (4--43) Since the velocity of the end conductors is given by v = rwm , Equation (4--43) can be rewritten as ei!>d = 2(rwm)B&I cos wmt = 2rlB~m cos wmt Finally, the flux passing through the coil can be expressed as <p = 2rlBm (see Problem 4- 7), while W m = W e = W for a two-pole stator, so the induced voltage can be expressed as I eind -¢w cos wi I (4-44) Equation (4--44) describes the voltage induced in a single-turn coil. If the coil in the stator has Nc turns of wire, then the total induced voltage of the coil will be I eind -Nc¢w cos wt I (4-45) Notice that the voltage produced in stator of this simple ac machine wind-ing is sinusoidal with an amplitude which depends on the flux <p in the machine, the angular velocity w of the rotor, and a constant depending on the construction of the machine (Nc in this simple case). This is the same as the result that we ob-tained for the simple rotating loop in Section 4.1. Note that Equation (4--45) contains the term cos wt instead of the sin wt found in some of the other equations in this chapter. 1lle cosine tenn has no spe-cial significance compared to the sine- it resulted from our choice of reference direction for 0: in this derivation. If the reference direction for 0: had been rotated by 90° we would have had a sin wt tenn. The Induced Voltage in a Three-Phase Set of Coils If three coils, each of Nc turns, are placed around the rotor magnetic field as shown in Figure 4-1 6, then the voltages induced in each of them will be the same in magnitude but wi II differ in phase by 120°. 1lle resulting voltages in each of the three coils are e .... .(t) = Nc <Pw sin wt V ew(r) = Nc <Pw sin (wt - 120°) ee,,,(t) = Nc <pw sin (wt- 2400) v V (4--46a) (4-46b) (4--46c) Therefore, a three-phase sct of currents can generate a unifonn rotating magnetic field in a machine stator, and a uniform rotating magnetic field can gen-erate a three-phase sct of voltages in such a stator. 254 ELECTRIC MACHINERY RJNDAMENTALS ~.M FIGURE 4- 16 The production of three-phase voltages from three coils spaced 120° apan. The RMS Voltage in a Three-Phase Stator TIle peak voltage in any phase of a three-phase stator of this sort is Since w = 2nf, this equation can also be written as Enuu = 27rNc <pf TIlerefore, the nns voltage of any phase of this three-phase stator is 2,,-EA = \lfNc<pf IE, - !2,,-Ncf I (4-47) (4-48) (4-49) (4- 50) TIle nns voltage at the terminnls of the machine will depend on whether the stator is Y- or .1.-connected. I fthe machine is V-connected, then the tenninal voltage will be V3 times EA; if the machine is .1.-connected, then the tenninal voltage will just be equal to EA. Example 4-2. The following information is known about the simple two-pole generator in Figure 4--16. The peak flux density of the rotor magnetic field is 0.2 T, and the mechanical rate of rotation of the shaft is 3600 r/min. The stator diameter of the machine is 0.5 m, its coil length is 0.3 m, and there are 151lU1ls per coil. The machine is V-connected. (a) What are the three phase voltages of the generator as a ftmction of time? (b) What is the nns phase voltage of this generator? (c) What is the nns tenninal voltage of this generator? Solutioll The flux in this machine is given by <P = 2rlB = dlB ACMACHINERYFUNDAMENTALS 255 where d is the diameter and I is the length of the coil. Therefore, the flux in the machine is given by <p = (0.5 mXO.3 m)(0.2 T) = 0.03 \Vb The speed of the rotor is given by w = (3600 r/minX27T radXI minl60 s) = 377 radls (a) The magnitudes of the peak. phase voltages are thus Emu = Nc<Pw = (15 turnsXO.03 Wb)(377 radls) = 169.7 V and the three phase voltages are e"".(t) = 169.7 sin 377t V ebb.(t) = 169.7 sin (377t -1200) V e,At) = 169.7 sin (377t - 240°) V (b) The nns phase voltage of this generator is Ell = E~x = 16~V = 120V (c) Since the generator is V-connected, VT = v'5EIl = 0(120 V) = 208 V 4.5 INDUCED TORQUE IN AN AC MACHINE In ac machines under nonnaI operating conditions, there are two magnetic fields present--.:1. magnetic field from the rotor circuit and another magnetic field from the stator circuit. The interaction of these two magnetic fields produces the torque in the machine, just as two pennanent magnets near each other will experience a torque which causes them to line up. Figure 4-1 7 shows a simplified ac machine with a sinusoidal stator flux dis-tribution that peaks in the upward direction and a single coil of wire mounted on the rotor. TIle stator flux distribution in this machine is Bs<,.a ) = Bs sin a (4- 51) where Bs is the magnitude of the peak flux density; B:!..a ) is positive when the flux density vector points radially outward from the rotor surface to the stator surface. How much torque is produced in the rotor of this simplified ac machine? To find out, we will analyze the force and torque on each of the two conductors separately. The induced force on conductor I is F = i(l x B) = ilBs sin a The torque on the conductor is "TiDd.] = (r x F) = rilBs sin a with direction as shown counterclockwise ( 1-43) 256 ELECTRIC MACHINERY RJNDAMENTALS a IUia)1 '" Bs sin a ""GURE4- 17 A simplified ac machine with a sinusoidal sta.tor flux distribution a.nd a single coil of wire mounted in the rotor. TIle induced force on conductor 2 is F = i(lxB) = ilBs sin a TIle torque on the conductor is 1";oo.t = (r x F) = rilBs sin a with direction as shown counterclockwise TIlerefore, the torque on the rotor loop is l1"ind = 2rilBs sin a counterclockwise I ( 1-43) (4- 52) Equation (4- 52) can be expressed in a more convenient fonn by examining Figure 4-1 8 and noting two facts: I. The current i flowing in the rotor coil produces a magnetic field of its own. The direction of the peak of this magnetic field is given by the right-hand rule, and the magnitude of its magnetizing intensity HR is directly propor-tional to the current flowing in the rotor: ACMACHINERYFUNDAMENT ALS 257 FIGURE 4- 18 B , , , , , , , , \ II .. , '-j<" _ f a _ "'-----l _ _ _ , , , , , , I I1~ H R r= 180" - a The components magnetic flux density inside Ihe machine of Figure 4--17. HR = Ci where C is a constant of proportionality. (4- 53) 2. 1lle angle between the peak of the stator flux density Bs and the peak of the rotor magnetizing intensity HR is y. Furthennore, y=180o-a sin y= sin ( 1 80° a) = sin 0: (4- 54) (4- 55) By combining these two observations, the torque on the loop can be expressed as "Tioo = KH I13s sin a counterclockwise (4- 56) where K is a constant dependent on the construction of the machine. Note that both the magnitude and the direction of the torque can be expressed by the equation I "Tind -KHR X Bs I (4- 57) Finally, since B R = /LHR, this equation can be reexpressed as (4- 58) where k = KIp. Note that in general k will not be constant, since the magnetic per-meability p varies with the amount of magnetic saturation in the machine. Equation (4--58) is just the same as Equation (4--20), which we derived for the case of a single loop in a unifonn magnetic field. It can apply to any ac machine, not 258 ELECTRIC MACHINERY RJNDAMENTALS just to the simple one-loop rotor just described. Only the constant k will differ from machine to machine. This equation will be used only for a qualitative study of torque in ac machines, so the actual val ue of k is unimportant for our purposes. TIle net magnetic field in this machine is the vector sum of the rotor and sta-tor fields (assuming no saturation): B ... , = BR + Bs (4- 59) TIlis fact can be used to produce an equivalent (and sometimes more useful) ex-pression for the induced torque in the machine. From Equation (4- 58) "TiDd = kB R X Bs But from Equation (4- 59), Bs = BDe, - BR , so "Tind = kBR X (BDe, - BR) = k(BR x B"",) - k(BR x BR) Since the cross prOOuct of any vector with itself is zero, this reduces to (4- 58) (4-60) so the induced torque can also be expressed as a cross product of BR and BDe, with the same constant k as before. The magnitude of this expression is (4-6 1) where /j is the angle between BR and B ... ,. Equations (4-58) to (4-6 1) will be used to help develop a qualitative un-derstanding of the torque in ac machines. For example, look at the simple syn-chronous machine in Figure 4-1 9. Its magnetic fields are rotating in a counter-clockwise direction. What is the direction of the torque on the shaft of the machine's rotor? By applying the right-hand rule to Equation (4- 58) or (4-60), the induced torque is found to be clockwise, or opposite the direction of rotation of the rotor. Therefore, this machine must be acting as a generator. 4.6 WINDING INSULATION IN AN ACMACHINE One of the most critical parts of an ac machine design is the insulation of its wind-ings. If the insulation of a motor or generator breaks down, the machine shorts out. The repair ofa machine with shorted insulation is quite expensive, ifit is even possible. To prevent the winding insulation from breaking down as a result of overheating, it is necessary to limit the temperature of the windings. TIlis can be partially done by providing a cooling air circulation over them, but ultimately the maximum winding temperature limits the maximum power that can be supplied continuously by the machine. w <8> 0, , , , , , , r ! B , , , , , ACMACHINERYFUNDAMENTALS 259 FlGURE 4- 19 A simplified synchronous machine showing its rotor and stator magnetic fields. Insulation rarely fails from immediate breakdown at some critical tempera-ture. Instead, the increase in temperature produces a gradual degradation of the in-sulation, making it subject to failure from another cause such as shock, vibration, or electrical stress. 1l1ere was an old rule of thumb that said that the life ex-pectancy of a motor with a given type of insulation is halved for each 10 percent rise in temperature above the rated temperature of the winding. This rule still ap-plies to some extent today. To standardize the temperature limits of machine insulation, the National Electrical Manufacturers Association (NEMA) in the United States has defined a series of insulation system classes. Each insulation system class specifies the maximum temperature rise pennissible for that class of insulation. 1l1ere are three common NEMA insulation classes for integral-horsepower ac motors: 8, F, and H. Each class represents a higher pennissible winding temperature than the one before it. For example, the annature winding temperature rise above ambient tem-perature in one type of continuously operating ac induction motor must be limited to 80°C for class 8, 105°C for class F, and 125°C for class H insulation. The effect of operating temperatu re on insulation life for a typical machine can be quite dramatic. A typical curve is shown in Figure 4-20. This curve shows the mean life of a machine in thousands of hours versus the temperature of the windings, for several different insulation classes. The specific temperature specifications for each type of ac motor and gen-erator are set out in great detail in NEMA Standard MG 1-1993, Motors and Gen-erators. Similar standards have been defined by the International Electrotechnical Commission (IEC) and by various national standards organizations in other countries. ~ < B u g ~ '--... i'-~ "" 8 ------"'-1 \ 0 ~ \ '--... i'-1 \ ~ 0 , N , • 0 ~ M • , ij ~ "" ~ ------1 \ ~ ~ g \ 0 00 \ ~ 8 0 ~ spuemoljlll! SJIloH 260 AC MACHINERY FUNDAMENTALS 261 4.7 AC MACHINE POWER FLOWS AND LOSSES AC generators take in mechanical power and produce electric power, while ac motors take in electric power and produce mechanical power. In either case, not all the power input to the machine appears in useful form at the other end- there is always some loss associated with the process. The efficiency of an ac machine is defined by the equation ?"UI " ~ - X 100% P in (4-62) The difference between the input power and the output power of a machine is the losses that occur inside it. lllerefore, (4-63) The Losses in AC Machines The losses that occur in ac machines can be divided into four basic categories: I. Electrical or copper losses (/ 2 R losses) 2. Core losses 3. Mechanical losses 4. Stray load losses ELECTRICAL OR COPPER LOSSES. Copper losses are the resistive heating losses that occur in the stator (annature) and rotor (field) windings of the machine. TIle sta-tor copper losses (SCL) in a three-phase ac machine are given by the equation (4-64) where IA is the current flowing in each annature phase and Rio. is the resistance of each armature phase. The rotor copper losses (RCL) of a synchronous ac machine (induction ma-chines wi ll be considered separately in Chapter 7) are given by (4-65) where IF is the current flowing in the field winding on the rotor and RF is the re-sistance of the field winding. The resistance used in these calculations is usually the winding resistance at nonnal operating temperature. CORE LOSSES. The core losses are the hysteresis losses and eddy current losses occurring in the metal of the motor. These losses were described in Chapter I. 262 ELECTRIC MACHINERY RJNDAMENTALS TIlese losses vary as the square of the flux density (8 2) and, for the stator, as the l.5th power of the speed of rotation of the magnetic fields (nI. 5) . MECHANICAL LOSSES. The mechanical losses in an ac machine are the losses associated with mechanical effects. There are two basic types of mechanical losses: friction and windage. Friction losses are losses caused by the friction of the bearings in the machine, while windage losses are caused by the friction between the moving parts of the machine and the air inside the motor's casing. TIlese losses vary as the cube of the speed of rotation of the machine. TIle mechanical and core losses ofa machine are often lumped together and called the no-load rotationnlloss of the machine. At no load, all the input power must be used to overcome these losses. Therefore, measuring the input power to the stator of an ac machine acting as a motor at no load will give an approximate value for these losses. STRAY LOSSES (OR MISCELLANEOUS LOSSES). Stray losses are losses that cannot be placed in one of the previous categories. No matter how carefully losses are accounted for, some always escape inclusion in one of the above categories. All such losses are lumped into stray losses. For most machines, stray losses are taken by convention to be I percent of full load. The Power-Flow Diagram One of the most convenient techniques for accounting for power losses in a ma-chine is the power-flow diagram. A power-flow diagram for an ac generator is shown in Figure 4-21 a. In this figure, mechanical power is input into the machine, and then the stray losses, mechanical losses, and core loses are subtracted. After they have been subtracted, the remaining power is ideally converted from me-chanical to electrical fonn at the point labeled P f!¥' TIle mechanical power that is converted is given by (4-66) and the same amount of electrical power is produced. However, this is not the power that appears at the machine's terminals. Before the tenninals are reached, the electrical12R losses must be subtracted. In the case of ac motors, this power-flow diagram is simply reversed. The power-flow diagram for a motor is shown in Figure 4- 21 b. Example problems involving the calculation of ac motor and generator effi-ciencies will be given in the next three chapters. 4.8 VOLTAGE REGULATION AND SPEED REGULATION Generators are often compared to each other using a figure of merit called voltage regulation. Voltage regulation (VR) is a measure of the ability of a generator to keep a constant voltage at its terminals as load varies. It is defmed by the equation Stray losses losses Pi. '" 3V.JA cos 6 '" ..fjVdL cos 6 I PR losses FIGURE 4- 21 C~ losses (a) Core ACMACHINERYFUNDAMENTALS 263 PR losses P 00' ",3V.1.ot cos 60r ..fjVdL cos 6 losses losses ,b, (a) The power-flow diagram of a three-phase ac generator. (b) The power-flow diagram of a three-phase ac motor. I VR = v..l Vfl Vfl X 1 00% I (4-67) where V. t is the no-load tenninal voltage of the generator and Vfl is the full-load tenninal voltage of the generator. It is a rough measure of the shape of the gener-ator's voltage-current characteristic-a positive voltage regulation means a drooping characteristic, and a negative voltage regulation means a rising charac-teristic. A small VR is "better" in the sense that the voltage at the tenninals of the generator is more constant with variations in load. Similarly, motors are often compared to each other by using a figure of merit called speed regulation. Speed regulation (SR) is a measure of the ability of a motor to keep a constant shaft speed as load varies. It is defined by the equation 100% 1 (4-68) x (4-69) 264 ELECTRIC MACHINERY RJNDAMENTALS It is a rough measure of the shape of a motor's torque-speed characteristic- , positive speed regulation means that a motor's speed drops with increasing load, and a negative speed regulation means a motor's speed increases with increasing load. TIle magnitude of the speed regulation tells approximately how steep the slope of the torque-speed curve is. 4.9 SUMMARY There are two major types of ac machines: synchronous machines and induction machines. The principal difference between the two types is that synchronous ma-chines require a dc field current to be supplied to their rotors, while induction ma-chines have the field current induced in their rotors by transfonner action. TIley will be explored in detail in the next three chapters. A three-phase system of currents supplied to a system of three coi Is spaced 120 electrical degrees apart on a stator wi ll produce a unifonn rotating magnetic field within the stator. The direction of rotation of the magnetic field can be re-versed by simply swapping the connections to any two of the three phases. Con-versely, a rotating magnetic field wi II produce a three-phase set of voltages within such a set of coils. In stators of more than two poles, one complete mechanical rotation of the magnetic fields produces more than one complete electrical cycle. For such a sta-tor, one mechanical rotation produces PI2 electrical cycles. Therefore, the electri-cal angle of the voltages and currents in such a machine is related to the mechan-ical angle of the magnetic fields by P (J~ = "2(Jm TIle relationship between the electrical frequency of the stator and the mechanical rate of rotation of the magnetic fields is ".p f~ = 120 TIle types of losses that occur in ac machines are electrical or copper losses (PR losses), core losses, mechanical losses, and stray losses. E.1.ch of these losses was described in this chapter, along with the definition of overall machine effi-ciency. Finally, voltage regulation was defined for generators as 1 VR = Vol V Il V Il x 1 00% 1 and speed regulation was defined for motors as ISR = nal nn nil x 100% 1 ACMACHINERYFUNDAMENTALS 265 QUESTIONS 4-1. What is the principal difference between a synchronous machine and an induction machine? 4-2. Why does switching the current flows in any two phases reverse the direction of ro-tation of a stator's magnetic field? 4-3, What is the relationship between electrical frequency and magnetic field speed for an ac machine? 4-4. What is the equation for the induced torque in an ac machine? PROBLEMS 4-1. The simple loop rotating in a lUlifonn magnetic field shown in Figure 4--1 has the following characteristics: B =0.5Ttotheright r = O.lm l = 0.5 m w = 103 radls (a) Calculate the voltage elOl(f) induced in this rotating loop. (b) Suppose that a 5-0 resistor is COIlllected as a load across the terminals of the loop. Calculate the current that would flow through the resistor. (c) Calculate the magnitude and direction of the induced torque on the loop for the conditions in b. (d) Calculate the electric power being generated by the loop for the conditions in b. (e) Calculate the mechanical power being consumed by the loop for the conditions in b. How does this nwnber compare to the amount of electric power being gen-erated by the loop? 4-2. Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14 poles operating at frequencies of 50, 60, and 400 Hz. 4-3. A three-phase, four-pole winding is installed in 12 slots on a stator. There are 40 turns of wire in each slot of the windings. All coils in each phase are cOIlllected in series, and the three phases are connected in.6.. The flux per pole in the machine is 0.060 Wh, and the speed of rotation of the magnetic field is 1800 rhnin. (a) What is the frequency of the voltage produced in this winding? (b) What are the resulting phase and tenninal voltages of this stator? 4-4. A three-phase, Y-COIlllected, 50-Hz, two-pole synchronous machine has a stator with 2()(x) turns of wire per phase. What rotor flux would be required to produce a tenni-nal (line-to-line) voltage of 6 kV? 4-5. Modify the MATLAB problem in Example 4--1 by swapping the currents flowing in any two phases. What happens to the resulting net magnetic field? 4-6. If an ac machine has the rotor and stator magnetic fields shown in Figure P4--1, what is the direction of the induced torque in the machine? Is the machine acting as a mo-tor or generator? 4-7. The flux density distribution over the surface of a two-pole stator of radius rand length l is given by B = BMCOs(W.,f-a) Prove that the total flux lUlder each pole face is (4--37b) 266 ELECTRIC MACHINERY RJNDAMENTALS H , B •• 0 , 0 , 0 , , , , y , , w B, o ""GURE 1)~- 1 The ac machine of Problem 4--6. 4--8. In the early days of ac motor development. machine designers had great difficulty controlling the core losses (hysteresis and eddy currents) in machines. They had not yet developed steels with low hysteresis. and were not making laminations as thin as the ones used today. To help control these losses. early ac motors in the United States were run from a 25-Hz ac power supply. while lighting systems were run from a separale 60-Hz ac power supply. (a) Develop a table showing the speed of magnetic field rotation in ac machines of 2.4.6.8. 10. 12. and 14 poles operating at 25 Hz. What was the fastest rota-tional speed available to these early motors? (b) For a given motor operating at a constant flux density B, how would the core losses of the motor flmning at 25 Hz compare to the core losses of the motor nmning at 60 Hz? (c) Why did the early engineers provide a separate 60-Hz power system for lighting? REFERENCES I. Del Toro. Vincent. Electric Machines and Po .... er Systetru. Englewood Cliffs. N.J.: Prentice-Halt. 1985. 2. Fitzgerald. A. E.. and Charles Kingsley. Electric Machinery. New Y ork: McGraw-Hill. 1952. 3. Fitzgerald. A. E.. Charles Kingsley. and S. D. Umans. Electric Machinery. 5th ed .. New Y ork: McGraw-Hill. 1990. 4. International Electrotechnical Commission. Rotating Electrical Machines Part I: Rating and Perfortnllnce. IEC 34-1 (RI994). 1994. 5. Liwschitz-Garik. Michael. and Clyde Whipple. Altunating-Current Machinery. Princeton. N.J.: Van Nostrand. 1961. 6. McPherson. George. An Introduction to Electrical Machines and Transformers. New Y ort: Wiley. 1981. 7. National Electrical Manufacturers Association. Motors and Gl'nerators, Publication MG 1-1993. Washington. D.C.. 1993. 8. Werninck. E. H. (ed.). Electric Motor Handbook. London: McGraw-Hill. 1978. CHAPTER 5 SYNCHRONOUS GENERATORS S ynchronous generators or alternntors are synchronous machines used to con-vert mechanical power to ac electric power. This chapter explores the opera-tion of synchronous generators, both when operating alone and when operating to-gether with other generators. 5.1 SYNCHRONOUS GENERATOR CONSTRUCTION In a synchronous generator, a de current is applied to the rotor winding, which produces a rotor magnetic field. The rotor of the generator is then turned by a prime mover, producing a rotating magnetic field within the machine. This rotat-ing magnetic field induces a three-phase set of voltages within the stator windings of the generator. Two terms commonly used to describe the windings on a machine arefield windings and armature windings. In general, the tenn "field windings" applies to the windings that produce the main magnetic field in a machine, and the term "armature windings" applies to the windings where the main voltage is induced. For synchronous machines, the field windings are on the rotor, so the tenns "rotor windings" and "field windings" are used interchangeably. Similarly, the terms "stator windings" and "annature windings" are used interchangeably. The rotor of a synchronous generator is essentially a large electromagnet. The magnetic poles on the rotor can be of either salient or nonsalient construction. The tenn salient means "protruding" or "sticking out," and a salient pole is a mag-netic pole that sticks out from the surface of the rotor. On the other hand, a 267 268 ELECTRIC MACHINERY RJNDAMENTALS o 0, s End View Side View fo'IGURE 5-1 A non salient two-pole rotor for a synchronous machine. nonsalient pole is a magnetic pole constructed flush with the surface of the rotor. A nonsalient-pole rotor is shown in Figure 5-1 , while a salient-pole rotor is shown in Figure 5- 2. Nonsalient-pole rotors are nonnally used for two- and four-pole ro-tors, while salient-pole rotors are nonnally used for rotors with four or more poles. Because the rotor is subjected to changing magnetic fields, it is constructed of thin laminations to reduce eddy current losses. A de current must be supplied to the field circuit on the rotor. Since the ro-tor is rotating, a special arrangement is required to get the de power to its field windings. There are two common approaches to supplying this dc power: I. Supply the dc power from an external dc source to the rotor by means of slip rings and brushes. 2. Supply the dc power from a special de power source mounted directly on the shaft of the synchronous generator. Slip rings are metal rings completely encircling the shaft of a machine but in-sulated from it. One end of the dc rotor winding is tied to each of the two slip rings on the shaft of the synchronous machine. and a stationary brush rides on each sli p ring. A "brush" is a block of graphitelike carbon compound that conducts electric-ity freely but has very low friction. so that it doesn't wear down the slip ring. If the positive end of a dc voltage source is connected to one brush and the negative end is connected to the other, then the same dc voltage wi II be applied to the field wind-ing at all times regardless of the angu lar position or speed of the rotor. Slip rings and brushes create a few problems when they are used to supply dc power to the field windings of a synchronous machine. TIley increase the amount of maintenance required on the machine, since the brushes must be checked for wear regularly. In addition, brush voltage drop can be the cause of significant power losses on machines with larger field currents. Despite these problems, slip rings and brushes are used on all smaller synchronous machines, because no other method of supplying the dc field current is cost-effective. On larger generators and motors, brnshless exciters are used to supply the dc field current to the machine. A brushless exciter is a small ac generator with its (a) HGURE 5-2 Slip rings (a) A salient six-pole rotor for a synchronous ntachine. (b) Photograph of a salient eight-pole synchronous ntachine rotor showing the windings on the individual rotor poles. (Courtesy of Geneml Electric Company. ) (e) Photograph of a single S3.lient pole front a rotor with the field SYNCHRONOUS GENERATORS 269 ,b, ,d, windings not yet in place. (Courtesy ofGeneml Electric Company.) (d) A single salient pole shown after the field windings are installed but before it is mounted on the rotor. (Courtesy ofWestinglwuse Electric Company.) field circuit mounted on the stator and its armature circuit mounted on the rotor shaft. The three-phase output of the exciter generator is rectified to direct current by a three-phase rectifier circuit also mounted on the shaft of the generator, and is then fed into the main dc field circuit. By controlling the small dc field current of the exciter generator (located on the stator), it is possible to adjust the field current on the main machine without slip rings and brushes. This arrangement is shown schematically in Figure 5-3, and a synchronous machine rotor with a brushless exciter mounted on the same shaft is shown in Figure 5-4. Since no mechanical contacts ever occur between the rotor and the stator, a brushless exciter requires much less maintenance than slip rings and brushes. 270 ELECTRIC MACHINERY RJNDAMENTALS Exciter Exciter armature :L Three-phase rectifier r , , , , I" ---,-Synchronous machine Main Field -----------------~-------------+----------------N ~h Three-phase input (low current) ""GURE 5-3 fu , citer ,ld fi ,------, Three-phase output Maln annature A brush less exciter circuit. A small thrre-phase current is rectified and used to supply the field circuit of the exciter. which is located on the stator. The output of the armature cirwit of the exciter (on the rotor) is then rectified and used to supply the field current of the main machine. ""GURE 5-4 Photograph of a synchronous machine rotor with a brush less exciter mounted on the same shaft. Notice the rectifying electronics visible next to the armature of the exciter. (Courtesy of Westinghouse Electric Company.) Pilot exciter Pilot exciter field Permanent magnets I I I SYNCHRONOUS GENERATORS 271 Exciter , I Synchronous : generator , , Exciter armature ----I- Main field , , , , Th~ , ph~ , , rectifier , , , , --r-, --~-----------~----------~----------r--------------------I Three-phase" Pllot eXCl1er annature FIGURE 5-5 Threo· ph~ rectifier , rO~"='~P"~'~~==+ ' I ~\ -4 R, , , -t-ExcIter field Main armature A brushless excitation scheme that includes a pilot exciter. The permanent magnets of the pilot exciter produce the field current of the exciter. which in turn produces the field current of the main machine. To make the excitation of a generator completely independent of any exter-nal power sources, a small pilot exciter is often included in the system. Apilot ex-citer is a small ac generator with permanent magnets mounted on the rotor shaft and a three-phase winding on the stator. It produces the power for the field circuit of the exciter, which in turn controls the field circuit of the main machine. I f a pilot exciter is included on the generator shaft, then no external electric power is required to run the generator (see Figure 5-5). Many synchronous generators that include brushless exciters also have slip rings and brushes, so that an auxiliary source of dc field current is available in emergencies. The stator of a synchronous generator has already been described in Chap-ter 4, and more details of stator construction are found in Appendix B. Synchro-nous generator stators are nonnally made of prefonned stator coils in a double-layer winding. The winding itself is distributed and chorded in order to reduce the hannonic content of the output voltages and currents, as described in Appendix B. A cutaway diagram of a complete large synchronous machine is shown in Figure 5-6. This drawing shows an eight-pole salient-pole rotor, a stator with dis-tributed double-layer windings, and a brushless exciter. 272 ELECTRIC MACHINERY RJNDAMENTALS ""GURE 5-6 A cutaway diagram of a large synchronous machine. Note the saliem·pole construction and the on· shaft exciter. (Courtesy ofGl'neral Electric Company.) 5.2 THE SPEED OF ROTATION OF A SYNCHRONOUS GENERATOR Synchronous generators are by defmition synchronous, meaning that the electrical frequency produced is locked in or synchronized with the mechanical rate of rotation of the generator. A synchronous generator's rotor consists of an electro-magnet to which direct current is supplied. TIle rotor's magnetic field points in whatever direction the rotor is turned. Now, the rate of rotation of the magnetic fields in the machine is related to the stator electrical frequency by Equation (4-34): (4- 34) where !. = electrical frequency, in Hz nm = mechanical speed of magnetic field, in r/min (equals speed of rotor for synchronous machines) P = number of poles Since the rotor turns at the same speed as the magnetic field, this equation relates the speed of rotor rotation to the resulting electrical frequency. Electric power is generated at 50 or 60 Hz, so the generator must turn at a fixed speed depending on the number of poles on the machine. For example, to generate 60-Hz power in a two-pole machine, the rotor must turn at 3600 r/min. To generate 50-Hz power in a four-pole machine, the rotor must turn at 1500 rIm in. TIle required rate of rota-tion for a given frequency can always be calculated from Equation (4- 34). ; , ,' FIGURE 5-7 SYNCHRONOUS GENERATORS 273 '" '" "")'DC (constant) ..--,b , (a) Plot of flux versus field current for a synchronous generator. (b) The magnetization curve for the synchronous generator. 5.3 THE INTERNAL GENERATED VOLTAGE OFASYNCHRONOUSGENERATOR In Chapler 4, the magnitude of the voltage induced in a given stator phase was found to be (4- 50) This voltage depends on the flux ~ in the machine, the frequency or speed of ro-lation, and the machine's construction. In solving problems with synchronous ma-chines, this equation is sometimes rewritten in a simpler fonn that emphasizes the quantities that are variable during machine operalion. This simpler form is I EA ~ Kw I (5-1 ) where K is a constant representing the construction of the machine. If w is ex-pressed in electrical radians per second, then Nc K ~ V2 (5- 2) while if w is expressed in mechanical radians per second, then Nc P K ~ V2 (5- 3) The internal generated voltage EA is directly proportional to the flux and to the speed, but the flux itself depends on the current fl owing in the rotor field cir-cuit. The field circuit IF is related to the flux ~ in the manner shown in Fig-ure 5- 7a. Since EA is directly proportional to the flux, the internal generated volt-age EA is related to the field current as shown in Figure 5- 7b. lllis plot is called the magnetization cUIYe or the open-circuit characteristic of the machine. 274 ELECTRIC MACHINERY RJNDAMENTALS 5.4 THE EQUIVALENT CIRCUIT OF A SYNCHRONOUS GENERATOR The voltage EA is the internal generated voltage produced in one phase ofa syn-chronous generator. However, this voltage EA is not usually the voltage that ap-pears at the terminals of the generator. In fact, the only time the internal voltage EA is the same as the output voltage Vo/> of a phase is when there is no annature current fl owing in the machine. Why is the output voltage Vo/> from a phase not equal to EA , and what is the relationship between the two voltages? TIle answer to these questions yields the model of a synchronous generator. TIlere are a number of factors that cause the difference between EA and Vo/>: I, The distortion of the air-gap magnetic field by the current flowing in the sta-tor, called annature reaction. 2, The self-inductance of the annature coils. ), The resistance of the armature coils. 4, The effect of salient-pole rotor shapes. We will explore the effects of the first three factors and derive a machine model from them. In this chapter, the effects of a salient-pole shape on the operation of a synchronous machine will be ignored; in other words, all the machines in this chapter are assumed to have nonsalient or cylindrical rotors. Making this assump-tion will cause the calculated answers to be slightly inaccurate if a machine does indeed have salient-pole rotors, but the errors are relatively minor. A discussion of the effects of rotor pole saliency is inc1 uded in Appendix C. TIle first effect mentioned, and nonnally the largest one, is armature reac-tion. When a synchronous generator's rotor is spun, a voltage EA is induced in the generator's stator windings. If a load is attached to the terminals of the generator, a current flows. But a three-phase stator current flow will produce a magnetic field of its own in the machine. This stator magnetic field distorts the original ro-tor magnetic field, changing the resulting phase voltage. This effect is called armature reaction because the annature (stator) current affects the magnetic field which produced it in the first place. To understand annature reaction, refer to Figure 5--8. Figure 5--8a shows a two-pole rotor spinning inside a three-phase stator. There is no load connected to the stator. The rotor magnetic field DR produces an internal generated voltage EA whose peak value coincides with the direction of DR. As was shown in the last chapter, the voltage will be positive out of the conductors at the top and negative into the conductors at the bottom of the figure. With no load on the generator, there is no annature current flow, and EA will be equal to the phase voltage Vo/>. Now suppose that the generator is connected to a lagging load. Because the load is lagging, the peak current wi ll occur at an angle behind the peak voltage. TIlis effect is shown in Figure 5--8b. TIle current flowing in the stator windings produces a magnetic field of its own. This stator magnetic field is called Ds and its direction is given by the right-o E ' ",IOIX I , , B, , ,' E ' ",IDiX I , B, , , o o , , 0 , , , , , FIGURE 5-8 , , D, • 'e' ". '. • '0 E ,., • '. SYNCHRONOUS GENERATORS 275 o o , D, w , , , , , ,b, , , , , , v, , I I ... JII>X , ' o .' I 'AJIIH , , ' --:::::]'0 " " , , D, , , , , ,d, , o ~ E "" • The development of a model for armature reaction: (a) A rotating magnetic field produces the internal generated voltage EA' (b) The resulting voltage produces a lagging currentflow when connected to a lagging load. (e) The stator current produces its own magnetic field BS' which produces its own voltage E in the stator windings of the machine. (d) The field Us adds to "I/" distorting it into H .... The voltage E ... adds to EA. producing v . at the output of the phase. hand rule to be as shown in Figure 5--8c. The stator magnetic field Bs produces a voltage of its own in the stator, and this voltage is called E .... , on the figure. With two voltages present in the stator windings, the total voltage in a phase is just the sum of the internal generated voltage EA. and the annature reaction volt-age E"a,: (5-4) The net magnetic field 8 ... , is just the sum of the rotor and stator magnetic fields: (5- 5) 276 ELECTRIC MACHINERY RJNDAMENTALS v, FlGURES-9 A simple cirwit (see text). Since the angles of EA and BR are the same and the angles of E"a. and Bs, are the same, the resulting magnetic field Boe. will coincide with the net voltage Vo/>. The resulting voltages and currents are shown in Figure 5--8d. How can the effects of armature reaction on the phase voltage be modeled? First, note that the voltage E"a. lies at an angle of 90° behind the plane of maxi-mum current IA . Second, the voltage E."" is directly proportional to the current IA . If X is a constant of proportionality, then the armature reaction voltage can be ex-pressed as E"., = - jXIA (5-6) TIle voltage on a phase is thus '"I Vc-.---o Eo-A ---c j X ccI" , 1 (5- 7) Look at the circuit shown in Figure 5- 9. The Kirchhoff's voltage law equa-tion for this circuit is (5- 8) TIlis is exactly the same equation as the one describing the annature reaction volt-age. Therefore, the annature reaction voltage can be modeled as an inductor in series with the internal generated voltage. In addition to the effects of armature reaction, the stator coils have a self-inductance and a resistance. I f the stator self-inductance is called LA (and its cor-responding reactance is called XA) while the stator resistance is called RA , then the total difference betwccn EA and Vo/> is given by (5- 9) TIle annature reaction effects and the self-inductance in the machine are both rep-resented by reactances, and it is customary to combi ne them into a single reac-tance, called the synchronous reactance of the machine: XS= X + XA Therefore, the final equation describing Vo/> is I V4> - EA - jXS IA - RAIA I (5- 10) (5- 11 ) SYNCHRONOUS GENERATORS 277 I" + jXs R, + EA] ""' \' f] I, + I" R., + R, jXs R, v, + EA2 ""' \' f2 (&) L, FIGURE 5-10 The full equivalent circuit of a three-phase synchronous generator. II is now possible 1 0 sketch the equivalent circuit of a three-phase synchro-nous generator. The full equivalent circuit of such a generator is shown in Fig-ure 5- 10. This figure shows a dc power source supplying the rotor field circuit, which is modeled by the coil 's inductance and resistance in series. In series with RF is an adjustable resistor R adj which controls the flow of field current. The rest of the equivalent circuit consists of the models for each phase. E:1.ch phase has an internal generated voltage with a series inductance Xs (consisting of the sum of the armature reactance and the coil 's self-inductance) and a series resistance RA. TIle voltages and currents of the three phases are 120° apart in angle, but other-wise the three phases are identical. TIlese three phases can be either Y- or Ii-connected as shown in Figure 5- 11. If they are Y-connected, then the tenninal voltage VT is related to the phase voltage by (5- 12) 278 ELECTRIC MACHINERY RJNDAMENTALS E.n + v, v, + jXs (a) + ~ _ C=~ --Q + jXs v, jXs 'bJ ""GURE 5- 11 The generator equivalent circuit connected in (a) Y and (b) 8. I f they are a-connected, then (5- 13) TIle fact that the three phases of a synchronous generator are identical in all respects except for phase angle nonnally leads to the use of a per-phase equiva-lent circuit. The per-phase equivalent circuit of this machine is shown in Fig-SYNCHRONOUS GENERATORS 279 v, FIGURE 5-12 The per-phase equivalent circuit of a synchronous generator. The internal field circuit resistance and the external variable resistance have been contbined into a single resistor Rr . FIGURE 5-13 The phasor diagrant of a synchronous generator at unity power factor. ure 5- 12. One important fact must be kept in mind when the per-phase equivalent circuit is used: The three phases have the same voltages and currents only when the loads attached to them are balanced. If the generator's loads are not balanced, more sophisticated techniques of analysis are required. 1l1ese techniques are be-yond the scope of this book. 5.5 THE PHASOR DIAGRAM OF A SYNCHRONOUS GENERATOR Because the voltages in a synchronous generator are ac voltages, they are usually expressed as phasors. Since phasors have both a magnitude and an angle, the re-lationship between them must be expressed by a two-dimensional plot. When the voltages within a phase (E,t, V 4n jXSIA, and RAIA) and the current IA in the phase are plotted in such a fashion as to show the relationships among them, the result-ing plot is called a phasor diagram. For example, Figure 5- 13 shows these relationships when the generator is supplying a load at unity power factor (a purely resistive load). From Equation (5- 11 ), the total voltage E,t differs from the tenninal voltage of the phase V 4> by the resistive and inductive voltage drops. All voltages and currents are referenced to V4n which is arbitrarily assumed to be at an angle of 0°. This phasor diagram can be compared to the phasor diagrams of generators operating at lagging and leading power factors. 1l1ese phasor diagrams are shown 280 ELECTRIC MACHINERY RJNDAMENTALS V , jXSIA lARA ,,' E, jXSIA lARA V , ,b, ""GURE 5-14 The phasor diagram of a synchronous generator at (3) lagging and (b) leading power factor. in Figure 5- 14. Notice that, for a given phase voltage and armnture current, a larger internal generated voltage EA is needed for lagging loads than for leading loads. Therefore, a larger field current is needed with lagging loads to get the same tenninal voltage, because (5- 1) and w must be constant to keep a constant frequency. Alternatively, for a given field current and magnitude of load current, the terminal voltage is lower for lagging loads and higher for leading loads. In real synchronous machines, the synchronous reactance is nonnally much larger than the winding resistance RA, so RA is often neglected in the qualitative study of voltage variations. For accurate numerical results, R A must of course be considered. 5.6 POWER AND TORQUE IN SYNCHRON OUS GEN ERATORS A synchronous generator is a synchronous machine used as a generator. It con-verts mechanical power to three-phase electrical power. The source of mechanical power, the prime mover, may be a diesel engine, a stearn turbine, a water turbine, or any similar device. Whatever the source, it must have the basic property that its speed is almost constant regardless of the power demand. If that were not so, then the resulting power system's frequency would wander. Not all the mechanical power going into a synchronous generator becomes electrical power out of the machine.llle difTerence between input power and output power represents the losses of the machine. A power-flow diagram for a synchro-SYNCHRONOUS GENERATORS 281 p~= foppw.. Stray losses FIGURE 5-15 windage losses , , find w.. I , losses , , , (copper losses) The power-flow diagram of a synchronous generntor. nous generator is shown in Figure 5- 15. The input mechanical power is the shaft power in the generator fln = "Tappwm , while the power converted from mechanical to electrical fonn internally is given by = 3E,./1t cos "y (5- 14) (5- 15) where 'Y is the angle between Elt and lit- TIle difference between the input power to the generator and the power converted in the generator represents the mechan-ical, core, and stray losses of the machine. TIle real electrical output power of the synchronous generator can be ex-pressed in line quantities as and in phase quantities as ?"UI = 3'4,IA cos (J The reactive power output can be expressed in line quantities as Q,UI = ~VTIL sin (J or in phase quantities as (5- 16) (5- 17) (5- 18) (5- 19) If the annature resistance RIt is ignored (since Xs» RIt), then a very useful equation can be derived to approximate the output power of the generator. To de-rive this equation, examine the phasor diagram in Figure 5- 16. Figure 5- 16 shows a simplified phasor diagram of a generator with the stator resistance ignored. No-tice that the vertical segment be can be expressed as either Elt sin /j or X s lit cos (J. Therefore, EA sin /j lA cos (J = X , 282 ELECTRIC MACHINERY RJNDAMENTALS , o r ""GURE 5-16 """, " , , , , ............ 1:;' , , , , , , " , , jXs l,t I , , , V , • ___ L.L " b , Simplified phasor diagram with armature resistance ignored. and substituting this expression into Equation (5- 17) gives E,t sin .s =Xs l,t cos(} (5- 20) Since the resistances are assumed to be zero in Equation (5- 20), there are no elec-trical losses in this generator, and this equation is both PCOII¥ and Pout. Equation (5- 20) shows that the power produced by a synchronous genera-tor depends on the angle 8 between Vq,and EA. The angle 8 is known as the torque angle of the machine. Notice also that the maximum power that the generator can supply occurs when 8 = 900. At 8 = 90°, sin 8 = I , and (5- 21) TIle maximum power indicated by this equation is called the static stability limit of the generator. Nonnally, real generators never even come close to that limit. Full-load torque angles of 15 to 20° are more typical of real machines. Now take another look at Equations (5- 17), (5- 19), and (5- 20). IfVq, is as-sumed constant, then the real power output is directly prop011ionni to the quanti-ties J,t cos () and E,t sin 8, and the reactive power output is directly proportional to the quantity J,t sin (). These facts are useful in plotting phasor diagrams of syn-chronous generators as loads change. From Chapter 4, the induced torque in this generator can be expressed as (4- 58) or as (4-60) SYNCHRONOUS GENERATORS 283 The magnitude of Equation (4--60) can be expressed as Tind = kB,/J"", sin /j (4-6 1) where /j is the angle between the rotor and net magnetic fields (the so-called torque angle). Since BR produces the voltage E ... and BOel produces the voltage Vo/>. the angle /j between E ... and V 0/> is the same as the angle /j between BR and B. An alternative expression for the induced torque in a synchronous generator can be derived from Equation (5- 20). Because P C equal to zero in Figure 5-1 8, the internnl mnchine impedance is given by EA Zs = VR A 2 + X2 = -(5- 25) , IA Since Xs» RIl , this equation reduces to (5- 26) If Ell and III are known for a given situation, then the synchronous reactance Xs can be found. Therefore, an approximate method for detennining the synchronous reac-tance Xs at a given field current is I. Get the internal generated voltage Ell from the ace at that field current. 2. Get the short-circuit current now l,o.,sc at that field current from the Sec. 3. Find Xs by applying Equation (5- 26). 286 ELECTRIC MACHINERY RJNDAMENTALS Air-gap line ___ ---ace sec x, o o ""GURE 5- 19 A sketch of the approximate synchronous reacl3.nce of a synchronous generator as a function of the field current in the machine. The constant value of reactance found at low values of field current is the uns(J/umted synchronous reactance of the machine. TIlere is a problem with this approach, however. The internal generated voltage Ell comes from the acc, where the machine is partially saturated for large field currents, while III is taken from the sec, where the machine is unsatu-rated at all field currents. TIlerefore, at higher field currents, the Ell taken from the aec at a given field current is not the same as the Ell at the srune field current un-der short-circuit conditions, and this difference makes the resulting value of Xs only approximate. However, the answer given by this approach is accurate up to the point of saturation, so the unsaturated synchronous reactance Xs.~ of the machine can be found simply by applying Equation (5- 26) at any field current in the linear por-tion (on the air-gap line) of the acc curve. TIle approximate value of synchronous reactance varies with the degree of saturation of the ace, so the val ue of the synchronous reactance to be used in a given problem should be one calculated at the approximate load on the machine. A plot of approximate synchronous reactance as a function of field current is shown in Figure 5- 19. To get a more accurate estimation of the saturated synchronous reactance, refer to Section 5- 3 of Reference 2. If it is important to know a winding's resistance as well as its synchronous reactance, the resistance can be approximated by applying a dc voltage to the windings while the machine is stationary and measuring the resulting current flow. TIle use of dc voltage means that the reactance of the windings will be zero during the measurement process. SYNCHRONOUS GENERATORS 287 This technique is not perfectly accurate, since the ac resistance will be slightly larger than the dc resistance (as a result of the skin effect at higher fre-quencies). The measured value of the resistance can even be plugged into Equa-tion (5- 26) to improve the estimate of Xs, if desired. (Such an improvement is not much help in the approximate approach-saturation causes a much larger error in the Xs calculation than ignoring Rio. does.) The Short-Circuit Ratio Another parameter used to describe synchronous generators is the short-circuit ra-tio. 1lle short-circuit ratio of a generator is defined as the ratio of the field current requiredfor the rated voltage at open circuit to the field current required for the rated armature current at short circuit. It can be shown that this quantity is just the reciprocal of the per-unit value of the approximate saturated synchronous re-actance calculated by Equation (5- 26). Although the short-circuit ratio adds no new information about the genera-tor that is not already known from the saturated synchronous reactance, it is im-portant to know what it is, since the tenn is occasionally encountered in industry. Example 5-1. A 2oo-kVA, 480-y' 50-Hz, V-connected synchronous generator with a rated field current of 5 A was tested, and the following data were taken: 1. VT,OC at the rated h was measured to be 540 V. 2. h,se at the rated If was found to be 300 A. 3. When a dc voltage of 10 V was applied to two of the tenninals, a current of 25 A was measured. Find the values of the armature resistance and the approximate synchronous reactance in ohms that would be used in the generator model at the rated conditions. Solutioll The generator described above is V-connected, so the direct current in the resistance test flows through two windings. Therefore, the resistance is given by V 2R -= 10. -loe Voe 10 V Rio. = 2/0e = (2)(25 A) = 0.2 n The internal generated voltage at the rated field current is equal to V, EIo. = V as before. Therefore, the annature reaction voltage jXs IIl is larger than before but at the same angle. Now since Ell = Vo/> + jXsIIl j Xs III must stretch between Vo/> at an angle of 0° and Ell, which is constrained to be of the same magnitude as before the load increase. If these constraints are plotted on a phasor diagram, there is one and only one point at which the annature reac-tion voltage can be parallel to its original position while increasing in size. The re-sulting plot is shown in Figure 5- 22a. If the constraints are observed, then it is seen that as the load increases, the voltage V 0/> decreases rather sharply. Now suppose the generator is loaded with unity-power-factor loads. What happens if new loads are added at the same power factor? With the same con-straints as before, it can be seen that this time Vo/> decreases only slightly (see Fig-ure 5- 22b). 290 ELECTRIC MACHINERY RJNDAMENTALS E'A E, jXslA jXSIA , " 0 V' // V , . ~ . I, I', ~ ~ ~ ~ , ~ , ~ , ~ , ~ , ~ ',,/'y~ v , , , (a) " , ~ ~ ~ ~ E'A , ~ ~ I', ~ ~ I, " , ,,' ""GURE 5-11 " 1 11. I'll. jXSIA E, V. V; 'h' jXs IA V' V • • The elIect of an increase in generator loads at constant power factor upon its terminal voltage. (a) Lagging power factor; (b) unity power factor; (c) teading power factor. Finally, let the generator be loaded with leading-power-factor loads. If new loads are added at the same power factor this time, the annature reaction voltage lies outside its previous value, and Vo/> actually rises (see Figure 5- 22c). In this last case, an increase in the load in the generator produced an increase in the tenninal voltage. Such a result is not something one would expect on the basis of intuition alone. General conclusions from this discussion of synchronous generator behav-ior are I. If lagging loads (+ Q or inductive reactive power loads) are added to a gen-erator, Vo/> and the terminal voltage Vrdecrease significantly. 2. If unity-power-factor loads (no reactive power) are added to a generator, there is a slight decrease in V 0/> and the tenninal voltage. 3. If leading loads (--Q or capacitive reactive power loads) are added to a gener-ator, V 0/> and the tenninal voltage will rise. A convenient way to compare the voltage behavior of two generators is by their voltage regulation. The voltage regu lation (VR) of a generator is defined by the equation SYNCHRONOUS GENERATORS 291 Vn1 -Va I VR = Va x 100% (4-67) where V ol is the no-load voltage of the generator and V fl is the full-load voltage of the generator. A synchronous generator operating at a lagging power factor has a fairly large positive voltage regulation, a synchronous generator operating at a unity power factor has a small positive voltage regu lation, and a synchronous gen-erator operating at a leading power factor often has a negative voltage regulation. Normally, it is desirable to keep the voltage supplied to a load constant, even though the load itself varies. How can tenninal voltage variations be corrected for? The obvious approach is to vary the magnitude of E), to compensate for changes in the load. Recall that E), = Kcpw. Since the frequency should not be changed in a nonnal system, E), must be controlled by varying the flux in the machine. For example, suppose that a lagging load is added to a generator. Then the terminal voltage will fall, as was previously shown. To restore it to its previous level, decrease the field resistor RF" If RF decreases, the field current wil I increase. An increase in IF increases the flux, which in turn increases E)" and an increase in E), increases the phase and terminal voltage. nlis idea can be summarized as follows: I. Decreasing the field resistance in the generator increases its field current. 2. An increase in the field current increases the flux in the machine. 3. An increase in the flux increases the internal generated voltage E), = Kcpw. 4. An increase in E), increases Vo/> and the terminal voltage of the generator. The process can be reversed to decrease the tenninal voltage. It is possible to regulate the tenninal voltage of a generator throughout a series of load changes simply by adjusting the field current. Example Problems The following three problems illustrate simple calculations involving voltages, currents, and power flows in synchronous generators. The first problem is an ex-ample that includes the armature resistance in its calculations, while the next two ignore R),. Part of the first example problem addresses the question: How must a generator s field current be adjusted to keep VT constant as the load changes? On the other hand, part of the second example problem asks the question: lfthe load changes and the field is left alone, what happens to the terminnl voltage? You should compare the calculated behavior of the generators in these two problems to see if it agrees with the qualitative arguments of this section. Finally, the third example illustrates the use of a MATLAB program to derive the terminal charac-teristics of synchronous generator. Example 5-2. A 480-V, 6()"Hz, ~ -co lUlected, four-pole SynChroflOUS geflerator has the OCC shown in Figure 5--23a. This geflerator has a synchronous reactaflce of 0.1 n afld 292 ELECTRIC MACHINERY RJNDAMENTALS > • ~ • " ~ .. . § § ., , •• y 0 ~ 600 500 400 300 200 100 I I o 0.0 V / I / / 1.0 2.0 3.0 I,t '" 692.8 L - 36.87° A ""GURE 5-23 / /' / / 4.0 5.0 6.0 Field current. A v • , ., ,b, 7.0 8.0 9.0 10.0 (a) Open-drwit characteristic of the generator in Example 5- 2. (b) Phasor diagram of the generator in Example 5- 2. an annature resistance of 0.015 n. At fullload, the machine supplies 1200 A at 0.8 PF lag-ging. Under full-load conditions. the friction and windage losses are 40 kW. and the core losses are 30 kW. Ignore any field circuit losses. SYNCHRONOUS GENERATORS 293 (a) What is the speed of rotation of this generator? (b) How much field current must be supplied to the generator to make the terminal voltage 480 V at no load? (c) If the generator is now cOlUlected to a load and the load draws 1200 A at 0.8 PF lagging, how much field current will be required to keep the terminal voltage equal to 480 V? (d) How much power is the generator now supplying? How much power is supplied to the generator by the prime mover? What is this machine's overall efficiency? (e) If the generator's load were suddenly disconnected from the line, what would happen to its terminal voltage? (jJ Finally, suppose that the generator is cOIUlected to a load drawing 1200 A at 0.8 PF leading. How much field current would be required to keep Vrat 480 V? Solutioll This synchronous generator is .d.-connected, so its phase voltage is equal to its line voltage V. = Vr, while its phase current is related to its line current by the equation I L = ~/ • . (a) The relationship between the electrical frequency produced by a synchronous generator and the mechanical rate of shaft rotation is given by Equation (4--34): Therefore, ".p fe = 120 12!X60 Hz) _ 4 poles 1800 r/ min (4--34) (b) In this machine, Vr = V • . Since the generator is at no load, IA = 0 and EA = V •. Therefore, Vr = V . = EA = 480 V, and from the open-circuit characteristic, I" = 4.5 A. (c) If the generator is supplying 1200 A. then the armature current in the machine is 1..1 = 1 2~A = 692.8 A The phasor diagram for this generator is shown in Figure 5- 23b. If the terminal voltage is adjusted to be 480 V, the size of the internal generated voltage EA is given by EA = V . + RAIA + jXsI,\ = 480 LO° V + (0.015 n X692.8 L -36.87° A ) + (j0.1 0)(692.8 L -36.87° A ) = 480 LO° V + 10.39 L -36.87° V + 69.28 L53.13° V = 529.9 + j49.2 V = 532 L5.JO V To keep the tenninal voltage at 480 V, E,\ must be adjusted to 532 V. From Fig-ure 5- 23, the required field current is 5.7 A. (d) The power that the generator is now supplying can be found from Equation (5-16): (5--1 6) 294 ELECTRIC MACHINERY RJNDAMENTALS = VJ(480 VXI200 A) cos 36.87° = 798 kW To detennine the power input to the generator, use the power-flow diagram (Fig-ure 5-15). From the power-flow diagram, the mechanical input power is given by The stray losses were not specified here, so they will be ignored. In this genera-tor, the electrical losses are P olo< 10 .. = 311RA = 3(692.8 A)2(0.015 f.!) = 21.6 kW The core losses are 30 kW, and the friction and windage losses are 40 kW, so the total input power to the generator is P in = 798kW + 21.6kW + 30kW + 40kW = 889.6kW Therefore, the machine's overall efficiency is Pout 798 kW 7f = p x 100% = 8896 kW x 100% = 89.75% rn . (e) If the generator's load were suddenly disconnected from the line, the current IA would drop to zero, making EA = V •. Since the field current has not changed, lEAl has not changed and V. and Vr must rise to equal EA' Therefore, if the load were suddenly dropped, the terminal voltage of the generator would rise to 532 V. (f) If the generator were loaded down with 1200 A at 0.8 PF leading while the ter-minal voltage was 480 V, then the internal generated voltage would have to be EA = V. + RAIA + jX s I,\ = 480LO° V + (0.015 n)(692.8L36.87° A) + (j0.1 nX692.8L36.87° A) = 480 LO° V + 10.39 L36.87° V + 69.28 L 126.87° V = 446.7 + j61.7 V = 451 L7.10 V Therefore, the internal generated voltage EA must be adjusted to provide 451 V if Vr is to remain 480 V. Using the open-circuit characteristic, the field current would have to be adjusted to 4.1 A. Which type of load (leading or lag ging) needed a larger field current to maintain the rated voltage? Which type of load (leading or lagging) placed more thermal stress on the generator? Why? Example 5-3. A 480-V, 50-Hz, Y -connected, six-pole synchronous genera-tor has a per-phase synchronous reactance of 1.0 n. Its full-load armature current is 60 A at 0.8 PF lagging. This generator has friction and windage losses of 1.5 kW and core losses of 1.0 kW at 60 Hz at full load. Since the armature resistance is being ig-nored, assrune that the j 2R losses are negligible. The field current has been adjusted so that the terminal voltage is 480 V at no load. (a) What is the speed of rotation of this generator? (b) What is the terminal voltage of this generator if the following are true? SYNCHRONOUS GENERATORS 295 I. It is loaded with the rated current at 0.8 PF lagging. 2, It is loaded with the rated current at 1.0 PF. 3, It is loaded with the rated current at 0.8 PF leading. (c) What is the efficiency of this generator (ignoring the lUlknown electrical losses) when it is operating at the rated current and 0.8 PF lagging? (d) How much shaft torque must be applied by the prime mover at full load? How large is the induced cOlUltertorque? (e) What is the voltage regulation of this generator at 0.8 PF lagging? At 1.0 PF? At 0.8 PF leading? Solution This generator is V-connected, so its phase voltage is given by V. = Vr/ v'J. That means that when Vr is adjusted to 480 V, V. = 277 V. The field current has been adjusted so that V r ... = 480 V, so V. = 277 V. At no load, the armature current is zero, so the armature re-action voltage and the I},R}, drops are zero. Since I}, = 0, the internal generated voltage E}, = V. = 277 V. The internal generated voltage E},( = Kq,w) varies only when the field current changes. Since the problem states that the field current is adjusted initially and then left alone, the magnitude of the internal generated voltage is E}, = 277 V and will not change in this example. (a) The speed of rotation of a synchronous generator in revolutions per minute is given by Equation (4-34): Therefore, 1201. " = --m p _ limP Ie -120 = 120(50 H z) _ 1000 rlmin 6 poles Alternatively, the speed expressed in radians per second is Wm = (1000r/min)e6~~n)e~~ad) = 104.7 radl s (4-34) (b) I. If the generator is loaded down with rated current at 0.8 PF lagging, the re-sulting phasor diagram looks like the one shown in Figure 5- 24a. In this phasor diagram, we know that V. is at an angle of 0°, that the magnitude of E}, is 277 V, and that the quantity jXsI}, is jXsI}, = j(1.0 nX60 L - 36.87° A) = 60 L53.13° V The two quantities not known on the voltage diagram are the magnitude of V. and the angle 0 of E},. To find these values, the easiest approach is to con-struct a right triangle on the phasor diagram, as shown in the figure. From Figure 5- 24a, the right triangle gives E1 = (V. + Xsl}, sin 9)2 + (Xsl}, cos 9)2 Therefore, the phase voltage at the rated load and 0.8 PF lagging is 296 ELECTRIC MACHINERY RJNDA MENTALS 60 L 53.13° v, ,b , v, " , ""GURE 5-14 Generator phasor diagrams for Example 5- 3. (a) Lagging power factor; (b) unity power factor; (c) leading power factor. (277 vi = [V. + (1 .0 0)(60 A) sin 36.87°]2 + [( 1.0 n X60 A) cos 36.87°]2 76,729 = (V . + 36)1 + 2304 74,425 = (V . + 36)2 272.8 = V,., + 36 V,., = 236.8 V Since the generator is Y-colUlected, Vr = V3V. = 410 V. 2. If the generator is loaded with the rated current at unity power factor, then the phasor diagram wi11look like Figure 5- 24h. To find V . here the right tri-angle is SYNCHRONOUS GENERATORS 297 £1 = vi + (XsIA)2 (277V)2 = vi + [(1.00)(60A)]2 76,729 = Vi + 3600 Vi = 73,129 V. = 270.4 V Therefore, Vr = V3"V. = 46S.4 V. 3. When the generator is loaded with the rated current at O.S PF leading, the re-suiting phasor diagram is the one shown in Figure 5- 24c. To find V. in this situation, we construct the triangle OAB shown in the figure. The resulting equation is £1 = (V. - XsIA)2 + (XsI./t cos (/)2 Therefore, the phase voltage at the rated load and O.S PF leading is (277 V)2 = [V. - (l.OnX60A) sin 36.S7°f + [(1.0 0)(60 A) cos 36.S7o]2 76,729 = (V. - 36)2 + 2304 74,425 = (V. - 36)2 272.S = V. - 36 V. = 30S.S V Since the generator is Y-cOIUlected, Vr = V3"V. = 535 V. (c) The output power of this generator at 60 A and O.S PF lagging is POU,=3V.IAcos () = 3(236.S VX60AXO.S) = 34.1 kW The mechanical input power is given by = 34.1 kW + a + 1.0kW + 1.5kW = 36.6kW The efficiency of the generator is thus Pout 34.1 kW 7f = P, x 100% = 366 kW x 100% = 93.2% rn . (d) The input torque to this generator is given by the equation T = P in = -W. 36.6 kW 125.7 radls = 291.2 N · m The induced cOlUltertorque is given by P.:oov = Tind = Wv 34.1 kW 125.7 radls = 271.3 N · m (e) The voltage regulation of a generator is defined as 298 ELECTRIC MACHINERY RJNDA MENTALS v - " VR = nl () X 100% V, (4--67) By this defmition, the voltage regulation for the lagging, lUlity, and leading power-factor cases are 1. Lagging case: VR = 480 ~I~ ~ 1O V x 100% = 17.1% 2. Unity case: VR = 480 ~6; ~68 V x 100% = 2.6% 3. Leading case: VR = 480 ~3; ~35 V x 100% = -10.3% In Example 5- 3, lagging loads resulted in a drop in terminal voltage, unity-power-factor loads caused little effect on VT, and leading loads resulted in an in-crease in tenninal voltage. Example 5-4. Assume that the generator of Example 5- 3 is operating at no load with a tenninal voltage of 480 V. Plot the tenninal characteristic (terminal voltage versus line current) of this generator as its armature ClUTent varies from no-load to full load at a power factor of (a) 0.8 lagging and (b) 0.8 leading. Assume that the field current remains constant at all times. Solutioll The terminal characteristic of a generator is a plot of its tenninal voltage versus line cur-rent. Since this generator is Y-cOIlllected. its phase voltage is given by V. = VTIV'3 . If Vr is adjusted to 480 V at no-load conditions. then V . = fA = 277 V. Because the field C lUTent remains constant, fA will remain 277 V at all times. The output current h from this gener-ator will be the same as its armature current IA because it is V-connected. (a) If the generator is loaded with a 0.8 PF lagging clUTent. the resulting phasor di-agram looks like the one shown in Figure 5- 24a. In this phasor diagram. we know that V. is at an angle of 0°. that the magnitude of EA is 277 V. and that the quantity jXSIA stretches between V. and EA as shown. The two quantities not known on the phasor diagram are the magnitude of V. and the angle 0 of EA. To find V .. the easiest approach is to construct a right triangle on the phasor dia-gram. as shown in the figure. From Figure 5--24a. the right triangle gives ~ = (V. + XSIA sin (J)2 + (XSIA cos (J)2 This equation can be used to solve for V., as a flUlction of the current IA: V. = JE1 (XiA cos 0)2 - XiA sin 0 A simple MATLAB M-filecan be used to calculate V .(and hence VT) as a func-tion of current. Such an M-file is shown below: ~ M-file : t e rm_cha r a .m ~ M-file t o p l ot the t e rmina l cha r act eri s ti cs o f the ~ gene r a t or o f Examp l e 5-4 with a n 0.8 PF l agging l oad . ~ Firs t , initia lize the c urrent a mp litudes (2 1 values ~ in the r a nge 0- 60 A) i _a = (0, 1: 20) .. 3; SYNCHRONOUS GENERATORS 299 % Now initia liz e a ll other va lues v-ph ase = zeros( 1 ,2 1 ) ; e_a = 277.0; x_s = 1. 0; theta = 36 .S7 .. (p i / 1 SO ) ; % Converted t o radians % Now ca l c ulate v-ph ase f or each c urrent l eve l f or ii = 1: 2 1 e od (x_s .. i _a( ii ) .. cos(the ta ))" 2 ) (x_s .. i _a( ii ) .. s in (the ta )) ; % Ca l c ulate terminal voltage from the phase voltage v_t = v-ph ase .. sqrt (3) ; % Plot the terminal ch aracteri sti c, remembering the % the line c urre nt i s the same as i _a p l ot ( i _a, v_t , 'Col or' , 'k' , 'Linewi dth ' ,2.0) ; x l abe l ( 'Line CUrre nt (A) ' , 'Fontwei ght ' , 'Bo l d ' ) ; y l abe l ( 'Te rmina l Voltage (V) ' , 'Fontwei ght' , 'Bo l d ' ) ; title ( 'Te rmina l Ch a r acter i stic f o r O.S PF l agging l oad ' , ... 'Fontwei ght' , 'Bo l d ' ) ; gr i d on ; ax i s( [ O 60 400 550 ] ) ; The plot resulting when this M-file is executed is shown in Figure 5- 25a. (b) If the generator is loaded with a 0.8 PF leading current, the resulting phasor di-agram looks like the one shown in Figure 5- 24c. To fmd V.' the easiest ap-proach is to construct a right triangle on the phasor diagram, as shown in the fig-ure. From Figure 5- 24c, the right triangle gives E1 = (V. - XsfA sin 9)2 + (XsfA cos 9)2 This equation can be used to solve for V ", as a ftmction of the current fA: V. = JE l (XsfA cos (J)l + XsfA sin 9 This equation can be used to calculate and plot the terminal characteristic in a manner similar to that in part a above. The resulting tenninal characteristic is shown in Figure 5- 25b. 5.9 PARALLEL OPERATION OF AC GENERATORS In today's world, an isolated synchrono us generator supplying its own load inde-pendently of other generators is very rare. Such a situation is found in only a few out-of-the-way applications such as emergency generators. For all usual genera-tor applications, there is more than one generator operating in parallel to supply the power demanded by the loads. An extreme example of this situation is the U.S. power grid, in which literally thousands of generators share the load on the system. 300 ELECTRIC MACHINERY RJNDAMENTALS 550 > • 500 00 • r----I------------I------------~ .. " ., ~ 450 4000 \0 20 30 40 50 Line current. A ,,' 550 ,----,---,--,---,----,----, > ~ 500 • ~ e-o " § 450 ~ 4OO0· '----" IO c---C 20 ~---" 30 c----4O ~---c 50 '---~ 60 Line current. A ,b, ""GURE 5-25 (a) Thrminal characteristic for the generator of Example 5-4 when loaded with an 0.8 PF lagging loo.d. (b) Thrminal characteristic for the generator when loaded with an 0.8 PF leading load. Why are synchronous generators operated in parallel? There are several ma-jor advantages to such operation: I. Several generators can supply a bigger load than one machine by itself. 2. Having many generators increases the reliability of the power system, since the failure of anyone of them does not cause a total power loss to the load. 3. Having many generators operating in parallel allows one or more of them to be removed for shutdown and preventive maintenance. SYNCHRONOUS GENERATORS 301 '\ Generator I Lood / s, . -Generator 2 HGURE 5-26 A generator being paralleled with a running power system. 4. If only one generator is used and it is not operating at near full load, then it will be relatively inefficient. With several smaller machines in parallel, it is possible to operate only a fraction of them. The ones that do operate are op-erating near full load and thus more efficiently. This section explores the requirements for paralleling ac generators, and then looks at the behavior of synchronous generators operated in parallel. The Conditions Required for Paralleling Figure 5- 26 shows a synchronous generator G t supplying power to a load, with another generator Gl about to be paralleled with Gt by closing the switch St. What conditions must be met before the switch can be closed and the two generators connected? If the switch is closed arbitrarily at some moment, the generators are liable to be severely damaged, and the load may lose power. I f the voltages are not ex-actly the same in each conductor being tied together, there will be a very large cur-rent flow when the switch is closed. To avoid this problem, each of the three phases must have exactly the same voltage magnitude and phase angle as the con-ductor to which it is connected. In other words, the voltage in phase a must be ex-actly the same as the voltage in phase a' , and so forth for phases b-b' and c-c'. To achieve this match, the following paralleling conditions must be met: I. 1lle rms line voltages of the two generators must be equal. 2. 1lle two generators must have the same phase sequence. 3. 1lle phase angles of the two a phases must be equal. 4. 1lle frequency of the new generator, called the oncoming generator, must be slightly higher than the frequency of the running system. These paralleling conditions require some explanation. Condition I is obvi-o",- in order for two sets of voltages to be identical, they must of course have the same rms magnitude of voltage. The voltage in phases a and a' will be completely 302 ELECTRIC MACHINERY RJNDAMENTALS v, • v, • v, v, abc phase sequence acb phase sequence ,,' Generator I Lo"" Generator 2 Switch S[ ,b, ""GURE 5-27 (a) The two possible phase sequences of a three-phase system. (b) The three-light-bulb method for checking phase sequence. identical at all times if both their magnitudes and their angles are the same, which explains condition 3. Condition 2 ensures that the sequence in which the phase voltages peak in the two generators is the same. Ifthe phase sequence is different (as shown in Fig-ure 5- 27a), then even though one pair of voltages (the a phases) are in phase, the other two pairs of voltages are 120 0 out of phase. If the generators were connected in this manner, there would be no problem with phase a, but huge currents would fl ow in phases band c, damaging both machines. To correct a phase sequence problem, simply swap the connections on any two of the three phases on one of the machines. If the frequencies of the generators are not very nearly equal when they are connected together, large power transients will occur until the generators stabilize at a common frequency. The frequencies of the two machines must be very nearly equal, but they cannot be exactly equal. 1lley must differ by a small amount so SYNCHRONOUS GENERATORS 303 that the phase angles of the oncoming machine will change slowly with respect to the phase angles of the running system. In that way, the angles between the volt-ages can be observed and switch SI can be closed when the systems are exactly in phase. The General Procedure for Paralleling Generators Suppose that generator Gl is to be connected to the running system shown in Fig-ure 5- 27. TIle following steps should be taken to accomplish the paralleling. First, using voltmeters, the field current of the oncoming generator should be adjusted until its tenninal voltage is equal to the line voltage of the running system. Second, the phase sequence of the oncoming generator must be compared to the phase sequence of the running system. TIle phase sequence can be checked in a number of different ways. One way is to alternately connect a small induction motor to the terminals of each of the two generators. I f the motor rotates in the same direction each time, then the phase sequence is the same for both generators. If the motor rotates in opposite directions, then the phase sequences differ, and two of the conductors on the incoming generator must be reversed. Another way to check the phase sequence is the three-light-bulb method. In this approach, three light bulbs are stretched across the open terminals of the switch connecting the generator to the system as shown in Figure 5- 27b. As the phase changes between the two systems, the light bulbs first get bright (large phase difference) and then get dim (small phase difference). If all three bulbs get bright and dark together, then the systems have the same phase sequence. If the bulbs brighten in succession, then the systems have the opposite phase sequence, and one of the sequences must be reversed. Next, the frequency of the oncoming generator is adjusted to be slightly higher than the frequency of the running system. TIlis is done first by watching a frequency meter until the frequencies are close and then by observing changes in phase between the systems. TIle oncoming generator is adjusted to a slightly higher frequency so that when it is connected, it will come on the line supplying power as a generator, instead of consuming it as a motor would (this point will be explained later). Once the frequencies are very nearly equal, the voltages in the two systems will change phase with respect to each other very slowly. TIle phase changes are observed, and when the phase angles are equal, the switch connecting the two sys-tems together is shut. How can one tell when the two systems are finally in phase? A simple way is to watch the three light bulbs described above in connection with the discussion of phase sequence. When the three light bulbs all go out, the voltage difference across them is zero and the systems are in phase. This simple scheme works, but it is not very accurate. A better approach is to employ a synchroscope. A synchro-scope is a meter that measures the difference in phase angle between the a phases of the two systems. The face of a synchroscope is shown in Figure 5- 28. TIle dial shows the phase difference between the two a phases, with 0 (meaning in phase) 304 ELECTRIC MACHINERY RJNDAMENTALS \... S C , " _ ,h ,;,'C '"C"" _ J FIGURE 5-28 A synchrosrope. at the top and 180 0 at the bottom. Since the frequencies of the two systems are slightly different, the phase angle on the meter changes slowly. If the oncoming generator or system is faster than the running system (the desired situation), then the phase angle advances and the synchroscope needle rotates clockwise. If the oncoming machine is slower, the needle rotates counterclockwise. When the syn-chroscope needle is in the vertical position, the voltages are in phase, and the switch can be shut to connect the systems. Notice, though, that a synchroscope checks the relationships on only one phase. It gives no infonnation about phase sequence. In large generators belonging to power systems, this whole process of par-alleling a new generator to the line is automated, and a computer does this job. For smaller generators, though, the operator manually goes through the paralleling steps just described. Frequency-Power and Voltage-Reactive Power Characteristics of a Synchronolls Generator All generators are driven by a prime mover, which is the generator's source of mechanical power. TIle most common type of prime mover is a steam turbine, but other types include diesel engines, gas turbines, water turbines, and even wind turbines. Regardless of the original power source, all prime movers tend to behave in a similar fashion--.:1.s the power drawn from them increases, the speed at which they turn decreases. The decrease in speed is in general nonlinear, but some form of governor mechanism is usually included to make the decrease in speed linear with an increase in power demand. Whatever governor mechanism is present on a prime mover, it will always be adjusted to provide a slight drooping characteristic with increasing load. The speed droop (SD) ofa prime mover is defined by the equation I SO = nnl nn nil x 100% I (5- 27) where n o] is the no-load prime-mover speed and no is the full-load prime-mover speed. Most generator prime movers have a speed droop of 2 to 4 percent, as de-fined in Equation (5- 27). In addition, most governors have some type of set point .5 , I " J o ", o ,b , SYNCHRONOUS GENERATORS 305 Power. kW Power. kW HGURE 5-29 (a) The speed-versus-power curve for a typical prime mover. (b) The resulting frequency-versus-power curve for the generator. adjustment to allow the no-load speed of the turbine to be varied. A typical speed-versus-power plot is shown in Figure 5- 29. Since the shaft speed is related to the resulting electrical frequency by Equa-tion (4- 34), (4- 34) the power output of a synchronous generator is related to its frequency. An exam-ple plot of frequency versus power is shown in Figure 5- 29b. Frequency-power characteristics of this sort play an essential role in the parallel operation of syn-chronous generators. The relationship between frequency and power can be described quantita-ti vely by the equation where P = power output of the generator Jot = no-load frequency of the generator !.y. = operating frequency of system sp = slope of curve, in kW/Hz or MW/Hz (5- 28) A similar relationship can be derived for the reactive power Q and terminal voltage VT. As previously seen, when a lagging load is added to a synchronous 306 ELECTRIC MACHINERY RJNDAMENTALS VTo] Q, 0 kYAR consumed ""GURE 5-30 Qn Q (reactive power). kYAR supplied The curve of terminal voltage (Vr) versus reactive power (Q) for a synchronous generator. generator, its tenninal voltage drops. Likewise, when a leading load is added to a synchronous generator, its tenninal voltage increases. It is possible to make a plot oftenninal voltage versus reactive power. and such a plot has a drooping charac-teristic like the one shown in Figure 5-30. This characteristic is not intrinsically linear, but many generator voltage regulators include a feature to make it so. The characteristic curve can be moved up and down by changing the no-load tenninal voltage set point on the voltage regulator. As with the frequency-power character-istic, this curve plays an important role in the parallel operation of synchronous generators. TIle relationship between the terminal voltage and reactive power can be expressed by an equation similar to the frequency-power relationship [Equation (5-28)] if the voltage regulator produces an output that is linear with changes in reacti ve power. It is important to realize that when a single generator is operating alone, the real power P and reactive power Q supplied by the generator will be the amount demanded by the load attached to the generator- the P and Q supplied cannot be controlled by the generator's controls. Therefore, for any given real power, the governor set points control the generator's operating frequency Ie and for any given reactive power, the field current controls the generator's tenninal voltage VT. Example 5-5. Figure 5-31 shows a generator supplying a load. A second load is to be connected in parallel with the first one. The generator has a no-load frequency of 61.0 Hz and a slope sp of I MWlHz. Load I consumes a real power of I()(x) kW at 0.8 PF lag-ging. while load 2 consrunes a real power of 800 kW at 0.707 PF lagging. (a) Before the switch is closed. what is the operating frequency of the system? (b) After load 2 is cOIUlected. what is the operating frequency of the system? (c) After load 2 is cOIUlected. what action could an operator take to restore the sys-tem frequency to 60 Hz? SYNCHRONOUS GENERATORS 307 y "-Lo'" 1 /' Turbine generator I I Lo'" 2 FIGURE 5-31 The power system in Example 5- 5. Solutioll This problem states that the slope of the generator's characteristic is I MW/Hz and that its no-load frequency is 61 Hz. Therefore, the power produced by the generator is given by P = sl--JnJ - J.y. ) p f. y• = Jo l -sp (a) The initial system frequency is given by = 61 Hz -lOOOkW I MW/Hz = 61 Hz -I Hz = 60 Hz (b) After load 2 is connected, = 61 Hz -1800 kW I MW/Hz = 61 Hz -1.8 Hz = 59.2 Hz (5--28) (c) After the load is connected, the system frequency falls to 59.2 Hz. To restore the system to its proper operating frequency, the operator should increase the gov-ernor no-load set points by 0.8 Hz, to 61.8 Hz. This action will restore the sys-tem frequency to 60 Hz. To summarize, whe n a generator is operating by itself supplying the system loads, then I . 1lle real and reactive power supplied by the generator will be the amount de-manded by the attached load. 2. 1lle governor sct points of the generator will control the operating frequency of the power syste m. 308 ELECTRIC MACHINERY RJNDAMENTALS /, -p o Consumed ,,' ""GURE 5-32 p. kW supplied v, - Q o Consumed Q. kVAR supplied ,b, Curves for an infinite bus: (a) frequency versus power and (b) tenninal voltage versus reactive power. 3. The field current (or the field regulator set points) control the terminal volt-age of the power system. nlis is the situation found in isolated generators in remote field environments. Operation of Generators in Parallel with Large Power Systems When a synchronous generator is connected to a power system, the power system is often so large that nothing the operator of the generator does will have much of an effect on the power system. An example of this situation is the connection of a single generator to the U.S. power grid. 1lle U.S. power grid is so large that no reasonable action on the part of the one generator can cause an observable change in overall grid frequency. nlis idea is idealized in the concept of an infinite bus. An infinite bus is a power system so large that its voltage and frequency do not vary regardless of how much real and reactive power is drawn from or supplied to it. The power-frequency characteristic of such a system is shown in Figure 5- 32a, and the reac-tive power-voltage characteristic is shown in Figure 5- 32b. To understand the behavior of a generator connected to such a large system, examine a system consisting of a generator and an infinite bus in parallel supply-ing a load. Assume that the generator's prime mover has a governor mechanism, but that the field is controlled manually by a resistor. lt is easier to explain gener-ator operation without considering an automatic field current regulator, so this dis-cussion will ignore the slight differences caused by the field regulator when one is present. Such a system is shown in Figure 5- 33a. When a generator is connected in parallel with another generator or a large system, the frequency and terminnl voltage of all the mnchines must be the same, SYNCHRONOUS GENERATORS 309 Infinite bus ;-Generator )::::::::::::::::::::~U ,,' f. Pjmoo,' kW PiJJ. bw Pc· kW ,b, FIGURE 5-33 (a) A synchronous generator operating in parallet with an infinite bus. (b) The frequency-versus-power diagram (or lwuse diagmm) for a synchronous generator in parallel with an infinite bus. since their output conductors are tied together. Therefore, their real power-frequency and reacti ve power- voltage characteristics can be plotted back to back, with a common vertical axis. Such a sketch, sometimes infonnally called a house diagram, is shown in Figure 5- 33b. Assume that the generator has just been paralleled with the infinite bus ac-cording to the procedure described previously. Then the generator will be essen-tially "floating" on the line, supplying a small amount of real power and little or no reactive power. nlis situation is shown in Figure 5- 34. Suppose the generator had been paralleled to the line but, instead of being at a slightly higher frequency than the running system, it was at a slightly lower fre-quency. In this case, when paralleling is completed, the resulting situation is shown in Figure 5- 35. Notice that here the no-load frequency of the generator is less than the system's operating frequency. At this frequency, the power supplied by the gen-erator is actually negative. I n other words, when the generator's no-load frequency is less than the system's operating frequency, the generator actually consumes elec-tric power and runs as a motor. It is to ensure that a generator comes on line sup-plying power instead of consuming it that the oncoming machine's frequency is ad-justed higher than the running system's frequency. Many real generators have a 310 ELECTRIC MACHINERY RJNDAMENTALS !.. Hz P. k:W P. k:W ""GURE 5-34 The frequency-versus-power diagram at the moment just after paralleling. !.. Hz P. k:W Pc is constant, the angle of jXsllo. changes as shown, and therefore the angle and magnitude of 1 10. change. Notice that as a result the distance proportional to Q (110. sin ()) increases. In other words, increasing the field current in a synchronous generator operating in parallel with an infinite bus increases the reactive power output of the generator. To summarize, when a generator is operating in parallel with an infinite bus: I. The frequency and tenninal voltage of the generator are controlled by the sys-tem to which it is connected. 2. The governor set points of the generator control the real power supplied by the generator to the system. 3, The field current in the generator controls the reactive power supplied by the generator to the system. lllis situation is much the way real generators operate when connected to a very large power system. Operation of Generators in Par allel with Other Generators of the Same Size When a single generator operated alone, the real and reactive powers (P and Q) supplied by the generator were fixed, constrained to be equal to the power de-manded by the load, and the frequency and tenninal voltage were varied by the SYNCHRONOUS GENERATORS 313 governor set points and the field current. When a generator operated in parallel with an infinite bus, the frequency and tenninal voltage were constrained to be constant by the infmite bus, and the real and reactive powers were varied by the governor set points and the field current. What happens when a synchronous gen-erator is connected in parallel not with an infinite bus, but rather with another gen-erator of the same size? What wi ll be the effect of changing governor set points and field currents? If a generator is connected in parallel with another one of the same size, the resulting system is as shown in Figure 5- 38a. In this system, the basic constraint is that the sum of the real and reactive powers supplied by the two generators must equal the P and Q demanded by the load. The system freq uency is not con-strained to be constant, and neither is the power of a given generator constrained to be constant. The power-frequency diagram for such a system immediately after Gl has been paralleled to the line is shown in Figure 5- 38b. Here, the total power P,,,, (which is equal to PJood) is given by (5- 29a) and the total reactive power is given by (5- 29b) What happens if the governor set points of Gl are increased? When the gov-ernor set points of G2 are increased, the power-frequency curve of G2 shifts up-ward, as shown in Figure 5- 38c. Remember, the total power supplied to the load must not change. At the original frequency fj, the power supplied by G J and Gl will now be larger than the load demand, so the system cannot continue to oper-ate at the same frequency as before. In fact, there is only one freq uency at which the sum of the powers out of the two generators is equal to PJood ' That frequency fl is higher than the original system operating frequency. At that freq uency, Gl sup-plies more power than before, and G J supplies less power than before. Therefore, when two generators are operating together, an increase in gov-ernor set points on one of them I. Increases the system frequency. 2. Increases the power supplied by that generator. while reducing the power supplied by the other one. What happens if the field current of G2 is increased? TIle resulting behavior is analogous to the real-power situation and is shown in Figure 5- 38d. When two generators are operating together and the field current of Gl is increased, I. The system terminal voltage is increased. 2. The reactive power Q supplied by that generator is increased, while the re-active power supplied by the other generator is decreased. kW kVAR 314 t, 601h P~ ,b, Generator I t, h -----------j--------"'~ : II pc. , , , , , , pis) p,. P~ P~ ,,' Generator I V, -------r------, V" , , , , , , Qe. QGI {b Q,. Q. ,d , P" ~ Vn Qm P~ Generntor 2 kW Generator 2 kVAR HGURE 5-38 (a) A generator COIloected in parallel with another machine of the same size. (b) The corresponding house diagram at the moment generator 2 is paralleled with the system. (e) The effect of increasing generator 2's governor set points on the operation of the system. (d) The effect of increasing generator 2's field current on the operation of the system. Generator I Slope = I MW/Hz kW P t =1.5 MW 61.5 Hz 60 H, /= 60 Hz SYNCHRONOUS GENERATORS 315 Generator 2 P2= 1.0MW kW FIGURE 5-39 The llOuse dia.gram for the system in Example 5- :5. If the slopes and no-load frequencies of the generator's speed droop (frequency-power) curves are known, then the powers supplied by each generator and the resulting system frequency can be determined quantitatively. Example 5--6 shows how this can be done. EXllmple 5-6. Figure 5- 38a shows two generators supplying a load. Generator I has a no-load frequency of 61.5 Hz and a slope SpI of I MW/Hz. Generator 2 has a no-load frequency of 61.0 Hz and a slope sn of I MWlHz. The two generators are supplying a real load totaling 2.5 MW at 0.8 PF lagging. The resulting system power-frequency or house diagram is shown in Figure 5- 39. (a) At what frequency is this system operating, and how much power is supplied by each of the two generators? (b) Suppose an additional I-MW load were attached to this power system. What would the new system frequency be, and how much power would GI and G2 supply now? (c) With the system in the configuration described in part b, what will the system frequency and generator powers be if the governor set points on G2 are in-creased by 0.5 Hz? Solutioll The power produced by a synchronous generator with a given slope and no-load frequency is given by Equation (5- 28): P I = SPI(foJ.l - l>y.) P2 = sn(foJ2 - l>y.) Since the total power supplied by the generators must equal the power consumed by the loads, P loo.d = P I + P2 These equations can be used to answer all the questions asked. 316 ELECTRIC MACHINERY RJNDAMENTALS (a) In the first case, both generators have a slope of I MW/Hz, and GI has a no-load frequency of 61 .5 Hz, while G2 has a no-load frequency of 61.0 Hz. The total load is 2.5 MW. Therefore, the system frequency can be found as follows: P load = P I + P l = Spt(f ol.1 - I.Y ' ) + sn(f ol.l - I.Y ' ) 2.5 MW = (1 MW/Hz)(61.5 Hz - I.Y' ) + (1 MW/HzX61 Hz - f.y.) = 6 1.5 MW -(I MW/Hz)l.y• + 61 MW -(1 MW/Hz)f.ys = 122.5 MW - (2 MW/Hz)f.y• therefore (" = 122.5 MW - 2.5 MW = 60 0 H Jsy. (2MW/Hz) . z The resulting powers supplied by the two generators are P I = spl(fnJ .1 - f.y. ) = (1 MW/HzX61.5 Hz - 60.0 Hz) = 1.5 MW P2 = sn(fnJ.2 - f .y. ) = (1 MW/HzX61.0 Hz - 60.0 Hz) = I MW (b) When the load is increased by I MW, the total load becomes 3.5 MW. The new system frequency is now given by Pload = Spt(f ol.1 - I.Y ' ) + sn(f ol.l - f.y.) 3.5 MW = (1 MW/Hz)(61.5 Hz - I.y.) + (1 MW/HzX61 Hz - f.ys) = 6 1.5 MW -(I MW/Hz)l.y• + 61 MW -(1 MW/Hz)f.y. = 122.5 MW - (2 MW/Hz)f.y• therefore (" = 122.5 MW - 3.5 MW = 595 H Jsy. (2MW/Hz) . z The resulting powers are P I = spl(fnJ .1 - f.y.) = (1 MW/HzX61.5 Hz - 59.5 Hz) = 2.0 MW P2 = sn(fnJ.2 - f.y.) = (1 MW/HzX61.0 Hz - 59.5 Hz) = 1.5MW (c) If the no-load governor set points of Gl are increased by 0.5 Hz, the new system frequency becomes P load = Spt(f ol.1 - I.y.) + sn(f ol.l - I.Y ') 3.5 MW = (1 MW/Hz)(61.5 Hz - f.y.) + (1 MW/HzX61.5 Hz - f.y.) = 123 MW - (2 MW/Hz)f.y• 123 MW - 3.5 MW f.y• = (2MW/Hz) = 59.75 Hz The resulting powers are P I = P l = Spt(f ol.l - I.y.) = (1 MW/HzX61.5 Hz - 59.75 Hz) = 1.75 MW SYNCHRONOUS GENERATORS 317 Notice that the system frequency rose, the power supplied by G2 rose, and the power supplied by G1 fell. When two generators of similar size are operating in parallel, a change in the governor set points of one of them changes both the system frequency and the power sharing between them. It would nonnally be desired to adjust only one of these quantities at a time. How can the power sharing of the power system be ad-justed independently of the system frequency, and vice versa? The answer is very simple. An increase in governor set points on one gen-erator increases that machine's power and increases system frequency. A decrease in governor set points on the other generator decreases that machine's power and decreases the system frequency. Therefore, to adjust power sharing without changing the system frequency, increase the governor set points of one generator and simultaneously decrease the governor set points of the other generator (see Figure 5-40a). Similarly, to adjust the system frequency without changing the power sharing, simultaneously increase or decrease both governor set points (see Figure 5-40b). Reactive power and tenninal voltage adjustments work in an analogous fashion. To shift the reactive power sharing without changing Vn simultaneously increase the field current on one generator and decrease the field current on the other (see Figure 5-40c). To change the tenninal voltage without affecting the re-active power sharing, simultaneously increase or decrease both field currents (see Figure 5-4(kl). To summarize, in the case of two generators operating together: I. TIle system is constrained in that the total power supplied by the two genera-tors together must equal the amount consumed by the load. Neither/.y• nor VT is constrained to be constant. 2. To adjust the real power sharing between generators without changing/.y" simultaneously increase the governor set points on one generator while de-creasing the governor set points on the other. TIle machine whose governor set point was increased will assume more of the load. 3. To adjust /.Y' without changing the real power sharing, simultaneously in-crease or decrease both generators' governor set points. 4. To adjust the reactive power sharing between generators without changing VT, simultaneously increase the field current on one generator while decreas-ing the field current on the other. The machine whose field current was in-creased will assume more of the reactive load. 5. To adjust VTwithout changing the reactive power sharing, simultaneously in-crease or decrease both generators' field currents. It is very important that any synchronous generator intended to operate in par-allel with other machines have a drooping frequency-power characteristic. If two generators have flat or nearly flat characteristics, then the power sharing between Genemtor I ,-kW P, P" , kW P, Generator I -kVAR Q, Q, Generator I kVAR Q, H GURE 5-40 /. '" Generator 2 ------\ , I=constant l , , , , , , , , , -' P, P' , , ,' I. Hz ,b , v, Generator 2 , V T = con5lant I , ,,' v, ,d , , , , ,-Generator 2 Qi kW kW kVAR kVAR (a) Shifting power sharing without affecting system frequency. (b) Shifting system frequency without affecting power sharing. (c) Shifting reactive power sharing without affecting temJinal voltage. (d) Shifting terminal voltage without affecting reactive power sharing. 318 SYNCHRONOUS GENERATORS 319 t, p, ------------~==========;---------------- p, FIGURE 5-41 Two synchronous generators with flat frequency-power characteristics. A very tiny change in the no-load frequency of either of these machines could cause huge shifts in the power sharing. Ihem can vary widely with only the tiniest changes in no-load speed. 1llis problem is illustraled by Figure 5-41. Notice that even very tiny changes in InJ in one of the generators would cause wild shifts in power sharing. To ensure good control of power sharing between generators, they should have speed droops in the range of 2 to 5 percent. 5.10 SYNCHRONOUS GENERATOR TRANSIENTS When the shaft torque applied to a generator or the output load on a generator changes suddenly, there is always a transient lasting for a finite period of time be-fore the generator returns to steady state . For example, when a synchronous gen-erator is paralleled with a running power system, it is initially turning faster and has a higher frequency than the power system does. Once it is paralleled, there is a transient period before the generator steadies down on the line and runs at line frequency while supplying a small amount of power to the load. To illustrate this situation, refer to Figure 5-42. Figure 5-42a shows the magnetic fields and the phasor diagram of the generator at the moment just before it is paralleled with the power system. Here, the oncoming generator is supplying no load, its stator current is zero, E ... = V q., and RR = Ro ... At exactly time t = 0, the switch connecting the generator to the power sys-tem is shut, causing a stator current to fl ow. Since the generator's rotor is still turning faster than the system speed, it continues to move out ahead of the sys-tem's voltage Vo/>. The induced torque on the shaft of the generator is given by (4-60) The direction of this torque is opposite to the direction of motion, and it increases as the phase angle between DR and D ... ,Cor E ... and Vo/» increases.nlis torque opposite w 320 ELECTRIC MACHINERY RJNDAMENTALS ", ~ jXSIA Ill. \~ ,b, ""GURE 5-42 w Bs Tind '" k BR x" .. , Tind is clockwise D, (a) The phasor diagram and magnetic fields of a generator at the momem of paralleling with 3. large power system. (b) The phasor diagram and house diagram shortly after a. Here. the rotor has moved on ahead of the net nt3.gnetic fields. producing a clockwise torque. This torque is slowing the rotor down to the synchronous speed of the power system. the direction of motion slows down the generator until it finally turns at synchronous speed with the rest of the power system. Similarly, if the generator were turning at a speed lower than synchronous speed when it was paralleled with the power system, then the rotor would fall be-hind the net magnetic fields, and an induced torque in the direction of motion would be induced on the shaft of the machine. This torque would speed up the rotor until it again began turning at synchronous speed. Transient Stability of Synchronolls Generators We learned earlier that the static stability limit of a synchronous generator is the maximum power that the generator can supply under any circumstances. The maximum power that the generator can supply is given by Equation (5- 21): _ 3'"4,EA P max -X , and the corresponding maximum torque is _ 3'"4,EA T=-X W , (5- 21) (5- 30) In theory, a generator should be able to supply up to this amount of power and torque before becoming unstable. In practice, however, the maximum load that can be supplied by the generator is limited to a much lower level by its dynamic stability limit. To understand the reason for this limitation, consider the generator in Figure 5-42 again. If the torque applied by the prime mover (Tapp) is suddenly increased, the shaft of the generator will begin to speed up, and the torque angle [) will increase as described. As the angle [) increases, the induced torque Tind of the generator will SYNCHRONOUS GENERATORS 321 120 11Xl -- ----- -- -- -- -- -- --, E 80 -0 fin>'aDtaooow '-• , (j(] ! • , 40 .l " f\ / [\ V " V ~ 20 0.5 LO Time,s FIGURE 5-43 The dynamic response when an applied torque equal to 50% of Tmu is suddenly added to a synchronous generator. increase until an angle [) is reached at which T;Dd is equal and opposite to T opp' TIlis is the steady-state operating point of the generator with the new load. However, the ro-tor of the generator has a great deal of inertia, so its torque angle /) actually over-shoots the steady-state position, and gradually settles out in a damped oscillation, as shown in Figure 5-43. TIle exact shape of this damped oscillation can be detennined by solving a nonlinear differential equation, which is beyond the scope of this book. For more information, see Reference 4, p. 345. The important point about Figure 5-43 is that ifat any point in the transient response the instantaneous torque exceeds T"""" the synchronous generator will be unstable. The size of the oscillations depends on how suddenly the additional torque is applied to the synchronous generator. If it is added very gradually, the machine should be able to almost reach the static stability limit. On the other hand, if the load is added sharply, the machine will be stable only up to a much lower limit, which is very complicated to calculate. For very abrupt changes in torque or load, the dynamic stability limit may be less than half of the static sta-bility limit. Short-Circuit Transients in Synchronous Generators By far the severest transient condition that can occur in a synchronous generator is the situation where the three terminals of the generator are suddenly shorted out. Such a short on a power system is called afaull. There are several compo-nents of current present in a shorted synchronous generator, which will be de-scribed below. TIle same effects occur in less severe transients like load changes, but they are much more obvious in the extreme case of a short circuit. 322 ELECTRIC MACHINERY RJNDAMENTALS o Time Phase a DC component -~ o Time Phase b -0 ~ Time DC component Phase c ""GURE 5-44 The total fault currents as a function of time during a three-phase fault at the terminals of a synchronous generator. When a fault occurs on a synchronous generator, the resulting current flow in the phases of the generator can appear as shown in Figure 5-44. The current in each phase shown in Figure 5-42 can be represented as a dc transient component added on top of a symmetrical ac component. 1lle symmetrical ac component by itself is shown in Figure 5-45. Before the fault, only ac voltages and currents were present within the gen-erator, while after the fault, both ac and dc currents are present. Where did the dc currents come from? Remember that the synchronous generator is basically inductive-it is modeled by an internal generated voltage in series with the syn-chronous reactance. Also, recall that a current cannot change instantaneously in an inductor. When the fault occurs, the ac component of current jumps to a very SubtraDsient period Subtransient period ,I Transient period Transient plll (5- 35) If the rated voltage is known, then the maximum acceptable armature current de-tennines the rated kilovoltamperes of the generator: Srated = 3 '"4..rated I A max Srated = V3V L.rated l L.max (5- 36) (5- 37) 328 ELECTRIC MACHINERY RJNDAMENTALS ""GURE 5-47 How the rotor field current lintit sets the rated power factor of a generator. It is important to realize that, for heating the annature windings, the power factor of the armature current is irrelevant. The heating effect of the stator copper losses is given by (5- 38) and is independent of the angle of the current with respect to Vo/>. Because the cur-rent angle is irrelevant to the annature heating, these machines are rated in kilo-voltamperes instead of kilowatts. 1lle other wi nding of concern is the field winding. 1lle field copper losses are given by (5- 39) so the maximum allowable heating sets a maximum field current for the machine. Since Ell = K4>w this sets the maximum acceptable size for Ell. 1lle effect of having a maximum IF and a maximum E). translates directly into a restriction on the lowest acceptable power factor of the generator when it is operating at the rated kilovoltamperes. Fig ure 5-47 shows the phasor diagram of a synchronous generator with the rated voltage and armature current. The current can assume many different angles, as shown. The internal generated voltage Ell is the sum of V 0/> and jXs Ill. Notice that for some possible current angles the required E). exceeds E A,mu. If the generator were operated at the rated annature current and these power factors, the field winding wou ld burn up. TIle angle of III that requires the maximum possible Ell while V 0/> remains at the rated value gives the rated power factor of the generator. It is possible to op-erate the generator at a lower (more lagging) power factor than the rated value, but only by cutting back on the kilovoltamperes supplied by the generator. • • FIGURE 5-48 • 3Vl X, v • (a) ,b, 0 Volts kW " .;" , , E, ' B .' , , , , , SYNCHRONOUS GENERATORS 329 A Volts B , .' , , p 3V. l.ot cosO , , , k:VAR A Q=3V.l.ot sinO Derivation of a synchronous generator capability curve. (a) The generator phasor diagram; (b) the corresponding power units. Synchronous Generator Capability Curves The stator and rotor heat limits, together with any external limits on a synchro-nous generator, can be expressed in graphical fonn by a generator capability dia-gram. A capability diagram is a plot of complex power S = P + j Q. It is derived from the phasor diagram of the generator, assuming that Vo/> is constant at the ma-chine's rated voltage. Figure 5-48a shows the phasor diagram of a synchronous generator operat-ing at a lagging power factor and its rated voltage. An orthogonal set of axes is drawn on the diagram with its origin at the tip of V 0/> and with units of volts. On this diagram, vertical segment AB has a length Xs/). cos (), and horizontal segment OA has a length XsI). sin (). The real power output of the generator is given by P = 3'4,IA cos () (5- 17) 330 ELECTRIC MACHINERY RJNDAMENTALS the reactive power output is given by Q =3V 4 ,lA sine and the apparent power output is given by S=3V 4 ,lA (5-1 9) (5- 35) so the vertical and horizontal axes of this figure can be recalibrated in terms of real and reactive power (Figure 5--48b). The conversion factor needed to change the scale of the axes from volts to voJtamperes (power units) is 3 ~ 1 Xs: ,nd 3 ~ P = 3V4 ,lA cos e = X (XsIA cos e ) (5--40) , . 3 V.p . Q = 3 ~IA Sin e = X (XsIA Sin e ) , (5-41) On the voltage axes, the origin of the phasor diagram is at -Vo/,on the hori-zontal axis, so the origin on the power diagram is at (5-42) TIle field current is proportional to the machine's flux, and the flux is proportional to Elt = Kcf>w. TIle length corresponding to Elt on the power diagram is 3EA V.p X, (5-43) TIle annature current lit is proportional to Xsllt' and the length corresponding to Xsflt on the power diagram is 3Vq,11l.-1lle final synchronous generator capability curve is shown in Figure 5-49. It is a plot of P versus Q, with real power P on the horizontal axis and reactive power Q on the vertical axis. Lines of constant armature current lit appear as lines of constant S = 3Vq,IIt, which are concentric circles around the origin. Lines of constant field current correspond to lines of constant EIt, which are shown as cir-cles of magnitude 3EIt Vq,IXs centered on the point 3V ' - "-'-' X, (5-42) TIle armature current limit appears as the circle corresponding to the rated lit or rated kilovoJtarnperes, and the field current limit appears as a circle corre-sponding to the rated IF or Ell.- Any point that lies within both circles is a safe op-erating point for the generator. It is also possible to show other constraints on the diagram, such as the max-imum prime-mover power and the static stability limit. A capability curve that also reflects the maximum prime-mover power is shown in Figure 5- 50. FIGURE 5-49 Q. k:VAR 3.' • X, SYNCHRONOUS GENERATORS 331 Rotor current limit P. k:W Stator current limit The resulting generator capability curve. FIGURE 5-50 Q. k:VAR P. k:W Prime-mover power limit Origin of rotor current circle: 3.' Q= - ~ ' X, A capability dia.gram showing the prime-mover power limit. 332 ELECTRIC MACHINERY RJNDAMENTALS Example 5-8. A 480-V, 50-Hz, Y -connected, six-pole synchronous generator is rated at 50 kVA at 0.8 PF lagging. It has a synchronous reactance of 1.0 n per phase. As-swne that this generator is cOIUlected to a steam turbine capable of supplying up to 45 kW. The friction and windage losses are 1.5 kW, and the core losses are 1.0 kW. (a) Sketch the capability curve for this generator, including the prime-mover power limit. (b) Can this generator supply a line current of 56A at 0.7 PF lagging? Why or why not? (c) What is the maximwn amOlUlt of reactive power this generator can produce? (d) If the generator supplies 30 kW of real power, what is the maximum amount of reactive power that can be simultaneously supplied? Solutioll The maximum current in this generator can be fOlUld from Equation (5--36): s...'ed = 3 VoI!.l1IIod IA .max (5- 36) The voltage Vol! of this machine is VT 4S0 V Vol! = V3" = ~ = 277 V so the maximum armature ClUTent is ~ 50kVA (".max = 3 Vol! = 3(277 V) = 60 A With this infonnation, it is now possible to answer the questions. (a) The maximum permissible apparent power is 50 kVA, which specifies the max-imum safe armature current. The center of the EA circles is at Q= _ 3Vl X, = _ 3(277V)2 = - 230kVAR 1.0 n The maximum size of EA is given by EA = Vol! + jXslA = 277 LO° V + (i1.0 nX60 L - 36.87° A) = 313 + j48 V = 317 LS.7° V Therefore, the magnitude of the distance proportional to EA is 3EA VoI! DE: = X , = 3(317 VX277 V) = 263 kVAR I.on (5-42) (5-43) The maximum output power available with a prime-mover power of 45 kW is approximately SYNCHRONOUS GENERATORS 333 Q. k:VAR Stator currenl "-50 limit , -- _ / Field current limit --- 25 FIGURE 5-51 " , , \ --150 75 , P.k:W - 75 - \00 - 125 - 150 - 175 - 200 , Maximum printe-mover power - 225 ~-- -Origin of maximum I =::- rotor current - 250 circle The capability diagram for the generator in Example 5--8. P DIU_ = P IIIU';' - P mech I.,.. - P axe ..,.. = 45 kW - 1.5 kW - 1.0 kW = 42.5 kW (This value is approximate because the / 2R loss and the stray load loss were not considered.) The resulting capability diagram is shown in Figure 5--51. (b) A current of 56 A at 0.7 PF lagging produces a real power of P = 3Vo/J,I cos () = 3(277 VX56 AXO.7) = 32.6kW and a reactive power of Q = 3V4>/,Isin () = 3(277 V)(56AXO.714) = 33.2kVAR 334 ELECTRIC MACHINERY RJNDAMENTALS " < > " , , & .~ ! " 200 \00 0 - \00 - 200 r-- 300 r-, -------_ , ________ L ,; _ __ -' : L_ >-----;-----:1.0 PF -----'------------- ----------.... -.. ~ .. ~ ~ ~~~~~~~~~~-=~~~~--~~ o 50 100 150 200 250 300 350 400 450 500 Real power. kW ""GURE 5-52 Capability curve for a real synchronous generator rated at 470 kVA. (Courtesy of Maratlwn Electric Company.) Plotting this point on the capability diagram shows that it is safely within the maximum (It curve but outside the maximrun I" curve. Therefore, this point is not a safe operating condition. (e) When the real power supplied by the generator is zero, the reactive power that the generator can supply will be maximwn. This point is right at the peak of the capability curve. The Q that the generator can supply there is Q = 263 kVAR - 230 kVAR = 33 kVAR (d) If the generator is supplying 30 kW of real power, the maximum reactive power that the generator can supply is 31.5 kVAR. This value can be fOlUld by entering the capability diagram at 30 kW and going up the constant-kilowatt line lUltii a limit is reached. The limiting factor in this case is the field clUTe nt- the anna-ture will be safe up to 39.8 kVAR. Fig ure 5- 52 shows a typical capability for a real synchronous generator. Note that the capability boundaries are not a perfect circle for a real generator. nlis is true because real synchronous generators with salient poles have additional effects that we have not modeled. These effects are described in Appendix C. SYNCHRONOUS GENERATORS 335 Short-Time Operation and Service Factor The most important limit in the steady-state operation ofa synchronous generator is the heating of its armature and field windings. However, the heating limit usu-ally occurs at a point much less than the maximum power that the generator is magnetically and mechanically able to supply. In fact, a typical synchronous gen-erator is often able to supply up to 300 percent of its rated power for a while (un-til its windings burn up). This ability to supply power above the rated amount is used to supply momentary power surges during motor starting and similar load transients. It is also possible to use a generator at powers exceeding the rated values for longer periods of time, as long as the windings do not have time to heat up too much before the excess load is removed. For example, a generator that could supply 1 MW indefinitely might be able to supply 1.5 MW for a couple of minutes without serious hann, and for progressively longer periods at lower power levels. However, the load must finally be removed, or the windings will overheat. The higher the power over the rated value, the shorter the time a machine can tolerate it. Figure 5- 53 illustrates this effect. This figure shows the time in seconds re-quired for an overload to cause thennal damage to a typical electrical machine, whose windings were at nonnal operating temperature before the overload oc-curred.ln this particular machine, a 20 percent overload can be tolerated for JOOO seconds, a 100 percent overload can be tolerated for about 30 seconds, and a 200 percent overload can be tolerated for about 10 seconds before damage occurs. The maximum temperature rise that a machine can stand depends on the in-sulation class of its windings. There are four standard insulation classes: A, B, F, and H. While there is some variation in acceptable temperature depending on a machine's particular construction and the method of temperature measurement, these classes generally correspond to temperature rises of 60, 80, 105, and 125°C, respectively, above ambient temperature. 1lle higher the insulation class of a given machine, the greater the power that can be drawn out of it without over-heating its windings. Overheating of windings is a very serious problem in a motor or generator. It was an old rule of thumb that for each 1 DoC temperature rise above the rated windings temperature, the average lifetime of a machine is cut in half (see Figure 4- 20). Modern insulating materials are less susceptible to breakdown than that, but temperature rises still drastically shorten their lives. For this reason, a syn-chronous machine should not be overloaded unless absolutely necessary. A question related to the overheating problem is: Just how well is the power requirement of a machine known? Before installation, there are often only ap-proximate estimates of load. Because of this, general-purpose machines usualJ y have a sef>!ice factor. The service factor is defined as the ratio of the actual max-imum power of the machine to its nameplate rating. A generator with a service factor of 1.15 can actually be operated at 115 percent of the rated load indefinitely without harm. The service factor on a machine provides a margin of error in case the loads were improperly estimated. 336 ELECTRIC MACHINERY RJNDAMENTALS w' w' w , lei' o \ , 1.2 ""GURE 5-.53 \ '" , 1.4 ~ -~ , , 1.6 1.8 2 Per-unit current , , • , , , 2.2 2.4 2.6 2.8 ThemJal damage curve for a typical synchronous machine. assuming that the windings were already at operational temperature when the overload is applied. (Courtesy of Maratlwn Electric Company.) 5.12 SUMMARY A synchronous generator is a device for converting mechanical power from a prime mover to ac electric power at a specific voltage and frequency. The term synchronous refers to the fact that this machine's electrical frequency is locked in or synchronized with its mechanical rate of shaft rotation. 1lle synchronous gen-erator is used to produce the vast majority of electric power used throughout the world. TIle internal generated voltage of this machine depends on the rate of shaft rotation and on the magnitude of the field nux. The phase voltage of the machine differs from the internal generated voltage by the effects of annature reaction in the generator and also by the internal resistance and reactance of the annature wind-ings. The tenninal voltage of the generator will either equal the phase voltage or be related to it by V3, depending on whether the machine is ,6,- or V-connected. TIle way in which a synchronous generator operates in a real power system depends on the constraints on it. When a generator operates alone, the real and 3 SYNCHRONOUS GENERATORS 337 reactive powers that must be supplied are detennined by the load attached to it, and the governor set points and field current control the frequency and terminal voltage, respectively. When the generator is connected to an infinite bus, its fre-quency and voltage are fixed, so the governor set points and field current control the real and reactive power flow from the generator. In real systems containing generators of approximately equal size, the governor set points affect both fre-quency and power flow, and the field current affects both tenninal voltage and re-active power flow. A synchronous generator's abilit.y to produce electric power is primarily limited by heating within the machine. When the generator's windings overheat, the life of the machine can be severely shortened. Since here are two different windings (armature and field), there are two separate constraints on the generator. The maximum allowable heating in the armature windings sets the maximum kilovoltamperes allowable from the machine, and the maximum allowable heat-ing in the field windings sets the maximum size of E),- The maximum size of Elt and the maximum size of lit together set the rated power factor of the generator. QUESTIONS 5-1. Why is the frequency of a synchronous generator locked into its rate of shaft rotation? 5-2. Why does an alternator's voltage drop sharply when it is loaded down with a lag-ging load? 5-3. Why does an alternator's voltage rise when it is loaded down with a leading load? 5-4. Sketch the phasor diagrams and magnetic field relationships for a synchronous gen-erator operating at (a) unity power factor, (b) lagging power factor, (c) leading power factor. 5-5. Explain just how the synchronous impedance and annature resistance can be deter-mined in a synchronous generator. 5-6. Why must a 60-Hz generator be derated if it is to be operated at 50 Hz? How much derating must be done? 5-7. Would you expect a 400-Hz generator to be larger or smaller than a 6O-Hz genera-tor of the same power and voltage rating? Why? 5-8. What conditions are necessary for paralleling two synchronous generators? 5-9. Why must the oncoming generator on a power system be paralleled at a higher fre-quency than that of the nmning system? 5-10. What is an infinite bus? What constraints does it impose on a generator paralleled with it? 5-11. How can the real power sharing between two generators be controlled without af-fecting the system's frequency? How can the reactive power sharing between two generators be controlled without affecting the system's terminal voltage? 5-12. How can the system frequency of a large power system be adjusted without affect-ing the power sharing among the system's generators? 5-13. How can the concepts of Section 5.9 be expanded to calculate the system frequency and power sharing among three or more generators operating in parallel? 5-14. Why is overheating such a serious matter for a generator? > • 00 • ~ • 0 ., ;; '5 , ., , ~ 338 ELECTRIC MACHINERY RJNDA MENTALS 5-15. Explain in detail the concept behind capability curves. 5-16. What are short-time ratings? Why are they important in regular generator operation? PROBLEMS 5-1. At a location in Europe. it is necessary to supply 300 kW of 60-Hz power. The only power sources available operate at 50 Hz. It is decided to generate the power by means of a motor-generator set consisting of a synchronous motor driving a syn-chronous generator. How many poles should each of the two machines have in or-der to convert 50-Hz power to 60-Hz power? 5-2. A 23OO-V. lOOO-kV A. O .S-PF-Iagging. 60-Hz. two-pole. V-connected synchronous generator has a synchronous reactance of 1.1 0 and an armature resistance of 0.1 5 o. At 60 Hz. its friction and windage losses are 24 kW. and its core losses are IS kW. The field circuit has adc voltage of 200 V. and the maximum I" is IDA. The resistance of the field circuit is adjustable over the range from 20 to 200 O. The OCC of this generator is shown in Figure P5- 1. 3(XX) 2700 2400 2100 1800 1500 1200 900 600 300 / o 0.0 / / 1.0 fo'IGURE " 5- 1 V / 2.0 /' /" V V / 3.0 4.0 5.0 6.0 Field current. A The open-circuit characteristic for the generator in Problem 5- 2. 7.0 8.0 9.0 10.0 (a) How much field current is required to make Vr equal to 2300 V when the gen-erator is mlUling at no load? (b) What is the internal generated voltage of this machine at rated conditions? SYNCHRONOUS GENERATORS 339 (c) How much field current is required to make V r equal to 2300 V when the gen-erator is rulUling at rated conditions? (d) How much power and torque must the generator's prime mover be capable of supplying? (e) Construct a capability curve for this generator. 5-3. Assume that the field current of the generator in Problem 5- 2 has been adjusted to a value of 4.5 A. (a) What will the tenninal voltage of this generator be if it is connected to a 6.-colUlected load with an impedance of 20 L 30° O ? (b) Sketch the phasor diagram of this generator. (c) What is the efficiency of the generator at these conditions? (d) Now asswne that another identical 6.-colUlected load is to be paralleled with the first one. What happens to the phasor diagram for the generator? (e) What is the new tenninal voltage after the load has been added? (f) What must be done to restore the terminal voltage to its original value? 5-4. Assume that the field current of the generator in Problem 5- 2 is adjusted to achieve rated voltage (2300 V) at full-load conditions in each of the questions below. (a) What is the efficiency of the generator at rated load? (b) What is the voltage regulation of the generator if it is loaded to rated kilo-voltamperes with 0.8-PF-Iagging loads? (c) What is the voltage regulation of the generator if it is loaded to rated kilo-voltamperes with 0.8-PF-Ieading loads? (d) What is the voltage regulation of the generator if it is loaded to rated kilo-voltamperes with lUlity power factor loads? (e) Use MATLAB to plot the terminal voltage of the generator as a flUlction of load for all three power factors. 5-5. Assume that the field current of the generator in Problem 5- 2 has been adjusted so that it supplies rated voltage when loaded with rated current at unity power factor. (a) What is the torque angle 0 of the generator when supplying rated current at unity power factor? (b) When this generator is running at full load with lUlity power factor, how close is it to the static stability limit of the machine? 5-6. A 480-V, 4oo-kVA, 0.85-PF-Iagging, 50-Hz, four-pole, 6.-connected generator is driven by a 500-hp diesel engine and is used as a standby or emergency generator. This machine can also be paralleled with the normal power supply (a very large power system) if desired. (a) What are the conditions required for paralleling the emergency generator with the existing power system? What is the generator's rate of shaft rotation after paralleling occurs? (b) If the generator is cOIUlected to the power system and is initially floating on the line, sketch the resulting magnetic fields and phasor diagram. (c) The governor setting on the diesel is now increased. Show both by means of house diagrams and by means of phasor diagrams what happens to the genera-tor. How much reacti ve power does the generator supply now? (d) With the diesel generator now supplying real power to the power system, what happens to the generator as its field current is increased and decreased? Show this behavior both with phasor diagrams and with house diagrams. 5-7. A 13.8-kV, IO-MVA, 0.8-PF-Iagging, 60-Hz, two-pole, Y-COlUlected steam-turbine generator has a synchronous reactance of 12 n per phase and an armature resistance 340 ELECTRIC MACHINERY RJNDAMENTALS of 1.5 n per phase. This generator is operating in parallel with a large power system (infinite bus). (a) What is the magnitude of EA at rated conditions? (b) What is the torque angle of the generator at rated conditions? (c) If the field current is constant, what is the maximwn power possible out of this generator? How much reserve power or torque does this generator have at full load? (d) At the absolute maximum power possible, how much reactive power will this generator be supplying or consruning? Sketch the corresponding phasor dia-gram. (Assrune h is still unchanged.) 5-8. A 480-V, lOO-kW, two-pole, three-phase, 60-Hz synchronous generator's prime mover has a no-load speed of3630 rfmin and a full-load speed of3570 rfmin.1t is op-erating in parallel with a 480-V, 75-kW, four-pole, 60-Hz synchronous generator whose prime mover has a no-load speed of 1800 rhnin and a full-load speed of 1785 rfmin. The loads supplied by the two generators consist of 100 kW at 0.85 PF lagging. (a) Calculate the speed droops of generator I and generator 2. (b) Find the operating frequency of the power system. (c) Find the power being supplied by each of the generators in this system. (d) If Vr is 460 V, what must the generator's operators do to correct for the low ter-minal voltage? 5-9. TIrree physically identical synchronous generators are operating in parallel. They are all rated for a full load of 3 MW at 0.8 PF lagging. The no-load frequency of gen-erator A is 61 Hz, and its speed droop is 3.4 percent. The no-load frequency of generator B is 61.5 Hz, and its speed droop is 3 percent. The no-load frequency of generator C is 60.5 Hz, and its speed droop is 2.6 percent. (a) If a total load consisting of 7 MW is being supplied by this power system, what will the system frequency be and how will the power be shared among the three generators? (b) Create a plot showing the power supplied by each generator as a function of the total power supplied to all loads (you may use MATLAB to create this plot). At what load does one of the generators exceed its ratings? Which generator ex-ceeds its ratings first? (c) Is this power sharing in a acceptable? Why or why not ? (d) What actions could an operator take to improve the real power sharing among these generators? 5-10. A paper mill has installed three steam generators (boilers) to provide process steam and also to use some its waste products as an energy source. Since there is extra ca-pacity, the mill has installed three 5-MW turbine generators to take advantage of the situation. Each generator is a 4160-V, 6250-kVA, 0.85-PF-Iagging, two-pole, Y-cOIlllected synchronous generator with a synchronous reactance of 0.75 n and an annature resistance of 0.04 n. Generators I and 2 have a characteristic power-frequency slope sp of 2.5 MWlHz, and generators 2 and 3 have a slope of3 MW/Hz. (a) If the no-load frequency of each of the three generators is adjusted to 61 Hz, how much power will the three machines be supplying when actual system fre-quency is 60 Hz? (b) What is the maximum power the three generators can supply in this condition without the ratings of one of them being exceeded? At what frequency does this limit occur? How much power does each generator supply at that point? SYNCHRONOUS GENERATORS 341 (c) What would have to be done to get all three generators to supply their rated real and reactive powers at an overall operating frequency of 60 Hz? (d) What would the internal generated voltages of the three generators be under this condition? Problems 5- 11 to 5- 21 refer to a four-pole, Y-connected synchronous generator rated at 470 kVA, 480 V, 60 Hz, and 0.85 PF lagging. Its armature resistance RA is 0.016 n. The core losses of this generator at rated conditions are 7 kW, and the friction and windage losses are 8 kW. The open-circuit and short-circuit characteristics are shown in Figure P5- 2. 5- 11. (a) What is the saturated synchronous reactance of this generator at the rated conditions? (b) What is the unsaturated synchronous reactance of this generator? (c) Plot the saturated synchronous reactance of this generator as a function of load. 5-12. (a) What are the rated current and internal generated voltage of this generator? (b) What field current does this generator require to operate at the rated voltage, current, and power factor? 5-13. What is the voltage regulation of this generator at the rated current and power factor? 5-14. If this generator is operating at the rated conditions and the load is suddenly re-moved, what will the tenninal voltage be? 5-15. What are the electrical losses in this generator at rated conditions? 5-16. If this machine is operating at rated conditions, what input torque must be applied to the shaft of this generator? Express your answer both in newton-meters and in polUld-feet. 5-17. What is the torque angle 0 of this generator at rated conditions? 5-18. Assume that the generator field current is adjusted to supply 480 V under rated con-ditions. What is the static stability limit of this generator? (Note: You may ignore RA to make this calculation easier.) How close is the full-load condition of this genera-tor to the static stability limit? 5-19. Assume that the generator field current is adjusted to supply 480 V under rated con-ditions. Plot the power supplied by the generator as a function of the torque angle o. (Note: You may ignore RA to make this calculation easier.) 5-20. Assume that the generator's field current is adjusted so that the generator supplies rated voltage at the rated load current and power factor. If the field current and the magnitude of the load current are held constant, how will the terminal voltage change as the load power factor varies from 0.85 PF lagging to 0.85 PF leading? Make a plot of the tenninal voltage versus the impedance angle of the load being supplied by this generator. 5-2 1. Assrune that the generator is connected to a 480-V infinite bus, and that its field cur-rent has been adjusted so that it is supplying rated power and power factor to the bus. You may ignore the annature resistance RA when answering the following questions. (a) What would happen to the real and reactive power supplied by this generator if the field flux is reduced by 5 percent? (b) Plot the real power supplied by this generator as a function of the flux cp as the flux is varied from 75 percent to 100 percent of the flux at rated conditions. Open Circuit Characteristic 1200 1100 , , , , , , , , , , , , , '-c > ~ • • , '§ , ., , ~ < • ~ a , 3 • ~ lllOO 9lXl c ,/ C /' 800 700 C /' I c 6lXl 500 c / c 400 c / / 300 200 V 100 c °0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Field current. A Shon Circuit Characteristic 16lXl 1400 C 1200 c lllOO c 800 c 6lXl c 400 c 200 V o o / 0.2 / / , , 0.4 0.6 / , 0.8 HGURE )'5- 2 Field current. A ,b, I. I 1.2 1.3 1.4 1.5 /' / / , , , 1.2 1.4 (a) Open-cirwit characteristic curve for the generator in Problems 5- 11 to 5- 21. (b) Short-cin:uit characteristic curve for the generator in Problems 5- 11 to 5- 21. 342 SYNCHRONOUS GENERATORS 343 (c) Plot the reactive power supplied by this generator as a function of the flux cp as the flux is varied from 75 percent to 100 percent of the flux at rated conditions. (d) Plot the line current supplied by this generator as a function of the flux cp as the flux is varied from 75 percent to lDO percent of the flux at rated conditions. 5-22. A lDO-MVA. 12.S- kV. 0.8S-PF-Iagging. SO-Hz. two-pole. Y -cOlUlected synchronous generator has a per-unit synchronous reactance of 1.1 and a per-lUlit annature resis-tance of 0.012. (a) What are its synchronous reactance and annature resistance in oluns? (b) What is the magnitude of the internal generated voltage E./t at the rated condi-tions? What is its torque angle 0 at these conditions? (c) Ignoring losses in this generator. what torque must be applied to its shaft by the prime mover at full load? 5-23. A three-phase Y-cOIUlected synchronous generator is rated 120 MVA. 13.2 kV. 0.8 PF lagging. and 60 Hz. Its synchronous reactance is 0.9 n. and its resistance may be ignored. (a) What is its voltage regulation? (b) What would the voltage and apparent power rating of this generator be if it were operated at 50 Hz with the same annature and field losses as it had at 60 Hz? (c) What would the voltage regulation of the generator be at 50 Hz? 5-24. Two identical 600-kVA. 480-V synchronous generators are connected in parallel to supply a load. The prime movers of the two generators happen to have different speed droop characteristics. When the field currents of the two generators are equal. one delivers 4DO A at 0.9 PF lagging, while the other delivers 3DO A at 0.72 PF lagging. (a) What are the real power and the reactive power supplied by each generator to the load? (b) What is the overall power factor of the load? (c) In what direction must the field current on each generator be adjusted in order for them to operate at the same power factor? 5-25. A generating station for a power system consists offour 120-MVA, IS-kV, 0.85-PF-lagging synchronous generators with identical speed droop characteristics operating in parallel. The governors on the generators' prime movers are adjusted to produce a 3-Hz drop from no load to full load. TIrree of these generators are each supplying a steady 7S MW at a frequency of 60 Hz, while the fourth generator (called the swing generator) handles all increme ntal load changes on the system while main-taining the system's frequency at 60 Hz. (a) At a given instant, the total system loads are 260 MW at a frequency of 60 Hz. What are the no-load frequencies of each of the system's generators? (b) If the system load rises to 290 MW and the generator's governor set points do not change, what will the new system frequency be? (c) To what frequency must the no-load frequency of the swing generator be ad-justed in order to restore the system frequency to 60 Hz? (d) If the system is operating at the conditions described in part c, what would hap-pen if the swing generator were tripped off the line (disconnected from the power line)? 5-26. Suppose that you were an engineer plaMing a new electric cogeneration facility for a plant with excess process steam. You have a choice of either two IO-MW turbine-generators or a single 20-MW turbine-generator. What would be the advantages and disadvantages of each choice? 344 ELECTRIC MACHINERY RJNDAMENTALS 5-27. A 25-MVA. three-phase. 13.8-kV. two-pole. 60-Hz Y-connected synchronous gen-erator was tested by the open-circuit test. and its air-gap voltage was extrapolated with the following results: Open-circuit test Field current. A Line voltage. t V Extrapolated air-gap voltage. tV 320 13.0 15.4 365 13.8 17.5 380 14.1 18.3 475 15.2 22.8 The short-circuit test was then peIfonned with the following results: Short-circuit test Field current. A Affilature current. A 320 \040 The armature resistance is 0.24 n per phase. 365 1190 380 1240 475 1550 570 16.0 27.4 570 1885 (a) Find the unsaturated synchronous reactance of this generator in oluns per phase and per unit. (b) Find the approximate saturated synchronous reactance Xs at a field current of 380 A. Express the answer both in ohms per phase and per lUlit. (c) Find the approximate saturated synchronous reactance at a field current of 475 A. Express the answer both in ohms per phase and in per-unit. (d) Find the short-circuit ratio for this generator. 5-28. A 20-MVA, 12.2-kY, 0.8-PF-Iagging, Y-connected synchronous generator has a neg-ligible annature resistance and a synchronous reactance of 1.1 per lUlit. The gener-ator is connected in parallel with a 60-Hz, 12.2-kV infinite bus that is capable of supplying or consuming any amOlUlt of real or reactive power with no change in frequency or tenninal voltage. (a) What is the synchronous reactance of the generator in oluns? (b) What is the internal generated voltage EA of this generntor lUlder rated conditions? (c) What is the annature current IA in this machine at rated conditions? (d) Suppose that the generator is initially operating at rated conditions. If the inter-nal generated voltage EA is decreased by 5 percent, what will the new annature current IA be? (e) Repeat part d for 10, 15, 20, and 25 percent reductions in EA. (j) Plot the magnitude of the annature current 1..1 as a function of EA. (You may wish to use MATLAB to create this plot.) SYNCHRONOUS GENERATORS 345 REFERENCES 1. Chaston. A. N. Electric Machinery. Reston. Va.: Reston Publishing. 1986. 2. Del TOTO. V. Electric Machines and Po· .... er Systelll!i. Englewood ClilTs. N.J.: Prentice-Hall. 1985. 3. Fitzgerald. A. E., and C. Kingsley. Jr. Electric Machinery. New Yor(: McGraw-Hill. 1952. 4. Fitzgerald. A. E., C. Kingsley, Jr., and S. D. Umans. Electric Machinery. 5th ed., New York: McGraw-Hill. 1990. 5. Kosow. Irving L. Electric Machinery and Transformers. Englewood ClilTs. N.J.: Prentice-Hall. 1972. 6. Liwschitz-Garik. Michael. and Clyde Whipple. AlteflUlting-Current Machinery. Princeton. N.J.: Van Nostrand. 1961. 7. McPherson. George. An Introduction to Electrical Machines and Traruformers. New Yor(: Wiley. 1981. 8. Siemon. G. R., and A. Straughen. Electric Machines. Reading, Mass.: Addison-Wesley. 1980. 9. Werninck. E. H. (ed.). Electric Motor Hatufbook. London: McGraw-Hill. 1978. CHAPTER 6 SYNCHRONOUS MOTORS S ynchronous motors are synchronous machines used to convert electrical power to mechanical power. This chapter explores the basic operation of synchronous motors and relates their behavior to that of synchronous generators. 6.1 BASIC PRINCIPLES OF MOTOR OPERATION To understand the basic concept of a synchronous motor, look at Figure 6-1 , which shows a two-pole synchronous motor. 1lle field current IF of the motor produces a steady-state magnetic field HR. A three-phase set of voltages is applied to the stator orthe machine, which produces a three-phase current flow in the windings. As was shown in Chapter 4, a three-phase sct of currents in an annature winding produces a uniform rotating magnetic field Bs. Therefore, there are two magnetic fields present in the machine, and the rotor field will tend to line up with the stator field, just as two bar magnets will tend to line up if placed near each other. Since the stator magnetic field is rotating, the rotor magnetic field (and the rotor itself) will constantly try to catch up. TIle larger the angle between the two magnetic fields (up to a certain maximum), the greater the torque on the rotor of the machine. The basic principle of synchronous motor operation is that the rotor "chases" the rotating stator magnetic field around in a circle, never quite catching up with it. Since a synchronous motor is the same physical machine as a synchronous generator, all of the basic speed, power, and torque equations of Chapters 4 and 5 apply to synchronous motors also. 346 SYNCHRONOUS MOTORS 347 o /, 11, 0, o Tind ",k HRx nS '" counterclockwise o FIGURE 6-1 A two-pole synchronous motor. The Equivalent Circuit of a Synchronolls Motor A synchronous motor is the same in all respects as a synchronous generator, except that the direction of power flow is reversed. Since the direction of power fl ow in the machine is reversed, the direction of current fl ow in the stator of the motor may be expected to reverse also. Therefore, the equivalent circuit of a synchronous motor is exactly the same as the equivalent circuit of a synchronous generator, except that the reference direction of IA. is reversed. 1lle resulting full equivalent circuit is shown in Figure 6- 2a, and the per-phase equivale nt circuit is shown in Figure 6- 2b. As be-fore, the three phases of the equivalent circuit may be either Y- or d-connected. Because of the change in direction of lA., the Kirchhoff's voltage law equa-tion for the equivalent circuit changes too. Writing a Kirchhoff's voltage law equation for the new eq uivalent circuit yields I V4> - EA + jXS IA + RAIA I lEA -V4> - jXS IA - RAIA I (6-1) (6-2) This is exactly the srune as the equation for a generator, except that the sign on the current term has been reversed. The Synchronolls Motor from a Magnetic Field Perspective To begin to understand synchronous motor operation, take another look at a syn-chronous generator connected to an infinite bus. The generator has a prime mover 348 ELECTRIC MACHINERY RJNDAMENTALS I" j Xs R, )v., E" I, R .. I" R, j Xs ) V .' R, V, + E" "-' L, I" -j Xs R, )v" E" (a) I, RJ,{ I, --/' j Xs R, V, L, E, V . ,b, ""GURE 6-2 (a) The full equivalent circuit of a three-phase synchronous motor. (b) The per-phase equivalent circuit. turning its shaft, causing it to rotate. The direction of the applied torque Tapp from the prime mover is in the direction of motion, because the prime mover makes the generator rotate in the first place. The phasor diagram of the generator operating with a large field current is shown in Figure 6-3a, and the corresponding magnetic field diagram is shown in Figure 6- 3b. As described before, RR corresponds to (produces) EA , Rnet corre-sponds to (produces) Vo/>, and Rs corresponds to E"at (= -jXsIA). TIle rotation of both the phasor diagram and magnetic fie ld diagram is counterclockwise in the figure, following the standard mathematical convention of increasing angle. TIle induced torque in the generator can be found from the magnetic field diagram. From Equations (4-60) and (4-6 1) the induced torque is given by , FIGURE 6-3 (a) • , SYNCHRONOUS MOTORS 349 B , w.~ -rlf, =------- B~ ,b , (a) Phasor diagram ofa synchronous generator operating at a lagging power factor. (b) The corresponding magnetic field diagram. ~/~ --, , 8 , FIGURE 6-4 u , , 'V , ,,~ w.~ "cr,-------'" ,b , B , (a) Phasor diagram ofa synchronous motor. (b) The corresponding magnetic field diagram. (4-60) (4-61) Notice that from the magnetic field diagram the induced torque in this machine is clockwise, opposing the direction of rotation. In other words, the induced torque in the generator is a countertorque, opposing the rotation caused by the external applied torque "Taw Suppose that, instead of turning the shaft in the direction of motion, the prime mover suddenly loses power and starts to drag on the machine's shaft. What happens to the machine now? The rotor slows down because of the drag on its shaft and falls behind the net magnetic field in the machine (see Figure 6-4a). As the ro-tor, and therefore BR, slows down and falls behind Bne, the operation of the machine suddenly changes. By Equation (4--60), when BR is behind B ... " the induced 350 ELECTRIC MACHINERY RJNDAMENTALS torque's direction reverses and becomes counterclockwise. In other words, the ma-chine's torque is now in the direction of motion, and the machine is acting as a mo-tor. The increasing torque angle 8 results in a larger and larger torque in the direc-tion of rotation, until eventually the motor's induced torque equals the load torque on its shaft. At that point, the machine will be operating at steady state and syn-chronous speed again, but now as a motor. TIle phasor diagram corresponding to generator operation is shown in Fig-ure 6-3a, and the phasor diagram corresponding to motor operation is shown in Figure 6-4a. TIle reason that the quantity jXsI), points from V o/>, to E), in the gen-erator and from E), to V o/> in the motor is that the reference direction of I), was re-versed in the definition of the motor equi valent circuit. The basic difference be-tween motor and generator operation in synchronous machines can be seen either in the magnetic field diagram or in the phasor diagram. In a generator, E), lies ahead of V o/>, and BR lies ahead of 8 0 ... In a motor, E), lies behind V o/>' and BR lies behind Boe, . In a motor the induced torque is in the direction of motion, and in a generator the induced torque is a countertorque opposing the direction of motion. 6.2 STEADY-STATE SYNCHRONOUS MOTOR OPERATION TIlis section explores the behavior of synchronous motors under varying condi-tions of load and field current as well as the question of power-factor correction with synchronous motors. The following discussions will generally ignore the ar-mature resistance of the motors for simplicity. However, R), will be considered in some of the worked numerical calculations. The Synchronous Motor Torque-Speed Characteristic Curve Synchronous motors supply power to loads that are basically constant-speed de-vices. They are usually connected to power systems very much larger than the in-dividual motors, so the power systems appear as infinite buses to the motors. TIlis means that the terminal voltage and the system frequency will be constant regard-less of the amount of power drawn by the motor. 1lle speed of rotation of the mo-tor is locked to the applied electrical frequency, so the speed of the motor will be constant regardless of the load. The resulting torque-speed characteristic curve is shown in Figure 6- 5. The steady-state speed of the motor is constant from no load all the way up to the maximum torque that the motor can supply (called the pull-out torque), so the speed regulation of this motor [Equation (4-68)] is 0 percent. 1lle torque equation is (4-<>1 ) (5- 22) SYNCHRONOUS MOTORS 351 fpullou1 -----------------n_. - n, SR= on xlOO% '" SR=O% f",,0<1 -----------------L-___ ~-------- ,. '.~ FIGURE 6-S The torque-speed characteristic of a synchronous motor. Since the speed of the motor is oonstam. its speed regulation is zero. The maximum or pullout torque occurs when /j = 900. Nonnal full-load torques are much less than that, however. In fact, the pullout torque may typically be 3 times the full-load torque of the machine. When the torque on the shaft of a synchronous motor exceeds the pullout torque, the rotor can no longer remain locked to the stator and net magnetic fields. Instead, the rotor starts to slip behind them. As the rotor slows down, the stator magnetic field "laps" it repeatedly, and the direction of the induced torque in the rotor reverses with each pass. The resulting huge torque surges, first one way and then the other way, cause the whole motor to vibrate severely. The loss of syn-chronization after the pullout torque is exceeded is known as slipping poles. The maximum or pullout torque of the motor is given by (6-3) (6-4) These equations indicate that the larger the field current (and hence E,...), the greater the maximum torque of the nwtor. There is therefore a stability advantage in operating the motor with a large field current or a large E,.,. The Effect of Load Changes on a Synchronous Motor If a load is attached to the shaft of a synchronous motor, the motor will develop enough torque to keep the motor and its load turning at a synchronous speed. What happens when the load is changed on a synchronous motor? 352 ELECTRIC MACHINERY RJNDAMENTALS (a) , , , , IIA2 ilAl ""GURE 6-6 , , , , , , , IV, EA4 "j' __ CC __ _ (b) (a) Pltasor diagram of a motor operating at a leading power factor. (b) The effect of an increase in load on the operation of a synchronous motor. To find oul, examine a synchronous motor operating initially with a leading power factor, as shown in Figure 6--6. If the load on the shaft of the motor is in-creased, the rotor wi ll initially slow down. As it does, the torque angle 8 becomes larger, and the induced torque increases. The increase in induced torque eventu-ally speeds the rotor back up, and the motor again turns at synchronous speed but with a larger torque angle 8. What does the phasor diagrrun look like during this process? To find out, ex-amine the constraints on the machine during a load change. Figure 6--6a shows the motor's phasor diagram before the loads are increased. The internal generated volt-age EA is equal to K, and therefore the annature current IA also increases. Notice that the power-factor angle () changes too, becom-ing Jess and less leading and then more and more lagging. Example 6-1. A 20S-V, 4S-kVA, O.S-PF-Ieading, a-connected, 60-Hz synchro-nous machine has a synchronous reactance of 2.5 0 and a negligible armature resistance. Its friction and windage losses are 1.5 kW, and its core losses are 1.0 kW. Initially, the shaft is supplying a IS-hp load, and the motor's power factor is O.SO leading. (a) Sketch the phasor diagram of this motor, and find the values of lA, fL' and EA. (b) Assume that the shaft load is now increased to 30 hp. Sketch the behavior of the phasor diagram in response to this change. (c) Find lA, fL' and EA after the load change. What is the new motor power factor? Solutioll (a) Initially, the motor's output power is 15 hp. This corresponds to an output of POOl = (15 hp)(0.746 KWlhp) = 11.19 kW Therefore, the electric power supplied to the machine is Pin = P out + P""",blo - jXsIA = 20S L 0° V - (j2.5 0)(27.4 L 36.S7° A) = 20S L 0° V - 6S.5 L 126.S7° V =249.I -jS4.SV =2SSL - 12.4° V The resulting phasor diagram is shown in Figure 6-7a. (b) As the power on the shaft is increased to 30 hp, the shaft slows momentarily, and the internal generated voltage EA swings out to a larger angle /j while maintain-ing a constant magnitude. The resulting phasor diagram is shown in Figure 6-7b. (c) After the load changes, the electric input power of the machine becomes Pin = P out + Pmoc.b lo, and the motor looks purely resistive. As the field current is increased further, the annature current becomes leading, and the motor becomes a capacitive load. 11 is now acting like a capacitor-resistor combination, consuming negative reactive power -Q or, alternatively, supplying reacti ve power Q to the system. A plot of IA versus IF for a synchrono us motor is shown in Figure 6- 9. Such a plot is called a synchronous motor V cu",e, for the obvious reason that it is shaped like the letter V. There are several V curves drawn, corresponding to dif-ferent real power levels. For each curve, the minimum armature current occurs at unity power factor, when only real power is being supplied to the motor. At any other point on the curve, some reactive power is being supplied to or by the mo-tor as well. For field currents less than the value giving minimum lA, the annature current is lagging, consuming Q. For field currents greater than the value giving the minimum lA, the annature current is leading, supplying Q to the power system as a capacitor would. 111erefore, by controlling the field current of a synchronous motor, the reactive power supplied to or consumed by the power system can be controlled. When the projection of the phasor EA onto V 0/> (EA cos 8) is shoner than V 0/> itself, a synchronous motor has a lagging current and consumes Q. Since the field current is small in this situation, the motor is said to be underexcited. On the other hand, when the projection of EA onto Vo/> is longer than Vo/> itself, a synchronous SYNCHRONOUS MOTORS 357 FIGURE 6-10 (a) The phasor diagram of an underexcited synchronous motor. (b) The phasor diagram of an overexcited synchronous motor. motor has a leading current and supplies Q to the power system. Since the field current is large in this situation, the motor is said to be overexcited. Phasor dia-grams illustrating these concepts are shown in Figure 6-10. EXllmple 6-2. The 20S-V, 45-kVA, O.S-PF-Ieading, 8-cOIUlected, 60-Hz synchro-nous motor of the previous example is supplying a 15-hp load with an initial power factor of 0.85 PF lagging. The field current I" at these conditions is 4.0 A. (a) Sketch the initial phasor diagram of this motor, and fmd the values IA and EA. (b) If the motor's flux is increased by 25 percent, sketch the new phasor diagram of the motor. What are EA , lA, and the power factor of the motor now? (c) Assume that the flux in the motor varies linearly with the field current I". Make a plot of 1..1 versus I" for the synchronous motor with a IS-hp load. Solutioll (a) From the previous example, the electric input power with all the losses included is p~ = 13.69 kW. Since the motor's power factor is 0.85 lagging, the resulting annature current flow is I - n,R" ,~"::-;; A -3VoIIcos(J 13.69 kW = 3(20S V)(0.S5) = 25.8 A The angle (J is cos- 1 0.85 = 31.8°, so the phasor current 1..1 is equal to 1 ..1 = 25.8 L -31.So A To find EA , apply Kirchhoff's voltage law [Equation (6--2)]: EA = V oII -jXSIA = 20S L 0° V -(j2.5 0)(25.8 L - 31.So A ) =20SLOo V -64.5L5S.2° V = 182L - 17.5° V The resulting phasor diagram is shown in Figure 6- 11, together with the results for part b. 358 ELECTRIC MACHINERY RJNDAMENTALS , , , I;' I , I" , , , fV~"'208LOOV EA",182L - 17.5° Y.J '- E;' '" 227.5 L _ 13.9° Y ""GURE 6-11 The phasor diagram of the motor in Example 6--2. (b) If the flux cp is increased by 25 percent, then EA = Kcpw will increase by 25 per-cent too: EA2 = 1.25 EAI = 1.25(182 V) = 227.5 V However, the power supplied to the load must remain constant. Since the dis-tance EA sin /) is proportional to the power, that distance on the phasor diagram must be constant from the original flux level to the new flux level. Therefore, EA] sin 8] = EA2 sin ~ ~ = sin- t(EAt sin 8]) E" The annature current can now be found from Kirchhoff's voltage law: _ VI/! -EA2 1 ..1.2 -·X J , I _ 208 LO° V - 227.5 L - 13.9° V ..1. -j2.50 = 56.2 ~.~OA2° V = 22.5 L 13.2° A Finally, the motor's power factor is now PF = cos (13.2°) = 0.974 leading The resulting phasor diagram is also shown in Figure 6-11. (e) Because the flux is assumed to vary linearly with field current, EA will also vary linearly with field current. We know that EA is 182 V for a field current of 4.0A, so EA for any given field current can be fOlUld from the ratio ~ - ~ 182V -4.0A (6-5) SYNCHRONOUS MOTORS 359 The torque angle lj for any given field current can be found from the fact that the power supplied to the load must remain constant: EA I sin 01 = EA2 sin ~ ~ = sin- I ( EA I sin 01 ) E" (6-6) These two pieces of infonnation give us the phasor voltage EA. Once EA is avail-able, the new armature current can be calculated from Kirchhoff's voltage law: _ V</I -EAl IAl -'X J , (6- 7) A MATLAB M-file to calculate and plot IA versus IF using Equations (6- 5) through (6- 7) is shown below: % M-fil e : v_curve. m % M-fil e c reat e a p l o t o f a rmatur e curre nt ver s u s fi e l d % current f o r the syn chro nou s mo t o r of Exampl e 6- 2 . % Firs t , initia liz e the fi e l d curre nt values (2 1 va lues % in the range 3 . S- 5.S A) i _ f = (3S : 1 :5S ) / 1 0; % Now initia liz e a ll o ther values i _a = z e r os( 1 ,2 1 ) ; x_s = 2.5; v_phase = 20S; del tal = -1 7 .5 .. p i / 1 SO; % Pre - a llocate i _a array % Synchro no u s reac tance % Phase voltage at 0 degrees % delt a 1 in radian s e_a l = l S2 .. (cos (delta l ) + j .. s in (delt a l )) ; % Ca l culat e the armature current f o r each value f or ii = 1: 21 ond % Ca l culat e magnitude of e _a2 e_a2 = 45.5 .. i _ f (ii ) ; % Ca l culat e delta2 delta2 = asin ( abs (e_a l ) / abs (e_a2 ) .. s in (deltal ) ) ; % Ca l culat e the phasor e_a2 e_a2 = e_a2 " (cos (delt a2 ) + j" s in (delt a2 )) ; % Ca l culat e i _a i _a( ii ) = ( v_phase % Plot the v - curve p l ot (i _ f. abs (i _a) , ' Co l or ' , 'k' , 'Linewi dth' , 2.0 ) ; x l abel ('Fie l d Current (A) ' , 'Fo ntwe i ght' , 'Sol d ' ) ; y l abel ( ' Armature Current (A) ' , 'Fontwe i ght ' , 'Sold' ) ; titl e ( ' Sync hro no u s M o tor V- CUrve ' , 'Fontwe i ght ' , 'Sol d ' ) ; grid on; The plot produced by this M-flle is shown in Figure 6-12. Note that for a field current of 4.0 A, the annature current is 25.8 A. This result agrees with part a of this example. 360 ELECTRIC MACHINERY RJNDAMENTALS 30 29 28 < 27 " ~ 26 ! 25 ~ 24 23 22 \ \ \ \ / / / / / / " 21 3.' 4.0 4.5 5.0 ,.5 6.0 Field current. A ""GURE 6- 12 V curve for the synchronous motor of Example 6--2. The Synchronolls Motor and Power-Factor Correction Figure 6-13 shows an infinite bus whose OUlpUI is connected through a transmis-sion line 1 0 an industrial plant at a distant point. The industrial plant shown con-sists of three loads. Two of the loads are induction motors with lagging power fac-tors, and the third load is a synchronous motor with a variable power factor. What does the ability to set the power factor of one of the loads do for the power system? To find out, examine the following example problem. (Note: A re-view of the three-phase power equations and their uses is given in Appendix A. Some readers may wish to consult it when studying this problem.) Example 6-3. The infinite bus in Figure 6-13 operates at 480 V. Load I is an in-duction motor consruning 100 kW at 0.78 PF lagging, and load 2 is an induction motor con-sruning 200 kW at 0.8 PF lagging. Load 3 is a synchronous motor whose real power con-srunption is 150 kW. (a) If the synchronous motor is adjusted to operate at 0.85 PF lagging, what is the transmission line current in this system? (b) If the synchronous motor is adjusted to operale at 0.85 PF leading, what is the transmission line current in this system? (c) Assrune thai the transmission line losses are given by line loss where LL stands for line losses. How do the transmission losses compare in the two cases? p,. -Infinite bus Transmission line -Q,. Plant FIGURE 6-13 SYNCHRONOUS MOTORS 361 ,----------------, P, --Q, P, --'" P, --Q, Ind. motor Ind. motor Synchr. motor lOO kW 0.78 PF lagging 200 kW 0.8 PF lagging 150 kW PF = ? L __ ~ A simple power system consisting of an infinite bus supplying an industrial plant through a transmission line. Solutioll (a) In the first case, the real power of load I is 100 kW, and the reactive power of load I is Ql = P t tan () = (1 00 kW) tan (cos-l 0.7S) = (100 kW) tan 3S.7° = SO.2 kVAR The real power of load 2 is 200 kW, and the reactive power of load 2 is Q2= P2tan() = (200 kW) tan (cos-l O.SO) = (200 kW) tan 36.S7° = 150 kVAR The real power load 3 is 150 kW. and the reactive power of load 3 is Q]= p] tan() = (150 kW) tan (cos-l 0.S5) = (150 kW) tan 3 1.So = 93 kVAR Thus, the total real load is P'o< = Pt + P2 + p] = lOO kW + 200 kW + 150 kW = 450 kW and the total reactive load is Q,o<=Qt+ Q2+Q] = SO.2 kVAR + 150 kVAR + 93 kVAR = 323.2 kVAR The equivalent system power factor is thus PF = cos (J = cos (tan- I.2) = cos (tan- l 323.2 kVAR) P 450 kW = cos 35.7° = 0.Sl2 lagging 362 ELECTRIC MACHINERY RJNDAMENTALS Finally, the line current is given by P trX 450 kW IL = v'JVLcos 0 = v'J(480V)(0.812) = 667 A (b) The real and reactive powers of loads I and 2 are unchanged, as is the real power of load 3. The reactive power of load 3 is Q3 = PJ tan() = (150 kW) tan (cos- l 0.85) = (150 kW) tan (_31.8°) = -93 kVAR Thus, the total real load is P'fJI = PI + P2 + PJ = lOOkW + 200kW + 150kW = 450kW and the total reactive load is Q'fJI= QI+Q2+ Q3 = 80.2 kVAR + 150 kVAR - 93 kVAR = 137.2 kVAR The equivalent system power factor is thus PF = cos O = cos (tan- l Q) = cos (tan- l 137.2kVAR) P 450kW = cos 16.96° = 0.957 lagging Finally, the line current is given by P'fJI 450 kW IL = V3V L cos 0 = v'3(480 VXO.957) = 566 A (e) The transmission losses in the first case are The transmission losses in the second case are Notice that in the second case the transmission power losses are 28 percent less than in the first case, while the power supplied to the loads is the same. As seen in the preceding example, the abi lity to adjust the power factor of one or more loads in a power system can significantly affect the operating effi-ciency of the power system. TIle lower the power factor of a system, the greater the losses in the power lines feeding it. Most loads on a typical power system are induction motors, so power syste ms are almost invariably lagging in power factor. Having one or more leading loads (overexcited synchronous motors) on the sys-tem can be useful for the fo llowing reasons: I. A leading load can supply some reactive power Q for nearby lagging loads, instead of it coming from the generator. Since the reactive power does not have to travel over the long and fairly high-resistance transmission lines, the SYNCHRONOUS MOTORS 363 transmission line current is reduced and the power system losses are much lower. (nlis was shown by the previous example.) 2. Since the transmission lines carry less current, they can be smaller for a given rated power flow. A lower equipment current rating reduces the cost of a power system significantly. 3. In addition, requiring a synchronous motor to operate with a leading power factor means that the motor must be run overexcited. nlis mode of operation increases the motor's maximum torque and reduces the chance of acciden-tally exceeding the pullout torque. The use of synchronous motors or other equipment to increase the overall power factor of a power system is called power-factor correction. Since a syn-chronous motor can provide power-factor correction and lower power system costs, many loads that can accept a constant-speed motor (even though they do not necessarily need one) are driven by synchronous motors. Even though a synchro-nous motor may cost more than an induction motor on an individual basis, the ability to operate a synchronous motor at leading power factors for power-factor correction saves money for industrial plants. This results in the purchase and use of synchronous motors. Any synchronous motor that exists in a plant is run overexcited as a matter of course to achieve power-factor correction and to increase its pullout torque. However, running a synchronous motor overexcited requires a high field current and flux, which causes significant rotor heating. An operator must be careful not to overheat the field windings by exceeding the rated field current. The Synchronolls Capacitor or Synchronous Condenser A synchronous motor purchased to drive a load can be operated overexcited to supply reactive power Q for a power system. In fact, at some times in the past a synchronous motor was purchased and run without a load, simply for power-factor correction. nle phasor diagram of a synchronous motor operating overex-cited at no load is shown in Figure 6- 14. Since there is no power being drawn from the motor, the distances propor-tional to power (Ell sin /j and III cos () ) are zero. Since the Kirchhoff's voltage law equation for a synchronous motor is I, I (6-1) ""GURE 6- 14 The phasor diagram of a synchronous Cllpacitor or synchronous condenser. 364 ELECTRIC MACHINERY RJNDAMENTALS Lagging PF (+ Q consumed) Saturation Leading PF (+ QsuppIied) L-________ ~ _________ ~ (a) ""GURE 6-15 ,b, (a) The V curve of a synchronous capacitor. (b) The corresponding machine phasor diagram. the quantity jXSIA. points to the left, and therefore the armature current IA. points straight up. If V 4> and IA. are examined, the voltage-current relationship between them looks like that of a capacitor. An overexcited synchronous motor at no load looks just like a large capacitor to the power system. Some synchronous motors used to be sold specifically for power-factor cor-rection. 1llese machines had shafts that did not even come through the frame of the motor- no load could be connected to them even if one wanted to do so. Such special-purpose synchronous motors were often called synchronous condensers or synchronous capacitors. (Condenser is an old name for capacitor.) 1lle V curve for a synchronous capacitor is shown in Figure 6-15a. Since the real power supplied to the machine is zero (except for losses), at unity power factor the current fA. = D. As the field current is increased above that point, the line current (and the reactive power supplied by the motor) increases in a nearly linear fashion until saturation is reached. Figure 6-I 5b shows the effect of increasing the field current on the motor's phasor diagram. Today, conventional static capacitors are more economical to buy and use than synchronous capacitors. However, some synchronous capacitors may still be in use in older industrial plants. 6.3 STARTING SYNCHRONOUS MOTORS Section 6.2 explained the behavior of a synchronous motor under steady-state conditions. In that section, the motor was always assumed to be initially turning at synchronous speed. What has not yet been considered is the question: How did the motor get to synchronous speed in the first place? To understand the nature of the starting problem, refer to Figure 6- I 6. nlis figure shows a 6D-Hz synchronous motor at the moment power is applied to its stator windings. The rotor of the motor is stationary, and therefore the magnetic D , B, f=O S 'find = 0 ,,' FIGURE 6-16 SYNCHRONOUS MOTORS 365 D , D , • f=I1240s w 'find = Counterclockwise '{.u-/I= 111205 'find = 0 B , ,b, ,,' B , B, w 't-t-- B, w 1 = 31240 5 1=I/60s 'f;nd = clockwise 'f;nd = 0 ,d, ,., Staning problems in a synchronous motor---the torque alternates rapidly in magnitude and direction. so that the net 5taning torque is zero. field DR is stationary. The stator magnetic field Ds is starting to sweep around the motor at synchronous speed. Figure 6-1 6a shows the machine at time t = 0 s, when DR and Ds are exactly lined up. By the induced-torque equation (4- 58) the induced torque on the shaft of the rotor is zero. Figure 6--16b shows the situa-tion at time t = 1 1240 s. In such a short time, the rotor has barely moved, but the stator magnetic field now points to the left. By the induced-torque equation, the torque on the shaft of the rotor is now counterclockwise. Figure 6-1 6c shows the situation at time t = 1/120 s. At that point DR and Ds point in opposite direc-tions, and TiDd again equals zero. At t = 1160 s, the stator magnetic field now points to the right, and the resulting torque is clockwise. Finally, at t = 1/60 s, the stator magnetic field is again lined up with the ro-tor magnetic field, and T iDd = O. During one electrical cycle, the torque was first counterclockwise and then clockwisc, and the average torque over the complete 366 ELECTRIC MACHINERY RJNDAMENTALS cycle was zero. What happens to the motor is that it vibrates heavily with each electrical cycle and finally overheats. Such an approach to synchronous motor starting is hardly satisfactory-managers tend to frown on employees who burn up their expensive equipment. So just how can a synchronous motor be started? TIuee basic approaches can be used to safely start a synchronous motor: I. Reduce the speed of the stator mngneticfield to a low enough value that the rotor can accelerate and lock in with it during one half-cycle of the magnetic field 's rotation. This can be done by reducing the frequency of the applied electric power. 2. Use an extenwl prime mover to accelerate the synchronous motor up to syn-chronous speed, go through the paralleling procedure, and bring the machine on the line as a generator. TIlen, turning off or disconnecting the prime mover wil I make the synchronous machine a motor. 3. Use damper windings or amortisseur windings. The function of damper windings and their use in motor starting will be explained below. Each of these approaches to synchronous motor starting will be described in turn. Motor Starting by Reducing Electrical Frequency I f the stator magnetic fields in a synchronous motor rotate at a low enough speed, there will be no problem for the rotor to accelerate and to lock in with the stator magnetic field. TIle speed of the stator magnetic fields can then be increased to operating speed by gradually increasingf .. up to its normal 50- or 6O-Hz val ue. TIlis approach to starting synchronous motors makes a lot of sense, but it does have one big problem: Where does the variable electrical frequency corne from? Regular power systems are very carefully regulated at 50 or 60 Hz, so un-til recently any variable-frequency voltage source had to come from a dedicated generator. Such a situation was obviously impractical except for very unusual circumstances. Today, things are different. Chapter 3 described the rectifier-inverter and the cycloconverter, which can be used to convert a constant input frequency to any de-sired output frequency. With the development of such modern solid-state variable-frequency drive packages, it is perfectly possible to continuously control the elec-trical frequency applied to the motor all the way from a fraction of a hertz up to and above full rated frequency. If such a variable-frequency drive unit is included in a motor-control circuit to achieve speed control, then starting the synchronous motor is very easy- simply adjust the frequency to a very low value for starting, and then raise it up to the desired operating frequency for normal running. When a synchronous motor is operated at a speed lower than the rated speed, its internal generated voltage Ell = Kcpw will be smaller than normal. If Ell is reduced in magnitude, then the terminal voltage applied to the motor must be SYNCHRONOUS MOTORS 367 reduced as well in order to keep the stator current at safe levels. The voltage in any variable-frequency drive or variable-frequency starter circuit must vary roughly linearly with the applied frequency. To learn more about such solid-state motor-drive units, refer to Chapter 3 and Reference 9. Motor Starting with an External Prime Mover The second approach to starting a synchronous motor is to attach an external start-ing motor to it and bring the synchronous machine up to full speed with the ex-ternal motor. 1l1en the synchronous machine can be paralleled with its power sys-tem as a generator, and the starting motor can be detached from the shaft of the machine. Once the starting motor is turned off, the shaft of the machine slows down, the rotor magnetic field BR falls behind B ... " and the synchronous machine starts to act as a motor. Once paralleling is completed, the synchronous motor can be loaded down in an ordinary fashion. This whole procedure is not as preposterous as it sounds, since many syn-chronous motors are parts of motor-generator sets, and the synchronous machine in the motor-generator set may be started with the other machine serving as the starting motor. Also, the starting motor only needs to overcome the inertia of the synchronous machine without a load-no load is attached until the motor is par-alleled to the power system. Since only the motor's inertia must be overcome, the starting motor can have a much smaller rating than the synchronous motor it starts. Since most large synchronous motors have brushless excitation systems mounted on their shafts, it is often possible to use these exciters as starting motors. For many medium-size to large synchronous motors, an external starting motor or starting by using the exciter may be the only possible solution, because the power systems they are tied to may not be able to handle the starting currents needed to use the amortisseur winding approach described next. Motor Starting by Using Amortisseur Windings By far the most popular way to start a synchronous motor is to employ anwrtisseur or damper windings. Amortisseur windings are special bars laid into notches carved in the face of a synchronous motor's rotor and then shorted out on each end by a large shoT1ing ring. A pole face with a set of amortisseurwindings is shown in Figure 6-17, and amortisseur windings are visible in Figures 5- 2 and 5-4. To understand what a set of amortisseur windings does in a synchronous motor, examine the stylized salient two-pole rotor shown in Figure 6- 18. This ro-tor shows an amortisseur winding with the shorting bars on the ends of the two ro-tor pole faces connected by wires. (This is not quite the way nonnal machines are constructed, but it will serve beautifully to illustrate the point of the windings.) Assume initially that the main rotor field winding is disconnected and that a three-phase set of voltages is applied to the stator of this machine. When the 368 ELECTRIC MACHINERY RJNDAMENTALS o o Shorting "'" Shorting "'" o o o o FIGURE 6-17 A rotor field pole for a synchronous machine showing amortisseur windings in the pole face. (Courtesy ofGeneml Electric Company.) fo'IGURE 6- 18 A simplified diagram of a salient two-pole machine showing amortisseur windings. power is first applied at time t = a s, assume that the magnetic field Bs is vertical, as shown in Figure 6- 19a. As the magnetic field Bs sweeps along in a counter-clockwise direction, it induces a voltage in the bars of the amortisseur winding given by Equation (1-45): where ei!>d = (v x B) • I v = velocity of the bar relative to the magnetic field B = magnetic nux density vector I = length of conductor in the magnetic field ( 1-45) eind and i out of page ®® w 1-+'-- 8 . 00 00 00 eind and i into page I find = counterclockwise Shorting b= (a) 1=05 eind and i into page 00 00 00 '" I find = counterclockwise ". • • d . eind an J ® ® out of page (c) 1= 11120& FIGURE 6-19 SYNCHRONOUS MOTORS 369 "0 0" (b) 1=112405 0 0 00 +t)--u, I find=O 0 0 0 0 (d) 1=312405 The development of a unidirectional torque with synchronous motor amonisseur windings. The bars at the top of the rotor are moving to the right relative to the magnetic field, so the resulting direction of the induced voltage is out of the page. Similarly, the induced voltage is into the page in the bottom bars. These voltages produce a current fl ow out of the top bars and into the bottom bars, resulting in a winding magnetic field Bw pointing to the right. By the induced-torque equation 370 ELECTRIC MACHINERY RJNDAMENTALS the resulting torque on the bars (and the rotor) is counterclockwise. Figure 6-1 9b shows the situation at t = 1 1240 s. Here, the stator magnetic field has rotated 90° while the rotor has barely moved (it simply cannot speed up in so short a time). At this point, the voltage induced in the amortisseur windings is zero, because v is parallel to B. With no induced voltage, there is no current in the windings, and the induced torque is zero. Figure 6-1 9c shows the situation at t = 11120 s. Now the stator magnetic field has rotated 900, and the rotor still has not moved yet. TIle induced voltage [given by Equation (1-45)] in the amortisseur windings is out of the page in the bottom bars and into the page in the top bars. The resulting current flow is out of the page in the bottom bars and into the page in the top bars, causing a magnetic field Bw to point to the left. 1lle resulting induced torque, given by T;Dd = k Bw x Bs is counterclockwise. Finally, Figure 6-1 9d shows the situation at time t = 31240 s. Here, as at t = 1 1240 s, the induced torque is zero. Notice that sometimes the torque is counterclockwise and sometimes it is essentially zero, but it is always unidirectional. Since there is a net torque in a sin-gle direction, the motor's rotor speeds up. (1llis is entirely different from starting a synchronous motor with its normal field current, since in that case torque is first clockwise and then counterclockwise, averaging out to zero. In this case, torque is always in the same direction, so there is a nonzero average torque.) Although the motor's rotor will speed up, it can never quite reach synchro-nous speed. This is easy to understand. Suppose that a rotor is turning at synchro-nous speed. Then the speed of the stator magnetic field Bs is the same as the ro-tor's speed, and there is no relative motion between Bs and the rotor. If there is no relative motion, the induced voltage in the windings will be zero, the resulting current fl ow wi ll be zero, and the winding magnetic field will be zero. Therefore, there will be no torque on the rotor to keep it turning. Even though a rotor cannot speed up all the way to synchronous speed, it can get close. It gets close enough to n'YD< that the regular field current can be turned on, and the rotor will pull into step with the stator magnetic fields. In a real machine, the field windings are not open-circuited during the start-ing procedure. If the field windings were open-circuited, then very high voltages wou ld be produced in them during starting. I f the field winding is short-circuited during starting, no dangerous voltages are produced, and the induced field current actually contributes extra starting torque to the motor. To summarize, if a machine has amortisseur windings, it can be started by the following procedure: I. Disconnect the field windings from their dc power source and short them out. SYNCHRONOUS MOTORS 371 2. Apply a three-phase voltage to the stator of the motor, and let the rotor accel-erate up to near-synchronous speed. The motor should have no load on its shaft , so that its speed can approach n.ync as closely as possible. 3. Connect the dc field circuit to its power source. After this is done, the motor wi ll lock into step at synchronous speed, and loads may then be added to its shaft. The Effect of Amortisseur Windings on Motor Stability If amortisseur windings are added to a synchronous machine for starting, we get a free bonus-an increase in machine stability. The stator magnetic field rotates at a constant speed n.YD<, which varies only when the system frequency varies. I f the rotor turns at n,YD<, then the amortisseur windings have no induced voltage at all. If the rotor turns slower than n,YD<, then there wi ll be relative motion between the rotor and the stator magnetic field and a voltage wi ll be induced in the windings. nlis voltage produces a current fl ow, and the current fl ow produces a magnetic field. The interaction of the two magnetic fields produces a torque that tends to speed the machine up again. On the other hand, if the rotor turns faster than the stator magnetic field, a torque will be produced that tries to slow the rotor down. Thus, the torque produced by the anwrtisseur windings speeds up slow mnchines and slows down fast machines. These windings therefore tend to dampen out the load or other transients on the machine. It is for this reason that amortisseur windings are also called damper windings. Amortisseur windings are also used on synchronous generators, where they serve a similar stabilizing function when a generator is operating in parallel with other generators on an infinite bus. If a variation in shaft torque occurs on the generator, its rotor wi ll momentarily speed up or slow down, and these changes wi ll be opposed by the amortisseur windings. Amortisseur windings improve the overall stability of power systems by reducing the magnitude of power and torque transients. Amortisseur windings are responsible for most of the subtransient current in a faulted synchronous machine. A short circuit at the terminals of a generator is just another fonn of transient, and the amortisseur windings respond very quickly to it. 6.4 SYNCHRONOUS GENERATORS AND SYNCHRONOUS MOTORS A synchronous generator is a synchronous machine that converts mechanical power to electric power, while a synchronous motor is a synchronous machine that converts electric power to mechanical power. In fact, they are both the same physical machine. 372 ELECTRIC MACHINERY RJNDAMENTALS Supply Consume reactive power E" cos {j > V6 reactive power E" cos {j < V. Q Q Supply pow~ P E, E, " ~. ~\V • , , o • ~ I, E" leads V. Consume pow~ p I, " ~ ~V' ~ E , E" Jags E, V , ""GURE 6-10 Phasor diagrams showing the generation and consumption of real power P and reactive power Q by synchronous generators and motors. A synchronous machine can supply real power to or consume real power from a power system and can supply reactive power to or consume reactive power from a power system. All four combinations of real and reactive power flows are possible, and Figure 6-20 shows the phasor diagrams for these conditions. Notice from the figure that I. The distinguishing characteristic of a synchronous generator (supplying P) is that E" lies ahead o/V", while for a motor E" lies behind V",. 2. The distinguishing characteristic of a machine supplying reactive power Q is that E" cos lj > V", regardless of whether the machine is acting as a generator or as a motor. A machine that is consuming reactive power Q has E" cos lj < V",. 6.5 SYNCHRONOUS MOTOR RATINGS Since synchronous motors are the same physical machines as synchronous genera-tors, the basic machine ratings are the same. The one major difference is that a large SYNCHRONOUS MOTORS 373 '" GENERAL@ ELECTRIC " SYNCHRONOUS MOTOR FIGURE 6-21 A typical nameplate for a large synchronous motor. (Courtesy o!General Electric Company.) Ell gives a leading power factor instead of a lagging one, and therefore the effect of the maximum field current limit is expressed as a rating at a leading power factor. Also, since the output of a synchronous motor is mechanical power, a synchronous motor's power rating is usually given in horsepower rather than kilowatts. TIle nameplate of a large synchronous motor is shown in Figure 6-21. In addition to the information shown in the figure, a smaller synchronous motor would have a service factor on its nameplate. In general, synchronous motors are more adaptable to low-speed, high-power applications than induction motors (see Chapter 7). They are therefore commonly used for low-speed, high-power loads. 6.6 SUMMARY A synchronous motor is the same physical machine as a synchronous generator, except that the direction of real power fl ow is reversed. Since synchronous motors are usually connected to power systems containing generators much larger than the motors, the frequency and tenninal voltage of a synchronous motor are fixed (i.e., the power system looks like an infinite bus to the motor). The speed of a synchronous motor is constant from no load to the maximum possible load on the motor. The speed of rotation is _ _ 120..r.: nm -n sync -p The maximum possible power a machine can produce is _ 3V1>EA Pm:u.-X , (5- 21) 374 ELECTRIC MACHINERY RJNDAMENTALS If this value is exceeded, the rotor will not be able to stay locked in with the sta-tor magnetic fields, and the motor will slip poles. I f the field current of a synchronous motor is varied while its shaft load re-mains constant, then the reactive power supplied or consumed by the motor will vary. If Ell cos 8 > ~, the motor will suppl y reactive power, while if Ell cos 8< Vo/» the motor will consume reactive power. A synchronous motor has no net starting torque and so cannot start by itself. TIlere are three main ways to start a synchronous motor: I. Reduce the stator frequency to a safe starting level. 2. Use an external prime mover. 3. Put amortisseur or damper windings on the motor to accelerate it to near-synchronous speed before a direct current is applied to the field windings. If damper windings are present on a motor, they will also increase the sta-bility of the motor during load transients. QUESTIONS 6-1. What is the difference between a synchronous motor and a synchronous generator? 6-2. What is the speed regulation of a synchronous motor? 6-3. When would a synchronous motor be used even though its constant-speed charac-teristic was not needed? 6-4. Why can't a synchronous motor start by itself? 6-5. What techniques are available to start a synchronous motor? 6-6. What are amortisseur windings? Why is the torque produced by them unidirectional at starting, while the torque produced by the main field winding alternates direction? 6-7. What is a synchronous capacitor? Why would one be used? 6-8. Explain, using phasor diagrams, what happens to a synchronous motor as its field current is varied. Derive a synchronous motor V curve from the phasor diagram. 6-9. Is a synchronous motor's field circuit in more danger of overheating when it is op-erating at a leading or at a lagging power factor? Explain, using phasor diagrams. 6-10. A synchronous motor is operating at a fixed real load, and its field current is in-creased. If the armature current falls, was the motor initially operating at a lagging or a leading power factor? 6-11. Why must the voltage applied to a synchronous motor be derated for operation at frequencies lower than the rated value? PROBLEMS 6-1. A 480-V, 60 Hz four-pole synchronous motor draws 50 A from the line at unity power factor and full load. Assuming that the motor is lossless, answer the follow-ing questions: (a) What is the output torque of this motor? Express the answer both in newton-meters and in pound-feet. SYNCHRONOUS MOTORS 375 (b) What must be done to change the power factor to 0.8 leading? Explain your an-swer, using phasor diagrams. (c) What will the magnitude of the line current be if the power factor is adjusted to 0.8 leading? 6-2. A 480-V, 60 Hz 4OO-hp, 0.8-PF-Ieading, six-pole, ~-connected synchronous motor has a synchronous reactance of 1.1 {} and negligible annature resistance. Ignore its friction, windage, and core losses for the purposes of this problem. (a) If this motor is initially supplying 400 hp at 0.8 PF lagging, what are the mag-nitudes and angles of EA and IA? (b) How much torque is this motor producing? What is the torque angle O? How near is this value to the maximum possible induced torque of the motor for this field current setting? (c) If lEAl is increased by IS percent, what is the new magnitude of the armature current? What is the motor's new power factor? (d) Calculate and plot the motor's V curve for this load condition. 6-3. A 2300-V, I()(X)-hp, 0.8-PF leading, 60-Hz, two-pole, Y-cotulected synchronous mo-tor has a synchronous reactance of2.8 n and an annature resistance of 0.4 n. At 60 Hz, its friction and windage losses are 24 kW, and its core losses are 18 kW. The field circuit has a dc voltage of2oo V, and the maximwn IF is 10 A. The open-circuit characteristic of this motor is shown in Figure P6-1. Answer the following ques-tions about the motor, assuming that it is being supplied by an infinite bus. (a) How much field current would be required to make this machine operate at tulity power factor when supplying full load? (b) What is the motor's efficiency at full load and unity power factor? (c) If the field current were increased by 5 percent, what would the new value of the annature current be? What would the new power factor be? How much re-active power is being consumed or supplied by the motor? (d) What is the maximrun torque this machine is theoretically capable of supplying at tulity power factor? At 0.8 PF leading? 6-4. Plot the V curves (fA versus IF) for the synchronous motor of Problem 6- 3 at no-load, half-load, and full-load conditions. (Note that an electronic version of the open-circuit characteristics in Figure P6-1 is available at the book's website. It may simplify the calculations required by this problem. Also, you may assrune that RA is negligible for this calculation.) 6-5. If a 60-Hz synchronous motor is to be operated at 50 Hz, will its synchronous reac-tance be the same as at 60 Hz, or will it change? (Hint: Think about the derivation of Xs.) 6-6. A 480-V, lOO-kW, 0.85-PF-Ieading, 50-Hz, six-pole, V-connected synchronous mo-tor has a synchronous reactance of 1.5 n and a negligible annature resistance. The rotational losses are also to be ignored. This motor is to be operated over a continu-ous range of speeds from 300 to 1000 rlmin, where the speed changes are to be ac-complished by controlling the system frequency with a solid-state drive. (a) Over what range must the input frequency be varied to provide this speed con-trol range? (b) How large is EA at the motor's rated conditions? (c) What is the maximwn power that the motor can produce at rated speed with the EA calculated in part (b)? (d) What is the largest EA could be at 300 r/min? 376 ELECTRIC MACHINERY RJNDAMENTALS 3(XX) 2750 2500 2250 2(XX) 1750 1500 1250 IlXXl 750 250 / / o 0.0 ""GURE 1'( ;-1 / / / 1.0 2.0 /" / / / / 3.0 4.0 5.0 6.0 Field current. A The open-circuit characteristic for the motor in Problems 6-3 and 6-4. 7.0 8.0 9.0 10.0 (e) Assuming that the applied voltage V. is derated by the same amOlUlt as EA. what is the maximwn power the motor could supply at 300 r/min? if) How does the power capability of a synchronous motor relate to its speed? 6-7. A 20S-V. Y-connected synchronous motor is drawing 40 A at unity power factor from a 20S-V power system. The field current flowing under these conditions is 2.7 A. Its synchronous reactance is O.S n. Assume a linear open-circuit characteristic. (a) Find the torque angle o. (b) How much field current would be required to make the motor operate at O.S PF leading? (c) What is the new torque angle in part b? 6-8. A synchronous machine has a synchronous reactance of 2.0 n per phase and an ar-mature resistance of 0.4 n per phase. If EA = 460 L -80 V and V. = 4S0 L 0° V, is this machine a motor or a generator? How much power P is this machine consum-ing from or supplying to the electrical system? How much reactive power Q is this machine consuming from or supplying to the electrical system? SYNCHRONOUS MOTORS 377 6-9. Figure P6--2 shows a synchronous motor phasor diagram for a motor operating at a leading power factor with no R". For this motor. the torque angle is given by ~X~,J ~'!;; CO~' ~''-co tan 0 = -;-VI/! + Xsl" sin () , ( Xsl" cos () ) • - tan -V I/! + Xi " sin () Derive an equation for the torque angle of the synchronous motor if the amlature re-sistance is included. , , , , FIGURE P6-2 , , , , . , Xsl" Sin 0 V . \,"---" --'1 , jXsl" 0: Xsl" cos () , Phasor diagram of a motor at a Jeading power factor. 6-10. A 4S0-V, 375-kVA, O.S-PF-Iagging, V-connected synchronous generator has a syn-chronous reactance of 0.4 n and a negligible armature resistance. This generator is supplying power to a 4S0-V, SO-kW, 0 .8-PF-Ieading, V-connected synchronous mo-tor with a synchronous reactance of 1.1 n and a negligible annature resistance. The synchronous generator is adjusted to have a terminal voltage of 480 V when the mo-tor is drawing the rated power at unit y power factor. (a) Calculate the magnitudes and angles of E" for both machines. (b) If the flux of the motor is increased by 10 percent, what happens to the tenni-nal voltage of the power system? What is its new value? (c) What is the power factor of the motor after the increase in motor flux ? 6- 11, A 4S0-V, lOO-kW, 50-Hz, four-pole, V-connected synchronous motor has a rated power factor of 0.S5 Ieading. At full load, the efficiency is 9 1 percent. The annature resistance is O .OS n, and the synchronous reactance is 1.0 n. Find the following quantities for this machine when it is operating at full load: (a) Output torque (b) Input power (c) n .. (d) E" (e) 1 1 ,,1 if) P coov (g) P mocb + P core + P ""'Y 378 ELECTRIC MACHINERY RJNDAMENTALS 6-12. The V-connected synchronous motor whose nameplate is shown in Figure 6-21 has a per-unit synchronous reactance of 0.9 and a per-unit resistance of 0.02. (a) What is the rated input power of this motor? (b) What is the magnitude of EA at rated conditions? (c) If the input power of this motor is 10 MW. what is the maximum reactive power the motor can simultaneously supply? Is it the annature current or the field current that limits the reactive power output? (d) How much power does the field circuit consume at the rated conditions? (e) What is the efficiency of this motor at full load? if) What is the output torque of the motor at the rated conditions? Express the an-swer both in newton-meters and in pound-feet. 6-13. A 440-V. three-phase. V-connected synchronous motor has a synchronous reactance of 1.5 n per phase. The field C lUTent has been adjusted so that the torque angle 0 is 28° when the power supplied by the generator is 90 kW. (a) What is the magnitude of the internal generated voltage EA in this machine? (b) What are the magnitude and angle of the armature current in the machine? What is the motor 's power factor? (c) If the field current remains constant. what is the absolute maximum power this motor could supply? 6-14. A 460-V, 200-kVA. 0.80-PF-Ieading. 400-Hz. six-pole. V-connected synchronous motor has negligible armature resistance and a synchronous reactance of 0.50 per unit. Ignore all losses. (a) What is the speed of rotation of this motor? (b) What is the output torque of this motor at the rated conditions? (c) What is the internal generated voltage of this motor at the rated conditions? (d) With the field C lUTent remaining at the value present in the motor in part c. what is the maximwn possible output power from the machine? 6-15. A lOO-hp. 440-V. 0.8-PF-Ieading. 6.-cormected synchronous motor has an annature resistance of 0.22 n and a synchronous reactance of 3.0 O. Its efficiency at full load is 89 percent. (a) What is the input power to the motor at rated conditions? (b) What is the line C lUTent of the motor at rated conditions? What is the phase cur-rent of the motor at rated conditions? (c) What is the reactive power consumed by or supplied by the motor at rated conditions? (d) What is the internal generated voltage EA of this motor at rated conditions? (e) What are the stator copper losses in the motor at rated conditions? if) What is P OOIIV at rated conditions? (g) If EA is decreased by 10 percent. how much reactive power will be consumed by or supplied by the motor? 6-16. Answer the following questions about the machine of Problem 6-1 5. (a) If EA = 430 L 13.5° V and V . = 440 L 0° V. is this machine consuming real power from or supplying real power to the power system? Is it consuming re-active power from or supplying reactive power to the power system? (b) Calculate the real power P and reactive power Q supplied or consumed by the machine under the conditions in part a. Is the machine operating within its rat-ings nnder these circumstances? SYNCHRONOUS MOTORS 379 (c) If E,\ = 470 L 1 20 V and V. = 440 L 00 V, is this machine consuming real power from or supplying real power to the power system? Is it consuming re-active power from or supplying reactive power to the power system? (d) Calculate the real power P and reactive power Q supplied or consruned by the machine WIder the conditions in part c. Is the machine operating within its rat-ings under these circumstances? REFERENCES 1. Chaston. A. N. Electric Machinery. Reston. Va.: Reston Publishing. 1986. 2. Del Toro. V. Electric Machines atuf Pov.·er Systems. Englewood Cliffs. NJ : Prentice-Hall. 1985. 3. Fitzgerald. A. E.. and C. Kingsley. Jr. Electric Machinery. New York: McGraw-HilL 1952. 4. Fitzgerald. A. E.. C. Kingsley. Jr .. and S. D. Umans. Electric Machinery, 5th ed. New York: McGraw-Hill. 1990. 5. Kosow. lrving L. Control of Electric Motors. Englewood Cliffs. N.J.: Prentice-Hall. 1972. 6. Liwschitz..Garik. MichaeL and Clyde Whipple. Alternating-Current Machinery. Princeton. N.J.: Van Nostrand. 1961. 7. Nasar. Syed A. (ed.). Handbook of Electric Machines. New York: McGraw-Hill. 1987. 8. Siemon. G. R., and A. Straughen. Electric Machines. Reading. Mass.: Addison-Wesley. 1980. 9. Vithayathil. Joseph. PO'we, Electronics: Principles and Applications. New YorK: McGraw-Hill. 1995. 10. Werninck. E. H. (ed.). Electric MOlOr Handaook. London: McGraw-Hill. 1978. CHAPTER 7 INDUCTION MOTORS I n the last chapter, we saw how amortisseur windings on a synchronous motor cau Id develop a starting torque without the necessity of supplyi ng an external field current to them. In fact, amortisscur windings work so well that a motor could be built without the synchronous motor's main de field circuit at all. A ma-chine with only amortisseur windings is called an induction machine. Such ma-chines are called induction machines because the rotor voltage (which produces the rotor current and the rotor magnetic field) is induced in the rotor windings rather than being physically connected by wires. The distinguishing feature of an induction motor is that no de field current is required to run the machine. Although it is possible to use an induction machine as either a motor or a generator, it has many disadvantages as a generator and so is rarely used in that manner. For this reason, induction machines are usually referred to as induction motors. 7.1 INDUCTION MOTOR CONSTRUCTION An induction motor has the same physical stator as a synchronous machine, with a different rotor construction. A typical two-pole stator is shown in Figure 7-1. It looks (and is) the same as a synchronous machine stator. TIlere are two different types of induction motor rotors which can be placed inside the stator. One is called a cage rotor, while the other is called a wound rotor. Figures 7- 2 and 7- 3 show cage induction motor rotors. A cage induction motor rotor consists ofa series of conducting bars laid into slots carved in the face of the rotor and shorted at either end by large shoT1ing rings. This design is re-ferred to as a cage rotor because the conductors, if examined by themselves, would look like one of the exercise wheels that squirrels or hamsters run on. 380 ,ore FIGURE 7-2 Condoctor rings Rotor rotor conductors INDUCTION MOTORS 381 FIGURE 7-1 The stator of a typical induction motor. showing the stator windings. (Courlesy of MagneTek, Inc.) ,., ,b , (a) Sketch of cage rotor. (b) A typical cage rotor. (Courtesy ofGeneml Electric Company.) 382 ELECTRIC MACHINERY RJNDAMENTALS ,,' ,b, ""GURE 7-3 (a) Cutaway diagram of a typical small cage rotor induction motor. (Courtesy of Ma8neTek. Inc.) (b) Cutaway diagram of a typical large cage TOIor induction motor. (Counesy ofGeneml Electric Company.) TIle other type of rotor is a wound rotor. A wound rotor has a complete set of three-phase windings that are mirror images of the windings on the stator. The three phases of the rotor windings are usually V-connected, and the ends of the three rotor wires are tied to slip rings on the rotor's shaft. TIle rotor windings are shorted through brushes riding on the slip rings. Wound-rotor induction motors therefore have their rotor currents accessible at the stator brushes, where they can be examined and where extra resistance can be inserted into the rotor circuit. It is possible to take advantage of this feature to modify the torque- speed characteris-tic of the motor. Two wound rotors are shown in Figure 7-4, and a complete wound-rotor induction motor is shown in Figure 7- 5. INDUCTION MOTORS 383 (a) ,b , FIGURE 7-4 Typical wound rotors for induction motors. Notice the slip rings and the bars connecting the rotor windings to the slip rings. (Courtel)' ofGeneml Electric Company.) FIGURE 7-5 Cuta.way diagram of a. wound-rotor induction motor. Notice the brushes and slip rings. Also notice that the rotor windings are skewed to eliminate slot h3.ITllonics. (Courtesy of MagneTe/:. Inc.) 384 ELECTRIC MACHINERY RJNDAMENTALS Wou nd-rotor induction motors are more expensive than cage induction mo-tors, and they require much more maintenance because of the wear associated with their brushes and slip rings. As a result, wound-rotor induction motors are rarely used. 7.2 BASIC INDUCTION MOTOR CONCEPTS Induction motor operation is basically the srune as that of amortisseur windings on synchronous motors. That basic operation will now be reviewed, and some im-portant induction motor tenns will be defined. The Development of Induced Torque in an Induction Motor Figure 7--6 shows a cage rotor induction motor. A three-phase set of voltages has been applied to the stator, and a three-phase set of stator currents is flowing. These currents produce a magnetic field Bs, which is rotating in a counterclockwise direction.1lle speed of the magnetic field's rotation is given by (7-1 ) where Ie is the system frequency in hertz and P is the number of poles in the ma-chine. This rotating magnetic field Bs passes over the rotor bars and induces a voltage in them. 1lle voltage induced in a given rotor bar is given by the equation eioo = (v x H) • I where v = velocity of the bar relative to the magnetic field B = magnetic flux density vector I = length of conductor in the magnetic field ( 1-45) It is the relative motion of the rotor compared to the stator magnetic field that produces induced voltage in a rotor bar. The velocity of the upper rotor bars relative to the magnetic field is to the right, so the induced voltage in the upper bars is out of the page, while the induced voltage in the lower bars is into the page. nlis results in a current flow out of the upper bars and into the lower bars. How-ever, since the rotor assembly is inductive, the peak rotor current lags behind the peak rotor voltage (see Figure 7--6b). The rotor current flow produces a rotor mag-netic field H R. Finally, since the induced torque in the machine is given by (4- 58) the resulting torque is counterclockwise. Since the rotor induced torque is coun-terclockwise, the rotor accelerates in that direction. Maximum induced voltage , , , @ @ @ 0 @ • 0 0 0 " " " " ,., II, Net voltage , I ER , , I, , , , , , @ 0 @ • , 0 , , , IIi', 0 , , , , 0 , " " " ,,' 0 0 " 0 0 INDUCTION MOTORS 385 Maximum induced voltage Maximum , induced current , , " IR H , , @ 0 @ 0 • , 0 , 0 , , 0 , " , , , 0 , " " " ,b, ""CURE 7-6 The development of induced torque in an induction motor. (a) The rotating stator field lis induces a voUage in the rotor bars; (b) the rotor voltage produces a rotor currem flow. which lags behind the voUage because of the inductance of the rotor; (c) the rotor currem produces a rotor magnetic field liN lagging 90° behind itself. and liN imeracts with II ... to produce a counterclockwise torque in the machine. There is a fmite upper limit to the motor's speed, however. If the induction motor's rotor were turning at synchronous speed, then the rotor bars wou ld be sta-tionary relative to the magnetic field and there would be no induced voltage. If eioo were equal to 0, then there would be no rotor current and no rotor magnetic field. With no rotor magnetic field, the induced torque would be zero, and the rotor would slow down as a result of friction losses. An induction motor can thus speed up to near-synchronous speed, but it can never exactly reach synchronous speed. Note that in nonnal operation both the rotor and stator mngnetic fields BR and Bs rotate together at synchronous speed n,yDC' while the rotor itselftums at a slower speed. 386 ELECTRIC MACHINERY RJNDAMENTALS The Concept of Rotor Slip TIle voltage induced in a rotor bar of an induction motor depends on the speed of the rotor relative to the magnetic fields. Si nce the behavior of an induction motor depends on the rotor's voltage and current, it is often more logical to talk about this relative speed. Two tenns are commonly used to define the relative motion of the rotor and the magnetic fields. One is slip speed, defined as the difference be-tween synchronous speed and rotor speed: where n.up = slip speed of the machine n,yDC = speed of the magnetic fields nm = mechanical shaft speed of motor (7- 2) TIle other tenn used to describe the relative motion is slip, which is the rela-tive speed expressed on a per-unit or a percentage basis. That is, slip is defined as " " s = ~ (x 100%) n sync (7- 3) " - n s = 'YDC m(x 100%) n.". (7-4) lllis equation can also be expressed in terms of angular velocity w (radians per second) as W -W S = sync m(x 100%) w~oc (7- 5) Notice that if the rotor turns at synchronous speed, s = 0, while if the rotor is sta-tionary, s = 1. All normal motor speeds fall somewhere between those two limits. It is possible to express the mechanical speed of the rotor shaft in tenns of synchronous speed and slip. Solving Equations (7-4) and (7- 5) for mechanical speed yields I nm = ( 1 -s)n,yDC I (7-6) "' I wm ~ (I -s)w,ync I (7- 7) lllese equations are useful in the derivation of induction motor torque and power relationships. The Electrical Frequency on the Rotor An induction motor works by inducing voltages and currents in the rotor of the machine, and for that reason it has sometimes been called a rotating transformer. Like a transformer, the primary (stator) induces a voltage in the secondary (rotor), INDUCTION MOTORS 387 but unlike a transfonner, the secondary frequency is not necessarily the same as the primary frequency. If the rotor of a motor is locked so that it cannot move, then the rotor will have the same frequency as the stator. On the other hand, if the rotor turns at syn-chronous speed, the frequency on the rotor will be zero. What will the rotor fre-quency be for any arbitrary rate of rotor rotation? At nm = 0 rlmin, the rotor frequency fr = Jr, and the slip s = I. At nm = n,ync' the rotor frequency fr = 0 Hz, and the slip s = O. For any speed in between, the ro-tor frequency is directly proportional to the difference between the speed of the mag-netic field n.ync and the speed of the rotor nm . Since the slip of the rotor is defined as (7-4) the rotor frequency can be expressed as (7-8) Several alternative fonns of this expression exist that are sometimes useful. One of the more common expressions is deri ved by substituting Equation (7-4) for the slip into Equation (7--8) and then substituting for n,ync in the denominator of the expression: But n,yDC = 120 fr I P [from Equation (7- 1 )], so Therefore, (7- 9) Example 7-1. A 20S-V, lO-hp, four-pole, 60-Hz, V-connected induction motor has a full-load slip of 5 percent. (a) What is the synchronous speed of this motor? (b) What is the rotor speed of this motor at the rated load? (c) What is the rotor frequency of this motor at the rated load? (d) What is the shaft torque of this motor at the rated load? Solution (a) The synchronous speed of this motor is _ 120f,. n,ync - - p -= 120(60 Hz) _ 4 poles -ISOOr/ min (7- 1) 388 ELECTRIC MACHINERY RJNDAMENTALS (b) The rotor speed of the motor is given by n", = (I - s)n.yDC = (I - 0.95)(l800r/min) = 17lOr/min (c) The rotor frequency of this motor is given by Ir = s/e = (0.05)(60 Hz) = 3 Hz Alternatively, the frequency can be found from Equation (7-9): p /, = 120 (n,ync - nm) = lio(l800r/min -17IOr/min) = 3 Hz (d) The shaft load torque is given by (10 hpX746 W/hp) = (l7IOr/min)(2'lTrad/rXI min/60s) = 41.7N o m The shaft load torque in English units is given by Equation (1- 17): 5252P (7--6) (7--8) (7-9) where 'Tis in pOlUld-feet, P is in horser.ower, and n .. is in revolutions per minute. Therefore, 5252(10 hp) Tload = 17lOr/min = 30.71b o ft 7.3 THE EQUIVALENT CIRCUIT OF AN INDUCTION MOTOR An induction motor re lies for its operation on the induction of voltages and cur-rents in its rotor circuit from the stator circuit (transformer action). Because the in-duction of voltages and curre nts in the rotor circuit of an induction motor is es-sentially a transformer operation, the equivalent circuit of an induction motor will turn o ut to be very similar to the equivalent circuit of a transfonner. An induction motor is called a singly excited machine (as opposed to a doubly excited synchro-nous machine), since power is supplied to only the stator circuit. Because an in-duction motor does not have an independe nt field circ uit, its model will not con-tain an internal voltage source such as the internal generated voltage E,t in a synchronous machine. lt is possible to derive the equivalent circuit of an induction motor from a knowledge of transformers and from what we already know about the variation of rotor frequency with speed in induction motors. TIle induction motor model will be v, INDUCTION MOTORS 389 I, R, I, I, ---I. j + + Rc jXM E, -FIGURE 7-7 The transformer model or an induction motor. with rotor and stator connected by an ideal transformer of turns ratio a,/f" developed by starting with the transformer model in Chapter 2 and then deciding how to take the variable rotor frequency and other similar induction motor effects into account. The Transformer Model of an Induction Motor A transfonner per-phase equivalent circuit, representing the operation of an in-duction motor, is shown in Figure 7- 7. As in any transfonner, there is a certain re-sistance and self-inductance in the primary (stator) windings, which must be rep-resented in the equivalent circuit of the machine. The stator resistance will be called R1• and the stator leakage reactance will be called Xl. These two compo-nents appear right at the input to the machine model. Also, like any transformer with an iron core, the nux in the machine is re-lated to the integral of the applied voltage E l . TIle curve of magnetomotive force versus nux (magnetization curve) for this machine is compared to a similar curve for a power transfonner in Figure 7- 8. Notice that the slope of the induction mo-tor's magnetomotive force-nux curve is much shallower than the curve of a good transformer. TIlis is because there must be an air gap in an induction motor, which greatly increases the reluctance of the nux path and therefore reduces the coupling between primary and secondary windings. TIle higher reluctance caused by the air gap means that a higher magnetizing c urrent is required to obtain a given nux level. Therefore, the magnetizing reactance XM in the equivalent circuit will have a much smaller value (or the susceptance EM will have a much larger value) than it would in an ordinary transformer. The primary internal stator voltage El is coupled to the secondary EN by an ideal transformer with an effective turns ratio a.ff" The effective turns ratio a.ff is fairly easy to detennine for a wound-rotor motor- it is basically the ratio of the conductors per phase on the stator to the conductors per phase on the rotor, modi-fied by any pitch and distribution factor differences. It is rather difficult to see a.ff 390 ELECTRIC MACHINERY RJNDAMENTALS Transformer ""GURE 7-8 ;.Wb Induction motor ~, A-turns The magnetization curve of an induction motor compared to that of a transformer, clearly in the cage of a case rotor motor because there are no distinct windings on the cage rotor. In either case, there is an effective turns ratio for the motor, 1lle voltage ER produced in the rotor in turn produces a current fl ow in the shorted rotor (or secondary) circuit of the machine, TIle primary impedances and the magnetiwtion current of the induction mo-tor are very similar to the corresponding components in a transformer equivalent circuit. An induction motor equivalent circuit differs from a transfonner equiva-lent circuit primarily in the effects of varying rotor frequency on the rotor voltage ER and the rotor impedances RR and jXR' The Rotor Circuit Model In an induction motor, when the voltage is applied to the stator windings, a volt-age is induced in the rotor windings of the machine, In general, the greater the relative motion between the rotor and the stator magnetic fields, the greater the resulting rotor voltage and rotor frequency, The largest relative motion occurs when the rotor is stationary, called the locked-rotor or blocked-rotor condition, so the largest voltage and rotor frequency arc induced in the rotor at that condition, TIle smallest voltage (0 V) and frequency (0 Hz) occur when the rotor moves at the same speed as the stator magnetic field, resulting in no relative motion, The magnitude and frequency of the voltage induced in the rotor at any speed between these extremes is directly propoT1ional to the slip of the rotor, Therefore, if the magnitude of the induced rotor voltage at locked-rotor conditions is called EIlQ, the magnitude of the induced voltage at any slip will be given by the equation (7-1 0) INDUCTION MOTORS 391 + R, FlGURE 7-9 The rotor ci['(;uit model of an induction motor. and the frequency of the induced voltage at any slip will be given by the equation (7-8) This voltage is induced in a rotor containing both resistance and reactance. The rotor resistance RR is a constant (except for the skin effect), independent of slip, while the rotor reactance is affected in a more complicated way by slip. The reactance of an induction motor rotor depends on the inductance of the rotor and the frequency of the voltage and current in the rotor. With a rotor induc-tance of LR, the rotor reactance is given by XR = wrLR = 27rfrLR By Equation (7--8),/, = sf~, so XR -27rSfeLR -s(27rfeLR ) -sXRO (7-11 ) where XRO is the blocked-rotor rotor reactance. The resulting rotor equivalent circuit is shown in Figure 7- 9. The rotor cur-rent flow can be found as E, IR = RR + jsXRO (7-1 2) E", IR = R I + X R S } RO (7-1 3) Notice from Equation (7-1 3) that it is possible to treat all of the rotor effects due to varying rotor speed as being caused by a varying impedance supplied with power from a constant-voltage source ERO. The equivalent rotor impedance from this point of view is (7-1 4) and the rotor equivalent circuit using this convention is shown in Figure 7- 10. In the equivalent circuit in Figure 7-1 0, the rotor voltage is a constant EIiO V and the 392 ELECTRIC MACHINERY RJNDAMENTALS o 25 R, , I'IGURE 7-10 The rotor cirwit model with all the frequency (slip) effects concentrated in resistor RR. ~ ~ \ 100 nm. percentage of synchronous speed ""GURE 7-11 Rotor currem as a function of rotor speed. 125 rotor impedance ZR .• q contains all the effects of varying rotor slip. A plot of the current flow in the rotor as developed in Equations (7-1 2) and (7- 13) is shown in Figure7-11. Notice that at very low slips the resistive tenn RRI s» XRQ, so the rotor re-sistance predominates and the rotor current varies linearly with slip. At high slips, INDUCTION MOTORS 393 XRO is much larger than RRI s, and the rotor current approaches a steady-state value as the slip becomes very large. The Final Equivalent Circuit To produce the fmal per-phase equivalent circuit for an induction motor, it is nec-essary to refer the rotor part of the model over to the stator side. 1lle rotor circuit model that will be referred to the stator side is the model shown in Figure 7-1 0, which has all the speed variation effects concentrated in the impedance term. In an ordinary transformer, the voltages, currents, and impedances on the secondary side of the device can be referred to the primary side by means of the turns ratio of the transfonner: (7-1 5) Ip = I , _1: ,- a (7-1 6) and (7-1 7) where the prime refers to the referred values of voltage, current, and impedance. Exactly the same sort of transfonnati on can be done for the induction mo-tor's rotor circuit. If the effective turns ratio of an induction motor is a off, then the transfonned rotor voltage becomes the rotor current becomes and the rotor impedance becomes Ifwe now make the following definiti ons: R2 = a;ff RR (7-1 8) (7-1 9) (7- 20) (7- 21) (7- 22) then the final per-phase equivalent circ uit of the induction motor is as shown in Figure 7-1 2. The rotor resistance RR and the locked-rotor rotor reactance XIIQ are very dif-ficult or impossible to determine directly on cage rotors, and the effective turns ra-tio a off is also difficult to obtain for cage rotors. Fortunately, though, it is possible to make measurements that will directly give the referred resistance and reac-tance Rl and Xl, even though RR, XRO and aeff are not known separately. The mea-surement of induction motor parameters will be taken up in Section 7.7. 394 ELECTRIC MACHINERY RJNDAMENTALS I, R, I, --+ 1 .1 ~ v • Rc jXM E, -7 -""GURE 7-12 The per-phase equivalent ci['(;uit of an induction motor. Air-gap power : , , ::~~ ' "~oow. r Pu,,,,,f3vrhcos6 :: , R,~ P trictiOll '-L ODdwiD<!ay ""GURE 7-1J p= (Stator copper loss) (C~ losses) (Rotor 00 loss) The power-flow diagram of an induction motor. 7.4 POWER AND TORQUE IN INDUCTION MOTORS Because induction motors are singly exciled machines, their power and torque re-lationships are considerably different from the relationships in the synchronous machines previously studied. TIlis section reviews the power and torque relation-ships in induction motors. Losses and the Power-Flow Diagram An induction motor can be basically described as a rotating transfonner. Its input is a three-phase system of voltages and currents. For an ordinary transfonner, the output is electric power from the secondary windings. TIle secondary windings in an induction motor (the rotor) are shorted out, so no electrical output exists from normal induction motors. Instead, the output is mechanical. The relationship be-tween the input electric power and the output mechanical power of this motor is shown in the power-flow diagram in Figure 7- 13. INDUCTION MOTORS 395 The input powerto an induction motor fln is in the form of three-phase elec-tric voltages and currents. TIle first losses encountered in the machine are [ 2R losses in the stator windings (the stator copper loss P SCL) ' Then some amount of power is lost as hysteresis and eddy currents in the stator (P.:ore). The power re-maining at this point is transferred to the rotor of the machine across the air gap between the stator and rotor. This power is called the air-gap power PAG of the machine. After the power is transferred to the rotor, some of it is lost as / lR losses (the rotor copper loss P RCL), and the res.t is converted from electrical to mechani-cal form (P C<JII¥)' Finally, fri ction and windage losses PF&W and stray losses Pmlsc are subtracted. The remaining power is the output of the motor Pout. The core losses do not always appear in the power-flow diagram at the point shown in Figure 7- 13. Because of the nature of core losses, where they are ac-cou nted for in the machine is somewhat arbitrary. The core losses of an induction motor come partially from the stator circuit and partially from the rotor circuit. Since an induction motor nonnally operates at a speed near synchronous speed, the relative motion of the magnetic fields over the rotor surface is quite slow, and the rotor core losses are very tiny compared to the stator core losses. Since the largest fraction of the core losses comes from the stator circuit, all the core losses are lumped together at that point on the diagram. These losses are represented in the induction motor equivalent circuit by the resistor Rc (or the conductance Gd. If core losses are just given by a number (X watts) instead of as a circuit element they are often lumped together with the mechanical losses and subtracted at the point on the diagram where the mechanical losses are located. The higher the speed of an induction motor, the higher its friction, windage, and stray losses. On the other hand, the higherthe speed of the motor (up to n,ync)' the lower its core losses. Therefore, these three categories of losses are sometimes lumped together and called rotational losses. The total rotational losses of a mo-tor are often considered to be constant with changing speed, since the component losses change in opposite directions with a change in speed. Example 7-2, A 4S0-V, 60-Hz, SO-hp, three-phase induction motor is drawing 60 A at 0.S5 PF lagging. The stator copper losses are 2 kW, and the rotor copper losses are 700 W. The friction and windage losses are 600 W, the core losses are ISOO W, and the stray losses are negligible. Find the following quantities: (a) The air-gap power P AG (b) The power converted P """" (c) TheoutputpowerP OIIt (d) The efficiency of the motor Solutioll To answer these questions, refer to the power-flow diagram for an induction motor (Fig-ure 7- 13). (a) The air-gap power is just the input power minus the stator j 2R losses. The input power is given by 396 ELECTRIC MACHINERY RJNDAMENTALS Pin = V3"V Th cos () = V3"(480 V)(60 A)(O.8S) = 42.4 kW From the power-flow diagram, the air-gap power is given by P AG = Pin -PSCL. - P.:ore = 42.4kW - 2 kW -1.8kW = 38.6kW (b) From the power-flow diagram, the power converted from electrical to mechan-ical fonn is P C<JiIV = P AG -PRCL = 38.6 kW - 700 W = 37.9 kW (c) From the power-flow diagram, the output power is given by P out = P C<JiIV -P F& W -P mi«: = 37.9kW - 600W - OW = 37.3kW or, in horsepower, 1 hp Pout = (37.3 kW) 0.746 kW = SOhp (d) Therefore, the induction motor's efficiency is Power and Torque in an Induction Motor Figure 7- 12 shows the per-phase equivalent circuit of an induction motor. If the equi valent circuit is examined closely, it can be used to derive the power and torque equations governing the operation of the motor. TIle input current to a phase of the motor can be found by di viding the input voltage by the total equivalent impedance: V. I I -(7- 23) Z~ where Zeq = RI + JXI + . I Gc - )BM + V2/S + jX2 (7- 24) Therefore, the stator copper losses, the core losses, and the rotor copper losses can be found. The stator copper losses in the three phases are given by (7- 25) The core losses are given by (7- 26) INDUCTION MOTORS 397 so the air-gap power can be found as ICp -A -G -~ -R -;" ---p-,c -c ---P -,~ -' (7- 27) Look closely at the equivalent circuit of the rotor. The only element in the equivalent circuit where the air-gap power can be consumed is in the resistor Rl/S. Therefore, the air-gap power can also be given by I P AG = 3Ii~ 1 (7- 28) The actual resistive losses in the rotor circuit are given by the equation PRG- = 3 1~ RR (7- 29) Since power is unchanged when referred across an ideal transfonner, the rotor copper losses can also be expressed as I'P-R -cc -~ -3 1~ l C-R-, '1 (7- 30) After stator copper losses, core losses, and rotor copper losses are sub-tracted from the input power to the motor, the remaining power is converted from electrical to mechanical form. This power converted, which is sometimes called developed mechanical power, is given by = 3Ii R2 _ 312R , " = 31~ R2(~ - 1 ) (7- 31) Notice from Equations (7- 28) and (7- 30) that the rotor copper losses are equal to the air-gap power times the slip: (7- 32) Therefore, the lower the slip of the motor, the lower the rotor losses in the ma-chine. Note also that if the rotor is not turning, the slip S = 1 and the air-gap power is entirely consumed in the rotor. This is logical, since if the rotor is not turning, the output power Pout (= "Tload w",) must be zero. Since P.:<>D¥ = P AG - PRCL, this also gives another relationship between the air-gap power and the power con-verted from electrical to mechanical fonn: P.:onv = P AG -P RCL (7- 33) 398 ELECTRIC MACHINERY RJNDAMENTALS Finally, if the friction and windage losses and the stray losses are known, the output power can be found as 'I "p oo C-, - ~ ~ C O -", ---C p~, -& -W ---C P~ ~ -." -'1 (7- 34) TIle induced torque rind in a machine was defined as the torque generated by the internal electric-to-rnechanical power conversion. This torque differs from the torque actually available at the tenninals of the motor by an amount equal to the friction and windage torques in the machine. The induced torque is given by the equation (7- 35) TIlis torque is also called the developed torque of the machine. TIle induced torque of an induction motor can be expressed in a different fonn as well. Equation (7- 7) expresses actual speed in terms of synchronous speed and slip, while Equation (7- 33) expresses P"DDY in terms of P AG and slip. Substituting these two equations into Equation (7- 35) yields (1 -s)P A G r ind = ( 1 s)WS ytlC (7- 36) TIle last equation is especially usefu l because It expresses induced torque directly in tenns of air-gap power and synchronous speed, which does not vary. A knowl-edge of P AG thus directly yields r ind . Separating the Rotor Copper Losses and the Power Converted in an Induction Motor 's Equivalent Circuit Part of the power coming across the air gap in an induction motor is consumed in the rotor copper losses, and part of it is converted to mechanical power to drive the motor's shaft. It is possible to separate the two uses of the air-gap power and to indicate them separately on the motor equivalent circuit. Equation (7- 28) gives an expression for the total air-gap power in an in-duction motor, while Equation (7- 30) gives the actual rotor losses in the motor. TIle air-gap power is the power which would be consumed in a resistor of value Ris, while the rotor copper losses are the power which would be consumed in a resistor of value R2. TIle difference between them is P eDDY' which must therefore be the power consumed in a resistor of value Reonv = ~2 -R2 = R2(~ -1) (7- 37) INDUCTION MOTORS 399 I, R, I, --+ (SCL) 1 .1 + (RCL) (Core loss) R, jXM E, J ./ -FIGURE 7-14 The per-phase equivalent circuit with rotor losses and P CO« separated. Per-phase equivalent circuit with the rotor copper losses and the power con-verted to mechanical fonn separated into distinct e lements is shown in Figure 7- 14. Example 7-3. A 460-V. 25-hp. 6()"'Hz. four-pole. V-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: Rt = 0.641 n XI = 1.106 n Rl = 0.332 n Xl = 0.464 n XM = 26.3 n The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency. find the motor's (a) Speed (b) Stator current (c) Power factor (d) P OO 90°). Since the first effect is larger than the second one, the overall induced torque in-creases to supply the motor's increased load. When does an induction motor reach pullout torque? This happens when the point is reached where, as the load on the shaft is increased, the sin 8 tenn de-creases more than the BR tenn increases. At that point, a further increase in load decreases "TiDd, and the motor stops. It is possible to use a knowledge of the machine's magnetic fields to approx-imately derive the output torque-versus-speed characteristic of an induction motor. Remember that the magnitude of the induced torque in the machine is given by (4-61) Each tenn in this expression can be considered separately to derive the overall machine behavior. The individual tenns are L BR . TIle rotor magnetic field is directly proportional to the current fl owing in the rotor, as long as the rotor is unsaturated. TIle current flow in the rotor in-creases with increasing slip (decreasing speed) according to Equation (7-1 3). This current fl ow was plotted in Figure 7-11 and is shown again in Fig-ure 7-16a. 2. B"",. The net magnetic field in the motor is proportional to E] and therefore is approximately constant (E ] actually decreases with increasing current flow, but this effect is small compared to the other two, and it will be ignored in this graphical development). TIle curve for B"", versus speed is shown in Fig-ure 7-16b. 3. sin 8. TIle angle 8 between the net and rotor magnetic fields can be expressed in a very useful way. Look at Figure 7-15b. In this figure, it is clear that the angle 8 is just equal to the power-factor angle of the rotor plus 90°: (7-38) TIlerefore, sin 8 = sin (OR + 90°) = cos OR. TIlis tenn is the power factor of the rotor. The rotor power-factor angle can be calculated from the equation (7-39) 404 ELECTRIC MACHINERY RJNDAMENTALS J, oe I DRI r---~ ~---------7~-- '. n,ymc ,., L----------c-~- '. '.-,b, 0 '. ,e, '.-,~ ,d, '.-'. The resulting rotor power factor is given by PFR = cos (JR FlGURE 7- 16 Grapltical development of an induction motor torque-speed cltaT3cteristic. (a) Plot of rotor current (and titus IURI) versus speed for an induction motor; (b) plot of net magnetic field versus speed for the motor; (c) plot of rotor power factor versus speed for the motor; (d) the resulting torque-speed characteristic. PF = cos (tan-1SXRQ ) , R, (7-40) A plot of rotor power factor versus speed is shown in Figure 7-1 6c. Since the induced torque is proportional to the product of these three tenns, the torque-speed characteristic of an induction motor can be constructed from the INDUCTION MOTORS 405 graphical multiplication of the previous three plots (Figure 7-1 6a to c). 1lle torque-speed characteristic of an induction motor derived in this fashion is shown in Figure7-16d. This characteristic curve can be di vided roughly into three regions. The first region is the low-slip region of the curve. In the low-slip region, the motor slip in-creases approximately linearly with increased load, and the rotor mechanical speed decreases approximately linearly with load. In this region of operation, the rotor reactance is negligible, so the rotor power factor is approximately unity, while the rotor current increases Ii nearly with slip. The entire normnl steady-state operating range of an induction motor is included in this linear low-slip region. Thus in normal operation, an induction motor has a linear speed droop. The second region on the induction motor's curve can be called the moderate-slip region. In the moderate-slip region, the rotor freq uency is higher than before, and the rotor reactance is on the same order of magnitude as the rotor resistance. In this region, the rotor current no longer increases as rapidly as before, and the power factor starts to drop. TIle peak torque (the pullout torque) of the motor occurs at the point where, for an incremental increase in load, the increase in the rotor current is exactly balanced by the decrease in the rotor power factor. The third region on the induction motor's curve is called the high-slip re-gion. In the high-slip region, the induced torque actually decreases with increased load, since the increase in rotor current is completely overshadowed by the de-crease in rotor power factor. For a typical induction motor, the pullout torque on the curve will be 200 to 250 percent of the rated full-load torque of the machine, and the starting torque (the torque at zero speed) will be 150 percent or so of the full-load torque. Unlike a syn-chronous motor, the induction motor can start with a fuJI load attached to its shaft. The Derivation of the Induction Motor Induced-Torque Equation It is possible to use the eq uivalent circuit of an induction motor and the power-flow diagram for the motor to derive a general expression for induced torque as a function of speed. The induced torque in an induction motor is given by Equation (7- 35) or (7- 36): (7- 35) (7- 36) The latter equation is especially useful, since the synchronous speed is a constant for a given frequency and number of poles. Since w' ync is constant, a knowledge of the air-gap power gives the induced torque of the motor. The air-gap power is the power crossing the gap from the stator circuit to the rotor circuit. It is equal to the power absorbed in the resistance R2! s. How can this power be found? 406 ELECTRIC MACHINERY RJNDAMENTALS + v • -I, -""GURE 7-17 , R, I, -+ jXM E, -Per-phase equivalent circuit of an induction motor. ~ Refer to the equivalent circuit given in Figure 7-17. In this figure, the air-gap power supplied to one phase of the motor can be seen to be , R, P AG.t4> = 12s Therefore, the total air-gap power is _ 2 R2 P AG -312 , If /l can be determined, then the air-gap power and the induced torque will be known. Although there are several ways to solve the circuit in Figure 7-17 for the cur-rent ll, perhaps the easiest one is to detennine the lllevenin equivalent of the por-tion of the circuit to the left of the X's in the figure. Thevenin's theorem states that any linear circuit that can be separated by two tenninals from the rest of the system can be replaced by a single voltage source in series with an equivalent impedance. If this were done to the induction motor equivalent circuit, the resulting circuit would be a simple series combination of elements as shown in Figure 7-1 8c. To calculate the lllevenin equivalent of the input side of the induction motor equivalent circuit, first open-circuit the terminals at the X's and fmd the resulting open-circuit voltage present there. lllen, to find the Thevenin impedance, kill (short-circuit) the phase voltage and find the Zeq seen "looking" into the tenninals. Figure 7-1 8a shows the open tenninals used to find the Thevenin voltage. By the voltage divider rule, ZM V TH = V4> Z + Z M , = V jXM 4>R t + jX] + jXM llle magnitude of the Thevenin voltage Vru is (7-4 I a) INDUCTION MOTORS 407 jX, : , v '" jX/oI V TH R]+jX]+jX/oI • (-t)V, jX/oI Vrn XM VTH '" V. ~R]2+(X] +X/oI)2 ('J jX, R, (bJ jXrn (oj FIGURE 7-18 (a) The Thevenin equivalent voltage of an induction ntotor input circuit. (b) The Thevenin equivalent impedance of the input circuit. (c) The resulting simplified equivalent circuit of an induction motor. Since the magnetization reactance XM » X] and XM » RJ, the magnitude of the Thevenin voltage is approximately (7-4 (b) to quite good accuracy. Figure 7-1 8b shows the input circuit with the input voltage source killed. The two impedances are in parallel, and the TIlevenin impedance is given by ZIZM ZTH = Z l + ZM (7-42) This impedance reduces to 408 ELECTRIC MACHINERY RJNDAMENTALS (7-43) Because XM » Xl and XM + Xl »Rb the TIlevenin resistance and reactance are approximately given by (7-44) (7-45) TIle resulting equivalent circuit is shown in Figure 7-1 8c. From this circuit, the current 1 2 is given by VTH 1 2 = ZTH +2; _ ~~~~V~TH!lL~~~ Rrn + R2/ s + jXTH + jX2 The magnitude of this current is V TH /2 = Y(RTH + R2/sP + (Xrn + X2)2 TIle air-gap power is therefore given by R, P = 3[ 2 -AG 2 S = (Rrn + R2/si + (Xrn + X2)2 and the rotor-induced torque is given by P AG T;nd=w ~oc (7-46) (7-47) (7-48) (7-49) (7- 50) A plot of induction motor torque as a function of speed (and slip) is shown in Figure 7-1 9, and a plot showing speeds both above and below the normal mo-tor range is shown in Figure 7- 20. Comments on the Induction Motor Torque-Speed Curve TIle induction motor torque-speed characteristic curve plotted in Figures 7-1 9 and 7- 20 provides several important pieces ofinfonnation about the operation of induction motors. TIlis infonnation is summarized as follows: INDUCTION MOTORS 409 Pullout torque \ 400% " ~ = • • 300% • Starting • " torque § ( ~ 200% " ~ 100% ____________________ ~U~l~~~~~:~~ o Mechanical speed FIGURE 7-19 A typical induction motor torque-speed characteristic curve. I. 1lle induced torque of the motor is. zero at synchronous speed. 1llis fact has been discussed previously. 2. 1lle torque- speed curve is nearly linear between no load and full load. In this range, the rotor resistance is much larger than the rotor reactance, so the ro-tor current, the rotor magnetic field, and the induced torque increase linearly with increasing slip. 3. There is a maximum possible torque that cannot be exceeded. nlis torque, called the pullout torque or breakdown torque, is 2 to 3 times the rated full -load torque of the motor. The next section of this chapter contains a method for calculating pullout torque. 4. 1lle starting torque on the motor is slightly larger than its full-load torque, so this motor will start carrying any load that it can supply at fu ll power. 5. Notice that the torque on the motor for a given slip varies as the square of the applied voltage. nlis fact is useful in one fonn of induction motor speed con-trol that will be described later. 6. If the rotor of the induction motor is driven faster than synchronous speed, then the direction of the induced torque in the machine reverses and the ma-chine becomes a generator, converting mechanical power to electric power. 1lle use of induction machines as generators will be described later. 410 ELECTRIC MACHINERY RJNDAMENTALS 400 ] = 200 e '0 Braking If region ; § ] - 200 ~ -400 ""GURE 7-10 Tmn --........~ Pullout torque Motor region nsyDC..-/ Mechanical speed Generator region Induction motor torque-speed characteristic curve. showing the extended operating ranges (braking region and generator region). 7. If the motor is turning backward relative to the direction of the magnetic fields, the induced torque in the machine will stop the machine very rapidly and will try to rotate it in the other direction. Since reversing the direction of magnetic field rotation is simply a matter of switching any two stator phases, this fact can be used as a way to very rapidly stop an induction motor. The act of switching two phases in order to stop the motor very rapidly is called plugging. TIle power converted to mechanical fonn in an induction motor is equal to and is shown plotted in Figure 7- 21. Notice that the peak power supplied by the induction motor occurs at a different speed than the maximum torque; and, of course, no power is converted to mechanical fonn when the rotor is at zero speed. Maximum (pullout) Torque in an Induction Motor Since the induced torque is equal to PAG/w'Y"'" the maximum possible torque oc-curs when the air-gap power is maximum. Since the air-gap power is equal to the power consumed in the resistor R2 /s, the maximum induced torque will occur when the power consumed by that resistor is maximum. INDUCTION MOTORS 411 800 120 700 105 600 90 , 500 75 • Z ~ , ~ 400 60 c , ~ " , , " 300 45 0 200 30 100 15 0 0 250 500 750 1000 1250 1500 1750 2000 Mechanical speed. r/min FIGURE 7-21 Induced torque and power convened versus motor speed in revolutions per minute for an example four-pole induction motor. When is the power supplied to Rl/s at its maximum? Refer to the simplified equivalent circuit in Figure 7- 18c. In a situation where the angle of the load im-pedance is fixed, the maximum power transfer theorem states that maximum power transfer to the load resistor Rll s wil I occur when the magnitude of that im-pedance is equal to the magnitude of the source impedance. TIle equivalent source impedance in the circuit is Zsoun:c = RTH + jXTH + jX2 so the maximum power transfer occurs when R -i- = YRfH + (XTH + Xii (7- 51) (7- 52) Solving Equation (7- 52) for slip, we see that the slip at pullout torque is given by (7- 53) Notice that the referred rotor resistance Rl appears only in the numerator, so the slip of the rotor at maximum torque is directly proportional to the rotor resistance. 41 2 ELECTRIC MACHINERY RJNDAMENTALS 800 ,--------------------------------, R, I 700 600 R. ;"" z • ~400 "I ~ 300 200 100 R, R, o~~~~~~~~~~~~~~~~ o 250 500 750 1000 1250 1500 1750 2000 Mechanical speed. r/min ""GURE 7- 22 The effect of varying rotor resistance on the torque-.\peed characteristic of a wound-rotor induction motor. TIle value of the maximum torque can be found by inserting the expression for the slip at maximum torque into the torque equation [Equation (7- 50)]. TIle re-sulting equation for the maximum or pullo ut torque is (7- 54) TIlis torque is proportional to the square of the supply voltage and is also in-versely related to the size of the stator impedances and the rotor reactance. The smaller a machine's reactances, the larger the maximum torque it is capable of achieving. Note that slip at which the maximum torque occurs is directly propor-tional to rotor resistance [Equation (7- 53)], but the value of the maximum torque is independent of the value of rotor resistance [Equation (7- 54)]. TIle torque-speed characteristic for a wound-rotor induction motor is shown in Figure 7- 22. Recall that it is possible 1.0 insert resistance into the rotor circuit of a wound rotor because the rotor circuit is brought out to the stator through slip rings. Notice on the figure that as the rotor resistance is increased, the pullout speed of the motor decreases, but the maximum torque remains constant. INDUCTION MOTORS 413 It is possible to take advantage of this characteristic of wound-rotor induc-tion motors to start very heavy loads. If a resistance is inserted into the rotor cir-cuit, the maximum torque can be adjusted to occur at starting conditions. There-fore, the maximum possible torque would be available to start heavy loads. On the other hand, once the load is turning, the extra resistance can be removed from the circuit, and the maximum torque will move up to near-synchronous speed for reg-ular operation. Example 7-4. A two-pole, 50-Hz induction motor supplies 15 kW to a load at a speed of 2950 r/min. (a) What is the motor's slip? (b) What is the induced torque in the motor in Nom under these conditions? (c) What will the operating speed of the motor be if its torque is doubled? (d) How much power will be supplied by the motor when the torque is doubled? Solution (a) The synchronous speed of this motor is = 120f,. = 120(50 Hz) = 30CXl r/ min n.yDC P 2 poles Therefore, the motor's slip is = 3000r/min - 29.50 r / min(x 100%) 3(x)() rl mm = 0.0167 or 1.67% (7-4) (b) The induced torque in the motor must be assruned equal to the load torque, and P OO/IiV must be assumed equal to P load' since no value was given for mechanical losses. The torque is thus ~oov TiDd =W-m ~~~~~1 5~k~W~c.-~~-c = (2950 r/minX27Trad/rXI min/60 s) = 48.6N o m (c) In the low-slip region, the torque-speed curve is linear, and the induced torque is directly proportional to slip. Therefore, if the torque doubles, then the new slip will be 3.33 percent. The operating speed of the motor is thus 11m = (1 - s)n.yDC = (1 - 0.0333X3000r/min) = 2900 r / min (d) The power supplied by the motor is given by = (97.2 N 0 m)(2900 r/ minX27Trad/ rXI minI 60 s) = 29.5 kW 414 ELECTRIC MACHINERY RJNDAMENTALS Example 7-5. A460-V. 25-hp. 60-Hz. four-pole. Y-cOIlllected wOlUld-rotor induc-tion motor has the following impedances in ohms per phase referred to the stator circuit: Rl = 0.641 0 Xl = 1.106 0 R2 = 0.3320 X2 = 0.4640 XM = 26.3 0 (a) What is the maximrun torque of this motor? At what speed and slip does it occur? (b) What is the starting torque of this motor? (c) When the rotor resistance is doubled. what is the speed at which the maximum torque now occurs? What is the new starting torque of the motor? (d) Calculate and plot the torque-speed characteristics of this motor both with the original rotor resistance and with the rotor resistance doubled. Solutioll The Thevenin voltage of this motor is (7-4la) The Thevenin resistance is (7-44) J 26.3 n )' - (0.641 0-'\1.1060 + 26.3 0 = 0.590 0 The Thevenin reactance is XTH - Xl = 1.106 0 (a) The slip at which maximum torque occurs is given by Equation (7- 53): R, Smax = ~VijRijfu =cc + c=i (X~rn "=,, +=x~ ~' (7- 53) 0.3320 = =0.198 Y(0.590 0)1 + (1.106 0 + 0.464 0)2 This corresponds to a mechanical speed of nm = (I - S)n,yDC = (I - 0.198)(l800r/min) = 1444r/ min The torque at this speed is ~--,"--c-ce3,Vfl"~cc==C=~" Trrw; = 2W'YDC[RTH + YRfu + (XTH + X;l2 ] = 3(255.2 V)2 (7- 54) 2(188.5 rad/s)[O.590 0 + Y(0.590 O)l + (1.106 0 + 0.464 0)2] = 229N om INDUCTION MOTORS 415 (b) The starting torque of this motor is found by setting s = I in Equation (7- 50): 3ViHRl T,tan= W I(R +R)'+(X +X)'l 'YDC TH 2 rn 2 _ 3(255.2 V)1:0.3320) - (188.5 rad/s)[(0.590 0 + 0.3320)2 + (1.I()5 0 + 0.464 0)2] = I04N om (c) If the rotor resistance is doubled, then the slip at maximwn torque doubles, too. Therefore, Smax = 0.396 and the speed at maximum torque is nm = (I -s)n.yDC = (1 - 0.396)(1800 r /min) = 1087 r/min The maximum torque is still Trrw; = 229 Nom The starting torque is now _ 3(255.2 V)2(0.664 0) Tot"" -(188.5 rad/s)[(0.590.n + 0.6640)2 + (1.106 0 + 0.4640)2] = 170 Nom (d) We will create a MATLAB M-file to calculate and plot the torque-speed char-acteristic of the motor both with the original rotor resistance and with the dou-bled rotor resistance. The M-file will calculate the Thevenin impedance using the exact equations for V lll and Zrn [Equations (7-4la) and (7-43)] instead of the approximate equations, because the computer can easily perfonn the exact calculations. It will then calculate the induced torque using Equation (7- 50) and plot the results. The resulting M-file is shown below: % M-file, t orque_speed_curve.m % M-file c reate a p l ot o f the t orque- speed c urve o f the % induction mot or o f Exampl e 7- 5. % Fir s t , initia liz e the va lues needed in thi s program. rl = 0.641; % Stat or res i s tance xl = 1.10 6; % Stat or r eactance r2 = 0.332; % Rotor r es i s tance x2 = 0.464, xm = 26.3, v-ph ase = 460 / sqrt (3) , n_sync = 1800, w_sync = 188.5; % Ca l c ulate the Thevenin % 7- 41a a nd 7- 43. v_th 0 v-ph ase • ( = I % Rotor r eactance % Magne tizati on branch r eactance % Phase voltage % Syn chronou s speed (r / min ) % Sy nchronou s speed (rad /s) voltage and impedance from Equat i ons sqrt (rl " 2 • (xl • xm )" 2 ) I ; , " 0 ( ( jxm ) • (d • -jxl ) ) I (r1 • j (xl • xm ) ) , r -" 0 real ( z th ) , x -" 0 imag ( z_ th ) , 416 ELECTRIC MACHINERY RJNDAMENTALS ~ Now ca l c ulate the t orqu e- speed chara c teri s ti c f or many ~ s lips between 0 and 1. Not e that the fir s t s lip value ~ i s set t o 0.001 ins tead of exact l y 0 to avoi d d i v i de-~ by-zero probl ems. s = (0 ,1,50 ) / 50; s( l ) = 0 . 001; % Slip run = (1 -s) n_sync; % M ech a ni cal speed ~ Ca l c ulate t orque f or original rotor r es i s tance f or ii = 1,51 t _ indl (U ) = (3 v_th" 2 r2 / s (U )) / ... (w_sync (( r _ th + r2 / s (ii )) " 2 + (x_th + x2 ) " 2 ) ) ; ond ~ Ca l c ulate t orque f or doubl ed roto r r es i s tance f or ii = 1,51 t _ ind2 (U ) = (3 v_th" 2 (2r2 ) / s( U )) / ... (w_sync " (( r _ th + (2r2 )/s( U ))" 2 + (x_th + x2 )" 2 ) ) ; ond ~ Plot the t orque- speed c urve p l ot (run , t _ indl, ' Co l or' , 'k' , 'LineWi dth ' ,2 . 0); h o l d on; p l ot (nm , t _ ind2, ' Co l or', 'k' , 'LineWi dth ' ,2.0, 'LineStyl e ' , '-. ' ) ; x l abe l ( ' \ itn_( m)' , 'Pontwei ght' , 'Bo l d ' ) ; y l abe l ( ' \ tau_( ind) ' , 'Pontwei ght' , 'Bo l d ' ) ; title ('Induction mot or t orque- speed c harac teri s ti c ' , ... 'Pontwe i ght' , 'Bol d ' ) ; l egend ( ' Original R_(2) ' , 'Doubled R_(2} ' ) ; gri d on; h o l d o ff ; The resulting torque-speed characteristics are shown in Figure 7-23. Note that the peak torque and starting torque values on the curves match the calculations of parts (a) through (c). Also. note that the starting torque of the motor rose as R2 increased. 7.6 VARIATIONS IN INDUCTION MOTOR TORQUE-SPEED CHARACTERISTICS Section 7.5 contained the derivation of the torque-speed characteristic for an induction motor. In fact, several characteristic curves were shown, depending on the rotor resistance. Example 7- 5 illustrated an induction motor designer's dilemma-if a rotor is designed with high resistance, then the motor's starting torque is quite high, but the slip is also quite high at normal operating conditions. Recall that P C . -. , 200 E • Z 150 ] ./" ~ / ;? 1 - 100 50 Single-cage design .- . Double-cage design o o 200 400 600 800 1000 1200 1400 1600 1800 n",. rlmin ""GURE 7-29 Comparison of torque-speed characteristics for the single- and double-cage rotors of Example 7-6. The resulting torque-speed characteristics are shown in Figure 7-29. Note Ihal the double-cage design has a slightly higher slip in the normal operating range, a smaller maximum torque and a higher starting torque compared to Ihe corresponding single-cage rotor design. This behavior matches our theoretical discussions in this section. 7.7 TRENDS IN INDUCTION MOTOR DESIGN TIle fundamental ideas behind the induction motor were developed during the late 1880s by Nicola Tesla, who received a patent on his ideas in 1888. At that time, he presented a paper before the American lnstitute of Electrical Engineers [AlEE, predecessor of today's Institute of Electrical and Electronics Engineers (IEEE)] in which he described the basic principles of the wound-rotor induction motor, along with ideas for two other important ac motors-the synchronous motor and the re-luctance motor. Although the basic idea of the induction motor was described in 1888, the motor itself did not spring forth in full-fledged fonn. There was an initial period of rapid development, followed by a series of slow, evolutionary improvements which have continued to this day. TIle induction motor assumed recognizable modem form between 1888 and 1895. During that period, two- and three-phase power sources were developed to produce the rotating magnetic fields within the motor, distributed stator windings were developed, and the cage rotor was introduced. By 1896, fully functional and recognizable three-phase induction motors were commercially available. Between then and the early 1970s, there was continual improvement in the quality of the steels, the casting techniques, the insulation, and the construction INDUCTION MOTORS 427 1903 1910 1920 , 1940 19!54 1974 FIGURE 7-30 The evolution of the induction motor. The motors shown in this figure are all rated at 220 V and 15 hp. There has been a dramatic decrease in motor size and material requirements in induction motors since the first practical ones were produced in the 1890s. (Courtesy ofGeneml Electric Company.) FIGURE 7-31 Typical early large induction motors. The motors shown were rated at 2(xx) hp. (Courtesy ofGeneml Electric Company.) features used in induction motors. 1l1ese trends resulted in a smaller motor for a given power output, yielding considerable savings in construction costs. In fact, a modern 100-hp motor is the same physical size as a 7.S-hp motor of 1897. This progression is vividly illustrated by the IS-hp induction motors shown in Figure 7- 30. (See also Figure 7- 31.) 428 ELECTRIC MACHINERY RJNDAMENTALS However, these improvements in induction motor design did not necessar-ily lead to improvements in motor operating efficiency. The major design effort was directed toward reducing the initial materials cost of the machines, not toward increasing their efficiency. The design effort was oriented in that direction because electricity was so inexpensive, making the up-front cost of a motor the principal criterion used by purchasers in its selection. Since the price of oil began its spectacular climb in 1973, the lifetime oper-ating cost of machines has become more and more important, and the initial in-stallation cost has become relatively less important. As a result of these trends, new emphasis has been placed on motor e fficiency both by designers and by end users of the machines. New lines of high-efficiency induction motors are now being produced by all major manufacturers, and they are fanning an ever-increasing share of the in-duction motor market. Several techniques are used to improve the efficiency of these motors compared to the traditional standard-efficiency designs. Among these techniques are I. More copper is used in the stator windings to red uce copper losses. 2. The rotor and stator core length is increased to reduce the magnetic nux den-sity in the air gap of the machine. This reduces the magnetic saturation of the machine, decreasing core losses. 3. More steel is used in the stator of the machine, al lowing a greater amount of heat transfer out of the motor and reducing its operating temperature. The ro-tor's fan is then redesigned to reduce windage losses. 4. The steel used in the stator is a special high-grade electrical steel with low hysteresis losses. 5. The steel is made of an especially thin gauge (i.e., the laminations are very close together), and the steel has a very high internal resistivity. Both effects tend to reduce the eddy current losses in the motor. 6. The rotor is carefull y machined to produce a unifonn air gap, reducing the stray load losses in the motor. In addition to the general techniques described above, each manufacturer has his own unique approaches to improving motor efficiency. A typical high-efficiency induction motor is shown in Figure 7- 32. To aid in the comparison of motor efficiencies, NEMA has adopted a stan-dard technique for measuring motor efficiency based on Method 8 of the IEEE Standard 11 2, Test Procedure for Polyphase Induction Motors and Generators. NEMA has also introduced a rating called NEMA nominal efficiency, which ap-pears on the nameplates of design class A, 8 , and C motors. The nominal effi-ciency identifies the average efficiency of a large number of motors of a given model, and it also guarantees a certain minimum efficiency for that type of motor. The standard NEMA nominal efficiencies are shown in Figure 7- 33. INDUCTION MOTORS 429 FIGURE 7-32 A General Electric Energy Saver motor. typical of modem high-efficiency induction motors. (Courtesy ofGeneml Electric Company.) Nomin al Guanmh. 't!d minimum Nominal Guara nte ... >d minimum efficiency, o/~ efficiency, % efficiency, % efficiency, '7~ 95.0 94.1 SO.O 77.0 94.5 93.6 78.5 75.5 94.1 93.0 77.0 74.0 93.6 92.4 75.5 72.0 93.0 91.7 74.0 70.0 92.4 91.0 72.0 68.0 91.7 90.2 70.0 66.0 91.0 89.5 68.0 64.0 90.2 88.5 66.0 62.0 89.5 87.5 64.0 59.5 88.5 86.5 62.0 57.5 87.5 85.5 59.5 55.0 86.5 84.0 57.5 52.5 85.5 82.5 55.0 50.' 84.0 81.5 52.5 48.0 82.5 80.0 50.' 46.0 81.5 78.5 FIGURE 7-33 Table of NEMA nominal efficiency standards. The nominal efficiency represents the mean efficiency of a large number of sample motors. and the 8uar:mteed minimum efficiency represents the lowest permissible efficiency for any given motor of the class. (Reproduced by permission from Motors and GenemfOrs, NEMA Publication MG-I. copyright 1987 by NEMA.) 430 ELECTRIC MACHINERY RJNDAMENTALS Other standards organizations have also established efficiency standards for induction motors, the most important of which are the 8ritish (8S-269), IEC (IEC 34-2), and Japanese (JEC-37) standards. However, the techniques prescribed for measuring induction motor efficiency are different in each standard and yield dif-ferent results for the same physical machine. If two motors are each rated at 82.5 percent efficiency, but they are measured according to different standards, then they may not be equally efficient. When two motors are compared, it is important to compare efficiencies measured under the same standard. 7.8 STARTING INDUCTION MOTORS Induction motors do not present the types of starting problems that synchronous motors do. In many cases, induction motors can be started by simply connecting them to the power line. However, there are sometimes good reasons for not doing this. For example, the starting current required may cause such a dip in the power system voltage that across-the-line staning is not acceptable. For wound-rotor induction motors, starting can be achieved at relatively low currents by inserting extra resistance in the rotor circuit during starting. lllis extra resistance not only increases the starting torque but also reduces the starting current. For cage induction motors, the starting current can vary widely depending primarily on the motor's rated power and on the effecti ve rotor resistance at start-ing conditions. To estimate the rotor current at starting conditions, all cage motors now have a starting code letter (not to be confused with their design class letter) on their nameplates. The code letter sets limits on the amount of current the mo-tor can draw at starting conditions. 1l1ese limits are expressed in tenns of the starting apparent power of the motor as a function of its horsepower rating. Figure 7- 34 is a table containing the starting kilovoltamperes per horsepower for each code letter. To determine the starting current for an induction motor, read the rated volt-age, horsepower, and code letter from its nameplate. 1l1en the starting apparent power for the motor wi ll be S.1Mt = (rated horsepower)(code letter factor) and the starting current can be found from the equation S.tan IL = V3V T (7- 55) (7- 56) Example 7-7. What is the starting ClUTent of a 15-hp, 208-V, code-Ietter-F, three-phase induction motor? Solutio" According to Figure 7- 34, the maximwn kilovoltamperes per horsepower is 5.6. Therefore, the maximum starting kilovoltamperes of this motor is S,w<. = (15 hp)(5.6) = 84 kVA INDUCTION MOTORS 431 Nominal code Lock ... >d rotor, Nominal code Locked rotor, letter kYAJhp letter kVAJhp A 0--3.15 L 9.00-10.00 B 3.15--3.55 M 10.00-11.00 C 3.55-4.00 N 11.20-12.50 D 4.00-4.50 P 12.50-14.00 E 4.50-5.00 R 14.00-16.00 F 5.00-5.60 S 16.00-18.00 G 5.60--6.30 T 18.00-20.00 H 6.30--7.10 U 20.00-22.40 J 7.7- 8.00 V 22.40 and up K 8.00-9.00 FIGURE 7-34 Table of NEMA code letters. indicating the starting kilovolt amperes per horsepower of rating for a motor. Each code letter extends up to. but does not include. the lower bound of the next higher class. (Reproduced 17y permission from Motors and Generators. NEMA Publication MG-I. copyright 1987 byNEMA.) The starting current is thus (7- 56) 84kVA = vi3"(208 V) = 233 A If necessary, the starting current of an induction motor may be reduced by a starting circuit. However, if this is done, it will also reduce the starting torque of the motor. One way to reduce the starting current is to insert extra inductors or resis-tors into the power line during starting. While fonnerly common, this approach is rare today. An alternative approach is to reduce the motor's terminal voltage dur-ing starting by using autotransformers to step it down. Figure 7- 35 shows a typi-cal reduced-voltage starting circuit using autotransfonners. During starting, con-tacts 1 and 3 are shut, supplying a lower voltage to the motor. Once the motor is nearly up to speed, those contacts are opened and contacts 2 are shut. These con-tacts put fu JI line voltage across the motor. It is important to realize that while the starting current is reduced in direct proportion to the decrease in terminal voltage, the starting torque decreases as the square of the applied voltage. Therefore, only a certain amount of current reduc-tion can be done if the motor is to start with a shaft load attached. 432 ELECTRIC MACHINERY RJNDAMENTALS Line terminals 2 2 " " " " 3 3 Motor terminals Starting sequence: (a) Close I and 3 (b) Open I and 3 (c) Close 2 ""GURE 7-35 An autotransfonner starter for an induction motor. ~ ---1 1 F, M 'I II ~ F, M , ~ F, M, Disronnect switch Start Overload heaters Induction motor Stop OL LL---r ---"--;~M ""GURE 7- 36 A typical across-too-line starter for an induction motor. Induction Motor Starting Circuits A typical full-voltage or across-the-Iine magnetic induction motor starter circuit is shown in Figure 7- 36, and the meanings of the symbols used in the figure are ex-plained in Figure 7- 37. This operation of this circuit is very simple. When the start button is pressed, the relay (or contactor) coil M is energized, causing the normally open contacts M t , M2, and M) to shut. When these contacts shut, power is applied to the induction motor, and the motor starts. Contact M4 also shuts, ---11 If-0 e 0 II )( rx, OL )f FIGURE 7-37 Normally open Normally shut INDUCTION MOTORS 433 Disconnect switch Push button; push to close Push button; push to open Relay coil; contacts change state when the coil energizes Contact open when coil deenergized Contact shut when coil deenergized Overload heater Overload contact; opens when the heater gets too wann Typical components found in induction motor control circuits. which shorts out the starting switch, allowing the operator to release it without re-moving power from the M relay. When the stop button is pressed, the M relay is deenergized, and the M contacts open, stopping the motor. A magnetic motor starter circuit of this sort has several bui It-in protective features: I. Short-circuit protection 2. Overload protection 3. Undervoltage protection Short-circuit protection for the molor is provided by fuses F t , F2, and Fl. If a sudden short circuit develops within the motor and causes a current flow many limes larger than the rated current, these fuses will blow, disconnecting the motor from the power supply and preventing it from burning up. However, these fuses must not burn up during normal molor starting, so they are designed to require currents many times greater than the full -load current before they open the circuit. This means that short circuits through a high resistance and/or excessive motor loads will not be cleared by the fuses. Overload protection for the motor is provided by the devices labeled OL in the figure. These overload protection devices consist of two parts, an overload 434 ELECTRIC MACHINERY RJNDAMENTALS heater element and overload contacts. Under nonnal conditions, the overload con-tacts are shut. However, when the temperature of the heater elements rises far enough, the OL contacts open, deenergizing the M relay, which in turn opens the normally open M contacts and removes power from the motor. When an induction motor is overloaded, it is eventually drunaged by the ex-cessive heating caused by its high currents. However, this damage takes time, and an induction motor will not nonnally be hurt by brief periods of high currents (such as starting currents). Only if the high current is sustained will damage occur. The overload heater elements also depend on heat for their operation, so they wil l not be affected by brief perioos of high current during starting, and yet they wil I operate during long periods of high current, removing power from the motor be-fore it can be damaged. Undefi!oltage protection is provided by the controller as well. Notice from the figure that the control power for the M relay comes from directly across the lines to the motor. If the voltage applied to the motor falls too much, the voltage applied to the M relay will also fall and the relay will deenergize. TIle M contacts then open, removing power from the motor tenninals. An induction motor starting circuit with resistors to reduce the starting cur-rent flow is shown in Figure 7- 38. TIlis circuit is similar to the previous one, ex-cept that there are additional components present to control removal of the start-ing resistor. Relays lID, 2TD, and 3TD in Figure 7- 38 are so-called time-delay relays, meaning that when they are energized there is a set time delay before their contacts shut. When the start button is pushed in this circuit, the M relay energizes and power is applied to the motor as before. Since the 1 ID, 2TD, and 3ID contacts are all open, the full starting resistor is in series with the motor, reducing the start-ing current. When the M contacts close, notice that the 1 ID relay is energized. How-ever, there is a finite delay before the lTD contacts close. During that time, the motor partially speeds up, and the starting current drops off some. After that time, the 1 TO contacts close, cutting out part of the starting resistance and simultane-ously energizing the 2TD relay. After another delay, the 2TD contacts shut, cut-ting out the second part of the resistor and energizing the 3TD relay. Finally, the 3TD contacts close, and the entire starting resistor is out of the circuit. By a judicious choice of resistor val ues and time delays, this starting circuit can be used to prevent the motor starting current from becoming dangerously large, while still allowing enough current flow to ensure prompt acceleration to normal operating speeds. 7.9 SPEED CONTROL OF INDUCTION MOTORS Until the advent of modern solid-state drives, induction motors in general were not good machines for applications requiring considerable speed control. The normal operating range of a typical induction motor (design classes A, B, and C) INDUCTION MOTORS 435 Overload F M] heaters /~ I r~~~~ ------l..L....l..I-1 r---Resis]or 3TD Resis]or lTD 2TD 3TD Resistor lTD 2TD 3TD S<w Stop I OL lID lID 2ID 2ID 3ID FIGURE 7-38 A three-step resistive staner for an induction motor. Induction motor is confined to less than 5 percent slip, and the speed variation over that range is more or less directly proportional to the load on the shaft of the motor. Even if the slip could be made larger, the effi ciency of the motor would become very poor, since the rotor copper losses are directly proportional to the slip on the motor (remember that PRCL = sP AG). There are really only two techniques by which the speed of an induction motor can be controlled. One is to vary the synchronous speed, which is the speed of the stator and rotor magnetic fields, since the rotor speed always remains near n,ync. The other technique is to vary the slip of the motor for a given load. E:1.ch of these approaches will be taken up in more detail. The synchronous speed of an induction motor is given by 436 ELECTRIC MACHINERY RJNDAMENTALS 120f e p (7- 1) so the only ways in which the synchronous speed of the machine can be varied are (I) by changing the electrical frequency and (2) by changing the number of poles on the machine. Slip control may be accomplished by varying either the rotor re-sistance or the terminal voltage of the motor. Induction Motor Speed Control by P ole Changing TIlere are two major approaches to changing the number of poles in an induction motor: I. The method of consequent poles 2. Multiple stator windings TIle method of consequent poles is quite an old method for speed control, having been originally developed in 1897. It relies on the fact that the number of poles in the stator windings of an induction motor can easily be changed by a fac-tor of 2: I with only simple changes in coil connections. Figure 7- 39 shows a , Winding connections at back end of stator b fo'IGURE 7-39 \ d, ---b', p~= 60° 0, , " " , , / ' , b', " " , ---" b " ----, " /' a, A two-pole stator winding for pole changing. Notice the very small rotor pitch of these windings. INDUCTION MOTORS 437 simple two-pole induction motor stator suitable for pole changing. Notice that the individual coils are of very short pitch (60 to 90°). Figure 7-40 shows phase a of these windings separately for more clarity of detail. Figure 7-40a shows the current n ow in phase a of the stator windings at an instant of time during nonnal operation. Note that the magnetic field leaves the sta-tor in the upper phase group (a north pole) and enters the stator in the lower phase group (a south pole). This winding is thus pnxlucing two stator magnetic poles. a', I(t) I ,,' ) NtS ) N t S a] d] a2 d 2 S B -,b, FIGURE 7-40 Connections at far end of stator ,-" , , " " ,-" , ", , " " " B \' B /, " " B \' N a', B S B A close-up view of one phase of a pole-changing winding. (a) In the two-pole configuration. one coil is a north pole and the other one is a south pole. (b) When the connection on one of the two coils is reversed. they are both nonh poles. and the magnetic flux returns to the stator at points halfway between the two coils. The south poles are called consequent poles. and the winding is now a four-pole winding. 438 ELECTRIC MACHINERY RJNDAMENTALS Now suppose that the direction of current flow in the lower phase group on the stator is reversed (Figure 7--40b).1llen the magnetic field wi ll leave the stator in both the upper phase group and the lower phase group-each one will be a north magnetic pole. The magnetic fl ux in this machine must return to the stator between the two phase groups, producing a pair of consequent south magnetic poles. Notice that now the stator has four magnetic poles-twice as many as before. 1lle rotor in such a motor is of the cage design, since a cage rotor always has as many poles induced in it as there are in the stator and can thus adapt when the number of stator poles changes. When the motor is reconnected from two-pole to four-pole operation, the resulting maximum torque of the induction motor can be the same as before (constant-torque connection), half of its previous value (square-law-torque con-nection, used for fans, etc.), or twice its previous value (constant-output-power connection), depending on how the stator windings are rearranged. Figure 7--4 1 shows the possible stator connections and their effect on the torque-speed curve. 1lle major disadvantage of the consequent-pole method of changing speed is that the speeds must be in a ratio of 2: I. 1lle traditional approach to overcom-ing this limitation was to employ multiple stator windings with different numbers of poles and to energize only one set at a time. For example, a motor might be wound with a four-pole and a six-pole set of stator windings, and its synchronous speed on a 6O-Hz system could be switched from 1800 to 1200 r/min simply by supplying power to the other set of windings. Unfortunately, multiple stator wind-ings increase the expense of the motor and are therefore used only when ab-solutely necessary. By combining the method of consequent poles with multiple stator wind-ings, it is possible to build a four-speed induction motor. For example, with sepa-rate four- and six-pole windings, it is possible to produce a 6O-Hz motor capable of running at 600, 900, 1200, and 1800 r/min. Speed Control by Changing the Line Frequency I f the electrical frequency applied to the stator of an induction motor is changed, the rate of rotation of its magnetic fields "' )'DC will change in direct proportion to the change in electrical frequency, and the no-load point on the torque-speed characteristic curve will change with it (see Figure 7--42). The synchronous speed of the motor at rated conditions is known as the base speed. By using variable fre-quency control, it is possible to adjust the speed of the motor either above or be-low base speed. A properly designed variable-frequency induction motor drive can be very flexible. It can control the speed of an induction motor over a range from as little as 5 percent of base speed up to about twice base speed. However, it is important to maintain certain voltage and torque limits on the motor as the fre-quency is varied, to ensure safe operation. When running at speeds below the base speed of the motor, it is necessary to reduce the terminal voltage applied to the stator for proper operation. 1lle ter-minal voltage applied to the stator should be decreased linearly with decreasing INDUCTION MOTORS 439 T, T, T, T, T, T, T, T, T, Lines Lines S","" L, L, L, S"""d L, L, L, Low T, T, T, T4, T~ T6 Low T, T, T, Tt -T2-Tj High T, T, T, S","" L, Low T, High T, FIGURE 7-41 T, (a) T, T, Lines L, T, T, (,) T, T, L, T, T, 0",," T)-TrTJ together T, T4, T." T6 0",," T)-TrTJ together High T, T, T, (b) (b, High speed 1:: (all) , l""d(:"'::::1 r---' ~ (Cl Speed, rlmin (d) together T4, T~, T6 0",," Possible connections of the stator coils in a pole-changing motor. together with the resulting torque-speed characteristics: (a) Constant-torque connection-the torque capabilities of the motor remain approximately constant in both high-speed and low-speed connections. (b) Constant-lwrsepok'er connection---lhe power capabilities of the motor remain approximately constant in both h.igh-speed and low-speed connections. (el Fan torque connection---lhe torque capabilities of the motor change with speed in the same manner as f.an-type loads. 440 ELECTRIC MACHINERY RJNDAMENTALS 800,-------------------------------------, 700 600 , • Z 500 • ¥ 400 ~ I 300 200 lOOe--, 800 700 600 , • Z 500 • ~ 400 " , 300 0 ~ 200 100 0 0 ""GURE 7-42 Mechanical speed. r/min , ., 1000 1500 2000 2500 3000 3500 Mechanical speed. r/min ,b, Variable-frequency speed control in an induction motor: (a) The family of torque-speed characteristic curves for speeds below base speed. assuming that the line voltage is derated linearly with frequency. (b) The family of torque-speed characteristic curves for speeds above base speed. assuming that the line voltage is held constant. INDUCTION MOTORS 441 800,-------------------------------------, 700 600 , • Z 500 , ~ 400 ] 300 • 200P\-------\ o o 500 FIGURE 7-42 (roncluded) 1000 1500 2000 Mechanical speed. r/min I" (c) The torque-speed characteristic curves for all frequencies. 2500 3000 3500 stator frequency. This process is called derating. If it is not done, the steel in the core of the induction motor will saturate and excessive magnetization currents will flow in the machine. To understand the necessity for derating, recall that an induction motor is basically a rotating transfonner. As with any transfonner, the flux in the core of an induction motor can be found from Faraday's law: vet) = -N~ dl (1-36) If a voltage ve t) = VM sin wt is applied to the core, the resulting flux ~ is 'Wi ~ J h l)dl p = ~ !VM sinwtdt p I ~t) = -~ cos wt l (7- 57) Note that the electrical frequency appears in the denominator of this expression. Therefore, if the electrical frequency applied to the stator decreases by 10 percent while the magnitude of the voltage appl ied to the stator remains constant, the flux in the core of the motor wi I I increase by about 10 percent and the magnetization current of the motor wi ll increase. In the unsaturated region of the motor's 442 ELECTRIC MACHINERY RJNDAMENTALS magnetization curve, the increase in magnetization current will also be about 10 percent. However, in the saturated region of the motor's magnetiwtion curve, a 10 percent increase in flux requires a much larger increase in magnetization current. Induction motors are normally designed to operate near the saturation point on their magnetization curves, so the increase in flux due to a decrease in frequency will cause excessive magnetization currents to flow in the motor. (This same prob-lem was observed in transfonners; see Section 2.1 2.) To avoid excessive magnetization currents, it is customary to decrease the applied stator voltage in direct proportion to the decrease in frequency whenever the frequency falls below the rated frequency of the motor. Since the applied volt-age v appears in the numerator of Equation (7-57) and the frequency wappears in the denominator of Equation (7-57), the two effects counteract each other, and the magnetization current is unaffected. When the voltage applied to an induction motor is varied linearly with fre-quency below the base speed, the flux in the motor will remain approximately constant. TIlerefore, the maximum torque which the motor can supply remains fairly high. However, the maximum power rating of the motor must be decreased linearly with decreases in frequency to protect the stator circuit from overheating. TIle power supplied to a three-phase induction motor is given by P = v'JVLIL cos (J If the voltage VL is decreased, then the maximum power P must also be decreased, or else the current flowing in the motor wi II become excessive, and the motor will overheat. Figure 7-42a shows a family of induction motor torque-speed characteris-tic curves for speeds below base speed, assuming that the magnitude of the stator voltage varies linearly with frequency. When the electrical frequency applied to the motor exceeds the rated fre-quency of the motor, the stator voltage is held constant at the rated value. Al-though saturation considerations would pennit the voltage to be raised above the rated value under these circumstances, it is limited to the rated voltage to protect the winding insulation of the motor. The higher the electrical frequency above base speed, the larger the denominator of Equation (7-57) becomes. Since the nu-merator tenn is held constant above rated frequency, the resulting flux in the ma-chine decreases and the maximum torque decreases with it. Figure 7-42b shows a family of induction motor torque- speed characteristic curves for speeds above base speed, assuming that the stator voltage is held constant. If the stator voltage is varied linearly with frequency below base speed and is held constant at rated value above base speed, then the resulting family of torque-speed characteristics is as shown in Figure 7-42c. TIle rated speed for the motor shown in Figure 7-42 is 1800 r/min. In the past, the principal disadvantage of electrical frequency control as a method of speed changing was that a dedicated generator or mechanical fre-quency changer was required to make it operate. This problem has disappeared with the development of modern solid-state variable-frequency motor drives. In INDUCTION MOTORS 443 800 700 600 E "" Z ; ~ 400 ~ • , , • 300 • / -/ Lo,' 200 100 0 --------0 250 '00 "" 1000 1250 1500 1750 2000 Mechanical speed. r/min FIGURE 7-43 Variable-line-voltage speed control in an induction motor. fact, changing the line frequency with solid-state motor drives has become the method of choice for induction motor speed control. Note that this method can be used with any induction motor, unlike the pole-changing technique, which re-quires a motor with special stator windings. A typical solid-state variable-frequency induction motor drive wi ll be de-scribed in Section 7.10. Speed Control by Changing the Line Voltage The torque developed by an induction motor is proportional to the square of the applied voltage. Ifa load has a torque-speed characteristic such as the one shown in Figure 7-43, then the speed of the motor may be controlled over a limited range by varying the line voltage. This method of speed control is sometimes used on small motors driving fans. Speed Control by Changing the Rotor Resistance In wound-rotor induction motors, it is possible to change the shape of the torque- speed curve by inserting extra resistances into the rotor circuit of the ma-chine. The resulting torque- speed characteristic curves are shown in Figure 7-44. 444 ELECTRIC MACHINERY RJNDAMENTALS 800 700 R, R, R, (j()() E 500 Z • , ~ 400 g " , , ~ 300 R] '" 2Ro 200 R2", 3Ro RJ ",4Ro RJ '" 5Ro 100 R3", 6Ro 0 0 250 500 750 1000 1250 1500 17'" 2000 Mechanical speed. r/min fo'IGURE 7-44 Speed control by varying the rotor resistance of a wound-rotor induction motor. If the torque-speed curve of the load is as shown in the figure, then changing the rotor resistance wi ll change the operating speed of the motor. However, inserting extra resistances into the rotor circuit of an induction motor seriously red uces the efficiency of the machine. Such a method of speed control is nonnally used only for short periods because of this efficiency problem. 7.10 SOLID·STATE INDUCTION MOTOR DRIVES As mentioned in the previous section, the method of choice today for induction motor speed control is the solid-state variable-frequency induction motor drive. A typical drive of this sort is shown in Figure 7--45. TIle drive is very flexible: its in-put power can be either single-phase or three-phase, either 50 or 60 Hz, and any-where from 208 to 230 V. The output from this drive is a three-phase set of volt-ages whose frequency can be varied from 0 up to 120 Hz and whose voltage can be varied from 0 V up to the rated voltage of the motor. TIle output voltage and frequency control is achieved by using the pulse-width modulation (PWM) techniques described in Chapter 3. Both output frequency and output voltage can be controlled independently by pulse-width modulation. fig-ure 7--46 illustrates the manner in which the PWM drive can control the output fre-quency while maintaining a constant nns voltage level, while Figure 7--47 illustrates • • / Voltage. V Voltage. V 100 20 o 10 - 100 FIGURE 7-46 INDUCTION MOTORS 445 fo'IGURE 7-45 A typical solid-state variable-frequency induction motor drive. (Courtesy of MagneTek, Inc.) I. ms (a) 30 40 50 I. ms ,bl Variable-frequency control with a PWM waveform: (a) 6O-Hz.. 120-V PWM waveform: (b) 30-Hz. 12()' V PWM waveform. 446 ELECTRIC MACHINERY RJNDAMENTALS Voltage, V 100 t , ms - 100 "J Voltage, V 100 10 30 50 O ~ mf t , ms - 100 20 40 'bJ ""GURE 7-47 Variable voltage control with a PWM waveform: (a) 6O-Hz, 120-V PWM waveform: (b) 6().Hz, 6(). V PWM waveform. the manner in which the PWM drive can control the nns voltage level while main-taining a constant frequency. As we described in Section 7.9, it is often desirable to vary the output fre-quency and output nns voltage together in a linear fashion. Figure 7-48 shows typical output voltage wavefonns from one phase of the drive for the situation in which frequency and voltage are varied simultaneously in a linear fashion.· Fig-ure 7-48a shows the output voltage adjusted for a frequency of60 Hz and an nns voltage of 120 V. Figure 7-48b shows the output adjusted for a frequency of 30 Hz and an nns voltage of 60 V, and Figure 7-48c shows the output adjusted for a frequency of 20 Hz and an nns voltage of 40 V. Notice that the peak voltage out of the dri ve remains the same in all three cases; the nns voltage level is controlled by the fraction of time the voltage is switched on, and the frequency is controlled by the rate at which the polarity of the pulses switches from positive to negative and back again. TIle typical induction motor drive shown in Figure 7-45 has many built-in features which contribute to its adjustability and ease of use. Here is a summary of some of these features. The output waveforms in Figure 7-47 are actually simplified waveforms. The real induction motor drive has a much higher carrier frequency than that shown in the figure. INDUCTION MOTORS 447 Voltage. V P\VM waveform t. ms ,., Voltage. V P\VM waveform 100 20 30 o 10 40 50 t. ms - 100 ,b, Voltage. V PWM waveform 100 o WUlli:lm 30 , 40 t. ms - 100 10 20 ,< , FIGURE 7-48 Simultaneous voltage and frequency control with a P\VM wavefonn: (a) 6O-Hz. 120-V PWM waveform: (b) 30-Hz. 60-V PWM waveform: (c) 2O-Hz. 40-V PWM waveform. Frequency (Speed) Adjustment The output frequency of the drive can be controlled manually from a control mounted on the drive cabinet, or it can be controlled remotely by an external volt-age or current signal. The ability to adjust the frequency of the drive in response to some external signal is very important, since it permits an external computer or process controller to control the speed of the motor in accordance with the over-all needs of the plant in which it is installed. 448 ELECTRIC MACHINERY RJNDAMENTALS A Choice of Voltage and Frequency Patterns TIle types of mechanical loads which might be attached to an induction motor vary greatly. Some loads such as fans require very little torque when starting (or running at low speeds) and have torques which increase as the square of the speed. Other loads might be harder to start, requiring more than the rated full-load torque of the motor just to get the load moving. TIlis drive provides a variety of voltage-versus-frequency patterns which can be selected to match the torque from the in-duction motor to the torque required by its load. TIlree of these patterns are shown in Figures 7-49 through 7- 5 J. Figure 7-49a shows the standard or general-purpose voltage-versus-frequency pattern, described in the previous section. This pattern changes the out-put voltage linearly with changes in output frequency for speeds below base speed and holds the output voltage constant for speeds above base speed. (The small constant-voltage region at very low frequencies is necessary to ensure that there will be some starting torque at the very lowest speeds.) Figure 7-49b shows the resulting induction motor torque-speed characteristics for several operating fre-quencies below base speed. Figure 7- 50a shows the voltage-versus-frequency pattern used for loads with high starting torques. This pattern also changes the output voltage linearly with changes in output frequency for speeds below base speed, but it has a shal-lower slope at frequencies below 30 Hz. For any given frequency below 30 Hz, the output voltage will be higherthan it was with the previous pattern. This higher voltage will produce a higher torque, but at the cost of increased magnetic satura-tion and higher magnetization currents. The increased saturation and higher cur-rents are often acceptable for the short periods required to start heavy loads. Fig-ure 7- 50b shows the induction motor torque-speed characteristics for several operating frequencies below base speed. Notice the increased torque available at low frequencies compared to Figure 7-49b. Figure 7-51 a shows the voltage-versus-frequency pattern used for loads with low starting torques (called soft-start loads). TIlis pattern changes the output voltage parabolically with changes in output frequency for speeds below base speed. For any given frequency below 60 Hz, the output voltage will be lower than it was with the standard pattern. TIlis lower voltage will produce a lower torque, providing a slow, smooth start for low-torque loads. Figure 7- 5 I b shows the induction motor torque-speed characteristics for several operating frequencies below base speed. Notice the decreased torque available at low frequencies com-pared to Figure 7-49. Independently Adjustable Acceleration and Deceleration Ramps When the desired operating speed of the motor is changed, the drive controlling it will change frequency to bring the motor to the new operating speed. If the speed change is sudden (e.g., an instantaneous jump from 900 to 1200 rIm in), the drive v 8O\l 700 roo E • '00 Z • , 4O\l ~ ~ 300 WO 100 0 0 FIGURE 7-49 INDUCTION MOTORS 449 O ~----------~ ffi C----------C 1~ W c---- f H Z f_. (a) Torque--speed characteristic 200 4O\l roo 800 1000 12lXl 1400 Speed. rlmin ,b, 1roo 18O\l (a) Possible voltage-versus-frequency patterns for the solid-state variable-frequency induction motor drive: general-purpose paT/ern. This pattern consists of a linear voltage-frequency curve below rated frequency and a constant voltage above rated frequency. (b) The resulting torque-speed characteristic curves for speeds below rated frequency (speeds above rated frequency look like Figure 7-4lb). 450 ELECTRIC MACHINERY RJNDA MENTALS , • Z • , ~ ~ 8O\l 700 600 5O\l 2llll 100 v V .. ,od , , , , , , , , , o !------~ 60 ~----~ 1~ 2~O-- f. Hz f~. "I Torque-speed characteristic O~~~~~~~-o~L,,~-e~-+~-..~-.t , o 200 400 600 800 1000 1200 1400 1600 1800 Speed. r/min 'hI fo'IGURE 7-50 (a) Possible voltage-versus-frequency patterns for the solid-state variable-frequency induction motor drive: high-starting-torque patfem. This is a modified voltage-frequency pattern suitable for loads requiring high starting torques. lt is the same as the linear voltage- frequency pattern except at low speeds. The voltage is disproportionately high at very low speeds. which produces extra torque at the cost of a h.igher magnetization current. (b) The resulting torque-speed characteristic curves for speeds below rated frequency (speeds above rated frequency look like Figure 7--41b). INDUCTION MOTORS 451 v O ~----------iro '----------' I~W o---- f H z (a) 8O\l Torque--speed characteristic 700 roo E • '00 Z • , 4O\l ~ ~ 300 WO 100 0 0 200 4O\l 800 1000 12lXl 1400 lroo 18O\l Speed. r/min 'h ' FIGURE 7-51 (a) Possible voltage-versus-frequency patterns for the solid-state variable-frequency induction motor drive:fan torque pattern. This is a voltage-frequency pattern suitable for use with motors driving fans and centrifugal pumps. which have a very low starting torque. (b) The resulting torque-speed characteristic curves for speeds below rated frequency (speeds above rated frequency look like Figure 7-4lb). 452 ELECTRIC MACHINERY RJNDAMENTALS does not try to make the motor instantaneously jump from the old desired speed to the new desired speed. Instead, the rate of motor acceleration or deceleration is limited to a safe level by special circuits built into the electronics of the drive. TIlese rates can be adjusted independently for accelerations and decelerations. Motor Protection TIle induction motor drive has built into it a variety of features designed to protect the motor attached to the drive. The drive can detect excessive steady-state cur-rents (an overload condition), excessive instantaneous currents, overvoltage con-ditions, or undervoltage conditions. In any of the above cases, it will shut down the motor. Induction motor drives like the one described above are now so flexible and reliable that induction motors with these drives are displacing dc motors in many applications which require a wide range of speed variation. 7.11 DETERMINING CIRCUIT MODEL PARAMETERS TIle equivalent circuit of an induction motor is a very useful tool for detennining the motor's response to changes in load. However, if a model is to be used for a real machine, it is necessary to detennine what the element val ues are that go into the model. How can Rl> R2, Xl> X2, and XM be detennined for a real motor? These pieces of information may be found by perfonning a series of tests on the induction motor that are analogous to the short-circuit and open-circuit tests in a transfonner. TIle tests must be performed under precisely controlled conditions, since the resistances vary with temperature and the rotor resistance also varies with rotor frequency. The exact details of how each induction motor test must be performed in order to achieve accurate results are described in IEEE Standard 11 2. Although the details of the tests are very complicated, the concepts behind them are relatively straightforward and will be explained here. The No-Load Test TIle no-load test of an induction motor measures the rotational losses of the mo-tor and provides infonnation about its magnetization current. The test circuit for this test is shown in Figure 7- 52a. Wattmeters, a voltmeter, and three ammeters are connected to an induction motor, which is allowed to spin freely. The only load on the motor is the friction and windage losses, so all Peon v in this motor is consumed by mechanical losses, and the slip of the motor is very small (possibly as small as 0.00 1 or less). TIle equivalent circuit of this motor is shown in Figure 7- 52b. With its very small slip, the resistance corresponding to its power con-verted, Rl l - sys, is much much larger than the resistance corresponding to the rotor copper losses R2 and much larger than the rotor reactance X2. I n this case, the equivalent circuit reduces approximately to the last circuit in Figure 7- 52b. There, Initial Variable voltage, variable frequency, three-phase power 00.= I, -+ equivalent V. ( cin:uit: I, Since -R2('~S)>>R2 + v .( ,,' R2 (' - s)>>X2, , this cin:uit reduces to: + Combining V. ( RF&w and Reyields: FIGURE 7-52 R, jXt R, , R, INDUCTION MOTORS 453 I, -P, A I, -A No load I, -P, A lit + IB+ Ie IL = 3 , ., 12= 0 -jX2 R, " I. j R, jXM ~ R2(';S ) ~ Rfri<:tiOll, win<b!O, ~ &""'" »XM 'h ' The no-load test of an induction motor: (a) test circuit; (b) the resulting motor equivalent cin:uit. Note that at no load the motor's impedance is essentially the series combination of RI,jXj • andJXM . the output resistor is in parallel with the magnetization reactance XM and the core losses Re. In this motor at no-load conditions, the input power measured by the meters must equal the losses in the motor. The rotor copper losses are negligible because the current / l is extremely small [because of the large load resistance R2(1 - s)/s], so they may be neglected. The stator copper losses are given by 454 ELECTRIC MACHINERY RJNDAMENTALS so the input power must equal P;o = P SCL + P.:orc + PF& W + Pmisc = 3f r RI + Prot where P rot is the rotational losses of the motor: PM = P oore + PF&W + P mis.c (7- 25) (7- 58) (7- 59) TI1US, given the input power to the motor, the rotational losses of the machine may be detennined. TIle equivalent circuit that describes the motor operating in this condition contains resistors Re and R2( I - sys in parallel with the magnetizing reactance XM . The current needed to establish a magnetic field is quite large in an induction mo-tor, because of the high reluctance of its air gap, so the reactance XM will be much smaller than the resistances in parallel with it and the overall input power factor will be very small. With the large lagging current, most of the voltage drop will be across the inductive components in the circuit. The equivalent input impedance is thus approximately (7-60) and if Xl can be found in some other fashion, the magnetizing impedance XM will be known for the motor. The DC Test for Stator Resistance TIle rotor resistance R2 plays an extremely critical role in the operation of an in-duction motor. Among other things, Rl determines the shape of the torque-speed curve, detennining the speed at which the pullout torque occurs. A standard mo-tor test called the locked-rotor test can be used to detennine the total motor circuit resistance (this test is taken up in the next section). However, this test ftnds only the total resistance. To find the rotor resistance Rl accurately, it is necessary to know RI so that it can be subtracted from the total. TIlere is a test for Rl independent of Rl , Xl and X2. This test is called the dc test. Basically, a dc voltage is applied to the stator windings of an induction mo-tor. Because the current is dc, there is no induced voltage in the rotor circuit and no resulting rotor current now. Also, the reactance of the motor is zero at direct current. TIlerefore, the only quantity limiting current now in the motor is the sta-tor resistance, and that resistance can be detennined. TIle basic circuit for the dc test is shown in Figure 7- 53. This figure shows a dc power supply connected to two of the three terminals of a V-connected in-duction motor. To perfonn the test, the current in the stator windings is adjusted to the rated value, and the voltage between the terminals is measured. TIle current in Voc (variable) + '-/ FIGURE 7-53 Current-limiting resistor Test ci['(;uit for a dc resistance test. INDUCTION MOTORS 455 R, the stator windings is adjusted to the rated value in an attempl to heat the wind-ings to the same temperature they would have during nonnal operation (remem-ber, winding resistance is a function of temperature). The currenl in Figure 7- 53 flows through two of the windings, so the total resistance in the current path is 2Rt . Therefore, Voe 2Rt = -J-oe I R, Vee I 2loc (7-6 1) With this value of Rt the stator copper losses at no load may be detennined, and the rotational losses may be found as the difference between the input power at no load and the stator copper losses. The value of Rt calculated in this fashion is not completely accurate, since it neglects the skin effect that occurs when an ac voltage is applied to the wind-ings. More details concerning corrections for temperature and skin effect can be found in IEEE Standard 11 2. The Locked-Rotor Test The third test that can be perfonned on an induction motor to detennine its circuit parameters is called the locked-rotor test, or sometimes the blocked-rotor test. This test corresponds to the short-circuit test on a transformer. In this test, the ro-tor is locked or blocked so that it cannot move, a voltage is applied to the motor, and the resulting voltage, current, and power are measured. Figure 7- 54a shows the connections for the locked-rotor test. To perform the locked-rotor test, an ac voltage is applied to the stator, and the current flow is adjusted to be approximately full-load value. When the current is fu ll-load value, the voltage, current, and power flowing into the motor are measured. TIle equiva-lent circuit for this test is shown in Figure 7- 54b. Notice that since the rotor is not moving, the slip s = 1, and so the rotor resistance Ris is just equal to R2 (quite a 456 ELECTRIC MACHINERY RJNDAMENTALS I, -Adjustable-V voltage. I, adjustable-b -A frequency. three-phase power source I, , -A , ,' I, R, I, --v, 'h' I<'IGURE 7-54 p p /,= /,= It .. , XM »IR2+jX21 Rc »IR2 + jX21 So neglect Rc and XM The locked-rotor test for an induction motor: (a) test circuit: (b) motor equivalem circuit. small value). Since R2 and X2 are so small, almost all the input current will flow through them, instead of through the much larger magnetizing reactance XM . Therefore, the circuit under these conditions looks like a series combination of X ], R], X2 , and Rl . nlere is one problem with this test, however. In nonnal operation, the stator frequency is the line frequency of tile power system (50 or 60 Hz). At starting con-ditions, the rotor is also at line frequency. However, at nonnal operating conditions, the slip of most motors is only 2 to 4 percent, and the resulting rotor frequency is in the range of I to 3 Hz. nlis creates a problem in that the line frequency does not represent the nonnal operating conditions of the rotor. Since effecti ve rotor resis-tance is a strong function of frequency for design class Band C motors, the incor-rect rotor frequency can lead to misleading results in this test. A typical compro-mise is to use a frequency 25 percent or less of the rated frequency. While this approach is acceptable for essentially constant resistance rotors (design classes A and D), it leaves a lot to be desired when one is trying to find the nonnal rotor re-sistance of a variable-resistance rotor. Because of these and similar problems, a great deal of care must be exercised in taking measurements for these tests. INDUCTION MOTORS 457 After a test voltage and frequency have been set up, the current fl ow in the motor is quickly adjusted to about the rated value, and the input power, voltage, and current are measured before the rotor can heat up too much. 1lle input power to the motor is given by P = V3"V T IL cos (J so the locked-rotor power factor can be found as PF = cos (} = V3V, I " and the impedance angle (J is just equal to cos- l PF. (7-62) The magnitude of the total impedance in the motor circuit at this time is and the angle of the total impedance is O. Therefore, ~ = RLR + jXi.R ~ IZ"loo, e + jIZ"I'in e The locked-rotor resistance RLR is equal to I R" - R, + R, I while the locked-rotor reactance X~ is equal to X LR = X; + X; (7-63) (7-64) (7-65) (7-66) where X; and X; are the stator and rotor reactances at the test frequency, respectively. The rotor resistance Rl can now be found as (7-67) where Rl was detennined in the dc test. The total rotor reactance referred to the sta-tor can also be found. Since the reactance is directly proportional to the frequency, the total equivalent reactance at the nonnal operating frequency can be found as X LR = ~ra1.ed X LR = Xl + X 2 llesl (7-68) Unfortunately, there is no simple way to separate the contributions of the stator and rotor reactances from each other. Over the years, experience has shown that motors of certain design types have certain proportions between the rotor and stator reactances. Figure 7- 55 summarizes this experience. In nonnal practice, it really does not matter just how XLR is broken down, since the reactance appears as the sum XI + X2 in all the torque equations. 458 ELECTRIC MACHINERY RJNDAMENTALS Xl and Xl as functions of XLR Rotor Dcs i ~n X, X, Wound rotor 0.5 XU! 0.5 XU! Design A 0.5 XU! 0.5 XU! Design B 0.4 XU! 0.6 XU! Design C 0.3 XU! 0.7 XU! Design D 0.5 XU! 0.5 XU! H GURE 7-55 Rules of thumb for dividing rotor and stator ci["(;uit reactance. Example 7-S. The following test data were taken on a 7.S-hp, four-pole, 20S-V, 60-Hz, design A, Y-colUlected induction motor having a rated current of 28 A. DC test: No-load test: Voc = 13.6 V Vr = 20SV IA = S.12A la = S.20A le = S.18A loc = 28.0A f = 60 Hz P",,= 420W Locked-rotor test: Vr = 25 V IA = 28.IA fa = 28.0A Ie = 27.6A f = IS Hz P"" = 920W (a) Sketch the per-phase equivalent circuit for this motor. (b) Find the slip at the pullout torque, and find the value of the pullout torque itself. Solutio" (a) From the dc test, Voc 13.6 V R[ = 2/0c = 2(2S.0A) = 0.2430 From the no-load test, IL.av = S.12A + 8.2~A + 8.ISA = S.17A 20S V V</l,nl = ~ = 1lOV INDUCTION MOTORS 459 Therefore, 1 1 120 V ZnJ = 8.17 A = 14.70 = XI + XM When XI is known, XM can be fOlUld. The stator copper losses are PSCL = 3/ f RI = 3(8.17 A)2(0.243 n) = 48.7 W Therefore, the no-load rotational losses are = 420W - 48.7 W = 371.3 W From the locked-rotor test, I = 28.IA + 28.0A + 27.6A = 279A L.av 3 . The locked-rotor impedance is 1 1_1'. --"'- _ 25V -ZLR -110. -..;J/1o. -V3"(27.9 A) - 0.517 n and the impedance angle (J is _ _ I P,n () - cos V!VTh _ _ I 920W - cos v'3(25 VX27.9 A) = cos- t 0.762 = 40.4 0 Therefore, RLR = 0.517 cos 40.4° = 0.394 0 = RI + R2. SinceRI = 0.243 n, R2 must be 0.151 n. The reactance at IS Hz is X LR = 0.517 sin 40.4 0 = 0.335 0 The equivalent reactance at 60 Hz is X LR = ;: X LR = (~~~~) 0.3350 = 1.340 For design class A induction motors, this reactance is assumed to be divided equally between the rotor and stator, so XI = X2 = 0.67 n XM = I ZruI - XI = 14.7 n - 0.67 0 = 14.03 n The final per-phase equivalent circuit is shown in Figure 7- 56. (b) For this equivalent circuit, the Thevenin equivalents are fOlUld from Equations (7-41 b), (7-44), and (7-45) to be V lll = 114.6 V Rlll = 0.221 n Xlll = 0.67 n Therefore, the slip at the pullout torque is given by ,~~~R~, ~~~, ' -mox -v'R~ + (Xlll + X;ll (7- 53) 460 ELECTRIC MACHINERY RJNDAMENTALS R, ""GURE 7-56 jO.67 fl. , , , , Rc -s: (unknown) <-, , , , jXM=j14.03fl. Motor per-phase equivalent circuit for Example 7-8. = 0.1510 =0.111 = 11.1% V(0.243 0)2 + (0.67 0 + 0.67 0)2 The maximum torque of this motor is given by Tmox = 2 W'YDC[Rll{ + VRfu + (Xll{ + X2)] _ 3(114.6 V)l (7- 54) - 2(188.5 rad/s)[O.22 1 0 + V(0.221 Of + (0.670 + 0.67 Of] = 66.2N o m 7.12 THE INDUCTION GENERATOR TIle torque-speed characteristic curve in Figure 7- 20 shows that if an induction motor is driven at a speed greater than n.y "" by an external prime mover, the di-rection of its inducted torque will reverse and it will act as a generator. As the torque applied to its shaft by the prime mover increases, the amount of power pro-duced by the induction generator increases. As Figure 7- 57 shows, there is a max-imum possible induced torque in the generator mode of operation. This torque is known as the pushover torque of the generator. If a prime mover applies a torque greater than the pushover torque to the shaft of an induction generator, the gener-ator wil I overspeed. As a generator, an induction machine has severe limitations. Because it lacks a separate field circuit, an inducti on generator cannot produce reactive power. In fact, it consumes reactive power, and an external source of reactive power must be connected to it at all times to maintain its stator magnetic field. 1llis external source of reacti ve power must also control the terminal voltage of the generator-with no field current, an induction generator cannot control its own output voltage. Normally, the generator's voltage is maintained by the exter-nal power system to which it is connected . 1lle one great advantage of an induction generator is its simplicity. An in-duction generator does not need a separate field circuit and does not have to be driven continuously at a fi xed speed. As long as the machine's speed is some , 0 0 • - 1(XX) - 1500 0 FIGURE 7-57 ~ region Generator region ~ 1000 /2000 "~ Mechanical speed. r/min Pushover torque 3000 INDUCTION MOTORS 461 The torque-speed characteristic of an induction machine. showing the generator region of operation. Note the pushover torque. value greater than n.ync for the power system to which it is connected, it will func-tion as a generator. The greater the torque applied to its shaft (up to a certain point), the greater its resulting output power. TIle fact that no fancy regulation is required makes this generator a gooj choice for windmills, heat recovery systems, and similar supplementary power sources attached to an existing power system. In such applications, power-factor correcti on can be provided by capacitors, and the generator's tenninal voltage can be controlled by the external power system. The Induction Generator Operating Alone It is also possible for an induction machine to function as an isolated generator, in-dependent of any power system, as long as capacitors are available to supply the reactive power req uired by the generator and by any attached loads. Such an iso-lated induction generator is shown in Figure 7- 58. The magnetizing current 1M required by an induction machine as a function of tenninal voltage can be found by running the machine as a motor at no load and measuring its annature current as a function of terrninal voltage. Such a magneti-zation curve is shown in Figure 7- 59a. To achieve a given voltage level in an in-duction generator, external capacitors must supply the magnetization current cor-responding to that level. Since the reactive current that a capacitor can produce is directly proportional to the voltage applied to it, the locus of all possible combinations of voltage and current through a capacitor is a straight line. Such a plot of voltage versus current 462 ELECTRIC MACHINERY RJNDAMENTALS Terminals p I, --Three-phase induction generator --Q p I Q -Q To loads Capacitor bank ""GURE 7-58 An induction generator operating alone with a capacitor bank to supply reactive power. for a given frequency is shown in Figure 7-59b. If a three-phase set of capacitors is connected across the terminnls of an induction generator, the no-load voltage of the induction generator will be the intersection of the generator's magnetization CUlVe and the capacitor s load line. TIle no-load tenninal voltage of an induction generator for three different sets of capacit.:1.nce is shown in Figure 7-59c. How does the voltage build up in an induction generator when it is first started? When an induction generator first starts to turn, the residual magnetism in its field circuit produces a smal I voltage. TImt smal I voltage produces a capacitive current fl ow, which increases the voltage, further increasing the capacitive cur-rent, and so forth until the voltage is fully built up. If no residual flux is present in the induction generator's rotor, then its voltage will not build up, and it must be magnetized by momentarily running it as a motor. TIle most serious problem with an induction generator is that its voltage varies wildly with changes in load, especially reactive load. Typical tenninal char-acteristics of an induction generator operating alone with a constant parallel ca-pacitance are shown in Figure 7-60. Notice that, in the case of inductive loading, the voltage collapses very rapidly. This happens because the fixed capacitors must supply all the reactive power needed by both the load and the generator, and any reactive power diverted to the load moves the generator back along its magneti-zation curve, causi ng a major drop in generator voltage. It is therefore very diffi-cult to start an induction motor on a power system supplied by an induction gen-erator- special techniques must be employed to increase the effective capacitance during starting and then decrease it during nonnal operation. Because of the nature of the induction machine's torque-speed characteristic, an induction generator's frequency varies with changing loads: but since the torque-speed characteristic is very steep in the nonnal operating range, the total fre-quency variation is usually limited to less than 5 percent. This amount of variation may be quite acceptable in many isolated or emergency generator applications. INDUCTION MOTORS 463 Capacitor bank: voltage Vc- V (1 M" no-load armature current) (Lagging amperes) (a) Medium C Small C -------------;'---------------r ---'0' FIGURE 7-59 Small Medium capacitance C (mediumZd capacitance C / ~/ (large Zcl / ~ / / 1 / / / / ~ Large capacitance C / ~ ~ (Small Zcl 1/ , (Capacitor bank: current) (leading amperes) ,b, I Large C (a) The magnetization curve of an induction machine. It is a plot of the tenninal voltage of the machine as a function of its magnetization current (which lags the phase voltage by approximately 90°). (b) Plot of the voltage-.::urrent characteristic of a capacitor bank:. Note that the larger the capacitance. the greater its current for a given voltage. This current leads the phase voltage by approximately 90°. (c) The no-load terntinal voltage for an isolated induction generator can be found by ploning the generator terminal characteristic and the capacitor voltage-.::urrent characteristic on a single set of axes. The intersection of the two curves is the point at which the reactive power demanded by the generator is exactly supplied by the capacitors. and this point gives the no-load ferminall"Oltage of the generator. 464 ELECTRIC MACHINERY RJNDAMENTALS v, ""GURE 7-60 The terminal voItage--<:urrent characteristic of an induction generator for a load with a constant lagging power factor. Induction Generator Applications Induction generators have been used since early in the twentieth century, but by the 1960s and 1970s they had largely disappeared from use. However, the induc-tion generator has made a comeback since the oil price shocks of 1973. With en-ergy costs so high, energy recovery became an important part of the economics of most industrial processes. The induction generator is ideal for such applications because it requires very litt Ie in the way of control systems or maintenance. Because of their simplicity and small size per kilowatt of output power, in-duction generators are also favored very strongly for small windmills. Many com-mercial windmills are designed to operate in parallel with large power systems, supplying a fraction of the customer's total power needs. In such operation, the power system can be relied on for voltage and frequency control, and static ca-pacitors can be used for power-factor correction. 7.13 INDUCTION MOTOR RATINGS A nameplate for a typical high-efficiency integral-horsepower induction motor is shown in Figure 7--6 1. The most important ratings present on the nameplate are L Output power 2, Voltage ), Current 4, Power factor 5, Speed 6, Nominal efficiency INDUCTION MOTORS 465 I ..... NO " . Hf. LOUIS ALLIS FIGURE 7-61 The nameplate of a typical lIigh-efficiency induction motor. (Courtesy of MagneTek, Inc.) 7. NEMA design class 8. Starting code A nameplate for a typical standard-efficiency induction motor would be similar, except that it might not show a nominal efficiency. The voltage limit on the motor is based on the maximum acceptable mag-netization current flow, since the higher the voltage gets, the more saturated the motor's iron becomes and the higher its magnetization current becomes. Just as in the case of transformers and synchronous machines, a 60-Hz induction motor may be used on a 50-Hz power system, but only if the voltage rating is decreased by an amount proportional to the decrease in frequency. nlis derating is necessary be-cause the flux in the core of the motor is. proportional to the integral of the applied 466 ELECTRIC MACHINERY RJNDAMENTALS voltage. To keep the maximum nux in the core constant while the period of inte-gration is increasing, the average voltage level must decrease. TIle current limit on an induction motor is based on the maximum acceptable heating in the motor's windings, and the power limit is set by the combination of the voltage and current ratings with the machine's power factor and efficiency. NEMA design classes, starting code letters, and nominal efficiencies were discussed in previous sections of this chapter. 7.14 SUMMA RY TIle induction motor is the most popular type of ac motor because of its simplicity and ease of operation. An induction motor does not have a separate field circuit; in-stead, it depends on transfonner action to induce voltages and currents in its field circuit. In fact, an induction motor is basically a rotating transfonner. Its equivalent circuit is similar to that of a transfonner, except for the effects of varying speed. An induction motor nonnally operates at a speed near synchronous speed, but it can never operate at exactly n,yDC. There must always be some relative mo-tion in order to induce a voltage in the induction motor's field circuit. TIle rotor voltage induced by the relative motion between the rotor and the stator magnetic field produces a rotor current, and that rotor current interacts with the stator mag-netic field to produce the induced torque in the motor. In an induction motor, the slip or speed at which the maximum torque oc-curs can be controlled by varying the rotor resistance. The value of that maximum torque is independent of the rotor resistance. A high rotor resistance lowers the speed at which maximum torque occurs and thus increases the starting torque of the motor. However, it pays for this starting torque by having very poor speed reg-ulation in its normal operating range. A low rotor resistance, on the other hand, re-duces the motor's starting torque while improving its speed regulation. Any nor-mal induction motor design must be a compromise between these two conflicting requirements. One way to achieve such a compromise is to employ deep-bar or double-cage rotors. lllese rotors have a high effective resistance at starting and a low ef-fective resistance under normal running conditions, thus yielding both a high starting torque and good speed regulation in the same motor. The same effect can be achieved with a wound-rotor induction motor if the rotor field resistance is varied. Speed control of induction motors can be accomplished by changing the number of poles on the machine, by changing the applied electrical frequency, by changing the applied tenninal voltage, or by changing the rotor resistance in the case of a wound-rotor induction motor. TIle induction machine can also be used as a generator as long as there is some source of reacti ve power (capacitors or a synchronous machine) available in the power system. An induction generator operating alone has serious voltage reg-ulation problems, but when it operates in parallel with a large power system, the power system can control the machine's voltage. Induction generators are usually INDUCTION MOTORS 467 rather small machines and are used principally with alternative e nergy sources, such as windmills, or with energy recovery syste ms. A lmost all the really large generators in use are synchronous generators. QUESTIONS 7-1. What are slip and slip speed in an induction motor? 7-2. How does an induction motor develop torque? 7-3. Why is it impossible for an induction motor to operate at synchronous speed? 7-4. Sketch and explain the shape of a typical induction motor torque-speed characteris-tic curve. 7-5. What equivalent circuit element has the most direct control over the speed at which the pullout torque occurs? 7...(j. What is a deep-bar cage rotor? Why is it used? What NEMA design c1ass(es) can be built with it? 7-7. What is a double-cage cage rotor? Why is it used? What NEMA design class(es) can be built with it? 7-8. Describe the characteristics and uses of wound-rotor induction motors and of each NEMA design class of cage motors. 7-9. Why is the efficiency of an induction motor (wolUld-rotor or cage) so poor at high slips? 7-10. Name and describe four means of controlling the speed of induction motors. 7-11. Why is it necessary to reduce the voltage applied to an induction motor as electrical frequency is reduced? 7-12. Why is tenninal voltage speed control limited in operating range? 7-13. What are starting code factors? What do they say about the starting current of an in-duction motor? 7-14. How does a resistive starter circuit for an induction motor work? 7-15. What infonnation is learned in a locked-rotor test? 7-16. What infonnation is learned in a no-load test? 7-17. What actions are taken to improve the efficiency of modern high-efficiency induc-tion motors? 7-18. What controls the tenninal voltage of an induction generator operating alone? 7-19. For what applications are induction generators typically used? 7-20. How can a wOlUld-rotor induction motor be used as a frequency changer? 7-2 1. How do different voltage-frequency patterns affect the torque-speed characteristics of an induction motor? 7-22. Describe the major features of the solid-state induction motor drive featured in Sec-tion 7.10. 7-23. Two 480-V, lOO-hp induction motors are manufactured. One is designed for 50-Hz operation, and one is designed for 6O-Hz operation, but they are otherwise similar. Which of these machines is larger? 7-24. An induction motor is rlUlning at the rated conditions. If the shaft load is now in-creased, how do the following quantities change? (a) Mechanical speed (b) Slip 468 ELECTRIC MACHINERY RJNDAMENTALS (c) Rotor induced voltage (d) Rotor current (e) Rotor frequency (j) PRCL (g) Synchronous speed PROBLEMS 7-1. A dc test is performed on a 460-V. ~-connected. lOO-hp induction motor. If Voc = 24 V and foc = 80A. what is the stator resistance R]? Why is this so? 7-2. A 220-V, three-phase. two-pole. 50-Hz induction motor is ruooing at a slip of 5 per-cent. Find: (a) The speed of the magnetic fields in revolutions per minute (b) The speed of the rotor in revolutions per minute (c) The slip speed of the rotor (d) The rotor frequency in hertz 7-3. Answer the questions in Problem 7- 2 for a 480-V. three-phase. four-pole. 60-Hz in-duction motor running at a slip of 0.035. 7-4. A three-phase. 60-Hz induction motor runs at 890 rhnin at no load and at 840 rlmin at full load. (a) How many poles does this motor have? (b) What is the slip at rated load? (c) What is the speed at one-quarter of the rated load? (d) What is the rotor's electrical frequency at one-quarter of the rated load? 7-5, ASO-kW, 440-V, 50-Hz, six-pole induction motor has a slip of 6 percent when op-erating at full-load conditions. At full-load conditions, the friction and windage losses are 300 W, and the core losses are 600 W. Find the following values for full-load conditions: (a) The shaft speed n .. (b) The output power in watts (c) The load torque 1lood in newton-meters (d) The induced torque 11 ... in newton-meters (e) The rotor frequency in hertz 7-6. A three-phase, 60-Hz, four-pole induction motor runs at a no-load speed of 1790 rlmin and a full-load speed of 1720 r/min. Calculate the slip and the electrical fre-quency of the rotor at no-load and full-load conditions. What is the speed regulation of this motor [Equation (4-68)]? 7-7, A 208-V, two-pole, 60-Hz, V-connected woood-rotor induction motor is rated at IS hp. Its equivalent circuit components are Rl = 0.200 n Xl = O.4lOn P moc.b = 250 W For a slip of 0.05, find (a) The line C lUTent h R2 = 0.120 n X2 = 0.410 n P.u.c""O (b) The stator copper losses P SCL (c) The air-gap power PAG XM = IS.On P<Ote = 180W INDUCTION MOTORS 469 (d) The power converted from electrical to mechanical fonn P <oov (e) The induced torque Tm (f) The load torque Tload (g) The overall machine efficiency (h) The motor speed in revolutions per minute and radians per second 7-8. For the motor in Problem 7- 7. what is the slip at the pullout torque? What is the pullout torque of this motor? 7-9. (a) Calculate and plot the torque-speed characteristic of the motor in Problem 7- 7. (b) Calculate and plot the output power versus speed curve of the motor in Prob-lem 7- 7. 7-10. For the motor of Problem 7- 7. how much additional resistance (referred to the sta-tor circuit) would it be necessary to add to the rotor circuit to make the maximum torque occur at starting conditions (when the shaft is not moving)? Plot the torque-speed characteristic of this motor with the additional resistance inserted. 7-11. If the motor in Problem 7- 7 is to be operated on a 50-Hz power system. what must be done to its supply voltage? Why? What will the equivalent circuit component values be at 50 Hz? Answer the questions in Problem 7- 7 for operation at 50 Hz with a slip of 0.05 and the proper voltage for this machine. 7-12. Figure 7-1 8a shows a simple circuit consisting of a voltage source. a resistor. and two reactances. Find the Thevenin equi valent voltage and impedance of this circuit at the terminals. Then derive the expressions for the magnitude of Vrn and for Rrn given in Equations (7-4lb) and (7-44). 7-13. Figure P7-1 shows a simple circuit consisting of a voltage source. two resistors. and two reactances in series with each other. If the resistor RL is allowed to vary but all the other components are constant. at what value of RL will the maximum possible power be supplied to it? Prove your answer. (Hint: Derive an expression for load power in terms of V. Rs. Xs. RL • and XL and take the partial derivative of that ex-pression with respect to Rd Use this result to derive the expression for the pullout torque [Equation (7- 54)]. jXs v(Z) FlGURE 1'7-1 Circuit for Problem 7- 13. 7-14. A 440-V. 50-Hz, two-pole, V-connected induction motor is rated at 75 kW. The equivalent circuit parameters are R[ = 0.075 0 X[ = 0.170 PFAW = 1.0 kW For a slip of 0.04, find R2 = 0.065 n X2 = O.170 Pmioc = 150W XM = 7.20 P axe = 1.1 kW 470 ELECTRIC MACHINERY RJNDAMENTALS (a) The line clUTent h (b) The stator power factor (c) The rotor power factor (d) The stator copper losses PSCL (e) The air-gap power PAG (j) The power converted from electrical to mechanical fonn P OO/IiY (g) The induced torque 7; ... (h) The load torque Tlood (i) The overall machine efficiency 71 OJ The motor speed in revolutions per minute and radians per second 7-15. For the motor in Problem 7-14, what is the pullout torque? What is the slip at the pullout torque? What is the rotor speed at the pullout torque? 7-16. If the motor in Problem 7-14 is to be driven from a 440-V, 60-Hz power supply, what will the pullout torque be? What will the slip be at pullout? 7-17. Plot the following quantities for the motor in Problem 7-14 as slip varies from 0 to 10 percent: (a) T ind: (b) P OOGY: (c) P 00': (d) efficiency 71. At what slip does P 00i. equal the rated power of the machine? 7-18. A 20S-V, 60 Hz six-pole, V-connected, 25-hp design class B induction motor is tested in the laboratory, with the following results: No load: Locked rotor: OC test: 208 V, 22.0 A, 1200 W, 60 Hz 24.6 V, 64.5 A, 2200 W, 15 Hz 13.5 V, 64 A Find the equivalent circuit of this motor, and plot its torque-speed characteristic curve. 7-19. A 460-V, four-pole, 50-hp, 60-Hz, Y-colUlected, three-phase induction motor devel-ops its full-load induced torque at 3.S percent slip when operating at 60 Hz and 460 V. The per-phase circuit model impedances of the motor are RI = 0.33 n XI = 0.42 n XM =30 n X2 = 0.42 n Mechanical, core, and stray losses may be neglected in this problem. (a) Find the value of the rotor resistance R2. (b) Find T malI , s"""v and the rotor speed at maximum torque for this motor. (c) Find the starting torque of this motor. (d) What code letter factor should be assigned to this motor? 7-20. Answer the following questions about the motor in Problem 7-19. (a) If this motor is started from a 460-V infinite bus, how much current will flow in the motor at starting? (b) If transmission line with an impedance of 0.35 + jO.25 n per phase is used to connect the induction motor to the infinite bus, what will the starting current of the motor be? What will the motor's tenninal voltage be on starting? (c) If an ideal 1.4: I step-down autotransformer is connected between the transmis-sion line and the motor, what will the ClUTent be in the transmission line during starting? What will the voltage be at the motor end of the transmission line dur-ing starting? INDUCTION MOTORS 471 7-21. In this chapter. we learned that a step-down autotransformer could be used to reduce the starting current drawn by an induction motor. While this technique works. an au-totransfonner is relatively expensive. A much less expensive way to reduce the start-ing current is to use a device called y-~ starter. If an induction motor is nonnally ~-cOIUlected. it is possible to reduce its phase voltage V. (and hence its starting cur-rent) by simply reconnecting the swtor windings in Y during starting. and then restoring the cOlUlections to ~ when the motor comes up to s~ed. Answer the fol-lowing questions about this type of starter. (a) How would the phase voltage at starting compare with the phase voltage under normal lUlUling conditions? (b) How would the starting current of the Y-colUlected motor compare to the start-ing current if the motor remained in a ~-connection during starting? 7-22. A 460-V. lOO-hp. four-pole. ~-connected. 60-Hz. three-phase induction motor has a full-load slip of 5 percent. an efficiency of 92 percent. and a power factor of 0.87 lagging. At start-up. the motor develops 1.9 times the full-load torque but draws 7.5 times the rated current at the rated voltage. This motor is to be started with an auto-transformer reduced-volwge starter. (a) What should the output volwge of the starter circuit be to reduce the starting torque until it equals the rated torque of the motor? (b) What will the motor starting ClUTent and the current drawn from the supply be at this voltage? 7-23. A wound-rotor induction motor is operating at rated voltage and frequency with its slip rings shorted and with a load of about 25 percent of the rated value for the ma-chine. If the rotor resistance of this machine is doubled by inserting external resis-tors into the rotor circuit. explain what hap~ns to the following: (a) Slip s (b) Motor speed n .. (c) The induced voltage in the rotor (d) The rotor current (e) "rio<! (f) P out (g) PRCL (h) Overall efficiency 7f 7-24. Answer the following questions about a 460-V, ~-COlUlected. two-pole. 75-hp. 60-Hz. starting-code-Ietter-E induction motor: (a) What is the maximum current starting current that this machine's controller must be designed to handle? (b) If the controller is designed to switch the stator windings from a ~ connection to a Y connection during starting. what is the maximum starting current that the controller must be designed to handle? (c) If a 1.25: I step-down autotransfonner starter is used during starting. what is the maximum starting current that will be drawn from the line? 7-25. When it is necessary to stop an induction motor very rapidly. many induction motor controllers reverse the direction of rotation of the magnetic fields by switching any two stator leads. When the direction of rotation of the magnetic fields is reversed. the motor develops an induced torque opposite to the current direction of rotation. so it quickly stops and tries to start turning in the opposite direction. If power is re-moved from the stator circuit at the moment when the rotor s~ed goes through zero. 472 ELECTRIC MACHINERY RJNDAMENTALS then the motor has been stopped very rapidly. This technique for rapidly stopping an induction motor is called plugging. The motor of Problem 7- 19 is running al rated conditions and is to be stopped by plugging. (a) What is the slip s before plugging? (b) What is the frequency of the rotor before plugging? (c) What is the induced torque "ria<! before plugging? (d) What is the slip s inunediately after switching the stator leads? (e) What is the frequency of the rotor immediately after switching the stator leads? (j) What is the induced torque "ria<! immediately after switching the stator leads? REFERENCES I. Alger. Phillip. Induction Machines. 2nd ed. New York: Gordon and Breach. 1970. 2. Del Toro. V. Electric Machines and Power Systems. Englewood Cliffs. N.J.: Prentice-Hall. 1985. 3. Filzgerald. A. E. and C. Kingsley. Jr. Electric Machinery. New York: McGraw-Hilt. 1952. 4. Filzgerald. A. E .• C. Kingsley. Jr .• and S. D. Umans. Electric Machinery. 51h ed. New York: McGraw-Hilt. 1990. 5. Institute of Electrical and Electronics Engineers. Standard Test Procedure for Pol)phase Induction MOlOrs and Genemtors. IEEE Standaro 112-1996. New York: IEEE. 1996. 6. Kosow. Irving L. Control of Electric Motors. Englewood Cliffs. N.J.: Prentice-Hall. 1972. 7. McPherson. George. An Introduction 10 Electrical Machines and Tmnsfonners. New York: Wiley. 1981. 8. National Eleclrical Manufaclurers Association. MOlOrs and GenemlOrs. Publication No. MG I-1993. Washington. D.C.: NEMA.I993. 9. Siemon. G. R .• and A. Siraughen. Electric Machines. Reading. Mass.: Addison-Wesley. 1980. 10. Vithayalhil. Joseph. Pov.'ef Electronics: Principles and Applications. New York: McGraw-Hill. 1995. II. Weminck. E. H. (ed.). Electric MOlOr Handbook. London: McGraw-Hill. 1978. CHAPTER 8 DC MACHINERY FUNDAMENTALS DC machines are generators that convert mechanical energy to de electric energy and motors that convert de electric energy to mechanical energy. Most de machines are like ac machines in that they have ac voltages and currents within thern---dc machines have a de output only because a mechanism exists that converts the internal ac voltages to de voltages at their tenninals. Since this mechanism is called a commutator, de machinery is also known as commutating machinery. The fundamental principles invo lved in the operation of de machines are very simple. Unfortunately, they are usually somewhat obscured by the compli-cated construction of real machines. This chapter will first explain the principles of de machine operation by using simple examples and then consider some of the complications that occur in real dc machines. S.I A SIMPLE ROTATING LOOP BETWEEN CURVED POLE FACES The linear machine studied in Section 1.8 served as an introduction to basic ma-chine behavior. Its response to loading and to changing magnetic fields closely re-sembles the behavior of the real dc generators and motors that we will study in Chapter 9. However, real generators and motors do not move in a straight line-they rotate. The next step toward understanding real dc machines is to study the simplest possible example of a rotating machine. The simplest possible rotating dc machine is shown in Figure 8-1. It consists of a si ngle loop of wire rotating about a fixed axis. The rotating part of this ma-chine is called the rotor, and the stationary part is called the stator. TIle magnetic 473 474 ELECTRIC MACHINERY RJNDAMENTALS d N s (. ) N s r-= >s -f:::: = '---~ -~ -f---0 ' < b . . -( ): . • d " • " ',. (b) «) F" s • , N F " (d) ""GURE 8-1 A simple rotating loop between curved pole faces. (a) Perspective view; (b) view of field lines; (e) top view; (d) front view. OCMACHINERYFUNDAMENTALS 475 .. -" • FIGURE 8-2 Derivation of an equation for the voltages induced in the loop. field for the machine is supplied by the magnetic north and south poles shown on the stator in Figure 8- 1. Notice that the loop of rotor wire lies in a slot carved in a ferromagnetic core. 1lle iron rotor, together with the curved shape of the pole faces, provides a constant-width air gap between the rotor and stator. Remember from Chapter I that the reluctance of air is much much higher than the rei uctance of the iron in the ma-chine. To minimize the reluctance of the flux path through the machine, the mag-netic flux must take the shortest possible path through the air between the pole face and the rotor surface. Since the magnetic flux must take the shortest path through the air, it is per-pendicular to the rotor surface everywhere under the pole faces. Also, since the air gap is of unifonn width, the reluctance is the same everywhere under the pole faces. The uniform reluctance means that the magnetic flux density is constant everywhere under the pole faces. The Voltage Induced in a Rotating Loop If the rotor of this machine is rotated, a voltage will be induced in the wire loop. To detennine the magnitude and shape of the voltage, examine Figure 8- 2. 1lle loop of wire shown is rectangular, with sides ab and cd perpendicular to the plane of the page and with sides be and da parallel to the plane of the page. The mag-netic field is constant and perpendicular to the surface of the rotor everywhere un-der the pole faces and rapidly falls to zero beyond the edges of the poles. To determine the total voltage e,OI on the loop, examine each segment of the loop separately and sum all the resulting voltages. TIle voltage on each segment is given by Equation (1-45): eind = (v x B) • I ( 1-45) 476 ELECTRIC MACHINERY RJNDAMENTALS I. Segment abo In this segment, the velocity of the wire is tangential to the path of rotation. The magnetic field 8 points out perpendicular to the rotor surface everywhere under the pole face and is. zero beyond the edges of the pole face. Under the pole face, velocity v is perpendicular to 8, and the quantity v x 8 points into the page. TIlerefore, the induced voltage on the segment is positive into page under the pole face beyond the pole edges (8-1 ) 2. Segment be. In this segment, the quantity v x 8 is either into or out of the page, while length 1 is in the plane of the page, so v x 8 is perpendicular to I. Therefore the voltage in segment be will be zero: ecb = 0 (8- 2) 3. Segment ed. In this segment, the velocity of the wire is tangential to the path of rotation. The magnetic field 8 points in perpendicular to the rotor surface everywhere under the pole face and is. zero beyond the edges of the pole face. Under the pole face, velocity v is perpendicular to 8, and the quantity v x 8 points out of the page. 1llerefore, the induced voltage on the segment is positive out of page under the pole face beyond the pole edges (8- 3) 4. Segment da. Just as in segment be, v x 8 is perpendicular to I. Therefore the voltage in this segment will be zero too: ead = 0 (8-4) 1lle total induced voltage on the loop ei!>d is given by under the pole faces beyond the pole edges (8- 5) When the loop rotates through 180°, segment ab is under the north pole face in-stead of the south pole face. At that time, the direction of the voltage on the seg-ment reverses, but its magnitude remains constant. The resulting voltage e,,,, is shown as a function of time in Figure 8- 3. 1llere is an alternative way to express Equation (8- 5), which clearly relates the behavior of the single loop to the behavior of larger, real dc machines. To de-rive this alternative expression, examine Figure 8-4. Notice that the tangential ve-locity v of the edges of the loop can be expressed as v = rw IX: MACHINERY FUNDAMENTALS 477 2vBI C r----, ,81 0 r-----~.------r------~------T_------ , - vBI - 2vBI FIGURE 8-3 The output voltage of the loop. -v=rw FIGURE 8-4 w , Pole surface area Ap .. nrl • Rotor surface area A =2nrl Derivation of an alternative form of the induced voltage equation. 478 ELECTRIC MACHINERY RJNDAMENTALS where r is the radius from axis of rotation out to the edge of the loop and w is the angular velocity of the loop. Substituting this expression into Equation (8- 5) gives _ [2rWBI eioo -0 eioo = {2r~BW under the pole faces beyond the pole edges under the pole faces beyond the pole edges Notice also from Figure 8-4 that the rotor surface is a cylinder, so the area of the rotor surface A is just equal to 2nrl. Since there are two poles, the area of the rotor under each pole (ignoring the small gaps between poles) is Ap = nrl. TIlerefore, under the pole faces beyond the pole edges Since the nux density B is constant everywhere in the air gap under the pole faces, the total flux under each pole is just the area of the pole times its flux density: Therefore, the final form of the voltage equation is e. , ~ {~"'W '" 0 under the pole faces beyond the pole edges (8-6) TIms, the voltage generated in the machine is equnl to the product of the flux inside the mnchine and the speed of rotation of the machine, multiplied by a constant representing the mechanical construction of the machine. In general, the voltage in any real machine will depend on the same three factors: I. The flux in the machine 2. The speed of rotation 3. A constant representing the construction of the machine Getting DC Voltage out of the Rotating Loop Figure 8- 3 is a plot of the voltage e"", generated by the rotating loop. As shown, the voltage out of the loop is alternately a constant positive value and a constant negative value. How can this machine be made to produce a dc voltage instead of the ac voltage it now has? One way to do this is shown in Figure 8- 5a. Here two semicircular conduct-ing segments are added to the end of the loop, and two fixed contacts are set up at an angle such that at the instant when the voltage in the loop is zero, the contacts IX: MACHINERY FUNDAMENTALS 479 N Commutator s Brushes (.) - <'W (b) FIGURE 8-S Producing a dc output from the machine with a commutator and brushes. (a) Perspective view; (b) the resulting output voltage. short-circuit the two segments. In this fashion, every time the voltage of the loop switches direction, the contacts also switch connections, and the output of the con-tacts is always built up in the same way (Figure 8- 5b). This connection-switching process is known as commutation. TIle rotating semicircular segments are called commutator segments, and the fixed contacts are calJed brushes. 480 ELECTRIC MACHINERY RJNDAMENTALS N Commutator s (a) Current into cod page N ,...!..... Current out of page F cJ, imd .......... :::J w, S ~,Z, "-b (h) ""GURE 8-6 Derivation of an equation for the induced torque in the loop. Note that the iron core is not shown in part b for clarity. The Induced Torque in the Rotating Loop Suppose a battery is now connected to the machine in Figure 8- 5. The resulting configuration is shown in Figure 8-6. How much torque will be produced in the loop when the switch is closed and a current is allowed to fl ow into it? To deter-mine the torque, look at the close-up of the loop shown in Figure 8-6b. TIle approach to take in detennining the torque on the loop is to look at one segment of the loop at a time and then sum the effects of all the individual seg-ments. TIle force on a segment of the loop is given by Equation (1-43): IX: MACHINERY FUNDAMENTALS 481 F = i(lx8) ( 1-43) and the torque on the segment is given by T = rFsin () (1-6) where () is the angle between rand F. The torque is essentially zero whenever the loop is beyond the pole edges. While the loop is under the pole faces, the torque is I. Segmentab. In segment ab, the current from the battery is directed out of the page. TIle magnetic field under the pole face is pointing radially out of the ro-tor, so the force on the wire is given by Fab- i(lx8) -ilB tangent to direction of motion TIle torque on the rotor caused by this force is "Tab -rF sin () -r(iIB) sin 900 -rilB CCW (8- 7) (8-8) 2. Segment be. In segment be, the current from the battery is flowing from the upper left to the lower right in the picture. The force induced on the wire is given by Fbc -i(lx8) -0 since I is parallel to 8 (8- 9) TIlerefore, "Tb< = 0 (8-10) 3. Segment ed. In segment ed, the current from the battery is directed into the page. The magnetic field under the pole face is pointing radially into the ro-tor, so the force on the wire is given by Fed -i(l X 8 ) -ilB tangent to direction of motion TIle torque on the rotor caused by this force is "Ted -rF sin () -r(iIB) sin 900 -rilB CCW (8-11 ) (8-1 2) 4. Segment 00. In segment 00, the current from the battery is flowing from the upper left to the lower right in the picture. The force induced on the wire is given by 482 ELECTRIC MACHINERY RJNDAMENTALS Fda -i(I XB) -0 since I is parallel to B Therefore, Tdo = 0 TIle resulting total induced torque on the loop is given by under the pole faces beyond the pole edges (8-13) (8-14) (8-15) By using the facts that Ap ,., rrrl and cp = ApB, the torque expression can be re-duced to { , ~. - ~, Tind = : under the pole faces (8-16) beyond the pole edges TIlUS, the torque produced in the machine is the product of the flux in the machine and the current in the machine, times some quantity representing the me-chanical construction of the machine (the percentage of the rotor covered by pole faces). In general, the torque in any real machine will depend on the same three factors: L The flux in the machine 2, The current in the machine ), A constant representing the construction of the machine Example 8-1. Figure 8--6 shows a simple rotating loop between curved pole faces connected to a battery and a resistor through a switch. The resistor shown models the total resistance of the battery and the wire in the machine. The physical dimensions and charac-teristics of this machine are r = O.5m R = 0.3!l VB = 120 V I = 1.0m B = O.25T (a) What happens when the switch is closed? (b) What is the machine's maximum starting current? What is its steady-state angu-lar velocity at no load? (c) Suppose a load is attached to the loop, and the resulting load torque is 10 N· m. What would the new steady-state speed be? How much power is supplied to the shaft of the machine? How much power is being supplied by the battery? Is this machine a motor or a generator? IX: MACHINERY FUNDAMENTALS 483 (d) Suppose the machine is again unloaded, and a torque of 7.5 N • m is applied to the shaft in the direction of rotation. What is the new steady-state speed? Is this machine now a motor or a generator? (e) Suppose the machine is rulUling unloaded. What would the final steady-state speed of the rotor be if the flux density were reduced to 0.20 T? Solution (a) When the switch in Figure 8-6 is closed, a current will flow in the loop. Since the loop is initially stationary, e jod = O. Therefore, the current will be given by VB - eiod VB i = = R R This current flows through the rotor loop, producing a torque 2.; 7iod = -'I" ~ ccw This induced torque produces an angular acceleration in a counterclockwise di-rection, so the rotor of the machine begins to turn. But as the rotor begins to tum, an induced voltage is produced in the motor, given by so the current i falls. As the current falls, 7ind = (2hr)cpi.t.. decreases, and the ma-chine winds up in steady state with 7iod = 0, and the battery voltage VB = eind' This is the same sort of starting behavior seen earlier in the linear dc machine. (b) At starting conditions, the machine's current is . VB l20V I = J[ = 0.3fi = 400 A At no-load steady-state conditions, the induced torque 7ind must be zero. But 7ind = 0 implies that current i must equal zero, since 7ind = (2hr)cpi, and the flux is nonzero. The fact that i = 0 A means that the battery voltage VB = eind' There-fore, the speed of the rotor is _ VB _ ~ W -(2i1r)cp -2rlB l20V = 2(0.5 mXI.O mXO.25 T) = 480 rad/s (c) If a load torque of 10 N om is applied to the shaft of the machine, it will begin to slow down. But as w decreases, eind = (2hr)cpwJ. decreases and the rotor cur-rent increases [i = (VB -eind.t.. )/R]. As the rotor current increases, ITIOdI in-creases too, until I 7;",,1 = 171Nd1 at a lower speed w. At steady state, 171 .... 1 = 17indl = (2hr)cpi. Therefore, . 7jod 7ind 1= = --(2hr)cp 2rlB ION om = (2XO.5 m)(1.0 mXO.25 T) = 40 A 484 ELECTRIC MACHINERY RJNDAMENTALS By Kirchhoff's voltage law, eind = V B - iR, so eiDd = 120 V -(40 AXO.3 ll) = 108 V Finally, the speed of the shaft is ejDd eiDd w = (2/Tr)q, = 2rlB 108 V = (2)(0.5 mX1.0 m)(0.25 1) = 432 radls The power supplied to the shaft is P = TW = (10 N • mX432 rad/s) = 4320 W The power out of the battery is P = V Bi = (120 V)(40 A) = 4800 W This machine is operating as a motor, converting electric power to mechanical power. (d) If a torque is applied in the direction of motion, the rotor accelerates. As the speed increases, the internal voltage eind increases and exceeds Vs, so the current flows out of the top of the bar and into the battery. This machine is now a gen-emtor. This current causes an induced torque opposite to the direction of mo-tion. The induced torque opposes the external applied torque, and eventually I11NdI = IT UldI at a higher speed w. The current in the rotor will be . 7ind Tim I = (2hr)q, = 2rlB 7.5 N. m = (2)(0.5 mX1.0 mXO.25 T) = 30 A The induced voltage eind is eind -V B + iR -120 V + (30 AXO.3 !l) -129 V Finally, the speed of the shaft is eiod eiDd W = (2/Tr)q, = 2rlB 129 V = (2X0.5 m)(1.0 mXO.25 T) = 516 radls (e) Since the machine is initially lUlloaded at the original conditions, the speed w = 480 radls. If the flux decreases, there is a transient. However, after the transient is over, the machine must again have zero torque, since there is still no load on its shaft. If"TIDd = 0, then the current in the rotor must be zero, and VB = eind. The shaft speed is thus eiod eiDd w = (2/Tr)q, = 2rlB IX: MACHINERY FUNDAMENTALS 485 120 V = (2)(0.5 mX 1.0 m)(0.20 T) = 600 rad/s Notice that when the flux in the machine is decreased, its speed increases. This is the same behavior seen in the linear machine and the same way that real dc motors behave. 8.2 COMMUTATION IN A SIMPLE FOUR-LOOP DC MACHINE Commutation is the process of converting the ac voltages and currents in the rotor of a dc machine to dc voltages and currents at its tenninals. It is the most critical part of the design and operation of any dc machine. A more detailed study is nec-essary to determine just how this conversion occurs and to discover the problems associated with it. In this section, the technique of commutation will be explained for a machine more complex than the single rotating loop in Section 8. 1 but less complex than a real dc machine. Section 8.3 will continue this development and explain commutation in real dc machines. A simple four-loop, two-pole dc machine is shown in Figure 8- 7.lllis ma-chine has four complete loops buried in slots carved in the laminated steel of its rotor. TIle pole faces of the machine are curved to provide a uniform air-gap width and to give a uniform nux density everywhere under the faces. The four loops of this machine are laid into the slots in a special manner. The "unprimed" end of each loop is the outermost wire in each slot, while the "primed" end of each loop is the innennost wire in the slot directly opposite. The winding's connections to the machine's commutator are shown in Figure 8- 7b. Notice that loop 1 stretches between commutator segments a and b, loop 2 stretches between segments band c, and so forth around the rotor . • N s d (.J FIGURE 8-7 (a) A four-toop two-pole dc machine shown at time WI = 0°. (continues) 486 ELECTRIC MACHINERY RJNDAMENTALS Back side of coil I Back side of coil 4 3 I' 42' S Pol, faces Comrnutator _ segments -~-Brushes ""GURE 8-7 (roncluded) 3' 24' N I ' 2 , b 43' (bl Back side of coil 2 • , E=4~ Back side of coil 3 3 ] ' 42' S 3' 24' N (" (b) The voltages on the rotor conductors at this time, (c) A winding diagram of this machine showing the interconnections of the rotor loops. At the instant shown in Figure 8- 7, the \,2,3', and 4' ends of the loops are under the north pole face, while the \',2',3, and 4 ends of the loops are underthe south pole face. TIle voltage in each of the 1,2,3', and 4' ends of the loops is given by eind -(v x B) • I positive out of page ( 1-45) (8-1 7) TIle voltage in each of the 1 " 2', 3, and 4 ends of the ends of the loops is given by -vBI positive into the page ( 1-45) (8-1 8) IX: MACHINERY FUNDAMENTALS 487 " -----:7 ,,/ · ---,;..3,--., • .;:"",:: =.::45 ,-' __ I ' b N 3' I" 2 + e -I ' ov ov 4' + e -Ib, d + e -2' , d + e -4 4 S 3 ov ov 3' E=2e FIGURE 8-8 The same machine at time WI = 45°. showing the voltages on the conductors. The overall result is shown in Figure 8- 7b. In Figure 8- 7b, each coil represents one side (or conductor) of a loop. Ir the induced voltage on anyone side of a loop is called e = vBI, then the total voltage at the brushes or tile machine is 1£ -4, wt -O' I (8-1 9) Notice that there are two parallel paths for current through the machine. The ex-istence of two or more parallcl paths for rotor current is a common feature of all commutation schemes. What happens to the voltage E of the terminals as the rotor continues to rotate? To fmd out, examine Figure 8- 8. lllis figure shows the machine at time wt = 45°. At that time, loops 1 and 3 have rotated into the gap between the poles, so the voltage across each of them is zero. Notice that at this instant the brushes 488 ELECTRIC MACHINERY RJNDAMENTALS w • -wl= 90° " 3 , ~E~ N b ~f-I d s " 2 The same machine at time WI = 90°, showing the voltages on the conductors. of the machine are shorting out commutator segments ab and cd. This happens just at the time when the loops between these segments have 0 V across them, so shorting out the segments creates no problem. At this time, only loops 2 and 4 are under the pole faces, so the terminal voltage E is given by I E ~ " (8- 20) Now let the rotor continue to turn through another 45°. TIle resulting situa-tion is shown in Figure 8-9. Here, the 1',2,3, and 4' ends of the loops are under E. volts 5 , 4 , 3 , 2 , , o ° FIGURE 8-10 '-' 45 ° 90 ° ° IX: MACHINERY FUNDAMENTALS 489 ~ 180 ° 225 ° 270 ° ° 315 360 ° wI The resulting output voltage of the machine in Figure 8- 7. the north pole face, and the I, 2', 3', and 4 ends of the loops are under the south pole face. The voltages are still built up out of the page for the ends under the north pole face and into the page for the ends under the south pole face. 1lle re-sulting voltage diagram is shown in Figure 8-1 8b. There are now four voltage-carrying ends in each parallel path through the machine, so the terminal voltage E is given by wt - 900 1 (8- 21) Compare Figure 8- 7 to Figure 8- 9. Notice that the voltages on loops 1 and 3 have reversed between the two pictures. but since their connections have also reversed, the total voltage is still being built up in the same direction as before. This fact is at the heart of every commutation scheme. Whenever the voltage re-verses in a loop, the connections of the loop are also switched, and the total volt-age is still built up in the original direction. The terminal voltage of this machine as a function of time is shown in Fig-ure 8-1 0. It is a better approximation to a constant dc level than the single rotat-ing loop in Section 8.1 produced. As the number of loops on the rotor increases, the approximation to a perfect dc voltage continues to get better and better. In summary, Commutation is the process of switching the loop COIUlections on the rotor of a dc machine just as the voltage in the loop switches polarity, in order to maintain an es-sentially constant dc output voltage. As in the case of the simple rotating loop, the rotating segments to which the loops are attached are called commutator segments, and the stationary pieces that ride on top of the moving segments are called brushes. The commutator segments 490 ELECTRIC MACHINERY RJNDAMENTALS in real machines are typically made of copper bars. The brushes are made of a mix-ture containing graphite, so that they cause very little friction as they rub over the rotating commutator segments. 8.3 COMMUTATION AND ARMATURE CONSTRUCTION IN REAL DC MACHINES In real dc machines, there are several ways in which the loops on the rotor (also called the armature) can be connected to its commutator segments. nlese differ-ent connections affect the number of parallel current paths within the rotor, the output voltage of the rotor, and the number and position of the brushes riding on the commutator segments. We wil l now examine the construction of the coils on a real dc rotor and then look at how they are connected to the commutator to pro-duce a dc voltage. The Rotor Coils Regardless of the way in which the windings are connected to the commutator segments, most of the rotor windings themselves consist of diamond-shaped preformed coils which are inserted into the armature slots as a unit (see Figure 8-11 ). Each coil consists of a number of turns (loops) of wire, each turn taped and insulated from the other turns and from the rotor slot. Each side of a turn is called a conductor. 1lle number of conductors on a machine's annature is given by where I Z ~ 2eNc I Z = number of conductors on rotor C = number of coils on rotor Nc = number of turns per coil (8- 22) Nonnally, a coil spans 180 electrical degrees. nlis means that when one side is under the center of a given magnetic pole, the other side is under the cen-ter of a pole of opposite polarity. The physical poles may not be located 180 me-chanical degrees apart, but the magnetic field has completely reversed its polarity in traveling from under one pole to the next. The relationship between the electri-cal angle and mechanical angle in a given machine is given by where O e = electrical angle, in degrees 0", = mechanical angle, in degrees P = number of magnetic poles on the machine (8- 23) If a coil spans 180 electrical degrees, the voltages in the conductors on either side of the coil will be exactly the same in magnitude and opposite in direction at all times. Such a coil is called afull-pitch coil. Nc turns insulated from ,~h other FIGURE 8-11 OCMACHINERYFUNDAMENTALS 491 I", length of conductor (a) (b) (a) The shape of a typical prefornled rotor coil. (b) A typical coil insulation system showing the insulation between turns within a coil. (Courtesy ofGeneml Electric Company.) Sometimes a coil is built that spans less than 180 electrical degrees. Such a coil is called afractional-pitch coil, and a rotor winding wound with fractional-pitch coils is called a chorded winding. 1lle amount of chording in a winding is described by a pitch factor p, which is defined by the equation electrical angle of coil p = 180 0 x 100% (8- 24) Sometimes a small amount of chording will be used in dc rotor windings to im-prove commutation. Most rotor windings are hi!o-layer windings, meaning that sides from two different coils are inserted into each slot. One side of each coil will be at the bot-tom of its slot, and the other side will be at the top of its slot. Such a construction requires the individual coils to be placed in the rotor slots by a very elaborate 492 ELECTRIC MACHINERY RJNDAMENTALS , \ • , ---:.------""GURE 8-12 The installation of prefonned rotor coils on a dc machine rotor. (Courtesy of Westinghouse Electric Company.) procedure (see Figure 8-1 2). One side of each of the coils is placed in the bottom of its slot, and then after all the bottom sides are in place, the other side of each coil is placed in the top of its slot. In this fashion, all the windings are woven to-gether, increasing the mechanical strength and unifonnity of the final structure. Connections to the Commutator Segments Once the windings are installed in the rotor slots, they must be connected to the commutator segments. There are a number of ways in which these connections can be made, and the different winding arrangements which result have different advantages and disadvantages. TIle distance (in number of segments) between the commutator segments to which the two ends of a coil are connected is called the commutator pitch Ye. If the end of a coil (or a set number of coils, for wave construction) is connected to a commutator segment ahead of the one its beginni ng is connected to, the winding is called a progressive winding. If the end of a coil is connected to a commutator segment behind the one its beginning is connected to, the winding is called a ret-rogressive winding. If everything else is identical, the direction of rotation of a progressive-wound rotor will be oppos.ite to the direction of rotation of a retrogressive-wound rotor. Rotor (armature) windings are further classified according to the plex of their windings. A simplex rotor winding is a single, complete, closed winding wound on a rotor. A duplex rotor winding is a rotor with two complete and inde-pendent sets of rotor wi ndings. I f a rotor has a duplex winding, then each of the windings will be associated with every other commutator segment: One winding IX: MACHINERY FUNDAMENTALS 493 c+ I c C+I C-I C C+I I" Ib, FIGURE 8-13 (3) A coil in 3 progressive rotor winding. (b) A coil in 3 retrogressive rotor winding. will be connected to segments I, 3, 5, etc., and the other winding will be con-nected to segments 2, 4, 6, etc. Similarly, a triplex winding will have three com-plete and independent sets of windings, each winding connected to every third commutator segment on the rotor. Collectively, all armatures with more than one set of windings are said to have multiplex windings. Finally, annature windings are classified according to the sequence of their connections to the commutator segments. There are two basic sequences of arrna-ture winding connections-iap windings and wave windings. In addition, there is a third type of winding, called a frog-leg winding, which combines lap and wave windings on a single rotor. These windings will be examined individually below, and their advantages and disadvantages will be discussed. The Lap Winding The simplest type of winding construction used in modem dc machines is the sim-plex series or lap winding. A simplex lap winding is a rotor winding consisting of coils containing one or more turns of wire with the two ends of each coil coming out at adjacent commutator segments (Figure 8- 13). If the end of the coil is con-nected to the segment after the segment that the beginning of the coil is connected to, the winding is a progressive lap winding and Yc = I; if the end of the coil is connected to the segment before the segment that the beginning of the coil is con-nected to, the winding is a retrogressive lap winding and Yc = - I. A simple two-pole machine with lap windings is shown in Figure 8- 14. An interesting feature of simplex lap windings is that there are as many par-allel current paths through the machine as there are poles on the mnchine. If C is the number of coils and commutator segments present in the rotor and P is the 494 ELECTRIC MACHINERY RJNDAMENTALS 2 N I-oo--J s \ 8 5 ""GURE 8-14 A simple two-pole lap-wound dc machine. number of poles on the machine, then there will be c/P coils in each of the P par-allel current paths through the machine. The fact thai there are P current paths also requires that there be as many brushes on the machine as there are poles in order to tap all the current paths. This idea is illustrated by the simple four-pole motor in Figure 8-15. Notice that, for this motor, there are four current paths through the rotor, each having an equal voltage. The fact that there are many current paths in a multi pole machine makes the lap winding an ideal choice for fairly low-voltage, high-current machines, since the high currents required can be split among the several different current paths. This current splitting pennits the size of individual rotor conductors to remain reasonable even when the total current becomes ex-tremely large. TIle fact that there are many parallel paths through a multipole lap-wound machine can lead to a serious problem, however. To understand the nature of this problem, examine the six-pole machine in Figure 8-1 6. Because of long usage, there has been slight wear on the bearings of this machine, and the lower wires are closer to their pole faces than the upper wires are. As a result, there is a larger volt-age in the current paths involving wires under the lower pole faces than in the paths involving wires under the upper pole faces. Since all the paths are connected in par-allel, the result will be a circulating current flowing out some of the brushes in the machine and back into others, as shown in Figure 8-17. Needless to say, this is not gooj for the machine. Since the winding resistance of a rotor circuit is so small, a very tiny imbalance runong the voltages in the parallel paths will cause large cir-culating currents through the brushes and potentially serious heating problems. TIle problem of circulating currents within the parallel paths of a machine with four or more poles can never be entirely resolved, but it can be reduced somewhat by equalizers or equalizing windings. Equalizers are bars located on the rotor of a lap-wound dc machine that short together points at the same voltage N N lit · " " " " " I 13 2 14 3 IS ,16y IXX'>: yP , FIGURE 8-15 16 s 41651 ' 62'7 XXx 6 , d , IX: MACHINERY FUNDAMENTALS 495 s 4 s N IOL-__ II 13 12 s , ., ;x ;X N " " " " " " 3' 8 4' 9 5' 10 6' 117' I ;X :xxx :xxx 8 h ,b, s ,- i-" " " N " " I 8' 139' 1410 151116~ 13' ;:xxx :X14· • m '~ (a) A four-pole lap-wound de motor. (b) The rotor winding diagram of this machine. Notice that each winding ends on the commutator segment just after the one it begins at. This is a progressive lap winding. 496 ELECTRIC MACHINERY RJNDAMENTALS s N N s s N ""GURE 8-16 A six-pole dc motor showing the effects of bearing wear. Notice that the rotor is slightly closer to the lower poles than it is to the upper poles. level in the different parallel paths. The effect of this shorting is 1 0 cause any cir-culating currents that occur to now inside the small sections of windings thus shorted together and to prevent this circulating current from flowing through the brushes of the machine. TIlese circulating currents even partially correct the flux imbalance that caused them to exist in the first place. An equalizer for the four-pole machine in Figure 8-1 5 is shown in Figure 8-1 8, and an equalizer for a large lap-wound dc machine is shown in Figure 8-1 9. If a lap winding is duplex, then there are two completely independent wind-ings wrapped on the rotor, and every other commutator segment is tied to one of the sets. Therefore, an individual coil ends on the second commutator segment down from where it started, and Yc = ~2 (depending on whether the winding is progressive or retrogressive). Since each set of windings has as many current paths as the machine has poles, there are twice as many current paths as the ma-chine has poles in a duplex lap winding. In general, for an m-plex lap winding, the commutator pitch Yc is lap winding (8- 25) and the number of current paths in a machine is I a -mpl lap winding (8- 26) v, IX: MACHINERY FUNDAMENTALS 497 Cirwlating Current + .l. .l. ... + + + + + , , , " " -----+ + + + + , , , " " -----+ + + + + , , , " " -----+ + + + + , , , " " -----+ + + + + , , , " " -----T T T -e+ slightly greater voltage e- slightly lower voltage FIGURE 8-17 The voltages on the rotor conductors of the machine in Figure 8--16 are unequal. producing circulating currents flowing through its brushes. where a = number of current paths in the rotor m = plex of the windings ( I, 2, 3, etc.) P = number of poles on the machine The Wave Winding + + + + + The series or wave winding is an alternative way to connect the rotor coils to the commutator segments. Figure 8- 20 shows a simple four-pole machine with a 498 ELECTRIC MACHINERY RJNDA MENTALS Equalizer bars 1 ?< > X ------------Ii I r-------I " " " N I: I , I: I N ':1 , II N " " " II 1 13 2 14 3 5 4 6 5 I' 6::k 7 3' 8 4' 9 5' 10 6' II 7' 1 8' 13 9' 14 10 15~1612 XXX >< >< X XX< x,;x X x,;x x,;x >x: >< f'""" 'h ' "m"~ + ~ + + 6' , , , --+ + 6 , " , --+ + " , 8 , --+ + , , 8' , v, --+ + 4' , 9 , --+ + 4 , " , --+ + ,- , 10 , --+ + 3 , 10' , --~ -""GURE 8-1 8 ,,) + '" , -+ 14 , -Equalizers + 13' , -+ 13 , -+ 12' , -+ 12 , -Equalizers + 11' , -+ 11 , -(b) B=h " W 16 16' , , ' 2 ,-B=h + , + , + , + , + , + , + , + , (a) An equalizer connection for the four-pole machine in Figure 8--15. (b) A voltage diagram for the machine shows the points shoned by the equalizers. s 2 " , b + N • .-0 .1.-, h 7 s FIGURE 8-10 A simple four-pole wave-wound dc machine. IX: MACHINERY FUNDAMENTALS 499 FlGURE 8- 19 A closeup of the commutator of a large lap-wound de machine. The equalizers are moumed in the smaIl ring just in front of the commutator segments. (Courtesy ofGeneml Electric Company.) N 500 ELECTRIC MACHINERY RJNDAMENTALS simplex wave winding. In this simplex wave winding, every other rotor coil con-nects back to a commutator segment adjacent to the beginning of the first coil. TIlerefore, there are two coils in series between the adjacent commutator seg-ments. Furthennore, since each pair of coils between adjacent segments has a side under each pole face, all output voltages are the sum of the effects of every pole, and there can be no voltage imbalances. TIle lead from the second coil may be connected to the segment either ahead of or behind the segment at which the first coil begins. If the second coil is con-nected to the segment ahead of the first coil, the winding is progressive; if it is connected to the segment behind the first coil, it is retrogressive. I n general, if there are P poles on the machine, then there are PI2 coils in se-ries between adjacent commutator segments. If the (PI2)th coi I is connected to the segment ahead of the first coil, the winding is progressive. If the (PI2)th coil is connected to the segment behind the first coil, the winding is retrogressive. In a simplex wave winding, there are only two current paths. There are c/2 or one-half of the windings in each current path. The brushes in such a machine will be located a full pole pitch apart from each other. What is the commutator pitch for a wave winding? Figure 8- 20 shows a progressive nine-coil winding, and the end of a coil occurs five segments down from its starting point. In a retrogressive wave winding, the end of the coil occurs four segments down from its starting point. Therefore, the end of a coil in a four-pole wave winding must be connected just before or just after the point halfway around the circle from its starting point. TIle general expression for commutator pitch in any simplex wave winding is simplex wave (8- 27) where C is the number of coils on the rotor and P is the number of poles on the machine. The plus sign is associated with progressive windings, and the minus sign is associated with retrogressi ve windings. A simplex wave winding is shown in Figure 8- 2 1. Since there are only two current paths through a simplex wave-wound rotor, only two brushes are needed to draw off the current. TIlis is because the segments undergoing commutation connect the points with equal voltage under all the pole faces. More brushes can be added at points 180 electrical degrees apart if desired, since they are at the same potential and are connected together by the wires un-dergoing commutation in the machine. Extra brushes are usually added to a wave-wound machine, even though they are not necessary, because they reduce the amount of current that must be drawn through a given brush set. Wave windings are well suited to building higher-voltage dc machines, since the number of coils in series between commutator segments pennits a high voltage to be built up more easily than with lap windings. A multiplex wave winding is a winding with multiple independent sets of wave windings on the rotor. These extra sets of windings have two current paths each, so the number of current paths on a multiplex wave winding is IX: MACHINERY FUNDAMENTALS 501 8 9 2 3 4 5 6 7 8 9 2 9 7' 1 8' 2 9' 3 I' 4 2' 5 3' 6 4 7 5' 8 6' 9 7' 1 8' 2 9' d , FIGURE 8-21 The rotor winding diagram for the machine in fi gure 8-20. Notice that the end of every second coil in series connects to the segment after the beginning of the first coil. This is a progressive wave winding. I a -2m I rnul1iplex wave (8- 28) The Frog-Leg Winding Thefrog-leg winding or self-equaliZing winding gets its name from the shape of its coils, as shown in Figure 8- 22. It consists of a lap winding and a wave wind-ing combined. The equalizers in an ordinary lap winding are connected at points of equal voltage on the windings. Wave windings reach between points of essentially equal vol1age under successive pole faces of the same polarity, which are the same lo-cations that equalizers tie together. A frog-leg or self-equalizing winding com-bines a lap winding with a wave winding, so that the wave windings can function as equalizers for the lap winding. The number of current paths present in a frog-leg winding is I a - ' Pm,." I frog-leg winding (8- 29) where P is the number of poles on the machine and mJap is the plex of the lap winding. EXIllllpie 8-2. Describe the rotor winding arrangement of the four-loop machine in Section 8.2. Solutioll The machine described in Section 8.2 has four coils. each containing one turn. resulting in a total of eight conductors. It has a progressive lap winding. 502 ELECTRIC MACHINERY RJNDAMENTALS Coil ~~r,~ W ·d· / ave Win lOgS Fl GURE 8-22 A frog-leg or self-equalizing winding coil. 8.4 PROBLEMS WITH COMMUTATION IN REAL MACHINES TIle commutation process as described in Sections 8.2 and 8.3 is not as simple in practice as it seems in theory, because two major effects occur in the real world to disturb it: L Annature reaction 2, L dildt voltages TIlis section explores the nature of these problems and the sol utions employed to mitigate their effects. Annature Reaction If the magnetic field windings of a dc machine are connected to a power supply and the rotor of the machine is turned by an external source of mechanical power, then a voltage wi ll be induced in the cond uctors of the rotor. This voltage wil I be rectified into a dc output by the action of the machine's commutator. Now connect a load to the tenninaIs of the machine, and a current will flow in its armature windings. TIlis current flow wi ll produce a magnetic field of its own, which wi ll distort the original magnetic field from the machine's poles. TIlis distortion of the flux in a machine as the load is increased is called armature re-action. It causes two serious problems in real dc machines. TIle first problem caused by annature reaction is neutral-plane shift. The mngnetic neutral plane is defined as the plane within the machine where the N IX: MACHINERY FUNDAMENTALS 503 Magnetic neutral plane ;:,w-I-- New neutral plane w Old neutral plane y-t---(b) 0 " w N 0 " S ~ 0 " 0 " N 0 " S ( .) o (,) FIGURE 8-23 The development of annature reaction in a dc gelJerator. (a) Initially the pole flux is unifonnly distributed. and the magnetic neutral plane is vertical; (b) the effect of the air gap on the pole flux distribution; (c) the armature magnetic field resulting when a load is connected to the machine; (d) both rotor and pole fluxes are shown. indicating points where they add and subtract; (e) the resulting flux under the poles. The neutral plane has shifted in the direction of motion. velocity of the rotor wires is exactly parallel to the magnetic nux lines, so that eind in the conductors in the plane is exactly zero. To understand the problem of neutral-plane shift, exrunine Figure 8- 23. Fig-ure 8- 23a shows a two-pole dc machine. Notice that the nux is distributed uni-fonnly under the pole faces. The rotor windings shown have voltages built up out of the page for wires under the north pole face and into the page for wires under the south pole face. The neutral plane in this machine is exactly vertical. Now suppose a load is connected to this machine so that it acts as a genera-tor. Current will now out of the positive terminal of the generator, so current wi ll 504 ELECTRIC MACHINERY RJNDAMENTALS be flowing out of the page for wires under the north pole face and into the page for wires under the south pole face. This current fl ow produces a magnetic field from the rotor windings, as shown in Figure 8- 23c. This rotor magnetic field affects the original magnetic field from the poles that produced the generator's voltage in the first place. In some places under the pole surfaces, it subtracts from the pole flux, and in other places it adds to the pole flux. The overall result is that the magnetic flux in the air gap of the machine is skewed as shown in Figure 8- 23d and e. No-tice that the place on the rotor where the induced voltage in a conductor would be zero (the neutral plane) has shifted. For the generator shown in Figure 8- 23, the magnetic neutral plane shifted in the direction of rotation. If this machine had been a motor, the current in its ro-tor would be reversed and the nux would bunch up in the opposite corners from the bunches shown in the figure. As a result, the magnetic neutral plane would shift the other way. I n general, the neutral-plane shifts in the direction of motion for a generator and opposite to the direction of motion for a motor. Furthennore, the amount of the shift depends on the amount of rotor current and hence on the load of the machine. So what's the big deal about neutral-plane shift? It's just this: The commu-tator must short out commutator segments just at the moment when the voltage across them is equal to zero. Ifthe brushes are set to short out conductors in the vertical plane, then the voltage between segments is indeed zero until the machine is loaded. When the machine is loaded, the neutral plane shifts, and the brushes short out commutator segments with a finite voltage across them. The result is a current now circulating between the shorted segments and large sparks at the brushes when the current path is interrupted as the brush leaves a segment. The end result is arcing and sparking at the brushes. TIlis is a very serious problem, since it leads to drastically reduced brush I ife, pitting of the commutator segments, and greatly increased maintenance costs. Notice that this problem cannot be fixed even by placing the brushes over the full-load neutral plane, because then they wou Id spark at no load. In extreme cases, the neutral-plane shift can even lead to flashover in the commutator segments near the brushes. The air near the brushes in a machine is normally ionized as a result of the sparking on the brushes. Flashover occurs when the voltage of adjacent commutator segments gets large enough to sustain an arc in the ionized air above them. If flashover occurs, the resulting arc can even melt the commutator's surface. TIle second major problem caused by annature reaction is called flux weak-ening. To understand flux weakening, refer to the magnetization curve shown in Figure 8- 24. Most machines operate at flux densities near the saturation point. TIlerefore, at locations on the pole surfaces where the rotor magnetomotive force adds to the pole magnetomotive force, only a small increase in nux occurs. But at locations on the pole surfaces where the rotor magnetomotive force subtracts from the pole magnetomotive force, there is a larger decrease in flux. 1lle net result is that the total averageflux under the entire pole face is decreased (see Figure 8- 25). IX: MACHINERY FUNDAMENTALS 505 q,. Wb , , •• , 1 , , .. , , , L-------------cf---}--~--------------- ~·A . tums Pole mmf ,/ """ - annature Pole mmf Pole mmf + annature mmf mmf l1q,i - flux increase under reinforced sections of poles l1q,d - flux decrease under subtracting sections of poles FIGURE 8-24 A typical magnetization curve shows the effects of pole saturation where armature and pole magnetomotive forces add. Flux weakening causes problems in bolh generators and motors. In genera-tors, the effect of flux weakening is simply to reduce the voltage supplied by the generator for any given load. In motors, the effect can be more serious. As the early examples in this chapter showed, when the flux in a motor is decreased, its speed in-creases. But increasing the speed of a motor can increase its load, resulting in more flux weakening. It is possible for some shunt dc motors to reach a runaway condi-tion as a result offlux weakening, where the speed of the motor just keeps increas-ing until the machine is disconnected from the power line or until it destroys itself. L dildt Voltages The second major problem is the L dildt voltage that occurs in commutator seg-ments being shorted out by the brushes, sometimes called inductive kick. To un-derstand this problem, look at Figure 8- 26. This figure represents a series of com-mutator segments and the conductors connected between them. Assuming that the current in the brush is 400 A, the current in each path is 200 A. Notice that when a commutator segment is shorted out, the current flow through that commutator 506 ELECTRIC MACHINERY RJNDAMENTALS Stator Field § windings -S § N / ~ __ "'--'>. L-/ __ "'--'>. 1 Rotor -Motion of generator M fmotor ::f. A • turns -Oilon 0 Pole -/ magnetomotive force -", .............. ............... " --, ............... , ................ , . , Rotor magnetomotive force ...... ..-- Net 9' '!f. A· turns .Wb ... 1 -Note: Saturation at pole tips fo'IGURE 8- 25 --..........IP. Wb / Old neutral point New neutral poim The flux and magoetomotive force under the pole faces in a de machine. At those points where the magnetomotive forces subtract. the flux closely follows the net magoetomotive force in the iron; but at those points where the magnetomotive forces add, saturation limits the IOtal flux present. Note also that the neutral point of the rotor has shifted. segment must reverse. How fast must this reversal occur? Assuming that the ma-chine is turning at 800 r/min and that there are 50 commutator segments (a reason-able number for a typical motor), each commutator segment moves under a brush and clears it again in t = 0.00 15 s. 1l1erefore, the rate of change in current with re-spect to time in the shorted loop must average di 400 A dt - 0.00 15 s -266,667 A ls (8- 30) IX: MACHINERY FUNDAMENTALS 507 -200 A Direction of commutator motion 200 A -200 A 200 A -? 200 A (a) 1=0.0015 s 200 Ar----" --200 A 200 A 200 A 200 A \ r-------i-~,"i------------ , Brush reaches beginning of segment b , , , , , , ...-- Spark at trailing edge of brush Brush clears end of segment a ----- -Ideal conunutation 200 A -Actual commutation with inductance taken into account (b) FIGURE 8-26 (a) The revel"S3.1 of current flow in a coil undergoing commutation. Note that the current in the coil between segments a and b must reverse direction while the brush ShOMS together the two commutator segments. (b) The current reversal in the coil undergoing commutation as a function of time for both ideal commutation and real commutation. with the coil inductance taken into account. With even a tiny inductance in the loop, a very significant inductive voltage kick v = Ldildt will be induced in the shorted commutator segment. This high voltage naturally causes sparking at the brushes of the machine, resulting in the same arc-ing problems that the neutral-plane shift causes. 508 ELECTRIC MACHINERY RJNDAMENTALS Solutions to the Problems with Commutation TIlree approaches have been developed to partially or completely correct the prob-lems of armature reaction and L dildt voltages: I. Brush shifting 2. Commutating poles or interpoles 3. Compensating windings E:1.ch of these techniques is explained below, together with its advantages and dis-advantages. BRUSH SHIFTING. Historically, the first attempts to improve the process of commutation in real dc machines started with attempts to stop the sparking at the brushes caused by the neutral-plane shifts and L dildt effects. The first approach taken by machine designers was simple: If the neutral plane of the machine shifts, why not shift the brushes with it in order to stop the sparking? It certainly seemed like a good idea, but there are several serious problems associated with it. For one thing, the neutral plane moves with every change in load, and the shift direction reverses when the machine goes from motor operation to generator operation. lllerefore, someone had to adjust the brushes every time the load on the machine changed. In addition, shifting the brushes may have stopped the brush sparking, but it actually aggravated the flux-weakening effect of the armature reaction in the machine. TIlis is true because of two effects: I. The rotor magnetomotive force now has a vector component that opposes the magnetomotive force from the poles (see Figure 8- 27). 