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7100 | https://books.google.com/books/about/Inequalities.html?id=tXVowgEACAAJ | Inequalities - Godfrey Harold Hardy, John Edensor Littlewood, George Pólya - Google Books
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Inequalities
Godfrey Harold Hardy, John Edensor Littlewood, George Pólya
The University Press, 1934 - Calculus - 314 pages
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Contents
INTRODUCTION 1
ELEMENTARY MEAN VALUES 12
MEAN VALUES WITH AN ARBITRARY 65
11 other sections not shown
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1934
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1934
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a₁a₂ambnargumentbest possiblebilinearbilinear formsboundedchaptercoefficientsconsiderconstantcontinuous and strictlyconvergentconvex functioncurvedecreasesdeduce Theoremdefineddefinitiondenoteeffectively proportionalequalityequationexamplef and gfinite numberfollows from TheoremFourier seriesgeneralisationsgiveHardy and LittlewoodHenceHilbert'sHölder's inequalityhomogeneousincreasing functioninfiniteintervalLebesgueLebesgue integralslimitM₁meansMinkowski's inequalitymonotonic functionsmultilinear formsnecessary and sufficientnon-negativenul setobtainPólyaproof of Theoremprove Theoremproves the theoremquadratic formr+s+t=0replaceresultRieszsatisfiedSchurStieltjes integralstrictly monotonicsufficient conditionsummationSupposesymmetricalTheorem 13Theorem 326Theorem 9theorytrueunless fvaluesvariablesW. H. Youngweightswritex₁zeroαιακΣαΣΣ
Bibliographic information
Title Inequalities
Cambridge mathematical library
AuthorsGodfrey Harold Hardy, John Edensor Littlewood, George Pólya
Edition 10
Publisher The University Press, 1934
Original from the University of Michigan
Digitized Feb 5, 2010
Length 314 pages
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7101 | https://www.youtube.com/playlist?list=PL_3h4GvKc6GByotM_WpBP7v4OEbxLIQZQ | Chapter 7 | SOLUTION MANUAL for "College Physics" | OpenStax - YouTube
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Chapter 7 | SOLUTION MANUAL for "College Physics" | OpenStax
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Chapter 7 | SOLUTION MANUAL for "College Physics" | OpenStax
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Want all of the answers to Chapter 7 (Work, Energy, and Energy Resources) in the OpenStax "College Physics" textbook? Well now you can! In this learning playlist, Andrew, from The Glaser Tutoring Company, will walk you through every question found in Chapter 7 of College Physics! As always, Andrew will approach every question using the easiest, step-by-step approach so that you can learn efficiently! By going though all of the questions in Chapter 7, you will learn about: ● Work: The Scientific Definition ● Kinetic Energy and the Work-Energy Theorem ● Gravitational Potential Energy ● Conservative Forces and Potential Energy ● Nonconservative Forces ● Conservation of Energy ● Power ● Work, Energy, and Power in Humans ● World Energy Use Let's get learning!
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Chapter 7 | SOLUTION MANUAL for "College Physics" | OpenStax
by The Glaser Tutoring Company
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Want all of the answers to Chapter 7 (Work, Energy, and Energy Resources) in the OpenStax "College Physics" textbook?...more Want all of the answers to Chapter 7 (Work, Energy, and Energy Resources) in the OpenStax "College Physics" textbook?...more...more
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1
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7.16 | A hydroelectric power facility (see Figure 7.37) converts the gravitational potential energy
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7.26 | Using values from Table 7.1, how many DNA molecules could be broken by the energy carried by
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7.28 | If the energy in fusion bombs were used to supply the energy needs of the world, how many of
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7.29 | Use of hydrogen fusion to supply energy is a dream that may be realized in the next century.
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7.30 | The Crab Nebula (see Figure 7.40) pulsar is the remnant of a supernova that occurred in A.D.
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7.31 | Suppose a star 1000 times brighter than our Sun (that is, emitting 1000 times the power)
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7.32 | A person in good physical condition can put out 100 W of useful power for several hours at a
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7.33 | What is the cost of operating a 3.00-W electric clock for a year if the cost of electricity
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7.35 | What is the average power consumption in watts of an appliance that uses 5.00 kW ⋅ h of
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7.38 | How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp = 746 W) to
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7.39 | Find the useful power output of an elevator motor that lifts a 2500-kg load a height of 35.0
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7.40 | What is the available energy content, in joules, of a battery that operates a 2.00-W electric
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7.41 | How long would it take a 1.50×10^5-kg airplane with engines that produce 100 MW of power to
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7.42 | Calculate the power output needed for a 950-kg car to climb a 2.00º slope at a constant 30.0
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7.43 | Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take
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7.45 | What is the power output in watts and horsepower of a 70.0-kg sprinter who accelerates from
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7.47 | What is the efficiency of an out-of-condition professor who does 2.10×10^5 J of useful work
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7.48 | Energy that is not utilized for work or heat transfer is converted to the chemical energy of
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7.49 | Using data from Table 7.5, calculate the daily energy needs of a person who sleeps for 7.00 h
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7.50 | What is the efficiency of a subject on a treadmill who puts out work at the rate of 100 W
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7.51 | Shoveling snow can be extremely taxing because the arms have such a low efficiency in this
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7.52 | Very large forces are produced in joints when a person jumps from some height to the ground.
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7.53 | Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet
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7.54 | Calculate the energy in kJ used by a 55.0-kg woman who does 50 deep knee bends in which her
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7.55 | Kanellos Kanellopoulos flew 119 km from Crete to Santorini, Greece, on April 23, 1988, in
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7.56 | The swimmer shown in Figure 7.43 exerts an average horizontal backward force of 80.0 N with
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7.57 | Mountain climbers carry bottled oxygen when at very high altitudes. (a) Assuming that a
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7.58 | The awe-inspiring Great Pyramid of Cheops was built more than 4500 years ago. Its square base
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7.59 | (a) How long can you play tennis on the 800 kJ (about 200 kcal) of energy in a candy bar?
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7.60 | Calculate the force the woman in Figure 7.45 exerts to do a push-up at constant speed, taking
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7.61 | A 75.0-kg cross-country skier is climbing a 3.0º slope at a constant speed of 2.00 m/s and
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7.62 | The 70.0-kg swimmer in Figure 7.43 starts a race with an initial velocity of 1.25 m/s and
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7.63 | A toy gun uses a spring with a force constant of 300 N/m to propel a 10.0-g steel ball. If
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7.64 | What force must be supplied by an elevator cable to produce an acceleration of 0.800 m/s2
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7.65 | A car advertisement claims that its 900-kg car accelerated from rest to 30.0 m/s and drove
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7.66 | Body fat is metabolized, supplying 9.30 kcal/g, when dietary intake is less than needed to
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7.69 | A 105-kg basketball player crouches down 0.400 m while waiting to jump. After exerting a
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2.9K views • 5 years ago
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7102 | https://pubmed.ncbi.nlm.nih.gov/1246489/ | Rapid diagnosis of iron deficiency by measurement of free erythrocyte porphyrins and hemoglobin: the FEP/hemoglobin ratio - PubMed
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. 1976 Jan;57(1):136-41.
Rapid diagnosis of iron deficiency by measurement of free erythrocyte porphyrins and hemoglobin: the FEP/hemoglobin ratio
S Piomelli,A Brickman,E Carlos
PMID: 1246489
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Rapid diagnosis of iron deficiency by measurement of free erythrocyte porphyrins and hemoglobin: the FEP/hemoglobin ratio
S Piomelli et al. Pediatrics.1976 Jan.
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. 1976 Jan;57(1):136-41.
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S Piomelli,A Brickman,E Carlos
PMID: 1246489
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The concentration of free erythrocyte prophyrins (FEP) and of hemoglobin can be measured on blood samples spotted on filter paper. The FEP/hemoglobin ratio in iron deficiency increases exponentially with a decrease of both transferrin saturation and hemoglobin level. The FEP/hemoglobin is an indicator of imparied heme synthesis. In small children an elevation of the FEP/hemoglobin ratio is a better indicator of iron-deficiency anemia than low transferrin saturation. The FEP/hemoglobin ratio is normal in thalassemia trait and renal anemia but it may be elevated in sickle-cell anemia. Measurement of FEP and hemoglobin on filter paper provides a useful diagnostic tool for the diagnosis of iron deficiency, of anemia, and (in populations at risk) of leas intoxication.
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The measurement of free erythrocyte porphyrin (FEP) as a simple means of distinguishing iron deficiency from beta-thalassemia trait in subjects with microcytosis.Stockman JA 3rd, Weiner LS, Simon GE, Stuart MJ, Oski FA.Stockman JA 3rd, et al.J Lab Clin Med. 1975 Jan;85(1):113-9.J Lab Clin Med. 1975.PMID: 1141725
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Serum ferritin and free erythrocyte protoporphyrin in rheumatoid arthritis.Hannonen P, Möttönen T, Oka M.Hannonen P, et al.Rheumatol Int. 1986;6(4):185-8. doi: 10.1007/BF00541286.Rheumatol Int. 1986.PMID: 3787091
Lack of correlation between free erythrocyte porphyrin and serum ferritin values at birth and at 2 months of life in low birthweight infants.Faldella G, Alessandroni R, Salvioli GP, Capelli M, Paolini M, Minak G Jr, Tiraferri S.Faldella G, et al.Arch Dis Child. 1983 Mar;58(3):216-9. doi: 10.1136/adc.58.3.216.Arch Dis Child. 1983.PMID: 6838253 Free PMC article.
Iron nutritional status of preschool children.Raman L, Pawashe AB, Ramalakshmi BA.Raman L, et al.Indian J Pediatr. 1992 Mar-Apr;59(2):209-12. doi: 10.1007/BF02759985.Indian J Pediatr. 1992.PMID: 1398850
Iron deficiency: diagnosis and treatment.Dallman PR.Dallman PR.West J Med. 1981 Jun;134(6):496-505.West J Med. 1981.PMID: 7257364 Free PMC article.
Phototoxicity of protoporphyrin as related to its subcellular localization in mice livers after short-term feeding with griseofulvin.Sandberg S, Romslo I.Sandberg S, et al.Biochem J. 1981 Jul 15;198(1):67-74. doi: 10.1042/bj1980067.Biochem J. 1981.PMID: 7326002 Free PMC article.
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7103 | https://artofproblemsolving.com/wiki/index.php/Symmetric_sum?srsltid=AfmBOop-WNz5rV0brnfAy9da4j4DkLJBIZOMzYla2dttwMZSUySqifPT | Art of Problem Solving
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Symmetric sum
The symmetric sum of a function of variables is defined to be , where ranges over all permutations of .
More generally, a symmetric sum of variables is a sum that is unchanged by any permutation of its variables.
Any symmetric sum can be written as a polynomial of elementary symmetric sums.
A symmetric function of variables is a function that is unchanged by any permutation of its variables. The symmetric sum of a symmetric function therefore satisfies
Given variables and a symmetric function with , the notation is sometimes used to denote the sum of over all subsets of size in .
See also
Cyclic sum
Muirhead's Inequality
PaperMath’s sum
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7104 | https://oeis.org/A000975/internal | The OEIS is supported by the many generous donors to the OEIS Foundation.
a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).
%I #530 Aug 22 2025 13:35:21
%S 0,1,2,5,10,21,42,85,170,341,682,1365,2730,5461,10922,21845,43690,
%T 87381,174762,349525,699050,1398101,2796202,5592405,11184810,22369621,
%U 44739242,89478485,178956970,357913941,715827882,1431655765,2863311530,5726623061,11453246122
%N a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).
%C Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - _Andreas M. Hinz_, Feb 15 2017
%C Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)
%C Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.
%C Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 553 = 165 = 10100101 in binary. - _Antti Karttunen_, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - _Antti Karttunen_, Aug 31 2016
%C Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - _Bill Blewett_, Dec 21 2000
%C a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - _Sen-peng Eu_, Jun 15 2002
%C A087117(a(n)) = A038374(a(n)) = 1 for n > 1; see also A090050. - _Reinhard Zumkeller_, Nov 20 2003
%C Intersection of A003754 and A003714; complement of A107907. - _Reinhard Zumkeller_, May 28 2005
%C Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - _Peter Munn_, Oct 06 2022
%C a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - _Paul Barry_, Jul 18 2005
%C Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} kA119440(n+1, k). - _Emeric Deutsch_, May 19 2006
%C In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - _Hans Isdahl_, Jan 07 2008
%C Equals A002450 and A020988 interlaced. - _Zak Seidov_, Feb 10 2008
%C For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - _K.V.Iyer_, Apr 13 2009
%C Starting with 1 and convolved with [1, 2, 2, 2, ...] = A000295. - _Gary W. Adamson_, Jun 02 2009
%C a(n) written in base 2 is sequence A056830(n). - _Jaroslav Krizek_, Aug 05 2009
%C This is the sequence A(0, 1; 1, 2; 1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - _Wolfdieter Lang_, Oct 18 2010
%C From _Vladimir Shevelev_, Jan 30 2012, Feb 13 2012: (Start)
%C 1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).
%C 2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)
%C From _Hieronymus Fischer_, Nov 22 2012: (Start)
%C Represented in binary form each term after the second one contains every previous term as a substring.
%C The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2(2^(2n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.
%C Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n. (End)
%C Number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - _Patrick Labarque_, Feb 09 2013
%C A090079(a(n)) = a(n) and A090079(m) <> a(n) for m < a(n). - _Reinhard Zumkeller_, Feb 16 2013
%C a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - _Geoffrey Critzer_, Dec 15 2013
%C a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - _Ran Pan_, Apr 18 2015
%C a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))H(2), where is the Kronecker product. - _William P. Orrick_, Jun 28 2015
%C Conjectured record values of A264784: a(n) = A264784(A155051(n-1)). - _Reinhard Zumkeller_, Dec 04 2015. (This is proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - _Antti Karttunen_, Aug 31 2016
%C Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - _Robert Price_, Dec 05 2016
%C For n > 4, a(n-2) is the second-largest number in row n of A127824. - _Dmitry Kamenetsky_, Feb 11 2017
%C Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows: 2; 2'; 1,1; 1',1 = 1',1; 1',1'. - _Gregory L. Simay_, Sep 02 2017
%C Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2x^2 - 2x^3 - ...) = x/(1 - 2x - x^2 + 2x^3). - _Gregory L. Simay_, Sep 10 2017
%C a(n) = 2^(n-1) + 2^(n-3) + 2^(n-5) + ... a(2k -1) = A002450(k) is the sum of the powers of 4. a(2k) = 2A002450(k). - _Gregory L. Simay_, Sep 27 2017
%C a(2n) = n times the string in binary representation, a(2n+1) = n times the string followed with in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - _Ctibor O. Zizka_, Nov 06 2018
%C Except for 0, these are the positive integers whose binary expansion has cuts-resistance 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. Cuts-resistance 2 is A329862. - _Gus Wiseman_, Nov 27 2019
%C From _Markus Sigg_, Sep 14 2020: (Start)
%C Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4a(n-2)+1) = s(12a(n-2)+4)+1 = s(3a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.
%C Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)
%C From _Gary W. Adamson_, May 14 2021: (Start)
%C With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:
%C ..1 3 7 15...
%C ..1 0 1 1 1 0 1 0 1 0 1 1 1 0 1...
%C ..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)
%C .....1...........2.......................5...; as to zeros.
%C ..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:
%C ..0, 1, 2, 0, 1, 2, ... whereas even numbered disks move in the pattern:
%C ..0, 2, 1, 0, 2, 1, ... (End)
%C Except for 0, numbers that are repunits in Gray-code representation (A014550). - _Amiram Eldar_, May 21 2021
%C From _Gus Wiseman_, Apr 20 2023: (Start)
%C Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:
%C {1} {1} {1} {1}
%C {2} {2} {2}
%C {3} {3}
%C {1,3} {4}
%C {1,2,3} {1,3}
%C {2,4}
%C {1,2,3}
%C {1,2,4}
%C {1,3,4}
%C {2,3,4}
%C The complement is counted by A005578.
%C For mean instead of median we have A051293, counting empty sets A327475.
%C For normal multisets we have A056450, strongly normal A361202.
%C For partitions we have A325347, strict A359907, complement A307683.
%C (End)
%C In binary form, because the sequence 10101_2... keeps repeating, one can represent a(n) = floor(0.(10)_2 2^n). Because 0.(10)_2 = 0.(10)_2 4 - 2, then 0.(10)_2 = 2/3. Substituting this value in the first expression, we obtain a(n) = floor(2^(n+1)/3). This also agrees with the formula of Paul Barry, Oct 08 2005. - _Giovanni Ciriani_, Mar 31 2025
%D Thomas Fink and Yong Mao, The 85 Ways to Tie a Tie, Broadway Books, New York (1999), p. 138.
%D Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009.
%H G. C. Greubel, Table of n, a(n) for n = 0..3300 (terms 0..300 from T. D. Noe)
%H Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
%H Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
%H Sergei L. Bezrukov et al., The congestion of n-cube layout on a rectangular grid, Discrete Mathematics 213.1-3 (2000): 13-19. See Theorem 1.
%H F. Chapoton and S. Giraudo, Enveloping operads and bicoloured noncrossing configurations, arXiv:1310.4521 [math.CO], 2013. Is the sequence in Table 2 this sequence? - _N. J. A. Sloane_, Jan 04 2014. (Yes, it is. See Stockmeyer's arXiv:1608.08245 2016 paper for the proof.)
%H Ji Young Choi, Ternary Modified Collatz Sequences And Jacobsthal Numbers, Journal of Integer Sequences, Vol. 19 (2016), #16.7.5.
%H Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
%H Madeleine Goertz and Aaron Williams, The Quaternary Gray Code and How It Can Be Used to Solve Ziggurat and Other Ziggu Puzzles, arXiv:2411.19291 [math.CO], 2024. See pp. 1, 17, 42.
%H David Hayes, Kaveh Khodjasteh, Lorenza Viola and Michael J. Biercuk, Reducing sequencing complexity in dynamical quantum error suppression by Walsh modulation, arXiv:1109.6002 [quant-ph], 2011.
%H Clemens Heuberger and Daniel Krenn, Esthetic Numbers and Lifting Restrictions on the Analysis of Summatory Functions of Regular Sequences, arXiv:1808.00842 [math.CO], 2018. See p. 10.
%H Clemens Heuberger and Daniel Krenn, Asymptotic Analysis of Regular Sequences, arXiv:1810.13178 [math.CO], 2018. See p. 29.
%H Andreas M. Hinz, The Lichtenberg sequence, Fib. Quart., 55 (2017), 2-12.
%H Andreas M. Hinz, Sandi Klavžar, Uroš Milutinović, and Ciril Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 56. Book's website
%H Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
%H Jia Huang, Nonassociativity of the Norton Algebras of some distance regular graphs, arXiv:2001.05547 [math.CO], 2020.
%H Jia Huang, Norton algebras of the Hamming Graphs via linear characters, arXiv:2101.05711 [math.CO], 2021.
%H Jia Huang and Erkko Lehtonen, Associative-commutative spectra for some varieties of groupoids, arXiv:2401.15786 [math.CO], 2024. See p. 17.
%H Jia Huang, Madison Mickey, and Jianbai Xu, The Nonassociativity of the Double Minus Operation, Journal of Integer Sequences, Vol. 20 (2017), #17.10.3.
%H Ryota Inagaki, Tanya Khovanova, and Austin Luo, On Chip-Firing on Undirected Binary Trees, Ann. Comb. (2025). See p. 10.
%H INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 394
%H Hans Isdahl, "Knitwear" puzzle
%H D. E. Knuth and O. P. Lossers, Partitions of a circular set, Problem 11151 in Amer. Math. Monthly 114 (3) (2007) p 265, E_3.
%H S. Lafortune, A. Ramani, B. Grammaticos, Y. Ohta and K.M. Tamizhmani, Blending two discrete integrability criteria: singularity confinement and algebraic entropy, arXiv:nlin/0104020 [nlin.SI], 2001.
%H Robert L. Lamphere, A Recurrence Relation in the Spinout Puzzle, The College Mathematics Journal, Vol. 27, Nbr. 4, Page 289, Sep. 96.
%H Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
%H Georg Christoph Lichtenberg, Vermischte Schriften, Band 6 (1805). See chapter 6, p. 257.
%H Saad Mneimneh, Simple Variations on the Tower of Hanoi to Guide the Study of Recurrences and Proofs by Induction, Department of Computer Science, Hunter College, CUNY, 2019.
%H Richard Moot, Partial Orders, Residuation, and First-Order Linear Logic, arXiv:2008.06351 [cs.LO], 2020.
%H Gregg Musiker and Son Nguyen, Labeled Chip-firing on Binary Trees, arXiv:2206.02007 [math.CO], 2022.
%H Kival Ngaokrajang, Illustration of initial terms
%H Ahmet Öteleş, On the sum of Pell and Jacobsthal numbers by the determinants of Hessenberg matrices, AIP Conference Proceedings 1863, 310003 (2017).
%H Vladimir Shevelev, On the Basis Polynomials in the Theory of Permutations with Prescribed Up-Down Structure, arXiv:0801.0072 [math.CO], 2007-2010. See Example 3.
%H Chloe E. Shiff and Noah A. Rosenberg, Enumeration of rooted binary perfect phylogenies, arXiv:2410.14915 [q-bio.PE], 2024. See pp. 15, 17.
%H A. V. Sills and H. Wang, On the maximal Wiener index and related questions, Discrete Applied Mathematics, Volume 160, Issues 10-11, July 2012, Pages 1615-1623 - _N. J. A. Sloane_, Sep 21 2012
%H N. J. A. Sloane, Transforms
%H Elen Viviani Pereira Spreafico, Francisco Regis Vieira Alves, Paula Maria Machado Cruz Catarino, and Eudes Antonio Costa, A brief study of the one parameter Lichtenberg numbers, VI Int'l Conf. Math. Appl. Sci. Eng. (ICMASE 2025).
%H Paul K. Stockmeyer, An Exploration of Sequence A000975, arXiv:1608.08245 [math.CO], 2016; Fib. Quart. 55 (2017) 174.
%H Eric Weisstein's World of Mathematics, Baguenaudier
%H A. K. Whitford, Binet's Formula Generalized, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.
%H Index entries for sequences related to binary expansion of n
%H Index entries for linear recurrences with constant coefficients, signature (2,1,-2).
%F a(n) = ceiling(2(2^n-1)/3).
%F Alternating sum transform (PSumSIGN) of {2^n - 1} (A000225).
%F a(n) = a(n-1) + 2a(n-2) + 1.
%F a(n) = 22^n/3 - 1/2 - (-1)^n/6.
%F a(n) = Sum_{i = 0..n} A001045(i), partial sums of A001045. - _Bill Blewett_
%F a(n) = n + 2Sum_{k = 1..n-2} a(k).
%F G.f.: x/((1+x)(1-x)(1-2x)) = x/(1-2x-x^2+2x^3). - _Paul Barry_, Feb 11 2003
%F a(n) = 2a(n-1) + a(n-2) - 2a(n-3). - _Paul Barry_, Feb 11 2003
%F a(n) = Sum_{k = 0..floor((n-1)/2)} 2^(n-2k-1). - _Paul Barry_, Nov 11 2003
%F a(n+1) = Sum_{k=0..floor(n/2)} 2^(n-2k); a(n+1) = Sum_{k = 0..n} Sum_{j = 0..k} (-1)^(j+k)2^j. - _Paul Barry_, Nov 12 2003
%F (-1)^(n+1)a(n) = Sum_{i = 0..n} Sum_{k = 1..i} k!k Stirling2(i, k)(-1)^(k-1) = (1/3)(-2)^(n+1)-(1/6)(3(-1)^(n+1)-1). - Mario Catalani (mario.catalani(AT)unito.it), Dec 22 2003
%F a(n+1) = (n!/3)Sum_{i - (-1)^i + j = n, i = 0..n, j = 0..n} 1/(i - (-1)^i)!/j!. - _Benoit Cloitre_, May 24 2004
%F a(n) = A001045(n+1) - A059841(n). - _Paul Barry_, Jul 22 2004
%F a(n) = Sum_{k = 0..n} 2^(n-k-1)(1-(-1)^k), row sums of A130125. - _Paul Barry_, Jul 28 2004
%F a(n) = Sum_{k = 0..n} binomial(k, n-k+1)2^(n-k); a(n) = Sum_{k = 0..floor(n/2)} binomial(n-k, k+1)2^k. - _Paul Barry_, Oct 07 2004
%F a(n) = A107909(A104161(n)); A007088(a(n)) = A056830(n). - _Reinhard Zumkeller_, May 28 2005
%F a(n) = floor(2^(n+1)/3) = ceiling(2^(n+1)/3) - 1 = A005578(n+1) - 1. - _Paul Barry_, Oct 08 2005
%F Convolution of "Number of fixed points in all 231-avoiding involutions in S_n." (A059570) with "1-n" (A024000), treating the result as if offset was 0. - _Graeme McRae_, Jul 12 2006
%F a(n) = A081254(n) - 2^n. - _Philippe Deléham_, Oct 15 2006
%F Starting (1, 2, 5, 10, 21, 42, ...), these are the row sums of triangle A135228. - _Gary W. Adamson_, Nov 23 2007
%F Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n [1,0,0] = [A005578(n), A001045(n), a(n-1)]. - _Gary W. Adamson_, Dec 25 2007
%F 2^n = 2A005578(n-1) + 2A001045(n) + 2a(n-2). - _Gary W. Adamson_, Dec 25 2007
%F If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)stirling2(k,j)x^(m-k) then a(n-3) = (-1)^(n-1)f(n,3,-2), (n >= 3). - _Milan Janjic_, Apr 26 2009
%F a(n) + A001045(n) = A166920(n). a(n) + A001045(n+2) = A051049(n+1). - _Paul Curtz_, Oct 29 2009
%F a(n) = floor(A051049(n+1)/3). - _Gary Detlefs_, Dec 19 2010
%F a(n) = round((2^(n+2)-3)/6) = floor((2^(n+1)-1)/3) = round((2^(n+1)-2)/3); a(n) = a(n-2) + 2^(n-1), n > 1. - _Mircea Merca_, Dec 27 2010
%F a(n) = binary XOR of 2^k-1 for k=0..n. - _Paul D. Hanna_, Nov 05 2011
%F E.g.f.: 2/3exp(2x) - 1/2exp(x) - 1/6exp(-x) = 2/3U(0); U(k) = 1 - 3/(4(2^k) - 4(2^k)/(1+3(-1)^k - 24x(2^k)/(8x(2^k)(-1)^k - (k+1)/U(k+1)))); (continued fraction). - _Sergei N. Gladkovskii_, Nov 21 2011
%F Starting with "1" = triangle A059260 [1, 2, 2, 2, ...] as a vector. - _Gary W. Adamson_, Mar 06 2012
%F a(n) = 2(2^n - 1)/3, for even n; a(n) = (2^(n+1) - 1)/3 = (1/3)(2^((n+1)/2) - 1)(2^((n+1)/2) + 1), for odd n. - _Hieronymus Fischer_, Nov 22 2012
%F a(n) + a(n+1) = 2^(n+1) - 1. - _Arie Bos_, Apr 03 2013
%F G.f.: Q(0)/(3(1-x)), where Q(k) = 1 - 1/(4^k - 2x16^k/(2x4^k - 1/(1 + 1/(24^k - 8x16^k/(4x4^k + 1/Q(k+1)))))); (continued fraction). - _Sergei N. Gladkovskii_, May 21 2013
%F floor(a(n+2)3/5) = A077854(n), for n >= 0. - _Armands Strazds_, Sep 21 2014
%F a(n) = (2^(n+1) - 2 + (n mod 2))/3. - _Paul Toms_, Mar 18 2015
%F a(0) = 0, a(n) = 2(a(n-1)) + (n mod 2). - _Paul Toms_, Mar 18 2015
%F Binary: a(n) = (a(n-1) shift left 1) + (a(n-1)) NOR (...11110). - _Paul Toms_, Mar 18 2015
%F Binary: for n > 1, a(n) = 2a(n-1) OR a(n-2). - _Stanislav Sykora_, Nov 12 2015
%F a(n) = A266613(n) - 202^(n-5), for n > 2. - _Andres Cicuttin_, Mar 31 2016
%F From _Michael Somos_, Jul 23 2017: (Start)
%F a(n) = -(2^n)a(-n) for even n; a(n) = -(2^(n+1))a(-n) + 1 for odd n.
%F 0 = +a(n)(+2 +4a(n) -4a(n+1)) + a(n+1)(-1 +a(n+1)) for all n in Z. (End)
%F G.f.: (x^1+x^3+x^5+x^7+...)/(1-2x). - _Gregory L. Simay_, Sep 27 2017
%F a(n+1) = A051049(n) + A001045(n). - _Yuchun Ji_, Jul 12 2018
%F a(n) = A153772(n+3)/4. - _Markus Sigg_, Sep 14 2020
%F a(4k+d) = 2^(d+1)a(4k-1) + a(d), a(n+4) = a(n) + 2^n10, a(0..3) = [0,1,2,5]. So the last digit is always 0,1,2,5 repeated. - _Yuchun Ji_, May 22 2023
%e a(4)=10 since 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101, 1111 are the 10 binary strings switching 0000 to 1111.
%e a(3) = 1 because "lrc" is the only way to tie a tie with 3 half turns, namely, pass the business end of the tie behind the standing part to the left, bring across the front to the right, then behind to the center. The final motion of tucking the loose end down the front behind the "lr" piece is not considered a "step".
%e a(4) = 2 because "lrlc" is the only way to tie a tie with 4 half turns. Note that since the number of moves is even, the first step is to go to the left in front of the tie, not behind it. This knot is the standard "four in hand", the most commonly known men's tie knot. By contrast, the second most well known tie knot, the Windsor, is represented by "lcrlcrlc".
%e a(n) = (2^0 - 1) XOR (2^1 - 1) XOR (2^2 - 1) XOR (2^3 - 1) XOR ... XOR (2^n - 1). - _Paul D. Hanna_, Nov 05 2011
%e G.f. = x + 2x^2 + 5x^3 + 10x^4 + 21x^5 + 42x^6 + 85x^7 + 170x^8 + ...
%e a(9) = 341 = 2^8 + 2^6 + 2^4 + 2^2 + 2^0 = 4^4 + 4^3 + 4^2 + 4^1 + 4^0 = A002450(5). a(10) = 682 = 2a(9) = 2A002450(5). - _Gregory L. Simay_, Sep 27 2017
%p A000975 := proc(n) option remember; if n <= 1 then n else if n mod 2 = 0 then 2A000975(n-1) else 2A000975(n-1)+1 fi; fi; end;
%p seq(iquo(2^n,3),n=1..33); # _Zerinvary Lajos_, Apr 20 2008
%p f:=n-> if n mod 2 = 0 then (2^n-1)/3 else (2^n-2)/3; fi; [seq(f(n),n=0..40)]; # _N. J. A. Sloane_, Mar 21 2017
%t Array[Ceiling[2(2^# - 1)/3] &, 41, 0]
%t RecurrenceTable[{a == 0, a == 1, a[n] == a[n - 1] + 2a[n - 2] + 1}, a, {n, 40}] ( or )
%t LinearRecurrence[{2, 1, -2}, {0, 1, 2}, 40] ( _Harvey P. Dale_, Aug 10 2013 )
%t f[n_] := Block[{exp = n - 2}, Sum[2^i, {i, exp, 0, -2}]]; Array[f, 33] ( _Robert G. Wilson v_, Oct 30 2015 )
%t f[s_List] := Block[{a = s}, Append[s, If[OddQ@ Length@ s, 2a + 1, 2a]]]; Nest[f, {0}, 32] ( _Robert G. Wilson v_, Jul 20 2017 )
%t NestList[2# + Boole[EvenQ[#]] &, 0, 39] ( _Alonso del Arte_, Sep 21 2018 )
%o (PARI) {a(n) = if( n<0, 0, 2 2^n \ 3)}; / _Michael Somos_, Sep 04 2006 /
%o (PARI) a(n)=if(n<=0,0,bitxor(a(n-1),2^n-1)) \ _Paul D. Hanna_, Nov 05 2011
%o (PARI) concat(0, Vec(x/(1-2x-x^2+2x^3) + O(x^100))) \ _Altug Alkan_, Oct 30 2015
%o (PARI) {a(n) = (42^n - 3 - (-1)^n) / 6}; / _Michael Somos_, Jul 23 2017 /
%o (Haskell)
%o a000975 n = a000975_list !! n
%o a000975_list = 0 : 1 : map (+ 1)
%o (zipWith (+) (tail a000975_list) (map ( 2) a000975_list))
%o -- _Reinhard Zumkeller_, Mar 07 2012
%o (Magma) [(2^(n+1) - 2 + (n mod 2))/3: n in [0..40]]; // _Vincenzo Librandi_, Mar 18 2015
%o (GAP) List([0..35],n->(2^(n+1)-2+(n mod 2))/3); # _Muniru A Asiru_, Nov 01 2018
%o (Python)
%o def a(n): return (2(n+1) - 2 + (n%2))//3
%o print([a(n) for n in range(35)]) # _Michael S. Branicky_, Dec 19 2021
%Y Partial sums of A001045.
%Y Row sums of triangle A013580.
%Y Equals A026644/2.
%Y Union of the bijections A002450 and A020988. - _Robert G. Wilson v_, Jun 09 2014
%Y Column k=3 of A261139.
%Y Cf. A000295, A005578, A015441, A043291, A053404, A059260, A077854, A119440, A127824, A130125, A135228, A155051, A179970, A264784.
%Y Complement of A107907.
%Y Row 3 of A300653.
%Y Other sequences that relate to the binary representation of the terms: A003714, A003754, A007088, A022290, A056830, A104161, A107909.
%Y Cf. A000120, A001511, A003242, A027383, A070939, A164707, A295235, A319416.
%Y Cf. A005186, A033491, A153772.
%Y Cf. A014550, A035263
%Y Cf. A051293, A067659, A079309, A231147, A325347, A359893, A359907, A361801.
%K nonn,easy,nice,changed
%O 0,3
%A _Mira Bernstein_, _N. J. A. Sloane_, _Robert G. Wilson v_, Sep 13 1996
%E Additional comments from _Barry E. Williams_, Jan 10 2000 |
7105 | https://www.youtube.com/watch?v=YXWKpgmLgHk | Inverse Trigonometric Functions
Professor Dave Explains
3910000 subscribers
7314 likes
Description
471089 views
Posted: 17 Jan 2018
We know about inverse functions, and we know about trigonometric functions, so it's time to learn about inverse trigonometric functions! These are functions where you plug in valid values that trig functions can possess, and they spit out the angles that produce them. There's a little more to it than that, so let's get into it!
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186 comments
Transcript:
Professor Dave again, let’s talk about inverse
trig functions. We know about the six trigonometric functions,
and we know about inverse functions, so it’s time to learn about inverse trigonometric
functions. Let’s recall that when trying to find the
inverse of a function, like F of X equals root X, we change this into a Y, and then
we make X and Y swap places. Then, we solve for Y. In this case, that would involve squaring
both sides, which gives us Y equals X squared. Essentially, inverse functions sequentially
undo anything operating on X in the original function. Now let’s try Y equals sine X. If we swap the variables, how do we solve
for Y? The only way to do this is to take the inverse
sine of both sides. This will cancel out the sine operating on
Y, and we get Y equals inverse sine X. We must note that inverse sine X is not the
same thing as one over sine X. If we had the quantity sine X to the negative
one power, that would equal one over sine X, which we already know is cosecant X. Instead, when we have the negative one here,
it’s the inverse sine function, and rather than sine X, where you plug in an angle and
get a sine value, for inverse sine X, you plug in a valid sine value, and it gives you
the angle that generates it. Inverse sine can also be represented this
way, as arcsin X. We should quickly note that in order for a
function to have an inverse function, it must pass the horizontal line test, because when
we find the inverse, we are reflecting the function across the line Y equals X, and whatever
we get must pass the vertical line test to be considered a function. So sine X as a whole, does not have an inverse
function, but we will instead look at a restricted domain for this function, from negative half
pi to half pi. This section does pass the horizontal line
test, and it is the inverse of this section that gives us the inverse sine function, which
will look like this. The domain of this function will be negative
one to one, because those are the only sine values that are possible. The range will be negative half pi to half
pi, because those are the angles we get when we plug each possible unique sine value. So let’s make sure we know how to evaluate
these. What is the inverse sine of root two over
two? Well, if we recall our unit circle, the angle
that gives a sine of root two over two is quarter pi, so the inverse sine of root two
over two is quarter pi. Now let’s move on to inverse cosine. This will be the inverse of cosine X from
zero to pi. This is a different restriction than we had
for sine, because sine goes from negative one to positive one as it moves through quadrants
four and one. Cosine spans all the possible values from
one to negative one as it goes through quadrants one and two, so we take the graph of cosine
X from zero to pi, and we reflect it across the line Y equals X to get the inverse cosine
function. As we would expect, the domain is negative
one to one, and the range is zero to pi. What is the inverse cosine of negative one
half? Well that would be the angle in this interval
that gives a cosine value of negative one half, and if we remember the unit circle,
we know that must be two thirds pi. Now let’s look at inverse tangent. Starting with tangent X, we see that we will
want to restrict things to within negative half pi and half pi, so that it passes the
horizontal line test. Now to get the inverse tangent function, we
reflect across Y equals X, and we get this. Now the domain is all real numbers, while
the range is negative half pi to half pi, since those are the angles we get when we
plug in all the possible tangent values, which go towards positive and negative infinity
as cosine gets very small. These are trickier to evaluate in your head,
but sometimes we can do it, like the inverse tangent of root three. We might realize that the angle that produces
a tangent of root three must have a sine value over a cosine value that equals root three,
and of the common angles, that must be a third pi, since root three over two divided by one
half is the same as root three over two times two, which equals root three. Luckily, rather than having to evaluate inverse
trig functions by hand, we can always just plug them into a calculator. Just make sure it’s in radian mode, plug
in the sine, cosine, or tangent value, and use the buttons that say things like inverse
sine, or arcsin, to find out what the corresponding angle is. Just remember, you may get a messy answer
that will be hard to express as fractions of pi, but it will be the correct answer. Let’s check comprehension. |
7106 | https://www.youtube.com/watch?v=5qh1NDRis58 | High School Physics: Free Fall
Dan Fullerton (APlusPhysics)
32500 subscribers
709 likes
Description
135891 views
Posted: 20 Aug 2011
In this video Dan Fullerton provides a brief introduction to free fall problems for students in high school physics courses including Regents Physics, IB, AP Physics, and others.
APlusPhysics.com provides a variety of free educational resources to assist students in finding success in high school physics courses such as Honors Physics, Regents Physics, and AP Physics. Resources include video mini-lessons, online tutorials, help forums, problem sets, downloads, and review books.
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For more information, check out and associated books in the series specific to courses:
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AP Physics 1 Essentials:
AP Physics 2 Essentials:
238 comments
Transcript:
Introduction hi everyone today I'd like to talk about freefall and our goal here is going to be to use the kinematic equations we learn to solve problems for objects moving at a constant acceleration while they're in freef Fall well what is freef What is Free Fall fall now first thing freef fall is an object that is not being acted upon by any Force other than gravity so for example while I'm holding this pen it's not in free fall if I let go while it's in midair the only force acting on it is gravity well wait you might say what about air resistance you're absolutely right in reality air resistance is also acting on this pen when I let go of it however for the purposes of this course we're going to neglect air resistance so freef fall is an object is a situation where we do not have air resistance or air friction acting for example if we drop this ball in sheet of paper it's pretty obvious that the ball is going to drop fall fast F than the paper however if we could remove the air from the room we would find that they fall at the same rate and Commander David Scott actually did this on an Apollo Mission when he was on the moon he dropped a hammer and a feather at the same time and they both fell at the same rate they hit the ground at the same time so we're going to analyze the motion of objects while they're falling or being thrown up while their only force acting on them is gravity and will neglect air resistance or air friction for the purposes of this course later on in other physics courses you will go into considerably more depth on air Acceleration Due to Gravity resistance now the acceleration due to gravity is how much an object near the surface of the Earth accelerates toward the center of the earth we typically think of this as a constant at 9.8 m/s squar but in reality this acceleration due to gravity this constant that we give the symbol lowercase G it's a bit of a misnomer really what we should call it is the gravitational field strength near the surface of the Earth and it is that as well but what it's really trying to tell you is if we drop this ball it will accelerate toward the center of the earth at 9.8 m/s Square it'll accelerate 9.8 m/s every second as we move further away from the earth G decreases and as we get closer to the Center of the Earth or if we were on another planet that's more massive and we'll learn more about gravity later on but G can actually change on other planets so near the surface of the Earth the acceleration due to gravity the acceleration experienced by an object in freef fall is 9.8 m/s squared always toward the center of the Objects earth now we'll start off by talking about objects that are falling from rest anything that we drop or that falls from rest has an initial velocity of V KN of zero now when we talked about the kinematic equations we said we're going to call the direction of an object's initial motion positive so for an object dropped from rest since it's initially moving downwards we'll call that the positive direction then when we look at the problem we also realize that the acceleration which is down toward the center of the earth points in the same direction as our positive axis since they point in the same direction our acceleration is going to be positive G or positive 9.81 m/s squared other than that it's just a standard kinematics problem let's take a look at an Example example how far will a brick starting from rest fall freely in 3 seconds neglect air resistance well we can figure this out by first recognizing that this is a vertical motion problem since the object will go down first we'll call that the positive direction down and we'll set up our vertical motion table V KN V Delta y a and t and since it starts from rest V not is equal to zero Delta Y is what we're trying to find and we know the acceleration is positive G positive because acceleration points down and positive is down so that's going to be 9.8 m/s squ we don't know what our final velocity is we do know that time is 3 seconds so trying to find how far it falls that's Delta y we need to pick a kinematic equation that has most or all of our variables in it well the one that comes to mind very easily is deltay equals VT plus 12 a t^2 and a nice trick for simplification if you notice that V is 0 0 anything is 0 so that whole term becomes Zer this simplifies to Delta y = 12 a t^2 now we can substitute in our values with units to say that Delta y equal 1/2 our acceleration 9.8 m/s squar time the square of our time 3 seconds squared and when I do all this plug it into my calculator I should come up with a deltay or a vertical displacement of about 44 M and notice our answer is positive that means that it was displaced in the same direction that we called positive there so it was displaced 44 M down of course that makes sense you wouldn't drop a brick and have it fall upwards so let's take a look at a more complicated type of situation what happens when you launch an object upward or it starts starts with some initial velocity going up well we have to look at the motion of the object on the way up and on the way down now since the object's initial motion is upward we're going to call the up Direction our positive y we will call our acceleration which points down is in the opposite direction as our positive axis so our acceleration now is going to be negative G or 9.8 m/s squar now things to note here for objects that travel up and down there are a couple tricks that make this much much simpler at the highest point as an object goes up when it right before it comes down for an instant its velocity is zero so at its highest point we know V equals z that can be very very helpful in solving problems other things to notice is for an object that travels up first Split Second stops and then comes back down is the symmetry of motion however ever long it takes for it to come up if it comes back to the exact same altitude to the exact same height it'll take the exact same amount of time to come up as it does to come down whatever its initial velocity is as you launch it if you launch it at 50 m/s up it'll Slow Down slow down slow down stop then it will turn around start coming down faster faster faster and when it gets to that exact same height you launched it from it'll be going 50 m/s down as long as we neglect air resistance so we have symmetry of motion we can use that to help us solve some of these problems let's take a look at an Speed example a ball thrown vertically upward reaches a maximum height of 30 m at its maximum height what's the speed of the ball well the object travels up its highest point it stops comes back down V equals z at its highest point so what's the speed at its maximum height zero that Basketball simple how about if we look at a basketball player who jumps straight up for a rebound if she's in the air for8 seconds how high did she jump all right now we've got something that's a little bit more involved first off the basketball player starts on the ground jumps up so we're going to call up our positive y direction as we make our vertical motion table now we can also realize that the basketball player travels up and then comes back down and at the highest point velocity is Z and if it takes 8 seconds for the entire trip up and down it must take4 seconds to go up and4 seconds to come down so here's another trick to solving some of these problems often times it's a lot simpler to just solve half of the problem which is what we'll do here let's analyze the motion of the basketball player on the way up only well for a vertical motion table we have V KN V Delta y a and t we don't know the initial velocity but our final velocity if we're looking at just the motion on the way up we can say that that's zero acceleration if we called up the positive direction acceleration points down so positive Y is this way acceleration is that way they're in opposite directions therefore the acceleration must be G or 9.8 m/ second squ and finally if we're looking at just the motion on the way up the time involved is4 seconds not8 seconds got to be careful there we're looking for Delta Y how high she jumped and we don't know V KN yet well I know three things so I know I can solve this problem but if I'm looking for Delta y I don't see any equations that can give me Delta y directly so why don't we solve for V first and then we'll find VTA Delta y so we can start with v = v + a t or v = v minus a now we can substitute in our values V equal V 0 minus a which is 9.8 m/s SAR t.4 seconds for a total initial velocity of about 3.92 m/ second so we know V now it's no longer a question mark it's 3.92 m/s now we can solve for Delta y we can use any equation we want that has Delta Y in it and the easiest one that I can think of Delta y equal v t + a t^ 2 and we can substitute indirectly Delta y = 3.92 2 m/s our time of4 seconds + 12 a 9.8 m/s squar t^ 24 seconds squared for a total deltay of about 0.78 M that's a little less than 1 M about a yard which makes sense a reasonable vertical jump is somewhere in the two three to two to 3T range 3T is a really big jump or 1 meter would be a really great vertical leap so for a basketball player who's training 78% of a meter that's a pretty good jump that's reasonable so there's a problem where we had to go through a couple steps and also recognize the symmetry of motion as the basketball player jumped up and came back down let's take a look at another problem this time we have a ball that's thrown straight down W with the speed of5 m/s from a height of 4 M what is the speed of the ball 7 seconds after it is released well we recognize right right away it's a vertical motion problem and the initial velocity the initial motion of the ball is going to be down so let's call that our positive y now we can make our vertical motion table V KN V Delta y a and t we know the initial velocity is5 m/s down since it's down and we call down the positive direction V KN is going to be positive .5 m/s we're trying to find the final speed so there's our find Delta y oo this is tricky it says it's thrown from a height of 4 M but it doesn't ask us Delta Y and it doesn't say the total distance traveled by the ball in this time is 4 m so we really don't know deltay got to read that one very carefully acceleration well acceleration points down and positive y points down they're in the same direction therefore this must be our positive G or 9.8 m/s squared and we want to know the speed after seconds has elapsed so T is 7 Seconds now we can go and we can solve for our final velocity directly we know v v a and t we have a kinematic equation with those variables in it already so we can say that v = v + a t or V = .5 m/s plus 9.8 m/s squared times our time 7 Seconds handy dandy calculator should tell me this is about 7.4 m per Baseball second all right let's take a look at one last sample problem if we have a baseball player and she throws the ball upward with a speed of 30 m/ second and we're going to neglect friction find the maximum height reached by the baseball well for this sort of problem again it's a vertical motion problem and the ball starts going up so we'll call that positive Y and we can make our vertical motion table V KN V Delta y a and t our acceleration points down it always does and since that's in the opposite direction of positive y our acceleration is going to be 9.8 m/s squar what else do we know it starts with an initial velocity of 30 m/s and 30 m/s is upward in the same direction we called our positive y so that's a positive V kn we only have two things here so far we're trying to find the maximum height Delta y we need one more piece of information well if the ball is going up and then coming back down what we can do is we can realize that its highest point right there its velocity is zero so when its velocity is zero that's at the point where it's at its maximum displacement so if we want its maximum displacement will solve for this problem when that final velocity is equal to zero there's our third unknown right away we can jump into our kinematic equations with those four variables v^2 = v^ 2 + 2 a deltay rearrange this to get Delta y by itself Delta y = v 2 - v^ 2/ 2 a or Delta y equals as we substitute in with units v^2 that's 0^ 2 minus 30 m/s our V value squared all over 2 -9.8 m/s squar run that through the calculator and I should come up with a Delta y right around 45.9 or 46 m Next Steps all right that should give you a good idea of what sorts of problems you're going to see for next steps why don't you see if you can create your own dropped object problem see if you can figure out something interesting or fun to drop uh who knows maybe you drop your physics teacher off the third floor uh third floor roof of your school or whatever it happens to be how long does it take that teacher to hit the ground and then see if you can solve the problem then try your own vertically launched problem where you start with some initial velocity upward and see if you can solve that and of course if you need help want more information or extra practice you can always check out A+ physics.com thanks and have a terrific day |
7107 | https://dictionary.cambridge.org/us/dictionary/english-polish/egregious | EGREGIOUS definition | Cambridge Dictionary
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Translation of egregious – English-Polish dictionary
egregious
adjective
formal uk
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/ɪˈɡriːdʒəs/us
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extremelybad or shocking in an obvious way
rażący
an egregious example of racism
(Translation of egregious from the Cambridge English–Polish Dictionary © Cambridge University Press)
Examples of egregious
egregious
A more egregious misrepresentation involves the issue of content versus process.
From the Cambridge English Corpus
While public-sector size in general may not be associated with corruption, egregious forms of state intervention are another matter.
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Some of the most egregious violations of liberty involve not the intent to punish but rather the paternalistic intent to help.
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Were there particular instances or examples of "politicizing" science that you felt were most striking or egregious?
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But it is unfortunate that the very last sentence is marred by an egregious slip.
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It is an egregious mistake, however, to think that it is such "openness" that accounts for dominance.
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If they did not, then the decision-its substance aside-can be seen as illegitimate with respect to those millions of citizens in a rather egregious way.
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Now we resort to policing from outside agencies only to punish a few egregious violations.
From the Cambridge English Corpus
These examples are from corpora and from sources on the web. Any opinions in the examples do not represent the opinion of the Cambridge Dictionary editors or of Cambridge University Press or its licensors.
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Translations of egregious
in Chinese (Traditional)
（錯誤等）極其嚴重的，極壞的，令人震驚的…
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in Chinese (Simplified)
（错误等）极其严重的，极坏的，令人震惊的…
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in Spanish
mayúsculo, atroz…
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in Portuguese
terrível, flagrante…
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in more languages
in Turkish
in Russian
oldukça kötü, şok eden…
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вопиющий…
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7108 | https://www.chegg.com/homework-help/questions-and-answers/two-electrolytic-cells-one-containing-gallium-69723-g-mol-ga3-ions-containing-nickel-58693-q71456749 | Solved There are two electrolytic cells, one containing | Chegg.com
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There are two electrolytic cells, one containing gallium (69.723 g/mol) as Ga3+ ions and the other containing nickel (58.693 g/mol) as Ni2+ ions. They are connected in series (meaning one after the other, so that they both have the same current passing through them). Which cell has a larger mass of metal deposited after some time and why? The nickel (Ni)
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Question: There are two electrolytic cells, one containing gallium (69.723 g/mol) as Ga3+ ions and the other containing nickel (58.693 g/mol) as Ni2+ ions. They are connected in series (meaning one after the other, so that they both have the same current passing through them). Which cell has a larger mass of metal deposited after some time and why? The nickel (Ni)
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There are 2 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 Solution,
To solve this, we need to consider Faraday's laws of electrolysis, particularly the first l... View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question
Transcribed image text:
There are two electrolytic cells, one containing gallium (69.723 g/mol) as Ga3+ ions and the other containing nickel (58.693 g/mol) as Ni2+ ions. They are connected in series (meaning one after the other, so that they both have the same current passing through them). Which cell has a larger mass of metal deposited after some time and why? The nickel (Ni) cell because the lower valency of the ion allows for more atoms to be deposited than for the gallium. They both have the same mass because the current passing through the cells must be identical. The gallium (Ga) cell because the molar mass of gallium is greater than that of nickel. The gallium (Ga) cell because the +3 valence state of the ion allows for more atoms to be deposited for the same current.
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7109 | https://journals.tubitak.gov.tr/math/vol44/iss5/28/ | "Hypergeometric distribution of the number of draws from an urn with tw" by J. L. GONZALEZ-SANTANDER
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Home>Turkish Journal of Mathematics>Vol. 44 (2020)>No. 5
Turkish Journal of Mathematics
Hypergeometric distribution of the number of draws from an urn with two types of items before one of the counts reaches a threshold
Authors
J. L. GONZALEZ-SANTANDER
Abstract
We consider an urn with $R$ elements of one type and $B$ elements of other type. We calculate the probability distribution $P_{n_{R},n_{B}}^{R,B}\left( s\right) $ wherein the random variable $s$ is the number of draws from the urn until we reach $n_{R}$ elements of type $R$ or $n_{B}$ elements of type $B$. We calculate the mean value $\left\langle s\right\rangle $ and the standard deviation $\sigma $ of $P_{n_{R},n_{B}}^{R,B}\left( s\right) $ in terms of hypergeometric functions. For $n_{R}=n_{B}$ and $B=R$ , we reduce $\left\langle s\right\rangle $ and $\sigma $ in terms of elementary functions. Also, the normalization condition leads to a new hypergeometric summation formula involving ${3}F{2}$ terminating series with unity argument. For $n_{R}=n_{B}$, we provide an alternative proof of this summation formula using $q$-hypergeometric functions. As a consistency test, computer simulations have been performed to confirm the analytical results obtained.
DOI
10.3906/mat-1907-121
Keywords
Generalized hypergeometric function, hypergeometric probability distribution, $q$-hypergeometric function
First Page
1881
Last Page
1898
Recommended Citation
GONZALEZ-SANTANDER, J. L (2020). Hypergeometric distribution of the number of draws from an urn with two types of items before one of the counts reaches a threshold. Turkish Journal of Mathematics 44(5): 1881-1898.
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7110 | https://www.varsitytutors.com/practice/subjects/geometry/help/how-to-find-an-angle-in-a-rhombus | Geometry - How to find an angle in a rhombus | Practice Hub
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HomeGeometryLearn by ConceptHow to find an angle in a rhombus
How to find an angle in a rhombus
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Geometry › How to find an angle in a rhombus
Questions 1 - 9
1
In the rhombus shown above, angle has a measurement of degrees. Find the measurement of angle .
Explanation
The four interior angles in any rhombus must have a sum of degrees.
The opposite interior angles must be equivalent, and the adjacent angles have a sum of degrees.
Since angle is adjacent to angle , they must have a sum of degrees.
The solution is:
2
Angle has a measurement of degrees. Find the sum of angles and
Explanation
The four interior angles in any rhombus must have a sum of degrees.
The opposite interior angles must be equivalent, and the adjacent angles have a sum of degrees.
Since, both angles and are adjacent to angle --find the measurement of one of these two angles by: .
Angle and angle must each equal degrees. So the sum of angles and degrees.
3
Given Rhombus and diagonal .
Explanation
The rhombus referenced is below:
As a rhombus is a parallelogram, consecutive angles and are supplementary - that is,
.
Set and solve:
A diagonal of a rhombus bisects the angles at its endpoints, so, specifically, bisects . Therefore,
.
4
Given: Rhombus with diagonals and intersecting at point .
True or false: must be a right angle.
True
False
Explanation
One characteristic of a rhombus is that its diagonals are perpendicular. It follows that must be a right angle.
5
Given: Parallelogram such that .
True or false: Parallelogram cannot be a rhombus.
False
True
Explanation
A rhombus is defined to be a parallelogram with four congruent sides; there is no restriction as to the measures of the angles. Therefore, a rhombus can have angles of any measure. The correct choice is "false".
6
A rhombus has two interior angles with a measurement of degrees. What is the measurement of each of the other two interior angles?
Both of the remaining angles are degrees.
One angle is degrees and the other angle must be degrees.
All of the interior angles of the rhombus are degrees.
Not enough information is provided to solve this problem.
Explanation
The four interior angles in any rhombus must have a sum of degrees. The opposite interior angles must be equivalent, and the adjacent angles have a sum of degrees.
Thus, if a rhombus has two interior angles of degrees, there must also be two angles that equal:
Check:
7
In the above rhombus, angle degrees. Find the sum of angles and
Explanation
The four interior angles in any rhombus must have a sum of degrees. The opposite interior angles must be equivalent, and the adjacent angles have a sum of degrees.
Since ,
And
So,
8
Using the rhombus above, find the sum of angles and
Explanation
The four interior angles in any rhombus must have a sum of degrees.
The opposite interior angles must be equivalent, and the adjacent angles have a sum of degrees.
Since then,
9
In the rhombus shown above, angle has a measurement of degrees. Find the sum of angles and
Explanation
The four interior angles in any rhombus must have a sum of degrees.
The opposite interior angles must be equivalent, and the adjacent angles have a sum of degrees.
Angles and are opposite interior angles, so they must have equivalent measurements.
The sum is:
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7111 | https://www.skuola.net/fisica/concetti-generali/velocita-luce.html | Velocità della luce, una costante universale cenni di ottica geometrica e storia della misura della costante c
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Velocità della luce - Una costante universale
Velocità della luce - Una costante universale
Appunto di fisica sulla velocità della luce, i tentativi della sua misurazione da Galileo a Fizeau, brevi cenni ai fenomeni di diffusione, riflessione e diffrazione.
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La velocità della luce è la più alta conosciuta, pari a 300.000 km/s, ed è fondamentale per fenomeni quotidiani come la fotosintesi e la trasmissione di informazioni.
La natura della luce ha una doppia caratteristica: si comporta sia come onda che come particella, confermata da teorie e esperimenti di scienziati come Newton, Huygens, Young e Maxwell.
Galileo tentò di misurare la velocità della luce con un esperimento semplice, ma fu Roemer nel 1675 a ottenere una prima misura basata su osservazioni astronomiche.
Fizeau nel 1849 sviluppò un metodo per misurare la velocità della luce usando una ruota dentata e specchi, ottenendo risultati molto vicini al valore attuale accettato.
Fenomeni come la diffusione, la riflessione e la diffrazione della luce spiegano la visione dei colori, le immagini negli specchi e la deviazione delle onde luminose da ostacoli.
In questo appunto di fisica si affronta un argomento sicuramente affasciante: la velocità della luce. Una breve escursione storica sulle teorie riguardo la sua natura e sugli esperimenti finalizzati alla misurazione del suo valore da Galileo a Fizeau. I fenomeni quotidiani che dipendono dalla luce e brevi cenni di ottica geometrica su diffusione, riflessione e diffrazione, completano la trattazione.
Indice
Luce, più veloce non si può
Teorie sulla natura della luce
Misurare la velocità della luce, esperimento di Galilei
Roemer e Fizeau misurano il valore di c
Diffusione e riflessione della luce
Diffrazione della luce
Luce, più veloce non si può
La velocità della luce è veramente sbalorditiva ed è la più alta velocità che si conosca. In un secondo, un raggio di luce percorre la distanza di 300.000 km, la distanza media fra la Terra e la Luna. Questo valore è una costante fisica ed è indicato con c minuscolo, usando la notazione scientifica:
c=3,0⋅10 8 m/s
Il Sole dista in media dalla Terra 150 milioni di chilometri, alla luce occorrono 8 minuti e 13 secondi per giungere sulla Te4rra. Molti fenomeni della vita quotidiana dipendono dalle proprietà della luce. Ad esempio, le piante attraverso la fotosintesi convertono l’energia del sole in energia chimica. La luce è il mezzo principale per trasmettere e ricevere informazioni dagli oggetti che ci circondano, da quelli più vicini a quelli lontanissimi nell’universo.
Ma che cos’è la luce?
La luce è una radiazione elettromagnetica che si propaga anche nel vuoto, viaggiando alla velocità di 300.000 chilometri al secondo e trasporta energia. Il Sole la nostra stella ci illumina e ci riscalda emanando radiazione elettromagnetica.
Teorie sulla natura della luce
Prima dell’inizio del XIX secolo si pensava che la luce fosse costituita da un flusso di particelle emesse dall'oggetto che si stava osservando. Un’esponente di questa teoria, nota come teoria corpuscolare della luce, fu Newton, la sua idea era che delle particelle emesse da una sorgente di luce fossero in grado di entrare nell'occhio dell'osservatore e quindi di stimolare la sensazione visiva. Nel 1678 Huygens, si fece promotore di un’altra teoria quella ondulatoria con la quale riuscì a spiegare le leggi della riflessione e della rifrazione. Soltanto nel 1801 Thomas Young verificò che, in opportune condizioni, la luce dà origine a fenomeni di interferenza e quindi riuscì a dimostrare la natura ondulatoria della luce. Successivamente Maxwell nel corso dell’800 affermò che la luce era una forma di onda elettromagnetica di alta frequenza, teoria confermata sperimentalmente da Hertz, che riuscì a produrre e a rilevare le onde elettromagnetiche. Esiste però un fenomeno che è in contraddizione con la teoria ondulatoria della luce si tratta dell’effetto fotoelettrico, scoperto sempre da Hertz. Il fisico, nei suoi esperimenti osservo che quando la luce colpiva una superficie metallica, in determinate circostanze venivano emessi degli elettroni dalla superficie e che l'energia cinetica di un elettrone emesso era indipendente dall'intensità della luce. Questo fenomeno era in piena contraddizione con la teoria delle onde, la quale prevedeva che un fascio di luce più intenso dovesse cedere più energia all’elettrone.
Sulla base del modello dell’energia quantizzata si ipotizzava che l’energia di un’onda elettromagnetica fosse distribuita in pacchetti di energia chiamati fotoni e quindi in accordo con la teoria di Einstein l'energia di un fotone risultava proporzionale alla frequenza dell'onda elettromagnetica.
La luce va quindi intesa come caratterizzata da una doppia natura o natura duale, in alcuni casi si comporta come onda e in altri casi si comporta come particella.
Misurare la velocità della luce, esperimento di Galilei
Uno dei primi esperimenti per misurare la velocità della luce fu eseguito da Galileo Galilei. Egli fece posizionare due osservatori su due torri distanti 10 km, ciascun osservatore aveva una lanterna munita di un otturatore; nell’esperimento un osservatore avrebbe aperto la sua lanterna e il secondo osservatore avrebbe fatto lo stesso nel momento in cui avesse osservato la luce proveniente dalla prima lanterna. Galileo era convinto di poter ricavare la velocità misurando il tempo di transito del fascio di luce tra le due lanterne poiché la distanza era nota. L’esperimento non andò a buon fine perché, come Galileo stesso concluse, il tempo di transito della luce è molto più piccolo dei tempi di reazione degli osservatori.
Roemer e Fizeau misurano il valore di c
Nel 1675 l’astronomo danese Roemer, riuscì ad ottenere una prima misura realistica del valore della velocità della luce. I suoi calcoli si basavano sulle osservazioni astronomiche di un satellite di Giove, la luna Io. Il periodo di rivoluzione di Io intorno a Giove di circa 42,5 ore e Giove ha un periodo di rivoluzione intorno al sole di circa 12 anni. L'astronomo raccolse dati per più di un anno e notò che i periodi erano maggiori della media quando la Terra si allontanava da Giove minori quando la Terra si avvicinava. Attribuì questa variazione di periodo alia posizione dell’osservatore. Huygens, elaborò i dati di Roemer e stimò un valore della velocità c, prossimo a quello ufficialmente riconosciuto. L’esperimento ha una importanza storica perché stabilisce che la luce ha una velocità finita per quanto grande essa sia.
Il fisico Fizeau, nel 1849 ideò un apparato molto semplice costituito da una ruota dentata posta a distanza fissa da uno specchio; la ruota trasformava un fascio continuo di luce in una serie di impulsi luminosi, conoscendo la distanza, il numero di denti della ruota e la velocità angolare della ruota, misurò il tempo totale impiegato dalla luce per andare e tornare dallo specchio riflettente. Trovo un valore di c di 310.000 km al secondo. Altri sperimentatori con apparati simili stimarono valori della velocità della luce più accurati e prossimi al valore corrente accettato.
Diffusione e riflessione della luce
L’ottica geometrica è una parte della fisica che studia la propagazione delle onde luminose, e usa l’approssimazione dei raggi luminosi. Con questa approssimazione un raggio luminoso è praticamente un fascio di luce molto sottile che viene rappresentato con un tratto di retta.
Quando siamo in una stanza al buio, non siamo in grado di distinguere gli oggetti che ci circondano. È sufficiente la luce della fiamma di una candela per vedere gli oggetti che sono intorno a noi; noi vediamo gli oggetti quando la luce proveniente da essi colpisce i nostri occhi.
Il fenomeno della diffusione è responsabile della visione dei colori degli oggetti investiti da luce bianca cioè da una sovrapposizione di onde elettromagnetiche di tutte le lunghezze d'onda. Ad esempio noi percepiamo il colore rosso di una mela perché la sua superficie riflette in maniera diffusa la radiazione di lunghezza d'onda che corrisponde al colore rosso mentre assorbe tutte le altre lunghezze, la diffusione spiega anche il colore blu del cielo oppure Il rosso e l’arancio che sono i colori dell’alba e del tramonto.
La nostra immagine allo specchio o le immagini di altri veicoli nello specchietto retrovisore dell'automobile sono dovute al fenomeno della riflessione della luce. Quando i raggi luminosi colpiscono una superficie come quella di uno specchio vengono riflessi di un certo angolo secondo le leggi dell’ottica geometrica.
Diffrazione della luce
La diffrazione è un altro comportamento caratteristico delle onde, per il quale esse sono in grado di superare gli ostacoli deviando dalla linea retta. Per la luce questo fenomeno consiste nel produrre delle ombre dai contorni che non sono sempre netti, facendo passare un raggio di luce attraverso una sottilissima fenditura e posizionando uno schermo ad una certa distanza dalla fenditura si osserva una figura di diffrazione che è formata da frange luminose alternate a frange scure.
Per ulteriori approfondimenti sulla luce vedi anche qua.
Domande da interrogazione
Qual è la velocità della luce e perché è considerata una costante fisica?
La velocità della luce è di 300.000 km al secondo, ed è la velocità più alta conosciuta. È considerata una costante fisica, indicata con "c", perché rimane invariata nel vuoto.
Quali sono le principali teorie sulla natura della luce?
Le principali teorie includono la teoria corpuscolare di Newton, la teoria ondulatoria di Huygens e Young, e la teoria elettromagnetica di Maxwell. La luce ha una natura duale, comportandosi sia come onda che come particella.
Come è stata misurata per la prima volta la velocità della luce?
La prima misura realistica della velocità della luce fu ottenuta da Roemer nel 1675, basandosi sulle osservazioni astronomiche del satellite di Giove, Io. Fizeau nel 1849 utilizzò un apparato con una ruota dentata per misurare la velocità della luce.
Quali fenomeni ottici quotidiani dipendono dalla luce?
Fenomeni come la fotosintesi nelle piante, la visione degli oggetti, la diffusione dei colori e la riflessione delle immagini negli specchi dipendono dalle proprietà della luce.
Che cos'è la diffrazione della luce e come si manifesta?
La diffrazione è un fenomeno in cui le onde luminose superano ostacoli deviando dalla linea retta. Si manifesta producendo ombre dai contorni non netti e frange luminose e scure quando la luce passa attraverso una fenditura sottile.
Appunti correlati
#### Velocità della luce e misure fisiche
#### Lavoro di una forza costante
#### Doppia natura della luce
#### Propagazione della luce
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STRETTAMENTE NECESSARI |
7112 | https://www.youtube.com/watch?v=4FWtnwg1y3s | Solving Advanced Absolute Value Inequalities
GreeneMath.com
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Posted: 2 Dec 2019
In this lesson, we will learn how to solve more advanced absolute value inequalities. First, we will learn how to solve absolute value inequalities with two absolute value operations involved. To solve these types of problems, we will set the arguments of the absolute value operations equal to zero. We will then solve these equations and use the solutions to split up our number line into three intervals. From there, we will replace our absolute value operation with the expression itself if the argument is 0 or positive in the interval or the negative of the expression if the argument is negative in the interval. We will then go through each possible inequality and check to see if the solution lies in our given interval. Once we have verified all the correct solutions, we can report our answer. Our last example features an absolute value inequality with a rational expression involved as the argument of the absolute value operation.
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Transcript:
in this lesson we want to discuss how to solve more advanced absolute value inequalities so over the course of the last few lessons we've discussed some methods that can be used to solve some more advanced absolute value equations now again we're just going to go one step further and think about how to apply these same strategies when inequalities are involved so before we kind of jump into the examples I just want to take a minute to refresh your memory on absolute value inequalities just the basic processes that we use when we're solving these guys because it's been a while since we looked at them so I think it's important just to get a little quick refresher in so we should know that if a is greater than zero so a just represents some positive real number if we have the absolute value of X is greater than a okay greater than this positive real number this gives us in or write a compound inequality with or so X can be greater than a or X can be less than the negative of a so as a quick example let's say we had the absolute value of x and this is greater than seven we should know that again this sets up as X is greater than seven or X is less than negative seven graphically it's really easy to figure this out again I'm thinking about what values here for X have a distance from zero that is greater than seven well if I start at zero and I go seven units to the right it can't be seven itself but it can be anything larger right so I just put a parenthesis at seven to say it's not included can I shade everything to the right same thing goes in the other direction if I start at zero when I go seven units to the left I put a parenthesis at negative seven let me just shade everything to the left like that so this would be my solution graphically and we also have this solution here kind of in our standard notation and you can even use interval notation if you want you can say from negative infinity up to but not including negative seven and then the Union with seven is not included again out to positive infinity okay so let's take a look at kind of the other scenario that happens so if a is a positive real number again if a is greater than zero if the absolute value of x is less than this positive number a well then X is to be between negative a and a right X is greater than negative a and it's less than a so here we have a compound inequality with and so let's look at this quick example here suppose we have the absolute value of x and now it's less than 7 okay so we're gonna have a compound inequality with and so basically X is going to be greater than a negative of 7 and it's gonna be less than 7 itself again basically the thing about this graphically I'm thinking about all the values whose absolute value or whose distance from zero is less than 7 so starting at 0 I could basically travel up to but not including 7 so again at 7 I just put a parenthesis there and again I can go the opposite direction so starting at 0 I could go down to okay down to but not include negative 7 so I put a parenthesis there as well so that'll be my graphical solution again from negative 7 to 7 with neither of those included and then in interval notation I'd have negative 7 comma 7 again with a parenthesis next to each alright so two special cases that come up that we need to be aware of again if the absolute value operation in this case is the absolute value of X if we say this is less than 0 there's no solution here ok there's nothing I can plug in for X and take the absolute value of and say that it's going to be less than 0 because the result of this operation here is going to be 0 which is obviously not less than 0 or it's going to be some positive number which is also never gonna be less than 0 so in this situation there is no solution and our last special case we're going to come across if a now is a negative number if it's less than 0 so some negative real number then if we have the absolute value of X is greater than this negative real number well then we're gonna have a solution that's all real numbers right if I have the absolute value of X is greater than negative 5 well that's always true right no matter what I plug in for X the result of this will either be 0 which is greater than negative 5 or it'll be positive which is also greater than negative 5 so this is always true or you could say true for all real numbers ok so now let's get to kind of the first exam and hopefully you saw the lesson where we solved more advanced absolute value equations if you did that lesson already this is going to be a breeze for you we're basically using the same concepts okay so suppose you had something like the absolute value of x plus 3 is greater than the absolute value of 2x minus 1 again when you look at these what you want to do is you want to think about where the expressions inside the absolute value bars are going to be 0 okay you're thinking about kind of the turning point so where could this guy go from negative to positive or from positive to negative well that can only happen when that expression is equal to 0 okay so if I say X plus 3 is equal to 0 and I solve that I get that X is equal to negative 3 ok and then alternatively if I say that 2x minus 1 is equal to 0 if I add 1 each side of the equation I get 2x is equal to 1 divide both sides by 2 and I find that x equals 1/2 ok so we're going to use these numbers negative 3 and 1/2 to kind of split the number line up into 3 intervals ok and I'm not going to do it on this number line here I'm gonna use a table because I find it to be a little bit easier so the numbers we were working with again our negative 3 and positive 1/2 so we're using that to split the number line up into 3 intervals I just prefer to do this on a table because it's just easier for me to organize things so the first interval will be from negative infinity up to but not including negative 3rd okay so that's my first interval then my second interval negative 3 I'm going to include that okay and this is going to be out too but not including 1/2 then my third interval will be anything that includes 1/2 or larger okay so from 1/2 out to positive infinity now what we're gonna do here let's go back up so we have the absolute value of x plus 3 is greater than the absolute value of 2x minus 1 let's go ahead and write that real quick so the absolute value of x plus 3 is greater than the absolute value of 2x minus 1 okay so what we're doing here is we're looking at each interval we're thinking about is the expression inside the absolute value bars negative is it zero or is it positive okay that's what I want to consider so for this guy right here think about the fact that in this interval it's going to be negative right I know that X plus three is equal to zero when x equals negative three so anything less than negative 3 means it's going to be negative right so in this interval I will replace this expression here this absolute value of x plus 3 with the negative of X plus 3 okay so I just wrap it in parenthesis and put a negative out in front to express that okay for this guy right here this 2x minus 1 it's also gonna be negative right because 2x minus 1 is going to be 0 when X is 1/2 so if we're less than 1/2 which we are in this case it's going to be negative so again I just wrap it in parentheses so 2x minus 1 put a negative out in front okay and we'll deal with this scenario in a second let's just kind of move on or write everything so in this interval here we have from negative 3 up to but not including 1/2 so at negative 3 we know that this guy would be 0 but anything larger than that means it's going to be positive okay so whether it's 0 or some positive value I can just take the expression itself so I can just say it's X plus 3 here and again this is still in a format of where it's larger than negative 3 so in this interval it'll also be just X plus 3 okay then for 2x minus 1 again it's made 0 when X is 1/2 in this interval between negative 3 including negative 3 up to but not including 1/2 it's still going to be negative right because that's less than 1/2 so I would say the negative of 2x minus 1 and then here it could either be 0 if X is 1/2 or some positive value so I just drop the absolute value bars and write the expression so 2x minus 1 okay so a very simple process again if it's 0 or positive in that interval you replace the absolute value expression with just what's inside the absolute value bars okay that's all you need if it's a negative okay then what you want to do is drop those absolute value bars wrap it in parentheses put a negative out in front you're basically changing it into its opposite all right so now we've got a lot of work to do we've got to go through the different possibilities so the first possibility is that both of these are negative so let's kind of scroll down and crank that out so I'm gonna have the negative of packs plus 3 is greater than the negative of 2x minus 1 now if we want we can clear the negatives by just dividing both sides by negative 1 but again because we're working with inequality and we divided by a negative we've got to flip that guy okay so this is gonna cancel and this is gonna cancel you have X plus 3 now is less than 2x minus 1 let's go ahead and subtract 2x away from each side of the inequality and let's also subtract 3 away from each side of the inequality so this is gonna cancel this is gonna cancel we'll have X minus 2x which is negative x and this is lost then you've got negative 1 minus 3 which is negative 4 as a final step let's divide both sides by negative 1 and again we divided by a negative so flip this guy you're gonna get that X is greater than 4 okay so X is greater than 4 now we want to compare X is greater than 4 to the interval that we're working with the interval that we're working with specifically says that X is basically coming from negative infinity up to but not including negative 3 X is greater than 4 is not in that interval so this solutions going to be rejected okay let me erase all this and I'll kind of write it in the column so I'm just gonna write X is greater than 4 here and again this is not within our interval so we reject that solution okay now let's think about this interval here again where the X plus 3 is positive and we're going to use the negative of 2x minus 1 okay so let's scroll down and we're going to say that we have X plus 3 is positive and then this guy is going to be negative so the negative of 2x minus 1 okay so let's go ahead and solve this guy I'm going to distribute this this will be negative 2x plus 1 so if I add 2x to both sides of the inequality and I subtract 3 away from each side of the inequality this will cancel this will cancel X plus 2 X is 3 X this is greater than 1 minus 3 is negative 2 divide both sides by 3 and we get that X is greater than negative 2/3 okay let me erase this real fast X is greater than negative 2/3 let me write that up here so X is greater than negative 2/3 okay is that in our interval the interval is from negative 3 including negative 3 up to but not including 1/2 well yet is so this is a partial solution okay so for right now I'm just gonna write this guy as again not including negative 2/3 because it's strictly greater than up to but not including 1/2 okay and then we'll think about this solution and if we have a solution there we'll have two partial solutions we can put together as a full solution okay so in this case everything's positive so you just drop the absolute value bars and go with what we have so you'd have X plus 3 is greater than 2x minus 1 let's subtract 3 away from each side of the inequality so that would cancel let me subtract 2x away from each side of the inequality so that would cancel X minus 2x is going to give me negative X this is greater than negative 1 minus 3 is negative 4 let's divide both sides by negative 1 so we know that this is gonna flip right so this is gonna end up being a less than so you have X is less than positive 4 okay so again let me erase all this so we say X is less than 4 does that fit in our interval yeah it does right because we have from 1/2 including one half how to positive infinity X is less than 4 fits in there you can basically say it's another partial solution you would include one half and you would go up to but not include okay so if I take these two partial solutions and I just take the union of the two so again from negative 2/3 not including that up to but not including 1/2 and the union with you have 1/2 which is included up to 4 okay so this would be your solution set and basically when you combine the two you can get rid of this part right here and just kind of slide this down and that's what you're gonna get right you're gonna get from negative 2/3 not including that it's up to but not including four okay so let me kind of drag this up to our original page and we'll paste our solution here again from negative two-thirds not including that up to and not including 4 so negative 2/3 let's just go ahead and say that's about right here and I'm just gonna mark that and say this is negative 2/3 so I'd put a parenthesis there because it's not included before is not included just shade everything in between okay and then we could also write this saying that X is greater than negative 2/3 and less than 4 okay so this is your interval notation there's your standard notation and then this is graphically so those are the solutions for again the absolute value of x plus 3 is greater than the absolute value of 2x minus 1 all right let's go ahead and take a look at another example so this guy here is going to be a little bit more complicated but we're going to solve it using the same strategy so we have the absolute value of x plus 5 and then plus we have the absolute value of X minus 3 and then this is greater than 14 so the first thing we want to do is just take what's inside of the absolute value bars in each case so this expression here X plus 5 and this expression here X minus 3 we just want to set those guys equal to zero and see what the result is so if we have X plus 5 equals 0 very simple subtract 5 away from each side of the equation we get X is equal to negative 5 then for the other guy we have X minus 3 so X minus 3 equals 0 I had 3 to both sides of the equation we get X is equal to 3 so again these two numbers are gonna help me split the number line up into kind of three intervals we're gonna have from negative infinity to negative 5 with negative 5 not being included then we'll have negative 5 to 3 with negative 5 included and 3 not and then 3 included in anything larger okay so let's set this up using a table again I don't like to use the number line itself I like a little table it helps me organize things so again our numbers are just going to write them here we have negative 5 and positive 3 so again from negative infinity up to but not including negative 5 and then negative five will be included here so negative five is included and then up to but not including three and then three will be included and then out to positive infinity okay so let's rewrite our inequality again if we go back up we have the absolute value of x plus five then plus the absolute value of X minus three and this is greater than 14 so let's write that so we have the absolute value of x plus 5 and then plus we have the absolute value of X minus 3 and this is greater than 14 okay so we know in this first interval the 1 all the way to the left where we're coming from negative infinity going up to but not including negative 5 both of these guys are gonna be negative right this guy right here is gonna be negative because it's 0 or x equals negative 5 right so we're less than negative 5 so this guy's negative this guy is also negative because it's 0 when x equals 3 and we're definitely less than 3 in this case so we're gonna say that this guy is the negative of X plus 5 and this guy is going to be the negative of X minus 3 okay and then in this case right here now we're from negative 5 including that up to 3 but not including that well now this guy right here this X plus 5 that expression is going to be either 0 or positive so I can replace it with the expression itself so we'll put just X plus 5 and then this guy right here this X minus 3 is so gonna be negative right because again it's 0 when X is 3 we're less than 3 so it's still gonna be negative so we put the negative of X minus 3 and then in this last case where we're 3 or anything larger well then this X plus 5 is going to be positive and then for X minus 3 that could be 0 or positive so again we just put X minus 3 there ok so again if it's 0 or positive you just take what's inside the absolute value bars and you can write that down basically thinking about I just drop the absolute value bars if it's negative well then you drop the absolute value bars wrap it in parenthesis and then put a negative out in front okay so that's all we're doing all right so let's go through the possibilities now again this is the tedious part so in the first case both are negative so I have the negative of X plus 5 then plus the negative of X minus three and you can just put the minus there instead of plus negative doesn't really matter thank greater than fourteen okay so I'm going to distribute this negative to each term I'm gonna have negative X minus five and then minus X plus three and this is greater than 14 let me scroll down get a little room going so I know that minus X minus another X is negative 2x I know that minus 5 plus 3 is negative 2 this is greater than 14 let me go ahead and add 2 to both sides of the inequality we know this will cancel negative 2x is greater than 16 so then when I divide both sides by negative 2 remember we got to flip this so this becomes a less than now you'll have X is less than negative 8 okay so X has lost the negative 8 is that in the interval that we're working in right now well yeah we said that our current interval is anything less than negative 5 negative 8 is definitely less than negative 5 so we're good to go right this is gonna be a valid solution for us but again it's just part of our solution because we might have more to come okay so we just consider it as a partial solution for now all right so let me erase all this so again for this part I said that X was less than negative 8 and I'm gonna go ahead just put a check mark there so we know that we accepted that solution all right for the next interval again from negative 5 and including that up to but not including 3 this guy's positive this guy's negative so let's go ahead and do that so X plus 5 again that's positive and then plus this X minus 3 that expression is gonna be negative so I'm gonna put minus the quantity X minus 3 and before we can senior let's just go ahead just rib you this so this would be minus X and plus 3 make that a little quicker then this is greater than 14 okay so X minus X that's gonna drop out so this is gone right so you basically have that 8 is greater than 14 which is false okay that's never gonna be true so nothing's gonna work in that interval so you don't even have to put anything you just say it doesn't work all right so the next one we're gonna look at we have from 3 and including 3 out to positive infinity and again both expressions are positive here so we can just drop the absolute value bars in this case so we just say X plus 5 plus X minus 3 is greater than 14 so X plus X is 2x and then 5 minus 3 is 2 so plus 2 and this is greater than 14 all right so let's subtract 2 away from each side of the inequality and then we can cancel this will have 2x is greater than 12 let's divide both sides of the inequality by 2 we'll find that X is greater than 6 okay X is greater than 6 is that in our interval well yea this interval is from 3 out to positive infinity X is greater than 6 that is any interval so we're basically good to go with that so let's erase this we'll say X is greater than 6 let me put a checkmark there so basically we would just combine these two as one statement we would say that X is less than negative 8 or X is greater than 6 so let me just copy this real quick we'll go back up and let me just paste this in so again X is less than negative 8 so find negative 8 I'm gonna put a parenthesis there I'm just gonna shade everything to the left or also X could be greater than 6 so find 6 put a parenthesis there shade everything to the right okay in interval notation again I could just say from negative infinity up to but not including negative 8 and the Union with anything larger than 6 right so in both cases I'm using a parenthesis to show that it's not included all right so let's kind of change things up and look at another common type of problem that you'll see in this section so let's suppose we saw something like the absolute value of 3 over X minus 1 and this is greater than 5 so how can we solve this type of problem well one approach that we can use we realize that this guy right here this absolute value operation first and foremost there's one of them and it's already isolated for us on one side of the inequality so you have this absolute value operation and this is greater than 5 okay so for right now don't even worry about what's inside of there just think about the fact that earlier in the lesson we said if we had the absolute value of X this is greater than some positive real number a this led to X being greater than a or X is less than negative eight right we saw this with our example we said if the absolute value of X was greater than seven we said X was greater than seven or X was less than negative seven okay so this should be pretty crystal clear at this point now the thing is you can do this if you have a more complex expression inside of the absolute value bars okay so I can replace this X with something generic like you will just say u is some algebraic expression so it could be a rational expression like we have here it could be a quadratic expression it could be something more complicated than the typical linear expression that you're working with when you solve basic absolute value inequalities so what I can do is erase this and erase the X and just put au there okay so it's the same kind of concept it just takes more time to solve it it's just gonna be way more tedious okay so let's use this to set this up so basically I'm gonna take my rational expression inside the absolute value bars and I'm gonna say 3 over X minus 1 since I have a greater than I'm gonna say it's greater than this positive number 5 then or I'm gonna say 3 over X minus 1 I'm gonna say is less than the negative of 5 okay just use this rule here that's all I did okay so let me erase this we don't need this anymore so basically at this point all we need to do is solve two rational inequalities and then kind of combine our two solution sets that we have so I'm gonna start out with this one right here so I'm gonna copy this we're gonna go down to a fresh sheet all right so let me just paste this in here and I want you to recall that the very first thing you want to do when you're solving a rational inequality is get it in the format where you have a single rational expression on one side and 0 on the other okay so the way we're gonna do that here is we're just gonna subtract 5 away from each side of the inequality and so this would cancel and become 0 so essentially I would have 1 I would have 3 over X minus 1 and then we'd have minus 5 and this is greater than 0 okay so now I want to get a common denominator going let me just kind of slide this down a little bit and what I'm gonna do is just multiply this by the quantity X minus one and then over the quantity X minus one so now I'll have a common denominator negative five times X is negative 5x and then negative five times negative one is going to be plus five okay so kind of continuing now I have a common denominator so I can just combine the numerators so we have three and then let me make this more clear this is negative 5x and I'll just put a plus here so we have three plus five which is eight okay you have eight and you have your negative 5 X so let me write negative 5x plus 8 as the numerator this is over the common denominator which is X minus 1 and this is greater than zero okay so now we've accomplished count of the first task we have a single rational expression on the left side in this case and we have 0 on the other okay so the next thing we want to do is find out where the critical values are some people call them endpoints or boundary whatever you want to call them okay I'll just say critical values so that is going to occur where the numerator is going to be zero and where the denominator is going to be zero I would take this numerator which in this case is negative 5 X plus 8 and I would set it equal to zero so that's one of them then I would take the denominator which is X minus one set that equal to zero so both of those guys when we saw that the two solutions there are the critical values or again endpoints so let me add one to both sides of this equation that gives me x equals one so that's one of those guys let me subtract eight away from each side of this equation this is gonna cancel they'll have negative 5x is equal to negative eight divide both sides by negative five and you're gonna get that X is equal to 8/5 so these are our two critical values so we're going to use that to split the number line up into three intervals so basically you'll have from negative infinity up to but not including one and then between one and eight fifths right neither's included and then anything greater than eight fifths okay so those are the three intervals you're gonna have you're gonna pick a number in each interval okay you're gonna test that number in the original inequality you're gonna see if it satisfies it if it does that interval works it's part of your solution set if it doesn't it gets rejected it's not part of the solution set so you can use a table for this if you want or you could do a number line for these I prefer to actually use a number line so let's do that all right so the critical values again we have one and we have eight fifths so what is right here so let me just draw a nice little vertical line and it won't be perfect but we'll put that in there and then eight fifths is one point six now I've kind of divvy the number line up in a different way you'll see that the notches here are separated by 0.5 or 1/2 so from one we go to 1.5 or you can say three halves so eight fifths again is one point six if this is 1.5 1.6 let's just say it's about right there let's say this is 8/5 okay so let's go ahead and put another kind of vertical line there we're gonna divvy up the number line and just basically say that let me kind of make that a little better we'll say that this is interval a basically anything less than 1 anything between 1 and 8 fifths we'll call that interval B and anything larger than 8 fifths we'll call that interval C so we'll have a B and C so the original inequality was 3 over X minus 1 is greater than 5 so just take values in each interval check them see if they work again if they do it's part of your solution if they don't it's not part of your solution very very easy just a very tedious process so from interval a I'm gonna pick 0 plug that in there you would have 3 over 0 minus 1 is negative 1 so 3 over negative 1 is that greater than 5 no this would end up being negative 3 negative 3 is certainly not greater than 5 so this is false right in interval a nothing's gonna work so for interval B again I'm between 1 and 8 fifths or between 1 and 1.6 I think the easiest number to use there will be 1.5 so let's go ahead and plug in a 1.5 there 1.5 minus 1 would be 0.5 so you would have 3 divided by 0.5 is greater than 5 you know if you divide 3 by 0.5 it's the same thing as multiplying 3 by 2 right if you divide 3 by a half you could say when you crank that out you get 3 times 2 which is going to be 6 is 6 larger than 5 yes it is is true all right for the last one let's take a look at something in interval C I'm gonna pick two so 2 minus 1 is 1 3 over 1 is 3 so you basically have that 3 is greater than 5 which is false okay let me write that a little better so false okay so only numbers or only values and interval B are gonna satisfy this inequality it's a strict inequality so the endpoints we don't even need to consider okay they won't work they will be excluded so the solution for this part is going to be anything larger than 1 up to but not including 8 finn's so let's just copy this and we'll go back up and again this is just gonna be a partial solution so this is for this one and we could really write that X is greater than 1 and it's less than 8/5 okay so this is just a partial solution we're gonna work on this guy now and see what else we can find so let's copy this now let's go down all right so let's paste this other scenario here we have 3 over X minus 1 and this has less than negative 5 so again I want to get one rational expression on one side or one fraction and 0 on the other so same processes last time so I'm going to add 5 to both sides of the inequality I'm gonna have that 3 over X minus 1 then plus 5 is less than 0 I'm gonna get a common denominator so let me again move this down so I'm gonna multiply this by X minus 1 over X minus 1 and this is gonna lead to 3 over X minus 1 plus you have 5 times X which is 5 X and then minus 5 times 1 is 5 and this is over X minus 1 okay then this is less than 0 so let me erase this and we don't need this for right now we'll write it again when we need it so let me just drag this up and we make that 3 a little bit better okay so now we're just going to kind of combine the numerators we have a common denominator so we'd have 5 X and then 3 minus 5 is negative 2 so minus 2 this is over X minus 1 and this is less than 0 okay so at this point again I want to take my numerator set it equal to zero take my denominator set it equal to zero that will give me the two endpoints or critical values or boundaries okay so we would have 5 X minus 2 is equal to 0 add 2 to both sides of the equation I'm going to get that 5x is equal to 2 we would divide both sides by 5 and find that X is equal to 2/5 okay so that's one critical value and then X minus one we know if we set that equal to zero we would get that x equals one right we solved that earlier ok so let's go up and again the two critical values we have one and then we have 2/5 okay so one is right here let's just put a vertical line there 2/5 in decimal form is point 4 so I have a notch for 1/2 point 4 would be just a little bit less so let's just say it's right there so this is going to be my 2/5 let me just put a vertical line there okay so we have interval a interval B and interval C again interval A in this case is going to be anything that's less than 2/5 interval B will be between 2/5 and 1 and in turn will see will be anything larger than 1 so the inequality we're working with is 3 over X minus 1 and that's less than negative 5 so let's choose a number from interval a again I'm gonna choose 0 because it's easy to work with so 0 minus 1 is negative 1 you would have that 3 over negative 1 is less than negative 5 that's gonna be false right this would be negative 3 is less than negative 5 negative 3 is a smaller negative value than negative 5 so it's actually a larger number right so this is gonna be false okay what about for interval B well it's kind of a tight window but between 2/5 and one you can use 1/2 so if I do point 5 there I'm gonna have point 5 minus 1 which is negative point 5 so I would have 3 over a negative 0.5 and this is less than negative 5 of course if I divide 3 by negative 0.5 it's like multiplying 3 times negative 2 so this would basically be negative 6 here negative six is a larger negative the negative five so it is less than negative 5 so this is true then for interval C again I'm just gonna choose two nice and easy to work with two minus one gives me one so I would have that three is less than negative five which is obviously false okay so my solution here is gonna be for interval be anything between two-fifths and one so between two fifths and one okay so for this guy my solution is again between two-fifths and one so now it's going to be the union of these two solution sets so let me erase this I don't need any of this anymore because I'm thinking about the problem as a whole now so I'm just going to start with this once I'm going to say between two-fifths and one again neither is included and the union with this guy which is between one and eight fifths neither is included okay now you could also write this with our kind of standard way using that or we can say that X is going to be greater than two-fifths and less than one and then or you have this solution so you have that pax is gonna be greater than one and less than eight fifths okay so let's copy this and go to the number line we'll make a nice little graph to finish this off alright so the first thing I do is just make a little notch for two-fifths so again that's point four so I'm gonna put it right there and just say this is two-fifths again this value here for this notch is one half so two fifths would be less than that then for eight fifths I'm gonna say again that's one point six as a decimal so three halves is one point five so it's gonna be a little bit more than that so let's say it's right there so this will be my eight fifths okay so basically two fifths is not included so I'll put a parenthesis there we're gonna go up to but not including one so I'll put a parenthesis there but then again at one I'm not including it so I'm gonna put another parenthesis facing the opposite direction then I'm going to shade up to but not including eight fifths and I'll put a parenthesis there so that's a nice little graph you can make and in this case there might be a little easier to use open circles so what we might want to do to make the graph a little cleaner is put an open circle here at two fifths put an open circle here at one and put an open circle here at 8/5 okay so you could do it this way it's a little cleaner or the other way using the kind of parentheses for each it doesn't matter either way it's the same solution so we see that X is greater than two-fifths and less than 1 or X is greater than 1 and less than 8 fifths |
7113 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6660795/ | Protein synthesis rates and ribosome occupancies reveal determinants of translation elongation rates - PMC
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Proc Natl Acad Sci U S A
. 2019 Jul 10;116(30):15023–15032. doi: 10.1073/pnas.1817299116
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Protein synthesis rates and ribosome occupancies reveal determinants of translation elongation rates
Andrea Riba
Andrea Riba
a Institut de Génétique et de Biologie Moléculaire et Cellulaire, Université de Strasbourg, 67404 Illkirch Cedex, France;
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a,1, Noemi Di Nanni
Noemi Di Nanni
b Institute of Biomedical Technologies, National Research Council, 20090 Segrate (MI), Italy;
c Department of Industrial and Information Engineering, University of Pavia, 27100 Pavia, Italy;
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b,c, Nitish Mittal
Nitish Mittal
d Biozentrum, University of Basel, 4056 Basel, Switzerland
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d, Erik Arhné
Erik Arhné
d Biozentrum, University of Basel, 4056 Basel, Switzerland
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d, Alexander Schmidt
Alexander Schmidt
d Biozentrum, University of Basel, 4056 Basel, Switzerland
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d, Mihaela Zavolan
Mihaela Zavolan
d Biozentrum, University of Basel, 4056 Basel, Switzerland
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d,1
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a Institut de Génétique et de Biologie Moléculaire et Cellulaire, Université de Strasbourg, 67404 Illkirch Cedex, France;
b Institute of Biomedical Technologies, National Research Council, 20090 Segrate (MI), Italy;
c Department of Industrial and Information Engineering, University of Pavia, 27100 Pavia, Italy;
d Biozentrum, University of Basel, 4056 Basel, Switzerland
1
To whom correspondence may be addressed. Email: arriba87@gmail.com or mihaela.zavolan@unibas.ch.
Edited by Julia Bailey-Serres, University of California, Riverside, CA, and approved June 12, 2019 (received for review October 9, 2018)
Author contributions: A.R., A.S., and M.Z. designed research; A.R., N.D.N., N.M., E.A., and M.Z. performed research; N.M. and E.A. contributed new reagents/analytic tools; A.R., N.D.N., and E.A. analyzed data; and A.R. and M.Z. wrote the paper.
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Issue date 2019 Jul 23.
Copyright © 2019 the Author(s). Published by PNAS.
This open access article is distributed under Creative Commons Attribution-NonCommercial-NoDerivatives License 4.0 (CC BY-NC-ND).
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PMCID: PMC6660795 PMID: 31292258
Significance
Although sequencing of ribosome footprints has uncovered aspects of mRNA translation, the determinants of ribosome flux remain incompletely understood. Combining ribosome footprint data with measurements of protein synthesis rates, we inferred transcriptome-wide rates of translation initiation and elongation in yeast strains with varying translation capacity. We found that the translation elongation rate varies up to ∼20-fold among transcripts and is significantly correlated with the rate of translation initiation. Our data indicate that the amino acid composition of the synthesized proteins impacts the rate of translation elongation to the same extent as measures of codon and transfer RNA (tRNA) adaptation. Elongation is slow on transcripts encoding ribosomal proteins, which have a lower protein output compared with other transcripts with similar ribosome densities.
Keywords: translation, yeast, protein charge, TASEP, ribosomal proteins
Abstract
Although protein synthesis dynamics has been studied both with theoretical models and by profiling ribosome footprints, the determinants of ribosome flux along open reading frames (ORFs) are not fully understood. Combining measurements of protein synthesis rate with ribosome footprinting data, we here inferred translation initiation and elongation rates for over a 1,000 ORFs in exponentially growing wild-type yeast cells. We found that the amino acid composition of synthesized proteins is as important a determinant of translation elongation rate as parameters related to codon and transfer RNA (tRNA) adaptation. We did not find evidence of ribosome collisions curbing the protein output of yeast transcripts, either in high translation conditions associated with exponential growth, or in strains in which deletion of individual ribosomal protein (RP) genes leads to globally increased or decreased translation. Slow translation elongation is characteristic of RP-encoding transcripts, which have markedly lower protein output compared with other transcripts with equally high ribosome densities.
Gene expression analysis frequently relies on the high-throughput sequencing of cellular messenger RNAs (mRNAs). While the mRNA expression levels may be sufficient to decipher how cells respond to specific stimuli, they explain protein abundances only to a limited extent, with coefficients of determination R 2 in the range of 0.14 to 0.41 (1, 2). Protein levels vary over a much wider range than the levels of the corresponding mRNAs, indicating extensive regulation of protein metabolism, and especially synthesis (1). Translation is predominantly regulated at the initiation step (3), whose rate varies broadly between mRNAs, depending on the structural accessibility of the 5′-end to translation factors, and on the presence of upstream open reading frames. The latter generally hinder translation of the main open reading frame (ORF) (4). Translation elongation rates also differ between mRNAs, primarily due to codon biases and differences in the availability of cognate transfer RNAs (tRNAs). Whether and how the translation elongation rate is dynamically modulated is currently debated (2, 5–8). tRNA availability, translational cofolding of the polypeptide chain, and the presence of positively charged amino acids in the nascent peptide have all been linked to variation in elongation rate (5–8). Furthermore, it has been proposed that the codon usage is the substrate of “translational programs” that adjust the protein output of specific classes of mRNAs to the state (proliferation or differentiation) of the cell (9). However, explicit comparison of the coverage of 5′ and 3′ halves of ORFs by ribosome footprints did not reveal clear differences, indicating that bottlenecks in elongation along coding regions are uncommon (2).
Insights into the dynamics of translation and putative bottlenecks have emerged from theoretical studies, in particular of the totally asymmetric simple exclusion process (TASEP), introduced 5 decades ago (10). In a simple form of this model, ribosomes bind to mRNAs according to an initiation rate, move stochastically to downstream codons with an average elongation rate, if these codons are not already occupied by ribosomes, and are released at the end of the coding region with a given termination rate. The interplay of these rates gives rise to 3 distinct regimes. If initiation is infrequent, proteins are synthesized at a rate equal to the initiation rate and the ribosome density on ORFs is very low. As the initiation rate increases relative to the rate of elongation, the ribosome density on the ORF increases in parallel with the protein output. Finally, when the rate of initiation is too high, ribosomes start to “collide,” the ribosome density becomes very high, and the protein output drops markedly (11).
Currently available technologies enable predictions about the relationship between ribosome flux (corresponding to the protein synthesis rate) and ribosome density along ORFs to be tested. Ribosome density along ORFs can be studied with high resolution by sequencing of ribosome-protected mRNA footprints, a method known as ribosome footprinting or ribosome profiling (2). The approach has already uncovered novel principles of resource allocation and translation regulation (12, 13). Furthermore, model-based analyses of ribosome profiling data uncovered sources of local variation in ribosome densities and translation elongation along transcripts (14). However, the ribosome flux has rarely been measured directly, despite mass spectrometry-based methods being able to provide estimates of synthesis and degradation rates for a substantial fraction of eukaryotic proteomes (15). Direct measurement of protein synthesis rate is necessary to detect global changes in translation capacity between conditions (16) and for studying translation in an ORF-specific manner, because the protein synthesis rates can be inferred from ribosome profiles only up to a constant factor.
To fill this gap and further uncover factors that underlie variations in translation elongation rates between ORFs, we measured protein synthesis rates transcriptome-wide, by pulsed stable isotope labeling of amino acids in culture (pSILAC), in the widely studied experimental model of exponentially growing yeast cells. Combined analysis of pSILAC and ribosome footprinting data revealed the range of variation in translation elongation rates between yeast ORFs. Among broadly studied determinants of this rate, most indicative were the availability of cognate tRNAs and the frequency of positively charged amino acids in the synthesized protein. We found no evidence that translation is curbed by ribosome collisions either in exponentially growing wild-type yeast or in mutant strains with global alterations in translation. Rather, we found that translation elongation on mRNAs encoding positively charged proteins (particularly ribosomal proteins [RPs]) is slower compared with other mRNAs with similar ribosome densities.
Results
Ribosome Allocation Is Largely Explained by the Copy Number and Length of ORFs.
To uncover determinants of translation speed in exponentially growing yeast cells, we analyzed a recently published ribosome footprinting dataset obtained in this system (4), from the perspective of the TASEP model of translation (Fig. 1 A). Denoting by the number of ribosomes bound to a coding region of codons, and assuming that the rate with which a ribosome completes the polypeptide chain is given by the product of the frequency of finding a ribosome at the stop codon and the effective ribosome translocation (and termination) rate , the change in the number of ribosomes bound on the mRNA is given by the differential equation . Here, is the number of mRNA molecules, is the effective rate of translation initiation on an mRNA molecule, and we assume the broad region of parameter values where ribosome collisions are rare. This model predicts that the number of ribosome-protected fragments (RPFs) mapping to a specific mRNA is proportional to the mRNA abundance, the length of the ORF, and the ratio of the effective rates of initiation and elongation, .
Fig. 1.
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Predicted and observed relationships in gene expression in the BY4741 yeast strain. (A) Illustration of the classical totally asymmetric exclusion process (TASEP) with constant rates of initiation, elongation, and termination. (B) Relationship between protein abundance (ref. 18) and the density of RPFs on the ORF, or the mRNA abundance. (C) Relationship between the number of RPFs mapped to individual mRNAs and the corresponding ORF length, mRNA level, and both. p and s are Pearson’s and Spearman’s correlation coefficients, respectively.
Testing this prediction with the experimental dataset mentioned above (4), we found that the number of RPFs mapped to a specific mRNA indeed correlated very well with the relative abundance of the mRNA estimated by mRNA sequencing. However, further incorporating the ORF length slightly but significantly (z score = −4.74, Fisher z-transformation test) reduced the correlation, rather than improved it (SI Appendix, Fig. S1). As it was reported that the estimation of mRNA abundance by RNA sequencing is a critical aspect to control in the analysis of ribosome footprinting data (4), we repeated the analysis of the scaling behavior with estimates of mRNA abundance from another RNA-seq dataset, obtained by sequencing of RNAs purified directly with oligo(dT) from yeast cell lysates (17). The 3′-end bias in ORF coverage by RNA-seq reads, which strongly affects the accuracy of mRNA abundance estimates (4), was limited in this dataset, comparable with that inferred from the data obtained with a Ribo-zero protocol (SI Appendix, Fig. S2). We found that, when using the RNA samples obtained by oligo(dT)-based purification, both the mRNA level and ORF length contributed to the number of RPFs, as expected (Fig. 1 C). We therefore used this mRNA-sequencing dataset for the analyses described below, but present similar results with the RNA-seq data from reference 4 in SI Appendix.
The mRNA levels alone explained 65% of the variance in RPF numbers. Further taking into account the ORF length increased this number to 74% (Fig. 1 C), setting an upper bound of 25% on the variance in RPFs that could be due to differences in ribosome density along transcripts. From previously published measurements of protein levels in the same yeast strain (18), we further inferred that the number of RPFs explained ∼50% of the variance in protein levels, compared with only 38% explained by the mRNA abundance (Fig. 1 B).
Ribosome Allocation Predicts Protein Synthesis Rates.
Theoretical analysis of the TASEP model showed that the main dynamical regimes are defined by the density and the flux of ribosomes on mRNAs (11), the latter corresponding to the protein synthesis rate. To infer the translation regime of individual yeast mRNAs, we determined relative ribosome densities on individual ORFs from the RPF and RNA-seq data, knowing the ORF lengths. The estimates that we obtained here correlated quite well (Fig. 2 A; Spearman correlation coefficient, 0.46; P = 1.6e-194) with those from a much earlier study that determined the distribution of individual mRNA species across polysomal fractions corresponding to 1, 2, 3, etc., translating ribosomes, with microarrays (19). Having computed ribosome densities for each ORF, we used pSILAC to measure the corresponding protein synthesis rates.
Fig. 2.
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Analysis of protein synthesis rates. (A) Ribosome densities derived from the sequencing of RPFs (x axis) or estimated based on the relative abundance of RNAs across polysomal fractions in ref. 19 (y axis). (B) Protein synthesis rates s can be estimated from the dynamics of light peptide (P) accumulation within a short time interval t (in minutes) after medium change (Inset). Examples of linear fits to the peptide accumulation curves for the proteins indicated in the legend. (C) Histogram of R 2 values of the linear fit for all 1,616 measured proteins. (D) Relationship between ribosome allocation per codon and the protein synthesis rate. Highlighted in the red box are the proteins with highest synthesis rates. The orange box highlights the cluster of RPs. p and s are Pearson’s and Spearman’s correlation coefficients, respectively.
On a short timescale, upon shifting cells from a medium with heavy-isotope–containing amino acids to a medium with light-isotope–containing peptides, “light” peptides should accumulate proportionally to the protein synthesis rates (Fig. 2 B). Indeed, we found that the light peptide accumulation in the first 30 min after the medium change was very well described by a linear model (R 2 for the linear fit >0.8 for 1,114 of the 1,616 proteins; Fig. 2 C). Furthermore, the protein synthesis rates thus estimated correlated better with the density RPFs than the protein levels did (Pearson correlation coefficients of 0.81 and 0.7, respectively; Fig. 2 D and Fig. 1 C). This conforms to the expectation that RPFs reflect protein synthesis, while protein levels are set by the balance between synthesis and degradation.
Protein Synthesis Rates Are Not Limited by Ribosome Collisions in Exponentially Growing Yeast Cells.
Although protein synthesis rates increased linearly with the ribosome allocation over the entire range, a small cluster of ORFs did not conform to this relation but had distinctly lower protein output than other ORFs with similarly high numbers of allocated ribosomes (Fig. 2 D, orange box). These ORFs encoded almost exclusively RPs, while other highly translated ORFs, with a single exception (the translation elongation factor 2 [TEF2]), encoded genes involved in sugar metabolism (Fig. 2 D, red box). Indeed, Gene Ontology terms Glucose metabolic process and Gluconeogenesis and KEGG Glycolysis/Gluconeogenesis pathway were strongly enriched in this set (false discovery rate [FDR] for these biological processes: 1e-14 and 2e-12, respectively; Fig. 2 D). To understand the dynamics of translation on individual ORFs, we then sought to infer their absolute rates of translation initiation and elongation.
A yeast cell needs about 2 h to divide (20) and contains about 5 × 10 7 protein molecules (21), whose average half-lives are ∼8.8 h (22). These numbers define the total number of proteins produced by a yeast cell per unit time, allowing us to convert the relative protein synthesis rates inferred from the pSILAC time series to absolute rates of molecules per unit time. Further taking into account the estimated number of 40,000 (23, 24) mRNA molecules in a yeast cell, we obtained protein synthesis rates per mRNA (Methods). To directly compare the experimental data (Fig. 3 B) with predictions of the TASEP model (Fig. 3 A), we converted the relative ribosome densities that we obtained from sequencing of RPFs to absolute densities of ribosomes per codon (rpc) using the first principal component of the scatter of RPF-based densities as a function of absolute density measured in ref. 19 (Fig. 2 A). Results of protein synthesis rate and ribosome density for individual ORFs, computed using either oligo(dT) or Ribo-zero RNA-sequencing datasets, are given in Datasets S1 and S2.
Fig. 3.
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Predicted and observed relationship between the protein synthesis rate and the ribosome density on the corresponding ORF. (A) TASEP model predictions with isoclines corresponding to individual translation initiation (gray dotted lines; rate range, 0.01 to 1.9/s; increments of 0.1 starting from second line at 0.1; mean, 0.04/s) and elongation rates (colored lines; rate range, 1 to 20 aa/s). Superimposed is the first principal component of the experimental data shown in B, for which the mean initiation and elongation rates were 0.04/s and 2.63 aa/s, respectively. (B) Similar visualization of experimental results: protein synthesis rates were measured by pSILAC and converted to molecules per mRNA per second from the expected protein mass doubling time; ribosomes densities were obtained from the fit of ribosome footprint densities to numbers of ribosomes per codons (rpc) estimated by ref. 19. The black contour indicates 90% of the empirical distribution approximated through the 2D kernel density estimation from the R package “MASS.”
In the model, ORFs that initiate translation at very low rates have very low protein output and their ribosome coverage per codon reflects the rate of translation elongation (Fig. 3 A). Although our dataset contained only few proteins with very low synthesis rate, the ∼10-fold range in ribosome coverage that we infer this way is comparable to the 20-fold range that we observed for ORFs with the same RPF density (Fig. 3 B). The model also predicts that protein synthesis rate and ribosome coverage increase linearly with the initiation rate, as long as ribosome collisions do not halt elongation (Fig. 3 A). The ∼100-fold range of variation in protein output in the experimental data corresponds to a similar range of variation in translation initiation rate. Thus, our analysis indicates that translation is primarily regulated at the level of initiation, as reported before (4). The regime of high ribosome density and low protein output exhibited by the model was not observed in our data (Fig. 3 B). We also carried out simulations of the inhomogeneous TASEP model using the codon-specific speeds inferred from Ribo-seq codon densities (SI Appendix, Materials and Methods) and found that differences in elongation speed between codons are not sufficient to explain the observed variability in synthesis-density scatter (SI Appendix, Fig. S3). Furthermore, even in the inhomogeneous model, queued ribosomes only start to accumulate abruptly and limit protein output at very high ribosome densities (∼0.05 rpc).
An additional indication of evolutionary optimization of translation so as to avoid wasteful ribosome collisions comes from comparing the first principal component of the experimental data from Fig. 3 B with the equal elongation rate isoclines from the simulation (black line in Fig. 3 A). This comparison suggests that the rates of translation initiation and elongation are correlated, transcripts for which the initiation rate is high having also higher rate of elongation compared with transcripts for which the initiation rate and protein output are lower (SI Appendix, Fig. S3). Allowing for a 2-fold error in the estimation of protein synthesis rate or ribosome density, either due to experimental variability or to processes that we did not consider here such as sequestration or degradation of mRNAs with stalled ribosomes (25), does not destroy this correlation (SI Appendix, Fig. S4). Thus, although ribosome stalling may contribute to the variability in synthesis density scatter observed in Fig. 3 B, it is unlikely that those ribosome collisions curb the rate of protein synthesis in the high protein output regime of exponentially growing yeast.
Combining Protein Synthesis Rate and Ribosome Footprint Density Reveals the Speed of Translation Elongation.
If ribosome queueing events are rare, proteins should be synthesized at the rate at which new chains are initiated the ribosome density can be approximated as and the ratio of protein synthesis rate to ribosome density (SDR) will be the effective translation elongation rate, We used this relationship to uncover features of the mRNA or of the corresponding protein that most strongly affect elongation (Fig. 4). The positive correlation of the SDR with the average speed of ribosomes, calculated from the normalized ribosome densities at individual codons, served as control (Fig. 4 A). We found that all measures related to tRNA availability and codon usage [tRNA adaptation index (tAI) (26), “normalized tAI” (27), fraction of optimal codons (FOP) (28), and codon adaptation index (CAI) (29)] (see Methods for details) correlated positively with the SDR, as expected (30) (Fig. 4 A). Interestingly, the ribosome density was also positively correlated with the SDR (Fig. 4 A). This could be due to a correlation between initiation and elongation, but also to ribosomes promoting translation speed by resolving RNA secondary structure, consistent with the negative correlation between the average predicted propensity of 5-nt-long windows along the ORF to be in single-stranded conformation (“accessibility,” denoted as Acc5) and the SDR (see below). Participation of the newly incorporated amino acid in a protein domain was also associated with faster elongation, and this association was not due to a specific type of domain such as α-helix or β-sheet.
Fig. 4.
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Determinants of translation elongation rate in yeast. (A) Correlation coefficients (P values indicated on the bars) of SDR with features related to codon speed, biochemical properties of the encoded protein, and RNA secondary structure accessibility. (B) Correlation coefficients of the average accessibility of regions in the ORF of length indicated by the x axis with SDR. (C) Correlation coefficients of SDR with the probability of regions of 20 nt starting at the position indicated on the x axis relative to the A site to be in single-stranded conformation. Positions where the correlation coefficients are highly significant (P value of 0) are marked by asterisk. In all panels, Spearman and Pearson correlation coefficients are shown in blue and orange, respectively.
Strikingly, the feature most strongly anticorrelated with SDR was the estimated isoelectric point (pI), reflecting the charge of the encoded protein. Other global features of the encoded protein such as the proportion of aromatic amino acids (aromaticity), hydropathicity [measured by the grand average of hydropathy (GRAVY) index (31)], molecular weight (Mw), and instability [measured by the instability index (32)] had much smaller correlation with SDR and thus with translation speed. Incorporating all of the features shown in Fig. 4 A into a linear model predicted better the protein synthesis rate than the ribosome density alone (correlation coefficient, 0.69 vs. 0.49; Fisher’s Z test z score = −7.84). Using only the tAI from among the tRNA abundance-related features with high pairwise correlation (SI Appendix, Figs. S5 and S6) reduced the predictive power of the model only marginally (correlation coefficient, 0.65). The linear model also highlighted the most explanatory features, which were (in order of the significance of their correlation coefficient being different from zero) the ribosome density (P< 2e-16), isoelectric point pI (P = 3.64e-15), codon adaptation index (P = 1.22e-11), molecular weight of the encoded protein (P = 9.83e-05), ORF length (P = 0.000118), GRAVY index (P = 0.000256), tRNA adaptation index (P = 0.000977), domain coverage (P = 0.006493), and fraction of optimal codons (P = 0.009106). The weights of individual features in the linear model are shown in Dataset S3. Qualitatively similar but somewhat lower in magnitude correlation coefficients were obtained when the other RNA-seq dataset, from ref. 4, was used in the analysis (SI Appendix, Fig. S7).
Consistent with RPs being elongated relatively slowly (Fig. 2 D), the set of 50 genes with greatest distance from PC1 in the direction of reduced elongation rate was strongly enriched in RPs (15 genes are in GO:CC ribosome; hypergeometric test FDR = 3e-8) but also contained other positively charged proteins such as histones HTA1 and HHF1, the INH1 inhibitor of F1F0-ATP synthase, and the SEC62 component of the Sec63 complex for protein targeting to the endoplasmic reticulum. No specific biological process or cellular component was preferentially represented among the genes with the highest elongation rates.
Complex Effect of RNA Secondary Structure on Translation Elongation.
Although it did not significantly contribute to the linear model, the structural accessibility of translated RNAs—measured by the average probability of windows of n nucleotides along the ORF of being predicted in single-stranded conformation—was anticorrelated with SDR for n up to 20 to 40 nt (Fig. 4 B). This indicates that RNAs that are highly structured are also highly translated (33). Although this seems counterintuitive, a theoretical study proposed that structural rearrangements of the mRNA during translation may serve to maintain an optimal ribosomal flux for high protein output (34). On the other hand, structural accessibility of the RNA immediately ahead of the decoded codon was significantly anticorrelated with the ribosome density on the decoded codon, as found in vitro (35, 36) (Fig. 4 C). Our results thus indicate that, although ribosomes progress faster through unstructured regions of the ORFs, unstructured RNAs ultimately have lower translational output.
Influence of Incorporated Amino Acids on Translation Speed.
A functional analysis uncovered sequence-dependent rearrangements of the nascent polypeptide in the ribosomal exit tunnel, suggesting that side-chain size and charge of the incorporated amino acid impact the rate of polypeptide chain elongation, as do cotranslational protein folding and interaction with chaperones (37). Indeed, our analysis provides evidence for both size and charge of amino acids affecting translation speed; negatively charged proteins are synthesized at up to ∼2-fold higher rates, on average (comparing first last pI quantile bins; Fig. 5 A and B), compared with positively charged proteins. Furthermore, among nonpolar amino acids, those with small side chains are associated with faster elongation, whereas the more voluminous ones have the opposite effect (Fig. 5 C). The amino acid charge and relative abundance of cognate tRNAs impact translation elongation rate to a similar degree.
Fig. 5.
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Influence of encoded amino acids on the translation elongation rate. (A) Positively charged proteins have low synthesis rate for the density of ribosomes on their corresponding ORFs. Each point represents an mRNA, with x and y coordinates corresponding to the ribosome density and protein synthesis rate, respectively, both on a log 10 scale. The color indicates the isoelectric point of the encoded protein, red indicating proteins with high pI (positively charged) and blue indicating proteins with low pI (negatively charged). (B) SDR distributions for increasing isoelectric point quantiles (left–right bins, t test, P = 2e-8). (C) Spearman (dark shade) and Pearson (light shade) correlation coefficients of SDR with amino acid frequencies in the encoded proteins (Top; only values of P< 0.05 are shown) and respective amino acids sizes (Bottom).
The explanatory power of linear models using relative frequencies of encoded amino acids or features related to tRNA abundance along with the ribosome density on the ORF were very similar (Pearson’s R, 0.69 vs. 0.68). The most informative amino acids were Arg (P value of the coefficient being different from zero in the linear fit = 2.35e-10), Pro (P = 1.68e-07), Ala (P = 1.87e-07), Glu (P = 1.14e-06), and Ser (P = 0.00861). See Dataset S3 for the inferred weights of these features.
Ribosomal Protein mRNAs Are Translated Slowly for Their Ribosome Densities.
Assuming that all initiating ribosomes complete translation and that they elongate at similar rates across transcripts the ribosome footprint density is generally used as an estimate of the translation efficiency, defined as the number of protein molecules produced per mRNA molecule per unit time (12). As RPs represent a high translation burden for the cell, their transcripts should be highly optimized for translation. Indeed, RP-encoding genes have a significantly higher tRNA adaptation index than other genes (Fig. 6 A; t test, P = 9.7e-49). However, the high ribosome density of the RP-encoding transcripts is not a simple reflection of high translation efficiency, because RPs also have a much higher isoelectric point than other proteins (Fig. 6 B; t test, P = 2.3e-47). Visualizing the tAIs of the ORFs and the pIs of proteins along with the protein synthesis rate and ribosome allocation on individual ORFs, clearly illustrates that positively charged proteins stand out as having lower than expected protein output for the ribosome densities on the corresponding ORFs (Fig. 6 C and SI Appendix, Fig. S8), or in other words, the ribosome density on their ORFs is higher compared with ORFs encoding other proteins that are synthesized at the same rate. These results suggest that the interaction of positively charged proteins, and in particular of RPs, with the negatively charged exit tunnel, slows down translation elongation, increasing the ribosome density on the ORF without a corresponding increase in protein output.
Fig. 6.
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Properties of RPs (n = 122) compared with all other quantified yeast proteins (n = 992). Box plots of (A) tRNA adaptation index of individual ORFs and (B) isoelectric point for corresponding ribosomal and all other yeast proteins. (C) Protein synthesis rate [log 10(peptides/s/mRNA)] as a function of ribosome density [log 10(RPKM/TPM)] for transcripts encoding RPs (brown) and all other proteins (gray).
Perturbed Translation Dynamics in Ribosomal Protein Deletion Strains.
Deletion of specific RP genes has been associated with changes in translation and replicative life span (17, 38). In particular, deletion of rpl7a (Δrpl7a strain) led to ribosome assembly defects and overall decreased protein synthesis (measured by the incorporation of a methionine analog), whereas deletion of rpl6a (Δrpl6a strain) led to increased protein production (17). To determine how the translation parameters of individual ORFs are affected in these strains, we measured the protein synthesis rates by pSILAC and analyzed them jointly with ribosome profiling data obtained before (17). Results for individual ORFs are given in Datasets S4–S6, for the wild-type control, Δrpl6a, and Δrpl7a strains.
We found that accumulation of light peptides in the mutant strains was less well explained by a linear fit compared with the wild-type strain, especially for the high-translation Δrpl6a strain (SI Appendix, Fig. S9). In both mutant strains, the correlation between ribosome density and protein synthesis rates was lower (Fig. 7 A and B) compared with the wild type maintained in the same conditions (SI Appendix, Fig. S10). For the Δrpl7a strain, the decrease was due, in large part, to ORFs encoding proteins involved in starch and sucrose metabolism (FDR = 1.86e-6), glycolysis and gluconeogenesis (FDR = 0.00389), whose protein output was higher than expected for their observed ribosome densities in all of the strains (Fig. 7 B and SI Appendix, Fig. S10). Excluding the 31 ORFs with a log 10 ribosome density lower than −1 led to correlation coefficients comparable with those obtained for the Δrpl6a strain (both Pearson and Spearman correlation coefficients = 0.41). We then compared the synthesis rate–density relationship of these strains with that of the wild-type strain analyzed in the same study (17) (SI Appendix, Fig. S10). The ribosome density changed very little in the Δrpl6a strain (Fig. 7 C), and large, correlated changes in density and flux (more than 2-fold in either direction) were only observed for 11 ORFs. Furthermore, we did not find any ORF that was highly translated in the wild-type strain and whose protein output collapsed in the high-translation Δrpl6a strain, as would be expected if ribosome collisions occurred in this high-translation strain. This was not due to missing protein synthesis rate data, because the large majority (25 of 31) of ORFs with highest ribosome density and measured output in the wild type were also measured in the Δrpl6a strain.
Fig. 7.
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Translation parameters in yeast strains with deletions in RP genes, Δrpl6a and Δrpl7a. (A and B) Ribosome density–protein synthesis rate plots for the 2 strains; highlighted in red are outliers (having ribosome density of <0.1 RPKM/TPM; Dataset S7) in the Δrpl7a strain. (C and D) Change in the ribosome density vs. change in the synthesis rate in the Δrpl6a and Δrpl7a strains compared with wild type. The Insets show the number of transcripts in each of the 4 quadrants of the plots. In the violin plots, the distributions of density and synthesis rate changes are shown for 5 bins of SDR values (20% of transcripts in each bin), from the lowest SDR (left-most bin) to the highest (right-most bin). P values of the t test comparing the mean density and synthesis rate changes between 20% transcripts with highest and lowest SDR values, respectively, are shown.
In contrast, hundreds of proteins had reduced synthesis rates in the Δrpl7a strain relative to wild type, with correspondingly reduced ribosome densities along ORFs (Fig. 7 D). The change in ribosome density was well correlated with the change in protein output (correlation coefficients = 0.64 [Spearman, P< 2e-16] and 0.67 [Pearson, P< 2e-16]). However, ORFs with high SDR had a higher reduction in output compared ORFs with low SDR. This is indeed the behavior expected upon a global reduction in translation that comes with reduced ribosome biogenesis in the Δrpl7a strain. Namely, ORFs on which elongation is relatively slow and are in the elongation-limited regime of translation in the wild-type strain will not undergo as large a change in protein output upon the reduction of translation initiation rate as ORFs that are in the initiation-limited regime already in the wild-type strain. This can be inferred from the size of the intervals between 2 lines of distinct translation initiation rates (dashed lines in Fig. 3 B) along 2 lines of high and low elongation rates (colored lines in Fig. 3 B). These results demonstrate that the analysis of protein synthesis rates and ribosome densities enables the inference of translation initiation and elongation parameters for individual genes and that these parameters can be used to uncover elements that regulate translation in individual strains and conditions.
Discussion
Protein synthesis is a central activity in all cells, which has to be appropriately adjusted to resources and to the signals that a cell receives. The overall ribosome content of mammalian cells is strongly linked to their proliferation rate, in actively dividing cells ribosomal RNAs (rRNAs) taking up ∼80% of all nucleic acids and ∼15% of the biomass (39). Understanding how translation is regulated in relation to the cellular state is important, as changes in the protein synthesis capacity can lead to both cancers (40, 41) and changes in organism life span (17, 38, 42). Although theoretical models of biosynthetic processes have been proposed and studied for decades (10, 11, 43–46), measurements of translation dynamics across a large fraction of the transcriptome became possible only recently. Taking advantage of abundant data generated for the yeast Saccharomyces cerevisiae and measuring protein synthesis rates with the high transcriptome coverage afforded by currently available methods, we evaluated the translation initiation and elongation rates for individual yeast ORFs.
Using additional datasets to estimate absolute protein synthesis rates as well as ribosome densities per codon, we found that the translation initiation rate varies over a ∼100-fold range among yeast transcripts (Fig. 3). This is consistent with an initial estimation of translation efficiency based on ribosome profiling (2) as well as with the results of a study that used these data to parametrize a whole-cell model of translation, which found that the time between initiation events on individual mRNAs (5th to 95th percentile) is from 4 to 293 s (45). However, a narrower range of variation, ∼11-fold (1st to 99th percentile), was reported based on the initial analysis of the ribosome profiling data that we also used here (4), as well as in a subsequent study of a more limited set of proteins (14). It was suggested that inaccuracies in estimation of mRNA expression levels could account for discrepancies in estimates of translation efficiency from ribosome profiling (4). However, here we found that to explain the direct measurements of protein synthesis rates, the wider range of variation (∼150-fold; SI Appendix, Fig. S11) in translation initiation rates was indeed necessary. This was the case irrespective of the protocol used to prepare the mRNA-sequencing samples that were used in the analysis of ribosome densities. The similar results obtained based on mRNA level estimates with 2 sequencing protocols is perhaps not surprising, as the 3′-end bias of these mRNA-sequencing data was similar (SI Appendix, Fig. S2) and the transcript abundance estimates showed limited systematic differences between the 2 datasets (SI Appendix, Fig. S12). However, it is interesting to note that the data obtained with the optimized Ribo-zero protocol did not yield the expected scaling of RPFs with mRNA length and abundance, even though the mRNA abundances inferred from these data were very highly correlated with the number of RPFs. In the future, it will be interesting to determine the translation status of mRNA species that are preferentially enriched by different protocols. It is also unlikely that the wider range in translation initiation rate is due to error in estimating the protein synthesis rates because our analysis only included ORFs for which peptide accumulation was well described by a constant accumulation rate. The selection of transcripts for analysis in different studies may account for some of the reported differences in the range of rate variation, as the study of ref. 14, for example, used only ORFs of at least 200 codons and with a minimum ribosome density of 10 per site. This amounted to 894 ORFs, of which 826 are also covered by our analysis. However, our analysis includes 290 additional ORFs, some with relatively low translation. Despite this, the mean initiation and elongation rates in our data are quite close to those reported before, namely mean waiting time between initiation events of ∼25 s compared with a median of 8 s reported by ref. 14, and elongation rates of 2.63 aa/s compared with the 5.6 aa/s reported for mouse peptides by ref. 47, based on a ribosome runoff assay. More importantly, previous studies did not measure protein synthesis rates directly, but rather estimated initiation and elongation rates from ribosome densities. This can be done up to a constant scale factor, which was assumed to be identical between genes and set such as to achieve a specific target elongation rate toward the 3′-end of the ORF (14). Our data indicate, however, that there are substantial differences in protein output of ORFs with similar ribosome densities, underscoring the importance of direct measurements of protein synthesis rates to analyze the dynamics of translation.
The translation parameters of short ORFs, many of which encode RPs, have been the topic of much discussion (4, 45). The high ribosome density observed on short ORFs (19) has been attributed to their being evolutionarily optimized for protein output through high rate of translation initiation (45). As the high codon adaptation index exhibited by these ORFs would predict fast elongation and thereby low ribosome density (45), high ribosome density on short ORFs has also been interpreted as evidence for initiation being the main determinant of ribosome density. Consistently, we also found a small but significant correlation between ORF length and the principal component of the protein synthesis rate—ribosome density scatter, which is indicative of the translation initiation rate (Fig. 3 and SI Appendix, Fig. S13). However, our results reveal a more complex picture, which suggests that the charge of the encoded protein is an important determinant of ribosome flow. ORFs with similar overall ribosome density differ by up to ∼20-fold in protein output. This effect is not captured by models that assume that the rate of elongation depends only on the tRNA availability-dependent decoding speed at the A site of the ribosome. Indeed, we demonstrated that protein synthesis rates can be predicted with significantly higher accuracy when taking into account global features of the encoded protein such as the pI than when using solely the ribosome density.
Dissecting the independent contributions of various features to the rate of elongation is nontrivial because these features are not uniformly represented among various classes of proteins. RPs, in particular, tend to be short, positively charged, and enriched in amino acids such as lysine and arginine, which are targeted by enzymes such as trypsin, used during sample preparation for mass spectrometry. These are also the amino acids that are isotope labeled for pSILAC. Although we cannot completely exclude these confounding factors influencing our estimates of elongation rates, we did try to minimize their effect. In particular, we estimated the protein synthesis rates by measuring the accumulation of light peptides, after switching the cells from a medium with heavy isotopes to a medium with light isotopes so as to not impair protein synthesis. Furthermore, although the frequency of lysines and arginines in RPs is higher compared with other proteins, RPs do yield peptides that are sufficiently long and amenable to quantification. Finally, the estimated “rate” with which a given RP-derived peptide accumulates as a function of time should not be affected by the enzymatic digestion during samples preparation.
Overall, we found that the rate of elongation varies up to ∼20-fold among yeast ORFs, less than the rate of initiation. As the determinants of translation elongation rate are actively debated (9, 30, 45, 48, 49), we evaluated their relative contributions in our data. We further included in our study yeast strains with globally perturbed translation through RP gene deletions. We found no evidence that translation elongation severely curbs protein output, either in the exponentially growing BY4741 yeast strain, or in the Δrpl6a and Δrpl7a deletion strains, the first with higher and the second with lower overall protein synthesis rate compared with the BY4741 wild type. Rather, several lines of evidence point to evolutionary optimization of ORF sequences to maintain appropriate ribosome flux and minimize the chance of ribosome collision. For instance, ORFs with high protein output have high rates of translation initiation and at the same time a high codon adaptation index. This is predicted to enable fast elongation, as optimal codons will be rapidly found by cognate tRNAs that are in highest abundance. Our data provide transcriptome-wide evidence for the high elongation rates of highly expressed ORFs. Thus, although initiation rates vary over a wide range, the protein output increases in parallel with the ribosome density, without the latter reaching saturation. Moreover, the density of RNA secondary structure predicted in the ORF was positively correlated with the translation elongation rate, not negatively correlated, as would be expected if RNA structure were to hinder translation. This indicates that the RNA structure may also help maintaining the flux of ribosomes along the ORF to minimize ribosome collisions, as proposed in a previous study (34). Interestingly, we were able to confirm the positive influence of RNA secondary structure using dimethyl sulfate-sequencing–based measurements of secondary structure density (50) rather than computational predictions; despite the experimental dataset being sparser than our computational predictions (only 231 ORFs for which we had protein synthesis and ribosome density data also had experimental data on secondary structure), the density of secondary structure (measured by a Gini index; see ref. 50 and Dataset S8) correlated positively with our SDRs (Pearson correlation coefficient = 0.17; P = 0.007). Our results do not exclude “controlled” ribosome stalling at specific positions, such as on upstream ORFs (51), or at codons for which cognate tRNAs are limiting in specific conditions, where active regulatory mechanisms are used to modulate the output of specific ORFs (49). They also do not exclude that slow clearance of the ribosomes from the 5′-end of transcripts reduces the initiation rate to some extent [the concept of 5′-ramp (14)]. Rather, our data support the notion that ORFs have undergone evolutionary selection to minimize the chance of ribosome stalling due to imbalanced initiation and elongation rates.
That the charge of the translated protein affects the rate of translation elongation has been observed before (7, 8, 14, 52) and has been attributed to variation in the “friction” of the polymeric chain with the ribosomal exit tunnel. This effect is most marked for the positively charged RPs, whose elongation rate is low relative to other proteins whose transcripts have similar ribosome densities, and also in comparison with negatively charged RPs (SI Appendix, Fig. S14). Furthermore, even considering only transcripts whose codon usage is not optimal (CAI < 0.5), the vast majority of which encode non-RPs, the SDR is significantly lower when comparing encoded proteins with high predicted pI (>7.5) with those with low predicted pI (<7.5) (SI Appendix, Fig. S14). Our results thus provide a rationale for the previous observation that the ratio of protein to mRNA molecules is lower for RP-encoding compared with other genes (53). It is important to note that establishing a causal role of protein charge on elongation rate remains a big challenge. RPs are unusual in many respects that could affect or feedback on translation. They are very small, very abundant, under very strong selection, etc. Among all of the features of transcripts and proteins that we have tested, the pI had one of the highest correlations with the elongation rate, which argues for a more direct contribution of this parameter to the elongation rate. However, a causal effect will need to be established through additional experiments. A very exciting possibility would be to apply the recently developed nascent polypeptide chain tracking technique to a variety of constructs, engineered to vary in one specific aspect such as the pI. Although the data available to date are very limited, one study reported average translation elongation rates of ∼8, 10, and 12 aa/s for 3 very distinct proteins, histone H2B, lysine demethylase KDM5B, and actin (54), whose pIs (from GeneCards, are 10.32, 6.26, and 5.29, respectively. Thus, albeit extremely sparse, these data are consistent with our finding that the pI of the protein is anticorrelated with the average speed with which the ribosome elongates the polypeptide chain, a finding that extends beyond the unusual class of RPs.
The interaction of RPs with the negatively charged rRNAs likely imposes a strong selection pressure for positive charge on RP genes (55), which in turn sets an upper bound on the rate of translocation of the polypeptide chain through the ribosome channel. It will be interesting to explore whether this slower elongation rate may have as side effect an increased translation fidelity of these very abundant proteins (56).
Over 10 y ago it was discovered that protein folding takes place already cotranslationally and that helices can fold within the ribosome exit tunnel (57). A recent study further suggested that nonoptimal codons drive effective cotranslational folding of α-helices and β-sheets (27). Although our results are consistent with these conclusions, they indicate that the positive correlation of translation speed with high density of protein domains is not limited to particular secondary-structure elements.