2. The change in annature current distribution causes the flux to bunch up even more at the saturated parts of the pole faces. Another slightly different approach sometimes taken was to fix the brushes in a compromise position (say, one that caused no sparking at two-thirds of full load). In this case, the motor sparked at no load and somewhat at full load, but if it spent most of its life operating at about two-thirds of full load, then sparking was minimized. Of course, such a machine could not be used as a generator at ,II-the sparking would have been horrible. By about 1910, the brush-shifting approach to controlling sparki ng was al-ready obsolete. Today, brush shifting is only used in very small machines that al-ways run as motors. TIlis is done because better solutions to the problem are sim-ply not economical in such small motors. COMMUTATING POLES OR INTERPOLES. Because of the disadvantages noted above and especially because of the requirement that a person must adjust the brush positions of machines as their loads change, another solution to the problem of brush sparking was developed. TIle basic idea behind this new approach is that New neutral plane Brushes Jc---r::..;O~d neutral plane o N o o FIGURE 8-27 Net magnetomotive force ::f ... s Rotor magnetomotive force ::fR (a) IX: MACHINERY FUNDAMENTALS 509 N New neutral plane Old neutral plane ---f=--- w o ~, o o New net magnetomotive , f_ " (h) s Original net I magnetomotive f=, 1 , : ::fR , (a) The net magnetomotive force in a dc machine with its brushes in the vertical plane. (b) The net magnetomotive force in a dc machine with its brushes over the shifted neutral plane. Notice that now there is a component of armature magnetomotive force directly oppOiling the poles' magnetomotive force. and the net magnetomotive force in the machine is reduced. if the voltage in the wires undergoing commutation can be made zero, then there will be no sparking at the brushes. To accomp lish this, small poles, called com-mutating poles or interpoles, are placed midway between the main poles. These commutating poles are located directly over the conductors being commutated. By providing a flux from the commutating poles, the voltage in the coils undergoing commutation can be exactly canceled. If the cancellation is exact, then there will be no sparking at the brushes. The commutating poles do not otherwise change the operation of the ma-chine, because they are so small that they affect only the few conductors about to undergo commutation. Notice that the armature reaction under the main pole faces is unaffected, since the effects of the commutating poles do not extend that far. nlis means that the flux weakening in the machine is unaffected by commutating poles. How is cancellation of the voltage in the commutator segments accom-plished for all values of loads? nlis is done by simply connecting the interpole 510 ELECTRIC MACHINERY RJNDAMENTALS windings in series with the windings on the rotor, as shown in Figure 8- 28. As the load increases and the rotor current increases, the magnitude of the neutral-plane shift and the size of the L dildt effects increase too. Both these effects increase the voltage in the conductors undergoing commutation. However, the interpole flux increases too, producing a larger voltage in the conductors that opposes the volt-age due to the neutral-plane shift. The net result is that their effects cancel over a broad range of loads. Note that interpoles work for both motor and generator op-eration, since when the machine changes from motor to generator, the current both in its rotor and in its interpoles reverses direction. Therefore, the voltage effects from them still cancel. What polarity must the flux in the interpoles be? The interpoles must induce a voltage in the conductors undergoing commutation that is opposite to the voltage caused by neutral-plane shift and L dildt effects. In the case of a generator, the neu-tral plane shifts in the direction of rotation, meaning that the conductors undergoing commutation have the same polarity of voltage as the pole they just left (see Figure 8- 29). To oppose this voltage, the interpolcs must have the opposite flux, which is the flux of the upcoming pole. In a motor, however, the neutral plane shifts opposite to the direction of rotation, and the conductors undergoing commutation have the same flux as the pole they are approaching. In order to oppose this voltage, the in-terpoles must have the same polarity as the previous main pole. Therefore, I. The interpoles must be of the same polarity as the next upcoming main pole in a generator. -v, N s R -----d+ -I, ""GURE 8-28 A de machine with imerpoles. IX: MACHINERY FUNDAMENTALS 511 2. The interpoles must be of the same polarity as the previous main pole in a motor. The use of commutating poles or interpoles is very common, because they correct the sparking problems of dc machines at a fairly low cost. TIley are almost always found in any dc machine of I hp or larger. It is important to realize, though, that they do nothing for the flux distribution under the pole faces, so the flux-weakening problem is still present. Most medium-size, general-purpose mo-tors correct for sparking problems with interpoles and just live with the flux-weakening effects. N New neutral plane Now neutral plane FIGURE 8-29 u s n (a) (b) Determining the required polarity of an interpole. The flux from the interpole must produce a voltage that opposes the existing voltage in the conductor. 512 ELECTRIC MACHINERY RJNDAMENTALS -- Rotor (amlature) flux - - - Aux from compensating windings o , , " \ , N' , , , , IS , , I o , ,' N " ""GURE 8-30 (, , o ,b, Neutral plane no/ shifted with load The effect of compensating windings in a dc machine. (a) The pole flux in the machine; (b) the fluxes from the armature and compensating windings. Notice that they are equal and opposite; (c) the net flux in the machine. which is just the original pole flux. , , , , COMPENSATING WINDINGS. For very heavy, severe duty cycle motors, the flux-weakening problem can be very serious. To completely cancel armature re-action and thus eliminate both neutral-plane shift and flux weakening, a different technique was developed. This third technique involves placing compensating windings in slots carved in the faces of the poles parallel to the rotor conductors, to cancel the distorting effect of annature reaction. These windings are connected in series with the rotor windings, so that whenever the load changes in the rotor, the current in the compensating windings changes, too. Figure 8- 30 shows the ba-sic concept. In Figure 8- 30a, the pole flux is shown by itself. In Figure 8- 30b, the rotor fl ux and the compensating winding fl ux are shown. Figure 8- 3Oc represents the sum of these three fluxes, which is just equal to the original pole flux by itself. Figure 8- 3 \ shows a more careful development of the effect of compensat-ing windings on a dc machine. Notice that the magnetomotive force due to the IX: MACHINERY FUNDAMENTALS 513 Stator ~~~i"" -L---~ s------'>.~ L---~ N------'>.~ j ! ! Rotor ~~~ClIT0J:]0~· ~0c·IT0J:]6 ~. ~" []"'iJ,,~®~~,,~,, ~~ -Motionof ::f,A· turns Pole magnetomotive force Compensating ,,(!"inding /~ /\ r-~ Rotor genel1ltor -Motion of mmm magnetomotiveforce '-__ -':1..., =::fpole +::fR +::f"" ::f. A • turns FIGURE 8-3 1 I Neutral plane 00' shifted :1..., =::f pole The flux and magnetomotive forces in a de machine with compensating windings. compensating windings is equal and opposite to the magnetomotive force due to the rotor at every point under the pole faces. The resulting net magnetomotive force is just the magnetomotive force due to the poles, so the flux in the machine is unchanged regardless of the load on the machine. The stator of a large dc ma-chine with compensating windings is shown in Figure 8- 32. The major disadvantage of compensating windings is that they are expen-sive, since they must be machined into the faces of the poles. Any motor that uses them must also have interpoles, since compensating windings do not cancel L dildt effects. The interpoles do not have to be as strong, though, since they are canceling only L dildt voltages in the windings, and not the voltages due to neutral-plane shifting. Because of the expense of having both compensating wind-ings and interpoles on such a machine, these windings are used only where the ex-tremely severe nature of a motor's duty demands them. 514 ELECTRIC MACHINERY RJNDAMENTALS ""GURE 8-32 The stator of a six-pole dc machine with imerpoIes and compensating windings. (Courtesy of Westinghouse Electric Company.) 8.5 THE INTERNAL GENERATED VOLTAGE AND INDUCED TORQUE EQUATIONS OF REAL DC MACHINES How much voltage is produced by a real dc machine? The induced voltage in any given machine depends on three factors: I. The flux = BAp The rotor of the machine is shaped like a cylinder, so its area is A = 27frl (8- 34) I f there are P poles on the machine, the n the portion of the area associated with each pole is the total area A divided by the number of poles P: A = A = 27frl p P P (8- 35) The total flux per pole in the machine is thus 4> = BAp = B(27frl) = 27fr1B p p (8- 36) lllerefore, the internal generated voltage in the machine can be expressed as Finally, where E _ ZrwBI A-a (8- 33) (8- 37) (8- 38) (8- 39) In modern industrial practice, it is common to express the speed of a ma-chine in revolutions per minute instead of radians per second. The conversion from revolutions per minute to radians per second is (8-40) so the voltage equation with speed expressed in tenns of revolutions per minute is 516 ELECTRIC MACHINERY RJNDAMENTALS I E, ~ K'" I (8-4 1) where I K'= ~ I (8-42) How much torque is induced in the annature of a real dc machine? The torque in any dc machine depends on three factors: I. The flux <p in the machine 2. The armature (or rotor) current I ... in the machine 3. A constant depending on the construction of the machine How can the torque on the rotor of a real machine be determined? The torque on the armature of a real machine is equal to the number of conductors Z times the torque on each conductor. TIle torque in any single conductor under the pole faces was previously shown to be (8-43) If there are a current paths in the machine, then the total annature current I ... is split among the a current paths, so the current in a single conductor is given by I, leo"" = a and the torque in a single conductor on the motor may be expressed as dAIB Tcood = - a -(8-44) (8-45) Since there are Z conductors, the total induced torque in a dc machine rotor is (8-46) The flux per pole in this machine can be expressed as J.. = BA = B(2 7rrD = 27rr1B '+' p P P (8-47) so the total induced torque can be reexpressed as (8-48) Finally, I Tind = Kn = (288XO.OS Wb)(200 r/min) = 2880 V (e) There are two parallel paths through the rotor of this machine, each one consist-ing of 712 = 1440 conductors, or 720 turns. Therefore, the resistance in each current path is Resistancelpath = (720 turnsXO.OII !lIturn) = 7.92 n Since there are two parallel paths, the effective armature resistance is R}, = 7.9; n = 3.96 n (d) If a lOOO~ load is connected to the tenninals of the generator, and if R}, is ig-nored, then a current of 1 = 2880 V/HXX) n = 2.88 A flows. The constant K is given by K = ZP = (2880)(12) = 2750:2 27TO" (27T)(2) . Therefore, the countertorque on the shaft of the generator is "Tind = Kt$/}, = (27S0.2XO.05 Wb)(2.88 A) = 396 Nom 8.6 THE CONSTRUCTION OF DC MACHINES A simplified sketch of a dc machine is shown in Figure 8- 33, and a more detailed cutaway diagram of a dc machine is shown in Figure 8- 34. TIle physical structure of the machine consists of two parts: the stator or sta-tionary part and the rotor or rotating part. TIle stationary part of the machine con-sists of the frame, which provides physical support, and the pole pieces, which project inward and provide a path for the magnetic flux in the machine. TIle ends of the pole pieces that are near the rotor spread out over the rotor surface to dis-tribute its flux evenly over the rotor surface. TIlese ends are called the pole shoes. 1lle exposed surface of a pole shoe is called a pole face, and the distance between the pole face and the rotor is calJed the air gap. IX: MACHINERY FUNDAMENTALS 519 Field pole and r.J "Oid Nameplate -+--,'1--'<:. Yoke ----j'!-Frame FIGURE 8-33 A simplified diagram of a de machine. (a) FIGURE 8-34 (a) A cutaway view of a 4O(X)..hp, 700, V. 18-pole de machine showing compensating windings. interpoles. equalizer. and commutator. (Courtesy ofGeneml Electric Company.) (b) A cutaway view of a smaller four'pole de motor including interpoles but without compensating windings. (Courtesy of MagneTek lncorpomted.) 520 ELECTRIC MACHINERY RJNDAMENTALS TIlere are two principal windings on a dc machine: the annature windings and the field windings. The armature windings are defined as the windings in which a voltage is induced, and the field windings are defined as the windings that produce the main magnetic flux in the machine. In a nonnal dc machine, the annature windings are located on the rotor, and the field windings are located on the stator. Because the annature windings are located on the rotor, a dc machine's rotor itself is sometimes called an armature. Some major features of typical dc motor construction are described below. Pole and Frame Construction TIle main poles of older dc machines were often made of a single cast piece of metal, with the field windings wrapped around it. TIley often had bolted-on lami-nated tips to reduce core losses in the pole faces. Since solid-state drive packages have become common, the main poles of newer machines are made entirely of laminated material (see Figure 8- 35). This is true because there is a much higher ac content in the power supplied to dc motors driven by solid-state drive pack-ages, resulting in much higher eddy current losses in the stators of the machines. TIle pole faces are typically either chamfered or eccentric in construction, mean-ing that the outer tips of a pole face are spaced slightly further from the rotor's surface than the center of the pole face is (see Figure 8- 36). This action increases the reluctance at the tips of a pole face and therefore reduces the flux-bunching ef-fect of annature reaction on the machine. ""GURE 8-35 Main field pole assembly for a de motor. Note the pole laminations and compensating windings. (Courtesy of General Electric Company.) IX: MACHINERY FUNDAMENTALS 521 The poles on dc machines are called salient poles, because they stick out from the surface of the stator. The interpoles in dc machines are located between the main poles. TIley are more and more commonly of laminated construction, because of the same loss problems that occur in the main poles. Some manufacturers are even constructing the portion of the frame that serves as the magnetic flux 's return path (the yoke) with laminations, to further re-duce core losses in electronically driven motors. Rotor or Armature Construction The rotor or armature of a dc machine consists of a shaft machined from a steel bar with a core built up over it. The core is composed of many laminations stamped from a stccl plate, with notches along its outer surface to hold the arrna-ture windings. TIle commutator is built onto the shaft of the rotor at one end of the core. TIle annature coils are laid into the slots on the core, as described in Section 8.4, and their ends are connected to the commutator segments. A large dc machine rotor is shown in Figure 8- 37. Commutator and Brushes The commutator in a dc machine (Figure 8- 38) is typically made of copper bars insulated by a mica-type material. TIle copper bars are made sufficiently thick to pennit normal wear over the lifetime of the motor. The mica insulation between commutator segments is harder than the commutator material itself, so as a ma-chine ages, it is often necessary to undercut the commutator insulation to ensure that it does not stick up above the level of the copper bars. The brushes of the machine are made of carbon, graphite, metal graphite, or a mixture of carbon and graphite. They have a high conductivity to reduce elec-tricallosses and a low coefficient of friction to reduce excessive wear. They are N s ,., ,b, FIGURE 8-36 Poles with extra air-gap width at the tips to reduce armature reaction. (a) Chamfered poles; (b) eccentric or uniformly graded poles. s 522 ELECTRIC MACHINERY RJNDA MENTALS ""GURE 8-37 Photograph of a dc machine with the upper stator half removed shows the construction of its rotor. (Courtesy of General Electric Company.) ""GURE 8-38 Close-up view of commutator and brushes in a large dc machine. (Courtesy of General Electric Company.) IX: MACHINERY FUNDAMENTALS 523 deliberately made of much softer material than that of the commutator segments, so that the commutator surface will experience very little wear. The choice of brush hardness is a compromise: If the brushes are too soft, they will have to be replaced too often ; but if they are too hard, the commutator surface will wear ex-cessively over the life of the machine. All the wear that occurs on the commutator surface is a direct resu It of the fact that the brushes must rub over them to convert the ac voltage in the rotor wires to dc voltage at the machine's terminals. If the pressure of the brushes is too great, both the brushes and commutator bars wear excessively. However, if the brush pressure is too small, the brushes tend to jump slightly and a great deal of sparking occurs at the brush-commutator segment interface. This sparking is equally bad for the brushes and the commutator surface. TIlerefore, the brush pressure on the commutator surface must be carefully adjusted for maximum life. Another factor which affects the wear on the brushes and segments in a dc machine commutator is the amount of current fl owing in the machine. llle brushes normally ride over the commutator surface on a thin oxide layer, which lubricates the motion of the brush over the segments. However, if the current is very small, that layer breaks down, and the friction between the brushes and the commutator is greatly increased. lllis increased friction contributes to rapid wear. For maximum brush life, a machine should be at least partially loaded all the time. Winding Insulation Other than the commutator, the most critical part of a dc motor's design is the in-sulation of its windings. I f the insulation of the motor windings breaks down, the motor shorts out. The repair of a machine with shorted insulation is quite expen-sive, if it is even possible. To prevent the insulation of the machine windings from breaking down as a result of overheating, it is necessary to limit the temperature of the windings. This can be partially done by providing a cooling air circulation over them, but ultimately the maximum winding temperature limits the maximum power that can be supplied continuously by the machine. Insulation rarely fails from immediate breakdown at some critical tempera-ture. Instead, the increase in temperature produces a gradual degradation of the in-sulation, making it subject to failure due to another cause such as shock, vibration, or electrical stress. There is an old rule of thumb which says that the life ex-pectancy of a motor with a gi ven insulation is halved for each JO percent rise in winding temperature. This rule still applies to some extent today. To standardize the temperature limits of machine insulation, the National Electrical Manufacturers Association (NEMA) in the United States has defined a series of insulation system classes. Each insulation system class specifies the max-imum temperature rise permissible for each type of insulation. lllere are four stan-dard NEMA insulation classes for integral-horsepower dc motors: A, 8, F, and H. Each class represents a higher pennissible winding temperature than the one before it. For example, if the annature winding temperature rise above ambient tempera-ture in one type of continuously operating dc motor is measured by thermometer, 524 ELECTRIC MACHINERY RJNDAMENTALS it must be limited to 70°C for class A, WO°C for class B, \30°C for class F, and ISSoC for class H insulation. TIlese temperature specifications arc set out in great detail in NEMA Stan-dard MG I -1993, Motors and Generators. Similar standards have been defined by the International Electrotechnical Commission (lEC) and by various national stan-dards organizations in other countries. 8.7 POWER FLOW AND LOSSES IN DC MACHINES DC generators take in mechanical power and produce electric power, while dc motors take in electric power and pnxluce mechanical power. In either case, not all the power input to the machine appears in useful fonn at the other end-there is always some loss associated with the process. TIle efficiency of a dc machine is defined by the equation (8- 50) TIle difference between the input power and the output power of a machine is the losses that occur inside it. Therefore, P oot -~oss .,, = x 100% P;.o (8- 51) The Losses in DC Machines The losses that occur in dc machines can be divided into fi ve basic categories: I. Electrical or copper losses (l lR losses) 2. Brush losses 3. Core losses 4. Mechanical losses 5. Stray load losses ELECTRICAL OR COPPER LOSSES. Copper losses are the losses that occur in the armature and field windings of the machine. TIle copper losses for the anna-ture and field windings are given by Armature loss: Field loss: where P), -annature loss P F -field circuit loss PA = li RA P F = I} RF (8- 52) (8- 53) IX: MACHINERY FUNDAMENTALS 525 I, -annature current I, -field current R, -annature resistance R, -field resistance The resistance used in these calculations is usually the winding resistance at nor-mal operating temperature. BRUSH LOSSES. 1lle brush drop loss is the power lost across the contact poten-tial at the brushes of the machine. It is given by the equation Ip sD -VsDIA I where PBD = brush drop loss VBD = brush voltage drop IA = annature current (8- 54) The reason that the brush losses are calculated in this manner is that the voltage drop across a set of brushes is approximately constant over a large range ofarrna-ture currents. Unless otherwise specified. the brush voltage drop is usually as-sumed to be about 2 V. CORE LOSSES. The core losses are the hysteresis losses and eddy current losses occurring in the metal of the motor. These losses are described in Chapter 1. lllese losses vary as the square of the flux density (B 2) and. for the rotor, as the 1.5th power of the speed of rotation (nI. 5) . 1\" IECHANICAL LOSSES. 1lle mechanical losses in a dc machine are the losses associated with mechanical effects. There are two basic types of mechanical losses: friction and windage. Friction losses are losses caused by the friction of the bearings in the machine, while windage losses are caused by the friction between the moving parts of the machine and the air inside the motor's casing. These losses vary as the cube of the speed of rotation of the machine. STRAY LOSSES (OR MISCELLANEOUS LOSSES). Stray losses are losses that cannot be placed in one of the previous categories. No matter how carefully losses are accounted for, some always escape incl usion in one of the above categories. All such losses are lumped into stray losses. For most machines, stray losses are taken by convention to be I percent of full load. The Power-Flow Diagram One of the most convenient techniques for accounting for power losses in a ma-chine is the power-flow diagram. A power-flow diagram for a dc generator is shown in Figure 8- 39a. In this figure, mechanical power is input into the machine, 526 ELECTRIC MACHINERY RJNDAMENTALS Stray losses , , , , , , Mechanical losses EA IA '" iDd Ul m I'R losses ""GURE 8-39 Core losses p~ , Core losses , ,' Mechanical losses ,b, I'R losses Stray losses Power-flow diagrams for de machine: (a) generator: (b) motor. and then the stray losses, mechanical losses, and core losses are subtracted. After they have been subtracted, the remaining power is ideally converted from me-chanical to electrical fonn at the point labeled P"ODV. TIle mechanical power that is converted is given by I P"onv -Tindw m I (8- 55) and the resulting electric power produced is given by (8- 56) However, this is not the power that appears at the machine's tenninals. Be-fore the terminals are reached, the electricalllR losses and the brush losses must be subtracted. In the case of dc motors, this power-now diagram is simply reversed. The power-now diagram for a motor is shown in Figure 8- 39b. IX: MACHINERY FUNDAMENTALS 527 Example problems involving the calculation of motor and generator effi-ciencies will be given in the next two chapters. S.S SUMMARY DC machines convert mechanical power to dc electric power, and vice versa. In this chapter, the basic principles of dc machine operation were explained first by looking at a simple linear machine and then by looking at a machine consisting of a single rotating loop. The concept of commutation as a technique for converting the ac voltage in rotor conductors to a dc output was introduced, and its problems were explored. The possible winding arrangements of conductors in a dc rotor (lap and wave windings) were also examined. Equations were then derived for the induced voltage and torque in a dc ma-chine, and the physical construction of the machines was described. Finally, the types of losses in the dc machine were described and related to its overall operat-ing efficiency. QUESTIONS 8-1, What is conunutation? How can a commutator convert ac voltages on a machine's armature to dc voltages at its terminals? 8-2, Why does curving the pole faces in a dc machine contribute to a smoother dc output voltage from it? 8-3, What is the pitch factor of a coil? 8-4, Explain the concept of electrical degrees. How is the electrical angle of the voltage in a rotor conductor related to the mechanical angle of the machine's shaft? 8-5, What is conunutator pitch? 8-6, What is the plex of an armature winding? 8-7, How do lap windings differ from wave windings? 8-8, What are equalizers? Why are they needed on a lap-wound machine but not on a wave-wolUld machine? 8-9, What is annature reaction? How does it affect the operation of a dc machine? 8-10, Explain the L dildt voltage problem in conductors lUldergoing commutation. 8-11, How does brush shifting affect the sparking problem in dc machines? 8-12, What are conunutating poles? How are they used? 8-13, What are compensating windings? What is their most serious disadvantage? 8-14, Why are laminated poles used in modem dc machine construction? 8-15, What is an insulation class? 8-16, What types of losses are present in a dc machine? PROBLEMS 8-1, The following infonnation is given about the simple rotating loop shown in fig-ure 8--6: 528 ELECTRIC MACHINERY RJNDA MENTALS B = O.S T l = O.5 m r = O.I25 m VB = 24 V R = 0.4 0 w = 250 radls (a) Is this machine operating as a motor or a generator? Explain. (b) What is the current i flowing into or out of the machine? What is the power flowing into or out of the machine? (c) If the speed of the rotor were changed to 275 rad/s, what would happen to the current flow into or out of the machine? (d) If the speed of the rotor were changed to 225 rad/s, what would happen to the current flow into or out of the machine? &-2. Refer to the simple two-pole eight-coil machine shown in Figure PS- I. The follow-ing information is given about this machine: \ ~ad I I __ wne __ ' , , , w , I .. I ---------:::/~~'',. , -5' -i/-"'=:::---------N , , , I I , , I I , , , , , I I , , , I I , , , I I ' , , , , , 7""~-~-~ -:= -~-~~8~. ~,~ O~6~ . ~5~~ -~-~------,£~7 3 3' - -- --I 4 ---- -, , I I , I , I , , , , , , , , / 1' : 5 \ , , , r----'-----"' I 20" I 20° I , , , , , , , , , , , , s Given: II '" 1.0 T in the air gap I '" 0.3 m (length of sides) r '" 0.08 m (radius of coils) n '" 1700 r!min ---- Lines on this side of rotor ----Lines on other side of rotor Jo'IGURE 1'8-1 The machine in Problem 8- 2. IX: MACHINERY FUNDAMENTALS 529 B = 1.0T in air gap 1= 0.3 m (length of coil sides) r = 0.08 m (radius of coils) n = 1700 rlmin ccw The resistance of each rotor coil is 0.04 n. (a) Is the armature winding shown a progressive or retrogressive winding? (b) How many current paths are there through the armature of this machine? (c) What are the magnitude and the polarity of the voltage at the brushes in this machine? (d) What is the annature resistance RA of this machine? (e) If a 1O~ resistor is connected to the tenninals of this machine, how much cur-rent flows in the machine? Consider the internal resistance of the machine in de-tennining the current flow. (f) What are the magnitude and the direction of the resulting induced torque? (g) Assuming that the speed of rotation and magnetic flux density are constant, plot the terminal voltage of this machine as a flUlction of the current drawn from it. 8-3. Prove that the equation for the induced voltage of a single simple rotating loop 2 , . , = - "'w In 7T '+' (8-6) is just a special case of the general equation for induced voltage in a dc machine (8--38) 8-4. A dc machine has eight poles and a rated current of 100 A. How much current will flow in each path at rated conditions if the armature is (a) simplex lap-wound, (b) duplex lap-wound, (c) simplex wave-wound? 8-5. How many parallel current paths will there be in the armature of a 12-pole machine if the armature is (a) simplex lap-wolUld, (b) duplex wave-wound, (c) triplex lap-wound, (d) quadruplex wave-wolUld? 8-6. The power converted from one fonn to another within a dc motor was given by Use the equations for EA and "Tiod [Equations (8--38) and (8-49)] to prove that EAtA = "Tiodw .. ; that is. prove that the electric power disappearing at the point of power conversion is exactly equal to the mechanical power appearing at that point. 8-7. An eight-pole, 25-kW, 120-V dc generator has a duplex lap-wound armature which has 64 coils with 16 turns per coil. Its rated speed is 2400 r/min. (a) How much flux per pole is required to produce the rated voltage in this genera-tor at no-load conditions? (b) What is the current per path in the annature of this generator at the rated load? (c) What is the induced torque in this machine at the rated load? (d) How many brushes must this motor have? How wide must each one be? (e) If the resistance of this winding is 0.011 0: per turn, what is the armature resis-tance RA of this machine? 8-8. Figure PS- 2 shows a small two-pole dc motor with eight rotor coils and four turns per coil. The flux per pole in this machine is 0.0125 Wh. 530 ELECTRIC MACHINERY RJNDAMENTALS 2 3 N s ""GURE 1'8-2 The machine in Problem 8--8. (a) If this motor is cOIUlected to a 12-V dc car battery, whal will the no-load speed of the motor be? (b) If the positive terminal of the battery is connected to the rightmost brush on the motor, which way will it rotate? (c) If this motor is loaded down so that it consrunes 50 W from the battery, what will the induced torque of the motor be? (Ignore any internal resistance in the motor.) &-9. Refer to the machine winding shown in Figure P8-3. (a) How many parallel current paths are there through this annature winding? (b) Where should the brushes be located on this machine for proper commutation? How wide should they be? (c) What is the plex of this machine? (d) If the voltage on any single conductor lUlder the pole faces in this machine is e, what is the voltage at the lenninals of this machine? 8-10. Describe in detail the winding of the machine shown in Figure PS-4. If a positive voltage is applied to the brush under the north pole face, which way will this motor rotate? REFERENCES I. Del Toro. V. Electric Machines and Pov.·er Systems. Englewood Cliffs. N.J.: Prentice-Ha.lt. 1985. 2. Fitzgerald. A. E., C. Kingsley. Jr .. and S. D. Umans. Electric Machinery. 5th ed. New York: McGraw-Hilt. 1990. 3. Hubert. Charles I. Preventative Maintenance of Electrical Equipment. 2nd ed. New York: McGraw-Hilt. 19ff}. 4. Kosow. Irving L. Electric Machinery and Transfanners. Englewood Cliffs. N.J.: Premice-HatJ. 1972. 5. Na.tional Electrical Manufacturers Association. Motors and Generators, Publication MG 1-1993. Washington, D.C.. 1993. 6. Siskind. Charles. Direct Current Machinery. New York: McGraw-Hilt. 1952. 7. Werninck. E. H. (ed.). Electric Motor Handbook.. London: McGraw-Hilt. 1978. IX: MACHINERY FUNDAMENTALS 531 9 10 8 11 / 16' I 3' 7 I , I / 12 , I / -15' I , , I / ---6 I I g h 13 j , N , I , , , , 6' 5 " m , 14 , P " " I -/ I , I ---/ , -/ , 8' I 11' / \0'/ 3 16 2 ('1 :x ~~ :XXX -------, , , , , " , N , , , , , , , , , , , , ------~ m " " p " b , d , f g h ; j , , m " (hI FIGURE P8-J (a) The machine in Problem 8--9. (b) The annature winding diagram of this machine. 532 ELECTRIC MACHINERY RJNDAMENTALS ' __ 10 N s 14 2 ""GURE 1'8-4 The machine in Problem 8- 10. CHAPTER 9 DC MOTORS AND GENERATORS D c motors are de machines used as motors, and de generators are de machines used as generators. As noted in Chapter 8, the same physical machine can op-erate as either a motor or a generator-it is simply a question of the direction of the power now through it. This chapter will examine the different types of de mo-tors that can be made and explain the advantages and disadvantages of each. It will include a discussion of de motor starting and solid-state controls. Finally, the chapter will conclude with a discussion of de generators. 9.1 INTRODUCTION TO DC MOTORS The earliest power systems in the United States were de systems, but by the 1890s ac power systems were clearly winning out over de systems. Despite this fact, de motors continued to be a significant fraction of the machinery purchased each year through the 1960s (that fraction has declined in the last 40 years). Why were dc motors so common, when dc power systems themselves were fairly rare? There were several reasons for the continued popularity of dc motors. One was that dc power systems are still common in cars, trucks, and aircraft. When a vehicle has a dc power system, it makes sense to consider using dc motors. An-other application for dc motors was a situation in which wide variations in speed are needed. Before the widespread use of power electronic rectifier-inverters, dc motors were unexcelled in speed control applications. Even ifno dc power source were available, solid-state rectifier and chopper circuits were used to create the necessary dc power, and dc motors were used to provide the desired speed control. 533 534 ELECTRIC MACHINERY RJNDA MENTALS , ., ,b , ""GURE 9- 1 Early de motors. (a) A very early de motor built by Elihu Thompson in 1886. It was rated at about \oj hp. (Courtesy ofGeneml Electric Company.) (b) A larger four-pole de motor from about the turn of the century. Notice the h.andle for shifting the brush.es to the neutral plane. (Courtesy ofGeneml Electric Company. ) (Today, induction motors with solid-state drive packages are the preferred choice over dc motors for most speed control applications. However, there are still some applications where dc motors are preferred.) DC motors are often compared by their speed regulations. TIle speed regu-lation (SR) of a motor is defined by 'I S -R -~ --: W~ " -' Wn =~W~ "-X -l-()() -% '1 W-l ) rx: MmDRS AND GENERATORS 535 I SR = nul ~ nfl x 100% I (9-2) It is a rough measure of the shape of a motor's torque- speed characteristic-a positive speed regulation means that a motor's speed drops with increasing load, and a negative speed regulation means a motor's speed increases with increasing load. The magnitude of the speed regulation tells approximately how steep the slope of the torque- speed curve is. DC motors are, of course, driven from a dc power supply. Unless otherwise specified, the input voltage to a de motor is assumed to be constant, because that assumption simplifies the analysis of motors and the comparison between differ-ent types of motors. There are five major types of dc motors in general use: I. 1lle separately excited dc motor 2. 1lle shunt dc motor 3. 1lle pennanent-magnet dc motor 4. 1lle series dc motor 5. 1lle compounded dc motor Each of these types will be examined in turn. 9.2 THE EQUIVALENT CIRCUIT OFADC MOTOR The equivalent circuit of a dc motor is shown in Figure 9-2. In this figure, the ar-mature circuit is represented by an ideal voltage source E), and a resistor R),. This representation is really the Thevenin equivalent of the entire rotor structure, in-cluding rotor coils, interpoles, and compensating windings, if present. The brush voltage drop is represented by a small battery V bruwJ.), so the ar-mature current in the motor IA = (VT - EA J.)/RA increases. As the annature current rises, the induced torque in the motor increases (1]00 = K4> IA i ), and finally the in-duced torque will equal the load torque at a lower mechanical speed of rotation w. TIle output characteristic of a shunt dc motor can be derived from the in-duced voltage and torque equations of the motor plus Kirchhoff's voltage law. (KVL) TIle KVL equation for a shunt motor is VT = EA + lARA The induced voltage EA = K4>w, so VT = K¢w + lARA Since "Tind = K4>IA, current IA can be expressed as "Tind IA = K4> Combining Equations (9-4) and (9-5) prOOuces Finally, solving for the motor's speed yields (9- 3) (9-4) (9- 5) (9-6) (9- 7) This equation is just a straight line with a negative slope. TIle resulting torque- speed characteristic of a shunt dc motor is shown in Figure 9--6a. 540 ELECTRIC MACHINERY RJNDAMENTALS "---------------------------- '00 (a) ------WithAR ---NoAR "---------------------------- '00 ,b, ""GURE 9--6 (a) Torque-speed characteristic of a shunt or separately excited dc motor with compensating windings to eliminate armature reaction. (b) Torque-speed characteristic of the motor with annature reaction present. II is important to realize that, in order for the speed of the motor to vary lin-early with torque, the other terms in this expression must be constant as the load changes. TIle tenninal voltage supplied by the dc power source is assumed to be constant- if it is not constant, then the voltage variations will affect the shape of the torque- speed curve. Another effect internal to the motor that can also affect the shape of the torque-speed curve is armature reaction. If a motor has annature reaction, then as its load increases, the flux-weakening effects reduce its flux. As Equation (9-7) shows, the effect ofa reduction in flux is to increase the motor's speed at any given load over the speed it would run at without armature reaction. The torque-speed characteristic of a shunt motor with annature reaction is shown in Figure 9--6b. If a motor has compensating windings, of course there will be no flux-weakening problems in the machine, and the flux in the machine will be constant. rx: MmDRS AND GENERATORS 541 " R, " --+ 0.060 R;). ~ I', son + R, < ~ E, VT",250V L, NF '" 1200tuTns FIGURE 9- 7 The shunt motor in Example 9--1. Ifa shunt dc motor has compensating windings so that its flux is constant regardless of load, and the motor's speed and armature current are known at any one value of load, then it is possible to calculate its speed at any other value of load, as long as the armature current at that load is known or can be detennined. Example 9- 1 illustrates this calculation. Example 9-1. A 50-hp, 250-V, 1200 r/min dc shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 n. Its field circuit has a total resistance Rodj + RF of 50 fl, which pro-duces a no-load speed of 1200 r/min. There are 1200 tlU1lS per pole on the shunt field wind-ing (see Figure 9-7). (a) Find the speed of this motor when its input current is 100 A. (b) Find the speed of this motor when its input current is 200 A. (c) Find the speed of this motor when its input current is 300 A. (d) Plot the torque-speed characteristic of this motor. Solutioll The internal generated voltage of a dc machine with its speed expressed in revolutions per minute is given by (8-41) Since the field current in the machine is constant (because Vr and the field resistance are both constant), and since there are no annature reaction effects, the flux in this motor is constant. The relationship between the speeds and internal generated voltages of the motor at two different load conditions is thus (9-8) The constant K ' cancels, since it is a constant for any given machine, and the flux t " ., § ." " .~ , !. a 300 250 233 200 150 100 50 I I o 0.0 I I / I I 1.0 2.0 rx: MmDRS AND GENERATORS 545 ,/ V / / V / 3.0 4.0 4.3 5.0 6.0 7.0 8.0 9.0 10.0 Field current, A FIGURE 9-9 The magnetization curve of a typical 25()' V dc motor. taken at a speed of 1200 r/min. EXllmple 9-2. A 5O-hp, 250-V, 1200 r/min dc shlUlt motor without compensating windings has an armature resistance (including the brushes and interpoles) of 0.06 n. Its field circuit has a total resistance RF + Radj of 50 n, which produces a no-load speed of 1200 r/min. There are 1200 turns per pole on the shunt field winding, and the armature re-action produces a demagnetizing magnetomotive force of 840 A • turns at a load current of 200 A. The magnetization curve of this machine is shown in Figure 9-9. (a) Find the speed of this motor when its input current is 200 A. (b) This motor is essentially identical to the one in Example 9- 1 except for the ab-sence of compensating windings. How does its speed compare to that of the pre-vious motor at a load current of 200 A? (c) Calculate and plot the torque-speed characteristic for this motor. Solutioll (a) If IL = 200 A, then the armature current of the motor is V, lit = IL - IF = h - R , - 2ooA -250 V -195A -50n -546 ELECTRIC MACHINERY RJNDA MENTALS Therefore, the internal generated voltage of the machine is EA = Vr -lARA = 250 V - (195 A)(O.()5 fi) = 238.3 V At h = 200 A, the demagnetizing magnetomoti ve force due to armature reac-tion is 840 A · turns, so the effective shunt field current of the motor is "'" r,,= I F - N , (9-12) From the magnetization curve, this effective field current would produce an in-ternal generated voltage EAO of 233 Vat a speed 110 of 1200 r/min. We know that the internal generated voltage EAO would be 233 V at a speed of 1200 r/min. Since the actual internal generated voltage EA is 238.3 V, the ac-tual operating speed of the motor must be (9-13) EA 238.3 V (1200 I . ) n= £ no = 233V rmm = " 1 227 r/min (b) At 200 A of load in Example 9--1, the motor's speed was n = 1144 r/min. In this example, the motor's speed is 1227 r/min. Notice that the speed of the motor with armature reaction is higher than the speed of the motor with no armature reaction . This relative increase in speed is due to the flux weakening in the ma-chine with armature reaction. (c) To derive the torque- speed characteristic of this motor, we must calculate the torque and speed for many different conditions of load. Unfortunately, the de-magnetizing armature reaction magnetomotive force is only given for one con-dition of load (200 A). Since no additional information is available, we will as-srune that the strength of '.JAR varies linearly with load current. A MATLAB M-file which automates this calculation and plots the resulting torque-speed characteristic is shown below. It peIfonns the same steps as part a to detennine the speed for each load current, and then calculates the induced torque at that speed. Note that it reads the magnetization curve from a file called f i g9 _ 9 . ma t. This file and the other magnetization curves in this chapter are available for download from the book's World Wide Web site (see Preface for details). ~ M-file: s hunt_ t s_curve. m ~ M-file c r eat e a p l ot o f the t o r qu e - speed cu rve o f the ~ the s hunt dc mot or with a r mature r eaction in ~ Exampl e 9- 2. ~ Get the magneti zation c u rve. Thi s fil e conta ins the ~ three variabl es if_va lue, ea_value, a nd n_O. l oad fig9_9. ma t ~ Firs t , initia li ze the values n eeded in thi s p r ogr am. v_t = 250; % Te rmina l voltage (V) rx: MmDRS AND GENERATORS 547 c - f 0 50 ; • Fi e l d r es i s t a nce (ohms) c -" 0 0 . 06 ; • Arma tur e r es i s t a nce (ohms) i 1 0 10, 10,300 ; • Li ne curre nt s (A ) 0 - f 0 1200 ; • Number of turns 0 0 fi e l d f - " c O 0 84 0 ; • Arma tur e r eact i on , 200 A % Ca l c u l a t e the a rma ture c urre nt f or each l oad . i a = i _ l -v_t I r _ f ; % Now ca l c u l a t e the i nt e rna l gene r a t ed vo l t age f or % each a rma ture c urre nt. e_a = v_t -i _a r _a; (A- t i m) % Ca l c u l a t e the a rma ture r eact i on MMF f or each a rma ture % c urre nt. f _ar = ( i _a I 200) f _ar O: % Ca l c u l a t e the e ffect i ve fi e l d c urre nt. i f = v_t I r _ f -f _ar I n f : % Ca l c u l a t e the r esu l t i ng i nt e rna l gene r a t ed vo l t age a t % 1200 r / mi n by i nt e r po l a t i ng the mot or 's magne t i za t i on % c urve . e_aO = i nt e r p l (if_va l ues, ea_ va l ue s, i _ f , ' spline ' ) ; % Ca l c u l a t e the r esult i ng speed f r om Equa t i on (9 - 13 ) . n = ( e_a . 1 e_aO ) .. n_O: % Ca l c u l a t e the i nduced t or que correspond i ng t o each % speed f r om Equa t i on s ( 8- 55 ) a nd (8- 56 ) . t _ i nd = e_a . i a .1 (n .. 2 .. p i I 60): % Pl ot the t or que - speed c urve p l ot (t _ i nd, n , 'Col or' , 'k' , 'LineWi dth' ,2 . 0 ) ; h o l d on ; x l abe l ( ' \ t au( i nd} (N- m) ' , ' Fontwei ght' , 'Bo l d ' ) ; y l abe l ( ' \ i tn_( m} \ rm\ b f (r / mi n ) ' , 'Fontwei ght' , 'Bo l d ' ) : ( ' \ b f Shunt OC mot or t or que - speed c h a r act e r i s t i c ' ) ax i s( [ 0 600 1100 1300 ] ) ; g r i d on ; h o l d o ff ; The resulting torque-speed characteristic is shown in Figure 9-10. Note that for any given load. the speed of the motor with armature reaction is higher than the speed of the motor without armature reaction. Speed Control of Shunt DC Motors How can the speed of a shunt dc motor be controlled? There are two common methods and one less common method in use. 1lle common methods have already been seen in the simple linear machine in Chapter 1 and the simple rotating loop in Chapter 8. The two common ways in which the speed of a shunt dc machine can be controlled are by 548 ELECTRIC MACHINERY RJNDAMENTALS 1300 ,------,--,----,------,--,------, 1280 1260 1240 1220 t--~---'---: ~ 1200 ./"80 ]]60 ]]40 ]]20 ]]00 '-----c'cC_-~-C'c---c'cC_-~-~ o 100 200 300 400 500 600 'tiD<! N·m ""GURE 9- 10 The torque--speed characteristic of the motor with armature reaction in Example 9--2. I. Adjusting the field resistance RF (and thus the field nux) 2. Adjusting the tenninal voltage applied to the annature. The less common method of speed control is by 3. Inserting a resistor in series with the armature circuit. E:1.ch of these methods is described in detail below. CHANGING THE FIELD RESISTANCE. To understand what happens when the field resistor of a dc motor is changed. assume that the field resistor increases and observe the response. If the field resistance increases, then the field current de-creases (IF = Vr lRF i ), and as the field current decreases, the nux w. Therefore, EA will drop to EA2 = 0.99 EAt = 0.99(245 V) = 242.55 V The annature current must then rise to f = 250 V -242.55 V = 298 A A 0.25 n . Thus a I percent decrease in flux produced a 49 percent increase in armature current. So to get back to the original discussion, the increase in current predomi-nates over the decrease in flux, and the induced torque rises: U Tjnd = K4> fA Since Tind > Tto"," the motor speeds up. However, as the motor speeds up, the internal generated voltage EA rises, causing fA to fall. As fA falls, the induced torque Tind falls too, and fmally T;Dd again equals Ttood at a higher steady-state speed than originally. To summarize the cause-and-effcct behavior involved in this method of speed control: I. Increasing RF causes fF(= VT IRF i ) to decrease. 2. Decreasing IF decreases 4>. 3. Decreasing 4> lowers EA (= K4>J..w). 4. Decreasing EA increases fA (= VT -EA J..)IRA-5. Increasing fA increases T;od(= K4>UA fI), with the change in fA dominant over the change in flux). 6. Increasing Tind makes T;od > Ttood, and the speed w increases. 7. Increasing to increases EA = Kcf>wi again. 550 ELECTRIC MACHINERY RJNDAMENTALS (a) ,b, 8. Increasing Elt decreases lit-FlGURE 9-12 The effect of field resistance speed control on a shunt motor's torque-speed characteristic: (a) over the motor's normal operating range: (b) over the entire range from no-load to stall conditions. 9. Decreasing lit decreases "Tind until "Tind = "TJoad at a higher speed w. The effect of increasing the field resistance on the output characteristic of a shu nt motor is shown in Figure 9-1 2a. Notice that as the flux in the machine decreases, the no-load speed of the motor increases, while the slope of the torque-speed curve becomes steeper. Naturally, decreasing RF would reverse the whole process, and the speed of the motor would drop. A WARNING ABOUT FIELD RESISTANCE SPEED CONTROL. TIle effect of in-creasing the field resistance on the output characteristic of a shunt dc motor is shown in Figure 9-1 2. Notice that as the flux in the machine decreases, the no-load speed of the motor increases, while the slope of the torque-speed curve be-comes steeper. This shape is a consequence of Equation (9- 7), which describes the tenninal characteristic of the motor. In Equation (9- 7), the no-load speed is R, I, Variable -+ voltage controller + E, V, --FIGURE 9-13 /-rx: MmDRS AND GENERATORS 551 I, -y,,! I, ~ R, L, VTis oonstant VA is variable + V, Armature voltage control of a shunt (or separately excited) dc motor. proportional to the reciprocal of the nUX: in the motor, while the slope of the curve is proportional to the reciprocal of the flux squared. Therefore, a decrease in flux causes the slope of the torque- speed curve to become steeper. Figure 9- I 2a shows the tenninal characteristic of the motor over the range from no-load to full-load conditions. Over this range, an increase in field resis-tance increases the motor's speed, as described above in this section. For motors operating between no-load and full-l oad conditions, an increase in RF may reliably be expected to increase operating speed. Now examine Figure 9- 12h. This figure shows the tenninal characteristic of the motor over the full range from no-load to stall conditions. It is apparent from the figure that at very slow speeds an increase in field resistance will actually de-crease the speed of the motor. TIlis effect occurs because, at very low speeds, the increase in annature current caused by the decrease in Ell. is no longer large enough to compensate for the decrease in flux in the induced torque equation. With the flux decrease actually larger than the armature current increase, the in-duced torque decreases, and the motor slows down. Some small dc motors used for control purposes actually operate at speeds close to stall conditions. For these motors, an increase in field resistance might have no effect, or it might even decrease the speed of the motor. Since the results are not predictable, field resistance speed control should not be used in these types of dc motors. Instead, the annature voltage method of speed control shou ld be employed. CHANGING THE ARMATURE VOLTAGE. TIle second form of speed control in-volves changing the voltage applied to the armature of the motor without chang-ing the voltage applied to the field. A connection similar to that in Figure 9- \3 is necessary for this type of control. In effect, the motor must be separately excited to use armature voltage control. If the voltage VA is increased, then the annature current in the motor must rise [Ill. = (VA i - EA)IRAl As /11. increases, the induced torque "Tind = K4>IA i in-creases, making "Tind > "TJoad, and the speed w of the motor increases. 552 ELECTRIC MACHINERY RJNDAMENTALS v" '----------------f ind ""GURE 9-14 The effect of armature voltage speed control on a shunt motor's torque-speed characteristic. Bul as the speed w increases, the internal generaled voltage EA(= K4>wi) increases, causing the armature current to decrease. This decrease in J A decreases the induced torque, causing Tind to equal Ttoad at a higher rotational speed w. To summarize the cause-and-effect behavior in this method of speed control: I . An increase in VA increases JA [= (VA i - EA)/RA]. 2. Increasing J A increases Tind (= K4>JA i ). 3. Increasing Tind makes TiDd > TJoad increasing w. 4. Increasing w increases EA(= K4>wi). 5. Increasing EA decreases JA [= (VA i - EA)/RAl 6. Decreasing JA decreases Tind until Tind = TJoad at a higher w. TIle effect of an increase in VA on the torque-speed characteristic of a sepa-rately excited motor is shown in Figure 9- 14. Notice that the no-load speed of the motor is shifted by this method of speed control, but the slope of the curve re-mains constant. INSERTING A RESISTOR IN SERIES WITH THE ARl\l ATURE c m.CUIT. If a resistor is inserted in series with the annature circuit, the effect is to drastically in-crease the slope of the motor's torque-speed characteristic, making it operate more slowly if loaded (Figure 9- 15). lllis fact can easily be seen from Equation (9- 7). The insertion of a resistor is a very wasteful method of speed control, since the losses in the inserted resistor are very large. For this reason, it is rarely used. It will be found only in applications in which the motor spends almost all its time operating at fuJI speed or in applications too inexpensive to justify a better form of speed control. TIle two most common methods of shunt motor speed control- field resis-tance variation and armature voltage variation- have different safe ranges of operation. rx: MmDRS AND GENERATORS 553 '------------------'00 FIGURE 9-15 The effect of armature resistance speed control on a shunt motor's torque-speed characteristic. In field resistance control, the lower the field current in a shunt (or sepa-rately excited) dc motor, the faster it turns: and the higher the field current, the slower it turns. Since an increase in field current causes a decrease in speed, there is always a minimum achievable speed by field circuit control. This minimum speed occurs when the motor's field circuit has the maximum permissible current flowing through it. If a motor is operating at its rated terminal voltage, power, and field current, then it will be running at rated speed, also known as base speed. Field resistance control can control the speed of the motor for speeds above base speed but not for speeds below base speed. To achieve a speed slower than base speed by field cir-cuit control would require excessive fie ld current, possibly burning up the field windings. In armature voltage control, the lower the armature voltage on a separately excited dc motor, the slower it turns; and the higher the armature voltage, the faster it turns. Since an increase in annature voltage causes an increase in speed, there is always a maximum achievable speed by armature voltage control. This maximum speed occurs when the motor's armature voltage reaches its maximum permissible level. If the motor is operating at its rated voltage, field current, and power, it will be turning at base speed. Annature voltage control can control the speed of the motor for speeds below base speed but not for speeds above base speed. To achieve a speed faster than base speed by armature voltage control would require excessive annature voltage, possibly damaging the annature circuit. These two techniques of speed control are obviously complementary. Ar-mature voltage control works well for speeds below base speed, and field resis-tance or field current control works well for speeds above base speed. By com-bining the two speed-control techniques in the same motor, it is possible to get a range of speed variations of up to 40 to I or more. Shunt and separately excited dc motors have excellent speed control characteristics. 554 ELECTRIC MACHINERY RJNDAMENTALS Maximum torque fmax Maximum powerPmu fmax constant P mu constant VA control RFcontrol "------cC------ ,-,~ ""GURE 9-16 fmu constant so P rmx constant P IOU '" f mn w..,-_____ _ V" control VA control <-----~------- ,. ,~ Power and torque limits as a function of speed for a shunt motor under annature volt and field resistance control. TIlere is a significant difference in the torque and power limits on the ma-chine under these two types of speed control. The limiting factor in either case is the heating of the annature conductors, which places an upper limit on the mag-nitude of the annature current Ill.. For annature voltage control, the flux in the motor is constant, so the maxi-mum torque in the motor is (9-1 4) This maximum torque is constant regardless of the speed of the rotation of the mo-tor. Since the power out of the motor is given by P = "TW, the maximum power of the motor at any speed under annature voltage control is (9- 15) TIlUS the maximum power out of the motor is directly proportional to its operat-ing speed under armature voltage control. On the other hand, when field resistance control is used, the flux does change. In this form of control, a speed increase is caused by a decrease in the ma-chine's flux. In order for the annature current limit not to be exceeded, the in-duced torque limit must decrease as the speed of the motor increases. Since the power out of the motor is given by P = "TW, and the torque limit decreases as the speed of the motor increases, the maximum power out of a dc motor under field current control is constant, while the maximum torque varies as the reciprocal of the motor's speed. TIlese shunt dc motor power and torque limitations for safe operation as a function of speed are shown in Figure 9-16. TIle following examples illustrate how to fmd the new speed of a dc motor if it is varied by field resistance or annature voltage control methods. rx: MaJ"ORS AND GENERATORS 555 RA '" 0.03 n --", +~-------, ,-----v./v--~ + + ,b, FIGURE 9-17 (3) The shunt motor in Example 9--3. (b) The separately excited de motor in Example 9--4. Example 9-3. Figure 9--17a shows a 1000hp. 250-V, 1200 rhnin shlUlt dc motor with an armature resistance of 0.03 n and a field resistance of 41.67 O. The motor has compen-sating windings. so armature reaction can be ignored. Mechanical and core losses may be as-sumed to be negligible for the purposes of this problem. The motor is assmned to be driving a load with a line current of 126 A and an initial speed of 1103 r/min. To simplify the prob-lem. assmne that the amolUlt of armature current drawn by the motor remains constant. (a) If the machine's magnetization curve is shown in Figure 9-9. what is the mo-tor's speed if the field resistance is raised to 50 n1 (b) Calculate and plot the speed of this motor as a ftmction of the field resistance RI' assuming a constant-current load. Solutioll (a) The motor has an initial line current of 126 A, so the initial armature current is 150 V IA I = lu - 1Ft = 126A - 41.67 n = 120A Therefore, the internal generated voltage is EAI = VT -IA tRA = 250 V - (120 A)(0.03fi) = 246.4 V After the field resistance is increased to 50 n, the field current will become 556 ELECTRIC MACHINERY RJNDAMENTALS I -VT 2S0V -SA F2 -RF -50 n -The ratio of the internal generated voltage at one speed to the internal generated voltage at another speed is given by the ratio of Equation (&--41) at the two speeds: Elll _ K'q,l n2 Elll -K'q,n[ (9- 16) Because the armature current is assumed constant, EIlI = Elll, and this equation reduces to ~, n2=q,lnl (9- 17) A magnetization curve is a plot of Ell versus IF for a given speed. Since the val-ues of Ell on the curve are directly proportional to the flux, the ratio of the inter-nal generated voltages read off the curve is equal to the ratio of the fluxes within the machine. At IF = 5 A, Ello = 250 V, while at IF = 6 A, Ello = 268 V. There-fore, the ratio of fluxes is given by q,[ 268 V = = 1.076 q,l 250 V and the new s~ed of the motor is n2 = !:nl = (1.076XII03r/min) = 1187r/min (b) A MATLAB M-file that calculates the s~ed of the motor as a ftmction of RF is shown below. ~ M -file, r f speed_contro l .m ~ M -file c reate a p l ot o f the speed of a s hunt dc ~ mot or as a f unc t i on o f f i e l d r es i s tance, assumi ng ~ a con s tant armature c urrent (Exampl e 9- 3). ~ Get the magnet i zat i on c urve. Thi s f il e conta i n s the ~ three var i abl es if_va l u e, ea_va l u e, and n_O. l oad fig9_9.mat ~ Fir s t , i nit i a li ze v_t = 250; c - f 0 40,1:70; c -" 0 0.03; i -" 0 12 0; the va l ues n eeded i n thi s program. ~ Termi na l vo l tage (V) ~ Fie l d re s i s tance (ohms) ~ Armature re s i s tance (ohms) ~ Armature c urrent s (A) ~ The approach here i s t o ca l c u l ate the e_aO at the ~ re f erence fi e l d c urrent , and then to ca l cu l ate the ~ e_aO f or every fi e l d c urrent. The r e f e rence speed i s ~ 1103 r / mi n , so by knowi ng the e_aO and r e f e rence ~ speed, we will be abl e to ca l cu l ate the speed at the ~ other fi e l d c urrent. rx: MmDRS AND GENERATORS 557 % Ca l c ula t e the int e rna l gen e r a t ed voltage a t 1200 r / min % f or the r e f e r e nce fi e l d c urre nt (5 A) by int e r po l a ting % the mot or 's magnetiz ation c urve. The r e f e r e nce speed % correspond ing t o thi s fi e l d c urre nt i s 1103 r / min . e_aO_r e f = int e r p l (if_va lues, ea_values, 5, ' spline ' ) ; n r e f = 1103; % Ca l c ula t e the fi e l d c urre nt f or each value o f fi e l d % r es i s t a nce. i _ f = v_t . / r _ f ; % Ca l c ula t e the E_aO f or eac h fi e l d c urre nt by % int e r po l ating the mot or 's magneti zation c urve. e_aO = int e r p l (if_va lues, ea_values, i _ f , ' spline ' ) ; % Ca l c ula t e the r esulting speed from Equation (9-17): % n2 = (phil / phi2 ) .. nl = (e_aO_l / e_aO_2 ) .. nl n2 = ( e_aO_r e f . / e_aO ) .. n_ r e f ; % Plot the speed ver s u s r _ f c urve. p l ot (r _ f , n2, 'Col or' , 'k' , 'LineWi dth' ,2.0 ) ; h o l d on ; x l abe l ( 'Fi e l d r es i s t a nce, \Omega ' , 'Fontwei ght' , 'Bo l d ' ) ; y l abe l ( ' \ itn_( m} \ rm\ b f (r / min ) ' , 'Fontwei ght' , 'Bo l d ' ) ; title ( ' Speed vs \ itR_( F ) \ rm\ b f f or a Shunt IX: M ot or' , 'Fontwei ght' , 'Bo l d ' ) ; ax i s( [40 70 0 1400 ] ) ; grid on ; h o l d off ; The resulting plot is shown in Figure 9-1 8. 1400 1200 , , IlXXl .§ 800 ~ • 600 0 400 200 0 40 45 60 65 70 Field resistance. n FIGURE 9- 18 Plot of speed versus field resistance for the shunt dc motor of Example 9-3. 558 ELECTRIC MACHINERY RJNDAMENTALS Note that the assumption of a constant annature current as RF changes is not a very go<Xl one for real loads. The current in the annature will vary with speed in a fashion dependent on the torque required by the type of load attached to the mo-tor. These differences wi ll cause a motor 's speed-versus-RF curve to be slightly different than the one shown in Figure 9-1 8, but it will have a similar shape. Example 9-4. The motor in Example 9- 3 is now cotUlected separately excited, as shown in Figure 9-17b. The motor is initially flmning with VA = 250 V, IA = 120 A, and n = 1103 rlmin, while supplying a constant-torque load. What will the speed of this motor be if VA is reduced to 200 V? Solutio" The motor has an initial line current of 120 A and an armature voltage VA of 250 V, so the internal generated voltage EA is EA = Vr -lARA = 250 V -(l20AXO.03!l) = 246.4 V By applying Equation (9-16) and realizing that the flux ~ is constant, the motor's speed can be expressed as To find EJa use Kirchhoff's voltage law: ", = EJa = Vr -lJaRA (9-16) Since the torque is constant and the flux is constant, IA is constant. This yields a voltage of EJa = 200 V -(120 AXO.03 !l) = 196.4 V The final speed of the motor is thus E A2 196.4V,'03 /· 879/· ~ = EAt nt = 246.4 V r mill = r min The Effect of an Open Field Circuit TIle previous section of this chapter contained a discussion of speed control by varying the field resistance of a shunt moto r. As the field resistance increased, the speed of the motor increased with it. What would happen if this effect were taken to the extreme, if the field resistor really increased? What would happen if the field circuit actually opened while the motor was running? From the previous dis-cussion, the flux in the machine would drop drastically, all the way down to ~res, and EA(= K~w) would drop with it. This would cause a really enonnous increase in the armature current, and the resulting induced torque would be quite a bit higher than the load torque on the motor. TIlerefore, the motor's speed starts to rise and just keeps going up. rx: MmDRS AND GENERATORS 559 The results of an open field circuit can be quite spectacular. When the au-thor was an undergraduate, his laboratory group once made a mistake of this sort. The group was working with a small motor-generator set being driven by a 3-hp shunt dc motor. The motor was connected and ready to go, but there was just one little mistake- when the field circuit was connected, it was fused with a O.3-A fuse instead of the 3-A fu se that was supposed to be used. When the motor was started, it ran nonnally for about 3 s, and then sud-denly there was a flash from the fuse. ]mmediately, the motor's speed skyrock-eted. Someone turned the main circuit breaker off within a few seconds, but by that time the tachometer attached to the motor had pegged at 4000 r/min. TIle mo-tor itself was only rated for 800 rim in. Needless to say, that experience scared everyone present very badly and taught them to be most careful about fie ld circuit protection. In dc motor starting and protection circuits, afield loss relay is nonnally included to disconnect the motor from the line in the event of a loss of field current. A similar effect can occur in ordinary shunt dc motors operating with light fields if their annature reaction effects are severe enough. If the annature reaction on a dc motor is severe, an increase in load can weaken its flux enough to actually cause the motor's speed to rise. However, most loads have torque-speed curves whose torque increases with speed, so the increased speed of the motor increases its load, which increases its annature reaction, weakening its flux again. TIle weaker flux causes a further increase in speed, further increasing load, etc., until the motor overspeeds. nlis condition is known as runaway. In motors operating with very severe load changes and duty cycles, this flux-weakening problem can be solved by installing compensating windings. Unfortunately, compensating windings are too expensive for use on ordinary run-of-the-mill motors. TIle solution to the runaway problem employed for less-expensive, less-severe duty motors is to provide a turn or two of cumulative compounding to the motor's poles. As the load increases, the magnetomotive force from the series turns increases, which counteracts the demagnetizing magnetomotive force of the annature reaction. A shunt motor equipped with just a few series turns like this is calJed a stabilized shunt motor. 9.5 THE PERMANENT-MAGNET DC MOTOR A permanent-magnet de (PM DC) motor is a dc motor whose poles are made of pennanent magnets. Permanent-magnet dc motors offer a number of benefits compared with shunt dc motors in some applications. Since these motors do not require an external field circuit, they do not have the field circuit copper losses as-sociated with shunt dc motors. Because no field windings are required, they can be smaller than corresponding shunt dc motors. PMDC motors are especially common in smaller fractional- and subfractional-horsepower sizes, where the ex-pense and space of a separate field circuit cannot be justified. However, PMDC motors also have disadvantages. Pennanent magnets can-not produce as high a flux density as an externally supplied shunt field, so a 560 ELECTRIC MACHINERY RJNDAMENTALS Residual flux density n ... ------------------.f~t-----------------H ~r~) ""GURE 9-19 Coercive Magnetizing intensity He -----(.j (a) The magnetization curve of a typical ferromagnetic material. Note the hysteresis loop. After a large ntagnetizing intensity H is applied to the core and then removed. a residual flux density n ... rentains behind in the core. This flux can be brought to zero if a coercive magnetizing intensity He is applied to the core with the opposite polarity. In this case. a relatively sntall value of it will demagnetize the core. PMDC motor will have a lower induced torque "rind per ampere of annature cur-rent lit than a shunt motor of the same size and construction. In addition, PMDC motors run the risk of demagnetization. As mentioned in Chapler 8, the annature current lit in a dc machine produces an annature magnetic field of its own. The ar-mature mlllf subtracts from the mmf of the poles under some portions of the pole faces and adds to the rnrnfofthe poles under other portions of the pole faces (see Figures 8- 23 and 8- 25), reducing the overall net nux in the machine. This is the armature reaction effect. In a PMDC machine, the pole nux is just the residual flux in the pennanent magnets. If the armature current becomes very large, there is some risk that the armature mmf may demagnetize the poles, pennanenlly re-ducing and reorienting the residual flux in them. Demagnetizalion may also be caused by the excessive heating which can occur during prolonged periods of overload. Figure 9- 19a shows a magnetization curve for a typical ferromagnetic ma-terial. II is a plot of flux density B versus magnelizing intensity H (or equivalently, a plot of flux J • 00 • ~ " • , , • 00 • 0 ~ = (246 VX50 A) _ • (3690r/minXlmin/60sX27Tradlr) - 31.8N m (b) To calculate the complete torque-speed characteristic, we must repeat the steps in a for many values of armature current. A MATLAB M-file that calculates the torque-speed characteristics of the series dc motor is shown below. Note that the magnetization curve used by this program works in terms of field magnetomo-tive force instead of effective field current. % M-fi l e : seri es_t s_curve .m % M-fi l e c r eat e a p l ot of the t orque- speed c urve o f the % the seri es dc mot or with armat ure reaction in % Exampl e 9- 5. % Get the magne tizati on c urve . Thi s fil e cont a in s the % three variabl es mmf_values, ea_values, a nd n_O. l oad fig 9_22 .mat 300 2'" ./" .....-/ II .. '" 1200 rhnin / / 200 / / I'" / II 100 I I 1000 2000 3000 4(xx) 5000 6(XXl 7000 g(XX) 9000 1O.(xx) Field magnetomotive force 3'. A . turns HGURE 9-12 The magnetization curve of the motor in Example 9-5. This curve was taken at speed II,. = 1200 r/min. rx: MmDRS AND GENERATORS 567 % Firs t , i nitia liz e v_t = 250; the values needed in thi s program. % Te rminal voltage (V) r _ 0 i 0 -n_ o 0 0.08; 0 10, 10,300; 0 25; % Armature + fi e l d r es i s tance (ohms) % Armature (line) c urre nt s (A) % Numbe r of series turn s on fi e l d % Ca l culat e the MMF f o r each l oad f=n_s i_a; % Ca l culat e the int e rnal generated voltage e_a . e_a = v_t -i _a r _a; % Ca l culat e the r esulting interna l generated voltage at % 1 200 r / min by int e rpolating the mot or 's mag ne tization % curve. e_aO = int e rpl (mmf_va lues,ea_values, f ,'spline'); % Ca l culat e the motor's speed fr om Equat i on (9 -1 3) . n = (e_a . 1 e_aO) n_O; % Ca l culat e the induced t orque corresponding t o each % speed from Equations (8 - 55 ) and (8 - 56 ) . t _ ind = e_a . i a . 1 (n 2 p i I 60); % Plo t the t o rque - speed curve p l o t (t _ ind, n , ' Co l or' , 'k' , 'LineWi dth', 2 . 0) ; ho l d on; x l abel ( ' \ tau {ind) (N-m) ' , ' Fontwe i ght ' , 'Sol d' ) ; y l abel ( ' \ itn_ {m) \ nn\bf (r I min ) , , 'Fontwe i ght ' , 'Sol d' ) ; titl e ( ' Serie s IX: M ot or To rque - Speed Chara c t e ri s ti c ' , 'Font we i ght , , 'So l d' ) ; axi s( [ 0 700 0 5000 ] ) ; gri d on; ho l d o ff; The resulting motor torque-speed characteristic is shown in Figure 9- 23. Notice the severe overspeeding at very small torques. Speed Control of Series DC Motors Unlike with the shunt dc motor, there is only one efficient way to change the speed of a series dc motor. That method is to change the terminal voltage of the motor. If the terminal voltage is increased, the first term in Equation (9-23) is in-creased, resulting in a higher speed for any given torque. The speed of series dc motors can also be controlled by the insertion of a se-ries resistor into the motor circuit, but this technique is very wasteful of power and is used only for intennittent periods during the start-up of some motors. Until the last 40 years or so, there was no convenient way to change VT, so the only method of speed control available was the wasteful series resistance method. That has all changed today with the introduction of solid-state control circuits. Techniques of obtaining variable terminal voltages were discussed in Chapter 3 and will be considered further later in this chapter. 568 ELECTRIC MACHINERY RJNDAMENTALS 5(XX) 4500 4(xx) 3500 " 3(xx) ~ 2500 • " 2(xx) 1500 I(xx) 500 00 100 2lXl 3O\l 400 '00 6\lll 700 fiDd· N · m ""GURE 9-23 The torque-speed characteristic of the series dc motor in Example 9--5. 9.7 THE COMPOUNDED DC MOTOR A compounded dc motor is a motor with both a shunt and a seriesfield. Such a mo-tor is shown in Figure 9- 24. The dots that appear on the two field coils have the same meaning as the dots on a transfonner: Cu"ent flowing into a dot produces a positive magnetonwtive force. If current fl ows into the dots on both field coils, the resulting magnetomotive forces add to produce a larger total magnetomotive force. This situation is known as cumulative compounding. If current flows into the dot on one field coil and out of the dot on the other field coil, the resulting magnetomotive forces subtract. In Figure 9-24 the round dots correspond to cumulative compound-ing of the motor, and the squares correspond to differential compounding. 1lle Kirchhoff's voltage law equation for a compounded dc motor is Vr = E), + f),(R), + Rs) 1lle currents in the compounded motor are related by fA= IL -1F V T f F = -RF (9- 24) (9- 25) (9- 26) 1lle net magnetomotive force and the effective shunt field current in the com-pounded motor are given by ,nd ~-~~-~ I 9i'oet -9i'F ~ 9i'SE 9i'AR I (9- 27) (9- 28) rx: MmDRS AND GENERATORS 569 (b' FIGURE 9-24 The equiva.lent circuit of contpounded dc motors: (a.) Ions-shunt connection: (b) shon-shunt connection. where the positive sign in the equations is associated with a cumulatively com-pounded motor and the negative sign is associated with a differentially com-pounded motor. The Torque-Speed Characteristic of a Cumulatively Compounded DC Motor In the cumulatively compounded dc motor, there is a component of flux which is constant and another component which is proportional to its annature current (and thus to its load). TIlerefore, the cumulatively compounded motor has a higher starting torque than a shunt motor (whose flux is constant) but a lower starting torque than a series motor (whose entire flux is proportional to armature current). In a sense, the cumulatively compounded dc motor combines the best fea-tures of both the shunt and the series motors. Like a series motor, it has extra torque for starting; like a shunt motor, it does not overspeed at no load. At light loads, the series field has a very small effect, so the motor behaves approximately as a shunt dc motor. As the load gets very large, the series flux 570 ELECTRIC MACHINERY RJNDAMENTALS "m' rhnin Shunt Cumulatively compounded ,~- Series L _ ~ _ fjnd ,,' ,~ Shunt Cumulatively compounded ~----=:::::::=L._ ,~ ,b, ""GURE 9-25 (a) The torque-speed characteristic of a cumulatively compounded dc motor compared to series and shunt motors with the same full-load rating. (b) The torque-speed characteristic of a cumulatively compounded dc motor compared to a shunt motor with the same no-lood speed. becomes quite important and the torque-speed curve begins to look like a series motor's characteristic. A comparison of the torque-speed characteristics of each of these types of machines is shown in Figure 9- 25. To detennine the characteristic curve ofa cumulatively compounded dc mo-tor by nonlinear analysis, the approach is similar to that for the shunt and series motors seen before. Such an analysis will be illustrated in a later example. The Torque-Speed Characteristic of a Differentially Compounded DC Motor In a differentially compounded dc motor, the shunt magnetomotive force and se-ries magnetomotive force subtract from each other. This means that as the load on the motor increases, lit increases and the flux in the motor decreases. But as the flux decreases, the speed of the motor increases. This speed increase causes an-other increase in load, which further increases lit, further decreasing the flux, and increasing the speed again. The result is that a differentially compounded motor is rx: MmDRS AND GENERATORS 571 H GURE 9-26 The torque-speed characteristic of a L _______ fjmd differentially compounded dc motor. unstable and tends to run away. This instability is much worse than that of a shunt motor with armature reaction. It is so bad that a differentially compounded motor is unsuitable for any application. To make matters worse, it is impossible to start such a motor. At starting con-ditions the annature current and the series field current are very high. Since the se-ries flux subtracts from the shunt flux, the series field can actually reverse the mag-netic polarity of the machine's poles. The motor will typically remain still or turn slowly in the wrong direction while burning up, because of tile excessive annature current. When this type of motor is to be started, its series field must be short-circuited, so that it behaves as an ordinary shunt motor during the starting perioo. Because of the stability problems of the differentially compounded de motor, it is almost never intentionally used. However, a differentially compounded motor can result if the direction of power flow reverses in a cumulatively compounded generator. For that reason, if cumulative ly compounded dc generators are used to supply power to a system, they will have a reverse-power trip circuit to disconnect them from the line if the power flow reverses. No motor- generator set in which power is expected to fl ow in both directions can use a differentially compounded motor, and therefore it cannot use a cumulatively compounded generator. A typical terminal characteristic for a differentially compounded dc motor is shown in Figure 9- 26. The Nonlinear Analysis of Compounded DC Motors The determination of the torque and speed of a compounded dc motor is ill us-trated in Example 9--6. EXllmple 9-6. A lOO-hp, 250-V compounded dc motor with compensating wind-ings has an internal resistance, including the series winding, of 0.04 O. There are J(X)O turns per pole on the shunt field and 31lU1lS per pole on the series winding. The machine is shown in Figure 9-27, and its magnetization curve is shown in Figure 9-9. At no load, the field resistor has been adjusted to make the motor run at 1200 r/min. The core, mechanical, and stray losses may be neglected. 572 ELECTRIC MACHINERY RJNDAMENTALS + O.04n < L, N F = " RkJ! lmIlJ. (9--34) Solving for n yields (9--35) where n must be rounded up to the next integer value, since it is not possible to have a fractional number of starting stages. If n has a fractional part, then when the final stage of starting resistance is removed, the armature current of the mo-tor will jwnp up to a value smaller than lmu. In this particular problem, the ratio lminllr=. = 0.5, and RIOt is VT 250 V R~ - -, -= 700 A = 0.357 n m n log (RAiR,,J log (0.05 fIA).357 fi) n= = =2 84 log (lmin1Imax) log (350 MOO A) . The number of stages required will be three. (b) The annature circuit will contain the annature resistor RA and three starting re-sistors RI, R2, and RJ. This arrangement is shown in Figure 9--28. At first, EA = 0 V and IA = 700 A, so _ Vr _ IA - R + R + R + R - 700 A A I 2 J Therefore, the total resistance must be (9--36) This total resistance will be placed in the circuit WItii the ClUTent falls to 350 A. This occurs when EA = Vr -IAR,,,, = 250 V -(350 AX0.357 !l) = 125 V When EA = 125 V,IA has fallen to 350 A and it is time to cut out the first starting resistor RI . When it is cut out, the ClUTent should jump back to 7ooA. Therefore, R + R + R = Vr -EA = 250 V -125 V = 0 1786 n A 2 J I 700 A . mn (9--37) This total resistance will be in the circuit untillA again falls to 350A. This occurs when EA reaches EA = Vr -IAR,,,, = 250 V -(350A)(0.1786!l) = 187.5 V 578 ELECTRIC MACHINERY RJNDAMENTALS When EA = 187.5 V, fA has fallen to 350 A and it is time to cut out the second starting resistor R2. When it is cut out, the current should jump back to 700 A. Therefore, R +R = VT -EA =250V-187.5V =008930 A 1 I 700 A . ~ (9- 38) This total resistance will be in the circuit until fA again falls to 350 A. This occurs when EA reaches EA = VT -fAR,o< = 250V -(350A)(0.08930) = 218.75V When EA = 218.75 V, fA has fallen to 350 A and it is time to cut out the third starting resistor R1. When it is cut out, only the internal resistance of the motor is left. By now, though, RA alone can limit the motor's current to VT -EA 250 V - 218.75 V lAo = R = 0.050 , = 625 A (less than allowed maximrun) From this point on, the motor can speed up by itself. From Equations (9- 34) to (9- 36), the required resistor values can be calculated: RJ = R'OI.3 -RA = 0.08930 - 0.05 0 = 0.03930 R2 = R'0I2 - RJ -RA = 0.1786 0 - 0.0393 0 - 0.05 0 = 0.0893 0 R] = R"".l - R2 - R3 - RA = 0.357 0 - 0.1786 0 - 0.0393 0 - 0.05 0 = 0.1786 0 And RJ, Rl., and RJ are cut out when EA reaches 1 25, 187.5, and 218.75 V, respectively. DC Motor Starting Circuits Once the starting resistances have been selected, how can their shorting contacts be controlled to ensure that they shut at exactly the correct moment? Several dif-ferent schemes are used to accomplish this switching, and two of the most com-mon approaches will be examined in this section. Before that is done, though, it is necessary to intrrxluce some of the components used in motor-starting circuits. Figure 9- 30 illustrates some of the devices commonly used in motor-control circuits. 1lle devices illustrated are fuses, push button switches, relays, time delay relays, and overloads. Figure 9- 30a shows a symbol for a fuse. The fuses in a motor-control cir-cuit serve to protect the motor against the danger of short circuits. They are placed in the power supply lines leading to motors. If a motor develops a short circuit, the fuses in the line leading to it wi ll burn out, opening the circuit before any damage has been done to the motor itself. Figure 9-30b shows spring-type push button switches. There are two basic types of such switches-normally open and nonnally shut. Nonnally open con-tacts are open when the button is resting and closed when the button has been rx: MmDRS AND GENERATORS 579 -~o O~--0 1 0 Normally open Normally closed ,,' FIGURE 9-30 1 T ,b, Normally Nonnally open closed ,d, OL Heater , ., (a) A fuse. (b) Normally open and normally closed push button switches. (c) A relay coil and comacts. (d) A time delay relay and contacts. (e) An overload and its normally closed contacts. pushed, while normally closed contacts are closed when the button is resting and open when the button has been pushed. A relay is shown in Figure 9-30c. It consists of a main coil and a number of contacts.1lle main coil is symbolized by a circle, and the contacts are shown as par-allel1ines. TIle contacts are of two types- nonnally open and nonnally closed. A normally open contact is one which is open when the relay is deenergized, and a normally closed contact is one which is closed when the relay is deenergized. When electric power is applied to the relay (the relay is energized), its contacts change state: 1lle nonnally open contacts close, and the nonnally closed contacts open. A time delay relay is shown in Figure 9- 3Od. It behaves exactly like an or-dinary relay except that when it is energized there is an adjustable time delay be-fore its contacts change state. An overload is shown in Figure 9- 30e. It consists ofa heater coil and some nonnally shut contacts. The current flowing to a motor passes through the heater coils. If the load on a motor becomes too large, then the current flowing to the mo-tor wi 11 heat up the heater coils, which will cause the normally shut contacts of the overload to open. TIlese contacts can in turn activate some types of motor protec-tion circuitry. One common motor-starting circuit using these components is shown in Fig-ure 9- 31. In this circuit, a series of time delay relays shut contacts which remove each section of the starting resistor at approximately the correct time after power is 580 ELECTRIC MACHINERY RJNDAMENTALS + 1 1 F, F, R .. R, L, FL E, M R"., M + -OL F, lID 2TD 3TO F, SO", ~ Srop I 0 J , FL OL lID M 2ID ITO 3ID 2TO ""GURE 9-31 A dc motor starting circuit rising time delay relays to cut out the starting resistor. applied to the motor. When the start button is pushed in this circuit, the motor's ar-mature circuit is connected to its power supply, and the machine starts with all re-sistance in the circuit. However, relay ITO energizes at the same time as the motor starts, so after some delay the ITO contacts will shut and remove part of the start-ing resistance from the circuit. Simultaneously, relay 2m is energized, so after an-other time delay the 2TD contacts wil I shut and remove the second part of the tim-ing resistor. When the 2TD contacts shut, the 3TD relay is energized, so the process repeats again, and finally the motor runs at full speed with no starting resistance present in its circuit. I fthe time delays are picked properly, the starting resistors can be cut out at just the right times to limit the motor's current to its design values. rx: MmDRS AND GENERATORS 581 + I I Radj OL S"rt FL OL ~ Stop ~ J~I O r------{0 M IA }---1 IAR ~-j~---------{ 2A }---1 2AR 3AR ~-Ie---------~ 3A }--~ FIGURE 9-32 (a) A de motor starting circuit using oountervoltage-sensing relays to eut out the starting resistor. Another type of motor starter is shown in Figure 9- 32. Here, a series of re-lays sense the value of Ell in the motor and cut out the starting resistance as Ell rises to preset levels. This type of starter is better than the previous one, since if the motor is loaded heavily and starts more slowly than normal, its armature re-sistance is still cut out when its current falls to the proper value. Notice that both starter circuits have a relay in the field circuit labeled FL. This is afield loss relay. If the field current is lost for any reason, the field loss 582 ELECTRIC MACHINERY RJNDAMENTALS I, I A 2A 3 A 700 A k----'T'----~T----"-''------" I, I, ,b , ""GURE 9- 32 (conclud ... >d) (b) The armature current in a dc motor during starting. relay is deenergized, which turns off power to the M relay. When the M relay deenergizes, its nonnally open contacts open and disconnect the motor from the power supply. This relay prevents the motor from running away if its field current is lost. Notice also that there is an overload in each motor-starter circuit. If the power drawn from the motor becomes excessive, these overloads will heat up and open the OL nonnally shut contacts, thus turning off the M relay. When the M re-lay deenergizes, its nonnally open contacts open and disconnect the motor from the power suppl y, so the motor is protected against damage from prolonged ex-cessive loads. 9.9 THE WARD·LEONARD SYSTEM AND SOLID·STATE SPEED CONTROLLERS TIle speed of a separately excited, shunt, or compounded dc motor can be varied in one of three ways: by changing the field resistance, changing the annature volt-age, or changing the armature resistance. Of these methods, perhaps the most use-ful is annature voltage control, since it permits wide speed variations without af-fecting the motor's maximum torque. A number of motor-control systems have been developed over the years to take advantage of the high torques and variable speeds available from the anna-ture voltage control ofdc motors. In the days before solid-state electronic compo-nents became available, it was difficult to produce a varying dc voltage. In fact, the nonnal way to vary the armature voltage of a dc motor was to provide it with its own separate dc generator. An armature voltage control system of this sort is shown in Figure 9- 33. TIlis figure shows an ac motor serving as a prime mover for a dc generator, which rx: MmDRS AND GENERATORS 583 IX generator rx: motor R" I " I " R" + + wm E" v" Vn E" Three-phase motor (induction or synchronous) Rn Ln Rn Ln In I In I Three-phase rectifier Three-phase rectifier and control circuit and control circuit (a) + -Switch to , , ' IX "" reverse power >' connections ' ' , ' weroutput + -D, D, D, D, D, D. Three-phase input power ,b , FIGURE 9-33 (a) A Ward-Leonard system for dc motor speed control. (b) The circuit for producing field current in the dc generator and dc motor. in turn is used to supply a dc voltage to a dc motor. Such a system of machines is called a Ward-Leonard system, and it is extremely versatile. In such a motor-control system, the annature voltage of the motor can be controlled by varying the field current of the dc generator. This annature voltage 584 ELECTRIC MACHINERY RJNDAMENTALS Generator operation (r reversed and ro normal) - rind Motor operation (both rand ro reversed) ""GURE 9-34 Motor operation (normal r andro) Torque-speed curves Generator operation (r normal and ro reversed) The operating range of a Ward-Leonan:l motor-control system. The motor can operate as a motor in either the forward (quadram 1) or reverse (quadrant 3) direction and it can also regenerate in quadrams 2 and 4. allows the motor's speed to be smoothly varied between a very small value and the base speed. The speed of the motor can be adjusted above the base speed by reducing the motor's field current. With such a flexible arrangement, total motor speed control is possible. Furthermore, if the field current of the generator is reversed, then the polar-ity of the generator's annature voltage will be reversed, too. This will reverse the motor's direction of rotation. Therefore, it is possible to get a very wide range of speed variations in either direction of rotation out of a Ward-Leonard dc motor-control system. Another advantage of the Ward-Leonard system is that it can "regenerate," or return the machine's energy of motion to the supply lines. If a heavy load is first raised and then lowered by the dc motor ofa Ward-Leonard system, when the load is falling, the dc motor acts as a generator, supplying power back to the power system. In this fashion, much of the energy required to lift the load in the first place can be recovered, reducing the machine's overall operating costs. TIle possible modes of operation of the dc machine are shown in the torque- speed diagram in Figure 9- 34. When this motor is rotating in its nonnal direction and supplying a torque in the direction of rotation, it is operating in the first quadrant of this figure. If the generator's field current is reversed, that will re-verse the tenninal voltage of the generator, in turn reversing the motor's annature voltage. When the armature voltage reverses with the motor field current remain-ing unchanged, both the torque and the speed of the motor are reversed, and the machine is operating as a motor in the third quadrant of the diagram. If the torque or the speed alone of the motor reverses while the other quantity does not, then the machine serves as a generator, returning power to the dc power system. Because rSCR] Th~ r ph= SCR., input FIGURE 9-35 lr SCR2 l,-SCR~ I I I, Operation "0' possible lr SCRJ V- SC~ (., (b, rx: MmDRS AND GENERATORS 585 ,Free-wheeling diode) D, v, K. Operation "oj possible + I v, E,C) -(a) A two-quadrant solid-state dc motor controller. Since current cannot flow out of the positive terminals of the armature. this motor cannot act as a generator. returning power to the system. (b) The possible operating quadrants of this motor controller. a Ward-Leonard system pennits rotation and regeneration in either direction, it is called afour-quadrant control system. The disadvantages of a Ward-Leonard system should be obvious. One is that the user is forced to buy three full machines of essentially equal ratings, which is quite expensive. Another is that three machines will be much less efficient than one. Because of its expense and relatively low efficiency, the Ward-Leonard sys-tem has been replaced in new applications by SCR-based controller circuits. A simple dc armature voltage controller circuit is shown in Figure 9- 35. The average voltage applied to the armature of the motor, and therefore the aver-age speed of the motor, depends on the fraction of the time the supply voltage is applied to the armature. This in turn depends on the relative phase at which the 586 ELECTRIC MACHINERY RJNDAMENTALS -. Th ph inp '" m t I " -Generator (regeneration) -... ""GURE 9-36 (a) ,b, V +\ V, E, _ I Motor Generator __ (regeneration) -I' + -(a) A four-quadrant solid-state dc motor controller. (b) The possible operating quadrants of this motor controller. SCRs in the rectifier circuit are triggered. This particular circuit is only capable of supplying an annature voltage with one polarity, so the motor can only be re-versed by switching the polarity of its field connection. Notice that it is not possi-ble for an annature current to now out the positive terminal of this motor, since current cannot now backward through an SCR. nlerefore, this motor cannot re-generate, and any energy supplied to the motor cannot be recovered. This type of control circuit is a two-quadrant controller, as shown in Figure 9- 35b. A more advanced circuit capable of supplying an annature voltage with ei-ther polarity is shown in Figure 9- 36. n lis annature voltage control circuit can rx: MmDRS AND GENERATORS 587 (a) FIGURE 9-37 (a) A typical solid-state shunt dc motor drive. (Courtesy of MagneTek, Inc. ) (b) A close-up view of the low-power electronics circuit board. showing the adjustmems for current limits. acceleration rate. deceleration rate. minimum speed. and maximum speed. (Courtesy of MagneTel;. Inc.) pennit a current fl ow out of the positive terminals of the generator, so a motor with this type of controller can regenerate. If the polarity of the motor field circuit can be switched as well, then the solid-state circuit is a full four-quadrant con-troller like the Ward-Leonard system. A two-quadrant or a full four-quadrant controller built with SCRs is cheaper than the two extra complete machines needed for the Ward-Leonard system, so solid-state speed-control systems have largely displaced Ward-Leonard systems in new applications. A typical two-quadrant shunt dc motor drive with armature voltage speed control is shown in Figure 9- 37, and a simplified block diagram of the drive is shown in Figure 9- 38. nlis drive has a constant field voltage supplied by a three-phase full -wave rectifier, and a variable annature terminal voltage supplied by six SCRs arranged as a three-phase full-wave rectifier. The voltage supplied to the ar-mature of the motor is controlled by adjusting the firing angle of the SCRs in the bridge. Since this motor controller has a fixed field voltage and a variable anna-ture voltage, it is only able to control the speed of the motor at speeds less than or equal to the base speed (see "Changing the Armature Voltage" in Section 9.4). The controller circuit is identical with that shown in Figure 9- 35, except that all of the control electronics and feedback circuits are shown. ~ ~ : Pr~-;ct~~ci~i~ -----;l~~ ---: t-: (Protective devices Current-1 : tied to fault relay) In;~~ng ! : Sample foc 1 1 trips J L___ _ 341 P()',\o"er 3-phase , I Full-wave bridge ~I --, (diodes) Current feedback Voc Sync voltage T 341 P()',\o"er 1-------Speed<: i Low-pow~;I.;;:;~~;T -~ adj ~ Accelerutionl I I : deceleration r--s",,,, regulator Current ---, I I Firing I ! • circuit r-r-: I---I OC~~ rx: motor 1 Three-phase 1 full wave SCR-bridge Main H V limiter I L --,-FlGURE 9-38 I I _ I T achometer speed feedback contacts -----------------4---, r-t~--r-~--Start stop circuit Fault relay Run -.l .:L I R" I -l---colf-t-T ~ re~a! I , 1 _ -.l L _____ _ I Tachometer Shunt field A simplified block diagram of the t)pical solid-state shunt dc motoc drive shown in Figure 9--37. (Simplified/rom a block diagmm provided l7y MagneTek, Inc.) rx: MmDRS AND GENERATORS 589 The major sections of this dc motor drive include: I. A protection circuit section to protect the motor from excessive armature cur-rents. low tenninal voltage. and loss of field current. 2. A start/stop circuit to connect and disconnect the motor from the line. 3. A high-power electronics section to convert three-phase ac power to dc power for the motor's annature and field circuits. 4. A low-power electronics section to provide firing pulses to the SCRs which supply the annature voltage to the motor. This section contains several major subsections, which will be described below. Protection Circuit Section The protection circuit section combines several different devices which together ensure the safe operation of the motor. Some typical safety devices included in this type of drive are I. Current-limiting fuses, to disconnect the motor quickly and safely from the power line in the event of a short circuit within the motor. Current-limiting fuses can interrupt currents of up to several hundred thousand amperes. 2. An instantaneous static trip, which shuts down the motor if the annature cur-rent exceeds 300 percent of its rated value. If the armature current exceeds the maximum allowed value, the trip circuit activates the fault relay, which deenergizes the run relay, opening the main contactors and disconnecting the motor from the line. 3. An inverse-time overload trip, which guards against sustained overcurrent conditions not great enough to trigger the instantaneous static trip but large enough to damage the motor if allowed to continue indefinitely. TIle term in-verse time implies that the higher the overcurrent flowing in the motor, the faster the overload acts (Figure 9- 39). For example, an inverse-time trip might take a full minute to trip if the current flow were 150 percent of the rated current of the motor, but take 10 seconds to trip if the current now were 200 percent of the rated current of the motor. 4. An undefi!oltage trip, which shuts down the motor if the line voltage supply-ing the motor drops by more than 20 percent. 5. Afield loss trip, which shuts down the motor if the field circuit is lost. 6. An overtemperature trip, which shuts down the motor if it is in danger of overheating. StartlStop Circuit Section The start/stop circuit section contains the controls needed to start and stop the mo-tor by opening or closing the main contacts connecting the motor to the line. The motor is started by pushing the run button, and it is stopped either by pushing the 590 ELECTRIC MACHINERY RJNDAMENTALS I 3/",oed I",oed ----------------------------"--':10'---'2"0--30 "'---'40"- ': 50 '---'60"-- Trip time. s ""GURE 9-39 An inverse-time trip characteristic. stop button or by energizing the fault relay. In either case, the run relay is deener-gized, and the main contacts connecting the motor to the line are opened. High-Power Electronics Section The high-power electronics section contains a three-phase full-wave diode recti-fier to provide a constant voltage to the fie ld circuit of the motor and a three-phase full-wave SCR rectifier to provide a variable voltage to the annature circuit of the motor. Low-Power Electronics Section TIle low-power electronics section provides firing pulses to the SCRs which sup-ply the annature voltage to the motor. By adjusting the firing time of the SCRs, the low-power electronics section adjusts the motor's average armature voltage. TIle low-power electronics section contains the following subsystems: I. Speed regulation circuit. TIlis circuit measures the speed of the motor with a tachometer, compares that speed with the desired speed (a reference voltage level), and increases or decreases the annature voltage as necessary to keep the speed constant at the desired value. For exrunple, suppose that the load on the shaft of the motor is increased. If the load is increased, then the motor will slow down. TIle decrease in speed wi II reduce the voltage generated by the tachometer, which is fed into the speed regulation circuit. Because the voltage level corresponding to the speed of the motor has fallen below the reference voltage, the speed regulator circuit will advance the firing time of the SCRs, producing a higher annature voltage. The higher armature voltage will tend to increase the speed of the motor back to the desired level (see Figure 9-40). rx: MmDRS AND GENERATORS 591 Speed adjustment potentiometer +~,J 1--------1 -----' C' -_0 + + I I I Voltage I V .. r I I I regulator I t--'~ c-' -~O -, , l ____ -.l (V,oo:b ex speed) Tachometer rx: motor , ., 2 " , ,b, FIGURE 9-40 (a) The speed regulator cin:uit produces an output voltage which is proportional to the difference between the desired speed of the motor (set by V tor (measured by Vtoob). This output voltage is applied to the firing cin:uit in such a way that the larger the output voltage becomes, the earlier the SCRs in the drive turn on and the higher the average terminal voltage becomes. (b) The effect of increasing load on a shunt dc motor with a speed regulator. The load in the nx>tor is increased. If no regulator were present. the motor would slow down and operate at point 2. When the speed regulator is present. it detects the decrease in speed and boosts the armature voltage of the motor to compensate. This raises the whole torque-speed characteristic curve of the motor. resulting in operation at point 2'. With proper design, a circuit of this type can provide speed regulations of 0. 1 percent between no-load and full-load conditions. 1lle desired operating speed of the motor is controlled by changing the ref-erence voltage level. The reference voltage level can be adjusted with a small potentiometer, as shown in Figure 9-40. 592 ELECTRIC MACHINERY RJNDAMENTALS 2. Current-limiting circuit. This circuit measures the steady-state current flow-ing to the motor, compares that current with the desired maximum current (set by a reference voltage level), and decreases the annature voltage as nec-essary to keep the current from exceeding the desired maximum value. The desired maximum current can be adjusted over a wide range, say from 0 to 200 percent or more of the motor's rated current. This current limit should typically be set at greater than rated current, so that the motor can accelerate under full-l oad conditions. 3. Acceleration/deceleration circuit. TIlis circuit limits the acceleration and de-celeration of the motor to a safe val UC. Whenever a dramatic speed change is commanded, this circuit intervenes to ensure that the transition from the orig-inal speed to the new speed is smooth and does not cause an excessive anna-ture current transient in the motor. TIle acceleration/deceleration circuit completely eliminates the need for a starting resistor, since starting the motor is just another kind of large speed change, and the acceleration/deceleration circuit acts to cause a smooth increase in speed over time. TIlis gradual smooth increase in speed limits the current flow-ing in the machine's annature to a safe value. 9.10 DC MOTOR EFFICIENCY CALCULATIONS To calcu late the efficiency of a dc motor, the following losses must be detennined: I. Copper losses 2. Brush drop losses 3. Mechanical losses 4. Core losses 5. Stray losses TIle copper losses in the motor are the { 2R losses in the armature and field circuits of the motor. These losses can be found from a knowledge of the currents in the machine and the two resistances. To determine the resistance of the anna-ture circuit in a machine, block its rotor so that it cannot turn and apply a small dc voltage to the armature tenninals. Adjust that voltage until the current flowing in the annature is equal to the rated annature current of the machine. TIle ratio of the applied voltage to the resulting armature current fl ow is Rio.' The reason that the current should be about equal to full-load value when this test is done is that Rio. varies with temperature, and at the full-l oad value of the current, the annature windings will be near their normal operating temperature. TIle resulting resistance will not be entirely accurate, because I. The cooling that normally occurs when the motor is spinning will not be present. rx: MmDRS AND GENERATORS 593 2. Since there is an ac voltage in the rotor conductors during nonnal operation, they suffer from some amount of skin effect, which further raises armature resistance. IEEE Standard 11 3 (Reference 5) deals with test procedures for dc machines. It gives a more accurate procedure for determining RA , which can be used if needed. The field resistance is detennined by supplying the full-rated field voltage to the field circuit and measuring the resulting field current. TIle field resistance RF is just the ratio of the field voltage to the field current. Brush drop losses are often approximately lumped together with copper losses. If they are treated separately, they can be detennined from a plot of contact potential versus current for the particular type of brush being used. The brush drop losses are just the product of the brush voltage drop V BO and the annature current IA. The core and mechanical losses are usually detennined together. If a motor is allowed to tum freely at no load and at rated speed, then there is no output power from the machine. Since the motor is at no load, IA is very small and the annature copper losses are negligible. Therefore , if the field copper losses are subtracted from the input power of the motor, the remaining input power must consist of the mechanical and core losses of the machine at that speed. TIlese losses are called the no-load rotational losses of the motor. As long as the motor's speed remains nearly the same as it was when the losses were measured, the no-load rotational losses are a gD<Xl estimate of mechanical and core losses under load in the machine. An example of the detennination of a motor's efficiency is given below. Example 9-8. A SO-hp. 2S0-V. 1200 rlmin shunt dc motor has a rated armature current of 170 A and a rated field current of 5 A. When its rotor is blocked. an armature voltage of 10.2 V (exclusive of brushes) produces 170 A of current flow. and a field volt-age of2S0 V produces a field current flow of S A. The brush voltage drop is assumed to be 2 V. At no load with the terminal voltage equal to 240 V. the annature current is equal to 13.2 A. the field current is 4.8 A. and the motor's speed is IISO r/min. (a) How much power is output from this motor at rated conditions? (b) What is the motor's efficiency? Solutioll The armature resistance of this machine is approximately RA = 10.2 V =006 0 170A . and the field resistance is R ~=2S0 V =50 0 , 5 A Therefore, at full load the armature [2R losses are P A = (l70A)2(0.06!l) = 1734 W and the field circuit [ 2R losses are 594 ELECTRIC MACHINERY RJNDAMENTALS P F = (SA:Y(500) = 12S0W The brush losses at full load are given by P btuob = VBofA = (2 V)(170A) = 340W The rotational losses at full load are essentially equivalent to the rotational losses at no load, since the no-load and full-load speeds of the motor do not differ too greatly. These losses may be ascertained by detennining the input power to the armature circuit at no load and assuming that the annature copper and brush drop losses are negligible, meaning that the no-load armature input power is equal to the rotationa1losses: P,ot. = POOle + P = (240 VX13.2 A) = 3168W (a) The input power of this motor at the rated load is given by P~ = VrlL = (250VXI7SA) = 43,7S0W Its output power is given by P 00' = Pm - Pb<wb - P <:U -P , + R, v, A simplified equivalent ein:uit of a de generator. with RF eombining the resistances of the field coils and the variable control resistor. 9.12 THE SEPARATELY EXCITED GENERATOR A separately excited dc generator is a generator whose field current is supplied by a separate external dc voltage source. The equivalent circuit of such a machine is shown in Figure 9-44. In this circuit, the voltage VT represents the actual voltage measured at the tenninals of the generator, and the current II. represents the cur-rent flowing in the lines connected to the terminals. The internal generated volt-age is Ell, and the armature current is lit- It is clear that the annature current is equal to the line current in a separately excited generator: The Terminal Characteristic of a Separately Excited DC Generator (9-40) TIle terminnl characteristic of a device is a plot of the output quantities of the de-vice versus each other. For a dc generator, the output quantities are its tenninal volt-age and line current. llle tenninal characteristic of a separately excited generator is rx: MmDRS AND GENERATORS 597 I, -,-----".fII'V--~+ + v, FIGURE 9-44 A separately excited de generator. thus a plot of VT versus IL for a constant speed w. By Kirchhoff's voltage law, the terminal voltage is (9-41) Since the internal generated voltage is independent of lA, the tenninal characteris-tic of the separately excited generator is a straight line, as shown in Figure 9-45a. What happens in a generator of this sort when the load is increased? When the load supplied by the generator is increased, IL (and therefore IA) increases. As the annature current increases, the lARA drop increases, so the terminal voltage of the generator falls. This tenninal characteristic is not always entirely accurate. In generators without compensating windings, an increase in IA causes an increase in armature reaction, and armature reaction causes flux weakening. This flux weakening causes a decrease in EA = K<p J..w which further decreases the tenninal voltage of the generator. TIle resulting tenninal characteristic is shown in Figure 9-45b. In all future plots, the generators will be assumed to have compensating windings unless stated otherwise. However, it is important to realize that annature reaction can modify the characteristics if compensating windings are not present. Control of Terminal Voltage The tenninal voltage of a separately excited dc generator can be controlled by changing the internal generated voltage EA of the machine. By Kirchhoff's volt-age law VT = EA -lARA, so if EA increases, VT will increase, and if EA decreases, VT will decrease. Since the internal generated voltage EA is given by the equation EA = K<pw, there are two possible ways to control the voltage of this generator: I. Change the speed of rotation. If w increases, then EA = K in the machine increases. As the nux rises, EA = K\iw must rise too, so VT = EA i -lARA increases. In many applications, the speed range of the prime mover is quite limited, so the tenninal voltage is most commonly controlled by changing the field cur-rent. A separately excited generator driving a resistive load is shown in Figure 9-46a. Figure 9-46b shows the effect of a decrease in field resistance on the ter-minal voltage of the generator when it is operating under a load. Nonlinear Analysis of a Separately Excited DC Generator Because the internal generated voltage of a generator is a nonlinear function of its magnetomotive force, it is not possible to calculate simply the value of EA to be expected from a given field current. TIle magnetization curve of the generator must be used to accurately calculate its output voltage for a given input voltage. In addition, if a machine has armature reaction, its nux will be reduced with each increase in load, causing EA to decrease. TIle only way to accurately deter-mine the output voltage in a machine with annature reaction is to use graphical analysis. TIle total magnetomotive force in a separately excited generator is the field circuit magnetomotive force less the magnetomotive force due to annature reac-tion (AR): + v, FIGURE 9-46 I, v, v' , R, L, ( ?-1 (a) VT -------- "~"-\ 'bJ rx: MmDRS AND GENERATORS 599 R, I, + E, v, R~ (a) A separately excited de generator with a resistive load. (b) The effect of a decrease in field resistance on the output voltage of the generator. (9-42) As with de motors, it is customary to define an equivalent field current that would produce the same output voltage as the combination of all the magnetomotive forces in the machine. The resulting voltage EAO can then be detennined by locat-ing that equivalent field current on the magnetization curve. The equivalent field current of a separately excited de generator is given by • "'AR IF=IF-N F (9-43) Also, the difference between the speed of the magnetization curve and the real speed of the generator must be taken into account using Equation (9- 13): EA _!!.. EAO no (9- 13) The following example illustrates the analysis of a separately excited de generator. 600 ELECTRIC MACHINERY RJNDA MENTALS + v" .. I, 0.05.n I, 0-300.\1 ~ + 20fi R, R, VF", 430 V (+) E, V, NF",!()(Xlturns L, ""GURE 9-47 The separately excited dc generator in Example 9--9. Example 9-9. A separately excited dc generator is rated at 172 kW. 430 V. 400 A. and 1800 r/min.1t is shown in Figure 9-47. and its magnetization curve is shown in Fig-ure 9-48. This machine has the following characteristics: RA = 0.05 n RF = 20 n Rodj = Ot03oo n VF = 430V NF = J()(X) turns per pole (a) If the variable resistor Rodj in this generator's field circuit is adjusted to 63 n and the generator's prime mover is driving it at 1600 rlmin, what is this generator's no-load tenninal voltage? (b) What would its voltage be if a 360-A load were connected to its tenninals? As-swne that the generator has compensating windings. (c) What would its voltage be if a 360-A load were connected to its terminals but the generator does not have compensating windings? Asswne that its annature reaction at this load is 450 A · turns. (d) What adjustment could be made to the generator to restore its tenninal voltage to the value found in part a? (e) How much field current would be needed to restore the terminal voltage to its no-load value? (Assume that the machine has compensating windings.) What is the required value for the resistor R ..Jj to accomplish this? Solutio" (a) If the generator's total field circuit resistance is RF + R ..Jj = 83 n then the field current in the machine is V F 430 V IF = RF = 83 n = 5.2 A From the machine's magnetization curve, this much current would produce a voltage EAO = 430 V at a speed of 1800 r/min. Since this generator is actually turning at n .. = 1600 rlmin, its internal generated voltage EA will be E, " = (9- 13) 500 450 430 410 400 300 200 100 o I I rx: MmDRS AND GENERATORS 601 /' / V 0.0 1.0 2.0 3.0 4.0 /5.0 6.0 7.0 8.0 9.0 10.0 4.75 5.2 6.15 Field current. A Note: When the field current is zero, E" is about 3 V. FIGURE 9-48 The magnetization curve for the generator in Example 9--9. E = 1600 r/m~n 430 V = 382 V " 1800 r/mm Since Vr = E" at no-load conditions, the output voltage of the generator is Vr = 382 V. (b) If a 360-A load were cotUlected to this generator's terminals, the tenninal volt-age of the generator would be Vr = E" -I"R" = 382 V - (360 AXO.05 0 ) = 364 V (c) If a 360-A load were connected to this generator's terminals and the generator had 450 A ° turns of armature reaction, the effective field current would be 1 = I _ ~AR = S.2 A _ 450A oturns = 4.75A F F N F J(X)() turns 602 ELECTRIC MACHINERY RJNDAMENTALS From the magnetization curve, EAO = 410 V, so the internal generated voltage at 1600 rlmin would be = EAO no E = 1600 r/m~n 410 V = 364 V A 1800 r/mlll Therefore, the tenninal voltage of the generator would be Vr = EA -lARA = 364 V - (360 AXO.05 0) = 346 V It is lower than before due to the annature reaction. (9-13) (d) The voltage at the tenninals of the generator has fallen, so to restore it to its original value, the voltage of the generator must be increased. This requires an increase in EA, which implies that R..tj must be decreased to increase the field current of the generator. (e) For the terminal voltage to go back up to 382 V, the required value of EA is EA = Vr + lARA = 382 V + (360 AXO.05 0) = 400 V To get a voltage EA of 400 V at n .. = 1800 rlmin would be 1600 rlmin, the equivalent voltage at EA _ .!!. (9-13) E = 18oor/m~n 400 V = 450 V AO 1600 r/mlll From the magnetization curve, this voltage would require a field current of IF = 6.15 A. The field circuit resistance would have to be V, RF + Radj = T , 430V 200 + Radj = 6.15A = (:f).90 Radj = 49.90 - 500 Notice that, for the same field current and load current, the generator with annature reaction had a lower output voltage than the generator without annature reaction. The annature reaction in this generator is exaggerated to illustrate its ef-fects- it is a good deal smaller in well-designed modem machines. 9.13 THE SHUNT DC GENERATOR A shunt dc generator is a de generator that supplies its own field current by hav-ing its field connected directly across the terminals of the machine. The equi va-lent circuit of a shunt dc generator is shown in Figure 9-49. In this circuit, the ar-mature current of the machine supplies both the field circuit and the load attached to the machine: (9-44) + E, I, " -R, "1 /. IA = IF+h Vr = EA - lARA V, IF = -R, I, -v., L, rx: MmDRS AND GENERATORS 603 + v, HGURE 9-49 The equivalent circuit of a shunt de generator. The Kirchhoff's voltage law equation for the armature circuit of this machine is (9-45) This type of generator has a distinct advantage over the separately excited dc generator in that no external power supply is required for the field circuit. But that leaves an important question unanswered: If the generator supplies its own field current, how does it get the initial field nux to start when it is first turned on? Voltage Buildup in a Shunt Gener ator Assume that the generator in Figure 9-49 has no load connected to it and that the prime mover starts to turn the shaft of the generator. How does an initial voltage appear at the terminals of the machine? The voltage buildup in a dc generator depends on the presence ofa residual flux in the poles of the generator. When a generator first starts to turn, an internal voltage will be generated which is given by EA = K\ Rl (the critical resistance). then the generator's voltage will never build up. 3. The field resistance may be adjusted to a value greater than the critical re-sistance. To understand this problem, refer to Figure 9- 51. Nonnally, the shunt generator will build up to the point where the magnetization curve in-tersects the field resistance line. If the field resistance has the value shown at Rl in the figure, its line is nearly parallel to the magnetization curve. At that point, the voltage of the generator can fluctuate very widely with only tiny changes in RF or lit. nlis value of the resistance is called the critical resis-tance. If RF exceeds the critical resistance (as at R3 in the figure), then the steady-state operating voltage is essentially at the residual level, and it never builds up. The solution to this problem is to reduce Rr. Since the voltage of the magnetization curve varies as a function of shaft speed, the critical resistance also varies with speed. In general, the lower the shaft speed, the lower the critical resistance. The Terminal Characteristic of a Shunt DC Generator The terminal characteristic of a shunt dc generator differs from that of a separately excited dc generator, because the amount of field current in the machine depends on its terminal voltage. To understand the terminal characteristic of a shunt gen-erator, start with the machine unloaded and add loads, observing what happens. As the load on the generator is increased, II- increases and so lit = IF + IL i also increases. An increase in lit increases the annature resistance voltage drop IItRIt, causing Vr = Elt - lit i RIt to decrease. This is precisely the same behavior observed in a separately excited generat.or. However, when Vr decreases, the field current in the machine decreases with it. This causes the flux in the machine to 606 ELECTRIC MACHINERY RJNDAMENTALS v, ---=::---===--------------}- :~A --- I -;i:Id weakening effect L-____________________________ ~ ""GURE 9-52 The terminal characteristic of a shunt dc generator. decrease, decreasing EA- Decreasing EA causes a further decrease in the terminal voltage Vr = EA J.. - lARA' TIle resulting terminal characteristic is shown in Figure 9-52. Notice that the voltage drop-off is steeper than just the lARA drop in a sepa-rately excited generator. In other words, the voltage regulation of this generator is worse than the voltage regulation of the same piece of equipment connected sep-aratelyexcited. Voltage Control for a Shunt DC Generator As with the separately excited generator, there are two ways to control the voltage of a shunt generator: I. Change the shaft speed W m of the generator. 2. Change the field resistor of the generator, thus changing the field current. Changing the field resistor is the principal method used to control tenninal voltage in real shunt generators. If the field resistor RF is decreased, then the field current IF = VrIRFJ.. increases. When IF increases, the machine's flux \6/ -r~ , , EA versus IF '-________________ -.L-________ ~ 11'01 FIGURE 9-53 Graphical analysis of a shunt dc generator with contpensating windings. Figure 9- 53 shows a magnetization curve for a shunt dc generator drawn at the actual operating speed of the machine. The field resistance RF> which is just equal to VTIIF> is shown by a straight line laid over the magnetization curve. At no load, VT = E A and the generator operates at the voltage where the magnetization curve intersects the field resistance line. The key to understanding the graphical analysis of shunt generators is to re-member Kirchhoff's voltage law (KVL): (9-45) (9-46) The difference between the internal generated voltage and the tenninal voltage is just the lARA drop in the machine. The line of a 11 possible values of EA is the mag-netization curve, and the line of all possible tenninal voltages is the resistor line (IF = VT IRF)· Therefore, to find the tenninal voltage for a given load, just deter-mine the lARA drop and locate the place on the graph where that drop fits exactly between the E A line and the V T line. There are at most two places on the curve where the lARA drop will fit exactly. If there are two possible positions, the one nearer the no-load voltage will represent a normal operating point. A detailed plot showing several di fferent points on a shunt generator's char-acteristic is shown in Figure 9- 54. Note the dashed line in Figure 9- 54b. lllis line is the terminal characteristic when the load is being reduced. llle reason that it does not coincide with the line of increasing load is the hysteresis in the stator poles of the generator. 608 ELECTRIC MACHINERY RJNDAMENTALS , L V, 1. lARA drop ---I V VIT> - ~ , :; , ,". , , , I , 1 ".// VV / /,/ /// , , ~ ,/ , , I, ,,' ,b, ""GURE 9-54 Graphical derivation of the terminal characteristic of a shunt dc generator. If annature reaction is present in a shunt generator, this process becomes a little more complicated. The armature reaction produces a demagnetizing magne-tomotive force in the generator at the same time that the lARA drop occurs in the machine. To analyze a generator with annature reaction present, assume that its ar-mature current is known. Then the resistive voltage drop lARA is known, and the demagnetizing magnetomotive force of the annature current is known.1lle tenni-nal voltage of this generator must be large enough to supply the generator's nux after the demagnetizing effects of armature reaction have been subtracted. To meet this requirement both the annature reaction magnetomotive force and the lARA drop must fit between the Ell. line and the VT line. To detennine the output voltage for a given magnetomotive force, simply locate the place under the mag-netization curve where the triangle formed by the armature reaction and lARA ef-fects exactly fits between the line of possible VT values and the line of possible Ell. values (see Figure 9-55). 9.14 THE SERIES DC GENERATOR A series dc generator is a generator whose field is connected in series with its ar-mature. Since the annature has a much higher current than a shunt field, the series field in a generator of this sort will have only a very few turns of wire, and the wire used will be much thicker than the wire in a shunt field. Because magneto-motive force is given by the equation'?} = NI, exactly the same magnetomotive force can be produced from a few turns with high current as can be produced from many turns with low current. Since the fu ll-load current flows through it, a series field is designed to have the lowest possible resistance. 1lle equivalent circuit of a series dc generator is shown in Figure 9-56. Here, the annature current, field rx: MmDRS AND GENERATORS 609 Ell. '" Vr at no load r ---------:;::'J ......... C lARA drop Ell. with load r---"-''----'-''<;::/lr( Ell. versus IF Vr with load f------y"-"'k;Y Demagnetizing mmf (converted to an equivalent field current) '-------------------------- ~ FIGURE 9-55 Graphical analysis of a shunt dc generator with annature reaction. (NSF. turns) + v, HGURE 9-S6 111. "' ls"'IL Vr",EA -IA(RA+Rs) The equivalent circuit of a series dc generator. currenl, and line currenl all have the same value. The Kirchhoff's voltage law equation for this machine is (9-47) The Terminal Characteristic of a Series Generator The magnetization curve of a series dc generator looks very much like the magne-tization curve of any other generator. At no load, however, there is no field current, so Vr is reduced to a small level given by the residual nux in the machine. As the load increases, the field current rises, so E ll. rises rapidly. The iA(RA + Rs) drop goes up too, but at first the increase in Ell. goes up more rapidly than the iA(RA + Rs) drop rises, so Vr increases. After a while, the machine approaches saturation, and Ell. 610 ELECTRIC MACHINERY RJNDAMENTALS / / b / / / ~~---T)-/","'" I 111. (RA + Rs) drop / / V, '--------------- h (= Is= 111.) ""GURE 9-57 Derivation of the terminal characteristic for a series dc generator. V, Armature reaction '--__ -'L-~ ""GURE 9-58 A series generator tenninal characteristic with large armature reaction effects. suitable for electric welders. becomes almost constant. At that point, the resistive drop is the predominant effect, and VT starts to fall. lllis type of characteristic is shown in Figure 9- 57. It is obvious that this machine would make a bad constant-voltage source. In fact, its voltage regulation is a large negative number. Series generators are used only in a few specialized applications, where the steep voltage characteristic of the device can be exploited. One such application is arc welding. Series generators used in arc welding are deliberately designed to have a large annature reaction, which gives them a terminal characteristic like the one shown in Figure 9- 58. Notice that when the welding electrodes make contact with each other before welding commences, a very large current flows. As the op-erator separates the welding electrodes, there is a very steep rise in the generator's voltage, while the current remains high. This voltage ensures that a welding arc is maintained through the air between the electrodes. OCMmDRSANDGENERATORS 611 + FIGURE 9-59 IIt "'h+IF vT '" Elt -11t(RIt + Rsl v, IF"'1fF :¥ .... '" NF IF + NSFJIt - ::fAR • v, The equivalent cirwit of a cumulatively compounded dc generator with a long-shunt connection. 9.15 THE CUMULATIVELY COMPOUNDED DC GENERATOR A cumulatively compounded dc generator is a dc generator with both series and shunt fields, connected so that the magnetomoti ve forces from the two fields are additive. Figure 9- 59 shows the equivalent circuit ofa cumulatively compounded dc generator in the "long-shunt" connection. TIle dots that appear on the two field coils have the same meaning as the dots on a transfonner: Current flowing into a dot produces a positive magnetomotive force. Notice that the annature current flows into the dotted end of the series field coil and that the shunt current IF flows into the dotted end of the shunt field coil. Therefore, the total magnetomotive force on this machine is given by (9-48) where 'ifF is the shunt field magnetomotive force, 'ifSE is the series field magneto-motive force, and 2FAR is the armature reaction magnetomotive force. The equiva-lent effective shunt field current for this machine is given by NFl; = NFIF + NsEJA -2FAR The other voltage and current relationships for this generator are IIA - iF + I, I (9-49) (9- 50) (9- 51) 612 ELECTRIC MACHINERY RJNDAMENTALS • v, ""GURE 9-60 The equivalent cin;uit of a cumulatively compounded dc generator with a short-shunt connection. (9- 52) TIlere is another way to hook up a cumulatively compounded generator. It is the "short-shunt" connection, where the series field is outside the shunt field circuit and has current IL flowing through it instead of I .... A short-shunt cumula-tively compounded dc generator is shown in Figure 9-60. The Terminal Characteristic of a Cumulatively Compounded DC Generator To understand the terminal characteristic of a cumulatively compounded dc gen-erator, it is necessary to understand the competing effects that occur within the machine. Suppose that the load on the generator is increased. Then as the load in-creases, the load current IL increases. Since IA = IF + IL i , the annature current IA increases too. At this point two effects occur in the generator: I. As IA increases, the IA(RA + Rs) voltage drop increases as well.1l1is tends to cause a decrease in the terminal voltage VT = EA - IA i (RA + Rs). 2. As IA increases, the series field magnetomotive force 91'SE = NSEIA increases too. This increases the total magnetomotive force 91"0' = NFIF + NSEIA i which increases the flux in the generator.1l1e increased flux in the generator increases EA, which in turn tends to make VT = EA i - IA(RA + Rs) rise. TIlese two effects oppose each other, with one tending to increase VT and the other tending to decrease VT. Which effect predominates in a given machine? It all depends on just how many series turns were placed on the poles of the ma-chine. The question can be answered by taking several individual cases: I. Few series turns (NSE smnll). If there are only a few series turns, the resistive voltage drop effect wins hands down. The voltage falls off just as in a shunt OCMmDRSANDGENERATORS 613 v, Undercompounded Shum L-__ /~-------~ " FIGURE 9-61 Terminal characteristics of cumulatively compounded dc generators. generator, but not quite as steeply (Figure 9--61). This type of construction, where the full-load tenninal voltage is less than the no-load tenninal voltage, is called undercompounded. 2. More series turns (NSE larger). If there are a few more series turns of wire on the poles, then at first the flux-strengthening effect wins, and the terminal voltage rises with the load. However, as the load continues to increase, mag-netic saturation sets in, and the resistive drop becomes stronger than the flux increase effect. In such a machine, the terminal voltage first rises and then falls as the load increases. If VT at no load is equal to VTat full load, the gen-erator is called flat-compounded. 3. Even more series turns are added (NSE large). If even more series turns are added to the generator, the flux-strengthening effect predominates for a longer time before the resistive drop takes over. The result is a characteristic with the fu ll-load tenninal voltage actua lly higher than the no-load tenninal voltage. If VTat a full load exceeds VTat no load, the generator is called over-compounded. All these possibilities are illustrated in Figure 9--61. lt is also possible to realize all these voltage characteristics in a single gen-erator if a diverter resistor is used. Figure 9--62 shows a cumulatively com-pounded dc generator with a relatively large number of series turns NSE. A diverter resistor is connected around the series field. If the resistor Rdh is adjusted to a large value, most of the annature current flows through the series field coil, and the generator is overcompounded. On the other hand, if the resistor RcJjy is adjusted to a small value, most of the current flows around the series field through RcJjy, and the generator is undercompounded. lt can be smoothly adjusted with the resistor to have any desired amount of compounding. 614 ELECTRIC MACHINERY RJNDAMENTALS "/. Y I, R, R, L, I, -• -• • I, I y.: (+)£, /-R, • L, ""GURE 9-62 A cumulatively compounded dc generator with a series diverter resistor. Voltage Control of Cumulatively Compounded DC Generators + v, -TIle techniques available for controlling the tenninal voltage of a cumulatively compounded dc generator are exactly the same as the techniques for controlling the voltage of a shunt dc generator: I. Change the speed of rotation. An increase in w causes Ell = KtPwi to in-crease, increasing the terminal voltage VT = Ell i - IIl(RIl + Rs). 2. Change the field current. A decrease in RF causes IF = VT I RFJ.. to increase, which increases the total magnetomotive force in the generator. As ?ft", increases, the nux -' f ~ ~ !, " E • • 320 300 Speed 1200r/min 280 260 V /' 240 220 V 200 180 / 160 140 / 120 100 / 80 60 / 40 20 / ,II o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 l.l 1.2 1.3 1.4 Shum field currem. A ""GURE 1 '9- 1 The masnetization curve for the dc motor in Problems 9--1 to 9- 12. This curve was made at a constam speed of 1200 r/min. 9-7. Plot the torque-speed characteristic of this motor assuming no armature reaction. and again assuming a full-load armature reaction of 1200 A ollU1lS. For Problems 9--8 and 9-9. the shunt dc motor is reconnected separately excited. as shown in Figure P9- 3. It has a fixed field voltage V I' of 240 V and an annature voltage VA that can be varied from 120 to 240 V. 9-8. What is the no-load speed of this separately excited motor when R>dj = 175 nand (a) VA = 120 V. (b) VA = 180 V. (c) VA = 240 V? 9-9. For the separately excited motor of Problem 9--8: (a) What is the maximwn no-load speed attainable by varying both VA and R>dj? (b) What is the minimwn no-load speed attainable by varying both VA and R>dj? rx: MmDRS AND GENERATORS 623 --0.4On ~ IF I ~ Rodj + lOon Vr = 240 V FIGURE P9-2 The equiva.lent circuit of the shunt ntotor in Problems 9- 1 to 9- 7. I, I, R, I, --" -+ + 0.40 n R., + VF= 240 V RF = 100 n E, V A =120to240V FIGURE P9-J The equiva.lent circuit of the separately excited motor in Problems 9--8 and 9--9. 9-10. If the motor is connected cumulatively compOlUlded as shown in Figure P9-4 and if R adj = 175 O. what is its no-load speed? What is its full-load speed? What is its speed regulation? Calculate and plot the torque-speed characteristic for this motor. (Neglect annature effects in this problem.) 9- 11. The motor is connected cumulatively compounded and is operating at full load. What will the new speed of the motor be if Radj is increased to 250 O ? How does the new speed compare to the full-load speed calculated in Problem 9--1O? 9-12. The motor is now connected differentially compounded. (a) If Radj = 175 O. what is the no-load speed of the motor? (b) What is the motor's speed when the annature current reaches 20A? 40 A? 60A? (c) Calculate and plot the torque-speed characteristic curve of this motor. 9-13. A 7.5-hp. l20-V series dc motor has an armature resistance of 0.2 0 and a series field resistance of 0.16 O. At full load. the current input is 58 A. and the rated speed is 624 ELECTRIC MACHINERY RJNDAMENTALS O.4411",RA +RS . '" Cumulatively compounded • '" Differentially compounded + 100 n VT", 240 V •• ""GURE 1'9-4 The equivalent cin:uit of the compounded motor in Problems 9- 10 to 9--12. 1050 r/min. Its magnetization curve is shown in Figure P9-5. The core losses are 200 W, and the mechanical losses are 240 W al full load. Assume that the mechanical losses vary as the cube of the speed of the motor and that the core losses are constant. (a) What is the efficiency of the motor at fullload1 (b) What are the speed and efficiency of the motor ifit is operating at an annature current of 35 A 1 (c) Plot the torque-speed characteristic for this motor. 9-14. A 20-hp, 240-V, 76-A, 900 r/min series motor has a field winding of 33 turns per pole. Its annature resistance is 0.09 n, and its field resistance is 0.06 n. The mag-netization curve expressed in terms of magnetomotive force versus EA at 900 r/min is given by the following table: 95 188 212 229 243 :Ji. A • turns '00 1500 2000 2500 3000 Annature reaction is negligible in this machine. (a) CompUle the motor's torque, speed, and output power a133, 67,100, and 133 percent of full-load armalure ClUTent. (Neglect rotational losses.) (b) Plot the torque-speed characteristic of this machine. 9-15. A300-hp, 44O-V, 560-A, 863 r/min shunt dc motor has been tested, and the follow-ing data were taken: Blocked-rotor test: VA = 16.3 V exclusive of brushes 110. = 500 A No-load operation: VA = 16.3 V including brushes 110. = 23.1 A V F = 440 V IF = 8.86A IF = 8.76A n = 863 r/min > J • ~ • " ~ • • , , a 1! ~ 160 150 140 130 120 110 100 90 80 70 60 50 / 40 30 20 / / / 10 / o o 10 / / rx: MmDRS AND GENERATORS 625 5,.., 1200r~ /" / / / / / 20 30 40 60 70 Series field current. A FIGURE 1'9-5 The magnetization curve for the series moior in Problem 9--13. This curve was taken al a constant speed of 1200 r/min. What is this motor's efficiency at the rated conditions? [Note: Assrune that ( 1) the brush voltage drop is 2 V, (2) the core loss is to be determined at an armature volt-age equal to the armature voltage lUlder full load, and (3) stray load losses are 1 per-cent of full load.] Problems 9- 16 to 9--19 refer to a 240-V, IOO-A de motor which has both shunt and series windings. Its characteristics are RA =O.14f.! Rs = 0.04 n R" = 200 n Radj = 0 to 300 n, currently set to 1 20 n N" = 1500 turns Nsf', = 12 turns nO. = 1200 r/min 626 ELECTRIC MACHINERY RJNDAMENTALS 300 250 > 200 j "' ~ / ~ 150 , ~ • E • • 100 / / o 0.0 ""GURE 1'9-6 0.25 S"",d 1200 r/min / / / 0.50 0.75 Field current. A 1.00 The masnetization curve for the dc motor in Problems 9--16 to 9--19. 1.25 1.50 This motor has compensating windings and interpoles. The magnetization curve for this motor at 1200 rfmin is shown in Figure P9--6. 9-16. The motor described above is connected in shunt. (a) What is the no-load speed of this motor when R odj = 120 0 ? (b) What is its full-load speed? (c) Under no-load conditions. what range of possible speeds can be achieved by adjusting Rodj? 9-17. This machine is now connected as a cumulatively compounded dc motor with Rodj = 120 n. (a) What is the full-load speed of this motor? (b) Plot the torque-speed characteristic for this motor. (c) What is its speed regulation? 9-18. The motor is recotulected differentially compolUlded with R adj = 120 O. Derive the shape of its torque-speed characteristic. rx: MmDRS AND GENERATORS 627 9-19. A series motor is now constructed from this machine by leaving the shlUlt field out entirely. Derive the torque-speed characteristic of the resulting motor. 9-20. An automatic starter circuit is to be designed for a shlUlt motor rated at 15 hp. 240 V. and 60 A. The annature resistance of the motor is 0.15 n. and the shunt field re-sistance is 40 n. The motor is to start with no more than 250 percent of its rated ar-mature current. and as soon as the current falls to rated value. a starting resistor stage is to be cut out. How many stages of starting resistance are needed. and how big should each one be? 9-2 1. A 15-hp. 230-V. 1800 rlmin shunt dc motor has a full-load armature current of 60 A when operating at rated conditions. The annature resistance of the motor is RA = 0.15 n. and the field resistance R" is 80 n.The adjustable resistance in the field cir-cuit R>dj may be varied over the range from 0 to 200 n and is currently set to 90 n. Annature reaction may be ignored in this machine. The magnetization curve for this motor. taken at a speed of 1800 r/min. is given in tabular fonn below: 0.80 1.00 1.28 242 8.' I 150 I 180 0.00 2.88 (a) What is the speed of this motor when it is ruIllling at the rated conditions spec-ified above? (b) The output power from the motor is 7.5 hp at rated conditions. What is the out-put torque of the motor? (c) What are the copper losses and rotational losses in the motor at full load (ignore stray losses)? (d) What is the efficiency of the motor at full load? (e) If the motor is now unloaded with no changes in tenninal voltage or R>dj' what is the no-load speed of the motor? (f) Suppose that the motor is running at the no-load conditions described in part e. What would happen to the motor if its field circuit were to open? Ignoring ar-mature reaction. what would the final steady-state speed of the motor be under those conditions? (g) What range of no-load speeds is possible in this motor. given the range offield resistance adjustments available with Radj? 9-22. The magnetization curve for a separately excited dc generator is shown in Figure P9- 7. The generator is rated at 6 kW, 120 V. 50 A. and ISOO rlmin and is shown in Figure P9-8. Its field circuit is rated at SA. The following data are known about the machine: RA = O.ISO ~j = Ot030n N" = 1000 turns per pole V,, = 120V R,, =24n Answer the following questions about this generator. assruning no armature reaction. (a) If this generator is operating at no load. what is the range of voltage adjustments that can be achieved by changing Radj? (b) If the field rheostat is allowed to vary from 0 to 30 n and the generator's speed is allowed to vary from 1500 to 2()(x) r/min. what are the maximwn and mini-mum no-load voltages in the generator? 628 ELECTRIC MACHINERY RJNDAMENTALS 160 150 140 a ---V-a a /" 13 12 II / / I I a a I / / I I 40 / a / a / ,II 30 2 a 2 3 4 , 6 7 Shunt field current. A a 1000 2000 3000 4000 5000 6000 7000 Field mmf. A· turns ""GURE 1'9-7 The magnetization curve for Problems 9--22 to 9--28. This curve was taken at a speed of 1800 r/min. 9-23. If the armature current of the generator in Problem 9--22 is 50 A. the speed of the generator is 1700 r/min. and the tenninal voltage is 106 V, how much field current must be flowing in the generator? 9-24. Assuming that the generator in Problem 9- 22 has an annature reaction at full load equivalent to 400 A • turns of magnetomotive force. what will the terminal voltage of the generator be when I" = 5 A. n .. = 1700 r/min. and IA = 50 A? 9-25. The machine in Problem 9--22 is reconnected as a shunt generator and is shown in Figure P9-9. The shunt field resistor R>dj is adjusted to 10 n. and the generator's speed is 1800 r/min. rx: MmDRS AND GENERATORS 629 I, R , I, I, -- -+ O.~8n + :/ R~ 120 V V, RF=24f1 ~ c,JE ' V, L, FIGURE 1'9-8 The separately excited de generator in Problems 9--22 to 9--24. R, ,-~-----"V'VV'--------~e--C----~ + 0.18 n i'" J IF R-». + 24 n ~ RF V, FIGURE 1'9-9 The shunt de generator in Problems 9- 25 and 9--26. (a) What is the no-load lennina! voltage of the generator? (b) Assruning no armature reaction, what is the terminal voltage of the generator with an armature current of 20 A? 40 A? (c) Assruning an annature reaction equal to 300 A • turns at [unload, what is the lennina! voltage of the generator with an armature current of 20 A? 40 A? (d) Calculate and plot the terminal characteristics of this generator with and with-out armature reaction. 9-26. If the machine in Problem 9--25 is running at 1800 r/min with a field resistance Rodj = 10 n and an annature current of 25 A, what will the resulting terminal voltage be? If the field resistor decreases to 5 n while the armature current remains 25 A, what will the new terminal voltage be? (Assrune no armature reaction.) 9-27. A 120-V, 50-A cumulatively compounded dc generator has the following characteristics: RA + Rs = 0.2 1 n RF = 20 n R>dj = 0 to 30 n, set to 10 n N F = J()(X) turns NSE = 20 turns n .. = 1800 r/min 630 ELECTRIC MACHINERY RJNDA MENTALS 0.21 n + 200 v, • L" N,,= 1000 turns ""GURE 1'9- 10 The compounded dc generator in Problems 9--27 and 9- 28. The machine has the magnetization curve shown in Figure P9- 7. Its equivalent cir-cuit is shown in Figure P9--1O. Answer the following questions about this machine, assuming no annature reaction. (a) If the generator is operating at no load, what is its terminal voltage? (b) If the generator has an armature current of 20 A, what is its tenninal voltage? (c) If the generator has an armature current of 40 A, what is its tenninal voltage? (d) Calculate and plot the tenninal characteristic of this machine. 9-28. If the machine described in Problem 9- 27 is reconnected as a differentially com-pounded dc generator, what will its teoninal characteristic look like? Derive it in the same fashion as in Problem 9-27. 9-29. A cumulatively compounded dc generator is operating properly as a flat-compounded dc generator. The machine is then shut down, and its shunt field con-nections are reversed. (a) If this generator is turned in the same direction as before, will an output voltage be built up at its tenninals? Why or why not? (b) Will the voltage build up for rotation in the opposite direction? Why or why not? (c) For the direction of rotation in which a voltage builds up, will the generator be cumulatively or differentially compolUlded? 9-30. A three-phase synchronous machine is mechanically connected to a shlUlt dc ma-chine, fonning a motor-generator set, as shown in Figure P9--ll. The dc machine is connected to a dc power system supplying a constant 240 V, and the ac machine is connected to a 480-V, 60-Hz infinite bus. The dc machine has four poles and is rated at 50 kW and 240 V. It has a per-unit annature resistance of 0.04. The ac machine has four poles and is V-connected. It is rated at 50 kVA, 480 V, and 0.8 PF, and its saturated synchronous reactance is 2.0 n per phase. All losses except the dc machine's annature resistance may be neglected in this problem. Assume that the magnetization curves of both machines are linear. (a) Initially, the ac machine is supplying 50 kV A at 0.8 PF lagging to the ac power system. rx: MmDRS AND GENERATORS 631 MGset rx: machine AC machine 1'1 R, 6 R, ~ AC power , V, E, system , (infinite bus) & L, ~ R, L, + V, FIGURE 1'9-11 The motor-generator set in Problem 9- 30. I. How much power is being supplied to the dc motor from the dc power system? 2. How large is the internal generated voltage EA of the dc machine? 3. How large is the internal generated voltage EA of the ac machine? (b) The field current in the ac machine is now increased by 5 percent. What effect does this change have on the real power supplied by the motor- generator set? On the reactive power supplied by the motor- generator set? Calculate the real and reactive power supplied or consumed by the ac machine under these condi-tions. Sketch the ac machine's phasor diagram before and after the change in field current. (c) Starting from the conditions in part b. the field current in the dc machine is now decreased by I percent. What effect does this change have on the real power supplied by the motor-generator set? On the reactive power supplied by the motor- generator set? Calculate the real and reactive power supplied or con-swned by the ac machine lUlder these conditions. Sketch the ac machine's pha-sor diagram before and after the change in the dc machine's field current. (d) From the above results. answer the following questions: I. How can the real power flow through an ac-dc motor- generator set be controlled? 2. How can the reactive power supplied or consumed by the ac machine be controlled without affecting the real power flow? REFERENCES 1. Chaston. A. N. Electric Machinery. Reston. Va.: Reston Publications. 1986. 2. Fitzgerald. A. E.. and C. Kingsley. Jr. Electric Machinery. New YorK: McGraw-Hill. 1952. 3. Fitzgerald. A. E.. C. Kingsley. Jr .• and S. D. Umans. Electric Machinery. 5th ed. New York: McGraw-Hill. 1990. 4. Heck. C. Magnetic Materials and Their AppliCllTions. London: Butterwonh & Co .. 1974. 632 ELECTRIC MACHINERY RJNDAMENTALS 5. IEEE Standard 113-1985. Guide on Test Procedures for DC Machines. Piscataway. N.J.: IEEE. 1985. (Note that this standard has been officially withdrawn but is still available.) 6. Kloeftler. S. M .• R. M. Ken;hner. and J. L. Brenneman. Direct Current Machinery. Rev. ed. New York: Macmillan. 1948. 7. Kosow. Irving L. Electric Machinery atuf Tmnsjormers. Englewood ClilTs. N.J.: Prentice-Hall. 1972. 8. McPherson. George. An Introduction 10 Electrical Machines and Tmnsfonners. New York: Wiley. 1981. 9. Siskind. Charles S. Direct-Current Machinery. New York: McGraw-Hill. 1952. 10. Siemon. G. R .• and A. Straughen. Electric Machines. Reading. Mass.: Addison-Wesley. 1980. II. Werninck. E. H. (ed.). Electric MOlOr Handbook. London: McGraw-Hili. 1978. CHAPTER 10 SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS C hapters 4 through 7 were devoted to the operation orthe two major classes of ac machines (synchronous and induction) on three-phase power systems. Motors and generators of these types are by far the most common ones in larger commercial and industrial settings. However, most homes and small businesses do not have three-phase power available. For such locations, all motors must run from single-phase power sources. nlis chapter deals with the theory and operation of two major types of single-phase motors: the universal motor and the single-phase induction motor. The universal motor, which is a straightforward extension of the series de motor, is descri bed in Section 10.1. The single-phase induction motor is described in Sections 10.2 to 10.5. The major problem associated with the design of single-phase induction motors is that, unlike three-phase power sources, a single-phase source does not produce a rotat-ing magnetic field. Instead, the magnetic field produced by a single-phase source remains stationary in position and pulses with time. Since there is no net rotating magnetic field, conventional induction motors cannot function, and special de-signs are necessary. In addition, there are a number of special-purpose motors which have not been previously covered. These include reluctance motors, hysteresis motors, stepper motors, and brushless dc motors. 1lley are included in Section 10.6. 633 634 ELECTRIC MACHINERY RJNDAMENTALS v, ""CURE 10-1 Equivalent cin;uit of a univeTS3.1 motor. 10.1 THE UNIVERSAL MOTOR Perhaps the simplest approach to the design of a motor that will operate on a single-phase ac power source is to take a dc machine and run it from an ac supply. Recall from Chapter 8 that the induced torque of a dc motor is given by (8-49) If the polarity of the voltage applied to a shunt or series dc motor is reversed, both the direction of the field flux and the direction of the armature current reverse, and the resulting induced torque continues in the same direction as before. Therefore, it should be possible to achieve a pulsating but unidirectional torque from a dc motor connected to an ac power supply. Such a design is practical only for the series dc motor (see Figure 10- 1), since the annature current and the field current in the machine must reverse at ex-actly the same time. For shunt dc motors, the very high field inductance tends to delay the reversal of the field current and thus to unacceptably reduce the average induced torque of the motor. In order for a series dc motor to function effectively on ac, its field poles and stator frame must be completely laminated. If they were not completely lam-inated, their core losses would be enonnous. When the poles and stator are lami-nated, this motor is often called a universal motor, since it can run from either an ac or a dc source. When the motor is running from an ac source, the commutation will be much poorer than it would be with a dc source. The extra sparking at the brushes is caused by transfonner action inducing voltages in the coils undergoing com-mutation. These sparks significantly shorten brush life and can be a source of radio-frequency interference in certain environments. A typical torque-speed characteristic of a universal motor is shown in Fig-ure 10- 2. It differs from the torque-speed characteristic of the same machine op-erating from a dc voltage source for two reasons: I. The armature and field windings have quite a large reactance at 50 or 60 Hz. A significant part of the input voltage is dropped across these reactances, and therefore Ell is smaller for a given input voltage during ac operation than it is during dc operation. Since Ell = Kcpw, the motor is slower for a given SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 635 " Series IX motor """""" Universal motor ......... (AC supply) ........................ ... ""GURE 10-2 Comparison of the torque-speed characteristic of a universal motor when operating from ac and dc L ___ Tind power supplies. annature current and induced torque on alternating current than it wou ld be on direct current. 2. In addition, the peak voltage of an ac system is V2 times its nns value, so magnetic saturation could occur near the peak current in the machine. This saturation could significantly lower the nns nux of the motor for a given cur-rent level, tending to reduce the machine's induced torque. Recall that a de-crease in flux increases the speed of a dc machine, so this effect may partially offset the speed decrease caused by the first effect. Applications of Universal Motors The universal motor has the sharply drooping torque- speed characteristic of a dc series motor, so it is not suitable for constant-speed applications. However, it is compact and gives more torque per ampere than any other single-phase motor. It is therefore used where light weight and high torque are important. Typical applications for this motor are vacuum cleaners, dri lls, similar portable tools, and kitchen appliances. Speed Control of Universal Motors As with dc series motors, the best way to control the speed of a universal motor is to vary its nns input voltage. The higher the rms input voltage, the greater the re-sulting speed of the motor. Typical torque-speed characteristics of a universal mo-tor as a function of voltage are shown in Figure 10- 3. In practice, the average voltage applied to such a motor is varied with one of the SCR or TRIAC circuits introduced in Chapter 3. Two such speed control circuits are shown in Figure 10-4. TIle variable resistors shown in these figures are the speed adjustment knobs of the motors (e.g., such a resistor would be the trigger of a variable-speed dri II). 636 ELECTRIC MACHINERY RJNDA MENTALS v, v, v, '--------''--------------',,------ Tjoo ""GURE 10-3 The effect of changing teffilinal voltage on the torque-speed characteristic of a universal motor. + /. ~c 0, ,n D, = = C ~DIAC (a) L, (series field) C;) -s CR D2 is a free-wheeling diode to control inductive k:ick: effects. +~----~------------, Rc + v (t) TRIAC C ,b , ""GURE 10-4 Sample universal motor speed-control ci["(;uits. (a) Half-wave; (b) full-wave. o o o SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 637 Stator o o FlGURE 10-5 Construction of a single-phase induction motor. The rotor is the same as in a three-phase induction motor. but the stator has only a single distributed phase. 10.2 INTRODUCTION TO SINGLE-PHASE INDUCTION MOTORS Another common single-phase motor is the single-phase version of the induction motor. An induction motor with a squirrel-cage rotor and a single-phase stator is shown in Figure 10-5. Single-phase induction motors suffer from a severe handicap. Since there is only one phase on the stator winding, the magnetic field in a single-phase induc-tion motor does not rotate. Instead, it pulses, getting first larger and then smaller, but always remaining in the same direction. Because there is no rotating stator magnetic field, a single-phase induction motor has no staning torque. This fact is easy to see from an examination of the motor when its rotor is stationary. The stator flux of the machine first increases and then decreases, but it always points in the same di rection. Si nce the stator magnetic field does not ro-tate, there is no relative motion between the stator field and the bars of the rotor. Therefore, there is no induced voltage due to relative motion in the rotor, no rotor current flow due to relative motion, and no induced torque. Actually, a voltage is induced in the rotor bars by transfonner action (df/JIdt), and since the bars are short-circuited, current flows in the rotor. However, this magnetic field is lined up with the stator magnetic field, and it produces no net torque on the rotor, "rind = kBR X Bs = kBRBS sin "y = kBRBS sin 180 0 = 0 (4- 58) At stall conditions, the motor looks like a transformer with a short-circuited sec-ondary winding (see Figure 10-6). 638 ELECTRIC MACHINERY RJNDAMENTALS II, FlGURE 10-6 The single-phase induction motor at starting conditions. The stator winding induces opposing voltages and currents into the rotor cirwit. resulting in a rotor magnetic field lined up with the stator magnetic field. 7;M = O. TIle fact that single-phase induction motors have no intrinsic starting torque was a serious impediment to early development of the induction motor. When in-duction motors were first being developed in the late l880s and early 1890s, the first available ac power systems were 133-Hz, single-phase. With the materials and techniques then available, it was impossible to build a motor that worked well. The induction motor did not become an off-the-shelf working product until three-phase, 25-Hz power systems were developed in the mid- 189Os. However, once the rotor begins to tum, an induced torque will be produced in it. TIlere are two basic theories which explain why a torque is produced in the rotor once it is turning. One is called the double-revolving-field theory of single-phase induction motors, and the other is called the cross-field theory of single-phase induction motors. E.1.ch of these approaches will be described below. The Double-Revolving-Field Theory of Single-Phase Induction Motors TIle double-revolving-field theory of single-phase induction motors basically states that a stationary pulsating magnetic field can be resolved into two rotating magnetic fields, each of equal magnitude but rotating in opposite directions. The induction motor responds to each magnetic field separately, and the net torque in the machine will be the sum of the torques due to each of the two magnetic fields. Figure 10- 7 shows how a stationary pulsating magnetic field can be re-solved into two equal and oppositely rotating magnetic fields. The flux density of the stationary magnetic field is given by , Bs (t) = (Bmn cos wt) J A clockwise-rotating magnetic field can be expressed as ( 10-1 ) 0, 0-lie ... /' " W W (:1) " , ,d, FIGURE 10-7 SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 639 "-W II, ,b, " , ,., "-W " - \ ) o-W W , " ( \ "ow 1--------1 " ,f , The resolution of a single pulsating magnetic field into two magnetic fields of equal magnitude by rotation in opposite directions. Notice that at all times the vector sum of the two magnetic fields lies in the vertical plane. Bew(r) = ( t Bmax cos wt)i - ( t Bmax sin wi fi ( 10-2) and a counlerclockwise-rotating magnetic field can be expressed as Bcew(t) = ( 1 Bmax cos wt)i + ( 1 Bmax sin wi fi ( 10-3) Notice that the sum of the clockwise and counterclockwise magnetic fields is equal to the stationary pulsating magnetic field Bs: 8,(1 ) ~ B~(I) + BccwCt) (IO-d) TIle torque-speed characteristic of a three-phase induction motor in re-sponse to its single rotating magnetic field is shown in Figure 10--8a. A single-phase induction motor responds to each of the two magnetic fields present within it, so the net induced torque in the motor is the difference between the two torque- speed curves. This net torque is shown in Figure lO--8b. Notice that there is no net torque at zero speed, so this motor has no starting torque. TIle torque- speed characteristic shown in Figure 1 0--8b is not quite an ac-curate description of the torque in a single-phase motor. It was formed by the 640 ELECTRIC MACHINERY RJNDAMENTALS ----------------------t----------------'------'. -----------""GURE 10-8 ,,' --, " ' , , , , , , -' , " , ---, , ---------------,b, (a) The torque-speed characteristic of a three-phase induction motor. (b) The torque-speed characteristic curves of the two equal and oppositely rota.ting stator magnetic fields. superposition of two three-phase characteristics and ignored the fact that both magnetic fields are present simultaneously in the single-phase motor. If power is applied to a three-phase motor while it is forced to turn back-ward, its rotor currents will be very high (see Figure 10--9a). However, the rotor frequency is also very high, making the rotor's reactance much much larger than its resistance. Since the rotor's reactance is so very high, the rotor current lags the rotor voltage by almost 900, pnxlucing a magnetic field that is nearly 180 0 from the stator magnetic field (see Figure 10- 10). The induced torque in the motor is proportional to the sine of the angle between the two fields, and the sine of an an-gle near 180 0 is a very small number. TIle motor's torque would be very small, ex-cept that the extremely high rotor currents partially offset the effect of the mag-netic field angles (see Figure 10- 9b). On the other hand, in a single-phase motor, both the forward and the reverse magnetic fields are present and both are produced by the same current.llle forward SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 641 - n,j'DC ,,' PF =cosfl =sinll '. '.~ ,b, fo'IGURE 10-9 The torque-speed characteristic of a three-phase induction motor is proportional to both the strength of the rotor magnetic field and the sine of the angle between the fields. When the rotor is turned bad:waro. IN and Is are '. very high. but the angle between the '.~ fields is very large. and that angle '0' limits the torque of the motor. and reverse magnetic fields in the motor each contribute a component to the total voltage in the stator and, in a sense, are in series with each other. Because both magnetic fields are present, the forward-rotating magnetic field (which has a high effective rotor resistance Ris) will limit the stator current now in the motor (which produces both the forward and reverse fields). Since the current supplying the re-verse stator magnetic field is limited to a small value and since the reverse rotor magnetic field is at a very large angle with respect to the reverse stator magnetic field, the torque due to the reverse magnetic fields is very small near synchronous speed. A more accurate torque-speed characteristic for the single-phase induction motor is shown in Figure 10-11. In addition to the average net torque shown in Figure 10-11, there are torque pulsations at twice the stator frequency. 1l1ese torque pulsations are caused when the forward and reverse magnetic fields cross each other twice each cycle. Al-though these torque pulsations pnxluce no average torque, they do increase vibra-tion, and they make single-phase induction motors noisier than three-phase motors of the same size. 1l1ere is no way to eliminate these pulsations, since instantaneous power always comes in pulses in a single-phase circuit. A motor designer must al-low for this inherent vibration in the mechanical design of single-phase motors. 642 ELECTRIC MACHINERY RJNDAMENTALS Direction of magnetic w 0, field rotation Plane of maximum EI/ , • ~ Direction of • rotor rotati0/-j 0 0 Voltage • , @ polarity " , , " II .. , @ • Plane of , , maximum 1 1/ , Current , Current , polarity " , @ polarity , X , 0 , @ , X , 0 X X Voltage polarity 0, ""GURE 111-10 When the rotor of the motor is forced to turn backwmd. the angle r between 111/ and Bs approaches 180°. 1";00 Forward .... ".".". ".. , , ( CUn-ll ,-, , , , , , , , , , , / ' ----------------.--+=~-~¥-=----+.c_ '. - n,}'DC _ __ _ n.y"" /'"..". .... --, , , , , , , , ' \ , , , , '-' Reverse curve ""GURE III-II The torque--speed characteristic of a single-phase induction nx>toT. taking into account the current limitation on the backward-rotating magnetic field caused by the presence of the forward-rotating magnetic field. SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 643 The Cross-Field Theory of Single-Phase Induction Motors The cross-field theory of single-phase induction motors looks at the induction mo-tor from a totally different point of view. 1l1is theory is concerned with the volt-ages and currents that the stationary stator magnetic field can induce in the bars of the rotor when the rotor is moving. Consider a single-phase induction motor with a rotor which has been brought up to speed by some external method. Such a motor is shown in Figure 1O-1 2a. Voltages are induced in the bars of this rotor, with the peak voltage oc-curring in the windings passing directly under the stator windings. 1l1ese rotor voltages produce a current now in the rotor, but because of the rotor's high reac-tance, the current lags the voltage by almost 90°. Since the rotor is rotating at nearly synchronous speed, that 90° time lag in current produces an almost 90° an-gular shift between the plane of peak rotor voltage and the plane of peak current. The resulting rotor magnetic field is shown in Figure 1O-1 2b. The rotor magnetic field is somewhat smaller than the stator magnetic field, because of the losses in the rotor, but they differ by nearly 90° in both space and current FIGURE 10- 12 Plane of max IR voltage-----.... \ ' , Plane of maximum . I E, (a) The development of induced torque in a single-phase induction nx>tor. as explained by the cross-field theocy. If the stator field is pulsing, it will induce voltages in the rot"--r'-' Rmm Rotor volta.ges ""GURE 10-12 (concluded) -~ . voltages (b' (b) This delayed rotor current produces a rotor magoetic field at an angle different from the angle of the stator magnetic field. I I . I , / "'\ I URI /'\ / , , , , ""GURE 10-13 '\ " I , , , , , , , , , , , , 9d) SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 645 II, II, ,b, wt=45° wl= 135° 0, "'t=225° ",t=315° II .. , (b) The vector sum of the rotor and stator magnetic fields at various times. showing a net magnetic field which rotates in a counterclockwise direction. motor wi ll develop a net torque in the direction of motion, and that torque wi ll keep the rotor turning. If the motor's rotor had originally been turned in a clockwise direction, the resulting torque would be clockwise and would again keep the rotor turning. 646 ELECTRIC MACHINERY RJNDAMENTALS 10.3 STARTING SINGLE-PHASE INDUCTION MOTORS As previously explained, a single-phase induction motor has no intrinsic starting torque. TIlere are three techniques commonly used to start these motors, and single-phase induction motors are classified according to the methods used to pro-duce their starting torque. These starting techniques differ in cost and in the amount of starting torque produced, and an engineer normally uses the least ex-pensive technique that meets the torque requirements in any given application. TIle three major starting techniques are I. Split-phase windings 2. Capacitor-type windings 3. Shaded stator poles All three starting techniques are methods of making one of the two revolv-ing magnetic fields in the motor stronger than the other and so giving the motor an initial nudge in one direction or the other. Split-Phase Windings A split-phase motor is a single-phase induction motor with two stator windings, a main stator winding (M) and an auxiliary starting winding (A) (see Figure 10- 1 4). TIlese two windings are set 90 electrical degrees apart along the stator of the motor, and the auxiliary winding is designed to be switched oul of the circuit at some set speed by a centrifugal switch. TIle auxiliary winding is designed to have + 1 ·1 R. < ,. VAC , ~. jXM a 5· ~ ~ I. I ""GURE 10-14 Auxiliary winding jX" R, (a) ,b, -I, V Centrifugal switch (a) A split-phase induction motor. (b) The currents in the motor at starting conditions. SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 647 a higher resistance/reactance ratio than the main winding, so that the current in the auxiliary winding leads the current in the main winding. This higher RIX ratio is usually accomplished by using smaller wire for the auxiliary winding. Smaller wire is pennissible in the auxiliary winding because it is used only for starting and therefore does not have to take full current continuously. To understand the function of the auxiliary winding, refer to Figure 10-15. Since the current in the auxiliary winding leads the current in the main winding, 300% ® ® ® ® 0 Line fvoItage /,,-, ~ I,'/ / /" \ "-, / , "-, , , , Main plus starting winding , 200% ~""---100% Main winding alone FIGURE 10-15 ". ® 0 0 ,., '- , , ,b, , ,., Auxiliary winding Main winding 0 II, 0 Centrifugal switch , , , , , (a) Relationship of main and auxiliary magnetic fields. (b) I .. peaks before 1 M • producing a net counterclockwise rotation of the magnetic fields. (c) The resulting torque-speed characteristic. 648 ELECTRIC MACHINERY RJNDAMENTALS ""GURE 10-16 Cutaway view of a split-phase motor. showing the main and auxiliary windings and the centrifugal switch. (Courtesy of Westinghouse Electric Corpomtion. ) the magnetic field BA peaks before the main magnetic field B M . Since BA peaks first and then BM , Ihere is a net counterclockwise rotation in the magnetic field. In other words, the auxiliary winding makes one of the oppositely rotating stator magnetic fields larger than the other one and provides a net starting torque for the motor. A Iypicallorque-speed characteristic is shown in Figure 1O-I Sc. A culaway diagram of a split-phase motor is shown in Figure 10-16. It is easy to see the main and auxiliary windings (the auxiliary windings are the smaller-diameler wires) and the cenlrifugal switch that cuts the auxiliary windings oul of the circuit when the motor approaches operating speed. Split-phase motors have a moderate starting torque with a fairly low start-ing current. They are used for applications which do not require very high starting torques, such as fans, blowers, and cenlrifugal pumps. 1lley are available for sizes in the fractional-horsepower range and are quite inexpensive. I n a split-phase induction motor, the current in the auxiliary windings always peaks before the current in the main winding, and therefore the magnetic field from the auxiliary winding always peaks before the magnetic field from the main wind-ing. The direction of rotation of the motor is detennined by whether the space an-gie of the magnetic field from the auxiliary winding is 90° ahead or 90° behind the angle of the main winding. Since that angle can be changed from 90° ahead to 90° behind just by switching the connections on the auxiliary winding, the direction of rotation of the nwtor can be reversed by switching the connections of the auxiliary winding while leaving the main winding's connections unchanged. SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 649 +o---~--~-------------------, 1 ·1 Centrifugal R. i sWItch C Auxiliary winding , ,' I. 'h' FIGURE 10- 17 (a) A capacitor-start induction motor. (b) Current .angles at starting in this motor. Capacitor-Start Motors For some applications, the starting torque supplied by a split-phase motor is insuf-ficient to start the load on a motor's shaft. In those cases, capacitor-start motors may be used (Figure 10-1 7). In a capacitor-start motor, a capacitor is placed in se-ries with the auxiliary winding of the motor. By proper selection of capacitor size, the magnetomotive force of the starting current in the auxiliary winding can be ad-justed to be equal to the magnetomotive force of the current in the main winding, and the phase angle of the current in the auxiliary winding can be made to lead the current in the main winding by 90°. Since the two windings are physically sepa-rated by 90°, a 90° phase difference in current will yield a single unifonn rotating stator magnetic field, and the motor will behave just as though it were starting from a three-phase power source. In this case, the starting torque of the motor can be more than 300 percent of its rated value (see Figure 10-18). Capacitor-start motors are more expensive than split-phase motors, and they are used in applications where a high starting torque is absolutely required. Typi-cal applications for such motors are compressors, pumps, air conditioners, and other pieces of equipment that must start under a load. (See Figure 10-1 9.) 650 ELECTRIC MACHINERY RJNDAMENTALS ,~ 400% Main and aUXilip winding V 300% , , ~ , 200% '\ , , , , , 100% ~ , , ~ , -(' , --------Main winding only Switch ,~ ""GURE 10-18 Torque-speed characteristic of a capacitor-start induction motor. Permanent Split-Capacitor and Capacitor-Start, Capacitor-Run Motors The starting capacitor does such a good job of improving Ihe torque-speed char-acteristic of an induction motor that an auxiliary winding with a smaller capacitor is sometimes left pennanently in the molor circuit. If the capacitor's value is cho-sen correctly, such a motor will have a perfectly unifonn rotating magnetic field at some specific load, and it will behave just like a three-phase induction motor at that point. Such a design is called a pennanent split-capacitor or capacitor-start-and-run motor (Figure 10- 20). Pennanent split-capacitor motors are simpler than capacitor-start motors, since the starting switch is not needed. At normal loads, Ihey are more efficient and have a higher power factor and a smoother torque than ordinary single-phase induction motors. However, pennanent split-capacitor molors have a lower starting torque Ihan capacitor-start motors, since the capacitor must be sized to balance lhe currents in the main and auxiliary windings at normal-load conditions. Since the starting cur-rent is much greater than the nonnal-load current, a capacitor that balances the phases under nonnal loads leaves them very unbalanced under starting conditions. If both the largest possible starting lorque and the best running conditions are needed, two capacitors can be used with the auxiliary winding. Motors with two capacitors are called capacitor-stan, capacitor-run, or two-value capacitor motors (Figure 10- 21). TIle larger capacitor is present in the circuit only during starting, when it ensures that Ihe currents in the main and auxiliary windings are roughly balanced, yielding very high starting torques. When Ihe motor gets up 1 0 speed, the cenlrifugal switch opens, and the pennanent capacitor is left by itself in the auxiliary winding circuit. TIle pennanent capacitor is just large enough to bal-ance the currents at nonnal motor loads, so the motor again operates efficienlly with a high torque and power factor. TIle pennanent capacitor in such a motor is typically about 10 1 0 20 percent of the size of the starting capacitor. Sh," !(oy Rotor Assembly Rotllling } Staning Switch FIGURE 10- 19 SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 651 ('J Shaft End (Rear) End Bracket I Phase T.E.F.e. motor capa.citor start exploded view general purpose 56 frame Stator Assembly Starting Capacitor Capacitor Cover / -- Capacitor Cover Mounting Screws _ ~::::~;:-~Stationary Starting Switch ;: Tenninat Board Front End Bmcket Fan Shroud (bJ Fan Shroud Mounting Screws (a) A capa.citor-start induction ntotor. (Courtesy of Emerson Electric Company.) (b) Exploded view of a capacitor-start induction motor. (Courtesy of Westinghouse Electric Corpomtion.) The direction of rotation of any capacitor-type motor may be reversed by switching the connections of its auxiliary windings. 652 ELECTRIC MACHINERY RJNDAMENTALS I. j R. ~ ~ ~ < ,. , ~. 0 ~ C ~ , ~':l:/ jX. ~ Auxiliary win~i~ g jX, R, I, (a) '00 400 % 300 % 100 % /' '" / \ % /' \ 200 '.~ 'h' ""GURE 10-20 (a) A permanent split-capacitor induction ntotor. (b) Torque-speed characteristic of this motor. Shaded-Pole Motors A shaded-pole induction motor is an induction molor with only a main winding. Instead of having an auxiliary winding, it has salient poles, and one portion of each pole is surrounded by a short-circuited coil called a shading coil (see Figure 1O- 22a). A time-varying flux is induced in Ihe poles by the main winding. When the pole flux varies, it induces a voltage and a current in the shading coil which opposes Ihe original change in flUX.1llis opposition retards the flux changes un-der the shaded portions of the coils and therefore produces a slighl imbalance be-tween Ihe two oppositely rotating stator magnetic fields. TIle net rolation is in Ihe direction from the unshaded to the shaded portion of the pole face. The torque-speed characteristic of a shaded-pole molar is shown in Figure IO- 22b. Shaded poles produce less starting torque than any other type of induction motor starting system. TIley are much less efficient and have a much higher slip SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 653 -~ '). I RM < I' 7 '\ '\ ,. " ~. := " c ~ " "" 5/ jXM 00 , ( ., fjmd 400 % Starting and running capacitors % / Y" \ % , , , , , , , , , , , , , , , , 300 2llll (00 % , _---<- Runmng , , , , ---capacitors , , , Switch "~ (b' FIGURE 10- 21 (a) A capacitor-start. capacitor-run induction motor. (b) Torque-speed characteristic of this motor. than other types of single-phase induction motors. Such poles are used only in very small motors (Y..o hp and less) with very low starting torque requirements. Where it is possible to use them, shaded-pole motors are the cheapest design available. Because shaded-pole motors rely on a shading coil for their starting torque, there is no easy way to reverse the direction of rotation of such a motor. To achieve reversal, it is necessary to install two shading coils on each pole face and to selectively short one or the other of them. See Figures 10-23 and 10- 24. Comparison of Single-Phase Induction Motors Single-phase induction motors may be ranked from best to worst in tenns of their starting and running characteristics: I. Capacitor-start, capacitor-run motor 2. Capacitor-start motor 654 ELECTRIC MACHINERY RJNDA MENTALS 300% 200% Stator winding ,,' "-------------------------~--- "m "."" ,b , ""GURE 10-22 (a) A basic shaded-pole induction motor. (b) The resulting torque-speed characteristic. I Phase. shaded pole special purpose 42 frame motor Self-aligning sleeve bearing HFR -~~ lubricant ""GURE 10-23 Shading coil "V" skew rotor Cutaway view of a shaded-pole induction motor. (Courtesy ofWestin8lwuse Electric Corporation.) SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 655 Slot cell insulation Automatic reset thermal overload protector I Phase. shaded pole special purpose 42 frame stator assembly + rotor assembly ,',moo.';",~, thennal overload protector Shading coil Shaded portion of pole face ( ., I Phase shaded pole 42 frame wound stator Slot cell insulation Shading coil (5) ___ ~ (b' FIGURE 10-14 Close-up views of the construction of a four-pole shaded-pole induction motor. (Courtesy of Watinghouse Electric Corporation.) 3. Pennanent split-capacitor motor 4. Split-phase motor S. Shaded-pole motor 656 ELECTRIC MACHINERY RJNDAMENTALS Naturally, the best motor is also the most expensive, and the worst motor is the least expensive. Also, not all these starting techniques are available in all motor size ranges. It is up to the design engineer to select the cheapest available motor for any given application that will do the job. 10.4 SPEED CONTROL OF SINGLE-PHASE INDUCTION MOTORS In general, the speed of single-phase induction motors may be controlled in the same manner as the speed of polyphase induction motors. For squirrel-cage rotor motors, the following techniques are available: I. Vary the stator frequency. 2. Change the number of poles. 3. Change the applied tenninal voltage VT. In practical designs involving fairly high-slip motors, the usual approach to speed control is to vary the terminal voltage of the motor. 1lle voltage applied to a motor may be varied in one of three ways: I. An autotransformer may be used to continually adjust the line voltage. nlis is the most expensive methexl of voltage speed control and is used only when very smooth speed control is needed. 2. An SCR or TRIAC circuit may be used to reduce the nns voltage applied to the motor by ac phase control. nlis approach chops up the ac wavefonn as described in Chapter 3 and somewhat increases the motor's noise and vibra-tion. Solid-state control circuits are considerably cheaper than autotransfonn-ers and so are becoming more and more common. 3. A resistor may be inserted in series with the motor's stator circuit. This is the cheapest methexl of voltage control, but it has the disadvantage that consider-able power is lost in the resistor, reducing the overall power conversion efficiency. Another technique is also used with very high-slip motors such as shaded-pole motors. Instead of using a separate autotransformer to vary the voltage ap-plied to the stator of the motor, the stator winding itself can be used as an auto-transfonner. Figure 10--25 shows a schematic representation of a main stator winding, with a number of taps along its length. Since the stator winding is wrapped about an iron core, it behaves as an autotransfonner. When the full line voltage V is applied across the entire main winding, then the induction motor operates normally. Suppose instead that the full line voltage is applied to tap 2, the center tap of the winding.1llen an identical voltage will be in-duced in the upper half of the winding by transformer action, and the total winding Tap I + + V -2V + V oltage /" V apphed 300% 2llll% 100% FIGURE 10-26 T", 2 SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 657 Ferromagnetic core Jo'IGURE 10-25 The use of a stator winding as an autotransformer. If voltage V is applied to the winding at the center tap. the total winding voltage will he 2V. ~~ _ _ ~ Vo Vo Vo The torque-speed characteristic of a shaded-pole induction motor as the terminal voltage is changed. Increases in VTmay he accomplished either by actually raising the voltage across the whole winding or by switching to a lower tap on the stator winding. voltage will be twice the applied line voltage.1lle total voltage applied to the wind-ing has effectively been doubled. Therefore, the smaller the fraction of the total coil that the line voltage is applied across, the greater the total voltage will be across the whole winding, and the higher the speed of the motor will be for a given load (see Figure 10- 26). This is the standard approach used to control the speed of single-phase mo-tors in many fan and blower applications. Such speed control has the advantage that it is quite inexpensive, since the only components necessary are taps on the main motor winding and an ordinary multiposition switch. It also has the advan-tage that the autotransformer effect does not consume power the way series resis-tors would. 658 ELECTRIC MACHINERY RJNDAMENTALS 10.5 THE CIRCUIT MODEL OF A SINGLE-PHASE INDUCTION MOTOR As previously described, an understanding of the induced torque in a single-phase induction motor can be achieved through either the double-revolving-field theory or the cross-field theory of single-phase motors. Either approach can lead to an equivalent circuit of the motor, and the torque-speed characteristic can be derived through either method. TIlis section is restricted to an examination of an equivalent circuit based on the double-revolving-field theory- in fact, to only a special case of that theory. We will develop an equivalent circuit of the main winding of a single-phase in-duction motor when it is operating alone. The technique of symmetrical compo-nents is necessary to analyze a single-phase motor with both main and auxiliary windings present, and since symmetrical components are beyond the scope of this book, that case will not be discussed. For a more detailed analysis of single-phase motors, see Reference 4. TIle best way to begin the analysis of a single-phase induction motor is to consider the motor when it is stalled. At that time, the motor appears to be just a single-phase transfonner with its secondary circuit shorted out, and so its equiva-lent circuit is that of a transformer. This equivalent circuit is shown in Figure 100027a. In this figure, Rl and Xl are the resistance and reactance of the stator winding, XM is the magnetizing reactance, and Rl and X2 are the referred values of the rotor's resistance and reactance. The core losses of the machine are not shown and will be lumped together with the mechanical and stray losses as a part of the motor's rotational losses. Now recall that the pulsating air-gap flux in the motor at stall conditions can be resolved into two equal and opposite magnetic fields within the motor. Since these fields are of equal size, each one contributes an equal share to the resistive and reactive voltage drops in the rotor circuit. It is possible to split the rotor equivalent circuit into two sections, each one corresponding to the effects of one of the magnetic fields. The motor equivalent circuit with the effects of the forward and reverse magnetic fields separated is shown in Figure 10--27b. Now suppose that the motor's rotor begins to turn with the help of an auxil-iary winding and that the winding is switched out again after the motor comes up to speed. As derived in Chapter 7, the effective rotor resistance of an induction motor depends on the amount of relati ve motion between the rotor and the stator magnetic fields. However, there are two magnetic fields in this motor, and the amount of relative motion differs for each of them. For the fOlward magnetic field, the per-unit difference between the rotor speed and the speed of the magnetic field is the slip s, where slip is defined in the same manner as it was for three-phase induction motors. The rotor resistance in the part of the circuit associated with the forward magnetic field is thus O.5R.Js. TIle forward magnetic field rotates at speed n.ync and the reverse magnetic field rotates at speed -n.ync. TIlerefore, the total per-unit difference in speed (on a base of n.ync) between the forward and reverse magnetic fields is 2. Since the rotor SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 659 I, I, --+ R, j X ] + jX2 V E , jXM R, (a) I, -+ R, jX] + E" jO.5XM 0.5R2 Forward -V + jO.5X2 E" jO.5XM ~ 0.5R2 Reverse -,b, FIGURE 10-27 (a) The equivalent circuit of a single-phase induction motor at standstill. Only its main windings are energized. (b) The equivalent circuit with the effects of the forward and reverse magnetic fields separated. is turning at a speed s slower than the forward magnetic field, the total per-unit dif-ference in speed between the rotor and the reverse magnetic field is 2 - s. There-fore, the effective rotor resistance in the part of the circuit associated with the re-verse magnetic field is 0.SRI(2 - s). The final induction motor equivalent circuit is shown in Figure 10- 28. Circuit Analysis with the Single-Phase Induction Motor Equivalent Circuit The single-phase induction motor equivalent circuit in Figure 10- 28 is similar to the three-phase equivalent circuit, except that there are both forward and back-ward components of power and torque present. 1lle same general power and torque relationships that applied for three-phase motors also apply for either the forward or the backward components orthe single-phase motor, and the net power and torque in the machine is the difference between the forward and reverse components. The power-now diagram of an induction motor is repeated in Figure 10- 29 for easy reference. 660 ELECTRIC MACHINERY RJNDAMENTALS R, + " + E lF O.5ZF jO.5XM 0.5 R2 Forward , -V + jO.5X2 E IB 0.5ZB jO.5XM 0.5 R2 Reverse 2-> -""GURE 10-28 The equivalent cin:uit of a single-phase induction motor running at speed with only its main windings energized. Slator 00 losses ""GURE 10-29 Rotor copper losses Core Mechanical losses losses Stray losses Rotational losses The power-flow diagram of a single-phase induction motor. To make Ihe calculalion of the input current flow into the motor simpler, it is customary to define impedances ZF and ZB, where ZF is a single impedance equivalent to all the forward magnetic fie ld impedance elements and ZB is a sin-gle impedance equivalent to all the backward magnetic field impedance elements (see Figure 10-30). These impedances are given by . (Ris + jX.0(jXM) ZF = RF + }XF = (Ris + jX2) + jXM ( 10-5) -+ z, v -+ z, V D SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 661 Fl G URE 10-30 A series oombination of RF andJXF is the jXp Thevenin equivalent of the forwaJd-field impedance elements. and therefore Rr must consume the same power from a given current as Rlls would. . RI(2 - s) + jX2 ZB = RB + JXB = [R2/(2 s) + jX21 + jXM ( 10-<5) In tenns of Zp and ZB, the current flowing in the induction motor's stator winding is ( 10-7) The per-phase air-gap power of a three-phase induction motor is the power consumed in the rotor circuit resistance O.5RJs. Similarly, the forward air-gap power of a single-phase induction motor is the power consumed by 0.5R2/s, and the reverse air-gap power of the motor is the power consumed by O.5RJ(2 - s). Therefore, the air-gap power of the motor could be calculated by detennining the power in the forward resistor O.5RJs, detennining the power in the reverse resis-tor 0.5Ri(2 - s), and subtracting one from the other. TIle most difficult part of this calculation is the determination of the sepa-rate currents fl owing in the two resistors. Fortunately, a simplification of this cal-culation is possible. Notice that the only resistor within the circuit elements com-posing the equivalent impedance Zp is the resistor Ris. Since Zp is equivalent to that circuit, any power consumed by ZF must also be consumed by the original cir-cuit, and since Ris is the only resistor i n the original circuit, its power consump-tion must equal that ofimpedancc Zp. Therefore, the air-gap power for the forward magnetic field can be expressed as 662 ELECTRIC MACHINERY RJNDAMENTALS ( 10-8) Similarly, the air-gap power for the reverse magnetic field can be expressed as PAGJJ = n(0.5 RB) ( 10--9) 1lle advantage of these two equations is that only the one current / 1 needs to be calculated to detennine both powers. TIle total air-gap power in a single-phase induction motor is thus (10-10) TIle induced torque in a three-phase induction motor can be found from the equation P AG Tind = --w ,"" where P AG is the net air-gap power given by Equation (10-10). (10-11 ) 1lle rotor copper losses can be found as the sum of the rotor copper losses due to the forward field and the rotor copper losses due to the reverse field. (10- 12) TIle rotor copper losses in a three-phase induction motor were equal to the per-unit relative motion between the rotor and the stator field (the slip) times the air-gap power of the machine. Similarly, the forward rotor copper losses of a single-phase induction motor are given by P RCL•F = SPAG•F (10- 13) and the reverse rotor copper losses of the motor are given by (10-14) Since these two power losses in the rotor are at different frequencies, the total ro-tor power loss is just their sum. TIle power converted from electrical to mechanical fonn in a single-phase induction motor is given by the same equation as P.:OD¥ for three-phase induction motors. TIlis equation is Since Wm = (1 - s)w.ync, this equation can be reexpressed as PCRf(2 - s) + jX2 ZB - RB + }XB -[Rf(2 s) + jX2] + jXM (3.13 fl/1.95 + jl.56 0)(j58.2 0) = (3.13 fl/1.95 + jl.56 0) + j58.2 0 (2.24L44.2° OXj5S.2 0) = (1.61 0 + jl.56 0) + j5S.2 0 = 2.ISL45.9° 0 = 1.51 + jl.56 0 These values will be used to detennine the motor clUTent, power, and torque. (a) The synchronous speed of this motor is II = 12Qf , = 120(60 Hz) = .ync P 6 pole 1200 r/min Since the motor is operating at 5 percent slip, its mechanical speed is 11m = (I -s)n~yDC (10-5) (10--6) 664 ELECTRIC MACHINERY RJNDAMENTALS n .. = (I - 0.05X 1200 r/min) = 1140 r/min (b) The stator current in this motor is "">CCC,"or,",,<o"".i~i~O~L~O, ""V~"'C-"'7<""C-"'CC"" = 1.52 n + j2.10 n + 0.5(25.4 n + j30.7 n) + 0.5(1.51 n + j1.56 n) = .""i'iii'0,;L,O" " ,,V Ccon -lIOLO° V _ 466L 50 6° A 14.98 n + jI8.23 n - 23.6L50.6° n -. -. (c) The stator power factor of this motor is PF = cos (- 50.6°) = 0.635 lagging (d) The input power to this motor is Pm = VI cos () = (110 VX4.66 AXO.635) = 325 W (e) The forward-wave air-gap power is PAGE = l i(o.5 RF) = (4.66 A)2(12.7 0) = 275.8 W and the reverse-wave air-gap power is PAG.B = l i(o.5 RB) = (4.66 A)2(0.755 V) = 16.4 W Therefore, the total air-gap power of this motor is PAG = PAG.F -PAG.B = 275.8 W -16.4 W = 259.4 W if) The power converted from electrical to mechanical fonn is P OO<IV = (I - s) P AG = (I - 0.05)(259.4 W) = 246 W (g) The induced torque in the motor is given by = 259.4 W = 2(x;N o (1200r/min)(lminJ60sX27Tradlr) . m (h) The output power is given by POlS. = Pooov -P~ = Pooov - POOle - PIDOCb - P_ y = 246 W - 35 W -16 W = 195 W (i) The load torque of the motor is given by p ~ T_ = w". (10--7) (10--8) (10-9) (10--10) (10--17) (10--11 ) SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 665 = 195W = 163N o (1140 r/minXI min/60 s)(21Tradlr) . m OJ Finally, the efficiency of the motor at these conditions is 195 W 100% = 325 W x 100% = 60% 10.6 OTHER TYPES OF MOTORS Two other types of motors- reluctance motors and hysteresis motors-are used in certain special-purpose applications. These motors differ in rotor construction from the ones previously described, but use the same stator design. Like induction motors, they can be built with either single- or three-phase stators. A third type of special-purpose motor is the stepper motor. A stepper motor requires a polyphase stator, but it does not require a three-phase power supply. The final special-purpose motor discussed is the brushless dc motor, which as the name suggests runs on a dc power supply. Reluctance Motors A reluctance motor is a motor which depends on reluctance torque for its opera-tion. Reluctance torque is the torque induced in an iron object (such as a pin) in the presence of an external magnetic field, which causes the object to line up with the external magnetic field. TIlis torque occurs because the external field induces an internal magnetic field in the iron of the object, and a torque appears between the two fields, twisting the object around to line up with the external field. In or-der for a reluctance torque to be produced in an object, it must be elongated along axes at angles corresponding to the angles between adjacent poles of the external magnetic field. A simple schematic of a two-pole reluctance motor is shown in Figure 1O-3 . It can be shown that the torque applied to the rotor of this motor is pro-portional to sin 20, where 0 is the electrical angle between the rotor and the stator magnetic fields. TIlerefore, the reluctance torque ofa motor is maximum when the angle between the rotor and the stator magnetic fields is 45°. A simple reluctance motor of the sort shown in Figure 10-3 I is a synchro-nous motor, since the rotor will be locked into the stator magnetic fields as long as the pullout torque of the motor is not exceeded. Like a nonnal synchronous mo-tor, it has no starting torque and will not start by itself. A self-starting reluctance motor that will operate at synchronous speed until its maximum reluctance torque is exceeded can be built by modifying the rotor of an induction motor as shown in Figure 10- 32. In this figure, the rotor has salient poles for steady-state operation as a reluctance motor and also has cage or amortisseur windings for starting. TIle stator of such a motor may be either of single- or three-phase construction. The torque- speed characteristic of this mo-tor, which is sometimes called a synchronous induction motor, is shown in Fig-ure 10-33. 666 ELECTRIC MACHINERY RJNDAMENTALS Single-phase 0' three-phase stator ""GURE 10-31 , "iDd""sin2/i The basic concept of a reluctance motor. o FIGURE 10-32 The rotor design of a "synchronous induction" or self-starting reluctance motor. An inleresting variation on Ihe idea of Ihe reluctance motor is Ihe Syn-crospeed motor, which is manufactured in the United States by MagneTek, Inc. TIle rotor of this motor is shown in Figure 10- 34. It uses "flux guides" to increase Ihe coupling between adjacenl pole faces and therefore to increase the maximum-reluctance lorque of the motor. With these flux guides, Ihe maximum-reluctance torque is increased to about 150 percent of the rated torque, as compared to just over 100 percent of the rated torque for a conventional reluctance motor. Hysteresis Motors Another special-purpose motor employs the phenomenon of hysteresis to produce a mechanical lorque. TIle rotor of a hysteresis motor is a smoolh cylinder of mag-netic material with no teeth, protrusions, or windings. TIle stator of the motor can be either single- or three-phase; but if it is single-phase, a permanent capacitor SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 667 ~ ,-------------~~---, 500 , 400 Main and auxiliary , , winding I I "', , , , , , ' ~ f--;;:; I I \ , , , , , , , , , Varies with / I \ , , starting I I \ position " ::? I \ of rotor " :c: : al \ I . .e! 1 2!.. \ I OJ I '"' \ " '" I \ , , , , , , " Main winding I \ , , " only I 200 100 , , O ~--~--~--~--~--~ o 20 40 60 80 100 Percentage of synchronous speed (a) FIGURE 10-34 ""GURE 10-33 The torque-speed characteristic of a single-phase self-starting reluctance ntotor. 'h' (a) The aluminum casting of a Synchrospeed motor rotor. (b) A rotor lamination from the motor. Notice the flux guides connecting the adjacent poles. These guides increase the reluctance torque of the motor. (Courtesy of MagneTek, Inc.) should be used with an auxiliary winding to provide as smooth a magnetic field as possible, since this greatly reduces the losses of the motor. Figure 10-35 shows the basic operation of a hysteresis motor. When a three-phase (or single-phase with auxi liary winding) current is applied to the sta-tor of the motor, a rotating magnetic fie ld appears within the machine. This rotat-ing magnetic field magnetizes the metal of the rotor and induces poles within it. When the motor is operating below synchronous speed, there are two sources of torque within it. Most of the torque is produced by hysteresis. When the 668 ELECTRIC MACHINERY RJNDAMENTALS Stator - "6-} ""GURE 10-35 , , II, , 0, 1 -6---..1 , , , , ~777 ' , Rotor The construction of a hysteresis motor. The main component of torque in this motor is proponional to the angle between the rotor and stator magnetic fields. magnetic field of the stator sweeps around the surface of the rotor, the rotor flux cannot follow it exactly, because the metal of the rotor has a large hysteresis loss. TIle greater the intrinsic hysteresis loss of the rotor material, the greater the angle by which the rotor magnetic field lags the stator magnetic field. Since the rotor and stator magnetic fields are at different angles, a fmite torque will be produced in the motor. In addition, the stator magnetic field will produce eddy currents in the rotor, and these eddy currents produce a magnetic field of their own, further increasing the torque on the rotor. TIle greater the relative motion between the ro-tor and the stator magnetic field, the greater the eddy currents and eddy-current torques. When the motor reaches synchronous speed, the stator flux ceases to sweep across the rotor, and the rotor acts like a pennanent magnet. The induced torque in the motor is then proportional to the angle between the rotor and the stator mag-netic field, up to a maximum angle set by the hysteresis in the rotor. TIle torque-speed characteristic of a hysteresis motor is shown in Figure 10--36. Since the amount of hysteresis within a particular rotor is a function of only the stator flux density and the material from which it is made, the hysteresis torque of the motor is approximately constant for any speed from zero to n,ync. The eddy-current torque is roughly proportional to the slip of the motor. TIlese two facts taken together account for the shape of the hysteresis motor's torque-speed characteristic. FIGURE 10-37 SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 669 ""GURE 10-36 n.. The torque--speed characteristic ofa motor. A small hysteresis motor with a shaded-pole stator. suitable for running an electric clock. Note the shaded stator poles. (Stt'phen J. Chapman) Since the torque of a hysteresis motor at any subsynchronous speed is greater than its maximum synchronous torque, a hysteresis motor can accelerate any load that it can carry during normal operation. A very small hysteresis motor can be built with shaded-pole stator con-struction to create a tiny self-starting low-power synchronous motor. Such a mo-tor is shown in Figure 10- 37. It is commonly used as the driving mechanism in electric clocks. An electric clock is therefore synchronized to the line frequency of the power system, and the resulting clock is just as accurate (or as inaccurate) as the frequency of the power system to which it is tied. 670 ELECTRIC MACHINERY RJNDAMENTALS Stepper Motors A stepper nwtor is a special type of synchronous motor which is designed to ro-tate a specific number of degrees for every electric pulse received by its control unit. Typical steps are 7.5 or 15° per pulse. 1llese motors are used in many control systems, since the position of a shaft or other piece of machinery can be controlled precisely with them. A simple stepper motor and its associated control unit are shown in Figure 1 0- 3S. To understand the operation of the stepper motor, examine Figure 10-39. TIlis figure shows a two-pole three-phase stator with a pennanent-magnet rotor. If a dc voltage is applied to phase a of the stator and no voltage is applied to phases band c, then a torque wi ll be induced in the rotor which causes it to line up with the stator magnetic field Bs, as shown in Figure 10--39b. Now assume that phase a is turned off and that a negative dc voltage is ap-plied to phase c. TIle new stator magnetic field is rotated 60° with respect to the previous magnetic field, and the rotor of the motor follows it around. By continu-ing this pattern, it is possible to construct a table showing the rotor position as a function of the voltage applied to the stator of the motor. If the voltage produced by the control unit changes with each input pulse in the order shown in Table 10-1, then the stepper motor will advance by 60° with each input pulse. It is easy to build a stepper motor with finer step size by increasing the num-ber of poles on the motor. From Equation (4- 31) the number of mechanical de-grees corresponding to a given number of electrical degrees is (10- 18) Since each step in Table 10-1 corresponds to 60 electrical degrees, the number of mechanical degrees moved per step decreases with increasing numbers of poles. For example, if the stepper motor has eight poles, then the mechanical angle of the motor's shaft will change by 15° per step. 1lle speed of a stepper motor can be related to the number of pulses into its control unit per unit time by using Equation (10-IS). Equation (10-1S) gives the mechanical angle of a stepper motor as a function of the electrical angle. If both sides of this equation are differentiated with respect to time, then we have a rela-tionship between the electrical and mechanical rotational speeds of the motor: 2 (1O-19a) wm = pW~ 2 (IO-19b) "' nm = p n ~ Since there are six input pulses per electrical revolution, the relationship between the speed of the motor in revolutions per minute and the number of pulses per minute becomes I nm = -iP npulses (10- 20) where n pulse< is the number of pulses per minute. SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 671 b " + b Control "' ( ) Voc unit , d B, +~--~ ( a) 1 2 3 4 , 6 7 8 , Voc 1---, Phase voltages. V ,"1~ number '. '. " 1 Voc 0 0 '. 2 0 0 - Voc 3 0 Voc 0 4 - Voc 0 0 5 0 0 Voc 6 0 - Voc 0 (b' ( " FIGURE 10-38 (a) A simple three-phase stepper motor and its associated control unit. The inputs to the control unit consist of a de power source and a control signal consisting of a train of pulses. (b) A sketch of the output voltage from the control unit as a series of control pulses are input. (e) A table showing the output voltage from the control unit as a function of pulse number. 672 ELECTRIC MACHINERY RJNDAMENTALS b o B, 0, " . b' ('J (bJ b c' 0 0 0, B , .' ~ O· b' (cJ ""GURE 10-39 Operation of a stepper motor. (a) A voltage V is applied to phase a of the stator. causing a current to flow in phase a and producing a stator magnetic field n s. The interaction of nR and n s produces a counterclockwise torque on the rotor. (b) When the rotor lines up with the stator magnetic field. the net torque falls to zero. (c) A voltage - V is applied to phase c of the stator. causing a current to flow in phase c and producing a stator magnetic field ns. The interaction of DR and Ds produces a counterclockwise torque on the rotor. causing the rotor to line up with the new position of the magnetic field. • TIlere are two basic types of stepper motors, differing only in rotor con-struction: permanent-magnet l)pe and reluctance l)pe. TIle pennanent-magnet type of stepper motor has a permanent-magnet rotor, while the reluctance-type stepper motor has a ferromagnetic rotor which is not a pennanent magnet. (The rotor of the reluctance motor described previously in this section is the reluctance type.) In general, the pennanent-magnet stepper motor can produce more torque SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 673 TABLE 10-1 Rotor position as a function of voltage in a two-pole stepper motor Input pulse number 2 3 4 5 6 Phase ,·oltaj!cs " b V 0 0 0 0 V - V 0 0 0 0 - v , Rotor position 0 0° - V 60° 0 120° 0 1SO° V 240° 0 3lX1° than the reluctance Iype, since the permanent-magnel stepper motor has lorque from both the pennanent rotor magnetic field and reluctance effects. Reluctance-type stepper motors are often built with a four-phase stator winding instead of the three-phase stator winding described above. A four-phase stator winding reduces the steps belween pulses from 60 electrical degrees to 45 electrical degrees. As menlioned earlier, the torque in a reluctance motor varies as sin 20, so the reluctance torque between steps will be maximum for an angle of 45°. Therefore, a given reluctance-type stepper motor can produce more torque with a four-phase stator winding than with a three-phase stator winding. Equation (10- 20) can be generalized to apply to all stepper motors, regard-less of the number of phases on their stator windings. In general, if a stator has N phases, it takes 2N pulses per electrical revolution in thai motor. 1l1erefore, the re-lationship between the speed of the motor in revolutions per minute and the num-ber of pulses per minute becomes (10- 21) Stepper motors are very useful in control and positioning systems because the compuler doing the controlling can know both the exact speed and position of the stepper motor without needing feedback infonnation from the shaft of the mo-tor. For example, if a control system sends 1200 pulses per minute 1 0 the two-pole stepper motor shown in Figure 10--38, then the speed of the motor will be exaclly -3(2 ; oles}l 200 pulses/min) = 200 r/rnin (10- 20) Furthennore, if the initial position of the shaft is known, then the computer can de-tennine the exact angle of the rotor shaft at any fulure time by simply counting the total number of pulses which it has sent to the control unit of the stepper motor. 674 ELECTRIC MACHINERY RJNDAMENTALS Example 10-2. A three-phase permanent-magnet stepper motor required for one particular application must be capable of controlling the position of a shaft in steps of 7.5°, and it must be capable of running at speeds of up to 300 r/min. (a) How many poles must this motor have? (b) At what rate must control pulses be ra:eived in the motor's control unit if it is to be driven at 300 r/min? Solutioll (a) In a three-phase stepper motor, each pulse advances the rotor's position by 60 electrical degrees. This advance must correspond to 7.5 ma:hanical degrees. Solving Equation (10--18) for P yields '. (60") P =2 0m =2 7.5° = 16 poles (b) Solving Equation (10--21) for npuhe. yields npul ... = NPn .. = (3 phasesX 16 poles)(300 r/min) = 240 pulsesls Brushless DC Motors Conventional dc motors have traditionally been used in applications where dc power sources are available, such as on aircraft and automobiles. However, small de motors of these types have a number of disadvantages. nle principal disad-vantage is excessive sparking and brush wear. Small, fast dc motors are too small to use compensating windings and interpoles, so armature reaction and L dildt ef-fects tend to produce sparking on their commutator brushes. In addition, the high rotational speed of these motors causes increased brush wear and requires regular maintenance every few thousand hours. I f the motors must work in a low-pressure environment (such as at high altitudes in an aircraft), brush wear can be so bad that the brushes require replacement after less than an hour of operation! In some applications, the regular maintenance required by the brushes of these dc motors may be unacceptable. Consider for example a dc motor in an ar-tificial heart-regular maintenance would require opening the patient's chest. In other applications, the sparks at the brushes may create an explosion danger, or unacceptable RF noise. For all of these cases, there is a need for a small, fast dc motor that is highly reliable and has low noise and long life. Such motors have been developed in the last 25 years by combining a small motor much like a pennanent magnetic stepper motor with a rotor position sensor and a solid-state electronic switching circuit. These motors are called brushless de nwtors because they run from a dc power source but do not have commutators and brushes. A sketch of a small brushless dc motor is shown in Figure 10-40, and a photograph of a typical brushless dc motor is shown in Figure 10-41. The rotor is similar to that of a pennanent magnet stepper motor, except that it is nonsalient. nle stator can have three or more phases (there are four phases in the exmnple shown). SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 675 " + ~ d' H , I , Control unit "' H , ~' , , d' d /1" "' Position sensor input (a) r-~-----------,--,_---------L---L-- , 1 1 ,- ___ I , 1 ____ 1 , r---------~---L----------,--,---- , 1 1 (b, FIGURE 10-40 (a) A simple brushless dc motor and its associated control unit. The inputs to the control unit consist of a dc power source and a signal proportional to the current rotor position. (b) The voltages applied to the stator coils. 676 ELECTRIC MACHINERY RJNDAMENTALS ,,' (h, ""GURE 10-41 (a) Typical brushless dc motors. (b) Exploded view showing the permanent magnet rotor and a three-phase (6-pole) stator. (Counesy of Carson Technologies. Inc.) TIle basic components of a brushless dc motor are I. A rennanent magnet rolor 2. A stator with a three-. four-, or more phase winding 3. A rotor position sensor 4. An electronic circuit 1 0 control the phases of the rotor winding A brushless dc motor functions by energizing one stator coil at a time with a constanl dc voltage. When a coil is turned on, it produces a stator magnetic field Bs, and a lorque is produced on the rotor given by which lends 1 0 align the rolor with the stator magnelic field. At the time shown in Figure 1O--40a, the stalor magnelic field Bs points to the left while the pennanent magnet rotor magnetic field BR points up, prOOucing a counterclockwise lorque on the rotor. As a resuli Ihe rotor will tum to the left. SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 677 If coil a remained energized all of the time, the rotor would turn until the two magnetic fields are aligned, and the n it would stop, just like a stepper motor. The key to the operation of a brushless dc motor is that it includes a position sen-sor, so that the control circuit will know when the rotor is almost aligned with the stator magnetic field. At that time coil a will be turned off and coil b will be turned on, causing the rotor to again experience a counterclockwise torque, and to con-tinue rotating. nlis process continues indefinitely with the coils turned on in the order a, b, C, d, - a, - b, -C, - d, etc., so that the motor turns continuously. The electronics of the control circuit can be used to control both the speed and direction of the motor. The net effect of this design is a motor that runs from a dc power source, with full control over both the speed and the direction of rotation. Brushless dc motors are available only in small sizes, up to 20 Wor so, but they have many advantages in the size range over which they are available. Some of the major advantages include: I. Relatively high efficiency. 2. Long life and high reliability. 3. Little or no maintenance. 4. V ery little RF noise compared to a dc motor with brushes. 5. V ery high speeds are possible (greater than 50,000 r/min). The principal disadvantage is that a brushless dc motor is more expensive than a comparable brush dc motor. 10.7 SUMMARY The ac motors described in previous chapters required three-phase power to func-tion. Since most residences and small businesses have only single-phase power sources, these motors cannot be used. A series of motors capable of running from a single-phase power source was described in this chapter. The first motor described was the universal motor. A universal motor is a se-ries dc motor adapted to run from an ac supply, and its torque-speed characteris-tic is similar to that of a series dc motor. The universal motor has a very high torque, but its speed regulation is very poor. Single-phase induction motors have no intrinsic starting torque, but once they are brought up to speed, their torque-speed characteristics are almost as good as those of three-phase motors of comparable size. Starting is accomplished by the addition of an auxiliary winding with a current whose phase angle differs from that of the main winding or by shading portions of the stator poles. The starting torque of a single-phase induction motor depends on the phase angle between the current in the primary winding and the current in the auxiliary winding, with maximum torque occurring when that angle reaches 90°. Since the split-phase construction provides only a small phase difference between the main and auxiliary windings, its starting torque is modest. Capacitor-start motors have 678 ELECTRIC MACHINERY RJNDAMENTALS auxiliary windings with an approximately 90° phase shift, so they have large start-ing torques. Permane nt split-capacitor motors, which have smaller capacitors, have starting torques intennediate between those of the split-phase motor and the capacitor-start motor. Shaded-pole motors have a very small effecti ve phase shift and therefore a small starting torque. Reluctance motors and hysteresis motors are special-purpose ac motors which can operate at synchronous speed without the rotor field windings required by synchronous motors and which can accelerate up to synchronous speed by themselves. These motors can have either single- or three-phase stators. Stepper motors are motors used to advance the position of a shaft or other mechanical device by a fixed amount each time a control pulse is received. 1lley are used extensively in control systems for positioning objects. Brushless dc motors are similar to stepper motors with pennanent magnet rotors, except that they include a position sensor. 1lle position sensor is used to switch the energized stator coil whenever the rotor is almost aligned with it, keep-ing the rotor rotating a speed set by the control electronics. Brushless dc motors are more expensive than ordinary dc motors, but require low maintenance and have high reliability, long life, and low RF noise. They are available only in small sizes (20 W and down). QUESTIONS 10- 1. What changes are necessary in a series dc motor to adapt it for operation from an ac power source? 10-2. Why is the torque-speed characteristic of a lUliversal motor on an ac source dif-ferent from the torque-speed characteristic of the same motor on a dc source? 10-3. Why is a single-phase induction motor lUlable to start itself without special auxil-iary windings? 10-4. How is induced torque developed in a single-phase induction motor (a) according to the double revolving-field theory and (b) according to the cross-field theory? 10-5. How does an auxiliary winding provide a starting torque for single-phase induc-tion motors? 10-6. How is the current phase shift accomplished in the auxiliary winding of a split-phase induction motor? 10-7. How is the current phase shift accomplished in the auxiliary winding of a capacitor-start induction motor? 10-8. How does the starting torque of a pennanent split-capacitor motor compare to that of a capacitor-start motor of the same size? 10-9. How can the direction of rotation of a split-phase or capacitor-start induction mo-tor be reversed? 10-10. How is starting torque produced in a shaded-pole motor? 10-11. How does a reluctance motor start? 10-12. How can a reluctance motor run at synchronous speed? 10-13. What mechanisms produce the starting torque in a hysteresis motor? 10-14. What mechanism produces the synchronous torque in a hysteresis motor? SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS 679 10-15. Explain the operation of a stepper motor. 10-16. What is the difference between a permanent-magnet type of stepper motor and a reluctance-type stepper motor? 10-17. What is the optimal spacing between phases for a reluctance-type stepper motor? Why? 10-18. What are the advantages and disadvantages of brush less de motors compared to or-dinary brush dc motors? PROBLEMS lO-1. A 120-V, ~ hp, 60-Hz, four-pole, split-phase induction motor has the following impedances: RI = 1.80n R2 = 2.50 n XI = 2.40 n X2 = 2.40 n At a slip of 0.05, the motor's rotational losses are 51 W. The rotational losses may be assumed constant over the normal operating range of the motor. If the slip is 0.05, find the following quantities for this motor: (a) Input power (b) Air-gap power (c) P coov (d) POOl (e) "T"md (j) "", (g) Overall motor efficiency (h) Stator power factor 10-2. Repeat Problem 10-1 for a rotor slip of 0.025. 10-3. Suppose that the motor in Problem 10-1 is started and the auxiliary winding fails open while the rotor is accelerating through 400 r/min. How much induced torque will the motor be able to produce on its main winding alone? Assuming that the ro-tationallosses are still 51 W, will this motor continue accelerating or will it slow down again? Prove your answer. 10-4. Use MATLA8 to calculate and plot the torque-speed characteristic of the motor in Problem 10--1, ignoring the starting winding. 10-5. A 220-V, 1.5-hp, SO-Hz, two-pole, capacitor-start induction motor has the follow-ing main-winding impedances: RI = 1.40n R2 = 1.50n Xl = 1.90n X2 = 1.90n At a slip of 0.05, the motor's rotational losses are 291 W. The rotational losses may be assumed constant over the normal operating range of the motor. Find the fol-lowing quantities for this motor at 5 percent slip: (a) Stator ClUTent (b) Stator power factor (c) Input power (d) PAG (e) P coov 680 ELECTRIC MACHINERY RJNDAMENTALS (j) P-(g) Tind (h) Tlood (i) Efficiency 10-6. Find the induced torque in the motor in Problem 10--5 ifit is operating at 5 percent slip and its terminal voltage is (a) 190 V, (b) 208 V, (c) 230 V. 10-7. What type of motor would you select to peIform each of the following jobs? Why? (a) Vacuum cleaner (b) Refrigerator (c) Air conditioner compressor (d) Air conditioner fan (e) Variable-speed sewing machine (j) Clock (g) Electric drill 10-8. For a particular application, a three-phase stepper motor must be capable of step-ping in 10° increments. How many poles must it have? 10-9. How many pulses per second must be supplied to the controllUlit of the motor in Problem 10-8 to achieve a rotational speed of 600 r/min? 10-10. Construct a table showing step size versus number of poles for three-phase and four-phase stepper motors. REFERENCES I. Fitzgerald. A. E.. and C. Kingsley. Jr. Electric Machinery. New Yort: McGraw-Hill. 1952. 2. National Electrical Manufacturers Association. Motors and Generators. Publication No. MG I-1993. Washington. D.C.: NEMA. 1993. 3. Veinott. G. C. Fractional and Sulifractional Horsepov."er Electric Motors. New Y ork:: McGraw-Hill. 1970. 4. Werninck:. E. H. (ed.). Electric Motor Handbook.. London: McGraw-Hill. 1978. APPENDIX A THREE-PHASE CIRCUITS A lmost all electric power generation and most of the power transmission in the world looay is in the form of three-phase ac circuits. A three-phase ac power system consists of three-phase generators, transmission lines, and loads. AC power systems have a great advantage over de systems in that their voltage levels can be changed with transfonners to reduce transmission losses, as described in Chapter 2. Three-phase ac power systems have two major advantages over singlc-phase ac power systems: (I) it is possible to get more power per kilogram of mctal from a three-phase machine and (2) the power delivered to a three-phase load is constant at all times, instead of pulsing as it does in single-phase systems. lluce-phase systems also make the use of induction motors easier by allowing them to start without special auxiliary starting windings. A.I GENERATION OF THREE-PHASE VOLTAGES AND CURRENTS A three-phase generator consists of three single-phase generators, with voltages equal in magnitude but differing in phase angle from the others by l20?E.1.ch of these three generators could be connected to one of three identical loads by a pair of wires, and the resulting power system would be as shown in Figure A-l c. Such a system consists of three single-phase circuits that happen to differ in phase an-gle by I 20?The current flowing to each load can be found from the equation (A-I ) 681 682 ELECTRIC MACHINERY RJNDA MENTALS «) ""GURE A- I VA(t) ",,fi Vsin w/V V A ",VLOoV VB(t) '" ,fi V sin (wt _ 120°) V VB "'VL - 1200V Vdt) '" ,fi V sin (wt _ 240°) V Vc '" V L 240° V (.) (b) Z Z ",ZL(J Z Z ",ZL(J Z Z ",ZL(J (a) A three-phase generator. consisting of three single-phase sources equal in magnitude and 120° apart in phase. (b) The voltages in each phase of the generator. (c) The three phases of the generator connected to three identicalloods. lHREE-PHASE CIRCUITS 683 v, v, '" F'IGUREA- l (concluded) (d) Pltasor dia.gram showing the voltages in each phase. .... x V , C , '.... Ve + F'IGUREA- 2 " -z 'N z The three ci["(;uits connected together with a. common neutral. Therefore, the currents fl owing in the three phases are I = VLO° = iL- (J A ZL 6 1 = VL-120° = iL- 120 o -(J B ZL 6 1 = VL-240° =/L-240o -(J e ZL 6 (A- 2) (A- 3) (A-4) It is possible to connect the negati ve ends of these three single-phase gener-ators and loads together, so that they share a common return line (called the neu-tral). The resulting system is shown in Figure A- 2; note that now only four wires are required to supply power from the three generators to the three loads. How much current is fl owing in the single neutral wire shown in Figure A- 2? The return current will be the sum of the currents flowing to each individ-ual load in the power system. This current is given by 684 ELECTRIC MACHINERY RJNDAMENTALS IN = IA + IB + Ie = /L- O + /L- O -120 0 + /L- O -240 0 = / cos (- 0) + jf sin (- 0) + I cos (- () -120°) + jf sin (- () -120°) + l cos (- () - 240°) + jlsin (- () - 240°) = f[eos (- (}) + cos(- () -120°) + cos (- (} - 240°)] + jf [sin (- 0) + sin (- (} -120°) + sin (- (} - 240°)] Recall the elementary trigonometric identities: cos (a - m = cos a cos {3 + sin a sin {3 sin (a - m = sin a cos {3 - cos a sin (3 Applying these trigonometric identities yields (A- 5) (A-6) (A- 7) IN= [[cas (- 0) + cos (- U) cos 120 0 + sin (- 0) sin 120 0 + cos (- 0) cos 240° + sin (- (f) sin 240°] + j/[sin (- 0) + sin (- 0) cos 120° - cos (- 0) sin 120 0 + sin (- () cos 240° - cos (- () sin 240°] [ 1 0. 1 0.] IN = I cos (- 0) -"2 cos (- 0) + 2 Sin (- (}) -"2 cos (- (}) - - ,-Sin (- (}) +jIsin C- O ) l sin c-O )- 0"COS(-O) l Sill C- O) + 0 cos C-0)] lL 2 2 2 2 As long as the three loads are equal, the return current in the neutral is zero! A three-phase power system in which the three generators have voltages that are exactly equal in magnitude and 120° different in phase, and in which alJ three loads are identical, is called a balanced three-phase system. In such a system, the neutral is actually unnecessary, and we could get by with only three wires instead of the original six. PHASE SEQUENCE. The phase sequence of a three-phase power system is the order in which the voltages in the individual phases peak.1lle three-phase power system illustrated in Figure A-I is said to have phase sequence abc, since the volt-ages in the three phases peak in the order a, b, c (see Figure A-I b). TIle phasor di-agram of a power system with an abc phase sequence is shown in Figure A-3a. It is also possible to connect the three phases of a power system so that the voltages in the phases peak in order the order a, c, b. This type of power system is said to have phase sequence acb. 1lle phasor diagram of a power system with an acb phase sequence is shown in Figure A- 3b. TIle result derived above is equally valid for both abc and acb phase sequences. In either case, if the power system is balanced, the current flowing in the neutral will be O. lHREE-PHASE CIRCUITS 685 v c v , FIGUREA-3 v, v , v c v, (b) (a) The phase voltages in a power system with an abc phase sequence. (b) The phase voltages in a power system with an acb phase sequence. A.2 VOLTAGES AND CURRENTS IN A THREE-PHASE CIRCUIT A connection of the sort shown in Figure A- 2 is called a wye (Y) connection be-cause it looks like the letter Y. Another possible connection is the delta (.6.) con-nection, in which the three generators are connected head to tail. The 11 connec-tion is possible because the sum of the three voltages VA + VB + V c = 0, so that no short-circuit currents will flow when the three sources are connected head to tail. Each generator and each load in a three-phase power system may be either Y- or l1-connected. Any number of Y- and l1-connected generators and loads may be mixed on a power system. Figure A-4 shows three-phase generators connected in Y and in 11. 1lle volt-ages and currents in a given phase are called phase quantities, and the voltages be-tween lines and currents in the lines connected to the generators are called line quantities. The relationship between the line quantities and phase quantities for a given generator or load depends on the type of connection used for that generator or load. These relationships wi ll now be explored for each of the Y and 11 connections. Voltages and Currents in the Wye (Y) Connection A Y-connected three-phase generator with an abc phase sequence connected to a resistive load is shown in Figure A- 5. The phase voltages in this generator are given by V = V LO° - . V",. = Vo/>L- 120° Ven = Vo/>L-240° (A-8) V co / 686 ELECTRIC MACHINERY RJNDAMENTALS '. -"+ C'BV~ V~ , ).' " ". V M " '. -b ~) v. , -,.) ""GURE A-4 (a) Y connection. (b) ;l. connection. . , --'.1 • V '" '. -b ""GURE A-S Y-connected generator with a resistive load. V . '0 \100 , ~ ~ ~ '2· '. Resistive load V . (h) '. V oo '. --" Since the load connected to this generator is assumed to be resistive, the current in each phase of the generator will be at the same angle as the voltage. Therefore, the current in each phase will be given by Ia = JLO° I b = JL- 120° (A- 9) v • lHREE-PHASE CIRCUITS 687 FlGUREA-6 Line-to-line and phase (line-to-neutral) voltages for the Y connection in Figure A- 5. From FigureA- 5, it is obvious that the current in any line is the same as the current in the corresponding phase. TIlerefore, for a Y connection, Y connection I (A-I 0) llle relationship between line voltage ,md phase voltage is a bit more complex. By Kirchhoff's voltage law, the line-to-line voltage Vah is given by Vah = Va -Vb = Vo/> LO° -Vo/>L- 120° = Vo/> - (-~ Vo/> - J V; Yo/»~ =~ Vo/> + J V; Yo/> -v'3Vo/>(~ + J1) -v'jV~30° Therefore, the relationship between the magnitudes of the line-to-line voltage and the line-to-neutral (phase) voltage in a V-connected generator or load is Y connection I (A-II ) In addition, the line voltages are shifted 30° with respect to the phase voltages. A phasor diagram of the line and phase voltages for the Y connection in Figure A- 5 is shown in Figure A-6. Note that for Y connections with the abc phase sequence such as the one in Figure A- 5, the voltage of a line leads the corresponding phase voltage by 30°. For Y connections with the acb phase sequence, the voltage of a line lags the cor-responding phase voltage by 30°, as you will be asked to demonstrate in a prob-lem at the end of the appendix. 688 ELECTRIC MACHINERY RJNDAMENTALS '. " " -A -""GURE A-7 ~-oonnected generator with a resistive load. Resistive Lo .. Although the relationships between line and phase voltages and currents for the Y connection were derived for the assumption of a unity power factor, they are in fact valid for any power factor. The assumption of unity-power-factor loads simply made the mathematics slightly easier in this development. Voltages and Currents in the Delta (d ) Connection A .6.-connected three-phase generator connected to a resistive load is shown in Figure A-7. The phase voltages in this generator are given by Vab = V~LO ° V".. = V~L -1 20 ° Yea = V~L-240 ° Because the load is resistive, the phase currents are given by lab = 1 4> LO° 1 "..= I4>L-120° t, = 1 4> L -240 0 (A-1 2) (A- 13) In the case of the .6. connection, it is obvious that the line-to-line voltage between any two lines will be the same as the voltage in the corresponding phase. In a .6. connection, .6. connection I (A-1 4) 1lle relationship between line current and phase current is more complex. It can be found by applying Kirchhoff's current law at a node of the .6.. Applying Kirch-hoff's current law to node A yields the equation la = lab - lea = 1 4>LO° - 14>L-240° ( I .0 ) 3 .0 = 14>--"2 /4> +} T I4> ="2 /4> -} TI4> I. I. TablcA- l I. lHREE-PHASE CIRCUITS 689 FIGURE 2-8 Line and phase currents for the;l. cOlloection in Figure A- 7. Summary of relationships in Y and &. connections \' connection ;l. connl.>etion Voltage magnitudes Vu = v'3V. Vu = V. Current magnitudes it = I. it = 0 /. abc phase sequence \' ... leads \'. by 30° I. lags I ... by 30° acb phase sequence \' ... lags \'. by 30° I. leads lob by 30° - 0 /. (': - jk) -y'3J",L-30° Therefore, the relationship between the magnitudes of the line and phase currents in a .6.-connected generator or load is .6. connection I (A-I S) and the line currents are shifted 30° relative 1 0 the corresponding phase currents. Note that for .6. connections with the abc phase sequence such as the one shown in Figure A- 7, the current of a line lags the corresponding phase current by 30° (sec Figure A- 8). For .6. connections with the acb phase sequence, the current of a line leads the corresponding phase current by 30°. TIle voltage and current relationships for Y- and .6.-connected sources and loads are summarized in Table A- I. 690 ELECTRIC MACHINERY RJNDAMENTALS ,----y I ;.'" ,-L'-, + Z ~ v",,(t) b --------------~ A.3 POWER RELATIONSHIPS IN THREE-PHASE CIRCUITS FlGUREA- 9 A balanced Y-connected load. Figure A- 9 shows a balanced V-connected load whose phase impedance is Z", = ZL (f'. If the three-phase voltages applied to this load are given by van(t) = V2V sin wt Vbn(t) = V2V sin(wI' -120°) vao(t) = y2V sin(wI' - 240°) then the three-phase currents fl owing in the load are given by iit) = V'Ll sin(wt -()) ib(t) = v'2Isin(wt -120° -()) ( (t) = v'2I sin(wt - 240° -()) (A-1 6) (A-1 7) where I = VIZ. How much power is being supplied to this load from the source? TIle instantaneous power supplied 1.0 one phase of the load is given by the equation I pet) = v(t)i(t) I lllerefore, the instantaneous power supplied to each of the three phases is PaCt) = van(tKlt) = 2VI sin(wI') sin(wI' -()) Pb(t) = v",,(t)ib(t) = 2 VI sin(wI' -120°) sine wt -120° -()) pJt) = vao(t)iJt) = 2Vl sin(wt - 240°) sin(wI' - 240 0 -()) A trigonometric identity states that sin a sin (3 = k[cos(a -(3) - cos(a - (3)1 (A- 18) (A-1 9) (A- 20) Applying this identity to Equations (A-1 9) yields new expressions for the power in each phase of the load: lHREE-PHASE CIRCUITS 691 p I-____ PhaseA -··-PhaseB -------. Phase C Total power -" , \ I \ ,'. ~. I \ " 1 \' \I \ ,' , I. 'i ti. 1\ ~ . ~ /. " / ~ ; , ; , , , , " I " .' , , , , " I \ I \ " , , , , , , '\ I \ ./ \ ,' , I 'i .. , , , , , I ' , -\ , , \ , " • , \ , , , -I :', , , ' , ' , " " , , , , , , , , I ' ,', " I \ I \ : ' , ; " ' " • \ I ' / i , " , I ' I \ , , ; , , ; , ; , ; , , , , , , , , , , , ; , ; , ; , , , , , , , , , , \ , , , , , , \ ; , ; , , " , F'IGUREA- 1O , , " I , ' , , " I , , , , " , , " , , , , , , , , , , ' , , \ ' , , , , ; , , " I , ;\ " '; I " I , , I, , , , , , , , , '" 8 , 2 Instantaneous power in phases a, b, and c, plus the total power supplied to the load. PaCt) = V/[COS () -cos(2wl -())] Pb(t) = V/[COS () -cos(2wl - 240 0 -())] p/t) = V/[COS () -cos(2wl - 480 0 -())] WI (A- 2I) TIle total power supplied to the e ntire three-phase load is the sum of the power supplied to each of the individual phases. The power supplied by each phase consists of a constant component plus a pulsing component. However, the pulsing components in the three phases cancel each other out since they are 12(? out of phase with each other, and the final power supplied by the three-phase power system is constant. This power is given by the equation: p,jl) = Plt(t) + Ps(t) + pel,t) = 3Vl cos () (A- 22) llle instantaneous power in phases a, b, and c are shown as a function of time in Figure A-I O. Note that the total power supplied to a balanced three-phase load is constant at all times. TIle fact that a constant power is supplied by a three-phase power system is one of its major advantages comp.:1.red to single-phase sources. Three-Phase Power Equations Involving Phase Quantities The single-phase power Equations (1-60) to (1-66) apply to each phase ofa Y- or a-connected three-phase load, so the real, reactive, and apparent powers supplied to a balanced three-phase load are given by 692 ELECTRIC MACHINERY RJNDAMENTALS p = 3 V",1 ,/> COS () (A- 23) Q = 3V",I", sin () (A- 24) S =3V",I", I (A- 2S) p = 3 / ~ Z cos () (A- 26) Q = 3 / ~ Z sin () (A- 27) S = 3/~Z (A- 2S) 1lle angle () is again the angle between the voltage and the current in any phase of the load (it is the same in all phases), and the power factor of the load is the cosine of the impedance angle (). The power-triangle relationships apply as well. Three-Phase Power Equations Involving Line Quantities It is also possible to derive expressions for the power in a balanced three-phase load in tenns of line quantities. This derivation must be done separately for Y- and ~-c onnec ted loads, since the relationships between the line and phase quantities are different for each type of connection. For a Y-connected load, the power consumed by a load is given by p = 3Vo/>lo/> cos () (A- 23) For this type of load, II- = 10/> and Vu = V3V 0/>, so the power consumed by the load can also be expressed as (A- 29) For a l1-connected load, the power consumed by a load is given by (A- 23) For this type of load, II- = V3/0/> and Va = V 0/>, so the power consumed by the load can also be expressed in tenns of line quantities as P = 3Vu ( ~) cos () = V3Vull- cos () (A- 29) lHREE-PHASE CIRCUITS 693 This is exactly the same equation that was derived for a V-connected load, so Equation (A-29) gives the power of a balanced three-phase load in tenns of line quantities regardless of the connection of the load, The reactive and apparent powers of the load in terms of line quantities are (A-3D) (A-31) It is important to realize that the cos () and sin ()terms in Equations (A-29) and (A-30) are the cosine and sine of the angle between the phase voltage and the phase current, not the angle between the line-to-line voltage and the line current. Remember that there is a 30 0 phase shift between the line-to-line and phase volt-age for a Y connection, and between the line and phase current for a .1.. connection, so it is important not to take the cosine of the angle between the line-to-line volt-age and line current. A.4 ANALYSIS OF BALANCED THREE-PHASE SYSTEMS If a three-phase power system is balanced, it is possible to determine the voltages, currents, and powers at various points in the circuit with a per-phase equivalent circuit, This idea is illustrated in Figure A-II. Figure A-II a shows a V-connected generator supplying power to a V-connected load through a three-phase trans-mission line. In such a balanced system, a neutral wire may be inserted with no effect on the system, since no current flows in that wire. nlis system with the extra wire in-serted is shown in Figure A-II b. Also, notice that each of the three phases is iden-tical except for a 120 0 shift in phase angle. Therefore, it is possible to analyze a circuit consisting of one phase and the neutral, and the results of that analysis wil I be valid for the other two phases as well if the 120 0 phase shift is included. Such a per-phase circuit is shown in Figure A-Il c. There is one problem associated with this approach, however. It requires that a neutral line be available (at least conceptually) to provide a return path for current fl ow from the loads to the generator. nlis is fine for V-connected sources and loads, but no neutral can be connected to .1..-connected sources and loads. How can .1..-connected sources and loads be included in a power system to be analyzed? The standard approach is to transfonn the impedances by the Y--h. transfonn of elementary circuit theory. For the special case of balanced loads, the Y - .1.. transformation states that ad-con nected load consisting of three equal im-pedances, each of value Z, is totally equivalent to a V-connected load consisting of three impedances, each of value Z /3 (see Figure A-1 2). This equivalence means that the voltages, currents, and powers supplied to the two loads cannot be distinguished in any fashion by anything external to the load itself. 694 ELECTRIC MACHINERY RJNDA MENTALS Transmission line (a) Transmission line Neutral ,b, Transmission line • • 1 -II • I, + "--Z. -" , H GURE A- l1 (a) A Y -connected generator and load. (b) System with neutral insened. (c) The per-phase equivalent circuit. lHREE-PHASE CIRCUITS 695 , -------------, ,-----------------, ~ 3 L ___ -.l L __ ~ FIGUREA- 12 Y-;I. transformation. A Y-connected impedance of 713 n is totally equivalent to a ;I.-connected impedance of Z n to any circuit connected to the load's terminals. 0.06 0 jO.120 0.06 0 jO.12 n + Vc~ '" l20L- 240o v"" '" l20LO" 208 V 0.060 jO.120 FlGUREA- 13 The three-phase circuit of Example A- I. If a.-connected sources or loads include voltage sources, then the magni-tudes of the voltage sources must be scaled according to Equation (A- II ), and the effect of tile 30° phase shift must be included as well. Example A- I. A 208-V three-phase power system is shown in Figure A- 13. It consists of an ideal 208-V V-connected three-phase generator cOlUlected through a three-phase transmission line to a V-connected load. The transmission line has an impedance of 0.06 + jO.12 n per phase, and the load has an impedance of 12 + j9 n per phase. For this simple power system, find (a) The magnitude of the line current IL (b) The magnitude of the load's line and phase voltages V u and V#. 696 ELECTRIC MACHINERY RJNDAMENTALS + "-' -0.06 n jO.l2 n -I, V . '" 120 L 0° V ,,( Z, I 2+fJ n Fl GUREA- 1 4 Per-phase circuit in Example A- I. (e) The real, reactive, and apparent powers consruned by the load (d) The power factor of the load (e) The real, reactive, and apparent powers conswned by the transmission line if) The real, reactive, and apparent powers supplied by the generator (g) The generator's power factor Solutioll Since both the generator and the load on this power system are V-connected, it is very sim-ple to construct a per-phase equivalent circuit. This circuit is shown in Figure A- 14. (a) The line current flowing in the per-phase equivalent circuit is given by v 1 1' = ,--:;"-"'" Zl ... + Zl .... 120 L O° V = m (oCi. 06 ;C:;: +-' j" O.'C12 "'1l") :o'+ ~(I" 2C; + :C j "'9mll ) 120LO° 120 L O° = ~~"' - ~~'" 12.06 + j9.12 -IS.12L 37.1 0 = 7.94L -37.lo A The magnitude of the line current is thus 7.94 A. (b) The phase voltage on the load is the voltage across one phase of the load. This voltage is the product of the phase impedance and the phase current of the load: V . L = 1 .t.Z#, = (7.94L -37.l o AXl2 + j9 0 ) = (7.94L -37.1 ° AXI5L36.9° 0) = 119.IL - 0.2° V Therefore, the magnitude of the load 's phase voltage is V . L = 119.1 V and the magnitude of the load's line voltage is Vu.= V3V#-=206.3 V (e) The real power consumed by the load is P=3 VV.cos (J = 3(119.1 VX7.94 A) cos 36.9° = 2270 W lHREE-PHASE CIRCUITS 697 The reactive power consumed by the load is QIood = 3 V.I. sin 0 = 3(119.1 VX7.94 A) sin 36.90 = 1702 var The apparent power consumed by the load is Slood = 3VJ. (d) The load power factor is = 3(119. 1 VX7.94A) = 2839 VA PFIood = cos 0 = cos 36.90 = 0.8 lagging (e) The current in the transmission line is 7 .94L -37.1 A, and the impedance of the line is 0.06 + jO.12 n or O.134L63.4° n per phase. Therefore, the real, reac-tive, and apparent powers consumed in the line are PliDe = 31iZ cos 0 = 3(7.94 A? (0.1340) cos 63.4 0 = 11.3 W Q1ine = 31lZ sin 0 = 3(7.94 A? (0.1340) sin 63.40 = 22.7 var S1ine = 31iZ = 3(7.94A?(0.1340) = 25.3 VA (A- 26) (A- 27) (A- 28) (jJ The real and reactive powers supplied by the generator are the swn of the pow-ers consumed by the line and the load: P sea = P1iDe + Plood = I1.3W+2270W =2281 W Q8«1 = Q1ine + QIood = 22.7var + 1702var = 1725var The apparent power of the generator is the square root of the sum of the squares of the real and reactive powers: s = yp2 + Q2 = 2860 VA to" t"" t"" (g) From the power triangle, the power-factor angle 0 is _ _ IQ8"' _ _ 11725VAR _ 0 08", - tan p - tan 2281 W - 37.1 ." Therefore, the generator's power factor is PF gen = cos 37. 10 = 0.798 lagging 698 ELECTRIC MACHINERY RJNDAMENTALS 0.060 p.120 IL 0.06 0 p.12 0 V en ", 120 L - 240° Y V ",,'" 120LOoy Vu ",208 V V ",,'" 120L - 120oV '..... Z~ + 0.06 0 p.12 0 Jo'IGURE A- IS Three-phase circuit in Example A- 2. 0.06 0 +jO.1 2 0 .1 -I" " + • + + V . '" 120 L 0" -V . v ' " ---Jo'IGURE A- 16 Per-phase circuit in Example A- 2. Example A-2. Re~at Example A- I for a ~ -connected load, with everything else unchanged. Solutioll This power system is shown in FigureA- 15. Since the load on this power system is ~ con-nected, it must first be converted to an equivalent Y form. The phase im~dance of the ~ connected load is 12 + j9 n so the equivalent phase impedance of the corresponding Y fonn is z, . ZY = T = 4+}3n The resulting per-phase equivalent circuit of this system is shown in Figure A- 16. (a) The line current flowing in the per-phase equivalent circuit is given by lHREE-PHASE CIRCUITS 699 120L O° V = " (O". 06 = + c) "· O ".1C: 2 ~1l~ ) :-: + ;',; ( 4OC+ ~ j 3 'Om ) 120L O° 120L O° = = 4.06 + j3.12 5.12L37.5° = 23.4L -37.5° A The magnitude of the line current is thus 23.4 A. (b) The phase voltage on the equivalent Y load is the voltage across one phase of the load. This voltage is the product of the phase impedance and the phase cur-rent of the load: V, - I' Z' .L-.L.L = (23 .4L -37.5° A X4 + j3 fl) = (23.4L -37.5" A X5L36.9° n ) = 117L - 0.6° V The original load was ~ cOlUlected, so the phase voltage of the original load is V¢L= V3(117 V)=203 V and the magnitude of the load's line voltage is Va = V.u, = 203 V (c) The real power consumed by the equivalent Y load (which is the same as the power in the actual load) is P_=3V.,t.cos (J = 3(117 VX23.4 A) cos 36.9° = 657 1 W The reactive power consumed by the load is Q 1Nd = 3V.,t.sin (J = 3(11 7 V)(23.4 A) sin 36.9° = 4928 var The apparent power consumed by the load is SI""" = 3V.,t. (d) The load power factor is = 3(117 V)(23.4A) = 82 13 VA PF_ = cos (J = cos 36.9° = 0.8 lagging (e) The current in the transmission is 23.4L -37.5° A, and the impedance of the line is 0.06 + j O.12 n or O.134L63.4° n per phase. Therefore, the real, re-active, and apparent powers consruned in the line are PliDe = 31l Z cos (J = 3(23.4A)2(O.134 n ) cos 63.4° = 98.6 W (A-26) 700 ELECTRIC MACHINERY RJNDAMENTALS Qti ... = 3/lZ sin (J = 3(23.4 A:Y(0.1 34 f.!) sin 63.4 0 = 197var Sti ... = 3/lZ = 3(23.4 A:Y(0.1 34 f.!) = 220 VA (A- 27) (A- 28) (f) The real and reactive powers supplied by the generator are the sums of the pow-ers consumed by the line and the load: P800 = PbJte + P_ = 98.6W + 6571 W = 6670W Q gen = Qlime + QIoad = 197 var + 4928 VAR = 5125 var The apparent power of the generator is the square root of the SlUll of the squares of the real and reactive powers: S = yp2 + Q2 = 8411 VA ~ 800 800 (g) From the power triangle, the power-factor angle (J is _ _ IQ8ea _ _ 15125var _ 0 (J800 - tan p - tan 6670 W - 37.6 ,-Therefore, the generator's power factor is PF seo = cos 37.6° = 0.792 lagging A.S ONE-LINE DIAGRAMS As we have seen in this chapter, a balanced three-phase power system has three lines connecting each source with each load, one for each of the phases in the power system. The three phases are all similar, with voltages and currents equal in amplitude and shifted in phase from each other by 120°. Because the three phases are all basically the same, it is customary to sketch power systems in a simple fonn with a single line representing all three phases of the real power system. These one-line diagrams provide a compact way to represent the interconnections of a power system. One-line diagrams typically include all of the major compo-nents of a power system, such as generators, transformers, transmission lines, and loads with the transmission Jines represented by a single line. The voltages and types of connections of each generator and load are usually shown on the diagrrun. A simple power system is shown in Figure A-I 7, together with the corresponding one-line diagram. A.6 USING THE POWER TRIANGLE I f the transmission lines in a power system can be assumed to have negligible im-pedance, then an important simplification is possible in the calculation of three-lHREE-PHASE CIRCUITS 701 Generator Load I Lood2 + t' + '~ • 7 ..,1 ('j Bus I L",", ( A connected "-G, Y connected '"'''''2 Y connected (bj FIGURE 2-17 (a) A simple power system with a V-connected generator. a A-connected load. and a V-connected load. (b) The corresponding one-line diagram. 7 -'2 phase currenls and powers. This simplificalion depends on the use of the real and reactive powers of each load to detennine the currents and power factors at vari-ous points in the system. For example, consider the simple power system shown in Figure A-1 7. If the transmission line in that power system is assumed to be lossless, the line volt-age at the generatorwilJ be the same as the line voltage at the loads. If the gener-ator voltage is specified, then we can find the current and power factor at any point in this power system as follows: I. Detennine the line voltage at the generator and the loads. Since the transmis-sion line is assumed to be lossless, these two voltages will be identical. 2. Determine the real and reactive powers of each load on the power system. W e can use the known load voltage to perfonn this calculation. 3. Find the total real and reactive powers supplied to all loads "downstream" from the point being examined. 702 ELECTRIC MACHINERY RJNDAMENTALS BusA , , I" ~ , , 480 V three-phase HGURE A- IS The system in Example A- 3. Lo,' I Lo,' 2 Delta connected Z , '" IOL30on Wye connected Z, '" 5L- 36.87°n 4. Determine the system power factor at that point, using the power-triangle relationships. 5. Use Equation (A- 29) to detennine line currents, or Equation (A- 23) to de-tennine phase currents, at that point. nlis approach is commonly employed by engineers estimating the currents and power fl ows at various points on distribution systems within an industrial plant. Within a single plant, the lengths of transmission Ii nes will be quite short and their impedances will be relatively small, and so only small errors will occur if the impedances are neglected. An engineer can treat the line voltage as constant, and use the power triangle method to quickly calculate the effect of adding a load on the overall system current and power factor. Example A-3. Figure A- 18 shows a one-line diagram of a sma11480-V industrial distribution system. The power system supplies a constant line voltage of 480 V, and the impedance of the distribution lines is negligible . Load I is a ~ -co nnected load with a phase impedance of IOL30° n, and load 2 is a V-connected load with a phase impedance of 5L -36.87° n. (a) Find the overall power factor of the distribution system. (b) Find the total line current supplied to the distribution system. Solutioll The lines in this system are assumed impedanceless, so there will be no voltage drops within the system. Since load I is ~ cormected, its phase voltage will be 480 V. Since load 2 is Y connected, its phase voltage will be 4801\13 = 277 V. The phase current in load I is Therefore, the real and reactive powers of load I are P I = 3V'I/'1 cos (J = 3(480 V)(48 A) cos 30° = 59.9 kW lHREE-PHASE CIRCUITS 703 QI = 3V. I/. I sin (J = 3(480 VX48 A) sin 30° = 34.6 kvar The phase C lUTent in load 2 is I~ = 2~7 0V = 55.4 A Therefore, the real and reactive powers of load 2 are P2 = 3 V.2/~cos (J = 3(277 V)(55.4 A) cos( - 36.87°) = 36.8 kW Q2 = 3 V~/.2 sin (J = 3(277 V)(55.4 A) sin( -36.87°) = -27.6 kvar (a) The total real and reactive powers supplied by the distribution system are P,ot = PI + P2 = 59.9 kW + 36.8 kW = 96.7 kW Q,ot = QI + Q2 = 34.6 kvar - 27.6 kvar = 7.00 kvar From the power triangle, the effective impedance angle (J is given by (J = tan- I ~ _ - I 7.00 kvar _ 4140 - tan 96.7 kW -. The system power factor is thus PF = cos (J = cos(4.14°) = 0.997 lagging (b) The total line current is given by QUESTIONS h = ~o-"P- V3V Lcos () 1 -96.7 kW = 1l7 A L -V3(480 V)(0.997) A-I. What types of cOlUlections are possible for three-phase generators and loads? A-2. What is meant by the tenn "balanced" in a balanced three-phase system? A-3. What is the relationship between phase and line voltages and currents for a wye (Y) cOlUlection? A-4. What is the relationship between phase and line voltages and currents for a delta (.6.) cOlUlection? A-5. What is phase sequence? A-6. Write the equations for real, reactive, and apparent power in three-phase circuits, in tenns of both line and phase quantities. A-7. What is a Y-6. transform? 704 ELECTRIC MACHINERY RJNDAMENTALS PROBLEMS A- I. TIrree impedances of 4 + j3 n are.6. connected and tied to a three-phase 208-V power line. Find I., IL' P, Q, S, and the power factor of this load. A-2. Figure PA- I shows a three-phase power system with two loads. The a-connected generator is producing a line voltage of 480 V, and the line impedance is 0.09 + fJ.16 n. Load I is Y connected, with a phase impedance of2.5L36.87° n and load 2 is a connected, with a phase impedance of 5L -200 n. I" 0.090 jO.l60 -V""",480L- 2400Y • '".1 N Vah '" 480LO" Y Vbe ",480L- 1200 V 0.09 n jO.l6 n 0.09 n jO.l6 n Generator Loodl fo'IGURE I'A-1 The system in Problem A- 2. (a) What is the line voltage of the two loads? (b) What is the voltage drop on the transmission lines? (c) Find the real and reactive powers supplied to each load. Lood2 Z.t '" 2.5L36.87°n Z.2'" 5L- 20on (d) Find the real and reactive power losses in the transmission line. (e) Find the real power, reactive power. and power factor supplied by the generator. A-3. Figure PA- 2 shows a one-line diagram of a simple power system containing a sin-gle 480-V generator and three loads. Assume that the transmission lines in this power system are lossless, and answer the following questions. (a) Assrune that Load I is Y cOlUlected. What are the phase voltage and currents in that load? (b) Assrune that Load 2 is.6. connected. What are the phase voltage and currents in that load? (c) What real, reactive, and apparent power does the generator supply when the switch is open? (d) What is the total line current IL when the switch is open? (e) What real, reactive, and apparent power does the generator supply when the switch is closed? lHREE-PHASE CIRCUITS 705 Bus I I, -Lood I lOO kW 0.9 PF lagging 480 V 1.0"'2 SO kVA Y connected O.S PF lagging SO kW 0.S5 PF leading -1 1.0'" 3 I FIGURE PA- 2 The power system in Problem A- 3. (f) What is the total line current h when the switch is closed? (g) How does the total line current h compare to the sum of the three individual currents It + 1 2 + 1 3? If they are not equal, why not? A-4. Prove that the line voltage of a Y -connected generator with an acb phase sequence lags the corresponding phase voltage by 30°. Draw a phasor diagram showing the phase and line voltages for this generator. A-5. Find the magnitudes and angles of each line and phase voltage and current on the load shown in Figure PA-3. " -'« FIGURE PA-3 The system in Problem A- 5. A-6. Figure PA-4 shows a one-line diagram of a small 480-V distribution system in an industrial plant. An engineer working at the plant wishes to calculate the current that will be drawn from the power utility company with and without the capacitor bank switched into the system. For the purposes of this calculation, the engineer will as-sume that the lines in the system have zero impedance. (a) If the switch shown is open, ftnd the real, reactive, and apparent powers in the system. Find the total current supplied to the distribution system by the utility. 706 ELECTRIC MACHINERY RJNDAMENTALS SOV , , , , , ~ I, ' , ""GURE P A- 4 The system in Problem A--6. Lood I Lood 2 ~ T Delta connected z. ",IOL30o n Wye connected z. '" 4L36.87° n Capacitor "',' Wye connected z. '" 5L- 90on (b) Repeat part (a) with the switch closed. (c) What happened to the total current supplied by the power system when the switch closed? Why? REFERENCE I. Alexander. Charles K., and Matthew N. O. Sadiku: Fundamentals of Electric Circuits, McGraw-Hill. 2CXXl. APPENDIX B COIL PITCH AND DISTRIBUTED WINDINGS A s mentioned in Chapter 4, the induced voltage in an ac machine is sinusoidal only if the harmonic components of the air-gap flux density are suppressed. This appendix describes two techniques used by machinery designers to suppress harmonics in machines. B.1 THE EFFECT OF COIL PITCH ON AC MACHINES In the simple ac machine design of Section 4.4, the output voltages in the stator coils were sinusoidal because the air-gap flux density distribution was sinusoidal. If the air-gap flux density distribution had not been sinusoidal, then the output voltages in the stator would not have been sinusoidal either. They would have had the same nonsinusoidal shape as the flux density distribution. In general, the air-gap flux density distribution in an ac machine will not be sinusoidal. Machine designers do their best to pnxluce sinusoidal flux distributions, but of course no design is ever perfect. llle actual flux distribution will consist of a fundamental sinusoidal component plus hannonics. TIlese hannonic components of flux will generate hannonic components in the stator's voltages and currents. The hannonic components in the stator voltages and currents are undesir-able, so techniques have been developed to suppress the unwanted hannonic com-ponents in the output voltages and currents ofa machine. One important technique to suppress the harmonics is the use offractionnl-pitch windings. 707 708 ELECTRIC MACHINERY RJNDAMENTALS o o N r-, o fo'IGURE 8- 1 s 0 Pp'" 90° mechanical Il 180° electrical ~Nl---e!l-s The pole pitch of a four-pole machine is 90 mechanical or 180 electrical degrees. The Pitch of a Coil The pole pitch is the angular distance between two adjacent poles on a machine. TIle pole pitch of a machine in mechanical degrees is I pp=~ 1 (8- 1) where Pp is the pole pitch in mechanical degrees and P is the number of poles on the machine. Regardless of the number of poles on the machine, a pole pitch is always 1 80 electrical degrees (see Figure B- 1 ). I f the stator coil stretches across the same angle as the pole pitch, it is called afull-pitch coil. If the stator coil stretches across an angle smaller than a pole pitch, it is called afractional-pitch coil. The pitch of a fractional-pitch coil is of-ten expressed as a fraction indicating the portion of the pole pitch it spans. For ex-ample, a 5/6-pitch coil spans fi ve-sixths of the distance between two adjacent poles. Alternatively, the pitch of a fractional-pitch coil in electrical degrees is given by Equations (B- 2): e m P ~ - x 180 0 P , (B- 2a) where Om is the mechanical angle covered by the coil in degrees and Pp is the ma-chine's pole pitch in mechanical degrees, or (8- 2b) COIL PITCH AND DISTRIBlJfED WINDINGS 709 Air-gap flUl( density: B(a ):. BM cos (wl - a) , -d Rotor Air gap Stator +-+f----- p 90" P - ~ , - b '., --~CC~e-------R , " .. ccw~'---VoltageiSreallYintothepage. oorroalonlS. . B· . h II smce IS negatIVe ere. FIGURE B-2 A fractional-pitch winding of pitch p. The vector magnetic flux densities and velocities on the sides of the coil. The velocities are from a frame of reference in which the magnetic field is stationary. where e ", is the mechanical angle covered by the coil in degrees and P is the num-ber of poles in the machine. Most practical stator coils have a fractional pitch, since a fractional-pitch winding provides some important benefits which will be explained later. Windings employing fractional-pitch coils are known as chorded windings. The Induced Voltage of a Fractional-Pitch Coil What effect does fractional pitch have on the output voltage of a coil? To find out, examine the simple two-pole machine with a fractional-pitch winding shown in Figure 8-2. TIle pole pitch of this machine is 180°, and the coil pitch is p. 1lle voltage induced in this coil by rotating the magnetic field can be found in exactly the same manner as in the previous section, by detennining the voltages on each side of the coil. The total voltage wil l just be the sum of the voltages on the indi-vidual sides. 710 ELECTRIC MACHINERY RJNDAMENTALS As before, assume that the magnitude of the flux density vector B in the air gap between the rotor and the stator varies sinusoidally with mechanical angle, while the direction of B is always radial ly outward. If a is the angle measured from the direction of the peak rotor flux density, then the magnitude of the flux density vector B at a point around the rotor is given by B = BM cosa (8-3a) Since the rotor is itself rotating within the stator at an angular velocity W m , the magnitude of the flu x density vector B at any angle a around the stator is given by I B - BM cos (wt a) I The equation for the induced voltage in a wire is eiD<! = (v x B) - 1 where v = velocity of the wire relative to the magnetic field B = magnetic flux density vector 1 = length of conductor in the magnetic field (B-3 b) ( 1-45) lllis equation can only be used in a frame of reference where the magnetic field appears to be stationary. If we "sit on the magnetic field" so that the field appears to be stationary, the sides of the coil wi ll appear to go by at an apparent velocity vre], and the equation can be applied. Figure 8-2 shows the vector magnetic field and velocities from the point of view of a stationary magnetic field and a moving wire. I. Segment abo For segment ab of the fractional-pitch coil, a = 90° + pl2. As-suming that B is directed radially outward from the rotor, the angle between v and B in segment ab is 900, while the quantity v x B is in the direction of I, so eba = (v x B) -I = vBI directed out of the page - -vBM cos [w,,! - (900 + ~)] I - -vBt.I cos (wmt - 90° - ~) (B-4) where the negative sign comes from the fact that B is really pointing inward when it was assumed to point outward. 2. Segment be. The voltage on segment be is zero, since the vector quantity v x B is perpendicular to I, so ecb = (v x B). 1 = 0 (8-5) COIL PITCH AND DISTRIBlJfED WINDINGS 711 3. Segment cd. For segment cd, the angle a = 90° - pI2. Assuming that B is di-rected radially outward from the rotor, the angle between v and B in segment cd is 90°, while the quantity v x B is in the direction ofl, so edc = (v x B) · I = vBI directed out of the page = -vBt.! cos (W "I - 90° + ~) (8-6) 4. Segment da. TIle voltage on segment da is zero, since the vector quantity v x B is perpendicular to I, so ead = (v x B) • I = 0 (8-7) Therefore, the total voltage on the coil will be By trigonometric identities, cos (W "I - 90° - ~) = cos (wm t -90°) cos ~ + sin (wmt - 90°) sin ~ cos (W "I - 90° + ~) = cos (wm f -90°) cos ~ - sin (wmt - 90°) sin ~ sin (wm f - 90°) = -cos wmf Therefore, the total resulting voltage is eind = V BMI[-cos(wmt -900)COS ~ - sin (w"I- 90 0)Sin~ + cos (wmt - 90°) cos ~ - sin(wmt - 90°) sin ~] Since 2vB&l is equal to f TIlerefore, the nns voltage of any phase of this three-phase stator is 2" EA = V2 Nc kp4>f = V'17rNc kp4>f (8-9) (8-10) (8-11 ) (8-12) (8-13) (8-14) (8-15) Note that for a full-pitch coil, p = 180 0 and Equation (8-15) reduces to the same result as before. For machines with more than two poles, Equation (B-9) gives the pitch fac-tor if the coil pitch p is in electrical degrees. If the coil pitch is given in mechani-cal degrees, then the pitch factor can be given by Harmonic Problems and Fractional-Pitch Windings I kp = sin¥1 (8-16) TIlere is a very good reason for using fractional-pitch windings. It concerns the ef-fect of the nonsinusoidal flux density distribution in real machines. nlis problem can be understood by examining the machine shown in Figure 8-3. This figure shows a salient-pole synchronous machine whose rotor is sweeping across the sta-tor surface. 8ecause the reluctance of the magnetic field path is much lmverdi-rectly under the center of the rotor than it is toward the sides (smaller air gap), the flux is strongly concentrated at that point and the flux density is very high there. TIle resulting induced voltage in the winding is shown in Figure B-3. Notice that it is not sinusoidal-it contains many harmonic frequency components. Because the resu lting voltage wavefonn is symmetric about the center of the rotor flux, no even harmonics are present in the phase voltage. However, all COIL PITCH AND DISTRIBlJfED WINDINGS 713 N B, ,,' 181 ,b, , " FIGURE B-3 (a) A ferromagnetic rotor sweeping past a stator conductor. (b) The flux density distribution of the magnetic field as a function of time at a point on the stator surface. (c) The resulting induced voltage in the conductor. Note that the voltage is directly proponional to the magnetic flux density at any given time. the odd harmonics (third, fifth, seventh, ninth, etc.) are present in the phase volt-age to some extent and need to be dealt with in the design of ac machines. In gen-eral, the higher the number of a given hannonic frequency component, the lower its magnitude in the phase output voltage; so beyond a certain point (above the ninth harmonic or so) the effects of higher hannonics may be ignored. 714 ELECTRIC MACHINERY RJNDAMENTALS When the three phases are Y or 6. connected, some of the hannonics disap-pear from the output of the machine as a result of the three-phase connection. The third-harmonic component is one of these. If the fundamental voltages in each of the three phases are given by ea(t) = EM] sin wt V e,,(t) = EM] sin (wt - 120°) eJ t) = EM] sin (wI - 240°) v V then the third-hannonic components of voltage will be given by e,,3(t) = EM) sin 3wl V eblt) = EM) sin (3wl - 360°) ec3(t) = EM) sin (3wl - 720°) V V (8- 17a) (8- 17b) (8- 17c) (8- 18a) (8- 18b) (8- 18c) Notice that the third-harmonic components of voltage are all identical in each phase. If the synchronous machine is V-connected, then the third-harmonic voltage beMeen any Mo terminnls will be zero (even though there may be a large third-hannonic component of voltage in each phase). If the machine is 6.-connected, then the three third-hannonic components all add and drive a third-harmonic current around inside the 6.-wi ndi ng of the machine. Since the third-harmonic voltages are dropped across the machine's internal impedances, there is again no significant third-hannonic component of voltage at the tenninals. This result applies not only to third-harmonic components but also to any multiple of a third-harmonic component (such as the ninth hannonic). Such spe-cial harmonic frequencies are called triplen harmonics and are automatically sup-pressed in three-phase machines. 1lle remaining hannonic frequencies are the fifth, seventh, eleventh, thir-teenth, etc. Since the strength of the hannonic components of voltage decreases with increasing frequency, most of the actual distortion in the sinusoidal output of a synchronous machine is caused by the fifth and seventh harmonic frequencies, sometimes called the belt harmonics. If a way could be found to reduce these components, then the machine's output voltage would be essentially a pure sinu-soid at the fundamental frequency (50 or 60 Hz). How can some of the harmonic content of the winding's terminal voltage be eliminated? One way is to design the rotor itself to distribute the flux in an approxi-mately sinusoidal shape. Although this action will help reduce the hannonic con-tent of the output voltage, it may not go far enough in that direction. An additional step that is used is to design the machine with fractional-pitch windings. 1lle key to the effect of fractional-pitch windings on the voltage prrxluced in a machine's stator is that the electrical angle of the nth hannonic is n times the elec-trical angle of the fundrunental frequency component. In other words, if a coil spans 150 electrical degrees at its fundamental frequency, it wil I span 300 electrical de-grees at its second-hannonic frequency, 450 electrical degrees at its third-hannonic frequency, and so forth. If p represents the electrical angle spanned by the coil at its COIL PITCH AND DISTRIBlJfED WINDINGS 715 f undamental freq uency and v is the number of the hannonic being exmnined, then the coil will span vp electrical degrees at that harmonic frequency. Therefore, the pitch factor of the coil at the hannonic frequency can be expressed as I kp = sinT I (8-1 9) The important consideration here is that the pitch factor of a winding is different for each harnwnic frequency. By a proper choice of coil pitch it is possible to almost eliminate harmonic freq uency compone nts in the output of the machine. We can now see how hannonics are suppressed by looking at a simple example problem. Example B-1. A three-phase, two-pole stator has coils with a 5/6 pitch. What are the pitch factors for the harmonics present in this machine's coils? Does this pitch help sup-press the harmonic content of the generated voltage? Solution The pole pitch in mechanical degrees of this machine is 360° P = -- = 180° , P (B-1) Therefore, the mechanical pitch angle of these coils is five-sixths of 180°, or 150 0 • From Equation (B-2a), the resulting pitch in electrical degrees is p = £1m X 180° = 150° X 180° = 150° P p 1 80 0 (B-2a) The mechanical pitch angle is equal to the electrical pitch angle only because this is a two-pole machine. For any other number of poles, they would not be the same. Therefore, the pitch factors for the fundamental and the higher odd harmonic fre-quencies (remember, the even hannonics are already gone) are Fundamental: Third harmonic: Fifth hannonic: Seventh hannonic: Ninth harmonic: 150" kp = sin - 2- = 0.966 k = sin 3(150°) = -0.707 , 2 k = . 5(150°) = 0259 pS1ll 2 . k = . 7(150°) = 0259 pS1ll 2 . k = sin 9(150°) = -0.707 , 2 (This is a triplen hannonic not present in the three-phase output.) (This is a triplen hannonic not present in the three-phase output.) The third- and ninth-hannonic components are suppressed only slightly by this coil pitch, but that is unimportant since they do not appear at the machine's terminals anyway. Between the effects of triplen hannonics and the effects of the coil pitch, the third, fifth, seventh, and ninth hannonics are suppressed relative to the futuiamentalfrequency. There-fore, employing fractional-pitch windings will drastically reduce the harmonic content of the machine's output voltage while causing only a small decrease in its ftmdamental voltage. 716 ELECTRIC MACHINERY RJNDAMENTALS 300 200 > 100 ~ ~ > 0 ;; 0.10 0 ! ~ - 100 - 200 - 300 fo'IGURE 8-4 ~-I , ~- Fractional pitch \ ....-- Full Y pitch , , , 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Time. cycles \ , \ , , \ \ , I I I I I 1.00 I I I I The line voltage out of a three-phase generator with full-pitch and fractional-pitch windings. Although the peak voltage of the fractional-pitch winding is slightly smaller than that of the full-pitch winding. its output voltage is much purer. TIle tenni nal voltage of a synchronous machine is shown in Figure 8-4 both for full-pitch windings and for windings with a pitch p = 150°. Notice that the fractional-pitch windings produce a large visible improvement in waveform quality. It should be noted that there are certain types of higher-frequency hannon-ics. called tooth or slot hamwnics. which cannot be suppressed by varying the pitch of stator coils. These slot harmonics will be discussed in conjunction with distributed windings in Section B.2. 8.2 DISTRIBUTED WINDINGS IN A C MA CHINES In the previous section. the windings associated with each phase of an ac machine were implicitly assumed to be concentrated in a single pair of slots on the stator surface. In fact, the windings associated with each phase are almost always dis-tributed among several adjacent pairs of slots, because it is simply impossible to put all the conductors into a single slot. TIle construction of the stator windings in real ac machines is quite compli-cated. Normal ac machine stators consist of several coils in each phase, distributed in slots around the inner surface of the stator. In larger machines, each coil is a preformed unit consisting of a number of turns, each turn insulated from the oth-ers and from the side of the stator itself (see Figure 8-5). The voltage in any COIL PITCH AND DISTRIBlJfED WINDINGS 717 FIGURE B-S A typical preformed stator coil. (Courtesy ofGeneml Electric Company.) ,., ,b, FIGURE B-6 (a) An ac machine stator with preformed stator coils. (Courtesy of Westinghouse Electric Company.) (b) A close-up view of the coil ends on a stator. Note that one side of the coil will be outennosl in its slot and the other side will be innermost in its slol. This shape permits a single standard coil form to be used for every slot on the stator. (Courtesy ofGeneml Electric Company.) single turn of wire is very small, and it is only by placing many of these turns in series that reasonable voltages can be produced. 1lle large number of turns is nor-mally physically divided among several coils, and the coils are placed in slots equally spaced along the surface of the stator, as shown in Figure 8-6. 718 ELECTRIC MACHINERY RJNDAMENTALS Phase belt or ph= =gro ~" ~P _ _ _ --I FIGURE 11-7 A simple double-layer full-pitch distributed winding for a two-pole ac machine. TIle spacing in degrees between adjacent slots on a stator is called the slot pitch r of the stator. The slot pitch can be expressed in either mechanical or elec-trical degrees. Except in very small machines, stator coils are normally fonned into double-layer windings, as shown in Figure 8-7. Double-layer windings are usu-ally easier to manufacture (fewer slots for a given number of coils) and have sim-pler end connections than single-layer windings. They are therefore much less ex-pensive to build. Figure 8-7 shows a distributed full -pitch winding for a two-pole machine. In this winding, there are four coils associated with each phase. All the coil sides ofa given phase are placed in adjacent slots, and these sides are known as aphase belt or phase group. Notice that there are six phase belts on this two-pole stator. In general, there are 3P phase belts on a P-pole stator, P of them in each phase. Figure 8-8 shows a distributed winding using fractional-pitch coils. Notice that this winding still has phase belts, but that the phases of coils within an indi-vidual slot may be mixed. The pitch of the coils is 5/6 or 150 electrical degrees. The Breadth or Distribution Factor Dividing the total required number of turns into separate coils permits more effi-cient use of the inner surface of the stator, and it provides greater structural strength, since the slots carved in the frame of the stator can be smaller. However, the fact that the turns composing a given phase lie at different angles means that their voltages will be somewhat smaller than would otherwise be expected. COIL PITCH AND DISTRIBlJfED WINDINGS 719 Phase belt FIGURE B-8 A double-layer fractional-pitch ac winding for a lI'.o-pole ac machine. To illustrate this problem, examine the machine shown in Figure 8-9. This machine has a single-layer winding, with the stator winding of each phase (each phase belt) distributed among three slots spaced 20° apart. If the central coil of phase a initially has a voltage given by Ea2 = E LOoy then the voltages in the other two coils in phase a will be E,,]=EL-20o y E,,3= EL200y The total voltage in phase a is given by E" = Ea ] + Ea2 + E,,) = EL-20° + ELO° + EL20° = E cos (_20°) + jEsin (_20°) + E + E cos 20° + jEsin 20° = E + 2Ecos 20° = 2.879E nlis voltage in phase a is not quite what would have been expected if the coils in a given phase had all been concentrated in the same slot. nlen, the volt-age Ea would have been equal to 3E instead of2.879E. The ratio of the actual volt-age in a phase of a distributed winding to its expected value in a concentrated winding with the same number of turns is called the breadth factor or distribution factor of winding. The distribution factor is defined as _ V.p actual kd -V4>expected with no distribution (8-20) 720 ELECTRIC MACHINERY RJNDAMENTALS Phase belt ""GURE 11- 9 A two-pole stator with a single-layer witxling cOI\Sisting of three coils per phase, each separated by 20". TIle distribution factor for the machine in Figure 8 - 9 is thus (B- 21) TIle distribution factor is a convenient way to summarize the decrease in voltage caused by the spatial distribution of the coils in a stator winding. It can be shown (see Reference I, page 726) that, for a winding with n slots per phase belt spaced ydegrees apart, the distribution factor is given by sin (nyf2) k ~ ="'f'2c d n sin (yf2) (B- 22) Notice that for the previous example with n = 3 and y= 20°, the distribution fac-tor becomes sin (ny!2) k --d -n sin (yf2) -which is the same result as before. sin[(3)(200)!2] 3 sin(200!2) = 0.960 (B- 22) COIL PITCH AND DISTRIBlJfED WINDINGS 721 The Generated Voltage Including Distribution Effects The nns voltage in a sing le coil of Nc turns and pitch factor kp was previously de-tennined to be (8-15) If a stator phase consists of i coils, each containing Nc turns, then a total of Np = iNc turns will be present in the phase. The voltage present across the phase will just be the voltage due to N p turns all in the same slot times the reduction caused by the distribution factor, so the total phase voltage will become (8-23) The pitch factor and the distribution factor of a winding are sometimes combined for ease of use into a sing le winding factor k.,. The winding factor of a stator is given by I k. - k,k, I (8-24) Applying this defmition to the equation for the voltage in a phase yields I EA ~ V2wNpk.f I (8-25) Example B-2. A simple two.JXlle, three.phase, Y·connected synchronous machine stator is used to make a generator. It has a double-layer coil construction, with four stator coils per phase distributed as shown in Figure 8-8. Each coil consists of 10 turns. The windings have an electrical pitch of 150°, as shown. The rotor (and the magnetic field) is rotating at 3000 rlmin, and the flux per pole in this machine is 0.019 Wb. (a) What is the slot pitch of this stator in mechanical degrees? In electrical degrees? (b) How many slots do the coils of this stator span? (c) What is the magnitude of the phase voltage of one phase of this machine's stator? (d) What is the machine's tenninal voltage? (e) How much suppression does the fractional-pitch winding give for the fifth-harmonic component of the voltage relative to the decrease in its fundamental component? Solutioll (a) This stator has 6 phase belts with 2 slots per belt, so it has a total of 12 slots. Since the entire stator spans 360 0 , the slot pitch of this stator is 360" ")' = ---u- = 30° This is both its electrical and mechanical pitch, since this is a two-pole machine. (b) Since there are 12 slots and 2 poles on this stator, there are 6 slots per pole. A coil pitch of 150 electrical degrees is 150°1180° = 5/6, so the coils must span 5 stator slots. 722 ELECTRIC MACHINERY RJNDAMENTALS (e) The frequency of this machine is f = n"'p = (3000 r/min)(2 poles) = 50 H 120 120 z From Equation (8-19), the pitch factor for the fundamental component of the voltage is k = sin vP = sin (1)(150°) = 0966 p 2 2 . (B-19) Although the windings in a given phase belt are in three slots, the two outer slots have only one coil each from the phase. Therefore, the winding essentially oc-cupies two complete slots. The winding distribution factor is sin (n),12) sin[(2)(300)12] kd = n sin (),12) = 2 sin (30°12) = 0.966 Therefore, the voltage in a single phase of this stator is E}, = V2" 7rNpkpkd~f (B-22) = V2" 7r(40 tums)(O.966)(0.966XO.019 WbX50 Hz) = 157 V (d) This machine's tenninal voltage is VT = v'5E}, = V3"( 157 V) = 272 V (e) The pitch factor for the fifth-harmonic component is k =sinvP _ . (5XI500)_0259 p 2- S1ll 2-' (B-19) Since the pitch factor of the fundamental component of the voltage was 0.966 and the pitch factor of the fifth-harmonic component of voltage is 0.259, the fundamental component was decreased 3.4 percent, while the fifth-hannonic component was decreased 74.1 percent. Therefore, the fifth-hannonic compo-nent of the voltage is decreased 70.7 percent more than the fundamental com-ponent is. Tooth or Slot Harmonics Although distributed windings offer advantages over concentrated windings in terms of stator strength, utilization, and ease of construction, the use of distributed windings introduces an additional problem into the machine's design. The pres-ence of uniform slots around the inside of the stator causes regular variations in reluctance and flux along the stator's surface. These regular variations produce harmonic components of voltage called tooth or slot harmonics (see Figure B-JO). Slot harmonics occur at frequencies set by the spacing between adjacent slots and are given by (B-26) B COIL PITCH AND DISTRIBlJfED WINDINGS 723 FIGURE B-1O Stator with slots Flux density variations in the air gap due to the tooth or slot hannonics. The reluctance of each slot is higher than the reluctance of the metal surface between the slots. so flux densities are lower directly over the slots. where volo< = number of the hannonic component S = number of slots on stator M = an integer P = number of poles on machine The valueM = 1 yields the lowest-frequency slot harmonics, which are also the most troublesome ones. Since these hannonic components are set by the spacing between adjacent coil slots, variations in coil pitch and distribution cannot reduce these effects. Re-gardless ofa coil's pitch, it must begin and end in a slot, and therefore the coil 's spacing is an integral multiple of the basic spacing causing slot hannonics in the first place. For example, consider a 72-s10t, six-pole ac machine stator. In such a ma-chine, the two lowest and most troubles.ome s.tator hannonics are 724 ELECTRIC MACHINERY RJNDAMENTALS = 2(1)(72) + 1 = 23 25 6 ' These hannonics are at 1380 and 1500 Hz in a 6O-Hz machine. Slot harmonics cause several problems in ac machines: I. They induce harmonics in the generated voltage of ac generators. 2. The interaction of stator and rotor slot harmonics produces parasitic torques in induction motors. 1llese torques can seriously affect the shape of the mo-tor's torque-speed curve. 3. They introduce vibration and noise in the machine. 4. They increase core losses by introducing high-frequency components of volt-ages and currents into the teeth of the stator. Slot hannonics are especially troublesome in induction motors, where they can induce harmonics of the same frequency into the rotor field circuit, further re-inforcing their effects on the machine's torque. Two common approaches are taken in reducing slot hannonics. They are fractional-slot windings and skewed rotor conductors. Fractional-slot windings involve using a fractional number of slots per rotor pole. All previous examples of distributed windings have been integral-slot wind-ings; i.e., they have had 2, 3, 4, or some other integral number of slots per pole. On the other hand, a fractional-slot stator might be constructed with 2~ slots per pole. TIle offset between adjacent poles provided by fractional-slot windings helps to reduce both belt and slot harmonics. This approach to reducing hannonics may be used on any type ofac machine. Fractional-slot hannonics are explained in de-tail in References 1 and 2. 1lle other, much more common, approach to reducing slot hannonics is skewing the conductors on the rotor of the machine. nlis approach is primarily used on induction motors. llle conductors on an induction motor rotor are given a slight twist, so that when one end of a conductor is under one stator slot, the other end of the coil is under a neighboring slot. This rotor construction is shown in Fig-ure 8-1 I. Since a single rotor conductor stretches from one coil slot to the next (a distance corresponding to one full electrical cycle of the lowest slot hannonic fre-quency), the voltage components due to the slot hannonic variations in flux cancel. 0,3 SUMMARY In real machines, the stator coils are often of fractional pitch, meaning that they do not reach completely from one magnetic pole to the next. Making the stator wind-ings fractional-pitch reduces the magnitude of the output voltage slightly, but at the same time attenuates the hannonic components of voltage drastically, result-ing in a much smoother output voltage from the machine. A stator winding using fractional-pitch coi Is is often cal led a chorded winding. Certain higher-frequency hannonics, called tooth or slot harmonics, cannot be suppressed with fractional-pitch coils. These harmonics are especially trouble-COIL PITCH AND DISTRIBlJfED WINDINGS 725 FIGURE 0-11 An induction motor rotor exhibiting conductor skewing. The skew of the rotor conductors is just equal to the distance between one stator slot and the next one. (Courtesy of MagneTek, Inc.) some in induction motors. They can be reduced by employing fractional-slot windings or by skewing the rotor conductors of an induction motor. Real ac machine stators do not simply have one coil for each phase. In or-der to get reasonable voltages out of a machine, several coils must be used, each with a large number of turns. This fact requires that the windings be distributed over some range on the stator surface. Distributing the stator windings in a phase reduces the possible output voltage by the distribution factor kd' but it makes it physically easier to put more windings on the machine. QUESTIONS B-1. Why are distributed windings used instead of concentrated windings in ac machine stators? B-2. (a) What is the distribution factor of a stator winding? (b) What is the value of the distribution factor in a concentrated stator winding? B-3. What are chorded windings? Why are they used in an ac stator winding? B-4. What is pitch? What is the pitch factor? How are they related to each other? B-5. Why are third-hannonic components of voltage not fOlUld in three-phase ac machine outputs? B-6. What are triplen harmonics? B-7. What are slot harmonics? How can they be reduced? B-8. How can the magnetomotive force (and flux) distribution in an ac machine be made more nearly sinusoidal? PROBLEMS B-1. A two-slot three-phase stator armature is wOlUld for two-pole operation. If fractional-pitch windings are to be used. what is the best possible choice for wind-ing pitch if it is desired to eliminate the fifth-hannonic component of voltage? B-2. Derive the relationship for the winding distribution factor kd in Equation (B- 22). 726 ELECTRIC MACHINERY RJNDAMENTALS B-3. A three-phase four-pole synchronous machine has 96 stator slots. The slots contain a double-layer winding (two coils per slot) with four turns per coil. The coil pitch is 19124. (a) Find the slot and coil pitch in electrical degrees. (b) Find the pitch, distribution, and winding factors for this machine. (c) How well will this winding suppress third, fifth, seventh, ninth, and eleventh harmonics? Be sure to consider the effects of both coil pitch and winding distri-bution in your answer. B-4. A three-phase four-pole winding of the double-layer type is to be installed on a 48-slot stator. The pitch of the stator windings is 5/6, and there are 10 tlU1lS per coil in the windings. All coils in each phase are cOIUlected in series, and the three phases are cormected in 6.. The flux per pole in the machine is 0.054 Wb, and the speed of rotation of the magnetic field is 1800 r/min. (a) What is the pitch factor of this winding? (b) What is the distribution factor of this winding? (c) What is the frequency of the voltage produced in this winding? (d) What are the resulting phase and tenninal voltages of this stator? B-5. A three-phase, Y-cOIUlected, six-pole synchronous generator has six slots per pole on its stator winding. The winding itself is a chorded (fractional-pitch) double-layer winding with eight tlU1lS per coil. The distribution factor kd = 0.956, and the pitch factor kp = 0.98 1. The flux in the generator is 0.02 Wb per pole, and the speed of ro-tation is 1200 r/min. What is the line voltage produced by this generator at these conditions? B-6. A three-phase, Y -cormected, 50-Hz, two-pole synchronous machine has a stator with 18 slots. Its coils form a double-layer chorded winding (two coils per slot), and each coil has 60 turns. The pitch of the stator coils is 8/9. (a) What rotor flux would be required to produce a terminal (line-to-line) voltage of 6 kV? (b) How effective are coils of this pitch at reducing the fifth-hannonic component of voltage? The seventh-hannonic component of voltage? B-7. What coil pitch could be used to completely eliminate the seventh-harmonic com-ponent of voltage in ac machine armature (stator)? What is the minimum number of slots needed on an eight-pole winding to exactly achieve this pitch? What would this pitch do to the fifth-harmonic component of voltage? B-8. A 13.8-kV, V-connected, 60-Hz, 12-pole, three-phase synchronous generator has 180 stator slots with a double-layer winding and eight tlU1lS per coil. The coil pitch on the stator is 12 slots. The conductors from all phase belts (or groups) in a given phase are connected in series. (a) What flux per pole would be required to give a no-load terminal (line) voltage of 13.8 kV? (b) What is this machine's winding factor kO'? REFERENCES 1. Fitzgerald, A. E.. and Charles Kingsley. Electric Machinery. New York: McGraw-Hili, 1952. 2. Liwschitz-Garik, Michael. and Clyde Whipple. A lternating-Current Machinery. Princeton. N.J.: Van Nostrand. 1961. 3. Werninck. E. H. (ed.). Electric Motor Handbook.. London: McGraw-Hill. 1978. APPENDIX C SALIENT-POLE THEORY OF SYNCHRONOUS MACHINES T he equivalent circuit for a synchronous generator derived in Chapter 5 is in fact valid only for machines built with cylindrical rotors, and not for machines built with salient-pole rotors. Likewise, the expression for the relationship be-twccn the torque angle 8 and the power supplied by the generator [Equation (5- 20)] is valid only for cylindrical rotors. In Chapter 5, we ignored any effects due to the saliency of rotors and assumed that the simple cylindrical theory ap-plied. This assumption is in fact not too bad for steady-state work, but it is quite poor for examining the transient behavior of generators and motors. The problem with the simple equivalent circuit of induction motors is that it ignores the effect of the reluctance torque on the generator. To understand the idea of reluctance torque, refer to Figure C-l. nlis figure shows a salient-pole rotor with no windings inside a three-phase stator. I f a stator magnetic field is produced as shown in the figure, it will induce a magnetic field in the rotor. Since it is much easier to produce a flux along the axis of the rotor than it is to produce a flux across the axis, the flux induced in the rotor will line up with the axis of the rotor. Since there is an angle between the stator magnetic field and the rotor magnetic field, a torque will be induced in the rotor which will tend to line up the rotor with the stator field. The magnitude of this torque is proportional to the sine of twice the angle between the two magnetic fields (sin 20). Since the cyli ndrical rotor theory of synchronous machines ignores the fact that it is easier to establish a magnetic field in some directions than in others (i.e., ignores the effect of reluctance torques), it is inaccurate when salient-pole rotors are involved. 727 728 ELECTRIC MACHINERY RJNDAMENTALS c' o b' o FIGURE C-l o A salient-pole rotor, illustrating the idea of reluctance torque, A magnetic field is induced in the rotor by the stator magnetic field, and a torque is pnxluced on the rotor that is proportional to the sine of twice the angle between the two fields, CI DEVELOPMENTOFTHE EQUIVALENT CIRCUIT OF A SALIENT-POLE SYNCHRONOUS GENERATOR As was the case for the cylindrical rotor theory, there are four elements in the equivalent circuit of a synchronous generator: I. The internal generated voltage of the generator E), 2. The armature reaction of the synchronous generator 3. The stator winding's self-inductance 4. The stator winding's resistance TIle first, third, and fourth elements are unchanged in the salient-pole theory of synchronous generators, but the annature-reaction effect must be modified to ex-plain the fact that it is easier to establish a flux in some directions than in others, TIlis modification of the armature-reaction effects is accomplished as ex-plained below, Figure C- 2 shows a two-pole salient-pole rotor rotating counter-clockwise within a two-pole stator, TIle rotor flux of this rotor is called GR, and it points upward, By the equation for the induced voltage on a moving conductor in the presence of a magnetic field, ei!>d = (v x B) • I ( 1 -45) the voltage in the conductors in the upper part of the stator wi ll be positive out of the page, and the voltage in the conductors in the lower part of the stator will be into the page, The plane of maximum induced voltage will lie directly under the rotor pole at any given time, o o , , FIGURE C-2 , , , , SALIENT-POLE THEORY OF SYNCHRONOUS MACHINES 729 , '" I "A,1nH , , II, o • c-~ Pl~ne of max 1 ..1 -~, , , (b, , , , , o o : Plane of Ii!:A,1nH , B, o ( ., o o , '" I "A,1nH B , o , , Plane of lA, max ~~: , , , , , , o "'l=~. '(''----f-+- Plane of ~ Id, InH o o (,' -= " Magnetomotive fo= '3is" Stator magnetomotive fo= o :J d" Direct axis component of magnetomotive f=. :J~ .. Quadrature axis component of magnetomotive f=. The effects of annalure reaction in a salient-pole synchronous generator. (a) The rotor magnetic field induces a voltage in the stator which peaks in the wires directly under the pole faces. (b) If a lagging load is connected to the generator. a stator current will flow that peaks at an angle behind EA' (c) This stator current 1 ..1 produces a stator magnetomotive force in the machine. 730 ELECTRIC MACHINERY RJNDAMENTALS EA. "'"" Plane of Bs I •. ~~ • • • • / 0 B. Plane of ++--. Hd-'-T'----- ++- IJ.",ox • • ,d, , , Bs fornonsalient pole, Bs with ,alient poles " ?d" -~, • ?" ..:...J... . ~. ~J< m.,since it is easierto establish nux alons the di=t axis. ""GURE C-l (concluded) 0 ® • • • • • 'A ~ : Plane of I max I. , • • • • ". • • • " ~ ,., " . • Plane of I, ~ 0 '\ '.-. Plane of max IJ ® V ~ " EA+Ed +E. (d) The stator magnetomotive force produces a stator flux lis. However. the direct-axis component of magnetomotive force produces more flux per ampere-turn than the quadrature-axis component does. since the reluctance of the direct-axis flux path is lower than the reluctance of the quadrature-axis flux path. (e) The direct- and quadrature-axis stator fluxes produce armature reaction voltages in the stator of the machine. I f a lagging load is now connected to the tenninals of this generator, then a currefll will flow whose peak is delayed behind the peak voltage. lllis current is shown in Figure C- 2b. llle stator currefll flow produces a magnetomotive force that lags 900 be-hind the plane of peak stator current, as shown in Figure C- 2c. In the cylindrical theory, this magnetomotive force then produces a stator magnetic field Bs that lines up with the stator magnetomotive force. However, it is actually easier to pro-duce a magnetic field in the direction of the rotor than it is to produce one in the direction perpendicular to the rotor. Therefore, we will break down the stator magnetomotive force into componeflls parallel to and perpendicular to the rotor's axis. Each of these magnetomotive forces produces a magnetic field, but more flux is produced per ampere-turn along the axis than is produced perpendicular (in quadrature) to the axis. I I I E, E, v, ---- E • SALIENT-POLE THEORY OF SYNCHRONOUS MACHINES 731 FIGURE C-3 The phase voltage of the generator is just the sum of its internal generated voltage and its armature reaction voltages. The resulting stator magnetic field is shown in Figure C- 2d, compared to the field predicted by the cylindrical rotor theory. Now, each component of the stator magnetic field produces a voltage of its own in the stator winding by annature reaction. These annature-reaction voltages are shown in Figure C- 2e. The total voltage in the stator is thus (C-l ) where EJ is the direct-axis component of the annature-reaction voltage and Eq is the quadrature-axis component of annature reaction voltage (see Figure C- 3). As in the case of the cylindrical rotor theory, each armature-reaction voltage is di-rectly proportional to its stator current and delayed 90° behind the stator current. Therefore, each armature-reaction voltage can be modeled by EJ = -jxJIJ Eq = -jxqIq and the total stator voltage becomes Vo/> = E ... - jxJIJ - jxqIq (C- 2) (C- 3) (C-4) The annature resistance and self-reactance must now be included. Since the annature self-reactance X ... is independent of the rotor angle, it is nonnally added to the direct and quadrature annature-reaction reactances to produce the direct syn-chronous reactance and the quadrature synchronous reactance of the generator: IXrx,+ X AI IXq -Xq + X A I (C- 5) (C-6) 732 ELECTRIC MACHINERY RJNDAMENTALS ""GURE C-4 The phasor diagram of a salient-pole synchronous generator. o 0' I, b ""GURE C-S Constructing the phasor diagram with no prior knowledge of 8. E; lies at the same angle as EA. and E; may be determined exclusively from information at the terminals of the generator. Therefore. the angle 6 may be found. and the current can be divided into d and q components. TIle annature resistance voltage drop is just the armature resistance times the ar-mature current IA . lllerefore, the final expression for the phase voltage of a salient-pole syn-chronous motor is (C- 7) and the resulting phasor diagram is shown in Figure C-4. Note that this phasor diagram requires that the annature current be resolved into components in parallel with EA and in quadrature with EA. However, the an-gie between EA and IA is lj + (), which is not usually known before the diagram is constructed. Normally, only the power-factor angle () is known in advance. It is possible to construct the phasor diagram without advance knowledge of the angle 0, as shown in Figure C- S. TIle solid lines in Figure C- S are the same as the lines shown in Figure C-4, while the dotted lines present the phasor diagram as though the machine had a cylindrical rotor with synchronous reactance Xd . SALIENT· POLE THEORY OF SYNCHRONOUS MACHINES 733 The angle 0 of EA can be found by using infonnation known at the tenninals of the generator. Notice that the phasor E;' which is given by I EA = V4> + RAIA + jXqIA I (C-8) is collinear with the internal generated voltage EA' Since E; is determined by the current at the terminals of the generator, the angle {) can be detennined with a knowledge of the annature current. Once the angle 0 is known, the annature cur· rent can be broken down into direct and quadrature components, and the internal generated voltage can be detennined. Example C- 1. A 480-V, 60-Hz, d-cOIlllected, four-pole synchronous generator has a direct-axis reactance of 0.1 n, and a quadrature-axis reactance of 0.075 n. Its anna-ture resistance may be neglected. At fun load, this generator supplies 1200 A at a power factor of 0.8 lagging. (a) Find the internal generated voltage EA of this generator at full load, assruning that it has a cylindrical rotor of reactance Xd . (b) Find the internal generated voltage EA of this generator at fun load, assuming it has a salient-pole rotor. Solutioll (a) Since this generator is d-connected, the armature current at fun load is IA = 12~A = 693 A The power factor of the current is 0.8 lagging, so the impedance angle () of the load is () = cos-l 0.8 = 36.87° Therefore, the internal generated voltage is EA = V4> + jXSIA = 480 L 0° V + j(O.1 nX693 L - 36.87° A) = 480 L 0° + 69.3 L 53.13° = 524.5 L 6.1° V Notice that the torque angle 8 is 6.1°. (b) Asswne that the rotor is salient. To break down the current into direct- and quadrature-axis components, it is necessary to know the direction of EA. This di-rection may be detennined from Equation (C- 8): = 480L 0° V + 0 V + j(0JJ75 nX693L -36.87° A) = 480LO° + 52L53.13° = 513L4.65" V (C-8) The direction of EA is 8 = 4.65". The magnitude of the direct-axis component of C lUTent is thus Id = IA sin ({) + /)) = (693 A) sin (36.87 + 4.65) = 459 A 734 ELECTRIC MACHINERY RJNDAMENTALS and the magnitude of the quadrature-axis component of current is Iq = IA cos «() + Ii) = (693 A) cos (36.87 + 4.65) = 519 A Combining magnitudes and angles yields Id = 459 L -85.35° A Iq = 519 L 4.65° A The resulting internal generated voltage is EA = Vo/> + RAIA + jXdld + jX,}-q = 480 L 0° V + 0 V + j(O.1 0)(459 L -85.35" A) + j(0.075 OX519 L 4.65" A) = 524.3 L 4.65° V Notice that the magnitude O/ EA is not much affected by the salient poles, but the angle of EA is considerably different with salient poles than it is without salient poles. C.2 TORQUE AND POWER EQUATIONS OF SA LIENT-POLE MACHINE TIle power output of a synchronous generator with a cylindrical rotor as a function of the torque angle was given in Chapter 5 as _ 3~ V.~E -; A ,"' _ i_ " ~S p = -X, (5- 20) TIlis equation assumed that the annature resistance was negligible. Making the same assumption, what is the output power of a salient-pole generator as a func-tion of torque angle? To find out, refer to Figure C--6. TIle power out of a syn-chronous generator is the sum of the power due to the direct-axis current and the power due to the quadrature-axis current: v ~ cos {j I, , 90 - ' , , IA I, ""GURE C-6 , , v • Determining the power output of a salient-pote synchronous generator. Both I~ and I. contribute to the output power. as shown. SALIENT-POLE THEORY OF SYNCHRONOUS MACHINES 735 P = Pd+ Pq (C- 9) = 3 V4>ld cos (900-8) + 3 V ¥q cos 8 = 3V4>ld sin8+ 3 V4> lq cos 8 From Figure C-6, the direct-axis current is given by _ EA -\j, cos 5 ld -X , (C-I O) and the quadrature-axis current is given by _ V1> sin 5 lq -X , (C-II ) Substituting Equations (C- IO) and (C-ll ) into Equation (C- 9) yields ( EA -V1> cos 8 ) ( V1> sin 8 ) P = 3V1> X; sin 8 + 3V1> Xq cos 8 3V1>EA (I 1) = X sin 5 + 3 vt X - X sin 5 cos 5 , " Since, sin 8 cos 8 = ~s in 28, this expression reduces to P = 1> Asin5+~ d q sin25 3VE 3 V 2 (X - X) Xd 2 J<JXq (C-1 2) The first tenn of this expression is the same as the power in a cylindrical ro-tor machine, and the second tenn is the additional power due to the reluctance torque in the machine. Since the induced torque in the generator is given by T iD<! = P.:ODv1wm, the in-duced torque in the motor can be expressed as (C- 13) TIle induced torque out of a salient-pole generator as a function of the torque angle {j is plotted in Figure C- 7. PROBLEMS C-1. A 4S0-V, 200-kVA, O.S-PF-Iagging, 6O-Hz, four-]Xlle, Y-connected synchronous gen-erator has a direct-axis reactance of 0.25 n, a quadrature-axis reactance ofO.IS n and an annature resistance of 0.03 n. Friction, windage, and stray losses may be assruned negligible. The generator's open-circuit characteristic is given by Figure P5-I. (a) How much field current is required to make Vrequal to 480 V when the gener-ator is flmning at no load? (b) What is the internal generated voltage of this machine when it is operating at rated conditions? How does this value of E,\ compare to that of Problem 5- 2b? 736 ELECTRIC MACHINERY RJNDAMENTALS f ind> N·m Total torque Electrical -~~----'''----~----~-----'~-- angle 0. 90" ""GURE C-7 Reluctance torque 1800 degrees Plot of torque versus torque angle for a salient-pole synchronous generator. Note the component of torque due to rotor reluctance. (c) What fraction of this generator's full-load power is due to the reluctance torque of the rotor? C-2. A l4-pole, Y -connected, three-phase, water-turbine-driven generator is rated at 120 MVA, 13.2 kV, 0.8 PF lagging, and 60 Hz. Its direct-axis reactance is 0.62 n and its quadrature-axis reactance is 0.40 n. All rotational losses may be neglected. (a) What internal generated voltage would be required for this generator to operate at the rated conditions? (b) What is the voltage regulation of this generator at the rated conditions? (c) Sketch the power-versus-torque-angle curve for this generator. At what angle 0 is the power of the generator maximum? (d) How does the maximum power out of this generator compare to the maximum power available if it were of cylindrical rotor construction? C-3. Suppose that a salient-pole machine is to be used as a motor. (a) Sketch the phasor diagram of a salient-pole synchronous machine used as a motor. (b) Write the equations describing the voltages and currents in this motor. (c) Prove that the torque angle obetween E,\ and V. on this motor is given by C-4. If the machine in Problem C- l is flmning as a motor at the rated conditions, what is the maximrun torque that can be drawn from its shaft without it slipping poles when the field current is zero? Constants Charge of the electron Permeability of free space Permittivity of free space COD"crsion fa ctors Length F~ Torque Energy Power Magnetic flux Magnetic flux density Magnetizing intensity APPENDIX D TABLES OF CONSTANTS AND CONVERSION FACTORS e = - 1.6X to- 19 C tto = 47'1 X 10- 7 Him £0 = 8.854 X 10- 11 F/m I meter (m) I kilogram (kg) I newton (N) I newlon-meter (N • m) I joule (1) I watt (W) I horsepower I weber (Wb) I tesla (f) I ampere ' turn/m = 3.281 ft = 39.37 in = 0.0685 slug = 2.205 lb mass (Ibm) = 0.22481b force (lb ' n = 7.233 poundals = 0.102 kg (force) = 0.738 pound-feet (lb ' ft) = 0.738 foot-pounds (ft ' lb) = 3.725 X 10-1 horsepower-hour (hp ' It) = 2.778 X 10-1 kilowatt-hour (kWh) = 1.341 X IO-J hp = 0.7376 ft · lbf Is = 746W = ]Oi maxwells (lines) = ]Wb/m1 = 10,000 gauss (G) = 64.5 kilolineslinl = 0.0254 A • turns/in = 0.0126 oersted (Oe) 737 oOC p.-.. "que""e, 684, 68~, 689 -..-, iDcbon ll"DOH'or, 461J.-464 .oymcJ.""""" ll"DOH'or, 267~.Su abo SJ'DC1ronow gonmtOor >c machine, 231J.-472 ""'lpitch. 707_716 do machine.<, ~ 681 <le6DO<l 2JO di,,,ibolod wiDdiDgS. 716-72~ efficiOJlC)'.261 iDcbon motor, 380-472. SN abo [""moIor 100 ..... 261_262 _tomooive for<:elflux di,,,ibo1ioo, _m power tbw di.uam- 262, 263 .~ reguh. ioD, 26J.-264 .oymcJ.ODOOIS !l"DOH,orS, 267_34.'1. Sa abo SJ'DC1mnow S.,...,.... symcJ.ODOUII """on, 346-379. Su abo SJ'DC1moow mol" .. '(II" .... , 237_2J8, ~~~8 von!", 2B, 2~.'!.'! v()hlIS·ro~OD,26l-261 wiDdiD,l! imulotioo:>, 2'!8-260 >C Olochine ""'orl, 716 >C lDOIor. Sa lnWctioo moIor, SyncIroDow IOOIor >c phase ""810 COIIIro!. ] W-H16 >c phase OJIItrol 177_186 ocb p.-.. "que""e, 684, 68.'1, 689 Ac<:elentiOlO'<IocolontiOll circui~ m Acro .. _ the-liDe .wting, 430, 4]2 Air gap,.'In Air_ SlIP flux deruity dihe ll"DOntioa ciJ<:uiu. 171 Ana]yoi,. Sn abo Noali ..... analyst. <:ww lati""ly~dc genontOlll.614-61.'! diJJe ... ti. Uy ~dc gmont<n.618-619 _ de S .... ""or. 6()6...«)Il Aoplar oe<:elontioa. 4 Angular position. J Aoplar velocity, l-4 Apporo .. r. 4<J-j() oymcJ.""""" ll"DOH'or. 327 tnmsfonool. ]]8-lJ9 Apporo .. , rating odvaot"8". 112-113 ~xi ...... tnmsfonoe:r models, 89 Anna, ... , 490 • .'120 Anna,,,,,, Irus, .'124 Anna,,,,,, .. ""ion do machine.<,~ • .'120 pole foe ... .'120 ool"",.ly exci,ed de geJ.""'or, ~98 de.-or,~ oymcJ.""""" l!,,,,on'or, 274 AJmatore reoctioa vonge, 276 AJmatore re,si""""" vong. drop, 732 AJmatore winding,,- 26/, 492-493, ~20 A01""" .. fonne:r,I09-H6 'ppare1I' power nlilll adv""ge, H2-H3 di>adv"'''ge, H ~ inler .. li~,m variable-voltage H~ voltay/""""'" rel" iorubips. III A01""" .. fonne:r , taner (indUCIioo .-or~ 432 Av<Rf-O flux per ,~ .. , 73 Av<Rf-O power, 48.. 49 Balanced thru-p..... "'Y"em<. 69J-700 Balanced Y-<:O<IDOCOed Iood, 690 Base , pee<l438, ~~3 Belt lwlDOlliCl, 714 Blockod_ roIor CODditioo, 390 Blockod_ roIor ""or reac1aD::e, 391 Blockod_ roIor Ie", 4.l~ Bre:dh factor, 718.-720 BreakOOw. 'orque, 409. Sa aIu! Pullou, 'orqu. Breakover voltay (V""~ I~~ Br",b,268 Br",b drop 10< .... ~~, ~93 Br",b kwe., ~~ Br",b shifting, ~ Br",b voltage , ~~ Br",be., 489, ~21--,'!2J Br",bl .. , de !DOlor, 674--(,77 Br",bl .. , exciters, 268-271 Cage iooOC1ioo mol'l- 329 Copacitive ~ ~I Copaci!or-Qar', capaci'Q1_ JILII !DOlor, MI),M3 Copacitor-Qar'motIn, 64~~3 C .. lnt .... ,28 Ownferod pole .. ~20, ~21 Ow). of !be eleC1rolll, 737 Qde,171 a.opper, 186-193 forcedcomm01"ioo, 187, 18S-189 ponllel-<:","""or <:<>m.un"ion circuits, 191_193 series-apaci'or cornmu,,,ion cir<:uiu, 189-191 a.or[lI>01<>DlO1ive force, 27 ~Ipitcb, 707_716 fno::1ioruol_ pitcb coil. 709-712 fno::1ioruol_ pitcb wiDdi"W', 712-716 barmooicpob ...... , 712-716 pitch of oooil708-700 Commoa "'""'" <£.0), 109 INDEX Cormnoo voltage (V d, 100 Cormnoo winding, 100 CormnutatiD,l! ouochlnery. SN de ~ CormnutatiD,l! pole.<fIO'e:rpoI .... ~II CormnutatiOll choppe:r circui,., 18S-189 cyclocoovetter,210 defined, 479, 4&'!, 439 ew nW COI1llW1lItioa i.verler, 194 fcreed, 187, US-189, 210 four_ loop de oucl1ine, ~90 panllel-<:apOcitor commOO " ioo circui" 191_193 .. If-<:Ol1lDllltatiOll iovene:r, I~ =ie.-<:apOcitor commU1ll'ioo circui" 189-191 CormnutatQ1, 473. 499, ~19, ~21--,'!23 Cormnutat<>1 pi,cb, 492 CormnutatQ1 "[lIBOJI,., 479, 489 Coq>ar:o'or, 202 Coq>e1Wlting wiDdi"W', ~IWlJ Coq>lex power, ~ I ~oded de [lOJ>On1Q1 CUDaIlatively compout><lood, 611--t;1~ dilJeTe.otially ~d, 61~19 ~oded de !DOlor, ~73 oqui .. leD1 circuit, ~ Kirchbo/J', voltage law, ~ ... g .... omotive force, S68-569 oooli ..... aruolysit, HI--,'!73 , peed coDtrol. ~73 'orqu~ chlInocleru1ioc (<le """"'~ nO_ .,'I71 CoqJu1er ...... = program. Sa MATI.AB Condenser, J64 ConWctance, 12 ConWctor, 490 ConWctor .kewlog, 724, 7~ Conoeque'" pole., 437 Coruta1Jl-frequeJJC)' cyclocoovener, 209 Coruta1Jl _ borsepowe:r conoec1iOll, 439 Coruta1Jl-<lU1pII1_ power conoec1iOll, 438 Coru ....... 737 Coruta1Jl _ torqu. <:O<IDOCIiOll, 438, 439 CorrveniOll faC1<n, 737, Sa aIu! U .. ,. of ... 1!UJO -~. '" machioe .. 261 demodJj .... ,~24--.,'I~ de lOOIor" ~92 i_OIl motors, 39~ modeHD,I!,86 RCL. 261. 662 =-~, . ingle-pbafonne:r, 86, 102 ~'"'. '" machioe .. 261 262 demodJj .... ,~~ 739 740 INDEX Core 10 .. _ _ C""'-demo«n, ~3 inWctioa .-or.<, 3~ Core form tnmsforme:r, til Core_ Jo... ~ 83 Cou.o'OJvoItlIge-seruing reI. ys, ~81 Cross_ field 1boory, 64l-64~ CSlI%-I99 euo..Jati"" ~g, ~ euo..Jati""ly <:<>mJlOlUlded de l"DOH1Or, 611~1~ euo..Jati""ly <:<>mJlOlUlded de motor, ~" ~"' <:OIIll1lOII, 100 <:WIll, ~I Ie><li.og, ~2 magnetiuion. 81--1l2 ..n...,I09 _ _ phase circuit, ~9 Y <:oaDOCIion. 68~--688 eurr..t u.u.b, )]9-140 eurr..t source invo:r1OJ (CSI). 1~199 eurr..t lnASforlDOr, 68, 14 1_142 eurr..t - ~mitilll circuit, ~92 eurr..t - ~mitilll flUeS, ~89 Cyclooor!verto .. , 209-21 8 basic coacoplS, 209-14 circu1atilll Cirwbting \C)', 11:» CyHt>drical_ lDOIor ouocmn., 247 ~ wiOOings. >67-371 de ...... , ... voltage coruoller ciJ<:uit, ~&'! de genenton, ~94-619 ""mub'ively cornp:>Illlded, 611~1~ "'f"",d,~ diff ... "'i.uyc~61~19 oquival ... , ciJ<:uit, ~~, ~96 lan,y, ~~ prjlDO JOOVer, m >epar1IIoly excited. ~. Su abo Sepanoely exciled do:: genentor series, ti08--filO shlLll~ 602--6()Ij. Su also ShILll' do:: .--types, ~94 voltage .. guI"'ion. ~9~ do:: machine, 473_1;]2 >C modlioo., compand, 681 ..,...."'" reacti"", ~ brush .biftilll, ~ brushe.!, ~21....,'!2J OOIIllDItatiog poIe ... inlOJp:> .... , ~" OOIIllDItation. 4~89 OOIIllDIta1Or, 473, ~19, ~21....,'!23 OOIIlJ>O"SIIIilll winding., ~ IW 13 coastn>ction. ~ IW2J efficiency, ~24 fum loop mxbillO, 48~-4\1O froJI-Ie! wit>di,,!!, ~1....j(J2 Il"lIOJaton. SN de go,.,,"'01'1 iMul" ioo, ~21....,'!2J LJiIdI vohgo, ~ lap wit>di,,!!, 49l-497 Io.<oo.,~~~ moIOI'I. Sa de moton poIoIfnme <:oo>1rI><:ti0ll, ~2(h'!21 power_ llowdiogom. ~~....,'!26 proI>le,,",,~14 roIlIting loop between curved pole face., 473-4&'! ""orlarmature ..,...uuctioo, ~21 ""or coil., 49()....492 ""or (armature) ...mdings, 492-493 torque, ~2, ~ I(h'! 17 voltage, 47~80, ~ I WI6 .... ve winding. 497....j(J1 de machine tintiOll CIlJ""" ~36-,'!38 de motor, UJ,'!94 ~de.-or.,~Tl COII1JoI and P"'octior> oquipme.t, ~73 Off OCieDCY, ~92-,'!94 "'lui ...... circuit, ~3~....,'!>6 10 ..... m-,w3 DIlIg...,; .. tioa curve, Hli-,'!38 motor.otarting circuit!. HW82 PMDC moton, ~~2 J>OP"Iacity, ~34 >epanoely oxciled ..... or, ~JWJ9, ~~1. SN 000 S_ de~. ie. de motor .. ~2--,'!68 • t.rnI de mot"", ~3B-.'!39. SN abo Shu.t de motor. loIid ... ,e de motor "","roller, ~8~....,'!92. SN also Sclid_ ..... do:: motor<:Olltroller .peed .. guI"'ioo (SR~ ~3W3~ .tarting proI>lemslsoJutiOD!l, ~'73--,'!ll2 type" ~3~ Ward_ l«>nard 'Y""'" ~83-,'!&,! de motor .tarting circllib, SlB-.'!82 de Ie", oB4-4.l~ dc-to-do:: <:OIIvo:r1ers, 1S6-193. S" aI.!o a.,.,., D<ep-bar rotor, 420-422 DeIt. (.1) conDectioo, ~8--U9 DeIt.-dek. (.1-.l.) OOIIDOcti"", 12J DeIt. _ wye (.1_ y) <:Ooctioo, 121_122 Denti,,!!, 134, 441 o..ig. da .. A lDOIor .. 422 o..ig. dass B motors, 422 o..ig. dass C mo«n, 422 o..ig. clas. D lDOIon, 422-423 o..ig. dass F inWctioa motor, 42J o..ig. clas .... , 421-42J Develq>ed mod>anicaI power, 397 Develq>ed torquo, 398 DlAC. 1~8 Differenlially 00IIlJ>0Il1"I0 de 8"DOH'or, 61~19 Differenlially OOIIlpoundood de motor, ~" Digital pulse genentiOA circuit., 171 Diode, I~J-I~ defillOd, m fru-wboelins, 186 PN,I54 PNPN, I~I~~ . .. itching time, I~ "iger, 1~4 D;"ct .oyo:bromu ... act ance, 73 I Dj,,,ibuted wit>dioy, 716-72'! bRodtWdistributioo foetor, 718-720 dOl bar!DOllioc .. 722-724 volt"!!,,,721 Dj,,,ibutioo factor, 71 !!-720 Dj,,,ibutioo tnmsforme:r, til, 98 Dj,,,ibutioo tnmsforme:r oamepb ,o, 141 Di""rter re,si"or, 6)], 614 DooW .. 1:1 Dot COO"".tiOll, 70, 84 Double-<:"!!,, ""or, 4:D-422 Double_ laye:r wit>di"!!,,, 718 Double_ .. voh ing_ field tbeory, 6]8-642 Doubly excited .!J"I"-' machine,31S Duplex lap winding. 496 Duplex NOor winding. 492 Dynamioc " ability Hmit, 3D Ecc ... tric pol .. , ~20, ~21 Eddy cmrenI !two" 28 Fancby'. b w,31 ferromap>etic ctioo motors, 428-430 nominal, 428-429 tnmsforme:r, 102 ElectJic ciJ<:uit, II ElectJic.l degru .. 490. ~ ElectJic.lloose.!, 261. Su aI.!o Cq>pe:r b •• ElectJic.l macbiDO, I ElectJooic.s circuit 00ard, ~87 EIIOlJ!J< 737 EIIOlJ!Y Jo... IOC macbine.s, 261_262 bub !two., ~~ copper 10<00 •. SN Cq>pe:r 10< ... ""'" ...... SN Cae Jo... do:: machine .. ~24-~m do:: IDOIOI'I, ~W93 oddy """""'10< .... SN Eddy """".t b _ ferromap>etic ctioo motor<, 394-396 mocba.oical 10.<00., 262 00- roIlItiOlllll Jo..., 262 rotatiooW 10.<00., 39~ .tr>y los .... 262 tnmsfonoon, 102 wit>date Ioose.!, 262 Eopisb 'Y'tem of unit, 2. ~3, 737. Su aI.!o Units of 1D01!UrO Equalizers, 494. 496, 498 Equalirilll woo"!!,,, 494, 496. 498 Equi ...... circuit cornp:>Illldod do:: IDOIOI'I, ~ \fonne:r, 73 illlh>ctioo motor,3!8--.J94 per ~. SN Pe:r-pbaoe oquivole'" circuit .uJj"'-poIe .yo:/>roIIt>w l"lIOJator, 728-734 "pon'ely exci,ed de go .. ""or, ~ "pon'ely exci,ed de~, ~3l! .. rie. do:: genen<or, ti09 ..w.. de 8"oo",'or, 603 ..w.. de motor, ~38 'J'~' 8"DOH'or, 1:14-1:19 'J'~' motor, 347 tnmsforme:r, 86-94 Equi ...... field CIlJre.t, ~, ~99 Excitati ... ""' ...... BJ Ex .................. _ i __ •. 194 """ """"" ,.; • ~J9 """ """"" ........ «8. (51 1'010",.2&-32 F ... -i",.op<ed dio:Ioe.s. IS. F •• I,. l21.J26 FerrornaS .... i< mllori .... delined, 8 ... ..,. _ .. U--2B ~ __ 21_Z mopoti...o.c. """,,.lO6O porroeotQty.21 Fiold emil. llol H.1d IpJ!.171 Firioe;...po. 179. l!1_IM _05 .... r .. ld, Flosboo .... 5(M FIat~d.61J ~, '" ....:hi .... l46--ZO .... pt> flu.< ....... y diooib.nioo, 701 Ie..".. 0 . 7l. 79 ~ cimHI. 12-13 III. 7l. 79 _idalll.27 .. ne.dc_56l FIo .. <k";'y. 10. 19-11 Flo .. de";'y vector. 247 Flo .. linb, .. ll, T1 Flu _ok""" ~ F<rco. 737 """"""' ....... lin. 113-169. 210 f<lrm.w<Ulld .. .-or ~ 717 m..1oop ~ dc'-DO. 48j m..paIo """wound dc ..-.. 495 Fow.poIo iIIo<Ied.p<>1e iDaioo m,," ... 6;H Fow.poIo$II"" wiDdi.D,. 243 Fwr'!"'io ........... 0<1 dc _ ... 499 m...,.odn. oow.-dc ___ conIrOlIef. 516 F .............. ail. 491. 11)8, 709-712 F ........... piIdI wi!ldi_&,. m. 712-716 Froctiooal .. ..,. ..........., .. 724 F ....... 518 Flftow .... li.' diode. 1!6 --, illlMctioro -.. 386-J8II. 43&-40 ",..326-311 ", __ 373 1nB1annor. 1J4 Froq ... ""y.powe. d>ono::t<rilllX. .104-J05. ll7.119 Froq"'''''''''_or diogml. JO'l.110 Frielioo _ .. 262 5Z mlOPJ·fJea. 14 """.Ie, wi!ldi"", 1001 • .'JO:Z hII-lood "Oa.c ~t.ioG. 100 FoIl--piIdI ail. (90. 1011 Full· ..... rr<'\if.rr ... "' ... ph .... 167_168 ph.169--17O F ........... 1IIaJ (roqtoeO<)'. :1).1 F ... d7l.579 G .. "'""'"' (GTO) tII)'Rroar. 157 60 __ ElKtric.~ ___ .• :/9 6oDOnl."...,.- pol ..... 4411 • .w9 ~-dc • .s<J4-619. Su /JbtJ dc ....... _ ...... , i ......... 46Cl-46I b .... cklDOlOr • • 1-4241 .JII<bronoo •• 267-3<15, SHabo Sy.:trDnou. II""R"" 0. ..... '01 ocb .... 8. l5 GbosI pb .... 126 00_ mocllanis .... )Of tnpbiclll 11 .... SH JuWy"a GTO m,n-t. 157 lWf--onve....,;r .. r .... &Ie~ 163-166 tJ. ... ~ 168-169 H""",,,"c problems. 218-221. 712-116 Ho.ting limiI. )35 Hip ......... uanomia"" Ii . .. 16 Hi~.IocIfoooi<:o...ruor.. ~ Hi~ .oIid-.-<kn-. Sa _ .1ecUooIi", Hi&fw~p rqioo. oIQ5 Hi~.torq ... pol ..... 4411. 4,;) HoIdi", """ ... (I~~ 155 H<n<p<>W .... 737 H""", diogml. 309. III Hyo&eH!i .. 26 Hyo&eH!io """". n Hy~io ....,... 26.. XO Hpocraio -. 2S H,.-n ..-... 666-669 Ie. 109 [.' ]]0. I~' Ie.. llO [.~ 109 I'R -. Saeq,p. bun IdcodlrWlO. 67_76 cin:ui ... 72-16 .". 01 .. 01 ...... formor to, 56 doo COIIYemioo. 70 iJI¥dmoo Ir'USformo'ioa, 71 _72 magnetizatioo CUtVO. 85 .wI, 71 __ c.ymboI.~ twim ...;o.@ IGBT. [61_162 21» ........... oJ.,..,..... ll5 <kfuood. 71 ideal Ir'USformer. 71_72 ........ 1 .....,hitIo. 2&S rotor circo~ ",0<101.)91 Tbov ..... 4111 ~ • ..p . .tS. 49. ~l. 51 __ ,,-'12-33 _ """",. Su ~ !Jduoced YOl~ . SN Volt. InduoctiOll ....... or. bIo ...,~-.,. ....... 448. 451 OfF""" 380-312.<117-419 cin:ui, _I por_1tn. ~5:z...16O ...... ~n~466 de t ... ~ (S<l-4H deep.bor roICJf. (20-422 <k .. !" clawa. 421-423 diOling&iollod re ....... 31t'J ~ ......... 42n-422 ,..;........,2 cfficic...,.. '~30 . _ frequon<,'" . 3116·JU equiyolc.t cin:ui~ 113-l94 INDEX 741 bistorical .............. (~211 iO<b:ed Iorque. :l84-JSS. <lOI--IOII liK~ . • )S."'O liK "Oltqe..uJ Io<:to<I-...--. (jj..(~ ~.~ ... ,"";c field, <102 Dl>!"";c.-or M •• r ciKuit. 432-4:14 ...!..,.;mioot ........ :189. 390 -mod of ~ pole>. 43&-431 ....... ......., ___ • .02-454 "'too.t' • ......-. '". ___ • . ~ 4]),-4]4 pc • .pw.e eqWyoi..- ciKui~ 194 pole clwJsin& 4:16-4)8 _or. 39t.J98. <166 _or limio. 4tS6 _ ... _di ....... J94 po """", .• I0-411 P"o\'M ctTacof_. ,~, _.cin:uit ..... ___ .·~ m .illlgle-plwe __ 637-665, Sa abo 50", ... "" ... iD\aioo -, >peed <:am>1. 434-444 ...... ,, 430-4l4 ..... ,code ......... 30. 431 a , ___ ope«!. Jas. 435-436 ""'fK. :l84-JSS. )91, 401..«JII. 410-411 ~ -".ri.oUc. 401-416 tnn>fonneJ _I. l89-l90 YOl, •• 44.1. 465--166 voLtaso ", .. s. 46S-466 ·oItase· ...... ·f~ po"" .... ~, ..... 0<1 ..-. llW50I bdocWrI __ "'"'" cimoi<x OJ bdocWrI...- tod.g>te bipol .. tnruiotor ( IGBT~ 161_H'i21JYj ~ oc ...-bi"a 259.260 dc __ (91. 52l-514 "-;0s. 335. )]6 rroIows • __ . ll5 l .. orruol rnxbi .. lmpcoon. 371 Kirdobotr', """ ... 1. 688 Kir"""""', ""hp la~ e<>cnpo .. dod doc moI<Jr. 568 O<I"iv ... ., cu.:.~ ('j'Dcluonow 81').216 ""'"v ..... cu.:.~ ('j'Dcluonow ~ :141 r .. 1d ~nt __ (,~ ~ l:!1 ~_de-,J7 MJIOI"Iy ""ciood de ........... WI MJIOI"/y ""ciood de --. 519 ...... de~ . 1iO\I ....... de.~ ...... de ........... 6W ..... 'de_.339 Inn .. onn .. pho .... di.lOO V ......... icJn.OO Knee. 21 KVL .q ... "i .... 5<. Kirchb:>If', volt .. .. .. L dildt ""~ op. ~-501 l..o"'''1~''' 51 l..o","1 po..... r....". .~ , _ ..... 289. 290 .)'IIdIIi0chiDe.:I(;....47 I ..... boo;" "'!u .. .,.".l6-17 ge-.to<. .. 41 -42. 44 ""-.... 19-41.44 ......... .... ..-.... J7-39. 4l-47 .......... .,...".,.,..,.. __ 239-299 .,...".,.,..,.. ""-. 351-J.S5 ~~.I\14 Lo:""""""",, __ J90 Lo:I:ockooor .... 455-451 u., ...... , -.......ioD. 611. 616 Lou.. .s.. Enav lou Low.po......1octroDic.s .. eli .... 59()....S92 Low .... '" n:]!ioon. 4m M.,..ti. ci"",i~ 11_21 M.,..ti. ca .. 9 M.,..ti. don>aiu.. V M.,..1ic f>tld. 8-21 bolie Fri""iples. I .ifm 0I.........up. 32. n FIIMIoy"' .... 28-32 rri ...... 1Joa. 14 itoioced ........ . l4-J.'I ~0I-,,402 .-..pel;" .m.;~ 11_21 ~oJ.8-IO "" .. i., field. 1J1-246. .s.. _ Roari", ... pic f .. 1d .impIo loop. 2.JO...ZJ8 "os"·ph iD<!ioo onodIiDe. 46J illll><!ioo _ .lIl9. J90 .,.-.................,..Vl ..--.fomw. n 136. J90 lJIYllS.V Inaofunnoo-. 136 2»-V de ............ SCI MopeIili .. "'1Uq.. 21. m MopeIili .. .........". 319 MopIoo>oIi .. 'on:. Ie ....... ~ In ... ,,"fti<'6- 5()@.. ;'A)9 """'I'"",at"''''"''np. 51WLJ ~de_or.568 ~1ati .. ly """'pollDdod de 80" .... ""'. (;11 deon""""". 305. ~ . 51W13 dift" ... .,i .. ly~ de 80 ........... 616 _I"";. ci""';~ II ..,... .. ly . ........ dc geDOnl<><. 598 ..... d_1 1 Mope....,. .. 'on:. (....t). II Moi.IrMdormor. III t.'-al de ........... " '" '75 ...... m MATI.AO fiolll......,.(.~ O>OIor).m fuing IIlJIe. 111-113 I ..... de or. 46 ... S .... iccircuil. I6-11 ... S .... i .. 'i". OIl ....... 537...,'138 ripple 10C!<f. 165-166 roIO!l",,,,....k: field. 245....246 .pe«l (ohurl de 1JIOIUt> SO.&Je-tltI- ioGICIi"" -.-=iol.putpOOO. 665.-667 >IO"ppOf. 6~74 .I}"~" 346-31'9. 5<. abo Sy __ 00l ...... 6J4...6J(; ... ples 1'9 ",ndlo,. 496 Multiple.o!Olor ... Indlo, .. 4J8 Multiple ........ ";r><Ii"lo 500 Multiple .... ,JodI .... 491 M .... 01 fluJ. 78. 19 _pIm. 1010. 141 ildottiorr ......... 4601-46.5 .,.do __ .173 .............. 1.0.141 Ntpr; .. FOIIp. 214 NEMA""""_n.4]1 NEldAi ....... ,;OI ....... i .. ognoI.lIoDoroo-' Ie nocbi .... 259 i .. ognoI.IIoDoJlO"'<' de moI<Jn. 523 ...... , ... Iotion lif. yo. ",odiOS ~roIUI •• :zro NEMA nomio" .ff .:",1IC)" .tondonls. 429 NEMAS_MGI ·I991/M,.. 4 1fd Gt., ,,,,,,,,,.l :tl9. 524 &, Ntooonll'lano .... n. SOWOt __ (N). 737 ___ ...... (N . ... ~ 7)1 __ · ... wof~6 N<>.1ood to!lIionoI -. 262. ~3 ND.1ood ....... .ol ""bs .. W ND.1ood ..... 4}2-4$4 NomiDoI 011" .:",,,,,,. 425-429 Noocircu .............. "FIDe ............ 214-215 NorIU .... ",oly'; •. S .. abo ....... Iy .... -'P""odocl doc """"". 571...,'173 "ponlOly .. doed de , ......... ....." """" de . 54)...,'147 ~poIo .26! ~-poIe _ ..... 247 -.um-pole ......... 26! NamotlY ."""' m Namotly operr .......... ~7a...,'l7Il -, ooc.= QIun·.I ..... 12. SO Om<>omi.,.. )01 Obo·~ .. di ......... XIO Opo...,it .. ~ _m>ric (OCC). 2U_214 Opoo..citco.n 1011 .I}"~' ."""'''''. 23.J ....... 1 ............ 90-91 Opeo.4 """"".,iOll. 126-129 OpeI."OI)'I ... -<lella .......... ion. 1)()..]Jl Oulplll "OIlodios. 66 Ow«oonpoo-. 61 3 Owrbe","", d windinp. 1l:!. 136 0-'" 0-.... ," ... .. ,,,..." o-..~ trip. ~19 Pw:ollol .,.......,.. of .. ..-..-"'-'" odvulO,U. lOO-lOl ~ . po'I'"<f . h.-rUli ..... 311. 319 , .... toton d oomo Ii .... 112-J18 inf,DiIO 100 .. J08...J 12 lars.P"'""'rsySlOlW, lO8-312 panlleling <>JDdj,ions, JOI--J03 Slep-by->'ep procew.., lO3-J04 symcJ.rucq>e, JOJ--J()ol ~gbt-bulb metbo<l 302, JOJ Parallel-ap.dtor <:OIIllIUI,,,,joo circui~ 191_193 Peal illvon>e vohg. (PlY), I~ P=e-..;o of rippl., 163 Per ........ , spH'-<:,.,acitor moton, "" M' Per ........ , mapIO, de (PMOC) .-or, ~~9-j62 Per ........ , mapIO, Slepper motor, ~~; Per .... bility ferromagnetic ... ,ori. ls, 21 fne spac., 10, 737 mapIOtic, 9,21 rei",;"", 10 ... tsof ......... , to Per ... """", 12 Permittivity of fne space, 737 Per_ phase .q.;vole'" circui, balao::ed p.. .. sySlOlW. 693,694 iDdoction moIor, 394 symcJ.ODOOIS ll"DOH'or, !l8-!l9 symcJ.ODOUII """or, 348 Per omit system of .... ......"..,.., ';D,I;le-phase tnmsforme:r, 94 pbaoetnmsfonoor,12l-124 Phasebek,718 Phasegroop,718 Ph ... q.aotitie.!l, 6&l Phase .. qoeoce, 6&4, 6&'! Pbasor diagnm ..tienl pole ,ynd>romo. gonontor, 732 .ymcJ.OGOWi geDmlior, !l9-280 symcJ.ODOOIS motor, 349 tnmsfonoor,IOO-101 Pilot .xciler, 271 Pitch foetor, 491, 712 Pitch of . ooil, 708-700 PlY, 154 Plex (armature windiog.). 492 PloW"l,41O PMOC motar, ~~9-j62 PNPN diode, I~m Pol. dwipD,I;, 436-438 PoI.faee,~18 PoI.piece .. ~18 Pol. pitch, 701! PoI.oboe.,~18 I'ooition """", 6/7 I'ooitiv.groop, 21 4 Pot •• tiol tnrufor ... r, 68, 141 r,7--11 oir gap, 3~, 397. 4()Ij "', 49-j(]. Su aIu> A".... <: Cyclocollvon .... de_ to-\q>per DIAC , I~8 diode, 1~3_ 154 010 tbyJUtor, I S7 hormonic poblem.!, 2U_221 IOBT,161_162 in_oro. 193--10} PNPN diode, I~m power tnmsUtor, 160 powerlsJ-d coonparUoru, 162 I"lsecircoit.,111_ln I"Ise . yr>chnJDizatioo, In rectifier circuits. 163_170. Su aI>o Roctifio:r circuits rectifier i""" ... r. 193--109. Su aoo ~". SCR 1~~ lS7 TRIAC, 1~8-1~9 voltay voriatioo (II<: pbaoe coruol), In l86 P<>wer factor, ~2 .yr>chrnnow g""""or, 328 .yr>chrnnow ImIor, 360-J63 P<>wer ~mits, 327 P<>werOll',71 P<>wer tnmsfor ... "" 6/ P<>wer tnmsis'or (PTR~ 160 P<>wer triangle, ~I -.B. 700-703 P<>wer_ foetor correc:tiOll, 360-.J63 P<>wer_ no... diogom, 262 "" g."""or, 262, 263, 280-Z81. ~26 ""motor, 263, ~26 . ingle_ pbose induoctioo motor, 660 J>r.for ... d ",It or <:Oils, 716. 717 Pri.....,. windi"l, 66 Prime IDOVOJ de ll"DOH'or, m .yr>chrnnow g""""or, 304, JO~ . yr>chrnnow ImIor. 367 Prime-mover P"'""'r limit, 331 Progro ... "" lap ...mding, 4~ Progro ... "" rotor ...mding, 493 Progro ... "" .... ve .. inding, ~I Progro ... "" windi.S, 492 Protoction cir<:oit .. ctiOll, ~89 P1R,I60 Pollou, 'orque induction motor, 41(]...413 .yr>chrnnow motor, 3~1 Polo. circuits, 171_ ln Polo • • yr>clmniuion, In Pol_width 1IlOWhtti0ll, 202 Pol_width 1IlOWhtti0ll (indooctior> motor,~ 444--447 Pol_width 1IlOWhtti0ll (PWM) i""""',",202-209 Posh bot'OIl.wi,d>e., ~78. S79 Posho"", 'orq •• , 460. 461 PWM drive (indoclion 1OOI0IlI~ 444--447 PWM illvonon, 202-209 R"'ing. induction motor, 464-466 . yr>chrnnow g""""OII, 326-J36. Su abo SyAClrODOUS l"oontor rating • . yr>chrnnow ImIor, 372-373 tnrufor ... r, 134-139 RQ., 261. 662 ,.-direct .yr>chrnnolU, 731 induction motors , 4.17. 4~8 leak"8",417 magnetizing, 389 quachronoIU. 731 INDEX 743 NOor,391 , ubtnmsie"',3l'! ,YnChrODOUS. 276. 28~. 286 tnnoienl, 32~ Reoctive power, 49 Reocti"" power_ voltay cbaracleristic, ~ Real p<>Wer, 48 Real tnMformor, 76-36 """"""ion of, '0 ide. l1nn>fo:r, 8.'! """, to.. <:mreJII, 83 """"'" ",,;ado! COII"".tiOc., 12, 13 Reluctaoc. 1OOI0Ill, 66~ Reluctar>c.,orque, 66.'! Reluctaoce-'ype ""opper mot"'", 6/3 Re .. Wol flux. 27 Re .. W"" .tanor circoi~ 434. 433 Re"Slive .tarler circoit (i_OIl IDOIOII). 434. 43~ ~ve rotor ... indiog, 493 ~ve ...mding, 492 Reverse blocking diode ~ype tbyJisIor, 154 Ripple foctor, 163, 164-166 IOU voltay (~phase ",,"'or~ l'!4 R0001iog loop be1wuo CIlJvM pole bee.!!, 47J-.U~ R0001iog .... goe1ic C .. ld, 2J8-246 .lec1ricrol frequoer>:y1s!-d of rotatiOll, U~ MATLAB prognm, 24~246 "''''''''ing direction. ~6 R0001iog 1nn>formo:r, 386 Roootiomllosse.r, 3~ Roootiomi motion. 3-,'1 Rotor, 231. 473 Rotor coils, 490-492 Rotor _ 1o .. (R(1). 261. 662 Rotor reactao::e. 391 Rotor .Up. 386 Rotor windiog. , 26/. 492-493 Runa .... y, ~~ Salie., pole, 26/, ~21 Salie., poI. machine., 247 Salient pol. rotor, 268, 269 Salient_ pole symcJ.OIIOUS ll"DOH'or, 728-734 Salient_ pole 1beory of .yr>chrnnow machine .. 727_736 S"' ..... od .ynd>romo. ",octane., 286 S"' ..... ion curv., 21 S"' ..... ion regiOll, 21 sec, 284 SQ.,261 744 INDEX Soou-T~ BI.1l2 SCR.I»- I ~ SCR.- 110< dmoi ... s.r Solid. >l1II.do_OOIIIroIIo< .,. '" .sec-:I.y windi,,&- 66 Solf.a>mllllltllliooo ",..run. 19~ Solf-eq .. lillnl windill,l;. ~1. ~2 Solf •• tanial ,duc\ao;:e ~. 66.'1. 666 s.,..r.ltly .""iled do 8"",,,. 11,.. ~ oquivol..., c"""~. m _u moIy.lli .. ~'111-602 IOnniul ~Iic. ~%-.'I91 .. nniul~. '91-.j911 s.por.lyc.cllOdde -.... 5l11--5l9, 551.Sn_Sbo .. de-. Series ....... (1.), 100 Series de .--. 6OS-tiiO SoriOIde~~ ~ •• 564 "', '" ICVLcq.roliaa. ~ .".d 00IIIr0I. 561 .. nnioaJ clIonocIeri.rtia. 563--567 tr>rq .... 561-UJ torq ... ...".d choncle:ristic. 563-j6, s.. ... 1Ii_Io, , ... ilUJr. 6lJ. 61 4 s.. ... volt. (Val. 100 s.. ... ..,IIIIi." 100 Setin (lop) wiDenlioo (.,.~ ........ ). lJ!l SM .. de ........ , Ii02.-alII ,..u,6Ci6-MI oqui. .u.:.;,. 6W ICVLoquotK.. 6W IOrmioai ~c, ~ ""'"II" tUlpeed c"""oI. '52. HJ """lOut. vall • .".d II1roI. "I~n. HJ-.j~ oquivolun c"""~. 538 field mblu<z ..- <>JDroI, "~I,HW~ iaotrtirls _ ... i • ..;.. ..m cirait, 552, 5.'13 ICVLoquoo;... H8 ,.....".... S(3-5(7 ""'. fodd dmoi~ ~ ..-. "1--555 .'''''-.1 __ 3.W IOrmioai ~ 53'1-.S43 lI>rq'M...,...d chooracleriolic. ,S4(I SI ~oill. 2, 54, 731. Su abo UDiLt ot ~.-SiIi-m Siaoplo.lop windi,,&- 493 Si.",Io. roIOr windi"ll, 492 Si"",lo. wov ... indi"!l:. ~ Si.!Io-ph .... """.n1 """"'" i""""or. 19S-191 Si.!Io-ph .... I.....,.'" _ .. 637-66.'1 w-pp JIO""OI. 661-M2 copocilor ... .-t ~ 649--6!11 cimr. model. 6S1-6/i!1 atIIII·fodd -,.. 60. 645 Io motI'III'0l. ~ 7 lpIiI.pt> .iodi.lS. 646-648 .lOtIi.l. 646--656 Sin!Io-ph'" 0'I0I0r>, 6JJ Si.!ly .. cited, 311S SiJ.pole de """"'. 496 S ....... d ""'" ~or .. 724. 72'1 SL;p, }86 SI;p rinp. :za SI;p rillP 00<1 tn ...... 268 Slip .pted. JU Slippiaspola. )$1 SIoI iIInI>ooiao. T.ll-724 SIoI pildL, 11' Sol-JWI iIIIioooioro -.. 423 Sol-JWI -.. <t48 Soft ...... propwA Sa MAJLYI SoIicl-flMo do ..,.." <:OOOoJIo<. ~ block iii ...... 3101] 1O""power .lo<.-IroDic.o 0<0CIi0D, ~ Jow'J'O"'or doolroaiCi JeClioo. 59O-.!l92 p«>I'<'Iioa ci.",,;, .. cIiOll, ~89 • ...v.1Op dlWit oecIiooo. ~90 ''''''"'l''- v •. ftur.q."d"" .. Ior.~ SoIOcWWI in<h<:tlorr __ driVCl, ~" SoW-mIO v.n.t.Je.h""""1 irrdoctioo _dri..,.4t4. 0145 SpoQ~n.bmon. 611 Spoo<I. Sn_T~ -..:. .... K rnodoi 26J-.264 m C\IIIW.JMi..,ey ......,.,~odod de rnoIOf.573 de n1OIor •• 5~J5 iO<kl.lnI de mo«:n. 341-.j~ ....... pIwo IndlOCIX>D. "'-'" .~ .......... Z72,JZ7 .~ """""- J7l .... ..,1'$Il moron. 635 Spot<l drop (SO~ )()l Spot<l rqulllliorr (SR~ 2M-264. "'-'" Spood .. pIIoIioa cirait. 5\11hS91 Splil.phuo ...,.",.. 646-6411 Spri",.I",. pooh batto • ..;tcbos, 571. 579 Sqo ... low-lOtqooe C<>.lnI _. 5.'19 s-io)', Wolli-. ~ ...... de --. 51]..3&2 iao.h><liooo rnoIOI • .tJO-U4 00. do rnaohi ... 37-39. ~2---I7 ""I"<'Iic """'" .!arIo. circoir. 432---1:1-1 ... Uti •• IW'I ... in-u;,. 434. 4J~ oi..p.1"'- i""oC'Iiooo roo!<Jr. ...." .,.oo:t.oaooo. _ . J64..J71 $wt"'s cocIo lotIOn, 4]0, 411 S .............. n • S .............. 0105 SwWtop ciJaril ..."iOG, ~ Srmc ac &e , :) "",,--.., 19J SboIi.c .111.,,1)' limit. 2Sl. l<n St.Ior. 231. ' n, 116 Stoia- coil. 1 11 St.Ior <appor _ (SCl.~ 261 St.Ior wifllli .... 261 S .. ody-flOlO f .. _ oarrero:. 124 S .. ody-,oWl period. 32) S"ody-,oWl .,. __ ""' ....... 350-J64 S .. p.do...., MII_f",,,,,,,,. 100. 110 S .. ppo. """"'. 670-674 S"p."I' ,"""""",for",,, •• 109. 110 Sin)' 1Dooo •• 262, '25 S.,-iooo InnOtorr... 67 S"""" ..... , .... rot. 11l S"""" .... '" ponod, 12.l S"""" .... '" .. ..,,. 12'1 S,_i clCillllioa. 2(il..271 o;::opobiJi!y """"". l29-lM MOWO Y eli.,..... 271. 712 p<nIioa of K 1· ... nIIDrO .,...pIwo <;.i""Io., cirelli,. 278--27'9 p/Ia>. 179-280 pik!Ir .. cit ... 271 po ....... 280-lIJ power.f1Dw llioparn. 280-2111 "'in,o. 126-JJ6. s.. _ Sr-F-nIiIrp .-.;.., power ....... """'--iolic. """'" .ce.,.. ~ ... io.,211 ohr:IIt-ararilleSt. 2&01 ohr:IIt-ararii tna.IIiau. 12J---326 .IIi ..... _ opmoIUo,', "'-m .Iip ri ... ond 1INoha. :za "'" ot .01 ...... m ""ie "abi~1)' ~mil. 282 ay~.......,..a. 311-312 ay~ ... octanu. 216., 2M. 286 ~2Sl-m .......... 319-326 ."ty &coor. 2ti. 290 '""lias" 113. 321 '""lias' ~ 19O-191 5~..,....wcopobi~I)''''''', ,"-'" 5~..,....w fIIj .... l26-lJ6 '" J'C""Of. m ~.J26...J21 kilo~.ll7 _f""""'.328 .u.,.o..polo Iheofy. 127_136 .. moo fo ~atioc>. J1J oqoivu.," c"","~ 3017 r .. 1d ....... , uc'" J5S-360 "w._f""""'.loI9.J~ .. 0<1; .. ..,...., r"""",. J52. 362 Imd~ l!II-l!I5 "'"P""'" r .. 1d, :U1-3~ """"Platt. 313 ""' ............. 01 . l49 po ...... ~«mCIiOi. 36O-36l puIlouI """'"- l!Il rIliap.. l1W13 opo«IoI "" .... 313 IP""'I ...as. m --.. )6.1-311 .....,.. "",1'IItio». J».J6.I ay""'-'- apcilor. )6.1 ay""'-'-.-. ~ 171-312 ~~~.utic...,... .. "..", .odofOu:i .. dIovuucited, l56-l!l7 V < ...... J56 ."""so. 313 S~"""" flljj,,,.11l-l73 S~ _ V C'UtYO. 156 Syacbr""""" 10'''''''Il0l. 276., 28.5 Sya.d>raooopo. MJ-104 Syacbr_d . 666. 667 Sy"~"" J"' ...... """" (51). 2. 54. 737. Sr. oIs<a!ioN.~. Su tJI", ..... 1.;0" T ......... ~ri,.;< cumliloti ... ty............- ............. 611-613 do&DOd., '19 ~ .. .......,. «IIIl""'- .-.-..617 __ I)" aci .. d poonkO". ""'" ..... "".6()9...610 _.~~1 ..... de paonoor. 005--W6 ..... de -. 539-.S4J Tormiat .... ~ ~I ....... 10S-J06 yn.,.,66 Tall (T). 10. 1n T .... 1'1-. 426 lbonDO! ....,.. ......... 136 n.. ...... iq>tdaa: .. 4(J7 n.. ...... ...a..-:. oad 1n-4(17 n.c....- Elibu. ~ n.-~JIb<.bolIb IDO!bod, .lO2. Jro n.-pt..<...rc.iu. 631_106 ........... 631 bot_ !hr .. ~ ' )"10 ..... M~~ del .. (.1) <>JODOCtiooo. 688-689 FDOnIioon of ""ttav ... """" ..... M'~ ~ooIptwe ",mj,io •• 00 "",,·Iine di.~. 700 fb- OOIJOeDOe. 684. 00 r ,elOliomltipo. 69()...691 _rlri .. sle,.1OO-103 ""1Ia,\IO ... """" ..... 63~ Y COIIDe<:!ion. 68J-688, 689 Y·lo n..f~ i\93. 69S 11fte.pbooe.......- ..,."" i""" ..... 197_199 n.-.,t.-fuU-"a", lKlif. 1m-ITO 'tlfte.pbao paon __ 131.132 llfte.pb-. .... .II6--I26 deb."""""Clioo., III dek .. ..,o ~ 121_122 ..... I)' ... "'ofmo......,... ... 12J-124 e"""SO IOII(ce invenor. '~m n.-"." .. ,i"ive " "'or (in<b:ti"" .-or). 4JS n.-.. ;.. tbyri'<OO. ISS--IS6 ThyTi"<JI deflDOd.l54 010. 1ST ....... ;... 1S4--1~ T ......... .,. .. Ia,.. ~loI. 51'9. SAO Tl:.JOb...-.... 722--124 .W '" modIi ..... 211-2lo11. 2'lS--2511 ~......,..~ loop. 2J4.-23II .., __ ~:z. 511h'111 '"'""' , <k~Joped. JII8 ~.400.~1 INDEX 745 ildKtioo. _ . 184-38!1. 39&, «11,,(13 ... Ilooot. Su ......,.. .....,... ...-~ 0160, <161 rnJ "" . 2J1_llS ~~~121 .aIioOI-f".....::biDe. 7J.t-13' ...... '" .,---11" 2I2-2IJ ., ____ .331 _01",,-..5.137 T<Y<pIO ..... lSl. 213 T~~" _iIor ...... ~1Or ..... .M1 capoci<oo ..... . 650 rumnIoIivolJ.",..",..odod de .... Of. ....,'" diffeuolioJly """"""",d de moIOI. 57O.,S11 by......,;. moIOI. 668. 669 io<ti"" moclIi ... 461 io<ti"" m<It<f, 401-4)6 1 """WI<o, .1p~'-<,opoci,,,,,...-.. 651 ..If"'''''i,. .. 11I<\aIQ """"". 66/ ",ie. de IrO)IOr" ~ .bodod.poIo m<It<f. 6;S4 .b"", de moon. 540 .i.&lt.p/lAoo i_oo . 639. ""M' . pli<.pt..< . 6017 . ytd>rooolw . )~51 ~i-"'.640 uniwroal . 616 Too .. aalMioo """"at. 8) Trudamos,65-151 .......... _ ....... I)8..J19 __ ~ 109-II6.,Su.,J,., ~ ""'"' f ...... 6/ ...,., ... 141_142 ~'"imlHb.ll9-loIO <kr,.", diMibuoioo., 6/ offia...cy. 100 eqoi.uOl...rc.it. IJ6..9.I ~""'I-I~ idoooL 61-'M. Sff ... kIoal ....... i~66 i ..... _ ... 1 41)..1.2 .102 "";n. III umopb ... 14Q 141 " . ,.iI ... " 9(1..9) por.uni! .y ... m of ............... "'. 94 phaooo diosr- 100-10; p" •• Ii.lI41 1"".66 raliOjS, 134-139 ... l 16-M. Su ~..., R.oJ InIUfor .... ' """i." :lS6 .boll f<JIm. 6/. 68 .t.<.t.ciKui, ..... 91-92 "tp-op. 109, 110 . ubo!o!ioo., 6/ ' ..... 108-109 =C," .1lI ~ 116-126. SfftJI",n.r.. - .... -.....~ (,-......ror-..). 116-13). s.....s.o ~ ..-...f<noIMj<ll (r.o ......... , ....,~ .............. 134 746 INDEX Trmsf",,....-Con< voIIage .. gul"'ion. 100-101 Trmsf",,.... actiO<l, 8 Trmsf",,.... ID<XIeI of i_OIl moIor, ,~~ Trmsf",,.... ...... pl"'., 140, 141 Trmsf",,.... opoD_ drcuil' •• ~ 90---91 Trmsf",,....pbasordiagnm, 100-101 Trmsf",,.... "lings, 134-139 Trmsf",,.... ohon-Mion1 <UrrOD1, 324 T ...... .,. period, 323 T ...... .,. re:o::w.::o, 3~ T ...... " or IOBT, 161_162 PTR, 160 TRIAC. 1~8-1~9 Trigor dioo.. 154 Tr~e ideotiti ... , 6&4, 690 Trip"" barmoni<:.r, 714 Triplex willdi"-l, 493 Tor", ratio, 69 Two-byer...u.dings.491 Two-polo bp-do:: rnachlno, 494 Two-polo .oy~. molar, 347 T~ ootid ..... do:: """or COf"'''''>r, 6/ Uni .. ot .... :.s".. angular :lCcolonilioo, 4 angular voloc!'y, 3---4 ' WM01I' er, 50 llvorsioo (""or. , 737 .locIrical qw.owic., 54 English .nits, 2, ~3 llux linbS" 31 magnetic field doruiry, 10 magnetic flux doruiry, 10 magnetomotivo foR:o, II por ..... biliry, 10 l""""'r,7 reocrivo l""""'r, 49 real power, 48 SI .nits. 2, 54 ,oblo of """" .... , 737 ' orquo, ~ •• u Uni'y power (ocror, 2l!9, 290 Univor..J """or, 634--f;36 U ... 1l1r1IIod rogiOll, 21 U ... 1l1r1IIod .oy~ ... actaoc., 286 VI<> I~~ Ve,l09 V .. 110 V,,110 V .. ' 109 "--. )'1>CIror>ow eapociro, 364 . )'1>CIror>ow """or, 3~ Varioble-freqo>on::y Cfclooorrv.:" .r, 200 Varioble-line-vol1loS' • .,-1 <:O<I1ro1 (inWcriorr motor). 443 Varioble-vol1loS' aur"" .... formor, II~ "oo~ II<: rnachlno .. 233, 230-~~ II<:p/>asoconroL 177_186 . mat .... ">C1ioo, 276 coil orr tw<>-poIo . tator, 250-~3 ~'OO C<>Dth>ctor movloS i. magDOlic f .. ld, ~, cwmJatjvoly compoW><lod de !l"DOn'or, 611. 614 dcmoo:bi ..... 47~---480, ~IWI6 --y cl1lonocleruric, W~ ... Ioe machino.<, 233 rms vohg. (rhne phaso " "'or). 254 >epanIOly .xciled de SOllOrator, ~97~98 soric., 100 .t.rnr de SOllOrator, 603---60:'!, 606 .~Io """rioS loop. 2J 1_233 . )'1>CIror>ow gonontor, 273, 327 . )'1>CIror>ow """or, 3Tl Thovenio, 406-407 three--pIw>o circuil, ~ three--pIw>o .., of coil~ ~3 rnnsform ... ,I34 Y oonoocUOII, 685---68Il Vo/rage Wi]Wp (shlUl' S.,..nllor). ~ Vo/rage dividoor rul., 406 Vo/r age nllio ocross . lnMf",,...., 78-80 Vol,,,!!,, .. gub'ioo (VR) oe Dl:lCbino.s, 262-263 full_ lood, 100 .oy~. !l"DOn,or. , 290-291 1r1mSf"""",,", 100 Volr"!!,, "gub ,or, 109 Volr"!!""",,,,co illVOnor (VSl), I~, 196, 199-202 VR. Su Vo/rage rogulariO<l (VR) VSI. I~, 196, 199--W2 V_ VoonooC1ioo, 126-129 Word_ l=nard 'y'lOm, ~83~8~ W.D (W). 737 W.v • .,;00;"8, 497---5()1 W.ber (Vr'b), 737 Wlodago los .... 262, ~2'! Wllldi"-l oe macbino.s, 250, 2~~ omorIir..-or, 36/--.l71 lIrDIlI'uro, 267, 492-493 <:borded, 491. 700 0JIIIID0Il,109 <:OIIIpO""ory. ~IWlJ darnpor,36/--.l71 di.rriOOrod, 716-7~ doublo-l' yo:r,491. 718 'quolizio.'!, 494. 496, 498 f .. ld,267 fr>Ctioool_ pircb,249,712-716 froS-IoS, 501---502 b p,493---497 rwlripl. ""or, 438 pimory, 66 pogre"ivo,492 ogre .. ivo. 492 rot",, 492---493 .. cor><lory, 66 .. rio .. 109 '011i.,., 66 'wo-b yor, 491. 718 W . vo, 497..j()1 Wllldi"-l foctor, 721 Wllldi"-l ;""'b riOll, ~8-260 -, wo.u.d NOor, 382, 383 wo.u.d rotor inWcrioa .-or, 382--384 Wyo (Y) oonoocUorr,~, 689 Wyo-dol,. (Y-.1J <:OIIDOCI:iOll, 120-121 Wyo-wyo (Y-Y) COIIDOCIion. 118-120 Y oormocUorr, 685---68Il, 689 Y_ -<I. <:OIIDOCI:iOll, 120-121 Y_ -<l.1r1mSformatioo, 693, 69~ Yoke. ~19, ~21 Y-Y COIIDOCIion. 118-120 Higher Education The Online Learning Center (OLC) provides students with activities to further explore topics presented within the book and ensure a successful grade for the course. Activities may include: • interactive quizzes • Internet links to related materials • animations • flashcards • additional readings • answers to selected end-of-chapter questions • a career center Instructors will also appreciate how the OLC's library of teaching aids and compatibility with course management systems, including WebCT, BlackBoard, PageOut, and eColiege, make life a little easier. The McGraw'HiII Companies
13311	https://matheducators.stackexchange.com/questions/20709/phase-shift-vs-horizontal-shift-and-frequency-vs-angular-frequency-in-sinusoi	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Mathematics Educators Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Phase shift vs. horizontal shift, and frequency vs. angular frequency in sinusoidal functions Ask Question Asked Modified 4 years, 5 months ago Viewed 654 times 7 $\begingroup$ TL;DR version: It seems to me that high school curricula no longer distinguish between "horizontal shift" and "phase shift", or between "frequency" and "angular frequency", when teaching sinusoidal functions. Is this perception accurate, and if so is this a recent change in usage? Long version: When not at my day job, I tutor a few high school students, and I recently I have noticed that some of them seem to be learning to describe sinusoidal functions in a way that differs from how I understand important words. In particular, the distinction between "horizontal shift" and "phase shift" seems to be elided, as is the distinction between "frequency" and "angular frequency". Let me use an example to illustrate what I mean. $$H(t) = 8 \sin \left( \frac{2\pi}{5}(t - 2) \right) + 9$$ which can be written in the equivalent form $$H(t) = 8 \sin \left( \frac{2\pi}{5}t - \frac{4\pi}5 \right) + 9$$ Here let's assume that $t$ is time, measured in seconds, and $H(t)$ is the height of something, measured in meters. My understanding is that the correct way to describe the features of this function is: The amplitude is $8$ meters. The midline is $y = 9$ meters (this describes the equilibrium position of the oscillating object). The period is $5$ seconds. The frequency is $\frac 15$ cycles/second, i.e. $\frac 15$ Hz. The angular frequency is $\frac{2\pi}5$ radians/second; note that this is the same as the coefficient of $t$ inside the argument of the sine function. The horizontal shift is $t = 2$ seconds; this means that the graph has been shifted $2$ seconds to the right, and therefore, when graphing the ``first'' period of the wave, we would begin at $t = 2$, rather than at $t = 0$. The phase shift is $t = \frac{4\pi}5$ radians; the interpretation of this is that the graph has been shifted $\frac 25$ of a cycle to the right (because a full cycle is $2\pi$ radians). What I am seeing: Students are being taught that in an expression of the form $A \sin (bt)$, the constant $b$ is called the "frequency", and that the period is computed as $T = \frac{2\pi}{b}$. I agree with the latter formula, but not the word attached to the constant $b$. Frequency is measured in Hz, so it must be the reciprocal of the period -- no? Students are also being taught that in an expression of the form $A \sin (b(t-h))$, the quantity $h$ is called both "horizontal shift" and "phase shift" -- apparently these are synonyms now? The quantity $bh$, which is what I would call the "phase shift", does not seem to have a name, and expressions like $\sin(3t - \pi)$ must first be rewritten as $\sin\left( 3 \left(t - \frac{\pi}3 \right) \right)$, so that the "phase shift" of $\pi/3$ can be identified. I would chalk this up to careless teachers, but I'm finding this in published (online) materials as well. For example, here is a CK-12 textbook that explicitly says "Phase shift is the horizontal shift left or right for periodic functions", and here is the same book defining frequency as "the number of cycles that occur in $2\pi$". Now I am not by any means a language prescriptivist; it's fine with me if the meaning of words changes, or if words are used in different ways by different communities. (All definitions are conventions, and all conventions are local.) But I am curious where the boundaries of these communities are, and in who uses words in specific ways. So now, my questions: In what curricula are the distinctions "frequency"/"angular frequency" and "horizontal shift"/"phase shift" preserved, and in which are they elided? (Answers might start with: "I teach X in country Y, and the convention here is...") Have these usages changed over time -- and in particular, is my sense that this is a recent shift accurate? To a lesser extent I am interested in discussion of the pros and cons of both approaches, but primarily I am looking not for opinions, but for factual information about how these words are used, and by whom. secondary-education trigonometry vocabulary Share edited Apr 12, 2021 at 15:37 Peter Mortensen 16955 bronze badges asked Apr 9, 2021 at 2:26 mweissmweiss 17.8k11 gold badge4343 silver badges9696 bronze badges $\endgroup$ 1 3 $\begingroup$ Interesting. I graduated hs in '74 and college (with BA in math) in '79. I started teaching trig at a university in '86. I don't recall ever seeing these distinctions. $\endgroup$ Sue VanHattum – Sue VanHattum ♦ 2021-04-09 14:27:34 +00:00 Commented Apr 9, 2021 at 14:27 Add a comment \| 2 Answers 2 Reset to default 8 $\begingroup$ These books are simply reflecting the longstanding and universal usage in physics and engineering, which is that these words can have either meaning, and any ambiguity is normally either resolved by context or unimportant. Share answered Apr 9, 2021 at 13:58 user507user507 $\endgroup$ Add a comment \| 4 $\begingroup$ In engineering and physics, in the English literature, angular frequency is oftentimes abbreviated in frequency. Sometimes this is explicitly stated at the beginning, but mostly it is given implicitly. However, the intended audience is expected to not make any confusion. The usage is certainly not recent, and I think I've seen it already about 40 years ago in even older books. It's worth noting that other languages employ instead two very distinct terms for the two concepts, and there's less risk of confusing abbreviations. For instance, in Italian and French the angular frequency is called pulsazione and pulsation, respectively. Similar terms are probably used also in Spanish and German. Actually, English has a similarly distinct term too, pulsatance, but this has never become widespread, and it'd probably understood by very few people. For a list of these terms in various languages see the entry at the The World's Online Electrotechnical Vocabulary. For what concerns phase shift and horizontal shift, I've never found authors who would not distinguish between the two, even though what you call horizontal shift is more commonly called time shift, or similarly for other quantities. I'd say that your example can be an idiosyncrasy limited to few people. Finally, in engineering, what you call midline is more commonly called DC component, DC offset or vertical offset. Share answered Apr 9, 2021 at 19:05 Massimo OrtolanoMassimo Ortolano 50833 silver badges1212 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions secondary-education trigonometry vocabulary See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 8 Unit circle vs. ratios of right triangles vs. wave functions for introducing trig functions 6 Proving trigonometric identities 18 Why do we conventionally treat trig functions as going anti-clockwise from the right? 7 Rules to eliminate erroneous solutions in Trig equations? Why do we state the antiderivative of $\sec x$ as $\ln \|\sec x +\tan x\|+C$? Is $180^\circ = \pi$? 8 "Amplitude" of Tan and Cot functions How to formulate this type of arcsin problem? 6 Is Plane Trigonometry by S. L. Loney still good as a textbook today? 3 Geographic differences in the way in which trigonometry is taught Hot Network Questions Is there a specific term to describe someone who is religious but does not necessarily believe everything that their religion teaches, and uses logic? How long would it take for me to get all the items in Bongo Cat? Numbers Interpreted in Smallest Valid Base Clinical-tone story about Earth making people violent Proof of every Highly Abundant Number greater than 3 is Even Verify a Chinese ID Number How to use cursed items without upsetting the player? The altitudes of the Regular Pentagon Is existence always locational? Is this commentary on the Greek of Mark 1:19-20 accurate? Passengers on a flight vote on the destination, "It's democracy!" how do I remove a item from the applications menu Are there any world leaders who are/were good at chess? Do sum of natural numbers and sum of their squares represent uniquely the summands? Is direct sum of finite spectra cancellative? Determine which are P-cores/E-cores (Intel CPU) Matthew 24:5 Many will come in my name! Are there any alternatives to electricity that work/behave in a similar way? Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results Find non-trivial improvement after submitting Who is the target audience of Netanyahu's speech at the United Nations? Calculating the node voltage How to sample curves more densely (by arc-length) when their trajectory is more volatile, and less so when the trajectory is more constant Exchange a file in a zip file quickly more hot questions Question feed
13312	https://www.scholarhat.com/tutorial/cpp/if-else-statements-in-cpp	By DotNetTricks 0 View All Browse Tutorials 01 Career Opportunities Top 50 Mostly Asked C++ Interview Questions and Answers 02 Beginner Understanding Friend Function in C++ A Beginner's Guide to C++ Language How to Write Your First C++ Program: Detailed Explanation What are Keywords in C++ \| List of all keywords in C++ ( Full Explanation ) Identifiers in C++: Differences between keywords and identifiers Data Types in C++: Primitive, Derived and User-defined Types Variables in C++ Programming Operators in C++: Arithmetic, Relational, Logical, and More.. What is Expressions in C++ \| Types of Expressions in C++ ( With Examples ) Conditional Statements in C++: if , if..else, if-else-if and nested if Switch Statement in C++: Implementation with Examples Loop in C++: A Detailed Discussion with Examples Understanding While Loop in C++ Understanding Do...While Loop in C++ Jump statements in C++: break statement Continue statement in C++: Difference between break and continue statement Goto and Return Statements in C++ What is a Function in C++? Explore Type of Function with Example What is Arrays in C++ \| Types of Arrays in C++ ( With Examples ) 03 Intermediate Strings in C++: String Functions In C++ With Example Pointers in C++: Declaration, Initialization and Advantages Call by Value and Call by Reference in C++ ( With Examples ) Function Overloading in C++ (Using Different Parameters) What is Recursion in C++ \| Types of Recursion in C++ ( With Examples ) Storage Classes in C++: Types of Storage Classes with Examples Multidimensional Arrays in C++ 04 Advanced Object Oriented Programming (OOPs) Concepts in C++ Access Modifiers in C++: Public, Private and Protected Constructors and Destructors in C ++ Inheritance in C++ with Modifiers Types of Inheritance in C++ with Examples Polymorphism in C++: Types of Polymorphism Function Overriding in C++: (Function Overloading vs. Overriding) Understanding Virtual Functions in C++: A Comprehensive Guide Interfaces and Data Abstraction in C++ ( With Examples ) Exception Handling in C++: Try, Catch and Throw Keywords 05 Training Programs Data Structures & Algorithms Course For Beginners C Programming Course For Beginners C++ Programming Course For Beginners Home Tutorials Cpp Conditional Statements In.. Conditional Statements in C++: if , if..else, if-else-if and nested if 26 Jul 2025 Beginner 29.1K Views 14 min read Learn with an interactive course and practical hands-on labs Free C++ Online Course With Certificate Start Learning Free Free Interview books Skill Test Conditional Statements in C++: An Overview The flow of a program is managed using conditional statements in C++. If a condition is true inside the ifstatement, the body of the statement executes. If false, elseis used to specify an alternate course of action. You can check several criteria sequentially using the else-ifstatement and learn C++ from scratch. For multi-way branching based on the result of an expression, use the switch statement. You can also check our Free C++ Course for in-depth knowledge of C++. Conditional Statements in C++ programming? In C++, conditional statements are control structures that let code make decisions. These statements are also known as decision-making statements. They are of the type if statement, if-else, if else-if ladder, switch, etc. These statements determine the flow of the program execution. Types of Conditional Statements in C++ In this article, we'll discuss the first kind of conditional statement i.e. if...else statements in C++. We'll discuss the switch statement in the next tutorial switch statement in C++. if … else statements if-else is the first kind of control statement in C++. In this, the operations are performed according to some specified condition. The statements inside the body of the if block get executed if and only if the given condition is true. Read More - Advanced C++ Interview Interview Questions and Answers Variants of if-else statements in C++ We have four variants of if-else statements in C++. They are: if statement in C++ if-else statement in C++ if else-if ladder in C++ nested if statement in C++ 1. if statement in C++ if statements in C++ are one of the most simple statements for deciding in a program. When you want to print any code only upon the satisfaction of a given condition, go with the simple if statement. Syntax ``` if(testcondition) { // Statements to execute if // condition is true } ``` In the if statement, the test condition is in (). If the testcondition becomes true, the body of the if block executes else no statement of the if block gets printed. Example ``` // C++ program to understand if statement include using namespace std; int main() { int i = 11; if (i > 15) { cout << "11 is greater than 15"; } cout << "I am Not in if"; return 0; } ``` This C++ code in C++ Compiler initializes the integer ito 11 and then checks to see if it is greater than 15 using an ifstatement. It outputs "I am Not in if" to the console because 11 is not greater than 15. Output ``` I am Not in if ``` 2. if-else statement in C++ It is an extension of the if statement. Here, there is an if block and an else block. When one wants to print output for both cases - true and false, use the if-else statement. Syntax if (testcondition) { // Executes this block if // condition is true } else { // Executes this block if // condition is false } Example ``` // C++ program to understand the if-else statement include using namespace std; int main() { int A = 16; if (A < 15) cout << "A is smaller than 15"; else cout << "A is greater than 15"; return 0; } ``` This C++ program initializes the integer variable Ato 16 and then checks to see if it is less than 15 using the if-elsestatement. It outputs "A is greater than 15" to the console because it isn't. Output ``` A is greater than 15 ``` 3. nested if statement in C++ We can include an if-else block in the body of another if-else block. This becomes a nested if-else statement. Syntax ``` if(test condition1) { //if condition1 becomes true if(test condition1.1) { //code executes if both condition1 and condition 1.1 becomes true } else { //code executes if condition1 is true but condition 1.1 is false } } else { //code executes if condition1 becomes false } ``` Example ``` // C++ program to understand nested-if statement include using namespace std; int main() { int a = 10; int b = 5; if (a > b) { // Control goes to the nested if if (a % 2 == 0) { cout << a << " is even"; } else { cout << a << " is odd"; } } else { cout << a << " is not greater than " << b; } return 0; } ``` In the above code in C++ Editor, the first if statement checks whether a>b. If a is greater than b, the nested if statement is checked. If the nested if condition is false, the else statement in the nested if block gets executed. If the first if statement evaluates to false, the nested if block gets skipped and the else block executes. Output ``` 10 is even ``` 4. if else-if ladder in C++ It is an extension of the if-else statement. If we want to check multiple conditions if the first ifcondition becomes false, use the if else-if ladder. If all the if conditions become false the else block gets executed. Syntax if (testcondition1) statement; else if (testcondition2) statement; . . else statement; Example ``` // C++ program to understand if-else-if ladder include using namespace std; int main() { int A = 20; if (A == 10) cout << "A is 10"; else if (A == 15) cout << "A is 15"; else if (A == 20) cout << "A is 20"; else cout << "A is not present"; return 0; } ``` Theif else-ifladder is shown in this section of C++ code. It sequentially compares the value of Ato several conditions. It outputs "A is 20"as the condition matched and comes out of the ladder. Output 20 Summary This article gives a vast idea about if-else conditional statement in C++ language, including its various types with syntax and examples. If you've understood everything we went over in this blog post, congratulations! Now you can check out for a C++ certificationto get some more insights and practice on such an important programming concept. FAQs Incorrect logic: Double-check your conditions for errors and ensure they accurately represent your desired behavior. Missing braces: Braces are crucial for defining the code block associated with each condition. Infinite loops: Be cautious with conditions that might never become false, leading to an endless loop. Dangling else: Ensure an else statement belongs to the correct if condition. The switch statement is useful when you have multiple conditions that check for different values of the same variable. It allows for more efficient comparisons and clearer code organization compared to nested if statements for multiple values. Share Article Similar Articles Interfaces and Data Abstraction in C++ ( With Examples ) Types of Inheritance in C++ with Examples Continue statement in C++: Difference between break and continue statement Function Overloading in C++ (Using Different Parameters) A Beginner's Guide to C++ Language Inheritance in C++ with Modifiers Previous Tutorial What is Expressions in C++ \| Types of Expressions in C++ ( With Examples ) Next Tutorial Switch Statement in C++: Implementation with Examples About Author View Profile Shailendra Chauhan (Microsoft MVP, Founder & CEO at ScholarHat) Shailendra Chauhan, Founder and CEO of ScholarHat by DotNetTricks, is a renowned expert in System Design, Software Architecture, Azure Cloud, .NET, Angular, React, Node.js, Microservices, DevOps, and Cross-Platform Mobile App Development. His skill set extends into emerging fields like Data Science, Python, Azure AI/ML, and Generative AI, making him a well-rounded expert who bridges traditional development frameworks with cutting-edge advancements. Recognized as a Microsoft Most Valuable Professional (MVP) for an impressive 9 consecutive years (2016–2024), he has consistently demonstrated excellence in delivering impactful solutions and inspiring learners. Shailendra’s unique, hands-on training programs and bestselling books have empowered thousands of professionals to excel in their careers and crack tough interviews. A visionary leader, he continues to revolutionize technology education with his innovative approach. Our Courses Data Structures & Algorithms Course For Beginners Free Free DSA Online Course with Certification, Dsa Course Free, Free Dsa Course With Certificate, Free Dsa Course View More C Programming Course For Beginners Free Enroll in our free C Programming online course and earn a certificate in just 21 days! Master fundamentals like variables, loops, functions, and pointers with hands-on exercises. Perfect for beginners—start coding today! View More C++ Programming Course For Beginners Free Free C++ Online Course with Certification in 2025! Designed for beginners, learn and certify in 21 days. Don’t miss out—register now and start coding! View More Live Training - Book Free Demo Azure AI & Gen AI Engineer Certification Training Program 28 Aug • 08:30PM - 10:30PM IST Get Job-Ready • Certification TRY FREE DEMO .NET Solution Architect Certification Training 30 Aug • 10:00AM - 12:00PM IST Get Job-Ready • Certification TRY FREE DEMO .Net Software Architecture and Design Training 30 Aug • 10:00AM - 12:00PM IST Get Job-Ready • Certification TRY FREE DEMO Advanced Full-Stack .NET Developer with Gen AI Certification Training 31 Aug • 08:30PM - 10:30PM IST Get Job-Ready • Certification TRY FREE DEMO ASP.NET Core Certification Training 31 Aug • 08:30PM - 10:30PM IST Get Job-Ready • Certification TRY FREE DEMO Free Master Classes & Live Batches Schedule 28 Aug Career launchpad Expert Q&A session Register Now 28 Aug Career launchpad Expert Q&A session Register Now 29 Aug Build & Deploy a .NET Application with Azure and GenAI Register Now 30 Aug Build & Deploy a Spring Boot Application on AWS like a Professional Register Now 05 Sep Become Azure AI-900 Certified Kickstart Your AI Career with Microsoft Azure Register Now Live Training - Book Free Demo Azure AI & Gen AI Engineer Certification Training Program 28 Aug • 08:30PM - 10:30PM IST Get Job-Ready • Certification TRY FREE DEMO .NET Solution Architect Certification Training 30 Aug • 10:00AM - 12:00PM IST Get Job-Ready • Certification TRY FREE DEMO .Net Software Architecture and Design Training 30 Aug • 10:00AM - 12:00PM IST Get Job-Ready • Certification TRY FREE DEMO Advanced Full-Stack .NET Developer with Gen AI Certification Training 31 Aug • 08:30PM - 10:30PM IST Get Job-Ready • Certification TRY FREE DEMO ASP.NET Core Certification Training 31 Aug • 08:30PM - 10:30PM IST Get Job-Ready • Certification TRY FREE DEMO A world-class EdTech platform Unlimited live sessions Quick notes Hands-on labs Real-world projects Interview prep Accept cookies & close this
13313	https://math.stackexchange.com/questions/4272137/functional-equation-fx-xf1-x	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Functional equation: $f(x)=xf(1/x)$ Ask Question Asked Modified 3 years, 11 months ago Viewed 1k times -1 $\begingroup$ I'm trying to find out the function $f$ that satisfies the functional equation $f\biggl(x\cdot f(\frac yx)\biggr)=x\cdot f\biggl(\frac{f(y)}{x}\biggr),\forall x \in \mathbb R\setminus{0},\ \forall y \in \mathbb R$, where $f:\mathbb R \to \mathbb R$ is a continuous monotonically increasing function, with $f(-1)=0$, $f(0)=1$, and $f(x)>x,\forall x \in \mathbb R$. Besides, $f$ satisfies this two functional equations: $-f(-f(x))=x, \forall x \in \mathbb R$ (It may be useful to consider the function $g(x)=-f(x)$, so $g(g(x))=x$). $f(x)=x\cdot f(1/x),\forall x \in \mathbb R\setminus{0}$. Thanks Note: I've found out that using the three equations there are 2 new relationships: $f(-x)=f(1)\cdot f\biggl(-\frac{f(x)}{f(1)}\biggr),\forall x \in \mathbb R$, with $f(1)$ unkown, but satisfying $f(f(f(1)))=(f(1))^2$. $f(-x^2)=f(x)\cdot f(-x), \forall x \in \mathbb R$. functional-equations Share edited Oct 16, 2021 at 22:33 Javiershadow1Javiershadow1 asked Oct 9, 2021 at 17:13 Javiershadow1Javiershadow1 744 bronze badges $\endgroup$ 5 $\begingroup$ Lets see. A differential equation can be made with:$$f(x)=f\left(\frac 1x\right)-\frac {f\left(\frac 1x\right)}x $$ Also let $y=\frac1x$:$$ f\left(\frac1y\right)=\frac{f(y)}y$$ $\endgroup$ Тyma Gaidash – Тyma Gaidash 2021-10-09 17:17:39 +00:00 Commented Oct 9, 2021 at 17:17 4 $\begingroup$ The title doesn't really match the question. $\endgroup$ Greg Martin – Greg Martin 2021-10-09 17:37:47 +00:00 Commented Oct 9, 2021 at 17:37 1 $\begingroup$ How far have you come yourself? Any thoughts of how to solve it? $\endgroup$ md2perpe – md2perpe 2021-10-09 18:17:06 +00:00 Commented Oct 9, 2021 at 18:17 1 $\begingroup$ @md2perpe The bullets are the OP's attempt. They are consequences of setting $ y = 0 $ and $ y = - 1 $ in the original functional equation. $\endgroup$ Mohsen Shahriari – Mohsen Shahriari 2021-10-09 18:26:23 +00:00 Commented Oct 9, 2021 at 18:26 $\begingroup$ Sorry, I'm a new member and my first attempt was to put the original functional equation in the title but I couldn't do that. As @Mohsen Shahriari says, the bullets are deduced from the original functional equation. $\endgroup$ Javiershadow1 – Javiershadow1 2021-10-09 19:14:30 +00:00 Commented Oct 9, 2021 at 19:14 Add a comment \| 1 Answer 1 Reset to default 2 $\begingroup$ New attempt: I don't know if this problem should be solved by calculus as per @TymaGaidash's comment suggests, but I attempted solve it with more elementary logic, i.e. to show that $x + 1$ is the only expression suitable: We know that the function is monotonically increasing. This means that $f(x + a) = f(x) + b \tag{1}$ and $f( \frac{1}{x + a}) = f(1/x) - c$, where $a, b, c \in \mathbb R^+ \cup {0 }$, except that $a \ne 0$ as with that we want to change the value of $x$ and see the function values. Using $f(x)=x\cdot f(1/x) \tag{2}$ we get $f(x + a) = (x + a) \cdot f( \frac{1}{x + a})$ $f(x) + b = (x + a) \biggl(f(1/x) - c \biggr)$ $f(x) + b = x \cdot f(1/x) + a \cdot f(1/x) - xc - ac$ $b = a \cdot f(1/x) - xc - ac$ $f(1/x) = \frac {b + xc}{a} + c \tag{3}$ Using $(3)$ and $f(-1)=0$ we get $f(-1) = \frac {b - c}{a} + c = 0$ $b - c + ac = 0$ $b = c - ac \tag{4}$ Using $(3)$ and $(4)$ we get $f(1/x) = \frac {c - ac + xc}{a} + c = \frac {c + xc}{a} = \frac {1 + x}{a} c \tag{5}$ Using $(2)$ and $(5)$ we get $f(x) = x \cdot f(1/x) = x \frac {1 + x}{a} c \tag{6}$ We have $f(x + a) = f(x) + b$ from $(1)$, and using $(6)$ for $f(x + a)$ and $f(x)$, and $(4)$ for $b$ we get $(x + a) \frac{1 + x + a}{a} c = x \frac{1 + x}{a} c + c - ac$ $(x + a)(1 + x + a)c = x(1 + x)c + ac - a^2c$ $x(1 + x)c + a(1 + x)c + (x + a)ac = x(1 + x)c + ac - a^2c$ $ac + axc + axc + a^2c = ac - a^2c$ $2axc = -2a^2c \tag{7}$ before division we need to consider, that by definition $a$ can not be zero, however $c$ might be, so let $c = 0$ for $(6)$ and using $f(0)=1$ we get $f(0) = 0 \frac {1 + 0}{a} 0 = 1$ and this is impossible so $c \ne 0$ so we divide $(7)$ by $2ac$ to get $x = -a \tag{8}$ Substituting $(8)$ into $(6)$ we get $f(x) = x \frac {1 + x}{-x} c = -(1 + x)c \tag{9}$ and using $f(0)=1$ with $(9)$ we get $f(0) = -(1 + 0)c = 1$ $c = -1$ and substituting it into $(9)$ we get $f(x) = -(1 + x)(-1) = 1 + x$ Original attempt: From $f(1)=0, f(0)=1$ it is suspicious to check $f(x) = x + 1$: For $f(x)>x,\forall x \in \mathbb R$: $x + 1 > x$ true. For $-f(-f(x))=x, \forall x \in \mathbb R$: $-f(-f(x)) = -f(-(x + 1)) = -f(-x - 1) = - (-x) = x$ true. For $f(x)=x\cdot f(1/x)$: $x + 1 = x\cdot \frac{x + 1}{x}$ $x + 1 = x + 1$ true. For $f\biggl(x\cdot f(\frac yx)\biggr)=x\cdot f\biggl(\frac{f(y)}{x}\biggr),\forall x \in \mathbb R\setminus{0}, \forall y \in \mathbb R$: $f\biggl(x\cdot \frac {y + x}{x}\biggr)=x\cdot f\biggl(\frac{y + 1}{x}\biggr)$ $f(y + x) = x\cdot \biggl(\frac{y + 1 + x}{x}\biggr)$ $y + x + 1 = y + 1 + x$ true. To show that $f$ can not be any other polinomial expression, one can reason like @MohsenShahriari did in the below comment. My version is to consider $f(x) = P(x)$ and the condition $f(1)=0$: the latter means that $-1$ is a root of $P(x) \Rightarrow P(x) = (x + 1) \cdot Q(x)$. Using $f(x)=x\cdot f(1/x)$ we get $(x + 1) \cdot Q(x) = x \cdot \biggl( \frac{1}{x} + 1 \biggr) \cdot Q \biggl( \frac{1}{x} \biggr)$ $(x + 1) \cdot Q(x) = (1 + x) \cdot Q \biggl( \frac{1}{x} \biggr)$ At this point we can exclude the values $\pm 1$ of $x$, because we want to confirm $\forall x$, so we can divide by $(x + 1)$: $Q(x) = Q \biggl( \frac{1}{x} \biggr)$ If $Q(x)$ is of degree $> 0$ then $(x - x_0)(x - x_1)...(x - x_n) = \biggl( \frac{1}{x} - x_0 \biggr) \biggl( \frac{1}{x} - x_1 \biggr) ... \biggl( \frac{1}{x} - x_n \biggr)$ Again, as we want to confirm $\forall x$, we can use $\lim_{x \rightarrow \infty}(x - C) = \infty$ and $\lim_{x \rightarrow \infty}(\frac{1}{x} - C) = -C$ where C is a real constant and with the Extended Sum Rule we get $\lim_{x \rightarrow \infty} \biggl((x - x_0)(x - x_1)...(x - x_n) \biggr) = \infty \ne \lim_{x \rightarrow \infty} \biggl( \biggl( \frac{1}{x} - x_0 \biggr) \biggl( \frac{1}{x} - x_1 \biggr) ... \biggl( \frac{1}{x} - x_n \biggr) \biggr) = (-x_0)(-x_1)...(-x_n)$ therefore $Q(x)$ is of degree $0$. Using the condition $f(0)=1$ we have $(0 + 1) \cdot Q(0) = 1 \Rightarrow Q(0) = 1, \forall x$, therefore the only polinomial expression suitable is $x + 1$. I didn't take the time to show if there is any solution as a non-polinomial expression. Share edited Oct 20, 2021 at 19:19 answered Oct 9, 2021 at 18:23 Dávid LaczkóDávid Laczkó 88566 silver badges1717 bronze badges $\endgroup$ 13 $\begingroup$ Until now that's the only solution I've found out. $\endgroup$ Javiershadow1 – Javiershadow1 2021-10-09 19:33:32 +00:00 Commented Oct 9, 2021 at 19:33 $\begingroup$ See my addition for $f$ as polinomial expression. $\endgroup$ Dávid Laczkó – Dávid Laczkó 2021-10-09 20:39:12 +00:00 Commented Oct 9, 2021 at 20:39 2 $\begingroup$ For showing that $ f ( x ) = x + 1 $ is the only polynomial solution, note that $ - f \bigl ( - f ( x ) \bigr ) = x $ implies that $ f $ cannot have a degree more than $ 1 $, i.e. it must be linear. $ f ( - 1 ) = 0 $ and $ f ( 0 ) = 1 $ then imply $ f ( x ) = x + 1 $ for all $ x \in \mathbb R $. $\endgroup$ Mohsen Shahriari – Mohsen Shahriari 2021-10-10 01:11:08 +00:00 Commented Oct 10, 2021 at 1:11 $\begingroup$ @MohsenShahriari Ah, yes, I wanted to simplify it so much that I oversimplified it. I took a different approach then yours which is more complicated at the end. $\endgroup$ Dávid Laczkó – Dávid Laczkó 2021-10-10 19:53:57 +00:00 Commented Oct 10, 2021 at 19:53 $\begingroup$ @Javiershadow1 I think I managed to prove that that is the only solution. $\endgroup$ Dávid Laczkó – Dávid Laczkó 2021-10-20 19:20:31 +00:00 Commented Oct 20, 2021 at 19:20 \| Show 8 more comments You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functional-equations See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related Can we give a definition of the cotangent based on a functional equation? 1 A logarithm-like functional equation Functional equation involving gamma function Functional Equation with Inverse Find $f'(0)$ if $f(x)+f(2x)=x\space\space\forall x$ 1 Is there any functional equation $f(ab+cd)= f(a)+f(b)+f(c)+f(d)$? 4 Which functions share a certain property of sinusoids? 4 Integral functional equation Hot Network Questions How to sample curves more densely (by arc-length) when their trajectory is more volatile, and less so when the trajectory is more constant I have a lot of PTO to take, which will make the deadline impossible Is it ok to place components "inside" the PCB Calculating the node voltage How to understand the reasoning behind modern Fatalism? Two calendar months on the same page Why do universities push for high impact journal publications? Do sum of natural numbers and sum of their squares represent uniquely the summands? Help understanding moment of inertia Can a state ever, under any circumstance, execute an ICC arrest warrant in international waters? How long would it take for me to get all the items in Bongo Cat? How to use cursed items without upsetting the player? How do trees drop their leaves? Why is the fiber product in the definition of a Segal spaces a homotopy fiber product? What is the name of the 1950’s film about the new Scots lord whose relative is a frog like creature living in the ancestral home? Can a Box have a lifetime less than 'static? How to home-make rubber feet stoppers for table legs? Bypassing C64's PETSCII to screen code mapping Copy command with cs names Passengers on a flight vote on the destination, "It's democracy!" How do you emphasize the verb "to be" with do/does? Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? Why is the definite article used in “Mi deporte favorito es el fútbol”? On the Subject of Switches more hot questions Question feed
13314	https://www.statlect.com/fundamentals-of-probability/conditional-probability	StatLect Index > Fundamentals of probability Conditional probability by Marco Taboga, PhD A conditional probability is a probability assigned to an event after receiving some information about other relevant events. Table of contents The mathematical setting A gradual approach The case of equally likely sample points Counting favorable cases First proof of conditional probability formula Comments about the formula A more general approach Mathematical properties Second proof of conditional probability formula Tackling division by zero The law of total probability Solved exercises Exercise 1 Exercise 2 Exercise 3 Aims The aims of this lecture are to: introduce the concept of conditional probability in a mathematically sound manner; provide two proofs of the formula used to compute conditional probabilities; discuss some of the subtleties involved in the formula, including potential division-by-zero problems. The mathematical setting Let be a sample space and let denote the probability assigned to the events . Suppose that, after assigning probabilities to the events in , we receive new information about the things that will happen (the possible outcomes). In particular, suppose that we are told that the realized outcome will belong to a set . How should we revise the probabilities assigned to the events in , to properly take the new information into account? The answer to this question is provided by the conditional probability , the revised probability assigned to an event after learning that the realized outcome will be an element of . A gradual approach Despite being an intuitive concept, conditional probability is quite difficult to define in a mathematically rigorous way. We take a gradual approach in this lecture: we first discuss conditional probability for the very special case in which all the sample points are equally likely; we then give a more general definition; finally, we refer the reader to other lectures where conditional probability is defined in even more abstract ways. The case of equally likely sample points Suppose that a sample space has a finite number of sample points: Suppose also that each sample point is assigned the same probability: Counting favorable cases In such a simple space, the probability of a generic event is obtained aswhere denotes the cardinality of a set, that is, the number of its elements. In other words, the probability of an event is obtained in two steps: counting the number of "cases that are favorable to the event ", that is, the number of elements belonging to ; dividing the number thus obtained by the number of "all possible cases", that is, the number of elements belonging to . For example, if , then First proof of conditional probability formula When we learn that the realized outcome will belong to a set , we still apply the rule However, the number of all possible cases is now equal to the number of elements of because only the outcomes belonging to are still possible. Furthermore, the number of favorable cases is now equal to the number of elements of because the outcomes in are no longer possible. As a consequence, By dividing numerator and denominator by , we obtain Therefore, when all sample points are equally likely, conditional probabilities are computed as Example Let us make an example. Suppose that we toss a die. Six numbers (from to can appear face up, but we do not yet know which one of them will appear. The sample space is Each of the six numbers is a sample point and is assigned probability . Define the event as follows:where the event could be described as "an odd number appears face up". Now define the event as follows:where the event could be described as "a number greater than three appears face up". The probability of isSuppose we are told that the realized outcome will belong to . How do we have to revise our assessment of the probability of the event , according to the rules of conditional probability? First of all, we need to compute the probability of the event : Then, the conditional probability of given is Comments about the formula In the next section, we will show that the conditional probability formulais valid also for more general cases (i.e., when the sample points are not all equally likely). However, this formula already allows us to understand why defining conditional probability is a challenging task. In the conditional probability formula, a division by is performed. This division is impossible when is a zero-probability event (i.e., ). If we want to be able to define also when , then we need to give a more complicated definition of conditional probability. We will return to this point later. A more general approach In this section we give a more general definition of conditional probability, by taking an axiomatic approach. First, we list the mathematical properties that we would like conditional probability to satisfy. Then, we prove that the conditional probability formula introduced above satisfies these properties. Mathematical properties The conditional probability is required to satisfy the following properties: Probability measure. has to satisfy all the properties of a probability measure. Sure thing. . Impossible events. If (, the complement of with respect to , is the set of all elements of that do not belong to ), then . Constant likelihood ratios on . If , and , then These properties are very intuitive: Property 1 requires that also conditional probability measures satisfy the fundamental properties that any other probability measure needs to satisfy. Property 2 says that the probability of a sure thing must be : since we know that only things belonging to the set can happen, then the probability of must be . Property 3 says that the probability of an impossible thing must be : since we know that things not belonging to the set will not happen, then the probability of the events that are disjoint from must be . Property 4 is a bit more complex: it says that if is - say - two times more likely than before receiving the information , then remains two times more likely than , also after receiving the information because all the things in and remain possible (can still happen) and, hence, there is no reason to expect that the ratio of their likelihoods changes. Second proof of conditional probability formula When the above properties are satisfied, then the conditional probability formula can be used. Proposition Whenever , satisfies the four above properties if and only if Proof We first show thatsatisfies the four properties whenever . As far as property 1) is concerned, we have to check that the three requirements for a probability measure are satisfied. The first requirement for a probability measure is that . Since , by the monotonicity of probability we have thatHence,Furthermore, since and , also The second requirement for a probability measure is that . This is satisfied becauseThe third requirement for a probability measure is that for any sequence of disjoint sets the following holds:But,so that also the third requirement is satisfied. Property 2) is trivially satisfied:Property 3) is verified because, if , thenProperty 4) is verified because, if , and , thenSo, the "if" part has been proved. Now we prove the "only if" part. We prove it by contradiction. Suppose there exist another conditional probability that satisfies the four properties. Then, there exists an event , such thatIt can not be that , otherwise we would havewhich would be a contradiction, since if was a conditional probability it would satisfy:If is not a subset of then implies also becauseandbut this would also lead to a contradiction because . Tackling division by zero In the previous section we have generalized the concept of conditional probability. However, we have not been able to define the conditional probability for the case in which . This case is discussed in the lectures: Conditional probability with respect to a sigma-algebra; Conditional probability distributions. The law of total probability We end this lecture by stating an important formula that allows us to write the probability of an event as a weighted sum of conditional probabilities. Let , ..., be events having the following characteristics: they are mutually disjoint: whenever ; they cover all the sample space: they have strictly positive probability: for any . The events , ..., are called a partition of . The law of total probability states that, for any event , the following holds:which can, of course, also be written as Proof The law of total probability is proved as follows: Solved exercises Some solved exercises on conditional probability can be found below. Exercise 1 Consider a sample space comprising three possible outcomes , , : Suppose the three possible outcomes are assigned the following probabilities: Define the events and denote by the complement of . Compute , the conditional probability of given . Solution We need to use the conditional probability formula The numerator isand the denominator is As a consequence, Exercise 2 Consider a sample space comprising four possible outcomes , , , : Suppose the four possible outcomes are assigned the following probabilities: Define two events Compute , the conditional probability of given . Solution We need to use the formula Butwhile, by using additivity, we obtain Therefore, Exercise 3 The Census Bureau has estimated the following survival probabilities for men: probability that a man lives at least 70 years: 80%; probability that a man lives at least 80 years: 50%. What is the conditional probability that a man lives at least 80 years given that he has just celebrated his 70th birthday? Solution Given an hypothetical sample space , define the two events We need to find the following conditional probability: The denominator is known: As far as the numerator is concerned, note that (if you live at least 80 years then you also live at least 70 years). But implies Therefore, How to cite Please cite as: Taboga, Marco (2021). "Conditional probability", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. The books Most of the learning materials found on this website are now available in a traditional textbook format. Probability and statisticsMatrix algebra Featured pages Characteristic function Convergence in distribution Bayes rule Uniform distribution Multivariate normal distribution Set estimation Student t distribution Gamma function Poisson distribution Main sections Mathematical tools Fundamentals of probability Probability distributions Asymptotic theory Fundamentals of statistics Glossary About Statlect Contacts Cookies, privacy and terms of use Glossary entries Estimator Continuous random variable Power function Probability mass function Probability space Loss function To enhance your privacy, we removed the social buttons, but don't forget to share.
13315	https://mmerevise.co.uk/a-level-maths-revision/differentiating-trig-functions/	Differentiating Trig Functions \| Revision \| MME Revise##### A Level MathsBiologyChemistryPhysicsFurther MathsA Level Predicted Papers 2026 ##### GCSE MathsEnglish LanguageEnglish LiteratureCombined ScienceBiologyChemistryPhysicsGCSE CoursesGCSE Predicted Papers 2025/26 ##### Key Stage KS3 MathsKS3 EnglishKS3 ScienceAll KS2All KS1 ##### 11 Plus 11 Plus11 Plus Exam Papers11 Plus Practice Papers11 Plus Mock Exams11 Plus Schools ##### Functional Skills Functional Skills Maths Level 2Functional Skills Maths Level 1Functional Skills English Level 2Functional Skills Maths PapersFunctional Skills Maths Cards Resit your GCSEs with confidence Expert support to help you pass first time Find out more ##### Past Papers GCSE Maths Past PapersGCSE Science Past PapersGCSE Biology Past PapersGCSE Chemistry Past PapersGCSE Physics Past PapersGCSE Combined Science Past PapersGCSE English Language Past PapersGCSE English Literature Past PapersA Level Maths Past PapersA Level Physics Past PapersA Level Biology Past PapersA Level Chemistry Past PapersView All Resit your GCSEs with confidence Expert support to help you pass first time Find out more Shop##### A Level MathsChemistryPhysicsBiologyFurther MathsView All A Level ##### GCSE MathsScienceEnglishView All GCSE ##### Other Functional SkillsKS3 MathsKS3 EnglishKS2KS1 ##### Courses & Exams Functional SkillsView All Courses & Exams ##### By Type Revision CardsPractice Exam PapersWorksheetsGuides & WorkbooksBundles & Study KitsCourses & Exams ##### Special Offers BundlesSale Not sure what you're looking for? View all products For Teachers GCSE Exams & Courses##### GCSE GCSE MathsGCSE EnglishGCSE ScienceGCSE BiologyGCSE ResitsGCSE Courses ##### Functional Skills Maths Level 2English Level 2 Predicted Papers##### Predicted Papers GCSE Predicted Papers 2025/26A Level Predicted Papers 2026Free Predicted Papers Tuition##### Tuition Book a Tutor11 Plus TutorFor Tutors Sign UpLog InLearn MoreFor Schools Sign UpLog InLearn MoreFor Schools Shop Account Login Best viewed on desktop, tablet or in landscape mode. Revise Topic Home Differentiating Trig Functions Revision Try it yourself Example Questions Worksheets 1 Supercharge your learning Learn More Differentiating Trig Functions A Level AQA Edexcel OCR Differentiating Trig Functions We’re now going to look at how to differentiate the simple trig functions – that’s \textcolor{blue}{\sin}, \textcolor{limegreen}{\cos} and \textcolor{red}{\tan}. Make sure you are happy with the following topics before continuing. Inverse Trig Functions Reciprocal Trig Functions A Level AQA Edexcel OCR Product A Level Maths Predicted Papers 2026 118 From:~~£24.99 ~~£19.99 The MME A level maths predicted papers are an excellent way to practise, using authentic exam style questions that are unique to our papers. Our examiners have studied A level maths past papers to develop predicted A level maths exam questions in an authentic exam format. The profit from every pack is reinvested into making free content on MME, which benefits millions of learners across the country. View Product Product A-A A Level Maths Practice Papers 5 From:~~£26.99 ~~£21.59 The MME A-A A level maths practice papers are excellent for those top achieving students to practise for their exams, using authentic exam style questions that are unique to our practice papers. Our content experts have studied A level maths past papers and specifications to develop specific A-A A level maths exam questions in an authentic exam style. The profit from every pack is reinvested into making free content on MME, which benefits millions of learners across the country. View Product Simple Trig Results Below is a diagram showing the derivative and integral of the basic trigonometric functions, \sin x and \cos x. A Level AQA Edexcel OCR General Trig Differentiation Below is a table of three derivative results. A Level AQA Edexcel OCR Using the Chain Rule with Trig Functions We can also use the Chain Rule to differentiate more complex trig functions. For example, say we have f(x) = \textcolor{blue}{\sin} (x^2 - x + 1). Then we can set \textcolor{red}{u} = x^2 - x + 1 and f(x) = g(\textcolor{red}{u}) = \textcolor{blue}{\sin} \textcolor{red}{u}. Then \dfrac{d\textcolor{red}{u}}{dx} = 2x - 1 and \dfrac{df(x)}{d\textcolor{red}{u}} = \textcolor{limegreen}{\cos} u. So, \dfrac{df(x)}{dx} = (2x - 1)\textcolor{limegreen}{\cos} (x^2 - x + 1) A Level AQA Edexcel OCR From First Principles We can also find the derivatives from first principles. For example, let f(x) = \textcolor{limegreen}{\cos} x. Then f'(x) = \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{\textcolor{limegreen}{\cos} (x + \textcolor{purple}{h}) - \textcolor{limegreen}{\cos} x}{\textcolor{purple}{h}} \right) = \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{\textcolor{limegreen}{\cos} x \textcolor{limegreen}{\cos} \textcolor{purple}{h} - \textcolor{blue}{\sin} x \textcolor{blue}{\sin} h - \textcolor{limegreen}{\cos} x}{\textcolor{purple}{h}}\right) = \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{\textcolor{limegreen}{\cos} x(\textcolor{limegreen}{\cos} \textcolor{purple}{h} - 1) - \textcolor{blue}{\sin} x \textcolor{blue}{\sin} \textcolor{purple}{h}}{\textcolor{purple}{h}}\right) Using small angle approximations, we have f'(x) = \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{-\dfrac{1}{2}\textcolor{purple}{h}^2 \textcolor{limegreen}{\cos} x - \textcolor{purple}{h}\textcolor{blue}{\sin} x}{\textcolor{purple}{h}}\right) = \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{-1}{2}\textcolor{purple}{h}\textcolor{limegreen}{\cos} x - \textcolor{blue}{\sin} x\right) = -\textcolor{blue}{\sin} x A Level AQA Edexcel OCR Differentiating Trig Functions Example Questions Question 1: Give an expression for \dfrac{dy}{dx} in terms of y, when x = \tan y. [2 marks] A Level AQA Edexcel OCR x = \tan y gives \dfrac{dx}{dy} = \sec ^2 y Then \dfrac{dy}{dx} = \cos ^2 y. Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Question 2: For \tan x^2, find the derivative with respect to x. [2 marks] A Level AQA Edexcel OCR Let u = x^2 and y = \tan u. Then \dfrac{dy}{dx} = 2x \times \sec ^2 u = \dfrac{2x}{\cos ^2 x^2} Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Question 3: Prove that the derivative of \sin kx is k\cos kx, using the first principles technique. [4 marks] A Level AQA Edexcel OCR f'(x) = \lim\limits_{h \to 0}\left( \dfrac{\sin (kx + kh) - \sin kx}{h} \right) = \lim\limits_{h \to 0}\left( \dfrac{\sin kx \cos kh + \sin kh \cos kx - \sin kx}{h}\right) = \lim\limits_{h \to 0}\left( \dfrac{\sin kx(\cos kh - 1) + \sin kh \cos kx}{h}\right) Using small angle approximations, we have f'(x) = \lim\limits_{h \to 0}\left( \dfrac{-\dfrac{1}{2}(kh)^2 \sin kx + kh\cos kx}{h}\right) = \lim\limits_{h \to 0}\left( \dfrac{-1}{2}kh\sin kx + k\cos kx\right) = k\cos kx Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Additional Resources Exam Tips Cheat Sheet A Level Open Formula Booklet A Level Open Specification Points Covered G1 – Understand and use the derivative of f(x) as the gradient of the tangent to the graph of y=f(x) at a general point (x,y); the gradient of the tangent as a limit; interpretation as a rate of change; sketching the gradient function for a given curve; second derivatives; differentiation from first principles for small positive integer powers of x and for \sin(x) and \cos(x) G2 – Differentiate e^{kx} and a^{kx}, \sin(kx), \cos(kx), \tan(kx) and related sums, differences and constant multiples Differentiating Trig Functions Worksheet and Example Questions Differentiation A Level Exam QuestionsMark Scheme Related Topics Inverse Trig Functions A Level Revise Reciprocal Trig Functions A Level Revise Secure Shopping Ways to Pay Spread the Cost Socials Follow our socials for revision tips and community support Contact Details 020 3633 5145 revise@mmerevise.co.uk Mon - Fri: 08:45 - 19:00, Sat 09:30-16:00 Evans Business Centre, Hartwith Way, Harrogate HG3 2XA Information About Us Contact Us MME Blog Reviews Safeguarding Modern Slavery Statement Careers For Teachers Become a Tutor MME Affiliates Help Articles Pass Browse Subjects A Level Maths Revision A Level Biology Revision A Level Chemistry Revision A Level Physics Revision A Level English Language Revision GCSE Maths Revision GCSE English Language Revision GCSE Science Revision KS3 Revision KS2 Revision KS1 Revision 11 Plus Numerical Reasoning Popular Products GCSE Predicted Papers A Level Predicted Papers GCSE Maths Books Functional Skills Level 2 Course in a Box – GCSE Maths (Guaranteed Pass) GCSE Maths Revision Guide Privacy & Policy / Terms / Site Map Providing Ofqual Approved Qualifications © 2025 MME When you visit or interact with our sites, services or tools, we or our authorised service providers may use cookies for storing information to help provide you with a better, faster and safer experience and for marketing purposes. By clicking continue and using our website you are consenting to our use of cookies in accordance with our Cookie Policy Continue Report a Question Question: Your Report Submit You must be logged in to vote for this question. Username or Email Address Password Don't have an account?Signup Log In Forgot Password? Email address Password Confirm Password Your personal data will be used to support your experience throughout this website, to manage access to your account, and for other purposes described in our privacy policy. Already have an account?Login Get Started
13316	https://www.geeksforgeeks.org/python/python-program-to-get-all-unique-combinations-of-two-lists/	Python program to get all unique combinations of two Lists Last Updated : 15 Jul, 2025 Suggest changes 4 Likes The goal is to combine each item from first list with each item from second list in every possible unique way. If we want to get all possible combinations from two lists. Python’s itertools library has a function called a product that makes it easy to generate combinations of items from multiple lists. Python ```` import itertools a = ["a", "b"] b = [1, 2] It generate all possible combinations of elements from both lists Convert the result into a list of tuples combinations = list(itertools.product(a, b)) print(combinations) ```` import itertools import itertools a = ["a", "b"] a = "a" "b" b = [1, 2] b = 1 2 ``` It generate all possible combinations of elements from both lists It generate all possible combinations of elements from both lists ``` ``` Convert the result into a list of tuples Convert the result into a list of tuples ``` combinations = list(itertools.product(a, b)) combinations = list itertools product a b print(combinations) print combinations Output ``` [('a', 1), ('a', 2), ('b', 1), ('b', 2)] ``` itertools.product automatically creates the combinations of elements from both lists. We use list() to convert the result into a list of tuples. There are other methods which can be used that are both fast and efficient : Table of Content Using itertools.permutations Using a Nested Loop Using List Comprehension Using itertools.permutations itertools.permutations function creates all possible permutations of a given list. We can combine this with zip to create unique combinations of two lists. Python ```` import itertools a = ["a", "b"] b = [1, 2] Generate all 2-length permuations of pairs by zipping lists combinations = list(itertools.permutations(zip(a, b), 2)) print(combinations) ```` import itertools import itertools a = ["a", "b"] a = "a" "b" b = [1, 2] b = 1 2 ``` Generate all 2-length permuations of pairs by zipping lists Generate all 2-length permuations of pairs by zipping lists ``` combinations = list(itertools.permutations(zip(a, b), 2)) combinations = list itertools permutations zip a b 2 print(combinations) print combinations Output ``` [(('a', 1), ('b', 2)), (('b', 2), ('a', 1))] ``` The zip(a, b) combines the two lists element-wise into pairs, and itertools.permutations then creates all possible orders of these pairs. This method returns all unique permutations, including reversed combinations. Using a Nested Loop We can get all combinations is by using two loops. We can loop through each element in both lists and combine them. Python ```` X = ["a", "b"] Y = [1, 2] Initialize an empty list combinations = [] Loop through each element in List_1 for x in X: # Loop through each element in List_2 for y in Y: combinations.append((x, y)) print(combinations) ```` X = ["a", "b"] X = "a" "b" Y = [1, 2] Y = 1 2 ``` Initialize an empty list Initialize an empty list ``` combinations = [] combinations = ``` Loop through each element in List_1 Loop through each element in List_1 ``` for x in X: for x in X ``` # Loop through each element in List_2 Loop through each element in List_2 ``` for y in Y: for y in Y combinations.append((x, y)) combinations append x y print(combinations) print combinations Output ``` [('a', 1), ('a', 2), ('b', 1), ('b', 2)] ``` Using List Comprehension If we have to write cleaner and shorter code, we can use list comprehension to achieve the same result. This is essentially the same thing as using loops but in a single line. It’s a concise way to combine elements. Python ```` X = ["a", "b"] Y = [1, 2] list combination of all possible elemenst of two lists(X and Y) combinations = [(x, y) for x in X for y in Y] print(combinations) ```` X = ["a", "b"] X = "a" "b" Y = [1, 2] Y = 1 2 ``` list combination of all possible elemenst of two lists(X and Y) list combination of all possible elemenst of two lists(X and Y) ``` combinations = [(x, y) for x in X for y in Y] combinations = x y for x in X for y in Y print(combinations) print combinations Output ``` [('a', 1), ('a', 2), ('b', 1), ('b', 2)] ``` V vanshgaur14866 Improve Article Tags : Python Python Programs Python list-programs Python-itertools Practice Tags : python Similar Reads Python Fundamentals Python Introduction Python was created 1991 with focus on code readability and express concepts in fewer lines of code.Simple and readable syntax makes it beginner-friendly.Runs seamlessly on Windows, macOS and Linux.Includes libraries for tasks like web development, data analysis and machine learning.Variable types ar 3 min readInput and Output in Python Understanding input and output operations is fundamental to Python programming. With the print() function, we can display output in various formats, while the input() function enables interaction with users by gathering input during program execution. Taking input in PythonPython's input() function 7 min readPython Variables In Python, variables are used to store data that can be referenced and manipulated during program execution. A variable is essentially a name that is assigned to a value. Unlike many other programming languages, Python variables do not require explicit declaration of type. The type of the variable i 6 min readPython Operators In Python programming, Operators in general are used to perform operations on values and variables. These are standard symbols used for logical and arithmetic operations. In this article, we will look into different types of Python operators. OPERATORS: These are the special symbols. Eg- + , , /, 6 min readPython Keywords Keywords in Python are reserved words that have special meanings and serve specific purposes in the language syntax. Python keywords cannot be used as the names of variables, functions, and classes or any other identifier. Getting List of all Python keywordsWe can also get all the keyword names usin 2 min readPython Data Types Python Data types are the classification or categorization of data items. It represents the kind of value that tells what operations can be performed on a particular data. Since everything is an object in Python programming, Python data types are classes and variables are instances (objects) of thes 9 min readConditional Statements in Python Conditional statements in Python are used to execute certain blocks of code based on specific conditions. These statements help control the flow of a program, making it behave differently in different situations.If Conditional Statement in PythonIf statement is the simplest form of a conditional sta 6 min readLoops in Python - For, While and Nested Loops Loops in Python are used to repeat actions efficiently. The main types are For loops (counting through items) and While loops (based on conditions). In this article, we will look at Python loops and understand their working with the help of examples. For Loop in PythonFor loops is used to iterate ov 9 min readPython Functions Python Functions is a block of statements that does a specific task. The idea is to put some commonly or repeatedly done task together and make a function so that instead of writing the same code again and again for different inputs, we can do the function calls to reuse code contained in it over an 9 min readRecursion in Python Recursion is a programming technique where a function calls itself either directly or indirectly to solve a problem by breaking it into smaller, simpler subproblems.In Python, recursion is especially useful for problems that can be divided into identical smaller tasks, such as mathematical calculati 6 min readPython Lambda Functions Python Lambda Functions are anonymous functions means that the function is without a name. As we already know the def keyword is used to define a normal function in Python. Similarly, the lambda keyword is used to define an anonymous function in Python. In the example, we defined a lambda function(u 6 min read Python Data Structures [Python String A string is a sequence of characters. Python treats anything inside quotes as a string. This includes letters, numbers, and symbols. Python has no character data type so single character is a string of length 1.Pythons = "GfG" print(s) # access 2nd char s1 = s + s # update print(s1) # printOut 6 min read]( Lists In Python, a list is a built-in dynamic sized array (automatically grows and shrinks). We can store all types of items (including another list) in a list. Can contain duplicate items.Mutable : We can modify, replace or delete the items. Ordered : Maintains the order of elements based on how they are 6 min readPython Tuples A tuple in Python is an immutable ordered collection of elements. Tuples are similar to lists, but unlike lists, they cannot be changed after their creation (i.e., they are immutable). Tuples can hold elements of different data types. The main characteristics of tuples are being ordered , heterogene 6 min readDictionaries in Python Python dictionary is a data structure that stores the value in key: value pairs. Values in a dictionary can be of any data type and can be duplicated, whereas keys can't be repeated and must be immutable. Example: Here, The data is stored in key:value pairs in dictionaries, which makes it easier to 7 min readPython Sets Python set is an unordered collection of multiple items having different datatypes. In Python, sets are mutable, unindexed and do not contain duplicates. The order of elements in a set is not preserved and can change.Creating a Set in PythonIn Python, the most basic and efficient method for creating 10 min readPython Arrays Lists in Python are the most flexible and commonly used data structure for sequential storage. They are similar to arrays in other languages but with several key differences:Dynamic Typing: Python lists can hold elements of different types in the same list. We can have an integer, a string and even 9 min read[List Comprehension in Python List comprehension is a concise and powerful way to create new lists by applying an expression to each item in an existing iterable (like a list, tuple or range). It helps you write clean, readable and efficient code compared to traditional loops.Syntax[expression for item in iterable if condition]P 4 min read]( Advanced Python Python OOP Concepts Object Oriented Programming is a fundamental concept in Python, empowering developers to build modular, maintainable, and scalable applications. OOPs is a way of organizing code that uses objects and classes to represent real-world entities and their behavior. In OOPs, object has attributes thing th 11 min readPython Exception Handling Python Exception Handling handles errors that occur during the execution of a program. Exception handling allows to respond to the error, instead of crashing the running program. It enables you to catch and manage errors, making your code more robust and user-friendly. Let's look at an example:Handl 6 min readFile Handling in Python File handling refers to the process of performing operations on a file, such as creating, opening, reading, writing and closing it through a programming interface. It involves managing the data flow between the program and the file system on the storage device, ensuring that data is handled safely a 4 min readPython Database Tutorial Python being a high-level language provides support for various databases. We can connect and run queries for a particular database using Python and without writing raw queries in the terminal or shell of that particular database, we just need to have that database installed in our system.A database 4 min readPython MongoDB Tutorial MongoDB is a popular NoSQL database designed to store and manage data flexibly and at scale. Unlike traditional relational databases that use tables and rows, MongoDB stores data as JSON-like documents using a format called BSON (Binary JSON). This document-oriented model makes it easy to handle com 2 min readPython MySQL MySQL is a widely used open-source relational database for managing structured data. Integrating it with Python enables efficient data storage, retrieval and manipulation within applications. To work with MySQL in Python, we use MySQL Connector, a driver that enables seamless integration between the 9 min readPython Packages Python packages are a way to organize and structure code by grouping related modules into directories. A package is essentially a folder that contains an __init__.py file and one or more Python files (modules). This organization helps manage and reuse code effectively, especially in larger projects. 12 min readPython Modules Python Module is a file that contains built-in functions, classes,its and variables. There are many Python modules, each with its specific work.In this article, we will cover all about Python modules, such as How to create our own simple module, Import Python modules, From statements in Python, we c 7 min readPython DSA Libraries Data Structures and Algorithms (DSA) serve as the backbone for efficient problem-solving and software development. Python, known for its simplicity and versatility, offers a plethora of libraries and packages that facilitate the implementation of various DSA concepts. In this article, we'll delve in 15 min readList of Python GUI Library and Packages Graphical User Interfaces (GUIs) play a pivotal role in enhancing user interaction and experience. Python, known for its simplicity and versatility, has evolved into a prominent choice for building GUI applications. With the advent of Python 3, developers have been equipped with lots of tools and li 11 min read Data Science with Python NumPy Tutorial - Python Library NumPy is a core Python library for numerical computing, built for handling large arrays and matrices efficiently.ndarray object €“ Stores homogeneous data in n-dimensional arrays for fast processing.Vectorized operations €“ Perform element-wise calculations without explicit loops.Broadcasting €“ Apply 3 min readPandas Tutorial Pandas (stands for Python Data Analysis) is an open-source software library designed for data manipulation and analysis. Revolves around two primary Data structures: Series (1D) and DataFrame (2D)Built on top of NumPy, efficiently manages large datasets, offering tools for data cleaning, transformat 6 min readMatplotlib Tutorial Matplotlib is an open-source visualization library for the Python programming language, widely used for creating static, animated and interactive plots. It provides an object-oriented API for embedding plots into applications using general-purpose GUI toolkits like Tkinter, Qt, GTK and wxPython. It 5 min readPython Seaborn Tutorial Seaborn is a library mostly used for statistical plotting in Python. It is built on top of Matplotlib and provides beautiful default styles and color palettes to make statistical plots more attractive.In this tutorial, we will learn about Python Seaborn from basics to advance using a huge dataset of 15+ min readStatsModel Library- Tutorial Statsmodels is a useful Python library for doing statistics and hypothesis testing. It provides tools for fitting various statistical models, performing tests and analyzing data. It is especially used for tasks in data science ,economics and other fields where understanding data is important. It is 4 min readLearning Model Building in Scikit-learn Building machine learning models from scratch can be complex and time-consuming. Scikit-learn which is an open-source Python library which helps in making machine learning more accessible. It provides a straightforward, consistent interface for a variety of tasks like classification, regression, clu 8 min readTensorFlow Tutorial TensorFlow is an open-source machine-learning framework developed by Google. It is written in Python, making it accessible and easy to understand. It is designed to build and train machine learning (ML) and deep learning models. It is highly scalable for both research and production.It supports CPUs 2 min readPyTorch Tutorial PyTorch is an open-source deep learning framework designed to simplify the process of building neural networks and machine learning models. With its dynamic computation graph, PyTorch allows developers to modify the network€™s behavior in real-time, making it an excellent choice for both beginners an 7 min read Web Development with Python Flask Tutorial Flask is a lightweight and powerful web framework for Python. It€™s often called a "micro-framework" because it provides the essentials for web development without unnecessary complexity. Unlike Django, which comes with built-in features like authentication and an admin panel, Flask keeps things mini 8 min readDjango Tutorial \| Learn Django Framework Django is a Python framework that simplifies web development by handling complex tasks for you. It follows the "Don't Repeat Yourself" (DRY) principle, promoting reusable components and making development faster. With built-in features like user authentication, database connections, and CRUD operati 10 min readDjango ORM - Inserting, Updating & Deleting Data Django's Object-Relational Mapping (ORM) is one of the key features that simplifies interaction with the database. It allows developers to define their database schema in Python classes and manage data without writing raw SQL queries. The Django ORM bridges the gap between Python objects and databas 4 min readTemplating With Jinja2 in Flask Flask is a lightweight WSGI framework that is built on Python programming. WSGI simply means Web Server Gateway Interface. Flask is widely used as a backend to develop a fully-fledged Website. And to make a sure website, templating is very important. Flask is supported by inbuilt template support na 6 min readDjango Templates Templates are the third and most important part of Django's MVT Structure. A Django template is basically an HTML file that can also include CSS and JavaScript. The Django framework uses these templates to dynamically generate web pages that users interact with. Since Django primarily handles the ba 7 min readPython \| Build a REST API using Flask Prerequisite: Introduction to Rest API REST stands for REpresentational State Transfer and is an architectural style used in modern web development. It defines a set or rules/constraints for a web application to send and receive data. In this article, we will build a REST API in Python using the Fla 3 min readHow to Create a basic API using Django Rest Framework ? Django REST Framework (DRF) is a powerful extension of Django that helps you build APIs quickly and easily. It simplifies exposing your Django models as RESTfulAPIs, which can be consumed by frontend apps, mobile clients or other services.Before creating an API, there are three main steps to underst 4 min read Python Practice Python Quiz These Python quiz questions are designed to help you become more familiar with Python and test your knowledge across various topics. From Python basics to advanced concepts, these topic-specific quizzes offer a comprehensive way to practice and assess your understanding of Python concepts. These Pyt 3 min readPython Coding Practice This collection of Python coding practice problems is designed to help you improve your overall programming skills in Python.The links below lead to different topic pages, each containing coding problems, and this page also includes links to quizzes. You need to log in first to write your code. Your 1 min readPython Interview Questions and Answers Python is the most used language in top companies such as Intel, IBM, NASA, Pixar, Netflix, Facebook, JP Morgan Chase, Spotify and many more because of its simplicity and powerful libraries. To crack their Online Assessment and Interview Rounds as a Python developer, we need to master important Pyth 15+ min read We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy Improvement Suggest Changes Help us improve. Share your suggestions to enhance the article. Contribute your expertise and make a difference in the GeeksforGeeks portal. Create Improvement Enhance the article with your expertise. Contribute to the GeeksforGeeks community and help create better learning resources for all. Suggest Changes min 4 words, max Words Limit:1000 Thank You! Your suggestions are valuable to us. Interview Experiences Admission Experiences Career Journeys Work Experiences Campus Experiences Competitive Exam Experiences
13317	https://www.naspghan.org/files/documents/pdfs/position-papers/Guideline_for_the_Evaluation_of_Cholestatic.23.pdf	Copyright © ESPGHAL and NASPGHAN. All rights reserved. Guideline for the Evaluation of Cholestatic Jaundice in Infants: Joint Recommendations of the North American Society for Pediatric Gastroenterology, Hepatology, and Nutrition and the European Society for Pediatric Gastroenterology, Hepatology, and Nutrition Rima Fawaz, yUlrich Baumann, zUdeme Ekong, §Bjo ¨rn Fischler, jjNedim Hadzic, Cara L. Mack, #Vale ´rie A. McLin, Jean P. Molleston, yyEzequiel Neimark, zzVicky L. Ng, and §§Saul J. Karpen ABSTRACT Cholestatic jaundice in infancy affects approximately 1 in every 2500 term infants and is infrequently recognized by primary providers in the setting of physiologic jaundice. Cholestatic jaundice is always pathologic and indicates hepatobiliary dysfunction. Early detection by the primary care physician and timely referrals to the pediatric gastroenterologist/hepatologist are important contributors to optimal treatment and prognosis. The most common causes of cholestatic jaundice in the ﬁrst months of life are biliary atresia (25%–40%) followed by an expanding list of monogenic disorders (25%), along with many unknown or multifactorial (eg, parenteral nutrition-related) causes, each of which may have time-sensitive and distinct treatment plans. Thus, these guidelines can have an essential role for the evaluation of neonatal cholestasis tooptimize care. The recommendationsfromthisclinicalpractice guideline are based upon review and analysis of published literature and the combined experience of the authors. The committee recommends that any infant noted to be jaundiced after 2 weeks of age be evaluated for cholestasis with measure-ment of total and direct serum bilirubin, and that an elevated serum direct bilirubin level (direct bilirubin levels >1.0mg/dL or >17mmol/L) warrants timelyconsideration for evaluationand referraltoa pediatric gastroenterologist or hepatologist. Of note, current differential diagnostic plans now incorporate consideration of modern broad-based next-generation DNA sequencing tech-nologies in the proper clinical context. These recommendations are a general guideline and are not intended as a substitute for clinical judgment or as a protocol for the care of all infants with cholestasis. Broad implementation of these recommendations is expected to reduce the time to the diagnosis of pediatricliverdiseases,includingbiliaryatresia,leadingtoimprovedoutcomes. Key Words: biliary atresia, hepatoportoenterostomy, Kasai, liver biopsy, neonatal cholestasis, neonatal jaundice, radionuclide scan (JPGN 2017;64: 154–168) PREAMBLE C holestatic jaundice in infancy is an uncommon but poten-tially serious problem that indicates hepatobiliary dysfunc-tion. Early detection of cholestatic jaundice by the primary care physician and timely, accurate diagnosis by the pediatric gastro-enterologist are important for successful treatment and an optimal prognosis. The Cholestasis Guideline Committee consisted of 11 members of 2 professional societies: the North American Society for Gastroenterology, Hepatology and Nutrition, and the European Society for Gastroenterology, Hepatology and Nutrition. This committee has responded to a need in pediatrics and developed an updated clinical practice guideline for the diagnostic evaluation of cholestatic jaundice in the infant. There is an obligate focus upon identifying infants with cholestasis due to biliary atresia (BA), but also incorporating the recognition that most forms of cholestasis in this age group are due to non-BA causes. Thus, a structured and broad-based diagnostic approach is required. The recommendations presented in this clinical practice guideline are based on review and analysis of published literature as well as the experience of the authors and colleagues. The quality of evidence supporting the recommendations is based on the Grading of Recommendation, Assessment, Development, and Evaluation workgroup. Each recommendation is assigned a class (reflecting benefit vs risk) and level (assessing strength or certainty). Using these approaches, the recommendations presented herein provide an approach to diagnose infants with cholestasis. These guidelines are intended to be flexible and tailored to the individual patient and local practice and are not meant to determine standards of care for all infants. This guideline has been approved both by the North American Society Received April 20, 2016; accepted July 6, 2016. From the Division of Gastroenterology, Hepatology and Nutrition, Boston Children’s Hospital, Harvard Medical School, Boston, MA, the yDivision Paediatric Gastroenterology and Hepatology, Department of Paediatric Kidney, Liver and Metabolic Diseases, Hannover Medical School, Hann-over, Germany, the zYale New Haven Hospital Transplantation Center, Yale University School of Medicine, New Haven, CT, the §Department of Pediatrics, Karolinska University Hospital, CLINTEC, Karolinska Institute, Stockholm, Sweden, the jjPaediatric Centre for Hepatology, Gastroenterology and Nutrition King’s College Hospital, London, UK, the Section of Pediatric Gastroenterology, Hepatology and Nutrition, Children’s Hospital Colorado, University of Colorado School of Medicine, Aurora, CO, the #Swiss Center for Liver Disease in Children, University Hospitals Geneva, Geneva, Switzerland, the Indiana Uni-versity School of Medicine/Riley Hospital for Children, Indianapolis, IN, the yyDivision of Pediatric Gastroenterology, Hepatology and Nutrition, Hasbro Children’s Hospital, The Warren Alpert School of Medicine at Brown University, Providence, RI, the zzDivision of Pediatric Gastro-enterology, Hepatology and Nutrition, The Hospital for Sick Children, University of Toronto, Toronto, Canada, and the §§Department of Pediatrics, Emory University School of Medicine/Children’s Healthcare of Atlanta, Atlanta, GA. CLINICAL GUIDELINES 154 JPGN Volume 64, Number 1, January 2017 Copyright © ESPGHAL and NASPGHAN. All rights reserved. for Gastroenterology, Hepatology and Nutrition and the European Society for Gastroenterology, Hepatology and Nutrition after an extensive review. LITERATURE SEARCH A systematic literature search was performed using acces-sible databases of relevance: PubMed, MEDLINE from 2002 until 2015 for targeted topics and keywords (see Supplementary Digital Content 1, Table, The search involved only articles published in English and involving human subjects. GRADES OF EVIDENCE Grades of evidence for each statement were based on the grading of the literature and were assigned using the American Association for the Study of Liver Diseases Practice Guidelines method: Grading of Recommendation Assessment, Development, and Evaluation workgroup with minor modifications (1). The strength of recommendations in the Grading of Recommendation Assessment, Development, and Evaluation system was classified as outlined in Supplementary Digital Content 2, Table, http:// links.lww.com/MPG/A734. BACKGROUND Cholestasis is defined as reduced bile formation or flow resulting in the retention of biliary substances within the liver normally excreted into bile and destined for elimination into the intestinal lumen. Cholestasis is generally recognized by evaluation of serum studies, with elevation of serum conjugated (or direct) bilirubin and bile acids as central readily identified features of hepatobiliary dysfunction. Although cholestasis and hyperbilirubi-nemia are not synonymous, during cholestasis normal bile acid flux and conjugated bilirubin excretion into bile are both impaired and frequently linked. Hence, a central feature of conjugated (or direct) hyperbilirubinemia is a practical clinical marker and surrogate of cholestasis. Distinguishing jaundice caused by cholestasis from noncholestatic conditions (such as physiologic jaundice of the newborn) is critical because cholestatic jaundice is likely patholo-gic, and therefore patients with cholestatic jaundice will benefit from prompt diagnosis and institution of specific therapy. Choles-tasis can be classified into biliary (obstructive, large extrahepatic, or small intrahepatic bile ducts) or hepatocellular (defect in membrane transport, embryogenesis, or metabolic dysfunction) in origin. Cholestatic jaundice affects approximately 1 in every 2500 term infants and is thus infrequently seen by most providers taking care of infants (2). The most common causes of cholestatic jaundice in the first months of life are BA (25%–40%) and an array of individually uncommon genetic disorders (25%). Often, however, the etiology is unknown. It may be associated with prematurity or intravenous soy lipid infusions (see following sections) (3). The rate of patients designated by the descriptive term, ‘‘idiopathic neonatal hepatitis’’ as the cause of neonatal cholestasis, continues to decline with advancements in diagnostic evaluation and discovery of new etiologies, now clinically discoverable with the use of available next-generation DNA sequencing technologies (see following sections). Other causes of neonatal cholestasis include extrahepatic obstruction from common duct gallstones or choledochal cyst; metabolic disorders such as tyrosinemia type I, galactosemia, and inborn errors of bile acid metabolism; panhypopituitarism; Alagille syndrome (ALGS); infection; parenteral nutrition (PN)-associated liver disease and a broad array of generally rare disorders (Table 1) (4–42). The common clinical feature of impaired bile flow resulting from either biliary obstruction or hepatocellular metabolic derangements requires a broad-minded approach to the individual cholestatic infant—without which opportunities for pro-viding effective therapeutic interventions may be overlooked. The incidence of neonatal cholestasis is increased in pre-mature infants, more so in those born at the limits of viability than those born closer to term. PN-related cholestasis is present in up to one-fifth of neonates receiving PN for >2 weeks (43). Longer duration of PN and intestinal failure are independent risk factors for the development of PN cholestasis in infants and has led to the consideration for reducing exposure to soy lipids wherever appro-priate (43,44). In addition, small for gestational age is a strong independent risk factor for neonatal cholestasis (45). This clinical guideline is not meant to address cholestasis in the preterm infant on PN, but close follow-up and serial measurements of fractionated bilirubin levels early in life are important, alongside monitoring growth and tolerance of enteral feedings. Persistent cholestasis in any infant should, however, be considered pathologic and identifi-able causes of cholestasis, including BA should be ruled out in a timely fashion, because another cholestatic condition can certainly be present in an infant who requires PN. It should be noted that the incidence of BA or genetic forms of cholestasis is the same in premature as in term infants; thus, premature infants warrant consideration for the same evaluation of neonatal cholestasis as do full-term infants. Several studies demonstrate a higher incidence of BA in preterm infants compared with term infants, and delayed diagnosis results in poorer outcome (46,47). Biliary Atresia BA is the most frequent identifiable cause of obstructive jaundice in the first 3 months of life. The prevalence of BA varies according to location around the globe: 1 in 6000 live births in Taiwan, 1 in 12,000 in the United States, 1 in 19,000 in Canada, and 1 in 18,000 in Europe (48–50). There are 3 classifications of BA: the nonsyndromic form (84%), which is the most common; BA with at least 1 malformation but without laterality (eg, situs inversus) defects (6%); and the syndromic BA with laterality defects (10%). The latter 2 groups have other associated anomalies predominantly in the cardiovascular (16%) and gastrointestinal (14%) systems, but the group without laterality defects has more frequent genitourinary anomalies. Patients with BA with laterality defects more commonly have splenic anomalies (51). The etiology of BA is unknown and theories of pathogenesis include genetic contributions to bile duct dysmorphogenesis, viral infection, toxins, chronic inflammatory or Address correspondence and reprint requests to Saul J. Karpen, MD, PhD, Department of Pediatrics, Emory University School of Medicine/ Children’s Healthcare of Atlanta, 1760 Haygood Dr, HSRB E204, Atlanta, GA 30322 (e-mail: skarpen@emory.edu). Supplemental digital content is available for this article. Direct URL citations appear in the printed text, and links to the digital ﬁles are provided in the HTML text of this article on the journal’s Web site (www.jpgn.org). C.L.M., J.P.M., V.L.N., and S.J.K. are members of the NIH-supported ChiLDReN network, which focuses upon pediatric cholestatic diseases; N.H. is a consultant for Alnylam Pharmaceuticals and Alexion UK; E.N. is currently employed by Vertex Pharmaceuticals; S.J.K. is an unpaid consultant from Intercept Pharmaceuticals. The authors report no conﬂicts of interests. Copyright # 2016 by European Society for Pediatric Gastroenterology, Hepatology, and Nutrition and North American Society for Pediatric Gastroenterology, Hepatology, and Nutrition DOI: 10.1097/MPG.0000000000001334 JPGN Volume 64, Number 1, January 2017 Guideline for the Evaluation of Cholestatic Jaundice in Infants www.jpgn.org 155 Copyright © ESPGHAL and NASPGHAN. All rights reserved. TABLE 1. Anatomic and monogenic disorders of neonatal cholestasis Disease Presentation Radiology Gene (s) Gene function References Multisystem disease Alagille syndrome GGTP, cholesterol often elevated, eye, and cardiac ﬁndings, LB not always clearly diagnostic when performed early in life Vertebral anomalies JAG1; NOTCH2 Signaling ligand; receptor for Jagged 1 (5,6) ARC syndrome Lax skin, limb contractures, renal tubular acidosis VPS33B; VIPAR Membrane protein recycling; basolateral sorting of canalicular proteins involved in bile secretion (7–9) Congenital disorders of glycosylation Multisystemic Numerous genes coding for glycosylation enzymes N- and O- protein glycosylation leading to impaired function (10,11) Cystic ﬁbrosis Elevated sweat chloride; possible ductular proliferation on LB Cystic ﬁbrosis trans-membrane receptor (CFTR) Chloride channel (12,13) Mitochondrial disorders Multisystemic Nuclear genes; mitochondrial genes May impact mtDNA replication, protein translation, electron transport (14–16) Neonatal ichthyosis sclerosing cholangitis syndrome Hypotrichosis, alopecia, cholestasis CLDN1 Claudin-1: tight junctions (17) Panhypopituitarism LB: duct paucity, low pituitary hormones on stimulation, adrenal insufﬁciency MRI may reveal microadenoma or absent sella Trisomy 21 Typical stigmata Unknown Unknown (18) Extrahepatic bile duct abnormalities Biliary atresia LB diagnostic of obstruction with bile duct proliferation and bile duct plugs; acholic stools Situs or vascular anomalies in 5% to 10%; possible absence of gallbladder Choledochal cyst Abdominal mass along with features that overlap with BA (see below) Cyst seen by US Choledocholithiasis Acholic stools US and IOC diagnostic ABCB4 Multidrug resistance P-glycoprotein, MDR3 (19) Congenital perforation of the common bile duct Ascites without liver disease Echogenic ascites Neonatal sclerosing cholangitis GGTP often >800 IU/L; LB shows small duct destruction IOC shows pruning of small bile ducts Hepatocellular diseases Alpha-1-antitrypsin deﬁciency GGTP often high, a-1-antitrypsin level low (often falsely low in neonates), Pi type ZZ or SZ SERPINA1 Anti-protease (20–23) Bile acid synthesis defects GGTP normal, FABMS of urinary bile acids, may present with cirrhosis, fat-soluble vitamin deﬁciencies CYP7B1; AKR1D1 (SRD5B1); HSD3B7 Oxysterol 7a-hydroxylase D4–3-oxosteroid-5b-reductase deﬁciency 3b-hydroxy-D5-C27-steroid dehydrogenase deﬁciency (24–26) Bile acid conjugation defects FABMS of urinary bile acids BAAT; BAL Absence of conjugation (27,28) Fawaz et al JPGN Volume 64, Number 1, January 2017 156 www.jpgn.org Copyright © ESPGHAL and NASPGHAN. All rights reserved. Disease Presentation Radiology Gene (s) Gene function References PFIC1 GGTP low or normal; diarrhea and FTT; LB/EM helpful ATP8B1 FIC1 translocates phospholipids from outer to inner canalicular membrane (ﬂoppase)—also expressed in intestine: considered multisystem disease (29) PFIC2 Low or normal GGTP; LB/EM helpful ABCB11 Bile salt export pump (30) PFIC3 Elevated GGTP ABCB4 Phospholipid ﬂippase responsible for phosphatidylcholine transport into bile (31) Tight junction protein 2 mutations Severe cholestasis TJP2 Failure of tight junctions and protein localization (32) Transient neonatal cholestasis (neonatal hepatitis) GGTP and AP 200 to 400 IU/L, ALT and AST 80 to 200 IU/L, LB negative for obstruction ATP8B1; ABCB11; ABCB4 FIC1 polymorphisms; MDR3 polymorphisms (19,33) Inborn errors of metabolism Urea cycle defects Citrin deﬁciency Normal liver enzymes or slightly elevated SLC25A13 Mitochondrial aspartate-glutamate carrier (34,35) Ornithine trans-carbamylase deﬁciency Neonatal hyperammonemia with/ without cholestasis and with/ without liver failure OTC Mitochondrial enzyme of urea cycle (36) Carbohydrate metabolism Galactosemia Cholestasis and liver dysfunction GALT Galactose-1-phosphate uridyltransferase (37–39) Amino acid metabolism Tyrosinemia type 1 May present with liver failure, Fanconi-related nephropathy, or seizures FAH Fumarylacetoacetate hydrolase (40) Lipid metabolism Niemann-Pick type C Splenomegaly NPC1 acid sphingomyelinase (41) Lysosomal acid lipase deﬁciency (Wolman disease) Hepatomegaly, features suggesting NAFLD (neonatal liver failure) Hyperechoic liver LIPA Lysosomal acid lipase (42) When multiple mutations have been identiﬁed, the original paper is referenced. This list is not exhaustive; rather it is an overview of the most characterized genetic diseases and congenital conditions which may present as neonatal cholestasis. ALT ¼ alanine aminotransferase; AP ¼ alkaline phosphatase; ARC ¼ arthrogryposis-renal dysfunction-cholestasis syndrome; AST ¼ aspartate aminotransferase; CT ¼ computed tomography; EM ¼ electron microscopy; FABMS ¼ fast atom bombardment mass spectroscopy; FTT ¼ failure to thrive; GGTP ¼ gamma-glutamyl transferase; IOC ¼ intraoperative cholangiogram; LB ¼ liver biopsy; MDR3 ¼ multidrug resistance 3 gene; MRI ¼ magnetic resonance imaging; mtDNA ¼ mitochondrial DNA; NAFLD ¼ nonalcoholic fatty liver disease; PFIC ¼ progressive familial intrahepatic cholestasis; US ¼ ultrasound. TABLE 1. (Continued) JPGN Volume 64, Number 1, January 2017 Guideline for the Evaluation of Cholestatic Jaundice in Infants www.jpgn.org 157 Copyright © ESPGHAL and NASPGHAN. All rights reserved. autoimmune-mediated bile duct injury (52–55). Direct hyperbilir-ubinemia is identified sooner after birth in patients with BA compared with normal (control, noncholestatic) infants, suggesting that the initiation of the biliary injury occurs before, or soon after birth (ie, perhaps due to intrauterine insult or genetic etiology), thus minimizing the likelihood of biliary tract disease acquired after birth (56). Timely diagnosis is important to optimize the response to a Kasai hepatic portoenterostomy (HPE) aimed at reestablishing bile flow (57). If the HPE is performed within the first 60 days of life, 70% of patients will establish bile flow; after 90 days of life <25% of patients will have bile flow (3). Late diagnosis of BA however remains a problem worldwide for a variety of reasons including the obligate visual overlap with normal physiologic jaundice and the lack of a readily applicable newborn screening. The average age at HPE in the United States is 61 days and 44% of patients still undergo HPE after 60 days of life (58). In Europe, late diagnosis is also a challenge and average age at HPE has been reported between 57 and 68 days (59–61). In the largest outcome series from Canada, medium age at HPE was 55 days but late referral was still problematic (49). Although not systematically evaluated, surgical outcome has been associated with the expertise in performing HPE in Europe, with improved outcome seen with centralized care models (49,62,63). In the United States this may be more challenging, but referral to a specialized center with expertise in performing HPE remains crucial. The optimal management of infants with delayed presentation of BA remains controversial. Some series report successful HPE drainage beyond 90 days of life reaching 13% to 35% (49,58,64). In a large series that examined outcomes in 743 infants with BA, 2-, 5-, 10-, and 15-year survival rates with native liver were 57.1%, 37.9%, 32.4%, and 28.5%, respectively (57). Moreover, survival rates with native liver decreased as the age at surgery increased from <45 to 90 days. The investigators in this study estimated that if every patient with BA underwent the Kasai operation before 46 days of age, 5.7% of all liver transplantations performed annually in France in patients younger than 16 years could be avoided. These studies highlight the importance of early detection of cholestasis by providers that can improve outcomes (57). These also indicate a need for unbiased screening for cholestasis and BA, perhaps via yet-to-be discovered newborn screening or the application of stool color cards as success-fully used in Taiwan (65). Non-BA Etiologies of Neonatal Cholestasis Treatable conditions that can present with cholestatic jaun-dice include bacterial sepsis, galactosemia, tyrosinemia, panhypo-pituitarism, bile acid synthetic defects, or obstructive gallstones. These infants often appear acutely ill and early diagnosis will enable timely initiation of directed treatment. Conversely, infants with BA usually appear otherwise healthy and grow normally which may deceive the parent or physician into believing that the jaundice is physiologic or caused by breast-feeding (4). It is important to note that medical management and optimization of nutrition to prevent complications of neonatal cholestasis is beneficial even when specific treatment is not available or curative. The differential diagnoses include a variety of anatomic, infectious, autoimmune, genetic, metabolic, and congenital conditions. This list is not meant to be exhaustive but rather an overview to help orient the reader (see Table 1). EVALUATION OF THE JAUNDICED INFANT Jaundice or icterus is clinically evident when the total serum bilirubin level exceeds 2.5 to 3.0 mg/dL (42–51 mmol/L). Visual determinations of bilirubin levels are inherently problematic. Several studies confirm the inability of even experienced caregivers to accurately estimate an infant’s total serum bilirubin level (66) and this visual assessment cannot determine if the jaundice is due to indirect or direct hyperbilirubinemia. The most important initial step in evaluating a jaundiced infant is measuring serum total and direct (or conjugated) bilirubin. Jaundice at 2 weeks of age is a relatively common finding, observed in 2.4% to 15% of newborns (67,68); however, it should alert the provider of the possibility of cholestasis, although testing of all jaundiced newborns at the 2-week visit will detect cholestasis in relatively few (4). Providers have several options: the most direct is to test serum for total and direct bilirubin at the 2-week visit (or if concerned at any age), but in the absence of any significant ‘‘red Flags,’’ the infants can follow one of several paths. These ‘‘Flags’’ are detailed in Tables 2 and 3 (69). If the 2-week-old infant is breast-fed and has a normal physical examination, no history of dark urine or acholic stool, another option is to see the infant back for follow-up in 1 week according to local practice and caregiver/parental comfort with the plan. If this course is taken, and the jaundice persists at 3 weeks of age, laboratory evaluation is recommended (4). If a 2-week-old icteric infant is bottle-fed then fractionation of bilirubin is recom-mended. If the infant’s first visit is at 4 weeks of age as is common practice in many European countries, then any jaundiced infant should be investigated promptly by measurement of total and direct bilirubin. The actual age of the infant when measurement of a fractionated bilirubin is performed is dependent upon several factors and not meant to be proscribed—but practically the measurement should coincide with the clinical status of the infant in the context of accepted local practice. The earlier measurements of fractionated bilirubin are performed, the earlier a diagnosis of cholestasis can be made or excluded and thus help direct optimal and timely clinical care plans. The most commonly used laboratory determination, the diazo or van den Bergh method, does not specifically measure conjugated bilirubin but reports direct bilirubin, which includes both conjugated bilirubin and delta bilirubin (conjugated bilirubin covalently bound to albumin). For methodological reasons, the higher the total bilirubin (TB) (even if nearly all unconjugated bilirubin) the higher the reported direct bilirubin; hence, specific measurements of conjugated bilirubin are optimal if available (70,71). Because canalicular excretion of bilirubin can be rate-limiting to overall clearance, infants with elevated unconjugated bilirubin may retain some conjugated bilirubin, but this is variable because of the distinct canalicular transporters for conjugated bilirubin (Abcc2) and bile acids (Abcb11), and their differential expression in the setting of cholestasis and age. Therefore, in the presence of elevated TB, direct/conjugated bilirubin levels are considered abnormal when values are >1.0 mg/dL (17 mmol/L) regardless of the TB (72). Thus, for this guideline, an abnormal direct/conjugated bilirubin is defined as a serum value >1.0 mg/dL (17 mmol/L), because it is physiologically and clinically complex to consider incorporating consideration of whether or not the direct fraction exceeds 20% of the TB level as mentioned in some publications (4,73). In a healthy newborn baby with indirect/unconjugated hyper-bilirubinemia, the most common causes of jaundice are physiologic jaundice and breast milk jaundice. Both are self-limited matura-tional disorders characterized by an elevation of serum indirect/ unconjugated bilirubin. Infants who are breast-fed are more susceptible to neonatal jaundice because maternal milk contains b-glucuronidase that breaks down conjugated bilirubin to form unconjugated bilirubin and hence increases the enterohepatic cir-culation of bilirubin (4,74,75). Expressed breast milk also contains factors that may inhibit the conjugating enzyme in hepatocytes (76). Fawaz et al JPGN Volume 64, Number 1, January 2017 158 www.jpgn.org Copyright © ESPGHAL and NASPGHAN. All rights reserved. Please refer to the American Academy of Pediatrics guidelines for the management of unconjugated hyperbilirubinemia in the new-born infant 35 or more weeks of gestation (77). Recommendations: 1. Any formula-fed infant noted to be jaundiced after 2 weeks of age should be evaluated for cholestasis with measurement of total and conjugated (direct) serum bilirubin (1A). Depending upon local practice, breast-fed babies that appear otherwise well may be followed clinically until 3 weeks of age, at which time if they appear icteric should then undergo serum evaluation of total and conjugated (direct) serum bilirubin. 2. Measurements of serum bilirubin should always be fractionated into unconjugated (indirect) or conjugated (direct) hyperbilirubinemia (1A). 3. Conjugated (direct) hyperbilirubinemia (>1.0 mg/dL, 17 mmol/ L) is considered pathological and warrants diagnostic evaluation (1A). TABLE 2. Parameters of clinical interest in the history of the cholestatic infant Family history Consanguinity Increased risk of autosomal recessive disorders Neonatal cholestasis in the parents or siblings Cystic ﬁbrosis, a-1-antitrypsin deﬁciency, progressive familial intrahepatic cholestasis, Alagille syndrome are all genetic conditions causing neonatal cholestasis History of repeated fetal loss or early demise Gestational alloimmune liver disease Spherocytosis and other hemolytic diseases Known to aggravate conjugated hyperbilirubinemia Prenatal history Prenatal ultrasonography ﬁndings Presence of choledochal cyst, cholelithiasis, bowel anomalies or concern for syndrome Cholestasis of pregnancy May be seen in heterozygotes for PFIC gene mutations; mitochondrial disorder Acute fatty liver of pregnancy Neonatal long-chain 3-hydroxyacyl-coenzyme A dehydrogenase (LCHAD) deﬁciency Maternal infections TORCH infections Infant history Gestational age Prematurity as a risk factor for neonatal hepatitis SGA Increased risk of neonatal cholestasis, congenital infections Alloimmune hemolysis; glucose-6-P-dehydrogenase deﬁciency; hydrops fetalis Increased risk of neonatal cholestasis Neonatal infection Urinary tract infection, sepsis related cholestasis, CMV, HIV, syphilis, etc Newborn screen Panhypopituitarism galactosemia, fatty acid oxidation defects, cystic ﬁbrosis Source of nutrition: breast milk, formula, PN Galactosemia, hereditary fructose intolerance, PN-associated liver disease Growth Genetic and metabolic disease Vision Septo-optic dysplasia Hearing PFIC1, TJP2 Vomiting Metabolic disease, bowel obstruction, and pyloric stenosis Stooling Delayed stooling: CF, panhypopituitarism; diarrhea: infection, metabolic disease Stool color Acholic stools: cholestasis, biliary obstruction Urine characteristics: smell and color Dark urine (conjugated hyperbilirubinemia), metabolic disease Excessive bleeding May indicate coagulopathy, vitamin K deﬁciency Disposition: irritability, lethargy Metabolic disease or sepsis, panhypopituitarism Abdominal surgery Necrotizing enterocolitis, intestinal atresia CF ¼ cystic ﬁbrosis; CMV ¼ cytomegalovirus; HIV ¼ human immunodeﬁciency virus; PFIC ¼ progressive familial intrahepatic cholestasis; PN ¼ parenteral nutrition; TJP ¼ tight-junction protein; TORCH ¼ Toxoplasma gondii, other viruses, rubella, cytomegalovirus, and herpes simplex virus. TABLE 3. Physical ﬁndings in children with neonatal cholestasis Assessment of general health Ill appearance may indicate infection or metabolic disease, infants with biliary atresia typically appear well General appearance Dysmorphic features: Alagille syndrome in the neonate rarely exhibits characteristic facial appearance with a broad nasal bridge, triangular facies, and deep-set eyes. Typical facial features may appear at around 6 months of age, but are often nonspeciﬁc (69) Vision/slit lamp examination Hearing Congenital infection, storage disease, septo-optic dysplasia, posterior embryotoxon, cataracts Congenital infections, PFIC1, TJP2, mitochondrial Cardiac examination: murmur, signs of heart failure Congenital heart disease: Alagille syndrome, biliary atresia splenic malformation syndrome Abdominal examination Presence of ascites; abdominal wall veins, liver size and consistency, spleen size and consistency (or absence thereof), abdominal masses, umbilical hernia Stool examination (crucial—the primary physician should make every effort to view stool pigment) Acholic or hypopigmented stools suggest cholestasis or biliary obstruction Neurologic Note overall vigor and tone PFIC ¼ progressive familial intrahepatic cholestasis; TJP ¼ tight-junction protein. JPGN Volume 64, Number 1, January 2017 Guideline for the Evaluation of Cholestatic Jaundice in Infants www.jpgn.org 159 Copyright © ESPGHAL and NASPGHAN. All rights reserved. HISTORY Obtaining a detailed prenatal and infant history is funda-mental and should include details of the neonatal screening and any medication including vitamin K supplementation. Details of feed-ing history should be noted as well as the timing of the first bowel movement, because delayed passage of meconium can be seen in patients with cystic fibrosis (CF). The history should system-atically collect information about the onset of jaundice, changes in stool pigmentation, and urine color. It is important to identify history of pale or acholic stools and it is highly recommended to observe the stool pigment (see below). It is well recognized that parents and health care professionals assess stool pigmentation subjectively and abnormally pale stools are frequently misinter-preted as normal. Acholic stools were correctly identified only by 63% of health care providers (78). Stool color charts may be helpful in review of history and ascertaining lack of pigmentation of stools in children with suspected liver disease. In Taiwan, use of a stool color card proved to be effective with 95.2% sensitivity for pale stools (79). A large prospective cohort study using home-based screening for BA with a stool card proved cost effective in Canada (80). Use of the stool card has been piloted in some European countries, such as Switzerland (81) but has not been systematically implemented across the United States or Europe. Many efforts are being investigated to increase awareness and recognition of acholic stool. In addition, the common intersection of prematurity, inability to advance enteral feedings, and use of prolonged soy lipid–based PN leads to cholestasis, commonly known as parenteral nutrition–associated cholestasis (PNAC) (82). This is a major confounder in the evaluation of the cholestatic infant, and it is often worthwhile for caregivers to note the timing and initiation of PN in relation to serial measurements of fractionated bilirubin levels, especially if direct hyperbilirubinemia precedes the initiation of PN. Details in the family history including previous and current pregnancy such as miscarriages, pruritus, or overt liver dysfunction in maternal history should be noted; history of maternal fever, rash, adenopathy, or medication intake can be helpful. The family history should not only focus on known liver conditions but also on hemolysis and/or cardiac and vascular anomalies. A detailed overview of noteworthy features is given in Tables 2 and 3. PHYSICAL EXAMINATION The clinician performing a physical examination should not only focus on the abdomen but should also consider extrahepatic signs, such as: dysmorphic features, poor growth, dermatologic, neurologic, or pulmonary symptoms (Table 3). Palpation of the abdomen may reveal firm hepatomegaly suspicious for the diag-nosis of BA, often with a prominent middle or left lobe. Spleno-megaly in BA appears after the newborn period, and if present at a young age of 2 to 4 weeks should point toward other diseases such as storage or hematologic disorders. Cardiac examination is the key, as discovery of a murmur may suggest ALGS or cardiac anomalies associated with BA (eg, septal defects). For a variety of causes, right heart failure may lead to impaired hepatic venous outflow, hepa-tomegaly, and cholestasis. Hypoplastic (male) genitalia may be a feature of panhypopituitarism, but normal genitalia does not exclude this condition. Confirming whether the infant can visually fix and follow is helpful to rule out septo-optic dysplasia, but often cross-sectional brain imaging is required for this diagnosis (83,84). Direct observation of urine color, and most importantly stool color, is a necessary component of the assessment of the jaundiced infant, as acholic stools and dark urine often indicates the presence of cholestasis and conjugated hyperbilirubinemia. It is important to note that there are no findings obtained by a careful history or a detailed physical examination that are unique to BA patients. Recommendations: 4. A thorough physical examination is crucial to the proper evaluation of the jaundiced infant. Attention to hepatomegaly, splenomegaly, and ill appearance warrants special consider-ations (1A). 5. Direct visualization of stool pigment is a key aspect of a complete evaluation of the jaundiced infant (1A). DIAGNOSTIC EVALUATION This section is devoted to the diagnostic approach to the infant with cholestasis. In addition to laboratory studies, imaging and liver histopathology are important to evaluate for bile duct patency because cholestatic infants must be evaluated promptly to exclude treatable surgical conditions. As noted above, performance of the Kasai HPE for BA is much less likely to benefit infants if performed after 3 months of age (85), hence the importance of an expedient and efficient evaluation. LABORATORY EVALUATION During the evaluation of the infant with cholestasis, labora-tory investigations will help define the etiology, the severity of the liver disease and detect treatable conditions. A critical and important initial blood test is the measurement of serum conjugated (direct) bilirubin (DB), which, if elevated, is a reliable laboratory indicator of cholestasis at this age. Accompany-ing evaluation of DB levels are standard biochemical and synthetic liver tests to assess the severity of the liver disease to include TB, alanine aminotransferase (ALT), aspartate aminotransferase (AST), alkaline phosphatase (AP), gamma glutamyl transpeptidase (GGTP), prothrombin time (PT) with the international normalized ratio (INR), glucose, and albumin. An elevated serum AST without substantial increase in ALT, TB, or DB may point to a hematologic or muscular process, because AST is an enzyme present in red blood cells and myocytes. GGTP value is typically higher in neonates than older children (86) and is generally elevated during cholestasis (87). Some cholestatic diseases, however, present with normal or low GGTP, including progressive familial intrahepatic cholestasis (PFIC) type 1 (ATP8B1 deficiency) and 2 (ABCB11 deficiency), bile acid synthesis disorders (BASDs) and tight-junction protein (TJP) type 2 deficiency (32,88). Other conditions including ALGS, PFIC3 (due to ABCB4 deficiency), and often, but not always, BA frequently present with a high GGTP. Serum AP levels are gener-ally less helpful than serum GGTP in the evaluation of cholestatic infants since the normal range of serum AP levels varies greatly in growing infants. Bacterial cultures of blood, urine, and other fluids should be obtained as dictated by the clinical assessment. Severe coagulopathy unresponsive to parenteral vitamin K administration and out of proportion to the liver injury may indicate gestational alloimmune liver disease, metabolic disease, or sepsis. When evaluating a patient with cholestasis, it is crucial to review the standard local newborn screening as many diseases that cause cholestasis are tested such as hypothyroidism, galactosemia, tyr-osinemia, and CF. Some countries have extended newborn screens that can be performed upon request. The minimum evaluation for any health care professional encountering an infant with jaundice present after the age of 14 days should include a full history including family history and gestational history of the mother, physical examination, inspection of stool color, and obtaining a fractionated bilirubin measurement. When cholestasis is suspected, expedited focused Fawaz et al JPGN Volume 64, Number 1, January 2017 160 www.jpgn.org Copyright © ESPGHAL and NASPGHAN. All rights reserved. investigations (Table 4, Tier 1) are recommended. A disciplined and stepwise approach to the infant with confirmed cholestasis in concert with a pediatric gastroenterologist/hepatologist can then follow in the ordering of laboratory tests appropriate in each situation, and enabling a targeted workup (Table 4, Tier 2). Some local variation is unavoidable because of available expertise (Table 4). ‘‘Red flags,’’ which mandate evaluation for BA include acholic stools, high GGT cholestasis without alternative etiology, and abnormal or absence of gallbladder on ultrasound. Conditions that mimic BA such as a-1-antitrypsin deficiency, CF, ALGS, and others should be excluded early on in the evaluation process. DIAGNOSTIC IMAGING A fasting abdominal ultrasound is an easy and noninvasive first diagnostic imaging investigation to assess for visible obstruct-ing lesions of the biliary tree or identification of choledochal cyst, and to assess for signs of advanced liver disease or vascular and/or splenic abnormalities (89). Several hepatic sonographic parameters such as the triangular cord sign, abnormal gall bladder morphology, lack of gall bladder contraction after oral feeding, nonvisualization of the common bile duct, hepatic artery diameter, and hepatic artery diameter to portal vein diameter ratio, subcapsular blood flow have been suggested to aid in the diagnosis of BA (90–94), although none can singularly confirm a diagnosis of BA. It is useful, however, to know that many, but not all, infants with BA have a small or undetectable gall bladder (95). In addition, findings such as abdominal heterotaxy, midline liver, polysplenia, asplenia, and preduodenal portal vein increase the concern for BA with malfor-mations. It is imperative to remember that a normal ultrasonography (US), however, does not rule out nonsyndromic BA. Hepatobiliary scintigraphy (HBS) has been used to confirm biliary tract patency, but can be limited by its low specificity (range 68.5%–72.2%), and a nondiagnostic result when bile flow is limited as a result of a wide variety of etiologies (96). Patients with interlobular bile duct paucity, idiopathic neonatal hepatitis, low birth weight, and those on PN may have nonexcreting scans (97). This limited accuracy of HBS in differentiating idiopathic neonatal hepatitis from BA was demonstrated in a study by Yang et al (98) in which magnetic resonance cholangiopancreatography (MRCP), US, technetium 99m-iminodiacetic acid HBS, HBS single photon emission computed tomography (HBS SPECT), and liver biopsy were compared. The goal of this study of 69 infants with cholestatic jaundice and a final diagnosis of idiopathic neonatal hepatitis, and BA was to determine which modality may help distinguish between these 2 diagnoses. All of the 69 infants underwent MRCP, US, HBS, SPECT, and liver biopsy. HBS had sensitivity and a specificity of 88.2% and 45.7% for detecting BA, respectively, with an accuracy of 66.7%. Scintigraphy adds little to the routine evaluation of the cholestatic infant, but may be of value in determining patency of the biliary tract, thereby excluding BA. In this study, liver biopsy had the highest sensitivity in detecting BA at 100%, a specificity of 94.3% and an accuracy rate of 96.9%. A recent meta-analysis addressing the utility of HBS yielded a pooled sensitivity of 98.7% (98.1–99.2%) and a specificity of 70.4% (range 68.5%–72.2%) of a nondraining HBS for excluding BA. This shows that false negative results (excretion of the tracer into the bowel despite BA) are extremely rare (96). Limited reports describe infants with apparently initially excreting HBS and a subsequent diagnosis of BA, although the technical limitations of the study may have been a factor in its utility (100,101). Many clinicians and radiologists administer phenobarbital for 5 days before the study, in an attempt to enhance biliary excretion of the isotope and increase its discriminatory value (99), which often unnecessarily delays the diagnosis of BA and the necessary HPE (57,89). Further work is necessary to assess the utility of premedication for HBS (100,101). Despite the use of the diagnostic tests described above, it is still not easy to discriminate between BA and other causes of neonatal cholestasis. As detection of patency of the extrahepatic biliary tree is the primary goal of diagnostic evaluations in infants with cholestasis, the role of endoscopic retrograde cholangiopan-creatography (ERCP) in the diagnosis of BA has been studied by various groups (102,103). Although ERCP has proved effective with high positive and negative predictive values for BA (sensitivity 86%–100%, specificity 87%–94%, positive predictive value 88%– 96%, negative predictive value 100%) (102,104), ERCP requires an TABLE 4. Targeted investigations of the persistently cholestatic infant Tier 1: Aim to evaluate after cholestasis has been established in order to both identify treatable disorder as well as to deﬁne the severity of the liver involvement Blood—CBC þ differential, INR, AST, ALT, AP, GGTP, TB, DB (or conjugated bilirubin), albumin and glucose. Check a-1-antitryspin phenotype (Pi typing) and level, TSH, T4 if newborn screen results not readily available Urine—urinalysis, culture, reducing substances (rule out galactosemia) Consider bacterial cultures of blood, urine and other ﬂuids especially if infant is clinically ill. Verify results of treatable disorders (such as galactosemia and hypothyroidism) from newborn screen Obtain fasting ultrasound Tier 2: Aim to complete a targeted evaluation in concert with pediatric gastroenterologist/hepatologist General—TSH and T4 values, serum bile acids, cortisol Consideration of speciﬁc etiologies Metabolic—serum ammonia, lactate level, cholesterol, red blood cell galactose-1-phosphate uridyltransferase, urine for succinylacetone and organic acids. Consider urine for bile salt species proﬁling ID—direct nucleic acid testing via PCR for CMV, HSV, listeria Genetics—in discussion with pediatric gastroenterologist/hepatologist, with a low threshold for gene panels or exome sequencing Sweat chloride analysis (serum immunoreactive trypsinogen level or CFTR genetic testing) as appropriate Imaging CXR—lung and heart disease Spine—spinal abnormalities (such as butterﬂy vertebrae) Echocardiogram—evaluating for cardiac anomalies seen in Alagille syndrome Cholangiogram Liver biopsy (timing and approach will vary according to institution and expertise) Consideration for consultations Ophthalmology Metabolic/Genetic (consider when to involve, especially when there is consideration for gene panels or whole exome sequencing) Cardiology/ECHO (if murmur present or has hypoxia, poor cardiac function) General pediatric surgery Nutrition/dietician ALT ¼ alanine aminotransferase; AP ¼ alkaline phosphatase; AST ¼ aspartate aminotransferase; CBC ¼ complete blood count; CFTR ¼ cystic ﬁbrosis trans-membrane receptor; DB ¼ conjugated (direct) bilirubin; ECHO ¼ echocardiogram; GGTP ¼ gamma-glutamyl transferase; HSV ¼ herpes simplex virus; ID ¼ infectious diseases; INR ¼ international normal-ized ratio; PCR ¼ polymerase chain reaction; TB ¼ total bilirubin; TSH ¼ thyroid-stimulating hormone. JPGN Volume 64, Number 1, January 2017 Guideline for the Evaluation of Cholestatic Jaundice in Infants www.jpgn.org 161 Copyright © ESPGHAL and NASPGHAN. All rights reserved. experienced endoscopist, specific infant endoscopy equipment not readily available at many centers, and a general anesthetic. The superiority of ERCP compared with other types of cholangio-grams has not been demonstrated (105). A few reports have suggested that MRCP is a well-estab-lished noninvasive modality for visualizing the biliary system, including the first-order branches of the intrahepatic bile ducts, extrahepatic bile ducts, and gallbladder (106). The diagnostic value of 3-dimensional MRCP for BA in a large cohort of cholestatic infants and neonates was therefore recently evaluated, with a reported specificity of 36% and sensitivity of 99% (107). Recent case series have documented the technique and feasibility of percutaneous transhepatic cholecysto-cholangiogra-phy (PTCC) to exclude BA (108,109). In the largest series reported (109), PTCC was performed in combination with simultaneous liver biopsy. Although this was reported to effectively exclude BA with a lower negative laparotomy rate, there is a genuine concern that PTCC may be used unnecessarily in infants in whom a liver biopsy alone would have excluded biliary obstruction. Moreover, a PTCC may not be able to demonstrate retrograde patency of the biliary tree into the liver and may miss proximal obstruction, thereby obviating a surgical cholangiogram in a patient who may have BA. Impor-tantly, the specificity of liver biopsy in diagnosing biliary obstruc-tion in this case series was much lower than frequently reported values (89). Taken together, the use of MRCP, ERCP, and PTCC has a limited role in the general guidance to caregivers toward diagnosing BA in the present era. HISTOPATHOLOGY Liver biopsy often remains the cornerstone of the diagnostic workup of infants with cholestatic jaundice as interpretation by an experienced pathologist will provide the correct diagnosis in 90% to 95% of cases and avoid unnecessary surgery in patients with intrahepatic disease (89,110,111). Pathologists participating in the National Institutes of Health–supported Biliary Atresia Research Consortium (BARC, currently the ChiLDReN consortium [childrennetwork.org]) have developed and evaluated a standar-dized system for reporting of liver biopsies from infants with cholestasis. Overall, the pathologists diagnosis of obstruction in clinically proven cases of BA ranged from 79% to 98%, with a positive predictive value of 90.7%. The group diagnosed BA with a high level of sensitivity and identified infants with biliary obstruc-tion with reasonable interobserver agreement (110). Of note, a diagnosis of BA or obstruction other than BA was made in 14 of 15 cases of PN-associated liver disease and all 3 cases of a-1-antitrypsin deficiency. Conversely, a majority of the pathologists favored a nonobstructive diagnosis in the 3 cases of progressive familial intrahepatic cholestasis and 1 case of bile acid synthetic disorder. In cases of idiopathic neonatal hepatitis, the percentage of cases read by each pathologist as nonobstructive ranged from 57% to 93%. The classic histologic features of biliary obstruction are bile duct proliferation, bile plugs, portal or perilobular fibrosis, and edema, with preservation of the basic hepatic lobular architecture (Fig. 1). In idiopathic neonatal hepatitis, lobular disarray and inflammatory cells are seen within the portal areas, and the bile ductules show little or no alteration (Fig. 2). Giant cell transform-ation can be seen in 20% to 50% of patients with BA (112,113); however, it is not as prominent as that seen in idiopathic neonatal hepatitis (111). Some disorders that can mimic BA histologically are PN-associated cholestasis, CF and a-1-antitrypsin deficiency. They may show variable ductular reaction and may be impossible to distinguish from BA without clinical data (110,111). It is however important to recognize that the earliest histologic changes of BA may be relatively nonspecific, and biopsies performed too early in the course of the disease may result in a falsely negative diagnosis (110,111,114). The proper use of liver biopsy therefore remains a central component of the diagnostic evaluation of infants with cholestatic jaundice as the differential diagnosis is perhaps the broadest of any age group and encompasses obstructive as well as, more commonly, nonobstructive disorders. In addition to its role in diagnosis, the liver biopsy may also reveal histologic features of significant prognostic value, such as the degree of fibrosis, which may help predict outcome following HPE and the decision to proceed with HPE (115,116). Although sonography-guided percutaneous core liver biopsy is considered to be a safe and effective procedure in children with a low complication rate of 1.7% (117), the overall A B C FIGURE 1. Liver biopsies from 2 individuals with biliary atresia (BA). A, H&E stain of a 3-month-old infant with BA, highlighting peribiliary ﬁbrosis, ductular proliferation, bile duct plugs. B, High-power view of (A), emphasizing bile duct plugs and damaged cholangioles. C, Liver biopsy from a 6-week-old infant with BA, highlighting peribiliary ﬁbrosis, disordered cholangiocyte proﬁles and scattered inﬂammatory inﬁltrate. FIGURE 2. Idiopathic neonatal hepatitis: lobular disarray with giant cell transformation. Fawaz et al JPGN Volume 64, Number 1, January 2017 162 www.jpgn.org Copyright © ESPGHAL and NASPGHAN. All rights reserved. complication rate in infants, even in the hands of an experienced physician, was reported in 1 small series to be 4.6% (3/65 infants had a bleeding event that required an intervention) (118). Recommendations: 6. The abdominal ultrasound is useful in excluding choledochal cyst or gallstone disease causing extrahepatic bile duct obstruction. It may demonstrate an absent or abnormal gallbladder, or other features suggestive, but not diagnostic, of BA (1A). 7. Limited speciﬁcity precludes the use of the HBS scan as a stand-alone test in making a deﬁnitive diagnosis of BA (1B). Deﬁnitively demonstrated bile ﬂow by selective use of HBS may be of value in excluding BA (1B). 8. Limited speciﬁcity of MRCP, ERCP, PTCC provides a limited role in the general guidance to caregivers toward diagnosing BA in the present era. 9. In the hands of an experienced pediatric pathologist, histopathological ﬁndings of bile duct proliferation, bile plugs, and ﬁbrosis in an appropriately timed liver biopsy is the most supportive test in the evaluation of the infant with protracted conjugated hyperbilirubinemia (1B). Diseases other than BA that cause cholestasis can be determined via histologic examination of the liver. INTRAOPERATIVE CHOLANGIOGRAM The intraoperative cholangiogram and histologic examin-ation of the duct remnant is considered the criterion standard to diagnose BA (89,119,120). Interestingly, in up to 20% of cases, even a cholangiogram can suggest an incorrect diagnosis—cases with a hypoplastic biliary tree, ALGS and CF being confounding diagnostic conditions (121). Hence, preoperative testing for CF and ALGS is helpful in assisting in the interpretation of the cholangio-gram and decreasing false-positive results. Intraoperative cholan-giogram is typically performed after biliary obstruction is suggested in a liver biopsy or if sufficient clinical indications suggest direct referral to the surgeon for the procedure. If BA is confirmed (ie, nonvisualization of a patent extrahepatic biliary tree), a (Kasai) HPE is usually performed immediately, unless there are consider-ations made by the team that it would be in the best interest of the infant to proceed to transplant evaluation and not undergo the HPE. No effective diagnostic tools currently determine whether a patient should proceed to HPE, and it is up to the team of specialists to determine whether the patient would be better served without the HPE. It is important to note that diagnostic evaluation to rule out BA should be expedited especially when the infant is above 6 weeks of age. The younger the age at diagnosis of BA, the more likely that the HPE will be successful (at least in the short term, see above). Recommendation: 10. Evaluation by intraoperative cholangiogram and histological examination of the duct remnant is considered the gold standard to diagnose biliary atresia (1A). OTHER CAUSES OF NEONATAL CHOLESTASIS Structural Abnormalities Choledochal Cyst Patients with choledochal cysts present with laboratory findings suggestive of cholestasis. Sometimes patients have cho-langitis and present with fever, elevation of the GGTP, and direct hyperbilirubinemia. Ultrasonography can often differentiate between choledochal cyst and BA as the bile ducts are typically dilated or cystic and the gallbladder is not atretic (122). A diagnosis of choledochal cyst in a cholestatic neonate should, however, always prompt careful evaluation for BA (atresia of the distal common bile duct accompanied by cystic dilation: type 1 BA). In a few studies, cyst size appeared to decrease between prenatal diagnosis and birth in patients with BA but did not change in patients with choledochal cyst (123,124). Moreover, choledochal cysts can coincide with BA. Select Genetic/Metabolic Disorders Alagille Syndrome ALGS is an autosomal dominant multisystem disorder characterized by paucity of interlobular ducts. It is the most common form of familial intrahepatic cholestasis occurring in 1 in 30,000 live births. Diagnosis is usually made by the clinical findings, laboratory, and diagnostic evaluation, and confirmed by sequencing of JAG1 and NOTCH2 genes, with mutations found in 95% and 5% of patients with ALGS, respectively (125). Clinical criteria for the diagnosis of ALGS includes ductopenia on liver biopsy and a characteristic Alagille facies (broad forehead, small pointy chin, but is often difficult to recognize in the neonatal period), posterior embryotoxon, butterfly vertebrae, renal disease, and a variety of developmental cardiac defects (most commonly peripheral pulmonic stenosis) (126) or tetralogy of Fallot. Direct hyperbilirubinemia and occasional acholic stool that may improve with age can be present (126). Serum ALT and bile acids are usually elevated. The GGTP is an important test in orienting the practitioner toward this disorder because it is often disproportionately elevated, sometimes up to 20 times the normal value. Practitioners are encouraged to evaluate for associated clinical abnormalities found in patients with ALGS when the diagnosis is suspected either on liver biopsy or by characteristic facies, before proceeding with intraoperative cholangiogram. Cystic Fibrosis Some infants with CF present with abnormal liver tests, suggestive of biliary obstruction because of the presence of abnor-mal bile with plugging of the common bile duct (127). Checking the newborn screen for immunoreactive trypsinogen is helpful. The criterion standard remains sequencing of the cystic fibrosis trans-membrane receptor (CFTR) gene or a positive sweat chloride test, but this is sometimes not possible as infants may not produce enough sweat (128). Progressive Familial Intrahepatic Cholestasis PFIC is a group of unrelated monogenic disorders in which mutations in one of the genes involved in canalicular hepatobiliary transport results in progressive cholestasis and liver injury (Table 1). Patients with PFIC1–3 have a significant elevation of the total serum bile acids. A important clinical finding in individuals with PFIC types 1 and 2 (due to ATP8B1 and ABCB11 gene deficiencies, respectively) is the presence of a normal or low GGTP out of proportion to the degree of cholestasis (129), associated with normal or low serum cholesterol.SomeofthepatientswithnormalGGTPPFICwerefound to have a mutation in the TJP2 gene, which causes failure of protein localization and disruption of tight-junction structure, leading to severe cholestatic liver disease that can present early in life (32). Patients with PFIC type 3 (ABCB4 deficiency) have elevated GGTP and a variable degree of cholestasis—typically presenting later in infancy or in early childhood (130,131). JPGN Volume 64, Number 1, January 2017 Guideline for the Evaluation of Cholestatic Jaundice in Infants www.jpgn.org 163 Copyright © ESPGHAL and NASPGHAN. All rights reserved. Alpha-1-Antitrypsin Deﬁciency This is the most common cause of inherited neonatal cho-lestasis. Approximately 10% to 15% of neonates with this condition will present with cholestasis and a combined picture of hepatocel-lular injury and obstruction with elevation of the ALT, AST, GGTP, and AP. The cholestasis is usually severe and the presence of acholic stools may present a challenge because of the resemblance to BA. Although some patients may develop cirrhosis early on, jaundice clears in most patients by 4 months of age (132). The diagnosis is made based on the phenotype (normal: MM; abnormal: ZZ or SZ; heterozygous: MZ, MS) (133). It is important to note that neonates with ZZ phenotype may have no biliary excretion on scintigraphic studies (134) and liver biopsy may appear obstructive (110). Hence, obtaining the phenotype early in the evaluation of cholestasis could avoid unnecessary biopsy in this condition. Checking for serum levels of a-1-antitrypsin could be helpful if used along with the phenotype to distinguish patients who are homozygous for the Z allele or SZ compound heterozygotes, both of whom may develop liver disease. Patients with MZ, MS, SZ, or homozygous SS A1 phenotypes do not present with neonatal cholestasis unless associated with another cause (135). Serum a-1-antitrypsin concentrations alone are an insufficient test since a-1-antitrypsin is an acute phase reactant and during illnesses may be elevated (136,137). Of note, there have been few case reports of the concurrence of a-1-antitrypsin deficiency and BA (138,139). Bile Acid Synthesis Disorders More than 14 enzymes are involved in the synthesis of bile acids from cholesterol precursor molecules. Bile acid synthetic disorders (BASDs) are rare, but in many cases are treatable forms of cholestasis. Not all of the infants with the genetic abnormalities leading to BASD present with cholestasis and jaundice; some may have a more indolent presentation later during childhood. These conditions often present with normal or low GGTP. Total serum bile acids are usually low, in contrast to other cholestatic disorders. Fast atom bombardment mass spectrometry of urine should be con-sidered as a screening tool before starting ursodeoxycholic acid; it is possible to perform rapid diagnosis of potential inborn errors in bile acid synthesis from urinary bile acid analysis. Molecular techniques then identify the specific mutations in genes encoding the enzymes responsible for bile acid synthesis (25,26,140). Treatment with the end products of bile acid synthesis, cholic acid and chenodeoxy-cholic acid, is often curative for several of the BASDs, prompting directed evaluations. Select Inborn Errors of Metabolism A group of metabolic conditions classified as inborn errors of metabolism can present with cholestasis, and as in the other cholestatic disorders, the practitioner has to have a high index of suspicion. The initial laboratory testing recommended when the clinical picture is compatible with an inborn errors of metabolism includes blood gases, electrolytes, glucose, ammonia, uric acid, lactic acid, pyruvic acid (L:P ratio), ketone bodies; and in urine, ketone bodies, 2-keto acids, reducing substances, acylglycines, pH; and sulfites (141–144). Newborn screening for galactosemia and tyrosinemia are performed in some countries to identify infants before they are symptomatic. Infants with tyrosinemia can present with mild cholestasis, although more typically present with coa-gulopathy disproportionate to other biochemical abnormalities (145). Diagnosis via serum fumarylacetoacetate hydrolase enzyme determination or urine succinylacetone is vital in this treatable but life-threatening condition. With the incorporation of gene panels and exome sequencing, these disorders may have more precise and timely means of genetic investigations in the near future. Infections Cytomegalovirus Cytomegalovirus (CMV) the most common congenital infec-tion, affects 1% to 2% of newborns. Most infected newborns are asymptomatic; unfortunately 5% to 10% of the patients have a myriad of clinical symptoms that include low birth weight, micro-cephaly, periventricular calcifications, chorioretinitis, and deafness. Hepatosplenomegaly and direct hyperbilirubinemia are the most prominent liver-related problems (146,147). The diagnosis of con-genital CMV is confirmed by culture or PCR from the nasopharynx, saliva, blood, or urine soon after birth. Urine CMV culture or CMV-DNA detection by PCR is presently used for the diagnosis (148,149). The immunoglobulin M (IgM) CMV-specific antibodies can be monitored but are of limited value and may be less sensitive. Evidence of recent CMV infection at the time of diagnosis of BA has been reported by multiple investigators, but a role for CMV in the etiology of BA remains unproven (150–152). Viral Hepatitis A, B, and C In general, these hepatotropic viruses do not cause neonatal cholestasis. Single case reports document special circumstances where these infections present with neonatal cholestasis. Specific studies for these infectious agents in the evaluation of neonatal cholestasis are generally unwarranted. Other Infections Syphilis, rubella, toxoplasmosis, and herpes virus can present with neonatal cholestasis, coagulopathy, and growth restriction. Obtaining a good maternal history and discussing with the obste-trician and neonatal intensive care team about placental abnorm-alities can help with directing the workup for an infection. Typically, infants with these infections present with jaundice within first 24 hours of life. Congenital syphilis incidence is rising in the United States (153). Urinary tract infections present with choles-tasis in the neonatal period and a urine culture should be obtained early on in the diagnostic evaluation of cholestasis (154). Endocrine Disorders Thyroid Disorders Few reports in the pediatric literature describe cholestatic liver disease in infants born to mothers with Graves disease (155). The newborn screen is designed to detect high levels of thyroid-stimulating hormone (TSH); hence, in cases of central hypothyr-oidism, this can be missed and repeating a blood TSH, free T4, and T3 may be helpful (156,157). Panhypopituitarism Pituitary hormones are involved in the regulation of bile synthesis and excretion and bile flow. The neonates with this condition present with elevation of the total and direct bilirubin; they may have hypoglycemia and even shock from adrenal insuf-ficiency. Some infants have associated septo-optic dysplasia and on physical examination will lack the ability to focus or track. Diag-nostic evaluation includes TSH, total and free T4, early morning cortisol level, and a brain magnetic resonance imaging. In these patients, a nonfasting ultrasound should be requested as prolonged Fawaz et al JPGN Volume 64, Number 1, January 2017 164 www.jpgn.org Copyright © ESPGHAL and NASPGHAN. All rights reserved. fasting can lead to devastating complications related to severe hypoglycemia. The cholestasis resolves with the correction of the pituitary hormone insufficiency (83,84,158–160). Rare Diseases and Idiopathic Neonatal Cholestasis The neonatal cholestasis ‘‘black box’’ of unidentified etiol-ogies continues to shrink but is still a substantial group of disorders (131). The identifiable causes of neonatal cholestasis have grown more numerous largely because of application of modern tech-niques of clinical genetics. Reducing costs and easy access to genetic testing, including exome and genome sequencing, and targeted gene panels, have facilitated diagnosis. Among the most studied causes of neonatal cholestasis in recent years are metabolic diseases and disorders of bile transport and bile acid synthesis. It is, however, important to stress that the meaning of a gene mutation or polymorphism is dependent on clinical context. Advanced sequen-cing methods promise to further increase the diagnostic yield of genetic approaches. Table 1 summarizes known genetic findings presenting as neonatal cholestasis. CONCLUSIONS Cholestatic jaundice in an infant is a typical presenting feature of neonatal liver disease and is frequently clinically con-fused with the more common prolonged unconjugated hyperbilir-ubinemia. Identification of infants with cholestasis remains crucial and is in the domain of the primary care provider, generally uncovered with measurement of a serum-fractionated bilirubin. Hence a careful history, thorough physical examination, and frac-tionation of serum bilirubin are recommended in any infant with jaundice seen after 2 weeks of life. The relative rarity of cholestatic jaundice in contrast to unconjugated hyperbilirubinemia in this age range dictates that many jaundiced infants will be tested to detect those with elevated direct bilirubin levels. Investigation for neonatal cholestasis in these settings is, however, highly beneficial, despite its rarity, because of the gravity of the consequences of missing BA and monogenic diagnoses that have specific, and often life-saving, interventions. This guideline has been developed to assist in this process and is not intended as a substitute for clinical judgment or as a protocol for the care of all infants with cholestasis. Vigilance is crucial in detecting these infants and referring them to a pediatric gastroenterologist or hepatologist who can provide the essential diagnostic and treatment modalities to optimize outcome. REFERENCES 1. Guyatt GH, Oxman AD, Vist GE, et al. GRADE: an emerging consensus on rating quality of evidence and strength of recommenda-tions. BMJ 2008;336:924–6. 2. Dick MC, Mowat AP. Hepatitis syndrome in infancy—an epidemio-logical survey with 10 year follow up. Arch Dis Child 1985;60:512–6. 3. Balistreri WF, Bezerra JA. Whatever happened to ‘‘neonatal hepati-tis’’? Clin Liver Dis 2006;10:27–53v. 4. Moyer V, Freese DK, Whitington PF, et al. Guideline for the evaluation of cholestatic jaundice in infants: recommendations of the North American Society for Pediatric Gastroenterology, Hepatology and Nutrition. J Pediatr Gastroenterol Nutr 2004;39:115–28. 5. Li L, Krantz ID, Deng Y, et al. Alagille syndrome is caused by mutations in human Jagged1, which encodes a ligand for Notch1. Nat Genet 1997;16:243–51. 6. McDaniell R, Warthen DM, Sanchez-Lara PA, et al. NOTCH2 muta-tions cause Alagille syndrome, a heterogeneous disorder of the notch signaling pathway. Am J Hum Genet 2006;79:169–73. 7. Ackermann MA, Patel PD, Valenti J, et al. Loss of actomyosin regulation in distal arthrogryposis myopathy due to mutant myosin binding protein-C slow. FASEB J 2013;27:3217–28. 8. Cullinane AR, Straatman-Iwanowska A, Zaucker A, et al. Mutations in VIPAR cause an arthrogryposis, renal dysfunction and cholestasis syndrome phenotype with defects in epithelial polarization. Nat Genet 2010;42:303–12. 9. Gissen P, Johnson CA, Morgan NV, et al. Mutations in VPS33B, encoding a regulator of SNARE-dependent membrane fusion, cause arthrogryposis-renal dysfunction-cholestasis (ARC) syndrome. Nat Genet 2004;36:400–4. 10. Bjursell C, Stibler H, Wahlstro ¨m J, et al. Fine mapping of the gene for carbohydrate-deﬁcient glycoprotein syndrome, type I (CDG1): linkage disequilibrium and founder effect in Scandinavian families. Genomics 1997;39:247–53. 11. Tan J, Dunn J, Jaeken J, et al. Mutations in the MGAT2 gene controlling complex N-glycan synthesis cause carbohydrate-deﬁcient glycoprotein syndrome type II, an autosomal recessive disease with defective brain development. Am J Hum Genet 1996;59:810–7. 12. Cutting GR, Kasch LM, Rosenstein BJ, et al. Two patients with cystic ﬁbrosis, nonsense mutations in each cystic ﬁbrosis gene, and mild pulmonary disease. N Engl J Med 1990;323:1685–9. 13. Cutting GR, Kasch LM, Rosenstein BJ, et al. A cluster of cystic ﬁbrosis mutations in the ﬁrst nucleotide-binding fold of the cystic ﬁbrosis conductance regulator protein. Nature 1990;346:366–9. 14. Carroll CJ, Isohanni P, Po ¨yho ¨nen R, et al. Whole-exome sequencing identiﬁes a mutation in the mitochondrial ribosome protein MRPL44 to underlie mitochondrial infantile cardiomyopathy. J Med Genet 2013;50:151–9. 15. Holt IJ, Harding AE, Morgan-Hughes JA. Deletions of muscle mi-tochondrial DNA in patients with mitochondrial myopathies. Nature 1988;331:717–9. 16. Wallace DC, Zheng XX, Lott MT, et al. Familial mitochondrial encephalomyopathy (MERRF): genetic, pathophysiological, and bio-chemical characterization of a mitochondrial DNA disease. Cell 1988;55:601–10. 17. Hadj-Rabia S, Baala L, Vabres P, et al. Claudin-1 gene mutations in neonatal sclerosing cholangitis associated with ichthyosis: a tight junction disease. Gastroenterology 2004;127:1386–90. 18. Arnell H, Fischler B. Population-based study of incidence and clinical outcome of neonatal cholestasis in patients with Down syndrome. J Pediatr 2012;161:899–902. 19. Gonzales E, Davit-Spraul A, Baussan C, et al. Liver diseases related to MDR3 (ABCB4) gene deﬁciency. Front Biosci (Landmark Ed) 2009;14:4242–56. 20. Kidd VJ, Wallace RB, Itakura K, et al. Alpha 1-antitrypsin deﬁciency detection by direct analysis of the mutation in the gene. Nature 1983;304:230–4. 21. Kueppers F, Briscoe WA, Bearn AG. Hereditary deﬁciency of serum alpha-L-antitrypsin. Science 1964;146:1678–9. 22. Sharp HL, Bridges RA, Krivit W, et al. Cirrhosis associated with alpha-1-antitrypsin deﬁciency: a previously unrecognized inherited disorder. J Lab Clin Med 1969;73:934–9. 23. Talamo RC, Feingold M. Infantile cirrhosis with hereditary alpha 1 -antitrypsin deﬁciency. Clinical improvement in two siblings. Am J Dis Child 1973;125:845–7. 24. Setchell KD, Schwarz M, O’Connell NC, et al. Identiﬁcation of a new inborn error in bile acid synthesis: mutation of the oxysterol 7alpha-hydroxylase gene causes severe neonatal liver disease. J Clin Invest 1998;102:1690–703. 25. Cheng JB, Jacquemin E, Gerhardt M, et al. Molecular genetics of 3beta-hydroxy-Delta5-C27-steroid oxidoreductase deﬁciency in 16 patients with loss of bile acid synthesis and liver disease. J Clin Endocrinol Metab 2003;88:1833–41. 26. Schwarz M, Wright AC, Davis DL, et al. The bile acid synthetic gene 3beta-hydroxy-Delta(5)-C(27)-steroid oxidoreductase is mutated in progressive intrahepatic cholestasis. J Clin Invest 2000;106: 1175–84. 27. Falany CN, Johnson MR, Barnes S, et al. Glycine and taurine con-jugation of bile acids by a single enzyme. Molecular cloning and expression of human liver bile acid CoA:amino acid N-acyltransferase. J Biol Chem 1994;269:19375–9. 28. Falany CN, Xie X, Wheeler JB, et al. Molecular cloning and expres-sion of rat liver bile acid CoA ligase. J Lipid Res 2002;43:2062–71. JPGN Volume 64, Number 1, January 2017 Guideline for the Evaluation of Cholestatic Jaundice in Infants www.jpgn.org 165 Copyright © ESPGHAL and NASPGHAN. All rights reserved. 29. Bull LN, van Eijk MJ, Pawlikowska L, et al. A gene encoding a P-type ATPase mutated in two forms of hereditary cholestasis. Nat Genet 1998;18:219–24. 30. Strautnieks SS, Bull LN, Knisely AS, et al. A gene encoding a liver-speciﬁc ABC transporter is mutated in progressive familial intrahe-patic cholestasis. Nat Genet 1998;20:233–8. 31. Deleuze JF, Jacquemin E, Dubuisson C, et al. Defect of multidrug-resistance 3 gene expression in a subtype of progressive familial intrahepatic cholestasis. Hepatology 1996;23:904–8. 32. Sambrotta M, et al. Mutations in TJP2 cause progressive cholestatic liver disease. Nat Genet 2014;46:326–8. 33. Liu LY, Wang XH, Lu Y, et al. Association of variants of ABCB11 with transient neonatal cholestasis. Pediatr Int 2013;55:138–44. 34. Tamamori A, Okano Y, Ozaki H, et al. Neonatal intrahepatic choles-tasis caused by citrin deﬁciency: severe hepatic dysfunction in an infant requiring liver transplantation. Eur J Pediatr 2002;161:609–13. 35. Kobayashi K, Sinasac DS, Iijima M, et al. The gene mutated in adult-onset type II citrullinaemia encodes a putative mitochondrial carrier protein. Nat Genet 1999;22:159–63. 36. Russell A, Levin B, Oberholzer VG, et al. Hyperammonaemia: a new instance of an inborn enzymatic defect of the biosynthesis of urea. Lancet 1962;2:699–700. 37. Reichardt JK. Molecular analysis of 11 galactosemia patients. Nucleic Acids Res 1991;19:7049–52. 38. Reichardt JK, Packman S, Woo SL. Molecular characterization of two galactosemia mutations: correlation of mutations with highly con-served domains in galactose-1-phosphate uridyl transferase. Am J Hum Genet 1991;49:860–7. 39. Reichardt JK, Woo SL. Molecular basis of galactosemia: mutations and polymorphisms in the gene encoding human galactose-1-phos-phate uridylyltransferase. Proc Natl Acad Sci U S A 1991;88:2633–7. 40. Phaneuf D, Labelle Y, Be ´rube ´ D, et al. Cloning and expression of the cDNA encoding human fumarylacetoacetate hydrolase, the enzyme deﬁcient in hereditary tyrosinemia: assignment of the gene to chromo-some 15. Am J Hum Genet 1991;48:525–35. 41. Vance JE. Lipid imbalance in the neurological disorder, Niemann-Pick C disease. FEBS Lett 2006;580:5518–24. 42. Schiff L, Schubert WK, McAdams AJ, et al. Hepatic cholesterol ester storage disease, a familial disorder. I. Clinical aspects. Am J Med 1968;44:538–46. 43. Lauriti G, Zani A, Auﬁeri R, et al. Incidence, prevention, and treatment of parenteral nutrition-associated cholestasis and intestinal failure-associated liver disease in infants and children: a systematic review. JPEN J Parenter Enteral Nutr 2014;38:70–85. 44. Jolin-Dahel K, Ferretti E, Montiveros C, et al. Parenteral nutrition-induced cholestasis in neonates: where does the problem lie? Gastro-enterol Res Pract 2013;2013:163632. 45. Champion V, Carbajal R, Lozar J, et al. Risk factors for developing transient neonatal cholestasis. J Pediatr Gastroenterol Nutr 2012;55:592–8. 46. Chiu CY, Chen PH, Chan CF, et al. Biliary atresia in preterm infants in Taiwan: a nationwide survey. J Pediatr 2013;163:100–3e1. 47. Fischler B, Haglund B, Hjern A. A population-based study on the incidence and possible pre- and perinatal etiologic risk factors of biliary atresia. J Pediatr 2002;141:217–22. 48. The NS, Honein MA, Caton AR, et al. Risk factors for isolated biliary atresia, National Birth Defects Prevention Study, 1997-2002. Am J Med Genet A 2007;143A:2274–84. 49. Schreiber RA, Barker CC, Roberts EA, et al. Biliary atresia: the Canadian experience. J Pediatr 2007;151:659–65665 e1. 50. McKiernan PJ, Baker AJ, Kelly DA. The frequency and outcome of biliary atresia in the UK and Ireland. Lancet 2000;355:25–9. 51. Schwarz KB, Haber BH, Rosenthal P, et al. Extrahepatic anomalies in infants with biliary atresia: results of a large prospective North American multicenter study. Hepatology 2013;58:1724–31. 52. Sokol RJ, Mack C. Etiopathogenesis of biliary atresia. Semin Liver Dis 2001;21:517–24. 53. Mack CL. The pathogenesis of biliary atresia: evidence for a virus-induced autoimmune disease. Semin Liver Dis 2007;27:233–42. 54. Schreiber RA, Kleinman RE. Genetics, immunology, and biliary atresia: an opening or a diversion? J Pediatr Gastroenterol Nutr 1993;16:111–3. 55. Bezerra JA. Potential etiologies of biliary atresia. Pediatr Transplant 2005;9:646–51. 56. Harpavat S, Finegold MJ, Karpen SJ. Patients with biliary atresia have elevated direct/conjugated bilirubin levels shortly after birth. Pedia-trics 2011;128:e1428–33. 57. Serinet MO, Wildhaber BE, Broue ´ P, et al. Impact of age at Kasai operation on its results in late childhood and adolescence: a rational basis for biliary atresia screening. Pediatrics 2009;123:1280–6. 58. Shneider BL, Brown MB, Haber B, et al. A multicenter study of the outcome of biliary atresia in the United States, 1997 to 2000. J Pediatr 2006;148:467–74. 59. Hoerning A, Raub S, Deche ˆne A, et al. Diversity of disorders causing neonatal cholestasis—the experience of a tertiary pediatric center in Germany. Front Pediatr 2014;2:65. 60. Wildhaber BE, Majno P, Mayr J, et al. Biliary atresia: Swiss national study, 1994-2004. J Pediatr Gastroenterol Nutr 2008;46:299–307. 61. Grizelj R, Vukovic ´ J, Novak M, et al. Biliary atresia: the Croatian experience. Eur J Pediatr 2010;169:1529–34. 62. Davenport M, Caponcelli E, Livesey E, et al. Biliary atresia in England and Wales: results of centralization and new benchmark. J Pediatr Surg 2011;46:1689–94. 63. Lampela H, Ritvanen A, Kosola S, et al. National centralization of biliary atresia care to an assigned multidisciplinary team provides high-quality outcomes. Scand J Gastroenterol 2012;47:99–107. 64. Davenport M, Puricelli V, Farrant P, et al. The outcome of the older (> or ¼ 100 days) infant with biliary atresia. J Pediatr Surg 2004;39:575– 81. 65. Lien TH, Chang MH, Wu JF, et al. Effects of the infant stool color card screening program on 5-year outcome of biliary atresia in Taiwan. Hepatology 2011;53:202–8. 66. Moyer VA, Ahn C, Sneed S. Accuracy of clinical judgment in neonatal jaundice. Arch Pediatr Adolesc Med 2000;154:391–4. 67. Winﬁeld CR, MacFaul R. Clinical study of prolonged jaundice in breast- and bottle-fed babies. Arch Dis Child 1978;53:506–7. 68. Kelly DA, Stanton A. Jaundice in babies: implications for community screening for biliary atresia. BMJ 1995;310:1172–3. 69. Kamath BM, Loomes KM, Oakey RJ, et al. Facial features in Alagille syndrome: speciﬁc or cholestasis facies? Am J Med Genet 2002;112:163–70. 70. Lo SF, Kytzia HJ, Schumann G, et al. Interlaboratory comparison of the Doumas bilirubin reference method. Clin Biochem 2009;42:1328– 30. 71. Lo SF, Doumas BT. The status of bilirubin measurements in U.S. laboratories: why is accuracy elusive? Semin Perinatol 2011;35:141– 7. 72. Rosenthal P, Blanckaert N, Kabra PM, et al. Formation of bilirubin conjugates in human newborns. Pediatr Res 1986;20:947–50. 73. Kefﬂer S, Kelly DA, Powell JE, et al. Population screening for neonatal liver disease: a feasibility study. J Pediatr Gastroenterol Nutr 1998;27:306–11. 74. Gourley GR. Breast-feeding, neonatal jaundice and kernicterus. Semin Neonatol 2002;7:135–41. 75. Maisels MJ, Gifford K. Normal serum bilirubin levels in the newborn and the effect of breast-feeding. Pediatrics 1986;78:837–43. 76. Arias IM, Gartner LM, Seifter S, et al. Prolonged neonatal unconju-gated hyperbilirubinemia associated with breast feeding and a steroid, pregnane-3(alpha), 20(beta)-diol, in maternal milk that inhibits glu-curonide formation in vitro. J Clin Invest 1964;43:2037–47. 77. American Academy of Pediatrics Subcommittee on Hyperbilirubine-mia. Management of hyperbilirubinemia in the newborn infant 35 or more weeks of gestation. Pediatrics 2004;114:297–316. 78. Bakshi B, Sutcliffe A, Akindolie M, et al. How reliably can paediatric professionals identify pale stool from cholestatic newborns? Arch Dis Child Fetal Neonatal Ed 2012;97:F385–7. 79. Chen SM, Chang MH, Du JC, et al. Screening for biliary atresia by infant stool color card in Taiwan. Pediatrics 2006;117:1147–54. 80. Schreiber RA, Masucci L, Kaczorowski J, et al. Home-based screening for biliary atresia using infant stool colour cards: a large-scale pro-spective cohort study and cost-effectiveness analysis. J Med Screen 2014;21:126–32. 81. Wildhaber BE. Screening for biliary atresia: Swiss stool color card. Hepatology 2011;54:367–8. Fawaz et al JPGN Volume 64, Number 1, January 2017 166 www.jpgn.org Copyright © ESPGHAL and NASPGHAN. All rights reserved. 82. Carter BA, Karpen SJ. Intestinal failure-associated liver disease: management and treatment strategies past, present, and future. Semin Liver Dis 2007;27:251–8. 83. Ellaway CJ, Silinik M, Cowell CT, et al. Cholestatic jaundice and congenital hypopituitarism. J Paediatr Child Health 1995;31:51–3. 84. Karnsakul W, Sawathiparnich P, Nimkarn S, et al. Anterior pituitary hormone effects on hepatic functions in infants with congenital hypopituitarism. Ann Hepatol 2007;6:97–103. 85. Sokol RJ, Shepherd RW, Superina R, et al. Screening and outcomes in biliary atresia: summary of a National Institutes of Health workshop. Hepatology 2007;46:566–81. 86. Hirfanoglu IM, Unal S, Onal EE, et al. Analysis of serum gamma-glutamyl transferase levels in neonatal intensive care unit patients. J Pediatr Gastroenterol Nutr 2014;58:99–101. 87. Lu FT, Wu JF, Hsu HY, et al. gamma-Glutamyl transpeptidase level as a screening marker among diverse etiologies of infantile intrahepatic cholestasis. J Pediatr Gastroenterol Nutr 2014;59:695–701. 88. van Mil SW, Houwen RH, Klomp LW. Genetics of familial intrahe-patic cholestasis syndromes. J Med Genet 2005;42:449–63. 89. Balistreri WF. Neonatal cholestasis. J Pediatr 1985;106:171–84. 90. Mittal V, Saxena AK, Sodhi KS, et al. Role of abdominal sonography in the preoperative diagnosis of extrahepatic biliary atresia in infants younger than 90 days. AJR Am J Roentgenol 2011;196:W438–45. 91. Humphrey TM, Stringer MD. Biliary atresia: US diagnosis. Radiology 2007;244:845–51. 92. Lee HJ, Lee SM, Park WH, et al. Objective criteria of triangular cord sign in biliary atresia on US scans. Radiology 2003;229:395–400. 93. Kim WS, Cheon JE, Youn BJ, et al. Hepatic arterial diameter measured with US: adjunct for US diagnosis of biliary atresia. Radiology 2007;245:549–55. 94. Tan Kendrick AP, Phua KB, Ooi BC, et al. Biliary atresia: making the diagnosis by the gallbladder ghost triad. Pediatr Radiol 2003;33:311–5. 95. Farrant P, Meire HB, Mieli-Vergani G. Ultrasound features of the gall bladder in infants presenting with conjugated hyperbilirubinaemia. Br J Radiol 2000;73:1154–8. 96. Kianifar HR, Tehranian S, Shojaei P, et al. Accuracy of hepatobiliary scintigraphy for differentiation of neonatal hepatitis from biliary atresia: systematic review and meta-analysis of the literature. Pediatr Radiol 2013;43:905–19. 97. Gilmour SM, Hershkop M, Reifen R, et al. Outcome of hepatobiliary scanning in neonatal hepatitis syndrome. J Nucl Med 1997;38:1279– 82. 98. Yang JG, Ma DQ, Peng Y, et al. Comparison of different diagnostic methods for differentiating biliary atresia from idiopathic neonatal hepatitis. Clin Imaging 2009;33:439–46. 99. Majd M, Reba RC, Altman RP. Hepatobiliary scintigraphy with 99mTc-PIPIDA in the evaluation of neonatal jaundice. Pediatrics 1981;67:140–5. 100. Shneider BNK, Superina R, Erlichman J, et al. Diagnostic imaging in neonatal cholestasis: a multi-center prospective analysis. Hepatology 2006;44:437A. 101. Williamson SL, Seibert JJ, Butler HL, et al. Apparent gut excretion of Tc-99m-DISIDA in a case of extrahepatic biliary atresia. Pediatr Radiol 1986;16:245–7. 102. Shanmugam NP, Harrison PM, Devlin J, et al. Selective use of endoscopic retrograde cholangiopancreatography in the diagnosis of biliary atresia in infants younger than 100 days. J Pediatr Gastro-enterol Nutr 2009;49:435–41. 103. Shteyer E, Wengrower D, Benuri-Silbiger A ˜ , et al. Endoscopic retro-grade cholangiopancreatography in neonatal cholestasis. J Pediatr Gastroenterol Nutr 2012;55:142–5. 104. Keil R, Snajdauf J, Rygl M, et al. Diagnostic efﬁcacy of ERCP in cholestatic infants and neonates—a retrospective study on a large series. Endoscopy 2010;42:121–6. 105. Chardot C. Endoscopic retrograde cholangiopancreatography in patients with neonatal cholestasis: an additional diagnostic tool for selected indications. J Pediatr Gastroenterol Nutr 2009;49: 380–1. 106. Metreweli C, So NM, Chu WC, et al. Magnetic resonance cholangio-graphy in children. Br J Radiol 2004;77:1059–64. 107. Liu B, Cai J, Xu Y, et al. Three-dimensional magnetic resonance cholangiopancreatography for the diagnosis of biliary atresia in infants and neonates. PLoS One 2014;9:e88268. 108. Meyers RL, Book LS, O’Gorman MA, et al. Percutaneous cholecysto-cholangiography in the diagnosis of obstructive jaundice in infants. J Pediatr Surg 2004;39:16–8. 109. Jensen MK, Biank VF, Moe DC, et al. HIDA, percutaneous transhe-patic cholecysto-cholangiography and liver biopsy in infants with persistent jaundice: can a combination of PTCC and liver biopsy reduce unnecessary laparotomy? Pediatr Radiol 2012;42:32–9. 110. Russo P, Magee JC, Boitnott J, et al. Design and validation of the biliary atresia research consortium histologic assessment system for cholestasis in infancy. Clin Gastroenterol Hepatol 2011;9:357–62e2. 111. Morotti RJD. Pediatric Cholestatic Disorders. Approach to pathologic diagnosis. Surg Pathol Clin 2013;6:205–390. 112. Davenport M, Tizzard SA, Underhill J, et al. The biliary atresia splenic malformation syndrome: a 28-year single-center retrospective study. J Pediatr 2006;149:393–400. 113. Rastogi A, Krishnani N, Yachha SK, et al. Histopathological features and accuracy for diagnosing biliary atresia by prelaparotomy liver biopsy in developing countries. J Gastroenterol Hepatol 2009;24:97– 102. 114. Azar G, Beneck D, Lane B, et al. Atypical morphologic presentation of biliary atresia and value of serial liver biopsies. J Pediatr Gastroenterol Nutr 2002;34:212–5. 115. Weerasooriya VS, White FV, Shepherd RW. Hepatic ﬁbrosis and survival in biliary atresia. J Pediatr 2004;144:123–5. 116. Shteyer E, Ramm GA, Xu C, et al. Outcome after portoenterostomy in biliary atresia: pivotal role of degree of liver ﬁbrosis and intensity of stellate cell activation. J Pediatr Gastroenterol Nutr 2006;42:93–9. 117. Govender P, Jonas MM, Alomari A, et al. Sonography-guided percu-taneous liver biopsies in children. AJR Am J Roentgenol 2013;201:645–50. 118. Amaral JG, Schwartz J, Chait P, et al. Sonographically guided percu-taneous liver biopsy in infants: a retrospective review. AJR Am J Roentgenol 2006;187:W644–9. 119. el-Youssef M, Whitington PF. Diagnostic approach to the child with hepatobiliary disease. Semin Liver Dis 1998;18:195–202. 120. Hays DM, Woolley MM, Snyder WH Jr et al. Diagnosis of biliary atresia: relative accuracy of percutaneous liver biopsy, open liver biopsy, and operative cholangiography. J Pediatr 1967;71:598–607. 121. Altman RP, Abramson S. Potential errors in the diagnosis and surgical management of neonatal jaundice. J Pediatr Surg 1985;20:529–34. 122. Kim WS, Kim IO, Yeon KM, et al. Choledochal cyst with or without biliary atresia in neonates and young infants: US differentiation. Radiology 1998;209:465–9. 123. Matsubara H, Oya N, Suzuki Y, et al. Is it possible to differentiate between choledochal cyst and congenital biliary atresia (type I cyst) by antenatal ultrasonography? Fetal Diagn Ther 1997;12:306–8. 124. Casaccia G, Bilancioni E, Nahom A, et al. Cystic anomalies of biliary tree in the fetus: is it possible to make a more speciﬁc prenatal diagnosis? J Pediatr Surg 2002;37:1191–4. 125. Hartley JL, Gissen P, Kelly DA. Alagille syndrome and other heredi-tary causes of cholestasis. Clin Liver Dis 2013;17:279–300. 126. Emerick KM, Rand EB, Goldmuntz E, et al. Features of Alagille syndrome in 92 patients: frequency and relation to prognosis. Hepa-tology 1999;29:822–9. 127. Lykavieris P, Bernard O, Hadchouel M. Neonatal cholestasis as the presenting feature in cystic ﬁbrosis. Arch Dis Child 1996;75:67–70. 128. Wagener JS, Zemanick ET, Sontag MK. Newborn screening for cystic ﬁbrosis. Curr Opin Pediatr 2012;24:329–35. 129. Lu FT, Wu JF, Hsu HY, et al. Gamma-glutamyl transpeptidase level as a screening marker among diverse etiologies of infantile intrahepatic cholestasis. J Pediatr Gastroenterol Nutr 2014;59:695–701. 130. Pawlikowska L, Strautnieks S, Jankowska I, et al. Differences in presentation and progression between severe FIC1 and BSEP deﬁ-ciencies. J Hepatol 2010;53:170–8. 131. Hoerning A, Raub S, Deche ˆne A, et al. Diversity of disorders causing neonatal cholestasis—the experience of a tertiary pediatric center in Germany. Front Pediatr 2014;2:65. 132. Suchy FJ. Neonatal cholestasis. Pediatr Rev 2004;25:388–96. JPGN Volume 64, Number 1, January 2017 Guideline for the Evaluation of Cholestatic Jaundice in Infants www.jpgn.org 167 Copyright © ESPGHAL and NASPGHAN. All rights reserved. 133. Ovchinsky N, Moreira RK, Lefkowitch JH, et al. Liver biopsy in modern clinical practice: a pediatric point-of-view. Adv Anat Pathol 2012;19:250–62. 134. Johnson K, Alton HM, Chapman S. Evaluation of mebrofenin hepa-toscintigraphy in neonatal-onset jaundice. Pediatr Radiol 1998; 28:937–41. 135. Pittschieler K, Massi G. Liver involvement in infants with PiSZ phenotype of alpha 1-antitrypsin deﬁciency. J Pediatr Gastroenterol Nutr 1992;15:315–8. 136. Topic A, Ljujic M, Nikolic A, et al. Alpha-1-antitrypsin phenotypes and neutrophil elastase gene promoter polymorphisms in lung cancer. Pathol Oncol Res 2011;17:75–80. 137. Lang T, Mu ¨hlbauer M, Strobelt M, et al. Alpha-1-antitrypsin deﬁ-ciency in children: liver disease is not reﬂected by low serum levels of alpha-1-antitrypsin—a study on 48 pediatric patients. Eur J Med Res 2005;10:509–14. 138. Nord KS, Saad S, Joshi VV, et al. Concurrence of alpha 1-antitrypsin deﬁciency and biliary atresia. J Pediatr 1987;111:416–8. 139. Tolaymat N, Figueroa-Colon R, Mitros FA. Alpha 1-antitrypsin deﬁ-ciency (Pi SZ) and biliary atresia. J Pediatr Gastroenterol Nutr 1989;9:256–60. 140. Setchell KD, Heubi JE. Defects in bile acid biosynthesis—diagnosis and treatment. J Pediatr Gastroenterol Nutr 2006;43(suppl 1):S17–22. 141. Hommes FA, Varghese M. High-performance liquid chromatography of urinary oligosaccharides in the diagnosis of glycoprotein degrada-tion disorders. Clin Chim Acta 1991;203:211–24. 142. Blau N, Beck M, Matern D. Tetrahydrobiopterin induced neonatal tyrosinaemia. Eur J Pediatr 1996;155:832. 143. Wraith JE. Diagnosis and management of inborn errors of metabolism. Arch Dis Child 1989;64 (10 Spec No):1410–5. 144. Mak CM, Lee HC, Chan AY, et al. Inborn errors of metabolism and expanded newborn screening: review and update. Crit Rev Clin Lab Sci 2013;50:142–62. 145. Crofﬁe JM, Gupta SK, Chong SK, et al. Tyrosinemia type 1 should be suspected in infants with severe coagulopathy even in the absence of other signs of liver failure. Pediatrics 1999;103:675–8. 146. McCracken GH Jr, Shineﬁeld HM, Cobb K, et al. Congenital cytomegalic inclusion disease. A longitudinal study of 20 patients. Am J Dis Child 1969;117:522–39. 147. Watkins JB, Sunaryo FP, Berezin SH. Hepatic manifestations of congenital and perinatal disease. Clin Perinatol 1981;8:467–80. 148. Plosa EJ, Esbenshade JC, Fuller MP, et al. Cytomegalovirus infection. Pediatr Rev 2012;33:156–63. 149. Wieringa JW, de Vries JJ, Murk JL. Congenital CMV infections. Ned Tijdschr Geneeskd 2013;157:A6250. 150. Xu Y, Yu J, Zhang R, et al. The perinatal infection of cytomegalovirus is an important etiology for biliary atresia in China. Clin Pediatr (Phila) 2012;51:109–13. 151. Jevon GP, Dimmick JE. Biliary atresia and cytomegalovirus infection: a DNA study. Pediatr Dev Pathol 1999;2:11–4. 152. Fischler B, Ehrnst A, Forsgren M, et al. The viral association of neonatal cholestasis in Sweden: a possible link between cytomegalo-virus infection and extrahepatic biliary atresia. J Pediatr Gastroenterol Nutr 1998;27:57–64. 153. Mattei PL, Beachkofsky TM, Gilson RT, et al. Syphilis: a reemerging infection. Am Fam Physician 2012;86:433–40. 154. Seeler RA, Hahn K. Jaundice in urinary tract infection in infancy. Am J Dis Child 1969;118:553–8. 155. Regelmann MO, Miloh T, Arnon R, et al. Graves’ disease presenting with severe cholestasis. Thyroid 2012;22:437–9. 156. Mitchell ML, Hsu HW, Sahai I. Changing perspectives in screening for congenital hypothyroidism and congenital adrenal hyperplasia. Curr Opin Endocrinol Diabetes Obes 2014;21:39–44. 157. Minamitani K, Inomata H. Neonatal screening for congenital hy-pothyroidism in Japan. Pediatr Endocrinol Rev 2012;10(suppl 1):79–88. 158. Geffner ME. Hypopituitarism in childhood. Cancer Control 2002;9:212–22. 159. DeSalvo D, Pohl JF, Wilson DP, et al. Cholestasis secondary to panhypopituitarism in an infant. J Natl Med Assoc 2008;100:342–4. 160. Kaufman FR, Costin G, Thomas DW, et al. Neonatal cholestasis and hypopituitarism. Arch Dis Child 1984;59:787–9. Fawaz et al JPGN Volume 64, Number 1, January 2017 168 www.jpgn.org
13318	https://arxiv.org/pdf/2405.00884	%PDF-1.5 % 96 0 obj << /Filter /FlateDecode /Length 4622 >> stream xÚí;ÉvãÈ÷ú ©÷$¹"ñ\|Émwùµ»z{üÜ>@$$�£¿X2±KÑã9Ì#2ÖÌ¼xZÄß}Ê~PÆ ±°.¥Y$ZGVÛÅjûá¯khûý"4Y¨çv!E è/ÊÅÿñáöþÃ¯~ÚE¥I,÷$b(Å&YÜ¯]ÞeÍÕßî¿¸Xª7.Jå¦\|STkjU\²f1j»««.o»¬ê¨]GZÉàÕ¦®Û¼åVYÿ?\_Ý(/¿ÏÊz¥qn2G½Ë«~§g6p#"LwÐnµ-Ð2]Í¹H¥ö+ænÌÛ}I\|øî~ð6é1áuêÞ@x' ªAë~¼ºÑ2q\ ½¼å}Ö wPHç¨±fùaOÆTl°QZP£àXþy<&¹ócnçÆÌÆº4F¿02©\XàL' oéÖoé#n Ch¸àÒ¸Y4hRÜÆ\|!&4{ÉÔÍ3ÌxçwñÑf²y$\.BDÀÕÌðGd¹õd¹#a²L¸;CÊ»ñ9RÞzRÞ!å3D19¾ Íh#X6¥Á V.ÿ@¿?Á¯�t°'#è§+bOId¤Ö0ë°Om#£0ïõìºjX)\_Wùê¿á?ÓÇ3ë(Ù~»e ð x®t¼üU?½8±Tpý@ \|/cúô«é7¢ÓF¦HÜÒ7"}8»¸ííbÏ^Ùzåûé²Ûä 8UUTO mê+ÐröÛë=Wòw+ JÕ[ZOt¡¢TJ^Ç2¢H?ÆàÄcFT#y1eÃéza&èzCödh�<r¥À:ÚÀ FÝÌ!ãÈH1FB"!ád,ØÚ÷#ÑÏ4FB½\_3þäÒeY·þà\pÞ\£bW\ù±�Xab²Ë¶ùh{¤ímJTnA 2ï¡F 8)Ó×¨éOq=QØÃ°b-¡W,ô%è§c¡^EÄXüXw¹?MÖ1A?Ç±lÚÎ#¼-Öë2¿æ¦¬òGZfmÇ )#$&"ñ\|à×ÌVÝ?ðµÓËuóW9u+Ö?ê}·ª·èÉáW6×}¸ñrßù!?v×ÔÙCQ]ÆÔÁÒ73æMl§æhs®-¬\|ì;qj#'_;bf'gTL¶ð=G¦s§o9kÅ,£>Õ×°�K®õe°S±°oÀ¢ûóýÖXÖäÇmÙ6¸z»ñÄJO¹»º¨:7l AÔ{\J´ITÉ6y¶öÃ ¦ÍË\_®YæmÄ p 'x ÞGhXãå= L�áÉ®®)Oi5Ø(Jë¼+ÄükÀOl(®Ajb»ÜYQqCÖ2ÉÛzöÐ&¿æ÷àÐÆí¾Ù5Eÿ!¿aJmè¸ÊöØM «hP«º)WÅªãVÑ6ÜÒ úûÿRÔû[²ª=äo$ýÀ¦)P¿xÊrkÖÍIù®-Ü2ÛíÊg¶S"YH¿!¼Íó-Ãð@òó÷CÖ+ÕªØ´\|ãùbA×<Ï XÕÍó5¨%¸r\,g+À¶U6WbÙqu]ç¾OU{Ð6ë:¤G.É7Fx¯ yO¶¢úI2 ²¦ °ëºnZnÙd¨X¶ ¡ãÁ^(K £NMH]¬qÒyr�Ù¬i® EzÀMµµwáIíYËßuYz«1É=ú½bèÓ>oÉ´Ì3;Z ×7ÆjÀ¼åJ·iò«0±4þvaM½½æ\A¬´µ<¸´Û@<ÄS!&°¢¨ëNº¨+nËH÷Õû\Î'v¹æ¶eXÃírB&ó65:rìnÎè\| ~ÖÑ1jÎ¾ËßôS)ÏX®°5ÝL¸.Ym2dhbt²ÖPzF#~geÀµðës?¨ï°©[/M\|¿4ÞæÀë @I®çy¹Þ?¡´)K£ÕQÖ²ùAX]¡kCÁÁvÏM/óGò¦p9P�CÎ±"mÈ_h@åçebÛæ;Oä~UÊèÀUÊ(ï=pÕÌ±ç~¯Ñïpæ;ßÑ3\| êyWCøVC¿wq]É@êQ\ul:v4Ã!Þ}_DÆMù 0! N9Öo�¾àÊ$Qo4owäC§AAÕ#¬ Àò2)g ÎõM¿'ö4¸ÅüúÙZ°tWËj¿}` öëe³�+dE=¾+±t4·¿�ãnB%¬×^Xä/ =ÄØ8!Ü3ðÞÕ ÍE¾ös÷±�~´õÜnÚ±ñ± Ö1ï À+O¨ >ÈÐRøEÃÐ å ýð±9¡qWH\}dÁ¨o?póüm8Ûp�GþN¼÷TïárRHX¢ç"Uêý({Ç`ÁoBháþ`?ùÚ+waÒ¬?zÀ¬oÄCÇài?bÏ-v^»"ÈkWêÈm> cHWÆlaIÊ,vGÊ\f×Êb÷²cáÐe+YlñÑÓÎ-QýÞå;Kl¦ÃwÉ0xtº1bçqª +¨ßlaf03.^V$»£ÁºæQ=SÓ¨n4%¬Ys7rK�®ì¬r4;g ëidØ7 È«®É«Õ¹ ÒãØµbÈ !U ÷[çpêëüßs`¯¥L\|\|~Þ:Ï¸Æµ±ªëZùÈNÖñ½÷Uª Ru¹4»ùXå<@`ñ7:X¢°4\|\_aõXX®C1÷ÃÊÔ0¬nð%jlãð Ô¯`£¥GåHÑ ÚñþPj\_òwGtc-[¶bÇ¼;ää[ÀÇà¸@1=þ�±ÈIWd%o\¬ò"1°Çy¨ êò¤Y']z £à@" }î»Ìj¨ÛC¶B\|\_Ò®xôFF/ªBO¬¬2ßi :7p¤PÊÚá+?ÄàÙjïºSÍjGC ó]¹·äV°§=ýõÀ¾jaGíã3yú²GºQ/OÅ¼å¬?lsôÕñ¤Ý¶ ò© ábË5" 6ãT3rß\ü°Í'ZÇå$LQJD±J/¡´Dþ{Ò3ä¿>;C^2;£§ü%Ò3\_êó3Fïí¡õ ñ;?ævnLò Yö³6Á¬ºî½3tÁ`8rÚ}CÞ=ÉÛ8IPx`1ZL¦.ÍÑL¥ôtuÉ4Jä)£Ãdt ÿ'c´pÏó3ø8þÞD-cIÌ%Lô\|¢Ïô° ýtÓ¿\ÓZÿ<Ê¸§ßï¹6þk2ª?\ >¯ã'}¥}\|M}KO>gcºdçÇÎY¬éá�=¥ÀMxçÎ1î>ÃÍÐC1 èÍÄÉÿ·T#Ku#ádOÅ#�;7÷Ð°ø~ X¼Á`ýê·xn0Ob9å³¤ÜBE <^¼Ø%2Jj2rñéu,Øß°õè{$×û©DBëI'ÜLÒ±êM ñûú>µA1N"ã+PoðÎàÁ1%DL,#¥Î¡BªÎ¢È3iPõRÈãl ¼³9 Ïz/ßxa\_¾ßÌÅy±Ö\|ÈSª¾]¬[èp¯r«N¨¥}>ÜkøA'¡V"÷õYàU^à7¢È²¨´¸XÃÆ"&\_ki®W9gPüºæÒÆéJ/¢eÅ7¢xÖÇþF=wi`T S¾B¬çBb-]dBVÞ«ÿ÷QKÅ°¸µ¤¡tåWõB÷adÍE0`\_tòFÅõªÈê}K`®Ýe¤¯jñHð]8ªT]GÌ-±ÁZ3órÀCÐ§rý¶8H4²(=í 1ªRñPþQ9æü#øÌþÏ·£a?Ì¯ÅÇ<6ø\_À!³Þ?¤·¤Ì?Þ(ÌÔä¬aËÁ¡'9~�[ïW\|29~hê�ó×èÝ] \|§Kô>(öÍÐøÈ9¯'Ïô²mwôÜ<)uÚÆËCÎßM¾eòÞàóÌ¦Áí(ýz·»låÁt åýÃ´ÐãCçòìþ]s)pæÎ\UHc¦}AÛª.Ël×æëþ[ülìÓçgcÓ = 6¹&^øõZ'-¬ð³>¶rRæ på°^Ô 9NÚ¿}QÓ¨ôéqîá@oÈzÙqJ¸rÿ2»±2< 1 ~~2Cû(}Á¤×§)AgÎà¢:¦8z´§³ÎóçØÄ#Ùø@ÁÍL¢]b5 Qæ ^Vq> stream xÚí<Ùä¶ïóõæên¸IEìÃÃ:¼öÆîNGøAÖ»ÓE©,,µúïÀ«QÝ5Æ»Ò#ÈLd&®îWéêW©ÿ¾¹}õÇ¯E\Wy[iW·ïV:¤^9¥4\_ÝnWß¯\_'W7&M×·»òêFtý¶/êmÑ^)·ÞzH³?õUS\_ýpû?~ÛIR«a 7ø] \»®ún½)ê¦®6ÅáïöpÍÉ]µÙq¥ªcPYtÕþÓÛ²®¤Yÿ\|%ÌºÜrÅwmsàF #ZÝA$7~Å/UsèxøÍ;þÛæ®¸«öUÿxeÍú¡}7$¶PèéÔ,ü=ÏýgøCÏIAWd\{Ï?ÁWL.ÝHmÝk \¢-¤A%³sT&Qv$E`R%y¦¢Ã¦£3Yb¬}`³©Äk`9Ô»Ñ\_Þ\_»Ë¾Iqc\_íPEt+ ËN¥Ù\|ùEI\f[ïY"¤}¯-TÖê%\Ïwt¤°~þ½Ù "ÅLú ÅÜ¨ø´\|\_84¬òÄ-Ï7IÏn¦<¿S%CàÉX-c «T%Vø%¢¢ xb9:[XöTóï¤ZT¬ãq\_ÑÑ2¨~Ò¥,&þÖÔ8 L~Kz¦pûøÓºoGPOúüý¨+hg±MYn;ÎÝ=zèþtOj'æ\TW¿üOÇ²åôÏ¸ý©ôMîìì9c<ý;ª\s ²ãÉk¼.èrÐw[v§}ß% þÎ×nN-'&ºÛõøg¬ÒôaÒæ(K» xÊí :Îl3 ÀL×ÿÒÝ n¼m¶1¶ËÈÜ\_Ç$k3¬ØøÎxä6Ý¢\SC78:Tµú¸H®ä\ÙÜÀ AO%åPf&3 }Éù4Å"d q´!xÄ%.=+BÐÃ®"Üfg©¸råb ª&u5´ ÆÄÀtkÑ,áöð&ÊScG¡H÷ö}GbqÐ¤²·M¨ü¹B9ØÑv>¡}ÙÇFb>Õ# o}ØWòç\_ M©¹u£XMjY;YðÃúýS¼\_ BiR«ëyçÖis°i+IÍ èµp[Q¢ºëqeÔÅ µaÏ))Ïî9ÌÕ¾°åì¨ÛÝr3E"IG©�¼éXnúDK0XÆ&¿ïY&©L²t"ôo2Ø]!È2&9¿èa¶©e)Á;(( ý©;¨A$4î¿?\³F\fp\,'S%éÿaêBsÛTÒ§ôsûåÃ1 U\_Tµï·ànîØAÆ& 1BQ¼¢ÀOë�ïýmÞ¦ à?Oë"Òð·kö<¿Ø±Ô&Z@J%Z\î-Hß CÒ«¤ëSåª#3ö3û\ªe>¥^þA[~¦ÅÛÄ¥3Ns&ÿ²u³NßD¦i ¼J\_ÀtþA¨[ÌbºgíúSRýRÆ1§ÀìqïEöÏ¯A?Kag»lÿD~õA\_�Ð&Æ¢/Nô ôEPÄòOx8ºtK6yÁÃèd'À³ú@JßúO"[Y##6/3¬O±tÍGéF«³üwu!ê³Ðùo\_óqËò XsrÊxba9½À²RåeY ý®D(«>¥ ü×äY0b=Ï÷T9üÊÆlô"`É¦iQÛnj´X`)^÷H `[nª4Rûèü@zeóCÜBBG{8´e±í¾b'çàäp¦Õ¡3 &v0�ìn¢(O/?é\_ûê±oèG{¡Öo¦StçµågèØjÙèòòÇpLÓá� ûYèÌÚÉÕi SN}ßñý©ìÐ?ËJ<Rg£¶Ô®\_sgÊñ,C5~ÏIRáÓ`;:sS`:SÆ:á:ip8n¦g)¶0f9( 9Ú§ÉÓ6Zð£ÙbK.<(!êô¹Jþa²Y fÍ Y"&¤\Ø �{Ã"AMVNü&tÍÿç{\ÐÉÇ\Näëã®eBp ¨c6C1Nªh¤,P»îzà¦\D®é\_§w¯oõ[»¾rd~Õ±!6D0¯×¦õ©æ®ÛBÛæÃnm&Yc\Lh[iø»ÙmõîqÑô¾¹¯6ImË°Rw%vÄéQi4fA5:åeX-¹b0· t"Ò´·×O§¶lC0È awÚp¤H¨öÍøÒÛ=ðÛ¼æ ³mHlKÉuéÿËBêî¹ÕmR¿ýM:x~íoÎJ\%ntÒ¼î~íäÔ¸; 8lÚ]F<]S QD9zøPyc$ÿþ\_³àmÅº«(¨C[ÉñB�Ä V @ÑyAqÜùº¡3¨oÿá\|\|%ÅÓhaÉAôÈInX¸u³%¿ Z4aqwMËë¸Ù°69þàëÈOÉr\_�;är¬¦uYnç&²M±ß`TY¹À7ºéD6uÓ=±j÷IOèç¹qÚ/8`ªi cçu\Xµô DCs5©q2ÉBoYEË±÷þÛñ·n< .7 0í£×þßûoUÈ¶<ö;NwÍÉÿ¦½Kîs¸4ÛPPÚôÜ`XsX »Cº!,$¤Á#19v¼eðvó¥Yx{¼·aìíD ÷v& ÿ9û$Yá³9\|ÚT¾Õ`åNgþ ¬Ôû¤�Gm³=¡á^P\9ªÑa `§»ËMÍà®ôà¿à{Û ñ¬uðaäN'N êðyøóa~Åí½fÈ-6sêeñÞÁUÄ¿Ó4\ðìA¯ ÐèÉú>Ùr¯a>b¼ÚÓ':³kÉT(¶]à¿ßg?$³%¾ü>Ñs2ï[ °\ýê¶ sqÑ\|täóg1nKæµÏ¯¹eCe"qÎd1Ë³\|åP�½`{?sì'±eÇÌÿUãû5l CEÓ CP{6}{±DAËÐ"¬�>Ä0µ}Y vGqïTüà«ÿxCBX¯¡¦Q K5lTuR¬jü«Ö%Å× da{ÐhÚÑ© ªU}À[Tæãd5FñFéÑI±0À¤¿®ZôWÁÆÀõfÆaK:Uó¤ã d&©øþ vy[»KÄ=ûïì§ðÔAa½:ËT8¶fø ëÕÀVÅñÎöBS £hÔ3al6u¨N«Ä¦!¦ÏG2DècÔÔüdH%d%ª·Z¶¹aÉVåFSÐÁo>`î!TeÏm×¾øÉ¯)¯ Ã#Ü¦`ðÐ<T5ÏA ô± \`Øò/w¾¶SÓ#uÌ q,k2H¥g tÕTNçÙ I5YKH-aéåÁW[)Æp¬l¶ãgÅ/´ãk`ÚoÉä$]&gp1á9¼åE B,"/fùt m¨ªv O\_sápVÀzòÈyX: A90}¼õJrÈßKÙp æóXÈCiéÀ5üYÈùH$jOÑXÿ�4ÞQÈ@ç%'F¨ßWì {{Ù'¦SX`äÆ27\|fé:t8ØgGlË¢jK³\_Àÿ²WíK»D§Æk\_&f�C=\Ê«AÄ'½¶Bv,\_ëÁÃfåTÿoý·à~ï«löMÐÐë¬Q1îN{SçüÑ69ºþßÒ8ùì®s+'Ûh(Ûz¨ì¡d/ìb¶1?OT«=]¼1±E\_ ÉVÒ[ôtWt\|¿ZÁ¿;¾=ÞÛÛ{ÿ#Î~kúÝØfDÖÐe¼(XfZL°ãÄgEìJÀTðÀQdÝ¬Á2Ä@ÙRë&=ÃÝv©Ø$¥÷ûHï6 "�"D¡VØ <øzÑwv×\_gçlÎHÈ}@·RsQÇ\_ÊcßjÔ{tò±5¦&·ËUAÝÀÒ\ª >#o²¨×ÿ\´ûP§ØÊ~½ºéÉJ¦äàõ<@òo<ôÁä×ÿè+êî¡lÏîªSçU¦WiµAÓt¾p\_9{ß§ yð}há×rSº®z]w:]F¤f\ÜÔz[ZÇ¢d È ûA 4v³@Úhã¶n[Ö~Äx2½.+ZåGK¬ø8£ÕbdFÎ¢wãÞQtíë5æLXfr oãáÍ37çU>FÍûtÚ[ÐRgÌcV÷8\-ÏÅw¾ãäW}À9´á õ$]AýðØÁ¸çIÑèª;]ø¾#w$i:Ê»v¸oíÄ¹$ÝýìÙvËGUkc1TÝûcW Úã¾IÒê¤ q ÒzÑn@Þïö ðPªì«^¤ Ñ 2[ÚYtæwqcèf]÷Õ%È7Òû\|C §¡TÌT@êB! Öý#ÆyÑ¹B¤ß,Ä(©5ÈáÊ3¦FF#ð1#B¦4½!XªÑ2à]xýKT× n8#ï4ñKNèøÝô}áá+K>ðÔ$@Jfê% Àhn·RVÃ7û(Ph ¯§>B±Y²Ù!äBk;J@rç¿\_"Ðì\¾)wÈÏÔ> Ó:«&$«ÐÊ¬¿åÖDÓ Ö8,%µ^yn¤éQ õ6«0á%3.o4Ã pÄ$8'«,²Q««ZÚýÿ²º~þ»SÊV3+d�$Ñ\ÕS¿Ký3Rãm4JÙLùï3Éì .>XåR]D.AÿÏ¹øBSjÅH»üÉ/}m¦Á ud"qBNf(#3tybõ%\|Ñé[ÌO<3ºÈ .³±þýéXMk/¢Fêß-9·y@½J(Óáþ5j¡Ï e+çúÛÝxlÆØQ¦éÁ¢fèÁø:^MAû}íëp¸/G²i/t±æ®iº�Ý1Í+fC¼äÎNñÁòt#jö"Ð?d± ´ ì¢H{9¬UúÄ$ºe!°ÝêIdíM^Fá¤2zÁa~s ð!ÌÅÓwOç/ÎÌÜº\õm¡rX¯ÀO$è}FQê¦£¾ÞÇ±Ó\_añ§ZxÕ7þmÛ¿VªÞ¦O(¯G\,6¨4ìÎÖsá]b\ q ¾~À6]·ôÔÄì ÃA¦ÜuÏ9t@1Ê? àe ¿®ïÛ²ð îÔ«7¿'?>$Ê-¾£XWLÝ\_É!6¿m¾iÇÚçd\Û(1÷½üyOîÎDÏßÁÉ»ïÅ`( Ã»ùdÍ#¡¿AC&O-:ºãwm¿C\ÔÕmÚê-ü\_+²ÞÃG j²Ãu º2ÃI¤Á¬¬ ï^3¸#Ý³×Kz: ;EÛWÁ Ú1¥ qý}y×äåJý#X]p¥!úJÐuæWä5Õ¶W²1Éïe#+ù±9µ°Nª9ñËò3í8{¹Ó¦l\_nKÿë$ØÞgMë ¥=ä9Å¬7ØTÑÜ£ã.<È ¬1þ.N` 1øô1ø\-²56Ì ª:¼®Ë¾cz³ø6N×6>\_ö\ÎnjMQü½üð®TÁ[6U·¥¯7øHt×6\_ÕÐG2J¹¡ØNþ÷ endstream endobj 124 0 obj << /Filter /FlateDecode /Length 5646 >> stream xÚÍ;Ùë¶ï÷+ôfuCl\R\_;vîTÜ;rüÀØ§)Ræö×ÏÙ@jª}íöçE88Xà�gW¼Ú¯Õ×ïb)ßß½û·¯òdGy¢ÕÝÃÊf�ÚUjLç«»Ýê»õ¡¼¹µJ·EÃÀ=!ôº+Ou±-w}æ²àXlUSn¤^?Ü¨x]lAÖ]YìdXu<ÕU0×÷wÿK¼U\Ê] iÇS·.î«º¹2´\öÕq¬×&ëV èý ô?JåÔµû®8¡Úuý\|¸uÄM.KX÷ìã?þÐv4ÕBGm³uûÀ%} mÛu%î +}[Øw3uê¥[7ÖeÏ`Ñ Ö2îpóÔ·ýÖéÒ g¬££6V6cµÛÕ%#·ðåcÃx\ã¬ãPÜ×l/ÚþÇºëÆúõsSÑìÖSasªdTÃ©·Ó!a -�¤Z×Õ#Þ²®m»ãÆÓ¼/Ë¾\üMr£ãõ8,\·ÆSÉ-ªÉÝº)o´[ÿMeÇmÇbW2D#þîÜ¶"îw¨znôeÁE }îËm1öx±¡m¶åòh4]ïÊ~ÛU'")O t\_·ÚÏnÜd®5íÀ@ÛÔÏ EÿÌÃz] DCQbáéÉiM>Õ,\_FZ){.OmßWtm°FÏÍÀ/=P \;Ah¹tD~E¸YóÕò©¬ëú»$J³º5IØ}¸¹>á/ÎAñ%!VFy+de:TÁwiènn½þ¿`µþ+a¾¤ßóýþ×4ã9:é,ñ3~»ðÍ$Mê;ÜÑ¤W×ÿ¦ff}ANãFeë=2M½,ºjÊúÈ;Õ0DLÙjû+jFÍú¶¸òyÏeÁEWöc=lx`N·KÜs\¡ãE }ð6!x\_2LòH}ÿêËiÚðºS&ä;²¤5ÐÊsx!Ï'Ê±<´~^Å0À;Ò°Ç] ]óõ®e,H'Êø^]h°ëHpE±©VÏ×±àËÅXQÜ, HÃú )}}?ErG;ç!Áq#¶ê/'] \£Å¢äë=ë$aMµnT@ªçi.îYìUè¼¯ö#\© x¢Nt´�Ë¢õ·»±ñZaL¨¾v;ùPæ?ÍYÔåI§vb"7ÞxêÇÅ®m»¥«û[ÌÅ¾-u ¹hþ%qpCÒ¶<øÛmÙ�kûÉ&rEëÈæçøK9mÉ%ÀGGhÝÛGX$Ù¥ ¨�([ð:ê§rÖ==«/d»ÔXyðÃóLÓÏ.HÁ»¸2Ï®å¨Ið$#L×ûv$GXOåY½#ú1sMµì 5T átNð0Ð¬a%³Ý¤d`]3¨ì½1Ññ\S�L¬ô&f}Þc8±¿úhÇ¡/ÉOýBòc½\¼[bÔ6 uKÜ V#µ?½s¦Z0rÍõØ0¢©jPÒ\|XY\_íX&Ã~aI ïY9QÉØÛªgæøÉ<¹æd¤·¬M(.8rðPL>¦°Û»?Ý½ûá¥Ç+µÒ\lfV:VX¿Ûã»ï¾W;hÍE ^®¨ççIqT½úöÝ°Gsö=?Ê]d2Ë\_4KÖ}¤¦³F^ñw°Ãü©Íæ\uQ%sÀg7`ÏÎÏ&ÖËgi�³ðhôË£Q¾RMeydo8óT°mÿ³Aû½j¶DV =pó2±\|GñmÆàîeàáñ«µjEÊº ÿoC·Ø=Dg FF©M&ºö1úo÷(a¬Ú±ü]Ø¥záüHfìuÉéæ ììL�c1Çµì7R%Ã:ËDÔàW�Ç¹ 1�x ¡Éº'tÃºËÑd{fd6`?>ljè^)eQ¤MCyO+É×%ØqÔîÈÞBg\_¢Qç^¤s0$(öX\ÑkøpÀ,é[.Y^¡8èUæõrð[K¢Ú¼/þögp8´øaè©pGüF#\_%¿)\÷mþ¬tênLÊ¾ÑdcÃZÐ0¬ñ×ßj#ëplêPídÔ´Á/]YÉÆ"teÿ-mþV¦ÀMêÏ¤ÞD±ÖìÅÌ³K3áì@¥Ð-Ê %Ô®ôy¿hvÜFXP=}¬mkf i.6²w_/Ow/ÃþR¶ ºO²2ßÔ-XèÅÅ¢ùð9oÎZØüt´w^(ÙÍcAß}�Ñþó&ùQ?.Sª£Ì©Ø«.ÖqC©xèwæûhFËòÕ]ä,FÎ¸@Y\Ã¨ÒYÆD¾ªA4ªUjúì-jIçQ^\&%0UÛCAB wB;ÀÅCJç0ÖF\ùù!a¥Ú±+aà,R}qá/íe89 Wh/»EÇ?tÃGÀ9)±É¦©Ð³(a¾ D¥2Å}\AÑ§d¾½ñ% æ[4é³ö3[Mt3b6r/¹1-¾ùA'Û'.Á¡GV;MäMé%;Ä¥ìßª,DMòÍR]H¸áÎ ¬Ùì[XLà;èçÆ7Üå>½Å3Ý±jÊ©·ô%KT¦Ó<+éH W¥xXsa«'hú�ìdçö}6õN~äü +ô\éºanrS%FJkW£=kX(7àä>àh%QO¨£#ï6¶@1YO2DH¤±lc"p:XËL¤\I;ôa²�"îÑõ]ÕoÇwó(<é±2W£bX6ü\¼{Z(ÿp[K ©BÉaVtw¾iQÁÕ ôA³Qýô©Ï:BÎ3p¬URRÎAÛÕS�½ûaÊø²y\_�¹h~Ô<"uY ½{.Iþ&ÖPH¬éÉ\_ùØqì\ÆÊNBÑ¤ÝHæH¢¯Èd:X0O¼Ïw 6ÈûQðÊN¡/ës'þbH·B P¸Ðökú§åè·/b <ÖùOë$Ë�Ðw üÐ ²Qç(TìÙÐyúÌPê}·Å#y:ÜptÝÍ8lÃûwÆdÏü99ÊêrÎÜÐ´Çg¸Y;uÀAöDß«\|òê¾6Ià& �c[Ì¡áJÁÅLcD¿·¦P9ò´ÜK%a/ù6½×ÐIÆþ,¹PX3IÇÉÐðª"ôÑô%/VrçIÞÒ ÷cÚ -&KKw¬åÛ(V¥1ÐÌv¶Ûx¦ùÐM9Q¹ÇÜ3ýxÆ^æúU¹¼ z7ÊÌÈ¥L@.\|¿ ìf Ç´°gî%z MÆz $Zð¼ÌX²h²éÌ¨2YaÒ?rÈßÉÎÄJpPu¨Í<)F®SõÕ×ÏoÌ;@gg@Ã\^ËªÆpèµhv .þ^¢q» g°ÔèjËs~ÝáúÓÁ#¾iwe}-J/%!Íým»aç9d2èb9·ªSðmÚ¶f14CÁ7Nä7$aªþ¾¥½¿½Ê ÍY°M÷mBüP=ÒÖ6\!6 åÇ¯¿éBÛsaXUeæV ¤üÎ}ÿý"å ¹ sÓÈ§ÅëGf¹4(i©n¥É1KEîQZäºäÁK/kó@@�¿¥è?²6ÒÔ±ÆßÀÑxë�MÜú55øqaz¬dôí-¾(QÒdh Où001²O1!Ù¤>×¶æ¯»=ûÙ·(cAçz¤kb.ö¤ïçIì´$îÎIý¡:õÜíÈÉg?w dT^ç°$1Ó»gì\kHk P¾ÈÎÛ0þÇªéRÃDÇr8´;Æ?,GòÏ<àcJ)&ç6Õ5¹)¾Ü1vvg³²È´¤µfzJtAð»¬é 3÷å}WàË¥NCÁ\¹.1Wúhéûü¤qV4xÅE3ÅÞK5Ýµræ2¶°Y/]{¢ê¹×9XbÓ}^ ¹imîÁ(�²Q³MiÊNbÙ ð\Ú6ÅÄ[x³[Aä!}W¿[ ¶yîG¾]ÙZ&[Ë{�±þzÎ<ÊÞ ôîXîñø?x?NæÛÈf2öåDÑÞ-îm]Ò�ðRcDÏx¿0�¾\|A¦áÎ¨W5ÜÞ$(G6NY:"ò(éÓJ¹YvyØ9äy½ïWÈ)óqÉ)7ù^½Å dK("ÏXãF(eÕ1Îs¾«,ç]"_ \8Nð¢ ¿¢ÄC,J(¿ªötÔ«ËÒ6HÜ¾ªÀt\|¬p> ÍºkOïY.ß¥½¤±çÅ:±!Û'>¦ «k1°Ôp/r±ÃÈ¹îá ö'uÎ·néDüÓáSI×ÛªÛ9æôlD·Ñ§Qr °¿wKêô»n¸3@}5ÕsÀ$Ò}M©³ü° ¾Õ 09ô±d1§)ç}æ4¦îe2ÉÝ§µ¹Bî3³p?ÛnúWU\ãnWIN0V¤;¥!#¢?äB´ngB¸× 1H>)²)$4?¤$>u3QS¦«òÚ!9}jþË ³^,?åËÚ$ñÎÇw¾¥/ckÓøï"i&NÝöÊCèAñôffiAVô<±P/.¥4KjsÛí\|^t0WåQ[à.\|óZ¨c(°E.Û¯Ey>R ÌjÒòbbÊ·ÅÆ'Nd²Ê³òç\|£Øj§)$¡´ç 21`vòúúie®Ç)Ã(«! FÚ)Ni\| Õå0 @t +Ë"\pÚåWS!lGZ¥+!¿¢ÀÐs%qfE}ïüøQzèÙJua\|ra8ß\| qs¤3ÒÃF¹GÆäÞß¸ôÉX«UÐUo»õ©QªÍëO 0ï¯c:1éÍ"Üý×Ï~¶ÁEx¹36$§äè ÐAáÌÒ<¼/ Öà?d¥phö-fIGI¼&lSáv4}Ç~é¤Ü¿µ^åö7eö8ÅÿéýXà'$¼ö·ßi¦¹/Ê·1E iOmðßÌhøo)±taÕ<¿ñq/è-¡Kº¦ÓÿI¿&BAAE¼p%eàã0èÌ2\|å»=z£¨áfÐm<ÍngaWÿ Wendstream endobj 142 0 obj << /Filter /FlateDecode /Length 5316 >> stream xÚÍ;]sä¶ïû+ôæQÕ »;YÇ{å\îÎ;UGÍPgccþýõÀÅHÚõVê^H°F£¿©®®ÔÕ_Þ(¹}ûæßùU¹Î¯nï¯lM{åITyu»½úiUWÝõ.íªêûvÓTC½ÅçlõØ ¾÷çãfhÚ£�2®¯´÷ hÏÃ¦=Ô=C·u¿é;õ·ÿxÝ¤iRf9ý®Þ·×¾õ%öÌýýßh7A_«Di 3$ÓÏð=¾¸Xfä hÀ×7&S«Û]Í~è×i¶zxbH}8©Z=áÒr'=iÓºªã¼ÑÆ\þ#h(\_BtÛ¶ Sï£ ö¸kpÃæØU2pWõþúÈÓ¾ÚÔÛ×aÍèÌ°þ c µt¦Àq.) ..±¥J#k³uW1÷p®ûÞþg¥t×iRµØoãQ¦/oDW#³þÌSWûÜ²kûáÕïÎÄ ·à·GÄFùN8Q£ÞïãÏ[1ë=u-þ°atwÕê ;0¯¿#EºÚÕûÓýy/À£Ü÷ÀTÍ±øôoqÁ¾ÚË{@ßóf8wuÏÈèDF²îÚtUÁ;ÒðÄzÃ®«Ã6§¶ïä(êlcS÷?kí~VJáÊ<@ ð u}A\¥wÜ²\|×ÜÑÒòxäÏÀØÉMÀá! í¡»%"ç»Ç½¿¯» ÓÓtWaQç0¨º[lúÎ¹Þ®«øöÐVÃZX¥CBÑáà¹iç¡cÁ¨rMXR»bä#\|÷úþ\|àq}æ¸i»®ÞÈ«=,×o !ÊF¾Ot¶¹c:c%Î^6»¶íQ\|áCïÇÜÉvÕ4§±0>WòRÛåÇ]= D>Ùqc±»Ù¤7ÜîÎÈ\Ø\|¬ù¾©ÜØÖÈVÇ0LNJPëÏCPÀiùx2pÎ][ðJÉIa}Sd ²^Váö/ù FMõlX¾pAÓ&eTå/ÀÁé Ïeº:ý®yÄÁ ]!T¹Êd§Èg]ã16sLqôqærÌ& j.ãÑÖíàP ~=að[ÿ&23ó#hlþÖí \|µ¸¿kØz³"íûQùÆ@´¥Ä_píeùsÒ!UI:ä^àlwò®%ñÓnRgçá%BµÐ\| úuÄîE;,+\_2·'1(Ì$1¬qÒ<ëó§º@~Ø@¿ÑX{ÑÄ^3êS+Æ¯«=?ì\|»öºîÛî ÉTR,£¨ÍÀ&%°öF3©d4³ÚG1``ÌÀQL[0I0å7ÒèbÏ4°H±¹%!Eiµ±o)û=,+Àáe~Ëb '5Ï L�ògÞFc6[¯Ö$ÞÉ®Àè~áØTgv±Ñà ÝçLbB<)Ì÷¾éú¸©ô¬ð8· Hc½'¢!ºuÍðt! ûëíñöðÎ¥JO¶NQÜµðã?¡®Ôê-×Å¦\|èð£t9Gl¾X ¤Ò¬ßÕA'[D¶åû40\²pÔwÏwú$6êÑØÝÆ&h[x:$)o¿Zðw8-Èp[Üe-160$×ù4È=Ä päõî ÍÞEÙl²Çæ!Þù£.¥bDd ¬ÃüsìÈ$3aÄ¿G$^æ yµpýJäè×È+ fü)2¯ÑNÝs´U¥§GLV/®HÍ")í"= Õd\|DÔÜ"ç6 !<îÃ¹Ñ·kn<ÊkÛöø:M?e3¤KLSõçªP3&#¾a"CXS+ Íæ²ð«=c§P}2¢gv²Õùtª»MÕK°t_¹«ÓPª÷±ÿÞÇR%¼àô"å[Fx'åMK¾àõ©"´ÁC¿©rÜ"8¾å»Ç,U³q. ¾zc%æ¢-S"%ÜúS½iÐÓÜðó�Û¯¹M3Ãý´'Oº\LéX ?}Ó) v(@ü×!±EËàE~!(yÉiÈö3-±bf(ÐwlQ¹ñTd�-RØ;òüYØÄÏÍ¼t!·äÍWwÍ¾²þÈ¹'MÊdõ ½å°àçvXGÌÇ×ÚìÛMì©ÍücÐg1îÄõ>1¶CûãN^"?Üwäj<7%´²³è¥w,1¬²¯À BG!ÒE:räPíFÅØ ¬'lHPdÔ¨P ÏçõX{ûæ×7(cÕUzeI´É®µ ò«ÍáÍOÿTW[è'®tW4ôp£@´¿úþÍpÆøÚ(»KbØ,QÉ!æ,G®UaX\|\_ÀÁ¨Äèl-À¤NÍÓhï Ï$Æ\£¬Ë¤y¢ýÔ( êÐ©,%ëÒ¬¾úú»·¼iï~ðqïD£yBóßÏãòý4 :àï×798pM¨Øþ]ÿJ×·ÿ#võÍ¤{Ìµ¨oà¥ùG{¢tÊpÍ;öI\ÃH¬ùX®ÉÁÝ0îÕ'qM^Rn?k@«õb~50{<ÃÒyi<ÏäI ËþòLaç<Ë2X<À9d%ÖT¹7© IÕ YZ3ÓIæÓR¢ä~á!>hÿF¾éÅ ,R ÁÀ\rû\ÞúÊ RUµH~#ÕÉ3Ç¼+V%ª\ä B KÈÝù2çSë.9\_B¶æþ§P>�¸½ Q[sOÍUY5û>$ãØùóÞÀ<v¬r Åp ä¦×q¡à û&oKBübìõ\æ¨ßP<Ûr ½%µgÊS;0²c-3µ2ÅÐÇQ¡s^ê&Õ äo{¥Lºà s¶@~~TÇÕÖþ~Üû<{±Ýxäñaà(éÇ4åJÃ»À�v\| Ðút�9÷OSê×îÚ!ôÉ[ñð÷Ñx¸)cP}^Ð4-î¼Ä\éååp:2¥9sJÎ{÷ eEw.Vûêøp®jsZi²iúÆ/Bª¹ºq \§\|~tJî9a .ûÁúje-Pé¥t/¶æ¼/<0]&hØeJfH9ulJè}5ç}L°ÀùQ/&³1Nõiü70û¸¯\|÷¿T¾ÀY¢^Ç¤Ö ÞÌÜøø6n\|Ó[QR¸y/¡ÚO¥F~°\|NPê$uAì¢O6J×¿y¼KËJ>ZG#×z°ö9 'ö÷§X=)²ì÷þåÏ±xøMZ£<+Ó ÍÄ§þ¼Ùq2pA§ :Ì^BNáiåjìä5¶cì Íëd9§2ÊÀ)ùTBCï¾s}Y09Ð¡úpGÙ%IÆGÝ Êö¼©£vÕÆZëÏwCË¢ø%�8D¨3AÑ£áô;4H¸e¶Ï3/ô<N?gôG$½åãGåKåx)ô À»'G3Æ-ãä^å«óQ+&BhV¬am¤HlÚÍô²¯s¬¶°%ÃEË_\ÔêÇ]HÏF£¼j¤¬4/¿éx-+²¹òëd1ÍM §WþW1JõÉAôe ±68êØ$æ²±ZÈknJÀT§[8øN%°PÊe'®vÄ=®0Á¥0ùzúbxÏMþP>¾Md'²"IGýrAæ_(C#Ëa<Úã^¾Ü=%Û"ùºýs.Ö©Ô';Ò)Ù¶$Ó¨rRºÀ¢~6÷þEAdÎ ïÝ5ñ> Ê9©G92¤¾Éy%¼rH0R^¥ò{Qòßt¯úq»k UéêB¡¢¯M3©ñÜêEûîàÊ&¸VndÝhõ!RôàGDª±ÙÕ'¬Uéºàôv5j°1áO(Ã5üÿ{,¾tÕ¹BcË÷pp5ÓáAÒã±Á{d\|:Ï &ÚúWJzïB2x\| ûl\|×úõJ#ò!á[?÷³Ù·.ü 7@=úá]óïügLåÅ4S+:Åí¹#"«1Ø}/ºx¦¯éd«t{äuÞ©(Èï~Ø-±(q1¿éÈ?d7xc¢;Ëk©Å·sAP ïg·N¬¾w×÷ÞÇøÊ¦Ñþ¡ IvÿÏÍèÇÒ9 =sÖc5t�/d/ÕÚÉ$³=Ãÿp=wÕÒ«¦Ub}Á{áO¦¯$¬ôoádp·üú.°n nl$±CÑ%Ö¸E9Ò\|E¦Kÿ¢Ì&ÿ½LËñä¨CêÌÇúsî>çº{®ý×x®óÀCô¬á/ìéò¯÷ßÏ�ò»Ö'Õ§ÈtZ'ÙX}!¼2Å Ä<ð2òEô9=v3µp>óÏFMúÅ°À_ Ó2üb8§'fZÒgéiÿÏTlrL³í ïÅsJ^(Ùì··oþÝìendstream endobj 151 0 obj << /Filter /FlateDecode /Length 2887 >> stream xÚíZYsã6~_ÁG©ÊBpÈÖ>Ìq6ÉÎxÙÊähTHjlÿûí@Ù;[ª-W8}�\_£¦ÉEBïÑ½/\_°D¦Äj(ÍÓ¡ÄX\¹¡ëÄ "\ù;S9 NR%Å8±©ö"^OgLi>¹@^tÃ£½N:Ú¨ñÃY6¢«¡_'QÏ(hë~¥¼tÆ£)áJî &ÁÇÙÊFHIÊä¸û¦À Åºµ¢©Ô+ ;Ãý¡³RÏyÊré¿zù»9f<çåÞæÐÛ¥t'í£¥W¤×æYzX rqT c¨ _á¥mê&&ã ëfyáS/ô¥cKI¨TO#¼DSåyzyô#ìÉñr ýrìiìå@ö^cÊªìôÁñe=Ñcé&«DRF¨N=WgS+&Ï\_üðÚCÍÏ§ó°ñ¡ÌïýaÀÑ$ìG ëük:h8\_BBù½û}ç~\_»ß° ì;ÞDàA ÈÛÈÝ/ù£sê~v ûe~Pæ(ýnæ"B0pöçÔßWq3©êFÇ�ö«8Ãû!¸Ðh®àtB àÄ<ÑCyCÜ¾Äý«ÔCð[wô± iD"à·VH¬C¤cÓH<Ã#ÇQpóh9Zp0ÜB±éq(ª6ú! \('Û'[�×ÔêPûêåà&%F>¡ 8×)¡ì6x øhá²0ôgáãA÷ÈaTTw(xx¦á´Z;:Ô¼ço¶�Ã#}Æ¤ònÀ¾¥lûÃõÀÕægÎEdAA, UóÑQüe°ìPo#kÀH©¬!£Po¹¯áé~påçS0öYoäþwL$ Ëõ"Ä\aw: nc�\|+vc¯sa#ïäè%;g®÷ÿ3i¤áxæN>SÊë¦õEÖæ¢Á¯d¾ÚäaÄù¶\´EUæb]¬²ÚWÚ\|/³0b$Ç:@äf R}xQØ ±pwfq¤²ÍfUäK9× \¬ ~Ä-¯uÈ ðeÊÕ$[msEÉ¦®ÛÐp½óo t·%>ExJïUV.cd�Ü;ã\±)0$ÎqK+÷÷Í$RÛd0xÞOË¨Ì³¨Ê X°!®r¯.·¼=Ô¦Û¬0^µÒ«ÎªÉïSF'y¾ :zEúê@0'®GÀ5ÿ³4�<8 Àê=íeQ/}p$7H\%¿)RLj0^µv:Ú²ªÐ3ÏÛ« óË {ó£aÅ¨KpCàÿUtg:Xþµ\|Ò^ ìòB'3oPí0cå!ZînTÑH@-Hjúe¯ÿæ®ÑlØGºtíØÑ»dê× 87éäÛhÞ0Ó\_rê¨Jôä¿ÇHÀYcû©AG)¸ßïôÀÙTîH×åg¬£1p/q¸ËªE¦3ôÆïðú øc[c�Sng]ÜDÓ6º"ÊÑú±j]6uYz{Z.³)§!ûàÊªõeÞ,êbîSÔ% ä(?5¸9èãI6êfÝQò'/Û¬X¡ÞpF\ªrå·×¾Ý=øæ° ÇowìèôU·¯3obY5°�3Ð�\ËI6÷¹sg«ßVÕv©A-Bº ó¦Én¬²Ìã.fQmë&÷}u¾ÊÚ.G#F¦Û¯93®ò5ñÍ§ou2¨çç¸_Ï²tnPæÇÌWÙwß2¯®}?uëÐÉf[oª&ªÆ§q/Ö¾Ð§×2Tñïª:è¬FÀ}%´ßÍÛÍ¾RÆÌà¬³¡1VÎ]Ú}ÑbîüíeºÇÍë�Ã«6W¡Õ¹ÙÉ¥Ç/hÞnpà�ß]oWaÛB®îß/{ÄyÎ~Ñ=Rtÿ«ÙïUÜ¶Û:LÍÝæÎmÝ«EP.ïÝÓ³b½Îátéq tüC¨íhR\|ëYäAthþS)³Þ³C Çµ·=¾nY¿íð[÷¼Zl> stream xÚìÝtTg~ÿ±FeE; ÔØ¬YÓbÖf±1c³^0Ä!ÔuãÇx0ºCL3=4$Xj(RP[¤¨ ¨F5 rF3õÿþõ=Üs=°� $ñ~Î«ájæÎÑ}ç3O)-��µTQQnÜýá®Óé ¹ýÑS��ª¢ ´,aº¡÷3��¥ÁèÑ£ÿøÇ?÷3��u÷»¾¾¾V«õÐ¡CzgII 9��¨¢Ò²,Xàæææáá!·åÀÅÅÅ¥÷FR¸d��ÀEIIIqqqQQÑG}d±XÜÜÜVZõþûïkàhìùã?Ò·��Ü_QQÑ¢E´Ã¿ýÛ¿eggçääÌ5ËÇÇGî<}ú´)Üýá.9��¨ödÐaÅ¹¹¹QQQ×¯_I&¹ùî»ï4U0rÒ²î =�� uãóÏ?wss³Z²vãÆÐÐÐððð©S§jqæÌùaa<}��Ýn/-ë°råJaùòå7nÜÛÐÐPÍ3aíÕìp8\��Üýáîºuë<<<ÜÝÝ/_f³Ù¯_¿{³Lddd\Ü¸qã\|}}ÝÜÜÎ9£¿®ëP2��±bÅ FÍ3'555666222,,LC¨¨¨ÐÐÐàà`ùQî7n®@qþüù¼¼<§ÓYZ3è O��Bw¸[\¼~ýzOOO9sæÄÅÅEGG_¿~=""BnoÞsÃDv;v¬Íf+)ãt:d ��ÀSEûlÛ¶MC fddDGGÛl6#R¨0gKJJ=z´»»{«V®]»VXXèt:É��xÚlÚ´I?,\¸0333..îúõë7nÜ°ÙlQQQr[YÎ9bÄyæÍË/ÊcjvÁé��à©R\¼çÛ=îîînnnsæÌIOO»yófxxøõë×£ÊT3è>¡¡¡ò_¯½ö<_ZZÚÝîrn��xÊ.F¹k×.OOO77·Y³fÝºu+66Öf³]¿~]²ÔlÁ3É>Ævbb¢ì6dÈooï.]ºddd Êàl��P¿cÇ¹»»{zzÎ5+))ÉAMá>"""233£¢¢úöíëææÖ¹sç¬¬,N5��õ¹kAaa¡Ãá r+3qÄäää¸¸¸°°0]]ÆOU%g ½\|ùrZZÚµk×ºuëæîî.·wîÜáÌ��PÿgeüñÇ/^¼èáá¡!CRRvH 3FL } ��úçîwüñÇÜÜ\¹ ÷õõuss=zt\\ttô7"""þ 1n¼VVÖ¥K:tèàíí=dÈ§ÓÉù�� >)..Ö¡111-Z´°Z£FÒ àºÉ£ä ºEPPP«VÜÜÜ^yåÎ?��õÓé¼ûÃÝ[·nuèÐÁ××÷õ×_¯_¿~óæM]¿Ò5m<ê:uêTy½.\SWÊz��ü¹BI¹§ÄÄ¥cJlÙ²EÚõnnn«VÊÎÎÆ¾´ëå¶kI>>+W®ÌÌÌÔ 111U_³òry®~ýúÉébðððýû÷ËKÐÁÓÓ³k×®·nÝÒáFuÓ¨a ú\òªõüC Ç��Pa¿J��P¥¤¤è,Ã_¿~ý+WÕ+ÌËìØ±C¿Ó_¼x±<´÷¥ .Mþ7oÊ#ÔP&M3æµ4hP6mäø5}N:UíI\\BBBjjê»ï¾+Ïèååe³ÙJ+o§ �GÏJï;Ó¥[æÝîUÙv::Î±\|vQa=§��<8¡ø)mµ®¹A£F ¶víZiPkY,¥sAÒ²Îr»mÛ6møòÉ'ÙÙÙºø£1!Ã#¶åï3xzz+WZVÍôNíÌðÚk¯?~<>>^s5@ãäääQ£FÉ¶lÙ2==]OTUª@��àÑÃ#Ð £ Ë {{t-lÙÈÿ)ý_ÝYÁ��¡RÔj¼/¤½¬uoooäÖ××·mÛ¶ÒlÿòË/¥¥on&9rDsE¥¦¦J;,,ÌeýÊÈäð\|\|\|F=\|øð#G¾óÎ;M6ÕrÛµk×3gÎÈkC_¹QäÀdCN±_þò¹¹¹¥esB3�@eßsUDYÁýKLýDj,RuÑ¸@Ó�¹5&z §ãÌ�PÓ¤i¬. Úa@6tZÅ6mÚ :tÓ¦M²óÑ£G¥½/÷Ï3'==]§JÔ\| Fs9W^yEªòtÚ}Ê+'NÔ£mÝºõ¬Y³¤í/;ÄÅÅÝ¨a!!!IIIro½õÎ;'''V>ÞOzÜZ+Þ}v0ºÒivãòH«°Þ\|�:d0÷´ñPN§3##ãÖ[¡¡¡/^üî»ï8°ÝdÕªU+V¬X¾\|ùÒ¥Kç6ùÉ ��ðÌ%éËehÌäNcÞÙîÐ¡»»û¸qãtGû1<<<,,LÓËäy iàëó¦¤¤N4IþKSn¥ !mÿ¨¨(Ù§¦sí5!'Ú«AÎÝn'g�ð ÅÅÅÆZEFëOï42mUuì¼DèÑ+-[ØkiÝú ³"ë;ëòf©yòó=å6###...88øÔ©S;vìøê«¯>øà)S¦9rðàÁÝ»wb·mÛ¶rÛ¦M¦M¿(ÑZQ1odsåÇ��T£D®,jÐÑ ^^^ò¿V«U'FØ»woZZZhh¨NV þðððÎ¿¿nn®ñ¼))©W¯^Æ±8qB~åþËkVãD JiÈ!r�ýúõ+Ãáx Î�P§:í¹iù¿ÿû¿'OüÃþ0hÐ i þ¿xæg´Ü[ÙnØ°a»víúñÇËå»@On9ÛuKãNc[îOOO ùî»ïÖ¬YóÑG½óÎ;òQiÑ¢E³fÍ7n,___©xxx4(c)cT]ÊWfi¦üüüÚ·oß©S§={J¹ úhÍ5òJ9¡+eþtBÈÊæ8ªÊûû@²5��5JªmZóòòA»víö\|»G¿ÈËËëØ±£Ô&O\|ëÖ<¡ãÊryRi{zzÊaøûûëpNóþ6-:::))éìÙ³ZÕ[µj+uW¹é9!5jÐ#ÃkÛ¶ÉÀ]ª4\|Ø�ÔuR4HK_·;&L>ýûï¿ªò8òhòúàÒP5ÒcÈ¿yõ æ\|ÐÁHÌõ¿÷ïßÿá¾õÖ[R0bòä êÛ·ïo¾9oÞ¼+VK)\úsKQÜ9��uWVVÕjõõõýøãÓÒÒN§ÞÑ¾}{©?=:11QòÒÖu¶ÙÍ9<<ïÕ«Wµnó/:Ë/»ü4ðcbbdO©áhGM¹ýôÓOå°#ÊÜ¨yFÔÜªU+9wÞyGê-�êÿùÿÑabº¬°\oÿú×¿FU7yLydy\|}¢^½zÝ¿eÄæ^ ä aß\|óÍ G©E¿w0£sßo~ó©0ìØ±ãÚµk:rAg%2¢ùÑeRP~IÕ3��PwIQ~ñER877·M6R©ééér7:à= 9¾.3Øíö)µ_Ëò_RnJ!~ðàÁÅ2Ä?ÁÑjµz{{7oÞ\ÞÙ±cÇÊnìÑÛàUÉ��Ô ZUÓ~R÷êÕKZÇ ÐuZò±9ÃÏz ôp/!::úèÑ£ÚãwüøñFµPÔ�Ô rÉB/Vÿùÿ©K4mÚtË-QO<¯<»~Ûb .M×ú3¸¼:óò6åçççääÜºuëÄK,yçwüüüÌËDêÀÆwèÐ¡k×®£Fúì³Ï\Vs¨6>9��O-GiÌ.·½{÷úÛ/ùË°°°ØØØõsu£¶c+((ú^£F¤v7mÚ4=çr'£Ô-þóµeÄ¹sç¢yv9='Oº!õþj :ÐÒ$..îÒ¥K6l5kV¯^½¼¼¼¬V«ÐYåÇg}¶{÷îoÿîíÏ>ûìàÁ6Íèüæ ç°°°ÐáphàÑ×ò g��ài&EvQQVZ ÕÈÐÐÐ¬¬,©½H{9>>ááçååíùvÔôäÄJ=PÃr�uÈ ´]?zôè]²&È1Èè!mÛ¶\|»µ¾¾·oß¶ÙlÇÛ¼yó;ï¼óì³Ïê¼ æ [µjÕ·oß'~òÉ'»wï0ÏÓhÞI{3ÙEUÖ$g��UaLÍ4xð©¢tëÖ-888==Ýèü¯Ó3<\Î"u¹4iÒÄËËköìÙ¥÷Vò2êKÔ�ÔBziúÃþ ØiÓ¦EÕ&r\ê0mÚ´9wî\ff¦´¥CÎðè¤,g/::ZPìÞ½[«ÄsæÌ1¾ÆÚF0Æ»^²&OU[É±é÷ûR<ÉÛívc$à-dµ@e-zã¥hbïÞ½~øá)Szõêe³Q§mì×¯ßÔ©S¿üòË#GÈË_1?±v¤Óé4fW0úK$��ÇL¿é6lÔg:tèpöìÙÜÜÜààà¤¤$]²¼B¤Ug³Ù"##cbbËHÕ}ûöí:õ÷²eËÙ� 60&)))z±1b\ÊjmÎ Ç&G(Çéíí p¹Ò\|Rg²²¡ "66vçÎ/3fL.]äÈ===5Us.%òK/½4oÞ¼M6I¹,/P ã}ÑñúF ÁB9j��ðØ8ÿ¼úê«¥M6RZeDDD\®TNÎðä\GFFÊvLLÌÍ7¯]»¹víZ©I6lØpåÊå{5ÐÀ¢=Ü4Õ9»uë&²¨ÚMPS¶ÿþÆ yzùå;uêdL° rÛ¡C)S¦HYXXX¨¿eüºËôä �� ÖÚÝn×í·ÞzKG?>!!!::ZÚä 5½ÍfÛÛ·ooß¾]§ _¾\|¹~ÿUz¯Ó2³\|x"4öR@¿j_½zuTÝ!G«5¤§§?y äå ¿k×®=z´lÙRO£®k,üýý.\xùòåÜÜÜÒ{ ³9:Ð¨Aß¼9��¨Í´KÃ»ï¾+5 >}:!!!44466V;ù9¹LVðèt@JDDmÙÈÉÉùúë¯õëU«Vé�[r�OÑù¿cÇriúíoU×È1Ë8ð¡søS§NX±bÈ!Í5Ó Ãb±H ëíí-÷¡C®_¿>&&¦ @;Þ½AKd½àçççÓs@Î��j9í©bÅùóç¥å©3ê69CÍõjÐzzÓÒÒ¤þ©ßvmÙ²¥ô§ýÌK>º�CÎàt:7nÜ(%OOÏ¿üå/u.g8}ú´N¨xåÊ; F ív»ËÒ?~\|Á;v4Vp/#-[¶ 8xðàíÛ·Ís6>hzL��j3í+têI&i¥((((===""¢\£$®?]Ï1...:::222,,,üADÔSÚWD6´ßÍfS¸jÕ\Ô]ª÷¥¦n´ôm�ðøé,³gÏªäÈåøÛ·oÿè§BgÅ4¶÷íÛ7a j?4Mc´<í×¯ß\_~\áº¥¦ÕÈ��@ýtv)m®Î;·A ÜÜ\iíJ·²Ìn©mPkËÁ3ëPØl¶ÔÔÔ5kÖè�Þ7:Ny×È�<þòbóæÍr!jÒ¤IXXXÍäÈåøåUH«¿Ô´Wög0®´'N6mZÃ ½¼¼¤èÔ zûöïÞÞ¾}{BBBeÝnòW»C¤zJOnëiÑ+ç\??ýúõRµZ;wî÷Kî!g�PÓåyCÆúeý¬Y³¢ê29~y¿úÕ¯î´H9eÊfÍi§ßÔ©S÷ïßït:®ºB´ù,?jV\>[ g��õÖs>üðC///%K:uj÷îÝ»ví:rälì5ÙW½hÛ¶m;vìX¾\|ù1cv= ÝõÔo÷È?ãÌ8p@¶å,ÉÆ¡C'N¨½äkU«¯ä �j4g(-[($$D®?RL\½zµNç rüò\äµÄÅÅÍ¨·ÛNgùÆ»1÷¥KæÏ¯!ÅbñôôìÜ¹ó9s.^¼h¥Á¥¿F F&LV��ê=ê>p+GQÚù³Á=Æ4Vfî&æ/wÌ=Hå¡¼¼¼¤l<î#wÀå^OO²ùT¿G8p@ß¾òSó©PiÃßÿýßËÅgôèÑQu¼ y-Ó¦M3Ïëh\Qµ+BXXØüùóÛ·oïáá¡©^~þyùÅ³gÏ¹ñ. Cù^g÷9É\Ã�@}ÊÖ¬YSac_ªUFbPõAwZµ···¯¯¯9dGÖjzÐ¨áiËäååuéÒ¥Ò{©£¨¹¡´l¢½øìØ±£ä ò\äµ<óÌ3òºty±²qãÆ vîÜÙ(¥+s·nÝäþ'Nh ¡ýÌçGó Ç²UåúÌ5��Ô?Ã¥#÷\_Ó¸,Ò]@.ß×U&íDª£PÃÃÃÇ×ÀDë«ú+.ø³ÿJê)óI®ì[0»Ý^j+¹:\ ×¹ïÝ»W.Ú[·¶ÙlÕÒÒ½1µïý÷4fÇÕµª¼ y-ò.]ºùå\_öë×OË&]Ò××÷ùç9sf`` \W5ÔÕÅ=u½Þhesæ3��§Qk2þIÓU[ôR³2îÔ¹~ªØÄðJ3!!aÿþýýû÷7¤0;uê<~aa¡óÕSåO¯Ë ×eä¼d:\êe¬ºèïï/ñI&Uã$ sæÌyï½÷¤-òäÉÊv;{ö¬ì »-X° 44´»48+OOO??¿\_~yöìÙ.]2\_«rýñçðY��¨Jõ© qqqGyýõ×õ+!cñ=9r¤6]VÇ£´� &./ÒÐÖkøÿüçêjæÿ×ý×#´øø {5\¿~}Ú´iºÏøñã¶çÃ×¢UÅ îæ W®\©ÑVÿ-[<<<ô¹7n¼gÏ¨&¯HÇ÷IIT\\Ì¸ ��Ú ¤¤¤´l¬Ä5k¤¶æåå5kÖ,ý/Ã¡Û¿ûÿ'\ÏÉÉ!g�º3üõ¯ÑVÿÉ'¡iÓ¦.\¨éA^>ÝÝîjq��'«¸Ýn?~¼Åbqwwß·o\_QQQé½Ù´ä¿¤ææt:¥º({3��9C"""¤(1¯R4{öìÁ3HÉUÅ2 �� z¹Ô¯cbb:vìèééi±XÒÓÓKËz8èr¥esjià U8Ù g�ºxå ã&V®\© Ö³gOÙhØ°á×\_ýxÆM¯Ôétj©¤(É?)¿¼ä>��ÕXÛÔÜ@oO8!õ@ÅòÒK/sJ3�@Ý¢ÍíòèÑ£;v§h×®ÝüÇüùÏÖe.\_\|ñÅ³gÏÖô<V«µüÌ M\_��5¡¸¸¸´¬Ê+¥zÖ A3f¸dä �Por¹Õ± 5´®eHHÈ#´@ùì³Ïäë×¯O4IÃ©S§ÖÜxÃë³´hÑßßæÌ7o,¿öÆì:å��@µ+((Ð©ÅÖ[5yøñpJ 6kÒ¤\ð7lØPí-}ÍöÿþßÿÓcÄ×®]3ztëÖM§^·n] å òÜîiØ°a ¬V«þ($?¶k×nÈ!Û¶mcs¡ü��¨^iii;w¶X,[·NMM--Cq¹È� ®ëÕ«4½.\Xí-ýo¿ý¶eËòà:tpé/±nÝ:mõwíÚ500°&ryEòøcÆL}Û¶müñØ±c{öìÙªU+#p0²9)5köòË/Ï7oóæÍaaaùùùÆéÒrÍétº Çp8Æ=Ú5BaÎ��ðô0�²-5=k >\êK:5Ö}D'�@]'ú)S¦ÈÔ¨QÕÛÌÿÏÿüÏAéåË»üodd¤®@!Þ}÷ÝðððjÏäÉoÞ¼Ù\äÉ¼d)Ú bbb;¶qãÆ Ê1ôîÝ»eË5Ò£òððpww·Z:¥ÅbiÚ´éK/½4cÆíÛ·§§§ggg;N-(u8F FtÏ��ÔKºH¥V-[¦ßìL>Ý<ù6ý� f¯´µåúß½{÷jlãß¸qcöìÙÚÿÝï~Z~ _ýêW²Ã3Ï<³råÊjÏäÉ_¾\|Y 5]Éáp%Iy'»É>ò¿111Çßºuë¢EÆ÷/¶iÓÆÏÏ¯\|Å¢ó?¼öÚk³fÍÚ°aÃÅããóòòÌ³@æ¨sQR�ºBk\RÛüüü#FHÅIêK{÷î5j =�¨7e°ÛíÚd6æOxt7onØ°¡<ìsÏ='möÊv[½zõ3Ï<#»uìØñØ±cÕ2ÈkÑ 77W 5{A2}tiË2Dè)..6Òxuûöíèèè ;w.Y²dÒ¤I záÚ·o¯¯´An?åçç7xðàÉ'oÜ¸ñüùóW¯^ÈÈÈb·RÕBhÏ:?��:G]qqq]»vêPçÎ¥%õã+¾U§!j[,ðç?ÿ¹ºùÛ¶m_fíÚµ÷ÙMÚÝ-Ò=«7g×"¯èÙg5¿L£h«0oq¡ý¢¢"Ùec¥ôôôë×¯÷Ýwß\|óÍòåËgÍ5lØ°^½zùøøxzzálX,¹³yóæ ÒþÇ?zô¨ääd)õ1ÍOáDh�b·Ûõ¥6ªø^W7Q¾�G¯U ©[º»»¿úê«wîÜ)-L¡ßø3�ÀÓPèU}äÈÒ4iRT}¡Kg~òÉ'¥¦¢ÓñÊècpFD{áÃ¯\¹rþüùrýýýÛ¶mëååea±Xtý-Z 0üøñ}öÙ ä/^¼¨] ôÙ+,Êuj)¾5yÐÃýeôûôQ¤\|��BûJÝC\$r;gÎ«Õ\µ èyyy?2PzÃhº={V¼[·¶Ùlõ dW!¯E^Qppð£ æ´ÁÌÜÆ×A:í¶ë}Ý¾lçääÜ¼yóäÉ»wï^·nÝÜ¹s'L0ÀvíÚéø MÌ£0Ú¶m;dÈÙmñÅ_}õÕ¾}ûN:f·ÛË'Æ1ëQi¡¯ä2(ÃP¾�Gýúë¯kÎãÇßï;� ÞÓÙ mmØîØ±£ä òäµøúúVKÈPÅ Lª ÃÑ ÒÞ¯lKùÔÔÔÐÐÐ'O~óÍ7þùøñã\_~ùåöíÛ{zzÊËñöööðð0ÄhÕªUÿþýG9cÆÅoÞ¼ùðáÃòPÆXóÎ0hß ãxÌKWóG��Î7zöì)UÞ½{iÅCW/,,üÙz�ÔÁÜþþá¤h=zt=ÈäUÈkY¶lÙc.³\+7Í}!Ê÷%(Ï\Úfgg'&&FFF^¾\|yûöí\_\|ñE@@À!CÚµkçåååããcµZîîîrgûöíýýý\_}õÕ3f\|üñÇëÖÛóí/¦¥¥UøÐi(ìv{~~¾ÔdÃétêÌºÈµùÈ]Æ\_TøÚùãB]dÌûj^)�P!¤iü¸iÓ&ìkäÈRu)¨/(9�<=eqmùï)¤ÝzõêÕ:2ÈñË«×róæÍÚõÜ¿ÕBÜèv¢½t¼¼¼ÔÔT}½GY»víôéÓ\_ýõ¶mÛz{{½°X,:#DÃ ;wîüòË/5jæÌ\_~ùåÎ;Ï;^Y£Ô4CÀétÚívaIùîÔP.?û· Ö» ´ ìCj Úårýúõ:Ô\|u%g�§Ëµ]zÿý÷ëtÎ Ç/¯bèÐ¡õ¦ù£+Ic¿ @Ên¦½ñ£.EsçÎë×¯=ztõêÕ¯¼òJëÖ¥¹0È7îÔ©SïÞ½W@ý&E¿T ºwïîããÓ¾}ûØØX¹S§§6\!��èèû¿ýÛ¿æìÙ³ëhÎ G.Çÿw÷wZÌiÇþzS~U¥\~òÚétfee%%%®\_¿~òäÉ¿ùÍo7o®áÕjÕõ/4ðòòm??¿6mÚtêÔièÐ¡~øá¦MÎ9 ¦3>oÇg¨0" HSóÔÔ=PÓP¥÷º1lÝºU>äRaÖ<NãÆ=<<ÆË ¨McüææÍ½½½å²ùæoVÖñV��Ãa·Û¿ÿþ{iKzzz>}ºÎ rÌúÕüÕ«Wµm{ÿ© Éî;éFBBØM6M6mðàÁZµ2wxæ:Cª5êØ±ã!CæÏ¿nÝºÀÀÀÄÄD÷atPg©©ºH¸ÌIÝ53èWo3øúúêeG/#GäÔWR¦?^§¡^±bè$g��Ü¯½ö¿ýíoë\Î Ç,G>nÜ8s{¶\|W=róÎFÿÒÊøÙ">>þØ±ck×®:uêàÁ5kf41Å/ô{a£õÑ®]»W^yeÁ6múî»ï´gµËsOÝ'g8xð \|P222bðP víÚ%øï¾û.0 ^ æ´ã¹sçôúIÎ��¸ý9==]¿¡^½zu ähumÌÌL+9CÔÑè¢ò-}ÿ2><º\|üøñuëÖM2EóÌß»LÑ¼yó~ýúÍ9sÕªU.]JJJrYtºCÎpèÐ!ù4N4);;; ÅétîÜ¹ÓÃÃc \|ÀúçÅ[·ºVÛ¶mÓÒÒÉ¨É��U¡mv\_\_ß :Q¹ãüÅ/~!ÇüÍ7ßK7¾ o:bccO8±nÝ:iÇ 8°E:ÿ±þ¦0Êðóóó÷÷>}úþô'ùÝó#ëÒ²!?å3üh ¬Ã<& ØÑlÿþýòÁrFFÆuø9rÃÃ¯]»vûöí¯¿þZþÉ:MWÒm-j¥�ý/täo¼Q¾/"Y� t B·nÝ¤YËC9B9N9Ú±cÇêJzµ1tEê'ÃÜÈ7nÜ8}úô!C<÷ÜsòVêÆä.7oÞ§O)S¦\|þùçÇÓGÓdIãM7úfÌ vðÙ g g¨ö¡Aä @]ÏJï¬Å«T±´;êÕ«+,î©q�ª"??¿aÃR¦1f³ÕÚAMPóoþæoîÜ¾S>ç\|Qyu ÍeÌoß/ÞÄÇÇëø©S§jþ 1í¤Õjõðð0¹§I&}úô>}úÊ+¥Í(ð³êEä ä ä �\¼BB///#GhJïp8È��JR{ÔFÜäÉkmÎ Ç¦\_\|KE×ñw°2O8YY\_6Ü¿sóæÍ'N¬\_¿~Ú´iC yþùç[¶léííí2¥ñ£®TØ£GI&-\_¾üøñãòÁ.((P/"g g g�fåËA½09sF¯ìÐ¡®^]áLÑä �ª8ÅÚX?~- ä¨ôð.\¸à²BïýPÝ¿ò µm'ÚíöèèècÇmØ°aÖ¬Y#GìÑ£GÛ¶m5jÔ°aCóú:(CÇe4iÒ¤wïÞ'O^¶lÙÁÃÒÓÓ³²²äÑN§9!1#î0Ï2ê²[é½¡D^ä ä �êJm ÔÀ_Åk¹)Ñh�ê-w´î-¦MV«B9=°½{÷V>×1ê·£io~÷5(ÈÏÏFÐáÃ×¯_?{öì·÷v÷îÝ»téÒºuk«ÕÚ 1ìÈ"4iÒ«W¯¥KÊ§ëÅqqqÉÉÉwîÜ§3&�Ñ'õËMtIæ$g gàÔ9Æ¼ÊRx½÷Þ{Z>®^½Ú½�P-¤´oß>-kF-UÊ'0È1È!4è¤9IÎ�mæË!77×eè¨ËG:;;;::úÈ#6múàÞ}÷Ý¡CúûûwìØ±yóæÆ £ç®ËÙªU«~ýúM÷écÞÏ-99ÚàÖ[ýû÷?dooï'NètÒ�T/ ±Ï9£\_õöîÝûÜ¹sO0dgcÐ/ ÿò¿è¥ôgxioOó[o 1g æÞFó\_gªÔíÜÜ\ÍvöìÙ;v,Z´( àíß½=dÈ.]º4kÖÌeÙ b±´hÑ¢O>&MZºtéÎ;;võêUù¬JKóÊ>Æê:yC~ÈÈ�<ÁK¢ÜîÚµK×îÙ³gdd¤JRt2�PCí8)\¤Ù´iS)zävË-O$dçÕchÙ²åµÿºær$ OíçóA?Ì÷çòY[i=ÅÄÄ9sfïÞ½+V¬X°ÁÄßzë^½zýÌùrZµò÷÷=çÏ¿aÃ=ßî9þ¼´Èrrròóó©´å8Ãáz]aa¡.ÌQ~Òûw0ÿEðñ g g�PA\|øáZÍ5ËXêÇt�ª½5ùÊ+¯h4räÈË/?¶AKQú7ÞHOO¯°UE!GÌÓ´kN³PaßØYöVU\|\|ü¥K>¼víÚÅ¿ÿþûòAíÚµk&Mi'Í)4ÁZ·ný/3fîÜ¹+W®ÜºuëÉ'¥VPP Ç®þkpè!é. j¤"FRQ>-á%g g�P¡Bí(¥Ï¤ItÎ¢7eî@�¨QÒ´gÍ5úumãÆ.]\5ÌMäñåYä¹´&Ï.%ÎSDÎ Ì$iê¼BÿÑö7ZôÆØÙ''''55õòåË§NÚ½{·\|hçÌ3zôè^½zµlÙÒÃÃÃ¼ §¡Cþþþo½õÖØ±c-Z´~ýú'OFDDdffã/ÌoéÁqD ä ä �îOË²7o8Ðb±xyy]¹rÅø\_ccJ�Àã8dÈmµiÓæÓO?ýë\_ÿZí <¦}äó/{~ðÁ«W¯>zôhHHHZZZ^^^ùµ\¬È£ú÷EÎ@Î�@þf¥ ±Zþþþ7oÞ,½®ë:.ã&8]�ÇàðáÃ;vÔÆßôéÓ¿ÿþûjIäqäÑä1õÁýë\_?~29m0~4ß\_õ\_¯lÎòu¶üüüÌÌÌ[·nIc-((èÀK,8qbïÞ½7o.Õ?óÁÜ B~ìÐ¡C×®] 4tèÐ ¬ZµêØ±c¡¡¡)))æF´Ë±9{§Óiy NóFGNÎ�r ~3g)\|ðói�jUìðáÃÒØ1?¿þõ¯çÍ·oß¾]SößßG0gÏÇ§8jOþP½9î¦C\biûÛíö;wî$$$ÄÆÆ^¾\|ùÀ\|òÉøñã{õêÕ¬Y3ooo·´hÑ¢cÇÝºu{é¥¤ê¸jÕªC]¹rEÚÚãBÕæi'u® c1óCe3d3¨ëeYnnîo¼¡ÃwìØQÙLÈ��²§ìoü®ÿüóØív}@DtÙ m°k�bD%z1=¦®jRSVg,Kj^ÄåIj[Î&Ï.·)TÈf³ÉdÄ=3mÐÃÐg 5ç ´V'Ë\|ÅÓkÜÎ=ÛËËK®u-ª°ìà¼��3TÌÌÌsçÎ-\_¾\|Ò¤I=zôðõõõôôÔÎ Ê§ ÛF 8Pv^¶lÙ#G¤m+ M{}áæ¥?+<3÷¶¢º^yùÀ¡öç ò¼rJÃÃÃ£¢¢nVîêÕ«Òº[# &?9�½ kSÙÈÉÉyýõ×å\çååµyóæÊÊ,Î��´é]¾ù¯\\Bn³³³Ï;·víÚ3g¾ø¾¾¾åG<é\Ú·mÛ¶ýúõ8qÅ÷îÝ{åÊÌÌLcöª5ùè ×Ú\òZ3Øl¶¸¸89ù)))E :×nlllZZZbbb\|\|\|LL6ù###É�èuO.t: §§çµk×\¼ ��]ít:ºäüÓùô /mï»ä ¥÷æ0î×ßMNN>uêÔÿøÇiÓ¦õë×¯EÆ´:ó¤KÿfÍùûûO2eÙ²eneëqTc¨¢\_}õÕ{ï½ç0ÏöíÛ'Ç9~üøôôôÈ2O$mÐçöööÖÙt+ ´ÃCvvvÿþýÛµk'G~ôèÑøøxmòÓaÝºu4(33ÓÆ¼¢1��ð(Ê/»ù³íýª·¥~þüùõë×Ï;WÚÈ-[¶Ô)ÍÍ/t JÝnÖ¬Y÷îÝÇ÷ñÇïÚµëÅ:æBèåÑ¼"§y[í³¡3èt3gÎÔ§7o<¦ñ8¥e]õ+¿'3ÜÆÆÆ¶jÕJSiïÉ ã#îÜ¹óÆoè<{ölBBæä ÀÓIUóEÓ§O×Ë¯Ë��¨åt¾GV¿yÕ´´´ .lÜ¸qá¯¾újÇ5\|0º@¸LÑ´iÓ>}ú,Y²DÇ\_äææêCi a\|§5jÙF®>èÿ>\|¸.÷©^®^½:??\_¬°öj8pàüïÄ333ÈX m³GFFÊm\|\|¼3Hû½\|Î íú°°° sÇÙ¨mtZqëÖ={êÔ¾»wï.ýé¬5��@í'ÍI¡?jwc±Nc "²²².\_¾¼nÝº >¼[·nÏ?ÿ\|Ó¦M-TåÖH!´ÑÝ¢EÞ½{O\|X/]ºt?Fý»Óå'~ü)Î��ÎÁ¥V¬#,ÌIyllk#\|ºlåÁï3nb±HC^îôõõõöö{7n¼jÕªÚ3´lÙRñãÇOþ©I&ICþÝwßýðÃ;tèµZeOæÆ:>Õ«WëõmÄ-Þ×Xý¿G��<åQCUªÖ÷ýoÙ²eºV¦.j©-w¹¶¹··÷¤Í¾råÊË/¯X±BþköìÙÅÅÅQQQµ!gÐ¹+µ£ò+q>}:>>^WÊ g�ê±»?Ü5¯t8r«\VZU¾W±è0�� gxD¹¹¹r;jÔ(c¥ i·k×nìØ±/>\|ø°´¥]\zo¹ù³gÏÊ>o¾ùfZZZmXo¢M6ºLö¸¨f²AÎ�sÆß#��ðÐFÝ¥KO>ùäÀ!!!©©©º¦±ÛÝîfggËÆþýû¥Q?~üøOKÎÐ¸qcíÌ°uëÖÝåìùvÜ~÷Ýw}úôÑVó3�OmÛ¶iÙ»wo¹n£$È��¢Õéôôô#y0Ç N§S§Ô;vìr@@@VVVmÈüüü´F\|\|\|V9ÙÙÙ:äÈ#å°V+9ðôX²döh6mù¢GÎ��ÔEEEw¸»{÷ni ?>55õºÉÒ¼¢¥¹];\|øpmn?>))Éhø3�õ&¢zóÆoh¹bÅ #>ý±Ê8��ÀcË,KíÉüüüÈ§öI0wÄ kÝºµNl{ðàAóä ��9CUú3!g�6ÅÅÅC'dÝºuy¸¿G��àiÎ´k´®QaÎpçÎ#g0ÏÏ@Î�ÔruÛ3gê_º\£äoJWÉÑÿ"g��Èª3HÝ%gÐiÃÃÃ¯]»FÎ�ÔK:\Bne[®H ²Zæ ��3;hg��ä Ui³ëL ZµÆGxx¸Ë²´ëÃ4gpww¦ýÙ³gãããÉz6ÈEI{+õìÙ333S{2èª��È(g¿Î)÷»ä ±±±ä @ý0cÆ ¹"ÉEÒ¤I-ÈeÊétÊõ¿)��¡êÍv½ÊÍÍ-..ÎËË)$5Øl¶¤¤$ÙS\_îf~ Îqù3ïÞ½»»»»×úõëõ¿tÊG��ä 3-wAò½ä6222,,LvÓØ¨tÊGqåÊ///77·Î;\_¾\|ÙápÈÿø��È´Ù®ÃÃ¯^½\·Îî¨#,"Ê<æ$g�jÓéüæot¬Äo¾yçÎÒ²n úä6�� gxÐÁè±Þ.ßÀ7w g�ê¨¹sçº>}ziY¿»Ý^XX(r��ä 3¸d?»OU~¨UÙBiYGèèè^½zé¼¯_}õUIþ��ú3hÿrr æØív¹=sæL£Fä¯¦{÷î.\k9��@Î@Î@Î�<M6YV77·Ñ£Ggeeéæ©ø ��êtÎ dÍtErr æ1BþXÜÝÝ-Z¤ËV:¿þ��ZE³ÛíR7þÉÆ¶Óé,ÒQÒûöí:ÿïÿû'3Ô9ä @éuFÿä¯¦{÷îòææ¶cÇ{¿þj��Z3§ÓYb"xóºgaaá3g¤ÚÿÞ{ïÙívÍFÎ@Î�T ÃQZ¶l¥þ-;v¬I&nnn=zôÖä��¨ý¤V_XXxôèÑÝ»w1{íîÛ·ïÐ¡C/^\¸p¡Ôü{öì¹zõjrr z/Gº±xñb///ww÷Ñ£Gëò"W#��Pké÷ÔiÖ£]§§§ÜúøølÞ¼¨ÞÁn·÷íÛ×b±¸»»üñÇú_ÅÅÅFþR��Z®°°Pn-Z¤aÎ·f(3èoÿîíc~ÊäoÚ´INé»ï¾+ (´¢ÜO¢¢"ÍöÜsÏéuæðáÃr§y¬îFÎ��ÔrÚ¼&pË-+ËÁËË«mÛ¶²1~üøÛ·o2<\ÎPPP°uëV9S§NÕæ¹<>ìëëk±X:uêtçÎÒ²n ÌÉ��Ô9:»»Tæxþ º½sçÎÓ§O{xxL0!33ááräää7J{êíß½-'¿ Ï!ZuN:U/5#FÐ{l±E��@-QÅ¹~cß¯_¿ûä ^^^;wîÿáÃ=<<ÆAðÐýtÜIÅ²ÁM{§N¼0hÐ ½Èüñ4.YæDä ��@-É\5ë4kÚ¤ÕinÝº 5\|kiöj¼àéé)·ò£ü··wûöísssõqÈ=g¸sçÎºuëÌ}Eä¿òÊ+qqq.-,cEQ >ë~ÈÛµk'M6=þ¼ÞY>O a��jOÎþQ¹ÍÊÊúè£tÖiäzyy5nÜxýúõÏ=÷ÅbÑÆ¯nÈíàÁu]{íÞ¿ÿ~¹Ì1)))3ÈùÜ¶mÆ9sæ$$$ôìÙÓÈÚ·oTjÏ¢ãÓig¡öáÃuýN:ÉõDï×��¨ån¤§§oÜ¸QZ²Æ\| Zµ }¤Ú¿yóf¹ÓÃÃ£A3,X°ÀhÒ¡:r»Ý®ëZN4IN¬Óé,((7¢aÃþÈíÒ¥K³²²JËfä3úgÖ¬YÚ]jäÈòaþ¨¬��µ\rrò¡Czöì)UzmÆ6iÒdøðáAAAZ×oÌug___ÝGÚ¼(-ûzQ[Ú#àÁº cZZCØµk»»ûèÑ£åäËés+¬üüüM6½ð FÚ0jÔ¨ÈÈÈû/yÉØ Ô-yyyÖHsùòå¥÷zï2��µTÑ Ëß)))§N:t¨¥Ôê;tèðÎ;ïÏF;ëÌ3#FhÒ¤jyï{î¹ýû÷gddßt};ä}gIÔZrQrIÉäA./òáïØ±ãÉ'YK��¨=´w-8N¨geeN0Ajò:c«V¾víZÝÁh¢ÏdcçÎÃ ÓmsÎ¤ ôg¨¹AÏ°F óæÍëÒ¥1]äÊ+###õ=µ9j'#:ÓËü¸eË¡\_¿~6Ô´9��ðÄ+ð.K$''Î=Ûh6nÜxøðá6mJMM5~QCr4XÐ¦«³y7Ýß[rêÍHÇÜÎó\_XXh¼Åò&1+R~ëìÙ³ùùùF@ µËGQ>¢v»];\M>]?À&M\\#÷3��µMzzzPPÐ\|Ð¶m[]R¼öÚk«V2¾øÖÚ¾Ôó¥/íÙ s£·æ .ß³k÷ Í, 9C5æ JN²vP1\²mïÅ3gÎ\|öÙg5GêÛ·ï¦Mä1KMëµ'gÐ¥\|z322 ¤W§mÛ¶éÿæååWr��àñÐ ÚÒw:R-7òã¥K,YÒ£GoµZ_yåuëÖ]»vÍ¨±ÿX s¶ð³ÎüÚÖ! @Ë1&Ñ¸/Íf^×rûöíÒæ5jñNû¨¸4ÜÌSd$&&®Y³fÈ!:Wd«V>þøã°°°ÿ½;êjÿ8.Ö@D4¸Å¢VVk´j¬ÖKÝãqkªÕX«Z£¨õkÜHÝÞº5®qk\ã¬DX(Yÿ·ÿyÉí ²ùýLa¹sçÎs~÷,zÏìììÌÌLy\|UÎ Ç.+×¯\¹òå_êü0W¯^-¢äaï��®¢nº_õOoß¾½xñ¼yó¾ùæ=©íááÑ¦MeËÉíJÈ}´_Dtð19Ã¾}ûä© $ ÞX»xÛ·oß¸qCZéééÛ¶msuu°&BEÖÙäÊÁuC Ñ>óòé\|4ßPÊÅWvv¶¢r%ÊÝÝÝÍÍcÇqqqÖ£��(eRW7uï+W®Ì7¯sçÎf~Ã ,XpçÎiQ¾}ûV»ÖK»R¯ëèþJY A÷ÿþ®Ú¤¤¤[7Q8Ù?òNáåË7nÝ8lØ0[~¢s'ë,ÚÓcCVròZµZ¸p¡<¾9tn q(5999zÈ1BË~øA 4øEur�� iU zûÚµkÛµkWµjU©¨»»»{xx;öúõëÉÉÉ¶Ò÷Ï ÜÐÛ·oMêÕ«×ÖÞ²eK¹þêÕ+[þ:?£HÖáéÓ§æF��n¼ý5//ïíÛ·¦ë5g°vr3fLNNN&M<==Íb"×î].dÍÈ�T/#��\|æ´£¾Ô²³³õz^^ÞêÕ«=<<¬!vcPýúõÓªTõêÕåWºú¤úõëËÞ~üø1£N�33��TôAÇJª£\®X±¢^½zÖÎ Ö¹ çÜ\_¼xÑ AZµj\_¿~óæÍómzOßSE\|«5kÖìÙ³§U«V²cãããÉ�33��TÒÍÊÊ5jTpp°´yÝÝÝåÒÃÃC®hO//¯~ýú¹^¿~íç£R=z¤µSê¨�Ê=fÌí½àîîÞ¨Q£ ¼zõJW¢ÌËË+ñÞ ä ä �>2ê}w ¦ú'O69û��ÏÚx!;;Û¸ÿþ%K7n¬«öë×/&&FÿZÄÝTu��ä %»µ'~GO4��[R-ykgn:Ì©S§´£ C&MfÌqíÚ5íº U]°��rÒù²ÖA¿¯õ;:::}��J¿ZRtÝÃTülö®uø_j�BÎðÈ��@éç ïÌfÌÑ´iSWWWooo©øøø,Z´èñãÇ¦h£��P÷° Öëú«Táòòò6mÚôÍ7ßøúúJÄÍÍM.GqçÎÛ¿çj�� @×®]ëÕ«²mÛ¶´´4kçÆG��J9ÃG"g��¥ÌLù7sæÌ rËþijtRÙËµ+Tó��ä ä ��à³ s·ÌÌÌ¨¨¨ÐÐPwwwÍÜÜÜëüf¬Ä?ùØÏ�Oá#3��¢s,hÃÔ4ÌerròÏ?ÿèåå%õ WWW¹üî»ïN8aËÒøÇúøìd��99��øÜêÖd !!aîÜ¹þþþfd««këÖ r]r��9C 6öËj³É��@ÑU÷º\ÜÌ3===¥F!u3¹tww¯W¯ÞÒ¥KuHENN\Onnnvv¶ùGù�@ÎP4åe ß¼yóàÁ«W¯;wîÅW®\yòäuãËpþdr��PDeF§\°¦ Zo1ÕÌÌÌë×¯O2E§\p·ó÷÷ÿæoVZõòåK¹³Ãc:' ú°��3(ùRÎÍÍ]³fMË-«8ùî»ï~ÿïïúµ[ÛLÎ��&õ'XãôôôÄÄÄxAGF #¼½½[·n=eÊsçÎiM,''G{uæÚY»n$3��È ü.6#FÐlßÏÏ¯gÏkÖ¬¹téÒùóçgÏÝ©S'ÙZùë¨Q£äû· 7��FjÏ??\|øpddd«VôDÔaäJ Õ7näÙé¿¤§§k¼PXÝ�@Îð¾Û#ÛpúôiwwwÙ$ù.Þ»w¯Ün~lö!óçÏ÷ôô;øøø$$$Èwq\|M"g��àóaf @ff¦lù?¾\|ùòÌ3k×®9rd@@ ¢zõê½{÷^°ÁñãÇµVSX¦ é�¡rsssrräÊæÍå»ØÓÓsãÆú=n~ÌÔÊ½zõûÈ÷Ü¹smÿÎðÉ��À§Ètþ(rAx0w¸wïÞ¶mÛ-Z^«V-3©««ëwß}7{öì;wÆÆÆêJÈãX£sçº 9�áÃrww÷}ûöegg[s³gÎm®Q£Æ%K8ðI1$��³gÏnÜ¸qÚ´i=zôÐY¤Ì¼RÍ5?~ü;nß¾4¦Ðy¥4©pòPÌ��ro& §OÖe£«W¯þìÙ3kÎ¢w¯é'O¾~ýúÍ7eõ]LÎ��ÀgE\çÎûå\_úß¿iÓ¦ÖA.ÃfÍµ{÷îë×¯§¦¦: KëÄ zÝLNEÎ�� gøôT¶J¾Äýüü4j¨VÚ®^½jÝ63!##Ãæ´f49��(°¦¡+U ³4¤\ÏÎÎÎÊÊÊÉÉ±°Î´°~ýúñãÇ÷èÑ£nÝºÞÞÞR?ÑÕ"\|\|\|ºvíuúôéøøxU¡=þ¶(lF33y�áV�ô9&&F¿Áuê$ÿÐÐÐ¦MFFF>zôÈ?2Q\Ú��]ÇÐ@§VÐ4À¡Ú ·ÄÆÆJ%dÚ´iÝºukÒ¤I téI¡]jÕª5fÌ;wÞ¼yóÙ³gÚ?ÁÌ éü°ÿ¼Þ&��9Ã'.]ºäêêêîî.ßé^^^¦k¢§§gõêÕ»té¢#&¿ÄÉ��@\_ÜÚÒxõêÕ¹sç-Z4lØ° iÅCWÐ¶N¶0gÎcÇ%%%¥¥¥>\¼³bCÎ�� g(Q~kçää9rDÛïòuïëëëééiÀÀÀ7æææq&MÝHÎ��@y#Õ¿þúëôéÓÛ¶m¸¹¹Õ«W¯GR·Ù°aÃ±cÇÍÚÖu$ öµ eQ?9��+gÐÅ#.\_¾¬0\|÷Ýwc¥9¯[#?xð¹[ppplllyÈ"""´ªcx�$_¬¾¾¾ÕªU ëÞ½ûÐ¡CgÎ)õ½{÷^¸páÙ³gR1p¨áX×}(°n ý'uù ³¥Ðå³ ì[HÎ��@ÅÊ´µ_·n]Ùooo¹n6Õ\|§ûwìØÑÓÓSwïÞ- 7ñÓO?ýõ×_ù>>5nÙ²%)))''G¶6;;ûÁÒ÷ööÖ;Ì;×VF}ut§ÉÆ+ôÎu-§Ë.vÝä~øáÊúSÄú¬Ü��9ÃGÒNÚf_ºt©©B£FFÕ®];wwwÙZ///¹ ×ÿ9ÃäÉ5gÐ!¦îD%�$ðæÛ³W¬÷�ð9ç :8Bkr}É%¦Wáææ¦³,zxx 8Ð?Sl¹VLÎ0eÊÙ0ÙÓW��@ÎPúÌ¤ÐÖÍØ¾}û7mÚ´iÛ¶mÿïû?{öÌfïa¦n(hÍ´?9��àóTsÓArÞ3³µ¥\_Výæ$g��\|ÎÊmÎnÑî ÒÏÎ'7fffj/Ýø2ìÏ@Î��Qn×(Pa;á63��T¬¡"g��À gøHä ��ä ��á#3��3\|$r�� rDÎ��AÎðÈ��0È>9��9ÃG"g��À gøH²»È��Pä ��áãw üØÈ�� g(HÎ��i&3\| æ��À gøHä ��ä ��á#3��3\|$r�� rDÎ��AÎðÈ��0È>9��ÎÍä±cÇ3\|Ì$g��àþùßÿýO®DDD3\|sèèhÝ«��\|n´EVÊnÃqh>SúñÑùU>eàég IçG®gffÚþyÉ+ÿXNÊOý@É&íÑÃRËysË Ç¥ù®/µh-ùnZÊ¡ëPÈóÖWãÙzÜjYjýSÑUÙÏæÐÍµ3CSÞÚòC]½p<¨LÍº¤¤¤ÄÇÇ'$$<{öìõë×RôIeXJ<ÓÖûÛBÊÉòY�ÊVÉf¿yóò¹¢Ó6Ö²³³ã>\|{ÿþýÇ÷7++Kîà\£0Ç5C+ïny®·vºZs¼¢ ¹ãÓ98´Cyë+D©µA=éëR~Þ»wO Ò/^Xã/8Ã7×Ôl56S3ºTèÓZ¼è-BoÝº%å§\|Å¿zõÊ4mÎ)´¼±Û#e©)ê)?+1mc ==ýùóçRÌÞ¸qCªýõÍrºÁÔWµúj¾aÏYþöËsjÏVPG½@¥©Há¼~ýúzõêy{{W©RÅÓÓÓÅN®O:Uª%ZÍ779@aÿ¼³r^ôf=ÃRt8%%¥zõêú×ëëëj§uþM69T\¨WT ¡ÕªUÓ·RD6è{·ïÙ³ÇtfPoÞ¼éÛ·¯ÜaçÎE{VæAÙê/úø´fR×ÍMM b¤wîÜ¡WAéiÜo¿ýÖ/XQ=JGí\|H8p@þ4vìØ¥ÓaL\_TçÎõ(ýñÇmÌ¸û´"::Zz9åRc©²j1«¤(~òäõïÖÇkêkú3¼ï!úÎò¶8e¾åå3%oÝl\|rc-ôÔ õ�]NNN ¤øÕ»M6 .ØþÝk×Ú¾ècµQ£FZ;ª^½z>}._¾¼nÝº ØA>qÔ+¢7oÞÈû(ï¬~~~rL®X±BÏÕ«W·k×.88Ø´ï>\|hËC®Ì1Cn¯Zµê¥K úÙìa¯uNçÔËáþ½ß+S¤¯Y³Æô6éÑ£G7oÞ¢.9h)H+®Q£FÉÁ©í²ºuë.Z´hÿþýGHMý9s¦?éééò_ÖVü¦M ;>Íü3Ö#Özê¯ÚíÇzø}@ùæ0,nñÅ²mòñËyóæñ.Wú¸L¢]»viï©yÖ«WoÈ!¿ýöÉsæÌiÒ¤ Íô[gÎÉÈÈ¿Ê>hÐ ÓõÑ¢v~p>>Ín-QÅlq_rHMÛÝî«¯¾ºÁAAA#F g�P9H]B gù¾¾zõªµ MÊºnÝºé¹ãfÍIq-E· =/ _ÙÛ·o7GÝ/üýýåö^½zÉýÓÒÒ¬ =HÌ{¥ÇVu¬ó:ê :{^G©Ö+¶lÙ¢ÍÐÐÐë§,22ROjÔªUzE#oÓ²eËôôóó³ÒÒ·>%%EJN-¥>©]ÒPÖ××Wõ6Ë8b%«u^Pk¦Ç¤)WånRüjùé#Ø\oÅ gÖÆ¦lä={´×ã¼ÿ÷ýõÕM4·»÷ñúõëÕ«W×ÄëÖf:½FDDèñéíí-Å¦Í>%¼õ/^¬Q£Ü¾páBÍÿmÿWË=ÓCÌª&yÐ-±Î5èìÓMá[Ù6ù÷;wîhªÉ43¤´R²VåhéÞ½»ÖWå 5G¾ïRîÜ¹SÏ.Iýúõr»¿6û ={öÔÕ R³Îj.ÿkb}L3+C¤ ¥wzzºÞSëÃÖùþ¾K©hµsçÎþ£Çü �ºÉ'k¹cËÏråRj¹ZÙRTZ@Ú»uë)cccu2=IgJZ3÷¯óu¼ÍiÆë×ßÙòÒ§ Ò+´6¥ãààùµ-EñI@=yÃL VN©hó_«¨¨(CÑ´iSëÁ#ÿUµjÕQ£FY#)k_ xtÃØzëù SK1ùXaçå²K.Ú#ýîÝ»ÚÔ¬C¢}ûö²ýõêÕÏ9CÅ-B/_¾,GÚM« ¯_¿vvôµÆV>&'';×«môÔµ®áh9Gv¸[ÇÕÁµ¿\|ÊL TN)ÞåÐÒÈ#9&+¢M6i{<::Ú{Úð×Æ\i×®öN²Ô¼Ë/_¾¯~9öìÙã\|ì9Ìi¬Ìá\µhµHÅO§¤LMMmÞ¼¹¶ÔüüüärîÜ¹¼Ñ;mññÿßÞ·ÀTufmgf2ç3Þ×¨H¼x×h¡X¢¢ñÆ£S[ktÔ¨-±^ñqªÄA,©c§:¢AÁ bð x E#Hã¨¥É\|/ùÿý?á +o÷Ôé\|S}Ã9{¿{÷³×~Öz×»ÖÝN:Ì&àìsÒÁ¼õÀ®]»ºK$p @ ¸ñ([Ë Åûø¨?aÏ©m&îMKÖðÐ÷îÝc\|lõêÕvv¿ö¤Aj\|úé§´xxvÃÝæìS¦L¡),,Ä.PÅ7o>}:ÕõÌ3wîÜi&»§¥¥AãÛ9sæ,^¼xéÒ¥> mpÐ+Wîß¿üÁ Øòã?^¸páÖ[_§jõ<Þ[ìæ\LYY®rð aà(\\:}7·4ü8ÆÁ"""¸W~~þªU«¸c¯^½vìØñõ×\_cF!ûäÉóæÍû³;wîø3è1mÚ4ü¹gÏ.^¼h!µºXÊ<¾}û2)Ý1ÁGY÷T!¸�"Í=»iÓ¦={ödH¾'pÍÕòª³d!Û·oOsôÅ piÒ¤I´½0zIIIgÎ![¨{1Èúõë]~Ìr4²¯gÏ;w.ÊnÜ¸±È8ÏÄ-S?¯²³³q2C á8®Æª'ùÉ'Ô$4XlÙ².9®2.+S\_\|KÙtìØ<Äååå�¯7oÆÄÄài¾wï^[Iî9rÄµpö±#IbsÐx³víZ<y¸7n\|òäÉëÊl>#{8[8¡¡¡0ò°öºÐÖºº¤¤¤sçÎ4¾ móìÙ3Öá ÐÜ^±bd�v °\³f è=¼{÷.Ø ÃÊÍ?úÓbk6>á?ÿüsØpXÅÔÔÔ/¾øbÖ¬YÐ8Øa7"QáÐ¡CTª¾£ø6S> �tÇ ßÿ}ÏYÆ!Á)cüyëÖ-hxUüÈ1ÓK;!H¸Î¯EEEÐÕÌv°ÒëÀX×Ú¼ÌÌLn0gÎ~ýú1áÿøP(((°N@frÝó·Dz<°KïÞ½-[F÷ÊGW9xKÙ©S'eKgß={pãAÿðáCRË-CmD½yó&Ý@ìµtéR#Õc $¶ã2ÎÖ¿Ð^$ä³ÒØkÚ´i®6¨+/ýÊ+Ðíp\QmsÖP8¬^'Qt9\pcÞ¼y6 ÆEî ( PXXxéÒ%X3FÉÒÓÓ]JÆÆÆz5SÆÇc ifÔQ£Fåææ<ÆÆl\|øá¸S\|ÛÃª3¦®\|@?[!VG±DeÅÁ0;´ux8VTTXÅ{v×¥«¬¬Ü³gOii)\YØRØ5+Ê§'N¯Z´hÁ H9±½{÷ºtxàWüñáÃã=Ëöoß¾}ZZÚM ãÖà¹öYqFZQ%hHÉÎÎöjÿàÆ^2¶¤¤ë»tébr�a?nãÉneoy§¼.Z´ÏÍîµk×ãôéÓgöìÙ&-¨ °=§ÙÏzì$Î6!!ag~{¤¸¸'lklKÙ[A/_¶j½xvÿý÷îÿ¬;vþüùéÓ§CÁr¯èèh(d&QQ[ùÜ-[.^¼811³ ;>ÏÉÉa)XcÆLÌ@¥lÜ¸qóæÍÌ´ÞØIÈuÅ¼yçS§NX±J~öÙgÜö¤5kÕXå«ç$ÏpØ}FÇÄÄ7Ä 3áá[ÊfÎÒr3gÎ¤Á+6æfÔÇä½;w.Y²ÄxËíë3øfëÿ,//ç}Ô[·GIT]é($CdddJJ óªZÌbØgeeEEE 22uÌ1 ö¾ïöq³£GZyhÚÕÕÀ+¤Ã¹3¦=¬_¿mIÑ÷Þ{/)) Û³Ä(ìgÏ=¹¤u ¶Ö¦}4ÔÔÔÿL+áß´iÓ¬_cß\|{Î;wîÌ3\^à¬Ü1/lß¯_?eÁKNNÁC¶oßîÕÌYÀ¦sGB®[·nÛ¶m£G¦Æö¬ªªWúñÒ¼mÛ¶3üÌ,8àÙMgöÊ-Pc+&°×-[FÉìV°eÚ´iøe»°}\|\|<¥&x5cÆ 6Ì?ÓgøONLp4 ²à-n%Pw Ûà+ìkKêZ×ÁCàÌ;wYïÞ½ñôïÐ¡Cxx8\|»ÈÞ Ôøè£¨iéyuîÜxÄ»wïkÆZd°áÌU0EÊLKØÜÜ\öh£ F�TTTX zùòåÖYÞìgff& ÁhwéÒåþýû4é8îÀ)Kàfr)G]\××½Å¦XP\_ËGii©%°µåÐ¡C#""ààL1óuñ BÏ\5OíÁ�Ð·oß .222àÁQ]Ðã\ÉÔ©Sÿ²eËiÐÆüG¹qãF=z+:q=t?ÇÍÊ»ï¾ë,^8H?\D+ Ó´iSü )·ï»}¬0Æeh 6nÜÈæäáË/I¹;gUËð¾'Ñ@+ð¸zõjUÜ#0ÚôaKJJ f]ÿ^'Î!K.áüccc~rý»ú¹øüMLLdæ Y:jÔ(°eëÖp¸ìÁJòXùå'O÷¯-F «5{ölR®O>ùùùÜ¯)))Ö Éc¤¸¸Ø8nVVM4NìOú7f°JÎõó ´d4ãB¶ä?ßt¥¼\ ÍpÅ)fY VnÄþùÅmc0Ùµì?þ8iÒ$XZ¼6;ÃW6\_êÕvdï¸w¬)®-Z´À¡×¬YcÃ¾ }à)¸õÄ�ìÎ>D¾¦\| 2¤¼¼ÜmÑ"{+Bð VZEOÀÊ§·k×.<<Ü,¹¡¦/B<>\|ØLÑzõêe¢ëaÀ§MÆPð-[hB:Ä#¶lÙòáÃÏWË+®¬¬¬5ÎM-ÝÖ\ôaëA¬¼¿¤ÀÅÅE0`�'ËðJ¥7xîC»¹ÌÊ¶=6'§LÅ \¹N:1+Ò Z2ð'Wmì«1cÆðvºöjÖrzeÓ¦MÜ>::º®8O»Xý(¯&%ÒùS[L]ô òãxMr½9¾vìØî«ìzNI¯º$ Ì,¨# rÒ%ä$/Ä³UÛ³½@Î¥¥¥qlªöÕHÿþûïy§´iÓæu ´¥8ØØX7ÍÌÖI÷¬T]ff&/¥ï)[×½{w<-Nkm ð-6c~©Î¼äó¯°oýúõ\|Î9 Ì±µfgÎáÉ@Hð®a� øÍ7ß¼²aÄ¾ïöÏpñ¸ûµk×¸VhéÒ¥Òèã 4PÅÅÅ¼è®N Ù[·ÈÈÈ7nXVãx?~üxðbÀ-ríQ&Bà[²_6 )-Ãhù3äù;6moq+ÕgÀ.Ï=³tJ¼&''çäädggÒÀùà<½ê#Ù[AVXXc;räH¦8rI&í!>CÇÄ�wE<üñãÇiº1$0³×B,vÇcñ0yòd~÷ù \|ò OÃùç?ÿ'súôi_¯ ²zGä4öU>CPþ»+W®äjsÏMiÌ;Å:øèòä É6a.FÆP³µIIIîüiT¼kÝº5ÁçÑÑÑ8<5_Òy¯(Ý½:RÜàÕFc%¸¸8üýë_ÒºèÁË¨¹zõxeEfË"¡¡¡ôÔlûM6Á¾5iÒþkðCèdRÚ»æ®¼¼!ÛñáÅ¹ýáÃq°räÔs³fÍXÕÁ¨ Ë-{ Ü-»ëA¤{n@iié×_¯;j&ü +Îðëµ~ <à ºBPW®\û Í¹9ÖÉg«ëº oU©2¶(oø ¡B½Í:QE §Ýç¾U{¨§ïªO'àÜ Ñ-s²®ÎVB°M¯¦{©ÙFz£ µè+³d½é¹ÄÄDðøÃ©S§l@JY3fH.àð322Z±b6ëØ±£1ßE~~>KlØ°ª¸eììäþ·�íö¢Egö°e�-åååp²l¼$»ú.\àômÛ¶Y XB¦8>-õ(\\b~õ«r³c«o¶oßÎõ5Ý»w÷jÖ/¬[·AÐê^uj.Øxù\_ÕÀ¶mÛB²Jd>}h¢aè¸.ÃØ¸ËÇòóó± vOIIñÚ§+ÊÕú´êx:ðæb´Mü "pEdZZXjÑ$\_=R¯º£0×¯\_'µÜ\|PñÁ x$�£ÚªU+ð´oß 6wîÜa=ÀÓ'¿Ì 4ÝéìIÿ ê¯ÖLÐlÄÉ·mÐhO>MOO?\|ø0Yy.f8p5"6cU^�\µºÐµÆ,ËÅ --Ñ®];&ÛlÚ´É3þ\µj[KÄÞ0ÁþáÃuÅxü9öÞäùóçCÅÄÅÅiÝ A §O¦¥MHH íu=HM÷ìÙ[BÚãÖAàcÇy5¹óæÍÃ=zôsL~h³À>» �ý IÃõ¦ø±¥Á¯cf]¼fÍ®.++Ó^ >×zÕMôjÚðÑÚµkKíÜ¹ºRä=z4¹çU)cð3g¨E}yæ&§ùÇÄÄ3gºÍU¸À}O¯ðåõðzíÚ5%81_d ¬_à ei ðÁ_~ù%e$æòHyÑ;wîLIÏ¿úê+OÈËËc\|Õ[70¯¾qÀ,0gÄØ½yóæ0¹\|ÛÅ\|[·ºÑc++oß¾å6ÓzÃ¤ãTGá©%JP6s²°T®³ããÁÜ+0ÁH2Î�¢Ú-b¤5Ï]§EuîÞ½K xz?_FÐzgû8 à¬¬3¦/IÒ%ñ(fÐ»ÔÕo4±22-½)ÁjØ%6è±',7IÜ½{7cPd0fÌ\|2aS¹·nÝãô¶Ê6Ô.yÜC0Î0cÆ ×òÙÍIÜ]ºt©¨¨¨ëÁÍqª<°¨ ;ólÁ9}útOù 9Ó+hÑ¢ëªS\sRõí·ßAîÚµ«Õs8räïÔ´¶<uíè@¹% rss~/ÁÑ!Ö[çUÏnðyÁB=³fÍ¢xIÅ~Õ2\ÊÕ«WÓír}dnÕêÀ%NuÅÅÅY}H G¬\¹Ò¦ (ùùùPªà å.vüf7ëúç¦úðZµªü£]¿~ÝT14sUU)üº Ñðñ÷¿ÿÙ)ÑÑÑ.1Ü%^uÚ©NVNÛ¶m£¯«Èu»0z:tà 6ùõq�ñÄW®\©¬¬d &Åü}~ß y_:uª."aÑ£G3ØOðjòðplöìÙ"dpáæÍìüW½´´´Ög%¯58\°Ûà ×MlÝºÕªÓ¤¥¥±XÇ¦M¼z ÀùóçA6¸ZüóÎ;äó \|Å 8Ñ@{¸xñb3c~ôÑGÃ1bDDDÄ{Õ?~üÀÙ±cÇÎ;«®áx5ÇË¬0fg¹q�wë,ñ 3ÀJ[IIX+æ6°7?>wîÜíÛ·¸4ÓAfÖëAQ÷Ù3±Þ¯¼Ñì }aðí§~ BÃÞÚÁôéÓ9a:Ù¦´Ö®]Ëm¦NjV1++kÌ¹îvûBpËØØX «¯þÃ?$%%ñ[v�0omÐ A ;ÌÀh[·æZ]KÅ~ Æ¹ 8ûPV5<ñUùùùäö/¤ií9g?44Ò´¸¸ØWÿáÃ¶N¹¨¨¬2!iÓÓÓ-ÛÉ¸ÁÄë:¨yóætHìn.V¶Ú\TiòLøì³Ï¨$}ufåË³5I\|\|¼ÅQ7lØ@!î_^åååW]ºtá\|+¶<{ölÛ¶mqÝSSS-Ë ng îûn.XÇµÖ×>¯Yýßîß¿ßsªöY ¬þø-Þ¿¼yóæz #Î@3Å^À¬þá9ùp¼¤8wîg�Ó<§¸q+äð[P«k×®drJJÅ(zsrr,@G°Ïn×ú[¸Öy\Äo###©@²³³Õ×RF�vÒ¡²}çwJKKÝÒ÷pÖfÎiUNÍaGýÛj°û0¬Ü¤IàÝ·iÓÆ²m¶VÕzû6Ì8m>SÓù!�¯2©Óº5b[¸p!ÈFÃ8hÐ 8þ§N>}ºI\p^¢³f+ÑËz§ém·Z»ÁJø._¾wsÛ2220%ó°c§tôàä'..Sn¢C?-²Ê«üÞ{ï¹qÇs©8ûT±§É+WÌ®ØÛ¶mã#ÞìÌ¼$¦MVTTpc[2¶wïÞÜÜ\ $çùdggûôg¨ÕòOø8«&MlÜ¸Q«Xu«\|3ñDN:A À$ÛËz5µ"Á v UhWÙæ _aK-X°�?ÌòÌîAÁÆx+mÚ´ ÆEÉ,EÓõÛÉ±cÇrÜ2üñ(ÞôôôîÝ»ÛOÃ bA×tÛHXiN·QñZö8ÓXoßt)G¤Ý6ðÑ£GÔÏIIIç lzÕºukë¿æUäø0à-øÀ%lë_ 0G¸{÷®®xPLcEÐ-[Ëfmþ¬w)ÝÕ¾ÖA211^¸pÓ",nÐÕÐ¶dc<ÇWÖ(=lØ°Wæë²]Km³\|~² ÁEN¾N0ÁÑò¯ZµÊg{oß¾m-Afæºà«Ù³gÃº¡YÆ'QQQn¶8¿êÐ¡¯± N�mêÊ±yd\|ûñÇ3öÅíL'Ý¼¡¯º¯>À·ò~óÏþ366{cxs5e~~>¬±ÍX»U¼6oÞ~ÈIúàÁ¦ú´iÓÆî=02^,Y×¬QÃ¸YAAÄ L2×ånbÕ¥D^^ä¢EhèHú]»v½~ý:»ôZ^,ûàa �ÉÁL³´6×F¬\¹Òsê²´È% W<¬=å+1tèPbà)ñFÃÍz¼0òZ $Bã3ÐÁý_±bE.]ºwï×^½zõìÙvóéÓ§4õlvfOs\|Ñ·oßöíÛÃl²W³lÊ)Ø½ÿþ',,¬[·nÇsÛ3Î�cRXXxþüy¨blÆ-[FeËdò7õ¹rss ßòèÑ#]ñà[:ìøñã#G £ð [ðIOOwEµýóªW¯cËwß}Úµk9V (Ø}Au Ò£G·ß~{îÜ¹^ÍÌgtÞc,XÀöéÓ'44Ç«é]È/=z4ÎyðO1hÐ hû&í6lØ°'Nü¿: &4L ÊÊÊ>üðCª_¿~´¢¸èàUbb¢ÑØwM¯^½ 3«F�lËfO<éÝ»7ÙW[ #iä´8ce9990ªw5¾úê+zj°íõ0ÁaÊãË/ÛÍ3ÝxfÙÞº: \JÝääd0¤S§N4zxQ###mÁxûÁ?�÷¯ðÎÏÏwë×¯Ç##ð0e ñÚµkéÜY^Tcn Ø¼°0æ?üð¯U=ðý40öþýûqqqÍ5KKKólÓØÅªUÁëéÓ§ñT¬Àd;¼OúpÉ.° <Ì,Ëö@4'aîzTï'OìÞDLÎá¢øøxFe1½ð¿pákì\|<3Ëg»ræÌ®öÐÕ¡qÄÖ^¾\|é¶Kó}ë:û¦Fð- ¨_6×úZ ÿøã1hë&ºwïÞÝÝs°:íµó+ BÁÆ(¾qÝ:W?'-ÈH}£c}ßÚÈ¤Cóûßÿ~ìØ±FNk¦é\ÞúýÐÀVjµO( ÁeHÍôÙZ =p¯©Uä³Ñ° ?d£òÐgW]õkë&Þzë-\_Ï »s³ºeEÿø+\Ï7n ^íÏO¹µøWÖ{5åA\| ±t«ËÄÏ¹xÜT¾h +..æDó5kÜ2í¯ðÅÄÉ\_X%øp·Å¿nß(×¦ýO ¬õa}$Õ¼GhØ9xXXôjhh¨WdkBn@ã¨gìöáýÅÇ]ØX¡Ñøtµ¢íËÖ½æ±8Í,JÎGØã#0yòJæÇ©vcÿ÷Mð3«)Ò;ÃfGEE1þà«Îñ\_cÿpðßu@Q õæ\|MCs¿kÝ×4óûi>Ã[o½ÅuCu±]¼ÞÔÞþqÌßºuËÕ?¦«BýÜøùúö\_Ö«þæ7¿éÕ«Ëpµ÷%ß¾¾=Aø5Xò×\énÿ¯YK ì#GpM\jj\¶ÆÞt\|YòÆª{Þô¸¿³\|÷ßßJ[¿¾ùK \¡Aõuáþójð²Éoû[²QAhq³%%%°¿ûÝï^g¨õ&Òþ-æëgêUæ3ôíÛÍ lÀºô ¿,ÁuìPl»?'Î (Îðóã 9nÜ8PtÄ/\_¾´<ùÿ½8 ¼&s»\|ùrV?«?Î ÿIm\|£8 4L0e1,, Æ¶gÏÞOgþóúDAx»½ÿ~&U¦¤¤àÅg°hCtt4¸ÈÈHöWUAh¨qàòåËÿB½\APAêpo÷îÝ\|Ëe3 4L×o¿ý¶ ¬¬ÏÌhÿkõáßk×®íØ±ãÅná)é¡!P¯´´ôÜ¹s°¢l+¬ÿ¡!À5Çß³gõúo¬Y2_¾\|yj-Å ÿ+A M³î4ánÍ Í§sßK] Ä¡3Cªÿ ¡ÆDN¡qh�Ïé1dóR S0Ààöõ~qua ëIÔðËR¯fQþOe?ÉI6d©l×üãl- ñ~^ÍvAøßså¼WUÐÿðËaý' ÉN AAvA!AAA¹u 4nüÃ~Ðendstream endobj 156 0 obj << /Type /XObject /Subtype /Image /BitsPerComponent 8 /ColorSpace /DeviceGray /Filter /FlateDecode /Height 491 /Length 689 /Width 1400 >> stream xÚíÁ1�� þ©g ? ��>Gendstream endobj 2 0 obj << /Type /ObjStm /Filter /FlateDecode /First 810 /Length 2491 /N 100 >> stream xÚÍZ[OÛL¾Ï¯Ëöf<çÃ }Z =IT«în>àÕ#Ç¡ûë÷yíàÄ!8.X«½Àc{ÞÌ<óæyÇH&eZ0Ç¤,2©é)aqËL¦ñ'=ÓýÉÕ¥õ)ËýÈ1¯SyçiÐ~Y´h0ðiÏ¤é6Xf$JQÏ{ ¥b0,iJë1 8üÞj RKM6¢ÅxÁhf ¢·W¯ÁSJKQ¥Z§q\üÎ}�bç;¢ßÓjñVÀç Ë0qik%%(Ï°jÀE'VEáW³¼°Ð%,<Ò(PPÀ"é¡ fÐÁ¢&uZ±©¢Æc y¥FhÝ£S tÒYDÐa !&ö²oÉ 1¬U¬ ¨úgÖá ö³Î! ëë±ÓKapEHRL$,ï%%Ù¸^ª±V)I@;B} ½A¡-R\|Ä¢5)RS\D=Á½§Iª¾væÆõ¡d2#\Æ@7øµSªöMOêXÊB1 t]Bxn¨Ãs[JølÈÏ\,XQ!WI7.+á²!:ÿ ',9eÉû¬É1{µÈ¦U^Ì¹\|Íþøcòê»öþ»#]æt©èRÒ¥ Ë%]t¶½yÛ;ýxåE;Ï®ÎèrM.ßÔ~éN{AY+u³g5xÓvY»²ß\<ÙIûfAÛõí»ìõf¾¨7tIérß1o;æò-Û®ÚÓÖÐ×½JP¿aÙÓVÅk ícÙ>f!µKsÚnÚ»5¨íñ[>ª\_ÿØ\|Sí^í3U»¾ÃµÛ)nÛ»µ×ÓÖü\_îeÛÑ¸K¯JõöþÕ.°3í´ nmêµ÷Í¸Ã»DÚÞåÏäýÖnb\|ß¶½Þ-ú¹ ³v}øwØòíVVë\©z¾×]íZÏ\|æ1¾'×· ¯zé®'$§Ëª~þÏN7Ey5&GxÀüç´iÅ¾áp\c¼ç Ü ¤35äÙ¶-¦yqñ÷Û,-g\_Ê"½Ègù¢Ê§'Yº(æùüê#t7åWP!Ô6®U\Ä5Z¸²¢g Wez{OÓÙ¨<¸h¦6k°6éÓÄ{éû-²Ù]>ÿUUV.Þ^æUQJpÇñ°À¥ªÈq÷WÎsPé>dwéÌûü+u}\åt¡P/F¿¦¡Hp\(´è²ÒÈ±2ì÷áGQT.[í[£èHEÅ5\|ªkÄa^æL]9IÓÁDE½2©«KdZû,O9\/]nE\|ðë$§¾õÕ2êU$õøÉa½©&§É?N>Òß«ëªº]ü-IÊbÇN§ÅeÆò\ùÿÌÏØïÿüÎfÞ6nôzwcKýç¿þ Í#ö4ï7ðùr6;J4Ô¢Þb±g5kýQ%ZÕ+ê7¨Ç¥PÜ Uöj úT¥8Þ÷\01:FA¶4ð¬à]tÌtVµº§ÃïÛP°z Cî6Äöô43Ã¿cÉYöWµÍ¶ÈMtÛä&çØà «úe$¥RÖNç2ZjaÖDèÏÊpïÿ&® ¹¯ãBhfà ÃàÑø#f8+-2l¤ãDhÕ<÷Æ4;ìK.z\5ÚÁÁÚ]«$è!¢ÁT¿èCXYq!²¹ á2DTÛfH ®#j+ìàð1¾$Ôè y+ÖèóYÁF«6í¨ñ¦(=³YYH!(Öb5ZîÙÐõÄ@¼nw5 ^:P~Z]½Iï³ÅÉr-A«ótB2"8+°ÑÓ§ÊN-hïw3Er?ÏÎ®³¢¼?»N«¯ÅrFQm)¥Ó®l}ÞÇ¶ý² Ú^ÙR¤Ý¸§bª;�ÛÜ¾è;ÌvEÛCüIµËIeRÏ =Ù¬¾A4[µ~ÕUÇ Il:ÚÑ&Bð{Åés6 ÚJí H3b@¢Rz÷ÇãÍ¬¸ÌçÅ¯ãr9ÿ\_ÓÙÏÅ/Ô#!Z-ÏËÈ#ß®Àý¼&n¼OK#Vk»#Õÿ"P $¶5¢ íþßU´²H÷v i?8îh§\[ &W÷åôuÍ \|^ZÎÝuùüjYn>Ê\_JÏ¡Ù5Ùñ\5ÔµWvowUÛ¢éÖ8\yìáUÉA;sqßéägÍwp<Z"D\_/Ûú(ù§éû®~Y¢6[éÛéóµ^%j½JÔÚZ(Áé/q$®ä (¬¡r¿ï¬Áñ¤@}/ hö&Xýuì«ìa\iá±>UtrlB\_¬¥hû@¸í85rxvdWµ>6G:qé}(OT&ë4dÝ ~wu¾Ín6yO·ÊèDÚFvÏ<½(Ð¬\|hñ·í}Y,oëdØ»Ò+"¤5Ð¿3lÄ\_s�Õ &ge:\_ÜÒ Ó{&ÇÙ]>ÍNÞ¿aÉGVËl¤úÓ?Ò@{Èè¡.¤²LZÚi\«úñÜIDÙB+@(I¬¾Å�~L§Yû1Ä1ÿÁ!Q5¤\»>û¨¡DïIØÎV Ï²Ê¨æ¬ÔÓÆJæ¿½Ù4endstream endobj 168 0 obj << /Filter /FlateDecode /Length 4466 >> stream xÚZ[sãÆ±~×¯à[È\ æ UyØ]{MÙÙ$VrRç"!1 Ð�hYþõçëî\(h³.»T¥éiÌ¥g¦ûënÆ«ÇU¼úê&¾\¸IPÆ«dhå«ÔÈ·ÚnþýxµG×WqæéêOt¡z}{ó··w7¿Uåi¬îVQNUc®îö«¯ÚÜ&Ö©õãæ?wÆzéÒÔÑæ¼M¢$öìÊ³ÍÆuèw« ×u&[¿'î/ï^nGÇVöz;ëÿÔvr7ÛNeiºJUeßÐÌVqfnPëuL+Á·Y»tÅë·(³x7òº9ï»%Þ¥º<J~åU²,J³DVòÖ¯Dvh?¹jMãtØá§¤x?ç¸ßfÖE.õ«~÷ vøîjÃûnI;Ô:Êú¥n¾CF©U+§tè<¼Lûæí×\_nnµ×ÿÅÞ§y2[òGyn$!<]Ç#BêVcFføÇæïÍ%´/Ô?òÿoøÿLÿ³¤qìþû3¶qd3þû0óÈ3RiØ., È¥}Å3}àÿÿ,ç/ sC86O?-î:·yxÃÃýûbY#¥ãVTbýR± ê'Ïq¹q\_pz8\Wc\|Qn]ÿ¸Iìº¬ûÎò?K´ ËøÓ°«#IøÚ{¢Þ/Íäi2JU84I$Õq¤Q¦y7&735=ËDÆå« Óvs«\¼~:T»ª6\_m)²ë°©ª8 KßùÜ6´ß«}U?JWQKWU÷ª¯FÈN(æºÅFskeÞK½/Û®Çne¼æÊdÝJ!¦ûcy\|·:h¤g_v»¶º÷\|ø +©ê¶B«j!´r×ûlÃkååkÐ³5à¹vR>ùÞC±QqØ+(jÝe½Æh=_ýÓ&ñ»Ðë®8VðÒzØõOç"{wMÝUX©´hÜ{¨<¦ÁR73î±fåë÷6¶ ¥TM×ûAºqè¾-ªºÜÏä¯ÀY?ç§-U:8 s\|&[?ÊUà&VÇ(¯+êî÷EÄ $á¨Ú}·$#Vg~ ªT§2ê·%ö³ß^1ÖtÊ½´O¼ TÊÓùØÐ¤¾]xcYt}¤-Ë¥eÔz{U0q ødPïeótWx{ îË]ÕUgaÁÒÅyªzzÄÚxÎºé#©}À«]ÕC²Évi¼KWXýîqò¦&nxSB¥³7þis»~ E2)í]v³_j'¥òÈéüJNGÎ ket¤ãìê[¹áÖß{th÷Çª;H³¢½K©=@b JZ3¢û<i?Gkz¦H$Ydø{ê°ü¯KÜéõ¥²nèlÆxM´j=èxíqL¬ÿTi)õ¹ # ©0ªÑ% @´wk{¨ÓkÁ«&j�ßãÿÎ Uo6È¶ì.GR\|4üpQ aÛù¾ªº¦çõòXuÅð tÌ¼ 6Ãí0ÉÊÀHæ®�@XíïßÍ¦ #0ªÔCt½°Ñ £& ¦×PJ³ Û íÌGÑ]1ÌO»OÀ¤v¸}]³y¹,M.¡»¿È]ÁMd÷o"wÒnFç?¤6å¯¦Ðj&ÿ<ÇÛw+å¡ú5òF6IìëòÏ#«Ëßò§È;MÑGòÕ#AoqìéÉôÃð{ÓzÎÊÒæðT¯T]Ù.XcgGûRéÁfÅ zËOb)hém%idñ:ÏÔðnÝï4'²ÖDÝpÕqù\L\L¢®Ì·[Ð<.¥·[Dù.¶ñÇº[ê¯NÅc¨"{T¦Ö¸¾c]Ïl@©¿X%»o#¼\|xùj¦\¤®}ÙíIfD×ºÎ\_u©óUàN¶ Ì[]fHýËHr_Iõ4IxºTý}¼ø&O8íÿëWßPÚþ +]ì[¡?ùþÇ².ÇaE,B¶G¥¹i#Ö V#ÓIõ¡mNõÒR%[?¶Åæ$bðPåS£ ÃíQ~ÙÛ²¿BáNw3#íW=u ~jöåqëYI¤Ú\|JýRïz¬Ûõß Ã° òß ÖnbßW@+tq¼V[)5àÜFî0ç8xXté½Üh� ÏS\|íãM¯1¿²ê?ÔR¡òõ®èHÿ$?íÄ¯¡Äöú¶Úùõå\|üY:'QKZ©Qð¶BaO�å}y¨ø@Pß7¯ +Ôµ]ñª"U½C£uU½»~0÷D-ê�àIìà{æQ.¾2Cún^¤OeÙú\|,vå\|îàW<õÓ^]GB&ñSYì÷]¶0yîKÛA=ùÍ -þá·Æ=î Ý Lí¤Ý\ú]sK[ßQ96´E®0/î«#VØ ýqmnX\7\|LxÆTeÑD§L¾ÊrPÙ5ÇËÉë$ß¾¯jöî¼6ô,¨åñTóLökÎ4äÐÑÒùÈ÷ÏRz5# íá¯½0¬;¼¦¥c¼rGI+ À½'°Á":5~àdñë#Ç¢@e<g!<ÈGvd¡s½è£ÔÓ0vj¹±¯ hàÓR )QYdÇxß7ï6·©Z\|v:Ê'ïÞ~æ{?ñ(ijÓÌ]6²øÊÚ' «3Q ¡©?, ¥¢Ôe!ÏtE:Ê ¢\çüòe KÖ3¤ég,Dý¯¨Ï^ý-äñ[ c?J±´-kJ}î¾gà(Ve»1u£AqÆoçjgÖÿú e0Ø4 O½NEû, G QMz½2¢¯p^ôª5CáNØ àO7¼óÐt\|çÔðØÊ9&&n HVJèÇjU/=Ù½í!°¬;_eý2h}¥G6XÞ¦%FMXb(øq!u 0GåÞsÿX±¥yb Atq ~DR¨Êþ>ì±(<"´%HÍ~ß I°üHnÀçåø!1eX¸fO:¼0%Êp#Q¤Ô\|yínÃ7s?ó[íáU&@ýM^¥Æ<@RñcÙ-i@16q±0ðWí§"g7Mý4øËÓ¶t§ChÔ¢îð×öÒA@¬4Õß^zaëG$)-yfBp¯ /ÅN¢°þ¥8ðóÙ(Lr¤rNÚC÷ pø ôÜí²à�m@\D>Ãö)jðê²ë¤íöÒSÐ&ßïåÑG&ábï½bf\|È=Íýq}¦1è0Á1'¥N¡µæP¤_0©ö¹CÃ1ùoT}W"i¼4Ù»FUÇ@Ö;E£D3ei¥ît§>íÎnàÑ% arwÊFZåã$&iÃ¯ùÿ?BpÈÜÎÄd(¬kùÔ&<Ù½å½AÛ#¥ ïÖ@ü¼»sYúïè²< mÎ>élÍq¤G3É¹à¸,À·BY¾»Ò£ðÊ¢ö£±Crw,ÚêálX¾øG(¿¼_ZÈêdåè©ÖíN7?Ü�yèÆw$:¹²ádêï?Ôêææoø £Þao'ã¾ÌEkç"ðýæÇ§ÍcÅ¾©ón¦ Î«[m÷,Õ?rÝ$ªãt}w(ý¥¾CyKÉÉ¶;þ#Özôb!�&)^µÞýbÀ±b»ÈñÈZ0[U> ÷Q7®ÿ)@Av AÛ¯²^Eòñî¹J$Ãk×ÈÕ:àþôÔM] YòÆHOBáÅIÖôÁT lõtýÔFÍò�8ÔÇ³ °b7$ . õy·ÏÒ[áþÃX2.ÖùúÐ0OM ³Ì7¥pÇH0ñ¥¤OÎ¶èNtá;Îy'Ç8[u4¢&V!¨ª8dcbýüp©ZY:ÈS =x^³T^TRÃ¿Ï¥ÛiD<è/3 ¶¯ÞÌE#JQ§¢ØJ÷¦+Ëå»FÓÛØÿPÅRyÐTD~hÇÁI¥nyñ³/p²\|Á8¹dÚ×]H2¿·]# Êè<ûîûæC¥Ëê\�Fo·B/¼'ÏÉ["�¼\_º ñ´æX-XêÇH]Ó>N¢4D?ãü9:éEüÎN�?HnÝ\_íKýô EP~®: ì ÁHÄÙäó·@Ýòh3E ÇMNX¾¹¾èÙÄ£Aã9~³pä±BØ&å¤8á4":JdDÎë³ Ê\_\|w)/ÉÍtx£G á"ËàåWá$løGÕ þFP2õ¡ÉÓ¯þðndMéÖðÿÜd0¹endstream endobj 162 0 obj << /Type /XObject /Subtype /Image /BitsPerComponent 8 /ColorSpace /DeviceRGB /Filter /FlateDecode /Height 744 /Length 150226 /Width 1300 >> stream xÚìÝUÕá?ðóöÈ°÷Ä-®ªú«mkm\8q¡.T´µUë¨Tp!¨¨ E + !{½}÷ùqßËKÃÑþkòýô6f¼pÎ»ç{Ï¢��àgÎGS]U×þí��ÙÙ��þÚ¦ü\|s�³Kù5 jèü'øû� ;ã ��?>t¤gg øygÓ>XjÖ4Ý¯H#%À÷ÎÎ¬Ä0»n@��Mó¿öhÔ§F«~ú½·æ´ivÞÊþ¹§]æÒ´%(-Ýiñ¶½ÿ³³ÙþøÏeÒu¯¿ÚîN\|é¯ç)��.Ø¡å5JÕÝîyk %Bkl³ÇØañ6Æ);[©6i¤nÀ\_²jñ>@çíY³CTT¦º®¬tËwËKJJV]³~ÃM²¤Ù÷±:dgÓ.uf[iÜ{NoE[ÿuzoRKço û¬ë¦©ûÈÈë iÿ×2t³.^L µ¢Ô°±×ì qr1Hµý©³³µËf²:'«ªv©ïYº¨ïüÑÄ×¢Èww,dg��.Ù Ò?ñB>!>)èÕocc¢Y4 :¿-k!°Ôª:8ÓGñf\|ÞJ@þ¿í Ñ¬7Ì¶o¨Ö\²z[Ìn·h]cÐ®NÅÃ0¦Á9¾léáÃ¯ÊÊÌ!ÇðN·}îrå ;ãï/TYaÒÒÇ°V[v¶ÌÔ7÷HKU+k¯}ßªv)àWøË45¢4QKÚh¨nòS{õïKÄéq°¿xöpØkS?$DãWÒªÅº4Gf³[Ç\3, j;à!»×ÿ zòm·~ÿ¹ÛB-{+fÇ´Ì¨.¶±F?&K©ÿXí¯euµôìÌN8f¸¦´Äïdï^Ê_²Aáo# ªÆDõÐh¶ÒÊÕ5òÚ-ÿ¨Ëw\Ë��èB uáôgz²ìÀ;Þ74®¦ÝfgÖBnÜún?¿yîÀóï.½ÏÿÅìlw%LL m¾éÆk£Oyï ¡ûe ;egÃ4C"2³ùÖ+y9Î´6¾SùËtL)$$Ûí.üøÓ%iy×jíÏöVÞ¬ÿùìlµeçdÔP¥ÊÎOß{1=¹}n7!~~ø¤Øéé7ç³!%nj×ù ©ì¢f%ÕVO~¬UV.%u'Ãþ¦Ëá$nñ Ï<Æ/²X»)%ÿùì^É¾wv6ë7}ËÞÛKÜ9¬Ú¤ðÈìlP¥©yõ£z7 ÷hEI^,ÓÅ;dg��.]y³î!;Ê³sÃÿ¥ìÜyX0eëÄ¨X³ô-~Xw ©h´_&²3´ËÎfR}5ªï1Ü#[ö"=»½~;2{\|_@ö;³Æ±8?÷9óÍÔSk÷å³Ð¸Íï]ÚÿkýÎ2ÙóT4ê ^Ìð½£~tñË üÿL~à'aï9Äû¯wçjé gg£kºê\|V gé¸=§8§8dlvedäÞ:áªèîÇ5uÓÛOgN¿¨µkJÙÖÒ³³ÕÌÎnÎø¬dbZ<³×¥Dßxø>øÈ3Ï¼)aÐD\ÙÝ@3~ÀXu��øßÍÎúécÙ9À[9®Vo¯ßkv®ßÌ²sÏ9ýÎýïwÞe±0ûû,ûèqwÄ7t[4á_vië¬©VýFeÚC,Äíwû3÷àÊÕk&Ç³«KV>zßÙ¼%ÌÃ³È¾×TV»ÿ±+EÿµìlOÔ4í ª6,ùýúÁ�û{9I x÷CÖoÛÊ;ÑÌxSYÉÿxT!�a±È[¸¶ZRt°uõìlE©^MµÕã¯<Ú-21!yÿ:§dåú»XS²Ë¹³_\|AÐápØ=Ñnëûþ¢û÷r\|ïnïÑ1M§)w®ë¸\XììñÍ/Y§¥»¡½ú¢bþ;>ê£:ÇãìdI Í0cÈÎ��]3;ó®ÿ+Êê÷2ß9N¶_äá]Pì¾çÝ¹¥}vVUU.Ä%gòqkº¡¡lìóXlïQ[.êÛö¥Ùnd±ÉÓµbÁ¢Ï#H'Tµ¤MHÅÎYÐÖ$W©]ýùB²¢¿,Øÿ_ ³òbÙG[ý?mÞxXÏìLÑ}F\ýÎ½lg»rnÉÈJ»±ûi ±ªºÖ¡éH$R³jÒ¡¬ò+T{Ìb5<þ/Û%ÈOö1;ìîKY©xµeuFë=¾W>Z$B"mWwíÙÛ®/gÞÞÚ°B×®G ª×Rmå¸+Èæ%ÃïTÝ°û{ÄUªëÕ«¾íåQ»s¯¸©¾5ÑiáÔ C®®ï²ý¾gy¢Ñö<ÚGäÙ'¹Ì$«¬&îZ)Ñ\N<®!%àêuíù ^ÿëõÔÐc®ü¥ÈÎüç°Þq4þ;ªF4®E��º^væCW]þY%ö-;ó Êê{î[¿O¿³¢({ýi°6í¦é¬W9ÅP#\|z¦±jßÎûX¾ÈéÌ9zC%µ× O>H þ»/K§Z mªÜ?/3Y ÿ°«ïÜiÐÎ³3_:+FmØ6¤w/\ÉéûÉH8³¶4Ë¿/-¶ûÊe·í[Ï,õî¦:ðªéJòóÎû ^á"ÑP§%\ÆÞ¶é^·'h¢"eßôr±(nß9¾¯VáfUGÃ¾y6ÙË'uNÁñØ¬ýç×=ÿÏÙEÈzª¿ìp±Ä^¦7ptuÝî?+z Æ[×Ì&N7a»(ÐcHcK¢³®t¹ãÑ¸]5avwBkËË©ë·æ.U/µî7;ÏèüTcízüÄhÜ¹aÈÎ. ~ºb]}±- c®¾PüAö'©Õ-5ñìÈÎ��Ý1;ëV[vÎæ+fõ?çíí³3k®Èæ:kÀ´´´ìØ±cíÚµK.]µjUEEìØÝkÑ CÑT(ÊËËËJ·ð£¬lSi9»ìS5+óUÔî¤ÆÆYo?Êø~·##oÿª&Þ+ µu=#;OÁÚE/Lèád±9ÀüÎC4guSVµµÉY¤UcT©Y6ï ýÍºþág#ö-E»[átbØ óÌnOl¤jSªöÈæTjf444~÷Ýw+WåEcÙ³¼ËØl+m&¯h¥6óCBÉ \vgtj'·TFfwillÜ´¹lyÉ V%YåÅb¦ËÙÎJê1Bëæ½EHÀç ®·WTTÙó8äfpüqÍH-U\¦Ý?¿oxÿØô;wíkPòÐÄ²ó¸Ë¨ë ªÝc2äóÅ¨¥wÐ!¢«ºÀ=h7ÙYe6oXµvÕò+W¬Û^^£jì¼@wÏ,ðòQÉ³Issseeå5k,YNápXæè´GOn^Èª½n\i§-±Åïòæ#¦L+íåðqÍ+�D·.Ïô:?xsç.k)åc%zÛ¥d°zàèyìè\|D§´V<Úù¦ð��ð3M\|°rj§ËÿÁ=egÕË.ñììãýÎÌ~g\ï0U áúÛË®¾úJ{ÅUÃårÉÏøíw%íê²µeL-ÔÒüäÓSø(Ñ¿m¥VÞUÝî[µn§jEKoýÝù~ö3úpú·ø»ûiz[vÆ¿p·/áÚ´öÜ¹¬qULÜ§Î/k+YrÓVK¶«v]Zì¿q+RuÐà"^Ù×ÝÿpD¤G»È°~ýòødhÿ]{Vë$f°6)îWàòÌÀ¨gÉj%v®¡2ØRúÅ'dø¼ÃåN\_¬xÜ÷Ó]lÝW¦G~µØé§y=~^5ÒîuÀ=ó÷ç[ãz\ôtó¯<"hZ¬½ðo¿Ü^¿Óév;.qßSO?kñu¬ GìÍæx"øzökùn±ZVÕuj¿'ØûÛÚ ÊüëqQAûÀIëêÿ óÁáÿ{vniý§ìØCvfõ!ÒôÇ³O/à#ü 3óaÙÙLOÍ\|ßd¾ø³Ï?íð9Ø¼1AÚãqØûÇóÅõµ¿#/°o¿õFïÞ½Ó)ËøÜNrÁyçÎ5;aXû¤¦D[O\|q·ëÞ§§ð¾cUEClzhï'¥[Ûÿ,x²zêÉùÑÍÕ¥K½ü±àûß¬nVM¥þØ~= ïPÏæµ1èv3ÐyÿWÍ}j#y��èm!#ñÅg YíhOpæ²ÕáÝ¯ùcÈKËös{xE'BF¿³oØ&'1-ÎÚ!z¸6½pïu¹>ò¯æô8nñ§öºq^tã5¢y°TÍS3LCUÊúo³]ì^{VöÚbK,ªÿí¥ÉöyýCW+tb·Ê²Sº8úà ÙRMìDÓú/{Æ-+By'^gè'×Ò >!@e¡T=ª©~"ÞÍk~³tñ¦²-M1¾ûY¯OÇd+¥C ùzòozJ[uJU8µBÇõÎÏó/Õ¾¬ü5Û«M#¤ EÓXOp[a0Y-¼~!.' ¸äRá\|üÊ&gkql¶Q}ëQ{f»rú²ÛÃ^8«,8\üx³/¸æÏÛ-º®U£US&ävù×HíKÕ0MTÜtÚa°/Öwó¥Éì%¾ýÁóoSÅ2¬)³³a$/,5×=rò?ÄäjyEH®ãÍS¼¨?¶¸S9w>ùb\þfÓÎMKøYÒé%Þü÷¿]©°ªiÔÐ3§¯xoÉà§·ÇÉ·&_Ï3»aw#R��àçÛ2c¿ýL¾ÌÎþìE+wÆªÒÚmÃsDC'Ðû¬?o¥´0?£c.;m?Ñøùßèæ[FÜ÷ÀÝ¹yAÑÌºeî£Î«Òå²×ÐMç ÈòÝuïzáïOüÛS7ÁÇc;e]3jà/<¼óß/ÚH9Nv_ÊG¼}ûq¶µ '¾ój\ ÏyõC/îEÄ5Epæ»Êh<>«"J»Î©4í<ÌM¡ ÔªZ,r±§è/O¼Ô>;Ë±Ú,6ß+W\ÕÉõä«ìStnim¬?¥0¯ ¹qy;þñ&Ý~ÓÈ\ÈÕÇéÏ!Á~·©U<\BP+²dîë>QæÄëðæ:úÛÆßõÀý7pXQPøzÕè»¨Uúà¨óûÄ¯Ï~Õç_\öääç&Þ7¾8+=ZÀí]iÞñ¯½_+f;³ì¬ÉAÚVûåÛý]ÌèêOóø¨ËÎOYµº=DUã/=¬½I»óç¨ÍUíkJú²u<²¢´óé ×á?ðäs¯ÉìÌÿ¯T¬üÏE>Ñ¥ëÊdÏ¿ä×MxåÅççxìG\|Ë8÷ÀòÔÜj>ND0·¹ò$ãÏfyvè¿0éÇ)ÎögË]¥YÎüà e×³I£z´êø¼\|Q+óo\|ìÍ^yRO©Pù2¥jhQN.Þ[úWr¶E;»7ûoVhüBZÝÜ öh=Ésò×êrgxÝÈ<áQrÕM5ªËD��.$6ïgòd Äyø)¿xcÎï¾ÿN'ÞûöG³ÞùÙô×Ü¢ÕïÉíqú_ÊdÃ µ5$¶Í{å[B²YË'ýý_QÖ6²TjF¨ÙúñÌW³äØmo¸{üeì+!»QÚºxk~ôÎ(ô:úþáæq¦\ôËT >óôµ c:ÄlgÆmÎ¨OµÑõÏß~I4¯²2ò¯ ó0®àj?¤3ÏÿKOÙvå½»\|[½ÈÎr+&ÏhTEË:!f7ªr9¬TXä½®f[vQUeÙÙÜsvNÐÚ2´gÏ,h G=ñZXLæSôðË.Î´7À"¼Á Wóûij$jß1qäð\;0d¬ÕD»>þÍødoÀQ°ß7&Ä\|2¦/ù0h'Í#9û¨3yøÕõúog\\úqþê4.~#±jrËGÓ){óº¼½ùeE\®k1är¸ì3RÇøQñÊ÷+Èæo>8þ\_Õ!;wF UÖ°ì,Ç>¸rö¯jí¬Î±ÍS3 W<<ææ+¹´¿ÇsS«èñJÕtòÀ =»øÆÚhXóõ&Z³îCzð+®qýæªÑQ9(ÊÒxÍþùnq¿ÀàcÎ\WÓÌ«fX\|8ãå§¸ëî1è íü JÝÉE=óÕÜzÅl ùö¬[jgÙÙjØQÊÎÁ9ß,çUBÈEî¹ò·¢W½ßg4Åcª¸vÖ$¯»��@öY=öÉ{Sr¼mãöåg6?\^g'Ü¼SX¤ìü >²ÎßëÌÛ·4¾ïshã}d÷³öÊ»ZÄèÏd¨M©ÈõùÀl~lRÚ"~hýÚ¸<>Z5X}ÔÎV>M,n/ÊÄ³m©8®Ðä«þú]ÇWÆ3ã füM§è³¥ Ýgg3C'õÊá½ÄE£ÆòÉCn^¿.N§¸$Ø«Rá¥]ÜOuÒº©OC)ò ø©Dí1hMÍÙYYù¼:ûñéóí=³d×°Ê>ûôÝ¿ºyÌðWÞÔÏ¾I6Ò¸i~!9,¨;_mQª( [ÉK,ÈÓ{n¸$3×O0c¶»MvþíùrPµ#óÿ~?ò/·4räÈ#F\wÝu¿~M~ È ç¸áèÁEe¼Ù3©¡÷îÔ¢ä3®¹¿VÌ³ajÔÒÐÖ¿\r:ßfÜCÜýÖï0x&·Z#Û¾ÍvÊüÞï«5-2�³Ðª\|9¿åñã\|í±/wg>åS{vÌÎrÍ}ÈÎ.§ÈÎ>ÿËóú¡rç»{>¿½³ÏÃ®c6Ø ±¢qM>6��tÉìLÜA>mk÷ÙÙ!¦³æCßOÜN-è}ß£J&ÙºõçÊ6Ï{Ô¯®0ù¼Ðwi?g6Îþ×1ÇÓKnü ßEDWæ¿ùdþÜ¹¯ØhN¶ÃMöe5}òÎs§<òÎ¼y+¶UU´FuùhJù¢·EvvçÈÊf>¥ÁÚôÈ}WV ~fö©1xÓ§A±5?xïªIãM\_ðæÜßy{ö;ï\|ðÖ{ìõÞ»3g½ûÞ3ßÿtö gÏ]Ti\g1Q¯ÞìÜrrQÏ\þ´Å7?9MìöÊ²@å³cÚò¦-X¶«%$Á"Jè~9Y>>ù !'þÎÞ·±æ$§ÄñªSã+¹¬\|(Ú1iÜ]¯N}íóU×ÕUñ»G6/æ«èñÿA9åS¡YkNè\|b#¾eåò{ÆLzsöùß :Ænr5þvv'_!È57ýî>ÑH<y¥uQÓ ¶n³ÉÍÏ#Þ\Éã3}\|b×2Ëåp8ìø&f\|ûã/ÿõN]»[¦XÞ½ïBÙÝ³o½N4¢).mTimZ6o�ßÀÎ<¹×ßÿrÏ[nÜYº( g:û¯ÎWkÒ¦¶k7¬ U¥î½í®7Þøñ·ë »ôilçIù\|¡3g!«¹ö¢}ÉÎ´¡²,=Á®à MyK¹÷Ò³ÅoÑû\_Ü &Y\|3öc+mï+��è bóÞg¯Êå"^Ö6ÏË{uÄû]~/qåR/ãW<ìæ2{móË9ÿÈûÚ¸³Þ]¾mgrå^Ùf3êÍºÒ^n·~wæõwñ¿ûúDôH zeæ'MÍa-ÆwÅÓRY{?Ú ÇCölCüEå²EïL¯$HÜÇV6ìÉeÐ®µí°b´Ã_°¹1ÏÊä«ÜÙ{·Ô Évådxí¿[Î±wòªÙnW±ÇW\J¬Î ¶sïÙÙl9µ(wÏ"Ãh+½öÔþb1w!C«M{¤µ´¼ÔÄG?rÇm9ü%÷#dÐfÑ ¯»¤ª 3?rÙ"¶ÐÚ÷J<= [o¶ª×gÚ+é}çO¯®¨d· mS4Yl+Q^tqhÉìldlÄ²Á =t«ºÅYA~©-«àñ·f4§S»Ov¾ô°<»ød_W0uÕÌöªb9áóß+YJ¾9+/f´\|PÀ!´û±cèüÐS¡/ÎW}R,^!{rÞ;EmÜ¾¼KLäqó:z÷ä§7U5ò\-?kXf²8Æu±æ¼¬Ñª¡ùybÙÀÿ)®/í³sTfçÝfg/ñd~´du[v¦ñ{/;CdçÎûÈÎ¿ÜÄ7@v��è\ÚÖ ZøÙA'ñeÌZRÒ´ýM±YgÕæS ¬äÍïwîå"#³ÎñËeúíà \ÉÊÅKíÌM´^;ìô,Öñì?äÜ?¾¥¦ry^@ô :ý\|0ÑþêØGz\|ùòobª¦tÁïÉ?{ì:ªþÝGE+5å«l q·ËÐZôì<éçôë×97lU¤O2¶ÖU- \Ä´³SNv¿£ÛÓ§&¤²ìlð6<;;ÏÎz²ß¹á´¢ ÈÎ¹7?ùo ÚºñÄÇ ¸Åç3g¾5gæ[¾ýîGÓÞ;í¹o~ðá»³æÎýáÜ=½ÄítÿÊ¨,òÛ½c8kÒ¿+)æ]fï·W\óêÔw7om2d6e ªeÆhëÎÓèÅñö³Û8uæäßòÀ¸é\|¶9¢ÕËõÃµäòöÊI%j©²ÒÆûî½<à#A«§,×yaf£¸{ÿn ]¿6ÉµÀAÇÞ:æõG&ÿó fòSìÃðáW¥íÀ+ÐÇËøàX£^ÑEØI+jØëi;2²òûÎ=ûÃÙ3?3kæ¸9Îþä3öùô×^ðÈO¾wo}¨;}¾Õ;yiÈçÍðåÜ5òîó¿jjUäÄHÍ½§ sEÍM³-".¯ÁüÌÎ<_;óG=ùJÔ÷Meg?ñä¸d=/ëÜ :\|Ïå§3äkÃrÃ-ªéìiåÚ��Ðe²³ûbÆB"9u¸?,Y_+þVg¨lÿ×o>-5Øísú »~ åc³õDtäUÃz<½¾C«ºn/m.æ[ jÆM=¡hª¥Äúýð¼3hð)WÕS£Eçí÷/>Éã©ÑÉObRÛå×Þ0û«å¡d·hÓ×QcÕ3?Ï'ãw6ÚãÃÑZ6flÒzÙÊûÁÒÒV1«9yFµû\Cçý\|^§fgzïËM Î^ÉìÜÊZàC{öäÙÙ[tã/)íÓµ¢bdtÝ©Å^Ñ¢ÎùÔTþj¶MdG°ßå±ÛüÎ5<~q{¬ øz]¿eÖÒÖÒÓúó}¬\|öÝÞGîÎd5/ÈøÌso®ß^×"²³ý+ZñðÖ5YAþøA?\|J\|"×X foÎg?mæpDUÔ²;å°mÃ¶ì]"Ú¸iøå§yÅØo'ûÈûüÛ;ÄZfêGvîÙïQup&/P¹Ä{ri%MèÔÐ-qÝ±só9çd× 4wÑºVFä&E¬EoV}äçýÆìÜárç¸Ã-ÉKÕ<·ß®>{m2wRÝ¹¬©[VmùD¾,^Å3Ù\|.7çÀãÎþùÒjÝ^Næ0ï8±PîWxýärm{ó5Ùï¢ê¡Ey{ÈÎî¢b¾3ÿÑz×§ùJYsUL.¯ÍûËùÞÈÎ��],Y,1¥Øø¸¼³V¬mÞS¿³Iµ(m.;§È¦Eï3oÞ"ºX{èÅIwNñÁG]GÚûùÚ¼:ß>WåsØ¨fÞÕýø- >óZ±,¦iS¾ùÖ+~ç&rB¨¯CöåT²ÃÙ¯¨£æú3ÝyÄu\U Î²ÿ @¶± g½/gô«ïZ-ºJùµ SLì½«\|\V ùðÍe9Rl$k)¼ÿw>·øý¾^Õ-lå¼Æc zfð9û½~ÿØs µw=6å5)3ZÚT¯8µNl?ñ¡ª4TuF\_,°åµ7.wîeïFæX.eKì7%RÏ÷É¢õ1y\Ö@«f½ð+¹y 3\\|L:Øì$÷<÷ÏFö\+Qµ)Z½yÔuyb3èL·; ²³7À÷´UÌç\²\|#\_3ØS5ErNQ¢e+ìÍ{\|/ Nÿ^´m¶ÜKK- ]÷\|ÁÞd¿cÙ9Ë^¡ñÍµbHË¾kZ 6%Í¿ûÓY"þ:ø»¸ªV¥aÃ¾ºÄ{ÕÒÖm_°õðf»IËëLÍG,ÁáLxÈ¼ºÏ"/ÜZ¾ò®¿ ÏñóÚ-önã Nò S/ß#Ñë'Nßüu{ÀR&Y.æ ñeºnÊkZêª?]é\|]J>¥uXÿ>pÞÈ'^¯ÙÜÇì\|ÈY×ÄE ë}[:ú��º#Á²³ Êß3W¬ßCv6yvÓ¦M¿( Ùb:X³nÙLiIãñÚÛ®<÷~yúr«ÆdC´©ä7}êëoòcêÔ©oÏxë½·ßþÆ{Uq9%4ÐD ËáË-zæ¡±røH ¿ÓÁ5K=ñïÿÒ4>lÆùf:¡F5ÚøÕ¢Ïî=ÊN&.GÀçq;IdÖêüj_GM¾ JlË¯< G8«ïÇ_®hUÅk²K§Mìð®£êu¯ìì9gçv·yC\ccï"ÇÎ¼Ù$çhvj+ã%X1B4²©qã9üÔÃoþ)g^2mÚ»¯O1íõÿ~}:;^{ãíé3ÞmÚ/N}aÚÛÿzçÝ7¦Íþï÷çÛ^LÊ¡¥óç}«Twvr¦uÀåpû$3iy}5¿CeçzæùY=q\_;ùQö;�¿æà³°[+ÈËõ9ÑûsDM?è¬?$Gm<>#;�� ;§²³gç2y¢õ¿.a®f¯psO2S}Xné\|mU#FµËN<7½¼ûÅïEÐÕù(¾ÜFWôWY_Û°úÛþúè¹Ö2Éõú+¸òþúÉ"> Ïh¦ËÎyvvf÷á2;£é¥6DëÎÜðêXñt~ã£,ú©r Ü]Ñõl·¤¦òÙÌg²Å¢ð? >Ò;ÑxLn6z,ñÌó-¼o%/\|±¹aÛ29øÜ0éE¯õ(m-Vìõó{ÿúUºý2t¹Q¥¦j¦Ü×&ONî?mÑ^1ÐFbüQ£;¾[4÷_%ûÝ~o}¹·ÚÿU¾äoònüa5èÃ5UsÞzí¬°ÃÇK<ý.üýÝqjïä,þñÅ_¼ïto0ÇåÎá=ïÁ«ËyMMÄv£eÌv·Àw(ãÙ9G&bïÑ[êåùB¾ëòk£xEëY\|ÞôÍôýåN84('C¸x\_ö/þ2v'¿¡fEËîÉë¥¯ø/±ã~êªûOcÓæe=ù}Äwl^Ê÷s§çÜ¥e¼ÜËlî¾òt1;£ï!Ãn5HÓùeÑö[��@ÉÎOÚÙÙéõÝ²3o¨í²sï³y¿s-oAÄæOÔÏ}æõÍ7bÉÏ¶xX±"To j-kwe9<ÄÑeg>5LD«vÔnÜ>÷Ã¥«××óvJD¤Õ4ÞmÅi¸æÕÉåy=¬O \;arÔÒh´je gNá=f®,Öªßn{åöìl�3BÕ²Ùÿ¸ËN$¸úìhûßtã-y£×Ô>ù\oHçºükÃFÔ³£ÍÇçû}ÊO¼ú&±í±IåÝÅI¥´bõN9ñ ûÀëú§ÈÎa¾z/_ÎÐ/>esFY0-FUC1X¾VãTQB-Y¿µ¢ª¾Eæ[Q·ÔT¬ünå¬>Û©ëu¢& ß4Uë+ÖåùÄÀ×¾ÄyÀúêD£Î¢H$Rµmé_Ï÷uj¡½Xs³½²w¬¾ä«Näk6ÜÎ'Y$¶®XÀ/L¹½Ä[Èªì¡§_^Ú"vÒÚ]"@ëÙùËdwÌÎj2ÊÑû ±Ù'Þ,övv.É~iöMC§±ÚÍßõ+}¹ ÎùÝír9 ¾Ø£\÷Z2çÊK¯X°¡¼¬ªYnÉÆªI]CEUÉÒù®µ ¶¦Z,W'"Tk Fcé×ózzåL@ßóþT÷=LÔ]Ì×ö ÅÇ\þçfbº¬øö V]ù¼ö:»ÉÎÞ^-Û,Þ[¬ÙùÐ3F-ð´«ü°u¶��ºmvæM-B[JÏ+$¼ßÁSØç[·QÊ÷iÛ¿\|¿XÎ&'_;¢ÞÞ»Jn½ÃJXñ´éçvÚD{<6uoÞSí¬>Ë9ð¼ÿ»Ý{é¼ÏZì£ñA§áÚ{øS0ê©çE¿Ë Û?>¯_äÊ nk¦rãgEÑLMTÖ¦òCúä²BîÉÎ%Á}<¿Vv[)º/·µ½aqUK(¦%¢J¤©á Nåël9ûø² ËþV-ã®¾@´É§ ¥¬Ù/¦LÆ¶>5nØêÖM\|ýoûë 1"Jµmüî ¾w!ýþõþªZ±ãZB¤ÞÌÖUª6ÒXÍýzurIÎÀM5,Ó¹o?í{>¬¨Ü\Ì²³«\ñÏÞ}ïçD\o ñrû¯Îàóý9ýC¶ ³.¦H°ÿÆn¿7\|vgæàCÎ¹¼^Q+DõºuèÎ^#@É;bS}rS6µÆvMz&µ(©jj²O4²>ð¨NT®É²oì{Êo[xL®¼oø©~qQ+Ó}Lýqµ«íÄeP½Æ¶¿ò\|~ÊÛkà¯Goõj-#;,Õ¢ÜAg^Ô$:³%uùW£vÊ¸zøÝüª§0m°¦ª-Ë<¼=Hü}æ}»µí¹ÌÐ=WÁkgàaÃþØbÑV#ÎÞX ag"\S��@vNÏÎöìcÞÜpæÁ½òrM¦ç>[Ì¾6iLã"º}ÅçyÖ(qg>É»WÅ«¶°¸nÐh( »²uåäÎ³~ísëi"±&¨YÞ¸öc¿ØIw=g°²²AR¾¹_Hë{ÿ³ýÝ=]qkµxQ¥Kæ²×ÍÐ¸äË©½s[mÕ[þøÛ32xº."Ù¯©æ"e��RÅ+ü�¾Wv��îYÁÝÝáßÚÐ üCÿÈ¿á®@$Çá Ì��@8Å6nÜ¸º´ 6ÔÖÖ¢ÛôÇWÍ7¯_¿~tik×® Ã3Ü��@£µC¡ÌV+W®,--]]ZA~ÊÎèMûLÓÙØõÐ%° >ÊZ#³³LÍÈÎ��ÊÎápE²²2äGèXû}dqãÆòc]]ÀÞïÌÒì£L]©/YkØ¿){3,--Meg Õ��Ùe«[·ªª?KËz³ýggV\_XõOLîØ?¥¢(,5Ëhdg��è#ìw®¨¨ÀÛ]lö§Æä#;ÿÙe+ö¥¼ÖÔÙe²3ÃjÊÆ�� \{egÖPd-F1¦·ýû²²+¦¦óüå²²²uëÖ±?) Îø{v± #ì# ÎìÍÙ��vÍÎ±X5Y@vFvdçn7ÈÎ��Ð!0ÈÎÈÎì ÈÎ��°çìÇån;ÈÎÈÎììì ��»¶åíU«V!;#;²3²3²3��ì®¹È²óÊ+ÙÙÙ��õ÷üû!;#;w« ý\_,óæ}>²huòMV<)Ê'��²óOÒ"jkcÔê- dgdçTkì/ K~Ý]ÚçÈÎÝ'-³"mQR]\|Ü%G[Ev1wÉÓ~Ì¿µÇ±\_XTL³ý��N©\7ÄgZl×vr»{eñM»Ì½µ»hvÖÅ!3¥¨"°Ä7Y\_3 úó¯ÈÎÝ¢ñb¬¼ô\VÌ Qñ¯®$S.J=µØÅ¥ÖÖ¸\sµbZÎ²°#®'øùÅj»Hkª¢%ùÛ~P+]<I5MKEôÔéJß0µPÄç¸ª°ä?QãÔîf\_hòÉìlR$h��dçYc#aÒhX2Kp)Z;¢©º¡n¦¦%¿#ÛEÈÎÐ-³sz\µ¡]]@vaá6 ª&¨¦zUC%RöO«¶ÏÎ©köµÒJµfj5S³òsIí¾ã]®¬ Çi"AV~Ò1#$\Q,Li8&O1zòÃÓ9»}¨¶´Ð(û6»O,=;ÛÉ?¾¼ZEMóãÍ´¹ÆTÓizGv��@vþ±5 3hÜnXº¥Dx ÈÒkwnÚ°nþ S²jÓMµJ²ébÕ´;îÍdgdçïÕd+\_£\ ÔHÄ ÚªÓûFMÝ.¹0²s>Í¸ZÛHËËé¶ ª¤´Öî6Ó/ þ¹BwiåfZµî\MËi,FÅUÕ´ì¼ÆÊlSÝ^A·£Õ;h´6QÚLiE««éöm´±ÒFwðKºh¢èÖm´¦&Ø¢©KT8}©ÚåÐ0¥U;iÅ&Z¹Æ©8±YÈÌ��ÈÎ?U 0 ÞIñ&ÞI¶]5áÑ. 8${ï¤§Ôú²ÈmÑ¸ì Ý2;Ã ÑhùùöÎ²ëKî¤'ÿÝµWDvî YWûÑ#6zPóØrû_Ý¾^dÏöQtÕlÝ©/»üÑÓ oêå»u@áØãä:°ûÍöÙY 'ÝºÿÀQ½ÞzÒ¥»5LÃ,5zÉ¸g/ëûÇ¼ýî>ùìîoJ{% ½±àácþTpÐýúÜuÈ^½ê-üûfr²³ýÞY²nÕ¤¿^{ðm?pÔ¡·ýjasU¸K��@vþ_ D©ePC¶èW_\| ÉnÓ¸ÛË?JÞÜ{¦¼Xmò>½K´I@\|¶x'XÄR«×Ú]Ù2óhTi³I»jBvîÊ56Åè¢ÿ{ »Ùcô÷ù´%Ì¾Íg«&Õ¶¹Ì Kê¦¾Ñ76×9©È1aÞ=ÇÐÆPúr^Ï:ïX¿ñõn æÝw÷q¾6¦÷×Qºj ç÷Ì{ÄÑîÛfæàtÚòwv§ÇX?å Ü{ì¨ÿ¬7ÍÊÎxÈçbÎX7ö©\_ô¼=×¿ï}ç}E[dvn·þ6�� ;ÿàÜ(:¨QµjA.!Ù,%Ëåõzüv¯³ÓéeÜ'!ìp:ËOüùCó9íóç¿vÖ(³ø@ií¤»®ÍvÉzë&Ú¶01²3ü\|°»ä¾Ýç!9î?éUuC#5L]v:ëìRÖ#V?ýÙ ï Üä± y,?ðÀç7-©NüTv¦1ª®{ÿÛÜIåºî<ü²¹uüÒÓNö\ãÿqRîh?hÐ¤;JÄ¨k¹üW6<óÅÜN²&9ÈÃ2aPþC§Ï7·ÕR2uÝ4 1hJå\|ß}ú¤»sbqáØs?§ÍÈÎ��ÈÎ?Uë?ªkü:cãò\»\_}Ètfö<ï«¾]±VÓ ÙæÔU,ÿtæá$¯ÛEÕïw?ÄÀmèÙYÎ®ë#&å4£ñ²¾Å§ÎA9úèýü?7ÉRiÕËìcìÜ¥E¨QöÑÒGzÜ$óÉ¸ã¯ÿfj,FG¿\³´8]ûÛG\|/É@ºzcü¿þ¾Yä\_8Ä¤^0XmXxû#úÜC2&e?0lm¬ã¡=ð×þ~,ËÎdB¿É#ò'ÙùEìu ûô°n¨ÑS³¾265Ê«£~«×7.;øP®ça¿÷½ïüãÊ/\Ò Êæhbôã3F» Ç:2'ö;dò+ÅmÄZaî¸w÷¤}ºmý}yçÖ§ßØÿ{üp<\Hî>ôîÕoÕ$G~hTeÙçt^'ãÔúrÜ\_dÙÙ9Qfçú{ Ldg��dçø¸»è¬)ö$$µÿÝ~gñÛ£mm~ûf¬ýcê¼MÅÛWék9r¼àýkÈÎðóÏÎf§Üÿ2JcT©>ì(ýC.þÍÜ¥"EûésË¸75Q5 ;P±c<~Þ}ÏÈÎ]»îZZ´ÉÚôËÇè5ÆåÓgàc¿ÜB1ÞÜ×WOìw#{\Æ~O½aÙÃî1¹Þúún?ëõ¦©MÅTÚº¸åýn%þ]Î{{5y+µø6ÌàEO<ÜÑ$û¡öü®qAÊ~[¦áçÞzàh1Þx\Lò8&es³ï?ü #Ôjoñ°Ç=ñì5~1îÙ# YXT8öÜ´1Í©��¬é¢-ÛÊv&;½¯¶:Ú@³-;òKÙúAËAXv&þ>·5!;CwÈÎ4ÙÛüeºmm/±¶ñf]óØvä!.?q&Ãêu;ð=ÒY)K.Ëì ÿÓ'Ci¥ÛïqÞAwòòï;r~¢¢NDÓäB^¬{éW½îñdësì£×Î¥[²nÏÏ}°ØÿÀÐQ\_½jç\_ßÐ2h6þsÕm}ï"®ñÁ±¼SýÜ=òå<ú·#ûÞB2Çì·ßÄ;VoÉ Úôìç#I&åú'ee?å¾ øÇùà6ýªLª¥ äQ:ÿ¡¿^tñL(dÙùKÚ¥Ù��Ùù§È1ªW®ùðe·½vÁÁg]K&gu8XÛ)¡óÏb¥×\_v\£3ðÊ[(©jú¹¯ìì¼ÙY¾Ø£¯UÕè',f¹d@Y+V³ê1}êä�¯PB \_·¬E�þ^¡\|a±.²x²s×®»ÐhÕ{¥c¹ñeÜ>øDèdå J]tÁøÅw îÙïÿ±wðQëßí=½A ÅõÚ®×v½«Wý«´@H/¤¨"¨( bAEPPzïBIïÍ³§Î;3çl _Ñç÷9Ý³g×9Ïw6úûy{ ãwuJ7ó]6çGñN!l²íuÀÒç íT1Fl !qïücÁÌKcÆ�Czb¤q[È¯XID6Ì\ùRf6rr§U9q/«"sMº«ut¿Ì óë38ßeð-ÊáÊÉoô¥ìLEEEEEEÙù'F'dLyî.Yÿ zÐeë"2z¤Ó±³b9 µãHCS³@ì»Dó©þFì¬\|Ç6CwõÐ8K &åuØz÷ M}WRvOÝ÷ök±mï"A¯J aÊÎTíÑº^]ìýxÈh l½îzwòQ45¦b×æÆ×fGG$é»¨\|ëqèNýò¡éFSv\|tþM¿ . lvÝ×åFtÎ2¨'öì=õ±R(É×TáêÙ£!«gb~Ò6ëIv'NÃ¬Fõ¥( Ëéô+\^´úÅ)æÎy!Q¹ñ®þV4à{ç ÊÉ~(z½/aç¨ìÛOdgòLEEEEEEÙù÷HnØ¼oP(î<ìh»þro {zÌxÑÁCÎ£lûÓ:ÛT výýnÐ?ºÒ¾°)ÉÓ^]JR;[ W:æ!J{ôØÕ¥Õ^å%ÊÎTí}ð Øa\Å\_½ôzßð \»OÏg·ÝerÀÆ·W>×g°¥wI,Q =-°õçò¹Çb'Ù )]Þ\ßR)ïIèI»/êZmJ¹rÌÚ5äT$ü»5s/ôz³3)ÒÝ4gõ¾c´¬Ð¨»áöfxèWúÈÑEäêÃr»\2õ¾b g$tØ!ü)gæ¨ÊÎTTTTTTÿpbtCîø51À¢&m{Ü\a&@yzv&væ9;èøegÊÎgcglÌÎ¼ÝOÞ£GSÇ¨Ö~¿§åê=+ÞÖÈì¬ SðZ]¸°$^ ír(;\_ässm-d~sé?C'ZUù½Ìi#WÁÖZeaé³o^ÕmÖÕsè;Of66ñ?ß116b"0çv¾bfÒNµ ÜÿÕ¯iQÉZuNL§¼{?wWÕ+ÑÜò:ÄÎ)@ß³káØS²³);4&÷=°Ä k×}éK K¾ÚR±föKÚ0·à3Rè×å¼uydªÎ«as+¤1ÛTTTTTTÿbtæýW£ Wîw÷ØÃtá·ÅÎ¸Wª$?zp¡TIÀnh©£&);¡ÿ?ÐCõVS©T½o{ê\W¸¹G¶:KûF7t® ±¤Þ¹ÏÊÎ÷Ü%\ÛùC¿íÕ%# äDÁÓ®RÇ±¦S"²C i}wÎ«Äw{ö[7$N¶Iq¹¯Wò£«¡°&gÞ°®ùcA×¸Ìÿl#ÓW&�fç¼y§MA÷Ä¢qr0_iúæ·×oÎ;d<î<øú×OõÊR36ÃzÙJg]áÒÆ7®TêlSv¦¢¢¢¢¢¢ìüG²³w÷n¸×�ÐÛJæ;ÜéØÙ×ÙV×Fà, ©þì¬üwÞa~^ðZg505@e±ì×JñmÑzfÚKá8Éòú²ïjHe�g±ÏÊÎõNþR}ã¦õ7djò<¼rê1¥KUùÚ½¯w¨7Dk/_ËðènìöÏWMÌ5ª £t#8Ëêñx¯eá²£õ½ynÁ .T~¿óØy¬Á{ùXãÀê+ìð×{_íÙ)u)®ülRJ�=o{îìëdvÌ¾ý'Øì¾hm:<©¨¨¨¨¨(;ÿ£?øk'dÜ«U�ÄtR a³7Ô9Û3qÀÄã¶Ï:üGAÙ²ó9°³l\|³xü3uWw°\pÕ·0[yÍ ;6EÉe@§Nßu·éQ¶XÒß¹£//fvV¼n<òäkzgEé2.IüÏ Èxy4Ò÷gÏ©Qv+üïFI7çVÄÔÅÍËz!ù&0þg×ÌkhÚ7loøph%ÃfKíóÚþ ¥Jo)Yu®Í{HÄD Îïz:v6gDEe_±Uªo!ÙB,,ÝeÿxHI3L½%_ÚÚ9ÍÊÅoÍ{"qua;_TTTTTTÿjbdËÑ!¢Öcv¾q\|Õ¹Ú$Yí"\½BÙ²ó¹°3ãirU{~ÕÙ¹õÙä&BÿÙ¬ºC»jX ë¹bkS\%ÆËI</rvfñ/@Î +güøLñ¨¢~soÚí<.¹ý\_]²U ·ëÝgí%Gbg½´éÞ7{M³Ô®¾öXä]°qÎú £4Æ iWüì©jöÏwMRÿ$9Ë;T^Èîq x§ª¾®½ûÕÄØ <ßlL<éÉÃø©ë§Ì»±³}oª£Üa©(;Svn§ìì®ÙÅkP p¸z(ògegdûp¥V×43¤7íQEõ·g,Vn$W Ó¨ÚSiÔ²t^«Aà¬%�£ ýt]oÚ+0Ö´MÙù¢¾ÕA(æ® éVëø»aøån <¬ú²8§\_2°æY£ ~q\|ÃqXÓÞñ@qÙ£ßÍh±-;½5$:è&uíQt:v6GåöÝË;{ToqÃCß4ÄM9Ñ!)We¯{ë2Þº&4 ³sXn©]EEEõwcçÓm¹èÙ.ö/(1º¡Xµdú=3SïË¶bCH<AÞ ÅºÍßçX»gÎxëèegÊÎçÄÎ"<û¿y¯ ÄNO í©T2C£C«²ïY£Á³ Í+�:¢§½÷C«å ì\|³3)(/@V5nÌxïzë Éºð=¶);;³8ìÚUûØBIT6P~ù¥¶±©¶¦¢¾úX}uEmuUuEeùñÊÚÆ¦Ê\_ÎÀf@ÕÆ º¯Æ'©ÚõøG3ÅlËïørM²m îÜcÚ#+ÅÀ×#ÒÕÚÔ¾£çACÙËIY´bðÝþÌ´K;h 9Æ>3³ê]NHë6é=$[iì,Ï© õgcg9#"\g°ìwfÉ¿¸ TÙ(n¹óÕË2Í¶t]çðÛ\_½ê\_ nÕf©);SQQuÖ ÜXå©$ ÈjúÃ-Àëïá\_$uïOÔ¡ì\|AÅ§\|Ý°øÐ09QÓòÎ7þn;'O-/½G¶Äjä'±óú#vÊÎTvvC¡Æyl«Yûø!Ão\ÉeÀDIV ¯ ¬ÞveÜÚ Ô~,²3U;O:3»ñÿN\uIýw=ÓÌya±S®u¸0<×nM¾î¶¶BÎ# ½,ö)7C¸÷ÝåÿËRë'zÎ¸ìåëòL %üÒ)Ïì!)ÿmo,§cgLVÌÎã,æôÎQYWoª[\|¥öxüK;>ð9.ÕbK³D¦gÙÀ¤ßÙáÛª¢£ªÝX¢¨t«ô§CòþIÁË4ÖÔíÞ±;wîÞ±kwmmÓí×1yáO³ªg<¿yÈÁìp;bÐ§çû�%\|Óà>(8J k>eç 7}%ô6lýhv�V�°õ³÷¿ÌÍpÀ\|ùÛ}qt©.2v$A>È, Pñõ{¹¡Ë¤÷¾ÁÙ K%á³¼s,rdqÀ¥�ê?È{!V4èñ ì¶ÇÆ ÄGÙªý³³²öó»°ää¾AØ¤8sn)WcIêýèN¹/ }ì,H¸ wýÞÅ½³ÌÆ<69Nª+´êRbó¶-«"§ÚD_4 v<ûÊ ÀØ}rÒ6 &41;÷M2v¶MÄ¨îggßÐ÷¬KÜÿöÊñÝG;M²ÓUIZ @^LhÖ°²3U{3/X¼pºø'd>pÄ,hmØ¾æ§'¼?L§5ÕN: Fß¨ßu÷×?o@Ë¼½ýåE&kÖ×'Y±f¼dÛé( $AfÅáfØs! xÃa×Úß \#0Ä2Ëß©¨83èh£ì\|¡Æ" Õ#»à¤gÜèÙ¦³tÝ\| Ù@vÌD¯JhÎy S'Ýõòý·hä¾µ¦P ³9sÐh¾3ÕÅÅÎY/à¢Iòd ·\|hWÿ ½bS-ôq�ºáúËÉ¡\|³÷ø¾P\4{ßÓèmíø{M/Þû\¥E$Aä]æÇw ¬Cµ9KZ»ßT qÛ5¹ü<£ÈègMüa±qY ¤Àª+ ×äw)º\|;¬wÈÁÏØÃçÇì<íÍ¡ñã9»W·É·n'N¤Ö¹kÇöo0fþÎÛ¤ÊÖ@Ì¶Äà_Å7?1ëª $è �ÈÎa#VÁZ?;SQQQµ ¯îxã+_2M?/ý¸GE E2ªT&^ë©¨åÊ£&³à]{ 0GSu;ª ¦oÉY!ÏÀÎ.rÈ[yÓ}["§äþxðÛð´NìjÐ[1öÅéïºÈXÊÎ/¤X{ ô¶´l\_IÌ{ª6 þÏï8\Úähkeñ®¥s\_ �³Q¬aÀ5zÊvyþQv¦ºHÙÙIéåÑ�àºôj[ÿÛYç=ÈGàîu@¯ýÆ^ñ! >kFO§<¸4egªvÈÎB (ª6,þyôU/.¹&c¡Mg Íè7uËÊ:d¢X$IT6ðÍCð[þ¼[zfªMY\ã+ª¬Øî®¯:¾UÄ vÞ8mæ5 cìì¯ª°óµc{%ë¸¿3éQÅà; .ÈypÑ0/k×WÌ»\|¬:<è¦�0I§ÏësÇ/°Å}û�TTTTíÁ:ò.¿µCñPÞ·ÉdSkÀ/®@j@¸¬îg-x3C±tü¸¡ B¹åÿ¨\;àcöGK\|Gg¹¥DL,ìÔk'n½,T.@ÕýùWß%Ä¡Ë>eç ¾_öAÜVmòüÕªAÔ¸×>7ÞÑõÏ%5¸ÿPv¦ì\|vZdÙäï ÇIþèÎùá·3ÊÖç Ôì»èÐùógï¥Su1CjØ§AÙùbfg)ÈïLÄB{ióÒácCbSÚ¼HU~bdÎ5[H.QRÆ¾J.? ÞßÓ{¢ÞLS" i}ïÿ8©zàì:DÚ9Äß§¾yMÜxÊéÑ£0u6%£GöY¿íl0dÆEd\_µV6 ÝÒØïÀÙ=èqtñ{«Æ%di@6�Ùjv×ÐaWN&R/U;agNY(Õ}PôdbdðÀü¥ì¼ÛÅ:9Cô4íÝ¶63ùÿBµ¸\|´m+iõa8Òãß> ò!p¢¯¶DÇ¹=ñ>wºvÃSUòÀÐ>Ýô¤v«Ù ,±OÍvëK¹Q\|¦ì\|æ¯ÌÍØæáXÈÖXóu×H\6ÌlÔúYg±�µ¨u@oÓ�@ÓD½ñîRd53Ä×ÐñEÙ²óYßdÜ]Ô ¨´Àf²ö¨j$vt{ò³3¹C:¹¦C]8áÙ µU·/~.¥ìLÕ®í¶6ÝÜýë7Ç1§w6åêsûqÈ9·³hQÞDro¬ÿ\|àèð %+&uè¬Ãß7±(ÏPRí¼ ÜYôæ-ñc5¡}.4q/Z}§õÀæYë&ôI ~Ã6©°³(³³�yzèqA [¡PÓ ÷þóÝkÉ�dÚ :§üsÄÚ[OÙªØÞÍ >óVüeq¤Ü¶C«&ëwÝ»¿E^YQbDÉÍ0w´ÞrElH¨Q¶¨Ã.ùPu¬Àúz$·'±3¶u9¯ty¼g6ý·{ùGtç8ÎívðG÷x½Þl�9Mül§}!§Ó Û6ç:9!0Ðe¯Ü¶¹»Å¥Òkh4Àÿô·v);Sv¾ W\|%{òLR·é×î»ýF�VV£×�ãÞ Åì>}ú'nº%;'ùüm¼Ê=egÊÎg!Û¶l¶éT¤îWÌ#c²^ÿÊ\|Æ¡Ï¢ÛJó«Y/ûv£b\_Î\|²3Uû¾9ðþNR¤ ¢#ó¾Éêÿbä±ã¯»bf=Nµó\|AÄdÊËhô8á±g^ÿÇåúuxyÖ½;SRÀoÇ}[¡Åo<4$¹Gü¸ÁÃÒÆJD GãÙ?X?yðó»4 éöb¡ÙoT¸ÀÏ0Bßà¢1è ÇÀºß^3!±Sêµ&<±KZµtxÊÎTTTíÄüÆ×ÎÒ¡qj fgìO¾ë]ÝI½´ÔñZyEöP#ÝÛëÆÛôÄeêðMnª,]PrRC£¸y(x0xx+½=ÎÈ7tÿm¡1²Îñ.÷R9Áë=»¥#¿´ÜëCñT/cùþ^K-#3Ð¨5Jó0föÇ4f²ó¼¾Cå;¢W^QÒ&xÑÛ\SydëÖÍûJJ~ÝÎ%Õ]h J-d; Òö(;Svîèìëfeã8CV:¬°I®Oy.k�niëÐ!çqºHL/³MÕ~ï 8ÆÝüÿ;%a4òkjay9¬«\|5Î5ÁYTqV¡£{áÑRØTE'ÉÆ7ÜÒJÞ´p¹/±VÕM°¾ 6g¼çÙÙÙKaI¬9ñfÃà,<È >ÈÊôbÈo¡×-åû ½LU±!µ£¨¨¨Ú×}ÔõóûÓãq ¶J«ìvkÓG½,Ë{¼ J¼dCÌr oeòìº/>ò±¢å¹© ª\'.Á5 K^fs¯¨,ynTè$B=dpïA Wzlååõoc³57µlÞ¼uÛÖ6mA\mm¢Oî=Íq &·¶ºjkê<¸mÛ¶Ý{÷WTU¿ g6¿íö¦Ã¶lÞ{UUÃá&á}b!ßP\|W¤+Ö"vÖêJe³à\|gSü·?¡µ(;_xC%}<Ý²5ÞWb=[&cÆÓ }cÏKZ§¹<ÏÇ!w^èèÔ(;Sv>ëÛQÖn¼½ë+ö%ÅåkVå%ñ^¿S:ô\|¡ì\|³3StGÄà¢Àsv¶TG½¤¸«ï6Oî vv¾Ënll¨ vèðèKÒlüÐ-¸½8ZÏÝÍÓ\_ILc¬y6z} \|½G¾Û(Ð&õ<è½ärÍ$3ÔD¿ùHEEEõ×ßAIym®ùên!aªûuÊ¦:¯©LÍ"ÁgÓy,-HÐY×#Ú@+ÕÏä¿YÎ% ¾2G2ÿ{1¤ÁzÕð»9ßB³ílIê3wñNozÒ¥3¬K#¡%Ô{ìÈAC®Ô-[¸ÍjµhtUÁ>t´¢=Ô!ºmszþùçå´m$FÐý<6)uqµ ,Ú :Ö¯ÿõÆá7@kÐ«µÒîôíÝsþü¸à à¢D¼{-ÇöþªGïQ¥:\iö'ó½û]L@ TQ/ÏøÀE{TQv¾Ð½ØòaxÜaN¡Å¬Á©¢y)0ayÝ);S]Ìì,g÷&\.@gdgGGJÉ7 ÆÃRv¦jÇìl"¦ ;ðäÑÃH-¬ØÊrþRñØº=¿Ðý7ZÝ,v«HNIvAÕÄ+ Ï1qNAòó¸ë)Y¹ÄI6>60Ä�;û¦ÏUÞ"1îÀ;uc¨¨¨¨þúuB}Ë¡õfÉÕ¹ÿ=udÍDq±ú"É /¸qàþÝÛöîÞÅ %½kî½9^©@cí;Ér\_+g\_RÓÛµ+óÑ¡6ÏèÃ\_>&á3hÍæH¼7WùÞÔVk© @WÚï¨pd4þÆ>mÉò t"Dálä ×¯ýyå\|àÃfüE«Wi45="\ èû¬ÞÕ"ggãj\FµòðYIàµJ¯ è>#þ¥1rÀ]G99NOïqAWmsù! AÔlxø±¥µÕ6@¡fhtH(>QÔØ7>r+w1ºäSv¾ ¶O\ò4ÊN÷þ\|dðD¾XÆ&eçsüpdI{9ùRRR>L4\_8ë¸%,Ñç;BRpéK±mùÊÎTíê@¶ÅÀn¿¼sg¿áûÔvó¾>wràï\Ì\_Þ÷ÀãO Ä>$RRPf ïkí// fzÉÕ¹oûrTTTTíduB¶l÷÷s 8ìØô½^Îý�#dAAÀQjåo, µZd!ËMC´0; {GÁ#Åb ë~ëKÅý7)ÕØ[ó[^E&H:ScÐé&õZÇ=\|SRL»V\ñÛlKIËJÏÌÇ=l±÷Yk¶ØÜ¿v¯¼WZ[²ÒLJG°F£·þß¨±S§M¿ç{ :ÜYKAgë"®(¸#ñ=°[¾¤áØkP«ûõþÆ«9ùYW NERºUÀ2àÊûÇ7²PYÎÝ.È;«>öß¶mh%)@ÌÑë¢ÌÄ¯°3eç¿èÝÆÜNsPvþ3BÇsîóå¢eç¶÷¯s�Pv¦êøìY²áuþ\|og~¼äïóz.ÙÖ nO-Ùêb0°íßùAÁã&ÄÎP éòåºAö0/¼AGý{Y¾îT¤@0;GÄÎØè¹¿~6×JzJcO°Ê0Ñ¸Próè!öOÍ²êq.BZcÔOÇÙJ%ïÇÓFÅË¨k~cþgleq-2¯Çíi®¿õê!x@eóÖTÈ»-å· ÅORºôüiëNbYè\ó¡Ýk;EÊïPuh÷V·]þÜpµäÏiáXFÄ%d »üºh[8~\|ô¸ :À(;ÿµì\|²¹AÙùo,»Ýî¯AÙùwÈ)ÊÎTíüÿÃì¼OÙª°3»%ç¿WZj¶ûö27G·Eö8+àÌB�M¶ì¼àÁUQ§bgþôìX¾\|hg9Ì;×îØíUÒE×!. ±gÝr3ÀÁ5ÈÎ<ç9{éßcºÍ·ÜûÃWhBI Û¬Je3õ}ið Ð^>$XHsÞ¼rm ÄãL-äë¾ùümR üuë}ÁF¤\«ßØÀQO·VrV õ±ó' ¤=egÊÎÿzÉE>÷Úk¯\¹²¾¾þÌS�ýUÎw¦ì, );SQv¦ìLEEõwèÌÜ§® ÁÙ0}Ø55ydvVÀÄboüfå?ÿî¯¡ãË/¿øbÙw\|¾¶ÅæhÑÀtÿÜò;Ãû;ËÑ'=0ä¯H¢ÅpZXxÑ3TYµ '5Àa½³Wßwe;¯IÍÇzPyî¼ªÑ3´n×üS®IAY.s p#0\«w{sùaçmÖªUç¾$öF¼®ê]kJpÝìVý¨ÜT��Üù±ð¯¯vº3¸3Ü¹·Òß3:q×ÿÝZtÚ¥ûWR£±;ÿ¾q\_6õxîu��î úpg¸óÃÍOö¯§ë5î þ3w>z��¸3;Ãapg0ÀÜ��¸3;¸óÿYÃ��ÜÀáÎ�î ��àÎ�î w!p1;��¸3Üî wpg��wpg¸3;��¸3;ÃÜ��ÀÜî àÎ��î àÎpg�w��ýÙ-ôz½}ösgéVè+Üî w��EaÇÃ:¬»xàÀ¸�ÜÀáÎpg�� Ð't»Ý¬¯øþûï¿úê«ë×¯gBwpg¸3Ü��0Zbf¾Ì¬=?öb¸3��# ¶º®K5´óòò;°ûàEÚ��À�tçÀãÆÆÆeËåææ2w.\\zã7zZ6��0�ÅÙår½òÊ+LKJJ²³³srr{ì±Í§y<iÍr3��0L9øFeeeyy9çüüü²²²íÛ· ³63Æ¼3��FQæ¿åÖæ£G>þøã999EEEsæÌaîüÒK/9Äû��0�Å9°ZZ°Ýnýõ×,çùY¼xquu5AÃàÎ��º ð#SàíÛ·Ï3ù2sç[·z½Þ3Ü��À@Àn·/]ºÉ2³æÜÜ\¦ÏkÖ¬aÏË½ÌrnÝÃ�� Ë¼é°×ëeÌ\|¹¼¼¼ é3èÚÚZ·ÛMýÛ��%ÎrõµüqÇeeeÌå"í9sælÝºµÛñ8i�� >³ûC-[¶saaaQQQvvöË/¿ìr¹ä1ª�É§�� LÚÚÚÖ®][^^^RRÏÜItMM ¼H{��<ùÕu]JñöíÛçÍWPPÀÄ9//ÝWVVD��ÈîxÜØØ¸lÙ²ÜÜÜbfÍL\_{íµ^��8»\®W^y¥°°°¤¤$;;;''gÁuuu2È¶çåå\¹Òívk>)³��º®».U°~ýúB&ÎÙÙÙ-:vìn��a»wï7o^yyy ´´tóæÍÁ6��P>räÈO>]PPWTTôòË/···QUU.ØÆ��\|¥ Ø¯¾újnn.³æ9sæ0}^¼xñ¡CE aÞ0(ÞÿáÙÃI@û�� {ú#½nLîiåååyyyÌóÛ·oG×%Í§±;ì^íÀ12ê8êW¿þÆ\_"¾ÊÿD¦ êx<vÝC��Ú°öNFÖOú! �Ò{~ÁÏÔÖÖ>õÔSÌ333srrß~ûmÃáv»qå¹B´¥~Á@\ÿmw��Ìôé@·énà\�rR [ æÀcËµvíÚ9sæ0q9sfYYÙ%K>L»N ðñ^NðÐa5]{½\2rê¿çÈu8(ó��¬xäÈ/NÈNÁÁvÐa\_þýû{&\_¹H»¨¨(777;;ûGÙºu«kv8Áp¹\{÷îÝuBzÖÐ¿8pà@cc#Üù?¯2Ü½{7JThÃ.wN§b ��ë:t5¬³gÏöx'èÿ°ÎÌngù { Ã\|¾wv\_]]½bÅ ¦Ì%%%yyyÌ \_zé¥ÖÖÖ~ w«µäÊÊJfXìÁn?ÒåÉ° ôdM\_å±cÇhÙR´\_µqî¼ÿ~Ö²°KÐIÇ@?"ÐÊ°²v ��yXcÇz5²ÏÏz8¬os555UUUìëäH¾,»7N§ó·Þ\(((dgg/\¸ut{-!p;³ó&ÅyÛ¶mìÜ>\|¸ÆOuuõÑ£GÙ9?D5è°o}¬üïÙ³gÇwF¬°ÿ°q Ì;³vì«d\_.ûfåh¡¬Øá��¸3s+Ö]TýÃPBödjkkåAõ5ûºçÎ\_\\\TTÄ zÃ Ô¿'î\|Rw>pà�SõûûWÜØØ<ïoókqgfÍííín·[F �¡ÀãñìÙ³gÿþýpg��ÍåbEy½Ð {¬oÃz8¬{Ãú®v»}ñÅ¹¹¹999%%%«W¯[ÕNPBàÉ@®W<\|øp¯HPÕ-D38ýëpç¯ÑÙ)u¹\e½ #±½\pg��ÊYÛ·cÇæ8!¡û~;WVV¾òÊ+sæÌ)\\bÖÌî.]ZSS#8E³sÅêÜÅ\|ôèÑ^~ p÷ìïÃÖÖÖ;w»3Jþ¿ ;{�^¯ÐC.È;��îÌzpçPug&ÎË/gÊWXXXRR²yóæ@Æ%$9uwv¹\pg¸3;¸3��¸3NHHº3ûr,æææ®Y³F.Òd\|Æ>÷SÜî àÎ�î ��î wUwfßoeeå'\|ËîÛí)«ät\ÎÒ)º3;oÒàÎpg�w;Ã��ÀA¹³L#¢ª\Öfÿ'5E®Ù9zug�wpg¸3��wpgTæÎ23Üî àÎpg¸3��¸3;îÎº¸ÁQìáÎ}W¸3Üy�UèÿÇ2¯ÿ»¿O§¬$½üÍ:+¨³��¸ó¿ó®«]®±pg0ÝÙÐ;{=o}ö t-ôÜî wþèSÍ0+ÆêÃ-³"mPR¸ïáÑF÷äµ¥×üOê=ûiq71ü?½TÌn-Ü��wþª0ëÑ¥9Ë¬êQÍKn¨ës7w±øNpçÿ´wä£è¹5ÃkP~£T¡>®ût=!}pUãÕÇg(ºá¡ÔÃ (¸3Üù«XOÜ8S^q(¿÷IÖçõ%´×oÀC¯Ê±Wã¥×k¸4Ú!.^ÿuÞ'J=5\|æðk[6#Ô«Ý WVvsûX«AÃ,$54Mñª§ÌØ÷zÄO>ñktÊú-EµN¡á ñ Y E\_ÎxÙòQÜÔf?¨ò¿;ëX1�;\| ¿ØTóøØõÛçP¸3Ü¹OÐ¨æcCã õVõmX·ÅçnS7Çú4npg0°Ü9¸HKèÒ;;~X¸YË x¨¤¾ª8(3RöÕ\]ÝYj�¼mTm¥F+Õ[)7dÝß.ônÖ¼Ôí¦5Úø�¬æÕ¢¹´¥:]ÅøkD«Äw´Q»vøÄ +ØM3æï/G«¨Î4ßÝJ[Û¨ÛCU vy¸3��î\|êS]ºÅ�ªÊTÙÏí\|î ¢;wí@¸E øo÷ÆJìK=PDÅ1ÔqóÖBÜîüÝYñ÷òUª0ih£µ0m>êPùr]Ô¾ÒC²ºÚÐL«ªè£ÔQMi9+,/ÿ¦~z=ü±Ö9iõAZ»Ö}N«¨ËÅõWvgÿÜ33ÙúåQzd¯¡m´ÒVJ[ Z_O¿Î �;Uª7SåøgB0"¦xB¸{w;÷íÎRÜÇjv²Åj!VÆo$t6yòä;þò×·Þû ¾ÍÝáßjÆwªXwÐÅÍpÐª& 7«KRñ¼'åîh´}WÃÚëî{îô±Ãïü½òìóiIõy}ÞVçûäÖ²kÒîj2:-÷ÒKKÖH=æ¼°ÞÕ½íNï+§Lsß°1S®üÅ)NêdÖ¬ûóûåÛÇg\uý¬ ¿D¥ï\tgê9£Ï8ûßì£>;Þ¿�J7ÿãSÐª!:tµÅþþ¬FuÁ¤nÞÐZë ~��;êçtQ½ª¾rM²DsAHÏ?ÖN;ÿáÎpç)�Ó^½½æàB"xÕãöL.ô9..Îjeu'¡'ßèòh¾�w;U}6ø$X»¡Ô6ï~g(!©L DÇ¥LlVh«NC;8#Ü94k.mqÑ?5jøèìÑ#ò~þµ;ÙÓ\|7±¢SÖ¹ùøæÆ§ÇÝV<ÈR4&9ó¢JÚìçeH}öQwÍî½ÏN¼?&9'99ã_<Ýç)ÝQôÄMÃ¦Ç$çwaQÖs1ÇGíË>xðÜ"Òs£È}Ñ3/¾oÃÒñ~þ}Ó¬Ìùøù»XÓvå>üýÁScbòFÈºñ#ÚÖ"ÝyÄ�À¿¾Æ}´ê%Ù¿ÅD!³\_·Öë£¡}î w>Ñbò«¹ê÷swäÃB5EX$xöÙ\³a#C3ç¯lVÍ7#$ôáÎpç¯èÎ\¥n¦6Ïø}}1«Jüª\[hgb¸3è?°»9kù·eEA+ÿ©ìi¦Î·2+U)"æcÝªúùëÿ:n+ 'å1¤<%zÖ9Köm®ü;SUvy{êèILaõÁÉ¥ÞläCOuìwå/º2i:±æ[<R¬ºA5<ôøî=Ä[HI4)Rxí{úJ:Ê¡ñÙ>ßä¯Q©V;ÿÊÔ$lö´ÜÞ¥pg��ÜùdÉßq îÎë.ZÿÙeÃ,ÓÆM<Q?¼yÖÈQn5Äè0!Î1ráH8¹êçi#KÁ³î N[Ú©víÒô2;ä]zÇÖ§Å¹ª\\U£.ÍwÌM¿¸¥ìq3I\| + ·EFçñ¿?~Pø¯{1¼Ê « ¦>rÞðL;{P¬o¯£Í\jÙozübæÎ¤häîÝ,å»;ÏÝxûÐ$ºyQ¤4Ìzá¿nçÛ$8ìïQLwÖä&lªoÊ[pqj±IÎ¿a=mn;¢áÎ��¸ó z7ÁÉø¾mû+òi³0xÁçÚ¬$Þu¸Õ%¶Î´»pg0àÜYe7tgNÌ ):$±Êì³òã¶·\1nhÜ 3ì{.jeæ�wÌ53tó6J½¿îÉh9¨N&]t¦Xa%c?¢ 8ÃZÐ-4 Ü9D:zª{kÄÔXRsù¹OüµÁß\(BWuî×UµïLºß:4$ç%Ä&EDEå>íö/DÅPÙ¼¶´¸è[×M·¦åZf<û¡\_mã®"VØÇs\_28þðÏË·Íÿø®QÄVBÈl~³¤I¯:æ\_ù¡R'Uä T( f\|÷èùÌÃfKwn²Ë î �;÷)Î<Í.®>§úÜüJì©þÍwÏ·ñai¯ýÑ¡í¥«O£î/]Ù(B {t·ØB×XæÎìûe=Ø@¨d�wî,çb[ýþ\|ÞÙj%áwÍ{®ZäéíøÚ¶ó[ùh¾ë!î²= Ô©SaÀÁ�qg7B]MT©ûí·Ï¾uÎþæÞýDîk ÃïÏ{N¸Ù®RÑL¦¸s6uìÞáÐöý°ìÌÁÙácÊxò¬É^^ø ±=¬y}sÙ°#ò']{÷È¬¤¨Y#b¦\|÷ùã[ýI ÙCÛ>¶?1öU 9øîW:L ²+íÆ¹K.=$N0gFe§w³W;orÆto~8GXÎ'%äLþ@s´ÊÅþV-ù7\ä=v~3ñÙÒrox6; ��w>q¿ù/-Ôhh=òQºÇ?"Q#ï5º/Iãa±dø·j¼ðE\æCd÷&Üî\|¯Ùö5ìû#,$"ùöG^8,æ{AsÒ¶Ê_^È¢IÄ%»êLË;àÎ2¨¯JÃ¬núåö¡"¸ÿ]ù#¬î{þÙ±Ö(>D×¤"àUt_Ð ¸38/¬¼zÝôèÌçnuþ®ºf3·²)e¦#9Ë>366ä%¥¿J\| %5HlöåS7üÓìMÊig>KÝüÄÍ a11yg¿T_!³GQ.àý²ç¸Äe?{Ú6ñ\àá¡-½{ï÷AÅIQÅñ ñá$¦<:{Èù:(Ú)ªZgu¥ï.O)În.züÑu÷»X O-úp]Þ¿YRó£#2î;+gó?{ø~gÅÃë«ëJ5î �;²;+!Ñ"@:¿¸b(ËL-Ég·ò±Îæö/7ÆÊeÛ¶!ç~ïWêðQéÎZ¨ômàÎpçûá¡º½aÿ£Êj!6æÎÏîÃ÷½î¥-U×¤EòY6KbÒ9?«½©vî ;bá£j3uW_6Ô6·-)aI<>¯ÖbØw¤¹¯!üÌïÿªA4+þµÞ?/-ÜÆÜÇ¼¶þçùóî'ñÓIböÿy¡1¦j°Ï¿gçæ§¯È8#=vÊ¤ç«¶ÕRWöê_7køë6«Ç]²Äó[ôó« RäÚHÆøå¿>D ¹Z3Xò~ñãç§O%¶YãÇM ´IÌWk¼ìÝ{'Ü2{PRÞÐèë%ïÿï¸a iùÃÇ\þ9Î\uQOÉÞ.¦pçAi¹7vwgly�ÀOèÎ"Dóè¦çÌº2ÈÏ 5zjþú£Ëãù²í¼ñ#»Î»D´ 5àÝpgêîÌö5»wfêL"îY¶²Ú\Mg"¸1ý¡n×#÷Þ5Ü¦jÖ·D¿ºoh\|¿pg¸óI3Òî£ÎºCï7¸ 9ïE±¥¹ªîÿÙ2"IþþÁf¯ùZÕ ·ÁÁé]ÌuVV)¾7î?))'Ö{ÆYÙ.cHòËzXMÖýïø{#£rÆ(½ñu·SçUO\t¯mX~\|ìôOT}ZoÎ,óDT;jPyÉ½<&û¦(ýpö¦Í ÝYDãhYüþýgÞ;+1-oüçt[+Ý÷³9gOÊHÉLÊqNùOñ[¡ép;¥ïæ=:)-î �;jÇ&°f[¥b»ïØÌ?Ü8Ø!²£Ai2õl\_³d!ñ6B¢þó·�j©Þpg0ÜúXq4ìÿ8FFÏ¶Åü!¿¸¢Õ¹¿¶î¨ªªª-[¶¼öÚkß½úêT3 têàKRååEïî w>©8ëS¨»¡ø?àÕ!ÊJÏøèüLëw¾µÄ\ðûo4MÁ 4¤úðpçug¾Øºz7<ú 1¢ 3~ð>íhàë'>\Jüyá#§Åç}Éßÿ\|¶j´¹M}ïGÓ§N'qyÃ.Y0uúeðíËûþõÑÌÔá$oÐà¢¿è®k2g7¨åÌS3µhü¨)½ºstn üsvÒí´áÃ£O¾( sÖÔÍOÈÅäééó]iºóû´ÕI±f��w>y÷FL+ºZÝÑ´cT%6ÌBHÜùßù©+ìq¶S¶HxÄ°HÛK÷µÑ6Á@qgÞë©ß¿5ÊbAX-¬VÓ¢¬"^Ü¸þú¹OTÍ¿ÛÙïî w>i1 ¼ªÐc;'ÅÄB&~ÿU2®pk+7öCg¦^§b¢ãÇÔµ¤ÞûáÎ!ëÎ\^ù \_,FrÓÈô ®þ Îtn9ÒúÖä´äÜ´°ìIÓ¶ÿ½p¦¼Ûò¿f\|q¼¥Hzî[DMcÅÿ£Ü¿_5r¶-²xÔàßUêàÇÝ¹péÉÄ:{ìd¬0húÖ% \ZC;Ú´©nÙe3Rscg1o÷æCfn)I#¥\2êËc²ÓÒr/þÌhÀ= =´ÃþÌS£3ÉyÖÌ [Í?þ³¼'®a&VäÎF4R��¸ó7&æ¼sËÎMCX_tÿuw»V;ÝºæõÊ$ ´õÈÄô0è9<õ5båÑàÎpç£ÉH_T¯:èwgb9÷'wítÑÚ6µÕO£}ð¿/ytP<³X 3è(1êÝc¬¥ø´Ðø~áÎpçãquðÜ J]ÝÎ÷[W¤ïÝ!×,y:õÚM] F°òÕÉ$rô[Õíf>¹³&.Ïpgpö¡ä(5ÚiÃÃëoð -¹ôÜÁ7ì¥íàÝþ«£Rs¬Ö¼Q7¿0'82ÄàP{#ýäG77d4ÿ×µÔã¦öE[¦iuñzO]K ¦ðRÃDù£Ù/HA Gõµf;&')-÷ ½É?am÷ÑÏ}÷®!Y[±fÅYrõfçQñO{ò]-Ý99ïÆõ´ÕNýÖ w�À{ÿHþK¿á;ãÏ£¢Í×ZdNñ("&"6)ÒnI ç{¡#±$,éûwæ·ÊíÒî BÞuÑW ýèmQÄ0¶Füuî3Õf\|ÔÎ.Cmk´ýðO¿5>@È¾÷6¶)>ä¨¡ïÎrÙ§æ¦îå÷üUKØèÀîùpö³lb¢Å\|sb9E¢Æ~ÿ×8EµRÄÜÖ Ê$:m§ÍÜOBbsbø¯Cûni~öûÉqÃù°V·ö\|¹J¿Ì\|ýC3HØ¬Ô!ßÚi´§ö;úÎÈ¬p9vò¢?UQÕKÍÑ=ÝyléÔJ>mv¿4jÃ©gO?©¨eîlw·k\|¤ªéÝpó¼1gä¢¨éÃÏ/øÃ~þÒË^ÁÜÙìÎþ��pçÞ»7F5;Õ' ³É@.¼ÃsTÍn ; y$ÖõI(4yG»0:Ü �wæk¶u½F¸s4« 1q·Ï{æ[ªûa%GUEgÇðRµÑuøÓXs5w=ã CqvzAh»3ÇGk¶+¢YI$¤g±I"¬ÈpÞ²[ kWxóO"µ¹iQãhhÌÃC´©ã¾;kë5×Ñw¾3;mXY\üÌ13ßZ³ÒÙÞ;1ÄfÆ%>0î Z#TÈT¡uÿ:wæ SV\|î[-;÷Ñ«0nî ð µöáfªzô½«;,º }:(5®¤/wIË?³V3wvkØ×æ¢û¶4/:ï~2\|vLt^zBÆ¥¹ç{èÆìEW$ÎäîßéÎ�;÷-Î"ªïØ¾×ÿ!3o²^ ïáÎÝ EðTµdxT¯Aµôõ6±'4\�î w>1¼¯éõû¶ñüÎ¬fDÇÜ±e¨½Û¶î£n×¬ßý:o#1i¯nß¡æàÎ ´ÝO={w¿¶b b-F$@,«D[xÙÀXÖp1ÇZÖ¼$?ô×fË"å îÊúÌË\ûF)Ýõ«£§;òiÌmöÏkg´³&=ôÃ¾[÷ùzhÃþöUWNI[8¸lÛ²7Ý'ç¦ØR"§]ôdsEu¨º«;\_t2wKË?»Vµ\_$°¼Ï ½{É;÷¤ÜKbóc-ScÇÎý¹oÍÔ§oÍ;�àÎ§Þc£Ôë¢®£¹ê ¾^ÎA,/ý¸æ½¶¦©©¡ãX£½¾±¾ºáËÆº¦æÆcµGV?óÖ½å=dÐ7îQÌ .pgòîÌLù\| {ýîó§GWéÓrÝábD%¬ßsÀw}\|§f{Ó¯¯½0Q,e"òöûí Çjê«êkêëêkj«Ö6kn©úå4BEà½!çÿàÒÅMàÎàtíFÉ\|%:\_í«\½azÌ6{ØèÒÛÞ¥õçKFl3ÎòÑR7»ö{½!²ò8xL\|·ýiîä!ÅáùQ?{áOóëþIrbIVÙ?ùªíTÎ£Z(ýødîlÉNIË;\_Î;³\_Ä~ßuP£¶Yÿôó.1,/ùÆyþÉïÏ ;¾%Á¼éÿAú2½Ë»®ê(epç~wo¼íû7óÊÖøäoik¦ Î¾Ïc?~VbÆg VlÝ<ïÖ»\|pg¸ó)ÙÊÝ9p wç?,\yX,.íµ¹P5§¡úç_Är×N ÃvU·)X³ ;39hlÿ4qbÒ_ðVôa¡"¡é"²¶ù<´þKËÕÝQÄ:lå»{ÌÞÜÆÝ(JÛufùÞçúMoÉJ³¦¥]2uÿ¸a1Ó"§]óRÏzªGõ±;êuó5~±ô;çEÙ&,¸øæ·c-&Y©çÍýËÎn-ËÜÙîìäîü@lLÖ°´YWõ¦;óKóñ^ky-/Oº+fHflüÌØÔäYñ¤ sÞÙáªBé %õõT@0o;+vl«¬¬Ü±mûcMÎY»güu¸3ÿ½fE;ÃûsbU©cåb³{02ïé 5r£Yça¤mø/ßÞÇ³2ÓåªîÈ¤þ8ípgvrXg�î BÖe.UiÜûéÎ¶®Ùö¾Ö{?²©a)$\|èÑ·ùÐáÎpç^¯¢þ?¹sÍ«Ëgý?6bQ°ü5>«j¢wæ4Ù3U ±i=SÿÙw¤Gð�ÄÝðûyoOG~gpº»³ /S»íÚîïMJÌOÎ4{xä¬°¸ÌÔ3fÝö¹ÌË¦²rÏ¬ÙëáÛ¤Ý6}QóüÜ¨BPN²H-lzzAåZ1À¤©A _³ýqéã zØòÇ-ZÁ3 P9« w>sj´pç++ôg;+<´ {¬tè»¬6öÞ¡ñQYkA8É'¤pP¬ë×Ñf¸suD°FVê¼¼ÏçÄ´ãº»eù&usb ³Gpd3Ößß½uWÍ1·pk®_uM,»ÓüZýU :È}ZçlJÜ¹^ó;¨rlòÈ$áÎI$åìOkìrM©ÞÅü9V{k÷®á<Î°¤¤¨5üº®(J ï¤ñ uÿëþÑÒ¸;×ïßÊúö±áá<zô¿¯ùèUÿZõüsò¶ò§^~ñ ¹öòKyÈVMxÖZ°³¦ÉG5\ùÐáÎpç¥"hÞY=´quºÈÎ@¬ÑgßôÇz¹PÏ/Ý;è'ÚN}öo1<ÚÔgë=ó}Ù5Üvî¬u.rÕxFçãÏ¿wß¥ÿKgGØfÇ%¦äN\ÿt]#«ªÙ2¤8óLL~«ïZúÝñ9a¶YD²ÒFç]¹[?æ1³"Ê\_$ÖÌjÌ·<ôè#§tº3\_õ¤î¼xÃ Ó#¢rRx~g£Êã\_¤ë¡ZUÝ®YzÑ°älQFHADdáØô¼> m:»$¸8ëþå\|@[ý#m~S¶k g¦l ¤Öáq-Q¶è¨dæÖ ©£÷6ñdjföò>ËÅÛ\_Ù×lkmkÅÞz¸s?¬].êþrú×FËú>aõ[ûíÐût^¥£\_í\_nznIï±Ê~f¥-ò\Í³ÚÁ.Învëw«íàÎpç}(ÞÛáãMQ¢M$ÑfµZ¼°&XHjL,HdÕ¤háSæÒnÃ w!éÎÁÅªé·ü0ýË$ñ©7?>ê©yô}<³§§±6ÄõÞKËÅ«", éÔ©¡î²îlÍ;Ënµj}ñ;SR¦[ ÃÇ.úÖ§" nµ¯Êó¯¼PèñæMiI"3Æÿø©ûyv\\æ¦Íº¡S¾OÇñ.\_xÅi$:oÜ¸ÌJ-ûí}4eÜt-gHJî¥[imwRÅ[úÜ\¢Üì&bXþÎ#gYI.!¹a$wLòÌ®§{§Ñ(P]Tµ±±Îz¸ÕL3jÒ·s£CcÝtEóz¨Çµ·²ò'7\|?ÁÆ:6VÞ»yÉë)u"ä1ÿëº9ã;©óöÑér%yPèx�wîî¬UmÇ{)K\xäª>ÎÙáóèU¾vê;N K&QéÌVnª8\êßÊÆÕ³SaN}pg¸sH¹³¦ÔøÌÆiE·Ò}¶ýD1ICW½ña:\|!õýáÎ'ø8¬?6" á¶èøQuXAªôèKÏa°Oíò5ÇemÖ$BÆ¯YD ¯î²ú,Ëogu©ôóÿÿí»££g)¸\áò ÙX'Iå®-M/{_rz-:+9uúÅîy£«2?ÀTlÕ¹óqJ·,üîð)ÖÄì3Î)±KØÄÃÜyãgLML1fHÖ5F½pg]57«úÄ&kwíP¨jÇôÿýª¨éäÄÛr& ËøùFC±ÓÀ2_¸s¿×fþ ¶×ÏÉ¹3/c(bM^ùÖÇ^¾}ÀëøM\\|mö×Á§¡yÀÑ3yÁ!Êºòªÿ^t»µVßÐÜ»;w¸½F Ì¶¦w'd¸¤®+�Üùôü<]W§±#òîÎn¢Ôn0éõõºµAtÕ½<¨ê]I"IxÒ_ª£´MÌ;ÃA¨º³eÐk¿üÈÄçR»gAvúÑ~øøãoþdk ²°Öå}àÎ dÝ]?+>Ý+§9"~yo&¯ºfèC°ö¼¥Gq±ä¡ÜûeZ72üsÀÁé\wý3q²ä - G¾6ë{Æqî ©WÞ÷æü&ªø:/úL9ygðÙîvZuûüM6lÈÔQåÿ\|;¤öñÈØ<Ê.vó7ðÙ©~ lám~r÷WÎ\| JìwÄÇiÃÒ ÎþSú¨ûÎ4õÆº]¬¶eÝ9-{ ¨ÔÃn\|]mÜÔüEc=8fhæUüýÝpRÿöêË1àé¨ÙùA\|ßÝEbHÊ'mlgEKsµÛåo½tHµë?MH-D%©gõòtí¢µ»ÛDAde¤ÃÐà«} ÁbW¯£Kü ^ht6ÁáàÎpçÓð]vV½¯ õþ¨,}»³.ú?r Tá»l¼ÔáÒÛÅ!§¼Ð²z5Û¨,!éÎþ¶U]Esû EÕÕçe7MõvÊ;vhª¾Àê>î BÖå åÉÜ¼ªÖÁzÿMbGËÒxÔ]ÕOäÎ>1ËÑ.¦Ùø{9ªÀéªÍbû§GçaýU@aùé±FZSCª©ZCÕÎ¡SM,wrq¬F8èÁítëAZu¶5PÍIuUJáÿjÎ´\w ÑxÖ¦ÍÕ"AÏ0;s.áàôp5m®£^UÌ-j¼ÖfM}f6îãÕ¶ûØï=@«wQûaJd~R£§ñþx-vQoõ/®è_ 78«p©H9ÛNU+¥\|æWNöÊÒÁ¡á¥>ç¥có.qé$2ýß±+rWÑ¨ªº;óãp8Xõ>ùäÖrÕÖÖ¶µµiwê®~du\;T³ø[·nÝòIEmý±@Ãw;÷'}6:· d^L&~SDH]h¶8¬Ã«ÈÝ²?Xa TÝYóbXµ×m±ÇcF6rûxoÊ+z#nU×µ.áXCàû;ÃOðqÌÃòÖÁ ¤ü¦÷Ò¶°J¡ªSk¾3H,y Õ�;¤; 9ui\õ@W}QíuÖêUêU¨c×ìÎíU v66S»:Úùj]DgÚ,ÝAö)TµS¯=PI:ÿfÊ k³ÑÛår'ó(ÕlºÄ:]7û½b ÇAÕV±Õ-zfî]Ïþª+TwU¤PÄ@R'ï7%°N,(Ð\ác4ï¯^#ÿä¿{}Ã§.Ññj]Ûë©¯õçW\uÙÅÑ±<ßHxdTx±úWÜÝtýõmÝÚyU7;=Êys sø¨Û÷lÚ´É±Ëÿ´¸÷6m9Ô¸<øI Üù´½ò3¿UøåYó%u>ØÓG'ßïÎpg\_ÕÜüi¯ªªrñrT^É§}@î,/è¢èZOÿ5d ^ö®-%¤KH¸ó�¨/=ßw§;³ÖÀÓÅÅ¥Þçqå·ÚP](9I-¤Xv¯AU^rSVÔÙ}{uz¨!Rùèþ\|Ûrû§sI_g)b=1¯Æ' uv \æÝH.dvß´àù§ÊÛÎ¿O ¼áÎý¼XüëÝñÒ¢1Ò£ã.¸súsei ûÒókcÅÝnoeÿwuxt\È3å«¶÷=(>F¾'ßC#=;Ø5Ä<´ø%·25Dày½mÖ$<ùî¼cÓFóXßáQ$'aH[²ze35¯ó1£á[;÷&ò&HåcPF+KøÒ%îÜk¯^ÀVPTUk ;ËÏ¥÷ßÐèíÖw"¬Ù�îÌX%²3#¯''XJk¶>cZ¢Ó^Ä¹Öå0 wUOÑÍÄ=fÉåC¥fÁækt½ôûºõÎ]bêBÌ/ëAtúó.%Fþ×Î_Þavä:¥Ù×½Í2:ûoZPÜdêÿ,!ÓN Ø2é5x¯ü_o,!Q N·¡b¯ÌÞ³uµVÇÐt¾ÍçáKûkA³ÿA×ªkûÒg> 9åÑþ0~Õ×ëõ ôÔiîAÙ{áÎýµ@²"ècºýÇï§ñ)Þh3ú¶9{\| ×pÏàËx\|\íkäµÀ¨Ã.¼ùöÏ5¹ãm§º³î³·ÏM& á"´eð¬§Ö9x¡RùV�5ï÷7±¿Äb#\|^:î¬!g]¿öÝ÷?ÚòÁ-¿ùå/þð[êw§øóaáwpçP¸ \_Þ¿]]ôèÞ\wî7c\|Å[( ¸óam;9Ô>Kw>å÷;þÝFüV·¯CÔÅ\|oÝôß}?/¥N\|Á½xêoÌ}7ºpg£ª-×OHMé«"]°«6ñíþ"§Ø©§^Úo!èx6ì÷· w6;Þþ?Ãå2ïä+zOß¼\|ÿ¾ÌNÎ3B&º;ÃDUÕç{¾´´tÙ²e7ov8§Îì\|áÎ«ë"Eyy9«//¾ø"{\|>-;~ïÞ½¬²ÀáÎ�îü_tg wî÷(<ZwÙÈä$îÎq¶ÄñÕm\|J·[êýkÌnF}\|Ã25Z.ÍSsÚ~Y,X{4U×:Ît]÷ìÙ# �Ü�¸3øêÈ0ë·~Þe³Å;r\ïð¯>uw{uÇdç!º ¯OEë.Cí Ïn·×ÔÕoÚøáK/½ô»?ÝK,6k!#.üÉ½3ôBÜß_Ããl']róÌ:DwÆ'ùõP3dÈ<½[IpçÿôSß)ðß-§k½;%w.///((È0/Í.¡ìÊå¼óiêÎ½öÕû©Ò[Uó÷8Numö¿ÓóÑO×½?pçSqgV/ë¬& VkÊÊÊ/\_¾eËÖýÌ úðáÃL¯Xãô='F°»ÙàÎ!"¯ÿ¥_Óe'ò×P;cÁCñ2ë£#óß)6[G\|Ñ¢{g½²'þ/3Fyx²¦ÀmÔÓèj<¸êþô¦ëDNç0ñîþìÎ$\ä®"¶±ãoø+È§Ýr\_³<ñ;¿iIÖd4yzÕïÎÆ74DàÎ'¿$ê'ÿwýkm´ lq§"\Ý\¨~Zõ¦àÎ\_ÉÌJKKsss°ÅÅÅÌ V¯^Í:r\?¸³$(3ÈIQýI´¯k\_s"A½ý´,ÒÀOÑY½2±YMÕgÅÌYeaU¦¿¯Ùî3>p¼mßÁùvàÎàZ£K¦§Î'õ¿SãU-PU~Þ;§'TZ ÛÕõàaúÉÜYG§þr6\|TiYs{ áK¨ ]øÆÎj±XZë~¢-GÕæc¶ û×ì:$£rûÌ¤enê=ôÒã3 èrÀ-<º6#ÑAØ&¸ú¶óýÙÙ>ó¶«ñ5Û©÷Ï{ÎÅ~ÏèçnYqCW(\Wôh$à×2·ºÚ£Òz<ÿ\_vgöånÝºõÉ'død<5ÀX¶lÙ³Ï>ûôÓO³®~A}ÀªnºÎ{wzïÃ±dC§wußîÜ½/ÙmÐêt^zWðÕ¸m²$3Õ &C¯üØÞ;³b¦ eÿà=´þÕ¡$0åNHsõ¯êyÑqÝ·.LN1ÇÚØÝ½¹eì9ì¢ZkG£Fm÷Ðv·á¼þË¹WG5þ;r«öpg·Aeí¯fÍpg¸ó×tª#zÏ¢Õ»Îk£¾Éáÿ$û$¨ASo´R®vRøoVæÎ{öìaî,'Xóû !- ¬¬LªñÝY®Ge.ÀD½Yó5k9r\|Ðÿw6:Å9ØõîU¬³ß ø'{ûU@ï};Q\_uIïoî¼jÕ\V6¤3p¤;fCI'E2±-[v:®Ó8iKÔW{atöÄ5±P;ÕE½¥ëlòä\|§;wp:wl\ºôÙ Ð\_ÑçQ©K»9õ¯Vyë£ª/¸d\_Åã vUV ÃoÝçY½ÓÌs Mã?Y=7óvÀJÀ@dã^ª¢.>ùIuÚm¥8/Ìº®ðQÄ#¹·};kl ö©vtgêßä.ló\ÌÎ+'ûId\|ÙªÍ"3DÓ+OÄ3·ð¸Jþ{çuy?þï]neo{ãÀUµ®ºë¬µÕ¶£ýÕ¶NYdDY (((8PQD( +�! IÈ^ßûÎçÿï].+b.ù¼\_g<É÷ñy?ãóàòdÓDB¢ÃNü EiÍqÁnÄ¨·ÑúTw&IÇÀÁñ\|Ü\ºÈÍ;>éÈ¾1²¿ N9Õ½ >Ux¤ºêù1¥ÖgÊ){BÏ²ù%Ý¹¨¨ësEEEuuuÕi¨îcTPêêêæÍÇ¦¡Ï¼f;-- ËÔÚµk7nÜXRRR[[Û#EÀûÜY:y·@¯zWèik¼O :£;CDýð/ä¿¡ç»3[¶]VVÛÆÊÊÊ\ ªU[·¥8§§§¯\¹×={öàÆYU@lyî¦¿P\´× §Òzå])ª-y:½;+Þª"vYìÝÍ Òe7CÄWàÎ²ß¥ Ú¨ÅZZQ[+²[¹ÙZÃx&R~ÜKZ\|±YXj³!§dX°$ D»Ny8t ¨ÃN~[ê¨Gõ¨¦Õ[QYòk ¾~ý®±¼¹NºÙÉÒÿ@;jq"ÔêÎ°èñòL\D[ËÁo-ØYÉ~düÿÛ\xr¡Å%ÊEO"C$'¿9ó¥r 3Çéã¹KµÓüÚÆ©Oh20}øð#ÇEOßEòÍ8ÔTùýú0¢ÜÁyÜÐkm×Î¢ ÇäGn%eõ{îåàÎàÎ¿Xµ q¾ÖÛÄzê·m]ýØ÷³AAúØ£Ï¬ýrWnÙSãHdlÜNù~×¦ý_\|þYNOjèt:³Ùxë7ðÁÓôþÔe+Zv}ýÑ#÷ßm b9ùp¥ ¾ï§?ú\|ò.ð£¦rÉAÁö;ãXE¬{êiè%ÝY¶lóSRRðW6SÆæÚØüZjjþçªU«>,BGGzFr;¢ÇN Ø¶ã¥yéIZ cÇï¼ÿÉ¥ï\|ikóv'ót×Ð«¾ån"²5~»òùs^sñ»o¾ùÖë¯¿>oÞ<\|=çù1þü9o¼ß¾·)ãÎ555Ð;tÛ°z9}úôÙ³g³Å¸Ö°9zü£EaÅfRùr±¦q+2=Å«5cR«v~IöÅÿ»à¿Íàé]0ªøízè¦>ªZyxr yZeÕ´ôÉ&sg^{îùÝÊk<\|-UwÎCÁÓù5Ü´7 ©'Æ ÌLIM2üqfþÆe8Ät©Ì;NÅ-3mIO?q!¤ý4ZÌ6m.öÚd Ó)8Á;Xüj¨/óqÌôÊ-ÒqV®?loÖãH>ö»u]7ì¯¯¸<ÚN¼;ãFýÍ,'î,Û'?rËÉî,;;ç\Ù6¦ Þ$þã.£³pF/c¼¤ÄÓÇô¿üö2còxØ3»³·]Åìi\|îÏ´Lôz]y"ô´ôzúÏ>ßµ_¤s@è@B³½ñð¥Ã£uAûeè/c"ÇÍqÑ¦°Q¥Çì^}{HN$pçtg¶»oý6K¼bÅ v:oÙf³aîî¬zEònÞoS<'¾id0-¹:#óÑ³ì>áûïªxÇ©Íxw =F#ñµWß0 DkÑ¬Óuëà««¸-k¾ÝÐNª§¶Ã¢ç¯Ùw>;³3ÙÈËªùõ×_ß¹sgcc£6¢¨îìÈ Ý¬/´ KcYÿ~ò%¸d¸µ]6Þ+ÎIaWÉ² r5)<ÄFEùÑ»î»dqu CÏþ÷æ±±Á©× Ï}¢ ¢"ÈtÞÙÉn9H2êPPÝÌO^ûaøTCH"<ÕhÎÎ0&GÆ&3åêÌ¯\_¯Ar£ 8%E}´ ÞÖÔV\|Ås <ß%:iÐ ñmÀ¤Á±S¾3düc¦üék¾Ù¥ 6ºFq"©u»">(DÎª2Gb9{á¯7}³·¹ÝÍË2¯¸í ×®Çjë9íÄºæ-u4£»xó{Ói4Êéb½öD»Ã©"Y.#[GÅöDDbqõºó¿y?ÝJÏ·¿¿;ÓÑTºµ@ÁÁÏg¼"ûrS(Xk\_}ÇØ¨8-I0:idbDLRX\zXdzxhFløK£^úðo¦$aNälÜ³sÙÅ/ÅnñéÑìo1±ñ)£N}²»a½v ³v¨ävä¨n9øvü²NOç¿b8½÷ô\2+FÇ:IÄª3ÄõñQ£«¬¨D 2)ryëñoúÅDDs(²Ïdz.yÛËÞz¥ gpt~¡x½%Ó3 10¸¦ ¡VºVÉîIÜC¾ðì 6ï,¨ßTÜÜùü¹³ÊvZ0·µ\_¿ß£Éô8%lÈÆ;Ù#wlýª¬Þl tq÷&N;K²Ê6?ÁÙ7\_¾Ik[µjÓíVT×T>>ÊBÚÄa·9j#m=Ï\hC¶¿Þ~ñf"×SÈo6ï&oê«í¼ûæË-äõB9nÔÑ÷7ËdOO",¸sÏwgl©©© ,Ø¾};;û¤ v»ñwAp¹\=ÜÎ(Î)X+¬3Ó>93,zû·Ó%»m5GþxÇuáA\´)ÒÌ¹¡Í¦eY9åè®M?;úAtò%[a(£^3\|Ä×\_ý½»½üî®;ÿü»û¾Ü¿è;<«,àÎgéÎìÐó?þÍ2ãW ÿî¬tÍV¤"Áæ)ßþøc"èú RL±O½²¢æÀ©;w+¬³.zØz¡¥¢t»p·háÛr® ªµ+¨GGÒß½ãò\L%bÖ\Æ £§ýñ>whf:bÕÐÞÙï>8b<Ê&["SF$ïµí>~Ê/téÍ£3#¹tN>pð¾en£P7ÙXç.ÿlíÔ±±Y±é£bR/¸(ïÒKr/·òÆ¸ßÈw²¯¿0õÁ¯ÜµHO5z¨;+ÞÙg ·«b}]å}¸×«Ê:½ÁÒ¹l+V ã\_ÔØÜîÍí.G)¶¯¾ü¬s½\ÜôÄÅMÒÞ´ùÖ2¿¬7r443/]ý-ÓÅ=3{ugÈFà)àËàÎç-jñ&á¥»Íu\|Ëe ´68]ì¨ÒÑí{¨èr×Ë6þsáÃ·6uD¸ÌJ3¬'\l% ·° !¬¬¦ÍÕÅ·ßÎ5¯g%hµ(ôùËØÞkONìÛ¬£k»MñWüþ)ÒÔ²ß´ÚN$ZûíoèD6ÖçÈµjëiáíàÎ=ÛãoóæÍûöí«=ëHIIIOvgöº B i-Z\_xê÷&+ ÓÇ}úÕ·/ïþ©Òò\_nªÇ¥×§ð·?Ù×¬%Ó²ªróBò [¼OÆ°úsÜ æ6(Ó=Úeô;]ÝTãés:½à:t÷àÎ=®Êàb¿fÍÒÒR»ÝÎVe¹~1wÆUý³çÙVº¦ À¿½îÀæa!\L^ÉÔÓõ{\è7úh©³Ræ5e;§yü#·k}KØ¥\_ï\|iúpÄ%Týû´mIºø÷¾Øg÷t$õàTFàÎ.È%¯¼ÿ·þ,ú)qÆÉ¿¹ïi'h@dCtvç~üèÀ¸è©¡æ¤K¯ý¯foT?¥é¾-¥ÝzØPõ¶¶ew¾<&\d Ç¥q\n.÷êÑù6!Y¦WÜ j<Ü¸è¦çu±Iqjø ¬;¶ keÕät ZíüÞ=ñQ\VlHò ÿoY¡fôDÛ:;ëö¡¡i#dþ}'B'ÈÀf\|kEmM$ç½µ7¶XæX'�ôÙïèQr«dGX\|ø@bR¢¦¿F£Wu!¡Éðì³ÏK<gÞÊïÿyøPñ]wÝÁ·Þ¶¿ä)ìmÅ#ÙÜ³>ì¹¸Î8Éï OþûT'§¼±Z[³ \|·óÎ�¸óyqç¯¿Ï¶JêoÛÑJoHDkÃÚëðÑÏ½ò>IDöK+§/¦Øo\_®cCRÁ3±¼¶àGEª5ï¿>I>fØo\g¡[RÓËItÕ«9Óúl@¾´Ep )ºÛvmÅ aÑá(ôñüEåtÉ·ÜBpç3÷,¤?Ë+=Ö»84Èol®Þk ®Kú~=> Ñl´¢mAíG/ %#>:Cäåwÿ§R@Ô%r¦ì=DsçÎÊäðßäçãH}8ü¢{xº¤ÊéÄ4Ë@L²p³HhóÎ2îÐ]VÎÎ(JéÎ¸¬»Édëì¨/kû;s±}ùý£tÙà;²ç´Ñïu/Ù@'¯q¤õûyÕlæçÎò Ôºç¢ 39´4lÄµwµ Ú�Õ)¹¹ÁsH[ÝòØ´±±É&ãÔáC³þ¾!Eè8bEÛwù£!uÐàiwîEÕé"3Ý4lª²jé×©£3Å§ê¢§Y :.ãòc¹«Æ¼òp µp¿S\:ÿÉCc'r CÄôv§£ÀÞZàF¡FuÃÐähcZtXÊÕgÿµ¥ó÷iGhË³/l M}óÜ9X~:ÈúªÔLSÅÖ, rÓàÎ×ÉØñM=t2ª®=QzäðÎÝ»mûöïw°d5]ÝÙ'¹3Ûw;yÞµÿàA\|+&7^öuURldí¡Ä ÞøÇ-ú¬Ùz<~ºÝïÜÝ(Òwã�ãÎÞCÙptoÿè}d>ÞÄ£69DQD÷(n75ìö;EÕ×º¯úcDZÌN«íúwòH¨þxÉtsÝñ¶«´ ?Ô!ù#ÛRúÀetnàÒkVãZ)¶=~× z²¦3nàME.É$ìµ&spø7høuL ´,1ÿ»¬ÇW=£V;Y~ÔßÝÙË(ÞêÛüãåÒåÚ®ßú].¶ yjg§;þuAqW¨'QñrwóÎ²÷"Ivüýæ«ÈùA±ü'§]¤øÜYõwgü#Áwzfîü×7 ¾g~0ÓdÜøàPóÎ¨ºæ(KM¬éîÅCã ì®!òÉÙ+h4®jÛ"~ÀE§ ¶MscØ¯¥ÿgÅµÍÚAWtÊM¡×ÄÓ¸ ÷ [nPÜìvÑ\_±g pçFFm{ë\ý\pL),uÔ.¬Ö¦:\_ADS@µK¦nýOuÈ%KªïðC:YQ»rmò ÑæÉCãRYcoãÉÛ¸xÔV×£#ÍèD²V£9rÚÑÜV$K,QµÅ{Æ\_0Y0u\ ª=Ê·ñ[ß+ëÃêwÞ+[þ¸Ú4Y:û¿vÝnU"8¥÷wþ!}þ)î,ü ;#©Kû î î\|>C-î?Vëªl åÍ±ÿª<\¢++höV®?¸7Ö»mÍ¶²æÎÓNSõ WÉR· Hæ%\|ÃMiCùÎ+ã#£ÉC8óÉÞ¸é)È¥÷ìô$Ôúf¹U?sÈ´9sÇ]0îÄiN³1rÀª ß·ÑSälÈo Ü¹Ï¸3n\|ÈEPT¾®áøcOýíHC]YP:P[ùýÌtp)üÇÌ;#vµÊr]lh4îÓ Ê[ìÝ7AÎî\|ìÎÖßÄpt¤+öé%eÔÙô4ëzÀsëÎTûÔÒk8KÒãò%j§a°½9»æ¦IÿýU¿©:cÎ¹ÿþAßìÃiºW'r>õêSK®h"»æÈiû×kLÑ2ÇÌ¨ :ÝYµ!åÈ[Ëþ1lB%eè¨Ü'·#õ8joûßÞqIêÈgB?ßïÄËS¾¾[:ÆÎnpËªÝæ!Ì»QyæºG&sq9ý§}aËÄÏ 9qpÿg#½3dÀçF^0åÖ÷«¶Ziýj'in°;{À¡ô7ùsº³[-"ÿE¾gy¿svçR;;ëª¡UíørvmhÄG{öåm =&YAÞ ¢ÓõÅ;c´ÓjÈ{÷Õ�ét}7Ûê=n>È¦¶ååß»íïÈL² s&CvöûS_aÀ=¢ KN²îÎ2°§«³¨);²úÃ5×_{þ5Ì:9cÝ³\|ý6ÁW]z@Àî\|Ý¹'çÙÖEkçé°º#c\0ÖùÇUµñ%ÁëÎ,?É-Wô§ÓjAñÜôÍí°û¶T2Äÿ¨àù07dÌpÎäH&oNo 2[8zÎvLÞrÃc{ä´Noã"÷ØÔàÎçöb²Æç%@æU·Û+G$[\ ²³Ñ¦¦²;bCÈÞCì³VÓ¨[ë~Àq;áj.bÔ¸N?xÒìÞ4ôZè¦Å]¢µÞÅÑÆ7þõ]B[)Üh÷ÐµîàÎèÎZ@ vç?¾>&63§\_0~ÊÅê>¯A¦Up £º;Pé\_æ\_Ò?ãR9sAdp¦1<Û¢K3É\| J : äÅaz ¡rÙUÕAj=�îÜ½;³Ür ug¿Mg +û¢/ª:óC;;¿ªÈcTùå{y$Ë<þÏbyÿ»ízr¬°r,·59¼}ÿ¶mÍvÔÿË\_~F§ÿ;µNCô¦Xü^MëV½¤ç4çÌØ1ôw>uÄ;SFþ#C!éPé,Jk§í?7]AÖ«ZèÓ&c§ {Ý·-ìd ¹\H}¸3¸óùqg6@JäUã¨;ë¸Çæ½·\_«#¢¹3¢Hi¹"ÁNj@ìÅ·<êOnáOFr#û±Ü¿ÞJO\_~1géGNg3]Ë\r y!3A#sÐæAo{¼¾sYHÏÝô î\|ÎÝ¹¬¬, Ü¹SQy7n+K¤;ÞB~¼;W·Ul¥:"8ÓÐfÍ£Qê°Ð&¼äÔzä:ïÌ ¹éUdK)I>C$§§¦�whwî@5·LæÌ{¸MøvØ4!ukök%ÓY¦Üðª³QaîLrA¶\|BîöÈVî W%§ÊÆ/ºfÔ$ÎyQwîÜ©W¬="<}TØÔþ¡©1aãð-<%\"#44Ë³R¢¢/±wu-Nô"[ \º\§\_h6g( 1F2ÃÃæ½ôÕKû%ÇG¼dÍ 5Má¢¦H»ôwofU°÷ Òû³ëfwnËÎý<õíýWnû?�w>UN WÍ&')ë \pÔ¹7é÷hÒÊ<ª)½5ÆBÅ5úÿæ,¯ð {W4wHLARYös÷ÒDª6MfäL>G.K¿."kMeßF3æîª¤ÅKä¨ºoD¸\w?+ÈlÀñQÐ×üö¿VoEþ?\ \_Yº ÊüO¹sOv+Å{þ&¾Ó\|¼1©Mñ\_ó×F$W;¾Öw D{ëGo'Y¿ q£nú RN wpÔ^?(2ìïÀÆpÑ}±£DEÚP¡åØ¹-Õ+ ñß5-î,Ã¼srgÿ\a=xÎç ÍñEf%Èá5b;vçßEÈpP¿;ÅUÀ½LõðÖjÄ\ÈÀ^]Bê/{«ª¢½hEMÅ½X¶.rø-/ð% S{n4îPÆásgv .sÕ·äIåL©c/Ìýs3-q¹éXT B»Ö¬ÿ{=zÃ&ÔÓVkkG)JÞýïàÝ( ¨ò¥¿5 N¿pøô¿Ôá ©¾ÜtÏ¼×®6ÅÊeEër2/xõÔ/fnmÞ~ßùö±ÏéÇdE¥q±R~ó¡ãPUEÜû×$LÑM3séQÆ ÇMÿÃÇö=ÇPm;joE U¼÷Ûô .LÌ ãRbCS¯_Ø¸£åðww>;½;;ox)v¤Ö\|¿na(Ð ãBúo8xG9{;xTø¦cÁÄ÷{jÉkí¦'¾ÙVwÆ,¯Zþ×?ÿù·VÎ.,eÉ8ÈIý¾vD¶ÈCðÈºI {Eú÷¥&\_üÞ»ã\_"¤õ\ÉLæ®õÑ³Ï-zUPÉÜùÜ^Ï�rg\|¹3Û-ÏÌý£j¯d1Eêÿ»í!Fßú°ë³,HvÔá&\cbûcHk9ÅSÍH>ñÑkQlEGp$;f£�îÜ×ÜùèÑ£ãÎ]õYñÎªawn:ò»hYe¡üÑî,6ÙJ·1^²j»vgÒûð¾cm¶ò©.ÄWg?tÑÃoXî üÌîÜªn>¤Z9eä¸¼É$¿^ #aÃ¶}ðÅa/pA¹B¯ÛN´#rv ÛÝYíbªîñº³Üu õWÄIaó a7ýÙ©>&èµØ'I8ù¢ÕP$ÿoCRÃáµï²ÜG\ððÙlwîk>\|8ðÝÙÝmb{Ü¼3.íb{ëïCØ¬!nòÌwÈq'¢ìÕ ÎäÀDÑ7ÅÐ3ªßö¯ZpgàgqçÎygªº7gÈðdiÊÐ ò¨#§ýÜYk§È³%çõëC\_L#'ÝµA%a¯¦¾£<Ï;¡ãrCôÙ#[?õ8É?ÜIÔAü÷2M?@gjNóyã~ñb&ün:Õ,Ó N¡É25¢ªÊzªæç?¹'!ã2õ!É?ªùÖî�tgµû¾ôÁ1!Ú·fû[ü¡u4Ç¶ÓÏ^¿·¹³$Ñ(~kPª[EÚRÓÃ?ï¬ íÙZ¶\_Ig²ta+$ùôZIÎz¿\|Có¨Í¯¾vð@²ýÁð«ïÐz¤Á $à7(åAZîlíÈ '®uBÍ6²ÿÅÅ\ôê{J§;+=³1wwîzÊÉîüø¾óÏªLØë¯$ÝUØWÞá}×%V/øêëÃ9êÎ!/¾º¢ªÓqÅ;çÈ¯bM®K]~Ç¹Ð£ ¨BvNìÍ¥úÜyïK ®M 2dþï=q iø;ÿ¹mwV[Ú9géqnz¤)ãòéGVÛ;ÿÉNrÊû-ûnqÕ° % 4aÉzÎ¸{sJ¨ÕtS$\wÈ éÈçÎ\²½h¨D.<3L \_¼ïÌ;�îLÛaÕä&$41é8cÜÿe½êFÝòäújÅËXuhÎ<¼ÆÊf»Êi¢ WksÉí[6ÎÿîÛïlw8éQU¼vLÊ1 ä[S¹m \ 7-wì ç£GlÝ´aZ~Á¦CÇ[\|¶IIã©¶ËEôHÃ¯¾éi2?