All of the distinct features that we analyzed here, namely tRNA/codon usage, structure accessibility of the RNA and protein charge, have small and comparable correlation with elongation rate. Altogether, they explain approximately one-half of the variance in elongation rate. This indicates that more detailed models that also include positional features (14) as well as more accurate ribosome coverage profiles (58) will be necessary to improve the prediction of translation dynamics. Our measurements of protein synthesis rates in multiple yeast strains with different translation capacity provide an ideal test bed for new models.
Materials and Methods
Simulations.
All simulations of the TASEP model have been performed with C++ code developed in-house and available in the github repository at the following link: The size of the ribosome footprint has been set to 10 codons.
Analysis of Poly-A Selected RNA-Seq.
Reads from the fastq files associated with the publication of ref. 17 have been trimmed with cutadapt (59) with parameters “–error-rate 0.1 –minimum-length 15 –overlap 1,” first of the 3′ adapter (TGGATTCTCGGGTGCCAAGG) and then of the poly-A tail [“adapter” = (A)50]. Resulting sequences of at least 15 nt were then mapped to yeast ORFs obtained from the yeastgenome.org (60) database ( with the bowtie2 (61) aligner, version 2.2.9, with parameter “-q” (fastq format). For each read, the best alignment reported by bowtie2 was used, and expression levels for each ORF, expressed in reads per kilobase per million (RPKM) were calculated by dividing the number of reads mapped to the ORF by library size and ORF length, then multiplying by 10 6. For each ORF, the expression level used in the analysis was the average computed from 3 replicates.
Analysis of Ribosome Profiling Data.
Ribo-seq data have been downloaded from the Gene Expression Omnibus database (62) (accession number GSE53313). Reads from raw fastq files were trimmed with cutadapt (3′ adapter: TCGTATGCCGTCTTCTGCTTG) with parameters “–error-rate 0.1 –minimum-length 15 –overlap 1.” The first 8 nt corresponding to random barcodes were then trimmed as well and the remainder of the sequence was first aligned to rRNAs with bowtie2, version 2.2.9, parameters “-q” and “–un” to indicate the fastq format of the input file and to obtain also the unmapped reads. The latter were then aligned to a database consisting of all yeast ORFs extended by 200 nt upstream and downstream, to be able to reconstruct full-length ribosome profiles along the ORFs. To allow for the possibility of closely spaced ORFs which would lead to reads mapping in an overlapping manner to the 2 genes, we extracted up to 2 best mappings, with bowtie2 parameter “-k 2.” The positions of the “A” site of ribosomes in individual reads were inferred as in ref. 4.
Codon densities have been estimated as in ref. 4 by considering only ORFs with more than 1 read per codon and removing the first 200 codons. For each ORF, deviations were computed relative to the average number of reads per codon. These relative densities of reads were then collected for individual codons and averaged to get the estimated residence time of the ribosome in each codon.
Analysis of Protein Synthesis Rates.
For the pSILAC experiment, the S. cerevisiae strain BY4741 was grown as described (22). Briefly, synthetic medium containing 2% glucose, yeast nitrogen base (6.7 g/L), and dropout medium (2 g/L) containing all of the amino acids except l-lysine was prepared. Initially, a preculture of yeast was grown at 30 °C, 200 rpm, in 3 biological replicates obtained by inoculating 3 different colonies in 5 mL of heavy SILAC synthetic medium containing heavy l-lysine-2HCl, 13 C 6, 15 N 2 (Thermo Fisher; 88209) at a final concentration of 30 mg/L. The preculture step was repeated one more time so that all of the proteins became tagged with heavy isotope. The preculture thus obtained was used to grow cells at optical density of A600 = 0.4 in 200 mL. At this point, cells were centrifuged, washed twice with light SILAC synthetic medium containing light l-lysine-2HCl (Thermo Fisher; 89987) at concentration of 30 mg/L and transferred to 200 mL of light SILAC media. Cells were harvested at 0, 5, 15, 30, 60, 120, and 180 min for the mass-spectrometric analysis.
Cells were lysed in a buffer containing 1% sodium deoxycholate, 0.1 M ammonium bicarbonate, and 10 mM TCEP using strong ultrasonication (Bioruptor; 10 cycles, 30 s on/off; Diagenode). Samples were heated to 95 °C for 10 min, and after cooling, the protein concentration was determined by the BCA assay (Thermo Fisher Scientific), using a small sample aliquot. Fifty micrograms of protein were alkylated with 15 mM chloroacetamide for 30 min at 37 °C and incubated with sequencing-grade modified trypsin (1/50 [wt/wt]; Promega) overnight at 37 °C. After acidification using 5% TFA, precipitated detergent was removed by centrifugation (14,000 rpm, 5 min). Peptides were desalted on C18 reversed-phase spin columns according to the manufacturer’s instructions (Microspin; Harvard Apparatus) and dried under vacuum.
The setup of the μRPLC-MS system was as described previously (63). Chromatographic separation of peptides was carried out using an EASY nano-LC 1000 system (Thermo Fisher Scientific), equipped with a heated reversed-phase–high-performance liquid chromatography column (75 μm × 37 cm) packed in-house with 1.9-μm C18 resin (Reprosil-AQ Pur; Dr. Maisch). Aliquots of 1-μg total peptides were analyzed per liquid chromatography–tandem mass spectrometry run using a linear gradient ranging from 95% solvent A (0.15% formic acid, 2% acetonitrile) and 5% solvent B (98% acetonitrile, 2% water, 0.15% formic acid) to 30% solvent B over 90 min at a flow rate of 200 nL/min. Mass spectrometry analysis was performed on Q-Exactive HF mass spectrometer equipped with a nanoelectrospray ion source (both Thermo Fisher Scientific). Each MS1 scan was followed by high-collision–dissociation of the 10 most abundant precursor ions with dynamic exclusion for 20 s. Total cycle time was ∼1 s. For MS1, 3 × 10 6 ions were accumulated in the Orbitrap cell over a maximum time of 100 ms and scanned at a resolution of 120,000 full width at half-maximum (FWHM) (at 200 m/z). MS2 scans were acquired at a target setting of 10 5 ions, accumulation time of 50 ms, and a resolution of 15,000 FWHM (at 200 m/z). Singly charged ions and ions with unassigned charge state were excluded from triggering MS2 events. The normalized collision energy was set to 27%, the mass isolation window was set to 1.4 m/z, and 1 microscan was acquired for each spectrum.
The acquired raw files were imported into the Progenesis QI software (version 2.0; Nonlinear Dynamics, Limited), which was used to extract peptide precursor ion intensities across all samples applying the default parameters. The generated mgf files were searched using MASCOT using the following search criteria: full tryptic specificity was required (cleavage after lysine or arginine residues, unless followed by proline); 3 missed cleavages were allowed; carbamidomethylation (C) was set as fixed modification; oxidation (M) and heavy SILAC (K8) were applied as variable modifications; mass tolerance of 10 ppm (precursor) and 0.02 Da (fragments). The database search results were filtered using the ion score to set the FDR to 1% on the peptide and protein level, respectively, based on the number of reverse protein sequence hits in the datasets. The relative quantitative data obtained were normalized and statistically analyzed using our in-house script (SafeQuant) as above (63). The mass spectrometry proteomics data have been deposited to the ProteomeXchange Consortium via the PRIDE (64) partner repository with the dataset identifier PXD014357.
The protein synthesis rates have been obtained from the slope of linear regression constrained to 0 at time point 0. This analysis was performed with the lm() function of R (version 3.4.2). For each regression, R 2 values have been recorded for further analysis (Fig. 2 C).
Scaling mRNA Copy Numbers, Protein Synthesis Rates, and Ribosome Densities.
To obtain absolute protein synthesis rates, the relative rates obtained from pSILAC were scaled, using the known values of the number of protein molecules per yeast cell (21), the doubling time of yeast, and the average half-life of yeast proteins (22). Furthermore, as the proteomics experiment does not capture all proteins, the uncaptured fraction had to be taken into account. The fraction of captured proteins has been approximated as follows: reads from ribosome footprints were used to compute normalized ORF abundances in the Ribo-seq data (RPKMs). The total abundance of translated ORFs that were not captured in the proteomics data, relative to all of the ORFs captured in Ribo-seq, was used as the fraction of uncaptured protein. The steady-state level of protein per cell should be given by the ratio of synthesis and degradation rates. The synthesis rate can thus be calculated as the product of the steady-state level of protein per cell and the degradation rate. The latter is the result of 2 processes, protein degradation and dilution due to cell growth. This leads to the following expression for the average synthesis rate of the captured fraction:
To obtain synthesis rates for individual ORFs, we multiplied the total synthesis rate of the captured fraction by the relative synthesis rate inferred by fitting the light peptide accumulation in pSILAC:
To infer absolute densities of ribosomes per codon, we determined the first principal component of the absolute ribosome densities from ref. 19 relative to our estimates based on RPFs (Fig. 2 A). We used this first principal component to map relative densities we obtained based on RPF and mRNA-seq reads to ribosomes densities per codon for each ORF.
Absolute abundances of mRNA molecules per cell were obtained by rescaling the relative numbers inferred from RNA-seq to obtain a total of 40,000 transcripts per cell, as found in previous work (23, 24).
Computation of mRNA/Protein Features.
Protein features used in the analysis of translation elongation rate have been downloaded from the yeast genome database at the following link:
The tRNA adaptation index (tAI) and normalized tAI (ntAI) have been computed as in refs. 26 and 27 with a custom Python script. The RNAplfold tool from ViennaRNA package (65) (version 2.1.8) was used with default parameters to estimate structural accessibility along ORFs. For each ORF, the average accessibility of windows of a specified size has been calculated.
We analyzed the following features:
•
Molecular weight (Mw) of the protein in daltons;
•
Isoelectric point (pI): the pH at which the protein does not carry net electric charge;
•
Grand average of hydropathicity (GRAVY score): the average of hydropathy values of all amino acids in the protein (31);
•
Aromaticity score: the frequency of aromatic amino acids, Phe, Tyr, and Trp;
•
Codon adaptation index (CAI): measure of the bias of codon usage in a coding sequence with respect to a reference set of genes (29). It is defined as the geometric mean of the weights over all codons in the sequence :
where the weight of each of codon is computed from the reference sequence set, as the ratio between the observed frequency of the codon and the frequency of the most frequent synonymous codon for that amino acid:
with synonymous codons.
•
tRNA adaptation index (tAI): measure of the adaptation of each transcript to the pool of tRNAs (26). Similarly to CAI, tAI is the geometric mean of weights associated to each codon:
where
and
with as number of tRNA isoacceptors, as gene copy number of tRNA j th recognizing i th codon, and as a selective constraint of codon–anticodon coupling.
•
Normalized tRNA adaptation index (ntAI): normalized version of tAI based on the codon usage in the transcriptome (27), it has a similar form to tAI, but with being scaled by the normalized codon expression in the following way: is the usage of codon taking into account the abundance of individual transcripts:
with being the transcript abundance of gene and as the number of occurrences of codon within the ORF of the gene The usage of codon is then defined as follows:
and finally the weights that used in the calculation of the tAI are defined as follows:
The factor substitutes in the formula for tAI.
•
Fraction of optimal codons (FOP): fraction of optimal codons in the ORF (28). The optimal codon for an amino acid is the codon most used to encode the amino acid in the ORFs encoding the top expressed proteins;
•
Instability index: measure of protein half-life estimated based on the dipeptide composition of the protein (32);
•
Domain coverage: fraction of protein covered by Pfam domains predicted by InterPROScan (66);
•
α-Helix, β-sheet, coil: fraction of the protein sequence involved in the indicated types of secondary structures predicted by PSIPRED (67);
•
Accessibility, 5 nt: average probability of finding a window of size 5 nt in an open conformation predicted with RNAplfold from ViennaRNA package (65).
Enrichment Tests.
Gene ontology and KEGG pathway enrichments have been performed through the STRING database (68).
Supplementary Material
Supplementary File
pnas.1817299116.sapp.pdf (2.2MB, pdf)
Supplementary File
pnas.1817299116.sd01.tsv (203.1KB, tsv)
Supplementary File
pnas.1817299116.sd02.tsv (181.5KB, tsv)
Supplementary File
pnas.1817299116.sd03.xlsx (6.3KB, xlsx)
Supplementary File
pnas.1817299116.sd04.tsv (203.1KB, tsv)
Supplementary File
pnas.1817299116.sd05.tsv (196.2KB, tsv)
Supplementary File
pnas.1817299116.sd06.tsv (166.3KB, tsv)
Supplementary File
pnas.1817299116.sd07.tsv (433B, tsv)
Supplementary File
pnas.1817299116.sd08.tsv (19.8KB, tsv)
Acknowledgments
A.R. thanks Joao C. Guimaraes for discussions and knowledge sharing that helped in developing the project.
Footnotes
The authors declare no conflict of interest.
This article is a PNAS Direct Submission.
Data deposition: The mass spectrometry proteomics data have been deposited in the ProteomeXchange Consortium (www.proteomexchange.org/) via the PRIDE partner repository (dataset identifier PXD014357). The C++ code developed for simulations of the TASEP model are available at
This article contains supporting information online at www.pnas.org/lookup/suppl/doi:10.1073/pnas.1817299116/-/DCSupplemental.
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Associated Data
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Supplementary Materials
Supplementary File
pnas.1817299116.sapp.pdf (2.2MB, pdf)
Supplementary File
pnas.1817299116.sd01.tsv (203.1KB, tsv)
Supplementary File
pnas.1817299116.sd02.tsv (181.5KB, tsv)
Supplementary File
pnas.1817299116.sd03.xlsx (6.3KB, xlsx)
Supplementary File
pnas.1817299116.sd04.tsv (203.1KB, tsv)
Supplementary File
pnas.1817299116.sd05.tsv (196.2KB, tsv)
Supplementary File
pnas.1817299116.sd06.tsv (166.3KB, tsv)
Supplementary File
pnas.1817299116.sd07.tsv (433B, tsv)
Supplementary File
pnas.1817299116.sd08.tsv (19.8KB, tsv)
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Startseite » Deutsch » Grammatik Deutsch » Zeitformen Deutsch » Präteritum (Imperfekt, Vergangenheit)
Präteritum (Imperfekt, Vergangenheit)
2,50€
Deutsche Grammatik
Erläuterung und Übungen zum Präteritum
48 Seiten, 19 Übungen (inkl. Abschlusstest und Lösungen)
Das Präteritum (Imperfekt, Vergangenheitsform) verwenden wir im Deutschen für Erzählungen und Berichte in der Vergangenheit, vor allem in der Schriftsprache. Im mündlichen Sprachgebrauch nehmen wir statt Präteritum häufig das Perfekt.
Erfahre in unserer Erläuterung alles zur Verwendung und Bildung und teste dein Können in den Übungen.
Erläuterung zur Bildung, Verwendung und Besonderheiten mit Beispielen
Liste mit starken und gemischten Verben (unregelmäßige Verben)
Übungen inkl. Test mit gemischtem Schwierigkeitsgrad
Lösungen zu allen Übungen
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Nutzungsbedingungen: Sie dürfen die Materialien für den eigenen Unterricht an einem Bildungsinstitut, zur Nachhilfe oder im privaten Umfeld beim schulischen Lernen verwenden und als Ausdruck an Ihre Schüler oder Kinder verteilen.
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Artikelnummer: PDF-DEU-0006 Kategorie: Zeitformen DeutschSchlagwörter: Vergangenheit, Deutsch, Grammatik, Übungen, Zeitformen, Präteritum
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Erläuterung und Übungen zum Präteritum
Inhalt
Präteritum/Imperfekt – deutsche Zeitform für die Vergangenheit
Übung Präteritum – haben/sein – 1./3. Person 1 (A1)
Übung Präteritum – haben/sein – 1./3. Person 2 (A1)
Übung Präteritum – Konjugationtabelle (haben/sein) (A2)
Übung Präteritum – Konjugation von haben (A2)
Übung Präteritum – Konjugation von sein (A2)
Übung Präteritum – haben/sein (A2)
Übung Präteritum – schwache Verben (Tabelle) (B1)
Übung Präteritum – schwache Verben (B1)
Übung Präteritum – starke Verben (Tabelle) (B1)
Übung Präteritum – starke Verben (Sätze) (B1)
Übung Präteritum – eingeschobenes e vor t/st (schwache Verben) (B1)
Übung Präteritum – starke und gemischte Verben 1 (B1)
Übung Präteritum – starke und gemischte Verben 2 (B1)
Übung Präteritum – eingeschobenes e vor t/st (starke Verben) (B2)
Übung Präteritum – Ereignisse ohne Bezug zur Gegenwart (B2)
Übung Präteritum – Chronologische Reihenfolge (B2)
Übung Präteritum – Unfallbericht (B2)
Übung Präteritum – Aschenputtel (B2)
Übung Präteritum – Übungen (Abschlusstest)
Starke und gemischte Verben im Deutschen (unregelmäßige Verben)
Lösungen
48 Seiten, 19 Übungen (inkl. Abschlusstest und Lösungen)
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К 407,9 Б 13,8 Ж 33,8 У 12,0
САЛАТ ОЛИВЬЕ
200 гр. 499 р.
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К 362,0 Б 10,0 Ж 30,1 У 12,5
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210 гр. 649 р.
Кальмар, куриное филе, микс салата, болгарский перец, черри, спаржа, кешью и паназиатский соус
К 263,3 Б 17,6 Ж 13,1 У 18,5
САЛАТ С БУЖЕНИНОЙ
200 гр. 549 р.
Буженина, микс салата, пармезан, черри, кедровые орехи, горчичный соус
К 354,2 Б 16,1 Ж 30,3 У 4,2
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230 гр. 499 р.
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170 гр. 549 р.
Лист салата, курица, крутоны, пармезан, соус Цезарь, черри
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195 гр. 549 р.
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К 323,7 Б 20,8 Ж 17,8 У 19,9
САЛАТ С МИНИ-КЕБАБАМИ
250 гр. 499 р.
Мини-кебабы из курицы, черри, сыр Креметта, микс салата, вареное яйцо, огурцы, оливковая заправка
К 342,9 Б 8,7 Ж 32,4 У 3,9
САЛАТ С СЕМГОЙ И АПЕЛЬСИНОВОЙ ЗАПРАВКОЙ
220 гр. 749 р.
Семга с/с, черри, огурцы, лук фри, микс салата, сыр Креметта, апельсиновая заправка
К 172,9 Б 13,0 Ж 12,2 У 2,6
САЛАТ РОДИЧЕ
230 гр. 599 р.
Ростбиф, запеченный картофель, бекон, грибы, хрустящее яйцо, домашний майонез
К 415,2 Б 16,7 Ж 32,7 У 13,3
ЗЕЛЕНЫЙ САЛАТ С ПЕСТО
160 гр. 499 р.
Микс салата, черри, сыр Креметта, кедровый орех, соус Песто
К 228,4 Б 5,5 Ж 20,9 У 4,4
САЛАТ С КОПЧЕНЫМИ МОРЕПРОДУКТАМИ
185 гр. 999 р.
Копченая креветка, копченый гребешок, микс салата, черри, перец халапеньо, апельсиновая заправка
К 129,7 Б 10,1 Ж 8,1 У 4,0
САЛАТ С ТРЕСКОЙ И ТОМАТАМИ
250 гр. 549 р.
Микс салата, томаты, треска в панировке, картофельные дольки, ореховый соус
К 395,1 Б 75,8 Ж 8,0 У 4,9
ХОЛОДНЫЕ ЗАКУСКИ
ПАШТЕТ ИЗ КУРИНОЙ ПЕЧЕНИ
160/30 гр. 399 р.
Куриная печень с багетом и сладким луковым джемом
К 386,0 Б 17,2 Ж 25,7 У 18,7
АССОРТИ БРУСКЕТТ
200 гр. 549 р.
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320 гр. 949 р.
Три вида сыра, мед, грецкий орех, виноград, груша
К 696,1 Б 34,2 Ж 42,1 У 45,0
МЯСНЫЕ ДЕЛИКАТЕСЫ
270/30 гр. 899 р.
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Семга, сельдь, копченая масляная рыба
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БАКЛАЖАНЫ С ГРИБАМИ ШИИТАКЕ
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Жареные баклажаны в соусе Унаги
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ШАМПИНЬОНЫ, ЗАПЕЧЕННЫЕ С СЫРОМ
200 гр. 499 р.
Свежие шампиньоны, запеченные с сыром Моцарелла
К 238,8 Б 16,0 Ж 17,9 У 3,3
СУПЫ
ТОМ ЯМ
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ЛЕПЕШКА ВОСТОЧНАЯ
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КЕСАДИЛЬЯ С ОХОТНИЧЬИМИ КОЛБАСКАМИ
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ГРЕНКИ
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ЛУКОВЫЕ КОЛЬЦА
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СЫРНЫЕ ПАЛОЧКИ
140/50 гр. 449 р.
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КУРИНЫЕ НАГГЕТСЫ
200/50 гр. 499 р.
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КАМЧАТСКАЯ КОРЮШКА ФРИ
330 гр. 999 р.
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ПАСТЫ
УДОН НА ВЫБОР: С КАЛЬМАРОМ ИЛИ КУРИЦЕЙ
280 гр. 529 р.
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ЗЕЛЕНАЯ ПАСТА
270 гр. 549 р.
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ФАРФАЛЛЕ С БЕКОНОМ И ПЕСТО
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К 530,2 Б 23,1 Ж 28,9 У 44,3
ПАСТА КАРБОНАРА
280 гр. 599 р.
Паста, мясные копчености, сливки, сыр Пармезан
К 936,9 Б 34,7 Ж 44,8 У 98,5
ГОРЯЧИЕ БЛЮДА ИЗ РЫБЫ И МОРЕПРОДУКТОВ
ФИШ ЭНД ЧИПС
370 гр. 799 р.
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ФИЛЕ ЧЕРНОЙ ТРЕСКИ
260 гр. 1299 р.
С брокколи и сырным муссом
К 365,3 Б 39,8 Ж 18,8 У 9,1
ШАШЛЫК ФИНСКИЙ
380 гр. 999 р.
Кета в беконе, картофельные дольки, соус барбекю
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КОТЛЕТЫ ИЗ ЧЕРНОЙ И БЕЛОЙ ТРЕСКИ
220 гр. 699 р.
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СТЕЙК ИЗ СЕМГИ
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ГОРЯЧИЕ БЛЮДА ИЗ МЯСА И ПТИЦЫ
ФАХИТОС С МЯСОМ НА ВЫБОР: СВИНИНА ИЛИ КУРИЦА
360/40 гр. 699 р.
Мясо в пикантном соусе с овощами, подается с лавашом, сметаной, аджикой и халапеньо
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ШАШЛЫК НА ВЫБОР: СВИНИНА ИЛИ КУРИЦА
315 гр. 799 р.
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ЖАРЕХА С МЯСОМ НА ВЫБОР: СВИНИНА ИЛИ ГОВЯДИНА
390 гр. 649 р.
Мясо, картофель, грибы шампиньоны, лук
К 521,4 Б 16,1 Ж 31,3 У 43,5
КОЛБАСКА НА ГРИЛЕ
340 гр. 799 р.
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ЖАРКОЕ ПО-ДОМАШНЕМУ
350 гр. 1099 р.
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МЕДАЛЬОНЫ ИЗ ГОВЯЖЬЕЙ ВЫРЕЗКИ
260 гр. 1499 р.
Подается с картофелем, луком гриль и перечным соусом
К 191,6 Б 21,0 Ж 9,8 У 4,6
СТЕЙКИ
СТЕЙК ИЗ СВИНИНЫ
345/40 гр. 749 р.
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К 1128,8 Б 32,6 Ж 90,4 У 21,1
СТЕЙК ИЗ КУРИЦЫ
120/100/30 гр. 599 р.
Подается с овощами на гриле и соусом Тар-тар
К 249,3 Б 23,7 Ж 16,5 У 1,4
СТЕЙК РИБАЙ
250/130/40 гр. 2499 р.
Подается с грибами на гриле, картофельными дольками и перечным соусом
К 1094,4 Б 61,8 Ж 82,7 У 25,6
СТЕЙК ФЛАНК
150/70/40 гр. 1799 р.
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К 360,5 Б 28,1 Ж 24,2 У 7,4
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Подается с картофельными дольками, перцем халапеньо и соусом Барбекю
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СТЕЙК ТРАЙ ТИП
170/140/40 гр. 1699 р.
Подается с картофельными дольками, маринованным огурцом и соусом Барбекю
К 360,5 Б 28,1 Ж 24,2 У 7,4
БЛЮДА НА КОМПАНИЮ
СЭТ №1
210/50 гр. 649 р.
Чипсы из лаваша, картофель фри, луковые кольца, кетчуп
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СЭТ №2
470/100 гр. 1449 р.
Сырные палочки, крылья барбекю, картофель фри, гренки, кетчуп, соус Барбекю
К 1439,2 Б 77,4 Ж 69,3 У 126,3
СЭТ №3
700/80 гр. 1899 р.
Сырные палочки, кольца кальмара, куриные наггетсы, картофельные дольки, салат Коул Слоу, соус Кимчи, кетчуп
К 1342,7 Б 96,4 Ж 40,0 У 149,1
БОКС ЧИКЕН
1100 гр. 2299 р.
Филе куриного бедра, жареное на гриле в азиатском соусе, мини-кебабы из мяса говядины и свинины, куриная голень в медово-соевом соусе, салат Греческий, соус аджика, горчичный соус
К 580,0 Б 36,2 Ж 22,6 У 58,1
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您当前位置：公务员考试网>国家公务员考试网> 2022国考行测备考：图形中的汉字字母规律例题解析
2022国考行测备考：图形中的汉字字母规律例题解析
2021-07-23 17:26:17 公务员考试网文章来源：华图教育
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公考路漫漫，华图常相伴，这篇文章主要是针对图形中的汉字、字母等规律进行介绍。汉字、字母在考场中是经常出现的一类图形，很多考生对于解答这一类题目需要花费很长的时间，甚至选不出来答案。在考场中要快速的解答这一类题目，首先我们要清楚，具体有哪些规律的存在才能更快的去思考具体考的是哪一类规律。华图通过对近十年所有考题进行定性定量的分析可知，主要是围绕以下几个方面去考查规律，一、汉字结构(上下、左右、半包围)，二、样式类(加、减、同、异)，三、数量类(笔画、封闭面、部分数)，四、属性类(对称、曲直、封闭)。接下来我们看看具体在题目中如何去运用。
【真题演练1】从所给的四个选项中，选择最合适的一个，使之呈现一定的规律性( )。
1.C【解析】图形中出现汉字，九宫格优先横看，观察发现，题干第一行汉字的结构为：左右、全包围、上下;第三行图形中的汉字结构为：全包围、上下、左右，三种结构均出现一次。第二行图形中的汉字结构为：上下、?、全包围，因此?处应为左右结构的汉字，只有C项符合。因此，选择C选项。
【真题演练2】从所给的四个选项中，选择最合适的一个，使之呈现一定的规律性( )。
2.C【解析】图形中出现汉字，每个汉字都出现了“斤”，考虑样式类整体求同，排除B、D。继续观察每个汉字都是左右结构，排除A。因此，选择C选项。
【真题演练3】从所给的四个选项中，选择最合适的一个，使之呈现一定的规律性( )。
3.D【解析】题干每个图形都是一个汉字，优先考虑笔画数。第一组三个图形笔画数分别是4、3、2，第二组前两个图形笔画数6，5，故“?”处应该是一个4笔画汉字。A项9笔，B项3笔，C项8笔，D项4笔。因此，选择D选项。
【真题演练4】从所给的四个选项中，选择最合适的一个，使之呈现一定的规律性( )。
4.B【解析】元素组成不同，优先考虑属性规律。从左到右观察发现图1、图3、图5为既轴对称图形，又中心对称图形，图2、图4为中心对称图形。问号处应为中心对称图形。因此，选择B选项。
通过以上的题目以及解析可以知道，对于汉字、字母这一类图形的规律，主要是围绕上述的四个方面来考查，另外针对这四类的规律，其实每一类也有着不同的题型特征。第一类：汉字结构，这一类题目的呈现特征是具有明显的结构特点，如上下、左右等结构;第二类：样式类，这一类题目的呈现特征是存在相似的特征，具有相同的地方但是也有一些不同的地方;第三类：数量类，这一类题目的呈现特征是比较凌乱，是一些不规则的汉字、字母;第四类：属性类，这一类题目的呈现特征是具有共性的特点，如全直线、全曲线等共性，所以大家在考场中要明确有哪几类规律的存在以及每一类规律的不同特征，只有掌握清楚，才能更快的找到图形的规律，在考场中游刃有余。
这篇文章主要针对判断推理中图形的相关内容进行介绍，在备考的路上还需要大家多做题多思考，成功是留给有准备的人。
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7117 | https://www.lessonplanet.com/article/elementary-math/making-sense-of-dollars-and-cents | Making Sense of Dollars and Cents | Lesson Planet
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Elementary Math
Making Sense of Dollars and Cents
Making Sense of Dollars and Cents
Teaching the money system to youngsters through games, literature, and real-world simulations
By Nicole Schon
It’s never too early to foster a love of math. What better way to learn than with something students have used themselves–money! Learning the money system offers children many opportunities for hands-on, interactive lessons.
Helping Learners Buy In
Part of the challenge in teaching our monetary system lies in the connection children are asked to make between an object (the coin) and the “value” of that object. Guiding learners to understand this abstract correlation takes time, patience, and a lot of practice. In my second grade classroom, I sought ways to actively engage students in thinking about money. This involved a variety of games, group activities, and, my personal favorite, literature.
Playing with Money
While most children at this age have already had some exposure to money, they are not truly aware that each different type of coin has a specific value. They do know that money buys them things they want, and so having the opportunity to play with real money immediately increases engagement across the board.
I introduce pennies first, since this value is the simplest to understand; one penny equals one cent. We then do a variety of activities to familiarize them with the values of every other coin. A simple game involves small groups; each playing with a spinner (mine are made from a laminated piece of construction paper, a brass fastener, and laminated card stock for the spinner). The spinner lands on random amounts of money, such as $0.75, $1.36, etc. The goal is to find as many ways to make the amount as possible. The game can be played either competitively, with each person finding his own methods, or collaboratively, with each group working together to beat the other teams in the class.
Literature Connection
I used the book Alexander Who Used to Be Rich Last Sunday as both an introduction to, and a formative assessment during the money unit. The book follows a young boy who doesn’t really understand the values of different coins as he spends and trades his money over the course of a week. On the first read through, most children do not understand the follies that young Alexander makes throughout the story. Once we’ve learned about the various values of coins, we read the book a second time, and kids groan when the main character makes his poor financial decisions.
I turn this into an assessment by having small groups work together to add and subtract each of Alexander’s financial transactions throughout the book. Those who do not fully understand how much each coin is worth do not end up at the correct sum, which is zero. This also lead to a graphing activity to chart how much money Alexander has at the end of each day.
Bringing in Real-World Experience
Nothing makes my class happier than the culminating activity of our unit on money: a classroom store. This activity serves as a summative assessment, allowing me to quickly see who has truly mastered the values of each different type of coin, and who needs more reinforcement and practice.
To set up the store, I had the class assist me in labeling each “inventory” item with a price I assigned. For the inventory, I use a variety of methods depending on the student population and funding. Sometimes a white elephant model was convenient, having class members bring in small trinkets and toys to sell. Other times, I bought things like pencils, stickers, erasers, and other small items at a 99 cent store. Another option is to only pretend to buy and sell items–the white elephant model works well with this. I also found that kindergarten teachers at my school were happy to provide a variety of play grocery items, dress up clothes, toys, and other items. I found that the children enjoyed the activity just as much whether or not the selling was real.
During the activity, every “shopper” is given a certain amount of money to spend, including two or more of each type of coin. At some point, each child also gets to act as cashier. To gain a better sense of each child’s understanding, I have the “customer” add up his or her expenses and the “cashier” double check the math. Once the amount is verified, the customer has to give the right amount, and the cashier has to give the right change. I sit next to the cashier and take notes on what each learner’s strengths and stretches are when it comes to using money.
The more experiences children have playing with and manipulating numbers early on, the stronger their number sense foundation. Below is a collection of resources geared to increase youngster’s understanding of the money system.
Right on the Money
An interactive introduction to American currency first uses the internet to help kids discover how much each coin is worth and read facts about the penny. Then children are lead through a video. After each segment, they practice combining different coins and bills to create the same value amount.
Counting Money
After a short review of the dollar value of each type of coin, learners engage in writing word problems that involve calculating money. They then use manipulatives (either real or fake coins) to solve problems written by their classmates.
Gazillions
Second graders engage in a video to better understand the dollar amounts of coins and bills. They create specified dollar amounts in a variety of contexts, including various web-based games and a worksheet. The culminating activity calls for learners to write a letter to a friend that includes a money-story problem and solution.
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7118 | https://mathshelper.co.uk/gcse_docs/Intersecting%20Chord%20Theorem.pdf | Intersecting Chord Theorem 1. Find the value of x in the diagram below.
2 6 x 3 x = 4 (only) 2. Find the value of x in the diagram below.
2 8 x x x = 4 (only) 3. Find the value of x in the diagram below.
2 4 x x + 7 x = 1 (only) 4. Find the value of x in the diagram below.
2 x + 7 x + 1 x + 2 1 J.M.Stone x = 3 (only) 5. Find the value of x in the diagram below (Hard factorisation! Eliminate fractions.
Best of luck.).
x + 1 2 12x x + 5 2 x + 3 2 x = 1 2 (only) 6. Find the value of x in the diagram below.
13 x 9 1 x = 5 (only) 7. Find the value of x in the diagram below.
1 7 x 10 x = 4 (only) 8. Find the value of x in the diagram below.
8 x 4 1 2 J.M.Stone x = 2 (only) 9. Find the value of x in the diagram below.
13 x 5 1 x = 2 (only) 10. Find the value of x in the diagram below.
15 5 x x = 10 (only) 11. Find the value of x in the diagram below.
x + 4 x 4 x = 2 (only) 12. Find the value of x in the diagram below.
x 3 2 4 3 J.M.Stone x = 1 (only) 13. Find the value of x in the diagram below.
x + 6 x 4 1 x = 2 (only) 14. Find the value of x in the diagram below.
4 x 2 14 x = 4 (only) 15. Find the value of x in the diagram below.
x + 6 x + 2 5 1 x = 1 (only) 16. Find the value of x in the diagram below.
20 x −3 x −1 x + 2 4 J.M.Stone x = 5 or x = 10 17. Find the value of x in the diagram below.
2x x + 1 2 18 x = 3 (only) 18. The chords AB and CD of a circle meet at X inside the circle. XA = 3, AB = 7, XC = 2.
Find CD.
5 J.M.Stone |
7119 | https://www.accessdata.fda.gov/drugsatfda_docs/label/2010/018662s060lbl.pdf | Company logo ACCUTANE® (isotretinoin capsules) Rx only Pregnancy Wa rning Log o PREGNANT CONTRAINDICATIONS AND WARNINGS Accutane must not be used by female patients who are or may become pregnant. There is an extremely high risk that severe birth defects will result if pregnancy occurs while taking Accutane in any amount, even for short periods of time. Potentially any fetus exposed during pregnancy can be affected. There are no accurate means of determining whether an exposed fetus has been affected. Birth defects which have been documented following Accutane exposure include abnormalities of the face, eyes, ears, skull, central nervous system, cardiovascular system, and thymus and parathyroid glands. Cases of IQ scores less than 85 with or without other abnormalities have been reported. There is an increased risk of spontaneous abortion, and premature births have been reported. Documented external abnormalities include: skull abnormality; ear abnormalities (including anotia, micropinna, small or absent external auditory canals); eye abnormalities (including microphthalmia); facial dysmorphia; cleft palate. Documented internal abnormalities include: CNS abnormalities (including cerebral abnormalities, cerebellar malformation, hydrocephalus, microcephaly, cranial nerve deficit); cardiovascular abnormalities; thymus gland abnormality; parathyroid hormone deficiency. In some cases death has occurred with certain of the abnormalities previously noted. If pregnancy does occur during treatment of a female patient who is taking Accutane, Accutane must be discontinued immediately and she should be referred to an Obstetrician-Gynecologist experienced in reproductive toxicity for further evaluation and counseling. Special Prescribing Requirements Because of Accutane’s teratogenicity and to minimize fetal exposure, Accutane is approved for marketing only under a special restricted distribution program approved by the Food This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 2 and Drug Administration. This program is called iPLEDGE™. Accutane must only be prescribed by prescribers who are registered and activated with the iPLEDGE program. Accutane must only be dispensed by a pharmacy registered and activated with iPLEDGE, and must only be dispensed to patients who are registered and meet all the requirements of iPLEDGE (see PRECAUTIONS). Table 1 Monthly Required iPLEDGE Interactions Female Patients of Male Patients, And Female Childbearing Potential Patients Not of Childbearing Potential PRESCRIBER Confirms patient counseling X X Enters the 2 contraception methods X chosen by the patient Enters pregnancy test results X PATIENT Answers educational questions before X every prescription Enters 2 forms of contraception X PHARMACIST Contacts system to get an authorization X X DESCRIPTION Isotretinoin, a retinoid, is available as Accutane in 10-mg, 20-mg and 40-mg soft gelatin capsules for oral administration. Each capsule contains beeswax, butylated hydroxyanisole, edetate disodium, hydrogenated soybean oil flakes, hydrogenated vegetable oil, and soybean oil. Gelatin capsules contain glycerin and parabens (methyl and propyl), with the following dye systems: 10 mg — iron oxide (red) and titanium dioxide; 20 mg — FD&C Red No. 3, FD&C Blue No. 1, and titanium dioxide; 40 mg — FD&C Yellow No. 6, D&C Yellow No. 10, and titanium dioxide. Chemically, isotretinoin is 13-cis-retinoic acid and is related to both retinoic acid and retinol (vitamin A). It is a yellow to orange crystalline powder with a molecular weight of 300.44. The structural formula is: Structural Formula CLINICAL PHARMACOLOGY Isotretinoin is a retinoid, which when administered in pharmacologic dosages of 0.5 to 1.0 mg/kg/day (see DOSAGE AND ADMINISTRATION), inhibits sebaceous gland function and keratinization. The exact mechanism of action of isotretinoin is unknown. 2 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 3 Nodular Acne Clinical improvement in nodular acne patients occurs in association with a reduction in sebum secretion. The decrease in sebum secretion is temporary and is related to the dose and duration of treatment with Accutane, and reflects a reduction in sebaceous gland size and an inhibition of sebaceous gland differentiation.1 Pharmacokinetics Absorption Due to its high lipophilicity, oral absorption of isotretinoin is enhanced when given with a high-fat meal. In a crossover study, 74 healthy adult subjects received a single 80 mg oral dose (2 x 40 mg capsules) of Accutane under fasted and fed conditions. Both peak plasma concentration (Cmax) and the total exposure (AUC) of isotretinoin were more than doubled following a standardized high-fat meal when compared with Accutane given under fasted conditions (see Table 2). The observed elimination half-life was unchanged. This lack of change in half-life suggests that food increases the bioavailability of isotretinoin without altering its disposition. The time to peak concentration (Tmax) was also increased with food and may be related to a longer absorption phase. Therefore, Accutane capsules should always be taken with food (see DOSAGE AND ADMINISTRATION). Clinical studies have shown that there is no difference in the pharmacokinetics of isotretinoin between patients with nodular acne and healthy subjects with normal skin. Table 2 Pharmacokinetic Parameters of Isotretinoin Mean (%CV), N=74 Accutane 2 x 40 mg Capsules AUC0-∞ (ng⋅hr/mL) Cmax (ng/mL) Tmax (hr) t1/2 (hr) Fed 10,004 (22%) 862 (22%) 5.3 (77%) 21 (39%) Fasted 3,703 (46%) 301 (63%) 3.2 (56%) 21 (30%) Eating a standardized high-fat meal Distribution Isotretinoin is more than 99.9% bound to plasma proteins, primarily albumin. Metabolism Following oral administration of isotretinoin, at least three metabolites have been identified in human plasma: 4-oxo-isotretinoin, retinoic acid (tretinoin), and 4-oxo-retinoic acid (4-oxo tretinoin). Retinoic acid and 13-cis-retinoic acid are geometric isomers and show reversible interconversion. The administration of one isomer will give rise to the other. Isotretinoin is also irreversibly oxidized to 4-oxo-isotretinoin, which forms its geometric isomer 4-oxo-tretinoin. After a single 80 mg oral dose of Accutane to 74 healthy adult subjects, concurrent administration of food increased the extent of formation of all metabolites in plasma when compared to the extent of formation under fasted conditions. 3 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 4 All of these metabolites possess retinoid activity that is in some in vitro models more than that of the parent isotretinoin. However, the clinical significance of these models is unknown. After multiple oral dose administration of isotretinoin to adult cystic acne patients (≥18 years), the exposure of patients to 4-oxo-isotretinoin at steady-state under fasted and fed conditions was approximately 3.4 times higher than that of isotretinoin. In vitro studies indicate that the primary P450 isoforms involved in isotretinoin metabolism are 2C8, 2C9, 3A4, and 2B6. Isotretinoin and its metabolites are further metabolized into conjugates, which are then excreted in urine and feces. Elimination Following oral administration of an 80 mg dose of 14C-isotretinoin as a liquid suspension, 14C activity in blood declined with a half-life of 90 hours. The metabolites of isotretinoin and any conjugates are ultimately excreted in the feces and urine in relatively equal amounts (total of 65% to 83%). After a single 80 mg oral dose of Accutane to 74 healthy adult subjects under fed conditions, the mean ± SD elimination half-lives (t1/2) of isotretinoin and 4-oxo-isotretinoin were 21.0 ± 8.2 hours and 24.0 ± 5.3 hours, respectively. After both single and multiple doses, the observed accumulation ratios of isotretinoin ranged from 0.90 to 5.43 in patients with cystic acne. Special Patient Populations Pediatric Patients The pharmacokinetics of isotretinoin were evaluated after single and multiple doses in 38 pediatric patients (12 to 15 years) and 19 adult patients (≥18 years) who received Accutane for the treatment of severe recalcitrant nodular acne. In both age groups, 4-oxo-isotretinoin was the major metabolite; tretinoin and 4-oxo-tretinoin were also observed. The dose-normalized pharmacokinetic parameters for isotretinoin following single and multiple doses are summarized in Table 3 for pediatric patients. There were no statistically significant differences in the pharmacokinetics of isotretinoin between pediatric and adult patients. 4 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 5 Table 3 Pharmacokinetic Parameters of Isotretinoin Following Single and Multiple Dose Administration in Pediatric Patients, 12 to 15 Years of Age Mean (± SD), N=38 Parameter Isotretinoin (Single Dose) Isotretinoin (Steady-State) Cmax (ng/mL) AUC(0-12) (ng⋅hr/mL) AUC(0-24) (ng⋅hr/mL) Tmax (hr)† Cssmin (ng/mL) T1/2 (hr) CL/F (L/hr) 573.25 (278.79) 3033.37 (1394.17) 6003.81 (2885.67) 6.00 (1.00-24.60) – – – 731.98 (361.86) 5082.00 (2184.23) – 4.00 (0-12.00) 352.32 (184.44) 15.69 (5.12) 17.96 (6.27) The single and multiple dose data in this table were obtained following a non-standardized meal that is not comparable to the high-fat meal that was used in the study in Table 2. †Median (range) In pediatric patients (12 to 15 years), the mean ± SD elimination half-lives (t1/2) of isotretinoin and 4-oxo-isotretinoin were 15.7 ± 5.1 hours and 23.1 ± 5.7 hours, respectively. The accumulation ratios of isotretinoin ranged from 0.46 to 3.65 for pediatric patients. INDICATIONS AND USAGE Severe Recalcitrant Nodular Acne Accutane is indicated for the treatment of severe recalcitrant nodular acne. Nodules are inflammatory lesions with a diameter of 5 mm or greater. The nodules may become suppurative or hemorrhagic. “Severe,” by definition,2 means “many” as opposed to “few or several” nodules. Because of significant adverse effects associated with its use, Accutane should be reserved for patients with severe nodular acne who are unresponsive to conventional therapy, including systemic antibiotics. In addition, Accutane is indicated only for those female patients who are not pregnant, because Accutane can cause severe birth defects (see Boxed CONTRAINDICATIONS AND WARNINGS). A single course of therapy for 15 to 20 weeks has been shown to result in complete and prolonged remission of disease in many patients.1,3,4 If a second course of therapy is needed, it should not be initiated until at least 8 weeks after completion of the first course, because experience has shown that patients may continue to improve while off Accutane. The optimal interval before retreatment has not been defined for patients who have not completed skeletal growth (see WARNINGS: Skeletal: Bone Mineral Density, Hyperostosis, and Premature Epiphyseal Closure). 5 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 6 CONTRAINDICATIONS Pregnancy: Category X. See Boxed CONTRAINDICATIONS AND WARNINGS. Allergic Reactions Accutane is contraindicated in patients who are hypersensitive to this medication or to any of its components. Accutane should not be given to patients who are sensitive to parabens, which are used as preservatives in the gelatin capsule (see PRECAUTIONS: Hypersensitivity). WARNINGS Psychiatric Disorders Accutane may cause depression, psychosis and, rarely, suicidal ideation, suicide attempts, suicide, and aggressive and/or violent behaviors. No mechanism of action has been established for these events (see ADVERSE REACTIONS: Psychiatric). Prescribers should read the brochure, Recognizing Psychiatric Disorders in Adolescents and Young Adults: A Guide for Prescribers of Isotretinoin. Prescribers should be alert to the warning signs of psychiatric disorders to guide patients to receive the help they need. Therefore, prior to initiation of Accutane therapy, patients and family members should be asked about any history of psychiatric disorder, and at each visit during therapy patients should be assessed for symptoms of depression, mood disturbance, psychosis, or aggression to determine if further evaluation may be necessary. Signs and symptoms of depression, as described in the brochure (“Recognizing Psychiatric Disorders in Adolescents and Young Adults”), include sad mood, hopelessness, feelings of guilt, worthlessness or helplessness, loss of pleasure or interest in activities, fatigue, difficulty concentrating, change in sleep pattern, change in weight or appetite, suicidal thoughts or attempts, restlessness, irritability, acting on dangerous impulses, and persistent physical symptoms unresponsive to treatment. Patients should stop Accutane and the patient or a family member should promptly contact their prescriber if the patient develops depression, mood disturbance, psychosis, or aggression, without waiting until the next visit. Discontinuation of Accutane therapy may be insufficient; further evaluation may be necessary. While such monitoring may be helpful, it may not detect all patients at risk. Patients may report mental health problems or family history of psychiatric disorders. These reports should be discussed with the patient and/or the patient’s family. A referral to a mental health professional may be necessary. The physician should consider whether Accutane therapy is appropriate in this setting; for some patients the risks may outweigh the benefits of Accutane therapy. Pseudotumor Cerebri Accutane use has been associated with a number of cases of pseudotumor cerebri (benign intracranial hypertension), some of which involved concomitant use of tetracyclines. Concomitant treatment with tetracyclines should therefore be avoided. Early signs and symptoms of pseudotumor cerebri include papilledema, headache, nausea and vomiting, and visual disturbances. Patients with these symptoms should be screened for papilledema and, if present, they should be told to discontinue Accutane immediately and be referred to a neurologist for further diagnosis and care (see ADVERSE REACTIONS: Neurological). 6 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 7 Serious Skin Reactions There have been post-marketing reports of erythema multiforme and severe skin reactions [eg, Stevens-Johnson syndrome (SJS), toxic epidermal necrolysis (TEN)] associated with isotretinoin use. These events may be serious and result in death, life-threatening events, hospitalization, or disability. Patients should be monitored closely for severe skin reactions, and discontinuation of Accutane should be considered if warranted. Pancreatitis Acute pancreatitis has been reported in patients with either elevated or normal serum triglyceride levels. In rare instances, fatal hemorrhagic pancreatitis has been reported. Accutane should be stopped if hypertriglyceridemia cannot be controlled at an acceptable level or if symptoms of pancreatitis occur. Lipids Elevations of serum triglycerides in excess of 800 mg/dL have been reported in patients treated with Accutane. Marked elevations of serum triglycerides were reported in approximately 25% of patients receiving Accutane in clinical trials. In addition, approximately 15% developed a decrease in high-density lipoproteins and about 7% showed an increase in cholesterol levels. In clinical trials, the effects on triglycerides, HDL, and cholesterol were reversible upon cessation of Accutane therapy. Some patients have been able to reverse triglyceride elevation by reduction in weight, restriction of dietary fat and alcohol, and reduction in dose while continuing Accutane.5 Blood lipid determinations should be performed before Accutane is given and then at intervals until the lipid response to Accutane is established, which usually occurs within 4 weeks. Especially careful consideration must be given to risk/benefit for patients who may be at high risk during Accutane therapy (patients with diabetes, obesity, increased alcohol intake, lipid metabolism disorder or familial history of lipid metabolism disorder). If Accutane therapy is instituted, more frequent checks of serum values for lipids and/or blood sugar are recommended (see PRECAUTIONS: Laboratory Tests). The cardiovascular consequences of hypertriglyceridemia associated with Accutane are unknown. Animal Studies: In rats given 8 or 32 mg/kg/day of isotretinoin (1.3 to 5.3 times the recommended clinical dose of 1.0 mg/kg/day after normalization for total body surface area) for 18 months or longer, the incidences of focal calcification, fibrosis and inflammation of the myocardium, calcification of coronary, pulmonary and mesenteric arteries, and metastatic calcification of the gastric mucosa were greater than in control rats of similar age. Focal endocardial and myocardial calcifications associated with calcification of the coronary arteries were observed in two dogs after approximately 6 to 7 months of treatment with isotretinoin at a dosage of 60 to 120 mg/kg/day (30 to 60 times the recommended clinical dose of 1.0 mg/kg/day, respectively, after normalization for total body surface area). 7 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 8 Hearing Impairment Impaired hearing has been reported in patients taking Accutane; in some cases, the hearing impairment has been reported to persist after therapy has been discontinued. Mechanism(s) and causality for this event have not been established. Patients who experience tinnitus or hearing impairment should discontinue Accutane treatment and be referred for specialized care for further evaluation (see ADVERSE REACTIONS: Special Senses). Hepatotoxicity Clinical hepatitis considered to be possibly or probably related to Accutane therapy has been reported. Additionally, mild to moderate elevations of liver enzymes have been observed in approximately 15% of individuals treated during clinical trials, some of which normalized with dosage reduction or continued administration of the drug. If normalization does not readily occur or if hepatitis is suspected during treatment with Accutane, the drug should be discontinued and the etiology further investigated. Inflammatory Bowel Disease Accutane has been associated with inflammatory bowel disease (including regional ileitis) in patients without a prior history of intestinal disorders. In some instances, symptoms have been reported to persist after Accutane treatment has been stopped. Patients experiencing abdominal pain, rectal bleeding or severe diarrhea should discontinue Accutane immediately (see ADVERSE REACTIONS: Gastrointestinal). Skeletal Bone Mineral Density Effects of multiple courses of Accutane on the developing musculoskeletal system are unknown. There is some evidence that long-term, high-dose, or multiple courses of therapy with isotretinoin have more of an effect than a single course of therapy on the musculoskeletal system. In an open-label clinical trial (N=217) of a single course of therapy with Accutane for severe recalcitrant nodular acne, bone density measurements at several skeletal sites were not significantly decreased (lumbar spine change >-4% and total hip change >-5%) or were increased in the majority of patients. One patient had a decrease in lumbar spine bone mineral density >4% based on unadjusted data. Sixteen (7.9%) patients had decreases in lumbar spine bone mineral density >4%, and all the other patients (92%) did not have significant decreases or had increases (adjusted for body mass index). Nine patients (4.5%) had a decrease in total hip bone mineral density >5% based on unadjusted data. Twenty-one (10.6%) patients had decreases in total hip bone mineral density >5%, and all the other patients (89%) did not have significant decreases or had increases (adjusted for body mass index). Follow-up studies performed in 8 of the patients with decreased bone mineral density for up to 11 months thereafter demonstrated increasing bone density in 5 patients at the lumbar spine, while the other 3 patients had lumbar spine bone density measurements below baseline values. Total hip bone mineral densities remained below baseline (range –1.6% to –7.6%) in 5 of 8 patients (62.5%). 8 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 9 In a separate open-label extension study of 10 patients, ages 13-18 years, who started a second course of Accutane 4 months after the first course, two patients showed a decrease in mean lumbar spine bone mineral density up to 3.25% (see PRECAUTIONS: Pediatric Use). Spontaneous reports of osteoporosis, osteopenia, bone fractures, and delayed healing of bone fractures have been seen in the Accutane population. While causality to Accutane has not been established, an effect cannot be ruled out. Longer term effects have not been studied. It is important that Accutane be given at the recommended doses for no longer than the recommended duration. Hyperostosis A high prevalence of skeletal hyperostosis was noted in clinical trials for disorders of keratinization with a mean dose of 2.24 mg/kg/day. Additionally, skeletal hyperostosis was noted in 6 of 8 patients in a prospective study of disorders of keratinization.6 Minimal skeletal hyperostosis and calcification of ligaments and tendons have also been observed by x-ray in prospective studies of nodular acne patients treated with a single course of therapy at recommended doses. The skeletal effects of multiple Accutane treatment courses for acne are unknown. In a clinical study of 217 pediatric patients (12 to 17 years) with severe recalcitrant nodular acne, hyperostosis was not observed after 16 to 20 weeks of treatment with approximately 1 mg/kg/day of Accutane given in two divided doses. Hyperostosis may require a longer time frame to appear. The clinical course and significance remain unknown. Premature Epiphyseal Closure There are spontaneous reports of premature epiphyseal closure in acne patients receiving recommended doses of Accutane. The effect of multiple courses of Accutane on epiphyseal closure is unknown. Vision Impairment Visual problems should be carefully monitored. All Accutane patients experiencing visual difficulties should discontinue Accutane treatment and have an ophthalmological examination (see ADVERSE REACTIONS: Special Senses). Corneal Opacities Corneal opacities have occurred in patients receiving Accutane for acne and more frequently when higher drug dosages were used in patients with disorders of keratinization. The corneal opacities that have been observed in clinical trial patients treated with Accutane have either completely resolved or were resolving at follow-up 6 to 7 weeks after discontinuation of the drug (see ADVERSE REACTIONS: Special Senses). Decreased Night Vision Decreased night vision has been reported during Accutane therapy and in some instances the event has persisted after therapy was discontinued. Because the onset in some patients was 9 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 10 sudden, patients should be advised of this potential problem and warned to be cautious when driving or operating any vehicle at night. PRECAUTIONS Accutane must only be prescribed by prescribers who are registered and activated with the iPLEDGE program. Accutane must only be dispensed by a pharmacy registered and activated with iPLEDGE, and must only be dispensed to patients who are registered and meet all the requirements of iPLEDGE. Registered and activated pharmacies must receive Accutane only from wholesalers registered with iPLEDGE. iPLEDGE program requirements for wholesalers, prescribers, and pharmacists are described below: Wholesalers: For the purpose of the iPLEDGE program, the term wholesaler refers to wholesaler, distributor, and/or chain pharmacy distributor. To distribute Accutane, wholesalers must be registered with iPLEDGE, and agree to meet all iPLEDGE requirements for wholesale distribution of isotretinoin products. Wholesalers must register with iPLEDGE by signing and returning the iPLEDGE wholesaler agreement that affirms they will comply with all iPLEDGE requirements for distribution of isotretinoin. These include: • Registering prior to distributing isotretinoin and re-registering annually thereafter • Distributing only FDA approved isotretinoin product • Only shipping isotretinoin to − wholesalers registered in the iPLEDGE program with prior written consent from the manufacturer or − pharmacies licensed in the US and registered and activated in the iPLEDGE program • Notifying the isotretinoin manufacturer (or delegate) of any non-registered and/or non activated pharmacy or unregistered wholesaler that attempts to order isotretinoin • Complying with inspection of wholesaler records for verification of compliance with the iPLEDGE program by the isotretinoin manufacturer (or delegate) • Returning to the manufacturer (or delegate) any undistributed product if registration is revoked by the manufacturer or if the wholesaler chooses to not re-register annually Prescribers: To prescribe isotretinoin, the prescriber must be registered and activated with the pregnancy risk management program iPLEDGE. Prescribers can register by signing and returning the completed registration form. Prescribers can only activate their registration by affirming that they meet requirements and will comply with all iPLEDGE requirements by attesting to the following points: • I know the risk and severity of fetal injury/birth defects from isotretinoin. 10 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 11 • I know the risk factors for unplanned pregnancy and the effective measures for avoidance of unplanned pregnancy. • I have the expertise to provide the patient with detailed pregnancy prevention counseling or I will refer her to an expert for such counseling, reimbursed by the manufacturer. • I will comply with the iPLEDGE program requirements described in the booklets entitled The Guide to Best Practices for the iPLEDGE Program and The iPLEDGE Program Prescriber Contraception Counseling Guide. • Before beginning treatment of female patients of childbearing potential with isotretinoin and on a monthly basis, the patient will be counseled to avoid pregnancy by using two forms of contraception simultaneously and continuously one month before, during, and one month after isotretinoin therapy, unless the patient commits to continuous abstinence. • I will not prescribe isotretinoin to any female patient of childbearing potential until verifying she has a negative screening pregnancy test and monthly negative CLIA-certified (Clinical Laboratory Improvement Amendment) pregnancy tests. Patients should have a pregnancy test at the completion of the entire course of isotretinoin and another pregnancy test 1 month later. • I will report any pregnancy case that I become aware of while the female patient is on isotretinoin or 1 month after the last dose to the pregnancy registry. To prescribe isotretinoin, the prescriber must access the iPLEDGE system via the internet (www.ipledgeprogram.com) or telephone (1-866-495-0654) to: 1) Register each patient in the iPLEDGE program. 2) Confirm monthly that each patient has received counseling and education. 3) For female patients of childbearing potential: • Enter patient’s two chosen forms of contraception each month. • Enter monthly result from CLIA-certified laboratory conducted pregnancy test. Isotretinoin must only be prescribed to female patients who are known not to be pregnant as confirmed by a negative CLIA-certified laboratory conducted pregnancy test. Isotretinoin must only be dispensed by a pharmacy registered and activated with the pregnancy risk management program iPLEDGE and only when the registered patient meets all the requirements of the iPLEDGE program. Meeting the requirements for a female patient of childbearing potential signifies that she: • Has been counseled and has signed a Patient Information/Informed Consent About Birth Defects (for female patients who can get pregnant) form that contains warnings about the risk of potential birth defects if the fetus is exposed to isotretinoin. The patient must sign the informed consent form before starting treatment and patient counseling must also be done at that time and on a monthly basis thereafter. 11 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 12 • Has had two negative urine or serum pregnancy tests with a sensitivity of at least 25 mIU/mL before receiving the initial isotretinoin prescription. The first test (a screening test) is obtained by the prescriber when the decision is made to pursue qualification of the patient for isotretinoin. The second pregnancy test (a confirmation test) must be done in a CLIA-certified laboratory. The interval between the 2 tests should be at least 19 days. − For patients with regular menstrual cycles, the second pregnancy test should be done during the first 5 days of the menstrual period immediately preceding the beginning of isotretinoin therapy and after the patient has used 2 forms of contraception for 1 month. − For patients with amenorrhea, irregular cycles, or using a contraceptive method that precludes withdrawal bleeding, the second pregnancy test must be done immediately preceding the beginning of isotretinoin therapy and after the patient has used 2 forms of contraception for 1 month. • Has had a negative result from a urine or serum pregnancy test in a CLIA-certified laboratory before receiving each subsequent course of isotretinoin. A pregnancy test must be repeated every month, in a CLIA-certified laboratory, prior to the female patient receiving each prescription. • Has selected and has committed to use 2 forms of effective contraception simultaneously, at least 1 of which must be a primary form, unless the patient commits to continuous abstinence from heterosexual contact, or the patient has undergone a hysterectomy or bilateral oophorectomy, or has been medically confirmed to be post-menopausal. Patients must use 2 forms of effective contraception for at least 1 month prior to initiation of isotretinoin therapy, during isotretinoin therapy, and for 1 month after discontinuing isotretinoin therapy. Counseling about contraception and behaviors associated with an increased risk of pregnancy must be repeated on a monthly basis. If the patient has unprotected heterosexual intercourse at any time 1 month before, during, or 1 month after therapy, she must: 1. Stop taking Accutane immediately, if on therapy 2. Have a pregnancy test at least 19 days after the last act of unprotected heterosexual intercourse 3. Start using 2 forms of effective contraception simultaneously again for 1 month before resuming Accutane therapy 4. Have a second pregnancy test after using 2 forms of effective contraception for 1 month as described above depending on whether she has regular menses or not. Effective forms of contraception include both primary and secondary forms of contraception: 12 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 13 Primary forms Secondary forms • tubal sterilization Barrier: • partner’s vasectomy • male latex condom with or without • intrauterine device spermicide • hormonal (combination • diaphragm with spermicide oral contraceptives, • cervical cap with spermicide transdermal patch, Other: injectables, • vaginal sponge (contains implantables, or vaginal spermicide) ring) Any birth control method can fail. There have been reports of pregnancy from female patients who have used oral contraceptives, as well as transdermal patch/injectable/implantable/vaginal ring hormonal birth control products; these pregnancies occurred while these patients were taking Accutane. These reports are more frequent for female patients who use only a single method of contraception. Therefore, it is critically important that female patients of childbearing potential use 2 effective forms of contraception simultaneously. Patients must receive written warnings about the rates of possible contraception failure (included in patient education kits). Using two forms of contraception simultaneously substantially reduces the chances that a female will become pregnant over the risk of pregnancy with either form alone. A drug interaction that decreases effectiveness of hormonal contraceptives has not been entirely ruled out for Accutane (see PRECAUTIONS: Drug Interactions). Although hormonal contraceptives are highly effective, prescribers are advised to consult the package insert of any medication administered concomitantly with hormonal contraceptives, since some medications may decrease the effectiveness of these birth control products. Patients should be prospectively cautioned not to self-medicate with the herbal supplement St. John’s Wort because a possible interaction has been suggested with hormonal contraceptives based on reports of breakthrough bleeding on oral contraceptives shortly after starting St. John’s Wort. Pregnancies have been reported by users of combined hormonal contraceptives who also used some form of St. John’s Wort. If a pregnancy does occur during isotretinoin treatment, isotretinoin must be discontinued immediately. The patient should be referred to an Obstetrician-Gynecologist experienced in reproductive toxicity for further evaluation and counseling. Any suspected fetal exposure during or 1 month after isotretinoin therapy must be reported immediately to the FDA via the MedWatch number 1-800-FDA-1088 and also to the iPLEDGE pregnancy registry at 1-866-495-0654 or via the internet (www.ipledgeprogram.com). 13 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 14 All Patients Isotretinoin is contraindicated in female patients who are pregnant. To receive isotretinoin all patients must meet all of the following conditions: • Must be registered with the iPLEDGE program by the prescriber • Must understand that severe birth defects can occur with the use of isotretinoin by female patients • Must be reliable in understanding and carrying out instructions • Must sign a Patient Information/Informed Consent (for all patients) form that contains warnings about the potential risks associated with isotretinoin • Must fill and pick up the prescription within 7 days of the date of specimen collection for the pregnancy test for female patients of childbearing potential • Must fill and pick up the prescription within 30 days of the office visit for male patients and female patients not of childbearing potential • Must not donate blood while on isotretinoin and for 1 month after treatment has ended • Must not share isotretinoin with anyone, even someone who has similar symptoms Female Patients of Childbearing Potential Isotretinoin is contraindicated in female patients who are pregnant. In addition to the requirements for all patients described above, female patients of childbearing potential must meet the following conditions: • Must NOT be pregnant or breast-feeding • Must comply with the required pregnancy testing at a CLIA-certified laboratory • Must fill and pick up the prescription within 7 days of the date of specimen collection for the pregnancy test • Must be capable of complying with the mandatory contraceptive measures required for isotretinoin therapy, or commit to continuous abstinence from heterosexual intercourse, and understand behaviors associated with an increased risk of pregnancy • Must understand that it is her responsibility to avoid pregnancy one month before, during and one month after isotretinoin therapy • Must have signed an additional Patient Information/Informed Consent About Birth Defects (for female patients who can get pregnant) form, before starting isotretinoin, that contains warnings about the risk of potential birth defects if the fetus is exposed to isotretinoin • Must access the iPLEDGE system via the internet (www.ipledgeprogram.com) or telephone (1-866-495-0654), before starting isotretinoin, on a monthly basis during therapy, and 1 14 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 15 month after the last dose to answer questions on the program requirements and to enter the patient’s two chosen forms of contraception • Must have been informed of the purpose and importance of providing information to the iPLEDGE program should she become pregnant while taking isotretinoin or within 1 month of the last dose Pharmacists: To dispense isotretinoin, pharmacies must be registered and activated with the pregnancy risk management program iPLEDGE. The Responsible Site Pharmacist must register the pharmacy by signing and returning the completed registration form. After registration, the Responsible Site Pharmacist can only activate the pharmacy registration by affirming that they meet requirements and will comply with all iPLEDGE requirements by attesting to the following points: • I know the risk and severity of fetal injury/birth defects from isotretinoin. • I will train all pharmacists, who participate in the filling and dispensing of isotretinoin prescriptions, on the iPLEDGE program requirements. • I will comply and seek to ensure all pharmacists who participate in the filling and dispensing of isotretinoin prescriptions comply with the iPLEDGE program requirements described in the booklet entitled Pharmacist Guide for the iPLEDGE Program. • I will obtain Accutane product only from iPLEDGE registered wholesalers. • I will not sell, buy, borrow, loan or otherwise transfer isotretinoin in any manner to or from another pharmacy. • I will return to the manufacturer (or delegate) any unused product if registration is revoked by the manufacturer or if the pharmacy chooses to not reactivate annually. • I will not fill isotretinoin for any party other than a qualified patient. To dispense isotretinoin, the pharmacist must: 1) be trained by the Responsible Site Pharmacist concerning the iPLEDGE program requirements. 2) obtain authorization from the iPLEDGE program via the internet (www.ipledgeprogram.com) or telephone (1-866-495-0654) for every isotretinoin prescription. Authorization signifies that the patient has met all program requirements and is qualified to receive isotretinoin. 3) write the Risk Management Authorization (RMA) number on the prescription. Accutane must only be dispensed: • in no more than a 30-day supply • with an Accutane Medication Guide • after authorization from the iPLEDGE program 15 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 16 • prior to the “do not dispense to patient after” date provided by the iPLEDGE system (within 30 days of the office visit for male patients and female patients not of childbearing potential and within 7 days of the date of specimen collection for female patients of childbearing potential) • with a new prescription for refills and another authorization from the iPLEDGE program (No automatic refills are allowed) An Accutane Medication Guide must be given to the patient each time Accutane is dispensed, as required by law. This Accutane Medication Guide is an important part of the risk management program for the patients. Accutane must not be prescribed, dispensed or otherwise obtained through the internet or any other means outside of the iPLEDGE program. Only FDA-approved Accutane products must be distributed, prescribed, dispensed, and used. Patients must fill Accutane prescriptions only at US licensed pharmacies. A description of the iPLEDGE program educational materials available with iPLEDGE is provided below. The main goal of these educational materials is to explain the iPLEDGE program requirements and to reinforce the educational messages. 1) The Guide to Best Practices for the iPLEDGE Program includes: isotretinoin teratogenic potential, information on pregnancy testing, and the method to complete a qualified isotretinoin prescription. 2) The iPLEDGE Program Prescriber Contraception Counseling Guide includes: specific information about effective contraception, the limitations of contraceptive methods, behaviors associated with an increased risk of contraceptive failure and pregnancy and the methods to evaluate pregnancy risk. 3) The Pharmacist Guide for the iPLEDGE Program includes: isotretinoin teratogenic potential and the method to obtain authorization to dispense an isotretinoin prescription. 4) The iPLEDGE program is a systematic approach to comprehensive patient education about their responsibilities and includes education for contraception compliance and reinforcement of educational messages. The iPLEDGE program includes information on the risks and benefits of isotretinoin which is linked to the Medication Guide dispensed by pharmacists with each isotretinoin prescription. 5) Female patients not of childbearing potential and male patients, and female patients of childbearing potential are provided with separate booklets. Each booklet contains information on isotretinoin therapy including precautions and warnings, a Patient Information/Informed Consent (for all patients) form, and a toll-free line which provides isotretinoin information in 2 languages. 6) The booklet for female patients not of childbearing potential and male patients, The iPLEDGE Program Guide to Isotretinoin for Male Patients and Female Patients Who Cannot Get Pregnant, also includes information about male reproduction and a warning not 16 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 17 to share isotretinoin with others or to donate blood during isotretinoin therapy and for 1 month following discontinuation of isotretinoin. 7) The booklet for female patients of childbearing potential, The iPLEDGE Program Guide to Isotretinoin for Female Patients Who Can Get Pregnant, includes a referral program that offers female patients free contraception counseling, reimbursed by the manufacturer, by a reproductive specialist; and a second Patient Information/Informed Consent About Birth Defects (for female patients who can get pregnant) form concerning birth defects. 8) The booklet, The iPLEDGE Program Birth Control Workbook includes information on the types of contraceptive methods, the selection and use of appropriate, effective contraception, the rates of possible contraceptive failure and a toll-free contraception counseling line. 9) In addition, there is a patient educational DVD with the following videos — “Be Prepared, Be Protected” and “Be Aware: The Risk of Pregnancy While on Isotretinoin” (see Information for Patients). General Although an effect of Accutane on bone loss is not established, physicians should use caution when prescribing Accutane to patients with a genetic predisposition for age-related osteoporosis, a history of childhood osteoporosis conditions, osteomalacia, or other disorders of bone metabolism. This would include patients diagnosed with anorexia nervosa and those who are on chronic drug therapy that causes drug-induced osteoporosis/osteomalacia and/or affects vitamin D metabolism, such as systemic corticosteroids and any anticonvulsant. Patients may be at increased risk when participating in sports with repetitive impact where the risks of spondylolisthesis with and without pars fractures and hip growth plate injuries in early and late adolescence are known. There are spontaneous reports of fractures and/or delayed healing in patients while on therapy with Accutane or following cessation of therapy with Accutane while involved in these activities. While causality to Accutane has not been established, an effect must not be ruled out. Information for Patients See PRECAUTIONS and Boxed CONTRAINDICATIONS AND WARNINGS. • Patients must be instructed to read the Medication Guide supplied as required by law when Accutane is dispensed. The complete text of the Medication Guide is reprinted at the end of this document. For additional information, patients must also be instructed to read the iPLEDGE program patient educational materials. All patients must sign the Patient Information/Informed Consent (for all patients) form. • Female patients of childbearing potential must be instructed that they must not be pregnant when Accutane therapy is initiated, and that they should use 2 forms of effective contraception simultaneously for 1 month before starting Accutane, while taking Accutane, and for 1 month after Accutane has been stopped, unless they commit to continuous abstinence from heterosexual intercourse. They should also sign a second Patient 17 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 18 Information/Informed Consent About Birth Defects (for female patients who can get pregnant) form prior to beginning Accutane therapy. They should be given an opportunity to view the patient DVD provided by the manufacturer to the prescriber. The DVD includes information about contraception, the most common reasons that contraception fails, and the importance of using 2 forms of effective contraception when taking teratogenic drugs and comprehensive information about types of potential birth defects which could occur if a female patient who is pregnant takes Accutane at any time during pregnancy. Female patients should be seen by their prescribers monthly and have a urine or serum pregnancy test, in a CLIA-certified laboratory, performed each month during treatment to confirm negative pregnancy status before another Accutane prescription is written (see Boxed CONTRAINDICATIONS AND WARNINGS and PRECAUTIONS). • Accutane is found in the semen of male patients taking Accutane, but the amount delivered to a female partner would be about 1 million times lower than an oral dose of 40 mg. While the no-effect limit for isotretinoin induced embryopathy is unknown, 20 years of postmarketing reports include 4 with isolated defects compatible with features of retinoid exposed fetuses; however 2 of these reports were incomplete, and 2 had other possible explanations for the defects observed. • Prescribers should be alert to the warning signs of psychiatric disorders to guide patients to receive the help they need. Therefore, prior to initiation of Accutane treatment, patients and family members should be asked about any history of psychiatric disorder, and at each visit during treatment patients should be assessed for symptoms of depression, mood disturbance, psychosis, or aggression to determine if further evaluation may be necessary. Signs and symptoms of depression include sad mood, hopelessness, feelings of guilt, worthlessness or helplessness, loss of pleasure or interest in activities, fatigue, difficulty concentrating, change in sleep pattern, change in weight or appetite, suicidal thoughts or attempts, restlessness, irritability, acting on dangerous impulses, and persistent physical symptoms unresponsive to treatment. Patients should stop Accutane and the patient or a family member should promptly contact their prescriber if the patient develops depression, mood disturbance, psychosis, or aggression, without waiting until the next visit. Discontinuation of Accutane treatment may be insufficient; further evaluation may be necessary. While such monitoring may be helpful, it may not detect all patients at risk. Patients may report mental health problems or family history of psychiatric disorders. These reports should be discussed with the patient and/or the patient’s family. A referral to a mental health professional may be necessary. The physician should consider whether Accutane therapy is appropriate in this setting; for some patients the risks may outweigh the benefits of Accutane therapy. • Patients must be informed that some patients, while taking Accutane or soon after stopping Accutane, have become depressed or developed other serious mental problems. Symptoms of depression include sad, “anxious” or empty mood, irritability, acting on dangerous impulses, anger, loss of pleasure or interest in social or sports activities, sleeping too much or too little, changes in weight or appetite, school or work performance going down, or trouble concentrating. Some patients taking Accutane have had thoughts about hurting themselves or putting an end to their own lives (suicidal thoughts). Some people tried to end their own 18 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 19 lives. And some people have ended their own lives. There were reports that some of these people did not appear depressed. There have been reports of patients on Accutane becoming aggressive or violent. No one knows if Accutane caused these behaviors or if they would have happened even if the person did not take Accutane. Some people have had other signs of depression while taking Accutane. • Patients must be informed that they must not share Accutane with anyone else because of the risk of birth defects and other serious adverse events. • Patients must be informed not to donate blood during therapy and for 1 month following discontinuation of the drug because the blood might be given to a pregnant female patient whose fetus must not be exposed to Accutane. • Patients should be reminded to take Accutane with a meal (see DOSAGE AND ADMINISTRATION). To decrease the risk of esophageal irritation, patients should swallow the capsules with a full glass of liquid. • Patients should be informed that transient exacerbation (flare) of acne has been seen, generally during the initial period of therapy. • Wax epilation and skin resurfacing procedures (such as dermabrasion, laser) should be avoided during Accutane therapy and for at least 6 months thereafter due to the possibility of scarring (see ADVERSE REACTIONS: Skin and Appendages). • Patients should be advised to avoid prolonged exposure to UV rays or sunlight. • Patients should be informed that they may experience decreased tolerance to contact lenses during and after therapy. • Patients should be informed that approximately 16% of patients treated with Accutane in a clinical trial developed musculoskeletal symptoms (including arthralgia) during treatment. In general, these symptoms were mild to moderate, but occasionally required discontinuation of the drug. Transient pain in the chest has been reported less frequently. In the clinical trial, these symptoms generally cleared rapidly after discontinuation of Accutane, but in some cases persisted (see ADVERSE REACTIONS: Musculoskeletal). There have been rare postmarketing reports of rhabdomyolysis, some associated with strenuous physical activity (see Laboratory Tests: CPK). • Pediatric patients and their caregivers should be informed that approximately 29% (104/358) of pediatric patients treated with Accutane developed back pain. Back pain was severe in 13.5% (14/104) of the cases and occurred at a higher frequency in female patients than male patients. Arthralgias were experienced in 22% (79/358) of pediatric patients. Arthralgias were severe in 7.6% (6/79) of patients. Appropriate evaluation of the musculoskeletal system should be done in patients who present with these symptoms during or after a course of Accutane. Consideration should be given to discontinuation of Accutane if any significant abnormality is found. 19 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 20 • Neutropenia and rare cases of agranulocytosis have been reported. Accutane should be discontinued if clinically significant decreases in white cell counts occur. • Patients should be advised that severe skin reactions (Stevens-Johnson syndrome and toxic epidermal necrolysis) have been reported in post-marketing data. Accutane should be discontinued if clinically significant skin reactions occur. Hypersensitivity Anaphylactic reactions and other allergic reactions have been reported. Cutaneous allergic reactions and serious cases of allergic vasculitis, often with purpura (bruises and red patches) of the extremities and extracutaneous involvement (including renal) have been reported. Severe allergic reaction necessitates discontinuation of therapy and appropriate medical management. Drug Interactions • Vitamin A: Because of the relationship of Accutane to vitamin A, patients should be advised against taking vitamin supplements containing vitamin A to avoid additive toxic effects. • Tetracyclines: Concomitant treatment with Accutane and tetracyclines should be avoided because Accutane use has been associated with a number of cases of pseudotumor cerebri (benign intracranial hypertension), some of which involved concomitant use of tetracyclines. • Micro-dosed Progesterone Preparations: Micro-dosed progesterone preparations (“minipills” that do not contain an estrogen) may be an inadequate method of contraception during Accutane therapy. Although other hormonal contraceptives are highly effective, there have been reports of pregnancy from female patients who have used combined oral contraceptives, as well as transdermal patch/injectable/implantable/vaginal ring hormonal birth control products. These reports are more frequent for female patients who use only a single method of contraception. It is not known if hormonal contraceptives differ in their effectiveness when used with Accutane. Therefore, it is critically important for female patients of childbearing potential to select and commit to use 2 forms of effective contraception simultaneously, at least 1 of which must be a primary form (see PRECAUTIONS). • Norethindrone/ethinyl estradiol: In a study of 31 premenopausal female patients with severe recalcitrant nodular acne receiving OrthoNovum® 7/7/7 Tablets as an oral contraceptive agent, Accutane at the recommended dose of 1 mg/kg/day, did not induce clinically relevant changes in the pharmacokinetics of ethinyl estradiol and norethindrone and in the serum levels of progesterone, follicle-stimulating hormone (FSH) and luteinizing hormone (LH). Prescribers are advised to consult the package insert of medication administered concomitantly with hormonal contraceptives, since some medications may decrease the effectiveness of these birth control products. • St. John’s Wort: Accutane use is associated with depression in some patients (see WARNINGS: Psychiatric Disorders and ADVERSE REACTIONS: Psychiatric). Patients should be prospectively cautioned not to self-medicate with the herbal supplement St. John’s Wort because a possible interaction has been suggested with hormonal contraceptives based on reports of breakthrough bleeding on oral contraceptives shortly after 20 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 21 starting St. John's Wort. Pregnancies have been reported by users of combined hormonal contraceptives who also used some form of St. John's Wort. • Phenytoin: Accutane has not been shown to alter the pharmacokinetics of phenytoin in a study in seven healthy volunteers. These results are consistent with the in vitro finding that neither isotretinoin nor its metabolites induce or inhibit the activity of the CYP 2C9 human hepatic P450 enzyme. Phenytoin is known to cause osteomalacia. No formal clinical studies have been conducted to assess if there is an interactive effect on bone loss between phenytoin and Accutane. Therefore, caution should be exercised when using these drugs together. • Systemic Corticosteroids: Systemic corticosteroids are known to cause osteoporosis. No formal clinical studies have been conducted to assess if there is an interactive effect on bone loss between systemic corticosteroids and Accutane. Therefore, caution should be exercised when using these drugs together. Laboratory Tests • Pregnancy Test: − Female patients of childbearing potential must have had two negative urine or serum pregnancy tests with a sensitivity of at least 25 mIU/mL before receiving the initial Accutane prescription. The first test (a screening test) is obtained by the prescriber when the decision is made to pursue qualification of the patient for Accutane. The second pregnancy test (a confirmation test) must be done in a CLIA-certified laboratory. The interval between the two tests must be at least 19 days. − For patients with regular menstrual cycles, the second pregnancy test must be done during the first 5 days of the menstrual period immediately preceding the beginning of Accutane therapy and after the patient has used 2 forms of contraception for 1 month. − For patients with amenorrhea, irregular cycles, or using a contraceptive method that precludes withdrawal bleeding, the second pregnancy test must be done immediately preceding the beginning of Accutane therapy and after the patient has used 2 forms of contraception for 1 month. − Each month of therapy, patients must have a negative result from a urine or serum pregnancy test. A pregnancy test must be repeated each month, in a CLIA-certified laboratory, prior to the female patient receiving each prescription. • Lipids: Pretreatment and follow-up blood lipids should be obtained under fasting conditions. After consumption of alcohol, at least 36 hours should elapse before these determinations are made. It is recommended that these tests be performed at weekly or biweekly intervals until the lipid response to Accutane is established. The incidence of hypertriglyceridemia is 1 patient in 4 on Accutane therapy (see WARNINGS: Lipids). • Liver Function Tests: Since elevations of liver enzymes have been observed during clinical trials, and hepatitis has been reported, pretreatment and follow-up liver function tests should be performed at weekly or biweekly intervals until the response to Accutane has been established (see WARNINGS: Hepatotoxicity). 21 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 22 • Glucose: Some patients receiving Accutane have experienced problems in the control of their blood sugar. In addition, new cases of diabetes have been diagnosed during Accutane therapy, although no causal relationship has been established. • CPK: Some patients undergoing vigorous physical activity while on Accutane therapy have experienced elevated CPK levels; however, the clinical significance is unknown. There have been rare postmarketing reports of rhabdomyolysis, some associated with strenuous physical activity. In a clinical trial of 217 pediatric patients (12 to 17 years) with severe recalcitrant nodular acne, transient elevations in CPK were observed in 12% of patients, including those undergoing strenuous physical activity in association with reported musculoskeletal adverse events such as back pain, arthralgia, limb injury, or muscle sprain. In these patients, approximately half of the CPK elevations returned to normal within 2 weeks and half returned to normal within 4 weeks. No cases of rhabdomyolysis were reported in this trial. Carcinogenesis, Mutagenesis and Impairment of Fertility In male and female Fischer 344 rats given oral isotretinoin at dosages of 8 or 32 mg/kg/day (1.3 to 5.3 times the recommended clinical dose of 1.0 mg/kg/day, respectively, after normalization for total body surface area) for greater than 18 months, there was a dose-related increased incidence of pheochromocytoma relative to controls. The incidence of adrenal medullary hyperplasia was also increased at the higher dosage in both sexes. The relatively high level of spontaneous pheochromocytomas occurring in the male Fischer 344 rat makes it an equivocal model for study of this tumor; therefore, the relevance of this tumor to the human population is uncertain. The Ames test was conducted with isotretinoin in two laboratories. The results of the tests in one laboratory were negative while in the second laboratory a weakly positive response (less than 1.6 x background) was noted in S. typhimurium TA100 when the assay was conducted with metabolic activation. No dose-response effect was seen and all other strains were negative. Additionally, other tests designed to assess genotoxicity (Chinese hamster cell assay, mouse micronucleus test, S. cerevisiae D7 assay, in vitro clastogenesis assay with human-derived lymphocytes, and unscheduled DNA synthesis assay) were all negative. In rats, no adverse effects on gonadal function, fertility, conception rate, gestation or parturition were observed at oral dosages of isotretinoin of 2, 8, or 32 mg/kg/day (0.3, 1.3, or 5.3 times the recommended clinical dose of 1.0 mg/kg/day, respectively, after normalization for total body surface area). In dogs, testicular atrophy was noted after treatment with oral isotretinoin for approximately 30 weeks at dosages of 20 or 60 mg/kg/day (10 or 30 times the recommended clinical dose of 1.0 mg/kg/day, respectively, after normalization for total body surface area). In general, there was microscopic evidence for appreciable depression of spermatogenesis but some sperm were observed in all testes examined and in no instance were completely atrophic tubules seen. In studies of 66 men, 30 of whom were patients with nodular acne under treatment with oral isotretinoin, no significant changes were noted in the count or motility of spermatozoa in the ejaculate. In a study of 50 men (ages 17 to 32 years) receiving Accutane (isotretinoin) therapy for nodular acne, no significant effects were seen on ejaculate volume, sperm count, total sperm motility, morphology or seminal plasma fructose. 22 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 23 Pregnancy: Category X. See Boxed CONTRAINDICATIONS AND WARNINGS. Nursing Mothers It is not known whether this drug is excreted in human milk. Because of the potential for adverse effects, nursing mothers should not receive Accutane. Pediatric Use The use of Accutane in pediatric patients less than 12 years of age has not been studied. The use of Accutane for the treatment of severe recalcitrant nodular acne in pediatric patients ages 12 to 17 years should be given careful consideration, especially for those patients where a known metabolic or structural bone disease exists (see PRECAUTIONS: General). Use of Accutane in this age group for severe recalcitrant nodular acne is supported by evidence from a clinical study comparing 103 pediatric patients (13 to 17 years) to 197 adult patients (≥18 years). Results from this study demonstrated that Accutane, at a dose of 1 mg/kg/day given in two divided doses, was equally effective in treating severe recalcitrant nodular acne in both pediatric and adult patients. In studies with Accutane, adverse reactions reported in pediatric patients were similar to those described in adults except for the increased incidence of back pain and arthralgia (both of which were sometimes severe) and myalgia in pediatric patients (see ADVERSE REACTIONS). In an open-label clinical trial (N=217) of a single course of therapy with Accutane for severe recalcitrant nodular acne, bone density measurements at several skeletal sites were not significantly decreased (lumbar spine change >-4% and total hip change >-5%) or were increased in the majority of patients. One patient had a decrease in lumbar spine bone mineral density >4% based on unadjusted data. Sixteen (7.9%) patients had decreases in lumbar spine bone mineral density >4%, and all the other patients (92%) did not have significant decreases or had increases (adjusted for body mass index). Nine patients (4.5%) had a decrease in total hip bone mineral density >5% based on unadjusted data. Twenty-one (10.6%) patients had decreases in total hip bone mineral density >5%, and all the other patients (89%) did not have significant decreases or had increases (adjusted for body mass index). Follow-up studies performed in 8 of the patients with decreased bone mineral density for up to 11 months thereafter demonstrated increasing bone density in 5 patients at the lumbar spine, while the other 3 patients had lumbar spine bone density measurements below baseline values. Total hip bone mineral densities remained below baseline (range −1.6% to −7.6%) in 5 of 8 patients (62.5%). In a separate open-label extension study of 10 patients, ages 13 to 18 years, who started a second course of Accutane 4 months after the first course, two patients showed a decrease in mean lumbar spine bone mineral density up to 3.25% (see WARNINGS: Skeletal: Bone Mineral Density). Geriatric Use Clinical studies of isotretinoin did not include sufficient numbers of subjects aged 65 years and over to determine whether they respond differently from younger subjects. Although reported 23 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 24 clinical experience has not identified differences in responses between elderly and younger patients, effects of aging might be expected to increase some risks associated with isotretinoin therapy (see WARNINGS and PRECAUTIONS). ADVERSE REACTIONS Clinical Trials and Postmarketing Surveillance The adverse reactions listed below reflect the experience from investigational studies of Accutane, and the postmarketing experience. The relationship of some of these events to Accutane therapy is unknown. Many of the side effects and adverse reactions seen in patients receiving Accutane are similar to those described in patients taking very high doses of vitamin A (dryness of the skin and mucous membranes, eg, of the lips, nasal passage, and eyes). Dose Relationship Cheilitis and hypertriglyceridemia are usually dose related. Most adverse reactions reported in clinical trials were reversible when therapy was discontinued; however, some persisted after cessation of therapy (see WARNINGS and ADVERSE REACTIONS). Body as a Whole allergic reactions, including vasculitis, systemic hypersensitivity (see PRECAUTIONS: Hypersensitivity), edema, fatigue, lymphadenopathy, weight loss Cardiovascular palpitation, tachycardia, vascular thrombotic disease, stroke Endocrine/Metabolic hypertriglyceridemia (see WARNINGS: Lipids), alterations in blood sugar levels (see PRECAUTIONS: Laboratory Tests) Gastrointestinal inflammatory bowel disease (see WARNINGS: Inflammatory Bowel Disease), hepatitis (see WARNINGS: Hepatotoxicity), pancreatitis (see WARNINGS: Lipids), bleeding and inflammation of the gums, colitis, esophagitis/esophageal ulceration, ileitis, nausea, other nonspecific gastrointestinal symptoms Hematologic allergic reactions (see PRECAUTIONS: Hypersensitivity), anemia, thrombocytopenia, neutropenia, rare reports of agranulocytosis (see PRECAUTIONS: Information for Patients). See PRECAUTIONS: Laboratory Tests for other hematological parameters. Musculoskeletal skeletal hyperostosis, calcification of tendons and ligaments, premature epiphyseal closure, decreases in bone mineral density (see WARNINGS: Skeletal), musculoskeletal symptoms 24 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 25 (sometimes severe) including back pain, myalgia, and arthralgia (see PRECAUTIONS: Information for Patients), transient pain in the chest (see PRECAUTIONS: Information for Patients), arthritis, tendonitis, other types of bone abnormalities, elevations of CPK/rare reports of rhabdomyolysis (see PRECAUTIONS: Laboratory Tests). Neurological pseudotumor cerebri (see WARNINGS: Pseudotumor Cerebri), dizziness, drowsiness, headache, insomnia, lethargy, malaise, nervousness, paresthesias, seizures, stroke, syncope, weakness Psychiatric suicidal ideation, suicide attempts, suicide, depression, psychosis, aggression, violent behaviors (see WARNINGS: Psychiatric Disorders), emotional instability Of the patients reporting depression, some reported that the depression subsided with discontinuation of therapy and recurred with reinstitution of therapy. Reproductive System abnormal menses Respiratory bronchospasms (with or without a history of asthma), respiratory infection, voice alteration Skin and Appendages acne fulminans, alopecia (which in some cases persists), bruising, cheilitis (dry lips), dry mouth, dry nose, dry skin, epistaxis, eruptive xanthomas,7 erythema multiforme, flushing, fragility of skin, hair abnormalities, hirsutism, hyperpigmentation and hypopigmentation, infections (including disseminated herpes simplex), nail dystrophy, paronychia, peeling of palms and soles, photoallergic/photosensitizing reactions, pruritus, pyogenic granuloma, rash (including facial erythema, seborrhea, and eczema), Stevens-Johnson syndrome, sunburn susceptibility increased, sweating, toxic epidermal necrolysis, urticaria, vasculitis (including Wegener’s granulomatosis; see PRECAUTIONS: Hypersensitivity), abnormal wound healing (delayed healing or exuberant granulation tissue with crusting; see PRECAUTIONS: Information for Patients) Special Senses Hearing hearing impairment (see WARNINGS: Hearing Impairment), tinnitus. Vision corneal opacities (see WARNINGS: Corneal Opacities), decreased night vision which may persist (see WARNINGS: Decreased Night Vision), cataracts, color vision disorder, conjunctivitis, dry eyes, eyelid inflammation, keratitis, optic neuritis, photophobia, visual disturbances 25 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 26 Urinary System glomerulonephritis (see PRECAUTIONS: Hypersensitivity), nonspecific urogenital findings (see PRECAUTIONS: Laboratory Tests for other urological parameters) Laboratory Elevation of plasma triglycerides (see WARNINGS: Lipids), decrease in serum high-density lipoprotein (HDL) levels, elevations of serum cholesterol during treatment Increased alkaline phosphatase, SGOT (AST), SGPT (ALT), GGTP or LDH (see WARNINGS: Hepatotoxicity) Elevation of fasting blood sugar, elevations of CPK (see PRECAUTIONS: Laboratory Tests), hyperuricemia Decreases in red blood cell parameters, decreases in white blood cell counts (including severe neutropenia and rare reports of agranulocytosis; see PRECAUTIONS: Information for Patients), elevated sedimentation rates, elevated platelet counts, thrombocytopenia White cells in the urine, proteinuria, microscopic or gross hematuria OVERDOSAGE The oral LD50 of isotretinoin is greater than 4000 mg/kg in rats and mice (>600 times the recommended clinical dose of 1.0 mg/kg/day after normalization of the rat dose for total body surface area and >300 times the recommended clinical dose of 1.0 mg/kg/day after normalization of the mouse dose for total body surface area) and is approximately 1960 mg/kg in rabbits (653 times the recommended clinical dose of 1.0 mg/kg/day after normalization for total body surface area). In humans, overdosage has been associated with vomiting, facial flushing, cheilosis, abdominal pain, headache, dizziness, and ataxia. These symptoms quickly resolve without apparent residual effects. Accutane causes serious birth defects at any dosage (see Boxed CONTRAINDICATIONS AND WARNINGS). Female patients of childbearing potential who present with isotretinoin overdose must be evaluated for pregnancy. Patients who are pregnant should receive counseling about the risks to the fetus, as described in the Boxed CONTRAINDICATIONS AND WARNINGS. Non-pregnant patients must be warned to avoid pregnancy for at least one month and receive contraceptive counseling as described in PRECAUTIONS. Educational materials for such patients can be obtained by calling the manufacturer. Because an overdose would be expected to result in higher levels of isotretinoin in semen than found during a normal treatment course, male patients should use a condom, or avoid reproductive sexual activity with a female patient who is or might become pregnant, for 1 month after the overdose. All patients with isotretinoin overdose should not donate blood for at least 1 month. DOSAGE AND ADMINISTRATION Accutane should be administered with a meal (see PRECAUTIONS: Information for Patients). 26 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 27 The recommended dosage range for Accutane is 0.5 to 1.0 mg/kg/day given in two divided doses with food for 15 to 20 weeks. In studies comparing 0.1, 0.5, and 1.0 mg/kg/day,8 it was found that all dosages provided initial clearing of disease, but there was a greater need for retreatment with the lower dosages. During treatment, the dose may be adjusted according to response of the disease and/or the appearance of clinical side effects — some of which may be dose related. Adult patients whose disease is very severe with scarring or is primarily manifested on the trunk may require dose adjustments up to 2.0 mg/kg/day, as tolerated. Failure to take Accutane with food will significantly decrease absorption. Before upward dose adjustments are made, the patients should be questioned about their compliance with food instructions. The safety of once daily dosing with Accutane has not been established. Once daily dosing is not recommended. If the total nodule count has been reduced by more than 70% prior to completing 15 to 20 weeks of treatment, the drug may be discontinued. After a period of 2 months or more off therapy, and if warranted by persistent or recurring severe nodular acne, a second course of therapy may be initiated. The optimal interval before retreatment has not been defined for patients who have not completed skeletal growth. Long-term use of Accutane, even in low doses, has not been studied, and is not recommended. It is important that Accutane be given at the recommended doses for no longer than the recommended duration. The effect of long-term use of Accutane on bone loss is unknown (see WARNINGS: Skeletal: Bone Mineral Density, Hyperostosis, and Premature Epiphyseal Closure). Contraceptive measures must be followed for any subsequent course of therapy (see PRECAUTIONS). Table 4 Accutane Dosing by Body Weight (Based on Administration With Food) Body Weight Total mg/day kilograms pounds 0.5 mg/kg 1 mg/kg 2 mg/kg 40 88 50 110 60 132 70 154 80 176 90 198 100 220 20 25 30 35 40 45 50 40 50 60 70 80 90 100 80 100 120 140 160 180 200 See DOSAGE AND ADMINISTRATION: the recommended dosage range is 0.5 to 1.0 mg/kg/day. INFORMATION FOR PHARMACISTS Access the iPLEDGE system via the internet (www.ipledgeprogram.com) or telephone (1-866 495-0654) to obtain an authorization and the “do not dispense to patient after” date. Accutane must only be dispensed in no more than a 30-day supply. 27 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 28 REFILLS REQUIRE A NEW PRESCRIPTION AND A NEW AUTHORIZATION FROM THE iPLEDGE SYSTEM. An Accutane Medication Guide must be given to the patient each time Accutane is dispensed, as required by law. This Accutane Medication Guide is an important part of the risk management program for the patient. HOW SUPPLIED Soft gelatin capsules, 10 mg (light pink), imprinted ACCUTANE 10 ROCHE. Boxes of 100 containing 10 Prescription Paks of 10 capsules (NDC 0004-0155-49). Soft gelatin capsules, 20 mg (maroon), imprinted ACCUTANE 20 ROCHE. Boxes of 100 containing 10 Prescription Paks of 10 capsules (NDC 0004-0169-49). Soft gelatin capsules, 40 mg (yellow), imprinted ACCUTANE 40 ROCHE. Boxes of 100 containing 10 Prescription Paks of 10 capsules (NDC 0004-0156-49). Storage Store at controlled room temperature (59° to 86°F, 15° to 30°C). Protect from light. REFERENCES 1. Peck GL, Olsen TG, Yoder FW, et al. Prolonged remissions of cystic and conglobate acne with 13-cis-retinoic acid. N Engl J Med 300:329-333, 1979. 2. Pochi PE, Shalita AR, Strauss JS, Webster SB. Report of the consensus conference on acne classification. J Am Acad Dermatol 24:495-500, 1991. 3. Farrell LN, Strauss JS, Stranieri AM. The treatment of severe cystic acne with 13-cis-retinoic acid: evaluation of sebum production and the clinical response in a multiple-dose trial. J Am Acad Dermatol 3:602-611, 1980. 4. Jones H, Blanc D, Cunliffe WJ. 13-cis retinoic acid and acne. Lancet 2:1048-1049, 1980. 5. Katz RA, Jorgensen H, Nigra TP. Elevation of serum triglyceride levels from oral isotretinoin in disorders of keratinization. Arch Dermatol 116:1369-1372, 1980. 6. Ellis CN, Madison KC, Pennes DR, Martel W, Voorhees JJ. Isotretinoin therapy is associated with early skeletal radiographic changes. J Am Acad Dermatol 10:1024-1029, 1984. 7. Dicken CH, Connolly SM. Eruptive xanthomas associated with isotretinoin (13-cis-retinoic acid). Arch Dermatol 116:951-952, 1980. 8. Strauss JS, Rapini RP, Shalita AR, et al. Isotretinoin therapy for acne: results of a multicenter dose-response study. J Am Acad Dermatol 10:490-496, 1984. OrthoNovum 7/7/7 is a registered trademark of Ortho-McNeil Pharmaceutical, Inc. Patient Information/Informed Consent About Birth Defects (for female patients who can get pregnant) To be completed by the patient (and her parent or guardian if patient is under age 18) and signed by her doctor. 28 This label may not be the latest approved by FDA. For current labeling information, please visit ______________ NDA 018662/S-060 Page 29 Read each item below and initial in the space provided to show that you understand each item and agree to follow your doctor's instructions. Do not sign this consent and do not take isotretinoin if there is anything that you do not understand. A parent or guardian of a minor patient (under age 18) must also read and initial each item before signing the consent. (Patient’s Name) 1. I understand that there is a very high chance that my unborn baby could have severe birth defects if I am pregnant or become pregnant while taking isotretinoin. This can happen with any amount and even if taken for short periods of time. This is why I must not be pregnant while taking isotretinoin. Initial: __ 2. I understand that I must not get pregnant 1 month before, during the entire time of my treatment, and for 1 month after the end of my treatment with isotretinoin. Initial: _ 3. I understand that I must avoid sexual intercourse completely, or I must use 2 separate, effective forms of birth control (contraception) at the same time. The only exceptions are if I have had surgery to remove the uterus (a hysterectomy) or both of my ovaries (bilateral oophorectomy), or my doctor has medically confirmed that I am post-menopausal. Initial: 4. I understand that hormonal birth control products are among the most effective forms of birth control. Combination birth control pills and other hormonal products include skin patches, shots, under-the-skin implants, vaginal rings, and intrauterine devices (IUDs). Any form of birth control can fail. That is why I must use 2 different birth control methods at the same time, starting 1 month before, during, and for 1 month after stopping therapy every time I have sexual intercourse, even if 1 of the methods I choose is hormonal birth control. Initial: 5. I understand that the following are effective forms of birth control: 29 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 30 Primary forms Secondary forms • tubal sterilization (tying Barrier: my tubes) • male latex condom with or without • partner’s vasectomy spermicide • intrauterine device • diaphragm with spermicide • hormonal (combination • cervical cap with spermicide birth control pills, skin Other: patches, shots, under-• vaginal sponge (contains the-skin implants, or spermicide) vaginal ring) A diaphragm and cervical cap must each be used with spermicide, a special cream that kills sperm I understand that at least 1 of my 2 forms of birth control must be a primary method. Initial: 6. I will talk with my doctor about any medicines including herbal products I plan to take during my isotretinoin treatment because hormonal birth control methods may not work if I am taking certain medicines or herbal products. Initial: 7. I may receive a free birth control counseling session from a doctor or other family planning expert. My isotretinoin doctor can give me an isotretinoin Patient Referral Form for this free consultation. Initial: 8. I must begin using the birth control methods I have chosen as described above at least 1 month before I start taking isotretinoin. Initial: 9. I cannot get my first prescription for isotretinoin unless my doctor has told me that I have 2 negative pregnancy test results. The first pregnancy test should be done when my doctor decides to prescribe isotretinoin. The second pregnancy test must be done in a lab during the first 5 days of my menstrual period right before starting isotretinoin therapy treatment, or as instructed by my doctor. I will then have 1 pregnancy test; in a lab. • every month during treatment • at the end of treatment • and 1 month after stopping treatment 30 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 31 I must not start taking isotretinoin until I am sure that I am not pregnant, have negative results from 2 pregnancy tests, and the second test has been done in a lab. Initial: 10. I have read and understand the materials my doctor has given to me, including The iPLEDGE Program Guide for Isotretinoin for Female Patients Who Can Get Pregnant, The iPLEDGE Birth Control Workbook and The iPLEDGE Program Patient Introductory Brochure. My doctor gave me and asked me to watch the DVD containing a video about birth control and a video about birth defects and isotretinoin. I was told about a private counseling line that I may call for more information about birth control. I have received information on emergency birth control. Initial: 11. I must stop taking isotretinoin right away and call my doctor if I get pregnant, miss my expected menstrual period, stop using birth control, or have sexual intercourse without using my 2 birth control methods at any time. Initial: 12. My doctor gave me information about the purpose and importance of providing information to the iPLEDGE program should I become pregnant while taking isotretinoin or within 1 month of the last dose. I also understand that if I become pregnant, information about my pregnancy, my health, and my baby’s health may be shared with the maker of isotretinoin and their authorized parties who maintain the iPLEDGE program and government health regulatory authorities. Initial: 13. I understand that being qualified to receive isotretinoin in the iPLEDGE program means that I: • have had 2 negative urine or blood pregnancy tests before receiving the first isotretinoin prescription. The second test must be done in a lab. I must have a negative result from a urine or blood pregnancy test done in a lab repeated each month before I receive another isotretinoin prescription. • have chosen and agreed to use 2 forms of effective birth control at the same time. At least 1 method must be a primary form of birth control, unless I have chosen never to have sexual contact with a male (abstinence), or I have undergone a hysterectomy. I must use 2 forms of birth control for at least 1 month before I start isotretinoin therapy, during therapy, and for 1 month after stopping therapy. I must receive counseling, repeated on a 31 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 32 monthly basis, about birth control and behaviors associated with an increased risk of pregnancy. • have signed a Patient Information/Informed Consent About Birth Defects (for female patients who can get pregnant) that contains warnings about the chance of possible birth defects if I am pregnant or become pregnant and my unborn baby is exposed to isotretinoin. • have been informed of and understand the purpose and importance of providing information to the iPLEDGE program should I become pregnant while taking isotretinoin or within 1 month of the last dose. • have interacted with the iPLEDGE program before starting isotretinoin and on a monthly basis to answer questions on the program requirements and to enter my two chosen forms of birth control. Initial: My doctor has answered all my questions about isotretinoin and I understand that it is my responsibility not to get pregnant 1 month before, during isotretinoin treatment, or for 1 month after I stop taking isotretinoin. Initial: I now authorize my doctor ___ to begin my treatment with isotretinoin. Patient Signature:_____ Date: __ Parent/Guardian Signature (if under age 18):__ Date:__ Please print: Patient Name and Address____ ____ Telephone ___ I have fully explained to the patient, ___, the nature and purpose of the treatment described above and the risks to female patients of childbearing potential. I have asked the patient if she has any questions regarding her treatment with isotretinoin and have answered those questions to the best of my ability. Doctor Signature: ____ Date: PLACE THE ORIGINAL SIGNED DOCUMENTS IN THE PATIENT’S MEDICAL RECORD. PLEASE PROVIDE A COPY TO THE PATIENT. 32 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 33 Patient Information/Informed Consent (for all patients): To be completed by patient (and parent or guardian if patient is under age 18) and signed by the doctor. Read each item below and initial in the space provided if you understand each item and agree to follow your doctor’s instructions. A parent or guardian of a patient under age 18 must also read and understand each item before signing the agreement. Do not sign this agreement and do not take isotretinoin if there is anything that you do not understand about all the information you have received about using isotretinoin. 1. I, _________, (Patient’s Name) understand that isotretinoin is a medicine used to treat severe nodular acne that cannot be cleared up by any other acne treatments, including antibiotics. In severe nodular acne, many red, swollen, tender lumps form in the skin. If untreated, severe nodular acne can lead to permanent scars. Initials: __ 2. My doctor has told me about my choices for treating my acne. Initials: _ 3. I understand that there are serious side effects that may happen while I am taking isotretinoin. These have been explained to me. These side effects include serious birth defects in babies of pregnant patients. [Note: There is a second Patient Information/Informed Consent About Birth Defects (for female patients who can get pregnant)]. Initials: 4. I understand that some patients, while taking isotretinoin or soon after stopping isotretinoin, have become depressed or developed other serious mental problems. Symptoms of depression include sad, “anxious” or empty mood, irritability, acting on dangerous impulses, anger, loss of pleasure or interest in social or sports activities, sleeping too much or too little, changes in weight or appetite, school or work performance going down, or trouble concentrating. Some patients taking isotretinoin have had thoughts about hurting themselves or putting an end to their own lives (suicidal thoughts). Some people tried to end their own lives. And some people have ended their own lives. There were reports that some of these people did not appear depressed. There have been reports of patients on isotretinoin becoming aggressive or violent. No one knows if isotretinoin caused these behaviors or if they would have happened even if the person did not take isotretinoin. Some people have had other signs of depression while taking isotretinoin (see #7 below). Initials: 33 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 34 5. Before I start taking isotretinoin, I agree to tell my doctor if I have ever had symptoms of depression (see #7 below), been psychotic, attempted suicide, had any other mental problems, or take medicine for any of these problems. Being psychotic means having a loss of contact with reality, such as hearing voices or seeing things that are not there. Initials: 6. Before I start taking isotretinoin, I agree to tell my doctor if, to the best of my knowledge, anyone in my family has ever had symptoms of depression, been psychotic, attempted suicide, or had any other serious mental problems. Initials: 7. Once I start taking isotretinoin, I agree to stop using isotretinoin and tell my doctor right away if any of the following signs and symptoms of depression or psychosis happen. I: • Start to feel sad or have crying spells • Lose interest in activities I once enjoyed • Sleep too much or have trouble sleeping • Become more irritable, angry, or aggressive than usual (for example, temper outbursts, thoughts of violence) • Have a change in my appetite or body weight • Have trouble concentrating • Withdraw from my friends or family • Feel like I have no energy • Have feelings of worthlessness or guilt • Start having thoughts about hurting myself or taking my own life (suicidal thoughts) • Start acting on dangerous impulses • Start seeing or hearing things that are not real Initials: 8. I agree to return to see my doctor every month I take isotretinoin to get a new prescription for isotretinoin, to check my progress, and to check for signs of side effects. Initials: 9. Isotretinoin will be prescribed just for me — I will not share isotretinoin with other people because it may cause serious side effects, including birth defects. Initials: 10. I will not give blood while taking isotretinoin or for 1 month after I stop taking isotretinoin. I understand that if someone who is pregnant gets my donated blood, her baby may be exposed to isotretinoin and may be born with serious birth defects. Initials: 34 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 35 11. I have read The iPLEDGE Program Patient Introductory Brochure, and other materials my provider gave me containing important safety information about isotretinoin. I understand all the information I received. Initials: 12. My doctor and I have decided I should take isotretinoin. I understand that I must be qualified in the iPLEDGE program to have my prescription filled each month. I understand that I can stop taking isotretinoin at any time. I agree to tell my doctor if I stop taking isotretinoin. Initials: I now allow my doctor ___ to begin my treatment with isotretinoin. Patient Signature: _____ Date: _ Parent/Guardian Signature (if under age 18): __ Date: _ Patient Name (print) _____ Patient Address ___ Telephone (..) _____ I have: • fully explained to the patient, ___, the nature and purpose of isotretinoin treatment, including its benefits and risks • given the patient the appropriate educational materials, The iPLEDGE Program Patient Introductory Brochure and asked the patient if he/she has any questions regarding his/her treatment with isotretinoin • answered those questions to the best of my ability Doctor Signature: ______ Date: _ PLACE THE ORIGINAL SIGNED DOCUMENTS IN THE PATIENT’S MEDICAL RECORD. PLEASE PROVIDE A COPY TO THE PATIENT. MEDICATION GUIDE ACCUTANE (ACK-u-tane) (isotretinoin capsules) Read the Medication Guide that comes with Accutane before you start taking it and each time you get a prescription. There may be new information. This information does not take the place of talking with your doctor about your medical condition or your treatment. 35 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 36 What is the most important information I should know about Accutane? • Accutane is used to treat a type of severe acne (nodular acne) that has not been helped by other treatments, including antibiotics. • Because Accutane can cause birth defects, Accutane is only for patients who can understand and agree to carry out all of the instructions in the iPLEDGE program. • Accutane may cause serious mental health problems. 1. Birth defects (deformed babies), loss of a baby before birth (miscarriage), death of the baby, and early (premature) births. Female patients who are pregnant or who plan to become pregnant must not take Accutane. Female patients must not get pregnant: • for 1 month before starting Accutane • while taking Accutane • for 1 month after stopping Accutane. If you get pregnant while taking Accutane, stop taking it right away and call your doctor. Doctors and patients should report all cases of pregnancy to: • FDA MedWatch at 1-800-FDA-1088, and • the iPLEDGE pregnancy registry at 1-866-495-0654 2. Serious mental health problems. Accutane may cause: • depression • psychosis (seeing or hearing things that are not real) • suicide. Some patients taking Accutane have had thoughts about hurting themselves or putting an end to their own lives (suicidal thoughts). Some people tried to end their own lives. And some people have ended their own lives. Stop Accutane and call your doctor right away if you or a family member notices that you have any of the following signs and symptoms of depression or psychosis: • start to feel sad or have crying spells • lose interest in activities you once enjoyed • sleep too much or have trouble sleeping • become more irritable, angry, or aggressive than usual (for example, temper outbursts, thoughts of violence) • have a change in your appetite or body weight • have trouble concentrating • withdraw from your friends or family • feel like you have no energy • have feelings of worthlessness or guilt • start having thoughts about hurting yourself or taking your own life (suicidal thoughts) • start acting on dangerous impulses • start seeing or hearing things that are not real 36 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 37 After stopping Accutane, you may also need follow-up mental health care if you had any of these symptoms. What is Accutane? Accutane is a medicine taken by mouth to treat the most severe form of acne (nodular acne) that cannot be cleared up by any other acne treatments, including antibiotics. Accutane can cause serious side effects (see “What is the most important information I should know about Accutane?”). Accutane can only be: • prescribed by doctors that are registered in the iPLEDGE program • dispensed by a pharmacy that is registered with the iPLEDGE program • given to patients who are registered in the iPLEDGE program and agree to do everything required in the program What is severe nodular acne? Severe nodular acne is when many red, swollen, tender lumps form in the skin. These can be the size of pencil erasers or larger. If untreated, nodular acne can lead to permanent scars. Who should not take Accutane? • Do not take Accutane if you are pregnant, plan to become pregnant, or become pregnant during Accutane treatment. Accutane causes severe birth defects. See “What is the most important information I should know about Accutane?” • Do not take Accutane if you are allergic to anything in it. Accutane contains parabens as the preservative. See the end of this Medication Guide for a complete list of ingredients in Accutane. What should I tell my doctor before taking Accutane? Tell your doctor if you or a family member has any of the following health conditions: • mental problems • asthma • liver disease • diabetes • heart disease • bone loss (osteoporosis) or weak bones • an eating problem called anorexia nervosa (where people eat too little) • food or medicine allergies Tell your doctor if you are pregnant or breastfeeding. Accutane must not be used by women who are pregnant or breastfeeding. 37 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 38 Tell your doctor about all of the medicines you take including prescription and non prescription medicines, vitamins and herbal supplements. Accutane and certain other medicines can interact with each other, sometimes causing serious side effects. Especially tell your doctor if you take: • Vitamin A supplements. Vitamin A in high doses has many of the same side effects as Accutane. Taking both together may increase your chance of getting side effects. • Tetracycline antibiotics. Tetracycline antibiotics taken with Accutane can increase the chances of getting increased pressure in the brain. • Progestin-only birth control pills (mini-pills). They may not work while you take Accutane. Ask your doctor or pharmacist if you are not sure what type you are using. • Dilantin (phenytoin). This medicine taken with Accutane may weaken your bones. • Corticosteroid medicines. These medicines taken with Accutane may weaken your bones. • St. John’s Wort. This herbal supplement may make birth control pills work less effectively. These medicines should not be used with Accutane unless your doctor tells you it is okay. Know the medicines you take. Keep a list of them to show to your doctor and pharmacist. Do not take any new medicine without talking with your doctor. How should I take Accutane? • You must take Accutane exactly as prescribed. You must also follow all the instructions of the iPLEDGE program. Before prescribing Accutane, your doctor will: • explain the iPLEDGE program to you • have you sign the Patient Information/Informed Consent (for all patients). Female patients who can get pregnant must also sign another consent form. You will not be prescribed Accutane if you cannot agree to or follow all the instructions of the iPLEDGE program. • You will get no more than a 30-day supply of Accutane at a time. This is to make sure you are following the Accutane iPLEDGE program. You should talk with your doctor each month about side effects. • The amount of Accutane you take has been specially chosen for you. It is based on your body weight, and may change during treatment. • Take Accutane 2 times a day with a meal, unless your doctor tells you otherwise. Swallow your Accutane capsules whole with a full glass of liquid. Do not chew or suck on the capsule. Accutane can hurt the tube that connects your mouth to your stomach (esophagus) if it is not swallowed whole. • If you miss a dose, just skip that dose. Do not take 2 doses at the same time. • If you take too much Accutane or overdose, call your doctor or poison control center right away. 38 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 39 • Your acne may get worse when you first start taking Accutane. This should last only a short while. Talk with your doctor if this is a problem for you. • You must return to your doctor as directed to make sure you don’t have signs of serious side effects. Your doctor may do blood tests to check for serious side effects from Accutane. Female patients who can get pregnant will get a pregnancy test each month. • Female patients who can get pregnant must agree to use 2 separate forms of effective birth control at the same time 1 month before, while taking, and for 1 month after taking Accutane. You must access the iPLEDGE system to answer questions about the program requirements and to enter your 2 chosen forms of birth control. To access the iPLEDGE system, go to www.ipledgeprogram.com or call 1-866-495-0654. You must talk about effective birth control methods with your doctor or go for a free visit to talk about birth control with another doctor or family planning expert. Your doctor can arrange this free visit, which will be paid for by the company that makes Accutane. If you have sex at any time without using 2 forms of effective birth control, get pregnant, or miss your expected period, stop using Accutane and call your doctor right away. What should I avoid while taking Accutane? • Do not get pregnant while taking Accutane and for 1 month after stopping Accutane. See “What is the most important information I should know about Accutane?” • Do not breast feed while taking Accutane and for 1 month after stopping Accutane. We do not know if Accutane can pass through your milk and harm the baby. • Do not give blood while you take Accutane and for 1 month after stopping Accutane. If someone who is pregnant gets your donated blood, her baby may be exposed to Accutane and may be born with birth defects. • Do not take other medicines or herbal products with Accutane unless you talk to your doctor. See “What should I tell my doctor before taking Accutane?” • Do not drive at night until you know if Accutane has affected your vision. Accutane may decrease your ability to see in the dark. • Do not have cosmetic procedures to smooth your skin, including waxing, dermabrasion, or laser procedures, while you are using Accutane and for at least 6 months after you stop. Accutane can increase your chance of scarring from these procedures. Check with your doctor for advice about when you can have cosmetic procedures. • Avoid sunlight and ultraviolet lights as much as possible. Tanning machines use ultraviolet lights. Accutane may make your skin more sensitive to light. 39 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 40 • Do not share Accutane with other people. It can cause birth defects and other serious health problems. What are the possible side effects of Accutane? • Accutane can cause birth defects (deformed babies), loss of a baby before birth (miscarriage), death of the baby, and early (premature) births. See “What is the most important information I should know about Accutane?” • Accutane may cause serious mental health problems. See “What is the most important information I should know about Accutane?” • serious brain problems. Accutane can increase the pressure in your brain. This can lead to permanent loss of eyesight and, in rare cases, death. Stop taking Accutane and call your doctor right away if you get any of these signs of increased brain pressure: • bad headache • blurred vision • dizziness • nausea or vomiting • seizures (convulsions) • stroke • skin problems. Skin rash can occur in patients taking Accutane. In some patients a rash can be serious. Stop using Accutane and call your doctor right away if you develop conjunctivitis (red or inflamed eyes, lik “pink eye”), a rash with a fever, blisters on legs, arms or face and/or sores in your mouth, throat, nose, eyes, or if your skin begins to peel. • stomach area (abdomen) problems. Certain symptoms may mean that your internal organs are being damaged. These organs include the liver, pancreas, bowel (intestines), and esophagus (connection between mouth and stomach). If your organs are damaged, they may not get better even after you stop taking Accutane. Stop taking Accutane and call your doctor if you get: • severe stomach, chest or bowel pain • trouble swallowing or painful swallowing • new or worsening heartburn • diarrhea • rectal bleeding • yellowing of your skin or eyes • dark urine • bone and muscle problems. Accutane may affect bones, muscles, and ligaments and cause pain in your joints or muscles. Tell your doctor if you plan hard physical activity during treatment with Accutane. Tell your doctor if you get: • back pain 40 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 41 • joint pain • broken bone. Tell all healthcare providers that you take Accutane if you break a bone. Stop Accutane and call your doctor right away if you have muscle weakness. Muscle weakness with or without pain can be a sign of serious muscle damage. Accutane may stop long bone growth in teenagers who are still growing. • hearing problems. Stop using Accutane and call your doctor if your hearing gets worse or if you have ringing in your ears. Your hearing loss may be permanent. • vision problems. Accutane may affect your ability to see in the dark. This condition usually clears up after you stop taking Accutane, but it may be permanent. Other serious eye effects can occur. Stop taking Accutane and call your doctor right away if you have any problems with your vision or dryness of the eyes that is painful or constant. If you wear contact lenses, you may have trouble wearing them while taking Accutane and after treatment. • lipid (fats and cholesterol in blood) problems. Accutane can raise the level of fats and cholesterol in your blood. This can be a serious problem. Return to your doctor for blood tests to check your lipids and to get any needed treatment. These problems usually go away when Accutane treatment is finished. • serious allergic reactions. Stop taking Accutane and get emergency care right away if you develop hives, a swollen face or mouth, or have trouble breathing. Stop taking Accutane and call your doctor if you get a fever, rash, or red patches or bruises on your legs. • blood sugar problems. Accutane may cause blood sugar problems including diabetes. Tell your doctor if you are very thirsty or urinate a lot. • decreased red and white blood cells. Call your doctor if you have trouble breathing, faint, or feel weak. • The common, less serious side effects of Accutane are dry skin, chapped lips, dry eyes, and dry nose that may lead to nosebleeds. Call your doctor if you get any side effect that bothers you or that does not go away. These are not all of the possible side effects with Accutane. Your doctor or pharmacist can give you more detailed information. Call your doctor for medical advice about side effects. You may report side effects to FDA at 1-800-FDA-1088 or Roche at 1-800-526-6367. How should I store Accutane? • Store Accutane at room temperature, between 59° and 86°F. Protect from light. • Keep Accutane and all medicines out of the reach of children. General Information about Accutane Medicines are sometimes prescribed for conditions that are not mentioned in Medication Guides. Do not use Accutane for a condition for which it was not prescribed. Do not give Accutane to other people, even if they have the same symptoms that you have. It may harm them. 41 This label may not be the latest approved by FDA. For current labeling information, please visit NDA 018662/S-060 Page 42 This Medication Guide summarizes the most important information about Accutane. If you would like more information, talk with your doctor. You can ask your doctor or pharmacist for information about Accutane that is written for health care professionals. You can also call iPLEDGE program at 1-866-495-0654 or visit www.ipledgeprogram.com. What are the ingredients in Accutane? Active Ingredient: Isotretinoin Inactive Ingredients: beeswax, butylated hydroxyanisole, edetate disodium, hydrogenated soybean oil flakes, hydrogenated vegetable oil, and soybean oil. Gelatin capsules contain glycerin and parabens (methyl and propyl), with the following dye systems: 10 mg — iron oxide (red) and titanium dioxide; 20 mg — FD&C Red No. 3, FD&C Blue No. 1, and titanium dioxide; 40 mg — FD&C Yellow No. 6, D&C Yellow No. 10, and titanium dioxide. This Medication Guide has been approved by the U.S. Food and Drug Administration. Dilantin is a registered trademark of Warner-Lambert Company LLC. Distributed by: Company logo AEC43780_PI_N(1) PI Revised: January 2010 Copyright © 2000-200x by Roche Laboratories Inc. All rights reserved. 42 This label may not be the latest approved by FDA. For current labeling information, please visit |
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7121 | https://archive.org/details/bouchersprosthod0000unse | Boucher's prosthodontic treatment for edentulous patients : Free Download, Borrow, and Streaming : Internet Archive
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Boucher's prosthodontic treatment for edentulous patients
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Boucher's prosthodontic treatment for edentulous patients
Publication date1985TopicsComplete dentures, Edentulous mouth, Denture, Complete, Mouth, EdentulousPublisher St. Louis : Mosby Collectioninternetarchivebooks; printdisabled; inlibraryContributorInternet ArchiveLanguageEnglishItem Size 1.8G
x, 646 p., p. of plates : 25 cm
Bibliography: p. 590-636
Includes index
Notes
some text are cut due to margins too tight inherent from the source.
cut off text tight binding
Access-restricted-item true Addeddate 2021-06-22 19:04:44 Associated-names Boucher, Carl C; Hickey, Judson C; Zarb, George A. (George Albert), 1938-; Bolender, Charles L Boxid IA40141410 Camera USB PTP Class Camera Collection_set printdisabled External-identifierurn:oclc:record:1280729605
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7122 | https://www.vedantu.com/question-answer/does-atmospheric-pressure-affect-boiling-point-class-12-chemistry-cbse-5ffb48e1bfdd3912f3df2724 | How does atmospheric pressure affect boiling points?
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How does atmospheric pressure ...
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How does atmospheric pressure affect boiling points?
Answer
Verified
476.7k+ views
Hint: We know that the temperature at which vapor pressure of a liquid becomes equal to the vapor pressure surrounding the liquid and the liquid changes to its vapor state is termed as boiling point of the substance.
Complete step by step answer:
Let’s first discuss atmospheric pressure in detail. Atmospheric pressure is also known by the name of air pressure. This is the force exerted by air on a surface. An instrument called a barometer is used to measure atmospheric pressure.
Let’s discuss the effect of atmospheric pressure on the boiling point. We know that boiling point is achieved when vapour pressure of liquid equals to the atmospheric pressure. That means, increasing the atmospheric pressure results in an increase of boiling point and lowering of atmospheric pressure results in lowering of boiling point of the liquid.
Let’s discuss the reason behind the above phenomenon. The above phenomenon occurs because vapor pressure of water is dependent on temperature. It rises as temperature increases and lowers as temperature drops. We can also assume atmospheric pressure as a pressure that pushes the liquid in a container back into the container (prevents the transition to the gaseous phase). So, greater the pressure means that more force is pushing back the particles down to the container. So, we need to heat the liquid more to achieve boiling point.
Note: It is to be noted that on increasing the temperature boiling point of the liquid rises because when temperature increases, energy available for liquid molecules to undergo expansion into the gaseous phase also increases.
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7123 | https://download.e-bookshelf.de/download/0000/0016/74/L-G-0000001674-0002331320.pdf | The Essence of Dielectric Waveguides C. Yeh • F. I. Shimabukuro The Essence of Dielectric Waveguides 123 C. Yeh California Advanced Studies 2432 Nalin Drive Los Angeles CA 90077 USA F. I. Shimabukuro California Advanced Studies 2432 Nalin Drive Los Angeles CA 90077 USA ISBN 978-0-387-30929-3 e-ISBN 978-0-387-49799-0 Library of Congress Control Number: 2008923746 c ⃝2008 Springer Science+Business Media, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden.
The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights.
Printed on acid-free paper 9 8 7 6 5 4 3 2 1 springer.com To Our Families Vivian, John, and Evelyn; Siblings-Dorothy, Richard, and Vicky Karen, Susan and Lee PREFACE “It is our responsibility as scientists, knowing the great progress which comes from a satisfactory philosophy of ignorance, the great progress which is the fruit of freedom of thought, to proclaim the value of this freedom, to teach how doubt is not to be feared but welcomed and discussed; and to demand this freedom as our duty to all coming generations” —— Richard Feynman, 1955 —— First, as students from Cal Tech and MIT and then as researchers and teachers from other universities and industry, we are benefited greatly from the philoso-phy of learning practiced by these and other distinguished universities in the US, namely, learn and teach the fundamentals and not the fashions. Under this guiding light, this comprehensive book was formed, covering the most important modern topics on guided waves. As such, it may be used as a research reference book or as a textbook for senior undergraduate students or first-year graduate students. The lectures for an one-semester or one-quarter course on guided waves along surface wave structures can begin with a review of EM fundamentals (Chap. 2), and then move on to a discussion on the general important and relevant characteristics of these guided surface waves (Chap. 3). Then follows the rigorous analytic treatment for canonical structures (planar, circular, and elliptical) (Chaps. 4–8). By the end of these lectures, the students would have gained a very solid theoretical foun-dation on this subject. Then the fun part starts. The students can now learn how they may make use of their fundamental knowledge to treat the many modern up-to-date applications: linear and nonlinear wave propagation in fibers, solitons in fibers and WDM beams propagation in fibers (Chaps. 9 and 10), plasmon (sub-wavelength) waves (Chap. 12), waves in periodic structures (photonic structures) (Chap. 13), surface waves on metamaterial (artificial material) and other exotic (moving medium) structures (Chap. 14). Finally, the students can now be intro-duced to the many numerical approaches (Chap. 15) that can be used on the vari-ous guided wave structures, with the comforting knowledge that they possess the necessary theoretical foundation to correctly interpret the numerical data.
Substantial amount of the material of the text appears in book form for the first time. References are given to the original sources. However, unintentional over-sight by us is unavoidable. For this the authors offer their apologies. It is curious to note that many popular references (with many citations in the literature) may not represent the papers published by the originators of the concepts. Special care has been taken by us not to follow this erroneous path. References are listed at the end of each chapter for clarity and ease of usage.
As far as nomenclatures and symbols are concerned, we have not been able to have a given symbol to represent a single unique entity throughout the whole book.
viii Preface Instead, we only make sure that a given symbol clearly and uniquely represents a single entity in that chapter. Whenever possible, universally accepted nomencla-tures are used to represent vector and scalar quantities.
It is with deep gratitude and great pleasure for us to acknowledge the sig-nificant guidance and encouragement given to us by Professors C. H. Papas, J. R. Whinnery, and R. W. Gould. We also wish to acknowledge with special thanks to Dr. Peter Siegel who introduced us to the field of terahertz research and who planted the seed for us to pursue the writing of this book. Throughout our professional careers, we benefited greatly from the many positive advice and en-couragement from our colleagues. We express our deepest thanks and gratitude to them. Finally, we express our sincerest thanks to Marshall Kwong for his dedicated professional graphic arts work for this book, without which this book would be incomplete.
We greatly appreciate the careful reading and constructive comments by the reviewers.