ü.¸ssgUÑæËääÞrß¨hb¯ý,×;äÒ»ë:ÖQ»t4JóÄCBÈ,XP)üWv6/hó¤ëW<ÈaCµþrK¶fûJ§Çäæ²8Eì¦Ç×ÜYØ(ÚeÛ«jt®÷xå-Oº_x!¸ssg>+Ôù}æ=A÷iê¹¨ ·5£v¶hÅ¤nÙ®OMãØ:Üàû[ì¢\|æ3kqµIö¡þ.(7·ãxC³æÎähiò<'Orùfeya µz_\|uÜÜ9 Ü7.ÔZy_\0;=ê+´5Ûg_å³I] 7Ç_VÒ,;YÒIFCA¼[´#Oë·¦á´ýÎ_©;ÁFºÞøÙÜYÆZZpÙ ºÐd%ulá µtöÖ/~\û'^;!(tª%,ã®[q;©¶@¤Çþ¢)¡Z¯;ÿ¿;dª¡êµ»ó#'p\¶%2ÿ?}8á(9^·!'ÝYu#Ùãæ;ZPÅìo\_º -,"'.:í²÷í¥y6\\|íçãM¾\Ê_w¥WÈ®Ê2}}¬¸r{PK:øï·¯Æ¦&L½twGEC �@�º³v(Ý®,¬×?©jðö5Ê4À¿ÓäFT·gH8§Ç1Ä>ðLå« Ù]ä]NÿPû»h$¯ç¶hòÈãô¦ÍPiv#w«\|k4=òÓ0îÞzDÖþt»®?ûuÉi;$Qem¼Õéb lû>{'NO3#<üýpç¾æÎoí$+R+jwEq!//YçûøÉüJ´½clËW9äÎZ;ä·9y;=ñMõxÜîÎícdìjÞüÉ#h¯×GMù^ÉY¯zÄ¾Þ«ÝmnNÃÅEûº,¢C=8w%¸3¸sWw¦A¢Újîe°zùJÜÖkéö=,H¢Sæh=¸9ô)FN÷òòoëé+¸H7"a'© w%U5Þó«:7CÞ+ ø9y0:³äL¿úÏV+R!¥SÝî#D )jç×Þ¶³ÿ¨°èjìÎô\_ Í¼ ugº)Y%[ÿ½\_ÍgGåÿªaS&õÅNÞ%ÙÃ=SãÃ³ooÒOqón©@ºwêÜ[bÇ/Þ/$ñª¹µªqt$Ye9Ü$g8ª.µ¾?l!,;K1nÆ#ÇdÝ!@ù�� ÝY ûi&9Ë¿æ¸p¢Æ!Üú×zºåÆÅ.'nAOL\|øfºàX#xÀG\]äy.¤à D NkwÒ´D ¼í¸¹J¾ûAe GP]qóïÛ]äM·+MÈYþÇkFðûZ"9ã°g3ÈÄWçþýÞ0l $µwLÊ«Ë<ÌUE¨+¥eS ç¸Ð¾?èòò�î î\|Þ\v�[ë\só¨è(ìÚ2 CðOÐÅ³ØQDöìÏlZ¢lÀNðn[Gpv¸Þy¸^YJ7%ÉÆþcg±ó¸Ããý©G FuåÔXíÄ·ëN/g§Èi$z¢;;wçÎ j¿'>ùmU_ifî\|åI)ãª¡ùOÖ! Û7]õ§ipåÐn¼ê_¾lÚï¾7þ}¡o'K@$'²;ÔÚ÷¾3üÙpãø}ÊåÌþGye\¥JßÝô\|Üx.,/:$gÌ;q-ox½mböU!/qÆì!é·?»iY3Pn�Ôfdå^6¿!>7SÜ.Ó·E¾~ÜÜù\|U#¹)Ð¨ài¹}d\+ó:3g{äáÇÿ;{ÖEáú\ H©å.X°úÛZdýEÖÁÎEÄoÄîDÖ\|7T;[wí!eqU!+· z=g yË5÷=rçß¸×bdO!UÓÇÎ\_µ¡Ý·cSÕFÉÀÁ{p¿èM ïx$ÔÐü@d,Iöþð;«4=vJSkéË-6jè vÓ¶ß%°¤d«ÑåÃHZÖÿøÛÞÈ9""&ì#ÒîC:oùúvî w~nw&ÛeRøéÆÇç§GG K¿ò?«þrgþ¨~Éæ°pCJqè\_Ïzú0=PKu+r~òefº´A¡ÙwÿiåzÕí{?º§Zõ®3R½ß¬°èJºfÌ; »Ý´tÓr9qÙçèÈ;;ó.IÒ/%:"5::eÐ¨ô\_ÝPx×åé¿¾8{\ÿK1dZtÈäÏ\|>· ¡f²ÐCÌjûøÈÇM¢ã¸ÔáaY×^6ãWæüöîù¿1,\+2(5Ø5,é×k[\hg¦¨àÎ��¢;k1¿¬MxIH²Ö\|;,¬Ü&ÄêBÌ-é)8ìAÑK>\o'Á¼ ±3 nTwøú²l:D?ÀtMÅqöú$]7q³ì,Þ³9 2²þ09"Èl1ÐsØ×Ífãà;+:�UeÃ^ýÉ²ÙáøFÎZ¼Êî1¿¹ÿHüBÐn2¬Ùw>¯1?ÛÇFs¯hG-«.$·=v×ålY¯'sÅAA¸ÈsÁ,E)á_¶ÓÅl¦7Öñ¤Q\|e¿à²þg¡À¨I3a Öl¬÷ÐÒeo¬Ùw�}f~£ûVëÝd32üÄì¥eÌ ´}27í?Ñxk¤Ö'Ä¿ñ!¤ýÇóäFÒuÔC§·zBªah¿¥M·ÑXoÕw¸3ð3rÃö\_zûÉ¦è)úÜ¸ÈiñéÁái¦ðÌ¨ic&qÿ.©ÖÓ%L+Ë>ü\|êÀ¢ Ã¸×Ý÷NÙË Ê~î\|rêFåÏ¼vÙuÆÍy¬Fb+}ãD¸÷êp¡Ú5%ó/~&apj^üD¦ÇÅ¥&DOMà&ÌYCC&]úeI'Éæªtê¹êà«WlsIRTÿ£>Ã¬Ë¶¤÷ Oí9,\uÜñ×¨ù¦îØ3��íÎB~~lC\YüøoñS¶r4Ä"{zJiØAåTFN¹e$îGhö§n¸ Ç0¡~QW×5ùr¶¦²ÌÌôåôQ þ3Cbç.XÁ Èãñï\TÊòq{$GõãEO>þP#1úQä]Òjî÷ùÆ}$ã6IÝYww>ïîÌò¤²Ùüìå-}9TÏfõ¸"Ñ(ßxéÕ·lÞ{L¡»¤M$Ý,ÒYùFz UÄ¾ißazb¹¬ußLµ"WýêùsÃH}0°5Ü4¥^éÏ¿t°¼³À\¸ /¬°P òl;�2É1$ ¦ÆúæÐé+?«f '$Á[åÜtýj§ñØ¨þf³3÷óûEí°{ËnÄàÜ±rÑ"g4ÛaF³ìô1\_wë}[öWXÙægÚqtuçÚn;:^náö½W{söÃ&õ28fJ\øKýÇ>ùúÔ=K¹$ø× \æùãßl[té c'^ ßI+$R'{½+?Rì%GéáEÅ¬öo¾ßuüD½ÒõÉ$=0Y¥ê!)\pù%ùç=nþ½göøW&ò$¨º¥ví;¸kOqÑ®ÇªÙ4è]!ñ$ugI;~ºç-àÎàÎþõî!HW²fÉC²q8¨Ûvý+Øh]ÇH]~ÝY¥ 6Çãy2,{²òÊ[7Õ5Uí,ú¾®½ÍIû2Îh;IÊKIaÆÜÇÀ{²S5¨¨}Í®ÕßÞÙ¸ÿóÒov(q f{;Ûp$Ö¹Ý-Æí®úvÔÚ¬ï8ÅOÚì±6+jm@-ÇPu-ÞGÒAõ»T[+[:Py-:º¾róúo6ÛoeõÈÕÔV$thÏÍ^Éûö¸ÃÁÞÃÙ¶vÔ¸³vû§¿\èíGÊJÝÞ·pz::»!w� PÝYò;UP$ºÈa¬µFo:lYE¼Cu-Xiáy«@þ¥ÙUí¼fý{´ÓÉzn9U[L¤ã ÖÞz$?×pôÜt®Á× ¨Ìq¸äTÁxhrnòæNÑê{Ïýz 2q ;;ï_½-CßDæ=AÝ²Ì©©¯Ûji±wÖYeêÎ¤®Ébçt>M CL¸& 9MÛÉrÏGØy^ð2©};SüôÜÜ¹G»3ùDDË·û#Yr{A¥.[8Ojúñ\_·<ÏªÈ©²\O{2^¡ b§@°(×5Y³ËCj DæÎÚ +JO=Ø Ü9Àã?¯9ÞñA9y¥µÄ¾ð6#¹ÛT¬vb \êÜïLêê]´ä áïhÏEÿ\ÞFÊ®nA%ó!4¼Ó~?Ô,ÉE ²ô¾NÅÃFwÙÍªpÀÛ3}{l~ÆÓù pg��Ò6ÞÙIÚH$RPw\~Ý(Hêé¦¥¦QOù6¸3¸sOGôCÉ÷H«H´»¯þàëh}P×ÖÈ'wjJg7¤¢[\_À{>w<¾òÜMiÙ3än£³¨I~%ËðpÕ;6º ¥6,ùHìÒËÐQå®ßìRSüëæÉ=Á6��ïÎ?Å7ÿ÷fðLÏô¹³z®ßÜÜ¹§"hDýyJwwJ î îÜsÇ NWb;ãïÿ9séW\| ÔnÄÜøå[ýqµégùB+(3��;÷®îÜó?\pgpg�ÜÜ¹§ºùý´g_-pgàuç³©��àÎ©Ïj�¸àÎàÎ�¸3¸3¸3¸3ð »3w��ÜÜÜÜ�wÀÏY;��¸3¸3¸3¸3�î î ;��¸3�î î ;;½pg��Ü�wÀÁá;àÎ��î î ;àÎàÎ�¸3��;àÎàÎ�¸3¸3�î ��àÎ�¸3¸3�î î ;��¸3�î î ;;àÎ��ð#üµ¢¢¹Ífc�î ;;àÎ�¸3��0Í5û~4cÆ 6àî¯££pg�ÜÜÜÜ�w��z=<Ïçççc#ÎÍÍÅ_SSSó(Ø§Lîÿ? 99?fóæÍX¯öîÝ WÜ�wîãî\|ðàAlµµµp5Î;ãë î î ��=Í¶}ûv¬ÏØsrr Ù\sØÙwð>ú¨¼¼w8Èrõ>ªªªÀÏ¹;8qérT ÐÀfkkë¾}û°îÕÕÕA?Wî±Ûí>½zìóÅì±cÇ<È¾ñ��pàÎ ë3VæÄÄÄôôôÓ¹3þºxñbÜåa ÀîÝª¶¶¶è-àO³±±±¹¹ùÐ¡CàÎçÖñÅÄ#®8MMM§¡ (ð§V^^¾{÷nìÕÕÕPàÏ;³ë/o{{{Ð+Ë¾}ûpÿ¿² ¥,é ��8\yåWX\°nÝ3wî\çq=¹3îK(Å@à?ÇÃ½{÷OÄb®,¾ygÿë\|¨;J�5àÎçÜY5ñ-áüQ=z}²0ï ��:6mÎ9iii,¥v·îÜÜÜÌt�÷¸+,//ÇÜ;8è °#kEQØæ¸3ð]êýÝq(ðgZTT¿~öÃ}574ì3eþêk!¡Ì��=9÷ÝWçyvÛ¶mìg;3}NLLÌ§ùoôc«)-ÖÖVv§@\|+QÙØÑÑáv»Ùmo~JÃ×Ðjµ6° :àÖl×ÕÕá¯¸úà;°Bãêêj\_Äª[ñ{RMF;°±A{{;Ø8<��o¾ù&Væ¼¼¼¤¤¤L?°;c¡Æw¶oß¼IÝ&Áo"^Øè³f_N$àbUãsé@êÀµulaó¹Õg_9©ñüF;°õø#r��VHÏóüêÕ«±5³ó±&\|÷ÝwsæÌ9É?ýôSß³@£úT!w�àoNYk��ÀÏuúMA¾ÿþûY³fú¦Z³fËå?©©ñwçùóç³vÎ¦ÿ¤³o\ pgàßþénpMÁVk°©4(äç§ó��øÀqÿCìlÙªªª ²éæôôôE544øÒòåË±5§¥¥Í7¯Û5¥0 ��ÿó@Ä ��àü³o ÌVë§~¥8??óÌ3Ù1ìä).+33Ë5vê³Ù W�À�� àªë·cG.,,ÌÈÈÀyúôéëÖÃÖìñxNbü7áBÌ�àÎ��Ðë±Z ,é²±5ggg/Y²{±obº[GÆßw¹\ó�¸3��ô²àÝg[ñ?¿üòË´´4õ+))iÖ¬YÅÅÅØÍýõcc~¸È��@ï@QlÇEEEl®955ÍÈÈøê«¯l6ç«äàÎ��@ïémyyù ³²²òòò²³³srrð?NzðYÎ;��@oãñ\|ôÑGØ§O^XXÄ2i#:èþe¦Ìø®��Ðñ-¢EyñÞ½{srrÒÓÓ±;ggggff®_¿Þjµvû\[.#��Ð»ÝÙw¿¹¹yÑ¢Eiiiyyy .ljjò?¦êTïw��ú8»\®õë×§¤¤¤§§³Ì3gÎÜ¿¿@a;ÕÁ��¾àÎ¢¢¢6Ý¿fggoÜ¸±££?Ü#K2¸3��Ð§RÜÒÒ²dÉÔÔÔxð Ë¤955Ý¶m~�+�� Z³Ïï¾ûnnnn^^^å>hmme?u»ÝpÑ��>¨Ï²$-[¶YÎÎÎf)6lp��ú&6mÕªUl\_sNN¶æ%K´´´ø2i��@ßDd\|Û´iS%2gÎCá¢î ��ôqjjj°)çååaqÎÌÌÄ\_7lØàp8üWh;��}U«V±ó§XN°¥K655ôpg�� ýWv\_ÅÝ»wc\_Æyúôé¯¿þúÑ£G=h2��Ð7Yd,Î<Ï+RVVMA¶aÃI£Pà¢��}Ó Ç\|¹°°»óÜ¹s}aÖÌq+��ôbG>õ ìÜ¹³yæÌ¥¥¥ìÁ= [3\@�� O¹3»\_UUµpáB\_B°/¿üREEQ ��ÐÁvìp8Ö[}9'''77»ó jkkÙOáìf�� ÏxçÎ3fÌ+´±5Ï3gÏ=Ý>®��Ðwé²kjjÞ}÷Ý vU~~þÚµkm6Ë�æev"¸3��Ðw%ÙívöÙgLÓÓÓçÍUútOq��ú¢(²;ûöí3gNvvvfffeÏ=,'\%�� /#KrCCÃë¯¿Îiw^½zµÝngÖ,\%�� /³aÃ¬¬¬ìììÜÜ\\|çÍ7ßd´aº��èã5>\|øp~~>¶fvjsaaáöíÛÙfË��èSøbce%Ùjµ.\¸ÐÎÍÍÅú¼råÊÖÖVün��ú,n·Ýùì³Ï°5gfffee¥§§/Z´¨ªª Ñ3§\|Æ~ W ��èÅîè¨3g²ÍXñ-[¶ ¿Yi¶TÄ�Õ¶�½õãOü§�ë¸�� G³¬³f,"áööö·Þz+333???+++//oùòå¾LÚ'½ÄK¾à�°z0¾ðA+^àjüÏ°þ×w$Ð×¸�p5Wz)>É=õ£?)ÔÁ}÷æÍ§OM={v]]ÏÞ0öú:¶~wûZwÀ£{ôÁuø} \¾SY ¾+ù?X°X¢oÆQ�ÐËà^ÝnommõíbîÒÒÒ7Þx[sff&¶flÐØ£AM=¾sRÛÞ×Ú\|\|ÙÍæñ¿¯ª«+n? ìh}ioo~á§à>Ã5l®¹X:::\|Í#ðSÀÕjúÒ»innfÓl@Ü�C@¯fÿþýEEE555Ý®©Ãõûï½¿bq~çwp'¼sÍ¸Uw8g÷î;5_ÊÊÊ/ÅÅÅÜß~HiiéáÃìØý °¦Ç+9 Ôt}illÊòÓ«LYYÙYö\|lhÜ�ËuôèQÜhãpÅ0%@¯�ûöíÛ³gÓçêêêZlA6oÞÅòiÏ5Yp:w.//ÇÝÓ« zG0ÃQ\YpeÁ.®,ÏüÏ°»3P×÷2gè\À@6T¸k þÊªokðÓÝWcÇák{Ó}Þ,ZÀ_Y$ï;@/pçR n·«ªªjÞe,wøcõwgÖDã;¸wÆ¦IÁõyýúõ8Ökg6ìÎ,ÔÇë@%Ð+ÀM®,¸Rø\öütî¼{÷nÜáËëÎ+ À%ë\Ê;CwÆÖ//Éð>Ýõð&àÀf²Ì!; �´;ãÀ·Ø8\_¾²ðæÔÜòÀ½<ÏãëKuu5\Ì^öá"gµµµpMÎ;:t÷Ë>á:n Ù:;Ã5vg¬Wp5z¥;=ztÏ=iiiyyyééé\|ðAKK/ îøÇº3®/àÎàÎ�¸3¸3¸3¸3�î �àÎ@orç¢¢¢½{÷®\¹2##ãµ×^Ã 8û)kÀá\;àÎàÎ�¸3¸3�î ��¸3¸3;v§ªªjË-¸ÑöíÒeKµÙh¸VàÎ�¸3¸3�î î ;��îÜÝ¹^]]ïåjWËVN#Î2û.î³d©¤ûGøÿì\Å¹¨g{vÕ-YrÁP!$¡ä%$÷%7ô$ä¦rrIHPvÉS lL10c0î¶\%Ëeõ.mï§ÿoÊÙÕJV1y¹ f/«Ý³gËæÿæoJ?-[úfª½³y)ù%U'PSU5Q ]$úQêðQÕ×,ÖÕaàXhÔkÎÎyªÇavÈ°3x<)±½e6[Ô¬8pvæ-ßÙY#³1è\|©ÐMÑú4)ÙÑØL.±©UK§ø;kyu}9;svWiÀHZ%ÛGËÕ\_Ò±°\§(+2UpäÌÓ¬sv¡áìÌ[Þ³³¾-Fux5wÕÀ4{K :>±³ÈÙ·O{¦Çw;c&H<"ÊNÎÎÚDì±Á9ILøV�¤@ AÜÑ~ôC°BadyQjÑ C{ªN×ÃñYÃø#ÿSðwC<±ÄcOA$ i ü²<ÝyhØEäC&5&!HÜ¦hG!ëqòõÃ"dhD.ÅÌ·æì×ìÇaÒ2) BTÿ²Be;çËxàìÌÙy\¹D2ºAë-¿ B6Y%f&S8Ù#AÒuÏ² ¾O(HFíù2yrvæì\|J¨@RÁóÓdM¦fM ô÷4îß·ö£7mÛµ§±¹±³\_ ÛRR5g¶þ§üÅåìÌÙùÙP«LÄ"a^HÒÄ$§DEH) {ÚO"næÊ:ATý/ÎÎ¼}ó¼ÄÔGb » » ÚÐ s?¡!¯è±'J¦UC%TLÙYC¨ ZwÀÖaÙ°òuÒß\|Þ~Zû¡/AÔ,îªH½Å3]eö ÐL¥#j?Ú µÖ¼«WÕðþ°j=4ô_ 6muñA7mÓÈÝlÁ>hl¶vhmCíÐÜNîw°Þ íäÏ¶Nhë&%�-!èÁ¿ÌGÉ4ñÙNzÆ#-Ô<ÇìYñFÏêµTD¹ûI\ggÞòñ�å(À ¨·ÿ'B&¤7#ûÞsXyçUóoÜþ×7ïiT)Gë7°5ÎÎ¼MvÖW¨.i%$F:ÚZ¿ç®ï2qÉHv#rxoºûÁÎ 8$Pi ÎggÞ¦;SWU9vÓ5¸©<àf¦ÝYhÌ-3æî¬oï÷wÝßÔ:HüYejÏ à¢è=fÎÎSfÜf9id§ ©Á-~}Muù'¾\|ì9àOîã°:Á E²E$!1�-ðÈ]oWØ·:öx\»¼î^7¾Ýæ+ZUT¾ÿß®W@qãM× fnÎ²3q\|R¢äôÜßw\tÆÇÕÞ}eu¥»+ªvÌ^W~Ì+sN'þ(DD]cÓ ÃÎºø\_º{å1UÛ«ª¶VTn+¯ÜYVY[VïoQ¹©²÷Í3p¯Ü\1ssEÍæò¹ËÙ\vÜ²y§B4ÍGÉ4ag2þà{ÏÞWP!½ñmÿîOoèWÉng:3¾óo9æìÌÙy¢È Ð@ÜyûÕ§Z\6£k7-ÇfBv yÄê)9æì\\ÎÎ@ó°ã¥ç#b³@Õj5sØYÇg&hd²!í¸¿ÚÕ«DãìõXÔ4hoù+³óTµ Ù¼in.\|+@¬³ñüZ<æA×ÜÝsÎîHmÑ¨»lãÀ@+<ºGeá~;êµe¯7nµÖAeÀjê3¡×Õì²m·>ª¯¼ Á8±ì¦N {5b»vÀ÷ß¯ªh,-n´¡H+dF J[migÀèpùYÌ-¥%[ÊªaéÛ0Ð}È5êm« ¹CzöÆ]Ëq»NW´¨hÐåê7[LV¿ÉÌ:¾?² E¨äã¢¹äqvñÎ"h!;¿2ËéAÈGìÎ&d\@Ö²82æ�5ïqvæì<~S$T?¤¶ßuÕIeVb23Yn/e k<&böº]úCä>¶+Iö78;svàKeæ´ï\|¿Ø\d3é0LC"µBY¢íE§~çêFÎËÙ³óÄBlÑ.=«"2²"Åêrå³9ÒT°bæux å³ û ggÞ>QK£8>2$ ÒÞsöqa+RMm\_®À'8Lý&¢ahkÜ{ñYÕEA»!,WÉ@yåþbï&u»×UWækQÔê3wÙQÌYØ³»æèðïþ ýU$ y&'M Çï}\QÑer¥Lî»¨Åîî1³µ´üPAaÓµ· Ùã9Làq&<%õÕÇÁ+ïB ±>« B\|+<õÛæh5Eí®oiÓÑ4£ ãv{qQWQN÷tù\º¼3» kº ç~8ëDsvÎ7vVGý¼dC¬á-êõu\|V©7MR[D{zÎ:!jF,k¯::ªNt²h'tìý¸¦¢³ÐvLÖvûQ»f}þçIèi'ôµBÛØ·îýÓRó Ï°¢·h§§î{zÃ øIcuM¤:·TÍØËñçI:goñ·ý êÚ µúºÉ ¶ÃcuÖÔ$í2÷»K7TÍuôÔóI@l;<ùæb[ØîÙã=O?,~{þ xþqXü8Xwr1²±®½ÿ%Þ¦þBvd·jg�³³ d_Å2ºçÃÀÀSt4§ «7$ñY³ó0v´,;»H¬³¹>ß$Ë¡Í p2ÝENÒñÆûÊ\|¥ ãõ#¤RÞ\_ÎÎÇY^¨óo>\|gß´¹l3OmK@:WR©ÇIE/ÁíåçYÙ_áúÜ÷(S¾îggÎÎG¬¤oú÷óËE¹ÙçvÅ?¤]1\|Ö¨È¨©§¼+ivÊ¯ÿÇkí4?@¶ôT××9;OINv¢ýýg±!Éìkyt a&Î±lX½¿Ê³[LöíÇÀ² / qK{"n© ÜÒrÕ·]¨ÁÖgñ8áKJPu ÈjÝÜSáHÛ\óaÙvèNBJe9Åô$gø>óHÚv¬¨×fÛqôçÀ¥yËDbOLì§on/v¥;ç\�{0 ÚøS)´ËÓÍØ'é©Ë³s²3ýÍªÈ¤y£ex29PlzgWä ÅÜ+Îû·s¡èAI\|ÈÁÎÙ³óDì ¤õÙ¦ììFÎS:ýlRÏ°svPã ì«x kõ>ÙÈ«ëËÙ³ó8B{"íÇyÍ,¶Ùõ½ý2 ÖvÏ# B¼cS¹yn;«zw³3oÓ5ÊÎß?¯xl{mnW#º/øl(ÀG-ÈXü¿-o¦ÚX$ÎÎ¼}ì,éìdì,X\|M#ØyüÑ¯&!ÒÛwÁÙ[ÒíóÕÀ' 5é1ècA¸zöí:ªÊï´³°£´-8)r HîÇ¯ïaJÚ]ÝÕgÃÄXÍQOÖk'Ò©¿nþYC©µ¡ ä½ÏÃî^)éxöö®2GÔ]¾ë¨o>ZTO&)]=LH0Ë´«ó2Þ]tè-²ªjrDwÿà¢Ó$U°×èÑ ¥!ñËËÎsUÆÌ3vô¤üÀlr¾TàäìÌÙù³3ò"ûd~ÕÙYÎÎBSdÿZõÅX3¼pQú)ùr}9;sv{uIÒUÿÎÓi.>ñ«ðø$Û 2@s£Æ÷ÿçÕç¡ñÔ\ö[jÛ¤©q9;sv>2vNÞôýs(;»}öHvv°Ô³³Ùl%×>üJ+@>MÁ§ÔÀeì,Nü'ÙOóP»²vFAÊQ2uÎút¥IqLsµÜE;àª=!«£ßU]{ò· 1�J íê¢ëû\,ÙÓæÃÊ&§4óT¦Î±!&£°ôÉ'WwáÛ¯ºA?DÌÇÃCìì±cÞ%°pyPÌÖÖFþ&ÙE³s¾åÙÎ^X è?ÁÆMf%°Týä×wÇIöÆðþwASaþõ½OöQÖ(¨ù28;sv>"vNdíÎÅÈ~v÷�_Gag% ¶¯.²JUEÈ^ýôÇudÆåìÌÛ4g%RÛ\_ñÿ£Z±Y væ±Ø½ .Hvô¢ D°:%ðxgÞ¦;cUJ³<ÛdGö®66;+½½Ìªçß.x~Cc+¨y¤¡svj ¡è8ðÏ±³Ô~xõ¯ûKìq£ÏXxü ÆÆ! URÄ5ZBïæ Ç{;Ì¿yÆF×ñðQÍÖR]òÉÛ+]Å0Ìîó5hlVÊ;!M 9¨lN!h±®Ù´JÑÇ%´8$ëáé?·ÛÎòsÿ4ÔI¦Ëi0½aßþ¡±ÎÙ9ÙyØ¨Aí]t÷¯°²c3;\~oK'½äæ@¥F Ô¨ø¤�;^Ãx©A¾°�ggÎÎ³³LØyÁ\_p:E&×{#UV¢!SúáìÌÙùÀdPÃ7]zF1^mÞ9átòv9%;ÒÚt Ìg±÷'·ìøVÔC ÁÙ·Oáfg%±Ý}¿ûVs!Ü¥ÍîÏÃ[ é$AÏl?l Nd6Y ÿ³ÅáLÚ\÷Í%¯C\J (ýÐ²}ÏÌAæAGE[å\|¸u!lÞ ¡ô 'ÕXôd°´I!P# DÛSdè¹ìv®¯<�BÚImàÅ+ÛSé¡ß·ücçO:\ãÙw´ÚTnä=®W¦Oj:þ¯\0aæóoïa/I&äË ggÎÎ°3Ë³ß³àSm´FrÌûx{_këG[;ë]ý½}6l\|ô¿Úh$ !õåww©hJ ÎÎ¼å;;+I;1xL4Ú¹æ¬v:þc¤þæXì¬£ú"ì, UÎÎ¼MvA ÞtéWØv«££=ûöõuõôöuu³~ nß®mµ ï¾KÓB¾eïlf=8;óöÙ²sxpð¬cV+ì°3Vÿ¶ýÏ9£ocÑé°~�>ÿÁÎiÍÓmEÅ£¢Õ3î~qR\_J!»oèªr÷"2ÛS¦¸ov]Ùæ~<�«?C!àH º:!6� ¦¤&%@3æãXöúÒÏÃª=°c?t÷�Ì{:¡öÒ;» £Àjî>?Ä0ó±¿vg½Å;Ö?7 ! ßþíùÝ¦¬@Øß¿ö='B+2Ï<ñüv+$V^¢59;ó6-ØXÄ¯[pÅiºÃÊif£Ñf±è}Òí~eíúß fg\|ÎÎ¼å9;Ç!¼ÿ4r1±Í»ø×�!hyeTviv?IìÖdÇþ©{4ggÞòÓ/;]ggS²»Ñ8bÁÃj#)h[úÆG"MØ\epåìÌÛTcg!ë7Í3+E®\lôÛÕ&æÆÇi½°ûýæbÉ\Úk'\_õGõ·LÆ Ö¿öÒÎÅîÌ6LÐ!«cÐSØTà¬÷ºöz=u5ÕðÓÃÛïB{ôõC4BÜ4E?ÎÎ7cv³%áòµ[ ZJÊöø¶zw4x xë}õEÅ-îòWÕ^×ÌÚÊc?"lÜÉ4"yÈÎM^Uðì¸þ²¯1&poêôg"HÍ21uÖ¬jn+B{ÓZ^ù#pvæì<¾n#\)P«[pùéfZZ=³$Mu³QÿËlÇ¢ä\|îÕIÏx!¼FoÓÅº÷æqÀÿ +ÏÿM;@hPÊXìL=ç)kF @«KÄÍÙ·iÅÎf¼¨v6Xa8ÑÈöò÷ ;ÃBÇÙ·)ÅÎéìo][Pít}jS0 û©c i-m[N§Ùu¯c#Ï ßwîa0 B!« Ð ý£®¼jÇÓìóõ9q+CAR-&Áfk³Ú÷Wµö/Àßô% «©©QÆÎ-¥6°ÒcÊjKz¹ÏaZÌa³5huDÌ. y$ä KZ·VÌ7P 8gçübçÌ¶ ÉÕ®øcûä 4øòEeD:mtàíE 1 ÈdþÃ[è¦Óµ"çÉÅåìÌÙy\|Y¨ßsïe§»1¸XÅAVs®EÀb2apf(]Bç ÿÆÏ[$2k¢[N9;;&ÆU¹MXÊ\.øÃAÌXÇ¤å>µ¡Å\I"h)+±4Ö8;óïì<ä³mCVdô!û0VFV«Õn·Û,VÁ4{Læ¯ÿÃFë Hü5ò9;ç5;k#;fçÆµ[j,«AvW5TuB^s+_3tÓ³�Ê l{·mF¹\|è¸þó~ iU^ÆÊ\|2ýaèÀÍá{¬;oGqI×Ýçq¤ÜÍîm=È³ÏRSwìÿú°b¤¾ó37¨°%í¤§¼ÓYÚXU]WS]_1÷U3wÎ¨ÚWZ} hf·æoöyUÏ}å¨9°z¤\|ä;ãùH¥Ã3 BÇª¥S,Ff{ÙËo³ ìè¶UZÉ0lv£ò¯ìé¥Ò8;ó6Øy×½VÀj@&×{vvu÷÷vu³ÞÓÙÕßÛw\_Ãï¾{õw¾cEÈibè¯\þ§d¦ ggÞòÓm?¥ÀPh²v>ëúîÜÒ£±sæ>ag¦åÍÅåìÌÙùØä ËÄ;VoÛÝ ôú2Íï÷ã/¾jÕªk®¾¢îÝ:l¥ó½BÃv¡8;óö±³ÿÌñÙY¥h]·óäb¿ËwÎØ=ã\ØÌTvs·\Úaë[uNd(ëE«îp\÷#ëYfQÙ!¾>ØSËí·ìQÚQêk·Ù»¬àÝo>aëÜoCoÆCbvV½©a-èö´O½ k·ÀÚõðáføh3lØ ë7ÁÍFx#\|°ÖoMaëf@ä>ÛyÆÎx2pLóÔiiPzÎïµÐM²¹=e49ñG5ÛõÌD323px°:ô·Ûy3SsvæìvóÉpR©«iZ«Éa@ÅAbmH Á ²1 f2#»9güÍ«4S;¯QÅÛ4g Cä Ûqï\_( òàE¦Ó{9#¥Ùoªñ®þ¥e3¸®øó£t¾\\_ÎÎÇcçTßE56Z¡ oì4ÆaiBv&¡B[{z)étÊ�ggÎÎ-È,åáôáØÄ1\ÌmÜzÝÅµ!SÕÜüLü¸Ä§;«mÙóÃSÚzÅ{½\_V=bLvÎà³�õ <ñ§OA�4ÚO§6Ò�IH 2B:VS2^4Vð$óaTL@)vCkÝàWíµ!(pºfl9õÐIê\1vV¾±©Ô2è(Û}Ô7aB zºdòAdSX6EÚß¼dgzÉiv9ø·ß^9Û@óº-ßHU}ý>áh¾i¶ ³³³Óýî\|aÎÎ'!"£ß~ßå'Rk2×Û7r¯eL¦íùã\_4ëë[ ³ÊÙ·0ëR¨RWCº½¥£ÌqÏØQù5hÐÉPÒ¿6+3MHYÖ5RQ¡]ÎÎùÄÎt¦ÖhÍî]ï,r2¶Ûu×_h¡@Y¥å5ÙÅHÐ-XSló"ïÈ¿ÎÎÇoYí ;M§õõV\|±³HkÞ/fYï,<ÄÙ·éÀÎdYH<ðÕj¯OOì\|êm) Ág${¶ÊVªQ0yT3ôÒµvd¯ÜÔæìÌgçÃFUÔÿÞ>&,ê þ:YáñÎ¼}6ì,ý°³B\|z[ÎÿRW±3Zàê«+ýá~>>ík·¶BÏêÎº-3KlVÁd?TP O¯�@T&Ìî÷¢ µÚ "¿ºü@«-k"·ä<2µ_ ýÐ¾µó¼ãC÷Í¾i°s²¹¥½Ü((ß8ë\8hHnöDZt½ÐU¿ïÛç¶ù<1£=©stFH'±± Ô}°usÐ¾u¹ðúFð' ?Á9!b9R$ñÙbì]í.³EÜAoÍºù3uÈaHÖavn-³û=e«Ï$Äñ×IWèú%©P ®O^1ÓS´Ë999;ç;kt?Iu®÷ê>¥ÆS.è£ñËè¦LÏ5H4+Ô»fùßzú0Ç §^ÔVlöàaþ SÉK³3gçØOàQHîøËU'ay1ÛfßÑa"Ì¦©Ù±¤õ"ÕqÛõË¬BUu!Oü48;sv>!Þ{a5ÉFsN8½s7Õwà%"k±´ ÊL$PBÐQÝw/ f4«9 Å³®±=2õóÒsvæì\|ØLÇ¹¾ñÒ³H¢Á5«9Q³ õTúíeHGÞðs,VD³º©\|åÞÎ 8ªr»3oú$'dìÜ~~\|·¡¨°¡(²ôÎ·RmüñÚ¯-¾µ¥Üñ\|gÙ¡Q®£ É:A $%ÉÀ]Õ{XX4¤ü ÅXÕæì<Å;uÊNúøëuz"mï½agm;+4}a é¯éF4 ¹Ý«¶5�¢º«µ#è/ìÌÙ9ïØ9µã®«N1ÿ6ßüí¹¿ðK¬¿øÒUï¬~nÉ7ßô;3-îl7R 5r]öãßÉh¢/×³3gçñì\_NsLH©d+>ëÛß¯Ý0J³ÍÖî¦=Ë dßC. ¹½ÈSrÝ½ÿÎøêqvæm°ó.#ìl'ÙY½¿_ø÷§V¬Z²\|ù²e/±þÜsÏ¾ðó×^ûË@SP vd+½øßv8;óöiÞ4h5-C¤Ûö¼¸ ©h°p.üãAõù¿iÏ?Ï=Ï>Ï,çq¿Ï.×!ùÒIÈuÄVèß]wúñÍnSÜmÙ\mÖyû¾�=÷CWémp¨A¼ù?×Cæ$Bi§gÝ ÷>FR{ ahVÄÁÁµëç´;PÙR¶z×÷>ß·@kô5AË~x÷è¯~Ô0»BðøätVo;æ è£ßMNCº{¤'oiòYbfÇ!Ï\¸ãïðôõ±Çà§µgÅÏÀ³áYü¥Às/¾xíËaÉÊM0 C\aÑJYkä´ Î/vAè?®Æc# àDö{»#AÝñb¨fLÍ¦Q¤À ?ºdÙ°>d/½èª?È¥ÏcÓ$@bc³3gç ØY�Ïí©ÚW@ê·IÕs£¥¤Ï4Ål3YHe§90<#·UºG©Sd~Ì9;Ø( &¶®\Rf@.£ÉbrlÎLÑC½ô¡Ñb° ÅF»Ïà"±÷wö[¿BT QÊËÙ³ó!õÙ®D¬º!ÑëE¹«K¦^¨Íjt°(æ äßJPË¢I ÎÎ¼}³>ÜÒ²sï¹ó#V[ýÞêªv_q»·¼Í7£ÍWÞ^\ÖYTÖS8³§°æ§ô-oiò¥%5E# Áý¦s?W_nô»PÊîUfµØ ö:]ûJKpß_RÒRìëó:¢N@}oÉÜäon®$rÑ,cÐ«½WU(,#kØ^Øl²7;=ÝÍî®Ò¶²NoqÛÔgÄ\|mµ®üTØÕG¬Æ¢H¿Y�Ò{à¹¾¤zÄyîLXw&mñþýûËÊÊJJJ-Z´xñ[·¢hb±+tgÝÝAwFwÆÏGwFÝy»3©®¼ùæË/'î¼téR"Ñ«W¯æW'¡;#èÎèÎº3º3~>º3 èÎèÎ$íèè ¡Î³Ï<»xñ;î¸Hô²eË~úiöàBwFøõmii!õ! _QÐtgtgüü³èÎ¸ÞAÆHuM¤@$l&på×1ÕÕÕëÛÕÕEþ\br¡zè!bÐ .,...++Ûºu+ye¢OçFNæÎä\|FÄØodD£Q¾¡ºóYtgV^¯Ô+M wæ oe'74GÉzÊ¯yäãÎxÃEwæ¥&OÍR{àçÝÏ'\Y~MIyÁýM$ ÜT\|¦PSS)×ceylsL"¡éÁî6v9ùî<°xñ>(!Î;;ósKÊ FÑICKK lønn\|ÜÇÑÎ;6·&Ü ÆÕÈ¤[@( á:ÅVzÈäÔ\|Nñ +VIôí{<¾ñÆ>úè]wÝÅù@3&-Z°yN úÐ¡C¡ÝAwFwFÐtgA$Yãÿ»iÔJWUUmÜ¸ñ¾ûîãÎDKJJJÄ yä¾kî ;£;ã9AwFÐA©�iÈÈø\æD'B;w¾ø«V\..æ¯¼ò :tgÝÝÝÝAwFAcEÒä·Ú¼y3ñèç{®ªª O º3îîçäÿtgAìîLÚ>ÞD£QlûÐtgtgtgtgÝAANèÎª¬¬$\_òM¢§NpÇtgÝyê tgtgÝyj¶eÇ\(ç¼}ÑH%ñAÛ©ÖR!2QÜ¹¡¡7S/Ï6«uõ$¿ÄèÎèÎïDi'9NZj45É"tgtçÓ»QØæ®1¦@%O´î<nZmô¦¥õ·>æFó÷¸·G¿«ÑÆC×NÚzèFpø)cÖ±î\|ªwà)ïYTvøÇ}8 È¸3w«h4\V×øOÖão¥ÿYI \_ÒL3úØª).ìÉ2û,±IÐ?´qÍÐqìxQ?N®A¦¶;Õ}d¤0ëô$u6y$=¨¬%í$oDwFÖI9Pe]¥®è,(Ñ + q>æÏÔFÞÆ[3º32EÜYf;)8bYRÅFxÀO[QV´ÑÑ.î !Y»³Htgä\ÆCäãÎÄIÁaw²cç7B}Rw&e@#¡ý¨äQóDAòAÈ ø=0ìß$ÑyJ5mn±^$Í°ëñú¬C$¤ÿQEðö@ÈA??1Ve¼;ÓqçÑÛ.0DÉCàÑÇ(;BÀÏýóG$ð)à×!d(8º3 SÙ :Qê´¾ØÄÖUÔWª=]søHeMc_HVÆ72ñZtg$ÝyüeCÉáØ?o{Çß¥ò¨E7ú·:Ù£ tgtçÓÕçÄ(2ÈjtÔ(è"Á·§©®¦¦º²²úðýÕí}%@yV´3 TDrhèÎÈ¹¬ì5&+DWA 7ôtCÇôiõO¥xtVô©îú\¢D³»[_3´};á¹õðÚKôxõ%Ø´ Ú=Ð0¥MLYb³Åãï"0�Ñ CG¡½Þ~6¿oo·¶Á» ®¼"Ï´QÇchÍêH öCc+íöhéÖú¼íÐÑN¿<ÚG{èïÖ6m>è%g@Á»A)íÎ æ5Þ/Êgjo¿´þ¦O\ëÒ-UL@RlÉþ¿ôê;[£,ã3\|Ôd¹ÄèÎèÎ'wg®ÑþîR³2ÑÉtÅWüW¿~sëö>4wné2=Ðdwg-1ì¦³Õ$\y xóåç?ûÑK,¤s )-f{Z¶Ù%ö¹×Þôô;»DÖ9Åûs%:g6ªbtçÉsçÝ=²î½åæ·g¼;ãÊç/¹¼Q¶¢5F2 6î !P!<�mpßÒMýÎê4wefjEfyÜýFvAýw~Ï¿Ã Ñ9Þ,ÊÃÍ w&j�¢\|àÐ¯Ý93óHaFc~FMa^UáôÜóv\üùWÀÚ§À�¿èÒiæÎDº¼Ï¼vñôÓ§ï+,Ú_PT_t0¿òª¶ (8î$»;!)&Z¨»e¯K }J6£\|X.$ô955ëßÖm óõ<Àäîî\|:e³²%ÿwûs¸RIpZM´ø¤ð¢bfÿ±ÐÃ!¸2§]qõ¾Ö¿±ÐÆ1Ñ'É}K,´ý ËÍEEv5ÞtU[eÐ=§êüÏAÿô3ÙÓd"#0ÔßupZzkºmÈmü<)5Õæ"Ï½ë ÝÒa:3ÝG²RwÎ_á°ô2= äîÌC1A)éÎ,øàßûÞ÷6:ÙBº ¤¿^Õ7B\|3)æi ;£;ðe ^ÐÝþ=7iTl6ÁBÊÃ7{íÓ{½þÄ=muwÏÿS.y i_ô5BÖgzXG;#>«t¨7¦B½½×_°Êïþqè÷Rí½>,ÐËçÎðäe3CVGcnåìÏCBo ÷ÀÀQèl ¸¿d{Óm9ÍRv^ez>ÜõvC"íF 1zxa kïôóýòûD\çíÍ¼Ý 5G¡½ú{ÀÛuÍÊ®Y³"4¿ñ¤æ½;}TÔ°ÏË¯Å!GÿÜcq¤UgÍkáñÇÕuëO§uÃìxká±'à±§àÑ§aísá¼ Q ÷FdJ»³6ðîû ³á�ûw?ê¥F áÑMø£{/Èv@Èb#ÎÇ¿üÓ(MCwFßiê»30ËÿÃýÏ·²b¢Çá!±,ËÑáOÏÉ;\E_üÙRR Ð©áÎìN� \|óÁ\¶Æ±3sÁFtêãöØydÿÅþËs2Æ¬ ®ýMî3wf3·cìa¤¿ÿº«¨;s;¦ôôÝ9 ¡æÞ\_~·ÝmUÎ~»«&£~·£Ôd5"³!P#ti3ye¨êöù¾z}AZ(%%[Yx lxÃ¤(<Çèw^éÈ+2£3ë£°µ:B´¢b¡3=°»¼¡hF· øÌæî¬ÜÈFçZÏ) û X ÝAÜyÐ{è¼OC£FàÑb~°#dG¦ïé)Q5tgAÎÈkH$yþ°þKÝ4%@qÏß·-iã»$äAW!ìó6ì)rf eÎóÿYÕ3ÙÝGªäú¶¶¶að-ðGwçÎ´Hh}Mûé¸³Ù,X2~³.f'z}ü5×l=gzM½ÄènoèÎHÒ»3iCú}®MuØV$\|\|ÉHñ¬xÖI½§r\_ X-fÁ%XÞ©ë $Kß,ºóäªä4à <ë¯ôÛÙÕ:ã\è8"xçÕ,GØe÷»2ªf^ÿxDL[ÕG·¹¢;²lÃè©=ôõk6¿ ¤çóèö@T$-Mmß[èÙÝMÎà¹Ð¨f'%Å5ªßuÙ¦Ðg·ºðrðXÞ2 ô�Ããó;r#Î¼ó¿Õhmºc;øÒáÎ<×åx©EwFdÊº³ùôï¥9TÍBJÖùWßÒ«Ð '±92ZQò¡ÈüÏÙèKpÎ¼ñÖß£;#ÉïÎ\|÷6Yñ4²õÎÖÁõóÏµ±qç Á_¹äÍ ë8õê#½e£;#ÉïÎz^ä4rhÌþèWÚ0�¼}Qø~µÜÙÏ²õ^ø«dBÉjudùçê56CwFÎ±;Ë;û¹;Ö¬æcÜù@/\|¶×b VOöl({ Z\|´èD/Wc äífå÷Û¬]Ð/ß»:ýT! jxøýÓÌ»gæçàvÈòÆI× w6~=bý½0ÿWuy¶ºôÜ-³>Uut7)vò0ÄêàïÅÝùÎ@jAåÜ¯Ã¶ HGyúøm¬ÿÐ\| õjtgAÎ bìèèH"whTëûìÙ©4¼q BúÆmýò8wÖÆt2Ò\IZ 4_{ÕvÙ>qóî]ÿOyò_YtgtçÓ8úòÖ\|éÖÿvVÛB©;º³{Ú¡ ¾ õ´ 5-±´~ì9èÎLYw&Ü)qÒ@Å"iîôF :ûxwVe¶³B!:ÐyTòç CwFw>-¿!ñ6iÚI÷¨2§vÎO×ÓÍ2Oør;®ÏµÑ õ)ýKÝãBJÝîÜVQDÚ³r;ËU¶?º¬©ñmÆº3uJÿêØ··¹²t¦\àzgäÃpg½wÖ£ ¶jëÿz$Û&93çÃ¯ÑHv}²7i9@ñAç®×NÕgq ë\(o¢ ÚÈ£ó;ÜªÕ=r^çù7Bct¶ÓTÞ1ºÿä#\_úX±î0í·%-¹ó$ôDjáñ¿täØ]ûæ\| êhãáZ¯Ç¿q#ûBçº3 SÖC 4¾¹a¹NØNÌÓ¾÷«;Åñ Zõ£fxl^íÛäaÕ0ê$¯IÑÑOÇot-êw&ê,XS»v}ÛgÔX9¢?¬ü¯ßLoSõ5ozÙJÏ¨$&ÁÅEwFw~OwÞþUl>áæ?vÆÓNª4qÊÄvº2ÝHTI{¢éF.kZtÜ£ ùðÜÞ·;ÕÖ Üþ¦4!êÊiÏ¸^o�ÈÒ)ÞH·BeCÆaTÊ%¿îLKÛ¦Õç^O¿± è#4ßdÇ¡Ê"w mÎ¼óáÏKÇAèð_¤I°uöË) ÄB4-æE¤[EÓuý ´ ëkç·gYìÙ»>Í�>}tHwèêèA©ëÎÖ/ßeaûR¹g~f3ÆªÚqî<êÐ£vÖ³y²W¢;£;';Ó uJÀÓ´ÇÅ³gÛ]?..9ä 6õôvÆ!UÄÞ½{_{íµ¯».ÇØ:§àßîé ³¤%ÉÅEwFw~¯?)¶}ýê\|îÎi®;_x³¯\Öuç¸>KDuú1YT>ÝùðÜydpðú+6+ì ÜHëÈ¡æo\ìÏsÓr> Ûû!hl°8û¹ü¿"ëáGºó ${ÁÑ¹Pö BôßÉ¯$ùaÉÛòì^«Y´¹#¦4_êôÆé¿3çrX¾ ^\|È ¡0ôõ@h2DTxt ø¤Õpç-µ&÷rx£ÕCO/Ê¼· ºØÑÛI®nèì¥µvA¿!RjñÞEdº³�½aóúE4°É&dn©iXww³FÓhúÊ¢;£;¿§:Sõc}Mû3vÁÀl6ÿl¶Ò\|z¦Ô9_ùõa£ij\|>é/.º3ºó{ýI-kKæ¤ðÿlîïãêEøw6é2ÿ}ÊFJë§äîLLw}Ð·gßÅv%ËÜ-y\_Qèïý5¶ÞGîotH¦\óbíÇwðÆ'nÄB0Òæûåwú§åBZ¦juÆ¬Îa«£×ájJs6æeU¦9+§i?ÿ%lz:{``�BPE6cP¥QZ�5ðØ¼£ÙVÝl:3ZÓÚró«s³äeÊM¯ËLoHÏ¬ÉÉªÍÎiK-hsO?ìq°èw.ºvïH oA¦®;Ëçÿêº,· Øm#ýÕzQS¸³fìÎ©#Ñ\|äÝÝù´)nðÑÓtÐpgC°£Ðc1Læãé\|ÀÃ Ä¢<òQÑ©àÎòµ"AÈ¡ÍKÚ?«[=¬ÁuçÑÖözb®¨Æ\_©Ð§;Ç\|ÐôÎîéBÐ"Ì¦/Bu4:;Ï?¼ý\|SNVPp{S.ö\|á0à%Eè ô5º¿7ÍSÕÓi1ÛCva0EÐíÑno1Yk§Í~ç¢½O@w�<ÙVôçîÜKÜ9%"!5:è´ö:-ýNË°Õ2b± Û~[Òd!kØÛdÍÙW8v¾KGÀA¦¬;kµ«}ÛeåãdoT7³]¡½ýqÆ5&ºHÀ(Ër,9g£;£;';»AÔIÙP:+ìtÓòÊ¬É1Öù«YpÚ¨TÌ©þÏ]G^VtgtçdpçîøùtAÈ%îlq½°¯² ¨Ä[ã!¾ÎR}²2(¢¦hÉ³¼Ý9©ÝY?ö îÜ¼}ç4!à0,ÙGn d{\È¤»KtåØûÏ¾ófIBöpéàM¿ïF ¾ó²,C$ C! ýþ²eo?·ºpZSNF_ª#$îVöÙÑ%¸XgÕ\ò/P-¢¥pwWCªÖW_Ûí~gNoÚ´¦³Î/"ÇÁiE äÍlÌÙ5»1ûüÃÓ.Ü1kîÌ·ÞhoA¦®;Ë[\_Xè¶±q3GöÛG:¢ Ec¾¸³ÞhÇ´I²o/º3ºó{¡òL\_ u´ÄÝYHùÈ·S¿ì3À¨ªªztÍ½ùiBjàN!ÅË!Xg½s¨æ@RÔä(/èÎèÎ§vçwY=KÒ;;]¯ikMìÔ\|RwV1KÔÙ HòI^îCwö^wjwÖOpHCÐ±«üÒÔ¡T³×UYø98ßùÄî,i@wNjwV=AÜUyãôÎ,k§)ã@îµP> >õ¤¯76CÏ"(-ðÏRÓD¡°Ór%,yFDêÎÄ©>-E1êb~<Ð\Ü»5/«×iM)>½Þ9Ú < \_Ôò®ÜªÏa< ui²î ¡(c ùä(æÄè³\Aò$¿GÝùL!LSË§ÜFBúOo\_æJ,£~BwVgRKCÝÝÍ5 ÀF@7\ÓI ;£;;Ó9ÛÖÍÜÙi#þúóÿhgICcYf£e:wi;à6ÊWà¾¨% EùînèÎHr»³é\©P S/!åëÿ{oÐUÿ'®WGÝ9za²;sgvæç9ßç9¥\|cÍùPGtÚFÉ¾^\Órmy- :Ùü =±áu³¼Bõí½¥5'vÆÕ� båÀØ'"¾¢+ª¡-¤ØiÇÊa"üà?ñ´»{ù^¡]»§ÀE³/Um ©öw6~v\_õÀó{¢mÃ¤S¼ü[níÜ¥¯mÝ&Ã¤/~ìì¼o©)èÜtD¶ª«ú+¡ÍÇÕÝ£\é,ù=¡ÏÌ\¥¼íòrñ·u)S£X²ó¤dgÝçÄGeg´Ã[·»zwP7> ]aHÇÝwVx,4áî0tZ1Ûô ÕµSaun 1xã¹-§¿²bÃgNþA^B#ÉÝ&l4z£;×aµÁÖ×¶T 10,¯õ h�Kv'ßVj#?ß^s!«c¨átþÓøÙCñ :¤ØYH(é¶+Î;¡B´gñ«kÙ/vggcA4£µ;ì´'´C°U æò¡P{Ü+@vFMIv¦M¦e¨1ËÎn÷wÿò\|ÇÙtÖ¬ilm²_ÚÐBvF"ìiûþGÑ¸&+±é^ÝÎ¢NÃ=;ÇåIy¨±°6yÅmvj[ÍGÈÎ¨ÆÍ Ë;¡Óþ vNB¦:W}Xíx ¢Þií³O¨DÇegIb§JC¼ ^þk}-ã+pNßvì7¡ {nè¨v'-BGa9Ô6ó7¨°¸í íß,iæyRt\|Ð{î'BæA¿qæ¥°î/Ó}çÔxúº®2gØS¹mæ©S¼uÅèb K4ÖÔ\cR òZ«à5B¡5væzõÇü EÓÿÊ0õñØ<<#Lðáy¿f8à\3NýßÅµ1²3jJ²3u,dÊÎ^B&ÊÎW>ø\|; ×¾d5®KÁk¾÷ em¯ÞÙ0fu(°³FÜùe/ÜG[ Ú©dÆ§Î3cQÇgg^Q ñç®.q2v6û>óåöå\!;£4;Ëÿ;³3Ä¾znW¡}Øné\þÕ0 p³Ñ»&;ÉÃ0P¿¶ºxØaOì îrxüeb6j( R þv×Î"!b:þèÏn0?.HÀ9 irK·6w }uï¹GýwÎº2r¼xö¨zêºrg° \|ýìó E#Æ§fëX¦]êÒ³?a½×§Ní{ ìüo³ éÞOÜ¬¯ ÿêÆÇ2:í»©g3Ù¡]ÈmxÿÕ2 }( CuÍ\ðaºÛìÆ2û;Sv7ììðí5f[Ñpkãz§µ6 Öªî¡´4ùë÷C(C öÌô2µ8 wÙ/ïx ÁXXÍYÎÖ¤ïOX[OZúÅ,Ø©Á8Ì¶/¯îèqßuàÇzÝgï:ý5kiaç®^ºÇKÜ!Å2;ºlÆ&ÁÕª¢C<Û¯("¥A·kËôcàÑa0I¶.-Ó}ãèçÖo^ØYTìagÑú£ ;\|G:ºe÷=B²¸\|¥o¼²"-ä%@Á9)QiúLB=°ié¶JgÄmðT¯8RàK½Rí;ßÀÙyÝagCc¢ ·VYµ{ÖéKitnËd ev?Ã«éÐ¢Q(Ô!ÇÎZ6Ø'ÔºÕ"6Óáo¸ã12dÆy;ÒÉ4¤4O o~ë9kiEû×Ú}sÏûßÁ©²/@¾ßööö:&Þ¨×<²óXv&jÚà5ÊþöeaM²«RÐ¼qu×a¶:W)>öô EÖg²Ó3²3²ó~\% (dÙòêCåà¦ÝTPvó\_H°å¦4ïõ¬²ÎoêèÕo¿Ie!¤m÷ìN¿<,¦Ô'×å+S\æì<:ý]Ð-=ÕÇChÒ½ @: É8=2 È¤(¹¦(¹òR®×bMðàuÍVÑåì¬-3àÇw2ch()zH)H!ÖßY7£h¤Ò'Ù-{¯½K7ù¡$pKý÷Îë®¶¦=Ö¯ª¥øHx5èMS;%Æ"hIãìeoµMæ¤ÉÙV2_O¶éÞ1açzíé?¶¹"î-'B½QÄhl¡È vd'Ô#])&Ö+jj2ÃÜH¼Q(Ô!ÄÎ¬M´(~íüb^×ÅfÜæY<úBW÷H\Ê:m^¿ç@\|¿qÕ,8½õa9Sd\_�ÙÙyÝhÈ 4op Çj¥k]³xåÃ/ýó¥E/ðãùÏ.YüÜwæÉ'ú Ä¶ ¦òà½ÅDvFMuvMO'!ÖRÇ/þçdcÅ¶:/}îõ@4Ã\ÒëlßþîÎ<u\|°ÐÃZ\Ç6¶ó¸(yj¤U";O.ß(C;%Ð¢}Á3ç$XþÃá{µÑü}þþ4<ý , Çxæøûbý¯Ï¥^xR§¢)º¤ª# wB¨nË s»J©7îð úY[~<ú ìÜ}]ôèéö¸ùW+K-!%-)§§ÖîÛf¥½rUb·BÛ5(êt PP´o¶V¯{ <þ¬X Ý0Ü ;ká×ß¿lcOòK&\_À]³ù¨3aýmJ2}Ø&?qCK-nuµ7? O-Ô{Lò õ©'à©'áé¿ÃßÉµ}Ï¼ÈÅ°ðÕÔË+ià· ZZ4ÜH'B¡P;ë ö}fz¹Ý,ÁdÌnÁdQÜ ½c¥QtêûÐ}òåíukc�AQÎ ²3²ó>þ(ê H}-¸a¸<~ÁÅ³Åbø×F à&,pëÏ¡Ýºì:ØK¢a#gÎ,ôÙ±Õ!IÛ-{ÎXìì®Åa§y@ö\A¨üpc¿ç¨¼øndgÔt3.7Ìu¢}guw¤s°Ð?\3«¨¤«°¢³¨ª³¨¢«¤¼§¸¼ß?½ß?£µ ì²Ô ¯ðT2Êí¼6R¿ñüã¦;n!$Ø%ÿô.OQKiYSU9++ºÊJ%ÞdC6 AukñÌÄÕÞ0Åp-["ÁN8o¿°lfù°Ç,§¿ËáVÓìõ7Øíþ6oñ@iùH{È"ÄsÈY³´äX¨ »Æ´-:mðô¼r·n·Np8ýÎRwsiAs±·µÈÛìw7ø<õnãðzùñañå§\|öº³ ô Ö·EÁËBì ,ÑLÍ@4 ©Øõ¿»;3Ù¦ÏíÍ-·c2R¢M1'~¦sh¯SÅ¹AvFvÞ7;«RoË&ºzdbin0Vùøle±{Ê^~wMT2u¾\dgdç 2¿¤øýÙÖ³à0[\|Ãli¬lz1÷¾°½½i~2íìì:Ðì,Éh$.8¶µXºí«sÄl!;bñ,¾Õµ9¢6WÜZý!sciyúÉXV?ù%rP®LöAßxð¦î!«=as'îÓC¸Ã¶»ûÇÝî+X[9½Fã¥EEãqÑ¼òÌrF:aõû+fVç-ì(ò÷¸ÃN§èôªÖà ¶¨:m½î:oyå' ¶B:¤Q¢AÝÒR½úìÛ§E}A·5RäöY=¦AyÐavÉ1ä¶cØc§ÛÉú¢[ÏûÝæ.0Åù+ÞCvF¡PSµl¢J¶}jð@oû£÷Ü6³Øã1øÃÞ±¹/þÿYºj¨1�]}Ê¸yÈÎÈÎûv«t¯³Nà¨mF4j¾\«Ê/¾ø?üðÚõxFÃòÚúä.Uìì¼¿sLõ Ý}K÷<óàeN{Ý±Ñôæ¬¯}céÆm4\6¯ÞR®«²3êÀ±³NShK5æ,ýÚ¯O/x¿¬ìÃêõe5kÊ¦¯©¨Y\_9}}ù´Meµe%ËKVW¯ª¨^SQüò¬Y¡çßÎ«LÝcHÉB°pù¬;Êg¶ÎÚ^XMþêÚ¢ê÷K\z/½ ^ 1R4Ï²J BÊ¾5-éô~åõö+®x½²b{ÍºÊ5 ewÍ³¶ ðÃsê/þ¼¿:»AgæDÝ65CS²Ã�Á§î¡fÆªªiËÊJÖL+ý°¢huYÉÊå%+«JWV¯ª,\_]ÁÊÜñÊOvÜùeeà½¶ÐRDdiÈÎ(êPag£ï>:³D1 ôÔÕ®ß°aC4Y»yûhÚh@%UGvFbì 9ÿ%£)\ñ~%5è¬I²"C¥ÅN³0DËìé´aÊêÇè£^²3jÊ³³6öP C\|l ÊZ4êélÚ^»~íªX2¼~Û¦MÛû4 ý¤RÄ¸°²Åì:à¥c0-l£¥·Úº¡u�: £ºz »:: ½Ú ³ ºÚ¡» :Û §Ô Û»\k¼L(3 ½=PÛ�/¿ï¥Ç;äv=´ÀpdÞÿÌI5kViÕlEíEî(,÷UtÖ6X»Þ\_ K×ÁMðÆÛ°\|Ô7C(N3¶É$»«Kôi æÑèê~èhÞ.ho§Gg'tuBg=:zÆ}ôè ó¦ªÌ'CÎíÊìB¡¦4;ë£¾mæ¬Ó~ ª\§éª¦1ìóÁP¥ @\4R6 HÙ~Æl£ vVpÍú3HÉ#CZøn!´¬i¹voú¤ïìì¼ÿàÌ.´ªUÌ°úBù$Ö¤E{i #Óò0ö IÀ:Û¨ Íivè¬b,iq£=#"DUHk@®át ÒqH xÞ&£HRM¨Óä$ç «©¸mdU&ø,gÄ¨Ð¿TÑ¥df´Ä=Q;óØ I4uwþÕÙR'ÇÚ<ê¹+Æ¦3²ó¡m;çSµ<ÉJ:»S@RqdgÔAºtÞn×Ò&Rd°× .\ä S(ÖJÙ\Y¼m!àìÈnì6È£GÈIÒ2íóÄñÙ¸ôÝø¢²2lH%sÙ¢jP6ðÎÎrÎ×Òó¥DÐÓì\RØ«RÔØÿ]»î©.Eírìþ°]V®§¢ïÍÄïðÆöK'¦C÷°ý3µlßï¯ëf}rmÿÎìúÏ/^5ó:]9¦ÌýNÍF>c/~vFÐv±ìÝ!h¹¥§]gqÏ¨ïç<Î¬´Ï§êcÞ6^Í(êPbgmì¿wítÎc¤Îæ;Oy¥ÓiIøGûº÷þÕOÍ}gãßOvÖÇû-²ó¡2Y\|T>ÙY÷y¹^±ãÈÎ¨dÄ{¦ó]¡M( ì;á&±\Ý$dgdçCw1n¸á®»îZ¸páÖ[ù$³ÆÆFWäõ¡úÉåØYCvF¡Q(²óÃÏØûO«ïwFdçÉÌÎ\|+Ü7oÞõ×_Çw;÷ßÿ[o½588Èc¶ LEvÞÅÎôyµáîU#;ÿìLåV¦o¼ñø¹C~øôÓOoÞ¼Ãrss3Ó 3u×ôý°¬=ÍlÈÎ(Öý+ÿöÛR+½( Ùy\gbÿÖåÇ>d¿Å=6Ù1¯ÛmÜ÷¥w'GìüØ9Ènº@¹åP°pá>øÍ5Sµ±ü~òf«åwçcú»?\;sq{á¶Cnod?þo¼±téRb,xÖÃXW×ôýÖö©w×nó²3ê�\ô:LOfä«æµL÷1ÈÎ(jJ±sÎàÃmb/ç·ÿZÐr«qì´= »8 m÷GðÎ¥Ò ¡¤íNÇ£Îqf2ì'N4ñs8ø4ggòåø··ÝvÛíyºíÖMLÄÃ7o¿¿ AsÈà�êë¯¿þÞ{ï}óÍ7ÁàÞë O\pØmÎÍPÇegn;ÊXzfvHãW»®:ñÞé9KÐØKIòÙhá\|ñäÇabwÜqÇÍûùèÝu×]äÁÄj^zé¥Ïö:\_é»çµ¦ýaj}·óû,r?ÍÙu@Í^ mP®^ý£F ®ìh0ÞjéËZ~fÃ7$Óac~ËmÛHÉîæ=$&BvF¡PSwçýgæ°³ý.{ôVvYÜß';$Ø!³¹¼+;§ vÏy$21Ø¹¾¾~óæÍ<ùÆ<Í;TEþvÞß~ûí¹-³½³3_và#÷ JÞsÏ=SÅOÚ/vñcO©ÊË2Ø9yÙyË-/¾ø"A?þ¥£¸ÈÅýõ×ÛÙÏó° ¾An/^<)^ö:_hûêuþÇæï qÈÎ¨> r+»�ÙKRË!m;Ëtð×óÁYÎöÊÆú¾MQßÅÄT2ÝÜ´MîÎij2:Àg"dg 5uØY6tyÏb66N;ohWvÇ,O[d{È'JYÐòpa·£>,Õy+¶c Äï°3ñpvÞo:E°èºë®#Á}î;óuþsrÿ©§Z·nÝ$,4Þl\öÊXç\|ÏÛ£Ë¨jÙ}gylQ$ö8}WYùìÜØØ¸páB^ 9·\|DìÇlìÏãYÛ»îºë¾ûî[²dI¿$Idök¾È^5üÊ×v5qhïÁ'@êxBvÒ ¶Ð©)tLWwª\=Ã¨6}¤ìm×³zþ¥=ºj´ç¢7Ädtàg Þmu¤Ó÷ìB¡¦;gÁ9GÁÛÕMbìØèncóA«ªåÜùð9¿ñ2î@Ë¢·{©_´«×#åì¡+cÞ=ÆOÖ>ÈìSÓÐÔI©á½_ÃÊ¨Tnº!f3NÒ=¤§µ½©¾¦fÅåË.]ÚÓ6øbïzf v±aå$äÎ5ÔêV~ºøóÚµrníéð »==])Ö»SûGÓ´Ì×é=!îîîlï'ºdûÃ��oéªÝ9C}yyGµ¨Os×¤M#A§èx9Üã·Øv66õ³" ÒÌêN[¦l©ÜËÎ2õL\KÍH¾å'É;h¼Z=iNËÖl¿SÍ®hò41x ïõK&DfY6ê¥3ÛLÊ¦¦ý¨u~,äüîºþ5_¡« ÍÚ¢Z²l1áÎîW6t#°QG èø¨fÃ/FV+¦m ¥FH²'ôG!:ëUÊIk¶ÇixaöÉà±U+ÓÇ²4.ùb, ¤4f;ÓpU6yIE!1ÍJð}oTk:ÓôZfæL{§¦A/GýeVÙYkòß½P¾2ô §ceË~Ú;åvxÙàZ@Ómµ :MðV²½il5%ÛfÛµ¡ó÷~=AìÁ¬¦Êóo ï½¢Ó\_N±ÒªMr]ÓC³iÄ4"B¡ºý\_Å"IM©ÑíJ û6KZ\|¾àJ¿4Ø+04QXggR jJ ¾°Ä«ØÛVB5lÊìS¢Á44Íub¿^ßIHXúVµÉ¦£ªy5¤Äø��¾)²Lá4¬¿¦§b{qû yýQ5·ÌjRMäñ¸hE£¥T7\_Og<öÌÏë ²Îê¢ópc¨b2% ²Æ§Öf¦ô¿!Tª½-H6¸¿¬1Ýd«ìºè}¹ö\¼Æ=+¸Ï»¦ÍWBPUØP±zOáÓ1EGgv§9W±FvXÈ®!t<ûH:©ÚåÎ0;Ì=G¢ÙB^9ÚÌ9tCÕd1¼5=m\uí~dÚvUf ¤©O >E¼ìx2Áü.QéÕöÇº¢n«=¨bTNRYzï}%ìÒf±×D¿1ØoMæöËÒÒ=¡Åß²:UªÉÖCXÓ$ÍU 4»Nv¤R3ñFÞ»ÈÈ¯£÷a#b1!¸>ù©ðêtë½ (Æ6¯Ä¨Ä¤WªÝÜ5:DÙ\|ÖgÍLa¼SÕL¿))Ê´H×2¢ç^>ÆÌü@³3øz¨ÑÝlT¶dÙ´Ë\³Å$Bï'ì]ÐS¦Ú3��ßÆû0¿n/Q÷Èi8~ß®÷þ÷Vï{1£76¥gÝ[ÓÞAdnÿê\|mÜÌ=}¸Ùÿ°úyðwåqÎÇ§iØ¥,²ÍðøÒoõ{Jvñ¡kíÐ´¾ÒÒ¾ÔðìÝ¬ÛJØÚ¯0(Òßs sE;2³¡1«ß{±MÖµ XQööþM°Ä¶rFS¢ÏÌB9£¹ïì3vÅ¸3úÏ2ý3ªû~¶aÄêMi½±½3äÞ Ì> Xý¯öy.,��øOTê×ð¿Ê ÿº®þwvÎ¯'«± ¾C¾áú®<Ôo{ooÝ¹ª¤ÿ'Ð\_ÄB¥î³ÅúO®ö fÓööÞÛ¬J}ÙUÌþ6��xL�ïÓMxç'Æ½´þ��¨T�¾·¹��ÀG¥¨T��P©��¾Ó4ñ!��Àw¦iø¾eºA?Àt:OcAQ\|��ÿ{LÓe94D,öÿP¥ ÙBj h¨É?:¬��à{ñº©K¶yóæ5@³víÚ¸Üÿ!ô¬««£çj0 Ù¸q#F;��À÷¨R#HMM tÜÀþÛÚÚ Rÿsjkk×[· hV¬X¡��ð}©TJ"X¹r%u¼7nÜØÓÓÓÒÒÒ¾ÚÀNýuvv655Q{õêÕ[¶lÁÿÏM¦¾¾ª~úÒÏ¶Ó8;9ô-Ðuu5U©ô~Ñ��ð}a¦P©Ô1£~pÛ@T¦Q1UWWú]ÙP© ªªlj28èaîä¡wBj,T¥?.��øUu¼[ZZ¾RõñWF£7nJýnUjuuu,¡OJ%<£þq¡R��~ ªÑhËÀ@ÄR»ººKýÎU5x<!3Àìþ}³±TôÆ��øÞUjMMMSS¼îG$eK¡R¿CZ]]-:iBË@¥��ÿJ¥^7TTJJJ��Jýo¼-~\|}÷tTØTê7´¨Tð½Öw2¶¿Wk@ó��ú_T©ÃJJý:Ö"»¦<JÝÙnÓÈQ©f?ÑjÙûéÎá_g¦øÊÉÅÊ9µ£k¬��TêNêqÛñ ¾Äg0·9\Jýo«Ômëï6cS%$Å]³TêN¨)Èb]EÜ¢ét kÔ(n¯¹dïíÖG¿Êö¥}ë·'«ÿb?À²1ìz;>ÎÓé³ÌA¿¶z¯Õ_¥ò·ÔgÉHü óÃÜÎ1àr��@¥,Äû R R·o6Ô× IB¥B¥î#©T«T3sç7ø0îU©zßÛø¨T]ÜòùÄafßþñK3G¥jüé;Ì?ÈÕ9BÕV©ìrTjÙ°¾ bdäoN¸¶JÝ\J��úCW©}ê¹§Èø»¦Jí èhvLÉÔ²±$¨TðCµSvTé¦e·oºo%ùÒ4Y~BÝÌÌ9æ ³ì0²«7X§)¾Öû¤éngfÙî-¿Y&r¨I¢Ë\«fµ¦¸.e¹ïb»þÕ1V��\õ¥Rõ¯;GÐÒIP©P©\_G¥ò8;Ìþ.T\øZ£w²qkäJT\¹ì¿b¹Ms}jÚTç¹ÄÄÈ§½QN³W¹¿ÝÖv&;w¸W¥ÆJ«fóÅÉÙCôJÝþ:ª¹+T\��¨Ô #¦i{!Ü´äÌ¯~N©¡¦e¿b¨T¨Ô¯´n#ôP¥X¦JÓ>tqP[¶¦+iílg¨ÔöÖ;é×V&×2eEd~ÛrÕÊF$·Õ»Yegô§ë$FÄ5¢tlð y3ÅvsóóÛYåV\|d~NIW¾\KÄNUbÐ3¨L¢R»&§3ªX$må¨ï£"~ ��\u'ª\w¼Ó\|§]BLí¡vvvÒ7;þüeËÕÕÕQí¦êÁg}ESÉã¡R¡R¿=Qj$%J·¥¦©°oõi¤ C¦ßÒûÛôôkMÓ£�T\ø~ô©Õûµ!¾¥#4 Ýí$ÜN]ÄHóìu+'´º#°B¥¦!réj%ímdm-ùðSòÉòÑòÿ·¥tI"Nmð½±ÄR¹Jí±åÁ\c$% õäóä3z,&\_,&KÖ6Ò!©ÄbêØÒìlVf¡UYÆo$MzÒ¤;Îp´ïÑ9º3ÿ6'��\õªRS&Q¡¦mÍkï½ïNI<ßï8Óév¹}Þ{î¿·¹½-¥)¹Æª4T\TêWaéÑÛ~sÖPävJÛíuI>ävIÎÌú\|ÖYgüéO÷PG¨q RÁ÷5nûwMuçW~÷ Ñ$ë-¯kWFúÊ G<×è!=HýRòàÝKV~QR¸vÐ µ%õy%Á ¡U¥C?;¼ú:im&é0 f¶Ì/a¶9°c]dÓ ò>bqqY]á Mù ÕúÌë\_TQ5ÄäÉgÉÚ:-ÅÏÀª[Yí"æÿhÝkK®..\_]Rjôëô±Ûîì�� Rp'\9ÜÓðóóÆHn)Ãá°¿r:NêK.·jÕ»\|$LH?\ì:\hÑ;~}ê jSrøÉåR¼^·ÏçÆràÁ6níf)ýt^ÉFßÙwsC¥î\ãVËVÁ1Þºõ§ý°·bð>¤¥÷Se¿å¾¤tÐ¼ÄÐIÏÝD^zrÎÈ²õAoØãíöz»þh0 d\_Hñdw~¿hmIÅûÃYï0uæ"ØêÕÎ6Ýa²\|ÑÌakKÜá7OvÊî¸·8/ìñÛ\îÎaCaÙç»íI¦Lc»TyLVc9ÀT¥¶ROþýPSE!q,É«;½i·7é¿÷pûK+pã��ÀNí¥R)êr×ÔÔlÞ¼y'tUv¼G·¡®[ñ×ñ³sÜîl85KÐ]B®%kêÉº:2÷Sòà}ëKIqIñ¢}$¬"ÉÿÏÞYÀÉU ü»í¬%ÝÁKqK <m±+õ¢!È#HBHp( B�Å4X<ÙìnÖ5ë2;n×ï÷Î9wfvv@hd÷\|¿avæÞ;w#ßÿ\|&fo7R£øÏüºÎõî³ûæÁÁOÀOÀãOÀcOÀíµç_dõ&L0a(uW¢Tª¨ HtºgËL6;BÅÞÃ++4ÍdhÌ¦ö¥[ñÁùÿ³.c'2"[Ñëµ½½TõaÊdÜSqm2jE6L©¤Jõ~Y%9FEZ#ó5¿ñ)cjÅÃÊ»±7H9jÊd7£T>.,BsÇ%'õZÓÈDÎß4áú¹Ð"atCd�RaR7&ÎÖÐy?ì-A))æÔXx,\GR cÄª°ziýTç²w"cûÌ@ýNñNÀPIèÚ >[9ÙÕeEq«ËZÊýïu$QRfÃ'rô-Íú ÉÞRv4Ô%aXQHFÉc¶H¶E³ _\s_Õ0a £ÔÝR1;Ä@\\|ßõ%Y¨ÝGoÝ¿³îK¤�øÿÆVøö_ýì§YóvWùÓnÔÊ(É¸§TU¡TÒÛÝÓ¢Gù?ÄG_}~X]u9~ÿàó}�ÑñsRÇ¥rQ¿ÌJp0 ï?¶©q.«hômuÎ?@Ò¢J¤ø�R¦æù49>õ±;\dãì°iææ²S M&åÉÏ[økEæá}à$ d'½$Û õôÃê÷;¦zSnkXç«)?$òhz^5£Ô!³»fúi°Eè×OÀ( &L0JÝ(+ �DÚìsæ4èÌËÞ[Ë©:"² sÑîL7ò¯ 1[F©-u(¡¤3ÚF«!4®+ G"¤w^÷Ð.ZíIÅ¥NPJ ôvth)6w»±zzTHgLiÆ]A"~<©nC¨ @wSõTWÌcÜÅÃ+!�¤¸Ä@l¿½Ñk7Sv4T »QiDÄ:«øY ÆÙáN¸üìp£èÚ¢=acJUg·_õô3¡VD^¶b5[,vd¨æWTeµS0a £Ô]R¤·¶ý{)5÷UÚzØ ç §kÑXIOß\±åÚ\_dÔ!ÍuÉõ7V÷RL8JµlR%EpÓR«öÔ®ò"ä¤[@¿{ðÅTORìJjCíç{$L:ÑQT[²?ÔÇû.©ÃJ(U7Y£TU¤¶H¹§³X¯zú½eúSB© @Gò[« (®÷o¶ì4BO\@ãH §¤©ÆLÃ÷±oqÏq³ªÏ»Úzh% e¾)¥¶fõô³¡N$ñY�Õ1»I2£T&L0a(u¥T9±Æ{~}hÑ¤+ÿÚG¤üåÚN½}ã=¥2JÝiJ(¥ÎÊ£Ti-U j>Zþ¢;@ÉùôèEÄqtR'¥ H×)n¯/ÖÇ,ÆG)üå>hÿëe4HÓ2i¢ª]_9ö5«+q>b³Ü ªz RR£ÏÝÒ]æ>n.k+;6¶@\ggö}d"Ñ Û#$j58"òþ HÜ¤<KÏ4TV[§ä/[Ú}y-«K3Ve &RwJíÝ4Ë&HçE}VÔ<õ³Rwxòp)d¯Ï(ÉD Ôx¥oK©NÌ É¶²BI dr#¯>(DµB ÒxØØa:á(U@ºªçús:PÐhh+ÚVÔc¤ÇKÙPæLY%UÎH<HÜzi¬Ä2P½§¾K2ôÊIPû ñÃ-SÝÛÎ¬ýwõä=áÞG¡¡¢IRn&A¸c&U$Ae'3i2ÊiÚé5ÇtÞÆ²ã¡2CdcJC2M1)§¢$S:H~WQ¦¥WF©L0aQê.D©úÓvbëñg5¾byV<÷¨ö¤MbÊdBQª!½¹d]mkKO_W/î2]Z«¯¯]¹òã¿ß¯Í®%$3 SÁ o\|S¨EH'îR'¥ò1è[sú~½q¸& <÷K$ó5I~³§¨54_T"éð£n »\|6Ñäi5WÀãoBÉ!Òwý®©L¶£BQ½ËiðOÙ8y&\<ö"ln¶a¤ Ê©¡^Ë@EÍß8b½òÜÍ. lýéðÆÇ°ilYÐP [¡·ò¤q-ië¾n¨ë¡4qUf &L¥îBº#<4» òrú-=<$¬[_¡ð0Je2)UQ<<¢7]xB©.ÇØTvZ)#VÉm±t:íUó÷VÇ4§EÀ2.,5R'¥¦cÐQµvßX¥ßhÝ2õTG ¨,}Y9e¥ >YÒás)zÿ io¸h6 ¦ÈºFxõô_~JÃÞ>nw(Ù²ÝÅe«½%ìyïµWA aÆ/ù64´4B=7r'¬]Þî÷ÈÈA3à[ ÏÔÎS9Réd?T C{¼÷><ðÀGÓ'×»]¾�2V;í{7ís6)Õ�ÒýÄF©-nëÎ½Ö1>¬5aýZ¨Ú ¸U¶©<¯Þ�5 z3TÕÀ6¦X\\&L0aºQj:×K5Ø§Õ(Í(ïR³"Q<À"VAxF©L&¥:õv»§\|ëPDØc9¢ ¿êWôY YýÈ{à«úÉÈùÚRì²ÊG sÓiû\Içä@=Y0P\ukWhJ%ü¡Huª·¦$rtêËÞRw¿NSfR$RL ¯MB $ BQè5¼G+¦õXqÀèk1Ãk!òñ¶Ýuô§ MRÀ§Hv&76ü)s à&ëËùUdõ&LÐ9P3~¿Eº(õ§TkOZkµ½©L}9U¦\oCÚ¹)^ 76× çiARL J%øé)UÞf¢È<cjsÝ9dx\|È²×oç.MÉ@J×0Je² PªÜÅB\_äôúJJ#¯©:÷àcÒXÚä9 V´Ìy êö(\&Hç:àÚmÖÑÞè.ùÏB4FKÃD!¾tab %e"åI2 É(¹B ¹ÿ>sB§ë2:øÐ I\%µ^¼³³Ø²t"l ¤,w²8ÛÇ¯ªøQU©¾:²È1ad\|QÍ\Ó .j¶eÇývêQM®£1!ÊÛïvd\øàZ;=®ï¥2 \ò7\x2¥T¯Á4ã+(:º:?Ìxü:ö=êç7 0NæsF©»1¥FcÀF² ÕL0J4\SÇ¸rc+Öô özp´ÐA¨ÏU¼æ~èT±>H"fJ%"¡kSÃ´¸Ã´¢Z Y}^Sîç4NBÃ ù& ×? [ESSBÒ##LF_tú×Õ0µË-EOû4~PªÛRä 8&9¤ ¥ò óCsiï/¥TuÄ\|Ì( q¦y£ ¦ÀÅ DÝ'ÿºúçy¨¹E¯×ÉéõÙý³Î9ÙÇkx²KÏL¤Rè'Ë(Gü{UÍýëýl¨ÿÎ+ÌGX6a2JýV:£�©ðÚEÏ!DëgØyÿö~ämµ²«r$ê/>ý�/=Zo)Úûøé(fÊdBP±¥jê4+ºÃ\zd17Yk- bëU¿8ÖOµ#kù_^$jaÊd T7 )ÚYqS§Øî¤ã×EÓ-!ÿ¤þþÈ@v¥2÷ª¨¤EôúÿUJÝ²Üt÷ß¾óîko¼¾üõ¥Z{å¥ç~rÁÕÛu }é1_ßAm>ã£r£ÔÝRkRi%óÁ÷< 'çÁÃùðèc°}ò<ú Ì\|{&glj?yß6[Mq«¯Ï¿ÇZßtxìi¨«æjh©j¨Ù¾zc=äwqG J÷;>:ëhëÊ5eûLÖ°EðÙÚ=¶ê)eäÞû�67@]¬¯/ª?=³«Øvèy£9d^5ýr©´¶ãT<={k3[úÝSàÖàïÿ'ùóàÑyð0móðóùöÈL[°® ±½¦9ê×F,Ç2-ÈÆº&»7¥j\|ö¶Ë qBz3Òa5Þrçý íýQA BOéXÝÚÕsÿv£9LÈª×ÙÑkpíùÉÚöb3A~:MQá3ª$?ÐpááÓÊ(ºÍÆVðyOe"Tæy(õÛTñÌX¾Ø¦CæÃÒ«Õn2e<õ H³¶?è¼ê�a3\|AGÑB"cÑw@uhÑ a\x0JÝÍ(j¦Rex-\XÑ+n)ò´¹z=þ>¿ÛçÙZàÛêÔé)ëõ}Q±OhÑ»V »/É u>èªj÷'¦?U¦F geÊ&»»ÈÔS"jüø YÌx¶ ÷À3~^3ÒÞ$YEÜ>«×ïç²Q¿A³Î<õ £Ï×.öUi¶ kßÑóÿpÿ× ü»$@MÞZ¹²Ä\yLDÏ!ögóÚù ÉHÇ&JF©ß¥f#q°ÞÀo©þb¯Rem2 ¶b\5ì:RÆO Cç?¾ÌD õ68y\tRF©R¿TÈRj@£TæPÝÔ±QØ,Ìä^ 32Ïûõ_ÆÕLÎ(u·§TõÌ6ÞìÌaãWR?rz×[:;=õöRéñ×!LºÊqøÿx Á!xäÞ lÝ.KØ¨ [ §5é²áÇ´´¨Ußg²v8§¬\|0¼ô>8$!Ée#_3!þÆ?[^îk,qÁiOZ,a!b´Æ-Ù2D)ntt¸+6t´H~Sn]WbI[u¼Åç\| ¼IY0%½7Ó¸¸¥\| nÎ²Ú³oAX4JIva%Í( Ý~êãh¬B ýèÓ¡Ó,¥g_~}Jêhm')LnXî©?¬¬ÔÕaC!2NZ]Ûç´¤eaÑÙÉl! yË§T&f3 -2<úæD_n�FMÍö¦µìY$«üu¸U±yÎ·ûY. +<5 Bðé»nði&TÌ ¡T¡Ôkæ?ÁÁ¨5L¥~[ªj}'=_ _±üóÏ2tRM&SVû6"dÕ'=¸QwBJf¯ÀTÓ¬^\ @© 1·¨á?\\|º\_Ëq¹ dÓe¢ÔSO>áªk¯zyÙÒÎáp\K/F-í4ü¥îfªZ¶² yn:ãøõÅÎúÍ>\_¯°¶°pK«¡ÈÕæóuæjñ{ü¥ þÉmVMÝW~eD¨µf"ITº ~=<\|ïæ²-¥Eµõ>\_£××äõÕøÖû}á3~ÿ\!w-ÙØJ$">¦<ÏÿïÚ§M«ñø6Úµþîé3ª\æIÑ~ ßpï©À×xé·ö¼qrq]EYuQá:£aÒ¤j¿\_¡Ñëoòøë}¸ù´Vçõjíß¥3úf?KnêJÅ÷£0[\&ãaê>¶áûýÅ<G¦Ö$#Iö ^¥ùäV.\_æÖéN,®\_ßõø�Ý¨£ÿÓüa�ºDHá)QÎä¦Èîb \$¥ªÃJ6ëÔìGªiàIç@¤a}ÝªÏ7ÝX5¦p<1 !$x9Ò?K$ñy6¾µqÓ¦ ëÖm¨¯k\|åtlµ ='f\_~/FÉ§¦©¡íõ7®]µiÃz¼zGx9�¸\|UÛðË�0>-î¬µiQ~½ÕaÁ?Þ¤=Ê=x§Ü¼ðÍäTðL¥~»\x®°Ñ9Eïéé\_ãp(ºqCe\_ßªï;ÞQ\£Ô¯äLå5²Rä¥3ªGµQòòXQ\ïeÏPª(@<áa C0i¡!Òi£O´×CA´Ê£mÄ¢"R$akÔÖÁÆÊLÛT [;apbÑá2Û3ôÌ1¶íuÌ\|QE!ê/ÛÒ Ý½× H"ÖÕT%ït¢[&ZÃ\_$Ûym8×Ç6üí7£2Í Ý}¡@×wÖTëê©tþE¶0DªÛê$\Ù¥"eUÅ$ð!ùPàÀi$În¿ìïÏôi»k\ÆÏÖ¿]v&¹¿sÿ¨ H¼G©)HvÝtù¹V\|Ã}Ø9Wr"¾z\fV}ö?à!4Y\øæÌFÑl=û+êúqY#GÍ<õEæbLQ"Ã=¸êì6dwX1@OQ,ngñsî Þä!ß´ò\_gþð ;B¢DÜ½¬6\|¢¾ì°£xåÍ(=AÌ0�¡[¦ Ìfd4Þýø#ÑtäàÒÒ!CÑ¥w?ÌÏÎQêwûesZÆÄÏ^R¥îÜ¸Qda¬Z;¦mK R\|ßúßþ¿Ïk¹IÎ¤~#+î]xÁý 71+J¨Tv46ÿÓß'0a2$j7×¶r2UÃR´ççÇ�b"«¶¥T ªüãðøÉæ/>®Û�%Z/ü/ Ù¿<Pª§lÓ¯m´ vÈKRnË-G¢}%Ó¿"}_x# t?1çJ'Bnd£6^b©´X b»Ôtvdp>·ü£A)ëN¬bdM×½~åR²c²Zr©cÌZÎb×ÄàDpB¤®÷üÞLÀì:×8#&ÜôÈQ\øsêµÏhµ®¤ú»7®4tÈd9õ;ãQ¿%FösYÈ/i)½æá¥2JÝ¥ÔrF©L&8¥þ÷¼É(ÉnO©cX#£L+ª~W/'ìÿ)&L¾tÖK�ôn^þ!IËïDºé¸}¡ÿHÎRª¶'6B©\æWÊ\�í@5'en$øTÄzÏ/"ù[S\N¹ë¡ôTêÃ nsé! E3Oùó0Ò^$´<üÇõDÕéÌ:B>d(øëõ7Üô·?TùÍ5!ó®'^Ih\x<j\|¿�3¥Çt½Ý«·º®úÝu÷?tßÙg³ ³1uKm%öK%ßLTpÝOàÏÒÈaD½gîýwÜ:÷÷ÙoÓ/é}¿£Ïû¥NpºH°¹êË/©êèÄ\?$r©tynÄ%¿½ÿyF©R¿/¾E1§h\èhQêDìÿ\_©:2JeÊd¢P\l½RÃUõ»è{R0a²³^ [^s1FBÑ {ükí¼Í\|¢¹m(@²©¢J;yI£T)¥¨¥Að¢bÒUwýxÏ"kÙôS¯íÓHMÊ¸öæ f1¦ §øç6Ò2KMö~üÒ~ÉRª³aD´¼¸øãG+¿È2%ÿ§Ï±C÷X6h>þ=k¹Mm¥¾ðzË©[±X¨åÓ~@KÌu/®ªë$¦)P¢'úIýW§Õ^ñÎ{«)RÕ ú[¾(÷SNE6¤+ÞLµ+JãeZI U!)#¿9¼ÔGl©¢ëæ-bÊ(uW¥F©±ÿç¼¾5»Ü[Ú ?Ñ(õF©L¾û%ÙR0aò5(U¨{ðÊSl¬\È¸ßç[vXJFyª¸a:ä¹/ÄÓqéMæ1ò{þ$íiò´S®éÕ²ñrz~ím?Aó5N:éÏÍÎ'9Ù"GWèhÉQbÃü°².-æc^ÄáÊUï:¦ZNþí_:µXQ~hÎ¯%õ^ÖCO¿8¢B\Íy Æ"½5í,ó'7ÝÖKîCñÀQnO2Ùõ%óyKËçx\1ðÉ²'½Äzé×¾äÓÕ=ôKI% Dp<$d5bèðR£TF©»þÁ(uê\|ðA$Ñp5_K¶TÖ%¥Nhío'#±¿Ñ.7FØÿ&L\|é<µvö/òM?²ÑÐ)ÇÈÜBk¼üë·Þ[ñÞo½ö&i¯öþo½¾\|ñò7.{ó_ ßZ=ÀKÀ§ ^Ç¹¸åÒ?õÄ«º)EJiqD=VÃ¬ºóü}¥Z½Å§þ® AÞIV½ÿøälñº+îûGö>4x&ÊRµÑÙ¿:Í¥£Å5\Ó;"öÝséYe4´ô Y TÓ%ITÅ¿gþSÏ¼¿¦²7®¤hN::¸ÀWJK^tÍìB2ó¼J"H)yhë}7þqùëË>Zµz KçOñj6§ªªjBQ@ê;¼ÔÅ(Q£TF©» ´··ßqÇsçÎ}'>ûì3 _¶83JU·S¯{"ge:nHóëöñiN]ùÆÅ7A©ã<§&LF±zö/ 6F2Ö% 8[�16<¸çÒ§E§G:Î £¹¬FêCKï"ëÌê^êy+qn¹óüÃ½" §\|Õvl©RSR©W°»øÔkê�äÝøó÷^éÓÕY¼1õ=9EwI=j2«¥!ÝT÷öc>èÈóîç ý.¿òGvGÖ÷ÖG_RFNÜ p"&åLìÐ{¡eoÒGO.¢·jG&×Iç]øyU>+CÔø£1\KriÊË¯=x·wk¬ê &ãG-Óäõ7·OÆÖPQÓÇD©[J)øàa\4Å.f@3²XÉôVäÐÕÛçàÎVú< Býí??ÈM«¯îqÊÕZ\ªFYÁt¼á¶ f:tÄZrÚÕ ´À@Ö O-£¹uÅ{ý½O¾ôÆÊÏV\|ñÎ¢%o-[²ì%¯,_ºhÙ³«Þ}ì7?9ÒÏuxuÊ:H1TÕ¯z¬ÈLÄk¡�«7;}G;ëçÖ4µEx§UgøL+ý~²øô/I LÈæþÉ%W>ûÊÛûÒqßÆ7'òZfQÉK<iS%Zñu¤þNrWgcñïÒieÍoLÀ½¶Ûôê{,®Ê3.£¦¥l¼ôgñHê²}79¦oz \|P AOØoØnLZ>ê DÕ(UC¬ø¯ 6SÐV¸zòþ°e+ôôC"ÃA Ð g Oµ>êp ¥!gîiu£2 ¦Cg/ àUðÅç°æSXó9à;ÅÊ( ST¶ÿÏÉDòí½¢ìàà¬Ëg 'òß-ÄÊØÊê¼Sò3&CP³ìÄ=(ZÊØòÖùEÞGÿZ%OE'µ´·§)A~PËÄ(õ?ÜM94T¬'üiv^þÐ«íÔó\|twÉb/ ^î§Ù¯}WÎyLë®;¤ ªÖBgóçK2ieÆ¹÷=°gÅ6ü ÙtÈì@f2íS_]Q�beE´ð]9ÎQê7¹4ì ]-ýöæ´0 ·îÑÛH¢>÷åó~[¾`XtìçsÛ1úZ&]ýà?Ó9_!ºÝC×)Á7´å' M5¹/½ki×HÖzF©L¾qJ¥Æp_üC»-Jáü?Bt2ÜíI\|}ìIbJíú¤ùÌ½½æ·\|ÓþçCG.I3åcv §pÒÛ¿ý¼»8¸§iB°PgèÔýcv]Ò¼o&.ÖKÍPªÈ&¸÷ÚAOÁ°©bãÌó îxùÅàq'uMß«¦À½Þï_7å�xt´ �'Ëq>û¨îÅ'ä1SÈ ÙKàÖÛ7ÕT¹ E®"_KaÅð¬sáÝCoMu$ &LZà¶Ý{ÿ.yúùÄ<åaTÊ»KÎ¿±ì,¨R´Ô#NQ¾R!ë<çnÌÚ/!$÷þzüæàqè]Õó? 8tÉ5T¤û§]S¾f·»¦ÀYSi+ñ X ¼×Ýk³¯tøá_«! B,N¿T m[+&¥>nqlµVy§¬)Ùyô±ëüîV§-ÌªÉ=dv7TìO½½ÜH>K&L0úÜ £TF©ß8¥Bª§í ¬ój%5JSgû\|J¥ä¾þ¢s®rà©Wð$ãËÅÄg!OÆåwÉ±ïkêbZ¾ët(Ò]}íÅgøhý§é=<³Ï£TyW%F©RI"Ù´?EO¥(& m=Ó£'¾¸Vç×¦T) U(jüÚYtýÜ!1Ï\®Ç"ÒF¯9×I©è«?Y©Áµ°ÔI«7-0fsi"Ëw<ÊKc¨¢X kq1'Ã8³$ÉéÜçþ÷ÇSzì\|bX\,õÌ¶ÔÌy©'²^\3Z=ÒsiYC²JÖ²ÿ¼ã±@2x8X(»Aú«¾:ß'VeýÊÞÝK=£?.Xêtì¿þÞ¾KÏ;²Ç¤yÏÿFUC¬ÇÏäiûhOû×//Îä/zïoÒ¡ÔR¾óQ3£©æ¤Ò}S:ø¹ GÝÇ²TfÄ5Õ_¹¤×#mö¼ÙôÊ. Z×ª1Qu«Ø�ý~ÍÀ)ä¶VçPu#E%¢¦þR²êl$¢Q,ÁcÃü-Ëq úèÖÿë¢ngKñ9ôN#%©��©§#I-¦ÑÊåJÙ6I²ztÛb H'\|ð%#"\:·L5~gýlíïTó½jéå}3DãË;«Ö:9Hea@89Éù©bF?j2ºVzÚë¹¥JÉ}þOV?¬¤-õLò K=-õè½íÔÃÝéy©Ç²ÔÉåcÒjðÀXùPQV^þ_·<5õUjtg0Ê R"4ÞrM¹È\|°äýhÕaE¯ÃRÁÉß©S¢ýy1½±¬"ÿ5ß¥1ñc ój&ÇHöÜA»ä+\PõÙ¥¾d ÿLîêÚmÒÞËBGÛÆ¯»(è²$küÑÛ5åqÔöüç»ó-~£:w6í\|júE§òØÙf^{¨¶P³³ë\%´§¿àè.Q%Ý£¤E)¡%&²ìbÌRÑS,%ìö¦ì9´§kb¸��¥ª(riÍs´¤ÀÞ¾ûQ¯E×eyW\|·=µÎ#OÍ5WDÕb¤ÆÚkÜ¢¯$?þz4 .¡©l®´¥j¼êi«®Þ¿óùÿÐoE1ÎjÈfÅ %jÈcCVÉír.øïÛ·òGêbðÈ¥R§¥¥þè'Eæ¼Î3Ã@Æ±Ð4ÙHþÏ÷®ã¹ YEe~Õ§ð®¢e<ÄÔ?Ba û©}ÿlI=Ñ}ç©àÔ\ÍaZõ\_^iÜæiý)µap#úüBJ8Hu¯TÎ\|.i¨hÝòm´T2MjO?º¥f¬ö+,µÃÿÙKÎcXª>N±ôDEwivÈÛ»Þj¢±HZÍÍ ]¦@/íx²¥Øc¯ÜYF{ZÈ§S AýTU¯?ÿíháIÍù>ÔE#¿èp?þ¯^$KshïäÚ��Xê©²Ô¨É¨hõ÷ÄL·\|©à/ sK&³µ!jñRtdï}·Í±d KÛ×¼^ëééKrêªxzH[sãrEDíqì8Ô7Àó~~Ä >}H5EtÀf£WöòL1!»3Í\ÊæO·T^p¾¿½6Ï²\ÒòÇi aLÔàåÙ:j<@#}tÓÿø¼¹ KNKá"ªï¤ñ0 õ?ué%ï-¼hWÁy».ûµó'g'#3YIþ »#»"ê¶'óZ\Cíì��©§!¨ÖD¡ÈGdÑFÇ@õÖB3ôµÍ»õáWÃÜR¤¥²ã¾<¯ÀÃ«»x . ERëêfî_DJEÌ\êT~óõET IVKÞ/nýOûhb8×aÎªS´ås,L}%k¶½×¡±°È¥Rg¥XV\_Ry>!WV~í'UÉE\ÑB¡ªSã¤ö\_ÝýÌ¸¸c$; Þ£FhàÍ.æØ>Ü«ñ JfZBò¢,tbãÔüÆ»çÙF ¤n¯7qýzêù Iª\ì¢\|e0G çØªËfS}+ÉF²°.®JFãQóÍµÈ"GhÕOØR$k]4Rµo;äÎ ;.è<÷Ë4 QØ¼£ë¼O)ÔG£õ³Ùù²§¼ùü¿ þ10ÞvðóWöäø³;.§RbÖï5g¯ð ´ò8íøCk±Óo±÷ÛÏÝÁ©×00 ��z:,5°91N¾«æ3õKîùûã=9zc,${õá[Ü\|J\oýÝKáÌ)£Fíêc\QÚIq¹ùuueqKìs÷uBÉÕTFHëÞòó¸üòyv×þðçý©ÉLü\»ªa,z KupKuæê:k §¡xdÌòé\cM,6F(vøo/Èæ,V)ûÃm£WfG³>zùÛrÌ¬©üÍ/ñkLZ/êqÝ N.ä¦ß¹¢«ØÎ÷ÖåÏ£vPÄ§kQ¾¿Ã¡í7{¼ñEÃ\_ýC2;þs[2Bã]ÔÕI4KMÌ²Å½"ÕOþ.:ÔX]^Þít´g¹k ?N[«(¤¨dÀR��ÀRO¥êÉoèµû×õh¬öb«{ÞÁCÃì¤Rr¤ë#wº$¯KLJõÎoôiäðy1=úóï\|=ÖévínéÕÌ'gÌè wm×[«ÕV.ÙÊwhFBcãõ7ÿèËä"8É^¸£¹s09Ï.UòX\,uª¥&TêMZª+ï7ÍRYøõsø$p«d)ýÒ?þ"Æ/O P¬c÷æÕ¬+óé©¥NïeU-}aEQbA¿ôð¯ Í+E)çC!QX ÕjD3 NZmôÎÃïI9R èÐóé¥çh´ÆFhx~po}Ìå¶¾Wx½UOq¦¨Ì\|dttÝuÓùÐç>þ+éí"W+JÑ3°ÔáÎñÏ\²[eÛÅG³ÔZ·ýÀù¹#9ö~»ç½ÙÒª ÔÙEaùh¸Vÿ{õÜ¬ ×Úïò¼=ï#Ô6J¨B£o¢ö7vÏÊ²º¢RÎxÎý×ýö ÓwP¨wVyòy ysßúÈW¨-§'g��°ÔSk©zò®Y7"Bþî/Í/bÎèÌrIYÙõÕøöú;~]Vm7'²ð»èÖ§ßð9I?qÇÿàYì÷<¶½V×àzkÎ6 ê#ÝæçpµóÕ,çÇ®¾ê_øÌ?ÿYí,dq»¥ì¢{½h\|Òrg&°TXjª·§,UIZjÎñ-B¡¶ª$,uÎù×ý 2ámÕ¯3=X¿õo÷;nY<·Pòð « ¬eÝÏ5cj.»´Ô3òxKæ°ý+DJ{ø_<8GÍ¶ô[] ÅåÃùÝü³á+ÎñççE%¹ç´{~±ú¢ìB(Û¥G)°O¾ïx]îçÒ}oXI4ÒéduíO²TUa@§±zìþê²¼üCîj«·ÚG_øBó¢Eæ{Æ²¥x§&¿x¬H %Æø4pm¹{oyQÈáMH«´-¯¾úî¿¸õEsíCiÜnëpì_)ÕSC> ¦��¥QMô5)Ä(0@_úøG<ýÖ"Yl\ ³$Õî´X³X¼ýÓU¿5£Õ] ¢¨\|Ñ²ýú»d·1íÌ§©+zòx"BrØßÔx·Ø+9<öì,ÅÙ1$ÄÓbÏ²¹Íþøëo(5¤Oö1,:ÕR5Æ?]ÃíÒó³ u84ÅRðPÓ{bæ)ëeE\|ñ¡xjwÌ\b8HÃM?ùÌG¼± ë/,³²6¿µóÿÿïÇÔd¥5µ¬ªvFÎ¥Î>j(õÐàÿ~¡²ØÙÓSØïõöØìQWnÈ^ÞÛ~íßQgÆt>kÕÜ ?Ä7ßP;;·« dOÙGè§hdTUZzRjæuI^nW¦ñöÎ/\Þîuw{/ëüôÉ¯æÊÜæô±¨¹:ªÆGr·Ü÷îÜüÒNoñxnþÅ6æÉñx»$ë!§³~V9Ýu7E 3¿GÑÅÖÐù{è÷÷äç:ývGÃÙîqôeKyÒINë¬Ù{Ï»\[ÈÖc¸ía©��©§ËR J-dXôôO11å1±ÈÄu ©üìg?}°©!ÄÌB b5Uå®ÛÄO#ÇûöþÊ^ÒHÖô§¶ü>?7ÞYV9³Îb.×¿ýì?Z;:tqEãH¥RÏhKÕÌ;ÑèKK]V«Õãùõ=µt]ÞIÛëþá¾YÙN^.Ûyåu_Ávæ//#äaÖ;X±Úí¬§°NãtÚ?uõUû÷ÐÒDê2qç ÌC¥Î³ñ³CbéÙ¹pÑÎ¦ù öïUú^áü³/¥g·Sç¯ç%VºÖL/äÙè£ê¶-/^tþ9ÚrÎ§´G¶QT¬6Æ}3£$\|²7(Úöö·¯~yaÙöó?Ý÷)j95!fÅ¦dpÑ;4HÍïøoþWÎ½gþÜÊÒÙU¥åûËKwÍÿú%¤;GÃ> ).ëºÊU×hdú��3ÉR§ZC$c¯þ·wvötîß³w ·/ýÖ&$B¤ìQqEM¨OÍÒ'Öàaw4ewZ½gÏº¦ jPÝªÓ.êQ-}¹°Ô3ÝOûÊ ilÆ(,Ujh:«ióm¯þÄ¾üòÃÞrvÛXXþ®ÝUY<ëÝùçÑmwSýaõSØO�¥~ \òCd¶Öm+rHVLHYN©äÒ^Ý¬ª.ÆÊþ¥Rÿ@[u ea¶]²8%5¨¼»¤îXÈfY7Ü}WºÏ1Ýa©°Ô<§á$tÔÄ³OmµYx$[dg·v÷dµ±±{ìcsÉGwÕ·¥~Å¬:}ÆXªÙ\_©gøN«¤IV/ ôæòqWÖ¸S=û"êcüZrÿ<^L¥\|=è ßÝþê¢cÜîs8ÆÜ®';ÎÙqG¶lËuå7n?^zÆÇ)&$ØRÊøkÕ¾çå5ÚÆ óÝ±ø]>wnÀé²XÆ<¹½Å»Tn}b2�UÍ±]Ö±Ik£Öä&É¡Z1#bgÍufóÃZeé\÷c?ZXê 8 ëGQYD¡woøï¿e¡¶ÓåtIEÞ®Jdl©N.0cLKMge¥ÍI8®ÌKÓô¾Îz§U²³(ÛÎê<É/Y]VCQµ8\<g½Èzó½w²ãEHK]þJK=é'vðIQyohè?¿ÿ÷^«=Ë¼x#å)µºÝÙæõ»S²:$§Û¼ÊÃówÞ®éì\_JEM{LK5/Ò ìþ«&/®¾8hbRnsÙ%Ô7Êá¥wÿH×fARX¦Cûþú/jómcù¹²Í«»gu9òZ ÷{÷TÎ.n..så0]³g äe7k?û%uú(,lÉÆ"!z{Ïþ¼ÜîK8Ï>îpºZòfµÎ¿¨aÁï·Ì\ëÌÍö¹m±§Ï[7w>mÞéãF²$"í°¾yigS²ÈSÐ/ÙÛÝVoî¡\|ÖçóÖæ-LµbÞòJYÛ¾à ±��K=Í:ñO=L#åJNÅ !\|¸PÊ®ùá?DØõÌíuX\8;,UÌEÒ;j«ø3\={öªyy>Iõ¸êòéÆeôÚ;Tsº©½ÞÜA«W¶ÎÎï²JqWV»$í,ÿ½ÛÄ¯&W·PöÀö\|«,YùhÉ´þ6ºm=Ýó[º÷·t·hwÝj÷M´ß=J(ö�ÎXK53BC¡Ð0T¦.7£æ¯y~C \ÙRÉË?q ´¥l·½uhx9:¡v²h>\ÃRA&ÑhtFYª!ª2jÊX{UNKµeýåí}¬G¼,IÚã¾ÑO/,bÁyiWrã§¿ ¯ÛKG¥¶¶Lºôt?ÁcQRÃû_ÞR"¬®©îÅ7dÇÄ6ÉClTLE¾õ[òÚ¨ûW;¢ÌîõXêYe©ñ¦©îøáµ½Å®dH9 ¿¬ q^(ä£òRÔGrR©ëðø×?ÑW&EÝöö¬æOÒÃûÈ/ ¿ö§;èí'[ç(ÙnÉÖ~Þ%ÔXG¾A>[6¡±� QÀÏ 4Õ¼³½ÔÕ\_àô¹\îrúÉ4KlR[õoî(°-9m³?A !RP¼¤ß¼ùEó±×já¯h�¥&KMÕÓ >³^gqÅø?~á$9½úÜW++wflÎª¸·ß¬-æèÌÙý°TpDgÊ¢¸¯2/Kd'zs~\|×Ã\_íNïH¦bð$4Pý\I¤1IÞOS4ÌK3ßRyÕñKË ì¦wZ¤ïÛûyµ³?G5~Ñd>T®ÿíÿÕçJ!c´øBÚòEB©ñ0buÁvªy«¶ÔÎsúlEÅñ¿+¢òà-i©c¶üú_ ºDø'P��Xê)UcRù4j_Ð>\ÏëZ¥ùÿ¶ê×ð]»¨ ÛÅ~äò.è§céa]\î×ºa©°Ôê/jHàp¥'%JRçûw?ÒHÌ(¤^5[j qÅ7¯%ÈÊ%ûÇÛx0×éÞ_©°Ô8ª² zü¥«øDÔlI\÷±/ST3[sö´Èÿøf¯r~åß5:ÃRÁt³Ô¾k?Úítw^»n{z %×NÕ(¡PBåóEã\|u¾j~êi©]ôÚö¼öËéî7iHÌ¶2$·h·åzÆ¤Ü¹WQ3OÄb©?ª#=Îc99H#ô/_,FÞÝçÓÞüOª6ZÚYè·\øWT¯R8£Z±15Çnr�ÀRO¥ò¥¾RªòkÚÚðó¿]]À®ìsö.Ç_»o¥Ú%KÉO¿¥R8;¬ñc+,jè<5Ñ×¾?m©ß½ûñ: YG·Ôóòï\|$¹Ï²?Õåã©3 ¿ÀRa©'°DÉßxÕlG>Ï&(¤[÷ ð%føõÐsP²LM\@Ö]\_5k\]qÝ·îyñ@KÓÐRÙn¥ú·\_¶g)Ù%õeSc\_Óäóµ1$ÁfZ\L-Õü>ºMW©Å(öåÏy÷òoÑ ;ó$ûIïßwsCº/¤Ô¢@,p :£ì ÃCôØo^PÔyÕ§k¾ö]jéfÝO¼òh¥qKmPÙÏSÄôèý�Xêé²T[jru05Aj?Yjãc©"©ðãm<þÝù®äÚ×üõ·Ò«¦ÒL\ K=Foa=$LíÜeZjÎwï~ªÉ×tå,;/ùk))¸ôoY¼r fª¥R4>°×dZ\\×Yk\|L®½Ç~¡ª¤\¤FzÌ>x3ÄF; ðEm¢ em/rìÕýw»òãÖª\_ z~r'íFªÔfÐZ�Xêô°T>(jûß}v¾$Ð"·®SËbÅÿå×ÚyTîlî}MÝfÖ9BÁRa©'Ò[Ü2Ã{sÌ /î¾óV±ôQ,Un¾åF¯B/9ü¿ [ÇÄ6í÷+X\,õD,õµgnËONI-øß;\®ç°HX?jyxC1ÅÅ%¬3çDK=»,UóS¬¦÷ÓQ"Ù¬m%¦W)dn¯¦\|RUvf0øÏmS¿nþ§s\ZrçÑÃ/STeçÒû©éµùyF®G±¹Ç=%us?DËÖSgù«Öy)2>Õ04UÖU,A£×zTÓRef©ôàn¯#whÎUT5JÃA ù¹ ³ Q4Äg¼FóñÆî°Ë éÔ,1X�°ÔSwääÄR>)5zóþÊËìRNye÷úTóSåb;Éûïe'¦0ÀRÁL·TñËBæØpg¥Ç"Y-<þ»7.;8nëèIÁÞleeåK/½tÝg¯Î53ò.þzÓ¸(cÊBo ÎK5¢¯nY/In~¥Fº÷åNGäà ¥R\|'êÌëb=bM5k ³;¦ñàEÕê a©g¥ÆÔµ÷à?ÜçüÙ¹M³¯¦ª0G=yª kÊ\ óib \_8ÁnØN¯>Ò]àVìÞÃ®tÿüäaN¶ PÅÏæXÆWvpáén¢MQÕ!j¢¡(Ë}´dÄ¬ñË_G³ÔÞ<"¹¼èÙ7èÀ{T·«¨©êª¨µTWCMûyÛ÷.uôQ}? ÇgH >�°Ôi!ªæü+ÓhËE\|ÁuIÊùØ_9jrN¾'_÷%^1dË ¯¨ ¦êX-ª'Å:kÜÉm¬NG2ÑÆüÝn·Ú²¬vö+çå_þIôÃ¼ÌÚÈøg¥_´¢Y¯×[ôBu ½çëS,UO[UC¦°óAq÷Xn RGÍÞJeÎ«nÁç©N%® z¼¦È\|äÓGz7í\|¢£ W³ :/¤.'¸ª³® Óà®éÜü·%äÉ6r UG~Ð3ìÎï)ónAÉÎó/þ×ÿ£WwÓ°ÉÑÕ¼�·ÔøXª.Ùü.oÇ[?«tgQöî%{çÔ45Ì©+unkµ%¥ûç^´õ«éa�XêéTÍ\°¯ÐIÞùØz/\_ÅÞ³î¦2,jñÎ¤tÜrýwÝ<êæ¿]óô{2e\¥³Re2b¦¥N²lÅeµZ-KsS3õ)Ëá¾K_¯OêþõJ1<<Ìn«««7ÝygUJÖÆÎñHÎ²·Q]Æ©àl°Ô¯+$'ë,¼çt ÅÍº{ÅRYä/ÇùØi\;GÕõtETX\8íÊùó-5æ§Ö=;çg÷»mvoãù£fÙôDRWaÒÔÿ»~ÒºhïsíE^EÊµKß½\|¡Ì®è2)£¢È�ùÈ7JímôÊ6ýÖõ/Ì)¨+óØ<²äQ$e8Ï«?ÿK4PØPôéÃÑïýES®\_ÊÞ}½VGïVÑþýT]KUµt .ÙØýÚtð�ÕÔRM=ÕuÐ,ê/�©§áP,¦%xY[ñÃIÊµ[ÅJ©^ÉáÄ(û7A$¯CÊw8]YVû£ðÊï/îK-q7Ý©°Ô²3¾G$-ÕjËlÿ¸ñÙ¦ä%ê$ ¿Ç ÿ·®ý¸ËÌýr?ùÕïÇfDd K¥À! úæõ9æ¾ïÈÛ¼øU{ÐÍ\'mÁÓIáÛ±èjÀLÚ¥ée©q? T·\|á#£¹ÙWPcÉiRGú© 6?E»h÷ Í.{DÊî°Í¥ [(Ê+ÄS"L²L1b2iÁU ëZQ¾SB¥qPí{´þÖÆ¼I KeyóèÙ\|Új1Jr¾é&¯³ÇZPµèóÔà¥Ø3ÇUÑtÑÄ}v"ãbËó>7×{Ï ÀRÿÌ3Nê\|°á#¥ya£¢øKrñaU/$C²YÌIv¼NdµZ%kTöÝC TmêÓ?Ra©'ô¦¦¥æüðö-DSGc±A UEúZò%³"ÅSº°Ã¯Æ¦ÿØ,z"Ú²ó÷e¦¥z<ßùwE\|,KeF ñ2Q÷¶lÁúÑ(Í¢°Ôén©w¸ô#Ô?&ê\|ý1KUü:XóµË;¶¼Åû1zµÍ,:©ËR sÙìH¸±Ýí·y æÒ÷RÀ§Æ$È7NC~ê&9àëÓÈSO)ñÇ©³ZÇ>ui ßµY»³Ëw]þ-ê' k\|^ªÞD-m/Îvï+¿WvÏèfy3E45=aß"}NOy:ö#æ1(ùmmêÌz£Ñ�zBïHÓXhðÀ+±@:/Ë%eYÒóë,V®¥SîÄ=»MJm1kÙ}/ÔT#ó¨ed¬ù<}N°TXê a©íUÌRÝN«]ò/·?ÆÞß %EZb�ia-ýÁß×,7ë»Æ§ÿØ,z{Iú«.+\>}Û>çQ³<²]ÚRcZôEä¯ør·Å.yrÏý×wtôÏZ-°Ôén©q)¯µlª¥N¾F¾£óõRû÷ÐCmÙNÅV4{ýj ¿æ"Æ'ªMHåOËkøF):JëßY8{¼0g$;wWñ"ªl!Öq~zë¹\_ÿü óÎyæ+ÉeýÆÐT?ÏÏ ÆãyúùêÜÁªíð<·?;k,öÛóÿÚD'4hÖXr¨Ð3h/}·èªIPP¸uúôi¬õ%ÈPÄB9)Å°Ô åZB\|´qE5¿¤ã¤úIau±üPòcSdÀRmì@7ðO\_üihV/s±ïàá¡±ÁÑ±Ññ±¾¾¾¡Ñá¡Ñî¾ÞÁ¡®áºç¶Þ[èöä¸J¤¬Ù]Å7 d¬x7%ka©°Ô·Ô!±^\¯÷/w>Ù19ã×H!òRÃç-ç9 =;ûµºÖñé?6K¥ÀQ5L#~üé¼"%O²Î{jÇ!!&®j¦ÎÑØë´îöäÒl©è#{Çx¥X\øÀ,5À-5Ä-5·eö¥48v3Eä\)Ó«Ï ¥ÓdPÓÛm³ódIÙë/§VË F-!s!e¢l¯hZtúÓstØ¢ÙÒwïåK 2OÁôÈëþußBkOaveÉ9ôV+ÛP4È<¯¸=¡xHQÍê®;0ÏîË·³è³T#ÞS4LCêýËÜÎ¹×QS¢¼LPYÔq2{jF3ßóamòafG>;á§Ã´¥òO%å9!üÇW�RCµUïÖ¨¬:p5&ª8F4QKgÊPÀR4úª $)zs.øË~-/yäÔäÏDP5}HvW/Ê:%{ñËU=Ã:iGZjú K¥ÎKUê¶óãÇ±Tßðÿ}-÷/¾lä¶>¥³Rõ(UmÝXÄ+0ï,ÿáÏ Æ(læýê)7¦bI Þð³%×!¶y/ûÎÿç#X\øÀ,Õ?2rÕ¥ÌRc·iÎe4è§TÖ®'mªJRN3UùøZ%ú?y Øê¤þÜyÿ5uLNa¾S'¦X\¿UhüuhZXÊ¶F\R}v.Ýö4õ2yR¬½¹©\¤¡üY\_n NBt¦:ìOrK(Í6L#U-×ÛcíÅ;ç}dS¨SÛ´ª¥(gÈ]ÊWÉiNP[j´8÷P&7g;¨Ráã³ÞRÓ~j~&1¶7ÄUÒCÔ]Þ·avaÛå²ÚV§] eÙ%kÞßþý¿î¬j#ÔÙW¦O§A�K==¯2£úÍMÜ9ùþa$uE(f©\|V!o[ú\_y\|m UrÕ÷cC¹/FæríÓ K=¡7Å¯"+\yþã7öM¢3,\|»j¶åMmÒ¬ìì ÛG#ÌKg¥¢ÒJ ÿ\|¯;WÜ¢2üÿÞúè0¢f©ü4Û¿k{ÛÐæàðÌzboóRÁa©q©´yÎ45ø %k\| UOÛ¸é\|©%³D³TªzöÀy±\Éos´äÎ£Û#Fa]ñËj\OÇMIS¡öªÈß\|ª×.ÅmÛ[½ðCÔ?\|I:ðì2É# HRSÙ"zt+D=%Íàa¯¤%³) á>Úõlï\WÌ%õ¸ó·\_ò jO%õ»éî5ÞüQÇì= ¿È×© Oþøþi¤ÍQ¸«"K9iÖZÆÛÖÏÖ¼\_ó°Hõ&ÁÛnþ÷lä´SÃs©l©ésYR[²HYÅøæ¿´Eùx8�z4TRÃ¤ù¯çñòã¶yÏ©ø2Rm¢ÚEêAÑáo%{[óñCb¦ÅÄ¤¦,:CNIºi©U$Î?®ÿ¹ý±äL£#,Õµxya±ÄWæ!zÙÅW\|Û'STD°T0Ã-Ä,-V½í©rI5Ä\Rîüîx[ªY¾@}(ºû¬fYùlvFò\zÍ×ú £Æ/ø -Ò»K?Jí£)äU\|#~ )à§@Ì7dòGùà¦^F^¡ÑÕ?íç¸¥ñ,G[éy¾ÿ^I#qQ(F(Å¢ü6¦¾AªÜ¿g^A;3b§;,jÌ\_HÏ¿ÁcòTý4ÖÐû½ë:ÝÑ<Ïx^NÃÜrzh3 ðuj1~ËºMz¨ñ@Ëlo8Ç¶Ù[òçÒïÞ ¿Åéì¤»Wuysöó®¦ú E(Ä^F7·p¢dcï='k×ªI9=Ë«×ê)é×Ìï:>B¾ö KÜbÐG²:Í¬hÜR-é:/¹^É+I^É{þ¡1òéGàl·TC&c´mÏ d·)¾òËÿ¢QÌsúù!×ü©Æº¿yÑ,Ùû,,^ûbHý4n>çni<9DÕ§>,:/ëðá!e «Êck,NKÉü»\|ág:ÅÖ[yæ 6üå'ÿ!¦£ÚøÆvÉZþFÍ�Ì§ÿÙ K=¡÷0%º~þÍkøà(Òl;ßf}Ïo¶vÇÌøVÇêv~áKêãë ^ð^{(rX\8íª&ÕAXªXÆ1l-¢5é®{éÞ;èÞ t×´ñ·t×Î}´ñ~ºó~º÷! ÉçQdn.c-í×]ÔæB.{ÈUÐ\_tîÞEôÛMÔpÕRk-5Õò5Iÿçß\ =ãE¹rvvPr VÜGý ^ýÚìFr/µxwvy¯Í1îBù®¶\WíÜ9ü¼²ª¨¡öUÓC¿Ó¿ù¥®Sãd±Í«YôiþTfy-JáFºoqG-fs äÍ¥o¡õ·Ó½÷ÐwÐÆ;hhw°ûw&ÛoîJ¶»¡'÷ÑïÙ1ÙÌ3DE%í¬MúÕ\_vÏvñKlYÉ1Ó^±}ÇÞBª&d-®¯ÞýwßúrúH'YÊæ\|ôoúB%�K=ýDIé]üÏ\_²&¯öÌh[°Jõ¨=qyø½Çîô&/ Íó^9¢ò¤lÖ+ç32t5>/ K!\r{xã@g'}EÔêì9YV%~~ÊJnÁþáq³Ý»fóÓCÄ{ÅLø(©°Ôì1¼h/Åû.]ê5V©XòlR~~N±ÓÎNÉbçÝF\ÏqHîmUA^03!YgB:!,u:YªÕd&XÊR5Ig¹}Åó[ J[K¼m%¹}Þ¢~oQO·³° ³VwN\_Áwæ\_8þøË\|x,¢QçPÌGÝ5ÍüPkIî ËÊÊñÙsÜÎÚÒ¼½³rY«\Ím,põd[F=6¿mSTã.¡¿º#4\ ÈN,w&/=û\Ýyt8íR@{W¡·yöìúYÌo)Óãõ]¸]RÝn\_á¼}¥§½ýP9 Z"õÆý7\æÛIÖA{vÓÑÛZ\Ô^TÔ\_ÄÞZWAQgaQgA kùÅù¥¬U^øúÕ?æ+ÚúLR515~ö¦ýê\/N®Ö\_ÿn8Å×Ç°å?ýFå¸JA±Ì¬$ñÄÏ=VòsÜRSrÎ¹þöGl\_°Ô3¶oEýU¤¬l©à£a3.vQ'ãzQbÃÆXÛ¢¬l«äqäKÎÒ§+ëÃæì«3èðlÎÅlA§5{ö{]Km>~úI¤f¿\Q6©×:")îtFv9;ÛgµÝ°:T#nsÄX³ÛXÚ¬lÖçÌ©ÿëëi"VTÓR^]X½jÄÉ 6íÎóø5jGîæ~JG®¢Éj(@òØ ÜÉ+ÿ»RV®7O>^¸\|yÕFªÚUjÛ]°TXêtë[rË®ç½æ:¨YïüÛ\RPÕc[ªbèQqí(´féñ¢¥¬,wîOWþjë+É³>\B(ê´¹K¥°¥\}õî,Énåk {ìN16ýÓëõÎ;çºÏþ×nÛSYÏGÍã¼/;aÈBý¦Xj"ÁPaÛôÐãÈÿðý%Åm Ôæå¿çòÔåv/\TÛR6ÏÿÙ¯Ð#/QW©Üy³x"S¬kø_?~ùþòùsjK÷eg7ÍU[TT_XÔ_Ô-j,Àl ùùfÛQ~NÿÍ\U´T^\ðlKeo8Î?Õ8ÅÛ~úÅf¯½ä?ýFT%WÃ©éÉø¼é ËÅe;·ä\ÚüìkAó"VÐM¯,,³;ñH ®þÀÞý{676nëêöGbé¢rÌ#BJPûBÉò¡ÀÓáýµUÕÝ~Y?ÒJq¥áqÌ[j½'å2é²ãZzþÆäÇNú-,:,UO®¬N$ÇÚÜ¬ 2Ê, zñJSkª©yßñùÆÞ}÷jÁÐè©©ùn¢Í ¡¥N»½7i©JB~òÒø('Ûø0o££¢;æÏGÇÉ¤8QLYq¨FùSé\|ÔÙAõ TYlª¨³)(STOeùGG ¾ÒY"Ì@ä÷QíÁÆ¶õ0õôQÿ0/èU' ¦kjJ)ÈÞHªe´ÑtÚØ» '>=ãdwv¤&ß³Sÿþy_pKräI×û&¢j®ü¼jàÑñç\|§[]ßøûzvWåXú± Å£Q^Ù²õÓý¨Ã¬)Ço,Î,aùßüÝ®=ujºÒrêbàÿÏÞyIU^}üÎìôÝíÖh4ÆÞ[Ô¿±kÄ D°+Æì½T^¶°Ëî²½M¯wn=ß[îÌÎ6DDØóî³ÌÎÎ½3ÜyËù½ï)÷\9\_\ºeóª;M<µ°Yò,}¹dsBFJEJEõ³º[íRênÎÅZÒ¹Ê0ÎÕQRR÷ØæçnT\ÔcÊ1ØÃ;°\_¤ÔþJ©?ÜJ©öl$gÒ÷DÚ·XÈ!'¥1½©§Sö ô £$$PÞ}Pj&cÉþ<¡©§Ìÿ©B:]ÃPyÎÑæzjÑD"ËZ!æ V¶ uffr:åÁfåc$Èi¢Õ½¬BfÞç^²zÉo"úÏë/ÎWNæÝ&.Lf¶ &òtÞSóßvSudU¤TR\RêAgã ¥¢R{¢Tæ'¨1àt=]H©¨R»°\]¤Ñ?vUÔ÷àJu;¬BFæ£ï.iê¥>/¡ÓU9ÕX¬ÓôXTK Þäh¬ÿY©ÓEAÔfµg &ÇÈøÏMûç?î9ë§Çf[y¹vu Ù¹Kê[$HUl O¼îjµZ 2ì mÉÏ¬Ü×xc"RR ); Rj¥H©aûGJíR]áGêO©_RQ,¥v¡®b+èç$®Æ@ ÝóûsJ(7Z2²ÝêÑ\|lmZï]\|dãEÙ©4>;k/Ýe-8éÒ¶´«4o£¬Ð=Ö~y\©ËÅwH³ ï}á#s)9YÝ÷Û3 ¦¥#È3êÄV.[µþÒ?\Ùµ7Jo«'(¤Ôþõ_ív ¥ÄJ$é÷³äÁ.li¤Ô´CaùND H©ÃLAn,Ëü^ñD$¤Ô¨¦ÿ¶ÇÚËB}ÿQo>Ø]î½æ\|7mÏSåÓ#=QË¡Ä¶P¹I2Ï yÆ³0§Ê¹µd îa¼àIlAn²¡ÉØgÉe÷MSÓB£ÓBÿ¸ôa¬´$Ñ¨3.¯MP¿r2àzUEîa4EJEJEJEJ=pæ»·ÞzkÉ%mmm@3ÑÉ©»×ÝGJíB©RêA#¾ÓØØøñÇöL¿iü¯ÑÏåRüäÊQ5¨ ÅOV°&¡BÒ¨ 8ùÑ! mQZtåÒÕQe¿üÒ\|pZ4êü«Éjôtõmy£OkSYi�]ÚeýTRêL©©é^ëSîûUH©H©iJù.Î?Pê&òúôé¯½úéñx))®'ò2iÒ$ÒYzê© ¶µµõ8¶ðª²Í¤T¤Ô~>¦§YYFÅ±Î°©ï ¥þ íWvô»ÎÓR»ÞÁ×\PÈsé²#$±Z=Z²öVÚ]å©ã(?iQÉ1yI¢,¸TU6�ENDÉKÍMúÚ%\_~öü'nøýM¬DkØ9uñµ ãPJ}àÊs(/;õÄÔ[«0+UöPH©ûRS+Õ=P\¶ÆÝ£Tb([pZÝÙäÅ\_Ø^~ùåµk×B¡Ô}#J 3ÒY¸ð=4ô¶ªÓÉQù=Rù÷VFñ TBuuu¤\_LNÓ4}úé§¤ðÎG2øPÝ°aÃwF°ö¶Fª¬¬V.Rê�$¬ óÅdu\_ZRrØú®r·Y/Ù}ôô»Úwà@b\|0z°©§sEû]ÓsÚ¶Þéî3\_å~6Ð¢cS®¿h À\fæ´ á¦Û¥rs@çiAUE ¡¶( ÖºT¦ %dk[í}ô²ÏÍu92-´¤Õ=uZ-ä§M%Od>çJ1é6 Zð¾ßUD>¥dÔÅ7JüVõ´dêißrRê¾^åK§T@JÝ3JÝ¼y31yÅ1´·Ó oræÌÓ¥Ã{ìØ±³fÍzûí·1H$ÁEEÅºuëø^jw¯¤ÔRIiooÇ.Ó§OÜzè¡©S§þïÿ#£ w�nhh(++#ý».ôo\_^(UEJEJíçªjØÖ¦P0°ð»a¾7¥ê'\©×÷î#ÜàLI=PªÞûÑ3 ¢ööéò3Þ6ÄÌZ\O~^VÏ@Ó¿´{Û1¦#6l\Ñ¶s3ù[L©4½ÍÅÿ>\|WN¦±émá©ÉbsedÐ7q y¼vªeÀ¨3~Ö@VªÑ;ùKrè+ÜùØô3Ð½S½SÉ/Rê~£ÔÎc6háÐò}(<¨®®®C1Õ0555Í?¿7J4i±ºYN~_~øá÷Þ{oáäfø¿1¥r§nÒ_PIÕ××Î²\|ùrÒz£TÒGÆÇ» é;Ï=÷ÜçN: ¹¥¢(öû<~ö½¤Ô~Í§Nj²jlUÉöRi¶U-P~øûvï_©Ed;¢5LøþÖñÙö$zk×è[Ýk¾ý¨ !AI»ûì;gTR÷Ý ïtJU5ðÔq Næ\|áµlÅ�yç3Snq6e4eoFféðQ7ÿéÎæ½²nÍzIã¯¹!O° ¡cÎ¾2Fwt%¼ &]{q>ÛK½}ö«Q¤T¤Ô~d vR+++7lØ0mÚ´ñãÇOBMÄ!³ç.4aòbb~S¦L!¯¿÷Þ{þè£¾óÎ;---rïwQªö=)µ[°éZUU5þ\|ÒY& "Ý´\|Ò î¿ÿ~Ò#R])¨ü1÷&ÉÝ#ÉOrÊÜ¹s ËD"ªS Rq>Q1]ÓÓý¢Ý\Ý ªõøLú®\ª¦A¥~~H©ÄXµjÕC=4®w?ÈDçLcÇídïºë.òÐ=ß!"ÈOþ'r:aÕuëÖõ§Ù'm3»ó¦Þskßå$¾yD¦¬@8Ôsï~us ÍãP_H§0¾pqÐRj:öØ_H·J$äþë}÷ÝG?üðÃ}ôQ0ì§ÍÅaI;EqÏ¸5 pÀD ¥«.)k_§9n(Èj".qs94^ñôC¼¨q¥°]0N'"'È#YMv<ÖÉõWTåá¤0D$Æ/òüû6RòÿÐTtËK¿qE@ò=pÃåt/U°ÎA¿¹kJ@;S §&´@ãIr QÚÉ)Bæä?òÓç½^aµ BU°e=ðèÜVZÀW-ã:+Q´Îq[vø¿U ª¡I×ÿ¢Roé¾fìçj{©H©}fQGÖ5QNDåD\|9 «Éó@&éÓ§÷H©]B¹ Ú¯Y³ômÛ¶þÒQûÌýÜ¸RÒe$EN°\Ktôù¡m÷ÈýAÈu¹S¶ô×)OE5U=ÝñÆÓ÷dÐdÁ&ÁÿåjÕTùÎ¼<ÕÇ Îvf»CÈx/oÑSÑtÀÄë.¤QpÇìù4÷¯®ÙQH©ûRÓÆB¤¤}¶¬ÿêúk/uqvÁ"³.¾ìº¬á èT#°H±ÁF1VÔ ·o»óÎ;yn1¶ö#~êÉ¯¿þz !Æ4Õ0?è ¢®ÉUzî/¿üÕ¯~e¶ØÈa±Ó+}õÕK¿].ÉºÚ¤»óh¾_)§á4ÝÏí¿ÿýï."XgÏýÞ{ï¥êø©¨¨ ¿655¥P·oZgL¤Ñj"¨rS]í¸qã2¨çbI N=ýôW\_;aÎ]tµºwÙOiKcß\|þá=òìssëEsóüÜy¾h\Ô /W4ëñÛÚÚJë §ßr7zÛK%"kîÜ¹N=Rø¹¥©¦oSªÖÙzç¥+=^½ôkÒeL.ûÇM^UvRÓ÷ª´¨.ÚZÇo·ÚÈa±9Èqæg¾üò«j?Ìz P\M¢Ú´ù×£>hqþeCF/u <þ\_Êª>2P²H§ÔÝÊF8&U¡!övXSþñQç/ý«7N¸IJU¨sH)UÑSfªy å««î¹ï£á,²¸täÇ³x àhlêJßG!ñ¾ð\_íÙ'µggÇü1Ìy\|§>3/ôÜër 2x:Ü¶A BpçqÅ,«µ V 1Ä-#úÅ_17BÓJxV,üè°¢¬<IÈ° BÎ°S®jº¼A× îã÷gSÊ´ÔuùÉÛý $-øÔ´níQÅ¹ô\Z¼ÅÐ#O¿ÆHÔEØø&]wn)µÓ3ïyx.s@×¹ýßRê>¦TV!XÔ¡PÍýfiÁÌä±-K°e ¦Ácþ¯ÖË3cÁS:R©÷lI 5ß}ëõÄR·Y¬ \|¦²¾úÅê?-j$ JXãôòb¼¹bÔBãÒiÅpÚø25»ty?êæz\_¡ÔmÛ¶s±Ï!Õþß\æ)K{ìØ±?üðË/¿Ll°v2«¦uÒGøÍ¬ít¯±Î£v$àeÙð< µ¿úøC©ÚÜÍ<ç;QV¶ûÕZD=éùÔS!:è(þÌôH Í9CÜÎd'JÉd2,&ÒIÈûÙ\_ûlI(5ôÉ¼µoú�mËñfxý©%5¥Î¦ÒÌ\»Ça fey óväæ ç\_TíÜÊµ.EHz\|É¨o©ÅÔ)Î·Ôcê Õ¸o<ÿNðI)&¢!$ ª\|#E q#üëï« ê j¶vwf(;ÏëÊ©s¹·ær8¼ô.´!¡;ÜÞÒÉe°uÍçCÔ (Ï±£'¹î¬Zwnm6ýuMÑ!or.TÓþR;ZF-bSGAö+V8§Yv»3¹¹C×ºùrwKDÆ59Ç9Ì6à'aºÝM¿[XYþæÖµr$²¬ÉñÛÿöú¼æ?5ghvk7\|04 0åáÕØHH?Fxüµg1ãüîÏòáQb»±ø!¥îË¡R5J£òloî¯»ldî@AÈ¦³¸Ýjq¦w 0ÈQx²_'á"² J¥¥ÌØxÛuçc=]<$ÆÉí0ÒQÛ!£àÏVk2èR8U¶DêÊd§xÖ.XÝ&»Ûv-nv Ylne6ÞÙLAJí{ÊoÈÜ¹sÓ÷^\|ñÅõë×¾À½=»tò\|EEE\_§T%ÝF©öÙÞMëÖml6GíTt]ÓBæ 2?ä BÉºm^cØ{^¦LRªdÔõV"àÝv¬UppÛÚ TYÌø×ùùºí¾Ô´ÚO ZM#Ez.©Y³f-X°ûç ÔdüJìäÚA \wîqY¼!;¬7=ùÚ"ÑH=ö¿ØÕö'±¸e#ÄÐDÓÓÿº\'##Í¾fD·SÞ¬%«k¼±þc}!¥ö{HUEjÞ¼:wÍ lo¾µÍ.Tæfî\º-/»&×Ñkj4 >«cóðñNié¨½µXÊ; %X ¢h[yð÷µ%B@üÈð©Ñ Tà¥jx�+ HD¡±róÇo),ûlÛ²³·(©\ZWÓè7;×$°º¿6Ç¡ï¼L¦¡êêEïp ù´SkÑïîú§cQe¥§52ôù?}ñ"¾ cw öì7>ú²ÙEàÛùÈ·ò%Á% ^ã7F5yã ïóN0nWIöÀc¬ªåÐ¾yÓ¯O=±Èd¢î ¦!GçÕ!ii ¥öuV%X:vìXªk×®õz½$'ÉÏî¡RËËËû.¥ÒvÎJwÔµÕoød% ¥3æ¼LgtÕ\_³í«\_>"SèRUzlÄ&ÅùÑzðéÒbïÔ6}UÊ nòã°CGuÖY^zégyaR\_pñ%g\_SúùMm¸ÚÅ\É2uêÔ3g¾óÎ;õõõ]n£ÔtVMÆt¤5N×aÚj®øÙhÅS¶ Ønzò N© ìN^PîÀöU ¡lÁùøËoSÿIUj¬Üô»sÏ&3¹~Ó1ôø�?Õ¨¿Ñëd ¥öW85sÒ£P½ý¼oN^SPQûû°£ªêëø çV ZmBkaÉáGÃF)Ð½}GûL<-P±n¸¸6OðÚ¯@C'ßMÉ I¿5þmÜV{ýe[ ìMskæ Åcàåw¡¢\|u°å«ª³· 3 n-òÕàa7é·¦S3/Ö¬?9m{#jq×9©tríÏ.Ù~Ê%å§]¸åôó7y~ÙYçn>óÌUg\_òÒI¿fjpÆtqìL©47Îs%Mkõ[ÖÝÿýQ9®Ò1ÑêÈÈpdgåýùÎ»·V7H©Aé Ä ÕêÊ·uÙ¥ØÎN¦UpZÍj÷¿YYUÿÎR3³U2úüN P"ÿ¸é<ê�f½ïág{R¼¢/~oH©ûR Z77ûã:¡Û ®QQjëW] ñÅ1Ë[têí¡í:GºD½MµÇueR\_G¿½:sJ\|\|S\| µ¿ýä¤bÞó¹ûV÷¼"WÖZ¿\0¶QÝ'^tmc�br×)9våi' ´:¦BÁ>äÕ¥Û×BJíÓZÁ0½ºDo©~8¥nfê£Êé:nÒ£ÃúÝn¾[cw¿¶%1°iR=Ðvþå?'r¦½@0Îýt·ÃW¸gJÕèÈ<Å¥¯=YJ7L çz¼A~7:åJ.qæð §ÆCJíw¦¦ùýþÊÊÊÃuy w?¤ÔÔ¤$ýÈh5êÛ¼ú¨ÁÇR0!ít³Æ5#Krï\a¥K%Iì¥òòí\_~¹°¾®qéÊU;[Ú¢1 :Hs)¤¦} af $Ê6Ü°iõºÍk×n^Ïì Ý¢)º"éä'}>Ä·Sõ(hQÊUÃ-s9¾¿úRê÷ê»c9÷Jô!ôVÍ·¹ÜÌv¸.úÓ"-¥¡öi¡ºVShÊ4ÙKé1ÞZÅóò+Z/JZUûå¹ûþ4mÐ>úÔkÝU-iúË»vCJí«:»î>]úQUUUìI}ïÕÑRXªtÅ?mì?¨EÈ6=@[¬D3[Ç=S9E¬_ðöÓBvñ[vPY}:7ÎÝ9öÒð=KáÈKÿ±èJhAðmýûÇÑÝwîÑù'Ã»B¡?äb±y®}{ÅQý³wzPS=b Týêô,2)kàåÜ¿O\|&êg&Üæ¦á!yBÎè!6¿HbªÐe 8+1§A4~í)ÇåÓÅ¤üs®ºCégT6Û±Nj_U¤Ô½Ò}ú%¥êÉ& C6ÜÒ^n6ÓµJ%CyñÉµà, ¥yÁÏìéÝq HÐåÿðû¯Owf^:oÕÔ³ÙAM&cHD çþóºBy,{ð¶úÞáîúq(5Òïýgk¾vØ«sGÀ µSc¦äiùS+³I¨Ì+/WR¸#æ¤&Ú;¥9ls®ÝïÌöf8ë³2Û ÜQ YÝaçÀ#§õAfÝÅýPµic¡%j'\»ì7à;s OÈª¤DÃ.¿pÙàa+^ J³HóÀe2Ûòü¡ðúRÚW¥Î=4µË³�R³ÛmJM+õañúéÙÚí«rLn¥Ó!1£J2!ý&ëÒDÓ¯U"H\ÔWWÙSBDnMQhÕ©Î÷R©3Í« H©ûN2·aõ@ÜW}xÛA³téM µ£ï$óþË þóÐÝtåÙdÅá ô÷Kce·¢-mË7\|ûògøéæ-»ÚQ\ZýOçqÓPð¼C.ûG9ßK¥{¼a46l\_ûõ¼÷Ù"5fWM®5Ü¿yh[ ÏÈùÛÜ÷ªs¢+ûËÃ-RËËË-\Õ:à)Uæm2¶}ÜïNdþ.aÀ);é%ÝßéLÐ¨µá¥¶rÍgNApYrç×lIèÿ\|O{©ÜêVA ÉmuGfÿÏÇæÓ³RK¬L¦NX\)ôSÞÖÇên ¥Ôªó9(J(µ¦~-O'¬«hÖSÿpÛéC LdJ2Üø¤A©»÷ÿÑJÌVwÿö<Á)i ÄWaU¾x(×}ýN© d³tño.Ûæ1bF±OÞ?¤Ô~ "lÏ\Ò�ÿ¾£Dð ¦Üa°#¨c2]õgÛañV(ûpu¡ f\mNx\|>Ý%ðÆqäÉ<;=½/´tà!^W×9bkþÞ;ÆBÅ¸øÔË¶ñt±ÚÑÎ£¨¯ß¨×j6o.(?HÇ´6n ºO@8 á0%YC¬>üï!d²Ö;)!Å5HH´<'AF9ªcnÉumëbz²µ0Ó[Õôx¨d×kÏbF\\oé»¨iß³ 6BU¨r\|\|ÏT¥®Oé×ÆÒoñ'uÊ(¢jH©H©ûR5PBõëiVjkà8ç½Å©fHzG<7mÔHËöÅÄ>¶9=Ã½ps£/½ ô¼:¤ÇuÄt]5\|±RifÈ ný6fúËÛ+iê¬$- Ô^]óEÚa$§Ï)¦-\|íñÞÉ@%Ó+õVº5ÒSjò½ÊìÖéÉ Ào )uÖ´å2º³#3 î½¿b£bÄ\_S·ûG·n\_c8Íüß~ ¢ïÒÑ8Éµ3I×ãZ<"±¥¦¶Í\_.Î£nY®\|ÁUøá¶º&cøã^¬$ù gZqX°:5¥ï¾iaÙ¹©qãºÕ¯z¥F¬7Rê¢H$²©Lv}ÙêÖYw{ÅÙÅfÊ¶¬f¿Y Fêé´Á]¥Ti;aHsHÌ»uüÓÉþ¦÷\´7+¹iÛWÍÓfÌ9tS°X»I°d 6+yìòæu�ZøLU5MJ[ÕÑã÷\ºgOêó µu7þñúJæ×Î¢9<;/È1ÓYÉé¾éÉ7¾Ï^¹ _ Ôç¦) l¥·Íú_Ì¨¤Àwd£TC"íOs y¤?e]=þ¥z£ RêÇ TæÔéo q¼ärúR¸o¤_iöJ²à ¥Ö/ÜqÑHO®%;hõQÓê-·tÀnº¨\|øÇ{á,üEúj}çr"ÖÑþß d=þÁø×ÀÌ;Ûr M×ül¦íðüsí§»sÄè²¼ÕÅ+ O¼Um×õpÜè"ÄBÓ#úÿ]°XÅ'ØaÒäù¹ëó³×dZ7å»6ågÕ ¼àÿàË%4 Oäg/í#Þ~T? Ôd ¥êíß¾J,]J©ÎVløÜ§1Bì ÔÄ5ý³®§š‡okÊw"†dšñxsõöÊ¥‹ÜûÇ+rM4[“Î²†\_\|ç„†™2Ï¬%Zô¤•.†êwlªÜ²aÁ»ïœ{ìQ.# ¤C0•¾öÉú(Ã>0+i{ H©H©ûÊðÖ@ŽAãÖc²º d¶\_ÿØ«•�J1Yã ´ž}øºåšuø˜³®åÛC½Q\\_¥±îƒGï%vû¡‡ 6±’gfÁdgÛ¶VÚuYùïÌ‚¹øÔ+nnVxyn•g?àN;H©H©}Õè=½‚"I\_ý%ù¶<º—šsëÓoó¿»9 J¶l’iœ‚½ô¶ÙsyŒ–H-\|æ¼¨¿\¿ Õã¶>ëÎH©¨E©ŠK"øšv ¨Ys}a¾Óo¯Ìü\_ÙP¨$ã@íÐüÍº Fl7 Á¬A+‡že2µ´JU;oxuHÒ (ƒ\_JºÈ¨òž92dæQ¾o«kÃã³)c¶B\|ËÆ F79íB üßdø¼qEáÐòü\|OvqÄ•ïwÙÛìŽfkÞÆÌÁŸv,,ZIûÝCÐë.-uCÅlY¢ó;\íÙÙž¼l95~V¬Àµ™Áštp(Ìy‘¾{HãnË¡PH©»g#ºnÙOv§Ô®îdZ¢¥\|µ±—jÏ¼sÎ;µ©9ý»Öá(ÑZí¢wŸ0[“ÃFÕÊ’Ì¸†üöÇš“ˆšÐô„.iééÊèÇŒÔ~ë™?¥õ‰iÎK†‰î 6ó¼¯h‰Æ¶ÆL êûŸRÉ—[VV–H$ÐŠ8à)µÃÓF…j×g,£—é¦Ç^«d+'J%ÝFÑzÛGSË–ÀÒoá‘YëKÜ-¹ö˜ÅÌ-YsØÏaS=(²ªh}§¯ß^1 Ç›íl+Îß˜—¾X¼ªš¡ºjwÀ{¯n\Ì®·!A úméø¶BJŸí\_(ª/Q\PO¨ØÖòo^aåbœ„R¿,«è•RÅö‘E«…ÖI¿ýé·vEÒ{3ÝÉ(W Á)5ÑŒ¾bõ'ÏÿûÊ\_ýö¹9/<:müaƒ²ˆaËb»ªö¢¿?üb¾‡<·B‹§'²‹‡&Ý\|ãØ»ÿùÒü×þ5ööB‡ÀKÜÐYÅï/ªP8 'Bûk×)õà¦T"±©fi–Ù4Ô\|óã¯W½t£\ÛÑÔuòœïÎ+. ÈiA(UÒ’^\=Q\m×šèiØÉø×b²Œ:þaîßN³ÃKÔ�ƒÒòú÷ä‘ãÙBÖ€\ŸÈÞ7Áü~û\;DJ=È)ºN4¼ ‚nPªÛ)õO¾Åë¥îÞU}ÞÊU6» Ø„Rï~êe±#‹f’R „=œþ›“¨¾)÷Ð ïÝÉÖ”:ê#¥¢~JÝ>¸H\ëðëR”ªð…Ä¥ú–Ác·UÛ…Pæ uÃ› c‹Qª² Jåí\êH›€öÊð£!"ê9þ:þŽ(µ}ÙÎËF·ä J†YÍ(Úé<ôãCÏ„ðø!€X\|~ØQ¶xtIs¡=^[™3&<aBöø›?œÿÙOÆ,4pÉ±ÇÀ£O@k˜~Ôé\_2y ø«®ÝqîOZ¬äUæ\ñwÒp@JE¡Rwgø €^³à•Vn\ØóvA©žë ÌF\êÏ¼ÝÎó"é½åÔUX:òx‡«a,H�9˜L¹åÉÇ0³kfØÜ‚cÀg›êé8ŽQDUy²R%gžÅjŒZ/ÆØëƒù95¯9+ËH&“%äSQÙ¥¤T¤Ôý@©šØ¼õë\Ap[hö¤ÔyL32ñ©ª95O4Ÿ3jP¡2vo$ævÐ#¥rß�r YÉj0ào©®©oŒRgû(µ{Ô8í'\«�ßº³wRk–5ø‹µœRyEc¤T¤ÔO©ÌVO$ µöì :¼Û\ßR_mÙ×l^p J©Ñnª&±ï¦«Gg²d Ù¹'ÜØ¢T)µ×)eÎð7¬ÍuHæ¬PÆhÿ7sf¤ì©k<~C«ënftfÁF)çël=Ðô.¾ê½Èî~zá¦Vb ÿ¥¢AGcÚ,Ï³3MØ2$hÆ4¯ù5B¼úç#çRJÍ·»¬kM¨ÆÀÇbó!¢³áR%¨z4ìuJ ;Û¾),s 8ë×ªA©èñº{5·cÐºå\|f÷Zuý^ÛKMpÁp#Tè.¨ê9}ä æ²^pé'³gPj²öYZÙ2Ö±Gd× ÉbTZõÕV±fB³g Îôå$¥&X^Ó¾e ¥"¥î¥¦Ç¥îN½TÒ¦´hýÑ%y6!C°ÞõÔë$$Íz>ÃwÙðËávÚ[]Å¿¸ãé¥j}9h)µJëðø 44}XéDÏ»<Î}:-$ÜðBóÒ¿<ÚWèôç[6æW°] ¹ÒÉ÷R;l-=ý4¹#A©-çò¸z¡TÙÁ²-WßcX¬;FÀµ Ê4äJîÈùD8ÜógÔQWÆÖ°r-ÄeUTÒÞþKm;C0¥T= ±(<0ûîÚ!b³lÍ ëkAAJE¡RwcìTã 4«²X\|§+ütþtÅH·57°s¼Äê²~°¥ÙÇÿ®H{D©¨^Z"1Ô$Æ?Óê\-íLX4®Ñ�\|%¸sÒ {®Í}lM{ç(´Ã bªB=VOR;Äç /ä ð£.¸1 ÛiÝ?7RêÁI©lJnG$mµ¾j+¥T:ûÐ?NANQ\½R\Rj?0èi®¼¶ºT\joÚ¹dÊX×) Èírèê»y_4Ú®Æ®ö}TÚ·V"Ð¢fÜ[þõj»°Oß1¤ÔþI©IÓ(ÐÒ\|òag×T8ý÷àeO,¥hÔôhZÖì8ëÐZ»Ð{éù_÷i\¤s·«+lÊj¡ÙÿàµçÑMÚë'Mlð¯¬rÝWÏ~h«S6-õg1(UB¾ØJbAöÓ_þÕOGÊý6ö!¥Ê 'HC\ñàïÏ±ëºÄ}Üï«uSèºìWopòìÖBá«_«´ÑLªs¢TCGo¬,Û°bÑ¿Û¬ÉqFW9¥ÒeoYÔEkÍf»EÈÈÊ²ûû#o°@n))µÿRêÍ¿¾yÎÓud&ÕÓºPjÚ8¬ºä¹ëêshÀ©¹pàq¿ (4hDëöÖ x¬ ³ d>õáº¤TÔ ª�¡6ò×Ágqmx ´±¦[%SIãÜüéòáBKP[2H}ôUºFÃC¡ÔT§ÓIW¿ônu~9¥Î><Ò¥jR7À+Sê3ÙCà2hSß5v$kx«¯á«w¶¸¢&Çf[!¬ÜÎB²¨¨ à¯m ³¦j)^çAÁ×ÿº«Î!Ä¬m9GÀz¸B!¥~¢ÒùZ¦N¿ª÷çE(5ÃÄI¶ûÑ%Tø¦hÔý¾òB¢Ð2øøf)=i¿1u2Ô´lº(5Æm¦¼ Û½F]u /U%Z{n½Å» îã+[ÙéõMC) ÓqæÕ!HÁ&2\|_4 ÏÛxRñ0/dýäÐ oîWG-¤ÔR }óÓ®v1Ç�!ïµmtÞÒmêÒEÚ?µFbí+þ3{ÙÄ(upca{°L<\|Id¸ªÕÞÙÛÌ¸ïÆ[Ètd.Ü¶³Á©á$ º¨ª¼+-}çiQÔxèõ-l×ýIGJEJ= (5Ý&ïê4öêCnÌf)÷°í00<&qZùòé)I²ºkÉê¦úíÚÏç3¼ùDyÍ±uåeQcinh4ï@>zr]±ÐìÛ-ðÅ'h8IÊr£ÊL3= Jõg§)5Í $v\5¼ýpÛ·æ´;¯B»f¼fÔ2Ó¥8´µÿð»óDÁµ-(lÝIÓÿÖW¿vÄáß[=ðG^Ô¹^2ÖZ©__]äì¢v»1lÇ!çC½ÜÛF¡PH©(X2döÖïþgBoî¸×UÄÉ 86ÆAoµq¨ÉBöðKþ86¨è²§åç¨·jäQd½ÿÖÙ·\K½³ó¡ß¬ö§Ã1B¡S¢ö»ðÕopZu8.->ü²(¿¤¯î¬¡nòiÌ9Bááî0 +È¡µï¼Ë24 À«&½Ø¯ß,RêAH©»{m5Ô4¯+×æLÁ2dÖ¼ÚÊuR¹ïn¸ù!y.ê<àÊÏ;65,cEg¾_�ZªFJ¾óÒ×.<Ú÷>ü¹©<(n&@ @´ê¤A+Ïn5dIE{m¹Ò+iH©H©¥ÎEuV¾Ã¨VÆ:@Çj8¼î£ÁsSç¼ë37JlþXÛ¹#IÊ0ÙjÔhÕýz)õ QÜÛ¾ZvÅ'ÔåäH÷ÍzFu(\�Q¶ZøóeÕy/×²fÀ@ØR¢È²MjÔ MN@<Ñ8ä±Ì>õÚI»·fl'´®Y9À)ºó5\: \e3 l##Òí [2óãÎm£V/õ Ú¹¶ÒÓv8[iûøp4®n\|¡.ÐÛaÁù.¿àh0\;æWÐ\©¸B!¥î¥rî½søäY·ÔÆYv7½BjDù?®=ØÞv«C°}º±..+É\|.:xE,êiYU¹¯Gü«>øOXìÄ(<î¬ß�Z4R'aERB4FO^qQ.ëawÿë ]fAª\|ë%4½9Ó7ü¯ÓÿhOYôSÇ£Û7±iÍ-döÜ· Mñý9e#¥´j\|ñÄf\|m$óº°h5kbq¿/!l¥;ËòÄê.XJVU"ÝA¢ÕÕÕWû§¥72²Î¾þÏMFµVæ~¯F¨8RRjÿ T¥T»SwðéðÕáu;($¢^Ø¯/yD@oø+Fefñ¬®1Û[pªQ2E?7+×Û6 ýü'~wïXwBJEí]ÉË7þö¸ÚHnÎ¦Ü!ðáBúU-B\,�r3\|9[® ºmAK¯oÑ(Ë·ð·ÈTXU-~$ à§¸mzø ÷1ÐEh¯ðÕµM�¥ <á¾« L!se�L}êü4]\!ÞÁmðÌ·MÎÊ+·Ø dá,ðü¸ª<8óBÖ £ÎJêx ªµPµcãÀaM®ìF{á¬eê{¹üP(T_£T0Æ1[¹ró¯:?¸0tô7x¸�©ùïWZ$n±ßyÄ¿ör"£Ñ¥:÷7YÌìLtP¥EReUtêÌàÆ³ä¨¯¨Ð7c¢\E+6;ú0§AÇÖîðAÆ1x VÜñÜf»á¾eÒìF^¡Øü¡¶-ß.dÏ ©-ÁsþQ ¼X´ÙR÷¥j[¿øïC¬Mµ0Ã9dcEia9$FÞÿx¾CÈq0¿¡[ýj»zé Ûôýå·§ä¿ÙÈÉöo¶Õ´ª©@ëÀ¿9·ámFFQøÕê!8µ4¼ÛþyÓe.£M°æ.Ü^ÍSnhÜ¤×ú\¼R\RjÏ\)ÐÚPªÃ}ã=QªFæhçæA4\ÄhÙÞÕïK©Ì^7Áê÷7QÐemsº×9ÆÏÚZÆç¦J\|ÛÖC\§ÉãÎÿvè±°¥-úA¯ú¯×Èo³ÚE!+sxÓ¹7ÃÒz¨Ay9ø¶ÃëJ =ÙÅ^WÉö¼¾ª$æîB!¥~·=ÀllVóEñøë6+7Çê~RÄ@Èüë?¼ïï·;X\|(±s¹)kYUsÄ®XË[ ¹G4i¡,>"×4nYg¼«#GÈÌýÃÍw<2û©«.¿ä,Í´s¾XS#u2hþ·Jl sc çí÷Ü?mÆÌn·E06mÃ¬Q[Eh&¦Ñhº\_, ¤ÔRÓ\³,¾º ^d79S¦ .½ê73ÿwIAn¦ÞP0ëí¯üFfDÐYGUl{àÚÓKÈ ôemÜ xWcÐÖpl;ùåMäò¶ãO9íüsÏ¼áê_hxí)fg¦Yðüû ÈG%ôôê«}îÖ!¥"¥v[5e{©¥fíR!®ZgfK æA#Ï¹6Ú9O¶£zÉç�ßmª^uíïgÏ\|ôËDú¦wSÆÌ>h'×R!.+r:¥öI!¥ösIÔAF®Ü{ÁÆA'ÓÜáØRXÚræ)0éö©/q·Ø]q÷ M¹ÃÚø74Óªðò{mÿçÁC¬ysÆÀÓoC»OGSÆì'4;¥g×ªÆÉE¬8ÈÏ?µiP~{nV3¿Ì·£xHðøªF±màÀZ·5c9íåÅáÍÏÁKf-I¢®:>H4Ï~[ä»òTÁå±åU¹áò_7t\øÌ1uÃì6ÁguÔØ,òSØágy3EJE¡RwR5£-3Ú²}S¡Ýl#èh¶f8ÝVgVN¦ÃÊ6hL4p4ãOzuð«©F¦1tÊ£ ']Ò¶¶;Ú©áW¤ä¸ÙüÍ6º+jf[£6ÆêLÑ§Ù;xùÊíku.ÒeqbÕ¼ôPvõ6óÍÚ42ÌÕ}ÌÉ¿%\ìa¾Ô©¤T¤Ô}ÜøCÅC¦a2ã7×\xñ6ÚbIÛµð¶ïÈ°ÚÍ&³ÊÎ¼uÊ¿=ÜjQ¸÷STRé®èøÛ.@N#].3gqùÎPòr"ñXpëÖ£sòsÓ ^Ì)\¾IÅá,öW¿ZfyLS;Øn¯Ö×0))µgJUð¶E×6íY·?úb-ó .ªGZËWNoØg³ýj·Sx½çú�ÚUîSÄQ¦ùH©¨½<ìérõÐ²0pçùk íÕYY Yùõy9;Jmµ¥voVNkà÷Ð² ¯]aK¥Æ ¼9>w\|ÙF÷ 5?§ÞV%3=ÜNUÈ)þÕSc¯rYuêMèB©z@b!Ý mÍðÆóK²6æå· %Äïpy]9-v·7+¯:Ó½eÀ xh6Ý EjbÆV¬h üçyÙ-YÙÍVK«Ó^YmoÉZsMyÛ,þXQQÑÒ~¢P(¤ÔÞ=5h(Æ¤"ð>º<Á ¨Úlf3±3®¹á¯/ªèÝ,tBrÍçiæÛnkYVuêBýµ¨rÐã:y\¹çR# ©Õaµ>úøcÔ¸WÓ £ Û×¢;¶ÝvãvsQ3Ùy~º³d¸.+É\|îzOR\RêI©z²+©=�Þ~ãM¦,3¶ÑàIÓ?ë¬36oI°F=çUjaËOÍjã6µ+{ÉÊÕQEëÔc5j¼ùÊk¹Ùn¶td$ô%ÔápÜvû_kvÊF_èhò}ÓäFJEJíN©Q\ì Ëõï9sÕéZ q@^&ì?;çlH¡JóF¯ÎçrÑDJdºáÓÍfùé'¬^·9%Sóé®]¤TÔÞlçdvZ ´Þ÷Îè_m6lùÀåJW ÿ\|ÈÑðÞPï¥+ïl0JýjäOÿäµÏG¾xàá¯8QÿÄ"lI2¼bMßH¥j¾ùÍ ôñðSëîzÍ <¨§#KÆkÓÌ~QÔÀÎõÍûÓçÃ.+u¯ÉwÏË[SôõÐÃ«¯¾ 6næµIe_:3ôT·@ù7í÷ÿñ½Á W®.,^Q¿pÀÀÏF ?ò<4y!¢ PÉ}³¡PhHôJíÙ¦Ä£Aßêåß®]½ª¶¾±lË¶vX7V©JRç4ÔDÎ"'%ZçYíZlÝ8Pa·5kÖ¬b"³íw!tò\E7¬ûfÙ7³¦l}M}ÌÆc:r)ÓÓ ¥"¥îJí bÑx¼lÓ¦Åß\|½³~çªå+Sã@J u¢ Ò$I!yRë¼jcäAeeå Ö®]KnN}}=il)ôëBJEJíRé$AÖtµ£5ó¨»ôÿN(7~U%QÑE dYÎ%$SÝ!m«¨Xºby]cÃõëZÚÛM#G\J¨ÉÔÁjª¦RêG0£zÀ×�\|o¿ÖÁ×Kà«åP^G3è±jl ¨¼i÷hlh @(b%ÔqN ý4FÛS m Pí =Æ½n\Xçoùö¢Ð¼\ËÙç°è#Xø%-³¹üÑqIÓez¤önõÄk!²<õðÕWðîÇðÑXü-¬ZGß×Kkj<Ñ¡[WC¡PH©»;¼WNÄeµÓ$H~K0u¼^Wø8Iþ¦jtäëÒ2¤Bä¿c"¦(ÌÀèR5rÝ½ÑX¨ÐhuI>l«zúxÚW(U$ÒR CbJ5b9wbÕäJNêõéeuÒ¥QñwI½ ¿49¤T¤ÔN=tPj²è¼°ÖS#¿Êª2ªÔy9ëÞTùM ×QXiã'¾GJíBJ=PJRzÌÆüõAÌKS{"Ä´Ü¢t´b£\3·üm EA óv×Ã ÈQ²Æ·QN Ó< ªº¥I2ÐÄð;ªÔª7èªèDÖn{ oü<.ÖØUu^ÙÕ\¿I%ÿòIÄi½×8B@\¸Jú¦ÄFuÏbËE¡R÷Ô ÀInö0± Ò59r YÑ$IQËx3 Ør5=z4¹9¢v²¥2Ôåv¬è 70Tý;í¤T¤ÔýF©¼=$m [¯éØåI§TÒøó»ßlº¼¾ÃÏ))µæø¥Ù ´î=¥·&Da¶ì=«%É´Ë¯H©¨}=e$æ= ºUÄh××jjG\yêÐ®j¬L¦¯$@KU/ªIL·yºJëÉLÒØðôÀ VÉUÉ¡)yN%Ôä ÉX%ÇÚReû½tLCäÞmtêø ¥¢PH©{ü µÎ×zÇ Ô)û½ëlTµ=òBì.w:RR÷½ÉBJEJý!ÿ´4J»MµÉðnÈnöß¾ìåz Ái÷6Ö¥½iFn¸×,Ùë¼"ÇôH+¿»é=¼ 3ôTÃk·sÈõfï®É,ÜU2~èWër¤srç#8M¢PH©{m%£ØÒf}>.êZjDå«Ó{j)õBú.G<)))))µ?Qj§¤u$¦î=ÑQJWö; ¥¢ö ¥jÉð(¹3ë¥láz'\|Ma\Ü TÙÈd¤ zî½§ôü¼Öés¤B!¥îñ(ÚiQÎ Tµ·ôüìod,S1HK/~Ñ5æîûiÇÑé]õtÖ{>»×K!¥"¥¢RRû¨ ß¥êßqFJÕöè}RQûXI\_Y@ ÞÃÖªÞÑ¾ LÓ)üðC^5L7¶¤i+S÷gPýE\| üÏ¯Ú3qxr7lØ°y7´Õ¯Þk¥644ðï)uEîÁrWÓ{¿Ïå¨~.þµÁpÝºuT±w P(ê;A/iVTTÌ1cÚ´i<{RJ³fÍnÀsæÌEEftk-ÝÒÞ¦Í¨þ£.æ7ù¹mÛ¶@ ¶ÄgU0vCØûcáyÀ=~õÎöÿ½D&£ÊÊJî¾z:0tx Uò/cA¡P(Ô.ÔÚÚúÞ{ï7îÁ$@Jð\|¿<Ò}÷Ý7a®ó¬,Ëé¾:d!¿rROÑ êC¤»Ê»ÌwJGõ+¥©\_Ñã÷tHÖGK¿Ã¨~T\_ ÉÍýLÈè®B¡P¨Þ´jÕª'>}:¡TòsÖ¬YN§NJÈü<9mÚ´úúú]\_gVª ¿Ç½u¿õ P(ÔÁ36Ý@¡P(TºdY®ªª={6!P¤ãÆ# ºpáp8üÖ[oÝ{ï½'O&ºiÓ¦T+R\R\ )B¡RQ( µwÅ^yå)S¦Üwß}Ü¿wîÜ¹^¯ÿ5L>@ë3zè¡%Kì±A·)o R\ B!¥¢P(j×Z°ÁOsJ<úè£ååå¼,fjÃtñÅ'N$Jv7'´ºRQH©( B¡P¨]#'Og¤iáÐÆÆÆ'\|òþûïç¨R-ZÄu·¨gÍõôÓOG"JP( B¡P( õÃÅó%r>Ção¼ñÆÔ©SgÎ9mÚ´±cÇÎ7¯¥¥%³· ¢nªªòz½¸×B¡P( B¡P¨½ª$}óÍ7'N2eÊøñã ¨>ñÄeeeü¼ ÐEy¾B¡P( B¡P¨®ººº¹sçò,¾cÇ4iÒ§~Ê÷O r¢È Myööx¤T B¡P( B}/I"Jòq<3<Ï=÷ï=ç;}:¢nÞ¼¯N%dÊa{ @ ]uqä$úé§¦§§8MeZºtiii)°z7BÅñ @ @ n·6¯¢(ÊÌÌ´Z<~:gÎ @ @ Ð5(ONgEEÅÅ³³³ &''§¥¥}úé§p@ @ tÍÄ~üñÇNÉ#Õ%Kòè\ßn@ @ k Î7/-- ªÕjÍÌÌ,,Éä.@ ºIí²sËÊÊ-[fµZ ¥&&&®]»Ön·{a®@ ®¨666îÚµ+---===)))##cîÜ¹%%%ü§<\K²ïBT@ ®¢¼±ÑÎ9Ójµæää¤¦¦fffîÞ½»ÉBe$@ u·xß¬¬¬¤¤¤äääµk×666¶£Q T@ u·6mÚ4}út¨ÙÙÙ¼/¾±P@ ºUzíµ×RSS ¥ZVB©»wïæ~;<(@ @WK²${WWW¿õÖ[N³222Þ}÷ÝúúzÜQ@ @ «.QN§Íf[¿~}rrrvvvzzzZZÚ òóóÉÈ ¢P@ @ P÷É»wÌþýûgÎIàtÖ¬YäqÎ97oæ¼äS@ ]UUU½ûî»V«MMMMfZ²dIYY)@ ®%«º\®Í78MHH V«uþüùæËÀ3 @ ên9E=räÈìÙ³ÓÒÒ²²²¢æää¬_¿Þétú @ @ îVccãÒ¥KùSò±páBÍ&K²( 8å¸êÍ @ @ èÊåÅL%\|ºqãÆÌÌLB¦¼JÒk¯½Æ«øúJõ\C@ ]¹¼¨òN .\¸ðiNNNzzzVVÖÚµk., @ @ «.§DK.MMMÍÎÎ&Æ¨äíÔys52yìËý^½]3Æv_õ\|×$%ÏøF>Pè!vÙ»({ï»o XËJJí¥ÊnöO××Ï tª@© ^B©ÞßH1P(µË¦ä@,(JMùRªÒ&:)ñXwP\|Nó¥Tþ!²oÀ«\|1Jé»$ïRJ%}TnaETòª;ÛRUw¨ÏJ(õÚ1êîDö@©@©]£TÚòI4±ý4oQ:<ÀEJJí2¥¶Õuñ J½'æãÆt Ð Êã¤N1¦ìöÚÍÝ»cÒÖHÇ"¼àêKË,¶é3±ãkMì$ªôÛ=Ë¾JU:,mC©ì\_m¥6¨=C½g]jÇÝcNã¥¥vRÝ\|¾ \÷¾©@©@©I©RW~ÀNJJ½zX»Õp@© ¿kÏ<áCÙ#y.±¼[å5ª:N²oØ´Cª0Õ6/²¯°Ì¸RªHIö~;×·CDmKÁj'ç¥^[¹NwôT¢°HJo© Ú·À;$IØ¬Z§¿²¤ ,q;òtrÛÃ¯¡((µk#y¦Íé¤¢:}Üö}· zà5 T Ô+y¨è°¢´iWÚß¥jS)=ª],Wò°ªE7Gªìrä '8Z¹´Ev:h"×eX«FÉÓÕfe¨"ûxW\÷Á$¶KÄ)#ßÅ¾-&u÷²êåU÷Ìª©=ÒÚ,eü9ù¬¶�¥^!Üºå~ k»l((µ+,#h"Ýý[ÚªÇHE±^ ]"vµ¾½cPJ¥bßä.n°e>Í^)))!ígÇÌËË#µ Û÷uHRA½RY¢ÜBGEq±nUä£¬¥üR'®kÆ¶l«Çì²a+ì»vJõ°¥]Àµ6\YmE¸¡þ ÁP·&8(Ñ^×WW-\_ÑJ÷æ ÞZ} n)Á åX´aÙNÇ#Õî² Jk\¸¢ÛJ°ÇHÖ[\Û.ÑRj0®mu X(ÇR-/T¢zßëR¯n/ÐAVO¨8ãÎú#CBQ! iúlÈ-Ú¶· jG¥ÚSêI&ã¶s¿þH©ì°XöÌ¸¨e°ÃWZ!Íþúù#Ysæí;yÒéáS TÐEÔÔÔ¸jWèÄN½K./.;iMK ¢Óé¸Éh4^§7É^RYá% ²R»oÈ>{ö,12¸Î':öµ 'n,¸ÿúhj¦pdêóæ¼ÕÖ!5Àc¥ìünÌ-ûøÞçã¼7eäYÏä³Iê_©²ê¥£&£FóÀî>¥OÈ+1}ïýQÎ3õJU°S¤÷EÈk[Ë±ÿ±<2yÜç¾Ô@GOO{í×7<(ñþ;3'}qt¾;qÍÛ_gÜñRÌI±o»cæ?µ/êb T½Kº0¢2à=?ÿç×½Ø7z ý'ýüklkÁØS£ ª'u3¥ª¾Ý#¿Ú®ï¾Ù¡xk®ÿÃOMmawÍ[q (Ô«(ÎÒv¬äÍ\|æ0F©J£ÑqV¥ÿ6»ut^iU3 &)@© ^D©\|dx±ÄÚºsýï©ÙhÜpz¡íèôª)Y3êO �¥z¥¶K äÉjcÝÑ2ó0!}¡ÿBh¡fT T¨¶oû]æÀ8«Fc ÎþÅfl¯òÜt ªì/Ô.yûô!SJÓé3LùQ1KûDc©R¬V(ù ßxCÐ´hyÏ _¬§±¶÷¥¿~ïÔ]Êõ³'Æ¸R À:pù²½GNF$¤Íê¦}yÏª Be(Jûõ\µï°¾~ìT#J³¤<ô\_\]{\]Y»ÝÎ;ÆBï:ÜzýÇCóÝ¨u? ^øáõ®ºÒ§5J ×hP82ô;^îªW±à û©,+[¦ý>ßÆRÉý%}ã÷tåêßOÑnÉÍ\|rLÌFd G°vU#«PpD2k¢æµàRæu×56ø»×íK©ÇÈ+@¦W(Þùø'¥^´BDÆ !ÿø>KÏ6hµÉd±X´ZúNCá5DFüîùT¡Æ/¨KÊ;À¡ÔÖ½= ªùSC(XËw ôáCgx?îF·Rj@TU«Ô±KùÍñ%(ã¬Å?49!hêðåÒ¡ ÞUfUÉ oÆåÙ_<7%8ÌP2f´/D-:»¯ãz9[Xº¯2<¥\|³zË+ý¦õqÆôä\|YGìH®ÆøëioÞ9"UoN2dú+0qÃT }oÕnJ@lfÑ´¡áSïýoÝ9ý@QêÝáÕç7¶ÿûÕ{O5kSûG¥ü\|3®«Ã2þömM©î}èÒ!±¼ìø®+ s¸!È¬ê3DoÍ\|ã3ÏÔÉ%a TP/£TÙ[dþéÖPê2èu¡Ãç¾¹îÝÕkW¯ZÍ÷ßÿÓO?:ujÿ¸{ëÝñ"ÓÝ¿¾S±ÿÇRR»B©îJ/.§íøO!³ÞhùÕ¯~µoß>n,³ WUUmÙøß!}ÍdÐ1²}ÃÃ) Ãý>H Ú%Lñf²"JV«±P\|S$5 r£ÆÜt=1ÛóLMdÚPYl Ô6ÝFì:²é«I}' ÓÌPdöRîBz.,Ñ°\|;-ÊRS >÷è{û&jôñ$MÁÙ($µÏ_OÊc±4Ìü1Oñ#^¸àþØd){H¤õáÏòzb r%ÆÓÜ<4éúg¾¼ã",9±]\¾hë?GNDæ,B©H4³ÿ^D-PÄÞ_ÅRísæSJÕ¥ôïôm¸(µû)ÕöÝoHvb©4ëéÍ¤cÔ Ô[4& ÒDüáùZz®Ó!x(UõDçRAPê)¦¦¦¦� T¶{s¶ïÊüÓ-RI:Êfo¦×kÚñÉ ²ôLÒÅ>WC¹Héç~7Pj÷ .D©ÌuVqsñC#¢ÌZ ÄÑ ºóÈs¼wu:àv@ V-Ëòäë%öe²Û'Jù??ÿèÑ£äz&ÄÀÕ:uIK:°óüþæ°WöÀèÃÈÿÆÖzw ?¥Ýº- TPO¾»NßÚ6ÈÐXo¹géß°Ã¥ «C$QJu¦}¦zóW¢MEIAÈ´©¡±¿=ÌÊQz)µøzºyÜä°dÙ:ô¶¹ãO²%©,nû¥uáCµºR±äR)¥Ú=¡ÑäèÌ°>SÆää¾o#V%ø°»éÕce{Î{Æ·§TJínPõøá\£Ô²w!eÔý-yÒÊëé8éÑÐ \_4¸Ytù,ÞJ>¥2!ÿUbûW¾)úÑÈ4úLQÛÝºZÍ\_ÆbÕÊù)¼¦ÒGÏY¾¾\:RAKÀQª@§¸åÊÕ3¦Ðú4Dª1)¯Ávv8í;XZR¾» ùÃÜÕÿÁGKØ êê[1¬ÙîÄªÛÏüõÇ#,tQªñþû=q¸<,T5áÉcc%³Õóf5pjb¥®hVZ½zæ3oéÖ% õ³r.2ÙAî´+ëvÿ;vR¨ÎÑ'gìöM4fë¬aÉw¯JØIZs°jùm¡ÉzSüÇW¿ï±_kkæ£®K5R Ìxã z@·¼qõ;Æß2YaÔgdCÓ´û Mùá®ÒF¾áªä¦TSª¼=mÁÝR¯±_áûJwGuÚ\|0k_?d1¯Úµ%_¨u/&2^÷£G'7¾NV]_¼Onèï]#PPj ç+ÇömÓþÄ( ÃèÊÎZ¿\\|ä¿nÛýü¿IqgRR/M©twÔZÜøí¨Ø0 jMÑoÜa&gÔ]Bq"K¶ ÒÈ¢vUA,Ô({ÖÔ½&ÐTþu4]xE]±)Óß%èÝÃo Ö¤³ô1}NÀ5ìGN©lï ¿ßÜ(5Ð)øFå/,72QoJ u÷ÆºµØîÄ.àU%Ü¤H¹ oüb@F_uèÈÿõmõTd SB%íîIÞ& jdJÜHãëÛ^>A~ÇÒbþºlÃxÛ´Å·M3êÒ>é0ÆôÄÈ$ ×,Ý1aÔc÷A]zPpv9;"$eØ½sÿþ ÆlgÉC©I·¥,¸«_"Pêµi( ;ø®µÞu÷º\ô§G$ Aq°Üñ¯û áT½!¡¸½çêI#QùÎª.z�¥RÛËs;,îyÑÏf½á] pJUìX,=¿} uªé¢:óØqÖØµ£¿çäR#9ìúæù'ÆYLò\rjnY¹ ê Ê+~0ÚN\ýÎÒd3 ¤ê»no1ñÄÞãP¢m?Pq©ÝK©¤_¦Ûs�¥z¸¸rÕáÄ^F¡¦¾é7ÎÊ]SLñÃ%cþ´Áwÿ&õºðpÝS-9Åq¯Þ?lZ¬nê ûý¹»ÇÚÃ _z÷×%å Í\|àÒÒÌm¢¥ãMin¢C)CúgM8q9ñ_°S}U¾±í¥&êL)æ°¹Ê3û¦öES1-<,õþÃk Y®-Õau[ú»$R¯W¡>M¦±o~#d6yðzkXïCþ÷¥´ZVåùë «itUPÌ³SW)tó]EU[hãJ>¥òßjÜ²ÓC©d¼¹¤ªÖ¯¸°lË[³yéd¾ôãÃ¢7 dJ±³ú3XÒF¿¾v«È²o:¡T»¶ÑÞ\Z\r>n.P\PjHpW³dÎ&JoI×Zú£ ëêe§pÕæº½3eäc§iwcAÁÍ2ÝXòw{J Ü¡+Àuçõ÷M¬öÿÿûø]RZ·¿zéé¦Í×/þvËY\>åGFL@¡SL×§Ý{B°5É,ÝæÛªûïNjLñÿüµ»ÜuÈ\¥o¶.¼9.¡äý¦½\|îD#]hõ¤×·üóÆ %¡Ïa\|0c{@khUo¶ñÐzÇyKóuHýµe¾6ÿöñ@©×R:í&xEº,ç½þÚßõ´Ò¢Øp¤¶î$GñÍá¬ú¢6ÅÝWèbÊSh8Õ¿¦¥¥^¥îöä-RMÈt}IugºNºÕ"éô9(äÆÑ"¸ êZuì¡>¨?-"núòT]³ «Ø9¥²MDQ]vÓ=¬¨öÿØPPê¥Heõc¶kO\|ýQ¨Ð¨i½õËìmÆÙgw-ÈgAèM ¼Ø¯Ê)Õ.C,ÔÃ)µd;Þ÷ûúNÒ§ öÜV¨êÄÕ¯nùÇÈIÝ¤ÐØ±gp] .Þp.áÎ¡è©(zÊÐÅ'¶ÖPUéøí·µ5ñc¦YU -~¤²=··¦,C)5¥ÿ¬]K½y¡OfLäÔÁÇðÉ\øÈü11ñKF!aÈ³<-Öáa¿y-·f½zPê5óÓ-1@bÐ¸h3±æ3u&M¿mÆÍ\|¾B8;{üoÙë¤ï»\|Óm@ô@©@©]j&X¢Õ÷d?1¦/-.¦ÓÜ¾/ï\|QE!SIIIAAñBw~µmÎÌl«8ÆtÖ\|þ±1» ¸ëcøï~4@©@©]¢Ô²cEk,dÐ@Q¦Ø\|Ûä©ôÒÙEà£Éì çÉÌÍhþJJíÏÙÎMùÓOBy}xËõ; (Ä1¯8¸a¹Ù½3÷ðRV3JÝµd¨òFNSï{sÃ3qñf5ã·a:ßEÖÎ=»ÜÐ)Ón¿mái½½Ê ×ÿ/)4&g\w÷üÄMDæÝ9%\^R£öÒXëÝ¶fîHYp{\%÷5þ]ªÈØE0VÀUonyÔDcxrô ôÛsñù:\³§êÛ&¦"d IýÊÞ×«èpD7ýfi ;îèz(Ue];7'Û×qrûçî\|°9«ªÝ5Èí®=ú¹;Õ$¬ïÍ?ûµ¯¼Ì lJe±TÙd?~;ó4H¨®Em¥cÛitÈh OÞÿô¿M230¶ ×ÇJ4¥Úqñ;CYx}ûÏªT\g'ÔuJUq êªøÎðç^o¶£ ùÐÎÉÜ'±XÏ_¥gfnüVÄ"_§'D_ Ô@]ÙA�ØÎ¼Ù±AÓeÿð5ÛÞ2Y óø¸¥¢ú«±/ÇE2&Þ\|öórìrÒÜÝãSþOØ¤µö8ã»é¦d äx&}ÙO£S AýRþ\|m?ã1:¿JZ0¦_%õkG©"®Z²íå[_1FRJ½óZÚ@qø<=²ÞÕ¡äÈ(ëí;qI æuæ FïNY4(õZ÷ §N:e-Ö¾ðÇGÙ¦ê¦à¾7(ià;9]X(ÃÍçÇ óê'Î×³FàÞG(øJÜ&ÜýøÑF¤ÑéÆô!4ÁÃ){[µ(.+B+?û²Qj½³:@©@©¦TÕÙXx¸é´Ä0bü4õ[LóÁ ªü MÊAw%_P SªBëÚÙªR'V·.>ñÐÿ\|¶·Ò9¬Ló¼äZÈO5F¤ÿÑþZ¾ {iªCÆ/¨'ß\_ÆÊªwfÆ¦¡Äþøùô2ò¢HZoñ¯g\÷RHp@ãä½¯V×Ñ7æ\_°òëàjQJ9éÞ-GëISWj]øÀ£IÃbSÌ©þ÷Ãyù,¨æ1�òôëäwöOÖ£¤s&Æ¸íDCÐRÕK·=Ñ>(cÌA\\G ÿÚìøôfÝù"RcÆ¼ùÌ,Ò½oTB©{ ï¿×¶¹pª+õ¸ªÀB½k-ÒÄ{øÉImí.ü¨Õþ\ûFOFÐÇæ²P\¨P\±ÑJeÓ5ZMP4Ò§zwTMn7hºã¡ÇjÙ¢Óÿí((µ+j;»7ªØGLÔ&³);\_ãÓ¥ú\+ü�ûq¥¥vRj"µX®ùñ> ×[Ô ÔGç©8í{ÅC[¢i\Ý³;êæ\_÷È Ï8Ù TPv£è¢Qã&tøÉå÷ôI4ûlTòúñYoÿ¾ß+A)Lû{.Æ¥\|k&\q°üãþ-¦ôÈ;'ì]RKzíá²÷îbrÝ¢üm óúPjàÁÙ0JÅm(Õ: óÖýR \_À¶¿øÛ^2Df 4-M½ñÜ%nRÞ¶,Ôø½¦êÙc³mûÏ\|÷®ÆÈ×¯i(J.E{N\ìJ[ù³M¤1£§cYPPjíEpqJ5³µÛE~½ÿ\|euCÍf«¨¨ gÏÝ¾ù¿/=õ(9MOÕ@öäóRý}g T ÔKSªlÇ²í>,5ÑwÓOÏØ©Óàì\|é§ )GËå Ú5#°Tù¶ãÍîe$?<þwÅììà½àÆcÅT=2Øz´¼.ÊîÿÛ Pj�SÛD"'áòY[2¡ñ!?8&Q¬Ç»=¶ï$}püÐÇ?Y~Ím²½HõøÛ HE(yøM³,%ZºúèWÌ¡)C#îÙ+5yÌ ê})uh¥ÔjF©äwðRjhª96sÔ>µ@±³³³×ýu@¢FÓúN¹íK±¼ÚÖæy· H4tD©2PjwRªBg jÏ<8,"\t:# éC4:'uÑ Ñ1F)ºÝÅ,[»·Ö êÊýK=ýøí´2!2<[ÇV¸/êUÅì¨Üºn-¥ jüühq=æÔRAM©Í¸äà}!¡ÁÃ~ï j¿DówÇÚc.ìrÍJJ½¤¹Ô¯n>µ4é/dÔ0Ðå$4nV4Aôy;Ñ&Û¼óÛ§Eæóõ~@© ÝÒeF8×ízã¦a¢6éÏ«sm¸y¯í½û§FE'O¼þÕ{ X$¾p=.û×úE'#ÔwÐÌ_À¶&lðÞOÿ E¤ß:jÎKqSvÉ¬:\|gj£¯SJUpõòmãod0[Í}¦: TÔ»ÓKÉ»k]øôo^Ó?UgH2¤ ûïçÎÒPÝ×ÖE£ûµRjSi@©ÝM© w s?#ýíi¤]vz-2u&£¸Øf¨ãëïÈ ;~ô{ù ¸@©@©]5©7íË~ÖP¶<[«SZF3Ló²oQÀ&¿¥ÚÏX4\|Ùjð_§ÿ§Ø½(ÐªÔcáØ¸!æçè£PØõ{KíJQAíäsì.±îtÞÑÊêV»J8¥r5ácwÅ"B§1ÈøT´ä0ÒCCg=Ã(¯aEOã¡È8äTìà+JhÊ ¬ROÍ_ÿôðAÙ±!IÃ¥Ý²WYÿÙO#SM@©×R©gÐì,éßÇDû¿ 8CÔÈ¢ÒºÒÊÒÊÒ¢²Ê²rWUÖÔU×VgOy.e Ñ¿{iNö®x¼¥uÏ" T Ô@ªñÛ;í;é.0é¶r^¦Ë·[cDªwÞÆU»5e) MhÐ¸çO¸w¯J ¥vø'9±ZS°cC,7yÿOÿSh ö¸ :)\_È{áæ¢4\8wÆßÍ(fò×Eÿ¿¹@©@©@ãúÙ·ÆGÑYq¡Ö[r®ûé{÷Y2-kÿ¨xOÆ¯J@(µJ'\ØQøõÇohÜvþòºË×-÷ÓIGÝ±}¥Ò°»7ðù ½½ê/ô ÚEJ(¥>~£ÔCÐmÕQª¤ÒÒåªc_õáòÁ±#Ê< êÑ\Ñqo¿!}- úps.ñK ÷Ät'öÐÍÊwhùjV#²ôÛW.øÿÍJJ½sUVóÍö"ép1pä}vºi²w%x×È\~bt8Oz3#ýuöW j ì Úk(ÕÅsGJþ3<)(4#2:õ¯ä/ÞÏatÏòÜj,¹(¢ 2Qêäî3)TÒçY=úÁsaÉ"Òn¿gáô\VWõì:ì¡ÔmRS ¥ÆÏë¥ÒÅ§Rîø¡ÔÈÄ~¬JmjC©M¿Çe{jÞ39<.3$tFHxVdÈ´CR[êhËJíÎ¦ÕÿøOFR/Z«GÁ#þ[íÛoCRØ>\¹;qÃùG¯ µ¡H3$mù&öMà¢9Ä<¤/ûI Ú¥fBºJ¥7h¥îR,T¹ìéÿû©{ ÈõCÆ/¨7P¯&Úüõ;ð/d¼0Þr®¬M ßkB«Ú¥bÜtÿwGËk4coùñãPm ((µ Äå5s'÷7F5!Ëi×ÊnCêR,ûOæ_Ø^Ý:ôÁÇÇV¤@XÇ Ú+X.#©þzÎÈÁÉËÄ cbXôÜ(ý¾ÃSþzÅ=G(ÊNPþæë£MVc54,«OpfxÄ9»Ö±MÅ_¸c©R&¥ßð±p,ñ×ÀoÅ_ýãæIú°ä~Ò8¤T6xJd3×n}R 3çMÕd S¶Ö0Ý n$9õ ÔkÔN$GÓ7;x¬Gk ¾õ¡"·(ô¢CUvõéAþZ'uÇEÜ\´oÅüP÷Æ©úÞò½Mýs¾Ø(ï�¦ÔÛ)¥j-ËÈ ñÛ!¥ª®Êì´M&7£"üþ¾Õ,(èJFÎÛTüãÑ¾kb ¾ãðn&ö'GT®ànÀu§^\|ìXß{ØÌý¿8\Øìÿ;7¥¥^ÚQì¸1ï¾~¬ÚqÇBoØS¦BlÛ·æ%·-7cùÛæsÜ[rëQøSùÎÔøõôöîYH#¤.LXýÈðI(2gÌ9ÐÔÍvÅF±9¥ ØU\_ï:r{RLXÒd ¦Ó¦ÆM½eOS-àÕRk¥Þæ¡Ô©ù¾6\[s«ÞØÁ)u@ÿô¶£TZIhU7á£Î½epº%"49:]J\Ä·æZj¼Pjw6;kß>¯ÖG((iÑê\¶nëÖêìpVéë«²L<Õ¸Që÷mîuË,®øKÝ$ T ÔË2,5áæÜÌ'ï ñÔµH6ëý?ùðýÈñÉ5ï¾ýÎê+ßþÏ^\|8æzwÍ132 {ì¹Ìz6Ý'¥z¥ºýiÙ·kS¸§öÆÐW£ ùÓÏíÞuÐårçý/üæÝÿ¼ÆÙs·;$ê9s\y{J(¥º\. ¡¬Øëów¬íËëìéCÿü¹nÎf¬´Rª÷NÖcGÙOG rW¹ì?5km½³:@©=Ú±)Y¢»#Ðr¿u«Y¯ú$étI]fdXò-oÜUË²hdÚÙÁMYÃtëW=Ø/Ò´úý4i7 IûY>n&?p¨ì-¿´zRÚ;¥h4ÉFLO8ì¥T¸æÍí/Ü<Ñ4huÜ!¥º}#ç6r±_ÎU©^9vbpL:ábB©/¥Ö1JTAR%¾!PêU,éÎ\n.lH^Ôè¹SR9HF,Û^ùã/I:mß¾ÎÙmkÃ(UJÇ;P)õ¡ÔOÉk¤AZ½N¯ÕÃd0t:^±µD6ÉDc¾m¤åRoX@©@©]´gÖ¨®/>^id@êèfMÙ]ÈÔ¨7ð}^ô·ýù¯ùtvÁ[Ø¢$ TP�Rf5áùÿ'ÿå÷Á\|ì°Ä,þòx s§UW³o,ÕÇSi(JlÚóÁ÷@¥{¨ 12äR{¥ EÒëO5!ã¡TºFµðÍÝ¿n"2§é4YýuÖ{y/±À]æê¤z(µQêmJzØ'7¥&öÃ(UgÉç4+8oáÆûOFæ ¤ËBÚ>áÚjÉééR»©©Ø±óÛãßeý[(BýGýä.> vB©ÍAÈ"Ë÷näöÒ#PèÈÝÈ'ÏIÏ©rT Ô©dÃöýOÔ!% ô:aS\Ëî%ZX®ÑtOý3µF Ó8.LCü½P\PjíÅãfH\v9ôÕXs{W¦V1\5²5¢hE>sø«+ß'CCt±¹TÙÏí((õ""¿?õ§kª¢,z6f ða%Ít¼p÷·n^JU°ûáª7±\8-1Aí> òOwZñ±ç²î<ÕlÎA #Ç¼öL±oõíôB� ª\¸à´cÅñútJ Oy0çôÆ2l'ÓÄrÕDwUoI[4jiRú 9ñ§Tj@¢ËÞØñìÍQXrlÿôû©¥-^³özªHÿ )Õ}ãdL<Ñí?ÞÇÌÞtq«¾Ì¥î9EJÇ;�)üW'þü»åkoÔTÃÄØÔxß÷¾0þùwÖ\|TR';Y]S/é¥z¥²¸Si¡ ²Uû6~üäï54F :E« ÓûÍ}ãý2ÊqÇ© ,À~© �¦TþËïÙ³KÇmBùÄ > HbDmK©¼fhB\Ü¯ð«C4ËF8¬´Ã@ Ô«ät;éA ®¤wç~[õLßèä¼LÊövWOò#×((õ2&�¥z¬4w¥Do±kß ãï�((õ; 6tÁ-ÒHhAtÊ8räÍV·ÿÁÒJìq v \JõA0wêx txx~îòÁØÕ@~1"xVÒjº nlÁµµ¸± õ´Rug\),Úà¶l«ÅMõ´iö ¡y]2wIlô[Õ«l¸±m(òÁ»t¸Ökêqc=v:Ù\Ü¿§¦Ê\_$çÔªÅ5üÓ<;×´µ¾gZßï2¾¾ßÿU@©@©éQ\à-ô"¥¥^¡ ÊZ 5îèmQ¸Õ3ÀÖc�¥¥~w¯ërÀè"R{ªv½tÚ´ÜÁ8µZÁq·ð¯çð,}UR{b¿ êÍÚuÏ(zI\|»ðe TPªÏX£´sRA ïÃßkK©Ø¨@©@©@©@©@©@©@©þo8"¢·µð¶Ô¶ñ�¥RRRA@©@©@©@©]¢^-ßS3ß}Ñ¸Ãéës¥¥\.·Îêk/]¥ÞK½hkJí]ó3¾6íI8ªxwvèÝ ¿@© Ñ6/T^Å¾çY@©@©½YË/ß»wocccWÆ T Ô^(¯Ï¼zÕêO?ý¸ÓÄv8Ü(:¼V½3ã(äÕÙ³g-[FÀÓÎ gzÇ\_\_tíª�¥¥RRÛ^Ú.ôfWH©~Ök¥vqì T2sæÌôôôéÓ§/^¼8//ã§oÌÈ/(µmö§ªÑÙ�¥¥^DÜ"Þ{ï=n/ ,ø/êêêøUºÐ^R/F©í\_ð3;Jí 233ÉÈýÎ;ï>\|XÅE©jÛã»ò)P\È'cºy¹ ¼ø@©ÝÞ^UJíðsRJ< Ä¬¬,«ÕCþÁìß¿ß?)Õ;Í¢¸Ã]�((õJ\eËe0«ñÚËúõëm6[»Ë Ú{ï·ª:N2X>1cyÌdJOO\_»v-¡×®düú÷xvì§yw:Jñ¥PêÕmÅjG×¾ ãw¥vWï§øøÞ>½ßUk j^wA±ÿQ\ø\++¡ãhçeÿ\½ïàj^^×E'WäÈrI].u{¥² ¶:¡T/RÁèU/\_Æäk)[W¬X±ÿ~2 p{áK¼ÉÈrúôéÒÒR ÔK ÷4ß JTUUåq@\|¤àÑ®Þû$�9s \_È{yy¿oòùirWc åo�¥ö2FåñI¡;rvP»4\Ê¨ÐCØxAü<ÛXì8Û$Ý»Øó ªÔê&º·JínJR;¢Ô£G:t¨ pønexJ%ÊÊÊ>}ú¬Y³>ûì3/ ~ñ%Éô(Dõë¾R¿¥8q ï÷:k?J/÷B-[¶,%$$$&&«!$KÒó455+I÷úX[ S±êêê }Þ,AÈcQQQfffgÊyóæmØ°>aRb&G¸ÅµÝr((äG=J«×½PïxAùÔÅàä¢Ú:Ê(ôðÀ¶ ¥^=ì(ã·Ûz7?ÎøÍe"w E ¦ÒÒRÎ{öìéOyN#y$î·Õj%ORSS-Z´nÝ:òb/ÄCs¹\=¬WlÛ@ùôXë<ä{wRÉý-(( öRØz½K!Wæüùó¡Tb Ädsrr¸Y³xà$ï½ø< ÔVñù ¥òÜüü\|Ò©H×AwíÚÑ¹¸¥L><Áå£>Ú»w/¬{Úè5÷ÓRAÃªíS=+·,ßU¯Ê/ Ú Ï êR¹AÊ\_ç´Áõë×÷¨\Ë·y1ò¹Õ´TÜ={ òíý¡?+äTÅ)äø +¢Úúx°ÇÇã © år°aKeÑÅ8K ÜäJ½ZÞ¹O$]%^g±¿ø2kßR½S ¸MéW=/¥9rt;wî\|ï½÷Þ1^µzÅäÉë¯¿Þ¥ò×É#îæñ(Þzë7~ßIYRVIw(ºZDÃ%J¢e¥õÇWÒÚum\|.L·+Æ9·ª]Æ/Ñ²eË/_þ±Ò¼öÚkQ\\_ÇÍ£¨ÄR²æÎKümþ¼Íø©cÖ®ý·ü]¢ä"cJSsùTË¡ªr»Y&¿£T2¾¬Y³4b/oÈøB.\Ø¥ò¿cyË3ÈeÜ´iSW6ïîá¾WËÉÿ!Êr{JU/ctkWSBEÿBx ÔÞ#oÊ ;p}-®Ã ¸Ùñ»R ¢ñ©ìÄvòXkÈj±P-Z¹ä«E\ïÀöFÜR6·\À<\|R¯Ô;}ÖW%úËN2ò;EQ&ûÏ¹ÌþÐýY\q%QrTWW\²'JÅãzO£TrçææòÍ ®·=óIiiéEBfÞÜ-ò8êyyyçÏç\|ñ÷ZÝ¢]}UÞ½ lÄj"9é¢}»Dì°ÐÁÀ]j¹¼R+¿h¼ì¤ÏÉË555+ÆëtØ½¢/¥>\|Ü\_ÿ.°Ù=TB´råÊ)¼È7Ú ÷ôéÓ,X°k×®«!¿=Ó1S.Ê¼þ\ÁI§ÓNZEë¢ÈÑèiT«x!5Ùëé4ìéÀ¾Jîouu5yÅ¯cÝÑðêIR\Ï=àÉ9ÙÙÙëÖ;sæÌÙ³gIÏC¨·Y=©K«ëH'/H¢¥ Ìoj-pÚ¥j~îXª :ù{.¬\|èµD TPUÒ6ãó¸ð4>{ãªZLSp\HúU;^ÑÒ·¹T½µÕ¸¬Å'ØYq=uûTáRis¥^Äeø38ÿ.?m§°½ E¢OY3^Ø[ (õJ(µµDL¼p9KýZ¤ïk«·³sÔ6·½cñ\¿\°ÜÆ#v=ñäï ¢Ò \_?úûµë6\ N)m{l¥çqj»X\+1BïqáU\((èRù\7Sb Ü ¥dú^c©ÞÚq>íV²c©ñdî§ÿüK5[¤GÚG~õÔç\_¨·à¶ÏBþynÏY°B'óóòÆ¿ü²Ng¤Æ ÓÇq?üÁ\|«kp©>%Ð[ývÙÓùá7?öØcZ½zý'xb××{Q} ÓueðPàÉÌ0Ó®ÿ'æpß9sælØ°Ø?P A£G«ñöH=RÏHÑÖf7¢Ô?ú«GÌæKHleÝç§"C¢\|¾vÍØ;oÖR3B&si{öÅ¼3¥Ç÷/J%aUUßÆßþ¤¢¢¢\|¾"ÖD¬Ã;¸Ã!ÿ$\_Ìi,@R=øª¤:ìÚ®Ñ"ÅôB²ºÐ²ÔUJmõGbwUåÉÉ©&zc9Æ÷Þ{«d¿ÚM(µ×Lä ¹ÜË÷îHù(yXüuýS I¸nä¤[~òÓ¬óO 2~Ø;m·P\}tÔà¢søøË=7äÅ¸!b¼28æ¥QC\_ùÑßßOÞ\´Æ'ðsëªËELPªÃe[?}öÍÇnóRÿ¡écûL;:ó×\_->m6,µ°Ìdö{>Ú#¨¯KuÉî>M°zk6ñu(ToèWROW¶bìÅ:²Ö²ÌìB#nÊÿ×#w 6 zBZÌF3u)}GÞþÌùr,¨<ûþU-4}õ)Ü\2öqAKKKù6aRorïôéÓ\_ýõÃ·Ûd¼ ¯FE.ì÷gRe,lßOÔÿõ½aÄ´ôàÒC1ibÞô»ÌÝkg+±vÚhn 5®/OøÇ\_B 51S0hÍÁfÁýÈ°jÓVÞ)·ÔÑî¦\Ò/:ÊÏìã>QoD ¤×::÷£1ÆíÉ¯sg¶¹çíz¥K ¥vx.]ÊË$'''$$2%OfÏ½råÊ s´£Ô×¸[Rùô yêÎ\|ö\¶öí2 32(®ÆN_KÏRRo[¥½(ÍuåyÇnìÓXµbLZ×È?úçñ ÌñM¦ïµ«ñÛëwj/¾ < .?oy1kÏ²wÅÒYaRÏG)2MW[°p~Å¼øiHã6ÏÈ>tç¢»wà÷¸ä ÔuÃ¨êx÷«[ºeRßØHSi I3EgÅeö 4ê?u[ª¼¼O\| I¾à m¼²þõý&jÃPpªÖª·¤eERÃP\|hDÚ¬°Ç¸]ÅE&/5á#/\|ðëSM1Éæ0k°ÙjHÔ%Dk§Dã£¢' ~ð½jú5<Æ»ÿjOUñ+³áßáÀ«Í'ö}iÖÐ~L¢µ!çl-Í^ÇC¹4¥¶Nb¸ª½!\|¥]¤3Yx ÕÝ-"æ¥hF#Ç7zZÊA(Õ)¸ÙLäÉòåËwíÚUWW×áµ"/öJeû¢JìPØ ]+øÕÈþ%i{5ôtRÎf3\%@Ð èqu m³ûÄU;µ ýi-U}p,1fä=Y ³ÐEÿçvfX°« »cowåÅe2iLat¦ñOÑ#m v¢ùê÷ì[�¥^òúpX²d q¼³M%''gÍ5äu8^ø¥Ê¸5Êgfs©Ù·y© }4 ¾«´¡M.Ò9¥rD:«©ÊRcñH4},FíPTÞ¢ÑhÄÿùÓSj x?¡.@©]U2æò#¼¦Á vïÞ]VV/¨å·Ê:M\aÁ«±RÿÔ·ð^?Èðôü÷Î°U(·@B§n¯gIbwb#\fü\|æÜþÒé &Ï ýoè8, ºþ¸a2A½óñïþèÎdö 2'é&ê¢ûL6ÇÅbSµA(dæÑÚNSJæÚ9¥ÇíÌôÏ¼5ÓbG9!Ó¢C'õÕ7>4xªÁi0þ{°&ù¾ÇyÖ°<²ë\|ìo è3EÌÐÁQýbúG¾Ù/>::Þ¦ Né;ñöý«ÝfÃfjU¹çìæ¯êv!°P±mÝJÒi^CÆýXqHQ£ìò¥^ì×ÝË[ãßRüù[3¢5(úzChÜ=·Ð"ØwnÿbXLp0ñ1LÑHßï÷/fÔò·ÑïRýRçÎ»aÃk]üõ$JdêÖA\|ñÎÌ bÁÎ(cÐðuëwtbO>°mÝua(óFÒnÿ0af Ä»D;P ûº°½nóò¹ýÜ35¤1ðÉÆÚf"9UÛùÿBÌÁH\|÷ 0ÔgÐ¡zWwiçÖ[~óÐX7ÈZbCûÙ±¿~²«Oÿßîê£ÑèÈ§jbÝòMõ&À�¥öh ðÞ{ï¥§§GßRª"{§4é¸¤ð\_±rÓÊy¡,5>5a(è¶ÃþÅ»¤3·XÆg±6ÿó÷¿¦):P¤ë;ð¦å©&>Ö½5×v¼ÈýæÚ¯l ö\_@©þ(¾¹N-Z´yóf§¼¬JøéêËª¼U ì -WÿÍÝ7Z(r¬4þyÎ{y¤P+N±·m ±PåÆ¼îe+Aáç¿û¡U+9sì÷?¦L'5 Aw¯óxÖ=8û(5°)Õí"&Lpfß·oÞÒ7Mj¼5û8Îsáêý«3mðð(ØjËyCúï÷©Mu²½»³µ Äv+wT/ºq©5NÑ¼4Ï«)ù ØVOÎÝ0"9.(Á¡)·Ì<¼µY[dµVÁE¯nùÛÈÉ(,U£M°DµûÍ2\éÄ§lë',~dÔÔà¨É:K\|lÌ¤;wºjÊÝh¥ö¨Å'þI© ÍvTÄÎüt¤5ÜÖ ZÓÐÑÙ5JU¼EHiÎ¿!Ò.qÈ¹ïÑsÕ;.ÏÖ¾Vå ÕxüÈ2lw!m^¬3ï®RýR+++yÝ_0é°Í÷DJ¸êÈíýy.éSvWTªÄ·F£>tÚ¢Æì)§M×ïÔ.T«¸¾pì�Kw´Á§óéºæf»Sv4b© å«æ&ÅpÞ9éÏº³MTì(,ÚýI¯<ì®ÿ¹´79Y÷G ¥©öÿûÁ=ý AfM 2 ZµëDæ¹&@©=u öìÆBiÃAìÅÛÃðBþN©4UµÓüyg³P_ñïø¿Fkiô_ÃtQ(hlÅFUõJ=n¾6D®o.?íÎI»åGOÔ³¥"´nÀjÕÑÅqZ%-AQé}÷Jõ£=p{¦¯ÞC(U¤Ó4Û3Ëµó£¹ÑîÔôw¿ïl¥Ò ÏhÒí.>Æb¥0þ/?¡Á%S ¿s_ý>E>»ð-ëW.pOê"·ªªRÜ>ëRË(:ýR[kÅ%4)¸YVëU©îßéIÁ¨öìü5ß°Èw¸8ú29¾Ýj SCd@üåä¬VÅÔÝóDÄæÑq,Zk½ïgÏ9ºÖD¹pPèøcåÍC5KÆN2ôÉ Ñ¥ 5ã'1Ý-FrKÁuß=÷%E O7$Þpýô?ÐÅZXäÍUòDð$ÒË2 )ªó´ãðÐÄ(¤Eñnû«2#áf,InÆ¥4zpÒ²ûû§ê,iqÛ¢Ø+b\°uGê ,1Ó¢BR®lÕê:zÍ[ ÕÒ¦%Ç¥éL9#nöÊQmØ³Ké)ó=þI©ÎÌü_¬× +I¤§BÚÁù6¡¥ Ê¼fM°k&üñAæí¡ÐëK[psk^ +©^i.wëu #ú}¤Ôi·SÇ\[ó³RR»©èÔ(QgÙ®ÊÿüÍÒÐv{®¥WµvD\|º»ùáCè¨Þ/jÔõª»ÐèRU'vm\|omêF# ýú"÷ëªä®!7bñlcwÐ%H¦Ø¿RÀÖQzµÿË=ÃchþbpXÄÍNj!Kb%funüöÞ>²üuÞsì½É)ãï§}¥v¥æååõ\|Ju§øª¬JRýúgúÓaÀèdq×:êô·TaÈò..M©ª"9éÖvõGSÍ4è¦_N¬ac+~&ªÜ5í¸d_º OÌ}M[TÓaM Ô�¥TþOB©d¤îéªzCÕ©¦êÓ¤Íu»Ç!¬eÒ èR/ê15öá,£;>òÃcEeÞ,vHÎ¬6,IúS8O7,¨PZ÷%T1PèÚË+_ÿú^DAifMüÈ%¥ç¹Gä3qBzùÄUÜ8UgNrÿNGU¥B$^¶O¾ë¾]bÈèÒ²¿x÷ðïÿãñ_,hÀõ4ÛGôPªÚñÕÛÿ68ÉNMûå¸±½KqÉæM9£\_óRÿ¸îúÔvÄy!3¯½xrüwE§jLÓ\_3é�ÆÌe¤^¬¥^Oi·>ùp?ZÌë¡CGßWX´cóûl6/éÔvoAN±R½äD«îÜ8#ºüñ¥F©µ]¬ðz3³\_7±U«17(µµ¸Ó¥Ö-¥0¥\_DT F+uy7ÄQÜÓGßó«§\ï¯ºëË§O å'ÏV NþCQî¬«^no¨ùvïþíï¾¾VÙát5·:¤lþæïÞI³býjJ>÷%X53ÉVPwÏîík¿ØF9ÙÁúSbD OÌrºu´ÖÒ_¶¶MJªô}µO ÔîpÌxÆo¦Tw/oÞRñ§DÒ¤2 qÇO+ ¶y+OÐ]Åµ~ìB,ÕEÕç¶yL9oo©é2PábsÛÌÇ©;õüÃ·±jØ7þü/g ÔÞE©D¤óñJe¿;ñ ¥/>Äòm4È¢½¶äml¯ýÁ(6ÃóÌkA©î¸T?åÿ~ÎÞ¢o¬Ø°vä$÷";¶<Acëö¸¦©! ê.IÒ ÿ¹þ~ã>%rÀÿ99wªî¤àW÷ß3YÃ_DáÖÓÔrnÎgy£ ýôK½IÀçªOm9¹ãõÏ>:P^Ô%fÛ~qDQk1>8ÿ? H12ã¢2~¹××{§jû½[¾Ù¸à³·=3;fï\|vÎÈØ$ChæMÃ²'íÃ¸SªÊó\|zDäêÿãÃ´¢©)øí¿¨±CV×®Fg¼C~èùzr»D©ªÌÚTÔRr8NÌY1vÞG]ôÄq»Ô¦Kùä²³_öàÙmyåuÌßT((õZJTxÍoKPL4"sÜë¶µUÃ¡ð[n®øfWmÞz¤Úz¼¼Ö×ëè¬gp5®WÉ×4 N¢äò,ñ'¯Õ×ÛB [tH6ì×ÿüï8 ªØB3Y¤:74+^é6.7¥)Û·®~@h YbSßÝPVCo(5À(õôéÓþB©¨²öY9ñ?fëWn¡ÝY²gíVì;eüºv\|þf¸õ}Äà£åbµ»ê)§Tý~W vÝ²ÄJÓQÌmGlîÕ\|@©½Rùxçã'±TÌ7Z(9AW}ø¿¿?ec³¶ûÌ¼n^ß§^ý ùáj(Ö«ÇÎêögYm1¿x>½^Á®ÖMØzþ{T«g Ý?g¯9ÏçH½JRA×pQ¡L8þÿþ=4f2f 6û×ÌÝq!¡0Ipi9þôg1)´]1ÿ Ë°ýÿì\|TUÖÀßÌdJ&½Ó{Q, ØVÝÕu]u»º»®®ºkÙÏtHO&$U¤(Ø¨ ¨H7�$¤>ÉL¦¼z¿[ÞL&! Épþ¿1ÉäÍ»ïÿ½çKçÀºrPô=3D¯ßïbA$Þ{hXl(Ú¡ã´:ì©Aoí<?Ð^F/ÅÄ×R½\ýÆRÉH]9yúà:R:LÈc¾Ø}HT£rb·´\Ù]u<'LÌ Y±õh0ÅóÇíl ATx^Q×MÈAr±ýÙë w\¬7rú .0ëgèÚRIpõ°dÄ ÕAÓðv2 (ºw¬þÄÀ9#ØçhTK-¶Y°T¿²T\|$>\|êÔ©Îú["O-×îæ35uÍê]4Ä ÜÊ¶}6ÎÚDp>s©®¤ÖIÜµfa É 6\_SÑLÒ-s©n²WäDBá¯æ¢±\øµ9»s\_ ñW«Õ#üù>}NøKo/ÞÏógWXêJO<ûÿ>Q\àf4Îþ&æ¹yãÃb©dï'Å^vMXp §ç´1/üw©s"Ï=Ö£ðÍ¨îèm&.b¦¨&¿\_FG~Ùã¥÷éH¨oBÇî58(-K¾î×&AäÏ:ákíèëGÒc¢S Ú¤#ß°¦\ªêöÙXnõÌ6sm ÛPAÐ£§ÜÞo²!$-44íºg¤-Nj1^²q!] C3ÚwÉ¸ÿ©kBÇ·[ßÿeV·î=754ññgh;I3nHXêeÀÅ.²mëçó~º¥ÒÿýØÀ$2è¤\j©í$¶ÍéÒÖ%gÓYs p÷¶ ¯¯zjÀ¨ q\TÝ;ý×Gè Ò.Fb7RþÞX4ÖP¨tÓ{?ÎýÛÌ_]¯!u2 Yúd¹úDVùÄ&ÑeX©\|rx.~íY\]¬øLÃR%Oíþ\|µÔf©Nß}·=1?K-Ü¥¥s©Ú ßúªUy°T°Ô°ÛÚùåü³-µ 8>¿GK5Züy©ZØÿÜWBºÁ¥îÄÛRR¶\|5/@Çip$@f¤XYÄ^Y¨ Ze!ÓúÃ>eÈ]óÒ¯FÒ$ëX§á züvå÷¹5¬²¬n¼')©þuÛl¶.b©-¿sËø'Éÿt\|ûçÓI9m8g¾å´ÕÛs\_¥¥=1ÞÙ¬¡ Qçb§º<çqÿAê ;¿\|CG7vÍjõ%.°Ô«¬s)((èBê½fú´%Îý63ûSæRZÊÞÁÌb\_±Ôå©~êµTÉ[ì¼'#©ùaXÊÖ(9$°T #ED¬=TöeÜë Yj½[þúoÉQÁ8múÞãö#;¼Íö-Uõg;'2#ê®9\²ñWuÜ#d¢14M<3eÆÎøõ6d«eu¶t¶¥&Rüù§c¼Ö#-6 Þ°p&eàðw_ßû9är°D0Ö=ÁºÔK;3x÷&m_ÀRö>}OÆ¯Gèôw+§²\_cìú£NßK¯¥"WÙÎhbxfîÇElÇjÁ ÚÑªàó¶ü÷²Iü¬1s¦ÈóX²UÔcÇät¦ÿÌ~¯®\|é)ÏÔ¬ZnBÕHÊKxé>OMmÏÍ¡éqÇ{?ÕÉd(ÛE-U"ûlx Ó½lg~,Yª×Ó=7úÝùmõ<>¹Uº/XªXêaJW±Tß°yçRù#[>ÉÖ©6#ZÖ¥^Ô\©zmyît]°N?ät=I¥je©7N5\|Û.3 ø&¯,õ´Tf1ßRÙFÉÞQÝ~çúß-UÆ½G}]A]n³TIiÏRGÓ¾ÌØ<0._K®¤¥¬ÿ¾Bøy,ªÄ'éî1oÇ0åî)£rhë ~q^Kuxç H%ÇGkgõz±OtJ¸è$cÄ,.Ã:¥ïT!¶vUÁíÅ¥ð¾¢JÞÝyè¹íÿr91RªI áRÌºä©NÜ·ærYÙïI{6Yì,¦vé¹TÙ[ËT±o_=?ÌÔj N5!ìÖÕdY(Ä¶ ñ=ÍÇOnûØ×RùsYªû4¶TÉÔ?;ïÖ9»<°T¿¶TÚ"ÇÑümïØü)rãÁç´TWÍà>ã_]øiZÛÃRq¸wák¡i=a;rXýfÆþ}ùó²ö"ïZ=UÖ±së¯M½(ä\|gå¹gÇ&¿¿òÃ¬ÄWcMZòìûÅæ¬nºR9XêUj©çèVè0 É0¤¯¥úVO¢] xKÅLýòÔz¾)Ê\|CM3[Í¥3¿¤,wµAÃôþtçI°T°ÔNÞÀ}H\|cK\~0DNW<ÿÆG?¡Æ/npb]}á2þ©7HlìüwêÖ¿õH(MûÓZ 8°÷¯\_?®¥WÒROÔ\|ÛwRhK%ïtø]ÄÑÀS]ØÅ,U1´wÆ¸Ý,nzY·:\uQBcîî ¿zñw©\_ÏHÿ½cö7E[ ¦Éá×Ý0ýÚ¥6K.KÅ]õ¥',ÿÿfú¨õFÇ&pºTN6Ud?Aõ\¼"¹dØ¦³XªGò©-ïÏT3~MqßìÉoPeÅ×ROîÒÒY$ÎôÜüO+<å¹:ó~©©þi©¢)Åß¾?UÏö#Îc©µ¹ä¼¥Ãk>«QÏ~Fg·qBÛdç=ûdà #Yxê¬ò\Tõæüï~Å)f}~9QZÞ¹èÕNr:if±è@¼'èëQcþ'ïUW§s77"Ëå³,,µÓXª¿í³it<&L¤xô!Çÿ0240ñÚFdSZ[qïûæM)åz\|´í«e×c°T°ÔÎo© D·Uþ¯"±þÔ¡hbà ÄRígYÙæÛeEMùíN7¸¿æññ:q t¬¥æ¬zU{K¥Ä§ó÷OeöÊ¥ºëJ¦UY3rÙï°"©5»P-ÚpÍ¤àSÌA8nÒÝ¿\B6fRxY¨¥JDX%ÁF ùJu¨þÚ3jÃÃ½R9ã \|gvþºÖLi¡2¥sô<~b©È¾uÍ,Õ+ª ¥çZFf¹Oïüt^-7wûvï±¢[[jEÞ&Oí Ç²ß>Ìæë ÚÑ«¡Ò-ÎºK=uhzÞÃþ=ýíu¿i5º åd¢m/,ýßIf\|D\èÈL¬ É+æÅFqq·üöIgÜVBrÊÈ¨ÜNÙ"ÁÐ¤w \_üÝM¤ÁÆp¦þIo~¦Æê©©×RC.ÎR½8ýÊ×$>y\_wKQGvÈÿýGÓÈ\\¶Ô A\_í;ín±T±sö/©©b©mÇ\_HX\\_UCç >vôwmKÕM"¿b3ªË»ÛL+ k=¤Î¥¥WÈRWÓgb±Ô[Ïm©MnùëG#4º¡ý³&æ\ÀRÏ+õA\_I&SÍ¤)@çgk7¿Ùg¢>(ÕÈMtë1¹)«+Y¢JíôSÊÊñýsác¸ÀÓZ)ÃmXé\o~©òÅXªÒæªèYJ®i.$Õü6]é\ªì9�te+ùJ®À»±dOT�§!±yÖÆ<Ü7W¹y°T°Ô+ðùÊÍH(.?¸fÏj9±Tº[j=à{ú¢CÑ4ÀºpS^YÆ¯-Y(ïµTÅ§ÉÐ{$µô? ¤ùfäjF¼+¹6V#¡ä^q$s7\:¬¨Ê%©¯¦«ZI¾0ÝÒ&Õ\È%²Õ¢9KÜÇ¶ÅáÀ]ÃqÝúßûWIµTÈøKíªê3jªøãîmïÏ¡©fé²FÔDoÈîÀøWDÅæ¤C@ù?\|4¬KÅqÓ®ÙÕ¢¨©©]ÒRÏ¹.Uñíd¼RÏs«(D",mìèùÚÏ³fP"JëbÁê4ê /x~o¦f$¹ÙD·QþÓïÝÓmAbLðNîlIOkæÈê«È$),¿ ÷>$ÏvüiÁÀ°xNÐ#lìíEÈeólÞIZßXêöÕó#/RÛ²)lÍþ¹27DMAÞ[$û~DÙqÕg®:¹[ªA6zÝ kÈ]v°T°Ô¤H,ËséúN²ÏïNÚý)¢¤Æä¼iÕÜÜ¡l!¨.bu^y%½jÉ" 1KU\|,@ÖÏ±-%µÐºJ¤&�¾üÖÖüÔ§&¯Ã¬8P ËYñ6!Ù\¸Ù5XPÄFRIæÕ¾9÷÷¼6Ã¶ßõYD[ \|ÚNçRéØ'y8IÁBÈiCM¹;>ÊÐ³5æ!w®e©À¼Z ,,µ+ ï GÕ§;ÑËR[)Ùæej¾zP8DkbGÏüÜ¥2]@û&¹x»#÷t×Gé^Êþíó(Ò©+°T c?]\|Þ¹øC÷¦÷K Õ%\|dM¶GX.X¹$µ®¤È#ÙÊÐ,á)1±×_ÞW¨VÁ±Ãè[YªRQR-mÛç Éj²gßmçQÉÒÜqq£¹À)7Ê¼ôÈ7VoGD³GëÝ.vVÅNgå< MD;q0xìµe7ôÃEgôsS»Öál�Kí8K<¢ê[:Ëû=)öRl'Åjèö[-Ô×Ôh<ÖýÍ§ "È\@ïÜZ©È¿`©}aÄgµ5Èz\|D2º>q¦ OêìA~~~~~~~~
13319	https://www.inchcalculator.com/pentagon-calculator/	Inch Calculator Skip to Content Popular Searches calculate body fat what is today's date? calculate yards of concrete how long until 3:30? calculate a pay raise 30 minute timer calculate board and batten wall layout how to do long division calculate the best TV size how many days until Christmas? Pentagon Calculator Calculate any part of a pentagon using the pentagon calculator below. Have a Question or Feedback? Solution: \| \| \| --- \| \| Side (a): \| 3 \| \| Area (A): \| 15.484 \| \| Perimeter (P): \| 15 \| \| Diagonal (d): \| 4.854 \| \| Height (h): \| 4.617 \| \| Circumradius (R): \| 2.552 \| \| Inradius (r): \| 2.065 \| Learn how we calculate this below Add this calculator to your site On this page: Calculator How to Calculate the Properties of a Pentagon How to Calculate the Area of a Pentagon How to Calculate the Perimeter of a Pentagon How to Calculate the Diagonal of a Pentagon How to Calculate the Height of a Pentagon How to Calculate the Circumradius and Inradius of a Pentagon By Joe Sexton Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications. Full bio Reviewed by Pateakia Heath, PhD Pateakia has worked in education for 15 years and has three degrees, including a PhD, Master's degree, and Bachelor's degree. She specializes in mathematics. Full bio Share on Facebook Share on X Share on Pinterest Cite This Page Sexton, J. (n.d.). Pentagon Calculator. Inch Calculator. Retrieved September 22, 2025, from How to Calculate the Properties of a Pentagon A pentagon is a two-dimensional geometric shape with five straight sides. The properties of a pentagon include the length of the sides a, the area A, the perimeter P, the diagonal d, the height h, the radius (circumradius) R, and apothem (inradius) r. A pentagon is a polygon with five sides and five angles. When all its sides and angles are equal, it’s referred to as a ‘regular’ pentagon. A regular pentagon is symmetric, with all its internal angles measuring 108 degrees. It can be divided into five isosceles triangles of equal size. How to Calculate the Area of a Pentagon To find the area of a regular pentagon, you can use the pentagon area formula: A = \frac{1}{4} \sqrt{5 \left ( 5 + 2 \sqrt{5} \right ) } \times a^{2} The area A of a pentagon is equal to 1/4 times the square root of 5 times 5 plus 2 times the square root of 5, times the length of the sides a squared. How to Calculate the Perimeter of a Pentagon Since a regular pentagon has five equal sides, to find the perimeter, you can multiply the length of the sides by 5. P = 5a The perimeter P of a pentagon is equal to 5 times the length of the sides a. How to Calculate the Diagonal of a Pentagon In a regular pentagon, the diagonal connects two non-adjacent vertices. You can calculate the length of the diagonal using a formula: d = \frac{1 + \sqrt{5}}{2} \times a The diagonal d of a pentagon is equal to 1 plus the square root of 5, divided by 2, times the side length a. How to Calculate the Height of a Pentagon In a regular pentagon, the height extends from the center of one of the sides to the opposite vertex. You can calculate the height of the pentagon using a formula: h = \frac{\sqrt{5 + 2 \sqrt{5}}}{2} \times a The height h is equal to the square root of 5 plus 2 times the square root of 5, divided by 2, times the side length a. How to Calculate the Circumradius and Inradius of a Pentagon Just like a hexagon or an octagon, a pentagon has a circumradius and an inradius. The circumradius of a pentagon is the radius of the circumscribed circle that passes through all the vertices of the pentagon. The circumradius is often referred to as just the radius. The inradius, or the apothem, of a pentagon is the perpendicular distance from the center of the pentagon to one of its sides. In a regular pentagon, this is also the height of one of the isosceles triangles when split. You can calculate the circumradius and apothem using a few formulas. Circumradius (Radius) The formula to calculate the circumradius of a pentagon is: R = \frac{\sqrt{50 + 10 \sqrt{5}}}{10} \times a The circumradius R is equal to the square root of 50 plus 10 times the square root of 5, divided by 10, times the side length a. Inradius (Apothem) The formula to calculate the apothem of a pentagon is: r = \frac{\sqrt{25 + 10 \sqrt{5}}}{10} \times a The apothem r is equal to the square root of 25 plus 10 times the square root of 5, divided by 10, times the side length a. Similar Geometry Calculators Slope Calculator Volume Calculator - Find Volume Area Calculator Angle Conversion Calculators Area Conversion Calculators See All
13320	https://doctorabad.com/uptodate/d/topic.htm?path=therapeutic-apheresis-plasma-exchange-or-cytapheresis-complications	Therapeutic apheresis (plasma exchange or cytapheresis): Complications Therapeutic apheresis (plasma exchange or cytapheresis): Complications Drug Interactions Calculators PCUs What's New Patient Education Contents Therapeutic apheresis (plasma exchange or cytapheresis): Complications Topic Outline SUMMARY & RECOMMENDATIONS INTRODUCTION OVERVIEW OF COMPLICATIONS ANY REPLACEMENT FLUID Citrate-induced hypocalcemia Citrate-induced metabolic alkalosis Inadvertent removal of a medication Vascular catheter complications NON-PLASMA REPLACEMENT FLUIDS Hypokalemia Hypocalcemia Coagulation factor depletion Immunoglobulin depletion Angiotensin converting enzyme (ACE) inhibitor-related complications DONOR PLASMA OR RED BLOOD CELL EXPOSURE Anaphylactic reactions Hives Transfusion-related acute lung injury (TRALI) Infectious risks of plasma or RBC EVALUATION FOR SUSPECTED COMPLICATIONS MORTALITY SUMMARY AND RECOMMENDATIONS REFERENCES GRAPHICS View All ALGORITHMS - Initial approach transfusion reaction TABLES - Infection risk of blood products - Transfusion reactions findings RELATED TOPICS Approach to the patient with a suspected acute transfusion reaction Blood donor screening: Laboratory testing Clinical manifestations and treatment of hypokalemia in adults Clinical manifestations of hypocalcemia Complications of central venous catheters and their prevention Definition, classification, etiology, and pathophysiology of shock in adults Immunologic transfusion reactions Initial evaluation and management of suspected acute coronary syndrome (myocardial infarction, unstable angina) in the emergency department Major side effects of angiotensin-converting enzyme inhibitors and angiotensin II receptor blockers Overview of intravenous immune globulin (IVIG) therapy Reflex syncope in adults and adolescents: Clinical presentation and diagnostic evaluation Selective IgA deficiency: Clinical manifestations, pathophysiology, and diagnosis Selective IgA deficiency: Management and prognosis Therapeutic apheresis (plasma exchange or cytapheresis): Indications and technology Transfusion-related acute lung injury (TRALI) Therapeutic apheresis (plasma exchange or cytapheresis): Complications Authors:Andre A Kaplan, MDJoy L Fridey, MDSection Editor:Arthur J Silvergleid, MDDeputy Editor:Jennifer S Tirnauer, MD Contributor Disclosures All topics are updated as new evidence becomes available and our peer review process is complete. Literature review current through:Feb 2018.\|This topic last updated:Apr 27, 2017. INTRODUCTION—Therapeutic apheresis is an extracorporeal treatment that separates blood components (plasma and/or cellular components) from the patient's blood for the treatment of conditions in which a pathogenic substance in the blood is causing morbidity. Therapeutic plasma exchange (TPE) denotes the selective removal of a patient’s plasma and replacement with another fluid; cytapheresis refers to selective removal of abnormal or excessive numbers of blood cells. This topic review will discuss the complications of therapeutic apheresis. An overview of the terminology used to describe apheresis procedures; the types of indications for which therapeutic apheresis is effective, including American Society for Apheresis (ASFA) therapeutic categories; and practical issues in apheresis techniques are discussed separately. (See "Therapeutic apheresis (plasma exchange or cytapheresis): Indications and technology".) An overview to the patient with a suspected acute transfusion reaction is also presented separately. (See "Approach to the patient with a suspected acute transfusion reaction".) OVERVIEW OF COMPLICATIONS—The basic premise of therapeutic apheresis is that removal of certain pathologic substances (or cells, in cytapheresis) will reduce organ or tissue damage and may permit reversal of a pathologic process. In order to prevent volume depletion during therapeutic plasmapheresis, the volume of plasma removed must be replaced by plasma, colloid, or crystalloid. In some cases, use of allogeneic (donor) plasma is preferred because it provides needed proteins or other factors; however, donor plasma should be avoided if possible, other than when it is indicated as replacement fluid, such as for thrombotic thrombocytopenic purpura, in which it provides needed proteins or other factors. (See "Therapeutic apheresis (plasma exchange or cytapheresis): Indications and technology", section on 'Replacement fluids'.) The frequency and types of complications from therapeutic apheresis depends upon the overall condition of the patient, the number of procedures, the replacement fluid, and the venous access device. A review of the reported complications from over 15,000 therapeutic plasma exchange (TPE) treatments found that adverse reactions were substantially more common with plasma than with albumin replacement (20 versus 1.4 percent) . Patients undergoing therapeutic cytapheresis will only have selected cells removed. For erythrocytapheresis, red blood cells (RBC) are used as replacement. Most other types of cytapheresis require minimal or no replacement, depending on the volume of cells removed and the patient's ability to tolerate the procedure. For all therapeutic apheresis, patients may be at risk for citrate-induced hypocalcemia, metabolic alkalosis, or complications related to the vascular catheter. When blood products (either plasma or RBCs) are used as replacement, the risk for citrate-related complications is higher due to citrate in blood components. In addition, patients who receive blood products as replacement are at risk for transfusion reactions and transfusion transmitted diseases. For therapeutic plasma exchange, when albumin and/or crystalloid is used as the replacement fluid, patients may also be at risk for depletion of coagulation factors or immunoglobulins. ●Any replacement fluid – A common issue is citrate-induced symptoms of hypocalcemia, due to binding of ionized (free) calcium by citrate, which is used as an anticoagulant during the procedure. Symptoms include paresthesias, nausea and vomiting, muscle cramps, chest pain, hypotension, and, in the extreme, tetany or arrhythmias such as QT prolongation. (See 'Any replacement fluid' below.) ●Non-plasma replacementfluids for plasma exchange– Non-plasma replacement fluids may cause hypokalemia, depletion of coagulation factors or immunoglobulins, depletion of potassium, or hypotension if the patient is taking an angiotensin converting enzyme (ACE) inhibitor. (See 'Non-plasma replacement fluids' below.) ●Donor plasma or RBC exposure – Exposure to donor plasma or RBCs may cause more serious complications, such as severe anaphylactic reactions or transfusion-related acute lung injury (TRALI), and may increase the risk of transfusion-transmitted disease . (See 'Donor plasma or red blood cell exposure' below.) Any change in the patient's status (eg, dyspnea, seizures, chest pain, hypotension not responsive to one or two fluid boluses) should prompt discontinuation of the TPE procedure and evaluation for the cause of the symptoms. (See 'Evaluation for suspected complications' below.) The overall incidence of death from TPE is 0.03 to 0.05 percent [1,2]. (See 'Mortality' below.) ANY REPLACEMENT FLUID Citrate-induced hypocalcemia—Generally, a solution of sodium citrate is infused either as the anticoagulant for the extracorporeal system. Citrate binds to ionized calcium to form soluble calcium citrate, thereby lowering the ionized (and possibly total) serum calcium concentrations. In patients with normal liver function, citrate is generally metabolized within approximately 1.5 hours. However, because of the duration of most therapeutic plasma exchange (TPE) or RBC exchange procedures, symptoms of citrate toxicity frequently occur. Potential early symptoms of hypocalcemia include perioral and distal extremity paresthesias or numbness. Severe reactions can include tetany, prolongation of the QT interval, arrhythmias, or hypotension. (See "Clinical manifestations of hypocalcemia", section on 'Acute manifestations'.) Electrocardiogram monitoring and intra-procedural evaluation of ionized calcium levels may be appropriate in patients who are receiving intravenous calcium infusions, become symptomatic, or are at risk of significant citrate sensitivity, including patients with altered mental status or pediatric patients . Strategies to reduce citrate-induced hypocalcemia include the following: ●Slowing the rate of exchange, which may cause resolution of symptoms. ●A successful prophylactic or treatment regimen consists of infusing one 10 mL ampule of 10 percent calcium chloride intravenously over 15 to 30 minutes, beginning 15 minutes after the start of TPE [1,4]. Calcium infusion can be repeated if the therapy will last substantially longer than an hour. •In one of the authors' experience (AAK), this regimen reduced the incidence of symptomatic hypocalcemia from 9.1 percent to 1 percent of treatments . •The other author (JLF) generally uses oral calcium supplements, reserving intravenous calcium infusion therapy for patients who cannot take oral calcium or the occasional patient who develops severe hypocalcemia or symptoms that do not improve with oral calcium. Five to 10 cc of 10 percent calcium gluconate administered intravenously over 10 to 15 minutes is effective in reducing symptoms when oral calcium is not effective; muscle contractions can develop if calcium chloride or calcium gluconate is administered too rapidly . ●As an alternative to the above approaches, intravenous calcium gluconate, administered as a continuous infusion of 10 mL of 10 percent calcium gluconate per 1000 mL, has been shown to prevent citrate toxicity . Citrate-induced metabolic alkalosis—Another complication of citrate administration is the development of metabolic alkalosis in patients with concurrent renal failure such as in anti-glomerular basement membrane (anti-GBM]) antibody disease (Goodpasture's disease) . Metabolism of the excess citrate generates bicarbonate, the excretion of which is limited by the renal failure. Citrate-induced metabolic alkalosis in individuals with renal insufficiency can be managed by hemodialysis. Patients with renal failure requiring therapeutic apheresis may require hemodialysis following the procedure to restore acid-base, electrolyte, and fluid balance as needed. Inadvertent removal of a medication—Substantial drug removal by TPE should be anticipated for medications that are highly protein-bound and have a small volume of distribution and are therefore primarily limited to the vascular space, and for therapeutic antibodies. ●Prednisone and prednisolone are not substantially removed, whereas cyclophosphamide and azathioprine are removed to some extent . ●Therapeutic antibodies such as rituximab and eculizumab are generally removed. This potential problem can be minimized by administering the drug after the exchange. ●Total parenteral nutrition (TPN) solutions should be infused after the exchange. Vascular catheter complications—Potential complications of central venous catheters include infection, pain, nerve damage, thrombosis, perforation, dissecting hematomas, air embolism, or arteriovenous fistulas. Many of these complications can also occur if peripheral veins are used. Confirmation of the proper placement of the vascular access device is standard practice intended to minimize some of these complications and is also required by the AABB cellular therapy standards . (See "Complications of central venous catheters and their prevention".) NON-PLASMA REPLACEMENT FLUIDS Hypokalemia—Non-plasma replacement fluids can induce a 25 percent reduction in the plasma potassium concentration in the post-apheresis period, due to dilution . If potassium depletion is suspected, the potassium level should be checked immediately post-procedure and deficiency treated if necessary. (See "Clinical manifestations and treatment of hypokalemia in adults".) Hypocalcemia—Infusion of large volumes of albumin may play a role in the development of hypocalcemia. This was shown in a study that randomly assigned 32 patients to receive 10 percent pentastarch (a type of hydroxyethyl starch [HES]) versus 5 percent albumin during the first half of an apheresis procedure . The risk of hypocalcemia was lower with pentastarch than with albumin (hypocalcemic symptoms in 8 versus 33 percent). Coagulation factor depletion—Therapeutic plasma exchange (TPE) with albumin or other non-plasma replacement fluid produces a predictable decrease in blood coagulation factors that may predispose to bleeding. Immediately after a single plasma volume exchange, the prothrombin time (PT) may increase by 30 percent and the activated partial thromboplastin time (aPTT) may double; these changes tend to return towards normal within four hours [4,11]. However, more severe and long-lasting changes can be induced when multiple exchanges are performed over a short period (eg, three or more treatments per week) . If a patient is expected to require multiple TPE procedures, a baseline assessment of coagulation status by measuring the PT, aPTT, and/or fibrinogen level should be done prior to initiating TPE; this may be repeated daily or every other day thereafter. For patients at risk of bleeding (eg, immediately after renal biopsy for anti-GBM disease), one or more liters of plasma (three to four units of plasma per liter) may be substituted as the replacement fluid; the plasma should be administered at the end of the procedure. Immunoglobulin depletion—Repeated TPE with a non-plasma replacement fluid will predictably deplete the patient's immunoglobulins, resulting in low serum IgG levels. The need for monitoring serum IgG levels depends upon a number of factors, including how frequently the exchanges are performed, how many blood volumes are exchanged with each procedure, the immune status of the patient, and the patient's ability to produce immunoglobulins. Removal of immunoglobulins and complement may lead to an immunodeficient state, leaving the patient susceptible to infection. Reports of patients treated with TPE for glomerulonephritis suggested an increase in opportunistic infections; however, these patients also received immunosuppressive agents and were often granulocytopenic . In a trial that randomized 86 patients with lupus nephritis to TPE or immunosuppressive therapy, TPE was not associated with a greater risk of infection and patients were not more prone to infection . For a patient undergoing aggressive TPE (eg, two to three plasma volumes replaced per procedure and/or consecutive daily procedures) without plasma replacement, a baseline IgG level should be obtained [15,16]. For patients whose IgG level falls below 500 mg/dL who have a systemic infection, or for whom concerns about infection exist, we use a single infusion of intravenous immune globulin (IVIG, 100 to 400 mg/kg), although IgG replacement has not been tested in this setting. (See "Overview of intravenous immune globulin (IVIG) therapy".) Angiotensin converting enzyme (ACE) inhibitor-related complications—Rarely, symptoms resembling anaphylaxis, such as flushing, hypotension, abdominal cramping, and other gastrointestinal symptoms, have been reported during TPE in patients receiving ACE inhibitors. These symptoms have been observed when albumin was used as the replacement fluid and the patient had taken the medication within 24 to 30 hours of TPE [17,18]. In one report of 299 consecutive patients undergoing TPE, these symptoms occurred in all 14 patients receiving an ACE inhibitor versus only 7 percent of those not receiving an ACE inhibitor . The etiology of these symptoms is unclear; one possibility is increased kinin generation, which has been thought to account for the angioedema that can occur with ACE inhibitors. (See "Major side effects of angiotensin-converting enzyme inhibitors and angiotensin II receptor blockers".) Based on these observations, it has been recommended that ACE inhibitors be withheld for 24 hours prior to TPE if the patient can tolerate discontinuation. Management is supportive. If future TPE is required, a different lot of albumin should be used, as kinin concentrations vary among albumin lots. DONOR PLASMA OR RED BLOOD CELL EXPOSURE Anaphylactic reactions—Anaphylactic reactions to plasma are the most common cause of serious complications and death associated with therapeutic plasma exchange (TPE) using plasma as a replacement fluid [2,19]. These reactions have been reported in up to 21 percent of patients and are most often characterized by fever, rigors, urticaria, wheezing, and hypotension [20,21]. Cardiopulmonary collapse is rare. (See "Immunologic transfusion reactions", section on 'Anaphylactic transfusion reactions'.) Management or prevention of anaphylactic reactions may include antihistamines, epinephrine, glucocorticoids, or more aggressive measures, depending upon the severity of the symptoms. (See "Immunologic transfusion reactions", section on 'Treatment and prevention of anaphylactic reactions'.) Anaphylaxis may be due to the presence of anti-IgA antibodies in an IgA-deficient patient that react with IgA in the donor plasma. In cases of anaphylaxis due to anti-IgA, it is necessary to obtain plasma from IgA-deficient donors. This product is usually only available from a registry of rare donors, and it may become difficult to obtain enough units to perform multiple TPE procedures on a given patient. (See "Selective IgA deficiency: Clinical manifestations, pathophysiology, and diagnosis" and "Selective IgA deficiency: Management and prognosis".) If plasma from an IgA-deficient donor is unavailable, we try to use a non-plasma replacement fluid such as albumin. (See "Therapeutic apheresis (plasma exchange or cytapheresis): Indications and technology", section on 'Replacement fluids'.) In rare cases, one of the authors (AAK) has been able to use pretreatment with ephedrine, corticosteroids, and diphenhydramine in a patient with anaphylaxis to plasma and thus to successfully perform TPE for thrombotic thrombocytopenic purpura during two pregnancies . Hives—In contrast to anaphylactic reactions, allergic reactions (eg, development of hives) can often be treated by interrupting the infusion temporarily, giving diphenhydramine or steroids, if warranted, and monitoring the patient for worsening symptoms. The infusion can be restarted if additional symptoms are absent. Allergic reactions, such as hives, are thought to be due to preformed IgE antibodies in the recipient or donor that react with a substance in the donor or recipient, respectively (eg, donor peanut allergy, with hives triggered due to recent peanut ingestion by the recipient) [22,23]. (See "Immunologic transfusion reactions", section on 'Urticarial (allergic) reactions'.) Transfusion-related acute lung injury (TRALI)—TRALI is a form of acute lung injury usually caused by antibodies in donor plasma that react with neutrophil or HLA antigens in the recipient, leading to leukoagglutination in the pulmonary circulation and non-cardiogenic pulmonary edema. TRALI typically presents as sudden-onset of hypoxic respiratory insufficiency with an abnormal chest radiograph within the first six hours of blood product exposure. Treatment is supportive; in severe cases the patient may need ventilatory support. The pathogenesis, diagnosis and management of TRALI is discussed in detail separately. (See "Transfusion-related acute lung injury (TRALI)".) Infectious risks of plasma or RBC—Estimates of the per-unit risk of transfusion-acquired viral disease in the United States are shown in the table (table 1). (See "Blood donor screening: Laboratory testing".) EVALUATION FOR SUSPECTED COMPLICATIONS—Symptoms occurring during apheresis may be due to volume shifts, metabolic complications, or reactions to plasma or red blood cells (RBC). Blood pressure should be measured often during apheresis (eg, every 10 to 15 minutes depending on the hemodynamic stability of the patient). We consider the following in evaluating the patient who develops symptoms: ●Shortness of breath – Causes of shortness of breath include the following: •Transfusion-related acute lung injury, which is the most likely cause of dyspnea if plasma is used as the replacement fluid. (See "Transfusion-related acute lung injury (TRALI)".) •Dyspnea, which could suggest the presence of pulmonary edema due to fluid overload, although this is less common. •Bronchospasm, which can be due to an anaphylactic-like reaction to plasma or RBC •Air emboli caused by air bubbles in the line, or pulmonary emboli due to inadequately anticoagulated blood that contains clots, are exceptionally unlikely because sensors in cell separators can detect air bubbles and clots and would trigger alarms. Anticoagulant volumes are controlled by state-of-the-art cell separators. •Very rarely, dyspnea, wheezing, chest pain, and hypotension unresponsive to fluid boluses may be produced by complement-mediated membrane bioincompatibility or sensitivity to the ethylene oxide used as a membrane sterilant [24,25]. ●Hypotension – Causes of hypotension include the following: •Citrate-induced hypocalcemia can occur with any replacement fluid. (See 'Citrate-induced hypocalcemia' above.) •Decreased intravascular volume may occur due to extracorporeal volume during the procedure, which is generally around 200 mL, but depends on the configuration of the apheresis kit. The extracorporeal volume of each kit type can be found in the manufacturer's information. With continuous flow technology, the extracorporeal volume is usually no more than 15 percent of the patient's intravascular volume; however, instrumentation that utilizes discontinuous flow technology may have higher extracorporeal volumes. Infusing additional intravascular fluid, slowing the procedure, or increasing the return rate can return the blood pressure toward the baseline level if hypotension is related to fluid shifts. •An acute coronary syndrome may present with hypotension and chest pain. If the patient has chest pain, the procedure should be stopped and an electrocardiogram obtained to evaluate for an acute coronary syndrome. (See "Initial evaluation and management of suspected acute coronary syndrome (myocardial infarction, unstable angina) in the emergency department".) •Vasovagal reactions are not common; these may present as hypotension accompanied by a decrease in pulse rate, diaphoresis, and/or syncope. Lowering the patient's head, using ammonium salts, and stopping the procedure temporarily are the appropriate in this setting. Modest fluid boluses with normal saline can also improve the patient's symptoms. (See "Reflex syncope in adults and adolescents: Clinical presentation and diagnostic evaluation", section on 'Vasovagal syncope'.) •Additional evaluation for hypotension is discussed in more detail separately. (See "Definition, classification, etiology, and pathophysiology of shock in adults".) A general approach to the suspected acute transfusion reaction is presented in detail separately (algorithm 1 and table 2). (See "Approach to the patient with a suspected acute transfusion reaction".) MORTALITY—The reported case fatality rate for therapeutic plasma exchange (TPE) is 3 to 5 per 10,000 (0.03 to 0.05 percent) [1,2]. In more than 50 TPE-associated deaths since 1989, respiratory or cardiac complications were most common . Cardiac arrhythmias were frequently observed, especially in those receiving plasma; the suspected etiology was a reduced ionized calcium concentration but a cause-and-effect relationship has not been proven. Among respiratory deaths, acute respiratory distress and symptoms of noncardiogenic pulmonary edema were observed shortly before death; these patients were also being infused with plasma. Anaphylaxis, vascular complications, hepatitis, sepsis, and thrombosis were other, less common causes of death. SUMMARY AND RECOMMENDATIONS ●In order to prevent volume depletion during therapeutic apheresis, the removed plasma must be replaced by a replacement fluid, which may be allogeneic plasma, colloid, or crystalloid. The frequency and types of complications from therapeutic apheresis depend on the overall condition of the patient, the number of plasma exchanges, the replacement fluid, and the venous access device. Adverse reactions are more common when plasma is used as the replacement fluid. (See 'Overview of complications' above.) ●Citrate causes hypocalcemia by binding to ionized calcium. Citrate-induced hypocalcemia can occur with any replacement fluid in therapeutic plasma exchange (TPE) and red blood cell (RBC) exchange, and symptoms of citrate toxicity can occur frequently. Potential early symptoms include perioral and distal extremity paresthesias or numbness; severe reactions can include tetany, prolongation of the QT interval, arrhythmias, or hypotension. A variety of strategies exist to reduce this complication. (See 'Citrate-induced hypocalcemia' above.) ●Other complications that can occur with any replacement fluid include metabolic alkalosis, inadvertent removal of a medication, or complications related to the vascular access catheter. (See 'Citrate-induced metabolic alkalosis' above and 'Inadvertent removal of a medication' above and 'Vascular catheter complications' above.) ●Non-plasma replacement fluids may cause hypokalemia or depletion of coagulation factors or immunoglobulins. Baseline assessment of the PT, aPTT, and/or fibrinogen level should be done prior to initiating apheresis for patients expected to undergo multiple procedures, and a baseline IgG level should be obtained for those expected to undergo aggressive plasma exchange. (See 'Coagulation factor depletion' above and 'Immunoglobulin depletion' above.) ●Flushing, hypotension, abdominal cramping, and other gastrointestinal symptoms have been reported, although rarely, during TPE in patients receiving an angiotensin converting enzyme (ACE) inhibitor. To reduce this complication, it has been recommended that ACE inhibitors be withheld for 24 hours prior to TPE if the patient can tolerate a short-term interruption. (See 'Angiotensin converting enzyme (ACE) inhibitor-related complications' above.) ●Exposure to donor plasma or RBC may cause anaphylactic reactions, allergic reactions (eg, hives), transfusion-related acute lung injury (TRALI), or exposure to infectious pathogens (see 'Donor plasma or red blood cell exposure' above). A general approach to evaluating and managing an acute transfusion reaction is presented in detail separately (algorithm 1 and table 2). (See "Approach to the patient with a suspected acute transfusion reaction".) ●Blood pressure should be measured often during therapeutic apheresis; hypotension and/or symptoms of shortness of breath or hypotension should be evaluated promptly. (See 'Evaluation for suspected complications' above.). ●The overall mortality of TPE is 0.03 to 0.05 percent; death is most commonly due to respiratory or cardiac complications. (See 'Mortality' above.) ●An overview of the terminology used to describe apheresis procedures; the types of indications for which therapeutic apheresis is effective, including American Society for Apheresis (ASFA) therapeutic categories; and practical issues in apheresis techniques are discussed in detail separately. (See "Therapeutic apheresis (plasma exchange or cytapheresis): Indications and technology".) Use of UpToDate is subject to the Subscription and License Agreement. REFERENCES Mokrzycki MH, Kaplan AA. Therapeutic plasma exchange: complications and management. Am J Kidney Dis 1994; 23:817. Huestis DW. Mortality in therapeutic haemapheresis. Lancet 1983; 1:1043. Apheresis Principles and Practice, 2nd ed, McLeod BC (Ed), AABB, Bethesda 2003. Kaplan AA, Halley SE. Plasma exchange with a rotating filter. Kidney Int 1990; 38:160. Weinstein R. Prevention of citrate reactions during therapeutic plasma exchange by constant infusion of calcium gluconate with the return fluid. J Clin Apher 1996; 11:204. Pearl RG, Rosenthal MH. Metabolic alkalosis due to plasmapheresis. Am J Med 1985; 79:391. Stigelman WH Jr, Henry DH, Talbert RL, Townsend RJ. Removal of prednisone and prednisolone by plasma exchange. Clin Pharm 1984; 3:402. Standards for Cellular Therapy Product Services, 6th ed, AABB, Bethesda 2013. Orlin JB, Berkman EM. Partial plasma exchange using albumin replacement: removal and recovery of normal plasma constituents. Blood 1980; 56:1055. Goss GA, Weinstein R. Pentastarch as partial replacement fluid for therapeutic plasma exchange: effect on plasma proteins, adverse events during treatment, and serum ionized calcium. J Clin Apher 1999; 14:114. Chirnside A, Urbaniak SJ, Prowse CV, Keller AJ. Coagulation abnormalities following intensive plasma exchange on the cell separator. II. Effects on factors I, II, V, VII, VIII, IX, X and antithrombin III. Br J Haematol 1981; 48:627. Gelabert A, Puig L, Maragall S, et al. Coagulation alterations during massive plasmapheresis. In: Plasma Exchange, Sieberth HG (Ed), Schattauer Verlag, Stuttgart 1980. p.71. Wing EJ, Bruns FJ, Fraley DS, et al. Infectious complications with plasmapheresis in rapidly progressive glomerulonephritis. JAMA 1980; 244:2423. Pohl MA, Lan SP, Berl T. Plasmapheresis does not increase the risk for infection in immunosuppressed patients with severe lupus nephritis. The Lupus Nephritis Collaborative Study Group. Ann Intern Med 1991; 114:924. Kaplan AA. Towards a rational prescription of plasma exchange: The kinetics of immunoglobulin removal. Semin Dial 1992; 5:227. Keller AJ, Urbaniak SJ. Intensive plasma exchange on the cell separator: effects on serum immunoglobulins and complement components. Br J Haematol 1978; 38:531. Owen HG, Brecher ME. Atypical reactions associated with use of angiotensin-converting enzyme inhibitors and apheresis. Transfusion 1994; 34:891. Brecher ME, Owen HG, Collins ML. Apheresis and ACE inhibitors. Transfusion 1993; 33:963. Sutton DM, Nair RC, Rock G. Complications of plasma exchange. Transfusion 1989; 29:124. Ring J, Messmer K. Incidence and severity of anaphylactoid reactions to colloid volume substitutes. Lancet 1977; 1:466. Bambauer R, Jutzler GA, Albrecht D, et al. Indications of plasmapheresis and selection of different substitution solutions. Biomater Artif Cells Artif Organs 1989; 17:9. Apter AJ, Kaplan AA. An approach to immunologic reactions associated with plasma exchange. J Allergy Clin Immunol 1992; 90:119. Technical Manual, 17th ed, Roback JD (Ed), American Association of Blood Banks, Bethesda 2011. Jørstad S. Biocompatibility of different hemodialysis and plasmapheresis membranes. Blood Purif 1987; 5:123. Nicholls AJ, Platts MM. Anaphylactoid reactions due to haemodialysis, haemofiltration, or membrane plasma separation. Br Med J (Clin Res Ed) 1982; 285:1607. Topic 7942 Version 14.0
13321	https://us.sofatutor.com/math/videos/from-ratio-tables-to-equations	From Ratio Tables to Equations – Made Easy Skip to main content Login Subjects Math English Language Arts Science Social Studies How it works Fill knowledge gaps without pressure Practice anywhere, anytime Learn by playing Prepare at your own pace Nurture a Love of Reading Try sofatutor for 30 Days Pricing Subjects All school subjects Math English Language Arts Science Social Studies How it works Fill knowledge gaps without pressure Our tutorial videos allow students to learn at their own pace without any pressure or stress. They can watch, pause or rewind the videos as often as they need until they understand the content. Practice anywhere, anytime Our interactive exercises come in a variety of formats so students can practice in a playful way. They get feedback and hints while solving the exercises. As a result, they learn and grow from their mistakes instead of feeling frustrated. Learn by playing Children in elementary school can use Sofahero to review independently and stay motivated. They master topics while going on exciting adventures, without the help of adults. Prepare at your own pace Students can use the worksheets to prepare themselves for mock tests: Simply print them out, fill them out, and check the answers using the answer key. Nurture a Love of Reading With interactive e-books children can playfully train their active listening and comprehension skills. Pricing Try sofatutor for 30 Days Login Do you already have an account? Log in: Your e-mail address Your password Password forgotten Try sofatutor for 30 Days Discover why over 1.6 MILLION students choose sofatutor! Try it for 30 Days Math Middle School Collections of Equivalent Ratios From Ratio Tables to Equations From Ratio Tables to Equations Watch videosStart exercisesShow worksheets Start exercises! Improve your grades now while having fun Unlock this video in just a few steps, and benefit from all sofatutor content Videos for all grades and subjects that explain school material in a short and concise way. Boost your confidence in class by studying before tests and mock tests with our fun exercises. Learn on the go with worksheets to print out – combined with the accompanying videos, these worksheets create a complete learning unit. 24-hour help provided by teachers who are always there to assist when you need it. Try it for 30 Days Cancel at any time Rating Ø 5.0 / 1 ratings You must be logged in to be able to give a rating. Wow, thank you! Please rate us on Google too! We look forward to it! Go to Google The authors Erin K. Basics on the topicFrom Ratio Tables to Equations ----------------------------------------------------- After this lesson, you will be able to transform ratio tables into equations. The lesson begins by teaching you how to identify patterns in a ratio table and to determine the constant rate. It leads you to learn how to assign variables and set up the proportion to get the equation. It concludes with isolating the desired variable in the equation for ease of manipulation. Learn about changing ratio tables to equations by understanding the balance of processes in the human body! This video includes key concepts, notation, and vocabulary such as ratio (a relationship between two non-negative numbers, both of which are not zero); ratio table (a structured set of equivalent ratios); constant rate (the rate at which a quantity changes in relation to another is fixed or unchanging); proportion (two or more equivalent ratios); variable (a symbol, usually a letter, that is used to represent one or more values) and equation (a statement of equality between two quantities). Before watching this video, you should already be familiar with ratio, ratio table, constant rate, variable, and equation. After watching this video, you will be prepared to learn to solve real-world and mathematical problems involving unit rate, percent, and measurement conversion. Common Core Standard(s) in focus: 6.RP.A.3 A video intended for math students in the 6th grade Recommended for students who are 11 - 12 years old ### TranscriptFrom Ratio Tables to Equations When you listen to your favorite song, the music can move you! While dancing the robot, you're running on the electric energy of quality beats! In fact, our bodies work like machines. There are moving parts, and lots of inputs and outputs. We eat, drink, and breathe to keep our bodies in balance. Changing ratio tables to equations, gives us a way to understand the balance of processes that happen in our bodies. Let's start by examining our breathing process. The air we breathe in and out is composed of many gases. These gases are mostly nitrogen, and some parts oxygen and carbon dioxide. The nitrogen is stable as it is and doesn't change within our bodies. On the other hand, the inhaled oxygen is used to perform many important functions, like fueling our muscles, while the carbon dioxide is exhaled as a waste gas. Let's think about the breath you just took. Only 60 milliliters of that air is oxygen. We can use "ml" to represent mililiters. 60 mililiters is about the volume of two ping pong balls. Now breath out. With that exhale, only 15 milliliters of that is carbon dioxide. That’s about half a ping pong ball. So, what is the ratio of oxygen inhaled to carbon dioxide exhaled? 60 to 15. Here we have a ratio table. Now it shows what happens when we start to dance faster and our breathing volume increases. Do you notice any patterns in the table? For every 4 milliliters of increase in oxygen inhaled, there is a corresponding 1 milliliter increase in carbon dioxide exhaled. The 4 to 1 ratio is preserved in every breath! From this ratio table we can write an equation that examines the relationship between oxygen and carbon dioxide. An equation gives us a tool to find any gas amounts we want. We can use the 4 to 1 pattern, ratio, we marked earlier to help us establish the relationship. We see that 60 equals the pattern of 4 times 15, 64 equals 4 times 16, and so on, and so forth. With this relationship identified, it is easier to write an equation. When writing an equation, we should should identify our variables. We want to know about the oxygen and carbon dioxide relationship for any amount. Therefore, let’s call oxygen, 'x', and carbon dioxide, 'c'. In general, we can see oxygen, 'x', equals 4 times the carbon dioxide, 'c'. That's a useful equation and will give us how much oxygen we inhaled when we know how much carbon dioxide was exhaled. If you want to solve for carbon dioxide you can use the same equation and just isolate the c variable. You can’t live on oxygen alone. To move with fluidity, you need water, too. Even though different people have different masses, they all have a similar water to other bio-material components. This table shows the mass of water to other bio-materials found in different guys in kilograms. Do you see any patterns in this table? As the water mass increases by 3, the other bio-materials mass increases by 2, maintaining a nice, hydrated, 3 to 2 ratio. This ratio could give us our water mass when we know the mass of our other biomaterials. Do you know the mass of your other bio-materials? However, most people only know their total mass. How can we calculate the total mass of the given information? Just add the water and other bio-materials together to get the total mass for each person. Notice anything interesting in the ratio table? Because the water increased by 3 each time and the bio-material increased by 2 combined they increased by 5. This constant increase can be nicely translated into equations using ratios. Remember to first identify your variables. Let's define water as 'w', other bio-materials as 'b,' and total mass as 't.' So, we can write the proportion of other bio-materials, 'b', to total, 't', as 2 to 5, and the proportion of water,'w', to total, 't' as 3 to 5. We can solve for whichever variable we want to dance with. For example, let's pick water, 'w'! Isolate 'w' by multiplying both sides by 't'. This gives us water is equal to three-fifths 't.' For example, if your total mass is 55 kilograms, 33 of those kilograms are from water. So, we've learned a lot. Can you summarize the steps needed to go from ratio tables to equations? First, identify patterns. There should always be a constant ratio. Second, identify your variables and set-up the equivalent ratios. Lastly, isolate the variable you want to find. Smooth dancers have to breathe deep, stay hydrated, and understand their bodies. Oh? Looks like we've overlooked some gases. His machine is also running on fumes. But, that's another story. READ MORE 0 comments From Ratio Tables to Equations exercise Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video From Ratio Tables to Equations. ### Decide which statements are true given the ratio. Hints When ratios are written as words the order is important. The first word corresponds to the first number, the second word corresponds to the second number. If you have a ratio of 7 7 7 oranges to 6 6 6 apples. That means for every 7 7 7 oranges you have 6 6 6 apples. So if you have 21 21 2 1 oranges that means you have 18 18 1 8 apples. The ratio 21 18=7 6\frac { 21 }{ 18 } = \frac { 7 }{ 6 } 1 8 2 1=6 7 Solution When ratios are written as words, order is important! The first word corresponds to the first number and the second word corresponds to the second number. In this problem, it is given that the ratio of milliliters of oxygen inhaled to carbon dioxide exhaled is 4 to 1. This means that the 4 4 4 corresponds to oxygen and the 1 1 1 corresponds to carbon dioxide. In order for a statement below to be true, this oxygen 4 4 4 to carbon dioxide 1 1 1 ratio needs to there. For every 4 4 4 ml of oxygen inhaled, 1 1 1 ml of carbon dioxide is exhaled is true. This statement holds the oxygen 4 4 4 to carbon dioxide 1 1 1 ratio given. For every 1 1 1 ml of oxygen inhaled, 4 4 4 ml of carbon dioxide is exhaled is not true. In this statement the ratio of oxygen to carbon dioxide is 1 1 1 to 4 4 4, not 4 4 4 to 1 1 1 from the given statement. The ratio is backwards in this answer. When 80 80 8 0 ml of oxygen is inhaled, 20 20 2 0 ml of carbon dioxide is exhaled is true. It is given that the ratio of oxygen to carbon dioxide is 4 4 4 to 1 1 1. In this statement, the ratio is 80 80 8 0 to 20 20 2 0. 80 80 8 0 to 20 20 2 0 can be simplified by dividing both numbers by 20 20 2 0, resulting in 4 4 4 to 1 1 1. The amount of oxygen inhaled is 4 4 4 times the amount of carbon dioxide exhaled is true. From the given statement we know that for every 4 4 4 ml of oxygen inhaled, 1 1 1 ml of carbon dioxide is exhaled. To test if this statement is true, multiply the amount of carbon dioxide exhaled (1 1 1) by 4 4 4, which gives us 4 4 4, which is the amount of oxygen inhaled in the example. If 40 40 4 0 ml of carbon dioxide was exhaled, then 10 10 1 0 ml of oxygen was inhaled is not true. From the given statement we know oxygen to carbon dioxide is 4 4 4 to 1 1 1. In this statement we are given carbon dioxide to oxygen is 40 40 4 0 to 10 10 1 0. We can simplify this ratio by dividing both numbers by 10 10 1 0, which gives us a ratio of 4 4 4 to 1 1 1. This may seem correct, but in this statement we are given carbon dioxide first, not oxygen like the given statement. The numbers may be the same, but the order of the ratio is not, so this is not a true statement. ### Writing equations given a ratio table Hints In equivalent ratios, the numerator of the first fraction is related to the numerator of the second fraction and the denominator of the first fraction is related to the denominator of the second fraction. When solving for a variable, perform the opposite operation to move any other variable or constant to the other side of the equation. Remember that addition and subtraction are opposites, while multiplication and division are opposites. If asked to solve a d=5 7\frac { a } { d } = \frac { 5 } { 7 } d a=7 5 for a a a in terms of d d d, multiply both sides of the equation by d d d to get a=5 7 d a= \frac { 5 } { 7 } d a=7 5d. Solution The first step in writing an equation is always to define the variables. Once variables have been defined, then you can follow the following steps: Set up equivalent ratios. Remember that in equivalent ratios the numerators all must represent the same thing, and the denominators all must represent the same thing. The correct equivalent ratios are : b t=24 60=2 5 w t=36 60=3 5 w b=36 24=3 2\frac{b}{t} = \frac{24}{60} = \frac{2}{5} \ \frac{w}{t} = \frac{36}{60} = \frac{3}{5} \ \frac{w}{b} = \frac{36}{24} = \frac{3}{2} t b=6 0 2 4=5 2t w=6 0 3 6=5 3b w=2 4 3 6=2 3 To identify the equation that compares water to other bio-materials, look for the equation with both the variable that represents water, w w w, and the variable that represents other bio-materials, b b b. Also make sure that this equation has the correct ration of water to other bio-materials, which is 3 3 3 to 2 2 2. The equation is: w b=2 3\frac{w}{b} =\frac{2}{3}b w=3 2. When asked to isolate a variable, that means to solve for that variable and get everything else on the other side of the equations. In this step we must solve w b=2 3\frac{w}{b} =\frac{2}{3}b w=3 2 for w w w. To solve for w w w, we need to use the opposite operation to move the b b b to the other side. The fraction w b\frac { w }{ b }b w means w w w divided by b b b, to move b b b we must do the opposite of dividing, which is to multiply both sides by b b b. w b=3 2 b⋅w b=3 2⋅b w=3 2 b\begin{array}{rcl} \frac{w}{b} & = & \frac{3}{2} \ \color{#669900}{b~ \cdot} \frac{w}{b} & = & \frac{3}{2} \color{#669900}{ \cdot b} \ w & = & \frac{3}{2} b\ \end{array}b wb⋅b ww===2 32 3⋅b 2 3b The resulting equation is w=3 2 b w = \frac{3}{2} b w=2 3b ### Complete the ratio table and find an equation Hints Find the constant increase in each column by looking for the pattern. What is being added as you go down each column? Identify the ratio of femur length to body height for the table. Look across each row. What is the relationship between femur length and body height? Solution When analyzing a table, it is a good idea to look for patterns in the table. You should notice that the Femur Length column is increasing by 1 1 1 as the Body Height column is increasing by 4 4 4. This gives us a Femur Length to Body Height Ratio of 1 1 1 to 4 4 4. This ratio of 1 1 1 to 4 4 4 can also help find the relationship across each column. In this table, the Femur Length, f f f, multiplied by 4 4 4 results in the Body Height, h h h. It is a good idea to check to make sure this works for every row. 15⋅4=60 15\cdot 4=60 1 5⋅4=6 0 ✓ 16⋅4=64 16\cdot 4=64 1 6⋅4=6 4 ✓ 17⋅4=68 17\cdot 4=68 1 7⋅4=6 8 ✓ 18⋅4=72 18\cdot 4=72 1 8⋅4=7 2 ✓ Since this relationship works for each row in the table, a general equation can be written using f f f for Femur Length and h h h for Body Height. Following the relationship in each of the above examples, we get f⋅4=h f\cdot 4=h f⋅4=h or simplified algebraically 4 f=h 4f=h 4 f=h This equation can be used to find Body Height, h h h, given and Femur Length, f f f. In this question you were asked to determine the Body Height given a Femur Length of 30 inches. Using the equation created, 4 f=h 4f=h 4 f=h, and given f=30 f=30 f=3 0, you will have 4(30)=h 4(30)=h 4(3 0)=h, when simplified gives you 120=h 120 =h 1 2 0=h. Therefore when a Femur Length is 30 30 3 0 inches, the Body Height is 120 120 1 2 0 inches. ### Match ratio table to equation Hints Identify the constant increase in each column. Set up equivalent ratios, x y\frac{x}{y}y x and solve for x. Look for the relationship across each row. What operation do you have to perform on the value in y y y to get the value in x x x? Solution One way to write equations for ratio tables is to set up equivalent fractions. To write equivalent fractions you will need to find the constant increase for each variable by looking for the pattern down each column. The constant increase, equivalent fractions, and equations for each table are listed below. 1. The constant increase for x x x is 1 1 1. The constant increase for y y y is 2 2 2. The equivalent ratio equation is x y=1 2\frac{x}{y} = \frac{1}{2}y x=2 1. To solve for x x x, multiply both sides by y y y to get the equation x=1 2 y x= \frac{1}{2}y x=2 1y. 1. The constant increase for x x x is 2 2 2. The constant increase for y y y is 10 10 1 0. The equivalent ratio equation is x y=2 10\frac{x}{y} = \frac{2}{10}y x=1 0 2. To solve for x x x, multiply both sides by y y y to get the equation x=2 10 y x= \frac{2}{10}y x=1 0 2y. The fraction 2 10\frac{2}{10}1 0 2 can be simplified by dividing the numerator and denominator by 2 to get 1 5\frac{1}{5}5 1. The final equation for table 2 2 2 is x=1 5 y x= \frac{1}{5}y x=5 1y . 1. The constant increase for x x x is 5 5 5. The constant increase for y y y is 1 1 1. The equivalent ratio equation is x y=5 1\frac{x}{y} = \frac{5}{1}y x=1 5. To solve for x x x, multiply both sides by y y y to get the equation x=5 1 y x= \frac{5}{1}y x=1 5y. The fraction 5 1\frac{5}{1}1 5 simplifies to 5 5 5, so the final equation for table 3 3 3 is x=5 y x=5y x=5 y. 1. The constant increase for x x x is 12 12 1 2. The constant increase for y y y is 3 3 3. The equivalent ratio equation is x y=12 3\frac{x}{y} = \frac{12}{3}y x=3 1 2. To solve for x x x, multiply both sides by y y y to get the equation x=12 3 y x= \frac{12}{3}y x=3 1 2y. The fraction 12 3\frac{12}{3}3 1 2 simplifies to 4 4 4, so the final equation for table 4 4 4 is x=4 y x=4y x=4 y. ### Identify patterns in ratio tables. Hints To find the ratio x x x to y y y, you must first find the constant increase for each column. To identify the constant increase, look for a pattern in either the x x x- or y y y- column. Look at one column at a time. Start from the first number and ask yourself "what number am I adding to get the number below?" Repeat this for the entire column, if this value is the same for each then it is the the constant increase for that column. Look at the table below. x y 1 5 2 10 3 15 4 20\begin{array}{\|c\|c\|} \hline x & y \ \hline 1 & 5 \ 2 & 10 \ 3 & 15 \ 4 & 20 \ \hline \end{array}x 1 2 3 4y 5 1 0 1 5 2 0 Look at the numbers in the x x x-column: 1 1 1, 2 2 2, 3 3 3, 4 4 4. Ask yourself what number do can I add to 1 1 1 to get to 2 2 2? Repeat this for each number in the x x x-column: 2+?=3 3+?=4 2 + ? = 3 \ 3 + ? = 4 2+?=3 3+?=4. If you added the same number each time, then that is the constant increase for the x x x. Do the same for the y y y-column: 5+?=10 10+?=15 15+?=20 5 + ? = 10 \ 10 + ? = 15 \ 15 + ? = 20 5+?=1 0 1 0+?=1 5 1 5+?=2 0 The number you added every time should have been the same, this is the constant increase for y y y. Look at the table below. x y 1 5 2 10 3 15 4 20\begin{array}{\|c\|c\|} \hline x & y \ \hline 1 & 5 \ 2 & 10 \ 3 & 15 \ 4 & 20 \ \hline \end{array}x 1 2 3 4y 5 1 0 1 5 2 0 You can see that the x x x increases by 1 1 1 and the y y y increases by 5 5 5, this gives an x x x to y y y ratio of 1:5 1:5 1:5. Solution First identify the constant increase in both the x x x and y y y column for each table. Then, set up your x x x to y y y ratio using the constant increases you found. For the first table, _constant increase_ for x x x is 4 4 4 and for y y y is 3 3 3. The ratio x x x to y y y for this table is 4:3 4:3 4:3. For the second table, the _constant increase_ for x x x is 1 1 1 and for y y y is 7 7 7. The ratio x x x to y y y for this table is 1:7 1:7 1:7. For the third table, the _constant increase_ for x x x is 3 3 3 and for y y y is 4 4 4. The ratio x x x to y y y for this table is 3:4 3:4 3:4. For the fourth table, the _constant increase_ for x x x is 7 7 7 and for y y y is 1 1 1. The ratio x x x to y y y for this table is 7:1 7:1 7:1. ### Given an equation fill out the ratio tables Hints To find a missing value in a table, substitute in a number you are given into the equation and simplify. To decide which ratio equation to use, look at the value you are given. Which equation can you substitute that value into and simplify without having to rearrange the formula? Let's look at an example: Given the two equations a=3 4 b a=\frac{3}{4}b a=4 3b and b=4 3 a b=\frac{4}{3}a b=3 4a , find a a a if b=8 b=8 b=8. I would choose the equation that has a a a isolated and b b b on the other side: a=3 4 b a=\frac{3}{4}b a=4 3b . Then, substitute b=8 b=8 b=8 and simplify. a=3 4(8)a=24 4 a=6 a=\frac{3}{4}(8)\ a=\frac{24}{4} \ a=6 a=4 3(8)a=4 2 4a=6 Solution In this problem, three equivalent equations are given. That means that you can use whichever equation makes the most sense based on whether you are given d d d or t t t. You may choose to use one equation to complete the table, or use all of them. There is not a rule as to which equation you have to use. Some equations may take less steps, but all of them will arrive at the same answer. When solving this problem, I chose to use the equation with the variable isolated that I was not given. For example, if I was given d d d and needed to find t t t, I would use the equation t=7 2 d t=\frac{7}{2}d t=2 7d. Let's start by substituting in all d d d-values given into t=7 2 d t=\frac{7}{2}d t=2 7d. d=2 d=2 d=2, t=7 2(2)t=\frac{7}{2}(2)t=2 7(2), t=14 2 t=\frac{14}{2}t=2 1 4, so t=7 t=7 t=7. d=6 d=6 d=6, t=7 2(6)t=\frac{7}{2}(6)t=2 7(6), t=42 2 t=\frac{42}{2}t=2 4 2, so t=21 t=21 t=2 1 d=10 d=10 d=1 0, t=7 2(10)t=\frac{7}{2}(10)t=2 7(1 0), t=70 2 t=\frac{70}{2}t=2 7 0, so t=35 t=35 t=3 5 Now, let's substitute all of the t t t values into d=2 7 t d=\frac{2}{7}t d=7 2t. t=14 t=14 t=1 4, d=2 7(14)d=\frac{2}{7}(14)d=7 2(1 4), t=28 7 t=\frac {28}{7}t=7 2 8, so t=4 t=4 t=4 t=70 t=70 t=7 0, d=2 7(70)d=\frac{2}{7}(70)d=7 2(7 0), t=140 7 t=\frac {140}{7}t=7 1 4 0, so t=20 t=20 t=2 0 More videos in this topic Collections of Equivalent Ratios Determining Equivalent Ratios Ratio Tables - Additive and Multiplicative Structure Comparing Ratios Using Ratio Tables From Ratio Tables to Double Number Lines From Ratio Tables to Equations Plotting Ratios on the Coordinate Plane Reviewing Representations of Ratios What are Congruence and Similarity? Company Our Team Prices Jobs Platform My Account Enter code Proven learning success Gamification How it works Tutorial Videos Exercises Sofahero Worksheets Interactive books Help FAQ Give Us Feedback Legal Terms & Conditions Privacy Policy Cookie Details Cookie Preferences Contact Us Do Not Sell My Personal Information Any questions? Contact us! help@sofatutor.com sofatutor.com sofatutor.ch sofatutor.at sofatutor.com sofatutor.co.uk Any questions? Contact us! help@sofatutor.com Terms and Conditions › Privacy Policy › Cookie Details › Cookie Preferences › Contact Us › Jobs › Do Not Sell My Personal Information ›
13322	https://courses.lumenlearning.com/suny-biology2xmaster/chapter/nitrogenous-wastes/	The Excretory System Nitrogenous Wastes Learning Objectives By the end of this section, you will have completed the following objectives: Compare and contrast the way in which aquatic animals and terrestrial animals can eliminate toxic ammonia from their systems Compare the major byproduct of ammonia metabolism in vertebrate animals to that of birds, insects, and reptiles Of the four major macromolecules in biological systems, both proteins and nucleic acids contain nitrogen. During the catabolism, or breakdown, of nitrogen-containing macromolecules, carbon, hydrogen, and oxygen are extracted and stored in the form of carbohydrates and fats. Excess nitrogen is excreted from the body. Nitrogenous wastes tend to form toxic ammonia, which raises the pH of body fluids. The formation of ammonia itself requires energy in the form of ATP and large quantities of water to dilute it out of a biological system. Animals that live in aquatic environments tend to release ammonia into the water. Animals that excrete ammonia are said to be ammonotelic. Terrestrial organisms have evolved other mechanisms to excrete nitrogenous wastes. The animals must detoxify ammonia by converting it into a relatively nontoxic form such as urea or uric acid. Mammals, including humans, produce urea, whereas reptiles and many terrestrial invertebrates produce uric acid. Animals that secrete urea as the primary nitrogenous waste material are called ureotelic animals. Nitrogenous Waste in Terrestrial Animals: The Urea Cycle The urea cycle is the primary mechanism by which mammals convert ammonia to urea. Urea is made in the liver and excreted in urine. The overall chemical reaction by which ammonia is converted to urea is 2 NH3 (ammonia) + CO2 + 3 ATP + H2O → H2N-CO-NH2 (urea) + 2 ADP + 4 Pi + AMP. The urea cycle utilizes five intermediate steps, catalyzed by five different enzymes, to convert ammonia to urea, as shown in Figure 1. The amino acid L-ornithine gets converted into different intermediates before being regenerated at the end of the urea cycle. Hence, the urea cycle is also referred to as the ornithine cycle. The enzyme ornithine transcarbamylase catalyzes a key step in the urea cycle and its deficiency can lead to accumulation of toxic levels of ammonia in the body. The first two reactions occur in the mitochondria and the last three reactions occur in the cytosol. Urea concentration in the blood, called blood urea nitrogen or BUN, is used as an indicator of kidney function. Figure 1. The urea cycle converts ammonia to urea. Evolution Connection Excretion of Nitrogenous Waste The theory of evolution proposes that life started in an aquatic environment. It is not surprising to see that biochemical pathways like the urea cycle evolved to adapt to a changing environment when terrestrial life forms evolved. Arid conditions probably led to the evolution of the uric acid pathway as a means of conserving water. Nitrogenous Waste in Birds and Reptiles: Uric Acid Birds, reptiles, and most terrestrial arthropods convert toxic ammonia to uric acid or the closely related compound guanine (guano) instead of urea. Mammals also form some uric acid during breakdown of nucleic acids. Uric acid is a compound similar to purines found in nucleic acids. It is water insoluble and tends to form a white paste or powder; it is excreted by birds, insects, and reptiles. Conversion of ammonia to uric acid requires more energy and is much more complex than conversion of ammonia to urea Figure 2. Figure 2. Nitrogenous waste is excreted in different forms by different species. These include (a) ammonia, (b) urea, and (c) uric acid. (credit a: modification of work by Eric Engbretson, USFWS; credit b: modification of work by B. “Moose” Peterson, USFWS; credit c: modification of work by Dave Menke, USFWS) Everyday Connection Gout Mammals use uric acid crystals as an antioxidant in their cells. However, too much uric acid tends to form kidney stones and may also cause a painful condition called gout, where uric acid crystals accumulate in the joints, as illustrated in Figure 3. Food choices that reduce the amount of nitrogenous bases in the diet help reduce the risk of gout. For example, tea, coffee, and chocolate have purine-like compounds, called xanthines, and should be avoided by people with gout and kidney stones. Figure 3. Gout causes the inflammation visible in this person’s left big toe joint. (credit: “Gonzosft”/Wikimedia Commons) Section Summary Ammonia is the waste produced by metabolism of nitrogen-containing compounds like proteins and nucleic acids. While aquatic animals can easily excrete ammonia into their watery surroundings, terrestrial animals have evolved special mechanisms to eliminate the toxic ammonia from their systems. Urea is the major byproduct of ammonia metabolism in vertebrate animals. Uric acid is the major byproduct of ammonia metabolism in birds, terrestrial arthropods, and reptiles. Candela Citations CC licensed content, Shared previously Biology. Authored by: OpenStax. Provided by: OpenStax College. Located at: License: CC BY: Attribution Licenses and Attributions CC licensed content, Shared previously Biology. Authored by: OpenStax. Provided by: OpenStax College. Located at: License: CC BY: Attribution
13323	https://www.sciencedirect.com/topics/medicine-and-dentistry/saxitoxin	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Marine Toxin Attack Saxitoxin Saxitoxin is one of the family of neurotoxins that cause paralytic shell fish poisoning (PSP). This toxin is soluble in water and stabile at high temperatures.1 The toxin is made by small organisms called dinoflagellates, which contaminate shellfish (clams, scallops, oysters). The toxin is also produced by blue green algae. The algae may grow rapidly, producing blooms called „red tides.” Human ingestion of filter-feeding crabs and lobsters or shellfish results in an intoxication.1–4 Saxitoxin binds to voltage-gated sodium channels on nerve fibers and muscle cells. The binding of the toxin blocks conduction of the nerve impulse. General symptoms of saxitoxin poisoning are neurologic and respiratory paralysis.1–3 Clinical Features General: There is an initial latent period varying from 30 minutes to several hours following ingestion, before the onset of neurological symptoms. : Cardiovascular: There are no specific cardiovascular effects although in laboratory animals, saxitoxin caused hypotension and conduction defects. : Respiratory: Respiratory distress from muscular paralysis may occur up to 12 hours after intoxication. The respiratory paralysis may lead to death.1,5 : Gastrointestinal: Gastrointestinal symptoms may appear hours to days after ingestion. These symptoms may include nausea, vomiting, abdominal pain, and diarrhea.1,5 : Other: A tingling and burning sensation, initially occurring around the mouth and lips, is usually the first symptom. The numbness may be on the hands and spread over the chest and abdomen.1,5 These symptoms may progress, with difficulty walking and arm and leg weakness.1,5 Involuntary movements and tremors may occur.1 There are no specific antidotes for saxitoxin poisoning.1 Treatment is symptomatic. If oral ingestion is suspected, emptying the stomach via emetic or gastric lavage is recommended. Intubation and mechanical ventilation with monitoring to support respiration may be necessary.1 Routine laboratory studies are not helpful. Diagnosis is confirmed by detection of the toxin in food, water, or environmental samples.5,6 Saxitoxin is water soluble and therefore can be easily aerosolized. It is toxic by both inhalation and ingestion.1 View chapterExplore book Read full chapter URL: Book2006, Disaster MedicineWende R. Reenstra Chapter PHYCOTOXINS 1999, Encyclopedia of Food MicrobiologyArun Sharma Saxitoxin Saxitoxin is one of the major causes of paralytic poisoning occurring after consumption of seafood such as mussels, clams or oysters. Saxitoxin is a neurotoxin named after the butter clam (Saxidomus giganteus). The toxin is produced by dinoflagellates such as Protogonyaulax sp., Gymnodium catenatum, Alexandrium catenella and Alexandrium minutum. Mussels, clams and oysters feed on these dinoflagellates or red algae with which they may be associated and thus become toxic. The carnivorous fish that feed on these organisms will also become toxic. Thus human consumption of seafoods harvested from areas where these dinoflagellates thrive in abundance, i.e. algal blooms, can lead to outbreak of paralytic poisoning. Saxitoxin is a sodium channel blocker. All paralytic shellfish toxins contain a guanidino group. The positively charged guanidino group interacts with a negatively charged carboxyl group at the mouth of the sodium channel on the extracellular side of the plasma membrane of nerve and muscle cells. Blocking of the sodium ion transport through nerve and muscle cell membranes results in paralysis. Saxitoxin is also produced by a genus of cyanobacteria Aphanizomenon. View chapterExplore book Read full chapter URL: Reference work1999, Encyclopedia of Food MicrobiologyArun Sharma Chapter Saxitoxin 2014, Encyclopedia of Toxicology (Third Edition)S.E. Gad Abstract Saxitoxin is one of only two naturally occurring Schedule 1 chemical warfare agents, and is a toxic compound found naturally in dinoflagellates. Its toxicity stems from its ability to bind to voltage-gated sodium channels found in the membranes of excitable cells such as neurons and muscle cells, blocking synaptic transmission. The toxic effects of saxitoxin primarily affect the peripheral nervous system and can alter cardiac chronotrophy. In humans, the estimated lethal dose is 0.3–1 mg, while a single contaminated shellfish may contain 50 lethal doses. Bioaccumulation up the food chain may occur. View chapterExplore book Read full chapter URL: Reference work2014, Encyclopedia of Toxicology (Third Edition)S.E. Gad Chapter Saxitoxin 2014, Encyclopedia of Toxicology (Third Edition)S.E. Gad Background Information First recognized in 1957 by Shantz et al. in the butter clam Saxidomus giganteus, saxitoxin is a naturally occurring toxin that is synthesized by various marine dinoflagellates and cyanobacteria. It is used in neurochemical and molecular biology research, but regulatory concerns have focused on its weaponization and potent toxicological effects on animals and humans. Saxitoxin causes paralytic shellfish poisoning (PSP) in humans and other animals; whales having ingested organisms contaminated with the toxin have died just hours after exposure. In humans, PSP can occur as a result of consuming shellfish or other seafood in which saxitoxin has accumulated. While effects have been documented worldwide, the sources of contaminated seafood have been identified as primarily the west and east coasts of the United States. To date, more than 30 different saxitoxin analogs have been identified. These include pure saxitoxin (STX), neosaxitoxin (neoSTX), the gonyautoxins (GTX), and decarbamoylsaxitoxin (dc-STX); of these, STX, 6NeoSTX, GTX1, and dc-STX seem to be the most toxic. The term saxitoxin typically refers to this collection of compounds produced naturally by cyanobacteria. Saxitoxin is far more potent than the classic puffer fish toxin tetrodotoxin, and is one of only two naturally occurring Schedule 1 chemical warfare agents (the other is ricin). In 1970, President Nixon ordered stocks of the toxin destroyed in compliance with U.N. agreements on biological weapons; however, the Central Intelligence Agency revealed in 1975 that there was a remaining supply, which was distributed to research facilities by the National Institutes of Health in order to study neurological diseases. View chapterExplore book Read full chapter URL: Reference work2014, Encyclopedia of Toxicology (Third Edition)S.E. Gad Chapter Marine Toxin Attack 2006, Disaster MedicineWende R. Reenstra MEDICAL TREATMENTS OF CASUALTIES Saxitoxin Saxitoxin is one of the family of neurotoxins that cause paralytic shell fish poisoning (PSP). This toxin is soluble in water and stabile at high temperatures.1 The toxin is made by small organisms called dinoflagellates, which contaminate shellfish (clams, scallops, oysters). The toxin is also produced by blue green algae. The algae may grow rapidly, producing blooms called „red tides.” Human ingestion of filter-feeding crabs and lobsters or shellfish results in an intoxication.1–4 Saxitoxin binds to voltage-gated sodium channels on nerve fibers and muscle cells. The binding of the toxin blocks conduction of the nerve impulse. General symptoms of saxitoxin poisoning are neurologic and respiratory paralysis.1–3 Clinical Features General: There is an initial latent period varying from 30 minutes to several hours following ingestion, before the onset of neurological symptoms. : Cardiovascular: There are no specific cardiovascular effects although in laboratory animals, saxitoxin caused hypotension and conduction defects. : Respiratory: Respiratory distress from muscular paralysis may occur up to 12 hours after intoxication. The respiratory paralysis may lead to death.1,5 : Gastrointestinal: Gastrointestinal symptoms may appear hours to days after ingestion. These symptoms may include nausea, vomiting, abdominal pain, and diarrhea.1,5 : Other: A tingling and burning sensation, initially occurring around the mouth and lips, is usually the first symptom. The numbness may be on the hands and spread over the chest and abdomen.1,5 These symptoms may progress, with difficulty walking and arm and leg weakness.1,5 Involuntary movements and tremors may occur.1 There are no specific antidotes for saxitoxin poisoning.1 Treatment is symptomatic. If oral ingestion is suspected, emptying the stomach via emetic or gastric lavage is recommended. Intubation and mechanical ventilation with monitoring to support respiration may be necessary.1 Routine laboratory studies are not helpful. Diagnosis is confirmed by detection of the toxin in food, water, or environmental samples.5,6 Saxitoxin is water soluble and therefore can be easily aerosolized. It is toxic by both inhalation and ingestion.1 Conotoxin The venom of the cone snail is composed of small substances termed conotoxins. There are more than 2000 peptides identified7 that lead to a complex set of symptoms. The toxins are heat stable but are inactivated by the disinfectants glutaraldehyde and formaldehyde.7 The conotoxins mechanism of action can be divided into pre- and post-synaptic pathways. The presynaptic conotoxin blocks the release of acetylcholine.5,8 The postsynaptic conotoxin inhibits sodium, potassium, and calcium channels and blocks muscular contraction.5 The toxicity of the venom is thought to result from the additive effects and not the concentration of the toxin. Clinical Features General: The onset of symptoms is almost immediate upon injection. Common symptoms include localized pain, swelling, numbness, and ischemia at the injection site.5,7 The numbness, swelling, and tingling may spread rapidly from the injection site to involve the entire body.5,7 : Cardiovascular: No specific cardiac effects are seen. : Respiratory: Progressive weakness, droopy eyelids, headache, abdominal pain, and difficulty breathing, may occur. Death results from respiratory paralysis.5,7 : Gastrointestinal: Stomach cramps and nausea are common effects. : Other: The clinical course is characterized by rapid onset and deterioration for the first 6 to 8 hours.7 This is followed by improvement, and complete recovery may take 4–6 weeks.5,7 Diagnosis is by clinical signs and symptoms, and there are no laboratory tests available. Treatment is to immobilize the limb or site of envenomation. Pressure dressings should be applied and pain medication and tetanus prevention provided.7 Intubation and mechanical ventilation may be necessary to support breathing.5,7 Conotoxins are very small, stable toxins that theoretically may be weaponized and disseminated as aerosols. A search of the open scientific literature regarding their inhalation toxicity found no publications.7 They are poisonous by injection. Tetrodotoxin Tetrodotoxin is one of the best characterized marine toxins because of its involvement in fatal food poisoning. The toxin is named from the pufferfish family (Tetraodontidae), where it has been found to be concentrated in the liver and other organs.9,10 The toxin can also be found in the Blue Ringed Octopus, Parrot fish, crabs, newts, and algae.2,5,9–11 The toxin is made by a bacterium that forms a symbiotic relationship with the animals.5,12 Tetrodotoxin is a neurotoxin that interferes with transmission of the nerve impulse at the nerve-muscle junction.5,12 The toxin is heat stable and can be soluablized in acetic solutions.5,12 This toxin specifically blocks sodium channels on the nerve cell and inhibits transmission of the impulse.2,5,12 The target molecular channels are thought to be very similar to saxitoxin.9 Clinical Features General: The first symptom is increasing numbness and tingling in the face and around the mouth.12 These may extend to the extremities or become generalized.5,12 : Cardiovascular: Hypotension and cardiac arrhythmias may occur.5,12 : Respiratory: There is increasing respiratory distress. The victim usually exhibits difficulty breathing and cyanosis. Paralysis increases and convulsions, mental impairment, and cardiac arrhythmia may occur.5,12 : Gastrointestinal: Autonomic effects such as headache, diaphoresis, and chest pain may occur and gastrointestinal symptoms such as nausea, diarrhea, and/or vomiting may develop.5,12 : Other: A coagulation disturbance, which is an occasional complication, may lead to bleeding into the skin and mucosa, formation of blood blisters, and peeling of the skin. The neurologic involvement may start as muscular twitching and proceed to complete skeletal muscle paralysis, interfering with speech and swallow.10,12 The pupils, after initially constricting, may become fixed and dilated.12 The victim may be completely paralyzed but conscious. Untreated, the death rate is 50–60% in some studies.5,12 Death usually occurs within 4 to 6 hours, with a known range of about 20 minutes to 8 hours.10,12 Management is supportive and standard management of poison ingestion should be employed if intoxication is by the oral route. These include gastric lavage or emetics, particularly after control of the airway has been obtained. Intubation and mechanical ventilation may be required in severe intoxication.5,12 After weakness has become apparent, the treatment is symptomatic (e.g., maintenance of respirations, monitoring of vital signs and electrolytes).5,10,12 Because of the likelihood of consciousness being maintained with complete paralysis, periodic administration of a tranquillizer is recommended along with continuous reassurance.10,12 Relatively little is known about TTX as a possible toxin weapon. A company in Japan is known to produce the toxin.12 It is not known to be made in large quantities that could be used in weapons, and little or nothing has been published about its inhalational toxicity.12 Palytoxin Palytoxin is one of most potent marine toxins known. It was isolated first from corals located in the South Pacific.13,14 Originally it was thought that the toxin was made by the corals; now, however, it is known that the toxin is made by a dinoflagellate (a small single-celled organism) and the corals concentrate the toxin.2,8,13 It is estimated that the lethal dose for a human is less than 5 μg.3,10,13,14 Palytoxins are stable in seawater and alcohols. Extensive pharmacological research has determined that palytoxin is not a neurotoxin.13,15 It instead acts at the cellular membranes to make them porous to charged molecules such as sodium, potassium, and calcium.13,15 Without the gradients of these ions, the cells are unable to function or maintain the cell shape.14,15 Clinical Features General: Symptoms are rapid, with death occurring within minutes.13 : Cardiovascular: Initial symptom may be chest pain from constriction of the cardiac blood vessels. This may lead to cardiac ischemia and death of cardiac tissue. This may be seen on the EKG as peaked T waves or ST segment elevation.13 The next symptom may be the loss of consciousness as unstable blood pressure, particularly episodes of low blood pressure, reduce blood flow to the brain.13 : Respiratory: There may be difficulty breathing, with symptoms of wheezing. This again may be due in part to constriction of blood vessels in the lungs. : Gastrointestinal: There are no specific gastrointestinal effects. : Other: Hemolysis (breakdown of blood cells) may occur as the cell membranes become permeable to various ions, the red blood cells swell and the membranes rupture. This results in decreased oxygen carrying capacity. Death is thought to result from decreased oxygenation.13 There are no known therapeutics for Palytoxin poisoning. Relatively is little is known about Palytoxin as a possible toxin weapon. It is not known to be made in large quantities that could be used in weapons and little or nothing has been published about its inhalational toxicity.13 In summary these toxins act on a variety of sites. Table 134-1 summarizes their specific effects. View chapterExplore book Read full chapter URL: Book2006, Disaster MedicineWende R. Reenstra Chapter Natural Products Structural Diversity-II Secondary Metabolites: Sources, Structures and Chemical Biology 2010, Comprehensive Natural Products IIJun’ichi Kobayashi, Takaaki Kubota 2.09.3.2 Saxitoxin Saxitoxin (120), a paralytic shellfish poisoning alkaloid, was obtained from cultures of a red tide dinoflagellate Alexandrium sp. The toxin is also used for studies on ion channels. The biosynthesis of saxitoxin (120) involves arginine as a precursor of the guanidinio groups, whereas the perhydropurine skeleton is derived from Claisen condensation of acetate with arginine. The carbon atom in the side chain is derived from S-adenosyl methionine.142 View chapterExplore book Read full chapter URL: Reference work2010, Comprehensive Natural Products IIJun’ichi Kobayashi, Takaaki Kubota Chapter WATER ANALYSIS \| Algal and Microbial Toxins 2005, Encyclopedia of Analytical Science (Second Edition)C. Robillot, L.E. Llewellyn Saxitoxins Saxitoxin (C10H17N7O4; MW=299) was first isolated from the marine environment, namely from the clam Saxidomus giganteus from whence it derives its name. Saxitoxin (Figure 1E) has more than 25 naturally occurring chemical variants, all of which are tricyclic molecules with the 1,2,3- and 7,8,9-guanidino groups of STX itself possessing pKas of 11.3 and 8.2, respectively. They are all substituted tetrahydropuric bases, water soluble, thermostable and stable in acidic conditions. However, they are unstable in alkaline conditions and sensitive to oxidative conditions. Most derivatives present a common backbone, saxitoxin or N-hydroxysaxitoxin, and only differ on the location and number of sulfate substitutions. However, a new class of toxins has recently been discovered, with a phenolic group in place of the typical sulfate substitution. In the freshwater environment, they can be produced by the cyanobacteria Aphanizomenon flos-aquae, Anabaena circinalis, Lyngbya wollei, and Cylindrospermopsis raciborskii. As expected from their chemical structure, these toxins have very different polarities and their lack of chromophore complicates their analysis. The longest used bioassay for the saxitoxin family is intraperitoneal injection of mice, with toxin quantitated by comparing the time to death from respiratory paralysis to a standard series of mouse units. Saxitoxin is highly toxic, being lethal to guinea pigs at only 5 μg kg−1 when injected intramuscularly and at similar doses when injected intraperitoneally into mice. Saxitoxin blocks sodium ion entry into nerves and muscle by occlusion of the voltage gated sodium channel. This prevents the conduction of a cellular action potential, paralyzing any victim, sometimes fatally. A variety of in vitro assays have been developed that minimize or avoid live animal experimentation. Several capitalize on the sodium channel's affinity for these toxins. Neuronal cell lines lyse in the presence of veratridine, a sodium channel activator, and ouabain, which prevents removal of the excessive sodium ions allowed in by veratridine. In the presence of both these drugs, a sodium channel blocker such as saxitoxin rescues the cells. Cellular viability can then be measured by adding tetrazolium salts that are metabolized by living cells to a colored product. Alternatively, isolated cellular membranes, typically from brain tissue, are used to bind radiolabeled saxitoxin. After incubating receptor and radioligand in the presence of a test sample, any radiolabeled saxitoxin bound to the cell membranes are deposited onto filters by vacuum pressure. Radioactivity from the labeled saxitoxin is then measured with a signal reduction indicating the presence of saxitoxin. Saxitoxin is also bound by receptors unrelated to the sodium channel. The first is saxiphilin, a protein found in the circulatory fluid of many animals used in a receptor binding assay much like that described above for the sodium channel. An alternative to natural receptors is to generate antibodies to saxitoxin. Saxitoxin itself does not elicit an immune response from mammals and one does not want the animal used to generate the antibodies to die from the toxin. The toxin then is linked to a carrier protein to create an antigenic epitope and reduce its toxicity. An example carrier is the enzyme horseradish peroxidase, which can also be used to generate an assay signal from the enzyme converting a colorless substrate to a visible product. Saxitoxin will interfere with the binding between the saxitoxin–horseradish peroxidase conjugate and the antibody generated to the toxin–protein complex and reduce any assay signal. Such assays, however, are servant to the fact that antibodies raised toward one member of the saxitoxin family may not detect other chemical relatives. View chapterExplore book Read full chapter URL: Reference work2005, Encyclopedia of Analytical Science (Second Edition)C. Robillot, L.E. Llewellyn Review article Chemical casualties – a clinical overview 2022, Anaesthesia & Intensive Care MedicineJonathan L. Begley, Elissa M. Milford Saxitoxin Saxitoxin is a potent paralytic shellfish toxin present in a variety of sea life including puffer fish. It can theoretically be delivered by ingestion, air, injection, but is most commonly ingested accidently. Mechanism of action: Saxitoxin causes sodium channel blockade. Signs and symptoms: Ingestion causes perioral numbness spreading to the face and neck, and in severe cases the limbs in association with peripheral muscle weakness, cranial nerve weakness, and respiratory failure. Gastrointestinal symptoms and headache are common. Symptoms peak at 12 hours. Management: Supportive management (often including intubation) facilitates complete recovery in the absence of complications. View article Read full article URL: Journal2022, Anaesthesia & Intensive Care MedicineJonathan L. Begley, Elissa M. Milford Chapter Saxitoxin 2005, Encyclopedia of Toxicology (Second Edition)Samantha E. Gad, Shayne C. Gad Mechanism of Toxicity Saxitoxin binds to the sodium channels in the membranes of excitable cells (neurons and muscle cells) blocking synaptic transmission. Saxitoxin is connected to red tides. Saxitoxin reduces nerve conduction velocities. View chapterExplore book Read full chapter URL: Reference work2005, Encyclopedia of Toxicology (Second Edition)Samantha E. Gad, Shayne C. Gad Review article Medical Toxicology 2007, Emergency Medicine Clinics of North AmericaChristopher P. Holstege MD, ... Stephen G. Dobmeier BSN Saxitoxin Saxitoxin (STX) is associated with the syndrome known as paralytic shellfish poisoning and it poses a worldwide health problem . STX is formed by dinoflagellates, which cause a phenomenon known as a red tide. Marine life (mollusks, crabs, and fish) may feed on these and bioaccumulate the dinoflagellate toxins. Humans may inadvertently consume intoxicated seafood [43,44]. Numerous outbreaks of paralytic shellfish poisoning have been reported worldwide. Paralytic shellfish poisoning is caused by not only saxitoxin, but also other chemical variations of saxitoxin, for example decarbamoyl STX (dc-STX) and N-sulfocarbamoyl (B1) toxin. Governments reportedly began experimenting with saxitoxin in the 1950s. In 1969, President Richard Nixon banned biological weapons. Subsequently, nearly all the United States STX produced was destroyed. However, in 1975, approximately 10 g of STX was discovered in a storage facility, triggering a US Senate investigation and a redistribution of the remaining STX to universities for research purposes. In recent years, terrorist events have resulted in increased regulations of STX . STX has been listed in the “Select Agent Program” by the United States (Box 2). The US Department of Health and Human Services and the US Department of Agriculture published final rules, which implement the provisions of the USA Patriot Act and Public Health Security and Bioterrorism Preparedness and Response Act of 2002. These rules set forth the requirements for possession, use, and transfer of select agents and toxins. The select chemical toxins identified in the final rules have the potential to pose a severe threat to public health and safety, to animal and plant health, or to animal and plant products. The CDC regulates the possession, use, and transfer of these select agents and toxins that have the potential to pose a severe threat to public health and safety. The CDC Select Agent Program oversees these activities and registers all laboratories and other entities in the United States that possess, use, or transfer a select agent or toxin. Box 2 Health and human services select agent toxins : Abrin : Conotoxins : Ricin : Shigatoxin : T-2 toxin : Botulism toxins : Diacetoxyscirpenol : Saxitoxin : Staphylococcal enterotoxins : Tetrodotoxin Saxitoxin is also currently listed in schedule 1 of the Chemical Weapons Convention as one of the most potent toxins known. Saxitoxin and ricin are the only two naturally occurring toxins classified as schedule 1 of the Chemical Weapons Convention. Properties STX is a naturally occurring toxin. STX has also been synthesized using various different methods . STX is water-soluble, heat stable, and unaffected by cooking . The LCt50 of STX is 5 mg/min/m3 and is reportedly 2000 times more toxic than sodium cyanide by weight . Routes of exposure STX toxicity can occur by either ingestion or inhalation. Contamination of food or water with STX are viewed as viable concerns for mass human exposure . In animal experiments, inhalational routes of administration are more potent than oral routes, causing death within minutes compared with hours for oral . Pathophysiology STX is a specific high-affinity blocking ligand of voltage-dependent sodium channels . STX binds competitively to a site on the external surface of the channel, named toxin site 1. This binding inhibits sodium flux through these ion channels rendering excitable tissues such as nerves and muscle nonfunctional. Clinical manifestations There are no published reports of saxitoxin being used by terrorists, though concern remains high pertaining to its potential use, its marked toxicity, and its natural availability. There are numerous reports in the literature of saxitoxin being ingested in contaminated food, which sheds light on the clinical manifestations of STX toxicity. In two separate outbreaks reported in 1990, nine fisherman developed symptoms within 2 hours following consumption of STX-contaminated shellfish . Reported symptoms included numbness of the mouth (six of nine), vomiting (four of nine), paresthesias of the extremities (seven of nine), numbness and tingling of the tongue (two of nine), numbness of the face (five of nine), low-back pain (six of nine), and periorbital edema (one of nine). Two hours following the onset of symptoms, one of the fishermen suffered a “cardiopulmonary arrest” and died. Of the remaining eight, only two required hospitalization. The duration of neurologic symptoms was less than 24 hours, and those who had low back pain approximately 3 days. Of those who survived, all recovered uneventfully. In 1994, nine people, 38 to 80 years old, presented to the ED 6 to 18 hours after the first symptoms of STX toxicity occurred . In all cases, symptoms began 60 to 90 minutes after ingestion of contaminated shellfish and consisted of dizziness, ataxia, paresthesias (oral–facial and extremity), but no gastrointestinal complaints. Six people had progressive impairment of gait, confining them to a wheelchair after 6 to 8 hours. All had normal vital signs on arrival with axial ataxia and bilateral dysmetria. Four had bilateral nystagmus, and three had dysarthria. All had distal stocking and glove superficial impairment and bilateral moderate position and vibratory sense impairment, with preservation of tendon jerks. Hematology and biochemistry routine testing were normal in all patients. Cerebral spinal fluid from two patients was normal. None developed respiratory involvement, and all were discharged from the hospital within 3 days. Within 2 weeks, all recovered but still complained of fatigue, paresthesias, and memory loss, which persisted up to 3 months. There were no long-term sequelae. In 2000, a 65-year-old female reportedly ingested STX-contaminated blowfish and within minutes developed tingling of her lips and tongue, which intensified over the ensuing 2 hours . She developed increasing chest pain and had mild tachycardia and hypertension (160/70 mm Hg) requiring treatment with topical nitroglycerin . Six to 8 hours after ingestion, she developed ascending paralysis and declining pulmonary function requiring intubation. Over the following day she regained reflexes, and voluntary movement and was extubated 72 hours later. In 2002, two fishermen died following STX-contaminated shellfish ingestion . Symptoms before demise included lip paresthesias, nausea, extremity weakness, and “tongue immobilization.” The forensic examination of both victims did not show pathologic abnormalities with the exception of the lungs, which revealed pulmonary edema. STX was detected in gastric contents, body fluids, and tissue samples. In summary, following oral exposure, STX causes prominent paresthesias, often beginning circumorally and spreading to the limbs. This can then progress to paralysis with retention of reflexes . Cranial nerve dysfunction, hypersalivation, diaphoresis, respiratory failure, hypertension, and hypotension have all been reported. Laboratory testing Measuring levels of saxitoxin in human samples requires early acquisition of samples. Saxitoxin undergoes minimal metabolism, but is rapidly excreted into the urine. Saxitoxin can also concentrate in liver, spleen, and central nervous system tissues . Because of the extremely rapid excretion profile of saxitoxin compounds, urine samples are preferred rather than serum samples. All saxitoxin testing should be performed by a state health department laboratory where high performance LCMS/MS is used. These LCMS/MS methods can provide saxitoxin fingerprint analyses to determine if the saxitoxin agent was derived from an organic source (shellfish ingestion) or a purified saxitoxin source (biological warfare agent). Any cases suspected of saxitoxin exposure should collect urine specimens within 24 hours of exposure. Higher levels of saxitoxin exposure can extend detection times for several days . Alternatively less sensitive methods are available for detecting saxitoxin in human samples, such as high-performance liquid chromatography and receptor binding assays [57,58]. State facilities may request submission of shellfish samples, if the physician suspects saxitoxin exposure due to ingestion of contaminated shellfish. Current United States Food and Drug Administration (USFDA) guidelines require all shellfish sold in the United States be tested for paralytic shellfish toxins. The mouse bioassay is the current gold standard method approved by the USFDA for detection and quantitation of paralytic shellfish toxins in shellfish marketed to human consumers . This method identifies paralytic shellfish toxins by injecting mice with 1 ml of an acidic extract, then measuring time of death as a measurement of toxicity (5–15 min). The toxicity of the sample is then calculated with reference to dose–response curves established with saxitoxin standards and expressed in mouse units. Specificity for saxitoxin is based on the extremely rapid toxicity profile compared with other paralytic shellfish toxins. This method is not used to determine levels in humans. Treatment/disposition There is no known antidote for STX toxicity. Most patients will recover if they receive adequate and timely supportive care. View article Read full article URL: Journal2007, Emergency Medicine Clinics of North AmericaChristopher P. Holstege MD, ... Stephen G. Dobmeier BSN Related terms: Transcription Factors Guanosine Triphosphate Tetrodotoxin Shellfish Poisoning Cell Membrane Neurotoxin Intoxication Phosphotransferase Shellfish Toxin Sodium Channel View all Topics
13324	https://www.booksamillion.com/product/9789810238629	Statistical Mechanics : Rigorous Results by David Ruelle Shop Everything You Need for Your Faith Journey, In Stores & Online 15% off ALL Manga Online - Over 10,000 Items! Get FREE SHIPPING Every Day, Every Order! Join Our Millionaire's Club! Click Here to Start Saving Today! Get It Today! Same-Day Delivery or Buy Online, Pick Up In Stores Options Available in All Stores Shop Everything You Need for Your Faith Journey, In Stores & Online 15% off ALL Manga Online - Over 10,000 Items! Get FREE SHIPPING Every Day, Every Order! Join Our Millionaire's Club! Click Here to Start Saving Today! Get It Today! Same-Day Delivery or Buy Online, Pick Up In Stores Options Available in All Stores  Set My Store Millionaire's Club Help Gift Cards Events Educators menu close Books ALL Books Favorites Bestsellers Limited Editions AutographedBooks.com Cranford Classics Summer Says Boxed Sets Sales & Promotions What We Recommend Fiction Picks Nonfiction Picks New York Times Bestsellers Award Winners Staff Picks Book Clubs Shop All Featured New Releases Coming Soon & Pre-Orders Success Magazine Bestsellers Blog Shop by Category Art & Photography Banned Books Bibles Classics Calendars & Planners Cookbooks Current Events Debut Authors Diverse Voices Faithpoint Exclusives Graphic Novels Home & Design LGBTQ+ Voices Manga Mythology Poetry Roleplaying Guides Spanish Titles Browse More Subjects Fiction #BookTok Cozy Reading Fantasy Horror Historical Literary Mystery Romance Science Fiction Thrillers Nonfiction Biography Business Entertainment Health & Wellness History Science True Crime Shop By Format Audiobooks Coloring Books eBooks Large Print Paperback More Information Millionaire's Club Educators Business Services BookPage Authors & Publishers #Booktok ALL #Booktok Favorites Bestsellers New & Coming Soon Beautiful Books Romantasy Cranford Classics Dark Romance #BookTok Gifts Young Adult Autographed Copies Shelf Stoppers Rom Com Spicytok FYP Cozy Romance Featured Books-A-Million Exclusives Special Editions Illustrated Editions Fantasy & Sci-Fi Shop By Author A-G Sarah Adams Navessa Allen Jennifer L. Armentrout Lauren Asher Victoria Aveyard Tessa Bailey Holly Black Olivie Blake Pierce Brown H. D. Carlton Kiera Cass Sara Cate Cassandra Clare Blake Crouch Ruby Dixon Alice Feeney Tana French Stephanie Garber Rachel Gillig Laurie Gilmore Shop by Author G-L Hannah Grace Jasmine Guillory John Gwynne Callie Hart Ali Hazelwood Grady Hendrix Emily Henry Matt Haig Talia Hibbert Ana Huang Holly Jackson Simone St. James Danielle L Jensen Abby Jimenez Stephen Graham Jones Toshikazu Kawaguchi Rina Kent T Kingfisher TJ Klune R. F. Kuang Shop By Author L-R Christina Lauren E. Lockhart Sarah J. Maas Tahereh Mafi Kerri Maniscalco Freida McFadden Emily McIntire Karen M. McManus Silvia Moreno-Garcia Amber Nicole Alice Oseman Lynn Painter Caroline Peckham Devney Perry Terry Pratchett Nita Prose Ava Reid Taylor Jenkins Reid Leigh Rivers Katee Robert Shop by Author R-Z Lauren Roberts Jeneva Rose Kennedy Ryan Lyla Sage Riley Sager Brandon Sanderson V. E. Schwab Lucy Score Samantha Shannon Neal Shusterman Elsie Silver Sierra Simone Kaylie Smith Jesse Q. Sutanto Shantel Tessier Suzanne Valenti Chloe Walsh Ruth Ware Brynne Weaver Rebecca Yarros Special Editions Manga ALL Manga Shop By Series A-G Ao Haru Ride Attack on Titan Beastars Berserk Bleach Blue Period Call of the Night Chainsaw Man Death Note Demon Slayer Don't Toy With Me, Miss Nagatoro Given Goodnight Punpun Shop By Series H-M Haikyu!! Hell's Paradise Horimiya Jojo's Bizarre Adventure Jujutsu Kaisen Komi Can't Communicate Legend of Zelda Mob Psycho Monster My Dress-Up Darling My Hero Academia Shop By Series N-Z Naruto One Piece One-Punch Man Persona 5 Pokemon The Promised Neverland RWBY Splatoon Spy X Family Toilet-Bound Hanako-Kun Wotakoi : Love Is Hard for Otaku Yarichin Bitch Club Customers Also Liked Avatar the Series Collectible Plush Con Collectibles Funko Gaming Gifts Graphic Novels Horror Teen Books & Gifts Trading Cards More in Manga Bestsellers New Coming Soon Boxed Sets Manga to Anime Manga & Anime Collectibles Junji Ito Web Toons A uto g rap h ed Bibles Kids ALL Kids Favorites Bestsellers New Releases Coming Soon Comic & Graphic Novels 50 Books to Read Before 5 Coloring Page to Screen Boxed Sets Classics Favorite Characters Bluey Disney Dog Man Harry Potter Junie B. Jones Magic Tree House Percy Jackson Star Wars Kids Fiction Action & Adventure Board Books Fantasy Horror Mysteries Science Fiction Sports Stories Christian Fiction Shop All Fiction Nonfiction Activity Books Animals History Science & Nature Shop All Nonfiction Shop by Age Ages 0-2 Ages 3-5 Ages 6-8 Ages 9-12 Top Authors Eric Carle Erin Hunter Ann M. Martin Dav Pilkey Rick Riordan J. K. Rowling Rachel Renee Russel Dr. Seuss Tui T. Sutherland Mo Willems Educational Resources Educator Discount Card Study Guides Tools for Educators Preschool Kindergarten Elementary School Middle School High School Anti-Racism & Social Justice First Readers National Geographic S.T.E.M. More Categories Faithpoint Board Books Autographed Young Adult ALL Young Adult Favorites YA Bestsellers YA Coming Soon BookTok YA New Releases Boxed Sets Action & Adventure Fantasy Fiction Historical Fiction Romance Science Fiction Thrillers Popular Authors Jennifer L. Armentrout Victoria Aveyard Leigh Bardugo Jennifer Lynn Barnes Holly Black Kiera Cass Cassandra Clare Stephanie Garber Jenny Han Holly Jackson E. Lockhart Tahereh Mafi Laura Nowlin Lynn Painter Lauren Roberts Chloe Walsh Bestselling Series A-K Avatar: The Last Airbender Boys of Tommen Crave Crimson Moth Disney Twisted Tales Ember in the Ashes Good Girls Guide to Murder Harry Potter Heartstopper Hunger Games If You Liked The Hunger Games The Princes of Sin Bestselling Series L-Z Legendborn Cycle Lightlark Saga Once Upon A Broken Heart Powerless Trilogy Shadow and Bone Shatter Me The Arc of a Scythe The Inheritance Games The Summer I Turned Pretty To All the Boys I've Loved Before Twilight We Were Liars Vinyl Records Adult Collectibles ALL Adult Collectibles Favorites POP MART Powerless Collection Blind Boxes LEGO Loungefly Funko Collectible Plush Manga #BookTok Gifts Music BAM! Exclusive Vinyl Upcoming Releases New Releases Vinyl Accessories Best Selling CD's Shop Music Shop Featured Notable Personalities Behind-The-Scenes Home Decor Trading Cards Coloring Books Shop All Pop Culture Page to Screen Film Television Anime Kids Tabletop Role-playing Dungeons & Dragons Pathfinder Dice & Accessories Shop All Role-playing Games Fandoms Hello Kitty & Friends Lilo & Stitch Star Wars Disney Pokemon Batman Harry Potter Teenage Mutant Ninja Turtles Taylor Swift Five Nights at Freddy's Wicked How To Train Your Dragon One Piece Fallout Minecraft Toys & Games ALL Toys & Games LEGO Bestsellers Art and Architecture Icons LEGO Creator $25 and Under $50 and Under $100 and Under $200 and Under Shop All LEGO Bestsellers POP MART Trading Cards Top Toys Plush Blind Boxes Games & Puzzles Top Games Trading Cards Puzzles Ravensburger Puzzles Bookshelf Builds Roleplaying and D&D Gift Shop ALL Gift Shop Favorites Special Editions #BookTok Wicked Autographed Books Boxed Sets Gift Cards Books You Need to Read in 2025 Special Editions Best in Fiction Manga Teens & Young Adults Memoirs & Biographies Notable Nonfiction Kids Shop All Best Gifts for Kids Bluey Comics & Graphic Novels Top Toys Wish Lilo & Stitch Harry Potter Gaming Winnie the Pooh Disney Shop by Age 0-2 3-5 6-8 9-12 Shop by Series Shop by Subject Pop Culture & Fandoms Anime & Manga Funko Loungefly Plush Trading Cards Squishmallows #BookTok Gifts Horror Gifts Adult Coloring Books Godzilla Vinyl Shop Shop All Faith Gifts Beautiful Bibles Devotionals Gifts for Her Gifts for Him Gifts for Kids Prayer Books Shop All Sale ALL Sale Bargains in Fiction Action & Adventure Fantasy Historical Literary Mysteries Romance Sci-Fi Spies & Espionage Suspense Shop All Bargains in Nonfiction Biographies Body, Mind & Spirit Business Cooking Health & Wellness History Humor Music & Entertainment Personal Care Political Science Science Social Science Bargains in Young Adult Books Action & Adventure Fantasy Historical Fiction Romance Thrillers Shop All Bargains in Kids Fiction Action & Adventure Activity Books Comics & Graphic Novels Fantasy Horror Mysteries Sci-Fi Sports Stories Shop All Bargains in Kids Nonfiction Activity Books Animals Biography History Science & Nature Shop All Bargains in Faith & Inspiration Christian Fiction Christian Living Shop All Bargain Favorites Bestselling Fiction Myths & Lore True Crime Classics Body, Mind & Spirit Engage Your Brain Gothic Fiction Mysteries & Thrillers Kids Under $15 #BookTok keyboard_arrow_down  Sign In person Sign In Order Status Gift Card Balance Cart  0 Books Favorites Bestsellers Limited Editions AutographedBooks.com Cranford Classics Summer Says Boxed Sets Sales & Promotions What We Recommend Fiction Picks Nonfiction Picks New York Times Bestsellers Award Winners Staff Picks Book Clubs Shop All Featured New Releases Coming Soon & Pre-Orders Success Magazine Bestsellers Blog Shop by Category Art & Photography Banned Books Bibles Classics Calendars & Planners Cookbooks Current Events Debut Authors Diverse Voices Faithpoint Exclusives Graphic Novels Home & Design LGBTQ+ Voices Manga Mythology Poetry Roleplaying Guides Spanish Titles Browse More Subjects Fiction #BookTok Cozy Reading Fantasy Horror Historical Literary Mystery Romance Science Fiction Thrillers Nonfiction Biography Business Entertainment Health & Wellness History Science True Crime Shop By Format Audiobooks Coloring Books eBooks Large Print Paperback More Information Millionaire's Club Educators Business Services BookPage Authors & Publishers #Booktok Favorites Bestsellers New & Coming Soon Beautiful Books Romantasy Cranford Classics Dark Romance #BookTok Gifts Young Adult Autographed Copies Shelf Stoppers Rom Com Spicytok FYP Cozy Romance Featured Books-A-Million Exclusives Special Editions Illustrated Editions Fantasy & Sci-Fi Shop By Author A-G Sarah Adams Navessa Allen Jennifer L. Armentrout Lauren Asher Victoria Aveyard Tessa Bailey Holly Black Olivie Blake Pierce Brown H. D. Carlton Kiera Cass Sara Cate Cassandra Clare Blake Crouch Ruby Dixon Alice Feeney Tana French Stephanie Garber Rachel Gillig Laurie Gilmore Shop by Author G-L Hannah Grace Jasmine Guillory John Gwynne Callie Hart Ali Hazelwood Grady Hendrix Emily Henry Matt Haig Talia Hibbert Ana Huang Holly Jackson Simone St. James Danielle L Jensen Abby Jimenez Stephen Graham Jones Toshikazu Kawaguchi Rina Kent T Kingfisher TJ Klune R. F. Kuang Shop By Author L-R Christina Lauren E. Lockhart Sarah J. Maas Tahereh Mafi Kerri Maniscalco Freida McFadden Emily McIntire Karen M. McManus Silvia Moreno-Garcia Amber Nicole Alice Oseman Lynn Painter Caroline Peckham Devney Perry Terry Pratchett Nita Prose Ava Reid Taylor Jenkins Reid Leigh Rivers Katee Robert Shop by Author R-Z Lauren Roberts Jeneva Rose Kennedy Ryan Lyla Sage Riley Sager Brandon Sanderson V. E. Schwab Lucy Score Samantha Shannon Neal Shusterman Elsie Silver Sierra Simone Kaylie Smith Jesse Q. Sutanto Shantel Tessier Suzanne Valenti Chloe Walsh Ruth Ware Brynne Weaver Rebecca Yarros Special Editions Manga Shop By Series A-G Ao Haru Ride Attack on Titan Beastars Berserk Bleach Blue Period Call of the Night Chainsaw Man Death Note Demon Slayer Don't Toy With Me, Miss Nagatoro Given Goodnight Punpun Shop By Series H-M Haikyu!! Hell's Paradise Horimiya Jojo's Bizarre Adventure Jujutsu Kaisen Komi Can't Communicate Legend of Zelda Mob Psycho Monster My Dress-Up Darling My Hero Academia Shop By Series N-Z Naruto One Piece One-Punch Man Persona 5 Pokemon The Promised Neverland RWBY Splatoon Spy X Family Toilet-Bound Hanako-Kun Wotakoi : Love Is Hard for Otaku Yarichin Bitch Club Customers Also Liked Avatar the Series Collectible Plush Con Collectibles Funko Gaming Gifts Graphic Novels Horror Teen Books & Gifts Trading Cards More in Manga Bestsellers New Coming Soon Boxed Sets Manga to Anime Manga & Anime Collectibles Junji Ito Web Toons A uto g rap h ed Bibles Kids Favorites Bestsellers New Releases Coming Soon Comic & Graphic Novels 50 Books to Read Before 5 Coloring Page to Screen Boxed Sets Classics Favorite Characters Bluey Disney Dog Man Harry Potter Junie B. Jones Magic Tree House Percy Jackson Star Wars Kids Fiction Action & Adventure Board Books Fantasy Horror Mysteries Science Fiction Sports Stories Christian Fiction Shop All Fiction Nonfiction Activity Books Animals History Science & Nature Shop All Nonfiction Shop by Age Ages 0-2 Ages 3-5 Ages 6-8 Ages 9-12 Top Authors Eric Carle Erin Hunter Ann M. Martin Dav Pilkey Rick Riordan J. K. Rowling Rachel Renee Russel Dr. Seuss Tui T. Sutherland Mo Willems Educational Resources Educator Discount Card Study Guides Tools for Educators Preschool Kindergarten Elementary School Middle School High School Anti-Racism & Social Justice First Readers National Geographic S.T.E.M. More Categories Faithpoint Board Books Autographed Young Adult Favorites YA Bestsellers YA Coming Soon BookTok YA New Releases Boxed Sets Action & Adventure Fantasy Fiction Historical Fiction Romance Science Fiction Thrillers Popular Authors Jennifer L. Armentrout Victoria Aveyard Leigh Bardugo Jennifer Lynn Barnes Holly Black Kiera Cass Cassandra Clare Stephanie Garber Jenny Han Holly Jackson E. Lockhart Tahereh Mafi Laura Nowlin Lynn Painter Lauren Roberts Chloe Walsh Bestselling Series A-K Avatar: The Last Airbender Boys of Tommen Crave Crimson Moth Disney Twisted Tales Ember in the Ashes Good Girls Guide to Murder Harry Potter Heartstopper Hunger Games If You Liked The Hunger Games The Princes of Sin Bestselling Series L-Z Legendborn Cycle Lightlark Saga Once Upon A Broken Heart Powerless Trilogy Shadow and Bone Shatter Me The Arc of a Scythe The Inheritance Games The Summer I Turned Pretty To All the Boys I've Loved Before Twilight We Were Liars Vinyl Records Adult Collectibles Favorites POP MART Powerless Collection Blind Boxes LEGO Loungefly Funko Collectible Plush Manga #BookTok Gifts Music BAM! Exclusive Vinyl Upcoming Releases New Releases Vinyl Accessories Best Selling CD's Shop Music Shop Featured Notable Personalities Behind-The-Scenes Home Decor Trading Cards Coloring Books Shop All Pop Culture Page to Screen Film Television Anime Kids Tabletop Role-playing Dungeons & Dragons Pathfinder Dice & Accessories Shop All Role-playing Games Fandoms Hello Kitty & Friends Lilo & Stitch Star Wars Disney Pokemon Batman Harry Potter Teenage Mutant Ninja Turtles Taylor Swift Five Nights at Freddy's Wicked How To Train Your Dragon One Piece Fallout Minecraft Toys & Games LEGO Bestsellers Art and Architecture Icons LEGO Creator $25 and Under $50 and Under $100 and Under $200 and Under Shop All LEGO Bestsellers POP MART Trading Cards Top Toys Plush Blind Boxes Games & Puzzles Top Games Trading Cards Puzzles Ravensburger Puzzles Bookshelf Builds Roleplaying and D&D Gift Shop Favorites Special Editions #BookTok Wicked Autographed Books Boxed Sets Gift Cards Books You Need to Read in 2025 Special Editions Best in Fiction Manga Teens & Young Adults Memoirs & Biographies Notable Nonfiction Kids Shop All Best Gifts for Kids Bluey Comics & Graphic Novels Top Toys Wish Lilo & Stitch Harry Potter Gaming Winnie the Pooh Disney Shop by Age 0-2 3-5 6-8 9-12 Shop by Series Shop by Subject Pop Culture & Fandoms Anime & Manga Funko Loungefly Plush Trading Cards Squishmallows #BookTok Gifts Horror Gifts Adult Coloring Books Godzilla Vinyl Shop Shop All Faith Gifts Beautiful Bibles Devotionals Gifts for Her Gifts for Him Gifts for Kids Prayer Books Shop All Sale Bargains in Fiction Action & Adventure Fantasy Historical Literary Mysteries Romance Sci-Fi Spies & Espionage Suspense Shop All Bargains in Nonfiction Biographies Body, Mind & Spirit Business Cooking Health & Wellness History Humor Music & Entertainment Personal Care Political Science Science Social Science Bargains in Young Adult Books Action & Adventure Fantasy Historical Fiction Romance Thrillers Shop All Bargains in Kids Fiction Action & Adventure Activity Books Comics & Graphic Novels Fantasy Horror Mysteries Sci-Fi Sports Stories Shop All Bargains in Kids Nonfiction Activity Books Animals Biography History Science & Nature Shop All Bargains in Faith & Inspiration Christian Fiction Christian Living Shop All Bargain Favorites Bestselling Fiction Myths & Lore True Crime Classics Body, Mind & Spirit Engage Your Brain Gothic Fiction Mysteries & Thrillers Kids Under $15 #BookTok { "item_title" : "Statistical Mechanics", "item_author" : [" David Ruelle "], "item_description" : "This classic book marks the beginning of an era of vigorous mathematical progress in equilibrium statistical mechanics. Its treatment of the infinite system limit has not been superseded, and the discussion of thermodynamic functions and states remains basic for more recent work. The conceptual foundation provided by the Rigorous Results remains invaluable for the study of the spectacular developments of statistical mechanics in the second half of the 20th century.", "item_img_path" : " "price_data" : { "retail_price" : "95.00", "online_price" : "95.00", "our_price" : "95.00", "club_price" : "95.00", "savings_pct" : "0", "savings_amt" : "0.00", "club_savings_pct" : "0", "club_savings_amt" : "0.00", "discount_pct" : "10", "store_price" : "" } } Save Statistical Mechanics : Rigorous Results by David Ruelle 0.0 No RatingsWrite the First Review local_shippingShip to Me In Stock. FREE Shipping for Club Membershelp storeIn-Store Pickup Hardcover $95.00 Add to Cart  "Add Statistical Mechanics to your Cart")+ Add to Wishlist Zero interest 4 payments of $24 with Affirm. Learn more Overview This classic book marks the beginning of an era of vigorous mathematical progress in equilibrium statistical mechanics. Its treatment of the infinite system limit has not been superseded, and the discussion of thermodynamic functions and states remains basic for more recent work. The conceptual foundation provided by the Rigorous Results remains invaluable for the study of the spectacular developments of statistical mechanics in the second half of the 20th century. This item is Non-Returnable )x The Electromagnetic Origin of Quantum Theory and Light Dale M. Grimes$118.00 Hardcover )x Transverse Spin Physics Vincenzo Barone$128.00 Hardcover )x Inverse Semigroups, the Theory of Partial Symmetries Mark V. Lawson$155.00 Hardcover )x Fundamentals of Equations of State Shalom Eliezer$138.00 Hardcover )x Gauge Fields in Condensed Matter (in 2 Volumes)Hagen Kleinert$297.00 Hardcover )x Introduction to the Theory of Critical Phenomena Dimo I. Uzunov$158.00 Hardcover )x Lectures on Non-Equilibrium Theory of Condensed Matter Ladislaus Alexander Banyai$118.00 Hardcover )x Quasimolecular Modelling Donald Greenspan$65.00 Hardcover )x A Survey of Nonlinear Dynamics (Chaos Theory)Richard L. Ingraham$56.00 Hardcover )x Coherent States John R. Klauder$114.00 Hardcover )x Adiabatic Thermodynamics of Fluids Christian Fronsdal$128.00 Hardcover )x Diagrammatics Michael V. Sadovskii$148.00 Hardcover )x The Electromagnetic Origin of Quantum Theory and Light Dale M. Grimes$118.00 Hardcover )x Transverse Spin Physics Vincenzo Barone$128.00 Hardcover )x Inverse Semigroups, the Theory of Partial Symmetries Mark V. Lawson$155.00 Hardcover )x Fundamentals of Equations of State Shalom Eliezer$138.00 Hardcover )x Gauge Fields in Condensed Matter (in 2 Volumes)Hagen Kleinert$297.00 Hardcover )x Introduction to the Theory of Critical Phenomena Dimo I. Uzunov$158.00 Hardcover )x Lectures on Non-Equilibrium Theory of Condensed Matter Ladislaus Alexander Banyai$118.00 Hardcover )x Quasimolecular Modelling Donald Greenspan$65.00 Hardcover 1 )x A Survey of Nonlinear Dynamics (Chaos Theory)Richard L. Ingraham$56.00 Hardcover )x Coherent States John R. Klauder$114.00 Hardcover )x Adiabatic Thermodynamics of Fluids Christian Fronsdal$128.00 Hardcover )x Diagrammatics Michael V. Sadovskii$148.00 Hardcover )x The Electromagnetic Origin of Quantum Theory and Light Dale M. Grimes$118.00 Hardcover )x Transverse Spin Physics Vincenzo Barone$128.00 Hardcover )x Inverse Semigroups, the Theory of Partial Symmetries Mark V. Lawson$155.00 Hardcover )x Fundamentals of Equations of State Shalom Eliezer$138.00 Hardcover Details ISBN-13: 9789810238629 ISBN-10: 9810238622 Publisher: World Scientific Publishing Company Publish Date: April 1999 Dimensions: 9 x 6.2 x 0.9 inches Shipping Weight: 1.2 pounds Page Count: 236 Related Categories Books Science Mechanics - General Books Science Physics - General )x Phoebe Berman's Gonna Lose It Autographed Brooke Averick$28.00 Hardcover )x Merry Christmas, You Filthy Animal Autographed (Exclusive Deluxe Edition)Meghan Quinn$18.99 Paperback )x The College Scam Charlie Kirk$27.89 Hardcover )x Alchemised SenLinYu$31.36 Hardcover )x 107 Days Kamala Harris$26.40 Hardcover )x Becoming an Artist Autographed Scott Christian Sava$24.99 Hardcover )x Stop, in the Name of God Charlie Kirk$28.49 Hardcover )x Pagan Threat Lucas Miles$27.99 Hardcover )x Right Wing Revolution Charlie Kirk$26.99 Hardcover )x Phoebe Berman's Gonna Lose It Autographed Brooke Averick$28.00 Hardcover )x Merry Christmas, You Filthy Animal Autographed (Exclusive Deluxe Edition)Meghan Quinn$18.99 Paperback )x The College Scam Charlie Kirk$27.89 Hardcover )x Alchemised SenLinYu$31.36 Hardcover )x 107 Days Kamala Harris$26.40 Hardcover )x Becoming an Artist Autographed Scott Christian Sava$24.99 Hardcover )x Stop, in the Name of God Charlie Kirk$28.49 Hardcover )x Pagan Threat Lucas Miles$27.99 Hardcover 1 )x Right Wing Revolution Charlie Kirk$26.99 Hardcover )x Phoebe Berman's Gonna Lose It Autographed Brooke Averick$28.00 Hardcover )x Merry Christmas, You Filthy Animal Autographed (Exclusive Deluxe Edition)Meghan Quinn$18.99 Paperback )x The College Scam Charlie Kirk$27.89 Hardcover )x Alchemised SenLinYu$31.36 Hardcover )x 107 Days Kamala Harris$26.40 Hardcover )x Becoming an Artist Autographed Scott Christian Sava$24.99 Hardcover )x Stop, in the Name of God Charlie Kirk$28.49 Hardcover 1 BAM Customer Reviews Write the First Review Questions? Reach us via Live Chat Get the latest on trends, deals, and promotions Enter your email address Stay Connected with us when you sign up for text alerts. Enter phone number Related Links Top Toys Journals, Stationery & Notebooks Harry Potter SHOP arrow_drop_down Books eBooks Kids-A-Million Teen Bargain Books Entertainment Toys Fandom SERVICES arrow_drop_down Millionaire's Club Affiliate Program Business Services Education Services Gift Cards COMPANY arrow_drop_down Find a Store About Us Career Opportunities For Authors & Publishers Contact Us Read Our Blog Digital Accessibility HELP arrow_drop_down Help Desk Manage My Account Reset My Password Shopping Cart Shipping Info Where's My Order? Returns JOIN THE MILLIONAIRE'S CLUB Join the Millionaire's Club and receive FREE SHIPPING, plus tons of exclusive benefits and offers. JOIN THE CLUB NOW > © 1996–2025 Books-A-Million, Inc. All rights reserved. BOOKSAMILLION.COM \| Joemuggs.com \| Terms of Use \| CA Disclosures \| Privacy Policy store zip code Close In-Store Availability Close
13325	https://fr.scribd.com/document/225979790/Hyperbolic-Function-wikiPEDIA	Hyperbolic Function - wikiPEDIA \| PDF \| Trigonometric Functions \| Mathematical Analysis Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 353 views 11 pages Hyperbolic Function - wikiPEDIA Hyperbolic functions are analogs of trigonometric functions that are defined in terms of exponential functions rather than circular functions. The basic hyperbolic functions are hyperbolic s… Full description Uploaded by Amiya Ranjan Malik AI-enhanced title and description Go to previous items Go to next items Download Save Save Hyperbolic Function.wikiPEDIA For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Hyperbolic Function.wikiPEDIA For Later You are on page 1/ 11 Search Fullscreen Hyperbolic function From Wikipedia, the free encyclopedia Jump to:navigation,searchA ray through the origin intercepts the hyperbola in the point , where is twice the area between the ray and the -axis. For points on the hyperbola bel ow t he -axis, the area is considered negative (seeanimated versionwith comparison with the trigonometric (circular) functions). Inmathematics, hyperbolic functions are analogs of the ordinarytrigonometric,or circular, functions. The basic hyperbolic functions are the hyperbolic sine "sinh" (/ ˈ s ɪ n tʃ /or/ ˈʃaɪ n/), and the hyperbolic cosine "cosh" (/ ˈ k ɒʃ /), from which are derived the hyperbolic tangent "tanh" (/ ˈ tæn tʃ /or/ ˈθ æn/ [ citation need ed []]( ), and so on, corresponding to th e derived trigonometric functions. Theinverse hyperbolic functionsare the area hyperbolic sine "arsinh" (also called "asinh" or sometimes "arcsinh") and so on. Just as the points (cos t , sin t ) form a circle with a unit radius, the points (cosh t , sinh t ) form the right half of the equilateralhyperbola.Hyperbolic functions occur in the solutions of some important lineardifferential equations,for example the equation defining acatenary,and Laplace's equationinCartesian coordinates.The latter is important in man y areas ofphysics,includingelectromagnetic theory,heat transfer,fluid dynamics,andspecial relativity.The hyperbolic functions take real values for a real argument called ahyperbolic angle.In complex analysis, they are simplyrational functionsofexponentials,and so aremeromorphic.Hyperbolic functions were introduced in the 1760s independently byVincenzo Riccatiand Johann Heinrich Lambert. Riccati used Sc. and Cc. ( [co]sinus circulare ) to refer to circular functions and Sh. and Ch. ( [co]sinus hyperbolico ) to refer to hyperbolic functions. Lambert adopted the names but altered the abbreviations to what they are today. The abbreviations sh and ch are still used in some other languages, like French and Russian. adDownload to read ad-free Contents cosh( x ) is theaverageof e x and e −x (b) sinh( x ) is half thedifferenceof e x and e −x Hyperbolic functions (a) cosh and (b) sinh obtained using exponential functions e x and e − x The hyperbolic functions are:  Hyperbolic sine: adDownload to read ad-free  Hyperbolic cosine:  Hyperbolic tangent:  Hyperbolic cotangent:  Hyperbolic secant:  Hyperbolic cosecant: Hyperbolic functions can be introduced viaimaginary circular angles:  Hyperbolic sine:  Hyperbolic cosine:  Hyperbolic tangent:  Hyperbolic cotangent:  Hyperbolic secant: adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Dostojewski Notatki Z Podziemia (Całość) No ratings yet Dostojewski Notatki Z Podziemia (Całość) 102 pages Manual ACU802 100% (1) Manual ACU802 100 pages Hyperbolic Functions Overview No ratings yet Hyperbolic Functions Overview 10 pages Online Credit Risk Analytics and Modeling 0% (2) Online Credit Risk Analytics and Modeling 7 pages Hyperbolic Function: From Wikipedia, The Free Encyclopedia No ratings yet Hyperbolic Function: From Wikipedia, The Free Encyclopedia 9 pages Bifold SH06 PDF 50% (2) Bifold SH06 PDF 8 pages Standard Algebraic Expressions: Average No ratings yet Standard Algebraic Expressions: Average 4 pages Inverse Hyperbolic Functions Guide No ratings yet Inverse Hyperbolic Functions Guide 10 pages Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions No ratings yet Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 17 pages Lecture1 Calculus 2 No ratings yet Lecture1 Calculus 2 48 pages Hiperbolic Funtion No ratings yet Hiperbolic Funtion 8 pages NOTES 10.6 Hyperbolic Functions No ratings yet NOTES 10.6 Hyperbolic Functions 4 pages Hyperbolic Functions No ratings yet Hyperbolic Functions 9 pages Hyperbolic Function: Hyperbolic Tangent "Tanh" (/ Tæntʃ/ or No ratings yet Hyperbolic Function: Hyperbolic Tangent "Tanh" (/ Tæntʃ/ or 10 pages Prog 3 Hyperbolic Functions 1 No ratings yet Prog 3 Hyperbolic Functions 1 25 pages Hyperbolic Functions - Wikipedia No ratings yet Hyperbolic Functions - Wikipedia 18 pages IFPS User Manual 100% (1) IFPS User Manual 678 pages Hyperbolic Function - Wikipedia No ratings yet Hyperbolic Function - Wikipedia 11 pages Hyperbolic Function No ratings yet Hyperbolic Function 8 pages Hyperbolic Function No ratings yet Hyperbolic Function 9 pages Ia Lol XDDDDD No ratings yet Ia Lol XDDDDD 13 pages Sec67 PDF No ratings yet Sec67 PDF 12 pages ICSE VII Maths Ratio and Proportion 67% (3) ICSE VII Maths Ratio and Proportion 12 pages Hyperbolic Functions Explored No ratings yet Hyperbolic Functions Explored 19 pages Honeycomb FP2 No ratings yet Honeycomb FP2 165 pages Hyperbolic Functions No ratings yet Hyperbolic Functions 11 pages 5 8 No ratings yet 5 8 12 pages TX 9 FV 1 J CQUO4 Sy FTe ZCC No ratings yet TX 9 FV 1 J CQUO4 Sy FTe ZCC 21 pages Electron Mi Kro Skop No ratings yet Electron Mi Kro Skop 4 pages Calculus II: Hyperbolic Functions No ratings yet Calculus II: Hyperbolic Functions 18 pages Hyperbolic Function MTS 201 No ratings yet Hyperbolic Function MTS 201 13 pages Hyperbolic Functions Overview No ratings yet Hyperbolic Functions Overview 4 pages Scan 16 Apr 23 12 36 15 No ratings yet Scan 16 Apr 23 12 36 15 59 pages Hyperbolics and Inverse Differentiation No ratings yet Hyperbolics and Inverse Differentiation 10 pages Hyperbolic Function No ratings yet Hyperbolic Function 1 page Manual Instruction CPAM-EKA AIR C16 EKA KOOL V2 No ratings yet Manual Instruction CPAM-EKA AIR C16 EKA KOOL V2 8 pages Smithsonian Hyperbolic Fun CT I 020205 MBP No ratings yet Smithsonian Hyperbolic Fun CT I 020205 MBP 375 pages Hyperbolic Functions No ratings yet Hyperbolic Functions 51 pages Infectious Smile Gui No ratings yet Infectious Smile Gui 4 pages Chapter3 HyperbolicFunctions No ratings yet Chapter3 HyperbolicFunctions 16 pages Beyond Sin and Cos No ratings yet Beyond Sin and Cos 13 pages ATI FT Sensor Catalog 2005 No ratings yet ATI FT Sensor Catalog 2005 32 pages Hyperbolic Functions Guide No ratings yet Hyperbolic Functions Guide 11 pages 40 ICAGRI SlsiRevisi LukmanHakim No ratings yet 40 ICAGRI SlsiRevisi LukmanHakim 10 pages Hyperbolic No ratings yet Hyperbolic 23 pages Search No ratings yet Search 6 pages Soa PDF No ratings yet Soa PDF 64 pages EMC Engineering Exam Insights No ratings yet EMC Engineering Exam Insights 3 pages Hyperbolic Functions MST No ratings yet Hyperbolic Functions MST 12 pages Amlocor Steel Grade: Aml C No ratings yet Amlocor Steel Grade: Aml C 8 pages Unit 1 - What Kind of Movies Have You Been Watching Recently No ratings yet Unit 1 - What Kind of Movies Have You Been Watching Recently 12 pages Mast10006 Calculus 2 Notes - Compress No ratings yet Mast10006 Calculus 2 Notes - Compress 59 pages Hyperbolic Functions No ratings yet Hyperbolic Functions 5 pages Hyperbolic Functions Stewart No ratings yet Hyperbolic Functions Stewart 31 pages Introduction and Course Roadmap: Zicklin School of Business, Baruch College, CUNY No ratings yet Introduction and Course Roadmap: Zicklin School of Business, Baruch College, CUNY 4 pages Hyperbolic Functions No ratings yet Hyperbolic Functions 26 pages Socket Base Connections With Precast Concrete Columns PDF 100% (5) Socket Base Connections With Precast Concrete Columns PDF 11 pages Recognize A Potential Market No ratings yet Recognize A Potential Market 50 pages 4A Lesson Plan in English Grade 2: Valencia City Central School No ratings yet 4A Lesson Plan in English Grade 2: Valencia City Central School 3 pages Hyperbolic Functions Guide No ratings yet Hyperbolic Functions Guide 22 pages English MAINS Practice Shot 200 No ratings yet English MAINS Practice Shot 200 4 pages 29 Mathematics Hyperbolic Functions No ratings yet 29 Mathematics Hyperbolic Functions 14 pages Osborn Rule No ratings yet Osborn Rule 3 pages Fpure ch2 No ratings yet Fpure ch2 22 pages Hyperbolic Functions Norveen S. CAMPANILLA No ratings yet Hyperbolic Functions Norveen S. CAMPANILLA 15 pages Dorothy Allison No ratings yet Dorothy Allison 2 pages Hyperbolic Identities No ratings yet Hyperbolic Identities 4 pages M4 L1Assessment For Learning Using Assessment To Classify Learning and Understanding No ratings yet M4 L1Assessment For Learning Using Assessment To Classify Learning and Understanding 5 pages CP2 Chp6 HyperbolicFunctions No ratings yet CP2 Chp6 HyperbolicFunctions 46 pages Hyperbolic Functions Calculus PDF No ratings yet Hyperbolic Functions Calculus PDF 5 pages FSU MATH2400 Project1 No ratings yet FSU MATH2400 Project1 2 pages Transcript No ratings yet Transcript 3 pages Uninterruptible Power Supply (UPS) No ratings yet Uninterruptible Power Supply (UPS) 11 pages Decision Theory for Leaders No ratings yet Decision Theory for Leaders 12 pages Bộ đề kiểm tra định kì - lớp 6 - global success No ratings yet Bộ đề kiểm tra định kì - lớp 6 - global success 38 pages pdf24 Merged No ratings yet pdf24 Merged 27 pages Gingrich 2016 No ratings yet Gingrich 2016 5 pages SAEJ435 CV 001 100% (1) SAEJ435 CV 001 13 pages G1 Lecture 1st B Hyperbolic Functions No ratings yet G1 Lecture 1st B Hyperbolic Functions 42 pages EIL Participates in India Energy Week 2024 No ratings yet EIL Participates in India Energy Week 2024 9 pages Hyperbolic Functions - Lecture Notes No ratings yet Hyperbolic Functions - Lecture Notes 86 pages Ebook Monitoring Can Help Make Tailings Dams Safer No ratings yet Ebook Monitoring Can Help Make Tailings Dams Safer 17 pages 13.0 Hiperbolic Functions Theory No ratings yet 13.0 Hiperbolic Functions Theory 4 pages Hyperbolic Trigonometry No ratings yet Hyperbolic Trigonometry 20 pages Math223 LEC08 Hyperbolic FunctionsX No ratings yet Math223 LEC08 Hyperbolic FunctionsX 34 pages Documents Teaching Methods & Materials Mathematics ad Footer menu Back to top About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Support Help / FAQ Accessibility Purchase help AdChoices Legal Terms Privacy Copyright Cookie Preferences Do not sell or share my personal information Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Legal Terms Privacy Copyright Cookie Preferences Do not sell or share my personal information Support Help / FAQ Accessibility Purchase help AdChoices Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps Documents Language: English Copyright © 2025 Scribd Inc. We take content rights seriously. Learn more in our FAQs or report infringement here. We take content rights seriously. Learn more in our FAQs or report infringement here. Language: English Copyright © 2025 Scribd Inc. 576648e32a3d8b82ca71961b7a986505 scribd.scribd.scribd.scribd.scribd.scribd.scribd.scribd.