C. Yeh F. I. Shimabukuro Los Angeles CONTENTS 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Brief Historical Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Scope of this Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2. Fundamental Electromagnetic Field Equations . . . . . . . . . . . . . . . . . . . . . . . . 11 2.1 Maxwell Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.2 The Constitutive Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2.1 Simple Medium (Linear and Isotropic) . . . . . . . . . . . . . . . . . . . . . . . . 14 2.2.2 Anisotropic Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2.3 Left-Handed Medium (Metamaterial) . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.2.4 Conducting Medium. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16 2.2.5 Dielectric Medium with Loss. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17 2.2.6 Nonlinear Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.3 Boundary Conditions, Radiation Condition, and Edge Condition. . . . . . .20 2.3.1 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.3.2 Radiation Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3.3 Edge Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3.4 Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.4 Energy Relations: Poynting’s Vector Theorem . . . . . . . . . . . . . . . . . . . . . . . 29 2.5 Classification of Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.5.1 The Debye Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.5.2 Basic Wave Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.5.3 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 2.5.3.1 Rectangular Coordinates (x, y, z) . . . . . . . . . . . . . . . . . . . . . . 39 2.5.3.2 Circular Cylinder Coordinates (r, θ, z) . . . . . . . . . . . . . . . . . 40 2.5.3.3 Elliptical Cylinder Coordinates (ξ, η, z) . . . . . . . . . . . . . . . . 41 2.5.3.4 Parabolic Cylinder Coordinates (ξ, η, z) . . . . . . . . . . . . . . . . 42 2.6 Polarization of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.6.1 Linearly Polarized Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.6.2 Circularly Polarized Waves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .44 2.6.3 Elliptically Polarized Waves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .44 2.7 Phase Velocity and Group Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.8 The Impedance Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.9 Validity of the Scalar Wave Approach. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .47 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 x Contents 3. Propagation Characteristics of Guided Waves Along a Dielectric Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.1 Typical Surface Waveguide Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.2 Formal Approach to the Surface Waveguide Problems . . . . . . . . . . . . . . . 57 3.3 The ω-β Diagram: Dispersion Relations. . . . . . . . . . . . . . . . . . . . . . . . . . . .59 3.4 Geometrical Optics Approach. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .62 3.5 Attenuation Constant. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .65 3.5.1 Single Mode Case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .66 3.5.2 Multimode Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 3.6 Signal Dispersion and Distortion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 3.7 α and Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 3.8 Excitation of Modes on a Dielectric Waveguide . . . . . . . . . . . . . . . . . . . . . 79 3.8.1 Excitation Through Direct Incidence . . . . . . . . . . . . . . . . . . . . . . . . . 79 3.8.1.1 Incident Plane Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.8.1.2 Incident Gaussian Beam. . . . . . . . . . . . . . . . . . . . . . . . . . . . .82 3.8.2 Excitation Through Efficient Transitions. . . . . . . . . . . . . . . . . . . . . .85 3.9 Coupled Mode Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 3.10 Bends and Corners for Dielectric Waveguides . . . . . . . . . . . . . . . . . . . . . . 89 3.11 Systems and Noise. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .92 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 4. Planar Dielectric Waveguides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .99 4.1 Fundamental Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .99 4.2 Dielectric Slab Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 4.2.1 The TM Surface Wave Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 4.2.1.1 Cutoff Conditions for TM Modes. . . . . . . . . . . . . . . . . . . . .103 4.2.1.2 Distribution of Guided Power . . . . . . . . . . . . . . . . . . . . . . . . 105 4.2.1.3 Attenuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 4.2.2 The TE Surface Wave Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 4.2.3 Special Cases and Numerical Examples . . . . . . . . . . . . . . . . . . . . . . 109 4.3 Leaky Wave in a Heteroepitaxial Film Slab Waveguide . . . . . . . . . . . . . . 112 4.3.1 Leaky Modes along an Asymmetric Dielectric Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 4.3.2 Approximate Solutions of the Characteristic Equations. . . . . . . . .115 4.4 Multilayered Dielectric Slab Waveguides. . . . . . . . . . . . . . . . . . . . . . . . . . .118 4.5 Coupling Between Two Parallel Dielectric Slab Waveguides . . . . . . . . . 122 4.6 The Sommerfeld–Zenneck Surface Impedance Waveguide . . . . . . . . . . . 131 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Contents xi 5. Circular Dielectric Waveguides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .137 5.1 Fundamental Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 5.2 Modes on Uniform Solid Core Circular Dielectric Cylinder . . . . . . . . . . 139 5.2.1 Dispersion Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 5.2.2 Cutoff Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 5.2.3 Attenuation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .147 5.2.3.1 The Exact Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 5.2.3.2 The Perturbation Approach . . . . . . . . . . . . . . . . . . . . . . . . . . 148 5.2.4 Field Configurations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .150 5.3 The Sommerfeld–Goubau Wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 5.4 Modes on Radially Inhomogeneous Core Circular Dielectric Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 5.4.1 Formulation of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 5.4.2 Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 5.4.3 Hollow Cylindrical Dielectric Waveguide. . . . . . . . . . . . . . . . . . . . .165 5.5 Experimental Determination of Propagation Characteristics of Circular Dielectric Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 5.5.1 Ultrahigh Q Dielectric Rod Resonant Cavity. . . . . . . . . . . . . . . . . .167 5.5.2 Measured Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 5.6 Summary and Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 6. Elliptical Dielectric Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 6.1 Formulation of the Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .180 6.2 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 6.3 Mode Classifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 6.4 The Dispersion Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 6.4.1 Cutoff Frequencies of Modes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .197 6.4.2 Transition to Circular Cross-Section . . . . . . . . . . . . . . . . . . . . . . . . . 199 6.4.3 Approximate Characteristic Equations . . . . . . . . . . . . . . . . . . . . . . . 201 6.4.4 Propagation Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 6.4.4.1 The Even Dominant eHE11 Mode . . . . . . . . . . . . . . . . . . . . 204 6.4.4.2 The Odd Dominant oHE11 Mode . . . . . . . . . . . . . . . . . . . . . 205 6.4.4.3 Higher Order e,oHEn′m′ Modes . . . . . . . . . . . . . . . . . . . . . . 206 6.4.5 Field Configurations of the Dominant Modes . . . . . . . . . . . . . . . . . 207 6.4.6 Attenuation Calculation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .209 6.5 Weakly Guiding Elliptical Dielectric Waveguides . . . . . . . . . . . . . . . . . . . 210 6.6 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 6.7 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 xii Contents 7. Approximate Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 7.1 Marcatili’s Approximate Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 7.1.1 Approximate Solution for a Rectangular Dielectric Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 7.1.1.1 The Ey nm Modes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .223 7.1.1.2 The Ex nm Modes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .229 7.1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 7.2 The Circular Harmonics Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 7.3 Experimental Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 8. Inhomogeneous Dielectric Waveguides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .241 8.1 Debye Potentials for Inhomogeneous Medium . . . . . . . . . . . . . . . . . . . . . . 241 8.1.1 Rectangular Coordinates (x, y, z) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 8.1.2 Spherical Coordinates (r, θ, φ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 8.1.3 Circular Cylindrical Coordinates (p, θ, z). . . . . . . . . . . . . . . . . . . . .244 8.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 8.2.1 Structures with Transverse Inhomogeneity. . . . . . . . . . . . . . . . . . . .246 8.2.1.1 Wave Propagation along a Dielectric Slab with ϵ(x) and µo Immersed in Free-space . . . . . . . . . . . . . . . . . . . . . . 246 8.2.1.2 Waves in Metallic Rectangular Waveguide Filled with Transversely Inhomogeneous Dielectrics . . . . . . . . . 249 8.2.1.3 Circularly Symmetric Waves along a Cylindrical Radially Inhomogeneous Dielectric Cylinder . . . . . . . . . . 252 8.2.2 Structures with Longitudinal Inhomogeneity . . . . . . . . . . . . . . . . . . 255 8.2.2.1 Longitudinal Periodic Medium . . . . . . . . . . . . . . . . . . . . . . . 256 8.2.2.2 Solutions to the Hill Equation . . . . . . . . . . . . . . . . . . . . . . . . 259 8.2.2.3 Propagation Characteristics of Type (II) (TM) Waves in Periodic Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264 9. Optical Fibers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 9.1 Weakly Guiding Optical Fibers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .265 9.2 Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 9.2.1 Material Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 9.2.2 Waveguide Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 9.2.3 Total Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 9.3 Attenuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 9.4 The Propagation Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 9.5 Selected Solutions to the Propagation Equation . . . . . . . . . . . . . . . . . . . . . 282 Contents xiii 9.6 Wavelength Division Multiplexed Beams (WDM). . . . . . . . . . . . . . . . . . .284 9.6.1 Bit-Parallel WDM Single-Fiber Link. . . . . . . . . . . . . . . . . . . . . . . . .286 9.6.2 Elements of a 12-Bit Parallel WDM System . . . . . . . . . . . . . . . . . . 286 9.6.2.1 The Transmitter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 9.6.2.2 The Single-Mode Fiber. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .287 9.6.2.3 The Receiver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 9.6.3 Design Considerations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .289 9.6.3.1 Wavelength Spacing Considerations . . . . . . . . . . . . . . . . . . 289 9.6.3.2 Skew and Walk-off Considerations . . . . . . . . . . . . . . . . . . . 289 9.6.3.3 Loss Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 9.6.4 Experimental Demonstration of a Two Wavelength BP-WDM System. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .289 9.7 Concluding Remarks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .290 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 10. Solitons and WDM Solitons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 10.1 Nonlinear Refractive Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 10.2 The Nonlinear Pulse Propagation Equation . . . . . . . . . . . . . . . . . . . . . . . 298 10.3 Solution of the Nonlinear Pulse Propagation Equation . . . . . . . . . . . . . 305 10.4 Nonlinear Pulse Propagation for WDM Beams (Cross-Field Modulation Effects) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 10.4.1 Self-Phase Modulation (SPM) and Cross-Phase Modulation (CPM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 10.4.2 Normalized Nonlinear Propagation Equations for WDM Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 10.5 Soliton on a Single Beam. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .311 10.5.1 Bright Solitons. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .311 10.5.2 Dark Solitons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 10.6 Applications of Nonlinear Cross-Field Modulation (CPM) Effect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .313 10.6.1 Pulse Shepherding Effect (Dynamic Control of In-Flight Pulses with a Shepherd Pulse) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 10.6.1.1 Without Shepherd Pulse . . . . . . . . . . . . . . . . . . . . . . . . . 315 10.6.1.2 With Shepherd Pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 10.6.2 Enhanced Pulse Compression in a Nonlinear Fiber by a WDM Optical Pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 10.6.2.1 Shepherding and Primary Pulses are all in the Anomalous Dispersion Region . . . . . . . . . . . . . . . . . . . 320 10.6.2.2 The Shepherd Pulse is in the Normal Dispersion Region and the Primary Pulse is in the Anomalous Dispersion Regime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 xiv Contents 10.6.2.3 The Shepherd Pulse and Primary Pulses are all in the Normal Dispersion Region. . . . . . . . . . . . . . . . .326 10.6.2.4 Additional Simulation Study on WDM Copropagating Pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 10.6.3 Generation of Time-Aligned Picosecond Pulses on Wavelength-Division-Multiplexed Beams in a Nonlinear Fiber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 10.6.3.1 Generation of Time-Aligned Pulses . . . . . . . . . . . . . . . 329 10.6.3.2 Computer Simulation Results . . . . . . . . . . . . . . . . . . . . 329 10.6.3.3 Experimental Setup and Results . . . . . . . . . . . . . . . . . . 330 10.6.4 Bit Parallel WDM Solitons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 11. Ultra Low-Loss Dielectric Waveguides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .339 11.1 Theoretical Foundation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .339 11.1.1 Normal Mode Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 11.1.2 Geometrical Loss Factor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .340 11.1.3 Relationship between Geometrical Loss Factors for TE-Like Mode and for TM-Like Mode . . . . . . . . . . . . . . . . . 343 11.1.4 External Field Decay Consideration . . . . . . . . . . . . . . . . . . . . . . . 343 11.2 Experimental Verification. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .345 11.3 Example of Low-Loss Terahertz Ribbon Waveguide. . . . . . . . . . . . . . .350 References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .356 12. Plasmon (SubWavelength) Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 12.1 TM Wave Guidance Along a Metallic Substrate. . . . . . . . . . . . . . . . . . .360 12.2 TM Wave Guidance Along a Metallic Film . . . . . . . . . . . . . . . . . . . . . . . 365 12.3 Wave Guidance by Metal Ribbons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 12.4 SPP Waves Along Cylindrical Structures . . . . . . . . . . . . . . . . . . . . . . . . . 373 12.4.1 TM Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 12.4.2 HE Waves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .381 12.5 Nanofibers (Subwavelength Guiding Structures) . . . . . . . . . . . . . . . . . . 382 12.6 Conclusions and Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385 References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .387 13. Photonic Crystal Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 13.1 Fundamental Properties of Guided Waves in Periodic Structures. . . .389 13.2 Stop-Band and Pass-Band Property. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .391 13.3 Dielectric-Rod Array Waveguide. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .393 Contents xv 13.4 Band Gap and Waveguide Bends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 13.5 Photonic Bandgap Fiber. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .396 13.6 Analytic Study of Surface Wave Propagation Along a Periodic Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .406 14. Metamaterial and Other Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 14.1 Moving Dielectric Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 14.1.1 Relativity, Lorentz Transformation, and Minkowski Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 14.1.2 Reflection and Transmission of Electromagnetic Waves by a Moving Plasma Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 14.1.3 Mode Propagation Along Moving Dielectric Slabs. . . . . . . . . .418 14.1.3.1 TE Modes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .419 14.1.3.2 TM Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 14.1.4 Mode Propagation Along a Moving Dielectric Cylinder . . . . . 421 14.1.5 Wave Propagation on a Moving Plasma Column. . . . . . . . . . . .425 14.2 Anisotropic Medium Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 14.3 Metamaterial Artificial Dielectric Waveguides . . . . . . . . . . . . . . . . . . . . 435 14.3.1 Some Special Properties of Metamaterial . . . . . . . . . . . . . . . . . . 436 14.3.1.1 If ϵ < 0 and µ < 0, Then n < 0 . . . . . . . . . . . . . . . . . . . 436 14.3.1.2 Snell’s Law for n < 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 14.3.1.3 Poynting’s Vector and Wave Vector in Metamaterial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 14.3.1.4 Fresnel Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439 14.3.1.5 Formation of Metamaterials. . . . . . . . . . . . . . . . . . . . . .441 14.3.1.6 Cloaking with Metamaterial . . . . . . . . . . . . . . . . . . . . . 441 14.3.2 Metamaterial Surface Waveguides . . . . . . . . . . . . . . . . . . . . . . . . 442 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449 15. Selected Numerical Approaches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451 15.1 Outer Radiation Boundary Condition (ORBC) for Computational Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 15.2 Finite Element Method (FEM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 15.2.1 Circular Fiber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 15.2.2 Rectangular Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 15.2.3 Triangular Dielectric Guides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466 15.2.4 Elliptical Dielectric Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 15.2.5 Single Material Fiber Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 15.2.6 Concluding Remarks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .470 xvi Contents 15.3 Beam Propagation Method (BPM) or Forward Marching Split-Step Fast Fourier Transform Method . . . . . . . . . . . . . . . . . . . . . . . 470 15.3.1 Formulation of the Problem and the Numerical Approach. . .471 15.3.2 Gaussian Beam Propagation in a Radially Inhomogeneous Fiber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474 15.3.3 Fiber Couplers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478 15.3.4 Fiber Tapers and Horns. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .485 15.3.5 ω-β Diagram From BPM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 15.3.5.1 The Step-Index Circular Fiber. . . . . . . . . . . . . . . . . . . .491 15.3.5.2 Graded-Index Circular Fiber . . . . . . . . . . . . . . . . . . . . . 492 15.3.5.3 Rectangular Fiber. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .493 15.3.5.4 Elliptical Fiber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 15.3.5.5 Triangular Fiber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 15.3.5.6 Diffused-Channel Rectangular Waveguide. . . . . . . . .496 15.3.5.7 Non-Axisymmetric Graded-Index Fiber . . . . . . . . . . . 496 15.4 Finite Difference Time Domain Method (FDTD) . . . . . . . . . . . . . . . . . 498 15.4.1 Excitation of a Ribbon Dielectric Waveguide . . . . . . . . . . . . . . 498 15.4.2 Ribbon Waveguide Assembled from Dielectric Rods . . . . . . . 499 15.4.3 Dielectric Waveguide Transitions. . . . . . . . . . . . . . . . . . . . . . . . .500 15.5 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504 References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .506 Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 Author Index. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .517 1 INTRODUCTION The increasing capabilities of digital computation have altered the way electromag-netic problems are being solved. It is no longer necessary that analytical solutions be obtained. Many practical problems with complicated geometries for which there are no closed form analytic solutions can now be solved numerically. Nevertheless, understanding the fundamental behavior (the essence) of the solutions must still be gained from analytic solutions of canonical problems. In other words, correct inter-pretation of the numerical results must depend on knowing the essence of guided waves on certain related canonical structures. Therefore, the primary goal of this book is to provide an insight into this essence.
Review of the wave guiding structures over the whole electromagnetic spec-trum shows that, for frequencies below 30 GHz, mostly metal-based structures are used, and for frequencies above 30 GHz, increasing skin-depth losses in metal re-quires that low-loss structures be made without the use of any metallic material.
Hence, the importance of pure dielectric waveguides for carrying large bandwidth signals is established. See Fig. 1.1 for a display of spectral regions in which cer-tain guiding structures are useful. It is seen that the useful spectrum for dielectric waveguides can span more than seven decades, from 109 to 1016 Hz.
1.1 Brief Historical Background The concept of guiding electromagnetic waves either along a single conducting wire with finite surface impedance or along a dielectric rod/slab has been known for a long time. As early as 1899, Sommerfeld conceived the idea of guid-ing a circularly symmetric TM1 wave along a conducting wire with small surface resistivity. However, because of the large field extent outside the wire, this “open-wire” line remained a novelty with limited practical applications. In 1909, Som-merfeld treated the problem of an oscillating dipole above a finitely conducting plane . He found theoretically that there existed not only a radiated wave due 1This notation and classification will be discussed in detail later.
2 The Essence of Dielectric Waveguides Figure 1.1. Spectral regions for various waveguides to the oscillating dipole, but also a surface wave that traveled along the lossy sur-face . This is the well-known “Sommerfeld Problem”. In 1910, Hondros and Debye demonstrated analytically that it was possible to propagate a TM wave along a lossless dielectric cylinder. Zahn in 1915 and his two students, Ruter and Schriever , confirmed the existence of such a TM wave experimentally. Not until 1936 were the propagation properties of asymmetric waves on a round di-electric rod obtained by Carson et al. , who provided a complete mathematical analysis of this problem. It was noted in their paper that in order to satisfy the boundary conditions for the general case, a hybrid wave (i.e., the coexistence of longitudinal electric and magnetic fields) must be assumed. In other words, asym-metric TE and TM modes were inextricably coupled to each other along a circular dielectric rod. They also showed that (1) pure TE and TM waves could only exist in the circularly symmetric case and (2) there existed one and only one mode, namely 1 Introduction 3 the lowest order hybrid wave called the HE11 mode, which possessed no cutoff frequency2 and could propagate at all frequencies. All other circularly symmetric or nonsymmetric modes had cutoff frequencies. The dispersion relations of these modes were also obtained in their paper, but no numerical results were given. In their paper they also mentioned that Southworth in 1920 accidentally observed a guided TM wave in a trough of water. Later, in 1936, Southworth described more detailed experimental measurements on the phase velocity and attenuation of the circularly symmetric TM wave on a circular dielectric guide. Soon afterwards, in 1938, Schelkunoff wrote a paper on the coupled transmission line represen-tation of the waves and the impedance concept, which became the foundation for the development of microwave circuits.
In 1943, Mallach published his results on the use of the dielectric rod as a directive radiator. He showed experimentally that the radiation pattern obtained by the use of the asymmetric HE11 mode produced only one lobe in the principal direc-tion of radiation. Soon after Mallach’s paper, Wegener presented a dissertation in which the asymmetric HE11 mode, together with the lowest order circularly sym-metric TE and TM modes, were analyzed in detail. Not only were the numerical results of the propagation constants for these waves obtained, but also their atten-uation characteristics. He also obtained a few experimental points substantiating his theoretical results. Apparently, he was not aware of the earlier Carson, Mead, and Schelkunoff’s work. Elsasser in 1949 , independent of Wegener’s work, published his computation on the attenuation properties of these three lowest-order modes. In a companion paper, Chandler verified experimentally Elsasser’s re-sults on the dominant HE11 mode. He found that the guiding effect was retained even when the rod was only a fraction of a wavelength in diameter. For this case, since the greater part of the guided energy was outside the dielectric rod, very little loss was observed. This was also the first time the cavity resonator technique for open dielectric structure was introduced to measure the attenuation constant of the HE11 mode. It should be noted that the formula used by Chandler to obtain the at-tenuation constant, α, from the measured Q value was approximately correct, since it was derived assuming that the propagating mode was a TEM mode. The correct formula relating α and Q for the hybrid HE11 mode was given by Yeh in 1962 .
In the mid-1940s, Brillouin summarized his research on wave propagation in periodic structures in a book (1946) . In 1951, Sensiper wrote his thesis on wave propagation on a helical wire waveguide, a periodic structure waveguide.
In 1954, Pierce also provided the results on the interaction of electron beam 2The cutoff frequency does not have the conventional definition as that for the metal waveguides.
Here it is defined that at the cutoff frequency the open dielectric waveguide structure ceases to act as a binding medium for the guided surface wave, and the wave can no longer be guided by the structure.
4 The Essence of Dielectric Waveguides with slow waves guided by a periodic structure. The fundamental theory on wave propagation in periodic media is well established by these publications. At about the same time, the increasing demand for higher bandwidth low-loss transmis-sion lines for transcontinental television and long-distance phone transmission pro-vided the incentive to find new ways to transmit microwaves efficiently. King and Schlesinger studied the dielectric image line (1954), while Goubau experi-mented with a conducting wire coated with a thin sheath of dielectric material, a modified version of the Sommerfeld line (1950) . High loss or instability of the guided field due to the large field extent hampered further development of these approaches. During the 1950s, significant amount of research on the excitation of surface wave problem was carried out (See Collin’s book ). These investiga-tions together with Sommerfeld’s research provided the basic understanding of the problem of wave excitation on a dielectric structure. Another notable effort was the concentrated research undertaken by the Bell Laboratory investigators on the trans-mission of millimeter wave in a oversized circular conducting tube supporting the low-loss circularly symmetric TM wave. (This approach turned out to be not very fruitful due to high loss caused by the modal instability of the low-loss circularly symmetric TM mode in an oversized waveguide.) At that time, the Bell Laboratory group chose not to investigate dielectric fiber as a viable optical waveguide due to its high dielectric losses. History tells us that this was an unfortunate decision.
Observation of waveguide modes in optical fibers was first reported by Snitzer and Hicks in 1959, then later in 1961 by Snitzer and Osterberg, and by Kapany and Burke in 1961 . In 1961, Snitzer restudied the problem of wave guidance along an optical fiber (a circular dielectric cylinder). He provided detailed numer-ical computations on several lower-order modes and obtained field configurations for these modes . In 1962, Yeh solved the unique-canonical problem of surface wave propagation on an elliptical dielectric waveguide. Unlike the circular cylinder case where each mode can be described by a single order of Bessel func-tion, each surface wave mode for an elliptical dielectric cylinder would require infi-nite sums of all orders of Mathieu functions. In other words, the dispersion relation of each mode on an elliptical dielectric cylinder must be represented by an infi-nite determinant of all orders of Mathieu functions. In this case no pure TE or TM mode can exist on an elliptical dielectric cylinder; all modes must be of a hybrid type, that is, the HE type. Yeh not only provided the complete analytical solution to this problem, he also obtained numerical solutions on the propagation constants as well as the attenuation constants for the dominant modes. Experimental verifica-tions were also obtained by him. Independently, at about the same time, Lynkimov et al. gave an analytic solution to this problem, but no experimental or detailed numerical results were given. One notes that the use of elliptical fiber is one way of making a polarization-preserving fiber . In 1965, Bloembergen wrote a 1 Introduction 5 book summarizing his research on wave propagation in nonlinear dielectrics. His work on nonlinear dielectric became the backbone of the later discovery of solitons in optical fibers.
Two events changed the tempo and direction of research on the optical fiber as a viable information transmission link: (1) Kao and Hockham in 1964 recog-nized that if the impurities in optical fiber can be eliminated, the fiber may be-come a very low-loss transmission waveguide for optical signals; and (2) Kapron et al. in 1970 minimized these impurities in fused silica, resulting in the successful making of optical fiber with optical transmission losses of approxi-mately 20 dB km−1. These events awakened the researchers in the communica-tion communities throughout the world. Major efforts were started in the U.S.
(Bell Telephone Laboratories, Corning Glass Works, the Naval Electronics Lab-oratory Center in San Diego, and the Naval Research Laboratory), in the United Kingdom (Standard Telecommunications Laboratories and the British Post Of-fice), in Japan (Nippon Electric Company and the Nippon Sheet Glass Company), and in Germany (AEG-Telefunken, Schott Glass Company, and the Siemens Com-pany). Three major types of optical glass fibers were in contention: the solid core single-mode fiber, the liquid core fiber, and the solid core parabolic-index-variation multimode fiber. Because of the superior dispersion property (i.e., high bandwidth behavior) of the solid core single-mode fiber, it is now universally accepted as the long-distance fiber. At that time, the hope for an all-optical communication system also ignited a significant amount of research in integrated optical circuits, that is, planar imbedded optical dielectric waveguides . Because the index of refraction of the core region and that of the cladding region of an optical fiber are quite close, Snyder in 1969 and Gloge in 1971 provided a new look on the modes that can exist in this so-called “weakly guiding” fiber. Since the 1980s, the emphasis of the research communities has been towards finding ways to increase the band-width capacity and to decrease the loss behavior of a single mode fiber. The use of WDM (Wavelength Division Multiplexed) scheme and solitons [33,34] has provided the much sought-after improvement. From the 1990s until now, we find an explosion of novel dielectric waveguides due to the discovery of new materials.
The revolution in digital processing started in the 1950s finally took flight in the 1960s due to the rapid advances in the use of integrated circuits in dig-ital computers. The impact has been incredible and far-reaching. Many hereto-fore unsolvable engineering or scientific problems could now be solved using a relatively straightforward numerical computational approach. Much advances in the application of numerical techniques to guided wave problems were therefore developed in the period from 1965 to 1980. For example, Yee in 1966 devel-oped the FDTD (Finite Difference Time Domain) algorithm to solve the Maxwell equations numerically; Mur in 1981 developed an effective absorbing bound-6 The Essence of Dielectric Waveguides ary condition for FDTD; Yeh and Wang in 1972 made use of the two-point boundary value numerical approach to solve the graded-index fiber problem; Yeh and Lindgren also found an efficient numerical way to solve the many lay-ered guided wave structure problem; Yeh et al. in 1978 and, few months af-terwards, Feit and Fleck applied the beam propagation method to treat the problem of wave propagation in single-mode or multimode fibers; Yeh et al. in 1975 became the first group who successfully adopted the finite element tech-nique to solve a large variety of single-mode optical waveguides; and Mariki and Yeh in 1985 perfected the 3D TLM (Transmission-Line Matrix) technique based on the Schelkunoff’s impedance concept to solve the arbitrarily shaped di-electric waveguide problem. Several numerical approaches (e.g., FDTD, Finite Element Method, Beam Propagation Method) have already been developed into commercial software packages where a given problem is viewed as a “blackbox” having input data (that specify the problem parameters) and output data (that pro-vide numerical results). There is no need to understand the physics or engineering aspects of the problem. The increasing importance of these numerical approaches to treat guided waves in complex dielectric structures in such a mechanical manner is the reason why there is a need to write this book on the essence of dielectric waveguides. A more thorough discussion of these numerical techniques will be given in the chapter on numerical methods.
Although by the mid-1960s, most of the fundamental concepts of guided wave propagation on linear dielectric structures have been uncovered and understood, it is the explosive revolutionary applications of these concepts in the modern world that establish the importance of understanding the essence of dielectric (surface) waveguides. Optical fibers, which are basically dielectric waveguides, are now routinely used as high-bandwidth communication links. Integrated optical circuits, also basically dielectric waveguides, are in the process of being used exclusively for super-speed computers. Recently, the pursuit of high data rate optical integrated circuits that are compatible with electronic integrated circuits has succeeded in the development of silicon based optical integrated circuits, sources, modulators, and detectors. The only remaining unexploited spectral region is the terahertz band.
This is now being actively explored. It appears that, because of the high loss of metallic material in this spectrum, dielectric waveguides may be the only viable option for terahertz links. It should be pointed out that low-loss material in the ter-ahertz region has yet to be found. Lack of suitable low-loss material in the terahertz spectrum means that the traditional optical fiber approach cannot be used to design a low-loss terahertz waveguide. Yeh and Shimabukuro in 2000 found that the configuration of a high dielectric constant waveguide structure could affect greatly the loss behavior of the dominant TM-like mode. Hence, very low loss terahertz 1 Introduction 7 waveguide may be designed using this discovery. Other modern application areas for dielectric waveguides include the photonic crystal waveguide [44–47], basi-cally an air or dielectric core surrounded by periodic dielectric structures; surface plasmon polaritons guides [48–50], basically a type of Sommerfeld guide; left-handed material (metamaterial) waveguide, that is a dielectric waveguide whose core region is made with artificial dielectrics with negative permittivity and nega-tive permeability [51,52]. The surface plasmon waveguide is of special interest in nanostructure research because of the subwavelength property of its guided wave.
The peculiar behavior of waves guided by artificial metamaterial structure provides unique opportunity to invent new applications.
1.2 Scope of this Book The plethora of dielectric waveguides and its vast modern applications mean that it is not possible to write an all-encompassing book on dielectric waveguides. There-fore, our goal is to write a “back to the basics” book that provides the foundation of dielectric waveguides that is useful, clear, and easy to understand.
Chapter 2 presents the fundamental electromagnetic equations with new in-sight in boundary conditions, classification of fields, the impedance concept, and the scalar-wave approach. Then, an over-all view of dielectric waveguides without delving directly into the specific solution of a given dielectric guided wave struc-ture is presented in Chap. 3. The concepts given there are universally applicable to any dielectric waveguide. New and unique treatment on attenuation has been included.
Specific canonical dielectric guided wave structures will be treated in Chaps. 4–6. They are the planar, circular cylindrical, and elliptical cylindrical structures. Classical analytic modal solutions will be given and explained. The emphasis is to show how one may understand the wave guiding characteristics of a complex, perhaps more practical, dielectric structure from the knowledge of the fundamental solutions from these canonical structures. Approximate ap-proaches for the rectangular dielectric waveguide structure and other structures with no known analytic solutions and inhomogeneous dielectric waveguides are considered in Chaps. 7 and 8.
Subsequent chapters (Chaps. 9–14) will deal with modern applications.
Chapters 9 and 10 deal with linear or nonlinear optical fiber structures, where WDM propagation and WDM solitons will be emphasized. Chapter 11 deals with low-loss structures in the terahertz/millimeter wave region. Plasmon (subwave-length) waveguides are treated in Chap. 12. Chapter 13 deals with photonic crystal waveguides. Other uncommon structures, such as metamaterial structure, moving medium waveguide, and anisotropic material structures, are treated in Chap. 14.
8 The Essence of Dielectric Waveguides Finally, a brief description of several important numerical techniques with ex-amples will be given in Chap. 15.
References 1. A. Sommerfeld, “ ¨ Uber die fortplanzung elektrodynamisches wellen langes eines drahtes,” Ann. der Phys. Chem. 67, 233 (1899) 2. A. Sommerfeld, “An oscillating dipole above a finitely conducting plane,” Ann. der Physik, 28, 665 (1909); Ann. der Physik 81, 1135 (1926) 3. J. Zenneck, “Propagation of plane EM waves along a plane conducting surface,” Ann.
der Physik 23, 846 (1907) 4. D. Hondros and P. Debye, “Elektromagnetische wellen in dielektrischen drahtes,” Ann. der Physik 32, 465 (1910); D. Hondros, “Elektromagneticsche wellen in drahtes,” Ann. der Physik 30, 905 (1909) 5. H. Zahn, “Detection of electromagnetic waves along dielectric wires,” Ann. der Physik 49, 907 (1916) 6. H. Ruter and O. Schriever, “Elektromagnetische wellen an dielektrischen drahten,” Schriften des Naturalwissenschaftlichen vereines fur Schleswig-Holstein 16, 2 (1916) 7. J. R. Carson, S. P. Mead, and S. A. Schelkunoff. “Hyperfrequency waveguides-mathematical theory,” Bell Syst. Tech. J. 15, 310 (1936) 8. G. C. Southworth, “Hyperfrequency waveguides – general considerations and experi-mental results,” Bell Syst. Tech. J. 15, 284 (1936) 9. S. A. Schelkunoff, “The impedance concept and its application to problems of reflec-tion, refraction, shielding and power absorption,” Bell Syst. Tech. J. 17, 17 (1938) 10. Mallach, “Dielektrische richtstrahler,” Bericht des V. I. F. S (1943) 11. G. F. Wegener. “Ausbreitungsgeschwindigkeit wellenwiderstand und dampfung elek-tromagnetischer wellen an dielektrischen Zylindern,” Dissertation, Air Material Com-mand Microfilm ZWB/FB/RE/2018, R8117F831 (1946) 12. W. M. Elsasser, “Attenuation in a dielectric circular rod,” J. Appl. Phys. 20, 1193 (1949) 13. C. H. Chandler. “An investigation of dielectric rod as waveguides,” J. Appl. Phys. 20, 1188 (1949) 14. C. Yeh, “A relation between α and Q,” Proc. IRE, vol. 50, 2143 (1962) 15. L. Brillouin, “Wave Propagation in Periodic Structures,” Dover, New York (1953) 16. S. Sensiper, “Electromagnetic wave propagation on helical conductors,” Research Laboratory for Electronics, Mass. Inst. of Tech., Tech. Rept No. 194, May 16 (1951); Proc. IRE 43, 149 (1955) 1 Introduction 9 17. J. R. Pierce, “Theory and Design of Electron Beams,” D. Van Nostrand, Princeton (1950) 18. D. D. King, “Dielectric image line,” J. Appl. Phys. 23, 699 (1952); D. D. King and S. P. Schlesinger, “Losses in dielectric image lines,” IRE Trans. Microw. Theory Tech.
MTT-5, 31 (1957) 19. G. Goubau, “Surface waves and their application to transmission lines,” J. Appl. Phys.
21, 1119 (1950); G. Goubau, “Single conductor surface wave transmission lines,” Proc. IRE 39, 619 (1951) 20. R. E. Collin, “Field Theory of Guided Waves,” McGraw-Hill, New York (1960) 21. E. Snitzer and J. W. Hicks, “Optical wave-guide modes in small glass fibers. I. The-oretical,” J. Opt. Soc. Am. 49, 1128 (1959); E. Snitzer and H. Osterberg, “Observed dielectric waveguide modes in the visible spectrum,” J. Opt. Soc. Am. 51, 499 (1961); N. S. Kapany and J. J. Burke, “Fiber optics IX. Waveguide effects,” J. Opt. Soc. Am.
51, 1067–1078 (1961) 22. E. Snitzer, “Cylindrical dielectric waveguide modes,” J. Opt. Soc. Am. 51, 491 (1961) 23. C. Yeh, “Elliptical dielectric waveguides,” J. Appl. Phys. 33, 3235 (1962); C. Yeh, “Attenuation in a dielectric elliptical cylinder,” IEEE Trans. Antenn. Propag. AP-11, 177 (1963) 24. L. A. Lynbimov, G. I. Veselov, and N. A. Bei, “Dielectric waveguide with elliptical cross-section,” Radio Eng. Electron. (USSR) 6, 1668 (1961) 25. R. B. Dyott, “Elliptical Fiber Waveguides,” Artech House, Boston (1995) 26. N. Bloembergen, “Non-linear Optics,” W. A. Benjamin, New York (1965) 27. K. C. Kao and G. A. Hockham, “Dielectric fiber surface waveguides for optical fre-quencies,” Proc. IEEE 113, 1151 (1966) 28. F. P. Kapron, D. B. Keck, and R. D. Maurer, “Radiation losses in glass optical waveguides,” Appl. Phys. Lett. 17, 423 (1970) 29. D. Marcuse, Ed., “Integrated Optics,” IEEE Press, New York (1973) 30. A. W. Snyder, “Asymptotic expressions for eigenfunctions and eigenvalues of a dielec-tric or optical waveguide,” IEEE Trans. Microw. Theory Tech. MTT-17, 1130 (1969) 31. D. Gloge, “Weakly guiding fibers,” Appl. Opt. 10, 2252 (1971) 32. G. P. Agrawal, “Fibre-Optic Communication Systems,” Wiley, New York (2002) 33. A. Hasegawa and T. Tappert, “Transmission of stationary nonlinear optical pulses in dispersive dielectric fibers. I. Anomalous dispersion,” Appl. Phys. Lett. 23, 142 (1973) 34. C. Yeh and L. A. Bergman, “Existence of optical solitons on wavelength division multiplexed beams,” Phys. Rev. E 60, 2306 (1999) 35. K. S. Yee, “Numerical solution of initial boundary value problems involving Maxwell’s equations in isotropic media,” IEEE Trans. Antenn. Propag. 14, 302 (1966) 10 The Essence of Dielectric Waveguides 36. G. Mur, “Absorbing boundary conditions for finite-difference approximation of the time-domain electromagnetic field equations,” IEEE Trans. Electromag. Compat. 23, 1073 (1981) 37. C. Yeh and P. Wang, “Scattering of obliquely incident waves by inhomogeneous fibers,” J. Appl. Phys. 43, 3999 (1972) 38. C. Yeh and G. Lindgren, “Computing the propagation characteristics of radially strat-ified fibers – An efficient method,” Appl. Opt. 16, 483 (1977) 39. C. Yeh, L. Casperson, and B. Szejn, “Propagation of truncated gaussian beams in multimode or single-mode fiber guides,” J. Opt. Soc. Am. 68, 989 (1978) 40. M. D. Feit and J. D. Fleck, “Light propagation in graded-index optical fibers,” Appl.
Opt. 17, 3990 (1980) 41. C. Yeh, S. B. Dong, and W. Oliver, “Arbitrarily shaped inhomogeneous optical fiber or integrated optical waveguides,” J. Appl. Phys. 46, 2125 (1975); C. Yeh, K. Ha, S. B. Dong, and W. P. Brown, “Single-mode optical waveguides,” Appl. Opt. 18, 1490 (1979) 42. G. E. Mariki and C. Yeh, “Dynamic 3D TLM analysis of microstrip-lines on anisotropic substrates,” IEEE Trans. Microw. Theory Tech. MTT-33, 789 (1985) 43. C. Yeh, F. Shimabukuro, and P. H. Siegel, “Low-loss terahertz ribbon waveguides,” Appl. Opt. 44, 5937 (2005); C. Yeh, F. Shimabukuro, P. Stanton, V. Jamnejad, W. Imbriale, and F. Manshadi, “Communication at millimeter–submillimeter wave-lengths using a ceramic ribbon,” Nature 404, 584 (2000) 44. E. Yablonovitch, “Inhibited spontaneous emission in solid-state physics and electron-ics,” Phys. Rev. Lett. 58, 2059 (1987) 45. S. John, “Strong localization of photon in certain disordered dielectric superlattices,” Phys. Rev. Lett. 58, 2486 (1987) 46. J. Joannopoulos, R. Meade, and J. Winn, “Photonic Crystals,” Princeton Press, New Jersey (1995) 47. J. C. Knight, T. A. Birks, P. J. Russell, and D. M. Atkins, “All-silica single-mode optical fiber with photonic crystal cladding,” Optics Lett. 21, 1547 (1996) 48. R. H. Ritchie, “Plasma losses by fast electrons in thin films,” Phys. Rev. 106, 874 (1957); H. Raether, “Surface Plasmons,” Springer, Berlin Heidelberg New York (1988) 49. S. A. Maier, “Plasmonics: Fundamentals and Applications,” Springer, Berlin Heidelberg New York (2007) 50. H. A. Atwater, “The promise of plasmonics,” Verlag Scientific Am. 50, 56 (2007) 51. V. G. Veselago, “The electrodynamics of substances with simultaneously negative val-ues of permittivity and permeability,” Soviet Phys. Uspekhi 10, 509 (1968) 52. J. B. Pendry, D. Schurig, and D. R. Smith, “Controlling electromagnetic fields,” Sci-ence 312, 1780 (2006) 2 FUNDAMENTAL ELECTROMAGNETIC FIELD EQUATIONS All large-scale electromagnetic wave phenomena are governed by the Maxwell equations and the appropriate boundary conditions. In this chapter we shall dis-cuss the fundamental equations and relations dealing with electromagnetic waves [1–3].
2.1 Maxwell Equations On the basis of the established experimental laws, Maxwell postulated that the electromagnetic field vectors are subject to the following equations: ∇× E(r, t) = −∂B(r, t) ∂t , (2.1) ∇× H(r, t) = J(r, t) + ∂D(r, t) ∂t , (2.2) where E(r, t) = Electric field intensity (V m−1) H(r, t) = Magnetic field intensity (A m−1) D(r, t) = Electric displacement vector (C m−2) B(r, t) = Magnetic induction vector (Wb m−2) J(r, t) = Electric current density (A m−1) 12 The Essence of Dielectric Waveguides These vectors are functions of space, r (in meters), and time, t (in seconds). The mks or Giorgi system of units will be used throughout. On a macroscopic scale, the conservation of charge law can be expressed as follows: ∇· J(r, t) + ∂ρ ∂t = 0, (2.3) here, ρ(r, t) = Electric charge density (C m−3).
This is the equation of continuity. Faraday’s Law, Ampere’s Law, Gauss’ Law, and Coulomb’s Law are included or can be derived from the Maxwell equations and the equation of continuity. For example, (2.1) is a statement of Faraday’s Law, while (2.2), without the displacement current term, ∂D(r, t)/∂t, is a statement of Ampere’s Law. Maxwell postulated the existence of the displacement current term in (2.2) to express the wave nature of the electromagnetic fields. Since the divergence of the curl of any vector vanishes identically, taking the divergence of (2.1) yields ∇· ∂B(r, t) ∂t = ∂ ∂t (∇· B(r, t)) = 0 (2.4) or ∇· B(r, t) = 0.
(2.5) This is Gauss’ Law. From (2.5), the field of the magnetic induction vector B(r, t) is solenoidal. The divergence of (2.2) gives ∇· J(r, t) + ∇· ∂D(r, t) ∂t = 0 (2.6) and, from (2.3), one obtains ∂ ∂t [∇· (D(r, t) −ρ(r, t)] = 0 (2.7) or ∇· D(r, t) = ρ(r, t).
(2.8) This is Coulomb’s Law. One notes that these divergence equations (2.5) and (2.8) are not independent relations of Maxwell equations (2.1) and (2.2) and the equa-tion of continuity, (2.3). Limiting our investigation to linear phenomena, fields of arbitrary time variation can be constructed from harmonic solutions through the Fourier Transform Method, and there is no loss of generality with the assumption that the time-dependent variation of the fields may be factored out as follows: 2 Fundamental Electromagnetic Field Equations 13 E(r, t) = Re[E(r) ejωt], (2.9) where ω is the harmonic frequency of the wave, Re means the real part of, and E(r), the electric field vector, is a spatially dependent, complex function. Simi-lar time variations are assumed for the other field and source quantities, such as D(r, t) = Re[D(r) ejωt], H(r, t) = Re[H(r) ejωt], . . . , etc. The time-harmonic Maxwell equations and the continuity equation now take the forms ∇× E(r) = −jωB(r), (2.10) ∇× H(r) = J(r) + jωD(r), (2.11) ∇· J(r) = −jωρ(r).
(2.12) The associated divergence equations are ∇· B(r) = 0, (2.13) ∇· D(r) = ρ(r).
(2.14) The field vectors E(r), D(r), H(r), and B(r) are now spatially dependent com-plex functions. It is seen from (2.10) and (2.11) that given the source function J(r), there are four unknown quantities, E, B, D, and H, and two independent equa-tions (2.10) and (2.11). Two additional independent equations relating the field quantities, E, B, D, and H are needed in order that deterministic solutions for these quantities may be found. The needed equations are obtained from the consti-tutive relations.
2.2 The Constitutive Relations The constitutive relations are derived from the description of the macroscopic prop-erties of the medium in the immediate neighborhood of the specified field point. In general, we shall assume that, at any given point in a given medium, the vector D and H may be represented as a function of E and B.
D = F1(E, B), (2.15) H = F2(E, B).
(2.16) The functional dependencies of these functions are obtained from the macroscopic physical properties of the medium . The behavior of a material medium in an electromagnetic field can be described in terms of distributions of electric and 14 The Essence of Dielectric Waveguides magnetic dipoles. The medium can be characterized by two polarization density functions: P, the electric dipole moment per unit volume, and M, the magnetic dipole moment per unit volume. The polarization may be induced under the action of the field from other sources, or it may be virtually permanent and independent of external fields. The permanent polarizations will be designated by P0 and M0.
A few examples are given below.
2.2.1 Simple Medium (Linear and Isotropic) A simple medium is taken to be (a) linear, where D is a linear function of E and H is a linear function of B, and (b) isotropic, where D is parallel to E and H is parallel to B. In this simple medium, D = ϵE, H = 1 µB.
(2.17) The parameters ϵ and µ, which represent the macroscopic electromagnetic prop-erties, are, respectively, the permittivity and permeability of the medium. For isotropic inhomogeneous media, ϵ and µ may be functions of positions. For free-space, ϵ = ϵ0, µ = µ0, (2.18) where ϵ0 = 8.854 × 10−12 (F m−1) and µ0 = 4π × 10−7 (H m−1) are, respectively, the free-space permittivity and free-space permeability. The relationships between the field vectors and the polarization vectors are defined as follows: P + P0 = D −ϵ0E = (ϵ −ϵ0)E = χeϵ0E, (2.19) M + M0 = 1 µ0 B −H = µ µ0 −1 H = χmH, (2.20) where χe and χm are called the electric and magnetic susceptibilities. The elec-tric and magnetic polarization vectors are zero in free-space. Strictly speaking, the relations (2.19) and (2.20) are definable only for time-periodic phenomena, since in general ϵ and µ are functions of the frequency. The frequency dependence of the constitutive parameters is known as the dispersive property of the medium.
Hence, these relations are applicable to other than time-periodic, time-varying fields only when, over the significant part of the frequency spectrum covered by the Fourier components of the time dependence, the constitutive parameters ϵ and µ are sensibly independent of frequency.
2 Fundamental Electromagnetic Field Equations 15 2.2.2 Anisotropic Medium [5–7] In an anisotropic material medium, the electromagnetic properties are functions of the field directions about a point. Thus, in general, D = ϵ ·E ϵ = ⎡ ⎣ ϵ11 ϵ12 ϵ13 ϵ21 ϵ22 ϵ23 ϵ31 ϵ32 ϵ33 ⎤ ⎦, (2.21) B = µ ·H µ = ⎡ ⎣ µ11 µ12 µ13 µ21 µ22 µ23 µ31 µ32 µ33 ⎤ ⎦.
(2.22) Here, ϵij and µij are elements of the permittivity matrix and the permeability ma-trix describing the anisotropic characteristics of the medium. For inhomogeneous and anisotropic medium, ϵij and µij are functions of positions. For anisotropic and dispersive medium, ϵij and µij are functions of the frequency. The electromag-netic properties of a few common anisotropic material media are characterized as follows: (a) Magnetized Ferrite Medium ϵ = ⎡ ⎣ ϵ1 0 0 0 ϵ1 0 0 0 ϵ1 ⎤ ⎦= a scalar µ = ⎡ ⎣ µ11 µ12 0 µ21 µ22 0 0 0 µ33 ⎤ ⎦ (2.23) with an impressed static magnetic field along the axial z-axis.
(b) Crystalline Medium ϵ = ⎡ ⎣ ϵ11 ϵ12 ϵ13 ϵ21 ϵ22 ϵ23 ϵ31 ϵ32 ϵ33 ⎤ ⎦ µ = ⎡ ⎣ µ0 0 0 0 µ0 0 0 0 µ0 ⎤ ⎦= a scalar.
(2.24) (c) Uniaxial Medium ϵ = ⎡ ⎣ ϵ1 0 0 0 ϵ2 0 0 0 ϵ3 ⎤ ⎦ µ = ⎡ ⎣ µ0 0 0 0 µ0 0 0 0 µ0 ⎤ ⎦= a scalar.
(2.25) 16 The Essence of Dielectric Waveguides (d) Cold Plasma with Impressed Static Magnetic Field B0 ϵ = ⎡ ⎣ ϵ11 ϵ12 0 ϵ21 ϵ22 0 0 0 ϵ33 ⎤ ⎦ µ = ⎡ ⎣ µ0 0 0 0 µ0 0 0 0 µ0 ⎤ ⎦= a scalar.
(2.26) Here ϵij is a function of frequency as follows: ϵ11 = ϵ0 ⎡ ⎣1 − ω2 p (ω −jν) ω (ω −jν)2 −ω2 c ⎤ ⎦, ϵ22 = ϵ11, ϵ12 = jϵ0 ω2 pωc ω (ω −jν + ωc) (ω −jν −ωc) , ϵ21 = −ϵ12, ϵ33 = ϵ0 1 − ω2 p ω (ω −jν) , where ωp = nee2/meϵ0 1/2 is the electron plasma frequency, ne is the electron number density, e is the electronic charge, me is the electron mass, and ωc = eB0/me is the electron cyclotron frequency, where B0 is the impressed static mag-netic induction along the z-axis. The term ν is the collision frequency of the elec-trons with the heavier particles.
2.2.3 Left-Handed Medium (Metamaterial) [8–10] A class of artificial media can be characterized as follows: D = −ϵE, H = −1 µB.
(2.27) A left-handed material medium (usually artificially made) is one with negative permittivity and negative permeability in the frequency range of interest. The index of refraction in such a metamaterial medium is also negative. The permittivity and the permeability of the medium are usually frequency dependent and lossy. In other words, they are complex quantities.
2.2.4 Conducting Medium According to Ohm’s law J = σE, (2.28) |
7124 | https://testinar.com/operations/gcf_of_152_and_8991 | GCF of 152,8991
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What is the Greatest Common Factor of 152 and 8991?
Greatest Common Factor of 152 and 8991
GCF(152, 8991) = 1, Greatest common factor of 152 and 8991 is 1. Greatest Common Factor or Greatest Common Divisor of two numbers is the largest integer by which both the numbers can be divided. There are two different methods to calculate Greatest Common Factor of 152 and 8991. Greatest Common Factor by prime factorization method and Greatest Common Factor by matching factors method.
Greatest Common Factor of 152 and 8991 by prime factorization method
We will first find the prime factorization of 152 and 8991.
Prime Factorization of 152 is 1, 2, 2, 2, 19 and Prime Factorization of 8991 is 1, 3, 3, 3, 3, 3, 37.
Factorize(152)=(152)=1×2×2×2×19 1×2×2×2×19
Factorize(8991)=(8991)=1×3×3×3×3×3×37 1×3×3×3×3×3×37
Now we need to find any which are common for each number (1, 1) and multiply these numbers together.
G C F(152,8991)=1×1=1 G C F(152,8991)=1×1=1.
Greatest Common Factor of 152 and 8991 by matching factors method
List of positive integers factors of 152 leaving a remainder zero is 1, 2, 4, 8, 19, 38, 76, 152
List of positive integers factors of 8991 leaving a remainder zero is 1, 3, 9, 27, 37, 81, 111, 243, 333, 999, 2997, 8991
As you can see, 1 is the greatest and common number that 152 and 8991 divides into.
So the greatest common factor 152 and 8991 is 1.
G C F(152,8991)=1 G C F(152,8991)=1.
If you want to learn more about greatest common divisor, take a look at the Wikipedia page.
Greatest common factor of:
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7125 | https://www.worddb.com/opposite-word-for/allure | All Tools
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| Top 20 Top 10 Top 20 (Default) Top 50 Top 100 | All All (Default) Words Phrases | | | | | All All (Default) Verb Noun Adjective Adverb |
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| | Word | Save | Syns.. | | Usage | Type |
| | deter | | | | | verbv |
| | verb • try to prevent; show opposition to • turn away from by persuasion | | | | | | | |
| | discourage | | | | | verbv |
| | verb • try to prevent; show opposition to • deprive of courage or hope; take away hope from; cause to feel discouraged • advise or counsel in terms of someone's behavior | | | | | | | |
| | dissuade | | | | | verbv |
| | verb • turn away from by persuasion | | | | | | | |
| | repel | | | | | verbv |
| | verb • cause to move back by force or influence • be repellent to; cause aversion in • force or drive back • reject outright and bluntly • fill with distaste | | | | | | | |
| | repulse | | | | | verbv |
| | noun • an instance of driving away or warding off verb • force or drive back • be repellent to; cause aversion in • cause to move back by force or influence | | | | | | | |
• try to prevent; show opposition to
• turn away from by persuasion
• try to prevent; show opposition to
• deprive of courage or hope; take away hope from; cause to feel discouraged
• advise or counsel in terms of someone's behavior
• turn away from by persuasion
• cause to move back by force or influence
• be repellent to; cause aversion in
• force or drive back
• reject outright and bluntly
• fill with distaste
• an instance of driving away or warding off
• force or drive back
• be repellent to; cause aversion in
• cause to move back by force or influence
Alternatives for DETER
Alternatives for DISCOURAGE
Alternatives for DISSUADE
Alternatives for REPEL
Alternatives for REPULSE |
7126 | http://search-library.ucsd.edu/discovery/fulldisplay?docid=cdi_proquest_miscellaneous_733181221&context=PC&vid=01UCS_SDI:UCSD&lang=en&search_scope=ArticlesBooksEtc&adaptor=Primo%20Central&tab=ArticleBooksEtc&query=creator%2Cexact%2CMarcus%2C%20Yizhak%2CAND&facet=creator%2Cexact%2CMarcus%2C%20Yizhak&mode=advanced&offset=0 | Effect of Ions on the Structure of Water: Structure Making and Breaking - UC San Diego
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; Effect of Ions on the Structure of Water: Structure Making and Breaking
; Marcus, Yizhak;
; United States: American Chemical Society;
; Chemical Reviews, 2009-03, Vol.109 (3), p.1346-1370;
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Title
Effect of Ions on the Structure of Water: Structure Making and Breaking
Effect of Ions on the Structure of Water: Structure Making and Breaking
Effect of Ions on the Structure of Water: Structure Making and Breaking
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Marcus, Yizhak
Marcus, Yizhak
Marcus, Yizhak
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Chemical Reviews, 2009-03, Vol.109 (3), p.1346-1370
Chemical Reviews, 2009-03, Vol.109 (3), p.1346-1370
Chemical Reviews, 2009-03, Vol.109 (3), p.1346-1370
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Description
The structural nature of water, particularly with regard to its ionic content, is discussed. The effects of ions on viscosity, temperature and entropy are examined.
The structural nature of water, particularly with regard to its ionic content, is discussed. The effects of ions on viscosity, temperature and entropy are examined.
The structural nature of water, particularly with regard to its ionic content, is discussed. The effects of ions on viscosity, temperature and entropy are examined.
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United States: American Chemical Society
United States: American Chemical Society
United States: American Chemical Society
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Chemistry
Chemistry
Chemistry
Entropy
Entropy
Entropy
Ions
Ions
Ions
Molecular structure
Molecular structure
Molecular structure
Temperature
Temperature
Temperature
Water
Water
Water
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ISSN: 0009-2665
ISSN: 0009-2665
ISSN: 0009-2665
ISSN: 1520-6890
ISSN: 1520-6890
ISSN: 1520-6890
EISSN: 1520-6890
EISSN: 1520-6890
EISSN: 1520-6890
DOI: 10.1021/cr8003828
DOI: 10.1021/cr8003828
DOI: 10.1021/cr8003828
PMID: 19236019
PMID: 19236019
PMID: 19236019
CODEN: CHREAY
CODEN: CHREAY
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7127 | https://spines.com/paradox-definition-literature/ | Paradox Definition in Literature: Meaning, Examples & Usage Explained
If literature were a dinner party, paradox would be the charming guest who says something utterly confusing—and somehow, it makes perfect sense. A paradox is a statement that seems self-contradictory on the surface, but once you sit with it (preferably with a strong coffee), it reveals a deeper truth.
From Shakespeare’s “I must be cruel to be kind” to Orwell’s “War is peace”, paradoxes have long been used to challenge our logic and stir our imaginations. Paradoxes add depth to major literary themes by highlighting contradictions that reflect the complexity of real life. But what is the definition of paradox in literature, exactly? And how can such contradictions be so… insightful?
In this article, we’ll unpack the paradox definition in literature, explore famous examples, and show you why writers love to bend reality just enough to make you think twice.
What Is a Paradox?
Let’s start with the basics: a paradox is a statement that contradicts itself — yet somehow rings true. It’s the literary equivalent of saying, “Less is more” and actually meaning it.
In general, paradoxes expose a truth by turning logic on its head. In literature, they’re used to create depth, tension, irony, or even a bit of intellectual mischief. Think of a paradox as a brain teaser wrapped in poetic flair — it makes you pause and say, “Wait… what?”, followed quickly by, “Oh… I get it.”
Paradox Definition in Literature:
A paradox in literature is a seemingly contradictory statement or situation that reveals a deeper truth or insight when examined closely.
Paradoxes often contribute to plot twists or deeper thematic arcs and writers use paradoxes to explore complex ideas, challenge assumptions, or add a layer of sophistication to their work. It’s not about confusing the reader — it’s about inviting them to think deeper.
Paradox vs. Oxymoron: What’s the Difference?
At first glance, a paradox and an oxymoron might seem like literary twins — both involve contradiction, both sound a little puzzling — but they’re actually quite different in how they operate.
An oxymoron is a short phrase, usually two or three words, that combines opposing ideas. Think of expressions like “deafening silence,” “bittersweet,” or “seriously funny.” These mini-contradictions are often used for dramatic or poetic effect — they catch your attention, make you pause, and sometimes even make you smile.
A paradox, on the other hand, is more like a full thought experiment. It can be a sentence, a situation, or even an entire theme that appears illogical at first but reveals a deeper truth when you take a moment to reflect. For example, “The only constant is change” seems contradictory — how can change be constant? But once you sit with it, it makes perfect sense.
So while oxymorons play with words, paradoxes play with ideas. While they’re not the same as irony, paradoxes can have a similar effect, often making readers question what’s real, what’s expected, and what lies beneath the surface.
Why Writers Use Paradoxes in Literature
Writers don’t just toss paradoxes into their work to sound clever (well, not only for that reason). Paradoxes serve a powerful purpose in literature: they make readers think, question, and dig deeper. They’re the literary equivalent of planting a seed that blossoms the more you reflect.
So why are paradoxes so popular among authors, poets, and playwrights? Here are a few reasons:
1. To Challenge the Reader’s Perspective
A good paradox forces us to pause and rethink what we believe to be true. It introduces complexity where things seem simple, and simplicity where things seem complex. It encourages critical thinking — something literature does best when it’s not spoon-feeding the reader.
2. To Add Depth and Layers to a Theme
Themes like love, time, identity, or truth often resist black-and-white thinking. Paradoxes help writers explore these gray areas with more nuance. For example, in 1984, the slogan “Freedom is slavery” might sound absurd at first — but in context, it reveals the twisted logic of a totalitarian regime.
3. To Reflect Inner Conflict
Characters are rarely one-dimensional. They have contradictions, doubts, and clashing desires — just like us. Paradoxes work especially well in introspective first-person narratives, where a character’s inner turmoil is front and center. They often appear in dialogue making it more realistic, adding realism and internal conflict, helping readers connect with a character’s emotional or philosophical struggle. A character who says, “I feel most alone in a crowd,” reveals something deep and relatable in just one sentence.
4. To Spark Emotional or Philosophical Engagement
Paradoxes are memorable. They echo in your mind long after you’ve turned the page. They create that “Wait… huh?” moment — followed by a “Whoa.” That emotional or intellectual pause is exactly what great literature aims for.
In short, paradoxes are not just pretty word puzzles — they’re storytelling tools that push boundaries and reveal truth in unexpected ways.
Famous Examples of Paradox in Literature
Now that we’ve defined what a paradox is (and made peace with the fact that it does make sense eventually), let’s look at how some of the world’s most celebrated authors have used paradoxes to elevate their writing.
These literary paradoxes aren’t just clever lines — they’re powerful tools that reveal character depth, expose societal contradictions, or leave readers wonderfully unsettled.
“I must be cruel to be kind.” — William Shakespeare, Hamlet
On the surface, this statement sounds like emotional whiplash. How can cruelty be kind? But within the context of Hamlet, it reflects the painful truth that sometimes difficult actions are necessary for the greater good. Welcome to moral complexity, courtesy of Shakespeare.
“War is peace. Freedom is slavery. Ignorance is strength.” — George Orwell, 1984
Orwell didn’t just write paradoxes — he weaponized them. These slogans of the oppressive regime in 1984 reveal how language can be twisted to control thought. Each statement contradicts itself, yet in the warped reality of the novel, they serve a chilling logic.
“I can resist everything except temptation.” — Oscar Wilde, Lady Windermere’s Fan
Wilde was a master of paradox wrapped in wit. This line is humorous, yes — but also deeply human. It captures the essence of internal conflict: knowing better, yet doing otherwise. (We’ve all been there. Probably with snacks.)
“Much madness is divinest sense.” — Emily Dickinson
Dickinson flips the definition of sanity on its head. What society deems “madness” may actually be the truest form of insight — a paradox that celebrates individuality and questions conformity.
“It was the best of times, it was the worst of times…” — Charles Dickens, A Tale of Two Cities
This iconic opening line throws you straight into contradiction. How can both extremes exist at once? But Dickens isn’t confused — he’s capturing the turbulent, conflicting reality of a revolutionary era.
Paradoxes like these prove that literature doesn’t shy away from contradictions — it embraces them. Through these elegant contradictions, authors explore truths too complex for straight answers.
How to Identify a Paradox in a Text
So, you’re reading a novel, a poem, or maybe even a Shakespeare monologue, and suddenly a line makes your brain short-circuit. That might just be a paradox waving at you. But how can you tell the difference between a deep literary paradox and a sentence that just doesn’t make sense?
Here’s how to spot one in the wild:
Look for a Self-Contradictory Statement
Paradoxes often sound like they’re arguing with themselves. Phrases like “less is more” or “the beginning of the end” make you pause because they seem logically impossible. But if the contradiction feels intentional, and it nudges you to think deeper — you’ve got a paradox on your hands.
Ask: “Does This Reveal a Hidden Truth?”
A true literary paradox isn’t just a contradiction for contradiction’s sake. It should lead somewhere — to a philosophical insight, a character’s internal conflict, or a theme that challenges simple logic. If the line makes more sense the more you think about it, it’s likely a paradox.
Consider the Context
Paradoxes don’t exist in a vacuum — the surrounding story, tone, and theme matter. In a dystopian novel, paradoxes might reveal societal hypocrisy. In a love poem, they might express emotional tension. Always ask: what’s the author trying to show by turning logic on its head?
Watch for Tone: Is It Ironic or Philosophical?
Paradoxes often carry a serious or reflective tone, sometimes with a hint of irony or dry humor. If a statement makes you laugh uncomfortably and rethink your worldview, you’re probably dealing with a paradox — not a typo.
Quick Practice Example:
“The more you learn, the less you know.” At first glance, it sounds wrong. But dig deeper and it’s a statement about humility — the more you study, the more you realize how vast your ignorance really is. Classic paradox.
With a bit of practice, identifying paradoxes becomes second nature — like spotting plot twists or overused metaphors. And once you start noticing them, you’ll find they’re everywhere, quietly adding layers of meaning beneath the surface.
Using Paradox in Your Own Writing
Now that you’re well-versed in spotting paradoxes, why not try writing your own? Whether you’re working on your first book, a poem, or a late-night journal entry that feels way too deep, paradoxes can be powerful tools for adding depth, tension, and sophistication.
But like any literary device, paradox is best used with intention — not just because it sounds fancy (although let’s be honest, it usually does).
1. Start with a Contradiction
Begin by identifying two opposing ideas. Think about emotional opposites (love and hate), logical contradictions (freedom and control), or character traits (strength and vulnerability). Your goal is to find a pair that, while conflicting, can reveal something true when combined.
Example:
“The closer I get to you, the further I feel from myself.”
It’s contradictory, yes. But it also says something real about identity and relationships.
2. Make It Meaningful, Not Just Clever
A good paradox shouldn’t be a party trick — it should serve your story. Does it reflect your character’s inner struggle? Does it capture a theme you’re exploring? The best paradoxes make the reader stop and say, “Oh wow… that’s true.”
Bad paradoxes just make them say, “…Huh?” (and not in a good way).
3. Keep It Clear — Even When It’s Complex
Paradox isn’t an excuse for confusing writing. In fact, the magic lies in taking something complicated and expressing it in a surprisingly simple way. Aim for clarity within the contradiction — that’s what makes it hit hard.
4. Practice with Prompts
Try finishing these sentences with your own paradoxical twist:
Let your brain have a little fun bending the rules — that’s where great paradoxes are born.
Paradoxes invite your readers to wrestle with ideas, not just read them. So whether you’re writing the next great literary masterpiece or just want your Instagram captions to sound profound, a well-placed paradox might be your secret weapon.
Final Thoughts
Paradoxes are proof that literature doesn’t need to make immediate sense to be meaningful — in fact, sometimes the most powerful truths are hidden inside contradictions. From Shakespeare’s conflicted characters to Orwell’s chilling slogans, paradoxes force us to pause, think, and see the world from a different angle.
Let’s recap what we’ve covered:
In literature, as in life, things aren’t always black and white. Paradoxes help us live in the gray areas — where the most interesting stories usually happen.
FAQs – Paradox Definition Literature
Q1: What is a paradox in literature?
A paradox in literature is a statement or situation that appears to contradict itself but reveals a deeper truth upon reflection. Writers use paradoxes to highlight complexity, irony, or hidden meaning within characters, themes, or dialogue. These contradictions often make the reader pause and reconsider the ideas being presented. Paradoxes can add depth and provoke thought, making them a powerful literary device.
Q2: What is the best example of a paradox?
One of the most famous examples is Shakespeare’s line from Hamlet: “I must be cruel to be kind.” At first glance, it seems contradictory—how can cruelty be kindness? But within the story, it makes sense: Hamlet believes that by taking harsh action, he will ultimately do good. It’s a classic example of a literary paradox revealing emotional and moral conflict.
Q3: What is the difference between a paradox and an oxymoron?
An oxymoron is a short phrase (usually two words) that combines contradictory terms, like “bittersweet” or “deafening silence.” A paradox is broader—it’s often a full statement or concept that appears contradictory but holds a deeper meaning. While oxymorons are more stylistic, paradoxes are usually conceptual and philosophical. In short: oxymorons play with words, paradoxes play with ideas.
Q4: How do you explain a paradox to a child?
You can explain a paradox to a child by using a simple example like, “The more you learn, the less you know.” Tell them it’s something that sounds confusing at first but actually teaches you something interesting. You might say, “It’s like when something feels wrong but turns out to be right.” Use everyday situations to help them grasp the idea that some truths are hidden inside contradictions.
Q5: Why do authors use paradoxes in writing?
Authors use paradoxes to challenge readers, express inner conflict, or explore complex ideas that aren’t black and white. Paradoxes make writing more thought-provoking and emotionally layered. They can reflect how real-life situations are full of contradictions, making stories feel more human and relatable. In essence, paradoxes add richness and nuance to literary work.
Q6: What is the closest meaning of paradox?
The closest meaning of “paradox” is a contradiction that reveals a hidden truth. It’s something that seems to go against logic, but when you think about it, it actually makes sense. In literature, paradoxes help express truths that aren’t easy to say in straightforward terms. They invite readers to look beyond the obvious.
Q7: How can you identify a paradox in a text?
To identify a paradox, look for statements that seem self-contradictory or confusing at first glance. Ask yourself if the contradiction serves a purpose—does it reveal something deeper about the character, theme, or situation? If a line makes more sense the longer you think about it, it’s likely a paradox. Context and tone are key clues to spotting them.
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7128 | https://winter.group.shef.ac.uk/vsepr/radicals.html | | |
| You are at: University of Sheffield » Chemistry » Mark Winter » VSEPR |
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| | VSEPR home | | | | Introduction Assumptions Geomtries Geometry prediction Examples + CH4 + NH3 + OH2 + BF3 + H3N-BF3 + [PF6]- + [AsPh4]+ + Propene + SF3N Isomers + ClF3 Multiple bonds + [ClO4]- + [MeCO2]- Unpaired electrons + NO2 Beyond six electron pairs d-block systems Exercises Online VSEPR calculator | | | | | The University of Sheffield | | Department of Chemistry | VSEPR | Radicals and VSEPR calculation for nitrogen dioxide, NO2 Another complication crops up when there are unpaired electrons. This is well illustrated for nitrogen dioxide. So, for NO2 there is an integral number of electrons but a non-integral number of electron pairs. Since any orbital can accommodate 0, 1, or 2 electrons, 21/2 electron pairs must be placed into three orbitals. Therefore the geometry is based upon a trigonal-planar arrangement of electron pairs. Since the lone-pair orbital is only half filled, it demands less space, and the O-N-O angle opens out a little (to 134.1°) from the ideal trigonal angle of 120°. Addition of one electron to NO2 gives the nitrite anion NO2-. This last electron completes the half-occupied lone-pair orbital and this filling of the orbital causes it to fill out, and so close the O-N-O bond angle to 115°. Nitrogen dioxide, NO2 | Lewis structure: | | Central atom | nitrogen | | Valence electrons on central atom: | 5 | | 2 terminal oxygens each contribute 1 electron in the two σ bonds: | 2 | | Subtract two for the two electrons contributed by N to the two π bonds: | -2 | | Total: | 6 | | Divide by 2 to give electron pairs | 21/2 | | 21/2 electron pairs: | trigonal geometry for the 21/2 shape-determining σ-framwork orbitals | | | | The geometry of nitrogen dioxide, NO2. You can use your mouse to manipulate the molecule in the right hand "Jmol" image. | |
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A VSEPR tutorial on the WWW
VSEPR tutorial on the WWW, URL: Copyright 1996-2015 Prof Mark Winter [The University of Sheffield]. All rights reserved. Document served: Sunday 28th September, 2025 |
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A beam broadening method for phased arrays in wireless communications
EURASIP Journal on Wireless Communications and Networking
volume 2022, Article number: 91 (2022)
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Abstract
5G and IEEE 802.11ay introduce the use of the millimeter band as one promising solution to provide broadband wireless communication at multi-Gb/s user data rate. Due to the severe path-loss at such frequencies, it is generally assumed that large antenna arrays are used at the base station to steer narrow beams and build highly directional communication links towards the terminal points. However, broader and less directional beams are also of high interest in some of the steps involved in the establishment or the maintenance of the communication links . Indeed, search of a large area by narrow beams becomes too time consuming and link outage becomes more critical, thus affecting the latency and the robustness of the communications. A method enabling an adaptation of the beam widths is then worthwhile to consider. In this article, we investigate how narrow beams naturally produced by large antenna arrays can be broadened to adapt the beam width to a desired angular sector. We consider that the multi-antenna processing is performed by phase shifters on the radio-frequency stage since fully digital beamforming is hardly feasible in practice at such high frequencies. The main idea of our systematic phase-only beam broadening technique relies on the determination of a quadratic phase excitation law from a desired beam width and steering angle. We first lead a thorough analysis of the radiation behavior regarding the coefficients of such quadratic excitation. We then propose a calculation method for determining the polynomial coefficients as a function of the desired beam width and steering angle. This non-iterative beam broadening method is described for boresight and non-boresight directions and is intended for discrete antenna arrays.
1 Introduction
The recent introduction of the millimeter band in the last versions of the prevailing wireless communication standards, namely 5G NR and IEEE 802.11ay, is considered as one major enabler for the enhancement of the capacity of wireless networks. Working with carrier frequencies of several tens of GHz is indeed highly attractive owing to the very large bandwidth available in this portion of the radio spectrum. However, millimeter waves suffer from much higher propagation losses compared to lower frequencies. In addition to the strong path loss given by the well-known Friis transmission equation, signals at millimeter wave penetrate less easily through buildings, solid materials or even human bodies [1,2,3]. A convenient way to combat such drawbacks is to establish directional communications towards users or terminals by means of adaptive beamforming techniques.
Forming directional beams can for instance be easily implemented using a linear antenna array controlled by a linear phase excitation. The 3dB beam width obtained in this way has the remarkable property of being inversely proportional to the antenna array length while the maximum gain is proportional to (10\log _{10}(M))dB where M denotes the number of antennas composing the array . Although those properties are theoretically beneficial in order to combat strong path-losses and increase the received power, beam misalignment may occur in practical scenarios, especially with large antenna arrays, thus leading to poor link quality . On one other hand, very narrow beams are costly regarding beam scanning latency time and not well-suited for broadcast channels that have to be received by several users [6, 7]. Finally, in regard to the penetration problems of millimeter waves, it has been shown that less-directional beams can improve link resilience since the energy from non-line-of-sight paths is retrieved [8, 9]. As depicted throughout these scenarios, being able to adapt and increase the width of the formed beams becomes essential at various levels of the communication link management.
Beam broadening techniques have always been a subject of research for radar applications [10, 11] and has more recently become a topic of interest for mobile wireless communications as the community started looking at the millimeter band . Generally speaking, a broadened beam can be designed and controlled using adequate amplitude and phase excitations . Such an approach is well suited to the fully digital beamforming implementation which is however hardly applicable to the millimeter wave context [13, 14]. Pure analog or at least hybrid analog-digital beamforming architectures have rather to be considered for millimeter wave front-ends [13, 15]. This implies that beam direction and width have to be managed at the analog stage. On that basis, amplitude excitation may be achieved by controlling the gains of the power amplifiers while phase excitation may be obtained through phase shifters. For power efficiency reasons, it is however recommended for millimeter wave applications that power amplifiers operate at maximum power rather than tuning their gains [16, 17]. Consequently, phase-only element weights are preferable in practical millimeter wave beamformers. Finding the phase excitation that ensures a given beam width then becomes a nonlinear and non-convex optimization problem .
Optimization of beam broadening techniques constrained to unit amplitude weights have already been studied in many papers [10, 18,19,20]. The obtained methods yield interesting results but lack of flexibility since the optimization process has to be done offline regarding particular predefined configurations. Contrary to these methods, authors in provide a systematic approach for beam broadening. The idea consists in dividing the array into multiple logical subarrays, each being controlled by an independent linear phase excitation and being responsible for an elementary beam associated with a predefined direction. The resulting broadened beam is obtained by summation of the elementary ones. The major drawback of this method is that the number of broadened beams that can be formed is limited by the number of antenna elements comprised in the array. Indeed, each subarray has to be composed of the same number of antennas with the constraint that the number of subarrays should not exceed the number of antennas per subarray. For example, only three different beams are configurable for a 128-element antenna array. Authors in proposed a similar approach for which the steering directions of the elementary beams are refined through an optimization process. This solution appears to be more flexible but also more computationally expensive. Other beam broadening strategies can finally be found as in , where the broadened beam patterns are obtained through nonlinear parametric phase excitations. However, no other method than an exhaustive search is suggested for tuning the phase law parameters regarding the desired beam width. Besides, such an approach can be quite tricky since different quadratic coefficients give the same beam width.
In this paper, we introduce a beam broadening control method that is based on a quadratic phase excitation and that involves two parameters. Our method requires neither an exhaustive search nor an iterative one to find the adequate control parameters. Throughout the article, we establish a bijective function linking these two parameters with the beam width and the steering angle of the power pattern produced by the phased array. To that purpose, an analysis of the far-field radiation pattern of the array is first led to identify how the proposed quadratic phase law governs the beam shape. It is then demonstrated that the coefficients of the quadratic phase excitation can be expressed according to a new variable that has a near linear relationship with the beam width, for boresight and non-boresight directions. In this work, the beam width is characterized from the beam power efficiency. This choice has been made since the conventional half power beam width is not relevant in some configurations for which the ripples can exceed 3 dB. On that basis, we design our proposed beam broadening control method which can be implemented for various antenna array sizes, boresight and non-boresight directions.
The rest of the paper is organized as follows. In Sect. 2 some fundamental expressions about line-source radiation are reminded and the far-field radiation pattern for a quadratic phase excitation is derived. In Sect. 3, a beam control method that relies on the Fresnel functions is derived for that type of excitation. The beam width definition is then given in Sect. 4 and the relation between the Fresnel functions and the beam width is established for boresight and non-boresight directions. Finally, the accuracy of the proposed beam width control method is evaluated in Sect. 5 before concluding our work in Sect. 6.
2 Far-field beam pattern of line-sources
In this section, the general principles yielding the far-field radiation of a continuous line source are reminded and specifically derived and analyzed in case of a quadratic phase excitation. The obtained expressions serve as basis to the beam width control method developed in the sequel of the paper.
2.1 Radiation pattern of a continuous line-source
Following the conventional definition of the spherical coordinate system, the space factor (SF(\theta )) for a continuous line-source of length L placed symmetrically along the z-axis is given by ,
where (\theta) is the elevation angle, (k_0=\frac{2\pi }{\lambda }) represents the wave number with (\lambda) the wavelength, while I(z) and (\Phi (z)) correspond, respectively, to the amplitude and phase distributions along the source. Following the common assumption that the source operates at maximum power rate without any amplitude change at millimeter wave frequencies, a uniform amplitude distribution is considered, i.e. (I(z)=\frac{I_0}{L}). It is then noticeable that Eq. (1) relates the far-field pattern of the source to its excitation distribution through the Fourier transform of a complex exponential function f(z) as,
with,
A classic excitation of the form (e^{j\Phi (z)}) is the linear phase distribution defined as,
with,
Such linear phase excitation has the interesting property of steering the maximum radiation towards the direction (\theta _{\mathrm {max}}). One of the main drawbacks of this method is that the 3dB beam width is not tunable for a given direction. Indeed, the half power beam width is inversely proportional to the source length L and to (\sin \left( \theta _{\mathrm {max}}\right)).
Since the far-field pattern is related to its excitation through the fourier transform (2), temporal waveforms of the form (e^{j\Phi (t)}) with easily tunable power spreading properties in the frequency domain are good candidates for beam widening. The linear chirp is one of them as a configurable bandwidth can be swept by introducing quadratic variations on the instantaneous phase . We thus propose to conduct a deeper study on the influence of a quadratic phase excitation on the far-field radiation pattern.
2.2 Radiation pattern with a quadratic phase excitation
Let us consider a quadratic phase distribution of the form,
in which (B_1) given by Eq. (5) is the coefficient that controls the linear phase shift between antenna elements, i.e the beam direction, while (B_2) adds a quadratic phase shift that is expected to generate a broadened beam owing to the known spectrum shape of chirp signals mentioned above. Note that no constant coefficient is considered in the quadratic law since it would not affect the beam pattern. By integrating Eq. (6) into Eq. (1), we obtain,
with the vertex form of (\phi (z)) being,
Substituting variable z by (u=\sqrt{\frac{2B_2}{\pi }}\left( z+\frac{B_1+k_0\cos (\theta )}{2B_2}\right)), we get,
where,
Eq. (9) can be rewritten using the normalized cosine Fresnel integral (C(u)=\int _{0}^{u}{\cos {\left( \frac{\pi }{2}t^2\right) }}\mathrm{d}t) and the normalized sine Fresnel integral (S(u)=\int _{0}^{u}{\sin {\left( \frac{\pi }{2}t^2\right) }}\mathrm{d}t) as ,
Finally, the radiated power of a continuous line-source for a quadratic phase excitation can readily be expressed as follows,
This expression gives insight on the fact that the radiated power is governed by the behavior of Fresnel functions evaluated on (\underline{u}_\theta) and (\overline{u}_\theta), for (\theta \in [0,\, \pi ]). To intuitively understand how the quadratic phase variation is responsible for the beam widening phenomenon, it is interesting to express the difference (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }) between the integral bounds, or equivalenlty the Fresnel function evaluation points, that is,
This integration interval remarkably depends on coefficient (B_2), i.e the quadratic phase variation, for a given length source L. (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }) is depicted in Figs. 1 and 2, on which the left-hand plots correspond to the Fresnel functions, with a visualization of the interval range covered by variables (\underline{u}_\theta) and (\overline{u}_\theta) when (\theta) goes from 0 to (\pi). As observed when making the link with the right-hand figures, the beam shape results from the traveling of (\underline{u}_\theta) and (\overline{u}_\theta) points on the Fresnel functions. From Figs. 1 and 2, it is observed that (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }), i.e the distance between (\overline{u}_\theta) and ({\underline{u}}_\theta), drives the angular distance between the couple of Fresnel integrals ((C(\overline{u}_\theta ),S(\overline{u}_\theta ))) and ((-C(\underline{u}_\theta ),-S(\underline{u}_\theta ))). Thus, the angular distance between the couple of Fresnel integrals can be tuned through (B_2)for a given length source L. More precisely, the beam formation is triggered when (\underline{u}_\theta) pass by the minima of the normalized cosine Fresnel function and released when (\overline{u}_\theta) pass by the maxima of the normalized cosine Fresnel function. As can be observed comparing Figs. 1 and 2, the larger (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }), the higher the angular distribution of the radiated power will be.
Continuous line-source of length (L=16\lambda) ((\lambda =0.01\text {m})) with input parameters (B_1=0) and (B_2=600)
Continuous line-source of length (L=16\lambda) ((\lambda =0.01\text {m})) with input parameters (B_1=0) and (B_2=3000)
From this analysis, it turns out that (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }) plays a central role in the adaptation of the beam width. However, its influence through the Fresnel functions and far-field radiation pattern is highly nonlinear, which prevents a straightforward usage of Eq. (13) as a way to adapt parameter (B_2) and tune the beam width. For example, although it is clear that the lowest (B_2) value is zero, leading to a pure linear phase excitation (coming back to Eq. (6)), the upper limit is harder to define since (\underline{u}_\theta) and (\overline{u}_\theta) depend both on the length source L and the steering angle (\theta _{\mathrm {max}}). Hence, a deeper analysis of the influence of parameter (B_2) has to be led to properly control the broadening effect.
Highlight of the match between the Fresnel angular distance and the beam width
3 Beam broadening control
From the previous section, it is understood that the beam shape obtained from a quadratic excitation of a line-source depends on the evaluation intervals of the Fresnel integrals. Those intervals correspond to the ranges of (\underline{u}_\theta) and (\overline{u}_\theta) functions, both depending on the coefficients (B_1) and (B_2). Our goal is now to establish a formal expression making the link between such coefficients and the beam width. Due to the complexity of the manipulation of the transcendent Fresnel integrals, a complete closed form derivation is intractable to find such an expression. Some convenient approximation is however possible by considering the localization of the cosine Fresnel functions maxima.
3.1 Fresnel angular distance
As depicted in Fig. 3, the studied beam width is highly correlated to the spacing between the maxima of the cosine Fresnel functions (C(\overline{u}_\theta )) and (-C(\underline{u}_\theta )). In contrast, the influence of the sine Fresnel functions seems to remain less significant. In the sequel, the spacing between the maxima of the cosine Fresnel functions (C(\overline{u}_\theta )) and (-C(\underline{u}_\theta )) is referred to as the Fresnel angular distance (\Delta _F). The Fresnel angular distance is defined as,
where,
(\Delta _F) is expected to serve as a comprehensive dimensioning parameter, contrary to the function (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }) previously introduced. As further illustrated in 4, one additional motivation for introducing (\Delta _{F}) comes from the near linear relationship between that variable and the beam width of the phased array exploiting a quadratic phase excitation.
As cosine Fresnel functions C(u) and (-C(u)) take their global maxima at (u=1) and (u=-1), the analytical derivation of (\Delta _F) can be obtained from the expression of variables (\underline{u}_\theta) and (\overline{u}_\theta) introduced in Eq. (10) by simply solving (\underline{u}_\theta =-1) and (\overline{u}_\theta =1). Equation (14) is then rewritten,
with, (x_0=-B_1/k_0=\cos (\theta _{\mathrm {max}})) and,
It appears that the Fresnel angular distance depends on both the targeted steering angle (\theta _{\mathrm {max}}), through parameter (B_1), and the length variation (\Delta _x) as a function of the quadratic parameter (B_2). One may notice that interval (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }) is involved in the control of the Fresnel angular distance. However, the beam width adaptation through (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }) is not straightforward since this function is related to (\Delta _F) by a difference of nonlinear (\arccos) functions. Hence, as already discussed in the previous section, adapting the beam width upon (\Delta _{\overline{u}_\theta ,\underline{u}_\theta }) only is not practical, whereas using (\Delta _F) is of better convenience.
As additional comments, it is noticeable that the beam shape is symmetrical with respect to the boresight direction ((\theta _{\mathrm {max}}=90^{\circ })) for (B_1=0) and (\Delta _F) expression reduces to (\Delta _F=2\arccos (\Delta _x)-\pi). In other configurations however, i.e. for (B_1\ne 0), the beam shape is no longer symmetrical about the steering angle (\theta _{\mathrm {max}}). Remarkably, (\Delta _F=0) for (B_2=\frac{2\pi }{L^2}). This configuration corresponds to the situation for which the maxima of the Fresnel functions coincides. In that case, the beam width is limited, but not strictly equivalent to the nominal beam width obtained with a pure linear phase excitation. This latter case is rather obtained when (B_2) tends to 0. In that case, (\Delta _F) becomes negative, i.e. the maxima of the cosine Fresnel functions come in the opposite order than the one displayed in Fig. 3. To summarize, we have the following definition range of (\Delta _{F}),
where the first interval is obtained for (\Delta _x>0) and the second for (\Delta _x\le 0).
On one other hand, it is important to keep in mind that Eq. (16) assumes that the evaluated maxima in Eq. (15) correspond to the global maxima of the C(u) and (-C(u)) functions, which amounts to saying that the variables (\underline{u}_\theta) and (\overline{u}_\theta) span a range of values such that (\exists \, \theta \in [0,\, \pi ], \underline{u}_\theta =-1) and (\overline{u}_\theta =1). These constraints are implicitly embedded in Eq. (16) by the definition range of the (\arccos) functions that gives the range of values (\Delta _F). Hence, for a given desired steering angle (\theta _{\mathrm {max}}), (\Delta _x) should be such that (-1\le x_0+\Delta _x\le 1) and (-1\le x_0-\Delta _x\le 1) are both verified. After further analyzing how these inequalities restrain the range of values for (\Delta _F), it is determined that the maximum configurable value (\Delta _F^{\mathrm {max}}) is expressed as,
with,
Those two last equations may also be written as,
This drives us to an important result that the Fresnel angular distance is limited to a maximum value directly depending on the desired steering angle (\theta _{\mathrm {max}}). More specifically, we can conclude that the maximum Fresnel angular distance decreases as the steering angle increases.
3.2 (B_2) parameter adaptation
The objective is now to express the coefficient (B_2) according to the variable (\Delta _{F}), in such a way that, in the end, (B_2) can be computed from the desired Fresnel angular distance directly. To that purpose, one may rewrite (16) using the arccosine difference property defined as [24, p. 80, Eq. (4.4.33)],
where ({{\,\mathrm{Sign}\,}}(x)) denotes the sign function. This way, it is possible to express (\Delta _x) as a function of (\Delta _F). After simple manipulations we obtain,
One may verify that the constraint on the maximum value of (\Delta _F) given in Eq. (21) ensures that the argument of the square root is always positive, i.e. (\Delta _x) always exists.
It is then straightforward to combine equations (13) and (17) to extract (B_2) from a given (\Delta _x). Indeed, (B_2) merely consists in getting the roots of a quadratic form, which yields,
Through such formula, parameter (B_2) is obtained from a computed (\Delta _x) given by Eq. (22) for a desired (\Delta _F), with again (\Delta _{F}) satisfying (21). These last equations represent the mathematical statements upon which the proposed beam broadening design method presented in the next section is built.
Before getting further, it is however important to pay attention to the fact that additional constraints apply on (\Delta _x) or equivalently on (\Delta _F), to make Eq. (23) yield real values of (B_2). Even if condition (21) is sufficient for (\Delta _F\ge 0) since (\Delta _x\le 0) in that case, for (\Delta _F<0) however, we have (\Delta _x>0), and then we should impose,
After a complete derivation of such inequation involving Eq. (22), it is possible to identify the various conditions on (\Delta _F<0) for Eq. (23) to hold whatever the selected steering angle (\theta _{\mathrm {max}}). This is solved after classical though quite long mathematical derivations not detailed here. The results can be summarized as follows. Parameter (B_2) computed for negative (\Delta _F) is consistent if and only if,
such that,
where (\theta _0=\arccos (1-\frac{\lambda }{4L})), and,
Interestingly, for a source of large dimension compared to the wavelength, i.e. (L>>\lambda), (\frac{\lambda }{4L}\rightarrow 0) and (\theta _0 \rightarrow 0). Consequently, we have (\Delta _F^{\mathrm {min}} = -\arccos (c_1c_2+c_3),\ \forall \theta _{\mathrm {max}}\in \left[ 0,\pi \right]). Meanwhile, if (L>>\lambda), (c_2\rightarrow c_1), and hence one can easily find that (c_1c_2+c_3 \rightarrow 1). This finally leads to (\Delta _F^{\mathrm {min}}\rightarrow 0). This analysis allows to anticipate and give insight to a later observation showing that for large antenna arrays, the beam broadening effect is essentially achievable for positive (\Delta _F).
4 Practical beam broadening method for antenna arrays
From the previous developments, we have demonstrated that the (B_2) coefficient of a quadratic phase excitation applied to a line source can be calculated according to the newly defined (\Delta _{F}) variable whatever the source length L and the steering angle (\theta _{\mathrm {max}}). We now intend to exploit such results to propose a simple way for controlling the beam width of discrete linear antenna arrays.
Remember that (\Delta _F) is not proven to be strictly equal to the beam width, but has rather been proposed as a practical and representative parameter of it. In this section then, our goal is to study in which extent some calibration or correction process may be elaborated to make (\Delta _F) be an actual tuning parameter of the beam width, and propose hereby a systematic method for beam broadening. In particular, as shown in the sequel, (\Delta _F) turns out to be almost linearly depending on the so-called beam efficiency used as a reference metrics to characterize the broadening of the steered beams.
4.1 Discrete-element array excitation
As a preliminary step, let us remind that the mathematical proofs derived in Sect. 3 about the Fresnel angular distance control through parameter (B_2) consider the radiation characteristics of a continuous source as established in Sect. 2. In practical scenarios however, discrete-element arrays controlled by an integer number of phase shifters are rather used. Nevertheless, the radiation characteristics of a discrete-element array can be approximated by those of a continuous source, making the beam broadening phenomenom defined in the previous sections still valid in spite of the discretization [12, Chapter 7.2]. In the sequel, we consider a linear antenna array, placed along the z-axis, that is composed of M antenna elements and controlled by a quadratic phase excitation. The array factor (AF(\theta )), that is the twin of the space factor (SF(\theta )) for discrete arrays, is then given by,
with
being the quadratic phase excitation. The position of the mth antenna element is given by (z_m=(m-1)d), d being the inter-element spacing. Without loss of generality, we will consider (d=\frac{\lambda }{2}) in the sequel.
As depicted in Fig. 4, the line-source discretization mainly induces higher sidelobes but the beam width is kept the same. The coefficient parameterization derived for line-sources is thus kept unchanged in the case of linear antenna arrays.
Comparison of space factor ((L=32\lambda)) and array factor ((M=64)) for the same quadratic phase excitation
4.2 Beam width characterization
To go further, we need an appropriate metrics to measure the width of the beam produced by the antenna array. As mentioned in the introduction, ripples can exceed 3 dB when using a pure phase excitation. Thus, relying on the conventional 3 dB beam width definition is not convenient in our case. We then propose to characterize the beam width from the so-called beam efficiency of the array [12, Chapter 2.10]. The beam efficiency of an antenna may be defined as,
It represents the ratio of the power radiated within a solid angle (\Omega _b) around a main direction (\theta _b) to the total power radiated. Strictly speaking, the radiated power should be measured around the barycenter of the power pattern, i.e at the angular value that splits the power pattern in such a way that 50% percent of the power is radiated on each angular sector. Hence, the actual steering direction (\theta _b) to consider is such that,
The barycenter is an essential metrics for ensuring that the radiated power is focused in the right direction, which may be slightly different from the initial steering angle (\theta _{\mathrm {max}}) as discussed later on.
From these considerations, it is then possible to define the beam width (\Omega _b^X) of the array for a target beam efficiency of (X\%), that is,
It simply represents the angular range around (\theta _b) within which (X\%) of the total power is radiated. Hence, the beam width has not a unique value, but rather depends on the targeted beam efficiency. Note that it is convenient to choose a high value of X since the width of the main lobe is close to the targeted beam efficiency (\Omega _b^X) for such a configuration. In the following sections, we consider that (X=80) but the described method could be repeated for other X values.
In the next sections, further details are provided regarding (\theta _b) and the relationship between (\Delta _{F}) and (\Omega _b^{X}) for boresight and non-boresight directions.
4.3 Beam width control for boresight direction
(\Delta _{F}) versus (\Omega _{b}^{80}) for (M\in {12, 24, 48, 128})
For the boresight direction, i.e (\theta _{\mathrm {max}}=\pi /2), the first thing to notice is that (\theta _b=\pi /2) regardless of the (\Delta _{F}) value. Indeed, since (\theta _{\mathrm {max}}=\pi /2), (B_1) is null and (\phi _n) is an even function resulting in a symmetrical power pattern whose barycenter coincides with the steering angle (\theta _{\mathrm {max}}).
As already discussed, (\Delta _F) is a practical parameter strongly linked to the width of the beam but is not formally expressed as a function of it. Therefore, we propose to investigate about the possible relationship between (\Delta _{F}) and the previously defined beam width (\Omega _b^{X}) for an antenna array composed of M antennas. The following procedure was used :
Set the variation range of (\Delta _{F}) for M antennas from Eqs. (21) and (25) as,
Compute (B_2) coefficient using Eq. (23) for each (\Delta _F).
Compute the phase law (\phi _m) from Eq. (28) for each (\Delta _F).
Compute the array factor (AF(\theta )) for each phase law using Eq. (27).
Find (\Omega _b^{X}) by solving Eq. (29) numerically for each (AF(\theta )) obtained from each (\Delta _F).
In Fig. 5, we plot (\Delta _F(\Omega _b^{80},M)) as a function of the obtained beam width (\Omega _b^{80}) for various array sizes M. An almost linear relationship is observed between the two parameters, at least until a maximum value (\Omega _{\mathrm {max}}^{80}(M)) of the beam width above which (\Delta _F(\Omega _b^{80},M)) rapidly goes to (\pi) rad. Interestingly, the larger the antenna array, the more linear the relationship. More precisely, we can state that,
meaning that (\Delta _F) asymptotically matches the beam width.
Indeed the (I(\theta )) function tends to be rectangular as M grows due to the compression of the Fresnel functions (C(\overline{u}_\theta )) and (-C(\underline{u}_\theta )) . Consequently, for large M values, (\underline{\theta }_{\mathrm {max}}) and (\overline{\theta }_{\mathrm {max}}) perfectly coincide with the edge of a rectangle which width directly corresponds to the beam width (\Omega _b^{100}).
Moreover, (\Delta _F^{\mathrm {min}}) tends to zero (recall discussion after Eq. (26)) as it is depicted in Fig. 5 where the range of values for (\Delta _F(\Omega _b^{80},M)) progressively becomes strictly positive when M grows.
As we may apply our approach for millimeter transmission scenario, we are interested in the situation where M is not too high. It is hence important to study the beam width control for (\Delta _{F}(\Omega _b^{X},M)\ge 0) as well as (\Delta _F(\Omega _b^{X},M)<0). As the relationship is asymptotically linear, we suggest to approximate it by a linear polynomial even at small M. As may be noticed from Fig. 5, the average slope of the function is slightly different for positive and negative values of (\Delta _F(\Omega _b^{80},M)). Hence, each linear approximation may be studied individually as follows.
4.3.1 Linear approximation for (\Delta _F(\Omega _b^{X},M)<0)
In this case we have (\Delta _F\in [\Delta _F^{\mathrm {min}}(M),\, 0[), with (\Delta _F^{\text {min}}(M)=-\arccos {\left( 1-\frac{1}{2M^2}\right) }). Accordingly, let (\Omega ^X_{0}(M)) and (\Omega ^X_{\mathrm {min}}(M)) denote the beam width values, respectively, associated with (\Delta _F=0) and (\Delta _F=\Delta _F^{\mathrm {min}}(M)). The linear relationship then writes,
It follows that if (\Omega _{0}^X(M)) and (\Omega ^X_{\mathrm {min}}(M)) are known (\forall M), then Eq. (33) directly gives the adequate (\Delta _F) for a targeted beam width (\Omega _b^X) and a given array size M. A numerical study of the variation of (\Omega _0^{80}) and (\Omega _{\mathrm {min}}^{80}) versus M yields the curves depicted in Fig. 6. It is observed that both functions may be approached by the multiplicative inverse of a linear polynomial, that is,
with (\alpha) and (\beta) some real scalar values, depending on X. Through a least square fitting to such model for (X=80\%) for instance, we get the following approximations,
(\Omega _{\mathrm {min}/0}^{80}(M)) and (\tilde{\Omega }_{\mathrm {min}/0}^{80}(M)) for (M\in [12,128])
The tightness of the proposed model is validated in Fig. 6 where we observe that the beam width values (\tilde{\Omega }_{\mathrm {min}/0}^{80}(M)) generated by Eq. (35) are perfectly matching the actual ones obtained by simulations. We then conclude that (\Delta _F(\Omega ^{80}_b,M)) can accurately be approximated using Eq. (33) by substituting (\Omega ^X_0(M)) and (\Omega ^X_{\mathrm {min}}(M)) by the approached values given by Eq. (35). This methodology may be repeated for an other (\Omega _b^X), e.g. (X=75\%) or (X=90\%).
4.3.2 Linear approximation for (\Delta _F(\Omega _b^{X},M)\ge 0)
For (\Delta _{F}(\Omega _b^{X},M)\ge 0), we assume that (\Omega _{F}^X \in [\Omega _0^X(M),\, \Omega ^X_{\mathrm {max}}]) with (\Omega _{0}^X(M)) already given by Eq. (35) and (\Omega ^X_{\mathrm {max}}) chosen equal to (90^{\circ }). As observed in Fig. 5, the latter corresponds to a reasonable maximum value that enables the configuration of a large beam width while still ensuring a near linear relationship between (\Delta _{F}(\Omega _b^{80},M)) and (\Omega _b^{80}) whatever M. We may choose a higher value of (\Omega ^X_{\mathrm {max}}), possibly reaching a sector of (120^{\circ }), for arrays of several hundreds of antenna elements. On this basis, we may approximate (\Delta _F(\Omega ^X_b,M)) as a linear function as,
where (p^X(M)) is the slope of the linear polynomial for a given array size M. (p^X(M)) is determined numerically by means of linear regression analysis and is depicted in Fig. 7 for (M\in \left[ 12,128\right]). This function may be approximated by (\tilde{p}^{80}(M)) which is expressed as,
(p^{80}(M)) and (\tilde{p}^{80}(M)) for (M\in [12,128])
To summarize, the beam width (\Omega _b^X) ranges from (\Omega _{\mathrm {min}}^X(M)) to (\Omega _{\mathrm {max}}^X=90^{\circ }) and is related to (\Delta _F(\Omega _b^X,M)) through the function (\tilde{\Delta }_F(\Omega _b^X,M)) by means of two linear approximations given by Eq. (33) and (36). Interestingly, the terms composing Eq. (33) and (36), i.e (\Omega ^X_{\mathrm {min}}(M)), (\Omega ^X_{0}(M)) and (p^X(M)) can be accurately approximated by Eq. (35) and (37) for (M\in [12,128]) and (X=80\%). Therefore, a systematic relation connect the beam width (\Omega _b^{80}) to the Fresnel angular distance (\Delta _F). As a final step, the quadratic phase law can be computed systematically according to the desired (\Omega _b^{80}) since the quadratic coefficient (B_2) is related to (\Delta _F), recalling Eqs. (22) and (23). As for (\Delta _F(\Omega _b^{X},M)<0), this methodology may be repeated for an other (\Omega _b^X). Note that the near linear relationship between (\Omega _b^{X}) and (\Delta _F) is obtained for a larger range of (\Delta _F) values when X is large, e.g (X=70\%) or (X=80\%). Therefore, a high value of X is advised as the operating range of the proposed beam broadening method is larger for such a configuration.
4.4 Non-boresight directions
(\theta _b(\theta _{\text {max}},\Delta _F)) for an antenna array composed of (M=32) antenna elements and a maximum drift (\theta _{\mathrm {lim}}=2.5^{\circ })
For non-boresight directions, i.e (\theta _{\mathrm {max}}\ne \pi /2), (\phi _m) is no longer an even function since (B_1) introduces a linear phase shift. The resulting power pattern is asymmetric and the barycenter (\theta _b) depends on the (\Delta _{F}). As previously discussed, the barycenter is an essential aspect so it is important to ensure that the drift between the desired steering angle (\theta _{\mathrm {max}}) and the effective one, i.e the barycenter (\theta _{b}), is not too high.
For simplification purposes, we consider an antenna array comprising (M=32) antenna elements. In addition, we assume that (\theta _{\mathrm {max}}) ranges from 40(^\circ) to 140(^\circ) with an angular resolution of 5(^\circ). Considering those parameters, the barycenter (\theta _b(\theta _{\text {max}},\Delta _F)) is evaluated numerically for (\Delta _F\in \left[ \Delta _F^{\mathrm {min}}(\theta _{\text {max}}),\Delta _F^{\mathrm {max}}(\theta _{\mathrm {max}})\right]), recalling Eq. (21) and (25). As depicted in Fig. 8, the drift between (\theta _{\mathrm {max}}) and (\theta _b(\theta _{\text {max}},\Delta _F)) increases with (\Delta _{F}). Consequently, a trade-off exists between the maximum desired drift (\theta _{\mathrm {lim}}) and the maximum configurable beam width (\Omega _{\mathrm {max}}^X(\theta _{\mathrm {max}})). In order to ensure that the maximum effective drift doesn’t exceed (\theta _{\mathrm {lim}}), we determine numerically the maximum Fresnel angular distance (\Delta _F^{\text {max}}(\theta _{\text {max}},\theta _{\text {lim}})) which ensures that,
In the suggested implementation, a maximum drift (\theta _{\mathrm {lim}}) of 2.5(^\circ) is chosen owing to the 5(^\circ) angular resolution of (\theta _{\mathrm {max}}). The values taken by (\Delta _F^{\text {max}}(\theta _{\text {max}},\theta _{\text {lim}}=2.5^{\circ })) for (M=32) are illustrated in Fig. 8 and given in Table 1.
In order to determine the relationship between (\Delta _F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) and (\Omega ^X_b) for non-boresight directions, the 5-step procedure presented in Sect. 4.3 is applied, with respect to (\Delta _F\in \left[ \Delta _F^{\text {min}}(\theta _{\mathrm {max}}),\Delta _F^{\text {max}}(\theta _{\text {max}},\theta _{\text {lim}}=2.5^{\circ })\right]). In a similar manner, the relation between (\Delta _F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) and (\Omega ^X_b) is approximated by means of two linear approximations.
4.4.1 Linear approximation for (\Delta _F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})<0)
(\Omega _{\mathrm {min}/0}^{80}(\theta _{\mathrm {max}})) and (\tilde{\Omega }_{\mathrm {min}/0}^{80}(\theta _{\mathrm {max}})) for (\theta _{\mathrm {max}}\in [40,140]) and (M=32)
(\Delta _{F}) and (\tilde{\Delta }_{F}) versus (\Omega _{b}^{80}) for (M\in {12,16,32,64,128})
Following the same approach as the one described in Sect. 4.3 for (\Delta _F(\Omega _b^{X},M)<0) estimate, (\Delta _F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) may be approximated by,
A numerical study of the variation of (\Omega _0^{80}) and (\Omega _{\mathrm {min}}^{80}) versus (\theta _{\mathrm {max}}) yields the curves depicted in Fig. 9. It is observed that both functions may be approached by quadratic functions, that is,
with (\alpha), (\beta) and (\gamma) some real scalar values, depending on X. Through a least square fitting to such model for (X=80\%) for instance, we get the following approximations,
The tightness of the proposed model is validated in Fig. 9 where we observe that the beam width values (\tilde{\Omega }_{\mathrm {min}/0}^{80}(\theta _{\mathrm {max}})) generated by Eq. (41) match the ones obtained by simulations with fair accuracy. We then conclude that (\Delta _F(\Omega ^{80}_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) can be approximated using Eq. (39) by substituting (\Omega ^X_0(\theta _{\mathrm {max}})) and (\Omega ^X_{\mathrm {min}}(\theta _{\mathrm {max}})) by the approached values given by Eq. (41). This methodology may be repeated for an other (\Omega _b^X), e.g. (X=75\%) or (X=90\%).
Mean relative and absolute error for (\Omega _{b}^{80}\in \left[ \Omega _{\mathrm {min}}^{80}(M),\Omega _{\mathrm {max}}^{80}\right]) and (M\in [12,128])
(\Delta _{F}) and (\tilde{\Delta }_{F}) versus (\Omega _{b}^{80}) for (M=32), (\theta _{\mathrm {lim}}=2.5^{\circ }) and (\theta _{\mathrm {max}}\in {95^{\circ },110^{\circ },125^{\circ },140^{\circ }})
4.4.2 Linear approximation for (\Delta _F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}}) \ge 0)
The general idea described in Sect. 4.3 for (\Delta _F(\Omega ^X_b,M) \ge 0) is followed. The main difference is that the maximum beam width (\Omega _{\mathrm {max}}^{X}(\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) depends on the steering angle (\theta _{\mathrm {max}}) and on the maximum desired drift (\theta _{\mathrm {lim}}), while (\Omega _{\mathrm {max}}^{X}) is constant (M-invariant) in Sect. 4.3. Hence, (\Delta _F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) may be approximated by,
where (p^X(\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) is the slope of the linear polynomial for a given steering angle (\theta _{\mathrm {max}}) and a desired maximum drift (\theta _{\mathrm {lim}}). In contrast with the method described in Sect. 4.3, ({p^X}(\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) is not approximated by means of a linear regression analysis. Indeed, the following estimate,
is more appropriate to ensure that (\tilde{\Delta }_F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) doesn’t exceed (\Delta _F^{\text {max}}(\theta _{\text {max}},\theta _{\text {lim}})) and thus respect the expected maximum drift (\theta _{\mathrm {lim}}). Note that (\Omega _{\mathrm {max}}^{80}(\theta _{\mathrm {max}},\theta _{\mathrm {lim}}=2.5^{\circ })) values have been estimated numerically for (M=32) and are given in Table 1.
To summarize, the beam width (\Omega _b^X) ranges from (\Omega _{\mathrm {min}}^X(\theta _{\mathrm {max}})) to (\Omega _{\mathrm {max}}^X(\theta _{\text {max}},\theta _{\text {lim}})) and is related to (\Delta _F(\Omega _b^X,\theta _{\text {max}},\theta _{\mathrm {lim}})) through the function (\tilde{\Delta }_F(\Omega _b^X,\theta _{\text {max}},\theta _{\mathrm {lim}})) by means of two linear approximations given by Eq. 39 and 42. The maximum Fresnel angular distance (\Delta _F^{\text {max}}(\theta _{\text {max}},\theta _{\text {lim}})) in Eq. 43 limits the maximum drift to (\theta _{\text {lim}}) and has to be determined numerically in conjunction with (\Omega _{\mathrm {max}}^{X}(\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) for the steering angles (\theta _{\text {max}}) of interest. Remarkably, the terms (\Omega ^X_{\mathrm {min}}(\theta _{\mathrm {max}})) and (\Omega ^X_{0}(\theta _{\mathrm {max}})) can be approximated with fair accuracy by Eq. 41 for (\theta _{\mathrm {max}}\in [40^{\circ },140^{\circ }]) and (X=80\%), considering an antenna array composed of (M=32) antenna elements. Therefore, a systematic relation connect the beam width (\Omega _b^{80}) to the Fresnel angular distance (\Delta _F) for a given steering angle (\theta _{\mathrm {max}}) and desired maximum drift (\theta _{\mathrm {lim}}). As a final step, the quadratic phase law can be computed systematically according to the desired (\Omega _b^{80}) since the quadratic coefficient (B_2) is related to (\Delta _F) by means of Eq.22 and 23.
5 Results and discussion
In the last section, a 5-step procedure that enables to determine the relation between (\Delta _F) and (\Omega _b^X) has been described for boresight and non-boresight directions. From this procedure, the functions (\Delta _F(\Omega ^X_b,M)) and (\Delta _F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) have been determined numerically and approximated by (\tilde{\Delta }_F(\Omega ^X_b,M)) and (\tilde{\Delta }_F(\Omega ^X_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}})) in a systematic manner for (X=80\%). Consequently, the quadratic phase law (\phi _m) can be determined systematically too for an aimed beam width (\Omega ^{80}_b) and steering angle (\theta _{\mathrm {max}}). The proposed beam broadening method has the advantage of being lowly complex compared to existing solutions since neither the computation of the array factor nor any iterative process is needed to obtain the coefficients that drive the quadratic phase law (\phi _m). In this section, the accuracy of the proposed systematic beam broadening method is assessed by comparing the aimed beam width (\Omega _b^{80}) with the effective one. This evaluation actually measures the accuracy of the approximation (\tilde{\Delta }_F). The precision of the suggested beam broadening method is evaluated regarding the antenna array size M for the boresight direction and regarding the steering angle (\theta _{\text {max}}) for non-boresight directions. In the latter case, the maximum drift between the desired steering angle (\theta _{\mathrm {max}}) and the barycenter (\theta _{b}) is also measured.
5.1 Boresight direction
Mean relative error and maximum absolute drift for (\Omega _{b}^{80}\in \left[ \Omega _{\mathrm {min}}^{80}(\theta _{\mathrm {max}}),\Omega _{\mathrm {max}}^{80}(\theta _{\text {max}},\theta _{\text {lim}}=2.5^{\circ })\right]) and (\theta _{\mathrm {max}}\in [95^{\circ },140^{\circ }])
The mean relative error and the mean absolute error between the aimed beam width (\Omega _{b}^{80}) and the effective one are the metrics chosen for assessing the accuracy of the proposed systematic beam broadening method. The simulations are performed for (\Omega _b^{80}\in \left[ \Omega _{\mathrm {min}}^{80}(M),\Omega _{\mathrm {max}}^{80}=90^{\circ }\right]) and the obtained functions (\Delta _F(\Omega ^{80}_b,M)) and (\tilde{\Delta }_F(\Omega ^{80}_b,M)) are depicted in Fig. 10 for (M\in {12,16,32,64,128}). It is observed that those two functions are close to each other whatever the array size M, which is confirmed by the results obtained for the mean relative error and the mean absolute error (Fig. 11). The approximations suggested in Sect. 4.3 shouldn’t lead to a mean relative/absolute error that exceeds (2.5\%/0.8^{\circ }) for an antenna array size (M\in [12,128]). Note that the mean absolute error is higher for small M values since the relation between (\Delta _F(\Omega ^X_b,M)) and (\Omega ^X_b) is not strictly linear as observed in Fig. 10.
5.2 Non-boresight directions
The mean relative error between the aimed beam width (\Omega _{b}^{80}) and the effective one is the metrics chosen for assessing the accuracy of the proposed systematic beam broadening method for non-boresight directions, considering an antenna array composed of (M=32) antenna elements. In addition, the maximum drift between the desired steering angle (\theta _{\mathrm {max}}) and the barycenter (\theta _b) is calculated numerically to check that the (\theta _{\mathrm {lim}}) constraint introduced in Sect. 4.4 is verified. The simulations are performed for (\Omega _{b}^{80}\in \left[ \Omega _{\mathrm {min}}^{80}(\theta _{\mathrm {max}}),\Omega _{\mathrm {max}}^{80}(\theta _{\text {max}},\theta _{\text {lim}}=2.5^{\circ })\right]) and (\theta _{\mathrm {max}}\in [95^{\circ },140^{\circ }]) with an angular resolution of 5(\circ). As the array factors obtained for a steering angle (\theta _{\mathrm {max}}=\theta _0) and (\theta _{\mathrm {max}}=\pi -\theta _0) are symmetrical about (\theta =90^{\circ }), the interval (\theta _{\mathrm {max}}\in [40^{\circ },85^{\circ }]) is not evaluated since the performance would be exactly the same than the one obtained for (\theta _{\mathrm {max}}\in [95^{\circ },140^{\circ }]). The obtained functions (\Delta _F(\Omega ^{80}_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}}=2.5^{\circ })) and (\tilde{\Delta }_F(\Omega ^{80}_b,\theta _{\mathrm {max}},\theta _{\mathrm {lim}}=2.5^{\circ })) are depicted in Fig. 12 for (\theta _{\mathrm {max}}\in {95^{\circ },110^{\circ },125^{\circ },140^{\circ }}). As expected, it is observed that the width of (\Omega _b^{80}) range decreases as (\theta _{\mathrm {max}}) increases so as to limit the maximum drift to (\theta _{\mathrm {lim}}=2.5^{\circ }). The red curve depicted in Fig. 13 validates the approach described in Sect. 4.4 since the maximum drift between the desired steering angle (\theta _{\mathrm {max}}) and the barycenter (\theta _b) equals (\theta _{\mathrm {lim}}) whatever (\theta _{\mathrm {max}}). In addition, the mean relative error between the aimed beam width (\Omega _{b}^{80}) and the effective one doesn’t go beyond 3%, which seems fairly accurate. Note that the lower precision of the suggested method for steering angles close to the boresight directions is attributable to the greater width of (\Omega _b^{80}) range.
6 Conclusion
In this paper, the far-field radiation pattern produced by a continuous line-source for a quadratic phase excitation has been derived. An intuitive explanation of the beam broadening phenomenon, that involves the Fresnel functions, has also been provided for a better understanding of the quite complex expression that characterized the space factor for that type of excitation. It has also been shown that the quadratic coefficient of the phase law can be expressed as a function of the angular distance between the cosine Fresnel functions. As discussed in the paper, this angular distance precisely lead the beam width for very large antenna arrays but not for common size ones. Consequently, a procedure that enables to tune systematically the quadratic coefficient according to the desired beam width has been designed for boresight and non-boresight directions. In addition to being systematic, the proposed solution provides an accurate beam width control as shown by the simulation results. From the perspective of authors, the solution detailed in this paper has the benefit of being easily implementable considering both complexity aspect and hardware constraints, making it well-suited for mobile wireless communications in the millimeter band.
Further improvements could still be envisioned. Indeed, the current procedure enable to determine the quadratic coefficient whatever the array size for the boresight direction and whatever the steering angle for non-boresight directions. The next step would be to generalize the approach whatever the array size and the steering angle for non-boresight directions. Moreover, it would be interesting to extend the concept to planar arrays as 3D-beamforming is a key concept for the new generation of mobile communication systems.
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Authors and Affiliations
Orange Labs, Rennes, France
Corentin Fonteneau & Bruno Jahan
Univ Rennes INSA Rennes, IETR - UMR 6164, 35000, Rennes, France
Corentin Fonteneau & Matthieu Crussière
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Contributions
CF came up with the initial idea of beam broadening control by means of Fresnel functions, completed the analytical derivations, and conducted the simulations. MC proposed to characterize the beam width from the beam efficiency of the array. BJ suggested the barycenter metrics for non-boresight directions. CF and MC wrote this paper. All authors read and approved the final manuscript.
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Corentin Fonteneau received M.S. degree in electrical engineering from INSA Rennes, Rennes, France, in 2018. He is currently pursuing the Ph.D degree in Orange Labs, Rennes. His current research interests include massive MIMO, millimeter wave communication, ultra-reliable and low latency communication.
Matthieu Crussière received the M.S. and Ph.D. degrees in electrical engineering from the National Institute of Applied Sciences, France, in 2002 and 2005, respectively. Since 2005, he has been an Associate Professor affiliated to the Research Institute of Electronics and Telecommunications, Rennes, France. In 2014, he started collaborations as an Associate Researcher with the Institute of Research and Technology B-COM, Rennes. He has authored or co-authored over 80 technical papers in international conferences and journals. He has been involved in several European and French national research projects in the field of powerline, broadcasting, ultrawideband, and mobile radio communications. His main research interests lie in digital communications and signal processing, with a particular focus on multicarrier and multiantenna systems.
Bruno Jahan received the M.S. degree in optical and photonics and the M.S. degree in electronic systems from the University of Paris-Sud, Orsay, France, in 1989 and 1990, respectively. In 1991, he was with Télédiffusion de France as a Research Engineer. He joined Orange Labs (formerly France Telecom), Rennes, in 1998. His research interests include digital signals processing for wire and wireless communications.
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Fonteneau, C., Crussière, M. & Jahan, B. A beam broadening method for phased arrays in wireless communications.
J Wireless Com Network 2022, 91 (2022).
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This chapter focuses on Paramyxoviridae family whose member genuses include Rubulavirus , Avulavirus , Respirovirus , Pneumovirus , and Metapneumovirus. The virions of this family consist of a lipid envelope surrounding a nucleocapsid, and the envelope is derived directly from the host cell plasma membrane by budding, and contains two or three transmembrane glycoproteins. The virions contain a single molecule of linear, negative sense, ssRNA that is not infectious alone, but is infectious in the form of the nucleocapsid. Genome lengths for all viruses in the subfamily Paramyxovirinae are multiples of 6, which is a requirement for the efficient replication of the members of the subfamily Paramyxovirinae, but does not apply to the members of the subfamily Pneumovirinae. Some virions may contain positive sense RNA and thus, partial self-annealing of extracted RNA may occur. Members of the subfamily Paramyxovirinae encode 7–10 proteins (5–250 kDa) of which 2–4 (or more) are derived from the 2–3 overlapping ORFs in the P locus. Pneumoviruses encode 9–11 proteins of 4.8–250 kDa, including two proteins encoded by overlapping ORFs in the M2 locus. The virions are composed of 6% carbohydrate by weight which is again dependent on the host cell. Fusion and attachment proteins are glycosylated by N-linked carbohydrate side chains and in the subfamily Pneumovirinae, the attachment protein (G) is heavily glycosylated by O-linked as well as N-linked carbohydrate side chains.
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Copyright © 2012 International Committee on Taxonomy of Viruses. Published by Elsevier Inc. All rights reserved.
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How to find the values of 2 unknown variables in data set when its mean and standard deviation are known?
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Suppose I have a set of values like:
2,4,5,6,8,x,y.
The mean(arithmetic mean) and standard deviation of this data set are already known. Also the mean value of x and y are also known.
How can we determine the values of x and y?
statistics
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asked Feb 22, 2016 at 11:35
MaxoodMaxood
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$\begingroup$ Can we find from simultaneous equations? I tried that but still could not figure them out. I am looking for a definite solution. $\endgroup$
Maxood
– Maxood
2016-02-22 11:42:04 +00:00
Commented Feb 22, 2016 at 11:42
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We have two variables ($x,\ y$), so we need two equations. Luckily, we have two additional items of data, with which we can build our two equations: We'll denote the mean $\mu$ and the standard deviation $\sigma$. Let's use the information regarding the mean first: $$ \mu = \frac{\sum z_i}{n} = \frac{25 + x +y}{7} $$ So, our first equation is: \begin{equation} x+y = 7\mu - 25 \end{equation}
Now let's use the information regarding the standard deviation: $$\sigma^2 = \frac{\sum(z_i-\mu)^2}{n}$$ After some arithmetic this can be written as: $$x^2 + 2x\mu + y^2+ 2y\mu = 7(\mu^2- \sigma^2) -50 \mu + 145$$
So now we have our two equations for two variables, and we can solve via our favorite method. If you need more help, let me know.
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answered Feb 22, 2016 at 12:12
gordonjogordonjo
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This will result in a simple set of equations. $$ 2+4+5+6+8+x+y=7\mu $$ with $\mu$ the mean of the set and 7 the number of datapoints in the set.
$x+y=2\mu_2$ with $\mu_2$ the mean of $x$ and $y$. The set of this two equations should be redundant (infinite solutions) If it has no solutions, there is something wrong with the data of the question.
The last equation you need to be able to solve the problem is: $$ (\mu-2)^2 + (\mu-4)^2 + (\mu-5)^2 +(\mu-6)^2 + (\mu-8)^2 + (\mu-x)^2 + (\mu-y)^2 = \sigma^2 $$ whereby $\sigma$ is the standard deviation of the dataset.
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answered Feb 22, 2016 at 11:57
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$\begingroup$ Fine. But how are we going to find the values of x and y? $\endgroup$
Maxood
– Maxood
2016-02-22 12:01:38 +00:00
Commented Feb 22, 2016 at 12:01
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You do not actually need to know the mean value of $x$ and $y$ in order to solve this problem. It is enough to know either the mean value of $x$ and $y$ or the mean value of all the data taken together. Just one of those mean values, plus the standard deviation, is sufficient.
Here's one specific way to find $x$ and $y.$
First, you have to determine which "standard deviation" is meant. One possibility is that the given data are considered to be the entire population whose standard deviation is measured, that is, the "standard deviation" is $$\sigma = \sqrt{\frac 1N \sum (x_i - \mu)^2},\tag1$$ where $N$ is the number of data values. Another possibility is that the given data are considered to be a sample taken from a larger population, and the "standard deviation" is defined by the usual formula for the standard deviation of a sample (which is an estimate of the standard deviation of the entire population), $$s = \sqrt{\frac 1{N-1} \sum (x_i - \mu)^2}.\tag2$$
In the following, I will suppose that the "standard deviation" was defined by Equation $(1).$ If it actually was defined by Equation $(2),$ the following equations can be adapted to that definition by replacing $\sigma$ by $\left(\sqrt{\frac{N-1}{N}}\right)s$ wherever $\sigma$ appears.
In order to make this a general solution, not just applicable to the particular five given numbers in the question, let the list of data values be $$ x_1,\ x_2,\ x_3,\ \ldots,\ x_{N-2},\ x,\ y.$$ Then the mean of all the data is $$ \mu = \frac1N\left(x_1 + x_2 + x_3 + \ldots + x_{N-2} + x + y\right).$$ Therefore $$ x + y = N\mu - \sum_{i=1}^{N-2} x_i. \tag3$$ In the case where the other data and the mean of all the data are known, everything on the right-hand side of Equation $(3)$ is known, so we can set $A = N\mu - \sum_{i=1}^{N-2} x_i$ and compute the value of $A.$ But if we know the mean of $x$ and $y$ then we can just set $A$ to twice that mean. Either way, we have $$ x + y = A, \tag4 $$ where $A$ is a value we know how to compute. If we were given the mean of $x$ and $y$ but not $\mu,$ we can now find $\mu$ by using the equation $$\mu = \frac1N\left(A + \sum_{i=1}^{N-2} x_i\right).$$ From now on, we suppose that $\mu$ was either given or computed by this method.
For the standard deviation, there is a well-known computational trick: $$ \frac 1N \sum_{i=1}^N (x_i - \mu)^2 = \left(\frac 1N \sum_{i=1}^N x_i^2\right) - \mu^2. $$ We can therefore write $$ \sigma^2 = \frac 1N\left(x_1^2 + x_2^2 + x_3^2 + \ldots + x_{N-2}^2 + x^2 + y^2\right) - \mu^2, $$ which is equivalent to $$ x^2 + y^2 = N\sigma^2 - N\mu^2 - \sum_{i=1}^{N-2} x_i^2. \tag5 $$ Let $B$ be equal to the right-hand side of Equation $(5)$, which we can compute from quantities that are all known. That is, we compute the quantity $B$ and then we know that $$ x^2 + y^2 = B. \tag6$$
Square Equation $(4)$: $$ x^2 + 2xy + y^2 = A^2.$$ Subtract Equation $(6)$: $$ 2xy = A^2 - B. \tag7$$ If the right-hand side of Equation $(7)$ turns out to be zero, then one of the variables $x$ or $y$ is zero and the other is equal to $A.$ Otherwise, Equation $(7)$ implies $$ y = \frac{A^2 - B}{2x}.$$ Plugging this back into Equation $(4)$ gives us $$ x + \frac{A^2 - B}{2x} = A,$$ and since $x \neq 0$ this is equivalent to the quadratic equation $$ x^2 - Ax + \frac12(A^2 - B) = 0. \tag8$$
Use the standard quadratic formula to solve for $x$ in Equation $(8).$ The two solutions are the values of $x$ and $y.$ (If there is only one solution then $x$ and $y$ are both equal to that value; if there are no solutions then the values given in the problem statement are impossible.)
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answered Jul 11, 2018 at 0:58
David KDavid K
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Exp-10: Determination of Zinc Ions via Precipitation Titration
...............
Original title:
Exp-10 - ...............
Course
Environmental Engineering (ES 412)
70 documents
University
University of Southern Mindanao
Academic year:2015/2016
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Kathleen Suanzo
University of Southern Mindanao
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EXPT. 10 DETERMINATION OF ZINC BY
PRECIPITATION WITH
POTASSIUM FERROCYANIDE
USING INTERNAL INDICATOR
# Structure
10 Introduction Objectives 10 Principle 10 Requirements 10 Solutions Provided 10 Procedure 10 Observations and Calculations 10 Result
10 INTRODUCTION
In the previous experiment you learnt and performed precipitation titrations called argentometric titrations. These involved the determination of chloride ions in a given solution by precipitating them as silver chloride using silver nitrate as the titrant. The determination was done by two different methods, based on the type of indicator used for the determination of the end point. While the Mohr’s method used a precipitation indicator, the Fajan’s method employed an adsorption indicator. In this last experiment of this course you will learn and perform the determination of zinc ions in a given solution, by precipitating the ions with potassium ferrocyanide. Interestingly, you would be using a redox indicator for the determination of the end point.
# Objectives
After studying and performing this experiment, you should be able to:
state the principle of the determination of zinc ions in a given solution by precipitating the ions with potassium ferrocyanide,
explain various basis of end point determination in the precipitation titrations,
prepare a standard solution of zinc sulphate and use it for the standradisation of potassium ferrocyanide,
titrate the given solution of zinc ions with potassium ferrocyanide by using diphenylamine as internal indicator, and
calculate the amount of zinc ions in the given aqueous solution.
10 PRINCIPLE
The determination of Zn++ ions in neutral or weakly acidic solution by precipitation titration using standard K 4 [Fe(CN) 6 ] solution is based on the following reaction:
3ZnSO 4 + 2K 4 [Fe(CN) 6 ] →K 2 Zn 3 [ Fe(CN) 6 ] 2 ↓+3K 2 SO 4
In the titration of zinc ions with potassium ferrocyanide, the continuous addition of the titrant precipitates the analyte ions and immediately after the equivalence point there would be an excess of the titrant. This stage could be ascertained with the help of a suitable indicator. As mentioned above, in this titration you are going to use a redox indicator i., diphenylamine (transition range 0.73-0 V) for the purpose.
In order to use the redox indicator we need to create a redox system. You know that potassium ferrocyanide K 4 [Fe(CN) 6 ] is a complex salt of ferrous ions and like other ferrous compounds, it has reducing properties. Therefore, on oxidation, K 4 [Fe(CN) 6 ] forms the corresponding ferric salt; that is potassium ferricyanide, K 3 [Fe(CN) 6 ]. The presence of these two moieties in solution constitutes a redox system whose potential is given by the following equation: 3 6 4 6
0 [ ( ) ] 0 log 1 [ ( ) ]
Fe CN E Fe CN
− = + −
In order to create the above mentioned redox couple we need to add a little of potassium ferricyanide in the analyte solution. Therefore, in one of the ways of performing this determination, a small volume of the solutions of, potassium ferricyanide, K 3 [Fe(CN) 6 ] is added to the titrand ( zinc ion solution) along with diphenylamine indicator. This imparts a blue-violet colour to the solution. This can then be titrated with a solution of potassium ferrocyanide ions that precipitate the zinc ions as per the equation
3ZnSO 4 + 2K 4 [ Fe(CN) 6 ] →K 2 Zn 3 [Fe(CN) 6 ] 2 ↓+3K 2 SO 4
Since the added titrant is used up in precipitating the zinc ions, there are no free 4 [ Fe CN( ) ] 6 − ions, the potential of the solution does not change and the colour of the solution remains the same till all the zinc ions are precipitated out. Thereafter, first drop of K 4 [Fe(CN) 6 ] in excess lowers the oxidation potential of the solution to an extent that the indicator becomes colourless. This marks the end point of the titration. The precipitation of zinc with ferrocyanide is a little slow therefore the titration should not be performed rapidly.
10 REQUIREMENTS
Apparatus Chemicals Volumetric flask (100 cm 3 ) – 1 Burette (50 cm 3 ) – 1 Pipettes (10 cm 3 ) – 1 Weighing bottle – 1 Burette stand with clamp – 1 Conical flasks (100 cm 3 ) – 2 Funnel – 1 Beakers (250 cm 3 ) – 2
Zinc sulphate heptahydrate Potassium ferrocyanide Potassium ferricyanide Diphenylamine Ammonium sulphate Sulphuric acid
10 SOLUTIONS PROVIDED
0 M potassium ferrocyanide, K 4 [Fe(CN) 6 ]: It is prepared by dissolving about 1 of potassium ferrocyanide in about 20 cm 3 of distilled water taken in a 100 cm 3 volumetric flask and making up the volume with more distilled water.
Diphenylamine indicator: It is prepared by dissolving about 1 g of diphenylamine in 100 cm 3 of concentrated sulphuric acid.
Potassium ferricyanide, K 3 [Fe(CN) 6 ]: It is prepared by dissolving about 1 g of potassium ferrocyanide in about 20 cm 3 of distilled water taken in a 100 cm 3 volumetric flask and making up the volume with more distilled water.
Earlier, the equivalence point in this titration was determined generally by means of an external indicator, uranyl nitrate, which forms a brick-red precipitate of a uranyl salt with K 4 [Fe(CN) 6 ].
10 OBSERVATIONS AND CALCULATIONS
a) Preparation of standard solution of zinc sulphate
Mass of weighing bottle + zinc sulphate= m 1 g = ............. Mass of weighing bottle (after transferring zinc sulphate) = m 2 g = .............. g Amount of zinc sulphate transferred = m 1 – m 2 = m g = ............. g Molar mass (Mm) of zinc sulphate heptahydrate = 287 g mol− 1 Volume of zinc sulphate prepared = 100 cm 3 Molarity of standard zinc sulphate solution =
4
# 1000 10
# .........
# ZnSO 100 287 287.
# m m
# M M
# ×
# = = =
# ×
b) Standardisation of potassium ferrocyanide solution
Volume of standard ZnSO 4 solution taken in conical flask, VZnSO4 = .. 3 Solution in the burette: Potassium ferrocyanide Indicator used: Diphenylamine
Observation Table 10: Standardisation of potassium ferrocyanide
S. Volume of zinc sulphate (in cm 3 )
Burette reading Initial Final
Titre value (in cm 3 ) (Final-initial reading)
1
2
3
Concordant reading
The concentration of the given potassium ferrocyanide solution can be determined as follows.
The reactions involved:
3ZnSO 4 + 2K 4 [Fe(CN) 6 ] →K 2 Zn 3 [ Fe(CN) 6 ] 2 ↓+3K 2 SO 4
Molarity equation: 4 4
2 M ZnSO VZnSO = 3 M Potassium ferrocyanide VPotassium ferrocyanide
# 2
# 3
Zinc sulphate Zinc sulphate Potassium ferrocyanide Potassium ferrocyanide
# M V
# M
# V
# =
Substituting the values, the molarity of potassium ferrocyanide =
The molarity of given potassium ferrocyanide solution is = .........
c) Determination of zinc ions in the given solution by titrating against standaradised solution of potassium ferrocyanide. Volume of standard ZnSO 4 solution taken in conical flask, VZnSO4 = .. 3 Solution in the burette: Potassium ferrocyanide
Indicator used: Diphenylamine
Observation Table 10: Titration of given zinc sulphate solution with standradised solution of potassium ferrocyanide
S. Volume of zinc sulphate (in cm 3 )
Burette reading Initial Final
Titre value (in cm 3 ) (Final-initial reading)
1
2
3
Concordant reading
The concentration of the given zinc sulphate solution can be determined as follows.
The reactions involved:
3ZnSO 4 + 2K 4 [Fe(CN) 6 ] →K 2 Zn 3 [ Fe(CN) 6 ] 2 ↓+3K 2 SO 4
Molarity equation: 4 4
2 M ZnSO VZnSO = 3 M Potassium ferrocyanide VPotassium ferrocyanide
# 3
# 2
Potassium ferrocyanide Potassium ferrocyanide Zinc sulphate Zinc sulphate
# M V
# M
# V
# =
Substituting the values, the molarity of given zinc sulphate = The molarity of given zinc sulphate solution is = .........
10 RESULTS
The molarity of zinc ions in the given solution is found to be = ..... M
Exp-10: Determination of Zinc Ions via Precipitation Titration
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Exp-10: Determination of Zinc Ions via Precipitation Titration
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74
EXPT. 10 DETERMINATION OF ZINC BY
PRECIPITATION WITH
POTASSIUM FERROCYANIDE
USING INTERNAL INDICATOR
Structure
10.1 Introd uction
Objectives
10.2 Principle
10.3 Requirements
10.4 Solutions Provided
10.5 Procedure
10.6 Observations a nd Calcula tions
10.7 Result
10.1 INTRODUCTION
In the previous experiment you le arnt and perform ed precipita tion titrations called
argentometric titrations. These invol ved the de termination of chloride ions in a given
solution by precipitating them as silver chloride using s ilver nitrate as the titrant. The
determination was done by two d ifferent methods, based on the type of indicator used
for the determination of the end point. While the Mohr’s method used a precipitation
indicator, the Fajan’s method employ ed an ad sorption indicator. In this last experiment
of this course you will lear n and perform the determinatio n of zinc ions in a given
solution, by precipita ting the ions with potassium fe rro cyanide. Inte restingly, you
would be usin g a redox indi cator for the determ ination of the end point.
Objectives
After studying and perform ing this e xperim ent, you should be able to:
•state the principle of th e determination of zinc io ns in a given solution by
precipitating the ions with potassium ferrocyan ide,
•explain various basis of e nd point determination in th e precipitation titrations,
•prepare a standard solution of zinc su lp hate and us e it for th e standradisation of
potassium ferr ocyanide,
•titrate the given solution of zinc ions with potassi um ferrocyanide by using
dipheny lamine as internal ind icator, and
•calcula te the a mount of zinc io ns in the given a queous s olution.
10.2 PRINCIPLE
The determination of Zn++ ions in neutral or weakly acidic solution by precipitation
titration using standard K 4 [Fe(CN)6] solution is based on the following reaction:
[
]
[
]
4 2
2
6 3 2 6 4 4 SO 3K Fe(CN)Zn K Fe(CN)2K 3ZnSO+↓→+
In the titration of zinc ions with potassium ferrocyanide, the con tinuous addition of the
titrant precipitates the analyte ions and imme diately after the equivalence point there
would be an excess of the titrant. This stag e could be ascertained with the help of a
suitable indicator. As mentione d above, in this titration you are going to use a redox
indicator i.e., diphenylamin e (transition range 0.73-0.7 9 V) for the purpose.
75
In order to use the redox indicator w e nee d to create a redox sy stem. You know that
potassium ferr ocyanide K 4 [Fe(CN)6] is a com plex salt of ferrous ions and like other
ferrous compounds, it has red ucing properties. Therefo re, on oxidation, K 4 [Fe(CN)6]
forms the corresponding ferric salt; that is potassium ferricyanide, K 3 [Fe(CN)6]. The
presence of these two moieties in solution constitutes a redox system whose potential
is given by the following equation:
3
6
4
6
[()]
0.058
0.36 log
1[()]
Fe CN
E Fe CN
−
−
=+
In order to create the a bove mentioned redox couple we need to add a little of
potassium ferricyanide in the analyte solu tion. Therefore, in one of the ways of
performing this determ inatio n, a small volume of th e solutions of, potassium
ferricyanide, K 3 [Fe(CN)6] is added to the titrand ( zi nc ion solution) along with
dipheny lamine indicator. This imparts a blue-violet col our to the solution. This can
then be titrated with a solution of potassium ferrocyanide ions that precipitate the zinc
ions as per the equation
[]
[
]
4 2
2
6 3 2 6 4 4 SO 3K Fe(CN)Zn K Fe(CN)2K 3ZnSO+↓→+
Since the added titrant is used up in precipitating the zi nc ions, there are no free
4
6
[()]Fe CN−
ions, the potential of the solution does not change a nd the colour of the
solution remains the same till all the zinc io ns are precipitated out. Thereafter, first
drop of K 4 [Fe(C N)6] in excess lowers the oxidation potential of the solution to an
extent that the indica tor becomes colourless. This mark s the end point of the titration.
The precipitation of zinc with ferrocyanide is a little slow therefore the titr ation should
not be pe rforme d ra pidly.
10.3 REQUIREMENTS
Apparatus Chemicals
Volumetric flask (100 cm 3) – 1
Burette (50 cm 3) – 1
Pipettes (10 cm 3) – 1
Weighing bottle – 1
Burette stan d with clam p – 1
Conical flasks (100 cm 3) – 2
Funnel – 1
Beakers (250 cm 3) – 2
Zinc sulphate heptahy drate
Potassium ferrocyanide
Potassium ferricyanide
Diphenyl amine
Ammoniu m sulphate
Sulphuric acid
10.4 SOLUTIONS PROVIDED
1.0.05 M potas sium ferr ocyanid e, K 4 [Fe(C N)6]: It is prepa red by dissolving
about 1.85g of pota ssium fe rrocyanide in about 20 cm 3 of distilled water taken
in a 100 cm 3 volumetric flask and making up the volume with more distilled
water.
2.Diphenylamine indicato r: It is p repare d by dissolving abou t 1.0 g of
dipheny lamine in 10 0 cm 3 of concentr ated sulphur ic acid.
3.Potassium ferricyanide, K 3 [Fe(CN)6]: It is prepared by dissol ving about 1.65 g
of potassium ferro cyan ide in about 20 cm 3 of distilled water taken in a 100 cm 3
volumetric flask and m aking up th e volume with more distilled water.
Earlier, the equi valence point
in this titration was
determined generall y by
means of an external
indicator, uran yl nitrate,
which forms a brick-red
precipitate of a uranyl s alt
with K 4 [Fe(CN)6].
76
10.5 PROCEDURE
The determination of zin c ions in a gi ven solution using potass ium fer rocyanide
consists of the following steps:
a)Preparation of standard so lution of zinc sulphate
b)Standradisation of potassium ferrocy anide solution
c)Determination of zinc ions in th e given solution by titrating against
standaradised solutio n of potassium ferrocyanide.
Follow the instructions given below in sequential manner to carry out the
determ ination of zin c ions by using pota ssium ferrocya nide.
a)Preparation of standard solution of zinc sulphate
•Accurate ly weigh a bout 2.87 g of zi nc sulpha te heptahydr ate in a clean
dry weighing bottle, and tr ansfer the same to a clean volumetric flask of
100 cm 3 capacity th rough a glass funnel.
•Add about 20 cm 3 of distilled water and swirl the contents of the flask
until all the z inc sulphate he ptahydrate is dissolved
•Make the volume upto the mark by adding more distilled water
b)Standradisation of potassium ferrocyanide solution
•Pipette 10 cm 3 of the s tandard z inc sulphate into a 100 cm 3 conical flask
and add 2 g of (NH 4)SO 4, about 20 cm 3 of 3 M H 2 SO 4 solution, two or
three drops of K 3 [Fe(CN)6] solution, a nd 3 drops of diphenylam ine
indicator to it.
•The solution acquires a blue colour, titrate the solution with potassium
ferrocyanide solution taken in a burette.
•Continue the titration till the colour of the solution changes sharply (it
becomes colourless) after ad dition of a single drop.
•Repeat the standard ization procedur e at least three times and reco rd your
observations in Observat ion Table 10.1.
•Use t he data so obtaine d to dete rm ine the m olar con centration of the
potassium ferr ocyanide solution.
c)Determination of zinc io ns in the given so lution by titrating against
standaradised solution of pot assium ferrocyanide.
•Pipette 10 cm 3 of the given zinc sulpha te solution into a 100 cm 3 conical
flask and add 2 g of (NH 4)SO 4, about 20 cm 3 of 3 M H 2 SO 4 solution, two
or three dr ops of K 3 [Fe(CN)6] solution,& 3 drops of diphenyla mine
indicator to it,
•The solution acquires a blue colour, titrate the solution with potassium
ferrocyanide solution taken in a burette,
•Continue the titration till the s olution colour changes sharply (it becomes
colourless) after additio n of a single drop.
•Repeat the proced ure at least three times and r ecord you r observation in
Observa tion Table 10.2.
•Use t he data so obtaine d to dete rmine the molar concen tra tion of the zinc
ion solution
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77
10.6 OBSERVATIONS AND CALCULATIONS
a)Preparation of standard solution of zinc sulphate
Mass of weighing bottle + zinc sulph ate= m 1 g = ..............g
Mass of weighing bottle (after transferring zinc sulp hate) = m 2 g = .............. g
Amount of zinc sulphate transferred = m 1 – m 2 = m g = ............. g
Molar mass (M m) of zinc sulphate heptahydrate = 2 87.53 g mol−1
Volume of zinc sulphate prepared = 100 cm 3
Molarity of standard zinc sulphate solution =
4
1000 10.........
100 287.53 287.53
ZnSO
mm
M
M
×
×
b)Standardisation of potassium ferrocyanide solution
Volume of standard Zn SO 4 solution taken in conical flask, V ZnSO4 = …cm 3
Solution in the burette: Potassium ferrocyanide
Indicator used: Diphenylamine
Observation Table 10.1: Standardis ation of potassium f errocyanide
S.No. Volume of zinc
sulphate (in cm 3)
Burette reading
Initial Final
Titre value (in cm 3)
(Final-initial reading)
1
2
3
Concordant reading
The concentration of the given pota ssium ferrocyanide so lution can be
determined as follows.
The reactions involved:
[]
[
]
4 2
2
6 3 2 6 4 4 SO 3K Fe(CN)Zn K Fe(CN)2K 3ZnSO+↓→+
Molarity equation:
44
23
ZnSO ZnSO Potassi um ferrocyanid e Potassi um ferrocyanide
M
VM V=
2
3
Z
inc sulphate Zinc sulphate
Potassium ferrocyanide
Potassium ferrocyanide
M
V
M V
=
Substituting the values, the molarity of potassium ferrocyanide =
The molarity of g iven potassium ferrocyanid e solution is = ……….M
c)Determination of zinc io ns in the given so lution by titrating against
standaradised solution of po tassium ferrocyanide.
Volume of standard Zn SO 4 solution taken in conical flask, V ZnSO4 = …cm 3
Solution in the burette: Potassium ferrocyanide
78
Indicator used: Diphenylamine
Observation Table 10.2: Titration of given z inc sulphate solution with
standradised solution of potassium ferrocyanide
S.No. Volume of zinc
sulphate (in cm 3)
Burette reading
Initial Final
Titre value (in cm 3)
(Final-initial reading)
1
2
3
Concordant reading
The concen tration of the g iven zinc sulp hate solution can be determined as
follows.
The reactions involved:
[
]
[
]
4 2
2
6 3 2 6 4 4 SO 3K Fe(CN)Zn K Fe(CN)2K 3ZnSO+↓→+
Molarity equation:
44
23
ZnSO ZnSO Potassium ferrocyanide Potassium ferrocyanide
M
VM V
=
3
2
P
otassium ferrocyanide Potassium ferrocyanide
Zinc sulpha te
Zinc sulphate
M
V
M V
=
Substituting the values, the molarity of g iven zinc sulphate =
The molarity of given zinc sulphate solution is = ……….M
10.7 RESULTS
The molarity of zinc ions in the gi ven s olution is found to b e = ….. M
1 out of 5
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7137 | https://ndnhistoryresearch.com/2016/07/05/dentalia-shell-money-hi-qua-alika-chik/ | Dentalia Shell Money: Hi-qua, Alika-chik – The Quartux Journal
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Dentalia Shell Money: Hi-qua, Alika-chik
California Natives, Oregon indians, Uncategorized
July 5, 2016
—
3 minutes
Dentalia shells, a mollusk, is collected from the sea floor off of Vancouver Island and is used by native peoples in a broad region of North America. They grow up to 3 inches (or more) long for the North West Coast dentalia, while 2 inch dentalia are not uncommon and 1-1.5 inches are the most common. Tribes of the North West Coast and California used the dentalia for money, and for decorative and ceremonial artwork.
Measuring the Dentalia money string using the tattoo on his arm- Alikachik, Curtis image
The shells are tusk shaped, and when the animal dies, the tusk is hollow, which allow for easy stringing. The tusks are formed in layers and is easy to flake the shells, or cut off the narrow tip if the hole for the string is not large enough. The dentalia is normally strung with glass and lithic beads in contemporary use. The Native vulgare species has a glossy shell that varies in color from grey to yellow and white and may have banded color lines. Other varieties from throughout the world are in various colors.
Chinook, Columbia River Canoe, Elk Horn Purse with incised Hi-qua string
Yates, 1904, Prehistoric California, Bulletin of the Southern California Academy of Sciences
Further inland dentalia was used to adorn Indian regalia as far east as the Great Lakes. The Northwest trade networks, that extended as far as the Rockies along the principal river systems, Columbia, Fraser, Nass, Klamath, carried dentalia shells far inland. The Northwest Network would connect with that of the American Plains, and Ojibwe (Chippewa), Cree, and other tribes would trade the valuable shells to Plains and Great lakes tribes.
Overland Monthly vol 11
Mass of Dentalia shell necklaces, Northern California
Tribes in the inland region would make hats, capes and adorn their shirts and dresses with the shells by the thousands. Tribes in Washington, Oregon, British Columbia and California would make jewelry, and long dentalia shell necklaces. Many of the tribes would show their wealth at ceremonial events by heaping hundreds of shells necklaces. They would dance the necklaces in ceremonies all along the coasts and in the interiors.
Dentalium gathering, Image borrowed from online
The dentalia strings of Northern California were called Alikachik, (various spellings) and many were kept in elk horn money purses. In the north the strings were called Hiaqua or Hi-qua (various spellings). In northern California some tribes even began incising the dentalia with graphic designs, rubbing dyes and charcoal into the ridges to show geometric designs.
The dentalia strings or Alikachik, in the colonial period would fetch a good price, several gold pieces per string. Dentalia were even utilized, on a limited basis, in the fur trade. The dentalia shells have maintained their value to today and a single 2 inch vulgare dentalia can be worth as much as $3 per shell. The tribes on Vancouver island continue to harvest the shells and own the harvest rights to areas off the coast. They use rakes to gather the shells from the sea floor.
There are hundreds of types of dentalia that grow around the world. The green dentalia are said to grow in the Mediterranean, while very large giant dentalia, up to a foot long, grow off the coast of Vietnam. The Philippines serves as the principal trading areas for the shells, where they are processed and bleach before being shipped to worldwide markets.
Tribes in the North West Coast region still use the dentalia extensively and with the revival of Native dancing and ceremonial traditions, there is a revival in the arts and crafts around the use of dentalia.
Additional article on Trade Networks on this Blog.
A related article.
2 responses to “Dentalia Shell Money: Hi-qua, Alika-chik”
Dentalium | Shichils's Blog August 3, 2016 […] month my friends David Robertson (linguist) and David Lewis (historian & anthropologist) wrote about dentalium. I thought I’d write a bit more about dentalium among our tribes and […] Log in to Reply
Giving It to You Straight – Toothshell Hermit Crabs and Wampum Tuskshells – Walrus Neat April 30, 2023 […] Quartux Journal – Dentalia Shell Money: Hi-qua, Alika-chik […] Log in to Reply
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Review of Guidelines on Clonidine for Various Indications
Rapid Review
CADTH Health Technology Review
Ottawa (ON): Canadian Agency for Drugs and Technologies in Health; 2024 Aug.
Report No.: RC1543
Copyright and Permissions
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Key Messages
What Is the Issue?
Clonidine is an antihypertensive medication that has been used for a range of health conditions including hypertension, substance use disorders, menopause, restless leg syndrome, migraines, attention-deficit/hyperactivity disorder (ADHD), and Tourette syndrome. The role of clonidine in the treatment of these health conditions is unclear.
What Did We Do?
To inform decisions around the use of clonidine in various health conditions, we sought to identify and summarize recommendations from evidence-based guidelines.
We searched key resources, including journal citation databases, and conducted a focused internet search for relevant evidence published since 2014. One reviewer screened articles for inclusion based on predefined criteria, critically appraised the included guidelines, and narratively summarized the findings.
What Did We Find?
We identified 12 evidence-based guidelines that included recommendations on the use of clonidine. We identified 1 guideline on hypertension, 4 guidelines on substance use disorders, 4 guidelines on menopause, 2 guidelines on restless leg syndrome, and 1 guideline on Tourette syndrome. We did not identify any evidence-based guidelines that included recommendations on the use of clonidine for the treatment of ADHD or migraine prophylaxis.
The included guidelines recommend clonidine for hypertension in pregnant women, management of opioid withdrawal and alcohol withdrawal, and Tourette syndrome.
The recommendations in the guidelines for menopause were mixed. Two guidelines do not recommend clonidine and 2 guidelines recommend clonidine for the treatment of vasomotor symptoms (i.e., hot flashes) of menopause.
One guideline does not recommend the use of clonidine for restless leg syndrome in people who are pregnant or lactating and 1 guideline states that there is insufficient evidence to support or refute the use of clonidine in restless leg syndrome.
What Does It Mean?
The use of clonidine is recommended for some health conditions and is not recommended for others. Due to the inconsistency in recommendations on the use of clonidine for the control of hot flashes in menopause, decision-makers may wish to consider other factors such as patient preferences and availability of other treatment options.
Future evidence-based guidelines that include recommendations on the use of clonidine for the prevention of migraines, treatment of ADHD and the treatment of hypertension in a broader population would help fill the gaps identified in this report.
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Context and Policy Issues
What Are the Health Conditions Included in This Review?
Guidelines on hypertension, substance use disorders, menopause, restless leg syndrome, migraine prevention, ADHD, and Tourette syndrome were eligible for inclusion in this review. Hypertension (high blood pressure) is when the pressure in your arteries is consistently too high.1 Treatment can include lifestyle changes and medications (e.g., diuretics, angiotensin-converting enzyme inhibitors, angiotensin II receptor blockers, and calcium channel blockers).2 Substance use disorder is a condition in which there is a problematic pattern of substance use that causes distress or impairs day-to-day functioning.3 Treatment for substance use disorder can include detoxification, cognitive and behavioural therapies, and medications.3 Menopause is a time that marks the end of menstrual cycles and is diagnosed after a person has gone 12 months without a menstrual period.4 A potential symptom of menopause is hot flashes (also known as vasomotor symptoms) which are a sudden feeling of warmth that spreads over the body.5 Treatment for menopause can include hormonal (i.e., estrogen or estrogen and progesterone) and nonhormonal treatments (e.g., diet, exercise, medications).5
Restless leg syndrome is a condition which causes a strong urge to move the legs and usually occurs in the evening or at night when sitting or lying down.6 Treatment for restless leg syndrome can include at-home therapies (e.g., exercise, warm baths, stress reduction) or medications (e.g., iron supplements, antiseizure medications, dopamine agonists, benzodiazepines, opioids).7 A migraine is a headache that can cause severe throbbing pain or a pulsing sensation and is often accompanied by nausea, vomiting, and extreme sensitivity to light and sound.8 Medications aimed at preventing migraines can include blood-pressure lowering medications, antidepressants, antiseizure drugs, and calcitonin gene-related peptides monoclonal antibodies.9 ADHD is a mental health disorder that include symptoms such as difficulty paying attention, hyperactivity, and impulsive behaviour.10 Treatment for ADHD includes medication (stimulants and other medications), education, skills training, and psychological counselling.11 Tourette syndrome is a neurologic disorder that may cause sudden unwanted and uncontrolled repetitive movements or vocal sounds called tics.12 Treatment for Tourette syndrome can include behavioural treatments, psychotherapy, and medications (e.g., dopamine blocking agents, alpha-adrenergic agonists, stimulants, antidepressants).12
What Is Clonidine?
Clonidine is an alpha-2 adrenergic agonist that has antihypertensive (blood pressure lowering) effects.13 Clonidine lowers blood pressure by relaxing the arteries and increasing the blood supply to the heart.13 Common side effects of clonidine may include abdominal pain, headache, hypotension, fatigue, nausea, constipation, dry mouth, sexual dysfunction, dizziness, and sedation.13 There is also potential for rebound hypertension and withdrawal symptoms if clonidine is discontinued abruptly.13 Clonidine is indicated for the treatment of hypertension and should normally be used in patients in whom treatment with a diuretic or beta-blocker was ineffective or associated with unacceptable adverse effects.14 Clonidine is available as tablets for oral administration and the initial dose is 0.1 mg twice daily.14 After 2 to 4 weeks, further increments of 0.1 mg per day may be added until the desired response is achieved.14 The common therapeutic dose ranges from 0.2 mg to 0.6 mg per day.14 When discontinuing clonidine, the dosage should be reduced gradually.14
Why Is It Important to Do This Review?
In addition to its use in hypertension, clonidine has also been used for a range of other health conditions including ADHD, Tourette syndrome, managing withdrawal from opioids, benzodiazepines, and alcohol, restless leg syndrome, migraine prophylaxis, and control of hot flashes in menopause.13 A review of guidelines of clonidine can help decision-making around which indications clonidine should be used for as well as provide guidance on potential dosing and safety considerations.
Objective
The purpose of this report is to summarize and critically appraise evidence-based guidelines regarding the use of clonidine for hypertension, substance use disorders, control of hot flashes in menopause, restless leg syndrome, migraine prophylaxis, ADHD, and Tourette syndrome.
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Research Questions
What are the evidence-based guidelines regarding the use of clonidine in adults with hypertension?
What are the evidence-based guidelines regarding the use of clonidine in adults with substance use disorders (i.e., opioids, benzodiazepines, alcohol)?
What are the evidence-based guidelines regarding the use of clonidine for the control of hot flashes in adults with menopause?
What are the evidence-based guidelines regarding the use of clonidine in adults with restless leg syndrome?
What are the evidence-based guidelines regarding the use of clonidine for prevention of migraines in adults?
What are the evidence-based guidelines regarding the use of clonidine in adults with ADHD?
What are the evidence-based guidelines regarding the use of clonidine in adults with Tourette syndrome?
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Methods
Literature Search Methods
An information specialist conducted a literature search on key resources including MEDLINE, Embase, PsycInfo, the Cochrane Database of Systematic Reviews, the International HTA Database, the websites of major international health technology agencies, and those in Canada, as well as a focused internet search. The search approach was customized to retrieve a limited set of results, balancing comprehensiveness with relevancy. The search strategy comprised both controlled vocabulary, such as the National Library of Medicine’s MeSH (Medical Subject Headings), and keywords. Search concepts were developed based on the elements of the research questions and selection criteria. The main search concept was clonidine. Search filters were applied to limit retrieval to guidelines. Conference reviews and conference abstracts were excluded. The search was completed on June 12, 2024 and limited to English-language documents published since January 1, 2014.
Selection Criteria and Methods
One reviewer screened citations and selected studies. In the first level of screening, titles and abstracts were reviewed and potentially relevant articles were retrieved and assessed for inclusion. The final selection of full-text articles was based on the inclusion criteria presented in Table 1.
Table 1
Selection Criteria.
Exclusion Criteria
Articles were excluded if they did not meet the selection criteria outlined in Table 1, they were duplicate publications, or were published before 2014. Guidelines with unclear methodology were also excluded.
Critical Appraisal of Individual Studies
The included guidelines were critically appraised by 1 reviewer using the Appraisal of Guidelines for Research and Evaluation (AGREE) II instrument15 as a guide.
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Summary of Evidence
Quantity of Research Available
A total of 76 citations were identified in the literature search. Following screening of titles and abstracts, 48 citations were excluded and 28 potentially relevant reports from the electronic search were retrieved for full-text review. Ten potentially relevant publications were retrieved from the grey literature search for full-text review. Of these potentially relevant articles, 25 publications were excluded for various reasons, and 13 publications met the inclusion criteria and were included in this report. These comprised 13 reports pertaining to 12 unique evidence-based guidelines. Appendix 1 presents the Preferred Reporting Items for Systematic reviews and Meta-Analyses (PRISMA)16 flow chart of the study selection.
Additional references of potential interest are provided in Appendix 2.
Summary of Guideline Characteristics
This report included 12 evidence-based guidelines17-28 that included recommendations regarding the use of clonidine for various indications.
The Hypertension Canada (2020)17 guideline included recommendations on hypertension. The Canadian Research Initiative in Substance Misuse (CRISM) (2023),18 Veterans Affairs (VA)/Department of Defense (DoD) (2021),19 American Society of Addiction Medicine (ASAM) (2020),20 and Commonwealth of Australia (2014)21 guidelines included recommendations on substance use disorder. The North American Menopause Society (NAMS) (2023),22 Society of Obstetricians and Gynaecologists of Canada (SOGC) (2021),23 National Institute for Health and Care Excellence (NICE) (2019),24 and Endocrine Society (2015)25 guidelines included recommendations on menopause. The American Academy of Neurology (AAN) (2016)26 and International Restless Legs Syndrome Study Group (IRLSSG) (2015)27 guidelines included recommendations on restless leg syndrome. The AAN (2019)28 guideline included recommendations on Tourette syndrome. Two of the guidelines on substance use disorder included recommendations on the management of opioid withdrawal (VA/DoD 19 and Commonwealth of Australia 21) and 2 of the guidelines included recommendations on alcohol withdrawal (CRISM 18 and ASAM 20). We did not identify any evidence-based guidelines that included recommendations on the use of clonidine for the prevention of migraines or treatment of ADHD that met our inclusion criteria.
The guideline development groups were from Canada,17,18,23 the US,19,20 Australia,21 the UK,24 North America,22 or were international.25-28
Additional details regarding the characteristics of included guidelines are provided in Appendix 3.
Summary of Critical Appraisal
All the included guidelines had clear objectives, guideline questions, and target populations. Only 3 of the guidelines18,19,24 specifically sought the views and preferences of the target populations and 1 of the guidelines20 posted the recommendations for public feedback. Therefore, the recommendations in some of the guidelines may not adequately reflect the values and preferences of patients. Systematic methods were used to search for evidence in 7 of the included guidelines.18-20,24-26,28 In 4 of the guidelines,17,21-23 they state that literature searches were conducted, however, limited details on the search (e.g., databases searched, time frame, screening methods, and so forth.) were provided. The IRLSSG (2015)27 guideline only performed a search in a single database and therefore relevant evidence may have been missed due to lack of a comprehensive search strategy.
There was an explicit link between the recommendations and supporting evidence in 9 of the guidelines.17-19,22,24-28 The SOGC (2021)23 and ASAM (2020)20 guidelines did not include discussion of the evidence that supports the recommendations. The Commonwealth of Australia (2014)21 guideline stated that the approach of using clonidine and other medications to control symptoms of opioid withdrawal is well supported by evidence, however, do not include a description of this evidence. Providing clear descriptions of the evidence used to inform the recommendations increases transparency in the recommendation development process.
The competing interests of the guideline development group were disclosed in all the guidelines. The SOGC (2021)23 and ASAM (2020)20 guidelines did not include statements about any funding received for the development of the guideline.
Additional details regarding the strengths and limitations of included guidelines are provided in Appendix 4.
Summary of Findings
An overview of the recommendations and supporting evidence regarding the use of clonidine for the health conditions covered in the included guidelines is provided in Table 2.
Appendix 5 presents the recommendation in the included guidelines.
Table 2
Overview of Included Guidelines.
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Limitations
Some of the included guidelines are limited by the evidence identified to inform the recommendations. The supporting evidence for some of the recommendations was rated as low quality by the guideline authors or the recommendations were based on expert opinion. Additionally, we cannot determine whether the literature searches conducted to inform 4 of the guidelines17,21-23 were comprehensive as very limited details were provided. The IRLSSG (2015)27 guideline only searched a single database and may have missed relevant evidence.
We did not identify any evidence-based guidelines on the use of clonidine for the prevention of migraines or treatment of ADHD that met our inclusion criteria. However, the AAN (2019)28 guideline included a recommendation on the use of clonidine for patients with comorbid ADHD and tics. Most of the included guidelines did not include recommendations or guidance on clonidine dosing. The Hypertension Canada (2020)17 guideline included a recommendation for the use of clonidine in pregnant women however, it did not include recommendations on the use of clonidine for the treatment of hypertension in other populations.
There were inconsistencies in some of the recommendations in the included guidelines on menopause. The NAMS (2023)22 and NICE (2019)24 guidelines do not recommend clonidine for the treatment of vasomotor symptoms of menopause, whereas the SOGC (2021)23 and Endocrine Society (2015)25 guidelines included positive recommendations for clonidine for the treatment of vasomotor symptoms.
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Conclusions and Implications for Decision- or Policy-Making
We included 12 evidence-based guidelines on the use of clonidine for various health conditions in this report. These comprised 1 guideline on hypertension,17 4 guidelines on substance use disorders (2 on opioid withdrawal19,21 and 2 on alcohol withdrawal18,20), 4 guidelines on the treatment of vasomotor symptoms of menopause,22-25 2 guidelines on restless leg syndrome,26,27 and 1 guideline on Tourette syndrome.28
The Hypertension Canada (2020)17 guideline recommends considering clonidine as a second-line option for pregnant women with chronic hypertension, gestational hypertension, or preeclampsia.
The 4 guidelines on substance use disorders all included positive recommendations for the use of clonidine. The CRISM (2023)18 guideline recommends offering clonidine for withdrawal management in an outpatient setting for patients at low risk of severe complications of alcohol withdrawal. The ASAM (2020)20 guideline recommends clonidine for the control of autonomic hyperactivity and anxiety when symptoms are not controlled by benzodiazepines alone. The ASAM (2020)20 guideline does not recommend clonidine be used alone to prevent or treat withdrawal-related seizures or delirium. The VA/DoD (2021)19 guideline suggests clonidine as a second-line drug for opioid withdrawal management when methadone and buprenorphine are contraindicated, unacceptable, or unavailable. The Commonwealth of Australia (2014)21 guideline recommends abrupt cessation of opioid use and control of symptoms using non-opioid drugs such as clonidine as an option for the management of opioid withdrawal.
The recommendations on the use of clonidine for the treatment of vasomotor symptoms of menopause were mixed. The NAMS (2023)22 guideline does not recommend clonidine for the treatment of vasomotor symptoms of menopause. The NICE (2019)24 guideline does not recommend routinely offering clonidine as first-line treatment for vasomotor symptoms alone. The SOGC (2021)23 guideline recommends clonidine as a nonhormonal option for refractory vasomotor symptoms. The Endocrine Society (2015)25 guideline suggests a trial of clonidine for women with moderate to severe vasomotor symptoms who do not respond to or are intolerant of SSRIs/SNRIs, gabapentin, or pregabalin.
Neither of the guidelines for restless leg syndrome recommend the use of clonidine. The AAN (2016)26 guideline states that there is insufficient evidence to support or refute the use of clonidine for the treatment of restless leg syndrome. The IRLSSG (2015)27 guideline states that clonidine should probably not be considered for people with restless leg syndrome who are pregnant or lactating.
The AAN (2019)28 guideline includes several recommendations on the use of clonidine for Tourette syndrome. The guideline recommends the use of alpha2 adrenergic agonists (e.g., clonidine) for the treatment of tics when the benefits outweigh the risks.28 The guideline also recommends counselling patients on common side effects and monitoring heart rate and blood pressure of patients treated with alpha2 adrenergic agonists.28
Future evidence-based guidelines that include recommendations on the use of clonidine for the prevention of migraines, treatment of ADHD and the treatment of hypertension in a broader population would help fill the gaps identified in this report. Due to the inconsistency in recommendations on the use of clonidine for the treatment of vasomotor symptoms of menopause, decision-makers may wish to consider other factors such as patient preferences and availability of other treatment options when making decisions around the use of clonidine for this indication.
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Abbreviations
AAN
American Academy of Neurology
ADHD
attention-deficit/hyperactivity disorder
ASAM
American Society of Addiction Medicine
CRISM
Canadian Research Initiative in Substance Misuse
DoD
Department of Defense
IRLSSG
International Restless Legs Syndrome Study Group
NAMS
North American Menopause Society
NICE
National Institute of Health and Care Excellence
SOGC
Society of Obstetricians and Gynaecologists of Canada
VA
Veterans Affairs
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References
1.
Mayo Clinic. High blood pressure (hypertension). 2024; Accessed 2024 Jul 8.
2.
Mayo Clinic. High blood pressure (hypertension): Diagnosis & treatment. 2024; Accessed 2024 Jul 8.
3.
Cleveland Clinic. Substance Use Disorder (SUD). 2022; Accessed 2024 Jul 8.
4.
Mayo Clinic. Menopause. 2023; Accessed 2024 Jul 8.
5.
Cleveland Clinic. Menopause. 2024; Accessed 2024 Jul 8.
6.
Mayo Clinic. Restless legs syndrome. 2024; Accessed 2024 Jul 8.
7.
Cleveland Clinic. Restless legs syndrome. 2023; Accessed 2024 Jul 8.
8.
Mayo Clinic. Migraine. 2023; Accessed 2024 Jul 8.
9.
Mayo Clinic. Migraine: Diagnosis & treatment 2023; Accessed 2024 Jul 8.
10.
Mayo Clinic. Adult attention-deficit/hyperactivity disorder (ADHD). 2023; Accessed 2024 Jul 8.
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Mayo Clinic. Adult attention-deficit/hyperactivity disorder (ADHD): Diagnosis & treatment. 2023; Accessed 2024 Jul 8.
12.
National Institute of Neurological Disorders and Stroke. Tourette syndrome. Accessed 2024 Jul 8.
13.
Yasaei R, Saadabadi A. Clonidine. Treasure Island (FL): StatPearls Publishing; 2023: Accessed 2024 Jul 8.
14.
Clonidine (clonidine hydrochloride): 0.1 mg and 0.2 mg tablets [product monograph]. Saint-Laurent (QC): Sivem Pharmaceuticals ULC; 2023 Jun 1.
15.
Agree Next Steps Consortium. The AGREE II Instrument. Hamilton (ON): AGREE Enterprise; 2017: Accessed 2024 Jul 3.
16.
Liberati A, Altman DG, Tetzlaff J, et al. The PRISMA statement for reporting systematic reviews and meta-analyses of studies that evaluate health care interventions: explanation and elaboration. J Clin Epidemiol. 2009;62(10):e1-e34. [PubMed: 19631507]
17.
Rabi DM, McBrien KA, Sapir-Pichhadze R, et al. Hypertension Canada's 2020 comprehensive guidelines for the prevention, diagnosis, risk assessment, and treatment of hypertension in adults and children. Can J Cardiol. 2020;36(5):596-624. [PubMed: 32389335]
18.
Wood E, Bright J, Hsu K, et al. Canadian guideline for the clinical management of high-risk drinking and alcohol use disorder. CMAJ . 2023;195(40):E1364-E1379. [PMC free article: PMC10581718] [PubMed: 37844924]
19.
U.S. Department of Veterans Affairs. VA/DoD Clinical Practice Guidelines: Management of substance use disorder (SUD). 2021; Accessed 2024 Jul 3.
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American Society of Addiction Medicine. The ASAM clinical practice guideline on alcohol withdrawal management. 2020; Accessed 2024 Jul 3. [PubMed: 32909985]
21.
Australian Government Department of Health and Aged Care. National guidelines for medication-assisted treatment of opioid dependence. 2014; Accessed 2024 Jul 3.
22.
“The Nonhormone Therapy Position Statement of The North American Menopause Society” Advisory Panel. The 2023 nonhormone therapy position statement of The North American Menopause Society. Menopause. 2023;30(6):573-590. [PubMed: 37252752]
23.
Jacobson M, Mills K, Graves G, Wolfman W, Fortier M. Guideline no. 422f: Menopause and breast cancer. J Obstet Gynaecol Can. 2021;43(12):1450-1456 e1451. [PubMed: 34895583]
24.
National Institute for Health and Care Excellence. Menopause: Diagnosis and management. (NICE guideline NG23). 2019: Accessed 2024 Jul 3. [PubMed: 33141539]
25.
Stuenkel CA, Davis SR, Gompel A, et al. Treatment of symptoms of the menopause: An Endocrine Society Clinical Practice Guideline. J Clin Endocrinol Metab. 2015;100(11):3975-4011. [PubMed: 26444994]
26.
Winkelman JW, Armstrong MJ, Allen RP, et al. Practice guideline summary: Treatment of restless legs syndrome in adults: Report of the Guideline Development, Dissemination, and Implementation Subcommittee of the American Academy of Neurology. Neurology. 2016;87(24):2585-2593. [PMC free article: PMC5206998] [PubMed: 27856776]
27.
Picchietti DL, Hensley JG, Bainbridge JL, et al. Consensus clinical practice guidelines for the diagnosis and treatment of restless legs syndrome/Willis-Ekbom disease during pregnancy and lactation. Sleep Med Rev. 2015;22:64-77. [PubMed: 25553600]
28.
Pringsheim T, Okun MS, Muller-Vahl K, et al. Practice guideline recommendations summary: Treatment of tics in people with Tourette syndrome and chronic tic disorders. Neurology. 2019;92(19):896-906. [PMC free article: PMC6537133] [PubMed: 31061208]
29.
Pringsheim T, Holler-Managan Y, Okun MS, et al. Comprehensive systematic review summary: Treatment of tics in people with Tourette syndrome and chronic tic disorders. Neurology. 2019;92(19):907-915. [PMC free article: PMC6537130] [PubMed: 31061209]
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Appendix 1. Selection of Included Studies
Figure 1
Preferred Reporting Items for Systematic reviews and Meta-Analyses PRISMA Flow Chart of Study Selection.
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Appendix 2. References of Potential Interest
Note this appendix has not been copy-edited.
Guidelines (Unclear Methodology)
1. Guidelines BC. High-Risk Drinking and Alcohol Use Disorder. 2024;
2. Canadian ADHD Resource Alliance. Canadian ADHD Practice Guidelines, 4.1 Edition, Toronto (ON); CADDRA, 2020. Available from
3. BC Guidelines. Opioid Use Disorder - Diagnosis and Management in Primary Care. 2018;
Review Articles
1. Mattes JA. Treating ADHD in Prison: Focus on Alpha-2 Agonists (Clonidine and Guanfacine). J Am Acad Psychiatry Law. 2016;44(2):151-157. [PubMed: 27236168]
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Appendix 3. Characteristics of Included Publications
Table 3
Characteristics of Included Guideline – Hypertension.
Table 4
Characteristics of Included Guidelines – Substance Use Disorders.
Table 5
Characteristics of Included Guideline – Menopause.
Table 6
Characteristics of Included Guidelines – Restless Leg Syndrome.
Table 7
Characteristics of Included Guidelines – Tourette Syndrome.
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Appendix 4. Critical Appraisal of Included Publications
Note this appendix has not been copy-edited.
Table 8
Strengths and Limitations of Guidelines Using AGREE II – Part 1.
Table 9
Strengths and Limitations of Guidelines Using AGREE II – Part 2.
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Appendix 5. Recommendations in Included Guidelines
Note this appendix has not been copy-edited.
Table 10
Summary of Recommendations – Hypertension.
Table 11
Summary of Recommendations – Substance Use Disorders.
Table 12
Summary of Recommendations – Menopause.
Table 13
Summary of Recommendations – Restless Leg Syndrome.
Table 14
Summary of Recommendations – Tourette Syndrome.
Disclaimer: The information in this document is intended to help Canadian health care decision-makers, health care professionals, health systems leaders, and policy-makers make well-informed decisions and thereby improve the quality of health care services. While patients and others may access this document, the document is made available for informational purposes only and no representations or warranties are made with respect to its fitness for any particular purpose. The information in this document should not be used as a substitute for professional medical advice or as a substitute for the application of clinical judgment in respect of the care of a particular patient or other professional judgment in any decision-making process. The Canadian Agency for Drugs and Technologies in Health (CADTH) does not endorse any information, drugs, therapies, treatments, products, processes, or services.
While care has been taken to ensure that the information prepared by CADTH in this document is accurate, complete, and up-to-date as at the applicable date the material was first published by CADTH, CADTH does not make any guarantees to that effect. CADTH does not guarantee and is not responsible for the quality, currency, propriety, accuracy, or reasonableness of any statements, information, or conclusions contained in any third-party materials used in preparing this document. The views and opinions of third parties published in this document do not necessarily state or reflect those of CADTH.
CADTH is not responsible for any errors, omissions, injury, loss, or damage arising from or relating to the use (or misuse) of any information, statements, or conclusions contained in or implied by the contents of this document or any of the source materials.
This document may contain links to third-party websites. CADTH does not have control over the content of such sites. Use of third-party sites is governed by the third-party website owners’ own terms and conditions set out for such sites. CADTH does not make any guarantee with respect to any information contained on such third-party sites and CADTH is not responsible for any injury, loss, or damage suffered as a result of using such third-party sites. CADTH has no responsibility for the collection, use, and disclosure of personal information by third-party sites.
Subject to the aforementioned limitations, the views expressed herein are those of CADTH and do not necessarily represent the views of Canada’s federal, provincial, or territorial governments or any third-party supplier of information.
This document is prepared and intended for use in the context of the Canadian health care system. The use of this document outside of Canada is done so at the user’s own risk.
This disclaimer and any questions or matters of any nature arising from or relating to the content or use (or misuse) of this document will be governed by and interpreted in accordance with the laws of the Province of Ontario and the laws of Canada applicable therein, and all proceedings shall be subject to the exclusive jurisdiction of the courts of the Province of Ontario, Canada.
The copyright and other intellectual property rights in this document are owned by CADTH and its licensors. These rights are protected by the Canadian Copyright Act and other national and international laws and agreements. Users are permitted to make copies of this document for noncommercial purposes only, provided it is not modified when reproduced and appropriate credit is given to CADTH and its licensors.
About CADTH: CADTH is an independent, not-for-profit organization responsible for providing Canada’s health care decision-makers with objective evidence to help make informed decisions about the optimal use of drugs, medical devices, diagnostics, and procedures in our health care system.
Funding: CADTH receives funding from Canada’s federal, provincial, and territorial governments, with the exception of Quebec.
Questions or requests for information about this report can be directed to Requests@CADTH.ca
Key Messages
Context and Policy Issues
Research Questions
Methods
Summary of Evidence
Limitations
Conclusions and Implications for Decision- or Policy-Making
Abbreviations
References
Appendix 1. Selection of Included Studies
Appendix 2. References of Potential Interest
Appendix 3. Characteristics of Included Publications
Appendix 4. Critical Appraisal of Included Publications
Appendix 5. Recommendations in Included Guidelines
Copyright Notice
Copyright © 2024 - Canadian Agency for Drugs and Technologies in Health. Except where otherwise noted, this work is distributed under the terms of a Creative Commons Attribution-NonCommercial- NoDerivatives 4.0 International licence (CC BY-NC-ND).
Bookshelf ID: NBK607348 PMID: 39312608
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Context and Policy Issues
Research Questions
Methods
Summary of Evidence
Limitations
Conclusions and Implications for Decision- or Policy-Making
Abbreviations
References
Selection of Included Studies
References of Potential Interest
Characteristics of Included Publications
Critical Appraisal of Included Publications
Recommendations in Included Guidelines
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CADTH Health Technology Review
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7139 | https://math.libretexts.org/Bookshelves/PreAlgebra/Prealgebra_(Arnold)/03%3A_The_Fundamentals_of_Algebra/3.02%3A_Evaluating_Algebraic_Expressions | Skip to main content
3.2: Evaluating Algebraic Expressions
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: Aug 30, 2024
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3.1: Mathematical Expressions
3.3: Simplifying Algebraic Expressions
Page ID
: 22474
David Arnold
College of the Redwoods
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In this section we will evaluate algebraic expressions for given values of the variables contained in the expressions. Here are some simple tips to help you be successful.
Tips for Evaluating Algebraic Expressions
Replace all occurrences of variables in the expression with open parentheses. Leave room between the parentheses to substitute the given value of the variable.
Substitute the given values of variables in the open parentheses prepared in the first step.
Evaluate the resulting expression according to the Rules Guiding Order of Operations.
Let's begin with an example.
Example 1
Evaluate the expression x2−2xy+y2 at x=−3 and y=2.
Solution
Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression x2 − 2xy + y2 with open parentheses.
x2−2xy+y2=( )2−2( )( )+( )2
Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.
x2−2xy+y2 =(−3)2−2(−3)(2)+(2)2 =9−2(−3)(2)+4 =9−(−6)(2)+4 =9−(−12)+4 =9+12+4 =25 Original expression. Substitute −3 for xand 2 for y. Evaluate exponents first. Left to right, multiply 2(−3)=−6. Left to right, multiply: (−6)(2)=−12. Add the opposite. Add.
Exercise
If x = −2 and y = −1, evaluate x3 − y3.
Answer
: −7
Example 2
Evaluate the expression (a − b)2 If a = 3 and b = −5, at a = 3 and b = −5.
Solution
Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression (a − b)2 with open parentheses.
(a−b)2=(()−())2
Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.
(a−b)2=((3)−(−5))2 =(3+5)2 =82 =64 Substitute 3 for a and −5 for b. Add the opposite: (3)−(−5)=3+5 Simplify inside parentheses: 3+5=8 Evaluate exponent: 82=64
Exercise
If a = 3 and b = −5, evaluate a2 − b2.
Answer
: −16
Example 3
Evaluate the expression |a|−|b| at a = 5 and b = −7.
Solution
Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression |a|−|b| with open parentheses.
|a|−|b|=|( )|−|( )|
Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.
|a|−|b|=|(5)|=|(−7)| =5−7 =5+(−7) =−2 Substitute 5 for a and −7 for b. Absolute values first: |(5)|=5 and |(−7)|=7| Add the opposites: 5−7=5+(−7). Add: 5+(−7)=−2.
Exercise
If a = 5 and b = −7, evaluate 2|a| − 3|b|.
Answer
: −11
Example 4
Evaluate the expression |a − b| at a = 5 and b = −7.
Solution
Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression |a − b| with open parentheses.
|a−b|=|( )−( )|
Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.
|a−b|=|(5)−(−7)| =|5+7| =|12| =12 Substitute 5 for a and −7 for b. Add the opposite: 5−(−7)=5+7. Add: 5+7=12. Take the absolute value: |12|=12.
Exercise
If a = 5 and b = −7, evaluate |2a − 3b|.
Answer
: 31
Example 5
Evaluate the expression
ad−bca+b
at a = 5, b = −3, c = 2, and d = −4.
Solution
Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression with open parentheses.
ad−bca+b=( )( )−( )( )( )+( )
Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.
ad−bca+b=(5)−(−3)(2)(5)+(−3) =−20−(−6)2 =−20+62 =−142 =−7 Substitute: 5 for a, −3 for b, 2 for c, −4 for d. Numerator: (5)(=4)=−20, (−3)(2)=−6. Denominator: 5+(−3)=2. Numerator: Add the opposite. Numerator: −20+6=−14.Divide.
Exercise
If a = −7, b = −3, c = −15, 15, and d = −14, evaluate:
a2+b2c+d
Answer
: −2
Example 6
Pictured below is a rectangular prism.
The volume of the rectangular prism is given by the formula
V=LWH,
where L is the length, W is the width, and H is the height of the rectangular prism. Find the volume of a rectangular prism having length 12 feet, width 4 feet, and height 6 feet.
Solution
Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of of L, W, and H in the formula
V=LWH
with open parentheses.
V=( )( )( )
Next, substitute 12 ft for L, 4 ft for W, and 6 ft for H and simplify.
V=(12ft)(4ft)(6ft)=288ft3
Hence, the volume of the rectangular prism is 288 cubic feet.
Exercise
The surface area of the prism pictured in this example is given by the following formula:
S=2(WH+LH+LW)
If L = 12, W = 4, and H = 6 feet, respectively, calculate the surface area.
Answer
: 288 square feet
Exercises
In Exercises 1-12, evaluate the expression at the given value of x.
−3x2 − 6x + 3 at x = 7
7x2 − 7x + 1 at x = −8
−6x − 6 at x = 3
6x − 1 at x = −10
5x2 + 2x + 4 at x = −1
4x2 − 9x + 4 at x = −3
−9x − 5 at x = −2
−9x + 12 at x = 5
4x2 + 2x + 6 at x = −6
−3x2 + 7x + 4 at x = −7
12x + 10 at x = −12
−6x + 7 at x = 11
In Exercises 13-28, evaluate the expression at the given values of x and y.
|x|−|y| at x = −5 and y = 4
|x|−|y| at x = −1 and y = −2
−5x2 + 2y2 at x = 4 and y = 2
−5x2 − 4y2 at x = −2 and y = −5
|x|−|y| at x = 0 and y = 2
|x|−|y| at x = −2 and y = 0
|x − y| at x = 4 and y = 5
|x − y| at x = −1 and y = −4
5x2 − 4xy + 3y2 at x = 1 and y = −4
3x2 + 5xy + 3y2 at x = 2 and y = −1
|x − y| at x = 4 and y = 4
|x − y| at x = 3 and y = −5
−5x2 − 3xy + 5y2 at x = −1 and y = −2
3x2 − 2xy − 5y2 at x = 2 and y = 5
5x2 + 4y2 at x = −2 and y = −2
−4x2 + 2y2 at x = 4 and y = −5
In Exercises 29-40, evaluate the expression at the given value of x.
9+9x−x at x = −3
9−2x−x at x = −1
−8x+9−9+x at x = 10
2x+41+x at x = 0
−4+9x7x at x = 2
−1−9xx at x = −1
−12−7xx at x = −1
12+11x3x at x = −6
6x−105 + x at x = −6
11x+11−4 + x at x = 5
10x+115 + x at x = −4
6x+12−3 + x at x = 2
The formula
d=16t2
gives the distance (in feet) that an object falls from rest in terms of the time t that has elapsed since its release. Find the distance d (in feet) that an object falls in t = 4 seconds.
The formula
d=16t2
gives the distance (in feet) that an object falls from rest in terms of the time t that has elapsed since its release. Find the distance d (in feet) that an object falls in t = 24 seconds.
The formula
C=5(F−32)9
gives the Celcius temperature C in terms of the Fahrenheit temperature F. Use the formula to find the Celsius temperature (◦ C) if the Fahrenheit temperature is F = 230◦ F.
The formula
C=5(F−32)9
gives the Celcius temperature C in terms of the Fahrenheit temperature F. Use the formula to find the Celsius temperature (◦C) if the Fahrenheit temperature is F = 95 ◦F.
The Kelvin scale of temperature is used in chemistry and physics. Absolute zero occurs at 0◦ K, the temperature at which molecules have zero kinetic energy. Water freezes at 273◦ K and boils at K = 373◦ K. To change Kelvin temperature to Fahrenheit temperature, we use the formula
F=9(K−273)5+32.
Use the formula to change 28◦ K to Fahrenheit.
The Kelvin scale of temperature is used in chemistry and physics. Absolute zero occurs at 0◦ K, the temperature at which molecules have zero kinetic energy. Water freezes at 273◦ K and boils at K = 373◦ K. To change Kelvin temperature to Fahrenheit temperature, we use the formula
F=9(K−273)5+32.
Use the formula to change 248◦ K to Fahrenheit.
A ball is thrown vertically upward. Its velocity t seconds after its release is given by the formula
v=v0−gt,
where v0 is its initial velocity, g is the acceleration due to gravity, and v is the velocity of the ball at time t. The acceleration due to gravity is g = 32 feet per second per second. If the initial velocity of the ball is v0 = 272 feet per second, find the speed of the ball after t = 6 seconds.
A ball is thrown vertically upward. Its velocity t seconds after its release is given by the formula
v=v0−gt,
where v0 is its initial velocity, g is the acceleration due to gravity, and v is the velocity of the ball at time t. The acceleration due to gravity is g = 32 feet per second per second. If the initial velocity of the ball is v0 = 470 feet per second, find the speed of the ball after t = 4 seconds.
Even numbers. Evaluate the expression 2n for the following values:
i) n = 1
ii) n = 2
iii) n = 3
iv) n = −4
v) n = −5
vi) Is the result always an even number? Explain.
Odd numbers. Evaluate the expression 2n + 1 for the following values:
i) n = 1
ii) n = 2
iii) n = 3
iv) n = −4
v) n = −5
vi) Is the result always an odd number? Explain.
Answers
−186
−24
7
13
138
−134
1
−72
−2
1
69
0
9
36
−6
−71
1
5
46
−29
256 feet
110 degrees
−409◦ F
80 feet per second
i) 2
ii) 4
iii) 6
iv) −8
v) −10
vi) Yes, the result will always be an even number because 2 will always be a factor of the product 2n.
3.1: Mathematical Expressions
3.3: Simplifying Algebraic Expressions |
7140 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4103412/ | Developmental and Individual Differences in Understanding of Fractions - PMC
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Dev Psychol
. Author manuscript; available in PMC: 2014 Oct 1.
Published in final edited form as: Dev Psychol. 2012 Dec 17;49(10):1994–2004. doi: 10.1037/a0031200
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Developmental and Individual Differences in Understanding of Fractions
Robert S Siegler
Robert S Siegler
1 Carnegie Mellon University
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1, Aryn A Pyke
Aryn A Pyke
1 Carnegie Mellon University
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1 Carnegie Mellon University
✉
Correspondence concerning this article should be addressed to Robert S. Siegler or Aryn A. Pyke, Psychology Department, Carnegie Mellon University, Pittsburgh, PA 15213. rs7k@andrew.cmu.edu or apyke@andrew.cmu.edu
Issue date 2013 Oct.
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PMCID: PMC4103412 NIHMSID: NIHMS598075 PMID: 23244401
The publisher's version of this article is available at Dev Psychol
Abstract
We examined developmental and individual differences in 6 th and 8 th graders’ fraction arithmetic and overall mathematics achievement and related them to differences in understanding of fraction magnitudes, whole number division, executive functioning, and metacognitive judgments within a cross sectional design. Results indicated that the difference between low achieving and higher achieving children’s fraction arithmetic knowledge, already substantial in 6 th grade, was much greater in 8 th grade. The fraction arithmetic knowledge of low achieving children was similar in the two grades, whereas higher achieving children showed much greater knowledge in 8 th than 6 th grade, despite both groups having been in the same classrooms, using the same textbooks, and having the same teachers and classmates. Individual differences in both fraction arithmetic and mathematics achievement test scores were predicted by differences in fraction magnitude knowledge and whole number division, even after the contributions of reading achievement and executive functioning were statistically controlled. Instructional implications of the findings are discussed.
Keywords: fractions, numerical magnitude representations, arithmetic, mathematics difficulties, mathematics achievement
Research on numerical development has focused on the development of whole numbers. This knowledge is obviously important, but numerical development also involves understanding of other types of numbers, such as fractions. Fractions are crucial not only because they express values that cannot be expressed with whole numbers but also because they provide most children’s first opportunity to learn that many properties of whole numbers are not properties of numbers in general. All whole numbers can be represented by a single numeral, have unique successors, never decrease with multiplication, never increase with division, and so on. None of these properties, however, is true of fractions or of numbers in general. Instead, the only property that all real numbers have in common is that they have magnitudes that can be located and ordered on number lines.
Current Understanding of the Development of Fraction Knowledge
Many children have great difficulty acquiring fraction knowledge. Although children receive substantial fraction instruction beginning in 3rd or 4th grade (National Council of Teachers of Mathematics (NCTM), 2006), a recent National Assessment of Educational Progress revealed that 50% of 8th graders could not correctly order three fractions (2/7, 1/12, and 5/9) from least to greatest. The difficulty continues in high school and college; for example, on another NAEP item, fewer than 30% of 11 th graders translated .029 into the correct fraction (Kloosterman, 2010). The pattern is longitudinally stable; children who have early difficulties with fractions tend to have later difficulties as well (Hecht & Vagi, 2010; Mazzocco & Devlin, 2008).
Failure to master fractions has large consequences. It impedes acquisition of more advanced mathematics and precludes participation in many occupations (McCloskey, 2007; National Mathematics Advisory Panel (NMAP), 2008). A nationally representative sample of 1,000 U.S. Algebra 1 teachers rated fractions as one of the two largest weaknesses in students’ preparation for their course (National Opinion Research Center (NORC), 2007). Consistent with these informed opinions, analyses of two large, longitudinal data sets, one from the U.S. and one from the UK, indicated that fifth graders’ knowledge of fractions uniquely predicted those students’ knowledge of algebra and overall mathematics achievement in high school 5–6 years later, even after statistically controlling for other types of mathematical knowledge, verbal and non-verbal IQ, reading comprehension, working memory, and family income and education (Siegler et al., 2012).
Understanding fractions requires not only knowledge of procedures for solving fraction arithmetic problems but considerable conceptual knowledge as well. Conceptual knowledge of fractions involves knowing what fractions are: that they are numbers that stretch from negative to positive infinity; that between any two fractions are an infinite number of other fractions; that the numerator-denominator relation, rather than either number alone, determines fraction magnitudes; that fraction magnitudes increase with numerator size and decrease with denominator size; that fractions can be represented as points on number lines; and so on.
Understanding magnitudes appears to be a particularly important aspect of conceptual understanding of fractions. Several measures of fraction magnitude representations, including number line estimation, magnitude comparison, and ordering of multiple fractions, correlate highly with knowledge of fraction arithmetic and overall mathematics achievement from 5 th to 8 th grade (Bailey, Hoard, Nugent, & Geary, 2012; Mazzocco & Devlin, 2008; Siegler, Thompson, & Schneider, 2011). This relation between fraction magnitude knowledge and overall math achievement remains strongly present even when fraction arithmetic competence is statistically controlled.
Acquiring conceptual and procedural knowledge of fractions poses several serious challenges. One challenge in acquiring conceptual knowledge is that children’s massive prior experience with integers leads to a whole number bias, in which properties of positive integers are assumed to extend to fractions (Ni & Zhou, 2005). For example, even high school students often claim that there are no numbers between fractions such as 5/7 and 6/7 (as there are no integers between 5 and 6), and that multiplication cannot yield products smaller than both multiplicands (Vamvakoussi & Vosniadou, 2004).
Acquiring procedural knowledge of fraction arithmetic poses different difficulties. Components of different fraction arithmetic procedures are complexly related. Addition and subtraction of fractions require common denominators, but multiplication and division of fractions do not. Fraction multiplication problems can be solved by applying the arithmetic operation independently to the numerators and to the denominators, but fraction addition and subtraction problems cannot be solved in this way. In fraction division, but in no other operation, answers can be obtained through the invert and multiply procedure.
This analysis suggests that conceptual and procedural knowledge of fractions should be related, because without conceptual understanding, the procedures are highly confusable and difficult to remember. Consistent with this reasoning, conceptual and procedural knowledge of fractions correlate substantially over a wide age range (Hecht, 1998; Hecht, Close, & Santisi, 2003; Hecht & Vagi, 2010; Siegler, et al., 2011).
Goals of the Present Study
Describing Fraction Development among Lower and Higher Achieving Children
The present study was designed to provide a more nuanced description of developmental and individual differences in fraction arithmetic knowledge than currently available. To pursue this goal, we investigated differences between 6 th and 8 th graders in conceptual and procedural knowledge of fractions among low achieving (LA) and higher achieving (HA) children. We examined three tasks to assess conceptual knowledge of fractions (magnitude comparison and 0–1 and 0–5 number line estimation), five aspects of fraction arithmetic (accuracy, strategy use, strategy variability, confidence, and calculation errors), and two types of processes that seemed likely to underlie fraction proficiency (whole number division and executive functioning).
This design allowed comparison of HA and LA children on all of the measures, as well as comparing the types of errors made by the two groups, which have been found to vary in previous studies (Mazzocco & Devlin, 2008). Of special interest was whether increases in fraction knowledge between 6 th and 8 th grade would differ for HA and LA children.
One plausible position was that low achieving children generally benefit less from instruction than typically achieving children (Fuchs et al., 2005); they might be expected to make less progress in this domain too. Consistent with this argument, Hecht and Vagi (2010) and Mazzocco and Devlin (2008) found poorer fraction learning among LA than HA children between4 th and 5 th grade, and 7 th and 8 th grade, respectively. The LA and HA children in our sample were drawn from the same classrooms; thus, they were exposed to the same teachers, classmates, and textbooks and to very similar instruction. However, this does not imply that their learning was similar.
Another logic yielded the opposite prediction. Having learned less from initial instruction in fractions, LA children would have more to learn later. Such patterns of change have been observed at times with whole number procedures (Chong & Siegel, 2008). Instruction during middle school might help LA children catch up to classmates whose fraction understanding was already relatively good.
A third possibility was that patterns for conceptual and procedural knowledge would differ in ways partially consistent with both of the other hypotheses. Conceptual knowledge of fractions is usually taught before fraction arithmetic (National Council of Teachers of Mathematics (NCTM), 2006). If HA children leave elementary school with greater conceptual understanding of fractions, they might make greater strides in learning fraction arithmetic during middle school. In contrast, LA children might make as great or greater progress in conceptual understanding during middle school, due to their having more to learn in that area, but their arithmetic learning might be poorer, due to their not initially possessing adequate conceptual understanding to understand the arithmetic procedures. The present design allowed us to determine which of these scenarios most accurately described changes in fraction knowledge during middle school
Identifying Sources of Individual Differences in Fraction Knowledge
A second goal of the study was to examine the contributions of four potential sources of individual differences in fraction knowledge: weak executive functioning, inadequate understanding of whole number division, limited knowledge of fraction magnitudes, and inaccurate metacognitive assessments of one’s fraction arithmetic knowledge.
Contributions of executive functioning
Executive functioning, which involves control of working memory and attention, was one likely contributor to individual differences in fraction arithmetic and mathematics achievement. Inhibition and updating, two processes identified as central to executive functioning by Miyake et al. (2000), have been found to be related to whole number arithmetic; moreover, children with LA tend to exhibit executive functioning deficits relative to typically achieving peers (Bull & Scerif, 2001; Mazzocco & Kover, 2007; Passolunghi & Pazzaglia, 2004, 2005; Swanson & Beebe-Frankenberger, 2004).
Although most research relating executive functioning to math knowledge has been done with whole numbers, executive functioning might be at least as strongly related to fraction knowledge. Conceptual knowledge of fractions requires inhibiting the tendency to view fractions as two independent whole numbers and instead focusing on the relation between numerator and denominator. Procedural knowledge of fractions requires frequent updating of working memory to keep track of completed and yet to-be-completed sub-goals and procedures (mentally add 2/3 + 4/5 to get a sense of the demands of fractions arithmetic on updating).
Contributions of fraction magnitude knowledge
Logically, fraction magnitude knowledge could be unrelated to fraction arithmetic. Children could learn such procedures through rote memorization, without reference to the numerical magnitudes that the fractions represent. However, consistent relations have emerged, probably for several reasons. Fraction magnitude knowledge allows children to check the plausibility of their answers and to reject procedures that generate implausible answers (Hiebert & LeFevre, 1986). Magnitude knowledge also might inhibit use of flawed arithmetic procedures that treat numerators and denominators as independent whole numbers, rather than as parts of a single magnitude that can be combined with another magnitude. A third reason involves motivation. Without knowledge of fraction magnitudes, fraction arithmetic, pre-algebra equation solving, and other aspects of middle school mathematics become arbitrary procedures that generate meaningless outcomes. When material is meaningless, people have little motivation to learn it (Reyna, Chapman, Dougherty, & Confrey, 2011). For example, without understanding what 1/3 means, there is no particular reason why on 1/3X = Y, X must be larger than Y. As this example illustrates, fraction knowledge is necessary for subsequent mathematics, including pre-algebra and algebra, to make sense. Thus, fraction magnitude knowledge was expected to be related to overall math achievement as well as to fraction arithmetic, even after other aspects of intellectual functioning and mathematical knowledge were statistically controlled.
Contributions of whole number division
From one perspective, the relation of whole number division to fractions is obvious. Any fraction can be viewed as a division problem (N/M); in this sense, fractions are division. However, this logical relation does not imply that the two are closely related psychologically. Indeed, knowledge of whole number division and of fractions in 5 th grade independently predicted the children’s mathematics achievement in 10 th grade; neither predictive relation explained the other (Siegler, et al., 2012).
Whole number division might be important for understanding fractions for several reasons. Alone among whole number operations, division often yields fractional answers (e.g., 14 ÷ 3 = 4 2/3). Experience with whole number division also could provide knowledge of the magnitude of some fractions (e.g., 8/2) and a rough sense of the magnitudes of others (e.g., 10/3 is between 3 and 4). In contrast, whole number addition, subtraction, and multiplication never yield fraction answers and thus do not present opportunities to learn about them.
Contributions of metacognition
A fourth potential source of individual differences involves assessments of one’s own knowledge. Inadequate metacognitive knowledge might produce either overconfidence in arithmetic answers across the board or lack of differences in confidence in correct and incorrect answers. If children are confident that incorrect procedures and the answers they yield are correct, they might not check their work or consider whether the answers are plausible (Mazzocco, 2007).
Previous findings indicate that children who are more accurate on a given math task tend to be more confident in the correctness of their answers, and that typically achieving children are more confident in their answers than children with math difficulties (Desoete, Roeyers, & Buysse, 2001; Garrett, Mazzocco, & Baker, 2006; Luncangeli & Cornoldi, 1997). However, the children in the current study were considerably older than those in most previous studies (middle school vs. early elementary school). At these older ages, both high and low achieving children might accurately evaluate the correctness of their answers.
Thus, the main purposes of the present study were to describe the development of fraction knowledge during middle school among low achieving and higher achieving children and to test four potential sources of individual differences in fraction arithmetic and overall math achievement: knowledge of fraction magnitudes, knowledge of whole number division, executive functioning, and metacognitive awareness of one’s own fraction competence.
Method
Participants
Participants were 60 6 th graders (M CA = 12.04 years, SD = 0.61; 50% girls; 68% Caucasian, 25% African American, 2% Asian, 5% Biracial) and 60 8 th graders (M CA = 14.08 years, SD = 0.49; 53% girls; 67% Caucasian, 30% African American, 2% Asian, 1% Biracial) recruited from three low-income public school districts near Pittsburgh, PA. All students who returned consent forms were included in the study. Testing was done by two female research associates and the second author (one Asian, two Caucasians).
The math section of the previous year’s Pennsylvania System of School Assessments (PSSA), the standardized achievement test used in Pennsylvania, provided a measure of overall mathematics knowledge. Children whose scores were in the bottom 35% were classified as low achieving (LA,) and those in the top 65% were classified as high achieving (HA). This threshold has been used to good effect in previous studies (Hanich, Jordan, Kaplan, & Dick, 2001; Jordan, Hanich, & Kaplan, 2003). PSSA results for 8 children were unavailable because they had transferred from another state or because their parents did not release their scores. Among children whose scores were known, 33% (15 6 th graders, and 22 8 th graders) were LA. Test scores of 6 th and 8 th graders were comparable: for HA students, means = 66 th and 67 th percentiles, p =.916; for LA students, means = 18 th and 17 th percentiles, p =.894.
Tasks
Number line estimation
On the 0–1 task, children were presented number lines on a computer screen, each with 0 at the left end, 1 at the right, and the fraction to be estimated above the line. Children responded by moving the cursor to the position on the line that they thought corresponded to the number being estimated and clicking the track pad. Then, a new number line with a different fraction appeared, and the process repeated. To acquaint children with the track pad, the experimenter asked them to first locate the practice fraction 1/4.
After the practice trial, children estimated the positions of the 10 fractions that provided the experimental data: from smallest to largest, 1/19, 2/13, 1/5, 1/3, 3/7, 7/12, 5/8, 3/4, 7/8, and 13/14. Two of these fractions were drawn from each fifth of the number line. Here as on all experimental tasks, presentation order of items was random, and no feedback was provided.
The 0–5 number line task was identical, except that the right endpoint was labeled “5,” the practice fraction was 7/2, and the 10 fractions to be estimated were 1/5, 7/8, 11/7, 9/5, 13/6, 7/3, 13/4, 10/3, 9/2, and 19/4.
Magnitude comparison
Children were asked to compare 3/5 to a series of other fractions shown one at a time in random order on a computer screen: 2/7, 1/3, 5/11, 4/7, 2/3, 3/4, 7/9, and 5/6. The task was to press the “a” key if the fraction was smaller than 3/5 and to press the “l” key if the fraction was larger than 3/5.
Fraction arithmetic
Children were presented 16 problems, 4 for each of the 4 arithmetic operations, one at a time on a computer screen. For each operation, the four problems combined the operand 3/5 with 1/5, 1/4, 2/3, and 4/5. Thus, half of the problems for each operation had operands with equal denominators (e.g., 3/5 + 1/5; 4/5 + 3/5) and half had operands with unequal denominators (e.g., 3/5 + 1/4; 3/5 + 2/3). For subtraction and division problems, the larger operand was always listed first. On each item, children received a sheet of paper that stated the problem at the top and afforded ample space for paper-and-pencil calculations. After generating an answer, children were asked to rate their confidence in it on a 5-point scale (1=not confident at all, 5=extremely confident) and then to describe how they solved the problem. These explanations were audio recorded for later coding.
Whole number division
Children were presented three whole number division problems: 56 ÷ 8, 306 ÷ 9, and 91÷ 4. Pencil and paper for calculations were provided. For the two problems with whole number quotients, children received 1 point for being correct; for the problem whose quotient included a fraction, children received 1 point for a completely correct answer (e.g., 22.75 or 22 ¾), 0.75 point for an answer with a correct remainder (e.g., 22, R 3), and 0.5 point for answers where only the whole number was correct (e.g., 22 or 22.3).
Standardized math and reading achievement tests
We obtained children’s scores from the PSSA mathematics and reading sections given toward the end of 5 th and 7 th grades, about half a year before the study began. The PSSA math section examined a range of skills, including knowledge of whole number and fraction arithmetic; probability and statistics; interpretation of tables, graphs, and figures; pre-algebra; geometry; and series extrapolation. The PSSA reading section examined vocabulary, general comprehension, and inferential reasoning skills. For example, children read about a baker who argued that a man enjoying smells from his shop should have to pay for the pleasure; the children then were asked if the baker lacks: A. talent, B. popularity, C. success, or D. generosity.
Executive function
Children were presented two EF tasks adapted from Miyake, et al. (2000), one measuring updating of working memory and one measuring inhibition.
Working memory updating
Letters were presented one at a time on a computer screen for 2750 ms each. The task was to recall, in order, the three most recently presented letters when “???” appeared on the screen. For example, after seeing “Q, D, X, R, M, ???”, the child was to respond “XRM.” Children were encouraged to rehearse aloud the three most recent letters as each new letter appeared. This task required updating one’s memory by adding the most recently presented letter and dropping the 4 th most recently presented. After three practice items, children performed nine trials on which five, seven, or nine letters appeared before the recall request. The nine trials included three of each length, and were ordered randomly. The dependent measure was percent recall of the 27 letters from the nine trials.
Inhibition
A fixation cross was presented at the center of the computer screen for a period varying from 1500 to 3500 ms. A black square (0.4 degrees wide) then appeared for 200 ms on a randomly determined side of the screen. Next, a target letter appeared, either in the square’s previous location (pro-saccade trials) or on the opposite side of the screen (anti-saccade trials); the task was to identify the letter. Because the letter was present for only 150 ms before being masked by a gray square, initially looking toward the wrong side generally ruled out correct letter identification.
Participants first completed 26 practice trials: 18 pro-saccade trials followed by 8 anti-saccade trials. Then they completed the 36 anti-saccade trials that provided the experimental data. These anti-saccade trials were the trials of interest, because they required participants to inhibit the inclination to first look toward the cue that appeared in their periphery (the black square) and instead to attend to the opposite side of the screen. The dependent measure was the percent of anti-saccade target trials on which children correctly identified the letter.
Design and Procedure
Children were tested individually in a quiet room in their school and completed all tasks on a laptop computer during a single 45-minute session. Both the items within each mathematics task and the tasks themselves were ordered randomly. The two executive function tasks (updating and inhibition) were presented in random order following the math tasks.
Results
We describe, in order, findings on development of fraction magnitude representations (number line and magnitude comparison tasks), fraction arithmetic, whole number division, executive functioning, and predictors of individual differences in fraction arithmetic and overall math achievement. Pairwise comparisons using the Bonferroni correction were used to interpret statistical interactions. Mauchly’s test indicated that the assumption of sphericity was violated in a few cases; when that occurred, Greenhouse-Geisser corrections to the degrees of freedom were used. Performance of the eight children whose PSSA scores were unavailable was excluded from all analyses that included LA/HA status as a variable.
Fraction Magnitude Tasks
Number line estimation
Accuracy of number line estimation was indexed by percent absolute error (PAE), defined as:
PAE = (|Child’s Answer – Correct Answer|) / Numerical Range 100%
For example, if a child was asked to locate 9/2 on a 0–5 number line, and the child marked the location corresponding to 7/2, the PAE would be 20% ((ú3.5 – 4.5ú) / 5 100%). PAE is an index of error; the higher the PAE, the less accurate the estimate.
A 2 (grade: 6 th or 8 th) X 2 (achievement status: HA or LA) X 2 (number line: 0–1 or 0–5) ANOVA revealed that 8 th graders’ estimates were more accurate than those of 6 th graders (PAEs = 17% and 22%), F(1, 108) = 7.31, p = .008, η p 2 = .06; HA students’ estimates were more accurate than those of LA students (PAEs = 13% and 26%), F(1,108) = 63.62, p< .001, η p 2 = .37; and estimates were more accurate on 0–1 than 0–5 number lines (PAEs = 16% and 23%), F(1,108) = 23.17, p< .001, η p 2 = .18. No interactions were significant.
Magnitude comparison
A grade X achievement status ANOVA indicated that comparison accuracy increased marginally from 6 th to 8 th grade (69% vs. 75% correct), F (1,108) = 3.17, p = .078, η p 2 = .03, and was higher among HA than LA students (81% vs. 63%), F (1,108) =24.05, p< .001, η p 2 = .18. In both grades, the greater the distance of the comparison fraction from 3/5, the greater the number of correct comparisons: r (6) = .73, p = .039 for 6 th graders, and r (6) = .66, p =.076 for 8 th graders.
Fraction Arithmetic
To obtain a nuanced understanding of fraction arithmetic development, we examined accuracy, strategy use, strategic variability, calculation errors, and confidence ratings.
Accuracy
A 2 (grade) X 2 (achievement status) X 4 (arithmetic operation) X 2 (denominator: equal or unequal) ANOVA on number of correct answers yielded main effects for all four variables: grade (41% correct for 6 th graders vs. 57% for 8 th graders), F(1,108) = 15.57, p<.001, η p 2 =.13; achievement status (36% correct for LA students vs. 61% for HA students), F(1,108) = 37.97, p<.001, η p 2 =.26; arithmetic operation (60% correct for addition, 68% for subtraction, 48% for multiplication and 20% for division), F(1.93, 208.3) = 45.82, p<.001, η p 2 =.30; and denominator equality (55% correct for problems with equal denominators vs. 43% for those with unequal ones), F(1,108) = 48.87, p<.001, η p 2 =.31. Interactions were present for grade X achievement status, F(1,108) = 4.24, p =.042, η p 2 =.04; achievement status X arithmetic operation, F(1.93, 208.3) = 8.89, p<.001, η p 2 =.08; achievement status X denominator equality, F(1,108) = 34.34, p< .001, η p 2 =.24; denominator equality X arithmetic operation, F(2.13, 230.0) = 39.15, p<.001, η p 2 =.27; and among arithmetic operation, denominator equality, and achievement status, F(2.13, 230.0) = 5.85, p =.003, η p 2 =.05.
The three-way interaction and several of the two-way interactions reflected differences between HA and LA children in the effects of denominator equality on the four arithmetic operations (Table 1). Among HA children, accuracy was higher on equal than on unequal denominator problems for addition (87% correct on equal vs. 74% on unequal denominator problems, p = .013), subtraction (91% vs. 77%, p = .002), and division problems (36% vs. 27%, p = .002). In contrast, HA children’s multiplication accuracy was lower on equal than on unequal denominator problems (34% vs. 61%, p< .001). This difference reflected children often extending equal denominators in the operands to the answer (e.g., 1/5 × 3/5 = 3/5), an approach that was correct for addition and subtraction but incorrect for multiplication.
Table 1.
Percent Correct on Fraction Arithmetic by Problem Type for Low Achieving (LA) and Higher Achieving (HA) Students
| Operation | Denominators | Achievement Status |
:---
| | | LA | HA |
| Addition | Equal | 63 | 87 |
| | Unequal | 12 | 74 |
| Subtraction | Equal | 75 | 91 |
| | Unequal | 28 | 77 |
| | | | |
| Multiplication | Equal | 40 | 34 |
| | Unequal | 55 | 61 |
| Division | Equal | 12 | 36 |
| | Unequal | 6 | 27 |
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LA children exhibited effects of denominator equality that were similar but much larger than those of their HA peers for both addition (63% correct on equal denominators problems vs. 12%, on unequal denominator ones, p<.001) and subtraction (75% vs. 28%, p<.001). For multiplication, the effect of denominator equality was somewhat smaller among children with LA than among their HA peers, but in the same direction (40% on equal vs. 55% on unequal, p = .050). For division, the accuracy of children with LA was low regardless of whether denominators were equal or unequal (12% and 6% correct, p = .114).
The forest and the trees
It is easy to get lost in the details of these data and to miss the most striking finding: the differing pattern of change during middle school in fraction arithmetic knowledge of HA and LA children. As shown in Table 2, increases in percent correct between 6 th and 8 th grade were greater for HA than LA children on all eight types of fractions arithmetic problems. HA children’s accuracy was significantly higher among 8 th than 6 th graders on 6 of the 8 types of problems, compared to 0 of 8 for LA children. The LA children were already less accurate as 6 th graders, but the gap between their and the HA children’s accuracy widened considerably between 6 th and 8 th grade.
Table 2.
Gains in Fraction Arithmetic Accuracy from 6 th to 8 th Grade Among Low Achieving (LA) and Higher Achieving (HA) Children
| Operation | Denominators | % Accuracy Gain (8 th minus 6 th) |
:---
| | | LA | HA |
| Addition | Equal | 6 | 12 |
| | Unequal | 4 | 22 |
| Subtraction | Equal | 10 | 13 |
| | Unequal | 3 | 20 |
| Multiplication | Equal | 6 | 20 |
| | Unequal | 9 | 31 |
| Division | Equal | 11 | 36 |
| | Unequal | 11 | 40 |
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Note: Gains = 8 th grade – 6 th grade percent correct.
indicates significance at p<.05
Strategies
Children’s fraction arithmetic strategies fell into four categories. Correct strategies (54% of trials) were generally the standard fraction arithmetic algorithms. Independent whole numbers strategies (14%) involved performing the arithmetic operation on numerators and denominators separately, as if they were unrelated whole numbers (e.g., 3/5 + 1/4 = 4/9). Wrong fraction operation strategies (26%) involved importing components of procedures that are correct for another fraction arithmetic operation but inappropriate for the requested operation (e.g., leaving a common denominator unchanged on a multiplication problem, as in 3/5 × 4/5 = 12/5). Finally, the None/Other classification (7%) included trials on which children used idiosyncratic strategies (e.g., inverting the wrong fraction on a division problem) or refused to try any approach.
The analysis of frequency of correct strategies yielded highly similar results to the analysis of frequency of correct answers, despite correct strategies yielding incorrect answers on 9% of trials, usually because of whole number arithmetic errors (e.g., 5 × 3 = 20). Therefore, we only report analyses of frequencies of the two common incorrect strategies.
Independent whole numbers strategy
Treating numerators and denominators as independent whole numbers leads to incorrect answers for addition and subtraction, and to unnecessarily complex answers for division. We excluded multiplication from analyses of this strategy, due to it and the correct strategy being identical.
A grade X achievement status X arithmetic operation X denominator equality ANOVA indicated that the independent whole numbers strategy was more frequent among 6th than 8th graders (25% vs. 16%), F(1, 108) = 3.09, p = .082, η p 2 = .03; among LA than HA children (32% vs. 10%), F(1, 108) = 17.66, p< .001, η p 2 = .14; on addition than on subtraction or division problems (26%, 20%, and 15%, respectively), F(1.79, 193.4) = 10.37, p< .001, η p 2 = .09; and on problems with unequal than equal denominators (25% vs. 16%), F(1, 108) = 16.58, p< .001, η p 2 = .13.
These effects were qualified by interactions between arithmetic operation and denominator equality, F(1.74, 188.1) = 11.48, p< .001, η p 2 = .10; between arithmetic operation and achievement status, F(1.79, 193.4) = 9.49, p< .001, η p 2 = .08; and among arithmetic operation, denominator equality, and achievement status, F(1.74, 188.1) = 4.04, p = .024, η p 2 = .04. The three-way interaction and one two-way interaction reflected differing strategy use by HA and LA children. Among HA children, use of the independent whole numbers strategy was comparable for equal and unequal denominator problems for addition (10% and 12%, p =.643), subtraction (6% and 10%, p =.209), and division (12% and 8%, p =.306). Among LA students, the same was true for division (20% and 21%, p = .735), but these children used the independent whole numbers strategy twice as often on problems with unequal as equal denominators for addition (55% vs. 28%, p< .001) and subtraction (45% vs. 21%, p< .001). This suggests that LA children sometimes used the whole number strategy because they did not know how to generate a common denominator, rather than because they thought it was correct.
Wrong fraction operation strategies
Children often applied components of procedures that were correct for another fraction arithmetic operation but incorrect for the operation specified in the problem. A grade X achievement status X arithmetic operation X denominator equality ANOVA indicated that such wrong fraction operation strategies were more common among 6 th than 8 th graders (30% vs. 22% of problems), F(1, 108) = 5.31, p = .023, η p 2 = .05, and much more common on division and multiplication than on addition and subtraction problems (55%, 46%, 1%, and 1%, respectively), F(1.77, 190.0) = 130.98, p<.001, η p 2 = .55. The disproportionate frequency of these errors on multiplication and division problems seems attributable to children transferring components of earlier learned fraction arithmetic operations (addition and subtraction) to later learned ones.
Interactions were present for achievement status and grade, F(1, 108) = 5.55, p = .020, η p 2 = .05, achievement status and arithmetic operation, F(1.77, 191.0) = 3.30, p = .045, η p 2 = .03, and achievement status, grade, and arithmetic operation, F(1.77, 191.0) = 5.12, p = .009, η p 2 = .05. The three-way interaction reflected different patterns of differences between 6 th and 8 th grade for HA and LA children. Among HA children, wrong fraction operation strategies were more common in 6 th than 8 th grade on multiplication and division problems (65% to 34% and 65% to 29%, p’s < .001). Similar decreases on addition and subtraction were impossible, because wrong fractions operation strategies were already uncommon in 6 th grade (0% and 3%). In contrast, among LA children, use of the strategy was comparable for 6 th and 8 th graders for multiplication (47% and 40%, p =.596) and division (57% and 68%, p =.385). As with HA children, wrong operation errors were already rare in 6 th grade (3% and 1%), so no decrease was possible. Thus, another way that the arithmetic performance of HA children improved between 6 th and 8 th grade, and that of LA children did not, was in reduction of use of wrong fraction operation errors for multiplication and division.
Variability of fraction arithmetic strategies
Highly consistent strategy use might have been expected within the pairs of virtually identical problems (e.g., multiplication problems with common denominators, such as 3/5 1/5 and 3/5 4/5). However, most participants used different strategies on at least one pair of highly similar problems. This variability was present among a greater percentage of LA than HA children, 79% vs. 51%, F(1,108) = 7.94, p =.006, η p 2 = .07. LA children also generated a greater number of pairs of trials that showed such strategic variability (21% for LA and 11% for HA children), F(1,108) = 13.10, p< .001, η p 2 = .11.
This variability often did not reflect children lacking knowledge of the correct strategy and therefore vacillating between different incorrect strategies. Children used one correct and one incorrect strategy on 65% of the inconsistent pairs of trials (67% for HA children, 64% for LA children, n. s.).
Calculation errors during fraction arithmetic strategies
Incorrect answers sometimes resulted from whole number fact errors, such as writing “5 × 3 = 20.” A grade X achievement status ANOVA indicated that 6 th graders made more such errors than 8 th graders (25% vs. 16%), F(1,108) = 9.76, p =.002, η p 2 =.08, and that LA children made more of them than did HA children (28% vs. 13% of trials), F(1,108) = 28.60, p<.001, η p 2 =.21.
Confidence in fraction arithmetic answers
A grade X achievement status X arithmetic operation X denominator equality ANOVA indicated that confidence in the correctness of answers was greater among HA than LA children (means of 4.1 vs. 3.4 on the 1–5 rating scale), F(1, 108) = 20.17, p< .001, η p 2 = .16; on addition than subtraction, subtraction than multiplication, and multiplication than division problems (means = 4.2, 3.9, 3.7 and 3.1, respectively), F(2.23, 240.7) = 75.56, p< .001, η p 2 = .41; and on problems with equal than unequal denominators (3.8 vs. 3.6), F(1, 108) = 19.63, p< .001, η p 2 = .15.
Denominator equality and arithmetic operation interacted, F(3, 324) = 20.20, p< .001, η p 2 = .16. Confidence was higher on equal than unequal denominator problems for addition (4.4 vs. 4.1, p< .001), subtraction (4.1 vs. 3.7, p< .001), and division (3.3 vs. 2.9, p< .001), but lower on equal than unequal denominator multiplication problems (3.6 vs. 3.8, p = .002). This interaction for confidence ratings paralleled the one for accuracy; just as children erred more often when multiplication problems had equal denominators, they were also less confident in their answers on such problems. A three-way interaction of denominator equality, arithmetic operation, and grade was also present, F(3, 324) = 4.19, p=.006, η p 2 = .04, though the reasons for it were unclear.
Among the 92% of children who generated at least one correct and one incorrect answer on the 16 arithmetic problems, a within-subjects analysis revealed that children were more confident in their correct than incorrect answers (mean confidence rating = 4.2 vs. 3.5), F(1,109) = 126.61, p< .001, η p 2 = .54. Mean confidence in correct versus incorrect answers was higher among most children (79%-87%) in all four grade X achievement status groups.
Whole Number Division
A grade X achievement status ANOVA revealed that HA students were much more accurate than students with LA at whole number division (89% vs. 53% correct), F(1,108) = 43.26, p<.001, η p 2 =.29. No other effects were significant (F s ≤ 1).
Executive Functioning
A grade X achievement status ANOVA indicated higher memory updating scores for HA than LA children (78% vs. 69% correct), F(1,108) = 8.54, p =.004, η p 2 =.07. A grade X achievement status interaction, F(1,108) = 5.26, p =.024, η p 2 =.05, reflected higher updating scores among 8 th than 6 th grade HA children (82% vs. 74%, p =.042), but not among LA peers (65% for 8 th graders, 73% for 6 th graders, p =.176).
A grade X achievement status ANOVA indicated that HA children were more successful than LA ones in inhibiting inappropriate responses (40% vs. 34% correct), F(1,108) = 4.07, p =.046, η p 2 =.04. No effect for age or interaction was present.
Individual Differences
Predictors of fraction arithmetic performance. I
In both grades, almost all measures were related to fraction arithmetic accuracy (Table 3). To examine predictors of fraction arithmetic accuracy in a more structured way, we constructed a hierarchical regression model. We first entered the child’s PSSA reading score, to control for general intellectual ability. Next, we entered measures of performance on the experimental tasks in the order in which the competencies seem likely to be acquired during the developmental/educational process: 1) executive functions; 2) whole number division; and 3) fraction magnitude measures. The two EF measures were entered in a single step of the hierarchical regression, as were the three measures of fraction magnitude knowledge.
Table 3.
Correlations among Experimental Tasks and Math Achievement Tests for Grade 6 (above diagonal) and
| | | | | | | Fraction Arithmetic | | Executive Function |
:---: :---: :---: :---:
| | Math Achievement (PSSA) | Reading Achievement (PSSA) | Fraction Magnitude Comparison | Number line 0–.1 (PAE) | Number line 0–.5 (PAE) | Correct Strategy Use | Answer Accuracy | Basic Calculation Errors | Whole Number Division | Update (Letter Memory) | Inhibit (Anti-Saccade) |
| Math Achievement (PSSA) | 1 | .600 | .456 | −.477 | −.499 | .164 | .234 | −.470 | .466 | .010 | .345 |
| Reading Achievement (PSSA) | .745 | 1 | .460 | −.315 | −.361 | .351 | .369 | −.405 | .299 | .206 | .149 |
| Fraction Magnitude Comparison | .567 | .427 | 1 | −.493 | −.318 | .524 | .559 | −.519 | .418 | .088 | .123 |
| Number line 0–1 (PAE) | −.763• | −.651• | −.568 | 1 | .527• | −.266 | −.295 | .271 | −.341 | −.295 | −.109 |
| Number line 0–5 (PAE) | −.628 | −.507 | −.476 | .457 | 1 | −.325 | −.343 | .368 | −.484 | −.357 | −.190 |
| Fraction Arithmetic Correct Strategy | .621• | .579 | .455 | −.483 | −.507 | 1 | .960 | −.586 | .393 | .256 | .137 |
| Fraction Arithmetic Answer Accuracy | .615• | .577 | .405 | −.433 | −.540 | .917 | 1 | −.626 | .422 | .243 | .076 |
| Fraction Arithmetic Calculation Errors | −.540 | −.561 | −.383 | .470 | .353 | −.493 | −.635 | 1 | −.348 | −.136 | −.230 |
| Whole Number Division | .703• | .603• | .421 | −.708• | −.488 | .656• | .628 | −.562 | 1 | .287 | .231 |
| EF: Update (Letter Memory) | .507• | .569 | .306 | −.550 | −.457 | .363 | .446 | −.300 | .392 | 1 | .140 |
| EF: Inhibit (Anti-Saccade) | .306 | .277 | .228 | −.381 | −.354 | .248 | .265 | −.332 | .387 | .364 | 1 |
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Notes: N=60 per grade (N=56 per grade for PSSA achievement correlations)
p<.05;
p<.01.
Reading scores accounted for 14% of the variance in 6 th graders’ fraction arithmetic scores, F(1,54) = 8.53, p =.005, and 33% of the variance in 8 th graders’ scores, F(1,54) = 26.98, p<.001. The two executive function measures explained an additional 2% of variance for 6 th graders (n.s.), and 1% for 8 th graders (n.s.). Whole number division accounted for an additional 8% of variance for 6 th graders, F(1,51) =5.43, p =.024, and 6% for 8 th graders, F(1,51) = 5.31, p =.025. The fraction magnitude tasks contributed an additional 12% of variance for 6 th graders, F(3,48) = 3.04, p =.038, and 8% for 8 th graders, F(3,48) = 2.37, p =.082. In all, these predictors accounted for 36% of variance in 6 th graders’ fraction arithmetic scores, F(7,48) = 3.78, p =.002, and 48% of variance in the 8 th graders’ scores, F(7,48) = 6.36, p<.001.
This ordering of variables might be viewed as underestimating the contribution of executive functioning, due to it sharing variance with reading achievement. To evaluate this hypothesis, we re-ordered the predictors so that executive functioning was entered before reading achievement (the order of the other variables, and their contribution, remained constant). In this analysis, EF accounted for 4% of variance in 6 th graders’ fraction arithmetic accuracy, F(2,53) = 1.10, p = .342 (n.s.) but 17% of the variance in 8 th graders’ accuracy, F(2,53) = 5.28, p = .008. Reading achievement accounted for an additional 11% of variance in 6 th graders’ fraction arithmetic accuracy, F(1,52) = 6.83, p = .012, and 18% of variance in 8 th graders’ performance, F(1,52) = 13.91, p<.001.
Predictors of mathematics achievement
Bivariate correlations indicated that among both 6 th and 8 th graders, all three fraction magnitude tasks, whole number division, calculation errors and at least one EF measure were related to overall math achievement test scores. The relations tended to be stronger for the older children (Table 3). The relations of PSSA scores to fraction performance were not due to fraction questions being major parts of the test. Analysis of test guidelines suggested that only 10%-15% of PSSA mathematics test items are dedicated to fractions for the grades in question
To examine predictors of individual children’s PSSA math scores more systematically, we conducted a hierarchical regression analysis that paralleled the one for fraction arithmetic. Because instruction in fraction arithmetic generally comes after instruction in fraction magnitudes, we entered fraction arithmetic after fraction magnitudes.
PSSA reading scores accounted for 36% of the variance in 6 th graders’ PSSA math scores, F(1,54) = 30.40, p<.001, and 56% of the variance in 8 th graders’ scores, F(1,54) = 67.44, p<.001. The two executive function measures explained 8% of variance beyond that explained by the reading scores for 6 th graders, F(2,52) = 3.93, p =.026, and 2% for 8 th graders, (n.s.). Whole number division accounted for an additional 9% of variance for both 6 th graders, F(1,51) = 9.38, p =.004, and 8 th graders, F(1,51) = 14.13, p<.001. Fraction magnitude tasks contributed an additional 10% of variance for 6 th graders, F(3,48) = 4.10, p =.011, and 9% for 8 th graders, F(3,48) = 5.50, p<.002. The fraction arithmetic measure added a non-significant 1% to the variance explained by the other measures for both 6 th and 8 th graders. Together, these predictors accounted for 64% of variance in 6 th graders’ PSSA math scores, F(8,47) = 10.33, p<.001, and 76% of variance in 8 th graders’ scores, F(8,47) = 18.83, p<.001.
When entered first, EF accounted for 12% of variance in 6 th graders’ math achievement scores, F(2,53) = 3.59, p = .035, and 28% in 8 th graders’ scores, F(2,53) = 10.05, p<.001. Reading achievement explained 33% additional variance in 6 th graders’ math achievement scores, F(1,52) = 30.40, p<.001, and 30% in 8 th graders’ scores, F(1, 52) = 36.10, p<.001.
Fraction magnitudes, fraction arithmetic, and mathematics achievement
Fazio, Bailey, Thompson, and Siegler (under review) found that children who based number line estimates predominantly on fraction magnitudes did better on other mathematics tasks and on a math achievement test than children whose number line estimation was based predominantly on numerators or denominators alone. To examine whether this relation was again present, we followed Fazio et al.’s procedure of computing gamma correlations between each child’s 0–1 number line estimates and the order of the fractions if ranked by numerator, denominator, and fraction magnitude. The predictor corresponding to the largest of these three correlations was interpreted as the child’s predominant approach, unless all relations were weak (|gamma|<.25), which resulted in a strategy classification of “unknown”. Two-thirds of children (69%) relied predominantly on fraction magnitudes, 17% on numerators, and 12% on denominators.
As in Fazio et al., children who relied mainly on fraction magnitudes on the 0–1 number line task generated much better performance on all measures of mathematical knowledge than children who did not. This was true on 0–5 number line estimation among 6 th graders (PAE = 18 vs. 30), t(58) = −3.73, p<.001, d = −1.03, and 8 th graders (PAE = 17 vs. 25), t(58) = −2.75, p = .008, d = −.82; on magnitude comparison among 6 th graders (82% vs. 60% correct), t(58) = 4.64, p<.001, d = 1.28, and 8 th graders (83% vs. 55%), t(58) = 5.48, p<.001, d = 1.63; on fraction arithmetic among 6 th graders (50% vs. 39% correct), t(58) = 1.88, p = .066, d = .52, and 8 th graders (64% vs. 40%), t(58) = 3.13, p = .003, d = .93; on whole number division among 6 th graders (83% vs. 62% correct), t(30.81) = 2.55, p = .016, d = .78, and 8 th graders (87% vs. 39%), t(58) = 5.83, p<.001, d = 1.73; and on overall math achievement test scores among 6 th graders (PSSA = 1597 vs. 1320), t(54) = 4.39, p<.001, d = 1.26, and 8 th graders (1547 vs. 1213), t(54) = 5.91, p<.001, d = 1.86. Thus, the differentiation based on 0–1 number line estimation was strong not only on other measures of numerical magnitude knowledge but also on fraction arithmetic, whole number division, and mathematics achievement.
Focusing on the numerator alone or the denominator alone seemed likely to be related to LA/HA status, and it was. Reliance on fraction magnitudes was much higher among HA than LA children in both 6 th grade (81% versus 27%) and in 8 th grade (94% versus 45%).
Discussion
Development of Fraction Knowledge in LA and HA Children
The most striking finding from the study was that differences between the fraction arithmetic knowledge of HA and LA children widen substantially between 6 th and 8 th grade. HA children’s fraction arithmetic accuracy was much higher in 8 th than 6 th grade, but LA children’s accuracy was similarly low in both grades.
At least three factors contributed to this pattern. One was differentiating between properties of whole numbers and fractions. Among LA children, whole number strategies were equally common on fraction arithmetic problems in 6 th and 8 th grade; in contrast, among HA children, whole number strategies were fairly common in 6 th grade but almost nonexistent in 8 th grade. A second likely contributor to the differential improvement involved establishing common denominators. On operations that require equal denominators (addition and subtraction), LA 6 th and 8 th graders showed equally poor performance, but HA 8 th graders were considerably more successful than HA 6 th graders. A third likely contributor involved variability of strategies. Both 6 th and 8 th grade LA children and 6 th grade HA children fairly often used a correct and an incorrect procedure to solve pairs of virtually identical problems, but HA 8 th graders consistently used correct procedures.
Sources of Individual Differences in Fraction Arithmetic and Overall Math Achievement
Fraction magnitude knowledge
As found previously (Siegler, et al., 2011), individual differences in fraction magnitude knowledge were closely related to individual differences in fraction arithmetic and mathematics achievement. The present results showed that this relation remains present even after controlling for reading achievement and executive functioning.
The relation to math achievement test scores seems likely to be due in part to the importance of fractions for solving pre-algebra and algebra problems. Consistent with this interpretation, in both Siegler et al. (2011) and the present study, the relations of fraction magnitude knowledge to mathematics achievement test scores were stronger in 8 th grade, when children would have taken or be taking pre-algebra and algebra courses, than in 6 th grade. The fact that in both studies the relation was also present, albeit weaker, in 6 th grade suggests that fraction magnitude knowledge also facilitates learning of less advanced mathematics.
Whole number division
Whole number division accounted for substantial variance in 6 th and 8 th graders’ fraction arithmetic and overall math achievement test scores, even after controlling for reading achievement and executive function. This predictive relation probably had several sources: Whole number division and fractions are logically equivalent (N/M); whole number division is assessed on middle school mathematics achievement tests; and it is essential to solving many pre-algebra, algebra, and statistics problems that also appear on the tests. Consistent with this analysis, HA and LA children’s whole number division accuracy differed greatly (89% versus 53% correct.)
In contrast to this large difference between LA and HA children’s division accuracy, no difference was present between 6 th and 8 th graders in either group. A likely reason is that whole number division is not a focus of middle school math instruction. The lack of improvement during middle school helps explain why 5 th graders’ whole number division uniquely predicts 10 th graders’ mathematics achievement (Siegler, et al., 2012). If children do not learn whole number division when it is taught, they may be unlikely to learn it later.
Executive functioning
Our findings regarding relations of executive functioning to fraction arithmetic and overall mathematics achievement were complex. Eighth graders’ working memory updating and inhibitory skills correlated with their fraction arithmetic accuracy and math achievement. These relations made sense; success at mixed series of problems, like the present ones, requires inhibiting inappropriate procedures, such as maintaining equal denominators in the product during multiplication, and it also requires continuously updating working memory contents, such as which steps in a computation have been completed and which remain. In contrast, 6 th graders’ fraction arithmetic accuracy was not related to either executive function, and their mathematics achievement was related to inhibitory skill but not working memory updating. Adding to the complexity, when reading achievement was statistically controlled, the relation between executive functioning and fraction arithmetic disappeared at both grade levels, and the relation of executive functioning to overall math achievement was present for 6 th but not 8 th graders. These results were less consistent than those obtained in some previous studies (e.g., Bailey, et al., 2012) but similar to those obtained in others (e.g., Mazzocco & Kover, 2007). The reasons for this inconsistent pattern, both within the present study and in the literature in general, remain to be established.
Metacognitive judgments
Previous studies have not tested whether children who use inappropriate arithmetic procedures believe those procedures are correct or whether they suspect that the procedures are incorrect but use them anyway because they cannot recall a correct strategy. The present study provided evidence for both interpretations.
Consistent with the view that children often believe that incorrect procedures are correct, HA children were wrong on 26% of answers in which they indicated maximal confidence, and LA children were wrong on 46%. Such overconfidence may have negative consequences for learning. Children who are confident that their incorrect answers are correct might not check them or attend to subsequent instruction.
On the other hand, the present results also reflected considerable metacognitive knowledge. In both grades, children were more confident in their correct than in their incorrect answers, and their mean confidence on each problem correlated positively with their overall accuracy on the problem. The overall pattern suggests that both overconfidence in incorrect strategies and uncertainty about the correct strategy contribute to use of incorrect approaches.
Several limitations of this study should be noted. The study was cross-sectional, so the stability of individual differences could not be examined. The sample was relatively small and included few Hispanic or Asian students. Results regarding unique contributions of executive functioning did not yield clear conclusions.
Instructional Implications
A general instructional implication of these findings is that skills that are prerequisite for subsequent mathematics need to be taught much more effectively than they are at present. Illustratively, the recently adopted U. S. Common Core State Standards prescribe that whole number division be taught and learned in 4 th grade, but the present results demonstrate that many 8 th graders have not learned it. The strong relations of whole number division to both fraction arithmetic and math achievement test scores found in this study, as well as the inherently foundational role of division for later mathematics, show the necessity of improving teaching and learning in this area.
Another instructional implication of the present findings involves a target for early intervention that might avoid many subsequent difficulties that children have with fractions. Roughly one-third of children in the present study, and a similar percentage in Fazio et al. (under review), based their fraction magnitude estimates primarily on the numerator alone or the denominator alone. This tendency was associated with many negative outcomes, including poor performance on fraction arithmetic, whole number division, and overall mathematics achievement test scores. Relying on the numerator or denominator alone also overlapped with LA classification and with the tendency to treat numerators and denominators in fraction arithmetic problems as independent whole numbers.
Viewing fraction magnitudes as independent whole numbers is understandable, because fractions with larger numerators do tend to be larger, as do fractions with smaller denominators. Moreover, the approach is appropriate for fraction multiplication and yields correct answers to some other classes of problems that are often presented in textbooks, for example comparing fraction magnitudes with equal denominators (e.g., 1/5 versus 3/5) or equal numerators (e.g., 3/5 versus 3/7). However, it is not the right way to view fractions, and does not yield consistently correct performance across unrestricted sets of fraction comparison or arithmetic problems. Placing greater instructional emphasis on the need to view each fraction as an integrated magnitude that expresses the relation between its numerator and its denominator might avoid subsequent difficulties not only in fraction arithmetic but in learning of mathematics more generally.
Acknowledgments
This research was funded by Department of Education Grant R324C10004, as well as by the Teresa Heinz Chair at Carnegie Mellon University. Charles Vukotich, Jr. provided assistance with school recruitment with support from Grant Number UL1 RR024153 from the National Center for Research Resources (NCRR). We thank Pragna Prahalad and Megan Isserman for their assistance with data collection, coding, and analyses, the administrators, teachers, parents and students of our three participating middle schools; and Nancy Jordan and Lynn Fuchs for helpful comments on the manuscript.
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Current Understanding of the Development of Fraction Knowledge
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A strip of width w is the set of points in the plane which are on, or between, two parallel lines distance w apart. Let S be a finite set of n(n≥3) points in the plane, such that any three different points from S can be covered by a strip of width 1 . Prove that S can be covered by a strip of width 2.
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Question: A strip of width w is the set of points in the plane which are on, or between, two parallel lines distance w apart. Let S be a finite set of n(n≥3) points in the plane, such that any three different points from S can be covered by a strip of width 1 . Prove that S can be covered by a strip of width 2.
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A strip of width w is the set of points in the plane which are on, or between, two parallel lines distance w apart. Let S be a finite set of n(n≥3) points in the plane, such that any three different points from S can be covered by a strip of width 1 . Prove that S can be covered by a strip of width 2.
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7142 | https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.01%3A_Binary_Operations | a∈S and b∈S⟹a∗b∈S
∗
a,b,c,d
S
a=c and b=d⟹a∗b=c∗d.
a,b,c,d
S
a=b⟹a∗c=b∗c
a,b,c,d
S
c=d⟹a∗c=a∗d
2+3=1
2+2=0
0+3=3
2⋅3=2
2⋅2=0
1⋅3=3
2+3=0
2+2=4
0+3=3
2⋅3=1
2⋅2=4
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1: Binary Operations
Last updated
: Mar 13, 2022
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2: Introduction to Groups
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: 74638
W. Edwin Clark
University of South Florida
( \newcommand{\kernel}{\mathrm{null}\,})
The most basic definition in this course is the following:
Definition 1.1: Binary Operation
A binary operation ∗∗ on a set SS is a function from S×SS×S to SS. If (a,b)∈S×S(a,b)∈S×S then we write a∗ba∗b to indicate the image of the element (a,b)(a,b) under the function ∗∗.
The following lemma explains in more detail exactly what this definition means.
Lemma 1.1 A binary operation ∗∗ on a set SS is a rule for combining two elements of SS to produce a third element of SS. This rule must satisfy the following conditions:
(a)
: a∈S and b∈S⟹a∗b∈S.
[S is closed under ∗.]
(b)
: For all a,b,c,d in S
a=c and b=d⟹a∗b=c∗d.
[Substiution is permissible.]
(c)
: For all a,b,c,d in S
a=b⟹a∗c=b∗c.
(d)
: For all a,b,c,d in S
c=d⟹a∗c=a∗d.
Proof Recall that a function ff from set AA to set BB is a rule which assigns to each element x∈Ax∈A an element, usually denoted by f(x)f(x), in the set BB. Moreover, this rule must satisfy the condition x=y⟹f(x)=f(y)
x=y⟹f(x)=f(y)(1.1)
On the other hand, the Cartesian product S×SS×S consists of the set of all ordered pairs (a,b)(a,b) where a,b∈Sa,b∈S. Equality of ordered pairs is defined by the rule a=c and b=d⟺(a,b)=(c,d).
a=c and b=d⟺(a,b)=(c,d).(1.2)
Now in this case we assume that ∗∗ is a function from the set S×SS×S to the set SS and instead of writing ∗(a,b)∗(a,b) we write a∗ba∗b. Now, if a,b∈Sa,b∈S then (a,b)∈S×S(a,b)∈S×S. So the rule ∗∗ assigns to (a,b)(a,b) the element a∗b∈Sa∗b∈S. This establishes (a). Now implication ([eqn1.1]) becomes (a,b)=(c,d)⟹a∗b=c∗d.
(a,b)=(c,d)⟹a∗b=c∗d.(1.3)
From ([eqn1.2]) and ([eqn1.3]) we obtain a=c and b=d⟹a∗b=c∗d.
a=c and b=d⟹a∗b=c∗d.(1.4)
This establishes (b).
To prove (c) we assume that a=ba=b. By reflexivity of equality, we have for all c∈Sc∈S that c=cc=c. Thus we have a=ba=b and c=cc=c and it follows from part (b) that a∗c=b∗ca∗c=b∗c, as desired. The proof of (d) is similar. ◼■
Remark
In part (a) the order of aa and bb is important. We do not assume that a∗ba∗b is the same as b∗ab∗a. Although sometimes it may be true that a∗b=b∗aa∗b=b∗a, it is not part of the definition of binary operation.
Statement (b) says that if a=ca=c and b=db=d, we can substitute cc for aa and dd for bb in the expression a∗ba∗b and we obtain the expression c∗dc∗d which is equal to a∗ba∗b. One might not think that such a natural statement is necessary. To see the need for it, see Problem 1.7 below.
Part (c) of the above lemma says that we can multiply both sides of an equation on the right by the the same element. Part (d), says that we can multiply both sides of an equation on the left by the same element.
Binary operations are usually denoted by symbols such as +,⋅,∗,×,∘,⋆,∙,⋄,⊡,⊠,⊗,⊕,⊙,∨,∧,∪,∩,⋯
+,⋅,∗,×,∘,⋆,∙,⋄,⊡,⊠,⊗,⊕,⊙,∨,∧,∪,∩,⋯
Just as one often uses ff for a generic function, we use ∗∗ to indicate a generic binary operation. Moreover, if ∗:S×S→S∗:S×S→S is a given binary operation on a set SS, we write a∗ba∗b instead of ∗(a,b)∗(a,b). This is called infix notation. In practice, we abbreviate even more; just as we use abab instead of a⋅ba⋅b or a×ba×b in high school algebra, we will often use abab instead of a∗ba∗b for a generic binary operation.
Notation. We denote the natural numbers, the integers, the rational numbers, and the real numbers by the symbols NN, ZZ, QQ, and RR, respectively.
Recall that N={1,2,3,…}Z={…,−3,−2,−1,0,1,2,3,…}Q={nm:n,m∈Z and m≠0}
NZQ==={1,2,3,…}{…,−3,−2,−1,0,1,2,3,…}{nm:n,m∈Z and m≠0}
For now, we assume that students have a basic knowledge of all these number systems. Later in the course, we will give a list of axioms from which all properties of these number systems can be derived. See Appendix C for some basic properties of NN and ZZ that we will need from time to time.
We now list some examples of binary operations. Some should be very familiar to you. Some may be new to you.
Example 1.1 Ordinary addition on NN, ZZ, QQ and RR.
Example 1.2 Ordinary multiplication on NN, ZZ, QQ and RR.
Example 1.3 Ordinary subtraction on ZZ, QQ and RR. Note that subtraction is not a binary operation on NN since, for example, 1−2∉N1−2∉N.
Example 1.4 Ordinary division on Q−{0}Q−{0} and R−{0}R−{0}. Note that division is not a binary operation on NN and ZZ since, for example, 12∉N12∉N and 12∉Z12∉Z. Also note that we must remove 0 from QQ and RR since division by 0 is not defined.
Example 1.5 For each integer n≥2n≥2 define the set Zn={0,1,2,…,n−1}.
Zn={0,1,2,…,n−1}.
For all a,b∈Zna,b∈Zn let
a+b=a+b= remainder when the ordinary sum of aa and bb is divided by nn, and
a⋅b=a⋅b= remainder when the ordinary product of aa and bb is divided by nn.
The binary operations defined in Example 1.5 are usually referred to as addition modulo nn and multiplication modulo nn. The integer nn in ZnZn is called the modulus. The plural of modulus is moduli.
In Example 1.5, it would be more precise to use something like a+nba+nb and a⋅nba⋅nb for addition and multiplication in ZnZn, but in the interest of keeping the notation simple we omit the subscript nn. Of course, this means that in any given situation, we must be very clear about the value of nn. Note also that this is really an infinite class of examples: Z2={0,1}Z2={0,1}, Z3={0,1,2}Z3={0,1,2}, Z4={0,1,2,3}Z4={0,1,2,3}, etc. Just to be clear, we give a few examples of addition and multiplication:
In Z4Z4:
: 2+3=1, 2+2=0, 0+3=3, 2⋅3=2, 2⋅2=0 and 1⋅3=3.
In Z5Z5:
: 2+3=0, 2+2=4, 0+3=3, 2⋅3=1, 2⋅2=4 and 1⋅3=3\
Example 1.6 For each integer n≥1n≥1 we let [n]={1,2,⋯,n}[n]={1,2,⋯,n}.
A permutation on [n][n] is a function from [n][n] to [n][n] which is both one-to-one and onto. We define SnSn to be the set of all permutations on [n][n]. If σσ and ττ are elements of SnSn we define their product στστ to be the composition of σσ and ττ, that is, στ(i)=σ(τ(i))for all ∈ n.
στ(i)=σ(τ(i))for all ∈ n.
See Appendix B if any of the terms used in this example are unfamiliar.
Again, we have an infinite number of examples: S1S1, S2S2, S3S3, S4S4, etc. We discuss this example as well as the other examples in more detail later. First, we give a few more examples:
Example 1.7 Let KK denote any one of the following: Z,Q,R,ZnZ,Q,R,Zn. Let M2(K)M2(K) be the set of all 2×22×2 matrices [abcd]
[acbd]
where a,b,c,da,b,c,d are any elements of KK. Matrix addition and multiplication are defined by the following rules:
[abcd]+[a′b′c′d′]=[a+a′b+b′c+c′d+d′]
[abcd]⋅[a′b′c′d′]=[aa′+bc′ab′+bd′ca′+dc′cb′+dd′]
for all a,b,c,d,a′,b′,c′,d′∈K.
Example 1.8 The usual addition of vectors in Rn, n∈N. More precisely
Rn={(x1,x2,…,xn) | xi∈R for all i}.
Addition is defined by the rule: (x1,x2,…,xn)+(y1,y2,…,yn)=(x1+y1,x2+y2,…,xn+yn).
where xi+yi denotes the usual addition of the real numbers xi and yi.
Example 1.9 Addition modulo 2 for binary sequences of length n, n∈N. (This example is important for computer science.) In this case the set is Zn2={(x1,x2,…,xn) | xi∈Z2 for all i}.
Recall that Z2={0,1}. Addition is defined by the rule: (x1,x2,…,xn)+(y1,y2,…,yn)=(x1+y1,x2+y2,…,xn+yn).where xi+yi denotes addition modulo 2 (also called exclusive or) of xi and yi. More precisely 0+0=0, 0+1=1, 1+0=1 and 1+1=0.
Example 1.10 The cross product u×v of vectors u and v in R3. Recall that if u=(u1,u2,u3)v=(v1,v2,v3)
then u×v is defined by the formula u×v=(|u2u3v2v3|,−|u1u3v1v3|,|u1u2v1v2|),where |abcd|=ad−bc.
Example 1.11 The set operations ∪ and ∩ are binary operations on the set P(X) of all subsets of X. Recall that the set P(X) is called the power set of X; and, if A and B are sets, then A∪B is called the union of A and B and A∩B is called the intersection of A and B.
Definition 1.2:
Assume that ∗ is a binary operation on the set S.
We say that ∗ is associative if
x∗(y∗z)=(x∗y)∗zfor all x,y,z ∈ S.
2. We say that an element e in S is an identity with respect to ∗ if
x∗e=x and e∗x=xfor all x in S.
3. Let e∈S be an identity with respect to ∗. Given x∈S we say that an element y∈S is an inverse of x if both x∗y=e and y∗x=e.
4. We say that ∗ is commutative if x∗y=y∗x for all x,y∈S.
5. We say that an element a of S is idempotent with respect to ∗ if a∗a=a.
6. We say that an element z of S is a zero with respect to ∗ if z∗x=z and x∗z=zfor all x∈S.
Problem 1.1 Assume that ∗ is a binary operation on the set S. Prove the following statements.
(i) If e and e′ are identities with respect to ∗ on S then e=e′. [Hint: What is e∗e′?]
(ii) If z and z′ are zeros with respect to ∗ on S then z=z′. [Hint: What is z∗z′?]
Problem 1.2 Go through all of the above examples of binary operations and determine which are not associative. Show non-associativity by giving three specific elements a,b,c such that a∗(b∗c)≠(a∗b)∗c.
Problem 1.3 Go through all of the above examples of binary operations and determine which are not commutative. Show non-commutativity by giving two specific elements a,b such that a∗b≠b∗a.
Remark
A set may have several binary operations on it. For example, consider the set R of real numbers. We write (R,⋅), (R,+), and (R,−) to indicate the set R with the binary operations multiplication, addition and subtraction, respectively. Similarly, we use this notation for other sets such as the set M2(R), of 2×2 matrices over the real numbers R. We use (M2(R),⋅) and (M2(R),+) to denote matrix multiplication and matrix addition, respectively, on M2(R).
Problem 1.4 Determine which of the examples (R,⋅), (R,+), (M2(R),⋅), and (P(X),∪) have identities. If there is an identity, determine the elements which do not have inverses.
Problem 1.5 Determine which of the examples (R,⋅), (R,+), (M2(R),⋅), and (P(X),∪) have zeros. If there is a zero, determine whether or not there are non-zero elements whose product is zero.
Problem 1.6 Determine which of the examples (R,⋅), (R,+), (M2(R),⋅), and (P(X),∪) have idempotents. Try to find all idempotents in each case.
Problem 1.7 Here we give an example of a rule that appears to define a binary operation, but does not, since substitution is not permissible. Let a,b,c,d be integers with b≠0 and d≠0. Then ab∈Q and cd∈Q.
Define ∗ on Q by: ab∗cd=a+cb2+d2.Show that ab∗cd∈Q,so Q is closed under ∗. Show by specific example that this rule does not permit substitution.
2: Introduction to Groups |
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Acidic Amino Acids
In subject area:Neuroscience
Acidic amino acids are a category of amino acids characterized by having a negatively charged side chain at physiological pH levels.
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Amino Acids, Peptides and Proteins
2010, Comprehensive Natural Products IIP.J. Milne, G. Kilian
5.20.7Relevance of Amino Acids
5.20.7.1What Is an Amino Acid?
Amino acids are bifunctional compounds and the basic structural units/building blocks of proteins. In chemistry, an α-amino acid consists of an amino group, a carboxyl group, an R-group, and a hydrogen atom which is bonded to the α-carbon. The R-group represents a side chain specific to each amino acid and 20 different kinds of side chains (varying in shape, size, charge, hydrogen bonding capacity, and chemical reactivity) are commonly found in proteins. Although amino acids exist in either the l- or d-stereoisomer most of the amino acids found in nature are of the l-type and are isolated from proteins that contain only l-amino acids. l- and d- refer to the absolute configuration of optically active compounds.11,150
For all other amino acids, with the exception of glycine (Gly), the α-carbon is bonded to four different groups, and the two stereoisomers are mirror images that cannot be superimposed. Eukaryotic proteins are always composed of l-amino acids although d-amino acids are found in certain peptide antibiotics and some peptides of bacterial cell walls.10,150 The physical properties of amino acids are influenced by the degree of ionization at different pH values.
Amino acids are not only important for protein synthesis but also serve as precursors for hormones, coenzymes, alkaloids, cell wall polymers, porphyrins, antibiotics, nucleotides, pigments, and neurotransmitters.150 A deficiency of one amino acid may result in a negative nitrogen balance (Figure 9).11
5.20.7.2Classification of Amino Acids
There are several possible ways of classifying amino acids. From their isoelectric points, their basic, acidic, or neutral character can be distinguished. Alternatively, we can consider in turn those with aromatic or aliphatic side chains.151 Therefore, based on the properties of their R-groups (polarity and charge) amino acids are grouped into five main classes: (1) nonpolar (hydrophobic), aliphatic; (2) aromatic; (3) polar (hydrophilic), uncharged; (4) positively charged (basic); and (5) negatively charged (acidic).150 On the basis of the properties of their side chains, Stryer11 divided them into seven groups: (1) aliphatic side chains, (2) hydroxyl aliphatic side chains, (3) aromatic side chains, (4) basic side chains, (5) acidic side chains, (6) amide side chains, and (7) sulfur side chains. In biochemical terms they may be divided into either glucogenic (catabolized to pyruvate, α-ketoglutarate, succinyl CoA, fumarate, or oxaloacetate) or ketogenic (gives rise to ketone bodies) amino acids depending on the metabolism of the carbon chains.11 Alanine (Ala), cysteine (Cys), Gly, serine (Ser), threonine (Thr), and Trp are all glucogenic and are converted into either acetyl-CoA or oxaloacetate.150 Five amino acids (Phe, Tyr, isoleucine (Ile), Thr, and Trp) are both glucogenic and ketogenic. Only leucine (Leu) and lysine (Lys) are exclusively ketogenic. Both asparagine (Asn) and aspartate are degraded or converted into oxaloacetate whereas the three branched-chain amino acids (Leu, Ile, and valine (Val)) are oxidized as fuels only in extrahepatic tissues, for example, brain, muscle, kidney, and adipose.
Finally, they may be classified on the basis of whether they are nonessential, essential (exogenous), and conditionally essential amino acids.150–152 Food and tissue proteins contain 20 amino acids of nutritional importance. Nine of these amino acids (histidine (His), Ile, Leu, Lys, methionine (Met), Phe, Thr, Trp, and Val) cannot be synthesized by the body and they are therefore essential or indispensable nutrients that must be obtained from the diet. The other 11 amino acids (Ala, arginine (Arg), aspartic acid (Asp), Asn, Cys, glutamic acid (Glu), glutamine (Gln), Gly, Pro, Ser, and Tyr) are also ordinarily obtained from the diet, but the body can synthesize them. They are therefore nonessential nutrients or nutritionally dispensable, but are equally as important as the indispensable amino acids for the nutrition of cells and for normal cell and organ function.153 For the conditionally essential amino acids, there is either an increased demand of certain amino acids, as seen in specific diseases or a decreased endogenous synthesis. There are only a few diseases that may cause isolated amino acid deficiencies, for example, in catabolic diseases, some of the amino acids (Arg, Cys, Gln, Gly, Pro, histidine, Ser, and Tyr) can become conditionally essential.150,152,154
5.20.7.3Choice of Amino Acid
Knowledge of how each individual amino acid and the manner in which they are combined affects important biochemical/pharmacological systems is important when considering their potential as novel agents. Understanding of the conformation, which an amino acid residue adopts during its interaction with either receptors or enzymes, is equally important.155 Thus, the choice of an amino acid plays a very important role in biological and structural relevance.
Alanine (abbreviated Ala or A) ((S)-2-aminopropanoic acid; α-aminopropionic acid) is a nonpolar, neutral, aliphatic amino acid with the formula HOOCCH(NH2)CH3. Ala plays a major role in the transport of nitrogen from skeletal muscles to the liver.150
Arginine (Arg or R) (2-amino-5-(diaminomethylidene amino)-pentanoic acid) is a polar, basic, positively charged amino acid with the formula HOOCCH(NH2)(CH2)3NH(CN+H2)NH2 and consists of a four-carbon aliphatic straight chain. The outer part of the side chain consisting of three nitrogens bonded to a carbon atom is called a guanidinium group. Owing to the conjugation between the double bond and the nitrogen lone pairs, the side chain of Arg can serve as hydrogen bond donor only.11 Arg is the precursor of nitric oxide (NO), an very important biological messenger.150 Arg-containing peptides can result in the formation of piperidones.73 The side chains of Arg and Trp can serve as hydrogen donors only, while the side chains of Ser, Thr, Asn, and Gln can serve as hydrogen bond acceptors and donors.11
Asparagine (Asn or N) ((2S)-2-amino-3-carbamoyl-propanoic acid) is a polar, uncharged amino acid with the formula HOOCCH(NH2)CH2CONH2. It has a carboxamide as the side chain’s functional group. Asx or B represent either Asn or Asp. Asn are often found near the beginning and end of alpha-helices, and in turn motifs in beta sheets.11,150
Aspartic acid (Asp or D) ((2S)-2-aminobutanedioic acid) is a polar, acidic, negatively charged amino acid with the formula HOOCCH(NH2)CH2COO−. The side chains of Asp and Glu are nearly always negatively charged at physiological pH; therefore, these amino acids are usually called aspartate (Asp) and glutamate (Glu). Asp and Glu play important roles as general acids in enzyme active centers, as well as in maintaining the solubility and ionic character of proteins.11,150
Cysteine (Cys or C) ((2R)-2-amino-3-sulfanyl-propanoic acid) is a polar, uncharged amino acid with the formula HOOCCH(NH2)CH2SH. It is an important thiol-containing (sulfur) amino acid and is classified as hydrophilic. The sulfhydryl of Cys is highly reactive and plays a crucial role in shaping proteins by forming disulfide links.11 The Cys thiol group is nucleophilic and easily oxidized to the disulfide cystine. Owing to the ability of thiols to undergo redox reactions, Cys has antioxidant properties. Cys residues play a valuable role by crosslinking proteins. Insulin is an example of a protein with cystine crosslinking. Cys is useful to detoxify the body from harmful toxins.11,150
Glutamic acid (Glu or E) ((2S)-2-aminopentanedioic acid) is a polar, amino acid with the formula HOOCCH(NH2)CH2CH2COOH. It has an acidic side chain. Glutamate is the precursor of Gln, Pro, and Arg, while Ser is the precursor of Gly and Cys.11
Glutamine (Gln or Q) ((2S)-2-amino-4-carbamoyl-butanoic acid) is a polar, uncharged amino acid with the formula HOOCCH(NH2)(CH2)2CONH2. The abbreviation Glx or Z represents either Gln or Glu. The side chain contains an amide group in place of the carboxylate and can be considered the amide of the acidic amino acid Glu. Gln is the most abundant naturally occurring, nonessential amino acid in the human body. It becomes conditionally essential in states of illness or injury. Both glutamate and Gln play key roles in nitrogen metabolism.150
Glycine (Gly or G) (aminoacetic acid, aminoethanoic acid) is a nonpolar, neutral, aliphatic amino acid with the formula HOOCCH(NH2)H.151 Gly is the simplest amino acid and plays important roles in peptide and protein chains. It does not contain a side chain and can thus fit into secondary structures where larger amino acids cannot.156 Gly acts as a transmitter in the CNS where it accomplishes several functions.157 Gly is a precursor of porphyrins.150 Gly, Pro, aspartate, Ser, and Asn enable reverse turns.150 The acylated amino group of Gly can accept a second acyl group to give rise to a diacylamide.72
Histidine (His or H) (2-amino-3-(3H-imidazol-4-yl) propanoic acid) is a polar, basic amino acid with the formula HOOCCH(NH2)CH2 (CCHNHCHN+H).151 His can be uncharged or positively charged, and is often found in active sites of enzymes where its imidazole ring can readily switch between these states resulting in the making and breaking of bonds.11 The unprotonated imidazole is nucleophilic and can serve as a general base, while the protonated form can serve as a general acid. The imidazole moiety can play a role in stabilizing the folded structures of proteins and is useful as coordinating ligand in metalloproteins. Intramolecular nucleophilic attack by the imidazole nitrogen can cause lactam formation158 as well as fission of peptide bonds.159
Isoleucine (Ile or I) ((2S,3S)-2-amino-3-methylpentanoic acid) is a nonpolar, neutral, aliphatic amino acid with the formula HOOCCH(NH2)CH(CH3)CH2CH3. Having a hydrocarbon side chain, Ile is classified as a hydrophobic amino acid. Together with Thr, Ile is one of the two common amino acids that have a chiral center. Although four stereoisomers of Ile are possible, Ile present in nature exists in only one enantiomeric form, that is, (2S,3S)-2-amino-3-methylpentanoic acid.11,150
Leucine (Leu or L) ((S)-2-amino-4-methyl-pentanoic acid) is a neutral, aliphatic amino acid with the formula HOOCCH(NH2)CH2CH(CH3)2 and with a hydrocarbon side chain, Leu is classified as a hydrophobic (nonpolar) amino acid.11,150
Lysine (Lys or K) (2,6-diaminohexanoic acid) is positively charged, amino acid with the formula HOOCCH(NH2)(CH2)4N+H3. Lys, Arg, and His have basic side chains. The ε-amino group often participates in hydrogen bonding and as a general base in catalysis. Lys plays a role in the formation of collagen and is important for proper growth and bone development.11
Methionine (Met or M) ((S)-2-amino-4-(methylsulfanyl)-butanoic acid) is a nonpolar, neutral, amino acid with the formula HOOCCH(NH2)CH2CH2SCH3. Together with Cys, Met is one of the two sulfur-containing proteinogenic amino acids and a great antioxidant. Its derivative S-adenosyl methionine (SAM) serves as a methyl donor.11,150
Phenylalanine (Phe or F) (2-amino-3-phenyl-propanoic acid) is a neutral, aromatic amino acid with the formula HOOCCH(NH2)CH2C6H5. It is classified as nonpolar because of the hydrophobic nature of the benzyl side chain.16,160 Tyr and Phe play a significant role not only in protein structure but also as important precursors for thyroid and adrenocortical hormones as well as in the synthesis of neurotransmitters such as dopamine and noradrenaline.16,160 The genetic disorder phenylketonuria (PKU) is the inability to metabolize Phe. This is caused by a deficiency of phenylalanine hydroxylase with the result that there is an accumulation of Phe in body fluids. Individuals with this disorder are known as ‘phenylketonurics’ and must abstain from consumption of Phe. A nonfood source of Phe is the artificial sweetener aspartame (l-aspartyl-l-phenylalanine methyl ester), which is metabolized by the body into several by-products including Phe. The side chain of Phe is immune from side reactions, but during catalytic hydrogenations the aromatic ring can be saturated and converted into a hexahydrophenylalanine residue.72
Proline (Pro or P) ((S)-pyrrolidine-2-carboxylic acid) is a nonpolar, neutral amino acid and is regarded as a helix breaker. It has an aliphatic side chain but it differs from other amino acids in that its side chain is bonded to both the α-carbon and nitrogen. Pro contains a secondary rather than a primary amino group, which makes it an imino acid.11 Because Pro lacks a hydrogen on the amide group, it cannot act as a hydrogen bond donor, only as a hydrogen bond acceptor. Pro is the only residue, which leads to an N-alkyl amide bond when incorporated into a peptide. Pro is therefore unique in having no amide proton to participate in hydrogen bonding and in having a cyclic side chain, which establishes conformational restrictions (the pyrrolidine ring decreases the conformational mobility of the DKP moiety, so that the side-chain rotamers for the nonprolyl residue is well defined).77,161 Pro can cause reversal of direction of peptide chains in globular proteins162 and acts as a conformational determinant in structural proteins, for example, collagen.144 Brandl and Deber163 suggested that cis–trans isomerism of Pro residues might play a role in transduction of transmembrane proteins. Inclusion of Pro in DKPs increases the solubility of the compounds in chloroform (CHCl3).77 Futhermore, studies have showed that Pro has the ability to isomerize and undergoes a transition from all-cis form (polyproline 1) to the all-trans form (polyproline 11).164 Pro is found in naturally occurring biologically active peptides, including peptide hormones, for example, angiotensin, bradykinin, oxytocin, vasopressin, melanocyte stimulating hormone (MSH) thyroid-releasing factor (TRF); gramicidin S and actinomycin; and antamanide.165 Dipeptides containing a Pro or a hydroxyproline residue, exhibit a marked propensity or strong tendency for intramolecular cyclization and it is not surprising that several Pro-based DKPs are detected in fermented and thermally treated foods.126,166,167 In addition, the thermal treatment of food appears to favor the preservation of some amino acids compared with others. For example, during the roasting of coffee beans there is a reduction in the amount of amino acids by 20–40%. The content of reactive amino acids such as Lys is strongly decreased, whereas others such as Pro or Phe remain nearly unchanged.71 Both Pro- and Gly-containing peptides have a higher probability of cyclization than peptides containing other amino acids.168
Serine (Ser or S) ((S)-2-amino-3-hydroxypropanoic acid) is a polar, neutral, uncharged amino acid with the formula HOOCCH(NH2)CH2OH. It has an aliphatic hydroxyl side chain and can be seen as a hydroxylated version of Ala.11 Ser participates in the biosynthesis of purines and pyrimidines and is also the precursor to several amino acids including Gly, Cys, and Trp (in bacteria). In addition, it is the precursor to numerous other metabolites, including sphingolipids11 and is present in enzymes such as α-chymotrypsin.169 Ser, Asn, and aspartate disrupt α helices.150
Threonine (Thr or T) ((2S,3R)-2-amino-3-hydroxybutanoic acid) has an aliphatic hydroxyl side chain and is classified as a polar, uncharged amino acid with the formula HOOCCH(NH2)CHOHCH3. Together with Ser and Tyr, Thr is one of the three proteinogenic amino acids bearing an alcohol group. Thr can be seen as a hydroxylated version of Val.11 With two chiral centers, Thr can exist in four possible stereoisomers, or two possible diastereomers of l-Thr. However, the name l-Thr is used for one single enantiomer, (2S,3R)-2-amino-3-hydroxybutanoic acid. The second diastereomer (2S,3S), which is rarely present in nature, is called l-allo-Thr.
Both Ser and Thr are commonly considered to be hydrophilic due to the hydrogen bonding capacity of the hydroxyl group and play a dynamic role in the functioning of cellular processes. Thr has been observed to have an effect on the eating patterns of mammals.170
Tryptophan (Trp or W) ((S)-2-amino-3-(1H-indol-3-yl)-propionic acid, α-amino-β-[3-indolyl]propionic acid, 1-β-indolylalanine, 2-amino-3-indolylpropionic acid) is a nonpolar, neutral, aromatic amino acid and has a indole ring attached to the methylene group. Compounds containing an indole ring are found in many pharmacologically active compounds available in the market today. Their therapeutic uses range from antiemetics and anti-inflammatories to the treatment of hypertension, migraine, and Parkinson’s disease.9 Trp is also the precursor to 5-hydroxytryptamine (5-HT), also known as serotonin, which is an important neurotransmitter in the brain.171 Trp is very sensitive under acidic conditions and its side chain is susceptible to oxidative degradation, dimerization, alkylation, and substitution with sulfenyl chlorides.72
Tyrosine (Tyr or Y) (4-hydroxyphenylalanine ((S)-2-amino-3-(4-hydroxyphenyl)-propanoic acid)) is a polar, neutral, aromatic amino acid with the formula HOOCCH(NH2)CH2C6H5OH and is the precursor of thyroxin, dopamine, norepinephrine (noradrenaline), epinephrine (adrenaline), and the pigment melanin.11 Being the precursor amino acid for the thyroid gland hormone thyroxin, a defect in this may result in hypothyroidism. Tyr is extremely soluble in water, a property that has proven useful in isolating this amino acid from protein hydrolysates. The occurrence of tyrosine-O-sulfate as a constituent of human urine and fibrinogen has been reported.172
Although Phe, Trp, and Tyr are nonpolar, Trp and Tyr are relatively more ‘polar’ than Phe because of the nitrogen indole ring and the Tyr hydroxyl group.150
Valine (Val or V) ((S)-2-amino-3-methyl-butanoic acid) is a nonpolar, neutral, aliphatic amino acid with the formula HOOCCH(NH2)CH(CH3)2. Along with Leu and Ile, Val is a branched-chain amino acid and is found in high concentrations in the muscles. Val is needed for muscle metabolism and coordination, tissue repair, and for the maintenance of proper nitrogen balance in the body.150 The steric hindrance present in Val and Ile (caused by branching) lowers the rate of coupling reactions, resulting in an increase in side reactions.72
α-Helix formation is enhanced by glutamate, Leu, and Met whereas the formation of a β-sheet is favored by Val, Phe, and Ile.150
In conclusion, certain residues such as Gly, Pro, Val, and Ile enhance the tendency for cyclization. Aspartyl and Asn residues can result in the formation of aminosuccinimide derivatives. Alkylation of Tyr, Trp, and Met side chains are possible and can happen very quickly.72
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Reference work2010, Comprehensive Natural Products IIP.J. Milne, G. Kilian
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SMELL, TASTE, AND CHEMICAL SENSING | Neurophysiology of Olfaction
2011, Encyclopedia of Fish PhysiologyT.J. Hara
Amino Acids
At physiological pH values, amino acids exist naturally in a zwitterion state, with the carboxylic acid moiety ionized and the basic amino group protonated. The entire class of amino acids has a common backbone of an organic carboxylic acid group and an amino group attached to a saturated carbon atom. The simplest is glycine, in which the saturated carbon atom is unsubstituted, rendering it optically inactive. The rest of the 20 most common amino acids are optically active and exist as both d- and l-stereoisomers. Naturally occurring amino acids are, for the most part, the l-isomers. Amino acid side chains are polar, nonpolar, or practically neutral.
Electrophysiological and optical imaging studies in salmonids, channel catfish (Ictalurus punctatus), and zebrafish (Danio rerio) show that: (1) only α-amino acids are stimulatory at biologically relevant concentrations, (2) the natural l-isomer of an amino acid is significantly more stimulatory than its d-isomer, (3) ionized α-amino and α-carboxyl groups are required, (4) the α-hydrogen of an amino acid must be free, and (5) the size and polar nature of the fourth α-moiety are important factors for stimulatory effectiveness (Figure 2). The amino acids effective as odorants are thus characterized by being simple, short, and straight chained, with only certain substituent groups.
The olfactory spectrum of amino acids is generally similar across all fish species examined. An amino acid receptor model involving two charged subsites, one anionic and the other cationic, capable of interacting with ionized amino and carboxylic groups of stimulant amino acid molecule is hypothesized. Biochemical evidence for the existence of such binding sites came from studies on binding experiments using radioactively labeled amino acids on an isolated olfactory preparation from trout. Physical parameters such as saturability, reversibility, affinity, and quantity of binding sites are reasonably consistent with receptor criteria. In addition, good agreement between the rank order of binding and electrophysiological potency further supports the conclusion that binding data reflect the interaction of amino acids with physiologically relevant receptors. Amino acids assert their activities through multiple receptor systems; note, for example, that in Figure 2 cysteine, glutamate, and arginine evoke different maximal effect, suggesting the involvement of different receptor-mediated physiological functions.
All lines of experimental evidence indicate that in channel catfish, Arctic char (Salvelinus alpinus), and rainbow trout (Onchorynchus mykiss) at least three receptor types (cysteine- (or, neutral amino acid-), arginine- (or, basic guanidine-containing amino acid-), and glutamate- (or, acidic amino acid-) receptor types) exist, which play dominant roles in the discrimination of amino acids (Figure 3). Specific binding of alanine of the salmonid olfactory preparation is inhibited by neutral amino acids, but only partially by arginine or glutamate. Each receptor recognizes the characteristic features, or epitope, of an amino acid molecule, including (1) hydrocarbon chain length, (2) difference in functional groups, and (3) position of functional groups. Although the three receptor types are quite specific under the natural condition, because all amino acids have some of these features in common (generic structural plan), they could all function as partial agonists, activating all available receptors depending on their affinities and concentrations. For example, the cysteine receptor is activated not only by its primary ligand cysteine, but also by other neutral amino acids including alanine, glutamine, and methionine, with lesser affinities, and generates receptor potentials accordingly. Also, electrophysiological and binding analyses show that the cloned goldfish olfactory receptor 5.24 is preferentially tuned to recognize basic amino acids; of 20 amino acids tested for competition binding, arginine and lysine have the highest affinities, suggesting that this receptor may in fact represent an Arg-, or more generally guanidino-containing amino acid receptor for this species. This would verify the existence of the Arg-receptor predicted electrophysiologically earlier in salmonids. It should be noted that the 5.24 receptor shares amino acid sequence identity to mammalian pheromone receptors, the Ca-sensing receptors, the T1R taste receptors, and the metabotropic glutamate receptors (mGluRs).
Recently, kynurenine, a metabolite of tryptophan and technically an amino acid, released by ovulating female masu salmon has been found to be a potent olfactory stimulant for spermiating males, with an amazingly low detection threshold of 10−14 M. Even immature or sexually regressed males as well as ovulated females detect it at 10−12–10−10 M. The urine of the reproductively mature female masu salmon contains high levels of kynurenine that is released into the water at a rate of 0.40 × 10−6 mole/female per hour, where it is thought to act as a male-attracting pheromone (see also SMELL, TASTE, AND CHEMICAL SENSING | Chemosensory Behavior). Whether this is a reproductive pheromonal cue specific for this salmonid species and behaves like ordinary amino acid odorants await future study.
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AMINO ACIDS | Metabolism
2003, Encyclopedia of Food Sciences and Nutrition (Second Edition)R.M.B. Ferreira, A.R.N. Teixeira
Introduction
Naturally occurring amino acids may be conveniently grouped into three categories: protein amino acids (sometimes known as ‘standard,’ ‘primary,’ and ‘normal’), uncommon amino acids, and nonprotein amino acids. Protein amino acids are those that are coded for in the genes and incorporated directly into proteins. For some time it seemed well established that all proteins, whatever their origin, were constructed from the same set of 20 amino acids. Recent studies, however, have shaken the foundation of this classical dogma. It now seems that the genetic code may dictate the incorporation of more than 20 amino acids. Thus, for example, selenocysteine and phosphoserine, previously considered to be uncommon amino acids, can be directly incorporated into the polypeptide chain. All protein amino acids are α,l-amino acids. It is not clear why amino acids incorporated by organisms into proteins are of the l form, since l-amino acids have no obvious inherent superiority over their d isomers for biological function. Thus, in this article, unless otherwise stated, an l configuration is assumed.
The 20 classical protein amino acids may be grouped into several classes reflecting important characteristics of their side chains: straight aliphatic amino acids (glycine, alanine), branched-chain amino acids (valine, leucine, isoleucine), hydroxy amino acids (serine, threonine), sulfur-containing amino acids (cysteine, methionine), aromatic amino acids (phenylalanine, tyrosine), heterocyclic amino acids (tryptophan, histidine), basic amino acids (lysine, arginine), acidic amino acids and their amides (aspartate, glutamate, asparagine, glutamine), and imino acid (proline). Amino acids can also be classified on the basis of the polarity of their side chains. (See PROTEIN | Chemistry.)
Analyses of proteins have revealed that they contain well over 100 different amino acids. The occurrence of uncommon amino acids in proteins is the result of posttranslational, covalent modification of protein amino acids. Cystine, for example, is formed by the posttranslational cross-linking of two cysteine residues. Citrulline, N-formylmethionine, O-galactosylserine, and N-acetylthreonine constitute other examples of amino acids found in proteins.
The amino acids found in proteins are by no means the only ones to occur in living organisms. Thus the term ‘nonprotein amino acids’ is used to include those naturally occurring amino acids which are present in free or combined forms but not in proteins. Over 200 nonprotein amino acids are known, most of them occurring in plants and frequently limited, in each case, to certain taxonomic groups. Some, such as cystathionine and saccharopine, fulfill important roles in the primary metabolic pathways. However, the great majority of these compounds have obscure functions and are generally regarded as secondary products. Many of the nonprotein amino acids fromplants are known to be toxic to animals, plants, and microorganisms. Some accumulate to exceptionally high levels, as in the case of 5-hydroxytryptophan, canavanine, or 3,4-dihydroxy-phenylalanine, which may constitute up to 14% of the seed weight in some Leguminosae species. Storage and protection against predation are probably two of the many possible roles that these amino acids play in plants.