13326	https://www.doubtnut.com/qna/213352261	3x−y3=21 x=y+7 Which ordered pair (x,y) satisfies the system of equations shown above? (0,-7) (4,27) (7,0) (9,-16) More from this Exercise The correct Answer is:C To solve the system of equations given by: 1. 3x−y3=21 2. x=y+7 we will follow these steps: Step 1: Simplify the first equation Start with the first equation: 3x−y3=21 Multiply both sides by 3 to eliminate the fraction: 3x−y=63 Step 2: Rewrite the second equation The second equation is already in a simple form: x=y+7 We can also rewrite it as: x−y=7 Step 3: Set up the system of equations Now we have the following two equations: 1. 3x−y=63 (Equation 1) 2. x−y=7 (Equation 2) Step 4: Use elimination method We will eliminate y by subtracting Equation 2 from Equation 1. Subtract Equation 2 from Equation 1: (3x−y)−(x−y)=63−7 This simplifies to: 3x−y−x+y=63−7 Step 5: Simplify the equation Now combine like terms: 2x=56 Step 6: Solve for x Divide both sides by 2: x=28 Step 7: Substitute x back into the second equation Now substitute x=28 into the second equation x=y+7: 28=y+7 Step 8: Solve for y Subtract 7 from both sides: y=28−7=21 Step 9: Write the ordered pair Thus, the ordered pair (x,y) that satisfies both equations is: (x,y)=(28,21) Final Answer The ordered pair that satisfies the system of equations is (28,21). --- To solve the system of equations given by: 3x−y3=21 x=y+7 we will follow these steps: Step 1: Simplify the first equation Start with the first equation: 3x−y3=21 Multiply both sides by 3 to eliminate the fraction: 3x−y=63 Step 2: Rewrite the second equation The second equation is already in a simple form: x=y+7 We can also rewrite it as: x−y=7 Step 3: Set up the system of equations Now we have the following two equations: 3x−y=63 (Equation 1) x−y=7 (Equation 2) Step 4: Use elimination method We will eliminate y by subtracting Equation 2 from Equation 1. Subtract Equation 2 from Equation 1: (3x−y)−(x−y)=63−7 This simplifies to: 3x−y−x+y=63−7 Step 5: Simplify the equation Now combine like terms: 2x=56 Step 6: Solve for x Divide both sides by 2: x=28 Step 7: Substitute x back into the second equation Now substitute x=28 into the second equation x=y+7: 28=y+7 Step 8: Solve for y Subtract 7 from both sides: y=28−7=21 Step 9: Write the ordered pair Thus, the ordered pair (x,y) that satisfies both equations is: (x,y)=(28,21) Final Answer The ordered pair that satisfies the system of equations is (28,21). Topper's Solved these Questions Explore 38 Videos Explore 60 Videos Explore 37 Videos Similar Questions x+y=−6 y−4x=4 If the ordered pair (x, y) satisfies the system of equations shown above, what is the value of xy? y≤3x+1 x−y>1 Which of the following ordered pairs (x,y) satisfies the system of inequalities above? Knowledge Check x=y-3 x2+2y=6 Which ordered pair (x, y) satisfies the system of equations shown above? x=2y+5 y=(2x−3)(x+9) How many ordered pairs (x,y) satisfy the system of equations shown above ? x+y=0 3x−2y=10 Which of the following ordered pairs (x y) satisfies the system of equations above? y=x2−4x+4 y=4-x If the ordered pair (x, y) satisfies the system of equations above, what is one possible value of x ? 3x=5y+19 2x+y=−x+7 What ordered pair (x,y) represents the solution to the system of equations above? 13x−16y=7 15y−15x=8 Which of the following ordered pairs (x, y) satisfies the system of equations above? 58x+72y=32 16x−23y=1 If the ordered pair (x, y) satisfies the system of equations above, what is the sum of the values of x and y? y=15x+4 y=37x−4 If the ordered pair (x, y) satisfies the system of equations above, what is the value of y? PRINCETON-PRACTICE TEST 1-Math test-Calculator 3x-y/3=21 x=y+7 Which ordered pair (x,y) satisfies the system of ... A certain homeowner uses a gas edger to clean up his lawn every time h... The table above shows the number of students who chose to be grated on... (4-a^2)-(2a^2-6) Which of the following expressions is equivalent to... The ordered pairs (3,-1) satisfies which of the following inequalities... A psychology student randomly selected 300 people from a group of peop... The scatterplot above shows the pH of seven well water samples in West... The scatterplot above shows the pH of seven well water samples in West... 25=(ky-1)^2 In the equation above y=-2 is one solution. If K is a co... Andrew works out for 30 minutes every other days. If he spends 35% of ... If 8x-8yz+2=74. What is the value of x-yz? A chef trimmed fat off a steak and was left with a steak weighing 8.80... A backpacker is packing survival rations that consists of granola bar... Ten floorboards with equals widths laid down side to side cover a widt... George recorded his distance from home over a five-hour period, his di... In the figure above , what is the value of a? y=-75x+5,000 the equation above models the amount of money y, in dol... The scatterplot above shows data for ten accounts opened by a company ... If x/3=4 and x+y=32, what is the value of x-y? The scatterplot above shows the height in centimeters for both the dro... Formula A: BMI=w/h^2 Formula B: BMI=(4w-100)/5 The formulas above ... Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation Contact Us
13327	https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-7th-evaluating-expressions-word-problems/v/evaluating-exponential-expressions	Evaluating expressions with variables: exponents (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Math: Pre-K - 8th grade Pre-K through grade 2 (Khan Kids) 2nd grade 3rd grade 4th grade 5th grade 6th grade 7th grade 8th grade See Pre-K - 8th Math Math: Get ready courses Get ready for 3rd grade Get ready for 4th grade Get ready for 5th grade Get ready for 6th grade Get ready for 7th grade Get ready for 8th grade Get ready for Algebra 1 Get ready for Geometry Get ready for Algebra 2 Get ready for Precalculus Get ready for AP® Calculus Get ready for AP® Statistics Math: high school & college Algebra 1 Geometry Algebra 2 Integrated math 1 Integrated math 2 Integrated math 3 Trigonometry Precalculus High school statistics Statistics & probability College algebra AP®︎/College Calculus AB AP®︎/College Calculus BC AP®︎/College Statistics Multivariable calculus Differential equations Linear algebra See all Math Math: Multiple grades Early math review Arithmetic Basic geometry and measurement Pre-algebra Algebra basics Test prep SAT Math SAT Reading and Writing Get ready for SAT Prep: Math NEW Get Ready for SAT Prep: Reading and writing NEW LSAT MCAT Economics Macroeconomics AP®︎/College Macroeconomics Microeconomics AP®︎/College Microeconomics Finance and capital markets See all Economics Science Middle school biology Middle school Earth and space science Middle school chemistry NEW Middle school physics NEW High school biology High school chemistry High school physics Hands-on science activities NEW Teacher resources (NGSS) NEW AP®︎/College Biology AP®︎/College Chemistry AP®︎/College Environmental Science AP®︎/College Physics 1 AP®︎/College Physics 2 Organic chemistry Cosmology and astronomy Electrical engineering See all Science Computing Intro to CS - Python Computer programming AP®︎/College Computer Science Principles Computers and the Internet Computer science theory Pixar in a Box See all Computing Reading & language arts Up to 2nd grade (Khan Kids) 2nd grade 3rd grade 4th grade reading and vocab NEW 5th grade reading and vocab NEW 6th grade reading and vocab 7th grade reading and vocab NEW 8th grade reading and vocab NEW 9th grade reading and vocab NEW 10th grade reading and vocab NEW Grammar See all Reading & Language Arts Life skills Social & emotional learning (Khan Kids) Khanmigo for students AI for education Financial literacy Internet safety Social media literacy Growth mindset College admissions Careers Personal finance See all Life Skills Social studies US history AP®︎/College US History US government and civics AP®︎/College US Government & Politics Constitution 101 NEW World History Project - Origins to the Present World history AP®︎/College World History Climate project NEW Art history AP®︎/College Art History See all Social studies Partner courses Ancient Art Asian Art Biodiversity Music NASA Natural History New Zealand - Natural & cultural history NOVA Labs Philosophy Khan for educators Khan for educators (US) NEW Khanmigo for educators NEW Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever. Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. Skip to lesson content 6th grade math Course: 6th grade math>Unit 6 Lesson 7: Evaluating expressions word problems Evaluating expressions with variables: temperature Evaluating expressions with variables word problems Evaluating expressions with variables word problems Evaluating expressions with variables: cubes Evaluating expressions with variables: exponents Math> 6th grade math> Variables & expressions> Evaluating expressions word problems © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Evaluating expressions with variables: exponents TN.Math: 6.EE.A.1, 6.EE.A.2.c Google Classroom Microsoft Teams About About this video Transcript In this math lesson, we learn to evaluate an algebraic expression with exponents by following the order of operations (PEMDAS). We substitute a given value for the variable, calculate exponents, perform multiplication, and finally, subtraction. By applying these steps, we successfully find the value of the expression.Created by Sal Khan and Monterey Institute for Technology and Education. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted sukhmani 7 years ago Posted 7 years ago. Direct link to sukhmani's post “I need help with a questi...” more I need help with a question that doesn't have anything to do with topic but it would be nice if someone helped me. A wetsuit cost one third of the cost of diving gear. David hires a wetsuit and diving gear for the same time and length and pays $144 in total. How much money did david spend on the wetsuit in dollars. Answer Button navigates to signup page •1 comment Comment on sukhmani's post “I need help with a questi...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer sukhmani 6 years ago Posted 6 years ago. Direct link to sukhmani's post “thank you so much i got i...” more thank you so much i got i got $216 for the diving suit is that correct? 1 comment Comment on sukhmani's post “thank you so much i got i...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Sam Eaton 7 years ago Posted 7 years ago. Direct link to Sam Eaton's post “For quiz/practice questio...” more For quiz/practice questions students really should be told how Khan would like the answers. Do they want fractions or do they want decimal answers? After several "wrong" answers to a question we figured out they wanted a fraction. Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Eve Georgoulias 7 years ago Posted 7 years ago. Direct link to Eve Georgoulias's post “usually they want it as t...” more usually they want it as the fraction/decimal is displayed in the problem. For example if they use decimals in the equations or title of the practice then most likely it will want decimal answers. I know this can be frustrating especially the equations with both, but it will allow you to retake the quiz so you can get the right answers and a good score Hope this helps! 1 comment Comment on Eve Georgoulias's post “usually they want it as t...” (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more April 9 years ago Posted 9 years ago. Direct link to April's post “At 0:43, Why does he use...” more At 0:43 , Why does he use PEMDAS? Everyone I know uses GEMS. Grouping Exponents Multiply/Divide Subtract/Add In order from left to right for the letters M and S. Answer Button navigates to signup page •2 comments Comment on April's post “At 0:43, Why does he use...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer PantherasKing 9 years ago Posted 9 years ago. Direct link to PantherasKing's post “PEMDAS stands for Parenth...” more PEMDAS stands for Parenthesis, Exponents, Multiplying, Division, Addition. and Subtraction. It really is the same, but it has more steps. 2 comments Comment on PantherasKing's post “PEMDAS stands for Parenth...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Kartikeye 12 years ago Posted 12 years ago. Direct link to Kartikeye's post “So what is exactly pemdas...” more So what is exactly pemdas? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer wonderLand 12 years ago Posted 12 years ago. Direct link to wonderLand's post “P= Parenthesis E=Exponent...” more P= Parenthesis E=Exponents M=Multiplication D=Division A=Addition S=Subtraction A good way to remember this is Please Excuse My Dear Aunt Sally, or PEMDAS! Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... elim.jocelyn 12 years ago Posted 12 years ago. Direct link to elim.jocelyn's post “I learned the order of op...” more I learned the order of operations as BEDMAS? Answer Button navigates to signup page •1 comment Comment on elim.jocelyn's post “I learned the order of op...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jordan L 5 years ago Posted 5 years ago. Direct link to Jordan L's post “How did they solve the pr...” more How did they solve the problem? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Laurel Ross 10 years ago Posted 10 years ago. Direct link to Laurel Ross's post “I always thought it was P...” more I always thought it was PEMADS...... not pemDAS............ does it make a difference? Answer Button navigates to signup page •2 comments Comment on Laurel Ross's post “I always thought it was P...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer KG Dragon 8 years ago Posted 8 years ago. Direct link to KG Dragon's post “what are grouping symbols...” more what are grouping symbols? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 8 years ago Posted 8 years ago. Direct link to Kim Seidel's post “Grouping symbols are pare...” more Grouping symbols are parentheses ( ); square brackets [ ]; and curly brackets { }. 1 comment Comment on Kim Seidel's post “Grouping symbols are pare...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Ryan F 13 years ago Posted 13 years ago. Direct link to Ryan F's post “Are radicals the same as ...” more Are radicals the same as exponents? Where do radicals fit into pemdas? Or should it really be called peRmdas?! Answer Button navigates to signup page •Comment Button navigates to signup page (0 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer dragon these numbers across your mind 3 years ago Posted 3 years ago. Direct link to dragon these numbers across your mind's post “how do you multiply it” more how do you multiply it Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kenneth Marshall 3 years ago Posted 3 years ago. Direct link to Kenneth Marshall's post “You have 3 squared. You w...” more You have 3 squared. You would do3 times 3 Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Video transcript Evaluate the expression 5y to the fourth minus y squared when y is equal to 3. So every place we see a y here, we could just replace it with a 3 to evaluate it. So it becomes 5 times 3 to the fourth power minus 3 squared. All I did is every time we saw a y here, I put a 3 there. Every time we saw a y, I put a 3. So what does this evaluate to? And we have to remember our order of operations. Remember, parentheses comes first. Sometimes it's referred to as PEMDAS. Let me write that down. PEMDAS, PEMDAS. P is for parentheses. E is for exponents. M and D are for Multiplication and Division. They're really at the same level of priority. And then addition and subtraction are at the same level. If you really want to do it properly, it should be P-E, and then multiplication and division are really at the same level. And addition and subtraction are at the same level. But what this tells us is that we do parentheses first. But then after that, exponentiation takes priority over everything else here. So we have to evaluate these exponents before we multiply anything or before we subtract anything. So the one exponent we'd have to evaluate is 3 squared. So let's remember. 3 to the first is just 3. It's just 3 times itself once. So it's just 3. 3 squared is equal to 3 times 3, 3 multiplied by itself twice. That's equal to 9. 3 to the third power is equal to 3 times 3 times 3. Or you could view it as 3 squared times 3. So it'll be 9. 3 times 3 is 9. 9 times 3 is equal to 27. 3 to the fourth is equal to 3 times 3 times 3 times 3. So 3 times 3 is 9. 3 times 3 is 9. So it's going to be the same thing as 9 times 9. So this is going to be equal to 81. So we now know what 3 to the fourth is. We know what 3 squared is. Let's just put it in the expression. So this is going to be equal to 5 times 3 to the fourth. 3 to the fourth is 81. So 5 times 81 minus 3 squared. And we have 3 squared right over here. It is equal to 9. 5 times 81 minus 9. Let's figure out what 5 times 81 is. So 81 times 5. 1 times 5 is 5. 8 times 5 is 40. So this right over here is 405. So it becomes 405 minus 9. So that is going to be equal to-- if we were subtracting 10, it would be 395. But we're subtracting one less than that. So it's 396. And we're done. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: Quiz 1 Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Allow All Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies [x] Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies [x] Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies [x] Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Reject All Confirm My Choices Top Voted
13328	https://artofproblemsolving.com/wiki/index.php/2006_SMT/Advanced_Topics_Problems/Problem_10?srsltid=AfmBOop1g0Qz5SCBzo4KQZGAuVI07fF8hlZeqgUj-3s_gEvf5gdJpg5P	Page Toolbox Search 2006 SMT/Advanced Topics Problems/Problem 10 Problem 10 Evaluate: Solution First of all, remember that Therefore we want and such that: Noticing that , we can set , and . Substituting this into the formula from the beginning, we see that: Using the identity that This sum telescopes, leaving us with the first and last terms only; which equals , and . So our answer is: Something appears to not have loaded correctly. Click to refresh.
13329	https://mathmonks.com/prism/pentagonal-prism	Pentagonal Prism - Formulas, Examples and Diagram Shapes Rectangle Square Circle Triangle Rhombus Squircle Oval Hexagon Pentagon Trapezoid Kite Parallelogram Quadrilateral Polygon Nonagon Heptagon Decagon Octagon Ellipse Parallelepiped Tetrahedron Cylinder Prism Sphere Pyramid Frustum Polyhedron Dodecagon Dodecahedron Octahedron Torus Cube Cone Hyperbola Rectangular Prism Fibonacci Sequence Golden Ratio Parabola Worksheets Calculators Fraction Calculator Mixed Fraction Calculator Greatest Common Factor Calulator Decimal to Fraction Calculator Angle Arithmetic Whole Numbers Rational Numbers Place Value Irrational Numbers Natural Numbers Binary Operation Numerator and Denominator Decimal Order of Operations (PEMDAS) Scientific Notation Symmetry Fractions Triangular Number Complex Number Binary Number System Logarithm Binomial Theorem Quartic Function Mathematical Induction Group Theory Modular Arithmetic Euler’s Number Inequalities Sets De Morgan’s Laws Transcendental Numbers About Us Shapes Rectangle Square Circle Triangle Rhombus Squircle Oval Hexagon Pentagon Trapezoid Kite Parallelogram Quadrilateral Polygon Nonagon Heptagon Decagon Octagon Ellipse Parallelepiped Tetrahedron Cylinder Prism Sphere Pyramid Frustum Polyhedron Dodecagon Dodecahedron Octahedron Torus Cube Cone Hyperbola Rectangular Prism Fibonacci Sequence Golden Ratio Parabola Worksheets Calculators Fraction Calculator Mixed Fraction Calculator Greatest Common Factor Calulator Decimal to Fraction Calculator Angle Arithmetic Whole Numbers Rational Numbers Place Value Irrational Numbers Natural Numbers Binary Operation Numerator and Denominator Decimal Order of Operations (PEMDAS) Scientific Notation Symmetry Fractions Triangular Number Complex Number Binary Number System Logarithm Binomial Theorem Quartic Function Mathematical Induction Group Theory Modular Arithmetic Euler’s Number Inequalities Sets De Morgan’s Laws Transcendental Numbers About Us Search Table of Contents Right Pentagonal Prism Oblique Pentagonal prism Formulas Surface Area Volume Last modified on August 3rd, 2023 chapter outline Right Pentagonal Prism Oblique Pentagonal prism Formulas Surface Area Volume Home » Geometry » Prism » Pentagonal Prism Pentagonal Prism A pentagonal prism is a three-dimensional solid consisting of two identical pentagonal bases connected by five lateral faces. It is a heptahedron. It has 7 faces, 15 edges, 10 vertices. Pentagonal Prism next stay CC Settings Off Arabic Chinese English French German Hindi Portuguese Spanish Font Color white Font Opacity 100%Font Size 100%Font Family Arial Text Shadow none Background Color black Background Opacity 50%Window Color black Window Opacity 0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25% 200%175%150%125%100%75%50% Arial Georgia Garamond Courier New Tahoma Times New Roman Trebuchet MS Verdana None Raised Depressed Uniform Drop Shadow White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% A 2-dimensional net can be used to construct a pentagonal prism to understand its shape. Generally a net is made from paper and cut in a manner such that it can be folded and modified into 3-D shape. Pentagonal Prism Net A pentagonal prism can be can also berightoroblique, depending on the alignment of its bases. It can also be regularorirregularbased on the uniformity of its cross-section. Right Pentagonal Prism A right pentagonal prism is a prism where the lateral faces are perpendicular to the bases. Thus, the lateral faces are rectangular. Therefore, two bases appear above one another when the prism rests on its base. Right Pentagonal Prism A right pentagonal prism can be regular when the base is a regular pentagon with equal base edges as shown in the above diagram or irregular when its base is an irregular pentagon with unequal base edges. Oblique Pentagonal prism An oblique pentagonal prism is a slanted prism where the lateral faces are not perpendicular to the bases. So the lateral faces are parallelogram-shaped. Therefore, the two pentagonal bases do not appear above one another when the prism rests on its base. Oblique Pentagonal Prism Like all other polyhedrons, we can calculate the surface area and volume of a pentagonal prism. Formulas Surface Area The formula is given below: Total Surface Area (TSA) = 5 ab + 5 bh, here a = apothem, b = base edge, h = height, lateral surface area Also,since Lateral Surface Area (LSA) = Base perimeter x height Here, Base Perimeter = 5 x base edge x height = 5bh ∴Lateral Surface Area (LSA) = 5bh Thus, we can write Total Surface Area (TSA) = 5 ab + LSA Find the lateral and the total surface area of a pentagonal prism with a base edge of 7 cm, an apothem of 4.81 cm and a height of 3 cm. Solution: As we know, Lateral Surface Area (LSA) = 5bh, here b = 7 cm, h = 3 cm ∴ LSA = 5 × 7 × 3 = 105 cm 2 Total Surface Area (TSA) = 5 ab +LSA, here a = 4.81 cm, LSA = 105 cm 2 ∴TSA = 5 × 4.81 × 7 + 105 = 273.35 cm 2 Finding the surface area of a pentagonal prism when the BASE EDGE and HEIGHT are known Find the surface area of a pentagonal prism with a base edge of 10.5 in and height 7 in. Solution: Here we will use an alternative formula. Total Surface Area (TSA) = 5 b h+1 2 5(5+2 5)b 2 , here b = 10.5 cm, h = 7 in. ∴ TSA = 5×10.5×7+1 2 5(5+2 5)×(10.5)2 ≈ 746.9 in 2 Volume The formula is given below: Volume (V) =5 2 a b h,here a = apothem, b = base edge, h = height Let us solve some examples to understand the concept better. Find the volume of a pentagonal prism with an apothem of 5.5 cm, a base edge of 8 cm and a height of 6 cm. Solution: As we know, Volume (V) = 5 2 a b h , here a = 5.5 cm, b = 8 cm, h = 6 cm ∴V = 5 2×5.5×8×6 = 660 cm 3 More Resources: Volume of a Prism Surface Area of a Prism Right Prism Oblique Prism Rectangular Prism Volume of a Rectangular Prism Surface Area of a Rectangular Prism Triangular Prism Volume of a Triangular Prism Surface Area of a Triangular Prism Hexagonal Prism Volume of a Hexagonal Prism Surface Area of a Hexagonal Prism Pentagonal Prism Volume of a Pentagonal Prism Surface Area of a Pentagonal Prism Trapezoidal Prism Volume of a Trapezoidal Prism Surface Area of a Trapezoidal Prism Square Prism Volume of a Square Prism Surface Area of a Square Prism Octagonal Prism Heptagonal Prism Decagonal Prism Last modified on August 3rd, 2023 Leave a comment Cancel reply Your email address will not be published.Required fields are marked Comment Name Email Website [x] Save my name, email, and website in this browser for the next time I comment. Δ About us Contact us Privacy Policy Categories Algebra Arithmetic Geometry Statistics Trigonometry Grades 1st Grade2nd Grade3rd Grade4th Grade5th Grade6th Grade7th Grade8th Grade9th Grade10th Grade11th Grade12th Grade Join Our Newsletter Subscribe to our weekly newsletter to get latest worksheets and study materials in your email. Email Please leave this field empty. © 2025 Mathmonks.com. All rights reserved. Reproduction in whole or in part without permission is prohibited. ✕ Do not sell or share my personal information. You have chosen to opt-out of the sale or sharing of your information from this site and any of its affiliates. To opt back in please click the "Customize my ad experience" link. This site collects information through the use of cookies and other tracking tools. Cookies and these tools do not contain any information that personally identifies a user, but personal information that would be stored about you may be linked to the information stored in and obtained from them. This information would be used and shared for Analytics, Ad Serving, Interest Based Advertising, among other purposes. For more information please visit this site's Privacy Policy. CANCEL CONTINUE Your Use of Our Content ✕ The content we make available on this website [and through our other channels] (the “Service”) was created, developed, compiled, prepared, revised, selected, and/or arranged by us, using our own methods and judgment, and through the expenditure of substantial time and effort. This Service and the content we make available are proprietary, and are protected by these Terms of Service (which is a contract between us and you), copyright laws, and other intellectual property laws and treaties. This Service is also protected as a collective work or compilation under U.S. copyright and other laws and treaties. We provide it for your personal, non-commercial use only. You may not use, and may not authorize any third party to use, this Service or any content we make available on this Service in any manner that (i) is a source of or substitute for the Service or the content; (ii) affects our ability to earn money in connection with the Service or the content; or (iii) competes with the Service we provide. These restrictions apply to any robot, spider, scraper, web crawler, or other automated means or any similar manual process, or any software used to access the Service. You further agree not to violate the restrictions in any robot exclusion headers of this Service, if any, or bypass or circumvent other measures employed to prevent or limit access to the Service by automated means. Information from your device can be used to personalize your ad experience. Do not sell or share my personal information. Terms of Content Use A Raptive Partner Site
13330	http://mathcentral.uregina.ca/QQ/database/QQ.09.97/bokinac1.html	y=(-2)^x Date: Tue, 9 Dec 1997 10:49:02 -0600 (CST) Name: Ken Who is asking: Student Level: Secondary Question: During our math class last week we came accross an equation that we could not graph we were wondering if you could help us try and graph it. The question is: y=(-2)^x Hi Ken, To deal with the function y=(-2)^x you need to decide what values of x give a real number. Certainly if x is an integer then (-2)^x is real. Also if x=1/3 or x=1/5 then (-2)^x is real but if x=1/2 then (-2)^x is not real. In fact (-2)^x is real if and only if x can be written as p/q where p and q are integers with no common factors other than 1 and q is odd. If x=p/q is such a rational number then (-2)^x=2^x if p is even and (-2)^x=-(2^x) if p is odd. To see the graph of y=(-2)^x first consider the graphs of y=2^x and y=-(2^x). The graph of y=(-2)^x is then What you don't see in the picture is that the graph has a large number of "holes". The only points on the graph have first coordinate x=p/q where p and q are integers with no common factors other than 1 and q is odd. If p is even the point with first coordinate x=p/q is on the "upper" curve and is p is odd the point with first coordinate x=p/q is on the "lower" curve. Cheers, Harley Go to Math Central To return to the previous page use your browser's back button.
13331	https://www.khanacademy.org/math/algebra-home/alg-rational-expr-eq-func/alg-direct-and-inverse-variation/e/direct_and_inverse_variation	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
13332	https://www.statskingdom.com/two-proportions-ci-calculator.html	Statistics Kingdom Confidence Interval Calculator For Two Proportions Confidence interval calculator for the difference between two proportions with calculation steps, using the normal distribution approximation. Two proportions confidence interval calculator When using the sample data, we know the different between the sample proportions but we don't know the true value of the different between the population's proportions. Instead, we may treat the population's proportions as random variables and calculate the confidence interval. First, we need to define the confidence level which is the required certainty level that the true value will be in the confidence interval Researchers commonly use a confidence level of 0.95. We use the normal approximation. Two proportions confidence interval formula \| \| \| \| \| \| --- --- \| CI = p̂1 - p̂2 ± Z1-α/2 √( \| p̂1(1 - p̂1) \| + \| p̂2(1 - p̂2) \| ) \| \| n1 \| n2 \| Two proportions margin of error formula MOE = Z1-α/2√(σ2p̂1 + σ2p̂2) \| \| \| \| \| \| --- --- \| MOE = Z1-α/2√( \| p̂1(1 - p̂1) \| + \| p̂2(1 - p̂2) \| ) \| \| n1 \| n2 \| How to use the two proportion confidence interval calculator? Confidence level - The certainty level that the true value of the estimated parameter will be in the confidence interval, usually 0.95. Rounding - how to round the results? When a resulting value is larger than one, the tool rounds it, but when a resulting value is less than one the tool displays the significant figures. Sample sizes (n1, n2) - the number of subjects. Sample proportion (p̂1, p̂2) or #successes (x1, x2): If the value you entered is between 0 and 1 - the calculator assumed that you enter proportion (proabability). When the value is 1, or larger - the calculator assumed that you enter the number of successes. For example, when the sample size is 12 if you enter 3 the tool assumes 3 successes, x=3, and will calculate p̂ = 3/12 = 0.25. If you enter 0.25, the tool assumes p̂ = 3/12 = 0.25, and the confidence interval will be the same. Continuity correction - since we use the continuous normal distribution to approximate the discrete binomial distribution, the Continuity correction support results more similar to the binomial distribution. If you are a student, don't use the continuity correction. R Code The following R code should produce the same results:
13333	https://pdfs.semanticscholar.org/a93e/3b78707bf3985fb4a7f999b5c4c85edf599e.pdf	JNMA I VOL 48 I NO. 2 I ISSUE 174 I APR-JUN, 2009 135 ABSTRACT Introduction: Bone marrow culture is considered superior to blood culture in evaluation of FUO. The aim of this study was to compare the usefulness of these two cultures. Methods: A one year prospective cross sectional study was conducted to ﬁ nd out the usefulness of bone marrow culture and blood culture in the diagnosis of FUO. Marrow aspirates in each case were sent for bacterial, myocbacterial and fungal culture. Simultaneously venous blood was sent for bacterial culture. The results of BMCs and BCs were compared. Results: Total 57 cases of FUO were included in the study. Male female ratio was 1.22:1. Age range was ﬁ ve to 83 years (median 30). Duration of fever was 21 to 365 days. Bacterial growth was seen in nine cases (15.78%) of BMCs and in three cases (5.26%) of corresponding BCs. Fungal or myocbacterial growth was not seen. Salmonella typhi was the commonest organism isolated in BMCs (three cases) followed by Staphylococcus aureus (two cases), Escherichia coli, Non fermenting Gram negative bacilli, Enterococcus species and Salmonella paratyphi–A (one case each). Two cases of Salmonella typhi and one case of Salmonella paratyphi–A were isolated in BCs. Conclusions: BMCs are more useful than BCs in evaluation of patients with FUO, especially in cases of salmonella infection and are particularly important when the patient has already taken antibiotics. In immuno-competent patients presenting with FUO, BMCs for mycobacteria or fungi is unlikely to yield any growth. Key Words: blood culture, bone marrow culture, fever of unknown origin Bone Marrow Culture Vs Blood Culture in FUO Jha A,1 Sarda R,1 Gupta A,1 Talwar OP1 1Department of Pathology, Manipal College of Medical Sciences, Pokhara, Nepal Correspondence: Dr. Abhimanyu Jha Department of Pathology Institute of Medicine, TUTH Maharajgunj, Kathmandu, Nepal. Email: jhaabhimanyu@yahoo.com Phone: 9851011684 ORIGINAL ARTICLE J Nepal Med Assoc 2009;48(174):135-8 INTRODUCTION Fever of unknown origin (FUO) is a common clinical problem and an enormous number of conditions, either uncommon diseases or unusual presentations of common diseases are implicated as cause.1 Diagnostic spectrum is changing over time because of advent of sophisticated tests; however, it still poses diagnostic challenge to the physicians. The three major categories of diseases giving rise to FUO are infections, tumors, and connective tissue diseases.2 Bone marrow cultures (BMCs) are performed in the evaluation of FUO, usually in conjunction with blood cultures (BCs), cultures of body ﬂ uids and tissues.3 BMCs are often positive when BCs are negative, especially in cases of enteric fever.4 In cases of salmonella infection, BMCs increase the diagnostic yield by about one-third when compared with BCs.5 In the present study we compared the usefulness of BMCs with BCs in an attempt to detect systemic infection in patient with FUO. JNMA I VOL 48 I NO. 2 I ISSUE 174 I APR-JUN, 2009 136 METHODS This was a one year prospective cross sectional study. All the cases of FUO referred to department of pathology of Manipal Teaching Hospital, were included in the study. FUO was deﬁ ned by the criteria of Petersdorf and Beeson: temperature of 38.3oC (101oF) or above persisting or recurring during a period of two weeks and seven days investigation in hospital, or if three out patient visits fail to result in a diagnosis.6 Patients not fulﬁ lling the above criteria were excluded from the study. In all the cases clinical ﬁ ndings were recorded and Patients were informed about the procedure. Venous blood is drawn for bacterial culture. Bone marrow aspirations were performed under local anesthesia from posterior superior iliac crest using Salah needle attached to 20 cc syringe, after aseptic precaution. About 10 ml of marrow was aspirated in each case for bacterial, myocbacterial and fungal cultures. For bacterial culture aspirate were inoculated into biphasic brain heart infusion medium and biphasic MacConkey medium and incubated at 37°C for seven days with regular subculture on third and sixth days till the growth was observed. Subcultures were done and incubated at 37°C for 18-24 hours. The growths in the slant were Gram stained and the isolates were identiﬁ ed by biochemical and serological tests. For Mycobacteria, aspirates were inoculated into Lowenstain Jensen medium and incubated at 370 C till the appearance of the growth, maximum for eight weeks. For fungi aspirates were inoculated into Sabourad dextrose agar medium and incubated at room temperature till the appearance of the growth, maximum for 4 weeks. The obtained data were collated and results were tabulated. The data were analyzed by measures of central tendency. RESULTS Total numbers of patients with fever, sent for BMCs were 74, but only 57 patients were included in the study. 17 cases were excluded from the study. HIV test was negative in all the cases. 30 cases were male and 27 were female. Male to female ratio was 1.22:1. Youngest patient was of ﬁ ve year and oldest patient was 83 year old. Median age was 30 years. Six cases (10.5%) were children (<15 years) and 10 cases (17.5%) were elderly (>65 years). Table 1 shows the age distribution of 57 patients. Shortest duration of fever was 21 days and longest was 365 days with mean of 30 days (Table 2). BMCs in nine (15.78%) out of 57 patients exhibited bacterial growth and in three cases (5.26%) corresponding BCs also exhibited bacterial growth (Table 3 and 4). None of the BMCs showed mycobacterial or fungal growth. Frequencies of various bacteria isolated in BMCs are shown in table 4; the commonest was Salmonella typhi (33.5%). In two cases bacterial growth in BMCs and BCs were comparable (Table 3). Table 1. Age distribution of patients with FUO Age group (years) No. of cases % 1-14 6 10.5 15-44 32 56 45-64 9 16 >65 10 17.5 Total 57 100 Table 2. Distribution of duration of fever in patients with FUO Duration of fever (days) No. of cases % 21-30 43 75.44 31-60 5 8.77 61-90 6 10.5 91-180 1 1.75 181-365 2 3.5 Total 57 100 Table 3. Result of bone marrow culture S.N. Age (year) Sex Organisms isolated in BMCs Organism isolated in corresponding BCs 1 37 F Escherichia coli No growth 2 27 F Staphylococcus aureus No growth 3 20 F Non fermenting Gram negative bacilli No growth 4 76 F Salmonella typhi No growth 5 15 M Staphylococcus aureus No growth 6 38 M Salmonella typhi No growth 7 12 F Enterococcus sps. Salmonella typhi 8 23 M Salmonella para typhi-A Salmonella paratyphi -A 9 5 F Salmonella typhi Salmonella typhi Jha et al. Bone Marrow Culture Vs Blood Culture in FUO JNMA I VOL 48 I NO. 2 I ISSUE 174 I APR-JUN, 2009 137 Table 4. Frequency of various organisms isolated in bone marrow culture Organisms No. of cases % Salmonella typhi 3 33.5 Staphylococcus aureus 2 22.5 Escherichia coli 1 11 Salmonella paratyphi-A 1 11 Enterococcus sps 1 11 Non fermenting GNB 1 11 Total 9 100 Table 5. Comparison of 5 studies evaluating detection of organism by bone marrow culture in patients with FUO Studies No. of cultures No. (%) of specimens with positive BMC Volk et al3 215 1 (0.5) Marsh et al13 124 4 (3.2) Riley et al14 433 51 (11.8) Nicholas et al15 342 59 (17.3) Present study 57 9 (15.78) DISCUSSION Diagnosis of FUO requires multidisciplinary approach including clinical history. History of travel in recent past and drug history is of great importance. Bone marrow examination in FUO is an important diagnostic modality. In present study BMCs alone yielded accurate diagnosis in 15.78% of cases. 15.78% (nine cases) of BMCs and only 5.26% (three cases) of BCs showed bacterial growth (Table 3). BCs did not reveal bacterial growth in isolation without BMCs revealing bacterial growth. Except in one case out of nine cases, BMCs alone were sufﬁ cient for the diagnosis without any role of BCs. In that one case BC revealed growth of Salmonella typhi; BMC in this case revealed growth of Enterococcus species the clinical signiﬁ cance of which is doubtful as patient responded to treatment for salmonella infection, although Enterococcus species is known to cause septicemia and infective endocarditis, and the ubiquitous nature of this organism requires caution in establishing the clinical signiﬁ cance.7 In one case BMC exhibited growth of non-fermenting gram negative bacilli which was not further classiﬁ able. In this study salmonella was the commonest isolate in the BMCs, Salmonella typhi and Salmonella paratyphi-A together constituted 44.44 % of total positive bone marrow bacterial culture (Table 4). If one case of positive blood culture for Salmonella typhi is considered, enteric fever alone constituted 55.55% of total infections and 8.77% of cases of FUO. There was no mycobacterial and fungal growth in any of the cases. In a study by Haq et al, infectious disease accounted for 63.21% of cases of FUO. Tuberculosis was the commonest accounting for 24.53%, followed by enteric fever (12.74%) and visceral leishmaniasis (9.43%).8 In a study by Jung et al, commonest cause of FUO was infections (46.4%) and enteric fever was the commonest (29.6%) among that followed by malaria (9%), and tuberculosis (5.2%).9 But these studies were not based only on BMCs and BCs, as in the present study. Nonetheless in this study enteric fever was the commonest cause of infection related FUO. 9 In a study by Farooqui and colleague organism was isolated from the bone marrow in all the cases of salmonella infection and from blood in only 66% cases. In the same study further subtyping of the Salmonella species showed Salmonella typhi 83%, Salmonella paratyphi-A 11.8%, Salmonella paratyphi-B 4.1% and Salmonella paratyphi-C 0.9%.4 In present study Salmonella typhi accounted for 80% and Salmonella paratyphi-A accounted for 20% of total salmonella infections (Table 3 and 4). Out of ﬁ ve cases of salmonella infection four cases (80%) showed positive BMCs while two cases (40%) showed growth in BCs as well. In one case (20%) growth was seen only in BC. Findings in this study are in accordance to study of Farooqui and colleague. BMCs could conﬁ rm a diagnosis of enteric fever in patients with negative BCs.4 One of the reasons for BCs being negative and BMCs being positive in enteric fever may be antibiotic treatment before collection of blood for culture. An injudicious antibiotic regimen may diminish or eliminate the organism from blood, but salmonella being an intracellular organism, persists in the reticulo-endothelial system including bone marrow.10 The intracellular location of bacteria protects them from conventional chemotherapeutic measures.11,12 Enteric fever is the only bacterial infection in humans for which bone marrow examination is recommended.5 In a study by John et al, the number of bacteria in blood but not bone marrow was correlated inversely with the duration of preceding fever.5 Thus, with increasing duration of illness the ratio of bone marrow to blood bacterial concentrations increased; the median ratio was 4.8 during the ﬁ rst week compared with 158 during the third week. This study provided proof that the concentrations of Salmonella typhi in bone marrow are higher than in blood. Bone marrow contained over 10 times more bacteria than blood.5 Effective antibiotic pre-treatment had greater effect in reducing blood counts of bacteria compared to bone marrow count.5 How salmonella survives in the bone marrow despite the presence of high concentrations of bactericidal antibiotic which reaches high intracellular concentration is unclear. Jha et al. Bone Marrow Culture Vs Blood Culture in FUO JNMA I VOL 48 I NO. 2 I ISSUE 174 I APR-JUN, 2009 138 In a study by Volk et al only 0.5% of BMCs and corresponding BCs in the patients with FUO showed clinically signiﬁ cant growth in an immunocompromised patient and the organism isolated was Mycobacterium avium intracellularae.3 In contrast present study showed 15.78% of growth in BMCs. A summary of diagnostic yield of BMCs in ﬁ ve studies is shown in table 5. A population characteristic of Volk et al was non-restricted while a population characteristic of Marsh et al, Riley et al and Nicholas et al was HIV patients.3, 13-15 In present study the isolates were bacteria but in other four studies the isolate was Mycobacterium avium intracellularae. Yield of BMCs in present study was 32 times better (15.78% versus 0.5%) than the other study (Volk et al). In absence of hematological justiﬁ cation for bone marrow evaluation, BMCs and BCs in a work up of patient signiﬁ cantly increase the cost. It is difﬁ cult to justify performing BMCs on all patients with FUO. Neither BMCs nor the BCs will detect the causes of all systemic infections; an algorithm that conserves resources should be followed in which BMCs are performed if the BCs and other easily obtained tissues have been cultured, and the results are repeatedly inconclusive. CONCLUSIONS A bone marrow examination including BMCs is important part of investigation of FUO, but this alone is insufﬁ cient to trace speciﬁ c etiology in all the cases. A good clinical history, radiological and hematological evaluation including bone marrow coupled with BMC and serology would yield diagnosis in a high proportion of cases. Yield of BMCs is more than BCs especially in cases of suspected salmonella infection and in patients treated with antibiotics as was the case in present study; however, its role is insigniﬁ cant in isolating mycobacteria or fungi in immuno-competent patients. REFERENCES 1. Arnow PM, Flaherty J P. Fever of Unknown Origin. The Lancet. 1997;350:575-80. 2. Knockaert Daniel C, Vanneste LJ, Bobbaers HJ. Recurrent or Episodic Fever of Unknown origin. Review of 45 cases and Survey of the Literature. Medicine. 1993;72:184-96. 3. Volk EE, Miller ML, Kirkley BA, Washington JA. The diagnostic usefulness of bone marrow cultures in patients with Fever of Unknown Origin. Am J clin Pathol. 1998;110:150-3. 4. Farooqui BJ, Khursid M, Ashfaq MK, Khan MA. Comparative yield of Salmonella typhi from blood and bone marrow cultures in patients with Fever of Unknown origin. J Clin Pathol. 1991;44:258-9. 5. Wain J, Bay PVB, Vinh H, et al. Quantization of bacteria in bone marrow from patients with typhoid fever: Relationship between counts and clinical features. J Clin Microbiology. 2001;39:1571-6. 6. Petersdorf RG, Beeson PB. Fever of Unknown origin: report on 100 cases. Medicine. 1961;40:1-30. 7. Facklam RR, Teixeira LM. Enterococcus. In: Balows A, Duerden BI. editors. Systemic Microbiology. Vol 2. In: Leslie C, Albert B, Max S. Topley & Wilson’s Microbiology and Microbial infections. 9th ed. London: Arnold publication; 1998. P. 669-82. 8. Haq SA, Alam MN, Hossain SM, et al. A study of prolonged pyrexia in Dhaka. Bangladesh Med Res Counc Bull. 1996;22:33-42. 9. Jung A, Singh MM, Jajoo U. Unexplained fever-analysis of 233 cases in a referral hospital. Indian J Med Sci. 1999;53:535-44. 10. Suter E. Interaction between phagocytes and pathogenic microorganisms. Bacteriol Rev. 1956;20:94-132. 11. Mandell GL. Interaction of intraleukocytic bacteria and antibiotics. J Clin Invest. 1973;52:673-9. 12. Holmes B, Quie PG, Windhorst DP, Pollara B, Good RA. Protection of phagocytozed bacteria from killing action of antibiotics. Nature. 1966;210:1131-2. 13. Marsh RD, Paul M, Siddique T, et al. Bone marrow culture for diagnosis of mycobacterial and fungal infections in febrile patients. J Fla Med Assoc. 1991;78:357-60. 14. Riley UBG, Crawford S, Barrett SP, et al. Detection of myocbacterial in bone marrow biopsy specimens taken to investigate pyrexia of unknown origin. J Clin Pathol. 1995;48:706-9. 15. Nicholas L, Florentine B, Lewis W, et al. Bone marrow examination for the diagnosis of mycobacterial and fungal infections in the acquired immunodeﬁ ciency syndrome. Arch Pathol Lab med. 1991;115:1125-32. Jha et al. Bone Marrow Culture Vs Blood Culture in FUO
13334	https://bolanlearokoyo.com/types-of-morpheme/	Skip to content Types of Morphemes: free and bound Bolanle Elizabeth Arokoyo Morphology Lecture Series #III We continue our lecture from The Morpheme. The morpheme can broadly be divided into free morphemes and bound morphemes. One basic feature of the morpheme is that it must be meaning bearing. The tree diagram below shows the distribution of the morphemes. Free Bound Lex. Fun. Infl. Deriv. Free Morphemes The free morpheme is a morpheme that can exist in isolation. It is a morpheme that has independent existence. It can stand on its own. It is the core of the word. It is a morphological unit whose existence is not dependent on any other unit whether morphological, phonological or syntactic; it occurs unattached. The free morpheme is a word by itself, the meaning carrying unit. The free morpheme is also the base, root and at times, the stem. As the base, root and stem, it can accept other morphemes to be attached to it. Examples of free morphemes in Igala are: atá ‘father’ ọ́ma ‘child’ íye ‘mother’ agboji ‘leader’ Free morphemes can also broadly be divided into two namely lexical and functional morphemes as discussed below. a. Lexical Morphemes Lexical morphemes are the meaning bearers of sentences. They serve as carriers of the information being conveyed in a sentence. Lexical morphemes form the open class of words as words belonging to this group are opened to word formation rules, new members of each class can easily be created. They are nouns, adjectives, adverbs and verbs. Consider the following examples from English and Yoruba. Noun Verb Adj Adv man eat deep fairly woman sleep ugly along boy drink nice deeply stone buy fair easily table smile certain quietly Yoruba 6. Noun orúkọ ‘name’ ọkùnrin ‘man’ ilé ‘house’ ago ‘watch’ 7. Verb lọ ‘go’ tà ‘sell’ sà ‘pick’ gé ‘cut’ Adjective kpukpa ‘red’ dára ‘good’ dúdú ‘black’ wúwo ‘heavy’ Adverb kíákíá ‘quickly’ tètè ‘fast’ ibẹ̀‘there’ mànà ‘definitely’ b. Functional Morphemes Functional morphemes show relationship between or among lexical morphemes. Unlike lexical morphemes they do not convey lexical meaning and they are very few in number. They belong to the closed class as new members are not admitted. They have a high frequency of occurrence and they are reciprocally exclusive. Two members of a class cannot be used together. They include all functional words like prepositions (on, in, at, under, over, etc.), pronouns (he, she, it, they, her, us, etc.), conjunctions (and, or, but, etc.), determiners (a, an, the, these, etc.), and interjections (ah! oh!, eh!, ughh!). Following in (10) below are examples of functional morphemes from Owé: 10. a. àti ‘and’ b. àbí ‘or’ c. sugbọ̀n ‘but’ d. èmi ‘I’ e. àwa ‘we’ f. kan ‘one / a’ g. abẹ́‘under’ h. inú ‘inside’ Functional morphemes are however open to inflectional morphemes as they are capable of being inflected for number, person, gender and case. This will be discussed under inflectional morphemes later. Bound Morphemes In morphology, a bound morpheme is a dependent morpheme. It is a morpheme that cannot occur without being attached to a root. The bound morpheme does not occur as an independent word. It never never occurs in isolation. It is only meaningful when attached to a root. Their major functions are to derive new words in which case they may change the grammatical class of the word and to provide additional grammatical information. In Yorùbá, all vowels except ucan serve the derivational purpose of creating nouns from verbs. For example, the noun ẹ̀sẹ̀‘sin’ is derived from theverb sẹ̀‘to sin’ by the addition of the bound morpheme ẹ-which functions as a nominalizer here. The same explanation obtains for the nouns ikú‘death’ and ìfẹ́‘love’ derived from the verbs kú‘to die’ and fẹ́‘to love’ respectively. The bound morpheme in these examples is ‘i’’.The oní-‘owner of’ morpheme in Yorùbá also derives new nouns from nouns. Oní-‘owner of’ is attached to nouns like ilé‘house’, igi‘tree’, omi‘water’, ata‘pepper’,etc. to derive onílé‘landlord’, onígi‘owner of tree’, olómi‘water seller’, and aláta‘pepper seller’ respectively. The forms ólómi‘water seller’ and aláta‘pepper seller’ have passed through some morpho-phonological processes which be explained later in subsequent lectures. The bound morpheme –edin English provides information about tense and so it is attached to verbs. It is added to words like walk – walked, smoke- smoked, kill- killed, buy-bought, hit- hit, catch- caught,etc. The bound morpheme changed the form of the verb from present to past. Morphemes are divided into different structural types. They include prefixes, suffixes, infixes, circumfixes, and suprafixes. All these bound morphemes are regarded as affixes. Bound morphemes perform two basic functions; derivational and inflectional. They are referred to as derivational morphemes (Derivational Morphemes) and inflectional morphemes (Inflectional Morphology) respectively. These will be discussed in subsequent lectures. With ample illustrations, classify the morpheme. Describe free and bound morphemes. Be creative! What are the possible forms you can come up with using affixes like un-, in-, dis-, il-, -s, -ren, -ed,on these words. Can you explain why the derived or inflected forms sound the way they do? List of abbreviations Adj. Adjective Adv. Adverb Deriv. Derivational Func. Functional Infl. Inflectional Lex. Lexical Excepts are taken from Arokoyo (2017) Arokoyo, Bolanle Elizabeth. (2017). Unlocking morphology. Ilorin: Chridamel Publishing House.
13335	https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:unit-circle/v/unit-circle-definition-of-trig-functions-1	Unit circle (video) \| Trigonometry \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever. Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. Skip to lesson content Algebra 2 Course: Algebra 2>Unit 11 Lesson 1: Unit circle introduction Unit circle Unit circle The trig functions & right triangle trig ratios Trig unit circle review Math> Algebra 2> Trigonometry> Unit circle introduction © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Unit circle Google Classroom Microsoft Teams About About this video Transcript Learn how to use the unit circle to define sine, cosine, and tangent for all real numbers.Created by Sal Khan. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Bryan Bao 2 years ago Posted 2 years ago. Direct link to Bryan Bao's post “I'm a little confused. Ca...” more I'm a little confused. Can someone help explain how to use a unit circle to help solve sine, cosine, and tangent? Answer Button navigates to signup page •Comment Button navigates to signup page (14 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Aristotle 2 years ago Posted 2 years ago. Direct link to Aristotle's post “Determine angle(theta) yo...” more Determine angle(theta) you are finding the cosine, sine, or tangent of. Using SOHCAHTOA in relation to theta, the angle in the math equation you are searching for, then use SOHCAHTOA to search for the proper opposite, adjacent, or hypotenuse. SOHCAHTOA changes based on the location of the angle you are searching for. For example, if the angle was top right corner, instead of at the right angle, SOHCAHTOA measures and inputs would be different. Fast forward to 8:00 and watch from there. It's a little confusing, but I'll explain. Based on those definitions you'll observe from 8 minutes and onward, you can now apply SOHCAHTOA. In this situation, cosine is the x coordinate. sine is the Y coordinate, and tangent is y/x coordinate. based on this Terminal side is like saying, based on the location of theta(angle) in this "unit circle equation"....... what sides correspond to the adjacent side, the opposite side, and the hypotenuse side BASED ON the location of theta you are solving for? now what is the sine, cosine, and tangent? (SOHCAHTOA to solve for the missing sides) NOW here is the final part to understand. The length of the "cosine" side will now act as the "x-coordinate", the length of the "sine" side will act as the "y-coordinate", and the hypotenuse will act and be measured by the y/x coordinate value. (the side length is like measuring an x coordinate, counting from 0 to 8 for example on the x-axis. the side value now equals 8 and the "cosine" coordinate will also be equal to 8; represented as (8,0) SOHCAHTOA will help you connect the dots, if that makes sense? the dots change in location to the angle you are solving for. When I say change location, this is because of either a clockwise rotation or counterclockwise rotation. this means that the "terminal side" will change location based on the math question's rotation and must constantly be reevaluated to ensure you are solving correctly. if you're still confused, let this information soak for a day and revisit this information. Rewatch the video explanation a few times too. Google "terminal angle/side trigonometry" and look at the graphs in images. Practice the practice problems and then revisit this information and do this a few times. If still confused, ask and I'll explain more. 9 comments Comment on Aristotle's post “Determine angle(theta) yo...” (40 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... chasephuayoung88 2 years ago Posted 2 years ago. Direct link to chasephuayoung88's post “You should at least expla...” more You should at least explain what sin and cosine are. This is intro to trigonometry Answer Button navigates to signup page •1 comment Comment on chasephuayoung88's post “You should at least expla...” (18 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer ks 2 years ago Posted 2 years ago. Direct link to ks's post “sin is opposite/hypotnuse...” more sin is opposite/hypotnuse. cosine is adjacent/hypotnuse 1 comment Comment on ks's post “sin is opposite/hypotnuse...” (14 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... alexmirzabdullaev 2 years ago Posted 2 years ago. Direct link to alexmirzabdullaev's post “if someone is having hard...” more if someone is having hard time of understanding this, I would encourage you to get additional information from "organic chemistry tutor" YouTube channel Answer Button navigates to signup page •8 comments Comment on alexmirzabdullaev's post “if someone is having hard...” (18 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Ella 2 years ago Posted 2 years ago. Direct link to Ella's post “I'm still very confused. ...” more I'm still very confused. He didn't really explain anything that shows up in the corresponding exercise that I could tell. Maybe it's just me? Answer Button navigates to signup page •Comment Button navigates to signup page (12 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer cossine 2 years ago Posted 2 years ago. Direct link to cossine's post “So the first step would b...” more So the first step would be figure out how things are defined? How is the sine function defined? Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... 1abruh 2 years ago Posted 2 years ago. Direct link to 1abruh's post “What is the most effcient...” more What is the most effcient way to memorize this thing 😭 Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Venkata 2 years ago Posted 2 years ago. Direct link to Venkata's post “Don't. Understand the ide...” more Don't. Understand the idea behind it. The most you'll need to know is the first quadrant (you don't even need to memorize it. Practice enough problems with it and you'll be able to remember it) as every subsequent quadrant is just adding pi or subtracting from 2pi or some manipulation like that. Comment Button navigates to signup page (20 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... smarzo3 a year ago Posted a year ago. Direct link to smarzo3's post “Okay ladies and gentlemen...” more Okay ladies and gentlemen, buckle up now, this is where it begins to get harder. Answer Button navigates to signup page •1 comment Comment on smarzo3's post “Okay ladies and gentlemen...” (13 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer #ThugLife 2 years ago Posted 2 years ago. Direct link to #ThugLife's post “is there a reason that we...” more is there a reason that we have to go counterclockwise? Answer Button navigates to signup page •Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Venkata 2 years ago Posted 2 years ago. Direct link to Venkata's post “Just convention. There wi...” more Just convention. There will be cases where you're told to go clockwise, but if it isn't mentioned, you're supposed to go counterclockwise. Comment Button navigates to signup page (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Smit 2 years ago Posted 2 years ago. Direct link to Smit's post “Some questions: 1. What ...” more Some questions: What is the fancy zero (𝛳) and why is he using it? If you didn't have a calculator, how would you compute sin, cos, etc.? How is the unit circle used? What is its purpose? Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Aeternum 2 years ago Posted 2 years ago. Direct link to Aeternum's post “1. The fancy zero is thet...” more 1. The fancy zero is theta. It's a Greek letter used in mathematics to represent angles. Some values you will memorize and others you'll find with the help of special triangles (30-60-90 and 45-45-90) and unit circles. You'll learn all of that later on in the Precalculus/Algebra II/Trigonometry courses. But you should be provided with a calculator on particularly hairy angles. The unit circle is used to find sin/cos/etc. values of certain angles. It's also used to visualize trigonometric functions and is very helpful when you start to learn about radians. As again with question 2, you'll learn later on in Khan :) Comment Button navigates to signup page (9 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... gameboyaj94 2 years ago Posted 2 years ago. Direct link to gameboyaj94's post “Tbh, this course is aweso...” more Tbh, this course is awesome (Trigonometry Course), but I still don't know why this course lacks many videos in order... Answer Button navigates to signup page •Comment Button navigates to signup page (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer cossine 2 years ago Posted 2 years ago. Direct link to cossine's post “You may find some videos ...” more You may find some videos missing. In this case it is good idea to refer YouTube. There are numerous videos on trigonometry some with playlist available. Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Smit 2 years ago Posted 2 years ago. Direct link to Smit's post “Does the size of the circ...” more Does the size of the circle matter? If my circle has a radius of 3, how does the unit circle work? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Cedar 2 years ago Posted 2 years ago. Direct link to Cedar's post “By definition, the unit c...” more By definition, the unit circle has a radius of 1. A circle with radius 3 isn't a unit circle. 2 comments Comment on Cedar's post “By definition, the unit c...” (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Video transcript What I have attempted to draw here is a unit circle. And the fact I'm calling it a unit circle means it has a radius of 1. So this length from the center-- and I centered it at the origin-- this length, from the center to any point on the circle, is of length 1. So what would this coordinate be right over there, right where it intersects along the x-axis? Well, x would be 1, y would be 0. What would this coordinate be up here? Well, we've gone 1 above the origin, but we haven't moved to the left or the right. So our x value is 0. Our y value is 1. What about back here? Well, here our x value is -1. We've moved 1 to the left. And we haven't moved up or down, so our y value is 0. And what about down here? Well, we've gone a unit down, or 1 below the origin. But we haven't moved in the xy direction. So our x is 0, and our y is negative 1. Now, with that out of the way, I'm going to draw an angle. And the way I'm going to draw this angle-- I'm going to define a convention for positive angles. I'm going to say a positive angle-- well, the initial side of the angle we're always going to do along the positive x-axis. So you can kind of view it as the starting side, the initial side of an angle. And then to draw a positive angle, the terminal side, we're going to move in a counterclockwise direction. So positive angle means we're going counterclockwise. And this is just the convention I'm going to use, and it's also the convention that is typically used. And so you can imagine a negative angle would move in a clockwise direction. So let me draw a positive angle. So a positive angle might look something like this. This is the initial side. And then from that, I go in a counterclockwise direction until I measure out the angle. And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. So the first question I have to ask you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate of this point of intersection. If you were to drop this down, this is the point x is equal to a. Or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine-- let me use the same green-- what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our soh cah toa definition. That's the only one we have now. We are actually in the process of extending it-- soh cah toa definition of trig functions. And the cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side-- for this angle, the adjacent side has length a. So it's going to be equal to a over-- what's the length of the hypotenuse? Well, that's just 1. So the cosine of theta is just equal to a. Let me write this down again. So the cosine of theta is just equal to a. It's equal to the x-coordinate of where this terminal side of the angle intersected the unit circle. Now let's think about the sine of theta. And I'm going to do it in-- let me see-- I'll do it in orange. So what's the sine of theta going to be? Well, we just have to look at the soh part of our soh cah toa definition. It tells us that sine is opposite over hypotenuse. Well, the opposite side here has length b. And the hypotenuse has length 1. So our sine of theta is equal to b. So an interesting thing-- this coordinate, this point where our terminal side of our angle intersected the unit circle, that point a, b-- we could also view this as a is the same thing as cosine of theta. And b is the same thing as sine of theta. Well, that's interesting. We just used our soh cah toa definition. Now, can we in some way use this to extend soh cah toa? Because soh cah toa has a problem. It works out fine if our angle is greater than 0 degrees, if we're dealing with degrees, and if it's less than 90 degrees. We can always make it part of a right triangle. But soh cah toa starts to break down as our angle is either 0 or maybe even becomes negative, or as our angle is 90 degrees or more. You can't have a right triangle with two 90-degree angles in it. It starts to break down. Let me make this clear. So sure, this is a right triangle, so the angle is pretty large. I can make the angle even larger and still have a right triangle. Even larger-- but I can never get quite to 90 degrees. At 90 degrees, it's not clear that I have a right triangle any more. It all seems to break down. And especially the case, what happens when I go beyond 90 degrees. So let's see if we can use what we said up here. Let's set up a new definition of our trig functions which is really an extension of soh cah toa and is consistent with soh cah toa. Instead of defining cosine as if I have a right triangle, and saying, OK, it's the adjacent over the hypotenuse. Sine is the opposite over the hypotenuse. Tangent is opposite over adjacent. Why don't I just say, for any angle, I can draw it in the unit circle using this convention that I just set up? And let's just say that the cosine of our angle is equal to the x-coordinate where we intersect, where the terminal side of our angle intersects the unit circle. And why don't we define sine of theta to be equal to the y-coordinate where the terminal side of the angle intersects the unit circle? So essentially, for any angle, this point is going to define cosine of theta and sine of theta. And so what would be a reasonable definition for tangent of theta? Well, tangent of theta-- even with soh cah toa-- could be defined as sine of theta over cosine of theta, which in this case is just going to be the y-coordinate where we intersect the unit circle over the x-coordinate. In the next few videos, I'll show some examples where we use the unit circle definition to start evaluating some trig ratios. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: exercise Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Allow All Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies [x] Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies [x] Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies [x] Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Reject All Confirm My Choices
13336	https://pmc.ncbi.nlm.nih.gov/articles/PMC4670877/	Early Detection of Fetal Malformation, a Long Distance Yet to Cover! Present Status and Potential of First Trimester Ultrasonography in Detection of Fetal Congenital Malformation in a Developing Country: Experience at a Tertiary Care Centre in India - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice J Pregnancy . 2015 Nov 23;2015:623059. doi: 10.1155/2015/623059 Search in PMC Search in PubMed View in NLM Catalog Add to search Early Detection of Fetal Malformation, a Long Distance Yet to Cover! Present Status and Potential of First Trimester Ultrasonography in Detection of Fetal Congenital Malformation in a Developing Country: Experience at a Tertiary Care Centre in India Namrata Kashyap Namrata Kashyap 1 Department of Maternal and Reproductive Health, Sanjay Gandhi Post Graduate Institute of Medical Sciences (SGPGIMS), Lucknow 226 014, India Find articles by Namrata Kashyap 1,, Mandakini Pradhan Mandakini Pradhan 1 Department of Maternal and Reproductive Health, Sanjay Gandhi Post Graduate Institute of Medical Sciences (SGPGIMS), Lucknow 226 014, India Find articles by Mandakini Pradhan 1, Neeta Singh Neeta Singh 1 Department of Maternal and Reproductive Health, Sanjay Gandhi Post Graduate Institute of Medical Sciences (SGPGIMS), Lucknow 226 014, India Find articles by Neeta Singh 1, Sangeeta Yadav Sangeeta Yadav 1 Department of Maternal and Reproductive Health, Sanjay Gandhi Post Graduate Institute of Medical Sciences (SGPGIMS), Lucknow 226 014, India Find articles by Sangeeta Yadav 1 Author information Article notes Copyright and License information 1 Department of Maternal and Reproductive Health, Sanjay Gandhi Post Graduate Institute of Medical Sciences (SGPGIMS), Lucknow 226 014, India ✉ Namrata Kashyap: dr.nmrata@gmail.com Academic Editor: R. L. Deter Received 2015 Sep 4; Accepted 2015 Oct 27; Issue date 2015. Copyright © 2015 Namrata Kashyap et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC4670877 PMID: 26759727 Abstract Background. Early detection of malformation is tremendously improved with improvement in imaging technology. Yet in a developing country like India majority of pregnant women are not privileged to get timely diagnosis.Aims and Objectives. To assess the present status and potential of first trimester ultrasonography in detection of fetal congenital structural malformations.Methodology. This was a retrospective observational study conducted at Sanjay Gandhi Postgraduate Institute of Medical Sciences. All pregnant women had anomaly scan and women with fetal structural malformations were included.Results. Out of 4080 pregnant women undergoing ultrasound, 312 (7.6%) had fetal structural malformation. Out of 139 patients who were diagnosed after 20 weeks, 47 (33.8%) had fetal structural anomalies which could have been diagnosed before 12 weeks and 92 (66.1%) had fetal malformations which could have been diagnosed between 12 and 20 weeks.Conclusion. The first trimester ultrasonography could have identified 50% of major structural defects compared to 1.6% in the present scenario. This focuses on the immense need of the hour to gear up for early diagnosis and timely intervention in the field of prenatal detection of congenital malformation. 1. Introduction Fetal structural malformations are seen in 3 to 5% of all pregnancies . Detection of malformation is tremendously improved with improvement in imaging technology. In majority of countries worldwide, second trimester scan between 18 and 22 weeks remains the standard of care for fetal anatomical assessment; however, most recent literature shows a significant improvement in detection of fetal abnormalities in first trimester of pregnancy . Besides nuchal abnormalities a wide range of central nervous system, heart, anterior abdominal wall, urinary tract, and skeletal abnormalities can be diagnosed between 11 and 14 weeks of scan. The clear benefits of first trimester ultrasound are early detection and exclusion of major congenital anomalies (not compatible with life or followed by severe handicap), reassurance, and relatively easier pregnancy termination if required. Currently, the review of recent literature suggests classification of fetal abnormalities as always detectable, potentially detectable, and undetectable till first trimester and anomaly scan. The diagnostic efficacy of first trimester anomaly scan and echocardiography between 11 and 14 weeks has been assessed in medium risk population by Becker and Wegner . The prevalence of major anomalies in their study group was 2.8%. The overall detection rate of fetal anomalies including cardiac defects was 84% and increased with raised nuchal thickness particularly more than 2.5 mm. This highlights the scope of first trimester scan apart from its conventional role in detection of chromosomal abnormality. First trimester screening is now no more limited to detection of raised nuchal thickness (NT). Becker et al. analysed 6879 cases to assess the prevalence and detection rate of major anomalies by applying first trimester anomaly scan and fetal echocardiography. They concluded that a significant number of fetal anomalies occur with normal NT and more than half of them could be detected in first trimester. Hence, even fetuses with normal NT should be offered first trimester anomaly scan and fetal echocardiography considering the ethical principles of nonmaleficence, justice, and respect for autonomy of pregnant women. Even in this era the benefits of this established technology are not in the reach of all. A vast majority of patients in India are not yet undergoing anomaly scan. We frequently encounter malformations always or potentially detectable during first trimester scan at third trimester or in postnatal period. It depends on both the expertise and resources available along with the awareness and sensitization in general population. This fact of diagnosis is particularly more important in countries like India where medical termination of pregnancy is legally allowed up to 20 weeks of gestation irrespective of malformation being lethal. We see a fair number of patients who are diagnosed with fetal malformation beyond 20 weeks and in that situation they are forced to seek termination services at small substandard centres since they get refusals from all relatively good hospitals due to legal issues associated with termination. Many of such patients get deteriorated due to septic abortion and unnecessary hysterotomy and so forth. Question then arises that where lies the fault, the awareness of the patients or the expertise of the sonologist. Henceforth, the study was planned to assess the prevalence of fetal malformation in a tertiary care referral centre and to assess the present status of first trimester ultrasonography in the detection of fetal malformations in a tertiary care centre in India. 2. Materials and Methods This was a retrospective observational study conducted at Sanjay Gandhi Postgraduate Institute of Medical Sciences. All pregnant women attending Department of Maternal and Reproductive Health, OPD, from August 2009 till October 2013 were enrolled in the study. All pregnant women underwent ultrasound (General electrical Voluson S8) and those with fetal structural malformations were evaluated. Malformations were classified according to gestational age of diagnosis, system involved, and type of malformation. Descriptive proportions and frequencies have been used to depict the data. 3. Results A total number of 4080 pregnant women underwent USG and amongst them 312 (7.6%) patients had fetal structural malformation. The malformations were classified according to gestational age as shown in Table 1. Malformations were classified according to various systems as shown in Table 2. Table 1. Number of malformations at different gestational age. \| Gestational age \| Number \| :---: \| \| <12 weeks \| 5 \| \| 12–20 weeks \| 98 \| \| >20 weeks \| 209 \| Open in a new tab Table 2. Classification of malformation according to the system involved. \| CNS, brain \| Number \| Skeletal \| Number \| :---: :---: \| \| Neural tube defects \| 30 \| Achondroplasia \| 2 \| \| Porencephaly \| 1 \| Hypochondroplasia \| 2 \| \| Anencephaly \| 11 \| Osteogenesis imperfecta \| 3 \| \| Occipital encephalocoele \| 8 \| Short rib polydactyly \| 6 \| \| Iniencephaly \| 2 \| Thanatophoric dysplasia \| 1 \| \| Ventriculomegaly \| 14 \| Single forearm bone \| 1 \| \| Arachnoid cyst \| 2 \| Cooks syndrome \| 1 \| \| Holoprosencephaly \| 8 \| Respiratory \| \| \| Agenesis of corpus callosum \| 8 \| CCAM \| 7 \| \| Dandy-Walker malformation \| 16 \| Pleural effusion \| 1 \| \| Diastematomyelia \| 2 \| Congenital high airway obstruction \| 3 \| \| Vermian agenesis \| 2 \| Extralobar pulmonary sequestration \| 2 \| \| Genitourinary \| \| Heart \| \| \| ADPKD \| 3 \| Structural cardiac malformations \| 44 \| \| ARPKD \| 1 \| Congenital heart blocks \| 8 \| \| Megacystis \| 1 \| Pericardial effusion \| 2 \| \| Gonadal cyst \| 2 \| Structural and rhythmic \| 1 \| \| Hydronephrosis \| 12 \| AV malformation \| \| \| Lower urinary tract obstruction \| 9 \| Vein of Galen malformation \| 1 \| \| Horseshoe kidney \| 1 \| Klippel-Trenaunay-Weber syndrome \| 1 \| \| Unilateral multicystic kidney \| 17 \| Others \| \| \| Bilateral cystic kidney disease \| 9 \| Fetal goitre \| 1 \| \| Gastrointestinal \| \| Cystic hygroma \| 6 \| \| Fetal ascites \| 1 \| Multiple \| \| \| Meconium peritonitis \| 2 \| Limb body wall complex \| 2 \| \| Gastroschisis \| 3 \| VACTERL \| 1 \| \| Mesenteric cyst \| 1 \| Meckel-Gruber syndrome \| 6 \| \| Enteric duplication cyst \| 1 \| Multiple malformations \| 20 \| \| Hirschsprung disease \| 1 \| Fetal tumors \| \| \| Tracheoesophageal fistula \| 2 \| Adrenal neuroblastoma \| 1 \| \| Congenital diaphragmatic hernia \| 3 \| Teratoma \| 3 \| \| Duodenal atresia \| 4 \| Sacrococcygeal teratoma \| 1 \| \| Isolated fetal ascites \| 2 \| \| \| \| Omphalocele \| 6 \| \| \| Open in a new tab 3.1. Malformations Detected prior to 20 Weeks Out of total malformed fetuses, 103 (33%) were detected prior to 20 weeks of gestational age and 209 (66.9%) were detected after 20 weeks of gestational age. Out of 103 women who were diagnosed with fetal malformations before 20 weeks, only 5 (1.6%) were detected prior to 12 weeks of gestational age and the remaining 98 (31.4%) were diagnosed between 12 and 20 weeks. Six patients amongst them presented before 12 weeks but malformations were missed and diagnosed later between 12 and 20 weeks. These cases were omphalocele, osteogenesis imperfecta, harlequin ichthyosis, Stickler syndrome, Fraser syndrome, and Dandy-Walker malformation. These conditions, however, are known to present late. Out of 103 patients diagnosed to have malformation prior to 20 weeks, 80 patients willingly underwent termination of pregnancy in view of malformation being lethal like a fetus with occipital encephalocoele terminated at 20 weeks of gestational age (Figure 1). We had prescribed protocol of oral mifepristone (200 mg) followed by misoprostol induction after 48 hours of mifepristone. Three patients were lost to follow-up. Ten patients had nonlethal malformation and were willing to continue pregnancy. All of them had postpartum neonatal intervention in the Department of Pediatric Surgery, Neonatology and Plastic Surgery, respectively (for posterior urethral valve, extra lobar sequestration, tracheoesophageal fistula, anorectal malformation, congenital diaphragmatic hernia with good LH ratio, meningocele, polycystic kidneys, megacystis, vesicoureteral reflux, and cleft lip palate). Figure 1. Open in a new tab Occipital encephalocoele diagnosed at 20 weeks which was terminated. Ten patients refused to continue pregnancy despite malformation being lethal. They had obstetrical procedure at their convenient places. Four amongst them had preterm still birth and six babies died in neonatal period. Biggest agony is that two amongst those continuing pregnancies with known lethal malformations had hysterotomy and two had cesarean section for anomalous fetus which could have been avoided. We found that with the present available technology majority of malformation could be diagnosed before 20 weeks (Box 1). Box 1. Open in a new tab List of common malformations detected before 20 weeks. First trimester sonography has huge potential of diagnosing fetal anomalies. We found that there are few malformations which could be easily diagnosed before 12 weeks (Box 2). Box 2. Open in a new tab List of common malformations we found usually detectable before 12 weeks. Five patients were diagnosed prior to 12 weeks for neural tube defect, holoprosencephaly, gastroschisis, cystic hygroma, and anencephaly. All of them had easy termination of pregnancy. 3.2. Malformations Detected after 20 Weeks Out of 312 pregnant women with malformations, 209 (66.9%) were diagnosed after 20 weeks. 109 had their first USG after 20 weeks and 100 had USG prior to 20 weeks but malformations were missed. Out of those 100 patients, 6 patients presented to our institute before 20 weeks and malformations were not confirmed until 24 weeks. In 94 women, they went for USG prior to 20 weeks at some other centre and malformation was missed. Amongst those six patients who presented to SGPGI prior to 20 weeks but were missed, there were one case each of Dandy-Walker malformation, autosomal dominant polycystic kidneys, late onset hydrocephalus, and tetralogy of Fallot. All of these tend to be diagnosed late. Two fetuses, one with cleft lip and one with neural tube defect, could have been diagnosed but were missed. There exists a group of malformation which lies in the grey zone of diagnosis before 20 weeks (Box 3). Box 3. Open in a new tab List of malformations we found undetectable before 12 weeks and difficult between 12–20 weeks. Out of 209 detected cases after 20 weeks, 70 (33.4%) patients had malformations which were detected after 20 weeks and are acceptable because these include conditions which tend to be diagnosed late in gestation like hydrocephalus (Figures 2(a) and 2(b)), agenesis of corpus callosum (Figure 2(c)), congenital cystic adenomatoid malformation (Figure 2(d)), various cardiac structural malformations, cystic kidney diseases, horseshoe kidney, Dandy-Walker malformations and variants, vein of Galen aneurysm, duodenal atresia, fetal goitre, intra-abdominal tumours, gonadal cyst, Hirschsprung disease, and isolated fetal ascites. Figure 2. Open in a new tab (a and b) show late onset hydrocephalus diagnosed at 27 weeks and 32 weeks, noted to occur late as a part of MASA syndrome (patient had history of X-linked hydrocephalus). (c) shows agenesis of corpus callosum which is known to be said with confirmation at later gestation. (d) shows fetal micro cystic congenital cystic adenomatoid malformation (CCAM) which is notorious to be missed early. 3.3. Malformation Missed Even though missed in first trimester, in 139 (66.5%) patients, fetal malformations could have been diagnosed between 12 and 20 weeks as shown in Box 1. These included malformations like neural tube defect (Figures 3(a) and 3(b)), acrania-exencephaly-anencephaly sequence (Figures 4(a), 4(b), and 4(c)), skeletal dysplasia (Figures 5(a), 5(b), and 5(c)), multicystic dysplastic kidneys (Figure 5(d)), and limb body wall complex (Figures 6(a), 6(b), and 6(c)). Figure 3. Open in a new tab (a and b) show two cases of neural tube defect diagnosed at advanced gestational ages of 22 weeks and 39 weeks. Figure 4. Open in a new tab (a, b, and c) show fetal exencephaly that could be diagnosed at 14 weeks but was missed and diagnosed at 19 weeks and as late as 27 weeks. Figure 5. Open in a new tab (a, b, and c) show skeletal malformation achondrogenesis (3D), thanatophoric dysplasia, and osteogenesis imperfecta. All of them potentially diagnosable before 12 weeks and usually before 20 weeks were missed and were diagnosed late. (d) shows bilateral multicystic dysplastic kidney diagnosed at 31 weeks, with fetus being continued as oligohydramnios. Figure 6. Open in a new tab (a, b, and c) show gross fetal deformity limb body wall complex, diagnosis delayed till 21 weeks in one fetus (Figures 6(a) and 6(b)) and 29 weeks in other fetuses (Figure 6(c)). (d) shows acardiac twin (twin reverse arterial perfusion sequence) diagnosed at 25 weeks when the normal twin had already decompensated making any intervention difficult. Prenatal interventions in very unique complications of monochorionic twins have become the treatment of choice but diagnosis of acardiac twinning was delayed till 24 weeks (Figure 6(d)). This delayed pick-up of these potentially salvageable conditions leads to high likelihood of adverse pregnancy outcome. The patient had demise of normal cotwin also at 28 weeks. 4. Discussion The overall prevalence of severe and lethal fetal structural malformation in our study was 7.6% which was higher than that reported in the literature for general population (3–5%), possibly because it was a referral centre for high risk pregnancy and fetal medicine; there is overreporting of cases. Our study calculated that CNS malformations were most common in our study population. As such, preconceptional folic acid is not commonly practiced in our study population. We realize that almost half (52.1%) of our patients had their first USG for anomaly detection after 20 weeks. It reflects the existing darkness of unawareness and vacuum of knowledge in patients and also in basic health care that are first to encounter pregnant women. We found that, out of the total number of women with diagnosed fetal malformation, 203 (65%) presented before 20 weeks. Hence, equally important is the fact to realize that almost half of these patients who had malformations detected after 20 weeks had their obstetrical sonography before 20 weeks and were missed. This missing out of an anomaly may be because of scarcity of good resolution machines, busy schedules, and lack of expertise as well. For years together, there have been substantial advances in magnification imaging and signal processing which increased the ability to visualize fetal anatomy; there has been great concern on the possibility to diagnose a wide range of fetal anomalies at the time of nuchal translucency scan by transvaginal and transabdominal sonography [7–9]. Almost half of malformations in our study were amenable to be diagnosed in first trimester as reported in current literature. These fetuses were having malformations like neural tube defects, anencephaly, holoprosencephaly, and gastroschisis (Box 2). Castro-Aragon and Levine reported that 60–67% of malformations could have been diagnosed prior to 12 weeks. This is far away from our scenario where we found that only 1.6% (5/312) were diagnosed prior to 12 weeks. This is possibly due to the lack of awareness and lack of expertise as well. Fong et al. in their study scanned 8,537 women between 11 and 14 weeks of gestation (crown rump length, 45–84 mm); there were 175 fetuses with an increased NT. Besides nuchal abnormalities, a wide range of other congenital anomalies can be diagnosed with US at 11–14 weeks of gestation, including defects of the central nervous system, heart, anterior abdominal wall, urinary tract, and skeleton. Oztekin et al. analyzed 1085 pregnancies; 21 (1.93%) fetuses had at least one major structural defect considered detectable by routine ultrasound screening. 14 (1.29%) were identified at early (first trimester) screening and an additional 5 (0.47%) were identified at late (second trimester) USG. They found that majority of fetal structural abnormalities can be detected by sonographic screening at 11–14 weeks, but detailed fetal anatomic survey performed at 18–22 weeks should not be abandoned. Rossi and Prefumo also laid stress that first trimester ultrasound can detect half of fetal malformations. They included nineteen studies on 78,002 fetuses, with 996 with malformations that were confirmed by postnatal or postmortem examinations. USG at 11 to 14 weeks detected malformation in 472 of the malformed fetuses (51%). Detection rate was highest for neck anomalies (92%) and lowest for limbs, face, and genitourinary anomalies (34% for each). The presence of associated anomalies appears to increase the accuracy of early ultrasonography. Multiple defects were more likely to be identified than isolated malformations (60% versus 44%). Detection rates ranged from 1% to 49% for spina bifida or hydrocephalus, ranged from 50% to 99% for valvular disease and septal defects, were 100% for acrania and anencephaly, and were 0% for corpus callosum agenesis and bladder exstrophy. Combination of transabdominal and transvaginal techniques resulted in a 62% detection rate versus 51% for transabdominal technique only and 34% for transvaginal technique only. Although first trimester ultrasound can detect about 50% of fetal malformations, it cannot replace second trimester ultrasound because several malformations develop later than the first trimester. Also to be kept in mind is the fact that accuracy of early ultrasonography can be compromised by transient findings like midgut herniation, small septal defects, and hydronephrosis which might get resolved during intrauterine life. Iliescu et al. did a prospective two-centre 2-year study of 5472 consecutive unselected pregnant women examined at 12 to 13 + 6 gestational weeks. The first trimester scan identified 40.6% of the cases detected overall and 76.3% of major structural defects. Major congenital heart disease (either isolated or associated with extracardiac abnormalities) was 90%. Major central nervous system anomalies were 69.5%. Fetuses with increased nuchal translucency (NT), the first trimester DR for major anomalies, were 96% compared to 66.7% amongst those with normal NT. There have been several studies seeing application for an extended protocol in which first trimester sonography is supported by a second anomaly scan. The obvious advantage of an extended protocol is that parents are offered the option of earlier and safer termination of pregnancy for the large majority of severe/lethal abnormalities. Early ultrasound might be more accurate than second trimester ultrasonography for detection of malformations associated with oligohydramnios and anhydramnios which lead to poor visualization at later gestation necessitating amnioinfusion. We have applied this kind of protocol at our centre particularly in high risk women. First trimester sonography with targeted imaging for fetal malformation appeared particularly more helpful in high risk women with previous history of fetus/neonates with malformations, known risk factors, for example, type 2 diabetes, patients prone for teratogenicity, for example, thromboembolic and valve replacement patients on warfarin, methotrexate intake for connective tissue disorders, and so forth, antiepileptic, chemotherapeutic drugs, and history of infection exposure like rubella. We diagnosed and terminated patients before 16 weeks with rubella exposure and subsequent pulmonary stenosis, severe bony stippling and craniofacial malformation associated with high doses warfarin, VSD associated with type 2 diabetes, neural tube defect with antiepileptic and tetraphocomelia with chemotherapeutic agents, renal agenesis in previous history of Fraser's syndrome, encephalocoele in previous Meckel-Gruber syndrome, ARPKD and ADPKD, and so forth. However, a detailed first trimester examination protocol involves supplementary resources: additional examination time and specialized personnel for the abnormal suspected/detected cases. Healthcare systems are yet to determine whether early first trimester diagnosis of most major structural abnormalities is cost-effective. Previous research, albeit using inferior ultrasound technology and a less extended protocol, found that the first trimester anomaly scan was cost-efficient in terms of medical and economic expenses, although they obtained lower detection rates [15, 16]. The present research about the effectiveness of early ultrasonography in the diagnosis of structural defects does have some conflicts, which made it a challenge that to what extent structural congenital abnormalities could be detected by the routine scanning of fetal anatomy combined with nuchal translucency measurement . Few other basic prerequisites associated with early prenatal diagnosis consist of the high experience required and high costs in terms of time and equipment . Even with all these circumstances, the situation in our country is such that a huge number of patients, 209 (66.9%), were diagnosed after 20 weeks which shows the lacunae which need to be filled. 5. Conclusion In our study we realized that even in a tertiary care centre only 1.6% fetuses with malformation are identified in first trimester. In a way, it throws light on the importance of screening as well as an immense need for early diagnosis and timely intervention in the field of prenatal detection of congenital malformation. A detailed examination of fetal anatomy during the routine 11–14 weeks of gestation scan provides a comprehensive assessment of fetal anatomy and can detect approximately half of major structural defects in both low-risk and high-risk pregnancies. Detection rate increases markedly beyond 13 weeks of gestation compared with 11 weeks of gestation. We have seen to be better convinced to diagnose holoprosencephaly, achondrogenesis, osteogenesis imperfecta, and spondylocostal dysostosis at 14 weeks compared to 12 weeks. It is also expected that because of the late development of some organ systems and the delayed onset of a significant number of major anomalies in the second and third trimester it is very unlikely that the early scan may replace second trimester ultrasonography. We need to identify structural malformations before 20 weeks except those conditions which are said to appear further late or reported with confirmation at a later gestational age like few posterior fossa abnormalities, duodenal atresia, and few renal abnormalities. The most important implication is safe termination and avoiding maternal threat to life by forced termination at resourceless and substandard centres. There could be an option of incorporating anomaly scan between 18 and 20 weeks in our health plans and guides at well registered centres with expertise at reasonable cost. Focus and emphasis should aim at detection of malformation earlier than 12 weeks owing to the very unique and clear facts that first trimester detection leads to easy termination of pregnancy and lessening of women's mental, physical, and psychological trauma. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. References 1.Garne E., Dolk H., Loane M., Boyd P. A. EUROCAT website data on prenatal detection rates of congenital anomalies. Journal of Medical Screening. 2010;17(2):97–98. doi: 10.1258/jms.2010.010050. [DOI] [PubMed] [Google Scholar] 2.Syngelaki A., Chelemen T., Dagklis T., Allan L., Nicolaides K. H. Challenges in the diagnosis of fetal non-chromosomal abnormalities at 11–13 weeks. Prenatal Diagnosis. 2011;31(1):90–102. doi: 10.1002/pd.2642. [DOI] [PubMed] [Google Scholar] 3.Becker R., Wegner R.-D. Detailed screening for fetal anomalies and cardiac defects at the 11–13-week scan. Ultrasound in Obstetrics and Gynecology. 2006;27(6):613–618. doi: 10.1002/uog.2709. [DOI] [PubMed] [Google Scholar] 4.Becker R., Schmitz L., Kilavuz S., Stumm M., Wegner R.-D., Bittner U. ‘Normal’ nuchal translucency: a justification to refrain from detailed scan? Analysis of 6858 cases with special reference to ethical aspects. Prenatal Diagnosis. 2012;32(6):550–556. doi: 10.1002/pd.3854. [DOI] [PubMed] [Google Scholar] Medical termination of pregnancy act, 2002, Ministry of Health and Family Welfare, Government of India, 6.Pagani G., D'Antonio F., Khalil A., Papageorghiou A., Bhide A., Thilaganathan B. Intrafetal laser treatment for twin reversed arterial perfusion sequence: cohort study and meta-analysis. Ultrasound in Obstetrics and Gynecology. 2013;42(1):6–14. doi: 10.1002/uog.12495. [DOI] [PubMed] [Google Scholar] 7.Souka A. P., Nicolaides K. H. Diagnosis of fetal abnormalities at the 10–14-week scan. Ultrasound in Obstetrics and Gynecology. 1997;10(6):429–442. doi: 10.1046/j.1469-0705.1997.10060429.x. [DOI] [PubMed] [Google Scholar] 8.Dugoff L. Ultrasound diagnosis of structural abnormalities in the first trimester. Prenatal Diagnosis. 2002;22(4):316–320. doi: 10.1002/pd.309. [DOI] [PubMed] [Google Scholar] 9.Whitlow B. J., Economides D. L. The optimal gestational age to examine fetal anatomy and measure nuchal translucency in the first trimester. Ultrasound in Obstetrics and Gynecology. 1998;11(4):258–261. doi: 10.1046/j.1469-0705.1998.11040258.x. [DOI] [PubMed] [Google Scholar] 10.Castro-Aragon I., Levine D. Ultrasound detection of first trimester malformations: a pictorial essay. Radiologic Clinics of North America. 2003;41(4):681–693. doi: 10.1016/s0033-8389(03)00045-9. [DOI] [PubMed] [Google Scholar] 11.Fong K. W., Toi A., Salem S., et al. Detection of fetal structural abnormalities with US during early pregnancy. Radiographics. 2004;24(1):157–174. doi: 10.1148/rg.241035027. [DOI] [PubMed] [Google Scholar] 12.Oztekin O., Oztekin D., Tınar S., Adıbelli Z. Ultrasonographic diagnosis of fetal structural abnormalities in prenatal screening at 11–14 weeks. Diagnostic and Interventional Radiology. 2009;15:221–225. [PubMed] [Google Scholar] 13.Rossi A. C., Prefumo F. Accuracy of ultrasonography at 11–14 weeks of gestation for detection of fetal structural anomalies: a systematic review. Obstetrics and Gynecology. 2013;122(6):1160–1167. doi: 10.1097/aog.0000000000000015. [DOI] [PubMed] [Google Scholar] 14.Iliescu D., Tudorache S., Comanescu A., et al. Improved detection rate of structural abnormalities in the first trimester using an extended examination protocol. Ultrasound in Obstetrics and Gynecology. 2013;42(3):300–309. doi: 10.1002/uog.12489. [DOI] [PubMed] [Google Scholar] 15.Whitlow B. J., Chatzipapas I. K., Lazanakis M. L., Kadir R. A., Economides D. L. The value of sonography in early pregnancy for the detection of fetal abnormalities in an unselected population. British Journal of Obstetrics and Gynaecology. 1999;106(9):929–936. doi: 10.1111/j.1471-0528.1999.tb08432.x. [DOI] [PubMed] [Google Scholar] 16.Dolkart L. A., Reimers F. T. Transvaginal fetal echocardiography in early pregnancy: normative data. American Journal of Obstetrics and Gynecology. 1991;165(3):688–691. doi: 10.1016/0002-9378(91)90310-n. [DOI] [PubMed] [Google Scholar] 17.Saltvedt S., Almström H., Kublickas M., Valentin L., Grunewald C. Detection of malformations in chromosomally normal fetuses by routine ultrasound at 12 or 18 weeks of gestation—a randomised controlled trial in 39 572 pregnancies. BJOG. 2006;113(6):664–674. doi: 10.1111/j.1471-0528.2006.00953.x. [DOI] [PubMed] [Google Scholar] 18.Rustico M. A., Benettoni A., D'Ottavio G., et al. Early screening for fetal cardiac anomalies by transvaginal echocardiography in an unselected population: the role of operator experience. Ultrasound in Obstetrics and Gynecology. 2000;16(7):614–619. doi: 10.1046/j.1469-0705.2000.00291.x. [DOI] [PubMed] [Google Scholar] Articles from Journal of Pregnancy are provided here courtesy of Wiley ACTIONS View on publisher site PDF (3.1 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Materials and Methods 3. Results 4. Discussion 5. Conclusion Conflict of Interests References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
13337	https://ops.fhwa.dot.gov/publications/fhwahop08024/chapter4.htm	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Table of Contents Traffic Signal Timing Manual Contact Information: Operations Feedback at OperationsFeedback@dot.gov This publication is an archived publication and replaced with the Signal Timing Manual - Second Edition. CHAPTER 4TRAFFIC SIGNAL DESIGN TABLE OF CONTENTS 4.0 TRAFFIC SIGNAL DESIGN CONCEPTS 4.1 Overview + 4.1.1 Relationship between Signal Timing and Traffic Control Design + 4.1.2 Traffic Signal System Design 4.2 Phasing Overview + 4.2.1 Definitions and Terminology + 4.2.2 Ring-and-Barrier Diagrams 4.3 Left-Turn Display Options + 4.3.1 Permissive Only Left-Turn Phasing + 4.3.2 Protected Only Left-Turn Phasing + 4.3.3 Protected-Permissive Left-Turn Phasing + 4.3.4 Split Phasing + 4.3.5 Prohibition of Left-Turns as a Phasing Option + 4.3.6 Guidelines for Selecting Left-Turn Phasing 4.4 Left-Turn Phase Sequence Options + 4.4.1 Lead-Lead Left-Turn Phase Sequence + 4.4.2 Lag-Lag Left-Turn Phase Sequence + 4.4.3 Lead-Lag Left-Turn Phase Sequence 4.5 Pedestrian Phasing 4.6 Right-Turn Phasing 4.7 Detection Fundamentals + 4.7.1 Detection Design Objectives + 4.7.2 Detector Operating Modes + 4.7.3 Controller Memory Modes + 4.7.4 Detection Design for High-Speed Approaches + 4.7.5 Detection Design for Low-Speed Traffic Movements 4.8 Detection Applications + 4.8.1 Basic Fully-Actuated Design + 4.8.2 Volume-Density Design + 4.8.3 Multiple-Detector Design + 4.8.4 Left-Turn Movements + 4.8.5 Right-Turn Movements 4.9 References LIST OF TABLES Table 4-1 Recommended distance between stop line and detector Table 4-2 Recommended detector locations and timing settings for multiple detector technique LIST OF FIGURES Figure 4-1 Physical components of a signal system Figure 4-2 Phasing and movement diagram for the intersection of two one-way streets Figure 4-3 Typical vehicular and pedestrian movements at a four-leg intersection Figure 4-4 Standard ring-and-barrier diagram Figure 4-5 Ring-and-barrier diagram for intersection of two one-way streets Figure 4-6 Ring-and-barrier diagram showing permissive phasing Figure 4-7 Ring-and-barrier diagram showing protected phasing Figure 4-8 Ring-and-barrier diagram showing protected-permissive phasing Figure 4-9 Ring-and-barrier diagram showing split phasing Figure 4-10 Prohibited left turns by time of day Figure 4-11 Guidelines for determining the potential need for a left-turn phase Figure 4-12 Ring-and-barrier diagram showing protected lead-lag left turns Figure 4-13 Illustration of the yellow trap Figure 4-14 Ring-barrier diagrams showing a leading pedestrian interval and an exclusive pedestrian phase4-19 Figure 4-15 Phase diagram illustrating a right-turn overlap Figure 4-16 Examples of pedestrian push buttons Figure 4-17 Example of vehicle detector design (through lane on side street) Figure 4-18 Indecision zone boundaries on a typical intersection approach Figure 4-19 Distance to the beginning and end of the indecision zone Figure 4-20 Detector location for multiple detector technique (level approach) 4.0 TRAFFIC SIGNAL DESIGN CONCEPTS This chapter documents the concepts of traffic signal design as they apply to traffic signal timing. The principles related to geometric design and operation are addressed in the Signalized Intersections: Informational Guide (1) The elements addressed in this Chapter include: signal control type, signal phasing, detection layout, and how the decisions made during traffic signal design affect signal timing for isolated (described in Chapter 5) and coordinated (described in Chapter 6) operation. This chapter consists of four sections. The first section provides an overview of the objectives of traffic signal design. The second section summarizes the basic concepts associated with key signal design elements. The third section describes a procedure for determining appropriate signal design elements for a given intersection. The last section provides guidelines for selecting signal design values and choosing from among design options. 4.1 OVERVIEW The topic of traffic signal design broadly includes any of an intersection's traffic signal control elements that have a physical presence at the intersection. In this regard, the traffic control type (e.g., pre-timed, actuated) is implemented in the signal controller and cabinet. The signal phasing is implemented using signal heads, signal indications, and logic in the controller that governs their sequence. Detection is provided by various devices that identify users (such as vehicles and pedestrians) of the intersection. Other design elements exist (e.g., preemption); however, the aforementioned three elements are present at almost all intersections, and have a significant influence on intersection safety and efficiency. Hence, they are the subject of discussion in this chapter. The objective of signal design is to produce a design that yields safe and efficient operation for the prevailing conditions. This objective is accomplished by making design choices that are tailored to the specific facility conditions. Signal timing settings (as described in Chapter 5) can be changed as needed to accommodate changes in traffic demand, pattern, but signal design elements are relatively static and are typically more difficult (or costly) to change and thus are discussed as an introduction. The efficiency of an intersection is directly impacted by its signal design, and the detection layout can have a significant effect on the safety associated with high-speed intersection approaches (2). 4.1.1 Relationship between Signal Timing and Traffic Control Design The traffic signal design process should recognize and accommodate signal timing considerations to insure effective operation of the intersection. A robust detection system is needed for the traffic signal to be able to respond to changes in traffic conditions. Detection systems sense when pedestrians and vehicles are at a traffic signal and use that information to determine who will be served next and how long the phase is served. The quality of intersection operation is particularly dependent on the relationship between the detection layout and the signal controller settings. For optimum performance, the detector layout and signal settings should be “tuned” to the geometry of the intersection, its traffic volume, and the approach speed. The tuning process consists of finding a balance between detector location (relative to the stop line), detector length, passage time, and minimum green time for the prevailing conditions. However, there is no strong consensus in the industry with regard to what is the “best balance.” Use of an iterative process in the design process results in an intersection that can take advantage of signal timing techniques to provide a high level of service to all users. The signal design can also be influenced by the traffic control device layout. Specifically, for a safe and effective signal design, it must be possible to properly position signal heads for maximum visibility for all movements. The MUTCD (3) describes the minimum standards for traffic signal displays. These standards address the number, size, mounting alternatives, physical arrangement, and placement of the signal heads. They also include special considerations for left-turn phasing that address the number and arrangement of lenses in the left-turn signal head, the location of this head, and the display sequence it presents during the signal cycle. The reader is referred to the MUTCD for further information on this topic. The intent of this chapter is to highlight specific issues that affect signal timing and not replicate the information contained within the MUTCD. 4.1.2 Traffic Signal System Design Traffic signals may operate in a system of intersections. The application of timing plans depends on the infrastructure available in the signal system. The typical hardware components of a signal system are shown in Figure 4-1. These components are described below. Figure 4-1 Physical components of a signal system This figure illustrates the five typical hardware components of a signal system. These components include the detector loop on the approach, the local controller housed in a cabinet adjoining the intersection, a master controller typically remotely located, a traffic control center that observes data from an entire region, and the communication conduits that connect the previous four elements. These are described in further detail in the paragraphs that follow. 1. Detection. Detectors are used to gather information about the “conditions” used in the local traffic signal controller at each intersection and to allocate time at the intersection. Detection may also be used to collect data that can support monitoring, managing, and measuring performance. Few agencies achieve all three functions with their detection system. 2. Local Controller. (further described in Chapter 5 and 6) The local controller operates the displays through the load switches using the signal timing provided by the user. The local controller implements specific strategies from field inputs or directly from the central signal system operator. 3. Master Controller. The master controller is an optional component of a system that facilitates communication between the “central” signal system and the local controllers for control decisions. The primary functions of the on-street master controller are to select the timing plans for a group of intersections, to process and store detector count information, and to monitor equipment operation. 4. Traffic Control Center (Signal System). The traffic control center is used for a variety of functions. The central signal system is an alternative to using a master controller and contains the operational database that stores controller data, monitors the system, and allows timing and other parameters to be modified. It may also facilitate the implementation of the advanced concepts described in Chapter 9. 5. Communication. Several components of the system communicate through many forms. Communication facilitates coordination between controllers (see below). These system components works together to implement a control strategy, such as coordination. To understand the implementation of the strategy, one must have a rudimentary understanding of how a system is configured and how coordination is maintained at a system level. This is particularly important for troubleshooting. 4.2 PHASING OVERVIEW Phasing represents the fundamental method by which a traffic signal accommodates the various users at an intersection in a safe and efficient manner. This section first presents definitions and terminology and then discusses the variety of phasing techniques used in common practice. 4.2.1 Definitions and Terminology Definitions used in traffic signal timing have resulted in some confusion. In many cases, “the old way of doing things” confounds the ability of the industry to educate people new to the field. Over the years, the description of the “individual movements” of the dual-ring 8-movement controller as “phases” has blurred into common communicated terminology of “movement number” being synonymous as “phase number”. Most signal designs and all controllers sold today provide eight standard phases within the signal controller; however, a four-phase intersection is commonly referred to in the literature to represent a standard four-legged intersection with protected left turns on all approaches. It is the intent of this document to reduce the ambiguous use of the term “phase” and for the purposes of this document, a phase is defined as a controller timing unit associated with the control of one or more movements. Each phase at an intersection has a set of timing, possibly containing vehicle and pedestrian timing. A phase may control both a through movement and a right turn movement on an approach. The MUTCD defines a signal phase as the right-of-way, yellow change, and red clearance intervals in a cycle that are assigned to an independent traffic movement or combination of traffic movements. Two additional terms that are important for improving the use of terminology within the signal timing industry is to articulate is the use of movement and interval. A movement reflects the user perspective and is defined by the user type and the action that is taken (turning movement for a vehicle or pedestrian crossing). Two different types of movements include those that have the right of way and those that must yield consistent with the rules of the road or the Uniform Vehicle Code. A simple example of the concept of movements is the intersection of one-way streets shown in Figure 4-1. In this example, the intersection is operated by two phases (2 and 4) and pedestrians are accommodated as concurrent movements to the traffic. Phase 2 will include a through and a right turn movement, while phase 4 will have a through and a left turn movement (appropriate turning movements are omitted from the diagram for simplicity). The right turn on phase 2 must yield to pedestrian traffic crossing the west leg of the intersection. Figure 4-2 Phasing and movement diagram for the intersection of two one-way streets The figure above consists of two illustrations. The first illustrates a simple example of movements using the intersection of two one-way streets. The illustration consists of two diagrams, one for phasing and one for movement. The diagram to the left is a phasing diagram indicating which movement is given right-of way during each phase of the two phase cycle. The initial phase, labeled phase two, gives right of way to vehicles heading from north to south, and pedestrians crossing the west side of the intersection. The final phase, labeled phase four, gives right-of-way to vehicles heading from east to west, and pedestrians crossing the north leg of the intersection. The second diagram illustrates permitted movements during each phase. During the initial phase, permitted vehicles movements are north to south and north to west, the latter having to yield to pedestrian movement. Permitted pedestrian movements during the initial phase are all north-south crossings on the west side of the intersection. On the final phase, permitted vehicle movement is east to west and east to south. Permitted pedestrian movement during the final phase are all east-west crossings on the north leg of the intersection. An interval is a duration of time during which the signal indications do not change. For example, a pedestrian phase contains three intervals—Walk, Flashing Don’t Walk, and solid Don’t Walk—and within the Walk and Flashing Don’t Walk intervals, the corresponding through movement will remain green. The movements served at an intersection can be categorized by the various users: vehicles, pedestrians, bicyclists, and transit. Figure 4-2 illustrates the typical vehicle and pedestrian movements at a four-leg intersection. In the context of this figure, bicycle and transit movements track the same paths as vehicle movements. These movements are regulated by the signal controller through their allocation to one or more signal phases. The following list defines some of the terms used to describe vehicle and pedestrian phasing (4): A vehicular phase is defined as a phase that is allocated to one or more vehicular traffic movements, as timed by the controller unit. A pedestrian phase is defined as a traffic phase allocated to pedestrian traffic that may provide a pedestrian indication either concurrent with one or more vehicular phases, or separate from all vehicular phases. A traffic phase is defined as the green, change, and clearance intervals in a cycle assigned to specified movement(s) of traffic. A cycle is defined as the total time to complete one sequence of signalization for all movements at an intersection. In an actuated controller unit, the cycle is a complete sequence of all signal indications. The assignments shown are typical for eight-phase controller operation although other assignments are possible. As can be seen in Figure 4-3, left-turn movements are assigned odd number phases, while through movements are assigned to even number phases. In this example, the southbound left-turn movement is protected and is associated with phase 5. The right-turn movements are not typically assigned to separate phases. For example, the westbound right-turn movement is compatible with the westbound through movement and thus shares phase 4. Typically, a pedestrian movement is associated with the concurrent vehicular phase running parallel and adjacent to it. As indicated earlier, this phase may include the concurrent right-turn movement that is associated with the through movement. As illustrated in Figure 4-3, pedestrians crossing the northern leg of the intersection are assigned the concurrent westbound vehicular phase (phase 4), which conflicts with the eastbound left turn (phase 3), this is further described in the next section. The operation of a concurrent phase is influenced by both the vehicular and pedestrian movements that it serves. Figure 4-3 Typical vehicular and pedestrian movements at a four-leg intersection The diagram above illustrates how vehicle and pedestrian movements interact in a four-legged intersection in which the major road runs north and south and the minor road runs east and west. The phases assigned to each movement in a conventional eight phase control approach. Vehicle movements include through, right-turn, and left-turns on each approach. Pedestrian movements include crossings on each of the four legs. By convention, through movements are assigned even phase designations, and left-turn movements are assigned odd phases designations. In the illustrated example, left turn phases are numbered one, three, five, and seven for left turns originating from the south, west, north, and east respectively. The through movement phases are numbered two, four, six and eight for through movements originating from the north, east, south and west respectively. Right turn phases are combined with their respective through movement, but must yield to any permitted pedestrian movement. Pedestrian phases run concurrent with through phases. Pedestrian crossings on the west leg run concurrent with southbound vehicle through movements, crossing on the north leg are concurrent with westbound through movements, and so on. 4.2.2 Ring-and-Barrier Diagrams Modern U.S. practice for signal control organizes phases by grouping them in a continuous loop (or ring) and separating the crossing or conflicting traffic streams with time between when they are allowed to operate, either by making the movements sequential or adding a barrier between the movements. The ring identifies phases that may operate one after another and are typically conflicting phases organized in a particular order. For instance, it may be desirable to separate the traffic traveling through the intersection in the northbound direction from the southbound left turn movement. A change interval and clearance time is used to separate that movement in time. In Figure 4-4, a “barrier” would be used to separate the east-west movements from north-south movements to avoid operating conflicting movements at the same time. They are also used to define a relationship between the rings to assure compatible movements. The barrier represents a reference point in the cycle at which a phase in each ring has reached a point of termination; both rings must cross the barrier simultaneously. The time sequence of phases can be described using a ring and barrier diagram. Some common rules for number phases at an intersection (which are applicable to the diagram) are provided in the following list, which assumes leading left turns and separate left-turn phases. Phases 1, 2, 3, and 4 are assigned to Ring 1. Phases 5, 6, 7, and 8 are assigned to Ring 2. Phases 1, 2, 5, and 6 are assigned to Barrier 1. Phases 3,4,7, and 8 are assigned to Barrier 2. A phase pair contains two phases within the same ring and barrier that cannot be displayed concurrently. Examples of phase pairs include 1+2, 5+6, 3+4, and 7+8. Phases within a phase pair can be reversed (e.g. 2+1, 6+5, etc.) Phase pairs within the same barrier must end simultaneously (i.e., end at the barrier). For example, phase pairs 1+2 and 5+6 must end simultaneously at the end of barrier 1 and phase pairs 3+4 and 7+8 must end simultaneously at the end of barrier 2. Phase pair 1+2 can operate concurrently with phase pair 5+6. Phase pair 3+4 can operate concurrently with phase pair 7+8. These phase pairs are also known as concurrency groups because they can time together. One common practice is to assign phases 2 and 6 to the major street movements and the phases on the other side of the barrier to the minor road movements. Another practice is to define this by direction (phase 4 may be the most northerly pointed phase). An example ring diagram is shown in Figure 4-4. The sequence of phases is shown as they occur in time, proceeding from left to right. The figure illustrates a phase sequence with left-turn movements leading the opposing through movements on both the major and minor streets. The diagram shows phase 1 and 5 ending at the same time, but they operate independently and can end at different times. The subsequent phase (phases 2 and 6 respectively) may begin once the previous phase has used its time. Once the barrier is crossed phases 3 and 7 operate followed by phases 4 and 8. The cycle ends with the completion of phases 4 and 8. Figure 4-4 Standard ring-and-barrier diagram The figure above illustrates a ring and barrier diagram corresponding to a typical four-legged intersection previously illustrated in figure 4-3. The figure appears as a two row by four column matrix. Each row represents a sequential ring. Each column in each row represents a sequential operating phase of the respective ring. Two barriers, demarked by double parallel vertical lines, are shown after the second and fourth columns. These barriers represent where the rings must be synchronized to avoid conflicting movements. For a typical four-legged intersection as depicted in figure 4-3, the diagrams works as follows: The first entry in each ring are protected left turns from the major north-south street. The first entry in ring one is vehicle movement from south to west, and the first entry in ring two is vehicle movement from north to east. The transition from column one to column two in each ring is illustrated by a yellow-red stripe indicating a transition in a signal. The second column of each ring represents the through movement and permitted right turns for vehicle movements originating from the north and south legs respectively, as well as their corresponding pedestrian movements. Columns three and four are similar to columns one and two, but address vehicle east-west vehicle movement and their related pedestrian movements. Between the second and third columns is a barrier. This barrier indicates that both ring one and ring two must complete their phase pairs before either can proceed to the third column. When considering signal phasing it is helpful to start at the most basic, one where two one-way streets intersect. Figure 4-5 shows a ring and barrier structure that compliments the phase diagram described previously in Figure 4-3. All movements on the street are served (vehicles, bicycles, and pedestrians) in one direction during one phase, and all movements on the cross street during the other phase. It should be noted that it is common practice to assign pedestrians to a through traffic movement, with the underlying assumption that a motorist or bicyclist making a right turn must yield to pedestrians as per most Uniform Vehicle Code (5). However, care should be exercised with this concurrent pedestrian phasing approach when there are exclusive right-turn lanes with high right turn volumes (6]. Figure 4-5 Ring-and-barrier diagram for intersection of two one-way streets This diagram illustrates the ring-barrier diagram for the intersection of two one way streets. The diagram consists of a one row by two column matrix representing of one ring with two phases. The initial phase allows traffic from the east leg to pass through the intersection and pedestrians to cross on the north leg concurrently. The last phase allows traffic from the north leg to pass through the intersection and pedestrians to cross on the west leg. In this example, the left-turning movements are either non-existent or prohibited (phase 2) or are protected (phase 4). Application of this concept at a more typical intersection of two-way streets uses the standard ring and barrier structure described in Figure 4-4. Assignment of phase numbers to signalized intersections is somewhat arbitrary based on historical design principles, but there are some rules that have been applied to standardize operation. Depending on the complexity of the intersection, 2 to 8 phases are typically used, although some controllers can provide up to 40 phases to serve complex intersections or sets of intersections. Developing an appropriate phasing plan begins with determining the left-turn phasing type at the intersection. Section 4.4 presents some of the guidance related to selection of appropriate left turn phasing based on traffic volumes and safety experience. 4.3 LEFT-TURN DISPLAY OPTIONS There are five options for the left-turn phasing at an intersection: permissive only, protected only, protected-permissive, split phasing, and prohibited. Phasing can have a significant impact on signal system effectiveness for a number of reasons, including: Permissive only left turn operation may reduce delay for the intersection, but may adversely affect intersection safely, because it requires motorists to choose acceptable gaps. Protected only left-turn phases may reduce delay for turning vehicles but are likely to increase overall intersection delay. Protected-permissive left turn phases can offer a good compromise between safety and efficiency but could limit available options to maximize signal progression during coordination unless innovative displays are used. Split phasing may be applicable with shared lanes, but could increase coordinated cycle length if both split phases are provided a concurrent pedestrian phase. Prohibited left turns may be used selectively to reduce conflicts at the intersection. 4.3.1 Permissive Only Left-Turn Phasing Permissive only operation requires left-turning drivers to yield to the conflicting vehicle and pedestrian traffic streams before completing the turn. In the permissive mode, the left-turn movement is served concurrently with the adjacent through movement. Both the left turn and the opposing through movements are presented with a circular green indication. Thus, in this left turn display option, a green arrow is never provided. Permissive operation is primarily used when traffic is light to moderate and sight distance is adequate. This display option provides the most efficient operation for green allocation at the intersection. The efficiency of this mode is dependent on the availability of gaps in the conflicting streams through which the turn can be safely completed. This mode can have an adverse affect on safety in some situations, such as when the left-turn driver's view of conflicting traffic is restricted or when adequate gaps in traffic are not present. The yellow trap can occur for the “permissive only” left turn when the opposite direction has a lagging left turn movement. Figure 4-6 and the following figures are adapted from those presented in the Signalized Intersections: Informational Guide Report and thus, phase 2 is defined as the eastbound movement. Figure 4-6 Ring-and-barrier diagram showing permissive phasing This figure illustrates the ring-and-barrier diagram for an intersection when permissive left turns are allowed. The diagram consists of a single ring with two phases. The first phase allows east and west through movement with permissive left turns and east-west pedestrians movements on the north and south legs. The second phase is similar to the first, except it applies to north-south traffic and pedestrian movement. 4.3.2 Protected Only Left-Turn Phasing Protected only operation assigns the right-of-way to drivers turning left at the intersection and allows turns to be made only on a green arrow display. This operation provides for efficient left-turn movement service; however, the added left-turn phase increases the lost time within the cycle length and may increase delay to the other movements. An exclusive left-turn lane is typically provided with this phasing as shown in Figure 4-7. The left-turn phase is indicated by a green arrow signal indication. This type operation is recognized to provide the safest left-turn operation. \ Figure 4-7 Ring-and-barrier diagram showing protected phasing This figure illustrates a protected only left-turn phasing ring and barrier diagram. The two ring, four column phase diagram allows for protected left turns from the east and west legs in the first phase. The second phase in each ring allows for through movements and rights turns from the east and west legs and corresponding east-west pedestrian movement. The remaining two phases are separated by a barrier, and repeats the movements for the north-south traffic. 4.3.3 Protected-Permissive Left-Turn Phasing Protected-permissive operation represents a combination of the permissive and protected modes. Left-turning drivers have the right-of-way during the protected left-turn phase. They can also complete the turn "permissively" when the adjacent through movement receives its circular green indication as illustrated in Figure 4-8. This mode provides for efficient left-turn movement service, often without causing a significant increase in delay to other movements. This mode also tends to provide a relatively safe left-turn operation, provided that adequate sight distance is available and turns during the permissive component can be safely completed. Figure 4-8 Ring-and-barrier diagram showing protected-permissive phasing This figure illustrates a protected-permissive left-turn phasing ring and barrier diagram. The diagram is similar to the protected only diagram shown in the previous figure, with the exception that permissive left turns are allowed to occur with the through movement. Protected-permissive phasing should be used with caution when a phasing sequence other than lead-lead left-turn phasing is being deployed. Section 4.4.2 provides additional detail on this matter. 4.3.4 Split Phasing Split phasing represents an assignment of the right-of-way to all movements of a particular approach, followed by all of the movements of the opposing approach. This is depicted in Figure 4-9. Spilt phasing may be necessary when intersection geometry results in partially conflicting vehicle paths through the intersections or where the approaches are offset such that left turning vehicles would have to occupy the same space to complete their turns. Split phasing avoids the conflict of 10 opposing left turn vehicle paths. Similarly, if the intersection has high left turn and through volume, the traffic engineer may have to use shared left turn and through lanes to make efficient use of the approach which would also result in split phasing for the approach. Figure 4-9 Ring-and-barrier diagram showing split phasing This figure illustrates split phasing for the north – south traffic movement in the ring and barrier diagram. The first two phases of the two ring diagram are the same as that for the protected-permissive left turn phasing for east-west movement. The third phase allows vehicle traffic from the north leg and pedestrian crossing on the west leg. All other traffic movement is prohibited. Similarly, the fourth phase in the diagram allows all traffic movements from the south leg and its corresponding pedestrian movement on the east leg. Split phasing may be helpful if any of the following conditions are present (7): There is a need to accommodate one or more left-turn lanes on each approach, but sufficient width is not available to ensure adequate separation in the middle of the intersection. This problem may also be caused by a large intersection skew angle. The larger left-turn lane volume is equal to its opposing through lane volume during most hours of the day ("lane volume" represents the movement volume divided by the number of lanes serving it.) The width of the road is constrained such that an approach lane is shared by the through and left-turn movements yet the left-turn volume is sufficient to justify a left-turn phase. One of the two approaches has heavy volume, the other approach has minimal volume, and actuated control is used. In this situation, the phase associated with the low-volume approach would rarely be called and the intersection would function more nearly as a "T" intersection. Crash history indicates an unusually large number of sideswipe or head-on crashes in the middle of the intersection and involving left-turning vehicles. This phasing is typically less efficient than other types of left-turn phasing. It typically increases the cycle length, or if the cycle length is fixed, reduces the time available to the intersecting road. 4.3.5 Prohibition of Left-Turns as a Phasing Option Prohibition of left turns on an approach is an option that has been implemented in some cases to maintain mobility at an intersection. In this case, a supplemental sign may be provided that indicates “no left turn”. In some cases, these have been applied only during certain times of day, when gaps in traffic are unavailable and operation of permitted phasing may be unsafe. Figure 4-10 is an example from Toronto, Ontario, that prohibits left turns during the morning and evening periods. Figure 4-10 Prohibited left turns by time of day In general, the operational mode used for one left-turn movement on a road is also typically used for the other (opposing) left-turn movement. For example, if one left-turn movement is permissive, the opposing left turn is also permissive. However, this agreement is not required and the decision of mode should be movement-specific based on factors such as sight distance, volumes, number of turning lanes, number of opposing lanes, and leading vs lagging left turn operation. 4.3.6 Guidelines for Selecting Left-Turn Phasing A variety of guidelines exist that have been developed to indicate conditions where the benefits of a left-turn phase typically outweigh its adverse impact to intersection operation. Many of these guidelines indicate that a left-turn phase can be justified based on consideration of several factors that ultimately tie back to the operational or safety benefits derived. These factors include: Left-turn and opposing through volumes Number of opposing through lanes Cycle length Speed of opposing traffic Sight distance Crash history The flowchart shown in Figure 4-11 can be used to assist in the determination of whether a left-turn phase is needed and whether the operational mode should be protected or protected-permissive. These guidelines were derived from a variety of sources (8;9). Application of the flowchart requires the separate evaluation of each left-turn movement on the subject road. Figure 4-11 Guidelines for determining the potential need for a left-turn phase This flow chart shows how to assist in determining the type of left-turn phasing for each movement at an intersection. The flowchart considers such factors as accident history, sight distance, number and type of lanes, volumes, and speed. Source: Adapted from (Kell and Fullerton, 1998; Orcutt, 1993; Traffic Engineering Manual, 1999). The objective of the flow chart is to identify the least restrictive left-turn operational mode. A secondary objective is to provide a structured procedure for the evaluation of left-turn phasing for the purpose of promoting consistency in left-turn phase application. The critical left-turn crash counts identified in the figure are based on an underlying average critical crash frequency and recognize the inherent variability of crash data. The underlying averages are 1.3 crashes per year and 3.0 crashes per year when considering protected-permissive and protected only left-turn phasing, respectively. If the reported crash count for existing permissive operation exceeds the critical value, then it is likely that the subject intersection has an average left-turn crash frequency that exceeds the aforementioned average (5 percent chance of error) and a more restrictive operational mode would likely improve the safety of the left-turn maneuver. The flowchart has two alternative paths following the check of opposing traffic speed. One path requires knowledge of left-turn delay; the other requires knowledge of the left-turn and opposing through volumes. The left-turn delay referred to in the flowchart is the delay incurred when no left-turn phase is provided (i.e., the left-turn movement operates in the permissive mode). 4.4 LEFT-TURN PHASE SEQUENCE OPTIONS It may be advantageous under certain circumstances to change the sequence in which left turns are served relative to their complementary through movements. This is done by reversing the sequence of a pair of complementary phases, as is shown for phases 1 and 2 in Figure 4-4. In this example, phase 1 is said to “lag” phase 2. Specifically, Figure 4-12 shows phases 2 and 6 starting and ending at different times in the cycle. This independence between the through phases can be desirable under coordinated operations because it can accommodate platoons of traffic arriving from each direction at different times. Figure 4-12 Ring-and-barrier diagram showing protected lead-lag left turns Figure 4-12: This figure illustrates a protected only left-turn phasing ring and barrier diagram, similar to Figure 4-7 but with a change in the left-turn phase sequence. The cycle starts with concurrent through and left-turn phases at the same time. The opposing left-turn phase begins after the end of the through phase in Ring 1. 4.4.1 Lead-Lead Left-Turn Phase Sequence The most commonly used left-turn phase sequence is the "lead-lead" sequence which has both opposing left-turn phases starting at the same time. If a single ring structure is used, then the two phases also end at the same time. If an actuated dual ring structure is used, then each left-turn phase 14 is assigned to a different ring such that each can end when the left-turn demand is served (i.e., they can end at different times). The advantages of this phasing option are: (1) that drivers react quickly to the leading green arrow indication and (2) it minimizes conflicts between left-turn and through movements on the same approach, especially when the left-turn volume exceeds its available storage length (or no left-turn lane is provided). A more detailed discussion of the advantages of leading left-turn phases is provided in Chapter 13 of the Traffic Engineering Handbook (10). 4.4.2 Lag-Lag Left-Turn Phase Sequence This left-turn phase sequence is most commonly used in coordinated systems with closely spaced signals, such as diamond interchanges. It has both opposing left-turn phases ending at the same time. If it is implemented in a single ring structure, then the two phases also start at the same time. If a dual-ring structure is used, then each left-turn phase is assigned to a different ring such that each can start when the left-turn demand is served (i.e., they can start at different times). Lagging left-turn phasing is also recognized to offer operational benefits for the following special situations: At "T" intersections when the one left-turn phase that exists is combined with a protected-permissive mode. At the intersection of a two-way street and a one-way street where the one left-turn phase that exists is combined with a protected-permissive mode. At a n interchange, or a pair of closely-spaced interconnected intersections, where both intersections have a left-turn phase and each are combined with the protected-permissive mode. When used with protected phasing, this phase sequence provides a similar operational efficiency as a lead-lead or lead-lag phase sequences. However, differences emerge when they are used with protected-permissive mode. One disadvantage of lagging left-turn phases is that drivers tend not to react as quickly to the green arrow indication. Another disadvantage is that, if a left-turn bay does not exist or is relatively short, then queued left-turn vehicles may block the inside through lane during the initial through movement phase. When lag-lag phasing is used at a four-leg intersection where both phases are used with the protected-permissive mode, then both left-turn phases must start at the same time to avoid the "yellow trap" (or left-turn trap) problem, illustrated in Figure 4-13. This problem stems from the potential conflict between left-turning vehicles and oncoming vehicles at the end of the adjacent through phase. Of the two through movement phases serving the subject street, the trap is associated with the first through movement phase to terminate and occurs during this phase's change period. The left-turn driver seeking a gap in oncoming traffic during the through phase, first sees the yellow ball indication; then incorrectly assumes that the oncoming traffic also sees a yellow indication; and then turns across the oncoming traffic stream without regard to the availability of a safe gap. The "yellow trap" problem can be alleviated by using one of the following techniques: Use the protected-only mode for both left-turn movements. Use a single-ring structure to ensure that both through movement phases end at the same time (use with the protected-permissive mode). Use a flashing yellow arrow or"Dallas Display" for both left-turn signal heads and use the protected-permissive mode for both left-turn movements. Figure 4-13 Illustration of the yellow trap This diagram shows sequential phasing for a protected-permissive left-turn (PPLT) movement that experiences a yellow trap. The diagram shows a five-section signal head display for the PPLT movement, a three-section signal head display for an adjacent through lane, and a three-section signal head display for the opposing through movement. The six sequential indications shown for the PPLT display are: (1) red ball indications for the three depicted signal heads (label reads - All red); (2) a green-arrow indication for the PPLT head and red ball indications for all three signal heads (label reads - Protected left turn); (3) a yellow-arrow indication for the PPLT head and red ball indications for all three signal heads (label reads, Clearance interval (end protected left turn)); (4) green-ball indications for the three signal heads (label reads - Permissive phase); (5) yellow ball indication for the PPLT and adjacent through signal heads and a green ball indication for the opposing through signal head (label reads - Change interval (Yellow trap)); and (6) red ball indication for the PPLT and adjacent through signal heads and a green ball indication for the opposing through signal head (label reads, Opposing through phase indication still green). Source: FHWA Signalized Intersection Guide In fact, under at least one condition, the second technique can operate more efficiently than dual-ring lead-lead phasing. This condition occurs when the left-turn volume is moderate to heavy and relatively equal on both approaches. Regardless, a detailed operational evaluation should always be used to confirm that lag-lag phasing operates more efficiently than other phasing options. The third technique avoids the yellow trap by using an overlap in the controller and a five-section left-turn signal head. An overlap is a controller output (to the signal head load switch) that is associated with two or more phases. In this application, the left-turn green and yellow arrow indications are associated with the subject left-turn phase; and the left-turn green, yellow, and red ball indications are associated with the opposing through movement phase (as opposed to those of the adjacent through phase). Two types of protected-permissive displays have been developed to provide more operational flexibility and avoid the “yellow trap” problem: the flashing yellow arrow and the “Dallas Display”. The flashing yellow arrow is contained within a three-, four-, or five-section head and provides a permissive indication to the driver that operates concurrent with the opposing through movement rather than the adjacent through movement. Researchers for National Cooperative Highway Research Program (NCHRP) Project 3-54 studied alternatives to the green ball indication for permissive left turn movements. This study was conducted over a 7-year period and comprised a very comprehensive research process, including engineering analyses, static and video-based driver comprehension studies, field implementation, video conflict studies, and crash analyses. This study (11) recommended that a flashing yellow arrow be allowed as an alternative to the circular green for permissive left-turn intervals. The flashing yellow arrow was found to be the best overall alternative to the circular green as the permissive signal display for a left-turn movement. A flashing yellow arrow was found to have a high level of understanding and correct response by left-turn drivers, and a lower fail-critical rate than the circular green. The flashing yellow arrow display in a separate signal face for the left-turn movement offers more versatility in field application. It is capable of being operated in any of the various modes of left-turn operation by time of day, and is easily programmed to avoid the "yellow trap" associated with some permissive turns at the end of the circular green display. Subsequent to the publication of the NCHRP research, FHWA has issued an Interim Approval and no longer requires a Request to Experiment for this display. As of this writing, it has been proposed for inclusion in the next revision to the MUTCD. In the “Dallas Display”, the signal head uses louvers on the yellow and green ball indications to restrict visibility of the left-turn display to adjacent lanes. The louvered signal head is referred to as the "Dallas Display." With this display, both left-turn phases can operate in the protected-permissive mode and the trap is avoided. 4.4.3 Lead-Lag Left-Turn Phase Sequence This left-turn phase sequence is generally used to accommodate through movement progression in a coordinated signal system. The aforementioned "yellow trap" may occur if the leading left-turn movement operates in the protected-permissive mode and the two through movement phases time concurrently during a portion of the cycle. The "yellow trap" problem can be alleviated by using one of the following techniques: Use the protected only mode for the leading left-turn movement. Use a single ring structure that ensures that the two through movement phases do not time concurrently (use with the protected-permissive mode). Use a flashing yellow arrow or "Dallas Display" for the left-turn signal head associated with the leading left-turn phase and use the protected-permissive mode for both left-turn movements. The first two techniques will likely have an adverse effect on operations, relative to a dual ring implementation of lead-lag phasing with protected-permissive operation. However, they avoid the potential adverse effect a yellow trap would have on safety. The third technique avoids the yellow trap by using an overlap in the controller and a five section left-turn signal head. This use of overlap was described in the previous discussion of lag-lag left-turn 17 phasing. However, in practice, the Dallas Display is used for both the leading and the lagging left-turn signal heads because it improves operational performance (12). Lead-lag phasing is also recognized to offer operational benefits for the following special situations: Where there is inadequate space in the intersection to safely accommodate the simultaneous service of the opposing left-turn movements. However, in this application, a single ring structure (or equivalent functionality in a dual ring structure) should be used to ensure that the two left-turn phases never time concurrently. Intersections where the leading left-turn movement is not provided an exclusive lane (or the available left-turn storage is relatively small). 4.5 PEDESTRIAN PHASING Pedestrian movements are typically served concurrently with the adjacent through movement phase at an intersection. This is done to simplify the operation of the intersection primarily and is largely a legacy issue in our application of signal logic and control. Typical application of pedestrian operation puts pedestrians in conflict with right-turning vehicles and left-turning vehicles that operate in a permissive mode, by inviting their movement at the same time. There are specific measures that can be used to mitigate this potential conflict, three common options include: Leading pedestrian interval. As shown in Figure 4-14b, a leading pedestrian interval starts a few seconds before the adjacent through movement phase. This allows pedestrians to establish a presence in the crosswalk and thereby reduce conflicts with turning vehicles. This option supports improved safety for pedestrians by allowing them increased visibility within the intersection and is applicable to intersections where there are significant pedestrian-vehicle conflicts (13). Lagging pedestrian interval. A lagging pedestrian interval option operates similarly to a leading pedestrian interval, except that the pedestrian walk interval starts several seconds after the adjacent through movement phase. This option allows a waiting right-turn queue to clear before the pedestrian walk indication is presented and reduces conflicts with right-turning vehicles. It is applicable to intersections where there is: (1) a high right-turn volume and (2) either an exclusive right-turn lane (or lanes) or the two intersecting roads have one-way traffic (14). Exclusive pedestrian phase (also “pedestrian scramble” or “Barnes’ Dance”) As shown in Figure 4-14c, an exclusive pedestrian phase dedicates an additional phase for the exclusive use of all pedestrians. This additional phase is configured such that no vehicular movements are served concurrently with pedestrian traffic. During this phase, pedestrians can cross any of the intersection legs and may even be allowed to cross the intersection in a diagonal path. This type of phasing has an advantage of reducing conflicts between right-turning vehicles and pedestrians, but it comes at a penalty of reduced vehicular capacity and longer cycle lengths (which increases delay to all users). The exclusive pedestrian phase is not frequently used but can be found in the central business districts of several cities, including Denver and San Francisco. Figure 4-14 Ring-barrier diagrams showing a leading pedestrian interval and an exclusive pedestrian phase This figure illustrates the leading pedestrian interval and the exclusive pedestrian phase using the ring and barrier diagrams. The first option adds an interval before starting the through movement. The second option provides an additional phase following the normal ring-barrier structure for the pedestrians. The third option provides an exclusive pedestrian phase with no permitted vehicle traffic. 4.6 RIGHT-TURN PHASING Two types of right-turn phasing are addressed in this section. The first type is based on the addition of a phase to the signal cycle that exclusively serves one or more right-turn movements. This type of right-turn phasing is rarely used. If it is being considered, then its operational or safety benefits should be evaluated and shown to outweigh its adverse impact on the efficiency of the other intersection movements. The second type of right-turn phasing is based on the assignment of the right-turn movement to the phase serving the complementary left-turn movement on the crossroad. The following conditions should be satisfied before using this type of right-turn phasing: The subject right-turn movement is served by one or more exclusive right-turn lanes. The right-turn volume is high (300 vehicles per hour or more) and is a critical movement at the intersection (see Chapter 3 for more details). A protected left-turn phase is provided for the complementary left-turn movement on the intersecting road. U-turns from the complementary left-turn are prohibited. If the aforementioned conditions are satisfied, then the appropriate operational mode can be determined. If the through movement phase for the subject intersection approach serves a pedestrian movement, then the right-turn phasing should operate in the protected-permissive mode. As shown in Figure 4-15, the permissive right-turn operation would occur during the adjacent through movement phase, and the protected right-turn operation would occur during the complementary left-turn phase. Figure 4-15 Phase diagram illustrating a right-turn overlap A ring and barrier diagram illustrating the use of permissive right turns (on overlap) during the adjacent through movement phase, and protected right turns during the complimentary left-turn phase. If the through movement phase for the subject intersection approach does not serve a pedestrian movement, then the right-turn phasing should operate in the protected only mode during both the adjacent through movement phase and the complementary left-turn phase. A controller overlap may be used to provide this sequence. 4.7 DETECTION FUNDAMENTALS Detection at an intersection informs the signal controller that a user desires service (often called “demand” for service). Detectors place calls into the traffic signal controller. The controller uses this information and the signal timing to determine the display provided to the users. Detection for pedestrians is limited in most cases to push buttons as shown in Figure 4-16, although accessible 20 pedestrian signal detectors are increasing in their use. The FHWA’s Pedestrian Facilities User Guide-Providing Safety and Mobility (15) is a resource describing more detail related to other equipment. Figure 4-16 Examples of pedestrian push buttons A pedestrian push button sign. It includes information for the pedestrian such as: Start crossing (Watch for turning cars), Don’t Start (Finish Crossing if Started) and Pedestrians should not be in crosswalk Source: Fred Ranck,FHWA, Illinois, Naperville, intersection of Washington Street at Shuman Blvd There are various forms of vehicle detection technologies, and strengths and weaknesses of each are described in the Traffic Detector Handbook, 3rd edition (2006). The detection design for an intersection describes the size, number, location, and functionality of each detector. Most engineering drawings include the wiring diagram for how detectors are associated to phases. Signal timing settings such as the passage time, delay, extend, and other related parameters are described in more detail in Chapter 5. The size and location of detectors is an important element in traffic signal design. Detectors can consist of one 6-foot-by-6-foot inductive loop detector, a series of closely spaced 6-foot-by-6-foot loop detectors (may be circular in shape as shown in Figure 4-17), one long (6-foot-by-40-foot) loop detector, or alternative detection technology (e.g., video, microwave, etc.) that can monitor an 21 equivalent length of roadway. This detection zone can be used to meet the objectives described below. Figure 4-17 Example of vehicle detector design (through lane on side street) Figure 4-17a depicts an intersection design in which two circular inductive loops are located in the left turn lane, the first loop is located 16 feet from the stop bar, and the second loop is located four foot from the stop bar. Figure 4-17b is a photograph of the same intersection. In the photograph, the locations of the loop detectors are visible due to the sealing compound that protects the embedded loops. (a) Schematic design of loop detector design(b) Physical representation of loop detectors A “call” represents the controller’s registration of a request for service on a specified phase. A call can be triggered by an actuation from any detection, vehicular, pedestrian, or other or through a controller function. When triggered by a vehicle detection unit, the call’s start and end time can be equal to that of the actuation from the detection unit, or it can be modified using the controller’s detector parameters (e.g., delay, extend, call, queue, etc.). These parameters are described in Chapter 5. 4.7.1 Detection Design Objectives The objective of detection is to detect vehicle presence and identify gaps in vehicle presence that are sufficiently long to warrant terminating the phase. There are many objectives of detection design that can be characterized with the following statements: 1. Identify vehicle presence on a phase. 2. Extend the phase to serve queued traffic and that which is progressed from upstream traffic signals. 3. Identify gaps in traffic where the phase may be ended and extend the green. 4. Provide a safe phase termination for high-speed movements by minimizing the chance of a driver being in the indecision zone at the onset of the yellow indication. The first and fourth objectives are safety related. The first objective addresses expectancy, while the fourth specifically addresses the potential crashes as a result of phase termination. The fourth objective is achieved by using advance detectors on the approach. The location of these detectors can vary and depends on the detection technology used as well as intersection approach speed. The safety benefit of this design tends to be more significant on high-speed approaches. The other objectives focus on intersection efficiency. The second and third objectives are designed to address efficiency. During low volume (late night or off-peak) conditions, detection should seek to serve all traffic identified without stopping. In peak conditions, green allocation should seek to measure flows and maintain flows that are near saturation flow as described in Chapter 3. Detection timing to achieve this objective will be discussed in more detail in Chapter 5. 4.7.2 Detector Operating Modes The detection operating mode refers to the way the detection unit measures activity and is set in the detection unit. It also affects the duration of the actuation submitted to the controller by the detection unit. One of two modes can be used: presence or pulse. The presence mode is typically the default mode. Pulse Mode Pulse mode is used to describe a detector which detects the passage of a vehicle by motion only (point detection). This is characterized by a short “on” pulse sent to the controller of 0.10 to 0.15 seconds duration. The actuation starts with the arrival of the vehicle to the detection zone and ends after the pulse duration. This mode is typically used when the detectors are located upstream of the stop line and the associated detector channel operates in the locking mode. Presence Mode Presence mode is used to measure occupancy and the actuation starts with the arrival of the vehicle to the detection zone and ends when the vehicle leaves the detection zone. Thus, the time duration of the actuation depends on the vehicle length, detection zone length, and vehicle speed. Presence mode measures the time that a vehicle is within the detection zone and will require shorter extension (or gap) timing with its use. Presence mode is typically used with long-loop detection located at the stop line. In this application, the associated detector channel in the controller is set to operate in the non-locking mode (see next section). In this mode, the delay or extend parameters in the controller (described in detail in Chapter 5) can be used to modify the call start and end times. Alternatively, the delay or extend functions in the detection unit could also be used to adjust the start and end time of the actuation. The combination of presence mode operation and long-loop detection typically require a small passage time value to maintain efficiency. This characteristic tends to result in an operational benefit through efficient queue service. Modifiers to the detector settings are commonly handled in the controller. This has increased in popularity because it provides a single location for information on all phases at the intersection. 4.7.3 Controller Memory Modes Controller memory modes refer to the controller’s ability to “remember” (i.e., retain) a detector actuation. One of two modes can be used: non-locking or locking. This mode is set in the controller for each of its detector channel inputs. It dictates whether an actuation received during the red interval (and optionally, the yellow interval) is retained until the assigned phase is served by the controller. All actuations received during the green interval are treated as non-locking by the controller. The non-locking mode is typically the default mode. Non-locking Mode In the non-locking mode, an actuation received from a detector is not retained by the controller after the actuation is dropped by the detection unit. The controller recognizes the actuation only during the time that it is held present by the detection unit. In this manner, the actuation indicates to the controller that a vehicle is present in the detection zone and the controller converts this actuation into a call for service. This mode is typically used for phases that are served by stop line detection. It allows permissive movements (such as right-turn-on-red) to be completed without invoking a phase change. In doing so, it improves efficiency by minimizing the cycle time needed to serve minor movement phases. Non-locking mode is not typically used with pulse detection due to an inability to detect vehicle presence after the pulse duration elapses. Locking Mode In the locking mode, the first actuation received by the controller on a specified channel during the red interval is used by the controller to trigger a continuous call for service. This call is retained until the assigned phase is serviced, regardless of whether any vehicles are waiting to be served. This mode is typically used for the major-road through movement phases associated with a low percentage of turning vehicles (as may be found in rural areas). One advantage of using this mode is that it can eliminate the need for stop line detection, provided that advance detection is provided and that it is designed to ensure efficient queue service. 4.7.4 Detection Design for High-Speed Approaches Detection designs for high speed approaches (speeds greater than 35 mph) have the objective to not only service the queue at the beginning of green but also to safely terminate the phase in the presence of a conflicting call. Stop bar detection is usually used to clear the queues and the multiple upstream detectors are used to safely terminate the phase. For efficient operation, the stop bar detector should be programmed as a queue detector so that the stop bar detector is disconnected after the queue clears and only the upstream detectors are used to safely terminate the phase. When stop bar detectors are not used, volume density functions should be used to provide appropriate minimum green time to clear the queues. The design of advance detection on high-speed approaches requires special attention. Drivers within a few seconds travel time of the intersection tend to be indecisive about their ability to stop at the onset of the yellow indication. This behavior yields an "indecision zone" (also known as a “dilemma zone”) in advance of the stop line wherein some drivers may proceed and others may stop. The location of this zone is shown in Figure 4-18. Figure 4-18 Indecision zone boundaries on a typical intersection approach Figure 4-18: This figure illustrates the indecision zone. In this zone some drivers may stop at the transition to yellow while others proceed through the intersection. Upstream from the indecision zone, drivers are probable to stop, while downstream drivers are probable to proceed through the intersection. The indecision zone location has been defined in several ways. Distance from stop line. Some researchers have defined it in terms of distance from the stop line (16;17). They define the beginning of the zone as the distance beyond which 90 percent of all drivers would stop if presented a yellow indication. They define the end of the zone as the distance within which only 10 percent of all drivers would stop. The distance to the beginning of the zone recommended by Zegeer and Deen corresponds to about 5 seconds of travel time. That recommended by ITE increases exponentially with speed, ranging from 4.2 to 5.2 s travel time, with the larger values corresponding to higher speeds. Travel time. Another definition of the indecision zone boundary is based on observed travel time to the stop line. Chang et al. (18) found that 85 percent of drivers stopped if they were more than 3 s from the stop line, regardless of their speed. Similarly, they found that drivers less than 2 s from the stop line would almost always continue through the intersection. Stopping sight distance. A third definition of the beginning of the indecision zone is based on safe stopping sight distance (SSD). A method for computing this distance is described in Chapter 3 of the AASHTO document, A Policy on the Geometric Design of Highways and Streets (19). The zone boundaries obtained by these three definitions are compared in Figure 4-19. The boundaries based on distance typically have an exponential relationship. Those based on travel time have a linear relationship. Based on the trends shown in the figure, the beginning and end of the indecision zone tend to be about 5.5 and 2.5 seconds, respectively, travel time from the stop line. These times equate to about the 90th-percentile and 10th-percentile drivers, respectively. Figure 4-19 Distance to the beginning and end of the indecision zone For these types of designs, the furthest detector upstream of the stop bar is usually located at the beginning of the indecision zone of the approach design speed (85th-percentile approach speed). This is usually at a distance of 5 to 5.5 seconds of travel time. Subsequent detectors have a design speed of 10 mph lower than the upstream detector. Typically 3 to 4 detectors are used to enable safe termination of the high speed approach phase. The detectors are allowed to extend the phase by the passage time programmed in the controller or by the extension time on the detector itself (see Chapter 5). Although the concept of the indecision zone has been known for many years, comprehensive research has not been completed to conclude what the appropriate way to address the human factors associated with intersection design and signal timing display. The indecision should not be confused with the dilemma faced by drivers determining whether there is distance to stop and, if not, to travel through the intersection before a conflicting movement receives a green indication. 4.7.5 Detection Design for Low-Speed Traffic Movements Detection design for low speed traffic movements (speeds of 35 mph or less) has a different objective compared to the detection design for high speed traffic movements. The primary objective of detection on low speed approaches is to call a phase and clear the queue while minimizing delay. Due to lower speeds, there is less emphasis on protection from dilemma zone or indecision zone on the approach. Hence, some agencies use only stop bar detection for low speed approaches. Stop bar detectors are usually operating in the presence mode. This facilitates the primary objective of detector calling the phase and clearing the queues. 4.8 DETECTION APPLICATIONS This section presents a series of detection applications. 4.8.1 Basic Fully-Actuated Design This design is based on the following assumptions: A detection zone located at the stop line. The detectors are operating in the presence mode. Non-locking memory is used for the associated detector channel in the controller. No recall is used for the phase to which the detector is assigned. The key element of this design is the determination of detection zone length. The optimal length represents a trade off in the desire to avoid both premature gap out and excessive extension of green. According to Lin (20), the ideal length of the stop line detection zone is about 80 feet. This length allows the passage time setting to be small such that the design is very efficient in detecting the end of queue while minimizing the chance of a premature gap-out. However, the installation and maintenance of such a long detector is often cost prohibitive and multiple detectors of a shorter length are often used. The following guidelines should be used to determine the appropriate length of the stop line detection zone: The detection zone should not be smaller than 20 feet. The zone should be positioned such that a queued vehicle cannot stop between its trailing edge and the crossroad. The zone should consist of one long detection zone or a series of loops. Other sensors that can provide the equivalent length of detection can also be used. 4.8.2 Volume-Density Design This design is based on the following assumptions: One 6-foot detector is located upstream of the stop line. The detector unit is operating in the pulse mode. Locking memory is used for the associated detector channel in the controller. No recall is used for the phase to which the detector is assigned. This design is not well-suited to approaches with a significant percentage of turning vehicles because these vehicles may unnecessarily submit a call for the through movement phase. A key element of this design is the location of the detector. This location should be based on the desired maximum allowable headway for the design. Research has shown that maximum allowable headways in the range of 1.8 to 2.5 seconds yield the snappiest operation, values of 2.6 to 4.5 seconds will allow detected vehicles to use the green, but may result in extension of green during low flow periods (this is discussed further in Chapter 5). Lower values are more appropriate for higher volume conditions. The advance detector should be located such that the travel time from the detector to the stop line for a vehicle traveling at the 85th percentile speed is equal to the maximum allowable headway. In this manner, the green interval is not unnecessarily extended by a vehicle that has crossed the stop line (as it is with stop line detection). Based on this principle, the recommended location for the detector is listed in Table 4-1 for a range of approach speeds. With this design, the passage time setting is equal to the maximum allowable headway (which may vary if gap-reduction is used). Guidelines related to the detection design as it relates to signal timing are described in Chapter 5 for determining the appropriate minimum green interval. Gap-reduction is not essential with a maximum allowable headway of 3.0 s but should be considered for a maximum allowable headway of 4.0 s. If gap-reduction is used, then the minimum gap should be equal to 2.0 s. Guidelines for setting the time before reduction parameter and the time to reduce parameter are described in Chapter 5. Table 4-1 Recommended distance between stop line and detector \| 85th Percentile Approach Speed, mph \| Distance between Detector and Stop Line, ft \| \| 3.0 s Max. Allowable Headway \| 4.0 s Max. Allowable Headway1 \| \| 20 \| 90 \| 115 \| \| 25 \| 110 \| 145 \| \| 30 \| 130 \| 1752 \| \| 35 \| 155 \| 2052 \| \| 40 \| 180 \| 2352 \| \| 45 \| 200 \| 2652 \| \| Notes: 1. Use the controller’s gap-reduction feature with a minimum gap 2.0 s. 2. Use the controller’s variable initial feature to ensure the minimum green duration is not unnecessarily long during low-volume conditions (see Chapter 5, Table 5-3). \| 4.8.3 Multiple-Detector Design This design technique, used by the city of Portland, Oregon, is based on the following assumptions: One 6-foot detector is placed at 60 feet (because safe stopping distance ends at 15 mph), this is the last detector the vehicle will cross. One 6-foot detector is located at the safe stopping distance upstream of the stop line, based on the approach speed; this is the first detector the vehicle will cross. At higher speeds, one or two additional detector(s) is(are) placed between the first advance detector and the last detector (at 60 feet). The detectors operate in presence mode. Non-locking memory is used for the associated detector channel in the controller. No recall is used for the phase to which the detector is assigned. The first detector registers demand for the vehicles as they approach the stop bar. The second detector is used like a speed sieve to determine whether the vehicle can stop safely upon its approach to the intersection (based on the safe stopping distance from the second detector. This design is well-suited to approaches with a significant percentage of turning vehicles because it is likely these vehicles will gap out as the vehicle slows to make the turn at the intersection or upstream driveway. This is further illustrated in Figure 4-20. Figure 4-20 Detector location for multiple detector technique (level approach) This figure illustrates the differences in the beginning and end of the indecision zone based on the three definitions discussed in the text the location of the first advance detector and the corresponding safe stopping distance as a function of the approach speed. Source: City of Portland, Oregon A key element of this design is the use of detectors and extend values within the controller timing to reduce the time between successive vehicles. A sample table of extend values from the City of Portland is shown in Table 4-2. Table 4-2 Recommended detector locations and timing settings for multiple detector technique \| Approach Speed, mph \| Location of Advance Detector (feet) \| Location of Second Detector (feet) \| Location of Detector Nearest Stop Bar (feet) \| Extend (Carryover) Value from Upstream Detectors (secs)1 \| \| 25 \| 105 \| 60 0.8 \| \| 30 \| 140 \| 60 2.1 \| \| 35 \| 183 \| 115 \| 60 \| 1.0 \| \| 40 \| 230 \| 130 \| 60 \| 1.6 \| \| 45 \| 283 \| 190 \| 115 & 60 \| 1.0 \| \| Notes: 1. A minimum gap of 0.5 seconds is used to allow the vehicle to leave the last detector. Source: City of Portland \| 4.8.4 Left-Turn Movements The guidelines provided in this section can be used to design the left-turn movement detection when this movement has an exclusive lane (or lanes). In general, the detection design for a left-turn movement should follow the guidelines offered for through movements, as described previously for basic fully-actuated design in Section 4.8.1. Some agencies use an additional detector approximately 100 feet upstream of the stop bar to provide some advance notice of a vehicle approaching the intersection. In this case, an extension (carryover) setting is provided to allow a vehicle leaving the first advance detector to reach the detection zone at the stop bar and a short gap setting is used for the phase (assuming stop bar detection in presence mode). In cases where the left-turn movement operates in the protected-permissive mode and the conflicting through movement phase is on recall, then the stop line detection zone could be set back from the stop line one or two vehicle lengths. This technique assumes that one or two vehicles can be served at the end of the conflicting through phase. A call for the left-turn phase would only be registered if the left-turn queue extends back over the detection zone, in which case there would be more vehicles that could clear at the end of the through phase. If the left-turn movement operates in the protected-permissive mode and the conflicting through movement phase is likely to rest in recall, then the controller’s delay parameter can be used with the channel assigned to the left-turn lane detector to minimize unnecessary phase changes. The delay value used should range from 3 to 15 seconds, with the larger values used when higher speeds and volumes exist on the conflicting approach. Most controllers have a detector switching setting that can be used to send calls from left-turn detectors to the adjacent through phases to extend the permissive green phase for both movements. If the left-turn movement operates in the permissive or protected-permissive mode and the adjacent through movement phase is not on recall, then it may be desirable to extend the stop line detection zone beyond the stop line. This extension would be intended to minimize the potential for stranding a turning vehicle in the intersection at the end of the permissive period. It may also ensure that the left-turn movement is detected under low-volume conditions. 4.8.5 Right-Turn Movements The guidelines provided in this section can be used to design the right-turn movement detection when this movement has an exclusive lane (or lanes). In general, the detection design for a right-turn movement should follow the guidelines offered for through movements, as described previously for basic fully-actuated design in Section 4.8.1. If the right-turn volume is moderate-to-high and the volume on the intersecting road is relatively low (such that many gaps for right-turn-on-red exist), then the controller’s delay parameter can be used with the channel assigned to the right-turn lane detector to minimize unnecessary phase changes. The delay parameter should also be considered when a conflicting phase is on recall. The delay value used should range from 7 to 15 s, with the larger values used when higher speeds and volumes exist on the intersecting road. If the right-turn volume is high but there are few gaps for right-turn-on-red, then the use of the delay parameter may not be appropriate because it may only increase the delay to the right-turning vehicles. 4.9 REFERENCES 1. Rodegerdts, L.A., B. Nevers, B. Robinson, J. Ringert, P. Koonce, J. Bansen, T. Nguyen, J. McGill, D. Stewart, J. Suggett, T. Neuman, N. Antonucci, K. Hardy, and K. Courage. Signalized Intersections: Informational Guide. Report FHWA-HRT-04-091. Federal Highway Administration, Washington, D.C., 2004. 2. Wu, C-S., C.E. Lee, R.B. Machemehl, and J. Williams. “Effects of Multiple-Point Detectors on Delay and Accidents.” Transportation Research Record 881. National Research Council, Washington, D.C., 1982, pp. 1-9. 3. Manual on Uniform Traffic Control Devices. U.S. Department of Transportation, Washington, D.C., December 2003. 4. National Transportation Communications for ITS Protocol: Object Definitions for Actuated Traffic Signal Controller (ASC) Units – 1202 v01.07. National Electrical Manufacturers Association, Rosslyn, Virginia, January 2005 5. National Committee on Uniform Traffic Laws and Ordinances, February 26, 2007 6. Hubbard, S.M.L., R. Awwad, and D.M. Bullock, “A New Perspective on Assessing Impact of Turning Vehicles on Pedestrian Level of Service at Signalized Intersections," Transportation Research Record, TRB, National Research Council, Washington, DC, TRB Paper ID# 07-0473, in press. 7. Rodegerdts et al. 2004 8. Kell, J. H., and Fullerton, I. J., Manual of Traffic Signal Design. 2nd Edition. Institute of Transportation Engineers, Washington, D.C., 1998. 9. Orcutt, F.L. The Traffic Signal Book. Prentice Hall, Englewood Cliffs, New Jersey, 1993. 10. Traffic Engineering Manual. Section 3 - Signals. Florida Department of Transportation, Tallahassee, Florida, March 1999 11. NCHRP Report 493 12. De Camp, G., and R.W. Denny. "Improved Protected-Permitted Left-Turn Signal Displays -- The Texas Approach." ITE Journal. Institute of Transportation Engineers, Washington, D.C., October, 1992, pp. 21-24 13. Lalani, N. Alternative Treatments for At-Grade Pedestrian Crossings. Institute of Transportation Engineers, Washington, D.C., 2001 14. Lalani, N., 2004 15. Zegeer, C.V., C. Seiderman, P. Lagerwey, M. Cynecki, M. Ronkin, and R. Schenider. "Pedestrian Facilities Users Guide – Providing Safety and Mobility." Federal Highway Administration, Washington, D.C., March 2002. 16. ITE Technical Committee 18 (P.S. Parsonson, Chair). "Small-Area Detection at Intersection Approaches." Traffic Engineering. Institute of Transportation Engineers, Washington, D.C., February 1974, pp. 8-17 17. Zegeer, C.V., and R.C. Deen. "Green-Extension Systems at High-Speed Intersections." ITE Journal. Institute of Transportation Engineers, Washington, D.C., November, 1978, pp. 19-24 31 18. Chang, M.S., C.J. Messer, and A.J. Santiago. "Timing Traffic Signal Change Intervals Based on Driver Behavior." Transportation Research Record No. 1027. Transportation Research Board, National Research Council, Washington, D.C., 1985, pp. 20-30. 19. A Policy on Geometric Design of Highways and Streets. American Association of State and Highway Transportation Officials, Washington, D.C., 2004 20. Lin, F.B. “Optimal Timing Settings and Detector Lengths of Presence Mode Full-Actuated Control.” Transportation Research Record No. 1010. Transportation Research Board, National Research Council, Washington, D.C., 1985, pp. 37-45 32 Previous \| Table of Contents \| Next \| \| \| \| \| US DOT Home \| FHWA Home \| Operations Home \| Privacy Policy United States Department of Transportation - Federal Highway Administration \| \|
13338	https://ijccm.org/abstractArticleContentBrowse/IJCCM/21587/JPJ/fullText	Gastrointestinal Stomas and Fistulas: What is Lost and What to Do? INVITED REVIEW Indian Journal of Critical Care Medicine Volume 24 \| Issue Suppl 4 \| Year 2020 Gastrointestinal Stomas and Fistulas: What is Lost and What to Do? Rajesh K Pande1, Arpit Gupta2 1,2 Department of Critical Care, BLK Center for Critical Care, BLK Superspeciality Hospital, New Delhi, India Corresponding Author: Rajesh K Pande, Department of Critical Care, BLK Center for Critical Care, BLK Superspeciality Hospital, New Delhi, India, Phone: +91 9810536268, e-mail: rajeshmaitree2000@gmail.com How to cite this article Pande RK, Gupta A. Gastrointestinal Stomas and Fistulas: What is Lost and What to Do? Indian J Crit Care Med 2020;24(Suppl 4):S175–S178. Source of support: Nil Conflict of interest: None ABSTRACT Abnormal connections between gastrointestinal tract (GIT) and skin are called enterocutaneous fistulas (ECFs). Presence of ECF is associated with significant morbidity and mortality. A stoma refers to a surgically created opening in the abdomen to divert feces or urine to the outside of the body, to compensate for partial or complete loss of bowel function. Gastrointestinal (GI) stomas and postoperative ECFs present a unique challenge to the intensivist due to development of malnutrition, dehydration, and sepsis leading to high morbidity and mortality. This review focuses on the basic concepts about the type of fistula and stomas, their indications and complications, and management. Principles of clinical management include replacement of fluid and electrolyte losses, control of sepsis along with reducing fistula output, prevention of malnutrition and psychological support, and skin care. Keywords: Enteral nutrition, Enterocutaneous fistula, Parenteral nutrition, Sepsis, Surgical stomas. INTRODUCTION Abnormal connections between gastrointestinal tract (GIT) and skin are called enterocutaneous fistulas (ECFs). Presence of ECF is associated with significant morbidity and mortality. Seventy-five to eighty-five percent of ECFs are surgical in origin due to gut injury or significant manipulation and adhenolysis during surgery, or anastomotic leakage later, which is common in susceptible patients with cancer or inflammatory bowel disease.1,2 Mortality ranges from 6 to 33%, with sepsis, high initial output fistula, high acute physiology and chronic health evaluation II (APACHE II) scores, low albumin, and complications being the leading contributors and predictors of mortality.3,4 The classification is based on anatomy (site—gastrocutaneous, enterocutaneous, colocutaneous, simple vs complex, end or lateral, and distal obstruction) or based on output-high (>500 mL/day) vs low (%3C;200 mL/day). Another classification is based on the etiology—iatrogenic-percutaneous drainage, operative procedure; trauma, foreign body, Crohn’s disease, malignancy, infectious like tuberculosis (TB), actinomycosis.5 Stoma is a surgically created small opening on the surface of the abdomen to divert the flow of feces or urine from the bowel or bladder, which is collected in a waterproof pouch called stoma bag. Stomas are created to provide temporary or permanent bowel opening in the abdomen (interruption in gut continuity) for diverse disease conditions. Stomas can be classified on the basis of site, style, and duration.6 A colostomy is created when the colon is unable to function normally due to a disease or trauma, or needs functional rest, by making an opening in the colon, and exteriorizing it by bringing out on the left side of abdominal surface to provide an exit for feces and flatus, which are collected in a colostomy bag. An ileostomy involves exteriorizing the ileum onto the surface on the right side of the abdomen, to provide an exit for feces and flatus, which are collected in an ileostomy bag. The clinical indications for colostomy and ileostomy are listed in Table 1. Proper stoma creation and management improves the quality of life, but poor stoma care can pose significant challenges requiring multiple hospitalizations with significant financial burden and complications.7 PROBLEMS WITH STOMAS (ENTEROSTOMY) Ileostomy is associated with an increased fluid requirement and need for wearing of an appliance and skin barrier, due to liquid to semiliquid consistency of stool that contains proteolytic enzymes. Similarly, the more proximal colonic intestinal stomas are associated with semiliquid stool, whereas distal stoma like sigmoid colostomy is associated with formed stool consistency, not requiring any change in fluid requirements. Table 1: Indications for colostomy and ileostomy\| Colostomy \| Ileostomy \| --- \| \| Carcinoma of the colon, rectum, or anus \| Ulcerative colitis \| \| Diverticular disease \| Crohn’s disease \| \| Obstruction \| Colon cancer \| \| Crohn’s disease \| Rectal cancer \| \| Radiation enteritis \| Radiation enteritis \| \| Ischemic bowel \| Bowel ischemia \| \| Fecal incontinence \| Trauma \| \| \| Congenital abnormalities \| \| \| Familial adenomatous polyposis \| High-output Enterostomy Ileostomy generally starts functioning within 24 hours after creation, and the initial output may vary from 0.5 to 2 L/day. The output depends on the intake and the size of bowel left. The normal daily output varies depending on the amount of enteral intake and the remaining length of small bowel. Ileostomy adaptation occurs gradually and is considered complete when the output is controlled between 300 and 700 mL/day. There are three phases—hypersecretory, adaptive, and stabilization.7,8 After a stoma is created, there can be high amount of biliary fluid losses, lasting 1–3 days representing the hypersecretory phase, but can last up to 2 months. It is followed by adaptation and stabilization phases, which may take up to 2 years with stoma output finally declining to 300–700 mL/day. High-output fistulas (HOF) with output %3E;1.5 to 2 L/day result in severe malnutrition, dehydration, renal dysfunction, and low magnesium levels, resulting in frequent hospital admissions and longer hospital stays. The incidence of HOF is about 16% in jejunosotmies/ileostomies. Jejunostomy and intra-abdominal sepsis are the most common causes of HOF. Major causes of high-output enterostomies are jejunostomy/ileostomy, intra-abdominal sepsis, short bowel syndrome, Clostridium difficile enteritis, and obstructed bowel.8 Management of High-output Enterostomy Infection and development of dehydration and a catabolic state should be avoided in these patients. Decreasing ostomy output and fluid and electrolytes, sepsis control, and nutritional management are the cornerstone of therapy. Necessary investigation includes complete blood count (CBC), liver and kidney function tests (LKFT), serum electrolytes, albumin C-reactive protein, procalcitonin (PCT), and appropriate body fluid cultures.1 Stool may be analyzed for C. difficile infection and fecal calprotectin for ongoing colonic inflammation. X-ray, CT scan/MRI, and MR enterography may be helpful in identify any obstruction, intra-abdominal abscess, or any other pathology. Fluid and Electrolytes Initial management should focus on fluid and electrolyte balance and preventing further loss of fluid. Daily meals should also include about a liter each of hypo- as well as hypertonic fluids. If the output still remains high, free water and hypoosmolar solutions may be avoided to reduce further intestinal losses. Electrolyte replacements should take care of Na+, K+, Mg+, and Zn+ by oral or intravenous (IV) replacement depending on the extent of loss.8 If the output is still high, medications like antidiarrheal, antimotility, and antisecretory agents can be used before meals to reduce bowel motility and intestinal secretions. Nutrition Enteral nutrition (EN) is preferred over parenteral nutrition (PN) due to its beneficial effects and the problems of line-related sepsis with PN. Enteral nutrition should include high-calorie, low fiber diet for better absorption of nutrients. In situations where gut cannot be used, e.g., short or obstructed bowel, partial PN can be used along with EN.8 Details of nutrition are discussed in the section on management of ECF. Controlling the Cause Endoscopy and imaging play a vital role to find out the reason for a high-output enterostomy and help in distinguishing between infective and inflammatory causes. Antibiotics can be used to treat an intra-abdominal infection, and in addition, fecal transplant can also be tried for resistant C. difficile enteritis. The problem of steatorrhea can be managed with drugs like cholestyramine.8 PROBLEMS WITH ENTEROCUTANEOUS FISTULA Complications of ECF include electrolyte and fluid imbalance, malnutrition, and sepsis.9 The management requires a multidisciplinary approach. As an intensivist we are more concerned with the early and medical complications like sepsis, fluid and electrolyte disturbances, and nutrition, and this review we will focus mainly on these aspects. Acute intestinal failure (AIF) refers to a functional decline in absorptive capacity of the gut, necessitating PN supplementation. Intestinal insufficiency refers to a reduction in gut absorptive capacity that does not require PN supplementation to sustain growth and good health.3 Favorable outcome is associated with surgical etiologies like appendicitis, diverticulitis, absence of obstruction, infection or inflammation, bowel in continuity, transferrin >200 mg/dL, length >2 cm, end fistula, fistula output %3C;200 mL/day, electrolyte homeostasis, and absence of sepsis and good fistula care.10 Sepsis Control It may require immediate source control in abdominal cavity, and/or adequate drainage of an intra-abdominal abscess. Endogenous bacterial translocation can often give rise to sepsis as in subacute bowel ischemia, colitis, or distended bowel, necessitating targeted antimicrobial therapy.3 The choice may include a combination of beta-lactam and beta-lactamase inhibitor (BL/BLI) antibiotic or carbapenem with anaerobic ± antifungal cover, which can be modified later as per the culture results. In absence of any overt intra-abdominal cause, a secondary non-abdominal septic focus like a pneumonia or central venous catheter-related infection should be suspected. Image-guided drainage is helpful in avoiding a second hit of sepsis associated with undertaking major surgery in the sick, septic patient.11 Optimizing Fluid and Electrolytes and Prevention of Dehydration Although significant amount of secretions are produced by the gut (6–8 L), but most of them are reabsorbed and a small proportion reaches the colon, where it again is reabsorbed maintaining fluid homeostasis. Intestinal loss immediately after surgery is quite high due to decrease in gut motility and inflammation. Electrolyte (Mg+, K+, Na+) homeostasis should be achieved in all patients.9 The output may increase significantly if there is a preexisting abdominal disease or sepsis. Ileal resections are associated with more fluid losses due to higher malabsorption and diarrhea as the jejunum poorly adapts to the loss of ileum. Duodenal and pancreatic fistulas may require replacement with bicarbonates. A urine output of 1 mL/kg/hour should be the goal of fluid resuscitation. Intake–output fluid monitoring is essential in these patients to monitor fluid losses due to diarrheal, nasogastric and stomal losses, and their replacement. Urinary sodium below 20 mmol/L generally suggests fluid and sodium loss that needs to be replaced. Early monitoring of urinary sodium helps in maintaining hydration and prevents development of acute kidney injury (AKI).9 After the early resuscitation phase and control of sepsis, deresuscitation should be performed to achieve a temporary negative balance to prevent bowel and abdominal wall edema and subsequent intra-abdominal hypertension. Table 2: Calorie and nutrition requirement of patients with enterocutaneous fistulas\| \| Calorie (kcal/kg/day) \| Protein (g/kg/day) \| Vitamin C \| Other vitamins \| Elements (zinc, copper, selenium) \| --- --- --- \| \| Low output \| 20–30 \| 1–1.5 \| 5–10 times normal \| At least normal \| At least normal \| \| High output \| 25–35 \| 1.5–2.5 \| 10 times normal \| 2 times normal \| 2 times trace elements \| Nutritional Support9,10 Indirect calorimetry is usually unavailable, and nutrition evaluation scores like Nutritional Risk Screening 2002 (NRS 2002) and Nutritional Risk In Critically Ill (NUTRIC) or Subjective Global Assessment (SGA) scores can be used for nutritional assessment. Nutritional Risk Screening 2002 ≥5 or a NUTRIC score ≥5, %3E;10% ongoing weight loss and signs of fat and/or muscle loss indicate high nutrition risk and severe malnutrition. As these patients are in hypercatabolic state, they should be given 25–35 kcal/kg/day, with 1.5 g/kg of proteins. Amino acid solutions can be added to PN (Table 2).10 Lipids should constitute 20–30% of all calories. Diet should be supplemented with trace elements, vitamins, and electrolytes. Caution should be done to avoid refeeding syndrome in patients with long-term malnutrition. Enteral nutrition remains the preferential route of nutrition delivery either through nasogastric or nasojejunal tubes or through a surgical or endoscopically created gastrostomy or jejunostomy. At times, nutritional homeostasis is difficult to maintain with EN alone, and partial PN (PPN) may be used to augment the calorie–protein deficit. Enetral feeding is beneficial and provision of 20% of calories fed enterally with supplementation of vitamin C and zinc helps in engaging the bowel and helps in maintaining functional integrity of gut (immune, hormonal, mucosal barrier, bacterial translocation, and hepatic protein synthesis). However, enteral feeds are not recommended in bowel obstruction, perforation, or ischemia due to hemodynamic instability. Enteral feeding can be given through endoscopically put nasoenteral tubes bypassing the fistula opening and delivering the feed to the distal healthy bowel, thereby maintaining the bowel continuity and avoiding the need of PN. But, this route is usually successful in patients with proximal rather than distal fistulas. Distal feeding has a positive effect of controlling biliary and pancreatic secretions though inhibitory feedback. Enteral feeding can be sometimes administered through the fistula (fistuloclysis) or the chyme may be reinfused (enteroclysis). Reinfusion of collected proximal secretions and chyme through the fistula primes and prepares the distal bowel and improves the nutritional status. Parenteral nutrition with medium chain triglycerides (MCT), polyunsaturated fatty acids (PUFA), and fish oil derived omega-3 fatty acids is recommended.9 Decreasing Gastrointestinal Losses and Increasing Gastrointestinal Absorption Hypergastrinemia and gastric acid hypersecretion are seen with major small bowel resections. Antacids like PPI/H 2 blockers significantly reduce this response, decrease the distal output, and are the mainstay of therapy. Loperamide (up to 40 mg/day), an antimotility drug in high doses, can help in reducing the output. Other drugs include longer acting codeine phosphate (up to 240 mg/day) and cholestyramine—a bile acid adsorber given before food to control diarrhea due to excessive bile salts. Somatostatin (up to 250 μg/hour) or octreotide also reduces bowel secretions and may be used effectively. Their use is associated with hyperglycemia and there is rebound after stopping.1 Protection of Surrounding Skin Leakage of enteric fluid results in skin damage and breakdown. It can be minimized by diverting the enteral effluent.12 Vacuum-assisted closure systems isolate the fistula output from the wound by application of −25 mm Hg pressure and help in wound healing. Psychological Support Psychological support is very important, as these patients developed fistula as a complication of surgery. They may require prolonged hospitalization, and an open smelly wound with fistula effluents may be quite depressing. As the treatment may take a long time and the fistula may not heal completely at times, the patients would need encouragement and support. DEFINITIVE SURGICAL REPAIR Thirty to seventy-five percent of fistula get closed by conservative management.13 Inflammatory adhesions due to peritonitis associated with ECF may persist for up to a year, if the abdomen is left open.14 In the presence of a decreasing fistula output and wound healing, it is prudent to delay reconstruction surgery. Surgical reconstruction can be planned if the patient’s nutritional status is good and there is no sepsis. A contrast-enhanced CT scan of abdomen (CECT) may help in identifying the fistula tract. CONCLUSION Enterocutaneous fistula and stoma care require sepsis and output control, skin care, avoidance of dehydration, malnutrition, and other complications. Surgical repairs should only be contemplated after control of these issues and good nutritional build up have been successfully resolved. REFERENCES 1. Evenson AR, Fischer JE. Clinical management of enterocutaneous fistula. J Gastrointest Surg 2006;10(3):455–464. DOI: 10.1016/j.gassur.2005.08.001. 2. Berry SM, Fischer JE. Classification and pathophysiology of enterocutaneous fistulas. Surg Clin North Am 1996;76(5):1009–1018. DOI: 10.1016/s0039-6109(05)70495-3. 3. Campos AC, Andrade DF, Campos GM, Matias JE, Cohello JC. A multivariate model to determine prognostic factors in gastrointestinal fistulas. J Am Coll Surg 1999;188(5):483–490. DOI: 10.1016/s1072-7515(99)00038-1. 4. Haffejee AA. Surgical management of high output enterocutaneous fistulae: a 24-year experience. Curr Opin Clin Nutr Metab Care 2004;7(3):309–316. DOI: 10.1097/00075197-200405000-00011. 5. Schecter WP. Management of enterocutaneous fistula. Surg Clin N Am 2011;91(3):481–491. DOI: 10.1016/j.suc.2011.02.004. 6. Engida A, Ayelign T, Mahteme B, Aida T, Abreham B. Types and indications of colostomy and determinants of outcomes of patients after surgery. Ethiop J Health Sci 2016;26(2):117–120. DOI: 10.4314/ejhs.v26i2.5. 7. Tsujinaka S, Tan KY, Miyakura Y, Fukano R, Oshima M, Konishi F, et al. Current management of intestinal stomas and their complications. J Anus Rectum Colon 2020;4(1):25–33. DOI: 10.23922/jarc.2019-032. 8. Adaba F, Vaizey CJ, Warusavitarne J. Management of intestinal failure: the high-output enterostomy and enterocutaneous fistula. Clin Colon Rectal Surg 2017;30(03):215–222. DOI: 10.1055/s-0037-1598163. 9. Klek S, Forbes A, Gabe S, Holst M, Wanten G, Irtun Ø, et al. Management of acute intestinal failure: a position paper from the European Society for Clinical Nutrition and Metabolism (ESPEN) special interest group. Clin Nutr 2016;35(6):1209–1218. DOI: 10.1016/j.clnu.2016.04.009. 10. Gribovskaja-Rupp I, Melton GB. Enterocutaneous fistula: proven strategies and updates. Clin Colon Rectal Surg 2016;29(02):131–137. DOI: 10.1055/s-0036-1580732. 11. Kaushal M, Carlson GL. Management of enterocutaneous fistulas. Clin Colon Rectal Surg 2004;17(2):79–88. DOI: 10.1055/s-2004-828654. 12. Ashkenazi I, Fuentes FT, Olsha O, Alfici R. Treatment options in gastrointestinal cutaneous fistulas. Surg J (NY) 2017;3(01):e25–e31. DOI: 10.1055/s-0037-1599273. 13. Pritts TA, Fischer DR, Fischer JE. Postoperative enterocutaneous fistula ed. RG, Holzheimer JA, Mannick ed. Surgical Treatment: Evidence-Based and Problem-Oriented. Munich: Zuckschwerdt; 2001. Available from: 14. Lynch AC, Delaney CP, Senagore AJ, Connor JT, Remzi FH, Fazio VW. Clinical outcome and factors predictive of recurrence after enterocutaneous fistula surgery. Ann Surg 2004;240(5):825–831. DOI: 10.1097/01.sla.0000143895.17811.e3.
13339	https://www.quora.com/Can-a-straight-line-be-tangent-to-a-parabola-at-only-one-point-Are-there-any-examples-where-this-is-not-the-case	Can a straight line be tangent to a parabola at only one point? Are there any examples where this is not the case? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Lines (general) Tangent Space Parabolas (geometry) Conic Sections Curves Geometric Mathematics Parabola Precalculus Tangent Lines 5 Can a straight line be tangent to a parabola at only one point? Are there any examples where this is not the case? All related (32) Sort Recommended Gary Ward MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views ·1y Can a straight line be tangent to a parabola at only one point? Are there any examples where this is not the case? Yes, to the first, and no, to the second. A parabola always curves in the same direction making it impossible for a straight line to be tangent at more than one point. Continue Reading Can a straight line be tangent to a parabola at only one point? Are there any examples where this is not the case? Yes, to the first, and no, to the second. A parabola always curves in the same direction making it impossible for a straight line to be tangent at more than one point. Upvote · 9 8 Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) ·Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. Continue Reading This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Upvote · 999 485 999 103 99 17 Related questions More answers below "Line intersect parabola at single point." Does that mean the line is tangent? Can you give an example of a straight line and a parabola? Can you provide an equation for the tangent line to the parabola y = 4x^2 at the point (1,4)? Are property lines straight? How many tangent and normal can be drawn to a parabola? Woojun Jung Knows Korean ·Updated 2y Related Do all straight lines intersect a parabola in exactly one point? No. There are other cases. i) No intersection In this case, the discriminant of the equation a x 2+b x+c=m x+n,a x 2+b x+c=m x+n,D<0 D<0 However, only real intersections do not exist. The intersection point is imaginary. ii) One intersection in this case, D=0 D=0 Actually, a vertical line can have one intersection. iii) Two intersections in this case, D>0 D>0 Every parabola and line has a relationship among these 3 cases. ——————————————————————————————————————— [Why do we use a discriminant?] Intersections become roots when subtracting two functions (because they have same y-value), so the discriminant of f(x)−g(x)=0 f(x)−g(x)=0 indicates the Continue Reading No. There are other cases. i) No intersection In this case, the discriminant of the equation a x 2+b x+c=m x+n,a x 2+b x+c=m x+n,D<0 D<0 However, only real intersections do not exist. The intersection point is imaginary. ii) One intersection in this case, D=0 D=0 Actually, a vertical line can have one intersection. iii) Two intersections in this case, D>0 D>0 Every parabola and line has a relationship among these 3 cases. ——————————————————————————————————————— [Why do we use a discriminant?] Intersections become roots when subtracting two functions (because they have same y-value), so the discriminant of f(x)−g(x)=0 f(x)−g(x)=0 indicates the number of intersections for f(x)=g(x)f(x)=g(x) Upvote · 9 3 9 8 Devashish Rai Studied at St. Xavier's School, Raj Niwas, Delhi ·8y Related "Line intersect parabola at single point." Does that mean the line is tangent? No . Observe, All lines that are parallel to the axis of the parabola intersect the parabola once and aren't tangent to it. Continue Reading No . Observe, All lines that are parallel to the axis of the parabola intersect the parabola once and aren't tangent to it. Upvote · 9 3 Tony Fleet 5y Related Can a tangent line touch more than one point? The lines y= 1 and y = -1 are both tangents to y= sin(x), and touch at an infinite number of points. Continue Reading The lines y= 1 and y = -1 are both tangents to y= sin(x), and touch at an infinite number of points. Upvote · 9 5 Related questions More answers below Can we consider the axis of the parabola as a normal or tangent line of the Parabola or not? Can we define the tangent of a straight line? Can you give some examples of lines, circles, and parabolas? How do I prove that a straight line passing through a parabola is tangent to it? Can you identify the circle that is tangent to both a parabola and a line? Gerald Grenier Uses both American and Queen's English · Author has 1.4K answers and 358.4K answer views ·1y No, that is basically the definition of a tangent. A straight line the only touches a curve at a single point Upvote · Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Learn More 999 116 Daniel Sheedy Teacher Aide at Whitsunday Christian College (2023–present) · Author has 91 answers and 347K answer views ·8y Related Is there a name given for a line that touches a parabola (in geometry) only once but isn't a tangent? The answer to this question depends on how you want this line to ‘touch’ the parabola. Usually the world ‘touch’ is interpreted to mean “not surpass a given border”, or in this scenario; “not to intersect the parabola”. I will however briefly answer the latter ultimatum (encase you were not strictly suggesting that there could be no intersection). A: Vertical Intercept. For a line to intersect a parabola only once, and not be tangent to any point on the parabola, it must be a strait vertical line. e.g. x=-1. As you can see there are infinate solutions to this. I could have a line intersect at any Continue Reading The answer to this question depends on how you want this line to ‘touch’ the parabola. Usually the world ‘touch’ is interpreted to mean “not surpass a given border”, or in this scenario; “not to intersect the parabola”. I will however briefly answer the latter ultimatum (encase you were not strictly suggesting that there could be no intersection). A: Vertical Intercept. For a line to intersect a parabola only once, and not be tangent to any point on the parabola, it must be a strait vertical line. e.g. x=-1. As you can see there are infinate solutions to this. I could have a line intersect at any location on the x-axis, and it would satisfy your condition - given that it is completely vertical. Now, let’s say you did indeed intend for no intersection. Then the answer would be a range of different equations. For example: This image shows how y=-x^2 ‘touches’ the parabola at x=0. Similar results can be produced with all parabolas. Say you had a parabola; f(x). Then, I can tell you that the parabola that ‘touches’ the parabola at it’s turning point is given by g(x) such that: g(x)=−f(x)+2 c g(x)=−f(x)+2 c Where c is the distance the parabola is from the x-axis. See if you can work out how & why this works. Here is another parabola which only ‘touches’ y=x^2 once: It’s equation: y=-(x-1)(x-3)+1 We are not just limited to parabolas either: It’s equation: y=cos(x-pi)+1 There are infinitely many solutions to this problem. I also just realized that we can have cubic polynomials (and certainly other functions) intersect the parabola only once - which may be considered a solution, depending on your question. Regards, Daniel Upvote · 9 4 9 1 Sharanya Mathur 8y Related "Line intersect parabola at single point." Does that mean the line is tangent? No In fact, a tangent to a curve (parabola in this case) is defined as the line that intersects the curve at 2 infinitesimally close points. If I'm not wrong the only line that intersects the parabola at only 1 point is the axis of the parabola. Also if you define a tangent in the usual way (i.e. A line that touches a curve at a point) then it just touches the curve and not intersects it. So yeah, the axis of the parabola is the only line that intersects it at a single point EDIT : I just saw one of the collapsed answers and realised that any diameter to the parabola intersects it at only 1 point. Upvote · 9 1 9 2 Sponsored by Amazon Business Solutions and supplies to support learning. Save on essentials and reinvest in students and staff. Sign Up 99 85 Vipul Pal B.Tech Electrical Engineer @ VJTI ·5y Related What is the equation of tangent to a parabola at a given point? The above derivation is for a standard parabola but it can be done for any parabola using the same method. Happy Learning! Continue Reading The above derivation is for a standard parabola but it can be done for any parabola using the same method. Happy Learning! Upvote · 99 50 9 3 9 1 Makarand Apte Works on Computational Geometry · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 638 answers and 3M answer views ·10y Related If from a given point, two tangents can be drawn to a parabola, then how can three normals be drawn to that parabola from that given point? Let's take an example. Consider the parabola y=x 2 y=x 2 The slope of the line joining a point (a,b)(a,b) to a point (x,x 2)(x,x 2) on this parabola will be b−x 2 a−x b−x 2 a−x If this line is tangent to the parabola, then it's slope has to be equal to the derivative of parabola at that point. So, b−x 2 a−x=d(x 2)d x=2 x b−x 2 a−x=d(x 2)d x=2 x This gives us a quadratic equation, which tells us that you can draw up to 2 tangents to the parabola from a given point. If the line is normal to the parabola, then it's slope has to be equal to the negative reciprocal of the derivative of Continue Reading Let's take an example. Consider the parabola y=x 2 y=x 2 The slope of the line joining a point (a,b)(a,b) to a point (x,x 2)(x,x 2) on this parabola will be b−x 2 a−x b−x 2 a−x If this line is tangent to the parabola, then it's slope has to be equal to the derivative of parabola at that point. So, b−x 2 a−x=d(x 2)d x=2 x b−x 2 a−x=d(x 2)d x=2 x This gives us a quadratic equation, which tells us that you can draw up to 2 tangents to the parabola from a given point. If the line is normal to the parabola, then it's slope has to be equal to the negative reciprocal of the derivative of parabola at that point. So, b−x 2 a−x=−1 2 x b−x 2 a−x=−1 2 x This gives us a cubic equation, which tells us that you can draw up to 3 normals to the parabola from a given point. Also note that you can draw tangents from a point to a parabola only if your point is "outside" that parabola. Upvote · 99 14 9 1 Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Learn More 999 138 James Buddenhagen Lives in Xico,Veracruz.Mexico (2006–present) · Author has 2.6K answers and 4.2M answer views ·5y Related Is it possible for a quadratic and a line to only have one point of intersection? Is it possible for a quadratic and a line to only have one point of intersection? One point of contact, yes. But it is not a point of intersection because the line does not cross the curve, just touches and then stays on the same side of the curve. Reason: when you solve the system of equations, for the quadratic and the line, the result is a quadratic (with real coefficients), and every quadratic has two roots, if not both imaginary (no point of intersection) then both real so two points of intersection, unless a double root, which is a point of contact as described above. Example: the quadratic Continue Reading Is it possible for a quadratic and a line to only have one point of intersection? One point of contact, yes. But it is not a point of intersection because the line does not cross the curve, just touches and then stays on the same side of the curve. Reason: when you solve the system of equations, for the quadratic and the line, the result is a quadratic (with real coefficients), and every quadratic has two roots, if not both imaginary (no point of intersection) then both real so two points of intersection, unless a double root, which is a point of contact as described above. Example: the quadratic y=x^2 and the line y=0 (i.e. the x-axis) have a point of contact at the origin. Edit: Aaron Kelly points out that a vertical line through the origin will intersect the parabola y=x^2 in exactly one point. So, I stand corrected (see comments). Possibly one might argue that “the point at infinity” is the second intersection point in that case. Upvote · 9 7 Aaron Kelly B.S. in Mathematics, Dickinson State University (Graduated 2012) ·5y Related Is it possible for a quadratic and a line to only have one point of intersection? I would posit that any quadratic has infinitely many lines that only intersect it at one point. To prove, consider all the vertical lines where this is the case. Is it not true for all vertical lines? Quadratic functions have an infinite X range, so I believe it is. In addition, I also believe that this is ONLY true for vertical lines, assuming you do not count tangent lines as “intersecting”. If you consider a line with any non-vertical slope, because the quadratic is constantly getting “steeper”, it should eventually cross the line (even for incredibly “steep” lines). Upvote · 9 1 Max Orchard BAdvSc.(Hons) in Mathematics&Computer Science, University of Queensland (Graduated 2022) · Author has 83 answers and 449.5K answer views ·7y Related How do I prove that a straight line passing through a parabola is tangent to it? The equation of a line is of the form y=m x+d y=m x+d where m m is the gradient (slope) of the line and d d is the y-intercept (I’m using different letters so the c c in the parabola is not confused with the d d here). The equation of a parabola is of the form y=a x 2+b x+c y=a x 2+b x+c where a,b,c a,b,c are real numbers. A line is a tangent of another function if it intersects the function only once. This means that you should calculate the intersection point(s) of the two equations, and if only one intersection point exists, the line is a tangent. Now I should say that this doesn’t work for every function. It works for a parabola Continue Reading The equation of a line is of the form y=m x+d y=m x+d where m m is the gradient (slope) of the line and d d is the y-intercept (I’m using different letters so the c c in the parabola is not confused with the d d here). The equation of a parabola is of the form y=a x 2+b x+c y=a x 2+b x+c where a,b,c a,b,c are real numbers. A line is a tangent of another function if it intersects the function only once. This means that you should calculate the intersection point(s) of the two equations, and if only one intersection point exists, the line is a tangent. Now I should say that this doesn’t work for every function. It works for a parabola because there is only one local minima/maxima so a line with the same slope as the tangent would not have any other intersections anywhere else along the function. A more rigorous method would be to find the derivative of the parabola and compare to the line, but that is not necessary here. An intersection point occurs when both the x x and y y values are equal. This means we have a system of equations. Setting the equations as equal to each other, we get: m x+d=a x 2+b x+c m x+d=a x 2+b x+c 0=a x 2+(b−m)x+(c−d)0=a x 2+(b−m)x+(c−d) This is just another parabola, so we can use the quadratic formula x=−b±√b 2−4 a c 2 a x=−b±b 2−4 a c 2 a x=−(b−m)±√(b−m)2−4 a(c−d)2 a x=−(b−m)±(b−m)2−4 a(c−d)2 a This equation will produce two values (due to the ±±). It will produce one value, however, if the value of √(b−m)2−4 a(c−d)(b−m)2−4 a(c−d) is the same as −√(b−m)2−4 a(c−d)−(b−m)2−4 a(c−d). √(b−m)2−4 a(c−d)=−√(b−m)2−4 a(c−d)(b−m)2−4 a(c−d)=−(b−m)2−4 a(c−d) 2√(b−m)2−4 a(c−d)=0 2(b−m)2−4 a(c−d)=0 0=√(b−m)2−4 a(c−d)0=(b−m)2−4 a(c−d) 0=(b−m)2−4 a(c−d)0=(b−m)2−4 a(c−d) If you plug the respective variables into this equation and find it to be true, the line is a tangent. Let’s try an example: Here we have the parabola given by y=2 x 2+4 x−6 y=2 x 2+4 x−6 and the line given by y=−4 x−14 y=−4 x−14. Seeing this picture, it is obvious that the line is a tangent. Plugging numbers into our equation (using a=2,b=4,c=−6,m=−4,d=−14 a=2,b=4,c=−6,m=−4,d=−14), (4−−4)2−4(2)(−6−−14)(4−−4)2−4(2)(−6−−14) =8 2−8(8)=8 2−8(8) =0=0 This means that it is a tangent, as it equals zero. Upvote · 9 3 Deleted Account Lives in New York, NY · Author has 2K answers and 9.1M answer views ·5y Related Is it possible for a quadratic and a line to only have one point of intersection? Sure. A quadratic curve and any vertical line would intersect in only one point. Or a horizontal line for a horizontal quadratic in the form of x = ay²+ by + c Continue Reading Sure. A quadratic curve and any vertical line would intersect in only one point. Or a horizontal line for a horizontal quadratic in the form of x = ay²+ by + c Upvote · Related questions "Line intersect parabola at single point." Does that mean the line is tangent? Can you give an example of a straight line and a parabola? Can you provide an equation for the tangent line to the parabola y = 4x^2 at the point (1,4)? Are property lines straight? How many tangent and normal can be drawn to a parabola? Can we consider the axis of the parabola as a normal or tangent line of the Parabola or not? Can we define the tangent of a straight line? Can you give some examples of lines, circles, and parabolas? How do I prove that a straight line passing through a parabola is tangent to it? Can you identify the circle that is tangent to both a parabola and a line? Can tangents of circles and spheres ever be straight lines? Can a parabola have 2 or more parallel tangents? Is there a name given for a line that touches a parabola (in geometry) only once but isn't a tangent? What are some examples of where tangent circles and straight lines meet each other in geometry? What is the relationship between a straight line and a circle if the straight line is tangent to the circle? Related questions "Line intersect parabola at single point." Does that mean the line is tangent? Can you give an example of a straight line and a parabola? Can you provide an equation for the tangent line to the parabola y = 4x^2 at the point (1,4)? Are property lines straight? How many tangent and normal can be drawn to a parabola? Can we consider the axis of the parabola as a normal or tangent line of the Parabola or not? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
13340	https://www.kuh.ku.edu.tr/mayo-clinic-care-network/mayo-clinic-health-information-library/first-aid/how-to-get-pregnant	Mayo Clinic \| Koc University Hospital OUR SERVICES OUR PHYSICIANS CONTACT US CORPORATE MAYO CLINIC HEALTH LIBRARY EN Koc University Hospital Member Of Mayo Clinic Care Network Mayo Clinic Health Information Library First Aid How To Get Pregnant OUR CENTERS OUR SERVICES OUR PHYSICIANS NURSING DIRECTORATE KOÇ HEALTHCARE ONLINE SERVICES PATIENT EXPERIENCE OFFICE INTERNATIONAL PATIENTS GUEST GUIDE CORPORATE AGREEMENTS HUMAN RESOURCES CORPORATE OUR VALUES MAYO CLINIC HEALTH INFORMATION LIBRARY CONTACT US HOMEPAGE MEMBER OF MAYO CLINIC CARE NETWORK FIRST AID HOW TO GET PREGNANT How to get pregnant Last Updated on July 20, 2024 Some couples seem to get pregnant simply by talking about it. For others, it takes time. If you're looking for tips on how to get pregnant, here's what you need to know. How to predict ovulation Understanding when you're ovulating — and having sex regularly five days before and on the day of ovulation — can improve the odds of conceiving. Ovulation is the process in which a mature egg is released from the ovary. After it's released, the egg moves down the fallopian tube and stays there for 12 to 24 hours, where it can be fertilized. Sperm can live inside the female reproductive tract as long as five days after sexual intercourse under the right conditions. Your chance of getting pregnant is highest when live sperm are present in the fallopian tubes during ovulation. In an average 28-day menstrual cycle, ovulation typically occurs about 14 days before the start of the next menstrual period. However, each person's cycle length may be different, and the time between ovulation and the start of the next menstrual period may vary. If, like many people, you don't have a perfect 28-day menstrual cycle, you can determine the length and midpoint of your cycle by keeping a menstrual calendar. Beyond the calendar, you can also look for ovulation signs and symptoms, including: Change in vaginal secretions (cervical mucus). Just before ovulation, you might notice an increase in clear, wet and stretchy vaginal secretions. Just after ovulation, cervical mucus decreases and becomes thicker, cloudy and less noticeable. Change in basal body temperature. Your body's temperature at rest (basal body temperature) increases slightly during ovulation. Using a thermometer specifically designed to measure basal body temperature, take your temperature every morning before you get out of bed. Record the results and look for a pattern to emerge. You'll be most fertile during the 2 to 3 days before your temperature rises. You also might want to try an over-the-counter ovulation kit, which can help you identify when you're most likely to ovulate. These kits test your urine for the surge in hormones that takes place before ovulation. Ovulation occurs about 36 hours after a positive result. Maximizing fertility: What to do Follow these simple tips for how to get pregnant: Have sex regularly. The highest pregnancy rates occur in couples who have sex every day or every other day. Have sex near the time of ovulation. If having sex every day isn't possible — or enjoyable — have sex every 2 to 3 days a week starting soon after the end of your period. This can help ensure that you have sex when you are most fertile. Maintain a normal weight. Overweight and underweight women are at increased risk of ovulation disorders. Also, consider talking to your health care provider about preconception planning. He or she can assess your overall health and help you identify changes that might improve your chances of a healthy pregnancy. Your health care provider will recommend taking folic acid a few months before conception to reduce the risk of spina bifida and other neural tube defects. Maximizing fertility: What to avoid To improve your odds of conceiving: Don't smoke. Tobacco has multiple negative effects on fertility, not to mention your general health and the health of a fetus. If you smoke, ask your health care provider to help you quit before you start trying to conceive. Don't drink alcohol. Heavy alcohol use might lead to decreased fertility. Generally, it's best to avoid alcohol if you're hoping to conceive. Curb caffeine. Research suggests that fertility isn't affected by caffeine consumption of less than 200 milligrams a day. That's about 1 to 2 cups of 6 to 8 ounces of coffee a day. Don't overdo strenuous exercise. Strenuous, intense exercise of more than five hours a week has been associated with decreased ovulation. Also, talk to your health care provider about any medications you are taking. Certain medications — even those available without a prescription — can make it difficult to conceive. When to talk to a doctor With frequent unprotected sex, most healthy couples conceive within one year. If you're age 35 or older and you have been trying to conceive for six months or more, or if you or your partner has known or suspected fertility issues, consider consulting with a health care provider. Infertility affects both men and women — and treatment is available. Depending on the source of the problem, your gynecologist, your partner's urologist or your family doctor might be able to help. In some cases, a fertility specialist offers the best hope. Test diagnosis tab content Test doctor tab content ©1998-2024 Mayo Foundation for Medical Education and Research.All rights reserved Terms of Use OUR CENTERS OUR SERVICES OUR PHYSICIANS NEWS FINESSE AND PERFECTION CENTER CORPORATE The Vehbi Koç Foundation (VKV) and Board of Directors Our Values History Message Of CEO Organizatıon Chart and Management HUMAN RESOURCES GUEST GUIDE Corporate Agreements Patient Admission Guidelines Accompanying Policy Plan Your Visit Frequently Asked Questions ONLINE SERVICES Koç e-Health Koç Healthcare Web Baby Lab Results LINKS Koç University School of Medicine School of Nursing CONTACT FORM Our team of experts will answer your questions as soon as possible. [x] Site kullanım koşullarını ve gizlilik politikasını kabul ediyorum. I accept the privacy and useragreement SEND 2025, Copyright, Koç University Hospital. Contact : +90 (850) 250 8 250 Protection of Personal Data Information Society Services Manage Cookie Preferences Search Choose a Language ###### English
13341	https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch16/lechat.php	Le Chatelier's Principle Le Chatelier's Principle Nahkleh Group Robinson Group Weaver Group Bodner Group Gas-Phase Reactions Gas-Phase Reactions Equilibrium Expressions Le Chateliers Principle Le Chatelier's Principle Le Chatelier's PrincipleChanges in Concentration Changes in PressureChanges in Temperature Le Chatelier's Principle In 1884 the French chemist and engineer Henry-Louis Le Chatelier proposed one of the central concepts of chemical equilibria. Le Chatelier's principle can be stated as follows: A change in one of the variables that describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the effect of this change. Le Chatelier's principle describes what happens to a system when something momentarily takes it away from equilibrium. This section focuses on three ways in which we can change the conditions of a chemical reaction at equilibrium: (1) changing the concentration of one of the components of the reaction (2) changing the pressure on the system (3) changing the temperature at which the reaction is run. Changes in Concentration To illustrate what happens when we change the concentration of one of the reactants or products of a reaction at equilibrium, let's consider the following system at 500 o C. N 2(g)+3 H 2(g)2 NH 3(g)K c = 0.040 Initial:0.100 M 0.100 M 0 Equilibrium:0.100 -C 0.100 - 3 C 2 C We obtain the following results when we solve this problem. [NH 3] = 2 C 0.0020 M [N 2] = 0.100 - C 0.099 M [H 2] = 0.100 - 3 C 0.097 M The fact that C is small compared with the initial concentrations of N 2 and H 2 makes this calculation relatively easy to do. But it implies that very little ammonia is actually produced in the reaction. According to this calculation, only 1% of the nitrogen present initially is converted into ammonia. What would happen if we add enough N 2 to increase the initial concentration by a factor of 10? The reaction can't be at equilibrium any more because there is far too much N 2 in the system. Adding an excess of one of the reactants therefore places a stress on the system. The system responds by minimizing the effect of this stress by shifting the equilibrium toward the products. The reaction comes back to equilibrium when the concentrations of the three components reach the following values. [NH 3] = 2 C 0.0055 M [N 2] = 1.00 - C 1.00 M [H 2] = 0.10 - 3 C 0.092 M By comparing the new equilibrium concentrations with those obtained before excess N 2 was added to the system, we can see the magnitude of the effect of adding the excess N 2. Before After [NH 3] 0.0094 M[NH 3] 0.026 M [N 2] 0.095 M[N 2] 0.99 M [H 2] 0.29 M[H 2] 0.26 M Increasing the amount of N 2 in the system by a factor of 10 leads to an increase in the amount of NH 3 at equilibrium by a factor of about 3. Adding an excess of one of the products would have the opposite effect; it would shift the equilibrium toward the reactants. Changes in Pressure The effect of changing the pressure on a gas-phase reaction depends on the stoichiometry of the reaction. We can demonstrate this by looking at the result of compressing the following reaction at equilibrium. N 2(g)+3 H 2(g)2 NH 3(g) Let's start with a system that initially contains 2.5 atm of N 2 and 7.5 atm of H 2 at 500 o C, where K p is 1.4 x 10-5, allow the reaction to come to equilibrium, and then compress the system by a factor of 10. When this is done, we get the following results. Before Compression After Compression P NH3 = 0.12 atm P NH3 = 8.4 atm P N2 = 2.4 atm P N2 = 21 atm P H2 = 7.3 atm P H2 = 62 atm Before the system was compressed, the partial pressure of NH 3 was only about 1% of the total pressure. After the system is compressed, the partial pressure of NH 3 is almost 10% of the total. These data provide another example of Le Chatelier's principle. A reaction at equilibrium was subjected to a stress an increase in the total pressure on the system. The reaction then shifted in the direction that minimized the effect of this stress. The reaction shifted toward the products because this reduces the number of particles in the gas, thereby decreasing the total pressure on the system, as shown in the figure below. N 2(g)+3 H 2(g)2 NH 3(g) Changes in Temperature Changes in the concentrations of the reactants or products of a reaction shift the position of the equilibrium, but do not change the equilibrium constant for the reaction. Similarly, a change in the pressure on a gas-phase reaction shifts the position of the equilibrium without changing the magnitude of the equilibrium constant. Changes in the temperature of the system, however, affect the position of the equilibrium by changing the magnitude of the equilibrium constant for the reaction. Chemical reactions either give off heat to their surroundings or absorb heat from their surroundings. If we consider heat to be one of the reactants or products of a reaction, we can understand the effect of changes in temperature on the equilibrium. Increasing the temperature of a reaction that gives off heat is the same as adding more of one of the products of the reaction. It places a stress on the reaction, which must be alleviated by converting some of the products back to reactants. The reaction in which NO 2 dimerizes to form N 2 O 4 provides an example of the effect of changes in temperature on the equilibrium constant for a reaction. This reaction is exothermic. 2 NO 2(g)N 2 O 4(g)H o = -57.20 kJ Thus, raising the temperature of this system is equivalent to adding excess product to the system. The equilibrium constant therefore decreases with increasing temperature. Practice Problem 7: Predict the effect of the following changes on the reaction in which SO 3 decomposes to form SO 2 and O 2. 2 SO 3(g) 2 SO 2 (g) + O 2 (g) H o = 197.78 kJ (a) Increasing the temperature of the reaction. (b) Increasing the pressure on the reaction. (c) Adding more O 2 when the reaction is at equilibrium. (d) Removing O 2 from the system when the reaction is at equilibrium. Click here to check your answer to Practice Problem 7
13342	https://www.etymonline.com/word/scrumptious	Scrumptious - Etymology, Origin & Meaning Search Log in Columns Forum Apps Premium Log in Origin and history of scrumptious scrumptious(adj.) 1833, American English, in countrified humor writing of "Major Jack Downing" of Maine (Seba Smith), "stylish, splendid, fine;" probably a colloquial alteration (intensification) of sumptuous. By late 19c. especially of food, "delicious, delightful," and it was noted 1890s and early 20c. as a vogue word among college girls (also as scrum, scrummy). Related: Scrumptiously; scrumptiousness. OED (2nd edition, print) has scrumptious as probably identical with dialectal scrumptious "mean, stingy, close-fisted," and ultimately related to shrimp. The editors insist the sense transition "is not impossible," and they compare nice. also from 1833 Entries linking to scrumptious sumptuous(adj.) late 15c., "costly, expensive; luxurious, magnificent," from Old French sumptueux or directly from Latin sumptuosus "costly, very expensive; lavish, wasteful," from sumptus, past participle of sumere "to borrow, buy, spend, eat, drink, consume, employ, take, take up," contraction of sub-emere, from sub "under" (see sub-) + emere "to take, buy" (from PIE root em- "to take, distribute"). Related: Sumptuously; sumptuousness. Also as a noun, sumptuosity "lavishness in expenditure, costliness or magnificence in living" (1550s, from Late Latin sumptuositas). Chapman (1616) used sumpture. Advertisement Close Trends of scrumptious adapted from books.google.com/ngrams/ with a 7-year moving average; ngrams are probably unreliable. More to explore delectable c. 1400, "delightful to one of the senses, highly pleasing," from Old French delectable delitable and directly from Latin delectabilis "delightful," from delectare "to allure, delight, charm, please," frequentative of delicere "entice" (see delicious). The earlier form in English delicious c. 1300, "delightful to the senses, pleasing in the highest degree" (implied in deliciously), from Old French delicios (Modern French délicieux), from Late Latin deliciosus "delicious, delicate," from Latin delicia (plural deliciae) "a delight, allurement, charm," from delicere " luscious late 15c., according to The Middle English Compendium a variant of Middle English licius "delicious" (c. 1400), which is a shortening of delicious, with the variant form perhaps influenced by Old French luxure, lusure. But OED 2nd ed. and Century Dictionary are against all this a shrimp early 14c., "slender, long-tailed, ten-footed, edible marine crustacean," Middle English shrimpe, probably from or related to Old Norse skreppa "thin person," from Proto-Germanic skrimp- (see scrimp). This is related to Old English scrimman "to shrink." The connecting notion bet nice late 13c., "foolish, ignorant, frivolous, senseless," from Old French nice (12c.) "careless, clumsy; weak; poor, needy; simple, stupid, silly, foolish," from Latin nescius "ignorant, unaware," literally "not-knowing," from ne- "not" (from PIE root ne- "not") + stem of scire "to + stem of scire \"to ") cohort early 15c., "company of soldiers, band of warriors," from French cohorte (14c.) and directly from Latin cohortem (nominative cohors) "enclosure," with meaning extended to "infantry company" in the Roman army through the notion of "enclosed group, retinue;" from assimilated form o toast "to brown with heat," late 14c., from Old French toster "to toast, to grill, roast, burn" (12c.), from Vulgar Latin tostare (source of Italian tostare, Spanish tostar), frequentative of Latin torrere (past participle tostus) "to parch," from PIE root ters- "to dry." Related: To election c. 1300, eleccioun, "act of choosing" someone to occupy a position, elevation to office" (whether by one person or a body of electors); also "the holding of a vote by a body of electors by established procedure; the time and place of such a vote," from Anglo-French eleccioun, Old forsake Old English forsacan "object to, oppose, refuse, deny; give up, renounce" (past tense forsoc, past participle forsacen), from for- "completely" + sacan "to struggle, dispute, wrangle; accuse, blame" (see sake (n.1)). Related: Forsaking. Similar formation in Old Saxon farsakan, Du Eucharist "sacrament of the Lord's Supper, the Communion," mid-14c., from Old French eucariste, from Late Latin eucharistia, from Greek eukharistia "thanksgiving, gratitude," later "the Lord's Supper," from eukharistos "grateful," from eu "well" (see eu-) + stem of kharizesthai "show favor Share scrumptious ‘cite’ Page URL: Copy HTML Link: Etymology of scrumptious by etymonlineCopy APA Style: Harper, D. (n.d.). Etymology of scrumptious. Online Etymology Dictionary. Retrieved September 29, 2025, from Copy Chicago Style: Harper Douglas, "Etymology of scrumptious," Online Etymology Dictionary, accessed September 29, 2025, MLA Style: Harper, Douglas. "Etymology of scrumptious." Online Etymology Dictionary, Accessed 29 September, 2025.Copy IEEE Style: D. Harper. "Etymology of scrumptious." Online Etymology Dictionary. (accessed September 29, 2025).Copy Advertisement Remove Ads Trending 1.gooey 2.money 3.Christian 4.coach 5.love 6.science 7.rapture 8.weird 9.Dives 10.god Dictionary entries near scrumptious scrubbing scrubby scruff scruffy scrum scrumptious scrunch scruple scrupulous scrutable scrutinise Advertisement Close Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. ABCDEFGHIJKLMNOPQRSTUVWXYZ Quick and reliable accounts of the origin and history of English words. Scholarly, yet simple. About Who Did This Sources Introduction Links Support Premium Patreon Merch Apps Get Chrome Extension Get iOS App Get Android App Dark Auto Light Terms of ServicesPrivacy Policy English (English) © 2001 - 2025 Douglas Harper
13343	https://stackoverflow.com/questions/59347528/how-to-compare-two-ratios-of-rectangles-sizes-to-determine-their-percentages-o	math - How to compare two Ratios of Rectangles / Sizes to determine their percentages of differences? - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the companyVisit the blog Loading… current community Stack Overflow helpchat Meta Stack Overflow your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up How are we doing? Please help us improve Stack Overflow. Take our short survey Let's set up your homepage Select a few topics you're interested in: python javascript c#reactjs java android html flutter c++node.js typescript css r php angular next.js spring-boot machine-learning sql excel ios azure docker Or search from our full list: javascript python java c# php android html jquery c++ css ios sql mysql r reactjs node.js arrays c asp.net json python-3.x .net ruby-on-rails sql-server swift django angular objective-c excel pandas angularjs regex typescript ruby linux ajax iphone vba xml laravel spring asp.net-mvc database wordpress string flutter postgresql mongodb wpf windows xcode amazon-web-services bash git oracle-database spring-boot dataframe azure firebase list multithreading docker vb.net react-native eclipse algorithm powershell macos visual-studio numpy image forms scala function vue.js performance twitter-bootstrap selenium winforms kotlin loops express dart hibernate sqlite matlab python-2.7 shell rest apache entity-framework android-studio csv maven linq qt dictionary unit-testing asp.net-core facebook apache-spark tensorflow file swing class unity-game-engine sorting date authentication go symfony t-sql opencv matplotlib .htaccess google-chrome for-loop datetime codeigniter perl http validation sockets google-maps object uitableview xaml oop visual-studio-code if-statement cordova ubuntu web-services email android-layout github spring-mvc elasticsearch kubernetes selenium-webdriver ms-access ggplot2 user-interface parsing pointers c++11 google-sheets security machine-learning google-apps-script ruby-on-rails-3 templates flask nginx variables exception sql-server-2008 gradle debugging tkinter delphi listview jpa asynchronous web-scraping haskell pdf jsp ssl amazon-s3 google-cloud-platform jenkins testing xamarin wcf batch-file generics npm ionic-framework network-programming unix recursion google-app-engine mongoose visual-studio-2010 .net-core android-fragments assembly animation math svg session intellij-idea hadoop rust next.js curl join winapi django-models laravel-5 url heroku http-redirect tomcat google-cloud-firestore inheritance webpack image-processing gcc keras swiftui asp.net-mvc-4 logging dom matrix pyspark actionscript-3 button post optimization firebase-realtime-database web jquery-ui cocoa xpath iis d3.js javafx firefox xslt internet-explorer caching select asp.net-mvc-3 opengl events asp.net-web-api plot dplyr encryption magento stored-procedures search amazon-ec2 ruby-on-rails-4 memory canvas audio multidimensional-array random jsf vector redux cookies input facebook-graph-api flash indexing xamarin.forms arraylist ipad cocoa-touch data-structures video azure-devops model-view-controller apache-kafka serialization jdbc woocommerce razor routes awk servlets mod-rewrite excel-formula beautifulsoup filter docker-compose iframe aws-lambda design-patterns text visual-c++ django-rest-framework cakephp mobile android-intent struct react-hooks methods groovy mvvm ssh lambda checkbox time ecmascript-6 grails google-chrome-extension installation cmake sharepoint shiny spring-security jakarta-ee plsql android-recyclerview core-data types sed meteor android-activity activerecord bootstrap-4 websocket graph replace scikit-learn group-by vim file-upload junit boost memory-management sass import async-await deep-learning error-handling eloquent dynamic soap dependency-injection silverlight layout apache-spark-sql charts deployment browser gridview svn while-loop google-bigquery vuejs2 dll highcharts ffmpeg view foreach makefile plugins redis c#-4.0 reporting-services jupyter-notebook unicode merge reflection https server google-maps-api-3 twitter oauth-2.0 extjs terminal axios pip split cmd pytorch encoding django-views collections database-design hash netbeans automation data-binding ember.js build tcp pdo sqlalchemy apache-flex mysqli entity-framework-core concurrency command-line spring-data-jpa printing react-redux java-8 lua html-table ansible jestjs neo4j service parameters enums material-ui flexbox module promise visual-studio-2012 outlook firebase-authentication web-applications webview uwp jquery-mobile utf-8 datatable python-requests parallel-processing colors drop-down-menu scipy scroll tfs hive count syntax ms-word twitter-bootstrap-3 ssis fonts rxjs constructor google-analytics file-io three.js paypal powerbi graphql cassandra discord graphics compiler-errors gwt socket.io react-router solr backbone.js memory-leaks url-rewriting datatables nlp oauth terraform datagridview drupal oracle11g zend-framework knockout.js triggers neural-network interface django-forms angular-material casting jmeter google-api linked-list path timer django-templates arduino proxy orm directory windows-phone-7 parse-platform visual-studio-2015 cron conditional-statements push-notification functional-programming primefaces pagination model jar xamarin.android hyperlink uiview visual-studio-2013 vbscript google-cloud-functions gitlab azure-active-directory jwt download swift3 sql-server-2005 configuration process rspec pygame properties combobox callback windows-phone-8 linux-kernel safari scrapy permissions emacs scripting raspberry-pi clojure x86 scope io expo azure-functions compilation responsive-design mongodb-query nhibernate angularjs-directive request bluetooth reference binding dns architecture 3d playframework pyqt version-control discord.js doctrine-orm package f# rubygems get sql-server-2012 autocomplete tree openssl datepicker kendo-ui jackson yii controller grep nested xamarin.ios static null statistics transactions active-directory datagrid dockerfile uiviewcontroller webforms discord.py phpmyadmin sas computer-vision notifications duplicates mocking youtube pycharm nullpointerexception yaml menu blazor sum plotly bitmap asp.net-mvc-5 visual-studio-2008 yii2 electron floating-point css-selectors stl jsf-2 android-listview time-series cryptography ant hashmap character-encoding stream msbuild asp.net-core-mvc sdk google-drive-api jboss selenium-chromedriver joomla devise cors navigation anaconda cuda background frontend binary multiprocessing pyqt5 camera iterator linq-to-sql mariadb onclick android-jetpack-compose ios7 microsoft-graph-api rabbitmq android-asynctask tabs laravel-4 environment-variables amazon-dynamodb insert uicollectionview linker xsd coldfusion console continuous-integration upload textview ftp opengl-es macros operating-system mockito localization formatting xml-parsing vuejs3 json.net type-conversion data.table kivy timestamp integer calendar segmentation-fault android-ndk prolog drag-and-drop char crash jasmine dependencies automated-tests geometry azure-pipelines android-gradle-plugin itext fortran sprite-kit header mfc firebase-cloud-messaging attributes nosql format nuxt.js odoo db2 jquery-plugins event-handling jenkins-pipeline nestjs leaflet julia annotations flutter-layout keyboard postman textbox arm visual-studio-2017 gulp stripe-payments libgdx synchronization timezone uikit azure-web-app-service dom-events xampp wso2 crystal-reports namespaces swagger android-emulator aggregation-framework uiscrollview jvm google-sheets-formula sequelize.js com chart.js snowflake-cloud-data-platform subprocess geolocation webdriver html5-canvas centos garbage-collection dialog sql-update widget numbers concatenation qml tuples set java-stream smtp mapreduce ionic2 windows-10 rotation android-edittext modal-dialog spring-data nuget doctrine radio-button http-headers grid sonarqube lucene xmlhttprequest listbox switch-statement initialization internationalization components apache-camel boolean google-play serial-port gdb ios5 ldap youtube-api return eclipse-plugin pivot latex frameworks tags containers github-actions c++17 subquery dataset asp-classic foreign-keys label embedded uinavigationcontroller copy delegates struts2 google-cloud-storage migration protractor base64 queue find uibutton sql-server-2008-r2 arguments composer-php append jaxb zip stack tailwind-css cucumber autolayout ide entity-framework-6 iteration popup r-markdown windows-7 airflow vb6 g++ ssl-certificate hover clang jqgrid range gmail Next You’ll be prompted to create an account to view your personalized homepage. Home Questions AI Assist Labs Tags Challenges Chat Articles Users Jobs Companies Collectives Communities for your favorite technologies. Explore all Collectives Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Collectives™ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How are we doing? Take our short survey Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to compare two Ratios of Rectangles / Sizes to determine their percentages of differences? Ask Question Asked 5 years, 9 months ago Modified5 years, 9 months ago Viewed 888 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I have a ractangle whose ratio is 80:40 and other rectangles with similar aspect ratios within a humbral for example 80:35, 85:45, that unbral is a decimal or integer. My problem is that I need to compare other rectangles with the first one and determine a difference in their respective aspect ratios in percentages, for example: 80:30 is 20% different in aspect ratio than 80:40. (20% is not a calculated data, it is an example idea, because I don't know how to do it). Could it be that a totally opposite aspect ratio is 100% different? For example: 80:40 is 100% different than 40:80 Imagine that you have a collection of rectangles and a target rectangle and you have to filter the collection leaving only those rectangles that have an aspect ratio similar to the target rectangle. Sample: ``` private float GetDiffRatio(FloatSize size1, FloatSize size2) { float fixedR1 = size1.Width / size1.Height; if (fixedR1 >= 0) // Invert and negative. fixedR1 = -(size1.Height / size1.Width); float fixedR2 = size2.Width / size2.Height; if (fixedR2 >= 0) // Invert and negative. fixedR2 = -(size2.Height / size1.Width); float rDiff = fixedR1 - fixedR2; return rDiff 100f; } ``` Test: float diffRatio = GetDiffRatio(new FloatSize(100f, 50f), new FloatSize(50f, 100f)); Results = -100f Test2 (inverted order of parameters): float diffRatio = GetDiffRatio(new FloatSize(50f, 100f), new FloatSize(100f, 50f)); Results = 100f I am not sure that this is a valid or correct form, I do not know if it can generate any condition that returns a wrong percentage of similarity. math discrete-mathematics Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Dec 15, 2019 at 20:27 ChevaCheva asked Dec 15, 2019 at 19:51 ChevaCheva 360 5 5 silver badges 13 13 bronze badges 1 I edited the question and placed an example.Cheva –Cheva 2019-12-15 20:29:07 +00:00 Commented Dec 15, 2019 at 20:29 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. The answer to this question depends a lot on what exactly you're trying to do or you're planning to uses this similarity function for. I find it very unintuitive to say that opposite aspect ratio leads to 0% similarity. I think comparing two rectangles r1=(2.1,2) and r2=(2,2.1) should be a lot more similar to each other then for example r3=(1,5) r4=(5,1). This is not me saying that it couldn't be useful in some case to have a similarity function like your's, I just want to explain that it depends a lot on what you're doing ... I would say a very obvious solution would be to just divide width by height of every rectangle and take as similarity-function s1 the absolute value from both values subtracted. So for my provided examples the result would be: s1(r1,r2) = \| 2.1/2 - 2/2.1 \| = 0.0976... s1(r2,r3) = \| 1/5 - 5/1 \| = 4.8 If it is also important that you have values between 0 and 1 you could additionally plug this values in for example this function ... where b must be smaller than 0 and is a parameter with what you can controll how fast the funciton converges to 1. YOu can play around with it here: In the case you really want something as you suggested, i would simply do the following: You take your constraint that every rectangle is between the ratio a:b and c:d. Than you calculate x1=a/b and x2=c/d and then you interpolate the value from zero to one between those values so: h(x1) = 0 h(x2) = 1 if you need more details on how to do this look here but i think it's very straight forward. The similarity function s3 can then be build again with the absolute value of the difference s3=\| h(r1)-h(r2) \| Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Dec 16, 2019 at 11:25 man zetman zet 901 9 9 silver badges 31 31 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions math discrete-mathematics See similar questions with these tags. The Overflow Blog The history and future of software development (part 1) Getting Backstage in front of a shifting dev experience Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure New and improved coding challenges New comment UI experiment graduation Policy: Generative AI (e.g., ChatGPT) is banned Report this ad Report this ad Community activity Last 1 hr Users online activity 23122 users online 25 questions 32 answers 114 comments 428 upvotes Popular tags javac#phpjavascriptnode.jspython Popular unanswered question What is Sender Account ID in sqs message attributes? amazon-web-servicesamazon-sqs Cherry 33.9k 2,873 days ago Related 7How to find rectangle of difference between two images 0comparing rectangles 9Difference (XOR) between two rectangles, as rectangles? 1Math for counting the difference of concentric rectangles 0Calculating a ratio difference? 1Algorithm to compute the difference between a set of rectangles and a single rectangle 2The percentage of the rectangle in the other rectangle on leaflet map 5Resize a Rectangle in a Percent while maintaining its aspect ratio 0Program to calculate and compare the area of two rectangles 1Calculate overlap for two lists of rectangles Hot Network Questions What is the meaning and import of this highlighted phrase in Selichos? Vampires defend Earth from Aliens Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation The geologic realities of a massive well out at Sea Is encrypting the login keyring necessary if you have full disk encryption? What happens if you miss cruise ship deadline at private island? With line sustain pedal markings, do I release the pedal at the beginning or end of the last note? In the U.S., can patients receive treatment at a hospital without being logged? Making sense of perturbation theory in many-body physics How to fix my object in animation Is there a way to defend from Spot kick? Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? Direct train from Rotterdam to Lille Europe How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? Drawing the structure of a matrix What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? Does the curvature engine's wake really last forever? Does a Linux console change color when it crashes? Why are LDS temple garments secret? Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left? How to home-make rubber feet stoppers for table legs? Does "An Annotated Asimov Biography" exist? Is it possible that heinous sins result in a hellish life as a person, NOT always animal birth? Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? Probable spam. This comment promotes a product, service or website while failing to disclose the author's affiliation. Unfriendly or contains harassment/bigotry/abuse. This comment is unkind, insulting or attacks another person or group. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Stack Overflow Questions Help Chat Products Teams Advertising Talent Company About Press Work Here Legal Privacy Policy Terms of Service Contact Us Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings
13344	https://lab.corewellhealth.org/2018/10/23/helicobacter-pylori-diagnosis-serologic-testing-is-no-longer-recommended/	Corewell Health Laboratory Helicobacter pylori Diagnosis – Serologic Testing is no Longer Recommended Overview Clinical guidelines no longer recommend serologic testing as a method for the diagnosis of H. pylori infection. Rather than IgG serology testing, other non-invasive testing methods such as H. pylori stool antigen and urea breath tests may be used to both diagnose and monitor response to therapy for H. pylori infection. In anticipation of an increase in stool antigen testing, currently a reference lab send-out test, Spectrum Health Regional Laboratory (SHRL) will be implementing this test as of October 29, 2018. By offering this test in-house, results will typically be available one day faster than present state. Background Seroprevalence studies suggest that half of the world’s population is infected with or has been exposed to H. pylori at some point in time, although the majority of those infected are asymptomatically colonized. A minority of infected individuals will develop gastritis and dyspepsia, and a fraction of those will further develop peptic ulcer disease or gastric cancers (adenocarcinoma or mucosa-associated lymphoid tissue (MALT) lymphoma). Diagnostic Approach Established clinical guidelines from the American Gastroenterology Association (AGA) and American College of Gastroenterologists (ACG) provide recommendations regarding the preferred testing approach for H. pylori infection1, 2. The gold standard diagnostic remains invasive endoscopy with gastric tissue biopsy followed by bacterial culture, histology, or rapid urease testing. This approach is recommended for higher risk patients (i.e. GI bleeding, sudden weight loss, excessive vomiting, or anemia) or those over the age of 60 for which a clinician may wish to rule out malignancies. The recommended approach for low risk patients includes non-invasive methods such as the urea breath test or stool antigen test. These methods test for active disease and may also be used to monitor response to therapy following completion of an antibiotic treatment regimen. Infected individuals typically remain colonized for life unless effective therapy is used to eradicate the organism. Serology H. pylori IgG serologic testing is no longer recommended by clinical guidelines. Due to the high seroprevalence and since antibodies may persist for years beyond acute disease, this test is unable to accurately distinguish between past and current infection. IgG antibodies may be detected 2-3 weeks post-infection, yet they remain elevated for years after eradication. Therefore, serologic testing cannot be used to document eradication post-treatment. Sensitivity (85%) and specificity (79%) are low for both acute and chronic infection, and an increasing number of insurance providers are no longer reimbursing for serologic testing3. Stool Antigen and Urea Breath Test AGA and ACG clinical guidelines recommend stool antigen testing or urea breath testing to diagnose active H. pylori infection and guide therapy. Both tests offer high clinical sensitivity and specificity (>95%), and either test may be used to monitor resolution of infection in response to therapy. For both tests, antibiotics and proton pump inhibitors must be discontinued at least 2 to 4 weeks prior to testing to ensure H. pylori viability4. A drawback of the stool antigen test includes the possibility for two patient visits if the specimen is collected at home. Drawbacks of the urea breath test include increased cost and lack of FDA approval for pediatric patients below the age of 3 years. Comparison of H. pylori Stool Antigen and Urea Breath Test \| \| \| \| --- \| \| Stool Antigen \| Urea Breath Test \| \| Test code \| LAB880 \| LAB572 \| \| Patient expense \| $ \| $$$ \| \| Approved patient population \| All ages acceptable \| Not available for patients under 3 years of age. \| \| Specimen collection \| Fresh stool in sterile container or Cary Blair transport media. May require two patient visits (one to pick up the collection container and another to drop off the specimen). \| Bag of “breath”, transported to Blodgett Hospital for analysis. Single visit \| \| Ease of collection \| Non-invasive. Patient can collect at home. \| Non-invasive. Patient collects breath at a lab draw site (or provider’s office) before and after drinking a urea solution. \| \| Specimen transport temperature \| Ambient or refrigerated \| Ambient \| \| Turnaround time \| One day (testing performed 7 days/week) \| One to three days (testing performed Monday-Friday) \| May require two patient visits (one to pick up the collection container and another to drop off the specimen). Single visit (testing performed 7 days/week) (testing performed Monday-Friday) Note: if endoscopy with biopsy is performed, tissue may be sent for reference lab testing by culture (Mayo Medical Laboratories, test ID HELIS). References Laboratory Services You Might Also Like Ammonia, NH3 Reminder: Send Out/Referral Testing at Spectrum Health Respiratory Pathogen Testing Algorithm Categories Posts by Specialty Contact Us Please contact us with any questions or concerns. The Laboratory Call Center is available 24 hours a day, 7 days a week, including holidays, to help answer your questions and also to schedule specimen pick-ups. Please Call 616.774.7721. Additional Resources Archives
13345	https://brainly.com/question/35715380	[FREE] Write the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH). Write the - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +68,9k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +21,4k Ace exams faster, with practice that adapts to you Practice Worksheets +8,3k Guided help for every grade, topic or textbook Complete See more / Chemistry Expert-Verified Expert-Verified Write the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH). Write the equation for this process. 2 See answers Explain with Learning Companion NEW Asked by emilyR2839 • 08/13/2023 0:01 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 14945433 people 14M 0.0 0 Upload your school material for a more relevant answer The neutralization reaction between Hydrochloric acid (HCl) and sodium hydroxide (NaOH) can be represented by the equation: HCl + NaOH → NaCl + H2O. This reaction results in the formation of sodium chloride (NaCl), which is a salt, and water (H2O). Explanation Neutralization is a chemical reaction that occurs between an acid and a base, resulting in the formation of a salt and water. In the case of Hydrochloric acid (HCl) and sodium hydroxide (NaOH), the reaction can be represented by the following equation: HCl + NaOH → NaCl + H2O Hydrochloric acid is a strong acid, while sodium hydroxide is a strong base. When they react, the hydrogen ion (H+) from the acid combines with the hydroxide ion (OH-) from the base to form water (H2O). The remaining ions, sodium (Na+) from the base and chloride (Cl-) from the acid, combine to form sodium chloride (NaCl), which is a salt. Learn more about neutralization reaction between hydrochloric acid (hcl) and sodium hydroxide (naoh) here: brainly.com/question/35084543 SPJ14 Answered by arenasaleem890 •27.6K answers•14.9M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 14945433 people 14M 0.0 0 Upload your school material for a more relevant answer The neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) produces sodium chloride (NaCl) and water (H₂O), represented by the equation HCl + NaOH → NaCl + H₂O. In this process, the hydrogen ions from the acid react with the hydroxide ions from the base. This reaction results in the formation of a salt and water, illustrating how acids and bases interact chemically. Explanation The neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) can be represented by the chemical equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) In this reaction: Acid and Base: Hydrochloric acid (HCl) is a strong acid, and sodium hydroxide (NaOH) is a strong base. Reaction Process: When these two substances are mixed, they undergo a chemical reaction. The hydrogen ions (H⁺) from the acid combine with the hydroxide ions (OH⁻) from the base to form water (H₂O). Salt Formation: The remaining ions, sodium (Na⁺) from the sodium hydroxide and chloride (Cl⁻) from the hydrochloric acid, combine to create sodium chloride (NaCl), which is common table salt. Net Ionic Equation: The net ionic equation for this reaction highlights the essential ions involved: H⁺(aq) + OH⁻(aq) → H₂O(l) This reaction is important in many applications, including titrations in chemistry, where the goal is to determine the concentration of an unknown acid or base by neutralizing it with a standard solution. Examples & Evidence The reaction of hydrochloric acid and sodium hydroxide can be compared to other neutralization reactions, such as acetic acid (vinegar) reacting with sodium bicarbonate (baking soda), where sodium acetate and water are produced. This demonstrates the general principle of neutralization across different acid-base pairs. Neutralization reactions occur in various laboratory and industrial processes, confirming the established principles of chemistry that dictate how acids and bases interact to form salts and water. Thanks 0 0.0 (0 votes) Advertisement Community Answer This answer helped 15250430 people 15M 0.0 0 The neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is represented by the equation HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l), forming water and sodium chloride. The net ionic equation for this reaction is H+(aq) + OH-(aq) → H2O(l), with Na+ and Cl- being spectator ions. The neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is an important concept in chemistry. Neutralization results in the formation of water and a salt. The balanced chemical equation representing this process is: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) In this reaction, HCl is a strong acid and NaOH is a strong base. When they react, they form sodium chloride (NaCl) and water (H2O). Moreover, this reaction can also be expressed as a net ionic equation: H+(aq) + OH-(aq) → H2O(l) Here, the sodium (Na+) and chloride (Cl-) ions are spectator ions and do not participate directly in the reaction, hence their exclusion from the net ionic equation. Answered by Anshuyadav •15.8K answers•15.3M people helped Thanks 0 0.0 (0 votes) Advertisement ### Free Chemistry solutions and answers Community Answer write the net ionic equation for: the reaction of hydrochloric acid with sodium hydroxide and the reaction of acetic acid with sodium hydroxide Community Answer Write the balanced chemical equation for EACH of the two neutralization reactions. Include states (s), (1), (aq), etc. a) hydrochloric acid and sodium hydroxide solution. b) nitric acid and sodium hydroxide solution. Community Answer What is the balanced chemical reaction for the neutralization of sodium hydroxide with hydrochloric acid Community Answer Write the balanced molecular equation for the acid-base reaction that occurs when aqueous solutions of hydrochloric acid (HCl) and sodium hydroxide (NaOH) are mixed. Community Answer If you write the balanced reaction for the neutralization of sodium hydroxide and hydrochloric acid, you will see that the reaction proceeds in a 2:1 fashion True or False Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. New questions in Chemistry Complete the table by providing the working out and formula for each substance. \| Substance \| Working Out \| Formula \| :--- \| magnesium, chloride \| \| M g C l 2 \| \| \| C =\text{_} ratio \| C H 4 \| \| calcium, sulphide \| \| CaS \| \| potassium, oxide \| =\text{} ratio \| K_O \| \| carbon, oxide \| =\text{_}\text{} : ratio \| \| \| iron (III), bromide \| =\text{_}\text{_} : ratio \| \| \| lead (IV), oxide \| = \| \| Which chemical compound is organic? bi g c i rc H 2O bi g c i rc C 12H 22O 11bi g c i rc K N O 3bi g c i rc KOH A student places a marshmallow on a stick and cooks it over a campfire. Identify the type of change that occurs to the marshmallow AND describe the evidence that shows this type of change has occurred. Which element will form a covalent bond with iodine, I? A. chlorine, Cl B. beryllium, Be C. barium, Ba D. potassium, K Which element will form a covalent bond with iodine, I? A. chlorine, Cl B. beryllium, Be C. barium, Ba D. potassium, K Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
13346	https://www.sohu.com/a/393904684_374969	高中数学：四种条件的判断方法 2020-05-08 22:08 来源: 点知教育由“四种条件”的定义可知：判断条件p是结论q的什么条件，实际上就是判断或的正确与否。只要运用题目中所给的条件和相关的数学知识加以判断即可。而对于抽象命题的判断，则只有将题中所给的逻辑关系画出示意图，再利用定义进行判断。例1、“”是“”的 A. 充分不必要条件 B. 必要不充分条件 C. 充要条件 D. 既不充分也不必要条件分析：“若p则q”是原命题，可知：①原命题真而逆命题不真，则p是q的充分不必要条件；②原命题不真而逆命题真，则p是q的必要不充分条件；③原命题、逆命题都真，则p是q的充要条件；④原命题、逆命题都不真，则p是q的既不充分也不必要条件。解析：命题中条件p是“”，结论q是“”。若，则且（即），这说明“ 且 ”是“”的充分条件。若，则，适合上式，但，可见由且推不出，这说明“”不是“且”的必要条件。故应选A。如果从命题的条件和结论之间的关系来判断有困难时，有时可以从集合的角度来考虑，尤其是所研究的条件p与q表示两数集时，这种方法就更显优越性。记条件p、q对应的集合为A、B，即：，。 ①若，则p是q的充分条件，q是p的必要条件；②若，则p是q的充分不必要条件，q是p的必要不充分条件；③若A=B，则p是q的充要条件；④若，且，则p是q的既不充分也不必要条件。例2、是否存在实数m，使“”是“”的充分条件？如果存在，求出m的取值范围。是否存在实数m，使“”是“”的必要条件？如果存在，求出m的取值范围。分析：充要条件反映了命题间相互推导的逻辑关系，同时也是集合之间关系的一种反映。如，则A中的元素是属于B的充分条件，B中的元素是属于A的必要条件。本题将“若p则q”的判断转换成两集合之间的一种包含关系，从而使问题便于判断。解析：设p：，q：。平台声明：该文观点仅代表作者本人，搜狐号系信息发布平台，搜狐仅提供信息存储空间服务。首赞 +1 点赞失败阅读 (1436) 我来说两句 0人参与， 0条评论搜狐“我来说两句” 用户公约点击登录搜狐小编发布推荐阅读据说，欧洲的小偷居然达成了一种共识，那就是可以去偷中国人的钱财和物品，但是绝对不... 吃一颗苹果 · 昨天 10:41 0 为什么男人都喜欢微胖的女人？过来人说了大实话俺也一样 · 今天 06:00 0 大尺度电影推荐，部部都是经典，值得你熬夜观看！梦新看世界 · 09-26 00:52 0 梦见与异性发生关系，大都因为这2种原因，别傻傻不懂小布点娱乐 · 昨天 21:50 0 江西一女子考上大学，却交不起学费，向大舅借了4万，毕业后女子去还钱，谁料，大舅说钱... 骊歌声声慢 · 昨天 18:10 0 93受阅女兵参加婚礼被骂惨！网友：穿这套衣服去是几个意思？史之春秋 · 09-26 10:33 0 主办方拒绝挂中国国旗，中方果断退出，其他国家：没有中国不开赛！朔方瞭望 · 今天 03:20 0 太突然！清华赴美博士李昊然去世，年仅28岁，大学邮件曝光去世原因，我国大使馆紧急发声影像温度 · 昨天 23:56 0 9月28日特讯！洪森发表涉华言论，罕见言辞引爆国际舆论小笛点评 · 今天 03:09 0 全世界都被普京骗了 , 打击乌克兰只是一个幌子 , 真正目标已布局4年！史谭a · 昨天 07:28 0 我嫌老公买的维C太苦，便拿着药瓶去医院，医生看完神色凝重地说：这里面不是维C，是米... 瓜汁橘长Dr · 09-26 02:20 0 甘肃32岁美女博士，因无法接受丈夫的真实身份，从23楼一跃而下瓜汁橘长Dr · 昨天 03:02 0 今年国庆不一般！有1个坏消息，2个别大意，3个要谨慎，早做准备小娟教做菜 · 09-26 22:32 0 新中国成立后，只有3个人可以随时见毛主席，他们分别都是谁？南风一史 · 昨天 03:20 0 国庆黄金周还没到，就已经出现了3大反常现象，还有一个让人担忧平说财经张平 · 昨天 11:31 0 再这样吃南瓜，肝就废了？医生：吃南瓜不注意5点，是在给肝上刑健康小小讲堂 · 昨天 18:30 0 清华大学本科毕业生李昊然去世，年仅27岁非常优秀，同学曝是自杀八卦娱V · 今天 00:17 0 台湾的终极解决方案，土地归中国，人员来去自由大国迷雾 · 昨天 00:05 0 男驴友爬雪山坠亡后续：现场视频流出，死状极惨，目击者曝真相历史特别冷 · 09-26 06:50 0 一夜之间突然倒戈！3名师长带领5万士兵赴美，还顺走600辆坦克朔方瞭望 · 今天 02:24 0 港媒已经收到风声，中国在为国家紧急事态，提前做充分的准备澜风说 · 昨天 23:38 0 官方通报：作为“一把手”，他安排单位买500斤散装白酒存于办公楼，用于日常接待政事儿 · 昨天 09:23 0 韩国女主持嘲讽：中国制造就是廉价、仿品，教授反问了一句，让她瞬间脸红相框框一切 · 09-26 11:57 0 快报快报！泽连斯基愿承认乌俄新边界峰火人生 · 昨天 12:01 0 年轻人正在逃离上海！这事怪不了任何人，577万老人换谁也顶不住火火焱焱焱焱 · 昨天 03:45 0 对肺特好的“3种蔬菜”，建议秋天要多吃，人人吃得起，别不懂吃秀厨娘 · 昨天 21:45 0 台湾要员曾放话：满足这3条件，台湾自动回归，实行“一国一制” 柠重 · 09-26 01:06 0 警方通报“华南理工大学发生车祸致1死1伤” 环球时报 · 今天 11:57 0 男子直播开飞机坠亡，曝遗体烧的只剩一条腿，细节曝光，早有预兆青橘罐头 · 昨天 23:44 0 美媒无语了:美部长召集全球美军将领回国，原来就为了这点事？！多看看啊这个世界 · 昨天 23:26 0 周总理逝世，没人敢读悼词，叶剑英：论资格论威望，只有一人可以壹周历史 · 昨天 04:25 0 19000吨钢筋桥梁“瞬间崩塌”，大桥竣工前突然垮掉，75人当场身亡朔方瞭望 · 09-26 02:38 0 “禁毒女警”陈光明，将7000名毒贩送进牢房，下场极其悲惨微历史之中山古国 · 昨天 03:13 0 泽连斯基在联合国很没有面子，泽连斯基在联合国大会上发表演讲，直播显示，泽连斯基未... 小陈言社会 · 09-25 15:38 0 “一家几口都是公务员”，人口小县财政供养入不敷出！副部级退休官员建议按比例削减公... 有料新语 · 09-26 06:17 0 中共中央批准，开除段成刚、龙翔党籍长安街知事 · 今天 06:12 0 茶叶是血糖的“杀手”？医生建议：不想血糖飙升，最好少喝4种茶健康小小讲堂 · 09-26 23:35 0 早设能撑久一点？性治疗师童嵩珍4种控制法让它慢点设燕麦小姐 · 09-25 23:04 0 6个症状提示脑供血不足，医生提醒：不想脑梗或痴呆，做好4件事健康小小讲堂 · 昨天 22:28 0 喊话4天，泽连斯基决定离职，乌恳求中国做一件事，流放名单公开北极星哨 · 09-26 06:53 0 21岁女子独自赴埃及旅游失联，最新消息：人找到了深圳新闻网 · 今天 08:19 0 喝对水，养出易瘦体质注册营养师范英英 · 今天 00:10 0 女儿叛逆行为背后的秘密？这个被父母忽略的关键因素刘承洛记忆重组 · 今天 11:47 0 血液净化及连续性肾脏替代治疗（ CRRT）相关知识孙医生工作室 · 今天 10:58 0 早餐为何不建议吃面包？不止面包，这6种食物，尽量少吃经络技巧 · 今天 02:12 0 县长被开除党籍、撤职、降为正科11年后，涉嫌严重违法，再次被查观察者网 · 今天 04:16 0 被低估的散结食材！白萝卜这么吃，比吃药还省心，早看早受益王进学主任 · 昨天 21:05 0 忘恩负义！白拿中国100亿援助和战机，转头将中国武器送美国研究墨珑甲说 · 09-26 06:45 0 广东与广西的辖区调整，广西的1个县，为何划归了广东省？阿鑫历史说 · 09-26 06:00 0 普京心意已决：除了“消灭”乌克兰，还有一国将在2025不复存在？昭烈忠臣 · 09-26 00:36 0 美国总统特朗普在联合国大会上，第二次提到了中国，他声称：“我很佩服中国，我给予中... 吃一颗苹果 · 09-26 00:28 0 泰安中心医院“蒲公英”名医科普专家走进宁阳县河滨社区开展宣讲悦闻天下 · 今天 11:40 0 美国防长:我们可能要开战了！达文西看世界 · 09-26 06:56 0 脑梗高发季到了！有3种不舒服千万别大意环球时报新媒体 · 今天 10:07 0 68岁男子清晨去世？医生叹息：中老年人，早上千万牢记“5不要” 39健康网 · 昨天 08:31 0 医院各层干部形成权色钱交易网，招聘等十几种腐败触目惊心四季日常诠释社会 · 09-26 15:11 0 国际顶尖肺癌专家宣布已患肺癌三年：这是一种荣幸，可亲身体验患者经历的一切红星新闻 · 今天 05:07 0 秋季心血管病高发，恶心、心慌、后背疼痛警惕心肌梗死全科与心理 · 今天 08:36 0 “卖女孩的红烧肉”店名争议，官方回应深圳新闻网 · 今天 10:43 0 40岁上海男子“被患精神分裂”11年，政审时才发现？“600号”复核：精神分裂→急性应激障碍鲁网 · 今天 02:43 0 张召忠：错过统一的最好时机，该出手时没出手，看完恍然大悟小舟说史 · 昨天 02:58 0 两个草包，把西贝害惨了简十一 · 昨天 02:34 0 黄永胜晚年被儿子问：你最敬佩谁？他回答了三个人，却没有林帅苏苏的幸福生活 · 昨天 05:33 0 内心特别坚强的三大生肖旅行说事 · 昨天 18:16 0 解放军当前拥有13个新编集团军，一个集团军究竟有多少兵力？宏爽评历史 · 09-26 01:08 0 乳癌术后乳房再造，怎么让两边对称？医美大众科普 · 昨天 22:00 0 全国田径冠军赛：吴艳妮12秒90强势夺冠，林雨薇身兼多项获亚军全景体育 · 今天 09:19 0 2025亚太痛风联盟年会召开,成都风湿医院闪耀国际学术舞台角落光 · 今天 11:07 0 中建二局硬核科技闪耀服贸会，引领行业创新浪潮麻婆豆腐不要辣 · 今天 09:05 0 一个人观看的日本电影免费麻花说娱乐 · 昨天 23:10 0 毛主席给我们攒下的三大红利已经消耗殆尽，以后得靠我们自己了！游游历史 · 昨天 03:14 0 美国不担心稀土卡脖子了？一个华裔科学家，帮他们搞了无稀土磁铁乾凉散人 · 昨天 04:20 0 阅兵迎外宾的车，50辆来自河南车主，不是国家缺车，是有三大意义鲸探所 · 昨天 03:19 0 呕吐用多拉司琼：牢记“三要三不要”，规范用药止吐护健康家庭常用指南 · 今天 06:37 0 让人称绝的军购：中国引进瑞士“双35高炮”，换来30年技术飞跃明轩聊体育 · 昨天 23:16 0 《志愿军3》首映礼：朱亚文黑成炭，宋佳好惊艳，陈凯歌咋老成这猫叔聊电影 · 昨天 06:17 0 为何有些甲状腺癌不需要立即手术？乐飞说甲状腺 · 昨天 23:35 0 搜狐医药 \| WAO主席与中方权威专家共同警示：儿童过敏性鼻炎诊疗绝非“成人缩小版” 搜狐健康 · 昨天 22:51 0 你以为在减肥，其实很伤身！常见的 5 个“无效减肥”行为，比发胖还可怕全科与心理 · 今天 09:17 0 睡着了也能瘦！熬夜的3大减肥陷阱会让你白忙一场注册营养师范英英 · 09-26 23:30 0 已经到底了点知教育文章 0 总阅读 1547.3万+ 24小时热文 1 华南理工大学校内发生车祸，官方通报：1人经抢救无效死... 32万阅读 2 从教室到食堂，狐友校园百科一站式全导航！ 130万阅读 3 受贿2.68亿余元，唐仁健一审被判死缓 116万阅读普林斯顿中国博士后李昊然去世，毕业于清华，刚完成论... 319万阅读搜狐号不爱吃青菜的小白兔不宁腿有哪些表现？请输入正确的登录账号或密码安全提示暂不发送发送语音验证码
13347	https://sentence.yourdictionary.com/liaison	Examples of "Liaison" in a Sentence \| YourDictionary.com Dictionary Thesaurus Sentences Grammar Vocabulary Usage Reading & Writing Articles Vocabulary Usage Reading & Writing Sign in Menu Word Finder Words with Friends Cheat Wordle Solver Word Unscrambler Scrabble Dictionary Anagram Solver Wordscapes Answers Sign in with Google Dictionary Thesaurus Sentences Grammar Vocabulary Usage Reading & Writing Word Finder Word Finder Words with Friends Cheat Wordle Solver Word Unscrambler Scrabble Dictionary Anagram Solver Wordscapes Answers Sentences DictionaryThesaurusSentencesArticlesWord Finder Make Our Dictionary Yours Sign up for our weekly newsletters and get: Grammar and writing tips Fun language articles WordOfTheDay and quizzes Sign in with Google By signing in, you agree to our Terms and Conditions and Privacy Policy. Success! We'll see you in your inbox soon. Thank you! Undo Home Sentence Liaison Liaison Sentence Examples liaison Meanings Synonyms Sentences She became his mistress, and the liaison lasted till he died. 119 60 In any case, my husband, Mike—who was voted to be the liaison with the soldiers—isn't likely to let it happen. 79 62 The air cavalry can help effect vital liaison between the units. 19 8 A legal nurse consultant serves as a liaison between the medical system and legal system. 16 6 The confessor united his influence with that of Madame de Maintenon to induce the king to abandon his liaison with Madame de Montespan. 33 25 The program also includes a family liaison, a person who promotes parental participation in the children's education and in workshops on literacy, nutrition, budgeting, health, and other topics. 4 1 A thorough Environmental Impact Assessment, supplemented by further studies and liaison with key stakeholders, predicts no significantly adverse effects. 4 2 The County Council co-ordinates planned work on the highway through regular liaison with all parties involved. 4 2 The gold Anglo-Saxon pendent excites the local Finds Liaison Officers. 3 1 Call your local public school or your homeschooling liaison if you want your child to take one of the state's standardized tests. 3 1 Advertisement The subjects covered include induction, home/school liaison, combating racism, and the psychological adaption of refugee young people. 3 2 There has been liaison with LPO, the Birdlife International partner in France, concerning impact of Arab falconry on Sakers and bustard species. 2 1 I wondered, as a liaison psychiatrist, whether there might be a therapeutic place for this technique in hospitals. 3 2 Grocery Trade Liaison Committee consists of senior company representatives. 4 3 Large companies often employ regional operations managers to serve as an upper management liaison for general managers based at local sites. 4 3 Advertisement He is also remarkable as one of the few kings of the house of Oldenburg who had no illicit liaison. 3 4 In any case, my husband, Mikeâ€”who was voted to be the liaison with the soldiersâ€”isn't likely to let it happen. 3 4 The sked was setup earlier in the week via email with no liaison on any band including cellular! 2 3 But if you can supply a fairly cogent reason their special liaison unit will often sell you a copy ' at BBC cost ' . 2 3 Identification of individual patients was made by the district nurse liaison officer and referral made to the district continence adviser. 2 3 Advertisement Treasure finds can also be reported to your local Finds Liaison Officer who can inform the relevant coroner of the find on your behalf. 2 3 She had a dalliance with a stable boy, you are the product of that liaison. 2 3 We arrived at the site about 3 hours later than we intended to be met by a very flustered bands liaison guy. 2 3 A less satisfactory organization, from my point of view, set up a liaison group, which met roughly fortnightly. 2 3 More liaison with TECs and graduate recruiters would help rectify this. i. What are the purposes of graduate recruiters would help rectify this. i. What are the purposes of graduate schools? 2 3 Advertisement The naval commander is assisted in the control of naval gunfire by navy liaison representatives located with supported ground forces. 2 3 Liaison is essential in multinational, joint, interagency, and combined arms operations. 2 3 The site liaison COs were happy with the proposal but expressed concern about introducing the new arrangement before the availability of a school-wide inventory. 2 3 Liaison with Mendip DC Rights of Way officer, and permission sought from adjacent landowners to install some signs. 2 3 During the consumer choice study, the AEBC will need to maintain close liaison with the Food Standards Agency. 2 3 I am also happy to receive suggestions for other ways of facilitating regional liaison. 2 3 We were under the command of a Royal Navy Captain who was in close liaison with our own Captain. 2 3 Would they have done so if she had indeed been the partner in an adulterous liaison with the man they both venerated? 2 3 Roles of wildly varying quality followed, until in 71 Christie began a professional and romantic liaison with Warren Beatty. 2 3 Covering induction, home/school liaison, combating racism, and the psychological adaptation of refugee young people. 2 3 We take full responsibility for the client liaison throughout the purchase. 2 3 This depends on the liaison librarian receiving copies of reading lists. 2 3 The brigade for which I was liaison officer attacked with 75 tanks of which 3 tanks survived. 2 3 During the year there was much preparatory hard work in liaison with Richard Taylor of Leeds Planning Department and Leeds Leisure Services. 2 2 Indeed, no liaison psychiatry consultant should be expected to provide a service without junior medical staff. 2 2 Experts from the South Wales Police football liaison office in Swansea were trying to identify ringleaders. 2 2 The Group emphasizes the importance of liaison and team work in managing an incident or outbreak involving Cryptosporidium in the drinking water supply. 2 2 A great deal has been made of Peter's infidelity towards his consort; but the only one who really suffered from his liaison with the ugly, stupid and vixenish countess Elizabeth Vorontsova was the unfortunate emperor. 2 2 In most cases, it is best to stick with the actual companies themselves rather than go through a cyberspace liaison. 1 2 These companies act as a liaison between you and the seller. 3 3 It also reveals an organization that serves as a liaison with the American government while remaining an independent charitable organization. 2 2 It's hard enough to imagine mom and dad in a romantic liaison in their heyday. 2 2 Tom also has a daughter Lien from a liaison during the Vietnam War and son Daniel from an affair with Emily Stewart. 2 2 Henry and Alice overheard Madison and Lauren talking about Amy and Ricky's past sexual liaison and worry for their friend. 2 2 Lise Charmel - The Liaison Secrete collection from Lise Charmel is sexy and seductive. 2 3 Liaison Dangereuse is a daring lingerie retailer that's put out some thought-provoking ads in recent years for Muslim women. 3 3 She was the press liaison for Falcon Crest when she and Lorenzo met. 3 3 Captain Mike Donovan (Don Franklin) is the military liaison and Frank's old buddy. 3 3 Kira - A Bajoran colonel in the Bajoran Militia who serves on Deep Space 9 as a Bajoran liaison officer. 3 3 Major Kira Nerys - A member of the Bajoran resistance movement, Kira acted as first officer and political liaison aboard Deep Space Nine. 2 2 To the majority of English readers Guenevere is best known in connexion with her liaison with Lancelot, a story which, in the hands of Malory and Tennyson, has assumed a form widely different from the original conception, and at once more picturesque and more convincing. 9 12 Marie's one attempt to interfere in politics, an effort to prevent the disgrace of the duke of Bourbon, was the beginning of her husband's alienation from her; and after the birth of her seventh child Louise, Marie was practically deserted by Louis, who openly avowed his liaison with Louise de Nesle, comtesse de Mailly, who was replaced in turn by her sisters Pauline marquise de Vintimille, and Marie Anne, duchess de Chateauroux, and these by Madame de Pompadour. 12 15 The date of her birth was placed at about 1409, her liaison with the king dated from 1 433. 9 13 Charles de Flahaut was generally recognized to be the offspring of his mother's liaison with Talleyrand, with whom he was closely connected throughout his life. 13 17 Meanwhile the Countess Potocka had established herself in Paris, but Charles de Flahaut had by this time entered on his liaison with Hortense de Beauharnais, queen of Holland. 10 14 She was married, after a liaison with the duke of Guise, to Henry of Navarre, afterwards Henry IV., on the eve of St Bartholomew's Day. 9 14 He once lay in hiding for two months with the duchesse du Maine at Sceaux, where were produced the comedietta of La Prude and the tragedy of Rome sauvee, and afterwards for a time lived chiefly at Luneville; here Madame du Chatelet had established herself at the court of King Stanislaus, and carried on a liaison with Saint-Lambert, an officer in the king's guard. 12 17 The same year he formed a liaison with Marie Claire Deschamps de Marcilly, widow of the marquis de Villette, whom he married in 1720 after the death in 1718 of Lady Bolingbroke, whom he had treated with cruel neglect. 9 14 In 1859 he was tried on a charge of murder, having shot Philip Barton Key, U.S. attorney for the District of Columbia, whom Sickles had discovered to have a liaison with his wife; but was acquitted after a dramatic trial lasting twenty days. 10 15 The " Admiralty War Office and Press Committee " had been formed in 1911, mainly through the efforts of Mr. (afterwards Sir) Reginald Brade, to establish a permanent liaison in peace and war between the Admiralty and the War Office on the one hand and the Press on the other. 9 14 Within a few months of this culminating triumph, she was threatened with utter ruin by the discovery of a supposed liaison with her gentleman of the bedchamber, William Mons, a handsome and unscrupulous upstart, and the brother of a former mistress of Peter. 11 16 Tricked into a liaison with the Fisher King's daughter Elaine, he becomes the father of Galahad, the Grail winner, and, as a result of the queen's jealous anger at his relations with the lady, goes mad, and remains an exile from the court for some years. 8 15 After a careful education, completed by the usual grand tour, Magnus learned the art of war under Gustavus Horn, and during the reign of Christina (1644-16J4), whose prime favourite he became, though the liaison was innocent enough, he was raised to the highest offices in the state and loaded with distinctions. 9 16 Constant's political career was spoiled by his liaison with Madame de Stael, and at the Restoration was further disturbed by his unreturned passion for Madame Recamier. 15 22 At this time ended his liaison with Mme de Balbi, and the influence of d'Avaray reached its height. 11 18 Madame de Charriere, although twentyseven years older than Constant, became his mistress, and the liaison, an affair possibly more of the intellect than of the heart, lasted until 1796, when Constant became intimate with Madame de Stael. 9 18 About 1802 he contracted with Madame de Condorcet a liaison which lasted till her death (1822). 11 26 Browse other sentences examples The word usage examples above have been gathered from various sources to reflect current and historical usage. They do not represent the opinions of YourDictionary.com. Related Articles Aviation Abbreviations & Acronyms: List of Common TermsThere are hundreds of aviation abbreviations used by people in the aviation industry. Most are used by pilots in the cockpit. This list includes a selection of them. Why Did Martha Stewart Go to Jail?Home-decorating guru Martha Stewart is known for her crafty party planning tips and delicious recipes. However, her involvement in a 2001 insider trading scandal led to Stewart’s arrest and five-month prison sentence in December 2001. But it wasn’t the insider trading charge that led to her imprisonment. So why did Martha Stewart go to jail? Words near liaison in the Dictionary liableness liage liaise liaised liaises liaising liaison liaison aircraft liam liana liane liang Random Word Learn a new word now! Get a Random Word Copyright © 2025 LoveToKnow Media. All Rights Reserved Features Dictionary Thesaurus Sentences Grammar Vocabulary Usage Reading & Writing Company About Us Contact Us Privacy Policy Editorial Policy Cookie Settings Terms of Use Suggestion Box Do Not Sell My Personal Information Random Word Learn a new word now! Get a Random Word Follow Us Connect Contact Us Suggestion Box Follow Us LinkedIn Facebook Instagram TikTok Do Not Sell My Personal Information Copyright © 2025 LoveToKnow Media. All Rights Reserved
13348	https://deepblue.lib.umich.edu/bitstream/handle/2027.42/27367/0000393.pdf?sequence=1	Gene, 63 (1988) 331-336 Elsevier 331 GEN 02308 Short Communications Genetic basis of hypoxanthine guanine phosphoribosyltransferase deficiency in a patient with the Lesch-Nyhan syndrome (HPRT,,,,,) (Recombinant DNA; cDNA cloning; RNase A mapping; secondary structure prediction; hyperuricemia; gout; phage ngtll vector) Beverly L. Davidson, Mohammed Pashmforoush, William N. Kelley and Thomas D. Palella Departments of Internal Medicine and Biological Chemistry, and the Rackham ArthritisResearch Center, Universityof Michigan Medical School, 5522 Medical Science Research Building I, Ann Arbor, MI 48109 (U.S.A.) Received 9 November 1987 Accepted 11 November 1987 Received by publisher 7 December 1987 SUMMARY The molecular basis for complete hypoxanthine guanine phosphoribosyltransferase (HPRT) deficiency has been determined in a patient with Lesch-Nyhan syndrome. A B-lymphoblastoid cell line derived from this patient expresses normal amounts of HPRT mRNA yet no detectable immunoreactive protein as determined by radioimmunoassay. These findings suggest either a decreased rate of translation or accelerated degradation due to enhanced proteolytic susceptibility. cDNAs synthesized from this patient’s RNA have a single nucleotide (nt) substitution, a C --, A transversion at nt 222. RNase A cleavage analysis confirms the presence of a mutation at this position within mRNA isolated from lymphoblasts from patient A.C. This transversion predicts a phenylalanine to leucine replacement at amino acid position 73 in the translated protein. We have designated this mutant HPRT,,,,. The mutation in HPRTFlint disrupts a strongly conserved region among PRTases from Escherichiu coli, rodents and man, suggesting an important role for this region for the normal function of HPRT. Since it is unlikely that this amino acid substitution alters the translational rate, we hypothesize that disruption of the secondary structure within this region renders HPRTFli,t more susceptible to proteolysis. Correspondence to: Dr. Thomas D. Palella, Department of Inter-nal Medicine, University of Michigan Medical School, 1150 West Medical Center Drive, 5520 Medical Science Research Building I, Ann Arbor, MI 48109 (U.S.A.) Tel. (313)747-3413. Abbreviations: aa, amino acid(s); A.C., patient initials; bp, base pair(s); CRM, immunologically cross-reactive material; GMP, guanosine monophosphate; HPRT, hypoxanthine guanine phos-phoribosyltransferase; IMP, inosine monophosphate; nt, nucle-otide(s); RIA, radioimmunoassay. 0378-I 119/88/%03.50 0 1988 Elsevier Science Publishers B.V. (Biomedical Division) 332 INTRODUCTION Hypoxanthine guanine phosphoribosyltransferase (HPRT; EC 2.4.2.8) is a purine salvage enzyme that catalyzes the conversion of hypoxanthine and guanine to IMP and GMP, respectively. Human HPRT mutants were isolated over 25 years ago and became the basis for development of a DNA-mediated transformation system (Szybalska and Szybalski, 1962; Scangos and Ruddle, 1981). Partial deficiency of this enzyme results in hyperuricemia and a severe form of gout (Kelley et al., 1967). Complete HPRT deficiency causes the Lesch-Nyhan syndrome, a disease manifested by hyperuri-cemia, hyperuricaciduria, uric acid renal stones and severe neurological abnormalities (Lesch and Nyhan, 1964; Seegmiller et al., 1967). The determination of mutations responsible for HPRT deficiency in man has been limited in the past to the study of mutations which do not greatly affect the intracellular enzyme concentration. The identifi-cation of the complete amino acid sequence of the normal enzyme (Wilson et al., 1982a) and the deter-mination of amino acid substitutions in four mutant proteins (Wilson et al., 1986) have led to a prelimi-nary understanding of the roles of certain domains with respect to the catalytic function of HPRT (Argos et al., 1983). To extend our understanding of the molecular basis of HPRT-deficient states, we have undertaken the characterization of mutant forms of human HPRT. HPRT-deficient subjects are a phenotypi-tally heterogeneous population with regard to enzyme activity and concentration. This phenotypic variability reflects the occurrence of new and inde-pendent mutations at the HPRT locus (Yang et al., 1984; Wilson et al., 1986). More than 75% of these F H mutants have normal amounts of HPRT specific mRNA. Furthermore, in all but five cases described to date, transcript size is grossly normal (Wilson et al., 1986; Yang et al., 1984). These characteristics, together with available techniques for cDNA clon-ing, provide the necessary rationale and tools for the identification of mutations which lead to dys-functional HPRT. A subject (A.C.) with the Lesch-Nyhan syn-drome, has been previously characterized. Erythro-cytes and lymphoblasts from A.C. had undetectable HPRT activity (< 0.1 mU/mg or < 0.7 y0 of control values). Immunoreactive protein, as determined by RIA using polyclonal anti-human HPRT, was also undetectable (< 1.5 ng CRM/mg or < 0.5 y0 of con-trol values). HPRT-specific mRNA from A.C. lymphoblasts was grossly normal with respect to size and amount on Northern-blot analysis (Wilson et al., 1986). We have identified a single nucleotide substitution in HPRT cDNA clones derived from A.C. mRNA. The resultant amino acid change replaces phenylalanine with leucine in a region of the protein strongly conserved in E. coli, rodent and human phosphoribosyltransferases (Hershey and Taylor, 1986). These data suggest an important role for Phe-73 in maintaining an enzymatically active protein. EXPERIMENTAL AND DISCUSSION (a) cDNA cloning and recombinant analysis B-lymphoblastoid cells were established from lymphocytes isolated from patient A.C. as previously described (Wilson et al., 1982b). RNA was isolated Fig. 1. Strategy for sequencing mutant cDNA clones. The entire cDNA was sequenced in both directions using HPRT-specific oligomers (arrows 1 + 5) or the Ml3 universal primer (unnumbered arrows). The nucleotide sequences of HPRT-specific primers 1 through 5 are listed in Table I. The site of the mutation is indicated by a triangle. Stippled bar, HPRT-coding sequence; single line, HPRT-untranslated sequence; vertical dashes, vector sequence; B, B&I site; E, EcoRI site; H, Hind111 site. The direction of the arrows corresponds to sequencing direction. 333 TABLE IHPRT specific oligodeoxynucleotide primers Oligodeoxy-Sequence (5’ + 3’) Priming site a Strand’ nucleotide (nt) 1AGTGATGATGAACCAG 31-47 2CACTGAATAGAAATAGT 251-267 _3GATATAATTGACACTGG 403-419 _4CCCCTGTTGACTGGTCA 337-321 +5AGTCCTGTCCATAATTA 138-122 +a Corresponds to the nucleotides of sense strand of HPRT cDNA where the A of the ATG start codon is position 1. Note that primers 4 and 5 hybridize to the sense strand and are therefore the complement of the HPRT cDNA sequence. ’The strand to which the primer anneals is represented as either sense ( + ) or antisense ( - ). using guanidium isothiocyanate (Chirgwin et al., 1979) and then subjected to oligo-d(T) column chromatography to obtain poly(A) + RNA (Aviv and Leder, 1972). Approximately 10 pg of poly(A)’ RNA were used to synthesize an oligo-d(T)-primed library (Okayama and Berg, 1982; Gubler and Hoffman, 1983), which was cloned into &gtll (Young and Davis, 1983). Recombinant phages were screened with an [ a-32P]dCTP-labeled MspI-TuqI (160 nt) fragment of normal HPRT cDNA containing portions of exons 1 and 3 and all of exon 2 (Brennand, 1983). Four HPRT-positive recombinants were isolated: pHPAC1, pHPAC2, pHPAC8 and pHPAC12. Restriction endonuclease mapping and nucleotide sequencing revealed that pHPAC8 and pHPAC12 were full-length, containing 654 nt of coding se-quence, 590 nt of 3’-untranslated sequence and 118 and 120 nt of 5’-untranslated sequence, respectively. pHPAC1 contained only 52 nt of 5’-untranslated sequence. (b) Sequencing of mutants The dideoxynucleotide sequencing strategy for pHPAC1, pHPAC8, and pHPAC12 is depicted in Fig. 1, and used both HPRT-specific primers and the Ml3 universal primer (Bethesda Research Labo-ratories; Biggin et al., 1983; Bonthron et al., 1985). The positions of the HPRT nucleotide sequence to which these primers anneal and the strands to which they anneal are shown in Table I. Sequencing of pHPAC8 showed two differences from normal HPRT sequence, a C + A transversion at nt 222 and a G -+ A transition at nt 658. Se-quencing of clones pHPAC12 and pHPAC 1 revealed only the C + A transversion at 222, indicating that the G -+ A transition is probably a cloning artifact. Consistent with this conclusion, RNase cleavage analysis also failed to detect the transition at nt 658 (not shown). The C + A transversion at nt 222 predicts aPhe --f Leu change at aa 73 (Fig. 2). The normal abundance of HPRT mRNA in patient A.C. sug-gests that one effect of this nucleotide substitution is at the level of the primary structure of the protein since no immunoreactive protein is detectable in lymphoblasts derived from this patient (Wilson et al., 1986). Phe-73 is strongly conserved among various phosphoribosyltransferases when aligned to give maximal overlap of predicted secondary struc-tures (Table II). Analysis (Chou and Fasman, 1984) 73 ‘W LYS ILei Phe Ala HPRT Flint TAT AAA TTm TTT GCT 222 TAT AAA TTO TTT GCT Normal TYr LYS I Phe Phe Ala Fig. 2. Mutant nucleotide (nt 214-228) and aa (7 l-75) sequence of HPRTFlint compared to normal. The C -+ A transversion at nt 222 (box) causes a Phe +Leu substitution at aa 73 (box). 334 TABLE II Conservation of Phe-73 among various phosphoribosyltransferasesa Phospho~bosyl-transferases Amino acid residues from position 69 to 76b 69 70 71 12 Reference c 73 14 15 76 HPRTmnt Human HPRT Hamster HPRT Mouse HPRT Human APRT Mouse APRT E. coli APRT E. coli XGPRT E. co& OPRT GUY GUY Gb GUY GUY GUY GUY GUY - GUY GlY GUY GUY Phe Phe Phe Cly GIY Tyr TY~ ‘W Tyr Leu Leu Leu Leu Be LYS J-YS LYS LYS --- Val Glu Leu Phe Phe Phe Phe Phe Phe Pro Phe Phe Ala Phe Ala Phe Ala Phe Ala Gly pro Gly Pro Gly Ala Gly Ala - -Asp Asp Asp Asp Ser Ser pro Leu Asp a Sequences are aligned to give maximal overlap of predicted secondary structure (Chou and Fasman, 1978) b Residue numbers correspond to the amino acid sequence of the mature human enzyme isolated from erythrocytes (Wilson et al., 1983) c A, Wilson et al. (1982b); B,C, Konecki et al. (1983); D, Wilson et al. (1986); E, Dush et al. (1985); F, Hershey and Taylor (1986); G, Pratt and Subramani (1983). H, Poulsen et al. (1983). of the region spanning aa residues 69 to 76 predicts an a-helical region directly adjacent to a fi-turn. Leu, which has a stronger propensity toward a-helical formation than does Phe, could result in the ex-tension of a helix into what normally exists as a turn. Furthermore, it is likely that the replacement of the strongly conserved aromatic side chain at this po-sition afTects appropriate packing to accommodate pSPH 1 E HP ‘------H Antisense RNA probe Normal HPRT RNA 336 bp AC HPRT mRNA -- 72 bp 264 bp Fig. 3. Map of plasmid pSPH1 used in RNase A mapping analysis. The pSP64 plaamid was cleaved with Hind111 + Sal1 and ligated to the 336-bp HindIII-XhoI fragment of normal HPRT cDNA sequence. The resultant recombinant plasmid, pSPH1, was linearized with EcoRI (E). Transcription with SP6 RNA poiymerase using the SP6 promoter results in the production of the antisense strand of HPRT sequence. The probe, hybridized to normal RNA or to RNA of patient A.C., is cleaved with RNase A and RNase Tt, giving either a 336-bp fragment (completely protected) or 72- and 264-bp fragments, respectively. The actual cleavage products are shown in Fig. 4. HP, HPRT sequences; SP6, SP6 promoter; vertical dashes and single line, vector sequences; blackened bar, RNA probe; stippled bar, mRNA sequences. 335 the preceding turn. There are numerous examples of the impact of single amino acid substitutions on the rate of folding and/or assembly of proteins in vivo Fig. 4. RNase Amapping of HPRT,,;,,. RNase cleavage pro-ducts of approx. 72 bp and 264 bp (see Fig. 3) are seen in RNA from patient A.C., compared to the completely protected frag-ments in the adjacent lane (normal). RNase Aand RNase Tl digestion (Gibbs and Caskey, 1986) were followed by two ethanol precipitations. The RNase-digested RNA was then electro-phoresed in a 6% denaturing polyacrylamide gel containing 8 M urea. The gel was dried and autoradiographed for 12 to 16 h. (Beasty et al., 1986). Thus, the impact of this muta-tion may also be at the levels of both secondary and tertiary structures. (c) RNase A mapping RNase A mapping of A.C. mRNA was done using pSPH1 which is the pSP64 plasmid (Promega Biotech) into which the XhoI-Hind111 fragment representing bp 149 to 485 of normal human HPRT cDNA has been cloned (Fig. 3). RNA transcripts were made using [ a-32P]GTP and hybridized to 100 pg of total RNA, as described by Gibbs and Caskey (1987). A mismatch between the cDNA clones of HPRTFIi,, and normal HPRT mRNA was con-firmed using this approach (Figs. 3 and 4). The 336-bp fragment spanning the area of the mutation was cleaved when hybridized to RNA isolated from lymphoblasts derived from patient A.C., yielding the cleavage products of 72 and 264 bp, appropriate sizes for a mutation site at bp 222 (Fig. 4). RNase cleavage was incomplete as noted by the presence of the 336-bp fragment. No cleavage products were seen when a labeled RNA probe spanning bp 658 was used (not shown), providing further evidence that the transition at 658 in pHPAC8 is a cloning artifact. (d) Conclusions Cloning of mutant forms of HPRT cDNA from deficient subjects is a rapid method for determining those structural alterations which render the enzyme inactive and/or unstable. This is the first report of a mutation in this region of the HPRT molecule being responsible for a complete absence of enzyme protein. The high degree of evolutionary con-servation of amino acid sequence in this region, as well as the profound impact of this mutation within that region, suggest an important role for this se-quence in the maintenance of normal protein stability and activity. ACKNOWLEDGEMENTS The authors gratefully acknowledge Ms. Ardith Listeman for her expert secretarial assistance. This 336 work was supported by National Institutes of Health grant DK-19045. T.D.P. is the recipient of an Arthritis Investigator Award from the Arthritis Foundation. REFERENCES Argos, P., Hanei, M., Wilson, J.M. and Kelley, W.N.: A possible nucleotide binding domain in the tertiary fold of phos-phoribosyltransferases. J. Biol. Chem. 258 (1983) 6450-6457. Aviv, A. and Leder, P.: Purification of biologically active globin messenger RNA by chromatography on oligothymidylic acid cellulose. Proc. Nat]. Acad. Sci. USA 57 (1967) 1735-1739. Beasty, A.M., Hurle, M., Manz, J.T., Stackhouse, T. and Matthews, C.R.: The effect of point mutations on the folding of globular proteins. UCLA Symp. Mol. Cell. Biol. 39 (New Series) (1986) 259-268. Biggin, M.D., Gibson, T.J. and Hong, G.F.: Buffer gradient gels and 35S-label as an aid to rapid DNA sequence determi-nation. Proc. Natl. Acad. Sci. USA 80 (1983) 3963-3965. Bonthron, D.T., Markham, A.F., Ginsburg, D. and Orkin, S.H.: Identification of a point mutation in the adenosine deaminase gene responsible for immunodeficiency. J. Clin. Invest. 76 (1985) 894-897. Chirgwin, J.M., Przybyla, A.E., MacDonald, R.J. and Rutter, W.J.: Isolation of biologically active ribonucleic acid from sources enriched in ribonuclease. Biochemistry 18 (1979) 5294-5299. Chou, P.Y. and Fasman, G.D.: Empirical predictions of protein conformation. Annu. Rev. Biochem. 47 (1978) 25 1-276. Dush, M.K., Sikela, J.M., Sohaib, A.K., Tischtield, J.A. and Stambrook, P.J.: Nucleotide sequence and organization of the mouse adenine phosphoribosyltransferase gene: presence of a coding region common to animal and bacterial phospho-ribosyltransferases that has a variable intron/exon arrange-ment. Proc. Natl. Acad. Sci. USA 82 (1985) 2731-2735. Gibbs, R.A. and Caskey, C.T.: Identification and localization of mutations at the Lesch-Nyhan locus by ribonuclease Acleavage. Science 236 (1987) 303-305. Gubler, H. and Hoffman, B.J.: A simple and very efficient method for generating cDNA libraries. Gene 25 (1983) 263-269. Hershey, H.V. and Taylor, M.W.: Nucleotide sequence and deduced amino acid sequence of Escherichia coli adenine phosphoribosyltransferase and comparison with other analogous enzymes. Gene 43 (1986) 287-293. Jolly, D.J., Okayama, H., Berg, P., Esty, AC., Filpula, D., Bohlen, P., Johnson, G.G., Shively, J.E., Hunkapillar, T. and Friedmann, T.: Isolation and characterization of a full length expressible cDNA for human hypoxanthine phosphoribosyl-transferase. Proc. Natl. Acad. Sci. USA 80 (1983) 477-481. Konecki, D.S., Brennand, J., Fuscoe, J.C., Caskey, C.T. and Chinault, A.C.: Hypoxanthine guanine phosphoribosyltrans-ferase genes ofmouse and Chinese hamster: construction and sequence analysis of cDNA recombinants. Nucl. Acids Res. 10 (1982) 6763-6775. Kelley, W.N., Rosenbloom, F.M., Henderson, J.F. and Seegmiller, J.G.: A specific enzyme defect in gout associated with overproduction of uric acid. Proc. Natl. Acad. Sci. USA 57 (1967) 1735-1739. Lesch, M. and Nyhan, W.L.: A familial disorder of uric acid metabolism and central nervous system function. Am. J. Med. 36 (1964) 561-570. Okayama, H. and Berg, P.: High efficiency cloning of full-length cDNA. Mol. Cell Biol. 2 (1982) 161-170. Paulsen, P., Jensen, K.F., Valentin-Hansen, P., Carlsson, P. and Lundberg, L.G.: Nucleotide sequence of the Escherichia coli pyrE gene and of the DNA in front of the protein coding region. Eur. J. Biochem. 135 (1983) 223-229. Pratt, D. and Subramani, S.: Nucleotide sequence of the Escherichia coli xanthine-guanine phosphoribosyltransferase gene. Nucl. Acids Res. 11 (1983) 8817-8823. Scangos, G. and Ruddle, F.H.: Mechanisms and applications of DNA-mediated transfer in mammalian cells. Gene 14 (1981) l-10. Seegmiller, J.E., Rosenbloom, F.M. and Kelley, W.N.: Enzyme defect associated with a sex-linked human neurological dis-order and excessive purine synthesis. Science 155 (1967) 1682-1684. Szybalska, E.H. and Szybalski, W.: Genetics of human cell lines, IV. DNA-mediated heritable transformation of a biochemical trait. Proc. Natl. Acad. Sci. USA 48 (1962) 2026-2034. Wilson, J.M., Baugher, B.W., Mattes, P.M., Daddona, P.E. and Kelley, W.N.: Human hypoxanthine guanine phosphoribosyl-transferase. Demonstration of structural variants in lympho-blastoid cells derived from patients with a deficiency. J. Clin. Invest. 69 (1982a) 706-715. Wilson, J.M., Tarr, G.E., Mahoney, W.C. and Kelley, W.N.: Human hypoxanthine guanine phosphoribosyltransferase. Complete amino acid sequence of the erythrocyte enzyme. J. Biol. Chem. 257 (1982b) 10978-10985. Wilson, J.M., Stout, J.T., Palella, T.D., Davidson, B.L., Kelley, W.N. and Caskey, C.T.: A molecular survey of hypoxanthine guanine phosphoribosyltransferase deficiency in man. J. Clin. Invest. 77 (1986) 188-195. Yang, T.P., Patel, P.I., China&, A.C., Stout, J.T., Jackson, L.G., Hildebrand, B.M. and Caskey, C.T.: Molecular evidence for new mutation at the HPRT locus in Lesch-Nyhan patients. Nature 310 (1984) 412-414. Young, R.A. and Davis, R.W.: Eficient isolation of genes by using antibody probes. Proc. Natl. Acad. Sci. USA 80 (1983) 1194-l 198. Communicated by J.A. Engler.
13349	https://lightcolourvision.org/dictionary/definition/tangent/	Tangent \| lightcolourvision.org Skip to navigationSkip to content lightcolourvision.org Search for: Search Menu Home Cookie Policy (EU) Profile Account Login Logout Members Password Reset Privacy Policy Register UpDraftCentral View All Diagrams User Search All Diagrams My account Your downloads Complete download Download agreement Articles Search site Site index Ask a question Explore: Light Explore: Colour Dictionary of light, colour & vision Explore: Vision Explore: Seeing more Dictionary of light, colour & vision Dictionary of light, colour & vision Dictionary of light, colour & vision Questions and answers by page Questions and answers by topic Campaign page 2020 Site Newsletter 2020 Contact us About us It’s our launch party 2024-25 Budget (basic) 2022 -23 Budget (full) Sample Page Home Cookie Policy (EU) Profile Account Login Logout Members Password Reset Privacy Policy Register UpDraftCentral View All Diagrams User Search All Diagrams My account Your downloads Complete download Download agreement Articles Search site Site index Ask a question Explore: Light Explore: Colour Dictionary of light, colour & vision Explore: Vision Explore: Seeing more Dictionary of light, colour & vision Dictionary of light, colour & vision Dictionary of light, colour & vision Questions and answers by page Questions and answers by topic Campaign page 2020 Site Newsletter 2020 Contact us About us It’s our launch party 2024-25 Budget (basic) 2022 -23 Budget (full) Sample Page $0.00 0 items Home / Tangent Tangent A tangent to a circle is a straight line that touches but does not intersect the circle and is at right angles to a radial line drawn from the centre of the circle. There are two main contexts to consider: Circles: A tangent to a circle is a straight line that touches the circle at exactly one point, like a line just brushing against a ball. There’s also a special property – the radius drawn from the centre of the circle to the point of touch is always perpendicular (at a 90-degree angle) to the tangent line. General Curves:A tangent line can also be applied to any smooth, curved shape. Here, the concept gets a bit more mathematical. We can define a tangent as a straight line that intersects the curve at exactly one point, but if we could zoom in infinitely close to that point, the curve would begin to resemble a straight line, and the tangent line would become indistinguishable from the curve itself. Related diagrams Each diagram below can be viewed on its own page with a full explanation. No posts found. References Category: Definitions Tags: P-Q-R-S-T, T Post navigation Previous post: Subtractive colour model Next post: Temperature FILES READY TO DOWNLOAD Light, Colour, Vision &How To See More v4.0 : Copyright © 2015-2026: MediaStudies Trust :All rights reserved: Your Privacy My Account SearchSearch for: Search Basket 0 HOME DIAGRAMS Explore: Light Explore: Colour Explore: Vision DICTIONARIES Dictionary: Definitions Dictionary: Summaries Dictionary: About Dirctionary: Rainbows ARTICLES Contents, forward & introduction Light & preconditions of visual perception Introduction to electromagnetism Sense-making & sense maker Vision in detail Rainbows in detail SEARCH Site index MORE My account Contact us About us Site news Donate? We are currently upgrading our GoDaddy hosting from Managed WordPress to Web Hosting Plus. Functionality may be temporarily compromised. Sorry for the inconvenience Dismiss Manage Consent To provide the best possible visitor experience, we use a minimum set of technologies such as cookies to store and/or access device information. Your consent allow us to process data such as browsing behaviour or unique IDs during your visit. Set your preferences below. Denying consent may disable some site features and functions. Functional- [x] Functional Always active The technical storage or access is strictly necessary for the legitimate purpose of enabling the use of a specific service explicitly requested by the subscriber or user, or for the sole purpose of carrying out the transmission of a communication over an electronic communications network. Preferences- [x] Preferences The technical storage or access is necessary for the legitimate purpose of storing preferences that are not requested by the subscriber or user. Statistics- [x] Statistics The technical storage or access that is used exclusively for statistical purposes.The technical storage or access that is used exclusively for anonymous statistical purposes. Without a subpoena, voluntary compliance on the part of your Internet Service Provider, or additional records from a third party, information stored or retrieved for this purpose alone cannot usually be used to identify you. Marketing- [x] Marketing The technical storage or access is required to create user profiles to send advertising, or to track the user on a website or across several websites for similar marketing purposes. Manage optionsManage servicesManage {vendor_count} vendorsRead more about these purposes Accept Deny View preferences Save preferencesView preferences {title}{title}{title} Manage consent
13350	https://blog.prepscholar.com/algebra-operations-act-math-strategies-and-formulas	Variables, exponents, and _more_ variables, whoo! ACT operations questions will involve all of these (and so much more!). So if you ever wondered what to do with or how to solve some of those extra long and clunky algebra problems (“What is the equivalent to 2 3 a 2 b−(18 b−6 c)+ …” you get the picture), then this is the guide for you. This will be your complete guide to ACT operations questions—what they’ll look like on the test, how to perform operations with multiple variables and exponents, and what kinds of methods and strategies you’ll need to get them done as fast and as accurately as possible. You'll see these types of questions at least three times on any given ACT, so let's take a look. What Are Operations? There are four basic mathematical operations—adding, subtracting, multiplying, and dividing. The end goal for any particular algebra problem may be different, depending on the question, but the operations and the methods to solve them will be the same. For example, when solving a single variable equation or a system of equations, your ultimate objective is to solve for a missing variable. However,when solving an ACT operations problem, you must use your knowledge of mathematical operations to identify an equivalent expression (NOT solve for a missing variable). This means that the answer to these types of problems will always include a variable or multiple variables, since we are not actually finding the _value_ of the variable. Let’s look at two examples, side-by-side. This is a single variable equation. Your objective is to find x. If (9 x−9)=−11, then x=? A. −92 9 B. −20 9 C. −11 9 D. −2 9 E. 70 9 This is an ACT operations problem. You must find an equivalent expression after performing a mathematical operation on a polynomial. The product (2 x 4 y)(3 x 5 y 8) is equivalent to: F. 5 x 9 y 9 G. 6 x 9 y 8 H. 6 x 9 y 9 J. 5 x 20 y 8 K.6 x 20 y 8 _(We will go through exactly how to solve this problem shortly)_ _Let's break down each component of an operations problem, step-by-step. (Also, bonus French braid lesson!)_ Operation Question How-To's Let us look at how to identify operations questions when you see them and how to solve for your answer. How to Identify an Operations Problem As we said before, the end goal of an operations problem is not to solve for a missing variable. Because of this, you can identify an operations problem by looking at your answer choices. If the question involves variables (instead of integers) in the given equation _and_ in the answer choices, then it is likely you are dealing with an operations problem. This means that if the problem asks you to identify an “equivalent” expression or the “simplified form” of an expression, then it is highly likely that you are dealing with an operations problem. How to Solve an Operations Problem In order to solve these types of questions, you have two options: you can either solve your problems by using algebra, or by using the strategy of plugging in numbers. Let’s begin by looking at how algebraic operations work. First, you must understand how to add, multiply, subtract, and divide terms with variables and exponents. (Before we go through how to do this, be sure to brush up on your understanding of exponents and integers.) So let us look at the rules of how to manipulate terms with variables and exponents. Addition and Subtraction When adding or subtracting terms with variables (and/or exponents), you can only add or subtract terms that have the _exact_ same variable. This rule includes variables with exponents—only terms with variables raised to the _same_ power may be added together (or subtracted). For example, x and x 2 CANNOT be combined into one term (i.e. 2 x 2 or x 3). It can _only_ be written as x+x 2. To add terms with variables and/or exponents, simply add the numbers before the variable (the coefficients) just as you would add any numbers without variables, and keep the variables intact. (Note: if there is no coefficient in front of the variable, it is worth 1. x is the same thing as 1 x.) Again, if one term has an additional variable or is raised to a different power, the two terms cannot be added together. Yes: x+4 x=5 x 10 x y−2 x y=8 x y No: 6 x+5 y x y−2 x−y x+x 2+x 3 These expressions all have terms with different variables (or variables to different powers) and so CANNOT be combined into one term. How they are written above is as simplified as they can ever get. Multiplication and Division When multiplying terms with variables, you may multiply any variable term with another. The variables do _not_ have to match in order for you to multiply the terms—the variables instead are combined, or taken to an additional exponent if the variables are the same, after multiplying. (For more on multiplying numbers with exponents, check out the section on exponents in our guide to advanced integers) xy=x y a bc=a b c zz=z 2 The variables in front of the terms (the coefficients) are also multiplied with one another as usual. This new coefficient will then be attached to the combined variables. 2 x3 y=6 x y 3 a bc=3 a b c Just as when we multiplying variable terms, we must take each component separately when we divide them. This means that the coefficients will be reduced/divided with regard to one another (just as with regular division), as will the variables. (Note: again, if your variables involve exponents, now might be a good time to brush up on your rules of dividing with exponents.) 8 x y 2 x=4 y 5 a 2 b 3 15 a 2 b 2=b 3 30 y+45 5=6 y+9 _When working on operations problems, first take each component separately, before you put them together._ Typical Operation Questions Though there are several ways an operations question may be presented to you on the ACT, the principles behind each problem are essentially the same—you must manipulate terms with variables by performing one (or more) of the four mathematical operations on them. Most of the operations problems you’ll see on the ACT will ask you to perform a mathematical operation (subtraction, addition, multiplication, or division) on a term or expression with variables and then ask you to identify the “equivalent” expression in the answer choices. More rarely, the question may ask you to manipulate an expression in order to present your equation “in terms of” another variable (e.g. “which of the following expressions shows the equation in terms of x?”). Now let’s look at the different kinds of operations problems in action. The product (2 x 4 y)(3 x 5 y 8) is equivalent to: F. 5 x 9 y 9 G. 6 x 9 y 8 H. 6 x 9 y 9 J. 5 x 20 y 8 K.6 x 20 y 8 Here, we have our problem from earlier, but now we know how to go about solving it using algebra. We also have a second method for solving the question (for those of you are uninterested in or unwilling to use algebra), and that is to use the strategy of plugging in numbers. We’ll look at each method in turn. Solving Method 1: Algebra operations Knowing what we know about algebraic operations, we can multiply our terms. First, we must multiply our coefficients: 23=6 This will be the coefficient in front of our new term, so we can eliminate answer choices F and J. Next, let us multiply our individual variables. x 4x 5 x[4+5] x 9 And, finally, our last variable. yy 8 y[1+8] y 9 Now, combine each piece of our term to find our final answer: 6 x 9 y 9 Our final answer is H, 6 x 9 y 9 Solving Method 2: Plugging in our own numbers Alternatively, we can find our answer by plugging in our own numbers (remember—any time the question uses variables, we can plug in our own numbers). Let us say that x=2 and y=3 (Why those numbers? Why not! Any numbers will do—except for 1 or 0, which is explained in our PIN guide—but since we are working with exponents, smaller numbers will give us more manageable results.) So let us look at our first term and convert it into an integer using the numbers we selected to replace our variables. 2 x 4 y 2(2 4)(3) 2(16)(3) 96 Now, let us do the same to our second term. 3 x 5 y 8 3(2 5)(3 8) 3(32)(6,561) 629,856 And finally, we must multiply our terms together. (2 x 4 y)(3 x 5 y 8) (96)(629,856) 60,466,176 Now, we need to find the answer in our answer choices that matches our result. We must plug in our same values for x and y as we did here and then see which answer choice gives us the same result. If you are familiar with the process of using PIN, you know that our best option is usually to start with the middle answer choice. So let us test answer choice H to start. 6 x 9 y 9 6(2 9)(3 9) 6(512)(19,683) 60,466,176 Success! We have found our correct answer on the first try! (Note: if our first option had not worked, we would have seen whether it was too low or too high and then picked our next answer choice to try, accordingly.) Our final answer is again H, 6 x 9 y 9 Now let us look at our second type of problem. For all real numbers b and c such that the product of c and 3 is b, which of the following expressions represents the sum of c and 3 in terms of b? A. b+3 B. 3 b+3 C. 3(b+3) D. b+3 3 E. b 3+3 This question requires us to translate the problem first into an equation. Then, we must manipulate that equation until we have isolated a different variable than the original. Again, we have two methods with which to solve this question: algebra or PIN. Let us look at both. Solving Method 1: Algebra First, let us begin by translating our equation into an algebraic one. We are told that the product of c and 3 is equal to b. A “product” means we must multiply c and 3 and so our equation looks like this: 3 c=b Now we are asked to find the _sum_ of c and 3. This means we must isolate c so that we can add them together. So let us first isolate c by using our knowledge of algebraic operations. 3 c=b c=b 3 Now, we can sum c and 3 by replacing our c with b 3. c+3 b 3+3 Our answer matches answer choice E. Our final answer is E. Solving Method 2: Plugging in numbers Alternatively, we can use our technique of plugging in numbers. Because our question deals with variables, we can choose our own numbers (so long as they follow the rules of our given information.) We are told that the product of c and 3 is equal to b. So let us assign a value to c and use this information to find the value of b. So let us say that c=4. (Why 4? Why not!) If c=4, then the product of c and 3 is: 3 c=b 3(4)=b b=12 So, when c equals 4, b equals 12. Now we must find the sum of c and 3. 3+c 3+(4) 7 Now that we have found our sum, we must identify the answer choice that gives us this sum. All of our answer choices are presented to us in terms of b, so we will use our found value of 12 to replace b for each. As with all PIN questions, let us start with the middle answer option. Answer choice C gives us: 3(b+3) We can tell just by looking at it that this will be far larger than 7, but we can always test this out. 3(12+3) 3(15) 45 We can eliminate answer choice C. Just by glancing, we can see that answer choices A and B will also be larger than 12, which means we can eliminate them as well. Let us try answer choice D. b+3 3 12+3 3 15 3 5 Answer choice D did not match our sum, which means we can eliminate it as well. By process of elimination, we are left with answer choice E, but let us test it to be sure. b 3+3 12 3+3 4+3 7 Success! We have found the answer choice that matches the sum we found. Our final answer is, once again, E, b 3+3. As you can see, the answer to your operations questions will always be in variables and the problem will always require you to interpret and manipulate expressions with variables, but there are always multiple options for how to solve these types of problems. _You've got the power to decide how you would like to solve and manipulate your operations problems. Magic!_ Strategies for Solving Operations Questions Now that we’ve seen the types of operations questions you’ll see on the ACT, let’s review our solving strategies. #1: Use PIN when needed (or to double-check your answer) If you ever feel concerned that you may be going down the wrong path while manipulating your operations problems, or if you simply want to double-check your answer, it's never a bad idea to use the strategy of plugging in numbers. Although it can take a little longer plug in your own numbers for your variables, you'll never have to fear misremembering how to manipulate your exponents, your variables, or your equations as a whole. Once you're able to use real numbers for your variables, the math will be a piece of cake. #2: Focus on one aspect of the term at a time It can become all too easy to lose yourself when working with multiple variables at once, especially when it comes to multiplication and division. The test-makers know this and will provide bait answers for any number of common mistakes. In order to keep all your components organized, focus on just one piece of each expression at a time. First, look at the coefficients, then look at the variables. This will help keep all your moving pieces in order and lessen the odds of mix-ups and mistakes. #3: Eliminate your answer options as you go Operations problems can sometimes mess with your head, not because they are inherently difficult, but because the ACT is a marathon and your brain can get tired and confused (and lazy). This, combined with the fact that all the answer choices generally look quite similar, with only small differences—a minus sign instead of a plus sign, one coefficient difference, etc.—can lead you to select the wrong answer, even when you know what the correct one should be. To avoid this kind of careless error (the worst kind of error!), eliminate your answer choices as you go through your problem. Know that the coefficient for your y value must be 3? Immediately cross out any answer choices that give you anything other than 3 y. It may seem inefficient to solve problems this way, but it will keep your answers much more clear. #4: Keep careful track of your negatives Not only can it be difficult to keep track of multiple variables, but it's even easier to mix-up the proper negative and positive signs. Many students make careless errors with their negative signs and the ACT test-makers are all too aware of this. They will provide all manner of bait answers for anyone who misplaces even a single negative sign, so be very careful. (a+2 b+3 c)−(4 a+6 b−5 c) is equivalent to: A. −4 a−8 b−2 c B. −4 a−4 b+8 c C. −3 a+8 b−2 c D. −3 a−4 b−2 c E. −3 a−4 b+8 c For a problem like this, we are being asked to subtract the _entire expression_, 4 a+6 b−5 c, from the entire expression, a+2 b+3 c. This means that the negative sign will be negating _every_ term in the expression 4 a+6 b−5 c. So we must put a negative sign in front of _each_ term. 4 a becomes −4 a 6 b becomes −6 b −5 c becomes −−5 c or +5 c. Now let us put these pieces together with the first expression. a−4 a=−3 a 2 b−6 b=−4 b 3 c+5 c=8 c Our final expression will be: −3 a−4 b+8 c Our final answer is E, −3 a−4 b+8 c. [Note: many (many!) students put a negative sign _only_ in front of the first term in the parenthesis, which in this case the 4 a. If you had done this, you would have gotten: a−4 a=−3 a 2 b+6 b=8 b 3 c−5 c=−2 c. This would have given you answer choice C, −3 a+8 b−2 c. Again the test-makers know this is a common error and there will always be a bait answer to tempt anyone who makes this kind of mistake.] _Operations in the "real world." Hyuk, yuk, yuk._ Test Your Knowledge Now that we’ve gone through the tips and tricks of operations questions, it’s time to put your knowledge to the test with more real ACT math problems. 1. Which of the following is an equivalent simplified expression for 2(4 x+7)−3(2 x−4)? F. x+2 G. 2 x+2 H. 2 x+26 J. 3 x+10 K. 3 x+11 2.Which of the following expressions is equivalent to 1 2 y 2(6 x+2 y+12 x−2 y)? A. 9 x y 2 B. 18 x y C. 3 x y 2+12 x D. 9 x y 2−2 y 3 E. 3 x y 2+12 x−y 3−2 y 3.t 2−59 t+54−82 t 2+60 t is equivalent to: F. −26 t 2 G. −26 t 6 H. −81 t 4+t 2+54 J. −81 t 2+t+54 K. −82 t 2+t+54 4.The expression −8 x 3(7 x 6−3 x 5) is equivalent to: A. −56 x 9+24 x 8 B. −56 x 9−24 x 8 C. −56 x 18+24 x 15 D. −56 18−24 x 15 E. −32 x 4 Answers: H, A, J, A Answer Explanations: 1. As always, we can solve this question using algebra or using PIN. Let us look at both ways. Method 1: Algebra First, we must distribute out our terms. Only afterwards will we subtract them. Let us take each half of our expression by itself. 2(4 x+7) 8 x+14 −3(2 x−4) −6 x+12 (Note: keep careful track of your negatives here, especially in the second half of our expression.) Now, we can put the two together. 8 x+14−6 x+12 2 x+26 We cannot go any further, as we have combined all our like terms. Our final answer is H,2 x+26 Method 2: PIN As an alternative to algebra, we can always use plugging in numbers. So let us assign our own value to x, which we will call 3. (Why 3? Why not!) This means that we will replace any x in our given equation with a 3. 2(4 x+7)−3(2 x−4) 2(4(3)+7)−3(2(3)−4) 2(12+7)−3(6−4) 2(19)−3(2) 38−6 32 Now, let us find the answer choice that matches with our found answer of 32, once we replace the x with 3. As usual, when using PIN, let us start with the middle answer option. 2 x+26 2(3)+26 6+26 32 Success! We found our answer on the first try. But remember—when using PIN, always check your other answer options to make sure there are not repeat correct answers. We can see straightaway that answer choices F and G will be too small, since answer choice H was a match. So let us try answer choice J. 3 x+10 3(3)+10 9+10 19 This answer choice is too small and we can see just by looking that answer choice K will be too small as well (since they only differ by 1). This means we are safe with our answer choice H, as no others produced a match. Our final answer is H, 2 x+26. As we saw from earlier in the guide and from the example problem above, we can always use algebra or PIN for our operations problems. Knowing that, we will only go through one method each for the rest of our answer explanations. 2: For this problem, let us do our solve using algebra (again, we could also use PIN, but for the sake of brevity, we are only choosing one method for each problem). We are given the equation: 1 2 y 2(6 x+2 y+12 x−2 y) Now, let us first make life easier by combining the like terms in the parenthesis. (6 x+2 y+12 x−2 y) (6 x+12 x+2 y−2 y) (18 x) The y terms cancel one another out, so we are left with only 18 x in the parenthesis. Now, we must multiply our 18 x by 1 2 y 2. As always, when multiplying, we must multiply first the coefficients and then combine them with the combined variables. So: 1 2 y 218 x (1 2)18=9 y 2x=y 2 x Put the two together and we have: 9 y 2 x So our final answer is A, 9 x y 2 3: Because we used algebra last time, let us try our hand at solving this question using PIN. Because we are using our own numbers, we don’t have to worry about whether or not we are matching up the right terms, or if we are combining them incorrectly; we can bypass all the mess and use numbers instead. We have one variable, t, so let us say that t=2. (Why 2? As always, why not!) t 2−59 t+54−82 t 2+60 t (2)2−59(2)+54−82(2)2+60(t) 4−118+54−328+120 −268 Now, we must find the answer choice that matches our found answer of 102, once we replace t with 2. Let us start in the middle, with answer choice H. −81 t 4+t 2+54 −81(2)4+(2)2+54 −81(16)+4+54 −1296+58 −1238 We can see just by looking that answer choice G will be too small as well (−2616=−416), and answer choice F will be too large (-26 4 = -104). So let us try answer choice J. −81 t 2+t+54 −81(2)2+2+54 −81(4)+56 −324+56 −268 Success! And we can also see that the only difference between answer choices J and K are the coefficient in front of t 2 (-81 vs. -82), so we know that answer K would produce an incorrect and smaller number than answer choice J. Our final answer is J, −81 t 2+t+54 4: Because we used PIN last time, let us use algebra for this problem. Because we do not have like terms in the parenthesis, we must distribute out our expression using multiplication. −8 x 3(7 x 6−3 x 5) −8 x 3(7 x 6)−−8 x 3(3 x 5) And take each piece separately. −8 x 3(7 x 6) =>−87=−56 and x 3x 6=x[3+6]=x 9 (for more on this, look to the section on exponents in our advanced integers guide). So, combined, we have: −56 x 9 And the other half of our expression will be the same. −−8 x 3(3 x 5) 8 x 3(3 x 5) =>83=24 and x 3x 5=x[3+5]=x 8 So, combined, we have: 24 x 8 Now our equation looks like this: −56 x 9+24 x 8 Our final answer is A, −56 x 9+24 x 8 (Take care! The only difference between answer choice A and B is the negative sign. If you weren’t careful with your double negatives, you may have fallen for this bait answer.) _Ten thousand gold stars for solving your operations problems!_ The Take-Aways Though operations problems are easy to get wrong if you’re going too quickly through the test (or trying to solve them in your head), the basic elements are the same as any problem with variables—combine like terms, keep your work organized, and use PIN if you feel overwhelmed (or simply want to double-check your answer). You have a multitude of options for solving ACT algebra questions, so don’t be afraid to use them. What’s Next? Still in the mood for math? Well we've got you covered! First, take a gander at exactly what's tested on the ACT math section in order to get a feel for your strong and weak points. Next, dive right into our ACT math guides for any topic you feel you haven't quite mastered (or just any topic you want to refresh). From circles to ratios, slopes to polygons, we've got your back. Running out of time on the ACT math section? Check out our guide on how to help maximize your available time in order to get your best score possible. Nervous about test day? Ease your mind by taking a look at what to do the night before and the day of the test. Trying for a perfect score? Look no further than our guide to getting a perfect 36 on the ACT math, written by a perfect-scorer.