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Reference work2003, Encyclopedia of Food Sciences and Nutrition (Second Edition)R.M.B. Ferreira, A.R.N. Teixeira
Chapter
Structure and Properties of Biologic Molecules
2012, Elsevier's Integrated Review Biochemistry (Second Edition)John W. Pelley
Amino acids
An amino acid contains four functional groups organized around the α-carbon: the α-amino group, the α-carboxyl group, a unique side chain (hydrogen in glycine), and hydrogen. The asymmetry of the α-carbon gives rise to two optically active (chiral) isomers termed l- and d-amino acids (Fig. 2-1). The l-form is unique to proteins, while the d-form appears in bacterial cell walls and some antibiotics.
The genetic code in DNA specifies 20 amino acids for the construction of polypeptides. The most useful method for grouping or classifying amino acids is by their hydrophobicity and charge properties in order to understand their location in proteins and their influence on protein structure.
Hydrophobic and Hydrophilic Amino Acids
Hydrophobic amino acids have nonpolar side chains and are usually found at the interior of a protein or where the surface interfaces with lipids (Table 2-1).
•
: Alanine and glycine have the smallest side chains. Glycine lacks a side chain, which makes it compatible with hydrophobic environments. Alanine is prominent in the transport of nitrogen from muscle to liver during fasting (alanine cycle).
•
: Valine, leucine, and isoleucine are referred to as “branched-chain amino acids.” Their metabolism is altered in maple syrup urine disease.
•
: Proline has a cyclized side chain joined back to its α-amino group to form an “imino” acid. It functions as a helix breaker in the secondary structure of proteins (see later discussion). It is also hydroxylated to hydroxyproline after incorporation into collagen (requires ascorbic acid).
•
: Phenylalanine, tyrosine, and tryptophan are aromatic amino acids. Phenylalanine is increased in the serum and tissues of patients with phenylalanine hydroxylase deficiency (phenylketonuria; PKU), characterized by an inability to synthesize tyrosine from phenylalanine. Tyrosine is a precursor to dopamine and the catecholamines and, in proteins, can be phosphorylated by the action of tyrosine kinases. Tryptophan serves as a precursor for serotonin and melatonin and can be converted to niacin. The aromatic amino acids are the primary sites of chymotrypsin cleavage in proteins.
•
: Methionine is a sulfur-containing amino acid. It is always the first amino acid incorporated into polypeptides, but it may be removed afterward. S-adenosyl methionine serves as a single carbon donor. Methionine is the site of cyanogen bromide cleavage in proteins.
Hydrophilic amino acids have side chains that form hydrogen bonds and are found where the surface interfaces with the water (Table 2-2).
•
: Serine and threonine are the hydroxyl-containing amino acids. Both can be phosphorylated by the action of various kinases. Serine serves as a single carbon donor to tetrahydrofolate (THF) to produce N5,N10-methylene THF and glycine.
•
: Cysteine, like its hydrophobic counterpart methionine, is a sulfur-containing amino acid. Its thiol group can undergo enzyme-catalyzed oxidation, but it is also sensitive to oxidation by air, forming cystine. Cysteine is a component of glutathione, a recyclable antioxidant in cells. It can form covalent disulfide crosslinks (Fig. 2-2) that stabilize the structure of proteins, especially secreted proteins.
•
: Aspartate, asparagine, glutamate, and glutamine are the acidic amino acids and their amides. Both aspartate and glutamate carry a negative charge at a pH of 7; aspartate is interconverted with oxaloacetate by aspartate aminotransferase (AST), and glutamate is interconverted with α-ketoglutarate by alanine aminotransferase (ALT). Asparagine and glutamine are polar, neutral amino acids. Glutamine is formed by glutamine synthetase action in the brain and liver to detoxify ammonia, and it also serves as a donor of amide nitrogen in the biosynthesis of purines and pyrimidines.
•
: Lysine, histidine, and arginine are the basic amino acids, and they carry a positive charge at a pH of 7. Lysine and arginine are the site of trypsin cleavage in proteins; both are present at high concentration in histones. Histidine is only weakly basic and is uncharged at a pH of 7. Histidine forms one of the six coordination bonds with Fe++ in the heme prosthetic group of hemoglobin and myoglobin. Arginine (pKa ~ 14) always has a positive charge at neutral pH; it has an important role in the binding of anionic molecules, such as nucleic acids.
Pharmacology
Asparaginase Therapy
Asparagine is required in high amounts by some types of leukemia cells, making systemic administration of asparaginase (hydrolysis to aspartate and ammonia) an effective treatment.
Key Points About Amino Acids
▪
: The amino acids that comprise polypeptides have a consistent structure and differ only by their side chain; they can be classified by their degree of hydrophobicity and the nature of the side chain functional groups.
▪
: The sequence of amino acid side chains determines the native (tertiary) structure of the proteins that contain them.
▪
: Free amino acids also have biologic functions in intermediary energy metabolism, in the endocrine system, and in neuronal function.
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Amino Acids
2002, Medical Biochemistry (Fourth Edition)N.V. BHAGAVAN
2.3Electrolyte and Acid-Base Properties
Amino acids are ampholytes, i.e., they contain both acidic and basic groups. Free amino acids can never occur as neutral nonionic molecules:
Instead, they exist as neutral zwitterions that contain both positively and negatively charged groups:
Zwitterions are electrically neutral and so do not migrate in an electric field. In acidic solution (below pH 2.0), the predominant species of an amino acid is positively charged and migrates toward the cathode:
In basic solution (above pH 9.7), the predominant species is negatively charged and migrates toward the anode:
The isoelectric point (pI) of an amino acid is the pH at which the molecule has an average net charge of zero and therefore does not migrate in an electric field. The pI is calculated by averaging the pK′ values for the two functional groups that react as the zwitterion becomes alternately a monovalent cation or a monovalent anion.
At physiological pH, monoaminomonocarboxylic amino acids, e.g., glycine and alanine, exist as zwitterions. That is, at a pH of 6.9-7.4, the α-carboxyl group (pK′ 2.4) is dissociated to yield a negatively charged carboxylate ion (-COO−), while the α-amino group (pK‘ 9.7) is protonated to yield an ammonium group The pK′ value of the α-carboxyl group is considerably lower than that of a comparable aliphatic acid, e.g., acetic acid (pK′ 4.6). This stronger acidity is due to electron withdrawal by the positively charged ammonium ion and the consequent increased tendency of a carboxyl hydrogen to dissociate as an H+. The α-ammonium group is corre- spondingly a weaker acid than an aliphatic ammonium ion, e. g., ethylamine (pK′ 9.0) because the inductive effect of the negatively charged carboxylate anion tends to prevent dissociation of H+. The titration profile of glycine hydrochloride (Figure 2-7) is nearly identical to the profiles of all other monoaminomonocarboxylic amino acids with nonionizable R-groups (Ala, Val, Leu, Ile, Phe, Ser, Thr, Gln, Asn, Met, and Pro).
The titration of glycine has the following major features. The titration is initiated with glycine hydrochloride,CI-(NCH2COOH) which is the fully protonated form of the amino acid. In this form, the molecule contains two acidic functional groups; therefore, two equivalents of base are required to completely titrate 1 mol of glycine hydrochloride. There are two pK′ values: due to reaction of the carboxyl group and due to reaction of the ammonium group. Addition of 0.5 eq of base to 1 mol of glycine hydrochloride raises the pH 2.34 (pK;), whereas addition of 1.5 eq further increases the pH to 9.66 (). At low pH values (e.g., 0.4), the molecules are predominantly cations with one positive charge; at pH values of 5-7, most molecules have a net charge of zero; at high pH values (e.g., 11.7), all of the molecules are essentially anions with one negative charge. The midpoint between the two pK′ values [i.e., at pH = (2.34 + 9.66)/2 = 6.0] is the pI. Thus, pI is the arithmetic mean of and values and the inflection point between the two segments of the titration profile.
The buffering capacities of weak acids and weak bases are maximal at their pK values. Thus, monoaminomono-carboxylic acids exhibit their greatest buffering capacities in the two pH ranges near their two pK′ values, namely, pH 2.3 and pH 9.7 (Figure 2-7). Neither these amino acids nor the α -amino or α-carboxyl groups of other amino acids (which have similar pK′ values) have significant buffering capacity in the neutral (physiological) pH range. The only amino acids with R-groups that have buffering capacity in the physiological pH range are histidine (imidazole; pK′ 6.0) and cysteine (sulfhydryl; pK′ 8.3). The pK′ values for R-groups vary with the ionic environment. The titration profile of histidine is shown in Figure 2-8. The pI is the mean of and .
Titration profiles of the basic and acidic amino acids lysine and aspartic acid are shown in Figures 2-9 and 2-10. The R-groups are ionized at physiological pH and have anionic and cationic groups, respectively. The pI value for aspartic acid is the arithmetic mean of and , whereas for lysine and histidine the pI values are given by the arithmetic mean of and . The pK′ and pI values of selected amino acids are listed in Table 2-2.
TABLE 2-2. pIC and pl Values of Selected Free Amino Acids at 25°C
| Amino Acid | (α-COOH) | | | pI |
--- ---
| Alanine | 2.34 | 9.69 (α-,) | 6.00 |
| Aspartic acid | 2.09 | 3.86 (β-COOH) | 9.82 (α-) | 2.98 |
| Glutamic acid | 2.19 | 4.25 (γ-COOH) | 9.67 (α-) | 3.22 |
| Arginine | 2.17 | 9.04 (α-) | 12.48 (Guanidinium) | 10.76 |
| Histidine | 1.82 | 6.00 (Imidazolium) | 9.17 () | 7.59 |
| Lysine | 2.18 | 8.95 (α-) | 10.53 (ε-) | 9.74 |
| Cysteine | 1.71 | 8.33 (-SH) | 10.78 (α-) | 5.02 |
| Tyrosine | 2.20 | 9.11 (α-) | 10.07 (Phenol OH) | 5.66 |
| Serine | 2.21 | 9.15 (α-) | 13.6 (Alcohol OH) | 5.68 |
: The pK′ values for functional groups in proteins may vary significantly from the values for free amino acids.
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Book2002, Medical Biochemistry (Fourth Edition)N.V. BHAGAVAN
Chapter
Fundamentals of Genes and Genomes
2014, Bioinformatics for BeginnersSupratim Choudhuri
1.9.6Acidic and Basic Proteins
At physiological pH (7.4), acidic proteins tend to be negatively charged and have a higher proportion of acidic amino acids (e.g. aspartic acid, glutamic acid), whereas basic proteins tend to be positively charged and have a higher proportion of basic amino acids (e.g. arginine, lysine).
Hydrophilic and charged amino acids are frequently associated with antigenic determinants (epitopes), such as arginine, lysine, aspartic acid, glutamic acid, asparagine, glutamine, serine, and threonine.
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Review article
Review Article The family of sodium-dependent glutamate transporters: a focus on the GLT-1/EAAT2 subtype
1998, Neurochemistry InternationalM.B Robinson
The acidic amino acids, glutamate and aspartate, are the predominant excitatory neurotransmitters in the mammalian CNS. Under many pathologic conditions, these excitatory amino acids (EAAs) accumulate in the extracellular fluid in CNS and the resultant excessive activation of EAA receptors contributes to brain injury through a process known as excitotoxicity. Unlike many other neurotransmitters, there is no evidence for extracellular metabolism of EAAs, rather, they are cleared by Na+-dependent transport mechanisms. Therefore, this transport process is important for ensuring crisp synaptic signaling as well as limiting the excitotoxic potential of EAAs. With the cloning of five distinct EAA transporters, a variety of tools were developed to characterize individual transporter subtypes, including specific antibodies, expression systems, and probes to delete/knock-down expression of each subtype. These tools are beginning to provide fundamental information that has the potential to impact our understanding of EAA physiology and pathophysiology. For example, biophysical studies of the cloned transporters have led to the observation that some subtypes function as ligand-gated ion channels as well as transporters. With these reagents, it has also been possible to explore the relative contributions of each transporter to the clearance of extracellular EAAs and to begin to examine the regulation of specific transporter subtypes. In this review, an overview of the properties of the transporter subtypes will be presented. The evidence which suggests that the transporter, GLT1/EAAT2, may be sufficient to explain a large percentage of forebrain transport will be critically reviewed. Finally, the studies of regulation of GLT-1 in vitro and in vivo will be described.
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Fundamentals of Genes and Genomes
2014, Bioinformatics for BeginnersSupratim Choudhuri
1.9Protein Structure and Function
Proteins (polypeptides) are translated from the mRNA, which carries the amino acid sequence information for the polypeptide. Translation proceeds from the N-terminal to C-terminal direction of the polypeptide being synthesized. Proteins are made up of structural units called amino acids. All amino acids are α-amino acids. They are called α-amino acids because the amino group (NH2) is attached to the α-carbon atom—that is, the carbon atom linked to the carbonyl carbon of the carboxyl group (COOH). The basic formula of an amino acid is shown in Figure 1.6A.
1.9.1Configuration and Chirality of Amino Acids
All amino acids except glycine (R=H) are chiral because the α-carbon is chiral or asymmetric. So, except for glycine all amino acids can have two mirror-image stereoisomers (enantiomers). According to the DL system of Fischer, all natural amino acids are in L-configuration (as opposed to monosaccharides, which exist in D-configuration) (Figure 1.6B); according to the RS system of Cahn–Ingold–Prelog, all natural amino acids are in the S-configuration (Figure 1.6C). So, the S-form is analogous to the L-form (see Box 1.8). Located on the alpha carbon is the “R” group, called the side chain. The nature of this side chain determines the identity of a particular amino acid. Glycine is the simplest amino acid because R=H. Amino acid side chains can be polar or nonpolar. Polar side chains may be charged or neutral. For example, two negatively charged amino acids are aspartic acid and glutamic acid. Two positively charged (i.e. protonated) amino acids are lysine and arginine. Figure 1.6D shows the numbering of carbon atoms of lysine. A small fraction of histidine is also positively charged at physiological pH. Proline is the only amino acid that has an imino group rather than an amino group. Although there are many more amino acids known so far, only 20 of them are standard amino acids used by all organisms during translation to synthesize proteins because they are encoded by the genetic code.
Box 1.8
1.
: The DL system of denoting enantiomers, originally introduced by Emil Fischer, is an old way of denoting the chirality of biological macromolecules. A more recent system is the RS system introduced by Robert Cahn, Christopher Ingold, and Vladimir Prelog. Naturally occurring amino acids have L-configuration according to the DL system, and S-configuration according to the RS system. In the RS system, first the priority of the groups attached to the chiral center is established. Then the order from the highest priority group to the second highest priority group, and so on, is established. If the order is clockwise, the molecule is said to have the R- (rectus) configuration; if the order is anticlockwise, the molecule is said to have S- (sinistrus) configuration. In Figure 1.6, NH3+ has the highest priority (because the atomic number of N is 7), followed by COO− (because the atomic number of C is 6). If the first atom of two groups has the same atomic number, then the priority of the group is determined by the second atom and so on. Thus, COOH will have higher priority than CH2OH.
2.
: The presence of two H atoms makes the α-carbon of glycine achiral (not chiral) or symmetric. As a result, glycine does not have any enantiomer (D/R or L/S isomer) and has no optical activity (dextro or levo).
1.9.2Ionic Character of Amino Acids
In solution at physiological pH (7.4), amino acids exist as dipole ions or zwitterions, where the amino group (NH2) exists as an ammonium ion (NH3+) and the carboxyl group (COOH) exists as a carboxylate ion (COO−) (Figure 1.6A). An amino acid can therefore act as a base as well as an acid, and hence is an ampholyte (having amphoteric properties). In a zwitterion, the+and−charges cancel each other to give the molecule a net charge of zero. However, at pH that is significantly higher or lower than physiological pH, amino acids undergo ionization. At acidic pH that is significantly lower than 7.4, the amino group has a positive charge while the carboxyl is neutral. At alkaline pH that is significantly higher than 7.4, the amino group is neutral while the carboxyl has a negative charge.
Amino acids of proteins in solution accept or lose protons depending on the nature of the side chains. The pKa values of amino acids (i.e. the tendency of amino acids to lose protons) play an important role in determining the pH-dependent properties of a protein in solution. Internal ionizable groups in proteins are essential for catalysis. During a cycle of function, these internal ionizable groups can experience different microenvironments, and their pKa values and charged states adjust accordingly.29
1.9.3Relationship between Protein Function and the Location of Amino Acids in the Polypeptide Chain
The location of amino acids in the folded conformation of a protein is relevant for the protein’s function and its interaction with the environment. For example, proteins located in a hydrophobic environment, such as membrane, have nonpolar (hydrophobic) side chains on the surface interacting with the membrane lipids. In contrast, proteins located in an aqueous environment, such as cytosol, have polar side chains (hydrophilic) on the surface interacting with the aqueous environment.
Arginine and lysine carry positive charges, and are often located on the interacting surface of proteins that interact with negatively charged molecules. Predictably, arginine and lysine are found on the surface of DNA-binding proteins that interact with the negatively charged phosphate group of DNA.
Similarly, aspartic acid and glutamic acid carry negative charges, and are often located on the interacting surface of proteins that interact with positively charged molecules. Aspartic acid and glutamic acid in calmodulin bind Ca++ ions, which carry a complementary positive charge. Many proteins in halophilic archaebacteria, which live in an extremely salty environment, have high localized concentrations (high charge density) of acidic amino acids on the surface. Such high charge density of acidic amino acids very effectively sequesters sodium ions, thus preventing denaturation and precipitation of cellular proteins. In fact, these proteins are denatured if placed in low salt concentration because the removal of sodium ions leaves many closely placed negative charges exposed, which strongly repel each other.
Serine, threonine, and tyrosine have hydroxyl groups (−OH) in their side chains. These OH groups can serve as phosphate attachment sites during phosphorylation. Many receptors that are involved in signal transduction are phosphorylated for activation, and consequently have these amino acid residues in their active sites. Phosphorylation causes conformational change in these receptors.
The sulfhydryl (-SH) group in cysteine is ideal for binding metals through metal—thiolate bonds. Naturally, cysteines are prevalent in many storage proteins that bind heavy metals. For example, in metallothionein, the intracellular metal-binding protein, one third of the amino acid residues are cysteines. The -SH group is also ideal for forming strong covalent disulfide linkages that stabilize the conformation of proteins. Expectedly, cysteines are found in many enzymes that function in harsh conditions of salt and pH, such as digestive enzymes like pepsin and chymotrypsin. The structure of many small proteins, such as insulin and ribonuclease, is stabilized by cysteine disulfide linkages. Cysteine disulfide linkages also confer rigidity to protein tertiary structure and are found in proteins like keratin in hair.
Proline occurs near the bend of polypeptide chains, and its ring forms a useful kink in the protein chain. Therefore, proline helps redirect the protein chain back inwards or around a tight corner.
Glycine and alanine, being very small, are flexible and can easily fit into tight spots. For example, glycine is the most abundant amino acid in the tight triple helix of collagen (about one-third of all amino acids). Alanine, being small and chemically inconspicuous, can be accommodated on the inside as well as outside of proteins. Alanine residues are very common in proteins. Attempts to confirm the functional role of specific amino acid residues in proteins involve mutagenesis experiments, and oftentimes the target amino acid is replaced by alanine.
1.9.4Linkage between Amino Acids—The Peptide Bond
Amino acids are linked together by peptide bonds (alpha peptide bonds), which are simply amide linkages between the NH2 and COOH groups of neighboring amino acids. The peptide bond has unique characteristics, which contribute to the overall structure of proteins. The peptide bond has a partial double-bond character. Thus, it is rigid and planar and not free to rotate. The plane on which it lies is called the amide plane. Peptide bonds are generally trans bonds—that is, the carbonyl oxygen and amide hydrogen are in trans position (Figure 1.6E). The CαC bonds are not rigid and they can freely rotate, being only limited by the size and character of the R groups. In lysine, the ε-amino group (Figure 1.6D) also participates in the formation of a peptide bond, which is called an isopeptide bond because it does not involve the usual α-amino group.
1.9.5Four Levels of Protein Structure
Proteins have four levels of structure: primary, secondary, tertiary, and quaternary. Primary structure refers to the amino acid sequence of a protein. Secondary structure refers to the conformation of the polypeptide backbone. Examples of secondary structures are helices (α-helix), pleated sheets (β-pleated sheet), and bends or turns (β-bend). Tertiary structure of a protein refers to its three-dimensional structure—that is, further folding of the secondary structure in the three-dimensional space. Quaternary structure refers to a structure achieved by proteins composed of more than one polypeptide chain. Each polypeptide chain, called a subunit, has its own primary, secondary, and tertiary structure. In quaternary structure, protein chains (subunits) can associate with one another to form dimers, trimers, and other higher orders of oligomers. Recent studies have shown that despite having definitive structure, many proteins have specific regions that are intrinsically disordered (see Box 1.9).
Box 1.9
Intrinsically Disordered Proteins: The “Unstructural” Aspect of Structural Biology30
It has long been known that structural flexibility exists in proteins and aids in ligand binding. Nevertheless, the “structure–function paradigm”—that is, that proteins possess definitive three-dimensional structures in order to perform their function—has been the standard paradigm in protein biochemistry. Experimental evidence accumulating since the turn of the millennium has brought to light a unique aspect of protein structure that challenges this traditional structure–function paradigm once thought to be a universal theme applicable to all proteins. These findings demonstrate that under native functional conditions, many proteins or specific regions of some proteins are intrinsically disordered, existing as molten globules, collapsed or extended random coils, transiently structured forms, etc. These proteins are called intrinsically disordered proteins (IDPs). IDPs lack a unique three dimensional structure, either entirely or in part, when alone in solution. About 10–35% of prokaryotic and about 15–45% of eukaryotic proteins are estimated to contain disordered regions that are at least 30 amino acid residues in length. A significant number of IDPs are involved in regulatory and signaling functions; hence, IDPs are more prevalent in eukaryotes than in prokaryotes. IDPs and IDP databases are discussed in section 8.11 (Chapter 8).
1.9.6Acidic and Basic Proteins
At physiological pH (7.4), acidic proteins tend to be negatively charged and have a higher proportion of acidic amino acids (e.g. aspartic acid, glutamic acid), whereas basic proteins tend to be positively charged and have a higher proportion of basic amino acids (e.g. arginine, lysine).
Hydrophilic and charged amino acids are frequently associated with antigenic determinants (epitopes), such as arginine, lysine, aspartic acid, glutamic acid, asparagine, glutamine, serine, and threonine.
1.9.7Nonstandard Amino Acids in Polypeptide Chains
As indicated earlier, selenocysteine and pyrrolysine are the two nonstandard amino acids that are incorporated directly into the polypeptide chain during translation. Selenocysteine has been found in lower as well as higher organisms (including mammals), while pyrrolysine has so far been found in certain archaebacteria. However, their occurrence in proteins is not nearly as universal as the 20 standard amino acids.
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Chapter
Scientific Fundamentals of Biotechnology
2011, Comprehensive Biotechnology (Second Edition)P. Newsholme, ... M. Krause
1.02.2General Properties, Classification, and Structure of Amino Acids
All of the amino acids used for protein synthesis have the same general structure (Figure 1). They contain a carboxylic acid group (–COO−) and an amino group () attached to the alpha carbon (the carbon atom next to the carboxylate group) in an l-configuration, a hydrogen atom, and a chemical group called a side chain (–R) that is characteristic for each different amino acid. In solution, the free amino acids can exist as zwitterions (a chemical compound that carries a total net charge of zero and is thus electrically neutral, but carries formal charges on different atoms), in which the amino group is positively charged and the carboxylate group is negatively charged. In proteins, these amino acids are joined into linear polymers called polypeptide chains through chemical bonding between the carboxylic acid group of one amino acid and the amino group of the next amino acid (Figure 2). As mentioned above, the chemical properties of the side chain determine the types of chemical interactions and molecular functions. Thus, amino acids are often grouped by the polarity of the side chain (e.g., charged, nonpolar hydrophobic, or uncharged polar) or by structural features (e.g., aliphatic, cyclic, or aromatic). The side chains of the nonpolar hydrophobic amino acids cluster together to exclude water in the hydrophobic effect characteristic of the core of globular proteins. The uncharged polar amino acids participate in hydrogen bonding. Amino acids that contain a sulfhydryl group, such as cysteine, form disulfide bonds with other components. The negatively charged acidic amino acids form ionic (electrostatic) bonds with positively charged molecules, such as basic amino acids. The charge on the amino acid at a particular pH is determined by the pKa of each chemical group that is associated with a dissociable proton. A list of the 20 amino acids, their classification, isoeletric point (pI – the pH at which a particular molecule or surface carries no net electrical charge), and pKa is detailed in Table 1.
Table 1. Structure, name, abbreviation, pKa and isoelectric point of the common amino acids
| Side chain (-R) structure | Name | Abbreviation | pKa/α-COOH | pKa/α-NH2 | pI |
--- --- --- |
| Neutral amino acids: nonpolar, aliphatic |
| Glycine | Gly or G | 2.3 | 9.6 | 6.0 |
| Alanine | Ala or A | 2.3 | 9.7 | 6.0 |
| Valine | Val or V | 2.3 | 9.6 | 6.0 |
| Leucline | Leu or L | 2.4 | 9.6 | 6.0 |
| Isoeucine | Ile or I | 2.4 | 9.7 | 6.1 |
| Neutral amino acids: polar, aliphatic |
| Serine | Ser or S | 2.2 | 9.2 | 5.7 |
| Threonine | Thr or T | 2.6 | 10.4 | 6.5 |
| Aspargine | Asn or N | 2.o | 8.8 | 5.4 |
| Glutamine | Gln or Q | 2.2 | 9.1 | 5.5 |
| Neutral amino acids: aromatic |
| Phenylalanine | Phe or Fq | 1.8 | 9.1 | 5.5 |
| Tryptophan | Trp or W | 2.4 | 9.4 | 5.9 |
| Tyrosine | Tyr or Y | 2.2 | 9.1 | 5.7 |
| Neutral amino acids: sulfur-containing |
| Cysteine | Cys or C | 1.7 | 10.8 | 5.0 |
| Cystine | Cys-Cys | 2.3 | 9.7 | 5.1 |
| Methionine | Met or M | 2.3 | 9.2 | 5.8 |
| Neutral amino acids: containing secondary amino group |
| Proline | Pro or P | 10.6 | 6.3 | 6.3 |
| Acidic amino acids |
| Aspartate | Asp or D | 2.1 | 9.8 | 3.0 |
| Glutamate | Glu or E | 2.2 | 9.7 | 3.2 |
| Basic amino acids |
| Lysine | Lys or K | 2.2 | 9.0 | 9.8 |
| Arginine | Arg or R | 2.2 | 9.0 | 10.8 |
| Histidine | His or H | 1.8 | 9.2 | 7.6 |
Amino acids fulfill vital roles throughout biology. Not only do they function as building blocks for protein synthesis but they can also act as precursors to neurotransmitters, signaling molecules, and antioxidants. Amino acids can be combined in a large number of possible arrangements to form a variety of different proteins that can function as monomers or in combination with other peptide chains to form functional multisubunit structures .
The specific linear arrangement of amino acids is vital to the function and conformation of a protein. A slight change in the amino acid sequence can be detrimental. A single amino acid substitution in hemoglobin, the protein that transports oxygen via red blood cells, leads to sickle cell disease . Red blood cells are disk shaped in healthy individuals. However, abnormal hemoglobin molecules crystallize causing the red blood cells to appear sickle shaped in affected individuals. People with the disease experience sickle cell crises in which the sickle-shaped cells block small blood vessels and obstruct blood flow .
The R group distinguishes amino acids from one another and dictates the unique properties of each amino acid . The R group is important as it allows for posttranslational modifications that can modulate a protein’s function. Examples of such posttranslational modification include phosphorylation, nitrosylation, and acetylation.
Proline is the only cyclic amino acid. It is more conformationally restricted than other amino acids due to its ring structure and has a major influence on protein arrangement . The aromatic amino acids have distinct absorption spectra in the near-ultraviolet range due to the presence of the aromatic ring. This allows proteins to be characterized and quantified using a simple laboratory spectrophotometer . Amino acids that have side chains carrying a negative charge are classified as acidic as a result of the carboxyl group in the side chain. Amino acids containing amino groups in their side chains are designated basic .
All amino acids except glycine share an important characteristic – they show chirality or asymmetry . This chirality results from the asymmetric alpha carbon connecting to four different groups. Some amino acids such as isoleucine and threonine have additional chiral centers because each possesses an additional asymmetric carbon . Glycine is the only amino acid that lacks a chiral center as the alpha carbon is not asymmetric, binding to two hydrogen atoms.
The word chiral is derived from the Greek word meaning hand. An amino acid that possesses a chiral center is labeled either an l-(levo, left) or a d-(dextro, right) stereoisomer according to the direction it rotates polarized light. The l-stereoisomer of an amino acid is a mirror image of the d-form, just as the right hand is a mirror image of the left hand. The l- and d-amino acids of a stereoisomeric pair are known as enantiomers . Stereoisomers share the same molecular formula and bond sequence but differ by the arrangement of their atoms in three-dimensional space. Only l-amino acids occur in mammalian proteins. However, d-isomers can be detected in abundance in peptidoglycan bacterial cell walls and also in certain antibiotics .
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Volume 2
2015, Knobil and Neill's Physiology of Reproduction (Fourth Edition)Yoel Sadovsky, Thomas Jansson
Amino Acids
Amino acids are required for fetal protein synthesis, and they constitute critical precursors in many biosynthetic pathways, including the synthesis of porphyrins, nitric oxide, neurotransmitters, and nucleotides. Nonessential amino acids are also used as fetal energy substrates. It is estimated that 32% of the energy requirement of well-nourished fetal sheep is met by amino acid oxidation.255 For most amino acids, the concentrations in the umbilical vein are two- to threefold higher than in the uterine vein, demonstrating that the transfer of amino acids across the placental barrier is an active process (reviewed in Refs 256–258). This is also supported by the observation that placental concentrations of amino acids are, in general, much higher than in the blood on either side of the placental barrier (Figure 39.14).
There are two main types of amino acid transporters—accumulative transporters and amino acid exchangers—both of which are found in the MVM. Accumulative transporters mediate cellular uptake, resulting in increased intracellular amino acid concentrations. Amino acid exchangers, in contrast, exchange one amino acid for another, resulting in altered amino acid composition without changing total concentration.259 The primary driving forces for amino acid uptake across the MVM mediated by accumulative transporters are (1) the inwardly directed Na+ gradient (e.g., the System A transporter and the taurine transporter), and (2) the potential difference between the intracellular and extracellular environments, which is the driving force for uptake of the cationic amino acids (arginine, histidine, and lysine). The l-amino acid transporter system is an example of an exchanger, which uses the steep outwardly directed concentration gradient of some nonessential amino acids to drive the uptake of essential amino acids, such as leucine, against its concentration gradient. The energy for the uphill transport of amino acids is ultimately generated by the Na+K+–ATPase, which extrudes sodium in exchange for potassium, thereby maintaining a low intracellular Na+ concentration and creating a potential difference across the plasma membrane.174 Subsequently, amino acids are transferred across the BM by facilitated diffusion driven by the outwardly directed concentration gradient (Figure 39.14). While accumulative transporters and exchangers are sufficient to account for the uptake of all amino acids in most cells, in transporting epithelia, such as the syncytiotrophoblast, efflux transport mechanisms in the BM are needed to allow net transfer of all amino acids across the epithelium.258
The asymmetric distribution of amino acid transporters between the syncytiotrophoblast MVM and BM is critical in order to generate a net flux of amino acids from mother to fetus (Figure 39.14). The human syncytiotrophoblast expresses at least 20 different amino acid transporters (Table 39.2), with each transporter mediating the uptake of several amino acids, and each amino acid can be transported by multiple transport systems. Specifically, accumulative transporters and amino acid exchangers are present in the MVM, whereas amino acid exchangers and efflux transporters predominate in the BM. Figure 39.14 represents a simplified depiction of placental amino acid transport pathways. Notably, it does not take into account placental amino acid metabolism or complex interactions among transporters. A series of studies of amino acid metabolism in the sheep fetus revealed that there is an intricate cycling of amino acids between the placenta and the fetal liver (reviewed in Refs 260,261). Specifically, glutamate is taken up from the umbilical, rather than the uteroplacental circulation, and is oxidized or converted to glutamine. Glutamine is released from the placenta into the fetal circulation, taken up by the liver, and, in part, converted to glutamate. A similar placental–liver cycle exists for serine and glycine. The physiological function of these interorgan exchanges of some amino acids and their relevance to the human placenta remain to be established.258 A mathematical model, recently applied to amino acid transport in the human placenta and validated using experimental data, indicated that the flux of serine and alanine is particularly sensitive to changes in transporter abundance in the BM.262
Neutral Amino Acids
Classical studies have identified System A as a major sodium-dependent transporter that mediates the uptake of small zwitterionic nonessential neutral amino acids into the cell. System A is pH sensitive, and it displays extensive hormonal and adaptive regulation. System A activity establishes the high intracellular concentration of nonessential amino acids, which are used to exchange for extracellular essential amino acids via System L. Thus, System A activity is critical for cellular uptake of both nonessential and essential amino acids. There are three isoforms of System A: sodium-dependent neutral amino acid transporter 1 (SNAT1), SNAT2, and SNAT4 (Table 39.2). Of these isoforms, SNAT1 and SNAT2 share extensive similarities in substrate profiles and transport mechanism, although SNAT2 is more widely expressed than SNAT1. SNAT4 also transports cationic amino acids, independent of Na+, and is predominantly expressed in the liver. However, all SNAT isoforms, including SNAT4, are expressed in the human placenta.184 System A activity is present in both the MVM and BM; however, its activity in the BM is markedly lower.185,186 These observations are consistent with studies using immunohistochemistry, localizing SNAT isoforms to the MVM.183 Although SNAT1 was reported to exhibit the major System A activity in cultured primary human trophoblasts,263 these findings are not entirely consistent with recent studies showing that silencing of mammalian target of rapamycin complex 1 and 2 (mTORC1 and mTORC2, respectively) in primary human trophoblasts completely inhibits System A activity and markedly decreases SNAT2 expression in MVM without affecting SNAT1 or SNAT4 expression,264 implicating SNAT2 as the predominant contributor to System A activity. Additional work is needed to determine the relative contribution of each SNAT isoform to System A activity in vivo and in vitro.
System ASC transports alanine, serine, and cysteine in a Na+-dependent manner. While ASC activity has been demonstrated in the BM,186 ASC function in the MVM remains unclear.185 The expression of ASCT2, which has been implicated as the primary transporter responsible for ASC activity in the human, is very low in the placenta, suggesting that an as-yet-unidentified ASC-like transporter may be responsible for the ASC activity in the placenta. System B0 is another Na+-dependent amino acid transporter that is highly expressed at the mRNA level in the human placenta. Although B0 activity is clearly present in choriocarcinoma cell lines, it remains to be established if this transporter is active also in syncytiotrophoblast plasma membranes.192 The data also remain controversial with respect to System N activity.187,188 Na+-dependent glycine uptake mediated by System Gly has been reported in the MVM.189
The β-amino acid taurine (2-aminoethanesulfonic acid) is the most abundant free amino acid in many tissues. Taurine is not incorporated into proteins and plays an important role in bile acid conjugation, defense against oxygen free radicals, regulation of neuronal excitability, and cell volume regulation. Taurine deficiency in fetal life is associated with growth failure, abnormal cellular development, retinal degeneration, cardiac damage, and central nervous system (CNS) dysfunction. The capacity to synthesize taurine is low in the human fetus, and transplacental transfer of taurine is thus the primary source of this important amino acid for the fetus. System β mediates cellular uptake of taurine, Na+, and Cl−, and this transporter is highly active in the MVM. In contrast, System β activity in the BM is only 6% of the MVM’s activity.190,191
The System L amino acid transporter is a sodium-independent exchanger that mediates cellular uptake of essential amino acids, including leucine, methionine, and tryptophan. In fact, because no accumulative transporters that transport essential amino acids are expressed in the MVM, System L exchange of nonessential amino acids in the syncytiotrophoblast cytosol with extracellular essential amino acids is critical for the uptake of essential amino acids across the MVM. This transporter is a heterodimer, consisting of a catalytic light chain, typically LAT1 (large neutral amino acid transporter 1, also known as SLC7A5) or LAT2 (also known as SLC7A8), and a heavy chain, 4F2hc/CD98 (SLC3A2). The heavy chain is believed to be important for trafficking of the light chain to the plasma membrane, where the two subunits form disulfide-bound heterodimers. System L activity has been reported in both the MVM and BM.185,186,192 Kudo and Boyd reported that MVM System L activity is due to expression of LAT1,265 whereas other investigators have suggested that LAT2 is the predominant isoform in the MVM.266 Lastly, the efflux transporter LAT4 is expressed in the placenta, and is functional in isolated perfused human placental cotyledons, suggesting that LAT4 may also play a role in the efflux of certain essential amino acids across the BM.193
Cationic Amino Acids
Cellular uptake of the cationic (or basic) amino acids lysine, arginine, and histidine is driven by the electric potential difference (negative intracellularly) across the plasma membrane. System y+ is the main transporter for cationic amino acids in the MVM,194,195 whereas y+L is likely to be the primary transporter in the BM.195,196 However, System y+L activity also has been identified in the MVM,194,195 and System y+ activity has been detected in the BM by some but not all investigators.194,195 Whereas System y+ only accepts cationic amino acids, y+L transports neutral amino acids in the presence of Na+. Whether or not System b0,+ activity is present in syncytiotrophoblast membranes remains controversial.195,196
Anionic Amino Acids
The anionic (acidic) amino acids glutamate and aspartate are not transferred in the maternal–fetal direction in the in vitro perfused placenta. Furthermore, glutamate is rapidly oxidized by cultured primary human trophoblasts, and there appears to be no net flux of glutamate from the placenta into the umbilical circulation in pregnant women.267 Collectively, these findings suggest that glutamate is not maternally derived, but originates from the fetal liver. Glutamate may be taken up from both the fetal and maternal circulation into the syncytiotrophoblast, where glutamate is oxidized or converted to glutamine. Consistent with this model, activity for System X−AG transporters, which mediate cellular uptake of anionic amino acids driven by the Na+ gradient and associated with co-transport of H+, have been shown to be present in both the MVM and BM.171,197
Regulation of Amino Acid Transporters
The System A transporter has been the primary focus of research into mechanisms regulating placental amino acid transport. System A transporter activity in human trophoblasts is stimulated by insulin, IGF-I, and EGF, and by substrate concentrations.250 Leptin increases System A activity in primary villous fragments at term. The proinflammatory cytokines interleukin 6 (IL6) and TNFα also stimulate trophoblast System A activity, and the effect of IL6 was shown to be mediated by signal transducer and activator of transcription 3 (STAT3). Adiponectin decreased insulin-stimulated System A amino acid uptake in cultured human trophoblasts by modulating insulin receptor substrate phosphorylation. Furthermore, chronic administration of adiponectin to pregnant mice inhibits placental insulin and mTORC1 signaling, downregulates the activity and expression of System A and L isoforms, and decreases fetal growth.268 Dexamethasone, a synthetic glucocorticoid, stimulates System A activity in cultured primary human trophoblasts and in term villous explants. Notably, dexamethasone administered to pregnant mice downregulates placental System A amino acid transport,269 suggesting that the effects of glucocorticoids on System A activity in vitro and in vivo are distinct. Furthermore, in cultured primary human trophoblasts, hypoxia decreases System A activity, which could be explained by a decreased protein expression of the two System A transporter isoforms SNAT1 and SNAT2.270 Lastly, oleic acid activates System A activity in cultured primary human trophoblasts, an effect mediated by Toll-like receptor 4.271
The mTOR signaling pathway responds to changes in nutrient availability and growth factor signaling to control cell growth and metabolism. Rapamycin inhibits mTOR and markedly decreases System A activity in primary human trophoblasts without affecting global protein expression of SNAT isoforms. These findings are consistent with the possibility that mTOR regulates amino acid transporter activity at the posttranslational level. Indeed, inhibition of mTOR using gene-silencing approaches decreases the activity of key placental amino acid transporters in primary human trophoblasts, an effect mediated by modulating the trafficking of specific transporter isoforms between the cell interior and the plasma membrane.264
The regulation of placental amino acid transporters other than System A has been insufficiently explored. Studies of the effect of insulin on System L activity in primary human trophoblasts have produced inconsistent results. However, mTOR is a powerful positive regulator of trophoblast System L activity, mediated by influencing LAT1 trafficking to the plasma membrane.264 The nitric oxide donor SIN1 inhibits taurine transport in MVM vesicles and in villous explants. mTOR increases mRNA levels of the taurine transporter, and stimulates taurine transport in cultured trophoblasts.
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Book2015, Knobil and Neill's Physiology of Reproduction (Fourth Edition)Yoel Sadovsky, Thomas Jansson
Related terms:
Basic Amino Acids
Eicosanoid Receptor
Arginine
Peptidase
C-Terminus
Amino Acid
Aspartic Acid
Glutamic Acid
Glycoprotein
Amino Terminal Sequence
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Example
Midpoint rule (rectangle method)
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Simpson’s rule
Composite Simpson’s rule
Weddle’s rule
Implementation
Other rules
Numerical integration
Contents
Example
Midpoint rule (rectangle method)
Implementation
Trapezoid rule
Error analysis
Concave-down functions
Simpson’s rule
Composite Simpson’s rule
Weddle’s rule
Implementation
Other rules
Numerical integration#
Numerical Methods
Numerical integration involves finding the integral of a function. While SymPy can be used to do analytical integration, there are many functions for which finding an analytical solution to integration is very difficult, and numerical integration is used instead.
To understand how to perform numerical integration, we first need to understand what exactly is the purpose of integration. For a 1D function, integration means finding the area underneath the curve. However, integration can also be extended for a 2D function and even a 3D function.
Quadrature is the term used for numerical evaluation of a definite (i.e. over a range [a,b]) integral, or in 1D finding the area under a curve.
Source: Wikipedia
Wondered why the symbol of integration looks like this ∫?
The symbol of integration actually comes from the cursive of capital letter S, meaning summation. But why would it come from the idea of summation? Well, integration is very much related to summation.
We know that for a 1D function, which will be our primary topic of discussion today, if we evaluate an integral, we find the area under the function, as illustrated in the figure above. If your curve was a straight line (remember, straight lines are but a subset of curves which has no curves), then it would be rather straightforward to calculate the area under the curve, since you would either have a rectangle if your line was a horizontal straight line, or a trapezoid if your line was not a horizontal straight line. Yet, what happens if your line got a bunch of bend and curves?
For example, we could have a very simple equation like
f(x)=sin(x)+5.
This equation is a sinusoidal function, of much more complicated shape that is not as easy to get the area through a rectangle or a trapezoid. Therefore, integration comes in to help you find the area. But how does the integration actually get you the area underneath the curve?
Well, what integration essentially does is basically breaking the area into smaller and smaller parts, evaluating the area of each part, and then summing each small part together. The small part of the area can be approximated to be a rectangle, a trapezoid, or some other weird shape if you find it suitable.
A very simple example, using rectangles are shown below:
Source: Wikipedia
Note
It should also be noted that although the example shown uses rectangular slices, it is not neccesary to use rectangular slices. It is not even neccessary to use slices that are of the the same width. Even if you used slices with different width, and which are not rectangles, like trapezoids for example, you could still make the slices increasingly thinner so that the summation of the area under the slices comes closer and closer to the area underneath the curve, until the slices become infinitely thin, and the summed area of slices becomes essentially equal to the area under the curve.
As each rectangle slice becomes thinner and thinner, the summed area from the rectangles become more and more closely fitting to the area under the curve. It should be understood that if the rectangle slice becomes infinitely thin, then the summed area from the rectangle would become so close to the area underneath the curve that the two would be essentially the same. You may find the mathematics for the Riemann integral on Wikipedia. Of course, integration has come a long way since the Riemann integral, and other integrals were developed to deal with the deficiencies with the Riemann integral.
The choice of approximation method, as well as the size of the intervals, will control the error. Better methods as well as smaller (i.e. more to cover our total interval of interest: [a,b]) sub-intervals will lead to lower errors, but will generally cost more to compute.
Here the following quadrature methods will be covered in the context of a simple function:
Midpoint rule (also known as the rectangle method)
Trapezoid rule
Simpson’s rule
Composite Simpson’s rule
Weddle’s rule.
Example#
Let’s begin with a simple function to demonstrate some of the most basic methods for performing numerical integration:
f(x):=sin(x),
and assume that we want to know what the area under the, sin function between 0 and π, i.e. [a,b]=[0,π].
The indefinite integral (or anti-derivative) of sin(x) is of course −cos(x) (plus a constant of integration, C, which we can simply ignore as we saw above as it drops out as soon as we perform a definite integral).
Since we know the indefinite integral exactly in this case, we can perform the definite integration (i.e. find the area under the curve) ourselves exactly by hand:
I:=∫0 π sin(x)=[−cos(x)+C]0 π=−cos(π)−(−cos(0))=−cos(π)+cos(0)=−(−1)+1=2.
We included the constant C here to just to emphasise again the fact that it’s present doesn’t matter - we can just not write it down in this type of expression.
Let’s start by plotting the function between these points.
import numpy as np
import matplotlib.pyplot as plt
Set up the figure
fig = plt.figure(figsize=(10, 4))
ax1 = plt.subplot(111)
Get the value of pi from numpy and generate 100 equally spaced values from 0 to pi.
x = np.linspace(0, np.pi, 100)
Calculate sin at these points.
y = np.sin(x)
plot
ax1.plot(x, y, 'b')
Set x axis limits between 0 and pi.
ax1.set_xlim([0, np.pi])
ax1.set_ylim([0, 1.1])
Label axis.
ax1.set_xlabel('$x$', fontsize=14)
ax1.set_ylabel('$f(x)=\sin(x)$', fontsize=14)
ax1.set_title('An example function we wish to integrate', fontsize=14)
Overlay a grid.
ax1.grid(True)
plt.show()
Midpoint rule (rectangle method)#
The midpoint rule is perhaps the simplest quadrature rule. For reasons you will see below it is sometimes also called the rectangle method.
Consider one of the subintervals [x i,x i+1]. The midpoint rule approximates the integral over this (the i-th) subinterval by the area of a rectangle, with a base of length (x i+1−x i) and a height given by the value of f(x) at the midpoint of that interval (i.e. at x=(x i+1+x i)/2):
I M(i):=(x i+1−x i)×f(x i+1+x i 2),for 0≤i≤n−1.
The midpoint estimate of I then simply involves summing up over all the subintervals:
I M:=∑i=0 n−1 f(x i+1+x i 2)(x i+1−x i).
Midpoint rule:
Divide the interval you want to calculate the area under the curve for into smaller pieces, each will be called a subinterval.
Assume that the interval begins at x 0 and ends at x n. We can pick a random subinterval [x i,x i+1] where 0≤i≤n−1. For example, your 1st interval will be [x 0,x 1] and your i=0, and your last interval will be [x n−1,x n] and your i=n−1.
For every subinterval, we approxiamte the slice to be a rectangle. To find the area of the rectangle we need to find the width and the height. The width of the rectangle is simply the width of the subinterval. The height of the rectangle can be estimated as the value of the function at the midpoint of the subinterval, so f(x i+1+x i 2). To find the area we simply multiply the width of the rectangle by the height of the rectangle.
Width of the rectangle:
x i+1−x i.
Height of the rectangle:
f(x i+1+x i 2).
Area of the rectangle:
(x i+1−x i)f(x i+1+x i 2).
Generalizing the above for all slices, where I M(i) is simply the area of the subinterval [x i,x i+1]. M subscript here denotes the use of the midpoint method.
I M(i):=(x i+1−x i)×f(x i+1+x i 2),for 0≤i≤n−1.
4. To find the area under the curve, we need to sum up all of the areas from the subinterval, so we are going to use the summation symbol. We know that the subinterval index goes from the first subinterval where i=0 to the last subinterval where i=n−1, thus we arrive at
I M:=∑i=0 n−1 f(x i+1+x i 2)(x i+1−x i).
Note that we dropped i prefix from I because now it is the whole area under the curve, not just the area from one subinterval.
Let’s write some code to plot the idea as well as compute an estimate of the integral using the midpoint rule.
this is a matplotlib function that allows us to easily plot rectangles
which will be useful for visualising what the midpoint rule does
from matplotlib.patches import Rectangle
def f(x):
"""The function we wish to integrate"""
return np.sin(x)
Get the value of pi from numpy and generate equally spaced values from 0 to pi.
x = np.linspace(0, np.pi, 100)
y = f(x)
Plot
fig = plt.figure(figsize=(10, 4))
ax1 = plt.subplot(111)
ax1.plot(x, y, 'b', lw=2)
ax1.margins(0.1)
Label axis.
ax1.set_xlabel('x', fontsize=14)
ax1.set_ylabel('$f(x)=\sin(x)$', fontsize=14)
Overlay a grid.
ax1.grid(True)
number_intervals = 5
xi = np.linspace(0, np.pi, number_intervals+1)
I_M = 0.0
for i in range(number_intervals):
ax1.add_patch(Rectangle((xi[i], 0.0), (xi[i+1] - xi[i]),
f((xi[i+1]+xi[i])/2), fill=False, ls='--', color='k', lw=2))
I_M += f((xi[i+1]+xi[i])/2)(xi[i+1] - xi[i])
ax1.set_title('The sum of the areas of the rectangles is $I_M =$ {:.12f}.'.format(I_M),
fontsize=14)
plt.show()
A more complex example is shown below, where the red line shows the original function we wish to compute the integral of, and the blue rectangles approximate the area under that function for a number of sub-intervals:
Implementation#
Note
Note that the SciPy module features many different integration functions, and you can find thorough documentation for these functions (including methods not covered in this course) here. This library does not contain a function for the midpoint rule, but it is trivial to create our own.
Clearly the sum of the areas of all the rectangles provides an estimate of the true integral. In the case above we observe an error of around 1.5%.
As we are going to compare different rules below, let’s implement a midpoint rule function.
def midpoint_rule(a, b, function, number_intervals=10):
""" Our implementation of the midpoint quadrature rule.
a and b are the end points for our interval of interest.
'function' is the function of x \in [a,b] which we can evaluate as needed.
number_intervals is the number of subintervals/bins we split [a,b] into.
Returns the integral of function(x) over [a,b].
"""
interval_size = (b - a)/number_intervals
# Some examples of some asserts which might be useful here -
# you should get into the habit of using these sorts of checks as much as is possible/sensible.
assert interval_size > 0
assert type(number_intervals) == int
# Initialise to zero the variable that will contain the cumulative sum of all the areas
I_M = 0.0
# Find the first midpoint -- i.e. the centre point of the base of the first rectangle
mid = a + (interval_size/2.0)
# and loop until we get past b, creating and summing the area of each rectangle
while (mid < b):
# Find the area of the current rectangle and add it to the running total
# this involves an evaluation of the function at the subinterval midpoint
I_M += interval_size function(mid)
# Move the midpoint up to the next centre of the interval
mid += interval_size
# Return our running total result
return I_M
Check the function runs if it agrees with our first
version used to generate the schematic plot of the method above:
print('midpoint_rule(0, np.pi, np.sin, number_intervals=5) = ',
midpoint_rule(0, np.pi, np.sin, number_intervals=5))
midpoint_rule(0, np.pi, np.sin, number_intervals=5) = 2.033281476926104
Now let’s test the midpoint function:
print("The exact area found by direct integration = 2")
for i in (1, 2, 10, 100, 1000):
area = midpoint_rule(0, np.pi, np.sin, i)
print("Area %g rectangle(s) = %g (error=%g)"%(i, area, abs(area-2)))
The exact area found by direct integration = 2
Area 1 rectangle(s) = 3.14159 (error=1.14159)
Area 2 rectangle(s) = 2.22144 (error=0.221441)
Area 10 rectangle(s) = 2.00825 (error=0.00824841)
Area 100 rectangle(s) = 2.00008 (error=8.22491e-05)
Area 1000 rectangle(s) = 2 (error=8.22467e-07)
Exercise
Create a log-log plot of error egainst the number of subintervals:
Create a list of interval sizes to test
interval_sizes_M = [1, 2, 4, 8, 16, 32, 100, 1000]
Initialise an array to store the errors
errors_M = np.zeros_like(interval_sizes_M, dtype='float64')
Loop over the list of interval sizes, compute and store errors
for (i, number_intervals) in enumerate(interval_sizes_M):
area = midpoint_rule(0, np.pi, f, number_intervals)
errors_M[i] = abs(area-2)
Plot how the errors vary with interval size
fig = plt.figure(figsize=(5, 5))
ax1 = plt.subplot(111)
ax1.loglog(interval_sizes_M, errors_M, 'bo-', lw=2)
ax1.set_xlabel('log(no. of intervals)', fontsize=16)
ax1.set_ylabel('log(error)', fontsize=16)
ax1.set_title('Convergence plot for $\sin$ integration\nwith the midpoint rule',
fontsize=16)
from myst_nb import glue
glue("midpoint_conv_fig", fig, display=False)
plt.show()
Observations:
With one rectangle, we are simply finding the area of a box of shape π×1, where π is the width of the rectangle and 1 is the value of the function evaluated at the midpoint, π/2. So of course the result is π.
As we increase the number of subintervals, or rectangles, we increase the accuracy of our area. We can observe from the slope of the log-log plot of error against number of subintervals that the error is a quadratic function of the inverse of the number of subintervals (or equivalently is quadratically dependent on the spacing between the points - the interval size). This demonstrates that (for this particular example at least), the method demonstrates second-order accuracy - if we halve the interval size the error goes down by a factor of 4!
The simplicity of this method is its weakness, as rectangles (i.e. a flat top) are rarely a good approximation for the shape of a smooth function.
We want to use as few shapes as possible to approximate our function, because each additional rectangle is one extra time round the loop, which includes its own operations as well as an extra evaluation of the function, and hence increases the overall computational cost.
Trapezoid rule#
As previously stated, the slices we use do not have to be rectangles, they can also be trapezoids. Rectangle rule is very similar to the trapezoid rule except for one small difference.
For the trapezoid rule, the width of the subinterval will be multiplied by
f(x i)+f(x i+1)2.
For the trapezoid rule, the subscipt we will use the subscript T. If we change the shape of the rectangle to a trapezoid (i.e. the top of the shape now being a linear line fit defined by the values of the function at the two end points of the subinterval, rather than the constant value used in the midpoint rule), we arrive at the trapezoid, or trapezoidal rule.
The trapezoid rule approximates the integral by the area of a trapezoid with base (x i+1−x i) and the left- and right-hand-sides equal to the values of the function at the two end points.
In this case the area of the shape approximating the integral over one subinterval, is given by:
I T(i):=(x i+1−x i)×(f(x i+1)+f(x i)2)for 0≤i≤n−1.
The trapezoidal estimate of I then simply involves summing up over all the subintervals:
I T:=∑i=0 n−1(f(x i+1)+f(x i)2)(x i+1−x i).
Let’s write some code to plot the idea and compute an estimate of the integral.
This is a matplotlib function that allows us to plot polygons
from matplotlib.patches import Polygon
Get the value of pi from numpy and
generate equally spaced values from 0 to pi.
x = np.linspace(0, np.pi, 100)
y = f(x)
plot
fig = plt.figure(figsize=(10, 4))
ax1 = plt.subplot(111)
ax1.plot(x, y, 'b', lw=2)
ax1.margins(0.1)
Label axis.
ax1.set_xlabel('$x$', fontsize=14)
ax1.set_ylabel('$\sin(x)$', fontsize=14)
ax1.set_title('Approximating function with trapezoids', fontsize=14)
Overlay a grid.
ax1.grid(True)
number_intervals = 5
xi = np.linspace(0, np.pi, number_intervals+1)
I_T = 0.0
for i in range(number_intervals):
ax1.add_patch(Polygon(np.array([[xi[i], 0], [xi[i], f(xi[i])], [
xi[i+1], f(xi[i+1])], [xi[i+1], 0]]), closed=True, fill=False, ls='--', color='k', lw=2))
I_T += ((f(xi[i+1]) + f(xi[i]))/2)(xi[i+1] - xi[i])
ax1.set_title('The sum of the areas of the trapezoids is $I_T =$ {:.12f}.'.format(I_T),
fontsize=14)
plt.show()
For our pictorial example used above, the approximation looks like it should be more accurate than the midpoint rule:
The tops of the shapes (now trapezoids) are approximating the variation of the function with a linear function, rather than a flat (constant) function. This looks like it should give more accurate results, but see below.
Note that numpy has a function for the trapezoid rule, numpy.trapz, but we’ll make our own that works in a similar way to our midpoint rule function.
def trapezoidal_rule(a, b, function, number_intervals=10):
"""Our implementation of the trapezoidal quadrature rule.
Note that as discussed in the lecture this version of the implementation
performs redundant function evaluations - see the composite implementation
in the homework for a more efficient version.
"""
interval_size = (b - a)/number_intervals
assert interval_size > 0
assert type(number_intervals) == int
I_T = 0.0
# Loop to create each trapezoid
# note this function takes a slightly different approach to Midpoint
# (a for loop rather than a while loop) to achieve the same thing
for i in range(number_intervals):
# Set the start of this interval
this_bin_start = a + (interval_size i)
# Find the area of the current trapezoid and add it to the running total
I_T += interval_size \
(function(this_bin_start)+function(this_bin_start+interval_size))/2.0
# Return our running total result
return I_T
We can test the function in a similar way:
print("The exact area found by direct integration = 2")
for i in (1, 2, 10, 100, 1000):
area = trapezoidal_rule(0, np.pi, np.sin, i)
print("Area %g trapezoid(s) = %g (error=%g)"%(i, area, abs(area-2)))
The exact area found by direct integration = 2
Area 1 trapezoid(s) = 1.92367e-16 (error=2)
Area 2 trapezoid(s) = 1.5708 (error=0.429204)
Area 10 trapezoid(s) = 1.98352 (error=0.0164765)
Area 100 trapezoid(s) = 1.99984 (error=0.000164496)
Area 1000 trapezoid(s) = 2 (error=1.64493e-06)
Error analysis#
It is important to understand the errors of the numerical integration method we are using. If there are a limited number of slices, the area covered by the slices from trapezoid rule or the midpoint rule will not be slightly different from the actual area under the curve. Of course, if we increase the number of slices, the error becomes smaller; however, we also increase the computational power required.
A good numerical integration method should be able to have few slices, meaning using little computational power, but still be able to have a small error only. Thus, to know if the numerical integration method we used is a good or a bad method, we need to analyse the errors, or more exactly the change in the errors with numbers of slices, of our numerical integration method.
A good method will have a rapid decrease in the error when increasing the number of slices, while a bad method will have a slow decrease in the error when increasing the number of slices.
The accuracy of a quadrature, i.e. the mid point rule, the trapezoidal rule etc. is predicted by examining its behaviour in relationship with polynominals.
We say that the degree of accuracy or the degree of precision of a quadrature rule is equal to M if it is exact for all polynomials of degree up to and including M, but not exact for some polynomial of degree M+1.
Note
Test your own code on function x 2 to demonstrate that midpoint and trapezoid rule won’t give the exact solution.
Clearly both the midpoint and trapezoid rules will give the exact result for both constant and linear functions, but they are not exact for quadratics. Therefore, they have a degree of precision of 1 (remember that through the error analysis, we found that it is 2nd order accurate, which is a different term to degree of precision!!).
Concave-down functions#
The first half of a sine wave is concave-down and we notice from the plot that trapezoidal rule consistently underestimate the area under the curve as line segments are always under the curve.
In contrast, the mid-point rule will have parts of each rectangle above and below the curve, hence to a certain extent the errors will cancel each other out.
This is why, for this particular example, the errors in the mid-point rule turn out to be approximately half those in the trapezoidal rule.
While this result turns out to be generally true for smooth functions, we can always come up with (counter) examples where the trapezoid rule will win.
Taylor series analysis can be used to formally construct upper bounds on the quadrature error for both methods.
We know that the error when integrating constant and linear functions is zero for our two rules, so let’s first consider an example of integrating a quadratic polynomial.
We know analytically that
∫0 1 x 2 d x=1 3 x 3|0 1=1 3.
Numerically, the midpoint rule on a single interval gives
I M=1(1 2)2=1 4,
while the trapezoidal rule gives
I T=1 0+1 2 2=1 2.
The error for I M is therefore 1/3−1/4=1/12, while the error for I T is 1/3−1/2=−1/6.
Therefore, the midpoint rule is twice as accurate as the trapezoid rule:
|E M|=1 2|E T|,
where |E| indicates the error (the absolute value of the difference from the exact solution).
This is the case for this simple example, and we can see from the actual error values printed above that it also appears to be approximately true for the sine case (which is not a simple polynomial) as well.
Simpson’s rule#
For our half sine wave, the rectangle method overestimates it by about 0.4%, while the trapezoid method underestimates it for 0.9%. We notice that in this situation, the rectangle method overestimates while the trapezoid method underestimates about twice the overestimate of the rectangle method. Could we combine this to obtain something more accurate?
Knowing the error estimates from the two rules explored so far opens up the potential for us to combine them in an appropriate manner to create a new quadrature rule, generally more accurate than either one separately.
Suppose I S indicates an unknown, but more accurate, estimate of the integral over an interval. Then, as seen above, as I T has an error that is approximately −2 times the error in I M, the following relation must hold approximately:
I S−I T≈−2(I S−I M).
This follows from the fact that I−I T≈−2(I−I M), provided that I S is closer to I than either of the other two estimates. Replacing this approximately equals sign with actual equality defines I S for us in terms of things we know.
We can rearrange this to give an expression for I S that yields a more accurate estimate of the integral than either I M or I T:
I S:=2 3 I M+1 3 I T.
We combined twice the overestimate from the rectangle method and once the underestimate from the trapezoid method, and then divided everything by 3 to obtain something more accurate! What we’re doing here is using the fact that we know something about (the leading order behaviour of the) two errors, and we can therefore combine them to cancel this error to a certain extent.
This estimate will generally be more accurate than either M or T alone. The error won’t be zero in general as we’re only cancelling out the leading order term in the error, but a consequence is that we will be left with higher-degree terms in the error expansion of the new quadrature rule which should be smaller (at least in the asymptotic limit), and converge faster.
The resulting quadrature method in this case is known as Simpson’s rule. Let’s expand the Simpsons rule by substituting in what we know about the rectangle rule and the trapezoid rule:
I S=2 3 I M+1 3 I T
=2 3(b−a)f(a+b 2)+1 3(b−a)(f(a)+f(b))2
=(b−a)6(f(a)+4 f(c)+f(b))
where a and b are the end points of an interval and c=(a+b)/2 is the midpoint.
Note that an alternate derivation of the same rule involves fitting a quadratic function (i.e. P 2(x) rather than the constant and linear approximations already considered) that interpolates the integral at the two end points of the interval, a and b, as well as at the midpoint, c=(a+b)/2, and calculating the integral under that polynomial approximation.
Let’s plot what this method is doing and compute the integral for our sine case.
Get the value of pi from numpy and generate equally spaced values from 0 to pi.
x = np.linspace(0, np.pi, 100)
y = f(x)
plot
fig = plt.figure(figsize=(10, 4))
ax1 = plt.subplot(111)
ax1.plot(x, y, 'b', lw=2)
ax1.margins(0.1)
Label axis.
ax1.set_xlabel('x', fontsize=16)
ax1.set_ylabel('sin(x)', fontsize=16)
Overlay a grid.
ax1.grid(True)
number_intervals = 5
xi = np.linspace(0, np.pi, number_intervals+1)
I_S = 0.0
for i in range(number_intervals):
# Use a non-closed Polygon to visualise the straight sides of each interval
ax1.add_patch(Polygon(np.array(),
closed=False, fill=False, ls='--', color='k', lw=2))
# Add the quadratic top - fit a quadratic using numpy
poly_coeff = np.polyfit((xi[i], (xi[i] + xi[i+1])/2.0, xi[i + 1]),
(f(xi[i]), f((xi[i] + xi[i+1])/2.0), f(xi[i+1])), 2)
# Plot the quadratic using 20 plotting points within the interval
ax1.plot(np.linspace(xi[i], xi[i+1], 20),
f(np.linspace(xi[i], xi[i+1], 20)), ls='--', color='k', lw=2)
# qdd in the area of the interval shape to our running total using Simpson's formula
I_S += ((xi[i+1] - xi[i])/6.) (f(xi[i]) + 4
f((xi[i] + xi[i+1])/2.0) + f(xi[i+1]))
ax1.set_title("The Simpson's rule approximation is $I_s =$ {:.12f}.".format(I_S),
fontsize=14)
plt.show()
It looks much closer to the actual function:
Let’s make a function to test it out.
def simpsons_rule(a, b, function, number_intervals=10):
""" Function to evaluate Simpson's rule.
Note that this implementation takes the function as an argument,
and evaluates this at the midpoint of subintervals in addition to the
end point. Hence additional information is generated and used through
additional function evaluations.
This is different to the function/implementation available with SciPy
where discrete data only is passed to the function.
Bear this in mind when comparing results - there will be a factor of two
in the definition of "n" we need to be careful about!
Also note that this version of the function performs redundant function
evaluations - see the composite implementation below.
"""
interval_size = (b - a)/number_intervals
assert interval_size > 0
assert type(number_intervals) == int
I_S = 0.0
# Loop to valuate Simpson's formula over each interval
for i in range(number_intervals):
# Find a, c, and b
this_bin_start = a + interval_size (i)
this_bin_mid = this_bin_start + interval_size/2
this_bin_end = this_bin_start + interval_size
# Calculate the rule and add to running total.
I_S += (interval_size/6) (function(this_bin_start) +
4 function(this_bin_mid) + function(this_bin_end))
# Return our running total result
return I_S
Let’s test the function:
print("The area found by direct integration = 2")
for i in (1, 2, 10, 100, 1000):
area = simpsons_rule(0, np.pi, np.sin, i)
print("Area %g Simpson's interval(s) = %g (error=%g)"%(i, area, abs(area-2)))
The area found by direct integration = 2
Area 1 Simpson's interval(s) = 2.0944 (error=0.0943951)
Area 2 Simpson's interval(s) = 2.00456 (error=0.00455975)
Area 10 Simpson's interval(s) = 2.00001 (error=6.78444e-06)
Area 100 Simpson's interval(s) = 2 (error=6.76471e-10)
Area 1000 Simpson's interval(s) = 2 (error=6.79456e-14)
For this simple function you should find far smaller errors, and which drop much more rapidly with smaller h (or more sub-intervals).
Observations:
The errors are lower than for the midpoint and trapezoidal rules, and the method converges more rapidly - i.e. the relative improvement only gets better for more subintervals.
This expression now integrates up to cubics exactly (by construction), so it is of order 4 (if we halve the interval size, the error goes by a factor of 2 4=16).
The convergence can be confirmed in the plot below:
The degree of accuracy or precision of this method is 3.
Simpson’s rule integrates to cubics exactly, so since it’s integrating exactly, cubics cannot contribute to the error. Only the quartic (4th) order terms contribute to the error, so it’s 4th order accurate.
We’re getting down to errors close to the machine precision now when we use 1000 subintervals. Remember: your average consumer grade hardware can only handle that many decimal points, and you will need some rather expensive hardware to have even higher levels of precisions. Continuing with 1000 subintervals is actually not helpful as the error that you will get stops decreasing as it is so small that your computer stops being able to discriminate it from 0. Remember we may well either have a relatively small number of data points, or want to minimise the number of function evaluations well below this relatively high number. This will mean that for problems with lots of variation, and/or in higher dimensions, that we still work to do in improving our quadrature methods.
As was the case with our first trapezoidal implementation, we are performing unnecessary function evaluations here; we can fix this issue through the implementation of a so-called composite version of the rule, which still gives the same result as your Simpson’s rule, but makes it but easier for the computer. The composite Simpson’s rule still does many evaluations, but fewer evaluations than your standard Simpson’s rule.
Composite Simpson’s rule#
If we assume that our interval [a,b] has been split up into n intervals (or n+1 data points) we can save some function evaluations by writing Simpson’s rule in the following form:
I S=Δ x 3[f(x 0)+4 f(x 1)+2 f(x 2)+4 f(x 3)+⋯+2 f(x n−2)+4 f(x n−1)+f(x n)]
=Δ x 3[f(x 0)+2∑i=1 n/2−1 f(x 2 i)+4∑i=1 n/2 f(x 2 i−1)+f(x n)].
This is known as the composite Simpson’s rule, or more precisely the composite Simpson’s 1/3 rule.
You can find a version of Simpson’s rule implemented by SciPy - scipy.interpolate.simps.
Note that this way of formulating Simpson’s rule (where we do not allow additional function evaluations at the midpoints of intervals - we assume we are only in a position to use the given data points) requires that n be even.
This way of writing the composite form in the case of n=2 is equivalent to the formula over [a,b] that introduced the additional midpoint location c.
Let’s implement this rule:
def simpsons_composite_rule(a, b, function, number_intervals=10):
"""Function to evaluate the composite Simpson's rule only using
function evaluations at (number_intervals + 1) points.
This implementation requires that the number of subintervals (number_intervals) be even
"""
assert number_intervals % 2 == 0, "number_intervals is not even"
interval_size = (b - a) / number_intervals
# Start with the two end member values
I_cS2 = function(a) + function(b)
# Add in those terms with a coefficient of 4
for i in range(1, number_intervals, 2):
I_cS2 += 4 function(a + i interval_size)
# And those terms with a coefficient of 2
for i in range(2, number_intervals-1, 2):
I_cS2 += 2 function(a + i interval_size)
return I_cS2 (interval_size / 3.0)
Let’s test the rule:
print("The area found by direct integration = 2")
for i in (2, 10, 100, 1000):
area = simpsons_composite_rule(0, np.pi, np.sin, i)
print("Area %g rectangle(s) = %g (error=%g)"%(i, area, abs(area-2)))
The area found by direct integration = 2
Area 2 rectangle(s) = 2.0944 (error=0.0943951)
Area 10 rectangle(s) = 2.00011 (error=0.000109517)
Area 100 rectangle(s) = 2 (error=1.08245e-08)
Area 1000 rectangle(s) = 2 (error=1.07869e-12)
This is a slight improvement for a simple function like sin, but will be much more of an improvement for functions which oscillate more, in a relative sense compared to the size of our bins.
Weddle’s rule#
We noted above that Simpson’s rule is fourth-order accurate. Suppose we take an approximation to I using n subintervals with Simpson’s rule and call the result I S, and then apply Simpson’s rule with double the number of intervals (2 n) and call the result I S 2.
Then we have two estimates for the integral where we expect I S 2 to be approximately 2 4=16 times more accurate than S. In particular, we expect the lowest (i.e. the leading) order error term in I S 2 to be precisely one sixteenth that of I S.
Similar to how we derived Simpson’s rule by combining what we knew of the error for the midpoint and trapezoidal rules, with this knowledge we can combine the two estimates from Simpson’s rule to derive an even more accurate estimate of I.
Let’s call this more accurate rule I W, which we can find by solving:
I W−I S=16(I W−I S 2),
for I W.
With a bit of manipulation,
I W−I S=16(I W−I S 2)
⟹I W−I S=16 I W−16 I S 2
⟹15 I W=16 I S 2−I S
⟹15 I W=15 I S 2+(I S 2−I S)
we get this expression
I W=I S 2+(I S 2−I S)15.
This is known as Weddle’s rule, or the extrapolated Simpson’s rule because it uses two different values for the interval size and extrapolates from these two to obtain an even more accurate result.
Making a function for this rule is easy as we can just call our Simpson’s rule functions with two values for the number of intervals.
Implementation#
We can implement this by calling already created functions for composite Simpson’s rule:
def weddles_rule(a, b, function, number_intervals=10):
""" Function to evaluate Weddle's quadrature rule using
appropriate calls to the composite_simpson function
"""
S = simpsons_composite_rule(a, b, function, number_intervals)
S2 = simpsons_composite_rule(a, b, function, number_intervals2)
return S2 + (S2 - S)/15.
We can test it in a similar way:
for i in (2, 10, 100, 1000):
area = weddles_rule(0, np.pi, np.sin, i)
print("Area with %g Weddle's interval(s) = %g (error=%g)"%(i, area, abs(area-2)))
Area with 2 Weddle's interval(s) = 1.99857 (error=0.00142927)
Area with 10 Weddle's interval(s) = 2 (error=6.44164e-08)
Area with 100 Weddle's interval(s) = 2 (error=6.23945e-14)
Area with 1000 Weddle's interval(s) = 2 (error=8.88178e-16)
Our final result is much more accurate for fewer required bins:
Other rules#
Note that the above technique of using the same rule, but with different values for the interval size, h, to derive a more accurate estimate of the integral is an example of what is more generally called Richardson extrapolation. Performing this approach using the trapezoid rule as the starting point leads to what is termed Romberg integration.
Taking the idea behind Simpson’s rule which fits a quadratic Lagrange interpolating polynomial to equally spaced points in the interval, end extending to any order Lagrange polynomial leads to the Newton-Cotes family of quadrature rules.
Note finally, that even wider families exist where the function being integrated is evaluated at non-equally-spaced points. And of course for practical application these ideas need to be extended to more than one dimension.
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By Imperial College London
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7145 | https://30c6a7ec78.clvaw-cdnwnd.com/a57710a763bcf3e7657ca5fa4f6dee08/200000858-033d404368/relativita%CC%80%20ristretta.pdf | www.liceoweb.it teoria della relatività ristretta Esercizi Teoria relatività ristretta www.liceoweb.it teoria della relatività ristretta Esercizio Un’astronave diretta verso la Terra passa di fianco a un «faro» spaziale con velocità c/4 rispetto a esso. Nell’istante in cui sono affiancati, il faro lancia verso Terra un lampo di luce verde e l’astronave un lampo di luce blu. Fuori dall’atmosfera terrestre c’è un sensore in grado di rilevare i lampi di luce. Quale dei due lampi è segnalato per primo dal sensore situato vicino alla Terra? Soluzione I due lampi di luce viaggiano alla stessa velocità (secondo postulato) per cui giungono al sensore nello stesso istante. Esercizio Qual è la differenza tra il principio di relatività galileiana e il primo principio della relatività ristretta? Soluzione Il primo postulato della relatività ristretta estende il principio di relatività galileiana, valido per i fenomeni meccanici, a tutti i fenomeni fisici, in particolare a quelli elettromagnetici. Esercizio In che cosa differiscono le opinioni di Maxwell ed Einstein sul valore della velocità della luce nel vuoto? Soluzione La differenza è nell’esistenza o meno dell’etere. Secondo Maxwell, la velocità della luce è pari a c solo nel sistema di riferimento in cui l’etere è in quiete. Secondo Einstein, per il quale l’etere non esiste non solo fisicamente ma anche non necessario concettualmente, il valore di c è lo stesso in tutti i sistemi di riferimento inerziali. www.liceoweb.it teoria della relatività ristretta Esercizio Al centro di un vagone di un treno, che si muove di moto rettilineo uniforme con velocità v, si trova una lampadina che a un certo istante viene accesa. I raggi di luce che si propagano verso i due passeggeri di testa e di coda del vagone vi arrivano nello stesso istante? Soluzione La simultaneità, essendo un concetto relativo, comporta che eventi simultanei in un sistema di riferimento non lo sono in un altro. Quindi, quando la lampadina viene accesa al centro del vagone, i due passeggeri, seduti a uguale distanza dalla lampadina, percepiscono la luce nello stesso istante. Per i due passeggeri i due lampi di luce (i due eventi) arrivano simultaneamente. Invece, per i due osservatori che si trovano fuori dal vagone, solidali con il sistema terra, gli eventi (i due lampi di luce percepiti dai due viaggiatori) non sono simultanei. Infatti, per i due osservatori i due eventi sono simultanei, avendo la luce la stessa velocità c in tutti i sistemi di riferimento inerziali (secondo postulato), ma per essi il viaggiatore seduto a sinistra del vagone percepisce il lampo di luce prima del viaggiatore a destra, perché nel tempo che la luce impiega per propagarsi, il treno, e quindi i viaggiatori, si sono spostati verso destra con velocità v. Esercizio Un treno è lungo 200 m e si muove alla velocità costante di 300 km/h. In un tratto rettilineo, agli estremi del treno esplodono due petardi. Secondo un osservatore posto a terra, le due esplosioni sono simultanee. Quanto vale, per l’osservatore posto al centro del treno, l’intervallo di tempo che separa l’arrivo dei lampi di luce delle esplosioni? www.liceoweb.it teoria della relatività ristretta Soluzione Utilizziamo come sistema di assi cartesiani quello solidale con il terreno e disposto lungo il binario. Le posizioni dei due petard sono x=-L/2 e x=L/2. Con questa scelta, all’istante t=0 s in cui sono avvenute le esplosioni (che, secondo l’osservatore fermo rispetto al terreno, sono state simultanee) l’osservatore sul treno occupava il punto di ascissa x=0 m. Per la legge del moto rettilineo uniforme, le leggi del moto per l’osservatore sul treno e per i due raggi di luce sono: Ora possiamo trovare gli istanti di tempo ts e td a cui i due raggi di luce giungono all’osservatore. ts è determinato imponendo che l’osservatore e il raggio di luce che proviene da sinistra occupino la stessa posizione. In modo analogo si trova td : Quindi, l’intervallo di tempo che separa gli istanti in cui l’osservatore vede le esplosioni dei petardi è: (attenzione alle unità di misura). La simultaneità, essendo un concetto relativo, comporta che eventi simultanei in un sistema di riferimento non lo sono in un altro. Ovviamente, essendo il valore di Δt molto piccolo, l’osservatore sul treno non può rendersi conto del ritardo tra l’arrivo di un lampo di luce e l’altro. Ciò conferma che la relatività della simultaneità (che comunque esiste) non ha effetti rilevanti nei fenomeni quotidiani caratterizzati da velocità molto piccole rispetto a a quella della luce. x0 = vt (osservatore) x0 = −L 2 +ct (raggio di luce da sinistra) x0 = L 2 −ct (raggio di luce da destra) vts = −L 2 +cts ⇒ ts = L 2(c −v) vtd = −L 2 +ctd ⇒ td = L 2(c+ v) Δt = ts −td = L 2(c −v) − L 2(c+ v) = Lv c2 −v2 = 200⋅83,3 (3⋅108)2 −83,32 =1,85⋅10−13 s www.liceoweb.it teoria della relatività ristretta Esercizio L’astronauta Mario viaggia verso una stella lontana alla velocità pari al 95% di quella della luce, mentre il suo gemello Dario rimane sulla Terra. Mario raggiunge la stella dopo un tempo Δt=10 anni, misurato con gli strumenti di bordo. Calcolare il tempo dello stesso viaggio (durata dello stesso fenomeno) secondo l’orologio terrestre di Dario. Soluzione La durata Δt=10 anni è il tempo proprio del fenomeno perché è misurata nel sistema di riferimento in cui l’astronave risulta ferma (sistema solidale con il fenomeno). La durata del viaggio misurata da Dario è: dove: Esercizio Facciamo qualche conto sui muoni prodotti nell’alta atmosfera dai raggi cosmici. Soluzione Secondo la cinematica classica (tempo e spazio assoluti), un muone che viaggia quasi alla velocità della luce (99,92%), nel tempo t=2,2 µs percorrerebbe la seguente distanza prima di decadere: Ed ecco il problema: sulla Terra ne arrivano più della metà, e l’atmosfera ha uno spessore di circa 15 km. Domanda: come fanno ad arrivare sulla Terra se “muoiono” dopo 660 metri? Soluzione: dobbiamo utilizzare la teoria della relatività, in quanto le formule e i concetti della fisica classica non funzionano quando gli oggetti sono molto veloci: Δt' =γΔt = 3,2⋅10 = 32 anni γ = 1 1−β 2 = 1 1−0,952 = 3,2 β = v c = 0,95c c = 0,95 d = cΔt = 3⋅108 ⋅2,2⋅10−6 ≈660 m Δt' =γΔt = 25⋅2,2 = 55 µs γ = 1 1−β 2 = 1 1−0,99922 = 25 β = v c = 0,9992c c = 0,9992 www.liceoweb.it teoria della relatività ristretta In questo tempo, il muone percorrerà la distanza: Ecco perché circa il 40% dei muoni arrivano sulla Terra. Esercizio Poichè anche i processi biologici devono soddisfare i postulati della relatività, calcolare la velocità che deve avere una navicella affinchè il suo equipaggio invecchi della metà rispetto al personale di controllo rimasto a terra. Soluzione La dilatazione del tempo comporta che i due intervalli di tempo Δt e Δt’, misurati nei due sistemi di riferimento diversi (navicella e Terra) per la durata dello stesso fenomeno, sono differenti: Pertanto, affinchè il tempo Δt sulla navicella sia la metà di quello Δt’ sulla Terra , il fattore di dilatazione deve essere 2, quindi: Conoscendo β è possibile ricavare la velocità della navicella: che è circa l’87% di quella della luce. Esercizio Un’astronave si trova ferma a una distanza di 15 UA (1unità astronomica = 1,496x1011 m) dalla Terra. Viene inviato un messaggio alle ore 9:55 del mattino secondo i loro orologi. Ø A che ora arriva il messagio sulla Terra? d = cΔt = 3⋅108 ⋅55⋅10−6 ≈16 km Δt' =γΔt γ = 1 1−β 2 = 2 ⇒ β = γ 2 −1 γ 2 = 4−1 4 = 0,866 β = v c ⇒ v = βc = 0,866c www.liceoweb.it teoria della relatività ristretta Ø Appena inviato il messaggio, l’astronave inizia a muoversi per 30 minuti, secondo i loro orologi, con una velocità costante v=0,600c. Quanto tempo è passato per gli orologi sulla Terra? Soluzione Le onde elettromagnetiche viaggiano alla velocità costante della luce, per cui il messaggio arriva sulla Terra alle ore: La dilatazione del tempo comporta che i due intervalli di tempo Δt e Δt’, misurati nei due sistemi di riferimento diversi (navicella e Terra) per la durata dello stesso fenomeno, sono differenti, per cui sulla Terra è passato un tempo più lungo pari a: Esercizio Un elettrone si muove con velocità ve=0,98c all’interno di un acceleratore di particelle, in cui è presente una etichetta a forma di triangolo equilatero, di lato L=4 cm, con l’altezza nella direzione del moto dell’elettrone. Determina l’area del triangolo nel sistema di riferimento dell’elettrone. Soluzione L’altezza del triangolo equilatero, nel suo sistema di riferimento, vale (teorema di Pitagora): Per l’invarianza delle lunghezze perpendicolari al moto relativo, solo la componente orizzontale del triangolo, ossia l’altezza, subisce una contrazione. Pertanto, nel sistema di riferimento dell’elettrone l’altezza appare contratta e di valore pari a: Δt = D c = 15⋅1,5⋅1011 3,0⋅108 = 7,5⋅103s = 2h5' ora = 9h55'+ 2h5' =12h Δt' =γΔt =1,25⋅30 = 37,5 min γ = 1 1−β 2 = 1 1−0,62 =1,25 β = v c = 0,6c c = 0,6 H = L2 −L 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 3 2 L2 = L 3 2 www.liceoweb.it teoria della relatività ristretta dove: Quindi, l’area del triangolo è: Esercizio Un aereo lungo L0=80 m (lunghezza propria) viaggia alla velocità di crociera di 900 km/h rispetto alla Terra. A) Calcolare la lunghezza dell’aereo misurata da un osservatore solidale con la Terra. B) Con quale velocità deve muoversi rispetto all’osservatore affinchè appaia contratto di 40 cm. Soluzione A) Applichiamo la formula della contrazione delle lunghezze: comunque: dove: Essendo la velocità dell’aereo molto piccola rispetto a quella della luce, gli effetti della contrazione non sono rilevanti. Volendo fare i calcoli con estrema precisione, si otterrebbe una contrazione dell’ordine del raggio dell’atomo di idrogeno (53 pm). B) Dalla formula della contrazione delle lunghezze ricaviamo il fattore di contrazione: H ' = H γ = 3 2 L γ = 3 2 4 5,03 = 0,689 cm γ = 1 1−β 2 = 1 1−0,982 ≅5,03 β = 0,98c c = 0,98 A = L⋅H ' 2 = 4⋅0,689 2 =1,4 cm2 L = L0 γ ≅L0 L < L0 γ = 1 1−β 2 = 1 1−(8,33⋅10−7)2 ≅1 β = v c = 250 3⋅108 = 8,33⋅10−7 γ = L0 L = 80 79,6 =1,005 www.liceoweb.it teoria della relatività ristretta e quindi la velocità dell’aereo: Esercizio Un’asta di lunghezza a riposo L0=10m si muove a una velocità costante pari a 70% di quella della luce e forma un angolo di 45° con la direzione del moto. Qual è la sua lunghezza secondo un osservatore fermo al suolo? Soluzione Le componenti (verticale ed orizzontale) dell’asta a riposo sono: Per l’invarianza delle lunghezze perpendicolari al moto relativo, la componente verticale della lunghezza rimane invariata. La componente orizzontale, invece, subisce una contrazione pari a: dove: Pertanto, la lunghezza dell’asta vista dall’osservatore a terra è: Passiamo ai calcoli: γ = 1 1−β 2 ⇒ β = γ 2 −1 γ 2 = 1,0052 −1 1,0052 = 0,0996 β = v c ⇒ v = βc = 0,0996⋅3⋅108 = 0,3⋅108 m / s = 3⋅104 km / s Δx = L0 cosα = L0 cos45° = 2 2 L0 Δy = L0senα = L0sen45° = 2 2 L0 Δx' = Δx γ = 2L0 2γ γ = 1 1−β 2 β = v c L = Δx' ( ) 2 + Δy ( ) 2 = 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 L0 2 γ 2 + 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 L0 2 = 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 L0 2 1 γ 2 +1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟L0 1+ 1 γ 2 www.liceoweb.it teoria della relatività ristretta La lunghezza di un segmento misurata in un sistema di riferimento in cui esso è in movimento risulta sempre minore della lunghezza propria (lunghezza a riposo) del segmento stesso. Esercizio Un’auto imbocca un tunnel alla velocità 2c/3. Il tunnel è lungo 50 m nel suo sistema di riferimento S. L’ingresso dell’automobile è simultaneo all’accensione delle luci. Quanto tempo impiega l’auto nel suo sistema di riferimento S’ per uscire dal tunnel? Soluzione Per la contrazione delle lunghezze, il tunnel misurato nel sistema S’ è lungo: dove: Esercizio Un disco ha forma circolare e raggio r=32 cm quando è fermo nel suo sistema di riferimento S. a) Se si muove di moto rettilineo uniforme alla velocità di 2x105 km/s rispetto a un osservatore fermo al suolo (sistema di riferimento S’), la forma del disco cambia? B) Dimostra che la forma del disco in moto è ellittica. C) Calcola la lunghezza degli assi dell’ellisse nel sistema di riferimento S’ dell’osservatore. Soluzione a) Per l’invarianza delle lunghezze perpendicolari al moto relativo, la componente verticale della lunghezza (diametro del disco perpendicolare al moto) rimane γ = 1 1−β 2 = 1 1−0,72 =1,4 β = 0,7c c = 0,7 L = 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟L0 1+ 1 γ 2 = 2 2 ⋅10⋅1+ 1 1,42 = 8,7 m L = L0 γ = 50 1,35 = 37 m γ = 1 1−β 2 = 1 1−0,672 =1,35 β = v c = 2c / 3 c = 0,67 www.liceoweb.it teoria della relatività ristretta invariata. La componente orizzontale (diametro del disco nella direzione del moto), invece, subisce una contrazione, per cui cambia la forma del disco in moto rispetto a S’. b) Fissando il seguente sistema di assi cartesiani, la corda P1P2 è lunga L0 nel proprio sistema di riferimento S: Rispetto al sistema di riferimento S’ solidale con il suolo, per la contrazione delle lunghezze, questo segmento L0 si contrae: Elevando a quadrato primo e secondo membro si ottiene la seguente equazione: L’equazione ottenuta è un’ellisse, per cui, come si voleva dimostrare, la forma del disco, rispetto al sistema S’, è cambiata. c) Per quanto abbiamo detto in precedenza, la lunghezza 2b dell’asse verticale dell’ellisse non cambia per l’invarianza delle lunghezze perpendicolari al moto: mentre la lunghezza 2a dell’asse orizzontale lungo la direzione del moto, rispetto al sistema S’ subisce una contrazione, e vale: dove: L0 = P 1P 2 = x1 −x2 = 2x = teorema di Pitagora ⎯ → ⎯⎯⎯⎯⎯= 2 r2 −y2 L = L0 γ = L0 1 1−β 2 ⇒ x1 ' −x2 ' = 2x' = 2 r2 −y2 ⋅1−β 2 ⇒ x' 1−β 2 = r2 −y2 x'2 1−β 2 = r2 −y2 2b = 2r = 64 cm 2a = 2r γ = 64 1,35 = 47,4 cm γ = 1 1−β 2 = 1 1−0,672 =1,35 β = v c = 2⋅105 3⋅105 = 0,67 www.liceoweb.it teoria della relatività ristretta Esercizio Nel sistema di riferimento del laboratorio due eventi avvengono nello stesso luogo a distanza di 2,0 s l’uno dall’altro. In un secondo sistema di riferimento, i due eventi avvengono a distanza di 2,1 s l’uno dall’altro. Quanto distano spazialmente tra loro i luoghi dei due eventi nel secondo sistema di riferimento? Soluzione Nella teoria della relatività non si può parlare dello spazio senza parlare del tempo e viceversa. L’unificazione dello spazio e del tempo genera una nuova grandezza chiamata spaziotempo. Nello spaziotempo per descrivere un fenomeno, occorre sapere che è avvenuto a un dato istante e in un certo punto dello spazio. Quindi per rappresentare un evento abbiamo bisogno di quattro coordinate (t,x,y,z). Ora, dati due eventi, le coordinate spaziotemporali di ciascuno di essi cambiano da un sistema di riferimento inerziale a un altro, ma la loro “distanza spaziotemporale” non varia. Ossia, l’intervallo spaziotemporale fra i due eventi: è invariante, cioè dipende soltanto dagli eventi stessi e non dal particolare sistema di riferimento usato per descriverli. L’intervallo spaziotemporale nei due sistemi di riferimento inerziali sono: Poiché l’intervallo spaziotemporale è invariante, e tenendo conto che nel primo sistema di riferimento i due eventi avvengono nello stesso luogo (Δs1=0), si ha: Esercizio La stella Vega si trova a 25,3 a.l. (anni-luce) dalla Terra e la vi vuole raggiungere utilizzando un’astronave che viaggia alla velocità di 0,8c. Qual è la durata del viaggio misurata dal centro spaziale sulla Terra? Qual è la durata del viaggio misurata dagli orologi dell’astronave? Δs ( ) 2 = Δx ( ) 2 + Δy ( ) 2 + Δz ( ) 2 −cΔt ( ) 2 Δs1 ( ) 2 = cΔt1 ( ) 2 → cΔt1 ( ) 2 − Δs1 ( ) 2 = 0 Δs2 ( ) 2 = cΔt2 ( ) 2 → cΔt2 ( ) 2 − Δs2 ( ) 2 = 0 Δs1 ( ) 2 = Δs2 ( ) 2 cΔt1 ( ) 2 = cΔt2 ( ) 2 − Δs2 ( ) 2 → Δs2 ( ) 2 = cΔt2 ( ) 2 −cΔt1 ( ) 2 = 2,1c ( ) 2 −2c ( ) 2 = 0,41c2 Δs = 0,41c2 = c 0,41 =1,92⋅105 km www.liceoweb.it teoria della relatività ristretta Soluzione Con la scelta dei sistemi di riferimenti come in figura, si ha: Quindi, se dal sistema di riferimento Terra la velocità dell’astronave è: la distanza Δx=25,3 a.l. fino a Vega sarà percorsa nell’intervallo di tempo: Nel sistema di riferimento dell’astronave la partenza da Terra e l’arrivo a Vega avvengono nello stesso luogo, l’astronave appunto, per cui si ha Δx’=0. Ciò significa che, nel sistema di riferimento dell’astronave, la durata del viaggio è la durata propria Δτ dell’evento «l’astronave parte dalla Terra e giunge a Vega». Tenendo presente il sistema di riferimento fissato, e l’invarianza dell’intervallo spaziotemporale, si ottiene: All’arrivo dell’astronave su Vega, per le persone a Terra saranno trascorsi 31,6 anni, ma per l’equipaggio a bordo ne saranno passati meno di 19. Esercizio Il mesone µ è una particella che, a riposo, decade dopo un tempo di vita media di circa 2,15 µ s. La stessa particella, in moto rispetto a un riferimento terrestre a velocità molto elevata, percorre una distanza media di 6,40 km prima di decadere. Calcola: a) il tempo di vita media del mesone nel sistema di riferimento terrestre; b) la velocità con la quale si muove il mesone nel sistema di riferimento terrestre. Δy = Δy' = Δz = Δz' = 0 v = 0,8 a.l. / anni Δt = Δx v = 25,3 0,8 = 31,6 anni Δx' ( ) 2 = cΔτ ( ) 2 → cΔτ ( ) 2 = 0 Δx ( ) 2 = cΔt ( ) 2 → cΔt ( ) 2 − Δx ( ) 2 = 0 Δx' ( ) 2 = Δx ( ) 2 cΔτ ( ) 2 = cΔt ( ) 2 −Δx ( ) 2 → Δτ = cΔt ( ) 2 −Δx ( ) 2 c = (1⋅31,6)2 −(25,3)2 1 =18,9 anni www.liceoweb.it teoria della relatività ristretta Soluzione a) Nel sistema di riferimento della particella il decadimento avviene nello stesso luogo, per cui si ha Δx’=0. Ciò significa che, nel sistema di riferimento della particella, il tempo di decadimento coincide con la vita media Δτ (durata propria) del mesone. Quindi, attraverso l’invarianza dell’intervallo spaziotemporale, il tempo di vita media del mesone nel sistema di riferimento terrestre è: Nel sistema di riferimento terrestre la vita media del mesone è più lunga, di cira un ordine di grandezza. b) La velocità con la quale si muove il mesone nel sistema di riferimento terrestre, essendo un moto rettilineo uniforme, vale: Praticamente, si muove alla velocità della luce. Esercizio Nello spazio-tempo a due dimensioni (una dimensione temporale e una spaziale), considera gli eventi A(0,72 μs; 1,5 km) e B(0,95 μs; 1,7 km) rilevati in un riferimento S. In un riferimento S’ i due eventi risultano simultanei. Calcola la velocità del sistema S’rispetto al sistema S. Soluzione Grazie all’invarianza dell’intervallo spaziotemporale (tenendo presente che nel sistema S’ i due eventi sono simultanei Δx’=0): Δx' ( ) 2 = cΔτ ( ) 2 → cΔτ ( ) 2 = 0 Δx ( ) 2 = cΔt ( ) 2 → cΔt ( ) 2 − Δx ( ) 2 = 0 Δx' ( ) 2 = Δx ( ) 2 cΔτ ( ) 2 = cΔt ( ) 2 −Δx ( ) 2 → Δt = cΔτ ( ) 2 + Δx ( ) 2 c = Δτ ( ) 2 + Δx c ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 2,15⋅10−6 ( ) 2 + 640⋅103 3⋅108 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= 2,14⋅10−5s v = Δx Δt = 640⋅103 2,14⋅10−5 = 2,99⋅108 m / s www.liceoweb.it teoria della relatività ristretta e alla trasformazione di Lorentz: si ottiene la velocità del sistema S’rispetto al sistema S: Esercizio Nel sistema di riferimento S un raggio di luce si sposta nel verso delle x positive e un altro si muove nel verso opposto. Qual è la velocità del primo raggio di luce visto nel sistema di riferimento solidale con il secondo? Risolvi il problema prima nell’ambito della meccanica classica e poi in quello della meccanica relativistica. Soluzione Risolviamo il problema ponendo u=c (la velocità del raggio di luce rispetto a S è c) e v=-c (il sistema di riferimento S’ solidale con il secondo raggio di luce si muove nel verso negativo delle x con una velocità di modulo c). Nell’ambito della meccanica classica, dove valgono le trasformazioni di Galileo, si ottiene: Il primo raggio di luce si allontana dal secondo con una velocità doppia di quella che esso ha nel sistema S. Secondo la teoria della relatività (secondo postulato), invece, la velocità della luce deve essere la stessa in tutti i sistemi di riferimento inerziali. Infatti, grazie alle trasformazioni di Lorentz, si ottiene: Δs ( ) 2 = Δs' ( ) 2 ⇒ cΔt ( ) 2 −Δx ( ) 2 = −Δx' ( ) 2 → = Δx' ( ) 2 = Δx ( ) 2 −cΔt ( ) 2 x = x'−vt' 1−β 2 Δx = Δx'−vΔt' 1−β 2 Δt'=0 ⎯ → ⎯⎯ Δx = Δx' 1−β 2 → β = Δx ( ) 2 −Δx' ( ) 2 Δx = Δx ( ) 2 − Δx ( ) 2 −cΔt ( ) 2 ⎡ ⎣ ⎤ ⎦ Δx = cΔt Δx β = v c ⇒ v = cβ = c2Δt Δx = (3⋅108)2 ⋅(0,95⋅10−6 −0,72⋅10−6) (1,7⋅103 −1,5⋅103) =108m / s u' = u−v = c −(−c) = 2c www.liceoweb.it teoria della relatività ristretta Quindi, il risultato ottenuto mostra proprio che, anche vista da un altro raggio luminoso, la velocità della luce rimane uguale a c. Esercizio Un nucleo di uranio-238 (massa mu=238,0508 u) può decadere in modo spontaneo, dando origine a un nucleo di torio-234 (massa mTh=234,0436 u) e a un nucleo di elio-4 (massa mHe=4,0026 u). Quanto vale l’energia emessa nel corso di tale decadimento? Soluzione La somma delle masse dei nuclei di torio e di elio è: risulta minore della massa del nucleo originale di uranio mu=238,0508 u. La differenza tra la massa iniziale e quella finale del sistema è: dove l’unità di massa atomica u=1,6605x10-27 kg. Per l’equivalenza massa-energia, alla scomparsa di questa massa deve corrispondere l’emissione di una quantità E di energia pari a: motivo per cui la somma complessiva di tutte le energie e di tutte le masse in gioco nel processo di decadimento deve rimanere inalterata nel tempo (principio di conservazione della massa-energia). u' = u−v 1−uv c2 = c −(−c) 1−c(−c) c2 = 2c 2 = c mTH + mHe = 234,0436u+ 4,0026u = 238,0462u Δm = mU −(mTh + mHe) = 238,0508u−238,0462u = 0,0046u = 0,0046⋅1,6605⋅10−27 = 7,6⋅10−30 kg E = Δm⋅c2 = 7,6⋅10−30 ⋅(3⋅108)2 = 6,8⋅10−13 J www.liceoweb.it teoria della relatività ristretta Esercizio Un oggetto di alluminio di massa 3 kg viene riscaldato da 20,0 °C a 870 °C. Il calore specifico dell’alluminio è circa 0,9 kJ/(kg"K). Determina la variazione di massa. Soluzione Combiniamo la relazione massa-energia con la relazione fondamentale della termologia: Ciò che abbiamo ottenuto è la differenza di massa che si dovrebbe avere. Ossia, l’oggetto di alluminio a 870 °C “pesa di più” dello stesso oggetto a 20 °C. Esercizio Al CERN di Ginevra i protoni vengono accelerati fino ad una energia E=7,20x10-8J. Qual è la velocità dei protoni quando raggiungono tale energia? Soluzione L’energia totale E di un corpo di massa a riposo m0 (massa a riposo del protone m0=1,67x10-27, kg) in movimento con velocità di modulo v rispetto a un dato sistema di riferimento, è: Da questa formula ricaviamo il fattore ϒ, noto il quale sarà possibile calcolare la velocità dei protoni quando avranno raggiunto l’energia E: La velocità dei protoni è praticamente indistinguibile da quella della luce. ΔE = Δm⋅c2 ΔE = CΔT Δm⋅c2 = CΔT ⇒ Δm = CΔT c2 = 3⋅0,9⋅103 ⋅850 (3⋅108)2 = 3⋅10−11 kg E =γm0c2 γ = E m0c2 = 7,20⋅10−8 1,67⋅10−27 ⋅(3⋅108)2 = 479 γ = 1 1−v2 c2 ⇒ v = c γ 2 −1 γ 2 = c⋅ 4792 −1 4792 ≅c www.liceoweb.it teoria della relatività ristretta Esercizio L’energia delle particelle elementari viene spesso espressa in MeV (1eV=1,6x10-19J) e di conseguenza, la massa di riposo viene espressa in MeV/c2. Un elettrone ha una massa di riposo pari a 0,51 MeV/c2. A) Qual è il valore della massa dell’elettrone fermo, espressa in chilogrammi? B) Quanta energia è necessaria per creare un elettrone? C) Quando l’energia relativistica totale dell’elettrone è il doppio di quella a riposo, qual è la sua quantità di moto? Soluzione A) La massa dell’elettrone fermo, espressa in chilogrammi vale: B) Sfruttando l’equivalenza massa-energia, l’energia necessaria per creare un elettrone è pari a: C) L’energia relativistica totale di una particella, e quindi del nostro elettrone, è data dalla seguente relazione: dove: Pertanto, se vogliamo che la sua energia raddoppi, deve essere: e la velocità dell’elettrone dovrà assumere il seguente valore: In definitiva, la quantità di moto dell’elettrone corrispondente a questa energia relativistica doppia è: me = 0,51 MeV c2 = 0,51⋅1,6⋅10−13 (3⋅108)2 ≅9⋅10−31kg E0 = m0c2 = 0,51 MeV c2 ⋅c2 = 0,51 MeV = 0,51⋅106 ⋅1,6⋅10−19 = 0,82⋅10−13J E =γm0c2 γ = 1 1−v2 c2 γ = 1 1−v2 c2 = 2 γ = 1 1−v2 c2 = 2 ⇒ v = 3 2 c = 2,6⋅108m / s p =γm0v = 2⋅9,1⋅10−31 ⋅2,6⋅108 = 47,3⋅10−23kg⋅m / s www.liceoweb.it teoria della relatività ristretta Esercizio I muoni sono particelle elementari con massa di riposo m0=105,7 MeV/c2 che si creano quando raggi cosmici di alta energia entrano in atmosfera. La vita media dei muoni è pari a τ=1,56 μs nel sistema di riferimento in cui sono fermi, e decadono in altre particelle elementari. Considera un muone creato a 10 km dal livello del mare con velocità v=0,98c diretto verso il basso. Calcola: a) la quantità di energia necessaria per creare il muone; b) l’energia totale relativistica del muone; c) la distanza percorsa da un muone nel sistema di riferimento in cui è fermo, secondo la teoria classica e quella relativistica. Soluzione a) Sfruttando l’equivalenza massa-energia, l’energia necessaria per creare un muone è pari a: b) L’energia relativistica totale di una particella, e quindi del nostro muone, è data dalla seguente relazione: dove: c) La distanza percorsa dal muone nel sistema di riferimento in cui è fermo (tempo proprio τ=1,56 μs ), secondo la fisica classica, vale: Invece il muone, avendo una velocità relativistica, subisce una dilatazione temporale della sua vita media: per cui la distanza percorsa secondo la teoria relativistica, sarà maggiore rispetto a quella classica: E0 = m0c2 = 105,7 MeV c2 ⋅c2 =105,7 MeV =105,7⋅106 ⋅1,6⋅10−19 =169⋅10−13J E =γm0c2 = 5⋅169⋅10−13 = 845⋅10−13J γ = 1 1−β 2 = 1 1−0,982 = 5 β = v c = 0,98c c = 0,98 d = vτ = 0,98⋅3⋅108 ⋅1,56⋅10−6 = 459 m Δt' =γΔt = 5⋅1,56⋅10−6 = 7,8⋅10−6s = 7,8 µs d = vΔt' = 0,98⋅3⋅108 ⋅7,8⋅10−6 = 2293 m www.liceoweb.it teoria della relatività ristretta Esercizio Un oggetto celeste molto lontano, per esempio un quasar, emette onde luminose con frequenza f. La frequenza delle stesse onde misurata sulla Terra è f’=0,75 f. A) Il quasar si sta allontanando o avvicinando? B) Calcolare la velocità di allontanamento (o avvicinamento) relativo tra la Terra e il quasar; C) Quanto vale il suo redshift? Soluzione A) Poiché f’λ), il quasar e la Terra si stanno allontanando. B) La velocità di allontanamento relativo tra la Terra e il quasar la ricaviamo dalla formula dell’effetto Doppler relativistico nel caso in cui sorgente (quasar) e osservatore (Terra) si allontanano: C) In astrofisica, l’effetto Doppler viene sfruttato attraverso l’introduzione della grandezza adimensionale z, chiamata redshift (spostamento verso il rosso), che nel nostro caso vale: Quindi, nel caso in cui sorgente ed osservatore si allontanano (f’λ) si ha z<0 (il colore della luce ricevuto è spostato verso il rosso dello spettro). Esercizio Una stella emette luce di lunghezza d’onda λ=520 nm (corrispondente a luce verde). La sua velocità di avvicinamento alla Terra è pari a due terzi della velocità della luce nel vuoto. Calcola la lunghezza d’onda rilevata da un osservatore a Terra. Soluzione f ' = f 1−v c 1+ v c ⇒ v = c 1− f f ' ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 1+ f f ' ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 3⋅108 ⋅ 1−0,75 f f ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 1+ 0,75 f f ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 8,4⋅107m / s z = f f ' −1= f 0,75 f −1= 0,33 www.liceoweb.it teoria della relatività ristretta Nota la relazione tra lunghezza d’onda e frequenza: e utilizzando la legge dell’effetto Doppler relativistico (caso sorgente ed osservatore in avvicinamento): si ottiene: che corrisponde a luce (onde elettromagnetiche) nell’ultravioletto. c = λ' f ' ⇒ λ' = c f ' f ' = f 1+ β 1−β dove : β = v c = 2 3 c c = 2 3 λ' = c f 1+ β 1−β = λ 1−β 1+ β = 520⋅10−9 ⋅ 1−2 3 1+ 2 3 = 233⋅10−9 m = 233 nm |
7146 | https://math.libretexts.org/Bookshelves/Precalculus/Elementary_Trigonometry_(Corral)/02%3A_General_Triangles/2.01%3A_The_Law_of_Sines | Published Time: 2016-12-21T00:43:05Z
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2: General Triangles
Elementary Trigonometry (Corral)
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Thu, 17 Nov 2022 22:10:27 GMT
2.1: The Law of Sines
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Contents
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Precalculus & Trigonometry
Elementary Trigonometry (Corral)
2: General Triangles
2.1: The Law of Sines
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Elementary Trigonometry (Corral)
Front Matter
1: Right Triangle Trigonometry Angles
2: General Triangles
3: Identities
4: Radian Measure
5: Graphing and Inverse Functions
6: Additional Topics
Back Matter
2.1: The Law of Sines
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Nov 17, 2022
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2: General Triangles
2.2: The Law of Cosines
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3254
Michael Corral
Schoolcraft College
( \newcommand{\kernel}{\mathrm{null}\,}\)
Table of contents
Theorem 2.1.1: Law of Sines
Proof
Example 2.1.1: Case 1 - One side and two angles known
Solution
Example 2.1.2: Case 2 - Two sides and one opposite angle known
Solution
Example 2.1.3: Case 2 - Two sides and one opposite angle known
Solution
Proof
Theorem 2.1.12.1.1: Law of Sines
If a triangle has sides of lengths aa , bb , and cc opposite the angles AA , BB , and CC , respectively, then
asinA \= bsinB \= csinC .
Note that by taking reciprocals, Equation 2.1.1 can be written as
sinAa \= sinBb \= sinCc ,
and it can also be written as a collection of three equations:
ab \= sinAsinB ,ac \= sinAsinC ,bc \= sinBsinC
Another way of stating the Law of Sines is: The sides of a triangle are proportional to the sines of their opposite angles.
Proof
To prove the Law of Sines, let △ABC be an oblique triangle. Then ∠ABC can be acute, as in Figure 2.1.1(a), or it can be obtuse, as in Figure 2.1.1(b). In each case, draw the altitude from the vertex at C to the side ¯AB . In Figure 2.1.1(a) the altitude lies inside the triangle, while in Figure 2.1.1(b) the altitude lies outside the triangle.
Figure 2.1.1: Proof of the Law of Sines for an oblique triangle △ABC
Let h be the height of the altitude. For each triangle in Figure 2.1.1, we see that
hb \= sinA
and
ha \= sinB
in Figure 2.1.1(b), ha\=sin(180∘−B)\=sinB as shown previously). Thus, solving for h in Equation 2.1.3 and substituting that into Equation 2.1.2 gives
asinBb \= sinA ,
and so putting a and A on the left side and b and B on the right side, we get
asinA \= bsinB .
By a similar argument, drawing the altitude from A to ¯BC gives
bsinB \= csinC ,
so putting the last two equations together proves the theorem.
◻
Note that we did not prove the Law of Sines for right triangles, since it turns out (see Exercise 12) to be trivially true for that case.
Example 2.1.1: Case 1 - One side and two angles known
Solve the triangle △ABC given a\=10 , A\=41∘ , and C\=75∘ .
Solution
We can find the third angle by subtracting the other two angles from 180∘ , then use the law of sines to find the two unknown sides. In this example we need to find B , b , and c . First, we see that
B \= 180∘ − A − C \= 180∘ − 41∘ − 75∘⇒B \= 64∘ .
So by the Law of Sines we have
bsinB \= asinA⇒b \= asinBsinA \= 10sin64∘sin41∘⇒b \= 13.7 , andcsinC \= asinA⇒c \= asinCsinA \= 10sin75∘sin41∘⇒c \= 14.7 .
Example 2.1.2: Case 2 - Two sides and one opposite angle known
Solve the triangle △ABC given a\=18 , A\=25∘ , and b\=30 .
Solution
In this example we know the side a and its opposite angle A , and we know the side b . We can use the Law of Sines to find the other opposite angle B , then find the third angle C by subtracting A and B from 180∘ , then use the law of sines to find the third side c . By the Law of Sines, we have
sinBb \= sinAa⇒sinB \= bsinAa \= 30sin25∘18⇒sinB \= 0.7044 .
Using the sin−1 button on a calculator gives B\=44.8∘ . However, recall from Section 1.5 that sin(180∘−B)\=sinB . So there is a second possible solution for B , namely 180∘−44.8∘\=135.2∘ . Thus, we have to solve twice for C and c : once for B\=44.8∘ and once for B\=135.2∘:
B\=44.8∘B\=135.2∘C\=180∘−A−B\=180∘−25∘−44.8∘\=110.2∘C\=180∘−A−B\=180∘−25∘−135.2∘\=19.8∘csinC\=asinA ⇒ c\=asinCsinA\=18sin110.2∘sin25∘csinC\=asinA ⇒ c\=asinCsinA\=18sin19.8∘sin25∘⇒ c\=40⇒ c\=14.4
Hence, B\=44.8∘, C\=110.2∘,c\=40 and B\=135.2∘, C\=19.8∘, c\=14.4 are the two possible sets of solutions. This means that there are two possible triangles, as shown in Figure 2.1.2.
Figure 2.1.2: Two possible solutions
In Example 2.1.2 we saw what is known as the ambiguous case. That is, there may be more than one solution. It is also possible for there to be exactly one solution or no solution at all.
Example 2.1.3: Case 2 - Two sides and one opposite angle known
Solve the triangle △ABC given a\=5 , A\=30∘ , and b\=12 .
Solution
By the Law of Sines, we have
sinBb \= sinAa⇒sinB \= bsinAa \= 12sin30∘5⇒sinB \= 1.2 ,
which is impossible since |sinB|≤1 for any angle B . Thus, there is no solution.
There is a way to determine how many solutions a triangle has in Case 2. For a triangle △ABC , suppose that we know the sides a and b and the angle A . Draw the angle A and the side b , and imagine that the side a is attached at the vertex at C so that it can "swing'' freely, as indicated by the dashed arc in Figure 2.1.3 below.
Figure 2.1.3: The ambiguous case when A is acute
If A is acute, then the altitude from C to ¯AB has height h\=bsinA . As we can see in Figure 2.1.3(a)-(c), there is no solution when a<h (this was the case in Example 2.3); there is exactly one solution - namely, a right triangle - when a\=h; and there are two solutions when h<a<b (as was the case in Example 2.2). When a≥b there is only one solution, even though it appears from Figure 2.1.3(d) that there may be two solutions, since the dashed arc intersects the horizontal line at two points. However, the point of intersection to the left of A in Figure 2.1.3(d) can not be used to determine B , since that would make A an obtuse angle, and we assumed that A was acute.
If A is not acute (i.e. A is obtuse or a right angle), then the situation is simpler: there is no solution if a≤b , and there is exactly one solution if a>b (see Figure 2.1.4).
Figure 2.1.4: The ambiguous case when A≥90◦
Table 2.1 summarizes the ambiguous case of solving △ABC when given a , A , and b . Of course, the letters can be interchanged, e.g. replace a and A by c and C , etc.
Table 2.1 Summary of the ambiguous case
There is an interesting geometric consequence of the Law of Sines. Recall from Section 1.1 that in a right triangle the hypotenuse is the largest side. Since a right angle is the largest angle in a right triangle, this means that the largest side is opposite the largest angle. What the Law of Sines does is generalize this to any triangle:
In any triangle, the largest side is opposite the largest angle.
Proof
To prove this, let C be the largest angle in a triangle △ABC . If C\=90∘ then we already know that its opposite side c is the largest side. So we just need to prove the result for when C is acute and for when C is obtuse. In both cases, we have A≤C and B≤C . We will first show that sinA≤sinC and sinB≤sinC .
Figure 2.1.5
If C is acute, then A and B are also acute. Since A≤C , imagine that A is in standard position in the xy-coordinate plane and that we rotate the terminal side of A counterclockwise to the terminal side of the larger angle C , as in Figure 2.1.5. If we pick points (x1,y1) and (x2,y2) on the terminal sides of A and C , respectively, so that their distance to the origin is the same number r , then we see from the picture that y1≤y2 , and hence
sinA \= y1r ≤ y2r \= sinC .
By a similar argument, B≤C implies that sinB≤sinC . Thus, sinA≤sinC and sinB≤sinC when C is acute. We will now show that these inequalities hold when C is obtuse.
If C is obtuse, then 180∘−C is acute, as are A and B . If A>180∘−C then A+C>180∘ , which is impossible. Thus, we must have A≤180∘−C . Likewise, B≤180∘−C . So by what we showed above for acute angles, we know that sinA≤sin(180∘−C) and sinB≤sin(180∘−C) . But we know from Section 1.5 that sinC\=sin(180∘−C) . Hence, sinA≤sinC and sinB≤sinC when C is obtuse.
Thus, sinA≤sinC if C is acute or obtuse, so by the Law of Sines we have
ac \= sinAsinC ≤ sinCsinC \= 1⇒ac ≤ 1⇒a ≤ c .
By a similar argument, b≤c . Thus, a≤c and b≤c , i.e. c is the largest side.
◻
This page titled 2.1: The Law of Sines is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Michael Corral via source content that was edited to the style and standards of the LibreTexts platform.
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2: General Triangles
2.2: The Law of Cosines |
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7148 | https://www.youtube.com/watch?v=W80lgbWy85w | Ex: Find Parametric Equations For Ellipse Using Sine And Cosine From a Graph
Mathispower4u
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The video explains how to determine the parametric equations using the graph of an ellipse.
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Transcript:
we want to find the parametric equations for the graph of the ellipse in the form X of T = A cosine T and Y of tal B sin T to begin now we'll write the cartisian equation or rectangular equation for the ellipse in this form here the first thing to recognize here is that we have a horizontal major axis and a vertical minor axis remember the major axis is always the longer axis so again because we have a horizontal major axis a is greater than b and this will be the form of our equation where the point H comma K will be the center and notice how our Center is the origin 0 comma 0 and now for a quick review the length of the major axis is 2 a or we can say the length from the center to one end point of the major axis is a and the length of the minor axxis is 2b or we can say the length from the center to one end point of the minor axis is B so looking at the major axis notice how this means that a is equal to 5 and B is equal to three again we already notice at the center H comma K is the origin this is all we need in order to write the cartisian equation we'd have x^2 / a^ 2 which would be x^2 / by 5^ 2 or 25 plus y^2 / B ^2 which would be Y 2 / 3^2 or 9 equal 1 and now we need to find our equations for X of T and Y of T to do this we'll use the trig identity sin s thet + cosine s Theta = 1 and since X of T has to be in the form of some constant time cosine T notice how here if we had X of T equals 5 cosine T notice how x^2 would be 25 cosine 2qu T and the 25s would simplify out leaving us with cosine squ T and then looking at the second fraction notice how y of T has to be in the form of a constant time sin T so if we let y of T be equal to 3 sin T notice that y^ sare would be 9 sin^2 t the N9 would simplify out and therefore we'd have cosine s t+ sin squ t equal 1 which we know is true and therefore these would be our parametric equations let's go ahead and show the work to check this so we'll substitute 5 cosine t for X so for this first fraction we'd have 5 cine T SAR / 25 plus now for the second fraction we'd have y^2 ID 9 or 3 sin t^ 2ar / 9 = 1 so if we Square 5 cine T we'd have 25 cosine 2 T / 25 plus here we'd have 9 sin 2 T ID 9 equal 1 and now if we simplify 25 ID 25 is 1 9 ID 9 equal 1 so we've just shown that our parametric equations are correct since cine s t plus sin 2 T does equal 1 so we have found our parametric equations X of T and Y of T correctly but before we go when we do find parametric equations the graph is sketched out in a particular direction which we call the orientation of the graph so let's take a look at that before we go let's begin by finding the point on our ellipse when T equals 0 well X of 0 would be equal to 5 cosine 0 well here's the graph of our cosine function cosine 0 is equal to 1 so the x coordinate would be five and now for the y coordinate y of Z would be equal to 3 sin0 here's the graph of our sign function sin 0 is zero so when T is zero we would be at the point 5 Z on the ellipse which should be this point here now let's find one more Point let's find the X and Y coordinates when t equal Pi / 2 x of Pi / 2 would be equal to 5 cosine Pi / 2 cosine Pi / 2 is equal to 0 so the x coordinate would be zero and then y of Pi / 2 would be equal to 3 sin Pi / 2 well sin Pi / 2 is equal to 1 so the y coordinate is three so notice when X is pi over 2 would be at 03 at this point here so as says T increases the points would be plotted counterclockwise and therefore using our parametric equations we say the orientation would be in this direction here so we can say as T increases the ellipse is traced in this direction or has this orientation I hope you found this helpful |
7149 | https://louis.pressbooks.pub/trigonometry/chapter/9-3-dot-product/ | 9.3 The Dot Product – Trigonometry
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Contents
Introduction
Introduction
About This Book
About the Authors
Authors
Chapter 1: Triangles and Circles
Introduction to Triangles and Circles
Introduction
Class Activities
1.1 Triangles and Angles
Algebra Refresher
Triangles and Angles
Triangles
Angles
Section 1.1 Summary
1.2 Similar Triangles
Algebra Refresher
Congruent Triangles
Similar Triangles
Using Proportions with Similar Triangles
Similar Right Triangles
Overlapping Triangles
Section 1.2 Summary
1.3 Circles
Algebra Refresher
The Distance Formula
Equation for a Circle
Section 1.3 Summary
Chapter 1 Summary and Review
Exercises: 1.1 Triangles and Angles
Exercises: 1.2 Similar Triangles
Exercises: 1.3 Circles
Exercises: Chapter 1 Review Problems
Chapter 2: Trigonometric Ratios
Introduction to the Trigonometric Ratios
Introduction
Class Activities
2.1 Side and Angle Relationships
Algebra Refresher
Introduction to Triangles
The Triangle Inequality
Right Triangles: The Pythagorean Theorem
Section 2.1 Summary
2.2 Right Triangle Trigonometry
Algebra Refresher
The Sine of an Angle
Using the Sine Ratio to Find an Unknown Side
The Cosine and the Tangent
The Three Trigonometric Ratios
Section 2.2 Summary
2.3 Solving Right Triangles
Algebra Refresher
Finding an Angle
The Special Angles
Section 2.3 Summary
Chapter 2 Summary and Review
Exercises: 2.1 Side and Angle Relationships
Exercises: 2.2 Right Triangle Trigonometry
Exercises: 2.3 Solving Right Triangles
Exercises: Chapter 2 Review Problems
Chapter 3: Laws of Sines and Cosines
Introduction to Laws of Sines and Cosines
Introduction
Class Activities
3.1 Obtuse Angles
Algebra Refresher
Angles in Standard Position
Trigonometric Ratios for Obtuse Angles
Trigonometric Ratios for Supplementary Angles
Supplements of the Special Angles
The Area of a Triangle
Section 3.1 Summary
3.2 The Law of Sines
Algebra Refresher
Law of Sines
Finding a Side
Solving Triangles with the Law of Sines
Finding an Angle
Applications
Measuring Astronomical Distances
Small Angles: Minutes and Seconds
Section 3.2 Summary
3.3 The Law of Cosines
Algebra Refresher
Law of Cosines
Finding a Side
Finding an Angle
Using the Law of Cosines for the Ambiguous Case
Application: Navigation
Which Law to Use
Section 3.3 Summary
Chapter 3 Summary and Review
Exercises: 3.1 Obtuse Angles
Exercises: 3.2 The Law of Sines
Exercises: 3.3 The Law of Cosines
Exercises: Chapter 3 Review Problems
Chapter 4: Trig Functions
Introduction to Trigonometric Functions
Introduction
Class Activities
4.1 Angles and Rotation
Algebra Refresher
Angles in Standard Position
Trigonometric Ratios for All Angles
Reference Angles
Using Reference Angles
Coterminal Angles
Solving Trigonometric Equations
The Special Angles
The Unit Circle
Section 4.1 Summary
4.2 Graphs of Trigonometric Functions
Algebra Refresher
Location by Coordinates
Bearings
Function Notation
The Tangent Function
Angle of Inclination
Section 4.2 Summary
4.3 Periodic Functions
Algebra Refresher
Period, Midline, and Amplitude
Sinusoidal Functions
Other Periodic Functions
Section 4.3 Summary
Chapter 4 Summary and Review
Exercises: 4.1 Angles and Rotation
Skills
Exercises Homework 4.1
Exercises: 4.2 Graphs of Trigonometric Functions
Skills
Homework 4.2
Exercises: 4.3 Periodic Functions
Skills
Exercises Homework
Exercises: Chapter 4 Review Problems
Chapter 5: Equations and Identities
Introduction to Equations and Identities
Introduction
5.1 Algebra with Trigonometric Ratios
Algebra Refresher
Evaluating Trigonometric Expressions
Simplifying Trigonometric Expressions
Powers of Trigonometric Ratios
Products
Factoring
Section 5.1 Summary
5.2 Solving Equations
Algebra Refresher
Trigonometric Equations
Graphical Solutions
Equations with Squares of Trig Ratios
Snell’s Law
Section 5.2 Summary
5.3 Trigonometric Identities
Algebra Refresher
Using Trigonometric Ratios in Identities
Pythagorean Identity
Tangent Identity
Solving Equations
Proving Identities
Section 5.3 Summary
Chapter 5 Summary and Review
Exercises: 5.1 Algebra with Trigonometric Ratios
Skills
Exercises Homework 5.1
Exercises: 5.2 Solving Equations
Skills
Exercises Homework 5.2
Exercises: 5.3 Trigonometric Identities
Skills
Exercises Homework 5.3
Exercises: Chapter 5 Review Problems
Exercises for Chapter 5 Review
Chapter 6: Radians
Introduction to Trig Radians
Introduction
Class Activities
6.1 Arclength and Radians
Algebra Refresher
Arclength and Radians
Arclength
Measuring Angles in Radians
Converting between Degrees and Radians
Arclength Formula
Unit Circle
Section 6.1 Summary
6.2 The Circular Functions
Algebra Refresher
Trigonometric Functions of Angles in Radians
Values of Special Angles
Reference Angles in Radians
Sine and Cosine of Real Numbers
Coordinates on a Unit Circle
The Tangent Function
The Circular Functions
Section 6.2 Summary
6.3 Graphs of the Circular Functions
Algebra Refresher
The Sine and Cosine Functions
The Tangent Function
Solving Equations
Modeling with Circular Functions
Domain and Range
Section 6.3 Summary
Chapter 6 Summary and Review
Exercises 6.1 Arclength and Radians
Exercises 6.2 The Circular Functions
Exercises 6.3 Graphs of the Circular Functions
Exercises: Chapter 6 Review Problems
Chapter 7: Circular Functions
Introduction to Circular Functions
Introduction
Class Activities
7.1 Transformations of Graphs
Algebra Refresher
Transformations of Graphs
Graphs of Sinusoidal Functions
Modeling with Sinusoidal Functions
The Tangent Function
Section 7.1 Summary
7.2 The General Sinusoidal Function
Algebra Refresher
The General Sinusoidal Function
Combining Transformations
Modeling with Sinusoidal Functions
Section 7.2 Summary
7.3 Solving Equations
Algebra Refresher
Multiple Solutions
Using a Substitution
Applications
Section 7.3 Summary
Chapter 7 Summary and Review
Exercises 7.1 Transformations of Graphs
Exercises 7.2 The General Sinusoidal Function
Exercises 7.3 Solving Equations
Exercises: Chapter 7 Review Problems
Chapter 8: More Functions and Identities
Introduction to More Functions and Identities
Introduction
Class Activities
8.1 Sum and Difference Formulas
Algebra Refresher
The Sum of Angles Identities
The Difference of Angles Identities
Sum and Difference Identities for Tangent
Double Angle Identities
Solving Equations
Section 8.1 Summary
8.2 Inverse Trigonometric Functions
Algebra Refresher
Inverse of a Function
The Graph of the Inverse
The Inverse Cosine and Inverse Tangent Functions
Modeling with Inverse Functions
Simplifying Inverse Expressions
Section 8.2 Summary
8.3 The Reciprocal Functions
Algebra Refresher
Three More Functions
Application to Right Triangles
Graphs of the Reciprocal Functions
Solving Equations
Using Identities
Section 8.3 Summary
Chapter 8 Summary and Review
Exercises: 8.1 Sum and Difference Formulas
Exercises: 8.2 Inverse Trigonometric Functions Exercises
Exercises: 8.3 The Reciprocal Functions Exercises
Exercises: Chapter 8 Review Problems
Chapter 9: Vectors
Introduction to Vectors
Introduction
9.1 Geometric Form
Skills Refresher
Notation for Vectors
Scalar Multiplication of Vectors
Addition of Vectors
Velocity
Components of a Vector
Using Components
Section 9.1 Summary
9.2 Coordinate Form
Skills Refresher
Unit Vectors
Converting between Geometric and Coordinate Form
Scalar Multiples of Vectors in Coordinate Form
Adding Vectors in Coordinate Form
Force
Section 9.2 Summary
9.3 The Dot Product
Skills Refresher
Components
Coordinate Form for Components
The Dot Product
Geometric Meaning of the Dot Product
Section 9.3 Summary
Chapter 9 Summary and Review
Exercises: 9.1 Geometric Form
Exercises: 9.2 Coordinate Form
Exercises: 9.3 The Dot Product
Exercises: Chapter 9 Review Problems
Chapter 10: Polar Coordinates and Complex Numbers
Introduction to Polar Coordinates and Complex Numbers
Introduction
10.1 Polar Coordinates
Algebra Refresher
Plotting Points
Regions in the Plane
Converting between Polar and Cartesian Coordinates
Equations in Polar Coordinates
Section 10.1 Summary
10.2 Polar Graphs
Algebra Refresher
Graphing in Polar Coordinates
Sketching Familiar Equations
Finding Intersection Points
Section 10.2 Summary
10.3 Complex Numbers
Algebra Refresher
Imaginary Numbers
Complex Numbers
Arithmetic of Complex Numbers
Products of Complex Numbers
Quotients of Complex Numbers
Graphing Complex Numbers
Zeros of Polynomials
Section 10.3 Summary
10.4 Polar Form for Complex Numbers
Algebra Refresher
Polar Form
Products and Quotients in Polar Form
Powers and Roots of Complex Numbers
Section 10.4 Summary
Chapter 10 Summary and Review
Exercises: 10.1 Polar Coordinates
Exercises: 10.2 Polar Graphs
Exercises: 10.3 Complex Numbers
Exercises: 10.4 Polar Form of Complex Numbers
Exercises: Chapter 10 Review Problems
Glossary
Summary of Adaptations
Full Resource Adaptations
Chapter-Specific Changes
Trigonometry
Chapter 9: Vectors
9.3 The Dot Product
Skills Refresher
1.
Find the height h of the triangle.
How far is the foot of the altitude from the vertex of the 18° angle?
2.
Find the height h of the triangle.
Find the length of the third side of the triangle.
3.
How far north is the tower from the airport? How far east?
What is the distance from A to P?
4.
Find the distance from A to C.
How far north is point C from point B?
―
Skills Refresher Answers
1.
1.55
4.76
2.
19.57
81.83
3.
36.92 mi, 15.63 mi
16.08 mi
4.
21.87
5.95
Learning Objectives
Find the component of w in the direction of v.
Compute the dot product.
Find the angle between two vectors.
Resolve a vector into components in given directions.
Components
We have seen that it can be useful to resolve a vector into horizontal and vertical components. We can also break a vector into components that point in other directions.
Imagine the following experiment: Delbert holds a ball at shoulder height and then drops it so that it falls to the ground (Fig b.) Francine holds a ball at shoulder height on an inclined ramp, then releases it so that it rolls downhill (Fig a.) Which ball will reach the ground first?
Although gravity causes both balls to speed up, the free-falling ball will reach the ground first. The force of gravity pulls straight down, the same direction as the motion of the free-falling ball, but the rolling ball must move at an angle to the pull of gravity, along the surface of the ramp. Only part of the gravitational force accelerates the rolling ball, and the rest of the force is counteracted by the surface of the ramp. What fraction of the gravitational force causes the ball to roll?
In figure (a), the gravitational force F is resolved into the sum of two vectors, F=u+v, where v points down the ramp, and u is perpendicular to the ramp. The magnitude of v is called the component of F in the direction of motion, and is denoted by comp v F. This is the portion of the gravitational force that moves the ball. From figure (b), we see that comp v F=∥F∥cosθ, where θ is the angle between F and v.
Component of a Vector.
The component of a vector F in the direction of vector v is
comp v F=∥F∥cosθ, where θ is the angle between F and v.
With a little geometry, you can verify that in this example the angle θ is the complement of the angle of inclination of the ramp, α. (Think of similar triangles.) Now suppose that we increase the angle of inclination. As α increases, θ decreases, cosθ increases, and hence comp v F increases. This result agrees with our experience: as the ramp gets steeper, the ball rolls faster.
Example 9.40.
Delbert uses a sheet of plywood as a ramp for his wheelbarrow. The ramp is inclined at an angle of 15° to the horizontal. The plywood can support a maximum weight of 200 pounds. What weight will the ramp support?
Solution
The component of the weight perpendicular to the ramp cannot be more than 200 pounds. If the weight of the wheelbarrow is w, and v is a vector perpendicular to the ramp, as shown below, then the angle between v and w is 15°, by similar triangles. Thus, the component of w in the direction of w is
comp v w=∥w∥cos15°=200∥w∥(0.9659)=200∥w∥=207.06
The ramp can support no more than about 207 pounds.
Checkpoint 9.41.
Delbert pushes a lawn mower whose handle makes a 40° angle with the horizontal. If he applies a force of 30 pounds in the direction of the handle, what is the component of the force in the horizontal direction?
Solution
23 pounds
Coordinate Form for Components
In the examples above, we computed the component of a force F in the direction of a vector v by knowing the angle between F and v. If the vectors are given in coordinate form (that is, v=a i+b j), we may not know the angle between them. Can we compute the component of a vector w in the direction of v, in terms of the coordinates of v and w?
Suppose v=a i+b j and w=c i+d j, as shown below. We will need to compute the cosine of θ in terms of a,b,c, and d.
First verify that
cosα=a∥v∥and sinα=b∥v∥cosβ=c∥w∥and sinβ=d∥w∥
Because θ=β=α, we use the subtraction formula for cosine.
cosθ=cosα cosβ+sinα sinβ=a∥v∥c∥w∥+b∥v∥d∥w∥=1∥v∥∥w∥(a c+b d)
And finally,
comp v w=∥w∥cosθ=a c+b d∥v∥
Now we have a formula for the component of a vector w in the direction of a vector v.
Component of a Vector in Coordinate Form.
If v=a i+b j and w=c i+d j, then the component of w in the direction of v is given by
comp v w=a c+b d∥v∥
Caution 9.42.
Recall that the “component” of a vector is a scalar. If a vector w is resolved into two component vectors, w=u+v, its components are the lengths of the vectors u and v. The vectors u and v themselves are called the vector components of w.
Example 9.43.
Compute the component of w=6 i+2 j in the direction of v=4 i+3 j.
Find the vector component of w in the direction of v.
Find the vector component of w perpendicular to v.
Solution
1. We use the formula derived above.comp v w=a c+b d∥v∥=4(6)+3(2)4 2+3 2=30 5=6 Remember that the component of w in the direction of v is a scalar; it is the length of the projection of w onto v.
2. We would like to find a vector q of length 6 in the direction of v, as shown at right. Because ∥v∥=5, we scale the components of v by 6 5 to get q=6 5(4 i+3 j)=24 5 i+18 5 j
3. Because w=p+q, where p is perpendicular to q, we have p=w−q=(6 i+2 j)−(24 5 i+18 5 j)=6 5 i−8 5 j
Checkpoint 9.44.
Write the vector w=6 i+2 j as the sum of two components, one parallel to v=i+j and the other perpendicular to v.
Solution
w=(4 i+4 j)+(2 i−2 j)
The Dot Product
The expression a c+b d, which we encountered above as part of the formula for comp v w, is quite useful and is given a name; it is called the dot product of the vectors v=a i+b j and w=c i+d j.
It is easy to remember the formula for the dot product if we think of adding the product of the i-components and the product of the j-components of the two vectors.
Dot Product.
The dot product of two vectors v=v 1 i+v 2 j and w=w 1 i+w 2 j is the scalar
v⋅w=v 1 w 1+v 2 w 2
Note 9.45.
From the formula above, you can see that the dot product is commutative. That is,
v⋅w=w⋅v
You can check that this is true in the following example.
Example 9.46.