13351	https://www.gov.uk/drug-safety-update/over-the-counter-cough-and-cold-medicines-for-children	Over-the-counter cough and cold medicines for children - GOV.UK Cookies on GOV.UK We use some essential cookies to make this website work. We’d like to set additional cookies to understand how you use GOV.UK, remember your settings and improve government services. We also use cookies set by other sites to help us deliver content from their services. You have accepted additional cookies. You can change your cookie settings at any time. You have rejected additional cookies. You can change your cookie settings at any time. Accept additional cookies Reject additional cookiesView cookies Hide cookie message Skip to main content Navigation menu MenuMenu Services and information Benefits Births, death, marriages and care Business and self-employed Childcare and parenting Citizenship and living in the UK Crime, justice and the law Disabled people Driving and transport Education and learning Employing people Environment and countryside Housing and local services Money and tax Passports, travel and living abroad Visas and immigration Working, jobs and pensions Government activity Departments Departments, agencies and public bodies News News stories, speeches, letters and notices Guidance and regulation Detailed guidance, regulations and rules Research and statistics Reports, analysis and official statistics Policy papers and consultations Consultations and strategy Transparency Data, Freedom of Information releases and corporate reports Search GOV.UK × Search GOV.UK Search Search GOV.UK Search Home Drug Safety Update Over-the-counter cough and cold medicines for children Advice on how to use cough and cold medicines safely for children under 12 years. From:Medicines and Healthcare products Regulatory AgencyPublished 11 December 2014 Therapeutic area:Paediatrics and neonatology and Respiratory disease and allergy Contents Key information Timing CHM conclusions Article date: April 2009 The Commission on Human Medicines (CHM) has advised on measures to improve the safe use of cough and cold medicines for children under 12 years. This follows a thorough review by the MHRA of the benefits and possible risks of over-the-counter (OTC) cough and cold medicines for children under 12 years. OTC cough and cold medicines containing the following active ingredients are affected by the advice: antitussives (dextromethorphan and pholcodine); expectorants (guaifenesin and ipecacuanha); nasal decongestants (ephedrine, oxymetazoline, phenylephrine, pseudoephedrine, and xylometazoline); and antihistamines (brompheniramine, chlorphenamine, diphenhydramine, doxylamine, promethazine, and triprolidine). Overall these measures include changes to age ranges, introduces new advice on labelling, introduces child-resistant packaging (to help prevent overdoses), and recommends research into how effective the medicines are in children over six years. Colds and coughs occur frequently in children but they are self-limiting and rarely harmful if left untreated. Moreover, many medicines given to children have not been properly studied in this population. Specific paediatric studies are needed because of differences between adults and children in drug handling or drug effects, which may lead to different dose requirements. OTC cold and cough remedies were introduced when the requirement to demonstrate safety and efficacy was less robust compared to today’s standards. However, over the years, the products have raised no special concern about safety. Key information Cough and cold remedies containing the above ingredients should no longer be used in children under 6 years as the balance of benefits and risk has not been shown to be favourable. Products for children from 6 to 12 years will continue to be available in pharmacies where advice can be given. Medicines to treat cough and colds in older children (6 to 12 years) can be considered supplementary to basic principles of best care. Some combinations (such as cough suppressants and expectorants) are being phased out while all liquid products containing these ingredients will be in a child resistant container in future. Timing Newly packaged products reflecting the above advice will start to be introduced to pharmacies later this year in time for the 2009/10 winter cough and cold season. In the meantime medicines with the older labelling will continue to be available and can be supplied for use by older children and adults. Immediate withdrawal of products with older labelling is not necessary because of the absence of a safety issue. Products currently authorised with General Sales List (GSL) legal status may continue to be sold on open shelves and remain available through other retail outlets, such as supermarkets, until the new packaging reflecting Pharmacy (P) legal status becomes available. We expect the change to be complete by March 2010. CHM conclusions There is no robust evidence that cold and cough medicines containing the above ingredients work. Given that there have been some reports of harm with these ingredients, the risks of cough and cold medicines containing them outweigh the benefits. For children aged over 6 years, the risk from these ingredients is reduced because: they suffer from cough and cold less frequently and consequently require medicines less often; with increased age and size, the risk of toxicity is lower; and they can say if the medicine is working. For these reasons cold and cough medicines containing the above ingredients can continue to be available for these older children, but only through pharmacies where advice can be given. More research is required on how effective these products are in children over six years Helpful advice for parents and carers on the basic principles of best care for children of all ages with coughs and colds can be found in the Department of Health’s book “Birth to 5”. Key aspects of this advice will be reflected in new Patient Information Leaflets accompanying all licensed products containing the active substances included in the review. For more information see the medicines for children section of our website See the regulation of children’s medicines drug safety update January 2009. See the safety messages for medicines section of our websitefor further information for healthcare professionals Article citation: Drug Safety Update April 2009, vol 2 issue 9: 8. Updates to this page Published 11 December 2014 Contents Related content Legal requirements for children's medicines Explore the topic Alerts and recalls Is this page useful? Maybe Yes this page is useful No this page is not useful Thank you for your feedback Report a problem with this page Help us improve GOV.UK Don’t include personal or financial information like your National Insurance number or credit card details. This field is for robots only. Please leave blank What were you doing? What went wrong? Send Cancel Help us improve GOV.UK To help us improve GOV.UK, we’d like to know more about your visit today. Please fill in this survey (opens in a new tab). Cancel Services and information Benefits Births, death, marriages and care Business and self-employed Childcare and parenting Citizenship and living in the UK Crime, justice and the law Disabled people Driving and transport Education and learning Employing people Environment and countryside Housing and local services Money and tax Passports, travel and living abroad Visas and immigration Working, jobs and pensions Government activity Departments News Guidance and regulation Research and statistics Policy papers and consultations Transparency How government works Get involved Support links Help Privacy Cookies Accessibility statement Contact Terms and conditions Rhestr o Wasanaethau Cymraeg Government Digital Service All content is available under the Open Government Licence v3.0, except where otherwise stated © Crown copyright
13352	https://stackoverflow.com/questions/51045413/complexity-of-the-sum-of-the-squares-of-geometric-progression	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams complexity of the sum of the squares of geometric progression Ask Question Asked Modified 7 years, 3 months ago Viewed 581 times 1 I have a question in my data structure course homework and I thought of 2 algorithms to solve this question, one of them is O(n^2) time and the other one is: T(n) = 3 n + 11 + 22 + 44 + 88 + 1616 + ... + lognlogn And I'm not sure which one is better. I know that the sum of geometric progression from 1 to logn is O(logn) because I can use the geometric series formula for that. But here I have the squares of the geometric progression and I have no idea how to calculate this. time-complexity Share Improve this question asked Jun 26, 2018 at 14:31 Tá´á´á´Ê Wá´ÊÊá´ÊÉ¢Tá´á´á´Ê Wá´ÊÊá´ÊÉ¢ 10211 silver badge1111 bronze badges Add a comment \| 1 Answer 1 Reset to default 2 You can rewrite it as: log n log n + ((log n) / 2) ((log n) / 2) + ((log n) / 4) ((log n) / 4) ... + 1 if you substitute (for easier understanding) log^2 n with x, you get: x + x/4 + x/16 + x/64 + ... + 1 You can use formula to sum the series, but if you dont have to be formal, then basic logic is enough. Just imagine you have 1/4 of pie and then add 1/16 pie and 1/64 etc., you can clearly see, it will never reach whole piece therefore: x + x/4 + x/16 + x/64 + ... + 1 < 2x Which means its O(x) Changing back the x for log^2 n: T(n) = O(3n + log^2 n) = O(n) Share Improve this answer edited Jun 27, 2018 at 7:52 answered Jun 26, 2018 at 15:37 libiklibik 23.1k1010 gold badges5151 silver badges9494 bronze badges Comments Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions time-complexity See similar questions with these tags. The Overflow Blog The history and future of software development (part 1) Getting Backstage in front of a shifting dev experience Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure New and improved coding challenges New comment UI experiment graduation Policy: Generative AI (e.g., ChatGPT) is banned Related 6 Iterate sum of two squares 2 What is the complexity of an arithmetic progression? Complexity of a nested geometric sequence Algorithm for finding the smallest number of precise squares which amount is n 0 Time complexity of this series 2 algorithm complexity of a sqrt series 1 Sum of squares of distinct integers 1 What is the time complexity of below geometric series? 0 What is the time complexity of this code, where one loop grows geometrically and the other decays algebraically? 3 Bounded square sum algorithm Hot Network Questions What is this chess h4 sac known as? Where is the first repetition in the cumulative hierarchy up to elementary equivalence? Should I let a player go because of their inability to handle setbacks? Who is the target audience of Netanyahu's speech at the United Nations? Do sum of natural numbers and sum of their squares represent uniquely the summands? Checking model assumptions at cluster level vs global level? What were "milk bars" in 1920s Japan? What can be said? Two calendar months on the same page Change default Firefox open file directory Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator Matthew 24:5 Many will come in my name! Do we need the author's permission for reference Is existence always locational? How long would it take for me to get all the items in Bongo Cat? ICC in Hague not prosecuting an individual brought before them in a questionable manner? "Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf Gluteus medius inactivity while riding How to home-make rubber feet stoppers for table legs? The rule of necessitation seems utterly unreasonable Copy command with cs names Are there any world leaders who are/were good at chess? Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation Analog story - nuclear bombs used to neutralize global warming more hot questions Question feed
13353	https://openstax.org/books/college-physics-2e/pages/3-4-projectile-motion	Skip to ContentGo to accessibility pageKeyboard shortcuts menu College Physics 2e 3.4 Projectile Motion College Physics 2e3.4 Projectile Motion Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory. Determine the location and velocity of a projectile at different points in its trajectory. Apply the principle of independence of motion to solve projectile motion problems. Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible. The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. Figure 3.34 illustrates the notation for displacement, where is defined to be the total displacement and and are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation to represent a vector with components and . If we continued this format, we would call displacement with components and . However, to simplify the notation, we will simply represent the component vectors as and .) Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, . Both accelerations are constant, so the kinematic equations can be used. Review of Kinematic Equations (constant ) 3.31 3.32 Figure 3.34 The total displacement of a soccer ball at a point along its path. The vector has components and along the horizontal and vertical axes. Its magnitude is , and it makes an angle with the horizontal. Given these assumptions, the following steps are then used to analyze projectile motion: Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so and are used. The magnitude of the components of displacement along these axes are and The magnitudes of the components of the velocity are and where is the magnitude of the velocity and is its direction, as shown in Figure 3.35. Initial values are denoted with a subscript 0, as usual. Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms: 3.33 3.34 3.35 3.36 3.37 3.38 3.39 3.40 Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below. Step 4. Recombine the two motions to find the total displacement and velocity . Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing and in the following form, where is the direction of the displacement and is the direction of the velocity : Total displacement and velocity 3.41 3.42 3.43 3.44 Figure 3.35 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because and is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory. Example 3.4 A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of above the horizontal, as illustrated in Figure 3.36. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Strategy Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which and . We can then define and to be zero and solve for the desired quantities. Solution for (a) By “height” we mean the altitude or vertical position above the starting point. The highest point in any trajectory, called the apex, is reached when . Since we know the initial and final velocities as well as the initial position, we use the following equation to find : 3.45 Figure 3.36 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally. Because and are both zero, the equation simplifies to 3.46 Solving for gives 3.47 Now we must find , the component of the initial velocity in the y-direction. It is given by , where is the initial velocity of 70.0 m/s, and is the initial angle. Thus, 3.48 and is 3.49 so that 3.50 Discussion for (a) Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height. Solution for (b) As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use . Because is zero, this equation reduces to simply 3.51 Note that the final vertical velocity, , at the highest point is zero. Thus, 3.52 Discussion for (b) This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using , and solving the quadratic equation for .) Solution for (c) Because air resistance is negligible, and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by , where is equal to zero: 3.53 where is the x-component of the velocity, which is given by Now, 3.54 The time for both motions is the same, and so is 3.55 Discussion for (c) The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below. In solving part (a) of the preceding example, the expression we found for is valid for any projectile motion where air resistance is negligible. Call the maximum height ; then, 3.56 This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity. Defining a Coordinate System It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the and positions. Often, it is convenient to choose the initial position of the object as the origin such that and . It is also important to define the positive and negative directions in the and directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object’s motion. When this is the case, the vertical acceleration, , takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, takes a positive value. Example 3.5 Calculating Projectile Motion: Hot Rock Projectile Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle above the horizontal, as shown in Figure 3.37. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact? Figure 3.37 The trajectory of a rock ejected from the Kilauea volcano. Strategy Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain and at the final time determined in the first part of the example. Solution for (a) While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using 3.57 If we take the initial position to be zero, then the final position is Now the initial vertical velocity is the vertical component of the initial velocity, found from = ()() = . Substituting known values yields 3.58 Rearranging terms gives a quadratic equation in : 3.59 This expression is a quadratic equation of the form , where the constants are , , and Its solutions are given by the quadratic formula: 3.60 This equation yields two solutions: and . (It is left as an exercise for the reader to verify these solutions.) The time is or . The negative value of time implies an event before the start of motion, and so we discard it. Thus, 3.61 Discussion for (a) The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air. Solution for (b) From the information now in hand, we can find the final horizontal and vertical velocities and and combine them to find the total velocity and the angle it makes with the horizontal. Of course, is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore: 3.62 The final vertical velocity is given by the following equation: 3.63 where was found in part (a) to be . Thus, 3.64 so that 3.65 To find the magnitude of the final velocity we combine its perpendicular components, using the following equation: 3.66 which gives 3.67 The direction is found from the equation: 3.68 so that 3.69 Thus, 3.70 Discussion for (b) The negative angle means that the velocity is below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.37.) One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further. Figure 3.38 Trajectories of projectiles on level ground. (a) The greater the initial speed , the greater the range for a given initial angle. (b) The effect of initial angle on the range of a projectile with a given initial speed. Note that the range is the same for and , although the maximum heights of those paths are different. How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed , the greater the range, as shown in Figure 3.38(a). The initial angle also has a dramatic effect on the range, as illustrated in Figure 3.38(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately . Interestingly, for every initial angle except , there are two angles that give the same range—the sum of those angles is . The range also depends on the value of the acceleration of gravity . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range of a projectile on level ground for which air resistance is negligible is given by where is the initial speed and is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described. When we speak of the range of a projectile on level ground, we assume that is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.39.) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic. Figure 3.39 Hypothetical projectile to satellite. From this theoretical tower, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved. PhET Explorations Projectile Motion Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target. Access multimedia content PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Paul Peter Urone, Roger Hinrichs Publisher/website: OpenStax Book title: College Physics 2e Publication date: Jul 13, 2022 Location: Houston, Texas Book URL: Section URL: © Jul 9, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
13354	https://andreaknightteacherauthor.com/fun-activities-for-teaching-prepositions/	Andrea Knight Teacher · Learner · Author Fun Activities for Teaching Prepositions Teaching prepositions and positional words is fun when you use games and other engaging activities. Little ones will eventually put pencil to paper, but let’s begin with some out-of-your-seat fun first! From Simon Says to a fun twist on musical chairs, these ideas will really get the ball rolling. First though, let’s grab a quick library of books about prepositions. These can be used at any time throughout your study to help children identify positional words and understand how they’re used in longer contexts. Where’s Spotby Eric Hill – In this interactive flap book, the children follow along with Spot’s mom as she searches the house for her mischievous puppy. It’s a sweet little book for recognizing various prepositions like under, behind, and inside. Go, Dog. Go! by P.D. Eastman – This book takes me back to my own childhood, so we’ll just go ahead and call it a classic. It’s fun and perfectly simple, offering plenty of opportunities to identify beginning prepositions like in, out, and on. We’re Going on a Bear Hunt by Michael Rosen and Helen Oxenbury – Speaking of classics, this book is an adventure for readers! Read it once just for fun. Then, on the second reading, join the family as you journey across, through, and over various landscapes. PREPOSITIONS ACTIVITIES WITH GET UP AND GO-NESS Young children don’t sit still for long, nor should they. I, however, am pretty good at it. But even my watch reminds me to stand up every hour, so obviously get up and go-ness is important for everyone. Before you pass out the prepositions worksheets, find a spot for these activities in your lesson plans. The Simons Say– This simple twist on Simon Says gives every child in the class a chance to be Simon. Standing in a large circle, hold one of the Simon Cards👆 and give an instruction using a preposition such as “Simon says wave your hands above your head.” After the children have followed the prompt, pass the Simon card to the right so the next person can be Simon. Coach the children to give an instruction that includes a preposition such as, “Simon says put one foot behind the other.” Keep passing the Simon card to the right until every child who wants a turn has had one. Obstacle Course – Set up an obstacle course in the classroom or out on the playground that includes activity stations labeled with prepositions. For example, “Crawl through the tunnel.” “Jump over the blocks.” “Skip around the tree.” “Walk under the slide.” Musical Chairs with a Twist– Instead of just removing chairs, call out prepositions when the music stops. This might sound like, “When the music stops, sit behind a chair.” Change it up each time. “This time, when the music stops, walk around a chair.” Body Position Game– Call out different body positions using prepositions. Some examples include, “Stand on one foot.” “Kneel between two friends.” “Sit under the flag.” “Lie down beside the bookcase.” Depending on your prompts, this can either be done all together or with one student volunteer at a time. Outdoor Nature Walk – Take the children on a nature walk and encourage them to notice things in their surroundings by using prepositional prompts such as, “Walk out the door.” “Find a flower in the garden.” “Run on the court.” “Skip through the grass.” “Sit under the tree.” Note about The Simons Say Activity: I included four cards in the PDF so children can play in small groups. This version gives kids more opportunities to practice without having to wait so long for their turn. ENGAGING ACTIVITIES FOR POSITIONAL WORDS We like these GAMES and activities for learning preposition words. They aren’t necessarily out-of-your-seat activities, but they’re engaging in a way that keeps the children’s attention while reinforcing a wide variety of prepositions. Give these ideas a go! Building Blocks Challenge– Pull out your building blocks or LEGO pieces and challenge the children to follow the details of your positional directions. For example, “Put the red block on top of the blue block.” “Now, put the yellow block underneath the blue block.” “Put a green block next to the yellow block.” This is also a great exercise for developing active listening skills. Preppy Pix– Preppy Pix is our shortened name for preposition pictures. It’s sort of like a scavenger hunt meets the game of Pictionary. To play, give each child a plain piece of paper and have them fold it into fourths. Together, write a preposition at the top of each section. Depending on the age of your students, you may need to write these together. Then give them a direction for each word. For example, if you wrote beside in the first box, you might say something like, “Go find and draw something that is beside a window.” The kids can circulate around the room to find an object that fits your prompt and draw it. Then, for example, if you wrote above in the next box, you might say, “Go find and draw something above the door.” Keep going until the children have drawn something in each box. Don’t forget to save time for sharing. We Have – Who Has?– If your students have done an I Have – Who Has activity, then they’ll have no problem doing this MORE SUPPORTIVE VERSION.👇 We like to do this activity in pairs when there aren’t enough cards for each child, or when we need to pair a developing reader with a more proficient reader. That way, they can play together and coach each other as needed. SPEAKING OF READING We’re big fans of getting books into kids’ hands that they can actually read… even better if there’s an interactive component to each. These texts are perfect for K-1 children. Both are adorable, but WHERE IS MONKEY? is our favorite! On each page, students decide between two prepositions or positional words to tell where Monkey’s body is in relation to the box. His facial expressions are hilarious and the children love reading their book when it’s all finished. The other student text, Preposition Pairs, combines new content with what we already know about opposites. On each page, a pair of opposite prepositions match the illustrations, such as on & off, in & out, and above & below. It’s simple enough for the children to read and has been really helpful in working on new vocabulary with our ELLs as well. PRACTICE PREPOSITIONS AND ASSESS When your students are ready to do a little more writing, these WORKSHEETS can offer an opportunity for more practice or informal assessment. We’ve even used some of them during small group instruction so we can model more explicitly for those children who need a little extra support. Hopefully, you’ve found something new for your lesson plans! If you have a book, or song, or activity you always use when teaching prepositions, share it with us in the comments section. We’re always on the lookout for fresh ideas to try with our students! Happy teaching! RELATED BLOG POSTS In the Loop
13355	https://www.quora.com/When-the-discriminant-is-zero-then-are-the-roots-two-non-distinct-real-numbers	When the discriminant is zero, then are the roots two non-distinct real numbers? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Discriminant Quadratic Formula Roots (mathematics) Real Numbers Theory of Equations Solving Quadratic Equatio... Mathematical Sciences Algebra 5 When the discriminant is zero, then are the roots two non-distinct real numbers? All related (35) Sort Recommended Shivanshu Mishra Lives in Bihar, India ·3y When discriminant is zero of a quadratic equation in which Coffecient belongs to real number and coefficient of x squrare is not equal to zero then both roots are Real and equal to -b/2a and in this case where discriminant is equal to zero the graph of quadratic equation touches x- axis at one point. Continue Reading When discriminant is zero of a quadratic equation in which Coffecient belongs to real number and coefficient of x squrare is not equal to zero then both roots are Real and equal to -b/2a and in this case where discriminant is equal to zero the graph of quadratic equation touches x- axis at one point. Upvote · 9 4 Related questions More answers below When discriminant is zero, then what are the roots? When the discriminant is zero, then what is the nature of the roots? What is Jungkook's real phone number? How many real roots does an equation have when its discriminant is negative and equal to zero? Can a quadratic equation have two real roots if the discriminant is equal to zero? If so, what is the method for finding these roots? Alberto Cid M.S.E. in Telecommunications Engineering&Data Transmission, Technical University of Madrid (Graduated 2008) · Author has 2K answers and 3.8M answer views ·3y Well, “two non-distinct” is sometimes said as “only one root with multiplicity 2”. This is the case: K·(x-r)·(x-r) = K·(x-r)^2 Where r is the root, and it has multiplicity 2, the exponent. Expand it : K·(x-r)^2 = K·(x^2 -2·r·x + r^2) So: a = K b = -2·r·K c = K·r^2 Discriminant: b^2 - 4·a·c = 4r^2 K^2 - 4·K·K·r^2 = 0 In case it is not zero, then the “plus minus” in the formula gives you 2 different roots and the difference (substraction) is twice the square root of the discriminant, in case it’s positive… In case it’s negative the different roots would be complex, not real and it would not have rea Continue Reading Well, “two non-distinct” is sometimes said as “only one root with multiplicity 2”. This is the case: K·(x-r)·(x-r) = K·(x-r)^2 Where r is the root, and it has multiplicity 2, the exponent. Expand it : K·(x-r)^2 = K·(x^2 -2·r·x + r^2) So: a = K b = -2·r·K c = K·r^2 Discriminant: b^2 - 4·a·c = 4r^2 K^2 - 4·K·K·r^2 = 0 In case it is not zero, then the “plus minus” in the formula gives you 2 different roots and the difference (substraction) is twice the square root of the discriminant, in case it’s positive… In case it’s negative the different roots would be complex, not real and it would not have real roots. Upvote · Assistant Bot · Sep 6 Yes — for a quadratic ax^2 + bx + c, the discriminant Δ = b^2 − 4ac determines the nature of the roots: Δ > 0: two distinct real roots. Δ = 0: one real root of multiplicity two (a repeated or double root). Both roots are equal to −b/(2a); they are real and not distinct as separate values but count as two roots algebraically. Δ < 0: two complex conjugate nonreal roots. Remarks: Geometric view: when Δ = 0 the parabola touches the x-axis at a single point (tangent). Algebraic consequence: the quadratic factors as a(a x + b/ a?) more precisely a(x + b/(2a))^2 when Δ = 0. Upvote · Anonymous 3y If d=0, then there is one real root. If d<0, then there are two distinct imaginary solutions. If d>0, then there are two distinct real solutions. Upvote · Related questions More answers below Why does the discriminant tell us the number of real roots? How do I find the discriminant of x^2+8x+k=0 (the values of k for which the equation has 2 distinct real roots)? What is the relationship between the discriminant of an equation and its real roots? Is the discriminant always negative or zero if the equation has no real roots? What is the method for finding the value of discriminant when there are real roots? Let b be a non-zero real number. Suppose the quadratic 2x^2+bx+half b =0 has two distinct real roots, then what is b? Calvin L. 18, Mathematics & Statistics major · Author has 10K answers and 2.2M answer views ·2y No, when discriminant is zero, it's just one value since it would be a repeating root. Upvote · 9 1 Dean Rubine Author of: It's not just π; all of trig is wrong! · Upvoted by Terry Moore , M.Sc. Mathematics, University of Southampton (1968) · Author has 10.6K answers and 23.7M answer views ·1y Related I’m aware that if the discriminant of a quadratic is 0 then the roots should be real and equal, but I find that imaginary and equal roots also give 0 as the discriminant. Is this incorrect? If a polynomial with real coefficients has a zero discriminant, we know the roots are equal and real. A polynomial with equal and imaginary roots will also have a zero discriminant, but will have at least one complex (not real) coefficient. For example: 0=(x−i)2=x 2−2 i x−1 0=(x−i)2=x 2−2 i x−1 has an imaginary linear coefficient and a zero discriminant: Δ=(−2 i)2−4(−1)=−4+4=0 Δ=(−2 i)2−4(−1)=−4+4=0 Upvote · 9 5 Dean Rubine Author of: It's not just π; all of trig is wrong! · Upvoted by Terry Moore , M.Sc. Mathematics, University of Southampton (1968) and Michael Jørgensen , PhD in mathematics · Author has 10.6K answers and 23.7M answer views ·1y Related Is it true that all real numbers are roots of a polynomial equation? Can you provide a counter example to disprove this statement? Bot question. Trivially every real number r r is the root of x−r=0.x−r=0. So the answer is: TRUE. It’s not the case that every real number is the zero of a polynomial with integer coefficients. Those real numbers are called algebraic numbers. π π is real and transcendental, not the zero of a polynomial with integer coefficients. That’s the usual dogma. As I grow old and grumpy, I’m not really buying it anymore. If we have a number that we can’t really write as a number, but only write down as the root of an equation we can’t really solve, is that a number? Or is real the greatest marketing breakthrough ev Continue Reading Bot question. Trivially every real number r r is the root of x−r=0.x−r=0. So the answer is: TRUE. It’s not the case that every real number is the zero of a polynomial with integer coefficients. Those real numbers are called algebraic numbers. π π is real and transcendental, not the zero of a polynomial with integer coefficients. That’s the usual dogma. As I grow old and grumpy, I’m not really buying it anymore. If we have a number that we can’t really write as a number, but only write down as the root of an equation we can’t really solve, is that a number? Or is real the greatest marketing breakthrough ever, an ironic name for numbers, some of which can never be written down? Upvote · 9 7 9 7 Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·2y Related Why does the discriminant have to be positive in order for there to be two real solutions for a quadratic function? Clearly, you need to know all the cases possible in the discriminant to make any real sense of your question. This is the basic Quadratic Formula… To get a clear understanding of the use of the discriminant we need to see the relation between the discriminant and the types of solutions. It all depends on how the discriminant determines the position of the graph. The following examples will show this well. Continue Reading Clearly, you need to know all the cases possible in the discriminant to make any real sense of your question. This is the basic Quadratic Formula… To get a clear understanding of the use of the discriminant we need to see the relation between the discriminant and the types of solutions. It all depends on how the discriminant determines the position of the graph. The following examples will show this well. Upvote · 9 7 Max Gretinski Studied Mathematics · Author has 6.6K answers and 2.5M answer views ·1y Related Can a real number have five fifth roots, or does it only have one? Why? The function f(x) = x 5 x 5 is strictly increasing. Therefore, it is one-to-one, ranging over all of the real numbers. This means that every real number has exactly one real fifth root. However, over the complex numbers, every nonzero number has exactly two square roots, three cube roots, … , five fifth-roots, and so on. Every real number has one real fifth root. The other four complex fifth roots are pairs of complex conjugates. This can be demonstrated relatively easily using a graph in the complex plane, but since complex graphs are not part of early algebra courses, I have not used such a graph. Continue Reading The function f(x) = x 5 x 5 is strictly increasing. Therefore, it is one-to-one, ranging over all of the real numbers. This means that every real number has exactly one real fifth root. However, over the complex numbers, every nonzero number has exactly two square roots, three cube roots, … , five fifth-roots, and so on. Every real number has one real fifth root. The other four complex fifth roots are pairs of complex conjugates. This can be demonstrated relatively easily using a graph in the complex plane, but since complex graphs are not part of early algebra courses, I have not used such a graph. Similarly, I have not used the exponential form of a complex number. Upvote · 9 2 9 3 Paul Holloway Author has 2.1K answers and 3.3M answer views ·2y Related Can the number of real roots be equal to zero? If yes, then why and when would that happen? If no, then why not and when wouldn't it happen? A polynomial is an algebraic expression in the form of a 1 x n+a 2 x n−1+a 3 x n−2…+a n x+c a 1 x n+a 2 x n−1+a 3 x n−2…+a n x+c. The order of a polynomial is the value of n. Thus a x 2+b x+c a x 2+b x+c is a second order polynomial because the highest power x is raised to is 2. In the same way, a x 3+b x 2+c x+d a x 3+b x 2+c x+d is a third order polynomial because the highest power x is raised to is 3. The root of a polynomial is the point where a graph of the polynomial crosses the x axis. For example where a x 3+b x 2+c x+d=0 a x 3+b x 2+c x+d=0. An odd ordered polynomial is where the order of the polynomial is an odd number, for example, 1,3,5,7 etc. An even orde Continue Reading A polynomial is an algebraic expression in the form of a 1 x n+a 2 x n−1+a 3 x n−2…+a n x+c a 1 x n+a 2 x n−1+a 3 x n−2…+a n x+c. The order of a polynomial is the value of n. Thus a x 2+b x+c a x 2+b x+c is a second order polynomial because the highest power x is raised to is 2. In the same way, a x 3+b x 2+c x+d a x 3+b x 2+c x+d is a third order polynomial because the highest power x is raised to is 3. The root of a polynomial is the point where a graph of the polynomial crosses the x axis. For example where a x 3+b x 2+c x+d=0 a x 3+b x 2+c x+d=0. An odd ordered polynomial is where the order of the polynomial is an odd number, for example, 1,3,5,7 etc. An even ordered polynomial is where the order of the polynomial is an even number, for example, 2,4,6,8 etc. Now all odd ordered polynomials (third order, fifth order, seventh order etc…) must have at the very least one real root because a negative number raised to an odd number will always be negative and the highest ordered term will, eventually, always overpower all the other terms and thus dictate the overall shape of the graph. An x value that’s a long way south of zero raised to the highest ordered term in an odd ordered polynomial must at some point make the graph negative. In the same way a sufficiently large positive value for x will cause the highest ordered term to dictate that the graph is positive. This means that at some point the graph must at least once cross the x axis. If the highest order is an even number then a negative number raised to an even power will always be positive. This is because a negative times a negative is a positive. This means the sign of the coefficient of the highest ordered term will dictate the shape of the graph. If the co-efficient of the highest ordered term is negative the graph will go from negative infinity to negative infinity rising to a maximum point that may or may not cross the x axis. If the co-efficient of the highest ordered term is positive then the graph will go from positive infinity to positive infinity sinking to a minimum point and may or may not cross the x axis along the way. Thus, even ordered polynomials can have zero real roots, one real root, or multiple real roots. images courtesy of Desmos. Some functions are periodic in nature, meaning they can cross the x axis an infinite number of times and hence have an infinite number of roots. Image courtesy of Desmos. Upvote · Peter Shea B. Sc in Mathematics&Computer Science, Monash University (Graduated 1972) · Author has 5.2K answers and 1.2M answer views ·2y Related What is the definition of a cubic equation? Does every cubic equation have three real roots if its discriminant is negative or zero but not for positive discriminants? All cubic equations may be written in the form ax³ + bx² + cx + d = 0, a ≠ 0. Equivalently, they may be written in the form x³ + px² + qx + t = 0. The coefficients may all be complex but you are most likely to be presented with examples where they are all real. Naturally, just as quadratics are more complex that linear equations, so are cubics more complex than quadratics. There are 3 discriminants: Δ₀ = b²-3 ac, Δ₁ = 2b³−9abc+27a²d and Δ = (Δ₁²−4Δ₀³) / (-27a²). If Δ > 0, the equation has three real solutions. If Δ = 0, there are one or two real solutions, which may be duplicated. If Δ < 0, there Continue Reading All cubic equations may be written in the form ax³ + bx² + cx + d = 0, a ≠ 0. Equivalently, they may be written in the form x³ + px² + qx + t = 0. The coefficients may all be complex but you are most likely to be presented with examples where they are all real. Naturally, just as quadratics are more complex that linear equations, so are cubics more complex than quadratics. There are 3 discriminants: Δ₀ = b²-3 ac, Δ₁ = 2b³−9abc+27a²d and Δ = (Δ₁²−4Δ₀³) / (-27a²). If Δ > 0, the equation has three real solutions. If Δ = 0, there are one or two real solutions, which may be duplicated. If Δ < 0, there is only one real solution. Note: if the coefficients are all real, there is always at least 1 real solution,because the graph will always cross the x-axis at least once. BTW, this is not my own work, just my summary, for the original, see WikiHo.com is for sale \| HugeDomains Join thousands of people who own a premium domain. Affordable financing available. . Upvote · Adrian Giles Studied Physics&Mathematics (Graduated 1985) · Author has 3.5K answers and 807.8K answer views ·2y Related What is the proof that every positive real number has two distinct non-real square roots and vice versa? Every positive real number has two distinct REAL square roots. I don't know what you mean by vice versa in this question?? Let P be a positive real number. We are looking for the roots of the polynomial: x^2-P=0. The FTA assured us that there are two roots possibly repeated possibly complex but not in this case because the equation can be written: (x-√P)(x+√P)=0 so x=+/—√P. Really, this is more sleight of hand than anything as I could have just said that if we’re satisfied that a positive root of P exists call it √P then —√P squared is also P. It's when P Is a negative number N that the square roots Continue Reading Every positive real number has two distinct REAL square roots. I don't know what you mean by vice versa in this question?? Let P be a positive real number. We are looking for the roots of the polynomial: x^2-P=0. The FTA assured us that there are two roots possibly repeated possibly complex but not in this case because the equation can be written: (x-√P)(x+√P)=0 so x=+/—√P. Really, this is more sleight of hand than anything as I could have just said that if we’re satisfied that a positive root of P exists call it √P then —√P squared is also P. It's when P Is a negative number N that the square roots become complex. Upvote · 9 1 Bernard Leak Firmware Developer (2008–present) · Author has 5.8K answers and 5M answer views ·5y Related If the discriminant is a nonexistent number like I, will the sum of the roots be (-b/a)? What do you mean by “non-existent”? What can it possibly mean to talk about the discriminant being a non-existent thing, and then giving an example of a non-existent thing it might be? I can say there's no such thing, like a unicorn, but I can't then say “like the one in my back garden” without talking complete nonsense. There isn't something which a non-existent thing is. Don't try to think of “real” numbers as “really existing” while “imaginary” numbers “don't really exist”. The terminology is utterly unhelpful for this purpose — as unhelpful as trying to apply a Marxist analysis of the strug Continue Reading What do you mean by “non-existent”? What can it possibly mean to talk about the discriminant being a non-existent thing, and then giving an example of a non-existent thing it might be? I can say there's no such thing, like a unicorn, but I can't then say “like the one in my back garden” without talking complete nonsense. There isn't something which a non-existent thing is. Don't try to think of “real” numbers as “really existing” while “imaginary” numbers “don't really exist”. The terminology is utterly unhelpful for this purpose — as unhelpful as trying to apply a Marxist analysis of the struggle between classes in set-theory, or hoping to find the keys to a better existence in the ideals of ring-theory, or trying to grow spuds in skew fields. For what it's worth, yes, if you use complex numbers as the coefficients of a quadratic equation, the sum of the roots is −b a−b a. Indeed, this is true for all non-constant polynomials, though one normally only speaks of the discriminant of quadratic equations. Upvote · Related questions When discriminant is zero, then what are the roots? When the discriminant is zero, then what is the nature of the roots? What is Jungkook's real phone number? How many real roots does an equation have when its discriminant is negative and equal to zero? Can a quadratic equation have two real roots if the discriminant is equal to zero? If so, what is the method for finding these roots? Why does the discriminant tell us the number of real roots? How do I find the discriminant of x^2+8x+k=0 (the values of k for which the equation has 2 distinct real roots)? What is the relationship between the discriminant of an equation and its real roots? Is the discriminant always negative or zero if the equation has no real roots? What is the method for finding the value of discriminant when there are real roots? Let b be a non-zero real number. Suppose the quadratic 2x^2+bx+half b =0 has two distinct real roots, then what is b? Why does every positive non-zero number have two distinct real square roots? What is the definition of a square root? Does every non-zero real number have two different square roots, one positive and one negative? What does it mean when the discriminant is zero? Can every non-zero real number have two square roots, one positive and one negative? What is the relationship between the squares of two non-zero real numbers? Related questions When discriminant is zero, then what are the roots? When the discriminant is zero, then what is the nature of the roots? What is Jungkook's real phone number? How many real roots does an equation have when its discriminant is negative and equal to zero? Can a quadratic equation have two real roots if the discriminant is equal to zero? If so, what is the method for finding these roots? Why does the discriminant tell us the number of real roots? How do I find the discriminant of x^2+8x+k=0 (the values of k for which the equation has 2 distinct real roots)? What is the relationship between the discriminant of an equation and its real roots? Is the discriminant always negative or zero if the equation has no real roots? What is the method for finding the value of discriminant when there are real roots? Let b be a non-zero real number. Suppose the quadratic 2x^2+bx+half b =0 has two distinct real roots, then what is b? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
13356	https://sites.pitt.edu/~trenchea/MATH1070/Solutions_MATH1070_2022.pdf	Solutions for exercises 1. Section 1.1 THE TAYLOR POLYNOMIAL. pn(x;a) = n ∑ i=0 (x−a)n n! f (n)(a) (Taylor approximation polynomial) ≡f(a)+ 1 1!(x−a)f ′(a)+ 1 2!(x−a)2 f ′′(a)+ 1 3!(x−a)3 f ′′′(a)+···+ 1 n!(x−a)n f (n)(a), provided f, f ′,··· , f (n) are well-deﬁned at x = a. Exercise 2. Produce the linear and quadratic Taylor polynomials for the following cases. Graph the function and these Taylor polynomials. (a) f(x) = √x, a = 1 (b) f(x) = sin(x),a = π 4 (c) f(x) = exp(cos(x)), a = 0 (d) f(x) = log(1+ex), a = 0 Solution: (a) f(x) = x1/2, f ′(x) = 1 2x−1/2, f ′′(x) = −1 4x−3/2, hence f(1) = 1, f ′(1) = 1 2, f ′′(1) = −1 4 and p2(x;1) = 1+ 1 2(x−1)−1 8(x−1)2. (b) f(x) = sin(x), f ′(x) = cos(x), f ′′(x) = −sin(x), hence f( π 4 ) = √ 2 2 , f ′( π 4 ) = √ 2 2 , f ′′( π 4 ) = − √ 2 2 , and p2 x; π 4 = √ 2 2 + √ 2 2 x−π 4 − √ 2 4 x−π 4 2 . (c) f(x) = ecos(x), f ′(x) = −sin(x)ecos(x), f ′′(x) = −cos(x)+sin2(x) ecos(x), then f(0) = e, f ′(x) = 0, f ′′(x) = −e, therefore p2(x;0) = e−ex2 2 . (d) f(x) = log(1+ex), f ′(x) = ex 1+ex , f ′′(x) = ex (1+ex)2 , hence f(0) = log(2), f ′(0) = 1 2, f ′′(0) = 1 4, so p2(x;0) = log(2)+ 1 2x+ 1 8x2. Exercise 3. Produce a general formula for the degree n polynomials for the following functions, all using a = 0 as the point of approximation. Graph the function and these Taylor polynomials. (a) f(x) = 1 1−x (b) f(x) = sin(x) (c) f(x) = √1+x (d) f(x) = cos(x) (e) f(x) = (1+x)1/3 Solution: 1 (a) f(x) = 1 1−x = (1 −x)−1, f ′(x) = (1 −x)−2, f ′′(x) = 2(1 −x)−3, f ′′′(x) = 3!(1 −x)−4, ···, f (n)(x) = n!(1 − x)−(n+1) = n! 1 (1−x)n+1 , hence f (n)(0) = n!, so 1 1−x ≈pn(x) = 1+x+x2 +···+xn about x = 0. (b) f(x) = sin(x), f ′(x) = cos(x), f ′′(x) = −sin(x), f ′′′(x) = −cos(x), f iv(x) = sin(x), ..., f (4m)(x) = (−1)2m sin(x), f (4m+1)(x) = (−1)2m+1 cos(x), f (4m+2)(x) = (−1)2m+2 sin(x), f (4m+3)(x) = (−1)2m+3 cos(x), .... Therefore all even-order derivatives vanish when evaluated at a = 0, and the Taylor polynomial will contain only odd terms: p2n(x;0) = x−x3 3! + x5 5! −···+(−1)n−1 x2n−1 (2n−1)! ≈sin(x) ∀n ≥1. (1.1) (c) f(x) = √1+x (d) f(x) = cos(x) p2n(x;0) = 1−x2 2! + x4 4! −···+(−1)n x2n (2n)! ≈cos(x) ∀n ≥0. (1.2) (e) f(x) = (1+x)1/3 Exercise 4. Does f(x) = 3 √x have a Taylor polynomials approximation of degree 1, based on expanding about x = 0? x = 1? Explain and justify your answers. Solution: Since f(x) = 3 √x ≡x1/3, f ′(x) = 1 3x−2 3 , then f ′(0) is not well deﬁned, while f ′(1) = 1 3. Therefore p1(x;0) := f(0)+x f ′(0) does not exist, while p1(x;1) := f(1)+(x−1) f ′(1) ≡1+ 1 3(x−1). x=linspace(0.5,1.5,100); >> hold all >> plot(x,x.^(1/3),’b-’) >> plot(x,1+1/3(x-1),’r.-.’) Exercise 8.Let f(x) = ex; recall the formulae (1.3) for pn(x;0) and pn(x;a) = ea n ∑ j=0 (x−a)j j! . Compare pn(x;0) and pn(x,1) to ex on [−1,2] for n = 1,2,3. Solution: For p1 comparison: >> x=linspace(-1,2,100); >> hold all >> plot(x,exp(x),’b-’,x,1+x,’r.-.’); >> plot(x,exp(1).(1+(x-1)),’m.-.’); >> legend(’e^x’,’p_2(x,0)’,’p_2(x,1)’) For p2 comparison: >> clear all >> clc >> x=linspace(-1,2,100); >> hold all >> plot(x,exp(x),’b-’,x,1+x+1/2x.^2,’r.-.’); >> plot(x,exp(1).(1+(x-1)+1/2(x-1).^2),’m.-.’) >> legend(’e^x’,’p_2(x,0)’,’p_2(x,1)’) 2 For p3 comparison: >> clear all >> clc >> x=linspace(-1,2,100); >> hold all >> plot(x,exp(x),’b-’,x,1+x+1/2x.^2+1/6x.^3,’r.-.’); >> plot(x,exp(1).(1+(x-1)+1/2(x-1).^2+1/6(x-1).^3),’m.-.’) >> legend(’e^x’,’p_3(x,0)’,’p_3(x,1)’) Exercise 10. (a) Produce the Taylor polynomials of degree 1,2,3,4 for f(x) = ex2 with a = 0 the point of approximation. (b) Using the Taylor polynomials for et, substitute t = x2 to obtain polynomial approximations for ex2. Compare with the results in (a). Solution: (a) We have that f(x) = ex2, f ′(x) = 2xex2, f ′′(x) = (2 + 4x2)ex2, f ′′′(x) = (8x + 4x + 8x3)ex2 = (12x + 8x3)ex2, f (4)(x) = (12+24x2 +24x2 +16x4)ex2 = (12+48x2 +16x4)ex2 hence f(0) = 1, f ′(0) = 0, f ′′(0) = 2, f ′′′(0) = 0, f (4)(0) = 12, and p4(x) = 1+ 2 2!x2 + 12 4! x4 ≡1+x2 + 1 2x4. (b) As x ≈0, then also t = x2 ≈0, and the function f(t) = et has the Taylor approximation about t ≈0 being et ≈Tn(t) = 1+t + 1 2t2 +···+ 1 n!tn, ex2 ≈p2n(x) = 1+x2 + 1 2(x2)2 +···+ 1 n!(x2)n. The second degree Taylor polynomial approximation gives et ≈T2(t) = 1+t + 1 2t2, ex2 ≈T2(x2) = 1+x2 + 1 2(x2)2 ≡1+x2 + 1 2x4, Exercise 11. The quotient g(x) = ex −1 x is undeﬁned for x = 0. Approximate ex by using Taylor polynomials of degrees 1, 2, and 3, in turn to determine a natural deﬁnition of g(0). Solution: The Taylor polynomial of degree n for ex for x ≈0 is pn(x;0) = 1+x+ 1 2x2 +···+ 1 n!xn ≈ex, (1.3) hence g(x) ≈pn(x;0)−1 x = 1+x+ 1 2x2 +···+ 1 n!xn − 1 x = 1+ 1 2x+···+ 1 n!xn−1 := qn−1(x). 3 Using continuity we can deﬁne g(0) = 1. Since q(j) n−1(0) = 1 j+1,∀j = 0 : n−1, we can also deﬁne values for the derivatives of g at 0: g(j)(0) = 1 j +1, ∀j = 0 : n−1. Exercise 12. The quotient g(x) = log(1+x) x is undeﬁned for x = 0. Approximate log(1+x) by using Taylor polynomials of degrees 1, 2, and 3, in turn to determine a natural deﬁnition of g(0). Solution: The Taylor polynomial of degree n for log(1+x) for x ≈0 is pn(x;0) = x−1 2x2 + 1 3x3 +···+ (−1)n−1 n xn ≈log(1+x), (1.4) hence g(x) ≈pn(x;0) x = x−1 2x2 + 1 3x3 +···+ (−1)n−1 n xn x = 1−1 2x+ 1 3x2 +···+ (−1)n−1 n xn−1 := qn−1(x). Using continuity we can deﬁne g(0) = 1. Since q(j) n−1(0) = (−1)j j+1 j!,∀j = 0 : n−1, we can also deﬁne values for the derivatives of g at 0: g( j)(0) = (−1)j j +1 j!, ∀j = 0 : n−1. 4 2. Section 1.2 THE ERROR IN TAYLOR’S POLYNOMIAL. f(x) = pn(x;a)+Rn(x,a) where the remainder (in differential form) is Rn+1(x;a) := (x−a)n+1 (n+1)! f (n+1)(cx)/ ∈Pn+1 provided f, f ′,··· , f (n), f (n+1) are well-deﬁned in a vicinity of x = a, and cx is a point between a and x. Exercise 2. Find the degree 2 Taylor polynomial for f(x) = ex sin(x), about the point a = 0. Bound the error in this approximation when x ∈[−π 4 , π 4 ]. Solution: Since f ′(x) = exsin(x)+cos(x) , f ′′(x) = exsinx+2cosx−sinx = 2ex cos(x), f ′′′(x) = 2excos(x)− sinx , we have f(x) = p2(x;0)+R3(x;0), p2(x;0) = x+ x2 2!2 = x+x2, R3(x;0) = 1 3!x32ecxcos(cx)−sin(cx) where cx is between 0 and x. Then for x ∈[−π 4 , π 4 ] we have \|R3(x;0)\| ≤1 3! π3 43 2 ecxcos(cx)−sin(cx) \| {z } ≤1 ≤π3 192 ≈0.1615. clear all >> clc >> [x,c] = meshgrid(-pi/4:0.01:pi/4); >> r3 = 1/3x.^3.(exp(c).(cos(c)-sin(c))); >> r3bad = 1/3x.^3.(exp(x).(cos(x)-sin(x))); >> f = exp(x).sin(x); >> T2=x+x.^2; >> error = f-T2; >> figure(1) >> hold all; grid on; >> surf(x,c,r3,’FaceAlpha’,0.5) >> xlabel(’x’); ylabel(’c’);zlabel(’remainder’); >> figure(2) >> hold all; grid on >> plot(x,error,’bd’,x,r3bad, ’r’) >> legend(’true error in T2’,’Taylor approximation of the error in T2 w/ c_x=x’) Exercise 5. How large should the degree 2n−1 be chosen in sin(x) = x−x3 3! + x5 5! −···+(−1)n−1 x2n−1 (2n−1)! \| {z } p2n−1(x;0) +(−1)n x2n+1 (2n+1)! cos(cx) \| {z } R2n+1(x) (2.1) 5 to have \|sin(x)−p2n−1(x)\| ≤0.001 ∀x ∈ h −π 2 , π 2 i ? Check your results by evaluating the resulting p2n−1(x) at x = π 2 . Solution: The remainder R2n+1(x) ≡sin(x)−p2n−1(x;0) = (−1)n x2n+1 (2n+1)! cos(cx) (where cx is between x and 0, hence cx ∈ h −π 2 , π 2 i ). Since \|x\|2n+1 ≤ π 2 2n+1 ∀x ∈[π/2,π/2], \|cos(cx)\| ≤1 ∀cx ∈[π/2,π/2], then the remainder can be bounded above by R2n+1(x) ≤(π/2)2n+1 (2n+1)! . Therefore R2n+1(x) ≤0.001 provided R2n+1(x) ≤(π/2)2n+1 (2n+1)! ≤0.001. Since n 1000(π/2)2n+1 (2n+1)! o n=3:5 = {4.6818,0.1604,0.0036}, the error is satisﬁed for all n ≥4. Let us check if the value of \|R9(π/2)\| = \|p7(π/2)−sin(π/2)\| ≤0.001?. Indeed, \|p7(π/2)−sin(π/2)\| = π 2 −(π/2)3 3! + (π/2)5 5! −(π/2)7 7! −1 ≡0.0001569 < 0.001. Exercise 6. Let pn(x;0) be the Taylor polynomial of degree n of the function f(x) = log(1 −x) about a = 0. How large n should be chosen to have \|f(x)−pn(x;0)\| ≤10−4 for x ∈[−1 2, 1 2]? For x ∈[−1, 1 2]? Solution: Note that by the approximation in Exercise 16 below (setting t = −x) we have log(1−x) = −x−1 2x2 −1 3x3 −···−1 nxn − 1 n+1xn+1 1 (1−cx)n+1 , (with cx between 0 and x). We now remark that for 0 ≤cx ≤x ≤1 2 we have 0 ≥−cx ≥−x ≥−1 2, 1 ≥1−cx ≥1−x ≥1 2, 1 ≤ 1 1−cx ≤ 1 1−x ≤2, 6 so xn+1 ≤ xn+1 (1−cx)n+1 ≤ xn+1 (1−x)n+1 ≤1, which gives a ”not so good” approximation. So we need something else! Use formula (1.20) in the book: log(1−x) = − x+ 1 2x2 +···+ 1 nxn − 1 1−cx xn+1 n+1 (1.20) with cx between 0 and x, so the remainder is \|Rn+1(x)\| ≤ 1 \|1−cx\| \|xn+1\| n+1 (because \|x\| ≤1 2 implies \|xn+1\| ≤ 1 2n+1 , and for all cx ∈[−1/2,1/2]: 1 \|1−cx\| ≤2) \|Rn+1(x)\| ≤ 2 1 2n+ 1(n+1) = 1 2n(n+1) ≤10−4 which holds for n ≥10. For x ∈[−1, 1 2] (we have that \| 1 1−cx \| ≤2, \|xn+1\| ≤1) \|Rn+1(x)\| ≤2\|xn+1\| n+1 ≤ 2 (n+1) which holds for n ≥20,000. A better approximation comes from evaluating on x ∈[−1,−1 2] (we have that \| 1 1−cx \| ≤2 3, \|xn+1\| ≤1) \|Rn+1(x)\| ≤2\|xn+1\| n+1 ≤ 2 3(n+1) ≤10−4 which holds for n ≥6,667. Exercise 8. How large should n be chosen in ex = 1+x+ 1 2x2 +···+ 1 n!xn \| {z } pn(x;0) + xn+1 (n+1)!ec \| {z } Rn+1(cx) (2.2) (where c is between x and 0) to have \|ex −pn(x;0)\| ≤10−5, ∀x ∈[−1,1]. Solution: Since \|x\|n+1 ≤1 ∀x ∈[−1,1], ecx ≤e1 ∀cx ∈[−1,1], (exponential function is increasing) 7 we can bound the remainder on [−1,1] by xn+1 (n+1)!ecx ≤ 1 (n+1)!e (as the exponential function is increasing). Then, the error bound xn+1 (n+1)!ecx ≤ 1 (n+1)!e ≤10−5 will be satisﬁed provided 1 (n+1)!e ≤10−5, i.e., 105e ≤(n+1)!, which holds for all n ≥8. (271,828.1828459046 versus 8! = 40,320.00,9! = 362,880.00) Exercise 9. Use Taylor polynomials with remainder term to evaluate the following limits: (a) limx→0 1−cos(x) x2 (b) limx→0 log(1+x2) 2x (c) limx→0 log(1−x)+xex/2 x3 Hint: Use Taylor polynomials for the standard functions [e.g., cos(t),log(1 +t) and et] to obtain polynomial approximations to the numerators of these fractions; and then simplify the results. Solution: (a) Recall that for the cos(x) we have about 0 the following Taylor expansion cos(x) = 1−x2 2! + x4 4! −···+(−1)n x2n (2n)! +(−1)n+1 x2n+2 (2n+2)! cos(cx). (2.3) For n = 1 we then have 1−cos(x) x2 = 1− 1−x2 2! + x4 4! cos(cx) x2 = 1 2! −x2 4! cos(cx) →1 2, since lim→0 x2 4! cos(cx) = 0. (b) Using log(1+u) = u−1 2u2 1 (1+cu)2 with cu between u and 0, we have that log(1+x2) = x2 + 1 2x4 1 (1+c2 x)2 and lim x→0 log(1+x2) 2x = 0. (c) limx→0 log(1−x)+xex/2 x3 8 Exercise 10. Verify 1 1−x = 1+x+···+xn + xn+1 1−x. (2.4) Solution: This is a trivial identity: move the last therm from the right-hand side to the left-hand side, and then multiply by (1−x): 1−xn+1 = (1+x+···+xn)(1−x) Exercise 13. Evaluate I = Z 1 0 ex −1 x dx within an accuracy of 10−6. Hint: Replace ex by a general Taylor polynomial approximation plus its remainder. Solution: From (2.2) we have (with cx between 0 and x) ex = 1+x+ 1 2!x2 +···+ 1 n!xn + xn+1 (n+1)!ecx hence the integrand writes ex −1 x = 1+x+ 1 2x2 +···+ 1 n!xn + xn+1 (n+1)!ecx − 1 x = 1+ 1 2!x+···+ 1 n!xn−1 + xn (n+1)!ecx and the integral becomes I = Z 1 0 ex −1 x dx = Z 1 0 1+ 1 2!x+···+ 1 n!xn−1 + xn (n+1)!ecx dx = x+ 1 2·2!x2 +···+ 1 n·n!xn 1 0 + Z 1 0 xn+1 (n+1)·(n+1)!ecxdx = 1+ 1 2·2! +···+ 1 n·n! + Z 1 0 xn+1 (n+1)·(n+1)!ecxdx. The error term is then Z 1 0 xn+1 (n+1)·(n+1)!ecxdx ≤ Z 1 0 xn+1 (n+1)·(n+1)!edx = e (n+1)·(n+1)! (since cx is between 0 and x, is then also in the interval [0,1], and the exponential function being increasing). The error being less than 10−6 is guaranteed for n such that e (n+1)·(n+1)! ≤10−6, or e·106 (n+1)·(n+1)! ≤1, n e·106 (n+1)·(n+1)! o n=7;8;9 = n 8.43;0.83;0.07} hence for all n ≥8. This implies I ≈1+ 1 2·2! +···+ 1 8·8! ≈1.317901815239985 9 >> S = 0; >> for i=1:8 S = S + 1/(ifactorial(i)); end >> S S = 1.317901815239985 Exercise 14. (a) Obtain a Taylor polynomial with remainder for f(t) = 1 1+t2 about a = 0. Hint: substitute x = −t2 into (2.4) 1 1−x = 1+x+···+xn + xn+1 1−x. () (b) Obtain a Taylor polynomial withe remainder for g(x) = arctan(x). Do this by integrating the result in (a) and using arctan(x) = Z x 0 1 1+t2 dt. Solution: (a) Using the substitution x = −t2 in (2.4)/() we obtain 1 1+t2 = 1−t2 +t4 −t6 +···+(−1)nt2n +(−1)n+1t2(n+1) 1+t2 . (b) Integrating we obtain arctan(x) = Z x 0 1 1+t2 dt = Z x 0 1 1+t2 dt = Z x 0 1−t2 +t4 −t6 +···+(−1)nt2n +(−1)n+1t2(n+1) 1+t2 dt = x−x3 3 + x5 5 −x7 7 +···+(−1)n x2n+1 2n+1 +(−1)n+1 Z x 0 t2(n+1) 1+t2 dt. Therefore arctan(x) = p2n+1(x)+R2n+3(x), where the Taylor polynomial is p2n+1(x) = x−x3 3 + x5 5 −x7 7 +···+(−1)n x2n+1 2n+1, and the remainder (obtained by the mean value theorem for deﬁnite integrals) is R2n+3(x) = (−1)n+1 Z x 0 t2(n+1) 1+t2 dt = (−1)n+1 1 1+c2 x Z x 0 t2(n+1)dt = (−1)n+1 1 1+c2 x x2n+3 2n+3, (2.5) with cx between 0 and x. Exercise 15. Deﬁne f(x) = R x 0 1−e−t t dt. Find a Taylor polynomial approximation of degree 2 for f(x). Give a formula for the approximation error, bounding it on the interval 0 ≤x ≤δ with δ > 0. What is the error bound for δ = 0.1? 10 Solution: Using the Taylor approximation for ex, (see expression (2.2)), ex = 1+x+ 1 2x2 +···+ 1 n!xn \| {z } pn(x;0) + xn+1 (n+1)!ec \| {z } Rn+1(cx) with cx is between x and 0, substitute x = −t e−t = 1−t + 1 2t2 −···+ (−1)n n! tn \| {z } pn(t;0) +(−1)n+1 tn+1 (n+1)!ec t \| {z } Rn+1(ct) with ct is between −t and 0, and obtain f(x) = Z x 0 1−e−t t dt = Z x 0 1− 1−t + 1 2t2 −···+ (−1)n n! tn −(−1)n+1 tn+1 (n+1)!ec t t dt = Z x 0 t −1 2t2 +···−(−1)n n! tn +(−1)n+1 tn+1 (n+1)!ect t dt = Z x 0 1−1 2t +···−(−1)n n! tn−1 +(−1)n+1 tn (n+1)!ect dt = x−1 22 x2 +···−(−1)n n·n! xn + (−1)n+1 (n+1)! Z x 0 tnectdt. Therefore the Taylor approximation of degree 2 for f(x) is obtained for n = 2 p2(x;0) = x−1 22 x2, and the error term can be bounded as follows (for x ∈[0,δ]) R3(x;0) = −1 6 Z x 0 t2ectdt ≤1 6 Z x 0 t2etdt ≤1 6 Z x 0 t2exdt = ex 1 6 Z x 0 t2dt = x3 3 ex ≤δ 3 18eδ. When δ = 0.1, the error bound is R3(x;0) ≤δ 3 3 eδ ≈6.1398·10−5. Exercise 16. Deﬁne g(x) = Z x 0 log(1+t) t dt. (a) Give a Taylor polynomial approximation to g(x) about x = 0. (b) Bound the error in the degree n approximation for \|x\| ≤1 2. (c) Find n so as to have a Taylor approximation with an error at most 10−7 on [−1 2, 1 2]. Solution: The Taylor polynomial approximation with remainder for log(1+t) about a = 0 (see (1.4)) is log(1+t) = t −1 2t2 + 1 3t3 +···+ (−1)n−1 n tn + 1 (n+1)!tn+1{log(1+t)(n+1)}\|t=ct = t −1 2t2 + 1 3t3 +···+ (−1)n−1 n tn + 1 (n+1)!tn+1(−1)nn!(1+ct)−(n+1) = t −1 2t2 + 1 3t3 +···+ (−1)n−1 n tn + (−1)n n+1 tn+1 1 (1+ct)n+1 , (2.6) 11 (a) Then we can approximate g(x) by g(x) = Z x 0 log(1+t) t dt ≈ Z x 0 t −1 2t2 + 1 3t3 +···+ (−1)n−1 n tn t dt = Z x 0 1−1 2t + 1 3t2 +···+ (−1)n−1 n tn−1 dt = x−x2 4 + x3 9 +···+(−1)n−1 xn n2 . (b) The error in the degree n approximation, using (2.6), is Z x 0 (−1)n n+1 tn+1 1 (1+ct)n+1 t dt = (−1)n n+1 Z x 0 tn 1 (1+ct)n+1 dt, hence it can be bounded for \|x\| ≤1 2 as follows (−1)n n+1 Z x 0 tn 1 (1+ct)n+1 dt ≤ 1 n+1 Z x 0 tndt = 1 (n+1)2 \|x\|n+1 ≤ 1 2n+1(n+1)2 . (c) The estimate above yields that an error of 10−7 is obtained provided 1 2n+1(n+1)2 ≤10−7, 2n+1(n+1)2 ≥107 >> n=10:13; >> 2.^(n+1).(n+1).^2 ans = 247808 589824 1384448 3211264 which means any n ≥12. Exercise 19. Find the Taylor polynomial about x = 0 for f(x) = log 1+x 1−x , x ∈[−1,1]. Solution: First we rewrite f(x) as f(x) = log 1+x 1−x = log(1+x)−log(1−x) and use (2.6) with t = x and t = −x respectively to obtain f(x) = log(1+x)−log(1−x) = x− 1 2x2 + 1 3x3 −S S S 1 4x4 +···+ (−1)n−1 n xn + (−1)n n+1 xn+1 1 (1+cx)n+1 − −x− 1 2x2 + 1 3(−x)3 −S S S 1 4x4 +···+ (−1)n−1 n (−x)n + (−1)n n+1 (−x)n+1 1 (1+c−x)n+1 12 = 2 x+ 1 3x3 +···+ 1 2m+1x2m+1 + (−1)n n+1 xn+1 1 (1+cx)n+1 + (−1)n n+1 (−x)n+1 1 (1+c−x)n+1 . Therefore the Taylor approximation polynomial is pn(x;0) = 2 x+ 1 3x3 +···+ 1 2m+1x2m+1 . () and the error term can be bounded on [−1,1] as Rn+1(x;0) = (−1)n n+1 xn+1 1 (1+cx)n+1 + (−1)n n+1 (−x)n+1 1 (1+c−x)n+1 ≤ 1 n+1\|x\|n+1 1 \|1+cx\|n+1 + 1 n+1\|x\|n+1 1 \|1+c−x\|n+1 ≤ 1 n+1\|x\|n+1 1 \|1+cx\|n+1 + 1 n+1\|x\|n+1 1 \|1+c−x\|n+1 ≤ 1 n+1\|x\|n+1 1+ 1 \|1+x\|n+1 . () Exercise 20. (a) Using the previous problem give a Taylor polynomial approximation to log(2), and bound the error. Hint: what is the needed value of x in order to use the previous results. (b) Consider evaluating log(z) for z ∈[1/2,1]. Use the previous problem to bound the error. Solution: (a) Setting 1+x 1−x = 2 we obtain x = −1 3, and therefore log(2) ≈pn −1 3;0 = 2 −1 3 −1 2 1 32 −1 3 1 33 −···−1 n 1 3n , with the error bound Rn+1 −1 3;0 ≤ 1 (n+1)3n+1 1+ 3n+1 2n+1 = 1 (n+1) 1 3n+1 + 1 2n+1 . (b) As above, we set 1+x 1−x = z for z ∈[ 1 2,1], hence x = z−1 z+1 ∈[−1 3,0] and using () we obtain log(z) ≈pn z−1 z+1;0 = 2 z−1 z+1 −1 2 (z−1)2 (z+1)2 + 1 3 (z−1)3 (z+1)3 +···+ (−1)n−1 n (z−1)n (z+1)n , with the error bound given by () Rn+1(z;0) ≤ 1 n+1 \|z−1\|n+1 \|z+1\|n+1 1+ \|z+1\|n+1 (2z)n+1 = 1 n+1\|z−1\|n+1 1 \|z+1\|n+1 + 1 (2z)n+1 . Exercise 21. (a) Consider evaluating π by using π = 4arctan(1). Using the results of Exercise 14, how many terms would be needed on the Taylor approximation arctan(x) ≈p2n−1(x) to calculate π with an accuracy of 10−10? Is this a practical method of evaluating π? 13 (b) Suggest another computation of π using the series for arctan(x), giving a more rapidly convergent method. Solution: (a) In order to have the error in π = 4arctan(1) = 4p2n−1(1;0)+4R2n+3(1) ≈4p2n−1(1;0) to be 10−10, by (2.5) it is sufﬁcient to have (recall that cx is now between 0 and 1) 4 R2n+3(1) ≡4 (−1)n+1 1 1+c2 x 12n+3 2n+3 = 4 2n+3 1 1+c2 x ≤ 4 2n+3 ≤10−10 which holds for n ≥2·1010 −3 2, or n ≥2·1010, a very large number of terms!!! (b) Let us use similarly π = 6arctan 1 √ 3 = 6p2n−1 1 √ 3;0 +6R2n+3 1 √ 3 ≈6p2n−1( √ 3;0) with the error (cx between 0 and 1 √ 3) 6 R2n+3 1 √ 3 ≡6 (−1)n+1 1 1+c2 x 1 3n+3/22n+3 = 6 2n+3 1 3n+3/2 ≤10−10. Then the number of terms would be n ≥18. n=linspace(16,19,4); >> e=610^(10)./((2n+3).(3.^(n+3/2))) e = 7.6641 2.4166 0.7642 0.2423 14 3. Section 1.3 POLYNOMIAL EVALUATION. For a polynomial of degree n pn(x) = a0 +a1x+a2x2 +···+anxn, an ̸= 0 the nested multiplication method pn(x) = a0 +x a1 +x a2 +···x(an−1 +anx) ··· uses only n multiplications. Exercise 9. Evaluate p(x) = 1−1 3!x3 + 1 6!x6 −1 9!x9 + 1 12!x12 −1 15!x15 as efﬁciently as possible. How many multiplications are necessary? Assume all coefﬁcients have been computed and stored for later use. Solution: It takes two multiplications to evaluate x3. Then write p(x) = 1+x3n −1 3! +x3h 1 6! +x3 −1 9! +x3 1 12! −1 15!x15io which would take ﬁve more multiplications; hence for any x it takes seven multiplications to evaluate p(x). Exercise 10. Show how to evaluate the function f(x) = 2e4x −e3x +5ex +1 efﬁciently. Hint: consider letting z = ex. Solution: With z = ex, the function writes f(x) = 2z4 −z3 +5z+1 = 1+z 5+z2−1+2z 15 4. Section 2.1 FLOATING-POINT NUMBERS. “Scientiﬁc notation” of a non-zero real number (in base 10): y = σ ·y·10e where σ = +1 or −1 (sign) y ∈[1,10) (signiﬁcand) e is an integer (exponent) while real number in binary format (base 2) x = σ ·x·2e where σ = +1 or −1 (sign) x ∈[1,2) (signiﬁcand) e is an integer. (exponent) DEFINITION 4.1 (IEEE single precision ﬂoating-point representation). The set of single precision ﬂoating-point numbers is denoted by F(2,24,−126,127) = {0}∪ n x ∈R; x = (−1)σ2e1+ 23 ∑ i=1 ai2−io where • 2 represents the base 2, • 24 is the precision of x (i.e., x = 1.a1a2 ···a23, with 24 (!!!) positions represented exactly; note that here ai ∈{0,1} are arbitrary in base 2), • the exponent e has limited values: −126 ≤e ≤127. REMARK 1. • The single ﬂoating-point numbers occupy 32 bits of memory: 1 position for the sign σ, 23 by the signiﬁcand (mantissa), and 8 positions for the exponent e: 28 = 256 = all the exponents between -126 and 127. • The LARGEST INTEGER which can be represented exactly in single precision is M = 224 = 16,777,216 function [infinity] = bug_SeriesOf1 snew = 1; sold = 0; while snew - sold > eps sold = single(snew); snew = single(sold+1); end infinity = snew; fprintf(’The values of infinity in single precision is %d\n’,infinity); end 16 >> bug_SeriesOf1 The values of infinity in single precision is 16777216 ans = single 16777216 • The number of single precision ﬂoating point numbers (the cardinal of the set F(2,23,−126,127)) is 1+2∗222 ∗(254) = 2,130,706,433 • The smallest and the largest single precision numbers are xmin = 2−126 = 1.175494350822288e−38 xmax = 2128 ∗(1−2−23) = 3.402823263561193e+38 DEFINITION 4.2 (IEEE double precision ﬂoating-point representation). The set of double precision ﬂoating-point numbers is denoted by F(2,53,−1022,1023) = {0}∪ n x ∈R; x = (−1)σ2e1+ 52 ∑ i=1 ai2−io where • 2 represents the base 2, • 53 is the precision of x (i.e., x = 1.a1a2 ···a23 ···a52, with 53 (!!!) positions represented exactly; note that here ai ∈{0,1} are arbitrary in base 2), • the exponent e has limited values: −1022 ≤e ≤1023. REMARK 2. • The double ﬂoating-point numbers occupy 64 bits of memory: 1 position for the sign σ, 52 by the signiﬁcand (mantissa) [the ﬁrst position of 1 is not counted], and 8 positions for the exponent e: 211 = 2048 = all the exponents between -1022 and 1023. • The LARGEST INTEGER which can be represented exactly in double precision is M = 253 = 9007199254740992 (9.007199254740992e+15) >> factorial(18)+1-factorial(18) ans = 1 >> factorial(19)+1-factorial(19) ans = 0 >> factorial(18) 17 ans = 6.402373705728000e+15 >> factorial(19) ans = 1.216451004088320e+17 Check that: >> factorial(18)+ 100 - factorial(18) = 100 ans = 100 >> factorial(19)+ 100 - factorial(19) ans = 96 >> factorial(20)+ 100 - factorial(20) ans = 0 For n ≥19, n! has the correct order of magnitude, but only the ﬁrst 15 digits are accurate! • The number of double precision ﬂoating point numbers (the cardinal of the set F(2,53,−1022,1023)) is 1+2∗251 ∗(2046) = 9.214364837600035e+18 (The size of Universe - number of atoms - is about 1080.) • The smallest and the largest double precision numbers are xmin = 2−1022 = 2.225073858507201e−308 xmax = 21023 ∗(1−2−52) = 8.988465674311578e+307 MATLAB : 170! = 7.257415615307994e+306 < xmax, 171! = Inf DEFINITION 4.3 (machine epsilon). ε is the distance between 1 and the nearest ﬂoating-point number, i.e., is the smallest number in F(β,t,L,U) such that 1+ε > 1, namely ε = β 1−t (epsilon machine) >> eps ans = 2.220446049250313e-16 The machine epsilon (β −t) in single precision : 2−23 = 1.192092895507812e−07 in double precision : 2−52 = 2.220446049250313e−16 18 REMARK 3. In an interval of the form [2e,2e+1), the ﬂoating-point numbers are equally spaced, at absolute distance 2e+1−t, where t is the number of signiﬁcant digits. Therefore in single precision, the distance is 2e−23, with −127 ≤e ≤126, while in double precision, the distance is 2e−52, with −1022 ≤e ≤1023. Indeed, two consecutive numbers in this interval are of the form 2e ·1.a1 ···a52 and 2e ·1.a1 ···{a52 +1}, hence the difference 2e ·2−52. REMARK 4. Unlike the absolute distance, the relative distance between two consecutive ﬂoating-point num-bers has a periodic behaviour, and depends only on the mantissa (the integer formed by the signiﬁcant digits) 1a1a2 ···at−1. Indeed, relative distance between two consecutive numbers 2e+1−t ·1a1 ···at−1 and 2e+1−t ·1a1 ···{at−1 +1}, is ∆x x = 2e+1−t ·1a1 ···{at−1 +1}−2e+1−t ·1a1 ···at−1 2e+1−t ·1a1 ···at−1 = 2e+1−t 2e+1−t ·1a1 ···at−1 = 1 1a1 ···at−1 ∈ h 1 2t −1, 1 2t−1 i DEFINITION 4.4. A real number x = σ · 1.a1a2 ···at−1 ···2e (written in base β = 2) has a floating-point representation fl(x) =σ ·1.a1a2 ··· g at−1at ····2e ( fl(x)) x =σ ·1.a1a2 ···at−1at ····2e where g at−1 =        = at−1 if at < β 2 or at−1 +1 if at ≥β 2 , (rounding) alternatively g at−1 = at−1. (chopping) REMARK 5. • fl(x) ∈F, ∀x ∈R. • fl(x) = x, ∀x ∈F. • ∀x,y ∈R we have that fl(x) ≤fl(y). (monotonicity property) REMARK 6. Note that the real number x = 0.1 has a ﬁnite representation in base 10: (0.1)10 = 1·10−1, while in base 2 has in inﬁnite representation: x10 = (0.1)10 = 0.001100110011001100··· 2 = 1.100110011001100···∗2−3 =: x2, 19 and therefore its ﬂoating-point representation/approximation does not coincide with it: x2 :=1.100110011001100···at−2at−1 \| {z } t signiﬁcant digits \|ataa+1 ···001100···∗2−3 fl(x2) =1.100110011001100···at−2 g at−1 \| {z } t signiﬁcant digits \|ataa+1 ···001100···∗2−3, DEFINITION 4.5. We remark that everything written so far holds for numbers which have the exponent e within the range from the deﬁnition of F. • For R ∋x ∈(−∞,−xmax)∪( xmax \|{z} βU(1−β) ,∞), fl(x) is not deﬁned !!! (overﬂow) • If x ∈(−xmin, xmin \|{z} β L−1 ) the operation of rounding fl(x) is deﬁned (anyway). (underﬂow) MATLAB : 2−1074 = 4.940656458412465e−324, 2−1075 = 0 PROPOSITION 4.6. For a real number x ∈R such that xmin ≤x ≤xmax it can be shown that fl(x) = x(1+δ), with \|δ\| ≤u where u = 1 2β 1−t ≡1 2ε is the roundoff unit (or machine precision). Exercise 5. The following MATLAB program produced the given output. Explain the results. (4.1) x = 0.0; while x < 1.0 x = x + 0.1; %disp([x, sqrt(x)]); end disp([x(end), sqrt(x(end))]); >> Exercise2_2_5 1.100000000000000 1.048808848170151 Solution: The intention was to STOP at x = 1.0, but due to ﬂoating-point representation (the decimal number 0.1 is not represented exactly due to rounding) the number 1.0 is not obtained exactly by the incrementation. Therefore, the code stops only after the 11th round of the loop. If we try instead x = 0.0; while x <= 1.0-eps x = x + 0.1; %disp([x, sqrt(x)]); end disp([x(end), sqrt(x(end))]); we obtain only 10 iterations in the loop >> Exercise2_2_5 1.000000000000000 1.000000000000000 Recall that by Remark 6 , (0.1)10 has an inﬁnite representation in base 2 (0.1)10 = 0.001100110011001100··· 2, 20 hence its 64 bits ﬂoating-point approximation has a rounding error in the 53rd binary signiﬁcant digit. Exercise 6. Let x ∈R+ be a positive real number: x = 1.a1a2 ···at−1\|at ····2e Consider a computer using a positive binary ﬂoating-point representation with t bits of precisions in the signiﬁcand, i.e., fl(x) =1.a1a2 ··· g at−1at ····2e ( fl(x)) x =1.a1a2 ···at−1at ····2e. Assume that chopping (rounding towards zero) is used in going from x ∈R outside the computer to its ﬂoating-point representation fl(x) inside the computer. (a) Show that 0 ≤x−fl(x) ≤2e−t+1 (b) Show that x ≥2e, and use this to show x−fl(x) x ≤2−t+1. (c) Let x−fl(x) x = −ε and then solve for fl(x). What are the bounds on ε? (This result extends to x < 0, with the assumption of x > 0 being used to simplify the algebra.) Solution: If chopping is used, then fl(x) =1.a1a2 ···at−1at ····2e x =1.a1a2 ···at−1at ····2e, x−fl(x) = 1.a1a2 ···at−1at ···−1.a1a2 ···at−1 ·2e = 0.0102 ···0t−1\|at ····2e = 0.at ····2e−(t−1). () (a) Using () it is easy to see that 0 ≤x−fl(x) ≤2e−t+1. (b) Obviously x ≥2e, and therefore 0 ≤x−fl(x) x ≤2e−t+1 2e = 2−t+1. (c) This means fl(x) = x(1+ε) and from above we have x−fl(x) x = \|ε\| ≤uchopping = 2−t+1 which in double precision (t = 53) gives uchopping = 2.220446049250313e−16. 21 Exercise 7. Let x ∈R+ be a positive real number: x = 1.a1a2 ···at−1\|at ····2e Consider a computer using a positive binary ﬂoating-point representation with t bits of precisions in the signiﬁcand, i.e., fl(x) =1.a1a2 ··· g at−1at ····2e (fl(x)) x =1.a1a2 ···at−1at ····2e. Assume that rounding is used in going from x ∈R outside the computer to its ﬂoating-point approximation fl(x) inside the computer. (a) Show that −2e−t ≤x−fl(x) ≤2e−t (b) Show that x ≥2e, and use this to show x−fl(x) x ≤2−t. (c) Let x−fl(x) x = −ε and then solve for fl(x). What are the bounds on ε? (This result extends to x < 0, with the assumption of x > 0 being used to simplify the algebra.) Solution: If rounding is used, then we have two options • If at = 0, then g at−1 = at−1, and fl(x) =1.a1a2 ···at−1 at \|{z} 0 at+1 ····2e x =1.a1a2 ···at−1 at \|{z} 0 at+1 ····2e, x−fl(x) = 1.a1a2 ···at−10at+1 ···−1.a1a2 ···at−1 ·2e = 0.0102 ···0t−1\|0tat+1 ····2e = 0.at+1 ····2e−t ≤2e−t. () • If at = 1, then g at−1 = at−1 +1, and fl(x) =1.a1a2 ···at−1 +1 at \|{z} 1 at+1 ····2e x =1.a1a2 ···at−1 at \|{z} 1 at+1 ····2e, so fl(x)−x = 1.a1a2 ···at−1 +1−1.a1a2 ···at−11at+1 ··· ·2e = 0.0102 ···1t−1 −0.0102 ···0t−11at+1 ··· ·2e = (1.00−0.1at+1 ··· \| {z } ≤0.1 )·2e−(t−1) ≤2−1 ·2e−t+1 = 2e−t () 22 (a) Using () and () we see that −2e−t ≤x−fl(x) ≤2e−t. (b) Obviously x ≥2e, and therefore 0 ≤\|x−fl(x)\| \|x\| ≤2e−t 2e = 2−t. (c) This means fl(x) = x(1+ε) and from above we have x−fl(x) x = \|ε\| ≤urounding = 2−t which in double precision (t = 53) gives the machine precision in double precision on 64 bits urounding = 1.110223024625157e−16. 23 5. Section 2.2 ERRORS: DEFINITIONS, SOURCES AND EXAMPLES. DEFINITION 5.1 (Absolute and Relative Errors). Let xtrue or xT denote the true of a quantity x (with some measuring units), and by xapproximate or xA the approximate value (obtained by physical measurements or numerical approximations). Then we deﬁne the following errors in the approximate value xA: • Error(xA) := xT −xA ≡xtrue −xapproximate (absolute error) (Note that Error(xA) has the same units as xT and xA.) • Rel(xA) := Error(xA) xT = xT −xA xT (relative error) (Note that Rel(xA) does not have units!) DEFINITION 5.2 (Number of signiﬁcant digits). The approximate value xA has t signiﬁcant digits relative to the true value xA if the number of leading digits in xA are correct relative to the corresponding digits in xT. More precisely: if the (t +1)th digit in the magnitude of the absolute error \|xT −xA\| is less or equal than 5 (or half the base where the calculations are performed), then xA has at least t signiﬁcant digits of accuracy relative to xT. PROPOSITION 5.3. It can be shown that if xT −xA xT ≤5·10−(t+1), then xA has t signiﬁcant digits of accuracy relative to xT. 5.1. Subsection 2.2.1 SOURCES OF ERROR. (E1) Modelling errors “All models are wrong, but some are useful”, an aphorism attributed to George Box. (E2) Blunders and Mistakes: arithmetic and/or programming errors (E3) Physical Measurements errors (E4) Machine representation and Floating-point Arithmetic errors (E5) Mathematical Approximation errors: (local) truncation errors 5.2. Subsection 2.2.2 LOSS-OF-SIGNIFICANCE ERRORS. When two “nearly equal” quantities are subtracted, leading signiﬁcant digits are lost. For example, in MATLAB floating-point double precision we have that >> 1-(1+1.e-12-1)/(1.e-12) ans = -8.890058234101161e-05 while, in exact arithmetic, the answer is obviously 0 !!! EXAMPLE 1. When evaluating the function f(x) = 1−cos(x) x2 for values of x close to 0, we expect to obtain in the limit x →0 that lim x→0 f(x) = 1 2. Due to ﬂoating-point arithmetic, this is what we obtain in double precision 24 n=[-1:-1:-10]’; x=[1.010.^n]; xsquare=x.^2; cosx=cos(x); numerator=[1-cos(x)]; f=[numerator./x.^2]; T = table(x,xsquare,cosx,numerator,f) x xsquare cosx numerator f __ _ __ ____________________ _________________ 0.1 0.01 0.995004165278026 0.00499583472197429 0.499583472197429 0.01 0.0001 0.999950000416665 4.99995833347366e-05 0.499995833347366 0.001 1e-06 0.999999500000042 4.99999958325503e-07 0.499999958325503 0.0001 1e-08 0.999999995000000 4.99999996961265e-09 0.499999996961265 1e-05 1e-10 0.999999999950000 5.00000041370185e-11 0.500000041370186 1e-06 1e-12 0.999999999999500 5.00044450291171e-13 0.500044450291171 1e-07 1e-14 0.999999999999995 4.99600361081320e-15 0.499600361081320 1e-08 1e-16 1 0 0 1e-09 1e-18 1 0 0 1e-10 1e-20 1 0 0 Notice that the values of f grow towards 0.5, with the best value obtained for x = 1.0 ∗10−5, namely f(10−5) ≈ 0.500000041370186, while for x smaller than 10−8, f(x) ≈0!!! If we instead use (2.3) to rewrite f(x) as follows f(x) = 1−cos(x) x2 = 1− 1−x2 2! + x4 4! −···+(−1)n x2n (2n)! +(−1)n+1 x2n+2 (2n+2)! cos(cx) x2 = 1 2! −x2 4! +···−(−1)n x2n−2 (2n)! −(−1)n+1 x2n (2n+2)! cos(cx), then, in order to have an approximation of machine epsilon order, say 10−16 for any \|x\| < 0.1, we need to estimate the remainder \|R2n\| := (−1)n+1 x2n (2n+2)! cos(cx) ≤ 10−2n (2n+2)!≤10−16 which holds for n ≥5, (indeed 10−10 12! = 2.087675698786810∗10−19). f=[numerator./x.^2]; f_approx = 1/factorial(2) - x.^2 /factorial(4) + x.^4/factorial(6) - x.^6/factorial(8) + x.^10/fa TT = table(x,f,f_approx) x f f_approx ______ _________________ _________________ 0.1 0.499583472197429 0.499583472197421 0.01 0.499995833347366 0.499995833347222 0.001 0.499999958325503 0.499999958333335 0.0001 0.499999996961265 0.499999999583333 1e-05 0.500000041370186 0.499999999995833 25 1e-06 0.500044450291171 0.499999999999958 1e-07 0.499600361081320 0.5 1e-08 0 0.5 1e-09 0 0.5 1e-10 0 0.5 See also Exercise 5.4. 0.999985 0.99999 0.999995 1 1.000005 1.00001 1.000015 1.00002 -4 -3 -2 -1 0 1 2 3 4 10-15 Noise/Error in function evaluation: -1+x(3+x(-3+x)) (x-1)3 (x-1)3 in nested form: -1+x(3+x(-3+x)) 0.9997 0.9998 0.9999 1 1.0001 1.0002 1.0003 -1 0 1 2 3 4 5 6 7 8 9 10-15 Noise/Error in function evaluation: 1+x(-4+x(6+x(-4+x))) (x-1)4 (x-1)4 FIG. 5.1. Noise in function evaluations: (x−1)3 and (x−1)4, evaluated in their nested forms. 5.3. Subsection 2.2.3 NOISE IN FUNCTION EVALUATION. 5.4. Subsection 2.2.4 UNDERFLOW AND OVERFLOW ERRORS. Exercise 1. Calculate the error, relative error, and number of signiﬁcant digits in the following approximations xA ≈xT: (a) xT = 28.254,xA = 28.271 (b) xT = 0.0028254,xA = 0.028271 (c) xT = e,xA = 19 7 (d) xT = √ 2,xA = 1.414 (e) xT = ln(2),xA = 0.7 Solution: (a) Error(xA) = xT −xA ≡28.254−28.271 = −0.0170, Rel(xA) = xT −xA xA ≡−0.0170 28.271 = −6.0132·10−4. xA = 28.271 has 3 signiﬁcant digits of accuracy relative to xT = 28.254 since the error in the fourth digit (7 versus 5) is 2, less than 5. (b) xT = 0.0028254,xA = 0.028271 (c) xT = e,xA = 19 7 Error(xA) = xT −xA ≡e−19 7 = 0.0040, Rel(xA) = xT −xA xA ≡0.0040 e = 0.0015. 26 xA = 2.714285714285714 has 3 signiﬁcant digits of accuracy relative to xT = 2.718281828459046 since the error in the fourth digit (8 versus 4) is 4, less than 5. (d) xT = √ 2,xA = 1.414 (e) xT = ln(2),xA = 0.7 Error(xA) = xT −xA ≡ln(2)−0.7 = −0.0069, Rel(xA) = xT −xA xA ≡−0.0069 ln(2) = −0.0099. xA = 0.70 has 1 signiﬁcant digit of accuracy relative to xT = 0.693147180559945 since xT −xA xA ≡0.0099 = 0.99·10−2 ≤5·10−1−1, (see Proposition 5.3). Exercise 5. In some situations, loss-of-signiﬁcance errors can be avoided by rearranging the function being evaluated, as was done with f(x) in (2.23): f(x) = x √x+1−√x 1 · √x+1+√x √x+1+√x = x √x+1+√x. (2.23) Do something similar for the following cases, in some cases using trigonometric identities. In all but case (b), assume x is near 0. (a) 1−cos(x) x2 (b) log(x+1)−log(x), x large (c) sin(a+x)−sin(a) (d) 3 √x+1−1 (e) √4+x−2 x Solution: The idea is to use identities to “transform” the functional in such a way to avoid taking differences of terms of ‘approximately equal’ size. (a) Here we could use cos(x) = 2sin2(x/2) 1−cos(x) x2 = 1− 1−2sin2(x/2) x2 = 2sin2(x/2)) x2 or 1 = cos2(x)+sin2(x) 1−cos(x) x2 = 1−cos(x) x2 1+cos(x) 1+cos(x) = 1−cos2(x) x2(1+cos(x)) = sin2(x) x2(1+cos(x)). See also Example 1: f=[numerator./x.^2]; f_approx = 1/factorial(2) - x.^2 /factorial(4) + x.^4/factorial(6) - x.^6/factorial(8) + x.^10 f_ex5a = (sin(x)).^2./(x.^2.(1+cos(x))); T = table(x,xsquare,cosx,numerator,f) TT = table(x,f,f_approx,f_ex5a) 27 x f f_approx f_ex5a __ ___ __ __ 0.1 0.499583472197429 0.499583472197421 0.499583472197423 0.01 0.499995833347366 0.499995833347222 0.499995833347222 0.001 0.499999958325503 0.499999958333335 0.499999958333335 0.0001 0.499999996961265 0.499999999583333 0.499999999583333 1e-05 0.500000041370186 0.499999999995833 0.499999999995833 1e-06 0.500044450291171 0.499999999999958 0.499999999999958 1e-07 0.499600361081320 0.5 0.5 1e-08 0 0.5 0.5 1e-09 0 0.5 0.5 1e-10 0 0.5 0.5 (b) log(x+1)−log(x), x large (c) sin(a+x)−sin(a) (d) Use the identity y3 −1 = (y−1)(y2 +y+1) with y = 3 √x+1 to obtain 3 √ x+1−1 = ( 3 √ x+1−1)(x+1)2/3 +(x+1)1/3 +1 (x+1)2/3 +(x+1)1/3 +1 = (x+1)3/3 −1 (x+1)2/3 +(x+1)1/3 +1 = x (x+1)2/3 +(x+1)1/3 +1. (e) Use (a−b)(a+b) = a2 −b2 √4+x−2 x = √4+x−2 x √4+x+2 √4+x+2 = 4+x−4 x(√4+x+2) = 1 √4+x+2. Exercise 6. Use Taylor polynomial approximations to avoid the loss-of-signiﬁcance errors in the following formulas when x is near 0. (a) ex −1 x (b) 1−e−x x (c) ex −e−x 2x (d) log(1−x)+xe x 2 x3 (e) 1−(1−x) √ 2−1 x (f) x−sin(x) x3 (g) x−sin(x) tan(x) (h) x+log(1−x) x2 Solution: 28 (a) Use ex ≈1+x+ 1 2!x2 + 1 3!x3 +···+ 1 n!xn to obtain ex −1 x ≈ 1+x+ 1 2!x2 + 1 3!x3 +···+ 1 n!xn − 1 x = 1+ 1 2!x+ 1 3!x2 +···+ 1 n!xn−1 (b) Similarly use e−x ≈1−x+ 1 2!x2 −1 3!x3 +···+ (−1)n n! xn 1−e−x x ≈ 1− 1−x+ 1 2!x2 −1 3!x3 +···+ (−1)n n! xn x = 1−1 2!x+ 1 3!x2 −···+ (−1)n−1 n! xn−1 (c) As above ex −e−x 2x ≈ 1+x+ZZ 1 2!x2 + 1 3!x3 +···+ HHH H 1 (2m)!x2m + 1 (2m+1)!x2m+1 − 1−x+ZZ 1 2!x2 −1 3!x3 +···+ HHH 1 2m!x2m + (−1)2m+1 (2m+1)! x2m+1 2x = 1+ 1 3!x2 + 1 5!x4 ···+ 1 (2m+1)!x2m (d) Use formula (1.20) to approximate log(1−x) ≈−x−1 2x2 −1 3x3 −···− 1 n+1xn+1 and (1.3) to obtain log(1−x)+xe x 2 x3 ≈ − x−@ @ 1 2x2 −1 3x3 −···− 1 n+1xn+1 +x 1+HH H (x/2)+ (x/2)2 2! + (x/2)3 3! +···+ (x/2)n n! x3 = −1 3 + 1 22·2! x3 +···+ − 1 n+1 + 1 2n·n! xn+1 x3 = −1 3 + 1 22 ·2! +···+ − 1 n+1 + 1 2n ·n! xn−2, n ≥2. (e) 1−(1−x) √ 2−1 x (f) x−sin(x) x3 (g) x−sin(x) tan(x) (h) x+log(1−x) x2 Exercise 13. Find an accurate value of f(x) = r 1+ 1 x −1 for large values of x. Calculate lim x→∞x f(x). Solution: When x is very large (i.e., x →∞), or equivalently when 1 x →0, the two quantities in f(x) are “close” to 29 each other q 1+ 1 x ≈1 and there is a loss-of-signiﬁcance error. To avoid taking the difference, we can rewrite f(x) as follows: f(x) = r 1+ 1 x −1 = r x+1 x −1 = √x+1−√x √x = √x+1−√x √x √x+1+√x √x+1+√x = x+1− x √x(√x+1+√x) = 1 √x √x+1+√x = 1 √ x2 +x+x 30 6. Section 2.3 PROPAGATION OF ERROR. 6.1. Propagated error in multiplication. Compare xT ·yT versus xA ·yA. First let us rearrange the relative error deﬁnition: Rel(xA) = xT −xA xT , xA = xT(1−Rel(xA)), yA = yT(1−Rel(yA)), and now evaluate the relative error in xA ·yA relative to xT ·yT: Rel(xA ·yA) := xT ·yT −xA ·yA xT ·yT ≡xT ·yT −xT(1−Rel(xA))·yT(1−Rel(yA)) xT ·yT = xT ·yT −xT ·yT 1−Rel(xA)−Rel(yA)+Rel(xA)·Rel(yA) xT ·yT = Rel(xA)+Rel(yA)+Rel(xA)·Rel(yA) ≈Rel(xA)+Rel(yA), for Rel(xA),Rel(yA) ≪1. 6.2. Propagated error in division. Compare xT/yT versus xA/yA. Similarly we evaluate the relative error in xA/yA relative to xT/yT as follows: Rel(xA ·yA) := xT yT −xA yA xT yT ≡ xT yT −xT (1−Rel(xA)) yT (1−Rel(yA)) xT yT = 1−1−Rel(xA) 1−Rel(yA) = 1−Rel(yA)−( 1−Rel(xA)) 1−Rel(yA) = Rel(xA)−Rel(yA) 1−Rel(yA) ≈Rel(xA)−Rel(yA), for Rel(yA) ≪1. 6.3. Propagated error in addition and subtraction. Compare xT ±yT versus xA ±yA. We start as above Rel(xA ±yA) := (xT ±yT)−(xA ±yA) xT ±yT ≡xT ±yT −xT(1−Rel(xA))∓yT(1−Rel(yA)) xT ±yT = xT ±Z Z yT − xT +xT ·Rel(xA))∓Z Z yT ±yTRel(yA)) xT ±yT = xT ·Rel(xA))±yTRel(yA)) xT ±yT , which by taking absolute values yields \|Rel(xA ±yA)\| = xT ·Rel(xA))±yTRel(yA)) xT ±yT ≤\|xT\|·\|Rel(xA)\|+\|yT\|·\|Rel(yA)\| \|xT ±yT\| ≤(\|Rel(xA)\|+\|Rel(yA)\|)\|xT\|+\|yT\| \|xT±yT\| . 31 If xT,yT are “close” in magnitude and opposite sign, \|xT+yT\| ≡0 and the term in the right-hand side of the estimate above could be very large!!! \|Rel(xA ±yA)\| ≤(\|Rel(xA)\|+\|Rel(yA)\|)·O(∞) !?! We say that the “subtraction” operation is ill-conditioned: small errors in two real numbers of same signs yields (possibly) large errors when subtracting them. (This is related to the loss-of-signiﬁcance errors.) EXAMPLE 2. Let ε = eps = 2.220446049250313e−16 be the machine epsilon in the double-precision ﬂoating point arithmetic, and consider the following numbers. The true and approximate values are: xT = 1+500ε, xA = 1−500ε, yT = 1, yA = 1. Therefore the relative errors are respectively Rel(xA) = 1000ε 1+500∗ε = 2.220446049250066e−13, Rel(yA) = 0. Nevertheless, the relative error in the difference xA −yA is “quite large”: Rel(xA −yA) = (xT −yT)−(xA −yA) xT −yT ≡500ε −(−500∗ε) 500ε = 2 ̸≈ Rel(xA)−Rel(yA) !!! 6.4. Propagated error in function evaluation. Let f ∈C1[a,b], and xT,xA ∈[a,b], assumed to be ‘close’ to each other. Then by the Taylor approximation, the absolute error in f(xA) can be approximated as Error(f(xA)) := f(xT)−f(xA) = f ′(c)·(xT −xA), with c between xT and xA, hence ≈f ′(xT)·(xT −xA), and therefore the relative error in f(xA) is Rel( f(xA)) ≈f ′(xT) f(xT) ·(xT −xA) ≡f ′(xT) f(xT) xT · xT −xA xT = f ′(xT) f(xT) xT ·Rel(xA). (6.1) DEFINITION 6.1. We denote the quantity cond( f) := f ′(xT) f(xT) xT (condition number) to be the condition number of the evaluation of f(x). We note that cond(f) could be very large. For Example, if f(x) = 2x and xT = 1000. xA = 1000.0001 Rel(xA) = −9.999999997489795e−08; f(xT) = 1.071508607186267e+301 32 f(xA) = 1.071582881077358e+301 Rel(f(xA)) = −6.931231582904155e−05; f ′(xT) f(xT) xT= 6.931471805599452e+02. Rel( f(xA)) Rel(xA)) = 6.931231584644037e+02. Exercise 2.4.6 To illustrate (6.1) and (condition number), compare π100 to π100.1. π100 = 5.187848314319592e+49, π100.1 = 5.817042766090191e+49 Then calculate Rel(π100.1) directly: Rel(π100.1) := π100 −π100.1 π100 = −0.121282353231856. and using (6.1): Rel(xA) = 100−100.1 100 = −0.0009999999999999 Rel( f(xA)) ≡f ′(xT) f(xT) xT ·Rel(xA) = log(π) π100 π100 100·100−100.1 100 = −0.114472988584934 Also give the (condition number) cond( f) := f ′(xT) f(xT) xT ≡log(π)∗100 = 1.144729885849400e+02. 7. Section 2.4 SUMMATION. Recall that due to ﬂoating-point approximation, the addition operation in “computer arithmetic” is not associative. EXAMPLE 3. Let n ∈N be an integer, and consider the ﬁnite sum: S = 0.1+0.2+···+0.1·n ≡0.1∗(1+2+···n) = n(n+1) 2·10 . (Recall also Exercise 5 in Section 2.1 (4.1), about 0.1 not having a ﬁnite representation in base 2.) %Error is summation n=1.e+11; Sexact = n(n+1)/20 Sinc = 0; Sdec = 0; for i=1:n Sinc = Sinc + i/10; end for j=n:-1:1 Sdec = Sdec + j/10; 33 end Sinc Sdec Sexact = 5.000000000050000e+20 Sinc = 5.000000000031216e+20 Sdec = 5.000000000082726e+20 Exercise 1. Write a computer program to evaluate S = n ∑ j=1 a j for arbitrary n, and apply it to teh series given below. Do the calculation by both the methods LS and SL. Calculate a true value for S using the given answer, and compare it with the values obtained by LS and SL. (a) ∑n j=1 1 j(j+1) = n n+1. n = 1.e+9; >> SL = 0; LS = 0; >> for i=1:n LS = LS+ 1/(i(i+1)); end >> for i=n:-1:1 SL = SL+ 1/(i(i+1)); end >> S =n/(n+1) gives S = 0.999999999000000, SL = 0.999999999000000 LS = 0.999999993626404 (b) ∑n j=1 1 j(j+2) = 3 4 − 2n+3 2(n+1)(n+2). n = 1.e+9; >> SL = 0; LS = 0; >> for i=1:n LS = LS+ 1/(i(i+2)); end >> for i=n:-1:1 SL = SL+ 1/(i(i+2)); end >> S = 3/4 - (2n+3)/(2(n+1)(n+2)) gives S = 0.749999999000000, SL = 0.749999999000000 LS = 0.749999993626380 34 8. Section 4.1 POLYNOMIAL INTERPOLATION. The aim of polynomial interpolation is to have a compact representation of available (large set of) data. It could also be used to obtain an approximation of a function which is known only by a ﬁnite number of values, at a given number of points. For example, given a set of points {(xi, f(xi))}N i=1, we would want to approximate the given function f(·) by a polynomial (global or piecewise), which can be easier to manipulate (via integration, derivation, etcetera). More general, we can ask the following question: Given (n+1) pairs of points {(xi,yi)\|i = 0,1,··· ,n} {(xi,yi)}n i=0 = {(x0,y0),(x1,y1),(x2,y2),··· ,(xn,yn)} find a function Φ(·) such that Φ(xi) = yi, ∀i = 0,1,··· ,n saying that Φ(·) interpolates {yi}n i=0 at the nodes {xi}n i=0. We note that in the above, the function Φ could be a global polynomial given (n+1) pairs (xi,yi) ﬁnd Πn ∈Pn, an interpolating polynomial: Πn(xi) = yi, i = 0,1,...,n (polynomial interpolation) or a piecewise polynomial (spline functions). THEOREM 8.1. Given (n+1) distinct nodes {xi}n i=0: x0,x1,··· ,xn (nodes) and (n+1) corresponding values {yi}n i=0: y0,y1,··· ,yn (values) there exists a unique global (interpolating) polynomial Πn ∈Pn such that : Πn(xi) = yi, ∀i = 0,1,··· ,n. Proof. Existence: constructive. Let us deﬁne the following polynomials ℓi(x) = n ∏ j=0 j̸=i x−x j xi −x j ≡(x −x0)·(x −x1) ··· (x −x j−1)·(x −xj+1) ··· (x −xn) (xi −x0)·(xi −x1)···(xi −xj−1)·(xi −xj+1)···(xi −xn). (Lagrange characteristic polynomial) We note1 that the Lagrange2 characteristic polynomials {ℓi(·)}i=0:n form a basis of Pn, ℓi(x j) = δij = 1 if i = j 0 if i ̸= j 1{ℓi}i=0:n are also orthogonal, with the inner product ⟨f,g⟩= R ∞ −∞e−x2 f(x)g(x)dx. 2Joseph-Louis Lagrange (25 January 1736 – 10 April 1813), Lagrange 35 and therefore the interpolating polynomial is Πn(x) = n ∑ i=0 yiℓi(x) ≡y0ℓ0(x)+y1ℓ1(x)+···+ynℓn(x). (interpolating polynomial in Lagrange form) Uniqueness: by contradiction. DEFINITION 8.2 (NEWTON3 DIVIDED DIFFERENCES). Let x0,x1,x2,··· ,xn be n+1 distinct nodes f[x0,x1] = f(x1)−f(x0) x1 −x0 (1st-order divided difference) f[x0,x1,x2] = f[x1,x2]−f[x0,x1] x2 −x0 (2nd-order divided difference) f[x0,x1,x2,x3] = f[x1,x2,x3]−f[x0,x1,x2] x3 −x0 (3rd-order divided difference) f[x0,··· ,xn] = f[x1,··· ,xn]−f[x0,··· ,xn−1] xn −x0 (nth-order Newton divided difference) THEOREM 8.3. Let f ∈Cn(I) where x0,x1,··· ,xn ∈I are n+1 distinct nodes. Then there exists ∃c ∈I such that f[x0,x1,··· ,xn] = 1 n! f (n)(c). (8.1) Proof. This is a consequence of the Langrange Mean Value Theorem. PROPOSITION 8.4. f[x0,x1,··· ,xn] = f[xi0,xi1,··· ,xin], ∀permutations (i0,i1,··· ,in). Also, in case of “repeating nodes”: f[x0,x0] = lim x1→x0 f[x0,x1] = lim x1→x0 f(x1)−f(x0) x1 −x0 = f ′(x0), (8.2) f[x0,x0,x1] = f[x0,x1]−f[x0,x0] x1 −x0 (8.2) == f[x0,x1]−f ′(x0) x1 −x0 , (8.3) f[x0,x0,··· ,x0 \| {z } (n+1)terms ] = 1 n! f (n)(x0). (8.4) THEOREM 8.5. Πn(x) = n ∑ i=0 ωi(x) f[x0,··· ,xi] ≡f(x0)+ω1(x) f[x0,x1]+···+ωn(x) f[x0,x1,··· ,xn] ≡f(x0)+(x−x0) f[x0,x1]+···+(x−x0)···(x−xn−1)f[x0,x1,··· ,xn] (interpolating polynomial in Newton form) 3Isaac Newton (25 December 1642 – 20 March 1726/27), Newton 36 COROLLARY 8.6. Note that in the ”repeated nodes case”: x0 = x1 = ··· = xn, using property (8.4), the (interpolating polynomial in Newton form) gives Πn(x) = f(x0)+(x−x0) f ′(x0)+(x−x0)2 f ′′(x0) 2! +···+(x−x0)n f (n)(x0) n! (8.5) which is exactly the (Taylor approximation polynomial). Exercise 1. Given the data points (0,2),(1,1), ﬁnd the following: (a) The straight line interpolating this data. (b) The function f(x) = a+bex interpolating this data. (Hint: Find a and b so that f(0) = 2, f(1) = 1.) (c) The function g(x) = a b+x interpolating this data. In each instance, graph the interpolating function. Solution: (a) The divided differences are xi yi [yi,yj] 0 2 1 1 1−2 1−0 = −1 hence Π1(x) = 2−x. (b) Since f(0) = a+b ≡2, f(1) = a+be ≡1, we have that a = 1−2e 1−e ,b = 1 1−e, so f(x) = 1−2e 1−e + 1 1−eex. (c) Since g(0) = a b ≡2,g(1) = a b+x ≡1, we have that a = 2,b = 1, so g(x) = 2 1+x. x = linspace(0,1,100); x_nodes = [0 1]; y_values = [2 1]; P1 = 2 - x; f = (1-2exp(1) + exp(x))/(1-exp(1)); g = 2./(1+x); plot(x,P1,’b.-.’,x,f,’r.-.’,x,g,’k.-.’,x_nodes,y_values,’g’) legend(’P1’,’f’,’g’,’data points’) Exercise 2. (a) Find the function P(x) = a+bcos(πx)+csin(πx), which interpolates the data xi yi 0 2 0.5 5 1 4 37 (b) Find the quadratic polynomial interpolating this data. In each instance, graph the interpolating function. Solution: (a) Since P(0) = a+b ≡2,P(0.5) = a+c ≡5,P(1) = a−b ≡4, we have that a = 3,b = −1,c = 2. (b) The divided differences are xi yi [yi,y j] [y0,y1,y2] 0 2 0.5 5 5−2 0.5 = 6 1 4 4−5 1−0.5 = −2 −2−6 1 = −8 hence P 2(x) = 2+6x−8x(x−0.5) = −8x2 +10x+2. x = linspace(0,1,100); x_nodes = [0 0.5 1]; y_values = [2 5 4]; P = 3 - cos(pix) + 2sin(pix); P2 = -8x.^2 + 10x + 2; plot(x,P,’b.-.’,x,P2,’r.-.’,x_nodes,y_values,’g’) legend(’P’,’P2’,’data points’) Exercise 12. (a) For n = 3, explain why ℓ0(x)+ℓ1(x)+ℓ2(x)+ℓ3(x) = 1 ∀x. (Hint: It is unnecessary to actually multiply out and combine (Lagrange characteristic polynomial) ℓi(x). Use (interpolating polynomial in Lagrange form) and a suitable choice of {y0,y1,y2,y3}.) (b) Generalize part (a) to an arbitrary degree n > 0. Solution: Exercise 16. As a generalized interpolation problem, ﬁnd the quadratic polynomial q(x) with deg q(x) ≤2 which q(0) = −1, q(1) = −1, q′(1) = 4. Solution: Using (8.2) to substitute f[1,1] = q′(1), the Newton divided differences are and therefore the (interpolating polynomia xi f(xi) f[xi,x j] f[x1,x2,x3] 0 -1 1 -1 0 1 -1 4 4 writes q(x) = q(0)+(x−0)q[0,1]+(x−0)(x−1)q[0,1,1] = −1+0·x+4x(x−1) = 4x2 −4x−1. 38 Exercise 24. Given the data below, i xi f(xi) 1 0.1 0.20 2 0.2 0.24 3 0.3 0.30 ﬁnd f[x0,x1] and f[x0,x1,x2]. Then calculate Π1(0.15) and Π2(0.15), the linear and quadratic interpolates evaluated at x = 0.15. Solution: The Newton divided differences are i xi f(xi) f[xi,xj] f[x1,x2,x3] 1 0.1 0.20 2 0.2 0.24 0.24−0.20 0.2−0.1 = 0.4 3 0.3 0.30 0.30−0.24 0.3−0.2 = 0.6 0.6−0.4 0.3−0.1 = 1 and therefore the interpolating polynomials (in Newton form) of degree one and two are Π1(x) = 0.2+0.4(x−0.1), Π2(x) = 0.2+0.4(x−0.1)+1(x−0.1)(x−0.2). while the values are Π1(0.15) = 0.22, Π2(0.15) = 0.2175. Exercise 25’. Let f(x) = 1 1+x2 (Runge example) and let x0 = 1,x1 = 2,x2 = 3. Calculate the divided differences f[x0,x1] and f[x0,x1,x2]. Using these divided differences, give the quadratic polynomial P 2(x) that interpolates f(x) at the given node points {x0,x1,x2}. Graph the error on the interval [1,3]. Solution: The divided differences are hence the polynomial is xi f(xi) f[xi,x j] f[x1,x2,x3] 1 1/2 2 1/5 1/5−1/2 2−1 = −3 10 3 1/10 1/10−1/5 3−2 = −1 10 −1/10+3/10 3−1 = 1 10 P 2(x) = 1 2 +−3 10(x−1)+ 1 10(x−1)(x−2). 39 x = linspace(1,3,100) ; P2 = 1/2 - 3/10 (x-1) + 1/10 (x-1) . (x-2) ; f = 1 ./ (1 + x .^2) ; plot( x , f-P2 , ’r.-.’ ) Exercise 26. (a) By using the program divdiff, calculate the divided differences D0 = f(x0), D1 = f[x0,x1], ···, D5 = f[x0,x1,x2,x3,x4,x5], for f(x) = ex. Use x0 = 0,x1 = 0.2,x2 = 0.4,...,x5 = 1.0. (b) Using the results of (a), calculate Pj(x) for x = 0.1,0.3,0.5 and j = 1,··· ,5. Compare these results to the true values of ex. Solution: % Exercise 4.1.26 x_nodes = linspace(0.0,1.0,6); y_values = exp(x_nodes); x_eval = linspace(0.1,0.5,3); divdif_y = divdif(x_nodes,y_values); fprintf ( 1, ’ j %d %d %d %d %d %d \n’,1:length(x_nodes)); fprintf ( 1, ’ Djf %1.4f %1.4f %1.4f %1.4f %1.4f %1.4f \n\n’,divdif_y); p_eval = interp(x_nodes,divdif_y,x_eval); fprintf ( 1, ’ j P_j(%g) P_j(%g) P_j(%g)\n’,x_eval); for i = 1:length(x_nodes)-1 p_eval = interp(x_nodes(1:i+1),divdif_y(1:i+1),x_eval); fprintf ( 1, ’ %d %1.7f %1.7f %1.7f \n’, i, p_eval); end fprintf ( 1, ’True %1.7f %1.7f %1.7f\n’,exp(x_eval)); >> Exercise_4_1_26 j 1 2 3 4 5 6 Djf 1.0000 1.1070 0.6127 0.2261 0.0626 0.0139 j P_j(0.1) P_j(0.3) P_j(0.5) 1 1.1107014 1.3321041 1.5535069 2 1.1045740 1.3504863 1.6454179 3 1.1052523 1.3498080 1.6488094 4 1.1051584 1.3498643 1.6487156 5 1.1051730 1.3498581 1.6487218 True 1.1051709 1.3498588 1.6487213 Exercise 27. Repeat Problem 26 with f(x) = arctan(x). Solution: % Exercise 4.1.27 x_nodes = linspace(0.0,1.0,6); y_values = atan(x_nodes); 40 x_eval = linspace(0.1,0.5,3); divdif_y = divdif(x_nodes,y_values); fprintf ( 1, ’ j %d %d %d %d %d %d \n’,1:length(x_nodes)); fprintf ( 1, ’ Djf %1.4f %1.4f %1.4f %1.4f %1.4f %1.4f \n\n’,divdif_y); p_eval = interp(x_nodes,divdif_y,x_eval); fprintf ( 1, ’ j P_j(%g) P_j(%g) P_j(%g)\n’,x_eval); for i = 1:length(x_nodes)-1 p_eval = interp(x_nodes(1:i+1),divdif_y(1:i+1),x_eval); fprintf ( 1, ’ %d %1.7f %1.7f %1.7f \n’, i, p_eval); end fprintf ( 1, ’True %1.7f %1.7f %1.7f\n’,atan(x_eval)); >> Exercise_4_1_27 j 1 2 3 4 5 6 Djf 0.0000 0.9870 -0.1786 -0.1857 0.1698 -0.0572 j P_j(0.1) P_j(0.3) P_j(0.5) 1 0.0986978 0.2960933 0.4934889 2 0.1004834 0.2907366 0.4667050 3 0.0999263 0.2912936 0.4639197 4 0.0996717 0.2914464 0.4636651 5 0.0996116 0.2914722 0.4636393 True 0.0996687 0.2914568 0.4636476 Exercise 28. The following data are taken from a polynomial p(x) of degree ≤5. What is the polynomial and what is its degree? xi f(xi) -2 -5 -1 1 0 1 1 1 2 7 3 25 Solution: The Newton divided differences are and therefore the interpolating polynomial (in Newton form), of degree three, is: Π3(x) = −5+6(x+2)−3(x+2)(x+1)+(x+2)(x+1)x. 41 xi f(xi) f[xi,xj] f[xi,xj,xk] f[···] f[···] f[···] -2 -5 -1 1 1−(−5) −1−(−2) = 6 0 1 1−1 0−(−1) = 0 0−6 0−(−2) = −3 1 1 0 0 0−(−3) 1−(−2) = 1 2 7 6 6−0 2−0 = 3 3−0 2−(−1) = 1 0 3 25 18 18−6 3−1 = 6 6−3 3−0 = 1 0 0 9. Section 4.2 ERROR IN POLYNOMIAL INTERPOLATION. We are evaluating the error made when ”replacing” a function f by its interpolating polynomial Πn f, obtained by interpolation on the nodes x0,x0,··· ,xn and node values f(x0), f(x1),··· , f(xn). DEFINITION 9.1. Given a n+1 nodes x0,x1,xn, the nodal polynomial is the polynomial of degree n+1, deﬁned as the product of all monomials (x−xi), namely: ωn+1(x) := (x−x0)(x−x1)···(x−xn) ≡ n ∏ i=0 (x−xi) ∈Pn+1 (nodal polynomial) REMARK 7. We note that ω′ n+1(x) = n ∑ j=0 n ∏ i=0 i̸=j (x−xi), ω′ n+1(xκ) = n ∏ i=0 i̸=κ (xκ −xi), and therefore the (Lagrange characteristic polynomial), respectively the (interpolating polynomial in Lagrange form) write, respectively: ℓi(x) = n ∑ i=0 ωn+1(x) (x−xi)ωn+1(xi), (Πn f)(x) = n ∑ i=0 ωn+1(x) (x−xi)ωn+1(xi) f(xi). THEOREM 9.2. Let x0,x1,··· ,xn be distinct n+1 nodes, and an arbitrary value x in the domain of deﬁnition of function f. Let now denote by Ix the smallest interval containing the nodes x0,x1,··· ,xn and also x. Assuming that f ∈Cn+1(Ix), then there exists an arbitrary cx ∈Ix such that the interpolation error at x ∈Ix is given by En(x) := f(x)−(Πn f)(x) = f (n+1)(cx) (n+1)! ωn+1(x) (interpolation error) = f[x0,x1,··· ,xn,x]ωn+1(x), (interpolation error / Newton form) where ωn+1 is the (nodal polynomial) of degree n+1. REMARK 8 (Drawbacks of global interpolation on equally spaced nodes.The Runge phenomenon). Notice that the error is inﬂuenced by the behavior of the nodal polynomial ωn+1, which can highly ﬂuctuate on equally spaced nodes, for n large. See the Runge counterexample and interpolation runge.m. (See the runge phenomenon.m, runge example.m and interpolation runge.m.) 42 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 FIG. 9.1. Global interpolation polynomials on equally spaced nodes, x nodes = [-3 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3], y values = [0 0 0 0 0 0 1 0 0 0 0 0 0]. -6 -4 -2 0 2 4 6 0 0.5 1 1.5 2 Runge counterexample on equally spaced nodes Runge function 1./(1+x. 2) interpolation polynomial nodes -6 -4 -2 0 2 4 6 0 0.5 1 1.5 2 Runge counterexample on equally spaced nodes Runge function 1./(1+x. 2) interpolation polynomial nodes -6 -4 -2 0 2 4 6 0 1 2 3 4 5 106 Runge counterexample on equally spaced nodes Runge function 1./(1+x. 2) interpolation polynomial nodes FIG. 9.2. The Runge function f(x) = 1 1+x2 interpolated on 10, 15 and 50 equally spaced nodes. Exercise 2. Consider interpolating f(x) = arctan(x) from a table of values of the function f given at equally spaced values of x for 0 ≤x ≤0.8; the x entries are given in steps of h = 0.01. (a) Bound the error f(x)−P 1(x) of linear interpolation in this table. The value of x is to satisfy x0 < x < x1, with x0 and x1 adjacent x entries in the table. (b) Bound the error f(x) −P 2(x) of quadratic interpolation in this table. The value of x is to satisfy x0 < x < x2, with x0,x1 and x2 adjacent x entries in the table. Solution: f(x) = arctan(x), f ′(x) = 1 1+x2 ≡(1+x2)−1, f ′′(x) = −2x(1+x2)−2 ≡− 2x (1+x2)2 , f ′′′(x) = (6x2 −2)(1+x2)−3 ≡6x2 −2 (1+x2)3 , f (4)(x) = 24x−24x3 (1+x2)−4 . Using the (interpolation error) formula we then have: 43 -5 -4 -3 -2 -1 0 1 2 3 4 5 -400 -300 -200 -100 0 100 200 300 400 Nodal polynomial on 5 equally spaced nodes in [-5,5] nodes nodal polynomial -5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5 105 Nodal polynomial on 11 equally spaced nodes in [-5,5] nodes nodal polynomial -5 -4 -3 -2 -1 0 1 2 3 4 5 -1.5 -1 -0.5 0 0.5 1 1.5 1011 Nodal polynomial on 21 equally spaced nodes in [-5,5] nodes nodal polynomial FIG. 9.3. Three nodal polynomials in [−5,5] using 5, 11 and 21 equally spaced nodes. Notice the magnitude on the y-axis! (a) for n = 1 E1(x) := f(x)−(Π1 f)(x) = f ′′(cx) 2! ω2(x) ≡−1 2! 2cx (1+c2 x)2 (x−x0)(x−x1), where x,cx ∈[x0,x1]. Since max x∈x0,x1(x−x1) = (x1 −x0)2 4 , (9.1) max y∈[x0,x1] f ′′(y) = f ′′√ 3 3 = 3 √ 3 8 , and h ≡x1 −x0, we have that \|E1(·)\| ≤1 2! 3 √ 3 8 h2 4 ≈8.1190×10−6. () (b) Similarly, for n = 2 E2(x) := f(x)−(Π2 f)(x) = f ′′′(cx) 3! ω3(x) ≡1 3! 6c2 x −2 (1+c2 x)3 (x−x0)(x−x1)(x−x2) where x,cx ∈[x0,x2]. Since max x∈[x0,x2] (x−x0)(x−x1)(x−x2) = 2 √ 3 9 h3, (9.2) max y∈[x0,x2]\|f ′′′(y)\| = \|f ′′′(0)\| = 2, where h = x1 −x0 ≡x2 −x1. Then we have \|E2(·)\| ≤1 3! ·2· 2 √ 3 9 h3 ≈1.2830×10−7. () Exercise 6. Suppose a table of values of f(x) = arctan(x),x ∈[0,1] is to be constructed, with values of arctan(x) given with a spacing of h. 44 1. If linear interpolation is used in this table, how small should h be in order for the interpolation error to be less than 5×10−6? For notational assumptions, see Exercise 2 (a). 2. If quadratic interpolation is used in this table, how small should h be in order for the interpolation error to be less than 5×10−6? For notational assumptions, see Exercise 2 (b). Solution: 1. From () in Exercise 2 (a) we have that \|E1(x)\| ≤3 √ 3 64 h2 hence in order for the error to be less than 5×10−6 it sufﬁces to have h ≤ s 5×10−6 64 3 √ 3 ≈0.0078. 2. Similarly, from () in Exercise 2 (b) \|E2(x)\| ≤2 √ 3 27 h3 and h ≤ 3 s 5×10−6 9 √ 3 2 ≈0.0339. Exercise 7. Consider constructing a table of values of f(x) = √x for x ∈[1,100], with values of f(x) given for x = 0,h,2h,···. Choose h so that when the linear interpolation is used in this table, the error is bounded by 5×10−6. Discuss and compare the linear interpolation error near x = 1 and x = 100. Solution: Since f ′(x) = 0.5x−1/2, f ′′(x) = −1 4x−3 2 , on any interval [xi,xi+1], the interpolation error is bounded by (see (interpolation error)) we have that \|E1(x)\| = f ′′(x) 2! (x−xi)(x−xi+1) ≤1 2! 1 4 max x∈[1,100]\|x−3 2 \| \| {z } =1 h2 4 = 1 32h2, ∀x ∈[1,100]. The error is less than 5×10−6 provided h ≤0.0126. When x ≈1, \|E1(x)\| ≈1 2! 1 4 h2 4 = 1 32h2 ≈0.0312×h2, while when x ≈100 \|E1(x)\| ≤1 2! 1 410−3 h2 4 ≈0.00003125×h2. Exercise 8. Let f(x) = x4 + √ 2x3 +πx. Verify whether f[0,1,2,3,4] = f[0,1,π,e,−1]. Solution: Since f (4)(x) = 4! ∀x, 45 using (interpolation error / Newton form) we conclude that indeed f[0,1,2,3,4] = f[0,1,π,e,−1] = 1. Exercise 9. Let f(x) = a0 +a1x+···+anxn be a polynomial of degree less than or equal to n, and let {x0,x1,··· ,xn} be distinct points. What is the value of f[x0,x1,··· ,xn]? Solution: Similar to Exercise 8, since f (n)(x) = n!an, from (interpolation error / Newton form) we have that f[x0,x1,··· ,xn] = f (n)(cx) n! = an. 46 10. Piecewise Interpolations. REMARK 9. (i) As seen in Figure 9.2, on the very smooth Runge counter-example function f(x) = 1 1+x2 , global interpolation on equally spaced nodes does not guarantee uniform convergence Πn f →f as n →∞. (See also Figure 9.3 for the behavior of the nodal polynomial.) (ii) On the other hand, equally spaced nodes are computationally convenient. (iii) (Lagrange) interpolation of low degree κ is sufficiently accurate, provided sufﬁciently small inter-vals are considered. Indeed, by (interpolation error) we have that \|Eκ(x)\| ≤\|f (κ+1)(cx)\| (κ +1)! \|ωκ+1(x)\| \| {z } ≤Chκ+1 . for f ∈Cκ+1[a,b], (10.1) on any consecutive h-equally spaced nodes. (For example, see (9.1) for κ = 1: \|ω2(x)\| ≤1 4h2 and (9.2) for κ = 2: \|ω3(x)\| ≤2 √ 3 9 h3.) 10.1. Section 4.3 INTERPOLATION USING SPLINE FUNCTIONS. The goal is to approximate a given function with splines, allowing a piecewise interpolation with global smoothness. DEFINITION 10.1. Let x0,x1,··· ,xn be (n+1) distinct nodes in [a,b], a = x0 < x1 < ··· < xn = b. The function sk(x) on [a,b] is a spline of degree k relative to the nodes xj if (1) sk is piecewise polynomial: sk [x j,x j+1] ∈Pk, j = 0,1,··· ,n−1 (10.2) (2) global smoothness/continuity at nodes: sk ∈Ck−1[a,b] (10.3) We denote Sk the space of splines sk on [a,b] relative to n+1 distinct nodes. dim(Sk) = n+k. PROPOSITION 10.2. For a function f ∈Cκ+1([a,b]) we have the following uniform convergence resullt ∥f −sκ∥∞≤Chκ+1∥f (κ+1)∥∞. Proof. Use (10.1). PROPOSITION 10.3. For a function square integrable satisfying f (m) ∈L2(a,b) for all m = 0,···κ + 1, there exists a positive constant C > 0, independent of h such that ∥( f −sκ)(m)∥L2(a,b) ≤Chκ+1−m∥f (κ+1)∥L2(a,b). In particular, for κ = 1 (piecewise linear interpolation) and m = 0 or m = 1 we have that ∥f −s1∥L2(a,b) ≤C1h2∥f ′′∥L2(a,b), ∥(f −s1)′∥L2(a,b) ≤C2h∥f ′′∥L2(a,b). For κ = 3 (piecewise cubic interpolation) and m = 0,1,2,3 we have that ∥f −s3∥L2(a,b) ≤C1h4∥f (4)∥L2(a,b), ∥( f −s1)′∥L2(a,b) ≤C2h3∥f 4∥L2(a,b), ∥( f −s3)′′∥L2(a,b) ≤C3h2∥f (4)∥L2(a,b), ∥(f −s1)′′′∥L2(a,b) ≤C4h∥f 4∥L2(a,b). 47 -5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Piecewise (linear and cubic splines) interpolation of the "Runge" example data exact piecewise linear not-a-knot cubic spline -5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Piecewise (linear and cubic splines) interpolation of the "Runge" example data exact piecewise linear not-a-knot cubic spline -5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Piecewise (linear and cubic splines) interpolation of the "Runge" example data exact piecewise linear not-a-knot cubic spline FIG. 10.1. The Runge function f(x) = 1 1+x2 interpolated with piecewise linear and not-a-knot cubic splines, on 6, 8 and 10 equally spaced nodes. (Compare with Figure 9.2 - for global interpolation on equally spaced nodes.) Proof. Use (10.1), Rolle’s theorem and the Cauchy-Bunyakovsky-Schwarz inequality. Hence a small interpolation error can be obtained even for low κ, provided h is sufﬁciently “small”. REMARK 10. We remark that relations (10.2)-(10.3) do not sufﬁce to fully characterize a spline of degree k. Indeed, by condition (10.2) we see that (k +1)n coefﬁcients sij must be determined (k +1 coefﬁcients, n inter-vals): sk, j(x) := sk x j,x j+1 = k ∑ i=0 si, j(x−xj)i, if x ∈[xj,xj+1], j = 0,··· ,n−1. On the other hand, the continuity at the (interior) nodes condition (10.3) only provides k(n−1) conditions: s(m) k, j−1(x j) = s(m) k, j (xj), j = 1,··· ,n−1, m = 0,1,··· ,k −1. Therefore the deﬁnition leaves (k +1)n \| {z } unknowns −k(n−1) \| {z } conditions = n+k undetermined coefﬁcients. Assuming that the splines were interpolatory, i.e., sk(xj) = f j, j = 0,··· ,n where f0,··· , fn are given values, we would still have n+k −(n+1) = k −1 unsaturated degrees of freedom. Further constraints are imposed, leading to (i) periodic splines s(m) k (a) = s(m) k (b), m = 1,2,··· ,k −1 (ii) natural splines: for k = 2ℓ−1, with ℓ≥2 s(ℓ+j) k (a) = s(ℓ+j) k (b) = 0, j = 0,1,··· ,ℓ−2 48 10.2. Cubic spline interpolation: k = 3. The cubic splines are particularly signiﬁcant since they are • splines of minimum degree which yield C2 approximation, and are • sufﬁciently smooth in presence of small curvatures (see Proposition 10.5 for the minimum norm property of the curvature). In this case there are only k −1 ≡2 unsaturated degrees of freedom. So for the closure we have either • natural cubic splines: s′′ 3(a) = s′′ 3(b) = 0, • not-a-knot spline with – active nodes x0,x2,··· ,xn−2,xn (hence n−2 intervals, 4n−8 unknowns; 3(n−3) smoothness on interior nodes PLUS n+1 interpo-lation conditions; giving 4n−8 equations) and – x1,xn−1 as only interpolating nodes REMARK 11. MATLAB’s function spline uses not-a-knot cubic splines. spline(x_nodes,y_values,x_evaluatepoint) DEFINITION 10.4 (Natural cubic splines). There is a unique function s(x) such that (S1) s3(x) ∈P3 on each subinterval [xj−1,xj], j = 1,··· ,n; (S2) s3(x),s′ 3(x),s′′ 3(x) are continuous on [a,b]; (S3) s′′ 3(a) = s′′ 3(b) = 0, which also interpolates the data {xi,yi}i=0,···,n. 10.3. Construction of natural cubic splines. Let us denote Mi := s′′ 3(xi), i = 0,··· ,n, and according to (S3) set M0 = Mn = 0. Then, the cubic spline has the form s3(x) =(xj −x)3Mj−1 +(x−xj−1)3Mj 6(xj −x j−1) + (xj −x)yj−1 +(x−xj−1)yj xj −xj−1 (4.63) −1 6(xj −xj−1)[(xj −x)Mj−1 +(x−x j−1)Mj], x ∈[xj−1,x j], where M1,··· ,Mn−1 are solutions of the linear system xj −xj−1 6 Mj−1 + xj+1 −xj−1 3 Mj + xj+1 −x j 6 Mj+1 = yj+1 −yj xj+1 −xj −yj −yj−1 xj −xj−1 , j = 1,2,··· ,n−1. (4.64) PROPOSITION 10.5 (Minimum norm property). Let f ∈C2[a,b] and s3 be the natural cubic spline interpolating f. Then Z b a s′′ 3(x) 2dx ≤ Z b a f ′′(x) 2dx (10.4) where equality holds if and only if f = s3. REMARK 12. The property (10.4) holds even for splines that satisfy s′ 3(a) = f ′(a), s′ 3(b) = f ′(b). PROPOSITION 10.6. Let f ∈C4[a,b] and [a,b] = ∪N−1 i=0 Ii be a partition of width hi, with h = maxhi, and β = h/maxi hi. Then the s3 cubic spline interpolating f satisﬁes ∥f (r) −s(r) 3 ∥∞≤Crh4−r∥f (4)∥∞, r = 0,1,2,3 with C0 = 5 384, C1 = 1 24, C2 = 3 8 and C3 = (β +β −1)/2. Exercise 2. Consider the data 49 x 1 2 3 4 5 y 3 1 2 3 2 (a) Find the piecewise linear interpolating functions ℓ(x). (b) Find the cubic spline function s(x) that interpolates the data and satisﬁes the not-a-knot boundary conditions s(z1) = f(z1), s(z2) = f(z2). (4.73) Note in this case that n = 3,x1 = 1,x2 = 3,x3 = 5,z1 = 2,z2 = 4. Graph both s(x) and ℓ(x) for x ∈[0,5]. Solution: (a) The Newton divided differences are hence the piecewise linear interpolating polynomial is: x y [yi,yj] 1 3 2 1 1−3 2−1 = −2 3 2 2−1 3−2 = 1 4 3 3−2 4−3 = 1 5 2 2−3 5−4 = −1 ℓ(x) =        3−2(x−1), x ∈[1,2] 1+(x−2), x ∈[2,3] 2+(x−3), x ∈[3,4] 3−(x−4), x ∈[4,5] (b) Use (4.64) to ﬁnd the cubic spline by hand. xj −xj−1 6 Mj−1 + x j+1 −xj−1 3 Mj + xj+1 −xj 6 Mj+1 = y j+1 −yj x j+1 −xj −y j −yj−1 x j −xj−1 , j = 2,3,··· ,n−1. (4.64’) Then taking n = 3, hence j = 2 we get x2 −x1 6 M1 + x3 −x1 3 M2 + x3 −x2 6 M3 = y3 −y2 x3 −x2 −y2 −y1 x2 −x1 , and using the not-a-knot conditions s(2) = 1, s(4) = 3 in formula (4.63) s(x) = (xj −x)3Mj−1 +(x−xj−1)3Mj 6(x j −xj−1) + (xj −x)yj−1 +(x−xj−1)yj xj −x j−1 −1 6(x j −xj−1) (x j −x)Mj−1 +(x−xj−1)Mj], x ∈[x j−1,x j] with j = 2,x1 = 1,x2 = 3,y1 = 3,y2 = 2 and x = z1 = 2 gives M1 +M2 = 6, respectively, with j = 3,x2 = 3,x3 = 5,y2 = 2,y3 = 2 and x = z2 = 4 gives M2 +M3 = −4. 50 1 1.5 2 2.5 3 3.5 4 4.5 5 1 1.5 2 2.5 3 3.5 piecewise linear cubic spline data points FIG. 10.2. The piecewise linear and natural cubic spline for Exercise 2. Therefore M1 = 25 4 , M2 = −1 4, M3 = −15 4 . You could also use MATLAB’s built-in functions for the piecewise linear interpolant and the not-a-knot cubic spline: x_nodes = linspace(1,5,5); % x nodes y_values = [3 1 2 3 2]; % y values t = linspace(1,5,100); % values where the interpolants are evaluated linear = interp1(x_nodes,y_values,t); % the piecewise interpolant cubic = spline(x_nodes,y_values,t); % not-a-knot cubic spline plot(t,linear,’k.-.’,t,cubic,’b.-.’, x_nodes,y_values,’ro’) legend(’piecewise linear’,’cubic spline’,’data points’) Exercise 3. Consider the data x 0 1/2 1 2 3 y 0 1/4 1 -1 -1 1. Find the piecewise linear interpolating function for the data. 2. Find the piecewise quadratic interpolating function. 3. Find the natural cubic spline that interpolates the data. 4. Find the not-a-knot interpolating cubic spline. When using (4.73), let x1 = 0,x2 = 1,x3 = 3, and z1 = 1 2,z2 = 2. Solution: 51 0 0.5 1 1.5 2 2.5 3 -2 -1.5 -1 -0.5 0 0.5 1 1.5 Exercise 4.3.3 piecewise linear piecewise quadratic not-a-knot cubic spline natural cubic spline data points polyfit p3 0 0.5 1 1.5 2 2.5 3 -2 -1.5 -1 -0.5 0 0.5 1 1.5 Exercise 4.3.3, (c) and (d) not-a-knot cubic spline natural cubic spline FIG. 10.3. The piecewise linear, piecewise quadratic, not-a-knot cubic spline, natural cubic spline and the cubic least-squares polyﬁt. 1. The ﬁrst-order Newton divided differences are x y [yi,yj] 0 0 1/2 1/4 1/4−0 1/2 = 1/2 1 1 1−1/4 1−1/2 = 3/2 2 -1 −1−1 2−1 = −2 3 -1 −1−(−1) 3−2 = 0 and the piecewise linear polynomial is ℓ(x) =        0+1/2x, x ∈[0,1/2] 1/4+3/2(x−1/2), x ∈[1/2,1] 1−2(x−1), x ∈[1,2] −1, x ∈[2,3]. 2. The second-order Newton divided differences are x y [yi,yj] [yi,yj,yk] 0 0 1/2 1/4 1/4−0 1/2 = 1/2 1 1 1−1/4 1−1/2 = 3/2 3/2−1/2 1−0 = 1 2 -1 −1−1 2−1 = −2 3 -1 −1−(−1) 3−2 = 0 0−(−2) 3−1 = 1 and the piecewise quadratic polynomial is e Π2(x) = 0+1/2x+x(x−1/2) ≡x2, x ∈[0,1] 1−2(x−1)+(x−1)(x−2) ≡x2 −5x+5, x ∈[1,3]. 3. Find the natural cubic spline that interpolates the data. Let denote the nodes and the values x0 = 0,x1 = 1/2,x2 = 1,x3 = 2,x4 = 3, (nodes) 52 y0 = 0,y1 = 1/4,y2 = 1,y3 = −1,y4 = −1, (values) and use formula (4.64) to ﬁnd the coefﬁcients M0,M1,··· ,M4, by taking j = 1,2,3 xj −x j−1 6 Mj−1 + xj+1 −xj−1 3 Mj + xj+1 −x j 6 Mj+1 = yj+1 −yj xj+1 −xj −yj −yj−1 xj −xj−1 , x1 −x0 6 M0 + x2 −x0 3 M1 + x2 −x1 6 M2 = y2 −y1 x2 −x1 −y1 −y0 x1 −x0 , (j=1) x2 −x1 6 M1 + x3 −x1 3 M2 + x3 −x2 6 M3 = y3 −y2 x3 −x2 −y2 −y1 x2 −x1 , (j=2) x3 −x2 6 M2 + x4 −x2 3 M3 + x4 −x3 6 M4 = y4 −y3 x4 −x3 −y3 −y2 x3 −x2 , (j=3) hence 1/2−0 6 M0 + 1−0 3 M1 + 1−1/2 6 M2 = 1−1/4 1−1/2 −1/4−0 1/2−0, (j=1) 1−1/2 6 M1 + 2−1/2 3 M2 + 2−1 6 M3 = −1−1 2−1 −1−1/4 1−1/2, (j=2) 2−1 6 M2 + 3−1 3 M3 + 3−2 6 M4 = −1−(−1) 3−2 −−1−1 2−1 , (j=3) or equivalently 1 12M0 + 1 3M1 + 1 12M2 = 3 2 −1 2 ≡1, (j=1) 1 12M1 + 1 2M2 + 1 6M3 = −2−3 2 ≡−7 2, (j=2) 1 6M2 + 2 3M3 + 1 6M4 = −1−(−1) 3−2 −−1−1 2−1 ≡2. (j=3) From the deﬁnition of a natural cubic spline we have that M0 = M4 = 0, and solving the above system M0 + 1 3M1 + 1 12M2 = 1, (j=1) 1 12M1 + 1 2M2 + 1 6M3 = −2−3 2 ≡−7 2, (j=2) 1 6M2 + 2 3M3 + 1 6M4 = −1−(−1) 3−2 −−1−1 2−1 ≡2. (j=3) we obtain M0 = 0, M1 = 38 7 , M2 = −68 7 , M3 = 38 7 , M4 = 0. Now using deﬁnition (4.63) we obtain the expressions on each interval for s3(x): s3(x) =(xj −x)3Mj−1 +(x−x j−1)3Mj 6(xj −xj−1) + (xj −x)y j−1 +(x−xj−1)yj xj −xj−1 (4.63) −1 6(xj −xj−1)[(xj −x)Mj−1 +(x−xj−1)Mj], (on x ∈[xj−1,xj]) 53 by taking j = 1,2,3,4 respectively s3(x) = (x1 −x)3M0 +(x−x0)3M1 6(x1 −x0) + (x1 −x)y0 +(x−x0)y1 x1 −x0 −1 6(x1 −x0)[(x1 −x)M0 +(x−x0)M1], ((j=1)⇒x ∈[x0,x1]) = (x2 −x)3M1 +(x−x1)3M2 6(x2 −x1) + (x2 −x)y1 +(x−x1)y2 x2 −x1 −1 6(x2 −x1)[(x2 −x)M1 +(x−x1)M2], ((j=2)⇒x ∈[x1,x2]) = (x3 −x)3M2 +(x−x2)3M3 6(x3 −x2) + (x3 −x)y2 +(x−x2)y3 x3 −x2 −1 6(x3 −x2)[(x3 −x)M2 +(x−x2)M3], ((j=3)⇒x ∈[x2,x3]) = (x4 −x)3M3 +(x−x3)3M4 6(x4 −x3) + (x4 −x)y3 +(x−x3)y4 x4 −x3 −1 6(x4 −x3)[(x4 −x)M3 +(x−x3)M4]. ((j=4)⇒x ∈[x3,x4]) Substituting in the values for the nodes, values and the coefﬁcients Mi we obtain s3(x) = x3 ·38/7 6/2 + x/4 1/2 −1 61/2·x·38/7, (x ∈[0,1/2]) = (1−x)3 ·38/7−(x−1/2)3 ·68/7 6(1−1/2) + (1−x)/4+(x−1/2) 1−1/2 −1 6(1−1/2)[(1−x)·38/7−(x−1/2)68/7], (x ∈[1/2,1]) = −(2−x)3 ·68/7+(x−1)3 ·38/7 6(2−1) + (2−x)−(x−1) 2−1 −1 6(2−1)[−(2−x)·68/7+(x−1)·38/7], (x ∈[1,2]) = (3−x)3 ·38/7 6(3−2) + −(3−x)−(x−2) 3−2 −1 6(3−2)(3−x)38/7. (x ∈[2,3]) Finally s3(x) =                        38 21x3 + 1 21x, x ∈[0,1/2] −106 21 x3 + 72 7 x2 −107 21 x+ 6 7, x ∈[1/2,1] 53 21x3 −87 7 x2 + 370 21 x−47 7 , x ∈[1,2] −19 21x3 + 57 7 x2 −494 21 x+ 145 7 , x ∈[2,3] 4. Find the not-a-knot interpolating cubic spline. When using (4.73), let x1 = 0,x2 = 1,x3 = 3, and z1 = 1 2,z2 = 2. (See Exercise 4 3 3.m and Figure 10.3.) x_nodes = [ 0 1/2 1 2 3]; % x nodes y_values = [ 0 1/4 1 -1 -1]; % y values t = linspace(0,3,300); % values where the interpolants are evaluated t1 = linspace(0,1,100); t2 = linspace(1,3,200); linear = interp1(x_nodes,y_values,t); % the piecewise interpolant 54 quadratic1 = t1.^2; quadratic2 = t2.^2 - 5t2 + 5; cubic = spline(x_nodes,y_values,t); % not-a-knot cubic spline p3 = polyfit(x_nodes,y_values,3) % polynomial of degree 3 approximated in L2 sense p3eval = polyval(p3,t); %p3 polynomial evaluated figure() hold all plot(t,linear,’k.-.’,[t1 t2],[quadratic1 quadratic2],... ’m.-.’,t,cubic,’b.-.’, x_nodes,y_values,’g’) plot(t,p3eval,’r.-.’) legend(’piecewise linear’,’piecewise quadratic’,’cubic spline’,... ’data points’,’polyfit p3’) end Exercise 5. Use the MATLAB built-in function spline to interpolate (see also (Runge example)) f(x) = 1 1+x2 , x ∈[−5,5], from Example 4.2.5 (the Runge phenomenon) with the following sets of x nodes: (i) {−5,−2.5,0,2.5,5} (ii) {−5,−3.5,−2,0,2,3.5,5} (iii) {−5,−4.5,−4,−3,−2,−1,0,1,2,3,4,4.5,5} In each case, graph the spline function and the function f(x). Compare with the global polynomial interpolation. Solution: -5 -4 -3 -2 -1 0 1 2 3 4 5 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 cubic (i) cubic (ii) cubic (iii) Runge FIG. 10.4. Three not-a-knot cubic spline approximations of the Runge example for Exercise 5. (See Exercise 4 3 5.m.) x_nodes1 = [ -5, -2.5 , 0 , 2.5 , 5]; % x nodes x_nodes2 = [ -5, -3.5, -2 , 0 , 2, 3.5 , 5]; x_nodes3 = [ -5, -4.5, -4, -3, -2 , -1, 0 , 1, 2, 3, 4, 4.5 , 5]; y_values1 = 1./(1+x_nodes1.^2); % y values 55 y_values2 = 1./(1+x_nodes2.^2); y_values3 = 1./(1+x_nodes3.^2); t = linspace(-5,5,200); % values where the interpolants are evaluated cubic1 = spline(x_nodes1,y_values1,t); % not-a-knot cubic spline cubic2 = spline(x_nodes2,y_values2,t); cubic3 = spline(x_nodes3,y_values3,t); runge = 1./(1+t.^2); % figure(1) % plot(t,cubic1,’k.-.’,t, runge,’b-’) % legend(’cubic interpolant’,’Runge’) % figure(2) % plot(t,cubic2,’k.-.’,t, runge,’b-’) % legend(’cubic interpolant’,’Runge’) % figure(3) % plot(t,cubic3,’k.-.’,t, runge,’b-’) % legend(’cubic interpolant’,’Runge’) figure(4) plot(t,cubic1,’r.-.’,t,cubic2,’b.-.’,t,cubic3,’k.-.’,t,runge,’g-’) legend(’cubic (i)’,’cubic (ii)’,’cubic (iii)’,’Runge’) Exercise 10. Is the following function s(x) = (x−1)3, x ∈[0,1] 2(x−1)3, x ∈[1,2] a cubic spline on the interval x ∈[0,2]? Solution: Since s′(x) = 3(x−1)2, x ∈[0,1] 6(x−1)2, x ∈[1,2] , s′′(x) = 6(x−1), x ∈[0,1] 12(x−1), x ∈[1,2] we have that s,s′,s′′ are continuous at x = 1. Moreover, s is cubic on each subinterval, and therefore is a cubic spline. Exercise 12. Deﬁne s(x) =    2x3, x ∈[0,1] x3 +3x2 −3x+1, x ∈[1,2] 9x2 −15x+9, x ∈[2,3] Verify that s(x) is a cubic spline function on [0,3]. Is it a natural cubic spline function on this interval? Solution: Since s′(x) =    6x2, x ∈[0,1] 3x2 +6x−3, x ∈[1,2] 18x−15, x ∈[2,3] , s′′(x) =    12x, x ∈[0,1] 6x+6, x ∈[1,2] 18, x ∈[2,3] we see that s,s′,s′′ are continuous at x = 1 and x = 2. Being cubic on each subinterval, s is therefore a cubic spline. s′′(0) = 0, but s′′(3) = 18 ̸= 0, hence s is not a natural cubic spline. Exercise 13. Deﬁne s(x) = x3 −3x2 +2x+1, x ∈[1,2] −x3 +9x2 −22x+17, x ∈[2,3]. 56 Is s(x) a cubic spline on [1,3]? Is it a natural cubic spline function? Solution: Since s′(x) = 3x2 −6x+2, x ∈[1,2] −3x2 +18x−22, x ∈[2,3] , s′′(x) = 6x−6, x ∈[1,2] −6x+18, x ∈[2,3]. we see that s,s′,s′′ are continuous at x = 2. Being cubic on each subinterval, s is therefore a cubic spline. Also s′′(1) = 0, s′′(3) = 0, hence s is a natural cubic spline. Exercise 14. Is the following function a cubic spline on [0,3]? s(x) =    x3, x ∈[0,1] 2x−1, x ∈[1,2] 3x2 −9, x ∈[2,3] Solution: We have s′(x) =    3x2, x ∈[0,1] 2, x ∈[1,2] 6x, x ∈[2,3] , s′′(x) =    6x, x ∈[0,1] 0, x ∈[1,2] 6, x ∈[2,3] hence s(1) = 1, s(2) = 3, s′(1−) = 1 ̸= 2 = s′(1+), and therefore s is not a cubic spline. Exercise 15. Deﬁne s(x) = x3 +2x2 +1, x ∈[1,2] −2x3 +βx2 −36x+25, x ∈[2,3] For a special value of β, s(x) is a cubic spline function on [1,3]. Find that value of β and then verify that s(x) is a cubic spline function on [1,3]. Is it a natural cubic spline function on this interval? Solution: Since s(2−) = 17, s(2+) = −63+4β, in order s to be continuous we have β = 20, hence s(x) = x3 +2x2 +1, x ∈[1,2] −2x3 +20x2 −36x+25, x ∈[2,3] s′(x) = 3x2 +4x, x ∈[1,2] −6x2 +40x−36, x ∈[2,3] , s′′(x) = 6x+4, x ∈[1,2] −12x+40, x ∈[2,3] . We have that s′(2−) = 20 = s′(2+),s′′(2−) = 16 = s′(2+), 57 hence s is a cubic spline. Also, since s′′(1) = 10 ̸= 0 s is not a natural cubic spline. Exercise 16. Is there a choice of coefﬁcients {a,b,c,d} for which the following function is a cubic spline? s(x) =    (x+1)3, x ∈[−2,−1] ax3 +bx2 +cx+d, x ∈[−1,1] (x−1)2, x ∈[1,2] Solution: Since s′(x) =    3(x+1)2, x ∈[−2,−1] 3ax2 +2bx+c, x ∈[−1,1] 2(x−1), x ∈[1,2] , s′′(x) =    6(x+1), x ∈[−2,−1] 6ax+2b, x ∈[−1,1] 2, x ∈[1,2] by imposing the continuity condition, we obtain the following s(−1−) = 0 = −a+b−c+d = s(−1+), s(1−) = a+b+c+d = 0 = s(1+), s′(−1−) = 0 = 3a−2b+c = s′(−1+), s′(1−) = 3a+2b+c = 0 = s′(−1+), s′′(−1−) = 0 = −6a+2b = s′′(−1+), s′′(1−) = 6a+2b = 2 = s′′(−1+). This overdetermined homogeneous system does not have a solution, so s cannot be a cubic spline. 58 11. ORTHOGONAL POLYNOMIALS IN APPROXIMATION THEORY. 11.1. APPROXIMATION FUNCTIONS BY GENERALIZED FOURIER SERIES. DEFINITION 11.1. A function w = w(x) which is non-negative and integrable on (−1,1) is called a weight function. For example w(x) = 1 (for ‘orthogonality’ of Legendre polynomials) w(x) = 1 √ 1−x2 (for ‘orthogonality’ of Chebyshev polynomials) DEFINITION 11.2 (Orthogonal polynomials). We say that the family of polynomials {pk;k = 0,1,··· ,deg(pk) = k,}, i.e., pk ∈Pk are mutually orthogonal on (−1,1) with respect to the weight w(·) if Z 1 −1 pk(x)pm(x)dx = 0, ∀k ̸= m. (orthogonal polynomials) DEFINITION 11.3 (weighted scalar product and norm). The following deﬁne the weighted scalar product and weighted norm ⟨f,g⟩w := Z 1 −1 f(x)g(x)w(x)dx (weighted scalar product) ∥f∥2 w := ⟨f, f⟩ 1 2 w (weighted norm) for the (weighted) function space L2 w(−1,1) = { f : (−1,1) →R, Z 1 −1 f 2(x)w(x)dx < ∞}. For example, with w(x) = 1 √ 1−x2 , the functions f(x) = 1 and g(x) = x are mutually orthogonal with respect to the L2 w inner product Z 1 −1 1·x 1 √ 1−x2 = − p 1−x2 1 −1 = 0. DEFINITION 11.4 (Generalized Fourier series). For all f ∈L2 w, the series S f(x) := ∞ ∑ k=0 ˆ fk pk(x), (generalized Fourier series) where ˆ fk = ⟨f, pk⟩w ∥pk∥2 w (kth Fourier coefﬁcient) is called the generalized Fourier series of f, ˆ fk is the k-th Fourier coefﬁcient. PROPOSITION 11.5. S f convergens in average (or in L2 w sense) to f: lim n→∞∥f −fn∥w = 0, (11.1) 59 where fn is the truncation of order n fn(x) = n ∑ k=0 ˆ fkpk(x) ∈Pn. (truncation of order n) Moreover, the following Parseval’s equality holds ∥f∥2 w = ∞ ∑ k=0 ˆ f 2 k ∥pk∥2 w, (Parseval equality) and for any n ∥f −fn∥2 w = ∞ ∑ k=n+1 ˆ f 2 k ∥pk∥2 w n↗∞ − →0. PROPOSITION 11.6 (minimization property). fn ∈Pn satisﬁes the minimization property ∥f −fn∥w = min q∈Pn ∥f −q∥w. (11.2) i.e., fn is the orthogonal projection of f over Pn in the sense of L2 w: fn ≡ProjectionPn( f). Proof. We note that by the (truncation of order n) formula and the orthogonality deﬁnition we have ⟨f −fn,q⟩w = ⟨ ∞ ∑ k=n+1 ˆ fk pk(x) \| {z } ∈Pn+1 ,q(x) \|{z} ∈Pn ⟩w = 0. Then using this property and the Cauchy-Schwarz inequality we have ∥f −fn∥C 2 w = ⟨f −fn, f −q⟩+ ⟨f −fn \| {z } ∈Pn+1 ,q−fn \| {z } ∈Pn ⟩ = ⟨f −fn, f −q⟩≤XXXX X ∥f −fn∥w∥f −q∥w, ∀q ∈Pn, which concludes the proof. DEFINITION 11.7 (monic polynomials). The space of monic polynomials is P1 n = p(x) = n ∑ k=0 akxk, an = 1 . THEOREM 11.8 (recursive 3-term formula). For any family of monic orthogonal polynomials {pk}, the following recursive 3-term formula holds pk+1(x) = (x−αk)pk(x)−βkpk−1(x), k ≥0, p−1(x) = 0, p0(x) = 1 where αk = ⟨xpk, pk⟩w ⟨pk, pk⟩w , αk = ⟨pk+1, pk+1⟩w ⟨pk, pk⟩w , k ≥0. 60 11.2. THE CHEBYSHEV POLYNOMIALS. Consider the Chebyshev weight w(x) = 1 √ 1−x2 on x ∈(−1,1). (Chebyshev weight) The space of square integrable functions with respect to the (Chebyshev weight) w(x) is L2 w(−1,1) = { f : (−1,1) →R : Z 1 −1 f 2(x)(1−x2)−1/2dx < ∞}. The inner product and norm are ⟨f,g⟩w = Z 1 −1 f(x)g(x)(1−x2)−1/2dx, (11.3) ∥f∥w = Z 1 −1 f 2(x)(1−x2)−1/2dx 1/2 . DEFINITION 11.9 (Chebyshev polynomials). The (Chebyshev polynomials) are deﬁned as Tk(x) = cos(kθ), θ arccosx, k = 0,1,··· . (Chebyshev polynomials) PROPOSITION 11.10 (3-term recursion of (Chebyshev polynomials)). The (Chebyshev polynomials) can be generated recursively by Tk+1(x) = 2xTk(x)−Tk−1(x), k = 1,2,··· , T0(x) = 1, T1(x) = x. In particular, for any k ≥0 notice that Tk ∈Pk. Proof. Using the trigonometric identities cos(n+1)θ = cosnθ cosθ −sinnθ sinθ cos(n−1)θ = cosnθ cosθ +sinnθ sinθ and adding them we obtain cos(n+1)θ \| {z } Tn+1(x) = 2cosnθ \| {z } Tn(x) cosθ −cos(n−1)θ \| {z } =Tn−1(x) the 3-term recursion. COROLLARY 11.11. Tn(x) = 2n−1xn +terms of lower degree, in other words 1 2n−1 Tn is a monic polynomial. For example T2(x) = 2x2 −1, T3(x) = 4x3 −3x, T4(x) = 8x4 −8x2 +1. 61 REMARK 13. Using the well-known trigonometric relations ⟨Tk,Tn⟩w = 0 if k ̸= n, ⟨Tk,Tn⟩w = c0 = π, if n = 0, cn = π 2 , if n ̸= 0. it follows that the (Chebyshev polynomials) are mutually orthogonal with respect with the inner product (11.3). The Chebyshev series of f ∈L2 w(−1,1) takes the form C f = ∞ ∑ k=0 ˆ fkTk, with ˆ fk = 1 ck Z 1 −1 f(x)Tk(x)(1−x2)−1/2 dx. REMARK 14. From the deﬁnition, the (Chebyshev polynomials) satisfy the following bound ∥Tn∥∞= max x∈[−1,1] Tn(x) = 1, for every n. (11.4) Let us recall now the (interpolation error) formula for global polynomial interpolation is f(x)−(Πn f)(x) = f (n+1)(cx) (n+1)! ωn+1(x), x ∈[−1,1]. where ωn+1 = (x−x0)···(x−xn), a monic polynomial, is the (nodal polynomial). Let us ask the following question what are the nodes x0,··· ,xn which minimize the “size” of ωn+1(x)? Or, in other words, what is the nodal polynomial ωn+1(x) of minimal size? PROPOSITION 11.12. The nodal polynomial of minimal size is ωn+1(x) = 1 2n Tn+1(x), ∥ωn+1∥∞≥∥21−(n+1)Tn+1∥∞≡∥2−nTn+1∥∞, the minimal size is 1 2n , and the interpolation nodes are the zeros of Tn+1(x): xj = cos 2j +1 2n+2π , j = 0,1,··· ,n. THEOREM 11.13 (Min-max theorem of Chebyshev). 1 2n−1 Tn ∞ ≤min p∈P1 n ∥p∥∞. (11.5) 11.3. THE LEGENDRE POLYNOMIALS. The Legendre polynomials are (orthogonal polynomials) on (−1,1) w.r.t. the weight function w(x) = 1, forming the space L2 w(−1,1) ≡L2(−1,1). DEFINITION 11.14. The Legendre polynomials can be deﬁned by the 3-term recursion formula Lk+1(x) = 2k +1 k +1 xLk(x)− k k +1Lk−1(x), k = 1,2,··· (11.6) 62 L0(x) = 1, L1(x) = x (3-term Legendre recursion) or equivalently by L0(x) = 1, Ln(x) = 1 n!2n dn dxn (x2 −1)n . (Legendre polynomials) REMARK 15. For all k = 0,1,··· we have that Lk ∈Pk, ⟨Lk,Lm⟩= δk,m k + 1 2 −1 , k,m = 0,1,2,··· (orthogonality) For any f ∈L2(−1,1), its Legendre series is L f = ∞ ∑ k=0 ˆ fkLk with ˆ fk = k +1/2 Z 1 −1 f(x)Lk(x)dx. Examples: L0(x) = 1, L1(x) = x, L2(x) = 1 2(3x2 −1), L3(x) = 1 2(5x3 −3x), L4(x) = 1 8 35x4 −30x2 +3 . (11.7) 63 12. Chapter 5. Numerical integration and Differentiation. 12.1. Section 5.0 NUMERICAL INTEGRATION. Let f ∈L1(a,b) (a real integrable function). Computing explicitly the deﬁnite integral I( f) = Z b a f(x)dx may be difﬁcult or even impossible. DEFINITION 12.1 (Quadrature formula). Any explicit formula that is suitable for providing an approximation of I( f) is said to be a quadrature formula or numerical integration formula. EXAMPLE 4. For example, consider replacing f by an approximation fn (depending on n ∈N) and integrating fn exactly. We would expect to have I( f) ≈In(f) := R b a f(x)dx, namely f(x) ≈fn(x), I( f) := Z b a f(x)dx ≈ Z b a fn(x)dx := In( f). REMARK 16. Let f ∈C0[a,b]. Then the quadrature error En(f) = I( f)−In( f) ≡ Z b a f(x)−fn(x) dx (quadrature error) satisﬁes En(f) ≤ Z b a f(x)−fn(x) dx ≤(b−a) f −fn ∞. Therefore, if for some n we have that ∥f −fn∥∞≤ε, then En( f) ≤ε(b−a). REMARK 17. The approximating function fn should be easily integrable, as it would be the case if for example fn ∈Pn. Therefore we naturally consider now the following approach, i.e., we approximate f ≈(Πn f) by the Lagrange interpolating polynomial over n+1 distinct nodes x0,x1,··· ,xn ∈[a,b] fn(x) := (Πn f)(x) = n ∑ i=0 f(xi)ℓi(x), and where ℓi(x) is the characteristic Lagrange polynomial of degree n associated with the node xi (see e.g. (Lagrange characterist DEFINITION 12.2 (Lagrange quadrature formula). Integrating exactly Πn f we obtain the following quadrature formula In( f) = n ∑ i=0 f(xi) Z b a ℓi(x)dx. (Lagrange quadrature formula) REMARK 18. We note that (Lagrange quadrature formula) is a special instance of the following quadrature formula In( f) = n ∑ i=1 αi f(xi), (quadrature formula) 64 where αi = Z b a ℓi(x)dx, (quadrature coefficients or weights) x0,x1,··· ,xn (quadrature nodes) REMARK 19. The (Lagrange quadrature formula) can be generalized to the (Hermite quadrature formula) by approximating f with its Hermite interpolating polynomial Hn f, i.e. f ≈(Hn f), to obtain 1 ∑ κ=0 n ∑ i=0 αik f (κ)(xi) (Hermite quadrature formula) We say that both the (Lagrange quadrature formula) and (Hermite quadrature formula) are interpolatory quadra-ture formulae, as in both cases f ≈Πn f or f ≈Hn f are obtained by exact integration of interpolating polynomials. DEFINITION 12.3 (Degree of exactness / degree of precision). We deﬁne as the degree of exactness (D.O.E.) of a (quadrature formula) the maximum integer r ≥0 for which In( f) ≡I( f), ∀f ∈Pr i.e., n ∑ i=0 αi f(xi) = Z b a f(x)dx ∀f ∈Pr. PROPOSITION 12.4. Any interpolatory quadrature formula that uses n+1 distinct nodes has degree of exactness at least n. Proof. Let f ≡(Πn f), where Πn f is the interpolating polynomial on the n+1 nodes. Then I(Πn f) := Z b a (Πn f)(x)dx = Z b a n ∑ i=0 f(xi)ℓi(x) dx = n ∑ i=0 f(xi) Z b a ℓi(x)dx = In(f), so r ≥n. REMARK 20. The degree of exactness of a (Lagrange quadrature formula) can be as large as 2n+1, in the case of the so-called Gaussian quadrature formulae, where the quadrature nodes (also called Gauss nodes) are the zeros of Chebyshev polynomials. Interpolatory Quadrature: instances of formula (Lagrange quadrature formula) for n = 0,1,2. Particular cases of the Newton-Cotes formulae: The midpoint rule (n = 0), trapezoidal rule (n = 1), Cavalieri-Simpson rule (n = 2). 12.1.1. THE MIDPOINT OR RECTANGLE FORMULA. This is the case corresponding to (Lagrange quadrature formula) for n = i = 0, i.e. approximating f on the interval [a,b] by the constant value (polynomial of degree 0) f a+b 2 , and integrate exactly: f(x) ≈(Π0 f)(x) := f a+b 2 , I( f) := Z b a f(x)dx ≈ Z b a f a+b 2 dx ≡(b−a) f a+b 2 := I0( f), 65 giving I0( f) = (b−a) \| {z } :=α0 f a+b 2 , (midpoint quadrature rule) which in the language of (quadrature formula) has • weight α0 = b−a • quadrature node x0 = a+b 2 . THEOREM 12.5. If f ∈C2[a,b], the (midpoint quadrature error) is E0( f) = h3 24 f ′′(ξ), (midpoint quadrature error) where h = b−a, and ξ ∈(a,b) is an arbitrary point. COROLLARY 12.6. The degree of exactness of the (midpoint quadrature rule) is 1: D.O.E. =1. Proof. Since for any constant and afﬁne functions f (r = 0,r = 1) we have f ′′ = 0, we obtain by the (midpoint quadrature erro formula that the error is zero E0( f) = 0 ≡I( f)−I0(f), hence the degree of exactness is 1. REMARK 21. If h = b−a the length of the integration interval is not sufﬁciently small, the (midpoint quadrature error) can be quite large. This drawback is common to all numerical integration formulae and can be overcome by using composite rules. THE COMPOSITE MIDPOINT FORMULA In order to minimize the error in the midpoint rule on relatively large intervals [a,b], instead of replacing f over [a,b] by Π0 f, we replace it with a composite Lagrange polynomial of degree , on m ≥1 subintervals. Let us introduce the following quadrature nodes: xk = a+(2k +1)H 2 , k = 0,··· ,m−1, (quadrature nodes) where H = b−a m . (mesh size) x0 = a+ H 2 , x1 = a+H + H 2 , ··· , xm−1 = a+(m−1)H + H 2 . Then using the (midpoint quadrature rule) we obtain I0,m = H m−1 ∑ k=0 f(xk) (composite midpoint formula) and by (midpoint quadrature error) we have respectively THEOREM 12.7 (The composite midpoint error). E0,m = b−a 24 H2 f ′′(ξ), (composite midpoint error) provided f ∈C2[a,b], where ξ ∈(a,b). (Due to the dependence on H2 of the (composite midpoint error) formula, we say that the (composite midpoint formula) has the order of inﬁnitesimal = 2. 66 Proof. Using the additivity of the integrals, the (midpoint quadrature error) formula and the discrete mean value theorem we have E0,m = m−1 ∑ k=0 H3 24 f ′′(ξk) = H3 24 mf ′′(ξ) = b−a 24 H2 f ′′(ξ), which concludes the proof. REMARK 22 (asymptotic error formula). If instead of using the discrete mean value theorem we use the deﬁni-tion of the Riemann sum, from the proof of Theorem 12.7 we obtain E0,m( f) = m−1 ∑ k=0 1 24H3 f ′′(ξk) = 1 24H2 H m ∑ k=0 f ′′(ξk) \| {z } ≈ R b a f ′′(x)dx namely E0,m( f) ≈H2 24 f ′(b)−f ′(a) . (asymptotic midpoint error formula) Since the (asymptotic midpoint error formula) provides an estimate of the error E0,m( f) = I( f)−I0,m( f), we can then also write the true value I( f) as I( f) = I0,m( f)+E0,m(f) ≈I0,m( f)+ H2 24 f ′(b)−f ′(a) , which leads to the following deﬁnition. DEFINITION 12.8 (Corrected midpoint rule). Icorrected 0 ( f) := I0,m( f)+ H2 24 f ′(b)−f ′(a) . (corrected midpoint rule) REMARK 23 (order of inﬁnitesimal - ratio of consecutive errors). From Theorem 12.7 or by the (asymptotic midpoint error formula) we notice that the ratio of the errors corresponding to composite midpoint rule with respectively m and 2m nodes (”doubling the number of nodes, halving the mesh”) satisﬁes E0,m E0,2m = b−a 24 (b−a)2 m2 f ′′(ξ) b−a 24 (b−a)2 (2m)2 f ′′( ˜ ξ) ≈22, and log(E0,m/E0,2m)/log(2) ≈2. (order of inﬁnitesimal) Section 5.1 THE TRAPEZOIDAL AND SIMPSON RULES 67 12.1.2. THE TRAPEZOIDAL RULE. This is the case corresponding to (Lagrange quadrature formula) for n = 1,i = 0,1, i.e. approximating f on the interval [a,b] by the piecewise linear interpolating polynomial (of degree n = 1) Π1 f, and integrate exactly: f(x) ≈(Π1 f)(x) := f(a)x−b a−b + f(b)x−a b−a, I( f) := Z b a f(x)dx ≈ Z b a f(a)x−b a−b + f(b)x−a b−a dx := I1(f), giving I1( f) = Z b a f(a)x−b a−b + f(b)x−a b−a dx = f(a) Z b a x−b a−bdx+ f(b) Z b a x−a b−adx = f(a) a−b Z b a (x−b)dx+ f(b) b−a Z b a (x−a)dx = f(a) a−b 1 2(x−b)2 x=b x=a + f(b) b−a 1 2(x−a)2 x=b x=a = −a−b 2 f(a)+ b−a 2 f(b) = b−a 2 f(a)+ f(b) (trapezoidal quadrature rule) I1( f) = b−a 2 \| {z } α0 f(a)+ b−a 2 \| {z } α1 f(b) (trapezoidal quadrature rule) which in the language of (quadrature formula) has • weights: α0 = α1 = b−a 2 • quadrature nodes x0 = a,x1 = b. THEOREM 12.9. If f ∈C2[a,b], the (trapezoidal quadrature error) is E1(f) = −h3 12 f ′′(ξ), (trapezoidal quadrature error) where h = b−a, and ξ ∈(a,b) is an arbitrary point. COROLLARY 12.10. The degree of exactness of the (trapezoidal quadrature rule) is 1: D.O.E. =1, as the (midpoint quadrature rule). Proof. Since for any constant and afﬁne functions f (r = 0,r = 1) we have f ′′ = 0, we obtain by the (trapezoidal quadrature e formula that the error is zero E1( f) = 0 ≡I( f)−I1(f), hence the degree of exactness is 1. REMARK 24. If h = b−a the length of the integration interval is not sufﬁciently small, the (trapezoidal quadrature error) can be quite large. This drawback is common to all numerical integration formulae and can be overcome by using composite rules. THE COMPOSITE TRAPEZOIDAL FORMULA In order to minimize the error in the trapezoidal rule on relatively large intervals [a,b], instead of replacing f over [a,b] by Π1 f, we replace it with a composite Lagrange polynomial of degree 1, on m ≥1 subintervals. Let us introduce the following quadrature nodes: xk = a+kH, k = 0,··· ,m, (quadrature nodes) where H = b−a m . (mesh size) 68 Then using the (trapezoidal quadrature rule) we obtain I1,m = H 2 m−1 ∑ k=0 f(xk)+ f(xk−1) = H 1 2 f(x0)+ f(x1)+···+ f(xm−1)+ 1 2 f(xm) = H 2 f(x0)+H f(x1)+···+H f(xm−1)+ H 2 f(xm) (composite trapezoidal formula) and by (trapezoidal quadrature error) we have respectively THEOREM 12.11 (The composite trapezoidal error). E1,m = −b−a 12 H2 f ′′(ξ), (composite trapezoidal error) provided f ∈C2[a,b], where ξ ∈(a,b). (Due to the dependence on H2 of the (composite trapezoidal error) formula, we say that the (composite trapezoidal formula) has the order of inﬁnitesimal = 2. Proof. Using the additivity of the integrals, the (trapezoidal quadrature error) formula and the discrete mean value theorem we have E1,m = m ∑ k=0 −H3 12 f ′′(ξk) = −H3 12 mf ′′(ξ),= −b−a 12 H2 f ′′(ξ), which concludes the proof. REMARK 25 (asymptotic error formula). If instead of using the discrete mean value theorem we use the deﬁni-tion of the Riemann sum, from the proof of Theorem 12.11 we obtain E1,m( f) = m ∑ k=0 −1 12H3 f ′′(ξk) = −1 12H2 H m ∑ k=0 f ′′(ξk) \| {z } ≈ R b a f ′′(x)dx namely E1,m( f) ≈−H2 12 f ′(b)−f ′(a) . (asymptotic trapezoidal error formula) Since the (asymptotic trapezoidal error formula) provides an estimate of the error E1,m( f) = I( f)−I1,m( f), we can then also write the true value I( f) as I( f) = I1,m( f)+E1,m(f) ≈I1,m( f)−H2 12 f ′(b)−f ′(a) , which leads to the following deﬁnition. DEFINITION 12.12 (Corrected trapezoidal rule). Icorrected 1 (f) := I1,m(f)−H2 12 f ′(b)−f ′(a) . (corrected trapezoidal rule) 69 REMARK 26 (order of inﬁnitesimal - ratio of consecutive errors). From Theorem 12.11 or by the (asymptotic trapezoidal error formula) we notice that the ratio of the errors corre-sponding to composite trapezoidal rule with respectively m and 2m nodes (”doubling the number of nodes, halving the mesh”) satisﬁes E1,m E1,2m = −b−a 12 (b−a)2 m2 f ′′(ξ) −b−a 12 (b−a)2 (2m)2 f ′′( ˜ ξ) ≈22, and log(E1,m/E1,2m)/log(2) ≈2. (order of inﬁnitesimal) 12.1.3. THE CAVALIERI-SIMPSON RULE. This is the case corresponding to (Lagrange quadrature formula) for n = 2,i = 0,1,2, i.e. approximating f on the interval [a,b] by the piecewise quadratic interpolating polynomial (of degree n = 2) Π2 f on interpolating nodes x0 = a,x1 = a+b 2 ,x2 = b, and integrate exactly: f(x) ≈(Π2 f)(x) := f(a)+ f( a+b 2 )−f(a) (b−a)/2 (x−a)++ f(b)−2 f( a+b 2 )+ f(a) (b−a)2/2 (x−a)(x−a+b 2 ), I( f) := Z b a f(x)dx ≈ Z b a (Π2 f)(x)dx := I2( f), giving I2( f) = b−a 6 f(a)+4 f a+b 2 + f(b) (Cavalieri-Simpson quadrature rule) I2( f) = b−a 6 \| {z } α0 f(a)+4b−a 6 \| {z } α1 f( a+b 2 )+ b−a 6 \| {z } α2 f(b) (Cavalieri-Simpson quadrature rule) which in the language of (quadrature formula) has • weights: α0 = b−a 2 , α1 = 4 b−a 6 , α2 = b−a 2 . • quadrature nodes x0 = a,x1 = a+b 2 ,x2 = b. REMARK 27. From the error analysis of the midpoint and trapezoidal rules, we note that the error in the mid-point rule is half the size of the error in the trapezoidal rule and of opposite sign, hence a linear combination of the midpoint and trapezoidal rules should be a better approximation. Indeed, notice that the (Cavalieri-Simpson quadrature rule) can be written as the linear combination of the (midpoint quadrature rule) and (trapezoidal quadrature rule): I2( f) = 2 3I0( f)+ 1 3I1( f), with the error given in this next result. THEOREM 12.13. If f ∈C4[a,b], the (Simpson quadrature error) is E2( f) = −1 90 h 2 5 f (4)(ξ), (Simpson quadrature error) where h = b−a, and ξ ∈(a,b) is an arbitrary point. 70 COROLLARY 12.14. The degree of exactness of the (Cavalieri-Simpson quadrature rule) is 1: D.O.E. =3. Proof. Since for any cubic functions f (r = 0,1,2,3) we have f (4) = 0, we obtain by the (Simpson quadrature error) formula that the error is zero E2( f) = 0 ≡I( f)−I2( f), hence the degree of exactness is 3. REMARK 28. If h = b−a the length of the integration interval is not sufﬁciently small, also the (Simpson quadrature error) can be quite large. This drawback is common to all numerical integration formulae and can be overcome by using composite rules. THE COMPOSITE CAVALIERI-SIMPSON FORMULA In order to minimize the error in the Cavalieri-Simpson rule on relatively large intervals [a,b], instead of replacing f over [a,b] by Π2 f, we replace it with a composite Lagrange polynomial of degree 2, on m ≥1 subintervals. Let us introduce the following quadrature nodes: xk = a+kH 2 , k = 0,··· ,2m, (quadrature nodes) where H = b−a m . (mesh size) Then using the (Cavalieri-Simpson quadrature rule) we obtain I2,m( f) = H 6 h f(x0)+2 m−1 ∑ r=1 f(x2r)+4 m−1 ∑ s=0 f(x2s+1)+ f(x2m) i (composite Simpson formula) and by (Simpson quadrature error) we have respectively THEOREM 12.15 (The composite Simpson error). E2,m( f) = −b−a 180 H 2 4 f (4)(ξ), (composite Simpson error) provided f ∈C4[a,b], where ξ ∈(a,b). (Due to the dependence on H4 of the (composite Simpson error) formula, we say that the (composite Simpson formula) has the order of inﬁnitesimal = 4. Proof. Indeed, using the (Simpson quadrature error) formula and the discrete mean value theorem we have that E2,m( f) = m−1 ∑ k=0 −1 90 H 2 5 f (4)(ξk) = −1 180 H 2 4 H m−1 ∑ k=0 f (4)(ξk) \| {z } =mf (4)(ξ) = −b−a 180 H 2 4 f (4)(ξ), which concludes the proof. REMARK 29. Verify the (composite Simpson error) formula for f(x) = x4 on [0,1], and on two subintervals. I( f) = Z 1 0 x4dx = 1 5, I2,1( f) = 1 6 04 +4· 1 24 +14 = 5 24, (using (Cavalieri-Simpson quadrature rule)) 71 E2,1( f) = I( f)−I2,1(f) ≡1 5 −5 24 = −1 120, f (4)(x) = 4!, E2,1( f) = −b−a 180 H 2 4 f (4)(ξ) ≡−1 180 1 2 4 24 = −1 120. (using (composite Simpson error)) I2,2( f) = 1 12 04 +4· 1 44 + 1 24 + 1 24 +4· 34 44 +14 = 77 384, (using (composite Simpson formula), H = 1 2) E2,2( f) = I( f)−I2,2(f) ≡1 5 −77 384 = 384−385 1920 = − 1 1920, E2,2( f) = −b−a 180 H 2 4 f (4)(ξ) ≡−1 180 1 4 4 24 = − 1 1920. (using (composite Simpson error)) REMARK 30 (asymptotic error formula). If instead of using the discrete mean value theorem we use the deﬁni-tion of the Riemann sum, from the above we obtain E2,m( f) = m−1 ∑ k=0 −1 90 H 2 5 f (4)(ξk) = −1 180 H 2 4 H m−1 ∑ k=0 f (4)(ξk) \| {z } ≈ R b a f (4)(x)dx namely E2,m( f) ≈−1 180 H 2 4f (3)(b)−f (3)(a) . (asymptotic Simpson error formula) Since the (asymptotic Simpson error formula) provides an estimate of the error E2,m( f) = I(f)−I2,m(f), we can then also write the true value I(f) as I( f) = I2,m(f)+E2,m( f) ≈I2,m(f)−1 180 H 2 4f (3)(b)−f (3)(a) , which leads to the following deﬁnition. DEFINITION 12.16 (Corrected Simpson rule). Icorrected 2 (f) := I2,m( f)−1 180 H 2 4f ′′′(b)−f ′′′(a) . (corrected Simpson rule) REMARK 31 (order of inﬁnitesimal - ratio of consecutive errors). From Theorem 12.15 or by the (asymptotic Simpson error formula) we notice that the ratio of the errors correspond-ing to composite Cavalieri-Simpson rule with respectively m and 2m nodes (”doubling the number of nodes, halving the mesh”) satisﬁes E2,m E2,2m = −b−a 180 (b−a)4 24m4 f (4)(ξ) −b−a 180 (b−a)4 24(2m)4 f ′′( ˜ ξ) ≈24, and log(E2,m/E2,2m)/log(2) ≈4. (order of inﬁnitesimal) 72 12.1.4. RICHARDSON EXTRAPOLATION. “Richardson extrapolation is a sequence acceleration method, used to improve the rate of convergence of a sequence of estimates of some value I = lim m→∞Im. In essence, given the value of Im for several values of m, we can estimate I by extrapolating the estimates to m = ∞.” We recall that by (composite midpoint error), (composite trapezoidal error), (composite Simpson error), the er-rors in the midpoint, trapezoidal and Simpson rules are all of the form I −Im ≈c mp , (error form) namely: I −I0,m = E0,m = b−a 24 H2 f ′′(ξ) ≡(b−a)3 24 f ′′(ξ) 1 m2 , (midpoint: p = 2,c = (b−a)3 24 f ′′(ξ)) I −I1,m = E1,m = −b−a 12 H2 f ′′(ξ) ≡−(b−a)3 12 f ′′(ξ) 1 m2 , (trapezoidal: p = 2,c = (b−a)3 24 f ′′(ξ)) I −I2,m = E2,m = −b−a 180 H 2 4 f (4)(ξ) ≡−(b−a)5 24 ·180 f (4)(ξ) 1 m4 , (Simpson: p = 4,c = −(b−a)5 24·180 f (4)(ξ)) hence the constants c and p in (error form) differ with the method and the function. Using the general form of the error (error form) with 2m we have I −I2m ≈ c 2pmp , hence using again (error form) in the above we have I ≈I2m + c 2pmp ≈I2m + 1 2p I −Im , 2p −1 2p I = I2m −1 2p Im, I ≈R2m := 2p 2p −1I2m − 1 2p −1Im (Richardson extrapolation) I −I2m ≈I −R2m ≡ 2p 2p −1I2m − 1 2p −1Im −I2m = 1 2p −1 I2m −Im (Richardson’s error estimate in I2m) REMARK 32. From (error form) we see that Im is an approximation of I with the order of convergence p, while the (Richardson extrapolation) R2m is an approximation of I with order at least p+1. (See also Figure 12.1.) Im ↘ I2m → R2m ↘ ↘ I4m → R4m → REMARK 33. In particular, from (Richardson’s error estimate in I2m) we obtain that in the case of the midpoint (p = 2), trapezoidal (p = 2) and Simpson rules (p = 2), the errors in the quadrature rule using 2m points are I −I0,2m ≈1 3 I0,2m −I0,m (error in midpoint rule) I −I1,2m ≈1 3 I1,2m −I1,m (error in trapezoidal rule) 73 I −I2,2m ≈1 15 I2,2m −I2,m (error in Simpson rule) Example 5.1.3 , page 193; Example 5.1.5 , page 198. Compute the integral Z 1 0 e−x2dx ≈0.746824132812427 using the (composite) midpoint, trapezoidal, Simpson rule on 2 and 4 subintervals. Find the errors and the ratios by which the errors decreases (rates of convergence / order of inﬁnitesimal). Solution: First we compute the Trapezoidal integrals on m = 2 subintervals: I(e−x2) = Z 1/2 0 (e−x2)dx+ Z 1 1/2(e−x2)dx ≈I1,2(e−x2) H=1/2 = = = = H 2 f(0)+2f(1/2)+ f(1) = 1 4 e0 +2e−1/4 +e−1 = 0.731370251828563 and m = 4 subintervals: I(e−x2) = Z 1/4 0 (e−x2)dx+ Z 1/2 1/4 (e−x2)dx+ Z 3/4 1/2 (e−x2)dx+ Z 1 3/4(e−x2)dx ≈I1,4(e−x2) H=1/4 = = = = H 2 f(0)+2f(1/4)+2 f(1/2)+2 f(3/4)+ f(1) = 1 8 e0 +2e−1/16 +2e−1/14 +2e−9/16 +e−1 = 0.742984097800381. The errors are E1,2 = I −I1,2 = 0.746824132812427−0.731370251828563 = 0.015453880983864, E1,4 = I −I1,4 = 0.746824132812427−0.742984097800381 = 0.003840035012046, hence the computed order of inﬁnitesimal (rate of convergence) is log(E1,2/E1,4)/log(2) = 2.008777822283925 ≈2.0088. Similarly, we compute the Simpson integrals on m = 2 subintervals: I(e−x2) = Z 1/2 0 (e−x2)dx+ Z 1 1/2(e−x2)dx ≈I2,2(e−x2) H=1/2 = = = = H 6 f(0)+4 f(1/4)+ f(1/2)+ f(1/2)+4f(3/4)+ f(1) = 1 12 e0 +4e−1/16 +2e−1/4 +4e−9/16 +e−1 = 0.746855379790987 and m = 4 subintervals: I(e−x2) = Z 1/4 0 (e−x2)dx+ Z 1/2 1/4 (e−x2)dx+ Z 3/4 1/2 (e−x2)dx+ Z 1 3/4(e−x2)dx ≈I1,4(e−x2) H=1/4 = = = = H 6 f(0)+4 f(1/8)+2f(1/4)+4 f(3/8)+2 f(1/2)+4f(5/8)+2 f(3/4)+4 f(7/8)+ f(1) 74 = 1 24 e0 +4e−1/64 +2e−1/16 +4e−9/64 +2e−1/14 +4e−25/64 +2e−9/16 +4e−49/64 +e−1 = 0.746826120527467. The errors are E2,2 = I −I2,2 = 0.746824132812427−0.746855379790987 = −3.124697856005110×10−5, E2,4 = I −I2,4 = 0.746824132812427−0.746826120527467 = −1.987715040008275×10−6, hence the computed order of inﬁnitesimal (rate of convergence) is log(E2,2/E2,4)/log(2) = 3.974533843274072 ≈3.9745. 100 101 102 n 10-14 10-12 10-10 10-8 10-6 10-4 10-2 errors Midpoint, Corrected midopint and Richardson M-extrapolation Midpoint corrected Midpoint Richardson extrapolation 100 101 102 n 10-14 10-12 10-10 10-8 10-6 10-4 10-2 100 errors Trapezoidal, Corrected trapezoidal and Richardson T-extrapolation Trapezoidal corrected Trapezoidal Richardson extrapolation 100 101 102 n 10-16 10-14 10-12 10-10 10-8 10-6 10-4 10-2 errors Simpson, corrected Simpson and Richardson S-extrapolation Simpson corrected Simpson Richardson extrapolation 100 101 102 n 10-13 10-12 10-11 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 errors Corrected midpoint/Trapezoidal, and Richardson-extrapolated midpoint/Trapezoidal vs Simpson corrected midpoint corrected Trapezoidal Richardson-extrapolated midpoint Richardson-extrapolated Trapezoidal Simpson FIG. 12.1. . Example_5_1_3_1st Midpoint = 0.778800783071405 0.754597943772199 0.748747131891009 75 0.747303578730748 0.746943912516367 0.746854072623361 0.746831617445408 0.746826003950687 0.746824600595743 Trapezoidal = 0.683939720585721 0.731370251828563 0.742984097800381 0.745865614845695 0.746584596788222 0.746764254652294 0.746809163637828 0.746820390541618 0.746823197246153 Trapezoidal calculations for int_0^1 exp(-x^2) dx: 1st : I-I{1,m} 2nd 3rd : I-CI_{1,m} 4th : I-RI_{1,m} Tabular = 8×5 table n Error RichardsonErrorEstim ErrorCorrected ErrorRichardson ___ _ __ __ __ 2 0.015454 0.01581 0.00012557 -0.0003563 4 0.00384 0.0038713 7.9575e-06 -3.1247e-05 8 0.00095852 0.00096051 4.9859e-07 -1.9877e-06 16 0.00023954 0.00023966 3.118e-08 -1.2462e-07 32 5.9878e-05 5.9886e-05 1.949e-09 -7.7946e-09 64 1.4969e-05 1.497e-05 1.2182e-10 -4.8725e-10 128 3.7423e-06 3.7423e-06 7.6136e-12 -3.0454e-11 256 9.3557e-07 9.3557e-07 4.7562e-13 -1.9038e-12 5th : p=log(I-I_{1,m})/log(2) 6th : (I_{1,2m}-I_{1,m})/(I_{1,4m}-I_{1,2m}) 7th : p=log((I_{1,2m}-I_{1,m})/(I_{1,4m}-I_{1,2m}))/log(2) Tabular = 8×6 table 76 n OrderConvExact ApproxRatio OrderWithRatios OrderConvCorrected orderTRichar ___ __ _ __ ___ __ 2 2.0247 0 -Inf 3.6453 3.9745 4 2.0088 0 2.03 3.98 3.9955 8 2.0022 4.084 2.0109 3.9964 3.999 16 2.0006 4.0305 2.0028 3.9992 3.9997 32 2.0001 4.0078 2.0007 3.9998 3.9999 64 2 4.002 2.0002 4 3.9997 128 2 4.0005 2 4 -Inf 256 2 4.0001 2 4.0007 -Inf Simpson calculations for int_0^1 exp(-x^2) dx: 1st : I-I_{2,m} 2nd 3rd : I-CI_{2,m} 4th : I-RI_{2,m} Tabular = 8×9 table n Error RichardsonErrorEstim ErrorCorrected ErrorRichardson OrderConv ___ ___ ____ ___ ___ _ 2 -0.0003563 0.019415 0.00015465 -0.019771 9.673 4 -3.1247e-05 -2.167e-05 6.87e-07 -9.577e-06 3.511 8 -1.9877e-06 -1.9506e-06 8.1587e-09 -3.7097e-08 3.974 16 -1.2462e-07 -1.2421e-07 1.188e-10 -4.1719e-10 3.995 32 -7.7946e-09 -7.7886e-09 1.8234e-12 -5.9752e-12 3.99 64 -4.8725e-10 -4.8715e-10 2.8533e-14 -9.1149e-14 3.999 128 -3.0454e-11 -3.0453e-11 4.4409e-16 -1.4433e-15 3.999 256 -1.9035e-12 -1.9034e-12 -1.1102e-16 0 3.999 Exercises 2, 3 and 9, page 200. Compute the integral Z π 0 ex cos(4x)dx = eπ −1 17 using the (composite) midpoint, trapezoidal, Simpson rule. Find the errors and the ratios by which the errors de-creases (rates of convergence). Solution: % Exercise 5.1.2, 3 and 9 (a): % use the composite-trapezoidal rule, % Simpson rule and the midpoint rule % to approximate % I = int_0^pi exp(x)cos(4x) dx ==== (exp(pi)-1)/17 % with n=2^j, j=1:9 % Write the values of the errors and the ratio: % (I-In)/(I-I2n) clear all 77 101 102 103 n 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 errors Trapezoidal, Simpson and midpoint for 0 ex cos (4x) dx Trapezoidal Simpson midpoint FIG. 12.2. Log-log plot of errors for the midpoint, trapezoidal and Simpson rules. % Use the function "trapezoidal(a,b,n0,index_f)" to compute the integral n0=2; a=0;b=pi; [integralT,differenceT,ratioT]=trapezoidal(a,b,n0,12); % Evaluate errors trapezoidal ellT=length(integralT); errorsTrapezoidal = (exp(pi)-1)/17 - integralT; ratioTexact(2:ellT)=errorsTrapezoidal(1:ellT-1)./errorsTrapezoidal(2:ellT); ratioT; orderTexact=log(abs(ratioTexact(2:ellT)))/log(2); format long n = n02.^linspace(0,ellT-1,ellT); H=((b-a)./n)’; [integralS,differenceS,ratioS]=simpson(a,b,n0,12); % Evaluate errors Simpson ellS=length(integralS); errorsSimpson = (exp(pi)-1)/17 - integralS; ratioSexact(2:ellS)=errorsSimpson(1:ellS-1)./errorsSimpson(2:ellS); ratioS; orderSexact=log(abs(ratioSexact(2:ellS)))/log(2); 78 [integralM,differenceM,ratioM]=midpoint(a,b,n0,12); % Evaluate errors Midpoint ellM=length(integralM); errorsmidpoint = (exp(pi)-1)/17 - integralM; ratioMexact(2:ellM)=errorsmidpoint(1:ellM-1)./errorsmidpoint(2:ellM); ratioM; orderMexact=log(abs(ratioMexact(2:ellM)))/log(2); format short fprintf(’Trapezoidal calculations for int_0^pi e^x cos(4x) dx: \n’) Tabular = table(n(2:end)’,integralT(2:end),errorsTrapezoidal(2:end),... orderTexact’,’VariableNames’,{’n’ ’Trapezoidal’ ’Error in trapezoidal’ ’OrderConvExact’}) Tabular = table(n(2:end)’,integralS(2:end),errorsSimpson(2:end),... orderSexact’,’VariableNames’,{’n’ ’Simpson’ ’Error in Simpson’ ’OrderConvExact’}) fprintf(’midpoint calculations for int_0^pi e^x cos(4x) dx: \n’) Tabular = table(n(2:end)’,integralM(2:end),errorsmidpoint(2:end),... orderMexact’,’VariableNames’,{’n’ ’Midpoint’ ’Error in midpoint’ ’OrderConvExact’}) figure(1) %hold all loglog(n(2:end),abs(errorsTrapezoidal(2:end)),’-sr’,n(2:end),abs(errorsSimpson(2:end)),’-b’,... n(2:end),abs(errorsmidpoint(2:end)),’-og’); grid on xlabel(’n’) ylabel(’errors’) legend(’Trapezoidal’,’Simpson’,’midpoint’,’location’,’best’) title(’Trapezoidal, Simpson and midpoint for \int_0^\pi e^x cos (4x) dx’) >> Exercise_5_1_2and3and9_a Trapezoidal calculations for int_0^pi e^x cos(4x) dx: Tabular = 8×4 table n Trapezoidal Error in trapezoidal OrderConvExact ___ __ __ __ 4 3.2491 -1.9467 3.6952 8 1.6245 -0.32213 2.5953 16 1.3757 -0.073329 2.1352 32 1.3203 -0.017918 2.033 64 1.3068 -0.0044542 2.0082 128 1.3035 -0.001112 2.002 256 1.3027 -0.0002779 2.0005 79 512 1.3025 -6.9468e-05 2.0001 Tabular = 8×4 table n Simpson Error in Simpson OrderConvExact ___ _ ___ __ 4 -4.5067 5.8091 1.8821 8 1.083 0.21938 4.7268 16 1.2928 0.0096054 4.5134 32 1.3018 0.00055202 4.1211 64 1.3024 3.3796e-05 4.0298 128 1.3024 2.1015e-06 4.0074 256 1.3024 1.3117e-07 4.0019 512 1.3024 8.1956e-09 4.0005 midpoint calculations for int_0^pi e^x cos(4x) dx: Tabular = 8×4 table n Midpoint Error in midpoint OrderConvExact ___ ___ __ _____ 4 -3.9445e-15 1.3024 4.033 8 1.1269 0.17547 2.8918 16 1.2649 0.037492 2.2266 32 1.2934 0.0090098 2.057 64 1.3002 0.0022303 2.0143 128 1.3018 0.00055618 2.0036 256 1.3023 0.00013896 2.0009 512 1.3024 3.4735e-05 2.0002 Exercise 11 (a), page 202. Show that the midpoint rule I0(f) = (b−a)f a+b 2 for approximating I( f) = R b a f(x)dx has the degree of precision (exactness) 1. (Hint: Consider only I(f) = R b 0 f(x)dx for suitable b and f(x) = 1,x,x2,···) Solution: I( f) := Z b 0 dx = b, I0( f) := b·1 ( f(x) = 1 : r = 0) I( f) := Z b 0 xdx = b2 2 , I0(f) := b· b 2. ( f(x) = x : r = 1) 80 I( f) := Z b 0 x2dx = b3 3 , I0( f) := b· b 2 2 . ( f(x) = x2 : r = 2) Since for f(x) = x2 the integral and the quadrature do not coincide, we conclude that the midpoint rule has D.O.E. =1. Exercise 12, page 203. Determine the degree of precision of the approximation Z 1 0 f(x)dx ≈1 4 f(0)+ 3 4 f 2 3 . Solution: I( f) := Z 1 0 dx = 1, I0( f) := 1 4 ·1+ 3 4 ·1 ≡1 ( f(x) = 1 : r = 0) I( f) := Z 1 0 xdx = 1 2, I0( f) := 1 4 ·0+ 3 4 · 2 3 ≡1 2. ( f(x) = x : r = 1) I( f) := Z 1 0 x2dx = 1 3, I0( f) := 1 4 ·02 + 3 4 · 2 3 2 ≡1 3. ( f(x) = x2 : r = 2) I( f) := Z 1 0 x3dx = 1 4, I0( f) := 1 4 ·03 + 3 4 · 2 3 3 ≡2 9. ( f(x) = x3 : r = 3) Since for f(x) = x3 the integral and the quadrature do not coincide, we conclude that the midpoint rule has D.O.E. =2. Exercise 13, page 203. The degree of precision of a quadrature rule is deﬁned as follows: If the formula has zero error when integrating any polynomial of degree ≤r, and if the error is nonzero for some polynomial of degree r +1, then we say the formula has degree of precision equal to r. Let Ih = 3h 4 [f(0)+3f(2h)]. What is the degree of precision of the approximation Ih ≈ R 3h 0 f(x)dx? (Hint: Consider f(x) = 1,x,x2,x3,etc.) Solution: • f = 1 → R 3h 0 1dx = 3h ≈3h 4 (1+3) = 3h. • f = x → R 3h 0 xdx = x2 2 3h 0 = 9h2 2 ≈3h 4 (0+3(2h)) = 3h = 9h2 2 . • f = x2 → R 3h 0 x2dx = x3 3 3h 0 = 33h3 3 = 32h3 ≈3h 4 02 +3(2h)2 = 9h3. • f = x3 → R 3h 0 x3dx = x4 4 3h 0 = 34h4 4 ̸= 3h 4 03 +3(2h)3 = 36h4. The degree of precision is 2. Exercise 14. Approximate I( f) = R 1 −1 f(x)dx by replacing f(x) with P 1(x), the linear interpolant to f(x) at nodes x0 = −1 3 and x1 = 1 3. Give the resulting numerical integration formula. What is the degree of precision (exactness)? Solution: Using the Lagrange form of the interpolating polynomial, the interpolant is P 1(x) = f(x0) x−x1 x0 −x1 + f(x1) x−x0 x1 −x0 = f −1 3 x−1 3 −1 3 −1 3 + f 1 3 x+ 1 3 1 3 + 1 3 81 = −3 2 f −1 3 x−1 3 + 3 2 f 1 3 x+ 1 3 , and the quadrature rule is Ih(x) = −3 2 f −1 3 Z 1 −1 x−1 3 dx+ 3 2 f 1 3 Z 1 −1 x+ 1 3 dx = −3 4 f −1 3 x−1 3 2 1 −1 + 3 4 f 1 3 x+ 1 3 2 1 −1 = −3 4 f −1 3 4 9 −16 9 + 3 4 f 1 3 16 9 −4 9 = f −1 3 + f 1 3 . To compute the degree of exactness we evaluate I(f) := Z 1 −1 dx = 2, Ih( f) := 1+1 ≡2 (f(x) = 1 : r = 0) I(f) := Z 1 −1 xdx = 0, Ih( f) := −1 3 + 1 3 ≡0 (f(x) = x : r = 1) I(f) := Z 1 −1 x2dx = 2 3, Ih( f) := −1 3 2 + 1 3 2 ≡2 9. ( f(x) = x2 : r = 2) Since for f(x) = x2 the integral and the quadrature do not coincide, we conclude that the quadrature rule has D.O.E. =1. Exercise 15, page 203. Consider the approximation I( f) = Z 1 −1 f(x)dx ≈f(−β)+ f(β) for some β satisfying 0 < β ≤1. Show it has degree of precision greater than or equal to 1 for any such choice of β. Choose β to obtain a formula with degree of precision greater than 1. What is the degree of precision of this formula? Solution: To compute the degree of exactness we evaluate I(f) := Z 1 −1 dx = 2, Ih( f) := 1+1 ≡2 ( f(x) = 1 : r = 0) I(f) := Z 1 −1 xdx = 0, Ih( f) := −β +β ≡0 ( f(x) = x : r = 1) I(f) := Z 1 −1 x2dx = 2 3, Ih( f) := (−β)2 +(β)2 ≡2β 2. ( f(x) = x2 : r = 2) I(f) := Z 1 −1 x3dx = 0, Ih( f) := (−β)3 +(β)3 ≡0. ( f(x) = x3 : r = 3) I(f) := Z 1 −1 x4dx = 2 5, Ih( f) := (−β)4 +(β)4 ≡2β 4. ( f(x) = x4 : r = 4) For β = q 1 3, the D.O.E. = 3, otherwise 1. Exercise 16, page 203. Approximate I( f) = R 2h 0 f(x)dx by replacing f(x) with P 1(x), the linear interpolant to f(x) at x0 = 0 and x1 = h. 82 Give the resulting numerical integration formula. What is the degree of precision (exactness)? Solution: Using the Lagrange form of the interpolating polynomial, the interpolant is P 1(x) = f(x0) x−x1 x0 −x1 + f(x1) x−x0 x1 −x0 = f(0)x−h −h + f(h)x h, and the quadrature rule is Ih(x) = f(0) Z 2h 0 x−h −h dx+ f(h) Z 2h 0 x h dx = −f(0) 2h (x−h)2 2h 0 dx+ f(h) 2h x2 2h 0 = 2hf(h). To compute the degree of exactness we evaluate I( f) := Z 2h 0 dx = 2h, Ih( f) := 2h·1 (f(x) = 1 : r = 0) I( f) := Z 2h 0 xdx = 2h2, Ih( f) := 2h·h ( f(x) = x : r = 1) I( f) := Z 2h 0 x2dx = 8h3 3 , Ih( f) := 2h·h2. ( f(x) = x2 : r = 2) Since for f(x) = x2 the integral and the quadrature do not coincide, we conclude that the quadrature rule has D.O.E. =1. Exercise 1, page 215. Using the (composite trapezoidal error) formula , bound the error in the trapezoidal rule I1,m applied to the following integrals Z π/2 0 cosxdx, (a) Z 1 0 e−x2 dx, (b) Z √π 0 cos(x2)dx. (c) Solution: (b) To evaluate the error in the composite trapezoidal rule for Z 1 0 e−x2 dx (b) we compute f(x) = e−x2, f ′(x) = −2xe−x2, f ′′(x) = (4x2 −2)e−x2, and bound max x∈[0,1]\|(4x2 −2)e−x2\| = 2. Therefore \|E1,m(f)\| = −b−a 12 H2 f ′′(ξ) ≤1 6H2, where H = 1 m, and m is the number of subintervals. (c) To evaluate the error in the composite trapezoidal rule for Z √π 0 cos(x2)dx (c) 83 we compute f(x) = −2xsin(x2), f ′(x) = −2xe−x2, f ′′(x) = −2sin(x2)−4x2 cos(x2), and bound max x∈[0,√π]\|2sin(x2)+4x2 cos(x2)\| = \|2sin(π)+4π cos(π)\| = 4π. Therefore \|E1,m( f)\| = −b−a 12 H2 f ′′(ξ) ≤ √π 12 H24π = π√π 3 H2, where H = 1 m, and m is the number of subintervals. Exercise 5, page 215. Using the asymptotic formula (asymptotic trapezoidal error formula) for the trapezoidal rule, estimate the number of m subdivisions to evaluate the following integrals to the given accuracy ε: (a) I = Z 3 1 ln(x)dx, ε = 10−8 (b) I = Z 2 1 ex −e−x 2 dx, ε = 10−10 Solution: (a) By the (asymptotic trapezoidal error formula) we have that E1,m( f) ≈−H2 12 f ′(b)−f ′(a) ≡−H2 12 1 b −1 a = 4 m2 1 18≤10−8 ( f ′(x) = 1 x,H = 2 m) m2 ≥2 9 ∗108, m ≥4715. Exercise 6, page 215. Using the asymptotic formula (asymptotic Simpson error formula) for the Simpson rule, estimate the number of m subdivisions to evaluate the following integrals to the given accuracy ε: (a) I = Z 3 1 ln(x)dx, ε = 10−8 (b) I = Z 2 1 ex −e−x 2 dx, ε = 10−10 Solution: (a) By the (asymptotic Simpson error formula) we have that E2,m(f) ≈−1 180 H 2 4f (3)(b)−f (3)(a) ≡−1 180 H 2 4 2 b3 −2 a3 = −1 180 1 m4 −2·26 27 ( f ′′′(x) = 2 x3 ,H = 2 m) = 13 1215 1 m4 ≤10−8 m4 ≥ 13 1215 ∗108, m ≥33. Exercise 8, page 216. (a) Consider using the trapezoidal rule to estimate the integral I = Z 1 −1 dx 2+x ≡ln(3) ≈1.098612288668110. Give both a rigorous error bound found for I −I1,m and an asymptotic error estimate I −I1,m . Using the rigorous error bound, determine how large m should be in order that \|I −I1,m\| ≤5×10−8. (b) Repeat with Simpson’s rule. 84 Solution: (a) The rigorous error bound (composite trapezoidal error) for the trapezoidal rule gives E1,m = −b−a 12 H2 f ′′(ξ) ≡−2 12 4 m2 2 (2+ξ)3 , (H = 2 m, f ′′(x) = 2 (2+x)3 ) \|E1,m\| ≤4 3 1 m2 ≤5×10−8, m > 5163. The (asymptotic trapezoidal error formula) gives E1,m( f) ≈−H2 12 f ′(b)−f ′(a) . ≡−1 12 4 m2 8 9 = −8 27 1 m2 , (H = 2 m, f ′(x) = − 1 (2+x)2 ) \|E1,m\| = 8 27 1 m2 ≤5×10−8, m > 2434. Indeed, using the composite trapezoidal rule with m = 2436 intervals the error is \|E1,2436\| = \|−0.000000049931155\| ≈5×10−8. (b) The rigorous error bound (composite Simpson error) for the Simpson rule gives E2,m( f) = −b−a 180 H 2 4 f (4)(ξ) = −2 180 1 m4 24 (2+ξ)5 , ( f (4)(x) = 24 (2+x)5 ,H = 2 m) E2,m( f) = 4 15 1 m4 ≤5×10−8 m4 ≥4 75 ∗108, m ≥49. By the (asymptotic Simpson error formula) we have that E2,m( f) ≈−1 180 H 2 4f (3)(b)−f (3)(a) ( f ′′′(x) = − 6 (2+x)4 ,H = 2 m) ≡−1 180 H 2 4 2 b3 −2 a3 = 1 180 1 m4 160 27 = 8 243 1 m4 ≤5×10−8 m4 ≥ 8 1215 ∗108, m ≥29. Indeed, using the composite Simpson rule with m = 29 intervals the error is \|E2,29\| = \|−4.641445316977411e−08\| ≈5×10−8. Exercise 13, page 217. (a) From (error form) I −Im ≈c mp , derive I −Im I −I2m ≈2p for all m for which the approximation is valid. (Hint: Consider (error form) for m and 2m.) 85 (b) Also derive the computable estimate I2m −Im I4m −I2m ≈2p. This gives a practical means of checking the value of the value of p (order of convergence), using three consec-utive values Im,I2m,I4m. Using the log function, we get p = log I2m −Im I4m −I2m !. log(2). (12.1) (Hint: Write I2m −In = (I −Im)−(I −I2m), and do the same for the denominator.) Solution: Exercise 16. (a) Following is a table of numerical integrals Im for an integral whose true value is I = 0.3. Assuming that the error has an asymptotic formula of the form I −Im ≈c mp , for some p > 0 and c, estimate the order of convergence p. Estimate c. Estimate the size of m in order to have I −Im ≤10−10. m Im 8 0.2993331765 16 0.2997899139 32 0.2999338239 64 0.2999791556 128 0.2999934344 256 0.2999979320 (b) Assuming I is not known (as is usually the case), estimate p. Solution: (a) m Im I −Im I−Im I−I2m I2m −Im I2m−Im I4m−I2m 8 0.2993331765 0.6668 e-03 16 0.2997899139 0.2101 e-03 3.1740 4.5674e-04 32 0.2999338239 0.0662 e-03 3.1747 1.4391e-04 3.1738 64 0.2999791556 0.0208 e-03 3.1748 4.5332e-05 3.1746 128 0.2999934344 0.0066 e-03 3.1748 1.4279e-05 3.1747 256 0.2999979320 0.0021 e-03 3.1749 4.4976e-06 3.1748 We have that p is approximately p ≈log(3.1749)/log(2) = 1.6667, and c ≈mp(I −Im) ≡2561.6667(0.3−I256) = 2561.6667(0.3−0.2999979320) = 0.0213. (m = 256) Then for the error to be less than 10−10 I −Im ≈c mp ≡0.0213 m1.6667 ≤10−10, m ≥99295. 86 (b) Using the approximation of the ratios we have that p is approximately p ≈log(3.1748)/log(2) = 1.6667. Exercise 19, page 218. In the following table of numerical integrals and their differences, give the likely value of p if we assume the errors behave like I −Im ≈c/mp. Also, estimate the error in I64. m Im Im −Im/2 2 0.702877396 4 0.781978959 0.07910 8 0.804500932 0.02252 16 0.810303086 0.005802 32 0.811764354 0.001461 64 0.812130341 0.0003660 (Hint: Use (12.1) to evaluate p and (Richardson’s error estimate in I2m) to estimate the error in I64) . Solution: p ≈log(0.001461/0.0003660)/log(2) = 1.9970, I −I64 ≈ 1 21.9970 −10.0003660 = 1.2234e−04. We shall solve using the information for I16 and I32. p ≈log(0.005802/0.001461)/log(2) = 1.989594117394499 ≈1.9895, (order of convergence) I −I32 ≈ 1 21.989594117394499 −10.001461 ≈4.9171e−04. (Richardson’s estimated error in I32) 87 13. Section 5.3 GAUSSIAN NUMERICAL INTEGRATION. The (orthogonal polynomials) play an important role in devising quadrature formulae with maximal degree of exact-ness. Let x0,··· ,xn be n+1 distinct point in [−1,1]. To approximate the weighted integral Iw( f) = Z 1 −1 f(x)w(x)dx where f ∈C[−1,1], we consider quadrature rules of the following type In,w( f) = α0 f(x0)+···+αn f(xn), (Gaussian quadrature formula) where αi are to be suitably determined, in order to maximize the degree of exactness (w.r.t. weight w(x)). THEOREM 13.1. The (Gaussian quadrature formula) has degree of exactness n+m if and only if is of interpo-latory type and the (nodal polynomial) ωn+1(x) is orthogonal on all polynomials of degree m−1: Z 1 −1 ωn+1(x)p(x)w(x)dx = 0, ∀p ∈Pm−1. (13.1) COROLLARY 13.2. The maximum degree of exactness of the (Gaussian quadrature formula) is 2n+1. Proof. By contradiction, assume m ≥n+2 (i.e. the D.O.E. > 2n+2) and therefore we could choose p = ωn+1 in (13.1), which yields to the absurd result ωn+1 ≡0. Setting m = n+1 gives the result. 13.1. Gauss-Legendre quadrature. We shall use the orthogonal (Legendre polynomials) (the weight func-tion w(x) = 1) to approximate the (Gaussian quadrature formula). The quadrature nodes x j are then the zeros of (Legendre polynomials) Ln+1 (see also the examples (11.7)) , and the quadrature weights are αj = 2 (1−xj) Ln+1′(x) 2 , j = 0,··· ,n. (Gaussian quadrature weights) REMARK 34 (Integration over an arbitrary interval). The integral I( f) := Z b a f(x)dx = b−a 2 Z 1 −1 f b−a 2 y+ a+b 2 dy has the following Gaussian quadrature formula In( f) = b−a 2 α0 f b−a 2 x0 + a+b 2 +···+ b−a 2 αi f b−a 2 xi + a+b 2 +···+ b−a 2 αn f b−a 2 xn + a+b 2 . (Gauss-Legendre quadrature) Exercise 2. (d) Apply I2,I3 to the integral Z π −π ecosxdx ≈7.95492652101284. 88 Calculate the errors and compare them with the earlier results for the trapezoidal and Simpson rules. Solution: First, using formula (5.62) for an arbitrary interval we have I( f) = Z b a f(x)dx ≡ Z π −π ecosxdx= b−a 2 Z 1 −1 f b+a+t(b−a) 2 dt ≡π Z 1 −1 f(πt)dt. Secondly, we give the value computed with 20 Gaussian nodes I20 = 7.954926520986633, E20 = 7.95492652101284−7.954926520986633 = 2.620748063009160e−11. (Gaussian error) We report also the errors using 20 nodes for the midpoint, trapezoidal and Simpson errors: I(f)−I0,20( f) = 7.95492652101284−7.954926521012845 = −4.440892098500626e−15, (midpoint error) I(f)−I1,20( f) = 7.95492652101284−7.954926521012844 = −3.552713678800501e−15, (trapezoidal error) I( f)−I2,20( f) = 7.95492652101284−7.954926519859693 = 1.153147799470844e−09 (Simpson error) We note that the midpoint and trapezoidal rules are ”super-convergent” (f(x) is periodic on [a,b])!!! midpoint(-pi,pi,20,11) ans = 7.954926521012845 >> trapezoidal(-pi,pi,20,11) ans = 7.954926521012844 >> simpson(-pi,pi,20,11) ans = 7.954926519859693 Now we apply formulae (5.50) or (5.53), with the values for the nodes and weights from Table 5.7, to obtain I2( f) = π w1 f(π ∗x1)+w2 f(π ∗x2) ≡π ecos(−0.577350269189626·π) +ecos(−0.577350269189626·π) = 4.939472538370615, [val,bp,wf]=gaussint(-pi,pi,2,11) val = 4.939472538370615 bp = -0.577350269189626 0.577350269189626 wf = 1.000000000000000 1.000000000000000 and respectively I3( f) = π w1 f(π ∗x1)+w2 f(π ∗x2)+w3 f(π ∗x3) 89 ≡π 0.555555555555556·ecos(−0.774596669241484·π) +0.888888888888889·ecos(0∗π) +0.555555555555556·ecos0.774596669241484·π = 9.224020450374859. [val,bp,wf]=gaussint(-pi,pi,3,11) val = 9.224020450374855 bp = -0.774596669241484 0 0.774596669241483 wf = 0.555555555555556 0.888888888888889 0.555555555555556 Exercise 8, page 230. Determine constants c1 and c2 in the formula Z 1 0 f(x)dx ≈c1 f(0)+c2 f(1) so that it is exact for all polynomials of as large a degree as possible. What is the degree of precision of the formula. Solution: Exercise 10, page 230. (b) To ﬁnd a formula I(f) := Z 1 0 f(x)log 1 x dx ≈w1 f(x1)+w2 f(x2) =: I2( f) which is exact for all polynomials of degree ≤3, set up a system of four equations with unknowns w1,w2,x1,x2. Do not attempt to solve the system; instead show that x1 = 15− √ 106 42 , x2 = 15+ √ 106 42 , w1 = 21 √ 106 x2 −1 4 , w2 = 1−w2 is a solution of the system. Solution: Taking f(x) = 1,x,x2,x3 and enforcing I( f) = I2(f) we obtain Z 1 0 log 1 x dx =1 = w1 +w2, ( f = 1) 90 Z 1 0 xlog 1 x dx =1 4 = w1x1 +w2x2, ( f = x) Z 1 0 x2 log 1 x dx =1 9 = w1x2 1 +w2x2 2, ( f = x2) Z 1 0 x3 log 1 x dx = 1 16 = w1x3 1 +w2x3 2, ( f = x3) which is easy to check that hold for the given values. 91 14. Section 5.4 NUMERICAL DIFFERENTIATION. APPROXIMATION OF FUNCTION DERIVATIVES. The main goal of this section is to approximate the derivative of a given function f at the n + 1 distinct nodes {xk,k = 0,···n} in [a,b], namely x0 = a, xn = b, xk+1 = xk +h, k = 0,···n−1, h = b−a n . The efﬁcient way of computing/approximating the values of f ′(xi) ≈ui using the nodal values of f(xk) is to use the following compact formula h k=m ∑ k=−m αkui−k = k=m′ ∑ k=−m′ βk f(xi−k), (compact ﬁnite differences) where {αk}k,{βk}k ⊂R are 2(m + m′ + 1) coefﬁcients, to be determined, and uk are the desired approximations of f ′(xk). DEFINITION 14.1 (stencil). The set of nodes involved in constructing the derivative of f at a certain node is called stencil. 14.1. Classical Finite Differences method. The simplest way to generate a formula like (compact ﬁnite differences) is by using the deﬁnition of the deriva-tive f ′(xi) = lim h→0+ f(xi +h)−f(xi) h . (I) (forward ﬁnite difference) Replacing the limit with the incremental ratio, with h ﬁnite, gives uFD i = f(xi +h)−f(xi) h , i = 0,··· ,n−1. (forward Finite Difference) REMARK 35. (a) Note that (forward Finite Difference) is a special instance of (compact ﬁnite differences), with m = 0,α0 = 1,m′ = 1,β−1 = 1,β0 = −1,β1 = 0. (b) The approximation uFD i to f ′(xi) given by (forward Finite Difference) is the slope of the line passing through the points xi, f(xi) and xi+1, f(xi+1) . (c) Assuming that f ∈C2[a,b], and using Taylor’s expansion f(xi+1) = f(xi)+hf ′(xi)+ h2 2 f ′′(ξi), with ξi ∈(xi,xi+1), we obtain that the error made by the approximation is f ′(xi)−uFD i = −h 2 f ′′(ξi). (14.1) (II) (centered ﬁnite difference) Using a centered incremental ratio, with h ﬁnite, gives uCD i = f(xi +h)−f(xi −h) 2h , i = 1,··· ,n−1. (centered Finite Difference) REMARK 36. 92 (a) Note that (centered Finite Difference) is also a special instance of (compact ﬁnite differences), with m = 0,α0 = 1,m′ = 1,β−1 = 1/2,β0 = 0,β1 = −1/2. (b) The approximation uCD i to f ′(xi) given by (forward Finite Difference) is the slope of the line passing through the points xi−1, f(xi−1) and xi+1, f(xi+1) . (c) Assuming that f ∈C2[a,b], and using Taylor’s expansion f(xi+1) = f(xi)+hf ′(xi)+ h2 2 f ′′(ξi)+ h3 6 f ′′′(ξ + i ), with ξi ∈(xi,xi+1), f(xi−1) = f(xi)−hf ′(xi)+ h2 2 f ′′(ξi)−h3 6 f ′′′(ξ − i ), with ξi ∈(xi−1,xi), we obtain that the error made by the approximation is f ′(xi)−uCD i = −h2 6 f ′′′(ξi), ξi ∈(xi−1,xi+1). (14.2) (III) (backward ﬁnite difference) Replacing the limit with the incremental ratio, with h ﬁnite, gives uBD i = f(xi)−f(xi −h) h , i = 1,··· ,n. (backward Finite Difference) REMARK 37. (a) Note that (backward Finite Difference) is also special instance of (compact ﬁnite differences). (b) Assuming that f ∈C2[a,b], and using Taylor’s expansion we obtain that the error made by the approxi-mation is f ′(xi)−uBD i = h 2 f ′′(ξi). (14.3) 14.2. Differentiation using interpolation. Since we can use interpolation to approximate f(x) by polynomials deﬁned using nodal values of f(xi), i.e., f(x) ≈(Πn f)(x), it is only natural to ask if we could not approximate f ′(x) ≈(Πn f)(x)? THEOREM 14.2. Assume f ∈Cn+2[a,b], and x0,x1,xn are n + 1 distinct interpolatory nodes in [a,b], and x ∈[a,b] is an arbitrary value. Then the error in approximation f ′(x) ≈(Πn f)(x) is f ′(x)−(Πn f)′(x) = ωn+1(x) f (n+2)(c1) (n+1)! +ω′ n+1(x) f (n+1)(c2) (n+1)! , where c1,c2 are arbitrary points in the interval determined by the nodes {x0,··· ,xn}. Proof. Recall the (interpolation error / Newton form) formula En(x) := f(x)−(Πn f)(x) = f (n+1)(cx) (n+1)! ωn+1(x). Then the result yields after differentiation. 14.3. The Method of Undetermined Coefﬁcients. Higher-order derivatives. DEFINITION 14.3 (Central ﬁnite difference formula). Assume f ∈C4[a,b]. Then the approximate of f ′′(xi) denoted by u′′ i or D2 f(xi) is u′′ i = f(xi+1)−2 f(xi)+ f(xi−1) h2 , ∀i = 1,···n−1, (central ﬁnite difference formula) 93 with the error f ′′(xi)−u′′ i = −h2 24 f (4)(xi +θih)+ f (4)(xi −ωih) ≈−h2 12 f (4)(xi +ξih), (14.4) where θi,ωi ∈[0,1). The error formula (14.4) is a consequence of Taylor expansions f(xi+1) = f(xi)+hf ′(xi)+ h2 2 f ′′(xi)+ h3 6 f ′′′(xi)+ h4 24 f (4)(ξi +θih), f(xi−1) = f(xi)−hf ′(xi)+ h2 2 f ′′(xi)−h3 6 f ′′′(xi)+ h4 24 f (4)(ξi −ωih), −2 f(xi) = −2 f(xi), hence u′′ i = f ′′(xi)+ h2 4! f (4)(ξi +θih)+ f (4)(ξi −ωih) , and ﬁnally using the Mean Value Theorem we obtain (14.4). Let us now derive (central ﬁnite difference formula) by a more general technique. We start by asking the question of ﬁnding an approximation to f ′′(x) by a linear combination of nodal values f(x+h), f(x), f(x−h), with the smallest error term: f ′′(x) ≈Af(x+h)+Bf(x)+C f(x−h). Therefore we seek the undetermined coefﬁcients A,B,C so that the error made by the approximation is smallest. Proceeding as above, we write the Taylor expansions f(x+h) = f(x)+hf ′(x)+ h2 2 f ′′(xi)+ h3 6 f ′′′(x)+ h4 24 f (4)(x)+O(h5), (·A) f(x) = f(x), (·B) f(x−h) = f(x)−hf ′(x)+ h2 2 f ′′(x)−h3 6 f ′′′(x)+ h4 24 f (4)(x)+O(h5), (·C) Af(x+h)+Bf(x)+C f(x−h) = (A+B+C)f(x) (≡0· f(x)) +h(A−C)f ′(x) (≡1· f ′(x)) + h2 2 (A+C) f ′′(x) (≡0· f ′′(x)) + h3 6 (A−C) f ′′′(x) + h4 24(A+C) f (4)(x)+O(h5). Imposing A+B+C = 0, h(A−C) = 0, h2 2 (A+C) = 1, we obtain A = C = 1 h2 , B = −2 h2 , which conﬁrms the (central ﬁnite difference formula) and the error (14.4). 94 14.4. Effects of error in function values. We shall look now at the effect of small (measurements or round-off) errors in the functions values into numerical differentiation. Let say that the true values f(xi) are approximated by ˆ f(xi), i.e., f(xi) ≈ˆ f(xi). Then using the (central ﬁnite difference formula) as an example, the approximation D2 f(xi) of the second derivative f ′′(xi), which uses instead of f(xi) the approximate values ˆ f(xi) = f(xi)+εi would write \ D2 f(xi) = 1 h2 f(x+h)+ε1 −2 f(x)−2ε2 + f(x−h)+ε3 = D2 f(xi)+ ε1 −2ε2 +ε3 h2 . Therefore the true error would satisfy f ′′(x)−D2 f(x) ≤ f ′′(x)−\ D2 f(x) + \ D2 f(x)−D2 f(x) ≤h2 12\|f (4)(x)\|+ 4ε h2 . (14.5) This suggests that as h ↘0, the last term in the right hand side will increase. An optimal h should then satisfy h2 ≈ε h2 . For ε ≈10−16 this means h = ε1/4 = 10−4. Exercise 1,2,3 page 241. (b) Find the numerical (forward, backward and centered ﬁnite differences) derivatives D1 f(x) at x = 1 for f(x) = arctan(x2 −x + 1), using h = 0.1,0.05,0.025,0.0125,0.00625; calculate the error, the ratio and estimate the error by using x = 1 instead of the intermediate point. Solution: First, the exact value for the derivative is f ′(x) = 2x−1 1+(x2 −x+1)2 , f ′(1) = 2x−1 1+(x2 −x+1)2 x=1 = 1 2, f ′′(ξ) ≈f ′′(1) = 1 2, f ′′′(ξ) ≈f ′′′(1) = −2.45. Using the (forward Finite Difference) formula we obtain: h D1 f Error Ratio Error estimate≈−1 4h Order 0.1 0.5209 −0.0209 - 0.0250 0.05 0.5115 −0.0115 1.8199 - 0.0125 0.863854 0.025 0.5060 −0.0060 1.9132 -0.0063 0.936025 0.0125 0.5031 −0.0031 1.9575 - 0.0031 0.968992 0.00625 0.5015 −0.0016 1.9790 - 0.0016 0.984736 Using the (centered Finite Difference) formula we obtain: h D1 f Error Ratio Error estimate≈−2.45 6 h2 Order 0.1 0.495855700471995 0.004144299528005 0.004083333333333 0.05 0.498959737494895 0.001040262505105 3.9839 0.001020833333333 1.99418 0.025 0.499739671191277 0.000260328808723 3.9960 0.000255208333333 1.99854 0.0125 0.499934901325987 0.000065098674013 3.9990 0.000063802083333 1.99963 0.00625 0.499983724301654 0.000016275698346 3.9997 0.000015950520833 1.99991 (SEE FIGURE 14.1 FOR THE EFFECT OF ROUNDOFF ERROR IN FUNCTION VALUES !!! Exercise 4, page 241. Let h > 0,xj = x0 + jh for j = 0,1,2,··· ,n, and let (Πn f)(x) be the degree n interpolating polynomial of f(x) at x0,x1,··· ,xn. Use this polynomial to estimate f ′(x0). Produce the actual formulae involving f(x0), f(x1),··· , f(xn) for n = 1,2,3,4. Also produce their error formulae. 95 x0 f0 x1 f1 f1−f0 h x2 f2 f2−f1 h f2−2f1+f0 2h2 x3 f3 f3−f2 h f3−2f2+f1 2h2 f3−3f2+3f1−f0 6h3 Solution: For n = 3, the Newton divided differences give the interpolating polynomial (Π4 f)(x) = f0 + f1 −f0 h (x−x0)+ f2 −2 f1 + f0 2h2 (x−x0)(x−x1)+ f3 −3f2 +3 f1 −f0 6h3 (x−x0)(x−x1)(x−x2) hence (Π4 f)′(x0) = f1 −f0 h + f2 −2f1 + f0 2h2 (x0 −x1)+ f3 −3 f2 +3f1 −f0 6h3 (x0 −x1)(x0 −x2) = 1 h 1 3 f3 −3 2 f2 +3 f1 −11 6 f0 . The error term is 1 4! f (4)(ξ) d dx (x−x0)(x−x1)(x−x2)(x−x3) x=x0 \| {z } ≈−6h3 ≈−h3 4 f (4)(ξ), ξ ∈[x0,x3]. Exercise 9, page 242. Use the (central ﬁnite difference formula) formula to estimate f ′′(x) for the function f(x) = arctan(x2 −x + 1), at x = 1, using h = 0.1,0.05,0.025,0.0125,0.00625,0.003125,···; calculate the error, and the ratios by which they decrease. Solution: Using the (central ﬁnite difference formula) formula we obtain: h D2 f Error Ratio Order 0.1 0.4999958333958 0.0000041666042 0.05 0.4999997395836 0.0000002604164 15.9997 4.00001 0.025 0.4999999837240 0.0000000162760 16.0000 4.01356 0.0125 0.4999999989828 0.0000000010172 1.9132 7.11339 0.00625 0.4999999999370 0.0000000000630 1.9575 -6.95384 0.003125 0.4999999999995 0.0000000000005 1.9790 -0.50524 0.0015625 0.5000000000564 −0.0000000000564 1.9790 -2.54057 0.00078125 0.4999999999200 0.0000000000800 1.9790 0 (SEE FIGURE 14.1 FOR THE EFFECT OF ROUNDOFF ERROR IN FUNCTION VALUES !!!) 96 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 10-14 10-12 10-10 10-8 10-6 10-4 10-2 100 102 104 forward difference central difference second derivative FIG. 14.1. Errors in numerical differentiation: the forward and central ﬁnite difference approximation of f ′, and the centered ﬁnite difference approximation of f ′′, at x = 1, with f(x) = arctan(x2 −x + 1). Notice the effects of errors in function values for small values of the mesh size h ≪1. Exercise 6, page 241. The error formulae (14.1) for the (forward Finite Difference), (14.2) for the (centered Finite Differenc and (14.4) for the (central ﬁnite difference formula), are nearly proportional to a power of h. These justify the use of Richardson’s extrapolation as in Section 12.1.4 (see (Richardson extrapolation) for the quadrature rules). For the (forward Finite Difference), derive the extrapolation formula f ′(x) ≈2Dh f(x)−D2h f(x) (Richardson extrapolation of forward ﬁnite difference) and show its error convergences to zero more rapidly than does the error for (forward Finite Difference). Derive corresponding error formulae for (centered Finite Difference) and (central ﬁnite difference formula), based on (14.2) and (14.4). Solution: Using the error formula (14.1) for the with (forward Finite Difference) h and 2h we have f ′(x)−uFD h ≈−h 2 f ′′(x)+O(h2), f ′(x)−uFD 2h ≈− 2h 2 f ′′(x)+O(h2), hence f ′(x) ≈2uFD h −uFD 2h +O(h2). Similarly, using the error formula (14.2) for the with (centered Finite Difference) h and 2h we have f ′(x)−uCD h ≈−h2 6 f ′′′(x)+O(h3), 97 f ′(x)−uCD 2h ≈−4h2 6 f ′′′(x)+O(h3), hence f ′(x) ≈1 3 4uCD h −uCD 2h +O(h3). Exercise 7 (or similar). Use the extrapolation formula (Richardson extrapolation of forward ﬁnite difference) to improve the answers given in table in Exercise 1,2,3. Solution: The extrapolation formula gives the values in the table (notice that the extrapolated values are more accurate than the values given by the (forward Finite Difference) in Exercise 1,2,3): h D1 f Error Order Extrapolated values Error in extrapolated Order Xtrp 0.1 0.5209 −0.0209 0.502063969827183 -0.002063969827183 0.05 0.5115 −0.0115 0.8638 0.500519610991171 -5.196109911710378e-04 1.9899 0.025 0.5060 −0.0060 0.9360 0.500130131651430 -1.301316514300543e-04 1.9974 0.0125 0.5031 −0.0031 0.9689 0.500032547283285 -3.254728328494494e-05 1.9993 0.00625 0.5015 −0.0016 0.9847 0.500008137720619 -8.137720619028066e-06 1.9998 Exercise 14, page 242. Using the following table of rounded values of f(x), estimate f ′′(0.5) numerically with stepsizes h = 0.2,0.1. x f(x) 0.3 7.3891 0.4 7.4633 0.5 7.5383 0.6 7.6141 0.7 7.6906 Also estimate the possible size of that part of the error in your answer that is due to the rounding errors in the table entries. Is this a serious source of error in this case? Solution: Using h = 0.2 (and h = 0.1 respectively) and the (central ﬁnite difference formula) to approximate f ′′(0.5) we have f ′′(0.5) ≈7.3891−2·7.5383+7.6906 0.04 = 0.077500000000019, (h = 0.2) ≈7.4633−2·7.5383+7.6141 0.01 = 0.080000000000080. (h = 0.1) The rounding error in the table (only four significant digits) is of order ε = 0.00005, hence using a reasoning similar to (14.5), the error due to rounding values is of the form 4ε h2 ≈0.005 (h = 0.2) ≈0.02. (h = 0.1) Comparing these error values with the estimated values of f ′′(0.5), we conclude that when using h = 0.2, the round-ing error has less signiﬁcant inﬂuence than when using h = 0.1. 98 15. Section 3. ROOTFINDING FOR NONLINEAR EQUATIONS. MOTIVATION 1. Let’s consider solving the following Initial Value Problem (IVP) u′(t) = F(u), t ∈(0,T], u(0) = u0 (IVP) i.e., we are looking to ﬁnd the solution u(t) : [0,T] →R, a continuously differentiable function satisfying the equation and the initial condition. For a general right-hand side term, this could be very difﬁcult, so let’s again settle with an approximate solution, namely a sequence {un} of approximate values un ≈u(tn), tn = t0 +n∆t, ∆t = T N ,n = 0,··· ,N. These values could be then computed iteratively. Using a numerical approximation of the derivative, for example the (backward Finite Difference), we can approximate (IVP) by the (backward Euler) formula:    un+1 −un ∆t = F(un+1), n = 0,··· ,N −1, u(0) = u0. (backward Euler) We now point out that (for each n = 0,··· ,N −1) un+1 is a root of the function f(x) = 0, where f(x) = x−∆t F(x)−un. DEFINITION 15.1. Let f : I = [a,b] ⊂R →R be a scalar function. The complex number α ∈C is called a root or zero of f if f(α) = 0. REMARK 38. The methods for the numerical approximation of a zero α of f are usually iterative, i.e., (given an initial guess) generate a sequence of values {x(k)}k≥0 such that lim k→∞x(k) = α. (15.1) -6 -4 -2 0 2 4 6 -8 -6 -4 -2 0 2 4 6 8 FIG. 15.1. f(x) = 0.1−x+sin(x),α = 0.853750156640865 99 >> a = fzero(inline(’0.1-x+sin(x)’),’) a = 0.853750156640865 % verify your result >> 1.-a+sin(a) ans = 1.110223024625157e-16 % machine precision: eps/2 DEFINITION 15.2 (Order of convergence). The sequence {x(k)} generated by a numerical method converges to α with order p ≥1 • if there exists C > 0 such that \|x(k+1) −α\| \|x(k) −α\|p ≤C, ∀k ≥k0, (where k0 is a suitable constant) (15.2) • or lim k→∞ x(k+1) −α (x(k) −α)p = C, (asymptotic error constant). (15.3) REMARK 39. Unlike the case of linear systems, the convergence of iterative methods for rootﬁnding of nonlinear equations depends in general on the choice of (the initial ”guess”) x(0). In such a case we say that the method is locally convergent: x(0) ∈V (α). DEFINITION 15.3 (Global convergence). The iterative methods which converge for all x(0) are called globally convergent. REMARK 40. If p = 1 (linear convergence), from (15.2) we see that in order to have convergence x(k) →α, it is necessary to have the convergence factor C < 1: \|x(k+1) −α\| ≤C\|x(k) −α\| ≤C2\|x(k−1) −α\| ≤··· ≤Ck+1\|x(0) −α\| →0 if and only if C < 1. DEFINITION 15.4 (Efﬁciency factor). Consider an iterative method with the q ≥1 order of convergence. If each iteration requires m units of ”work” (work involved in computing a function value or the value of its derivatives), then E := q 1 m (efﬁciency index) is the efﬁciency index of the iteration. REMARK 41. • The methods which converge linearly all have the efﬁciency index Elinear = 1. (Indeed, E = 1m.) • We shall see that Newton’s method has ENewton = 2 1 2 , as q = 2 and m = 2 (each iteration involves evaluations of f and f ′). REMARK 42 (Errors in rootﬁnding). Let us consider the fourth-order polynomial p4(x) = (x−1)4, and its round-off error approximation b p4(x) = (x−1)4 −ε, 100 where ε = 10−16. It is easy to see that the roots of each polynomials are αi = 1, i = 1,2,3,4, b αi = 1± 4 √ε, 1±i 4 √ε, i = 1,2,3,4. So, to an ε = 10−16 error in the function evaluation it corresponds a much (four orders of magnitude) larger relative error in the roots: Relative error(α1) = 1−(1−4 √ε) 1 = 4 √ε = 10−4. 15.1. Subsection 3.1 THE BISECTION METHOD. A GEOMETRIC APPROACH TO ROOTFINDING. Let us recall ﬁrst the Intermediate Value Theorem. THEOREM 15.5 (Intermediate Value Theorem). Given f ∈C([a,b];R) such that f(a)· f(b) < 0, there exists α ∈(a,b) such that f(α) = 0. The theorem states that a continuous function which has opposite sign values on an interval, also has one root in that interval; and this is the basic fact on which we can construct an iterative algorithm to get as close as possible to the root: I0 := [a,b]; generate a sequence Ik = [a(k),b(k)] such that Ik ⊂Ik−1 with f(a(k))· f(a(k)) < 0. Result: Bisection Method initialization: set a(0) = a, b(0) = b, x(0) = a(0) +b(0) 2 ; for k ≥0 do a(k+1) = a(k), b(k+1) = x(k) if f(x(k))· f(a(k)) < 0 ; a(k+1) = x(k), b(k+1) = b(k) otherwise ; x(k+1) = a(k+1) +b(k+1) 2 Terminate when tolerance is reached: \|b(k) −a(k)\| < ε or k ≥max iterate. end REMARK 43. • At step m we have that the error is \|α −x(m)\| < \|Im\| 2 • Speed of convergence: \|I0\| = \|b−a\|, ··· \|Ik\| = b−a 2k • The absolute error satisﬁes \|e(k)\| = \|α −x(k)\| ≤b−a 2k+1 k↗∞ − − − →0 (global convergence of bisection) Hence the bisection method is globally convergent!!! QUESTION 1. 101 (1) How many iterates m are necessary to obtain tolerance ε? Using (global convergence of bisection) we have that e(m) = \|α −x(m)\| ≤b−a 2m+1 ≤ε, b−a ε ≤2m+1, log b−a ε log(2) ≤m+1, m ≥−1+ log b−a ε log(2) . (15.4) (2) How many iterates m are necessary to obtain one extra signiﬁcant digit? Similarly b−a 2k+1 ≈\|α −x(k)\| = \|α −x(j)\| 10 ≈ b−a 2 j+1 1 10, 2k+ 1−j− 1 ≈10, k −j ≈log(10) log(2) ≈3.3219. (≈3.3 bisection iterates for one signiﬁcant digit) REMARK 44. • (global convergence of bisection): The bisection method is globally convergent!!! • (≈3.3 bisection iterates for one signiﬁcant digit) makes the bisection method slow!!!! The fact that is slow but globally convergent makes the bisection method a perfect approaching techinique! The fzero function of MATLAB uses a combination of bisection, secant and inverse quadratic interpolation methods: fzero(inline(-4+x^2),1) EXAMPLE 5 (COUNTEREXAMPLE - f DISCONTINUOUS). Let f(x) = 5 + (3.5x + 1) · 1 + 1 (3.5x+1)2 + 1 x2 , which has discontinuities at x = −1 3.5 ≈−0.285714285714286 and at x = 0, -0.4 -0.35 -0.3 -0.25 -0.2 -0.15 -1.5 -1 -0.5 0 0.5 1 1.5 104 so the assumption of Intermediate Value Theorem 15.5 does not hold. Hence we should doubt the output of the bisection algorithm, which does not know of this fault! Indeed, if we run it on [a,b] ≡[−5,5], with tolerance ε = 10−10 and max iterate = 100 we obtain 102 bisect(-5,5,1.e-10,100,10) root = -0.285714285782888 and no indication of any errors! But if we check now if this α = −0.285714285782888 is indeed a root, by evaluating f(α) and expecting f(α) = 0, we are in for a big surprise: f(α) = −4.1648e+09 ̸= 0!!!! REMARK 45. The errors in the approximation of the root by the bisection method does not decay monotonically, and therefore is not of order 1: \|e(k+1)\| ≰\|e(k)\| \|α −x(k+1)\| ≰\|α −x(k)\| EXAMPLE 6. For the case f(x) = x2 −4,α = 2 we see that indeed the errors do not decay monotonically: k x(k) α −x(k) 1 2.250000 -0.2500 2 1.125000 0.8750 3 1.687500 0.3125 4 1.968750 0.0312 5 2.109375 -0.1094 6 2.039062 -0.0391 7 2.003906 -0.0039 8 1.986328 0.0137 9 1.995117 0.0049 1 2 3 4 5 6 7 8 9 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 FIG. 15.2. The error in the bisection method does not decay monotonically. 103 function [root,error_bound,it_count]=bisect(a0,b0,ep,max_iterate,index_f) >> bisect(0,4.5,1.e-2,100,12) Iteration number: 1 Approximation of the root: 2.250000 Iteration number: 2 Approximation of the root: 1.125000 Iteration number: 3 Approximation of the root: 1.687500 Iteration number: 4 Approximation of the root: 1.968750 Iteration number: 5 Approximation of the root: 2.109375 Iteration number: 6 Approximation of the root: 2.039062 Iteration number: 7 Approximation of the root: 2.003906 Iteration number: 8 Approximation of the root: 1.986328 Iteration number: 9 Approximation of the root: 1.995117 root = 1.995117187500000 Exercise 1, page 77. Use the bisection method with a hand calculator or computer to ﬁnd the indicated roots of the following equations. Use a tolerance of ε = 10−4. (a) The real root of x3 −x2 −x−1 = 0, f1(x) = x3 −x2 −x−1. (b) The root of x = 1+0.3cosx, f2(x) := x−1−0.3cos(x). (c) The smallest positive root of cos(x) = 1 2 +sinx, f3(x) := cos(x)−1 2 −sin(x). Solution: (a) With tolerance 10−15 it takes 52 iterations to obtain the root = 1.839286755214161; f1(1.839286755214161) = −4.4409e−16. With tolerance 10−4: bisect(0,3,1.e-4,100,16) Iteration number: 1 Approximation of the root: 1.500000 The error estimate : 1.500000 104 Iteration number: 2 Approximation of the root: 2.250000 The error estimate : 0.750000 Iteration number: 3 Approximation of the root: 1.875000 The error estimate : 0.375000 Iteration number: 4 Approximation of the root: 1.687500 The error estimate : 0.187500 Iteration number: 5 Approximation of the root: 1.781250 The error estimate : 0.093750 Iteration number: 6 Approximation of the root: 1.828125 The error estimate : 0.046875 Iteration number: 7 Approximation of the root: 1.851562 The error estimate : 0.023438 Iteration number: 8 Approximation of the root: 1.839844 The error estimate : 0.011719 Iteration number: 9 Approximation of the root: 1.833984 The error estimate : 0.005859 Iteration number: 10 Approximation of the root: 1.836914 The error estimate : 0.002930 Iteration number: 11 Approximation of the root: 1.838379 The error estimate : 0.001465 Iteration number: 12 Approximation of the root: 1.839111 The error estimate : 0.000732 Iteration number: 13 Approximation of the root: 1.839478 The error estimate : 0.000366 105 Iteration number: 14 Approximation of the root: 1.839294 The error estimate : 0.000183 Iteration number: 15 Approximation of the root: 1.839203 The error estimate : 0.000092 x(15) = 1.839203; f1(x(15)) = −4.581389615827547×10−4; The error: α −x(15) ≡1.839286755214161−1.839203 = 8.375521416104093×10−5 ≈10−4 (b) With tolerance 10−15 it takes 52 iterations to obtain the root = 1.128425092992224. f2(1.128425092992224) = −8.8818×10−16. With tolerance 10−4: >> bisect(0,3,1.e-4,300,17) Iteration number: 1 Approximation of the root: 1.500000 The error estimate : 1.500000 Iteration number: 2 Approximation of the root: 0.750000 The error estimate : 0.750000 Iteration number: 3 Approximation of the root: 1.125000 The error estimate : 0.375000 Iteration number: 4 Approximation of the root: 1.312500 The error estimate : 0.187500 Iteration number: 5 Approximation of the root: 1.218750 The error estimate : 0.093750 Iteration number: 6 Approximation of the root: 1.171875 The error estimate : 0.046875 Iteration number: 7 Approximation of the root: 1.148438 The error estimate : 0.023438 Iteration number: 8 Approximation of the root: 1.136719 The error estimate : 0.011719 Iteration number: 9 106 Approximation of the root: 1.130859 The error estimate : 0.005859 Iteration number: 10 Approximation of the root: 1.127930 The error estimate : 0.002930 Iteration number: 11 Approximation of the root: 1.129395 The error estimate : 0.001465 Iteration number: 12 Approximation of the root: 1.128662 The error estimate : 0.000732 Iteration number: 13 Approximation of the root: 1.128296 The error estimate : 0.000366 Iteration number: 14 Approximation of the root: 1.128479 The error estimate : 0.000183 Iteration number: 15 Approximation of the root: 1.128387 The error estimate : 0.000092 x(15) = 1.128387; f2(x(15)) = −4.842073720920381×10−05; The error: α −x(15) ≡1.128425092992224−1.128387 = 3.809299222390372×10−05 ≈10−4 (c) With tolerance 10−15 it takes 52 iterations to obtain the root = 0.424031039490741. f3(0.424031039490741) = −6.6613e−16. With tolerance 10−4: >> bisect(0,3,1.e-4,300,18) Iteration number: 1 Approximation of the root: 1.500000 The error estimate : 1.500000 Iteration number: 2 Approximation of the root: 0.750000 The error estimate : 0.750000 Iteration number: 3 Approximation of the root: 0.375000 The error estimate : 0.375000 Iteration number: 4 Approximation of the root: 0.562500 107 The error estimate : 0.187500 Iteration number: 5 Approximation of the root: 0.468750 The error estimate : 0.093750 Iteration number: 6 Approximation of the root: 0.421875 The error estimate : 0.046875 Iteration number: 7 Approximation of the root: 0.445312 The error estimate : 0.023438 Iteration number: 8 Approximation of the root: 0.433594 The error estimate : 0.011719 Iteration number: 9 Approximation of the root: 0.427734 The error estimate : 0.005859 Iteration number: 10 Approximation of the root: 0.424805 The error estimate : 0.002930 Iteration number: 11 Approximation of the root: 0.423340 The error estimate : 0.001465 Iteration number: 12 Approximation of the root: 0.424072 The error estimate : 0.000732 Iteration number: 13 Approximation of the root: 0.423706 The error estimate : 0.000366 Iteration number: 14 Approximation of the root: 0.423889 The error estimate : 0.000183 Iteration number: 15 Approximation of the root: 0.423981 The error estimate : 0.000092 root = 108 0.423980712890625 x(15) = 0.423981; f2(x(15)) = 6.619539810059916×10−05; The error: α −x(15) ≡0.424031039490741−0.423981 = 5.003949074100511×10−05 ≈10−4 Exercise 2, page 77. To help determine the roots of x = tan(x), graph both y = x and y = tan(x), and look at the intersection points of the two curves. -2 -1 0 1 2 3 4 5 6 7 8 -10 -8 -6 -4 -2 0 2 4 6 8 10 (a) Find the smallest nonzero positive root of x = tan(x), with an accuracy of ε = 10−4. Note: The desired root is greater than π/2. (b) Solve x = tan(x) for the root that is closest to x = 100. Solution: (a) From the graph we see that the ﬁrst zero of the function f(x) = x−tan(x) is between [π/2,3π/2]. With ε = 10−15, in 52 iterates we get α = 4.493409457909064. bisect(pi/2+0.01,3pi/2-0.01,1.e-4,100,19) Iteration number: 15 Approximation of the root: 4.493476239621677 The error estimate : 0.000095 (b) >> bisect(31.5pi,32.5pi,1.e-4,100,19) Iteration number: 15 Approximation of the root: 102.0918862403462 The error estimate : 0.000096 Exercise 4, page 77. Show that for any real constants c and d, the equation x = −c−d cos(x) has at least one root. Hint: Find an interval [a,b] on which f(x) = x−c−d cos(x) changes sign. Solution: Exercise 6, page 77. Use the bisection method and a graph of f(x), to ﬁnd all roots of f(x) = 32x2 −48x4 +18x2 −1. The true roots are cos (2j −1) π 12 , j = 1,2,··· ,6. Solution: 109 -10 -8 -6 -4 -2 0 2 4 6 8 10 -15 -10 -5 0 5 10 f(x)=x-c-dcos(x) FIG. 15.3. f(x) = x−c−d cos(x), with c = 2,d = −3. -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 FIG. 15.4. f(x) = 32x2 −48x4 +18x2 −1. bisect(-1,-0.8,1.e-6,100,20) Iteration number: 18 Approximation of the root: -0.965926361083984 >> bisect(-0.8,-0.6,1.e-6,100,20) Iteration number: 18 Approximation of the root: -0.707106781005859 The error estimate : 0.000001 >> bisect(-0.4,-0.2,1.e-6,100,20) Iteration number: 18 Approximation of the root: -0.258818817138672 >> bisect(0.2,0.4,1.e-6,100,20) Iteration number: 18 Approximation of the root: 0.258818817138672 >> bisect(0.6,0.8,1.e-6,100,20) Iteration number: 18 Approximation of the root: 0.707106781005859 >> bisect(0.8,1.0,1.e-6,100,20) 110 Iteration number: 18 Approximation of the root: 0.965926361083984 Exercise 7, page 78. Using the program of Problem 5, solve the equation f(x) ≡x3 −3x2 +3x−1 = 0 ( f(x) = (x−1)3) with an accuracy of ε = 10−6. Experiment with different ways of evaluating f(x); for example, use (i) the given form, (ii) reverse its order, and (iii) the nested form f(x) = −1+x(3+x(−3+x)). Try various initial intervals [a,b], for example [0,1.5], [0.5,2.0], and [0.5,1.1]. Explain the results. (Note that α = 1 is the only root of f(x) = 0.) Solution: [a,b] x3 −3x2 +3x−1 −1+3x−3x2 +x3 −1+x(3+x(−3+x)) [0.0,1.5] 1.000006914138794 (18 iter) 1.000005483627319 (21iter) 1.000005483627319 (21 iter) [0.5,2.0] 1.000007390975952 (21 iter) 1.000004529953003 (21 iter) 1.000004529953003 (21 iter) [0.5,1.1] 1.000006675720215 (20 iter) 1.000005531311035 (20 iter) 1.000004386901856 (20 iter) Exercise 8, page 78. The polynomial f(x) = x4 −5.4x3 +10.56x2 −8.954x+2.7951 has a root α in [1,1.2]. Repeat Problem 7, but vary the intervals [a,b] to reﬂect the location of the root α. Solution: With ε = 10−15, starting from [1.08,1.15], in 46 iterations, the nested form gives α = 1.100006916588192. With ε = 10−6: [a,b] x4 −5.4x3 +10.56x2 −8.95x+2.7951 2.7951+x(−8.95+x(10.56+x(−5.4+x))) [1.00,1.20] 1.100002288818359 (18 iter) 1.099996185302734 (18 iter) [1.08,1.15] 1.100002059936524 (17 iter) 1.100007400512696 (17 iter) Exercise 9, page 78. Let the initial interval used in the bisection method have length b−a = 3. Find the number of midpoints cn that must be calculated with the bisection method to obtain an approximate root within an error tolerance of ε = 10−9. Solution: Using estimate (15.4) ≡−1+ log 3 10−9 log(2) = 30.4823, hence 31 iterates. Exercise 10, page 78. Consider the equation e−x = sinx. Find an interval [a,b] that contains the smallest positive root. Estimate the number of midpoints c needed to obtain an approximate root that is accurate within an error tolerance of 10−10. Solution: The function f(x) = e−x −sinx has the smallest root on the interval [0,2]. With ε = 10−10, the bisection method gives 111 0 1 2 3 4 5 6 7 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 e-x-sin(x) (root,f(root)) >> bisect(0.0,2.0,1.e-10,100,8) Iteration number: 34 Approximation of the root: 0.588532744033728 Using estimate (15.4) m ≥−1+ log b−a ε log(2) ≡−1+ log 2 10−10 log(2) = 33.2193, hence 34 iterates. Exercise 11, page 78. Let α be the smallest positive root of f(x) ≡1−x+sinx = 0 Find an interval [a,b] containing α and for which the bisection method will converge to α. -8 -6 -4 -2 0 2 4 6 8 -6 -4 -2 0 2 4 6 8 f(x)=1-x-sin(x) (root,f(root)) FIG. 15.5. Exercise 11: f(x) ≡1−x+sinx = 0. Then estimate the number of iterates needed to ﬁnd α within an accuracy of 5×10−8. Solution: With ε = 10−15 the bisection method gives α = 1.934563210752025 in 53 iterations. With varepsilon = 5×10−8 we obtain: >> bisect(0,2pi,5.e-8,100,9) Iteration number: 26 Approximation of the root: 1.934563195716542 Using estimate (15.4) m ≥−1+ log b−a ε log(2) ≡−1+ log 2π 5∗10−8 log(2) = 25.9050, 112 hence 26 iterates. Exercise 12, page 78. Let α be the unique root of x = 3 1+x4 . Find an interval [a,b] containing α and for which the bisection method will converge to α. -3 -2 -1 0 1 2 3 -4 -3 -2 -1 0 1 2 3 x-3/(1+x 4) (root,f(root)) FIG. 15.6. Exercise 12: x− 3 1+x4 = 0. Then estimate the number of iterates needed to ﬁnd α within an accuracy of 5×10−8. Solution: With ε = 10−15 the bisection method gives α = 1.132997565885066 in 50 iterations. With ε = 5 × 10−8 we obtain: >> bisect(0,2,5.e-8,100,21) Iteration number: 25 Approximation of the root: 1.132997542619705 Using estimate (15.4) m ≥−1+ log b−a ε log(2) ≡−1+ log 2 5∗10−8 log(2) = 24.2535, hence 25 iterates. Exercise 13 page 79. Let α be the largest root of f(x) ≡ex −x−2 = 0. Find an interval [a,b] containing α and for which the bisection method will converge to α. Then estimate the number of iterates needed to ﬁnd α within an accuracy of 5×10−8. Solution: With ε = 10−15 the bisection method gives α = 1.146193220620583 in 51 iterations. With ε = 5 × 10−8 we obtain: >> bisect(-1.5,1.5,5.e-8,100,22) Iteration number: 25 Approximation of the root: 1.146193251013756 Using estimate (15.4) m ≥−1+ log b−a ε log(2) ≡−1+ log 3 5∗10−8 log(2) = 24.8385, hence 25 iterates. 113 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 ex-x-2 (root,f(root)) FIG. 15.7. Exercise 13: ex −x−2 = 0. 15.2. Subsection 3.3 NEWTON’S METHOD. The methods of chord, secant and Newton GOAL: for better convergence - include values of both f and f ′, or suitable approximations. Using Taylor series of f about the root α: 0 = f(α) = f(x)+ f ′(ξ)·(α −x), ξ between x and α equivalently α = x−f(x) f ′(ξ), if x is close to α, and f is C1 near α. Result: Algorithm “iterative method” for k ≥0, given x(k) do x(k+1) = x(k) −f(x(k)) q(k) where q(k) ≈f ′(x(k)) ; end As particular cases: q(k) ≡q := f(b)−f(a) b−a ⇒x(k+1) = x(k) − f(x(k)) f(b)−f(a)(b−a) (chord method) q(k) ≡f(x(k))−f(x(k−1)) x(k) −x(k−1) ⇒x(k+1) = x(k) − f(x(k)) f(x(k))−f(x(k−1))(x(k) −x(k−1)) (secant method, x(0),x(1)) q(k) ≡f ′(x(k)) ⇒x(k+1) = x(k) −f(x(k)) f ′(x(k)) (Newton’s method) THEOREM 15.6 (Newton’s method - quadratic convergence). If α is a root of multiplicity one, and the initial guess x(0) is sufﬁciently close to α so that the Newton method converges x(k) k↗∞ − − − →α, then Newton’s method has order of convergence two: lim k→∞ α −x(k+1) (α −x(k))2 = −f ′′(α) 2 f ′(α). (Newton’s method - quadratic convergence) 114 Proof. Using the Taylor expansion about x(k): 0 = f(α) = f(x(k))+(α −x(k)) f ′(x(k))+ 1 2(α −x(k))2 f ′′(ξ (k)). Dividing by f ′(x(k)) and using the Newton’s method (x(k+1) = x(k) −f(x(k)) f ′(x(k))): 0 = f(x(k)) f ′(x(k)) \| {z } =x(k)−x(k+1) +α −x(k) + 1 2(α −x(k))2 f ′′(ξ (k)) f ′(x(k)) = x(k) −x(k+1) +α − x(k) + 1 2(α −x(k))2 f ′′(ξ (k)) f ′(x(k)) . Equivalently we have α −x(k+1) = (α −x(k))2 −f ′′(ξ (k)) 2 f ′(x(k)) , α −x(k+1) (α −x(k))2 = −f ′′(ξ (k)) 2 f ′(x(k)) which gives (Newton’s method - quadratic convergence). REMARK 46 (Newton’s method is locally convergent). Convergence of Newton’s method is guaranteed provided the initial guess x(0) is close enough to the root α: \|α −x(0)\| < 2 f ′(α) f ′′(α) . (3.21 - Newton’s method - local convergence) Proof. Indeed form the previous proposition we have: \|α −x(k+1)\| f ′′(α) 2 f ′(α) ≈ \|α −x(k)\| f ′′(α) 2 f ′(α) 21 ≈ \|α −x(k−1)\| f ′′(α) 2 f ′(α) 22 ≈ ··· ≈ \|α −x(0)\| f ′′(α) 2 f ′(α) 2k+1 , hence for convergence \|α −x(0)\| f ′′(α) 2 f ′(α) < 1, which concludes the argument. COROLLARY 15.7 (Error estimation). Using Taylor expansion: α −x(k) = −f(x(k)) f ′(ξ (k)) (Taylor expansion: f(x(k)) = f(α)+ f ′(ξ (k))(x(k) −α)) ≈−f(x(k)) f ′(x(k)) (using the Newton iteration: x(k+1) = x(k) −f(x(k)) f ′(x(k))) 115 = x(k+1) −x(k). Exercise 1, page 88. Carry out the Newton iteration x(n+1) = x(n) − x6 (n) −x(n) −1 6x5 (n) −1 , n ≥0 (3.12) with the two initial guesses x0 = 1.5 and x0 = 2.0. Solution: n x(n) f(x(n)) x(n) −x(n−1) 0 1.5 8.8906e+00 -1.9951e-01 1 1.300490883590 2.5373e+00 -1.1901e-01 2 1.181480416403 5.3846e-01 -4.2025e-02 3 1.139455590276 4.9235e-02 -4.6780e-03 4 1.134777625237 5.5032e-04 -5.3480e-05 5 1.13472414531 7.1136e-08 -6.9147e-09 6 1.134724138402 1.5543e-15 -2.2204e-16 n x(n) f(x(n)) x(n) −x(n−1) 0 2.0 6.1000e+01 -3.1937e-01 1 1.680628272251 1.9853e+01 -2.4989e-01 2 1.430738988239 6.1468e+00 -1.7577e-01 3 1.254970956109 1.6517e+00 -9.3433e-02 4 1.161538432773 2.9431e-01 -2.5185e-02 5 1.136353274171 1.6826e-02 -1.6227e-03 6 1.134730528344 6.5738e-05 -6.3898e-06 7 1.134724138500 1.0154e-09 -9.8702e-11 8 1.134724138402 -8.8818e-16 0 Notice the quadratic convergence in the log-log plot e1 = [-1.9951e-01 -1.1901e-01 -4.2025e-02 ... -4.6780e-03 -5.3480e-05 -6.9147e-09 -2.2204e-16]; e2 = [ -3.1937e-01 -2.4989e-01 -1.7577e-01 ... -9.3433e-02 -2.5185e-02 -1.6227e-03 -6.3898e-06 -9.8702e-11 0]; figure(1) loglog(abs(e1) , ’o-’,’linewidth’,2) grid on figure(2) loglog(abs(e2) , ’d-’,’linewidth’,2) grid on 116 1 2 3 4 5 6 7 10-16 10-14 10-12 10-10 10-8 10-6 10-4 10-2 100 1 2 3 4 5 6 7 8 10-12 10-10 10-8 10-6 10-4 10-2 100 FIG. 15.8. log-log scale plot of the errors. Exercise 2, page page 88. Using Newton’s method, ﬁnd the roots of the equations. Use an error tolerance of ε = 10−6. (a) The real root of x3 −x2 −x−1 = 0. (f1(x) := x3 −x2 −x−1) (b) The root of x = 1+0.3cosx; (f2(x) := x−1−0.3cosx) (c) The smallest positive root of cosx = 1 2 +sinx, (f3(x) := cosx−1 2 −sinx). Solution: -1 -0.5 0 0.5 1 1.5 2 2.5 3 -2 0 2 4 6 8 10 12 14 x3-x 2-x-1 root (a) With ε = 10−15 the root approximation is 1.839286755214161. With ε = 10−6 we obtain: >> [root,error_bd,it_count] = newton(0,1.e-6,100,3) Iteration number: 13 Approximation of the root: 1.839286755214161 The error estimate: -0.000000007291 f1(1.839286755214161) = −4.440892098500626e−16 -1 -0.5 0 0.5 1 1.5 2 2.5 3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 x-1-0.3cos(x) root (b) With ε = 10−15 the root approximation is 1.128425092992225. 117 With ε = 10−6 we obtain: >> [root,error_bd,it_count] = newton(0,1.e-6,100,6) Iteration number: 3 Approximation of the root: 1.128425092992225 The error estimate: -0.000000061308 f2(1.128425092992225) = 5.273559366969494e−16 -1 -0.5 0 0.5 1 1.5 2 2.5 3 -2 -1.5 -1 -0.5 0 0.5 1 cos(x)-0.5-sin(x) root (c) With ε = 10−15 the root approximation is 0.424031039490741 in 5 iterations. With ε = 10−6 we obtain: >> [root,error_bd,it_count] = newton(0,1.e-6,100,4) Iteration number: 3 Approximation of the root: 0.424031039490745 The error estimate: -1.5940e-07 f3(0.424031039490745) = −5.939693181744587e−15 Exercise 4, page 88. Give Newton’s method for ﬁnding m √a, with a > 0 and m a positive integer. Apply it to ﬁnding m √ 2 for m = 3,4,5,6,7,8, say, to six signiﬁcant digits. Hint: Solve xm −a = 0. Solution: m ‘root’ f(x(m)) iterates 3 1.259921050018 5.8526e-10 4 4 1.189207115676 4.5301e-09 4 5 1.148698356620 1.4128e-08 4 6 1.122462051041 2.9198e-08 4 7 1.104089517486 4.8335e-08 4 8 1.090507737437 7.0019e-08 4 Deﬁning f(x) := xm −a, f ′(x) = mxm−1, the Newton’s iteration writes x(n+1) = x(n) −f(x(n)) f ′(x(n)) ≡x(n) − xm (n) −a mxm−1 (n) = (m−1)xm (n) +a mxm−1 (n) . With tolerance ε = 5×10−7, ··· (see the Table). 118 Exercise 5, page 88. To help determine the roots of x = tan(x), graph both y = x and y = tan(x), and look at the intersection points of the two curves. -2 -1 0 1 2 3 4 5 6 7 8 -10 -8 -6 -4 -2 0 2 4 6 8 10 (a) Find the smallest nonzero positive root of x = tan(x), using Newton’s method, with an error tolerance of ε = 10−6. Note: The desired root is greater than π/2. (b) Solve x = tan(x) for the root that is closest to x = 100. The root near 100 will be difﬁcult to ﬁnd using Newton’s method. To explain this, compute the quantity M of (3.21 - Newton’s method - local convergence): M := −f ′′(α) 2 f ′(α) , and use it in the condition \|α −x(0)\| < 1 M = 2f ′(α) f ′′(α) for x(0). Solution: f(x) = x−tan(x), f ′(x) = 1− 1 cos2 x, f ′′(x) = −2 sinx cos3 x, x(n+1) = x(n) − 1 sinx(n) cos3 x(n) (Newton iteration) M := − 2 sinα cos3 α 2 1 cos2 α = −sinα cosα , with sin(4.28) cos(4.28) ≈98.9501 ⇒\|α −x(0)\| < 1 98.9501 ≈0.0101, and with sin(102.1) cos(102.1) ≈567.7807 ⇒\|α −x(0)\| < 1 567.7807 ≈0.0018, 119 >> [root,error_bd,it_count] = newton(4.288,1.e-6,200,31) Iteration number: 13 Approximation of the root: 4.493409630214 The error estimate: -1.7231E-7 f(4.493409630214) = −3.478965012071455×10−6 >> [root,error_bd,it_count] = newton(102.1,1.e-6,200,31) Iteration number: 6 Approximation of the root: 102.091966495223 The error estimate: -3.0315E-8 f(102.091966495223) = −3.159710088880274×10−4 Exercise 6, page 88. The equation f(x) ≡x+e−Bx2 cosx = 0, B > 0 has a uniques root, and it is in the interval (−1,0). -1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 B=1 B=5 B=10 B=25 B=50 Use Newton’s method to ﬁnd it as accurately as possible. Use values of B = 1,5,10,25,50. Among your choices of x0, choose x0 = 0, and explain the behavior observed in the iterates for the larger values of B. Solution: B “Root” iterates f(“root”) α 1 -0.588401776501 5 -1.1102e-16 -0.5884 5 -0.404911548209 8 5.5511e-17 -0.4049 10 -0.9845988999155 200 9.8385e-01 -0.3264 25 -0.999999979887 200 1.0000e+00 -0.2374 50 -1.000000000000 200 1 -0.1399 f(x) = x+e−Bx2 cosx, f ′(x) = 1−2Bxe−Bx2 cosx−e−Bx2 sinx; f(0) = 1, f ′(0) = 1, f(x) ≈x, f ′(x) = 1, (for B ≫1) f(−1) ≈−1, f ′(−1) ≈1 (for B ≫1) xn+1 = xn −f(xn) f ′(xn) ≈xn −xn 1 ≈0 (for B ≫1) 120 xn+2 = xn+1 −f(xn+1) f ′(xn+1) ≈0−f(0) f ′(0) ≈0−1 1 = −1 (for B ≫1) xn+3 = xn+2 −f(xn+2) f ′(xn+2) ≈−1−f(−1) f ′(−1) ≈−1−−1 1 = 0 ≡xn+1 (for B ≫1) xn+4 = xn+2. Exercise 10, page 89. Solve the equation x3 −3x2 +3x−1 = 0 (multiply root: (x−1)3) on a computer and use Newton’s method. Experiment with the choice of initial guess x0. Also experiment with different ways of evaluating f(x) and f ′(x). Note any unusual behavior in the iteration. Solution: -100 -80 -60 -40 -20 0 20 40 60 80 100 -1.5 -1 -0.5 0 0.5 1 106 (x-1)3 root: (1,0) x(0) iterates Root -100 42 0.999994068953 -10 36 0.999997000377 0 45 0.999995652810 0.5 31 0.999999229233 1.5 27 1.000007639946 10 35 1.000007110518 100 41 1.000007209443 Exercise 11, page 89. Solve the equation x4 −5.4x3 +10.56x2 −8.954x+2.7951 = 0 on a computer and use Newton’s method. Look for the root α located in [1,1.2]. Experiment with the choice of initial guess x0. Solution: f ′(x) = 4x3 −16.4x2 +21.12x−8.954, f ′(1) ≈−0.0340, f ′(1.2) ≈−0.0260. Using tolerance ε = 10−6: 121 1.09 1.095 1.1 1.105 1.11 1.115 -1 -0.5 0 0.5 1 1.5 10-6 x4 - 5.4 x 3 + 10.56 x.2 - 8.954 x + 2.7951 FIG. 15.9. Exercise 11 x(0) iterates Root f(‘root’) -100 86 1.099987461861 1.3323e-15 -10 192 1.099987564173 1.3323e-15 0 90 1.099987355516 0.5 27 1.099986626985 1.5 91 1.099986589433 4.4409e-16 10 13 2.100000000021 2.0999e-11 100 21 2.100000013363 1.3363e-08 15.3. Subsection 3.3 SECANT METHOD. Recall the general iterative method: x(k+1) = x(k) −f(x(k)) q(k) , where q(k) ≈f ′(x(k)). Using q(k) := f(x(k))−f(x(k−1)) x(k) −x(k−1) , we obtain the following: given x(0),x(1) ﬁnd: x(k+1) = x(k) − f(x(k)) f(x(k))−f(x(k−1)) x(k) −x(k−1) , ∀k ≥1. (secant method) Recall (Newton’s method - quadratic convergence): lim k→∞ α −x(k+1) Newton α −x(k) Newton 2 = −f ′′(α) 2 f ′(α). (Newton’s method - quadratic convergence) 122 THEOREM 15.8 (Secant method - superlinear convergence). If α is a root of multiplicity one, and the initial guess x(0) is sufﬁciently close to α so that the Secant method converges x(k) k↗∞ − − − →α, then the Secant method has order of convergence p = 1+ √ 5 2 ≈1.618: lim k→∞ \|α −x(k+1) Secant\| \|α −x(k) Secant\|q = max x,y∈V (α) 1 2 f ′′(y) f ′(x) q−1 . (Secant method: order q = 1+ √ 5 2 convergence) The proof of Theorem 15.8 is based on the following result. PROPOSITION 15.9. The error in the (secant method) method ek := α −x(k) secant satisﬁes the following recursion relation: ek+1 = −f ′′(ξk) 2f ′(θk)ekek−1. Proof. Recall the (interpolating polynomial in Newton form) (Π2 f)(x) = f(x0)+(x−x0)f[x0,x1]+(x−x0)(x−x1) f[x0,x1,x2], and choose x0 = x(k), x1 = x(k−1), x2 = α, where f(α) = 0 (and consequently (Π2 f)(α) ≡f(α) = 0). Therefore 0 = f(α) ≡(Π2 f)(α) = f(x(k))+(α −x(k))f[x(k),x(k−1)]+(α −x(k))(α −x(k−1)) f[x(k),x(k−1),α]. Also using the (secant method): x(k+1) −x(k) = − f(x(k)) f[x(k),x(k−1)] 0 = f(x(k))+(x(k+1) −x(k)) f[x(k),x(k−1)], ( hence subtracting the above relations we obtain ) 0 = (α − x(k) −x(k+1) + x(k)) f[x(k),x(k−1)]+e(k)e(k−1) f[x(k),x(k−1),α]. = e(k+1) f[x(k),x(k−1)]+e(k)e(k−1) f[x(k),x(k−1),α]. Therefore e(k+1) = −f[x(k),x(k−1),α] f[x(k),x(k−1)] e(k)e(k−1). Finally, using (8.1) in Theorem 8.3 we obtain: e(k+1) = −f[x(k),x(k−1),α] f[x(k),x(k−1)] e(k)e(k−1) ≈−f ′′(ξk) 2 f ′(θk)e(k)e(k−1), which concludes the proof. Then Theorem 15.8 is an easy consequence of Proposition 15.9 and the following result. 123 THEOREM 15.10. Let f be a twice continuously differentiable function in a neighborhood of the root α such that 1 2 f ′′(y) f ′(x) ≤M, ∀x,y in a neighborhood of α. Let x(0),x(1),··· ,x(n) denote the (secant method) iterates, en = α −x(n) be the error in the iterates, and let K := max n M\|e0\|, M\|e1\| 1 q o , where q = 1+ √ 5 2 . (Note that q is a solution of q2 −q−1 = 0.) Then \|en\| ≤1 M Kqn, n = 0,1,··· , (15.5) i.e., the (secant method) converges with order q = 1+ √ 5 2 ≈1.618, since en \|en−1\|q ≈1 M Kqn · M Kqn−1 q = Mq−1 Kqn Kqn−1·q = Mq−1 = constant. Proof. \|en\| = −1 2 f ′′(ξn) f ′(ζn) en−1en−2 ≤Men−1en−2 (using (15.5)) ≤ M · 1 M Kqn−1 · 1 M Kqn−2 = 1 M Kqn−2(q+1) (using q2 = q+1) = 1 M Kqn−2q2 = 1 M Kqn which concludes the argument. Comparison of efﬁciency: (Newton’s method) versus (secant method) Recall the order of convergence for Newton’s and Secant methods’: qN = 2, (q Newton′s method) qS = 1+ √ 5 2 ≈1.618. (Secant method) Also recall the deﬁnition of (efﬁciency index) E := q 1 m , (efﬁciency index) where m = units of work required for each iteration. PROPOSITION 15.11. Assume the work to compute f ′(x) is θ times the amount of work required to compute f(x). Then if θ > 0.44, the secant method is asymptotically more efﬁcient than Newton’s method. Proof. To compare the amount of work done by the Newton and secant method, we have Newton work: (1+θ)m (Newton iteration: compute f and f ′ ≈θ f) secant work: m (secant iteration: compute f) 124 therefore to obtain the same amount work, it takes k Newton iterations while k(1+θ) secant iterations: k{Newton iteration}m ≈(1+θ)k{secant iteration}m. In terms of errors this means: (me(0))qNk \| {z } Newton ≈ (me(0))qsk(1+θ) \| {z } secant , (qN = 2,qs = 1.618) 2k = 1.618k(1+θ), klog2 = k(1+θ)log(1.618) θ = log(2) log(1.618) −1 = 0.4404 which concludes the argument. Exercise 1, page page 96. Using the secant method, ﬁnd the roots of the equations. Use an error tolerance of ε = 10−6. (a) The real root of x3 −x2 −x−1 = 0. (b) The root of x = 1+0.3cosx; f(x) := x−1−0.3cosx. (c) The smallest positive root of cosx = 1 2 +sinx, f(x) := cosx−1 2 −sinx. Solution: We use as initial guesses x0 = 0,x1 = 2: (a) With ε = 10−15 the root approximation is 1.839286755214161. With ε = 10−6 we obtain, using secant: >> [root,error_bd,it_count] = secant(0,2,1.e-6,100,3) Iteration number: 9 Approximation of the root: 1.839286754889028 The error estimate: 3.2513e-10 while we recall the Newton’s method results: >> [root,error_bd,it_count] = newton(0,1.e-6,100,3) Iteration number: 13 Approximation of the root: 1.839286755214161 The error estimate: -7.2910e-09 (b) With ε = 10−15 the root approximation is 1.128425092992225. With ε = 10−6 we obtain: >> [root,error_bd,it_count] = secant(0,2,1.e-6,100,7) Iteration number: 5 Approximation of the root: 1.128425092992225 The error estimate: 1.3526e-10 while we recall the Newton’s method results: >> [root,error_bd,it_count] = newton(0,1.e-6,100,6) Iteration number: 3 Approximation of the root: 1.128425092992225 The error estimate: -6.1308e-08s (c) With ε = 10−15 the root approximation is 0.424031039490741 in 5 iterations. With ε = 10−6 we obtain: 125 >> [root,error_bd,it_count] = secant(0,2,1.e-6,100,4) Iteration number: 4 Approximation of the root: 0.424031039649515 The error estimate: 9.7260e-07 while we recall the Newton’s method results: >> [root,error_bd,it_count] = newton(0,1.e-6,100,4) Iteration number: 3 Approximation of the root: 0.424031039490745 The error estimate: -1.5940e-07 126 15.4. Subsection 3.4 FIXED POINT ITERATION. Exercise 2, page 106. (a) Calculate the ﬁrst six iterates in the iteration xn+1 = 0.5 1+x2 n with x0 = 1. Chose another initial guesses x0 and repeat this calculation. (b) Find an interval [a,b] satisfying the hypotheses of Theorem ??. Hint: For g(x) = 0.5 1+x, let a = min x∈(−∞,∞)g(x), b = max x∈(−∞,∞)g(x). (c) Prepare a table in the same manner as Table 3.5 in the preceding discussion. The true solution is α ≈0.423853799069783. Solution: (a) >> root = fixedpoint(inline(’0.5./(1+x.^2)’),0,5) The iterations are x=0.0000000000000000 The iterations are x=0.5000000000000000 The iterations are x=0.4000000000000000 The iterations are x=0.4310344827586207 The iterations are x=0.4216595638004512 The errors are e=0.423854 The errors are e=-0.076146 The errors are e=0.023854 The errors are e=-0.007181 The errors are e=0.002194 The ratios are r=0.000000 The ratios are r=-0.179652 The ratios are r=-0.313263 The ratios are r=-0.301029 The ratios are r=-0.305575 The "Picard" approximation of the root, after 5 iterations is 0.4216595638004512, with error 0.00219424 root = 0.421659563800451 127 iteration n x0 = 0 x0 = 1.5 0 0.0000000000000000 1.5000000000000000 1 0.5000000000000000 0.1538461538461539 2 0.4000000000000000 0.4884393063583816 3 0.4310344827586207 0.4036903902830513 4 0.4216595638004512 0.4299352084038428 5 0.4245214498823317 0.4219963812920247 -50 -40 -30 -20 -10 0 10 20 30 40 50 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.5/(1+x2) (b) We have a = min x∈(−∞,∞) 0.5 1+x2 = 0, b = max x∈(−∞,∞) 0.5 1+x2 = 0.5. (c) We have g′(x) = − x (1+x2)2 , g′(α) = − α (1+α2)2 ≈−0.304584803720867. The computations are n xn error α −xn ratios rn = α−xn α−xn−1 0 1.0000000000000000 -5.761462e-01 1 0.2500000000000000 1.738538e-01 -3.017529e-01 2 0.4705882352941176 -4.673444e-02 -2.688146e-01 3 0.4093484419263456 1.450536e-02 -3.103783e-01 4 0.4282412618114397 -4.387463e-03 -3.024719e-01 5 0.4225147716747344 1.339027e-03 -3.051940e-01 6 0.4242613938187147 -4.075947e-04 -3.043961e-01 7 0.4237296285978768 1.241705e-04 -3.046420e-01 8 0.4238916173427634 -3.781827e-05 -3.045674e-01 9 0.4238422799977083 1.151907e-05 -3.045901e-01 Exercise 3, page 106. How many solutions are there to the equation x = e−x? Will the iteration xn+1 = e−xn converge for suitable choices of x0? 128 Calculate the ﬁrst six iterates when x0 = 0. Solution: There is only one solution (y = x is increasing, while y = e−x is decreasing). -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.5 0 0.5 1 1.5 2 2.5 3 y=x y=e-x Since a = min x∈(−∞,∞)e−x = 0, b = max x∈(0,∞)e−x = e0 = 1, g′(x) = −e−x ∈(−1,0], ∀x ∈(0,1] the iterations will converge if x0 ∈(0,1]. We used x0 = 0.5: >> fixedpoint(inline(’exp(-x)’),0.5,7) The iterations are x=0.5000000000000000 The iterations are x=0.6065306597126334 The iterations are x=0.5452392118926050 The iterations are x=0.5797030948780683 The iterations are x=0.5600646279389019 The iterations are x=0.5711721489772151 The iterations are x=0.5648629469803235 while in 60 iterates x60 = 0.5671432904097841. Exercise 6, page 107. Convert the equation x2 −5 = 0 to the ﬁxed-point problem x = x+c(x2 −5) ≡g(x) with c a nonzero constant. Determine the possible values of c to ensure convergence of xn+1 = xn +c(x2 n −5) to α = √ 5. Solution: We have g′(x) = 1+2cx hence for convergence we have to ensure that \|g′(x)\| = \|1+2cx\| < 1, −1 < 1+2cx < 1, −1 < cx < 0, ∀x ∈[a,b]. Since α = √ 5 > 0, we need c < 0. With c < 0 we have that max x∈R x+c(x2 −5) = −1 2c +c 1 4c2 −5 = −1+20c2 4c > 0. (g′(x∗) = 0,x∗= −1 2c) 129 Then, since g′(x) = 1+2cx is decreasing (for c < 0), we chose a = −1 4c, b = max −1 2c,−1+20c2 4c . Exercise 7, page 107. What are the solutions α, if any, of the equation x = √1+x? Does the iteration xn+1 = √1+xn converge to any of these solutions (assuming x0 is chosen sufﬁciently close to α)? Solution: There is only one solution. Squaring the equation: -2 0 2 4 6 8 10 -2 0 2 4 6 8 10 y=x y= 1+x x2 = 1+x, x1,2 = 1± √ 5 2 , with one root being negative. Hence α = 1+ √ 5 2 . With g(x) = √1+x,g′(x) = 1 2√1+x ∈(0,1) for x ∈[1,2]. Exercise 8, page 107. Which of the following iterations will converge to the indicated α, provided x0 is chosen sufﬁciently close to α? If it does converge, determine the convergence order. (a) xn+1 = 15x2 n−24xn+13 4xn , α = 1. (b) xn+1 = 3 4 + 1 x3 n , α = √ 2. Solution: Exercise 9, page 107. Consider the rootﬁnding problem f(x) = 0 with root α, with f ′(x) ̸= 0. Convert it to the ﬁxed-point problem x = x+cf(x) ≡g(x) with c a nonzero constant. How should c be chosen to ensure rapid convergence of xn+1 = xn +cf(xn) to α (provided that x0 is chosen sufﬁciently close to α)? Apply your way of choosing c to the rootﬁnding problem x3 −5 = 0. Solution: 130 Exercise 11, page 108. The iteration xn+1 = 2−(1+c)xn +cx3 n will converge to α = 1 for some values of c (provided that x0 is chosen sufﬁciently close to α). Find the values of c for which convergence occurs. For what values of c, ﬁany, will the convergence be quadratic? Solution: Exercise 13, page 108. Use Aitken’s error estimation formula (3.53) α −xn ≈ λn 1−λn (xn −xn−1) (Aitken’s error estimate) to estimate the error α −x2 in the following iterations: (a) xn+1 = e−xn, x0 = 0.57 (b) xn+1 = 0.5 1+x4 n , x0 = 0.48 (c) xn+1 = 1+0.5sin(xn), x0 = 1.5 Solution: Exercise 15, page 108. Show that (3.52) α ≈xn − λn 1−λn (xn −xn−1) (Aitken’s extrapolation) can be rewritten as α ≈xn − (xn −xn−1)2 (xn −xn−1)−(xn−1 −xn−2). Solution: Exercise 18, page 109. What is the order of convergence of the iteration xn+1 = xn(x2 n +3a) 3x2 n +a as it converges to the ﬁxed point α = √a? Solution: 131 15.5. Subsection 3.5 ILL-BEHAVING ROOTFINDING PROBLEMS. EXAMPLE 7. ■3.5.1, page 109. (a) f(x) = (x −1)2(x +2) = x3 −3x +2 has two roots. The root α = 1 has multiplicity 2, and α = −2 is a simple root. For α = 1: >> newton(1.1,1.e-14,50,26) root(26) = 0.999999994126396 >> adptnewt(1.1,1.e-14,50,’x^3 - 3x + 2’,’3x^2 - 3’) root(10) = 1.000000003434133 m = 1 1 1 1 1 1 2 2 2 2 while for α = −2 >> newton(-2.1,1.e-14,50,26) root(5) = -2 >> adptnewt(-2.1,1.e-14,50,’x^3 - 3x + 2’,’3x^2 - 3’) root(5) = -2 m = 1 1 1 1 1 1 (b) f(x) = x3 −3x2 +3x−1 has α = 1 as a root of multiplicity 3. newton(1.1,1.e-14,50,27) root(27) = 1.000003927633628 >> adptnewt(1.2,1.e-14,50,’x^3 - 3x^2+3x -1’,’3x^2 - 6x + 3’) root(5) = 1.000000000000249 m = 1 1 1 2.9 2.9 2.9 (c) f(x) = 1−cos(x) has α = 0 as a root of multiplicity 2. newton(0.5,1.e-14,15,28) root(15) = 0.000014833345929 >> adptnewt(0.5,1.e-14,15,’1-cos(x)’,’sin(x)’) root(8) = 0.000000004235853 m = 1 1 1 1 1 1.99 1.99 1.99 1.99 EXAMPLE 8. ■ ■3.5.2, page 110. To illustrate the effect of a multiple root on a rootﬁnding method, use Newton’s method to calculate the root of α = 1.1 of f(x) = (x−1.1)3(x−2.1) (3.61) = 2.7951+x(−8.954+x(10.56+x(−5.4+x))). [root,error_bd,it_count] = newton(1,1.e-8,100,29) root(39)= 1.099999985781220 >> [xvect,xdif,fx,nit]=adptnewt(1,1.e-8,20, ... 132 ’(x-1.1)^3(x-2.1)’,... ’3(x-1.1)^2(x-2.1) + (x-1.1)^3’) root(12)= 1.099999999998076 m= 1 1 1 1 1 1 1 1 3.015 3.015 3.015 3.015 3.015 EXAMPLE 9. ■ ■3.5.4, page 112. Deﬁne f(x) = (x−1)(x−2)(x−3)(x−4)(x−5)(x−6)(x−7) (3.66) = x7 −28∗x6 +322∗x5 −1960∗x4 +6769∗x3 −13132∗x2 +13068∗x−5040. Change the coefﬁcient of x6 from - 28 to - 28.002. The change is relatively small Rel(28.002) = 0.002 28 ≈7.14×10−5. See Exercise 3.5.5. newton(1.1,1.e-8,10,20) root(5) = 0.999999999999999 >> newton(2.1,1.e-8,10,20) root(4) = 2.000000000000040 >> newton(3.1,1.e-8,10,20) root(4) = 2.999999999999869 >> newton(4.1,1.e-8,10,20) root(3) = 4.000000000001791 >> newton(5.1,1.e-8,10,20) root(4) = 4.999999999998231 >> newton(6.1,1.e-8,10,20) root(5) = 5.999999999997954 >> newton(7.1,1.e-8,10,20) root(5) = 6.999999999999877 >> newton(1.1,1.e-8,10,22) root(5) = 0.99999 7222287418 >> newton(2.1,1.e-8,10,22) root(5) = 2.00 1071574417416 >> newton(3.1,1.e-8,10,22) root(4) = 2.9 71740865487905 >> newton(4.1,1.e-8,50,22) root(14) = 2.001071574417471 ???? newton(5.1,1.e-8,50,22) root(20) = 0.999997222287422 ???? newton(6.1,1.e-8,50,22) \|\|\|\|\| >> newton(6.01,1.e-8,100,22) >> newton(6.001,1.e-8,100,22) root(36) = 0.999997222287423 ???? root(64) = 0.999997222287420 root(45) = 2.001071574417492 >> newton(7.001,1.e-8,100,22) root(29) = 2.971740865488212 ??? EXAMPLE 10. ■ ■3.5.6, page 114. Approximate the evaluation of a function f(x) and introduced the perturbed 133 0 1 2 3 4 5 6 7 8 -300 -200 -100 0 100 200 300 400 500 function fε(x) = f(x)+εg(x), (3.68) assuming both f,g ∈C1(R). Denote α(0) the root of f α(0) = 0, and α(ε) the root of f α(ε) +εg α(ε) = 0. (3.71) Assume α(0) is a simple root, i.e., f ′α(0) ̸= 0, and that α(ε) ≈α(0)+εα′(0). (3.70) Taking the derivative, with respect to varepsilon, of (3.71), we have 0 = f ′α(ε) α′(ε)+g α(ε) +εg′α(ε) α′(ε) = 0, (3.72) which in the limit ε →0 gives f ′α(0) α′(0)+g α(0) = 0, hence α′(0) = −g α(0) f ′α(0) . Finally from (3.70) we have that α(ε) ≈α(0)−ε g α(0) f ′α(0) . (3.74) 134 Exercise 1, page 115. Use Newton’s method to calculate the roots of f(x) = x5 +0.9x4 −1.62x3 −1.458x2 +0.6561x+0.59049 Print out the iterates and the function values. Produce the ratios of (3.62) α −xn α −xn−1 (3.62) by using the approximation (3.65) λ ≈λn ≡ xn −xn−1 xn−1 −xn−2 . (3.65) Repeat the problem for choices of x0. Make observations that seem important relative to the rootﬁnding problem. Note: The above will ﬁrst approach λ = m−1 m , as in (3.63), but they will then depart from it because of the noise in the evaluation of f(x) as xn approaches the root. Solution: >> x=linspace(-1.5,1.5,200); >> plot(x,x.^5 + 0.9 x.^4 - 1.62 x.^3 - 1.458 x.^2 + 0.6561 x + 0.59049,’linewidth’,2) >> grid on -1.5 -1 -0.5 0 0.5 1 1.5 -2 -1 0 1 2 3 4 5 There seems to be a solution near x0 = −1. Run: >> [xvect,xdif,fx,nit]=adptnewt(-1.0,1.e-8,15,’x^5 + 0.9 x^4 - 1.62 x^3 ... - 1.458 x^2 + 0.6561 x + 0.59049’,... ’5x^4 + 0.94 x^3 - 1.62 3x^2 - 1.458 2x + 0.6561’) >>xvect = -1.000000000000000 -0.967796610169491 -0.945731676794779 -0.930735491427873 -0.920603726405928 -0.913787245565033 -0.909214668903794 -0.906153506650819 -0.904106986054098 -0.899983001501482 135 -0.900000175441066 -0.897759560109618 -0.900014560779769 -0.900000095426333 -0.900000095426333 xdif = 0.032203389830509 0.022064933374713 0.014996185366906 0.010131765021945 0.006816480840895 0.004572576661238 0.003061162252975 0.002046520596722 0.004123984552615 0.000017173939584 0.002240615331449 0.002255000670151 0.000014465353436 0 n xn xn −xn−1 ratio mn 0 -1.000000000000000 1 -0.967796610169491 0.032203389830509 2 -0.945731676794779 0.022064933374713 0.022064933374713 0.032203389830509 = 0.6851 3.17 3 -0.930735491427873 0.014996185366906 0.014996185366906 0.022064933374713 = 0.6796 3.12 4 -0.920603726405928 0.010131765021945 0.010131765021945 0.014996185366906 = 0.6756 3.08 5 -0.913787245565033 0.006816480840895 0.006816480840895 0.010131765021945 = 0.6728 3.05 Exercise 2, page 115. Repeat Problem 1 for f(x) = x4 −3.2x3 +0.96x2 +4.608x−3.456 Solution: >> x=linspace(-2,2.5,200); >> plot(x,x.^4-3.2x.^3 + 0.96x.^2 + 4.608x-3.456,’linewidth’,2) >> plot(x,x.^4-3.2x.^3 + 0.96x.^2 + 4.608x-3.456,’linewidth’,2) >> grid on 136 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 -10 -5 0 5 10 15 20 25 30 35 >> [xvect,xdif,fx,nit]=adptnewt(-1.0,1.e-8,15,... ’x^4-3.2x^3 + 0.96x^2 + 4.608x-3.456’,... ’4x^3-3.23x^2 + 0.962x + 4.608’) xvect = -1.000000000000000 -1.266129032258065 -1.204549309017986 -1.200023550054160 -1.200000000635462 -1.200000000000000 xdif = 0.266129032258065 0.061579723240079 0.004525758963825 0.000023549418698 0.000000000635462 n xn xn −xn−1 ratio mn 0 -1.000000000000000 1 -1.266129032258065 0.266129032258065 2 -1.204549309017986 0.061579723240079 0.061579723240079 0.266129032258065 = 0.2314 1.30 3 -1.200023550054160 0.004525758963825 0.004525758963825 0.061579723240079 = 0.0735 1.08 4 -1.200000000635462 0.000023549418698 = 5 -1.200000000000000 0.000000000635462 = >> [xvect,xdif,fx,nit]=adptnewt(1.0,1.e-8,55,... ’x^4-3.2x^3 + 0.96x^2 + 4.608x-3.456’,... ’4x^3-3.23x^2 + 0.962x + 4.608’) root(26) = 1.200000002120660 m(1:10) = 1; m(11:26) = 2.002 137 >> [xvect,xdif,fx,nit]=adptnewt(2.5,1.e-8,55,... ’x^4-3.2x^3 + 0.96x^2 + 4.608x-3.456’,... ’4x^3-3.23x^2 + 0.962x + 4.608’) root(7) = 2.000000000000001 m(1:7) = 1 Exercise 3, page 115. Use the ﬁxed point iteration theory of Section 3.4 to derive the results (3.63) λ = m−1 m (3.63) to (3.65) λ ≈λn ≡ xn −xn−1 xn−1 −xn−2 . (3.65) To aid with this, ﬁrst write xn+1 = g(xn) ≡xn −f(xn) f ′(xn) and use (3.59) f(x) = (x−α)mh(x) (3.59) to write g(x) = x− (x−α)h(x) h(x)+(x−α)h′(x). Apply Corollary 3.4.3. Solution: Exercise 4, page 115. Do the calculation of α(ε) for the roots α(0) = 3 and 7, continuing Example 3.5.6 Solution: Exercise 5, page 115. Do the perturbation calculation for another change in (3.66) f(x) = (x−1)(x−2)(x−3)(x−4)(x−5)(x−6)(x−7) (3.66) = x7 −28x6 +322x5 −1960x4 +6769x3 −13132x2 +13068x−5040, namely f(x) = (x−α)mh(x) (3.59) Change the coefﬁcient of x4 from −1960 to −1960.14. What is the relative perturbation in the coefﬁcient? Calculate α(ε) for α(0) = 3 and α(0) = 5. Solution: >> x=linspace(0.99,7.5,300); hold all >> plot(x,x.^7 - 28x.^6 + 322x.^5 - 1960x.^4 + 6769x.^3 - 13132x.^2 + 13068x - 5040,’linewi >> plot(x,x.^7 - 28x.^6 + 322x.^5 - 1960.14x.^4 + 6769x.^3 - 13132x.^2 + 13068x - 5040,’r-’ >> grid on 138 0 1 2 3 4 5 6 7 8 -400 -200 0 200 400 600 800 1000 1200 >> newton(3.1,1.e-8,10,20) iteration 3 root = 2.999999999999869 error = 1.3641e-09 >> newton(3.1,1.e-8,10,21) iteration 22 root = 7.286538669471431 error = -2.6290e-13 x = 7.286538669471431; x^7-28x^6 + 322x^5 - 1960.14x^4 + 6769x^3 - 13132x^2 + 13068x - 5040 >> ans = -1.047737896442413e-09 !!! Rel(−1960.14) = −1960+1960.14 −1960 = −7.1429e−05, Rel(7.286538669471431) = 2.999999999999869−7.286538669471431 2.999999999999869 = −1.4288. Using the adaptive Newton >> [xvect,xdif,fx,nit]=adptnewt(3.1,1.e-12,20,... ’x^7-28x^6 + 322x^5 - 1960x^4 + 6769x^3 - 13132x^2 + 13068x - 5040’,... ’7x^6-286x^5 + 3225x^4 - 19604x^3 + 67693x^2 - 131322x + 13068’) xvect = 3.100000000000000 2.990669955991597 2.999951644121190 2.999999998635778 2.999999999999869 3.000000000000020 >> [xvect,xdif,fx,nit]=adptnewt(3.1,1.e-12,20,... ’x^7-28x^6 + 322x^5 - 1960.14x^4 + 6769x^3 - 13132x^2 + 13068x - 5040’,... ’7x^6-286x^5 + 3225x^4 - 1960.144x^3 + 67693x^2 - 131322x + 13068’) 139 xvect = 3.100000000000000 3.449945676463090 3.088404113427837 3.433682595476009 3.045650354416620 3.384324331877695 2.832796151707533 3.278834762298465 5.439015617931766 3.732066752473070 3.396321716375667 2.902383739324706 3.292205606645318 9.402956726076429 8.757106439567847 8.232009854042712 7.821615857800884 7.527259022415715 7.355095286893262 7.293828793565238 7.286631330974876 “error(23)”≡xdif(23) = 2.6300e-13 Note that we obtain a similar result for α = 4 using the adaptive Newton >> [xvect,xdif,fx,nit]=adptnewt(4.1,1.e-12,20,... ’x^7-28x^6 + 322x^5 - 1960x^4 + 6769x^3 - 13132x^2 + 13068x - 5040’,... ’7x^6-286x^5 + 3225x^4 - 19604x^3 + 67693x^2 - 131322x + 13068’) xvect = 4.100000000000000 3.997178659995060 4.000000061137822 4.000000000000780 4.000000000001791 4.000000000000982 >> [xvect,xdif,fx,nit]=adptnewt(4.1,1.e-12,20,... ’x^7-28x^6 + 322x^5 - 1960.14x^4 + 6769x^3 - 13132x^2 + 13068x - 5040’,... ’7x^6-286x^5 + 3225x^4 - 1960.144x^3 + 67693x^2 - 131322x + 13068’) xvect = 4.100000000000000 3.510499507654773 3.199352115513752 3.708017383043476 3.381581719997233 140 2.814258589966431 3.279273032244989 5.485672302222047 3.980927405944043 3.497245793868069 3.179504263511301 3.627038192532888 3.323323364169225 1.483326530003837 2.091075593497805 1.963276018923747 1.981986424571477 1.982378517709796 1.982378692843498 1.982378692843554 Example 3.5.6, page 114. Consider the previous Exercise, with root α = 4 and perturb the coefﬁcient of x6 to be 28−0.002. Using the adaptive Newton >> [xvect,xdif,fx,nit]=adptnewt(4.1,1.e-12,20,... ’x^7-28x^6 + 322x^5 - 1960x^4 + 6769x^3 - 13132x^2 + 13068x - 5040’,... ’7x^6-286x^5 + 3225x^4 - 19604x^3 + 67693x^2 - 131322x + 13068’) xvect = 4.100000000000000 3.997178659995060 4.000000061137822 4.000000000000780 4.000000000001791 4.000000000000982 >>[xvect,xdif,fx,nit]=adptnewt(4.1,1.e-12,20,... ’x^7-(28-0.002)x^6 + 322x^5 - 1960x^4 + 6769x^3 - 13132x^2 + 13068x - 5040’,... ’7x^6-(28-0.002)6x^5 + 3225x^4 - 19604x^3 + 67693x^2 - 131322x + 13068’) >> xvect = 4.100000000000000 4.388310401634414 3.649056979199154 5.018616764750590 4.641752171791878 4.427144675872155 4.057687450309482 4.365131813622936 2.214475977863997 1.860135859029350 1.984532566918768 2.000735230129887 141 2.001071427920883 2.001071574417471 2.001071574417471 Exercise 6, page 115. Consider the problem of solving x 1+x −0.99 = 0, calling its root α. Then let α(ε) be the solution of x 1+x −0.99+ε = 0. (a) Using (3.74) α(ε) ≈α(0)−ε g(α(0)) f ′(α(0)) (3.74) estimate α(ε)−α. (b) Calculate α(ε) directly, compute α(ε)−α, and compare with (a). Comment on your results. Solution: Exercise 7, page 116. Consider the polynomial f(x) = x5 −300x2 −126x+5005, which has a root α = 5. Also consider the perturbed functions Fε = f(x)+εx5 = (1+ε)x5 −300x2 −126x+5005, with ε being a small number. Letting α(ε) denote the perturbed root of Fε(x) = 0 corresponding to α(0) = 5, estimate α(ε)−5. Is ﬁnding α = 5 for f(x) = 0 an unstable rootﬁnding problem? Solution: >> x=linspace(4.5,5.5,200); >> plot(x,x.^5-300x.^2-126x+5005,... x,(1-0.1)x.^5-300x.^2-126x+5005,... x,(1-0.01)x.^5-300x.^2-126x+5005,... x,(1-0.001)x.^5-300x.^2-126x+5005,’linewidth’,2) >> grid on >> legend(’f’,’f-0.1x^5’,’f-0.01x^5’,’f-0.001x^5’) 142 4.5 4.6 4.7 4.8 4.9 5 5.1 5.2 5.3 5.4 5.5 -400 -300 -200 -100 0 100 200 300 f f-0.1x 5 f-0.01x 5 f-0.001x 5 Choose ε = −0.01 >> newton(5.1,1.0E-8,20,23) root(12) = 5.001052340122435 Choose ε = −0.1 >> newton(5.1,1.0E-8,20,24) root(6) = 4.524542317488499 Rel(5) = 5−5.001052340122435 5 ≡−2.1047e−04 = −0.00021047, versus 0.01 Rel(5) = 5−5.059128224241503 5 ≡−0.0118, versus 0.001 Rel(5) = 5−4.524542317488499 5 ≡0.0951, versus 0.1 Exercise 8, page 116. Newton’s method is used to ﬁnd a root of f(x) = 0. The ﬁrst few iterates are shown in the following table, giving a very slow sped of convergence. n xn xn −xn−1 ratio 0 0.75 1 0.752710 0.00271 2 0.754795 0.00208 0.00208 0.00271 = 0.7675 3 0.756368 0.00157 0.00157 0.00208 = 0.7548 4 0.757572 0.00118 0.00118 0.00157 = 0.7516 5 0.758441 0.000889 0.000889 0.00118 = 0.7534 What can be said about the root α to explain this convergence? 143 Knowing f(x), how would you ﬁnd an accurate value for α Solution: Hint: evaluate λn+1 ≈m−1 m and evaluate the multiplicity m. 144
13357	https://www.khanacademy.org/test-prep/mcat/physical-processes/thin-lenses/a/thin-lens-sign-conventions	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
13358	http://www.historymuse.net/readings/SpencerMANvSTATE1884.htm	Herbert Spencer, Man Versus the State (1884) Herbert Spencer THE MAN VERSUS THE STATE Committed to a traditional laissez-faire policy, however, some liberals attacked state intervention as a threat to personal freedom and a betrayal of central liberal principles. In The Man Versns the State(1884), British philosopher Herbert Spencer (1820—1903) warned that increased government regulation would lead to socialism and slavery. The extension of this policy . . . (of government legislation) fosters everywhere the tacit assumption that Government should step in whenever anything is not going right. “Surely you would not have this misery continue!” exclaims some one, if you hint . . . (an objection) to much that is now being said and done. Observe what is implied by this exclamation. It takes for granted. . . . that every evil can be removed: the truth being that with the existing defects of human nature, many evils can only be thrust out of one place or form into another place or form.—often being increased by the change. The exclamation also implies the unhesitating belief, here especially concerning us, that evils of all kinds should be dealt with by the State. . . . Obviously, the more numerous governmental interventions become, the more confirmed does this habit of thought grow, and the more loud and perpetual the demands for intervention. Every extension of the regulative policy involves an addition to the regulative agents— a further growth of officialism and an increasing power of the organization formed of officials. . . . . . Moreover, every additional State-interference strengthens the tacit assumption that it is the duty of the State to deal with all evils and secure all benefits. Increasing power of a growing administrative organization is accompanied by decreasing power of the rest of the society to resist its further growth and control. . . . “But why is this change described as ‘the coming slavery’?” is a question which many will still ask. The reply is simple. All socialism involves slavery. . . . Evidently then, the changes made, the changes in progress, and the changes urged, will carry us not only towards State-ownership of land and dwellings and means of communication, all to be administered and worked by State-agents, but towards State-usurpation of all industries: the private forms of which, disadvantaged more and more in competition with the State, which can arrange everything for its own convenience, will more and more die away, just as many voluntary schools have, in presence of Board-schools. And so will be brought about the desired ideal of the socialists. . . . … It is a matter of common remark, often made when a marriage is impending, that those possessed by strong hopes habitually dwell on the promised pleasures and think nothing of the accompanying pains. A further exemplification of this truth is supplied by these political enthusiasts and fanatical revolutionists. Impressed with the miseries existing under our present social arrangements, and not regarding these miseries as caused by the ill- working of a human nature but partially adapted to the social state, they imagine them to be forthwith curable by this or that rearrangement. Yet, even did their plans succeed it could only be by substituting one kind of evil for another. A little deliberate thought would show that under their proposed arrangements, their liberties must be surrendered in proportion as their material welfares were cared for. For no form of co-operation, small or great, can be carried on without regulation, and an implied submission to the regulating agencies…. … So that each (individual] would stand toward the governing agency in the relation of slave to master. “But the governing agency would be a master which he and others made and kept constantly in check; and one which therefore would not control him or others more than was needful for the benefit of each and all.” To which reply the first rejoinder is that, even if so, each member of the community as an individual would be a slave to the community as a whole. Such a relation has habitually existed in militant communities, even under quasi-popular forms of government. In ancient Greece the accepted principle was that the citiZen belonged neither to himself nor to his family, but belonged to his city—the city being with the Greek equivalent to the community. And this doctrine, proper to a state of constant warfare, is a doctrine which socialism unawares re-introduces into a state intended to be purely industrial. The services of each will belong to the aggregate of all; and for these services, such returns will be given as the authorities think proper. So that even if the administration is of the beneficent kind intended to be secured, slavery, however mild, must be the outcome of the arrangement…. The function of Liberalism in the past was that of putting a limit to the powers of kings. The function of true Liberalism in the future will be that of putting a limit to the powers of Parliaments.