Compute the dot product of v=5 i−3 j and w=4 i+j.
Solution
We apply the formula above to find
v⋅w=5(4)+(−3)(1)=17
In the examples above, you can see that the dot product of two vectors is a scalar. For this reason, the dot product is also called the scalar product.
Checkpoint 9.47.
Compute the dot product of v=6 i+2 j and w=−2 i+3 j.
Solution
−6
We can now write the formula for comp v w using the dot product.
comp v w=∥w∥cosθ=a c+b d∥v∥=v⋅w∥v∥
We have derived an alternate formula for a component of a vector.
Component of a Vector.
The component of w in the direction of v is the scalar
comp v w=v⋅w∥v∥
Example 9.48.
Compute the component of v=5 i−3 j in the direction of w=4 i+j.
Solution
Using the formula above, we have comp w v=v⋅w∥w∥. We first compute v⋅w and ∥w∥.
v⋅w=5(4)+(−3)(1)=17 and∥w∥=4 2+1 2=17
Thus,
comp w v=v⋅w∥w∥=17 17=17
The length of the projection of v in the direction of w is 17 units, as shown at right.
Checkpoint 9.49.
Compute the component of u=2 i+3 j in the direction of v=6 i+5 j.
Solution
27 61
Geometric Meaning of the Dot Product
An even more important relationship, which gives geometric meaning to the dot product, follows from the formula for a component. We now have two ways to compute the component of w in the direction of v:
comp v w=∥w∥cosθ and comp v w=v⋅w∥v∥
Equating these two expressions, we find
∥w∥cosθ=v⋅w∥v∥
or ∥v∥∥w∥cosθ=v⋅w. This is a geometric formula for the dot product.
Geometric Form for the Dot Product.
The dot product of two vectors v and w is the scalar
v⋅w=∥v∥∥w∥cosθ
where θ is the angle between the vectors.
The dot product is a way of multiplying two vectors that depends on the angle between them.
If θ=0°, so that v and w point in the same direction, then cosθ=1 and v⋅w is just the product of their lengths, ∥v∥∥w∥.
If v and w are perpendicular, then cosθ=0, so v⋅w=0. (Two vectors v and w are said to be orthogonal if their dot product is zero.)
If θ is between 0° and 90°, the dot product multiplies the length of v times the component of w in the direction of v.
Example 9.50.
Show that the vectors v=2 i+6 j and w=−9 i+3 j are orthogonal.
Solution
We compute the dot product of the vectors.
v⋅w=2(−9)+6(3)=0
Thus, ∥v∥∥w∥cosθ=0, so v and w are orthogonal. Because neither ∥v∥=0 nor ∥w∥=0, it must be the case that cosθ=0, so θ=90° or 270°.
Checkpoint 9.51.
Show that the vectors v=a i+b j and w=−b i+a j are orthogonal.
Find a vector w perpendicular to v=−3 i−5 j.
Solution
1. u⋅v=−a b+a b=0
2. 5 i−3 j
Using the dot product, we can find the angle between two vectors.
Angle Between Two Vectors.
The angle θ between two vectors v and w is given by
cosθ=v⋅w∥v∥∥w∥
Example 9.52.
Find the angle between the vectors u=−3 i+2 j and v=5 i+3 j.
Solution
We first compute u⋅v and the magnitude of each vector.
u⋅v=−3(5)+2(3)=−9∥u∥=(−3)2+2 2=13∥v∥=5 2+3 2=34
Then we apply the formula for cosθ.
cosθ=v⋅w∥v∥∥w∥=−9 13 34=−0.4281 θ=cos−1(−0.4281)=115.35°
The angle between the vectors is 115.35°.
Checkpoint 9.53.
Find the angle between the vectors u=4 i−6 j and v=2 i+8 j.
Solution
132.27°
Section 9.3 Summary
Vocabulary
Dot product
Scalar product
Orthogonal
Concepts
Dot Product (Coordinate Formula).
The dot product of two vectors v=v 1 i+v 2 j and w=w 1 i+w 2 j is the scalar
v⋅w=v 1 w 1+v 2 w 2
The dot product is a way of multiplying two vectors that depends on the angle between them.
Dot Product (Geometric Formula).
The dot product of two vectors v and w is the scalar
v⋅w=∥v∥∥w∥cosθ
where θ is the angle between the vectors.
The component of a vector w in the direction of vector v is the length of the vector projection of w onto w.
Component of a Vector.
The component of w in the direction of v is the scalar
comp v w=v⋅w∥v∥
Angle Between Two Vectors.
The angle θ between two vectors v and w is given by
cosθ=v⋅w∥v∥∥w∥
Two vectors v and w are orthogonal if v⋅w=0
Study Questions
If u and v have the same direction, what is comp u v?
If u is perpendicular to v, what is comp u v?
What is the angle between u and v that makes their dot product as large as possible?
What does the dot product of two unit vectors tell you?
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Next: Chapter 9 Summary and Review
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Lippincott® Illustrated Reviews: Biochemistry
Edition:9
9781975220495
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Author(s) : Emine Ercikan Abali , Susan D. Cline , David S. Franklin , Susan M. Viselli Ph.D.
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2025-02-25
9781975220495
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7151 | https://math.gordon.edu/ntic/ntic2021/section-two-gcd-facts.html | NTIC Two facts from the gcd
Skip to main content
Number Theory:In Context and Interactive
Karl-Dieter Crisman
Contents
IndexPrevUpNext
ContentsPrevUpNext
Front Matter
Colophon
About the Author
Dedication
Acknowledgements
To Everyone
To the Student
To the Instructor
1 Prologue
A First Problem
Review of Previous Ideas
Where are we going?
Exercises
Using Sage for Interactive Computation
2 Basic Integer Division
The Division Algorithm
The Greatest Common Divisor
The Euclidean Algorithm
The Bezout Identity
Exercises
3 From Linear Equations to Geometry
Linear Diophantine Equations
Geometry of Equations
Positive Integer Lattice Points
Pythagorean Triples
Surprises in Integer Equations
Exercises
Two facts from the gcd
4 First Steps with Congruence
Introduction to Congruence
Going Modulo First
Properties of Congruence
Equivalence classes
Why modular arithmetic matters
Toward Congruences
Exercises
5 Linear Congruences
Solving Linear Congruences
A Strategy For the First Solution
Systems of Linear Congruences
Using the Chinese Remainder Theorem
More Complicated Cases
Exercises
6 Prime Time
Introduction to Primes
To Infinity and Beyond
The Fundamental Theorem of Arithmetic
First consequences of the FTA
Applications to Congruences
Exercises
7 First Steps With General Congruences
Exploring Patterns in Square Roots
From Linear to General
Congruences as Solutions to Congruences
Polynomials and Lagrange's Theorem
Wilson's Theorem and Fermat's Theorem
Epilogue: Why Congruences Matter
Exercises
Counting Proofs of Congruences
8 The Group of Integers Modulo n n
The Integers Modulo n n
Powers
Essential Group Facts for Number Theory
Exercises
9 The Group of Units and Euler's Function
Groups and Number Systems
The Euler Phi Function
Using Euler's Theorem
Exploring Euler's Function
Proofs and Reasons
Exercises
10 Primitive Roots
Primitive Roots
A Better Way to Primitive Roots
When Does a Primitive Root Exist?
Prime Numbers Have Primitive Roots
A Practical Use of Primitive Roots
Exercises
11 An Introduction to Cryptography
What is Cryptography?
Encryption
A Modular Exponentiation Cipher
An Interesting Application: Key Exchange
RSA Public Key
RSA and (Lack Of) Security
Other applications
Exercises
12 Some Theory Behind Cryptography
Finding More Primes
Primes – Probably
Another Primality Test
Strong Pseudoprimes
Introduction to Factorization
A Taste of Modernity
Exercises
13 Sums of Squares
Some First Ideas
At Most One Way For Primes
A Lemma About Square Roots Modulo n n
Primes as Sum of Squares
All the Squares Fit to be Summed
A One-Sentence Proof
Exercises
14 Beyond Sums of Squares
A Complex Situation
More Sums of Squares and Beyond
Related Questions About Sums
Exercises
15 Points on Curves
Rational Points on Conics
A tempting cubic interlude
Bachet and Mordell Curves
Points on Quadratic Curves
Making More and More and More Points
The Algebraic Story
Exercises
16 Solving Quadratic Congruences
Square Roots
General Quadratic Congruences
Quadratic Residues
Send in the Groups
Euler's Criterion
Introducing the Legendre Symbol
Our First Full Computation
Exercises
17 Quadratic Reciprocity
More Legendre Symbols
Another Criterion
Using Eisenstein's Criterion
Quadratic Reciprocity
Some Surprising Applications of QR
A Proof of Quadratic Reciprocity
Exercises
18 An Introduction to Functions
Three Questions for Euler phi
Three Questions, Again
Exercises
19 Counting and Summing Divisors
Exploring a New Sequence of Functions
Conjectures and Proofs
The Size of the Sum of Divisors Function
Perfect Numbers
Odd Perfect Numbers
Exercises
20 Long-Term Function Behavior
Sums of Squares, Once More
Average of Tau
Digging Deeper and Finding Limits
Heuristics for the Sum of Divisors
Looking Ahead
Exercises
21 The Prime Counting Function
First Steps
Some History
The Prime Number Theorem
A Slice of the Prime Number Theorem
Exercises
22 More on Prime Numbers
Prime Races
Sequences and Primes
Types of Primes
Exercises
23 New Functions from Old
The Moebius Function
Inverting Functions
Making New Functions
Generalizing Moebius
Exercises
24 Infinite Sums and Products
Products and Sums
The Riemann Zeta Function
From Riemann to Dirichlet and Euler
Multiplication
Multiplication and Inverses
Four Facts
Exercises
25 Further Up and Further In
Taking the PNT Further
Improving the PNT
Toward the Riemann Hypothesis
Connecting to the Primes
Connecting to Zeta
Connecting to Zeros
The Riemann Explicit Formula
Epilogue
Exercises
Back Matter
A List of Sage notes
B List of Historical Remarks
C Notation
D List of Figures
E References and Further Resources
Introduction to the References
General References
Proof and Programming References
Specialized References
Historical References
Other References
Useful Articles
Index
Authored in PreTeXt
🔗
Section 3.7 Two facts from the gcd
🔗
Here are two facts that seem really obvious but do need proofs. All can be done just with the gcd, using no facts about primes from Chapter 6 as would typically be done. Kudos go to users Math Gems and coffeemath at math.stackexchange.com for most of these clever arguments. See this question for Proposition 3.7.1 and this question for Proposition 3.7.2.
🔗
Proposition 3.7.1.When perfect squares divide each other.
For integers a,z a,z it is true that
a 2∣z 2⟹a∣z a 2∣z 2⟹a∣z
🔗
###### Proof.
First, let (d=\gcd(a,z)\text{.}) Then we can write (z^2=a^2 \cdot k) for some integer (k\text{,}) and immediately write
\begin{equation} (z')^2 d^2=(a')^2 d^2 k \end{equation}
for some integers (z') and (a'\text{,}) by definition of gcd. (That is, (z=z'd) and (a=a'd\text{.}) Also note that (z',a') are now relatively prime; it is not hard to prove using the techniques of the previous chapter, or see Exercise 6.6.7.)
Cancelling the (d^2) (yes, we do assume this property of integers) yields
\begin{equation} (z')^2=(a')^2 k\text{.} \end{equation}
Since (\gcd(a',z')=1\text{,}) we have (a'x+z'y=1) for some (x,y\in\mathbb{Z}\text{;}) now we substitute for (1) in (a'\cdot 1\cdot x) (!) to get
\begin{equation} a'(a' x+z'y)x+z'y=1 \end{equation}
Now we have that (a'^2 x^2+z'(a'xy+y)=1\text{,}) so that (\gcd((a')^2,z')=1) as well. But of course (a'\mid (z')^2\text{.}) Clearly if a positive number is a divisor, but their greatest common divisor is 1, then that number is going to have to be 1 by definition of divisors. So (a'=1\text{.}) (If (a') was negative, the same argument for (-a') shows (-a'=1\text{,}) so really (a'=\pm 1\text{.}))
Hence (a=a'd=\pm d\text{,}) which is a divisor of (z\text{,}) we have the desired result.
🔗
Proposition 3.7.2.When the product of coprime numbers is a square.
If we have integers m,n,j m,n,j such that m n=j 2 m n=j 2 and gcd(m,n)=1,gcd(m,n)=1, then m m and n n are also both perfect squares.
🔗
###### Proof.
First, we will need a general fact about gcds:
\begin{equation} \gcd(x,y)^2=\gcd(x^2,xy,y^2) \end{equation}
See Exercise 3.6.22.
We know that (1=\gcd(m,n)=\gcd(m,n,j)\text{,}) so
\begin{equation} m = m\cdot \gcd(m,n,j)=\gcd(m^2,mn,mj)=\gcd(m^2,j^2,mj) \end{equation}
Now we use the fact, so that
\begin{equation} m=\gcd(m,j)^2\text{.} \end{equation}
That's a perfect square.
The same argument with (n) and (j) yields (n = \gcd(n,j)^2\text{.})
🔗
(For more ‘traditional’ proofs, see Section 6.4.) |
7152 | https://www.youtube.com/watch?v=paQLJKtiAEE | How Symmetry works in Quantum Physics: Gauge Theory Simplified!
Arvin Ash
1080000 subscribers
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Description
230937 views
Posted: 18 Jun 2022
Signup for your FREE trial to Wondrium here:
BACKGROUND VIDEOS:
How All Fundamental Forces work:
All Fundamental Particles visualized:
Maxwell's Equations:
Quantum Electrodynamics (QED):
Quantum Chromodynamics (QCD):
CHAPTERS:
00:00 Symmetry - root of physics
01:31 What is symmetry?
03:24 Intro to Group Theory
06:04 Noether's Theorem
07:17 U(1) symmetry simplified
09:43 Dirac equation transformation
11:10 How QED comes from U(1) symmetry
12:47 U(1) SU(2) SU(3) explained simply
15:32 Symmetry is the foundation of the universe
15:54 Further study on Wondrium
SUMMARY:
If you ask a physicist, what is at the core of physics, you will hear symmetry. What is symmetry? Gauge theory explained simply.
Symmetry is about actions that don't change anything. If we take an equilateral triangle, and put a mirror from one corner to the middle of the opposite side, we will see that the whole triangle. This is a symmetry of the equilateral triangle. Similarly we can rotate the triangle by 120 degrees, and it will look identical to what it was before.
What we just did is a simple example of something more complex - group theory. Group theory is the math behind the symmetries.The mathematics behind the symmetries of the equilateral triangle is called the dihedral GROUP of degree 3, where 3 refers to the triangle having three corners. We can change the elements, or permutations, using two different operations, rotation, and reflection. These two operations are called generators. The result of applying a generator doesn’t change anything visible. This is symmetry.
Symmetries give us rules for how to transform something while conserving a quantity. For the triangle, that conserved quantity is its shape, and the generators are rotation and reflection.
This leads us to Noether’s theorem which states that “For every symmetry there is a corresponding conservation law.” This directly connects symmetries with conserved quantities.
What happens if we take the limit of a polygon with an infinite number of edges? We get a circle. A circle of some radius, r, can be described on a 2D plane using polar coordinates by two equations. If we use complex numbers to represent the circle, we can write it with just ONE equation. This allows us to write one complex equation that achieves the same mathematically as two real equations.
It turns out that there’s also a symmetry group associated with this circle of complex numbers with a radius or magnitude of 1. It is called the U(1) group. The elements of the group are all the infinite possible angles phi around the circle.
Quantum mechanics is built on complex numbers. We can apply the symmetry with the simple transformation of moving around the circle. Do described the movement of fermions, we can use the Dirac equation. It describes any matter particle, like an electron, with some mass m moving in space. It does not describe any forces.
If U(1) symmetry exists, it would mean that if we applied our transformation, the Lagrangian would not change. The problem is that the Lagrangian DOES change when we apply this transformation, so this tells us that no U(1) symmetry exists.
However, if we modify the equation, by adding a new quantum field to the theory, a gauge field, we can get a symmetry. Another name for a gauge field is a force. Our theory works, and obeys U(1) symmetry transformations if we add some new terms to the equation. It turns out that this new term describes the electromagnetic force. The entire theory of Quantum Electrodynamics can be derived by the new transformed equation.
So by taking a theory for fermions (Dirac equation) and demanding a U(1) transformation we got the theory of electromagnetism. Similarly, the standard model is constructed to respect three symmetries or special unitary groups. And each group leads to a symmetry resulting in a conservation law and a fundamental force.
The U(1) group gives us conservation of electric charge, and is associated with the electromagnetic force. The SU(2) group gives us conservation of weak isospin, or weak charge, and is associated with the weak force. The SU(3) group leads to conservation of color charge and is associated with the strong force. It leads to the theory of quantum chromodynamics.
In addition, the number of generators corresponds to the number of bosons involved with each force. U(1) has one generator and one photon. SU(2) has 3 generators and 3 W+, W-, and Z. SU(3) has 8 generators and 8 different gluons.
gaugesymmetry
grouptheory
noetherstheorem
Symmetries seem to be the foundation of the laws of physics. Why this is the case is something no one knows.
726 comments
Transcript:
Symmetry - root of physics This video is brought to you by Wondrium. You can sign up for free and support this channel by clicking the link the description.
Physicists study how the universe works at the most fundamental level. They’re like
Sherlock Holmes. They’ve just come to the scene of the crime which is the universe. And
they’re looking high and low trying to find clues that can tell them what happened, how
things came to be, but also how it all works. If you ask a physicist, what make the universe
tick? What’s at the core of physics theories? One of the common answers you’ll hear is
symmetry. Symmetry lies at the heart of physics, and it appears to be a fundamental property of
the universe. And it leads to something quite profound. You see, “Every symmetry leads
to a conserved quantity” – For example, it leads to the law of conservation of energy –
the idea that energy overall is neither created nor destroyed, it only changes form -- similarly,
conservation of momentum and electric charge. And it even leads to fundamental forces, such as the electromagentic force. So, what is symmetry? How does it play a role
in physics? How does it lead to some of the most fundamental rules the universe seems to obey?
The answer will probably shake your idea about what physics really is. And it’s coming up right now… What is symmetry? Symmetry is about actions
– actions that in some sense don’t change What is symmetry? anything. This might sound like a stupid
concept. If you don’t change anything, then what’s the point of doing anything?
Well, let’s look at that. You might have unknowingly studied symmetry in one
of your elementary school classes, for example if you played with mirrors and geometrical objects.
Consider an equilateral triangle. This is a triangle where all sides and
angles are equal. In this case we can put a mirror from one corner to the middle of
the opposite side. If we do this and look into the mirror, the reflection
of the image will complete the triangle. Each side is a mirror image of the other side.
This is a symmetry of the equilateral triangle! But we do more. There is another
operation we can do, which recovers the original triangle. Consider rotating the
triangle by 120 degrees. Again, the triangle will look identical to what it was before.
Now, the problem is that we can’t see the change. So to help us understand the symmetry of
this triangle, let us write the letter A in it. This will help us identify the permutations.
Let’s start from the beginning, the initial position is one permutation. Then we
can mirror it, this gives us another permutation. Then we can rotate by 120 degrees, this gives us
the 3rd permutation. Again, we can consider the reflection giving us the 4th permutation.
We can rotation with 120 degrees again to get the 5th permutation and consider the
reflection to get the 6th permutation. Now note the next time we rotate with
120 degrees we end up where we started. Conclusively there are 6 permutations we can do
with this triangle. Now you’re probably asking, why are you wasting time on elementary school math?
Well, let me reveal the secret of this exercise. Intro to Group Theory What we just talked about is a simple example of
something much more complex, namely group theory. Group theory is the math behind the symmetries.
The mathematics behind the symmetries of the equilateral triangle is called
the dihedral group of degree 3, where 3 refers to the triangle having three
corners. We can also say it’s of order 6 as there are 6 permutations, also called group elements.
As another example, if we considered the symmetries of a square, the group theory behind
it would be the dihedral group of degree 4. Going back to our triangle, notice how we
were able to change between the elements, or permutations, using two different
operations, rotation, and reflection. We can move between all elements using
combinations of these operations. We have thus two ways of describing this group.
Either by considering the list of all elements, or by the operation we performed, in this
case reflections or rotations. In mathematics we call the latter, “generators,” because from
the generators of the group we can produce all the elements of the group.
You might ask, well what’s the problem with having just a list of elements?
Well, it’s fine for a simple group. But consider how many elements there would be if we did this
for a hexagon, which is a more complex dihedral group of degree 6. Just like with the triangle,
the generators are reflections and rotations, though the rotation angle is now only 60 degrees.
The point is, it’s a lot shorter and simpler to describe a group in terms of the generators
than a potentially long list of elements. And to make matters worse. Groups can have an
infinite number of elements, making it impossible to describe them without generators.
Before moving on, let us consider the importance of group theory, because
this is where the magic begins. I said in the beginning that symmetry is about
actions that in some sense don’t change anything. What I meant with actions is applying the
generators to a group. And as we have seen, the result of applying a generator
doesn’t change anything visible. They key here is that this is not the same as
nothing changed. The object has been transformed, but it looks the same. This
is key; we can do something, but in the triangle, it holds its shape., It remains the same and in general we stay within the group elements. Noether's Theorem Overall, what I want to impart is that symmetries
give us rules for how to transform something while conserving a quantity.
In the case of the triangle, that conserved quantity is its shape, and the
transformation is rotation and reflection. This leads us to what is called Noether’s
theorem developed by Emmy Noether around 1915. This theorem states something profound
about nature. In its simplified form, it says that “For every symmetry there
is a corresponding conservation law.” This is huge! This directly connects
symmetries with conserved quantities. For example, time translation symmetry, which
is the idea that the laws of physics are the same today as they were yesterday, as they
will be tomorrow, leads to the principle of conservation of energy.
And space translation symmetry, which is the idea that the laws of physics
are the same here as they are 10 meters away, or on the moon or anywhere else in the universe,
leads to the law of conservation of momentum. Similarly the fundamental forces of
nature also arise from certain symmetries. So we looked at a triangles and
polygons earlier. But what happens U(1) symmetry simplified if we take the limit of having a polygon with
an infinite number of edges? We get a circle. A circle of some radius, r, can be described
on a 2D xy-plane using polar coordinates by these equations:
If we make the simplification that the radius is 1, we simply get the following,
Two equations which are functions of the angle ϕ. This is what we get using real numbers ,
but it limits us mathematically. If we allow ourselves to use complex numbers to represent the
circle, we can write it with just one equation: Where r is the radius, e is Euler’s number
representing a constant, and i is the square root of minus one. It’s called an imaginary number.
Since we chose the radius to be 1, we can simplify this to just this.
This allows us to write one complex equation that achieves the same
mathematically as two real equations. This single complex number Z can describe
any point on the circle of radius 1. All we had to do was change one axis to be
imaginary. So the circle is on a complex plane. This shows how complex numbers
can be powerful math tools. Notice the point on the circle is
completely determined by the angle phi. It turns out that there’s also a symmetry group
associated with this circle of complex numbers with a radius or magnitude of 1. It is called the
U(1) group. The elements of the group are all the infinite possible angles phi around the circle.
We can write the transformation of the U(1) group formally as follows,
Where n is the generator, meaning n represents how much we rotate around the circle.
Now, where have you seen complex numbers in physics? In quantum mechanics.
It’s built on complex numbers. Let’s try to apply the symmetry of just moving
around a circle with this simple transformation. To describe fermions, which are matter
particles, we need some equation of motion that describes their behavior. To do
this we can use the Dirac equation. This is a scary looking equation, but don’t be
intimidated by this. You don’t have to understand the math. I just want you to see the equation so
you can see conceptually what happens to the math when we do these transformations. L is the Lagrangian. A Lagrangian
is just the difference between Dirac equation transformation kinetic and potential energy of a particle.
Psi as the wavefunction for a matter particle, and psi bar is the wave function for
the equivalent antimatter particle. The wavefunction, if you recall, when squared
tells us the probability of finding the particle in a given location when measured.
The first part in the parenthesis describes the particle’s movement
in spacetime, and m is the mass. Ok so this equation describes some
matter particle, like an electron, with some mass m moving in space.
You might say, ok so what? I don’t see any forces here. This is where group theory comes in.
If U(1) symmetry exists, it would mean that if we applied our transformation this equation like so:
where n is -1 for matter particles, and 1 for antimatter particles, the
Lagrangian would not change . The problem is that when we do this, the Lagrangian DOES change, so this tells us that there's no U(1) symmetry,
and we wind up with a broken theory. However, if we modify the equation, by
adding a new quantum field to the theory, a so-called gauge field, we can get a symmetry.
Another name for a gauge field is a force. So when a force is added to the equation, we find
that a symmetry exists. Our theory then works, How QED comes from U(1) symmetry and obeys U(1) symmetry transformations.
So what is this force that we just added? It turns out that this new term that was added to
the equation describes the electromagnetic force. The “e” in this case is not euler’s number, it is
the electric charge, and A_μ represents the photon field. So this tells us that photons are the force
particles that mediate the electromagnetic force. So while we had electrons in the initial Dirac
equation with no interaction, what we have here is an equation with electron to electron
interactions, mediated by the photon field. The full equation looks like this:
This additional term in front contains what’s called the electromagnetic
field strength tensor, which is also defined from the photon field. This term
essentially contains all the Maxwell equations. Notice what happened here. We took a theory for
fermions and demanded a U(1) transformation. Adding the additional mathematical terms to make
the symmetry work, gave us the electromagnetic force mediated between fermions by photons.
This is the theory of electromagnetic force, also known as quantum electrodynamics, or QED.
We ended up with the entire theory for electromagnetism just by demanding
one relatively simple symmetry. Because of Noether’s theorem, we know something
is conserved. In this case of the electromagnetic force, electric charge is conserved.
As it turns out the standard model U(1) SU(2) SU(3) explained simply is constructed to respect these symmetries or special unitary groups, U(1), SU(2) and SU(3). Sometimes if you ask a mathematician, they will write the entire
standard model in terms of these groups. When scientists say they are
studying Group theory, they are basically studying the mathematical symmetry
transformations of objects or abstract things. It turns out each group leads to a
symmetry resulting in a conservation law and a fundamental force.
As we saw the U(1) group gives us conservation of electric charge. It’s a symmetry
associated with the electromagnetic force. The SU(2) group gives us conservation of
weak isospin, or weak charge. This symmetry is associated with the weak force.
And as you might guess. the SU(3) group leads to conservation of color charge and
is associated with the strong force. It leads to the theory of quantum chromodynamics.
In other words, the three symmetry groups that the standard model respects gives us three
of the fundamental forces of our universe. We can do a similar mathematical exercises as we did for
U(1) for the SU(2) and SU(3) groups, but the math is much more complex.
The U(1) group is like a one dimension sphere, or circle, but you need 2 dimensions to visualize
it because a circle transforms on a 2D plane. Similarly the SU(2) group is
like a 3 dimensional sphere that transforms in four dimensions,
which we can’t visualize And SU(3) is symmetry is even
more difficult to visualize, because it would be an eight dimensional sphere
that requires 9 dimensions to visualize it. Now, you might think this is pretty cool, but
it gets even cooler. Remember that the generators are like the rules for how to
transform, while respecting the symmetry? The U(1) group has 1 generator because
there’s just one rotation around a circle. Now ask yourself, how many bosons are responsible
for mediating the electromagnetic force, associated with this group? Well, it’s
just one single particle, the photon. It turns out, each generator of a gauge group
corresponds to a force mediating particle. So with the SU(2) group which results in
the weak force, there are 3 generators since there are three ways to transform a 3D sphere.
Well , it is mediated by a total of 3 bosons, the W+, W- and the Z bosons.
Likewise, the SU(3) group has 8 generators since we are dealing with an
8D sphere. So it should come as no surprise then that the strong force is mediated by 8 different
gluons of different combinations of color charges. Symmetry is the foundation of the universe Symmetries seem to be the foundation of the
laws of physics. Why this should be the case is something no one can really answer. Most
people subjectively find beauty in symmetries. Even people with more symmetrical
faces are considered more attractive. Maybe the universe itself
finds beauty in symmetry too. Further study on Wondrium If you want to go further into the details
of symmetry, and also how it is fundamentally at the root of physics, I highly recommend a
lecture available on Wondrium, today’s sponsor, called, “The Power of Symmetry.” It’s
part of a 12 chapter course called, “The Higss Boson and beyond” taught by someone you
might be familiar with because I interviewed him on this channel last year, and is one of my
favorite science educators, Sean Carroll. In an effort to explain our search for the Higgs Boson,
Sean reviews everything from quantum field theory, to particle physics, to symmetry.
It’s a must watch for physics enthusiasts. You’ll find Sean and many other fantastic educators on Wondrium. That’s why I have been a member of Wondrium for years. I couldn’t
recommend them more. And that’s why my testimonial is even at the bottom of Wondrium’s home page. It’s
really easy to sign up because they are offering a free trial, and you can cancel at anytime.
If you want in-depth college level knowledge about all the things I talked about on this channel,
then be sure to click the special link in the description to take advantage of the free offer.
That link is Wondrium dot com slash arvin, wondrium.com/arvin. And you’ll be supporting this
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sure to hit the subscribe button so you can be informed the minute I post a new video. And
I’d love to hear your comments or questions, so please leave them in the comment section.
I’ll see you in the next video my friend. |
7153 | https://edia.app/worksheets/algebra_1/systems_of_linear_equations | Numbers and counting
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Algebra I
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7154 | https://www.youtube.com/watch?v=YCM2JReWS10 | Octal to Decimal Conversion
The Organic Chemistry Tutor
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Posted: 23 May 2018
This video tutorial explains how to convert octal to decimal numbers. The octal system is a base 8 system while the decimal system is a base 10 system.
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Transcript:
in this video we're going to talk about how to convert an octo number into a decimal number so how can we do that well let's start with a simple example let's say we have 370 in the octal system so what we're going to do is we're going to multiply the zero by 8 to the 0 power and then we're going to multiply the 7 by 8 to the first Power and the 3 by 8 to the second power and we're going to add up all these numbers so this is going to be 3 times 8 squared plus 7 times 8 to the first power plus 0 times 8 to the 0 power 8 squared that's 8 times 8 which is 64. and 7 times 8 is 56. and zero times anything is zero now 3 times 64. that's 192. and then we have 192 Plus 56. and so two plus six is eight nine plus five is fourteen carry the one and so we get a total of 248 in the decimal system so that's how we can convert an octal number into a decimal number now let's try another example 1071 in the octal system let's convert that to a decimal number so the first one the one we're going to multiply that by 8 to 0. and then the seven we're going to multiply that by eight to the first Power and it's 0 by 8 squared and the one by eight to the third so this is going to be one times a to the third plus zero times H squared plus seven times a to the first power plus one times eight to the 0 power so what is a cubed so 8 times 8 times 8 that's going to be 512. 0 times 8 squared is zero 7 times 8 is 56 and 8 to the 0 power is one so one times one is one so we have 5 12 plus 56 plus one and that's going to be 569 in the base 10 system and so that's it for that one now what about this one sixteen thousand four hundred seventy-five and the octal system let's convert it to a decimal number in the base 10 system so the first number the 5 we're going to multiply by eighth to zero and the second number by a to the one and the third by eight squared and the fourth by 8 cubed and the last one by eight to the fourth so this is going to be one times a to the fourth plus six times eight cubed plus four times eight squared and then plus seven times eight to the first power and then plus five times eight to the zero power now 8 raised to the fourth power if we multiply 4 8 together eight times eight times eight times eight that's four thousand ninety six now eight to the Third we know it's 5 12. and then H squared is 64. 7 times 8 is 56. and I need to put a plus sign here and then 8 to the 0 is 1. so we have 4096 and then six times five twelve that's 3072 and 4 times 64 is 256 and then plus 56 Plus 5. so let's go ahead and add up these five numbers and so the sum comes out to 7485 in the base 10 system so this is the final answer now what if we have a fractional number let's say 425 .28 in the octal system how can we convert that to a base 10 number in the decibel system so we know that the 5 will need to multiply by eight to the zero and the two by eight to the one and four by eight squared so if we follow the pattern of exponents that we see here the next number has to be 8 to the negative 1. so the number to the right of the decimal we need to multiply by eight to the minus 1 and then the last number by eight to negative two so we're going to have 4 times 8 squared plus two times eight to the first power plus 5 times 8 to the 0 power plus two times eight and minus 1 and then plus eight times eight and minus two now we know that a squared is 64. and 2 times 8 is 16. and eight to the zero is one five times one is five now what about 2 times 8 to the minus 1. 8 to the negative 1 is the same as one over eight and so therefore this expression becomes two times one over eight or we could say it's basically two divided by eight now for this expression 8 to the negative 2 is 1 over 8 squared and so what we have is eight divided by 8 squared which I'll write that here so 4 times 64. that's 256. now we can add 16 plus 5. that's going to be 21. and then 2 over 8 or 2 divided by 8 that's 0.25 and then 8 divided by 8 squared or 8 divided by 64. that's point one two five now let's add up the numbers so we have 256 plus 21 plus 0.25 plus 0.125 and so our answer is going to be 277 point three seven five in the base 10 system and so now you know how to convert a fractional octal number into a decimal number thanks for watching |
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The fundamentals of security risk measurements
2.4 The exponential function ex
For some reason certain numbers and expressions occur more frequently than others in nature. One of those is the number “e” and is defined by the following expression (written to only four decimal places, e is a so-called irrational number and does not have a finite decimal representation), e = (1 + 1/n)n = 2.7183 … as n approaches infinity.
The exponential function is defined as f(x) = ex where x is some arbitrary exponent. This function appears in many contexts in the natural world and is the solution to a number of equations related to physical processes that characterize risk. The logarithm in base e has a special designation and is referred to as the “natural logarithm.” This does not imply that the regular logarithm is in some way unnatural. The natural logarithm is abbreviated as “ln.” We now know from the previous discussion on logarithms that ln (ex) = x. This is because the natural logarithm undoes exponentiation for functions where e is the base in exactly the same way that regular logarithms do for expressions in base 10 or any other base.
Figure 2.6 shows a plot of y = ex where the exponent varies from x = 1 to x = 5. It is immediately obvious from the graph that this is a nonlinear function. The exponential function is a solution to a common and relatively simple differential equation that governs a number of processes relevant to physical security. This statement applies equally to the decreasing exponential function y = e−x depicted in Figure 2.7. Because of its general applicability to physical processes related to security, this function will appear throughout the book. Understanding the general behavior of the exponential function can lead to important insights into the vulnerability component of risk.
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Book2010, Metrics and Methods for Security Risk ManagementCarl S. Young
Chapter
Ordinary Special Functions
2006, Encyclopedia of Mathematical PhysicsW. Van Assche
Introduction
The exponential function, the logarithm, the trigonometric functions, and various other functions are often used in mathematics and physics. They are transcendental functions in the sense that they cannot be obtained by a finite number of operations as a solution of an algebraic (polynomial) equation. Typically, they are obtained by a Taylor series expansion. Many other higher transcendental functions arise in mathematical physics, often as solutions of differential equations. A precise knowledge of the behavior of such functions, their relation with other functions, addition, multiplication and composition properties, representations as an infinite series, or as an integral, often shed a lot of light onto the problem in which they arise. If they are sufficiently useful to a large audience, then they usually get a name and they will be called special functions. In what follows, we describe a few of these special functions of one variable, but clearly this is just a tip of the iceberg. Many other special functions exist and we refer to the classical tables of Abramowitz and Stegun (1964) and the Bateman manuscript project (Erdélyi et al. 1953–55) for more special functions. Nowadays, there have been numerous q-extensions of special functions (see q-Special Functions).
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Chapter
The exponential function
2003, Mathematics for Electrical Engineering and ComputingMary Attenborough
8.6 Summary
1.
: Many physical situations involve exponential growth or decay where the rate of change of y is proportional to its current value.
2.
: All exponential functions, y = at, are such that d y/dt = ky, that is, the derivative of an exponential function is also an exponential function scaled by a factor k.
3.
: The exponential function y = ethas the property that dy/dt = y, that is, its derivative is equal to the original function:
where e ≈ 2.71828. The inverse function to et is loge(t), which is abbreviated to In(t). This is called the natural or Napierian logarithm.
4.
: The general solution to d y/dt = ky is y = y0ekt, where y0 is the value of y at t = 0
5.
6.
: The hyperbolic cosine (cosh) and hyperbolic sine (sinh) are the even and odd parts of the exponential function:
These functions get the name hyperbolic because of their relationship to a hyperbola. The hyperbolic tangent is defined byThere are various hyperbolic identities, which are similar to the trigonometric identities (Table 8.2).
7.
: The inverse hyperbolic functions cosh−1(x)(x ≥ 1), sinh−1(x), tanh −1 (x) (−1 < x < 1) have the following logarithmic identities:
cosh−1 (x) is the inverse of cosh(x) if the domain of cosh(x) is limited to the positive values of x and zero.
8.
: Adding the derivatives and integrals of the exponential, In, hyperbolic and inverse hyperbolic functions to the tables of standard derivatives and integrals gives Tables 8.3 and 8.4.
9.
: Partial fractions can be used to integrate fractional functions such as
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Chapter
Deep Architectures
2018, Machine LearningMarco Gori
5.1.2.5 Exponential Functions
Finally, let us consider the case the exponential functions . Like in the previous cases, there is no such that . This can promptly be seen when considering that this equation is equivalent to . This holds true in the general case of computation in , which corresponds with real exponentiation and with sinusoidal functions in the case of values in . More details on this issue are covered in Exercise 6.
From this preliminary analysis it is clear that the cascading of units typically enlarges the space of functions. Interestingly, the way the space is enriched very much depends on the choice of σ. Exercise 8 proposes a nice example that involves polynomials, where we can clearly see the limitation of the constructed space.
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Chapter
Modeling Information Security Risk
2016, Information Security ScienceCarl S. Young
The Exponential, e, and the Functions ex and e−x
The exponential “e” is known as “Euler’s Number.” It is defined as 2.7183.3 From the discussion on exponents it is apparent that e1 = 2.7183, e2 = 7.389, e3 = 20.086, etc. The exponential function should not be confused with exponents which were discussed earlier in this chapter.
Many processes can be modeled using the exponential function since it is the solution to a number of differential equations. The simplest of these are dx/dt = ax and dx/dt = −ax.4 The solutions to these equations are xo eat and xo e−at, respectively, where xo is the value of x at time t = 0. Graphs of these two functions are presented in Figs. 2.5 and 2.6. Specific examples of the applicability of exponentials to security scenarios will be provided as they arise in the text.
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Chapter
Linear Circuit Analysis
2005, The Electrical Engineering HandbookP.K. Rajan, Arun Sekar
1.9.2 Complex Exponential Function
Define as a complex exponential function. By Euler's theorem,
(1.69)
The term is called the phasor of the sinusoidal function x(t). For linear RLCM circuits, the forced response is sinusoidal at the input frequency. Since the natural response decays exponentially in time, the forced response is also the steady state response.
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Chapter
Relativistic Electronic Structure Theory
2002, Theoretical and Computational ChemistryD. Sundholm
3.1 ERA
The exponential regular approximation (ERA) Hamiltonian can be obtained from equation (20) by inserting the exponential function defined in equation (9) into the general ansatz for the small component (6). One important advantage with the exponential regular approximation is that the exponential function declines much faster with r than the ZORA ansatz. Therefore, the exponential ansatz function does not cause the same complications as one experiences when performing molecular structure optimizations using the ZORA Hamiltonian . It does not possess any long-range behaviour that causes difficulties in the calculation of the molecular gradients. In addition, the ansatz does not contain nuclear attraction potentials or other terms that introduce gauge dependencies. The exponent in equation (9) is for all atoms so large that the function f(r) is for typical bond distances practically equal to one.
Since most quantum chemistry program packages employ Gaussian basis functions, one would actually prefer to use Gaussian functions in the ansatz. However, Gaussian functions do not have the proper shape at the nucleus. Their first derivative is zero at the origin, and one single Gaussian function cannot be used in the ansatz. Gaussian functions could eventually be used in the evaluation of the relativistic correction integrals appearing with the ERA ansatz. This has not been tested since the methods have been implemented into a fully numerical program.
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Chapter
Signal Analysis and Measurement
2001, The Laboratory ComputerJOHN DEMPSTER
6.6.7 Discriminating between competing models
Although curve-fitting procedures can determine the best-fit parameters for a given mathematical model, they do not directly determine whether a better model might exist. One of the most common situations where choices have to be made between models occurs when analysing signals which exhibit multi-exponential time courses. Exponential models are generally chosen because they can be related to the theoretical bases of physiological signals under study, but usually there is no a priori knowledge of the number of exponential components to be expected. In fact, determining the number of exponentials needed to fit the data may be one reason why the experiments are being done. Debates have arisen in the past as to the appropriate number of exponential components required to describe experimentally observed signals. For instance, the number of exponentials (two or three) required to describe the decay of endplate currents in the presence of certain drugs has been questioned (Ruff, 1977; Beam, 1976). Such differences may be due to real differences in the experimental procedures, tissues or cells used by the particular experimenters, but it may also be due to differences in the curve-fitting procedures used and the criteria used to discriminate between competing models. Clearly, quantitative criteria are required to compare the relative merits of different models.
For example, returning to the signal with a two-exponential decay which was discussed earlier (Fig. 6.19), we know that this signal has a two-exponential decay because it is a simulated record. However, the aim here is to work backwards to the same conclusion using only the evidence derived from the curve fitting. The first step is the fit of a series of exponential functions of increasing order. A family of exponential functions can be defined as
[6.52]
Figure 6.21 shows the fits of the first three of these functions, a single exponential (a), the sum of two (b) and three (c) exponential components. The parameter estimates are shown in the table.
A model can be rejected because it does not fit the data or because another model fits it significantly better. This decision can be based upon the residual standard deviation or on the randomness within the distribution of residuals. The single-exponential model in our example, for instance, can be rejected on both of these counts. Firstly, the distribution of residuals is clearly non-random, indicating that the fitted curve underestimates the data both at the beginning and end of the fitted region and overestimates it in the middle. The residual standard deviation (0.0427) is also almost twice as large for the two- and three-exponential fits (0.0274, 0.0267).
The degree of randomness within a residual distribution can be quantified using the runs test. A run is defined as a series of residuals of the same sign. If the model provides a good fit to the data, the residuals should be distributed randomly, resulting in a large numbers of short runs, no more than a few data points in length. On the other hand, poor fits should result in a small number of long runs. The number of discrete runs, U (positive or negative), observed within the set of residuals is used as the test statistic. If the numbers of positive and negative residual values, n+ and n-, are both greater than 10, the expected number of runs that ought to be observed if the distribution is random is
[6.53]
with a standard deviation of
[6.54]
(Rawlings, 1988). The probability of observing U or more runs, can then be obtained by computing the probability p(z ≤ zU) where z is a normally distributed variable and zU is
[6.55]
The value p(z ≤ zU) can be obtained from the normal probability distribution tables found in most statistical textbooks, or from a statistical function within a spreadsheet program (e.g. the FNORM function in Microsoft Excel).
Applying the runs test to the residuals of the three different exponential fits confirms the decision to eliminate the one-exponential model. A total of 77 runs are observed for that model where 109 would have been expected if the distribution had been random. The probability of this occurring by chance is less than one in a million. By contrast, the numbers of runs observed in the residuals associated with the two- and three-exponential fits are close to the expected values (128 vs. 129, p = 0.42; 127 vs. 129, p = 0.36). It is worth noting, however, that the runs test is based upon the assumption that the residuals are independent of each other. This may not always be the case, especially when the data points are obtained from a signal record which has been low-pass filtered. In such circumstances, the runs test should be used with caution.
Discriminating between the two- and three-exponential functions is more difficult. Both appear to fit the data well, with a random distribution of residuals. Both have fairly similar residual standard deviations, with the three-exponential model producing the lowest value. This does not, however, automatically mean that three exponential components are actually required. The flexibility gained by adding the extra parameters to the equation would be expected to improve the fit to some extent, even though no such component actually existed. A comparison of the parameter estimates provides some evidence that this is the case. The parameter standard errors are substantially greater for the three-exponential fit, as might be expected to happen if the contributions of two actual well-defined components to the signal had to be split between three. The fact that the amplitude of one of the components (A2) is negative also supports this.
It is useful, however, to have some objective
Parameter estimates and statistics for each model
| | | | | | | |
--- --- ---
| Empty Cell | A0 (mV) | τ0 (ms) | A1 (mV) | τt (ms) | A2(mV) | τ2 (ms) |
| True | 8 | 2 | 2 | 10 | − | − |
| Best-fit: | | | | | | |
| (a) | 9.10±0.085 | 3.23±0.041 | − | − | − | − |
| (b) | 8.49±0.09 | 1.94±0.035 | 1.88±0.18 | 9.88±0.59 | − | − |
| (c) | 9.52±7.56 | 1.62± 0.31 | 2.25±1.29 | 8.89± 3.25 | −2.25±1.15 | 0.64±0.43 |
| | | | | | | |
--- --- ---
| Empty Cell | F-test | | | | Runs test | |
| Empty Cell | σres | Fn,n+1 | p(F ≤ Fn,n+1) | Ut | E(U) | p(U ≤ Ut) |
| True | − | − | − | − | − | − |
| Best-fit: | | | | | | |
| (a) | 0.0427 | − | − | 77 | 109 | 7.28 × 10−7 |
| (b) | 0.02741 | 35.3 | 0.0015 | 128 | 129 | 0.42 |
| (c) | 0.0261 | 2.10 | 0.17 | 127 | 129 | 0.36 |
criterion for choosing the two- rather than three-exponential model. As we have seen, adding extra parameters will usually reduce the residual standard deviation to some degree. The key question is whether that improvement is significant, given the additional flexibility added by the extra parameters. The exponential functions discussed here can be described as a set of nested models, in that the one- and two-exponential functions are subsets of the three-exponential function. For such models a variance ratio test can be applied to determine whether the residual sums of squares yielded by a pair of models are significantly different. For two models (a,b) from a nested set, where a is a subset of b, the ratio is computed as
[6.56]
where SSQa and SSQb are the residual sums of squares, ma and mb the numbers of parameters for each model, and n is the number points in the data set (Horn, 1987). The significance probability is determined from the F probability distribution with ma and n – mb degrees of freedom (like the normal, the F distribution can be obtained in statistical tables, or as a spreadsheet function). Computing the ratios F1,2 and F2,3 between the three models yields values of 35.3 and 2.1, respectively, with probabilities 0.0015 and 0.17 of observing these ratios purely by chance when no significant improvement in the fit has occurred. The two-exponential can thus be clearly seen to be a significantly better fit than the single, but three exponentials are not significantly better than two.
Again, it is important to emphasise that too much weight should not be given to the results of a single experiment. It is only when consistent results, using the above techniques, have been obtained from a series of experimental trials, and real parameter standard errors computed, that a confident choice of model can be made. Unfortunately, equation [6.56] is not applicable to nonnested models which cannot be simply expressed as a subsets of each other. Therefore it does not provide a universal means of discriminating between competing models. Discussion of some approaches that can be used in these circumstances can be found in Horn (1987), Rao (1973), or Leamer (1983).
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Mini review
Activation functions in deep learning: A comprehensive survey and benchmark
2022, NeurocomputingShiv Ram Dubey, ... Bidyut Baran Chaudhuri
5 Exponential Activation Functions
The exponential AFs tackle the gradient diminishing problem of ReLU. Table 4 lists the properties of the exponential AFs. The Exponential Linear Unit (ELU) is given as,
Table 4. Summary of Exponential Linear Unit based activation functions.
| Name | Parametric | Monotonic | Smooth | Bounded |
--- ---
| Exponential Linear Unit (ELU), 2016 | Yes | Yes | Yes | For negative inputs |
| Scaled ELU (SELU), 2017 | Yes | Yes | Yes | For negative inputs |
| Continuously Differentiable ELU (CELU), 2017 | Yes | Yes | No | For negative inputs |
| Parametric ELU (PELU), 2017 | Yes | Yes | No | For negative inputs |
| Multiple PELU (MPELU), 2018 | Yes | Yes | No | For negative inputs |
| Fast ELU (FELU), 2019 | Yes | Yes | No | For negative inputs |
| Parametric Rectified Exponential Unit (PREU), 2019 | Yes | No | Yes | For negative inputs |
| Elastic ELU (EELU), 2020 | Yes | Yes | No | For negative inputs |
| Parametric Deformable ELU (PDELU), 2020 | Yes | Yes | Yes | For negative inputs |
(34)
having the output range in where is a learnable parameter. The ELU function exhibits all the benefits of the ReLU function. The ELU is differentiable, saturates for large negative inputs and reduces the bias shift. The negative saturation regime of ELU adds some robustness to noise as compared to the Leaky ReLU and Parametric ReLU. The ELU is extended to Scaled ELU (SELU) by using a scaling hyperparameter to make the slope larger than one for positive inputs. The SELU can be defined as,
(35)
having the output range in where is a hyperparameter. Basically, the SELU induces self-normalization to automatically converge towards zero mean and unit variance. The Parametric ELU (PELU) changes the saturation point and exponential decay and also regulates the slope of the linear function for the positive inputs for differentiability. The PELU AF can be written as,
(36)
having output range, where a and b are the trainable parameters. The parametric ELU is also explored in Continuously differentiable ELU (CELU) for the negative inputs. The CELU is given as,
(37)
having the output range in where is a learnable parameter. The PELU is also extended to multiple PELU (MPELU) by using two learnable parameters to represent MPELU as either rectified, exponential or combined. The MPELU can be expressed as,
(38)
having the output range in , where and are the trainable parameters.
A soft exponential AF interpolates between the exponential, linear and logarithmic functions using the trainable parameter . A Shifted ELU (ShELU) AF is also explored as a locally optimal function . A Parametric Rectified Exponential Unit (PREU) is designed as,
(39)
having the output range in , where and are the trainable parameters. The PREU utilizes the negative information near to zero effectively. The efficiency of ELU is improved in Fast ELU (FELU) AF with the help of the simple displacement bits and integer algebra operations. The FELU is defined as,
(40)
having the output range in with as a learnable parameter. Recently, the properties of ELU and RELU have been utilized to design an Elastic ELU (EELU) AF . The EELU is defined as,
(41)
having the output range in where and are the trainable parameters. The EELU preserves a small non-zero gradient for the negative input and exhibits an elastic slope for the positive input. A Parametric Deformable ELU (PDELU) AF tries to shift the mean value of output closer to zero using the flexible map shape . The PDELU is defined as,
(42)
having the output range in where is a learnable parameter. A ReLU-Memristor-like AF (RMAF) uses two hyperparameters to have ReLU like shape for positive input and to give more importance to the negative values near to zero. An Exponential Linear Sigmoid SquasHing (ELiSH) is defined in as,
(43)
Moreover, it is also extended to HardELiSH which is a multiplication of HardSigmoid and Linear in the positive part and HardSigmoid and ELU in the negative part. Here, HardSigmoid is defined as,
(44)
The ELU based AFs exploit the negative inputs without compromising with the non-linearity. Some ELU variants also modify the function for positive inputs to make it bounded.
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Chapter
Introduction
2009, Measuring Academic ResearchAna Andrés
The ‘end’ of science and the obsolescence of literature
Given the exponential growth of publications and authors proposed by Price in his book, it would seem that science will never cease to grow. However, this assertion must be treated carefully, and it is necessary to analyse in detail the possibility that science will grow according to an exponential pattern. Firstly, we have to bear in mind the exponential function. Its profile is shown in Figure 1.1 and, applied to our subject, it will represent the growth in the number of publications or authors over time. As the function shows, the growing trend will always be increasing and, consequently, science never ceases to grow.
There is evidence that productivity can fit this pattern. One example is shown in the article by López-Muñoz et al. (2006), who carried out a bibliometric study of scientific publications related to bipolar disorder between 1980 and 2004. They identified over 4,000 articles on this topic and analysed their distribution over time; the frequency of studies over time was counted and represented graphically. However, in this article the authors go one step further and assess the nature of the growth profile in bipolar disorder research. Thus they calculated the fit of their data to both a linear and an exponential function. The results showed that the percentage of variance explained by the linear model was 73.67 per cent of data variability, while the exponential model explained 90.19 per cent of the variance. It therefore seems that scientific productivity in the field of bipolar disorder follows Price’s law of exponential growth. Figure 1.2 shows the evolution of productivity over time as well as the linear and exponential functions for these data.
Although the exponential growth of productivity seems to be a valid explanation from a theoretical point of view, it becomes almost absurd when applying it to the real world. In real data, there is no undefined growth until infinity. On the contrary, exponential growth reaches a certain limit, after which the process becomes weaker rather than continuing to grow to an absurd extent. In fact, if the number of scientists were to grow exponentially, it would grow faster than the general population, which is nonsense.
A more suitable explanation for scientific growth is thus to consider exponential growth as a phase in a more complex growth pattern. Thus accelerated growth may exist (although with fluctuations), but would be followed by the stabilisation of productivity. This perspective is more realistic and believable, and the pattern corresponds to the logistic curve. This function is represented by the shape shown in Figure 1.3.
The logistic curve is limited by a base that corresponds to the initial value of growth. This would mean that, at the beginning, no publications have been found. The curve also has a ceiling, after which growth is not possible in the usual way. As can be seen in the graphical representation, there is a phase in which growth is exponential but only up to a certain value, after which growth becomes stabilised.
Consequently, it can be assumed that the exponential growth described by Price’s law is acceptable within a logistic function, so this period of accelerated growth will be followed by a stabilisation phase. In the example described above (López-Muñoz et al., 2006), we can conclude that the exponential growth of productivity in the field of bipolar disorder is only the beginning of a logistic curve.
Within this context, the term ‘big science’ corresponds to this phase of exponential growth. However, it is not possible to know when this increase in productivity will reach the point of stabilisation. In any event, this will never mean the end, but simply the progression to a new phase.
Related to the ideas already discussed regarding the growth of science, it is also worth mentioning the concept of the obsolescence of literature. The term obsolescence of literature was introduced by Price (1963) and refers to the decline in the use of documents over time. This concept is closely related to citation analysis, as the use of a document is measured by the number of citations it receives. While the productivity of a given field is advancing, the citations received for a given document usually decline over time. Finally, if the document is no longer cited it will become obsolete. Only a small group of articles will remain frequently cited over time, and these will become classics due to the fact that authors continue to cite them in recognition of their contribution. However, very few documents will achieve the status of classic articles (Wolfram, 2003).
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Related terms:
Activation Function
Approximation (Algorithm)
Least Squares Method
Avionics
Field Programmable Gate Arrays
Fourier Transform
Case Study
Logarithmic Function
Complex Exponential
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From Wikipedia, the free encyclopedia
Wikimedia Commons has media related to Alkanes.
The main article for this category is Alkane.
An alkane is an acyclicsaturatedhydrocarbon. See Alkane. Alkanes as substituents are called alkyl groups.
| hide v t e Alkanes |
| Methane (CH 4) Ethane (C 2 H 6) Propane (C 3 H 8) Butane (C 4 H 10) Pentane (C 5 H 12) Hexane (C 6 H 14) Heptane (C 7 H 16) Octane (C 8 H 18) Nonane (C 9 H 20) Decane (C 10 H 22) Undecane (C 11 H 24) Dodecane (C 12 H 26) Tridecane (C 13 H 28) Tetradecane (C 14 H 30) Pentadecane (C 15 H 32) Hexadecane / Cetane (C 16 H 34) Heptadecane (C 17 H 36) Octadecane (C 18 H 38) Nonadecane (C 19 H 40) Eicosane (C 20 H 42) Heneicosane (C 21 H 44) Tetracosane (C 24 H 50) Nonacosane (C 29 H 60) Hentriacontane (C 31 H 64) Pentatriacontane (C 35 H 72) Hectane (C 100 H 202) |
| Higher alkanes List of alkanes |
Subcategories
This category has the following 6 subcategories, out of 6 total.
A
Alkane derivatives(8 C)
B
Butane(3 C, 11 P)
E
Ethane(1 C, 2 P)
I
Lists of isomers of alkanes(6 P)
M
Methane(3 C, 52 P)
P
Propane(4 C, 19 P)
Pages in category "Alkanes"
The following 70 pages are in this category, out of 70 total. This list may not reflect recent changes.
Alkane
A
Alkane stereochemistry
B
Butane
C
Crocetane
D
Decane
Dimethylbutane
2,2-Dimethylbutane
2,3-Dimethylbutane
2,3-Dimethylhexane
2,5-Dimethylhexane
2,2-Dimethylpentane
2,3-Dimethylpentane
2,4-Dimethylpentane
3,3-Dimethylpentane
Dodecane
E
Eicosane
Ethane
3-Ethylpentane
H
Hectane
Heneicosane
Hentriacontane
Heptadecane
Heptane
Hexadecane
Hexane
Higher alkane
Hydrocarbon mixtures
I
Isobutane
Isobutane (data page)
Isocetane
Isopentane
K
Kerosene
L
List of straight-chain alkanes
Lycopane
M
2-Methyldodecane
2-Methylheptane
3-Methylheptane
2-Methylhexane
3-Methylhexane
2-Methyloctane
2-Methylpentane
3-Methylpentane
2-Methyltridecane
N
Neopentane
Nonacosane
Nonadecane
Nonane
Nujol
O
Octadecane
Octane
P
Paraffin wax
Pentadecane
Pentane
Pentatriacontane
Phytane
Pristane
Propane
S
Squalane
T
Tetra-tert-butylmethane
Tetracosane
Tetradecane
Tetraethylmethane
Tetramethylbutane
Tridecane
2,3,3-Trimethylpentane
2,3,4-Trimethylpentane
2,2,4-Trimethylpentane
Triptane
U
Undecane
Utreloxastat
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7157 | https://madformath.com/calculators/basic-math/base-converters/octal-to-decimal-converter-with-steps/octal-to-decimal-converter-with-steps | Calculators
Basic Math
Base Converters
Octal to Decimal Converter
⇩ BASE N TO DECIMAL CONVERTERS ⇩
23456789111213141516
OCTAL TO DECIMAL CONVERTER (WITH STEPS)
Enter an octal number.
(125)8 = (85)10
SOLUTION
We convert the absolute value of the entered octal number and put a minus sign after the conversion.
We multiply each digit by its place value and add the products.
(125)8 = (1 × 82) + (2 × 81) + (5 × 80)
= (1 × 64) + (2 × 8) + (5 × 1)
= 64 + 16 + 5
= (85)10
Finally we restore the minus sign of the entered number.
(125)8 = (85)10
OTHER INFORMATION
Click here to see how (85)10 is converted to octal.
Click here to see the binary equivalent of 125.
Click here to see the hexadecimal equivalent of 125.
Download Solution
Copied to clipboard
Copy Text
© MadforMath
See the Solution
⇩ BASE N TO DECIMAL CONVERTERS ⇩
23456789111213141516
INFORMATION
HEXADECIMAL AND DECIMAL NUMBERS
Binary, octal, decimal and hexadecimal numbering systems are commonly used in mathematics, computer science and electrical engineering.
Octal Numbering System (Base 8): In the octal numbering system, numbers are represented by eight digits: 0, 1, 2, 3, 4, 5, 6 and 7. The ones digit has a place value
of 80 = 1, the next digit to the left has a
place value of 81 = 8, then 82 = 64, and so on. As we move one place to the left, the place value increases by a factor of 8.
Decimal Numbering System (Base 10): In the decimal system, numbers are represented using ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
Each digit has a place value of 10 raised to a power depending on its position in the number.
OCTAL TO DECIMAL CONVERSION
Each octal number has a unique representation in other numbering systems. Octal to decimal (oct to dec) conversion is the process of computing the equivalent
decimal representation of a base 8 (octal) number.
To convert an octal number to decimal we multiply each digit by its place value and add the products.
Each place value in octal representation corresponds to an exponent of 8. The exponent increases by 1 as we move one digit to the left, starting from zero for the ones digit.
BINARY TO DECIMAL CONVERSION EXAMPLES
EXAMPLE:
Find the decimal equivalent of 5038.
To find the decimal equivalent of this number, we multiply each digit by its place value and then sum up the products.
5038 = 5 · 82 + 0 · 81 + 3 · 80
= 5 · 64 + 0 · 8 + 3 · 1
= 320 + 0 + 3
= 32310
So, the decimal equivalent of 5038 is 32310.
EXAMPLE:
Find the decimal equivalent of the 47.28.
The place values of B0.416 are shown below.
Multiply those values with the corresponding digits and add the products.
47.28
=4⋅8+7⋅1+
2⋅81
=32+7+82
=39⋅41
=39.2510
OCTAL TO DECIMAL CONVERSION TABLE
The table below displays the decimal equivalents of the smallest non-negative octal numbers up to 100.
| | | |
---
| 08 = 010 | 18 = 110 | 28 = 210 |
| 38 = 310 | 48 = 410 | 58 = 510 |
| 68 = 610 | 78 = 710 | 108 = 810 |
| 118 = 910 | 128 = 1010 | 138 = 1110 |
| 148 = 1210 | 158 = 1310 | 168 = 1410 |
| 178 = 1510 | 208 = 1610 | 218 = 1710 |
| 228 = 1810 | 238 = 1910 | 248 = 2010 |
| 258 = 2110 | 268 = 2210 | 278 = 2310 |
| 308 = 2410 | 318 = 2510 | 328 = 2610 |
| 338 = 2710 | 348 = 2810 | 358 = 2910 |
| 368 = 3010 | 378 = 3110 | 408 = 3210 |
| 418 = 3310 | 428 = 3410 | 438 = 3510 |
| 448 = 3610 | 458 = 3710 | 468 = 3810 |
| 478 = 3910 | 508 = 4010 | 518 = 4110 |
| 528 = 4210 | 538 = 4310 | 548 = 4410 |
| 558 = 4510 | 568 = 4610 | 578 = 4710 |
| 608 = 4810 | 618 = 4910 | 628 = 5010 |
| 638 = 5110 | 648 = 5210 | 658 = 5310 |
| 668 = 5410 | 678 = 5510 | 708 = 5610 |
| 718 = 5710 | 728 = 5810 | 738 = 5910 |
| 748 = 6010 | 758 = 6110 | 768 = 6210 |
| 778 = 6310 | 1008 = 6410 | 1018 = 6510 |
| 1028 = 6610 | 1038 = 6710 | 1048 = 6810 |
| 1058 = 6910 | 1068 = 7010 | 1078 = 7110 |
| 1108 = 7210 | 1118 = 7310 | 1128 = 7410 |
| 1138 = 7510 | 1148 = 7610 | 1158 = 7710 |
| 1168 = 7810 | 1178 = 7910 | 1208 = 8010 |
| 1218 = 8110 | 1228 = 8210 | 1238 = 8310 |
| 1248 = 8410 | 1258 = 8510 | 1268 = 8610 |
| 1278 = 8710 | 1308 = 8810 | 1318 = 8910 |
| 1328 = 9010 | 1338 = 9110 | 1348 = 9210 |
| 1358 = 9310 | 1368 = 9410 | 1378 = 9510 |
| 1408 = 9610 | 1418 = 9710 | 1428 = 9810 |
| 1438 = 9910 | 1448 = 10010 |
WHAT IS OCTAL TO DECIMAL CONVERTER?
Octal to decimal converter,
Computes the decimal equivalent of the entered octal number,
Describes the solution step by step and
Illustrates the place values.
HOW TO USE OCTAL TO DECIMAL CONVERTER?
You can use octal to decimal converter in two ways.
USER INPUTS
You can enter an octal number into the input box and then click the 'CONVERT' button. The result and explanations will appear below the calculator.
#### RANDOM INPUTS
You can click on the DIE ICON next to the input box to generate a random octal number, which will be automatically entered into the calculator. The result and
explanations will then appear below the calculator. You can also create your own examples and practice using this feature.
#### CLEARING THE INPUT BOX
To find the decimal equivalent of another hexadecimal number, click on the CLEAR button to clear the input box.
#### COPYING & DOWNLOADING THE SOLUTION
You can copy the generated solution by clicking on the 'Copy Text' link located below the solution panel.
You can also download the solution as an image file with a .jpg extension by clicking on the 'Download Solution' link located at the bottom of the solution panel. You can then share the downloaded image file.
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7158 | https://www.ncbi.nlm.nih.gov/books/NBK8263/ | Antifungal Agents - Medical Microbiology - NCBI Bookshelf
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Baron S, editor. Medical Microbiology. 4th edition. Galveston (TX): University of Texas Medical Branch at Galveston; 1996.
Medical Microbiology. 4th edition.
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Baron S, editor.
Galveston (TX): University of Texas Medical Branch at Galveston; 1996.
Contents
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Chapter 76 Antifungal Agents
Dennis M. Dixon and Thomas J. Walsh.
Go to:
General Concepts
Definition
An antifungal agent is a drug that selectively eliminates fungal pathogens from a host with minimal toxicity to the host.
Polyene Antifungal Drugs
Amphotericin, nystatin, and pimaricin interact with sterols in the cell membrane (ergosterol in fungi, cholesterol in humans) to form channels through which small molecules leak from the inside of the fungal cell to the outside.
Azole Antifungal Drugs
Fluconazole, itraconazole, and ketoconazole inhibit cytochrome P 450-dependent enzymes (particularly C14-demethylase) involved in the biosynthesis of ergosterol, which is required for fungal cell membrane structure and function.
Allylamine and Morpholine Antifungal Drugs
Allylamines (naftifine, terbinafine) inhibit ergosterol biosynthesis at the level of squalene epoxidase. The morpholine drug, amorolfine, inhibits the same pathway at a later step.
Antimetabolite Antifungal Drugs
5-Fluorocytosine acts as an inhibitor of both DNA and RNA synthesis via the intracytoplasmic conversion of 5-fluorocytosine to 5-fluorouracil.
Go to:
Introduction
The development of antifungal agents has lagged behind that of antibacterial agents. This is a predictable consequence of the cellular structure of the organisms involved. Bacteria are prokaryotic and hence offer numerous structural and metabolic targets that differ from those of the human host. Fungi, in contrast, are eukaryotes, and consequently most agents toxic to fungi are also toxic to the host. Furthermore, because fungi generally grow slowly and often in multicellular forms, they are more difficult to quantify than bacteria. This difficulty complicates experiments designed to evaluate the in vitro or in vivo properties of a potential antifungal agent.
Despite these limitations, numerous advances have been made in developing new antifungal agents and in understanding the existing ones. This chapter summarizes the more common antifungal agents. Three groups of drugs are emphasized: the polyenes, the azoles, and one antimetabolite.Table 76-1 summarizes the most important antifungal agents and their most common uses.
Table 76-1
The Major Antifungal Agents and Their Common Users.
Go to:
Polyene Antifungal Drugs
The polyene compounds are so named because of the alternating conjugated double bonds that constitute a part of their macrolide ring structure (Fig. 76-1). The polyene antibiotics are all products of Streptomyces species. These drugs interact with sterols in cell membranes (ergosterol in fungal cells; cholesterol in human cells) to form channels through the membrane, causing the cells to become leaky (Fig. 76-2). The polyene antifungal agents include nystatin, amphotericin B, and pimaricin.
Figure 76-1
Structures of some common antifungal agents.
Figure 76-2
Generalized fungal cell depicting the sites of action of the common antifungal agents.
Amphotericin B is the mainstay antifungal agent for treatment of life-threatening mycoses and for most other mycoses, with the possible exception of the dermatophytoses. Discovered by Gold in 1956, it can truly be said to represent a gold standard. Its broad spectrum of activity includes most of the medically important moulds and yeasts, including dimorphic pathogens such as Coccidioides immitis, Histoplasma capsulatum, Blastomyces dermatitidis, and Paracoccidioides brasiliensis. It is the drug of choice in treating most opportunistic mycoses caused by fungi such as Candida species, Cryptococcus neoformans, Aspergillus species, and the Zygomycetes. Resistance to this agent is rare, but is noteworthy for Pseudallescheria boydii, Fusarium spp., Trichosporon spp., certain isolates of Candida lusitaniae and Candida guilliermondii.
The drug must be administered intravenously and is associated with numerous side effects, ranging from phlebitis at the infusion site and chills to renal toxicity, which may be severe. A major advance in the use of this agent has resulted from an understanding of the mechanism of its renal toxicity, which is presumed to involve tubuloglomerular feedback. The suppression of glomerular filtration can be reduced by administering sodium chloride.
Nystatin was the first successful antifungal antibiotic to be developed, and it is still in general use. It is representative of the polyene antifungal agents developed later. The promise of its broad-spectrum antifungal activity is offset by host toxicity. Therefore, it is limited to topical use, where it has activity against yeasts such as the Candida species.
Pimaricin (natamycin), another polyene, is used topically to treat superficial mycotic infections of the eye. It is active against both yeasts and moulds.
Go to:
Azole Antifungal Drugs
The azole antifungal agents have five-membered organic rings that contain either two or three nitrogen molecules (the imidazoles and the triazoles respectively). The clinically useful imidazoles are clotrimazole, miconazole, and ketoconazole. Two important triazoles are itraconazole and fluconazole. In general, the azole antifungal agents are thought to inhibit cytochrome P 450-dependent enzymes involved in the biosynthesis of cell membrane sterols.
Ketoconazole set the stage for the orally administered antifungal azoles. It can be administered both orally and topically and has a range of activity including infections due to H capsulatum and B dermatitidis, for which it is often used in nonimmunocompromised patients. It is also active against mucosal candidiasis and a variety of cutaneous mycoses, including dermatophyte infections, pityriasis versicolor, and cutaneous candidiasis. It is not indicated for treatment of aspergillosis or of systemic infections caused by yeasts.
The triazoles (fluconazole, itraconazole) have become the standard for the azoles, and have replaced amphotericin B for managing certain forms of the systemic mycoses. Fluconazole is now routinely used to treat candidemia in non-neutropenic hosts, and is gaining acceptance for use in cryptococcosis and selected forms of coccidioidomycosis. Itraconazole has proven to be effective for histoplasmosis, blastomycosis, sporotrichosis, coccidioidomycosis, consolidation treatment for cryptococcosis, and certain forms of aspergillosis. Fluconazole can be administered either orally, or intravenously. The licensed formulation for itraconazole is oral, but an intravenous formulation is under study, and could be a significant addition directed at bioavailability problems relating to absorption of the oral formulation.
Side effects are not as common with the azoles as with amphotericin B, but life-threatening liver toxicity can arise with long-term use. Liver toxicity noted with ketoconazole has been less problematic with the triazoles. Other side effects include nausea and vomiting. Drug interactions are a potential problem between azoles and other drug classes and include cyclosporin, certain antihistamines, anticoagulants, and antiseizure, oral hypoglycemic and other medications that are metabolized via similar pathways in the liver.
Go to:
5-Fluorocytosine
In contrast to the situation with antibacterial agents, few antimetabolites are available for use against fungi. The best example is 5-fluorocytosine, a fluorinated analog of cytosine. It inhibits both DNA and RNA synthesis via intracytoplasmic conversion to 5-fluorouracil. The latter is converted to two active nucleotides: 5-fluorouridine triphosphate, which inhibits RNA processing, and 5-fluorodeoxyuridine monophosphate, which inhibits thymidylate synthetase and hence the formation of the deoxythymidine triphosphate needed for DNA synthesis. As with other antimetabolites, the emergence of drug resistance is a problem. Therefore, 5-fluorocytosine is seldom used alone. In combination with amphotericin B it remains the treatment of choice for cryptococcal meningitis and is effective against a number of other mycoses, including some caused by the dematiaceous fungi and perhaps even by C albicans.
Go to:
Other Antifungal Agents
Griseofulvin is an antifungal antibiotic produced by Penicillium griseofulvum. It is active in vitro against most dermatophytes and has been the drug of choice for chronic infections caused by these fungi (e.g., nail infections with Trichophyton rubrum) since it is orally administered and presumably incorporated into actively growing tissue. It is still used in such instances but is being challenged by some of the newer azole antifungal agents. The drug inhibits mitosis in fungi.
Potassium iodide given orally as a saturated suspension is uniquely used to treat cutaneous and lymphocutaneous sporotrichosis. This compound, interestingly, is not active against Sporothrix schenckii in vitro. It appears to act by enhancing the transepidermal elimination process in the infected host.
Two other classes of antifungal agents represent new additions to topical treatment of the dermatomycoses in Europe. The two allylamines (naftifine and terbinafine) inhibit ergosterol synthesis at the level of squalene epoxidase; one morpholene derivative (amorolfine) inhibits at a subsequent step in the ergosterol pathway.
Go to:
Selection of Antifungal Agents
In vitro susceptibility testing with the fungi is not yet standardized, and the results of in vitro tests do not always compare to the results obtained in vivo. Therefore, preliminary selection of an antifungal agent for clinical use is made primarily on the basis of the specific fungal pathogen involved. The spectrum of activity for the licensed antifungal agents is well defined through the results of preclinical and clinical testing with the most common fungal pathogens. This approach is useful in avoiding selection of antifungals for species of fungi that are known to have primary resistance to the agent, but less useful in selecting antifungals for species that are known to develop secondary (drug induced) resistance to a particular agent.
Antifungal drug resistance has become an increasing problem with the development of a larger compendium of antifungal agents. Drug resistance to the polyene antifungals is almost always primary resistance rather than secondary resistance. That is, the susceptibility profiles for the species are characteristic and inherent, and rarely change in response to exposure to the agent. For example, amphotericin B-resistant species such as Pseudallescheria boydii and Candida lusitaniae are well known, and do not appear to have originated from exposure to the antifungal. Despite decades of widespread clinical use of amphotericin B in Candida albicans infections, the development of secondary resistance has been exceedingly rare. In contrast, both primary and secondary resistance to 5-fluorocytosine are known to occur for strains of Candida species, serving as the basis for restricting use of this agent to combination therapy with other antifungal drugs.
The question of fungal resistance to the azole drugs is considerably more complex and is currently under evaluation. Examples of both primary and secondary resistance are known for the medically important yeasts and selected azole antifungals. Candida krusei as a species is typically resistant to fluconazole. Candida albicans strains are typically susceptible to fluconazole and certain other azole antifungals, but there are increasing reports of resistance, especially in HIV-infected hosts having undergone repeated courses of azole antifungal therapy. The question of drug resistance is complicated by the limitations in the available susceptibility testing methodology and the ability to distinguish between microbiological and clinical drug resistance. The latter typically occurs when an inhibitory antifungal agent reaches the limits of its activity in a host with a decreasingly efficient immune system.
With the advent of the polyenes, azoles, and fluorocytosine, previously fatal infections can now be treated. However, as modern medicine continues to extend life through aggressive therapy of other life-threatening diseases such as cancer, there is an increasing population at risk for opportunistic fungal infections. Such patients represent a special challenge because they often are left with little host immune function. Therefore, chemotherapeutic agents should be fungicidal and not just fungistatic. The search continues for fungicidal agents that are nontoxic to the host. Research is also directed toward immunomodulating agents that can reverse the defects of native host immunity.
Go to:
References
Casadevall A, Scharff MD. Return to the past: The case for antibody-based therapies in infectious diseases. Clin Infect Dis. 1995;21:150–61. [PMC free article: PMC7197598] [PubMed: 7578724]
Como JA, Dismukes WE. Oral azole drugs as systemic antifungal therapy. New Engl J Med. 1994;330:263–272. [PubMed: 8272088]
Dixon DM. In vivo models: evaluating antifungal agents. Methods Find Exp Clin Pharmacol. 1987;9:729. [PubMed: 3448452]
Espinel-Ingroff A, Shadomy S. In vitro and in vivo evaluation of antifungal agents. Eur J Clin Microbiol. 1989;8:352. [PubMed: 2497014]
Francis P, Walsh TJ. Evolving role of flucytosine in immunocompromised patients: New insights into safety, pharmacokinetics, and antifungal therapy. Clin Infect Dis. 1992;15:1003–1018. [PubMed: 1457631]
Fromtling RA (ed): Recent Trends in the Discovery, Development and Evaluation of Antifungal Agents. Prous, Barcelona, 1987 .
Galgiani JN. Antifungal susceptibility tests. Antimicrob Agents Chemother. 1987;31:1867. [PMC free article: PMC175816] [PubMed: 3326524]
Graybill JR. New antifungal agents. Eur J. Clin Microbiol. 1989;8:402. [PubMed: 2546775]
Heidemann JF, Gerkens JF, Spickard WA. Amphotericin B nephrotoxicity in humans decreased by salt repletion. Am J. Med. 1983;75:476. [PubMed: 6614033]
Iwata K: Drug resistance in human pathogenic fungi. Eur J Epidemiol 8:407-421, 1992 Rinaldi MG, Dixon DM (eds): The evolving etiologies of invasive mycoses. Infect Dis Clin Practice 1994:3(suppl):S47-S112 . [PubMed: 1397205]
Vanden Bossche H. Molecular mechanisms of drug resistance in fungi. Trends in Microbiology. 1994;2:393–400. [PubMed: 7850208]
Walsh TJ. Recent advances in the treatment of fungal infections. Meth Find Exp Clin Pharmacol. 1987;9:769. [PubMed: 2834615]
General Concepts
Introduction
Polyene Antifungal Drugs
Azole Antifungal Drugs
5-Fluorocytosine
Other Antifungal Agents
Selection of Antifungal Agents
References
Copyright © 1996, The University of Texas Medical Branch at Galveston.
Bookshelf ID: NBK8263 PMID: 21413319
Contents
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In this Page
General Concepts
Introduction
Polyene Antifungal Drugs
Azole Antifungal Drugs
5-Fluorocytosine
Other Antifungal Agents
Selection of Antifungal Agents
References
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Review Antifungal drug resistance to azoles and polyenes.[Lancet Infect Dis. 2002]Review Antifungal drug resistance to azoles and polyenes.Masiá Canuto M, Gutiérrez Rodero F. Lancet Infect Dis. 2002 Sep; 2(9):550-63.
Review Fungal sphingolipids: role in the regulation of virulence and potential as targets for future antifungal therapies.[Expert Rev Anti Infect Ther. 2...]Review Fungal sphingolipids: role in the regulation of virulence and potential as targets for future antifungal therapies.Mota Fernandes C, Del Poeta M. Expert Rev Anti Infect Ther. 2020 Nov; 18(11):1083-1092. Epub 2020 Jul 16.
Review Antifungals: From Pharmacokinetics to Clinical Practice.[Antibiotics (Basel). 2023]Review Antifungals: From Pharmacokinetics to Clinical Practice.Carmo A, Rocha M, Pereirinha P, Tomé R, Costa E. Antibiotics (Basel). 2023 May 9; 12(5). Epub 2023 May 9.
Review Drug resistance in human pathogenic fungi.[Eur J Epidemiol. 1992]Review Drug resistance in human pathogenic fungi.Iwata K. Eur J Epidemiol. 1992 May; 8(3):407-21.
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Dixon DM, Walsh TJ. Antifungal Agents. In: Baron S, editor. Medical Microbiology. 4th edition. Galveston (TX): University of Texas Medical Branch at Galveston; 1996. Chapter 76. Available from:
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The Right to Personhood: One Professor’s Fight for Animals
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Author(s)
Emma Atkinson
Sturm College of Law professor Justin Marceau says he believes some animals can—and should—be designated as persons under the law. Here’s why.
News •
Faculty & Staff •
What if dogs were more than just “man’s best friend?”
That’s an over-simplified version of the question being asked by animal rights lawyers like Justin Marceau, the Brooks Institute Faculty Research Scholar of Animal Law and Policy at the Sturm College of Law.
Essentially, some activists and legal scholars are arguing that some animals—like dogs and elephants—should be granted “legal personhood.” A legal person is defined as “a human or a nonhuman legal entity that is treated as a person for legal purposes.” Some business organizations, like corporations, have been deemed legal persons, but generally, animals are not.
Marceau says while it might seem radical to say that animals should be granted legal personhood, most animal rights activists argue that animals should be treated with basic decency.
“They’re not actually trying to have them vote in the next presidential election,” he says. “They're just saying, with [legal personhood] comes certain obligations to treat them in a way that respects their rights and ability.”
Does it matter that ‘an elephant never forgets?’
In January, the Colorado Supreme Court ruled against an animal rights group’s attempt to compel the Cheyenne Mountain Zoo to transfer its five elderly elephants to a “suitable elephant sanctuary.” The group claimed that the zoo mistreated the elephants to the point of “chronic frustration, stress, physical disabilities and brain damage.”
“The basic claim was that the confinement of animals of this level of intelligence is not permissible, and they used a legal vehicle called habeas corpus,” Marceau says.
“Habeas corpus” is a legal framework used to determine if the detention of a person—usually a prisoner—is valid. Ultimately, the Colorado Supreme Court ruled that habeas corpus does not apply to “nonhuman animals,” only to a “person.”
This case, Marceau says, gets to the heart of animal rights groups’ argument that some animals should be designated as legal persons.
“What these lawyers were essentially saying is—for certain beings for whom the science is particularly strong—that they have autonomy in the truest sense,” Marceau says. “They have a theory of mind. They're able to kind of reflect on the past. I mean, one of the elephants passed the mirror test.”
Marceau is referring to a 2005 experiment, done on Happy the Asian elephant, which purported to prove that elephants are self-aware.
“Happy faced her reflection in an 8-by-8-foot mirror and repeatedly used her trunk to touch an "X" painted above her eye,” NBC News described. “The elephant could not have seen the mark except in her reflection. Furthermore, Happy ignored a similar mark, made on the opposite side of her head in paint of an identical smell and texture, that was invisible unless seen under black light.”
What about Fido and Mittens—or Charlotte’s friend Wilbur?
As for the animals we hold nearest and dearest—dogs, cats, birds, lizards and other “companion animals” in our lives—they are the most protected species under the law. The Preventing Animal Cruelty and Torture (PACT) Act of 2019 makes the “crushing, burning, drowning, suffocating, impaling or sexual exploitation” of animals a federal offense. And many states have state-level laws that protect companion animals specifically.
Not all species are protected, even if some are domesticated and considered companion animals. It differs by situation.
“You could take an animal—like a pig. A pig could be a pet, and if you were raising a pig, you would be fully required to follow all the laws just as though she was your dog,” Marceau says. “If she was a wild pig, it would be totally different.”
Wild pigs, in many states, can be legally hunted and killed.
The state-level animal welfare laws that apply to companion animals like dogs and cats often have exemptions. Colorado’s animal protection laws provide some exemptions for people participating in “accepted animal husbandry practices”—breeding—for both livestock and companion animals.
And some of the “standard practice” exemptions that apply to livestock can be extreme, Marceau alleges.
“You might have a pig that's small and use blunt force trauma to the head to kill the animal, because he's not going to grow into a big pig. It's not going to be profitable,” he says. “That would be cruel to do to a dog. But it's not to a pig, because it's standard practice.”
“Nobody would think this is okay.”
In a recent decision handed down by the 16th circuit court in the state of Wisconsin, a judge granted the request of animal rights activist groups to appoint a special prosecutor in a case against a local commercial dog breeding facility called Ridglan Farms.
“Petitioners have shown that there is probable cause to believe that Ridglan has committed crimes under Wisconsin’s animal cruelty laws,” the order reads.
Marceau is involved in the case by way of Sturm’s Animal Activist Legal Defense Project (AALDP), which is representing the animal rights groups in the case. Marceau co-founded and helps to direct the clinic.
The case in Wisconsin outlines graphic details about how AALDP claims the dog breeders have been treating the dogs, including confining them to small spaces and performing medical procedures without anesthesia.
“It's horrific,” Marceau says. “Nobody would think this is okay, but this happens hundreds and hundreds and hundreds of times. I think people don't understand that this kind of thing exists and is out there. We're trying to show that these dogs are a form of legal person.”
Marceau’s big-picture case for animal rights
While Marceau does love animals, he says his fight for the rights of nonhuman beings is about more than advocating for furry friends. It began with his work on death penalty cases after law school.
“I've always been interested in beings who have been at the bottom of the barrel in the legal system, and the legal system has, in many ways, either failed them, or there's nothing else for them to do,” he says. “Beings who are suffering and that the law seems to have turned its back on—that's always just been a draw for me.”
He remembers hearing people on death row being described in ways that did away with their humanity: “They always had to be othered in a way that animalizes; ‘They’re cockroaches. They’re rats. We have to treat this person like a beast.’’’
Marceau says he sees animal rights as inextricably linked to the rights of all beings—humans, too.
“I don't think we're going to achieve justice for humans if animals keep suffering,” he says.
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7161 | https://askfilo.com/user-question-answers-smart-solutions/distributive-property-of-multiplication-example-rewrite-the-3232373130383836 | Question asked by Filo student
Distributive property of multiplication Example : 3×23=(3×20)+(3×3)=60+9=69 Rewrite the equations using the distributive property and find the answer.
Views: 5,793 students
Updated on: Feb 6, 2025
Text SolutionText solutionverified iconVerified
Concepts:
Distributive property, Multiplication
Explanation:
To use the distributive property, we can break down the number 20 into 20 = 20 + 0. This allows us to rewrite the multiplication as follows: 20×4=(20×4)+(0×4). Now we can calculate each part: 20×4=80 and 0×4=0. Adding these together gives us 80+0=80. Therefore, 20×4=80.
Step by Step Solution:
Step 1
Rewrite 20 as (20 + 0).
Step 2
Apply the distributive property: 20×4=(20×4)+(0×4).
Step 3
Calculate: 20×4=80 and 0×4=0. Then add: 80+0=80.
Final Answer:
80
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| Question Text | Distributive property of multiplication Example : 3×23=(3×20)+(3×3)=60+9=69 Rewrite the equations using the distributive property and find the answer. 1. 20×4= |
| Updated On | Feb 6, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 6 |
| Answer Type | Text solution:1 |
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Class 8 (Old)14 units · 96 skillsUnit 1 Rational numbersUnit 2 Linear equations in one variableUnit 3 Understanding quadrilateralsUnit 4 Data handlingUnit 5 Squares and square rootsUnit 6 Cubes and cube rootsUnit 7 Comparing quantitiesUnit 8 Algebraic expressions and identitiesUnit 9 MensurationUnit 10 Exponents and powersUnit 11 FactorisationUnit 12 Introduction to graphsUnit 13 Playing with numbersUnit 14 Visualising solid shapes
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Unit test
Unit 1 Unit 1: Rational numbers
Finding reciprocal of a fraction
Additive and multiplicative inverse of a rational number
Visualize distributive property
Distributive property
Order rational numbers
Negative fractions on the number line
Rational numbers between two rational numbers
Rational numbers: Unit test
Unit 2 Unit 2: Linear equations in one variable
Two-step equations
Equations with variables on both sides: decimals & fractions
Equations with variables on both sides
Equations with parentheses: decimals & fractions
Linear equations in one variable: Quiz 1
Solving equations reducible to linear form
Word problems linear equations (basic)
Sums of consecutive integers
Linear equations word problems (advanced)
Linear equations in one variable: Unit test
Unit 3 Unit 3: Understanding quadrilaterals
Open and closed curves
Polygon types
Angles of a polygon
Interior and exterior angles of a polygon
Analyze quadrilaterals
Quadrilateral types
Understanding quadrilaterals: Quiz 1
Side and angle properties of a parallelogram (level 1)
Side and angle properties of a parallelogram (level 2)
Diagonal properties of parallelogram
Understanding quadrilaterals: Quiz 2
Understanding quadrilaterals: Unit test
Unit 4 Unit 4: Data handling
Read picture graphs (multi-step problems)
Read bar graphs (2-step problems)
Knowing frequency distribution table better
Constructing frequency table
Data handling: Quiz 1
Read histograms
Create histograms
Plotting pie charts
Data handling: Unit test
Unit 5 Unit 5: Squares and square roots
Square roots
Square roots using prime factorisation (advanced)
Simplify square roots
Multiply or divide to make a perfect square
Squares and square roots: Quiz 1
Number of digits in a square root of a number
Finding square roots using division method
Roots of decimals & fractions
Approximating square roots
Squares and square roots: Quiz 2
Squares and square roots: Unit test
Unit 6 Unit 6: Cubes and cube roots
Cube roots
Multiply divide to make it a perfect cube
Cubes and cube roots: Quiz 1
Units digit of a cube root
Cube root of a cube number
Cubes and cube roots: Unit test
Unit 7 Unit 7: Comparing quantities
Percent word problems
Percentage change word problems
Profit and loss percent problems
Profit and loss word problems
Tax and tip word problems
Discount, markup, and commission word problems
Comparing quantities: Quiz 1
Find compound interest
Word problems on compound interest
Comparing quantities: Unit test
Unit 8 Unit 8: Algebraic expressions and identities
Add polynomials (intro)
Subtract polynomials (intro)
Multiply monomials
Multiply monomials (advanced)
Multiply monomials by polynomials
Multiply monomials by polynomials challenge
Algebraic expressions and identities: Quiz 1
Multiply binomials intro
Multiply binomials
Multiply binomials by polynomials
Multiply difference of squares
Polynomial special products: perfect square
Algebraic expressions and identities: Quiz 2
Algebraic expressions and identities: Unit test
Unit 9 Unit 9: Mensuration
Area of trapezoids
Area of composite shapes
Mensuration: Quiz 1
Volume of rectangular prisms
Volume of cylinders
Mensuration: Quiz 2
Mensuration: Unit test
Unit 10 Unit 10: Exponents and powers
Multiply powers
Powers of powers
Divide powers
Powers of products & quotients
Negative exponents
Exponents and powers: Quiz 1
Multiply & divide powers (integer exponents)
Powers of products & quotients (integer exponents)
Properties of exponents challenge (integer exponents)
Scientific notation
Exponents and powers: Quiz 2
Exponents and powers: Unit test
Unit 11 Unit 11: Factorisation
Factor monomials
Greatest common factor of monomials
Factor polynomials: common factor
Factorisation: Quiz 1
Difference of squares
Perfect squares
Factor polynomials using structure
Factoring quadratics intro
Factor quadratics by grouping
Factorisation: Quiz 2
Divide polynomials by monomials (with remainders)
Divide polynomials by linear expressions
Divide polynomials with remainders
Factorisation: Quiz 3
Factorisation: Unit test
Unit 12 Unit 12: Introduction to graphs
Identify coordinates
Identify points
Distance between points in first quadrant
Coordinate plane word problems (quadrant 1)
Introduction to graphs: Unit test
Unit 13 Unit 13: Playing with numbers
Divisibility tests
Divisibility tests for 8 and 11
Divisibility
Divisibility by 6, 12, 18 and 20
Playing with numbers: Unit test
Unit 14 Unit 14: Visualising solid shapesThis unit's exercises do not count toward course mastery.
Course challenge Test your knowledge of the skills in this course.Start Course challenge
Class 8 (Old)
Old version of our Class 8 course (may include outdated content).
Get the updated version of this course (aligned with the latest syllabus).
Unit 1: Rational numbers ------------------------
The role of 0 and 1: Rational numbersDistributive property: Rational numbersRational numbers on the number line: Rational numbers
Rational numbers between two rational numbers: Rational numbers
Unit 2: Linear equations in one variable ----------------------------------------
Solving equations with variable on one side: Linear equations in one variableSolving equations with variables on both sides: Linear equations in one variableEquations reducible to linear form: Linear equations in one variable
Linear equations word problems: Linear equations in one variable
Unit 3: Understanding quadrilaterals ------------------------------------
Polygons: Understanding quadrilateralsAngle sum property: Understanding quadrilateralsKinds of quadrilaterals: Understanding quadrilaterals
Properties of a parallelogram: Understanding quadrilateralsSome special parallelograms: Understanding quadrilaterals
Unit 4: Data handling ---------------------
Looking for information (Recap): Data handlingOrganising data: Data handlingHistograms: Data handling
Pie charts: Data handling
Unit 5: Squares and square roots --------------------------------
Square roots using factorisation: Squares and square rootsNumbers that are not perfect squares: Squares and square rootsSquare roots using long division: Squares and square roots
Estimating square roots: Squares and square roots
Unit 6: Cubes and cube roots ----------------------------
Cube roots using factorisation: Cubes and cube rootsNumbers that are not perfect cubes: Cubes and cube rootsCube roots using long division: Cubes and cube roots
Unit 7: Comparing quantities ----------------------------
Percent change word problems: Comparing quantitiesProfit and loss: Comparing quantitiesTax and discount: Comparing quantities
Compound interest: Comparing quantities
Unit 8: Algebraic expressions and identities --------------------------------------------
Addition and subtraction: Algebraic expressions and identitiesMultiplying monomials by monomials: Algebraic expressions and identitiesMultiplying monomials by polynomials: Algebraic expressions and identities
Multiplying binomials: Algebraic expressions and identitiesStandard identities: Algebraic expressions and identitiesPolynomials word problems: Algebraic expressions and identities
Unit 9: Mensuration -------------------
Area of a trapezium: MensurationArea of a polygon: MensurationCube and cuboid: Mensuration
Cylinder: Mensuration
Unit 10: Exponents and powers -----------------------------
Laws of exponents (Recap): Exponents and powersNegative exponents: Exponents and powersLaws of exponents (More practice): Exponents and powers
Numbers in standard form: Exponents and powers
Unit 11: Factorisation ----------------------
Factors of a monomial: FactorisationFactorisation using common factors: FactorisationFactorisation using identities: Factorisation
Factors of the form (x+a)(x+b): FactorisationDividing polynomials by monomials: FactorisationDividing polynomials by polynomials: Factorisation
Unit 12: Introduction to graphs -------------------------------
Coordinates: Introduction to graphsDistance between points in the first quadrant: Introduction to graphs
Unit 13: Playing with numbers -----------------------------
Test for divisibility: Playing with numbersSome more divisibility tests: Playing with numbers
Unit 14: Visualising solid shapes ---------------------------------
Mastery unavailable
Faces, edges, and vertices: Visualising solid shapes
Course challenge Test your knowledge of the skills in this course.Start Course challenge
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7163 | https://www.quora.com/Why-is-variance-additive-in-statistics-Since-this-often-appears-to-be-described-in-terms-of-the-Pythagorean-Theorem-of-Statistics-what-is-the-hypotenuse-comparable-to-in-statistical-terms | Something went wrong. Wait a moment and try again.
Proofs (mathematics)
Population Variance
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About Statistics
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Probability Concept
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Major Concepts in Statist...
5
Why is variance additive in statistics? Since this often appears to be described in terms of the Pythagorean Theorem of Statistics, what is the hypotenuse comparable to in statistical terms?
Mac Tan
M.S. Applied Statistics, New York University (2019)
·
Author has 1.4K answers and 8.4M answer views
·
Updated 6y
Other answers have noted that the additivity of variances fits nicely into a picture where independent random variables are perpendicular line segments in space, their sum is the hypotenuse of the resulting right triangle, and so of course it makes sense that variances would be additive as per Pythagoras. I just wanted to extend and generalize this intuition a bit.
Instead of just line segments in space, imagine your random variables, X and Y, as two vectors x and y on an inner product space, and the sum of the two random variables as the vector sum of those two. Because we’re attempting to cha
Other answers have noted that the additivity of variances fits nicely into a picture where independent random variables are perpendicular line segments in space, their sum is the hypotenuse of the resulting right triangle, and so of course it makes sense that variances would be additive as per Pythagoras. I just wanted to extend and generalize this intuition a bit.
Instead of just line segments in space, imagine your random variables, X and Y, as two vectors x and y on an inner product space, and the sum of the two random variables as the vector sum of those two. Because we’re attempting to characterize the dispersion of the random variable, we can add a constant to either, and variance and covariance will be unaffected. So for mathematical convenience, let’s assume X and Y are mean-centered (that is, we’ve subtracted off their respective means so that both have mean zero).
I hope it’s not too much of a stretch if I tell you that the covariance of X and Y is their inner product—that is,
Cov(X,Y)=⟨x,y⟩.
After all, the inner product satisfies the properties we expect of covariance:
Covariance is symmetric: Cov(X,Y)=Cov(Y,X). If we’re talking about real-valued random variables—and almost always we are, and from here on out in this answer I’m going to be assuming that we are—so is the inner product: ⟨x,y⟩=⟨y,x⟩.
The inner product is also going to be bilinear: if you have a third real-valued random variable Z with corresponding vector representation z, and real numbers a,b, then ⟨ax+by,z⟩=a⟨x,z⟩+b⟨y,z⟩ and ⟨x,ay+bz⟩=a⟨x,y⟩+b⟨x,z⟩. (Note that this is equivalent to that shift invariance property I described earlier if Z is a constant-valued random variable.)
The inner product is also a quadratic form, which corresponds to the fact that a random variable’s covariance with itself (which is just the variable’s variance) is always nonnegative (and always positive if the random variable is non-constant).
What’s more, a lot of nice relationships between linear algebra and statistics fall out nicely if you consider samples of size n from the random variables to be vectors over Rn and the covariance to simply be the dot product divided by n. I mean, it’s just that natural given our definition of sample covariance:
Cov(X,Y)=1n∑ni=1(xi−¯x)(yi−¯y)
but since ¯x=¯y=0, this reduces to
Cov(X,Y)=1n∑ni=1xiyi=1n⟨x,y⟩.
And from there other neat properties follow: for example, orthogonality in Rn implies that X and Y are uncorrelated, since orthogonality implies that their dot product, and thus their covariance, are zero.
Where does the additivity of variance come into play here? As we’ve noted, the variance of a random variable is the covariance of that random variable with itself: Var(X)=Cov(X,X). Conveniently enough, ⟨x,x⟩ is defined to be the square of the length of x, which is denoted ∥x∥.
This is great, because we’ve essentially transformed the problem into a geometric one, which we’ve known how to do for hundreds of years. Variance is vector length squared (scaled down by a factor of n, but that’s not that important here). Now we’re being given vectors x and y and we want a formula for the squared length of their vector sum, x+y.
Pictorially, we want the length of the red vector:
The law of cosines, which I honestly thought I’d never use since I first forgot about it after 8th grade geometry, saves us here. Since vectors are added head-to-tail, we can draw a new line segment connecting the heads of x+y and x that is parallel to and has the same length as y:
Those labels now denote side lengths, not names of vectors.
We can denote the angle between the green and blue sides of the triangle as ϕ. The law of cosines then tells us that
∥x+y∥2=∥x∥2+∥y∥2−2∥x∥∥y∥cosϕ.
Great! So all we need is ϕ, which is the supplement of the angle between the original vectors y and x. (It’s not hard to see why if you imagine x and y as two sides of a parallelogram.) Call this angle θ, so that we now have the relationship ϕ=π−θ. One last fact you might remember from geometry is that, the cosine of an angle’s supplement is equal to the negative of the cosine of the original angle. That allows us to write:
∥x+y∥2=∥x∥2+∥y∥2−2∥x∥∥y∥cos(π−θ)=∥x∥2+∥y∥2+2∥x∥∥y∥cosθ
Finally we can translate this back into statistical terms that we recognize (again, all of this is up to a constant factor of 1n, which isn’t important here):
∥x+y∥2 is the variance of the sum, Var(X+Y).
∥x∥2 is the variance of X and ∥y∥2 is the variance of Y.
This 2∥x∥∥y∥cosθ nonsense is…. Well, it actually makes a lot more sense when we recognize that the dot product of two vectors is equal to the length of the projection of the first vector on the second, scaled up by the length of the second vector. It can be shown that this is equivalent to saying that ⟨x,y⟩=∥x∥∥y∥cosθ where θ is again the angle between x and y. But this product, multiplied by 2, is exactly the third term of our law of cosines expression. And since we’ve said the inner product—the dot product since we’re still just working with real numbers—is the covariance, this 2∥x∥∥y∥cosθ nonsense is twice the covariance of X and Y.
Substituting these things into our law of cosines expression, we finally find the following:
Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)
which is the true variance relationship between random variables and their sum. Call it the Statistical Law of Cosines if you like—it’s basically the same thing. Like the geometric law of cosines, the Statistical Law of Cosines boils down to the Pythagorean Theorem of Statistics in the special case where the random variables are perpendicular—excuse me, uncorrelated. When random variables are uncorrelated (or independent) their covariance is by definition zero, and so you have Var(X+Y)=Var(X)+Var(Y), as we’re familiar with. And since vectors add head to tail, the hypotenuse of the right triangle formed by these two random variables is their vector sum, and its length is the standard deviation of the sum of those two random variables.
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Related questions
What are the different formulas of empirical variance (variance, descriptive statistics, statistics)?
What is an explanation, not a proof of the Pythagorean theorem (I'm plenty convinced thanks), for the failure of the staircase approximation of a hypotenuse, or a solution to this pathological approach to the approximation?
Probability (statistics):
What is an intuitive explanation of variance?
What is the importance of variance in statistics?
What is the difference between descriptive and inferential statistics?
Sergio Alvarez
Ph.D. from University of Maryland, College Park
·
Author has 115 answers and 186.8K answer views
·
7y
Variance is additive for
independent
random variables, not in general. If
X
and
Y
are independent, then they can be viewed as mutually orthogonal vectors that have as their lengths the standard deviations (square roots of the variances) of the respective variables. If you place these vectors so that the front of
X
touches the rear of
Y
,
then the sum,
X
+
Y
,
is the hypotenuse of a right triangle with sides
X
,
Y
. Pythagoras' theorem then gives you the standard deviation of
X
+
Y
as the square root of the sum of the variances, and you can square it to show that the variance of
X
+
Y
is the sum of the var
Variance is additive for
independent
random variables, not in general. If
X
and
Y
are independent, then they can be viewed as mutually orthogonal vectors that have as their lengths the standard deviations (square roots of the variances) of the respective variables. If you place these vectors so that the front of
X
touches the rear of
Y
,
then the sum,
X
+
Y
,
is the hypotenuse of a right triangle with sides
X
,
Y
. Pythagoras' theorem then gives you the standard deviation of
X
+
Y
as the square root of the sum of the variances, and you can square it to show that the variance of
X
+
Y
is the sum of the variances of
X
,
Y
.
Eric Hortop
Union Steward at Professional Institute of the Public Service of Canada (2013–present)
·
Author has 112 answers and 179K answer views
·
7y
I think comparisons to the Pythagorean theorem would see standard deviations as the sides. Variances are additive, standard deviations are the square roots of variances, so if the standard deviations of independent random variables X and Y are the legs and the standard deviation of X+Y is the hypotenuse.
Variances add for independent random variables — the proof is readily find-able online but the crux of it is that the expectation of xy is equal to the product of the expectations of x and y when X and Y are independent. If you know this then plugging x+y or x-y into the variance formula expres
I think comparisons to the Pythagorean theorem would see standard deviations as the sides. Variances are additive, standard deviations are the square roots of variances, so if the standard deviations of independent random variables X and Y are the legs and the standard deviation of X+Y is the hypotenuse.
Variances add for independent random variables — the proof is readily find-able online but the crux of it is that the expectation of xy is equal to the product of the expectations of x and y when X and Y are independent. If you know this then plugging x+y or x-y into the variance formula expressed as E(x^2)-mu^2 will result in the variance of the sum being the sum of the variances.
Michael Lamar
PhD in Applied Mathematics
·
Author has 3.7K answers and 17.5M answer views
·
Updated 4y
Related
Why is variance the average of the square of the differences? This does not give an accurate description of the average distance between a point and the mean, but an average distance between a point squared and the mean. Why not use absolute value?
This question is a good one. It shows insight and careful thinking, and it deserves a good answer. And there are several good answers. Unfortunately, very few of them are accessible to a person who hasn’t studied quite a bit of math. Certainly more math than is required for an introductory level stats course, for example. I’ll try to give you a few of the good reasons, but my answer may not be helpful if you don’t have the necessary background.
Have you ever wondered what makes the mean (i.e. the expected value), itself, an interesting measure of the center of a distribution? Why not the median
This question is a good one. It shows insight and careful thinking, and it deserves a good answer. And there are several good answers. Unfortunately, very few of them are accessible to a person who hasn’t studied quite a bit of math. Certainly more math than is required for an introductory level stats course, for example. I’ll try to give you a few of the good reasons, but my answer may not be helpful if you don’t have the necessary background.
Have you ever wondered what makes the mean (i.e. the expected value), itself, an interesting measure of the center of a distribution? Why not the median? Why not some other number “in the middle?” One reason is that the mean has an important property. It is the unique number,
^
μ
, that minimizes the expected squared distance to
^
μ
. To see why, let
f
(
^
μ
)
=
E
(
(
X
−
^
μ
)
2
)
. Assuming we can interchange the derivative and the expectation operations, we see that
f
′
(
^
μ
)
=
−
2
E
(
X
−
^
μ
)
. Setting this derivative equal to zero gives
E
(
X
)
=
E
(
^
μ
)
. And since
^
μ
is a number, its expected value is itself and we conclude that
^
μ
=
E
(
X
)
.
As you suggest, we could try to minimize the absolute distance instead of the squared distance.
g
(
~
μ
)
=
E
(
|
X
−
~
μ
|
)
. This one is harder because the absolute value function isn’t always differentiable. It turns out that this derivative is
−
1
whenever
X
−
~
μ
0
and is
+
1
whenever
X
−
~
μ
<
0
. So the derivative is zero if we pick
~
μ
to be a number such that the probability that
X
−
~
μ
0
exactly equals the probability that
X
−
~
μ
<
0
. What do we call a number such that half the probability is on one side and half the probability is on the other? We call it the median. And it IS an important measure of the center of a distribution.
So already, we see one good reason why we care about variance. Once we decide to use
^
μ
=
E
(
X
)
as our measure of center, it is much more natural to measure the spread using the same criterion that lead us to the expected value in the first place: namely the expected squared distance to the center,
E
(
X
)
. This doesn’t make it “wrong” to use the expected absolute distance to
E
(
X
)
— just less natural. If absolute distance is what you want, you should probably be using the median instead of the mean as your center so that
E
(
|
X
−
~
μ
|
)
becomes the more natural choice.
2. There is some REALLY nice linear algebra with analogs in Euclidean geometry just waiting to be exploited when you use the variance. For example, we can consider the space of all random variables with finite variance and use the covariance as an inner product to get a really nice inner-product space. The variance is then the Euclidean norm. Boom! Hilbert space. Want a unit vector? Just take a random variable in this space and rescale it by its standard deviation. You’ve got yourself a unit vector.
3. OK, if we have this nice vector space, what about orthogonality? Zero covariance is orthogonality! And independent random variables always have zero covariance. So we can visualize two random independent variables as being vectors that form a right angle. Do you remember what happens when you add two vectors that form a right angle? Their sum is the hypotenuse of that right triangle. Do you remember how to find the length of the hypotenuse? Pythagorean Theorem:
a
2
+
b
2
=
c
2
What does that have to do with variance? Well,
a
2
and
b
2
are the variances of our two independent random variables and
c
2
is the variance of their sum. So our intuition from Euclidean geometry has led us to an extremely important property of statistics: the variance of the sum of independent random variables is the sum of their variances. And guess what… That result is ONLY true because we are using variance (i.e. the expected value of the squared distance to the mean) and not some other measure of spread (e.g. the expected value of the absolute distance to the mean).
4. Don’t even get me started on how great least squares regression is. I mean, it’s just so beautiful how well it fits with your intuitive understanding of linear algebra. It’s all I can do not to start on my soap box here. But I’ll skip to perhaps the most useful result. When we pick the regression line that minimizes the squared distance from the data to the line, a whole host of really nice properties seem to emerge out of nowhere. One of the most useful is that if we model the unexplained error using the normal distribution, then the regression coefficients of the least-squares regression line inherit that normal distribution as well. That is not true if we use a regression line that minimizes something other than the squared distance to the line. And because those coefficients inherit the normal distribution, it allows us to do all the statistical inference we might ever want to do very conveniently using t-distributions and chi-squared distributions. None of that works out the same way using some other kind of regression line.
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Related questions
What is differentiate between statistics and statistic?
Why is the Pythagorean Theorem not a law?
Why do we compute variance in statistics?
What does variance signify in statistics?
Is statistics considered facts?
Assistant
Bot
·
Sep 7
Variance is additive (for independent components) because variance measures the expected squared deviation from the mean, and when two independent random variables are summed their squared deviations add just like orthogonal squared lengths. The Pythagorean analogy makes that formal: variance plays the role of squared length, independence plays the role of orthogonality, and the sum of variances is the squared length of the sum.
Key points
Definition: For random variables X and Y with finite means,
Var(X+Y) = Var(X) + Var(Y) + 2 Cov(X,Y).
If Cov(X,Y)=0 (in particular for independence), Var(X
Variance is additive (for independent components) because variance measures the expected squared deviation from the mean, and when two independent random variables are summed their squared deviations add just like orthogonal squared lengths. The Pythagorean analogy makes that formal: variance plays the role of squared length, independence plays the role of orthogonality, and the sum of variances is the squared length of the sum.
Key points
Definition: For random variables X and Y with finite means,
Var(X+Y) = Var(X) + Var(Y) + 2 Cov(X,Y).
If Cov(X,Y)=0 (in particular for independence), Var(X+Y) = Var(X) + Var(Y).
Why squared deviations add: Write X̃ = X − E[X], Ỹ = Y − E[Y]. Then
Var(X+Y) = E[(X̃+Ỹ)²] = E[X̃²] + 2E[X̃Ỹ] + E[Ỹ²] = Var(X) + 2Cov(X,Y) + Var(Y).
When E[X̃Ỹ]=0 the cross term vanishes and the squared terms add exactly, identical to a² + b² in the Pythagorean theorem.
Correspondence in the Pythagorean metaphor
Hypotenuse: the standard deviation (√Var) of the sum plays the role of the hypotenuse length.
Legs: the standard deviations (√Var(X), √Var(Y)) of the orthogonal components play the role of the legs.
Orthogonality: zero covariance (Cov=0) corresponds to perpendicular legs; independence implies orthogonality but orthogonality does not always imply independence except in special cases (e.g., jointly Gaussian).
Squared lengths ↔ variances: variance is the expectation of the squared deviation, so it directly maps to "squared length" in geometry.
Examples and caveats
Independent coin flips: for independent Bernoulli trials summing counts, variances add because trials are independent; the standard deviation of the total is the hypotenuse length of the individual standard deviations.
Uncorrelated but dependent: Cov=0 is enough for additivity, but Cov=0 need not imply independence; treating "orthogonal" as exact perpendicularity only fully matches geometric intuition in linear/Gaussian settings.
Nonlinear combinations: the Pythagorean picture applies to sums and linear decompositions; for nonlinear functions of variables you must use different expansions (Delta method, ANOVA decompositions).
Summary sentence
Variance is additive because it is an expected squared length and, when components are uncorrelated (orthogonal), their squared lengths add; the hypotenuse in the Pythagorean analogy is the standard deviation (the square root of variance) of the combined variable.
Michael Lamar
PhD in Applied Mathematics
·
Upvoted by
Justin Rising
, PhD in statistics · Author has 3.7K answers and 17.5M answer views
·
5y
Related
Is variance a problem in statistics?
Variance is a property of distributions. It’s a measure of how “spread out” the distribution is away from its “center.” (The math makes the notions of “spread out” and “center” quite precise, but I’ll continue to write this answer in a more conceptual way.)
Variance can be large or it can be small.
Large variance can be good or it can be bad.
Small variance can be good or it can be bad.
Everything depends on context.
That’s a very important life lesson that’s true much more universally than just in statistics, but it’s particularly true in statistics. To illustrate the point, let’s think about som
Variance is a property of distributions. It’s a measure of how “spread out” the distribution is away from its “center.” (The math makes the notions of “spread out” and “center” quite precise, but I’ll continue to write this answer in a more conceptual way.)
Variance can be large or it can be small.
Large variance can be good or it can be bad.
Small variance can be good or it can be bad.
Everything depends on context. That’s a very important life lesson that’s true much more universally than just in statistics, but it’s particularly true in statistics. To illustrate the point, let’s think about some examples.
Suppose you want to predict, as accurately as possible how much money you should expect to make when you get a job after you graduate from college. You do some research and find starting salary data from recent graduates with the same major as you, from the same school as you, with about the same grades as you, etc. In this case, you’d probably prefer a small variance over a large variance because the smaller the variance, the more accurately you can predict what you can expect to earn.
Suppose you are a gambler going to a casino for a night of entertainment. You have your choice of a great many games to play. For simplicity, let’s suppose that they all have the same (fairly small) negative expected return. In other words, on average, you will, on average, lose the same amount per game in every game. Which game should you play? This question is more subtle, but in this case, picking the game with the smallest variance is probably a bad idea — after all, you came to the casino to be entertained. If you could play a “game” with zero variance, that would mean that every time you played, you lost exactly the same small amount with no chance of every winning a single game. That sounds terribly boring to me, and that’s why no casino would ever create such a game. No one would want to play it. Instead, you would pick a game with a larger variance — large enough that you actually win a great many of the games (even though, on average, you still lose). That’s why slot machines are so popular. There’s a tiny chance of winning a huge jackpot, a reasonably large chance of winning a small amount, and an even larger chance of winning nothing. This kind of game has a large variance to keep people entertained.
A a great many academic applications of statistics are like the first example. Statistics helps us “know,” and “knowing” involves certainty. Variance works against certainty. It is, in fact, a measurement of UNcertainty. Any time uncertainty is a major concern, large variance is a problem.
But a great many non-academic settings resemble the second example more closely. In 1776, American colonists declared independence from England. That was a VERY high variance decision. Winning the inevitable war that would result from this declaration meant autonomy from King George III and British rule while losing that war almost certainly meant execution for the signers of that declaration and, most likely, stricter governance from overseas. Yet, about this decision, Thomas Jefferson supposedly said, “With great risk comes great reward.” (I was unable to find the source of this quote, so it may be falsely attributed to him.)
In this context, variance is risk. Thomas Heywood (The Captives IV. 1624) coined the popular maxim “Nothing ventured, nothing gained” to express this same sentiment. Large variance can imply the potential for a large windfall, but may also leave open the possibility of a disastrous calamity. In situations when the potential for great reward matters more than the certainty of the outcome, variance is not a problem — rather, its the solution.
Ravi Shankar
Ph.D. in Electrical Engineering & Mathematics, Indian Institute of Science, Bangalore (IISc) (Graduated 1993)
·
Author has 441 answers and 1.2M answer views
·
9y
Related
Why do pythagorean theorem can be applied to right angle triangle only?
Actually, the Pythagorean Theorem is a special case of something called the Law of Cosines.
The Law of Cosines states that in any triangle ABC (see figure)
a2=b2+c2−2bccosA
b2=a2+c2−2accosB
c2=a2+b2−2abcosC
Now, if any angle, say C, is a right angle, then cosC=0. So, the expression for c2 reduces to c2=a2+b2, which is Pythagora’s theorem.
Actually, the Pythagorean Theorem is a special case of something called the Law of Cosines.
The Law of Cosines states that in any triangle ABC (see figure)
a2=b2+c2−2bccosA
b2=a2+c2−2accosB
c2=a2+b2−2abcosC
Now, if any angle, say C, is a right angle, then cosC=0. So, the expression for c2 reduces to c2=a2+b2, which is Pythagora’s theorem.
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968)
·
Author has 16.6K answers and 29.3M answer views
·
Updated 6y
Related
Why are data points squared before adding and dividing by the mean to get variance? What does squaring do mathematically that adds a benefit?
Why are data points squared before adding and dividing by the mean to get variance? What does squaring do mathematically that adds a benefit?
That’s a better question than the two answers so far give credit for. Yes squaring does make the deviations from the mean positive, but so does ignoring the sign (taking the modulus).
One benefit is that the theory is much simpler. There is an unbiased estimator for the population variance based on the sample variance. This does not depend on the distribution the data comes from (but it does depend on random sampling). There is no distribution free unbiase
Why are data points squared before adding and dividing by the mean to get variance? What does squaring do mathematically that adds a benefit?
That’s a better question than the two answers so far give credit for. Yes squaring does make the deviations from the mean positive, but so does ignoring the sign (taking the modulus).
One benefit is that the theory is much simpler. There is an unbiased estimator for the population variance based on the sample variance. This does not depend on the distribution the data comes from (but it does depend on random sampling). There is no distribution free unbiased estimator of the MAD (mean absolute deviation).
One disadvantage is that squaring emphasises large values, so the variance is strongly affected by the extremes. This is also true of the mean, but it is worse for the variance.
Rebecca Warner
Teacher and author of statistics textbooks
·
Author has 1.2K answers and 7.8M answer views
·
7y
Related
It is often said that you can think of independent variables the same way as magnitudes at right angles. Thus, the Pythagorean theorem can be applied to both. But why would we think of independent variables this way other than that it is convenient?
Thinking about angles between vectors is a very useful way to understand and visualize factor analysis and principal components analysis.
Factor analysis begins with the set of all correlations among k variables (this is summarized in a correlation matrix, R).
A goal of factor analysis is to see whether you can understand the variables in terms of a smaller number of underlying dimensions. For example, suppose you have k different assessments of intelligence. Can these all be understood as measures of a single kind of general intelligence? Or are they better understood as measures of two kinds o
Thinking about angles between vectors is a very useful way to understand and visualize factor analysis and principal components analysis.
Factor analysis begins with the set of all correlations among k variables (this is summarized in a correlation matrix, R).
A goal of factor analysis is to see whether you can understand the variables in terms of a smaller number of underlying dimensions. For example, suppose you have k different assessments of intelligence. Can these all be understood as measures of a single kind of general intelligence? Or are they better understood as measures of two kinds of intelligence (verbal versus quantitative)? Or do you need to imagine many kinds of intelligence?
At one extreme, each of the tests measures something completely different from the others. At the other extreme, all the tests can be thought of as measuring one “same” thing. In most empirical applications, analysts try to reduce variables to two or three or other small numbers of dimensions or factors.
To represent the correlation information geometrically, start with variables X1 and X2. You can draw these vectors on a flat piece of paper (2 dimensional space).
In lecture, I demonstrate this two dimensional space by drawing two vectors on the table top.
When you add a third variable, you need another dimension, a vector above or below the table top. I put my elbow on the table and move it around to demonstrate.
When we add a rourth variable, we need a fourth dimension (by that time we can’t visualize it well).
I think of these vectors as a bundle of umbrella spokes in a k dimensional space; the angles among these k vectors correspond exactly to the correlations in the R matrix.
For dimension reduction, we want to “flatten” this bundle of vectors into a smaller dimensional space with minimal distortion of all the angles between vectors.
The results of factor analysis are represented by new vectors called “factors”. We now represent the correlations among variables in a different way, by showing the correlation of each variable with each factor (factor loadings are correlations between variables and factors).
The factor loadings can be used to reproduce the correlation matrix R. If we use k factors to represent correlations among k variables, we can perfectly reproduce R. If we use fewer factors, the ability to reproduce R declines. There are informal rules and goodness of fit measures to decide if the loss of information is acceptable.
(I am skipping many steps here… an initial solution is obtained for k factors to represent correlations among k variables; the factors are listed in order of their ability to reproduce correlations (this is indexed by eigenvalues); and a decision is make to retain only the few factors with the largest eigenvalues).
Here is a hypothetical situation in which variables are represented relative to two retained or extracted factors.
These data come from the Bem Sex Role Inventory. Bem’s goal was to show that self ratings of masculine and feminine traits are two separate dimensions, not opposite endpoints of a single dimension. You can see that the masculine adjectives form one ‘bundle’ of vectors; the feminine adjectives form a second ‘bundle’ of vectors. The small angles within bundles indicate that the variables in that group have fairly high positive correlations with each others. The nearly 90 degree angle between the bundles indicates that masculine items have small correlations with feminine items and vice versa.
From linear algebra, a pair of orthogonal factors forms the “basis” for a two dimensional space. However, the location of this reference vectors (factors) is arbitrary. Factor rotation involves rotating the factors in space. The goal is to align the factor axes with the bundles.
After rotation, the new solution looks like this.
Based on this rotated pattern of loadings, we can say that Factor 1 has high positive correlations with ratings of affectionate, warm, etc. and that Factor 2 has high positive correlations with dominant, aggressive, forceful, etc. Factors are named by looking at their correlates. We could call Factor 1 a femininity factor and Factor 2 a masculinity factor.
This is a convenient way to understand factor analysis.
Refer to Justin Rising's answer to What is the geometric interpretation of variance? and you will realize that the lengths of vectors in these diagrams provide information about the proportion of variance that can be reproduced for each variable.
After a factor analysis of mental ability items, the analyst can make statements about: How many factors (different kinds of) mental ability are needed to do a reasonably good job of representing the correlations among measures in the data, and, how would you name or characterize these factors based on the mental ability measures that each one correlates with?
Note that factor analysis does not provide a final answer to the question: How many kinds of mental ability are there out in the world? But only to the question, how many kinds of mental ability can we infer from the hand picked measures and cases in our sample? We can only “find” factors for which we have included measures and as with all other analyses, violations of linearity and other assumptions, and sampling error, affect results.
Pretty neat, huh?
All figures (except the first) are from Warner (2012), Applied statistics: From bivariate through multivariate techniques. Thousand Oaks, CA: Sage. The factor analysis chapter provides details about the computations of factor loadings.
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Mac Tan
M.S. Applied Statistics, New York University (2019)
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Author has 1.4K answers and 8.4M answer views
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Updated 6y
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It is often said that you can think of independent variables the same way as magnitudes at right angles. Thus, the Pythagorean theorem can be applied to both. But why would we think of independent variables this way other than that it is convenient?
This answer
goes into detail on the intuition behind it, but the gist of it is that you treat your random variables
X
and
Y
as vectors
x
and
y
with some defined inner product (and for real-valued data, this is going to be the dot product). It’s then reasonable to treat the inner product of the vectors as the covariance of the two random variables. After all, the inner product and the covariance share a lot of properties: both are (over the real numbers) symmetric; both are quadratic forms; and it turns out that for samples from a population, mean-centering the variables—which doesn’t affect co
This answer goes into detail on the intuition behind it, but the gist of it is that you treat your random variables X and Y as vectors x and y with some defined inner product (and for real-valued data, this is going to be the dot product). It’s then reasonable to treat the inner product of the vectors as the covariance of the two random variables. After all, the inner product and the covariance share a lot of properties: both are (over the real numbers) symmetric; both are quadratic forms; and it turns out that for samples from a population, mean-centering the variables—which doesn’t affect covariance—the covariance is exactly equal to the dot product:
Cov(X,Y)=1n∑ni=1(xi−¯x)(yi−¯y)=1n∑ni=1xiyi
which is the dot product of the vectors x and y.
Two neat things fall out of this:
The variance of a random variable is the square of the magnitude of its vector representation.
Not only can you apply the Pythagorean theorem to both the vectors and to the random variables, you can also apply the law of cosines, the more general formulation of the Pythagorean theorem to non-right triangles.
From this second property we can actually show geometrically that the law of cosines has an analogy in statistics: it’s the fact that
Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)
In the orthogonal case, the inner product of X and Y is zero. And now, since the covariance of X and Y is the inner product of their vector representations x and y, in the orthogonal case the above equation reduces to
Var(X+Y)=Var(X)+Var(Y)
In words: “the length of x+y squared is equal to the length of x squared plus the length of y squared,” which is exactly the Pythagorean theorem.
So to finally answer the question, we don’t just think of independent variables as orthogonal out of convenience. We think of them that way because they actually are orthogonal.
Reuven Harmelin
Studied Mathematics at טכניון (Graduated 1978)
·
Author has 2.3K answers and 1.9M answer views
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Updated Aug 23
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What makes some prime numbers appear in the hypotenuse of a Pythagorean triple, and why are they called Pythagorean Primes?
According to
Euclid’s formulas
every
primitive
Pythagorean triple
a,b,c
is given by
for some pair of relatively prime natural number
s>t
with different parity
.
Hence, the hypothenuse term
c
would be a prime number if and only if it can be represented a sum of two perfect squares, e. g.
According to
Fermat’s
famous theorem about
sums of pairs of perfect squares
we know that prime numbers which
can be represented a sums of two perfect squares are exactly all those prime numbers
p
of the form
p=4n+1
(check the four Pythagorean triples in the examples above), and therefore, the hypotenuse of a Pytha
According to Euclid’s formulas every primitive Pythagorean triple a,b,c is given by
for some pair of relatively prime natural number s>t with different parity. Hence, the hypothenuse term c would be a prime number if and only if it can be represented a sum of two perfect squares, e. g.
According to Fermat’s famous theorem about sums of pairs of perfect squares we know that prime numbers which can be represented a sums of two perfect squares are exactly all those prime numbers p of the form p=4n+1 (check the four Pythagorean triples in the examples above), and therefore, the hypotenuse of a Pythagorean triple is a prime number if and only if it is of that form.
As for your last question I am not familiar with the term “Pythagorean Primes” but in view of the answer to your first question it is quite clear why that name is quite reasonable.
P.S, Every prime number p>2 is the first side of the Pythagorean triple
Adrian Giles
Studied Physics & Mathematics (Graduated 1985)
·
Author has 3.5K answers and 807.2K answer views
·
2y
Related
Why does the sum of the squares of the two shorter sides of a right triangle equal the square of the length of the hypotenuse?
The distance from (0,0) to (x,y) is r=√(x^2+y^2) using the usual metric. Defining a circle a the locus of points equidistant from a point enables us to write the equation of a circle as
Half of maths comes from the circle which might as well be the unit circle r=1. This is not the usual proof of Pythagoras but just pointing out that if you accept this as the equation for a circle then you must accept the case when r is called c, the hypotenuse of a right triangle.
The proof of Pythagoras's Theorem is often a visual one like a square nested in a larger square and is rotated such that i
The distance from (0,0) to (x,y) is r=√(x^2+y^2) using the usual metric. Defining a circle a the locus of points equidistant from a point enables us to write the equation of a circle as
x^2+y^2=r^2
Half of maths comes from the circle which might as well be the unit circle r=1. This is not the usual proof of Pythagoras but just pointing out that if you accept this as the equation for a circle then you must accept the case when r is called c, the hypotenuse of a right triangle.
The proof of Pythagoras's Theorem is often a visual one like a square nested in a larger square and is rotated such that it's corners touch the outer square on each side of the outer square dividing each side into x and y lengths. The area of the big square is (x+y)^2 and the area of the enclosed square is c^2 where c is the hypotenuse of the 4 right triangles each of area (1/2)xy.
Since the area must be equal:
Reconciling these results leads to
which has generality because no assumptions had to be made about x and y except that they are perpendicular. It is a case where a geometric truth and an algebraic truth are clearly being presented.
Terry Moore
PhD in statistics
·
Author has 16.6K answers and 29.3M answer views
·
9y
Related
What is Variance in Statistics?
A probability distribution can be partially described by a number of parameters. Two of the most used types of measure are measures of location amd measures of spread. You are probably familiar with the mean, median and mode as measures of location. Two measures of spread are semi-interquartile range and standard deviation.
The variance of the distribution is the square of the standard deviation. It is not a useful measure in its own right, but it is a step in calculating a standard deviation, and also has the important property that the variance of a sum of independent variables is equal to th
A probability distribution can be partially described by a number of parameters. Two of the most used types of measure are measures of location amd measures of spread. You are probably familiar with the mean, median and mode as measures of location. Two measures of spread are semi-interquartile range and standard deviation.
The variance of the distribution is the square of the standard deviation. It is not a useful measure in its own right, but it is a step in calculating a standard deviation, and also has the important property that the variance of a sum of independent variables is equal to the sum of the variances.
To each parameter, there is a corresponding sample measure corresponding to a set of observations from the distribution.
To calculate a variance of a set of observations: subtract the mean from each observation, square the result and add to the other squares. Then divide by n-1 where n is the sample size.
The variance of a discrete distribution is similar but each term is weighted by the probability of the measure. For a continuous distribution the variance is given by an integral.
Related questions
What are the different formulas of empirical variance (variance, descriptive statistics, statistics)?
What is an explanation, not a proof of the Pythagorean theorem (I'm plenty convinced thanks), for the failure of the staircase approximation of a hypotenuse, or a solution to this pathological approach to the approximation?
Probability (statistics):
What is an intuitive explanation of variance?
What is the importance of variance in statistics?
What is the difference between descriptive and inferential statistics?
What is differentiate between statistics and statistic?
Why is the Pythagorean Theorem not a law?
Why do we compute variance in statistics?
What does variance signify in statistics?
Is statistics considered facts?
How are inferential statistics most often used?
What is variance, and how is it different from standard deviation?
Is variance a problem in statistics?
Was the Pythagorean theorem thought to be useless at first?
Why do we use statistics?
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7164 | https://askfilo.com/user-question-answers-smart-solutions/what-is-the-sum-of-all-possible-three-digit-numbers-that-can-3336343638383430 | Question asked by Filo student
What is the sum of all possible three-digit numbers that can be formed using the digits 4, 6, and 9, with each digit used only once in each number?
Views: 5,729 students
Updated on: Aug 13, 2025
Text SolutionText solutionverified iconVerified
Concepts
Permutations, Place Value, Summation of Numbers
Explanation
We need to find the sum of all three-digit numbers formed by the digits 4, 6, and 9 without repetition in any number. Since we are using three digits and each digit should be used only once in any number, every permutation of these digits will form a valid three-digit number.
There are 3 digits: 4, 6, and 9.
Number of three-digit numbers from these digits without repetition = 3! = 6.
The numbers formed are: 469, 496, 649, 694, 946, 964.
To find the sum quickly, we use a positional approach. Each digit appears exactly the same number of times in each place (hundreds, tens, units) when all permutations are listed.
Number of permutations = 6.
Each digit appears in each place exactly 2 times because for each position, fixing a digit, the other two digits can be arranged in 2! = 2 ways.
So,
Sum contributed by hundreds place = 19 2 100 = 19 200 = 3800
Sum contributed by tens place = 19 2 10 = 19 20 = 380
Sum contributed by units place = 19 2 1 = 19 2 = 38
Total sum = 3800 + 380 + 38 = 4218
Step-By-Step Solution
Step 1
Calculate number of permutations of digits 4, 6, and 9:
Step 2
List the numbers to confirm:
Step 3
Calculate how many times each digit appears in each place:
Step 4
Sum of digits = 4 + 6 + 9 = 19
Step 5
Calculate contribution of each place:
Step 6
Add all contributions:
Final Answer
The sum of all possible three-digit numbers formed by 4, 6, and 9 using each digit once is 4218.
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| Question Text | What is the sum of all possible three-digit numbers that can be formed using the digits 4, 6, and 9, with each digit used only once in each number? |
| Updated On | Aug 13, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 9 |
| Answer Type | Text solution:1 |
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7165 | https://web.mnstate.edu/jasperse/online/Practice-Sets-Answers-All-Organic-Chemistry-1.pdf | Practice Set Answer Keys, Organic Chemistry I Table of Contents • Online Organic Chemistry I, Chem 350, • Dr. Craig P. Jasperse, Minnesota State University Moorhead • For full class website, see • The website will include explanatory videos for each practice set, videos in which I talk through the process and logic involved in determining the correct answers. • My recommendation would be to do a complete two-sided printout of all of the practice sets, and all of the practice set answer keys, at the start of the semester. Test Page Test 1 Test 1 PS#1: Arrow-Pushing/Mechanisms Practice Set 3 Test 1 PS#2: Acid Base Practice Set 7 Test 1 PS#3: 3-D Structure-Drawing Practice Set 11 Test 1 PS#4: Isomers Practice Practice Set 15 Test 1 PS#5: Newman Practice Practice Set 17 Test 1 PS#6: Cyclohexane Chair Practice Set 21 Test 2 Test 2 PS#1: PS1: Radical Bromination Practice Set 25 Test 2 PS#2: PS2: Stereochemistry Practice Set 27 Test 2 PS3: 2 Extra Mechanisms + Product Predict Practice 31 Test 3 Test 3 PS1: Miscellaneous and Mechanisms Principles 35 Test 3 PS2: Test 3 Extra Mechanisms Practice 39 Test 3 PS3: Test 3 Alkene Reactions Practice 43 Test 3 PS4: Test 3 Extra Synthesis Practice (6 pages) 45 Test 4 Test 4 PS1: Test 4 HBr Addn to Dienes; NBS Allylic Bromination 51 Test 4 PS2: Test 4 Conjugation-Allylic-Diels-Alder Practice 55 Test 4 PS3: Aromatic Substitution Mechanisms (Products Provided) 57 Test 4 PS4: Aromatic Substitution Product Prediction/Mechanisms/Synthesis Design Practice 59 1 2 1 Jasperse Arrow-Pushing Practice, Page 1: • Draw arrows for each of the steps in the following reactions. • I won’t require this on tests, but you may find it useful to include all lone-pairs on atoms that react. • I won’t require this on tests, but you may find it useful to draw in all hydrogens on atoms that react. (It is not useful to draw in all H’s on atoms that don’t react.) • Remember that arrows track the movement of electrons, so an arrow should go from the source of electrons and point directly to the atom that accepts them. Br H H Br E2 OH Br SN1 E1 NaI I + NaBr + SN2 H H H + NaOH + HOH + NaBr O + Br-H2O + H H H Br + Br H2O H H + H3O H H H H O H O H OH2 O H H + H-Br 2.
3.
4.
5.
Br Br H2O 3 Organic Chemistry Mechanisms Practice. See Page 3 for a summary of mechanisms principles. 1.