13359	https://www.ecdc.europa.eu/en/congenital-rubella-syndrome	Congenital rubella syndrome (CRS) Skip to main content An official website of the European Union An official EU website How do you know? All official European Union website addresses are in the europa.eu domain. See all EU institutions and bodies This site uses cookies. Visit our cookies policy page or click the link in any footer for more information and to change your preferences. Accept all cookiesAccept only essential cookies Global Navigation Other sites ECDC European Antibiotic Awareness Day ESCAIDE - Scientific conference Eurosurveillance journal EVIP - Vaccination portal European Centre for Disease Prevention and Control An agency of the European Union Search Search Search Translate this page Close Disclaimer This is a machine translation provided by the European Commission’s eTranslation service to help you understand this page. Please read the conditions of use. Select language below Accept and continue Main Navigation Home Infectious disease topics Infectious disease topics ABCDEFGHIJKLMNOPQRSTUVWXYZ Related public health topics Spotlight Antimicrobial resistance (AMR) Avian influenza COVID-19 Mosquito-borne diseases One Health Seasonal influenza Video: From abroad to your backyard - dengue in Europe Publications and data Publications and data Scientific and technical publications Dashboards and databases Spotlight Weekly threats reports (CDTR) Annual Epidemiological Reports (AERs) Epidemiological updates The European Respiratory Virus Surveillance Summary (ERVISS) Training and tools Training and tools Training programmes Surveillance and outbreak tools Preparedness, prevention and control tools Communication materials Spotlight EpiPulse - the European surveillance portal for infectious diseases The European Respiratory Diseases Forecasting Hub (RespiCast) ECDC Crowd Learning Portal About ECDC About ECDC Who we are What we do Partners and networks Work with ECDC Procurement and grants Media centre Spotlight Strategic documents Annual reports of the Director Governance ECDC: On Air Home Infectious disease topics Congenital rubella syndrome (CRS) Infectious disease topics Congenital rubella syndrome (CRS) Case definition Surveillance and updates Congenital rubella syndrome (CRS) Congenital rubella is the infection of a foetus with rubella virus following the infection of the mother during pregnancy. ‘Congenital’ indicates that the foetus also becomes infected during pregnancy. Congenital rubella is a rare disease in the EU but cases of Congenital Rubella Syndrome are still reported every year according to reports fromWHO. Symptoms Rubella is a mild febrile rash illness. Up to 50% of infections may not present symptoms. However, as many as 85 out of 100 babies born to mothers who had rubella shortly before or during the first three months of pregnancy may develop health problems, so-called congenital rubella syndrome. Risk for people Rubella infection of the foetus may have severe consequences on the development of baby’s organs, which may result in birth defects or problems at a later stage in life.The type of complications may vary in nature and severity, depending on the stage of pregnancy when the infection occurred. Complications affect 85 out of 100 babies whose mothers had rubella just before or at the beginning of the pregnancy. Complications include: deafness cataracts heart defects brain disorders mental retardation bone alterations liver and spleen damage. Furthermore, an infant infected with rubella during pregnancy can continue to shed the virus for about a year, sometimes longer.Symptoms can appear in the infant immediately at birth and up to the age of four years. Sometimes, even after a while, more issues might come up, like diabetes, problems with the thyroid, or troubles with their vision or how their brain works. In approximately 20% of the cases, rubella infections result in the death of the foetus. How does it spread? Rubella is transmitted from person to person via droplets (the virus is present in throat secretions). It can affect anyone who is not protected. If infection in an unprotected person occurs during pregnancy, placental infection occurs and may lead to transplacental foetal infection. Vaccination and treatment There is a vaccine against rubella which is included in the vaccine schedules of all EU member states (see the ECDC Vaccine Sheduler). Women who are planning to get pregnant should be vaccinated if not vaccinated or uncertain about their vaccination history as this protects the mother as well as the foetus. Women should avoid getting pregnant during four weeks following vaccination. Protective measures The best way to prevent rubella is through vaccination. Featured European Vaccination Information Portal The main purpose of this website is to provide accurate, objective, up-to-date evidence on vaccines and vaccination in general. ECDC Vaccine Scheduler The Vaccine Scheduler is an interactive tool that shows vaccination schedules for individual EU/EEA countries and specific age groups. Immunisation and vaccines Vaccines represent one of the most effective and cost-saving public health intervention. Latest outputs Infographic ### Editable social media cards: European Immunization Week 2023 21 Apr 2023 Surveillance and monitoring ### Retrospective surveillance and enhanced case-finding of congenital rubella syndrome cases 16 Nov 2018 Scientific journal article ### Surveillance of congenital rubella and rubella infections in pregnancy in EU/EEA countries, 2012: Current status and future perspective to monitor elimination 1 Jan 2015 Public health guidance ### Measles and rubella elimination: communicating the importance of vaccination 5 Aug 2014 Scientific journal article ### Ongoing outbreak of rubella among young male adults in Poland: increased risk of congenital rubella infections 1 May 2013 Load More Disease network ### EUVAC.Net Doormat About us What we do Who we are Governance Work with us Vacancies Fellowships Procurement and grants Contact ECDC Contact us Media centre Social media NewsletterRSS Follow us: facebook linkedin bluesky threads instagram vimeo Footer Navigation Legal notice Copyright Access to surveillance data Personal data protection Language policy Cookies Accessibility Sitemap © European Centre for Disease Prevention and Control (ECDC) 2025 Back to top Home Infectious disease topics Congenital rubella syndrome (CRS) Case definition Surveillance and updates Infectious disease topics ABCDEFGHIJKLMNOPQRSTUVWXYZ Related public health topics Spotlight Antimicrobial resistance (AMR) Avian influenza COVID-19 Mosquito-borne diseases One Health Seasonal influenza Stefan, a tuberculosis survivor journey Publications and data Scientific and technical publications Annual Epidemiological Reports Epidemiological updates Weekly threats reports (CDTR) Dashboards and databases Spotlight Weekly threats reports (CDTR) Annual Epidemiological Reports (AERs) Epidemiological updates The European Respiratory Virus Surveillance Summary (ERVISS) Training and tools Training programmes Scientific exchange programmes Food- and Waterborne Diseases Expert Exchange Programme Observership programme Continuous professional development Fellowships Fellowship programme: EPIET/EUPHEM About the fellowship At a glance Fellowship experience Work of graduated fellows cohort 2022 The work of graduated fellows cohort 2021 Work of graduated fellows cohort 2020 Team coordination Apply for fellowship Apply for EU-track Apply for MS-track ECDC Fellowship Programme documents Institutes Postcards from the field MediPIET MediPIET partner countries and training sites Multimedia Graduated Fellows Core competencies Contact us Surveillance and outbreak tools Information sources for epidemic intelligence and outbreak response Preparedness, prevention and control tools Communication materials Spotlight EpiPulse - the European surveillance portal for infectious diseases The European Respiratory Diseases Forecasting Hub (RespiCast) ECDC Crowd Learning Portal About ECDC Who we are Our guiding principles Annual Declarations of Interest and Declarations of Commitment Key documents Decisions and regulations Financial documents Strategic documents Planned scientific outputs Annual reports of the Director Public access to documents EU/EEA routine surveillance open data policy Governance Management Board Composition Minutes Advisory Forum Members and Alternates Observers Minutes Competent Bodies Audit Committee Director of ECDC Former Directors Celebrating 20 years of ECDC What we do Partners and networks Disease and laboratory networks European Antimicrobial Resistance Surveillance Network (EARS-Net) European COVID-19 surveillance network (ECOVID-Net) European COVID-19 reference laboratory network (ECOVID-LabNet) Emerging Viral Diseases-Expert Laboratory Network (EVD-LabNet) European Emerging and Vector-borne Diseases Network (EVD-Net) European Creutzfeldt-Jakob Disease Surveillance Network (EuroCJD) European Diphtheria Surveillance Network (EDSN) European Food- and Waterborne Diseases and Zoonoses Network (FWD-Net) European Gonococcal Antimicrobial Surveillance Programme (Euro-GASP) European Influenza Surveillance Network (EISN) European Invasive Bacterial Disease Surveillance Network (EU-IBD) European Legionnaires’ Disease Surveillance Network (ELDSNet) European Network for Hepatitis B and C Surveillance European Network for HIV/AIDS Surveillance About the network Euro-HIV project & past reports European Network for STI Surveillance About the network Historical network: ESSTI Project European Reference Laboratory Network for Human Influenza (ERLI-Net) About the network Biosafety Influenza laboratory quality control Laboratory surveillance of influenza Tuberculosis Disease Network European Reference Laboratory Network for TB (ERLTB-Net) European Tuberculosis Surveillance Network EUVAC.Net European Surveillance of Antimicrobial Consumption Network (ESAC-Net) Data collection and analysis Healthcare-associated Infections Surveillance Network (HAI-Net) European network for sharing data on the geographic distribution of arthropod vectors, transmitting human and animal disease agents (VectorNet) European Antimicrobial Resistance Genes Surveillance Network (EURGen-Net) National Immunisation Technical Advisory Groups (NITAG) collaboration Network for the Microbial Safety of Substances of Human Origin (SoHO-Net) Support and services to EU/EEA countries EU Health Task Force (EUHTF) European Commission funding opportunities Support for countries neighbouring Ukraine Sustainable Development Goals EU institutions and agencies Vaccine Monitoring Platform International cooperation Africa CDC – ECDC Partnership EU Initiative on Health Security ECDC Accession Support to the Western Balkans and Türkiye Health Resilience in the Eastern Partnership Work with ECDC Staff positions Recruitment process What we offer Staff categories Selection committees for ongoing recruitments Reserve list Vacancy translations Traineeships Short term interim support Seconded National Expert Fellowship Programme (EPIET and EUPHEM) External experts ECDC Crowd Procurement and grants Ex-post publicity Media centre Campaigns ECDC: On Air Infographics and videos ECDC picture gallery Social media Spotlight Strategic documents Annual reports of the Director Governance ECDC: On Air
13360	https://arxiv.org/pdf/2509.13125	PARITIES IN RANDOM LATIN SQUARES MATTHEW KWAN, KALINA PETROVA, AND MEHTAAB SAWHNEY Abstract. In a Latin square, every row can be interpreted as a permutation, and therefore has a parity (even or odd). We prove that in a uniformly random n × n Latin square, the n row parities are very well approximated by a sequence of n independent unbiased coin flips: for example, the total variation error of this approximation tends to zero as n → ∞ . This resolves a conjecture of Cameron. In fact, we prove a generalisation of Cameron’s conjecture for the joint distribution of the row parities, column parities and symbol parities (the latter are defined by the symmetry between rows, columns and symbols of a Latin square). Along the way, we introduce several general techniques for the study of random Latin squares, including a new re-randomisation technique via “stable intercalate switchings”, and a new approximation theorem comparing random Latin squares with a certain independent model. Introduction An n × n Latin square is an n × n array filled with n different “symbols” (usually taken to be the integers 1, . . . , n ), with the property that each symbol appears exactly once in each row and each column. For example, the multiplication table of a group is always a Latin square; in general, Latin squares can be interpreted as multiplication tables of a class of algebraic structures called quasigroups . See for example for an introduction to this vast subject. Each row or column of an n × n Latin square L can be interpreted as a permutation of order n,which can be either even or odd. Let Nrow (L) be the number of odd row permutations, and let Ncol (L) be the number of odd column permutations. If L is the multiplication table of a group, then either Nrow (L) = Ncol (L) = 0 or Nrow (L) = Ncol (L) = n/ 2. However, for general Latin squares, the row and column parities can have much richer behaviour, and much is still unknown. For example, one of the most important conjectures in this direction is the Alon–Tarsi conjecture , which (in probabilistic language) says that if n is even, and L is a uniformly random n × n Latin square, then PNrow (L) is even = PNrow (L) + Ncol (L) is even ̸ = 12 . (The first equality is not part of the original conjecture; it was observed by Huang and Rota , in a paper where they also observed that the Alon–Tarsi conjecture has a number of surprising consequences in seemingly unrelated areas of mathematics; see for a modern survey). It was first proved by Alpöge that PNrow (L) is even = 12 + o(1) . (1.1) as n → ∞ . In other words, if the Alon–Tarsi conjecture is true, then it is true “just barely”. Going far beyond (1.1), it has been suggested by Peter Cameron (in a variety of different sources; see for example [7, 10, 11, 12, 13, 14, 15]) that the row parities of a Latin square might be statistically completely unconstrained, in the sense that one can model the n row parities of a random Latin square by simply making n independent coin flips. Conjecture 1.1. Let L be a uniformly random n × n Latin square. Then the distribution of Nrow (L) is approximately the binomial distribution Bin( n, 1/2) , as n → ∞ . (Note that exchanging rows does not affect the distribution of L, so the sequence of row parities ⃗ξrow (L) ∈ (Z/2Z)n has a permutation-invariant distribution. This means that if we condition on Nrow (L),then ⃗ ξrow (L) ∈ (Z/2Z)n is a uniformly random sequence in (Z/2Z)n, constrained to have exactly Nrow (L) “1”s. That is to say, in order to understand the distribution of ⃗ ξrow (L), it is enough to understand the distribution of Nrow (L).) MK was supported by ERC Starting Grant “RANDSTRUCT” No. 101076777. KP was supported by the Euro-pean Union’s Horizon 2020 research and innovation programme under the Marie Skłodowska-Curie grant agreement No. 101034413 . This research was conducted during the period MS served as a Clay Research Fellow. 1 arXiv:2509.13125v2 [math.PR] 17 Sep 2025 To elaborate on the attribution/history of Conjecture 1.1: the starting point seems to have been the problem session at the British Combinatorial Conference in 1993, where Cameron asked a related question that was extended by Jeannette Janssen. However, at that time they did not seem to be very confident that the statement of Conjecture 1.1 was actually true (in the BCC problem list they phrase the question as “is it true that...”). Cameron later posed the problem more assertively in a 2002 survey on permutations and permutation groups , and it seems he first referred to it as a “conjecture” in a 2003 lecture on random Latin squares . However, as far as we can tell, he never stated the problem in a fully precise form (always using language like “approximately”). There are many different ways to compare distributions, to make rigorous sense of the word “approx-imately”; in this paper we confirm Conjecture 1.1 in many different senses. For example, one strong way to compare distributions is in terms of total variation distance : for two probability distributions μ, ν on the same space, write dTV (μ, ν ) = sup A \|μ(A) − ν(A)\| (so for any event A, the probabilities of A with respect to μ and ν differ by at most dTV (μ, ν )). The following theorem is a consequence of our main (technical) result. Theorem 1.2. Let L be a uniformly random n × n Latin square. Then lim n→∞ dTV Nrow (L), Bin( n, 1/2) = 0 . Equivalently, writing ⃗ ξrow (L) ∈ (Z/2Z)n for the sequence of parities of rows of L, and writing Unif(( Z/2Z)n) for the uniform distribution on (Z/2Z)n, we have lim n→∞ dTV ⃗ ξrow (L), Unif(( Z/2Z)n) = 0 . We can actually prove much more than Theorem 1.2; we will discuss this momentarily, but first it is worthwhile to briefly discuss why this conjecture took so long to be resolved (and, in our opinion, why it is so interesting). Generally speaking, it is quite easy to make plausible predictions about uniformly random Latin squares, by making various kinds of approximate independence assumptions (for example, Cameron’s conjecture can be justified from the point of view that there is “no obvious reason” for the parities of different rows to be correlated). However, it is surprisingly difficult to rigorously prove anything nontrivial about uniformly random Latin squares, or even to study them empirically. The main issues are that Latin squares do not enjoy any neat recursive structure, and they are very “rigid” objects, in the sense that there are only very limited ways to make a “local perturbation” to change a Latin square into another one. To highlight the difficulties here, we remark that (despite some very ambitious conjectures; see [1, Section 4.1]) we still have a rather poor understanding of the total number of n × n Latin squares (the best known upper and lower bounds differ by exponential factors; see [56, Section 17]), and there is no rigorously justified way to efficiently sample a uniformly random Latin square (there are certain ergodic Markov chains on the space of n × n Latin squares [31, 51], but these are not known to be rapidly mixing). By now, there are quite a few known results about random Latin squares; the proofs of many of these results have required fundamental new additions to a very limited toolbox of techniques, and Theorem 1.2 is no exception (we discuss existing and new techniques in Section 2). 1.1. Row, column and symbol permutations. Our proof techniques actually allow us to go beyond Cameron’s conjecture, to approximate not just the distribution of Nrow (L) but also the joint distribution between Nrow (L), Ncol (L) and a third parameter Nsym (L), which counts the number of odd symbol permutations . Perhaps the most natural way to define a symbol permutation is to reinterpret a Latin square in a way that emphasises the natural symmetry between rows, columns and symbols: indeed, observe that an n × n Latin square (with symbols 1, . . . , n ) can be interpreted as an n × n × n array in which every entry is “0” or “1”, such that every two-dimensional “slice” of this array is a permutation matrix 1. So, the row and column permutations correspond to slices in two of the three possible directions, and the symbol permutations correspond to slices in the third direction. The row, column and symbol parities cannot vary completely freely: it was proved by Janssen and Zappa that for any Latin square L of order n, we have Nrow (L) + Ncol (L) + Nsym (L) = f (n) (mod 2) , 1The correspondence is that if the Latin square has symbol sin the (i, j )-entry, then we put a “1” in the (i, j, s )-entry of the corresponding n×n×narray. 2 where f (n) = ( 0 if n = 0 or 1 (mod 4) 1 if n = 2 or 3 (mod 4) . Subject to this constraint, we are able to prove the natural extension of Cameron’s conjecture to row, column and symbol parities. Specifically, let μ∗ be the conditional distribution of three inde-pendent Bin( n, 1/2) random variables, given that their sum is f (n) mod 2. We are able to prove that (Nrow (L), N col (L), N sym (L)) approximately has the distribution μ∗ as n → ∞ , in many different senses. Theorem 1.3. Let L be a uniformly random n × n Latin square, and let ⃗ X = ( X1, X2, X3) = ( Nrow (L), N col (L), N sym (L)) . Then, the following hold as n → ∞ .(1) (Law of large numbers) We have the convergence in probability 1 n (X1, X2, X3) p → (1 /2, 1/2, 1/2) , i.e., ⃗ X satisfies the same law of large numbers as μ∗.(2) (Central limit theorem) We have the convergence in distribution 1 pn/ 4 (X1 − n/ 2, X2 − n/ 2, X3 − n/ 2) d → N (0 , I 3), i.e., ⃗ X satisfies the same trivariate central limit theorem as μ∗.(3) (Local central limit theorem) For all ⃗x = ( x 1, x 2, x 3) satisfying x1 + x2 + x3 = f (n) (mod 2), we have P[⃗ X =⃗x] = 2 · 1(2 π(n/ 4)) 3/2 exp − (x1 − n/ 2) 2 + ( x2 − n/ 2) 2 + ( x3 − n/ 2) 2 2( n/ 4) o(n−3/2), i.e., ⃗ X satisfies the same local central limit theorem as μ∗.(4) (Total variation convergence) We have dTV (⃗X, μ ∗) → 0. (5) (Large deviation principle) Let H2 : α 7 → − α log 2 α − (1 − α) log 2(1 − α) be the base-2 binary entropy function, and let I(x1, x 2, x 3) = 3 − H2(x1) − H2(x2) − H2(x3). We have − inf ⃗x∈E ◦ I(⃗x) ≤ lim inf n −1 log 2 P[n−1⃗ X ∈ E] ≤ lim sup n−1 log 2 P[n−1⃗ X ∈ E] ≤ − inf ⃗x∈ E I(⃗x) for all Borel E ⊆ R3, where E and E◦ denote the closure and interior of E. That is to say, ⃗ X satisfies the same large deviation principle as μ∗. Actually, all parts of Theorem 1.3 are really consequences of a single master theorem (Theorem 6.4), but we feel the statement of that master theorem is a little too technical for this introduction. 1.2. Previous work. We end this introduction with a brief discussion of previous work on Cameron’s conjecture. The first theorem in this area was due to Häggkvist and Janssen , who proved that P[Nrow (L) = 0] ≤ (p3/4 + o(1)) n ≈ 0.87 n (1.2) (note that Theorem 1.3(5) implies that 2 P[Nrow (L) = 0] ≤ (1 /2 + o(1)) n). Second, generalising (1.1), Cavenagh and Wanless proved that (Nrow (L), N col (L)) is asymptotically equidistributed modulo 2 (this is a special case of Theorem 1.3(4)). Third, Cavenagh, Greenhill and Wanless proved that the probability that the first two rows of L have the same parity lies between 1/4 − o(1) and 3/4 + o(1) (note that Theorem 1.3(1) implies that this probability is 1/2 + o(1) , using the invariance of the distribution of L under permutations of the rows). Until now, these were the only rigorous pieces of evidence towards Cameron’s conjecture. Surprisingly, empirical evidence for small Latin squares has run counter to Cameron’s conjecture (and therefore Theorem 1.3): it seems that the approximations in Theorem 1.3 only start to “kick in” for reasonably large n (though, there is a limit to what can be done empirically, because randomly 2Curiously, the methods in this paper do not directly imply the corresponding lower bound P[Nrow (L) = 0] ≥(1 /2 + o(1)) n. However, this does seem to be provable with some additional tricks; we intend to explore this in future work with Catherine Greenhill and Lenka Kopfová. 3 sampling Latin squares is a notoriously difficult problem). Cameron himself observed that “evidence supports this conjecture fairly well, but the tails of the distribution seem a little heavier than it would predict”, suggesting that there might be good reason that the Häggkvist–Janssen bound mentioned above is significantly larger than 2−n. Alimohammadi, Diaconis, Roghani and Saberi attempted to test Conjecture 1.1 using sequential importance sampling (in the same paper, they gave rigorous grounding to this method). They were not able to find any evidence of the total variation convergence 3 dTV (Nrow (L), Bin( n, 1/2)) → 0, with random Latin squares of size n ≤ 15 .1.3. Notation. We use standard asymptotic notation throughout: for functions f = f (n) and g = g(n),we write f = O(g) or g = Ω( f ) to mean that there is a constant C such that \|f (n)\| ≤ C\|g(n)\| for sufficiently large n, we write f = Θ( g) to mean that f = O(g) and f = Ω( g), and we write f = o(g) or g = ω(f ) to mean that f /g → 0 as n → ∞ . We will often want to treat certain quantities as constants, for the purpose of asymptotic notation; we will always make this clear with phrasing like “fix a constant α > 0”. Somewhat less standardly, for ε > 0 we write f ± ε to denote a quantity that differs from f by at most ε.Regarding basic mathematical notation: for a real number x, the floor and ceiling functions are denoted ⌊x⌋ = max( i ∈ Z : i ≤ x) and ⌈x⌉ = min( i ∈ Z : i ≥ x). We will however generally omit floor and ceiling symbols and assume large numbers are integers, wherever divisibility considerations are not important. We write [n] = {1, . . . , n }, and all logarithms are base e unless explicitly stated otherwise. Finally, since this paper features so many objects of different types, we adopt some typographical conventions: random objects are printed in bold (e.g., a random variable X or a random Latin square L), and “ordered” objects are decorated with an arrow (e.g., vectors ⃗x, and later in the paper we will consider ordered partial Latin squares ⃗ P ). 1.4. Acknowledgements. The authors would like to thank Lenka Kopfová for helpful feedback on an earlier draft of this paper. 2. Discussion of proof techniques In this section we give a high-level overview of the techniques which have previously been brought to bear on random Latin squares, and with this context we describe the new ideas that go into the proof of Theorem 6.4. (We end the section with an outline of the structure of the rest of the paper.) As far as we know, all the rigorous work on random Latin squares uses one of three general classes of techniques. (1) Permutation-invariance. The distribution of a uniformly random n × n Latin square is in-variant under random permutation of the rows, columns, and symbols. So, one can attempt to study random Latin squares purely by considering the effect of such random permutations. This type of reasoning is rather limited, but does have some applications (see e.g. [9, 47]). (2) Enumeration. The most obvious way to study the probability that a random Latin square satisfies a property P is to simply count the number of Latin squares satisfying P, and divide by the total number of Latin squares. Unfortunately, we only have very crude methods to enumerate n × n Latin squares (the best known upper and lower bounds differ by a factor of about exp( n log 2 n)), so generally speaking this can only be used to study properties that are extremely unlikely (i.e., that occur with probability less than about exp( −n log 2 n)). Even for such properties, actually proving the necessary estimates is often a highly nontrivial matter (and often involves consideration of auxiliary random models, as we discuss in Section 2.1). For some examples of properties of random Latin squares proved by enumeration, see e.g. [5, 8, 28, 40, 42, 43, 44, 46, 50]. (3) Cycle switching. A common technique in probabilistic combinatorics is to study a random object by studying the effect of local “switching” operations. Unfortunately, it is very difficult to make a controlled local change to a Latin square. To give some idea of the difficulty, it is an open question to understand the minimum number f (n) such that every n × n Latin square can be transformed into some different Latin square by changing f (n) entries (all we know is that f (n) has order of magnitude between log n and √n; see ). There is only really one type of switching operation that has successfully been used to study random Latin squares, namely cycle 3In fact, they did not even see any evidence of convergence with respect to the so-called Wasserstein metric (Wasserstein convergence is significantly weaker than total variation convergence). 4 switchings (and minor variants thereof). We will define cycle switchings properly in Section 2.3, but for now we just emphasise that they are very unwieldy; in general, if we want to make a particular local change using a cycle switching, we may be forced to make far-reaching changes to the rest of the Latin square. Nonetheless, since cycle switchings are one of very few available tools, they have played a crucial role in many of the known results about random Latin squares (see e.g. [2, 17, 18, 29, 44]). Permutation-invariance alone is certainly not enough to prove Cameron’s conjecture, since any permu-tation of the rows, columns and symbols affects all the row parities in the same way (though, permutation-invariance can be used to easily show that P[Nrow (L) is even ] = 1 /2 when n ≥ 3 is odd). At first sight, enumeration-based approaches also seem to be fundamentally unsuitable to prove Cameron’s conjecture, since we are concerned with events that occur with non-negligible probability, and a multiplicative error of exp( n log 2 n) would completely overwhelm the main term. So, cycle-switching techniques seem to be the most promising avenue towards Cameron’s conjecture; indeed, such techniques underpin the results of Häggkvist–Janssen, Cavenagh–Wanless and Cavenagh–Greenhill–Wanless mentioned in the introduc-tion. However, the usual type of switching analysis is far too crude to have any hope of proving anything like Theorem 1.2 (we discuss this further in Sections 2.3 and 2.4), and some fundamental new ideas are required. In this paper, we use a combination of cycle-switching and enumeration-based techniques. To very briefly summarise our approach: we use enumeration-based techniques to prove a new approximation theorem, comparing a random Latin square to a certain Erdős–Rényi-type random hypergraph (in a stronger sense than previously available). This allows us to show that random Latin squares are ex-tremely likely to have a very rich constellation of 2 × 2 subsquares with certain strong disjointness and canonicity/stability properties. The special properties of these 2×2 subsquares allow us to independently perform many cycle switchings, to “re-randomise” a random Latin square without biasing its distribution. We are then able to prove the desired results using the randomness of our independent switchings, via some linear algebra over the finite field F2.In the following subsections, we describe the above ideas in more detail. 2.1. Enumeration and approximation. Let \|L n\| be the number of n × n Latin squares. This number is known to be \|L n\| = ne2 + o(n) n2 , and there are two known ways to prove this. (1) We can build up an n × n Latin square in a row-by-row fashion: at each step we choose an option for the next row that does not conflict with previous rows. One can use celebrated permanent estimates of of Egorychev–Falikman [21,22] and Bregman to show that the number of choices for each row does not depend too strongly on previous choices; multiplying these bounds over all steps yields exp( −O(n log n)) ≤ \|L n\| (n/e 2)n2 ≤ exp( O(n log 2 n)) (see [56, Theorem 17.3] for the details). (2) We can instead build up an n×n Latin square in an entry-by-entry fashion: at each step we choose a row/column/symbol triple that does not conflict with the partial Latin square constructed so far, and add it. Unfortunately, the number of choices at each step can depend quite dramatically on previous choices; in particular it is possible to “get stuck”, and find oneself in a position where there are no legal choices for the next row/column/symbol triple. However, it turns out that “on average” (i.e., if one makes random choices at each step), this entry-by-entry process is quite well-behaved. Combining a number of very powerful tools, one can use this process to prove exp( −O(n2−ε)) ≤ \|L n\| (n/e 2)n2 ≤ exp( O(n3/2)) . for some tiny 4 constant ε > 0. The details appear in , but are really an adaptation of a general enumeration approach systematised by Keevash . Specifically, the upper bound is proved using the so-called entropy method of Radhakrishnan , in a form pioneered by 4It is unclear what is the optimal εwith this approach, but certainly εcannot be greater than 1/2(this is a general barrier for techniques based on the triangle removal process , which we intend to explore further in upcoming work). 5 Linial and Luria . For the lower bound, one combines an analysis of an instance of the so-called triangle removal process (first studied by Spencer and Rödl and Thoma ), with a completion theorem of Keevash (here one needs a “second generation” completion theorem, building on Keevash’s earlier ideas in the setting of the existence of designs conjecture ; see also [19, 27, 38]). The approaches in (1) and (2) suggest two different ways to study random Latin squares, by comparing them to two different auxiliary random models. First, a k × n Latin rectangle is a k × n array filled with n different symbols, such that each symbol appears exactly once in each row and at most once in each column. The ideas in (1) show that all k × n Latin rectangles can be completed to Latin squares in “about the same number of ways” (up to a multiplicative factor 5 of about exp( n log 2 n)), so if we could show that some property is overwhelmingly likely to hold in a random k × n Latin rectangle, then we could deduce that that property is also very likely to hold in a random n × n Latin square. Similarly, consider the random process in (2) (which can be interpreted as an instance of the so-called triangle removal process ), and suppose we run this process for just a few steps (until the partial Latin square is half-full, say). The ideas in (2) allow us to show that most outcomes of this random process can be completed to Latin squares in “about the same number of ways” (up to a multiplicative factor of about exp( n2−ε)), so if we had some way to show that the first few steps of this random process are overwhelmingly likely to satisfy some property, then we could deduce that that property is also very likely to hold in a random n × n Latin square. Both of these two methods have quite fundamental quantitative limitations (they can only be used to study properties that hold with overwhelming probability), but they have seen many applications. To compare the two methods: the first method has much better quantitative aspects, and it is “more elementary” (not depending on Keevash’s sophisticated completion machinery). On the other hand, the second method is generally much easier to apply: the entry-by-entry random process can be straight-forwardly coupled with an Erdős–Rényi random hypergraph, which allows one to take advantage of the huge body of techniques that have been developed to study random graphs 6. The only known way to directly study random Latin rectangles is via delicate switching arguments (taking advantage of the fact that it is much easier to make local changes to a Latin rectangle than a Latin square). 2.2. A new approximation lemma. One of our key contributions in this paper is a new approximation lemma (Lemma 4.7) that combines the advantages of both the above methods (and is proved by a delicate combination of both methods). Roughly speaking, this lemma says that if one can show that the entry-by-entry random process described in the last subsection (i.e., the triangle removal process ) satisfies a property with probability 1 − ε, then that property holds in a random subset of a random Latin square with probability at least 1 − ε exp( O(n log 2 n)) . The reader may also be interested in an easy-to-apply corollary of our main approximation lemma (Corollary 5.5) comparing random Latin squares to Erdős– Rényi random hypergraphs. For example, these tools could have been used to remove the switching analysis (or remove the dependence on general-purpose switching estimates) from [5,28,42,44,50], and remove the dependence on Keevash’s completion machinery from theorems in [40,42]. For the present paper, our new approximation lemma is rather crucial: quantitative aspects are very important (see Remark 2.1), and the properties we need to consider are so complicated that it would have been a herculean task to study them via switching on Latin rectangles. We hope that our approximation lemma will also be broadly useful in future work on random Latin squares. 2.3. Cycle switchings. A cycle switching is specified by a pair of rows, a pair of columns or a pair of symbols, together with an entry of the Latin square “belonging to that pair”. For example, consider two rows r1, r 2 and an entry (r1, c, s ) (meaning that symbol s appears in row r1 of column c). The first step of the cycle switching is to “move our entry into the other row”: we put the symbol s into cell (r2, c ).However, there was already another symbol s′ in that cell, which we need to move somewhere else; we move it to the only available cell, which is (r1, c ). But there is already an instance of s′ in row r1 (in some column c′), so we move that instance of s′ to the cell (r2, c ′). But there was already some symbol s′′ in that cell, which we move to (r1, c ′). At this point, if we are very lucky, we will have s = s′′ , in which case no further moves are necessary, and we have successfully switched to a new Latin square. If 5Of course, this multiplicative factor depends on k, but it turns out that most of the contribution comes from the last few rows, so one cannot gain much by taking kvery close to n. 6The second method is also much more robust, and can be applied in much more general settings than random Latin squares, though this is not relevant for the present paper. 6 not, we can keep going, repeatedly moving entries between r1 and r2 as long as it takes to resolve all conflicts and reach a Latin square. (In the worst case, we could end up exchanging the entirety of rows r1 and r2). The smallest possible cycle switching is called an intercalate switching . In the above example, it corresponds to the case where s = s′′ , so the entire switching affects just four entries of the Latin square. Alternatively, an intercalate switching can be viewed as “flipping a 2 × 2 Latin subsquare”. In general, the switchings described above (“row switchings”, where we are switching between two rows r1 and r2)can be viewed as “flipping a 2 × k Latin subrectangle”, for some k.Cycle switchings have an easy-to-describe impact on the parities of rows, columns and symbols. For example, a row switching changes the parity of every column and symbol involved in the switching, and it may or may not change the parity of the two involved rows, depending on whether the length of the cycle switching is even or odd. More subtly, cycle switchings also have an impact on the structure of other cycle switchings (e.g., switching in a pair of rows can affect the switchings that are possible in some pair of columns), so one can consider combinations of cycle switchings with more complicated effects. It is even possible to combine multiple “partial” cycle switchings of different types (cf. the “cross-switch” operations in [17, 29]). One can use these types of switchings to estimate the relative likelihoods of various events. Indeed, given two sets of Latin squares A, B, we can design a switching operation to transform a Latin square in A into a Latin square in B. Then, for L1 ∈ A we can try to estimate the number of Latin squares in B that can be reached by such a switching, and for L2 ∈ B we can try to estimate the number of Latin squares in A that can reach L2 by such a switching. Dividing these two estimates gives us an estimate on \|B\| /\|A\| .For example, for a sequence ⃗x ∈ { 0, 1} k, let L⃗x be the set of n × n Latin squares for which the parities of the first k rows are described by ⃗x. Using the above types of ideas, Häggkvist and Janssen managed to show that if k is significantly smaller than n/ 2, we have \|L ⃗x \| ≤ 3\|L ⃗y \| whenever ⃗x and ⃗y differ in a single coordinate, and they iterated this estimate to deduce (1.2). Obviously, this comes very far short of proving Cameron’s conjecture; the condition on k and the extraneous factor of 3 are basically due to the fact that one has very little control over the structure of cycle switchings in an arbitrary Latin square. To make a desired change to the row parities it is necessary to make case distinctions with different combinations of cycle switchings that interact in different ways, and it is necessary to reserve a large subset of “junk rows” which may be affected by the cycle switchings in unpredictable ways. 2.4. Individual intercalate switchings. An important point is that we do not actually need to con-sider cycle switchings in arbitrary Latin squares: we can use the enumeration/approximation methods described in Sections 2.1 and 2.2 to prove that Latin squares are very likely to satisfy certain properties, and then take advantage of these properties in cycle switching arguments. This sounds like a natural approach, but to our knowledge it has never been employed before, because until recently the properties that could be proved using enumeration/approximation were extremely limited. In particular, a very relevant direction is the study of intercalates ( 2 × 2 Latin subsquares) in random Latin squares. If an intercalate involves two even (respectively, odd) rows, then switching that intercalate reduces (respectively increases) the number of even rows by exactly two. So, if we had very tight control of the number of intercalates among the even rows, and the number of intercalates among the odd rows, we could hope to estimate the ratios P[Nrow (L) = x] P[Nrow (L) = x + 2] , (2.1) for all x ∈ N, by considering all the ways to switch a single intercalate. Note that these ratios fully determine the distribution of Nrow (L) (except for a possible bias mod 2, but this is handled by (1.1)). Recently, resolving a conjecture of McKay and Wanless , Kwan, Sah and Sawhney showed how to use enumeration/approximation techniques (together with techniques from large deviations theory) to prove that L has n2/4 + o(n2) intercalates with probability 1 − o(1) . These techniques can be adapted to show that L typically has k2/4 + o(n2) intercalates in any subset of k rows, which allows one to estimate the ratio in (2.1) up to a multiplicative factor of 1 + o(n/ (x(n − x))) . This is already enough to make new progress towards Cameron’s conjecture: since binomial tails decay very rapidly, it is straightforward to deduce that Nrow (L) = n/ 2 + o(n) with probability 1 − o(1) ; that is, Nrow (L) satisfies the same law of large numbers as Bin( n, 1/2) . This implies Theorem 1.3(1), and with a bit more work it is also possible to prove the large deviation principle in Theorem 1.3(5) using this type of idea. 7 That is to say, by studying the effect of switching an intercalate, one can estimate the ratios (2.1) tightly enough to study the tails of Nrow (L) (and Ncol (L) and Nsym (L)). Unfortunately, it is not feasible to study the bulk of the distribution of Nrow (L) this way: since the fluctuations of Bin( n, 1/2) have order of magnitude √n, one would need to control the ratios (2.1) up to a factor of 1 + o(1 /√n); roughly speaking, this is comparable to showing that the number of intercalates Y in L satisfies P[\|Y−EY\| > t ] = o(1) for some t = o(n3/2). While we believe this to be true, it is beyond the reach of enumeration-based techniques: due to clustering phenomena in the upper tail of Y, the relevant deviation probabilities are simply not small enough 7 to tolerate the super-exponential error terms described in Section 2.1. (This is an instance of Janson’s “infamous upper tail” , and is discussed further in .) 2.5. Multiple intercalate switchings. So, instead of studying the effect of a single intercalate switch-ing, our approach is to make many disjoint intercalate switches at the same time .A naïve approach for this is as follows. Using our new approximation theorem, together with tech-niques from [40,42], one can show that a random n×n Latin square L typically has a collection of disjoint intercalates I which “robustly span almost all the rows”, in the sense that for every set R0 of about 8 log 2 n rows, there is an intercalate in I which involves a row in R0 and a row outside R0. Some simple linear algebra over F2 then shows that if we were to randomly and independently switch all the intercalates in I (thereby obtaining a new random Latin square L′), the resulting distribution of the n row parities would be uniform on some affine-linear subspace of Fn 2 with codimension at most about log 2 n, and it is easy to deduce a central limit theorem for Nrow (L′). (With a more sophisticated argument, one can handle the joint distribution of (Nrow (L′), N col (L′), N sym (L′)) and upgrade the central limit theorem to a local central limit theorem.) Remark 2.1 . We are really proving a central limit theorem for Nrow (L′) conditioned on an outcome of L. So, it is important that for different outcomes of L, the corresponding conditional distributions of Nrow (L′) are statistically indistinguishable. This would not be true if the codimension above were greater than √n, since the fluctuations of Bin( n, 1/2) have order of magnitude √n. So, the quantitative aspects of our new approximation theorem are very important here. (E.g., the approximation theorem in would not have sufficed.) Unfortunately, this only characterises the distribution of Nrow (L′), which could in principle be very different to the distribution of Nrow (L). The problem is that switching some intercalates can destroy other intercalates, or create new ones, so the probability of switching from L to L′ could be different from the probability of switching from L′ to L (i.e., the switching could “push probability mass” towards a subset of Latin squares, introducing bias to the distribution of L′). Trying to quantify this bias takes us back to the issues described in Section 2.4. The only way we were able to resolve this issue was to execute the entire argument above with a stable /canonical collection of disjoint intercalates. Specifically, we associate a collection of disjoint intercalates I(L) to each Latin square L, in such a way that if we switch any subset of these intercalates to obtain a new Latin square L′, then I(L) and I(L′) are exactly the same (except that obviously some of these intercalates are switched). So, the probability of switching from L to L′, and the probability of switching from L′ to L, are both exactly 2−\|I (L)\| = 2 −\|I (L′)\|, and L′ is a uniformly random Latin square. In a typical Latin square, the intercalates heavily intersect each other (and therefore cannot be inde-pendently switched). Given a pair of intersecting intercalates, how do we decide which is “the canonical one” that we’re allowed to switch? Our approach is to randomly sparsify the situation, to reduce the intersections between intercalates. Namely, we consider a random “template” T, which describes a sparse subset of “switchable” row/column pairs. Then, given a Latin square L, we define a collection of inter-calates IT(L) by starting with the collection of intercalates in L which use only row/column pairs in T,then deleting the intercalates which are “non-canonical” (the precise definition is a little complicated, but e.g. we delete all intercalates in intersecting pairs, and we delete all intercalates that could introduce a new intercalate when switched). The idea is that this deletion step should not have a very severe effect, since T is sparse. We need to show that a typical outcome of T yields collections of intercalates IT(L) which “robustly span almost all the rows” for almost all L, with which we can implement the argument described above. 7For this to work, one would need to be able to estimate the number of n×nLatin squares \|L n\|up to a subexponential exp( o(n)) multiplicative error. 8We cannot go smaller than about log 2n, due to the multiplicative error of about exp( nlog 2n)in our enumeration estimates. 8 Actually proving the necessary properties of IT(L) is a very delicate matter (much more so than any previous work on random Latin squares), largely due to the aforementioned “infamous upper tail” issue. Specifically, because enumerative estimates have super-exponential error terms, most steps of the proof need to hold with overwhelming probability, so we constantly need to worry about large deviation behaviour. Whenever we need an upper bound on some quantity, we need to be extremely careful to avoid situations where clustering phenomena cause upper tails to be too heavy. 2.6. Outline of the paper. The structure of the proof of Theorem 1.3, as distributed over the rest of the paper, is as follows. First (after Section 3, in which we record some standard concentration inequalities that will be used throughout the paper), in Section 4 we prove our new approximation theorem (Lemma 4.7) relating random Latin squares to the triangle removal process. In Section 5 we prove some simple lemmas relating the triangle removal process to Erdős–Rényi random hypergraphs, which will be useful in combination with Lemma 4.7. In Section 6, we reduce all parts of Theorem 1.3 to a technical “master theorem” (Theorem 6.4) giving a precise description of the distribution of (Nrow (L), N col (L), N sym (L)) . Then, in Section 7, we introduce the notion of a “stable intercalate”, which is an intercalate which can be safely switched without affecting any other intercalates. In Section 8 we show how to use linear-algebraic arguments with stable intercalate switchings to prove the master theorem described above, given a key lemma (Lemma 8.3) on stable intercalates in random Latin squares. Roughly speaking, the key lemma says that there is a “template” T (obtained via a sparse random set of row/column pairs), such that for a uniformly random Latin square, it is very likely that for any large-enough sets of rows, columns and symbols, we can find a stable intercalate (involving only entries in T ) “inside” those sets. The rest of the paper is devoted to the proof of Lemma 8.3. In Section 9 we break down this proof into two lemmas. First, Lemma 9.11 says that (inside any large-enough sets of rows, columns and symbols) there are likely to be many disjoint intercalates (saying nothing about whether they are stable). Second, Lemma 9.13 says that there are unlikely to be many entries which lie in certain “bad” arrangements of intercalates which could lead to our intercalates being non-stable. (Here, due to “infamous upper tail” issues, we cannot hope to consider the number of bad arrangements themselves, only the number of entries in them!) The first of these lemmas (Lemma 9.11) is proved in Section 10, via a careful 2-step application of Freedman’s martingale concentration inequality (using a “maximum disjoint family” technique of Bollobás ). The second lemma (Lemma 9.13) is proved in Section 11, and is much more involved. Here, our job is to understand “bad” arrangements of intercalates; there is an enormous range of possibilities for the structure of such an arrangement, so the first step is a switching argument (unrelated to the main switching argument in Section 8), which shows that every bad arrangement of intercalates can be switched to obtain one of four “basic types”, and therefore it suffices to restrict our attention to these types. We then carefully study how the four basic types of bad intercalate arrangements can emerge in the triangle removal process. 3. Concentration inequalities Throughout this paper we will frequently use a general-purpose concentration inequality for functions of independent Bernoulli random variables. The following statement appears as [57, Corollary 6] 9; it follows from the martingale approach of Freedman . Theorem 3.1. Let ⃗ g = ( g1, . . . , gN ) be a sequence of independent Bernoulli random variables with P[gi = 1] = pi. Let f : {0, 1}N → R satisfy \|f (⃗g) − f (⃗g ′)\| ≤ Ki for all pairs ⃗g,⃗ g ′ ∈ { 0, 1}N differing only in the i-th coordinate, and suppose Ki ≤ K for all i ∈ [N ]. Then for all t ≥ 0, we have P h f (⃗g) − E[f (⃗g)] > t i ≤ 2 exp − t2 2 PNi=1 piK2 i 2 Kt/ 3 ! . Note that, in order to obtain a strong bound from Theorem 3.1, we need to know that f (⃗g) is not very sensitive to changes in any individual gi. In this respect Theorem 3.1 is similar to the much more well-known bounded differences inequality of McDiarmid (which is a consequence of the Azuma–Hoeffding martingale concentration inequality). However, we need the stronger statement of Theorem 3.1 for its essentially-optimal dependence on p1, . . . , p N . 9In , Theorem 3.1 is stated only for the upper tail of f(⃗g); applying it to −f(⃗g)yields the lower tail. 9 We also record a corollary of Theorem 3.1 in the special case of weighted sums of Bernoulli random variables (this is basically the standard Chernoff bound). Corollary 3.2. Let g1, . . . , gN be independent Bernoulli random variables, consider weights w1, . . . , w N ∈ [0 , ∆] , and let X = w1g1 + · · · + wN gN . Then, for any δ > 0 we have P\|X − EX\| ≥ δEX ≤ 2 exp − δ2EX (2 + 2 δ/ 3)∆ ! . Approximating random Latin squares with the triangle removal process In this section, we prove an approximation lemma that relates random Latin squares to the triangle removal process (which is a tractable probability distribution on partial Latin squares). To state this lemma, we need some preparations. First, it will be convenient for us to take an alternative perspective on Latin squares. Fact 4.1. An n × n Latin square can be interpreted as a tripartite 3-uniform hypergraph (specifically, a subgraph of the complete tripartite 3-uniform hypergraph K(3) n,n,n ), in which every pair of vertices in different parts features in exactly one hyperedge. To see the correspondence, think of the three parts as the set of rows, the set of columns and the set of symbols; a hyperedge (r, c, s ) means that symbol s appears in the c-th column of the r-th row. From now on, we interchangeably take the hypergraph point of view and the n × n array point of view, depending on whichever is most convenient at any given moment (generally speaking, the hypergraph point of view is more expressive, and highlights the symmetry between rows, columns and symbols, but it is easier to draw pictures with the n × n array point of view). Definition 4.2. Let Ln be the set of all n × n Latin squares. A partial Latin square is a tripartite 3-uniform hypergraph in which every pair of vertices in different parts features in at most one hyperedge (alternatively, one can also interpret this as an n × n array in which only some of the entries are filled with symbols). Let Pn,m be the set of all n × n partial Latin squares with m hyperedges, and let Pn = S m≤n2 Pn,m ⊇ L n be the set of all n × n partial Latin squares (with any number of hyperedges). An ordered partial Latin square is a partial Latin square together with an ordering on its hyper-edges. Let ⃗ Pn,m be the set of all n × n ordered partial Latin squares with m hyperedges, and let ⃗ Pn = S m≤n2 ⃗ Pn,m .Now, our approximation lemma will compare uniformly random Latin squares to a distribution on (ordered) partial Latin squares; for this to make sense we need a notion of inheritance from a property of Latin squares to a property of (ordered) partial Latin squares. Definition 4.3. Let ⃗ U ⊆ ⃗ Pn be a property of ordered partial Latin squares and let T ⊆ L n be a property of Latin squares. For m ≤ n2, we say ⃗ U is (ρ, m )-inherited from T if for any L ∈ T , taking ⃗ Pm(L) to be a uniformly random subset of m hyperedges of L, equipped with a uniformly random order, we have ⃗ Pm(L) ∈⃗ U with probability at least ρ. Remark 4.4 . It will often be quite obvious how to specify the property ⃗ U, for a desired event T (and often, the ordering will play no role). For example, if T is the property of having fewer than (1 − ε)n2/4 intercalates, it would make sense to take ⃗ U to be the property of having fewer than (1 −ε/ 2)( m/n 2)4n2/4 intercalates. However, as we will see later in the paper, there is sometimes reason to make more exotic choices for ⃗ U. Remark 4.5 . Throughout the paper we will need to consider many different properties of (ordered) (partial) Latin squares, and we have attempted to adopt some notational conventions to help the reader keep track of the “type” of each property. Specifically, for properties of partial Latin squares, we will usually use the letter U (think “unfinished”) and for properties of complete Latin squares we will usually use the letter T (think “total”). An arrow on top of a property indicates that it is a property of ordered (rather than unordered) partial Latin squares. Now, the main lemma of this section concerns the triangle removal process , which we now define. 10 Definition 4.6. Let TRP( n, m ) be the distribution over ⃗ Pn,m ∪ {⊥} obtained as follows. Start with the complete tripartite graph Kn,n,n . At each step, choose a uniformly random triangle (among all the triangles in the current graph), and remove it. After m steps of this process, the sequence of removed triangles can be interpreted as an ordered partial Latin square ⃗ R ∈⃗ Pn,m (unless we ran out of triangles at some point in the process, in which case we take ⃗ R =⊥). We are now ready to state the main result of this section, comparing a uniformly random Latin square to the triangle removal process. Lemma 4.7. Fix constants α ∈ (0 , 1) and ρ ∈ (0 , 1] , and let m = αn 2. Consider properties ⃗ U ⊆ ⃗ Pn,m and T ⊆ L n such that ⃗ U is (ρ, m )-inherited from T . Let ⃗ R ∼ TRP( n, m ) be an ordered partial Latin square obtained by m steps of the triangle removal process, and let L ∼ Unif( Ln) be a uniformly random n × n Latin square. Then P[L ∈ T ] ≤ exp n log 2 n 1 − α − o(1) ! P[⃗R ∈⃗ U]. Lemma 4.7 is a quantitative improvement to [42, Theorem 2.4] (which is itself an adaptation of [40, Theorem 2.4]). Our proof of Lemma 4.7 bears some similarities to the proofs in [40, 42], but incorporates various additional ideas (related to our use of enumeration results that are stronger but much less robust). Remark 4.8 . It may seem that Lemma 4.7 is not suitable for studying “local” properties (e.g., properties concerning a particular vertex v). Such properties are important for the so-called absorption method ,which (is not relevant for the main results in this paper but) has played a central role in previous work on random Latin squares. Indeed, it is not hard to see that with probability at least exp( −O(n)) , the random hypergraph ⃗ R has no edges at all containing v, and this probability seems too large to have any hope of applying Lemma 4.7 to any interesting local property about v. However, one can generally overcome this issue with a judicious choice of ⃗ U, cf. Remark 4.4. For example (very informally speaking), if T is the property that v does not participate in a desired local structure, we can take ⃗ U to be the property that v has high degree but still does not participate in the desired local structure. The rest of this section is devoted to the proof of Lemma 4.7. We start with some further definitions. Definition 4.9. For an (ordered) partial n × n Latin square P , let G(P ) ⊆ Kn,n,n be the graph which contains an edge uv (of Kn,n,n ) if and only if there is no hyperedge of P containing u and v. That is to say, G(P ) is the graph of pairs that would need to be covered to complete P to a full Latin square. For a graph G ⊆ Kn,n,n , let dens( G) = e(G)/(3 n2). Definition 4.10. We say that a graph G ⊆ Kn,n,n , is ε-triangle-typical if its number of triangles is (1 ± ε)n3 dens( G)3. Let P△:εn,m ⊆ P n,m be the set of partial Latin squares P ∈ P n,m such that G(P ) is ε-triangle-typical. Let ⃗ P△:εn,m ⊆⃗ Pn,m be the set of ordered partial Latin squares ⃗ P ∈⃗ Pn,m such that for each i ≤ m, if we consider the partial Latin square Pi consisting of the first i hyperedges of ⃗ P , then Pi ∈ P △:εn,m . Definition 4.11. We say that a graph G ⊆ Kn,n,n , is γ-quasirandom if for all pairs of vertices u, v in different parts, the number of common neighbours of u and v is (1 ± γ)n dens( G)2. Let Pγn,m be the set of partial Latin squares P ∈ P n,m such that G(P ) is γ-quasirandom. Remark 4.12 . It is not hard to see that if a graph is γ-quasirandom, then it is necessarily O(γ)-triangle-typical (this fact is used in the previous approximation lemmas in [40, 42]). However, in our proof of Lemma 4.7 (with its strong quantitative aspects), we will need to consider quasirandomness and triangle-typicality with quite different parameters. In the next two lemmas, we show that for any Latin square L ∈ L n, if we take a random ordering of a random subset of L, then the resulting random object is likely to satisfy certain triangle-typicality and quasirandomness properties. This will allow us to assume that various graphs we encounter later in the proof are triangle-typical/quasirandom, which will be necessary for our enumeration techniques. Lemma 4.13. Fix a constant α ∈ (0 , 1) . Consider any L ∈ L n and let ⃗ Pαn 2 (L) ∈⃗ Pn,αn 2 be a random ordering of a random set of αn 2 hyperedges of L. Then, with ε = n−1 log n, we have P[⃗ Pαn 2 (L) /∈⃗ P△:εn,αn 2 ] ≤ exp( −Ω(log 2 n)) . 11 Proof. Fix some m ≤ αn 2. We will show that G(⃗ Pm(L)) is ε-triangle-typical with probability at least 1 − exp( −Ω(log 2 n)) . The statement of the lemma will then follow by a union bound over all m ≤ αn 2.Note that ⃗ Pm(L) consists of a uniformly random set of m hyperedges of L. Let Bp(L) be a random subset of the hyperedges of L where each hyperedge is taken independently at random with probability p = m/n 2. By “Pittel’s inequality” (see e.g. page 17 in ), for any property Q of unordered partial Latin squares, P[⃗ Pm(L) /∈ Q ] ≤ 3√mP[Bp(L) /∈ Q ]. Thus, letting X be the number of triangles in G(⃗ Pm(L)) , it suffices to show that P[X /∈ (1 ±ε)n3(1 −p)3] ≤ exp( −Ω(log 2 n)) .For any v1 ∈ V1, v 2 ∈ V2, v 3 ∈ V3 such that v1v2v3 ∈ L, the probability that v1v2v3 is a triangle in G(Bp(L)) is precisely 1 − p = 1 − m/n 2. On the other hand, if v1v2v3 /∈ L, the probability v1v2v3 is a triangle in G(Bp(L)) is (1 − p)3, since each pair of vertices among v1, v 2, v 3 is in a distinct hyperedge of L, and all these hyperedges need to be absent from Bp(L) in order for v1v2v3 to be in G(Bp(L)) . Thus, μ := E[X] = n2(1 − p) + ( n3 − n2)(1 − p)3 = Θ( n3). There are n2 hyperedges in L, and adding or removing a hyperedge from Bp(L) can affect X by at most 1 + 3( n − 1) , since each pair of vertices in e can form a triangle with n − 1 other vertices in G(Bp(L)) .Thus, by Theorem 3.1, we have P[X /∈ (1 ± ε)n3(1 − p)3] ≤ P[\|X − μ\| ≥ εμ/ 2] ≤ 2 exp − ε2μ2/42n2p1 + 3( n − 1) 2 + 2 1 + 3( n − 1) εμ/ 6 ! = exp − Ω(log 2 n), where in the first inequality we used that μ − n3(1 − p)3 = oεn 3(1 − p)3. □ Lemma 4.14. Fix a constant α ∈ (0 , 1) . Consider any L ∈ L n and let Pαn 2 (L) ∈ P n,αn 2 be a random subset of αn 2 hyperedges of L. Let γ = ω(n−1/2 log 1/2 n). Then P[Pαn 2 (L) /∈ P γn,αn 2 ] ≤ exp( −Ω( γ2n)) . Proof. Let m := αn 2. We will show that G(Pm(L)) is γ-quasirandom with probability at least 1 − exp( −Ω( γ2n)) .As in the proof of Lemma 4.13, let Bp(L) be a random subset of the hyperedges of L where each hyperedge is taken independently at random with probability p = m/n 2 = α. Let Su,v be the set of common neighbours of u and v, with respect to the graph G(Bp(L)) . It suffices to show that for every u, v that are in different parts of G(Bp(L)) , it holds that P[\|Su,v \| /∈ (1 ± γ)n(1 − α)2] ≤ exp( −Ω( γ2n)) .A union bound over all pairs of vertices u, v and Pittel’s inequality would then complete the proof. Fix some u, v in different parts of G(Bp(L)) . Let w be the unique vertex such that uvw is a hyperedge of L. Then P[w ∈ Su,v ] = 1 − p. On the other hand, for all w′̸ = w that are in the same part as w, we have P[w′ ∈ Su,v ] = (1 − p)2 (since w′ participates in some hyperedge together with u and in another hyperedge together with v, and both of these need to not be selected in Bp(L) to have w′ ∈ Su,v ). Thus, for X = \|Su,v \|, μ := E[X] = (1 − p) + ( n − 1)(1 − p)2 = Θ( n). There are 2n − 1 hyperedges in L that affect X, and adding or removing such a hyperedge from Bp(L) can affect X by at most 1. Thus, by Theorem 3.1, we have P[X /∈ (1 ± γ)n(1 − p)2] ≤ P[\|X − μ\| ≥ γμ/ 2] ≤ exp − γ2μ2/42(2 n − 1) + γμ/ 3 = exp( −Ω( γ2n)) , where in the first inequality we used that μ − n(1 − p)2 = (1 − p)p = o(γμ ). □ We next prove an upper bound on the number of completions of a partial Latin square, under triangle-typicality and quasirandomness assumptions. Theorem 4.15. Fix a constant α ∈ (0 , 1) , let γ = o(1) and let ε = n−1 log n. For any partial Latin square P ∈ P γn,αn 2 ∩ P △:εn,αn 2 , the number \|L n(P )\| of completions of P to an n × n Latin square satisfies \|L n(P )\| ≤ (1 − α)2ne2 !(1 −α)n2 · exp n log 2 n 1 − α − o(1) . 12 Proof. For a row x (respectively, column y), we write rx (respectively, cy ) for the number of empty cells in that row, in P . Then for an empty cell e = ( x, y ), we write Qx,y for the set of symbols that do not appear in the row or column of e. Note that Qx,y is precisely the set of common neighbours of x and y in G(P ). Then, let L(P ) be a uniformly random completion of P . We will estimate the entropy H[L(P )] = log \|L n(P )\| of L(P ).Let E ⊆ [n]2 be the set of empty cells in P . For each e = ( x, y ) ∈ E, let ze be the symbol in cell e in L(P ). So the sequence (ze)e∈E determines L(P ). For any total ordering ≺ on E, we have H[L(P )] = X e∈E Hze (ze′ : e′ ≺ e). (4.1) Now, consider a pair of sequences μ = ( μx)x, ν = ( νy )y ∈ [0 , 1] n (of real numbers in the range [0 , 1] , which we assume to be all distinct from each other). Let λ = ( λx,y )x,y be the array defined by λx,y = ( μx, ν y ).We can use λ to define a total order ≺λ on E, via reverse lexicographic order on the pairs λx,y . To be precise, write (x′, y ′) ≺λ (x, y ) when μx′ > μ x or when x′ = x and νy′ > ν y . Let Re(λ) be an upper bound on the conditional support size supp ze (ze′ : e′ ≺λ e) defined as follows. Re(λ) for e = ( x, y ) is 1 plus the number of symbols z ∈ Qe \ ze for which μxz < μ x and νyz < ν y , where xz and yz are the row and column such that cells (xz , y ) and (x, y z ) contain symbol z in L(P ).Since Re(λ) is an upper bound on \| supp( ze \| (ze′ : e′ ≺λ e)) \|, we have Hze (ze′ : e′ ≺λ e) ≤ E[log Re(λ)] . (4.2) It follows from (4.1) and (4.2) that H[L(P )] ≤ X e∈E E[log Re(λ)] . This is true for any fixed λ, so it is also true if λ is chosen randomly: let λ ∈ [0 , 1] 2 n2 be obtained via 2n independent random variables μx, νy each uniformly distributed in [0 , 1] (with probability 1 they are all distinct). Then H[L(P )] ≤ X e∈E E[log Re(λ)] . Next, for any completion L of P , any e = ( x, y ) and any λe = ( μx, ν y ) ∈ [0 , 1] 2, let RL,λ e e = E[Re(λ) \| L(P ) = L, λe = λe]. (Note that λe = λe occurs with probability zero, so formally we should condition on λe = λe ± dλe and take limits in what follows, but there are no continuity issues so we will ignore this detail). By the definition of Re(λ) and linearity of expectation, we have RL,λ e e = 1 + ( \|Qe\| − 1) μxνy . By Jensen’s inequality, E[log Re(λ) \| L(P ) = L, λe = λe] ≤ log RL,λ e e , so E[log Re(λ) \| L(P ) = L] ≤ E h log RL, λe e i = Z [0 ,1] 2 log 1 + ( \|Qe\| − 1) μxνy dλe. For C > 0 we compute Z 10 Z 10 log(1 + Cts ) d t ds = log(1 + C) − 2 + log(1 + C) − Li 2(−C) C , where Li 2(C) = − Z C 0 log(1 − t) t dt is an evaluation of the polylogarithm function, so, as C → ∞ , − Li 2(−C) = Z −C 0 log(1 − t) t dt = Z C 0 log(1 + t) t dt 13 ≤ Z 10 log(1 + t) t dt + C−1 X t=1 log(2 + t) t ≤ log 2(C + 1) + O(1) . Therefore, substituting C = \|Qe\| − 1, we have E[log Re(λ) \| L(P ) = L] ≤ log \|Qe\| − 2 + log \|Qe\| + log 2 \|Qe\| + O(1) \|Qe\| = log \|Qe\| − 2 + (1 + o(1)) log 2 \|Qe\|\|Qe\| . Now recall that P ∈ P γn,αn 2 , and that \|Qe\| = \|Qx,y \| is precisely the number of common neighbours of x and y in G(P ). Since G(P ) is γ-quasirandom, we have \|Qe\| = (1 ± γ)n(1 − α)2 = n(1 − α − o(1)) 2. Thus, using \|Qe\| ≤ n, and noting that the expression above does not depend on the choice of L, we get E[log Re(λ)] ≤ log \|Qe\| − 2 + log 2 n (1 − α − o(1)) 2n , Next, since P ∈ P △:εn,αn 2 with ε := n−1 log n, we have that the number of triangles in G(P ) is X e∈E \|Qe\| = (1 ± ε)n3(1 − α)3. Since the natural logarithm is a concave function, Jensen’s inequality yields X e∈E log \|Qe\| ≤ \| E\| log P e∈E \|Qe\|\|E\| ! ≤ \| E\| log (1 + ε)n(1 − α)2 . We now get log \|L n(P )\| = H[L(P )] ≤ X e∈E E[log Re(λ)] ≤ X e∈E log \|Qe\| + \|E\| log 2 n (1 − α − o(1)) 2n − 2 ! ≤ n2(1 − α) log (1 + ε)(1 − α)2n − 2 ! n log 2 n 1 − α − o(1) . Thus, \|L n(P )\| ≤ (1 + ε)(1 − α)2ne2 !(1 −α)n2 · exp n log 2 n 1 − α − o(1) ≤ (1 − α)2ne2 !(1 −α)n2 · exp ε(1 − α)n2 + n log 2 n 1 − α − o(1) = (1 − α)2ne2 !(1 −α)n2 · exp n log 2 n 1 − α − o(1) , recalling that ε = n−1 log n and so exp( ε(1 − α)n2) = exp( o(n log 2 n)) . □ The next ingredient we will need is a lower bound on the total number of n × n Latin squares. The current state of the art is the following bound (obtained via a celebrated result of Egorychev– Falikman [21, 22] on permanents of doubly stochastic matrices). Lemma 4.16 ([56, Theorem 17.2]) . The number of n × n Latin squares is at least \|L n\| ≥ ne2 n2 · exp( −O(n log n)) . Finally, we need the following lemma, which shows that each triangle-typical ordered partial Latin square is roughly equally likely to be the outcome of the triangle removal process. Lemma 4.17. Fix a constant α ∈ (0 , 1) . Consider ⃗ P ∈⃗ P△:εn,αn 2 for some ε > 0. Then for ⃗ R ∼ TRP( n, m ), we have P[⃗ R =⃗ P ] = (1 ± 2ε)αn 2 exp O(log n) en 3αn 2 (1 − α)3n2(1 −α). 14 Proof. Let N = n2. Since ⃗ P ∈⃗ P△:εn,αN , P[⃗ R =⃗ P ] = αN −1 Y i=0 1(1 ± ε)(1 − i/N )3n3 = (1 ± 2ε)αN n−3αN exp − 3 αN −1 X i=0 log 1 − iN ! . Note that αN −1X i=0 1 N log 1 − i + 1 N ≤ Z α 0 log(1 − t)d t ≤ αN −1 X i=0 1 N log 1 − iN . We have αN −1 X i=0 log 1 − iN − log 1 − i + 1 N ! = αN −1 X i=0 log 1 + 1 N − (i + 1) ≤ αN −1 X i=0 1 N − (i + 1) = O(log n). Thus, using the fact that R log x = x(log x − 1) , we get 3 αN −1 X i=0 log 1 − iN = 3 N Z α 0 log(1 − t) d t ± O(log n)= 3 N Z 11−α log s ds ± O(log n)= 3 N 1(log 1 − 1) − (1 − α) log(1 − α) − 1 ± O(log n)= 3 N − α − (1 − α) log(1 − α) ± O(log n), so exp − 3 αN −1 X i=0 log 1 − iN ! = e3N α (1 − α)3N (1 −α) exp O(log n). Therefore, P[⃗ R =⃗ P ] = (1 ± 2ε)αN exp O(log n) en 3αN (1 − α)3N (1 −α). □ We are now ready to prove Lemma 4.7. Proof of Lemma 4.7. Let N := n2 and ε := n−1 log n and γ := n−1/4. By the fact that ⃗ U is (ρ, m )-inherited from T and by Lemmas 4.13 and 4.14, we have that for every Latin square L ∈ T , a random ordering ⃗ Pm(L) of a random set of m hyperedges of L satisfies P[⃗ Pm(L) ∈⃗ U ∩ ⃗ P△:εn,αN ∩ P γn,αN ] ≥ ρ − o(1) , where we say that an ordered partial Latin square is in Pγn,αN if its underlying unordered partial Latin square is in Pγn,αN . Thus, letting S := ⃗ U ∩ ⃗ P△:εn,αN ∩ P γn,αN , P[L ∈ T ] = X L∈T 1 \|L n\| ≤ X L∈T 1 \|L n\| · P[⃗ Pm(L) ∈ S ] ρ − o(1) = 1 ρ − o(1) · 1 \|L n\| X L∈T X⃗P ∈S P[⃗ Pm(L) = ⃗ P ]. Changing the order of the two sums, we get P[L ∈ T ] ≤ 1 ρ − o(1) X⃗P ∈S P[⃗ Pm(L) = ⃗ P ], where ⃗ Pm(L) is a uniformly random ordering of a uniformly random subset of m hyperedges of L. Letting \|L n(⃗P )\| denote the number of completions of the underlying unordered partial Latin square of ⃗ P , we have P[⃗Pm(L) = ⃗ P ] = \|L n(⃗ P )\| · 1 \|L n\| · (N − m)! N ! , since for each Latin square, there are N (N − 1) . . . (N − m + 1) ways to select an (ordered) sequence of m hyperedges. Therefore, by Theorem 4.15 and Lemma 4.16, P[L ∈ T ] ≤ X⃗P ∈S (ρ−1 + o(1)) · \|L n(⃗P )\|\|L n\| · (N − m)! N ! 15 ≤ X⃗P ∈S (ρ−1 + o(1)) · n(1 − α)2 e2 N −\| ⃗ P \| exp n log 2 n 1 − α − o(1) · nO(n) e2 n n2 · (N − m)! N ! ≤ X⃗P ∈S exp n log 2 n 1 − α − o(1) · e2 n αN · (1 − α)2N (1 −α) · (N − m)! N ! . By Stirling’s approximation, we deduce P[L ∈ T ] ≤ X⃗P ∈S exp n log 2 n 1 − α − o(1) · e2 n αN · (1 − α)2N (1 −α) · N N (1 −α)(1 − α)N (1 −α) eN (1 −α) · eN N N = X⃗P ∈S exp n log 2 n 1 − α − o(1) en 3αN (1 − α)3N (1 −α). On the other hand, for each ⃗ P ∈⃗ U ∩ ⃗ P△:εn,αN ∩ P γn,αN , by Lemma 4.17 we have P[⃗ R =⃗ P ] ≥ (1 − 2ε)αN e−Θ(log n) en 3αN (1 − α)3N (1 −α) ≥ exp − (2 + o(1)) αn log n en 3αN (1 − α)3N (1 −α). Therefore, P[L ∈ T ] ≤ X⃗P ∈S exp n log 2 n 1 − α − o(1) P[⃗ R =⃗ P ] ≤ exp n log 2 n 1 − α − o(1) P[⃗ R ∈⃗ U]. □ Approximating the triangle removal process In this section, we collect some simple lemmas that allow us to analyse the triangle removal process via Erdős–Rényi (binomial) random hypergraphs. The latter are defined in terms of independent choices, and are therefore much easier to analyse. We then use these lemmas to deduce a user-friendly corollary (Corollary 5.5) of Lemma 4.7. Definition 5.1. Let K(3) n,n,n be the complete 3-uniform 3-partite hypergraph on n + n + n vertices, and let G(3) (n, p ) be the probability distribution on subgraphs of K(3) n,n,n obtained by including each possible hyperedge with probability p independently. Note that G ∼ G(3) (n, p ) may not be a partial Latin square, as it may contain two hyperedges intersecting in more than one vertex. Call this a conflict .Now, our first lemma is similar to [24, Lemma 2.6]. It approximates the triangle removal process with a binomial random hypergraph, for properties U that are monotone increasing in the sense that P ∈ U and P ⊆ P ′ then P ′ ∈ U (i.e. “adding additional hyperedges doesn’t hurt”). For multiple-exposure arguments, it will be important that this lemma can handle the triangle removal process started from a proper subgraph of Kn,n,n . Definition 5.2. For a partial Latin square P , let TRP( P, m ) be the partial Latin square distribution ob-tained with m steps of the triangle removal process starting from G(P ). Thus, an outcome of TRP( P, m ) always has m hyperedges, and none of these hyperedges conflict with P (unless we run out of triangles and get the outcome “ ⊥”) Lemma 5.3. Fix constants γ, α 1, α 2 such that γ > 0, and α1, α 2 ∈ [0 , 1] are sufficiently small with respect to γ. Define m1 = α1n2 and m2 = α2n2 and p = (1 + γ)α2/n . Let P ∈ P n,m 1 be a partial Latin square, and consider a monotone increasing property U of subhypergraphs of K(3) n,n,n . Let R ∼ TRP( P, m 2) and G ∼ G(3) (n, p ). Then P[R ∈ U ] ≤ P[G ∈ U ] + exp( −Ω( n2)) . Proof. Let G∗ be obtained from G by deleting all hyperedges involved in conflicts. We say that the hyperedges of G∗ are isolated in G. Note that we can couple (G∗, G) and R such that R ⊆ G as long as G∗ has at least m2 = α2n2 hyperedges (to see this, randomly order the hyperedges in G and run the triangle removal process on that ordering). Let X be the number of hyperedges in G∗ (i.e., the number of isolated hyperedges in G). Note that each hyperedge in P conflicts with no more than 3n of the n3 potential hyperedges in G. Thus, there 16 are at least (1 − 3α1)n3 potential hyperedges that do not conflict with P . Each of these is present and isolated in G with probability at least p(1 − p)3n ≥ (1 + γ)α2 exp( −6(1 + γ)α2)/n . Thus, we have E[X] ≥ (1 − 3α1)n3 · (1 + γ)α2 exp( −6(1 + γ)α2)/n ≥ (1 + γ/ 2) α2n2. Adding a hyperedge to G can increase X by at most 1, and removing a hyperedge from G can increase X by at most 3, by making three hyperedges isolated. Thus, by Theorem 3.1, P[X < α 2n2] ≤ P[\|X − E[X]\| ≥ γα 2n2/2] ≤ exp( −Ω( n2)) . Therefore, P[R ∈ U ] ≤ P[G ∈ U ] + P[X < α 2n2] = P[G ∈ U ] + exp( −Ω( n2)) . □ The next lemma appears as [42, Lemma 5.2], and handles properties U that are monotone decreasing in the sense that if P ∈ U and P ′ ⊆ P then P ′ ∈ U (i.e., “removing hyperedges doesn’t hurt”). Lemma 5.4 ([42, Lemma 5.2]) . Fix a constant α ∈ (0 , 1) . Let U ⊆ P n be a monotone decreasing property of partial Latin squares, let 10 R ∼ TRP( n, αn 2), let G ∼ G(3) (n, α/n ), and let G∗ be obtained from G by deleting all hyperedges involved in conflicts. Then P[R ∈ U or R =⊥] = O(P[G∗ ∈ U ]) . For completeness, we reproduce the simple proof of Lemma 5.4 (from ). Proof. We can couple R and G∗ such that, if e(G) ≤ αn 2, then either G∗ ⊆ R or R =⊥ (to see this, randomly order the hyperedges of G and run the triangle removal process on that ordering). Since U is monotone decreasing, we have P[R ∈ U or R =⊥] ≤ P[G∗ ∈ U \| e(G) ≤ αn 2]. Since e(G) is binomial with mean αn 2, we have that P[e(G) ≤ αn 2] = Ω(1) . Thus, P[R ∈ U or R =⊥] ≤ P[G∗ ∈ U ]/P[e(G) ≤ αn 2] = O(P[G∗ ∈ U ]) . □ To illustrate the tools from this section, we conclude this section by stating a corollary of Lemma 4.7, which is an easy-to-apply tool for studying random Latin squares. We do not actually use this corollary in the present paper (due to some technical issues related to multiple-exposure arguments, we need the full power of Lemma 4.7), but we hope that it may be convenient in future work on random Latin squares. Note that we have defined (ρ, m )-inheritedness (in Definition 4.3) only for properties of ordered partial Latin squares, but in what follows we extend this definition to unordered partial Latin squares in the natural way. Corollary 5.5. Fix constants α ∈ (0 , 1) and ρ ∈ (0 , 1) , and let m = αn 2. Let T be a property of n × n Latin squares and let U be a property of subhypergraphs of K(3) n,n,n , such that U is (ρ, m )-inherited from T . Let L be a uniformly random n × n Latin square. (1) If U is a monotone decreasing property, we have P[L ∈ T ] ≤ exp O(n log 2 n)P[G∗ ∈ U ], where G ∼ G(3) (n, α/n ), and G∗ is obtained from G by deleting all hyperedges involved in conflicts. (2) If U is a monotone increasing property, then for any constant γ > 0, if α is sufficiently small with respect to γ, we have P[L ∈ T ] ≤ exp O(n log 2 n)P[G ∈ U ] + exp( −Ω( n2)) , where G ∼ G(3) (n, (1 + γ)α/n ). Part (1) is obtained by combining Lemma 4.7 with Lemma 5.3 (with P = ∅ and α1 = 0 and α2 = α), and Part (2) is obtained by combining Lemma 4.7 with Lemma 5.4. 10 We previously defined TRP( n, m )as a distribution on ordered partial Latin squares, but here we are somewhat abusively viewing it as a distribution on unordered partial Latin squares (we simply ignore the ordering). 17 The master theorem for parity distributions All parts of Theorem 1.3 will be deduced from a master theorem which describes the distribution of (Nrow (L), N col (L), N sym (L)) in somewhat technical terms. To state this theorem, we need some termi-nology to describe certain types of distributions that arise naturally from random intercalate switches. Definition 6.1. Consider a random vector (c1 +B1, c 2 +B2, c 3 +B3), where B1, B2, B3 are independent binomial random variables satisfying Bi ∼ Bin( ni, 1/2) . We say that this random vector is (n, d )-near-binomial if n − d ≤ ni ≤ n and ci ≤ d for each i.The reader should think of d as a “defect parameter”: if d is small, then (n, d )-near-binomial random vectors are approximately distributed like three independent copies of Bin( n, 1/2) . We remark that if we want this approximation to have o(1) total variation error, then we need d = o(√n). Indeed, for each i we need E[Xi] − n/ 2 to be negligible compared to the fluctuations of Bin( n, 1/2) (which have order of magnitude √n). Definition 6.2. Say that a random vector (X1, X2, X3) ∈ N3 is (n, d )-2-near-binomial if it can be obtained from an (n, d )-near-binomial random vector by conditioning on a particular parity (even or odd) for each of X1, X2, X3. Definition 6.3. Say that a random vector ⃗ X is a 2-near-binomial mixture with parameters (n, d, ε ) if one can define an auxiliary random object Q (on the same probability space as ⃗ X), and an event A depending only on Q, such that: • P[A] ≥ 1 − ε, and • For any of the possible outcomes Q of Q satisfying A, the conditional distribution of ⃗ X given Q = Q is (n, d )-2-near-binomial. If d is small, then the above definition says that ⃗ X is closely approximated by a sequence of three independent Bin( n, 1/2) random variables, except for possible biases mod 2. Note that the distribution μ∗ featuring in Theorem 1.3 is a 2-near-binomial mixture with parameters (n, 0, 0) .Now, our main technical theorem is as follows. Theorem 6.4. Let L be a uniformly random n × n Latin square. Then (Nrow (L), N col (L), N sym (L)) is a 2-near-binomial mixture with parameters n, O (log 11 n), exp( −ω(n log 2 n)) . 6.1. Corollaries of the master theorem. In this subsection we briefly explain how to deduce all parts of Theorem 1.3 from our master theorem (Theorem 6.4). This is all quite routine. Note that Theorem 1.2 is an immediate consequence of Theorem 1.3(4). We start with basic consequences of the de Moivre–Laplace local central limit theorem and a large deviation principle for binomial distributions. Lemma 6.5. Let ⃗ X = ( X1, X2, X3) be a (n, o (√n)) -2-near-binomial random vector. (i) Recall that (X1 mod 2 , X2 mod 2 , X3 mod 2 ) always takes a common value ⃗v ∈ { 0, 1} 3. For any ⃗x ∈ Z 3 satisfying ⃗x =⃗v mod 2, we have P[⃗ X =⃗x] = 8 · 1(2 π(n/ 4)) 3/2 exp − (x1 − n/ 2) 2 + ( x2 − n/ 2) 2 + ( x3 − n/ 2) 2 2( n/ 4) o(n−3/2). (ii) Let H2 : α 7 → − α log 2 α − (1 − α) log 2(1 − α) be the base-2 binary entropy function, and let I(x1, x 2, x 3) = 3 − H2(x1) − H2(x2) − H2(x3). We have − inf ⃗x∈E ◦ I(⃗x) ≤ lim inf n −1 log 2 P[n−1⃗ X ∈ E] ≤ lim sup n−1 log 2 P[n−1⃗ X ∈ E] ≤ − inf ⃗x∈ E I(⃗x) for all Borel E ⊆ R3.Proof. Recall that ⃗ X is obtained from a (n, d )-near-binomial random vector ⃗ Y by conditioning on the event that ⃗ Y =⃗v mod 2 (which occurs with probability exactly 11 1/8). Now, the de Moivre–Laplace local central limit theorem (see e.g. [23, Chapter VII, Section 3, Theorem 1]) tells us that P[⃗Y =⃗x] = 3 Y i=1 1 p2π(ni/4) exp − (x′ i − ni/2) 2 2( ni/4) o(n−1/2) ! 11 Here we are using that Bin( n, 1/2) is equally likely to be even and odd, which is easy to prove e.g. by induction. 18 for some n1, n 2, n 3 = n − o(√n), and some x′ 1 , x ′ 2 , x ′ 3 satisfying x′ i = xi − o(√n). It is easy to see that the error terms in the ni and x′ i can be absorbed into the main error term (i.e., it makes no difference to replace n1, n 2, n 3 with n, and replace each x′ i with xi). The desired result follows, noting that conditioning on an event that occurs with probability 1/8 introduces a factor of 8.We can similarly deduce (ii) from a large deviation principle for binomial distributions (e.g. Sanov’s theorem, see [20, Theorem 2.1.10] 12 ). Actually, this is even simpler than (i), because in this case the conditioning mod 2 has a negligible impact. □ Now we are ready to deduce Theorem 1.3. Proof of Theorem 1.3. We start with Theorem 1.3(3) (the local central limit theorem). Consider any ⃗x ∈ Z 3 satisfying x1 + x2 + x3 = f (n) (mod 2). By Theorem 6.4, together with Lemma 6.5(i) and the definition of a 2-near-binomial mixture, P[⃗ X =⃗x] is of the form P[⃗X =⃗x mod 2 ] · 8 · 1(2 π(n/ 4)) 3/2 exp (x1 − n/ 2) 2 + ( x2 − n/ 2) 2 + ( x3 − n/ 2) 2 2( n/ 4) o(n−3/2) exp( −ω(n log 2 n)) . By the Cavenagh–Wanless theorem (mentioned in the introduction), we have P[⃗X =⃗x mod 2 ] = 1/4 + o(1) , and Theorem 1.3(3) follows. We can then deduce Theorem 1.3(4) (total variation convergence) by summing over all ⃗x (using e.g. Chebyshev’s inequality to control the tails). Indeed, let ⃗ Z ∼ μ∗. For any N , the variance of Bin( N, 1/2) is N/ 4 = O(N ), so (recalling the definition of a 2-near-binomial mixture and noting that the means of all relevant binomials are within O((log n)11 ) = O(√n) of each other), the variance of each component of ⃗ X and ⃗ Z is O(n). Then, for any K ≥ 1, taking ⃗ Z ∼ μ∗ and writing (n/ 2) ⃗ 1 = ( n/ 2, n/ 2, n/ 2) , we have dTV (⃗ X, μ ∗) = 12 X⃗x∈Z 3 P[⃗ X =⃗x] − P[⃗ Z =⃗x] ≤ O(( K√n)3) · o(n−3/2) + P∥⃗X − (n/ 2) ⃗1∥∞ ≥ K√n + P∥⃗Z − (n/ 2) ⃗ 1∥∞ ≥ K√n ≤ o(1) · K3 + O(1 /K 2). Taking K → ∞ sufficiently slowly, Theorem 1.3(4) follows. It is then a near-trivial matter to deduce Theorem 1.3(1–2) (the central limit theorem and law of large numbers), since total variation convergence is much stronger than convergence in distribution (see e.g. [54, Proposition 1.2]). Finally, Theorem 1.3(5) (the large deviation principle) follows from Theorem 6.4 and Lemma 6.5(ii). Indeed, Theorem 6.4 says that ⃗ X is a mixture of (n, o (√n)) -2-near-binomial distributions except with an exceptional probability exp( −ω(n log 2 n)) (which is negligible). □ Remark 6.6 . In the above proof, it was convenient to use the Cavenagh–Wanless theorem as a black box. However, we remark that given the machinery we developed in Sections 4 and 5, it would be a very simple matter to re-prove the Cavenagh–Wanless theorem. Indeed, using this machinery it is easy to prove that L is very likely to have a 2 × 3 Latin subrectangle (in the first half of the rows, columns and symbols of L, say), and a 3 × 2 Latin subrectangle (in the second half of the rows, columns and symbols of L, say), and switching on these two Latin subrectangles mixes between all four different possibilities for (Nrow (L), N col (L), N sym (L)) mod 2. 7. Stable intercalates An intercalate in a Latin square is a 2 × 2 Latin subsquare. To switch an intercalate with symbols a and b means to replace both occurrences of a with b and vice versa. The result of this operation is always another Latin square. As discussed in Section 2, we will prove Theorem 6.4 via random intercalate switches. It is very important that if we start with a uniformly random Latin square, and perform our random switches, the resulting Latin square is still uniformly random. To ensure this, we need to restrict our attention to intercalates with a certain stability property. We define this property in such a way that switching stable 12 A simple approximation argument (approximating general open and closed sets with product sets) can be used to combine large deviation principles for three independent random variables, see e.g. [48, Lemma 2.8]. 19 intercalates can never create a new intercalate or destroy an existing intercalate (and, in particular, if we only ever switch stable intercalates, then the set of available stable intercalate switches never changes). Specifically, we define an intercalate to be stable if it does not intersect other intercalates, and if switching it and/or some other intercalates cannot make it intersect another intercalate, as follows. Definition 7.1. Let P be a partial Latin square. For an intercalate A, we denote by ¯A the outcome of switching A. • We say that an intercalate in P is isolated if it does not share an entry with any other intercalate. • Consider a set of isolated intercalates {A1, . . . , A t} in P . We say that this is a critical set of intercalates if it is possible to switch a subset of A1, . . . , A t (resulting in intercalates B1, . . . , B t where Bi = Ai or Bi = ¯Ai for each i ∈ [t]) to create a new intercalate that shares an entry with each of B1, . . . , B t. • We say that an isolated intercalate in P is stable if it is not contained in a critical set of inter-calates. Observe that a critical set {A1, . . . , A t} may be a subset of another critical set {A1, . . . , A t+1 } (e.g., if switching a subset of A1, . . . , A t creates a new intercalate that intersects At+1 ). Any critical set D has size at most 4, since its members A1, . . . , A t are pairwise entry-disjoint, and for each Ai, either Ai or ¯Ai shares an entry with the new intercalate that can be produced by switching a subset of D.Now, in the following lemma we record the fact that switching stable intercalates leaves the set of stable intercalates switches unchanged. Definition 7.2. For an intercalate A with rows r1, r 2, columns c1, c 2, and symbols s1, s 2, define the tuple σ(A) = ( {r1, r 2}, {c1, c 2}, {s1, s 2}) (i.e., this records the rows, columns and symbols of the intercalate, without recording which of the two possible “states” it is in). For a (partial) Latin square P , let S(P ) be the set of all tuples σ(A) for all stable intercalates A in P (i.e., this records the intercalate switches we can perform, without recording which state each of those intercalates is in). Lemma 7.3. Let P1 be a Latin square, and suppose that we can switch a single stable intercalate in P1 to obtain a second Latin square P2. Then S(P1) = S(P2).Proof. Let A be the stable intercalate in P1 which is switched to obtain P2. Then σ(A) ∈ S (P1). We start with a simple observation which will be used multiple times. Claim 7.4. If an intercalate B̸ = ¯A is isolated in P2, it is also present and isolated in P1.Proof of claim. Since B shares no entry with ¯A in P2, we see that B is also present in P1. Furthermore, if there were any intercalate C in P1 that shares an entry with B, that intercalate C would not be present in P2, so it would have to share an entry with A in P1. But that would imply A is not isolated in P1,which would contradict our assumption that A is stable in P1. ■ Step 1: reversibility of the A switch. We will first show σ(A) ∈ S (P2), i.e., we could equally well have switched ¯A in P2 to obtain P1. This amounts to showing that ¯A is isolated in P2, and not contained in a critical set of intercalates. First, note that if ¯A were not isolated in P2, then there would be some intercalate B in P2 that shares an entry with ¯A. Since that entry would not be in P1, we would have that B is not in P1, implying that {A} is a critical set in P1, contradicting the stableness of A in P1. Thus, ¯A is isolated in P2.Now, suppose for contradiction that ¯A is contained in a critical set { ¯A, A 2, . . . , A t} of isolated interca-lates in P2. If t = 1 , this would contradict the isolatedness of either A in P1 or ¯A in P2; thus t ≥ 2. Since the intercalates A2, . . . , A t are isolated in P2, they are also present and isolated in P1, by Claim 7.4. But then {A, A 2, . . . , A t} is a critical set in P1: if switching S ⊆ { ¯A, A 2, . . . , A t} creates a new intercalate in P2, then switching ¯S = S ∪ { A} (if ¯A / ∈ S) or ¯S = S \ { ¯A} (if ¯A ∈ S) creates a new intercalate in P1.This contradicts the stableness of A in P1. We conclude that ¯A is stable in P2, so σ(A) ∈ S (P2). Step 2: the intercalates other than A. Since σ(A) ∈ S (P1) ∩ S (P2), by symmetry it is enough to show that for every intercalate B that is stable in P1, we have σ(B) ∈ S (P2). Let B̸ = A be a stable intercalate in P1. Then B must be present and isolated in P2, by Claim 7.4 and the symmetry between P1 and P2.Suppose for the purpose of contradiction that B is not stable in P2. Then B is contained in a critical set Z = {B, A 2, . . . , A t} of isolated intercalates in P2. Since the intercalates in {A2, . . . , A t} \ { ¯A} are isolated in P2 and distinct from ¯A, they are also present and isolated in P1. Now, either Z \ { ¯A} ∪ { A} or Z \ { ¯A} is a critical set in P1. In particular, if the new intercalate that can be created in P2 by switching a subset of Z shares an entry with either A or ¯A, then Z \ { ¯A} ∪ { A} is a critical set in P1. Otherwise, 20 Z \ { ¯A} is a critical set in P1. In either case, B is contained in a critical set in P1, contradicting that B is stable in P1. Thus, B must be stable in P2, so σ(B) ∈ S (P2), concluding the proof. □ Deducing the master theorem from stable intercalate information As discussed in Section 2, the main challenge in this paper is to show that random Latin squares typically have very rich constellations of stable intercalates. These intercalates can then be randomly switched to prove our master theorem on parity distributions (Theorem 6.4). In this section, we state our key lemma about stable intercalates in random Latin squares (Lemma 8.3), and show how to use this key lemma (together with some linear-algebraic arguments) to prove Theorem 6.4. After this section, the rest of the paper will be devoted to the proof of Lemma 8.3. 8.1. The key lemma on stable intercalates. Roughly speaking, we need the property that for all “reasonably large” sets of rows, columns and symbols, we can find stable intercalates compatible with those choices. This will imply that the stable intercalates are so well-spread throughout the rows, columns and symbols that randomly switching them will thoroughly mix up the row, column and symbol parities. Definition 8.1. Let P be an n × n (partial) Latin square, and consider some 1 ≤ ℓ ≤ n. We say that P is an (ℓ, β )-intercalate-expander if the following property holds. For any sets of rows R, R∗, any sets of columns C , C ∗, and any sets of symbols S , S ∗, such that five out of these six sets have size βn and the last one has size ℓ, there is a stable intercalate in P with one row in R, the other row in R∗, one column in C , the other column in C ∗, one symbol in S , and the other symbol in S ∗.Note that the sets of rows, columns, and symbols above are not necessarily disjoint. Thus, the stable intercalate promised by Definition 8.1 may for instance have both its rows in R ∩ R∗.Random Latin squares themselves are unlikely to be good intercalate-expanders, because they have too many intercalates (that are too likely to intersect each other and form critical sets of intercalates). We are able to overcome this with a sparsification technique: instead of considering all the stable intercalates in a Latin square L, we consider only the stable intercalates in some sparsified partial Latin square (and we make sure to use the same “sparsification template” T for all Latin squares). To formalise this, we need another definition. Definition 8.2. An n × n template is a subset T ⊆ [n]2, which we interpret as a set of row/column pairs. For a Latin square L ∈ L n, we write T ∩ L ∈ P n for the partial Latin square containing just the entries of L in positions specified by T .Now, our key lemma is as follows. Lemma 8.3. Let β > 0 be a constant. There exists a template T ⊆ [n]2 such that the following holds. Let L ∼ Unif( Ln) be a uniformly random Latin square (interpreted as a subgraph of K(3) n,n,n ) and let ℓ = log 11 n. Then P[T ∩ L is an (ℓ, β )-intercalate-expander ] ≥ 1 − exp( −ω(n log 2 n)) . 8.2. Deducing the master theorem on parity distributions. To deduce Theorem 6.4 from Lemma 8.3, we will need some more notation and preliminaries. Definition 8.4. Consider an n × n Latin square L. For a set of rows R and a set of intercalates I in L,let MR, I ∈ { 0, 1}\|R\|×\|I\| be the “incidence matrix” defined as follows: MR, I (r, I ) = ( 1 if row r participates in intercalate I 0 otherwise . We can similarly define MC, I or MS, I for a set of columns C or a set of symbols S. Lemma 8.5. Consider an n × n partial Latin square P that is an (ℓ, 1/10) -intercalate-expander, where n is sufficiently large and 1 ≤ ℓ = o(√n). Then we can find sets R, C, S of rows, columns and symbols, satisfying \|R\|, \|C\|, \|S\| ≥ n − ℓ, with the following property. If we let I(R, C, S ) be the set of all stable intercalates in P which are contained in the rows in R and the columns in C, only using the symbols in S, then the matrix MR, I(R,C,S ) MC, I(R,C,S ) MS, I(R,C,S )  ∈ { 0, 1}(\|R\|+\|C\|+\|S\|)×\|I (R,C,S )\| 21 has rank exactly \|R\| + \|C\| + \|S\| − 3 over F2. The left kernel vectors of this matrix are those vectors of the form (⃗x R⃗x C⃗x S ) ∈ F\|R\|+\|C\|+\|S\| 2 where ⃗x Z ∈ F\|Z\| 2 is the all-zero vector (0 , . . . , 0) or the all-one vector (1 , . . . , 1) for each Z ∈ { R, C, S }.Proof. First, we translate the desired statement into more combinatorial language. Consider the auxiliary 6-uniform hypergraph Q(R, C, S ) on the vertex set 13 R ⊔ C ⊔ S, where for every intercalate in I(R, C, S ) we put a hyperedge consisting of the two rows, two columns and two symbols of that intercalate. We want to choose R, C, S to have the property that for any choices of vertex subsets R′ ⊆ R and C′ ⊆ C and S′ ⊆ S which are not all “trivial” (i.e., at least one of these three subsets is nonempty and proper), there is some hyperedge of Q(R, C, S ) which intersects R′ ⊔ C′ ⊔ S′ in an odd number of vertices. Now, we specify R, C, S as follows. Let R be the set of all rows r for which there exist at least 6ℓ different stable intercalates in P which all involve r and which otherwise involve pairwise disjoint rows, columns and symbols. Define C, S similarly. Claim. Each of R, C, S has size greater than n − ℓ.Proof of claim. Suppose for contradiction (and without loss of generality) that there is a set R0 of ℓ rows which are not in R. Then we can use the intercalate-expansion property to greedily find n/ 10 stable intercalates in P which all involve disjoint rows, columns and symbols (except for their rows in R0). So, there must be some r ∈ R0 involved in (n/ 10) /\|R0\| ≥ 6ℓ stable intercalates which involve disjoint rows, columns and symbols (except that they all involve r). This contradicts the choice of R0. ■ Recall that the defining properties of R, C, S were that they are included in many (at least 6ℓ)“externally disjoint” stable intercalates in P . The above claim shows that almost all of those interca-lates lie in I(R, C, S ), and therefore correspond to hyperedges of Q(R, C, S ). Specifically, each “bad” row/column/symbol (which is not in R, C, S ) kills at most one of our externally disjoint stable interca-lates, so each vertex of Q(R, C, S ) is contained in more than 6ℓ − 3ℓ ≥ 3ℓ externally disjoint 6-uniform hyperedges. Now, consider vertex subsets R′ ⊆ R and C′ ⊆ C and S′ ⊆ S which are not all trivial (at least one of these subsets is nonempty and proper). We need to prove there is some hyperedge intersecting R′ ⊔ C′ ⊔ S′ in an odd number of vertices. Case 1: First, suppose that for each Z ∈ { R′, C ′, S ′}, either Z or its complement has size less than ℓ.This gives us three “small sets” of size less than ℓ. At least one of our small sets must be non-empty; let v be a vertex in that set. Recall that v is contained in at least 3ℓ externally disjoint hyperedges of Q(R, C, S ), so at least one of these hyperedges (externally) avoids our three small sets. Any such hyperedge has odd intersection with R ⊔ C ⊔ S. Case 2: If Case 1 does not occur, we may assume without loss of generality that R′ and R \ R′ both have size at least ℓ. Then we apply the intercalate-expansion property directly. We take the smaller of R′ and R \ R′ as R∗ and the larger as R; we take the larger of C′ and C \ C′ as C = C ∗; and we take the larger of S′ and S \ S′ as S = S ∗. This again gives us a stable intercalate in P (corresponding to an edge of Q(R, C, S )) with the desired odd intersection. □ Now we can prove Theorem 6.4. Proof of Theorem 6.4. Let T be as given by Lemma 8.3, and let ℓ = log 11 n. For a Latin square L ∈ L n,we say an intercalate in L is T -stable if it is present and stable in T ∩ L. Recall the notation S(P ) from Definition 7.2 (recording the set of stable intercalates in a partial Latin square P , but without recording which of the two possible “states” each intercalate is in). We consider an auxiliary multigraph G, whose vertices are the Latin squares L ∈ L n, and where there is an edge between two Latin squares L1, L 2 if it is possible to switch some of the T -stable intercalates in L1 to transform it into L2. So, for every L ∈ L n, the degree of L in G is precisely 2\|S (T ∩L)\| (including one loop edge from L to itself). In fact, by Lemma 7.3, the connected component of L is a clique (with loops) of order 2\|S (T ∩L)\|, where each vertex L′ ∈ L n in this clique has S(T ∩ L′) = S(T ∩ L). In particular, each component is regular, so the uniform distribution Unif( Ln) is a stationary distribution for the random walk on G.Therefore, if we first sample L ∼ Unif( Ln), and then we “re-randomise” L by randomly switching each T -stable intercalate with probability 1/2 independently (this is the same as stepping to a random 13 Here we use “ ⊔” to indicate a disjoint union. 22 neighbour of L in G), then the resulting random Latin square L′ still has the same uniform distribution as L.Now, for the rest of the proof, our goal is to show that ⃗ X′ = ( Nrow (L′), N col (L′), N sym (L′)) is a 2-near-binomial mixture with parameters (n, ℓ, exp( −ω(n log 2 n)) . Recalling Definition 6.3, this means we need to show how to reveal certain information about L and L′, in such a way that, conditional on a typical outcome of this revealed information, ⃗ X′ is a (n, ℓ )-2-near-binomial random vector (where here “typical” refers to an event which occurs with probability 1 − exp( −ω(n log 2 n)) ). First, reveal an outcome of L such that T ∩ L is an (ℓ, 1/10) -intercalate-expander. By Lemma 8.3, this occurs with probability 1−exp( −ω(n log 2 n)) . From now on, we will view L as a non-random object (i.e., all probabilities will implicitly be with respect to the conditional probability space given our outcome of L). Let R, C, S be the sets of rows, columns and symbols from Lemma 8.5, and let I = I(R, C, S ) be as in the statement of Lemma 8.5. Recall that L′ is obtained from L via random T -stable intercalate switches; reveal any outcome of these random switches for T -stable intercalates which are not in I. We will show that, in the resulting conditional probability space (which can be described in terms of \|I\| random coin flips), ⃗ X′ has a 2-near-binomial distribution with parameters (n, ℓ ).To see this, let ⃗ x1 ∈ { 0, 1}n,⃗ x2 ∈ { 0, 1}n,⃗ x3 ∈ { 0, 1}n be the sequences of parities of rows, columns and symbols in L′. In our conditional probability space, the only parts of ⃗ x1,⃗ x2,⃗ x3 that remain random are the subsequences ⃗ xR ∈ { 0, 1}R,⃗ xC ∈ { 0, 1}C ,⃗ xS ∈ { 0, 1}S corresponding to our identified sets of rows, columns and symbols. We can describe their joint distribution as (⃗ xR⃗ xC⃗ xS ) = ( ⃗y R⃗y C⃗y S ) + MR, I MC, I MS, I ⃗r, where ⃗ y R,⃗ y C ,⃗ y S describe the parities of our identified rows, columns and symbols in L (which we are viewing as non-random), ⃗ r ∈ { 0, 1}\|I\| is a uniformly random zero-one sequence of length \|I\| , and arithmetic is mod 2. Now, recall the structure of the left kernel vectors from Lemma 8.5. We see that (⃗ xR⃗ xC⃗ xS ) is a uniformly random element of {0, 1}R × { 0, 1}C × { 0, 1}S , except that the values of ⃗ 1 ·⃗ xR,⃗ 1 ·⃗ xC and ⃗ 1 ·⃗ xS are constrained (i.e., the parities of the rows, columns and symbols in R, C, S are uniformly random, except that the number of odd rows, number of odd columns, and number of odd symbols are constrained mod 2). This implies the desired result. □ Setup for the proof of the intercalate-expander lemma In this section, we set the stage for the proof of Lemma 8.3. We first rephrase it by slightly changing the nature and role of the template. After that, we state two main lemmas, and show how to deduce the (rephrased) intercalate-expander lemma from them (we will then spend the following sections of the paper proving these two lemmas). Recall the sparsifying role of the template T in Lemma 8.3. To prove Lemma 8.3, we will take a slightly different view on T . Namely, for most of the proof it will be convenient to instead work with a template hypergraph H (which specifies a set of row/column/symbol triples, instead of a set of row/column pairs). Our template hypergraph will be obtained by sampling G(3) (n, ε ) with an appropriate ε, as specified in the following definition. Definition 9.1. Throughout this and the following sections, we fix a constant β > 0 (cf. Lemma 8.3). We also let ε = η log −1 n for some η = o(1) that slowly goes to 0 (concretely, take η = 1 / log log n) and let ℓ = log 11 n.We now state a version of the intercalate-expander lemma with a random template hypergraph. Lemma 9.2. Fix a constant β > 0. Consider a random hypergraph H ∼ G(3) (n, ε ) and an independent random Latin square L ∼ Unif( Ln) (both interpreted as subgraphs of K(3) n,n,n ). Then P[H ∩ L is an (ℓ, β )-intercalate-expander ] ≥ 1 − exp( −ω(n log 2 n)) . Remark 9.3 . Lemma 9.2 trivially implies that there is a hypergraph H ⊆ K(3) n,n,n such that P[H ∩ L is an (ℓ, β )-intercalate-expander ] ≥ 1 − exp( −ω(n log 2 n)) . However, this would not have sufficed to prove Theorem 6.4 (we would have no guarantee that if we switch an intercalate which lies in H, the resulting switched intercalate still lies in H). The slightly different notion of a template in Definition 8.2 was chosen to avoid this issue. 23 We first show the (simple) deduction of Lemma 8.3 from Lemma 9.2. For this, we record the following simple fact, which we repeatedly use throughout the rest of the paper. Fact 9.4. Suppose H1 ∈ H 1 and H2 ∈ H 2 are independent random objects, let P(2) ⊆ H 1 × H 2, and for any H1 ∈ H 1 and p ∈ [0 , 1] let P(p) be the set of all H1 ∈ H 1 such that P[( H1, H2) ∈ P (2) ] > p . Then (1) if P[( H1, H2) ∈ P (2) ] ≤ p, then P[H1 ∈ P (√p)] ≤ √p.(2) If P[H1 ∈ P (p2)] ≤ p1, then P[( H1, H2) ∈ P (2) ] ≤ p1 + p2. Proof of Lemma 8.3. Say that L ∈ L n is good if P[H ∩ L is an (ℓ, β )-intercalate-expander ] ≥ 1 − exp( −ω(n log 2 n)) . By Lemma 9.2 and Fact 9.4(1), L is good with probability 1 − exp( −ω(n log 2 n).Now, let T ⊆ [n]2 be a random subset of row/column pairs, where each is included with probability ε independently. Note that for each fixed outcome L of L, the two random partial Latin squares H ∩ L and T ∩ L have exactly the same distribution. Therefore, for each good L, P[T ∩ L is an (ℓ, β )-intercalate-expander ] ≥ 1 − exp( −ω(n log 2 n)) , and Fact 9.4(2) yields P[T ∩ L is an (ℓ, β )-intercalate-expander ] ≥ 1 − exp( −ω(n log 2 n)) . Applying Fact 9.4(1) again, we see that almost all outcomes T of T are suitable for the conclusion of Lemma 8.3 (namely, T satisfies the required property with probability 1 − exp( −ω(n log 2 n)) ). □ We now move on to reducing Lemma 9.2 to two main lemmas. Recall that the property of being an intercalate-expander says that (with respect to any appropriate 6-tuple (R, R∗, C , C ∗, S , S ∗)) there is at least one stable intercalate consistent with our 6-tuple. Our first lemma (Lemma 9.11) will say that there are likely to be many intercalates consistent with every appropriate 6-tuple (and says nothing about isolatedness or stability). Our second lemma (Lemma 9.13) will say that there is likely to be a much smaller number of entries which participate in “bad sets” of intercalates which violate isolatedness or stability (i.e., they participate in a critical set of intercalates or a pair of intersecting intercalates). Remark 9.5 . It is crucially important here that we consider the number of entries in bad sets of inter-calates (as opposed to the number of bad sets of intercalates themselves). This is due to the “infamous upper tail” problem which arises when trying to estimate the upper tail of subgraph counts in random graphs and hypergraphs. Due to clustering phenomena, these upper tails are ‘fatter’ than their lower counterparts (we need to pay very close attention to tail bounds due to the exponential error terms in Lemma 4.7). Before stating our two lemmas, we need some preparations. Definition 9.6. For sets R and R∗ of rows, sets C and C ∗ of columns and sets S and S ∗ of symbols, we say that an intercalate is (R, R∗, C , C ∗, S , S ∗)-split (or just split if the parameters are clear from context) if it has one row in R, the other row in R∗, one column in C , the other column in C ∗, one symbol in S , and the other symbol in S ∗.By symmetry between the rows, columns, and symbols of a Latin square, in the context of Defini-tion 8.1 it suffices to consider the case where \|R∗\| = ℓ and \|R\|, \|C \|, \|C ∗\|, \|S \|, \|S ∗\| ≥ βn . Definition 9.7. We say a 6-tuple (R, R∗, C , C ∗, S , S ∗) is (ℓ, β )-permissible if we have \|R∗\| = ℓ and \|R\|, \|C \|, \|C ∗\|, \|S \|, \|S ∗\| ≥ βn .Our goal is then to show that, with very high probability, all (ℓ, β )-permissible 6-tuples have a split stable intercalate. Definition 9.8. Let P be a partial Latin square, let {A1, . . . , A t} be a critical set of isolated intercalates in P , and let A′ be an intercalate that can be created by switching a subset of {A1, . . . , A t}. Then we say the entries in A1 ∪ · · · ∪ At ∪ (A′ ∩ P ) comprise a critical configuration with respect to P .We say that a set of entries in a partial Latin square P is a bad configuration with respect to P either if it is a critical configuration with respect to P , or if it comprises two intercalates that intersect in one entry. (We will from now on omit “with respect to P ” if P is clear from the context.) We say that a bad configuration is (R, R∗, C , C ∗, S , S ∗)-split if at least one intercalate in it (that is, one of the at most four intercalates in the critical set of intercalates or one of the two intersecting intercalates, depending on the type of bad configuration) is (R, R∗, C , C ∗, S , S ∗)-split. 24 Remark 9.9 . We will sometimes need to talk about sets of entries which make up some arrangement of intercalates, and we will sometimes need to talk about sets of intercalates themselves. To try to assist the reader to keep track of this distinction, we will reserve the word “configuration” for a set of entries in a partial Latin square (i.e. a set of hyperedges, in the hypergraph perspective in Fact 4.1). Note that pairs of intercalates in a partial Latin square can intersect in at most one entry, so if an intercalate in a partial Latin square does not intersect any other intercalate in one entry, then it is isolated. Thus, an intercalate in a (partial) Latin square is stable if and only if it is not a subset of any bad configuration. Instead of reasoning directly about the total number of split intercalates, it will be easier to show lower bounds on the size of the maximum family of disjoint split intercalates (as this random variable behaves better with respect to Theorem 3.1). The next lemma states that if we consider a random Latin square intersected with a random sparse template hypergraph, then there is very likely to be a large family of disjoint split intercalates. Definition 9.10. For c > 0 and a tripartite hypergraph H ⊆ K(3) n,n,n , let T int (H, c ) ⊆ L n be the set of Latin squares L such that for some (ℓ, β )-permissible 6-tuple (R, R∗, C , C ∗, S , S ∗), the maximum family of disjoint (R, R∗, C , C ∗, S , S ∗)-split intercalates in H ∩ L has size smaller than cε 4nℓ . Lemma 9.11. Fix a constant c > 0 which is sufficiently small with respect to β. Then, consider a random hypergraph H ∼ G(3) (n, ε ) and an independent random Latin square L ∼ Unif( Ln). We have P[L ∈ T int (H, c )] ≤ exp( −ω(n log 2 n)) . Next, the following lemma gives an upper bound on the number of entries in split bad configurations. Definition 9.12. We say that an entry in a row in R∗ is covered by some split bad configuration if it belongs to a split intercalate in it. We say an intercalate or a bad configuration is R∗-split if it is (R, R∗, C , C ∗, S , S ∗)-split for some (ℓ, β )-permissible 6-tuple (R, R∗, C , C ∗, S , S ∗) containing R∗.For C > 0 and a hypergraph H ⊆ K(3) n,n,n , let T bad (H, C ) ⊆ L n be the set of Latin squares L such that for some set of rows R∗ of size ℓ, the number of entries in rows in R∗ covered by R∗-split bad configurations in H ∩ L is more than Cε 5nℓ . Lemma 9.13. There is an absolute constant C > 0 such that the following holds. Consider a random hypergraph H ∼ G(3) (n, ε ) and an independent random Latin square L ∼ Unif( Ln). We have P[L ∈ T bad (H, C )] ≤ exp( −ω(n log 2 n)) . We can now deduce Lemma 9.2 from Lemmas 9.11 and 9.13. Proof of Lemma 9.2. By the union bound, Lemma 9.11, and Lemma 9.13, there are C, c > 0 such that we have P[L ∈ T int (H, c ) ∪ T bad (H, C )] ≤ exp( −ω(n log 2 n)) . Note that if L /∈ T int (H, c )∪T bad (H, C ), then for every (ℓ, β )-permissible 6-tuple (R, R∗, C , C ∗, S , S ∗),there are many (R, R∗, C , C ∗, S , S ∗)-split intercalates in H ∩ L, and only a small fraction of them can have entries covered by (R, R∗, C , C ∗, S , S ∗)-split bad configurations. Therefore, at least one of them must be stable. By symmetry between rows, columns, and symbols, and a union bound, we see that we have P[H ∩ L is an (ℓ, β )-intercalate-expander ] ≥ 1 − exp( −ω(n log 2 n)) . That is to say, almost all outcomes H ∩ L of H ∩ L satisfy the conclusion of Definition 8.1. □ It remains to prove Lemmas 9.11 and 9.13, which we will do in Sections 10 and 11, respectively. Both proofs will make crucial use of Lemma 4.7 (to deduce results about random Latin squares from very-high-probability results about the triangle removal process). Given Lemma 4.7, the proof of Lemma 9.11 is a rather quick (though somewhat delicate) consequence of Theorem 3.1. The proof of Lemma 9.13 is much longer, and requires several additional ideas. 10. Existence of many disjoint intercalates As already mentioned, to prove Lemma 9.11, we employ Lemma 4.7 and work in the setting of the triangle removal process. To do that, we define a property of ordered partial Latin squares that satisfies suitable inheritance properties with respect to T int (H, c ) (recall Definition 4.3). 25 Definition 10.1. Recall the definitions of ℓ and ε from Definition 9.1. For c > 0 and H ⊆ K(3) n,n,n ,let ⃗ Uint (H, c ) ⊆⃗ Pn be the set of ordered partial Latin squares ⃗ P such that for some (ℓ, β )-permissible 6-tuple (R, R∗, C , C ∗, S , S ∗), the maximum family of disjoint (R, R∗, C , C ∗, S , S ∗)-split intercalates in H ∩⃗ P has size less than cε 4nℓ .Now, the following lemma is a version of Lemma 9.11 for the triangle removal process. Lemma 10.2. Fix constants α, c > 0, such that α is sufficiently small and c is sufficiently small in terms of α and β. Consider independent random hypergraphs ⃗ R ∼ TRP( n, αn 2) and H ∼ G(3) (n, ε ).Then P[⃗R ∈⃗ Uint (H, c )] ≤ exp( −ω(n log 2 n)) . We first deduce Lemma 9.11 from Lemma 10.2 using Lemma 4.7. Proof of Lemma 9.11. Let α, c be small enough for Lemma 10.2 and let ρ = 1. By Lemma 10.2 and Fact 9.4(1), with probability 1 − exp( −ω(n log 2 n) our random hypergraph H ∼ G(3) (n, ε ) is such that P[⃗R ∈⃗ Uint (H, c ) \| H] ≤ exp( −ω(n log 2 n)) . Let H be an outcome of H for which this holds. Note that ⃗ Uint (H, c ) is (ρ, αn 2)-inherited from T int (H, c ): indeed, if L ∈ T int (H, c ), then there is some (ℓ, β )-permissible 6-tuple (R, R∗, C , C ∗, S , S ∗) such that the maximum family of dis-joint (R, R∗, C , C ∗, S , S ∗)-split intercalates in H ∩ L has size smaller than cε 4nℓ , so for any P ⊆ L,the maximum family of disjoint (R, R∗, C , C ∗, S , S ∗)-split intercalates in H ∩ P also has size smaller than cε 4nℓ . Thus, by Lemma 4.7, P[L ∈ T int (H, c )] ≤ exp(2 n log 2 n)P[⃗ R ∈⃗ Uint (H, c )] ≤ exp( −ω(n log 2 n)) . Recalling our choice of H, Fact 9.4(2) then implies that P[L ∈ T int (H, c )] ≤ exp( −ω(n log 2 n)) , as desired. □ Proof of Lemma 10.2. We will prove the desired statement with c = 10 −5β5α4.Note that the property ⃗ Uint (H, c ) is monotone decreasing. Let G ∼ G(3) (n, α/n ), and let G∗ be obtained from G by deleting all hyperedges involved in conflicts (recall that a conflict is a pair of hyperedges that intersect in more than one vertex). By Lemma 5.4, for each H ⊆ K(3) n,n,n we have P[⃗ R ∈⃗ Uint (H, c )] = O(P[G∗ ∈⃗ Uint (H, c )]) , so for our random hypergraph H we have P[⃗ R ∈⃗ Uint (H, c )] = O(P[G∗ ∈⃗ Uint (H, c )]) . Thus, from now on we work in G∗. Fix an (ℓ, β )-permissible 6-tuple (R, R∗, C , C ∗, S , S ∗). We will show that the probability that the maximum family of disjoint (R, R∗, C , C ∗, S , S ∗)-split intercalates in H ∩ G∗ has size smaller than cε 4nℓ is at most exp( −ω(n log 2 n)) .Our main tool will be Theorem 3.1, but to effectively apply this concentration inequality we need to be quite careful with the way we interpret H ∩ G∗ as a function of independent choices. Specifically, for each potential hyperedge e, let gh e and ge be two independent Bernoulli random variables with P[gh e = 1] = αε/n and P[ge = 1] = α(1 − ε)/(n − αε ). We generate G and H ∩ G by sampling gh e and ge for each potential edge e, setting e ∈ H ∩ G if gh e = 1 , and setting e ∈ G if gh e = 1 or ge = 1 (or both). To see that this correctly describes the joint distribution of G and H ∩ G, note that P[e ∈ G] = 1 − (1 − P[gh e = 1])(1 − P[ge = 1]) = 1 − 1 − αε n 1 − α(1 − ε) n − αε = αε n + α(1 − ε) n − αε − α2ε(1 − ε) n(n − αε ) = αn . Now, the plan is to reveal G and H (via the random variables gh e and ge) in two phases. First, we reveal all entries in rows outside of R∗, and then we reveal entries in rows in R∗.Recall that a conflict is a pair of hyperedges that intersect in more than one vertex. We say a pair of entries {(r, c 1, s 1), (r, c 2, s 2)} in H ∩ G on the same row r / ∈ R∗ has a second-order conflict if s1 appears in column c2 or if s2 appears in column c1 on a row outside of R∗, with respect to G. For r ∈ R \ R∗, c1 ∈ C , c2 ∈ C ∗, s1 ∈ S and s2 ∈ S ∗, we say a pair {(r, c 1, s 1), (r, c 2, s 2)} is a half-intercalate if both its hyperedges are present in H ∩ G and have no conflicts with entries of G outside of R∗, and the pair has no second-order conflicts (which would make it impossible to complete it to an intercalate in G∗). 26 The idea is that in the first phase, by revealing all the entries in the rows outside R∗, we reveal all the half-intercalates. In the second phase, each half-intercalate {(r, c 1, s 1), (r, c 2, s 2)} can then be completed to a split intercalate in H ∩ G∗ via any pair of entries (r′, c 1, s 1), (r′, c 2, s 2) (with r′ ∈ R∗) in G, as long as the four entries involved in the intercalate do not conflict with other entries in G on rows in R∗.Let X be the number of half-intercalates. We will apply Theorem 3.1 to show concentration of X.First note that E[X] ≥ βn 2 2 (βn − ℓ)( εα/n )2(1 − α/n )8n−8 ≥ 10 −1ε2α2(βn − ℓ)β4n2e−16 α ≥ 10 −2ε2α2β5n3. To see this, recall that \|R \ R∗\| ≥ βn − ℓ and each of R, C , C ∗, S , S ∗ have size at least βn ; also, each pair of entries is present in H ∩ G with probability (εα/n )2, it has no conflicts with other entries in G with probability at least (1 − α/n )6n−6, and it has no second-order conflicts with probability at least (1 − α/n )2n−2.Recall that we work with the independent random variables gh e and ge for each of the O(n3) many potential hyperedges e outside the rows in R∗, and each of these random variables is 1 with probability O(1 /n ). The appearance of a hyperedge in H ∩ G can increase X by at most n − 1, since each entry can be in at most n − 1 half-intercalates. The appearance of a hyperedge in G can decrease X by at most 4n, since it can destroy at most 3 hyperedges in G∗ due to conflicts (and each of them could be in at most n − 1 pairs counted by X), and can introduce at most n second-order conflicts. Thus, the effect of changing any individual random variable (of the form gh e or ge) is O(n). By Theorem 3.1, with probability 1 − exp( −Ω( ε4n2)) , we have X ≥ 10 −3ε2α2β5n3. For the rest of the proof, we condition on an outcome of the first phase such that this is the case (i.e., in the rest of the proof, all probabilistic considerations are implicitly with respect to this conditional probability space). Now, in the second phase we reveal G and H ∩ G in rows in R∗, and estimate the size Z of the maximum family of disjoint (R, R∗, C , C ∗, S , S ∗)-split intercalates in H ∩ G∗. Note that each half-intercalate counted by X forms an intercalate with two other entries on row r ∈ R∗ if and only if these two other entries are present in H ∩ G and no entry in G in a row in R∗ conflicts with any of the four entries of the intercalate. We can compute a lower bound on E[Z] by subtracting the expected number of pairs of intersecting split intercalates from the expected number of split intercalates. Indeed, E[Z] ≥ Xℓ(εα/n )2(1 − α/n )8n−8 − Xnℓ 2(εα/n )4 − Xnℓ (εα/n )3 ≥ X ℓα 2ε2 2n2 ≥ 10 −4α4β5ε4nℓ, where the term (1 − α/n )8n−8 accounts for the probability that some entry in a fixed potential split intercalate conflicts with some entry in a row in R∗. The second term bounds the expected number of pairs of split intercalates that intersect in an entry outside of R∗, since for a fixed half-intercalate, there are at most n choices for the other column of the other intercalate and at most ℓ choices for the other row of each intercalate, and for each such combination of choices, four entries need to be present in rows in R∗. The third term bounds the expected number of pairs of split intercalates that intersect in an entry in R∗, since for a fixed half-intercalate, there are at most ℓ choices for the common to both intercalates row in R∗ and at most n choices for the other row of the other intercalate, which is outside of R∗ (and this determines the entries of both intercalates completely). We again apply Theorem 3.1 with the remaining unrevealed random variables gh e and ge. Since we are interested in a family of disjoint intercalates, the appearance of a hyperedge can increase Z by at most 1 and can decrease it by at most 3 (as it can have a conflict with at most 3 edges in G∗). Crucially, there are now only O(n2ℓ) many unrevealed random variables. Thus, by Theorem 3.1, we have that Z ≥ 10 −5α4β5ε4nℓ with probability at least 1 − exp( −Ωα,β (ε8nℓ )) .We conclude the proof with a union bound over all (at most 26n) choices for an (ℓ, β )-permissible 6-tuple (R, R∗, C , C ∗, S , S ∗). □ Upper-bounding the number of entries in bad configurations We now turn to the proof of Lemma 9.13. For any set of ℓ rows R∗, our goal is to upper bound the probability that there are many entries in rows in R∗ covered by R∗-split bad configurations. 11.1. Setup. As in the proof of Lemma 10.2, the special role of the rows in R∗ requires us to consider two “phases”. In the first phase we reveal a subset of the entries of our random Latin square, and study configurations of entries which are “almost” bad configurations (they are just missing some entries from R∗). In the second phase we reveal the rest of our random Latin square, and we study which of our 27 “almost bad configurations” give rise to actual bad configurations. However, this breakdown into phases is much more complicated than for Lemma 10.2 (for several different reasons), and requires some setup. Definition 11.1. Given a R∗-split bad configuration F , we say that the special entries of F are its two entries which belong to rows in R∗. (Technically, it is possible that there is more than one split intercalate in a bad configuration, or that the bad configuration contains an intercalate with both its rows in R∗. In these cases we imagine multiple “copies” of the bad configuration, each time making a different choice for the two special entries). Also, we say that the intercalate in F which contains the special entries is the special intercalate . Definition 11.2. Given a partial Latin square P and a set of rows R∗, consider a pair {e1, e 2} of row/column/symbol triples, belonging to some row in R∗, which are not already present in P but which can be added to P without causing conflicts (i.e., P ∪ { e1, e 2} is a partial Latin square). Say that {e1, e 2} is a threatened pair for P if there is some split bad configuration F in P ∪ { e1, e 2} which has e1 and e2 as special entries. In this case, we say F ∩ P is a threat configuration in P .Informally speaking, threat configurations are the sets of entries which are “in danger of being com-pleted to a split bad configuration”, only requiring the addition of their two special entries. Each of the possible ways to complete a threat configuration to a split bad configuration is described by a threat-ened pair. A single threatened pair may complete multiple different threat configurations to split bad configurations, and a single threat configuration may be completeable by multiple different threatened pairs (to different split bad configurations). Now, given the above terminology, we can be a bit more concrete about the high-level plan to prove Lemma 9.13. In a first phase, we reveal part of our random Latin square and upper-bound the number of threatened pairs in it (we do not upper bound the number of threat configurations themselves, cf. Remark 9.5). Then, in a second phase, we reveal the rest of the Latin square and upper bound the number of threatened pairs from the first phase that actually appear as entries in the second phase. By an averaging argument (over all the ways to split into two phases) we are able to deduce the desired bound on the number of entries in split bad configurations. For the first phase, we consider the following property. Definition 11.3. Recall the definitions of ℓ and ε from Definition 9.1. For H ⊆ K(3) n,n,n and C, α, ϕ > 0,let T threat (H, C, α, ϕ ) ⊆ L n be the set of Latin squares L such that the following holds. If we take a random subset Pαn 2 (L) ∈ P n,αn 2 of αn 2 entries in L, then with probability at least ϕ, for some set of rows R∗ of size \|R∗\| = ℓ, the number of threatened pairs for H ∩ Pαn 2 (L) is more than Cε 3n3ℓ.Note that here we are defining a property of Latin squares in terms of how their random subsets of entries behave. This may seem rather unwieldy, but it is very convenient for the ρ-inheritedness assumption in Lemma 4.7 (when it comes time to apply Lemma 4.7, the ρ-inheritedness assumption will hold basically by definition). The issue with more straightforward properties is that bad configurations and threatened pairs do not behave well under subsampling entries, and therefore ρ-inheritedness would be quite difficult to prove. (The issue is that a bad configuration is defined in terms of isolated intercalates, and deleting entries can cause more intercalates to be isolated 14 ). Now, the following two lemmas correspond to the two “phases” informally discussed above. Lemma 11.4. There is an absolute constant C′ > 0 such that the following holds. Fix constants α, ϕ > 0, and consider a random hypergraph H ∼ G(3) (n, ε ) and an independent random Latin square L ∼ Unif( Ln). We have P[L ∈ T threat (H, C ′, α, ϕ )] ≤ exp( −ω(n log 2 n)) . Lemma 11.5. Fix constants α, C ′, C, ϕ > 0, such that α is sufficiently small, C is sufficiently large in terms of α and C′, and ϕ is sufficiently small in terms of α. Consider a random hypergraph H ∼ G(3) (n, ε ) and an independent random Latin square L ∼ Unif( Ln). We have P[L ∈ T bad (H, C ) \ T threat (H, C ′, α, ϕ )] ≤ exp( −ω(n log 2 n)) . Lemmas 11.4 and 11.5 are both proved using Lemma 4.7, comparing random Latin squares to the triangle removal process. In addition, the proof of Lemma 11.4 also involves a switching argument directly 14 The reader may then wonder why we insist on isolatedness in the definition of a bad configuration. The reason is that it is very convenient to restrict our attention to isolated intercalates in a switching argument which will appear later in this section. 28 on random Latin squares (to tame the complexity of all the different possibilities for the structure of a threat configuration). We conclude this subsection with the brief deduction of Lemma 9.13 from Lemmas 11.4 and 11.5. Proof of Lemma 9.13. Let C′ be as in Lemma 11.4, and then let α, C, ϕ be as in Lemma 11.5 (for the same value of C′). Then we have P[L ∈ T bad (H, C )] ≤ P[L ∈ T bad (H, C ) \ T threat (H, C ′, α, ϕ )] + P[L ∈ T threat (H, C ′, α, ϕ )] ≤ exp( −ω(n log 2 n)) + exp( −ω(n log 2 n)) by Lemmas 11.4 and 11.5. □ In the next three subsections, we will prove Lemmas 11.4 and 11.5 (in reverse order). 11.2. Few bad configurations or many threatened pairs. In this subsection we prove Lemma 11.5. As previously mentioned, we will work with the triangle removal process, via Lemma 4.7. In contrast to our previous application of Lemma 4.7 in Section 10, this time the order of the edges will be important. Definition 11.6. Let ⃗ P ∈⃗ Pn be an ordered partial Latin square, and write e1, . . . , e m for its entries (in order). For ι, κ ∈ [0 , 1] , we write ⃗ P [ι, κ ] ∈ P n, (κ−ι)m for the partial Latin square consisting of the edges ei with ιm < i ≤ κm .Recall the definitions of ℓ and ε from Definition 9.1. Definition 11.7. Let α ∈ (0 , 1) , C > 0, and H ⊆ K(3) n,n,n . Let ⃗ Uthreat (H, C ) ⊆⃗ Pn be the set of partial ordered Latin squares ⃗ P such that for some set of rows R∗ of size ℓ = \|R∗\|, the number of threatened pairs for H ∩⃗ P [0 , 1/2] is more than Cε 3n3ℓ. Definition 11.8. We call a bad configuration consistent with ⃗ P if its corresponding threat configuration is in ⃗ P [0 , 1/2] , one of its special entries is in ⃗ P [1 /2, 3/4] , and its other special entry is in ⃗ P [3 /4, 1] . Let ⃗ Ubad (H, C ) ⊆⃗ Pn be the set of partial ordered Latin squares ⃗ P such that for some set of rows R∗ of size ℓ = \|R∗\|, there are more than Cε 5nℓ entries in ⃗ P [3 /4, 1] that are in rows in R∗ and are covered by R∗-split bad configurations in H ∩⃗ P that are consistent with ⃗ P .The following lemma is a version of Lemma 11.5 for the triangle removal process. Lemma 11.9. Fix constants α, C ′, such that α is sufficiently small. Consider independent random hypergraphs ⃗ R ∼ TRP( n, αn 2) and H ∼ G(3) (n, ε ). Then P[⃗ R ∈⃗ Ubad (H, C ′) \⃗ Uthreat (H, C ′)] ≤ exp( −ω(n log 2 n)) . Before proving Lemma 11.9, we show how it implies Lemma 11.5, using Lemma 4.7 and an averaging argument. Proof of Lemma 11.5. Let m = 2 αn 2 and ρ = α16 /24. We assume that ϕ ≤ ρ and ρC ≥ C′.By Lemma 11.9 and Fact 9.4(1), with probability 1 − exp( −ω(n log 2 n)) over the randomness of H we have P[⃗ R ∈⃗ Ubad (H, C ′) \⃗ Uthreat (H, C ′) \| H] ≤ exp( −ω(n log 2 n)) . The desired result would therefore follow from Lemma 4.7 and Fact 9.4(2), if we knew that ⃗ Ubad (H, C ′) \⃗ Uthreat (H, C ′) is (ρ, m )-inherited from T bad (H, C ) \ T threat (H, C ′, α, ϕ ) for all outcomes H of H. For the rest of the proof, our goal is to prove that this is the case. Suppose L ∈ T bad (H, C ) \ T threat (H, C ′, α, ϕ ), and let ⃗ Pm(L) ∈⃗ Pn,m be a random ordering of m random entries of L. We need to prove that ⃗ Pm(L) ∈⃗ Ubad (H, C ′) \⃗ Uthreat (H, C ′) with probability at least ρ.By the definition of T threat (H, C ′, α, ϕ ) and the fact that ⃗ Pm(L)[0 , 1/2] is a uniformly random set of αn 2 entries of L, first note that P[⃗ Pm(L) ∈⃗ Uthreat (H, C ′)] ≤ ϕ. So, it suffices to show that ⃗ Pm(L) ∈⃗ Ubad (H, C ′) with probability at least ρ + ϕ.Since L ∈ T bad (H, C ), there is some R∗ such that there are X ≥ Cε 5nℓ many entries in rows in R∗ covered by R∗-split bad configurations in H ∩ L. Let Y ≤ X be the number of such entries in ⃗ Pm(L)[3 /4, 1] which are covered by R∗-split bad configurations in H ∩⃗ Pm(L) consistent with ⃗ Pm(L).For each covered entry (r, c, s ) in H ∩ L counted by X, the probability that its associated split bad 29 configuration with s ≤ 16 entries is present in ⃗ Pm(L) is n2−sm−s /n2 m = (1 + o(1))(2 α)s. Given that it is present in ⃗ Pm(L), the probability that it is consistent with ⃗ Pm(L) and that (r, c, s ) ∈⃗ Pm(L)[3 /4, 1] is m/ 4 m · m/ 4 m − 1 · m/ 2 s − 2 / m − 2 s − 2 = (1 + o(1))(1 /4) 2(1 /2) s−2. Note that each isolated intercalate in H ∩ L whose entries are in ⃗ Pm(L) is also isolated in H ∩⃗ Pm(L), so each bad configuration in H ∩ L that is present in H ∩⃗ Pm(L) is also a bad configuration in H ∩⃗ Pm(L).Therefore E[Y] ≥ (1 + o(1)) X(2 α)s(1 /4) 2(1 /2) s−2 ≥ Xα 16 /4. Recall our assumptions on ϕ, C from the start of the proof. Let δ = α16 /8 ≥ ρ + ϕ and note that δX ≥ C′ε5nℓ . By Markov’s inequality, P[⃗ Pm(L) ∈⃗ Ubad (H, C ′)] is at least P[Y ≥ δX ] = 1 − P[X − Y > (1 − δ)X] ≥ 1 − EX − YX ≥ 1 − 1 − 2δ 1 − δ ≥ ρ + ϕ, as desired. □ Now we prove Lemma 11.9. Proof of Lemma 11.9. First, we reveal H ∩⃗ R[0 , 1/2] . We assume that for each choice of R∗ of size ℓ,the number of threatened pairs for H ∩⃗ R[0 , 1/2] is at most C′ε3n3ℓ (otherwise ⃗ Uthreat (H, C ′) holds, and there is nothing to prove). From now on, when we talk about threatened pairs in this proof, we mean threatened pairs for H ∩⃗ R[0 , 1/2] .Given revealed information, our goal is to show that with probability 1 − exp( −ω(n log 2 n)) , for every R∗ with \|R∗\| = ℓ, there are at most C′ε5nℓ entries of ⃗ R[3 /4, 1] that are in rows in R∗ and are covered by R∗-split bad configurations in H ∩⃗ R that are consistent with ⃗ R. Actually, it suffices to show this for a fixed choice of R∗ (a union bound over choices of R∗ has a negligible impact). So, fix some R∗ with \|R∗\| = ℓ.The next step is to consider H ∩⃗ R[1 /2, 3/4] . We say a threatened pair survives if one of its entries is in H ∩⃗ R[1 /2, 3/4] . Claim. Given revealed information, the number of surviving threatened pairs is at most C′ε4n2ℓ with probability 1 − exp( −ω(n log 2 n)) .Proof of claim. Let U be the set of G ⊆ K(3) n,n,n which intersect more than C′ε4n2ℓ threatened pairs. Then, for each possible outcome H of H, the set {G ⊆ K(3) n,n,n : H ∩ G ∈ U} is a monotone increasing property of subgraphs of K(3) n,n,n , so by Lemma 5.3, the probability that there are more than C′ε4n2ℓ surviving threatened pairs is P[H ∩⃗ R[1 /2, 3/4] ∈ U ] ≤ P[H ∩ G′ ∈ U ] + exp( −Ω( n2)) = P[G ∈ U ] + exp( −Ω( n2)) , where G′ ∼ G(3) (n, (α/ 2) /n ) and G ∼ G(3) (n, (εα/ 2) /n ).From here, we just need to apply a concentration inequality in G, to upper bound P[G ∈ U ]. For each row/column/symbol triple e, let we be the number of threatened pairs that e is in, and let X be the sum of we over all hyperedges e of G. Note that (for small α > 0) we have E[X] ≤ 2 · C′ε3n3ℓ · εα/ 2 n ≤ C′ε4n2ℓ 2 . Then, for any row/column/symbol triple e, consider each of the threatened pairs that e is in. Each of these threatened pairs, if added to ⃗ R[1 /2, 1] , would complete a split bad configuration in which e would be in a different intercalate. Since a single entry can be in at most n many intercalates (in any Latin square), it follows that e is in at most n threatened pairs; that is, we ≤ n.We can therefore apply Corollary 3.2 with ∆ = n and δ = 1 , to obtain P[G ∈ U ] ≤ P X ≥ E[X] + C′ε4n2ℓ 2 ≤ exp( −Ω( ε4nℓ )) , and the desired result follows (recalling the relationship between ℓ and ε in Definition 9.1). ■ 30 Now, reveal H∩⃗ R[1 /2, 3/4] ; by the above claim we may assume that our revealed outcome is such that there are at most C′ε4n2ℓ surviving threatened pairs. Then, the final step is to consider H ∩⃗ R[3 /4, 1] .Each surviving threatened pair’s presence in H ∩⃗ R is determined by whether its second entry is in H ∩⃗ R[3 /4, 1] . Say a row/column/symbol triple e is a threatened entry if it is the second entry of some surviving threatened pair. Claim. Given revealed information, there are at most C′ε5nℓ threatened entries present in H∩⃗ R[3 /4, 1] ,with probability 1 − exp( −ω(n log 2 n)) .Proof of claim. We proceed very similarly to the last claim (in fact, the situation is even simpler). Let G ∼ G(3) (n, (αε/ 2) /n ), and let Y be the number of threatened entries present in G. Using Lemma 5.3 as before, the probability that there are more than C′ε5nℓ threatened entries present in H ∩⃗ R[3 /4, 1] is at most P[Y > C ′ε5nℓ ] + exp( −Ω( n2)) .We have E[Y] ≤ C′ε4n2ℓ · αε/ 2 n ≤ C′ε5nℓ 2 , so by Corollary 3.2 (with ∆ = 1 ), we have P[Y > C ′ε5nℓ ] ≤ exp( −Ω( ε5nℓ )) ; the desired result follows. ■ Now, if e is an entry of ⃗ R[3 /4, 1] that is in a row in R∗ and is covered by a split bad configuration in H ∩⃗ R consistent with ⃗ R, then e was a threatened entry which happened to be present in H ∩⃗ R[3 /4, 1] (the converse may not hold, as bad configurations in subsets of H ∩⃗ R are not always bad configurations in H ∩⃗ R itself). So, the desired result follows from the above claim. □ 11.3. Upper-bounding the number of threatened pairs: setup and switching. It remains to prove Lemma 11.4, which amounts to a careful study of threat configurations in a random subset of a random Latin square. Recall that threat configurations are sets of entries that are “two entries away” from a bad configu-ration, and there are a very large number of possibilities for the structure of a bad configuration (e.g., critical configurations can have different numbers of intercalates, they can require different subsets of intercalates to be switched to create a new intercalate, and they can have different rows in R∗). To tame this complexity, one crucial observation is that for every bad configuration in a partial Latin square, there is some way to switch isolated intercalates to obtain a pair of intersecting intercalates. So, with a switching argument (unrelated to the main switching argument in Section 8), we can reduce the number of cases significantly: it suffices to consider bad configurations that correspond to pairs of intersecting intercalates, and bad configurations that are “one switch away” from a pair of intersecting intercalates. Definition 11.10. Given a partial Latin square P ∈ P n and a set of rows R∗, we define a basic split bad configuration to be a set F of seven entries in P , such that F contains an intercalate A with a row in R∗, and either F or (F \ A) ∪ ¯A is a union of two intersecting intercalates. Then, say that a pair {e1, e 2}, belonging to some row in R∗, is a basic threatened pair for P if there is some basic split bad configuration F in P ∪ { e1, e 2} with e1, e 2 ∈ F . In this case, we say F \ { e1, e 2} is a basic threat configuration in P .There are four cases for the structure of a basic threatened pair/basic threat configuration; see Figure 1. Note that we make no isolatedness assumptions in the above definition, so technically it is possible that a basic split bad configuration is not actually a split bad configuration (or that a basic threatened pair is not actually a threatened pair, or that a basic threat configuration is not actually a threat configuration). However, we still borrow the same terminology (e.g., we may talk about the special entries of a basic split bad configuration). c1 c2 c3 r∗ 1 s1 s2 r2 s2 s1 s3 r3 s3 s2 c1 c2 c3 r∗ 1 s2 s1 r2 s1 s2 s3 r3 s3 s2 c1 c2 c3 r1 s1 s2 r∗ 2 s2 s1 s3 r3 s3 s2 c1 c2 c3 r1 s2 s1 r∗ 2 s1 s2 s3 r3 s3 s2 Figure 1. The four possibilities for the structure of a basic threat configuration, each illustrated with a basic threatened pair in orange . In each case, rows in R∗ are marked with a star. (Rows not marked with a star may or may not be in R∗.) 31 Fact 11.11. Consider any partial Latin square R ∈ P n and set of rows R∗, and let R′ be obtained from R by switching each isolated intercalate with probability 1/2, independently. Let {e1, e 2} be any threatened pair for R. Then, {e1, e 2} is a basic threatened pair for R′, with probability at least (1 /2) 5. Proof. There is some threat configuration F0 in R corresponding to the threatened pair {e1, e 2}. By the definition of a split bad configuration, in R ∪ { e1, e 2}, there is a way to switch isolated intercalates to introduce a pair of intersecting intercalates containing {e1, e 2}. If we do the same switches in R (except possibly the switch of the special intercalate, which may not actually exist in R), we make {e1, e 2} abasic threatened pair. See Figure 2 for an example. Now, F0 intersects at most five isolated intercalates in R (it is clear that F0 contains at most three iso-lated intercalates, but the other entries in F0 could be contained in totally different isolated intercalates). The desired result follows. □ We next record a lemma saying that there are typically not many entries covered by basic threatened pairs. Recall the definitions of ℓ and ε from Definition 9.1. Definition 11.12. For C > 0 and a hypergraph H ⊆ K(3) n,n,n , let T basic (H, C ) ⊆ L n be the set of Latin squares L such that for some set of rows R∗ of size \|R∗\| = ℓ, the number of basic threatened pairs in H ∩ L is more than Cε 3n3ℓ. Lemma 11.13. There is an absolute constant C > 0 such that the following holds. Consider a random hypergraph H ∼ G(3) (n, ε ) and an independent random Latin square L ∼ Unif( Ln). We have P[L ∈ T basic (H, C )] ≤ exp( −Ω( ε3n2)) . We defer the proof of Lemma 11.13 to Section 11.4. First, we show how to use a switching argument (with Fact 11.11) to deduce Lemma 11.4. Note that, as we will see, this deduction crucially requires that the error probability in Lemma 11.13 is exp( −Ω( ε3n2)) rather than our usual exp( −ω(n log 2 n)) . We also need an upper bound on the size of the largest family of pairwise disjoint intercalates, as follows. Definition 11.14. Let T int upper (H) ⊆ L n be the set of Latin squares L for which there is a family of more than 80 ε4n2 disjoint intercalates in H ∩ L. Lemma 11.15. Consider a random hypergraph H ∼ G(3) (n, ε ) and an independent random Latin square L ∼ Unif( Ln). We have P[L ∈ T int upper (H)] ≤ exp( −ω(n log 2 n)) . We prove Lemma 11.15 with an averaging argument and Lemma 4.7 (this is a subset of the ideas we have already seen in Sections 10 and 11.2). Indeed, the following lemma is a version of Lemma 11.15 for the triangle removal process. Definition 11.16. Let ⃗ Uint upper (H, α ) ⊆⃗ Pn be the set of partial ordered Latin squares ⃗ P for which there is a family of more than 20 α4ε4n2 disjoint intercalates in H ∩⃗ P . Lemma 11.17. Let α ∈ (0 , 1) be a sufficiently small constant. Consider independent random hypergraphs ⃗ R ∼ TRP( n, αn 2) and H ∼ G(3) (n, ε ). Then we have P[⃗ R ∈⃗ Uint upper (H, α )] ≤ exp( −ω(n log 2 n)) .c1 c2 c3 c4 c5 c6 r∗ 1 s2 s1 r2 s3 s4 r3 s1 s2 s4 s3 r4 s3 s5 s6 s2 r5 s5 s3 r6 s2 s6 → c1 c2 c3 c4 c5 c6 r∗ 1 s2 s1 r2 s4 s3 r3 s1 s2 s3 s4 r4 s5 s3 s2 s6 r5 s3 s5 r6 s6 s2 Figure 2. On the left, an example of a threat configuration in a partial Latin square (with an example threatened pair in orange ). If we switch the three highlighted in-tercalates, then we obtain the partial Latin square on the right, which contains a basic threat configuration (highlighted). The threatened pair on the left has now become a basic threatened pair on the right. 32 Proof of Lemma 11.15 given Lemma 11.17. Let α > 0 be small enough for Lemma 11.17, let m = αn 2 and let ρ = α4/4.By Lemma 11.17 and Fact 9.4(1), with probability 1 − exp( −ω(n log 2 n)) our random hypergraph H ∼ G(3) (n, ε ) satisfies P[⃗ R ∈⃗ Uint upper (H, α ) \| H] ≤ exp( −ω(n log 2 n)) . Let H be such an outcome of H.To show ⃗ Uint upper (H, α ) is (ρ, m )-inherited from T int upper (H), let L ∈ T int upper (H), so there is a family of X ≥ 80 ε4n2 disjoint intercalates in H ∩ L. Let Y be the number of these intercalates which lie in a random ordered m-subset ⃗ Pm(L) of L. We have E[Y] = X · (1 + o(1)) α4 ≥ α4X/ 2. By Markov’s inequality, P[⃗ Pm(L) ∈⃗ Uint upper (H, α )] ≥ P Y ≥ α4 4 X = 1 − P X − Y > 1 − α4 4 X ≥ 1 − 1 − α4/21 − α4/4 ≥ ρ. That is to say, ⃗ Uint upper (H, α ) is (ρ, m )-inherited from T int upper (H). Thus, by Lemma 4.7, P[L ∈ T int upper (H)] ≤ exp(2 n log 2 n)P[⃗ R ∈⃗ Uint upper (H, α )] = exp( −ω(n log 2 n)) . Recalling our choice of H, Fact 9.4(2) concludes the proof. □ Proof of Lemma 11.17. For each H ⊆ K(3) n,n,n , note that ⃗ Uint upper (H, α ) is a monotone increasing property. So, by Lemma 5.3 (applied to each possible outcome H of H), we have P[H∩⃗ R ∈⃗ Uint upper (H, α )] ≤ P[H∩G′ ∈⃗ Uint upper (H, α )]+exp( −Ω( n2)) = P[G ∈⃗ Uint upper (H, α )]+exp( −Ω( n2)) , where G′ ∼ G(3) (n, 2α/n ) and G ∼ G(3) (n, 2εα/n ). Let X be the size of the maximum family of disjoint intercalates in G. We have E[X] ≤ n6(2 αε/n )4 = 16 α4ε4n2, and changing an edge of G changes X by at most 1 (since an entry can be part of at most one intercalate in a maximum disjoint family). By Theorem 3.1, P[G ∈⃗ Uint upper (H, α )] is at most P[X > 20 α4ε4n2] ≤ exp − Ω ε8n4 n3 · ε/n + ε4n2 ! = exp( −Ω( ε7n2)) = exp( −ω(n log 2 n)) , and the desired result follows. □ We now deduce Lemma 11.4 from Lemmas 11.13 and 11.15 and Fact 11.11. Proof of Lemma 11.4. Let C be as in Lemma 11.13, and let C′ = 100 C.By Fact 9.4(1) and Lemmas 11.13 and 11.15, with probability 1 − exp( −ω(n log 2 n)) over the random-ness of H, we have P[L ∈ T basic (H, C ) \| H] ≤ exp( −Ω( ε3n2)) , P[L ∈ T int upper (H) \| H] ≤ exp( −ω(n log 2 n)) . Fix such an outcome H ∈ K(3) n,n,n of H, and let S = T threat (H, C ′, α, ϕ ) \ T int upper (H). By Fact 9.4(2), it suffices to show that P[L ∈ S ] ≤ exp( −ω(n log 2 n)) . Let P ∈ P n,αn 2 be a uniformly random subset of αn 2 edges of L and let P′ be obtained from P by switching each isolated intercalate in H ∩ P with probability 1/2 independently. Let L′ = P′ ∪ (L \ P).Let X be the maximum over all choices of R∗ of the number of threatened pairs in H ∩ P, and let Y ≤ X be the maximum over all choices of R∗ of the number of threatened pairs in H ∩ P which become basic threatened pairs in H ∩ P′. By Fact 11.11 with R = H ∩ P (and the fact that an expected maximum is at least the maximum expectation) we have E[Y \| X] ≥ (1 /32) X, so by Markov’s inequality P Y ≥ 1100 X X = 1 − P X − Y > 99 100 X X ≥ 1 − 31 /32 99 /100 ≥ 1100 . Also, by the definition of T threat (H, C ′, α, ϕ ) ⊇ S , we have P[X ≥ C′ε3n3ℓ \| L ∈ S ] ≥ ϕ. Recalling that C′ = 100 C, and letting T = T basic (H, C ), we deduce that P[L′ ∈ T \| L ∈ S ] ≥ ϕ/ 100 or in other words P[L ∈ S ] ≤ 100 ϕ P[L ∈ S and L′ ∈ T ]. 33 So, it suffices to prove that P[L ∈ S and L′ ∈ T ] ≤ exp( −ω(n log 2 n)) .We introduce the notation L1 → L2 to mean that we can obtain L2 from L1 by switching some disjoint intercalates in H ∩ L1. Recall that L′ is obtained from L by switching some intercalates which are isolated in H ∩ P, and these intercalates are certainly disjoint, so for any of the possible outcomes (L1, L 2) of (L, L′) we have L1 → L2. We therefore have P[L ∈ S and L′ ∈ T ] ≤ X L2∈T X L1∈S ,L 1→L2 P[L = L1]P[L′ = L2 \| L = L1] ≤ X L2∈T P[L = L2] X L1∈S ,L 1→L2 1 ≤ X L2∈T P[L = L2] n4 + 80 ε4n2 80 ε4n2 ≤ P[L ∈ T ] exp( O(80 ε4n2 log n)) ≤ exp( −Ω( ε3n2)) · exp( O(80 ε4n2 log n)) ≤ exp( −ω(n log 2 n)) , as desired. Here, in the second inequality we used the fact that P[L = L1] = P[L = L2] (since L has the same probability of being equal to any Latin square). In the third inequality we used the fact that every Latin square has at most n4 intercalates, and the difference between L1 and L2 is described by at most 80 ε4n2 disjoint intercalate switches (note that L1 ∈ S implies that L1 /∈ T int upper (H), i.e., H ∩ L1 has at most 80 ε4n2 disjoint intercalates). Given a set of size N , the number of subsets of size at most k is bounded by N +kk . The fifth inequality is by our choice of H, and the final one uses that ε = o(1 / log n) (recall Definition 9.1). It is vital here that the error probability in Lemma 11.13 is exp( −Ω( ε3n2)) . □ 11.4. Upper-bounding the number of basic threatened pairs. It remains to prove Lemma 11.13. This does not really involve any new ideas. Specifically, the approach is to carefully consider how basic threat configurations can emerge through the triangle removal process, using Lemma 4.7 to relate this to a random Latin square. There are four different cases for the structure of a basic threat configuration (cf. Figure 1), each of which needs to be treated in a slightly different way. We start by giving each of the four cases in Figure 1 a name. Definition 11.18. Consider the four cases for the structure of a basic threat configuration, as follows 15 . c1 c2 c3 r∗ 1 r2 s2 s1 s3 r3 s3 s2 c1 c2 c3 r∗ 1 r2 s1 s2 s3 r3 s3 s2 c1 c2 c3 r∗ 1 s3 r2 s1 s2 r3 s3 s2 c1 c2 c3 r∗ 1 s3 r2 s2 s1 r3 s3 s2 These four cases describe four partial Latin squares Q1, Q 2, Q 3, Q 4 ∈ P 3 (from left to right). Each Qt has three rows, three columns and three symbols (though in Q1 and Q2, one of the rows does not have any entries in it). For each type t, and any partial Latin square P ∈ P n, an embedding from Qt into P is an injective map from the 9 vertices of Qt into the vertices of P (where row-vertices are mapped to row-vertices, column-vertices are mapped to column-vertices, and symbol-vertices are mapped to symbol-vertices), such that the image of every hyperedge of Qt is a hyperedge of Pt.Then, for a partial Latin square P ∈ P n, note that a basic threatened pair (together with an associated basic threat configuration) is specified by an embedding of some Qt into P , where we demand that r∗ 1 is mapped into R∗. In this case we say that the basic threatened pair has type t.Note that a basic threatened pair can have multiple types (if it has multiple associated basic threat configurations). Definition 11.19. For some t ∈ { 1, 2, 3, 4}, let π = ( e1, . . . , e 5) be an ordering of the entries of Qt, and let ϕ(Qt) be an embedding of Qt into an ordered partial Latin square ⃗ P ∈⃗ Pn. We say that ϕ(Qt) is π-consistent with ⃗ P if ϕ(ei) ∈⃗ P [( i − 1) /5, i/ 5] for all i ∈ { 1, . . . , 5}. We say that a basic threatened pair of type t is π-consistent with ⃗ P if it is associated with some π-consistent embedding ϕ(Qt).Now, we state a version of Lemma 11.13 for the triangle removal process. Recall the definitions of ℓ and ε from Definition 9.1. 15 These are exactly as in Figure 1, but in the last two cases we have swapped the first and second rows. 34 Definition 11.20. For a hypergraph H, a type t ∈ { 1, 2, 3, 4} and an ordering π of the entries of Qt,let ⃗ Ubasic (H, t, π ) ⊆⃗ Pn be the set of partial ordered Latin squares ⃗ P such that for some set of rows R∗ of size \|R∗\| = ℓ, the number of type-t basic threatened pairs in H ∩⃗ P which are π-consistent with ⃗ P is more than ε3n3ℓ. Lemma 11.21. Fix a sufficiently small constant α > 0, and any t ∈ { 1, 2, 3, 4}. There is an ordering π of Qt such that for independent random hypergraphs ⃗ R ∼ TRP( n, αn 2) and H ∼ G(3) (n, ε ), we have P[⃗R ∈⃗ Ubasic (H, t, π )] ≤ exp( −Ω( ε3n2)) . Before proving Lemma 11.21, we show how to deduce Lemma 11.13. Proof of Lemma 11.13. Let α ∈ (0 , 1) be small enough for Lemma 11.21, let m = αn 2 and let ρ = α5/57,and assume C ≥ 4(5 7/α 5). For each t, let πt be the ordering in Lemma 11.21. By Lemma 11.21 and Fact 9.4(1), with probability 1 − exp( −Ω( ε3n2)) over the randomness of H, we have that for each t ∈ { 1, 2, 3, 4}, P[⃗ R ∈⃗ Ubasic (H, t, π t) \| H] ≤ exp( −Ω( ε3n2)) . (11.1) Fix such an outcome H ⊆ K(3) n,n,n of H.Let ⃗ Ubasic (H) = S4 t=1 ⃗ Ubasic (H, t, π t). Given an ordered partial Latin square ⃗ P and a set of rows R∗,we say a type-t basic threatened pair is consistent if it is πt-consistent. To show ⃗ Ubasic (H) is (ρ, m )-inherited from T basic (H, C ), let L ∈ T basic (H, C ), so there is R∗ of size \|R∗\| = ℓ for which there are X > Cε 3n3ℓ many basic threatened pairs in H ∩ L. Let Y be the number of basic threatened pairs in H ∩ L which are present and consistent in an ordered random m-subset ⃗ Pm(L) of L. We have E[Y] = X · (1 + o(1)) α5(1 /5) 5 ≥ α5 56 X. Now, if Y ≥ (α5/57)X > 4ε3n3ℓ then ⃗ Ubasic (H, t, π t) holds for some t. Using Markov’s inequality, it follows that P[⃗ Pm(L) ∈⃗ Ubasic (H)] ≥ P Y ≥ α5 57 X = 1 − P X − Y > 1 − α5 57 X ≥ 1 − 1 − (α5/56)1 − (α5/57) ≥ ρ, which means ⃗ Ubasic (H) is (ρ, m )-inherited from T basic (H, C ).Thus, we have P[L ∈ T basic (H, C )] ≤ exp(2 n log 2 n)P[⃗ R ∈⃗ Ubasic (H)] ≤ exp( −Ω( ε3n2)) , where the first inequality follows by Lemma 4.7, and the second inequality follows by (11.1). Recalling our choice of H, Fact 9.4(2) implies that P[L ∈ T basic (H, C )] ≤ exp( −Ω( ε3n2)) , as desired. □ Now we prove Lemma 11.21, separately considering types 1 and 2, and types 3 and 4. Proof of Lemma 11.21 for types t = 1 and t = 2 . Recall that Q1 and Q2 have the following forms. c1 c2 c3 r∗ 1 r2 s2 s1 s3 r3 s3 s2 c1 c2 c3 r∗ 1 r2 s1 s2 s3 r3 s3 s2 If t = 1 , we take our ordering π to be e1 = ( r3, c 3, s 2), e 2 = ( r2, c 3, s 3), e 3 = ( r3, c 2, s 3), e 4 = ( r2, c 2, s 1), e 5 = ( r2, c 1, s 2). If t = 2 , we take our ordering π to be e1 = ( r3, c 3, s 2), e 2 = ( r2, c 3, s 3), e 3 = ( r3, c 2, s 3), e 4 = ( r2, c 2, s 2), e 5 = ( r2, c 1, s 1). Fix t ∈ { 1, 2} and let R∗ be a set of rows with \|R∗\| = ℓ. Among all the π-consistent embed-dings ϕ(Qt) of Qt into H ∩⃗ R which satisfy ϕ(r∗ 1 ) ∈ R∗, let N be the number of possibilities for (ϕ(r∗ 1 ), ϕ (c1), ϕ (c2), ϕ (s1), ϕ (s2)) . Then, N is an upper bound on the number of type-t basic threatened pairs in H ∩⃗ R which are π-consistent with ⃗ R. 35 We first reveal ⃗ R[0 , 1/5] , which has αn 2/5 entries. Each of them is present in H with probability ε independently. Thus, by a Chernoff bound, with probability 1 − exp( −Ω( εn 2)) , there are at most εn 2 many viable choices for ϕ(e1) (and therefore (ϕ(r3), ϕ (c3), ϕ (s2)) ). Assume this is the case. Next, we reveal H ∩⃗ R[1 /5, 3/5] . For each choice of (ϕ(r3), ϕ (c3), ϕ (s2)) , there are at most n choices for ϕ(e2) in column ϕ(c3) of ⃗ R[1 /5, 2/5] . This determines ϕ(r2) and ϕ(s3), and then there is at most one occurrence of ϕ(s3) in row ϕ(r3) of ⃗ R[2 /5, 3/5] , determining ϕ(c2). This makes a total of at most εn 3 choices for (ϕ(r2), ϕ (r3), ϕ (c2), ϕ (c3), ϕ (s2), ϕ (s3)) .Let S be the set of all possible choices for (ϕ(r2), ϕ (c1), ϕ (c2), ϕ (s1), ϕ (s2)) at this stage, so we have \|S\| ≤ εn 5 (there are at most n2 choices for (ϕ(c1), ϕ (s1)) ). Let S4 ⊆ S be the set of all (ϕ(r2), ϕ (c1), ϕ (c2), ϕ (s1), ϕ (s2)) ∈ S for which ϕ(e4) is an edge of H ∩⃗ R[3 /5, 4/5] , and let S5 ⊆ S4 be the set of all such tuples for which ϕ(e5) is an edge of H ∩⃗ R[4 /5, 1] . Note that N ≤ ℓ\|S5\| (there are at most ℓ choices for ϕ(r∗ 1 ) ∈ R∗). We next consider H ∩⃗ R[3 /5, 4/5] . We claim that \|S4\| ≤ ε2n4 with probability 1 − exp( −Ω( ε2n2)) .To see this, first note that, by Lemma 5.3, it suffices to prove an analogous bound with H ∩ G ∼ G(3) (n, (2 εα/ 5) /n ) in place of H ∩⃗ R[3 /5, 4/5] , where G ∼ G(3) (n, 2( α/ 5) /n ). But this is a direct consequence of Corollary 3.2, since each potential entry ϕ(e4) is present in H ∩ G with probability at most 2εα/ (5 n), and appears in at most n2 tuples in S.Reveal an outcome of H ∩⃗ R[3 /5, 4/5] satisfying \|S4\| ≤ ε2n4. We next consider H ∩⃗ R[4 /5, 1] : we claim that \|S5\| ≤ ε3n3 with probability 1 − exp( −Ω( ε3n2)) . The proof is basically as before: first note that it suffices to prove an analogous bound with H ∩ G ∼ G(3) (n, (2 εα/ 5) /n ) in place of H ∩⃗ R[4 /5, 1] ,and then the desired result follows from Corollary 3.2 (noting that each potential entry ϕ(e5) appears in at most n tuples in S4, since such a tuple is determined by a choice of ϕ(e4) in the same row). We have proved that the number of type-t basic threatened pairs in H ∩⃗ R which are π-consistent with ⃗ R is at most N ≤ ℓ\|S5\| ≤ ε3n3ℓ, with probability at least 1 − exp( −Ω( ε3n2)) . A union bound over all (at most 2n) choices of R∗ finishes the proof. □ Proof of Lemma 11.21 for types t = 3 and t = 4 . Recall that Q3 and Q4 have the following forms. c1 c2 c3 r∗ 1 s3 r2 s1 s2 r3 s3 s2 c1 c2 c3 r∗ 1 s3 r2 s2 s1 r3 s3 s2 If t = 3 , we take our ordering π to be e1 = ( r3, c 3, s 2), e 2 = ( r∗ 1 , c 3, s 3), e 3 = ( r3, c 2, s 3), e 4 = ( r2, c 2, s 2), e 5 = ( r2, c 1, s 1). If t = 4 , we take our ordering π to be e1 = ( r3, c 3, s 2), e 2 = ( r∗ 1 , c 3, s 3), e 3 = ( r3, c 2, s 3), e 4 = ( r2, c 2, s 1), e 5 = ( r2, c 1, s 2). Fix t ∈ { 3, 4} and let R∗ be a set of rows with \|R∗\| = ℓ. Among all the π-consistent embed-dings ϕ(Qt) of Qt into H ∩⃗ R which satisfy ϕ(r∗ 1 ) ∈ R∗, let N be the number of possibilities for (ϕ(r∗ 1 ), ϕ (c1), ϕ (c2), ϕ (s1), ϕ (s2)) . Then, N is an upper bound on the number of type-t basic threatened pairs in H ∩⃗ R which are π-consistent with ⃗ R.The first part is exactly the same as the previous proof: we first reveal ⃗ R[0 , 1/5] , which has αn 2/5 entries. By a Chernoff bound over the randomness of H, with probability 1 − exp( −Ω( εn 2)) , there are at most εn 2 many viable choices for ϕ(e1) (and therefore (ϕ(r3), ϕ (c3), ϕ (s2)) ). Assume this is the case. Next, we reveal H ∩⃗ R[1 /5, 3/5] . The numbers are slightly different than in the last proof: for each choice of (ϕ(r3), ϕ (c3), ϕ (s2)) , there are at most ℓ choices for ϕ(e2) in column ϕ(c3) of ⃗ R[1 /5, 2/5] (in a row in R∗). This determines ϕ(r∗ 1 ) and ϕ(s3), and then there is at most one occurrence of ϕ(s3) in row ϕ(r3) of ⃗ R[2 /5, 3/5] . This makes a total of at most εn 2ℓ choices for (ϕ(r∗ 1 ), ϕ (r3), ϕ (c2), ϕ (c3), ϕ (s2), ϕ (s3)) .Let S be the set of all possible choices for (ϕ(r∗ 1 ), ϕ (r2), ϕ (c1), ϕ (c2), ϕ (s1), ϕ (s2)) at this stage, so we have \|S\| ≤ εn 5ℓ (there are at most n3 choices for (ϕ(r2), ϕ (c1), ϕ (s1)) ). Let S4 ⊆ S be the set of all (ϕ(r∗ 1 ), ϕ (r2), ϕ (c1), ϕ (c2), ϕ (s1), ϕ (s2)) ∈ S for which ϕ(e4) is an edge of H ∩⃗ R[3 /5, 4/5] , and let S5 ⊆ S4 be the set of all such tuples for which ϕ(e5) is an edge of H ∩⃗ R[4 /5, 1] . Note that N ≤ \| S5\|.The end of the proof is similar to before. We next consider H∩⃗ R[3 /5, 4/5] : we claim that \|S4\| ≤ ε2n4ℓ with probability 1−exp( −Ω( ε2n2)) . To see this, we first note that it suffices to prove an analogous bound 36 with H ∩ G ∼ G(3) (n, (2 εα/ 5) /n ) in place of H ∩⃗ R[3 /5, 4/5] , and then the desired result follows from Corollary 3.2, noting that each potential entry ϕ(e4) appears in at most n2ℓ tuples in S.Finally we consider H ∩⃗ R[4 /5, 1] : we claim that \|S5\| ≤ ε3n3ℓ with probability 1 − exp( −Ω( ε3n2)) .Again, we first note that it suffices to prove an analogous bound with H ∩ G ∼ G(3) (n, (2 εα/ 5) /n ) in place of H ∩⃗ R[4 /5, 1] , and then the desired result follows from Corollary 3.2 (noting that each potential entry ϕ(e5) appears in at most nℓ tuples in S4, since such a tuple is determined by a choice of ϕ(e4) in the same row, and a choice of ϕ(r∗ 1 )). We then finish with a union bound over choices of R∗, as before. □ Concluding remarks In this paper we have proved (a generalisation of) Cameron’s conjecture, describing the joint distri-bution of the number of odd row/column/symbol permutations in a random n × n Latin square. There are various directions for further research, as follows. • Can quantitative aspects be improved? For example, in Theorem 1.3(4), we see that the total variation error of our approximation is o(1) . Actually, our proof gives an error of n−1/2+ o(1) , and it seems this can be improved to n−1+ o(1) with some additional calculations using the fact that E[Nrow (L)] = E[Ncol (L)] = E[Nsym (L)] = n/ 2. However, it seems plausible that the true error of the approximation might be super-exponentially small. • What about constrained distributions on Latin squares (e.g. symmetric Latin squares)? We expect that similar results should hold, but the available enumeration estimates are much weaker. It might be possible to prove analogues of Theorem 1.3(1) and Theorem 1.3(5) (using the ideas described in Section 2.4, which have relatively weak quantitative requirements), but new ideas would be required for the other parts of Theorem 1.3. • What about analogous questions for higher-dimensional analogues of Latin squares (sometimes called high dimensional permutations , see e.g. )? Here it seems completely new ideas would be required, because (we predict that) small switchings become rarer as the dimension increases. References Y. Alimohammadi, P. Diaconis, M. Roghani, and A. Saberi, Sequential importance sampling for estimating expectations over the space of perfect matchings , Ann. Appl. Probab. 33 (2023), no. 2, 799–833. J. Allsop and I. M. Wanless, Subsquares in random Latin rectangles , arXiv:2409.08446. L. Alpoge, Square-root cancellation for the signs of Latin squares , Combinatorica 37 (2017), no. 2, 137–142. B. Bollobás, The chromatic number of random graphs , Combinatorica 8 (1988), no. 1, 49–55. C. Bowtell and R. Montgomery, Almost every Latin square has a decomposition into transversals , arXiv:2501.05438. L. M. Brègman, Certain properties of nonnegative matrices and their permanents , Dokl. Akad. Nauk SSSR 211 (1973), 27–30. P. J. Cameron, Permutations , Paul Erdős and his mathematics, II (Budapest, 1999), Bolyai Soc. Math. Stud., vol. 11, János Bolyai Math. Soc., Budapest, 2002, pp. 205–239. P. J. Cameron, Asymmetric Latin squares, Steiner triple systems, and edge-parallelisms , arXiv:1507.02190. P. J. Cameron, Almost all quasigroups have rank 2, Discrete Math. 106/107 (1992), 111–115. P. J. Cameron, BCC problem list , , 2001. P. J. Cameron, Random Latin squares , slides for a presentation at the RAND-APX meeting at Oxford in 2003, , 2003. P. J. Cameron, Problems from “Permutations” , permutations.html , 2005. P. J. Cameron, Peter Cameron’s BCC problems , ,2007. P. J. Cameron, Problems , , 2013. P. J. Cameron, A niggling problem , posted on Peter Cameron’s blog, 01/24/a-niggling-problem/ , 2015. N. Cavenagh and R. Ramadurai, On the distances between Latin squares and the smallest defining set size , J. Combin. Des. 25 (2017), no. 4, 147–158. N. J. Cavenagh, C. Greenhill, and I. M. Wanless, The cycle structure of two rows in a random Latin square , Random Structures Algorithms 33 (2008), no. 3, 286–309. N. J. Cavenagh and I. M. Wanless, There are asymptotically the same number of Latin squares of each parity , Bull. Aust. Math. Soc. 94 (2016), no. 2, 187–194. M. Delcourt and L. Postle, Refined absorption: A new proof of the existence conjecture , arXiv:2402.17855. A. Dembo and O. Zeitouni, Large deviations techniques and applications , Stochastic Modelling and Applied Proba-bility, vol. 38, Springer-Verlag, Berlin, 2010, Corrected reprint of the second (1998) edition. G. P. Egorychev, The solution of van der Waerden’s problem for permanents , Adv. in Math. 42 (1981), no. 3, 299–305. D. I. Falikman, Proof of the van der Waerden conjecture on the permanent of a doubly stochastic matrix , Mat. Zametki 29 (1981), no. 6, 931–938, 957. 37 W. Feller, An introduction to probability theory and its applications. Vol. I , third ed., John Wiley & Sons, Inc., New York-London-Sydney, 1968. A. Ferber and M. Kwan, Almost all Steiner triple systems are almost resolvable , Forum of Mathematics, Sigma, vol. 8, Cambridge University Press, 2020, p. e39. D. A. Freedman, On tail probabilities for martingales , Ann. Probability 3 (1975), 100–118. B. Friedman and S. McGuinness, The Alon-Tarsi conjecture: a perspective on the main results , Discrete Math. 342 (2019), no. 8, 2234–2253. S. Glock, D. Kühn, A. Lo, and D. Osthus, The existence of designs via iterative absorption: hypergraph F -designs for arbitrary F , Mem. Amer. Math. Soc. 284 (2023), no. 1406, v+131. S. Gould and T. Kelly, Hamilton transversals in random Latin squares , Random Structures Algorithms 62 (2023), no. 2, 450–478. R. Häggkvist and J. C. M. Janssen, All-even Latin squares , Proceedings of the 6th Conference on Formal Power Series and Algebraic Combinatorics (New Brunswick, NJ, 1994), vol. 157, 1996, pp. 199–206. R. Huang and G.-C. Rota, On the relations of various conjectures on Latin squares and straightening coefficients ,Discrete Math. 128 (1994), no. 1-3, 225–236. M. T. Jacobson and P. Matthews, Generating uniformly distributed random Latin squares , J. Combin. Des. 4 (1996), no. 6, 405–437. S. Janson, T. Łuczak, and A. Rucinski, Random graphs , Wiley-Interscience Series in Discrete Mathematics and Opti-mization, Wiley-Interscience, New York, 2000. S. Janson and A. Ruciński, The infamous upper tail , Random Structures Algorithms 20 (2002), no. 3, 317–342, Probabilistic methods in combinatorial optimization. J. C. M. Janssen, On even and odd Latin squares , J. Combin. Theory Ser. A 69 (1995), no. 1, 173–181. A. D. Keedwell and J. Dénes, Latin squares and their applications , second ed., Elsevier/North-Holland, Amsterdam, 2015, With a foreword to the previous edition by Paul Erdös. P. Keevash, The existence of designs , arXiv:1401.3665. P. Keevash, The existence of designs II , arXiv:1802.05900. P. Keevash, A short proof of the existence of designs , arXiv:2411.18291. P. Keevash, Counting designs , J. Eur. Math. Soc. (JEMS) 20 (2018), no. 4, 903–927. M. Kwan, Almost all Steiner triple systems have perfect matchings , Proc. Lond. Math. Soc. (3) 121 (2020), no. 6, 1468–1495. M. Kwan, A. Sah, and M. Sawhney, Note on random latin squares and the triangle removal process , arXiv:2109.15201. M. Kwan, A. Sah, and M. Sawhney, Large deviations in random Latin squares , Bull. Lond. Math. Soc. 54 (2022), no. 4, 1420–1438. M. Kwan, A. Sah, M. Sawhney, and M. Simkin, Substructures in Latin squares , Israel J. Math. 256 (2023), no. 2, 363–416. M. Kwan and B. Sudakov, Intercalates and discrepancy in random Latin squares , Random Structures Algorithms 52 (2018), no. 2, 181–196. N. Linial and Z. Luria, An upper bound on the number of Steiner triple systems , Random Structures Algorithms 43 (2013), no. 4, 399–406. N. Linial and Z. Luria, Discrepancy of high-dimensional permutations , Discrete Anal. (2016), Paper No. 11, 8. N. Linial and M. Simkin, Monotone subsequences in high-dimensional permutations , Combin. Probab. Comput. 27 (2018), no. 1, 69–83. J. Lynch and J. Sethuraman, Large deviations for processes with independent increments , Ann. Probab. 15 (1987), no. 2, 610–627. C. McDiarmid, On the method of bounded differences , Surveys in combinatorics, 1989 (Norwich, 1989), London Math. Soc. Lecture Note Ser., vol. 141, Cambridge Univ. Press, Cambridge, 1989, pp. 148–188. B. D. McKay and I. M. Wanless, Most Latin squares have many subsquares , J. Combin. Theory Ser. A 86 (1999), no. 2, 322–347. A. O. Pittenger, Mappings of Latin squares , Linear Algebra Appl. 261 (1997), 251–268. J. Radhakrishnan, An entropy proof of Bregman’s theorem , J. Combin. Theory Ser. A 77 (1997), no. 1, 161–164. V. Rödl and L. Thoma, Asymptotic packing and the random greedy algorithm , Random Structures Algorithms 8 (1996), no. 3, 161–177. N. Ross, Fundamentals of Stein’s method , Probab. Surv. 8 (2011), 210–293. J. Spencer, Asymptotic packing via a branching process , Random Structures Algorithms 7 (1995), no. 2, 167–172. J. H. van Lint and R. M. Wilson, A course in combinatorics , 2nd ed., Cambridge University Press, Cambridge, 2001. L. Warnke, On the method of typical bounded differences , Combinatorics, Probability and Computing 25 (2016), no. 2, 269–299. P. Zappa, Triplets of Latin squares , Boll. Un. Mat. Ital. A (7) 10 (1996), no. 1, 63–69. Institute of Science and Technology Austria (ISTA). Am Campus 1, 3400 Klosterneuburg, Austria Email address : matthew.kwan@ist.ac.at Institute of Science and Technology Austria (ISTA). Am Campus 1, 3400 Klosterneuburg, Austria Email address : kalina.petrova@ist.ac.at Department of Mathematics, Columbia University, New York, NY 10027 Email address : m.sawhney@columbia.edu
13361	https://drc.bmj.com/content/9/1/e002300	Published Time: 2021-07-26 Management of type 2 diabetes in chronic kidney disease \| BMJ Open Diabetes Research & Care Search Advanced search Latest contentTopic CollectionsArchiveFor authorsAbout Home/Archive/Volume 9, Issue 1 Email alerts Cardiovascular and metabolic risk•26 July 2021•Open access Request permission Cite this article Share Download PDF Management of type 2 diabetes in chronic kidney disease Author affiliations • Jefferson L Triozzi 1 , L Parker Gregg 2 3 4 , Salim S Virani 5 6 7 8 , Sankar D Navaneethan 2 3 4 9 . Article options Abstract The management of patients with type 2 diabetes and chronic kidney disease (CKD) encompasses lifestyle modifications, glycemic control with individualized HbA1c targets, and cardiovascular disease risk reduction. Metformin and sodium-glucose cotransporter-2 inhibitors are first-line agents. Glucagon-like peptide-1 receptor agonists are second-line agents. The use of other antidiabetic agents should consider patient preferences, comorbidities, drug costs, and the risk of hypoglycemia. Renin–angiotensin–aldosterone system inhibitors are strongly recommended for patients with diabetes, hypertension, and albuminuria. Non-steroidal mineralocorticoid receptor antagonists, which pose less risk of hyperkalemia than steroidal agents, are undergoing further evaluation among patients with diabetic kidney disease. Here, we discuss important advancements in the management of patients with type 2 diabetes and CKD. Back to top Introduction Diabetic kidney disease (DKD) is the most common cause of chronic kidney disease (CKD) globally.1 DKD is characterized by albuminuria and reduced estimated glomerular filtration rate (eGFR), which are independent risk factors for end-stage kidney disease (ESKD), cardiovascular events, and death.2 3 However, some patients with diabetes develop reduced eGFR with minimal or no albuminuria and retain the risks of microvascular and macrovascular complications.4 Moreover, kidney biopsy data among patients with DKD indicate variability in histopathological findings and overlap with non-diabetic disease processes.5 With advancements in diabetes care and therapeutics, the prevalence of kidney disease among patients with diabetes has stabilized around 35%,6 and the incidence of acute cardiovascular events among patients with diabetes has decreased by over 50%.7 Nevertheless, the absolute number of patients with diabetes is rising, with an anticipated global prevalence of 7.7% by 2030.8 The impending burden of diabetes, in the context of the obesity epidemic, will likely affect younger patients with more time at risk to develop kidney disease and its complications.9 In this review, we describe the management of type 2 diabetes in CKD, highlight important aspects of clinical practice guidelines, and discuss the evidence that supports modern practice.10 11 Back to top Management of type 2 diabetes in CKD Overview Glycemic control delays the development of albuminuria and improves clinical outcomes in those with diabetes and kidney disease.12 DKD care requires a multifaceted approach, encompassing lifestyle modifications, glycemic control, cardiovascular risk mitigation, and blood pressure regulation with a renin–angiotensin–aldosterone system (RAAS) inhibitor. ACE inhibitors (ACEi) and angiotensin II receptor blockers reduce kidney disease progression and incident ESKD and are recommended in clinical practice guidelines. Metformin is recommended alongside sodium–glucose cotransporter-2 inhibitor (SGLT2i) as first-line DKD agents. Glucagon-like peptide-1 receptor agonists (GLP-1 RAs) are second-line agents that may decrease albuminuria and cardiovascular risk in some patients. Evolving data regarding SGLT2i, GLP-1 RA, and their combinations will likely impact standards of DKD care. Mineralocorticoid receptor antagonists (MRAs) have potential benefits in DKD, but the use of steroidal agents is limited by the risk of hyperkalemia among patients with reduced kidney function. Emerging safety and efficacy data for non-steroidal MRAs may support their adjunctive use in certain populations. Lifestyle modifications Lifestyle modifications encompassing dietary changes, increased physical activity, and smoking cessation have potential cardiovascular benefits and are recommended for all adults with diabetes and CKD (table 1). Studies examining the benefits of intentional weight loss in CKD are limited. However, weight loss through caloric restriction and exercise modifies cardiovascular risk factors and improves glycemic control in those with diabetes and hence likely benefits those with DKD. Restricting sodium intake to <2 g/day may decrease albuminuria and maximize the effects of RAAS inhibition. Sodium-sensitive hypertension is common in DKD, and thus sodium restriction may also mediate blood pressure and reduce cardiovascular events.13 Experimental studies and clinical trials previously suggested that protein restriction reduces glomerular hyperfiltration and delays the progression of non-diabetic CKD, but there is a paucity of data in those with DKD.14 Advanced DKD is a catabolic state associated with low muscle mass, cachexia, and malnutrition, and thus very low-protein diets are potentially harmful. Protein intake of 0.8 g/kg body weight/day is recommended for non-dialysis-dependent patients, and protein intake of >1.0–1.2 g/kg body weight/day is recommended for the hemodialysis and peritoneal dialysis population.10 11 Table 1 View inline•Open as popup Clinical practice guidelines from KDIGO and ADA Glycemic control Assessment of glycemic control Glycated hemoglobin (HbA1c) is the preferred laboratory measure of glycemia in DKD, with certain limitations. Changes in red blood cell turnover in advanced DKD may decrease the validity of HbA1c.15 HbA1c reflects average blood glucose over 90 days and cannot adequately capture glycemic variability, which is associated with vascular complications, oxidative stress, and hypoglycemia.16 In these situations, continuous blood glucose monitoring using retrospective, real-time, or intermittent techniques provides a more accurate assessment of glycemic control. Other biomarkers, such as glycated albumin and fructosamine, have not demonstrated clear or consistent advantages over HbA1c in CKD. Both albumin-based assessments and fructosamine may be affected by hypoalbuminemia, a common finding among patients with advanced kidney disease.17 For most patients with DKD, HbA1c monitoring every 3–6 months is adequate to guide therapy. Glycemic target Individualized HbA1c targets balance the benefits (eg, reducing microvascular complications) and risks (eg, hypoglycemia) of glycemic control in those with CKD. Glycemic control decreases the incidence of kidney disease and other microvascular complications among patients with diabetes.18 This benefit is supported by clinical trial data using albuminuria and eGFR as outcome measures of DKD. The Diabetes Control and Complications Trial and the corresponding observational follow-up study, Epidemiology of Diabetes Interventions and Complications, enrolled patients with type 1 diabetes and reported a significant and durable reduction in microalbuminuria and albuminuria in the intensive glycemic control group, along with a reduction in the development of stage 3 CKD.19 The United Kingdom Prospective Diabetes Study (UKPDS) enrolled patients with type 2 diabetes and demonstrated a 33% risk reduction of incident microalbuminuria in the intensive glycemic control group.20 Other studies show that intensive glycemic control is associated with a lower risk of incident ESKD. Post-trial analysis by the Action in Diabetes and Vascular Disease: PreterAx and DiamicroN MR Controlled Evaluation study group reported a 65% reduction of ESKD in the intensive glycemic control group.21 A post hoc analysis of the Steno-2 study suggested that intensive glycemic control, as part of a multifactorial behavioral and pharmacological intervention, slows eGFR decline and reduces ESKD and death.22 However, aggressive glycemic control is associated with increased mortality in CKD. In a secondary analysis of the Action to Control Cardiovascular Risk in Diabetes trial, intensive blood glucose control in patients with CKD increased mortality.23 Hypoglycemia is the most important limiting factor for intensive glycemic control, as insulin clearance and gluconeogenesis are impaired in patients with reduced kidney function. Intensive glycemic control poses greater risks for patients receiving medications that cause hypoglycemia, such as insulin or insulin secretagogues. Other factors that support a less aggressive approach to glycemic control in CKD include shorter life expectancy, pre-existing macrovascular complications, greater comorbid conditions, hypoglycemic unawareness, and limited resources for self-care. Clinical practice guidelines emphasize individualized HbA1c targets (table 1). The Kidney Disease: Improving Global Outcomes Diabetes Work Group recommends a target HbA1c range from <6.5% to <8% in non-dialysis-dependent DKD. The American Diabetes Association does not make a specific HbA1c recommendation for patients with kidney disease but recommends a target of <7% for most patients and <8% for patients with a limited life expectancy or high risk of complications. For example, an intensive HbA1c target would benefit a younger patient with early-stage CKD and no cardiovascular complications. Whereas an intensive HbA1c target would pose greater risk than benefit for an elderly patient with advanced-stage CKD, cardiovascular complications, and risk of hypoglycemia. Pharmacological agents Metformin Metformin is a first-line antidiabetic agent that can be safely administered in most patients with baseline eGFR of >30 mL/min/1.73 m 2. Metformin is a biguanide medication with multiple mechanisms of action, including insulin sensitization in peripheral tissues and the reduction of hepatic gluconeogenesis. In the UKPDS study, patients treated with metformin demonstrated decreased diabetes-related, cardiovascular-related, and all-cause mortality when compared with insulin and sulfonylureas.24 Metformin is associated with a reduction in ESKD and adverse cardiovascular outcomes among patients with DKD when compared with other antihyperglycemic agents.25 Kwon et al reported lower all-cause mortality and ESKD progression in patients with eGFR OF >30 mL/min/1.73 m 2 prescribed metformin, with no increased incidence in all-cause lactic acidosis events.26 In separate studies, metformin was associated with decreased risk of cardiovascular events and heart failure readmissions among patients with DKD.27 28 Metformin is often underprescribed or prematurely discontinued among patients with reduced eGFR due to a perceived risk of lactic acidosis. Early biguanide medications were recalled due to life-threatening risks of lactic acidosis, but clinically significant lactic acidosis due to metformin is rare and often attributable to other acute illnesses (table 2). In a large retrospective cohort using national-level data, Lazarus et al reported no difference between hospitalization for lactic acidosis among patients with reduced kidney function taking metformin versus sulfonylureas.29 An analogous study using data from the Veteran’s Health Administration found no difference in lactic acidosis hospitalizations between metformin and sulfonylurea users who developed reduced kidney function.30 In a cohort study using national-level data from Sweden, metformin demonstrated less risk of a composite endpoint of acidosis, serious infection, and all-cause mortality compared with insulin and other oral antihyperglycemic agents in the subgroup of patients with eGFR of 45–60 mL/min/1.73 m 2.31 It is recommended to continue metformin in those with eGFR of ≥45 mL/min/1.73 m 2, titrate cautiously or halve the dose with eGFR of 30–44 mL/min/1.73 m 2, and to discontinue with eGFR of <30 mL/min/1.73 m 2 and the dialysis population. Holding metformin during acute illness or acute kidney injury is reasonable. Table 2 View inline•Open as popup Selected observational studies reporting the risk of acidosis among metformin users with reduced kidney function Sodium–glucose cotransporter-2 inhibitor SGLT2i blocks the reabsorption of glucose and sodium in the proximal convoluted tubule, producing natriuresis and glucosuria. There is substantial evidence to support a reduced risk of ESKD, cardiovascular death, and hospitalization for heart failure. The cardiovascular and kidney benefits of SGLT2i are independent of the antihyperglycemic effect, which attenuates with lower eGFR. SGLT2i may improve glomerular hemodynamics, reduce oxidative stress, and optimize tissue energetics.32 SGLT2i efficacy for kidney and cardiovascular outcomes Large cardiovascular safety trials of SGLT2i demonstrated favorable secondary kidney outcomes among patients with type 2 diabetes and variable baseline kidney function (table 3). Empagliflozin, Cardiovascular Outcomes, and Mortality in Type 2 Diabetes (EMPA-REG OUTCOME) enrolled patients with eGFR of ≥30 mL/min/1.73 m 2 and demonstrated a 46% risk reduction of the composite secondary kidney outcome of doubling of serum creatinine, initiation of kidney replacement therapy, or renal death.33 In a post hoc analysis of EMPA-REG OUTCOME, empagliflozin demonstrated improved kidney function regardless of the baseline eGFR or degree of albuminuria.34 Canagliflozin and Cardiovascular and Renal Events in Type 2 Diabetes (CANVAS) enrolled patients with eGFR of ≥30 mL/min/1.73 m 2 and reported a 40% reduction in the composite secondary kidney outcome.35 Dapagliflozin and Cardiovascular Outcomes in Type 2 Diabetes (DECLARE-TIMI 58) enrolled patients with eGFR of ≥60 mL/min/1.73 m 2 and reported a 47% risk reduction in the composite secondary kidney outcome.36 However, in the Cardiovascular Outcomes with Ertugliflozin in Type 2 Diabetes (VERTIS CV) study, the reduction in the secondary composite kidney outcome was not statistically significant. EMPA-REG OUTCOME and CANVAS both demonstrated a significant reduction in the primary cardiovascular safety outcome, and DECLARE-TIMI 58 and VERTIS CV reached cardiovascular non-inferiority. These data argued for dedicated SGLT2i trials in the DKD population. Table 3 View inline•Open as popup Selected clinical trials of SGLT2i and SGLT1/2i, empagliflozin, canagliflozin, dapagliflozin, ertugliflozin, and sotagliflozin The Canagliflozin and Renal Outcomes in Type 2 Diabetes and Nephropathy (CREDENCE) trial was a seminal study of SGLT2i in DKD with a dedicated primary composite kidney outcome.37 CREDENCE enrolled patients with type 2 diabetes, eGFR of 30–90 mL/min/1.73 m 2, and a urine albumin-to-creatinine ratio of 300–5000 mg/g. The prespecified enrollment strategy aimed to include at least 60% of patients with eGFR of <60 mL/min/1.73 m 2, a population at higher risk of ESKD than previously studied. In this double-blind, placebo-controlled trial, canagliflozin reduced the primary composite kidney endpoint by 31%, with a noteworthy benefit in a secondary heart failure outcome. The trial was prematurely discontinued after 2.6 years due to overwhelming efficacy. CREDENCE demonstrated kidney benefits independent of baseline HbA1c, extent of HbA1c reduction, and stage of CKD, leading to the first kidney-related indication for SGLT2i by the US FDA in 2019. There is evolving information about the use of SGLT2i among patients with eGFR of <30 mL/min/1.73 m 2 and non-DKD. Dapagliflozin in Patients with Chronic Kidney Disease included 4304 participants, two-thirds with diabetes, with eGFR of 25–75 mL/min/1.73 m 2. Dapagliflozin reduced the composite risk of ≥50% eGFR decline, ESKD, and renal or cardiovascular death by 39%. In subgroup analyses, the benefit of dapagliflozin was consistent regardless of HbA1c, eGFR, or presence of diabetes. The Study of Heart and Kidney Protection With Empagliflozin (EMPA-KIDNEY) is an ongoing study that will report the effect of empagliflozin on kidney progression or cardiovascular (CV) death.38 EMPA-KIDNEY will deliver important information among patients with or without diabetes, with or without albuminuria, and eGFR as low as 20 mL/min/1.73 m 2. Dual SGLT1 and SGLT2 inhibition Sotagliflozin is a dual SGLT1 and SGLT2 inhibitor with antihyperglycemic efficacy in both type 1 and type 2 diabetes.39 The SGLT1 cotransporter is also present in the gastrointestinal lumen, where it delays glucose absorption and reduces postprandial blood glucose levels.40 In the Sotagliflozin in Patients with Diabetes and Chronic Kidney Disease (SCORED) trial, sotagliflozin decreased a composite outcome of cardiovascular death, hospitalization for heart failure, and urgent heart failure among patients with DKD with or without albuminuria.41 Sotagliflozin in Patients with Diabetes and Recent Worsening Heart Failure (SOLOIST-WHF) enrolled patients with diabetes during acute heart failure hospitalization and demonstrated improved heart failure outcomes among a population with mean eGFR 50 mL/min/1.73 m 2.42 Both SCORED and SOLOIST-WHF reported several adverse events and prominent gastrointestinal upset with sotagliflozin, which may hinder its widespread use. SGLT2i safety SGLT2i safety data are extrapolated from trials that were not powered to detect individual adverse events. In a large meta-analysis, SGLT2i was associated with increased genital mycotic infections but not bacterial urinary tract infections.43 Genital mycotic infections are more common in women and manageable with antifungal therapy. Studies suggesting risks of Fournier’s gangrene are limited by low event rates.44 SGLT2i decreases available carbohydrate and drives metabolism towards more efficient ketone-based energy sources, leading to weight loss, decreased tissue adiposity, and increased serum concentrations of β-hydroxybutyrate.45 Diabetic ketoacidosis, including euglycemic ketoacidosis, is more likely to develop among patients with certain risk factors, such as insulin dependence or acute stress states.46 A cohort study in the USA identified a twofold increase in ketoacidosis among new users of SGLT2i compared with dipeptidyl peptidase-4 inhibitors (DPP4i).47 Amputation, fracture, and acute kidney injury have also been reported. A risk of leg and foot amputation reported in CANVAS (amputation event rate of 6.3 per 1000 patient-years, p<0.001) was ultimately not observed in CREDENCE.35 37 The US FDA subsequently recalled a black box amputation warning for canagliflozin. Similarly, a risk of fracture reported in CANVAS (15.4 fracture events per 1000 patient-years, p<0.001) was not observed in other SGLT2i trials.33 35 36 Finally, pooled analyses suggest SGLT2i is indeed associated with decreased risk of acute kidney injury.48 Volume depletion was proposed as a mechanism for amputation, fracture, and acute kidney injury, although the relative increase in urinary volume related to SGLT2i is transient.49 Clinical practice guidelines now recommend SGLT2i for patients with type 2 diabetes with DKD and eGFR of ≥30 mL/min/1.73 m 2. It is reasonable to temporarily hold SGLT2i when predisposed to volume depletion or ketoacidosis, such as surgery, fasting states, or acute illnesses. History of urinary tract infection is not a contraindication to SGLT2i. In the absence of clear risk factors, the potential cardiovascular and kidney benefits justify SGLT2i therapy for most patients with DKD. Glucagon-like peptide-1 receptor agonist Incretin-based therapies stimulate postprandial neuroendocrine pathways to increase pancreatic insulin secretion, suppress glucagon release, increase satiety, and delay gastric emptying. In addition to an antihyperglycemic effect, incretin-based therapies improve risk factors for cardiovascular and kidney disease by improving blood pressure, body weight, and lipid profile. Other anti-inflammatory properties are proposed.50 Whereas DPP4i transiently reduces the breakdown of physiological incretin hormones, GLP-1 RA with prolonged half-lives resists degradation by DPP4 and stimulate incretin receptors at supraphysiological levels. As a possible consequence, GLP-1 RA has a greater effect in DKD. Evidence for GLP-1 RA in DKD We lack dedicated DKD clinical trials of GLP-1 RA with predefined primary kidney endpoints. Nonetheless, some GLP-1 RA have demonstrated improved secondary microvascular outcomes in cardiovascular safety trials, driven by a reduction of albuminuria (table 4). Of the available GLP-1 RA, liraglutide, semaglutide, and dulaglutide have demonstrated both cardiovascular benefits and antialbuminuric effects. Exenatide and lixisenatide have demonstrated cardiovascular safety without a cardiovascular benefit. All therapies are administered as subcutaneous injection except for an oral formulation of semaglutide, which has demonstrated cardiovascular safety without a cardiovascular benefit. Table 4 View inline•Open as popup Selected clinical trials of GLP-1 RA, liraglutide, semaglutide, and dulaglutide Liraglutide and Cardiovascular Outcomes in Type 2 Diabetes (LEADER) was a placebo-controlled cardiovascular safety trial of liraglutide and included a secondary composite microvascular outcome. The reduction in the microvascular outcome, solely driven by reduced incidence of albuminuria, was independent of baseline HbA1c.51 Similarly, Semaglutide and Cardiovascular Outcomes in Patients with Type 2 Diabetes (SUSTAIN-6) demonstrated both cardiovascular safety and a decreased secondary microvascular outcome driven by reduced incidence of microalbuminuria. SUSTAIN-6 had a surprising increase in retinopathy events not previously reported in incretin trials.52 An oral formulation of semaglutide was subsequently studied in the Oral Semaglutide and Cardiovascular Outcomes in Patients with Type 2 Diabetes (PIONEER-6) trial. Oral semaglutide demonstrated cardiovascular safety without cardiovascular superiority, with no significant difference eGFR change from baseline.53 In LEADER, greater cardiovascular benefit was seen among patients with eGFR <60 mL/min/1.73 m 2. In SUSTAIN-6 and PIONEER, however, there was no difference in cardiovascular safety stratified by stage of CKD. Other than an apparent antialbuminuric effect, there is limited evidence that GLP-1 RA influences eGFR decline or other clinically meaningful kidney outcomes. In a prespecified kidney analysis of LEADER, there was a statistically less eGFR decline among patients receiving liraglutide versus placebo, only present in the eGFR of 30–60 mL/min/1.73 m 2 subgroup.51 In the Dulaglutide versus Insulin Glargine in Patients with Type 2 Diabetes and Moderate-to-Severe CKD (AWARD-7) study, dulaglutide demonstrated significantly less eGFR decline (−1.1 mL/min/1.73 m 2) than daily titrated insulin glargine (−2.9 mL/min/1.73 m 2) after a 52-week follow-up period.54 Although dulaglutide was associated with reduced albuminuria in the placebo-controlled Dulaglutide and Cardiovascular Outcomes in Type 2 Diabetes trial, there were no differences in albuminuria in comparison with insulin glargine in the AWARD-7 trial.55 A dedicated kidney outcomes trial for GLP-1 RA will clarify their use in DKD. A Research Study to See How Semaglutide Works Compared with Placebo in People With Type 2 Diabetes and Chronic Kidney Disease (FLOW) is a placebo-controlled, phase III trial of semaglutide with a primary kidney outcome set to complete in 2024 (NCT03819153). FLOW is enrolling adults with DKD and will report the effect of semaglutide on the primary renal composite outcome of eGFR decline, kidney failure, or death from kidney or cardiovascular disease. Combination of SGLT2i and GLP-1 RA SGLT2i and GLP-1 RA have distinct clinical effects. In a systematic review and trial-level meta-analysis, GLP-1 RA and SGLT2i reduced cardiovascular events to a similar degree among patients with established atherosclerotic cardiovascular disease, but SGLT2i had a greater impact on preventing heart failure hospitalizations and DKD progression.56 Further supported by dedicated heart failure studies, SGLT2i is now incorporated into guideline-directed strategies for congestive heart failure.57 GLP-1 RA has not demonstrated benefit in heart failure. GLP-1 RA have a class interaction with the sympathetic nervous system, reflected by increased heart rate, with some signal towards harm in heart failure.58 Combination of SGLT2i and incretin-based therapies may be synergistic, as both drug classes mitigate cardiovascular risk factors through improvements in weight, lipid profile, and blood pressure. Dapagliflozin and exenatide, studied alone and in combination, demonstrated improved glycemic control and cardiovascular risk in combination. The benefits of combined dapagliflozin and exenatide persisted through the 2-year follow-up, although no long-term kidney outcomes were reported.59 Clinical practice guidelines recommend GLP-1 RA in DKD based on potential cardiovascular risk reduction and antialbuminuric effects of certain agents. GLP-1 RA compounds with long-half-lives will require less frequent dosing and may have better compliance. Dulaglutide, exenatide, and semaglutide have extended-release formulations. Dulaglutide, liraglutide, lixisenatide, and semaglutide are not eliminated by the kidneys. The oral formulation of semaglutide is available for patients with an aversion to needles. Dedicated safety studies of GLP-1 RA have not been performed among patients with reduced kidney function. Slow titration to the maximally indicated dose is recommended to lower the risk of hypoglycemia and to mitigate gastrointestinal upset, the most reported side effect. Other antidiabetic agents We have discussed the roles of metformin, SGLT2i, and GLP-1 RA in detail. There are limited head-to-head efficacy data for other antidiabetic agents among patients with DKD.60 Selection of other antidiabetic agents should consider patient preferences, comorbidities, drug costs, and the risk of hypoglycemia. Any antidiabetic agent with a prolonged duration of action, active metabolites, and elimination by the kidneys poses greater risk of hypoglycemia among patients with reduced kidney function. Sulfonylureas stimulate endogenous insulin secretion and are predominately eliminated by the kidneys. The sulfonylureas glipizide, glicazide, and glimepiride can be prescribed in eGFR <60 mL/min/1.73m 2, but glyburide has active metabolites and should be avoided. Glinides exert similar effects as sulfonylureas but have a shorter onset and duration of action. Repaglinide is metabolized by both the liver and kidneys and can be prescribed in eGFR of <60 mL/min/1.73m 2, but nateglinide has active metabolites and should be avoided. Both sulfonylureas and glinides should be avoided in eGFR of <15 mL/min/1.73 m 2 and the dialysis population. Disaccharidase inhibitors, such as acarbose and miglitol, delay carbohydrate digestion in the gut. Although disaccharidase inhibitors pose less risk of hypoglycemia, they are generally not recommended in advanced CKD due to limited safety data. Thiazolidinediones increase insulin sensitivity in peripheral tissues, have less risk of hypoglycemia, and do not require dose adjustments for patients with reduced kidney function and the dialysis population. Thiazolidinediones may cause weight gain and fluid retention. Insulin may be required for patients with progressive insulin resistance and pancreatic dysfunction. Insulin is eliminated by the kidneys and has potent glucose lowering effects. Patient education, routine monitoring, and dose reductions based on the degree of kidney disease are important to reduce the risk of hypoglycemia. RAAS blockade RAAS inhibitors RAAS blockade delays DKD progression in patients with albuminuria and hypertension.61 RAAS activation, mediated by the bioactive end products angiotensin II and aldosterone, causes increased sodium avidity and vascular tone, glomerular injury, and proteinuria. RAAS activation also contributes to systemic inflammation, the production of reactive oxygen species, and fibrosis in the kidney and cardiovascular systems. ACEi and angiotensin receptor blockers (ARBs) remain the mainstay RAAS inhibitors since their implementation at the turn of the century. Clinical practice guidelines recommend ACEi or ARB monotherapy titrated to the maximally titrated dose. Patients should be monitored for hyperkalemia or acute kidney injury (>30% elevation of serum creatinine) after initiating or increasing the dose of RAAS inhibitors. Mineralocorticoid receptor agonists MRAs may be an important component of RAAS blockade. MRAs reduce albuminuria and secondary markers of fibrosis and inflammation in the kidney. MRAs are also indicated for many common DKD comorbidities, including resistant hypertension and congestive heart failure. Despite these potential benefits, the use of MRAs in DKD is limited by the risk of hyperkalemia posed by steroidal MRAs like spironolactone and eplerenone. In a meta-analysis of 19 trials, the addition of steroidal MRAs to RAAS inhibitors resulted in a threefold risk of hyperkalemia.62 Potassium binders, low potassium diets, or diuretics were often used to mitigate the hyperkalemia risk in these trials, although unsuccessfully. Adverse safety events precluded adequate follow-up periods to detect cardiovascular and kidney endpoints in these trials. Finerenone Finerenone is a non-steroidal, dihydropyridine-based MRA with high affinity for the mineralocorticoid receptor, posing less risk of hyperkalemia.63 The Mineralocorticoid Receptor Antagonist Tolerability Study–Diabetic Nephropathy (ARTS-DN) study assessed varied doses of finerenone versus placebo among patients with diabetes and eGFR of >30 mL/min/1.73 m 2. ARTS-DN resulted in a dose-dependent reduction in albuminuria at 90 days with no risk of hyperkalemia.64 Efficacy and Safety of Finerenone in Subjects With Type 2 Diabetes Mellitus and Diabetic Kidney Disease (FIDELIO-DKD) demonstrated that those treated with finerenone had decreased CKD progression and cardiovascular events in DKD, with an 18% reduction of the primary composite outcome of kidney failure, sustained decrease of eGFR by at least 40%, or renal death.65 This outcome was driven by the reduction in eGFR, with no difference in the rate of kidney failure. There were more hyperkalemia events in the finerenone group (15.8%) than placebo (7.8%), although the study was not powered to assess adverse events. FIGARO-DKD is an ongoing study of finerenone that will examine a primary cardiovascular outcome event with secondary kidney endpoints among a similar DKD population (NCT02545049). Back to top Conclusion There has been great progress in the management of patients with type 2 diabetes and CKD. DKD management is multifaceted and individualized. A more lenient HbA1c target is appropriate in older adults and patients with advanced DKD or increased risk of hypoglycemia. Metformin is an important first-line, cost-effective agent with significant data supporting efficacy and safety for patients with DKD. SGLT2i is a first-line agent based on substantial clinical trial data supporting a reduction in ESKD, cardiovascular death, and hospitalization for heart failure. GLP-1 RA is a second-line agent that may reduce albuminuria and cardiovascular disease risk. Other antidiabetic agents, such as sulfonylureas, glinides, disaccharidase inhibitors, thiazolidinediones, and insulin, may be used for glycemic control based on the stage of kidney disease, individual risks and benefits. ACEi or ARB monotherapy is strongly recommended for patients with diabetes, hypertension, and albuminuria. Mineralocorticoid receptor antagonism is an unmet need in DKD due to the risk of hyperkalemia with steroidal MRA agents. Non-steroidal MRAs like finerenone have better safety profiles and may prove beneficial for certain patients with DKD. Effective DKD therapies target the shared pathways of diabetes, cardiovascular disease, and kidney disease, and a comprehensive approach will improve global outcomes for patients with DKD. Back to top Footnotes Contributors: JLT and SDN planned the contents of the review. JLT and SDN drafted the review and tables. LPG and SSV critically reviewed and edited the manuscript, and all authors were involved in the revision of the manuscript. Funding: SDN is supported by research funding from the Department of Veterans Affairs Health Services Research & Development (1I01HX002917-01A1) and a grant from the National Institutes of Health (NIDDK-R01DK101500). SSV is supported by a grant from the Department of Veterans Affairs Health Services Research and Development (1I01HX002917-01A1), World Heart Federation, and Tahir and Jooma Family. This work was also supported in part by the Center for Innovations in Quality, Effectiveness and Safety (CIN 13-413), Michael E. DeBakey VA Medical Center, Houston, Texas. SDN, LPG, SSV are employees of the US Department of Veterans Affairs. The interpretation and reporting of these data are the responsibility of the authors and in no way should be viewed as official policy or interpretation of the Department of Veterans Affairs or the US government. Funding agencies did not have any role in study design, data collection, analysis, reporting, or the decision to submit for publication. The content is solely the responsibility of the authors and does not necessarily represent the official views of the National Institutes of Health or Veterans Administration. Competing interests: SDN reports personal fees from Bayer, Boehringer-Ingelheim, REATA, Tricida, and Vifor, and grants from Keryx outside the submitted work. SSV has received honorarium from the American College of Cardiology in his role as the associate editor for Innovations, ACC.org. All remaining authors have nothing to disclose. Provenance and peer review: Commissioned; externally peer reviewed. Data availability statement No data are available. This is not a clinical trial. Ethics statements Patient consent for publication: Not required. Back to top References Bikbov B, Purcell CA, Levey AS, et al. Global, regional, and national burden of chronic kidney disease, 1990–2017: a systematic analysis for the global burden of disease study 2017. The Lancet 2020; 395:709–33. doi:10.1016/S0140-6736(20)30045-3•Google Scholar Ninomiya T, Perkovic V, de Galan BE, et al. Albuminuria and kidney function independently predict cardiovascular and renal outcomes in diabetes. JASN 2009; 20:1813–21. doi:10.1681/ASN.2008121270•Google Scholar Berhane AM, Weil EJ, Knowler WC, et al. Albuminuria and estimated glomerular filtration rate as predictors of diabetic end-stage renal disease and death. Clin J Am Soc Nephrol 2011; 6:2444–51. doi:10.2215/CJN.00580111•Google Scholar•PubMed Porrini E, Ruggenenti P, Mogensen CE, et al. Non-proteinuric pathways in loss of renal function in patients with type 2 diabetes. Lancet Diabetes Endocrinol 2015; 3:382–91. doi:10.1016/S2213-8587(15)00094-7•Google Scholar Fiorentino M, Bolignano D, Tesar V, et al. Renal biopsy in patients with diabetes: a pooled meta-analysis of 48 studies. Nephrol Dial Transplant 2017; 32:97–110. doi:10.1093/ndt/gfw070•Google Scholar•PubMed de Boer IH, Rue TC, Hall YN, et al. Temporal trends in the prevalence of diabetic kidney disease in the United States. JAMA 2011; 305:2532–9. doi:10.1001/jama.2011.861•Google Scholar Gregg EW, Li Y, Wang J, et al. Changes in diabetes-related complications in the United States, 1990-2010. N Engl J Med 2014; 370:1514–23. doi:10.1056/NEJMoa1310799•Google Scholar•PubMed Shaw JE, Sicree RA, Zimmet PZ, et al. Global estimates of the prevalence of diabetes for 2010 and 2030. Diabetes Res Clin Pract 2010; 87:4–14. doi:10.1016/j.diabres.2009.10.007•Google Scholar Thomas MC, Cooper ME, Zimmet P, et al. Changing epidemiology of type 2 diabetes mellitus and associated chronic kidney disease. Nat Rev Nephrol 2016; 12:73–81. doi:10.1038/nrneph.2015.173•Google Scholar Kidney Disease: Improving Global Outcomes (KDIGO) Diabetes Work Group. KDIGO 2020 clinical practice guideline for diabetes management in chronic kidney disease. Kidney Int 2020; 98:S1–115. doi:10.1016/j.kint.2020.06.019•Google Scholar•PubMed American Diabetes Association. 11. microvascular complications and foot care: standards of medical care in Diabetes-2021. Diabetes Care 2021; 44:S151–67. doi:10.2337/dc21-S011•Google Scholar•PubMed Coca. Role of intensive glucose control in development of renal end points in type 2 diabetes mellitus: systematic review and meta-analysis. Arch Intern Med 2012; 172. doi:10.1001/archinternmed.2012.3617•Google Scholar•PubMed Jain N, Reilly RF. Effects of dietary interventions on incidence and progression of CKD. Nat Rev Nephrol 2014; 10:712–24. doi:10.1038/nrneph.2014.192•Google Scholar Ko GJ, Obi Y, Tortorici AR, et al. Dietary protein intake and chronic kidney disease. Curr Opin Clin Nutr Metab Care 2017; 20:77–85. doi:10.1097/MCO.0000000000000342•Google Scholar•PubMed Nathan DM, Kuenen J, Borg R, et al. Translating the A1c assay into estimated average glucose values. Diabetes Care 2008; 31:1473–8. doi:10.2337/dc08-0545•Google Scholar Papachristoforou E, Lambadiari V, Maratou E, et al. Association of glycemic indices (hyperglycemia, glucose variability, and hypoglycemia) with oxidative stress and diabetic complications. J Diabetes Res 2020; 2020:1–17. doi:10.1155/2020/7489795•Google Scholar•PubMed Zelnick LR, Batacchi ZO, Ahmad I, et al. Continuous glucose monitoring and use of alternative markers to assess glycemia in chronic kidney disease. Diabetes Care 2020; 43:2379–87. doi:10.2337/dc20-0915•Google Scholar•PubMed Nathan DM, Cleary PA, Backlund J-YC, et al. Intensive diabetes treatment and cardiovascular disease in patients with type 1 diabetes. N Engl J Med 2005; 353:2643–53. doi:10.1056/NEJMoa052187•Google Scholar•PubMed DCCT/EDIC Research Group, de Boer IH, Sun W, et al. Intensive diabetes therapy and glomerular filtration rate in type 1 diabetes. N Engl J Med 2011; 365:2366–76. doi:10.1056/NEJMoa1111732•Google Scholar•PubMed Bilous R. Microvascular disease: what does the UKPDS tell us about diabetic nephropathy? Diabet Med 2008; 25:25–9. doi:10.1111/j.1464-5491.2008.02496.x•Google Scholar Wong MG, Perkovic V, Chalmers J, et al. Long-term benefits of intensive glucose control for preventing end-stage kidney disease: ADVANCE-ON. Diabetes Care 2016; 39:694–700. doi:10.2337/dc15-2322•Google Scholar Oellgaard J, Gæde P, Rossing P, et al. Intensified multifactorial intervention in type 2 diabetics with microalbuminuria leads to long-term renal benefits. Kidney Int 2017; 91:982–8. doi:10.1016/j.kint.2016.11.023•Google Scholar•PubMed Action to Control Cardiovascular Risk in Diabetes Study Group, Gerstein HC, Miller ME, et al. Effects of intensive glucose lowering in type 2 diabetes. N Engl J Med 2008; 358:2545–59. doi:10.1056/NEJMoa0802743•Google Scholar•PubMed Holman RR, Paul SK, Bethel MA, et al. 10-Year follow-up of intensive glucose control in type 2 diabetes. N Engl J Med 2008; 359:1577–89. doi:10.1056/NEJMoa0806470•Google Scholar Schernthaner G, Schernthaner-Reiter MH. Therapy: risk of metformin use in patients with T2DM and advanced CKD. Nat Rev Endocrinol 2015; 11:697–9. doi:10.1038/nrendo.2015.132•Google Scholar•PubMed Kwon S, Kim YC, Park JY, et al. The long-term effects of metformin on patients with type 2 diabetic kidney disease. Diabetes Care 2020; 43:948–55. doi:10.2337/dc19-0936•Google Scholar•PubMed Roumie CL, Chipman J, Min JY, et al. Association of treatment with metformin vs sulfonylurea with major adverse cardiovascular events among patients with diabetes and reduced kidney function. JAMA 2019; 322:1167–77. doi:10.1001/jama.2019.13206•Google Scholar•PubMed Roussel R, Travert F, Pasquet B, et al. Metformin use and mortality among patients with diabetes and atherothrombosis. Arch Intern Med 2010; 170:1892–9. doi:10.1001/archinternmed.2010.409•Google Scholar•PubMed Lazarus B, Wu A, Shin J-I, et al. Association of metformin use with risk of lactic acidosis across the range of kidney function. JAMA Intern Med 2018; 178:903–10. doi:10.1001/jamainternmed.2018.0292•Google Scholar Chu PY, Hackstadt AJ, Chipman J, et al. Hospitalization for lactic acidosis among patients with reduced kidney function treated with metformin or sulfonylureas. Diabetes Care 2020; 43:1462–70. doi:10.2337/dc19-2391•Google Scholar Ekström N, Schiöler L, Svensson A-M, et al. Effectiveness and safety of metformin in 51 675 patients with type 2 diabetes and different levels of renal function: a cohort study from the Swedish national diabetes register. BMJ Open 2012; 2. doi:10.1136/bmjopen-2012-001076•Google Scholar Petrykiv S, Sjöström CD, Greasley PJ, et al. Differential effects of dapagliflozin on cardiovascular risk factors at varying degrees of renal function. Clin J Am Soc Nephrol 2017; 12:751–9. doi:10.2215/CJN.10180916•Google Scholar Wanner C, Inzucchi SE, Lachin JM, et al. Empagliflozin and progression of kidney disease in type 2 diabetes. N Engl J Med 2016; 375:323–34. doi:10.1056/NEJMoa1515920•Google Scholar Wanner C, Inzucchi SE, Zinman B, et al. Consistent effects of empagliflozin on cardiovascular and kidney outcomes irrespective of diabetic kidney disease categories: insights from the EMPA-REG outcome trial. Diabetes Obes Metab 2020; 22:2335–47. doi:10.1111/dom.14158•Google Scholar•PubMed Neal B, Perkovic V, Mahaffey KW, et al. Canagliflozin and cardiovascular and renal events in type 2 diabetes. N Engl J Med 2017; 377:644–57. doi:10.1056/NEJMoa1611925•Google Scholar•PubMed Wiviott SD, Raz I, Bonaca MP, et al. Dapagliflozin and cardiovascular outcomes in type 2 diabetes. N Engl J Med 2019; 380:347–57. doi:10.1056/NEJMoa1812389•Google Scholar•PubMed Perkovic V, Jardine MJ, Neal B, et al. Canagliflozin and renal outcomes in type 2 diabetes and nephropathy. N Engl J Med 2019; 380:2295–306. doi:10.1056/NEJMoa1811744•Google Scholar Herrington WG, Preiss D, Haynes R, et al. The potential for improving cardio-renal outcomes by sodium-glucose co-transporter-2 inhibition in people with chronic kidney disease: a rationale for the EMPA-KIDNEY study. Clin Kidney J 2018; 11:749–61. doi:10.1093/ckj/sfy090•Google Scholar Musso G, Gambino R, Cassader M, et al. Efficacy and safety of dual SGLT 1/2 inhibitor sotagliflozin in type 1 diabetes: meta-analysis of randomised controlled trials. BMJ 2019; 60:l1328. doi:10.1136/bmj.l1328•Google Scholar Powell DR, Zambrowicz B, Morrow L, et al. Sotagliflozin decreases postprandial glucose and insulin concentrations by delaying intestinal glucose absorption. J Clin Endocrinol Metab 2020; 105:e1235–49. doi:10.1210/clinem/dgz258•Google Scholar•PubMed Bhatt DL, Szarek M, Pitt B, et al. Sotagliflozin in patients with diabetes and chronic kidney disease. N Engl J Med 2021; 384:129–39. doi:10.1056/NEJMoa2030186•Google Scholar•PubMed Bhatt DL, Szarek M, Steg PG, et al. Sotagliflozin in patients with diabetes and recent worsening heart failure. N Engl J Med 2021; 384:117–28. doi:10.1056/NEJMoa2030183•Google Scholar•PubMed Liu J, Li L, Li S, et al. Effects of SGLT2 inhibitors on UTIs and genital infections in type 2 diabetes mellitus: a systematic review and meta-analysis. Sci Rep 2017; 7. doi:10.1038/s41598-017-02733-w•Google Scholar Bersoff-Matcha SJ, Chamberlain C, Cao C, et al. Fournier gangrene associated with sodium-glucose cotransporter-2 inhibitors: a review of spontaneous postmarketing cases. Ann Intern Med 2019; 170:764–9. doi:10.7326/M19-0085•Google Scholar•PubMed Ferrannini E, Baldi S, Frascerra S, et al. Shift to fatty substrate utilization in response to sodium-glucose cotransporter 2 inhibition in subjects without diabetes and patients with type 2 diabetes. Diabetes 2016; 65:1190–5. doi:10.2337/db15-1356•Google Scholar•PubMed Palmer BF, Clegg DJ. Euglycemic ketoacidosis as a complication of SGLT2 inhibitor therapy. Clin J Am Soc Nephrol 2021; 17621120. doi:10.2215/CJN.17621120•Google Scholar•PubMed Fralick M, Schneeweiss S, Patorno E, et al. Risk of diabetic ketoacidosis after initiation of an SGLT2 inhibitor. N Engl J Med 2017; 376:2300–2. doi:10.1056/NEJMc1701990•Google Scholar Menne J, Dumann E, Haller H, et al. Acute kidney injury and adverse renal events in patients receiving SGLT2-inhibitors: a systematic review and meta-analysis. PLoS Med 2019; 16. doi:10.1371/journal.pmed.1002983•Google Scholar•PubMed Heise T, Jordan J, Wanner C, et al. Pharmacodynamic effects of single and multiple doses of Empagliflozin in patients with type 2 diabetes. Clin Ther 2016; 38:2265–76. doi:10.1016/j.clinthera.2016.09.001•Google Scholar•PubMed Tanaka T, Higashijima Y, Wada T, et al. The potential for renoprotection with incretin-based drugs. Kidney Int 2014; 86:701–11. doi:10.1038/ki.2014.236•Google Scholar Mann JFE, Ørsted DD, Brown-Frandsen K, et al. Liraglutide and renal outcomes in type 2 diabetes. N Engl J Med 2017; 377:839–48. doi:10.1056/NEJMoa1616011•Google Scholar Marso SP, Bain SC, Consoli A, et al. Semaglutide and cardiovascular outcomes in patients with type 2 diabetes. N Engl J Med 2016; 375:1834–44. doi:10.1056/NEJMoa1607141•Google Scholar Husain M, Birkenfeld AL, Donsmark M, et al. Oral Semaglutide and cardiovascular outcomes in patients with type 2 diabetes. N Engl J Med 2019; 381:841–51. doi:10.1056/NEJMoa1901118•Google Scholar•PubMed Tuttle KR, Lakshmanan MC, Rayner B, et al. Dulaglutide versus insulin Glargine in patients with type 2 diabetes and moderate-to-severe chronic kidney disease (AWARD-7): a multicentre, open-label, randomised trial. Lancet Diabetes Endocrinol 2018; 6:605–17. doi:10.1016/S2213-8587(18)30104-9•Google Scholar•PubMed Gerstein HC, Colhoun HM, Dagenais GR, et al. Dulaglutide and cardiovascular outcomes in type 2 diabetes (REWIND): a double-blind, randomised placebo-controlled trial. The Lancet 2019; 394:121–30. doi:10.1016/S0140-6736(19)31149-3•Google Scholar Zelniker TA, Wiviott SD, Raz I, et al. Comparison of the effects of glucagon-like peptide receptor agonists and sodium-glucose cotransporter 2 inhibitors for prevention of major adverse cardiovascular and renal outcomes in type 2 diabetes mellitus. Circulation 2019; 139:2022–31. doi:10.1161/CIRCULATIONAHA.118.038868•Google Scholar•PubMed Packer M, Anker SD, Butler J, et al. Effect of Empagliflozin on the clinical stability of patients with heart failure and a reduced ejection fraction. Circulation 2021; 143:326–36. doi:10.1161/CIRCULATIONAHA.120.051783•Google Scholar Margulies KB, Hernandez AF, Redfield MM, et al. Effects of liraglutide on clinical stability among patients with advanced heart failure and reduced ejection fraction. JAMA 2016; 316:500–8. doi:10.1001/jama.2016.10260•Google Scholar Jabbour SA, Frías JP, Ahmed A, et al. Efficacy and safety over 2 years of exenatide plus dapagliflozin in the DURATION-8 study: a multicenter, double-blind, phase 3, randomized controlled trial. Diabetes Care 2020; 43:2528–36. doi:10.2337/dc19-1350•Google Scholar•PubMed Snyder RW, Berns JS. Use of insulin and oral hypoglycemic medications in patients with diabetes mellitus and advanced kidney disease. Semin Dial 2004; 17:365–70. doi:10.1111/j.0894-0959.2004.17346.x•Google Scholar•PubMed Zhang Y, He D, Zhang W, et al. Ace inhibitor benefit to kidney and cardiovascular outcomes for patients with non-dialysis chronic kidney disease stages 3–5: a network meta-analysis of randomised clinical trials. Drugs 2020; 80:797–811. doi:10.1007/s40265-020-01290-3•Google Scholar Currie G, Taylor AHM, Fujita T, et al. Effect of mineralocorticoid receptor antagonists on proteinuria and progression of chronic kidney disease: a systematic review and meta-analysis. BMC Nephrol 2016; 17. doi:10.1186/s12882-016-0337-0•Google Scholar Agarwal R, Kolkhof P, Bakris G, et al. Steroidal and non-steroidal mineralocorticoid receptor antagonists in cardiorenal medicine. Eur Heart J 2021; 42:152–61. doi:10.1093/eurheartj/ehaa736•Google Scholar•PubMed Bakris GL, Agarwal R, Chan JC, et al. Effect of finerenone on albuminuria in patients with diabetic nephropathy: a randomized clinical trial. JAMA 2015; 314:884–94. doi:10.1001/jama.2015.10081•Google Scholar•PubMed Bakris GL, Agarwal R, Anker SD, et al. Effect of finerenone on chronic kidney disease outcomes in type 2 diabetes. N Engl J Med 2020; 383:2219–29. doi:10.1056/NEJMoa2025845•Google Scholar•PubMed Back to top Publication history Received: 29 March 2021 Accepted: 4 July 2021 First published: 26 July 2021 Back to top Responses Compose a response to this article Back to top Overview Abstract Introduction Management of type 2 diabetes in CKD Conclusion References Footnotes Publication history Responses Article metricsAltmetricDimensions Overview Abstract Introduction Management of type 2 diabetes in CKD Conclusion References Footnotes Publication history Responses Article metricsAltmetricDimensions CONTENT Latest content Archive Browse by collection In the news Top cited articles Most read articles JOURNAL About Editorial board Thank you to our reviewers Sign up for email alerts AUTHORS Instructions for authors Submit an article Open access at BMJ HELP Contact us Reprints Permissions Advertising Feedback form Website Terms & ConditionsPrivacy & CookiesContact BMJCookie settingsOnline ISSN: 2052-4897 Copyright © 2025 BMJ Publishing Group Ltd. All rights, including for text and data mining, AI training, and similar technologies, are reserved. Cookies and privacy We and our 185 partners store and access personal data, like browsing data or unique identifiers, on your device. Selecting I Accept enables tracking technologies to support the purposes shown under we and our partners process data to provide. Selecting Reject All or withdrawing your consent will disable them. If trackers are disabled, some content and ads you see may not be as relevant to you. You can resurface this menu to change your choices or withdraw consent at any time by clicking the Cookie settings link on the bottom of the webpage . Your choices will have effect within our Website. For more details, refer to our Privacy Policy.Cookie policy We and our partners process data to provide: Use precise geolocation data. Actively scan device characteristics for identification. Store and/or access information on a device. Personalised advertising and content, advertising and content measurement, audience research and services development. List of Partners (vendors) I Accept Reject All Manage preferences About Your Privacy Your Privacy Strictly Necessary Cookies Performance Cookies Functional Cookies Targeting Cookies Google & IAB TCF 2 Purposes of Processing Store and/or access information on a device 130 partners can use this purpose Personalised advertising and content, advertising and content measurement, audience research and services development 171 partners can use this purpose Use precise geolocation data 56 partners can use this special feature Actively scan device characteristics for identification 30 partners can use this special feature Ensure security, prevent and detect fraud, and fix errors 105 partners can use this special purpose Deliver and present advertising and content 107 partners can use this special purpose Match and combine data from other data sources 77 partners can use this feature Link different devices 55 partners can use this feature Identify devices based on information transmitted automatically 99 partners can use this feature Save and communicate privacy choices 84 partners can use this special purpose Your Privacy We process your data to deliver content or advertisements and measure the delivery of such content or advertisements to extract insights about our website. We share this information with our partners on the basis of consent and legitimate interest. You may exercise your right to consent or object to a legitimate interest, based on a specific purpose below or at a partner level in the link under each purpose. These choices will be signaled to our vendors participating in the Transparency and Consent Framework. Privacy and cookie policies List of IAB Vendors‎ Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Cookies Details‎ Google & IAB TCF 2 Purposes of Processing [x] Google & IAB TCF 2 Purposes of Processing Allowing third-party ad tracking and third-party ad serving through Google and other vendors to occur. Please see more information on Google Ads Cookies Details‎ Store and/or access information on a device 130 partners can use this purpose [x] Store and/or access information on a device Cookies, device or similar online identifiers (e.g. login-based identifiers, randomly assigned identifiers, network based identifiers) together with other information (e.g. browser type and information, language, screen size, supported technologies etc.) can be stored or read on your device to recognise it each time it connects to an app or to a website, for one or several of the purposes presented here. List of IAB Vendors‎\|View Illustrations Personalised advertising and content, advertising and content measurement, audience research and services development 171 partners can use this purpose [x] Personalised advertising and content, advertising and content measurement, audience research and services development Use limited data to select advertising 134 partners can use this purpose [x] Switch Label Advertising presented to you on this service can be based on limited data, such as the website or app you are using, your non-precise location, your device type or which content you are (or have been) interacting with (for example, to limit the number of times an ad is presented to you). View Illustrations Object to Legitimate Interests Remove Objection Create profiles for personalised advertising 96 partners can use this purpose [x] Switch Label Information about your activity on this service (such as forms you submit, content you look at) can be stored and combined with other information about you (for example, information from your previous activity on this service and other websites or apps) or similar users. This is then used to build or improve a profile about you (that might include possible interests and personal aspects). Your profile can be used (also later) to present advertising that appears more relevant based on your possible interests by this and other entities. View Illustrations Use profiles to select personalised advertising 97 partners can use this purpose [x] Switch Label Advertising presented to you on this service can be based on your advertising profiles, which can reflect your activity on this service or other websites or apps (like the forms you submit, content you look at), possible interests and personal aspects. View Illustrations Create profiles to personalise content 42 partners can use this purpose [x] Switch Label Information about your activity on this service (for instance, forms you submit, non-advertising content you look at) can be stored and combined with other information about you (such as your previous activity on this service or other websites or apps) or similar users. This is then used to build or improve a profile about you (which might for example include possible interests and personal aspects). Your profile can be used (also later) to present content that appears more relevant based on your possible interests, such as by adapting the order in which content is shown to you, so that it is even easier for you to find content that matches your interests. View Illustrations Use profiles to select personalised content 39 partners can use this purpose [x] Switch Label Content presented to you on this service can be based on your content personalisation profiles, which can reflect your activity on this or other services (for instance, the forms you submit, content you look at), possible interests and personal aspects. This can for example be used to adapt the order in which content is shown to you, so that it is even easier for you to find (non-advertising) content that matches your interests. View Illustrations Measure advertising performance 159 partners can use this purpose [x] Switch Label Information regarding which advertising is presented to you and how you interact with it can be used to determine how well an advert has worked for you or other users and whether the goals of the advertising were reached. For instance, whether you saw an ad, whether you clicked on it, whether it led you to buy a product or visit a website, etc. This is very helpful to understand the relevance of advertising campaigns. View Illustrations Object to Legitimate Interests Remove Objection Measure content performance 74 partners can use this purpose [x] Switch Label Information regarding which content is presented to you and how you interact with it can be used to determine whether the (non-advertising) content e.g. reached its intended audience and matched your interests. For instance, whether you read an article, watch a video, listen to a podcast or look at a product description, how long you spent on this service and the web pages you visit etc. This is very helpful to understand the relevance of (non-advertising) content that is shown to you. View Illustrations Object to Legitimate Interests Remove Objection Understand audiences through statistics or combinations of data from different sources 97 partners can use this purpose [x] Switch Label Reports can be generated based on the combination of data sets (like user profiles, statistics, market research, analytics data) regarding your interactions and those of other users with advertising or (non-advertising) content to identify common characteristics (for instance, to determine which target audiences are more receptive to an ad campaign or to certain contents). View Illustrations Object to Legitimate Interests Remove Objection Develop and improve services 106 partners can use this purpose [x] Switch Label Information about your activity on this service, such as your interaction with ads or content, can be very helpful to improve products and services and to build new products and services based on user interactions, the type of audience, etc. This specific purpose does not include the development or improvement of user profiles and identifiers. View Illustrations Object to Legitimate Interests Remove Objection Use limited data to select content 41 partners can use this purpose [x] Switch Label Content presented to you on this service can be based on limited data, such as the website or app you are using, your non-precise location, your device type, or which content you are (or have been) interacting with (for example, to limit the number of times a video or an article is presented to you). View Illustrations Object to Legitimate Interests Remove Objection List of IAB Vendors‎ Use precise geolocation data 56 partners can use this special feature [x] Use precise geolocation data With your acceptance, your precise location (within a radius of less than 500 metres) may be used in support of the purposes explained in this notice. List of IAB Vendors‎ Actively scan device characteristics for identification 30 partners can use this special feature [x] Actively scan device characteristics for identification With your acceptance, certain characteristics specific to your device might be requested and used to distinguish it from other devices (such as the installed fonts or plugins, the resolution of your screen) in support of the purposes explained in this notice. List of IAB Vendors‎ Ensure security, prevent and detect fraud, and fix errors 105 partners can use this special purpose Always Active Your data can be used to monitor for and prevent unusual and possibly fraudulent activity (for example, regarding advertising, ad clicks by bots), and ensure systems and processes work properly and securely. It can also be used to correct any problems you, the publisher or the advertiser may encounter in the delivery of content and ads and in your interaction with them. List of IAB Vendors‎\|View Illustrations Deliver and present advertising and content 107 partners can use this special purpose Always Active Certain information (like an IP address or device capabilities) is used to ensure the technical compatibility of the content or advertising, and to facilitate the transmission of the content or ad to your device. List of IAB Vendors‎\|View Illustrations Match and combine data from other data sources 77 partners can use this feature Always Active Information about your activity on this service may be matched and combined with other information relating to you and originating from various sources (for instance your activity on a separate online service, your use of a loyalty card in-store, or your answers to a survey), in support of the purposes explained in this notice. List of IAB Vendors‎ Link different devices 55 partners can use this feature Always Active In support of the purposes explained in this notice, your device might be considered as likely linked to other devices that belong to you or your household (for instance because you are logged in to the same service on both your phone and your computer, or because you may use the same Internet connection on both devices). List of IAB Vendors‎ Identify devices based on information transmitted automatically 99 partners can use this feature Always Active Your device might be distinguished from other devices based on information it automatically sends when accessing the Internet (for instance, the IP address of your Internet connection or the type of browser you are using) in support of the purposes exposed in this notice. List of IAB Vendors‎ Save and communicate privacy choices 84 partners can use this special purpose Always Active The choices you make regarding the purposes and entities listed in this notice are saved and made available to those entities in the form of digital signals (such as a string of characters). This is necessary in order to enable both this service and those entities to respect such choices. List of IAB Vendors‎\|View Illustrations Cookie List Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Clear [x] checkbox label label Apply Cancel Confirm My Choices Reject All Allow All
13362	https://www.dayuwz.com/What-Can-Go-Wrong-with-Distillation-id41590656.html	Alcohol distillation--What Can Go Wrong with Distillation? - DAEYOO : All Product Name Product Keyword Product Model Product Summary Product Description Multi Field Search Pусский简体中文 Home Product List Distilling Equipment Brewing Equipment Pharmaceutical Equipment Chemical Equipment IBC Totes/Storage Tank ASME Pressure Vessel Accessory About us Company Profile Professional Equipment Certificate The Journey Of Struggle After Sales Project Case Domestic Case Abroad Case Video Catalogue Exhibition Media/Press Blog FAQ Company News Contact Us Pусский 简体中文 : All Product Name Product Keyword Product Model Product Summary Product Description Multi Field Search Home Product List Distilling Equipment Brewing Equipment Pharmaceutical Equipment Chemical Equipment IBC Totes/Storage Tank ASME Pressure Vessel Accessory About us Company Profile Professional Equipment Certificate The Journey Of Struggle After Sales Project Case Domestic Case Abroad Case Video Catalogue Exhibition Media/Press Blog FAQ Company News Contact Us What Can Go Wrong with Distillation? Home » Media/Press » FAQ » What Can Go Wrong with Distillation? What Can Go Wrong with Distillation? Views: 39 Author: Site Editor Publish Time: 2024-10-16 Origin: Site Inquire Distillation is an interesting method and enables one to convert the raw materials into quality spirits at all times. However, just like any other complicated process, it has its disadvantages. That is why understanding what can go wrong during distillation is a requirement for all those who will be using any distilling equipment or distillation equipment.In this article, we will explore common issues that can arise, focusing on the use of still equipment, including moonshine stills, alcohol distillers, and copper stills. 1.Insufficient Preparations Right preparatory work must be undertaken before moving to the distillation stage. Neglecting to prepare can cause problems of several kinds: ●Contamination: Any impurities in the ingredients or equipment will affect the taste and quality of the final product. Always use cleaned still equipment and fresh ingredients. ●Inaccurate Measurements:Some amounts used as ingredients may be too small or too large such that the fermentation process is ineffective. It is important to note that measurement is very important in this process as it affects the quality of the distillate. 2.Issues with Fermentation Fermentation is an important phase during distillation. If something goes wrong here, it can have significant repercussions. ●Stuck Fermentation:Sometimes fermentation can stall, often due to insufficient yeast activity. This can happen when the temperature is low or there is not enough sugar for the yeast. ●Off-Flavors: If the fermentation period is improperly controlled, the extraction will have off-flavors. Extreme adjustments in temperature or other contaminants may cause the growth of unwanted compounds in the distal alcohol. 3.Distillation Process Errors Upon completion of fermentation, the next step is distillation. This stage can proceed without disturbances; however, some of the problems are worth noting: ●Low or High Temperature: Low temperature makes the alcohol sluggish therefore there are minimal yields. High temperature succeeds in turning the alcohol, but the tips usually suffer a lot of overheating, giving a burnt taste or even damaging the distillation equipment. ●Temperature Control:Controlled conditions in terms of temperature are critical for effective distillation. Deviations from the set temperature can cause impediments in the distillation process and thus the separation of alcohol and other compounds. 4.Equipment Malfunctions It is worth noting that poor or faulty distilling equipment can also lead to devastating results in the distilling process. ●Loose Stills:The most perfidious leak is a lack of water more commonly known as a dry running still. In your stills, drying occurs primarily as a result of loss of alcohol vapors and low vapor power which reduces efficiency and it can be very dangerous if vapour escapes. ●Dirty Parts:Over a period of time, the distillation devices can get choked with excess residues that hinders the normal vapor flow and quality of the distillate. Cleaning and periodic maintenance are vital to avoid such a situation. 5.Safety Hazards Safety should be taken into account in any arbet. To be safe, there are some hazards to watch out for. ●Flammable Vapors:Distilling produces fire-hazardous vapors. These vapours, if not trapped or ventilated adequately, can lead to fire or an explosion. Work always in a well-ventilated workplace and keep whatever is fuelling the fire away from your distillation unit. ●Pressure Build-Up:In closed systems, pressure can cumulatively increase if no regulation is clearly observed. If too much pressure is maintained, opportunities for explosion or even the destruction of the equipment can be there. A pressure relief valve is a necessity for every alcohol distiller. 6. Quality Control Issues The final product should have a quality of some level. Operational concern of alcohol distillation allows for the expectations that there will be some operational concerns. ●Poor Tasting Spirits:Off-tastes from fermentation and the distillation processes also make unpleasant taking spirits. Factory tasting and quality check procedures as production goes on help solving problems like this in time. ●Alcohol Content Variability: Different levels of alcohol concentration are likely when distillation is not uniform, making it difficult producing the desired product specifications. Measure alcohol levels with hydrometers or other tools to guarantee the right quantities of alcohol content. 7. Environmental Factors. External factors are also important during the distillation process. ●Temperature and Humidity:Fermentation and distillation can suffer from certain external conditions. High humidity may cause problems with yeast performance while variance in temperatures may affect the processes of distillation. ●Altitude:For high altitude operations, the distillation process must be modified since the boiling points are lower. Not taking into account this adjustment can translate to poor distillation. 8.Regulatory Compliance For people distilling for the purpose of making profit, barely focused ought to be on local laws and regulation. ●Licensing Issues:Engaging in the art of distilling alcohol without the right licenses spells trouble. Make sure you know, and adhere, to the law wherever you are with a distillation process. ●Quality Standards: Qualities of spirits are expected to meet certain thresholds in most countries. Ignoring this may attract fines and recall of the products. Conclusions Distillation is a complex art and science that demands a meticulous approach. A wide range of factors intrude from the time of preparation to fermentation to the actual distillation process. Knowing areas of possible trouble happening e.g: equipment failure, fermentation failure, safety risks and others enables a people to the risk around.READ MORE. Investing in quality distilling equipment can really improve your distilling experience, for example, the reliable moonshine still and the copper still. Distillation also depends on proper equipment maintenance and temperature as well as safety measures that should be implemented. With proper attention, time and commitment, most of the common mistakes can be avoided and still great spirits can be made. You are going to need knowledge, regarless of the fact wether you are a hobbiest or running a business, if you are going to be distilling. Previous :Is Home Distilling Legal in Australia? Next :Essential Oil Distiller: Your Guide to Extraction Excellence distilling equipmentDistillation equipmentstill equipmentMoonshine Stillalcohol distillercopper stillCopper Distillerydistillery equipmentcommercial distillery equipmentAlcohol Distillation Equipment Daeyoo Tech. Co., LTD.Wenzhou is a comprehensive emerging enterprise that integrates product design, research and development, manufacturing and engineering. DYE is at forefront of the R & D and manufacturing of distillation equipment in China and also one of the first manufacturers of copper distillers to learn and produce red copper distillers. Product List Distillation Equipment Brewery Equipment Pharmaceutical Equipment IBC TOTES Pressure Vessel AMSE Pressure Vessel Quick Links Home Products About us Project case Media/Press Contact Us Privacy Policy Contact Us Contact Us Submit +86-577-8680 6088 sales@dayuwz.com +86-15757874062 dye.salesteam@gmail.com Copyright © 2025 DAEYOO TECH. CO., LTD.WENZHOU.All rights reserved. Sitemap ｜Privacy Policy\| Support By Leadong. top
13363	https://www.lesswrong.com/w/well-defined	Well-defined — LessWrong LESSWRONG Wikitags LW Login Discussion 0 3 Well-defined Patrick Stevens • Specific instances • Functions Well-defined Discussion 0 3 Edited by Patrick Stevenslast updated 7th Aug 2016 Name "Well-defined" is a slightly fuzzy word in mathematics. Broadly, an object is said to be "well-defined" if it has been given a definition that is completely unambiguous, can be executed without regard to any arbitrary choices the mathematician might make, or generally is crisply defined. (The Wikipedia page on well-definedness contains many examples for those who are more comfortable with mathematical notation.) Specific instances Functions One of the most common uses of the phrase "well-defined" is when talking about functions. A function is well-defined if it really is a bona fide function. This usually manifests itself as the following: Whenever x=y , we have f(x)=f(y) : that is, the output of the function doesn't depend on how we specify the input to the function, only on the input itself. This property is often pretty easy to check. For instance, the function from N to itself given by n↦n+1 is "obviously" well-defined: it's trivially obvious that if n=m then f(n)=f(m) . However, sometimes it is not so easy. The function N→N given by "take the number of prime factors" is not obviously well-defined, because it could in principle be the case that some number n is equal to both p 1 p 2 p 3 and q 1 q 2 for some primes p 1,p 2,p 3,q 1,q 2 ; then our putative function might plausibly attempt to output either 3 or 2 on the same natural number input n , so the function would not be well-defined. (It turns out that there is a non-trivial theorem, the Fundamental Theorem of Arithmetic, guaranteeing that this function is in fact well-defined.) Well-definedness in this context comes up very often when we are attempting to take a quotient. The fact that we can take the quotient of a set X by an equivalence relation ∼ is tantamount to saying: The function X→X∼ , given by x↦[x] the equivalence class of X , is well-defined. Another, different, way a function could fail to be well-defined is if we tried to take the function N→N given by n↦n−5 . This function is unambiguous, but it's not well-defined, because on the input 2 it tries to output −3 , which is not in the specified codomain. Summaries You can edit summaries by clicking on them, reorder them by dragging, or add a new one (up to 3). By default you should avoid creating more than one summary unless the subject matter benefits substantially from multiple kinds of explanation. Summary "Well-defined" is a slightly fuzzy word in mathematics. An object is said to be "well-defined" if it has been given a definition that is completely unambiguous, can be executed without regard to any arbitrary choices the mathematician might make, or generally is crisply defined. Tab title "Well-defined" is a slightly fuzzy word in mathematics. An object is said to be "well-defined" if it has been given a definition that is completely unambiguous, can be executed without regard to any arbitrary choices the mathematician might make, or generally is crisply defined. Cancel Submit To pick up a draggable item, press the space bar. While dragging, use the arrow keys to move the item. Press space again to drop the item in its new position, or press escape to cancel. Cancel Submit "Well-defined" is a slightly fuzzy word in mathematics. Broadly, an object is said to be "well-defined" if it has been given a definition that is completely unambiguous, can be executed without regard to any arbitrary choices the mathematician might make, or generally is crisply defined. (The Wikipedia page on well-definedness contains many examples for those who are more comfortable with mathematical notation.) Specific instances Functions One of the most common uses of the phrase "well-defined" is when talking about functions. A function is well-defined if it really is a bona fide function. This usually manifests itself as the following: Whenever x=y, we have f(x)=f(y): that is, the output of the function doesn't depend on how we specify the input to the function, only on the input itself. This property is often pretty easy to check. For instance, the function from N to itself given by n↦n+1 is "obviously" well-defined: it's trivially obvious that if n=m then f(n)=f(m). However, sometimes it is not so easy. The function N→N given by "take the number of prime factors" is not obviously well-defined, because it could in principle be the case that some number n is equal to both p 1 p 2 p 3 and q 1 q 2 for some primes p 1,p 2,p 3,q 1,q 2; then our putative function might plausibly attempt to output either 3 or 2 on the same natural number input n, so the function would not be well-defined. (It turns out that there is a non-trivial theorem, the Fundamental Theorem of Arithmetic, guaranteeing that this function is in fact well-defined.) Well-definedness in this context comes up very often when we are attempting to take a quotient. The fact that we can take the quotient of a setX by an equivalence relation∼ is tantamount to saying: The function X→X∼, given by x↦[x] the equivalence class of X, is well-defined. Another, different, way a function could fail to be well-defined is if we tried to take the function N→N given by n↦n−5. This function is unambiguous, but it's not well-defined, because on the input 2 it tries to output −3, which is not in the specified codomain. Parents: Mathematics Parents Mathematics 1355 Upload image from computer Insert table Horizontal line Insert display math block Insert media Collapsible Section Footnote Insert Claim Click to edit block Drag to move Previous Next Upload image from computer Insert table Horizontal line Insert display math block Insert media Collapsible Section Footnote Insert Claim Click to edit block Drag to move Previous Next
13364	https://www.quora.com/What-is-the-difference-between-a-fatty-acid-and-an-oil-What-is-the-difference-between-a-fatty-acid-and-a-fat	What is the difference between a fatty acid and an oil? What is the difference between a fatty acid and a fat? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In What is the difference between a fatty acid and an oil? What is the difference between a fatty acid and a fat? All related (33) Sort Recommended Lewis/Rosemary Moore Former Chemical engineer, general mgr, prev atomic energy at Retired (terminally unemployed) · Author has 2K answers and 793.3K answer views ·2y Fatty acids are the building blocks of fats. The main difference between fats and oils is that the former solidify at room temperature. Chemically, they are described as “saturated” as opposed to unsaturated and polyunsaturated. Upvote · Sponsored by Interactive Brokers Check out IBKR Desktop. Our next-gen trading platform combines streamlined design and powerful tools to help you succeed. Learn More 999 105 Related questions More answers below What is the difference between a fat and an oil? What is the difference between a lipid and a fatty acid? What is the difference between fat and fatty acid? What are the differences between fatty acids and oils? How do you determine free fatty acid in oil? What is everything you know about healthy fats and fatty acids? Mahadevappa K L Former Retired Professor (1975–2014) · Author has 4.9K answers and 1.9M answer views ·2y Fatty acid is s a carboxylic acid with an aliphatic chain, which is either saturated or unsaturated. Fats and oils are made from varying fatty acids. The difference between them is that the fats are composed of high amounts of saturated fatty acids. Whereas, oils are composed of mainly unsaturated fatty acids. Animals a... Upvote · 9 1 Bob Mouk Former Industrial Research Chemist / Chemistry Professor · Author has 2.1K answers and 787.3K answer views ·2y Fats and oils are triglycerides. Fatty acids make fats (But they make oils too. There aren’t any “Oily acids”). Fats are solids at room temperature, oils are liquids. Upvote · Subra Velu Passionate about Machine Learning and Cycling. · Author has 3.9K answers and 14.6M answer views ·8y Related What is the difference between fats and oils? Oil is fat but not the other way around. Fat derived from plant sources of seeds, nuts and fruits are called vegetable oils. Sesame oil (from seeds), almond oil (from nuts) and Olive oil (from fruit) are examples of vegetable oil. They are also plant sources fat. But animal source fat are not called oils but carry a unique name. Butter, Ghee and Lard are animal fats but not called as oils. In general animal fat contain more of saturated fatty acids, which are very stable with long shelf life and even at boiling or smoking. The vegetable oil has mix of polyunsaturated fat, monounsaturated fat an Continue Reading Oil is fat but not the other way around. Fat derived from plant sources of seeds, nuts and fruits are called vegetable oils. Sesame oil (from seeds), almond oil (from nuts) and Olive oil (from fruit) are examples of vegetable oil. They are also plant sources fat. But animal source fat are not called oils but carry a unique name. Butter, Ghee and Lard are animal fats but not called as oils. In general animal fat contain more of saturated fatty acids, which are very stable with long shelf life and even at boiling or smoking. The vegetable oil has mix of polyunsaturated fat, monounsaturated fat and less of saturated fat (except coconut oil). Though the unsaturated fat are believed to be healthy, they re unstable and have a short shelf life before turning rancid. They turn rancid on heating or cooking too. Rancid oils cause more health hazard and not worth the benefits of unsaturated fat. Upvote · 9 5 9 1 Promoted by Almedia Marc Hammes Personal finance journalist @ Almedia ·Updated Apr 4 What are some legit ways to make money online at home? Like many of you reading this, I’ve been looking for ways to earn money online in addition to my part-time job. But you know how it is – the internet is full of scams and shady-grady stuff, so I spent weeks trying to find something legit. And I finally did! Freecash surprised me in all the right ways. I’ve earned over $1,000 in one month without ‘living’ on the platform. I was skeptical right up until the moment I cashed out to my PayPal. What is Freecash all about? Basically, it’s a platform that pays you for testing apps and games and completing surveys. This helps developers improve their appl Continue Reading Like many of you reading this, I’ve been looking for ways to earn money online in addition to my part-time job. But you know how it is – the internet is full of scams and shady-grady stuff, so I spent weeks trying to find something legit. And I finally did! Freecash surprised me in all the right ways. I’ve earned over $1,000 in one month without ‘living’ on the platform. I was skeptical right up until the moment I cashed out to my PayPal. What is Freecash all about? Basically, it’s a platform that pays you for testing apps and games and completing surveys. This helps developers improve their applications while you make some money. You can earn by downloading apps, testing games, or completing surveys. I love playing games, so that’s where most of my earnings came from (oh, and my favorites were Warpath, Wild Fish, and Domino Dreams). There’s a variety of offers (usually, the higher-paying ones take more time). Some games can pay up to $1,000 for completing a task, but these typically require more hours to finish. On average, you can easily earn $30–50/day. You pick your options — you’re free to choose whatever apps, games, and surveys you like. Of course, it’s not like you can spend 5 minutes a day and become a millionaire. But you can build a stable income in reasonable time, especially if you turn it into a daily habit. Why did I like Freecash? It’s easy. I mean it. You don’t have to do anything complicated. All you need is to follow the task and have some free time to spend on it. For some reason, I especially enjoyed the game Domino Dreams. My initial goal was to complete chapter 10 to get my first $30, but I couldn’t stop playing and ended up completing chapter 15. It was lots of fun and also free money: $400 from that game alone. No experience needed. Even if you’ve never done any ‘testing’ before, you can do this. You get straightforward task descriptions, so it’s impossible to go wrong. A task you might expect is something like: Download this game and complete all challenges in 14 days. You can do it from anywhere. I was earning money while taking the bus, chilling on the couch, and during my breaks. Fast cashing out. I had my earnings in my PayPal account in less than 1 day. I’m not sure how long it takes for other withdrawal methods (crypto, gift cards, etc.), but it should be fast as well. You can earn a lot if you’re consistent. I’ve literally seen users in the Leaderboard making $3,000 in just one month. Of course, to get there, you need time, but making a couple of hundred dollars is really easy and relatively fast for anyone. Don’t miss these PRO tips to earn more: I feel like most users don’t know about these additional ways to make more money with Freecash: Free promo codes: You can follow Freecash on social media to get weekly promo codes for free coins, which you can later exchange for money. Daily rewards and bonuses: If you use the platform daily, you’ll get additional bonuses that help you earn more. In-app purchases to speed up processes: While playing, you can buy items to help speed up task completion. It’s optional, but it really saved me time, and I earned 4x more than I spent. Choose the highest-paying offers: Check New Offers and Featured Offers to get the best opportunities that pay the most. Honestly, I still can’t believe I was able to earn this much so easily. And I’ve actually enjoyed the whole process. So, if you’re looking for some truly legit ways to earn money online, Freecash is a very good option. Upvote · 15.3K 15.3K 1.6K 1.6K 999 363 Related questions More answers below What is a non-essential fatty acid? How are fatty acids recovered from fats and oils? How do the different types of fatty acids differ? What are the benefits of higher fatty acids? What are higher fatty acids? Thomas Carr PhD in Chemistry, Case Western Reserve University (Graduated 1981) · Author has 571 answers and 135K answer views ·2y Related What is the difference between a fat and an oil? What is the difference between a lipid and a fatty acid? A fat is a triglyceride that is solid at room temperature. The fatty acids comprising it are saturated (few if any double bonds between successive carbon atoms). Often, but not always, it is obtained from animal sources. An oil is a triglyceride that is liquid at room temperature. The fatty acids comprising it are unsaturated (double bonds, where hydrogen atoms can be added). They are often obtained from plant sources. Triglycerides arise from the esterification of three fatty acids with the tri-functional alcohol, glycerol. The fatty acids consist of long chains of carbon atoms. Examples of whi Continue Reading A fat is a triglyceride that is solid at room temperature. The fatty acids comprising it are saturated (few if any double bonds between successive carbon atoms). Often, but not always, it is obtained from animal sources. An oil is a triglyceride that is liquid at room temperature. The fatty acids comprising it are unsaturated (double bonds, where hydrogen atoms can be added). They are often obtained from plant sources. Triglycerides arise from the esterification of three fatty acids with the tri-functional alcohol, glycerol. The fatty acids consist of long chains of carbon atoms. Examples of which are oleum and stearic acids. All triglycerides are lipids; but not all lipids are triglycerides. An example of such is cholesterol. Upvote · Ahmad Shamim Self Employed ·3y Related What is the difference between fats and oils? To put very simply, there is no essential difference between the two except in terms of their melting point. Fats have a higher melting point of the two and hence remain solid at room temperature. This difference can be attributed to the fact that a fat molecules being saturated(with hydrogen bonds) and hence single bonded, have a molecular configuration that makes it easy for them to stack on top of each other and be held tightly together. As a result, Fats have a relatively high melting point, enabling them to stay solid at room temperatures as opposed to oils(having double bonded carbon atoms) Continue Reading To put very simply, there is no essential difference between the two except in terms of their melting point. Fats have a higher melting point of the two and hence remain solid at room temperature. This difference can be attributed to the fact that a fat molecules being saturated(with hydrogen bonds) and hence single bonded, have a molecular configuration that makes it easy for them to stack on top of each other and be held tightly together. As a result, Fats have a relatively high melting point, enabling them to stay solid at room temperatures as opposed to oils(having double bonded carbon atoms), whose molecules are not strongly bonded to each other and thus found in liquid state at room temperatures. Upvote · 9 1 Sponsored by Adsy Guest blogging platform. Explore over 100K websites ready to publish your articles. Sign up now. Sign Up 9 1 Avneendra Prasad Former Ex Consultant Pathologist, Medical Superintendent at Government of NCT of Delhi (1980–2019) · Author has 709 answers and 611.8K answer views ·5y Related What's the difference between a fat and a lipid? Long answer to explain : Lipids include a diverse group of compounds that are largely nonpolar in nature. They are hydrocarbons that include mostly nonpolar carbon–carbon or carbon–hydrogen bonds. Non-polar molecules are hydrophobic (“water fearing”), and insoluble in water. Lipids include many groups of biological chemicals like Fats / Oils, waxes, Phospholipids, Sterols and perform many different functions in body and a cell. Cells store energy for long-term use in the form of fats. Lipids also provide insulation from the environment for plants and animals (Figure 1). For example, they help keep Continue Reading Long answer to explain : Lipids include a diverse group of compounds that are largely nonpolar in nature. They are hydrocarbons that include mostly nonpolar carbon–carbon or carbon–hydrogen bonds. Non-polar molecules are hydrophobic (“water fearing”), and insoluble in water. Lipids include many groups of biological chemicals like Fats / Oils, waxes, Phospholipids, Sterols and perform many different functions in body and a cell. Cells store energy for long-term use in the form of fats. Lipids also provide insulation from the environment for plants and animals (Figure 1). For example, they help keep aquatic birds and mammals dry when forming a protective layer over fur or feathers because of their water-repellant hydrophobic nature. Lipids are also the building blocks of many hormones and are an important constituent of all cellular membranes. Whereas Fat / Oil is a subgroup of Lipids and fat molecule consists of two main components—glycerol and fatty acids. Glycerol is an organic compound (alcohol) with three carbons, five hydrogens, and three hydroxyl (OH) groups. Fatty acids have a long chain of hydrocarbons to which a carboxyl group is attached, hence the name “fatty acid.” The number of carbons in the fatty acid may range from 4 to 36; most common are those containing 12–18 carbons. In a fat molecule, the fatty acids are attached to each of the three carbons of the glycerol molecule with an ester bond through an oxygen atom During this ester bond formation, three water molecules are released. The three fatty acids in the triacylglycerol may be similar or dissimilar. Fats are also called triacylglycerols or triglycerides because of their chemical structure. Some fatty acids have common names that specify their origin. For example, palmitic acid, a saturated fatty acid, is derived from the palm tree. Arachidic acid is derived from Arachis hypogea, the scientific name for groundnuts or peanuts. Fatty acids in fat or triglycerides may be saturated or unsaturated. In a fatty acid chain, if there are only single bonds between neighboring carbons in the hydrocarbon chain, the fatty acid is said to be saturated. Saturated fatty acids are saturated with hydrogen; in other words, the number of hydrogen atoms attached to the carbon skeleton is maximized. Stearic acid is an example of a saturated fatty acid Stearic acid is a common saturated fatty acid. When the hydrocarbon chain contains a double bond, the fatty acid is said to be unsaturated. Oleic acid is an example of an unsaturated fatty acid (Figure 4). Oleic acid is a common unsaturated fatty acid. Most unsaturated fats are liquid at room temperature and are called oils. If there is one double bond in the molecule, then it is known as a monounsaturated fat (e.g., olive oil), and if there is more than one double bond, then it is known as a polyunsaturated fat (e.g., canola oil). A fat may contain similar or different fatty acids attached to glycerol. Long straight fatty acids with single bonds tend to get packed tightly and are solid at room temperature. Animal fats with stearic acid and palmitic acid (common in meat) and the fat with butyric acid (common in butter) are examples of saturated fats. In animals fat is stored in adipose tissue in fat cells (adipocytes) and in plants, fat or oil is stored in mainly seeds and is used as a source of energy. Unsaturated fats or oils are usually of plant origin and contain cis unsaturated fatty acids. Cis and trans indicate the configuration of the molecule around the double bond. If hydrogens are present in the same plane, it is referred to as a cis fat; if the hydrogen atoms are on two different planes, it is referred to as a trans fat. The cis double bond causes a bend or a “kink” that prevents the fatty acids from packing tightly, keeping them liquid at room temperature. Olive oil, corn oil, canola oil, and cod liver oil are examples of unsaturated fats. Unsaturated fats help to lower blood cholesterol levels whereas saturated fats contribute to plaque formation in the arteries. Saturated fatty acids have hydrocarbon chains connected by single bonds only. Unsaturated fatty acids have one or more double bonds. Each double bond may be in a cis or trans configuration. In the cis configuration, both hydrogens are on the same side of the hydrocarbon chain. In the trans configuration, the hydrogens are on opposite sides. Essential fatty acids are fatty acids required but not synthesized by the human body. Consequently, they have to be supplemented through ingestion through the diet. Wax are made up of long fatty acid chains esterified to long-chain alcohols and cover the feathers of some aquatic birds and the leaf surfaces of some plants. Because of the hydrophobic nature of waxes, they prevent water from sticking on the surface. Phospholipids are major constituents of the plasma membrane, the outermost layer of animal cells. Like fats, they are composed of fatty acid chains attached to a glycerol or sphingosine backbone. Instead of three fatty acids attached as in triglycerides (fat), however, there are two fatty acids forming diacylglycerol, and the third carbon of the glycerol backbone is occupied by a modified phosphate group. A phospholipid is a molecule with two fatty acids and a modified phosphate group attached to a glycerol backbone. The phosphate may be modified by the addition of charged or polar chemical groups. Two chemical groups that may modify the phosphate, choline and serine, are shown here. Both choline and serine attach to the phosphate group at the position labeled R. A phosphate group alone attached to a diaglycerol does not qualify as a phospholipid; it is phosphatidate (diacylglycerol 3-phosphate), the precursor of phospholipids. The phosphate group is modified by an alcohol. Phosphatidylcholine and phosphatidylserine are two important phospholipids that are found in plasma membranes.A phospholipid is an amphipathic molecule, meaning it has a hydrophobic and a hydrophilic part. The fatty acid chains are hydrophobic and cannot interact with water, whereas the phosphate-containing group is hydrophilic and interacts with water. The phospholipid bilayer is the major component of all cellular membranes. The hydrophilic head groups of the phospholipids face the aqueous solution. The hydrophobic tails are sequestered in the middle of the bilayer. The head is the hydrophilic part, and the tail contains the hydrophobic fatty acids. In a membrane, a bilayer of phospholipids forms the matrix of the structure, the fatty acid tails of phospholipids face inside, away from water, whereas the phosphate group faces the outside, towards aqueous side. Steroids: Unlike the fats/oils or waxes and phospholipids explained earlier, steroids have a fused ring structure. Though they do not resemble the other lipids, they are grouped with lipids because they are also hydrophobic and insoluble in water. All steroids have four linked carbon rings and several of them, like cholesterol, have a short tail. Many steroids also have the –OH functional group, which puts them in the alcohol classification (sterols). Steroids such as cholesterol and cortisol are composed of four fused hydrocarbon rings. Cholesterol is the most common steroid. Cholesterol is mainly synthesized in the liver and is the precursor to many steroid hormones such as testosterone and estradiol, which are secreted by the gonads and endocrine glands. It is also the precursor to Vitamin D. Cholesterol is also the precursor of bile salts, which help in the emulsification of fats and their subsequent absorption by cells. It is a component of the plasma membrane of animal cells and is found within the phospholipid bilayer. Being the outermost structure in animal cells, the plasma membrane is responsible for the transport of materials and cellular recognition and it is also involved in cell-to-cell communication. IN SUMMARY LIPIDS ARE BROAD GROUPS CONTAINING FATS/OILS, WAXES, PHOSPHOLIPIDS AND STEROIDS. Upvote · 99 27 9 3 9 1 Tahir Mehmood PhD from PMAS Arid Agriculture University Rawalpindi (Graduated 2018) ·4y Related What is the difference between fats and oils? Fats and oil are both chiefly comprised of triglycerides (glycerol + fatty acid). In fats, a major proportion of fatty acid comprised of saturated fatty acids (no double bond). These saturated fatty acids have a straight structure. Hence these are closely packed which makes them solid at room temperature. On the other end, oil has a higher proportion of unsaturated fatty acids which have a bent structure. Hence they can not closely pack and appear as a liquid at room temperature. In summary, fats are solid at room temperature and oil are liquid at room temperature Upvote · 9 1 Promoted by Grammarly Grammarly Knows English ·Updated 2y What are the different errors and mistakes in English? Grammatical mistakes can be easy to make. But fear not! Our team at Grammarly has compiled a handy list of common grammatical errors to help make your writing accurate, clear, and professional. Ambiguous (“Squinting”) modifiers Incorrect: Listening to loud music slowly gives me a headache. Correct: When I listen to loud music, I slowly develop a headache. A squinting modifier is a misplaced modifier that, because of its location in a sentence, could modify either the phrase that precedes it or the one that follows it. In the example sentence, is the subject listening to music slowly or slowly gett Continue Reading Grammatical mistakes can be easy to make. But fear not! Our team at Grammarly has compiled a handy list of common grammatical errors to help make your writing accurate, clear, and professional. Ambiguous (“Squinting”) modifiers Incorrect: Listening to loud music slowly gives me a headache. Correct: When I listen to loud music, I slowly develop a headache. A squinting modifier is a misplaced modifier that, because of its location in a sentence, could modify either the phrase that precedes it or the one that follows it. In the example sentence, is the subject listening to music slowly or slowly getting a headache? To correct a squinting modifier, move its position in the sentence to clarify to the reader which word you intend to modify. Misuse of lie/lay Incorrect: He was laying on the couch. Correct: He was lying on the couch. If you plan to place or put an object somewhere, such as a plate on a table, you should use “lay.” If you intend to stretch out on a bed for a nap, you should use “lie.” The verb “lie” is an intransitive verb, which means it does not need an object. The transitive verb “lay” requires an object. It may take some getting used to this “lay” or “lie” business; after all, misuse of these verbs is common. But if you remember to lay down your fork before you’re full, then you won’t have to lie down later from overeating. Comma splices Incorrect: He was very hungry, he ate a whole pizza. Correct: He was very hungry. He ate a whole pizza. He was very hungry, so he ate a whole pizza. To splice means to connect or join. When a writer joins two independent sentences with a comma instead of separating them with a period or coordinating conjunction, that’s a comma splice. The comma has its jobs to do, but connecting two independent sentences isn’t one of them. Besides, the period gets testy when his sister, the comma, steals his thunder. Periods have their jobs, and so do commas, but never the twain shall meet—unless it’s in the form of a semicolon. Semicolons can also take the place of a coordinating conjunction, such as “and,” “but,” or “so,” among others. Run-on sentences Incorrect: Lila enjoyed the bouquet of tulips John gave her on prom night however she prefers roses. Correct: Lila enjoyed the bouquet of tulips John gave her on prom night; however, she prefers roses. Run-on sentences, also known as fused sentences, occur when two complete sentences are squashed together without using a coordinating conjunction or proper punctuation, such as a period or a semicolon. Run-on sentences can be short or long. A long sentence isn’t necessarily a run-on sentence. To avoid run-on sentences, see if there is more than one idea communicated by two or more independent clauses. In our examples, there are two complete sentences: Example: Lily enjoyed the bouquet of tulips John gave her on prom night. Example: She prefers roses. Both sentences are complete ideas by themselves; therefore, use a semicolon or a period to indicate that they are separate independent clauses. Using “could of” instead of “could have” Incorrect: Sam could of received an A on his essay, but he made too many grammatical mistakes. Correct: Sam could have received an A on his essay, but he made too many grammatical mistakes. “Could have” is always correct; “could of” never is. Writers probably make this grammar gaffe because, when we speak, the contraction “could’ve” sounds an awful lot like “could of.” Tautologies Incorrect: Jack made a water pail with his own hands for Jill. Correct: Jack made a water pail for Jill. Tautologies express the same thing twice with different words. In our example, the word “made” implies that Jack used his own two hands to create the pail. The prepositional phrase “with his own hands” creates a redundancy. Once you know what they are, it’s fun to discover tautologies: dilapidated ruins, close proximity, added bonus, large crowd...The list goes on and on! After reading through this list of common grammatical mistakes, you might be wondering how to remember all these rules as you write. The free Grammarly for Windows and Mac is here to help. It provides a second set of eyes on your writing in real-time, so you can avoid everyday grammar and spelling errors. Moreover, Grammarly Premium offers features that evaluate conciseness and readability as well as vocabulary enhancement suggestions and genre-specific writing style checks. These tools can help you identify when you are making these common writing mistakes so you can proactively learn and improve your writing. Upvote · 1.5K 1.5K 999 520 99 55 Fiza Ammar 4y Related What is the difference between fats and oils? The main difference between fats and oils is: Oils are composed of unsaturated fatty acids. They take a liquid form at room temperature. Fats, on the other hand, are composed of high saturated fatty acids that take the form of solid at room temperature. Want to know more; read the article: Upvote · 9 2 Bart Loews Your body has no warranty, but I'll help you RTFM · Author has 13.6K answers and 64.6M answer views ·9y Related What is the difference between fats and oils? Fats are molecules made up of a fatty acid chain with a gycerol head. They're are several different kinds of fats determined by the number of carbon atoms in the fatty acid chain and how many double bonds there are in the chain and where the figure bonds are. Oils are made up of a combination of different fats in varying concentrations. Upvote · 9 3 Aman Puri Founder at Steadfast Nutrition (2017–present) · Author has 626 answers and 299.8K answer views ·Apr 16 Related What are the different types of fatty acids? Fatty acids are building blocks of lipids. They are mono-carboxylic acid having a terminal carboxylate group. There two types of fatty acids: Saturated fatty acids are generally solids at room temperature due to large number of carbon atoms. In plants occur in palm oil. They have hydrophobic tail of zig-zag configuration and because of single bond they are free to rotate and are thus more flexible and elongate. These are mostly solid and are found in animal fat like butter, meat, and whole milk. But some saturated fatty acids are also found in plant sources like vegetable oil, coconut oil, ghee, Continue Reading Fatty acids are building blocks of lipids. They are mono-carboxylic acid having a terminal carboxylate group. There two types of fatty acids: Saturated fatty acids are generally solids at room temperature due to large number of carbon atoms. In plants occur in palm oil. They have hydrophobic tail of zig-zag configuration and because of single bond they are free to rotate and are thus more flexible and elongate. These are mostly solid and are found in animal fat like butter, meat, and whole milk. But some saturated fatty acids are also found in plant sources like vegetable oil, coconut oil, ghee, and peanut oil. Unsaturated fatty acids due to the presence of a double bond. Lowers melting point, and therefore lipids with unsaturated acids are mostly oily and become liquid at room temperature.The number of double bond show the degree of unsaturation. Most vegetable oils and fish oils are some of the important sources of unsaturated fatty acids. Unsaturated fatty acids are mostly found via plant sources like vegetable oil, sunflower oil, mustard oil, avocado oil, etc. Monounsaturated fats contain one double bond between the carbon atoms in the fatty acid chain. Also they are found in high concentrations in olive, peanut, and canola oils, avocados, nuts such as almonds, hazelnuts, and pecans, seeds such as pumpkin and sesame seeds. Polyunsaturated fats containing two or more double bonds in their carbon chain structure. These fatty acids are essential for human health and are categorised into omega-3 and omega-6 type. They are found in high concentrations in sunflower, corn, soybean, and flaxseed oils, walnuts, flax seeds and fatty fish. Upvote · Jason Dean Lives in Toronto, ON (1985–present) · Author has 3.1K answers and 1.2M answer views ·3y Related What is the difference between fat and fatty acid? A fat ,also called a triglcyeride,is composed of 3 fatty acids esterified to the hydroxyl groups of a glycerol backbone (Phospholipids are diacylglycerols; where only two hydroxyls are esterified to fatty acids,and the third one is attached to a phosphate group). Fatty acids are carboxylic acids which vary ,in the diet, from 6 carbons up to 24 carbons. In the human diet,they will always have an even number of carbon atoms. They may be saturated (without double bonds) or unsaturated (with one or more double bonds). Trans-double bonds don’t change the overall geometry of the chain very much ; how Continue Reading A fat ,also called a triglcyeride,is composed of 3 fatty acids esterified to the hydroxyl groups of a glycerol backbone (Phospholipids are diacylglycerols; where only two hydroxyls are esterified to fatty acids,and the third one is attached to a phosphate group). Fatty acids are carboxylic acids which vary ,in the diet, from 6 carbons up to 24 carbons. In the human diet,they will always have an even number of carbon atoms. They may be saturated (without double bonds) or unsaturated (with one or more double bonds). Trans-double bonds don’t change the overall geometry of the chain very much ; however; cis-double bonds create kinks in the chain,and markedly change the packing characteristics of the fat. Hydrogenation is a process that changes cis double bonds into trans-double bonds,resulting in trans fats that have higher melting points due to closer packing ,and also contribute more to cardiovascular disease,by clumping together more in the arteries. Unsaturated fatty acids in fats; along with sterols,will not pack together as efficiently,and thereby increase the fluidity of cell membranes,and lessen the incidence of atherosclerotic plaques. Upvote · 9 1 Doug Freyburger 20 year hobbyist studying low carb metabolism · Author has 37.7K answers and 38.4M answer views ·3y Related Are fatty acids considered to be in the category of fat? “Are fatty acids considered to be in the category of fat?” Amino acids are the building blocks of protein. In diet when we refer to “protein” we actually mean the mixture of amino acids in the protein that we are eating. Fat an fatty acids work exactly the same way. Fatty acids are the building blocks of fat. In diet when we refer to “fat” we actually mean the mixture of fatty acids in the fatty that we are eating. Just as there are types of amino acids, there are types of fatty acids. Just as there are essential amino acids, there are essential fatty acids. Just as it takes a mixture of the essent Continue Reading “Are fatty acids considered to be in the category of fat?” Amino acids are the building blocks of protein. In diet when we refer to “protein” we actually mean the mixture of amino acids in the protein that we are eating. Fat an fatty acids work exactly the same way. Fatty acids are the building blocks of fat. In diet when we refer to “fat” we actually mean the mixture of fatty acids in the fatty that we are eating. Just as there are types of amino acids, there are types of fatty acids. Just as there are essential amino acids, there are essential fatty acids. Just as it takes a mixture of the essential amino acids to be a “complete” protein, it takes a mixture of the essential fatty acids to be a “complete” fat. Unlike the essential amino acids it takes little effort to get enough variety of fatty acids to cover our essential needs. Like the essential amino acids, once we get enough of the essential fatty acids the rest of the fatty acids we eat matters little. There are categories of fatty acids. That’s an accurate chemical description. Extending to call fats “saturated” or “polyunsaturated” is a crude approximation that is not chemically accurate. But for diet purposes those categories work roughly as well as calling foods “incomplete protein” versus “complete protein”. We don’t ever actually have foods with none of one specific amino acid but we do have foods deficient enough in one that we need other foods. We don’t ever actually have foods with none of one specific fatty acid but we do have foods deficient enough in one that we need other foods. It’s crude and inaccurate to put fats into categories like that. But that level of crude is good enough for dieting. Putting fats into categories is only worth the bother on low fat diets. That’s because low fat diets tell their folks to eat less fat so the essential portion is important for them. For the rest of us we eat enough fat that we never need to care about types. As long as we avoid transfats, all we need to do is push newbies to eat more fat to escape prior low fat programming. Upvote · Related questions What is the difference between a fat and an oil? What is the difference between a lipid and a fatty acid? What is the difference between fat and fatty acid? What are the differences between fatty acids and oils? How do you determine free fatty acid in oil? What is everything you know about healthy fats and fatty acids? What is a non-essential fatty acid? How are fatty acids recovered from fats and oils? How do the different types of fatty acids differ? What are the benefits of higher fatty acids? What are higher fatty acids? What's better: monounsaturated fatty acids or polyunsaturated fatty acids? What are the different types of fatty acids? What are the disadvantages of fatty acid? Is trans fatty acid bad? What is a good all unsaturated fatty acid? Related questions What is the difference between a fat and an oil? What is the difference between a lipid and a fatty acid? What is the difference between fat and fatty acid? What are the differences between fatty acids and oils? How do you determine free fatty acid in oil? What is everything you know about healthy fats and fatty acids? What is a non-essential fatty acid? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
13365	https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems?srsltid=AfmBOoph0WK-_Hx_3FBmpdC1WRqmbJIoXABF-WUDZSlxjLNZz5ERE-mO	Art of Problem Solving 2022 AIME II Problems - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2022 AIME II Problems Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2022 AIME II Problems 2022 AIME II (Answer Key) Printable version \| AoPS Contest Collections • PDF Instructions This is a 15-question, 3-hour examination. All answers are integers ranging from to , inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers. No aids other than scratch paper, rulers and compasses are permitted. In particular, graph paper, protractors, calculators and computers are not permitted. 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 Contents [hide] 1 Problem 1 2 Problem 2 3 Problem 3 4 Problem 4 5 Problem 5 6 Problem 6 7 Problem 7 8 Problem 8 9 Problem 9 10 Problem 10 11 Problem 11 12 Problem 12 13 Problem 13 14 Problem 14 15 Problem 15 16 See also Problem 1 Adults made up of the crowd of people at a concert. After a bus carrying more people arrived, adults made up of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. Solution Problem 2 Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is , where and are relatively prime positive integers. Find . Solution Problem 3 A right square pyramid with volume has a base with side length The five vertices of the pyramid all lie on a sphere with radius , where and are relatively prime positive integers. Find . Solution Problem 4 There is a positive real number not equal to either or such thatThe value can be written as , where and are relatively prime positive integers. Find . Solution Problem 5 Twenty distinct points are marked on a circle and labeled through in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original points. Solution Problem 6 Let be real numbers such that and . Among all such -tuples of numbers, the greatest value that can achieve is , where and are relatively prime positive integers. Find . Solution Problem 7 A circle with radius is externally tangent to a circle with radius . Find the area of the triangular region bounded by the three common tangent lines of these two circles. Solution Problem 8 Find the number of positive integers whose value can be uniquely determined when the values of , , and are given, where denotes the greatest integer less than or equal to the real number . Solution Problem 9 Let and be two distinct parallel lines. For positive integers and , distinct points lie on , and distinct points lie on . Additionally, when segments are drawn for all and , no point strictly between and lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when and . The figure shows that there are 8 regions when and . Solution Problem 10 Find the remainder whenis divided by . Solution Problem 11 Let be a convex quadrilateral with and such that the bisectors of acute angles and intersect at the midpoint of Find the square of the area of Solution Problem 12 Let and be real numbers with and such thatFind the least possible value of Solution Problem 13 There is a polynomial with integer coefficients such thatholds for every Find the coefficient of in . Solution Problem 14 For positive integers , , and with , consider collections of postage stamps in denominations , , and cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to cents, let be the minimum number of stamps in such a collection. Find the sum of the three least values of such that for some choice of and . Solution Problem 15 Two externally tangent circles and have centers and , respectively. A third circle passing through and intersects at and and at and , as shown. Suppose that , , , and is a convex hexagon. Find the area of this hexagon. Solution See also 2022 AIME II (Problems • Answer Key • Resources) Preceded by 2022 AIME IFollowed by 2023 AIME I 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions American Invitational Mathematics Examination AIME Problems and Solutions Mathematics competition resources These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
13366	https://www.quora.com/What-is-the-difference-between-a-local-minimum-and-a-global-minimum-Why-will-a-greedy-algorithm-get-stuck-when-it-reaches-a-local-maximum-Can-anybody-provide-a-simple-example	What is the difference between a local minimum and a global minimum? Why will a greedy algorithm get stuck when it reaches a local maximum? Can anybody provide a simple example? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Global Minimum The Greedy Method Optimization Problems Computer Science Local Minima Algorithms Optimization Algorithm Greedy Algorithms 5 What is the difference between a local minimum and a global minimum? Why will a greedy algorithm get stuck when it reaches a local maximum? Can anybody provide a simple example? All related (33) Sort Recommended Alexander Talalai Author has 655 answers and 423.1K answer views ·2y Global minimum is a minimal value on the whole area of the function definition (or location where that value is reached). Local minimum is when the function is reaching minimum in some vicinity of its location. Not sure what to you mean under “greedy algorithm”. “Greedy” defines a class of algorithms, that do not change decisions made during work. Also I think you mistaken maximum and minimum in the question. Still, an algorithm that finds minimum going down along the gradient will find, not global, but a local minimum. In other words it will stuck in local minimum. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · Promoted by Grammarly Grammarly Great Writing, Simplified ·Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do Continue Reading There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. AI Grader simulates feedback based on your rubric and course context, so you can get a realistic sense of how your paper measures up. It helps you catch weak points and revise with confidence before the official grade rolls in. Citation Finder – Your research sidekick Not sure if you’ve backed up your claims properly? Citation Finder scans your paper and identifies where you need sources—then suggests credible ones to help you tighten your argument. Think fact-checker and librarian rolled into one, working alongside your draft. Reader Reactions – Your clarity compass Writing well is one thing. Writing that resonates with the person reading it is another. Reader Reactions helps you predict how your audience (whether that’s your professor, a TA, recruiter, or classmate) will respond to your writing. With this tool, easily identify what’s clear, what might confuse your reader, and what’s most likely to be remembered. All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Upvote · 999 201 99 34 9 3 Related questions More answers below Why will a greedy algorithm get stuck when it reaches a local maximum? Can anybody provide a simple example? What is the difference between a local and global minimum in terms of algorithms? What is the difference between local and global in math? What is the difference between local and global optima in greedy algorithms? Can someone explain it with a definition and then an easy example? What is the most efficient algorithm for locating the global minimum of a function? Alan Idler Chief Software Architect, Idleswell Software Creations · Author has 2.3K answers and 1.6M answer views ·2y A local minimum is a minimum over values only within a specified range; a global minimum is the minimum over the entire set. A greedy algorithm will happily find a local minimum, but it becomes stuck (deceived) by the local minimum because it is a minimum compared to its most closely connected neighbours. A greedy algorithm never considers values that would lead it beyond a local minimum to a global minimum. If a greedy approach is used to find a route in a network, it would choose a short connection to an intermediate node even if it leads to a longer route overall (a local minimum) than a long Continue Reading A local minimum is a minimum over values only within a specified range; a global minimum is the minimum over the entire set. A greedy algorithm will happily find a local minimum, but it becomes stuck (deceived) by the local minimum because it is a minimum compared to its most closely connected neighbours. A greedy algorithm never considers values that would lead it beyond a local minimum to a global minimum. If a greedy approach is used to find a route in a network, it would choose a short connection to an intermediate node even if it leads to a longer route overall (a local minimum) than a longer connection that goes through a shorter route (the global minimum). Upvote · Lardo BS in Computer Science, Carnegie Mellon University (Graduated 2017) ·8y Originally Answered: Why will a greedy algorithm get stuck when it reaches a local maximum? Can anybody provide a simple example? · Some problems don’t admit efficient solutions for finding the globally optimal answer. In these cases, one way to achieve an efficient approximation is to use a greedy algorithm. Consider the following problem: Given an array of positive integers, we would like to partition them into two arrays such that the difference between the sums of the arrays is minimized. For example, one optimal partition of [1,8,9,4,5] is [1,8,4] and [9,5]. However, this problem is NP-hard in general, and thus we may need to use a greedy algorithm to approximate the optimal solution. One such greedy algorithm is to ad Continue Reading Some problems don’t admit efficient solutions for finding the globally optimal answer. In these cases, one way to achieve an efficient approximation is to use a greedy algorithm. Consider the following problem: Given an array of positive integers, we would like to partition them into two arrays such that the difference between the sums of the arrays is minimized. For example, one optimal partition of [1,8,9,4,5] is [1,8,4] and [9,5]. However, this problem is NP-hard in general, and thus we may need to use a greedy algorithm to approximate the optimal solution. One such greedy algorithm is to add each element of the input array to the side that currently has a lower sum. The result of running this algorithm on the example above would be [1,9,5] and [8,4], which is clearly not optimal. In general, greedy algorithms are not assumed to provide optimal solutions until they are proven to do so. Upvote · 9 1 Adrian Giles Studied Physics&Mathematics (Graduated 1985) · Author has 3.5K answers and 807.6K answer views ·10mo Related What is the difference between a global minimum and a minimum point? When can a point be considered a global minimum? f(x₀) is a global minimum point iff f(x₀)<f(x) for all x. In the example, (0,0) is a local minimum but not a global minimum. There appear to be two global minima which means my definition is unsatisfactory. Sorry about that! Perhaps, then, the point(s) (xᵢ, f(xᵢ)) are global minima iff f(xᵢ)=f(xⱼ)<f(x) for all x≠xᵢ,xⱼ. It’s a bit messy and wanting of refinement but the intention is there. Continue Reading f(x₀) is a global minimum point iff f(x₀)<f(x) for all x. In the example, (0,0) is a local minimum but not a global minimum. There appear to be two global minima which means my definition is unsatisfactory. Sorry about that! Perhaps, then, the point(s) (xᵢ, f(xᵢ)) are global minima iff f(xᵢ)=f(xⱼ)<f(x) for all x≠xᵢ,xⱼ. It’s a bit messy and wanting of refinement but the intention is there. Upvote · 9 4 Promoted by Webflow Metis Chan Works at Webflow ·Aug 12 What are the best AI website builders now? When it comes to AI website builders, there are a growing number of options, but a few stand out for their power, flexibility, and ability to grow with your needs. Webflow’s AI Site Builder is a top choice for small businesses, in-house teams, and agencies who want the speed of AI and the freedom to fully customize every part of their site. With Webflow, you can: Describe your business or idea and instantly generate a unique, production-ready website—no coding required. Edit visually in a powerful no-code canvas, customize layouts, and add advanced interactions. Collaborate with your team in real Continue Reading When it comes to AI website builders, there are a growing number of options, but a few stand out for their power, flexibility, and ability to grow with your needs. Webflow’s AI Site Builder is a top choice for small businesses, in-house teams, and agencies who want the speed of AI and the freedom to fully customize every part of their site. With Webflow, you can: Describe your business or idea and instantly generate a unique, production-ready website—no coding required. Edit visually in a powerful no-code canvas, customize layouts, and add advanced interactions. Collaborate with your team in real time, streamline feedback, and manage all your content in one place. Publish instantly on enterprise-grade hosting with built-in SEO, security, and the flexibility to scale as you grow. Many other tools offer AI-powered templates or quick site launches, but Webflow stands out by letting you take control—so your site never feels generic, and you can easily update, expand, or redesign as your needs change. Want to see what AI-powered site building can really do? Try Webflow AI Site Builder for free today. Upvote · 99 15 Related questions More answers below What is an intuitive explanation of greedy algorithms? Can you explain how to use an optimization problem and greedy algorithms to find the maximum capacity of a box with items of different sizes? What is an algorithm for maximum and minimum? What are the conditions that an algorithm must meet in order to be converted to a greedy algorithm? Can you explain the difference between local maximum, global maximum, local minimum, and global minimum? Sherwin Dsouza B.Tech in Computer Engineering, Vishwakarma Institute of Technology, Pune (Graduated 2021) · Author has 94 answers and 325.2K answer views ·7y Related What is the difference between the global maximum and minimum and local maximum and minimum? Local extremum is that location of a function where the function tends to get an extremum value only in a certain interval within the entire domain. Usually local extrema are easy to find by using the concept of derivatives. This is not always true for global extrema. Global extrema points are those points where the function gets extreme values(minimum or maximum) when considering the entire domain of the function. In simple words, highest hill point or lowest depth point of the function is the global maximum or minimum point respectively. So it's highly possible that in a function, global maxi Continue Reading Local extremum is that location of a function where the function tends to get an extremum value only in a certain interval within the entire domain. Usually local extrema are easy to find by using the concept of derivatives. This is not always true for global extrema. Global extrema points are those points where the function gets extreme values(minimum or maximum) when considering the entire domain of the function. In simple words, highest hill point or lowest depth point of the function is the global maximum or minimum point respectively. So it's highly possible that in a function, global maximum is greater than local maximum and similarly the global minimum is less than local minimum. The reason why the local extremum are easy to find is because the derivative of the function at that point is always zero. But not necessarily in global extrema. For example, consider this function: f(x)=(x−1)(x−2)(x−3)(3 x−4),1 2≤x≤5 2 f(x)=(x−1)(x−2)(x−3)(3 x−4),1 2≤x≤5 2 Maths professor: If you were told to find the maximum value of the function, what'd you do? Student A: Sir, ez pz, just find derivative of function, make it zero, find x values, get 2nd derivative, find where is 2nd derivative negative, so, that x is the point of maxima. Then we put in f(x) and we get our maximum value.(After solving) it's at x=1.7 and maximum value is 0.3. Student B: hmm, function looks cheesy. Lemme plot it and see how it looks. (After plotting the rough graph of it, finds out the function is like this:) (Source: graphing calculator GeoGebra) Student B: Sir, maximum value is at x=0.5 and the maximum value is 4.69. Now people, who was right in finding the global maximum? Ofcourse looking at the graph you'd guess it's student B. But if you still felt that student A was right, it's mandatory to read further. Did B use derivative to find the maximum point? No. He 1st confirmed how the graph looks like in the given domain and plotted it. With that, he found out that at the left most boundary of the domain, the function gets the highest value. So what we learn from this short lesson is that if you want to find the global extrema points, inspect the behavior of the function in the given domain, because using derivatives DOES NOT ALWAYS give us GLOBAL extrema. Upvote · 9 3 9 1 Marek Čtrnáct Translator From English · Author has 821 answers and 938.3K answer views ·2y Related Can a function have a local minimum but no global minimum/maximum? Sure! A good example is y=x sin(x)y=x sin⁡(x): Each “wave” is bigger than the one before. There are infinitely many minimums and maximums, but none of them is global. Continue Reading Sure! A good example is y=x sin(x)y=x sin⁡(x): Each “wave” is bigger than the one before. There are infinitely many minimums and maximums, but none of them is global. Upvote · 9 2 9 2 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 616 Francesco Amato Studied at Polytechnic University of Turin (Graduated 1977) · Author has 4.5K answers and 1M answer views ·1y Related Can you provide an example of a function that has both a local minimum and local maximum but no global minimum or global maximum? f(x)=x 3−x f(x)=x 3−x lim x→±∞=±∞lim x→±∞=±∞ The stationary points are where f′(x)=3 x 2–1=0→x±=±1√3 f′(x)=3 x 2–1=0→x±=±1 3 f(1√3)=−0.384 f(1 3)=−0.384 (relative minimum) f(−1√3)=0.384 f(−1 3)=0.384 (relative maximum) They are relative since there are smaller and greater function values respectively. Continue Reading f(x)=x 3−x f(x)=x 3−x lim x→±∞=±∞lim x→±∞=±∞ The stationary points are where f′(x)=3 x 2–1=0→x±=±1√3 f′(x)=3 x 2–1=0→x±=±1 3 f(1√3)=−0.384 f(1 3)=−0.384 (relative minimum) f(−1√3)=0.384 f(−1 3)=0.384 (relative maximum) They are relative since there are smaller and greater function values respectively. Upvote · 9 3 Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·5y Related Can we say that every absolute minimum or maximum is also a local maximum or minimum? Look at this function… It has 3 maximum points A, B, C and 2 minimum points P and Q. The gradient is zero at each of these points. You could say that point B is also a “local maximum” because it is the highest point in the region around B but it is a SPECIAL local maximum because it is the maximum point in the whole function so we call it the absolute maximum. Continue Reading Look at this function… It has 3 maximum points A, B, C and 2 minimum points P and Q. The gradient is zero at each of these points. You could say that point B is also a “local maximum” because it is the highest point in the region around B but it is a SPECIAL local maximum because it is the maximum point in the whole function so we call it the absolute maximum. Upvote · 9 3 9 1 Sponsored by LPU Online Career Ka Turning Point with LPU Online. 100% Online UGC-Entitled programs with LIVE classes, recorded content & placement support. Apply Now 999 261 Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·6y Related Why is a local maximum in f'(x) a local minimum in f(x)? I think you are a little confused here. I made the poster below, for my classroom wall. I think it will sort out your problem… In the above diagram, a local minimum on the gradient graph would indicate a point of inflection on the basic graph. Continue Reading I think you are a little confused here. I made the poster below, for my classroom wall. I think it will sort out your problem… In the above diagram, a local minimum on the gradient graph would indicate a point of inflection on the basic graph. Upvote · 9 5 Sayantan Maity Lives in Kolkata, West Bengal, India ·5y Related What is the difference between greedy and dynamic programming? Can you give simple examples? Thank you for asking me this question. To begin with I want to state that I am no expert in this, I’ll give my thoughts and the way I have perceived these 2 topics. Firstly, we must understand what Greedy and Dynamic Programming are. These are techniques we use to solve Algorithmic problems. Can Greedy be used to solve every problem? NO. To explain Greedy simply, it takes the best choice at every step to find the overall solution. But it has it’s limitations. Check the answer below for this. Sayantan Maity's answer to What are the disadvantages of greedy? And if you think of it, if one technique could Continue Reading Thank you for asking me this question. To begin with I want to state that I am no expert in this, I’ll give my thoughts and the way I have perceived these 2 topics. Firstly, we must understand what Greedy and Dynamic Programming are. These are techniques we use to solve Algorithmic problems. Can Greedy be used to solve every problem? NO. To explain Greedy simply, it takes the best choice at every step to find the overall solution. But it has it’s limitations. Check the answer below for this. Sayantan Maity's answer to What are the disadvantages of greedy? And if you think of it, if one technique could solve all problems, other techniques would not have existed. Can Dynamic Programming solve every problem? Yes, if it satisfies the criteria required. What are these criteria? Is there Optimal Substructures? What does this mean? Can you represent the problem into smaller versions of itself? Are there Overlapping Sub-problems? i.e These smaller versions of that you just found out, are these repeating multiple times? Can you write Recursive Equations for these? If the answer to these 3 questions is YES. Then you can apply Dynamic Programming. However there are some problems where even after applying DP, you will not be able to get a solution in Polynomial Time. e.g The Travelling Salesman problem. So if you ask me what the difference between Greedy and DP is, I’ll say the method which they follow. Greedy takes the best possible choice at any point and tries to get the overall best solution. Whereas, Dynamic Programming will see if a problem can be broken into smaller sub-problems, are these sub-problems repeating, if yes,it will save the value of these sub-problems so that you don’t have to compute them again,saving significant computational time. Upvote · 9 2 Michael Tam HACCP Coordinator · Author has 2.5K answers and 949.2K answer views ·4y Related What is the local maximum and local minimum of f(x) =x^4-2x^2+3? f ‘ (x) = 4x^3 - 4x Setting f ‘ (x) = 0 to find the local maximum and local minimum 4x^3 - 4x = 0 4x (x^2 - 1) = 0 4x (x + 1)(x - 1) = 0 x = 0, -1, 1 When x < - 1, x < 0, x + 1 < 0, and x - 1 < -2 < 0, or x - 1 < 0, which makes f’(x) < 0 When -1 < x < 0, x < 0, -2 < x - 1 < -1 < 0, or x - 1 < 0, 0 < x + 1 < 1, or x + 1 > 0, which makes f ‘ (x) > 0 When 0 < x < 1, x > 0, -1 < x - 1 < 0, or x - 1 < 0, 0 < 1 < x + 1 < 2, or x + 1 > 0, which makes f ‘ (x) < 0 When x > 1, x > 0, x - 1 > 0, x + 1 > 0, which makes f ‘ (x) > 0 There is a local minimum at x = -1 and x = 1 , and a local maximum at x = 0. f(-1) = (-1)^4 Continue Reading f ‘ (x) = 4x^3 - 4x Setting f ‘ (x) = 0 to find the local maximum and local minimum 4x^3 - 4x = 0 4x (x^2 - 1) = 0 4x (x + 1)(x - 1) = 0 x = 0, -1, 1 When x < - 1, x < 0, x + 1 < 0, and x - 1 < -2 < 0, or x - 1 < 0, which makes f’(x) < 0 When -1 < x < 0, x < 0, -2 < x - 1 < -1 < 0, or x - 1 < 0, 0 < x + 1 < 1, or x + 1 > 0, which makes f ‘ (x) > 0 When 0 < x < 1, x > 0, -1 < x - 1 < 0, or x - 1 < 0, 0 < 1 < x + 1 < 2, or x + 1 > 0, which makes f ‘ (x) < 0 When x > 1, x > 0, x - 1 > 0, x + 1 > 0, which makes f ‘ (x) > 0 There is a local minimum at x = -1 and x = 1 , and a local maximum at x = 0. f(-1) = (-1)^4 - 2(-1)^2 + 3 = 1 - 2 + 3 = 2 f(0) = (0)^4 - 2(0)^2 + 3 = 3 f(1) = (1)^4 - 2(1)^2 + 3 = 1 - 2 + 3 = 2 Local minima are at (-1, 2) and (1, 2). Local maximum is at (0, 3) Upvote · 9 1 Daniel R. Page Theoretical Computer Scientist, CS PhD · Author has 3K answers and 5.3M answer views ·2y Related What is the difference between a greedy algorithm and an approximation algorithm? How can one prove that a greedy algorithm is optimal (or suboptimal)? These are two different concepts, that can overlap in narrow circumstances. The former is an algorithm design technique, while the other encompasses algorithms with provably-close guarantees to the optimum (in polynomial time) for every feasible solution produced by the algorithm. Many approximation algorithms aren’t greedy algorithms, some are though. Some greedy algorithms can perform very badly when placed into the more rigorous context of approximation algorithms, others are fine. Some classic examples include: List Scheduling (Makespan Minimization on Identical Parallel Machines). Taking ed Continue Reading These are two different concepts, that can overlap in narrow circumstances. The former is an algorithm design technique, while the other encompasses algorithms with provably-close guarantees to the optimum (in polynomial time) for every feasible solution produced by the algorithm. Many approximation algorithms aren’t greedy algorithms, some are though. Some greedy algorithms can perform very badly when placed into the more rigorous context of approximation algorithms, others are fine. Some classic examples include: List Scheduling (Makespan Minimization on Identical Parallel Machines). Taking edges of a maximal matching (Minimum Cardinality Vertex Cover). First Fit (Bin Packing). To answer your second question: You prove the greedy algorithm satisfies the desired property via mathematical proof. For example, a common property (the greedy-choice property) to prove is to show that making the selection (the locally optimal choice) always leads to a globally optimal choice. Common ways to prove such a claim is using an exchange argument or structurally properties of the problem. Upvote · Amit Gupta Founder of actucation.com and CEO at Paravyom Education ·9y Related On a graph, what's the difference between global and local maximum/minimum values? On a graph, many local maximum/minimum values may be possible, but there will be only one global maximum / minimum value. Local Maximum / Minimum Value: Consider a graph to understand the local maximum and minimum values. Local Maximum Value: Local maximum value of a function f (x) on a graph, is a value at a point (like M in the graph) which is greater than the values at the nearest adjacent points on left and right sides (like L and N in the graph) Thus, f (M) is the local maximum value of the function f (x). Local Minimum Value: Local minimum value of a function f (x) on a graph, is a value at a p Continue Reading On a graph, many local maximum/minimum values may be possible, but there will be only one global maximum / minimum value. Local Maximum / Minimum Value: Consider a graph to understand the local maximum and minimum values. Local Maximum Value: Local maximum value of a function f (x) on a graph, is a value at a point (like M in the graph) which is greater than the values at the nearest adjacent points on left and right sides (like L and N in the graph) Thus, f (M) is the local maximum value of the function f (x). Local Minimum Value: Local minimum value of a function f (x) on a graph, is a value at a point (like Q in the graph). Which is lower than the values at the nearest adjacent points on left and right sides (like P and R in the graph). Thus, f (Q) is the local minimum value of the function f (x). Global Maximum and Minimum Values : Global maximum and minimum values are also known as the absolute maximum and minimum values. Let’s consider a graph to understand the global maximum and minimum values. Global Maximum Value: Global maximum value of a function f (x) on a graph, is a value at a point (say A) which is higher than the entire values of f (x) at any other point. Thus, f (A) is the global/absolute maximum value of the function f (x). Global Minimum Value: Global minimum value of a function f (x) on a graph, is a value at a point (say B) which is lower than the entire values of f (x) at any other point. Thus, f (B) is the global/absolute minimum value of the function f (x). Upvote · 99 20 9 3 9 1 Related questions Why will a greedy algorithm get stuck when it reaches a local maximum? Can anybody provide a simple example? What is the difference between a local and global minimum in terms of algorithms? What is the difference between local and global in math? What is the difference between local and global optima in greedy algorithms? Can someone explain it with a definition and then an easy example? What is the most efficient algorithm for locating the global minimum of a function? What is an intuitive explanation of greedy algorithms? Can you explain how to use an optimization problem and greedy algorithms to find the maximum capacity of a box with items of different sizes? What is an algorithm for maximum and minimum? What are the conditions that an algorithm must meet in order to be converted to a greedy algorithm? Can you explain the difference between local maximum, global maximum, local minimum, and global minimum? What is the difference between a local and global optimal solutions by real world example? Why do people often get stuck at a local maximum in their careers or personal goals? Is there a way to use a greedy algorithm on a non-convex function without running the risk of getting stuck at a local min/max? If every local optimum is not always a global optimum, then what is the advantage of greedy algorithms? What is the difference between local and global minimum? Related questions Why will a greedy algorithm get stuck when it reaches a local maximum? Can anybody provide a simple example? What is the difference between a local and global minimum in terms of algorithms? What is the difference between local and global in math? What is the difference between local and global optima in greedy algorithms? Can someone explain it with a definition and then an easy example? What is the most efficient algorithm for locating the global minimum of a function? What is an intuitive explanation of greedy algorithms? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
13367	https://www.angelfire.com/fl4/Kevin/physics2/Ch27F.pdf	CHAPTER 27 The Microscopic Theory of Electrical Conduction 1 ∙ In the classical model of conduction, the electron loses energy on average in a collision because it loses the drift velocity it had picked up since the last collision. Where does this energy appear? The energy lost by the electrons in collision with the ions of the crystal lattice appears as Joule heat (I2R). 2 ∙ A measure of the density of the free-electron gas in a metal is the distance rs, which is defined as the radius of the sphere whose volume equals the volume per conduction electron. (a) Show that rs = (3/4πn)1/3, where n is the free-electron number density. (b) Calculate rs for copper in nanometers. (a) The volume occupied by one electron is 1/n = (4/3)πrs 3. Thus, rs = (3/4πn)1/3. (b) n = 8.47×1028 m–3 (See Table 27-1). Evaluate rs rs = 1.41×10–10 m = 0.141 nm 3 ∙ (a) Given a mean free path λ = 0.4 nm and a mean speed vav = 1.17×105 m/s for the current flow in copper at a temperature of 300 K, calculate the classical value for the resistivity ρ of copper. (b) The classical model suggests that the mean free path is temperature independent and that vav depends on temperature. From this model, what would ρ be at 100 K? (a) Use Equ. 27-7 (b) vav ∝ T1/2 ρ = 10 4 ) 10 (1.6 10 8.47 10 1.17 10 9 10 2 19 28 5 31 − − − × × × × × × × × Ω.m = 0.123 µΩ.m ρ100 = (0.123 µΩ.m)(100/300)1/2 = 0.071 µΩ.m 4 ∙ Calculate the number density of free electrons in (a) Ag (ρ = 10.5 g/cm3) and (b) Au (ρ = 19.3 g/cm3), assuming one free electron per atom, and compare your results with the values listed in Table 27-1. (a), (b) n = ρNA/M; ρ = density, M = molar mass (a) nAg = 5.86×1022 el./cm3; (b) nAu = 5.90×1022 el./cm3. The results agree with Table 27-1 5 ∙ The density of aluminum is 2.7 g/cm3. How many free electrons are present per aluminum atom? ne = electrons/atom = nM/ρNA ne = 10 6.02 2.7 26.98 10 18.1 23 22 × × × × = 3.00 electrons/atom 6 ∙ The density of tin is 7.3 g/cm3. How many free electrons are present per tin atom? Chapter 27 The Microscopic Theory of Electrical Conduction See Problem 5 ne = 10 6.02 7.3 118.7 10 14.8 23 22 × × × × = 4.00 electrons/atom 7 ∙ Calculate the Fermi temperature for (a) Al, (b) K, and (c) Sn. (a), (b), (c) TF = EF/k; k = 8.625×10–5 eV/K (a) TF = (11.7 eV)/k = 1.36×105 K; (b) TF = 2.45×104 K (c) TF = 1.18×105 K 8 ∙ What is the speed of a conduction electron whose energy is equal to the Fermi energy EF for (a) Na, (b) Au, and (c) Sn? (a), (b), (c) m / E 2 = e F F u ; use Table 27-1 for EF (a) uF = 1.07×106 m/s; (b) uF = 1.39×106 m/s; (c) uF = 1.89×106 m/s 9 ∙ Calculate the Fermi energy for (a) Al, (b) K, and (c) Sn using the number densities given in Table 27-1. (a), (b), (c) Use Equ. 27-15b (a) EF = 0.365(181)2/3 eV = 11.7 eV; (b) EF = 2.12 eV; (c) EF = 10.2 eV 10 ∙ Find the average energy of the conduction electrons at T = 0 in (a) copper and (b) lithium. (a), (b) Eav = 0.6EF (a) Eav = 0.6×7.04 eV = 4.22 eV; (b) Eav = 2.85 eV 11 ∙ Calculate (a) the Fermi temperature and (b) the Fermi energy at T = 0 for iron. Use Equ. 27-15b for EF and TF = EF/k EF = 0.365×(170)2/3 eV = 11.2 eV; TF = 1.30×105 K 12 ∙∙ The pressure of an ideal gas is related to the average energy of the gas particles by , av E N = PV 3 2 where N is the number of particles and Eav is the average energy. Use this to calculate the pressure of the Fermi electron gas in copper in newtons per square meter, and compare your result with atmospheric pressure, which is about 105 N/m2. (Note: The units are most easily handled by using the conversion factors 1 N/m2 = 1 J/m3 and 1 eV = 1.6×10–19 J.) P = 2(N/V)Eav/3 = 2(N/V)EF/5; EF = EF(eV)×e P = 0.4(8.47×1028)(7.04)(1.6×10–19) N/m2 = 3.82×1010 N/m2; P = 3.78×105 atm 13 ∙∙ The bulk modulus B of a material can be defined by V P V - = B ∂ ∂ (a) Use the ideal-gas relation E N = PV 3 2 av and Equations 27-15 and 27-16 to show that V C = V 5 3 / 5 -F E 2N = P where C is a constant independent of V. (b) Show that the bulk modulus of the Fermi electron gas is therefore V 3 F E 2N = P 3 5 = B (c) Compute the bulk modulus in newtons per square meter for the Fermi electron gas in copper and compare your result with the measured value of 140×109 N/m2. Chapter 27 The Microscopic Theory of Electrical Conduction (a) From Problem 12, P = (2/3)(N/V)Eav = (2/5)(N/V)EF. But EF is proportional to V–2/3 so P = CV–5/3, where C is a constant. (b) B = –(1/V)(dP/dV) = (5/3)CV–5/3 = (5/3)P = (2/3)(N/V)EF. (c) B = 63.6×109 N/m2 ≅ 0.5Bexp 14 ∙ Thomas refuses to believe that a potential difference can be created simply by bringing two different metals into contact with each other. John talks him into making a small wager, and is about to cash in. (a) Which two metals from Table 27-2 would demonstrate his point most effectively? (b) What is the value of that contact potential? (a), (b) The largest contact potential is when potassium and nickel are joined. Then Vcontact = (5.2 – 2.1) V = 3.1 V. 15 ∙ (a) In problem 14, which choices of different metals would make the least impressive demonstration? (b) What is the value of that contact potential? (a), (b) Silver joined to gold gives the smallest contact potential. Vcontact = 0.1 V 16 ∙ Calculate the contact potential between (a) Ag and Cu, (b) Ag and Ni, and (c) Ca and Cu. (a), (b), (c) See Table 27-2 (a) Vcontact = 0.6 V; (b) Vcontact = 0.5 V; (c) Vcontact = 0.9 V 17 ∙ When the temperature of pure copper is lowered from 300 K to 4 K, its resistivity drops by a much greater factor than that of brass when it is cooled in the same way. Why? The resistivity of brass at 4 K is almost entirely due to the “residual resistance,” the resistance due to impurities and other imperfections of the crystal lattice. In brass, the zinc ions act as impurities in copper. In pure copper, the resistivity at 4 K is due to its residual resistance, which is very low if the copper is very pure. 18 ∙ The resistivities of Na, Au, and Sn at T = 273 K are 4.2 µΩ⋅cm, 2.04 µΩ⋅cm, and 10.6 µΩ⋅cm, respectively. Use these values and the Fermi speeds calculated in Problem 8 to find the mean free paths λ for the conduction electrons in these elements. ρ = meuF/nee2λ; λ = mevF/nee2ρ; evaluate λ for Na, Au, and Sn. See Problem 8 for uF. Note: 1 µΩ.cm = 10–8 Ω.m Na: λ = 10 4.2 ) 10 (1.6 10 2.65 10 1.07 10 9.11 8 2 19 28 6 31 − − − × × × × × × × × m = 34.2 nm; Au: λ = 41.1 nm; Sn: λ = 4.29 nm 19 ∙∙ The resistivity of pure copper is increased by about 1×10-8 Ω⋅m by the addition of 1% (by number of atoms) of an impurity throughout the metal. The mean free path depends on both the impurity and the oscillations of the lattice ions according to the equation λ λ λ i t 1 + 1 = 1 (a) Estimate λi from data given in Table 27-1. (b) If r is the effective radius of an impurity lattice ion seen by an electron, the scattering cross section is πr2. Estimate this area using the fact that r is related to λi by Equation 27-9. (a) ρi = meuF/nee2λi; λi = meuF/nee2ρi = ρi 2 e e F e n m E 2 (b) λi = 1/niπr2; πr2 = 1/niλi λi = 10 ) 10 (1.6 10 8.47 10 9.11 10 1.6 7.04 2 8 2 19 28 31 19 − − − − × × × × × × × × × m = 66.1 nm ni = 8.47×1026 m–3; πr2 = 1.8×10–20 m2 = 0.018 nm2 20 ∙ A metal is a good conductor because the valence energy band for electrons is (a) completely full. (b) full, but there is only a small gap to a higher empty band. (c) partly full. (d) empty. (e) None of these is correct. Chapter 27 The Microscopic Theory of Electrical Conduction (c) 21 ∙ Insulators are poor conductors of electricity because (a) there is a small energy gap between the valence band and the next higher band where electrons can exist. (b) there is a large energy gap between the full valence band and the next higher band where electrons can exist. (c) the valence band has a few vacancies for electrons. (d) the valence band is only partly full. (e) None of these is correct. (b) 22 ∙ You are an electron sitting at the top of the valence band in a silicon atom, longing to jump across the 1.14-eV energy gap that separates you from the bottom of the conduction band and all of the adventures that it may contain. What you need, of course, is a photon. What is the maximum photon wavelength that will get you across the gap? ∆E = hc/λ; λ = hc/∆E λ = 1240/1.14 nm = 1088 nm 23 ∙ Work Problem 22 for germanium, for which the energy gap is 0.74 eV. ∆E = hc/λ; λ = hc/∆E λ = 1240/0.74 nm = 1.676 µm 24 ∙ Work Problem 22 for diamond, for which the energy gap is 7.0 eV. ∆E = hc/λ; λ = hc/∆E λ = 1240/7.0 nm = 177 nm 25 ∙∙ A photon of wavelength 3.35 µm has just enough energy to raise an electron from the valence band to the conduction band in a lead sulfide crystal. (a) Find the energy gap between these bands in lead sulfide. (b) Find the temperature T for which kT equals this energy gap. (a) Eg = hc/λ (b) T = Eg/k; k = 8.62×10–5 eV/K Eg = 1240/3350 eV = 0.370 eV T = 4.29×103 K 26 ∙ (a) Use Equation 27-24 to calculate the superconducting energy gap for tin (Tc = 3.72 K) and compare your result with the measured value of 6×10–4 eV. (b) Use the measured value to calculate the wavelength of a photon having sufficient energy to break up Cooper pairs in tin at T = 0. (a) Eg = 3.5kTc (b) λ = hc/Eg Eg = 3.5×8.625×10–5 ×3.72 eV = 1.12 meV; about twice the measured value. λ = 1240/6×10–4 nm = 2.07×106 nm = 2.07 mm 27 ∙ Repeat Problem 26 for lead (Tc = 7.19 K), which has a measured energy gap of 2.73×10– 3 eV. (a) Eg = 3.5kTc (b) λ = hc/Eg Eg = 2.17 meV; this is about 80% of the measured value λ = 0.454 mm 28 ∙∙ The number of electrons in the conduction band of an insulator or intrinsic semiconductor is governed chiefly by the Fermi factor. Since the valence band in these materials is nearly filled and the conduction band is nearly empty, the Fermi energy EF is generally midway between the top of the valence band and bottom of the conduction band, i.e., at Eg/2, where Eg is the band gap between the two bands and the energy is measured from the top of the valence band. (a) In silicon, Eg ≈1.0 eV. Show that in this case the Fermi factor for electrons at Chapter 27 The Microscopic Theory of Electrical Conduction the bottom of the conduction band is given by exp(–Eg/2kT) and evaluate this factor. Discuss the significance of this result if there are 1022 valence electrons per cubic centimeter and the probability of finding an electron in the conduction band is given by the Fermi factor. (b) Repeat the calculation in (a) for an insulator with a band gap of 6.0 eV. (a) At the bottom of the conduction band, e = kT 2 / E /kT ) g F e E (E − >>> 1 for T near room temperature. We can then neglect the 1 in the denominator of the Fermi function and f(Eg/2) = e E kT 2 / g − . We take T = 300 K; then the Fermi factor is exp[–0.5/(8.625×10–5×300)] = 4.05×10–9. Given that low a probability of finding an electron in a state near the bottom of the conduction band, the exclusion principle has no significant impact on the distribution function. With 1022 valence electrons per cubic centimeter, the number of electrons in the conduction band will be about 4.05×1013 per cm3. (b) With Eg = 6 eV, the Fermi factor at the bottom of the conduction band is now 4.44×10–51, and the probability of finding even one electron in the conduction band is negligibly small (4×10–29). 29 ∙∙ Show that at E = EF, the Fermi factor is F = 0.5. For E = EF, 1 = = 0 /kT ) F e e E (E • . Consequently, f(EF) = 1/2 30 ∙∙ What is the difference between the energies at which the Fermi factor is 0.9 and 0.1 at 300 K in (a) copper, (b) potassium, and (c) aluminum. (a) Find E - EF for f(E) = 0.1 Find E - EF for f(E) = 0.9 ∆E = E(0.1) – E(0.9); independent of EF (b), (c) ∆E same as for (a) 0.1{exp[–(E – EF)/300k]} = 0.9; E – EF = 0.057 eV 0.9{exp[–(E – EF)/300k]} = 0.1; E – EF = –0.057 eV ∆E = 0.114 eV ∆E = 0.114 eV 31 ∙∙ What is the probability that a conduction electron in silver will have a kinetic energy of 4.9 eV at T = 300 K? Since EF – 4.9 eV >> 300k, f(4.9 eV) = 1 32 ∙∙ Show that E 2 / 1 2 / 3 F E 2) / (3N = g(E) (Equation 27-30) follows from Equation 27-28 for g(E), and Equation 27-15a for EF. From Equ. 27-15a, V =         EF e 2 / 3 m 8 h 3N 2 π . Substitute this expression for V in Equ. 27-28 and simplify. The result is Equ. 27-30. 33 ∙∙ Carry out the integration ( )∫ = F E av dE E Eg N E 0 ) ( / 1 to show that the average energy at T = 0 is . F E 5 3 F 2 / 5 F 2 / 3 F 0 0 2 / 3 2 / 3 F 5 3 5 2 2 3 2 3 ) ( 1 F F E E E dE E E dE E Eg N E E = = = − − ∫ ∫ 34 ∙∙ The density of the electron states in a metal can be written g(E) = AE 1/2, where A is a constant and E is measured from the bottom of the conduction band. (a) Show that the total number of states is . 2 / 3 F AE 3 2 (b) Approximately what fraction of the conduction electrons are within kT of the Fermi energy? (c) Evaluate this fraction for copper at T = 300 K. Chapter 27 The Microscopic Theory of Electrical Conduction (a) 2 / 3 F 0 2 / 1 3 2 F AE dE AE N E = = ∫ 0. (b) The fraction of N within kT of EF is kTg(EF)/N = 3kT/2EF. (c) For copper at T = 300 K, 3kT/2EF = 5.51×10–3. 35 ∙∙ What is the probability that a conduction electron in silver will have a kinetic energy of 5.49 eV at T = 300 K? (a) Find f(E) for E = 5.49 eV, EF = 5.50, T = 300 K f(5.49 eV) = [exp(–0.01/0.0259) + 1]–1 = 0.595 36 ∙∙ Use the density-of-states function, Equation 27-28, to estimate the fraction of the conduction electrons in copper that can absorb energy from collisions with the vibrating lattice ions at (a) 77 K and (b) 300 K. (a) Use the result of Problem 34(b); T = 77 K (b) Repeat (a) for T = 300 K; see Problem 34(c) Fraction = 3×77×8.625×10–5/2×7.04 = 1.42×10–3 Fraction = 5.51×10–3 37 ∙∙ In an intrinsic semiconductor, the Fermi energy is about midway between the top of the valence band and the bottom of the conduction band. In germanium, the forbidden energy band has a width of 0.7 eV. Show that at room temperature the distribution function of electrons in the conduction band is given by the Maxwell-Boltzmann distribution function. n(E) = g(E)f(E). In this case, exp[(E – EF)/kT] = exp[(E – Eg/2)/kT] >> 1 so the e E/kT kT 2 / g − e = (E) f E . So, using Equ. 27-30, we have e E e E/kT 2 / 1 kT 2 / E 2 / 3 F g − − E 2 3N = n(E) There is generally an additional temperature dependence that arises from the fact that EF depends on T. At room temperature, exp[(E – Eg/2)/kT] ≥ exp(0.35/0.0259) = 7.4×105, so the approximation leading to the Boltzmann distribution is justified. 38 ∙∙∙ (a) Show that for E ≥ 0, the Fermi factor may be written as 1 + Ce 1 = f(E) kT / E (b) Show that if C >> e–E/kT, f(E) = Ae–E/kT << 1; in other words, show that the Fermi factor is a constant times the classical Boltzmann factor if A << 1. (c) Use N = dE n(E) ∫ and Equation 27-28 to determine the constant A. (d) Using the result obtained in part (c), show that the classical approximation is applicable when the electron concentration is very small and/or the temperature is very high. (e) Most semiconductors have impurities added in a process called doping, which increases the free electron concentration so that it is about 1017/cm3 at room temperature. Show that for these systems, the classical distribution function is applicable. (a) Let C = e E /kT F − . Then Equ. 27-32 becomes 1 + e C 1 = (E) f E/kT . (b) For the condition stated, one may neglect the 1 in the denominator of the expression for f(E). Then f(E) = Ae–E/kT, where A = 1/C. (c) ∫ ∞ − = 0 / 2 / 1 3 2 / 3 2 8 de e E h V m A N kT E e π . The definite integral has the value π 2 ) (kT 3/2 . Solving for A one obtains ) (kT 1 V N 2 / 3 2 / 3 e       m 8 h 2 = A 3/2 3 π . Chapter 27 The Microscopic Theory of Electrical Conduction (d) The constant factor has the value 1.06×10–56 in SI units. At room temperature, kT ≅ 4×10–21 J, so A ≅ 4×10–26n, where n = (N/V) is the electron concentration. The valence electron concentration is typically about 1029 m–3. To meet the condition A << 1 at room temperature n should be less than 1023 m–3, or about one millionth of the valence electron concentration. Since A depends on T–3/2, the electron concentration may be greater the higher the temperature. (e) 1017 cm–3 = 1023 m–3. So according to the criterion in (d), the classical approximation is applicable. 39 ∙∙∙ Show that the condition for the applicability of the classical distribution function for an electron gas (A << 1 in Problem 38) is equivalent to the requirement that the average separation between electrons is much greater than their de Broglie wavelength. Note that d, the separation between electrons is approximately (V/N)1/3 = A ) (kT 1 3 / 1 2 / 1 2 / 1 e m 8 h 2 1/2 1/3 1/6 π . Now p ~ (2kTm)1/2 so d ~ 0.2h/pA1/3 = 0.2λ/A1/3. It follows that if A << 1 then d >> λ. 40 ∙∙∙ The root-mean-square (rms) value of a variable is obtained by calculating the average value of the square of that variable and then taking the square root of the result. Use this procedure to determine the rms energy of a Fermi distribution. Express your result in terms of EF and compare it to the average energy. Why do Eav and Erms differ? We shall do this problem for kT << EF, i.e., in the approximation T = 0. E dE E E dE E E g N E E E 7 3 ) ( 2 3 ) ( 1 2 / 1 0 2 / 5 2 / 3 F 2 / 1 0 2 rms F F =         =         = ∫ ∫ = 0.655 EF. Note that Erms > Eav because the process of averaging the square of the energy weighs larger energies more heavily. 41 ∙∙∙When a star with a mass of about twice that of the sun exhausts its nuclear fuel, it collapses to a neutron star, a dense sphere of neutrons of about 10 km diameter. Neutrons are spin– 2 1 particles and, like electrons, are subject to the exclusion principle. (a) Determine the neutron density of such a neutron star. (b) Find the Fermi energy of the neutron distribution. (a) Find N/V; N = M/mn; V = πD3/6 (b) Use Equ. 27-15a, replacing me by mn N/V = 4.57×1045 m–3 EF = 8.77×10–11 J = 548 MeV 42 ∙ True or false: (a) Solids that are good electrical conductors are usually good heat conductors. (b) The classical free-electron theory adequately explains the heat capacity of metals. (c) At T = 0, the Fermi factor is either 1 or 0. (d) The Fermi energy is the average energy of an electron in a solid. (e) The contact potential between two metals is proportional to the difference in the work functions of the two metals. (f) At T = 0, an intrinsic semiconductor is an insulator. (g) Semiconductors conduct current in one direction only. (a) True (b) False (c) True (d) False (e) True (f) True (g) False 43 ∙ How does the change in the resistivity of copper compare with that of silicon when the temperature increases? The resistivity of copper increases with increasing temperature; the resistivity of (pure) silicon decreases with increasing temperature because the number of charge carriers increases. Chapter 27 The Microscopic Theory of Electrical Conduction 44 ∙ The density of potassium is 0.851 g/cm3. How many free electrons are there per potassium atom? ne = electrons/atom = nM/ρNA ne = 1.07 electrons/atom 45 ∙ Calculate the number density of free electrons for (a) Mg (ρ = 1.74 g/cm3) and (b) Zn (ρ = 7.1 g/cm3), assuming two free electrons per atom, and compare your results with the values listed in Table 27-1. (a), (b) n = 2NAρ/M The calculated values agree well with Table 27-1 (a) n = 2×6.02×1023×1.74/24.3 = 8.62×1022 (b) n = 13.1×1022 46 ∙∙ Estimate the fraction of free electrons in copper that are in excited states above the Fermi energy at (a) room temperature of 300 K and (b) 1000 K. The fraction that can be excited thermally are those within about kT of EF. (a) See Problem 27-34 (b) Repeat (a) for T = 1000 K fraction = 5.51×10–3 fraction = 5.51×10–3(10/3) = 1.82×10–2 47 ∙∙∙ A 2-cm2 wafer of pure silicon is irradiated with light having a wavelength of 775 nm. The intensity of the light beam is 4.0 W/m2 and every photon that strikes the sample is absorbed and creates an electron–hole pair. (a) How many electron–hole pairs are produced in one second? (b) If the number of electron–hole pairs in the sample is 6.25×1011 in the steady state, at what rate do the electron–hole pairs recombine? (c) If every recombination event results in the radiation of one photon, at what rate is energy radiated by the sample? (a) Find the number of photons/s that hit the surface. Number of electron-hole pairs per second = N (b) In the steady state, rate of recombination = rate of generation (c) Power radiated = power absorbed N = 4.0×2×10–4/[1.6×10–19×(1240/775)] = 3.125×1015 Number of electron–hole pairs per second = 3.125×1015 Recombination rate = 3.125×1015 s–1 Prad = 8×10–4 W = 0.8 mJ/s
13368	http://cs.indstate.edu/~arash/adslec7.pdf	Recurrence Relations (review and examples) Arash Raﬁey September 29, 2015 Arash Raﬁey Recurrence Relations (review and examples) Homogenous relation of order two : C0an + C1an−1 + C2an−2 = 0, n ≥2. Arash Raﬁey Recurrence Relations (review and examples) Homogenous relation of order two : C0an + C1an−1 + C2an−2 = 0, n ≥2. We look for a solution of form an = crn, c ̸= 0, r ̸= 0. Arash Raﬁey Recurrence Relations (review and examples) Homogenous relation of order two : C0an + C1an−1 + C2an−2 = 0, n ≥2. We look for a solution of form an = crn, c ̸= 0, r ̸= 0. C0crn + C1crn−1 + C2crn−2 = 0. Arash Raﬁey Recurrence Relations (review and examples) Homogenous relation of order two : C0an + C1an−1 + C2an−2 = 0, n ≥2. We look for a solution of form an = crn, c ̸= 0, r ̸= 0. C0crn + C1crn−1 + C2crn−2 = 0. We obtain C0r2 + C1r + C2 = 0 which is called the characteristic equation. Arash Raﬁey Recurrence Relations (review and examples) Homogenous relation of order two : C0an + C1an−1 + C2an−2 = 0, n ≥2. We look for a solution of form an = crn, c ̸= 0, r ̸= 0. C0crn + C1crn−1 + C2crn−2 = 0. We obtain C0r2 + C1r + C2 = 0 which is called the characteristic equation. Let r1, r2 be the roots of C0r2 + C1r + C2 = 0. There are three cases : 1 r1, r2 are distinct real numbers 2 r1, r2 are complex numbers (conjugate of each other) 3 r1 = r2 is a real number In all cases r1, r2 are called the characteristic roots. Arash Raﬁey Recurrence Relations (review and examples) Distinct Real Values : an + an−1 −6an−2 = 0 where n ≥2 and a0 = −1, a1 = 8. Arash Raﬁey Recurrence Relations (review and examples) Distinct Real Values : an + an−1 −6an−2 = 0 where n ≥2 and a0 = −1, a1 = 8. an = crn and the characteristic equation is r2 + r −6 = 0. Arash Raﬁey Recurrence Relations (review and examples) Distinct Real Values : an + an−1 −6an−2 = 0 where n ≥2 and a0 = −1, a1 = 8. an = crn and the characteristic equation is r2 + r −6 = 0. Therefore (r + 3)(r −2) = 0 and hence r = 2, −3. Arash Raﬁey Recurrence Relations (review and examples) Distinct Real Values : an + an−1 −6an−2 = 0 where n ≥2 and a0 = −1, a1 = 8. an = crn and the characteristic equation is r2 + r −6 = 0. Therefore (r + 3)(r −2) = 0 and hence r = 2, −3. So both an = 2n and an = (−3)n are solutions. Since one is not a multiple of the other we can write an = c1(2n) + c2(−3)n. Arash Raﬁey Recurrence Relations (review and examples) Distinct Real Values : an + an−1 −6an−2 = 0 where n ≥2 and a0 = −1, a1 = 8. an = crn and the characteristic equation is r2 + r −6 = 0. Therefore (r + 3)(r −2) = 0 and hence r = 2, −3. So both an = 2n and an = (−3)n are solutions. Since one is not a multiple of the other we can write an = c1(2n) + c2(−3)n. −1 = a0 = c1(20) + c2(−3)0 = c1 + c2 8 = a1 = c1(21) + c2(−3)1 = 2c1 −3c2. Arash Raﬁey Recurrence Relations (review and examples) Distinct Real Values : an + an−1 −6an−2 = 0 where n ≥2 and a0 = −1, a1 = 8. an = crn and the characteristic equation is r2 + r −6 = 0. Therefore (r + 3)(r −2) = 0 and hence r = 2, −3. So both an = 2n and an = (−3)n are solutions. Since one is not a multiple of the other we can write an = c1(2n) + c2(−3)n. −1 = a0 = c1(20) + c2(−3)0 = c1 + c2 8 = a1 = c1(21) + c2(−3)1 = 2c1 −3c2. Therefore an = 2n −2(−3)n. Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition In general suppose C0an + C1an−1 + C2an−2 + · · · + Ckan−k = 0 where C ′ i s are constant and C0 ̸= 0 and Ck ̸= 0 and r is the characteristic root with multiplicity 2 ≤m ≤k. Then the part of the general solution involving root r has the following form : (A0 + A1n + A2n2 + · · · + Am−1nm−1)rn where Ai are arbitrary constant. Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition In general suppose C0an + C1an−1 + C2an−2 + · · · + Ckan−k = 0 where C ′ i s are constant and C0 ̸= 0 and Ck ̸= 0 and r is the characteristic root with multiplicity 2 ≤m ≤k. Then the part of the general solution involving root r has the following form : (A0 + A1n + A2n2 + · · · + Am−1nm−1)rn where Ai are arbitrary constant. Example : an −6an−1 + 9an−2 = 0, a0 = 5 and a1 = 12. Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition In general suppose C0an + C1an−1 + C2an−2 + · · · + Ckan−k = 0 where C ′ i s are constant and C0 ̸= 0 and Ck ̸= 0 and r is the characteristic root with multiplicity 2 ≤m ≤k. Then the part of the general solution involving root r has the following form : (A0 + A1n + A2n2 + · · · + Am−1nm−1)rn where Ai are arbitrary constant. Example : an −6an−1 + 9an−2 = 0, a0 = 5 and a1 = 12. r2 −6r + 9 = 0 and hence (r −3)2 = 0 and r1 = r2 = 3 (here m = 2). Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition In general suppose C0an + C1an−1 + C2an−2 + · · · + Ckan−k = 0 where C ′ i s are constant and C0 ̸= 0 and Ck ̸= 0 and r is the characteristic root with multiplicity 2 ≤m ≤k. Then the part of the general solution involving root r has the following form : (A0 + A1n + A2n2 + · · · + Am−1nm−1)rn where Ai are arbitrary constant. Example : an −6an−1 + 9an−2 = 0, a0 = 5 and a1 = 12. r2 −6r + 9 = 0 and hence (r −3)2 = 0 and r1 = r2 = 3 (here m = 2). So an = (A0 + A1n)3n. Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition In general suppose C0an + C1an−1 + C2an−2 + · · · + Ckan−k = 0 where C ′ i s are constant and C0 ̸= 0 and Ck ̸= 0 and r is the characteristic root with multiplicity 2 ≤m ≤k. Then the part of the general solution involving root r has the following form : (A0 + A1n + A2n2 + · · · + Am−1nm−1)rn where Ai are arbitrary constant. Example : an −6an−1 + 9an−2 = 0, a0 = 5 and a1 = 12. r2 −6r + 9 = 0 and hence (r −3)2 = 0 and r1 = r2 = 3 (here m = 2). So an = (A0 + A1n)3n. a0 = 5 = A0 and a1 = 12 = (5 + A1)3 and hence A1 = −1. Arash Raﬁey Recurrence Relations (review and examples) nonhomogeneous ﬁrst-order relation Deﬁnition Consider the nonhomogeneous ﬁrst-order relation (k constant) an + C1an−1 = krn When rn is not a solution (C1 ̸= −r) for an + C1an−1 = 0 then an = A(−C1)n + B(rn) for some constants A, B. When rn is a solution for the recurrence, i.e. (−C1 = r) then an = Arn + Bnrn for some constants A, B. Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition Consider the nonhomogeneous second-order relation (k constant) an + C1an−1 + C2an−2 = krn With homogeneous relation (h) : an + C1an−1 + C2an−2 = 0. If 1 rn is not a solution for (h) then an = Arn + B(r1)n + C(r2)n 2 rn is a solution for (h) and (h) has other solution rn 1 , (r ̸= r1) then an = (A + Bn)rn + C(r1)n. 3 the characteristic equation r2 + C1r + C2 = 0 has r1 = r2 = r solution then an = Arn + Bnrn + Cn2rn. Arash Raﬁey Recurrence Relations (review and examples) Theorem Consider the recurrence relation an = c1an−1 + c2an−2 + · · · + ckan−k + f (n). Suppose f (n) = (bsns + bs−1ns−1 + · · · + b1n + b0)λn. 1 If λ is not a characteristic root then ap n = (dsns + ds−1ns−1 + · · · + d1n + d0)λn (nonhomogeneous solution) 2 If λ is a characteristic root with multiplicity m then ap n = nm−1(dsns + ds−1ns−1 + · · · + d1n + d0)λn Arash Raﬁey Recurrence Relations (review and examples) Example : an = 4an−1 −4an−2 + 2nn, a0 = 1 and a1 = 2. Arash Raﬁey Recurrence Relations (review and examples) Example : an = 4an−1 −4an−2 + 2nn, a0 = 1 and a1 = 2. r2 = 4r −4 and hence (r −2)2 = 0 and r1 = r2 = 2. Arash Raﬁey Recurrence Relations (review and examples) Example : an = 4an−1 −4an−2 + 2nn, a0 = 1 and a1 = 2. r2 = 4r −4 and hence (r −2)2 = 0 and r1 = r2 = 2. λ = 2 and m = 2 therefore an = n(d1n + d0)2n + c2n. Arash Raﬁey Recurrence Relations (review and examples) Example : an = 4an−1 −4an−2 + 2nn, a0 = 1 and a1 = 2. r2 = 4r −4 and hence (r −2)2 = 0 and r1 = r2 = 2. λ = 2 and m = 2 therefore an = n(d1n + d0)2n + c2n. a0 = 1 = c and a1 = 2 = 2(d1 + d0) + 2 and hence d1 = −d0, a2 = 12 = 2(2d1 −d1)4 + 4 d1 = 1. Therefore an = n(n −1)2n + 2n. Arash Raﬁey Recurrence Relations (review and examples) Example : an+2 = an+1an, a0 = 1, a1 = 2. Arash Raﬁey Recurrence Relations (review and examples) Example : an+2 = an+1an, a0 = 1, a1 = 2. a2 = 2, a3 = 4, a4 = 8. Arash Raﬁey Recurrence Relations (review and examples) Example : an+2 = an+1an, a0 = 1, a1 = 2. a2 = 2, a3 = 4, a4 = 8. Therefore we may assume that an = 2bn. Arash Raﬁey Recurrence Relations (review and examples) Example : an+2 = an+1an, a0 = 1, a1 = 2. a2 = 2, a3 = 4, a4 = 8. Therefore we may assume that an = 2bn. 2bn+2 = 2bn+12bn and hence 2bn+2 = 2bn+1+bn. Arash Raﬁey Recurrence Relations (review and examples) Example : an+2 = an+1an, a0 = 1, a1 = 2. a2 = 2, a3 = 4, a4 = 8. Therefore we may assume that an = 2bn. 2bn+2 = 2bn+12bn and hence 2bn+2 = 2bn+1+bn. Thus bn+2 = bn+1 + bn and b0 = 0, b1 = 1. Arash Raﬁey Recurrence Relations (review and examples) Example : an+2 = an+1an, a0 = 1, a1 = 2. a2 = 2, a3 = 4, a4 = 8. Therefore we may assume that an = 2bn. 2bn+2 = 2bn+12bn and hence 2bn+2 = 2bn+1+bn. Thus bn+2 = bn+1 + bn and b0 = 0, b1 = 1. bn is the n-th Fibonacci’s number. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no ”100”. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no ”100”. Let an be the number of such sequences. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no ”100”. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no ”100”. Therefore we have an−1 of such sequences. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no ”100”. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no ”100”. Therefore we have an−1 of such sequences. If the last symbol is 0 and the (n −1)-th symbol is 1 then the ﬁrst n −2 symbols is a binary sequences of length n −2 with no ”100”. Therefore we have an−2 of such sequences. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no ”100”. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no ”100”. Therefore we have an−1 of such sequences. If the last symbol is 0 and the (n −1)-th symbol is 1 then the ﬁrst n −2 symbols is a binary sequences of length n −2 with no ”100”. Therefore we have an−2 of such sequences. If the last symbol is 0 and the (n −1)-th symbol is 0 then all the previous symbols must be 0 (one such sequence). Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no ”100”. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no ”100”. Therefore we have an−1 of such sequences. If the last symbol is 0 and the (n −1)-th symbol is 1 then the ﬁrst n −2 symbols is a binary sequences of length n −2 with no ”100”. Therefore we have an−2 of such sequences. If the last symbol is 0 and the (n −1)-th symbol is 0 then all the previous symbols must be 0 (one such sequence). Therefore an = an−1 + an−2 + 1 with a1 = 2 and a2 = 4. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no ”100”. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no ”100”. Therefore we have an−1 of such sequences. If the last symbol is 0 and the (n −1)-th symbol is 1 then the ﬁrst n −2 symbols is a binary sequences of length n −2 with no ”100”. Therefore we have an−2 of such sequences. If the last symbol is 0 and the (n −1)-th symbol is 0 then all the previous symbols must be 0 (one such sequence). Therefore an = an−1 + an−2 + 1 with a1 = 2 and a2 = 4. an = ah n + ap n where ap n = A (constant) and (ah n is the homogenous part) an = c1(1+ √ 5 2 )n + c2(1− √ 5 2 )n + A. Arash Raﬁey Recurrence Relations (review and examples) Example : Find the solution to the recurrence relation : an = −3an−1 −3an−2 −an−3, a0 = 1, a1 = −2, a2 = −1. Arash Raﬁey Recurrence Relations (review and examples) Example : Find the solution to the recurrence relation : an = −3an−1 −3an−2 −an−3, a0 = 1, a1 = −2, a2 = −1. r3 + 3r2 + 3r + 1 = 0 is the characteristic equation. Arash Raﬁey Recurrence Relations (review and examples) Example : Find the solution to the recurrence relation : an = −3an−1 −3an−2 −an−3, a0 = 1, a1 = −2, a2 = −1. r3 + 3r2 + 3r + 1 = 0 is the characteristic equation. Therefore (r + 1)3 = 0 and hence r = −1 is a root with multiplicity 3. So an = (A0 + A1n + A2n2)(−1)n Arash Raﬁey Recurrence Relations (review and examples) Example : Find the solution to the recurrence relation : an = −3an−1 −3an−2 −an−3, a0 = 1, a1 = −2, a2 = −1. r3 + 3r2 + 3r + 1 = 0 is the characteristic equation. Therefore (r + 1)3 = 0 and hence r = −1 is a root with multiplicity 3. So an = (A0 + A1n + A2n2)(−1)n a0 = 1 = A0 and a1 = −2 = (1 + A1 + A2)(−1), and a2 = −1 = (1 + 2A1 + 4A2) Arash Raﬁey Recurrence Relations (review and examples) Example : Find the solution to the recurrence relation : an = −3an−1 −3an−2 −an−3, a0 = 1, a1 = −2, a2 = −1. r3 + 3r2 + 3r + 1 = 0 is the characteristic equation. Therefore (r + 1)3 = 0 and hence r = −1 is a root with multiplicity 3. So an = (A0 + A1n + A2n2)(−1)n a0 = 1 = A0 and a1 = −2 = (1 + A1 + A2)(−1), and a2 = −1 = (1 + 2A1 + 4A2) an = (1 + 3n −2n2)(−1)n. Arash Raﬁey Recurrence Relations (review and examples) Solve the simultaneous recurrence relation. an = 3an−1 + 2bn−1 and bn = an−1 + 2bn−1, a0 = 1, b0 = 2. Arash Raﬁey Recurrence Relations (review and examples) Solve the simultaneous recurrence relation. an = 3an−1 + 2bn−1 and bn = an−1 + 2bn−1, a0 = 1, b0 = 2. an−1 = 3an−2 + 2bn−2 and bn−1 = an−2 + 2bn−2. Therefore Arash Raﬁey Recurrence Relations (review and examples) Solve the simultaneous recurrence relation. an = 3an−1 + 2bn−1 and bn = an−1 + 2bn−1, a0 = 1, b0 = 2. an−1 = 3an−2 + 2bn−2 and bn−1 = an−2 + 2bn−2. Therefore bn−1 = an−2 + an−1 −3an−2 = an−1 −2an−2 So Arash Raﬁey Recurrence Relations (review and examples) Solve the simultaneous recurrence relation. an = 3an−1 + 2bn−1 and bn = an−1 + 2bn−1, a0 = 1, b0 = 2. an−1 = 3an−2 + 2bn−2 and bn−1 = an−2 + 2bn−2. Therefore bn−1 = an−2 + an−1 −3an−2 = an−1 −2an−2 So an = 3an−1 + 2an−1 −4an−2 = 5an−1 −4an−2, a0 = 1, a1 = 7. Arash Raﬁey Recurrence Relations (review and examples) Solve the simultaneous recurrence relation. an = 3an−1 + 2bn−1 and bn = an−1 + 2bn−1, a0 = 1, b0 = 2. an−1 = 3an−2 + 2bn−2 and bn−1 = an−2 + 2bn−2. Therefore bn−1 = an−2 + an−1 −3an−2 = an−1 −2an−2 So an = 3an−1 + 2an−1 −4an−2 = 5an−1 −4an−2, a0 = 1, a1 = 7. r2 −5r + 4 = 0 and (r −4)(r −1) = 0. Arash Raﬁey Recurrence Relations (review and examples) Solve the simultaneous recurrence relation. an = 3an−1 + 2bn−1 and bn = an−1 + 2bn−1, a0 = 1, b0 = 2. an−1 = 3an−2 + 2bn−2 and bn−1 = an−2 + 2bn−2. Therefore bn−1 = an−2 + an−1 −3an−2 = an−1 −2an−2 So an = 3an−1 + 2an−1 −4an−2 = 5an−1 −4an−2, a0 = 1, a1 = 7. r2 −5r + 4 = 0 and (r −4)(r −1) = 0. an = c14n + c2(1)n Arash Raﬁey Recurrence Relations (review and examples) Solve the simultaneous recurrence relation. an = 3an−1 + 2bn−1 and bn = an−1 + 2bn−1, a0 = 1, b0 = 2. an−1 = 3an−2 + 2bn−2 and bn−1 = an−2 + 2bn−2. Therefore bn−1 = an−2 + an−1 −3an−2 = an−1 −2an−2 So an = 3an−1 + 2an−1 −4an−2 = 5an−1 −4an−2, a0 = 1, a1 = 7. r2 −5r + 4 = 0 and (r −4)(r −1) = 0. an = c14n + c2(1)n a0 = 1 = c1 + c2, a1 = 7 = 4c1 + c2. Therefore c1 = 2 and c2 = −1. Arash Raﬁey Recurrence Relations (review and examples) Example : We are given a 2 by n grid and we want to ﬁll it out with dominos (2 by 1 or 1 by 2 grid). What is the number of ways of doing this ? Arash Raﬁey Recurrence Relations (review and examples) Example : We are given a 2 by n grid and we want to ﬁll it out with dominos (2 by 1 or 1 by 2 grid). What is the number of ways of doing this ? n 2 Arash Raﬁey Recurrence Relations (review and examples) Example : We are given a 2 by n grid and we want to ﬁll it out with dominos (2 by 1 or 1 by 2 grid). What is the number of ways of doing this ? n 2 Let bn be the number of ways : Either we put the last domino vertically(2 by 1) and then in the remaining we have bn−1 ways or We put two dominos horizontally (2 of 1 by 2) and then in the remaining we have bn−2 ways. Therefore bn = bn−1 + bn−2, b1 = 1 and b2 = 2. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no consecutive 0′s. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no consecutive 0′s. Let an be the number of such sequences. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no consecutive 0′s. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no consecutive 0′s thus we have an−1 of such sequences. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no consecutive 0′s. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no consecutive 0′s thus we have an−1 of such sequences. If the last symbol is 0 then the (n −1)-th symbol is 1 and the ﬁrst n −2 symbols is a binary sequences of length n −2 with no consecutive 0′s thus we have an−2 of such sequences. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no consecutive 0′s. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no consecutive 0′s thus we have an−1 of such sequences. If the last symbol is 0 then the (n −1)-th symbol is 1 and the ﬁrst n −2 symbols is a binary sequences of length n −2 with no consecutive 0′s thus we have an−2 of such sequences. Therefore an = an−1 + an−2 with a1 = 2 and a2 = 3. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of binary sequences of length n with no consecutive 0′s. Let an be the number of such sequences. If the last symbol is 1 then the ﬁrst n −1 symbols is a binary sequences of length n −1 with no consecutive 0′s thus we have an−1 of such sequences. If the last symbol is 0 then the (n −1)-th symbol is 1 and the ﬁrst n −2 symbols is a binary sequences of length n −2 with no consecutive 0′s thus we have an−2 of such sequences. Therefore an = an−1 + an−2 with a1 = 2 and a2 = 3. Now the solution is the n + 2-th Fibonacci’s number. Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition : We say a sequence S of 0, 1 is nice if the number of ones and the number of zeros are the same and in every preﬁx of S the number of ones is not less than the number of zero. Problem : What is the number of nice sequences of length 2n ? Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition : We say a sequence S of 0, 1 is nice if the number of ones and the number of zeros are the same and in every preﬁx of S the number of ones is not less than the number of zero. Problem : What is the number of nice sequences of length 2n ? Let bn be the number of nice-sequences of length 2n. Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition : We say a sequence S of 0, 1 is nice if the number of ones and the number of zeros are the same and in every preﬁx of S the number of ones is not less than the number of zero. Problem : What is the number of nice sequences of length 2n ? Let bn be the number of nice-sequences of length 2n. Consider the ﬁrst index i that the number of 1’s and the number of 0’s (from 1 to 2i) are the same. Arash Raﬁey Recurrence Relations (review and examples) Deﬁnition : We say a sequence S of 0, 1 is nice if the number of ones and the number of zeros are the same and in every preﬁx of S the number of ones is not less than the number of zero. Problem : What is the number of nice sequences of length 2n ? Let bn be the number of nice-sequences of length 2n. Consider the ﬁrst index i that the number of 1’s and the number of 0’s (from 1 to 2i) are the same. Then we can write : bn = i=n X i=1 bi−1bn−i b0 = 1, b1 = 1. Arash Raﬁey Recurrence Relations (review and examples) bn+1 = b0bn + b1bn−1 + · · · + bn−1b1 + bnb0 ∞ X n=0 bn+1xn+1 = ∞ X n=0 (b0bn + b1bn−1 + · · · + bn−1b1 + bnb0)xn+1 Arash Raﬁey Recurrence Relations (review and examples) bn+1 = b0bn + b1bn−1 + · · · + bn−1b1 + bnb0 ∞ X n=0 bn+1xn+1 = ∞ X n=0 (b0bn + b1bn−1 + · · · + bn−1b1 + bnb0)xn+1 Let f (x) = ∞ P n=0 bnxn be the generating function for b0, b1, b2, . . . . Arash Raﬁey Recurrence Relations (review and examples) bn+1 = b0bn + b1bn−1 + · · · + bn−1b1 + bnb0 ∞ X n=0 bn+1xn+1 = ∞ X n=0 (b0bn + b1bn−1 + · · · + bn−1b1 + bnb0)xn+1 Let f (x) = ∞ P n=0 bnxn be the generating function for b0, b1, b2, . . . . (f (x) −b0) = x ∞ P n=0 (b0bn + b1bn−1 + · · · + bn−1b1 + bnb0)xn = x[f (x)]2. Arash Raﬁey Recurrence Relations (review and examples) bn+1 = b0bn + b1bn−1 + · · · + bn−1b1 + bnb0 ∞ X n=0 bn+1xn+1 = ∞ X n=0 (b0bn + b1bn−1 + · · · + bn−1b1 + bnb0)xn+1 Let f (x) = ∞ P n=0 bnxn be the generating function for b0, b1, b2, . . . . (f (x) −b0) = x ∞ P n=0 (b0bn + b1bn−1 + · · · + bn−1b1 + bnb0)xn = x[f (x)]2. x[f (x)]2 −f (x) + 1 = 0 and hence f (x) = [1 ± √1 −4x]/(2x). Arash Raﬁey Recurrence Relations (review and examples) bn+1 = b0bn + b1bn−1 + · · · + bn−1b1 + bnb0 ∞ X n=0 bn+1xn+1 = ∞ X n=0 (b0bn + b1bn−1 + · · · + bn−1b1 + bnb0)xn+1 Let f (x) = ∞ P n=0 bnxn be the generating function for b0, b1, b2, . . . . (f (x) −b0) = x ∞ P n=0 (b0bn + b1bn−1 + · · · + bn−1b1 + bnb0)xn = x[f (x)]2. x[f (x)]2 −f (x) + 1 = 0 and hence f (x) = [1 ± √1 −4x]/(2x). √1 −4x = (1 −4x)1/2 = 1/2 0 + 1/2 1 (−4x) + 1/2 2 (−4x)2 + . . . Arash Raﬁey Recurrence Relations (review and examples) The coeﬃcient of xn, n ≥1 is 1/2 n (−4)n = (1/2)(1/2 −1)(1/2 −2) . . . ((1/2) −n + 1) n! (−4)n Arash Raﬁey Recurrence Relations (review and examples) The coeﬃcient of xn, n ≥1 is 1/2 n (−4)n = (1/2)(1/2 −1)(1/2 −2) . . . ((1/2) −n + 1) n! (−4)n which is (−1) (2n−1) 2n n . Then f (x) = 1 2x [1 −[1 − ∞ P n=1 1 (2n−1) 2n n xn]], and bn the coeﬃcient of xn in f (x) is half of the coeﬃcient of xn+1 in ∞ P n=1 1 (2n−1) 2n n xn. Therefore Arash Raﬁey Recurrence Relations (review and examples) The coeﬃcient of xn, n ≥1 is 1/2 n (−4)n = (1/2)(1/2 −1)(1/2 −2) . . . ((1/2) −n + 1) n! (−4)n which is (−1) (2n−1) 2n n . Then f (x) = 1 2x [1 −[1 − ∞ P n=1 1 (2n−1) 2n n xn]], and bn the coeﬃcient of xn in f (x) is half of the coeﬃcient of xn+1 in ∞ P n=1 1 (2n−1) 2n n xn. Therefore bn = 1 2[ 1 2(n+1)−1] 2(n+1) n+1 = 1 (n+1) 2n n Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of sequences of length n on the alphabet {0, 1, 2} with no consecutive 0’s and no consecutive 1’s. Let’s call such a sequence a good sequence. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of sequences of length n on the alphabet {0, 1, 2} with no consecutive 0’s and no consecutive 1’s. Let’s call such a sequence a good sequence. Let an be the number of good sequences of length n. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of sequences of length n on the alphabet {0, 1, 2} with no consecutive 0’s and no consecutive 1’s. Let’s call such a sequence a good sequence. Let an be the number of good sequences of length n. Case 1. Suppose the last symbol is 2. Then the ﬁrst n −1 symbols is a good sequence of length n −1 and we have an−1 of such sequences. Arash Raﬁey Recurrence Relations (review and examples) Example : What is the number of sequences of length n on the alphabet {0, 1, 2} with no consecutive 0’s and no consecutive 1’s. Let’s call such a sequence a good sequence. Let an be the number of good sequences of length n. Case 1. Suppose the last symbol is 2. Then the ﬁrst n −1 symbols is a good sequence of length n −1 and we have an−1 of such sequences. Let b0 n be the number of good sequences ending with 0 and b1 n be the number of good sequences ending with 1. Note that b0 n = b1 n = bn and we have an = an−1 + 2bn. Arash Raﬁey Recurrence Relations (review and examples) Case 2. Suppose the last symbol is 0. Then symbol n −1 is 2 or 1. Therefore bn = an−2 + bn−1 (an−2 for when n −1-symbol is 2). Now we have : an = an−1 + 2bn and bn = bn−1 + an−2, (2bn = 2bn−1 + 2an−2) Arash Raﬁey Recurrence Relations (review and examples) Case 2. Suppose the last symbol is 0. Then symbol n −1 is 2 or 1. Therefore bn = an−2 + bn−1 (an−2 for when n −1-symbol is 2). Now we have : an = an−1 + 2bn and bn = bn−1 + an−2, (2bn = 2bn−1 + 2an−2) Therefore (an −an−1) = (an−1 −an−2) + 2an−2 and hence an = 2an−1 + an−2 Arash Raﬁey Recurrence Relations (review and examples) O, θ notations For a given function g(n), Θ(g(n)) denotes the set Θ(g(n)) = {f (n) : there exist positive constants c1, c2, n0 such that c1 · g(n) ≤f (n) ≤c2 · g(n) for all n ≥n0} Intuition: f (n) belongs to the family Θ(g(n)) if ∃constants c1, c2 s.t. f (n) can ﬁt between c1 · g(n) and c2 · g(n), for all n suﬃciently large. Correct notation: f (n) ∈Θ(g(n)) Usually used: f (n) = Θ(g(n)). We also say that “f (n) is in Θ(g(n))”. Arash Raﬁey Recurrence Relations (review and examples) Examples of Θ-notation: f (n) = 2n2 = Θ(n2) because with g(n) = n2 and c1 = 1 and c2 = 2 we have 0 ≤c1g(n) ≤f (n) = 2 · n2 ≤c2 · g(n). Arash Raﬁey Recurrence Relations (review and examples) Examples of Θ-notation: f (n) = 2n2 = Θ(n2) because with g(n) = n2 and c1 = 1 and c2 = 2 we have 0 ≤c1g(n) ≤f (n) = 2 · n2 ≤c2 · g(n). f (n) = 8n5 + 17n4 −25 = Θ(n5) because f (n) ≥7 · n5 for n large enough n 8n5 + 17n4 −25 n5 7n5 1 8 · 1 + 17 · 1 −25 = 0 1 7 2 8 · 32 + 17 · 16 −25 = 503 32 224 and f (n) ≤8n5 + 17n5 = 25n5, thus c1 = 7, c2 = 25 and n0 = 2 are good enough. Arash Raﬁey Recurrence Relations (review and examples) Big-O-notation When we’re interested in asymptotic upper bounds only, we use O-notation (read: “big-O”). For given function g(n), deﬁne O(g(n)) (read: “big-O of g of n” or also “order g of n”) as follows: O(g(n)) = {f (n) : there exist positive constants c, n0 such that f (n) ≤c · g(n) for all n ≥n0} We write f (n) = O(g(n)) to indicate that f (n) is member of set O(g(n)). Obviously, f (n) = Θ(g(n)) implies f (n) = O(g(n)); we just drop the left inequality in the deﬁnition of Θ(g(n)). Arash Raﬁey Recurrence Relations (review and examples) Big-Omega-notation Like O-notation, but for lower bounds For a given function g(n), Ω(n) denotes the set Ω(g(n)) = {f (n) : there exist positive constants c, n0 such that c · g(n) ≤f (n) for all n ≥n0} Saying T(n) = Ω(n2) means growth of T(n) is at least the of n2. Clearly, f (n) = Θ(g(n)) iﬀf (n) = Ω(g(n)) and f (n) = O(g(n)). Arash Raﬁey Recurrence Relations (review and examples) o-notation Similar to O f (n) = O(g(n)) means we can upper-bound the growth of f by the growth of g (up to a constant factor) f (n) = o(g(n)) is the same, except we require the growth of f to be strictly smaller than the growth of g: For a given function g(n), o(n) denotes the set o(g(n)) = {f (n) : for any pos constant c there exists a pos constant n0 such that f (n) < c · g(n) for all n ≥n0} Arash Raﬁey Recurrence Relations (review and examples) omega-notation For a given function g(n), ω(n) denotes the set ω(g(n)) = {f (n) : for any pos constant c there exists a pos constant n0 such that c · g(n) < f (n) for all n ≥n0} In other words: lim n→∞ f (n) g(n) = ∞ if the limit exists. I.e., f (n) becomes arbitrarily large relative to g(n). Arash Raﬁey Recurrence Relations (review and examples) Heuristics that can help to ﬁnd a good guess. One way would be to have a look at ﬁrst few terms. Say if we had T(n) = 2T(n/2) + 3n, then T(n) = 2T(n/2) + 3n = 2(2T(n/4) + 3(n/2)) + 3n = 2(2(2T(n/8) + 3(n/4)) + 3(n/2)) + 3n = 23T(n/23) + 223(n/22) + 213(n/21) + 203(n/20) We can do this log n times Arash Raﬁey Recurrence Relations (review and examples) Heuristics that can help to ﬁnd a good guess. One way would be to have a look at ﬁrst few terms. Say if we had T(n) = 2T(n/2) + 3n, then T(n) = 2T(n/2) + 3n = 2(2T(n/4) + 3(n/2)) + 3n = 2(2(2T(n/8) + 3(n/4)) + 3(n/2)) + 3n = 23T(n/23) + 223(n/22) + 213(n/21) + 203(n/20) We can do this log n times 2log n · T(n/2log n) + log(n)−1 X i=0 2i3(n/2i) = n · T(1) + 3n · log(n)−1 X i=0 1 = n · T(1) + 3n log n = Θ(n log n) After guessing a solution you’ll have to prove the correctness. Arash Raﬁey Recurrence Relations (review and examples) Theorem Let a, b, c be positive integers with b ≥2 and let f : Z + →R. If f (1) = c and f (n) = af (n/b) + c for n = bk then for all n = 1, b, b2, b3, . . . , 1 f (n) = c(logn b +1), when a = 1 2 f (n) = c(anloga b−1) a−1 , when a ≥2. Arash Raﬁey Recurrence Relations (review and examples) stepwise reﬁnement – guessing loose lower and upper bounds, and gradually taking them closer to each other For T(n) = 2T(⌊n/2⌋) + n we see Arash Raﬁey Recurrence Relations (review and examples) stepwise reﬁnement – guessing loose lower and upper bounds, and gradually taking them closer to each other For T(n) = 2T(⌊n/2⌋) + n we see T(n) = Ω(n) (because of the n term) T(n) = O(n2) (easily proven) Arash Raﬁey Recurrence Relations (review and examples) stepwise reﬁnement – guessing loose lower and upper bounds, and gradually taking them closer to each other For T(n) = 2T(⌊n/2⌋) + n we see T(n) = Ω(n) (because of the n term) T(n) = O(n2) (easily proven) From there, we can perhaps “converge” on asymptotically tight bound Θ(n log n). Arash Raﬁey Recurrence Relations (review and examples) A neat trick called “changing variables” Suppose we have T(n) = 2T(√n) + log n Arash Raﬁey Recurrence Relations (review and examples) A neat trick called “changing variables” Suppose we have T(n) = 2T(√n) + log n Now rename m = log n ⇔2m = n. We know √n = n1/2 = (2m)1/2 = 2m/2 and thus obtain T(2m) = 2T(2m/2) + m Arash Raﬁey Recurrence Relations (review and examples) A neat trick called “changing variables” Suppose we have T(n) = 2T(√n) + log n Now rename m = log n ⇔2m = n. We know √n = n1/2 = (2m)1/2 = 2m/2 and thus obtain T(2m) = 2T(2m/2) + m Now rename S(m) = T(2m) and get S(m) = 2S(m/2) + m Looks familiar. We know the solution S(m) = Θ(m log m). Arash Raﬁey Recurrence Relations (review and examples) A neat trick called “changing variables” Suppose we have T(n) = 2T(√n) + log n Now rename m = log n ⇔2m = n. We know √n = n1/2 = (2m)1/2 = 2m/2 and thus obtain T(2m) = 2T(2m/2) + m Now rename S(m) = T(2m) and get S(m) = 2S(m/2) + m Looks familiar. We know the solution S(m) = Θ(m log m). Going back from S(m) to T(n) we obtain T(n) = T(2m) = S(m) = Θ(m log m) = Θ(log n log log n) Arash Raﬁey Recurrence Relations (review and examples) The “Master Method” Theorem Let a ≥1 and b > 1 be constants, let f (n) be a function, and let T(n) be deﬁned on the nonnegative integers by the recurrence T(n) = aT(n/b) + f (n), where we interpret n/b to mean either ⌊n/b⌋or ⌈n/b⌉. Then T(n) can be bounded asymptotically as follows. 1 If f (n) = O(n(logb a)−ϵ) for some constant ϵ > 0, then T(n) = Θ(nlogb a). 2 If f (n) = Θ(nlogb a), then T(n) = Θ(nlogb a · log n). 3 If f (n) = Ω(n(logb a)+ϵ) for some constant ϵ > 0, and if a · f (n/b) ≤c · f (n) for some constant c < 1 and all suﬃciently large n, then T(n) = Θ(f (n)). Arash Raﬁey Recurrence Relations (review and examples) Notes on Master’s Theorem 2. If f (n) = Θ(nlogb a), then T(n) = Θ(nlogb a · log n). Note 1: Although it’s looking rather scary, it really isn’t. For instance, T(n) = 2T(n/2) + Θ(n) we have nlogb a = nlog2 2 = n1 = n, and we can apply case 2. The result is therefore Θ(nlogb a · log n) = Θ(n log n). Arash Raﬁey Recurrence Relations (review and examples) Notes on Master’s Theorem 2. If f (n) = Θ(nlogb a), then T(n) = Θ(nlogb a · log n). Note 1: Although it’s looking rather scary, it really isn’t. For instance, T(n) = 2T(n/2) + Θ(n) we have nlogb a = nlog2 2 = n1 = n, and we can apply case 2. The result is therefore Θ(nlogb a · log n) = Θ(n log n). 1 If f (n) = O(n(logb a)−ϵ) for some constant ϵ > 0, then T(n) = Θ(nlogb a). Note 2: In case 1, f (n) = n(logb a)−ϵ = nlogb a/nϵ = o(nlogb a) , so the ϵ does matter. This case is basically about “small” functions f . But it’s not enough if f (n) is just asymptotically smaller than nlogb a (that is f (n) ∈o(nlogb a), it must be polynomially smaller! Arash Raﬁey Recurrence Relations (review and examples) 3. If f (n) = Ω(n(logb a)+ϵ) for some constant ϵ > 0, and if a · f (n/b) ≤c · f (n) for some constant c < 1 and all suﬃciently large n, then T(n) = Θ(f (n)). Arash Raﬁey Recurrence Relations (review and examples) 3. If f (n) = Ω(n(logb a)+ϵ) for some constant ϵ > 0, and if a · f (n/b) ≤c · f (n) for some constant c < 1 and all suﬃciently large n, then T(n) = Θ(f (n)). Note 3: Similarly, in case 3, f (n) = n(logb a)+ϵ = nlogb a · nϵ = ω(nlogb a) , so the ϵ does matter again. This case is basically about “large” functions n. But again, f (n) ∈ω(nlogb a) is not enough, it must be polynomially larger. And in addition f (n) has to satisfy the “regularity condition”: af (n/b) ≤cf (n) for some constant c < 1 and n ≥n0 for some n0. Arash Raﬁey Recurrence Relations (review and examples) Using the master theorem Simple enough. Some examples: T(n) = 9T(n/3) + n We have a = 9, b = 3, f (n) = n. Thus, nlogb a = nlog3 9 = n2. Clearly, f (n) = O(nlog3(9)−ϵ) for ϵ = 1, so case 1 gives T(n) = Θ(n2). Arash Raﬁey Recurrence Relations (review and examples) Using the master theorem Simple enough. Some examples: T(n) = 9T(n/3) + n We have a = 9, b = 3, f (n) = n. Thus, nlogb a = nlog3 9 = n2. Clearly, f (n) = O(nlog3(9)−ϵ) for ϵ = 1, so case 1 gives T(n) = Θ(n2). T(n) = T(2n/3) + 1 We have a = 1, b = 3/2, and f (n) = 1, so nlogb a = nlog2/3 1 = n0 = 1. Apply case 2 (f (n) = Θ(nlogb a) = Θ(1), result is T(n) = Θ(log n). Arash Raﬁey Recurrence Relations (review and examples) T(n) = 3T(n/4) + n log n We have a = 3, b = 4, and f (n) = n log n, so nlogb a = nlog4 3 = O(n0.793). Clearly, f (n) = n log n = Ω(n) and thus also f (n) = Ω(nlogb(a)+ϵ) for ϵ ≈0.2. Also, a · f (n/b) = 3(n/4) log(n/4) ≤(3/4)n log n = c · f (n) for any c = 3/4 < 1. Thus we can apply case 3 with result T(n) = Θ(n log n). Arash Raﬁey Recurrence Relations (review and examples) Exercises True or False 1) (n4+10n) 2n2+4 = O(n2) Arash Raﬁey Recurrence Relations (review and examples) Exercises True or False 1) (n4+10n) 2n2+4 = O(n2) 2) n2 log n (5n+4) = O(n) Arash Raﬁey Recurrence Relations (review and examples) Exercises True or False 1) (n4+10n) 2n2+4 = O(n2) 2) n2 log n (5n+4) = O(n) 3) 3n+2 = Ω(3n) Arash Raﬁey Recurrence Relations (review and examples) Exercises True or False 1) (n4+10n) 2n2+4 = O(n2) 2) n2 log n (5n+4) = O(n) 3) 3n+2 = Ω(3n) 4) (log n)log n = Ω(n/ log n) Arash Raﬁey Recurrence Relations (review and examples) Exercises Problem Give asymptotic upper and lower bounds for T(n) in each of the following recurrences. Assume that T(n) is constant for n ≤2. (a) T(n) = 2T(n/2) + n3 (b) T(n) = T(n −1) + n (c) T(n) = T(√n) + 1 (d) T(n) = 16T(n/4) + n2 (e) T(n) = 2T(n −1) + log n Arash Raﬁey Recurrence Relations (review and examples)
13369	https://pmc.ncbi.nlm.nih.gov/articles/PMC5816248/	Kinesin and Dynein Mechanics: Measurement Methods and Research Applications - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. Learn more Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice J Biomech Eng . 2018 Jan 12;140(2):0208051–02080511. doi: 10.1115/1.4037886 Search in PMC Search in PubMed View in NLM Catalog Add to search Kinesin and Dynein Mechanics: Measurement Methods and Research Applications Zachary Abraham Zachary Abraham 1 Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106 Find articles by Zachary Abraham 1, Emma Hawley Emma Hawley 2 Department of Biomedical Engineering, , Case Western Reserve University, , Cleveland, OH 44106 Find articles by Emma Hawley 2, Daniel Hayosh Daniel Hayosh 3 Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106 Find articles by Daniel Hayosh 3, Victoria A Webster-Wood Victoria A Webster-Wood 4 Mem. ASME , Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , 10900 Euclid Avenue, , Cleveland, OH 44106 , e-mail: Webster-wood@case.edu Find articles by Victoria A Webster-Wood 4,1, Ozan Akkus Ozan Akkus 5 Mem. ASME , Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106 Find articles by Ozan Akkus 5 Author information Article notes Copyright and License information 1 Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106 2 Department of Biomedical Engineering, , Case Western Reserve University, , Cleveland, OH 44106 3 Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106 4 Mem. ASME , Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , 10900 Euclid Avenue, , Cleveland, OH 44106 , e-mail: Webster-wood@case.edu 5 Mem. ASME , Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106 1 Corresponding author. Manuscript received July 2, 2017; final manuscript received September 7, 2017; published online January 12, 2018. Editor: Victor H. Barocas. Received 2017 Jul 2; Revised 2017 Sep 7; Issue date 2018 Feb. Copyright © 2018 by ASME 0148-0731/2018/140(2)/020805/11/$25.00 PMC Copyright notice PMCID: PMC5816248 PMID: 28901373 Abstract Motor proteins play critical roles in the normal function of cells and proper development of organisms. Among motor proteins, failings in the normal function of two types of proteins, kinesin and dynein, have been shown to lead many pathologies, including neurodegenerative diseases and cancers. As such, it is critical to researchers to understand the underlying mechanics and behaviors of these proteins, not only to shed light on how failures may lead to disease, but also to guide research toward novel treatment and nano-engineering solutions. To this end, many experimental techniques have been developed to measure the force and motility capabilities of these proteins. This review will (a) discuss such techniques, specifically microscopy, atomic force microscopy (AFM), optical trapping, and magnetic tweezers, and (b) the resulting nanomechanical properties of motor protein functions such as stalling force, velocity, and dependence on adenosine triphosophate (ATP) concentrations will be comparatively discussed. Additionally, this review will highlight the clinical importance of these proteins. Furthermore, as the understanding of the structure and function of motor proteins improves, novel applications are emerging in the field. Specifically, researchers have begun to modify the structure of existing proteins, thereby engineering novel elements to alter and improve native motor protein function, or even allow the motor proteins to perform entirely new tasks as parts of nanomachines. Kinesin and dynein are vital elements for the proper function of cells. While many exciting experiments have shed light on their function, mechanics, and applications, additional research is needed to completely understand their behavior. 1. Introduction Motor proteins are the molecular motors that are responsible for transporting payloads, often referred to as cargo. The transport mechanism of motor proteins is made possible by the power derived from adenosine triphosophate (ATP). The cyclic hydrolysis of ATP allows the motor protein to repeatedly bind and unbind to a filament, producing a step-like motion. A specific subset of motor proteins, called cytoskeletal motor proteins, translate along cytoskeletal filaments. Cargo transport via cytoskeletal motor proteins is achieved by a mechanochemical cycle composed of cytoskeletal filament binding, conformational change, filament release, and conformational relaxation (Fig. 1) [1,2]. While the mechanochemical stepping cycles of kinesin and dynein have been reviewed individually in the past decade, a review of the mechanics of both kinesin and dynein as well as the related measurement methods has not been performed. As such, this review will focus on the techniques for measuring and implications of the force and velocity capabilities of the cytoskeletal motor proteins kinesin and dynein. Fig. 1. Open in a new tab The directional stepping process of kinesin (left) and dynein (right) As both kinesin and dynein are critical to many important cell functions, it is important to understand their mechanics. As such, many techniques have been developed to investigate the locomotion of kinesin and dynein, specifically the force they generate and the velocity they attain. These techniques provide both fundamental information about the function of the proteins and allow researchers to investigate how failings in the protein function relate to clinical pathologies. Furthermore, understanding the mechanics of cytoskeletal motor proteins not only provides information about their native function, but also allows researchers to re-engineer the protein structure to improve native functions or enable them to complete novel tasks. 2. Structure and Function Previous reviews have discussed the structure and function of kinesin and dynein [3,4]. However, we will provide a brief discussion to serve as background. The motor proteins kinesin and dynein exist to transport biological payloads, such as proteins, organelles, and vesicles, along microtubule pathways, and provide forces to drive motion of flagellar structures and cilia . Active transport by kinesin and dynein provides a faster and more efficient mode of intracellular transport than diffusion. Furthermore, large payloads may simply not translate by diffusion. As such, the motor proteins are essential for their translocation. The microtubule pathways provide direct routes for the locomotion of motor proteins. Transport can be thought of in two forms: anterograde transport and retrograde transport. Anterograde transport, also known as plus-ended, refers to the transport of cargo from the center of the cell to the periphery. Meanwhile, retrograde transport, also known as minus-ended, refers to the transport of cargo from the periphery to the center of the cell . Kinesin motor proteins are innately limited to unidirectional stepping, allowing for either anterograde or retrograde transport, with most kinesins performing anterograde transport. Meanwhile, dynein is more likely to perform retrograde transport, although dynein motor proteins are capable of bidirectional stepping. The kinesin superfamily (KIF) is comprised of fourteen large families . To be considered part of the superfamily group, a motor protein must have a particular motor domain. The motor domain specific to KIF molecules is a globular domain (Fig. 2) which undergoes a consistent ATP-binding and microtubule-binding sequence to allow for locomotion, thus the transport characteristic that makes the KIF molecules motor proteins. The cargoes transported by kinesin bind to the motor domain. Additionally, most kinesin families include two heavy chains, two light chains, and an elongated coiled coil. As the understanding of the evolutionary relation of each KIF family is improved, a new classification nomenclature is now used to distinguish the fourteen large families. This nomenclature is simply Kinesin-1 to Kinesin-14, and while they may have different structures, they are unified by their motor domain . Fig. 2. Open in a new tab A comparison of the structure of dynein and kinesin. The left-side image, modified from Ref. , depicts dynein next to the globular motor domain that is consistent between all kinesins. The binding of dynein is modulated during the ATPase cycle via primarily the AAA1 domain. On the right side is a detailed view of the kinesin motor domain (image from public domain). Native kinesin's motion is unidirectional . The directionality of a motor protein's motion is related to the structural position of the motor domain. In the case of kinesin, possible locations for the motor domain are N-terminal, C-terminal, and middle of the molecule. In the nomenclature, these can be differentiated as N-kinesin, C-kinesin, and M-kinesin, respectively. Functionally, the primary difference is the direction of motion, where N-kinesin and M-kinesin are plus-end directed while C-kinesin are minus-end directed . As with kinesin, the dynein family is identified by a specific motor domain. Furthermore, dyneins identified to date consist of two or three heavy chains and a variable number of light chains. The two main branches in the dynein family are cytoplasmic dyneins and axonemal dyneins. Cytoplasmic dyneins are found in all eukaryotic cells. Their functional purpose is to traffic cargo along vesicles and to localize the Golgi apparatus to the center of the cell. Meanwhile, axonemal dyneins are specialized to create the sliding movements of microtubules that power the beating of cilia and flagella. Therefore, axonemal dyneins are critical to the motility of single celled organisms. Dimensionally, dynein is the largest molecular motor known with a molecular mass of 2 million Da [1,6] (Fig. 2). Kinesin and dynein have many attributes in common, both being motor proteins dependent on microtubules and ATP to catalyze a cyclic stepping motion to transport cargoes. However, there are unique qualities of each family of motor protein. Dynein has a larger step size than that of kinesin, making dynein a faster motor than kinesin. Although dynein is larger and faster, kinesin is capable of transporting larger payloads. The kinesin superfamily is a more extensive and diverse group of motor proteins, while dynein is relatively limited with respect to number of species in the family and the number of tasks the type of motor protein completes. 3. Experimental Methods for Motor Protein Analysis As kinesin and dynein are critical to the normal function of cells, it is important to understand the mechanics of these motor proteins. To this end, researchers have implemented several methods to assess the force and motility capabilities of the proteins. Protein motility can be tracked using microscopy techniques such as differential interference contrast (DIC) and fluorescence microscopy. Additionally, atomic force microscopy (AFM) is a force-detection technique that can be modified to measured motility. While conventional AFM is not suited for imaging motility, time-lapsed lateral molecular-force mode has been successfully used to visualize motor protein locomotion and stall force . In combination with microscopy, force spectroscopy techniques such as optical trapping and magnetic tweezers are used to assess motor protein stall force [10–12]. With the wealth of knowledge that has been gained from experimental data, mathematical models have also been developed to predict the behavior of kinesin and dynein, taking into account environmental conditions such as ATP concentration. 3.1. Microscopy. Tracking the motility of kinesin has been feasible for several decades. Kinesin was initially discovered through the use of video-enhanced differential interference contrast (VE-DIC) microscopy (Fig. 3(b)). By enhancing the contrast of images and removing signal noise, VE-DIC can produce images at a much higher resolution than conventional microscopy. Subsequently, VE-DIC was used to detect the step size and velocity of kinesin-coated beads traveling on surface bound microtubules [15–17]. The bead's position was measured at 30 frames per second at a precision of 1–2 nm, allowing for subsequent calculation of velocity relative to ATP and step size. Additionally, DIC has been employed in a technique called centrifuge microscopy during which the microscope stage is spun to impart centrifugal forces on a sperm. Rotationally induced forces are countered by surface-bound kinesin molecules. Adjusting the rotational acceleration changed the movement of the sperm, giving a force–velocity relation for kinesin . Fig. 3. Open in a new tab (a) A diagram of the optical trapping measurement technique. A bead is captured by the focused optical trap such that either a force is applied, or the escape force (F k) generated by the protein to move the bead within the trap are met with a resistive force (F t). (b) The motion of the bead can be tacked with DIC (as shown, image from Ref. , CC BY 3.0) or with fluorescent microscopy and related to the applied force. While DIC microscopy was crucial to the early investigation of kinesin, it has lost its place to fluorescence microscopy in visualizing kinesin, dynein, and/or microtubules through the use of fluorescent beads [16,18,19]. Fluorescent markers to label these proteins are quantum dots or other commercially available fluorophores [20–26]. Additionally, the motility can be measured indirectly by labeling microtubules, which can then be propelled by surface bound proteins [12,27–30]. Quantum dots are useful for measurements involving total internal reflection fluorescence microscopy [9,20–23]. Total internal reflection fluorescence is a highly detailed and sensitive technique in which light is totally internally reflected at the interface of glass and buffer in order to reduce the noise in typical measurement. 3.2. Atomic Force Microscopy. Atomic force microscopy is a widely used technique across a broad range of disciplines for force measurement and motility imaging at a nanometer resolution. In AFM, a microcantilever interacts with a sample, and the deflection of the cantilever is measured via a reflected laser. This allows calculation of the force on the cantilever. Additionally, time-lapse measurements can provide a picture of the sample's motion. Atomic force microscopy is useful for measuring the motility and force capabilities of kinesin. Using AFM, Schaap et al. were able to determine the step size and the velocity of kinesin . They were also able to create three-dimensional representations of kinesin on a microtubule (Fig. 4). In their study, AFM was operated in dynamic tapping mode at a frequency of 7 kHz and the feedback signal was the amplitude of oscillations. Tapping mode as such allowed repeated scans without forcing the kinesin molecules off the microtubule or damaging the cantilever. Fig. 4. Open in a new tab Using AFM, the position of kinesin on a microtubule can be visualized in three-dimensional. This functions for high concentrations of motor proteins (a) or for single proteins (b). Image from Ref. . In order to measure the stall force of motor proteins using AFM, the technique has been modified such that the force measurements are made with a cantilever that is perpendicular to the sample. Instead of measuring reflected light, this method, known as lateral molecular force microscopy (LMFM), measures scattered evanescent electromagnetic waves . This allows the cantilever to be manufactured on a much smaller level and be mounted vertically over the sample rather than horizontally. In this AFM mode, the cantilever itself is exerting force directly on the sample, rather than indirectly through the tip. Using this technique, Scholz et al. measured the force capabilities of single kinesin molecules bound directly to the cantilever . When kinesin was constrained to move in a straight line along the microtubules, the knowledge on the displacement and stiffness of the cantilever allowed the measurements of velocity, step size, and stall force of the kinesin . 3.3. Optical Trapping. Optical trapping, also referred to as optical tweezers, is commonly used to measure forces at the cellular level and lower scales. Samples are prepared by functionalizing the surface of a bead and subsequently coating the surface with the motor proteins or attaching a single microtubule. A substrate is similarly prepared, coated with the opposite components (i.e., microtubules or motor proteins, respectively). The optical trap is then formed by focusing a laser at the edge of the sample, using the resulting laser force to stall or slow the sample's movement (Fig. 3(a)). By using optical trapping alongside a motility technique, it is possible to simultaneously measure the force and the velocity of kinesin to find their relation. A single experiment can provide the stall force, the unforced velocity, and the force–velocity relationship to give a more clear understanding of how external forces affect the motion of kinesin [18,33]. The force generated by kinesin and dynein is measured by multiplying the stiffness of the optical trap by the displacement of the motor protein or microtubule bound bead from the trap center. When the bead reaches a certain distance, it is observed to rapidly snap back to the center of the trap. This snap back indicates that the stall force has been reached, causing the protein to unbind from the microtubule. This method has been used to find the stall force of both kinesin [19,34] and dynein [20,25]. Alternatively, the stall force can be determined from zero-velocity plateaus prior to snap back [16,18,33]. In contrast to snap back studies, where the force is increased until snap back occurs, the stall force can also be determined by lowering the trapping force until kinesin begins to move . Optical trapping can also be directly combined with DIC microscopy techniques to create an optical trapping interferometer. Unlike most cases where the motility measurement technique is separate from the optical trap, this method involves using the same laser to both trap a kinesin-coated bead and allow for displacement measurements using DIC due to optical polarization. Using the same laser for measuring position and force facilitates system alignment and do not interfere with use of DIC. As such, optical trapping interferometry was used to measure the step size of kinesin [16,33]. Building on the interferometry platform, constant force optical traps have been developed, known as force clamps, which respond to changes in the position of a kinesin-coated bead in a feedback loop to maintain the constant applied force. Applying a constant external force simplifies step size and isometric stall force measurement, and can also be used to measure forced velocity [33,35]. Additionally, such force clamps can be used to provide an external force to induce stepping of the motor protein for both kinesin and dynein . As expected, such external forces applied to dynein can be used to induce bidirectional stepping. Interestingly, the same behavior is seen for kinesin, despite it only performing unidirectional stepping in its native environment. 3.4. Magnetic Tweezers. The magnetic tweezers method is another single-molecule technique which connects a sample to a magnetic bead and then moves the bead using a magnetic field. By adjusting the force of the magnetic field, it is possible to determine aspects of the sample's behavior such as the stall force and the force–velocity relation in the same way as it is done with optical trapping (Fig. 5). Magnetic tweezers are unique in their ability to rotate a sample, which cannot be accomplished in optical trapping or AFM. Fig. 5. Open in a new tab (a) Magnetic tweezers, similar to optical traps, trap a bead which is attached to either a motor protein or a microtubule. An external force (F m) can then be applied to the bead via a magnetic field. The effect of the force on the bead can be seen between (b) when no magnetic field is present and (c) when the field is applied. (b) and (c) adapted from Ref. . Magnetic tweezers have seen limited use as compared to optical trapping, and the commercial availability of paramagnetic beads is likely to raise the appeal of magnetic tweezers . In one application, Fallesen et al. attached a magnetic bead to a microtubule, which was propelled by surface bound kinesin [12,28,29]. The magnetic tweezers were then used to manipulate the bead, applying an external force to the microtubule. By varying the external force, the velocity of the microtubule changes, allowing determination of the force–velocity characteristics of the motor proteins. 3.5. Indirect Force Measurement. In addition to the single-molecule techniques that directly measure the forces of motor proteins, these forces can be inferred indirectly. Dynein's force has been inferred from measuring the flagella that it drives [37–39]. For example, Schmitz et al. were able to measure the force produced by a bull sperm flagellum, and as a result, the force produced by the dynein was derived . The force of a beating sperm was found using a microprobe, and a model was developed for the force of dynein. In addition to using this estimation for their own research, they applied the same model to blue mussel cilia force data previously reported by Yoneda in 1960 to estimate the dynein force produced in blue mussel . These indirect methods have mostly been replaced by direct techniques, which can precisely measure an individual motor protein. 4. Review of Motor Protein Mechanics as Measured by Experimental Methods A systematic review of the literature indicated that while a substantial body of work exists with relation to kinesin biomechanics, fewer studies exist on dynein. Of the 31 papers analyzed, 24 focused on kinesin while only seven focused on dynein (see Tables 1 and 2). Therefore, there seems to be more room for research on mechanics of dynein family proteins. Table 1. Values reported for kinesin motility and force generation by various techniques. Key: OT—optical trap, M—model, FM—fluorescence microscopy, MT—magnetic tweezer, DSEC—Drosophila sequenced E. Coli, D—Drosophila, and LMFM—lateral molecular force microscopy (see Supplementary Materials, which are available under the “Supplemental Materials” tab for this paper on the ASME Digital Collection, for the complete table). \| Paper \| Method \| Species \| Motor protein type \| Stalling Force (pN) mean ± std. dev \| Unloaded velocity (nm/s) \| ATP concentration (mM) \| Temp (°C) \| --- --- --- --- \| \| \| OT \| Bovine \| Kinesin \| 7.2 ± 1.3 \| 600 \| 1 \| \| \| OT \| Bovine \| Kinesin \| 7.2 ± 1.3 \| 200 \| 0.01 \| \| \| \| MT \| DSEC \| Kinesin-1 \| 6.1 ± 0.7 \| 675 \| 1 \| \| \| \| LMFM \| Porcine \| Kinesin-1 \| 4.0 ± 0.2 \| 460 ± 80 \| 1 \| \| \| LMFM \| Porcine \| Kinesin-1 \| 4.0 ± 0.2 \| 170 ± 30 \| 0.01 \| \| \| \| OT \| Bovine \| Kinesin \| 7.3 ± 0.33 \| 1400 \| 1 \| 35 \| \| OT \| 7.3 ± 0.33 \| 1000 \| 1 \| 30 \| \| OT \| 7.3 ± 0.33 \| 750 \| 1 \| 25 \| \| OT \| 7.3 ± 0.33 \| 550 \| 1 \| 20 \| \| OT \| 7.3 ± 0.33 \| 375 \| 1 \| 15 \| \| \| OT \| Bovine \| Kinesin \| 6 \| 190 \| 1 \| \| \| OT \| 5 \| 100 \| 0.04 \| \| \| OT \| 5 \| 30 \| 0.005 \| \| \| \| MT \| D \| Kinesin-1 \| 4 \| 660 \| 1 \| \| \| \| MT \| D \| Kinesin-1 \| 12.2 \| 670 \| 1 \| \| \| \| OT \| Squid \| Kinesin \| 5.7 ± 0.4 \| 700 \| 2 \| \| \| OT \| Squid \| Kinesin \| 5.1 ± 0.5 \| 70 \| 0.01 \| \| \| \| OT \| — \| Kinesin-1 \| \| 100 \| 0.25 \| 1.8 \| \| OT \| — \| Kinesin-1 \| 5.2 ± 0.2 \| 200 \| 0.25 \| 11.8 \| \| OT \| — \| Kinesin-1 \| 5.3 ± 0.2 \| 400 \| 0.25 \| 21.8 \| \| OT \| — \| Kinesin-1 \| \| 800 \| 0.25 \| 31.8 \| \| \| M \| — \| Kinesin-2 \| 5 \| 625 \| \| \| \| \| M \| — \| Kinesin \| 7.5 \| \| \| \| \| \| M \| — \| Kinesin \| 5 \| 800 \| \| \| Open in a new tab Table 2. Values reported for dynein motility and force generation by various techniques. Key: OT—optical trap, M—model, FM—fluorescence microscopy, MT—magnetic tweezer, DSEC—Drosophila sequenced E. Coli, D—Drosophila, and LMFM—lateral molecular force microscopy (see Supplementary Materials, which are available under the “Supplemental Materials” tab for this paper on the ASME Digital Collection, for the complete table). \| Paper \| Method \| Species \| Motor protein type \| Stalling force (pN) mean ± std. dev \| Unloaded velocity (nm/s) \| ATP concentration (mM) \| Temp (°C) \| --- --- --- --- \| \| \| M \| — \| Dynein \| 1.5 \| 50 \| 0.001 \| \| \| M \| — \| 1.5 \| 100 \| 0.1 \| \| \| M \| — \| 1.5 \| 800 \| 1 \| \| \| M \| — \| 1.5 \| 1000 \| 10 \| \| \| \| OT \| Yeast \| Dynein \| 7 \| — \| 0.01 \| \| \| OT \| 7 \| — \| 0.025 \| \| \| OT \| 7 \| — \| 0.05 \| \| \| OT \| 7 \| — \| 0.1 \| \| \| OT \| 7 \| 49 \| 1 \| \| \| \| OT \| Bovine \| Dynein \| 1.1 \| — \| 1 \| \| \| OT \| Bovine \| Dynein \| 0.5 \| — \| 0.4 \| \| \| OT \| Bovine \| Dynein \| 0.8 \| — \| 0.7 \| \| \| OT \| Bovine \| Dynein \| 0.25 \| — \| 0.1 \| \| \| \| FM \| Model 1 \| Dynein \| 1.1 \| — \| 1 \| \| \| FM \| Model 2 \| Dynein \| 1.1 \| 9.5 \| 1 \| \| \| FM \| Model 3 \| Dynein \| 1.1 \| 56 \| 1 \| \| \| OT \| Rat \| Mammalian Dynein \| — \| 75 \| 10 \| 1.8 \| \| OT \| Rat \| Mammalian Dynein \| 1.2 ± 0.1 \| 250 \| 10 \| 11.8 \| \| OT \| Rat \| Mammalian Dynein \| — \| 750 \| 10 \| 21.8 \| \| OT \| Rat \| Mammalian Dynein \| 1.2 ± 0.1 \| 850 \| 10 \| 31.8 \| \| OT \| \| Yeast Dynein \| — \| 10 \| 1 \| 1.8 \| \| OT \| \| Yeast Dynein \| — \| 20 \| 1 \| 11.8 \| \| OT \| \| Yeast Dynein \| — \| 40 \| 1 \| 21.8 \| \| OT \| \| Yeast Dynein \| — \| 80 \| 1 \| 31.8 \| Open in a new tab 4.1. Step Size. Extensive body of work on kinesin reveals that the step size is consistently reported at approximately 8 nm [9,49]. The step size of kinesin matches the size of a tubulin dimer. The step size for dynein decreases in the presence of an applied force, with reported values ranging from 8 to 32 nm [46,47]. 4.2. Stall Force. Stall force is an important mechanical property of motor proteins as it dictates the types of loads and environments with which the protein can interact. Additionally, stall force can be related to the velocity of the protein's locomotion (Fig. 6(c)). One of the most common methods to measure force is optical trapping in combination with fluorescence microscopy to track the displacement of either the payload or kinesin over time to determine the velocity. Using optical trapping, the reported stall forces for kinesin ranged from 4.7 to 7.5 pN. This variability in reported values likely stems from variations in environmental conditions and experimental design. For instance, the values reported in Gagliano et al. (stall force of 7 pN) differ from those found in Schroeder et al. (stall force of 5–5.3 pN). However, Gagliano et al. inferred the velocity of kinesin based on the motion of fluorescently labeled microtubules propelled by surface bound kinesin in a viscous fluid environment in which they varied the viscosity, whereas Schroeder et al. directly measured the velocity of motor bound fluorescent beads traveling on surface bound microtubules. Fig. 6. Open in a new tab (a) The relationship between ATP concentration and kinesin velocity as reported across the literature (see Table 1). (b) The effect of temperature on kinesin velocity at two ATP concentrations [34,43]. At both concentrations, the relationship is seen to follow a similar trend. (c) Approximate force–velocity relationships adapted from the existing literature [9,12,18,50]. Magnetic tweezers have been used to determine the stall force of kinesin. Fallesen et al. [12,28,29] investigated stall force by applying a magnetic field to a bead attached to the positive end of a microtubule. This microtubule was placed on a reversed gliding assay where surface bound kinesin moved microtubules along the length of the assay. This magnetic field produced a counteracting force against the kinesin generated forces. Intuitively, as the force on the bead increased, the velocity at which the kinesin was able to move the microtubule decreased. These studies were able to experimentally determine how groups of kinesin molecules work together against an opposing force, where they found that scaling down the number of motors did not affect the force output of each singular motor . The applicability of magnetic tweezers to the investigation of kinesin motility and force capabilities were first demonstrated by applying the magnetic force as a square wave from 0 to 12.2 pN (well above the accepted stall force values of 5–7 pN). At 12.2 pN, the microtubule was stalled, whereas at 0 pN, unloaded velocity of 670 nm/s was observed . Additionally, the behavior of many kinesin motor proteins working together has been investigated using this magnetic tweezer gliding assay . Fluorescently labeled microtubules were sent across varying concentrations of surface bound kinesin on a coverslip against a magnetic force that varied from 0 to 31.4 pN. When the magnetic force was absent, the microtubule moved at a velocity of 600–675 nm/s; however, when the magnetic force reached 31.4 pN (4.7 pN/motor), the velocity decreased to 220 nm/s. As a result of this assay, a stall force per motor of 6.1 ± 0.7 pN was calculated . Scholz et al. used LMFM where microfabricated cantilevers with a spring constant of 0.03 pN/nm recorded stall force measurements for kinesin (4.0 ± 0.2 pN). In this study, kinesin was attached to the cantilever and it was allowed to move along a microtubule. This movement created deflection in the cantilever and at the maximum deflection of 110 ± 5 nm locomotion ceased. The associated stall force is lower than the majority of optical trapping values. This outcome is hypothesized to be the result of kinesin proteins interacting with the microtubules individually, as opposed to multiple motors interacting with the microtubule as is prevalent in optical trapping setups. 4.3. Effects of ATP Concentration on Force and Velocity. The concentration of ATP strongly affects the velocity at which both kinesin and dynein are capable of moving. On the other hand, ATP concentration does not seem to have an effect on the stall force. For kinesin and dynein, decreasing ATP concentration results in decreased velocities. This is demonstrated for kinesin, by Kojima et al. where upon changing from 1 mM to 10 μ M of ATP the velocity drops from approximately 600 nm/s to about 225 nm/s. However, the stall force remains at approximately 7 pN. A similar relationship of ATP concentration and velocity is shown by Scholz et al. using LMFM, where the velocity was dependent on ATP concentration (at 1 mM the unloaded velocity was 460 ± 80 nm/s, and at 10 μ M 170 ± 30 nm/s); however, the stall forces remained unchanged at roughly 4.2 pN per motor. In the absence of ATP, kinesin cannot move, even under externally applied forces . For unmodified kinesin at room temperature, the effects of increasing ATP concentrations on velocity appear to saturate at 1 mM, at which point the unloaded velocities range from 390 to 760 nm/s (Fig. 6(a)). Increasing the concentration of ATP beyond 1 mM has minimal effect on velocity. As with kinesin, dynein exhibits lower velocities in environments with lower ATP concentrations. However, unlike kinesin, dynein has been reported to step bidirectionally in an ATP absent environment under an applied force. Interestingly, this effect is asymmetric, requiring threefold greater force to initiate positive directional stepping . 4.4. Effects of Temperature on Force and Velocity. Temperature also affects the velocity capabilities of both kinesin and dynein. Kawaguchi and Ishiwata demonstrated that, at a fixed ATP concentration (1 mM), temperature does not affect the force production capability of kinesin as shown by a consistent stall force of 7.3 ± 0.33 pN, which was independent of velocity, whereas velocity is affected by the temperature of the system with a range of unloaded velocities of 375–1400 nm/s at 15–35 °C, respectively. A similar relationship was observed between temperature and velocity for both kinesin and dynein by Hong et al. (Fig. 6(b)). With the ATP concentration fixed at 0.25 mM, the stall force for kinesin was found to be 5.2 ± 0.2 pN with unloaded velocities ranging from 100 to 800 nm/s at 1.8–31.8 °C, respectively. While the velocities reported by Hong et al. vary from those of Kawaguchi et al., this variation is likely due to the differences in ATP concentration. Additionally, Hong et al. investigated the effect of temperature on the velocity of dyneins from two sources—rat neurons and yeast. For the investigation of the dynein from rat, the ATP concentration was held at 10 mM, whereas a concentration of 1 mM was used for the yeast dynein. The rat dynein had an unloaded velocity range of 75–850 nm/s at 1.8–31.8 °C, whereas the yeast dynein had a range of 10–80 nm/s across the same temperature gradient. The rat dynein was also shown to have a stall force of 1.2 ± 0.1 pN at both 11.8 and 31.8 °C, continuing the relationship of stall force being unrelated to the temperature of the system. In summary, by comparing the varying methods used on kinesin it becomes abundantly clear that the stall force, regardless of method, falls between 4 and 7 pN. In dynein, the variance in stall force occurs throughout the collected data; however, the most common reported stall force lay between 1 and 1.5 pN. Meanwhile, the velocity understandably has the most variance regardless of the measurement technique as it has multiple variables, such as temperature, ATP concentration, fluid viscosity, opposing force, and payload type, that control how rapidly the kinesin or dynein is able to move. 5. Computational Modeling of Kinesin and Dynein Locomotion In addition to experimental techniques, mathematical models have been used to investigate and predict the force and motility of kinesin and dynein. These models have been developed based on experimental data as well as used predictively. The predictions of these models, both standalone and from experimental values, generally agree with the experimental data. To get this accuracy, particularly with respect to velocity, the ATP concentration had to be adequately accounted for in the models. In Ref. , the mechanical behavior of multiple kinesin proteins working together was investigated using a stochastic model of the mechanochemical cycle kinesin undergoes. To do this, the model accounted for a varying amount of ATP present in the given system among other factors and found a stall force of approximately 6 pN per motor and a maximum velocity of 1000 nm/s. This is remarkably close to the stall force of 6.1 pN/motor reported by Fallesen et al. by using magnetic tweezers. The model also was able to predict that as the load increased, synchronicity between motors also increased. Additionally, the relationships between force, ATP consumption, and velocity in a viscoelastic environment have been investigated through the use of models. For example, Holzwarth et al. developed a model, based on a Stokes-like relation to describe force in a viscoelastic fluid, to calculate the drag force and work required for transport of vesicles by kinesin through a viscoelastic cytoplasm which was compared to the generally accepted values kinesin generates in an optical trap experiment when traveling through a buffer. With the assumption that the drag force reduces to zero between steps, the model predicted a motor force of 16 pN being required to generate movement at a consumption rate of 1 ± 0.7 ATP per step. This is compared to the experimental report that 6–7 pN of force kinesin is able to generate at a consumption rate of 1 ATP per step. The differences between the values produced by the model and the experimental values are likely associated with many assumptions in constructing the model, such as the vesicle being rigid whereas in a cell the vesicle would be deformable. Shao and Gao developed a simple theoretical model of kinesin locomotion in order to examine the hand-over-hand motion of kinesin stepping in both forward and backward directions when an external load is placed on the kinesin molecule. The model predicts the motion of both heads, taking into account the ATP concentration and external force. This allowed them to investigate the relationship between locomotion velocity and ATP concentration to compare to experimental values. The resulting velocity values calculated at 0.01 and 1 mM of ATP (150 and 550 nm/s) correlate well with the experimental values from Carter and Cross (175 and 800 nm/s) . Models have also been developed to describe not only the locomotion, but also the behavior of kinesin. Specifically, Schroder et al. employed a steady-state binomial statistical model to capture the effect of the stochastic behavior of motor attachment and detachment on track switching of a transported cargo . This model is based on experiments in which actin filaments, along which myosin-V traveled, and microtubules, along which Kinesin-2 traveled, were placed perpendicularly to each other. The kinesin with a payload was then sent along the microtubule toward the actin filament where the payload was either taken by the myosin or kept by the kinesin to continue on its path. The model was used to explore how initial and environmental conditions affect the probability that the myosin would take the payload from the kinesin. Such information can be used to investigate how this switching behavior may be varied by modifying the proteins or their environment, pointing toward possible protein engineering applications. Dynein locomotion has also been modeled predictively. A simplified probabilistic model was done by Mukherji to examine the unidirectional motion of dynein with variable step sizes. The model assumes dynein to be single-headed and that ATP hydrolysis causes the dynein to initially move by a distance of 8 nm. Values for stall force and velocity at varying ATP concentrations were determined via the model with these assumptions in mind. At ATP concentrations of 0.005, 0.04, and 2 mM, stall force correlated with 10, 10, and 8 pN, respectively, while unloaded velocity correlated with 19, 80, and 400 nm/s, respectively. Similarly, a simple ratchet model developed by Bameta et al. gives a good general overview of the force velocity relationship dynein has along with its step sizes and velocity variation with ATP concentration. This model allowed for two different step sizes in response to an external force. In this case, it was shown that at higher loads, smaller step sizes are more probable, while larger step sizes are more likely at lower loads. This model also had a saturated stall force value of 1.5 pN, along with an unloaded velocity range from 50 to 1000 nm/s. These unloaded velocities correspond to an ATP concentration range of 0.001–10 mM. In comparison with Gennerich et al. , where a stall force of 7 pN and an unloaded velocity of 49 nm/s at 1 mM of ATP concentration show the variability currently found in the understanding of how the force velocity relationship in dynein works. 6. Associations Between Motor Protein Functions and Disease Processes Many neurodegenerative diseases, and some cancers, have been directly linked to the inability of kinesin or dynein to complete their primary function of cargo transport. This failure may be attributed to a dysfunction in the motor protein or a structural defect in the microtubules they walk on to transfer their cargo. In the case of motor protein dysfunction, there is a deficit in the ability to complete one or both of its major functions (i.e., transportation of cargo through the cytoplasm via microtubules or organization of the mitotic spindle ). On the other hand, defects arise during the polymerization process that occurs during the formation of microtubules . Potential microtubule defects include missing dimer units in the tubule as well as an inconsistent number of protofilaments. As the role of a microtubule is to provide a pathway for the motor proteins to travel along, such defects prevent the motor proteins from being able to locomote . Disruption of either of these functions can result in a cascade of downstream defects. These include failure of cargo trafficking over long distances, function of cilia and flagella due to improper formation during development, resistance to pathogens, elimination of environmental toxins, and negative feedback of cell proliferation . The downstream defects resulting from motor protein dysfunction show themselves symptomatically in the form of diseases. For example, long-distance trafficking is crucial to axonal transport. Specifically, KIF1B is instrumental in the transport of synaptic precursors. Additionally, the transport of cargoes such as peroxisomes, neurofilaments, and Golgi-derived vesicles into neurites such as axons and dendrites is prevented . Processes such as these need motor proteins for transport, as diffusion lacks the necessary efficiency. Interruptions of long-distance transport result in neurodegenerative diseases such as ALS and Alzheimer's disease . Additionally, ineffective function of cilia and flagella is related to a variety of diseases depending on the species of cilia that are defective. This can materialize in the form of polycystic kidney disease, due to the lack of retrograde transport, or laterally inverted positioning of organs resulting from improper development of a fetus in utero, among other symptoms . Another adverse effect of poorly functioning microtubule motor proteins is the interruption of the cell replication cycle, namely the negative feedback loop. This lack of feedback will allow cells to reproduce at an unusually high rate, which has been linked to cancer . Although microtubule polymerization in cells is closely regulated, defects in the microtubule structure may occur as confirmed by scanning force microscopy, cryo-electron microscopy, and mechanical measurements . Microtubule defects can cause a wide range of physiological conditions including Alzheimer's disease, frontotemporal dementia, respiratory dysfunction, abnormal cilia beating, motor neuropathy, lower motor neuron disease (which plays a role in amyotrophic lateral sclerosis), axonal neuronopathy, hereditary motor sensory disease, and autosomal dominant spastic paraplegia . The extent to which microtubule defects affect motor protein locomotion has been investigated in vitro . Liang et al. observed the locomotion of kinesin-coated beads along defect ridden microtubules. They varied the number of kinesin proteins coating the bead, as well as the number of defects in the microtubule, and found that the magnitude of the impact of microtubule defects was related to both the number of motor proteins and number of microtubule defects. However, the number of microtubule defects was found to be more impactful. In essence, the larger the number of defects formed during the polymerization process, the more likely an interruption of transport would occur . 7. Engineered Proteins Bioengineering aims to use components of existing biological systems and synthetically alters them to better meet a need or serve a purpose. The concept of bioengineering can be applied to motor proteins to obtain nanomachines in which kinesin and dynein would be indispensable components. Kinesin and dynein are particularly interesting, as they are highly specialized motor proteins and have relatively high motility and tunability . The re-engineering of the motor protein can serve to improve the control over the locomotion parameters such as speed, direction, photosensitivity, chemical sensitivity, cargo size, and protein size . Modification of these parameters allows motor proteins to better perform native tasks or be used for completely novel applications. Engineering of motor proteins can be separated into two categories. First is to use many of the existing motor proteins to act as a motor for a nanodevice . Second is the alteration of the protein itself to add or take away components so the new motor protein can transport a larger payload more efficiently . In either sense, the motor proteins are serving an alternative purpose after alterations are made to their natural structure. Regardless of the end purpose, engineered motor proteins typically are comprised of three major components: the functional elements, the platform, and the motors themselves (Fig. 7). The functional elements are the components that interface with the cargo and are task specific. The scaffolding platform is a connector between the functional element and the motor protein, and it merges the two elements and allows them to become part of one system. The motors are the proteins themselves, selected based on the end need being addressed. Fig. 7. Open in a new tab Engineered motor proteins offer exciting new possibilities in research and clinical applications. By modifying an existing motor protein, researchers can generate new structures capable of performing novel tasks or improving native protein function. One venue of engineering motor proteins is the use of natural motor proteins as a motor for a nanodevice. The most common example of these nanodevices is the creation of molecular shuttles that function as analyte identifiers or analyte concentrators in a microfluidic chip system [50,62–64]. Nanodevices operating based on the principle of motility assays are integrated where kinesin or dynein are immobilized as “feet” to move microtubules along a desired path. This movement of microtubules along patterned tracks or filaments allows for the analyte capture and identification process to happen [50,62,63]. The propelled filaments can be used to sort and transport a wide range of cargo from standard microtubules to ones coated in proteins or antibodies for analysis . So as to achieve this outcome, antibodies can be bound to the filaments allowing the molecular shuttles to selectively bind specific proteins or even viral remnants , from the surrounding solutions . Although microfluidic chip systems and motility assays are the most common examples of natural motor proteins being used in nanodevices, many theories for future applications also exist throughout literature. For example, motor proteins could be used to drive rotary devices, to improve immunoassays, such as an enzyme-linked immunosorbent assay, which are commonly used to analyze biological samples, or the creation of a micro- or nanoscale assembly line system [50,62–64]. In other cases, a simple addition to the protein allows for increased control over the normal function of the protein, essentially allowing the experimenter to direct the motor protein rather than alter the function itself . These controls can be mechanochemical [61,67] or, in other cases, optogenetic [61,68]. At times, the type of preferred control mechanisms can depend on the application and what other factors are at play, so the larger system is not impacted with the use of the control mechanism. Mechanochemical controls are linked to the sensitivity of a motor protein to calcium, an ion naturally appearing in the body. Experimenters are able to increase the calcium sensitivity in kinesin, allowing for an increase in motility and changing the direction of locomotion . For example, researchers have modified kinesin, to create molecular motors with Ca 2+ controllable motility, by fusing two monomeric kinesins to a M13 peptide and a CaM dimer . This allows Ca 2+ on–off control, essentially creating a Ca 2+ dependent accelerator and brake system. Optical controls can be used to trigger motor protein locomotion upon exposure to light of a specific wavelength range. The photosensitivity is not necessarily that of the kinesin, but that of the molecules that induce the release of ATP binding. As the locomotion of motor proteins is a mechanochemical process, the chemical processes are tools to influence the mechanical output. Specific to kinesin, photochromic substrates can be integrated into the motor domain. This impacts the mechanical capabilities of kinesin upon exposure to UV light, as the activity of ATPase becomes limited. The regulation of this chemical process then controls the mechanical activity of the motor protein . There are a number of additional examples of modifications to motor proteins in order to vary locomotion in vitro. Dynein, which natively moves along microtubules, has instead been engineered to move along actin filaments. Dynein has a larger, more modular structure than other motor proteins, enabling a modular engineering approach wherein the dynein motor domain can be coupled with actin-binding proteins . Additionally, external modifications can be applied to dynein using DNA origami to create a chassis that can bind to a variable number of dynein proteins. This chassis can be produced with varying rigidities and allows coupling of multiple motor proteins. Varying the rigidity of the engineered chassis revealed that the collective locomotion of dynein exhibits cargo rigidity-dependent velocity . Furthermore, motor proteins have also been modified to perform non-native functions in cells in vitro. Taking inspiration from viral behavior in which viruses hijack cytoskeletal motor proteins for transport to the nucleus, Toledo et al. modified the dynein light chain LC8 to fuse it to the DNA binding domain 4. Such modification enabled dynein to aid in nonviral transfection . These devices lay the foundation for future in vitro cellular and in vivo research. By using external stimuli to modify the motor protein, intracellular processes could be changed in response. A more detailed review of these engineered motor proteins has been provided by DelRosso et al. . The engineering techniques presented here allow for the limitations innate to the motor proteins to be synthetically eliminated. Moving forward, as the techniques are further explored, the repurposing of the motor proteins will become more precise. Many of the experiments run to evaluate the impact of parameters such as light exposure and calcium ion concentration have been explored in vitro. In order to transition these engineered proteins to clinical applications, the effects of such changes in vivo must be investigated. 8. Conclusions While many interesting studies have delved into the mechanics of kinesin and dynein, there remains ample room for study. For instance, in conducting this review the lack of mechanical analysis that has been performed on dynein became increasingly clear. This drought of information also sheds some light on the volatility of the reported values, where the force required to stall dynein ranged from 12 pN to 0.25 pN. These intensive gaps signal a need for more research to be done to better understand the forces and movements that dynein is capable of generating in different environments and ATP concentrations. In addition to further research being conducted on both kinesin and dynein, the methods used to measure force and velocity can also be expanded upon and improved. For instance, LMFM is not commonly utilized due to difficulties in fabricating nanoscale microcantilevers. However, its ability to effectively measure the stall force and velocity, when combined with a displacement tracking method, of a single molecule provides significant insight into the single molecule mechanics of both kinesin and dynein. This advantage over optical trapping, where protein bound beads impeded the motor protein's motion, is worth investigating and developing. Single molecule mechanics as well as group mechanics of these motor proteins would not only give insight into their natural functions, but also inform methods for researchers that are investigating re-engineering these proteins. Beyond research into the basic functions of these motor proteins, recent developments allow researchers to fundamentally change the function or behavior of the proteins themselves. This opens an exciting realm of research ventures in which, by modifying the very structure of the motor proteins, clinical pathologies may be corrected or novel bio-nanotechnological devices developed capable of functioning within cells themselves. Funding Data National Institutes of Health (Grant Nos. R01 AR063701 and T32 AR007505). National Science Foundation (Grant No. DMR-1306665). Supplementary Material Supplementary Material Click here for additional data file. (15KB, zip) Contributor Information Zachary Abraham, Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106. Emma Hawley, Department of Biomedical Engineering, , Case Western Reserve University, , Cleveland, OH 44106. Daniel Hayosh, Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106. Victoria A. Webster-Wood, Mem. ASME , Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , 10900 Euclid Avenue, , Cleveland, OH 44106 , e-mail: Webster-wood@case.edu. Ozan Akkus, Mem. ASME , Department of Mechanical and , Aerospace Engineering, , Case Western Reserve University, , Cleveland, OH 44106. References . Alberts, B. , 1970, “ Molecular Motors,” Molecular Biology of the Cell, 4th ed., National Library of Medicine, New York. [Google Scholar] . Goodman, B. S. , Derr, N. D. , and Reck-Peterson, S. L. , 2012, “ Engineered, Harnessed, and Hijacked: Synthetic Uses for Cytoskeletal Systems,” Trends Cell Biol., 22(12), pp. 644–652. 10.1016/j.tcb.2012.09.005 [DOI] [PMC free article] [PubMed] [Google Scholar] . Hancock, W. O. , 2016, “ The Kinesin-1 Chemomechanical Cycle: Stepping Toward a Consensus,” Biophys. J., 110(6), pp. 1216–1225. 10.1016/j.bpj.2016.02.025 [DOI] [PMC free article] [PubMed] [Google Scholar] . Gennerich, A. , and Vale, R. D. , 2009, “ Walking the Walk: How Kinesin and Dynein Coordinate Their Steps,” Curr. Opin. Cell Biol., 21(1), pp. 59–67. 10.1016/j.ceb.2008.12.002 [DOI] [PMC free article] [PubMed] [Google Scholar] . Hirokawa, N. , and Takemura, R. , 2013, Kinesin Superfamily Proteins, 2nd ed., Wiley, New York. [Google Scholar] . Levy, J. R. , and Holzbaur, E. L. F. , 2006, “ Cytoplasmic Dynein/Dynactin Function and Dysfunction in Motor Neurons,” Int. J. Dev. Neurosci., 24(2–3), pp. 103–111. 10.1016/j.ijdevneu.2005.11.013 [DOI] [PubMed] [Google Scholar] . Spundich, J. A. , 2011, “ Molecular Motors, Beauty in Complexity,” Science, 331(6021), pp. 1143–1145. 10.1126/science.1203978 [DOI] [PMC free article] [PubMed] [Google Scholar] .Impetux, 2015, “ Measurement of the Stall Force of Kinesins in Living Cells With Lunam™ T-40i by IMPETUX,” Impetux, Barcelona, Spain, accessed Sept. 19, 2017, . Scholz, T. , Vicary, J. A. , Jeppesen, G. M. , Ulcinas, A. , Hörber, J. K. H. , and Antognozzi, M. , 2011, “ Processive Behaviour of Kinesin Observed Using Micro-Fabricated Cantilevers,” Nanotechnology, 22(9), pp. 1–7. 10.1088/0957-4484/22/9/095707 [DOI] [PubMed] [Google Scholar] . Hall, K. , Cole, D. , Yeh, Y. , and Baskin, R. J. , 1996, “ Kinesin Force Generation Measured Using a Centrifuge Microscope Sperm-Gliding Motility Assay,” Biophys. J., 71(6), pp. 3467–3476. 10.1016/S0006-3495(96)79542-5 [DOI] [PMC free article] [PubMed] [Google Scholar] . Kuo, S. C. , Gelles, J. , Steuer, E. , and Sheetz, M. P. , 1991, “ A Model for Kinesin Movement From Nanometer-Level Movements of Kinesin and Cytoplasmic Dynein and Force Measurements.,” J. Cell Sci., 14, pp. 135–138. [DOI] [PubMed] [Google Scholar] . Fallesen, T. L. , Macosko, J. C. , and Holzwarth, G. , 2011, “ Force-Velocity Relationship for Multiple Kinesin Motors Pulling a Magnetic Bead,” Eur. Biophys. J., 40(9), pp. 1071–1079. 10.1007/s00249-011-0724-1 [DOI] [PubMed] [Google Scholar] . Salmon, E. D. , 1995, “ VE-DIC Light Microscopy and the Discovery of Kinesin,” Trends Cell Biol., 5(4), pp. 154–158. 10.1016/S0962-8924(00)88979-5 [DOI] [PubMed] [Google Scholar] . Bugiel, M. , Fantana, H. , Bormuth, V. , Trushko, A. , Schiemann, F. , Howard, J. , Schäffer, E. , and Jannasch, A. , 2015, “ Versatile Microsphere Attachment of GFP-Labeled Motors and Other Tagged Proteins With Preserved Functionality,” J. Biol. Methods, 2(4), pp. 1–8. 10.14440/jbm.2015.79 [DOI] [Google Scholar] . Gelles, J. , Schnapp, B. J. , and Sheetz, M. P. , 1988, “ Tracking Kinesin-Driven Movements With Nanometre-Scale Precision,” Nature, 331(6155), pp. 450–453. 10.1038/331450a0 [DOI] [PubMed] [Google Scholar] . Svoboda, K. , Schmidt, C. F. , Schnapp, B. J. , and Block, S. M. , 1993, “ Direct Observation of Kinesin Stepping by Optical Trapping Interferometry,” Nature, 365(6448), pp. 721–727. 10.1038/365721a0 [DOI] [PubMed] [Google Scholar] . Block, S. M. , Goldstein, L. S. B. , and Schnapp, B. J. , 1990, “ Bead Movement by Single Kinesis Molecules Studied With Optical Tweezers,” Nature, 348(6299), pp. 348–352. 10.1038/348348a0 [DOI] [PubMed] [Google Scholar] . Kojima, H. , Muto, E. , Higuchi, H. , and Yanagida, T. , 1997, “ Mechanics of Single Kinesin Molecules Measured by Optical Trapping Nanometry,” Biophys. J., 73(4), pp. 2012–2022. 10.1016/S0006-3495(97)78231-6 [DOI] [PMC free article] [PubMed] [Google Scholar] . Schroeder, H. W. , Hendricks, A. G. , Ikeda, K. , Shuman, H. , Rodionov, V. , Ikebe, M. , Goldman, Y. E. , and Holzbaur, E. L. F. , 2012, “ Force-Dependent Detachment of Kinesin-2 Biases Track Switching at Cytoskeletal Filament Intersections,” Biophys. J., 103(1), pp. 48–58. 10.1016/j.bpj.2012.05.037 [DOI] [PMC free article] [PubMed] [Google Scholar] . Gennerich, A. , Carter, A. P. , Reck-Peterson, S. L. , and Vale, R. D. , 2007, “ Force-Induced Bidirectional Stepping of Cytoplasmic Dynein,” Cell, 131(5), pp. 952–965. 10.1016/j.cell.2007.10.016 [DOI] [PMC free article] [PubMed] [Google Scholar] . Reck-Peterson, S. L. , Yildiz, A. , Carter, A. P. , Gennerich, A. , Zhang, N. , and Vale, R. D. , 2006, “ Single-Molecule Analysis of Dynein Processivity and Stepping Behavior,” Cell, 126(2), pp. 335–348. 10.1016/j.cell.2006.05.046 [DOI] [PMC free article] [PubMed] [Google Scholar] . Verbrugge, S. , Kapitein, L. C. , and Peterman, E. J. G. , 2007, “ Kinesin Moving Through the Spotlight: Single-Motor Fluorescence Microscopy With Submillisecond Time Resolution,” Biophys. J., 92(7), pp. 2536–2545. 10.1529/biophysj.106.093575 [DOI] [PMC free article] [PubMed] [Google Scholar] . Korten, T. , Nitzsche, B. , Gell, C. , Ruhnow, F. , and Diez, S. , 2011, “ Fluorescence Imaging of Single Kinesin Motors on Immobolized Microtubules,” Methods Mol. Biol., 783(8), pp. 121–137. 10.1007/978-1-61779-282-3 [DOI] [PubMed] [Google Scholar] . Yildiz, A. , Tomishige, M. , Vale, R. D. , and Selvin, P. R. , 2004, “ Kinesin Walks Hand-Over-Hand,” Science, 303(5658), pp. 676–678. 10.1126/science.1093753 [DOI] [PubMed] [Google Scholar] . Toba, S. , Watanabe, T. M. , Yamaguchi-Okimoto, L. , Toyoshima, Y. Y. , and Higuchi, H. , 2006, “ Overlapping Hand-Over-Hand Mechanism of Single Molecular Motility of Cytoplasmic Dynein,” Proc. Natl. Acad. Sci. U. S. A., 103(15), pp. 5741–5745. 10.1073/pnas.0508511103 [DOI] [PMC free article] [PubMed] [Google Scholar] . Vale, R. D. , Funatsu, T. , Pierce, D. W. , and Romberg, L. , 1996, “ Direct Observation of Single Kinesin Molecules Moving Along Microtubules,” Nature, 380(6573), pp. 451–453. 10.1038/380451a0 [DOI] [PMC free article] [PubMed] [Google Scholar] . Nicholas, M. P. , Berger, F. , Rao, L. , Brenner, S. , Cho, C. , and Gennerich, A. , 2015, “ Cytoplasmic Dynein Regulates Its Attachment to Microtubules Via Nucleotide State-Switched Mechanosensing at Multiple AAA Domains,” Proc. Natl. Acad. Sci., 112(20), pp. 6371–6376. 10.1073/pnas.1417422112 [DOI] [PMC free article] [PubMed] [Google Scholar] . Fallesen, T. L. , MacOsko, J. C. , and Holzwarth, G. , 2011, “ Measuring the Number and Spacing of Molecular Motors Propelling a Gliding Microtubule,” Phys. Rev. E, 83(1), pp. 1–8. 10.1103/PhysRevE.83.011918 [DOI] [PubMed] [Google Scholar] . Fallesen, T. , Hill, D. B. , Steen, M. , MacOsko, J. C. , Bonin, K. , and Holzwarth, G. , 2010, “ Magnet Polepiece Design for Uniform Magnetic Force on Superparamagnetic Beads,” Rev. Sci. Instrum., 81(7), p. 074303. 10.1063/1.3469792 [DOI] [PMC free article] [PubMed] [Google Scholar] . Gagliano, J. , Walb, M. , Blaker, B. , MacOsko, J. C. , and Holzwarth, G. , 2010, “ Kinesin Velocity Increases With the Number of Motors Pulling Against Viscoelastic Drag,” Eur. Biophys. J., 39(5), pp. 801–813. 10.1007/s00249-009-0560-8 [DOI] [PubMed] [Google Scholar] . Schaap, I. A. T. , Carrasco, C. , De Pablo, P. J. , and Schmidt, C. F. , 2011, “ Kinesin Walks the Line: Single Motors Observed by Atomic Force Microscopy,” Biophys. J., 100(10), pp. 2450–2456. 10.1016/j.bpj.2011.04.015 [DOI] [PMC free article] [PubMed] [Google Scholar] . Antognozzi, M. , Ulcinas, A. , Picco, L. , Simpson, S. H. , Heard, P. J. , Szczelkun, M. D. , Brenner, B. , and Miles, M. J. , 2008, “ A New Detection System for Extremely Small Vertically Mounted Cantilevers,” Nanotechnology, 19(38), p. 384002. 10.1088/0957-4484/19/38/384002 [DOI] [PubMed] [Google Scholar] . Visscher, K. , Schnitzer, M. J. , and Block, S. M. , 1999, “ Single Kinesin Molecules Studied With a Molecular Force Clamp,” Nature, 400(6740), pp. 184–189. 10.1038/22146 [DOI] [PubMed] [Google Scholar] . Kawaguchi, K. , and Ishiwata, S. , 2000, “ Temperature Dependence of Force, Velocity, and Processivity of Single Kinesin Molecules,” Biochem. Biophys. Res. Commun., 272(3), pp. 895–899. 10.1006/bbrc.2000.2856 [DOI] [PubMed] [Google Scholar] . Fehr, A. N. , Asbury, C. L. , and Block, S. M. , 2008, “ Kinesin Steps Do Not Alternate in Size,” Biophys. J., 94(3), pp. L20–L22. 10.1529/biophysj.107.126839 [DOI] [PMC free article] [PubMed] [Google Scholar] . Carter, N. J. , and Cross, R. A. , 2005, “ Mechanics of the Kinesin Step,” Nature, 435(7040), pp. 308–312. 10.1038/nature03528 [DOI] [PubMed] [Google Scholar] . Schmitz, K. A. , Holcomb-Wygle, D. L. , Oberski, D. J. , and Lindemann, C. B. , 2000, “ Measurement of the Force Produced by an Intact Bull Sperm Flagellum in Isometric Arrest and Estimation of the Dynein Stall Force,” Biophys. J., 79(1), pp. 468–478. 10.1016/S0006-3495(00)76308-9 [DOI] [PMC free article] [PubMed] [Google Scholar] . Kanimura, S. , and Takahashi, K. , 1981, “ Direct Measurement of the Force of Microtubule Sliding in Flagella,” Nature, 293(5833), pp. 566–568. 10.1038/293566a0 [DOI] [PubMed] [Google Scholar] . Oiwa, K. , and Takahashi, K. , 1988, “ The Force-Velocity Relationship for Microtubule Sliding in Demembranated Sperm Flagella of the Sea Urchin,” Cell Struct. Funct., 13(3), pp. 193–205. 10.1247/csf.13.193 [DOI] [PubMed] [Google Scholar] . Yoneda, M. , 1962, “ Force Exerted by a Single Cilium of Mytilus Edulis,” J. Exp. Biol., 39, pp. 307–317. [Google Scholar] . Coppin, C. M. , Pierce, D. W. , Hsu, L. , and Vale, R. D. , 1997, “ The Load Dependence of Kinesin's Mechanical Cycle,” Proc. Natl. Acad. Sci. U. S. A., 94(16), pp. 8539–8544. 10.1073/pnas.94.16.8539 [DOI] [PMC free article] [PubMed] [Google Scholar] . Svoboda, K. , and Block, S. M. , 1994, “ Force and Velocity Measured for Single Kinesin Molecules,” Cell, 77(5), pp. 773–784. 10.1016/0092-8674(94)90060-4 [DOI] [PubMed] [Google Scholar] . Hong, W. , Takshak, A. , Osunbayo, O. , Kunwar, A. , and Vershinin, M. , 2016, “ The Effect of Temperature on Microtubule-Based Transport by Cytoplasmic Dynein and Kinesin-1 Motors,” Biophys. J., 111(6), pp. 1287–1294. 10.1016/j.bpj.2016.08.006 [DOI] [PMC free article] [PubMed] [Google Scholar] . Hyeon, C. , and Onuchic, J. N. , 2007, “ Internal Strain Regulates the Nucleotide Binding Site of the Kinesin Leading Head,” Proc. Natl. Acad. Sci. U. S. A., 104(7), pp. 2175–2180. 10.1073/pnas.0610939104 [DOI] [PMC free article] [PubMed] [Google Scholar] . Gibbons, F. , Chauwin, J.-F. , Despósito, M. , and José, J. V. , 2001, “ A Dynamical Model of Kinesin-Microtubule Motility Assays,” Biophys. J., 80(6), pp. 2515–2526. 10.1016/S0006-3495(01)76223-6 [DOI] [PMC free article] [PubMed] [Google Scholar] . Bameta, T. , Padinhateeri, R. , and Inamdar, M. M. , 2013, “ Force Generation and Step-Size Fluctuations in a Dynein Motor,” J. Stat. Mech. Theory Exp., 2013(02), p. P02030. 10.1088/1742-5468/2013/02/P02030 [DOI] [Google Scholar] . Mallik, R. , Carter, B. C. , Lex, S. A. , King, S. J. , and Gross, S. P. , 2004, “ Cytoplasmic Dynein Functions as a Gear in Response to Load,” Nature, 427(6975), pp. 649–652. 10.1038/nature02293 [DOI] [PubMed] [Google Scholar] . Ikuta, J. , Kamisetty, N. K. , Shintaku, H. , Kotera, H. , Kon, T. , and Yokokawa, R. , 2014, “ Tug-of-War of Microtubule Filaments at the Boundary of a Kinesin- and Dynein-Patterned Surface,” Sci. Rep., 4(5281), pp. 1–8. 10.1038/srep05281 [DOI] [PMC free article] [PubMed] [Google Scholar] . Howard, J. , Hudspeth, A. J. , and Vale, R. D. , 1989, “ Movement of Microtubules by Single Kinesin Molecules,” Nature, 342(6246), pp. 154–158. 10.1038/342154a0 [DOI] [PubMed] [Google Scholar] . Van Den Heuvel, M. G. L. , and Dekker, C. , 2007, “ Motor Proteins at Work for Nanotechnology,” Science, 317(5836), pp. 333–336. 10.1126/science.1139570 [DOI] [PubMed] [Google Scholar] . Hendricks, A. G. , Epureanu, B. I. , and Meyhöfer, E. , 2009, “ Collective Dynamics of Kinesin,” Phys. Rev. E, 79(3), pp. 1–12. 10.1103/PhysRevE.79.031929 [DOI] [PubMed] [Google Scholar] . Holzwarth, G. , Bonin, K. , and Hill, D. B. , 2002, “ Forces Required of Kinesin During Processive Transport Through Cytoplasm,” Biophys. J., 82(4), pp. 1784–1790. 10.1016/S0006-3495(02)75529-X [DOI] [PMC free article] [PubMed] [Google Scholar] . Shao, Q. , and Gao, Y. Q. , 2006, “ On the Hand-Over-Hand Mechanism of Kinesin,” Proc. Natl. Acad. Sci. U. S. A., 103(21), pp. 8072–8077. 10.1073/pnas.0602828103 [DOI] [PMC free article] [PubMed] [Google Scholar] . Mukherji, S. , 2008, “ Model for the Unidirectional Motion of a Dynein Molecule,” Phys. Rev., 77(5), pp. 1–8. 10.1103/PhysRevE.77.051916 [DOI] [PubMed] [Google Scholar] . Mandelkow, E. , and Mandelkow, E. , 2002, “ Kinesin Motors and Disease,” Trends Cell Biol., 12(12), pp. 585–591. 10.1016/S0962-8924(02)02400-5 [DOI] [PubMed] [Google Scholar] . Gramlich, M. W. , Conway, L. , Liang, W. H. , Labastide, J. A. , King, S. J. , Xu, J. , and Ross, J. L. , 2017, “ Single Molecule Investigation of Kinesin-1 Motility Using Engineered Microtubule Defects,” Sci. Rep., 7, p. 44290. 10.1038/srep44290 [DOI] [PMC free article] [PubMed] [Google Scholar] . Liang, W. H. , Li, Q. , Faysal, K. M. R. , King, S. J. , Gopinathan, A. , and Xu, J. , 2016, “ Microtubule Defects Influence Kinesin-Based Transport In Vitro,” Biophys. J., 110(10), pp. 2229–2240. 10.1016/j.bpj.2016.04.029 [DOI] [PMC free article] [PubMed] [Google Scholar] . Zhao, C. , Takita, J. , Tanaka, Y. , Setou, M. , Nakagawa, T. , Takeda, S. , Yang, H. W. , Terada, S. , Nakata, T. , Takei, Y. , Saito, M. , Tsuji, S. , Hayashi, Y. , and Hirokawa, N. , 2001, “ Charcot-Marie-Tooth Disease Type 2A Caused by Mutation in a Microtubule Motor KIF1Bbeta,” Cell, 105(5), pp. 587–597. 10.1016/S0092-8674(01)00363-4 [DOI] [PubMed] [Google Scholar] . Yamada, M. D. , Nakajima, Y. , Maeda, H. , and Maruta, S. , 2007, “ Photocontrol of Kinesin ATPase Activity Using an Azobenzene Derivative,” J. Biochem., 142(6), pp. 691–698. 10.1093/jb/mvm183 [DOI] [PubMed] [Google Scholar] . Baird, F. J. , and Bennett, C. L. , 2013, “ Microtubule Defects & Neurodegeneration,” J. Genet. Syndr. Gene Ther., 4, p. 203 10.4172/2157-7412.1000203. [DOI] [PMC free article] [PubMed] [Google Scholar] . DelRosso, N. V. , and Derr, N. D. , 2017, “ Exploiting Molecular Motors as Nanomachines: The Mechanisms of de Novo and Re-Engineered Cytoskeletal Motors,” Curr. Opin. Biotechnol., 46, pp. 20–26. 10.1016/j.copbio.2016.10.011 [DOI] [PubMed] [Google Scholar] . Goel, A. , and Vogel, V. , 2008, “ Harnessing Biological Motors to Engineer Systems for Nanoscale Transport and Assembly,” Nat. Nanotechnol., 3(8), pp. 465–475. 10.1038/nnano.2008.190 [DOI] [PubMed] [Google Scholar] . Hess, H. , 2011, “ Engineering Applications of Biomolecular Motors,” Annu. Rev. Biomed. Eng., 13(1), pp. 429–450. 10.1146/annurev-bioeng-071910-124644 [DOI] [PubMed] [Google Scholar] . Diez, S. , Korten, T. , and Ma, A. , 2010, “ Towards the Application of Cytoskeletal Motor Proteins in Molecular Detection and Diagnostic Devices,” Curr. Opin. Biotechnol., 21(4), pp. 477–488. 10.1016/j.copbio.2010.05.001 [DOI] [PubMed] [Google Scholar] . Ramachandran, S. , Ernst, K. , Bachand, G. D. , Vogel, V. , and Hess, H. , 2006, “ Selective Loading of Kinesin-Powered Molecular Shuttles With Protein Cargo and Its Application to Biosensing,” Small, 2(3), pp. 330–334. 10.1002/smll.200500265 [DOI] [PubMed] [Google Scholar] . Bachand, G. D. , Rivera, S. B. , Carroll-Portillo, A. , and Hess, H. , 2006, “ Active Capture and Transport of Virus Particles Using a Biomolecular Motor-Driven Nanoscale ANtibody Sandwich Assay,” Small, 2(3), pp. 381–385. 10.1002/smll.200500262 [DOI] [PubMed] [Google Scholar] . Shishido, H. , and Maruta, S. , 2012, “ Engineering of a Novel Ca 2+-Regulated Kinesin Molecular Motor Using a Calmodulin Dimer Linker,” Biochem. Biophys. Res. Commun., 423(2), pp. 386–391. 10.1016/j.bbrc.2012.05.135 [DOI] [PubMed] [Google Scholar] . Hoersch, D. , and Kortemme, T. , 2016, “ A Model for the Molecular Mechanism of an Engineered Light-Driven Protein Machine,” Structure, 24(4), pp. 576–584. 10.1016/j.str.2016.02.015 [DOI] [PubMed] [Google Scholar] . Furuta, A. , Amino, M. , Yoshio, M. , Oiwa, K. , Kojima, H. , and Furuta, K. , 2016, “ Creating Biomolecular Motors Based on Dynein and Actin-Binding Proteins,” Nat. Nanotechnol., 12(3), pp. 233–237. 10.1038/nnano.2016.238 [DOI] [PubMed] [Google Scholar] . Driller-Colangelo, A. R. , Chau, K. W. L. , Morgan, J. M. , and Derr, N. D. , 2016, “ Cargo Rigidity Affects the Sensitivity of Dynein Ensembles to Individual Motor Pausing,” Cytoskeleton, 73(12), pp. 693–702. 10.1002/cm.21339 [DOI] [PubMed] [Google Scholar] . Toledo, M. A. S. , Janissen, R. , Favaro, M. T. P. , Cotta, M. A. , Montiero, G. A. , Prazeres, D. M. F. , Souza, A. P. , and Azzoni, A. R. , 2012, “ Development of a Recombinant Fusion Protein Based on the Dynein Light Chain LC8 for Non-Viral Gene Delivery,” J. Controlled Release, 159(2), pp. 222–231. 10.1016/j.jconrel.2012.01.011 [DOI] [PubMed] [Google Scholar] Associated Data This section collects any data citations, data availability statements, or supplementary materials included in this article. Supplementary Materials Supplementary Material Click here for additional data file. (15KB, zip) Articles from Journal of Biomechanical Engineering are provided here courtesy of American Society of Mechanical Engineers ACTIONS View on publisher site PDF (1.9 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Structure and Function 3. Experimental Methods for Motor Protein Analysis 4. Review of Motor Protein Mechanics as Measured by Experimental Methods 5. Computational Modeling of Kinesin and Dynein Locomotion 6. Associations Between Motor Protein Functions and Disease Processes 7. Engineered Proteins 8. Conclusions Funding Data Supplementary Material Contributor Information References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
13370	https://www.youtube.com/watch?v=f8qygGoLvQ0	Use u Substitution To Find The Derivative. 3 Examples. Part 2 Ms Shaws Math Class 51100 subscribers 35 likes Description 7596 views Posted: 1 Oct 2018 1 comments Transcript: everyone we're going to use u-substitution to find the derivative of these three problems so we can get started and the first one is this one so the first thing we're going to do is rewrite this as y equals Ln of U that means u equals x squared plus 1 therefore the derivative of U is 2x with respect to X so that's our little substitution part we're going to be doing and then all you do is take the derivative of Y which is 1 over u du u next thing you do is substitution so the derivative of y equals 1 over u u happens to be x squared plus 1 and D U happens to be 2x DX so I'm going to just clean this up now divide both sides by D D to the X so that's going to be dy derivative of Y with respect to x equals this is like over 1 I'm going to put that up there so that's 2x over x squared plus 1 and that would be your final answer and the next one is this one so the first thing I'm going to rewrite this as H of x equals I'm going to let this be my U and we write this in a rational form instead of radical form so my u equals x cubed plus 2x minus 1 therefore my derivative of U is 3 x squared plus 2 with respect to X that's going to be our substitution now we merely take the derivative of H which is 1/3 you ooh to the 1/3 minus 3/3 which is minus 1 so we get 1/3 you to the negative 2/3 now my you happens to be this so that's X cubed plus I've got to put my D u here X cubed plus 2x minus 1 and my d u here happens to be 3x squared plus 2 DX and then we merely have to simplify this this is like your Y so you divide that side by so H derivative of X divided by DX equals now I'm going to write this as 1 over 3 and I'm going to bring this in the denominator so this is going to be X cubed plus 2x minus 1 to the 2/3 I got rid of my negative exponent here and put it in the denominator now the numerator is going to get this part so it's 1 times 3x squared plus 2 so my derivative of H equals and the 1 1 times that we don't need the 1 really there so we're just gonna write 3x squared plus 2/3 times X cubed plus 2x minus 1 to the 2/3 power and that's it thank you have a very nice day bye bye
13371	https://www.sciencedirect.com/science/article/abs/pii/S0002962915353350	Band 4.2 Abnormalities in Human Red Cells - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract References (71) Cited by (23) The American Journal of the Medical Sciences Volume 307, Issue 3, March 1994, Pages 190-203 Reviews Band 4.2 Abnormalities in Human Red Cells Author links open overlay panel Yoshihito Yawata MD Show more Add to Mendeley Share Cite rights and content ABSTRACT Abnormalities of membrane protein band 4.2 in human red cells are reviewed from the standpoints of clinical hematology, protein chemistry, membrane functions, and gene expression. This article will help more extensive investigations in clarifying the physiologic significance of this protein, and to understand abnormalities of band 4.2 in clinical, biochemical, biologic, and genetic aspects. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Recommended articles References (71) J. Palek et al. Mutations of the red blood cell membrane proteins: From clinical evaluation to detection of the underlying genetic defect Blood (1992) S. Hayashi et al. Abnormality in a specific protein of the erythrocyte membrane in hereditary spherocytosis Biochem Biophys Res Commun (1974) Y. Nozawa et al. Erythrocyte membrane of hereditary spherocytosis: Alteration in surface ultrastructure and membrane proteins, as inferred by scanning electron microscopy and SDS-disc gel electrophoresis Clin Chim Acta (1974) P. Jarolim et al. Band 3 Tuscaloosa: Pro 327→Arg 327 substitution in the cytoplasmic domain of erythrocyte band 3 protein associated with spherocytic hemolytic anemia and partial deficiency of protein 4.2 Blood (1992) A.C. Rybicki et al. Human erythrocyte protein 4.2 deficiency associated with hemolytic anemia and a homozygous 40 glutamic acid→lysine substitution in the cytoplasmic domain of band 3 (band 3 Montefiore) Blood (1993) D. Branton et al. Interaction of cytoskeletal proteins on the human erythrocyte membrane (abstract) Cell (1981) C. Korsgren et al. Purification and properties of human erythrocyte band 4.2. Association with the cytoplasmic domain of band 3 J Biol Chem (1986) E. Dotimas et al. Human erythrocyte band 4.2: Structural domain mapping of the protein purified by a novel procedure Biochim Biophys Acta (1993) V. Bennett et al. Association between ankyrin and the cytoplasmic domain of band 3 isolated from the human erythrocyte membrane J Biol Chem (1980) C. Korsgren et al. Associations of human erythrocyte band 4.2. Binding to ankyrin and to the cytoplasmic domain of band 3 J Biol Chem (1988) M.A. Risinger et al. Human erythrocyte protein 4.2, a high copy number membrane protein, is N-myristylated J Biol Chem (1992) L.A. Sung et al. Human erythrocyte protein 4.2: Isoform expression, differential splicing, and chromosomal assignment Blood (1992) H. Wada et al. Expression of major blood group antigens on human erythroid cells in a two phase liquid culture system Blood (1990) M. Hanspal et al. Asynchronous synthesis of membrane skeletal proteins during terminal maturation of murine erythroblasts Blood (1992) M. Hanspal et al. Erythropoietin enhances the assembly of α, β spectrin heterodimers on the murine erythroblast membranes by increasing β spectrin synthesis J Biol Chem (1991) H. Iida et al. Biochemical studies on abnormal erythrocyte membranes protein abnormality of erythrocyte membrane in biliary obstruction Biochim Biophys Acta (1976) E.E. Bouhassira et al. An alanine-to-threonine substitution in protein 4.2 cDNA is associated with a Japanese form of hereditary hemolytic anemia (protein 4.2 Nippon) Blood (1992) J. Folch et al. A simple method for the isolation and purification of total lipides from animal tissues J Biol Chem (1957) G.R. Bartlett Phosphorus assay in column chromatography J Biol Chem (1959) V. Bennett et al. Association between ankyrin and the cytoplasmic domain of band 3 isolated from the human erythrocyte membrane J Biol Chem (1980) M. Inaba et al. Two novel molecular isoforms of band 4.2 in Japanese Sika deer (Cervus nippon yesoensis, Heude) erythrocytes Biochim Biophys Acta (1990) G. Fairbanks et al. Electrophoretic analysis of the major polypeptides of the human erythrocyte membrane Biochemistry (1971) T.L. Steck The organization of proteins in the human red blood cell membrane J Cell Biol (1974) J. Palek et al. Clinical expression and laboratory detection of red blood cell membrane protein mutations Semin Hematol (1993) C.M. Cohen et al. Human erythrocyte membrane protein band 4.2 (pallidin) Semin Hematol (1993) B.E. Burke et al. Erythrocyte membrane skeleton abnormalities in hereditary spherocytosis Br J Haematol (1983) M. Fukuda et al. Membrane protein band 4.2-deficient hereditary elliptocytosis with a marked drug-induced hemolytic crisis (abstract) Acta Haematologica Japonica (1987) M. Fukuda et al. Drug-induced severe hemolytic crisis in a patient with hemolytic hereditary elliptocytosis with complete deficiency of red cell membrane protein band 4.2 Japanese Journal of Clinical Hematology (1988) A.C. Rybicki et al. Deficiency of protein 4.2 in erythrocytes from a patient with a Coombs negative hemolytic anemia. Evidence for a role of protein 4.2 in stabilizing ankyrin on the membrane J Clin Invest (1988) F. Inoue et al. Alteration of erythrocyte membrane proteins in a family with hereditary spherocytosis Acta Haematologica Japonica (1989) F. Inoue et al. Band 4.2 protein of erythrocyte membranes with hereditary spherocytosis Medicine and Biology (1989) H. Ito et al. A case of spherocytosis with complete deficiency of erythrocyte membrane protein band 4.2, accompanied with arachnoid cyst and congenital ocular anomalies Nihon Naika Gakkai Zassi (1989) H. Ideguchi et al. A case of hereditary spherocytosis associated with band 4.2 protein deficiency (abstract) Acta Haematologica Japonica (1989) H. Ideguchi et al. A genetic defect of erythrocyte band 4.2 protein associated with hereditary spherocytosis Br J Haematol (1990) K. Ata et al. Deficiency of membrane protein band 4.2 in red cells of three cases of congenital hemolytic anemia (abstract) Acta Haematologica Japonica (1989) View more references Cited by (23) Refined views of multi-protein complexes in the erythrocyte membrane 2012, Blood Cells Molecules and Diseases Citation Excerpt : Given this, we concur with the view that band 3 dimers are spatially separated and ‘tetramerisation’ is only mediated via ankyrin. In addition to its interactions with band 3 and spectrin, there is evidence that ankyrin also binds protein 4.2 , the absence of protein 4.2 causes hereditary spherocytosis (HS) [6,25,26] and a reduction in the association of band 3 with ankyrin . Protein 4.2 binds to cdb3 and ANK D3-4 [28,29], strengthening their interaction. Show abstract The erythrocyte membrane has been extensively studied, both as a model membrane system and to investigate its role in gas exchange and transport. Much is now known about the protein components of the membrane, how they are organised into large multi-protein complexes and how they interact with each other within these complexes. Many links between the membrane and the cytoskeleton have also been delineated and have been demonstrated to be crucial for maintaining the deformability and integrity of the erythrocyte. In this study we have refined previous, highly speculative molecular models of these complexes by including the available data pertaining to known protein–protein interactions. While the refined models remain highly speculative, they provide an evolving framework for visualisation of these important cellular structures at the atomic level. ### Protein 4.2 : A complex linker 2009, Blood Cells Molecules and Diseases Show abstract The peripheral membrane protein, protein 4.2, is one of the most abundant protein components of the erythrocyte membrane. Protein 4.2 has an important role in red cell membrane structure, its absence due to natural mutations in humans or gene knockout in mice has a detrimental effect on membrane stability and results in hereditary spherocytosis. It is known to be a point of connection between the band 3 complex and the Rhesus protein complex, through its associations with band 3 and CD47 and also via interactions with the cytoskeletal protein ankyrin. Considering its relatively high abundance and importance in stability of the erythrocyte membrane, protein 4.2 has proved a somewhat neglected protein in recent years. In this review we will summarize our current understanding of protein 4.2, discuss its known interactions and describe the effects and implications of protein 4.2 deficiency. Based on protein 4.2's close homology with transglutaminase family proteins, we propose a new speculative “open” homology structure for protein 4.2 that may represent the active, membrane associated protein 4.2 molecule in red blood cells and also explain the dependence of protein 4.2 on band 3 binding for stability. ### A deletional frameshift mutation in protein 4.2 gene (allele 4.2 Lisboa) associated with hereditary hemolytic anemia 1995, Blood Show abstract We studied a 26-year-old Portuguese patient with recessively transmitted hereditary hemolytic anemia. Protein 4.2 was absent from red cell ghosts by Western blotting. Although the 4.2 mRNA was not detected in Northern blots, it was shown to be present by a procedure based on nested reverse transcription-polymerase chain reaction (RT-PCR). Partial nucleotide sequencing disclosed a one-nucleotide deletion at nt 264 (or 265): AAG GTG Å AAG TG, in codon 88 (or 89) belonging to exon 2. This change, defining allele 4.2 Lisboa, placed in frame the nonsense triplet that normally overlaps codons 136 and 137 (GTG ACQ. This mutation, which abolishes an Eco NI site, was also found in the gene of the proband (homozygous state), her parents, and her brother (heterozygous state). Apart from anemia, the patient was free of clinical manifestations. Platelet membranes were also investigated using Western blots. Antibodies to red cell protein 4.2 showed a doublet (72 and 70 kD) both in the controls and the patient. This finding raises an interesting question concerning the relationship between this doublet and erythroid protein 4.2. ### Cell Membrane: The Red Blood Cell as a Model 2004, Cell Membrane the Red Blood Cell as A Model ### Band 4.2 Shiga: 317 CGC → TGC in compound heterozygotes with 142 GCT → ACT results in band 4.2 deficiency and microspherocytosis 1995, British Journal of Haematology ### A point mutation in the protein 4.2 gene (allele 4.2 Tozeur) associated with hereditary haemolytic anaemia 1995, British Journal of Haematology View all citing articles on Scopus Supported in part by a research grant for Idiopathic Disorders of Hematopoietic Organs from the Japanese Ministry of Health and Welfare, by a Grant-in-Aid for Scientific Research (No. 03671201, 04671540, and 05670936) from the Ministry of Education, Science and Culture of the Japanese Government, by the Japan Society for the Promotion of Science (JSPS)-the Centre National de la Recherche Scientifique (CNRS) Joint Research Projects, and by research grants from Kawasaki Medical School (No. 3–305, 4–301, and 5–101) and Nakatani Electronic Measuring Technology Association of Japan. View full text Copyright © 1994 Southern Society for Clinical Investigation. Published by Elsevier Inc. All rights reserved. Recommended articles Factor productivity nexus economic growth in Sub-Saharan Africa: Symmetric and asymmetric panel approaches World Development Sustainability, Volume 5, 2024, Article 100169 Urgaia R.Worku ### Unveiling the environmental gains of biodegradable plastics in the waste treatment phase: A cradle-to-crave life cycle assessment Chemical Engineering Journal, Volume 487, 2024, Article 150540 Danbee Park, …, Wangyun Won ### The role of cardiac surgery and interventional cardiology in addressing cardiovascular risks associated with anabolic steroid use Steroids, Volume 209, 2024, Article 109472 Ignazio Condello, Giuseppe Nasso ### Evaluating underlying factor structures using novel machine learning algorithms: An empirical and simulation study Journal of Business Research, Volume 173, 2024, Article 114472 Zhenning“Jimmy” Xu, …, Gary L.Frankwick ### Validating the measurement of red blood cell diameter in fresh capillary blood by darkfield microscopy: A pilot study Advances in Integrative Medicine, Volume 3, Issue 1, 2016, pp. 11-14 Sheriden Keegan, …, Tini Gruner ### Membrane Proteins The Membranes of Cells, 2016, pp. 219-268 Philip L.Yeagle Show 3 more articles About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy Cookie Settings Accept all cookies Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. You may also be able to exercise your privacy choices as described in our Privacy Policy Allow all Manage Consent Preferences Strictly Necessary Cookies Always active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. Cookie Details List‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. Cookie Details List‎ Contextual Advertising Cookies [x] Contextual Advertising Cookies These cookies are used for properly showing banner advertisements on our site and associated functions such as limiting the number of times ads are shown to each user. Cookie Details List‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Confirm my choices
13372	https://www.verywellhealth.com/mimics-of-psoriasis-2788343	What Conditions Can Be Mistaken For Psoriasis? 8 Skin Issues That Can Mimic Psoriasis Psoriasis is challenging to diagnose. Healthcare providers primarily rely on the appearance of the lesions (plaques). These can resemble skin changes caused by other conditions, such as eczema, dermatitis, lupus, and skin cancer. Also, different types of psoriasis have variations in their appearance, location, and severity. This is why it's common to misdiagnose psoriasis or incorrectly diagnose another condition as psoriasis. Your symptoms may be due to psoriasis. But here are eight medical conditions your healthcare provider will likely consider before coming to that conclusion. What Can Be Mistaken for Psoriasis Conditions that may appear similar to psoriasis include: To avoid misdiagnosis, healthcare providers and dermatologists usually do a differential diagnosis to rule out conditions with symptoms similar to psoriasis. When in doubt, they can exclude other possible causes using blood tests, cultures, skin biopsies, and other tools. This allows them to accurately confirm the diagnosis and start the appropriate treatment. Eczema Eczema is the name for a group of conditions that cause red, itchy skin patches similar to psoriasis. As opposed to psoriasis, an autoimmune disorder, eczema is characterized by an overactive (rather than self-destructive) immune response. Eczema tends to be itchier than psoriasis and can cause oozing and crusting when scratched. With psoriasis, the plaques can easily bleed when scratched, leaving behind a peppered pattern of blood spots known as the Auspitz's sign. The differences are most apparent under the microscope: Eczema vs. Psoriasis Eczema tends to be itchier than psoriasis and can cause oozing and crusting when scratched. With psoriasis, the plaques can easily bleed when scratched, leaving behind a peppered pattern of blood spots known as the Auspitz's sign. Seborrheic Dermatitis Seborrheic dermatitis is a skin condition that mainly affects the scalp, causing scaly patches of red skin and stubborn dandruff. It's easily mistaken for scalp psoriasis and vice versa. As with eczema, the conditions can be differentiated under the microscope by their acanthotic or non-acanthotic appearance. Seborrheic Dermatitis vs. Psoriasis With seborrheic dermatitis, dandruff flakes are fine and slightly yellowish, while the scalp is oily. With psoriasis, the flakes are silvery-white with a lamellar (scale-like) appearance. Moreover, psoriatic skin will be dry. Pityriasis Rosea Pityriasis rosea is a benign skin condition. The name is Latin for "fine pink scale." It generally starts with a large, slightly raised, scaly patch on the back, chest, or abdomen. This is called a herald patch. The herald patch is typically followed by the appearance of smaller patches that sweep outward like the boughs of a pine tree. Pityriasis rosea usually resolves completely within six to eight weeks. Psoriasis is characterized by recurrent flares. Pityriasis Rosea vs. Psoriasis The characteristic rash pattern is usually enough to differentiate pityriasis rosea from psoriasis. Psoriasis plaques tend to be far more irregular in shape with a more pronounced scaling. Lichen Planus Lichen planus is a skin condition thought to be autoimmune. It causes swelling and irritation in the skin, hair, nails, and mucous membranes. Lichen planus can create lacy white patches when it appears in the mouth, vagina, or other mucosal tissues. Lichen planus will typically manifest with psoriasis-like lesions on the wrists and limbs. Under the microscope, psoriasis and lichen planus both have an acanthotic appearance. But with lichen planus, there will be a band-like area of damage between the upper layer of skin (epidermis) and the middle layer (dermis). Lichen Planus vs. Psoriasis Lichen planus can be differentiated by the appearance of the skin lesions. They are thick like in psoriasis but are more purple and lack the characteristic scales. Lichen planus can also affect mucosal tissues, whereas psoriasis won't. Onychomycosis Onychomycosis, also known as tinea unguium, is a fungal infection of the nails. Nail psoriasis is frequently mistaken for onychomycosis. They have similar symptoms, like the thickening, crumbling, and lifting of the nail plate, and the formation of abnormal ridges and dents. Onychomycosis vs. Psoriasis One of the first clues that a nail disorder is psoriatic is the appearance of skin plaques elsewhere on the body. A healthcare provider or dermatologist will take a scraping from the nail and examine it under the microscope to confirm the diagnosis. If there are no fungal spores, it can reasonably be assumed that psoriasis is the cause. Psoriasis Doctor Discussion Guide Get our printable guide for your next healthcare provider's appointment to help you ask the right questions. Sign up for our Health Tip of the Day newsletter, and receive daily tips that will help you live your healthiest life. Thank you, {{form.email}}, for signing up. There was an error. Please try again. Lupus Systemic lupus erythematosus, also known as lupus, is an autoimmune disease that affects multiple organ systems, including the skin. One of the tell-tale signs of lupus is a rash formation on the cheeks and nose. This is called a butterfly rash. Healthcare providers can usually confirm lupus with a series of blood tests, including the antinuclear antibody (ANA) test. Skin biopsy, including the use of direct and indirect immunofluorescent tests, can also confirm lupus as the cause. Lupus vs. Psoriasis When lupus affects the skin specifically—a condition known as discoid lupus erythematosus—the lesions will be far less scaly than psoriasis and lack the lamellar appearance. Squamous Cell Carcinoma Squamous cell carcinoma is the second most common type of skin cancer. It manifests with thick, flat, scaly nodules. It's usually found on areas of the body damaged by ultraviolet (UV) rays from the sun or tanning beds. Squamous cell carcinoma can readily be diagnosed with a skin biopsy: Squamous Cell Carcinoma vs. Psoriasis Squamous cell carcinoma can mimic psoriasis but tends to be less scaly and limited to areas of sun-damaged skin. Generally, there will only be a handful of cancerous lesions. Mycosis Fungoides Mycosis fungoides, also known as Alibert-Bazin syndrome, is the most common form of cutaneous T-cell lymphoma—a dermatological variation of blood cancer known as non-Hodgkin lymphoma. Mycosis fungoides manifests with rash-like patches of skin. The lesions will appear scaly and often be extremely itchy in the early stages. The buttocks are often the first part of the body affected. Over time, the lesions can develop elsewhere, causing widespread redness or discoloration (depending on one's skin tone) and itching but far less scaling. A skin biopsy can help differentiate the diseases. With mycosis fungoides, the biopsied tissue will have microscopic pus-filled cavities in the epidermal layer, known as Pautrier abscesses. Mycosis Fungoides vs. Psoriasis Mycosis fungoides is easily confused with psoriasis in the early stages. Unlike psoriasis, mycosis fungoides is often accompanied by persistently swollen lymph nodes. Pancreas and liver enlargement are also common. See a Healthcare Provider The problem with misdiagnosis is that it can expose you to unnecessary and unhelpful treatments. Worse, by assuming it's psoriasis without exploring other possible causes, you can miss the signs of a potentially more serious illness. This is why self-diagnosing psoriasis is never a good idea. If you're worried about a skin condition, ask your healthcare provider for a referral to a dermatologist. Mention both skin-related and non-skin-related symptoms. Doing so increases your chance of reaching the correct diagnosis. Rendon A, Schäkel K. Psoriasis Pathogenesis and Treatment. Int J Mol Sci. 2019;20(6):1475. doi:10.3390/ijms20061475 Asz-sigall D, Tosti A, Arenas R. Tinea Unguium: Diagnosis and Treatment in Practice. Mycopathologia. 2017;182(1-2):95-100. doi:10.1007/s11046-016-0078-4 Haneke E. Nail psoriasis: clinical features, pathogenesis, differential diagnoses, and management. Psoriasis (Auckl). 2017;7:51–63. doi:10.2147/PTT.S126281 Maidhof W, Hilas O. Lupus: an overview of the disease and management options. P T. 2012;37(4):240–249. Kim, W.; Jerome, D.; and Yeung, J. Diagnosis and management of psoriasis. Can Fam Physician. 2017;63(4):278-85. Young, M.; Aldredge, L.; and Parker, P. Psoriasis for the primary care practitioner. J Am Assn Nurse Pract. 2017;29(3):157-78. doi:10.1002/2327-6924.12443 By Dean Goodless, MD Dean R. Goodless, MD, is a board-certified dermatologist specializing in psoriasis. Thank you, {{form.email}}, for signing up. There was an error. Please try again.