T T Watch for: Changes in: 1. Bonds 2. Lone Pairs 3. Formal Charge Test 1 PS#1: Arrow-Pushing/Mechanisms Practice Set 2 Page 2: Draw the arrow(s) for each of these steps. 1. OH H+ OH2 2. OH2 + H2O 3. H H H + H2O + H3O+ 4. Ph Br Ph + Br + _ 5. Ph Br Ph + Br _ OH OH _ 6. Ph Ph + Br _ D-Br D + 7. Ph + Br _ D + Ph D Br 8. H O CH3 H O CH3 H H OH CH3 9. OCH3 _ O OCH3 O _ 10. Ph Ph + Cl _ Br-Cl Br + Ph Br Cl 11. H NEt3 + + Br _ H Br H H NEt3 + 4 T T Test 1 PS#1: Arrow-Pushing/Mechanisms Practice Set 4 Some Arrow-Pushing Guidelines 1. Arrows follow electron movement. 2. Some rules for the appearance of arrows • The arrow must begin from the electron source. There are two sources: a. An atom (which must have a lone pair to give) b. A bond pair (an old bond that breaks) • An arrow must always point directly to an atom, because when electrons move, they always go to some new atom. 3. Ignore any Spectator Atoms. Any metal atom is always a “spectator” • When you have a metal spectator atom, realize that the non-metal next to it must have negative charge 4. Draw all H’s on any Atom Whose Bonding Changes 5. Draw all lone-pairs on any Atom whose bonding changes 6. KEY ON BOND CHANGES. Any two-electron bond that changes (either made or broken) must have an arrow to illustrate: • where it came from (new bond made) or • an arrow showing where it goes to (old bond broken) 7. Watch for Formal Charges and Changes in Formal Charge • If an atom’s charge gets more positive ⇒ it’s donating/losing an electron pair ⇒ arrow must emanate from that atom or one of it’s associated bonds. There are two “more positive” transactions: • When an anion becomes neutral. In this case, an arrow will emanate from the atom. The atom has donated a lone pair which becomes a bond pair. • When a neutral atom becomes cationic. In this case, the atom will be losing a bond pair, so the arrow should emanate from the bond rather than from the atom. • If an atom’s charge gets more negative ⇒ it’s accepting an electron pair ⇒ an arrow must point to that atom. Ordinarily the arrow will have started from a bond and will point to the atom. 8. When bonds change, but Formal Charge Doesn’t Change, A “Substitution” is Involved • Often an atom gives up an old bond and replaces it with a new bond. This is “substitution”. • In this case, there will be an incoming arrow pointing directly at the atom (to illustrate formation of the new bond), and an outgoing arrow emanating from the old bond that breaks 5 T T Test 1 PS#1: Arrow-Pushing/Mechanisms Practice Set 6 Test 1 PS#1: Arrow-Pushing/Mechanisms Practice Set Organic Chemistry Jasperse Acid-Base Practice Problems A. Identify each chemical as either an “acid” or a “base” in the following reactions, and identify “conjugate” relationships. -You should have one acid and one base on each side -You should have two conjugate pairs 1. 2. 3. 4. 5. B. Choose the More Basic for Each of the Following Pairs (Single Variable). You can use stability to decide. 6. NH3 NaNH2 7. NaOH H2O 8. 9. 10. 11. CH3CH2OH + NaOH CH3CH2ONa + H2O CH3CH2NHLi + CH3OH CH3CH2NH2 + CH3OLi CH3CH2CO2H + CH3MgBr CH3CH2CO2MgBr + CH4 CH3OH + H3O+ H2O + CH3OH2+ CH3CH2NH3+ + CH3OH CH3CH2NH2 + CH3OH2+ NH O NHNa ONa Ph O Ph O O NHNa NHNa O 7 T T Keys: 1. Charge 2. Elecronegativity 3. Resonance Test 1 PS#2: Acid Base Practice Set C. Rank the basicity of the following sets: Multiple Variable Problems 12. CH3MgBr CH3NHNa CH3NH2 13. 14. 15. D. Choose the More Acidic for Each of the Following Pairs: Single Variable Problems 16. 17. 18. 19. 20. 21. O O O OH NHNa ONa O OH ONa NH O O OH O NH3 NH4 OH2 OH OH NH2 CH3 NH2 OH OH OH O NH2 NH2 O 8 T T Test 1 PS#2: Acid Base Practice Set E. Rank the acidity of the following sets: Multiple Variable Problems 22. 23. 24. 25. 26. F. Draw arrow to show whether equilibrium favors products or reactants. (Why?) 27. 28. G. For the following acid-base reaction, a. put a box around the weakest base in the reaction b. put a circle around the weakest acid c. draw an arrow to show whether the equilibrium goes to the right or left. (4pt) 29. OH2 O OH O OH HF NH2 NH3 O OH O H2O OH NH2 OH O OH NH2 NH3 O NH2 O OH NH2 HeO OH O H2O + OH + H OH O H O O O NH2 OH NH + + ONa NH2 NHNa OH + + 9 T T Keys: 1. Charge 2. Elecronegativity 3. Resonance Test 1 PS#2: Acid Base Practice Set Chem 341 Jasperse Ch. 1 Structure + Intro 12 Acid-Base Chemistry (Section 1.13-18) Acidity/Basicity Table Entry Class Structure Ka Acid Strength Base Base Strength 1 Strong Acids H-Cl, H2SO4 102 2 Hydronium H3O+, ROH+ cationic 100 H2O, HOR neutral 3 Carboxylic Acid 10-5 4 Ammonium Ion (Charged) 10-12 5 Water 10-16 6 Alcohol 10-17 7 Ketones and Aldehydes 10-20 8 Amine (N-H) (iPr)2N-H 10-33 9 Alkane (C-H) 10-50 Quick Checklist of Acid/Base Factors 1. Charge 2. Electronegativity 3. Resonance/Conjugation When neutral acids are involved, it’s best to draw the conjugate anionic bases, and then think from the anion stability side. • The above three factors will be needed this semester. The following three will also become important in Organic II. 4. Hybridization 5. Impact of Electron Donors/Withdrawers 6. Amines/Ammoniums Cl O S O O HO , R OH O R O O R N R H R Charged, but only weakly acidic!
R N R R Neutral, but basic!
HOH HO ROH RO O ! H O !
(iPr)2N Li RCH3 RCH2 10 T T T Base Stability T e x t 1. Cations more acidic than neutrals; anions more basic than neutrals 2. Carbanions < nitrogen anions < oxyanione < halides in stability 3. resonance anions more stable than anions without resonance Test 1 PS#2: Acid Base Practice Set Molecular Structure 1 MOLECULAR STRUCTURE For each of the following molecules, draw their 3-D structure. You will usually need to have converted the condensed structure into a Lewis structure. Draw in all hydrogens. - For molecules involving lone-pairs, draw them with the lone pairs shown. While this may not be required for test questions, VSEPR is impacted by lone pairs, so they indirectly impact where atoms are located. For this exercise, try to show where in space the lone pairs will be. To do so, put a “double dot” on the end of a stick (in place), or wedge (in front) or hash (in back). Guidelines for Drawing Models: A. 3-D Perspective 1. Keep as many atoms as possible in a single plane (plane of the paper) by zig-zagging. Connections within the paper are drawn with straight lines. 2. Use wedges to indicate atoms that are in front of the plane. 3. Use hashes to indicate atoms behind the plane. B. For any tetrahedral atom, only 2 attachments can be in the plane, 1 must be in front, and 1 behind. -if the two in the plane are “down”, the hash/wedge should be up -if the two in plane are “up”, the hash/wedge should be down. -the hash/wedge should never point in same direction as the in-plane lines, or else the atom doesn’t looks tetrahedral -for polyatomic molecules, it is strongly preferable to NOT have either of the in-plane atoms pointing straight up. Straight-up in-plane atoms do not lend themselves to extended 3-D structures. 1. ALKANE. butane, CH3CH2CH2CH3 -take the chain and wiggle around all the single bonds. -The most stable actual shape is the one with the carbons zig-zagged and co-planar. -Notice the symmetry possible. 2. ALKANE. Pentane, CH3CH2CH2CH2CH3 Bad! These don' t look tetrahedral!
Good! Look tetrahedral 11 T T 4 Targets: 1. 3-D Drawing 2. Condensed formula to Lewis structure 3. Functional Groups 4. Isomers, including structure versus stereoisomers (p 4) Test 1 PS#3: 3-D Structure-Drawing Practice Set Molecular Structure 2 3. HALOALKANE. 2-bromobutane, CH3CHBrCH2CH3 -notice that if the 4 carbons are co-planar zig-zagged, the attached Br can’t be in the same plane. -try to compare with a partner 2 cases in which Br is in front versus behind. Are they the same molecule, or isomers? 4. ALKENE. Draw both: a) trans-2-butene, CH3CH=CHCH3 and b) cis-2-butene (trans means the two CH3 groups are on the opposite sides of the double bond; cis means they are on same side) -notice that not only the 2 double-bonded C’s but also the four atoms directly attached are all co-planar. 5. ALKYNE. 2-butyne, CH3CCCH3 -draw Lewis structure first 6. WATER. H2O 7. ALCOHOL. Ethanol, CH3CH2OH 8. ETHER. Diethyl ether, CH3CH2OCH2CH3 12 T T Test 1 PS#3: 3-D Structure-Drawing Practice Set Molecular Structure 4 9. FORMALDEHYDE. CH2O. -for 9-16, make sure you draw the Lewis structure before you build models and draw the 3-D picture. If you don’t know the connectivity, you have no chance! 10. ALDEHYDE. CH3CH2CHO. 11. KETONE. CH3CH2C(O)CH2CH3. 12. ACID. CH3CH2CO2H. 13. ESTER. CH3CH2CO2CH3. 14. AMMONIA. NH3 15. AMINE. (CH3CH2)2NH 16. AMIDE CH3CONH2. 13 T T Test 1 PS#3: 3-D Structure-Drawing Practice Set Molecular Structure 5 SAME OR DIFFERENT? Classify the following pairs as “same” or “isomers”. Rules: 1. Structures which can be interchanged or made equivalent by rotations around single bonds are considered to be the same. 2. “Isomers” are things with the same formula that can’t be made superimposable by simple rotations around single bonds. (For class, we will eventually need to be able to distinguish “structural isomers” from “stereoisomers”, so try to do that if possible. And within stereoisomers, by test two we’ll need to distinguish between “enantiomers”, which are mirror image isomers; versus “diastereomers”, which are cis/trans isomers.) Br H H Br Br H Br H H CH3 Cl H H Cl H Cl Cl H Cl H Cl H H Cl Cl H Cl H Cl H H Cl H Cl 14 T T Test 1 PS#3: 3-D Structure-Drawing Practice Set 1 Organic Chemistry I Test 1 Isomers/Resonance Recognition Practice. Note: You should be able to practice the first page fairly early during the class lectures. • The second page you won’t be able to process until almost the end, after completion of the chapter dealing with Newman Projections and Cyclohexane Chair conformations. For the following pairs, classify the relationship between each pair as either: • same compound • structural isomers • resonance structures • stereoisomers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. H CH3 H H CH3 H OH OH OH OH OH OH OH O O O O O Br Br H H H H Br Br Br Br H H Br H H Br Br Br H H Br H H Br Br Cl H H Cl Br H H Br Cl H H Cl Br H H Br Cl H H H H Cl Br 15 T T T Different condensed formula, Different name No atoms move. Formal charge, double bonds, lone pairs may.
Same name, condensed formula. But atoms can't be superimposed, even by single bond rotation. Two families: 1. Unequal mirror images (enantiomers) 2. Cis/trans (diastereomers) Remember, single bonds can rotate, but double bonds can't. Single bond rotation OK Test 1 PS#4: Isomers Practice Practice Set 2 For the following pairs, classify the relationship between each pair as either: • same compound • structural isomers • resonance structures • stereoisomers 25. 26. 27. (Note: review video discussion of this problem in the context of the Newman projections.) 28. 29. 30. 31. 32. 33. 34. 35. 36. CH2CH3 H H H H CH3 CH2CH3 H H CH3 H H iPr H H3C H H H iPr H H CH3 H H H CH3 H H CH3 H H iPr H3C H H3C H iPr CH3 H CH3 H H iPr CH3 H CH3 H H iPr H H CH2CH3 H H Br H Br H Br H Br H Br H H Br Br H Br H Br H Br H H H Br Br Br H Br H H Br H Br Br H Br H Br H Br H Br H Br H Br H H Br 16 T T T Test 1 PS#4: Isomers Practice Practice Set Newman'Projection'Practice' 1' Organic Chemistry I – Jasperse Newman Projection Practice A. For each of the following, draw the best and worst Newman projection, relative to the bond indicated. 1. Butane, relative to the C2-C3 bond 2. 1-chloropropane, relative to the C1-C2 bond 3. 2-methylbutane, relative to the C2-C3 bond 4. 2,2-dimethylbutane, relative to the C2-C3 bond 5. 2-chloro-2-methylpentane, relative to the C2-C3 bond Note: Cl is smaller than methyl B. Rotation Barriers. 6. Rank the rotation barriers relative to the indicated bonds, with 1 have the largest barrier • For convenience, Et = ethyl and iPr = isopropyl • Assume that a halogen, OH, or NH2 is smaller than a CH3 or any other alkyl group. Et Et Cl iPr Et Et iPr Et Et 17 (See page 4 for some summary of operations/steps for handling Newman projections) T T T Test 1 PS#5: Newman Practice Practice Set Newman'Projection'Practice' 2' C. For each of the following, use the words torsional and/or steric to explain why the first conformation is more stable than the second. (The answer key and explaining video will be a bit more detailed as appropriate.) a. For each, note if any “total eclipse” steric interactions exist (two non-hydrogens eclipsing) b. For each, note if any “gauche” steric interactions exist (two non-hydrogens gauche) 7. H CH3 H H CH3 H H CH3 H CH3 H H 8. H CH3 H H CH3 H H CH3 H H CH3 H 9. H CH3 H H CH3 H H CH3 H CH3 H H 10. H H H H H H H H H H H H 11. H CH3 iPr H CH3 H H CH3 iPr H H H3C 12. H iPr CH3 CH3 H H H iPr CH3 H CH3 H 13. H CH3 CH3 CH3 H H H CH3 CH3 H H H3C 18 T T T Test 1 PS#5: Newman Practice Practice Set Newman'Projection'Practice' 3' D. Newman Projection Energy Diagrams. 14. Draw a qualitative energy diagram for CH3CH2CH2CH(CH3)2, relative to the bond between the two CH2 carbons. The Newman projections are drawn below, using “iPr” as an abbreviation for the isopropyl CH(CH3)2 group. Put “S” (for staggered) by any “staggered” conformation, and “E” (for eclipsed) by an eclipsed conformation. H iPr H CH3 H H H iPr H H CH3 H H iPr H CH3 H H H iPr H H H H3C H iPr H H CH3 H H iPr H CH3 H H H iPr H H H H3C 0º 60º 120º 180º 240º 300º 360º Relative Energy Lowest Energy Highest Energy 15. Draw a qualitative energy diagram for CH3CH2CH(CH3)2, relative to the C2-C3 bond. The Newman projections are drawn below. H CH3 CH3 CH3 H H H CH3 CH3 H CH3 H H CH3 CH3 CH3 H H H CH3 CH3 H H H3C H CH3 CH3 H CH3 H H CH3 CH3 CH3 H H H CH3 CH3 H H H3C 0º 60º 120º 180º 240º 300º 360º Relative Energy Lowest Energy Highest Energy 16. Draw a qualitative energy diagram for CH3CH2CH(CH3)CH(CH3)2, relative to the bond between the CH2 and CH carbons. The Newman projections are drawn below, using “iPr” as an abbreviation for the isopropyl CH(CH3)2 group. H iPr CH3 CH3 H H H iPr CH3 H CH3 H H iPr CH3 CH3 H H H iPr CH3 H H H3C H iPr CH3 H CH3 H H iPr CH3 CH3 H H H iPr CH3 H H H3C 0º 60º 120º 180º 240º 300º 360º Relative Energy Lowest Energy Highest Energy 19 T T T Test 1 PS#5: Newman Practice Practice Set !
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Organic Chemistry I Jasperse Newman Projections and Cyclohexane Chairs. Steps. Steps for processing a di-substituted cyclohexane chair: Summary:((Draw(chairs;(install(sticks;(install(substituents(appropriately( 1. Draw both “right-“ and “left-handed” chairs 2. Draw in "axial" sticks on the relevant carbons; then draw in "equatorial" sticks on the relevant carbons • Use the left-most carbon for your first substituted carbon 3. On the left-most carbon, put your first substituent in on both chairs. • It should be equatorial in the “right-handed” chair, and axial in the other. 4. Use "upper/downer" logic to decide whether the second substituent belongs eq or ax on the first chair (then make it the opposite on the second chair) • Draw in the H’s on the relavent carbons 5. Are the two substituents eq/eq, eq/ax, or ax/ax? This will help recognize relative stability 6. If one subst. is forced axial, the preferred chair has the bigger subst. equatorial 7. The best cis vs trans isomer has both substituents equatorial. 8. Note: To draw and identify the best cis versus trans, just draw a chair with both groups equatorial, and then identify whether that is cis or trans( ( ( Steps(for(Drawing(the(Best(Newman(projection( Summary:((Draw(staggered(sticks;(install(substituents(appropriately( 1. Draw a staggered Newman projection, with three sticks on the “back” carbon and three on the “front”. Have a stick up on the back carbon, and one down on the front. 2. Draw your biggest substituent on the back carbon on the “up” stick 3. Draw your biggest substituent on the front on the “down” “anti” stick 4. Fill in the other two back attachments on the other two back-carbon sticks. 5. Fill in the other two front attachments on the other two front-carbon sticks. ( Steps(for(Drawing(the(Worst(Newman(projection( Summary:((Draw(eclipsed(sticks;(install(substituents(appropriately( 1. Draw an eclipsed Newman projection, with three sticks on the “back” carbon and three on the “front”. Have a stick up on both the back and front carbons. 2. Draw your biggest substituent on the back carbon on the “up” stick 3. Draw your biggest substituent on the front on the “up” “totally eclipsed” stick 4. Fill in the other back and front attachments. Note: The more severe the eclipsing in the “worst” projection, the greater the rotation barrier Tips for creating a Newman Projection Energy Diagram 1. Use the “worst” (totally eclipsed version) as 0º and 360º. 2. 120º and 240º will be the other “eclipsed” conformations => energy crests. 3. 60º, 180º, and 300º will be the staggered conformations => energy valleys 4. 60º and 300º will be the other two staggered conformations (gauche) => energy valleys. 5. To compared the relative energies of the eclipsed crests, evaluate the sizes of the eclipsing substituents (when two non-hydrogens eclipse) and 6. To compare the relative energies of the staggered valleys, evaluate the number/severity of gauche interactions 20 T T T Test 1 PS#5: Newman Practice Practice Set Cyclohexane+Chair+Practice+ 1+ Organic Chemistry I – Jasperse Cyclohexane Chair Practice A. Draw the two chair conformations for each of the following di-substituted cyclohexanes. Circle the more stable one. • For convenience, you may abbreviate the substituents (Me, Et, Pr, Bu, iPr, tBu, or the like rather than drawing out methyl, ethyl, propyl, butyl, isopropyl, t-butyl….) • Assume that a halogen, OH, or NH2 is smaller than a CH3 or any other alkyl group. • Remember to draw in the hydrogens on each of the “substituted” carbons 1. Cis-2-bromo-1-methylcyclohexane 2. Cis-3-isopropyl-1-methylcyclohexane 3. Cis -4-ethyl-1-hydroxycyclohexane 4. trans-2-butyl-1-isopropylcyclohexane 5. trans-3-t-butyl-1-methylcyclohexane 6. trans -4-chloro-1-propylcyclohexane 21 T T T Test 1 PS#6: Cyclohexane Chair Practice Set Cyclohexane+Chair+Practice+ 2+ B. For each of the following, do two things: A. draw the most stable chair form for the more stable stereoisomer for the molecule B. identify whether the more stable stereoisomer is cis or trans. 7. 1-butyl-2-methylcyclohexane 8. 3-t-butyl-1-methylcyclohexane 9. 1,4-diethylcyclohexane C. For each of the following, do two things: A. draw the most stable chair form B. identify whether the more stable stereoisomer would be the cis or the trans stereoisomer 10. Cis-2-chloro-1-ethylcyclohexane 11. trans-3-butyl-1-isopropylcyclohexane 12. trans -4-hydroxy-1-t-butylcyclohexane 22 T T T Test 1 PS#6: Cyclohexane Chair Practice Set !
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Organic Chemistry I Jasperse Newman Projections and Cyclohexane Chairs. Steps. Steps for processing a di-substituted cyclohexane chair: Summary:((Draw(chairs;(install(sticks;(install(substituents(appropriately( 1. Draw both “right-“ and “left-handed” chairs 2. Draw in "axial" sticks on the relevant carbons; then draw in "equatorial" sticks on the relevant carbons • Use the left-most carbon for your first substituted carbon 3. On the left-most carbon, put your first substituent in on both chairs. • It should be equatorial in the “right-handed” chair, and axial in the other. 4. Use "upper/downer" logic to decide whether the second substituent belongs eq or ax on the first chair (then make it the opposite on the second chair) • Draw in the H’s on the relavent carbons 5. Are the two substituents eq/eq, eq/ax, or ax/ax? This will help recognize relative stability 6. If one subst. is forced axial, the preferred chair has the bigger subst. equatorial 7. The best cis vs trans isomer has both substituents equatorial. 8. Note: To draw and identify the best cis versus trans, just draw a chair with both groups equatorial, and then identify whether that is cis or trans( ( ( Steps(for(Drawing(the(Best(Newman(projection( Summary:((Draw(staggered(sticks;(install(substituents(appropriately( 1. Draw a staggered Newman projection, with three sticks on the “back” carbon and three on the “front”. Have a stick up on the back carbon, and one down on the front. 2. Draw your biggest substituent on the back carbon on the “up” stick 3. Draw your biggest substituent on the front on the “down” “anti” stick 4. Fill in the other two back attachments on the other two back-carbon sticks. 5. Fill in the other two front attachments on the other two front-carbon sticks. ( Steps(for(Drawing(the(Worst(Newman(projection( Summary:((Draw(eclipsed(sticks;(install(substituents(appropriately( 1. Draw an eclipsed Newman projection, with three sticks on the “back” carbon and three on the “front”. Have a stick up on both the back and front carbons. 2. Draw your biggest substituent on the back carbon on the “up” stick 3. Draw your biggest substituent on the front on the “up” “totally eclipsed” stick 4. Fill in the other back and front attachments. Note: The more severe the eclipsing in the “worst” projection, the greater the rotation barrier Tips for creating a Newman Projection Energy Diagram 1. Use the “worst” (totally eclipsed version) as 0º and 360º. 2. 120º and 240º will be the other “eclipsed” conformations => energy crests. 3. 60º, 180º, and 300º will be the staggered conformations => energy valleys 4. 60º and 300º will be the other two staggered conformations (gauche) => energy valleys. 5. To compared the relative energies of the eclipsed crests, evaluate the sizes of the eclipsing substituents (when two non-hydrogens eclipse) and 6. To compare the relative energies of the staggered valleys, evaluate the number/severity of gauche interactions 23 T T T Test 1 PS#6: Cyclohexane Chair Practice Set 24 Test 1 PS#6: Cyclohexane Chair Practice Set 1 Organic Chemistry I Jasperse Test 2, Radical bromination: Extra Radical Bromination Product Prediction and Mechanism Practice Problems Note: In each of the following, draw the MAJOR mono-brominated product, and/or draw the mechanism (full arrow-pushing) for the propagation steps in the radical mechanism. Initiation need not be illustrated.. 1. 2. 3. 4. Br2, hv Br Br Br2, hv O Br2, hv Br2, hv T T T Note: Was error in original draft of this problem.
And answer key is corrected, but the explanatory video will not really fit this specific structure. 25 !
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5. 6. 7. 8. Br2, hv Br2, hv Br2, hv Br2, hv 26 T T T Test 2 PS#1: PS1: Radical Bromination Practice Set 1 Organic Chemistry I Test 2 Extra Stereochemistry Practice Problems Page 1: Designate R/S Page 2: Chiral or Achiral? Page 3: Same, Enantiomer, or Diastereomer? A. Designate the R/S configuration for any chiral centers in the following molecules. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. Br H H Br H H2N Br H HO H OH OH H HO OH H HO O NH2 H OH O H OH OH Br H HO O O CH3 H O O CH3 H N H O CH3 H NH2 OH H HO OH OH H O OH H O H H3CO CH3 H OH H O NH2 H CH3 OH H H2N OH H CH3 Br CH3 CH3 H H HO O H NH2 H H NH2 H3C H CH3 CH3 H HO CH3 CH3 H H CH3 HO H CH3 H H OH O H H Cl HO O H3C H O CH2OH H H HO HO H H HO H OH 27 This one is very hard.
But for each of the 5 chiral C's, the tie-breaking processes do work. The upper-left S is harder than the others.
R,R R,R, error in movie which said R, S.
Priorities: 1. Heteroatom. 2. C-with-heteroatom. 3. C > CH > CH2 > CH3. 4. Proceed down chain until point of difference. Really Hard!
Test 2 PS#2: PS2: Stereochemistry Practice Set 2 B. Identify each of the following molecule as chiral or achiral. (By circling the chiral ones.) Write “meso” where it applies. (In other words, if it is achiral despite having chiral centers). 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. N H O CH3 H NH2 OH H HO OH OH H O H H3CO H H3CO CH3 H H H3C OH H OH H O Cl Cl H H Cl H H Cl Cl H H Cl Cl H Cl H Cl Cl H H Cl Br H H OH OH OH OH OH CH3 OH H OH H OH OH H H 28 T One chiral carbon => chiral molecule Two common attachments => achiral Plane of symmetry => achiral Two common attachments => achiral Be able to rotate in order to better visualize planes of symmetry or lack thereof Test 2 PS#2: PS2: Stereochemistry Practice Set 3 C. Mark the relationships between the following structures as either “same”, “enantiomers”, or “diastereomers”. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. Cl H Cl H Cl H Cl H Cl H H Cl Cl H H Cl Cl H Cl H H3C Cl H Cl H OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH Cl H Cl H H Cl H Cl Cl H Cl H H Cl Cl H Cl H H Cl H Cl Cl H Cl Cl H H Cl H H Cl Cl H H Cl Cl H H Cl Cl H H Cl Cl H H Cl OH OH OH H OH H OH OH H H OH OH 29 With just one chiral C, if you can assign R/S for both, you can tell if same or different.
chiral => all flipped => enant achiral => all flipped => same chiral => all flipped => enant With two chiral C's: 1. orient them the same; 2. look for all flipped (mirror) vs some flipped (not mirror, diastereomer) 3. If mirror, check for plane of symmetry (meso, achiral, mirror same) vs not (chiral, mirror is enantiomer) Test 2 PS#2: PS2: Stereochemistry Practice Set 30 Test 2 PS#2: PS2: Stereochemistry Practice Set !
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Organic Chemistry I Test 2 Extra Mechanism Practice Problems Note: In each of these cases, I am asking you to draw the mechanism for the product shown. In some cases where both elimination and substitution may occur, there may be another product in addition to the one shown. And in cases where elimination is happening, there may be an additional structural isomer that could form. Regardless, you should be able to draw the mechanism for how the product that IS shown would have actually formed. 1. 2. 3. 4. 5. 6. Br ONa O OK Br H H O O Br OH Heat HS Br S Br OK OK Br H 31 T T T SN2: No intermediates SN1: 3 steps 1. Cation formation 2. Cation Capture 3. Loss of proton Mech explains changes in: 1. Bonds 2. Lone pairs 3. Formal charges Test 2 PS3: 2 Extra Mechanisms + Product Predict Practice !
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7. 8. 9. 10. OH Br H Br OH Heat O Br2, hv O Br Br Br2, hv 32 T T T E1: 1. Cation formation 2. Loss of a neighbor H Radical Bromination: 1. Abstract H using Br radical 2. Attach Br to carbon radical Test 2 PS3: 2 Extra Mechanisms + Product Predict Practice !
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In these problems, both predict the major product and draw the mechanism for its formation. If you expect both substitution and elimination to occur, draw both (bit if there is more than one alkene isomer possible, just draw the one that would form to greater extent,) and draw the mechanism for both. ASSUME ANYTHING THAT STARTS CHIRAL IS OPTICALLY ACTIVE TO START. 1. 2. 3. 4. 5. 6. Br SNa OK Br H HO Br H NaO Br H3C D H HO Br H3C D H Br H D O NaOH 33 T T T Test 2 PS3: 2 Extra Mechanisms + Product Predict Practice !
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7. 8. 9. 10. 11. H Br H D O NaOH Br H D O H2O Br2, hv Br2, hv 34 T T T Test 2 PS3: 2 Extra Mechanisms + Product Predict Practice Organic Chemistry I Jasperse Some Chapter 7 Quiz-Like Practice, But NOT REQUIRED. Answer key available: 1. How many elements of unsaturation are present for a molecule with formula C5H5NO2? a. 0 b. 1 c. 2 d. 3 e. 4 f. 5 2. Provide the proper IUPAC name for the alkene shown below. 3. Which of the following is correct for the geometry of the double bond shown below? Br a. E b. Z c. Neither E nor Z 4. Draw and all structural and stereoisomeric alkenes (no alkanes or cyclic compounds) with the formula C4H8. (stereoisomers included) 5. Choose the most stable alkene among the following. (may help to draw each of them out first…) a. 1-methylcyclohexene b. 3-methylcyclohexene c. 4-methylcyclohexene d. They are all of equal stability 6. a) Draw and circle the major alkene product that would result from the following reaction. b) In addition, draw any other minor isomers that would form, but don’t draw the same isomer twice. Br NEt3 heat 35 Test 3 PS1: Miscellaneous and Mechanisms Principles 7. a) Draw and circle the major alkene product for the following reaction. (There may be a lot of SN2 product that forms as well, but you need not draw that.) b) In addition, draw any other minor isomers that would form, but don’t draw the same isomer twice. Br NaOH 8. a) Draw and circle the major alkene product for the reaction shown. (There may be some SN1 product that forms as well, but you need not draw that.) b) In addition, draw any other minor isomers that would form, but don’t draw the same isomer twice. c) Draw a detailed, step-by-step mechanism for the pathway to the major product. OH H2SO4 heat 9. Provide the chemicals necessary for transforming 2-methylheptane (A) into 2-methyl-1-heptene (C), and draw the structure for the chemical B which you can make from A and which serves as a precursor to C. Above the arrows write in recipes for the A !B transformation and for the B !C transformation. recipe 1 recipe 2 A B C 36 Test 3 PS1: Miscellaneous and Mechanisms Principles For each of the following reactions, write whether the mechanism would be radical, cationic, or anionic? 1. HNO3 NO2 2. Br O2N NaOH OH O2N 3. H O CH3OH, H+ H OCH3 H3CO 4. Br2, peroxides Br 5. O Br2, NaOH O Br 6. H2O, H+ O OH OH 7. peroxides etc etc 8. O OCH3 H3CO H2O, H+ 9. O LiCH3 OLi CH3 37 Cationic (The H+ is the active ion. Nitrate does nothing.) Anionic. The hydroxide is the active ion, sodium cation is spectator. Cationic Cationic. H+ Cationic. H+ Radical. Peroxides or hv is clue. Anionic. The hydroxide is the active ion, sodium cation is spectator. Br2 without hv or peroxides does NOT by itself cause radical chemistry. Radical. Peroxides is clue. Anionic. CH3 anion is active, highly unstable anion. Lithium cation is a metal cation, which serves as a do-nothing spectator. Test 3 PS1: Miscellaneous and Mechanisms Principles Draw the arrow(s) for each of these steps. 1. OH H+ OH2 2. OH2 + H2O 3. H H H + H2O + H3O+ 4. Ph Br Ph + Br + _ 5. Ph Br Ph + Br _ OH OH _ 6. Ph Ph + Br _ D-Br D + 7. Ph + Br _ D + Ph D Br 8. H O CH3 _ H O CH3 _ 9. OCH3 _ O OCH3 O _ 10. Ph Ph + Cl _ Br-Cl Br + Ph Br Cl 11. H NEt3 + + Br _ H Br H H NEt3 + 38 Test 3 PS1: Miscellaneous and Mechanisms Principles !
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Organic Chemistry I Test 3 Extra Mechanism Practice Problems Page 1: Eliminations to make Alkenes. Page 2+3: Reactions of Alkenes Note: In each of these cases, I am asking you to draw the mechanism for the product shown, even if in some cases there may be other products formed as well. In these problems I’m telling you what type of mechanism is involved; I won’t on a test! ! Ch. 7 Elimination Reactions 1. E2, Small/Normal Base 2. 3. E2, Bulky Base 4. E2, Bulky Base using Neutral NEt3 5. 6. H+-Catalyzed Dehydration 7. 8. Br NaOCH3 NaOCH3 Br KO Br Br NEt3 Br NEt3 OH H2SO4 H2SO4 HO H2SO4 HO 39 T Test 3 PS2: Test 3 Extra Mechanisms Practice !
2! Ch. 8 Reactions. 9. Ionic H-X Addition 10. 11. H+ catalyzed H2O Addition 12. 13. X2 addition !
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Br H Br H Br Br Ph H+, H2O Ph OH H+, H2O OH Cl2 Cl Cl 40 T Either the blue or the red mechanism could explain the product with the stereochemistry illustrated.
Test 3 PS2: Test 3 Extra Mechanisms Practice !
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14. X2 addition 15. X2/H2O addition 16. 17. H+ catalyzed H2O addition 18. Br2 Ph Ph H3C Br Br Cl2 H2O Cl OH Br2 H2O H3C Br HO O H2O H+ OH HO O CH3 H2O H+ HO OH CH3 41 T Not Responsible for this Year's Testing Not Responsible for this Year's Testing Test 3 PS2: Test 3 Extra Mechanisms Practice 42 Test 3 PS2: Test 3 Extra Mechanisms Practice 1 Organic Chemistry I Test 3 Extra Alkenes Reactions Practice Problems. (First half of the alkenes reactions only) 1. Draw the major product for the reaction shown. (There may be some side products or isomers formed in addition to the major products, but you don’t need to draw them.) Draw the mechanism. HCl 2. Draw the major product for the reaction shown. (There may be some side products or isomers formed in addition to the major products, but you don’t need to draw them.) No mechanism required. HBr, peroxides 3. Draw the major product for the reaction shown. (There may be some side products or isomers formed in addition to the major products, but you don’t need to draw them.) Draw the mechanism. H+, H2O 4. Draw the major product for the reaction shown. No mechanism required. H2, Pt 43 T 1. Protonate on less substituted end to make carbocation on more substituted end.
2. Capture the cation 1. Protonate on less substituted end to make carbocation on more substituted end.
2. Capture the cation. Capture by neutral water results in cationic product.
3. Deprotonate to get back to neutral. Test 3 PS3: Test 3 Alkene Reactions Practice 2 5. Draw the major product for the reaction shown. (There may be some side products or isomers formed in addition to the major products, but you don’t need to draw them.) No mechanism required. 1. Hg(OAc)2, H2O 2. NaBH4 6. Draw the major product for the reaction shown. (There may be some side products or isomers formed in addition to the major products, but you don’t need to draw them.) 1. BH3-THF 2. H2O2, NaOH 7. Draw the major product for the reaction shown. Include stereochemistry. Draw the mechanism, and make sure it accounts for the product stereochemistry. Br2 8. Draw the major product for the reaction shown. Include stereochemistry. Draw the mechanism, and make sure it accounts for the product stereochemistry. Also, make sure that your mechanism really gives the product that you show. (You may actually want to work the mechanism first, so you make sure you draw the product correctly.) Cl2, H2O 44 T 1. Stereochemistry must be designated 2. Either enantiomer is fine. 1. Stereochemistry must be designated 2. Either enantiomer is fine.
3. Really ugly to draw the stereochem of the 3-membered ring. But for the enantiomer I drew, you'd need to show the bromide anion attacking the more substituted end, and you'd need to the original bromine to be in front. 1. Stereochemistry must be designated 2. Either enantiomer is fine.
3. For the enantiomer I drew, you'd need to show the chloride on the front, and have the water attack the right carbon. Your mechanism and product stereochemistry must be internally consistent. Test 3 PS3: Test 3 Alkene Reactions Practice !
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Organic Chemistry I Test 3 Extra Synthesis Practice Problems Page 1: Synthesis Design Practice. Page 2+3: Predict the Product Practice (including some that involve stereochemistry). Page 4: Cis/trans Stereospecific reactions: which recipe to use; which E or Z alkene to use. Page 5: Recognizing cationic/anionic/radical reactions, and reasonable intermediates/first steps Page 6: Elements of unsaturation/hydrogenation problems; ozonolysis puzzle problems. A. Provide reagents for the following transformations. 1. 2. 3. 4. 5. 6. 7. OH Br Br OH O Br Cl OH Br HO Br OH OH O O HO HO 45 T T T Test 3 PS4: Test 3 Extra Synthesis Practice (6 pages) !
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B. Draw the major product for each of the following reactions or reaction sequences. You needn’t bother to show side products or minor products. For chiral molecules that are racemic , you needn’t draw both enantiomers. BE CAREFUL TO SHOW THE CORRECT ORIENTATION, AND THE CORRECT STEREOCHEMISTRY IN CASES WHERE STEREOCHEM IS FACTOR. (3 points each). 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. H2SO4, heat OH NaOH Br NEt3 Br HBr HBr, peroxides 1. HBr, peroxides 2. NaOCH3 H2O, H+ 1. Hg(OAc)2, H2O 2. NaBH4 1. Hg(OAc)2, H2O 2. NaBH4 3. H2SO4 1. BH3-THF 2. NaOH, H2O2 1. BH3-THF 2. NaOH, H2O2 1. Hg(OAc)2, CH3OH 2. NaBH4 46 T T T Note: explicit stereochemistry must be drawn. The enantiomer would have been equally acceptable.
Test 3 PS4: Test 3 Extra Synthesis Practice (6 pages) !
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20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. !
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1. HBr 2. NaOCH3 Br2 Br2, H2O Br2 Br2, H2O Cl2 Cl2 PhCO3H Ph PhCO3H Ph CH3CO3H, H2O CH3CO3H, H2O Ph OsO4, H2O2 OsO4, H2O2 1. O3 2. Me2S 1. O3 2. Me2S 47 T T T Note: explicit stereochemistry must be drawn. The enantiomer would have been equally acceptable. This principle will apply for any of the reactions producing two chiral centers. Problems 23-32 Test 3 PS4: Test 3 Extra Synthesis Practice (6 pages) !
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C. Draw the alkene that would product the products shown. Make sure to make your drawing clear whether the starting alkene was E or Z. 35. 36. 37. 38. 39. 40. 41. D. What reagent(s) would you use to conduct the following transformations? 42. 43. Cl2 Ph Cl Cl Br2, H2O OH Br CH3CO3H, H2O OH OH OsO4, H2O2 OH OH 1. BH3-THF 2. NaOH, H2O2 Ph H OH H3C PhCO3H O H3C CH2CH3 1. O3 2. Me2S Ph Ph OH OH Ph CH3 Ph OH OH H3C 48 T T T Test 3 PS4: Test 3 Extra Synthesis Practice (6 pages) !
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E. Recognizing whether reaction mechanisms should be cationic, anionic, or radical; whether intermediates should be cationic, anionic, or radical; and recognizing what could be reasonably involved in the initial reaction step. 44. The transformation shown is common in many biological systems. Which of the following statements is definitely, absolutely false? a. The first step in the mechanism probably involves protonation of the carbonyl oxygen. b. The overall reaction involves an addition reaction c. The mechanism is probably radical in nature 45. For the transformation shown, which of the following statements is definitely, absolutely false? a. The first step in the mechanism probably involves protonation of a carbonyl oxygen. b. The overall reaction involves a substitution reaction c. The mechanism is probably anionic in nature d. The first step in the mechanism involves ethoxide anion grabbing a hydrogen. 46. Shown is a reaction, and some possible intermediates along the mechanistic pathway. Given the reaction conditions shown, which of the following statements is true? a. Structures A and B might be plausible intermediates; structure C definitely isn’t b. Structures A and C might be plausible intermediates; structure B definitely isn’t c. Structures B and C might be plausible intermediates; structure A definitely isn’t d. Structure A might be a plausible intermediates; structures B and C definitely aren’t 47. Shown is a reaction, and some possible intermediates along the mechanistic pathway. Given the reaction conditions shown, which of the following statements is true? a. Structures A and B might be plausible intermediates; structure C definitely isn’t b. Structures A and C might be plausible intermediates; structure B definitely isn’t c. Structures B and C might be plausible intermediates; structure A definitely isn’t d. Structure A might be a plausible intermediates; structures B and C definitely aren’t HO O O OH H+ H2O OEt O O 1. NaOEt 2. BrCH3 OEt O O CH3 O NMe2 OH H+ H2O H + HNMe2 NMe2 O A NMe2 OH B H OH C OCH3 O Ph MeOH NaOMe O Ph OCH3 Ph O Ph OCH3 O Ph OCH3 O Ph O H3CO Ph OCH3 OH A B C 49 T T T Test 3 PS4: Test 3 Extra Synthesis Practice (6 pages) !
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F. Elements of Unsaturation/Hydrogenation Problems. For each problem there will be multiple satisfactory solutions. 48. Provide a possible structure for a compound with formula C5H8, given that it reacts with excess H2/Pt to give C5H10. 49. Provide a possible structure for a compound with formula C6H8, given that it reacts with excess H2/Pt to give C6H12. 50. Provide a possible structure for a compound with formula C8H10, given that it reacts with excess H2/Pt to give C8H14. 51. Provide a possible structure for a compound with formula C6H8, given that it reacts with excess H2/Pt to give C6H12. G. Ozonolysis: Draw starting chemicals that will undergo ozonolysis to produce the products shown. In some cases there may be more than one satisfactory answer. 52. 53. 54. 55. 1. O3 2. Me2S O O H 1. O3 2. Me2S H H O O 1. O3 2. Me2S O O O O H H H H + 1. O3 2. Me2S H H H O O O O + + 50 T T T Answer must show one alkene and one ring. (Other structures also meet that requirement). H2/Pt test proved 1 alkene.
EU=2 originally. So the other EU must be ring. Answer must show two alkene and one ring. (Other structures also meet that requirement). H2/Pt test proved 2 alkenes.
EU=3 originally. So the other EU must be ring. Answer must show two alkenes and two rings. (Other structures also meet that requirement). H2/Pt test proved 2 alkene.
EU=4 originally. So the other two EU must be two rings. Answer must show two alkene and one ring. (Other structures also meet that requirement). H2/Pt test proved 2 alkenes.
EU=3 originally. So the other EU must be ring. Any of three answers Test 3 PS4: Test 3 Extra Synthesis Practice (6 pages) Practice with HBr addition to Dienes. Review on predicting products when H-X adds to a diene. 1. Always protonate first on an outside rather than inside carbon. • This will give an allylic rather than isolated cation 2. Is the diene symmetric or asymmetric? • If it’s symmetric, it doesn’t matter which outside carbon you add to first. • If it’s asymmetric, then protonating at different ends will likely give allylic cations of unequal stability. Thus you should decide which protonation site will give the best allylic cation. 3. Is the allylic cation (once you have protonated ) symmetric or asymmetric? • If it’s symmetric, you’ll get one structural isomer. • Is it’s asymmetric, you’ll get two structural isomers. Predict products. Draw mechanisms. Identify 1,2 versus 1,4 addition products. Identify thermodynamic product. 1. 2. 3. H H H H H H Br Br H H H H H H Br Br H H H H Br Br H H 51 T T T Problems 1-3 involve symmetric dienes. I have drawn the allylic cations that would result from protonation on either end, but this is only for illustration sake only. On a real test question, either cation would be fine. Problems 1 and 2 represent cases where the diene is symmetric, but the resulting allylic cation is not symmetric. Thus two structure isomers would be produced. Problem 3 represents a case where not only is the diene symmetric, but so too is the allylic cation. Thus only one isomer will form. Test 4 PS1: Test 4 HBr Addn to Dienes; NBS Allylic Bromination 4. 5. 6. 7. H H H H H H Br Br H H + Br Br Ph Ph Ph Ph Ph Ph Ph Ph Ph + H H H H H H Br Br H H + + Br Br H H H H Br Br H H Br Br H H H H H H H H Br Br H H + Br Br + 52 T T T Problems 4-8 all involve asymmetric dienes. I have drawn the allylic cations that would result from protonation on either end, and have drawn the products that would result. But the yields of products resulting from the "minor" allylic cation may be low. In each case the major allylic cation is aymmetric, leading to two structural isomers. Problems 4 and 7 both illustrate cases where the 1,2 addition is the more stable "thermodynamic" product. Test 4 PS1: Test 4 HBr Addn to Dienes; NBS Allylic Bromination Practice with NBS bromination of Alkenes. Review on predicting products for NBS allylic radical bromination of an alkene. 1. Any allylic spot with an H could give up an H to product an allylic radical. How many allylic spots are there? 2. If there is more than one allylic spot, is the alkene symmetric or asymmetric? In other words, will the different allylic spots give the same allylic radical or unequal allylic radicals? • If there is more than one allylic radical, they may be of unequal stability. So one might lead to more product than the other. Still, you should expect to get at least some product from each of the allylic radicals. 3. One you have made an allylic radical, is it symmetric or asymmetric? • If it’s symmetric, you’ll get one structural isomer from it. • If it’s asymmetric, you’ll get two structural isomers out. • Note that if you your allylic radicals are asymmetric, you’ll get two bromide products for each one. So if you have two different allylic sites each offering asymmetric allylic radicals, you’ll get 2 x 2 = 4 isomer products. Draw products following NBS/peroxides bromination. Identify radicals, and draw all resonance structures for the radicals. 1. 2. 3. Br + Br Br Br 53 T T T This illustrates the simplest case. There is only one allylic position, and the allylic radical is symmetric leading to a single product. Problem 2 has only a single allylic position, but proceeds via an asymmetric radical which leads to two products. Problem 3 represents a symmetric alkene. It has two allylic positions, but due to symmetry they are both equivalent. The allylic radical formed is symmetric, thus only a single isomer is produced.
Test 4 PS1: Test 4 HBr Addn to Dienes; NBS Allylic Bromination 4. 5. 6. 7. + Br Br + Ph Ph Ph Ph Ph + Ph Ph Ph Ph Br Br Br Br + + Br Br Br Br Br Br Br Br Br Br + + + 54 T T T Problem 4 represents a symmetric alkene. It has four allylic positions, but due to symmetry they are all equivalent. The allylic radical formed is asymmetric, thus two isomers are produced.
Problem 5 represents an asymmetric alkene. It has two allylic positions, and they are not equivalent. Each allylic radical formed is asymmetric, thus each produces two isomers. 2 isomers x 2 products from each => 4 isomeric products.
The yield of the two products born from less stable "minor" allylic radical may be low, but is not zero. Problem 6 represents another asymmetric alkene. It has two allylic positions, and they are not equivalent. Each allylic radical formed is asymmetric, thus each produces two isomers. 2 isomers x 2 products from each => 4 isomeric products.
The yield of the two products born from less stable "minor" allylic radical may be low, but is not zero. Problem 7 represents another asymmetric alkene, with 3 non-equivalent allylic positions The above two are actually the same Above radical is symmetric.
Test 4 PS1: Test 4 HBr Addn to Dienes; NBS Allylic Bromination Organic Chemistry I Jasperse Extra Practice Problems: Conjugated Systems, Dienes, Allylic Systems and the Diels-Alder Reaction 1. Rank the heats of hydrogenation for the following, 1 being most heat released and 4 being least heat. (Think: will the more stable isomer release more heat or less heat when it is hydrogenated)? 2. Rank the rate of reaction of the following toward SN1 substitution (AgNO3/CH3CH2OH), 1 being most reactive and 4 being least reactive. (Think: what determines the rates for SN1 reactions?) 3. Products A and B combine to make up over 90% of the product mixture. a. For each of the structures A-D, attach an H atom to the carbon that in fact added an H. b. Classify each of the four structures as either a 1,2 or 1,4 addition product. c. Draw the resonance structures for the cation that leads to both products A + B, and also draw the resonance structures for the cation that leads to both products C+D. 4. Draw the mechanism for formation of products A and B above. Br Br Br Cl Cl Cl Cl Cl + + + H-Cl A B C D 55 Answers Test 4 PS2: Test 4 Conjugation-Allylic-Diels-Alder Practice 5. 1,4-pentanediene is much more acidic than pentane. Explain why. (Think: what determines aciditiy?) 6. Draw the two major products for the following reaction. • Identify each as either a 1,2 or 1,4 addition product. • Write either “thermodynamic” or “kinetic” underneath each one. • Draw the two resonance structures for the intermediate from which both form. 7. Draw the major product or products for the following reaction. Draw the resonance structures for the intermediate from which both form. 8. Give the reactants (including stereochemistry) that would give the following Diels-Alder product. 9. Draw the major Diels-Alder product. H H a million billion times more acidic vs.
H H low acidity + 1.0 H-Br Br2/hv or NBS/peroxides OCH3 O CN + O heat 56 The diene produces a much more stable ANION Test 4 PS2: Test 4 Conjugation-Allylic-Diels-Alder Practice !
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Organic Chemistry I Jasperse Test 4 Extra Practice: Drawing Mechanisms when the Reactants and Products are Given Aromatic)Substitution)Mechanism)Practice.))Product)given.))) • A!subsequent!practice!set!will!give!additional!mechanism!practice,!but!will!also!require!
you!to!predict!the!product!as!well.!!It!will!also!to!have!a!bunch!of!other!product!prediction!
problems!that!don’t!focus!on!mechanisms,!and!a!bunch!of!synthesis!design!practice!
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Cl2, AlCl3 Cl Br2, FeBr3 O O Br HNO3, H2SO4 O2N 57 Answers Standard Mechanism: 1. Create cationic electrophile. Use the acid to enable this.
2. Aromatic bonds to cation. Arrow from aromatic to cationioc electrophile.
3. Deprotonation. Two electrons in C-H bond go to cation. At least 3 resonance structures always available for the cation. Test 4 PS3: Aromatic Substitution Mechanisms (Products Provided) !
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HNO3, H2SO4 NO2 2-bromopropane, AlCl3 Br Br AlCl3 Cl O NO2 NO2 O 58 Test 4 PS3: Aromatic Substitution Mechanisms (Products Provided) !
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Aromatic)Substitution)Reaction)Mechanisms))) • Draw!the!major!product!and!the!mechanism!for!each!of!the!following!reactions,!using!
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• Draw!the!resonance!structures!for!carbocationic!intermediate.!!!
Note:!!See!pages!3!and!4!for!more!production!prediction!problems.!
See!page!5!for!some!synthesis!design!problems.!!!
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O Cl2, AlCl3 Br2, FeBr3 N H HNO3, H2SO4 HNO3, H2SO4 Cl 59 T T T Nitrogen substituent is a strong activator, takes preference over the weaker methyl group Test 4 PS4: Aromatic Substitution Product Prediction/Mechanisms/Synthesis Design Practice !
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2-bromopropane, AlCl3 (Show only one substitution) Cl bromocyclopentane, AlCl3 (Show only one substitution) HO2C AlCl3 HO2C OCH3 Cl O AlCl3 Cl Cl O 60 T T T Test 4 PS4: Aromatic Substitution Product Prediction/ Mechanisms/Synthesis Design Practice !
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Br2, FeBr3 Cl AlCl3 NO2 H2N Cl2 1. HNO3, H2SO4 2. Br2, FeBr3 3. Fe, HCl 1. HNO3, H2SO4 2. Fe, HCl 3. Br2, FeBr3 1. KMnO4 2. HNO3, H2SO4 3. 2-bromobutane, AlCl3 4. Fe, HCl 1. HNO3, H2SO4 2. 2-bromo-2-methylbutane, AlCl3 3. Fe, HCl 4. KMnO4 1. SO3, H2SO4 2. 2-bromo-2-methylbutane, AlCl3 3. H2O, H+ 1. SO3, H2SO4 2. Cl2, AlCl3 3. H2O, H+ 4. Fe, HCl NO2 61 T T T Test 4 PS4: Aromatic Substitution Product Prediction/Mechanisms/Synthesis Design Practice !
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bromocyclopentane, AlCl3 OCH3 O2N Cl O 1. AlCl3, 2. AlCl3, 2-bromopropane 3. Zn(Hg), HCl Cl Cl O 1. AlCl3, 2. Zn(Hg), HCl 3. AlCl3, 2-bromopropane Cl Cl O 1. AlCl3, 2. AlCl3, 2-bromopropane 3. Zn(Hg), HCl Cl Cl O 1. SO3, H2SO4 2. AlCl3, 3. H2O, H+ 4. Zn(Hg), HCl 1. NBS, peroxides (or Br2, hv) 2. NaOCH3 1. NBS, peroxides (or Br2, hv) 2. NEt3 1. NBS, peroxides (or Br2, hv) 2. NEt3 1. NBS, peroxides (or Br2, hv) 2. NaOCH3 62 T T T Test 4 PS4: Aromatic Substitution Product Prediction/Mechanisms/Synthesis Design Practice !
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H2N O2N Br CO2H Cl Cl NH2 CO2H H2N Br 63 T T T Test 4 PS4: Aromatic Substitution Product Prediction/Mechanisms/Synthesis Design Practice 64 Test 4 PS4: Aromatic Substitution Product Prediction/Mechanisms/Synthesis Design Practice |
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7167 | http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/betaex.html | | | |
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| Cobalt-60 The cobalt-60 isotope undergoes beta decay with a half-life of 5.272 years. Cobalt-60 decays to Nickel-60 plus an electron and an electron antineutrino. The decay is initially to a nuclear excited state of Nickel-60 from which it emits either one or two gamma ray photons to reach the ground state of the Nickel isotope. This particular radioisotope is historically important for several reasons. It is involved in the radioactive fallout from nuclear weapons. For many years, the gamma radiation from this decay was the main source for radiation therapy for cancer. This decay was used in the famous experiment by C. S. Wu in which she demonstrated the nonconservation of parity . | Index |
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7168 | https://pubs.rsc.org/en/content/articlelanding/2017/sc/c7sc03364h | Jump to main content
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Issue 11, 2017
From the journal:
Chemical Science
Effects of vibrational excitation on the F + H2O → HF + OH reaction: dissociative photodetachment of overtone-excited [F–H–OH]−†
Amelia W.
Ray,a
Jianyi
Ma,b
Rico
Otto,‡a
Jun
Li,
c
Hua
Guo
d
and
Robert E.
Continetti
a
Author affiliations
Corresponding authors
a
Department of Chemistry and Biochemistry, University of California, San Diego, 9500 Gilman Drive, La Jolla, California 92093-0340, USA
E-mail:
rcontinetti@ucsd.edu
b
Institute of Atomic and Molecular Physics, Sichuan University, Chengdu, Sichuan 610065, China
E-mail:
majianyi81@163.com
c
School of Chemistry and Chemical Engineering, Chongqing University, Chongqing 401331, China
d
Department of Chemistry and Chemical Biology, University of New Mexico, Albuquerque, New Mexico 87131, USA
Abstract
The reaction F + H2O → HF + OH is a four-atom system that provides an important benchmark for reaction dynamics. Hydrogen atom transfer at the transition state for this reaction is expected to exhibit a strong dependence on reactant vibrational excitation. In the present study, the vibrational effects are examined by photodetachment of vibrationally excited F−(H2O) precursor anions using photoelectron-photofragment coincidence (PPC) spectroscopy and compared with full six-dimensional quantum dynamical calculations on ab initio potential energy surfaces. Prior to photodetachment at hνUV = 4.80 eV, the overtone of the ionic hydrogen bond mode in the precursor F−(H2O), 2νIHB at 2885 cm−1, was excited using a tunable IR laser. Experiment and theory show that vibrational energy in the anion can be effectively carried away by the photoelectron upon a Franck–Condon photodetachment, and also show evidence for an increase of branching into the F + H2O reactant channel. The experimental results suggest a greater role for product rotational excitation than theory. Improved potential energy surfaces and longer wavepacket propagation times would be helpful to further examine the nature of the discrepancy.
1. Introduction
The effects of vibrational excitation on the rates and dynamics of chemical reactions in the gas phase are the focus of considerable interest. In the case of atom–diatom reactions, Polanyi's rules explain the differing effects of reactant vibration and translational energy as a function of reaction energetics,1 and recent work has extended these concepts toward polyatomic systems.2–4 Considerable experimental effort has been devoted toward exploration of mode-selective chemistry involving polyatomic molecules both in the gas phase5–10 and at surfaces.11,12 These studies indicate that excitation of vibrational modes with large displacements along a reaction coordinate can preferentially promote surmounting specific barriers, giving rise to mode-selective chemistry. The present work is a study of the effect of anion vibrational excitation on the dynamics of the neutral F + H2O → HF + OH reaction by means of direct IR absorption. This reaction is initiated near the transition state for the bimolecular reaction by photodetachment of F−(H2O).13 Excitation of a suitable mode in the precursor anion changes the Franck–Condon overlap with the neutral surface, and in favorable cases should provide considerable control over product energy disposal and dynamics. Here, we examine the effects of excitation of the overtone of the F–H–OH ionic hydrogen bond (IHB, 2νIHB) in F−(H2O) using photoelectron-photofragment coincidence (PPC) spectroscopy. This anion mode is particularly interesting because it is well aligned with the neutral reaction coordinate and the stretching vibration of the H2O reactant. The experiment was carried out by coupling an infrared laser system into the ion beam line of the PPC spectrometer.14 The experimental results are compared to extensive full-dimensional (6-D) quantum dynamics calculations using ab initio anionic and neutral potential energy surfaces (PESs).
Neutralization of ionic species provides a useful approach to examining neutral reaction dynamics. Photodetachment photoelectron spectroscopy of negative ions has been used as a spectroscopy of the neutral transition state in isomerization and chemical reactions.15–19 These experiments probe neutral configurations as determined by the ground state anion precursor. To go beyond this restriction, Lineberger and coworkers, using negative ion–neutral–positive ion (NeNuPo) charge reversal spectroscopy, probed a broad portion of the neutral PES of Ag3 by photodetachment of Ag3− followed by femtosecond photoionization to probe the time evolution of the neutral wavepacket.20 An alternative approach to sampling other configurations on the neutral surface is to vibrationally excite the anion precursor prior to photodetachment. Neumark and coworkers have demonstrated this approach by using stimulated Raman excitation of C2− as demonstrated by photoelectron spectroscopy.21,22 However, stimulated Raman pumping is limited to those anion systems with optically accessible excited states below the photodetachment threshold. Other approaches including stimulated Raman-adiabatic passage (STIRAP) may also be applied on systems that have been extensively spectroscopically characterized.23 Lacking such information in the case of F−(H2O) we have taken the more general approach of direct absorption of infrared radiation used in a number of crossed molecular beam studies with vibrationally excited reactants7,8 as well as anionic cluster vibrational predissociation and vibrationally mediated photodetachment.24
We have previously studied the prototypical hydrogen-abstraction reaction F + H2O → FH2O → HF + OH in a joint experimental/theoretical effort combining PPC spectroscopy with full six-dimensional (6D) quantum-dynamics simulations of the dissociative photodetachment process on ab initio PESs.13 This reaction provides an excellent case for studying the effects of vibrational excitation on the decay of neutral complexes near the transition state for a bimolecular reaction. The system is characterized by three low-lying PESs, shown schematically in Fig. 1. Asymptotically, the ground state F(2P3/2) correlates adiabatically to HF + OH(2Π3/2) along the ground (X) state and also accesses the higher-lying excited (A) state, leading to formation of spin-orbit-excited HF + OH(2Π1/2).25–28 The third surface corresponds to the endothermic reaction of spin-orbit-excited F(2P1/2) + H2O to adiabatically form electronically excited OH(2Σ+) + HF products. Reaction barriers separate reactant-channel and product-channel complexes on both the X and A states,29–31 and the pre-reactive F–H2O complex was recently shown to possibly exert a strong influence on the overall reactivity.32
Reaction rate measurements on this system suggested a tunneling-mediated reaction at low temperatures, based in part on the observed kinetic-isotope effect and the observation of no temperature dependence.33 Wang and coworkers reported the first photoelectron spectroscopy experiments on the F−(H2O) at 6.42 eV,34 and the resulting fragmentation processes were studied by direct molecular-dynamic simulations.35 Nesbitt and coworkers carried out an extensive series of crossed-beam experiments on the F + H2O/D2O reaction using a combination of laser-induced fluorescence26,27 and high-resolution IR-laser absorption techniques for product detection.28 Despite the collision energies (∼0.25 eV) of these experiments falling well below the adiabatic barrier to the higher-lying A state, both OH(2Π3/2) and OH(2Π1/2) products were observed with a branching ratio of 0.69 : 0.31, suggesting that nonadiabatic surface-hopping occurs in the exit channel.27 The HF products were found to exhibit a vibrational population inversion, with the OH product behaving essentially as a spectator to the reaction.
The structure and energetics of the F−(H2O) anion itself have garnered significant attention from both experimental and theoretical perspectives.36–49 Johnson and coworkers used vibrational predissociation of F−(H2O)·Ar complexes to measure infrared spectra over 600–3800 cm−1, producing spectra dominated by the ionic hydrogen bond (IHB) that shuttles the shared proton between the F and O atoms.38,41,42,44 Features attributed to the fundamental and overtone of this mode were found to be strongly blue-shifted upon complexation with Ar, while Cl−(H2O)·Ar and Br−(H2O)·Ar are slightly red-shifted for the comparable mode.42 The strong blue shift was attributed to charge localization on F− in the ground state, giving the anion F−⋯H–OH character. Excitation of this ionic OH stretch leads to vibrationally induced intracluster proton transfer, forming an FH⋯OH− structure and resulting in a large anharmonicity.44 The competing effects of these different charge-localization states were proposed to result in the increased complexity of the F−(H2O)·Ar spectrum, with the large blue shift contributing to the significant experimental uncertainty in the frequency of the overtone of the IHB in the binary anion complex, F−(H2O)(2νIHB).42
These phenomena have made theoretical treatments of this system challenging and highlighted the need for accurate PESs and a full treatment of coupled motions. One dimensional analysis of the intramolecular proton transfer coordinate in F−(H2O) shows this delocalized charge state as a shelf in the PES contributing strongly to large anharmonicities for this IHB mode. Although calculated vibrational frequencies using this reduced-dimensionality approach were found to yield reasonable results, it was determined that a 2D model based on coupling between the OF stretch and the motion of the shared proton gave the most accurate results compared to the available experimental data for the F−(H2O) system.44 A number of recent theoretical studies have examined the vibrational structure of F−(H2O) at a much higher level of theory,45,47,49 including recent full dimensionality simulations of the photodetachment spectra of vibrationally excited F−(H2O), showing that excitation of the IHB accesses the more charge delocalized FH⋯OH− structure of the anion resulting in a change in the photoelectron spectrum.48 The effects of excitation of this anion IHB mode on the dissociative photodetachment of F−(H2O) is the focus of the present study.
In the following, we present a combined experimental – theoretical study of the effect of vibrational excitation of precursor anions on neutral-dissociation dynamics using PPC spectroscopy and 6D quantum dynamics calculations on ab initio anionic and neutral PESs. As outlined above, the neutral reaction coordinate is essentially along the motion of IHB in the parent anion. Thus, excitation of 2νIHB in F−(H2O) is expected to have a significant impact on the reactivity and product energy disposal. The experimentally observed effects of precursor anionic vibrational excitation on the dissociation dynamics of neutral F·H2O are presented and discussed in light of the extensive theoretical predictions.
2. Experimental methods
Experiments were carried out on a PPC spectrometer, shown schematically in Fig. 2, and described in detail elsewhere.14,50,51 Precursor F−(H2O) anions at m/z = 37 were synthesized in a coaxial-pulsed plasma discharge, stabilized/activated by a 1 keV electron beam acting on a pulsed supersonic expansion of C2F6/HeNe (20%/80%) passed over a 10% NH4OH/H2O mixture operating at 20 Hz. The resulting ions were skimmed, accelerated to 7 keV, re-referenced to ground in a 30 cm long tube, and mass-selected by time-of-flight (TOF). The ion packet was then injected into a cryogenically cooled electrostatic ion-beam trap (EIBT), where it was phase-locked to the output from a Ti:sapphire regenerative amplifier (Clark MXR CPA-2000; λ = 775.5 nm, repetition rate 1037 Hz, pulse width 1.1 ps) using an RF oscillator. The ion packet was then repetitively probed perpendicularly using third-harmonic radiation (258.5 nm, hνUV = 4.80 eV) over a 48 ms trapping period. The resulting photoelectrons were collected on an event-by-event basis, extracted orthogonal to the ion- and laser-beam axis, and imaged onto a time- and position-sensitive detector.
Information on the center-of-mass (CM) electron kinetic energy (eKE) and laboratory-frame recoil angle were obtained from the arrival time and position for each photoelectron. When appropriate, optimal resolution was achieved by equatorially slicing the resulting photoelectron spectrum, ensuring selection of only those photoelectrons with minimal z-velocity. This imposed detector acceptance function (DAF) necessarily results in a reduction of signal intensity at larger photoelectron velocities, which can be corrected for by dividing the experimental intensity distribution, N(eKE), by the acceptance function of the z-velocity slice to provide DAF-corrected intensity distributions, P(eKE).52 Calibration of the photoelectron detector using F− resulted in a sliced z-velocity component resolution of ΔE/E ∼ 3.8% (unsliced ΔE/E ∼ 7.2%) at 1.4 eV electron kinetic energy.
Neutral photofragments recoil out of the EIBT in a cone along the propagation direction of the ion beam, and were detected in coincidence with the photoelectron using a four-quadrant time- and position-sensitive multiparticle detector. This allows calculation of the product masses and kinetic energy release (KER) for each event containing momentum-matched neutral particles. Calibration of the multiparticle neutral detector using the dissociative photodetachment of O4− yielded a kinetic energy release resolution of ΔE/E ∼ 10%.50 The mass resolution m/Δm ∼ 15 does not allow for a direct discrimination of F + H2O versus HF + OH products.
Observation of the effect of vibrational excitation of precursor ions was achieved by irradiating the ion packet with infrared (IR) light coaxially in a counter-propagating fashion prior to injection into the EIBT as described in ref. 14 and shown in Fig. 2. Due to limitations in the IR-laser power at the νIHB fundamental of the anion, these experiments were carried out at 2νIHB. A precise experimental number for this transition has not been measured, with reported values in the range 2815–2930 cm−1.38,42,44 Earlier theoretical values, which are highly sensitive to level of theory and anharmonic treatment, covered an even larger range: 2844–3055 cm−1.36,39,44,45,47,48 The data presented here were taken at hνIR = 2885 cm−1, extrapolated back to the uncomplexed F−(H2O) from the vibrational-predissociation spectra of F−(H2O)·Arn clusters.38,42 The extrapolated value of 2885 cm−1 for 2νIHB in the anion is slightly lower than the reported value of 2905(20) cm−1 for neat F−(H2O) given by Johnson and coworkers,42 but within the stated experimental uncertainty. Thus, although the clearest effects were observed at 2885 cm−1, additional data was collected both at 2900 cm−1 and at 2872 cm−1, as presented in the (ESI†). The anion packet was irradiated at 2885 cm−1, 2872 cm−1, or 2900 cm−1 using the tunable output of a 10 Hz Nd:YAG (Surelite III EX) pumped KTP/KTA optical-parametric oscillator/optical-parametric amplifier (OPO/OPA) system (Laser Vision, 100–300 mW, 5 ns FWHM, bandwidth ∼3 cm−1). The IR wavelength λ was set using the OPO signal and idler wavelength, measured using a spectrometer (Ocean Optics HR2000+) and independently calibrated using photoacoustic spectroscopy on CH4. Counter-propagation of the IR pulse relative to the fast-ion packet gives rise to a Doppler shift of 2 cm−1, corrected for prior to data collection. Source fluctuations were mitigated by data collection in an interleaved fashion, where ion source and PPC measurements were run at a 20 Hz duty cycle, all the while synchronized with the 10 Hz IR laser system such that every other 48 ms trapping cycle was carried out on an IR-irradiated anion packet. IR-irradiation of the anions occurred ∼10 μs before the first photodetachment laser shot in the EIBT. These interleaved IR on/off data sets were separated during post-processing, calibrated, and analyzed in the usual way. Effects of IR excitation are analyzed by examining difference spectra (IR–no-IR), and the experimental uncertainty for each is determined on a bin-by-bin basis, using
, where NIR and NnoIR correspond to the number of events in the IR and no-IR spectra. Based on analysis of the stable channel discussed in the ESI,† a best estimate for the excitation fraction of the anion packet is ∼4%. Although the best photoelectron resolution is typically achieved by slicing along the z-velocity component, the subsequent DAF correction resulted in increased noise at larger eKE when determining the difference spectra. Consequently, all difference spectra presented here use the full unsliced data set.
3. Theory
The theoretical results reported here were obtained from full-dimensional quantum dynamical calculations using the ab initio based global PESs of the lowest two electronic states of FH2O,30,31 and a semi-global PES for the anion F−(H2O).13 Both the X and A states of the neutral are responsible for the highly exothermic F + H2O ↔ HF + OH reaction, but the latter has a much higher barrier. Both PESs are dominated by a deep post-transition state well, featuring a hydrogen-bonded FH⋯OH complex. The equilibrium geometry of the anion is close to the transition-state geometry of the neutral PESs, providing an ideal system for transition-state spectroscopy.13 The quantum dynamical Hamiltonian is written in either the diatom–diatom Jacobi or atom–triatom Radau–Jacobi coordinates (Fig. S1 in ESI†) and the total angular momentum J is set to zero.23 The J = 0 assumption is reasonable as the experimental temperature of the anion is expected to be < 100 K based on measurements of the photodetachment of OH− under similar conditions,53 but there is no question that a range of rotational levels in the F−(H2O) anion will be populated. The photodetachment is modeled within the Condon approximation, in which the ejection of the electron from the anion is approximated by a vertical excitation to the neutral species, and the electron carries away no angular momentum in s-wave photodetachment. Quantum dynamics on the neutral PESs were followed using the Chebyshev propagator,54 and final state distributions were obtained by projecting the wavepacket onto the product states.51,55
In the process of carrying out the supporting theoretical results reported here, exact calculations of the low-lying vibrational energy levels were determined on the F−(H2O) anion PES. The results are listed in Table 1 with the corresponding normal mode vectors displayed in Fig. 3. The agreement with experimental44,45 and previous theoretical results47,49 is quite good. The vibrational state excited in the experiment is identified as the overtone of the ionic hydrogen bond at 2νIHB = 2838 cm−1. This is 47 cm−1 below the estimated experimental value of 2885 cm−1, but given the 50 cm−1 splitting observed in the IR absorption spectrum in the region of 2νIHB,42 this is reasonable agreement. Table 1 also shows that the fundamental frequencies for the out-of-plane wagging and water bending modes are such that a combination of these excitations is nearly degenerate with the first overtone of the ionic hydrogen bond, contributing to the complexity of the spectrum in this energy range.
Table 1
Comparison of calculated vibrational energy levels of three low-lying vibrational levels with experimental band origins for the F−(H2O) anion (see caption for Fig. 3)
| Mode | Expt.45 | Expt.44 | Theory (VCI on KTCB PES)47 | Theory (SLBCL PES)49 | Theory (KTCB PES, this work) | Theory (new PES, this work) |
--- --- ---
| νIW (F−–HOH) | 426.65 | — | 433.24 | 423.5 | 432.67 | 426.87 |
| νIP | 576.27 | — | 566.61 | 560.7 | 565.96 | 563.18 |
| 2νIW | 829.10 | — | 836.86 | 823.2 | 835.68 | 828.74 |
| νOOP | 1184.36 | 1083–1250 | 1146.62 | 1156.1 | 1154.85 | 1173.74 |
| νIHB | 1464.54 | 1430–1570 | 1456.71 | 1477.9 | 1464.01 | 1468.04 |
| νB | 1653.56 | 1650 | 1623.25 | 1716.8 | 1642.74 | 1663.28 |
| 2νIHB | 2915.91 | 2815–2930 | 2872.49 | 2837.3 | 2847.03 | 2838.01 |
The numerical parameters used in the dynamical calculations are collected in Table S1.† Due to the strong long-range interaction between HF and OH species, a large diatom–diatom distance must be used in the calculations. To reduce the size of the basis set, an L-shape grid and the vibrational basis were adopted. Due to exceedingly long lifetimes of the Feshbach resonances in the post-transition state well, the propagation was terminated at 20 000 to 25 000 steps, which roughly correspond to 1 ps. The energy spectrum of the final wave packet was analyzed by taking the Fourier transform of the autocorrelation function.54
4. Results
The most unbiased approach to examining the effects of vibrational excitation on the dynamics of the reaction in the present case of limited excitation of the parent anion beam is to analyze difference spectra from the interleaved IR laser-on/IR laser-off measurements. In the limit of no correlations between the subset of ions excited, their decay rates in the EIBT, the dissociation rate of the neutral complexes, and the subsequent dissociation dynamics, these difference spectra should correspond to the difference of the theoretical IR and no-IR spectra. If the F−(H2O) spectrum is a, and the F−(H2O)(2νIHB) spectrum is b, the difference spectrum for an excitation fraction of f will be given by the difference D = (((1 − f)a + fb) − a) = f(b − a) ∝ b − a within a normalization constant. The quantum dynamics calculations show that these results are compared to cover ∼1 ps wavepacket propagation, and are thus most sensitive to prompt dissociation processes, so the present work focuses on a comparison of theory with the two-dimensional PPC difference spectrum as well as the difference spectrum for the total kinetic energy release, ETOT, where ETOT,i = eKEi + KERi is determined for each event i and summed. Difference photoelectron spectra for stable and dissociative neutral products are provided in the ESI.† As noted in the experimental section, resolution has been sacrificed in favor of statistics in the work presented here, and no slicing of the photoelectron velocity distributions are used, so the resolution of the PPC and ETOT spectra are reduced relative to the study of F−(H2O) without IR excitation in ref. 13.
The difference PPC spectrum illustrating the effect on the dissociation dynamics of the nascent FH2O neutral complex following photodetachment of F–H–O overtone-excited anions at hνIR = 2885 cm−1 is shown in Fig. 4(a). For reference, the PPC spectrum of F−(H2O) without IR excitation taken simultaneously with this data is shown in Fig. 4(b). In the ESI, Fig. S6† shows PPC difference spectra for IR photon energies of 2885, 2872 and 2900 cm−1, as well as a null difference spectrum, to provide a measure of the magnitude of the effect. The 2885 cm−1 spectrum exhibits the strongest effect of vibrational excitation, although the 2872 cm−1 spectrum also shows statistically significant enhancement above a total energy of 1.0 eV. The 2885 cm−1 spectrum exhibits the strongest effect and is therefore the focus of this work. All diagonal energetic limits for the dissociative pathways for both the no-IR and IR cases are tabulated in Table S2,† with the limits in Fig. 4(a) shifted above those in Fig. 4(b) by the addition of hνIR such that hν = hνuv (4.80 eV) + hνIR (0.36 eV) = 5.16 eV is the total available photon energy. In both frames, the rightmost diagonal line corresponds to the energetic limit for the HF + OH products formed in their ground rotational and vibrational states, denoted KE
UV+IRMAX and KE
UVMAX for the IR-excited and no-IR cases, respectively. For convenience, product states are referenced herein in the general form (nHF, nOH). The energetics for formation of HF(nHF = 0) + OH(nOH = 1) and HF(nHF = 1) + OH(nOH = 0) products, denoted (0, 1) and (1, 0) respectively, are denoted in the difference spectrum by dotted blue lines. The black diagonal line at 0.26 eV in the no-IR PPC spectrum, and black diagonal at 0.62 eV in the difference spectrum correspond to formation of the reactants F + H2O. Both spectra reveal a strong preference toward vibrational excitation in the products, consistent with the HF product fragment being formed with considerable internal excitation, in agreement with previous experimental studies.13,28
The difference spectrum in Fig. 4(a) shows clear effects from IR excitation of 2νIHB in F−(H2O). The most obvious difference is the positive red band of events above the no-IR KE
UVMAX (1.03 eV, not shown on the difference spectrum for clarity, but shown in Fig. 4(b)) and below the IR-excited product KE
UV+IRMAX line (solid black diagonal line). This new band in the IR excitation difference spectrum is consistent with accessing the ground vibrational state of the HF + OH dissociation pathway, where the additional energy from the IR photon has been partitioned into translational motion of the resulting photoelectron, appearing at greater eKE but identical KER as the no-IR case. This is essentially a classic ‘hot-band’ in photoelectron spectroscopy, where vibrational energy in the anion is converted to eKE, leaving the dissociating molecular fragments with less total energy. This band consists of two broad features, one closer to the KE
UV+IRMAX limit at higher eKE and smaller KER, and a second, weaker feature appearing at lower eKE, just above the energetic limit for HF(nHF = 0) + OH(nOH = 1) (blue dashed line), with a long tail extending out to KER ∼0.65 eV. The first feature trails off at KER ∼0.40 eV, as in the no-IR PPC spectrum. The second feature, with the long KER tail, could be interpreted as rotational excitation in at least one of the HF and/or OH products. The large KER may result from increased Franck–Condon overlap with the more-repulsive A-state barrier that plays an important role in the (0, 1) and (1, 0) channels for the reaction with no vibrational excitation,13 but the low intensity means that these features are not as statistically significant.
The next prominent diagonal band of enhanced (red) signal in Fig. 4(a) appears below the energetic limit for (1, 0) excitation, lying just above the adjusted UV + IR energetic limit for the F + H2O reactant channel. As in the no-IR spectrum, this band covers the full range of allowed eKE and KER, notably without the distinct KER cutoff seen in the (0, 0) features. The peak enhancement signal falls below the energetic limits for the (1, 0) or (0, 1) product states by 0.2–0.3 eV, which can also be rationalized with production of these vibrationally excited products with significant rotational excitation.
The limited product mass resolution under the current experimental conditions prevents resolution of the F + H2O reactant channel from the HF + OH product channel. The energetic limit for formation of the F + H2O reactant-channel products is denoted by the solid white diagonal line at 0.62 eV in Fig. 4(a). The energetic limits for production of the (0, 2) and (2, 0) final states are denoted by black dashed diagonal lines. This convoluted portion of the difference spectrum is dominated by “depletion” (blue) signal. The depletion observed here corresponds to the most intense portions of the no-IR PPC spectrum, where the feature near eKE = 0.4 eV is attributed to the no-IR (1, 0) band and the long horizontal band covering the range KER = 0.0–0.4 eV has previously been observed in the no-IR spectrum as eKE = 0 eV is approached.13 IR excitation shifting some of these high intensity features seen in the no-IR spectrum makes it is plausible that depletion signal would be observed in these regions in the PPC difference spectrum. However, these features are also less significant statistically as will be further discussed in the next paragraph.
Integrating along the diagonal produces the ETOT difference spectrum, shown in Fig. 5, with the integrated IR and no-IR spectra providing a direct measure of the product state distributions prior to taking the difference. The energetic limits for dissociation to the respective pathways are the same as for the PPC spectra, and are indicated here with vertical lines. The top panel shows signal enhancement in red, with the vertical lines corresponding to the IR-excited energetics. The bottom panel shows signal depletion in the difference spectrum with the no-IR KEMAX limits indicated. Error bars represent the statistical uncertainty and are calculated for each bin using
. Again, the red enhancement data above the no-IR KE
UVMAX at ETOT = 1.03 eV provides unambiguous evidence for 2νIHB excitation of the precursor anions in a broad band, consistent with the possible rotational excitation in one or both diatomic products. This is also true for the enhancement data in the (1, 0) channel, spanning the (1, 0) to (2, 0) energy range and peaking at the energetic limit for appearance of the F + H2O reactant channel. This is consistent with vibrational excitation of the anion leading to enhanced production of the reactant channel following photodetachment. The trend of apparent rotational excitation in the products does not appear in the case of the (2, 0) channel, which peaks closer to the (2, 0) limit. At the threshold for production of OH(nOH = 1), the (0, 1) limit, there is a blue suppression feature, indicating that 2νIHB excitation does not favor production of vibrational excitation in the ‘external’ OH bond that becomes the nascent OH product after dissociation. Obviously, as eKE decreases, the product density of states increases and assignment of these overlapping spectral features becomes less conclusive. A second point is that the error bars below eKE = 0.5 eV become comparable to the observed differences, so these are not statistically significant.
5. Discussion
This system is an ideal one for studying the transition-state dynamics and the effect of vibrational excitation of the shared hydrogen atom on the neutral F + H2O → HF + OH reaction. This can be most clearly seen by examining the wavefunctions of ground state F−(H2O) and F−(H2O)(2νIHB) anions superimposed on the X state PES, as shown in Fig. 6. The two anion wavefunctions access significantly different coordinate spaces on the neutral PESs, providing an expectation that a significant vibration-induced effect on the reaction dynamics will be observed. In the anion ground state, the negative charge is located primarily on the F atom, giving the anion an F−⋯H⋯OH type structure. Photodetachment from this ground state probes near the transition-state geometry on the neutral PES. With additional internal energy in the IHB mode, the anion assumes a more charge-delocalized structure, FH⋯OH−.44,48 This increased FH⋯OH− character might be expected to provide better Franck–Condon overlap with the product-channel FH–OH complex well and the product-channel asymptote, but the quantitative effects are more complex. The additional available energy in the system is significantly above the X state barrier, reaching near the top of the A state barrier as shown in Fig. 1. As in the case of the earlier study of this system without vibrational excitation,13 the dynamical calculations revealed that the dissociation of the neutral FH2O complex prepared by photodetaching F−(H2O) produces both F + H2O and HF + OH products, with significant internal excitation. The branching ratio between reactants and products as well as the vibrationally resolved branching percentages (not energy resolved) for HF + OH are summarized in Table 2. These results show that dissociation to F + H2O reactants more than triples to 37% for photodetachment to the X state while the A state contribution increases from 0 to 14% upon excitation of 2νIHB.
Table 2
Calculated fractions of vibrational state-resolved channels in the photodetachment of ground state F−(H2O) anion and F−(H2O) with 2vIHB overtone excitation. The contributions of the X and A states are reported separately
| Channel | Fraction (%) | | (nHF, nOH) | Fraction (%) | |
--- --- --- |
| Ground state F−(H2O) | 2vIHB | Ground state F−(H2O) | 2vIHB |
| X: F + H2O | 13% | 37% | | | |
| X: HF + OH | 87% | 63% | (0, 0) | 9.4% | 19.3% |
| (0, 1) | 2.7% | 0.8% |
| (0, 2) | 1.6% | 0.3% |
| (0, 3) | 0.0% | 0.1% |
| (1, 0) | 60.3% | 13.4% |
| (1, 1) | 2.6% | 2.7% |
| (2, 0) | 9.6% | 26.3% |
| A: F + H2O | 0% | 14% | | | |
| A: HF + OH | 100% | 86% | (0, 0) | 24.7% | 60.7% |
| (0, 1) | 3.2% | 1.2% |
| (0, 2) | 0.0% | 0.1% |
| (0, 3) | 0.0% | 0.0% |
| (1, 0) | 71.8% | 18.0% |
| (1, 1) | 0.0% | 0.6% |
| (2, 0) | 0.2% | 5.0% |
The calculated X and A state photoelectron spectra for ground state F−(H2O) and F−(H2O)(2vIHB) are shown in Fig. 7. The energy zero is defined as the minimum of the FHOH product-channel H-bonded complex of neutral FH2O. The overall shapes of the spectra are similar, but the intensities vary considerably, reflecting the access of different regions of the neutral PESs. For the X state, the distinct bands separated by about 0.5 eV can be clearly assigned to different product HF vibrational levels. The fine structure within each band is due to Feshbach resonances, as pointed out in our earlier work.14,23 The A state spectra are much smoother, indicating fast and more direct dissociation, due to the more repulsive PES produced by the higher barrier in the A state shown schematically in Fig. 1, with the corresponding bands generally higher in energy than those on the X state spectrum. In the same figure, the photon energies of the two experiments (UV and UV + IR) are indicated by vertical lines. It is worth noting in passing that these calculated spectra are very different from the recent report using the reflection principle,48 which does not take into consideration the multidimensional neutral dynamics following the initial photodetachment.
A key element of the present comparison of experiment and theory is the explicit theoretical treatment of the neutral dynamics on the ps timescale, including the role of repulsive states, Feshbach resonances, and reactant and product complexes. The calculated PPC spectrum for F−(H2O)(2vIHB) shown in Fig. 8 captures the breadth of these results, resolved into contributions from the X and A state into the reactant F + H2O and product HF + OH channels. Integration over these spectra provides the product state distribution results in Table 2, showing that the HF + OH channel dominates, particularly in the A state, but more F + H2O flux is seen for the vibrationally excited anion. In the HF + OH channel, the HF product is highly excited while the OH product in mostly in its ground vibrational state. The larger population in the HF(nHF = 0) product channel stems from an improvement in the Franck–Condon overlap, but the enhancement of the HF(nHF = 2) product channel is in part due to the higher photon energy that accesses more final states in that manifold. The large excitation in the HF vibration is due to the fact that the HF vibrational mode is well aligned with the anion vibrational excitation.56 As shown in Fig. 6, the initial wavepacket for F−(H2O)(2vIHB) has a large span in the H–F coordinate, leading to significant vibrational excitation in the HF product.
Given the limitations in the experiment with respect to the fraction of IHB-excited anions (∼4%) and signal-to-noise in the difference spectra, it is invaluable to have these quantum dynamics predictions of the PPC and ETOT spectra to compare with the data. It is important to recognize two issues when comparing the experimental and theoretical PPC spectra. First, the theoretical results are obtained from a relatively short (∼1 ps) wavepacket propagation. As a result, they might differ from the experimental spectra that include events that occur over the flight time from the interaction region to the detector, 7 μs at the beam energy of 7 keV. As shown in the theoretical photoelectron spectra in Fig. S6,† dissociation on a 1 ps timescale typically involves high-energy short-lived resonances and direct processes. Indeed, an important assumption in comparing the theoretical results for the dissociation of FH2O and FH2O(2vIHB) is that the dissociation rate is equal on the 1 ps timescale. If the dissociation rate is equal and the photodetachment cross sections are identical, the experimental and theoretical results should be comparable as straightforward differences of the FH2O and FH2O(2vIHB) theoretical predictions versus the IR–no-IR difference spectra as discussed in the results section. Second, it is known that the anion has a complex vibrational structure,42 and may exhibit intramolecular vibrational energy redistribution (IVR), leading to mixing with other vibrational states nearby. In particular, as noted earlier, a combination band involving the out-of-plane wagging and water bending modes is nearly degenerate with 2vIHB, and excitation or IVR into this combination band could lead to product rotational excitation. As noted in the experimental section, the anions are excited 10 μs prior to trapping in the EIBT, and then the data is acquired over a period of 48 ms, so there is significant time for IVR to occur. In principle, this effect can be examined in the experimental data, but signal-to-noise limitations make that beyond the scope of the present work. The theoretical predictions will thus be compared directly to the experimental PPC and ETOT difference spectra shown in Fig. 4(a) and 5 using theoretical difference spectra assuming 5% F−(H2O)(2vIHB) excitation, shown in Fig. 9 and 10, respectively.
The enhancement signal observed in the PPC difference spectrum in Fig. 4(a), appearing below the IR-excited KE
UV+IRMAX and above the no-IR KE
UVMAX, can come only from ground-state HF + OH products accessed after vibrational excitation of the precursor anions. This shows that vibrational energy along the proton transfer coordinate in the precursor anion can be carried away by the photoelectron in a bound-free Franck–Condon photodetachment, and is seen in both the experimental and theoretical results. This is largely because the vibrational coordinate is well aligned with the neutral reaction coordinate near the transition state. Beyond serving as proof of concept for successful preparation of F−(H2O)(2νIHB), this band is largely consistent with the higher resolution data previously reported for the experiment with ground state F−(H2O).13 The signal grows in starting at the energetic limit for (0, 0) and appears as a fairly broad diagonal band. Both the X and A states contribute to the increased intensity in the (0, 0) band, with the larger KER contributions produced by dissociation on the excited A state. Dissociation on the A state yields products with a larger kinetic energy, which is consistent with the fact that the A state PES is energetically higher in the Franck–Condon region and more repulsive. The second feature appears just above the energetic limit for (0, 1) at a similar KER and has a weak and less statistically significant tail extending to larger KER. The larger KER tail is not seen in the theoretical PPC difference spectrum in Fig. 9. The fact that the second feature appears well below the product KE
UV+IRMAX in the experimental spectrum is consistent with formation of these products with considerable rotational excitation in at least one of the product fragments. One important difference between theory and experiment is that the experimental PPC difference spectrum does not indicate the significant formation of rotationally cold HF + OH products in the upper-left corner of the energetically allowed region, as seen in the theoretical PPC spectra in Fig. 8 as well as the high-resolution cold F−(H2O) spectra in ref. 13. Examination of the no-IR F−(H2O) spectrum in Fig. 4(b) also shows diminished intensity in that region, suggesting that perhaps the 20 Hz ion source used in these measurements did not cool parent anion rotational states as effectively as in the earlier experiment.
Vibrational excitation in the HF product is particularly sensitive to changes in reaction dynamics, with the theoretical results indicating a suppression of HF(nHF = 1) and enhancement of HF(nHF = 0, 2) in both the X and the A states, as shown in Table 2. In the experimental no-IR PPC spectrum (Fig. 4(b)), HF(nHF = 1) signal dominates, appearing as an intense spot near eKE = 0.4 eV, and as a band with a sharp onset at the threshold covering the full range of available KER as first reported in ref. 13. In the difference PPC spectrum, the enhancement signal for this channel falls noticeably below the (1, 0) limit, observed as a prominent red band that extends to lower energies beyond the F + H2O reactant KE
UV+IRMAX= 0.62 eV (annotated solid black lines in Fig. 4(a) and 5). As such, the enhancement-signal peaking at the F + H2O reactant-pathway energetic limit in the difference spectra was initially attributed to increased contributions from the (1, 0) channel where products are formed with substantial rotational excitation in at least one product-channel fragment. With the reduced resolution in the present experimental results relative to ref. 13, however, some of the observed enhancement near the F + H2O KE
UV+IRMAX limit may arise from the formation of the F + H2O products predicted to be significantly enhanced in the dynamical calculations for photodetachment of F−(H2O)(2νIHB). The product distributions in Table 2 show that the (2, 0) population is enhanced by anion vibrational excitation, particularly in the X state. This appears as a large feature in the calculated PPC spectrum for the HF + OH products formed on the X state in the (2, 0) band in the upper left frame of Fig. 8. On the other hand, in the experimental data, the HF + OH signal is superimposed on the F + H2O signal, which is also significantly enhanced. These overlapping features make interpretation of the experimental PPC spectrum at lower eKE and KER increasingly difficult.
The inference of enhanced rotational excitation in the HF + OH product channels for photodetachment is not supported by the theoretical predictions. In Fig. 11, the rotational state distribution of HF is displayed at several different eKE on the X and A states, respectively. Note that the rotational state distribution of OH is not shown as it is the same as that of HF because of the J = 0 constraint. All HF rotational state distributions indicate limited excitation in the HF product and this is true at all eKE values on both the X and A states. The strongest jHF excitation takes place at eKE = 0.20 eV (brown lines). The maximum rotational angular momentum quantum numbers of HF are close to 12 in all the panels, and the corresponding rotational energy in the HF + OH products is about 0.2 eV. The relatively small rotational excitation in both HF and OH is consistent with previous experimental26–28 and theoretical results4,30,32,57 examining the dynamics of F + H2O reaction as promoted by bimolecular collisions. It is interesting to note, however, that also in the ground state results for FH2O reported in ref. 13 one of the most significant differences between theory and experiment was a broad region of rotationally excited products in the (0,0) manifold. The eKE signature for these events was also shown in ref. 13 to correspond to a broad product channel Feshbach resonance as seen in the ‘stable’ photoelectron spectrum surviving after 2 ps wavepacket propagation. It is possible that some of the discrepancies in apparent rotational excitation seen in the present work likewise derive from the dissociation dynamics of long-lived resonance states that are not effectively captured in short wavepacket propagations, and the interactions of these states with the bending potential on the neutral surface. In addition, as noted earlier, IVR into states with significant bending excitation may also play a role given the fact that these PPC measurements are integrated over nearly 50 ms trapping times following IR excitation.
The role played by OH vibrational excitation is also worth mentioning. In the theoretical predictions for both F−(H2O) and F−(H2O)(2νIHB), vibrational excitation of OH is very low and actually reduced for F−(H2O)(2νIHB). In the experimental results in ref. 13 however, a clear feature at the threshold for (0, 1) OH-excited products was observed, with the full range of product KER, implying involvement of both the X and A states. This is also seen in the no-IR PPC spectra reported in Fig. 4(b). However, the difference spectra, particularly the ETOT spectrum in Fig. 5, shows that in accord with the theoretical predictions there is no enhancement of the (0, 1) channel with vibrational excitation. From a theoretical perspective, in general, the HF vibrational state distribution is inverted when excited state product channels open, while little vibrational excitation is found for the OH product. This trend is consistent with both experimental studies of full collisions26–28 and theoretical observations4,30,32,57 in the F + H2O bimolecular reaction. The experimentally observed (0, 1) products implies that there are likely some details of the anionic or neutral potential energy surfaces that need further examination. This includes the excited A state, since the large range of KERs observed at the energetic threshold for (0, 1) is consistent with direct dissociation involving repulsive regions of both the X and A states.
6. Conclusions
This study reports a joint experiment-theory examination of the effects of overtone excitation of the ionic hydrogen bond, F−(H2O)(2νIHB) on the dissociative photodetachment and thus the impact of parent anion vibrational excitation on the dynamics of the F + H2O → HF + OH reaction. Comparing the experimental and theoretical PPC difference spectra, the key features of enhancement and depletion are reproduced by theory quite well. These results offer insights into dynamics induced by photodetachment in a large and different Franck–Condon region on the neutral PES. There are, however, several outstanding differences between theory and experiment for both F−(H2O) and F−(H2O)(2νIHB). These include the role of rotational excitation in the HF + OH product channels and the level of OH vibrational excitation induced in the products. Extension of theoretical treatments to the decay of long-lived resonance states would be of great interest in resolving this question, as well as explicitly accounting for IVR prior to photodetachment. An improved A state PES is another area that deserves examination.31 In addition, at some level an explicit account for the photodetachment dynamics, beyond the s-wave approximation and taking into account the Wigner threshold law that governs the energy dependence of photodetachment cross sections will become important.58 Considerable improvements in the experiment would also be worthwhile, in particular increasing the excitation fraction for F−(H2O)(2νIHB) and an improvement in the eKE spectral resolution to the level achieved by Neumark and co-workers using the slow electron velocity map imaging (SEVI) technique.59 With sufficient resolution and signal-to-noise, the time-dependent eKE could provide a direct experimental probe of IVR in the excited F−(H2O) anion. SEVI-level resolution has yet to be achieved in three-dimensional coincidence photoelectron imaging measurements, however. Experiments with colder parent anions using a new cryogenically cooled octupole accumulator trap would also be useful in better understanding the role of parent anion rotational excitation on the observed dynamics.60 In summary, the present joint experiment-theory study of the effects of vibrational excitation on the F + H2O → HF + OH reaction has been an extensive undertaking at the ‘benchmarking Frontier’ for four atom reactions, and illustrates the state-of-the-art for full-dimensionality quantum dynamics of four atom reactions.
Conflicts of interest
There are no conflicts of interest to declare.
Acknowledgements
This material is based upon work supported by the U.S. Department of Energy, Office of Science, Office of Basic Energy Sciences under award number DE-FG03-98ER14879. J. Daluz contributed to the experimental measurements. We also thank the National Natural Science Foundation of China (91441107 to JM and 21573027 to JL) and the Air Force Office of Scientific Research (Grant No. FA9550-15-1-0305 to HG) for supporting the theoretical calculations.
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Footnotes
†
Electronic supplementary information (ESI) available. See DOI: 10.1039/c7sc03364h
‡
Present address: Illumina, Inc., 5200 Illumina Way, San Diego, CA 92122 USA.
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Effects of vibrational excitation on the F + H2O → HF + OH reaction: dissociative photodetachment of overtone-excited [F–H–OH]−
A. W. Ray, J. Ma, R. Otto, J. Li, H. Guo and R. E. Continetti,
Chem. Sci., 2017, 8, 7821
DOI: 10.1039/C7SC03364H
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7169 | http://ccc.chem.pitt.edu/wipf/Web/Degassing_Literature.pdf | THE SOLUBILITY OF GASES IN LIQUIDS RUBIN BATTINO' Department of Chewistry, Illinois Institute of Technology, Chicago, Illinois 60616 AXD H. LAWRENCE CLEVER Department of Chemistry, Etnory University, Atlanta, Georgia 30392 Received October 2.5, 1965 CONTENTS I. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 11. Nomencl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396 1 1 1 . Methods ratus. . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 A. Manome&-Volumetric Methods. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 1. 397 2. Degassing the Solvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398 3. The Effect of Temperature on Solubility Measurements . . . . . . . . . . . . . . . . . . 399 4. The Apparatus of Cook and Hanson. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 5. Saturation Metho . . . . . . . . . . . . . . . . . . . . 400 13. Mass Spectrometric b .
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. . . . . . . . . . . . . . . . . . . . . 407 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 s . . . . . . . . . . . . . . . . . . . . . . . . . 407 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 The Solubility of Oxygen in Rater as a Comparison Standard.. . . . . . . . . . . . . . . 1. Regular Solution Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 2. Cell Potential and Cavity Models.. . . . . . . . . . . . . . . . . . . . . . . 410 3. Ot,her Contributions to Theory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 4. Special Studies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 C. Partial Molal Volumes of Gases Dissolved in Liquids . . . . . . . . . . . . . . 413 D. GM ,Solubilities in Mixed Nonelectrolyte Solvents. . . . . . . . . . . . . . . . . . . 413 E. Solvent Surface Tension and Gas Solubility . . . . . . . . . . . . . . . . . . . . 415 F. SaltEffec ts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 G. Solubility of Gases in Biological Fluids. . . . . . . . . . . . . . . . . . . . . . . . . . 419 H. Effect, of Pressure on Gas Solubility. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 I. Solubility of Gases in Molten Salts and Glasses. . . . . . . . . . . . . . . . . . . 424 J. The Solubility of Gases in Molten Metals and Alloys.. . . . . . . . . . . . . . . . 426 VI. Solribilit,y Data. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. 453 VII. References.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453 B. Temperahre Coefficient of G a s Solubility. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412 I. IKTRODUCTION the solubility of gases in hunian tissues to the solubility The solubility of gases in liquids has been under quantitative investigation since the beginning of the nineteenth century. The last decade has seen some remarkable advances in theory, empirical correlations, systems studied, and apparatus. Much of the earlier work was more qualitative than quantitative. Gas solubilities have become increasingly more important for both the theoretical understanding of the liquid state and solutions, and for practical applications from (1) Department of Chemistry. Wright State College. Dayton, Ohio 4j431. of gases in molten salts and metals. This review brings up to date the earlier comprehen- sive review of Pllarkham and Kobe (393). An annotated bibliography (1907-1941) on the solubility of Ar, COZ, He, and N z in organic liquids was prepared by Croxton (115). The two books by Hildebrand and Scott (243, 244) contain chapters on gas solubility and many references. There are other papers containing either general correlations or many references (170, 206, 247, 248, 282, 353, 373, 485, 670). The review by Him- melblau (249) on the diffusion of dissolved gases in 395 396 RUBIN BATTINO AND H. LAWRENCE CLEVER liquids is of interest, and the review by Rowlinson and K Henry's law constant or Setschenow equation param- Richardson (509) on the solubility of solids in com- pressed gases may interest some readers. The limited literature on the solubility of liquids in compressed gases has yet to be reviewed. The literature since Varkham and Kobe was searched and critically evaluated. The principal medium of the search was Chemical Abstracts through Vol. 59, or the end of 1963. All pertinent articles abstracted through the end of 1963 were included, although many articles which were published in 1964 and 1965 were also included. Where the original paper was not examined, a Chemical Abstracts reference is supplied in the list of references. The emphasis in this review was placed on physical methods of determining solubilities and on reporting and discussing only equilibrium or saturation solubili- ties. Thus, the vast literature on the analysis of gases absorbed in liquids was ignored. Since most of these analyses involve Chemical methods, the latter was also ignored except for oxygen in water. The extensive literature on the solubility of gases in ponds, streams, and ground waters was not included because of the equilibrium criterion. There is an almost nonexistent dividing line between gas solubility (in liquids) and vapor-liquid equilibrium, especially at elevated pressures. The solubility o f the vapors of some substances was included where the experimental conditions were such that the pressure was lower than the normal vapor pressure. Also in- cluded are: (a) solubilities of gases in molten metals, alloys, salts, and glasses; (b) solubilities of gas mix- tures; (c) solubility in tissues and some biological systems; (d) solubility nomographs; and (e) partial molal volumes of gases in liquids. The large body of literature on the solubility of gases in plastic materials was excluded since much of this data is for plastic films. The solubility of gases in solids and substances whose composition would tend to be indeterminable (like molten slags) was also excluded. 1 1 . NOMENCLATURE The system of notation used in this review follows. Some specialized symbols which apply to a particular approach, and where we wished to use the author's own notation for clarity, are defined where they are used. Upper Case Symbols A arbitrary constant C 1 , C d c , C W AC, AEV molar energy o f vaporization Q molar Gibbs free energy H z AH, A B " AHy molar heat, of vaporization concentration of gas dissolved in the liquid phase concentration of gas dissolved in the gas phase weight solubility (see section IVD) heat capacity change on solution Henry's law constant (see section VD) molar heat of solution K i KP K O K' K, K", K." L M P P .
, Po P T Pi S Si", Si 3 2 S: AS" T To Ti v, V V S X P , XI X P i Y eter Henry's law constant (to fit Eq 5) Henry's law constant ( t o fit Eq 6) Henry's law constant ( t o fit Eq 7) Henry's law constant, modified Henry's law constant, reduced limiting Henry's law constants in water and in salt Ostwald coefficient (see section IVB) molecular weight pressure solvent vapor pressure total pressure partial pressure or vapor pressure Kuenen coefficient (see section IVA) solubility of gas in pure solvent, or s d t solution partial molar entropy of gas in solution molar entropy of gas in gas phase molar entropy of solution temperature, "K critical temperature reduced temperature volume, molar volume volume of solvent volume of gas molar volume of gas in cc/mole at 0' partial molar volume at infinite dilution of gas partial molar volume at infinite dilution of electro- mole fraction solubility of gas in solution ideal mole fraction solubility of gas in solution vapor phase mole fraction arbitrary constants salt concentration, moles per liter gas concentration, moles per liter activity coefficient of dissolved gas in salt-free solu- activity coefficient of dissolved gas in salt solution gravitational constant excess Gibbs free energy Boltzmann constant salting-out parameter salting-out parameter for salt x nonelectrolyte self-interaction parameter salt-effect parameter molality pressure radius of spherical gas atom temperature, "C decimal fraction of solute in solution volume solution lyte Lower Case Symbols tion Greek Letter Symbols Bunsen coefficient (see section IVA) interaction parameters absorption coefficient (see section IVB) compressibility of pure solvent activity coefficient Hildebrand solubility parameter LennardJones force constant volume fraction or vapor phase activity coefficient density surface tension SOLUBILITY OF GASES IN LIQUIDS 397 1 1 1 . METHODS AND APPARATUS The variety of approaches which have been used to determine the solubility of gases in liquids is an ade- quate testimonial to man’s ingenuity. The equipment used ranges in complexity and cost from mass spec- trometers to the simple van Slyke apparatus, in time from minutes to many hours, and in precision from the purely qualitative to the highly precise. Primarily physical methods will be discussed in this section, although, in specific instances chemical methods are sometimes more precise. There are several reasons for omitting chemical methods (except for oxygen in water). First, they are normally specific for a particular gas and thus do not show general applicabil- ity. Second, it is quite difficult to distinguish between “(chemical) absorption” methods and those which involve purely equilibrium solubilities. The literature for the analysis of dissolved gases was in general ignored. Third, gas absorption studies usually involve complex chemical equilibriums, and, since their purpose is to study the chemical equilibrium involved, the result frequently is that insufficient information is available to properly evaluate the study as an equilib- rium gas solubility. Physical methods can be divided into two broad clas- sifications : saturation methods in which a previously degassed solvent is saturated with a gas under condi- tions where appropriate volumes, pressures, and tem- peratures may be determined; and extraction methods where the dissolved gas in a previously saturated solu- tion is removed under conditions where appropriate P, V , and T values may be evaluated. Equilibrium between the gas and liquid phases has been obtained by shaking a mixture of the two, by flowing a fiIm or stream of the liquid through the gas, by bubbling the gas through the liquid, or by flowing the gas over the liquid held stationary on some supporting medium (as in gas-liquid partition chromatography). The de- termination of the quantities of the components in the gas and liquid phases has been carried out chemically, volurnetricalJy (and with the supporting use of manom- eters), by mass spectrometer, and by gas-liquid partition chromatography. These methods will be discussed below. A. MANOMETRIC-VOLUMETRIC METHODS First, some general considerations are discussed. Cook (103, also see 104) gives an excellent analysis of the problems involved in gas solubility determinations, the magnitudes and importance of contributing fac- tors, and his approach to achieving a truly high- precision (& 0.0570) gas solubility apparatus. He is to be commended for a major contribution to gas solubil- ity determinations. In gas solubility determinations contributing fac- tors like purity of materials and the measurement of the physical properties of pressure, volume, and tem- perature are usually adequately determined. The discrepancies, often large, between published values which appear in the literature (for example, the solu- bility of atmospheric gases in water and sea water con- tinues to be measured and debated) are most probably due to other factors. Cook and Hanson (104) list these as being one or more of the folIowing: (a) failure to attain equilibrium; (b) failure to completely degas the solvent; (c) failure to ascertain the true amount of gas dissolved; and (d) failure to make certain that the transfer of gas from a primary container to the ap- paratus does not involve contamination. The last two factors can be controlled by proper design, calibrations, and corrections. It is better to determine the initial quantity of gas in the dry state, Le., free of solvent, since P-V-T data can be used, especially if there is any uncertainty as to whether the gas is saturated with solvent vapor. Where known, the more accurate equations of state should be used since even at 1 atm pressure deviations from the ideal gas law can be signifi- cant. The first two factors will be discussed further. For equilibrium solubility determinations the at- tainment of equilibrium is of prinie importance. In flow systems the attainment of equilibrium is checked for by determining the solubility at several rates of flow. In nontlow systems the agreement between solubility determinations found v,-ith varying both the vigor of the stirring (or shaking) and the pressure of the gas above and below the equilibrium pressure serves as the main criterion. For each type of apparatus it is important to provide these checks by varying the conditions. 1. The Solubility of Oxygen in Water as a Comparison Standard It would be a boon to workers in the field to have a reliable standard for comparison. The criteria here mould be that the solvent and gas are readily available in high purity, and that at some convenient conditions a number of workers would have obtained substantially identical values having followed different approaches. The solubility of oxygen in water at 25” and 1 atm meets these criteria. There have been several recent determinations following different approaches, in- cluding both chemical and physical methods, which are in substantial agreement (152, 166, 231, 318, 424, 425, 581), and the value suggested as a standard is the Bunsen coefficient of 0.02847. Considering these seven values, the average deviation is 0.00006 in the Bunsen coefficient or 0.21y0. This deviation is just about within the independent experimental error of the seven de- ternlinations. The values are summarized in Table I along with other values of interest. As a further aid Table I1 contains smoothed values of the solubility of 02, Nz, and Ar in water as a function of temperature. 398 RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE I SOLUBILITY OF OXYGEN IN WATER' Esti- mated Bunsen error, coed Worker Date % at 25' Winkler 1889 0.02831 Fox 1909 0.02890 Morrison and Billett 1952 0.5 0.02848 Truesdale, et al. 1955 0.27 0.02800 Steen 1958 0.12 0.02863 Elmore and Hayes 1960 0.20 0.02824 Morris, et al. 1961 0.27 0.02850 Klots and Benson 1963 0.16 0.02848 Douglas 1964 0.25 0.02830 Montgomery, et al. 1964 0.26 0,02849 Ref 657 194 428 610 581 166 425 318 152 424 be initially gas free, and for extraction methods, where the gas is to be completely extracted. The single criterion for complete degassing which has been most widely used is the reproducibility of a measurement or its corollary, the agreement between several workers for the same measurement. This criterion is open to two criticisms: (a) the repetition of systematic errors, and (b) the perpetuating of older measurements as standards when they may have been superseded by more reliable measurements. Two additional criteria (104) for complete degassing should be mentioned. The first is a dynamic method in which a vacuum thermocouple gauge is located between Green 1965 0.27 0.02843 231 the last liquid nitrogen trap (on the apparatus) and before the vacuum pump. During the degassing pro- Av (all 11) = 0.02843 cedure in which the solvent is simu~taneous~y boiled tained in recent studies is 0.02847. and pumped on, the vacuum gauge reading slowly falls from greater than 1000 p to the base pressure of the pump. When the pressure is at the base pressure of the pump, this indicates that only negligible amounts Av dev (all 11) = 0.00016 Recommended standard value based on the agreement at- 2. Degassing the Solvent The complete removal of gas from a liquid is im- portant for saturation methods, where the solvent must TABLE 1 1 (Units: Bunsen Coefficient X 103) SOLUBILITY OF NITROGEN, OXYGEN, AND ARGON I N WATER .4T 1 ATM Temp, OC: 0 6 10 15 20 25 3 0 35 40 50 Nitrogen Hamburg. 23.79 21.05 18.81 17.03 15.70 14.68 . . . ... ... . . . Adeney and Beckera ... 21.22 18.70 16.96 15.55 14.35 13.27 ... hlorrison and Billettb (428) ... 19.25 17.36 15.86 14.63 13.64 12.82 12.17 11.23 Douglas (152) 20.91 18.75 17.05 15.57 14.41 13.45 . . , ... ... Klots and Benson (318) . . . 21.18 18.99 17.24 15.84 14.66 13.45 . . . . . . . . . Fox. 23.19 20.68 18.63 17.02 15.72 14.65 13.75 12.99 12.33 11.16 Winkler" 23.12 20.50 18.29 16.56 15.18 14.10 13.19 12.35 11.64 10.71 Oxygen Winklera 48.89 42.87 38.02 34.15 31.02 28.31 26.08 24.40 23.06 20.90 FOX. 49.24 43.21 38.37 34.55 31.44 28.90 26.65 24.85 23.30 20.95 Truesdale, Downing, and Lowden (610) 47.65 41.73 36.98 33.20 30.27 28.00 26.29 24.94 ... . . . Elmore and Hayes (166) 49.30 43.15 38.16 34.12 30.88 28.24 25.97 . . . ... ... Douglas (152) 37.97 34.03 30.95 28.30 26.20 . . . ... Steen' (581) . . . 42.80 38.37 34.38 31.06 28.65 . . . ... . . . . . . Klots and Benson (318) . . . 43.03 38.14 34.23 31.11 28.48 . . . Green (231) 49.43 43.31 38.39 34.39 31.12 28.43 26.23 2 . . . Morris, Stumm, and Galal" (425) 49.36 43.33 38.34 34.27 31.03 28.50 26.58 . . . ... Morrison and Billettb (428) 38.32 34.35 31.13 28.48 26.30 24.48 22.97 20.71 Montgomery, Thom, and Cockburn (424) 48.19 43.04 38.13 34.21 31.06 28.49 26.41 24.70 23.31 ... Argon Estreichera 57.80 50.80 45.25 40.99 37.90 34.70 32.56 30.54 28.65 25.67 Winklera 53.0 ... 42.0 35.0 30.0 27.0 Antropoff" 56.1 . . . 43.8 37.9 34.8 . . . ... Lannungn . . . . . . 41.1 37.1 33.6 31.4 28.9 27.1 25.3 . . . Eucken and Hertzberg (173) 52.6 . . . . . . . . . 33.6 . . . . . . . . . ... . . . Holland and Clever (260) . . . . . . . . . . . . . . . 31.5 28.8 26.9 25.2 ... Koenig (327) . . . 37.9 35.8 32.5 ... Douglas (152) . . . 46.89 41.80 37.53 34.05 31.23 28.88 ... ... . . . Ben-Naim and Baer" (38) 53.64 47.16 41.89 37.62 34.13 31.21 28.65 . . . ... ... Klots and Benson (318) . . . 47.13 41.82 37.59 34.21 31.37 . . . . . . ... ... Brothers Printing, Ann Arbor, Mich., 1943. by us using four-constant equation. . . . ... . . . ... . . . . . . . . . . . . . . . . . . Friedman (197) 38.2 31.1 Morrison and Johnstoneb (429) 41.7 36.7 33.5 30.7 28.1 26.4 24.7 22.1 ' Values as corrected for impurities and smoothed in LandoltrBornstein, "Physikalisch-Chemische Tabellen, 1936 Edition," Edwards Smoothed Calculated from Morrison's equations and converted to volume of water. SOLUBILITY OF GASES IN LIQUIDS 399 of noncondensible gases are present. After the base pump pressure has been reached, an additional quantity of solvent should be evaporated to be doubly certain that the solvent is completely degassed. The second criterion requires cushioning the degassed solvent between mercury. I f a bubble appears, the degassing operation should be repeated. This test is quite sensitive since gas bubbles much smaller than 0.001 cc can readily be detected. Since solution rates are slow, these gas bubbles will persist for several minutes. The most frequently used method of degassing a liquid is to boil away a portion of it under vacuum. This procedure can be considered to be a batch binary distillation. The Ramsey-Rayleigh equation for this type of distillation predicts that the evaporation of as little as 0.1% of the solvent should reduce the gas content by several 1000-fold. However, this equation assumes that equilibrium conditions prevail between the gas and the liquid and under actual degassing operations this is not the case. In practice l0-20?& of the solvent is evaporated, and one (or both) of the criteria mentioned in the previous paragraph should be employed. Another degassing procedure employs the method of pumping on the frozen solvent. This procedure gives good results when it is important to minimize the loss of solvent, but it is necessary to repeat the process at least three times and pumping on the frozen solvent for periods of an hour or longer in each cycle. It is more important to test for completeness of degassing when employing this method than when evaporating large amounts of the liquid. Clever, et al. (97), employed a method for degassing that operated in two stages. The first stage involved pumping on boiling solvent to evaporate a portion of it, and to remove perhaps 90% of the dissolved gas. Then in the second stage this preliminarily degassed liquid is sprayed through a fine nozzle into an evacuated flask. This procedure was found to give rapid and com- plete degassing. A similar method was employed by Baldwin and Daniel (22) where they permitted an oil sample to slowly drip into an evacuated vessel. They found that this removed 97-98% of the dissolved gas. The technique of removing a gas completely from a liquid by stripping the liquid through bubbling an inert gas through the previously saturated liquid will be discussed in the section on gas chromatographic methods where this technique is essential to the method. 3. The E$ect of Temperature on Solubility Measurements A full analysis of the effect of temperature control on gas solubility measurements is given by Cook (103) who points out that there are four factors to consider: (a) the temperature coefficient of the solvent vapor pressure ; (b) the temperature coefficient of solubility, or more realistically, the change in the equilibrium partial pressure of the dissolved gas with temperature at an approximately constant concentration; (e) the temperature level of the experiment; and (d) the pressure level of the experiment. The magnitude of these factors will be quite dependent on not only the system studied but also on the type of apparatus used. After a complete analysis of all contributing factors, Cook found that temperature control to 0.1 O was ade- quate for an over-all precision of O.OS~o for his ap- paratus for the system Hz-n-heptane in the range - 30 to 50". It would appear that temperature control of 0.1" should be more than adequate for most purposes. 4. The Apparatus of Cook and Hanson (103, 104) The apparatus and procedure of Cook and Hanson will be described in somewhat more detail than other methods since they achieved an unusually high level of precision for a physical method. This apparatus is also described by blader, Vold, and Vold (384). It was mounted on a steel plate which was shaken at a fre- quency of about 170 min-' and an amplitude of about 3/8 in. The shaking mechanism imparted a horizontal motion to minimize pressure disturbances due to vertical accelerations of the mercury present. The entire apparatus was housed in an air thermostat con- trolled to at least ~ 0 . 1 " . A millimeter scale mounted on the steel plate and observed with a cathetometer provided a reference point for appropriate readings. An auxiliary gas-charging apparatus was connected to the solubility apparatus at point 12. Once the appara- tus was set up and charged all manipulations were made remotely so as not to disturb the temperature equilibrium. Volumetric calibration was performed by weighing mercury displaced from pertinent sections of the apparatus which are the gas bulbs, A and C, the gas burets, B and D, and the solvent bulb, E, from point 19 down to and including the solvent buret. Also calibrated was the secondary gas buret and bulb from point 19 down to stopcock 2. The appropriate capaci- ties of the various parts of the apparatus are: solvent bulb, 200 cc; primary gas bulb, 22 cc; secondary gas bulb, 1 cc; solvent buret, 0.1 cc/cm; primary gas buret, 0.25 cc/cm; and secondary gas buret, 0.015 cc/cm. An outline of the procedure used follows. First, the gas and solvent bulbs are filled with mercury, and then by displacing mercury into the manometer a sufficient quantity of solvent is admitted through the joint at point 12 followed by about an inch of air to provide space for boiling the solvent. The solvent is degassed by pumping and boiling using the thermocouple gauge check described earlier, and also by checking for residual gas bubbles by confining the solvent between mercury admitted through stopcocks 3 and 6. About The basic apparatus is shown in Figure 1. 400 RUBIN BATTINO AND H. LAWRENCE CLEVER ~ l , 2 , 3 a 4 , 5 , B : 2mm. "HYVAG' sToPcoucs Y 6,?.9 : 4 mm.'nYWG STQPCOCKS S TRANSFER TUEL MARY GAS BULB. 20s. p A R Y GAS BURETTE.6mm.I.D. SECONDARY GAS BULB Figure 1.-The gas solubility apparatus of Cook and Hanson (103, 104). Reprinted from U. S. Atomic Energy Commission Report UCRL-2459 by permission of the author. l0-20% of the solvent is evaporated in the degassing process which takes several hours. The gas-charging system is connected at point 13, and after the system is purged and checked for tightness an appropriate quan- tity of gas is admitted through stopcock 1. The gas- charging system is removed and the thermostat closed and set to control at 25". By appropriate pressure, volume, and temperature readings the quantity of gas charged may be calculated from an applicable equation of state. An interesting point here is that a change of 1-2 ern in the gas pressure can cause adiabatic heating or cooling sufficient to require an additional half-hour to attain temperature equilibrium. The gas is then completely transferred into solvent bulb E through the gas transfer tube by alternately applying vacuum through 6 and pressure through 7. Stopcock 6 is left open during the dissolution process where the entire apparatus is shaken. The initial dissolution is hastened by applying about 10 psig to the manometer. When a residual gas bubble of about 1 cc remains, it is drawn over into the secondary gas bulb and buret and its volume (which can be measured to 0.005% as a check on the rate of solution and the attainment of equilib- rium) is measured. The residual gas is returned to the solvent bulb and shaking is repeated. The process is repeated until equilibrium is attained, and it is also repeated at two or more pressures as a further guarantee of attaining equilibrium. The equilibrium pressure is determined to 0.1-0.2 mm. The entire process is repeated at successive temperatures, thus yielding for a single charging a set of measurements for a range of temperatures and pressures. The solvent volume is determined at 35" by draining mercury between calibrated points on E out through 5, weighing it, and calculating the solvent volume as the difference between the calibrated volume of E and the volume of the mercury which was drained off. The authors estimate the reproducibility as better than =tO.l'% in ref 104 and better than 10.05% in ref 103. This is by far the most precise apparatus for determining gas solubilities by physical methods reported to date. A serious disadvantage is that a good estimate of the solubility must be known beforehand since the apparatus is designed such that 95% of the gas should be absorbed. The size of the solvent bulb E would have to be changed for different systems (or ranges of solubility). This disadvantage is partially offset by permitting the reasonably large pressure range of 0.5-1.5 atm to be available through the manometer. 5. Saturation Methods The apparatus (Figure 2) designed by Morrison and Billett (427) was based on attaining saturation by flowing a liquid film through the gas. A modification of this design (for full details see ref 32) was used by Clever and co-workers (95-98), Saylor and Battino (530), and Koenig (327). The degassed solvent in M is injected drop by drop through A and flows in a thin film down the absorption spiral B and into gas buret C. The absorption section is initially charged with gas. Saturated solvent flows out of E at such a rate that the levels in C and in the leveling buret D are kept the same. Solvent from E is collected and measured. Readings of C give the volume of gas dissolved, while the volume of solvent is the amount collected at E together with the volume accumulating in C and D. The absorption section is thermostated. For high solubility gases an additional gas buret is sealed into the system above spiral B. The variation of the flow rate within wide limits had no appreciable effect on the solubility. The reproducibility was f 0.5%. A disadvantage of this apparatus is that normally only SOLUBILITY OF GABES IN LIQUIDS 401 one gas in one solvent at one temperature and one pressure can be made on a single charging of the ap- paratus. The dissolution vessel and connections for the ap- paratus of Ben-Naim and Baer (38) is shown in Figure 3. Degassed solvent is transferred to the previously evacuated vessel A through point D. This vessel is connected to a gas buret and manometer at point C and all of these parts are immersed in a water thermo- stat (=!=0.03") with a window for viewing. The solvent fills A entirely and the liquid levels are in the capillaries h, 1, and k when the system is charged with solvent saturated gas. Appropriate readings on the manometer, gas buret, and dissolution vessel are recorded. Dis- solution of the gas is caused by switching on a magnetic stirrer which forces liquid up through capillaries h and k into bulbs a and b. The gas enters A through the capillary 1 and dissolves at the cone-shaped interface which is formed. At an initial high stirring rate some solvent is forced through m into A and within 2-3 min about 99% of the gas has dissolved. Equilibrium takes an additional 4 hr to achieve with gentle stirring. The over-all precision is estimated as &0.2%. A microgasometric technique (originally developed by Scholander (544) whose paper should be read for details on microgasometric analysis) embodying some modifications of Steen's apparatus (581 )was used by Douglas (152) to determine the solubility of oxygen, argon, and nitrogen in distilled water. This method takes advantage of the fact that the ratio of absorbed gas volume to liquid volume is constant at a given equi- librium pressure. Equilibration takes place within 30 min, small samples are used, and the estimated precision is f 0.25-0.50~0. The temperature was regulated to f 0.01 '. The procedure can be understood by referring to Figure 4. With G and D filled with mercury and H filled with water, degassed water is introduced into the side arm D with a 10-cc syringe and a blunt tipped needle. The plug E is replaced. With H half-filled with water and F filled with mercury to the top of the capillary, a conventional gas transfer pipet (filled with the pure gas) is seated on the capillary and the gas meniscus pulled down to the mark on the capillary. The micrometer is set (using the leveling bulb) to a zero reading and then some of the gas (using the micrometer) is pulled into the absorption chamber G. The water is removed from H except for an in- dicator drop in the capillary below H. When the sys- tem has equilibrated the gas volume is read. The gas- free water is tilted in from the side arm, mechanical shaking begun, and the indicator drop kept at its mark by adjusting the micrometer. When equilibrium is attained the final gas volume is read. Then by open- ing stopcock s-1 the liquid volume is read by screwing the micrometer in until the liquid level reaches the mark on the capillary. The procedure is said to be Figure 2.-The gas solubility apparatus of Morrison and Billett Reprinted from the Journal of the Chemical Society by (427). permission of the Chemical Society. I\ d Figure 3.-The gas solubility apparatu3 of Ben-Naim and Baer Reprinted from The Transactions of the Faraday Society by (38). permission of the Faraday Society. simple, precise and rapid. Another apparatus using small samples (5 cc) is described by Thomsen and Gjaldbaek (599). Burrows and Preece (85) used a manometric method to determine the solubility of helium in three low- pressure oils. Their mixing chamber is shown in Figure 5. After the vessel J was charged with the degassed liquid it was charged with gas by draining liquid from J and weighing it. The stirrer moves the magnet-containing ring R (stirring the liquid in the thermostated bath) which nioves the perforated steel disk (stirring the oil and gas in the mixing chamber). The temperature is varied in 20" intervals from 20 to 80". The precision appears to be of the order of 2%. Ridenour, et al. (501), used a manometric method to 402 RUBIN BATTINO AND H. LAWRENCE CLEVER k a , e -e Figure 4.-The gas Solubility apparatus of Douglas (152). Reprinted from The Journal of Physical Chem6rtSy by permission of the copyright owners, The American Chemical Society. Figure 5.-The mixing chamher of the gas solubility apparatus of Burrows and Preeee (85): B, three-way tap; J , mixing cham- ber; N, tap; Q, heating-bath liquid; R, ring of magnets; V, perforated steel disk; W, driving crank. Reprinted from The Journal o f Applied Chemistry by permission of the editor. determine gas solubilities (to about 3%) in molten paraffin and microcrystalline waxes. Baldwin and Daniel (22) describe a method for the determination of gas solubility (to 1%) which is particularly useful with viscous liquids. Yeh and Peterson's apparatus (667) was used for gas solubilities in liquids and they report their precision as 0.5Y0. Loprest (374) developed an apparatus for the rapid determination of the solubility of gases in liquids at various temperatures. The apparatus was designed for a precision of lye and it is possible to obtain solubility data over a wide temperature range and at several partial pressures of the gas with a single charging of the system. Solvent does not come into contact with mercury surfaces. Wheatland and Smith (642) de- scribe a simple gasometric method for the determination of dissolved oxygen in water and saline water, and their method is precise to 0.5%. Iflots and Benson (318) give details for an apparatus for determining the solu- bility of Nz, Oz, and Ar in distilled water in the tem- perature range 2-27" and with an estimated accuracy approaching 0.1%. Koonce and Kobayashi (330) designed an apparatus for the solubility of methane in n-decane for the pres- sure range 200-1000 psia and the temperature range of -20 to 40°F. They estimated their probable error as +1.5%. Zampachova (674) also described a simple manometric method for determining the solubil- ity of vapors in liquids at elevated pressures. Krichev- skii and Sorina (337) give details for an apparatus for determining hydrogen solubility in cyclohexane in the temperature range 20-60" and up,to 700 atm. Cox and Head (114) described a novel appa- ratus for determining (to l%) the solubility of COz in H F solutions. The solubility chamber was gold-plated. Karasz and Halsej paratus for measuring the solul neon in liquid argon over the ten 87.5"K, and with the general limits 01 error as 0 .
1 % . The apparatus described by Kobatake and Hildebrand (322) was used for determining (to 0.3% and the range 5-30') the solubility of a large number of gases in various solvents. Swain and Thornton (592) de- scribe an apparatus for measuring the solubility of the methyl halides in light and heavy water at 29 and 40'. The Tsiklis and Svetlova (615) w a s used for of HCl, Ch, NOCl, NO, and H2S in cycl 3 range 10-40". Kogan and Kol'tsov (328) aescrme an apparatus precise to 0.25% for the solubility of Clz in CCI, in the range - Bodor, et al. (54), describe an apparatus for < gas absorption in liquids in the range -80 to U-. B. MASS SPECTROMETRIC METHODS The mass spectrometer may be used for gas solubility determinations. Basically, the technique calls for apparatus of the solubility .ohexane in thi .\ . ., 1 (294) detail an ap- bility of helium and iperature range 83.9- .. " - _- -20 G-71'. ietermining ^ ^ SOLUBILITY OF GASES IN LIQUIDS 103 outgassing a sample of a gas-saturated solvent, trapping the gas, and then analyzing the gas by mass spec- trometry. Such equipment was described by Cseko (116-118) for the determination of the solubility of gases and gas mixtures in liquid ammonia. In this way he determined the solubility of argon in liquid ammonia at room temperature and the pressure range 20-100 atm. Cantone and Gurrieri (91) used mass spectrometry to analyze water samples for CH4, 02, N2, and Ar. Faulconer and co-workers (17714, 241A, 469A, 479) describe mass spectrometric techniques for the analysis of various gases in blood. A prime advantage of using the mass spectrometer as an analytical tool is the ability to determine the ratio of dissolved gases and isotope effects in dissolved gases. Benson and Parker (42) describe the technique they used for the determination of N2/Ar and N2/02 ratios in distilled water and sea water in some detail. The precison with which they determined these ratios was given as fl%. O f course, by using standard values for the solubility of any one of the gases in a pair, the solubility of the second gas may be determined. For a more accurate determination of the N2/Ar ratio the extracted gas was cycled through a "Vycor" furnace at 700" filled with copper turnings. Benson and Parker (41) used the sanie basic technique, modified for collecting samples at sea, to determine Nz/Ar and nitrogen isotope ratios in aerobic sea water. IClots and Benson (319) determined the isotope effect in the solution of oxygen (320z and 340z) and nitrogen (2sN, and 29N2) in distilled water in the temperature range 2-27'. They found the extrapolated values of the ratio of the Henry's law constants at 0 ' to be 1.00085 f 0.00010 for nitrogen and 1.00080 f 0.0001S for oxygen. Benson (42A) discusses applications of the mass spectrographically determined ratios to problems in oceanography. C. GAS CHROMATOGRAPHIC METHODS In gas-liquid partition chromatography (glpc) a liquid (normally high boiling) is supported in a column on an inert stationary phase. A carrier gas continually bathes the liquid, and one can assume an equilibrium exists between the passing carrier gas and the amount of this gas which has dissolved in the liquid. A third substance (vapor, gas, or a mixture) is transported in the carrier gas and is partitioned between the carrier gas stream and the stationary liquid phase. Partition coefficients are determinable from knowing the reten- tion volumes and column characteristics. By also knowing the column dimensions and the quantity of partitioning liquid it is possible to determine gas solubilities. Of course, gas solubilities measured in this may are measured for systenis under special con- straints: (1) the liquids are restricted to high boilers; (2) the solubility is for a gas or vapor in a film of liquid (supported on a solid phase) and in which the carrier gas is already equilibrated; (3) the process involves steady states and transient equilibriums as the carried component is swept through the column; (4) in the portion of the gas stream where the carried component is, the carrier gas concentration is less than normal and as this portion of the gas stream passes any given point some of the carrier gas must be out- gassed; and (5) it is difficult to ascertain the carried component partial pressure as it is swept along as a band which may or may not be symmetrical in its concentration distribution. Despite such difficulties Zorin, Ezheleva, and Devyatykh (685) were able to determine the soubility of CH4, CZHZ, C3H6, isobutaiie, isobutylene, and COz in certain solvents, and their results were compared with static methods of determin- ing solubility. O f further significance is the fact that partition co- efficients which are determinable from glpc are in a way gas solubilities, but the considerations mentioned in the last paragraph must be kept in mind. Two texts which give more details on determining partition coefficients (and other matters) are ref 124 and 80. Kurkchi and Iogansen (349) used glpc for determining the solubility of CZHZ, C3H4, and C4H4 in several sol- vents and found that their results agreed within 5% with literature values. They describe their all- paratus in detail. The major use of gas chromatography in gas solubility determinations has been as an analytical tool for the quantification of gases extracted from saturated solu- tions of liquids. In this sense the gas chromatographic methods are similar to the mass spectrometric methods described in the last section. The advantage of gas chromatography as an analytical tool is the relative simplicity, low cost, and rapidity of the rneasuremen ts. Reproducibility is of the order of 1-2%, but uncertainties in the extraction of the dissolved gas niakes the over-all precison a bit poorer. Some unique extraction methods have been devised. A particularly simple procedure was described by RIcAuliffe (400) who determined the solubility in water of CI-CS hydrocarbons. i i 0.05-0.10-cc sample of the hydrocarbon-saturated water was directly injected into a gas chromatograph fitted with a suitable fractionator containing a drying agent to absorb water, the released hydrocarbon passing directly into the chromatograph column. Hydrocarbon concentrations were determined by measuring areas under curves and comparing with calibrations arrived at by using known amounts of the pure hydrocarbons. The same author (401) used a similar technique to determine the solubilities in water of 65 (paraffin, cycloparaffin, olefin, acetylene, cycloolefin, and aromatic) hydrocarbons. Swinnerton, Linnenbom, and Cheek (593) determined the amount of dissolved gases in aqueous solutions by stripping the 404 RUBIN BATTINO AND H. LAWRENCE CLEVER LEAN HELIUM IN RICH HELIUM OUT MUM WATER LEVEL RICH HLO IN I LEAN Ht0 OUT Figure 6.-Detail of the stripper used by Williams and Miller (652) to remove dissolved gases from a saturated liquid. Re- printed from Analytical Chemistry by permission of the copyright owners, The American Chemical Society. gases from the solutions in an all-glass sample chamber which is divided into two parts by a coarse glass- fritted disk. A known quantity of the saturated liquid is admitted to the sample chamber through a rubber serum cap. The carrier gas coming up through the fritted glass disk in a stream of fine bubbles completely strips the solution of its dissolved gases quickly and effectively. The same authors (594) described an im- provement of their sampling procedure. Elsey (167) describes a similar procedure for the determination of dissolved oxygen in lubricating oil. Wilson and Jay, et aE. (656), used a fritted-glass sample chamber coupled with gas adsorption chromatography for the analysis of blood gases and found that their approach gave results of equal precision to the van Slyke technique. In a subsequent paper Jay and Wilson, et al. (278), utilized the same apparatus for determining absorption coefficients for nitrous oxide in distilled water and in whole blood. The gas chromatographic technique again gave results comparable to the van Slyke tech- nique, but the former method has some advantages in speed, the ability to resolve gas mixtures readily, and overcoming the disadvantage found in the van Slyke method of having to make a correction for the small but variable amounts of gas which are not extracted. The problem of stripping or extracting dissolved gas was solved by Williams and Miller (652) in an interest- ing way. Figure 6 shows the details for their device which is used for stripping on a continuous basis (the gas-saturated liquid flows countercurrent to the carrier gas which also serves as the inert stripping gas) with intermittent sampling and analysis by a commercial gas chromatographic unit,. They compared several common techniques for purging water: dynamic and static vacuum, with and without manual and ultra- sonic agitation; ultrasonic treatment alone; and purg- ing with an inert gas (argon and helium). The most effective system tested was inert gas purging at Aow rates of 500-1000 cc/niin of 100-cc water samples. This technique removed 9598% of the dissolved oxy- gen in 15-30 sec, where the next best technique of dynamic vacuum with agitation took 1-2 min to re- move the same quantity of gas. In the gas chromato- graphic approach the rapidity of removal of dissolved gas is extremely important. The stripping unit (Figure 6) is about 4 in. long and 2 in. in diameter. As the mylar disks rotate through the liquid phase (kept at an opti- mum level of about one-fourth of the stripper volume), a thin flm of liquid (which is being continuously re- newed) is spread over their surfaces and exposed to the gas phase. The rate of gaseous exchange is very rapid. They found essentially 100% gas removal for helium/ water ratios of 10: 1 through 1 : 2 for up to 100 ml/min flow rates. Since it is necessary to have both accurate knowledge and control of the gas and water flow rates, a ratio of 1: 1 was arbitrarily chosen for their work to simplify the calculations. This simple and efficient gas stripper should find wide use. Ikels (266, 267) used a gas chromatographic tech- nique to determine the solubility of nitrogen and neon in water and extracted human fat. Kruyer and Nobel (347) measured the solubility of hydrogen in five sol- vents by stripping the gas from the solvent and measur- ing areas under an expulsion curve. The method was said to be precise to 13%. D. CHEMICAL METHODS FOR DISSOLVED OXYQEN Chemical methods have been long used for the de- termination of dissolved oxygen in pure water, natural waters, and aqueous solutions. In recent years there has been much controversy over oxygen solubilities in water, and this has sparked many new studies. The manometric approaches were discussed earlier, and most workers take the results of Mots and Benson (318) to be the most reliable. The Winkler method (657) and modifications of it have been among the most popular and the most accurate. Briefly, the Winkler method involves the oxidation of freshly precipitated manganous hydroxide by the dissolved oxygen to form manganic hydroxide. This step is favored by high pH. The solution is then made acidic under which conditions the manganic ion oxidizes iodide. In the presence of excess iodide the iodine is largely present as the complex triodide. In the last step the jodine is titrated with thiosulfate which is oxidized to tetrathionate. Excess thiosulfate is back- titrated aniperometrically with standard potassium iodate reagent. The equations for these steps are Mn2+ + 20H- = Mn(OH)2 2Mn(OH)2 + l/202 + HzO = 2Mn(OH) 2Mn(OII)a + 6H+ + 31- = 2MnZ+ + IS- $. 6 H a I + I- = Ia- I + 2500- = 21- + s101- The method depends on strict control of pH and iodide concentration. Possible errors in the Winkler method have been recently and extensively discussed (91A, 91B, 424). SOLUBILITY OF GASES IN LIQUIDS 405 The work of Truesdale, et at. (610), began the modern determinations of dissolved oxygen. They critically evaluated earlier work and made new meas- urements but, unfortunately, their work contains a systematic error and is low by some 2.5% Dr. B. A. Southgate, who is Director of Water Pollution Research at the Water Pollution Research Laboarory, Steven- age, Herts., England, believes that the true story is contained in ref 424, although both studies were carried out at his institution. In part, a private communication from Dr. Southgate states, "One thing which has come out of this work (424) is that it is pretty obvious that a very large proportion of the determinations of dissolved oxygen made before about 1961 or 2 must have been incorrect for the same reason that the solubility values determined by Truesdale (610) were incorrect. The biggest source of possible error of course occurs if one uses the present-day accurate values for solubility with incorrect determinations of concentration in water, and from these two values calculates the oxygen deficit.'' The paper by Montgomery, Thom, and Cockburn (424) described their modification of the Winkler method, and with their improved procedure they determined the solubility of oxgyen in pure water (0.4-27') and in sea water (2-27'). The agreement hetween their results and that for other workers (for solubilities in pure water) is shown in Figure 7. The ordinate in this figure is mg 0 2 / 1 .