13373	https://courses.lumenlearning.com/suny-physics/chapter/30-9-the-pauli-exclusion-principle/	Physics Atomic Physics The Pauli Exclusion Principle Learning Objectives By the end of this section, you will be able to: Multiple-Electron Atoms Figure 1. The Austrian physicist Wolfgang Pauli (1900–1958) played a major role in the development of quantum mechanics. He proposed the exclusion principle; hypothesized the existence of an important particle, called the neutrino, before it was directly observed; made fundamental contributions to several areas of theoretical physics; and influenced many students who went on to do important work of their own. (credit: Nobel Foundation, via Wikimedia Commons) All atoms except hydrogen are multiple-electron atoms. The physical and chemical properties of elements are directly related to the number of electrons a neutral atom has. The periodic table of the elements groups elements with similar properties into columns. This systematic organization is related to the number of electrons in a neutral atom, called the atomic number, Z. We shall see in this section that the exclusion principle is key to the underlying explanations, and that it applies far beyond the realm of atomic physics. In 1925, the Austrian physicist Wolfgang Pauli (see Figure 1) proposed the following rule: No two electrons can have the same set of quantum numbers. That is, no two electrons can be in the same state. This statement is known as the Pauli exclusion principle, because it excludes electrons from being in the same state. The Pauli exclusion principle is extremely powerful and very broadly applicable. It applies to any identical particles with half-integral intrinsic spin—that is, having s = 1/2, 3/2, … Thus no two electrons can have the same set of quantum numbers. Pauli Exclusion Principle No two electrons can have the same set of quantum numbers. That is, no two electrons can be in the same state. Let us examine how the exclusion principle applies to electrons in atoms. The quantum numbers involved were defined in Quantum Numbers and Rules as n, l, ml, s, and ms. Since s is always 1/2 for electrons, it is redundant to list s, and so we omit it and specify the state of an electron by a set of four numbers (n, l, ml, ms). For example, the quantum numbers (2, 1, 0, −1/2) completely specify the state of an electron in an atom. Since no two electrons can have the same set of quantum numbers, there are limits to how many of them can be in the same energy state. Note that n determines the energy state in the absence of a magnetic field. So we first choose n, and then we see how many electrons can be in this energy state or energy level. Consider the n=1 level, for example. The only value l can have is 0 (see Table 1 in Quantum Numbers and Rules for a list of possible values once n is known), and thus ml can only be 0. The spin projection ms can be either +1/2 or −1/2, and so there can be two electrons in the n = 1 state. One has quantum numbers (1, 0, 0, +1/2), and the other has (1, 0, 0, −1/2). Figure 2 illustrates that there can be one or two electrons having n = 1, but not three. Figure 2. The Pauli exclusion principle explains why some configurations of electrons are allowed while others are not. Since electrons cannot have the same set of quantum numbers, a maximum of two can be in the n = 1 level, and a third electron must reside in the higher-energy n = 2 level. If there are two electrons in the n = 1 level, their spins must be in opposite directions. (More precisely, their spin projections must differ.) Shells and Subshells Because of the Pauli exclusion principle, only hydrogen and helium can have all of their electrons in the n = 1 state. Lithium (see the periodic table) has three electrons, and so one must be in the n = 2 level. This leads to the concept of shells and shell filling. As we progress up in the number of electrons, we go from hydrogen to helium, lithium, beryllium, boron, and so on, and we see that there are limits to the number of electrons for each value of n. Higher values of the shell n correspond to higher energies, and they can allow more electrons because of the various combinations of l, ml, and ms that are possible. Each value of the principal quantum number n thus corresponds to an atomic shell into which a limited number of electrons can go. Shells and the number of electrons in them determine the physical and chemical properties of atoms, since it is the outermost electrons that interact most with anything outside the atom. The probability clouds of electrons with the lowest value of l are closest to the nucleus and, thus, more tightly bound. Thus when shells fill, they start with l = 0, progress to l = 1, and so on. Each value of l thus corresponds to a subshell. The table given below lists symbols traditionally used to denote shells and subshells. Table 1. Shell and Subshell Symbols Shell \| Subshell n \| l \| Symbol 1 \| 0 \| s 2 \| 1 \| p 3 \| 2 \| d 4 \| 3 \| f 5 \| 4 \| g \| 5 \| h \| 6 \| i To denote shells and subshells, we write nl with a number for n and a letter for l. For example, an electron in the n=1 state must have l = 0, and it is denoted as a 1s electron. Two electrons in the n = 1 state is denoted as 1s2. Another example is an electron in the n = 2 state with l = 1, written as 2p. The case of three electrons with these quantum numbers is written 2p3. This notation, called spectroscopic notation, is generalized as shown in Figure 3. Figure 3. Counting the number of possible combinations of quantum numbers allowed by the exclusion principle, we can determine how many electrons it takes to fill each subshell and shell. Example 1. How Many Electrons Can Be in This Shell? List all the possible sets of quantum numbers for the n = 2 shell, and determine the number of electrons that can be in the shell and each of its subshells. Strategy Given n = 2 for the shell, the rules for quantum numbers limit l to be 0 or 1. The shell therefore has two subshells, labeled 2s and 2p. Since the lowest l subshell fills first, we start with the 2s subshell possibilities and then proceed with the 2p subshell. Solution It is convenient to list the possible quantum numbers in a table, as shown below. Figure 4 Discussion It is laborious to make a table like this every time we want to know how many electrons can be in a shell or subshell. There exist general rules that are easy to apply, as we shall now see. The number of electrons that can be in a subshell depends entirely on the value of l. Once l is known, there are a fixed number of values of ml, each of which can have two values for ms First, since ml goes from −l to l in steps of 1, there are 2l + 1 possibilities. This number is multiplied by 2, since each electron can be spin up or spin down. Thus the maximum number of electrons that can be in a subshell is 2(2l + 1). For example, the 2s subshell in Example 1 has a maximum of 2 electrons in it, since 2(2l + 1) = 2(0 + 1) = 2 for this subshell. Similarly, the 2p subshell has a maximum of 6 electrons, since 2(2l + 1) = 2(2 + 1) = 6. For a shell, the maximum number is the sum of what can fit in the subshells. Some algebra shows that the maximum number of electrons that can be in a shell is 2n2. For example, for the first shell n = 1, and so 2n2 = 2. We have already seen that only two electrons can be in the n = 1 shell. Similarly, for the second shell, n = 2, and so 2n2 = 8. As found in Example 1, the total number of electrons in the n = 2 shell is 8. Example 2. Subshells and Totals for [latex]n=3\[/latex] How many subshells are in the n = 3 shell? Identify each subshell, calculate the maximum number of electrons that will fit into each, and verify that the total is 2n2. Strategy Subshells are determined by the value of l; thus, we first determine which values of l are allowed, and then we apply the equation “maximum number of electrons that can be in a subshell = 2(2l + 1)” to find the number of electrons in each subshell. Solution Since n=3, we know that l can be 0, 1, or 2; thus, there are three possible subshells. In standard notation, they are labeled the 3s, 3p, and 3d subshells. We have already seen that 2 electrons can be in an s state, and 6 in a p state, but let us use the equation “maximum number of electrons that can be in a subshell = 2(2l + 1)” to calculate the maximum number in each: 3s has l = 0; thus 2(2l + 1) = 2(0 + 1) = 2 3p has l = 1; thus, 2(2l + 1) = 2(2 + 1) = 6 3d has l = 2; thus 2(2l + 1) = 2(4 + 1) = 10 Total = 18 (in the n = 3 shell) The equation “maximum number of electrons that can be in a shell = 2n2” gives the maximum number in the n = 3 shell to be: Maximum number of electrons = 2n2 = 2(3)2 = 2(9) = 18. Discussion The total number of electrons in the three possible subshells is thus the same as the formula 2n2. In standard (spectroscopic) notation, a filled n = 3 shell is denoted as 3s23p63d10. Shells do not fill in a simple manner. Before the n = 3 shell is completely filled, for example, we begin to find electrons in the n = 4 shell. Shell Filling and the Periodic Table Table 2 shows electron configurations for the first 20 elements in the periodic table, starting with hydrogen and its single electron and ending with calcium. The Pauli exclusion principle determines the maximum number of electrons allowed in each shell and subshell. But the order in which the shells and subshells are filled is complicated because of the large numbers of interactions between electrons. Table 2. Electron Configurations of Elements Hydrogen Through Calcium Element \| Number of electrons (Z) \| Ground state configuration H \| 1 \| 1s1 \| \| \| \| \| He \| 2 \| 1s2 \| \| \| \| \| Li \| 3 \| 1s2 \| 2s1 \| \| \| \| Be \| 4 \| “ \| 2s2 \| \| \| \| B \| 5 \| “ \| 2s2 \| 2p1 \| \| \| C \| 6 \| “ \| 2s2 \| 2p2 \| \| \| N \| 7 \| “ \| 2s2 \| 2p3 \| \| \| O \| 8 \| “ \| 2s2 \| 2p4 \| \| \| F \| 9 \| “ \| 2s2 \| 2p5 \| \| \| Ne \| 10 \| “ \| 2s2 \| 2p6 \| \| \| Na \| 11 \| “ \| 2s2 \| 2p6 \| 3s1 \| \| Mg \| 12 \| “ \| “ \| “ \| 3s2 \| \| Al \| 13 \| “ \| “ \| “ \| 3s2 \| 3p1 \| Si \| 14 \| “ \| “ \| “ \| 3s2 \| 3p2 \| P \| 15 \| “ \| “ \| “ \| 3s2 \| 3p3 \| S \| 16 \| “ \| “ \| “ \| 3s2 \| 3p4 \| Cl \| 17 \| “ \| “ \| “ \| 3s2 \| 3p5 \| Ar \| 18 \| “ \| “ \| “ \| 3s2 \| 3p6 \| K \| 19 \| “ \| “ \| “ \| 3s2 \| 3p6 \| 4s1 Ca \| 20 \| “ \| “ \| “ \| “ \| “ \| 4s2 Examining the above table, you can see that as the number of electrons in an atom increases from 1 in hydrogen to 2 in helium and so on, the lowest-energy shell gets filled first—that is, the n = 1 shell fills first, and then the n = 2 shell begins to fill. Within a shell, the subshells fill starting with the lowest l, or with the s subshell, then the p, and so on, usually until all subshells are filled. The first exception to this occurs for potassium, where the 4s subshell begins to fill before any electrons go into the 3d subshell. The next exception is not shown in Table 2; it occurs for rubidium, where the 5s subshell starts to fill before the 4d subshell. The reason for these exceptions is that l = 0 electrons have probability clouds that penetrate closer to the nucleus and, thus, are more tightly bound (lower in energy). Figure 5 shows the periodic table of the elements, through element 118. Of special interest are elements in the main groups, namely, those in the columns numbered 1, 2, 13, 14, 15, 16, 17, and 18. Figure 5. Periodic table of the elements (credit: National Institute of Standards and Technology, U.S. Department of Commerce) The number of electrons in the outermost subshell determines the atom’s chemical properties, since it is these electrons that are farthest from the nucleus and thus interact most with other atoms. If the outermost subshell can accept or give up an electron easily, then the atom will be highly reactive chemically. Each group in the periodic table is characterized by its outermost electron configuration. Perhaps the most familiar is Group 18 (Group VIII), the noble gases (helium, neon, argon, etc.). These gases are all characterized by a filled outer subshell that is particularly stable. This means that they have large ionization energies and do not readily give up an electron. Furthermore, if they were to accept an extra electron, it would be in a significantly higher level and thus loosely bound. Chemical reactions often involve sharing electrons. Noble gases can be forced into unstable chemical compounds only under high pressure and temperature. Group 17 (Group VII) contains the halogens, such as fluorine, chlorine, iodine and bromine, each of which has one less electron than a neighboring noble gas. Each halogen has 5 p electrons (a p5 configuration), while the p subshell can hold 6 electrons. This means the halogens have one vacancy in their outermost subshell. They thus readily accept an extra electron (it becomes tightly bound, closing the shell as in noble gases) and are highly reactive chemically. The halogens are also likely to form singly negative ions, such as C1−, fitting an extra electron into the vacancy in the outer subshell. In contrast, alkali metals, such as sodium and potassium, all have a single s electron in their outermost subshell (an s1 configuration) and are members of Group 1 (Group I). These elements easily give up their extra electron and are thus highly reactive chemically. As you might expect, they also tend to form singly positive ions, such as Na+, by losing their loosely bound outermost electron. They are metals (conductors), because the loosely bound outer electron can move freely. Of course, other groups are also of interest. Carbon, silicon, and germanium, for example, have similar chemistries and are in Group 4 (Group IV). Carbon, in particular, is extraordinary in its ability to form many types of bonds and to be part of long chains, such as inorganic molecules. The large group of what are called transitional elements is characterized by the filling of the d subshells and crossing of energy levels. Heavier groups, such as the lanthanide series, are more complex—their shells do not fill in simple order. But the groups recognized by chemists such as Mendeleev have an explanation in the substructure of atoms. PhET Explorations: Build an Atom Build an atom out of protons, neutrons, and electrons, and see how the element, charge, and mass change. Then play a game to test your ideas! Click to run the simulation. Section Summary Conceptual Questions Problems & Exercises Figure 6. This schematic shows the electron beam in a CRT passing through crossed electric and magnetic fields and causing phosphor to glow when striking the end of the tube. Figure 7. The Millikan oil drop experiment produced the first accurate direct measurement of the charge on electrons, one of the most fundamental constants in nature. Fine drops of oil become charged when sprayed. Their movement is observed between metal plates with a potential applied to oppose the gravitational force. The balance of gravitational and electric forces allows the calculation of the charge on a drop. The charge is found to be quantized in units of −1.6 × 10−19 C, thus determining directly the charge of the excess and missing electrons on the oil drops. Glossary atomic number: the number of protons in the nucleus of an atom Pauli exclusion principle: a principle that states that no two electrons can have the same set of quantum numbers; that is, no two electrons can be in the same state shell: a probability cloud for electrons that has a single principal quantum number subshell: the probability cloud for electrons that has a single angular momentum quantum number l Selected Solutions to Problems & Exercises 1. (a) 32. (b) 2 in s,6 in p,10 in d, and 14 in f, for a total of 32. 3. (a) 2; (b) 3d9 5. (b) n ≥ l is violated; (c) cannot have 3 electrons in s subshell since 3 > (2l + 1)=2; (d) cannot have 7 electrons in p subshell since 7 > (2l + 1) = 2(2 + 1) = 6 (a) The number of different values of ml is ±l, (l − 1), . . . , 0 for each l > 0 and one for [latex]l=0\Rightarrow \left(2l+1\right)\[/latex]. Also an overall factor of 2 since each ml can have ms equal to either +1/2 or [latex]-1/2\Rightarrow 2\left(2l+1\right)\[/latex]. (b) for each value of l, you get [latex]2\left(2l+1\right)=\text{0, 1, 2, ...,}\left(n\text{-1}\right)\Rightarrow[/latex] [latex]\begin{array}{l}2\left{\left[\left(2\right)\left(0\right)+1\right]+\left[\left(2\right)\left(1\right)+1\right]+\dots+\left[\left(2\right)\left(n - 1\right)+1\right]\right}=\\underset{n\text{ terms}}{\underbrace{2\left[1+3+\dots+\left(2n-3\right)+\left(2n-1\right)\right]}}\end{array}\[/latex] to see that the expression in the box is = n2, imagine taking (n − 1) from the last term and adding it to first term: [latex]\begin{array}{ll}=&2\left[1+\left(n-1\right)+3+\dots+\left(2n-3\right)+\left(2n-1\right)-\left(n-1\right)\right]\=&2\left[n+3+\dots+\left(2n-3\right)+n\right]\end{array}\[/latex] Now take (n − 3) from penultimate term and add to the second term [latex]2\underset{n\text{ terms}}{\underbrace{\left[n+n+\dots+n+n\right]}}={2n}^{2}\[/latex]. 9. The electric force on the electron is up (toward the positively charged plate). The magnetic force is down (by the RHR). 11. 401 nm 13. (a) 6.54 × 10−16 kg; (b) 5.54 × 10−7 m 15. 1.76 × 1011 C/kg, which agrees with the known value of 1.759 × 1011 C/kg to within the precision of the measurement 17. (a) 2.78 fm; (b) 0.37 of the nuclear radius. 19. (a) 1.34 × 1023; (b) 2.52 MW 21. (a) 6.42 eV; (b) 7.27 × 10−20 J/molecule; (c) 0.454 eV, 14.1 times less than a single UV photon. Therefore, each photon will evaporate approximately 14 molecules of tissue. This gives the surgeon a rather precise method of removing corneal tissue from the surface of the eye. 23. 91.18 nm to 91.22 nm 25. (a) 1.24 × 1011 V; (b) The voltage is extremely large compared with any practical value; (c) The assumption of such a short wavelength by this method is unreasonable. Candela Citations
13374	https://leverageedu.com/explore/learn-english/antonyms-of-reveal-with-meaning-and-examples/	45,000+ students trusted us with their dreams. Take the first step today! One app for all your study abroad needs Start your journey, track your progress, grow with the community and so much more Verification Code An OTP has been sent to your registered mobile no. Please verify Resend OTP Thanks for your comment ! Our team will review it before it's shown to our readers. 11+ Antonyms of Reveal, Meaning and Examples The antonyms of “reveal” are Conceal, Hide, Cover, Withhold or Suppress. To make an impression and get well-versed in vocabulary one should know at least a few antonyms of general words like “reveal.” “Reveal” is a word that is regularly used in general day-to-day communication. We come across this word at least once a day. Not just this we also use its antonyms quite frequently. Meaning of Reveal “Reveal” is a word that is used in everyday conversation. However, to make things simpler for you we will explain the meaning of this word in short. “Reveal” means to make something previously unknown or hidden known or visible. 10+ Opposites of Reveal Mentioned below are some words that can be used as antonyms to the word reveal: Also Read: 110+ Antonyms Usage with Examples Given below are some examples of how the antonyms of “reveal” can be used in a sentence: Antonyms of Reveal Quiz She decided to ____ her plans for a surprise party until the last moment. a) Reveal b) Conceal c) Share Answer: (B) This was all about the antonym of “reveal” meaning and examples. Hope you understood the concept where it’s used. For more such blogs, follow Leverage Edu. Digvijay Singh Having 2+ years of experience in educational content writing, withholding a Bachelor's in Physical Education and Sports Science and a strong interest in writing educational content for students enrolled in domestic and foreign study abroad programmes. I believe in offering a distinct viewpoint to the table, to help students deal with the complexities of both domestic and foreign educational systems. Through engaging storytelling and insightful analysis, I aim to inspire my readers to embark on their educational journeys, whether abroad or at home, and to make the most of every learning opportunity that comes their way. Leave a Reply Cancel reply Your contact details will not be published. Required fields are marked Name. Email Save my name, email, and website in this browser for the next time I comment. Contact no. Submit You May Also Like A Bird in the Hand is Worth Two in the Bush Meaning, Examples, Synonyms 11+ Antonyms of Sour, with Meanings & Example Idioms for Reading You Should Know! What are Conditional Sentences in English Grammar? Recent Posts Leverage Edu Offline Centers Learn English Subscribe to Our Newsletter Get notified about the latest career insights, study tips, and offers at Leverage Edu. General Enquiries [email protected] Press [email protected] Call Us 1800-572-000 Get the android app Get the iOS app Our Offices Leverage Edu Tower, A-258, Bhishma Pitamah Marg, Block A, Defence Colony, New Delhi, Delhi 110024 Leverage Edu 3rd floor, Plot number, 1- C, Raipur Khadar, Sector 126, Noida, Uttar Pradesh 201313 Devonshire House, 60 Goswell Road, United Kingdom, EC1M 7AD Leverage Edu Freshwater, Sydney, NSW 2096, Australia General Enquiries [email protected] Press [email protected] Call Us 1800-572-000 Get the android app Get the iOS app Our Offices Leverage Edu Tower, A-258, Bhishma Pitamah Marg, Block A, Defence Colony, New Delhi, Delhi 110024 Leverage Edu 3rd floor, Plot number, 1- C, Raipur Khadar, Sector 126, Noida, Uttar Pradesh 201313 Devonshire House, 60 Goswell Road, United Kingdom, EC1M 7AD Leverage Edu Freshwater, Sydney, NSW 2096, Australia Leaving already? 8 Universities with higher ROI than IITs and IIMs Grab this one-time opportunity to download this ebook Connect With Us Request 45,000+ students trusted us with their dreams. Take the first step today! Resend OTP in Continue 45,000+ students trusted us with their dreams. Take the first step today! Need help with? Study Abroad UK, Canada, US & More Test Prep IELTS, GRE, GMAT & More Finances Scholarship, Loans & Forex Continue 45,000+ students trusted us with their dreams. Take the first step today! Country Preference UK USA Canada Ireland Dubai Germany Australia New Zealand Others Continue 45,000+ students trusted us with their dreams. Take the first step today! Which English test are you planning to take? IELTS TOEFL PTE Not Yet Continue 45,000+ students trusted us with their dreams. Take the first step today! Which academic test are you planning to take? GRE GMAT Not Sure yet Continue 45,000+ students trusted us with their dreams. Take the first step today! When are you planning to take the exam? Already booked my exam slot Within 2 Months Want to learn about the test Submit 45,000+ students trusted us with their dreams. Take the first step today! Which Degree do you wish to pursue? Bachelor’s Master’s PhD Continue 45,000+ students trusted us with their dreams. Take the first step today! When do you want to start studying abroad? January 2026 May 2026 September 2026 January 2027 Later Submit 45,000+ students trusted us with their dreams. Take the first step today! What is your budget to study abroad? Submit How would you describe this article ? How useful was this post? Click on a star to rate it! Average rating 0 / 5. Vote count: 0 No votes so far! Be the first to rate this post.
13375	https://www.quora.com/How-can-I-prove-if-x-y-0-then-1-x-1-y	How to prove if x>y>0 then 1/x<1/y - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Linear Inequalities Proofs of Mathematics For... Mathematical Concepts in ... Mathematical Proof Algebraic Concepts Algebra Algebraic Proofs Basic Concepts of Mathema... 5 How can I prove if x>y>0 then 1/x<1/y? All related (35) Sort Recommended David Joyce Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by Jon Jones , B. A. Mathematics, University of Chicago (1974) · Author has 9.9K answers and 68.4M answer views ·1y Your proof will depend on what you already know about inequalities. The typical starting point are the axioms of positive numbers. Addition axiom. The sum of two positive numbers is positive. Multiplication axiom. The product of two positive numbers is positive. Trichotomy. For each number, exactly one of the following three possibilities is true: the number is positive, it is zero, or the negation of the number is positive (in which case it’s called negative). Then a>b a>b is defined as meaning a−b a−b is positive. The given statements are x>y x>y and y>0.y>0. The first says x−y x−y is positive, and the second says y y Continue Reading Your proof will depend on what you already know about inequalities. The typical starting point are the axioms of positive numbers. Addition axiom. The sum of two positive numbers is positive. Multiplication axiom. The product of two positive numbers is positive. Trichotomy. For each number, exactly one of the following three possibilities is true: the number is positive, it is zero, or the negation of the number is positive (in which case it’s called negative). Then a>b a>b is defined as meaning a−b a−b is positive. The given statements are x>y x>y and y>0.y>0. The first says x−y x−y is positive, and the second says y y is positive. It follows by the addition axiom that x x is also positive. You want to prove 1/x<1/y,1/x<1/y, that is, 1/y−1/x 1/y−1/x is positive. That’s equivalent to saying (x−y)/(x y)(x−y)/(x y) is positive. You’re given that the numerator x−y x−y is positive. Since both x x and y y are positive, by the multiplication axiom, therefore the denominator x y x y is positive. You would be done if you could just prove that the quotient of two positive numbers is positive. That’s not among the axioms, so that will have to be proved from them. Theorem. The quotient of two positive numbers is positive. Proof. Let a a and b b be positive. By trichotomy, their quotient a/b a/b is either positive, zero, or negative. If you can prove that it’s not zero or negative, then the only case left is that it’s positive. Case a/b a/b is zero. That implies a a is zero, but a a is positive, and by trichotomy, it can’t be both zero and positive. So this case doesn’t occur. Case a/b a/b is negative. That means −(a/b)−(a/b) is positive. Since b b is also positive, therefore by the multiplication axiom, −a−a is positive. But by trichotomy, not both a a and −a−a are positive. So this case doesn’t occur. Since the other two cases are excluded, therefore a/b a/b is positive. Q.E.D.Q.E.D. Upvote · 99 29 9 2 9 1 Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) ·Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. Continue Reading This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Upvote · 999 485 999 103 99 17 Related questions More answers below How will I prove that (1−x)y(1−x)(1−x y)1−x y<1(1−x)y(1−x)(1−x y)1−x y<1 if x > 0, y > 0 and xy < 1? Why does x y+x y 2+x y 3+⋯=x y−1 x y+x y 2+x y 3+⋯=x y−1? How can I prove that x > 2x - 1 for all x? What is lim(x,y)→(0,1)1+2(y−1)3+x 2(y−1+x 4)x 2+2(y−1)2 lim(x,y)→(0,1)1+2(y−1)3+x 2(y−1+x 4)x 2+2(y−1)2? Can you prove (y−y 1=m(x 1−x))=(y=m x+c)(y−y 1=m(x 1−x))=(y=m x+c)? Martyn Hathaway BSc in Mathematics, University of Southampton (Graduated 1986) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 4.7K answers and 6.7M answer views ·1y Simple, divide everything by x y x y: x x y>y x y>0 x y x x y>y x y>0 x y ⇒1 y>1 x>0⇒1 y>1 x>0 ⇒1 x<1 y⇒1 x<1 y [The ‘>0’ is crucial here: When you divide by a positive number, the inequality sign remains the same, whereas dividing by a negative number reverses the inequality. ] Upvote · 9 1 9 1 Diedrich Ehlerding Retired (2019–present) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 2.3K answers and 1.3M answer views ·1y Assume x>y>0 x>y>0. Then, x y>0 x y>0; and 1 x y>0 1 x y>0 as well. therefore x x y>y x y x x y>y x y cancel common factors from the fractions: 1 y>1 x 1 y>1 x Upvote · 9 1 Elaine Dawe BMath, in Mathematics&Computer Science, University of Waterloo (Graduated 1985) · Author has 5.4K answers and 6.6M answer views ·1y Since x,y>0 x,y>0, then x y>0 x y>0, and we can divide through by x y x y without changing direction of inequalities, then simplify each expression: x>y>0 x>y>0 x x y>y x y>0 x y x x y>y x y>0 x y 1 y>1 x>0 1 y>1 x>0 1 x<1 y 1 x<1 y Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · Sponsored by StealthGPT All your schoolwork, one app. Trusted by millions of students, writing countless essays and answering any question. Learn More 99 82 Related questions More answers below Why is √-1 valid, while a division by zero is invalid, when neither one truly makes sense? How can I show using direct proof that for all x, 0<x<1, we have have 1/5 < 1/ (3x^2 + 2) < 1/2? How can I prove X>y and Z<0 then xz<yz؟? If xyz = 1, then how can I prove that (1+x+y−1)−1+(1+y+z−1)−1+(1+z+x−1)−1=1(1+x+y−1)−1+(1+y+z−1)−1+(1+z+x−1)−1=1? Let x 1,x 2,x 3,y 1,y 2,y 3≥0 x 1,x 2,x 3,y 1,y 2,y 3≥0 such that x 1+x 2+x 3≥y 1+y 2+y 3,x 1+x 2+x 3≥y 1+y 2+y 3,x 1 x 2+x 2 x 3+x 3 x 1≥y 1 y 2+y 2 y 3+y 3 y 1,x 1 x 2+x 2 x 3+x 3 x 1≥y 1 y 2+y 2 y 3+y 3 y 1,x 1 x 2 x 3≥y 1 y 2 y 3.x 1 x 2 x 3≥y 1 y 2 y 3. How do you prove that √x 1+√x 2+√x 3≥√y 1+√y 2+√y 3?x 1+x 2+x 3≥y 1+y 2+y 3? Zakary Jay Nicholls Tutor in Economics/Engineering/Mathematics/Physics at Education Sector (2011–present) · Author has 646 answers and 3.8M answer views ·1y Here’s a quick and dirty method (as contrast to the more proper methods already listed), this is the kind of “proof” you would expect to see in a secondary school math textbook or classroom. if x>y>0 x>y>0 this means both x and y are positive. Now we have (we now know they’re both pos so we will ignore the zero): x>y x>y [divide both sides by x] 1>y/x 1>y/x [divide both sides by y] 1/y>1/x 1/y>1/x Upvote · Nirmalya Chakraborti Lives in West Bengal, India · Author has 124 answers and 219.8K answer views ·6y Related Given xy>x+y, x>0 and y>0, how do we prove x+y>4? Without the application of any standard inequality, one can notice that the inequality x y>x+y x y>x+y is equivalent to (x−1)(y−1)>1(x−1)(y−1)>1 and either both quantities must be positive or both must be negative to give a positive product. We can exclude the possibility that both x−1 x−1 and y−1 y−1 are negative. This is because they can be both negative only when 0<x<1 0<x<1 and 0<y<1 0<y<1 giving us 0<(x−1)(y−1)<1 0<(x−1)(y−1)<1 which is not true. So, both are positive numbers. Their product is a number greater than one so either both quantities are greater than one or one is greater and one is not. The second possibility can be excluded since Continue Reading Without the application of any standard inequality, one can notice that the inequality x y>x+y x y>x+y is equivalent to (x−1)(y−1)>1(x−1)(y−1)>1 and either both quantities must be positive or both must be negative to give a positive product. We can exclude the possibility that both x−1 x−1 and y−1 y−1 are negative. This is because they can be both negative only when 0<x<1 0<x<1 and 0<y<1 0<y<1 giving us 0<(x−1)(y−1)<1 0<(x−1)(y−1)<1 which is not true. So, both are positive numbers. Their product is a number greater than one so either both quantities are greater than one or one is greater and one is not. The second possibility can be excluded since when one quantity is less than one , taking away one from it will give us a negative quantity. Sticking with the first possibility , both are greater than one , we have , x−1>1 x−1>1 y−1>1 y−1>1 which on adding provides the required inequality. Upvote · 9 8 Sponsored by US Auto Insurance Now These companies are overcharging you for auto insurance! Say goodbye to high car insurance rates if you live in these ZIPs. Learn More 99 16 Gaurav Kumar Former Mathematics Learner · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 536 answers and 1.7M answer views ·5y Related If x<y<0, is y/x<1 always true? Upvote · 9 3 9 1 9 1 Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views ·2y Related If 0<x,y, z<1 Show that x(1-y) +y(1-z) +z(1-x) <1? How can I proof that Start from (1−x)(1−y)(1−z)=1−x−y−z+x y+y z+z x−x y z(1−x)(1−y)(1−z)=1−x−y−z+x y+y z+z x−x y z Thus you have 1−x−y−z+x y+y z+z x−x y z>0 1−x−y−z+x y+y z+z x−x y z>0 that you can rewrite as 1−x(1−y)−y(1−z)−z(1−x)−x y z>0 1−x(1−y)−y(1−z)−z(1−x)−x y z>0 and so x(1−y)+y(1−z)+z(1−x)<1−x y z x(1−y)+y(1−z)+z(1−x)<1−x y z Since x y z>0,x y z>0, we get 1−x y z<1 1−x y z<1 and you’re done. Upvote · 9 6 Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Learn More 999 116 Sohel Zibara Studied at Doctor of Philosophy Degrees (Graduated 2000) · Author has 5.1K answers and 2.6M answer views ·2y Related If 0<x,y, z<1 Show that x(1-y) +y(1-z) +z(1-x) <1? How can I proof that Hoping for the best but expecting the worst, the worst being quora collapsing the answer. Continue Reading Hoping for the best but expecting the worst, the worst being quora collapsing the answer. Upvote · 9 4 9 3 Jonathan Devor PhD in Astronomy, Harvard University (Graduated 2008) · Author has 3.7K answers and 19.4M answer views ·7y Related How do I prove this theorem: if a>0, then 1/a>0? I think the simplest approach is doing a proof by contradiction. Given a>0 a>0, let’s suppose that 1 a≤0.1 a≤0. If that were true then their multiplication must be: 1=a⋅1 a≤0 1=a⋅1 a≤0. We get the left-hand equality since we’re multiplying a number with its inverse. And we get the right-hand inequality since multiplying any number by a positive number (a>0 a>0), doesn’t change its sign. However this inequality is clearly false (1≰0 1≰0). So our proposition (1 a≤0 1 a≤0) must also be false, and in fact 1 a>0 1 a>0. QED Upvote · 99 27 9 1 Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views ·4y Related How can I prove that 3^x > x^3, where x>=0? You can’t, because it ain’t true. The inequality is certainly satisfied for x=0 x=0, so let’s assume x>0 x>0. Then the inequality is the same as x log 3>3 log x x log⁡3>3 log⁡x (natural logarithm) and we can study the function f(x)=x log 3−3 log x f(x)=x log⁡3−3 log⁡x The limits at 0 0 and ∞∞ are ∞∞. Moreover f′(x)=log 3−3 x f′(x)=log⁡3−3 x so the function has a minimum at 3/log 3 3/log⁡3. Since f(3/log 3)=3−3 log 3 log 3=3 log e log 3 3<0 f(3/log⁡3)=3−3 log⁡3 log⁡3=3 log⁡e log⁡3 3<0 we can say that there’s a whole interval where the function is negative. However, since f(3)=0 f(3)=0, you can deduce that, for x>3 x>3, you have 3 x>x 3 3 x>x 3. Upvote · 9 3 Martin Jansche works in any number system, as long as it's base 10 · Author has 3.7K answers and 3.7M answer views ·Updated 1y Related How can I prove that 1>0? I’m assuming we are working in an ordered field or at least in a non-zero ordered ring. Relevant properties include: 0≠1 0≠1 in any field or non-zero ring. << is a strict total order. In particular we have trichotomy: exactly one of a=b a=b or a<b a<b or b<a b<a is true. << is compatible with addition: If a<b a<b then a+c<b+c a+c<b+c. << is compatible with positive multiplication: If a<b a<b and 0<c 0<c then a c<b c a c<b c. In light of this, reason as follows: Because 0≠1 0≠1 (per 1 above) from trichotomy (2) we get either 1<0 1<0 or 1>0 1>0, but not both. Which one is it? Assume 1<0 1<0. From compatibility of << with addition (3) we conclude 1+(−1)<0+(−1)1+(−1)<0+(−1) Continue Reading I’m assuming we are working in an ordered field or at least in a non-zero ordered ring. Relevant properties include: 0≠1 0≠1 in any field or non-zero ring. << is a strict total order. In particular we have trichotomy: exactly one of a=b a=b or a<b a<b or b<a b<a is true. << is compatible with addition: If a<b a<b then a+c<b+c a+c<b+c. << is compatible with positive multiplication: If a<b a<b and 0<c 0<c then a c<b c a c<b c. In light of this, reason as follows: Because 0≠1 0≠1 (per 1 above) from trichotomy (2) we get either 1<0 1<0 or 1>0 1>0, but not both. Which one is it? Assume 1<0 1<0. From compatibility of << with addition (3) we conclude 1+(−1)<0+(−1)1+(−1)<0+(−1) or in other words 0<−1 0<−1 (using additive group axioms I didn't specifically enumerate). From compatibility of << with multiplication (4, using 0<−1 0<−1 twice) we have 0⋅(−1)<(−1)⋅(−1)0⋅(−1)<(−1)⋅(−1), which means 0<1 0<1 (using ring axioms), but because of trichotomy (2) this cannot happen, since we had assumed 1<0 1<0. Contradiction! Therefore 1≮0 1≮0 and hence 1>0 1>0, which is what was to be proven. Upvote · 9 1 Dewi Morgan Adobe ERT member until 2019 · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 2K answers and 6.6M answer views ·5y Related If x<y<0, is y/x<1 always true? x<y<0 x<y<0 and 0<y<x 0<y<x both give us: both x x and y y are the same sign. x x has a larger magnitude (or “absolute value”) than y y. Given 1), dividing y/x y/x or x/y x/y will always give a positive result as any negatives cancel. Given 2), y/x y/x will always be dividing one number by another with a larger magnitude, so it’ll always be less than one. So given either x<y<0 x<y<0 or 0<y<x 0<y<x, we should be able to say for certain 0<y/x<1 0<y/x<1… unless they’re both infinite, in which case all bets are off and at that point it depends which kind of math you’re talking about and what kinds of infinity. Upvote · 9 2 Related questions How will I prove that (1−x)y(1−x)(1−x y)1−x y<1(1−x)y(1−x)(1−x y)1−x y<1 if x > 0, y > 0 and xy < 1? Why does x y+x y 2+x y 3+⋯=x y−1 x y+x y 2+x y 3+⋯=x y−1? How can I prove that x > 2x - 1 for all x? What is lim(x,y)→(0,1)1+2(y−1)3+x 2(y−1+x 4)x 2+2(y−1)2 lim(x,y)→(0,1)1+2(y−1)3+x 2(y−1+x 4)x 2+2(y−1)2? Can you prove (y−y 1=m(x 1−x))=(y=m x+c)(y−y 1=m(x 1−x))=(y=m x+c)? Why is √-1 valid, while a division by zero is invalid, when neither one truly makes sense? How can I show using direct proof that for all x, 0<x<1, we have have 1/5 < 1/ (3x^2 + 2) < 1/2? How can I prove X>y and Z<0 then xz<yz؟? If xyz = 1, then how can I prove that (1+x+y−1)−1+(1+y+z−1)−1+(1+z+x−1)−1=1(1+x+y−1)−1+(1+y+z−1)−1+(1+z+x−1)−1=1? Let x 1,x 2,x 3,y 1,y 2,y 3≥0 x 1,x 2,x 3,y 1,y 2,y 3≥0 such that x 1+x 2+x 3≥y 1+y 2+y 3,x 1+x 2+x 3≥y 1+y 2+y 3,x 1 x 2+x 2 x 3+x 3 x 1≥y 1 y 2+y 2 y 3+y 3 y 1,x 1 x 2+x 2 x 3+x 3 x 1≥y 1 y 2+y 2 y 3+y 3 y 1,x 1 x 2 x 3≥y 1 y 2 y 3.x 1 x 2 x 3≥y 1 y 2 y 3. How do you prove that √x 1+√x 2+√x 3≥√y 1+√y 2+√y 3?x 1+x 2+x 3≥y 1+y 2+y 3? If xyz=1, and x, y, z > 0 then how can I prove that (1+x+y) ^-1 + (1+x+z) ^-1 + (1+y+z) ^-1 <=1? How do you prove (x y+y x)(1 x+1 y)≥4(x y+y x)(1 x+1 y)≥4? How can you prove that if x^2=1 then x's either -1 or 1? How do you prove that x^2+y^2>x+y for all x,y>0? How do you prove that if x+y<z, then x<z and y<z? Related questions How will I prove that (1−x)y(1−x)(1−x y)1−x y<1(1−x)y(1−x)(1−x y)1−x y<1 if x > 0, y > 0 and xy < 1? Why does x y+x y 2+x y 3+⋯=x y−1 x y+x y 2+x y 3+⋯=x y−1? How can I prove that x > 2x - 1 for all x? What is lim(x,y)→(0,1)1+2(y−1)3+x 2(y−1+x 4)x 2+2(y−1)2 lim(x,y)→(0,1)1+2(y−1)3+x 2(y−1+x 4)x 2+2(y−1)2? Can you prove (y−y 1=m(x 1−x))=(y=m x+c)(y−y 1=m(x 1−x))=(y=m x+c)? Why is √-1 valid, while a division by zero is invalid, when neither one truly makes sense? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
13376	https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node78.html	10.3 Application of the Momentum Equation to an Aircraft Engine Thermodynamics and Propulsion Next:11. Aircraft Engine PerformanceUp:10. Integral Momentum TheoremPrevious:10.2 Application of theContentsIndex 10.3 Application of the Momentum Equation to an Aircraft Engine ===================================================================================================================================================================================================================================================================================================== Consider the jet engine shown in Figure10.4. Figure 10.4: Control volume for application of momentum theorem to an aircraft engine So we have: Everything that relates to flow through the engine is conventionally called thrust. Everything that relates to the flow on the outside of the engine is conventionally call drag. Therefore, gathering only those terms that relate to the fluid that passes through the engine, we have: The thrust is largely composed of the net change in momentum of the air entering and leaving the engine, with a typically small adjustment for the differences in pressure between the inlet and the exit. We could have arrived at the same equation by considering only the streamtube that passes through the engine as shown in Figure10.5. Note that the static pressure along the curved control surfaces is different from ambient pressure due to streamline curvature. Figure 10.5: Alternate control volume for application of momentum theorem to an aircraft engine Next:11. Aircraft Engine PerformanceUp:10. Integral Momentum TheoremPrevious:10.2 Application of theContentsIndex UnifiedTP
13377	https://bestpractice.bmj.com/topics/en-us/112	Skip to main content Skip to search Overview of seizure disorder  Menu  Close Overview  Resources  Log in or subscribe to access all of BMJ Best Practice Last reviewed: 5 Aug 2025 Last updated: 02 Aug 2023 This page compiles our content related to seizure disorder. For further information on diagnosis and treatment, follow the links below to our full BMJ Best Practice topics on the relevant conditions and symptoms. Introduction A seizure is defined as "a transient occurrence of signs and/or symptoms due to abnormal excessive or synchronous neuronal activity in the brain."Fisher RS, Acevedo C, Arzimanoglou A, et al. ILAE official report: a practical clinical definition of epilepsy. Epilepsia. 2014 Apr;55(4):475-82. Epilepsy is considered to be a disease of the brain defined by any of the following conditions:Fisher RS, Acevedo C, Arzimanoglou A, et al. ILAE official report: a practical clinical definition of epilepsy. Epilepsia. 2014 Apr;55(4):475-82. ILAE: EpilepsyDiagnosis.org Opens in new window At least two unprovoked (or reflex) seizures occurring more than 24 hours apart One unprovoked (or reflex) seizure and a probability of further seizures similar to the general recurrence risk (at least 60%) after two unprovoked seizures, occurring over the next 10 years Diagnosis of an epilepsy syndrome. Epileptic seizures may be classified as focal, generalized, or unknown onset, with subcategories of motor, non-motor (absence), with retained or impaired awareness for focal seizures.Fisher RS, Cross JH, French JA, et al. Operational classification of seizure types by the International League Against Epilepsy: position paper of the ILAE Commission for Classification and Terminology. Epilepsia. 2017 Apr;58(4):522-30. Scheffer IE, Berkovic S, Capovilla G, et al. ILAE classification of the epilepsies: position paper of the ILAE Commission for Classification and Terminology. Epilepsia. 2017 Apr;58(4):512-21. Relevant conditions \| Condition \| Description \| --- \| \| Status epilepticus \| go to our full topic on Status epilepticus A life-threatening neurologic condition defined as 5 or more minutes of either continuous seizure activity or repetitive seizures without regaining consciousness.Lowenstein DH, Bleck T, Macdonald RL. It's time to revise the definition of status epilepticus. Epilepsia. 1999 Jan;40(1):120-2. Status epilepticus can equally occur in patients with an established diagnosis of epilepsy or as the first unprovoked seizure in patients with no known history of epilepsy. Generalized convulsive status epilepticus in both its subtle and overt subtypes constitutes the most frequent variant. Recognition is crucial because rapid termination helps prevent serious brain injury, especially in patients with impaired consciousness. \| \| Generalized seizures in adults \| go to our full topic on Generalized seizures in adults Generalized seizures (generalized tonic-clonic seizures [GTCS]) classically involve loss of consciousness and a phasic tonic stiffening of the limbs, followed by repetitive clonic jerking. Most GTCS are self-limiting without intervention. The electroencephalogram (EEG) shows bisynchronous epileptiform activity in both cerebral hemispheres. GTCS occur in many different types of epilepsy. GTCS may also be provoked by an insult in an individual who does not have epilepsy. \| \| Generalized seizures in children \| go to our full topic on Generalized seizures in children Seizures may occur as a stand-alone event or may be recurrent. Key risk factors include genetic predisposition or family history, perinatal asphyxia, metabolic/neurodegenerative disorders, head trauma, or structural abnormalities of the central nervous system.Vestergaard M, Pedersen CB, Sidenius P, et al. The long-term risk of epilepsy after febrile seizures in susceptible subgroups. Am J Epidemiol. 2007 Apr 15;165(8):911-8. Daoud AS, Batieha A, Bashtawi M, et al. Risk factors for childhood epilepsy: a case-control study from Irbid, Jordan. Seizure. 2003 Apr;12(3):171-4. Detailed history is of paramount importance in the diagnosis, as key diagnostic factors lie in the history as opposed to ancillary investigations. \| \| Febrile seizures \| go to our full topic on Febrile seizures Seizures occurring in a febrile child, between the ages of 6 and 60 months, without evidence of intracranial infection, metabolic disturbance, or history of afebrile seizures. Febrile seizures are common in childhood. The cumulative incidence is estimated to be between 2% and 5% in the US and Western Europe, between 6% and 9% in Japan, and 14% in India and Guam.Patel N, Ram D, Swiderska N, et al. Febrile seizures. BMJ. 2015 Aug 18;351:h4240. Most febrile seizures resolve spontaneously and do not require acute or long-term anticonvulsant treatment. \| \| Absence seizures \| go to our full topic on Absence seizures Typically characterized by abrupt cessation of activity and responsiveness, lasting between 5 to 10 seconds, with minimal associated movements and no aura/postictal state. Typical absence seizures may be precipitated by hyperventilation and photic stimulation. EEG is the definitive test. Most typical absence seizures are medically responsive, and childhood absence epilepsy (CAE) tends to remit by adulthood. Atypical absence seizures have a less distinct beginning and end, and are not usually precipitated by hyperventilation. Atypical absence seizures tend to be medically refractory and associated with intellectual disability. \| \| Focal seizures \| go to our full topic on Focal seizures Focal (partial) seizures are the electrical and clinical manifestations of seizures that arise from one portion of the brain. They may occur in the context of an underlying structural brain abnormality, or may be idiopathic. The EEG typically indicates a localized discharge over the area of onset, or regions beyond the initial onset as the abnormal electrical activity propagates. The temporal lobe is the most common area of onset for focal seizures, but they may arise from any lobe. Consciousness is preserved in focal aware seizures, whereas focal impaired awareness seizures feature memory loss for the duration of the clinical event or impaired responsiveness at the time of the event.Fisher RS, Cross JH, French JA, et al. Operational classification of seizure types by the International League Against Epilepsy: position paper of the ILAE Commission for Classification and Terminology. Epilepsia. 2017 Apr;58(4):522-30. Focal seizures may evolve into bilateral tonic-clonic seizures. \| \| Infantile spasms syndrome \| go to our full topic on Infantile spasms syndrome An epilepsy syndrome typically presenting in infancy, with a varying etiology. Spasms may be flexor, extensor, mixed flexor-extensor, symmetrical or asymmetrical, and typically occur in clusters. A hypsarrhythmic pattern on EEG is characteristic, but not universally present at all times. Infantile spasms are often associated with an underlying disorder. Where no underlying cause is detected and there is normal development prior to onset, the prognosis tends to be better.Riikonen R. Infantile spasms: outcome in clinical studies. Pediatr Neurol. 2020 Jul;108:54-64. \| \| Evaluation of syncope \| go to our full topic on Evaluation of syncope Syncope is a sudden and transient loss of consciousness that is associated with a loss of postural tone, and resolves spontaneously and completely without intervention. Differentiating between syncope and epileptic seizures can sometimes be challenging. Twitching and jerking are often seen with vasovagal or cardiac syncope, which can be differentiated from rhythmic jerking of all the limbs in tonic-clonic seizures. Loss of bowel and bladder control, commonly seen with seizures, is rare during syncope. The presence of an aura and unconsciousness for >5 minutes is more typical of a seizure. \| Contributors Authors Editorial Team BMJ Publishing Group Disclosures This overview has been compiled using the information in existing sub-topics. References Our in-house evidence and editorial teams collaborate with international expert contributors and peer reviewers to ensure that we provide access to the most clinically relevant information possible. Key articles Fisher RS, Cross JH, French JA, et al. Operational classification of seizure types by the International League Against Epilepsy: position paper of the ILAE Commission for Classification and Terminology. Epilepsia. 2017 Apr;58(4):522-30. Full text Abstract Scheffer IE, Berkovic S, Capovilla G, et al. ILAE classification of the epilepsies: position paper of the ILAE Commission for Classification and Terminology. Epilepsia. 2017 Apr;58(4):512-21. Full text Abstract Reference articles A full list of sources referenced in this topic is available to users with access to all of BMJ Best Practice. Patient information Epilepsy: questions to ask your doctor More Patient information Log in or subscribe to access all of BMJ Best Practice Use of this content is subject to our disclaimer Log in or subscribe to access all of BMJ Best Practice Log in or subscribe to access all of BMJ Best Practice Log in to access all of BMJ Best Practice person personal subscription or user profile Access through your institution OR SUBSCRIPTION OPTIONS Cookies and privacy We and our 225 partners store and access personal data, like browsing data or unique identifiers, on your device. Selecting I Accept enables tracking technologies to support the purposes shown under we and our partners process data to provide. Selecting Reject All or withdrawing your consent will disable them. If trackers are disabled, some content and ads you see may not be as relevant to you. You can resurface this menu to change your choices or withdraw consent at any time by clicking the Cookie settings link on the bottom of the webpage . Your choices will have effect within our Website. For more details, refer to our Privacy Policy.Cookie policy We and our partners process data to provide: Use precise geolocation data. Actively scan device characteristics for identification. Store and/or access information on a device. Personalised advertising and content, advertising and content measurement, audience research and services development. About Your Privacy Your Privacy Strictly Necessary Cookies Performance Cookies Functional Cookies Targeting Cookies Google & IAB TCF 2 Purposes of Processing Store and/or access information on a device 161 partners can use this purpose Personalised advertising and content, advertising and content measurement, audience research and services development 211 partners can use this purpose Use precise geolocation data 67 partners can use this special feature Actively scan device characteristics for identification 41 partners can use this special feature Ensure security, prevent and detect fraud, and fix errors 128 partners can use this special purpose Deliver and present advertising and content 128 partners can use this special purpose Match and combine data from other data sources 97 partners can use this feature Link different devices 66 partners can use this feature Identify devices based on information transmitted automatically 122 partners can use this feature Save and communicate privacy choices 112 partners can use this special purpose Your Privacy We process your data to deliver content or advertisements and measure the delivery of such content or advertisements to extract insights about our website. We share this information with our partners on the basis of consent and legitimate interest. You may exercise your right to consent or object to a legitimate interest, based on a specific purpose below or at a partner level in the link under each purpose. These choices will be signaled to our vendors participating in the Transparency and Consent Framework. Privacy and cookie policies Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Targeting Cookies These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Google & IAB TCF 2 Purposes of Processing Allowing third-party ad tracking and third-party ad serving through Google and other vendors to occur. Please see more information on Google Ads Store and/or access information on a device 161 partners can use this purpose Cookies, device or similar online identifiers (e.g. login-based identifiers, randomly assigned identifiers, network based identifiers) together with other information (e.g. browser type and information, language, screen size, supported technologies etc.) can be stored or read on your device to recognise it each time it connects to an app or to a website, for one or several of the purposes presented here. Personalised advertising and content, advertising and content measurement, audience research and services development 211 partners can use this purpose Use limited data to select advertising 173 partners can use this purpose Advertising presented to you on this service can be based on limited data, such as the website or app you are using, your non-precise location, your device type or which content you are (or have been) interacting with (for example, to limit the number of times an ad is presented to you). Create profiles for personalised advertising 126 partners can use this purpose Information about your activity on this service (such as forms you submit, content you look at) can be stored and combined with other information about you (for example, information from your previous activity on this service and other websites or apps) or similar users. This is then used to build or improve a profile about you (that might include possible interests and personal aspects). Your profile can be used (also later) to present advertising that appears more relevant based on your possible interests by this and other entities. Use profiles to select personalised advertising 127 partners can use this purpose Advertising presented to you on this service can be based on your advertising profiles, which can reflect your activity on this service or other websites or apps (like the forms you submit, content you look at), possible interests and personal aspects. Create profiles to personalise content 51 partners can use this purpose Information about your activity on this service (for instance, forms you submit, non-advertising content you look at) can be stored and combined with other information about you (such as your previous activity on this service or other websites or apps) or similar users. This is then used to build or improve a profile about you (which might for example include possible interests and personal aspects). Your profile can be used (also later) to present content that appears more relevant based on your possible interests, such as by adapting the order in which content is shown to you, so that it is even easier for you to find content that matches your interests. Use profiles to select personalised content 48 partners can use this purpose Content presented to you on this service can be based on your content personalisation profiles, which can reflect your activity on this or other services (for instance, the forms you submit, content you look at), possible interests and personal aspects. This can for example be used to adapt the order in which content is shown to you, so that it is even easier for you to find (non-advertising) content that matches your interests. Measure advertising performance 196 partners can use this purpose Information regarding which advertising is presented to you and how you interact with it can be used to determine how well an advert has worked for you or other users and whether the goals of the advertising were reached. For instance, whether you saw an ad, whether you clicked on it, whether it led you to buy a product or visit a website, etc. This is very helpful to understand the relevance of advertising campaigns. Measure content performance 84 partners can use this purpose Information regarding which content is presented to you and how you interact with it can be used to determine whether the (non-advertising) content e.g. reached its intended audience and matched your interests. For instance, whether you read an article, watch a video, listen to a podcast or look at a product description, how long you spent on this service and the web pages you visit etc. This is very helpful to understand the relevance of (non-advertising) content that is shown to you. Understand audiences through statistics or combinations of data from different sources 124 partners can use this purpose Reports can be generated based on the combination of data sets (like user profiles, statistics, market research, analytics data) regarding your interactions and those of other users with advertising or (non-advertising) content to identify common characteristics (for instance, to determine which target audiences are more receptive to an ad campaign or to certain contents). Develop and improve services 135 partners can use this purpose Information about your activity on this service, such as your interaction with ads or content, can be very helpful to improve products and services and to build new products and services based on user interactions, the type of audience, etc. This specific purpose does not include the development or improvement of user profiles and identifiers. Use limited data to select content 52 partners can use this purpose Content presented to you on this service can be based on limited data, such as the website or app you are using, your non-precise location, your device type, or which content you are (or have been) interacting with (for example, to limit the number of times a video or an article is presented to you). Use precise geolocation data 67 partners can use this special feature With your acceptance, your precise location (within a radius of less than 500 metres) may be used in support of the purposes explained in this notice.
13378	https://research.cm.utexas.edu/nbauld/lipids.htm	IV Lipids Introduction. Lipids are a diverse class of naturally occurring molecules which have in common essentially only the method of their isolation, which is by extraction of cells or tissues with a nonpolar organic solvent. They are therefore relatively nonpolar themselves. Our discussions will center upon two main sub-groups of lipids: (1) triglycerides and (2) terpenes and steroids. Triglycerides Triglycerides arenaturally occurring tri-esters of 1,2,3-propanetriol (glycerol; also called glycerin). The carboxylic acid portion of the ester is typically derived from a long chain carboxylic acid, such as stearic acid and palmitic acid. The generalized structure of triglycerides is therefore as shown below: q In most cases, the carboxylic acids found in triglycerides have from 10-20 carbons in the chain. q The number of carbons in the chain is virtually always an even number. We shall see why this is a little later. q The R groups may be the same or different. Unsaturated Acids Although the carbon chains in many of the carboxylic acids found in triglycerides are saturated (except for the carbon of the carboxyl function), many unsaturated acids are found, also. In fact there may be one, two, three, or even more C=C bonds in the chain. Two examples of the unsaturated acids which are commonly found in triglycerides are given below: q Note that in most triglycerides, the C=C’s have the thermodynamically less stable cis geometry. q When two or more C=C’s are present, they are non-conjugated. . Fats and Oils. Naturally occurring triglycerides may be either solid or liquid at room temperature. Commonly, solid triglycerides are called fats and liquid triglycerides are called oils. The carboxylic acid moieties involved in forming fats and oils are often referred to as fatty acids (even numbered, C12-C18 mostly). q Triglycerides which have all or most of their acidic moieties saturated are most often solid. q These solid triglycerides or fats are most often found in animal triglycerides. q Triglycerides which have most of their acidic moieties unsaturated are more often liquids. q These are referred to as oils, and they are more commonly found in plant triglycerides. q There are exceptions to these generalities, e.g., some plants have large amounts of saturated triglycerides. q A simple explanation of why unsaturated triglycerides have lower m.p.’s (i.e., are liquid at room temperature) is found in the ease of packing of the respective molecules into a snug crystal lattice. q The all anti saturated alkane linkages fit together nicely and provide close approach and favorable van der Waals attractive forces. q The cis double bonds hinder close approach and diminish the van der Waals interactions (see dotted lines for repulsions which impede close approach of the rest of the chain. They also prevent the formation of extended, linear lattice. q Interestingly, unsaturated triglycerides which have trans double bonds tend to have higher m.p.’s than the ones with cis double bonds, and thus to be solid at room temperature. They, like the saturated analogues, tend to pack densely into a rather linear, extended lattice. Hydrolysis of Triglycerides Since triglycerides are esters, they are readily hydrolyzed via either acid catalysis or base promotion. Base promoted hydrolysis is of special importance and will be considered exclusively here. q When 3 moles of hydroxide ion are used to hydrolyse a mole of triglyceride in aqueous solution, the products are one mole of glycerol and three moles of sodium salts of the carboxylic acids which were present in bound form in the triglyceride. q Acidic aqueous workup gives the carboxylic acids, themselves. These naturally occurring acids are often called “fatty acids”. q Since fats and oils are available in abundance, this hydrolysis can be used to produce glycerol and either the sodium salts of carboxylic acids or the fatty acids, themselves. q The sodium salts are especially useful because they constitute “soap”. Soaps Soaps exert their cleansing action in aqueous solution because of the formation of fundamental particles called micelles. q An illustration of the structure of a soap micelle is shown below. Note that the interior of the micelle is nonpolar because it contains all of the nonpolar hydrocarbon “tails” of the soap. q The exterior of the micelle is highly polar because it contains all of the ions: the negative carboxylate ions and the sodium ions. These are nicely solvated and stabilized by the surrounding solvent molecules. The nonpolar “tails” are protected from the water, which which they would not interact favorably. q The hydrocarbon tails interact with each other favorably by van der Waals attractions. q The result is that nonpolar substances like grease are able to enter the interior of the micelle and be favorably solvated by the nonpolar tails. In this way, substances which would not dissolve in water at all, are able to be included inside the aqueous phase. q Note that micelles are not considered to be dissolved in the ordinary sense, because they are rather large particles, but neither are they considered to be suspended particles. They are intermediate species. Typically, micelles might contain 50-200 molecules in a roughly spherical shape. q It turns out, interestingly, that the length of the hydrocarbon tails which are present in naturally occurring fatty acids, are just right for forming stable micelles, which are essential to cleansing action. q If the fatty acid salt has less than 12 carbons, the van der Waals attractions between the tails is not large enough to afford a stable interior for the micelle. If it is more than about 18 carbons, the van der Waals interactions are too strong, and they result in precipitation of the salt, with the tails interacting more strongly in the solid, insoluble lattice. q Potassium salts can also be used. They tend to be softer solids (or even liquids) than sodium salts. Detergents q The basic principle behind the cleansing action of soap is the ability to form micelles with a nonpolar interior. q This merely requires molecules which have a highly polar part and a nonpolar part.Of course, the nonpolar part must have a length and shape which will permit the formation of a stable micelle. q For example, the ionic part could be an anion other than a carboxylate anion. Synthetic cleansers, known as detergents, have been synthesized which use the sulfonate anion as the ionic part. q By doing this, it has been found that sulfonate based detergents are compatible with “hard water” (i.e., water which contains dissolved calcium and magnesium ions), whereas soap is not. q In the presence of either of these ions, soap exchanges its counterion and precipitates out of solution as the insoluble calcium or magnesium salt of the fatty acid. However, the sulfonate salts of calcium and magnesium are soluble, so these ions do not interfere with micelle formation. . q See if you can devise an efficient commercial synthesis of the detergent shown above, starting with benzene and using the typical electrophilic aromatic substitution reactions which we studied earlier in the semester. [How do you install an alkyl group? A sulfonic acid moiety? Which should go first, alkylation or sulfonation?] Terpenes and Steroids Terpenes are naturally occurring organic compounds which are ultimately formed from acetyl coenzme Aviaisopentenyl pyrophosphate, the structure of which is shown below. q Since isopentenyl pyrophosphate is the ultimate source of all the carbon atoms in terpenes, they typically have a total number of carbon atoms which is divisible by five. qThe simplest terpenes (we can call them monoterpenes), therefore have 10 carbon atoms. The next higher series have 15 carbons . These are called sesquiterpenes ( these have 1.5 terpene units). The next groups has 20 carbons and are called diterpenes. Terpenes which have 30 carbons are called triterpenes. q The mechanism for the formation of the basic terpene building block is shown below. The starting point is isopentenyl pyrophosphate. A molecule of this reactant is first isomerized (the disubstituted double bond is isomerized into a more stable trisubstituted double bond; detailed mechanism not given here). q At this point the pyrophosphate moiety, which is a good leaving group, is allylic and reactive. In an S N substitution, the double bond of a molecule of isopentenylpyrophosphate acts as a nucleophile (at its unhindered terminal carbon) to displace the pyrophosphate moiety from the allylic position, and generate a carbon-carbon bond which links the two molecules, forming a ten carbon pyrophosphate. q The 10 carbon molecule is called geranyl pyrophosphate. This molecule is considered to be the key intermediate for the formation of all monoterpenes, and by further reaction with another molecule of isopentenylpyrophosphate, as a precursor for forming sequiterpenes and diterpenes. We will discuss the mechanism of formation of triterpenes further along. q We might note that the pyrophosphate anion (P2 O 6-4 ) is generally considered to be nature’s choice of leaving group. THE ISOMERIZATION STEP THE CARB0N-CARBON BOND FORMING STEPS q As was noted above, geranyl pyrophosphate is considered to be the precursor for all monoterpenes. An example of the formation of limonene, an essential oil present in lemons, is shown below. q It was noted above, also, that the C15 pyrophosphate, farnesyl pyrophosphate,is the direct precursor of all sesquiterpenes. q Incidentally, the triterpenes, which are closely related to the steroids, are not formed by continuing to add isopentenyl pyrophosphate molecules one at a time. Instead, two farnesyl pyrophosphate molecules are coupled together to give a C30 molecule which is the precursor of triterpenes and steroids. This molecule is called squalene. q The structure of squalene is given below, but the biochemical mechanism for its formation is not covered here. STEROIDS The polyunsaturated molecule squalene undergoes a unique and impressive cyclization reaction which leads to polyclic triterpenes and steroids. We will not examine this mechanism in detail, although it has been elegantly worked out, especially by Professor E.J. Corey (Harvard), but it is a carbocation mechanism not unlike we have seen for the formation of limonene, except that several rings are formed consecutively. q Steroids are naturally occurring molecules which typically have the tetracyclic polycyclic skeleton shown below. It may be noted that cholesterol and other steroids do not have 30 carbon atoms; they have lost a few, but they are derived biochemically from the triterpenes. Also, steroids have additional functionalities which triterpenes lack. Cholesterol A conformational structure which better reveals the specific conformation and configuration of cholesterol is also given below:
13379	https://www.youtube.com/watch?v=zSpRgEbyiO4	Interval Notation to Set-Builder Notation \| Pre-Calculus The Organic Chemistry Tutor 9880000 subscribers 2003 likes Description 151955 views Posted: 16 Sep 2022 This Pre-Calculus video tutorial explains how to convert from interval notation to set-builder notation. Precalculus - Free Formula Sheets: 73 comments Transcript: in this video we're going to talk about how to convert a mathematical expression from interval notation to set builder notation so let's start with a graph the interval on a number line and express it in set builder notation so here's the number line here's negative two this is three now we have brackets so what that means is we're going to have a closed circle at negative 2 and 3 and x is going to be between these two values so x is less than or equal to three it includes three but it's also greater than or equal to negative two and this is how we can put it in set builder notation so this is the answer on a number line and this is the answer in set builder notation so that's it for part a now let's move on to part b so it's between negative three and five now we have a parenthesis at negative three so i'm going to put an open circle there because it does not include negative three but we have a bracket at five so we need a closed circle at five you could also do it this way too sometimes you'll see a bracket here and a parenthesis or parenthesis right there as well so you can do it like this or some textbooks may graph it this way a parenthesis at negative 3 and a bracket f5 basically exactly what you see here so you can graph it on a number line any one of those two ways now x is less than or equal to five it includes five it does not include negative three so it's simply greater than but not equal to negative three and that's how we can put it in set builder notation so this is the answer in set builder notation now let's move on to part c by the way if you want to pause the video and try part c have a go for it so let's start with the number line we're going to start at 2 and this is going to go to positive infinity which is all the way to the right negative infinity is all the way to the left now we have a bracket at 2 so i'm going to put a closed circle and it goes to infinity so there's no stopping point so we're just going to have an arrow here so we could simply write that x is equal to or greater than two it's to the right of two and so that's all we need to write to express this interval in set build notation therefore this is the answer right here well and this too so that's how we can graph it on a number line and that's how we can put it in set builder notation let's work on some more examples try this one let's say we have the interval negative infinity to negative 4 within a set of parentheses go ahead and graph it on a number line and express that interval in set builder notation so we have negative infinity to the left and we're going to stop at negative four now we have an open circle at negative four and so we're going to shade towards negative infinity that's we could say that x is less than but not equal to negative four and that's how we can express it in set build notation so this is the final answer here as you can see it's not that difficult to convert an expression from interval notation to set build notation with a little practice you can definitely do it now let's try a harder example so let's say we have the interval from negative 2 to 1 and then union this is going to start from 3 and go to infinity so this time we have two intervals go ahead and graph it on a number line and then express it in set builder notation so let's create a bigger interval we have negative infinity on the left positive infinity on the right we have the numbers negative 2 1 and 3 in increasing order now starting from the left and moving to the right we have a closed circle at negative two and an open circle at one so x is going to be between those two values then we're gonna pick up at three going from left to right so we have an open circle at three and we're gonna go towards infinity so let's start with the first interval x is less than one but doesn't include one but it's greater than or equal to negative two so it's between negative two and one or let's look at the other interval x is greater than but not equal to three so now let's write it in set build notation so this is the first part of my answer this is the answer on a number line and this is our answer expressed in set builder notation now for the sake of practice let's go ahead and work on a similar problem so this is going to be part f so we're going to start from negative infinity and we're going to go to negative 5. we're going to close this with a bracket connect it with another interval we're going to start back up at negative 2 and we're going to end at 4. so go ahead and try this example problem graph it on a number line and then express the mathematical expression in set builder notation so we have negative infinity on the left positive infinity on the right going from left to right the first number is negative five the next one is negative two and the next important number is four so let's start with this so we have a closed circle at negative five and it goes towards negative infinity so it's between negative infinity and negative five then we pick up again at negative two with an open circle and we have an open circle at four so that's how we can graph it on a number line now let's focus on writing it in set build notation so for the first interval we could say that x is less than or equal to negative five it includes negative five now let's connect it with the second interval using an or statement now x is less than four it doesn't include four but it's just less than four but it's greater than negative two it doesn't include negative two since we have an open circle and now let's complete up answer by putting it in set build notation and that's basically it so now you know how to convert a mathematical expression from interval notation to set builder notation
13380	https://www.youtube.com/watch?v=j_fLGDMIgUI	Lec23/Graph Theory/Counting Paths Between Vertices Engineering Mathematics djganit 4460 subscribers 55 likes Description 3802 views Posted: 3 Jun 2024 @djganit In this lecture, we discuss how to determine the number of paths between two different vertices of a graph with example. Transcript: Hello friends I welcome you in lecture number 23 on graph theory in this lecture we will discuss that how to count the number of paths between two vertices in a given graph suppose we are given a graph G in which we are we are having this type of listing of vertices V1 V2 VN and suppose we are interested in the total number of parts from vertex say VI I to v j so our problem is how to determine the exact number of paths from vertex VI to vertex VJ in a given graph G so for that we can use adjacency Matrix of a graph we have discussed about adjacency Matrix in previous lecture in lecture number 22 so if you have not seen that lecture first watch that lecture on adjacency Matrix so suppose I am interested in the number of paths from vertex say VI to vertex VJ so for that suppose the paths I am interested R of length R suppose we are interested in the number of Paths of length R from vertex VI to vertex VJ so for counting those number of paths we have to find out the adjacency Matrix of given graph and if we are interested in the number of Paths of length R then we have to find out earth power of adjacency Matrix and if we are interested in the number of paths from vertex VI to vertex VJ then we have to identify the entry in I row and jth column of this Matrix so we have to decide the I entry of AR Power of adjacency Matrix and that entry will give us the total number of Paths of length R from vertex VI to vertex VJ for example if I am interested in the total number of paths from vertex V3 to vertex V5 and suppose we are interested in the number of Paths of length say 4 then first we have to find out fourth power of adjacency Matrix and because here we are interested in the number of paths from vertex V3 to vertex V5 we have to find out the 35th entry of this Matrix that is entry in third row and fifth column that entry will give us the total number of Paths of length four from vertex V3 to vertex V5 so that is a theorem which provides this result that I have written here and uh let capital G be a graph with adjacency Matrix denoted by capital A with respect to the ordering V1 V2 V b n and this result is applicable for directed as well as undirected graphs for simple graphs and Loops are also allowed so the number of different Paths of length are from vertex VI to vertex VJ equals the I entry of a to the power R here R is a natural number or non- negative integer so to understand this theorem I have taken one example we are not gr for the proof of this result uh we will remember this result that earth power of adjacency Matrix will help us to determine the number of different paths from vertex VI to vertex VJ and that is given by I entry of a to the power R I have taken this example here I have taken this undirected graph and I have to find out that how many paths of length four are there from vertex a to vertex D in the graph G which is shown in this figure so I am interested in the number of Paths of length R = 4 from vertex a to vertex D so first I will decide what is the adjacency Matrix of this graph and uh for that I will list my vertices in particular order suppose I am using this listing of vertices to determine the adjacency Matrix and after finding adjacency Matrix because I am interested in the paths of length four I have to determine fourth power of adjacency Matrix and then uh if we consider this as vertex 1 2 3 and 4 then I am interested in the path from vertex a to vertex D that is from vertex 1 to 4 therefore I have to find out the entry in first row and fourth column of a to power 4 so first we have obtained The adjacency Matrix for this graph and we have discussed that how to write down the adjacency Matrix for a given graph you have to first write down your vertices in a particular order suppose we are using this order then entries of adjacency Matrix are nothing but suppose we are interested in this entry then this corresponds to vertex a and vertex B so here you have to write down the number of Ages from vertex a to vertex B so here there is only one is from vertex A to B similarly here you have to write down the number of edges from a to itself that is no loop at a here so we write zero then here we write number of edges from vertex a to c there is only one is and there is no is from a to d same pattern you have to follow for each row here we have one a from B to a there is no a from B to itself from B to C also there is no is and there is one is from B to D then from C to a there is one is there is no is from C to B from C to C also there is no a and from C to D there is one a then from D to a no is from D to a from D to B there is one is from D to C also there is one is and there is no a from D to D itself so this is our adjacency Matrix and we are interested in the number of paths from vertex a to d and lengths of those paths are four so I have to determine fourth power of a so for that I will use matrix multiplication so those things I have written here first I have written the adjacency metrix of a given graph by considering this order of vertices first I find out power of second power of a that is a square that is given by a a so I have written Matrix a in the multiplication now we are using simple multiplication of matri which we call as standard matrix multiplication so for that we start with first row of first Matrix we consider first row into First Column that is R1 into C1 so we multiply corresponding entries and add them 0 into 0 then 1 into 1 then 1 into 1 and then 0 into zero so this will give me two so first entry is two here after that we consider R1 into C2 so you can see R1 into C2 that is 0 into 1 which is 0 + 1 into 0 is 0 then 1 into 0 is 0 and then 0 into 1 is also zero so second entry is zero in a similar way I have obtain this second power of Matrix a you can verify your calculations by going in this way okay after that I will consider first row into third column then first row into fourth col column after completing task for first row I will go for second row so I will multiply second row with First Column second row with second column second row with third column and second row with fourth column that will give me second row of a square uh we can see for this particular R2 into C1 so I have to multiply 1 and 0er then that will give me 0 then 0 and 1 that is 0 then again 0 and 1 that is 0 and 1 into 0 that is 0 so this will give me 0 so in a similar way you can verify all these calculations and we obtain second power of a then we are interested in the fourth power of a so I have to decide a to the power 3 that I can write as a s a so this is my a square and I will multiply a square with a that I have written here this is a square and a again I use that standard multiplication R1 into C1 then R1 into C2 C3 and C4 for example if we consider R1 into C1 then 2 into 0 that is 0 then 0 into 1 that is 0 then 0 into 1 again that will 0 2 into 0 is 0 so that is the first entry here then we consider R1 into C2 so 2 into 1 that is 2 + 0 into 0 that is 0 + 0 into 0 again that will give me 0 + 2 into 1 that is 2 and I will have 4 here now going in a similar way I have written third power of a and finally we are interested in the fourth power of a that we write as a cub a so this is our a cub that we multiply with adjacency Matrix and again using standard matrix multiplication I obtain a to the power 4 we can check for first R1 into C1 so 0 into 0 that is 0 + 4 into 1 that is 4 + 4 into 1 that is 4 and + 0 into 0 that is 0 and 4 + 4 is 8 so in this way other entries are obtained so this is fourth power of a and we will stop here because we are interested in the paths of length four from vertex a to vertex D so this is vertex a and this is vertex D so that means we have to decide entry in first row and fourth column of a to the power 4 so I am interested in the 1/4th entry that is 8 here so that is nothing but total number of Paths of length four from vertex a to vertex D so the number of Paths of length four from vertex a to d is the 1/4 entry of a 4 which is 8 hence there are exactly eight paths of length four from vertex a to vertex B so this is very easy and uh comfortable way of determining the number of Paths of length are from one vertex to another vertex in a given graph but here if the number is larger then this process of considering powers of matrices becomes tedious suppose we are interested in the number of Paths of length say eight in a given graph from one vertex to another vertex then in that case we have to obtain eth power of adjacency Matrix but for smaller Powers we can easily find out powers of matrices and we can determine the number of paths between two vertexes in a given graph so sometimes they can ask you to find out the number of paths between two vertices in a given graph so for that remember this result suppose you are interested in the number of Paths of length R from vertex VI to vertex VJ then first you find out adjacency Matrix then find R power of the adjacency Matrix and then determine the entry in I row and jth column of earth power of adjacency Matrix that entry will give us the exact number of paths from vertex VI to vertex VJ so this is all about this session we will continue with graph Theory lectures in upcoming sessions thanks for watching
13381	https://artofproblemsolving.com/wiki/index.php/Polynomial_ring?srsltid=AfmBOoor5zq9FK5FRY6h0thnR2s1zw0gWZH0_LPq9wfCLT5uCrUKCRv2	Art of Problem Solving Polynomial ring - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Polynomial ring Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Polynomial ring Given a ring, the polynomial ring is, informally, "the ring of all polynomials in a commutative with coefficients in ." That is, it is the ring of all sums of the form where is a nonnegative integer that varies from sum to sum. The ring is also an -module. Contents 1 Formal Definition 2 Polynomials and Functions 3 Finitude of Degree 4 Differential operators 5 See also Formal Definition We can rigorously define to be the set of all sequences of elements of with only finitely many terms nonzero: The we call the elements of polynomials (over ). For a polynomial , the terms are called the coefficients of . For example, are polynomials, but is not a polynomial. At this point, our formal definition of a polynomial may seem unrelated to our intuitive notion of a polynomial. To relate these two concepts, we introduce the some notation. We denote the polynomial by . For instance, we write: Typically, we repress the terms with coefficient and we do not write the coefficient on terms with coefficient . We also do not care about the order in which the terms are written, and indeed often list them in descending order of power. So we would write: We can now define addition and multiplication in in the canonical way: It is now a simple matter to verify that indeed constitutes a ring under these operations, and that it is commutative when is commutative. This ring has additive identity and multiplicative identity . The ring can be thought of as a subring of via the embedding . For a nonzero polynomial , the greatest integer such that is called the degree of . It is often denoted . By convention, the degree of the zero polynomial (i.e., of the polynomial ) is either undefined, or , or depending on the author. Polynomials and Functions Polynomials are not functions. The symbol does not represent a variable (in the usual sense of this word), but rather a commutative indeterminate, that is, a formal symbol that commutes with the elements of and whose powers are independent of each other over . However, polynomials are associated with functions, called polynomial functions. This is a historically important association: originally, the two concepts were almost inseperable. Indeed, polynomial functions were almost certainly the first functions studied. The concept of "function" was not articulated until the 12th to 14th centuries. By Euler's time, "functions" were explicit rules of association built from elementary expressions, though Euler himself generalized the concept to what we now call continuous functions. This began a long debate over how "function" should be defined that did not resolve until the 20th century, when the modern, abstract definition of "function" became standard. (Although the classical concept of a function as a "deterministic rule to compute an output based on an input" has survived in constructive mathematics and functional programming!) The history of the concept of polynomial is more obscure, but they were almost certainly not divorced from their function roots until the beginnings of modern algebra in the 19th century. Specifically, each element in is associated with a function mapping into itself; this function is evaluated at a value by replacing the symbol with the element in . More, formally, we can prove by induction on the degree of the elements of that for any and any , there is a unique element of that is equivalent to modulo . This unique element is sometimes denoted . Thus we may associate each element with the mapping of into itself. (Alternatively, we can associate with each element a homomorphism of into that is the composition of the canonical homomorphism of into and the canonical homomorphism of into .) It is important to note that although each polynomial in is associated with a function mapping into itself, it is not always possible to uniquely reconstruct from this function. In particular, if is finite, then the set of functions mapping into itself is finite, whereas is infinite (unless is a trivial ring), so some functions must be associated with infinitely many different polynomials. (In fact, it follows from the theory of cosets, applied to the additive groups involved, that every function that is associated with a polynomial must be associated with infinitely many polynomials.) For example, if is the ring of integers modulo , for a prime, then Fermat's Little Theorem states that the polynomials and are associated with the same functions mapping into itself. Nevertheless, in many infinite rings (such as the ring of integers), this association of polynomials with functions is unique. In such contexts, the polynomials are often identified with their functions, by abuse of language. The association of polynomials with functions is an important one: polynomials were first studied as polynomial functions, and indeed it was not until recently that functions gained their modern definition, quite divorced from polynomials. There is yet another reason why polynomials should not be regarded as the functions they represent. Namely, if is a commutative ring, then a polynomial can be evaluated not only at an element of , but also at an element of any -algebra (by replacing every appearance of by in ). For instance, can be applied to any square matrix with entries in , or to any other polynomial over (this results in the composite of the two polynomials), or to a linear operator on an -module. This is in contrast to actual functions, which come with a predetermined domain and are not defined outside of it. Notice that the commonly used notation for a polynomial is a particular case of evaluation: When one evaluates a polynomial at the indeterminate , one gets back (because replacing all 's by in does not change the polynomial). Thus, (and not just by convention). There is a yet more general context in which polynomials are defined. Namely, one can define whenever is an additive abelian group (not necessarily a ring). Then will be just an additive group, not a ring; nevertheless such constructions are occasionally of use (see, e.g., the definition of a [loop algebra]). The usefulness mainly stems from the fact that if is an -module for some commutative ring , then becomes an -module. Here, again, trying to regard polynomials as functions leaves one hopelessly lost. The above discussion was concerned with univariate polynomials (i.e., polynomials in one variable). One can define polynomials in multiple (even infinitely many) variables. The definition is similar to the above, although it requires some more technical bookkeeping, at least if one wants to keep track of variable names. There is a way to avoid some of the technicalities for finitely many variables by an inductive construction (i.e., defining a polynomial in variables to be a polynomial in variable over a polynomial ring in variables), although here again some care has to be taken (when you define a polynomial in two variables, you do not want to call both of them ). Finitude of Degree "Polynomials of infinite degree" are properly called formal power series. The set of formal power series over a ring constitutes a ring, denoted , of which the ring of polynomials is a subring. In general, formal power series are not associated with mappings of into itself, as infinitely iterated addition is not generally well-defined unless the sum converges. Differential operators Given a commutative ring , one can define an -linear map as follows: This operator is called the (formal) derivative (with respect to ), and behaves much like the derivative of a function in analysis, and in fact commutes with the natural map from polynomials to polynomial functions (although, once again, polynomials are not per-se functions). For example, the Leibniz rule and the chain rule hold for any two polynomials and . Unlike the derivative in analysis, the formal derivative does not rely on any limits or topology (in particular, can be any commutative ring, not necessarily or ), although it does have a property mimicking the "difference quotient" definition of the analytic derivative: If , then the polynomial is divisible by , and evaluating the quotient at (that is, substituting for ) yields precisely . (The second variable here is the analogue of the infamous in analysis, but here we need not take any limits.) See also Polynomial Formal power series Ring Retrieved from " Categories: Ring theory Abstract algebra Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
13382	https://www.econgraphs.org/textbooks/intermediate_micro/math/optimization/unconstrained	Unconstrained Optimization - EconGraphs BETA _Note: This work is under development and has not yet been professionally edited. If you catch a typo or error, or just have a suggestion, please submit a note here. Thanks!_ Appendix B / Optimization B.2 Unconstrained Optimization Unconstrained optimization means finding the global maximum or minimum of a function over its entire domain. Critical points In the case of a continuous, smooth function (one which is both continuous and continuously differentiable), a critical point — that is, a local maximum or minimum — occurs at a point where the function is “flat”. For a univariate function y=f(x)y = f(x)y=f(x), this occurs where the derivative d y/d x dy/dx d y/d x is equal to zero: y y y x x x f(x)f(x)f(x) 0.66 0.66 0.6 6 f′(x)=9.08 f^\prime(x) = 9.08 f′(x)=9.0 8 d y d x\frac{dy}{dx}d x d y x x x f′(x)f^\prime(x)f′(x) 0.66 0.66 0.6 6 9.08 9.08 9.0 8 DERIVATIVE OF A UNIVARIATE FUNCTION At each of the critical points, the derivative of the function is equal to zero. x=x = x= f(x)=24 x−6 x 2−8 x 3+3 x 4=11.50\color{#9467bd}{f(x) = 24x-6x^2-8x^3+3x^4 = 11.50}f(x)=2 4 x−6 x 2−8 x 3+3 x 4=1 1.5 0 f′(x)=24−12 x−24 x 2+12 x 3=9.08\color{#1f77b4}{f^\prime(x) = 24 - 12x - 24x^2 + 12x^3 = 9.08}f′(x)=2 4−1 2 x−2 4 x 2+1 2 x 3=9.0 8 See interactive graph online here. Global maxima where the derivative is zero It’s clear from the above graph that just setting d y/d x=0 dy/dx = 0 d y/d x=0 and solving for x x x does not necessarily find you a global maximum or minimum. However, there are special cases in which it does. For example, consider the function y=f(x)=16+8 x−2 x 2 y = f(x) = 16 + 8x - 2x^2 y=f(x)=1 6+8 x−2 x 2 The derivative of this is d y d x=f′(x)=8−4 x{dy \over dx} = f^\prime(x) = 8 - 4x d x d y=f′(x)=8−4 x This has is equal to 0 at x=2 x = 2 x=2: y y y x x x f(x)f(x)f(x) 1.00 1.00 1.0 0 f′(x)=4.00 f^\prime(x) = 4.00 f′(x)=4.0 0 d y d x\frac{dy}{dx}d x d y x x x f′(x)f^\prime(x)f′(x) 1.00 1.00 1.0 0 4.00 4.00 4.0 0 UNCONSTRAINED MAXIMUM OF A UNIVARIATE FUNCTION At the maximum of this function, its derivative is equal to zero. Everywhere to the left of the maximum, the derivative is positive; to the right, the derivative is negative. x=x = x= f(x)=16+8 x−2 x 2=22.00\color{#9467bd}{f(x) = 16+8x-2x^2 = 22.00}f(x)=1 6+8 x−2 x 2=2 2.0 0 f′(x)=8−4 x=4.00\color{#1f77b4}{f^\prime(x) = 8-4x = 4.00}f′(x)=8−4 x=4.0 0 See interactive graph online here. Note that in this case, the derivative starts out positive, ends up negative, and is continuously and monotonically decreasing: that is, f′′(x)<0 f^{\prime \prime}(x) < 0 f′′(x)<0. In such a case, the function is increasing for low values of x x x and eventually decreases; so at some point it must reach a maximum, and that maximum must be a global maximum. Unconstrained maxima for multivariable functions With a multivariable function, critical points occur when all partial derivatives are zero. As with a univariate function, this is a “flat” point on the function, only now it’s the flat in both the x x x and y y y directions.is For example, we saw previously that the function y=f(x 1,x 2)=8 x 1−2 x 1 2+8 x 2−x 2 2 y = f(x_1,x_2) = 8x_1 - 2x_1^2 + 8x_2 - x_2^2 y=f(x 1,x 2)=8 x 1−2 x 1 2+8 x 2−x 2 2 had a maximum at (2,4)(2,4)(2,4). The gradient of this function, ∇f\nabla f∇f, is the vector of its partial derivatives: ∇f(x 1,x 2)=[∂f(x 1,x 2)∂x 1∂f(x 1,x 2)∂x 2]=[8−4 x 1 8−2 x 2]\nabla f(x_1,x_2) = \left[\begin{matrix}{\partial f(x_1,x_2) \over \partial x_1} \ \ {\partial f(x_1,x_2) \over \partial x_2}\end{matrix}\right] = \left[\begin{matrix}8 - 4x_1 \ \ 8 - 2x_2\end{matrix}\right]∇f(x 1,x 2)=⎣⎢⎢⎡∂x 1∂f(x 1,x 2)∂x 2∂f(x 1,x 2)⎦⎥⎥⎤=⎣⎢⎡8−4 x 18−2 x 2⎦⎥⎤ This can be interpreted as the slope of a plane tangent to the function at the point (x 1,x 2)(x_1,x_2)(x 1,x 2). Try changing x 1 x_1 x 1 and x 2 x_2 x 2 in the diagram below to see how this works, in particular at the maximum (2,4)(2,4)(2,4): ∂y∂x 1{\partial y \over \partial x_1}∂x 1∂y x 1 x_1 x 1 f(x 1∣x 2)f(x_1 \| x_2)f(x 1∣x 2) 1.00 1.00 1.0 0 GRADIENT OF A MULTIVARIABLE FUNCTION x 1=x_1 = x 1= x 2=x_2 = x 2= Show f(x 1∣x 2)\text{Show $f(x_1 \| x_2)$}Show f(x 1∣x 2) Show f(x 2∣x 1)\text{Show $f(x_2 \| x_1)$}Show f(x 2∣x 1) ∂f(x 1,x 2)∂x 1=4.00∂f(x 1,x 2)∂x 2=6.00\begin{aligned}{\partial f(x_1,x_2) \over \partial x_1} &= 4.00\ {\partial f(x_1,x_2) \over \partial x_2} &= 6.00\end{aligned}∂x 1∂f(x 1,x 2)∂x 2∂f(x 1,x 2)=4.0 0=6.0 0 See interactive graph online here. At the maximum of the function at (2,4)(2,4)(2,4), ∇f(x 1,x 2)=[8−4×2 8−2×4]=[0 0]\nabla f(x_1,x_2) = \left[\begin{matrix}8 - 4 \times 2 \ 8 - 2 \times 4 \end{matrix}\right] =\left[\begin{matrix}0 \ 0 \end{matrix}\right]∇f(x 1,x 2)=[8−4×2 8−2×4]=[0 0] In other words, at the maximum of the function, both partial derivatives are zero, so the gradient is flat. Previous: Local and Global Maxima and Minima Next: Constrained Optimization and the Lagrange Method Copyright (c) Christopher Makler / econgraphs.org
13383	https://www.biochemjournal.com/archives/2024/vol8issue4/PartF/8-3-90-541.pdf	~ 438 ~ ISSN Print: 2617-4693 ISSN Online: 2617-4707 IJABR 2024; 8(4): 438-442 www.biochemjournal.com Received: 16-01-2024 Accepted: 20-02-2024 K Manoj Kumar Pandit Jawaharlal Nehru College of Agriculture and Research Institute, Karaikal, UT of Pondicherry, Puducherry, India S Nadaradjan Pandit Jawaharlal Nehru College of Agriculture and Research Institute, Karaikal, UT of Pondicherry, Puducherry, India A Premkumar Pandit Jawaharlal Nehru College of Agriculture and Research Institute, Karaikal, UT of Pondicherry, Puducherry, India MK Sakthianand Pandit Jawaharlal Nehru College of Agriculture and Research Institute, Karaikal, UT of Pondicherry, Puducherry, India Corresponding Author: K Manoj Kumar Pandit Jawaharlal Nehru College of Agriculture and Research Institute, Karaikal, UT of Pondicherry, Puducherry, India Flood induced adventitious root formation and ethylene accumulation in Solanum species K Manoj Kumar, S Nadaradjan, A Premkumar and MK Sakthianand DOI: Abstract The frequency of extreme events such as droughts and floods has increased as a consequence of climate change. Many crops have not been improved to tolerate soil anoxia and, therefore, floods cause important economic losses. During waterlogging, Solanum species exhibits three distinct responses which are adventitious root production, ethylene production and aerenchyma formation. The development of a new adventitious root system is crucial as it can replace the original roots that succumb to the hypoxic environment. Solanum torvum is a wild Solanum species characterized by the presence of numerous adventitious root along with aerenchyma. In this work, we have analyzed the Solanum species to flooding to determine flood tolerance and to use them as a rootstock for future breeding programme. We have also examined them for other character viz., aerenchyma formation, ethylene, and adventitious roots plants to detect differential adaptations under flooding. Solanum torvum form an abundant adventitious root system faster than other Solanum species, which results in tolerant species accumulating as much biomass as susceptible plants. In addition, several ethylene-induced responses such as epinasty, aerenchyma production, confirms the presence of ethylene during the flooding. Therefore, our results indicate that the rapid formation of a new root system together with ethylene is responsible for faster adaptation to flooding stress in Solanum torvum. Keywords: Ethylene, arenchyma, Solanum, adventitious root Introduction Solanaceous vegetables are nutritionally rich and high valued crops. These vegetables are highly sensitive to climate change and any sudden variation in the weather condition at any growth stage will affect the normal vegetative flowering and reproductive stages and hence the yield of the crop. Changes in climatic conditions possess increase in the occurrence of heavy rains in tropical and subtropical regions leading to flooding of agricultural lands. Currently flooding has become an important global crop production constraint causing significant yield reduction in several crops. Flooding is one among the abiotic stresses causing crop loss due to excessive or uncertainty in rainfall (Oh et al., 2014) . The change in endogenous production of ethylene and nitric oxide are some of the indicators which express the efficiency of water logging in terms of production and formation of adventitious roots and aerenchyma. Flooding is one of the critical abiotic stress factors which has a considerable impact on crop growth, eventually leads to decline in the yield and production of various crop (Normile 2008) . Due to flooding plants does not receive little (hypoxia) and no oxygen (anoxia) leading to poor oxygen level that will affect the respiration and prevent the root growth that in turn reduce the shoot growth and plant productivity of vegetable crops particularly tomato (Solanum species). In tomato ethylene was accumulated during flooding condition causing severe damage to the crop (Drew et al., 1979) . It was indicated that the decrease in yield was attributed to decrease in nitrogen uptake and reduced photosynthesis under waterlogging. The most common morphological response of plants during flowering is the development of adventitious roots which may appear in the hypocotyl region of the plant. Besides the development of adventitious roots, the root aeration was channelized through the formation of aerenchyma for the sustenance of the plant (Zhang et al., 2017) . Tomato is considered to be one of the most sensitive vegetable species to excessive soil moisture. Speculating the problem of flooding condition in future, alternate tool to mitigate stress can achieve through grafting. International Journal of Advanced Biochemistry Research 2024; 8(4): 438-442 ~ 439 ~ International Journal of Advanced Biochemistry Research Grafting has evolved as an promising technique to withstand the global climate change generated by biotic and abiotic stresses. It offers an opportunity to use rootstocks of the specific traits to have its influence over the scions phenotype. Selection of tolerant plants under waterlogging condition is the indicator that can be used as the rootstock for the production of water logging resistant plants. Hence the present study undertaken to screen Solanum species for flooding tolerance through morphological characterization. Materials and Methods Nine Solanum species used for this study were the recommended dose of fertilizer and common package of practice were followed in timely fashion. Seeds of nine Solanum species were sown in portrays and transferred to submergence tank containing red soil. Environmental conditions, watering regime, and imposition of flood treatment were as described by McNamara and Mitchell (1989) . Flood treatment was imposed by placing seedling into submergence tank. Plants were immersed to 3 cm above the soil surface. Control plants were placed within similar containers containing bottom-drainage holes, and were watered to field capacity once daily. Ethylene were estimated for all the species by using gas chromatography and the instrument specialized with flame ionization detector and a 2-mm-diameter (I.d.)×182-cm- long column packed with alumina. chromatography oven temperature were 80 °C for column 120 °C for injector and 150 °C for detector and flow rate of nitrogen carrier gas was 40 ml min-1 By the way samples of 2.5 cm long hypocotyls excised after 48 h of treatment weighed and enclosed in 5 ml vials later the 1 ml gas samples was withdrawn and injected into gas chromatography as described by Mcnamara and Mitchell (1990) . Adventitious root formation was counted on the basis of number of roots produced above soil surface after imposing flooding for 7 days. The no of adventitious roots were recorded at the 7th day of flooding. For histological studies, hypocotyl was sectioned after 7 days of flooding and was fixed using a solution containing formalin, glacial acetic acid and 95% ethanol in the ratio 5:5:90 then it was passed through an ethanol-tertiary butanol dehydration series, and embedded in Paraplast (Sass, 1958) . Sections taken were 10 μm thick and were stained with safranin/ aniline blue. Results and Discussion Gas diffusion into the plant is impeded, leading to low oxygen partial pressure that stimulates biosynthesis of ethylene by increasing ACC (1-aminocyclopropane-1-carboxylic acid) synthase activity and further enriching the stem with ethylene (Sauter, 2013) and also inducing adventitious root and aerenchyma tissue formation. Hormonal balance is an adaptive response to flooding condition by the regulation of increase in ethylene concentration, which interacts with gibberellins (Musgrave et al., 1972) and auxins. In this study based on their performance in flooding and with their morpho physiological, biochemical parameters the most susceptible and the tolerant species were identified and ethylene was determined in those species. Flooding leads to the increase in the concentration of ethylene in the hypocotyl region. Ethylene estimation was done for the selected tolerant and the susceptible species and the amount of ethylene content varied between these species. The Solanum species recorded higher concentration of ethylene S. aculeatissimum (1.65 nl/ g of fresh weight/ hr) followed by S. torvum (1.012 nl/ g of fresh weight/ hr) and S. sisymbriifolium (0.986 nl/ g of fresh weight/ hr) under the flooding condition and the lowest ethylene content was recorded in S. viarum of (0.02 nl/ g of fresh weight/ hr) (Plate 2). The ethylene production in the shoot tissue will be stimulated because of the anoxia condition (Steffens et al., 2011) that will encourage cell separation and development of air spaces in their stem (Khah, et al., 2012) . The acceleration of air spaces formation in the newly produced nodal roots of zea maize has some feature in common with swelling of preformed stems in sunflower induced by ethyphon or flooding. The present study revealed that flooding affected the rate of ethylene evolution from hypocotyls of the different species. The hypocotyls of the flooded Solanum species Solanum aculeatissimum (1.652 nl/ g of fresh weight/ hr) (Fig 3) produced significantly higher ethylene when compared to control. Studies by Bellini et al., 2014 , indicated that the restricted capacity of susceptible species for Arenchyma and adventitious roots development limited by its ability to synthesize ACC or C2H4 during inundation. These studies support for the present findings where significant increase in the number of adventitious roots. In S. torvum than in other Solanum species were the ethylene correlate with adventitious root numbers (Plate.1) (Fig 1). As from the previous study adventitious roots development was not found in control condition and it was significantly higher under flood imposed treatment. The number of adventitious ranged from 17-52 per plant (fig 3). S. torvum recorded significantly higher number of adventitious roots 52 followed by S. aculeatissimum 36. In some of species like S. nigrum, S. macrocarpon, there was no development of adventitious root. The microscopic thin section of the hypocotyl tissue of different species revealed that hypocotyl porosity was increased under flooding condition. The aerenchyma development in terms of porosity was more in Solanum torvum 1(a) 2(a), Solanum macrocarpon, Solanum aculeatissimum and Solanum sisymbriifolium and there was no development of aerenchyma in Solanum nigrum 1(b), 1(b). ~ 440 ~ International Journal of Advanced Biochemistry Research 1(a) Hypocotyl cross-sections of Solanum torvum seedlings control 1 1(b) 7days after flooding 1 Fig 1(a-b): Red circle showing absence of adventitious or. Magnification. 2(a) Hypocotyl cross-sections of Solanum nigrum seedlings control 2(b) 7days after flooding Fig 2(a-b): Showing absence of adventitious root primordia. Magnification. Fig 3: Effect of flooding on ethylene and adventitious root formation ~ 441 ~ International Journal of Advanced Biochemistry Research Plate 1: Adventitious root formation in tolerant species. Plate 2: Ethylene peak observed in gas chromatography of tolerant species. Conclusion Clear evidence is emerging demonstrating that the adventitious root and arenchyma formation is regulated and responds to environmental cues. Because adventitious roots are important for tolerance to flooding. In this study the hormonal interaction and formation of adventitious roots and aerenchyma formation is more evident. The formation of aerenchyma in adventitious roots and in the hypocotyl, region is common in flood tolerant species. It is concluded S. torvum is suitable for mother stock flood susceptible solanaceous vegetables. In further Physiological and molecular analysis in combination with genetic modification are planned to shed further light on the roles of the various signalling pathways. Reference 1. Oh M, Nanjo Y, Komatsu S. Gel-free proteomic analysis of soybean root proteins affected by calcium under flooding stress. Frontiers in Plant Science. 2014;5:559. 2. Normile D. Reinventing rice to feed the world. Science. 2008;321(5887):330-333. 3. Drew M, Jackson M, Giffard S. Ethylene-promoted adventitious rooting and development of cortical air spaces (Aerenchyma) in roots may be adaptive responses to flooding in Zea mays L. Planta. 1979;147(1):83-88. 4. Zhang Q, Huber H, Beljaars SJ, Birnbaum D, De Best S, De Kroon H, et al. Benefits of flooding-induced aquatic adventitious roots depend on the duration of submergence: linking plant performance to root functioning. Annals of Botany. 2017;120(1):171-180. 5. Mcnamara S, Mitchell C. Differential flood stress resistance of two tomato genotypes. Journal of the American Society for Horticultural Science. 1989 Nov 1;114(6):976-980. 6. Mcnamara ST, Mitchell CA. Adaptive stem and adventitious root responses of two tomato genotypes to flooding. Hort Science. 1990;25(1):100-103. 7. Sass JE. Botanical microtechnique. Botanical microtechnique. 3rd ed; c1958. 8. Sauter M. Root responses to flooding. Current Opinion in Plant Biology. 2013;16(3):282-286. 9. Musgrave A, Jackson MB, Ling E. Callitriche stem elongation is controlled by ethylene and gibberellin. Nature New Biology. 1972;238(81):93. 10. Khah EM, Katsoulas N, Tchamitchian M, Kittas C. Effect of grafting on eggplant leaf gas exchanges under Mediterranean greenhouse conditions. International Journal of Plant Production. 2012;5(2):121-134. ~ 442 ~ International Journal of Advanced Biochemistry Research 11. Bellini C, Pacurar DI, Perrone I. Adventitious roots and lateral roots: similarities and differences. Annual Review of Plant Biology. 2014;65:639-666. 12. Steffen W, Persson Å, Deutsch L, Zalasiewicz J, Williams M, Richardson K, et al. The Anthropocene: From global change to planetary stewardship. Ambio. 2011 Nov;40:739-761.
13384	https://www.physicsclassroom.com/class/waves/lesson-0/properties-of-periodic-motion	Physics Tutorial: Properties of Periodic Motion My Account × Custom Search Sort by: Relevance Relevance Date TPC and eLearning What's NEW at TPC? Task Tracker Edit Profile Settings Classes Users Voice Tasks and Classes Webinars and Trainings Subscriptions Subscription Subscription Locator Products and Plans Request a Demo Manual Order Form Subscription Selection Ad Free Account My Cart Read Watch Interact Practice Review Test Teacher-Tools TPC and eLearningWhat's NEW at TPC?Task Tracker Edit Profile Settings Classes Users Voice Tasks and Classes Webinars and Trainings Subscriptions Subscription Locator Products and Plans Request a Demo Manual Order Form Subscription Selection Ad Free Account My Cart Read Watch InteractChemistry Tutorial Measurement and Calculations Matter Elements, Atoms, and Ions Compounds,Names, and Formulae The Modern Atomic Model Chemical Bonding The Mole and its Applications Chemical Reactions Stoichiometry Gases and Gas Laws Solids, Liquids, and Intermolecular Forces Thermochemistry Solutions Kinetics and Equilibrium Acids and Bases Solution Equilibria Reference Physics Tutorial 1-D Kinematics Newton's Laws Vectors - Motion and Forces in Two Dimensions Momentum and Its Conservation Work and Energy Circular Motion and Satellite Motion Thermal Physics Static Electricity Electric Circuits Magnetic Fields and Electromagnetism Vibrations and Waves Sound Waves and Music Light and Color Reflection and Mirrors Refraction and Lenses Video Tutorial Kinematics Newton's Laws Vectors and Projectiles Forces in Two Dimensions Momentum and Collisions Work, Energy, and Power Circular Motion and Gravitation Static Electricity Electric Circuits Vibrations and Waves Sound Waves Light and Color Reflection and Mirrors Refraction and Lenses Multimedia Studios 1-Dimensional Kinematics Newton's Laws Vectors and Projectiles Momentum and Collisions Work and Energy Circular, Satellite, and Rotational Motion Einstein's Theory of Special Relativity Static Electricity Waves, Sound and Light Ray Optics QuickTime Movies Interactives About the Physics Interactives Task Tracker Usage Policy Kinematics Newtons Laws Vectors and Projectiles Forces in 2D Momentum and Collisions Work and Energy Circular Motion and Gravitation Balance and Rotation Fluids Static Electricity Electric Circuits Electromagnetism Waves and Sound Light and Color Reflection and Mirrors Refraction and Lenses Atomic Physics Chemistry Practice Review TestConcept Builders About the Concept Builders Version 2 Pricing For Schools Directions for Version 2 Measurement and Units Relationships and Graphs Kinematics Newton's Laws Vectors and Projectiles Forces in 2-Dimensions Momentum and Collisions Work and Energy Circular and Satellite Motion Rotation and Balance Fluids Static Electricity Electric Circuits Vibrational Motion Waves and Sound Light and Color Reflection and Refraction Chemistry Minds On About MOPs Features Teacher Accounts Pricing For Schools Task Tracker Directions Kinematic Concepts Kinematic Graphing Newton's Laws Vectors and Projectiles Forces in 2-Dimensions Momentum and Collisions Work and Energy Circular and Satellite Motion Static Electricity Electric Circuits Wave Motion Sound and Music Light and Color Reflection and Mirrors Refraction and Lenses Calc Pad About CalcPad Teacher Accounts Pricing For Schools Directions 1D Kinematics Newton's Laws Vectors and Projectiles Vectors and Forces in 2D Momentum and Collisions Work, Energy, and Power Circular Motion and Gravitation Simple Harmonic Motion Rotational Kinematics Rotation and Torque Rotational Dynamics Fluids Static Electricity Electric Fields, Potential, and Capacitance Electric Fields, Potential, and Capacitance with Calculus Electric Circuits Transient RC Circuits Electromagnetism Vibrations and Waves Sound Waves Light Waves Reflection and Mirrors Refraction and Lenses Units and Measurement Mole Conversions Stoichiometry Molarity and Solutions Gas Laws Thermal Chemistry Acids and Bases Kinetics and Equilibrium Solution Equilibria Oxidation-Reduction Nuclear Chemistry Science Reasoning Task Tracker NGSS Alignments 1D-Kinematics Newton's Laws Projectiles Momentum Energy Circular Motion Static Electricity Circuits Magnetism and Electromagnetism Thermal Physics Waves Sound Waves Light Waves Reflection and Mirrors Refraction and Lenses Chemistry The Review Session 1-D Kinematics Newton's Laws of Motion Vectors and Projectiles Forces in Two Dimensions Momentum and Collisions Work and Energy Packet Circular Motion and Gravitation Static Electricity Review Electric Circuits Waves Sound and Music Light and Color Reflection and Mirrors Refraction and Lenses ACT Test Center About the ACT ACT Preparation ACT Tips For Teachers Other Resources Physics Help Graphing Practice Recognizing Forces Vector Direction Vector Addition Teacher-ToolsCurriculum Corner Solutions Guide Solutions Guide Digital Download Usage Policy Solution Guide Corrections Motion in One Dimension Newton's Laws Vectors and Projectiles Forces in Two Dimensions Momentum and Collisions Work, Energy and Power Circular Motion and Gravitation Static Electricity Electric Circuits Wave Basics Sound and Music Light and Color Reflection and Mirrors Refraction and Lenses Chemistry of Matter Measurement and the Metric System Early Atomic Models Names and Formulas The Modern Model of the Atom Chemical Bonding The Mole and Its Applications Chemical Reactions Stoichiometry Gases and Gas Laws Solids, Liquids, and Intermolecular Forces Thermal Chemistry Solutions Kinetics and Equilibrium Lesson Plans About Algebra Based On-Level Physics Honors Physics Conceptual Physics Task Tracker Other Tools Teacher Pres'n Pack Contents Frequently Asked Questions Purchasing the Download The Laboratory About Teacher Guide Using Lab Notebooks Share CP Course Pack Contents Our Flavor of Conceptual Physics Frequently Asked Questions Purchasing the Download Question Bank Contents Purchasing the Digital Download Teacher Toolkits About the Toolkits Position-Velocity-Acceleration Position-Time Graphs Velocity-Time Graphs Free Fall Newton's First Law Newton's Second Law Newton's Third Law Terminal Velocity Vectors Projectile Motion Forces in 2 Dimensions Impulse and Momentum Change Momentum Conservation Work-Energy Fundamentals Work-Energy Relationship Circular Motion Roller Coaster Physics Universal Gravitation Satellite Motion Charge and Charging Coulombs Law Electric Fields Circuit Concepts Series Circuits Parallel Circuits Vibrational Motion Describing-Waves Wave Behavior Toolkit Standing Wave Patterns Sound Waves Resonating Air Columns Wave Model of Light Color Plane Mirrors Curved Mirrors Snells Law Total Internal Reflection Lenses The Photo Gallery 1-D Kinematics Newton's Laws Vectors - Motion and Forces in Two Dimensions Momentum and Its Conservation Work, Energy, and Power Circular Motion and Satellite Motion Thermal Physics Static Electricity Current Electricity Waves Sound Waves and Music Light Waves and Color Reflection and Ray Model of Light Refraction and Ray Model of Light NGSS Corner About the NGSS Corner NGSS Search Force and Motion DCIs - High School Energy DCIs - High School Wave Applications DCIs - High School Force and Motion PEs - High School Energy PEs - High School Wave Applications PEs - High School Crosscutting Concepts The Practices Physics Topics NGSS Corner: Activity List NGSS Corner: Infographics TPC and eLearning What's NEW at TPC? Task Tracker Edit Profile Settings Classes Users Voice Tasks and Classes Webinars and Trainings Subscriptions Subscription Locator Products and Plans Request a Demo Manual Order Form Subscription Selection Ad Free Account My Cart Read Watch Interact Chemistry Tutorial Measurement and Calculations Matter Elements, Atoms, and Ions Compounds,Names, and Formulae The Modern Atomic Model Chemical Bonding The Mole and its Applications Chemical Reactions Stoichiometry Gases and Gas Laws Solids, Liquids, and Intermolecular Forces Thermochemistry Solutions Kinetics and Equilibrium Acids and Bases Solution Equilibria Reference Physics Tutorial 1-D Kinematics Newton's Laws Vectors - Motion and Forces in Two Dimensions Momentum and Its Conservation Work and Energy Circular Motion and Satellite Motion Thermal Physics Static Electricity Electric Circuits Magnetic Fields and Electromagnetism Vibrations and Waves Sound Waves and Music Light and Color Reflection and Mirrors Refraction and Lenses Video Tutorial Kinematics Newton's Laws Vectors and Projectiles Forces in Two Dimensions Momentum and Collisions Work, Energy, and Power Circular Motion and Gravitation Static Electricity Electric Circuits Vibrations and Waves Sound Waves Light and Color Reflection and Mirrors Refraction and Lenses Multimedia Studios 1-Dimensional Kinematics Newton's Laws Vectors and Projectiles Momentum and Collisions Work and Energy Circular, Satellite, and Rotational Motion Einstein's Theory of Special Relativity Static Electricity Waves, Sound and Light Ray Optics QuickTime Movies Interactives About the Physics Interactives Task Tracker Usage Policy Kinematics Newtons Laws Vectors and Projectiles Forces in 2D Momentum and Collisions Work and Energy Circular Motion and Gravitation Balance and Rotation Fluids Static Electricity Electric Circuits Electromagnetism Waves and Sound Light and Color Reflection and Mirrors Refraction and Lenses Atomic Physics Chemistry Practice Review Test Concept Builders About the Concept Builders Version 2 Pricing For Schools Directions for Version 2 Measurement and Units Relationships and Graphs Kinematics Newton's Laws Vectors and Projectiles Forces in 2-Dimensions Momentum and Collisions Work and Energy Circular and Satellite Motion Rotation and Balance Fluids Static Electricity Electric Circuits Vibrational Motion Waves and Sound Light and Color Reflection and Refraction Chemistry Minds On About MOPs Features Teacher Accounts Pricing For Schools Task Tracker Directions Kinematic Concepts Kinematic Graphing Newton's Laws Vectors and Projectiles Forces in 2-Dimensions Momentum and Collisions Work and Energy Circular and Satellite Motion Static Electricity Electric Circuits Wave Motion Sound and Music Light and Color Reflection and Mirrors Refraction and Lenses Calc Pad About CalcPad Teacher Accounts Pricing For Schools Directions 1D Kinematics Newton's Laws Vectors and Projectiles Vectors and Forces in 2D Momentum and Collisions Work, Energy, and Power Circular Motion and Gravitation Simple Harmonic Motion Rotational Kinematics Rotation and Torque Rotational Dynamics Fluids Static Electricity Electric Fields, Potential, and Capacitance Electric Fields, Potential, and Capacitance with Calculus Electric Circuits Transient RC Circuits Electromagnetism Vibrations and Waves Sound Waves Light Waves Reflection and Mirrors Refraction and Lenses Units and Measurement Mole Conversions Stoichiometry Molarity and Solutions Gas Laws Thermal Chemistry Acids and Bases Kinetics and Equilibrium Solution Equilibria Oxidation-Reduction Nuclear Chemistry Science Reasoning Task Tracker NGSS Alignments 1D-Kinematics Newton's Laws Projectiles Momentum Energy Circular Motion Static Electricity Circuits Magnetism and Electromagnetism Thermal Physics Waves Sound Waves Light Waves Reflection and Mirrors Refraction and Lenses Chemistry The Review Session 1-D Kinematics Newton's Laws of Motion Vectors and Projectiles Forces in Two Dimensions Momentum and Collisions Work and Energy Packet Circular Motion and Gravitation Static Electricity Review Electric Circuits Waves Sound and Music Light and Color Reflection and Mirrors Refraction and Lenses ACT Test Center About the ACT ACT Preparation ACT Tips For Teachers Other Resources Physics Help Graphing Practice Recognizing Forces Vector Direction Vector Addition Teacher-Tools Curriculum Corner Solutions Guide Solutions Guide Digital Download Usage Policy Solution Guide Corrections Motion in One Dimension Newton's Laws Vectors and Projectiles Forces in Two Dimensions Momentum and Collisions Work, Energy and Power Circular Motion and Gravitation Static Electricity Electric Circuits Wave Basics Sound and Music Light and Color Reflection and Mirrors Refraction and Lenses Chemistry of Matter Measurement and the Metric System Early Atomic Models Names and Formulas The Modern Model of the Atom Chemical Bonding The Mole and Its Applications Chemical Reactions Stoichiometry Gases and Gas Laws Solids, Liquids, and Intermolecular Forces Thermal Chemistry Solutions Kinetics and Equilibrium Lesson Plans About Algebra Based On-Level Physics Honors Physics Conceptual Physics Task Tracker Other Tools Teacher Pres'n Pack Contents Frequently Asked Questions Purchasing the Download The Laboratory About Teacher Guide Using Lab Notebooks Share CP Course Pack Contents Our Flavor of Conceptual Physics Frequently Asked Questions Purchasing the Download Question Bank Contents Purchasing the Digital Download Teacher Toolkits About the Toolkits Position-Velocity-Acceleration Position-Time Graphs Velocity-Time Graphs Free Fall Newton's First Law Newton's Second Law Newton's Third Law Terminal Velocity Vectors Projectile Motion Forces in 2 Dimensions Impulse and Momentum Change Momentum Conservation Work-Energy Fundamentals Work-Energy Relationship Circular Motion Roller Coaster Physics Universal Gravitation Satellite Motion Charge and Charging Coulombs Law Electric Fields Circuit Concepts Series Circuits Parallel Circuits Vibrational Motion Describing-Waves Wave Behavior Toolkit Standing Wave Patterns Sound Waves Resonating Air Columns Wave Model of Light Color Plane Mirrors Curved Mirrors Snells Law Total Internal Reflection Lenses The Photo Gallery 1-D Kinematics Newton's Laws Vectors - Motion and Forces in Two Dimensions Momentum and Its Conservation Work, Energy, and Power Circular Motion and Satellite Motion Thermal Physics Static Electricity Current Electricity Waves Sound Waves and Music Light Waves and Color Reflection and Ray Model of Light Refraction and Ray Model of Light NGSS Corner About the NGSS Corner NGSS Search Force and Motion DCIs - High School Energy DCIs - High School Wave Applications DCIs - High School Force and Motion PEs - High School Energy PEs - High School Wave Applications PEs - High School Crosscutting Concepts The Practices Physics Topics NGSS Corner: Activity List NGSS Corner: Infographics Student Extras Teacher's Guides The Physics Classroom » Physics Tutorial » Vibrations and Waves » Properties of Periodic Motion Vibrations and Waves - Lesson 0 - Vibrations Properties of Periodic Motion Vibrational Motion Properties of Periodic Motion Pendulum Motion Motion of a Mass on a Spring Getting your Trinity Audio player ready... Hold down the T key for 3 seconds to activate the audio accessibility mode, at which point you can click the K key to pause and resume audio. Useful for the Check Your Understanding and See Answers. A vibrating object is wiggling about a fixed position. Like the mass on a spring in the animation at the right, a vibrating object is moving over the same path over the course of time. Its motion repeats itself over and over again. If it were not for damping, the vibrations would endure forever (or at least until someone catches the mass and brings it to rest). The mass on the spring not only repeats the same motion, it does so in a regular fashion. The time it takes to complete one back and forth cycle is always the same amount of time. If it takes the mass 3.2 seconds for the mass to complete the first back and forth cycle, then it will take 3.2 seconds to complete the seventh back and forth cycle. It's like clockwork. It's so predictable that you could set your watch by it. In Physics, a motion that is regular and repeating is referred to as a periodic motion. Most objects that vibrate do so in a regular and repeated fashion; their vibrations are periodic. (Special thanks to Oleg Alexandrov for the animation of the mass on a spring. It is a public domain acquired from WikiMedia Commons. ) The Sinusoidal Nature of a Vibration Suppose that a motion detector was placed below a vibrating mass on a spring in order to detect the changes in the mass's position over the course of time. And suppose that the data from the motion detector could represent the motion of the mass by a position vs. time plot. The graphic below depicts such a graph. For discussion sake, several points have been labeled on the graph to assist in the follow-up discussion. Before reading on, take a moment to reflect on the type of information that is conveyed by the graph. And take a moment to reflect about what quantities on the graph might be important in understanding the mathematical description of a mass on a spring. If you've taken time to ponder these questions, the following discussion will likely be more meaningful. One obvious characteristic of the graph has to do with its shape. Many students recognize the shape of this graph from experiences in Mathematics class. The graph has the shape of a sine wave. If y = sine(x) is plotted on a graphing calculator, a graph with this same shape would be created. The vertical axis of the above graph represents the position of the mass relative to the motion detector. A position of about 0.60 m cm above the detector represents the resting position of the mass. So the mass is vibrating back and forth about this fixed resting position over the course of time. There is something sinusoidal about the vibration of a mass on a spring. And the same can be said of a pendulum vibrating about a fixed position or of a guitar string or of the air inside of a wind instrument. The position of the mass is a function of the sine of the time. A second obvious characteristic of the graph may be its periodic nature. The motion repeats itself in a regular fashion. Time is being plotted along the horizontal axis; so any measurement taken along this axis is a measurement of the time for something to happen. A full cycle of vibration might be thought of as the movement of the mass from its resting position (A) to its maximum height (B), back down past its resting position (C) to its minimum position (D), and then back to its resting position (E). Using measurements from along the time axis, it is possible to determine the time for one complete cycle. The mass is at position A at a time of 0.0 seconds and completes its cycle when it is at position E at a time of 2.3 seconds. It takes 2.3 seconds to complete the first full cycle of vibration. Now if the motion of this mass is periodic (i.e., regular and repeating), then it should take the same time of 2.3 seconds to complete any full cycle of vibration. The same time-axis measurements can be taken for the sixth full cycle of vibration. In the sixth full cycle, the mass moves from a resting position (U) up to V, back down past W to X and finally back up to its resting position (Y) in the time interval from 11.6 seconds to 13.9 seconds. This represents a time of 2.3 seconds to complete the sixth full cycle of vibration. The two cycle times are identical. Other cycle times are indicated in the table below. By inspection of the table, one can safely conclude that the motion of the mass on a spring is regular and repeating; it is clearly periodic. The small deviation from 2.3 s in the third cyle can be accounted for by the lack of precision in the reading of the graph. CycleLettersTimes at Beginning and End of Cycle (seconds)Cycle Time (seconds) 1st A to E 0.0 sto 2.3 s 2.3 2nd E tp I 2.3 s to 4.6 s 2.3 3rd I to M 4.6 s to 7.0 s 2.4 4th M to Q 7.0 s to 9.3 s 2.3 5th Q to U 9.3 s to 11.6 s 2.3 6th U to Y 11.6 s to 13.9 s 2.3 Students viewing the above graph will often describe the motion of the mass as "slowing down." It might be too early to talk in detail about what slowing down means. We will save the lengthy discussion of the topic for the page later in this lesson devoted to the motion of a mass on a spring. For now, let's simply say that over time, the mass is undergoing changes in its speed in a sinusoidal fashion. That is, the speed of the mass at any given moment in time is a function of the sine of the time. As such, the mass will both speed up and slow down over the course of a single cycle. So to say that the mass is "slowing down" is not entirely accurate since during every cycle there are two short intervals during which it speeds up. (More on this later.) Students who describe the mass as slowing down (and most observant students do describe it this way) are clearly observing something in the graph features that draws out the "slowing down" comment. Before we discuss the feature that triggers the "slowing down" comment, we must re-iterate the conclusion from the previous paragraphs - the time to complete one cycle of vibration is NOT changing. It took 2.3 seconds to complete the first cycle and 2.3 seconds to complete the sixth cycle. Whatever "slowing down" means, we must refute the notion that it means that the cycles are taking longer as the motion continues. This notion is clearly contrary to the data. A third obvious characteristic of the graph is that damping occurs with the mass-spring system. Some energy is being dissipated over the course of time. The extent to which the mass moves above (B, F, J, N, R and V) or below (D, H, L, P, T and X) the resting position (C, E, G, I, etc.) varies over the course of time. In the first full cycle of vibration being shown, the mass moves from its resting position (A) 0.60 m above the motion detector to a high position (B) of 0.99 m cm above the motion detector. This is a total upward displacement of 0.29 m. In the sixth full cycle of vibration that is shown, the mass moves from its resting position (U) 0.60 m above the motion detector to a high position (V) 0.94 m above the motion detector. This is a total upward displacement of 0.24 m cm. The table below summarizes displacement measurements for several other cycles displayed on the graph. CycleLettersMaximum Upward DisplacementMaximum Downward Displacement 1st A to E 0.60 m to 0.99 m 0.60 m to 0.21 m 2nd E to I 0.60 m to 0.98 m 0.60 m to 0.22 m 3rd I to M 0.60 m to 0.97 m 0.60 m to 0.23 m 6th U to Y 0.60 m to 0.94 m 0.60 m to 0.26 m Over the course of time, the mass continues to vibrate - moving away from and back towards the original resting position. However, the amount of displacement of the mass at its maximum and minimum height is decreasing from one cycle to the next. This illustrates that energy is being lost from the mass-spring system. If given enough time, the vibration of the mass will eventually cease as its energy is dissipated. Perhaps, this observation of energy dissipation or energy loss is the observation that triggers the "slowing down" comment discussed earlier. In physics (or at least in the English language), "slowing down" means to "get slower" or to "lose speed". Speed, a physics term, refers to how fast or how slow an object is moving. To say that the mass on the spring is "slowing down" over time is to say that its speed is decreasing over time. But as mentioned (and as will be discussed in great detail later), the mass speeds up during two intervals of every cycle. As the restoring force pulls the mass back towards its resting position (for instance, from B to C and from D to E), the mass speeds up. For this reason, a physicist adopts a different language to communicate the idea that the vibrations are "dying out". We use the phrase "energy is being dissipated or lost" instead of saying the "mass is slowing down." Language is important when it comes to learning physics. And sometimes, faulty language (combined with surface-level thinking) can confuse a student of physics who is sincerely trying to learn new ideas. Period and Frequency So far in this part of the lesson, we have looked at measurements of time and position of a mass on a spring. The measurements were based upon readings of a position-time graph. The data on the graph was collected by a motion detector that was capturing a history of the motion over the course of time. The key measurements that have been made are measurements of: the time for the mass to complete a cycle, and the maximum displacement of the mass above (or below) the resting position. These two measurable quantities have names. We call these quantities period and amplitude. An object that is in periodic motion - such as a mass on a spring, a pendulum or a bobblehead doll - will undergo back and forth vibrations about a fixed position in a regular and repeating fashion. The fact that the periodic motion is regular and repeating means that it can be mathematically described by a quantity known as the period. The period of the object's motion is defined as the time for the object to complete one full cycle. Being a time, the period is measured in units such as seconds, milliseconds, days or even years. The standard metric unit for period is the second. An object in periodic motion can have a long period or a short period. For instance, a pendulum bob tied to a 1-meter length string has a period of about 2.0 seconds. For comparison sake, consider the vibrations of a piano string that plays the middle C note (the C note of the fourth octave). Its period is approximately 0.0038 seconds (3.8 milliseconds). When comparing these two vibrating objects - the 1.0-meter length pendulum and the piano string which plays the middle C note - we would describe the piano string as vibrating relatively frequently and we would describe the pendulum as vibrating relatively infrequently. Observe that the description of the two objects uses the terms frequently and infrequently. The terms fast and slow are not used since physics types reserve the words fast and slow to refer to an object's speed. Here in this description we are referring to the frequency, not the speed. An object can be in periodic motion and have a low frequency and a high speed. As an example, consider the periodic motion of the moon in orbit about the earth. The moon moves very fast; its orbit is highly infrequent. It moves through space with a speed of about 1000 m/s - that's fast. Yet it makes a complete cycle about the earth once every 27.3 days (a period of about 2.4x10 5 seconds) - that's infrequent. Objects like the piano string that have a relatively short period (i.e., a low value for period) are said to have a high frequency. Frequency is another quantity that can be used to quantitatively describe the motion of an object is periodic motion. The frequency is defined as the number of complete cycles occurring per period of time. Since the standard metric unit of time is the second, frequency has units of cycles/second. The unit cycles/second is equivalent to the unit Hertz (abbreviated Hz). The unit Hertz is used in honor of Heinrich Rudolf Hertz, a 19th century physicist who expanded our understanding of the electromagnetic theory of light waves. The concept and quantity frequency is best understood if you attach it to the everyday English meaning of the word. Frequency is a word we often use to describe how often something occurs. You might say that you frequently check your email or you frequently talk to a friend or you frequently wash your hands when working with chemicals. Used in this context, you mean that you do these activities often. To say that you frequently check your email means that you do it several times a day - you do it often. In physics, frequency is used with the same meaning - it indicates how often a repeated event occurs. High frequency events that are periodic occur often, with little time in between each occurrence - like the back and forth vibrations of the tines of a tuning fork. The vibrations are so frequent that they can't be seen with the naked eye. A 256-Hz tuning fork has tines that make 256 complete back and forth vibrations each second. At this frequency, it only takes the tines about 0.00391 seconds to complete one cycle. A 512-Hz tuning fork has an even higher frequency. Its vibrations occur more frequently; the time for a full cycle to be completed is 0.00195 seconds. In comparing these two tuning forks, it is obvious that the tuning fork with the highest frequency has the lowest period. The two quantities frequency and period are inversely related to each other. In fact, they are mathematical reciprocals of each other. The frequency is the reciprocal of the period and the period is the reciprocal of the frequency. This reciprocal relationship is easy to understand. After all, the two quantities are conceptual reciprocals (a phrase I made up). Consider their definitions as restated below: period = the time for one full cycle to complete itself; i.e., seconds/cycle frequency = the number of cycles that are completed per time; i.e., cycles/second Even the definitions have a reciprocal ring to them. To understand the distinction between period and frequency, consider the following statement: According to Wikipedia (and as of this writing), Tim Ahlstrom of Oconomowoc, WI holds the record for hand clapping. He is reported to have clapped his hands 793 times in 60.0 seconds. What is the frequency and what is the period of Mr. Ahlstrom's hand clapping during this 60.0-second period? See Answer In this problem, the event that is repeating itself is the clapping of hands; one hand clap is equivalent to a cycle. Frequency = cycles per second = 793 cycles/60.0 seconds = 13.2 cycles/s = 13.2 Hz Period = seconds per cycle = 60.0 s/793 cycles = 0.0757 seconds Amplitude of Vibration The final measurable quantity that describes a vibrating object is the amplitude. The amplitude is defined as the maximum displacement of an object from its resting position. The resting position is that position assumed by the object when not vibrating. Once vibrating, the object oscillates about this fixed position. If the object is a mass on a spring (such as the discussion earlier on this page), then it might be displaced a maximum distance of 35 cm below the resting position and 35 cm above the resting position. In this case, the amplitude of motion is 35 cm. Over the course of time, the amplitude of a vibrating object tends to become less and less. The amplitude of motion is a reflection of the quantity of energy possessed by the vibrating object. An object vibrating with a relatively large amplitude has a relatively large amount of energy. Over time, some of this energy is lost due to damping. As the energy is lost, the amplitude decreases. If given enough time, the amplitude decreases to 0 as the object finally stops vibrating. At this point in time, it has lost all its energy. We Would Like to Suggest ... Why just read about it and when you could be interacting with it? Interact - that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of ourMass on a Spring Interactive. You can find it in the Physics Interactives section of our website. TheMass on a Spring Interactiveprovides the learner with a simple environment for exploring the properties of periodic motion. Visit: Mass on a Spring Interactive Check Your Understanding A pendulum is observed to complete 23 full cycles in 58 seconds. Determine the period and the frequency of the pendulum. See Answer The frequency can be thought of as the number of cycles per second. Calculating frequency involves dividing the stated number of cycles by the corresponding amount of time required to complete these cycles. In contrast, the period is the time to complete a cycle. Period is calculated by dividing the given time by the number of cycles completed in this amount of time. frequency = 23 cycles/58 seconds = 0.39655 Hz = ~0.40 Hz period = 58 seconds/23 cycles = 2.5217 sec = ~2.5 s A mass is tied to a spring and begins vibrating periodically. The distance between its highest and its lowest position is 38 cm. What is the amplitude of the vibrations? See Answer Answer: 19 cm The distance that is described is the distance from the high position to the low position. The amplitude is from the middle position to either the high or the low position. Next Section: Pendulum Motion Motion of a Mass on a Spring Jump To Next Lesson: Waves and Wavelike Motion Tired of Ads? Go ad-free for 1 year Privacy Manager privacy contact home about terms © 1996-2025 The Physics Classroom, All rights reserved. By using this website, you agree to our use of cookies. We use cookies to provide you with a great experience and to help our website run effectively. Freestar.comReport This Ad
13385	https://www.youtube.com/watch?v=FLM3xlqw3WY	Equations of Circles Marty Brandl 27100 subscribers 255 likes Description 57315 views Posted: 29 Feb 2012 This video looks at using equations of circles to find the center and the radius of a circle to graph it, as well as, writing an equation for a circle, given the center and radius, or a point on the circle. It includes four examples. 40 comments Transcript: in this video we're going to take a look at equations of circles first of all it would be helpful to know the general equation for circles and that is x - h^ 2ar + y bugger plus y - k^ squar wow this thing's really giving me trouble today = R2 okay so that's the general form where h k is the center of the circle HK is the center and the radius of the circle is R so R is the radius h k is the center so for this first uh couple let's pick out what the radius and the center of the circle would be and then we can talk about how we would sketch a graph of it so for this first one we have just X SAR hm so what would lead what would give us just X squar well zero for H so that means my H or my x coordinate of my Center is at zero then I look for my y-coordinate of the center this is y + 2 notice the general form is y - K so to get that minus it must be minus a -2 so the y coordinate of my Center is -2 so there's my center of my circle then I need to find the radius of the circle and notice here the r squared so we take the square root of this number the square root of 121 would be 11 so my radius of that Circle would be 11 so what I can do then to graph it I would just go to 0 -2 on my coordinate plane and then go 11 in go 11 up go 11 to the right go 11 to the left go 11 down from that Center and then connect those dots and I've got my circle okay let's pick out what the center and the radius for this one would be all right so the x coordinate of the center this is x + 5 so remember if it's plus the general form is minus so it must be minus a -5 so -5 would be my Center here then the y coordinate of the center well it's y - 10 so that would be 10 so then we have that for our Center then we need to find the radius well remember the radius is right here squared so I have to take the square root of that well the square root of 9 would just be three so my radius is three again to graph this I would first go to the point - 510 so so over five to the left up 10 then go three in each direction from that Center and connect the dots and I've got my circle now let's take a look at how we could write the equation for a circle given some information for this first one we're given the center and we're also given the radius we're asked to write an equation for that so the center well my x coordinate that's the H my y coordinate that's the K so I'm going to go ahead and fill those in so this is going to be X and since it's -11 and this is minus it be min-1 which would be + 11 squared then we have plus Y and then it's going to be Min -8 so that would be+ 8 squared and that's equal to the radius squared so 4 SAR be 4 4 which would be 16 so an a circle with a center here and a radius there would have the equation of that now let's look at the last one this one they don't give us the radius but they do give us a point that's on the circle if we know the center and a point that's on the circle could we figure out what the radius is well sure we could we could use the formula to figure out what the radius is so do from the center to that point on the circle that distance will give us what the radius is so let's I'm going to take that over here remember the distance formula is D equal the square root of x sub 2 - x sub 1^ 2ar plus y sub 2 - y sub 1 squared like so so we're going to fill those things in so let's call this our x sub 2 and Y sub 2 that will call our x sub 1 y sub 1 so we have the square root of x sub 2 we're going to call that -7 so -7 then minus 2 squared plus our y sub 2 is -1 so 1 - -5 would be + 5 squared okay then simplify some stuff here so -7 - 2 would be -9 9 squared would be 81 or excuse me it' be 81 negative negative is positive so 81 then -1 + 5 would be 4 4 SAR would be 16 so + 16 and I end up with the square root of 97 this would be 8197 okay so what's that mean the radius has a length of the square < TK of 97 okay we can't really simplify that so we'll just leave it as it is then I'm going to go ahead and write my equation so I have all the stuff I need I have the center and I have the radius so remember in our general form back up here we have x minus H which in this case is 2 so x - 2^ squared it's all about drawing straight lines here today x - 2^ squared what a bugger I might need to do a little troubleshooting here then Plus y times or excuse me y minus K in this case we've got -5 so Min -5 would be + 5^ squared and that's equal to my radius squared so if my radius is the square < TK of 97 if I square that I get just 97 so there's my equation hopefully this is helpful remember remember equations of circles the general form is like this where h k is your Center and R is your radius you can both pick out the center and the radius if the you're given the equation or if you're given a center and the radius or a point that's on the circle you can find the radius and then write our equation hope this video is helpful uh keep working hard on your math and I know you'll do great
13386	https://www.geeksforgeeks.org/videos/decimal-to-octal-conversion//	Decimal to Octal Conversion - GeeksforGeeks \| Videos Data Structure Java Python HTML Interview Preparation Sign In Tutorials Courses Tracks DSA Practice Problems C C++ Java Python JavaScript Data Science Machine Learning Courses Linux DevOps SQL Web Development System Design Aptitude Videos Video Player is loading. Play Video Play Mute Current Time 0:00 / Duration 0:00 Loaded: 0% 0:00 Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate 2x 1.5x 1x, selected 0.5x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected English Captions auto Quality Audio Track Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. June 08, 2022 \|5.7K Views Decimal to Octal Conversion CPP,base-conversion,CPP Programs Save Share Like Description Discussion Given a decimal number as input, we need to write a program to convert the given decimal number into an equivalent octal number. i.e convert the number with base value 10 to base value 8. The base value of a number system determines the number of digits used to represent a numeric value. For example, the binary number system uses two digits 0 and 1, the octal number system uses 8 digits from 0-7 and the decimal number system uses 10 digits 0-9 to represent any numeric value. Examples: Input : 16 Output: 20 Input : 10 Output: 12 Input : 33 Output: 41 Decimal to Octal Conversion: Read More Recommendations 10:14 67.3K Views \| 11/01/2025... Program to Find the Sum of All Digits of a Number 10:37 22.2K Views \| 09/01/2025... Graham Scan Algorithm 03:53 16.7K Views \| 08/01/2025... Primality Test (Introduction and School Method) 04:41 9.7K Views \| 08/01/2025... C++ Program to Find Roots of a Quadratic Equation 08:23 5.9K Views \| 03/01/2025... C++ Program on Multiplication of Two Matrices 08:00 3.8K Views \| 03/01/2025... JavaScript Program to Check if Two Strings are Anagrams 03:36 3.0K Views \| 16/08/2024... C++ Program to Return Maximum occurring Character in an Input String 03:12 109.9K Views \| 18/07/2024... C++ Program to Find GCD 10:49 43.4K Views \| 17/07/2024... C++ Program to Calculate Power of a Number 04:45 57.0K Views \| 12/07/2024... JavaScript Program to Find the Factorial of a Number Corporate & Communications Address: A-143, 7th Floor, Sovereign Corporate Tower, Sector- 136, Noida, Uttar Pradesh (201305) Registered Address: K 061, Tower K, Gulshan Vivante Apartment, Sector 137, Noida, Gautam Buddh Nagar, Uttar Pradesh, 201305 Advertise with us Company About Us Legal Privacy Policy Contact Us Advertise with us GFG Corporate Solution Campus Training Program Explore POTD Job-A-Thon Community Blogs Nation Skill Up Tutorials Programming Languages DSA Web Technology AI, ML & Data Science DevOps CS Core Subjects Interview Preparation GATE Software and Tools Courses IBM Certification DSA and Placements Web Development Programming Languages DevOps & Cloud GATE Trending Technologies Videos DSA Python Java C++ Web Development Data Science CS Subjects Preparation Corner Aptitude Puzzles GfG 160 DSA 360 System Design @GeeksforGeeks, Sanchhaya Education Private Limited, All rights reserved
13387	https://dictionary.law.com/Default.aspx?selected=1536	Menu Copyright Â© 2019 ALM Media Properties, LLC. All Rights Reserved. Search Legal Terms and Definitions \| \| \| \| --- \| Browse: A B C D E F G H I J K L M N O P Q R S T U V W Y Z \| \| \| \| \| \| \| \| Enter a Legal Term all wordsany wordsphrase \| \| Search the Definitions all wordsany wordsphrase \| all wordsany wordsphrase all wordsany wordsphrase \| \| \| \| \| \| --- --- \| \| \| \| \| \| --- \| \| \| \| --- \| \| plaintiff n. the party who initiates a lawsuit by filing a complaint with the clerk of the court against the defendant(s) demanding damages, performance and/or court determination of rights. \| \| See also: complaint defendant petitioner \| \| \| \| \| \| --- \| \| \| \| --- \| \| plaintiff n. the party who initiates a lawsuit by filing a complaint with the clerk of the court against the defendant(s) demanding damages, performance and/or court determination of rights. \| \| See also: complaint defendant petitioner \| \| \| \| --- \| \| plaintiff n. the party who initiates a lawsuit by filing a complaint with the clerk of the court against the defendant(s) demanding damages, performance and/or court determination of rights. \| \| n. the party who initiates a lawsuit by filing a complaint with the clerk of the court against the defendant(s) demanding damages, performance and/or court determination of rights. \| \| \| The People's Law Dictionary by Gerald and Kathleen Hill Publisher Fine Communications \| Publications Law Topics Rankings More Law.com Copyright © 2025 ALM Media Properties, LLC. All Rights Reserved.
13388	https://en.wikipedia.org/wiki/Vaginal_artery	Vaginal artery - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Structure 2 Function 3 Other animals 4 See also 5 References 6 External links Vaginal artery [x] 9 languages العربية Bosanski Deutsch Español فارسی Français Português Română Tiếng Việt Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations.(June 2015) (Learn how and when to remove this message) Blood vessel \| Vaginal artery \| \| Arteries of the female reproductive tract (posterior view): uterine artery, ovarian artery and vaginal arteries. \| \| Vessels of the uterus and its appendages, rear view. \| \| Details \| \| Source \| Internal iliac artery Uterine artery \| \| Vein \| Vaginal venous plexus \| \| Supplies \| Urinary bladder, ureter, vagina \| \| Identifiers \| \| Latin \| arteria vaginalis \| \| TA98 \| A12.2.15.035F \| \| TA2 \| 4336 \| \| FMA \| 18832 \| \| Anatomical terminology [edit on Wikidata] \| The vaginal artery is an artery in females that supplies blood to the vagina and the base of the bladder. Structure [edit] Vaginal artery The vaginal artery is usually a branch of the internal iliac artery. Some sources say that the vaginal artery can arise from the uterine artery, but the phrase vaginal branches of uterine artery is the term for blood supply to the vagina coming from the uterine artery. The vaginal artery is frequently represented by two or three branches. These descend to the vagina, supplying its mucous membrane. They anastomose with branches from the uterine artery. It can send branches to the bulb of the vestibule, the fundus of the bladder, and the contiguous part of the rectum. Function [edit] The vaginal artery supplies oxygenated blood to the muscular wall of the vagina, along with the uterine artery and the internal pudendal artery. It also supplies the cervix, along with the uterine artery. Other animals [edit] In horses, the vaginal artery may haemorrhage after birth, which can cause death. See also [edit] Uterine artery References [edit] This article incorporates text in the public domain from page 616 of the 20th edition ofGray's Anatomy(1918) ^ Jump up to: abKyung Won, PhD. Chung (2005). Gross Anatomy (Board Review). Hagerstown, MD: Lippincott Williams & Wilkins. p.290. ISBN0-7817-5309-0. ^ Jump up to: abŁaniewski, Paweł; Herbst-Kralovetz, Melissa (2018-01-01), "Vagina", in Skinner, Michael K. (ed.), Encyclopedia of Reproduction (Second Edition), Oxford: Academic Press, pp.353–359, ISBN978-0-12-815145-7, retrieved 2021-01-18 ^Graziottin, Alessandra; Gambini, Dania (2015-01-01), Vodušek, David B.; Boller, François (eds.), "Chapter 4 - Anatomy and physiology of genital organs – women", Handbook of Clinical Neurology, Neurology of Sexual and Bladder Disorders, 130, Elsevier: 39–60, doi:10.1016/B978-0-444-63247-0.00004-3, ISBN9780444632470, PMID26003238, retrieved 2021-01-18 ^Mahendroo, Mala; Nallasamy, Shanmugasundaram (2018-01-01), "Cervix", in Skinner, Michael K. (ed.), Encyclopedia of Reproduction (Second Edition), Oxford: Academic Press, pp.339–346, ISBN978-0-12-815145-7, retrieved 2021-01-18 ^McAuliffe, Siobhan B., ed. (2014-01-01), "Chapter 12 - Reproductive disorders", Knottenbelt and Pascoe's Color Atlas of Diseases and Disorders of the Horse (Second Edition), W.B. Saunders, pp.443–513, doi:10.1016/b978-0-7234-3660-7.00012-2, ISBN978-0-7234-3660-7, S2CID241150397, retrieved 2021-02-06 External links [edit] Anatomy photo:43:13-0206 at the SUNY Downstate Medical Center - "The Female Pelvis: Branches of Internal Iliac Artery" \| hide v t e Arteries of the abdomen and pelvis \| \| Abdominal aorta \| \| Inferior phrenic \| Superior suprarenal \| \| Celiac \| \| Left gastric \| Esophageal branches \| \| Common hepatic \| Proper hepatic cystic Right gastric Gastroduodenal right gastroepiploic superior pancreaticoduodenal supraduodenal \| \| Splenic \| Pancreatic branches greater dorsal Short gastrics Left gastroepiploic \| \| \| Superior mesenteric \| Inferior pancreaticoduodenal Intestinal jejunal ileal arcades vasa recta Ileocolic colic anterior cecal posterior cecal ileal branch appendicular Right colic Middle colic Marginal \| \| Suprarenal \| Middle suprarenal \| \| Renal \| Inferior suprarenal Ureteral \| \| Gonadal \| Testicular artery Ovarian artery \| \| Lumbar \| Lumbar arteries \| \| Inferior mesenteric \| Left colic Marginal Sigmoid Superior rectal \| \| Common iliac \| \| Internal iliac \| \| Posterior surface \| \| Iliolumbar \| Lumbar branch Iliac branch \| \| \| Anterior surface \| \| Superior vesical artery \| Umbilical artery Medial umbilical ligament to ductus deferens \| \| Obturator \| Anterior branch Pubic branch Posterior branch Acetabular branch Cruciate anastomosis Corona mortis \| \| Middle rectal \| Vaginal branch ♀ / Prostatic branch ♂ \| \| Uterine ♀ \| Arcuate Vaginal branches Ovarian branches Tubal branches Spiral \| \| Vaginal ♀ / Inferior vesical ♂ \| \| Inferior gluteal \| Accompanying of sciatic nerve Cruciate anastomosis \| \| Internal pudendal \| Inferior rectal Perineal posterior scrotal posterior labial Bulb of penis/vestibule Urethral Deep artery of the penis helicine Deep artery of clitoris Dorsal of the penis Dorsal of the clitoris \| \| \| \| External iliac \| Inferior epigastric Corona mortis Deep circumflex iliac Femoral see arteries of lower limbs \| \| \| Median sacral \| Coccygeal glomus \| \| Portal: Anatomy \| Authority control databases \| Terminologia Anatomica \| Retrieved from " Categories: Wikipedia articles incorporating text from the 20th edition of Gray's Anatomy (1918) Arteries of the abdomen Vagina Hidden categories: Articles lacking in-text citations from June 2015 All articles lacking in-text citations Articles with short description Short description is different from Wikidata Anatomy NAV infobox with use of other NAV parameters This page was last edited on 23 May 2025, at 22:20(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Vaginal artery 9 languagesAdd topic
13389	https://www.onlinemathlearning.com/isolate-the-variable.html	Isolate A Variable - Transposition (video lessons, examples, solutions) ×search Algebra: Isolate A Variable (Transposition) Related Pages Basic Algebra Algebra Lesson - Transposition Combining Like Terms Solving Equations More Algebra Lessons In these lessons, we will learn one of the basic techniques of simplifying an algebraic equation so that we can eventually solve the equation. Math online courses Share this page to Google Classroom How To Isolate A Variable? Transposition is a method to isolate the variable to one side of the equation and everything else to the other side so that you can solve the equation. Two methods are covered here: Method 1: Using inverse operations Method 2: a shortcut trick that allows you to work faster The following table gives the steps to Isolate a Variable. Scroll down the page for examples and solutions. Printable & Online Algebra Worksheets Method 1 A quick review of the basic principles - all equations have two sides: a Left Side (LS) and a Right Side (RS). The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolate the variable (or the unknown quantity). For Example: 5x + 8 = 3x – 6 We want to get rid of the number 8 from the left side. So we subtract 8 from both sides of the equation. 5x + 8=3x – 6original equation– 8=– 8subtract 8 from both sides5x=3x – 14resulting equation Next, we want to get rid of 3x from the right side. So, we subtract 3x from both sides of the equation. 5x=3x – 14result from above– 3x=– 3xsubtract 3x from both sides2x=– 14resulting equation Now, we want to get rid of the coefficient 2. So, we divide 2 from both sides of the equation. 2x=– 14result from above÷2=÷2divide both sides by 2x=– 7resulting solution Check the Answer Now check your answer x = –7 by plugging it back into the original equation. 5x + 8 = 3x – 6 LS: 5 × (– 7) + 8 = – 35 + 8 = – 27 RS: 3 × (– 7) – 6 = – 21 – 6 = – 27 LS = RS, ∴ answer is correct. Algebra: How to solve equations by isolating the variable? An equation is a symbolic statement that two algebraic expressions are equal. To solve an equation means to find the variable or unknown. The rule is that the same operation must be done on both sides of the equation to preserve equality. Examples: x + 4 = 20 x - 4 = 20 4x = 20 x/4 = 20 Isolate the variable: 5x - x = 20 + 4 9 - 4x = 8x 8 - 5x - x = 14 Show Video Lesson How to isolate a variable and solve linear equations (variables on both sides of the equation)? Example: Solve for y and check your solution 20 - 7y = 6y - 6 Show Video Lesson How to isolate the variable using inverse operations to solve fraction equations? Examples: a + 1/2 = 4 3/7 + n = 2 2/5 5/7 x = 4 2/3 3/4 b = -24 Show Video Lesson How to isolate a variable in a formula? 2p = kx - q, isolate for x (3b -4)/2 = C, isolate for b Show Video Lesson How to isolate a variable when the variable or expression is in the denominator? 8/y = 3 (x + 3)/(x + 5) = 20 5/(x + 3) = 4/(x + 4) Show Video Lesson Transposition (Rearranging Equations) - Introduction What is transposition? What is it used for? Show Video Lesson Method 2 In this method we isolate the variable by moving like terms to one side of the equation. To maintain the equality of the equation, when removing a term from one side of the equation we perform the opposite operation to the other side. For example: 5x + 8 = 3x – 6 To remove + 8 from the LS, we subtract 8 from the RS. To remove + 3x from the RS, we subtract 3x from the LS. To remove the coefficient 2, we divide 2 on the RS. When you have grasped Method 2, it is faster because it allows you to perform several steps at the same time to isolate the variable. For example: 5x + 8 = 3x – 6 Shortcut Trick to help you solve equations (Variable on one side) Examples: (3x + 5)/7 = 2 (5x - 8)/3 = 4 x2+7−−−−−√+5\=9x2+7+5\=9 2x√2\=22x2\=2 5x−6√×42\=65x−6×42\=6 Show Video Lesson Shortcut Trick to help you solve equations (Variables on both sides) Examples: 3x + 4 = 2x + 6 2x + 1 = 4x - 9 x - 3 = 4x + 6 Show Video Lesson Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. Math e-books ×search We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. ×search Back to Top \| Interactive Zone \| Home Copyright © 2005, 2025 - OnlineMathLearning.com. Embedded content, if any, are copyrights of their respective owners. × Home Math By Grades Back Pre-K Kindergarten Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grades 7 & 8 Grades 9 & 10 Grades 11 & 12 Basic Algebra Intermediate Algebra High School Geometry Math By Topics Back Arithmetic Algebra Geometry Trigonometry Statistics Probability Word Problems Pre-Calculus Calculus Set Theory Matrices Vectors Math Curriculum Back NY Common Core Illustrative Math Common Core Standards Singapore Math Free Math Worksheets Back Worksheets by Topics Worksheets by Grades Printable & Online Common Core Worksheets Interactive Zone Math Tests Back SAT Math ACT Math GMAT GRE Math GED Math High School-Regents California Standards Math Fun & Games Back Math Trivia Math Games Fun Games Mousehunt Guide Math Video Lessons Back College Algebra College Calculus Linear Algebra Engineering Math Test/Exam Prep Back SAT Preparation ACT Preparation GMAT Preparation GRE Preparation GED Preparation ESL, IELTS, TOEFL GCSE/IGCSE/A Level Back KS3/CheckPoint 1 Science KS3/CheckPoint 2 Science KS3/CheckPoint 3 Science GCSE/IGCSE Maths GCSE/IGCSE Biology CGSE/IGCSE Chemistry CGSE/IGCSE Physics A-Level Maths Others Back Biology Chemistry Science Projects High School Biology High School Chemistry High School Physics Animal Facts English Help English For Kids Programming Calculators Privacy Policy Tutoring Services What's New Contact Us
13390	https://math.stackexchange.com/questions/2659288/proof-that-ratio-of-vertices-to-edges-in-an-infinite-square-grid-is-12	Skip to main content Proof that ratio of vertices to edges in an infinite (square) grid is 1:2? Ask Question Asked Modified 7 years, 6 months ago Viewed 544 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I'm making the statement that the ratio of vertices to edges in an infinite square grid is 1:2. I need this fact for deducing further theorems specific for my problem, however I can't find any theory on infinite grids. I would prefer to cite some literature on that (maybe also for hexagonal and triangular grids). If no formal literature exists that covers that issue I would want to write the proof. My thinking so far is each vertex has 4 edges connected to it and each edge 2 vertexes, which makes for a 2 to 4 ratio −>1:2. However I feel this is not formal enough and the proof should involve infinity of the grid. Please post possible literature or a formal proof if you think mine isn't sufficient. geometry infinity Share CC BY-SA 3.0 Follow this question to receive notifications edited Feb 21, 2018 at 2:58 fitzmerl duron 1,60533 gold badges3232 silver badges5555 bronze badges asked Feb 20, 2018 at 21:23 SugarOnBaconSugarOnBacon 4544 bronze badges 5 4 There are infinitely many vertices and infinitely many edges in an infinite grid. Your ratio does not make sense as it stands, since the "ratio" of two infinite quantities is not a (real) number. You can talk about the ratio of vertices to edges in a square grid of size n, and show this goes to 2 as n goes to infinity, but this need not be the same thing that you are trying to get at. – Matthew Conroy Commented Feb 20, 2018 at 21:30 3 In addition to @MatthewConroy 's comment: For every ratio there is a way you can count the edges and vertices in the grid to get it. Therefore there is no "the" ratio. – SK19 Commented Feb 20, 2018 at 21:40 I understand what you mean and that is part of the problem and why i asked for literature on this subject. While it is true that you can make a bijection between edges and vertices similarly to proving the set of all natural even/odd numbers is as large as the set of all natural numbers, i'm asking for the ratio as described by @MatthewConroy when you send n into infinity (please pardon my inaccurate description). However if you limit a grid by n, depending on the definition of the grid, the ratio might not be 2:1 (to get this ratio we need some edges pointing out of the grid am i correct?) – SugarOnBacon Commented Feb 20, 2018 at 22:45 1 If you look at a finite n×n grid graph, you can count the number of edges and vertices. Call them e(n) and v(n). Then you can look at the limit of e(n)/v(n) as n goes to infinity. This limit is 2, at least with the definition of grid that I have in mind. I can post details in an answer if that would be helpful. But this does not yield the "ratio of edges to vertices in an infinite grid". – Matthew Conroy Commented Feb 20, 2018 at 23:19 What is your problem exactly? A useful answer depends on the answer to this. – Somos Commented Feb 21, 2018 at 2:30 Add a comment \| 3 Answers 3 Reset to default This answer is useful 6 Save this answer. Show activity on this post. If we're being "careful with ∞", then let's consider an n×n square grid (with n2 vertices). In this grid, there are (n−2)2 interior vertices, which have 4 edges out of them; 4(n−2) edge vertices, which have 3 edges out of them; 4 corner vertices, which have 2 edges out of them. By the degree sum formula, adding these numbers gives twice the number of edges (because each edge is counted twice, once by each of its endpoints). If we call the number of edges m, then 2m=4(n−2)2+3(4n−8)+2(4)=4n2−4n so there are 2n2−2n edges. The ratio mn of edges over vertices is 2n2−2nn2=2−2n, which approaches 2 as n→∞. We can be fully rigorous and still do less work with asymptotic analysis. If we were to claim that each vertex has degree 4, and therefore that 2m=4n2, we would be wrong only for edges around the border. But there are O(n) edges along each border, whose degree differs from 4 by a constant, so we have 2m=4n2+O(n), and therefore mn=2n2+O(n)n2=2+O(1n). Once again, as n→∞, this approaches 0. Also, we can consider a finite toroidal square grid, which "wraps around" at the borders, Pac-Man style. Then every single vertex has degree 4, and the ratio of vertices to edges is 1:2 with no approximation. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Feb 20, 2018 at 23:27 Misha LavrovMisha Lavrov 160k1111 gold badges167167 silver badges304304 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. The answer’s actually quite simple. Sure there’s a mathematical proof, given an n×n square, there will be (n+1)² vertices and 2n(n+1) edges, which equals ∞² to 2∞² as n goes to ∞. However, this is the less fun explanation. Let’s play a game, we construct a square grid completely out of vertices conjoined with edges. In the first image, we repeatedly add a line connected to a dot, in the second we add two lines connected to one dot, and on the far right we add 3 connected to one. As you can see, if there aren’t enough vertices, there will end up being some missing lines, where as if there are too many, you end up with missing dots. Now, granted, depending on how you play the game, there might be some missing lines or dots either way, but unless there’s a 2 to 1 ratio, there’s no way you can fill in all the spaces. Go ahead, try it out, it’s impossible, no matter how hard you try. I know that that explanation wasn’t as mathematical as one would hope, but I figured this would be a pretty one method problem, so I wanted to provide something unique from all the other answers. If, however, you want me to expand on my introductory explanation, just comment and if be glad to. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 21, 2018 at 3:03 fitzmerl duron 1,60533 gold badges3232 silver badges5555 bronze badges answered Feb 21, 2018 at 0:40 Math MachineMath Machine 44333 silver badges1010 bronze badges Add a comment \| This answer is useful -1 Save this answer. Show activity on this post. Consider that any vertex has four edges connected to it. Pick two edges that are adjacent, that is, are adjacent sides of a square. The vertex and the two edges can be translated to every other vertex and this partitions all vertices and edges. Thus the ratio of vertices to edges is 1:2. What is important is the local connection between vertices and edges despite the fact that there are many different ways of pairing edges and vertices. There is only one way that is translation invariant. A similar proof applies to an infinite one dimensional grid where the ratio is 1:1. A similar proof applies to an infinite (triangular) grid where the ratio is 1:3. In general, if you have an infinite homogeneous grid with the degree of each vertex being a constant D, then since each edge connects exactly two vertices, the ratio of vertices to edges is 2:D. This applies to infinite homogeneous trees as a special case. So in the case of an infinite hexagonal grid the ratio is 2:3. The same proofs apply to a finite cycle, infinite cylinder, finite torus by formally identifying edges of infinite grids. It is clear that for a finite cycle there is a 1:1 ratio of vertices to edges, and this also applies in the limit of an infinite cycle. The infinite total number of vertices or edges makes no difference. It is the local connections between them that counts. If you want to be really careful, consider the lattice of points in the plane with integer coordinates as vertices with horizontal edges {(n,y)} where m≤y<m+1, and vertical edges {(x,m)} where n≤x<n+1. Each period square is {(x,y)} where n≤x<n+1,m≤y<m+1 and each contains one lattice point (n,m), one horizontal edge, and one vertical edge. The ratio is 1:2 in any period square. Similar proofs apply to other cases. Read the Wikipedia article Fundamental pair of periods for the related idea of fundamental period parallelogram which is very important in elliptic function theory. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 21, 2018 at 2:44 answered Feb 20, 2018 at 21:52 SomosSomos 37.5k33 gold badges3535 silver badges8585 bronze badges 2 1 But you could also list the vetices, list the edges, and then for every n assign the n vertex to the n edge, proving the ratio is 1-to-1. Or, you could assign to vertex n edges 6n−5 through 6n, proving the ratio is 1-to-6. Or any other ratio you want. – Gerry Myerson Commented Feb 20, 2018 at 21:57 Your approach is similar to mine, however others have pointed out some problems with that approach in infinity. See comments under the question. – SugarOnBacon Commented Feb 20, 2018 at 22:47 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry infinity See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Related 2 Find the Outgoing Edge with the Smallest Angles, Given one Incident Edges and Multiple Outgoing Edges 8 How is it that this shape can converge to what looks like a triangle but has a different perimeter? 5 Triangular grids on a torus 14 Why does a convex polyhedron being vertex-, edge-, and face-transitive imply that it is a Platonic solid? 2 How to infinitely tile a cartesian plane with the least "mine-coverage" for various "8-proximity" values? Hot Network Questions Could you charge a battery using with a long radio aerial? Driven right leg common mode suppression How do you get rid of extra villagers in ACNH that have an account on the Nintendo Switch? What happens as you fall into a realistic black hole? Activate the Laser Gates How to balance research and teaching responsibilities? Chapter title formatting with TikZ graphic Should 1 Peter 1:5 read ‘a’ or ‘the’ before ‘salvation’, or should it just read ‘salvation’? Non-committing? Have we been using deniable authenticated encryption all along? Half-Life 1 launches on the wrong monitor and outside of the screen How can I separate the 2 wheels in this model into individual objects? Excessive flame on one burner Spectral sequences every mathematician should know Can a nozzle-less engine be made efficient by clustering? Would weekly seasonal changes still allow a habitable planet? Splitting a long introduction into collapsible elements easier to read and translate Can you remove a variable in the 8-bit Microsoft BASICs? How NOT to get hyper-inflation in a vassal-state that is printing/minting money at exorbitant speed? Solve the crossed ladders problem How to top align panels My Canadian employer is sending me to Germany to work on a project. Do I need a visa or a work permit? Why do these two lines have the same probability of intersecting the circle? Resonance in hexa-2,4-dienyl cation Since the universe is expanding and spacetime is a single fabric, does time also stretch along with the expansion of space? Question feed By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
13391	https://artofproblemsolving.com/wiki/index.php/Karamata%27s_Inequality?srsltid=AfmBOoqydhDQZ_csNxoApm4N0DLNwAae9Mukbby2_GqhR3pFbwOjyq_n	Art of Problem Solving Karamata's Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Karamata's Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Karamata's Inequality Karamata's Inequality states that if majorizes and is a convex function, then Proof We will first use an important fact: If is convex over the interval , then and , This is proven by taking casework on . If , then A similar argument shows for other values of . Now, define a sequence such that: Define the sequences such that and similarly. Then, assuming and similarily with the 's, we get that . Now, we know: . Therefore, Thus, we have proven Karamata's Theorem. This article is a stub. Help us out by expanding it. See also Retrieved from " Categories: Stubs Algebra Inequalities Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
13392	https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/6fe5ce4d56945ce7241679998922fb3a_MIT6_436JF18_lec09.pdf	MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2018 Lecture 9 PRODUCT MEASURE AND FUBINI’S THEOREM Contents 1. Product measure 2. Fubini’s theorem In elementary math and calculus, we often interchange the order of summa tion and integration. The discussion here is concerned with conditions under which this is legitimate. 1 PRODUCT MEASURE Consider two probabilistic experiments with probability spaces ( 1, F1, P1) and ( 2, F2, P2), respectively. We are interested in forming a probabilistic model of a “joint experiment” in which the original two experiments are carried out independently. 1.1 The sample space of the joint experiment If the ﬁrst experiment has an outcome !1, and the second has an outcome !2, then the outcome of the joint experiment is the pair (!1, !2). This leads us to deﬁne a new sample space = 1 × 2. 1.2 The ˙-algebra of the joint experiment Next, we need a ˙-algebra on . If A1 ∈F1, we certainly want to be able to talk about the event {!1 ∈ A1} and its probability. In terms of the joint experiment, this would be the same as the event A1 × 1 = {(!1, !2) \| !1 ∈ A1, !2 ∈ 2}. 1 Ω Ω Ω Ω Ω Ω Ω Ω Thus, we would like our ˙-algebra on to include all sets of the form A1 × 2, (with A1 ∈F1) and by symmetry, all sets of the form 1 ×A2 (with (A2 ∈F2). This leads us to the following deﬁnition. Deﬁnition 1. We deﬁne F1 × F2 as the smallest ˙-algebra of subsets of 1 × 2 that contains all sets of the form A1 × 2 and 1 × A2, where A1 ∈F1 and A2 ∈F2. Note that the notation F1×F2 is misleading: this is not the Cartesian product of F1 and F2! Since ˙-ﬁelds are closed under intersection, we observe that if Ai ∈Fi, then A1 × A2 = (A1 × 2) ∩ ( 1 ∩ A2) ∈F1 × F2. It turns out (and is not hard to show) that F1 × F2 can also be deﬁned as the smallest ˙-algebra containing all sets of the form A1 × A2, where Ai ∈Fi. Alternatively, suppose F1 and F2 are generated by algebras F0,1, F0,2. That is Fi = ˙(F0,i), i = 1, 2. Then F1 × F2 is also the smallest ˙-algebra containing all sets of the form A1 × A2, where Ai ∈F0,i. In the sequel, we will talk about g : 1 × 2 → R – measurable functions with respect to F1 × F2. Recall, this means that for any Borel set B ⊂ R, the set {(!1, !2) \| g(!1, !2) ∈ B} belongs to the ˙-algebra F1 × F2. As a practical matter, it is enough to verify that for any scalar c, the set {(!1, !2) \| g(!1, !2) ≤ c} is measurable. Other than using this deﬁnition directly, how else can we verify that such a function g is measurable? The basic tools at hand are the following: (a) continuous functions from R2 to R are measurable; (b) indicator functions of measurable sets are measurable; (c) combining measurable functions in the usual ways (e.g., adding them, mul tiplying them, taking limits, etc.) results in measurable functions. The following proposition gives further information about F1 ×F2 and func tions measurable with respect to it. Proposition 1. Let E ∈F1 × F2 then for every !1 ∈ 1 the set E!1 , {!2 \| (!1, !2) ∈ E} belongs to F2. Consequently, for every F1 × F2-measurable function f and every !1 the function f!1 (!2) , f(!1, !2) 2 Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω is F2-measurable. Remark: E!1 and f!1 are called slicec of E and f at !1, respectively. Proof. Fix some !1 and deﬁne a collection of sets L = {E ∈F1 × F2 \| E!1 ∈F2}. When E = A1 × A2 the set E!1 is either empty or equal to A2. Thus L contains all the rectangles. On the other hand, for any sequence Ej we have [ (∪j Ej ) = (Ej ) !1 !1 j and (Ec)!1 = (E!1 )c . Thus L is closed under countable unions and complements. Hence L is a ˙ algebra, which by minimality of F1 ×F2 must be equal to the latter. This shows these statement for sets. Next, a slice of a simple function N X f = ai1Ei i=1 at !1 is itself a simple (hence measurable) function on ( 2, F2). This fol lows from what was just shown for slices of sets. For the general f we have f = limr→∞ fr, where fr are simple functions. Since the slice of each fr is F2 measurable and the class of F2-measurable functions is closed under taking limits the result follows. 1.3 The product measure We now deﬁne a measure, to be denoted by P1 × P2 (or just P, for short) on the measurable space ( 1 × 2, F1 × F2). To capture the notion of independence, we require that P(A1 × A2) = P1(A1)P2(A2), ∀ A1 ∈F1, A2 ∈F2. (1) 3 Ω Ω Ω Theorem 1. There exists a unique measure P on ( 1 × 2, F1 ×F2) that has property (1). Furthermore, for every E ∈F1 × F2 measure P(E) satisﬁes Z P(E) = P2(E!1 ) P1(d!1) (2) Z = P1(E!2 ) P2(d!2). (3) Proof. Uniqueness follows from the fact that A1 × A2 is a generating p-system for F1 × F2 (see Proposition 1 in Lecture 2). We only need to show existence. We start by showing that for every E ∈F1 × F2 the function fE(!1) , P2(E!1 ) is F1-measurable. Note that P2(E!1 ) is well-deﬁned by Proposition 1. Deﬁne a collection L = {E : fE is F1-measurable}. When E = A1 × A2 the function fE(!1) = P2(A2)1A1 (!1), which is clearly measurable. Thus L contains all rectangles. Next, if E and F are disjoint then so are E!1 and F!1 . Consequently, fE∪F (!1) = fE(!1) + fF (!2) if E ∩ F = Ø . (4) This implies that L contains all ﬁnite unions of disjoint rectangles. The latter is an algebra of sets (since (A1 × A2)c can be written as disjoint union of 3 rectangles). Finally, if Ej ր E and Ej ∈L then fEj ր fE (5) and therefore fE is F1-measurable. Same argument applies to Ej ց E. All in all L is a monotone class, containing an algebra that generates F1 × F2. So L = F1 × F2. We now deﬁne for any E ∈F1 × F2 Z P(E) , fE(!1)P1(d!1) . (6) It is evident that this assignment satisﬁes (1). Finite additivity of P follows from (4). It remains to show ˙-additivity, which in turn is equivalent to continu ity. The latter follows from (5) and the MCT. 4 Ω Ω Thus, existence of P is established. Furthermore, deﬁnition (6) is just a restatement of (2). Regarding (3), construct another measure P ′ by exchanging roles of ( 1, F1, P1) and ( 2, F2, P2) in (6). So constructed P ′ automatically satisﬁes (3). Moreover, P ′ also veriﬁes (1) and hence coincides with P on a p-system of rectangles A×B. By Proposition 1 of Lecture 2 we have: P ′ = P. The above discussion extends to the case of any ﬁnite number of probability spaces ( i, Fi, Pi), i = 1, 2, . . . , k. In particular there exists a unique measure P on = 1 × · · · × k such that for every collection of sets Ai ∈Fi, P(A1 × · · · × Ak) = P(A1) × · · · × P(Ak). The corresponding ˙-algebra on is the smallest ˙-algebra containing all sets of the form A1 × · · · × Ak where Ai ∈Fi. Moreover, this extends to a count able collections of probability spaces ( i, Fi, Pi), i = 1, 2, . . . , but now the measure is only deﬁned when a ﬁnite collection of the {Ai} are not k, i.e. i = 1, 2, . . . , k P(A1 × · · · × Ak × k+1 × k+2 × · · · ) = P(A1) × · · · × P(Ak). 1.4 Beyond probability measures Everything in these notes extends to the case where instead of probability mea sures Pi, we are dealing with general measures µi, under the assumptions that the measures µi are ˙-ﬁnite. (A measure µ is called ˙-ﬁnite if the set can be partitioned into a countable union of sets, each of which has ﬁnite measure.) The most relevant example of a ˙-ﬁnite measure is the Lebesgue measure on the real line. Indeed, the real line can be broken into a countable sequence of intervals (n, n + 1], each of which has ﬁnite Lebesgue measure. 1.5 The product measure on R2 The two-dimensional plane R2 is the Cartesian product of R with itself. We endow each copy of R with the Borel ˙-ﬁeld B and one-dimensional Lebesgue measure. The resulting ˙-ﬁeld B × B is called the Borel ˙-ﬁeld on R2 . The resulting product measure on R2 is called two-dimensional Lebesgue measure, to be denoted here by λ2. The measure λ2 corresponds to the natural notion of area. For example, λ2([a, b] × [c, d]) = λ([a, b]) · λ([c, d]) = (b − a) · (d − c). More generally, for any “nice” set of the form encountered in calculus, e.g., sets of the form A = {(x, y) \| f(x, y) ≤ c}, where f is a continuous function, λ2(A) coincides with the usual notion of the area of A. 5 Ω Ω Ω Ω Ω Ω Ω Ω Ω 2 Remark for those of you who know a little bit of topology – otherwise ignore it. We could deﬁne the Borel ˙-ﬁeld on R2 as the ˙-ﬁeld generated by the collection of open subsets of R2 . (This is the standard way of deﬁning Borel sets in topological spaces.) It turns out that this deﬁnition results in the same ˙-ﬁeld as the method of Section 1.2. FUBINI’S THEOREM Fubini’s theorem is a powerful tool that provides conditions for interchanging the order of integration in a double integral. Given that sums are essentially special cases of integrals (with respect to discrete measures), it also gives con ditions for interchanging the order of summations, or the order of a summation and an integration. In this respect, it subsumes results such as Corollary 1 at the end of the notes for Lecture 12. Fubini’s theorem holds under two different sets of conditions: (a) nonnega tive functions g (compare with the MCT); (b) functions g whose absolute value has a ﬁnite integral (compare with the DCT). We state the two versions sepa rately, because of some subtle differences. The two statements below are taken verbatim from the text by Adams & Guillemin, with minor changes to conform to our notation. Theorem 2. Let g : 1 × 2 → R be a nonnegative measurable function. Let P = P1 × P2 be a product measure. Then, R (a) g(!1, !2) dP2 is a measurable function of !1. 2 R (b) g(!1, !2) dP1 is a measurable function of !2. 1 (c) We have Z Z Z Z h i h i g(!1, !2) dP2 dP1 = g(!1, !2) dP1 dP2 1 2 2 1 Z = g(!1, !2) dP. 1× 2 Note that some of the integrals above may be inﬁnite, but this is not a prob lem; since everything is nonnegative, expressions of the form ∞−∞ do not arise. Pn Proof. For simple functions g = , Ei ∈F1 × F2 statement i=1 ai1Ei 6 Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω (a) follows from measurability of !1 7→ P2(E!1 ) established in the proof of Theorem 1. For a general g consider a sequence of simple functions gr(!1, !2) ր g(!1, !2) ∀!1, !2 as r →∞. Then we have shown that Z fr(!1) = gr(!1, !2) dP2 2 are F1 measurable and monotonically increasing fr ր f. By the MCT Z f(!1) , lim gr(!1, !2) dP2 (7) r→∞ 2 Z = lim gr(!1, !2) dP2 (8) r→∞ 2 Z = g(!1, !2) dP2. (9) 2 Since f is a limit of measurable fr’s – f must be measurable. By (9) the integral over 2 is also F1 measurable. This establishes (a) and (b) by symmetry. Finally (c), for a simple function g is just (2)-(3), while for a general function g we just R need to integrate (7) interchanging and lim by the MCT at will. Recall now that a function is said to be integrable if it is measurable and the integral of its absolute value is ﬁnite. 7 Theorem 3. Let g : 1 × 2 → R be a measurable function such that Z \|g(!1, !2)\| dP < ∞, (10) 1× 2 where P = P1 × P2. (a) For almost all !1 ∈ 1, g(!1, !2) is an integrable function of !2. (b) For almost all !2 ∈ 2, g(!1, !2) is an integrable function of !1. R (c) There exists an integrable function h : 1 → R such that g(!1, !2) dP2 = 2 h(!1), a.s. (i.e., except for a set of !1 of zero P1-measure for which R g(!1, !2) dP2 is undeﬁned or inﬁnite). 2 R (d) There exists an integrable function h : 2 → R such that g(!1, !2) dP1 = 1 h(!2), a.s. (i.e., except for a set of !2 of zero P2-measure for which R g(!1, !2) dP1 is undeﬁned or inﬁnite). 1 (e) We have Z Z Z Z h i h i g(!1, !2) dP2 dP1 = g(!1, !2) dP1 dP2 1 2 2 1 Z = g(!1, !2) dP. 1× 2 Remarks: 1. Both Theorems remain valid when dealing with ˙-ﬁnite measures, such as the Lebesgue measure on R2 . This provides us with conditions for the familiar calculus formula Z Z Z Z g(x, y) dx dy = g(x, y) dy dx. 2. In order to apply Theorem 3, we need a practical method for checking the integrability condition (10). Here, Theorem 2 comes to the rescue. Indeed, by Theorem 2, we have Z Z Z \|g(!1, !2)\| dP = \|g(!1, !2)\| dP2 dP1, 1× 2 1 2 so all we need is to work with the right hand side, and integrate one vari able at a time, possibly also using some bounds on the way. 8 Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Proof. By now converting from a non-negative case to integrable case should be familiar. Theorem 3 is no exception: Given a function g, decom pose it into its positive and negative parts, apply Theorem 2 to each part, and in the process make sure that you do not encounter expressions of the form ∞−∞. We omit the details. 3 SOME CAUTIONARY EXAMPLES We give a few examples where Fubini’s theorem does not apply. 3.1 Nonnegativity and integrability Suppose that both of our sample spaces are the nonnegative integers: 1 = 2 = {1, 2, . . .}. The ˙-ﬁelds F1 and F2 consist of all subsets of 1 and 2, respectively. Then, ˙(F1 × F2) is composed of all subsets of {1, 2, . . .}2 . Let both P1 and P2 be the counting measure, i.e. P(A) = \|A\|. This means that Z Z X X g dP1 = f(a), h dP2 = h(b), A B a∈A b∈B and Z X f d(P1 × P2) = f(c). C c∈C Consider the function f deﬁned by f(m, m) = 1, f(m, m + 1) = −1, and f = 0 elsewhere. It is easier to visualize f with a picture: 1 −1 0 0 · · · 0 1 −1 0 · · · 0 0 1 −1 · · · 0 0 0 1 · · · . . . . . . . . . . . . . . . So, Z Z X X f dP2 dP1 = f(n, m) = 0 1 2 n m Z Z X X 6= 1 = f(n, m) = f dP1 dP2. 2 1 m n In this example, the conditions of Fubini’s theorem fail to hold: the function f is neither nonnegative nor integrable. 9 Ω Ω Ω Ω Ω Ω Ω Ω 3.2 ˙-ﬁniteness Let 1 = (0, 1), let F1 be the Borel sets, and let P1 be the Lebesgue measure. Let 2 = (0, 1) let F2 be the set of all subsets of (0, 1), and let P2 be the counting measure. In particular, for every inﬁnite (countable or uncountable) subset of (0, 1), P2(A) = ∞. Let f(x, y) = 1 if x = y, and f(x, y) = 0 otherwise. Then, Z Z Z f(x, y) dP2(y) dP1(x) = 1 dP1(y) = 1, 1 2 1 but Z Z Z f(x, y) dP1(x) dP2(y) = 0 dP2(y) = 0. 2 1 2 In this example, the conditions of Fubini’s theorem fail to hold: the measure on (0, 1) is not ˙-ﬁnite. 10 Ω Ω Ω Ω Ω Ω Ω Ω MIT OpenCourseWare 6.436J / 15.085J Fundamentals of Probability Fall 2018 For information about citing these materials or our Terms of Use, visit:
13393	https://web2.ph.utexas.edu/~vadim/Classes/2014s/TrigReview.pdf	Trigonometric Functions By Daria Eiteneer Topics Covered: Reminder: relationship between degrees and radians The unit circle Definitions of trigonometric functions for a right triangle Definitions of trigonometric functions for a unit circle Exact values for trigonometric functions of most commonly used angles Trigonometric functions of any angle θ' in terms of angle θ in quadrant I Trigonometric functions of negative angles Some useful relationships among trigonometric functions Double angle formulas Half angle formulas Angle addition formulas Sum, difference and product of trigonometric functions Graphs of trigonometric functions Inverse trigonometric functions Principal values for inverse trigonometric functions Relations between inverse trigonometric functions Graphs of inverse trigonometric functions Using trigonometric functions: components of a vector Using trigonometric functions: phase shift of a wave Derivatives of trigonometric functions Note: All figures, unless otherwise specified, have a permission to be copied, distributed and/or modified under the terms of the GNU Free Documentation License, Version 1.2 or later. Reminder: Relationship Between Degrees and Radians A radian is defined as an angle subtended at the center of a circle for which the arc length is θ equal to the radius of that circle (see Fig.1). Fig.1. Definition of a radian. The circumference of the circle is equal to 2 R, where R is the radius of the circle. Consequently, π 360°=2 radians. Thus, π 1 radian=360°/2 π ≈ 57.296° 1° = (2 /360) radians π 0.01745 radians ≈ The Unit Circle In mathematics, a unit circle is defined as a circle with a radius of 1. Often, especially in applications to trigonometry, the unit circle is centered at the origin (0,0) in the coordinate plane. The equation of the unit circle in the coordinate plane is x2 + y2 = 1. As mentioned above, the unit circle is taken to be 360°, or 2 radians. We can divide the π coordinate plane, and therefore, the unit circle, into 4 quadrants. The first quadrant is defined in terms of coordinates by x>0, y>0, or, in terms of angles, by 0°< <90°, or 0< < /2. The second θ θ π quadrant is defined by x<0, y>0, or 90°< <180°, or /2< < . The third quadrant is defined by θ π θ π x<0, y<0, or 180°< <270°, or < <3 θ π θ π/2. Finally, the fourth quadrant is defined by x>0, y<0, or 270°< <360°, or 3 /2< <2 . θ π θ π Trigonometric Functions Definitions of Trigonometric Functions For a Right Triangle A right triangle is a triangle with a right angle (90°) (See Fig.2). Fig.2. Right triangle. For every angle in the triangle, there is the side of the triangle adjacent to it (from here on θ denoted as “adj”), the side opposite of it (from here on denoted as “opp”), and the hypotenuse (from here on denoted as “hyp”), which is the longest side of the triangle located opposite of the right angle. For angle , the trigonometric functions are defined as follows: θ sine of = sin = θ θ opp hyp cosine of = cos = θ θ adj hyp tangent of = tan = θ θ sinθ cosθ = opp adj cotangent of = cot = θ θ 1 tanθ = cosθ sinθ = adj opp secant of = sec = θ θ 1 cosθ = hyp adj cosecant of = csc = θ θ 1 sinθ = hyp opp Definitions of Trigonometric Functions For a Unit Circle In the unit circle, one can define the trigonometric functions cosine and sine as follows. If (x,y) is a point on the unit cirlce, and if the ray from the origin (0,0) to that point (x,y) makes an angle θ with the positive x-axis, (such that the counterclockwise direction is considered positive), then, cos = x/1 = x θ sin = y/1 = y θ Then, each point (x,y) on the unit circle can be written as (cos , sin ). Combined with the θ θ equation x2 + y2 = 1, the definitions above give the relationship sin2 +cos θ 2 =1. In addition, other θ trigonometric functions can be defined in terms of x and y: tan = sin /cos = y/x θ θ θ cot = cos /sin = x/y θ θ θ sec = 1/cos = 1/x θ θ csc = 1/ θ sin = 1/y θ Fig.3 below shows a unit circle in the coordinate plane, together with some useful values of angle θ, and the points (x,y)=(cos , sin ), that are most commonly used (also see table in the following θ θ section). Fig.3. Most commonly used angles and points of the unit circle. Note: For in quadrant I, θ sinθ>0, cos θ >0; for in quadrant II, sin >0, cos <0; for in quadrant θ θ θ θ III, sin <0, cos <0; and for in quadrant IV, sin θ θ θ θ Exact Values for Trigonometric Functions of Most Commonly Used Angles in degrees θ in radians θ sinθ cosθ tanθ 0 0 0 1 0 30 π 6 1 2 3 2 3 3 45 π 4 2 2 2 2 1 60 π 3 3 2 1 2 3 90 π 2 1 0 undefined 180 π 0 -1 0 270 3 π 2 -1 0 undefined 360 2π 0 1 0 Note: Exact values for other trigonometric functions (such as cotθ, secθ, and cscθ) as well as trigonometric functions of many other angles can be derived by using the following sections. Trigonometric Functions of Any Angle θ' in Terms of Angle θ in Quadrant I θ' sinθ' cosθ' tanθ' θ' sinθ' cosθ' tanθ' 90°+θ π/2+θ cosθ -sinθ cotθ 90°-θ π/2θ cosθ sinθ cotθ 180°+θ π+θ sinθ cosθ tanθ 180°-θ π-θ sinθ cosθ tanθ 270°+θ 3π/2+θ cosθ sinθ cotθ 270°-θ 3π/2θ cosθ sinθ cotθ k(360°)+θ k(2π)+θ k=integer sinθ cosθ tanθ k(360°)-θ k(2π)-θ k=integer sinθ cosθ tanθ Trigonometric Functions of Negative Angles sin(- ) = -sin θ θ cos(- ) = cos θ θ tan(- ) = -tan θ θ Some Useful Relationships Among Trigonometric Functions sin2 + cos θ 2 = 1 θ sec2 – tan θ 2 = 1 θ csc2 – cot θ 2 = 1 θ Double Angle Formulas sin2 = 2 sin cos θ θ θ cos2 = cos θ 2θ – sin2θ = 1-2 sin2θ = 2 cos2θ -1 tan2θ = 2 tanθ 1−tan 2θ Half Angle Formulas Note: in the formulas in this section, the “+” sign is used in the quadrants where the respective trigonometric function is positive for angle /2, and the “-” sign is used in the quadrants where the θ respective trigonometric function is negative for angle /2. θ sin θ 2 = ±  1−cosθ 2 cos θ 2 = ±  1cosθ 2 tan θ 2 = ±  1−cosθ 1cosθ = sinθ 1cosθ = 1−cosθ sinθ Angle Addition Formulas Note: in this and the following section, letters A and B are used to denote the angles of interest, instead of the letter . θ sinA±B= sinAcosB±cosAsinB cosA± B= cosAcos B∓sinAsinB tanA±B= tanA±tanB 1∓tanAtanB cot(A±B)= cotAcotB∓1 cotB±cotA Sum, Difference and Product of Trigonometric Functions sinA + sinB = 2sinAB 2 cosA−B 2  sinA – sinB = 2sinA−B 2 cosAB 2  cosA + cosB = 2cosAB 2 cosA−B 2  cosA – cosB = −2sinAB 2 sinA−B 2  sinA sinB = 1 2 [cosA−B−cosAB] cosA cosB = 1 2 [cosA−BcosAB] sinA cosB = 1 2 [sinA−BsinAB] Graphs of Trigonometric Functions (Fig.4, a-f) Note: In each graph in Fig.4, the horizontal axis (x) is measured in radians. Ref. Weisstein, Eric W. “Sine.” “Cosine.” “Tangent.” “Cotangent.” “Secant.” “Cosecant.” From MathWorld – A Wolfram Web Resource: Fig.4a. Graph of sin(x). Fig.4b. Graph of cos(x). Fig.4c. Graph of tan(x). Fig.4d. Graph of cot(x). Fig.4e. Graph of sec(x). Fig.4f. Graph of csc(x). Inverse Trigonometric Functions Inverse Trigonometric Functions If x=sin(y), then y=sin-1(x), i.e. s is the angle whose sine is y. In other words, x is the inverse sine of y. Another name for inverse sine is arcsine, and the notation used is y=arcsin(x). Similarly, we can define inverse cosine, inverse tangent, inverse cotangent, inverse secant and inverse cosecant. All of the inverse functions are many-valued functions of x (for each value of x, there are many corresponding values of y), which are collections of single-valued functions (for each value of x, there is only one corresponding value of y) called branches. For many purposes a particular branch is required. This is called the principal branch and the values for this branch are called principal values. x = sin(y) y = sin1(x) = arcsin(x) x = cos(y) y = cos 1(x) = arccos(x) x = tan(y) y = tan1(x) = arctan(x) x = cot(y) y = cot 1(x) = arccot(x) x = sec(y) y = sec1(x) = arcsec(x) x = csc(y) y = csc 1(x) = arccsc(x) Principal Values for Inverse Trigonometric Functions Principal Values for x0 Principal Values for x0 0sin −1xπ/2 −π/2sin −1x0 0cos −1xπ/2 π/2cos −1xπ 0tan −1xπ/2 −π/2tan −1x0 0cot −1xπ/2 π/2cot −1xπ 0sec −1xπ/2 π/2sec −1xπ 0csc −1xπ/2 −π/2csc −1x0 Relations Between Inverse Trigonometric Functions Note: In all cases, it is assumed that principal values are used. sin-1x + cos-1x = /2 π tan-1x + cot-1x = /2 π sec-1x +csc-1x = /2 π csc-1x = sin-1(1/x) sec-1x = cos-1(1/x) cot-1x = tan-1(1/x) sin-1(-x) = -sin-1x cos-1(-x) = - cos π -1x tan-1(-x) = -tan-1x cot-1(-x) = – cot π -1x sec-1(-x) = – sec π -1x csc-1(-x) = -csc-1x Graphs of Inverse Trigonometric Functions (Fig.5, a-f) Note: In each graph in Fig.5, the vertical axis (y) is measured in radians. Only portions of curves corresponding to principal values are shown. Ref. Weisstein, Eric W. “Inverse Sine.” “Inverse Cosine.” “Inverse Tangent.” “Inverse Cotangent.” “Inverse Secant.” “Inverse Cosecant.” From MathWorld – A Wolfram Web Resource: Fig.5a. Graph of sin-1(x). Fig.5b. Graph of cos-1(x). Fig.5c. Graph of tan-1(x). Fig.5d. Graph of cot-1(x). Fig.5e. Graph of sec-1(x). Fig.5f. Graph of csc-1(x). Using Trigonometric Functions Resolving Vectors into Components: The geometric way of adding vectors is not recommended whenever great accuracy is required or in three-dimensional problems. In such cases, it is better to make use of the projections of vectors along coordinate axis, also known as components of the vector. Any vector can be completely described by its components. Consider an arbitrary vector A (from now on, the bold-cased letters are used to signify vectors, whereas the regular-font letter A signifies the length of the vector A) lying in the xy-plane and making an arbitrary angle θ with the positive x-axis, as shown in Fig.6: Fig.6. Arbitrary vector A in the xy-plane. This vector A can be expressed as the sum of two other vectors, Ax and Ay. From Fig.6, it can be seen that the three vectors form a right triangle and that A=Ax+Ay. It is conventional to refer to the “components of vector A,” written Ax and Ay (without the boldface notation). The component Ax represents the projection of A along the x-axis, and the component Ay represents the projection of A along the y-axis. These components can be positive or negative. From Fig.6 and the definition of sine and cosine for a right triangle, we see that cos =A θ x/A and sin =A θ y/A. Hence, the components of A are Ax=A cosθ Ay=A sinθ These components from the two sides of a right triangle with hypotenuse of length A. Thus, it follows that the magnitude and direction of A are related to its components through the expressions A=Ax 2Ay 2 θ=tan −1Ay Ax  Note that the signs of the comonents Ax and Ay depend on the angle . For example, if =120 θ θ °, then Ax is negative and Ay is positive. When solving problems, one can specify a vector A either with its components Ax and Ay or with its magnitude and direction A and . Furthermore, one can express a vector A as θ A=Ax  xAy  y=Acosθ  xsinθ  y, where  x and  y are unit vectors (length of one) in the direction of x- and y-axis, respectively. Writing a Phase Shift of a Wave Waves can have many different shape. One of the simplest to deal with and also one that is of a particular interest when it comes to simple harmonic motion is a sinusoidal wave. It is called “sinusoidal” because the shape that the wave takes either in space or in time is that of a sine curve (see Fig.7a,b): Fig.7.(a) A one-dimensional sinusoidal wave in space. Fig.7b. A one-dimensional sinusoidal wave in time. A function describing such a wave can be written as y=Asin2 π λ x±2 π T tφ, where A is the amplitude of the wave, is the wavelength of the wave, T is the period of the wave, λ and φ is the phase shift of the wave. The ± sign in front of the second term in the parentheses depends on whether the wave is moving to the right (-) or to the left (+). The entire expression in the parentheses can also be written as (x,t) and is called the total phase of the wave, and is a Φ function of position x and time t, in contrast to the phase shift , which is a constant φ . Then, the entire expression becomes y = A sin (x,t). Φ Because the sine function is periodic with period 2 , sin( +2 n)=sin( ) for any integer n. π Φ π Φ φ is referred to as a phase-shift, because it represents a "shift" from zero phase. But a change in φ is also referred to as a phase-shift. Fig.8 shows two curves: the red one with zero phase, and the blue one with a non-zero phase. Fig.8. Phase shift. Two oscillators that have the same frequency and same wavelength, will have a phase difference, if their phase shift is different. If that is the case, φ the oscillators are said to be out of phase with each other. The amount by which such oscillators are out of phase with each other can be expressed in degrees from 0° to 360°, or in radians from 0 to 2 (see Fig.9a,b). π Fig.9a. Waves that are in phase. Fig.9b. Waves that are out of phase. Derivatives of Trigonometric and Inverse Trigonometric Functions d dx sin x=cos x d dx cos x=−sin x d dx tan x=sec 2 x d dx cot x=−csc 2 x d dx sec x=sec x tan x d dx csc x=−csc x cot x d dx sin −1 x= 1 1−x 2 , −π 2 sin −1 xπ 2 d dx cos −1 x= −1 1−x 2 , 0cos −1 xπ d dx tan −1 x= 1 1x 2 , −π 2 tan −1 xπ 2 d dx cot −1 x= −1 1x 2 , 0cot −1 xπ d dx sec −1 x= ±1 xx 2−1 , + if 0sec −1 xπ 2 , - if π 2 sec −1 xπ d dx csc −1 x= ∓1 xx 2−1 , - if 0csc −1 xπ 2 , + if −π 2 csc −1 x0
13394	https://barryhdayton.space/theoryEquations/textpartX.pdf	Theory of Equations Lesson 10 by Barry H. Dayton Northeastern Illinois University Chicago, IL 60625, USA www.neiu.edu/˜bhdayton/theq/ These notes are copyrighted by Barry Dayton, 2002. The PDF ﬁles are freely available on the web and may be copied into your hard drive or other suitable electronic storage devices. These ﬁles may be shared or distributed. Single copies may be printed for personal use but must list the website www.neiu.edu/˜bhdayton/theq/ on the title page along with this paragraph. “Maple” and “MAPLE” represent registered trademarks of Waterloo Maple Inc. 4.6. NEWTON’S IDENTITIES 123 4.6 Newton’s Identities While it was noticed early that the roots of a polynomial are not rational expressions of the coefﬁcients, the mathematician Girard noticed in 1629 that certain expressions in the roots were rational expressions of the coefﬁcients. Suppose we consider the polynomial equation t3 + p1t2 + p2t + p3 = 0 Let x1, x2, x3 be the roots. Then Girard noted that x1 + x2 + x3 = −p1 x2 1 + x2 2 + x2 3 = p2 1 −2p2 x3 1 + x3 2 + x3 3 = −p3 1 + 3p1p2 −3p3 x4 1 + x4 2 + x4 3 = p4 1 −4p2 1p2 + 4p1p3 + 2p2 2 While it seems clear that we can continue, it was not clear to Girard what the general pattern was. In fact, the general pattern is quite complicated but Isaac Newton published in 1683 a simple set of recursive equations. Note that for convenience we are writing the coefﬁcients of the polynomial in a non-standard order. Theorem 4.6.1 (Newton’s Identities) Let f(t) = tn + p1tn−1 + p2tn−2 + · · · + pn−1t + pn be a polynomial with roots (counted according to multiplicity) x1, x2, . . . , xn. For j = 1, 2, 3, . . . let sj = xj 1 + xj 2 + · · · + xj n Set pk = 0 for k > n. Then for all j > 0 sj + p1sj−1 + p2sj−2 + · · · + pj−1s1 + jpj = 0 What this theorem says is that we have the equations s1 + p1 = 0 s2 + p1s1 + 2p2 = 0 s3 + p1s2 + p2s1 + 3p3 = 0 s4 + p1s3 + p2s2 + p3s1 + 4p4 = 0 124 The ﬁrst equation allows us to solve for s1 in terms of p1. Then since we know s1 and the pk the second equation allows us to solve for s2. Now we know s1, s2 and the pk so we can solve the third equation for s3. We can continue in this manner to ﬁnd s4, s5 and so on for as long as we like. The reader should check that solving the four equations gives Girard’s formulas above. Newton did not bother to supply a proof for these identities. We will sketch a proof modiﬁed from Uspensky. We will work with formal power series, i.e. expressions of the form P∞ i=0 aixi . By formal we mean that we won’t worry about convergence, we will only do algebraic operations. The operations of sum and product for formal power series are as deﬁned for polynomials in Chapter 1 and we obtain an integral domain. The main surprise is that formal power series with non-zero constant term have multiplicative inverses. In particular we have the geometric series 1 1 −x = 1 + x + x2 + x3 · · · Proof: We start with the factorization f(t) = tn + p1tn−1 + · · · + pn = (t −x1)(t −x2) · · · (t −xn) We then take the logarithmic derivative f ′(t) f(t) = 1 t −x1 + 1 t −x2 + · · · 1 t −xn Multiplying by t gives tf ′(t) f(t) = t t −x1 + t t −x2 + · · · t t −xn (4.10) Now if we expand each term on the right using the geometric series we have t t −xi = 1 1 −xi t = 1 + xi t + x2 i t2 + x3 i t3 + · · · Adding, we see that the right hand side of 4.10 is n + (x1 + · · · + xn)1 t + (x2 1 + · · · + x2 n) 1 t2 + (x3 1 + · · · + x3 n) 1 t3 + · · · or, alternatively (writing s0 = n) tf ′(t) f(t) = s0 + s1 t + s2 t2 + s3 t3 + · · · (4.11) 4.6. NEWTON’S IDENTITIES 125 Multiplying 4.10 by f(t) then gives tf ′(t) = (tn + p1tn−1 + · · · + pn)(s0 + s1 t + s2 t2 + · · · ) (4.12) Multiplying out the right hand side of 4.12 gives a power series bntn + bn−1tn−1 + · · · b0 + b−1t−1 + · · · where bn−j = sj + p1sj−1 + · · · + pjs0 On the other hand, tf ′(t) = ntn + (n −1)p1tn−1 + · · · + pn−1t This says that the coefﬁcient of tn−j on the left is (n −j)pj which makes sense for all j > 0 since we set pj = 0 for j > n. Finally (since we are using formal power series we may equate the coefﬁcients of tn−j on both sides of 4.12 to get (n −j)pj = sj + p1sj−1 + · · · + pjs0 Subtracting (n −j)pj from both sides of this equation and using the fact that s0 = n gives Newton’s identity. We give an application of the use of Newton’s Identities. This is the application that Newton had in mind and gives a method for ﬁnding the largest real root of a polynomial (assuming such a root exists, is not a multiple root, and is actually the root of largest modulus). We note that this method is of no practical value today. The method is based on the fact that if the real root xn is is of larger modulus than any other root then for large k, sk = xk 1 + · · · + xk n ≈xk n. Thus the sequence s1, √s2, 3 √s3, 4 √s4, . . . should converge to xn. Example 4.6.2 Let f(x) = x3 −5x2 + 6x −1. Then p1 = −5, p2 = 6 and p3 = −1. From the identities we get s1 = −p1 = 5 s2 = −p1s1 −2p2 = −(−5)(5) −2(6) = 13 s3 = −p1s2 −p2s1 −3p3 = −(−5)(13) −(6)(5) −3(−1) = 38 s4 = −p1s3 −p2s2 −p3s1 = −(−5)(38) −(6)(13) −(−1)(5) = 117 Thus our sequence is s1 = 5, √s2 = 3.6055, 3 √s3 = 3.36197, 4 √s4 = 3.2888, . . . The actual root is 3.2469 126 Maple Implementation For actually calculating the ghost coefﬁcients sj with Maple it is easier to use equation 4.11 directly than Newton’s identities. The idea is to expand the left hand side of this equation in an asymptotic series, that is a series in negative powers of t. Thus to ﬁnd the sj in the above example one might do f:= tˆ3 - 5tˆ2 + 6t-1; asympt(tdiff(f,t)/f, t, 5); and the result would look like 3 + 5 t + 13 t2 + 38 t3 + 117 t4 + O( 1 t5) where s0, s1, . . . , s4 are the numerators and the O-term at the end is to remind you that this is an inﬁnite series. Replacing the 5 with a larger number will get you as many terms as you like. Exercise 23a [10 points] Use the method just described to ﬁnd the largest real root of f(x) = x4 −2x3 −5x2 + 6x + 3 correct to 2 signiﬁcant digits. 4.7 More on Newton’s Identities In this optional section we consider some more applications of Newton’s identities. The material in this section will not be needed in the sequel, so you may wish to skip this section to maintain the continuity of our story. For our next application we note that if we know the sk we can then calculate the coefﬁcients pj of our polynomial. For solving Newton’s identities for pj in terms of the sk we have p1 = −s1 p2 = s2 1 −s2 2 p3 = −2s3 + 2s1s2 + s3 1 −s1s2 6 and so on. Evidently these formulas get very complicated quickly, but numerically it is easy to solve for the pj’s directly from Newton’s Identities. 4.7. MORE ON NEWTON’S IDENTITIES 127 This technique has been used to calculate the eigenvalues of a matrix. Recall that the eigenvalues of an n × n matrix A are the roots of the characteristic polynomial f(λ) = det(λI−A) = λn+p1λn−1+p2λn−2+· · ·+pn−1λ+pn. (Note that we are using det(λI −A) rather than the formula det(A−λI) found in some linear algebra books in order that our polynomial be monic, these formulas differ by a factor of (−1)n.) If the eigenvalues (complex and counted according to multiplicities) are x1, x2, . . . , xn then it is not too difﬁcult to see that the sum s1 = x1 + · · · + xn is the sum of the diagonal entries of A, in fact both are equal to −p1. The sum of the diagonal entries of a matrix A is called the trace of the matrix, trace(A). It is a bit harder to see that the eigenvalues of A2 are x2 1, x2 2, . . . , x2 n. When there are no repeated eigenvalues this follows from the fact that λ is an eigenvalue of A if there exists a vector v ̸= 0 so that Av = λv. Then A2v = A(Av) = A(λv) = λAv = λ2v so λ2 is an eigenvalue of A2. It follows that the trace of A2 is then x2 1 +x2 2 +· · ·+x2 n = s2. More generally the argument above suggests that sk = xk 1+· · ·+xk n is the trace of Ak for each k > 0. In fact this is true. Thus the method is to calculate Ak for k = 1, 2, . . . , n and set sk to be the trace of Ak. Then working backwards using Newton’s identities we can ﬁnd the coefﬁcients p1, p2, . . . , pn of the characteristic polynomial. Example 4.7.1 We wish to ﬁnd the eigenvalues of the matrix A =   1 2 0 3 −2 1 2 1 4   We ﬁrst calculate A2 =   7 −2 2 −1 11 2 13 6 17  , A3 =   5 20 6 36 −22 19 65 31 74   We then see that s1 = trace(A) = 3, s2 = trace(A2) = 35 and s3 = trace(A3) = 57. Now from the identity s1 + p1 = 0 we calculate p1 = −3. From the identity s2 + p1s1 + 2p2 = 0 we have 35 + (−3)(3) + 2p2 = 0 so p2 = −13. Finally from s3 + p1s2 + p2s1 + 3p3 = 0 128 we obtain 57 + (−3)(35) + (−13)(3) −3p3 = 0 so p3 = 29. We thus conclude that the characteristic polynomial of A is f(λ) = λ3 −3λ2 −13λ + 29 Using any method from Chapter 2 we see that the roots of f(λ) are −3.3813, 1.9244 and 4.4569, i.e. these are the eigenvalues of A. Exercise 23b [20 points] Find the characteristic polynomial and the eigenvalues of the matrix A =     1 2 0 −1 2 1 −2 0 0 −2 3 3 −1 0 3 1     Hint: The eigenvalues are real. There are similar identities for ﬁnding sums of negative powers of the roots of a polynomial. Theorem 4.7.2 Let f(x) = p0tn+p1tn−1+· · ·+pn be a polynomial with roots (counted according to multiplicity) x1, x2, . . . , xn. Assume that pn ̸= 0, i.e. that no xi = 0 and set pk = 0 for k < 0. Deﬁne for k ≥0 s−k = x−k 1 + x−k 2 + · · · + x−k n = 1 xk 1 + · · · + 1 xk n. Then for all j ≤n (n −j)pj + pj+1s−1 + · · · + pnsj−n = 0 The proof is similar to that of Theorem 4.6.1 by noting that expansion as a formal power series in positive powers of t gives −tf ′(t) f(t) = s−1t + s−2t2 + s−3t3 + · · · (4.13) Maple Implementation Again the most efﬁcient way to calculate the ghost coefﬁcients s−j is to expand the left hand side of equation 4.13 as a series, eg. series(-tdiff(f,t)/f,t,8); would give s−j as the coefﬁcient of tj for j = 1, 2, . . . , 7. 4.7. MORE ON NEWTON’S IDENTITIES 129 Often in the literature the Newton’s identity for positive and negative powers is combined by adding the two identities as follows: Corollary 4.7.3 With hypotheses as in the previous theorem, setting s0 = n we have for all integers (positive, negative and zero) j p0sj + p1sj−1 + p2sj−2 + · · · + pnsj−n = 0 Actually, the Newton’s Identity for negative powers looks nicer if we use our more standard notation for polynomials: Corollary 4.7.4 Let f(x) = a0 + a1t + a2t2 + · · · + antn be a polynomial of degree n with a0 ̸= 0. Let x1, . . . , xn be the roots counted according to multiplicity and let s−k = x−k 1 + · · · + x−k n for all k > 0. Then for all j > 0 jaj + aj−1s−1 + aj−2s−2 + · · · + a0s−j = 0 It should be noted from this result that alghough the number of roots and the actual roots depends on the degree and all the coeﬁcients, the sum of the jth powers of the reciprocals of the roots depends only on the coefﬁcients a0, a1, . . . , aj when j < n and is independent of the degree. This observation motivated the mathematician Euler to apply this last corollary to power series. Most modern mathematicians will say that Euler’s argument is wrong, but his results are correct. Euler wanted to calculate the number ζ(n) = ∞ X k=1 1 kn for n a positive even integer. To this end he started with the series sin(x) = x −x3 3! + x5 5! −x7 7! + · · · and noted that sin(x) has roots kπ for all integers k. Dividing by x eliminates the root at 0 and replacing x by √x eliminates the negative roots according to Euler. Thus Euler argued that f(x) = sin(√x) √x = 1 −x 3! + x2 5! −x3 7! + · · · 130 has roots at (kπ)2 for k = 1, 2, 3, . . .. This is certainly correct, however (here is the questionable step!) Euler then set s−j = ∞ X k=1 1 ((kπ)2)j = 1 π2j ∞ X k=1 1 k2j (4.14) and calculated the s−j using the Newton’s Identities of the last corollary. For j = 1 he had 1a1 + a0s−1 = 0 so s−1 = −a1 = 1 3! = 1/6. Multiplying equation (4.14) by π2 gives ζ(2) = ∞ X k=1 1 k2 = π2s−1 = π2 6 Next using 2a2 + a1s−1 + a0s−2 = 0 gives s−2 = ( 1 3!)2 −2 5! = 1 90. Multiplying (4.14) by π4 gives ζ(4) = ∞ X k=1 1 k4 = π4s−2 = π4 90 It should be mentioned that there is still no simple formula for odd values of ζ, for example ζ(3) = P∞ k=1 1 k3 is known only numerically and only in the last few years has even been shown to be irrational. Exercise 23c [10 points] Assuming that Euler’s calculation is correct, ﬁnd ζ(6) and ζ(8). 4.8 Symmetric Polynomials The expressions sk = xk 1+· · ·+xk n are examples of symmetric polynomials. Symmetric polynomials played a large role in the development of the modern theory of solvability of polynomials. The 18th century mathmatician Edward Waring is often credited for much of the development of the theory of symmetric polynomials. More generally we start with a polynomial f(x1, x2, . . . , xn) in n variables xj. This means that f(x1, x2, . . . , xn) is a sum of terms each one is a constant times positive powers of some or all of the variables xj. f(x1, x2, . . . , xn) is symmetric if any permu-tation of the variables leaves the result unchanged. More precisely, f(x1, x2, . . . , xn) is symmetric if for all 1 ≤j < k ≤n f(. . . , xj, . . . , xk, . . .) = f(. . . , xk, . . . , xj, . . .). For example f(x1, x2, x3) = x2 1x2x3 + x1x2 2x3 + x1x2x2 3 is a symmetric polynomial but g(x1, x2, x3) = x1x2 2x3 3 4.8. SYMMETRIC POLYNOMIALS 131 is not for g(x2, x1, x3) = x2 1x2x3 3 ̸= g(x1, x2, x3). Consider a polynomial f(t) = t3 + p1t2 + p2t + p3 of degree three with roots x1, x2, x3. Then f(t) factors as f(t) = (t −x1)(t −x2)(t −x3). Multiplying this last expression back out we get f(t) = t3−(x1+x2+x3)t2+(x1x2+x1x3+x2x3)t−x1x2x3. Thus we conclude that p1 = −(x1 + x2 + x3) p2 = x1x2 + x1x3 + x2x3 p3 = −x1x2x3 We note that the coefﬁcients p1, p2, p3 are symmetric functions of the roots x1, x2, x3 which should not be suprising since the order in which we listed the roots was clearly irrelevant. This pattern holds also for polynomials of higher degree and was well known to mathematicians of the 17th and 18th century. Theorem 4.8.1 Let f(t) = tn + p1tn−1 + · · · + pn−1t + pn be a monic polynomial of degree n with roots x1, x2, . . . , xn. Then p1 = −(x1 + x2 + · · · + xn) = − n X j=1 xj p2 = x1x2 + · · · + xn−1xn = X 1≤j<k≤n xjxk p3 = −(x1x2x3 + · · · + xn−2xn−1xn) = − X 1≤j<k<ℓ≤n xjxkxℓ . . . pn = (−1)nx1x2 · · · xn Example 4.8.2 Find a polynomial with roots 1,2,3,4. By the Theorem p1 = −(1 + 2 + 3 + 4) = −10 p2 = 1 ∗2 + 1 ∗3 + 1 ∗4 + 2 ∗3 + 2 ∗4 + 3 ∗4 = 35 p3 = −(1 ∗2 ∗3 + 1 ∗2 ∗4 + 1 ∗3 ∗4 + 2 ∗3 ∗4) = −50 p4 = 1 ∗2 ∗3 ∗4 = 24 so f(t) = t4 −10t3 + 35t2 −50t + 24. 132 Example 4.8.3 Suppose we know for some reason (eg. we started Graeffe’s method) that f(x) = x3 −x2 −x −15 has an imaginary root of modulus √ 5. We wish to ﬁnd the roots. Call this root x1 = a + bi. Then x2 = a −bi is also a root so \|x1\|2 = x1x2 = 5. Now p3 = −15 = −x1x2x3 = −5x3 so we see easily that x3 = 3. But p1 = −x1−x2−x3 = −2a−3 = −1 so −2a = 2 or a = −1. Since \|x1\|2 = 5 = a2+b2 it follows that b2 = 4 so b = ±2. Thus the roots are −1 + 2i, −1 −2i and 3. Since Theorem 4.8.1 gives the close relation between the roots and the coefﬁcients one might hope that these formulas might give a way to ﬁnd the roots given the coefﬁ-cients. Unfortunately this will not work, but to some extent these formulas are the basis for all attempts after the days of Cardano. In Theorem 4.8.1 we can view p1, p2, . . . as symmetric polynomials in the variables x1, . . . , xn. These symmetric polynomials are often known in the literature as the ele-mentary symmetric polynomials. Thus we can view Theorem 4.6.1 as saying that the symmetric functions sk = xk 1 + · · · + xk n are polynomials in the elementary symmetric polynomials, eg. as Girard noted s3 = −p3 1+3p1p2−3p3. What is much more suprising is the following: Theorem 4.8.4 (Fundamental Theorem on Symmetric Polynomials) Let f(x1, x2, . . . , xn) be a symmetric polynomial in n variables with coefﬁcients in an in-tegral domain R. Then f(x1, x2, . . . , xn) can be expressed as a polynomial in the n elementary symmetric functions p1, p2, . . . , pn with coefﬁcients in R. The coefﬁcient ring R will generally be Z or Q. Proofs of this theorem are con-tained in the books by Uspenski and Lang, in addition Adams and Loustanau sketch a modern proof as an exercise and details of this proof can be found in Fine and Rosen-berger. We will simply illustrate by some examples. Example 4.8.5 In the two variable case the elementary symmetric functions are p1 = −x1 −x2 and p2 = x1x2. Consider the symmetric polynomial D = (x1 −x2)2. Expanding, D = x2 1 −2x1x2 + x2 2 = s2 −2p2 where s2 = x2 1 + x2 2 as in §4.6. By Newton’s Identities s2 = p2 1 −2p2 so D = p2 1 −2p2 −2p2 = p2 1 −4p2 Note that this is just the discriminant of the quadratic polynomial f(t) = t2 + p1t + p2. Example 4.8.6 For a real cubic polynomial f(t) = t3+p1t2+p2t+p3 the discriminant is D = (x1 −x2)2(x1 −x3)2(x2 −x3)2 where x1, x2, . . . , x3 are the roots. It is a bit much work to put down here but it has been shown that D = 18p1p2p3 −4p3 1p3 + p2 1p2 2 −4p3 2 −27p2 3 If p1 = 0 as in §4 then we simply have D = −4p3 2 −27p2 3 as in Theorem 4.4.1. 4.8. SYMMETRIC POLYNOMIALS 133 More generally given a polynomial f(t) = tn + p1tn−1 + · · · + pn of degree n with roots x1, . . . , xn the discriminant is deﬁned by D = (x1 −x2)2(x1 −x3)2 · · · (xn−1 −xn)2 i.e. the product of all differences (xj −xk)2 for j < k. The general formula, which requires some advanced techniques to justify, is D = det        s0 s1 · · · sn−1 s1 s2 · · · sn s2 s3 · · · sn+1 . . . . . . ... . . . sn−1 sn · · · s2n−2        where as in §6 sk = xk 1 + · · · + xk n. Newton’s Identities then can be used to express D in terms of the pj’s. Note in particular that if the coefﬁcients pj are real then D is real, if the pj are all integers then D is an integer. The generalization of Theorem 4.4.1 is Theorem 4.8.7 Let D be the discriminant of f(t) as above, where the coefﬁcients pj are real. Then f(t) has multiple roots if and only if D = 0. Otherwise if D > 0 then f(t) has an even number of pairs of imaginary roots, if D < 0 then f(t) has an odd number of pairs of imaginary roots. It should go without saying that except for n = 2, 3 calculating the discriminant is a terrible way to tell if f(t) has multiple roots or to count real roots. We will need a few more calculations for the next section: Example 4.8.8 Let f(t) = t4 +p1t3 +p2t2 +p3t+p4 be a biquadratic polynomial with roots x1, x2, x3, x4. Consider A = x1 + x2 −x3 −x4 B = x1 −x2 + x3 −x4 C = x1 −x2 −x3 + x4 Let a = A2, b = B2 and c = C2. Then a + b + c, ab + ac + bc and ABC are all seen to be symmetric polynomials in x1, . . . , x4. We can calculate a + b + c = 3p2 1 −8p2 (4.15) ab + ac + bc = 3p4 1 −16p2 1p2 + 16p1p3 + 16p2 2 −64p4 (4.16) ABC = p3 1 −4p1p2 + 8p3 (4.17) 134 We leave the ﬁrst two calculations as rather hard exercises for the reader (see Uspensky for example) and tackle only the third. We note that multiplying out we can have 3 types of terms, x3 j, x2 jxk and xjxkxℓwhere in each term j, k, ℓare different. Note that x1 occurs only with “+” signs, but each of x2, x3, x4 occur exactly twice with −signs in the expressions A, B, C. Thus the expansion of ABC will contain the terms x3 j for each j so it will contain s3 = x3 1 + x3 2 + x3 3 + x3 4. To get a term of the form x2 jxk we must pick two of A, B, C to pick out the j and the k comes from the third. There are 3 ways to do this. It can be seen that two of the ways give a “−” sign but the third gives a “+” so each of the 12 terms x2 jxk occurs with a coefﬁcient of −1 in the expansion. Now note, for example when j = 1, x2 1x2 + x2 1x3 + x2 1x4 = x2 1x1 + x2 1x2 + x2 1x3 + x2 1x4 −x3 1 = x2 1(−p1) −x3 1. Repeating this for j = 2, 3, 4, adding and multiplying by the −1, we see that the contribution of the x2 jxk in ABC is (x2 1 + x2 2 + x2 3 + x2 4)p1 + (x3 1 + x3 2 + x3 3 + x3 4) = s2p1 + s3. Finally we have terms of the form xjxkxℓwhere we can take j < k < ℓ. Note that each such term can be generated 6 ways, i.e. we can choose the j from either factor A, B or C, then we have only two factors from which to choose the k and we must choose the ℓfrom the remaining factor. By careful inspection, we see that 4 ways give “+” signs and 2 ways give “−” signs, thus the contribution of the xjxkxℓterms in the product ABC is 2x1x2x3 + 2x1x2x4 + 2x1x3x4 + 2x2x3x4 = −2p3. Thus we can conclude that ABC = s3 + (s2p1 + s3) −2p3 = 2s3 + s2p1 −2p3 = 2(−p3 1 + 3p1p2 −3p3) + (p2 1 −2p2)p1 −2p3 = −p3 1 + 4p1p2 −8p3 as claimed. Exercise 23d[10 points] Write the symmetric polynomial f(x1, x2, x3) = x3 1x2x3 + x1x3 2x3 + x1x2x3 3 in three variables in terms of the elementary symmetric functions p1, p2, p3 of three variables. 4.9 Lagrange’s Solution of the Biquadratic After Cardano’s publication of del Ferro, Tartaglia and Ferrari’s solution of the cubic and biquadratic, many mathematicians tried to ﬁnd similar methods for solving the quintic (5th degree) and higher degree equations. They failed, as we now know they must, and so, for the most part, history has not recorded their efforts. Several of the attempts were more noteworthy than the others, for example Vandermonde’s almost correct solution (in 1770) of the cyclotomic equation xn −1 in radicals of degree less than n (Gauss ﬁlled in the details in 1801). However the most signiﬁcant attempt was made by Lagrange. In order to attack higher degree polynomials he started with a de-tailed analysis of the solution of the cubic and biquadratic. The information he gained 4.9. LAGRANGE’S SOLUTION OF THE BIQUADRATIC 135 was, of course, not enough to help him solve the quintic, but it laid the foundation for the proofs by Rufﬁni, Abel and Galois that the quintic and higher degree polynomials could not, in general, be solved by radicals. A good discussion of the history of these ideas can be found in B.L van der Waerden’s “A History of Algebra”. Lagrange devised solution methods for the cubic and the biquadratic, these are given in Chapter XI of Us-pensky. We give his solution of the biquadratic since, unlike the cubic where Lagrange re-derives Cardano’s equations (see Exercise 23e), the solution is given in a different, and more elegant, form than that of Ferrari. We start with a polynomial f(t) = t4 + pt3 + qt2 + rt + s where for notational simplicity we are using the letters p, q, r, s instead of p1, p2, p3, p4 respectively. Let x1, x2, x3, x4 be the roots of f(t). As in example 4.8.8 we let A = x1 + x2 −x3 −x4 B = x1 −x2 + x3 −x4 (4.18) C = x1 −x2 −x3 + x4 and a = A2, b = B2 and c = C2. We then have a + b + c = 3p2 −8q = u (4.19) ab + ac + bc = 3p4 −16p2q + 16pr + 16q2 −64s = v (4.20) abc = (p3 −4pq + 8r)2 = w (4.21) From Theorem 4.8.1 it follows that a, b, c are the roots of the resolvent cubic g(t) = t3 + ut2 + vt + w The cubic equation g(t) = 0 can be solved by Cardano’s method (or Lagrange’s method in Exercise 23e) so a, b, c can be calculated. Then A, B, C can be found by taking square roots of a, b, c, being careful only to select signs so that ABC = −p3 + 4pq −8r as required by Example 4.8.8. We then have equations (4.18) together with the equation −p = x1 + x2 + x3 + x4 Thus we have a system of 4 linear equations in the 4 unknowns x1, x2, x3, x4 which can 136 be solved once and for all by x1 = −p + A + B + C 4 x2 = −p + A −B −C 4 x3 = −p −A + B −C 4 x4 = −p −A −B + C 4 While this method is very elegant in theory, we warn the reader that in practice we may have u ̸= 0 in the resolvant cubic, and in any case the resolvant cubic may have a messy solution, the square roots of which must then be calculated! We end this section with an analysis of Lagrange’s method. It is based on the following theorem, which is a special case of a theorem proved by Lagrange. This theorem deals with what we might call somewhat symmetric polynomials in that some permutations, but perhaps not all, may leave the value unchanged. Theorem 4.9.1 Let g(x1, x2, . . . , xn) be a polynomial in n variables. Suppose that un-der all permutations of the variables the polynomial takes on exactly m different values, g1 = g, g2, . . . , gm then there is a polynomial f(t) of degree m whose coefﬁcients are polynomials in the elementary symmetric functions of x1, . . . , xn so that the roots of f(t) are g1, g2, . . . , gm. The idea of the proof is to construct the elementary symmetric functions P1, . . . , Pm of the gj, i.e. P1 = g1 + g2 + · · · + gm, Pm = g1g2 · · · gm etc. It can be seen that the Pj are actual symmetric functions and hence by Theorem 4.8.4 expressible in the elementary symmetric functions on x1, . . . , xn. But by Theorem 4.8.1 g1, . . . , gm are roots of f(t) = tm + P1tm−1 + · · · + Pm. In Lagrange’s solution of the biquadratic we took for g(x1, . . . , xn) the somewhat symmetric function a = (x1 + x2 −x3 −x4)2. It should be noted that permuting the variables gives exactly three different values, mainly a, b and c. Thus a, b, c are roots of a polynomial whose coefﬁcients are polynomials in p, q, r and s, mainly the resolvent polynomial. In connection with Theorem 4.9.1, Lagrange noted that if one multiplied the number of different values taken by g(x1, . . . , xn) by the number of permutations which left g(x1, . . . , xn) unchanged, the product will always be n!, the number of all permutations 4.10. INSOLVABILITY OF THE QUINTIC 137 of the n variables. For example, in the paragraph above, a takes on 3 distinct values under permutations, but there are 6 ways to permute the variables so that a remains unchanged. 4 ∗6 = 24 = 4!. At the time Lagrange lived, group theory had not yet been invented. When group theory was invented 100 years later the mathematician Camile Jordan named a now famous theorem of group theory after Lagrange because Lagrange’s observation was simply a special case of this general theorem. Exercise 23e [30 points] Derive Lagrange’s solution of the cubic. Let f(t) = t3+pt2+ qt + r have roots x1, x2, x3 and set A = x1 + ωx2 + ω2x3 and B = x1 + ω2x2 + ωx3 where ω = −1+ √ 3i 2 is a cube root of 1. Show that a = A3 takes on only 2 values under permutations of the roots, mainly a and b = B3 and thus these two values are roots of the resolvent quadratic. Find the coefﬁcients of the resolvent quadratic in terms of p, q and r. Note also that AB is a symmetric function of the roots so calculate AB in terms of p, q, r. Then solving the resolvent quadratic, taking appropriate cube roots of a, b to get the correct AB, the formulas for A, B and −p = x1 + x2 + x3 give three linear equations in 3 unknowns which can be solved to obtain Cardano’s Equations. 4.10 Insolvability of the Quintic Lagrange had hoped that his study of solution methods of the cubic and biquadratic would lead to a solution of the quintic (5th degree polynomial equation). After all, by his Theorem, all one needed to ﬁnd was a suitable somewhat symmetric function of 5 variables that took on exactly 4 different values under permutation of the variables. If this function was a power of a linear function of the variables, then with the additional equation −p1 = x1 + · · · + x5 he could then solve the “resolvent biquadratic” and take roots to obtain 5 equations in 5 unknowns which could be solved for the roots of the original polynomial. Unfortunately, such a function does not exist. While Lagrange was optimistic that the solution method would be found, the Ital-ian mathematian Paulo Rufﬁni realized that Lagrange’s analysis could lead instead to a proof that no solution could exist. Rufﬁni claimed that he had such a proof in 1798 but many mathematicians were skeptical. What Rufﬁni actually proved was that La-grange’s method would not lead to a solution, but there was a gap in his argument that if Lagrange’s method did not work, no method would work. In 1824 the 22 year old Niels Henrik Abel ﬁlled the gap in Rufﬁni’s proof. It is important to understand exactly what Rufﬁni and Abel proved. They started with variables x1, x2, . . . , x5 and deﬁned p1, p2, . . . , p5 as in 4.8.1. The polynomial 138 f(t) = t5 + p1t4 + · · · + p5 is called the general quintic. The goal was to solve this in terms of p1, . . . , p5 using only the algebraic operations of addition, subtraction, mul-tiplication, division and the taking of roots (square roots, cube roots, etc.). In other words, the goal is to recover the variables x1, . . . , x5 from the polynomials p1, . . . , p5 using only algebraic operations. Of course if this were possible then by replacing the pj by the coefﬁcients of an actual polynomial and then doing these algebraic operations then the numbers x1, . . . , x5 obtained would be the actual roots. For example, in the case of the quadratic, the general quadratic is f(t) = t2+pt+q where p = −(x1 + x2) and q = x1x2. The general quadratic formula x = −p±√ p2−4q 2 then recovers x1, x2 from p, q as follows: −p ± p p2 −4q 2 = x1 + x2 ± p (x1 + x2)2 −4x1x2 2 = x1 + x2 ± p x2 1 −2x1x2 + x2 2 2 = x1 + x2 ± p (x1 −x2)2 2 = x1 + x2 ± (x1 −x2) 2 = x1 or x2 A solution of this type is called a solution by radicals for the general equation. What Rufﬁni and Abel proved is that a solution by radicals did not exist for the general quintic. In particular, there is no single solution method which works for all quintic equations. Several questions still remain. First, perhaps while no one single solution method works for all quintics, maybe there are several methods one of which would work for any given quintic. For instance, Cardano thought, wrongly as we now know, that 13 different methods were necessary to solve all cubics. Perhaps we need 13 methods to solve all quintics? There certainly are some types of quintics which can be solved, for example the cyclotomic equation x5 −1 = 0 was solved by radicals by Vandermonde. Perhaps different methods would solve other types of quintics. The ﬁrst question then is whether this is actually true. Even if it was not possible to have a ﬁnite list of solution methods covering all quintics, one would surely expect that for any given quintic with rational coefﬁcients the roots would be algebraic expressions involving rational numbers, sums, differences, products, quotients and roots of various orders. So the second question is: “is this true?” 4.10. INSOLVABILITY OF THE QUINTIC 139 In 1831 Evariste Galois showed that in fact the answer to both questions is no! For example, not only is there no algebraic method to ﬁnd the roots of x5 −6x + 3 = 0 but the roots cannot be expressed in terms of radicals. Galois went much further than this, by showing that this negative result applies also to polynomials of degree higher than 5. More importantly, he gave a method for determining whether or not a given polynomial could be solved by radicals (at least in principle, if not in practice). Galois’ method, like the method of Rufﬁni and Abel following Lagrange, involves permuting the roots of the polynomial. But unlike Lagrange, Galois does not allow all permutations of the roots, only a group of permutations which somehow preserve the algebra of numbers which can be built up from the roots (the set of such numbers is called the root ﬁeld of the polynomial). Thus Galois replaces the polynomial by its root ﬁeld and then replaces the root ﬁeld by the abstract algebraic object now known as the Galois group and shows that solvability of the polynomial is equivalent to some facts about the structure of the group. Galois’ proof method, now known as Galois Theory remains this day as one of the most elegant theories in mathematics. As there are many good accounts in the mathematical literature we will not pursue this any further here. For the reader who wants a reasonably elementary introduction we recommend the account in Birkhoff and Mac Lane or the one in Fine and Rosenberger. The technique of replacing one mathematical object (eg. polynomials) by others easier to analyze (eg. ﬁelds, groups) has become central in modern mathematics. In addition there are many important direct applications of Galois theory. In a modern research journal such as the Bulletin of the American Mathematical Society the name “Galois” appears as often as the name of any other single mathematician. Yet Galois’ work did not bring him any fame, or even recognition, in his lifetime. Galois sent two papers to Cauchy, who lost them. He sent one paper to Fourier who promptly died, and this paper is also lost. His most important paper (1831) was given to the mathematicians Poisson and Lacroix to review, but they couldn’t understand it. A year later Galois was shot in a duel, not yet 21 and not yet known. Finally in 1846 the paper was published by Liouville in his journal. However the importance of Galois’ work did not become apparent to the mathematical public until 1870 when Jordan published his full account of Galois theory.
13395	https://en.wikipedia.org/wiki/Stoichiometry	Jump to content Search Contents 1 Etymology 2 Definitions 3 Converting grams to moles 4 Molar proportion 5 Determining amount of product 5.1 Further examples 6 Stoichiometric ratio 7 Limiting reagent and percent yield 7.1 Example 8 Different stoichiometries in competing reactions 9 Stoichiometric coefficient and stoichiometric number 10 Stoichiometry matrix 11 Gas stoichiometry 12 Stoichiometric air-to-fuel ratios of common fuels 13 See also 14 References 15 External links Stoichiometry Afrikaans العربية Asturianu বাংলা Беларуская Български Bosanski Català Čeština Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Gaeilge 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Italiano עברית Қазақша Kreyòl ayisyen Кыргызча Latviešu Magyar മലയാളം Bahasa Melayu Монгол Nederlands 日本語 Norsk bokmål پنجابی Polski Português Romnă Русский Shqip Simple English Slovenščina Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska తెలుగు Türkçe Українська اردو 粵語中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikibooks Wikidata item Appearance From Wikipedia, the free encyclopedia Calculation of relative masses of reactants and products in chemical reactions Stoichiometry (/ˌstɔɪkiˈɒmɪtri/ ⓘ) is the relationships between the quantities of reactants and products before, during, and following chemical reactions. Stoichiometry is based on the law of conservation of mass; the total mass of reactants must equal the total mass of products, so the relationship between reactants and products must form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of the products can be empirically determined, then the amount of the other reactants can also be calculated. This is illustrated in the image here, where the unbalanced equation is: : CH4 (g) + O2 (g) → CO2 (g) + H2O (l) : However, the current equation is imbalanced. The reactants have 4 hydrogen and 2 oxygen atoms, while the product has 2 hydrogen and 3 oxygen. To balance the hydrogen, a coefficient of 2 is added to the product H2O, and to fix the imbalance of oxygen, it is also added to O2. Thus, we get: : CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) Here, one molecule of methane reacts with two molecules of oxygen gas to yield one molecule of carbon dioxide and two molecules of liquid water. This particular chemical equation is an example of complete combustion. The numbers in front of each quantity are a set of stoichiometric coefficients which directly reflect the molar ratios between the products and reactants. Stoichiometry measures these quantitative relationships, and is used to determine the amount of products and reactants that are produced or needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry. In the example above, reaction stoichiometry measures the relationship between the quantities of methane and oxygen that react to form carbon dioxide and water: for every mole of methane combusted, two moles of oxygen are consumed, one mole of carbon dioxide is produced, and two moles of water are produced. Because of the well known relationship of moles to atomic weights, the ratios that are arrived at by stoichiometry can be used to determine quantities by weight in a reaction described by a balanced equation. This is called composition stoichiometry. Gas stoichiometry deals with reactions solely involving gases, where the gases are at a known temperature, pressure, and volume and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, because of the existence of isotopes, molar masses are used instead in calculating the mass ratio. Etymology [edit] The term stoichiometry was first used by Jeremias Benjamin Richter in 1792 when the first volume of Richter's Anfangsgründe der Stöchyometrie oder Meßkunst chymischer Elemente (Fundamentals of Stoichiometry, or the Art of Measuring the Chemical Elements) was published. The term is derived from the Ancient Greek words στοιχεῖον stoikheîon "element" and μέτρον métron "measure." Ludwig Darmstaedter and Ralph E. Oesper have written a useful account on this. Definitions [edit] Main article: Amount of substance A stoichiometric amount or stoichiometric ratio of a reagent is the optimum amount or ratio where, assuming that the reaction proceeds to completion: All of the reagent is consumed There is no deficiency of the reagent There is no excess of the reagent. Stoichiometry rests upon the very laws that help to understand it better, i.e., law of conservation of mass, the law of definite proportions (i.e., the law of constant composition), the law of multiple proportions and the law of reciprocal proportions. In general chemical reactions combine in definite ratios of chemicals. Since chemical reactions can neither create nor destroy matter, nor transmute one element into another, the amount of each element must be the same throughout the overall reaction. For example, the number of atoms of a given element X on the reactant side must equal the number of atoms of that element on the product side, whether or not all of those atoms are involved in a reaction. Chemical reactions, as macroscopic unit operations, consist of many elementary reactions, where a single molecule reacts with another molecule. As the reacting molecules (or formula units or ion pairs) consist of a definite set of atoms in an integer ratio, the ratio between reactants in a complete reaction is also in an integer ratio. A reaction may consume more than one molecule, and the stoichiometric number counts this number, defined as positive for products (added) and negative for reactants (removed). The unsigned coefficients are generally referred to as the stoichiometric coefficients. Each element has an atomic mass (usually given as an average in the form of the standard atomic weight), and considering molecules as collections of atoms, every compound has a molecular mass (if molecular) or formula mass (if non-molecular), which when expressed in daltons is numerically equal to the molar mass in g/mol. By definition, the atomic mass of carbon-12 is exactly 12 Da, making its molar mass 12 g/mol. The number of chemical entities per mole in a substance is given by the Avogadro constant, exactly 6.02214076×1023 mol−1 since the 2019 revision of the SI. Thus, to calculate the stoichiometry by mass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction. The mass ratios can be calculated by dividing each by the total in the whole reaction. Elements in their natural state are mixtures of isotopes of differing mass; thus, atomic masses and thus molar masses are not exactly integers. For instance, instead of an exact 14:3 proportion, 17.031 g of ammonia consists of 14.007 g of nitrogen and 3 × 1.008 g of hydrogen, because natural nitrogen includes a small amount of nitrogen-15, and natural hydrogen includes hydrogen-2 (deuterium). A stoichiometric reactant is a reactant that is consumed in a reaction, as opposed to a catalytic reactant, which is not consumed in the overall reaction because it reacts in one step and is regenerated in another step. Converting grams to moles [edit] Stoichiometry is not only used to balance chemical equations but also used in "conversions" between quantities of a substance by dimensional analysis, e.g., converting from grams to moles using molar mass as the "conversion factor", or from grams to milliliters using density. For example, to express 2.00 g of NaCl (sodium chloride) as an amount (in moles), one would do the following: In the above example, when written out in fraction form, the units of grams form a multiplicative identity, which is equivalent to one (g/g = 1), with the resulting amount in moles (the unit that was needed), as shown in the following equation, Molar proportion [edit] Stoichiometry is often used to balance chemical equations (reaction stoichiometry). For example, the two diatomic gases, hydrogen and oxygen, can combine to form a liquid, water, in an exothermic reaction, as described by the following equation: : 2 H2 + O2 → 2 H2O Reaction stoichiometry describes the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above equation. The molar ratio allows for conversion between moles of one substance and moles of another. For example, in the reaction : 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O the amount of water that will be produced by the combustion of 0.27 moles of CH3OH is obtained using the molar ratio between CH3OH and H2O of 2 to 4. The term stoichiometry is also often used for the molar proportions of elements in stoichiometric compounds (composition stoichiometry). For example, the stoichiometry of hydrogen and oxygen in H2O is 2:1. In stoichiometric compounds, the molar proportions are whole numbers. Determining amount of product [edit] Stoichiometry can also be used to find the quantity of a product yielded by a reaction. If a piece of solid copper (Cu) were added to an aqueous solution of silver nitrate (AgNO3), the silver (Ag) would be replaced in a single displacement reaction forming aqueous copper(II) nitrate (Cu(NO3)2) and solid silver. How much silver is produced if 16.00 grams of Cu is added to the solution of excess silver nitrate? The following steps would be used: Write and balance the equation Mass to moles: Convert grams of Cu to moles of Cu Mole ratio: Convert moles of Cu to moles of Ag produced Mole to mass: Convert moles of Ag to grams of Ag produced The complete balanced equation would be: : Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag For the mass to mole step, the mass of copper (16.00 g) would be converted to moles of copper by dividing the mass of copper by its molar mass: 63.55 g/mol. Now that the amount of Cu in moles (0.2518) is found, we can set up the mole ratio. This is found by looking at the coefficients in the balanced equation: Cu and Ag are in a 1:2 ratio. Now that the moles of Ag produced is known to be 0.5036 mol, we convert this amount to grams of Ag produced to come to the final answer: This set of calculations can be further condensed into a single step: Further examples [edit] For propane (C3H8) reacting with oxygen gas (O2), the balanced chemical equation is: : C3H8 + 5 O2 → 3 CO2 + 4 H2O The mass of water formed if 120 g of propane (C3H8) is burned in excess oxygen is then Stoichiometric ratio [edit] Stoichiometry is also used to find the right amount of one reactant to "completely" react with the other reactant in a chemical reaction – that is, the stoichiometric amounts that would result in no leftover reactants when the reaction takes place. An example is shown below using the thermite reaction,[citation needed] : Fe2O3 + 2 Al → Al2O3 + 2 Fe This equation shows that 1 mole of iron(III) oxide and 2 moles of aluminium will produce 1 mole of aluminium oxide and 2 moles of iron. So, to completely react with 85.0 g of iron(III) oxide (0.532 mol), 28.7 g (1.06 mol) of aluminium are needed. Limiting reagent and percent yield [edit] Main articles: Limiting reagent and Yield (chemistry) The limiting reagent is the reagent that limits the amount of product that can be formed and is completely consumed when the reaction is complete. An excess reactant is a reactant that is left over once the reaction has stopped due to the limiting reactant being exhausted. Consider the equation of roasting lead(II) sulfide (PbS) in oxygen (O2) to produce lead(II) oxide (PbO) and sulfur dioxide (SO2): : 2 PbS + 3 O2 → 2 PbO + 2 SO2 To determine the theoretical yield of lead(II) oxide if 200.0 g of lead(II) sulfide and 200.0 g of oxygen are heated in an open container: Because a lesser amount of PbO is produced for the 200.0 g of PbS, it is clear that PbS is the limiting reagent. In reality, the actual yield is not the same as the stoichiometrically-calculated theoretical yield. Percent yield, then, is expressed in the following equation: If 170.0 g of lead(II) oxide is obtained, then the percent yield would be calculated as follows: Example [edit] Consider the following reaction, in which iron(III) chloride reacts with hydrogen sulfide to produce iron(III) sulfide and hydrogen chloride: : 2 FeCl3 + 3 H2S → Fe2S3 + 6 HCl The stoichiometric masses for this reaction are: : 324.41 g FeCl3, 102.25 g H2S, 207.89 g Fe2S3, 218.77 g HCl Suppose 90.0 g of FeCl3 reacts with 52.0 g of H2S. To find the limiting reagent and the mass of HCl produced by the reaction, we change the above amounts by a factor of 90/324.41 and obtain the following amounts: : 90.00 g FeCl3, 28.37 g H2S, 57.67 g Fe2S3, 60.69 g HCl The limiting reactant (or reagent) is FeCl3, since all 90.00 g of it is used up while only 28.37 g H2S are consumed. Thus, 52.0 − 28.4 = 23.6 g H2S left in excess. The mass of HCl produced is 60.7 g. By looking at the stoichiometry of the reaction, one might have guessed FeCl3 being the limiting reactant; three times more FeCl3 is used compared to H2S (324 g vs 102 g). Different stoichiometries in competing reactions [edit] Often, more than one reaction is possible given the same starting materials. The reactions may differ in their stoichiometry. For example, the methylation of benzene (C6H6), through a Friedel–Crafts reaction using AlCl3 as a catalyst, may produce singly methylated (C6H5CH3), doubly methylated (C6H4(CH3)2), or still more highly methylated (C6H6−n(CH3)n) products, as shown in the following example, : C6H6 + CH3Cl → C6H5CH3 + HCl : C6H6 + 2 CH3Cl → C6H4(CH3)2 + 2 HCl : C6H6 + n CH3Cl → C6H6−n(CH3)n + n HCl In this example, which reaction takes place is controlled in part by the relative concentrations of the reactants. Stoichiometric coefficient and stoichiometric number [edit] In lay terms, the stoichiometric coefficient of any given component is the number of molecules and/or formula units that participate in the reaction as written. A related concept is the stoichiometric number (using IUPAC nomenclature), wherein the stoichiometric coefficient is multiplied by +1 for all products and by −1 for all reactants. For example, in the reaction CH4 + 2 O2 → CO2 + 2 H2O, the stoichiometric number of CH4 is −1, the stoichiometric number of O2 is −2, for CO2 it would be +1 and for H2O it is +2. In more technically precise terms, the stoichiometric number in a chemical reaction system of the i-th component is defined as or where is the number of molecules of i, and is the progress variable or extent of reaction. The stoichiometric number represents the degree to which a chemical species participates in a reaction. The convention is to assign negative numbers to reactants (which are consumed) and positive ones to products, consistent with the convention that increasing the extent of reaction will correspond to shifting the composition from reactants towards products. However, any reaction may be viewed as going in the reverse direction, and in that point of view, would change in the negative direction in order to lower the system's Gibbs free energy. Whether a reaction actually will go in the arbitrarily selected forward direction or not depends on the amounts of the substances present at any given time, which determines the kinetics and thermodynamics, i.e., whether equilibrium lies to the right or the left of the initial state, In reaction mechanisms, stoichiometric coefficients for each step are always integers, since elementary reactions always involve whole molecules. If one uses a composite representation of an overall reaction, some may be rational fractions. There are often chemical species present that do not participate in a reaction; their stoichiometric coefficients are therefore zero. Any chemical species that is regenerated, such as a catalyst, also has a stoichiometric coefficient of zero. The simplest possible case is an isomerization : A → B in which νB = 1 since one molecule of B is produced each time the reaction occurs, while νA = −1 since one molecule of A is necessarily consumed. In any chemical reaction, not only is the total mass conserved but also the numbers of atoms of each kind are conserved, and this imposes corresponding constraints on possible values for the stoichiometric coefficients. There are usually multiple reactions proceeding simultaneously in any natural reaction system, including those in biology. Since any chemical component can participate in several reactions simultaneously, the stoichiometric number of the i-th component in the k-th reaction is defined as so that the total (differential) change in the amount of the i-th component is Extents of reaction provide the clearest and most explicit way of representing compositional change, although they are not yet widely used. With complex reaction systems, it is often useful to consider both the representation of a reaction system in terms of the amounts of the chemicals present { Ni } (state variables), and the representation in terms of the actual compositional degrees of freedom, as expressed by the extents of reaction { ξk }. The transformation from a vector expressing the extents to a vector expressing the amounts uses a rectangular matrix whose elements are the stoichiometric numbers [ νi k ]. The maximum and minimum for any ξk occur whenever the first of the reactants is depleted for the forward reaction; or the first of the "products" is depleted if the reaction as viewed as being pushed in the reverse direction. This is a purely kinematic restriction on the reaction simplex, a hyperplane in composition space, or N‑space, whose dimensionality equals the number of linearly-independent chemical reactions. This is necessarily less than the number of chemical components, since each reaction manifests a relation between at least two chemicals. The accessible region of the hyperplane depends on the amounts of each chemical species actually present, a contingent fact. Different such amounts can even generate different hyperplanes, all sharing the same algebraic stoichiometry. In accord with the principles of chemical kinetics and thermodynamic equilibrium, every chemical reaction is reversible, at least to some degree, so that each equilibrium point must be an interior point of the simplex. As a consequence, extrema for the ξs will not occur unless an experimental system is prepared with zero initial amounts of some products. The number of physically-independent reactions can be even greater than the number of chemical components, and depends on the various reaction mechanisms. For example, there may be two (or more) reaction paths for the isomerism above. The reaction may occur by itself, but faster and with different intermediates, in the presence of a catalyst. The (dimensionless) "units" may be taken to be molecules or moles. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules. The Ns and ξs are reduced to molar units by dividing by the Avogadro constant. While dimensional mass units may be used, the comments about integers are then no longer applicable. Stoichiometry matrix [edit] Main article: Chemical reaction network theory In complex reactions, stoichiometries are often represented in a more compact form called the stoichiometry matrix. The stoichiometry matrix is denoted by the symbol N. If a reaction network has n reactions and m participating molecular species, then the stoichiometry matrix will have correspondingly m rows and n columns. For example, consider the system of reactions shown below: : S1 → S2 : 5 S3 + S2 → 4 S3 + 2 S2 : S3 → S4 : S4 → S5 This system comprises four reactions and five different molecular species. The stoichiometry matrix for this system can be written as: where the rows correspond to S1, S2, S3, S4 and S5, respectively. The process of converting a reaction scheme into a stoichiometry matrix can be a lossy transformation: for example, the stoichiometries in the second reaction simplify when included in the matrix. This means that it is not always possible to recover the original reaction scheme from a stoichiometry matrix. Often the stoichiometry matrix is combined with the rate vector, v, and the species vector, x to form a compact equation, the biochemical systems equation, describing the rates of change of the molecular species: Gas stoichiometry [edit] Gas stoichiometry is the quantitative relationship (ratio) between reactants and products in a chemical reaction with reactions that produce gases. Gas stoichiometry applies when the gases produced are assumed to be ideal, and the temperature, pressure, and volume of the gases are all known. The ideal gas law is used for these calculations. Often, but not always, the standard temperature and pressure (STP) are taken as 0 °C and 1 bar and used as the conditions for gas stoichiometric calculations. Gas stoichiometry calculations solve for the unknown volume or mass of a gaseous product or reactant. For example, if we wanted to calculate the volume of gaseous NO2 produced from the combustion of 100 g of NH3, by the reaction: : 4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (l) we would carry out the following calculations: There is a 1:1 molar ratio of NH3 to NO2 in the above balanced combustion reaction, so 5.871 mol of NO2 will be formed. We will employ the ideal gas law to solve for the volume at 0 °C (273.15 K) and 1 atmosphere using the gas law constant of R = 0.08206 L·atm·K−1·mol−1: Gas stoichiometry often involves having to know the molar mass of a gas, given the density of that gas. The ideal gas law can be re-arranged to obtain a relation between the density and the molar mass of an ideal gas: : and and thus: where: P = absolute gas pressure V = gas volume n = amount (measured in moles) R = universal ideal gas law constant T = absolute gas temperature ρ = gas density at T and P m = mass of gas M = molar mass of gas Stoichiometric air-to-fuel ratios of common fuels [edit] See also: Air–fuel ratio and Combustion In the combustion reaction, oxygen reacts with the fuel, and the point where exactly all oxygen is consumed and all fuel burned is defined as the stoichiometric point. With more oxygen (overstoichiometric combustion), some of it stays unreacted. Likewise, if the combustion is incomplete due to lack of sufficient oxygen, fuel remains unreacted. (Unreacted fuel may also remain due to physical rather than chemical factors, such as slow combustion or insufficient mixing of fuel and oxygen – this is not due to stoichiometry.) Different hydrocarbon fuels have different contents of carbon, hydrogen and other elements, thus their stoichiometry varies. Oxygen makes up only 20.95% of the volume of air, and only 23.20% of its mass. The air-fuel ratios listed below are much higher than the equivalent oxygen-fuel ratios, due to the high proportion of inert gasses in the air. (The ratio of 14.7 as listed for gasoline only applies for combustion in combination with a catalytic converter.) \| Fuel \| Ratio by mass \| Ratio by volume [full citation needed] \| Percent fuel by mass \| Main reaction \| --- --- \| Gasoline \| 14.7 : 1 \| — \| 6.9% \| 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O \| \| Natural gas \| 14.5 : 1 \| 9.7 : 1 \| 6.9% \| CH4 + 2 O2 → CO2 + 2 H2O \| \| Propane (LP) \| 15.67 : 1 \| 23.9 : 1 \| 6.45% \| C3H8 + 5 O2 → 3 CO2 + 4 H2O \| \| Ethanol \| 9 : 1 \| — \| 11.1% \| C2H6O + 3 O2 → 2 CO2 + 3 H2O \| \| Methanol \| 6.47 : 1 \| — \| 15.6% \| 2 CH4O + 3 O2 → 2 CO2 + 4 H2O \| \| n-Butanol \| 11.2 : 1 \| — \| 8.2% \| C4H10O + 6 O2 → 4 CO2 + 5 H2O \| \| Hydrogen \| 34.3 : 1 \| 2.39 : 1 \| 2.9% \| 2 H2 + O2 → 2 H2O \| \| Diesel \| 14.5 : 1 [citation needed] \| — \| 6.8% \| 2 C12H26 + 37 O2 → 24 CO2 + 26 H2O \| \| Methane \| 17.23 : 1 \| 9.52 : 1 \| 5.5% \| CH4 + 2 O2 → CO2 + 2 H2O \| \| Acetylene \| 13.26 : 1 [citation needed] \| 11.92 : 1 \| 7.0% \| 2 C2H2 + 5 O2 → 4 CO2 + 2 H2O \| \| Ethane \| 16.07 : 1 [citation needed] \| 16.68 : 1 \| 5.9% \| 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O \| \| Butane \| 15.44 : 1 [citation needed] \| 30.98 : 1 \| 6.1% \| 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O \| \| Pentane \| 15.31 : 1 [citation needed] \| 38.13 : 1 \| 6.1% \| C5H12 + 8 O2 → 5 CO2 + 6 H2O \| Gasoline engines can run at stoichiometric air-to-fuel ratio, because gasoline is quite volatile and is mixed (sprayed or carburetted) with the air prior to ignition. Diesel engines, in contrast, run lean, with more air available than simple stoichiometry would require. Diesel fuel is less volatile and is effectively burned as it is injected. See also [edit] Non-stoichiometric compound Biochemical systems equation Chemical reaction Chemical equation Molecule Molar mass Ideal gas law References [edit] ^ Richter, J.B. (1792). Anfangsgründe der Stöchyometrie ... (in 3 vol.s) [Rudiments of Stoichiometry ...] (in German). Vol. 1. Breslau and Hirschberg, (Germany): Johann Friedrich Korn der Aeltere. p. 121. From p. 121: "Die Stöchyometrie (Stöchyometria) ist die Wissenschaft die quantitativen oder Massenverhältnisse ... zu messen, in welchen die chemischen Elemente ... gegen einander stehen." (Stoichiometry (stoichiometria) is the science of measuring the quantitative or mass relations in which the chemical "elements" exist in relation to each other.) [On pp. 3–7, Richter explains that an "element" is a pure substance, and that a "chemical element" (chymisches Element (Elementum chymicum)) is a substance that cannot be resolved into dissimilar substances by known physical or chemical means. Thus, for example, aluminium oxide was a "chemical element" because in Richter's time, it couldn't be resolved further into its component elements.] ^ Sinnott, R. K. (2005). Coulson and Richardson's Chemical Engineering (4th ed.). Amsterdam Paris: Elsevier Butterworth-Heinemann. p. 36. ISBN 978-0-7506-6538-4. ^ L. Darmstaedter; R. E. Oesper (1928). "Jeremias Benjamin Richter". J. Chem. Educ. 5 (7): 785–790. Bibcode:1928JChEd...5..785D. doi:10.1021/ed005p785. ^ What's in a Name? Amount of Substance, Chemical Amount, and Stoichiometric Amount Carmen J. Giunta Journal of Chemical Education 2016 93 (4), 583-586 doi:10.1021/acs.jchemed.5b00690 ^ "Stoichiometry of Chemical Reactions" (PDF). ^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "stoichiometric number, ν". doi:10.1351/goldbook.S06025 ^ Nijmeh, Joseph; Tye, Mark (2 October 2013). "Stoichiometry and Balancing Reactions". LibreTexts. Retrieved 5 May 2021. ^ Prigogine & Defay, p. 18; Prigogine, pp. 4–7; Guggenheim, p. 37 & 62 ^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "extent of reaction, ξ". doi:10.1351/goldbook.E02283 ^ Ghaderi, Susan; Haraldsdóttir, Hulda S.; Ahookhosh, Masoud; Arreckx, Sylvain; Fleming, Ronan M.T. (August 2020). "Structural conserved moiety splitting of a stoichiometric matrix". Journal of Theoretical Biology. 499: 110276. Bibcode:2020JThBi.49910276G. doi:10.1016/j.jtbi.2020.110276. hdl:1887/3134882. PMID 32333975.{{cite journal}}: CS1 maint: article number as page number (link) ^ Hofmeyr, Jan-hendrik S. (2001). "Metabolic control analysis in a nutshell". In Proceedings of the 2 Nd International Conference on Systems Biology: 291–300. CiteSeerX 10.1.1.324.922. ^ Reder, Christine (21 November 1988). "Metabolic control theory: A structural approach". Journal of Theoretical Biology. 135 (2): 175–201. Bibcode:1988JThBi.135..175R. doi:10.1016/s0022-5193(88)80073-0. PMID 3267767. ^ "Universal Industrial Gases, Inc: Composition of Air - Components & Properties of Air - Answers to "What is air?" - "What is air made up of?" -" What are air products and what are they used for?"". ^ John B. Heywood: "Internal Combustion Engine Fundamentals page 915", 1988 ^ North American Mfg. Co.: "North American Combustion Handbook", 1952 ^ "Air-fuel ratio, lambda and engine performance". Retrieved 2019-05-31. Zumdahl, Steven S. Chemical Principles. Houghton Mifflin, New York, 2005, pp 148–150. Internal Combustion Engine Fundamentals, John B. Heywood External links [edit] Look up stoichiometry in Wiktionary, the free dictionary. Library resources about Stoichiometry Resources in your library Engine Combustion primer from the University of Plymouth Free Stoichiometry Tutorials from Carnegie Mellon's ChemCollective Stoichiometry Add-In for Microsoft Excel Archived 2011-05-11 at the Wayback Machine for calculation of molecular weights, reaction coëfficients and stoichiometry. Reaction Stoichiometry Calculator a comprehensive free online reaction stoichiometry calculator. \| v t e Branches of chemistry \| \| Glossary of chemical formulae List of biomolecules List of inorganic compounds Periodic table \| \| Analytical \| Calorimetry Characterization Chromatography + GC + HPLC Crystallography Electroanalytical methods Elemental analysis Instrumental chemistry Mass spectrometry + EI + ICP + MALDI Separation process Spectroscopy + IR + Raman + UV-Vis + NMR Titration Wet chemistry \| \| Theoretical \| Computational chemistry + Mathematical chemistry Molecular modelling Molecular mechanics Molecular dynamics Molecular geometry + VSEPR theory Quantum chemistry \| \| Physical \| Chemical kinetics Chemical physics + Molecular physics Chemical thermodynamics Cryochemistry Electrochemistry + Spectroelectrochemistry + Photoelectrochemistry Equilibrium chemistry Femtochemistry Interface and colloid science + Micromeritics Mechanochemistry Microwave chemistry Photochemistry Sonochemistry Spectroscopy Spin chemistry Structural chemistry Surface science Thermochemistry \| \| Inorganic \| Ceramic chemistry Cluster chemistry Coordination chemistry Magnetochemistry Organometallic chemistry + Organolanthanide chemistry Solid-state chemistry \| \| Organic \| Dynamic covalent chemistry Enantioselective synthesis Fullerene chemistry Organic reactions Organic synthesis Petrochemistry Physical organic chemistry Polymer chemistry Retrosynthetic analysis Stereochemistry + Alkane stereochemistry Total synthesis / Semisynthesis \| \| Biological \| Biochemistry + Molecular biology + Cell biology Bioinorganic chemistry Bioorganic chemistry Bioorganometallic chemistry Biophysical chemistry Chemical biology + Bioorthogonal chemistry Clinical chemistry Medicinal chemistry + Pharmacology Neurochemistry \| \| Interdisciplinarity \| Nuclear chemistry + Radiochemistry + Radiation chemistry + Actinide chemistry Cosmochemistry / Astrochemistry / Stellar chemistry Geochemistry + Biogeochemistry + Photogeochemistry Environmental chemistry + Atmospheric chemistry + Ocean chemistry Clay chemistry Carbochemistry Food chemistry + Carbohydrate chemistry + Food physical chemistry Agricultural chemistry + Soil chemistry Chemistry education + Amateur chemistry + General chemistry Clandestine chemistry Forensic chemistry + Forensic toxicology + Post-mortem chemistry Nanochemistry + Supramolecular chemistry Chemical synthesis + Green chemistry + Click chemistry + Combinatorial chemistry + Biosynthesis Chemical engineering + Stoichiometry Materials science + Metallurgy + Ceramic engineering + Polymer science \| \| See also \| History of chemistry Nobel Prize in Chemistry Timeline of chemistry + of element discoveries "The central science" Chemical reaction + Catalysis Chemical element Chemical compound Atom Molecule Ion Chemical substance Chemical bond Alchemy Quantum mechanics \| \| Category Commons Portal WikiProject \| \| Authority control databases \| \| National \| United States France BnF data Czech Republic Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Stoichiometry Chemical reaction engineering Hidden categories: Pages using the Phonos extension CS1 German-language sources (de) CS1 maint: article number as page number Articles with short description Short description is different from Wikidata Pages including recorded pronunciations Articles containing German-language text All articles with unsourced statements Articles with unsourced statements from February 2024 All articles with incomplete citations Articles with incomplete citations from December 2024 Articles with unsourced statements from December 2024 Webarchive template wayback links Pages that use a deprecated format of the math tags Add topic
13396	https://www.everlywell.com/blog/sti-testing/gonorrhea-and-chlamydia-knowing-the-difference/?srsltid=AfmBOorkiNjLKKkBG_SQJ3RhjejDdPjAMsaBZ6-NPVHNphASmSQ35qDc	Chlamydia vs. Gonorrhea \| Differences in Discharge \| Everlywell Opens in a new window Opens an external website Opens an external website in a new window We use cookies and other technologies to collect information about your use of our websites and online apps. Some of this collection is required for the operation of our platform and cannot be disabled. If you click “I Agree,” we may also share this information with 3rd-party advertising and analytics partners. It may indicate to partners the webpages, products and/or services that you have viewed and/or purchased. To learn about who we share your data with and why go to our Sharing Settings. Also, by accessing this website or online app, you agree to be bound by our Privacy Policy I Agree Deny Non-Essential Sharing Settings Daily HealthDigestive HealthSexual HealthHormone HealthShopLearn Register Kit Blog Topics Blog Topics: Virtual Care STI Testing Food Sensitivity Weight Management Colon Cancer More... STI Testing Gonorrhea and Chlamydia: Knowing the Difference Medically reviewed by Neka Miller, PhD on March 26, 2020. To give you technically accurate, evidence-based information, content published on the Everlywell blog is reviewed by credentialed professionals with expertise in medical and bioscience fields. Table of contents What Is Chlamydia and Gonorrhea? Chlamydia and Gonorrhea Symptoms The Health Complications How Gonorrhea and Chlamydia Are Tested Related Content Both gonorrhea and chlamydia are sexually transmitted infections (STIs) that share a number of similarities. They’re both very common STIs (commonly transmitted by having unprotected sex with a sex partner). Their symptoms often resemble each other. And they can both have long-term health consequences if they aren’t treated. So what’s the difference between these two? Keep reading for a closer look at the symptoms of chlamydia vs. gonorrhea, how they can affect your health, and how you can check for both STIs with a single chlamydia and gonorrhea test you can take at home. Test For Gonorrhea From Home What Is Chlamydia and Gonorrhea? Both gonorrhea and chlamydia are common sexually transmitted infections occurring in men and women. So how do you get gonorrhea and chlamydia? They are transmitted through vaginal, anal, or oral sex with someone who’s infected. Both infections are caused by bacteria—Chlamydia trachomatis in cases of chlamydia and Neisseria gonorrhoeae in cases of gonorrhea. Although gonorrhea is a common sexually transmitted infection, chlamydia has a higher prevalence—with over 1.7 million cases of chlamydia reported in the United States in 2017. Risk factors for getting gonorrhea and chlamydia are often identical and include: Having multiple sex partners. You're more likely to be exposed to someone with a sexually transmitted infection if you have multiple sex partners. Unprotected sex. Condom usage during sex substantially reduces the risk of getting a sexually transmitted infection, so your risk is higher if you have unprotected sex. Having other STIs: If you already have a sexually transmitted infection, you can be at a greater risk of getting another STI. For example, if you contract chlamydia, you could be more likely to contract gonorrhea. Check For Chlamydia & Gonorrhea Today! Chlamydia and Gonorrhea Symptoms The symptoms of chlamydia and gonorrhea overlap, so it can be difficult to differentiate between the two unless you visit your healthcare provider or take a test for chlamydia or gonorrhea. The overlapping symptoms for chlamydia and gonorrhea in men and women include: A burning sensation during urination Abnormal genital or rectal discharge Pain in the rectum Sore throat With both chlamydia or gonococcal infections (another name for gonorrhea infections), men might also experience swelling and pain in the testicles and/or scrotum. In women, both a gonorrhea and chlamydia infection might be mistaken for a yeast infection. Women may also experience painful periods, bleeding between periods, pain during sex, or abdominal pain. Although the symptoms overlap, the discharge caused by chlamydia vs. gonorrhea can vary slightly. For a chlamydia infection, a woman’s vaginal discharge might have a strong odor and yellowish tint. Men might have a cloudy or clear discharge. With gonorrhea, both women and men may experience green, yellow, or white discharge. If you're a woman experiencing abnormal vaginal discharge or a man with abnormal penile discharge, be sure to consult your healthcare provider as soon as possible as this is a common sign of an infection. Test For Chlamydia & Gonorrhea From Home The Health Complications If chlamydia and gonorrhea are not treated, serious health complications can develop. Untreated chlamydia and gonococcal infections in women can spread to the uterus and fallopian tubes. When untreated bacteria spread to these areas of the body, pelvic inflammatory disease (PID) can result. Often, there might not be any initial symptoms of PID other than pelvic or abdominal pain. Pelvic inflammatory disease can cause long-term harm to a woman’s reproductive system. Chronic pelvic pain can be caused by inflammation of the fallopian tubes and other areas of the reproductive system. Ultimately, infertility as well as ectopic pregnancy—which occurs when the fertilized egg cannot reach the uterus due to scarring caused by PID—can develop. In pregnant women, PID can also lead to premature birth. With both gonorrhea and chlamydia, infections can be transmitted to the newborn baby during birth as a result of infected vaginal tissue. This can lead to infant health complications like eye infections and pneumonia. Gonorrhea and chlamydia can lead to health problems and complications in men, as well. It is rare for these infections to cause infertility, though epididymitis—inflammation of the tubes next to the testicles—can occur when this part of the body has a chlamydial infection or is infected by gonorrhea. This can cause testicular pain and swelling. A prostate gland infection known as prostatitis can also occur in men if chlamydia and gonorrhea symptoms go untreated. In this case, the bacteria spread to the prostate gland and can make urination and ejaculation painful. The infection can also cause fevers or pain in the lower back. When it comes to men, one difference between chlamydia vs. gonorrhea health complications is that chlamydia can also spread to the urethra. When it does, it can cause non-gonococcal urethritis, an infection of the tube that carries the urine, which causes pain, fever, and inflammation. For both women and men, a gonorrhea and chlamydia infection can develop into forms of arthritis. This is called reactive arthritis, which means arthritis caused by the body’s reaction to the infection. It can affect joints, the urethra, and the eyes. If it isn’t treated, gonorrhea can also spread through the bloodstream and infect critical parts of the body, like the heart. This kind of gonorrhea infection is called a disseminated gonococcal infection. Disseminated gonorrhea can result in symptoms like joint pain, fever, skin rashes, and sores—as well as severe health complications involving the heart, brain, and/or spinal cord. Because of the risk of long-term health complications associated with untreated gonorrhea and chlamydia, it's essential that you seek medical treatment right away if you've had sex with an infected sexual partner and have tested positive for one or more of these sexually transmitted infections. If you think you may have been infected by a sexual partner, talk with your healthcare provider as soon as possible—and consider checking for STDs with our at-home STD Test for women or men. Buy At-home Chlamydia & Gonorrhea Testing How Gonorrhea and Chlamydia Are Tested Both gonorrhea and chlamydia can be diagnosed with similar methods. A healthcare provider might do a physical examination to look for symptoms, and they may do a urine test to check for the bacteria that cause chlamydia or gonorrhea. But screening for chlamydia and gonorrhea doesn’t have to take place at a clinic. With the Everlywell at-home Chlamydia & Gonorrhea Test, you can easily check for these STIs from the comfort and privacy of your home. The kit comes with easy-to-follow instructions and everything you need to collect your sample at home, and your physician-reviewed results can be conveniently and securely viewed online on your device. Plus, if you receive a positive result after you get tested for chlamydia and gonorrhea with the Everlywell at-home test, you’ll have the opportunity to connect with our independent physician network—and may be prescribed medication to treat the infection. You can also opt for sexual health testing year-round through the Everlywell+ STI testing membership, giving you easy access to a wide of variety of STI test options. Check For Gonorrhea At Home! Related Content Gonorrhea: Signs and Symptoms How Do You Get Gonorrhea? What Is Disseminated Gonorrhea? Gonorrhea in Women Improve Member Outcomes and Close HEDIS Gap in Care with the Everly Health Chlamydia and Gonorrhea Test for Health Plans References Chlamydia - CDC Fact Sheet. Centers for Disease Control and Prevention. Accessed March 26, 2020. Gonorrhea - CDC Fact Sheet. Centers for Disease Control and Prevention. Accessed March 26, 2020. New CDC Report: STDs Continue to Rise in the U.S. Centers for Disease Control and Prevention. Accessed March 26, 2020. Chlamydia - CDC Fact Sheet (Detailed). Centers for Disease Control and Prevention. Accessed March 26, 2020. Pelvic Inflammatory Disease (PID) - CDC Fact Sheet. Centers for Disease Control and Prevention. Accessed March 26, 2020. Berger RE, Alexander ER, Harnisch JP, et al. Etiology, manifestations and therapy of acute epididymitis: prospective study of 50 cases. J Urol. 1979;121(6):750–754. doi:10.1016/s0022-5347(17)56978-5 Sutcliffe S, Nevin RL, Pakpahan R, et al. Prostate involvement during sexually transmitted infections as measured by prostate-specific antigen concentration. Br J Cancer. 2011;105(5):602–605. doi:10.1038/bjc.2011.271 Urethritis. StatPearls. Accessed March 26, 2020. Denison HJ, Curtis EM, Clynes MA, et al. The incidence of sexually acquired reactive arthritis: a systematic literature review. Clin Rheumatol. 2016;35(11):2639–2648. doi:10.1007/s10067-016-3364-0 Holmes KK, Counts GW, Beaty HN. Disseminated gonococcal infection. Ann Intern Med. 1971;74(6):979–993. doi:10.7326/0003-4819-74-6-979 Everlywell makes lab testing easy and convenient with at-home collection and digital results in days. Learn More Shop Digestive HealthVitamins & SupplementsDaily HealthSexual HealthHormone HealthGift Cards About The ScienceArticlesReviews Support FAQsHelpCareersPressBecome an AffiliateRefer a Friend Get news, updates, and exclusive discounts Follow us Everlywell offers health and wellness solutions including laboratory testing for wellness monitoring, informational and educational use. With the exception of certain diagnostic test panels, list available here, the tests we offer access to are not intended to diagnose or treat disease. None of our tests are intended to be a substitute for seeking professional medical advice, help, diagnosis, or treatment. At-home lab tests are not available in NY. Accessibility StatementWebsite Privacy NoticeTerms of UseConsumer Health Data Privacy NoticeYour Privacy Choices © Everlywell 2025
13397	https://smartasset.com/estate-planning/community-property-vs-joint-tenancy	Community Property vs. Joint Tenancy Mortgage Taxes Retirement Financial Advisors Banking Investing Credit Cards Life Insurance Refinance Personal Loans Student Loans SmartReads SmartAsset AMP About Press Terms of Service Privacy Policy Back to SmartAsset Home Mortgage Calculators Mortgage Calculator Closing Costs Calculator Cost of Living Calculator How Much House Can I Afford? Mortgage Lender Reviews Rocket Mortgage Review Better Mortgage Review AmeriSave Mortgage Review More Lender Reviews Compare Rates Mortgage Rates 30-Year Mortgage Rates 15-Year Mortgage Rates 7/1 Arm Mortgage Rates 5/1 Arm Mortgage Rates 3/1 Arm Mortgage Rates Helpful Guides Home Buying Guide Veteran Home Buying Guide Learn More How Much You Should Charge for Rent Avoid These Early Mortgage Payoff Mistakes How to Buy Land Down Payment Gifts Taxes Calculators Paycheck Calculator Income Tax Calculator Property Tax Calculator Tax Return Calculator Retirement Taxes Calculator Helpful Guides Tax Guide Tax Software Reviews Best Free Tax Software H&R Block Review Jackson Hewitt Review FreeTaxUsa Review Learn More Federal Income Tax Brackets How to Fill Out W-4 State Capital Gains Taxes Gift Tax Explained Your Standard Deduction Tax Allowances Retirement Calculators Retirement Calculator 401(k) Calculator Social Security Calculator RMD Calculator Helpful Guides Retirement Guide Financial Advisor Guide Estate Planning Guide Learn More RMD Table How to Calculate Your RMD 401(k) Withdrawal The Rule of 55 401(k) Contribution Advice Financial Advisors Find a Financial Advisor Top Financial Advisors Financial Advisors in Houston, TX Financial Advisors in Charlotte, NC Financial Advisors in Atlanta, GA Financial Advisors in New York, NY Firm Reviews Advisory Firm Reviews Fisher Investments Review Merrill Lynch Review Helpful Tools Financial Advisor Value Calculator Working with a Financial Advisor Guide Financial Advisor Resources Guide to Financial Advisor Business Plans Client Acquisition Strategies Client Retention Strategies Cold Calling Scripts for Financial Advisors Marketing Tips About SmartAsset AMP More Resources for Advisors Learn More How to Choose a Financial Advisor Are Financial Advisors Worth it? What Is Conservatorship? Family Trusts CFA vs. CFP Best Financial Planning Software Wealth Managers vs. Financial Advisors Financial Advisor Cost Banking Calculators Savings Calculator CD Calculator Compare Bank Accounts Compare Savings Accounts Compare Checking Accounts Compare CD Rates Compare Money Market Accounts Helpful Guides Checking & Savings Guide Banking Products Best Savings Accounts Best Money Market Accounts Best CD Rates Best Checking Accounts Bank Reviews Ally Bank Review Capital One Review USAA Review Bask Bank Review Personal Finance Resources What Is the Best Credit Card for You? Personal Loan Calculator Student Loan Calculator Budget Calculator Learn More Withdrawal Limits How to Get a Bank Statement FDIC Insurance Investing Calculators Investment Calculator Capital Gains Tax Calculator Inflation Calculator Asset Allocation Calculator Helpful Guides Investing Guide Compare Accounts Brokerage Accounts Learn More What is a Fiduciary? How to Invest $100k What's a Good P/E Ratio? Types of Investments Tax Free Investments Credit Cards Best Credit Cards Best Credit Cards Helpful Guides Credit Cards Guide Compare Cards What is the Best Credit Card for You? Life Insurance Calculators How Much Life Insurance Do I Need? Compare Quotes Life Insurance Quotes Helpful Guides Life Insurance Guide Refinance Calculators Refinance Calculator Compare Rates Compare Refinance Rates Helpful Guides Refinance Guide Personal Loans Calculators Personal Loan Calculator Compare Rates Personal Loan Rates Helpful Guides Personal Loan Guide Student Loans Calculators Student Loan Calculator Compare Rates Student Loan Refinance Rates Helpful Guides Student Loan Guide SmartReads SmartAsset AMP About Press Terms of Service Privacy Policy More Best Credit Cards Best Credit Cards Helpful Guides Credit Cards Guide Compare Cards What is the Best Credit Card for You? Calculators How Much Life Insurance Do I Need? Compare Quotes Life Insurance Quotes Helpful Guides Life Insurance Guide Calculators Refinance Calculator Compare Rates Compare Refinance Rates Helpful Guides Refinance Guide Calculators Personal Loan Calculator Compare Rates Personal Loan Rates Helpful Guides Personal Loans Guide Calculators Student Loan Calculator Compare Rates Student Loan Refinance Rates Helpful Guides Student Loans Guide Back to SmartAsset Home I'm an AdvisorFind an Advisor Search SmartReads/Estate Planning/Community Property vs. Joint Tenancy Community Property vs. Joint Tenancy Written by Stephanie Colestock \| Edited by Mo Mozuch Edited by Arturo Conde, CEPF® Updated on September 7, 2025, 6:10pm ET \| Fact Checked SmartAsset maintains strict editorial integrity. It doesn’t provide legal, tax, accounting or financial advice and isn’t a financial planner, broker, lawyer or tax adviser. Consult with your own advisers for guidance. Opinions, analyses, reviews or recommendations expressed in this post are only the author’s and for informational purposes. This post may contain links from advertisers, and we may receive compensation for marketing their products or services or if users purchase products or services. \|Marketing Disclosure Share When it comes to sharing property, there are a few different forms of legal ownership people can choose from. Two common options are joint tenancy and community property. Though joint tenancy and community property are similar in some ways, there are a few important differences to note when it comes to estate planning. Let’s take a look at what each entails and which might be better for your specific situation. A financial advisor can offer valuable insight into property ownership arrangements that fit your goals, risk profile and timeline. Next Steps:Estate planning can be overwhelming. We recommend speaking with a financial advisor. This free tool will match you with vetted advisors who serve your area. Here's how it works: Answer a few easy questions, so we can find a match. Our tool matches you with vetted fiduciary advisors who can help you on the path toward achieving your financial goals. It only takes a few minutes. Check out the advisors' profiles, have an introductory call on the phone or introduction in person, and choose who to work with. Enter your ZIP code to find your matches: Find Your Advisor What Is Joint Tenancy? Joint tenancy is a legal arrangement between two or more people who wish to share ownership of real property. Each owner in a joint tenancy arrangement holds equal ownership and has equal rights to the property. In turn, each owner has full rights to the property and can make changes without the other owner’s permission. Most married couples hold their property (such as the family home, vehicles, and joint bank accounts) as joint tenants. It’s a simple ownership method and neither individual can leave their share of the property to anyone else in such an arrangement. Survivorship Right of survivorship is provided in joint tenancy. This means that if one party dies, their share will automatically go to the other owner(s) of the property. Because of this, probate is avoided and the property will pass on to the surviving owner(s) without delay or added expense. Debts When it comes to debt liability, joint tenancy does offer some protection. Creditors can make claims against the property regardless of borrower., But unless the debts are also shared, creditors can only make a claim against the portion of the property that the debtor in question owns. Taxes and Value Property held in joint tenancy will receive a step-up (or step-down) basis on one-half of the current fair market value when an owner passes away. This tax-free price adjustment will only be applied to the half that was owned by the now-deceased spouse.This means that if one spouse passes away and the other decides to sell the property, he or she will be responsible for capital gains taxes only on their share of the property, not the deceased spouse’s half. What Is Community Property? Community property is another form of shared property ownership, but it is only available between a husband and a wife. Both parties hold equal, shared ownership of the property, regardless of who contributed what to the purchase. Survivorship Right of survivorship is also offered with community property, meaning that if one spouse passes away, their share of the property’s ownership will pass to the surviving spouse. This allows spouses to avoid the probate process and retain control of the property at all times. Debts When it comes to community property, creditors can make claims against shared assets if a debt is owed by either party. Unlike with joint tenancy, the half of the property owned by the non-owing spouse is not protected.This means that even if your spouse is the one who owes said creditor, your community property is considered liable as a whole. Taxes and Value With community property, the step-up basis (or step-down) will be applied to the entire property upon one spouse’s death. This means that if the real estate has appreciated in value, and then one spouse passes away, the other spouse can sell the property shortly thereafter without being responsible for capital gains. Community Property States Currently, only nine states offer community property. They are: Arizona California Idaho Louisiana Nevada New Mexico Texas Washington Wisconsin Comparing Joint Tenancy and Community Property Choosing between joint tenancy and community property depends on your state of residence, your tax planning goals, and your need for creditor protection. Both offer shared ownership and survivorship rights, but they function differently in ways that can affect your estate plan and finances. State laws. You either live in a community property or common law state. Community property is recognized in only nine states: Arizona, California, Idaho, Louisiana, Nevada, New Mexico, Texas, Washington and Wisconsin. If you do not live in one of these states, joint tenancy will likely be your main option for shared ownership. Tax treatment. One of the most significant differences is how each form of ownership handles the stepped-up basis at death. With community property, the entire property receives a step-up in basis when one spouse dies. By contrast, in joint tenancy only the deceased spouse’s half of the property is adjusted to current market value. Creditor protection. Community property does not shield against a spouse’s creditors. If one spouse owes a debt, the entire property can be targeted to satisfy the claim. Joint tenancy offers a bit more separation. A creditor may pursue only the debtor spouse’s share, leaving the other spouse’s share intact. In short, joint tenancy offers simplicity and some separation of liability, while community property provides stronger tax benefits but greater exposure to creditors. The right choice depends on where you live, your family’s financial circumstances and how you want assets handled after death. Using Community Property and Joint Tenancy in Your Estate Plan Community property and joint tenancy both affect how assets move after death, so they should be considered in an estate plan. With joint tenancy, the right of survivorship means your share of the property automatically goes to the other owner. This bypasses probate, but it also prevents you from leaving your share to children or other heirs through a will or trust. For couples in blended families, this can create conflicts if one spouse wants to pass assets to children from a prior marriage. Community property also offers survivorship, but it applies only to married couples in a handful of states. At the first spouse’s death, the surviving spouse takes full ownership of the property. Unlike joint tenancy, the entire property receives a step-up in basis for tax purposes, which can reduce or even avoid capital gains for the surviving spouse. This tax treatment can be an advantage in estate planning, but it also means that neither spouse can redirect their share to someone else while both are alive. In either structure, creditors may still be able to make claims. Community property is more exposed to a spouse’s individual debts, while joint tenancy can shield the non-debtor spouse’s share. An estate plan that uses wills or trusts may add extra protection or control over how assets transfer, especially when multiple heirs are involved. When deciding how to title property, it is important to weigh tax benefits, probate, creditor protection, and family goals. Joint tenancy may be simple and automatic, while community property offers stronger tax advantages in certain states. Both choices should be viewed as part of your larger estate plan. Bottom Line Both joint tenancy and community property offer shared ownership of real property, such as land or structures on land, though community property is reserved for spouses. Community property is only available in select states, allowing them to hold a shared (rather than divided) interest in an asset. Joint tenancy divides a property’s ownership into equal shares. Each ownership option has its own rules regarding capital gains. Both ownership options are also unique in terms of creditor liability. It’s important to consider your priorities before choosing an ownership option. Tips on Estate Planning If you’re not sure how your shared property arrangement will affect your loved ones – in terms of taxes, creditor liability or ease of inheritance – consult with a financial advisor in your area. Finding a financial advisor doesn’t have to be hard. SmartAsset’s free tool matches you with vetted financial advisors who serve your area, and you can have a free introductory call with your advisor matches to decide which one you feel is right for you. If you’re ready to find an advisor who can help you achieve your financial goals, get started now. Another issue you may want to consider, besides whether to own a property through joint tenancy or in a community property arrangement, is whether to buy instead of just rent. Use our free calcuator to help you make that decision. Photo credit: ©iStock.com/Weekend Images Inc., ©iStock.com/ferrantraite, ©iStock.com/XiXinXing Stephanie ColestockWas this content helpful? Yes No Read More About Estate Planning Trusts What Does It Mean to Be a “Trust Fund Baby”? July 16, 2025 Read More Trusts What Rights Does a Trust Beneficiary Have? July 2, 2025 Read More Trusts Trustee vs. Beneficiary vs. Grantor: Estate Planning Guide August 14, 2025 Read More Trusts Trust vs. Trust Fund: Definitions, Purposes, Key Differences July 24, 2025 Read More Recent posts Revealed: Annual List of Top Financial Advisor Firms How to Potentially Reduce RMD Taxes After Age 73 How Irrevocable Trusts Can Help Eliminate Estate Taxes More from SmartAsset Compare Financial Advisor Matches Today How to Find and Choose a Financial Advisor The Top Financial Advisors in the U.S. How Much Do I Need to Save for Retirement? Mortgage Calculator Compare Mortgage Rates Subscribe to our Newsletter Join 200,000+ other subscribers Subscribe Categories Advisor Resources Auto Career Checking Account Credit Cards Credit Score Data Studies Debt Estate Planning Financial Advisor Insights Insurance Investing Life Insurance Mortgage Personal Finance Personal Loans Refinance Retirement Small Business Student Loans Taxes Get in touch Contact Careers SmartAsset Search About SmartReads Captivate SmartAsset AMP Advisor Testimonials Press SmartAdvisorMatch Get Social Like on Facebook Like on X Legal Stuff Terms of Service Privacy Policy AdChoices GLB Notice Online Tracking Opt-Out Guide SmartAsset Advisors, LLC ("SmartAsset"), a wholly owned subsidiary of Financial Insight Technology, is registered with the U.S. Securities and Exchange Commission as an investment adviser. SmartAsset's services are limited to referring users to third party advisers registered or chartered as fiduciaries ("Adviser(s)") with a regulatory body in the United States that have elected to participate in our matching platform based on information gathered from users through our online questionnaire. SmartAsset receives compensation from Advisers for our services. SmartAsset does not review the ongoing performance of any Adviser, participate in the management of any user's account by an Adviser or provide advice regarding specific investments. We do not manage client funds or hold custody of assets, we help users connect with relevant financial advisors. This is not an offer to buy or sell any security or interest. All investing involves risk, including loss of principal. Working with an adviser may come with potential downsides, such as payment of fees (which will reduce returns). Past performance is not a guarantee of future results. There are no guarantees that working with an adviser will yield positive returns. The existence of a fiduciary duty does not prevent the rise of potential conflicts of interest. © 2025 SmartAsset, all rights reserved.
13398	https://www.nagwa.com/en/videos/806194067190/	Question Video: Understanding How the Gravitational Potential Energy of a Pendulum Varies with Time Physics Which of the lines on the graph correctly shows how the gravitational potential energy of a pendulum compared to that at its equilibrium position varies with time? Video Transcript Which of the lines on the graph correctly shows how the gravitational potential energy of a pendulum compared to that at its equilibrium position varies with time? On this graph, we see gravitational potential energy in joules plotted against time in seconds. There are a number of different lines on the graph. There’s a black one here. That’s a flat line. Then, there’s a red one, a yellow one, a blue one, and here a purple one. We want to know which line correctly shows how the gravitational potential energy of a pendulum compared to that at its equilibrium position varies with time. So, say that this is our pendulum, and we see it here at three different snapshots in time. The center position of the pendulum here is what we call its equilibrium position. This is where the pendulum naturally moves if it’s not perturbed. But then, if we do move it to a side, say over here, and then release the pendulum, we know that it will start this back-and-forth swinging motion. And if we imagine no friction in our system, then this motion goes on indefinitely. If we imagine that all of the mass of the pendulum is at its end, called the pendulum bob, and none of it is in the arm that supports the bob as that swings back and forth, then in that case we can understand the gravitational potential energy of this pendulum by tracking the motion of the bob as it moves. In general, the GPE of an object in a uniform gravitational field is equal to the mass of that object multiplied by the acceleration due to gravity times the object’s height above some reference. For our pendulum, that reference level, we can say, goes right through the middle of the bob when it’s at its equilibrium position. We’ll say that this height corresponds to a height of zero. This definition is important because it means that at instants in time when our pendulum is at its equilibrium position, its gravitational potential energy is zero. And therefore, whatever line on our graph correctly shows GPE must reach the horizontal axis. We see that enforcing that condition eliminates two of the possible lines, the black line and the purple one. Neither of these lines crosses the horizontal axis. And we see that they fail for another reason. Note that they show us a constant gravitational potential energy over time, whereas really we know that the height of our pendulum bob, as it moves up and down, is changing the gravitational potential energy of this system. For a few reasons then, we won’t choose the black line or the purple line as our answer. Considering once more our bob in this equilibrium position, we said that at that point the gravitational potential energy of the pendulum is zero. The question is, how does that amount of GPE relate to the GPE of the system at other times? In other words, is it a minimum value, a maximum value, or somewhere in between? Our sketch shows us that at every instant other than times where the bob is at this equilibrium position, the height value of the pendulum, as we’ve defined it, will be positive. And therefore, since 𝑔 and 𝑚 are both positive as well, at all those instants the gravitational potential energy of the pendulum will be positive too. This shows us that the zero points we expect on our line should be minima, that is, the low points of the gravitational potential energy of this pendulum. Now, if we go and look at the blue line, we see that this has maximum values at zero, while the yellow line shows us zero values between the max and mid values of the line. It’s only the red line which does have zero values where those values correspond to the low points on the curve that satisfies this condition. And so, this is our answer. It’s the red line on this graph that correctly shows how the gravitational potential energy of a pendulum compared to that at its equilibrium position varies with time. Lesson Menu Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Nagwa is an educational technology startup aiming to help teachers teach and students learn. Company Content Copyright © 2025 Nagwa All Rights Reserved Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
13399	https://www.ck12.org/section/theoretical-and-experimental-probability/	Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? Grade K to 5 Earth Science Life Science Physical Science Biology Chemistry Physics Advanced Biology FlexLets Math FlexLets Science FlexLets English Writing Spelling Social Studies Economics Geography Government History World History Philosophy Sociology More Astronomy Engineering Health Photography Technology College College Algebra College Precalculus Linear Algebra College Human Biology The Universe Adult Education Basic Education High School Diploma High School Equivalency Career Technical Ed English as 2nd Language Country Bhutan Brasil Chile Georgia India Translations Spanish Korean Deutsch Chinese Greek Polski EXPLORE Flexi A FREE Digital Tutor for Every Student FlexBooks 2.0 Customizable, digital textbooks in a new, interactive platform FlexBooks Customizable, digital textbooks Schools FlexBooks from schools and districts near you Study Guides Quick review with key information for each concept Adaptive Practice Building knowledge at each student’s skill level Simulations Interactive Physics & Chemistry Simulations PLIX Play. Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up HomeMathematicsTheoretical and Experimental Probability Theoretical and Experimental Probability Difficulty Level: At Grade \| Created by: CK-12 Last Modified: Dec 26, 2014 Introduction Weekend Woes “I don’t want to work on the weekend,” Carey said to Telly at lunch one day. “But that was part of the deal. We both have to work one day out of the weekend,” Telly said. “Well, which day do you want?” Carey asked. “I don’t know. I haven’t really thought about it,” Telly said. “But we could make it really random.” “How?” Carey asked. Telly took two pieces of paper and wrote Saturday on one and Sunday on the other. “Now we can figure out the probability of you getting Saturday or Sunday,” she said. We can stop there. This lesson is all about probability. Telly’s experiment is an example of experimental probability. Let’s talk more about this at the end of the lesson. What You Will Learn In this lesson, you will learn how to demonstrate the following skills. Recognize the theoretical probability of an event as the ratio of favorable outcomes to possible outcomes. Recognize the experimental probability of an event as the ratio of successful outcomes to trials attempted. Write and compare probabilities as fractions, decimals and percents. Make and compare predictions based on theoretical and experimental probabilities, justifying the use of either. Teaching Time I. Recognize the Theoretical Probability of an Event as the Ratio of Favorable Outcomes to Possible Outcomes In this chapter we have explored different outcomes for events, but not probability itself. Probability is defined as a mathematical way of calculating how likely an event is to occur. The probability of an event occurring is defined as the ratio of favorable outcomes to the number of possible equally likely total outcomes in a given situation. In ratio form, the probability of an event is: Theoretical probability is probability that is based on an ideal situation. For example, since a flipped coin has two sides and each side is equally likely to land up, the theoretical probability of landing heads (or tails) is exactly 1 out of 2. Whether or not the coin actually lands on heads (or tails) 1 out of every 2 flips in the real world does not affect theoretical probability. The theoretical probability of an event remains the same no matter how events turn out in the real world. Example Find the probability of tossing a number cube and having it come up “4”. Step 1: Find the total number of outcomes Step 2: Find the number of favorable outcomes. Step 3: Find the ratio of favorable outcomes to total outcomes. While theoretical probability is based on the ideal, we can also figure out experimental probability. II. Recognize the Experimental Probability of an Event as the Ratio of Successful Outcomes to Trials Attempted Theoretical probability is based on an ideal situation. Since a flipped coin seems equally likely to land up or down, the theoretical probability of landing heads (or tails) is 1 out of 2. Whether or not the coin actually lands on heads (or tails) 1 out of every 2 flips in the real world is something you must determine with experimental probability. Experimental probability is probability based on doing actual experiments – flipping coins, spinning spinners, picking ping pong balls out of a jar, and so on. To compute the experimental probability of the number cube landing on 3 you would need to conduct an experiment. Suppose you were to toss the number cube 60 times. Favorable outcomes: Total outcomes: 60 tosses Experimental probability: Write this comparison down in your notebooks. Example What is the experimental probability of having the number cube land on 3? \| trial \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| Total \| --- --- --- --- \| \| raw data:3s \| \| \| \| \| \| \| \| \| favorable outcomes:3s \| 1 \| 3 \| 0 \| 1 \| 2 \| 2 \| 9 \| \| total tosses total outcomes \| 10 \| 10 \| 10 \| 10 \| 10 \| 10 \| 60 \| \| experimental probability: favorable outcomes to total outcomes \| x \| x \| x \| x \| x \| x \| \| The data from the experiment shows that 3 turned up on the number cube 9 out of 60 times. Simplified, this ratio becomes: You can see that it is only possible to calculate the experimental probability when you are actually doing experiments and counting results. Example A spinner was spun in a probability experiment 48 times. The results are shown in the table. Compute the experimental probability of the spinner landing on yellow. \| color \| red \| green \| yellow \| Total spins \| --- --- \| raw data \| \| \| \| 48 \| \| total from tally \| 16 \| 14 \| 18 \| 48 \| \| favorable outcomes:yellow \| x \| x \| 18 \| x \| \| experimental probability: \| x \| x \| \| x \| The data from the experiment above shows that the arrow landed on yellow 18 out of 48 times. Simplified, this ratio becomes: Notice that the ratio 18 out of 48 was simplified to 3 out of 8. We can simplify probabilities because they are written in ratio form. III. Write and Compare Probabilities as Fractions, Decimals and Percents You’ve seen how to compute probabilities in terms of ratios. Since any ratio can be turned into a fraction, decimal, or percent, you can also turn any probability into a fraction, decimal, or percent. For example, when you toss a number cube, the probability of rolling a “3” is: You can write the same probability as a fraction simply by rewriting the two numbers in the ratio as the numerator and denominator of a fraction. This is the answer in fraction form. How can we turn this fraction into a decimal? You can turn a fraction into a decimal by dividing the numerator by the denominator. How can we turn the decimal into a percent? We can turn a decimal into a percent by multiply the decimal by 100 since a percent is out of 100. Then we can move the decimal point two decimal places to show the percent. To summarize, the probability of rolling a 3 with a number cube is 1 out of 6, or: \| ratio \| fraction \| decimal \| percent \| --- --- \| \| 1:6 \| \| 0.167 \| 16.7% \| Please write these examples down in your notebook. IV. Make and Compare Predictions Based on Theoretical and Experimental Probabilities, Justifying the Use of Either A prediction is a reasonable guess about what will happen in the future. Good predictions should be based on facts and probability. There are two main types of predictions. Type 1: Predictions based on theoretical probability: These are the most reliable types of predictions, based on physical relationships that are easy to see and measure and that do not change over time. They include such things as: coin flips spinners number cubes Type 2: Predictions based on data and experimental probability: These predictions are often reliable, but subject to change depending on the situation. They include such activities as: batting averages, shooting percentages, and similar data from sports predicting the weather sales figures from such things as movies, TV shows, products polls and surveys that measure opinion historical data that measures past events The difference between the two types of prediction is best illustrated by the following examples. Type 1: Coin flip prediction: of 100 flips, 50 are predicted to turn up heads Type 2: Weather prediction: a 50 percent chance of rain tomorrow Note that both predictions are about the future – and you never know what might happen in the future. Though it is HIGHLY UNLIKELY to occur, a coin could land on heads 10, or 20, or even 50 times in a row. Similarly, a weather data might predict a zero percent chance of rain. But on a given day, the unexpected could happen. The winds could change clouds could unexpectedly move out or move in. That’s why predictions based on experimental probability are always less reliable than those based on theoretical probability. In general, the greater the number of outcomes you have, the closer a prediction based on probability is likely to be. Write this statement down in your notebooks. One hundred coin flips should turn out to be close to 50 percent heads; 1000 flips should be even closer to 50 percent; 10,000 flips should be closer yet. The same thing goes with weather forecasts. Over a year, forecasts are a lot more likely to be accurate than over a single day or a single week. To make predictions, use the following formulas. Or, for type 2 situations in which you do not have theoretical probability: Problem: Out of 150 number cube rolls, predict how many will turn up greater than 4. Step 1: Find the theoretical or experimental probability. Step 2: Multiply the theoretical probability by the number of total trials Based on these numbers, you would predict that 50 out of 150 rolls would land on numbers greater than 4. Now let’s apply what you have learned about predictions from the problem in the introduction. Real-Life Example Completed Weekend Woes Here is the original problem from the introduction. Think about how this is an example of experimental probability and then figure out the probability of Carey working Saturday in fraction, decimal and percentage form. “I don’t want to work on the weekend,” Carey said to Telly at lunch one day. “But that was part of the deal. We both have to work one day out of the weekend,” Telly said. “Well, which day do you want?” Carey asked. “I don’t know. I haven’t really thought about it,” Telly said. “But we could make it really random.” “How?” Carey asked. Telly took two pieces of paper and wrote Saturday on one and Sunday on the other. “Now we can figure out the probability of you getting Saturday or Sunday,” she said. Remember, there are three parts to your answer. Solution to Real – Life Example Carey has a chance of working Saturday or Sunday. There are two possible outcomes. She has a one out of 2 chance of working on Saturday and a one out of two chance of working on Sunday. .50 50% chance or probability for each outcome. Vocabulary Here are the vocabulary words that are found in this lesson. Probability : a mathematical way of calculating how likely an event is to occur. Favorable Outcome : the outcome that you are looking for Total Outcomes : all of the outcomes both favorable and unfavorable. Theoretical Probability : probability based on an ideal situation relating favorable to total outcomes Experimental Probability : probability based on doing actual experiments. Prediction : a reasonable guess based on probability Time to Practice Directions: Solve each problem. A spinner has five sections: purple, yellow, green, blue and red. Find the probability for the arrow landing on blue on the spinner: List each favorable outcome. Count the number of favorable outcomes. Write the total number of outcomes. Write the probability. Find the probability for the arrow landing on red or green on the spinner: List each favorable outcome. Count the number of favorable outcomes. Write the total number of outcomes. Write the probability. Find the probability for the arrow NOT landing on yellow on the spinner: List each favorable outcome. Count the number of favorable outcomes. Write the total number of outcomes. Write the probability. Find the probability for rolling 6 on the number cube: List each favorable outcome. Count the number of favorable outcomes. Write the total number of outcomes. Write the probability. Find the probability for rolling greater than 2 on the number cube: List each favorable outcome. Count the number of favorable outcomes. Write the total number of outcomes. Write the probability. Find the probability for rolling less than 4 on the number cube: List each favorable outcome. Count the number of favorable outcomes. Write the total number of outcomes. Write the probability. Find the probability for rolling 1 or 6 on the number cube: List each favorable outcome. Count the number of favorable outcomes. Write the total number of outcomes. Write the probability. A box contains 12 slips of paper numbered 1 to 12. Find the probability for randomly choosing a slip with a number less than 4 on it: List each favorable outcome. Count the number of favorable outcomes. Write the total number of outcomes. Write the probability. Directions: Use the table to answer the questions. Express all ratios in simplest form. Use the table to compute the experimental probability of flipping a coin and having it land on heads. \| trial \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| Total \| --- --- --- --- \| \| raw data(heads) \| \| \| \| \| \| \| \| \| number of heads \| 5 \| 6 \| 6 \| 3 \| 6 \| 5 \| 31 \| \| total number of flips \| 10 \| 10 \| 10 \| 10 \| 10 \| 10 \| 60 \| \| experimental probability \| x \| x \| x \| x \| x \| x \| 31:60 \| How many favorable outcomes were there in the experiment? How many total outcomes were there in the experiment? What was the experimental probability of the coin landing on heads? 31:60 Use the table to compute the experimental probability of a number cube landing on 6. \| trial \| 1 \| 2 \| 3 \| 4 \| 5 \| Total \| --- --- --- \| raw data \| \| \| \| \| \| x \| \| number of 6's \| 4 \| 1 \| 1 \| 2 \| 1 \| 9 \| \| total tosses \| 10 \| 10 \| 10 \| 10 \| 10 \| 50 \| \| experimental probability \| x \| x \| x \| x \| x \| 9:50 \| How many favorable outcomes were there in the experiment? How many total outcomes were there in the experiment? What is the experimental probability of the arrow landing on yellow? Directions: Use what you have learned about probability, fractions, decimals and percents to answer each question. A bag has 6 red marbles, 5 blue marbles, 7 green marbles, 2 white marbles, and 5 yellow marbles. Find the probability of randomly picking out one of the following. What is the probability in fraction form of choosing red marble? What is the probability in decimal form of choosing green marble? What is the probability in percent form of choosing a green or blue marble? Which 3 marbles together have a 60 percent chance of being chosen? Which 2 marbles together have a 40 percent chance of being chosen? Which 3 marbles together have a 0.72 chance of being chosen? Notes/Highlights \| Color \| Highlighted Text \| Notes \| \| --- --- \| \| \| Please Sign In to create your own Highlights / Notes \| \| \| Currently there are no resources to be displayed. Description No description available here... Difficulty Level At Grade Tags CK.MAT.ENG.SE.1.Middle-School-Math-Grade-8.11 Subjects mathematics Grades 8 Standards Correlations - Concept Nodes License CC BY NC Language English \| Cover Image \| Attributions \| --- \| \| \| License: CC BY-NC \| Date Created Jul 26, 2013 Last Modified Dec 26, 2014 . \| Image \| Reference \| Attributions \| --- \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| Show Attributions Show Details ▼ Related Content Definition of Probability Read + Basic Probability Video Reviews Was this helpful? Yes No 0% of people thought this content was helpful. 00 Back to the top of the page ↑ Oops, looks like cookies are disabled on your browser. Click on this link to see how to enable them. Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? No Results Found Your search did not match anything in .

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.