, where the oxygen solubility expressed in this unit goes from 14.63 at 0" to 6.47 at 40". They critically discuss earlier work, and in particular showed that the values of Truesdale, et al. (610), were low due to losses of iodine vapor. They found that the effect of initial concentration on the rate of loss of iodine outweighs that of initial tem- perature, so that the loss of iodine is greater for water samples saturated with oxygen at a low temperature than for those saturated at a high temperature. This conclusion was verified experimentally, although others (166, 231) and Truesdale, et al. (610), have speculated on the reasons for the "low" values. A description of the Montgomery, et al., procedure follows. The sample is collected by standard techniques in a 65-cc bottle with a well-fitting ground-glass stopper. Without delay, 0.4 ml of manganous sulfate solution (480 g MnS01.4Hz0/1.) is added below the surface, followed by 0.4 cc of alkaline iodine solution added at the surface. (The alkaline iodide solution is prepared according to the method of Pomeroy and Kirschman.) The bottle is stoppered so sts to exclude air bubbles and shaken by rapidly inverting 12-15 times. After the precipitate has settled to the lower third of the bottle, it is shaken again (essential) and allowed to settle completely. Sulfuric acid (1 cc of a solution containing 400 cc of the concentrated acid per liter) is added down the neck of the bottle, which ; +0.2, I I I d 5 +0.1 g . E 4 0 - 2 -0.1 a g :: A LI -0.2 3 4 -0.3 m ' 0 e - 0 -0.4 .
e + .- c a" -0.5 0 10 20 30 40 Temperature, W. Figure 7.-Comparison of oxygen solubility values in pure water obtained by various authors with the values of Mont- gomery, et al. (424). (1) Whipple, G. B., and Whipple, M. C., J . Am. Chem. Soc., 33,362 (1911); (2) Truesdale, et 01. (610); (3) Carlson, T., Akad. Afh. Stockholm, 1912; (4) ref 424; ( 5 ) Klots and Ben-on (318); (6) Winkler (657); (7) Elmore and Hayes (166). Root mean square deviation of the work in ref 424. Reprinted from The Journal of Applied Chemistw by permission of the editor. is restoppered (excluding air "bubbles) and the contents mixed by shaking. After 10 min an aliquot is removed by pipet and titrated without delay with 0.0025 N thiosulfate solution. The thiosulfate is standardized against an iodine solution which is prepared by adding 20.00 ml of 0.0025 N potassium iodate solution to a few milliliters of water containing a quantity of alkaline iodide reagent equivalent to that present in sample titrations. Iodine is liberated by addition of the ap- propriate quantity of 40% sulfuric acid, which must be mixed thoroughly with alkaline solution before titra- tion is begun, or too high a factor will be obtained. The calculated concentration of dissolved oxygen is multi- plied by 1.012 to allow for dilution by the manganous sulfate and alkaline iodide reagents. The end point of the titration is determined amperometrically. Ap- parently, one of the important factors in these chemical methods is technique, and some practice is required. Tables IV and V in Montgomery, et al., provide some interesting comparisons. The first table compares the results of their procedure with five others (showing excellent agreement with Elmore and Hayes (166)). The second table shows the effect of speed of manipula- tion and titration in standard modifications of the Winkler method. The work of Montgomery, et al., deserves careful study. Elmore and Hayes (166) undertook an independent check of the solubility of oxygen in water. They carried out some 260 determinations in 52 replicate groups of experiments in the range 1.8-29.3'. The standard deviation in the measurements is of the order of 0.2%, and the results are presented in a table from 0-30' in 406 RUBIN BATTIFO AND H. LAWRENCE CLEVER 0.1" intervals. They followed a modification of the Winkler method using an amperometric end point. This paper critically evaluates earlier work and they conclude that their values are "the most satisfactory of those presently available for application to natural stream conditions." Morris, Stumm, and Gala1 (425) determined the solubility of oxygen in water by both manometric (11-30" and a precision of about 0.5%) and chemical (5-29" and a precision of about 0.3%) methods. Their work is in good agreement with the best recent measure- ments. They suggest that a possible reason for the results of Elmore and Hayes being slightly low at 30" is that they may have omitted making the temperature correction for a mercury barometer. This amounts to a negative correction of about 0.4% at 25". The most recent determination of oxygen solubilities in pure water and sea water has been that of Green (231) in the range 0-35". This independent work is in excel- lent agreement with Klots and Benson (318) and with Rlontgomery, et al. (424). The error in this work is estimated at 3~0.27%. Green's thesis contains an excellent analysis of earlier work and sources of error in the Winkler method, a modification of which he used. It is gratifying to note the excellent agreement among recent workers, and it appears that with this agreement and the attendant explanations of earlier discrepancies that a truly definitive set of values for the solubility of oxygen in water has been attained. Wheatland and Smith (642) used both a gasometric method and the Winkler method and found that their results from the two approaches agreed within experi- mental error (about 0.2%). Czerski and Czaplinski (123) determined the solubility of oxygen in air- saturated liquids by stripping the liquids with a carrier gas and then passing this through an electrochemical detector. The probable error is estimated at 0.9 mg Ozb. E. iWSCELLAKEOUS METHODS Enns, Scholander, and Bradstreet (169) give details of a method they used for the determination of the solubility of Ozl nT2, Ar, He, and COz in water and sea water at hydrostatic pressures up to 102 atm. For all gases examined the equilibrium pressure increased about 14% per increase in hydrostatic pressure of about 100 atm. The method also permits the calculation of par- tial molal volumes of the dissolved gases. Buell and Eldridge (79) describe an apparatus for gas solubility at high pressures where glass systems cannot be used. Khiteev (309) describes a mercury-free, high-pressure solubility apparatus for gases in petroleum. Safronova and Zhuze (518) also describe an apparatus for high- pressure, high-temperature solubilities in crude oils. Miner's apparatus (421) was constructed for measure- ment of the solubility of O2 and N z in liquid C02 at pressures up to 1000 atm and in the range -40 to 32". At the low-temperature end Denton, Lucero, and Roellig (143) describe an apparatus for the solubility of He in liquid hydrogen. The apparatus of Hu and MacWoocl (265) was designed for determining the solubility of gases in liquids at 113-181°K and 0-40 atm. An apparatus for the solubility of He, Ne, Ar, and Xe in molten fluorides at 0.5-2 atm and 600-800" is described by Grimes, Smith, and Watson (232). The molten fluoride is first saturated with a gas. Then the dissolved gas (in a known volume of the molten fluo- ride) is stripped with a second inert gas and the sample collected and analyzed on a mass spectrometer. Ryabukhin (511) gives details for an apparatus for determining the solubility of Clz in fused chlorides at The most commonly used method for gas solubilities in molten materials is Sievert's method. This is basically a gas-handling method where the total number of moles of gas introduced into the system is determined by measuring P, V , and T; and then by measuring the equilibrium pressures in the calibrated system the quantity of absorbed gas may be calculated. This approach is also used for gas-solid absorption determinations. Gas extraction methods are also eni- ployed. Some recent papers describing apparatus for gas solubilities in molten metals are ref 537 and 472. Mulfinger and Scholze (435) describe an apparatus for solubility in molten glasses. Bar-Eli and Klein (28) describe a method for de- termining gas solubility by measuring the rate of re- action between a gas and its solvent. This method should prove useful in those systems where the gas reacts with the solvent. Alexander (10) describes a microcalorimeter which he used to directly determine heats of solution of Ne, Ar, Kr, and Xe in water. Other papers which give details on gas solubility ap- paratus are: ref 480-XO in aqueous solutions; ref 38-S02 in aqueous ammonia solutions; ref 470-CHzCHC1 in methanol and trichloroethylene ; ref 239, 512, 616-CzHz solubility in various solvents; ref 234-solubility in waxes in the range 200-400". In gcneral, most of the papers on the solubility of gases in liquids give detailed descriptions of the pro- cedure. The gas solubility tables can serve as a general guide for references to apparatus applicable to particular problems. 700-1050". IV. XETHODS OF EXPREBSIKG GAS SOLUBILITY Gas solubilities have been expressed in a great many ways. The more popular of these along with inter- conversion formulas are presented. Since there are so many methods of expressing solubility, it is extremely important that each paper present a careful exposition of the manner in which their solubilities were calculated SOLUBILITY OF GASES IN LIQUIDS 407 and to also include a sample calculation to be doubly certain. This was emphasized by Markham and Kobe (393) but bears repetition along with repeating much of what they said about definitions. A. T H E BUNSEN COEFFICIENT, CY The Bunsen coefficient, CY, is defined as the volume of gas, reduced to 0" and 760 mm pressure of mercury, which is absorbed by the unit volume of solvent (at the temperature of the measurement) under a gas pressure of 760 mm. When the partial pressure of the gas above the solvent differs from 760 mm, it is corrected to this pressure by Henry's law. By way of example an equa- tion which can be used to calculate the Bunsen co- efficient is a = [I (V, z T 3 $ ) ( i ) ] ( T) (Eq 1) where P, is the partial pressure (in mm Hg) of the gas above the solution, T is the absolute temperature, V, is the volume of gas absorbed (at T and the total pres- sure of the measurement), and V, is the volume of the absorbing solvent. I f the solvent has a nonnegligible vapor pressure, then P, = PT - P, where PT is the total pressure in the system and P, is the solvent vapor pressure. Equation 1 obviously reduces to V, 273.15 17, T a = - - The corrections to standard conditions assume ideal gas behavior. Since real gases do not follow the ideal gas law, it is extremely important to specify the gas equation of state used for the correction to standard conditions. For most gases and conditions the dif- ference is negligible, being less than 1%, but the method of correction should still be specified. The Bunsen coefficient is sometimes just referred to as the absorption coefficient or the coefficient of absorption. The Kuenen coefficient, X, is the volume of gas (in cubic centimeters) at a partial pressure of 760 mni re- duced to 0" and 760 mm, dissolved by the quantity of solution containing 1 g of solvent. Thus the Kuenen coefficient is proportional to gas molality. B. THE OSTWALD COEFFICIENT, L The Ostn-ald coefficient, L, is defined as the ratio of the volume of gas absorbed to the volume of the ab- sorbing liquid, all measured at the same teniperature. The Ostwald coefficient is then L = V,/V, (Eq 3) For the reaction Gas (in liquid phase; C1) = Gas (in gas phase; Cg), the Ostwald coefficient may be written as where Cl is the concentration of the gas in the liquid phase and C, is the concentration of the gas in the gas phase. The Ostwald coefficient is in reality an equilib- rium constant, and as such is independent of the partial pressure of the gas as long as ideality may be assumed. However, to fix the value of the Ostwald coefficient, the temperature and the total pressure must be designated. If the total pressure is kept at 760 mm, then the volume of gas absorbed, reduced to 0" and 760 mm by the ideal gas laws, per unit volume of liquid is fre- quently designated as p, an absorption coefficient. It is important to clearly specify the method of calculating the solubility since p sometimes gets confused with a. C. THE HENRY'S L.~W CONSTANT The equation for a gas in equilibrium with a liquid may be written as Gas (in liquid phase; X I or Cl) = Gas (in gas phase; P, or C,). Henry's law can then be presented as P, = KlXl (Eq 5 ) P, = K2CI (Eq 6) c, = K,Cl (Eq 7) or in the case of a dilute solution of the gas as From the last equation it is noted that L = l/Kc, In the above equations X is the mole fraction and, of course, solubility may be expressed in terms of mole fractions. The volume fraction, molarity, and molality can also be used to express solubility. The Henry's law constants, particularly Kz, can be satisfactorily used to express solubility, but it must be remembered from thermodynamics that Henry's law is applicable only over a restricted range for dilute solutions and that Henry's law in practice is frequently just a limiting law. The method of calculating the Henry's law constant must be specified. The practice of converting solubility data from the experimental pressure to a partial gas pressure of 760 inm by applying Henry's law usually introduces no errors if the pressure range is reasonably small. D. T H E WEIGHT SOLUBILITY, c, The weight solubility, C,, is recommended by Cook (103) as a more logical unit than either the Bunsen or Ostwald coefficients. C , is defined as the number of moles of gas, with the partial pressure of the gas being 760 mm, per gram of solvent. This unit has the ad- vantage of essentially being a ratio of weights, thus permitting easy conversions and making certain calcula- tions simpler. E. INTERCONVERSIOS OF THE SOLUBILITY EXPRESSIONS (a) From the Bunsen coefficient B = (~(760 - Ps)/760 (Eq 8) 408 RUBIN BATTINO AND H. LAWRENCE CLEVER 7 8 9 IO II 12 13 14 15 16 I ? 18 - I f ’ ’ ’ ’ ’ I ’ ’ I ’ -R I n x2 Figure 8.-Entropy of solution of gases. I, He; 2, H2; 3, Ne; 4, Nf; 5, CO; 6, 02; 7, Ar; 8, CH,; 9, COZ; 10, Kr; 11, Xe; 12, GHt; 13, C2H4; 14, C2Ha; 15, SFa. Reprinted from “Regular Solutions” (244) by permission of the authors and the copyright owners, Prentice-Hall, Inc. , Englewood Cliffs, N. J. L = a(T/273.15) = ,3(T/273.15)(760)/(760 - Ps) (Eq 9) c w = a/v’op (Eq 11) P . is the partial pressure of the solvent, pa is the density of the solution, u is the decimal fraction of solute in solution, Vo is the molal volume of the gas in cc/mole at O”, and p is the density of the solvent at the tempera- ture of the measurement. + 760 (Eq 12) 17.033 X 106p, QM, Ke = 17,033/~ = K1 = 22,414 X 760 lOOOQ (Eq 13) M, is the molecular weight of the solvent. The units of Kl are those of pressure (mm Hg), and only for the very soluble gases does the constant term of 760 mm alter significantly the value of K1 calculated from the first term in Eq 12. The units of K2 are (mm Hg) (liters of solvent)/mole of gas. (b) From the Ostwald coefficient Q = L(273.15/T) (Eq 14) K, = 1/L (Eq 16) c w = L/Vtp (Eq 17) /3 = L(273.15/T)(760 - PB)/760 (Eq 15) Vt is the molal volume of the gas in cc/mole at the temperature of the measurement. (c) From the Henry’s law constant, K1 17.033 X lO6p (KI - 760)M. C Y = The 760 mm in the denominator may be neglected unless it is appreciable with respect to K1. (d) From the weight solubility, C, x = CWMd(1 + CWM,) (Eq 19) V. SOLUBILITY THEORY AND RELATIONSHIPS A. GAS SOLUBILITY AND THEORIES OF SOLUTION 1. Regular Solution Theory Regular solution theory has been most consistently applied to gas solubilities by Hildebrand and co- workers. The two books by Hildebrand and Scott (243, 244) contain excellent individual chapters on gas solubilities, namely, Chapter XV in ref 243 and Chapter IV in ref 244. Hildebrand and Scott (243, p 4) define regular solutions with the following sentence: “A regular solution is one involving no entropy change when a small amount of one of its components is transferred to it from an ideal solution of the same composition, the total volume remaining unchanged.” There are two equations based on regular solution theory frequently used to calculate gas solubilities. The first is 0.4343 7 2 -log x 2 = -log xz‘ + --- (61 - 62)’ (Eq 20) RT In this equation X2 is the mole fraction gas solubility, X2’ is the ideal gas solubility (calculable from Raoult’s law), V2 is the partial molal volume of the gas in the solution, and the 6’s are solubility parameters where the subscript 1 refers to the solvent. The solubility parameter, 6, sometimes referred to as the cohesive energy density, is the square root of the energy of vaporization per cc or 6 = (hEv/v)”’, where hEv is the molar energy of vaporization and P is the molar volume. For solutions where the molecules differ in size (Flory-Huggins model) the following equation is used -log $2 = -log Xz’ + 0.4343 ( 1 - - ;)$l+ where I $ is the volume fraction and P 1 is the molar volume of the solvent. Upon introducing some sim- plifying approximations the previous equation becomes -log x 2 = -log x z ’ + log T P 2 + Vl ( 2) 0.434372 (61 - 62)’ (Eq 22) 0.4343 1 - T + - RT For gases above their critical temperature the terms X;, 62, and P2 are evaluated by various extrapolations and approximations. Gjaldbaek and Hildebrand (219) SOLUBILITY OF GASES IN LIQUIDS 409 - 4 5 in a paper on the solubility of chlorine in n-perfluoro- heptane and other liquids justified the substitution of the partial molal volume of the gas in the solution for for molal volume of the pure gas. Despite the many assumptions made in the derivation of Eq 20, 21, and 22 they have proved to be remarkably satisfactory for solutions where both solvent and solute are nonpolar, and have been shown to give a fair approximation for slightly polar solvents. In some cases experimentally determined gas solubilities have been used in conjunc- tion with the above equations to arrive at empirical values for the gas solubility parameters. Among others Clever, et al. (97), have done this and also applied regular solution theory to rare gas solubilities (95-98). Three recent papers by Thomsen and Gjaldbaek (599-601) have compared calculated with experimental solubilities for a variety of gases in a variety of solvents. In ref 600 they show an interesting correla- tion (a straight line) between liZ calculated empirically from solubility measurements and JZ calculated from [(AHv - RT)/a]‘/’ where AHv is the heat of vaporiza- tion. Gjaldbaek and co-workers in a series of papers (216-223, 358), which are most useful and interesting for the range of systems investigated and the reliability of the measurements, have supplied the most extensive testing of Eq 20, 21, and 22. Gjaldbaek and Anderson (221) included an additional term in Eq 20 to account for the dipole contribution to the energy of vaporization for polar solvents. Jolley and Hildebrand (282) critically reviewed the literature for reliable gas solubilities which suited their purpose and drew a number of conclusions which are quoted below and illustrated in Figures 8, 9, and 10 taken from ref 244: “(a) For a given series of gases at 1 atm and 25’ dissolving in a series of solvents, log X2 decreases with increasing solubility parameter, 6 1 , of the solvent. (b) For different gases in the same solvent, log X z increases linearly with increasing Len- nard-Jones force constant, E/&, of the gas. (c) The entropies of solution of different gases in the same sol- vent vary linearly with R In Xz, and extrapolate at X2 = 1 to the entropy of condensing to pure liquid the vapor of the solvent from a hypothetical pressure of 1 atni. The temperature coefficient of solubility may thus be obtained from its isothermal value. Solubility increases with temperature from common solvents when Xyz is less than about and vice versa. (d) The partial molal entropy of solution of any one gas from 1 atni to the same mole fraction (here is nearly bhe same in all solvents except fluorocarbons, where it, is a little greater. In any one solvent, it increasea in going to gases with smaller force constant. This is attributed mainly to increase in freedom of motion of the adjacent molecules of the solvent rather than to change in the behavior of the gas molecule in a ‘cage’.” Kobatake and Hildebrand (322) added a b c , d , e f ,9 h , i j h , 1 . 1 , , , , I , , , I -0 .
5 1 I I I I I I I Figure 9.-Gas solubilities at 25” and 1 atm in log X2 us. the square of the solubility parameters of the solvents, 81: (a) C,F,,, (b) (C4F9)3?J, (c) c-CBF~~CF~, (d) ~-C&S, (e) CC12F.CCLF2, (f) n-C-iH16, (g) CaHnCH3, (h) C -c 6 1 I 1 2 , (i) CCh, (i) Ce”,CH,, (k) C,,HB, (1) CS2. Unpublished results reprinted by special permis- sion of the author, Professor Joel H. Hildebrand. tlk Figure 10.-Solubility us. “force constant” of gases. Reprinted from “Regular Solutions” (244) by permission of the authors and the copyright owners, Prentice-Hall, Inc., Englewood Cliffs, N. J. data for many more systems, and Archer and Hilde- brand (15) added data on the solubility of CF, and SFa in nonpolar solvents to the “regular” solutions. Prausnits (483) applied regular solution theory to gas-liquid solutions. He considered a three-step 410 RUBIN BATTINO AND H. LAWRENCE CLEVER process: (a) isothermal compression of the pure gas from its partial pressure and the pure liquid from its vapor pressure to the isometric mixing pressure; (b) isothermal, isometric, and isopiestic mixing at the isonietric mixing pressure ; and (c) isothermal expansion of the solution from the isometric mixing pressure to the equilibrium pressure. The resulting equations gave reasonable estimates of the solubilities of gases and also of the temperature coefficient of solubility. Prausnitz and Shair (485) presented a thermodynamic correlation for gas solubilities based on the two-step process of condensing the gas isothermally to a hypothetical liquid at 1 atm and then dissolving this hypothetical liquid in the solvent. This paper contains much useful information including a semiempirical method for correlating the solubilities of gases in polar solvents. Sherwood and Prausnitz (566) derived a relationship for the accurate determination of heats of solution of gases at high pressure. Yen and McKetta (670) derived equations based on regular solution theory for the thermodynamic correlation of nonpolar gas solubilities in polar, nonassociated liquids. They were able to semienipirically correlate solubilities of nonpolar gases in both polar and nonpolar solvents. Lachowicz and Weale (353) also derived equations based on regular solution theory to predict gas solubility in nonpolar liquids, and their application of their equations to existing data resulted in useful correlations. Smith and Walkley (577) found that it was in general impossi- ble to obtain solubility parameters for gases that lead to acceptable values for both the partial molal volumes and the solubility. 2. Cell Potential and Cavity Models Uhlig (620) proposed a cavity model in which he considered the solubility process to take place in two steps: first, doing work on the solvent against the solvent surface tension to create a cavity, and, second, placing the gas molecule in this cavity and calculating the energy of interaction between the gas and solvent molecules. This is a simple theory but its prediction of a linear relationship between log L and the solvent surface tension has been borne out by many examples. Eley (161, 162) considered a two-step process similar to that of Uhlig's, but was able to more carefully evaluate the separate contributions of each step to the energy and the entropy changes involved. His ap- proach showed reasonable success with both water and organic solvents, although he shom that the case of water is more complicated due to the possibility of structural modifications. Reiss, et al. (496) , extending ideas previously applied to the statistical mechanical theory of hard-sphere fluids determined an expression for the work of creating a spherical cavity in a real fluid. Systems such as helium in benzene are sufficiently close to the model to permit an experimental test of the theoretical expres- sion, since Henry's law constants may be obtained froin the expression. In addition, the surface tension and the normal heats of vaporization of fluids may be evaluated, and the authors found satisfactory agreement between calculated and experimental properties. Pierotti (474) developed a method, using equations derived by Reiss, et al. (496), for calculating the reversi- ble work required to introduce a hard sphere into a fluid and for predicting the solubility, the heat of solution, and the partial molar volume of simple gases in non- polar solvents. The equations are derived for the two-step process of creating a cavity in the solvent of suitable size to accommodate the solute molecule (the reversible work or partial molar Gibbs free energy re- quired to do this being identical with that for introduc- ing a hard sphere of the same radius as the cavity into the solution), and then introducing into the cavity a solute molecule which interacts with the solvent ac- cording to some potential law, for instance, a Lennard- Jones (6-12) paimise potential (the reversible work in the second step being identical with that of charging the hard sphere or cavity introduced in the first step to the required potential). By plotting the Henry's law constant against the polarizability of the solute gases for experimental data and extrapolating to zero polariza- bility a hard-sphere solubility is obtained which can be compared with solubilities calculated from the theory. This was done for He, Ne, Ar, Kr, Kz, Hz, and CH4 in benzene and carbon tetrachloride at 29S"K, and He, Ne, Hf, and Dr in argon at 87°K; in all but one case the predicted solubility is well within a factor of 2 of the observed solubility, which is very good agreement considering the crude method used to obtain the inter- action energy. By plotting the collision diameters of the rare gases against their polarizabilities and extrapo- lating to zero polarizability a hard-sphere diameter (2.33 A) corresponding to the extrapolated hard-sphere Henry's law constant may be evaluated. The theory also yields heats of solutions and partial molal volumes of the gases in solution. The predicted heats were usually within experimental error for all solutes except methane. The agreement between the predicted and calculated partial molal volumes is good, being better than those calculated by Smith and Walkley (577). In a second paper (475) Pierotti developed a theory of gas solubility in water along lines similar to the earlier paper except for the introduction of a term involving the solvent dipole moment. Good agreement mas found between the experimental and Balculated heats, entropies, and molar heat capacities of solution, and for the partial molar volumes of tk solutes. The calculated and experimental Henry's ldw constants for 16 solutes in water at 25" show good agreement, in only one case being off by a factor of 2. The "ab- normal" thermodynamic properties of aqueous solu- SOLUBILITY OF GASES IN LIQUIDS 41 1 tions were discussed with respect to the enthalpy and entropy of cavity formation. The theory is promising as a method for the investigation of gas solubilities in molten salts and molten metals. The thermodynamic properties of gas solubility in mater and organic sol- vents were explained by one theory which involved no assumptions concerning the structure of the solvent. Kobatake and Alder (323) discuss cell potentials and gas solubility theory. They develop a two-parameter cell potentid in a free-volume-type theory which is determined from two experimentally obtained thernio- dynamic quantities. The cell potential for a gas dis- solved in a liquid yields values of the free volume of the gas molecule that are about ten times larger than in a typical liquid. These large free volumes make the calculations less sensitive to the assumptions of the geometric arrangements of the neighboring particles and less dependent on the uncertainties in the linowl- edge of the intermolecular cell potentials. The large free volume indicates, as a number of authors have pointed out, that the gas molecule almost digs a ‘(hole” in the liquid. Kobatake and Alder use this idea to calculate a reasonable value for the interfacial tension of carbon tetrachloride. For CH4, nT2, Ar, Oz, and C2H6 in CC1, it was found that the gas molecule is surrounded by about seven neighbors and that the solvent molecules surrounding the gas contribute importantly to the thermodynamic functions. Fur- ther, it was found that the gas molecules perturb the solvent significantly over several molecular layers, while this is not the case in dilute liquid mixtures. 3. Other Contributions to Theory Ridenour, et al. (501), derived an equation from a thermodynamic consideration of gas solubility as a special case of vapor-liquid equilibrium in which the system temperature may be higher than the critical temperature of the more volatile component of the mixture. The equation was applied with reasonable success to the solubility of C02, Ar, Oz, and N2 in paraffin wax and the solubility of air in kerosene. Klots and Benson (320) discuss the thermodynamic properties of the atmospheric gases in aqueous solu- tions. Mastrangelo (398) derived an equation by statis- tical methods using a “quasi-chemical” type of equation for a two-component system that possesses one or more similar interactions per molecule. The agreement between the equation and data on the solubility of chlorofluoromethanes and ethanes in tetraethylene glycol dimethyl ether and chloroform in acetone was good. The equation should prove especially useful for solubility in polymeric solvents. Himmelblau (247) uses a five-parameter equation for expressing the solubility of 0 2 , Kz, He, Ha, Xe, and CH, in water from the freezing point of water to near its critical point. The average deviation between the calculated and experimental Henry’s law constants is about 3%. The partial molal heats and entropies of solution are presented from the freezing point to near the critical point of water. These partial molal heats of solution appeared to correlate linearly best at 25” with the force constants of the gases, and at 4” with the polarizability of the gases. Himmelblau and Arends (250) used the same five-parameter equation to correlate the solubility of 0 2 , IV2, Hz, He, Xe, CH4, C2H4, C2Hs, C3H8, n-C4Hlo, and l-butene in water at high tempera- tures and pressures. ATaniiot (437) discusses the solubility under pressure of gases in water. In a paper on the solubility of nonpolar gases (He, Hz, Ar, Kr, Xe, CH4, C2Hs, C3Hs) in water Namiot (439) derives an equation for calculating the number of water molecules bonded to one gas molecule, and also the number of displaced water molecules. These values were calcu- lated for the above-mentioned gases. Amirkhanov (12) derived an equation for the theoretical calculation of the solubility in water of gases obeying Henry’s lam. The derivation assumes Maxwell’s law of the distribu- tion of molecular velocities and considers the thermal energy of the gaseous molecule, under equilibrium con- ditions. Using this equation the calculated solubility of Ne, Ar, Kr, and Xe was found to agree within ex- perimental error with the experimental solubility. Salvetti and Trevissoi (522) examined on the basis of irreversible thermodynamics the absorption of gases by liquids. Trevissoi and Ferraiolo (605) on a similar basis studied the absorption of gases by liquids con- sidering the effect of surface tension. 4. Special Studies The reasonably well-defined gaseous standard state and the low solubility of gases in liquids has attracted many workers to make studies in which the solubility of a gas is used as a “probe” to the understanding of a solvent property or to a gas-solvent molecular inter- action. The extensive studies of the effect of an elec- trolyte solution on a nonelectrolyte activity is discussed in the section on salt effects. Several other studies are worth special mention. Marvel, Copley, Zellhoefer, and co-workers (106-111, 677, 678) carried out an extensive study of hydrogen bonding. They determined the solubility of vapors of CC&F, CClzF2, C2Cl3F3, and CzClzF4 in several hundred oxygen-containing compounds including alcohols, alde- hydes, ketones, acids, ethers, and oximes as well as compounds containing nitrogen and sulfur functional groups. They concluded that the C-H group in the halocarbons can hydrogen bond and that ethers, alde- hydes, ketones, and trialkylamines being the best solvents form the strongest hydrogen bonds. The solubility of an acidic or basic gas can be cor- related with the basicity of the solvent. Brown, CHC13, CHZC12, CH3C1, CHzClF, CHCLF, CHClF2, 412 RUBIN BATTINO AND H. LAWRENCE CLEVER et al. (75-77), used the solubility of HC1 as a measure of the basic properties of aromatic nuclei. Gerrard, Macklen, and co-workers have used hydro- halide gas solubilities (202-205, 207-212) as a measure of the basic function of oxygen and more recently sulfur (196) in certain organic functional groups. There is a detailed review of most of their work (206). B. TEMPERATURE COEFFICIENT OF GAS SOLUBILITY Only one direct calorimetric study of the heats of solution of gases has been made (10). All other values of the heat of solution of a gas in a liquid were derived from phase-equilibrium data via the important tem- perature coefficient of gas solubility. Progress is continuing to be made on understanding the basis of the temperature coefficient of solubility; however, useful predictions of the sign and magnitude of the tempera- ture coefficient of solubility in all systems awaits a better understanding of intermolecular forces. The.rare gases are good examples of the types of gas solubility temperature dependence one observes. All the noble gases have a negative temperature coefficient of solubility in water around room temperature and atmospheric pressure which goes through a minimum and becomes positive at high temperatures and pres- sures. In hydrocarbon solvents at room temperature and atmospheric pressure helium and neon solubility increases with temperature, argon solubility is almost independent of temperature, and krypton and xenon solubilities decrease with temperature. In the molten salts studied to date all the rare gases have a positive temperature coefficient of solubility at temperatures between 600 and 900” and pressures ranging from 0.5 to 2 atm. For gases dissolved in nonpolar solvents Hildebrand and co-workers have done much to bring order and understanding to the problem of the temperature de- pendence of solubility. Their work is well summarized in Chapter 4 of Hildebrand and Scott (244) where it is pointed out that for a sparingly soluble gas that obeys Henry’s law the entropy of solution is where X2 is the mole fraction solubility. Experi- mentally it is observed that plots of log Xz against log T are essentially linear for gases dissolved in nonpolar liquids. Thus, there is a regular system of relationships which exists between entropy and solubility with the dividing line between positive and negative tempera- ture coefficient of solubility coming at $ 2 - Szg = 0. Plots of Sz - Szg against - R In Xz are linear for a series o f gases in a given solvent (Figure 8). Gases with solubility less than about mole fraction generalIy have positive temperature coefficient of solubility ; gases with greater than about lo- mole fraction generally have negative temperature coefficients. Hildebrand (244) has calculated the entropy of transferring gas at 1 atm to solution at mole fraction for solvents varying from 5.8 to 10.0 in solu- bility parameter and gases varying from 10 to 300 in “force constants.” The entropy increases moderately with decrease in solubility parameter and increases largely with decrease in force constant. Plots o f the logarithm of mole fraction solubility against solvent solubility parameter squared show a smooth near- linear relation for each gas. Plots of the logarithm of mole fraction solubility against the gas “force con- stants” are linear for each solvent tested (Figures 9 and 10). Hildebrand shows that, excepting cases o f specific interaction, the entropy of solution is the sum of two factors: dilution and expansion. These faetors are both determined by the interrelations of intermolecular forces as measured by solvent solubility parameter,. and gas “force constants.” The temperature dependence o f solubility has been used more often to get the heat of solution rather than the entropy of solution. Plots of log XZ against 1/T are usually linear to the accuracy with which gas solu- bility is commonly measured. A temperature inde- pendent AH is assumed and calculated from the slope which is equal - AH/2.303R. The choice of solubility unit is o f some importance for it determines the reference standard state change; mole fraction, Xz, Henry’s constant, K1, molarity, C, and Ostwald coefficient, L, are commonly used. Plots of log X, and log K1 against 1/T have slopes o f equal magnitude but opposite sign. The same is true o f log CZ and log KZ plots. Enthalpies from the concentration plots, X 2 or C2, represent the standard state change of gas to solution; the K plots give enthalpies of the re- verse reactions. Plots o f log L against 1/T are coni- monly linear. To put the AH from such a plot on a mole fraction basis one must add RT. Where exceptionally accurate solubiIity data from over an extended temperature range are available, an equation o f the type U 10gX - - + blogT - c 0 % 24) 2 - T can be fitted to the data where a, b, and c are the con- stants. Standard thermodynamic manipulations of this equation give the temperature-dependent heat o f solution as AH = -2.303Ra 4- bRT and the heat capacity change on solution as AC, = bR SOLUBILITY OF GASES IN LIQUIDS 413 The accuracy of the data is seldom good enough to attach any more than qualitative significance to the AC, value. Sherwood and Prausnitz (566) have carefully analyzed the factors important in getting the enthalpy of solution from phase-equilibrium data at high pres- sure. Their general expression for the partial molal heat of solution is where & is the vapor phase activity coefficient, y the liquid phase activity coefficient, and Y the vapor phase mole fraction. They used the equation to calculate the heat of solution o f methane in decane at lo00 psia by approximating the vapor phase equation of state of moderate vapor densities by a virial expression through the second virial coefficient. Liquid phase corrections for (b In y2/b In Xz),,p were obtained which required knowledge o f the solution compressibility, partial molal volume o f the gas in solution, and phase-equilb rium data. The calculated hRz had an uncertainty of 10% which mostly reflected the 1% uncertainty in the experimental liquid phase compositions. At higher vapor densities the third virial coefficient becomes im- portant. The approach, which includes the effect of vapor and liquid nonideality, permits a relatively accurate calculation of the heats of solution for systems for which the experimental data are sufficiently ac- curate to warrant its use. Namiot (438) gives a thermodynamic interpretation of the observed minimum in the solubility of hydro- carbons in water at about 70". The effect of temperature on salt effects and on the solubility of gases in molten salts is discussed in the appropriate sections. C. PARTIAL MOLAL VOLUMES OF GASES DISSOLVED I N LIQUIDS An understanding of the partial molal volume o f gases in solution is of importance in the study of solution thermodynamic properties. Smith and Walkley (577) have tested the predictions of various thermodynamic theories of solution for the gas partial molal volume. They used available partial molal volumes of gases in nonpolar liquids and showed that a simple free volume theory predicted the correct magnitude and order of partial molal volumes in the various nonpolar liquids. In general, regular solution theory did not predict acceptable values of both solubility and partial molal volumes of a gas from a single value of gas solubility parameter. Hillier and Walkley (2448) have used a quantum equation of state and found good agreement between their predicted and the experimental partial molal volumes for Hz and Dz in liquid argon. Hildebrand and Scott (244) discuss the contribution of volume expansion on mixing to the entropy of solu- tion. Although volume expansion has little effect on the free energy of mixing, it can have a marked effert on the entropy of mixing. The correction to the partial molal entropy of solution of a dilute solute needs values of solvent internal pressure, (bP/bT) B , and solute par- tial molal volume, V2. Table VI11 catalogs references to gas-liquid systems for which partial molal volumes are reported. Included is the extensive 1931 work of Horiuti (261); references to other pre-1940 data can be found in Kritchevsky and Ilinskaya (336). Two techniques have been used at atmospheric pressure. Most workers have used some modification of Horiuti's apparatus (261) to directly determine solution dilation on dissolving the gas to near saturation. Some have determined solution density (396). Details of the technique and apparatus are discussed in references listed in Table VIII. Gamburg (198A) and Connolly and Kandalic (101) describe apparatus for the determination of partial molal volumes at high pressure. Some o f the partial molal volumes at high pressure listed in Table VI11 were not determined directly but got by fitting the experimental gas solubility to some form of the Krichevsky and Kasarnovskey quation (335, 336) for gas solubility at at high pressure. D . GAS SOLUBILITIES IN MIXED NONELECTROLYTE SOLVENTS A solution o f a gas in a binary nonelectrolyte mixed solvent is a three-component system, with two gas- solvent interactions and a solvent-solvent interaction of importance. Ben-Naim and Baer (39) have determined the solu- bility of argon in water-ethanol mixtures at six tem- peratures and nine concentrations between 0.015 and 0.25 mole fraction ethanol (Figure 11). At low tem- peratures there is a maximum in the solubility at low ethanol concentrations. Both viscosity and water partial molal volumes show a similar concentration dependence in the ethanol-water system. These results are explainable in terms o f the influence of ethanol on the structure of water. Small amounts of ethanol increase the concentration o f the icelike form of water at low temperatures; at about 30" the icelike structure of water is breaking down anyway and the argon solubility tends to increase monotonically from its value in pure warer to its value in pure ethanol. Dissolved argon itself influences the amount of icelike water present. Plots of As" and A B o for the argon solutions show similar trends when plotted against the mole fraction of ethanol. The entropies of solution of argon in pure water are negative as compared to pure 414 55 5c P 45 X 4 I C 3? RUBIN BATTINO AND H. LAWRENCE CLEVER I I I I 1 J 0 9 % 010 O d 0.20 0.25 XCZHIOH Figure 11.-The solubility of argon in aqueous ethanol. Ost- wald coefficients as a function of mole fraction ethanol (39). Reprinted from The Transactions of the Faraday Society by per- mission of the Faraday Society. alcohol. The value for pure ethanol has already been attained at 0.2 mole fraction ethanol, and this sug- gested to Ben-Naim and Baer that the abnormally low entropy of solution of argon in pure water cannot be attributed to an active formation of icelike water but that the argon shifts the already existing equilibrium toward the icelike form. A similar study of argon solubility in water-p-dioxane system is reported by Ben-Kaim and Moran (40). The maximum in the argon solubility at low concentrations and low tem- peratures seen in the water-ethanol system was not found in the water-dioxane system. This is interpreted to mean p-dioxane has a destabilizing influence on the large compact clusters of water molecules at all p- dioxane concentrations and temperatures studied. Schlapfer, Audykowski, and Bukowiecki (539) de- termined the solubility of oxygen from air at 30" over the full concentration range of aqueous solutions with methanol, ethanol, l-propanol, 2-propanol, l-butanol, ethylene glycol, and glycerine. The oxygen solubility decreases almost linearly with increasing weight per cent ethylene glycol or glycerine. In the monohydric alcohols the solubility behavior is somewhat similar to the argon solubility in ethanol-water at 32" with the solubility going through a minimum then increasing to the solubility in pure alcohol. Some limited studies of gas solubility in aqueous alcohol mixtures include X 2 (168), Nz and 0 2 (342), and CO2 (503) in ethanol-water, 0 2 in methanol-water (478), and various gases in beers and wines (2, 168, 172, 313, 407). Studies in other aqueous mixed solvents include: oxygen (248, 285) and COz (379) in aqueous sugar solutions; He and Ar in mater-saturated nitromethane (197); C02, acetylene, and ethylene in water-??- methylpyrrolidone (565) ; C O Z in various aqueous mono-, di-, and triethanolamine solutions (381, 568, 596) ; acetylene in aqueous diniethylformamide, di- oxane, and acetone (387), acetylene, methylacetylene, vinylacetylene, and diacetylene in aqueous dimethyl- formamide and N-methyl-2-pyrrolidinone (72, 72A) ; hydrogen sulfide in aqueous phenol (226) and in aqueous monoethanolamine (283) ; C102 in aqueous acetic acid (305) ; and O2 in aqueous (236). Solubilities in aqueous strong acid solutions are discussed under salt effects. For less polar mixed solvents O'Connell and Prausnitz (459) have considered the thermodynamics of a solu- tion consisting of one supercritical component, a gas, and two or more subcritical components, the liquid solvent components. They treat the case of a gas dis- solved in two miscible solvents and generalize the result to a gas dissolved in a mixture of any number of miscible solvents. In their notation subscript 1 is the solvent of lower vapor pressure, 2 is the gas, and 3 the solvent of higher vapor pressure. The solvent activity coefficients are both referred to the saturation vapor pressure of solvent 1. The activity coefficient of the gaseous solute is also related to the saturation pressure of solvent 1 with the gas reference fugacity related to the Henry's constant in solvent 1 in the absence of solvent 3. Thus, they use an unsymmetrical convention for normaliza- tion of the activity coefficients y1 --t 1 as X I + 1, y2 + 1 as X 2 + 0 at Xa = 0, and y3 + 1 as Xa O'Connell and Prausnitz use the approach of Wold and write one-paramenter Jlargules expansions for the excess free energy which in the symmetric convention 1. E g13E 923 E 912 - - a 1 2 X 1 X 2 ~ = f f 1 3 x 1 x 3 - = a23x2x3 RT RT RT (Eq 26) Then the molar excess free energy of the ternary solu- tion is assumed to be SOLUBILITY OF GASES IN LIQUIDS 415 E Q l Z Z __ = a 1 2 X 1 X 2 + RT They differentiate and transform from the symmetric to the nonsymmetric convention to obtain expressions for Ti as a function of the CY'S and mole fractions. The parameters ~ ~ 1 2 , CYZ3, and CY13 are evaluated from the solubility data for the gas in pure solvent 1, the solu- bility data for the gas in pure solvent 3, and the vapor- liquid equilibrium data for the binary solution of 1 and 3, respectively. Finally Henry's law constant for the mixed solvent is In Hz, mixed solvent (Pl") = X 1 In H21(P19) + X 3 In HdP1S) - C Y l z l x 3 (Eq 28) The equation shows that even if the two solvents form an ideal mixture ( a 1 3 = 0), Henry's constant for the solute in the mixed solvent is an exponential rather than a linear function of the solvent composition. O'Connell and Prausnitz have calculated Henry's constant for Hz in toluene-heptane and for O2 in isooctane-perfluoro- heptane, but no direct experimental confirmation is available. Koudelka (333, 334) has determined the solubility of COz in the six possible binary systems formed from methanol, acetone, chloroform, and benzene. He finds deviations for the activity coefficient of the gas in the binary solvent from the linear relation of activity coefficients in the pure liquids log ~ 2 , m i x t u r e = XI log Y? in 1 + X 3 log YZ in 3 (Eq 29) The deviations may be expressed in terms of an excess function which requires only one constant, a, in a term aX1X3 for simple nonpolar solutions added to Eq 29. Clever (95) determined the solubility of argon and krypton in binary p-xylene-p-dihalobenzene systems at 30". The results fit regular solution theory with mixed-solvent solubility parameters obtained from mixed-solvent surface tensions. Kruyer and Nobel (347) report that hydrogen solu- bility is a linear function of composition in benzene- cyclohexane mixtures. Hydrogen solubilities are also reported (9) for the aliphatic olefin mixtures hexane- hexene, hep tane-hep tene, and oc tane-oc tene. Other studies in mixed solvents are CO, and N 2 in 50:50 decanol-dodecanol (378)) Xz and O2 in 50:50 acetone-ethanol and isooctane-ethanol (342), acetylene in dioxane-dimethylformaniide (387) and in various binary systems of water, methanol, CH3C1, and (CH30CHzCHz)zO with the donor-type solvents di- niethylformanzide, dimethyl sulfoxide, and ethylene glycol (517). E. SOLVENT SURFACE TENSION AND GAS SOLUBILITY A particularly successful gas solubility correlation is the linear relationship between log L and solvent I n 4 m A N E ISOOCTANE p-xrms BENZENE CHLOHOBENZWE CARBON DISUVIDE BROYIRMZWE - YITROMZ?"AM IOMBENZWIB NITROBENZEIIE WATEZ surface tension, u, for a gas in a series of solvents, Schlapfer, et al. (539), show that such a plot for oxygen solubility in about 20 solvents including alcohols, hydrocarbons, ketones, esters, and halocarbons holds well except for the solvents ethylene glycol, glycerine, and water. Baldwin and Daniel (23) correlate Nz, Oz, and air solubilities with the surface tension of lubricating oils and fuels. The solubility of the five gases He, Ne, Ar, Kr, and Xe fit the linear log L against u plot almost within experimental error for 13 hydrocarbon solvents (96, 97) and for various substituted benzene solvents (530). Figure 12 shows such a plot for argon in 25 solvents. Uhlig (620) assumed the energy of forming a solvent cavity for the gas molecule is equal to the cavity sur- face area times the liquid surface tension to derive -4rr2u + E 2.303kT log L = (Eq 30) which predicts the linear log L against a plots at con- stant temperature. The criticism that the bulk surface tension is not appropriate to calculate the energy of formation of a molecular-sized cavity seems justified. Rather the surface tension is probably proportional to some solvent property that determines the gas dissolv- ing power of the solvent. Hildebrand and Scott (243, Chapter XXI) show that a plot of U/V~''~ agaiiiit AEv,,/V for 19 liquids is linear with slope 0.86. T1 is suggests the Hildebrand solubility parameter, 6, is proportional to ( u / ~ , ~ ' ~ ) 0 . 4 3 . 416 RUBIN BATTINO AND H. LAWRENCE CLEVER In mixed solvents the surface tension may not repre- sent such a property of the bulk solvent because of Gibbs enrichment of the surface by the component of lower surface tension. Gjaldbaek (217) showed that decreasing water surface tension 57y0 by adding a trace of aeresol decreased carbon dioxide solubility by only o.4yO at 25" and 1 atm. Similar results were ob- tained for NZ solubility in water (169). Although a trace of surface-active material does not change the equilibrium solubility of a gas in water, it may affect the rate of attaining equilibrium. The report of Yeh and Peterson (667) that COz, Kr, and Xe solubilities in lipids do not obey the Uhlig plot may be due to the presence of variable amounts of surface-active compo- nets in the lipids. The Uhlig relationship is approxi- mately obeyed for the solubility of Ar and Kr in the binary pxylene-p-dihalobenzene systems (95). In these binary systems the solvent molecules are of similar size, shape, and surface tension. The surface enrichment of one component is minimal so the solution surface tension still represents a bulk property. Burrows and Preece (85) base a derivation on the ideas of Eley (161, 162), and Uhlig, and the empirical relationship between surface tension and temperature, u = K/T" that predicts a linear relationship between log L (T/U)''(~+') and a/T, where n = or 1. The solubility of helium in petroleum and silicone oils and o f Ht, He, Nt, and methane in terphenyls (234) obeys the relationship. Plots of log L against 1/T were not linear for these systems. Plots of the solubility of Oz, Nz, and air in six kerosene jet fuels against l/a0.75 are linear (144). Justification of the plot is the postulation that gas solubility is proportional to the solvent compressibility (p) and the relationship ab'" equals a constant. The relationship of solvent surface tension and gas solublity in molten salts and in molten metals is dis- cussed in the appropriate sections. F. SALT EFFECTS The activity coefficients of nonelectrolytes solutes in aqueous salt solution were well reviewed by Long and McDevit in 1952 (373). The activity coefficient is a function of the concentra- tion of a,ll solute species and at a given temperature log fi can be represented by a power series in C., the elec- trolyte concentration, and Ci the nonelectrolyte solute gas concentration. It is usually assumed that for low C , and Ci where there is no chemical interaction between solute species only the linear terms are important and log f i = k,C, + kiCi (Eq 32) The last term can be ignored if ki, which results from the interaction of the nonelectrolyte with itself, is small, or if Ci is very small. In gas solubility studies Ci is often small enough to justify ignoring the ki term. Most theories are concerned with the calculation o f k,. Experimental measurements of the solubility of a gas in pure solvent and in a salt solution give the activity coefficient of the dissolved gas directly. The gas solute activity is the same in pure solvent and salt solution so and fiSi = fiOSiO f - f .
0 2 S." s i 1 - 1 where Sio and Si are gas solubility in salt solution, respectively. Thus 0 0 Since log fi" = kioSio pure solvent and + kiSi (Eq 34) - Si") (Eq 35) and if Si and Si" are low the last term can be ignored which is the same form as the well-know empirical Setschenow equation, log Sio/Si = KC.. However, in systems where the ki(Si - Si") term cannot be ignored K and k, are not the same. The distinction becomes important when comparing salting out of a non- electrolyte of low solubility with one of high solubility. A salt that increases the activity coefficient o f the dis- solved gas salts out and a salt that decreases the activ- ity coefficient o f the dissolved gas salts in. Long and McDevit (373) point out that the theories of the salt effect all have common underlying aspects but emphasize different approaches to the problem. They classify the theoretical approaches as emphasizing (1) hydration, (2) electrostatic, (3) van der Waals, and (4) internal pressure effects. The hydration theories propose that salting out results from the effective removal o f water molecules from their solvent role owing to the hydration o f the ions. The theories provide no explanation o f salting in. The electrostatic theories relate salt effects to the influence of nonelectrolyte on the dielectric constant o f the solvent. A nonelectrolyte that increases the di- electric constant o f the solvent will be salted in, one that decreases the dielectric constant will be salted out. The van der Waals theories extend the electrostatic theories by taking into account short-range forces, primarily dispersion forces, that may play an appreci- SOLUBILITY OF GASES IN LIQUIDS 417 able role in the specific effects of ions. The most de- tailed attempt to include dispersion forces is that of Bockris, Bowler-Reed, and Kitchener (53). The interal pressure of a salt solution can be related to changes in both volume and compressibility o f a solverit due to dissolved salts. Both changes have been correlated with salt effects. RIcDevit and Long (408) have calculated the free energy of transfer of a nonpolar electrolyte from pure water to salt solution assuming the neutral molecules modify the ion-water interaction in a simple manner and get a limiting law for k, which is (Eq 37) Pi"(VS - VBO) 2.3PoRT k, = where Via and V,' are, respectively, partial molal volumes at infinite dilution of nonelectrolyte solute and salt, V, i s the molar volume of pure (liquid) electrolyte, and Po is the compressibility of pure water. Both the van der Waals and internal pressure approaches ex- plain both salting-in and salting-out effects. KO new basic theories of salt effects have appeared since 1952, but several experimental tests based on gas solubility have been carried out. Morrison and co-workers (426,428,430) have studied the salting out of H2, He, Ne, K2, 02, Kr, Xe, SFB, CHI, C2H4, C2H6, C3HB, n-C4Hlo, and benzene vapor in some or a l l of the aqueous solutions of HC1, XH4CI, several alkali halides, BaC12, Lac&, and several tetralkyl- ammonium chlorides. Electrostatic effects were tested by assuming IC, proportional to a coulombic term, Zce2/r. Using crystallographic radii and using IiaC1 as a reference salt it was found that HC1, LiC1, BaC12, and LaCla cause a salting out less than expected from ionic size. IiCl behaves normally. KI is normal for low molecular weight gases, but causes a specific decrease in salting out which is proportional to the molecular size for the hydrocarbons and heavier organic gams. Morrison and Johnstone (430) have calculated theo- reticd ratios of k,/kKaCl from electrostatic and internal pressure theories, where k, represents the salting-out constant of HC1, LiC1, KC1, KHIC1, 1/2BaC12, KMe4C1, NEt4C1, NaBr, KaI, NaK03, or 1/2Na2S04. The in- ternal preasure results accord better with the observed order of ratios than the electrostatic approach but does not fit the NMe4Cl and NEt4C1 salt ratios for the more "inert" gases. The kz/kNaC1 ratio is linear for a series o f solute gases in a given salt solution except the salts KaI, NMe4C1, and NEt4C1. With these salts a specific effect is observed. When a correction for the van der Waals forces between ions and neutral molecules is applied the k,/kNaC1 ratios become linear for these salts. The salt effect constant, k,, can be referred to either unit volume of electrolyte solution or the unit weight of solvent. Morrison points out that the weight basis is more closely related to the theoretically significant mole fraction and reports his results on a solvent weight (molality) basis. On this basis nitric acid and tetra- alkylammonium halides salt in He, We, n-CrHlo, and benzene vapor but salt out SF6. The salting out of 02, Ar, Xe, CHr, and C2He by LiC1, NaC1, KC1, and MgClz (173), CzH6 by NaCl and CaCl2 (123), and C2Hz by numerous halide, nitrate, and sulfate salts (188) has been explained by hydration theories. Eucken and Hertzberg (173) have derived an expression for the hydration number of an ion based on an equilib- rium association o f water molecule clusters of one to eight molecules, the displacement of the equilibrium by the ions, and the competition of the ions and dis- solved gas molecules for water of hydration. They get ion hydration numbers around 10. Flid and Golynets (188) point out that salting out increases in the order the cations increase in ease o f hydration between 0 and 25", but that the order differs in the 50-70" range. In general, as pointed out by McDevit and Long, the hydration numbers got ten by gas solubility measure- ments do not correspond with degrees of hydration ob- tained from other experiments. Namiot (439) discusses aqueous gas solubilities in terms o f a two-structure model o f water. The dis- solved gas molecules transform some "liquid" water molecules to "icelike" molecules. An equation is given for calculating the number of water molecules bonded to one gas molecule and the number of dis- placed water molecules. The Setschenow constant, k ' , is related to the number of bonded water molecules. Clever and Reddy (99) have obtained salting-out constants for helium and argon by NaI in both methanol and water. The ratio is less than expected from the dielectric constant difference of the solvents. The van der Waals approach of Bockris, Bowler-Reed, and Kitchener (53) was not sufficiently sensitive to explain the k s ~ e O & , ~ n O ratio for either gas. Many studies have been made on the effect o f aqueous electrolyte solutions on the activity coefficients o f dissolved hydrocarbon gases. With one exception the studies of Table 1 1 1 were carried out in aqueous solution. Salting out is the general rule. Exceptions include Na dodecyl sulfate and K oleate, where micelle formation and increased interaction energy between the hydrocarbon gas and the hydrocarbon-like micelle interior may explain the enhanced solubility over that in pure water. Guanidine hydrochloride, nitric acid, and tetralkylammonium halides salt in hydrocarbons. The increased solubility of ethylene in silver nitrate solutions is certainly due to formation of the Agf.C2H4 complex ion. It is suggested (387) that the increased solubility o f acetylene in acetone in the presence of NaI is because acetylene is more soluble in an acetone.Na1 418 RUBIK BATTINO AND H. LAWRENCE CLEVER TABLE I11 SALT EFFECTS ON HYDROCARBON GASES Ethyl acetylene Benzene vapor, C4H10 Salts NaC1, CaC12, MgCl? NaCl BaC1, LiC1, KI Alkali halides Guanidine hydrochloride NaC1, CaC12 Xa dodecyl sulfate NaCl "Neutral salts" NaCl "Neutral salts" K oleate NaC1 "Keutral Salts" XaI (in acetone and di- methylformamide) NaC1, NaOH HN03, tetralkyl am- monium halides KC1, AgN03 Ref 422 153 428 173 640 123 660 621 268 364 260 403 402 303 188 387 571 430 complex than in pure acetone. NaI has little effect on acetylene solubility in dimethylforniamide. The Setschenow equation does not apply to iso- butylene solubilities in aqueous NaCl at temperatures near 0" (303). The interest in oceanography has resulted in studies of nitrogen, oxygen, and noble gas solubilities in sea water and saline solutions (41, 42, 153, 327, 611). These systems salt out. Green (231) has made a careful study of oxygen solubility and Douglas (152) has determined nitrogen and argon solubility as a function of chlorinity (halide as g of chlorine/kg of sea water) and temperature. Green shows the oxygen solubility obeys a Setschenow-like equation with chlo- rinity used in place of salt molality. The solubility of a gas over the range of 0 to 100% aqueous strong acid has been studied. The solubility of COz initially decreases (salts out), goes through a minimum, increases to a maximum at a composition corresponding to HzS04 .4Hz0, goes through a second minimum at a composition of HzS04.Hz0, then in- creases until pure HzS04 is reached (392, 555, 556). The minimums become less pronounced as the tem- perature increases (5X5). NazS04 in aqueous HzS04 solutions of various composition salts out (556). COZ is salted in as the concentration of HC104 increases from 0 to 50 wt. yo; it is salted out from 50 to 70 wt. % (392). The solubility of chlorine in aqueous 0 to 50 wt. %; HC104 decreases sharply up to 5 M and then stays constant to higher HClO4 concentrations (551). Oxygen solubility decreases to a minimum at about 80% HzS04 and then increases sharply as 100% H2S04 is approached. Oxygen solubility decreases steadily as H3P04 concentration increases (235). Solubilities of oxygen in various nitric acid (502) and in white and red fuming nitric acid (579) are reported. CIOz solubilities in aqueous HzS04 and aqueous acetic acid obey Henry's law (305). The neutral nature of PH3 has been deduced from its similar solubility in aqueous NaOH, NaC1, and HzS04 (639). The solubility of KZ in buffered solutions of various transition metal acetates indicated no unusual association of Nz and transition ion (70). Chlorine is salted out by BaCL (51), LEI, SrClz and BaClz (292), and LiClOk and NaC104 (291) in aqueous solution between 10 and 50". The temperature dependence of the salting-out constant, k,, is small and negative at least in aqueous solutions at atmospheric pressure and temperatures below 70". This is true of NzO and COZ in several chlo- rides, nitrates, and sulfates between 0 and 40" (391), Clz between 10 and 50" (51, 291, 292), and COz in sulfuric acid between 20 and 60" (555). The salting out of acetylene is more pronounced between 0 and 25" than at higher temperatures up to 70" for 23 salts. The acetylene solubility goes through a minimum between 25 and 70" with the minimum becoming weaker at higher concentrations of the salt; no mini- mum is observed for NaC1, ZnClz, ZnSO4, MgS04, S S 0 4 , CaS04, and Alz(S04)3 (188). Sulfur dioxide is salted out at low temperatures but salts in with NaHS03 at 90" (348). Long and McDevit (373) differentiate with respect to temperature their internal pressure expression for k, Eq 37, to get where Vi" and P," are, respectively, the partial molar volumes of nonelectrolyte and of electrolyte at infinite dilution. The relation predicts dk,/dT to be small and negative, to be smaller in the 25-50" range than the 0-25" range, and that dk,/dT will be small for LiCl and relatively large for KNO3 with salts such as NaC1, KCI, KBr, KI, WaOH, and 1/zNazS04 being intermediate in value. The predictions agreed well with the salt effect data of Alarkhani and Kobe (391) on NzO and COz when reasonable partial molal volumes of the gases were used. Morrison (426) gets approximate values for the difference in the heat capacity of solution between water and salt solution from the temperature dependence of gas solubility in water and in salt solution. He shows that ions that are thought to have a structure-building effect in water, such as Li+, decrease the heat capacity of solution much more than salts thought to have a "structure-breaking" effect on water, such as I-, which suggests that modification of the solvent structure in the vicinity of the nonelectrolyte molecules cannot be neglected in theories of salting out. Salt effect studies at elevated temperatures and pres- sures are technically difficult. There is a problem in SOLUBILITY OF GASES IN LIQUIDS 419 TABLE IV VALUES of K = (l/m) LOG K,'/K' __ Temp, O C 50 75 100 0.5 m NaCl 0.096 0.084 0.076 1 m NaCl 0.095 0.088 0.078 2 m NaCl 0.091 0.084 0.080 determining the partial pressure of the gas; a common practice is to simply subtract the steam table value from total gauge pressure to get the gas partial pressure. KO correction is made for the effect of salt or dissolved gas on the steam table pressure. Pray and co-workers (583) have determined the solubility of Hz and Oz in water and in about 0.17, 0.42, and 1.02 M solutions of both uranyl sulfate and uranyl fluoride between 100 and 280" and at pressures up to 3000 psi. He and Xe solubilities mere determined in uranyl sulfate solutions up to 500 psi. With the excep- tion of O2 in uranyl fluoride solution, for which salting out is alniost independent of temperature, salting out appears to increase with increasing temperature. Henry's law is applicable over only part of the pressure range; it fails at lower pressures as the uranyl salt concentration and temperature increase. Anderson, Keeler, and Klach (14) have determined Br and O2 solubilities in aqueous uranyl sulfate solu- tions between 100 and 300". Krypton obeyed Henry's law over the 10-4 to 10-1 psi partial pressure studied. Kr solubility was the same for water and in a solution which is 0.02 M in uranyl sulfate, 0.005 M in CuSO4, and 0.005 M in H2S04; when these concentrations were doubled Kr solubility increased. Oxygen solubility over the partial pressure range of 50-1500 psi was independ- ent of the solvent. Studies of COz solubility in CaCL solution to 700 atm (487) and air solubility in brine to 3500 psig (157) are reported. Ellis and Golding (164, 165) have carefully studied the solubility of COz in 0.5, 1.0, and 2.0 ;zi7 NaCl up to 330". They determined the density of KaCl solutions for temperatures to 350" and salt concentrations to 3 M to be able to interpret gas solubilities as volume distributions (Ostwald). They corrected water vapor pressures for dissolved COz and NaCl assuming Raoult's law. They estimate maximuin solubility errors may be as high as 5-1070 in the 2 M NaCl at 300" because of uncertainties in the water vapor pres- sure. Their plot of Henry's law constant against temperature is Figure 13. The Setschenow salting-out K is calculated as (llnz) log K,"/K" and results are in Table IV. Salting out decreases with rising temperature but passes through a minimum of about 150" and then increases as the critical temperature of the solution is raised by the dissolved salt. The temperature of the 150 200 250 300 350 0.070 0.090 0.128 0.172 0.376 0.076 0.089 0.128 0.176 0.318 0.073 0.084 0,111 0.151 0.244 I i I 10 000 8000 6000 X . c X 4000 2000 A 0 5 m NaCl A 0 3 r n LIHCO) (Marshc,i !I 01) D 1 rn NoCi x 2 m NaCl 1 1 I I 1 1 J 100 200 300 T c Figure 13.--T'alues of the Henry's law constant for the ~olution of carbon dioxide in water and in sodium chloride solutions be- tween 10 and 335' (165). Reprinted from The dmerican Journal of Science by permission of the editor. minimum is similar to that for the minimum solubility of COZ in water. Smith, Nagy, and co-workers (153, 578) have studied the effect of gas pressure on gas-aqueous salt systems near room temperature. They have determined the solubility of methane in aqueous NaCl, CaCI2, and mixtures of CaClz + NaCl and the solubility of nitrogen in ISaC1, CaC12, NazS04, and MgSOr solutions as a function of pressure up to 1000 psia. Salting out occurs at all pressures, but there are negative departures from Henry's law for any given salt concentration as the SZ pressure increases. Methane obeys Henry's law up to 200 psia. They estimate from their data that 1 ft' of sedimentary rock of 20% porosity, saturated with 50,000 ppm NaCl brine with the gas at the pressure at 1000-ft depth, can accommodate either 0.15 mole of Nz or 0.30 mole of methane. G. SOLUBILITY OF GASES I N BIOLOGICAL FLUIDS The solubility of gases in biological fluids has been studied as an aid in understanding respiration in plants and animals, the anesthetic properties o f various gases, the action of poisonous gases, and the unusual "salting- in" properties of detergent, denaturing, and protein solutions. The solubility in various vegetable and animal fats and oils of the gases H, (619), Ra (452), Nz and S e (266, 267), cyclopropane (52, 364), HS, 02, and n ' 2 420 RUBIN BATTINO AND H. LAWRENCE CLEVER 4 .
0 i I I I 30 L o . 20 i 1 0 7 0 5 7 9 ti PH Figure 14.-Butane binding by bovine serum albnmiri as a function of pH. The ordinate is the average number o f butane molecules bound per bovine serum albumin molecule, obtained from the excess solubility of butane in the protein solutions com- pared with the solubility in the protein-free solvent (641). Ite- printed from The J o u m l o f Biological Chemistry by permission of the copyright owners, The American Society of Biological Chemists, Inc. (127), Xe (360), CHCla and CHaNO2 (383), Hz, 02, Xn, and COz (532) have been measured. The solubility o f the gases NH3 (275), CS, (409), acetylene (486), cyclo- propane (179, 268, 479), N2O (17, 122, 567), COz (18, 567, 570), Xe-OZ mixtures (178), and Nz (176, 567) has been reported for blood, various serums, and homogenized tissues. Leonard (363) analyzed the interstitial gases in fruits and, assuming these gases to be in equilibrium with dissolved gases, estimated the C02 and O2 solubility in the fruit tissues. Gas solubility studies in biological systems are frequently small parts of larger studies and as a consequence overlooked in abstracting and indexing. Thus the listings above and in Table V may not be as complete as for some other fields. The recent valuable studies of the solubility of simple hydrocarbons in protein, detergent, and denaturating solutions by Wishnia (659, 661) and Wetlaufer, et al. (640, 641), deserve special mention for the insight they give to hydrophobic bonding. Some aqueous solutions of proteins and detergents show an increase in the solubility o f a gas over its solubility in water alone. Findlay noted such increases in C02 solubility in gelatin, hemoglobin, and methyl orange solutions before 1914 (see Markham and Kobe (393)). McBain (402- 404) reported propylene solubilities in various detergents, and suggested that the enhanced solubility was due to the hydrocarbon gas entering inside of the hydrocarbon-like detergent micelle. Butadiene (507) is also “salted-in’’ by a detergent. The increased solubil- ity of hydrocarbons in biological fluids was suggested to be due primarily to the high solubility of the gas in lipids (fats), but considerable evidence has accumulated to suggest that proteins have a special affinity for hydrocarbon and other gases (179, 268, 364, 567, 667). Wishnia (659) determined the solubility of ethane, propane, and butane in water and in aqueous solutions of bovine serum albumin, human hemoglobin, lysozyme, and sodium lauryl sulfate at several temperatures be- tween 10 and 35”. The increased solubility due to the protein in the solution was almost independent of temperature, and demonstrated that the transfer of dissolved gas from water to protein has a small enthalpy change and a large positive entropy change. Imai (268) has observed small enthalpies of transfer for cyclopropane in similar systems. Wetlaufer, et al. (640, Ml), studied the solubility of the hydrocarbon gases ethane, propane, butane, iso- butane, pentane, isopentane, and neopentane in aqueous bovine serum albumin (BSA) and the denaturating solutions 7 N urea and 5 N guanidinum chloride. They point out that these eight gases represent six of the common amino acid side chains. The binding of butane by BSA is a function of pH (Figure 14) and falls off rapidly in pH regions where the protein goes to a random-coiled structure. Wetlaufer and Lovrien (641) suggest four possible mechanisms for the protein hydrocarbon interaction : (a) a dissolved nonpolar molecule could attach itself to one accessible surface of a nonpolar cluster, perhaps with partial penetration; (b) one nonpolar molecule might penetrate into one relatively nonpolar interior of the protein and lodge there; (c) a nonpolar molecule could be bound to a protein in the immediate neighborhood of an ion pair so as to strengthen this ion-pair interaction; and (d) hydrogen bonds and other polar interactions may be facilitated by placing a nonpolar group close to the interacting groups. Wishnia’s (660) study of ethane, propane, butane, and pentane in water and in aqueous sodium ddecyl sulfate as a model protein system appears tu fit best with mechanism a or b, with (possibly) b being favored. Using the solubility and the temperature dependence of the hydrocarbon gas solubility in water, in aqueous detergent, and in hydrocarbons to obtain the free energy, enthalpy, and entropy changes in solution, he shows that the transfer of dissolved gas from water to detergent solution and from water to hydrocarbon are similar in thermodynamic behavior. The transfers have small, usually positive, enthalpy changes and a large positive entropy change. This suggests partial pene- tration, if not complete solution, of the low molecular weight hydrocarbon in the detergent micelle. Wetlaufer and Lovrien (64) calculate similar values of enthalpy and entropy change in the transfer of dissolved hydro- carbon gases from water to denaturing solutions of urea and guanidinium chloride. SOLUBILITY OF GASES IN LIQUIDS 42 1 + u a % 2 MOL PERCENT CARBON DIOXIDE Figure 15.-lowv-temperature phase equilibria. Premire-com- position diagrams for methane-propane system (left) and carbon dioxidepropane system (right) (5). Reprinted from Industrial and Engineering Chemistry by permission of the copyright owners, The Bmerican Chemical Society. Featherstone, De Bon, el al. (179, 434), determined the solubility of the anesthetic gases NzO, Xe, and cyclo- propane in aqueous bovine serum albumin, hemoglobin, and y-globulin. Only the hydrocarbon solubility increases as the albumin and hemoglobin concentration increases. Although the Xe solubility is not increased by the presence of hemoglobin, dielectric studies of the aqueous hemoglobin solutions (547) show that the non- rotating bound water is increased 17% by the presence of dissolved Xe. This evidence supports Pauling’s hypothesis (471) that protein side chains, water, and $ inert gas interact to cause the formation of microcrys- H 3 .
0 talline hydrates i n vivo, and that these hydrates in- 5 terfere with the normal electrical oscillations of the 2 nervous system and thus lead to anesthesia. H. EFFECTS O F PRESSURE ON GAS SOLUBILITY Since 1940 so much good gas solubility n-ork at high pressure has appeared and the understanding of gas solubility of both low and high pressures has so im- 0 200 300 4 0 0 500 600 650 proved that references to both low- and high-pressure TEMPERATURE OF solubilities are included. However, the special tech- niques and apparatus needed for high-pressure solubil- ity measurements will not be discussed. Krichevskii (3388) in a book (Russian) and Lachowicz (351) in a review have discussed solubility of gases at high pressure. In this review gas solublity is taken as a special case of a vapor-liquid phase equilibrium where the gas phase is principally one component and the liquid phase principally the second component. Two component vapor-liquid systems in which both components have an appreciable concentration in both phases are not classed as gas solubility. In general, systems classed as gas solubility have components with a greater dif- ference in critical temperature than those not classed as gas solubility. Examples of the two cases are shown in Figure 15. The methane-propane system (component’s critical temperatures differ by 179”) is 80 mole % or Figure 16.-I,sohars of the water solubility of oxygen gas as a function of temperature. Total gauge pressure (upper) and “hypothetical” oxygen part,ial pressure (lower) (686). Reprinted from The Transactions o f the Society o f Mechanical Engineers by permisaion of the copyright owners, The American Society of Mechanical Engineers. more methane in the gas phase and is classed as gas solubility, but the carbon dioxidepropane system (critical temperatures differ by 66’) is not classed as gas solubility. For systems where the information left some doubt the data were usually included as gas solubility. The reporting of high-pressure gas solubility presents difficulties when only the total pressure is directly measured. The effect of the high-pressure gas and dissolved gas on the true vapor pressure of the solvent is not known nor is it easily measured. Figure 16 shows isobars of oxygen solubility in water at total gauge 422 RUBIN BATTINO AND H. LAWRENCE CLEVER pressure and at a fictitious oxygen partial pressure obtained by subtracting the saturation vapor pressure of water from the total pressure. The usual negative temperature coefficient of gas solubility in water is noted to about 100°, but at higher temperatures the solubility increases with temperature. Henry’s law is apparently obeyed to only moderate pressures (686). Krichevsky and Kasarnovsky (335) have developed a thermodynamic equation for calculating the solubility of slightly soluble gases at high pressure in solvents of low vapor pressure. Michels, Gerver, and Bijl (417) deduce a similar equation directly from the general equations for the equilibrium of a binary mixture. Sattler (526) gives a thermodynamic derivation for cases with both a negligible and a finite solvent vapor pressure. Needed for the calculation is low-pressure gas solubility data of the pure gas at low pressure and thermodynamic data of the pure gas at high pressure. Kobayashi and Kats (324) use the rigorous thermo- dynamic relation [dG]T,X, = RTd In32 = V2dP (Eq 39) and Henry’s law fz” = KX2 (Eq 40) where G2 is the partial molal free energy of the dis- solved gas, and P, T, and R are pressure, temperature, and gas constant, respectively. Vz and Xz are partial molal volume and mole fraction of the dissolved gas, andfZ and fzo are partial molal fugacities of the solute gas at the total pressure and at the solvent saturation pressure, respectively. Integration of Eq 39 between the limits of the solvent vapor pressure Po and the total pressure, P, assuming Cz to be independent of pressure and changes in solution concentration, followed by eliminating fzo by Henry’s law gives the Krichevsky- Kasarnovsky type equation (Eq 41) f Z Vz(P - P O ) x z RT In- = In K + The equation is sometimes used in the form below, where a modified Henry’s constant K’ is defined In K’ - VzPo/RT. The equation has been successfully used to fit high- pressure gas solubility in water, methanol, and hydro- carbons (324, 335, 339, 420, 441, 443). Themodified Henry’s constant, K’, shows a family relationship for hydrogen dissolved in paraffin and in olefin solvents as a complicated function of temperature. It can be used to calculate hydrogen solubility in binary and ternary mixed hydrocarbon solvents with fair success (37). Kritchevsky and Ilinskaya (336) point out the em- pirical nature of Eq 41. The partial molal volumes from the slope of a plot of In fz/xz against P seldom agree with the experimentally determined partial molal volumes. The difference is due to the concentration dependence of partial molal volumes and partial molal heats of solvent and solute in the dilute binary mixture. De- parture of the Vz in Eq 41 and 42 from the experimental Vz is taken as a sensitive test for discovery of deviations from Henry’s law. An equation for a slightly soluble gas in a liquid under pressure is derived that takes into account the concentration dependence of partial molal volume. It adds a term -(A/RT)(l - xlZ) to the right-hand side of Eq 41. The equation is generalized to fit the solubility of mixtures of gases. Goniltberg (226A) showed that his modification of regular solution theory could be used to calculate the constant A . The Kritchevsky-Ilinskaya equation has been applied to ethylene solubilities in methanol, acetone, methyl ethyl ketone, and toluene (563, 564), to C O Z solubilities in C1 to C q hydrocarbons (606A), and to the H z solubility in cyclohexane (337) where the Hz partial molal volume is strongly affected by pressure. Efremova (156) dis- cusses the pressure dependence of partial molal volumes. Orentlicher and Prausnitz (463) have extended the treatment of hydrogen solubilities in cryogenic liquids at high pressure by taking into account the effect of composition on the activity coefficients. For the small concentration range common to gas solubility they assume the activity coefficient of the solvent is given by the oneparameter expression (Eq 43) A RT In y1 = -Xz2 and they use the modified Henry’s lavi equation 3 2 = YZKXZ (Eq 44) This leads to an equation 3 z A V2(P - P O ) l n - = l n K + - ( X 1 2 - l ) + RT 2 2 RT (Eq 45) which is siniilar in form to the Kritchevsky-Ilinskaya equation. Orentlicher and Prausnitz show that for small solubilities the equation can be put in the form where d20 is the vapor phase fugacity coefficient. Thus at small values of x 2 it predicts a linear In 32/x2 against P relationship. Henry’s law constant, K , the constant, -4, and 7 2 are all temperature dependent. Orentlicher and Prausnitz estimate Vz assuming hydrogen to behave as a hard- sphere gas. They get K and A from the fit of the equa- tion to the solubility of hydrogen in the solvents Ar, Go, N 2 , CH4, C2H4, CZH~, C3Hsl C3H6, and n-hexane. No significant variation of A with temperature was SOLUBILITY OF GASES IN LIQUIDS 423 found. They developed a simplified solution theory that suggests a plot of a reduced Henry's constant, K, (K divided by the solvent solubility parameter squared, evaluated at 0.7 To), against reduced tempera- ture, T, (TI = T/T,), should give a universal curve for all solvents. Their plot of K, against TI falls on at least two lines, one for the solvents ethane, ethylene, propane, and propylene and another for the inorganic liquids Ar, X2, and CO. The difference in the lines is taken to indicate a qualitative difference between hy- drogen-hydrocarbon and hydrogen-inorganic molecule intermolecular forces. Himmelblau and -4rends (250) have developed a five-constant equation to correlate the literature high- pressure solubility data of gases that do not react with water. Prausnitz, Edminster, and Chao (484) develop a solubility parameter expression for the vaporization equilibrium ratio that has application to high-pressure solubilities. Some other equations that have been used to correlate high-pressure gas solubility include for COz in methyl ethyl ketone, ethyl acetate, and toluene (559) A 'XZ lnf2jxz = In I i - __ RT for C02 in methanol (558) x = 0.425(P/P0) (Eq 48) Iiiyama and Hiraoka (314) fit the solubility of acetylene in benzene with regular solution theory in- cluding Flory-Huggins mixing (Eq 21) when the com- pressibility factor, 2, is included in the solubility pa- rameter calculation Lachowicz (351A) analyzed regular solution theory (Eq 20) and predicted deuterium should be more soluble than hydrogen. However, high-pressure solubility studies of Hz and D, in heptane and octane showed no significant difference (352). Namiot and Bondareva (440) compared simple hydro- carbon solubilities in water. At 200 atm and 40" the ratio of solubility for CH4/CzHS/C3Hj/C4H10 is 1 :0.44: 0.20 : 0.073. JIcKetta and eo-workers have studied the solubility of methane (120, l2l), ethane (119, l20), propane (20), ethylene (140), propylene (21), cyclo- propane (260), 1-butene (74), 1,3-butadiene (493), and propyne (270) in water. The "normal" behavior of a solubility minimum at some temperatures is observed for methane, ethane, propane, propylene, ethylene, and 1-butene, but in the regions of temperature and pres- sure studied cyclopropane, propyne, and 1,3-butadiene show no minimum. Above the condensation pressure of the gas there is a three-phase system of gas-water- rich liquid-hydrocarbon-rich liquid. Figure 17 corn- .. MOLE FRACTION 110' Figure 17.-Comparison of the solubility of various hydrocar- bons in liquid water at 100°F (37.8') as a function of pressure (493 . Reprinted from the Journal o f Chemical Engineering and Data by permission of the copyright owners, The American Chemical Society. pares the pressure dependence of hydrocarbon solubility in water at 38". Hydrocarbon solubility in water in- creases with unsaturation. 1,3-Butadiene is four times more soluble than n-butane, and acetylene is more soluble than the other hydrocarbon gases. The solubility of ethylene up to 120 atm is in the order hexane > cyclohexane > benzene (683). Hydrogen containing freons are much more soluble in water than other freons (469). The solubility of acetylene in acetone goes through a pronounced minimum at 3" and 15 atni. At higher temperatures the minimum tends to disappear (259). A test for the thermodynamic consistency of the solubility data for the hydrogen-helium system has been made (69A). The surface tension lowering due to dissoved Sz and argon up to 120 atm in water, hexane, octane, and methanol has been examined (3978). Hiraoka and Kiyama (251-253, 313-315) have studied the pressure dependence of acetylene and ethylene solubility. Acetylene shows positive devia- tions from Raoult's lam in water, methanol, and benzene, negative deviations in tetrahydrofuran. Heats of solution are exothermic in water and methanol, endo- thermic in benzene. Ryntani (512-517) finds that acetylene solubilities at high pressure show negative deviations from Raoult's law in electron-donating solvents. The acetylene solubilities are proportional to the 0 atoms/mole in polyethylene glycols unless steric factors n-ere present. The acetylene solubility was less than expected from the pure solvent solubili- lies in binary mixtures of donor-type molecules with either chloroform, water, or methanol. Some studies o f the solubility of gas mixtures in- clude observations that the presence of COZ decreases the solubility of iYz more than Nz decreases GO2 solu- bility (679). Methane decreases COZ but ethane in- creases COz solubility (3948). The solubility of CO Z from a Nz-H2 mixture is proportional to its fugacity calculated from the Beattie-Bridgman equation (673). 424 RUBIN BATTINO AND H. LAWRENCE CLEVER The solubility of CH4-N2-C02 mixed gas in water can be calculated from the solubility of the single gas and the activity coefficient of the components of the gas phase (437). The Krichevsky-Kasarnovsky equation holds ap- proximately for Ar-CH4 (443) and for He-CH4 (441) mixed gas solutes. Trivus (607) has tested the Kritchevsky-Ilinskaya equation for mixed gases and finds it good to no better than 12% at low gas con- centrations. The effect of hydrostatic pressure on the solubility of a gas is of interest to oceanographers. Klots (317) has used a simple thermodynamic approach to calculate the change in Henry’s law constant with depth. K (P) P Vz log- - - -- (Eq 50) K(0) 2.303RT K(P) and K(0) are Henry’s law constant at 0 and P hydrostatic pressure; V2 is the gas partial molal volume, assumed independent of pressure. His calculation shows that with surface water in equilbrium with gas at 1 atm, equilibrium solubility decreases (Henry’s constant increases) with water depth for both nitrogen and oxygen. Klotz (321) also assumed gas equilibrium exists between surface and ocean depths. He takes into ac- count the dependence of Gibbs free energy on depth (the position of the solute in the gravitational field), on pressure, and on the amount of solute dissolved in a specified amount of solvent at a particular depth. His thermodynamic derivation gives mdepth maurtaoe RT I n - = (I - V2p)d (Eq 51) where mdepth and msurfaea are molalities; PI2 is the gas molecular weight, g acceleration due to gravity, 7 2 the gas partial molal volume in cma/g, p the solvent density, and d the solvent depth in cm. Both Vz and p are assumed independent of pressure. The sign of the term (1 - 7 2 p ) determines whether the gas will increase or decrease in solubility with depth. The equation predicts H2, CO, and Nz solubilities to decrease with depth, CO2 solubility to increase with depth, and 0 2 solubility to be almost independent of depth. An experimental study of the effect of hydrostatic pressure on gases dissolved in water by Enns, Schol- ander, and Bradstreet (169) indicated He, Nz, 0 2 , Ar, and C02 all decrease in equilibrium solubility with depth. Actually, they determined the gas equilibrium pressure necessary to maintain the atmospheric gas pressure saturation value as the hydrostatic pressure increased from 0 to 1500 psig. The required equilib- rium gas pressure increased 13% for He, 14% for Nz, 02, and Ar, and 16% for C02. This would presumably imply the same percentage decrease in solubility in order to maintain a gas equilibrium pressure of 1 atm. I. SOLUBILITY OF GASES I N MOLTEN SALTS AND GLASSES The determinations of gas solubility in molten salts and glasses has been done primarily by workers in- terested in molten salt reactors, industrial processes using molten salt solvents as the aluminum industry, or glass technology. The most extensive work on gas solubilities in molten salts was carried out by Grimes and co-workers at Oak Ridge. They have determined the solubility of helium, neon, argon, and xenon in molten fluoride salt mixtures LiF-NaF-KF (50), NaF-ZrF4 (232), and LiF-BeFz (635). They have also determined solubili- ties of H F in NaF-ZrF4 mixed melts (553) and BF3 in a LiF-BeF2-ZrF4-ThF4-UF4 melt (554). The noble gas solubilities are of the order of to lo-’ mole of gas per cc o f melt with the solubility decreasing in the order He > Ne > Ar > Xe. The enthalpies of solution for all four gases are positive and become more positive as the gas atomic weight in- creases. Solubility increases with temperature. Henry’s law is obeyed over the 0.5-2-atm range studied. Grimes and co-workers have picked a standard state change for calculation of the thermodynamic properties of the solution so that the entropy of solution is purely a function of solvent-gas interactions. The standard state change is X(B,Cd) x(d.c,d) where X represents 1 mole of gas. The subscripts g and d denote the gas and liquid phases and c d is the con- centration of the gas dissolved in the liquid which is in equilibrium with the gas at concentration C,. They assume the change from C, to c d in the gas phase is ideal. Thus, in the desired standard state A B 0 T AS” = - R In c d / c g (Eq 52) where hRo is the heat of solution gotten from a plot of log Henry’s law constant against 1/T (50). The en- tropies of solution at 1000°K in NaF-KF-LiF and NaF-ZrF4 melts vary from - 0.1 to - 1.5 cal/deg/mole with no regular pattern for the four gases. In the LiF-BeF2 melts the entropiesare more negative, varying over the range -3.1 to -4.2 for the four gases. Grimes and co-workers (50, 635), using a model simi- lar to that of Uhlig (620), correlate the noble gas solubilities by equating the free energy o f solution of the gas to the free energy of formation of holes which were assumed to be of the same size as the gas molecule in a continuous fluid having the same surface tension as the solvent. The approach gives an expression for the Henry’s law constant as a function of the liquid surface tension, u, the radius of a spherical gas atom, T , and the temperature, T. SOLUBILITY OF GASES IN LIQUIDS 425 Although admittedly a naive model, it does remarkably well at correlating the noble gas solubility in molten salts. Taking the noble gas atomic radii as the same as in the solid and assuming the values to be the lower limit of the hole radius, the correct magnitude and order of solubility are predicted for the He, Ne, Ar, and Xe series. This is not the same order observed in non- electrolyte solvents at room temperatures where the normal surface tension range is 15-40 ergs/cm2. The molten salts studied have surface tensions between 112 and 180 ergs/cm2 and the free energy of forming holes is presumably a predominant term. The calculated Henry's law constant for helium and neon is high. A larger radius for helium and neon can be justified on the basis of thermal motion of these less polarizable atoms in the melt, and this would give better agreement between the calculated and experimental constants. The HI? (553) and BF3 (554), solubilities in molten fluorides obey Henry's law. The heats of solution are exothermic and the solubilities of both gases decrease with temperature. The entropies of solution at 1000°K are about -6 for H F and about -22 for BF3, both values being more negative than values for the noble gases in similar mixed melts. BF3 is much more soluble than the noble gases. The solubility of HF increases tenfold aa the NaF concentration increases from 45 to 80.5 mole % in the NaF-ZrF4 system. The increase is thought to be related to the high stability of HF-NaF compounds. Woelk (662) reports argon solubilities and discusses the relation of molten salt free volume to gas solubility. Scholze and Mulfinger (545) report the solubility of helium decreases when the concentration of lithium in a mixed silicate melt increases from 20 to 25 mole as Li20. However, the glass molar volume per two oxide ions shows negligible change from that of pure Si02. Assuming that lithium also enters cavities where the helium atom fits, the decrease in helium solubility was used to calculate the lithium ion radius, and a reason- able value of 0.76 A was obtained. Scholze and Mulfinger (435, 546) describe an appara- tus sensitive to lo- cc of helium and report the solu- bility of helium in various binary and ternary silicate melts between 1200 and 1480". The helium solubility increases with temperature and there is a relation between the temperature dependence of helium solu- bility and the thermad coefficient of expansion of the glasses investigated. In lithium silicate melts the amount of dissolved helium depends on the volume of interstices in the atomic constitution of the melt. Conditions for a purely physical solubility are fulfilled. The solubility of helium in Pyrex (506) is reported to be independent of temperature between 25 and 515". The solubility of hydrogen in molten alkali metal hydroxides is stated to be less than 60 mg of hydrogen/ 100 g of alkali hydroxide unless corrosion products are present. Corrosion products increase the solubility o f hydrogen (589). In Pyrex glass at 1170" hydrogen at 10 mm pressure has a solubility of 0.060 ml (STP)/g, but there was no measureable solubility of helium, oxygen, or argon (444). An observed increase in hy- drogen solubility in Ca0-AlzO8--SiOz slags as the CaO content increased was explained on the basis that the Ca-O-H grouping was more stable than the Si+-H grouping (451). Ryabukhin (511) finds the solubility of chlorine in melts of NaC1, KC1, and MgCl2 and their 1 : 1 mixtures show little salt dependence. Plots of log S T against 1/T are linear and the solubility, S, can be represented b s where A = (Q + QIT)/RT and Q and QI are work functions. Oxygen and nitrogen have a solubility of less than 10-4 mole/mole of salt at the freezing point in sodium, potassium, and cesium nitrates (195). Oxygen has a "negligible" solubility in glass ( 4 4 4 ) . Mahieux (385) determined the solubility of carbon dioxide in molten glass by a I4CO2 tracer technique. He found the solubility decreased with temperature over the 1100-1300" range. Krohn, Grjotheim, and co-workers (233, 364) observe that the solubility of carbon dioxide at 1 atm in molten NaCl and molten KC1 increases with increasing temperature with the heat of solution being twice as great in KC1 as in NaCI. The solubility is greater in NaCl at 800" but greater in KC1 at 950". They correlate the crossover in solubility with temperature to a similar crossover in free volume with temperature for the two salts. The solubility of water vapor in molten salts and glasses has received considerable attention. Scholze and Rlulhger (546) determined the solubility of water vapor in various binary and ternary silicate melts between 1250 and 1750". They found the water vapor solubility increases with increasing alkali content o f the glass, with increasing radius of the alkali ions, and with increasing temperature. The solubility depends on the square root of the water vapor pressure and can be used as a measure of melt basicity. Tomlinson (602) reports the solubility of water in sodium silicate decreases 10% on going from 900 to 1100". The solubility is not proportional to the square root of water vapor pressure especially at the lower vapor pressures possibly because some sodium vapor is formed which reacts and removes water vapor in cool parts of the solubility apparatus. Duke and Doan (154) find water vapor solubility is negligible in either NaNOa or KNOa until LiN03 is 426 RUBIN BATTINO AND H. LAWRENCE CLEVER added. In the mixtures containing LBO8 the solu- bility is measurable and linear with temperature. The solubility varies with [Li+l2 which suggest two Lit ions are involved for each water molecule. It is pointed out that lithium has a greater ion-dipole interaction with water than the other alkali metal ions. The solubility of water vapor at 20 mm in NaKOa, KNOs, and CsN03 at their freezing point is found to be near mole/mole of salt by Frame, Rhodes, and Ubbelohde (195). The solubility is proportional to the free volume in the melt. When divalent cations are added, e.g., Ba+2, there is a slight increase in water solubility. There is probably only a slight increase in vacancies because of the considerable electrostriction of the melt that takes place when a divalent ion is added. The solubility of water vapor in LiCl-KC1 melts was determined between 3 and 26 mm at 390 and 480". Henry' law is obeyed up to 10 mm at 390 and up to 18 mm at 480" according to Burkhard and Corbett (82, 83). At the higher pressures HC1 is detected and it is thought that lithium hydrolyzes. The solubility of HC1 has been determined in the same melts. Henry's law is obeyed up to 90 mm for HC1 but the straight line does not go through the origin at zero pressure indicat- ing a residue of hydroxide was initially present in the melts. The hydroxide residue was not thought to have affected the water solubility. Sulfur trioxide in glass is in the form of a SO2-02 mixture (45). J. THE SOLUBILITY OF GASES IN MOLTEN METALS AND ALLOYS Table VI1 lists references to the solubility of gases in molten metals and alloys. The solubility of a gas in a molten metal may represent a physical equilibrium be- tween the gas and metal alone, or it may represent an equilibrium between metal and an intermediate oxide, nitride, or hydride phase for which at constant tem- perature there is a fixed pressure of the corresponding gas. Especially helpful introductions to the subject of gases in liquid metals are two papers from a 1954 Australian symposium on gases in metals by Willis (655) and by Jenkins (278A). These papers, which also discuss adsorption on metals and gases in solid metals, summarize the theoretical approaches and list references to earlier review papers. The books of Turovtseva and Kunin (618A) and of Smithells (5788) contain infonna- tion on the solubility of gases in liquid metals. The diatomic gases usually are in the liquid metal in the atomic state. Their solubility obeys Sieved's square root of pressure law S = K d g It can be shown (2788) that Sievert's experimental law is an expression for Henry's law when the solution equilibrium for the atomic solution o f the gas is taken into account Hdg) F? 2H (soh) A detailed discussion of He, X z , and 0 2 gas solubilities in liquid metals is beyond the scope of this review. For details see references listed in Table VII. The solubility of the rare gases in liquid metals is quite low. No absorption of helium by mercury was detected at 25 atm and room temperature (431). &lcRlilIan (410A) calculated the solubility of xenon in liquid bismuth to be 3 X atomic fraction at 300" and 1 atm pressure by a free volume model for liquid bismuth. Experimental solubility determinations give 3 X lo- atomic fraction at 540" and 1 atm (171), 4 X (422A) and 2 X 10-lo (2428) atomic fraction at 500" and 1 atm pressure. The lowest value seems most reliable at this time. Surface adsorption and occlusion may be responsible for the higher results. McMillan also predicted high surface adsorption which has been confirmed (242A). Xenon solubilities in mercury and in sodium are aIso reported (4228). Johnson and Shuttleworth (281) and Johnson (280) review briefly earlier work on rare gas solubility in liquid metals and report measurements of krypton solubility in liquid cadmium, indium, lead, tin, and silver by a technique using radioactive %r. Their lower limit of measurement was an Ostwald coefficient of 10-7 atomic fraction). No solubility was detected in silver, but Ostwald coefficients in the other metal melts varied from lo-' to The temperature dependence of the Ostwald coefficient was used to get the energy, H , needed to transfer an atom at rest from the gas into the solution of the metal and the vibrational entropy, S, of the krypton atom dissolved in the liquid metal from h3 L = (2 .rrm K T ) where m is the mass of the krypton atom and l/a3 is the number of metal atoms in unit volume of the liquid metal. The experimental H was compared with a calculated H which considered the energy of transfer to be made up of three terms: (1) the energy to make a hole in the liquid the size of the krypton atom estimated from the hole area of krypton atom radius and the surface energy per unit area (surface tension) of the liquid metal extrapolated to 0°K; (2) the vibrational energy of a krypton atom in the liquid approximated as being the same as the vibrational energy of a metal atom as obtained from heat capacity data; and (3) the negative van der Waals interaction energy between the krypton atom and the surrounding metal atoms estimated from adsorption potentials of inert gases on metals. The estimated values of H agreed within a few kilocalories SOLUBILITY OF GASES IN LIQUIDS 427 TABLE V fbLUBlLITY Di~lI.1 Gases Helium Neon Argon Solvent Inert Gases Water Water (hydrostatic pressure) D20 Sea water Methanol n-Hexane. n-heptane, n-octane, n-nonane, n-decane, n-dode- cane, n-tetradecane, 2,3-dimethylhexane, 2,4-dimethyl- hexane, 3-methylheptane, isooctane, cyclohexane, benzene hfethylcyclohexane, perfluoromethylcyclohexane n-Perfluoroheptane Fluorobenzene, chlorobenzene, bromobenzene, iodobenzene, Nitromethane Polyethylene terephthalate-amorphous, glassy crystalline, rubbery crystalline Poly(viny1 acetate) Polyethylene, hydropol Diester, phosphate ester, dimethyl silicone, methyl phenyl Apiezon GW oil, silicone DC 702 oil, silicone DC 200 oil Esso synthetic oil, castor oil, DC 200/200 silicone fluid, blown rapeseed oil, shell rotary vacuum-pump oil Santovax R Lung tissue (blood-free hornogenates) Pentaborane Liquid CH, Liquid Nz Liquid argon hlercury Liquid Hz Aqueous solutions: toluene, nitrobenzene silicone, paraffin base oil, aromatic base oil Solute: 1 N NaC1. LiC1, HC1, KC1, NHaC1, BaClz, NanSOa, KI, Water satd with nitromethane Nitromethane satd with water Uranyl sulfate (40-243 g/L) NaBr, HNOa, NMeaI, NEtrBr Methanol, 0-4 M in IiaI Kate1 Sea water n-Hexane, n-heptane, n-octane, n-nonane, n-decsne, n-dode- cane, n-tetradecane, 3-methylheptane, 2,3-dimethylhex- ane, 2,4-dimethylhexane, isooctane, cyclohexane, benzene Methylcyclohexane, perfluoromethylcyclohexane Fluorobenzene, chlorobenzene, bromobenzene, iodobenzene, Nitromethane Poly(viny1 acetate) Olive oil, human fat (pooled) Liquid argon 1 N NaC1, LiCl, KI toluene, nitrobenzene Water Water (hydrostatic pressure) Sea water Methanol 1-Propanol, 1-butanol, 1-pentanol, ethylene glycol Pressure, atm 1 1 7-34 1-102 3.4-78 1 1 1 1 1 1 1 1 To 200 mm 1 34-68 1 High pressures 1 1 34-100 25-157 1-295 To 160 mm High pressures 2-7 I 1 1 1 1 To 200 mm 1 To 140 mm 1 1 1 1 1 To 700 1 1 1 1 1 1 1 1-102 1 1 1 1 Temp, O C 5-73 25 163-3 16 25 50-2S0 1-20 30 15-42 16-43 18-30 15-45 25 25-130 8-40 25 24-177 20-83 20-140 233-406 37 30-150 90-106'K 78-109'K 84-87.5' K 20-140 16-29"K 25 25 25 162-300 30 9-74 1-16 38 0-25 14-39 16-43 15-55 25 8-40 38 83.9-87.5'K 25 0-20 15-25 11-74 25-40 10-50 3-28 30 2-27 10-20 30 3-30 5-25 25 1-20 30 30 25-35 Measure- ment value (see section VI) Ref 2 1 1 1 1 1 2 1 2 2 1 1 1 2 1 1 1 1 1 1 1 1 0 2 1 1 1 1 1 0 2 2 1 2 1 1 2 1 1 2 2 2 1 2 1 2 2 0 2 2 2 1 1 2 1 1 1 1 2 2 429 195 481 169 585 327 99 97 98 322 530 197 415 411 414 61 85 431 234 89 67 229 228 294 431 505 430 197 197 583 99 4 29 327 267 327 97 98 530 197 411 267 293, 294 430 173 197 429 260 443 38 99 318 327 260 152 40 169 327 260 99 223 428 Gases Krypton RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Solvent Methanol, ethanol, 1-propanol, 1-butanol 1-pentanol, l-hex- p-Dioxane Nonpolar solvents n-Hexane, n-heptane, n-octane, n-nonane, n-decane, n-dode- cane, n-tetradecane, 3-methylheptane, 2,3-dimethylhex- ane, Z,&dirnethylhexane, isoootane, cyclohexane, bensene Fluorobenzene, chlorobenzene, bromobenzene, iodobencene, toluene, nitrobenzene Methylc yclohexane Toluene pXylene CCl4 Perfluorometh yloyclohexane n-Perfluoroheptane (C4Fo)aN Nitromethane Carbon disulfide Olive oil Paraffin wax Polyethylene, hydropol, and natural rubber Polyethylene terephthalate-amorphous, glassy crystalline, Eel blood Hydrated Fe and Al oxides; soh of egg albumin, gelatin, serum, and serum albumin Liquid NHI anol, 1-heptanol, 1-octanol rubbery crystalline Ammonia Krypton-bulk and absorbed in layers Xenon-absorbed in layers Aqueous solutions: Solute: NaCl up to 3.46 M NaI, NaCl, NaBr, KCl, LiCl NaCl, CaClz, MgCla 1 N NaC1, LiCI, KI Water-NaI s o h Water, 0-7 M in NaI Water-ethanol mixture (XE~OH = 0.015-0.25) Water satd with nitromethane Nitromethane satd with water Water-pdioxane (all compositions) Methanol, 0 -4 M in NaI Methanol soh of tetramethylammonium iodide, tetra- p-Xylene-pdichlorobenzene (3 mixtures) pXylene-pdibromobenzene (2 mixtures) p-Xylene-p-diiodobenzene (1 mixture) Nonaqueous solutions: methylammonium bromide, CaClx, NaI Water Sea water n-Hexane n-Heptane, n-dodecane n-Hexane, n-heptane, n-octane, n-nonane, n-decane, n-dode- cane, n-tetradecane, 3-methyllreptane, 2,3-dimethylhex- ane. 2,4-dimethylhexane, isooctane, cyclohexane, benzene pXylene pXylene Mesitylene Methylcyclohexane, perfluoromethylcyclohexane Fluorobenzene, chlorobenzene, bromobenzene, iodobenzene, Nitromethane Olive oil Amsco 123-15 Ultrasene toluene, nitrobenzene Presaure , atm 1 1 To 300 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 292-767 mm 1 1 1 1 25-100 To 800 25100 0-7800 h t n point O-eatn point 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 To 41,000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Temp, O C 25-35 5-25 1 5 4 1 15-55 - 10 to 25 16-43 15-30 30 -20 to 10 5-35 16-43 25 4-32 25 -20 to 25 25 22-37 72 25 25-130 7 10-40 0-60 25-100 0-25 70-150 77OK 78' K 0-20 20-40 5-90 25 30 30 4-30 25 25 5-2 5 30 30 30 30 30 7-75 1-24 30 25-45 100-300 0-24 -90 to 25 25 16-41 30 30 26 6-43 15-55 25 22-37 25-45 -55 to 150 23 Measure- ment value (see section VI) Ref 1 1 0 2 2 2 2 2 2 2 2 2 2 2 1 2 2 0 1 1 1 1 0 2 2 2 1 0 0 2 2 1 1 1 2 2 1 1 1 2 1 2 2 2 2 1 1 2 1 1 1 1 2 2 1 1 2 2 1 0 2 1 1 66 40 230 97 530 494 98 494 95 494 494 98 222 322 197 494 222 360 501 414 415 580 567 116 413 117, 118 614 573 5i2 173 260 422 4s0 260 99 39 197 197 40 99 260 95 95 95 429 327 582 668 14 327 582 682 97 95 582 582 98 530 197 360 667 582 582 SOLUBILITY OF GASES IN LIQUIDS TABLE V (Continued) 429 Ganes Xenon Radon Hydrogen Pressure, Solvent atm Terphenyl 1 Dog fat, human fat, rat-pooled fat 1 Aq 0-20% bovine serum albumin, 0-20% bovine hemoglo- 1 Aqueous solutions: bin, 04% bovine y-globulin Solute: 1 N NaCl, LiCl, HCl, KCl, "&I, BaClr, NazS04, KI, 0.066 M phosphate buffer, pH 7.0 1 KBr, KNOI, HNOa, NMaI, NEtrBr 0.9% NaCl Aq soh uranyl aulfate Methanol-water (83% methanol) pXylene-pdichlorobenzene (3 mixtures) p-Xylen4-pdibromobensene (2 mixtures) pXylene-pdiiodobensene (1 mixture) Water DsO 9ea water n-Hexane, n-dodecane, isooctane, benzene, cyclohexane n-Heptane Acetic acid Xylene (tech) pXylene Meaitylene Methylcyclohexane, perfluoromethylcyclohexane Toluene Celt Fluorobenzene, chlorobenzene, bromobenrene, iodobenrene, Nitromethane Aniline Dow Corning silicone oils Ultrasene, halocarbon 437, olive oil, pine oil, Dowtherm A, Dow Corning Anti-Foam A, Koppers Emulsion K-900 Amsco 123-15 Olive oil Dog fat, human fat, rat-pooled fat, olive oil Aq soh human hemoglobin Aq 0-20% bovine serum albumin, 0 4 0 % bovine hemoglo- Aqueous solutions: 1 N NaCl, KI 0-2.68 M NaCl 0.066 M phosphate bu5er, pH 7.0 0.9% NaCl Olive oil in water emulsion Uranyl sulfate (40 g/l.) Toluene (40%)-pine oil mixture Formic acid, acetic acid, propionic acid, butyric acid, hex- anoic acid, heptanoic acid, octanoic acid, nonanoic acid, valeric acid, oleic acid toluene, nitrobenzene bin, 0-8% bovine ?-globulin Solute: Decanoic acid, undecanoic acid, acrylic acid Lauric acid, tridecanoic acid Tributyrin, triacetin, trihexanoin, linoleic acid Olive oil, olive oil (Italian) Rat fatty acids (extracted), human fat (extracted), butterfat Petroleum oils Elementary Gases Water Met hanol Alcohol 2-Butanol 1 1 To 50,OOO 1 1 1 1 1 1 1 1 1 7-2 1 8-14 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 12-19 1 1 1 1 1 1 1 Very low 1 1 0-72 6&204 7-24 To 50 To 100 85-800 10-80 1 Temp, C 25 2 5 4 5 37 25 25-45 25-45 100-300 -120 to - 30 30 30 0-20 13-72 5-25 30 25-45 100-300 162-300 0-25 16-43 22 28 32 0-29 20 16-43 24 15-55 25 30 24-95 19-32 24-150 22-37 25-45 20 37 25 0-20 25-45 25-45 27 1OC-260 25 25-50 37 50 25-37 25-37 37 Room 25 12-71 217 2-335 52-343 100-162 24-99 0-45 20 80-150 Measure- ment value (see section VI) Ref 1 582 2 667 1 434 1 2 2 1 1 2 2 2 2 2 1 1 2 0 0 1 a 1 1 1 1 1 2 1 0 2 1 1 1 1 1 0 2 1 1 1 2 2 2 1 0 1 2 2 2 2 2 2 1 0 2 2 1 1 0 0 2 2 2 0 430 668 668 14 582 95 95 95 173 429 327 682 668 583 683 321 96 582 582 582 582 582 98 582 302 530 197 582 582 582 582 360 667 547 434 430 173 668 668 582 583 582 452 452 452 452 452 452 200 174 428 29 588 481 24 683 416 560 378 9 430 Gases RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Cyclohexanol Phenol Liquid methane Ethane n-Propane n-Butane Isobutane n-Hexane n-Heptane Solvent n-Octane Octane (tech) n-Nonane n-Decane 2,2,4-Trimethylpentane Ethylene Propylene Cyclohexane Benzene Toluene m-Xylene n-Perfluoroheptane CClr Carbon disulfide Cottonseed oil, lard Butter oil Tetralin, olive oil 2 oils (mol wt 400 and 610) Hydrocarbon mixture (av mol wt 250) Hydrocarbon (slack wax mol rvt 345) Para511 oil Various fats Poly(viny1 acetate) Santoaax R Decalin (t7an.s) Hydrated Fe and A 1 oxides: soln o f egg albumin, gelatin, Liquid argon Liquid nitrogen serum, and serum albumin Liquid diborane Liquid ammonia (nomograph) Pentaborane Octamethylcyclotetrrsiloxane Aqueous solutions: Solute: NaC1, LiC1, LaCls 1% gelatin in water Vranyl sulfate, uranyl fluoride (40-243 g/L) tions Ammoniacal cuprous carbonate and cuprous formate solu- Xaphtha mixture Isomeric dodecane mixture Hexene-hexane, heptene-heptane, octene-octane Cyclohexane-benzene mixtures Methane-propane mixtures Pressure, atm 1 1 34-272 180-230 17-545 7-540 20-103 20-545 34-200 48-145 120-680 50-300 1 1 1 1 1 50-300 1 20-40 1 12-342 17-545 17-545 48-145 To 700 1 48-145 240-2900 50-487 1 1 1 1 48-145 1 1 1 1 1 1 1 1 1 10-968 2-968 1 To 200 mm 1 1 1 10-100 3.5-46 1-900 5-45 0-40 34-100 8-2 1 1 1 To 100 To 4 30-85 34-342 1 34-68 Measure- ment value Temp, (see OC section VI) Ref 25-140 1 40-140 1 -157 to -101 0 90-127'K - 184 to 24 4-88 24-116 - 130 to 24 38-121 35 4-204 25-50 - 35 t o 35 35 -30 t o 50 35 -25 t o 35 25-50 80-150 25-30 40-80 -25 t o 35 38-150 -157 to -18 - 157 t o 24 35 20-60 25-60 35-73 25-150 25 10-45 10-35 25-60 - 15 to 35 80-150 35 -25 to 50 -25 to 35 25-50 0-3 5 -25 to 25 25 40 40-60 25 0-100 106-200 106-300 20 50-220 8-40 241-404 20-140 10-40 8i-140°K 90-9B°K 79-109"K 63-75'K 113-1 81 O K 30-50 30 13-72 25 100-162 100-200 93-149 80-150 40 -129 to -18 2 2 0 1 0 2 1 C 2 2 1 2 2 2 2 0 2 1 2 2 2 2 1 2 1 1 2 2 2 2 1 2 0 1 2 2 2 2 2 2 1 1 I 1 2 2 2 0 2 1 1 0 2 0 2 c 2 1 1 2 0 0 0 0 2 0 1 0 347 347 36 177 653 84 44.4 167 141 527 447 352 105 3i4 103 104 105 352 9 599 445 105 141 653 653 527 337 347 527 272 340 103 105 347 103,105 9 527 103 105 216 103,105 103,105 216 532 532 127 23 473 473 3;s 619 411 234 347 567 630 388 227 460 265 131 67 90 428 174 583 24 299 141 9 347 36 G8BW Deuterium Nitrogen SOLUBILITY OF GASES IN LIQUIDS TABLE v (Continued) Solvent DrO Heptane, octane, 2,2,4-trimethylpentane, benzene, toluene, n-Heptane, n-octane Liquid argon Liquid nitrogen KNHz in liquid NHI Kater CCh, n-perfluoroheptane, CSz Water (hydrostatic pressure) U-ater, Nz isotopes in -4naerobic seawater, Nz isotopes in 1-Propanol, 1-pentanol, cyclohexanol, ethylene glycol Methanol, ethanol, 1-propanol, 1-butanol, 1-pentanol, 1- Methanol, ethanol, 2-propanol, 1-butanol, ethanol (95%) Methanol hexanol, 1-heptanol, 1-octanol Alcohol Liquid methane Liquid propane Butane n-Hexane n-Heptane n-Heptane, n-octane, n-nonane 2,2,4-Trimethylpentane Cyclohexane Benzene Toluene Nonpolar solvents Freon-114 n-Perfluoroheptane Perfluoromethylcyclohexane Perfluorodimethylcyclohexane Acetone Ximomethane Dimethylformamide n-Propyl nitrate (C4Fs)aN Carbon disulfide Human fat (pooled), human fat (individual), dog fat, olive oil Olive oil, tetralin, oleic acid, ethyl palmitate Cottonseed oil, lard Butter oil Paraffin oil Santowax R 6 oils (av mol wt 400-670) Electrical insulating oil 4 crude oils 4 Russian crude oils Baku crude oils Crude oils, kerosine, and gasoline Petrowax A Paraffin wax Polyethylene terephthalate-amorphous, glassy crystalline, rubbery crystalline Pressure, atm 4-28 1 50-300 10-100 3-69 1 34-204 97 1 10-27 1 08-204 1 To 50 1 11-58 1 1 1 1 1 100-500 1-102 1 1 1 1 1 48-280 10-80 1 1 100-750 mm 100-750 mm 34-286 1 68-680 1 1 1 1 60-300 27-775 1 15-400 To 300 1-9 1 1 1 1 1 400-900 mm 1 1 1 1 1 1 1 400-760 mm 200-770 mm 1 1 1 0-300 1-300 50-300 1 246-790 mm 265-774 mm 1 Temp, O C 162-300 -35 to 35 25-50 87-12OoK 90-95'K -64 to -42 0-260 0-240 25 260-316 12-73 18 25 Room 30 150-200 2-27 3-37 30 3-30 38 0-25 25 2-27 5-29 25 2 5-3 5 -25 t o 50 0-75 0 to -45 25-35 20 90°K 90°K 38 25 32-182 25-35 20 -25 to 50 25 30-150 25 25 200-275 20-60 0-50 25 25-50 -25 to 50 25 Room 11-30 25 6-31 37 25 40 40-60 20-82 20 238-409 0-100 26-66 20 20-100 20-100 20-60 82 76 25-130 5-40 431 Measure- ment value (see section VI) Ref 0 2 2 2 0 1 1 0 2 2 2 0 2 0 1 1 0 2 2 1 2 2 1 1 2 2 2 1 2 1 2 1 2 1 1 0 2 0 2 2 2 2 2 2 2 1 0 1 2 2 2 2 1 1 1 2 2 2 2 1 1 1 2 1 1 2 2 0 0 0 0 1 1 1 583 105 352 630 388 28 588 369 427 481 428 157 70 24 593 578 442 318 176 200 152 260 317 169 319 499 223 66 342 339 560 260 378 626 626 4 218 3 599 23 342 218 420 340 218 617 230 654 218 218 218 342 197 237 13 322 218 322 266 127 532 532 378 377 234 23 181 519 518 682 604 501 501 415 432 Gases Oxygen RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Solvent Polyethylene, hydrogenated polybutadiene, and natural rub- 4 aircraft fuels 100 octane fuel (av mol wt 100) Kerosene (av mol wt 165) Kerosene (jet fuel) Human blood and plasma, human urine Blood Eel blood Hydrated Fe and Al oxides, s o h of egg albumin, gelatin, Liquid 601 Liquid SOr Liquid NHa Liquid NH: (nomograph) Liquid Cor Pentaborane Aqueous solutions: Solute: NaI ber serum, and serum albumin 1 N NaC1, LiCl, KI 0.5-4.0 N NaOH 1.38 N in NaCl and 5.41 N in CaClr, 2.5-5.0 N MgS04, 1-5.5 N NaCI, 1-4.2 N NarSO4, 1-11 N CaClr 0.15 M Zn(OAc)r, 0.05415 M Ni(OAc)n, 0.05-0.15 M Hg(OAc):, 0.03415 M Co(OAc)r, 0.06-0.15 M Mn(OAc)t, and0.05 MCu(0Ac)zin 1 MHOAc + 0.5 M NaOAc soln Detergent solution (hydrostatic pressure) Ammoniacal cuprous carbonate and cuprous formate NaCl, CaClr, MgCla NaI-methanol solutions Acetone + ethanol (50 vol. %), ethanol+ 2,2,4-trimethyl- pentane (50 vol. %) 50% Decanol-dodecanol soh Nonaqueous solutions: Water Water (hydrostatic pressure) Water ( 0 2 isotopes in) Water (nomograph) Sea water Sea water (hydrostatic pressure) Saline water Ethanol Ethanol, ethanol (95%), methanol, %propanol, 1-butanol Methanol, 1-propanol, %propanol, 1-butanol, ethylene gly- Methanol Alcohol Methyl acetate, ethyl acetate n-Heptane, n-octane n-Nonane col, glycerine, risinusal Pressure, atm 1 1 1 1 1 1 1 1 11-35 50-101 50-100 34-100 1 1 97 12-75 1 1-102 To 4 1 1 400 mm 97 1 1 1 7-2 I 1 To 100 1 68-136 1 1 1 1 1 14-23 1 1 1 1 1 1 To 200 100-500 1-102 1 690-780 mrn 1 1 1-102 1 1 1 1 1 1 1 1 1 Temp, "C 25 20 0-20 16 37 37.5 7-20 10-40 25 0-50 15-30 30-150 -32 to 28 30 13-72 0-240 30 25 25 25-35 - 25 to 50 20-82 0-240 25 20 0-20 218-343 13-75 100-163 3-39 0-330 0-36 25 6-24 2-29 20 125-200 Room 2-27 0-35 5-30 0.4-37 100-290 25 0.5-25 2-27 0-35 2-27 0-35 25 0-40 0-70 -25 to 50 20 20 20 20 25-35 25 9-30 Measure- ment value (see section VI) Ref 1 0 2 2 2 2 0 1 0 0 1 2 0 1 1 2 0 1 2 1 0 0 1 2 1 0 2 1 2 1 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 0 0 1 1 2 0 2 2 1 2 1 2 1 2 2 1 2 2 414 183 23 23 144 176 122 580 567 613 142 116 131 338 67 260 428 369 578 70 169 24 422 260 342 378 369 427 539 173 481 428 482 609 686 610 11 581 166 478 14 593 318 231 152 425 424 583 317 169 319 136 424 231 169 611 539 342 539 478 378 539 599 ,509 433 Gases SOLUBILITY OF GASES IN LIQUIDS TABLE V (Continued) Solvent 2,2,4-Trimethylpentane Benzene Toluene p-Xylene Xylene (tech) Carbon tetrachloride, tetrachloroethylene n-Perfluoroheptane Acetone Acetone, diethyl ether, paraldehyde Aniline n-Propyl nitrate (C4Hg)sN Carbon disulfide hlethylal Dioxane Liquid paraffin, olive oil, silicone fluid (1 and 5 cs.) Polydimethylsiloxane oil Polyethylene, hydropol. and natural rubber Polyethylene terephthalate-amorphous, glassy crystalline, Petroleum ether, paraffin oil, benzol, cracked gasoline Paraffin wax Petrowax A Olive oil, tetralin 100 octane (av mol wt 100) Kerosene (jet fuel) Kerosene (av mol wt 165) Ligroin, white spirit fraction, cracking gasoline 6 oils, kerosene, 100 octane aviation fuel Cottonseed oil, lard Butter oil 6 oils (av mol wt 400-670) Paraffin oil Electrical insulating oil Hydrated Fe and A 1 oxides; soln of egg albumin, gelatin, serum, and serum albumin Whole blood Blood Eel blood Liquid Clz Nitric acid (100%) Aqueous solutions: rubbery crystalline Solute: NaC1, LiC1. KCI, MgCh NaCl (15, 30 g/kg) NaC1, CaCIz, MgCh 0.5-4.0 N NaOH Uranyl sulfate, uranyl fluoride (40-240 g of U/L) KOH-water solutions Pure NHa solutions, pure NHs + (NH4)zSO4 solutions 0-100 a t % HzSO4, 0-100 wt % HsPOa, 0-40 wt % KOH 94% "01 + 6% HzO Nitric acid solutions (6-31'% H20) Uranyl sulfate solutions (40-243 g o f U/L) Uranyl sulfate Cane sugar, corn sugar, levulose, dextrose, corn syrup, invert syrup Dextrose, sucrose, levulose, cane sugar, cerelose, bakers glucose, sweetose, puritose, invert Methanol, ethanol, 1-propanol, 2-propanol, 1-butanol, ethylene glycol, glycerine (12-100 wt y o alcohol) 1 N NEhBr Acetone + ethanol (50 vol. %) 2,2,4-Trimethylpentane + ethanol (50 vol. %) KOH-methanol solutions 85% "0s + 15% NiOc KOH in methanol-water solutions Nitric acid solutions (7.5-20% NtO4) (25" only) Nonaqueous solutions: Nitrose and HB04 solution Pressure, atm 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4-2 1 1 1 1 238-808 mm 290-787 mm 1 1 1 1 1 1 1 1 200-770 mm 1 1 1 1 1 1 5-12 8-22 1 1 1 97 170 1 10-30 1 8-2 2 100 4-102 1 1 1 1 1 1 1 8-22 1 Temp, OC 20 -25 to 50 10-30 20 0-70 20 20 30-80 23-100 20 25 -25 to 50 20 20 Room 6-32 25 20 20 38 30 25 25-130 20 72 82 25 20 16 0-20 19 20 40 40-60 0-100 20 20 22-66 10-40 Room 37.5 8-20 25 25 38-88 0-20 0-36 5-90 0-240 100-290 20 110-130 -30 to 25 38-88 25 100-163 100-300 22 19-24 30 25 -25 to 50 -25 to 50 20 3S-88 20 25 19 Measure- ment value (see section VI) Ref 1 2 2 2 1 1 1 2 1 1 2 2 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 2 2 2 1 2 1 1 2 1 2 2 0 1 0 1 2 0 0 2 2 1 0 0 2 1 1 0 0 2 1 1 1 1 1 2 2 2 0 2 0 1 539 342 322 22,23 539 497 539 187 497 539 216 342 539 539 13 322 216 539 539 504 90 414 415 539 501 50 1 127 23 144 23 497 22 532 532 23 377 378 181 567 656 122 580 345 579 502 173 610 422 369 583 478 446 235 502 579 482 14 284 285 539 430 342 342 478 502 478 579 612 434 Gases Ozone Chlorine Air Methane RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Solvent Water Chloroform Freon 11-CClrF Freon 12-CClzFz Freon 13-CClFa Freon 22-CHClFz Freon 113-CClzFCClFz Freon 114-CCIFL!ClFz HzSOa (5-70%) Water Water (nomograph) Cyclohexane Benzene Chlorobenzene n-Perfluoroheptane Hexachlorobutadiene Carbon tetrachloride Liquid TiClr Aqueous solutions: Solute: 0.1-4 M LiClOa, 0.1-8 M NaClO4 NaCl (210-300 g/l.) 0.1-3.0 N BaClz LiCl, BaClz, SrCla HC104 (up to 50 wt %) Water Ice Sea water (several salinities) 2,2 ,4-Trimethylpentane Freon 12, Freon 22 n-Propyl nitrate Butter oil Cottonseed oil, lard 6 oils (petroleum), kerosene. 100-octane aviation fuel 2 100-octane fuels (av mol wt 100) Kerosene (av mol wt 165) 6 Oils (av mol wt 400-670) 8 jet fuels (kerosene) 2 kerosenes 9 heavy lube Oils, 4 light lube oils, diesel fuel, 3 aviation fuels 4 aircraft fuels Electrical insulating oil Beer Aqueous solutions: Solute: NaCl (30-309 gjl.) 0.5-4.0 N NsOH Ethanol (4%) Acetone (satd with CzHz) Compound Gases Water Water (nomograph) Methanol, ethanol, cyclohexanol Methanol, ethanol, 1-propanol, 1-butanol, 1-pentanol, 1- hexanol. 1-heptanol, 1-octanol Pressure, atm 0.06-1 .O 100-800 mm 1 1 1 1 1 2.0-2.7 1.5-6 1 0.1-1 1 97 1 1 68-238 1 1 1 1 To 7 1 1 1 1 1 1 1 1 1 1 68-238 97 1 1 1 20-660 20-680 1 1 1 11-51 1 1 1 7-815 1 1 Temp, O C 20 15 10-39 -25 to -39 20 -110 to -23 -110 to -94 - 110 -110 - 23 - 63 20 10-25 Measure- ment value (=e section VI) 0 1 1 1 1 1 1 1 1 1 1 0 2 10-40 1 10-50 2 25-68 2 25-65 2 0-2 5 2 - 20 to 92 1 40-90 2 20-30 2 10-50 0 30-70 1 10-50 0 10-50 0 0-50 0 0-240 0 21 1 20 2 20 2 25-65 2 0 0-30 1 0 0 0-30 1 20 2 -40 to 24 1 Room 1 40-60 1 40 1 20 2 20 2 0-20 2 0-100 2 16 2 - 18 to 49 2 21 1 0 22-66 2 0 25-65 2 0-240 0 0 040 0 25 2 0-20 2 25 1 25-171 2 2-40 2 12-75 2 18-37 2 25-30 1 25 1 5-45 1 Room 1 10-38 0 18-37 2 25-35 1 Ref 297 255 490 60 386 93 93 93 93 93 93 297 645-647 128 615 341 92 92 219 329 576 345 29 1 672 51 292 551 369 22 23 157 168 42 543 42 22,23 468 13 532 532 22 23 23 23 144 144 549 183 181 168 549 , 157 369 168 69 427 173 120 121 94 428 358 153 400 640 401 139 358 66 Gases Ethane SOLUBILITY OF GASES IN LIQUIDS TABLE V (Continued) Solvent Methanol Propane n-Hexane n-Heptane n-Octane n-Dodecane 2,2,4-Trimethylpentane Cyclohexane n-Decane Benzene Toluene Xylene n-Perfluoroheptane Acetone Dichloroethyl ether, ethylene glycol monoethyl ether Furfural Dimethylformamide Nitrobenzene Carbon disulfide Ethyl Cellosolve, chlorex, PFhlC-4F, paraffin oil Crude oils, kerosene, and gasoline Rurakhany crude oil Petroleum Baku crude oils 4 Russian crude oils Baku and American crude oils 4 crude oils Polyethylene, hydropol, and natural rubber Polyethylene terephthalate-amorphous, glassy crystalline, Paraffinic, naphthenic, and aromatic lean oil Santow-ax R Paraffin wax Liquid 602 con Aqueous solutions: rubbery crystalline Solute: NaCl (0-2.8 .
l f ) NaC1, LiC1, KI (1 m) NaC1, CaClz 1.53 N NaCl + 6.0 N CaClz NaCl CaClz NaCl, CaClz, hfgC1z 4.9 M guanidinium chloride 7 h f urea Water Methanol, ethanol, 1-butanol 1-Propanol, 1-pentanol. oyclohexanol, ethylene glycol Methanol, ethanol, 1-propanol, 1-butanol, 1-pentanol, 1- n-Heptane, n-nonane, n-octane 2,2,4-Trimethylpentane Cyclohexane n-Perfluoroheptane hexanol, 1-heptanol, 1-octanol Pressure, atm 40 7-98 1 41-198 1 1 1 1 41-198 1 14-68 0-238 103-350 30-150 7-326 1 1 7-360 1 1 1 1 200-900 mm 0.1-35 mm 1 0.1-35 mm I Up to 300 1-40 50-300 1-300 0-300 1 1 9-210 1 218-776 mm 12-34 7-68 1 1 25-51 14-61 11-74 1 1 1 Up to 93 4-680 1 4-82 1 1 1-16 1 1 1 1 1 1 1 1 I 1 7-68 1 Temp, "C -50 to 20 -25 to -60 -115 to 0 18-37 38-104 0-2 1 8-35 0-32 3-33 38-104 18-37 -29 to 4 21-121 100 100-250 66 7-21 18-37 66 5-28 3-30 18-30 18-37 30-70 30-70 5-40 30-70 5-50 15-35 5-50 20-60 20-70 66-84 20-100 20-100 40-80 20 25 25-130 450 237-407 72 -32 to 28 -40 to -54 0-20 13-72 4-45 30 30 25-30 5-90 5-45 5-45 2.5 38-171 0-20 38-lil 2-40 12-72 0 10-35 15-3R 25 5-45 Room 25 25-35 25-35 25 14-32 10-ti6 25 435 Measure- ment value (see section VI) Ref 0 2 2 2 2 2 2 2 2 2 2 2 0 2 1 0 2 2 0 2 2 2 2 0 0 1 0 1 2 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0 2 2 0 1 1 1 1 1 1 2 1 2 1 2 2 1 2 2 1 1 1 2 2 1 2 2 0 2 46 560 5 358 542 461 461 46 1 461 541 358 330 520 540 273 160 461 358 160 461 461 322 358 1 T 5 175 237 175 145 322 145 604 308 310 682 518 606 519 414 415 312 234 501 142 149 173 428 3i6 153 153 153 422 640 640 427 119 li3 120 94 428 123 659 6 ti0 400 640 401 223 223 66 599 322 301 222.599 436 Gases Propane Cyclopropane RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Solvent Benzene Dichloroethyl ether, ethylene glycol monoethyl ether Furfural (OFdrN Nitrobenzene Carbon disulfide Polyethylene, hydropo Polyethylene terephthalate 4 aircraft fuels Para5nic, naphthenic, and aromatic lean o i l 4 Surakhany crude oils Petroleum 4 Russian crude oils Baku and American crude oils Lysozyme (lo%), hemoglobin (5%), serum albumin solu- Aqueous solutions: tion (5%) Solute: 0-3 M NaCl 1 m NaCl, LiCl, KI 0.5-1.5 M CaC11, 0.5-2.0 M NaCl Sodium lauryl sulfate (1.8%) Sodium dodecyl sulfate 7 M urea, 4.9 M guanidinium chloride Water Water (nomograph) Methanol Ethanol 2 -Prop an o 1 n-Hexane, n-octane Benzene n-Perfluoroheptane Dichloroethyl ether, ethylene glycol monoethyl ether Furfural 1,4-Dioxane Nitrobenzene Paraffinic, naphthenic, and aromatic lean oil 3 Surakhany crude oils Petroleum Paraffin oil 4 Russian crude oils Baku crude oils Baku and American crude oils Polyethylene, hydropol, and natural rubber Bovine serum albumin solution (5%), hemoglobin (5761, lysozyme (10%) Aqueous solutions: Solute: NaC1, LiC1, KI (1 m) XaC1 (0-24 wt %) Sodium lauryl sulfate (1.8%) Sodium dodecyl sulfate 7 M urea 4.9 M guanidinium chloride Water n-Hexane Benzene n-Perfluoroheptane Dioxane Dog hemoglobin (9.5%), homogenized rabbit muscle, homog enized rabbit liver Olive oil, pooled rat fat Blood Bovine whole blood, plasma, olive o i l Pressure, atm 14-97 1 1 1 1 9-210 0-100 1-40 1-300 1 1 1 1-16 1 1 1 1 1 0-26 1-35 1 1 1 1 1 1-7 30-760 mm 12-760 mm 8-760 mm 1 10-50 1 1 1 8-204 0-2 5 1-40 1-300 50-300 1 1 1 0.1-1 1 1 1 1 1 1 1-41 1 1 1 1 1 1 1 1 1 Temp, O C 0-290 30-70 30-70 14-31 30-70 25 25 25 29 56-86 66-84 20-100 40-80 10-35 0-20 12-72 0 10-35 15-35 5-45 12-74 20-30 12-149 16-138 10-35 15-35 25 5-45 Room 16-54 0-50 0-50 0-50 25 150-210 25 25 30-70 30-70 25 30-70 29 40-85 66-84 20-100 20-100 40-80 25 10-35 12-72 0-20 10-35 15-3 5 5-45 5-45 35 25-38 21-104 25-34 25 25 25 25 35 25-38 27-37 25-38 Measure- t ment value (see section VI) Ref 1 0 0 2 0 2 1 1 0 1 0 0 0 0 2 2 2 1 2 2 1 2 2 0 2 2 2 1 1 1 0 2 2 2 2 2 2 2 0 0 2 0 1 0 0 0 0 0 0 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 2 300 175 175 322 175 222 414 415 183 312 307 310 518 606 659 173 428 123 659 660 640 428 94 324 20 659 660 400 640 401 138 344 343 344 600 273 600 600 175 175 600 1 i 5 312 307 310 377 518 682 606 414 659 428 621 659 660 640 640 364 268 269 60 1 601 601 601 60 1 364 52 479 268 Gases n-Butane Isobutane n-Pentane Neopentane Ethene SOLUBILITY OF GASES IN LIQUIDS TABLE V (Continued) Solvent Aqueous solutions: Solute: NaCl (0.9 wt %), Na phosphate buffer Bovine serum albumin (10%) Bovine serum albumin (0-25%) Bovine serum albumin, hemoglobin, y-globulin, &globulin Water Methanol Ethanol 2-Propanol Para5nic, naphthenic, and aromatic lean oil Aqueous solutions: Solute: NaC1, KCl, LiCl, HCl, KI, BaClz, LaClr ( 1 m) NaCl (0-24 wt %) 4.9 M guanidinium chloride 7 M urea 1 N KBr, NEbBr Hemoglobin (5%), bovine serum albumin solution, Na lauryl sulfate (1.8%). lysozyme (10%) Bovine serum albumin solutions with 0.15 M NaCl Bovine serum albumin (5%) Sodium dodecyl sulfate Water Methanol Ethanol 2-Propanol Dichloroethyl ether, ethylene glycol monoethyl ether Furfural Nitrobenzene Aqueous solutions: 7 M urea 4.9 M guanidinium chloride 1% potassium oleate Solute: Water P a r a 5 i c , naphthenic, and aromatic lean oil Bovine serum albumin in 0.15 M NaCl Sodium dodecyl sulfate Water 4.9 M guanidinium chloride 7 M urea Water Methanol Methanol, ethanol, 1-propanol, 1-butanol, 1-pentanol, 1- n-Decane n-Hexane, cyclohexane n-Hexane n-Heptane Meth ylc yolohexane Benzene hexanol, 1-heptanol, 1-octanol Pressure, atm 1 1 1 1 1 1 T o 200 1 1 1 1 1 200-760 mm 60-740 mm 80-760 mm 9-210 1 0.1-1 1 1 1 1 0.2-1 1 1 1 1 1 364-760 mm 23-760 mm 355-760 mm 1 1 100-650 mm 1 9-210 1 1 1 1 1 1 1 4.5-520 1 1 To 193 1-34 1 To 193 1-18 1 1 20-40 To 120 1 1 1 40-190 To 193 To 120 Temp, O C 35 35 37 37 20-30 11-76 40-100 10-35 15-35 25 5-45 Room 25-50 25-50 25-50 29 11-76 0-20 5-45 5-45 25 10-35 25 25 15-35 25 5-45 Room 25-50 10-50 25-50 30-70 30-70 30-70 5-45 5-45 25 15-35 29 25 15-35 15-45 Room 15-45 15-45 25 35-100 14-73 30 25-150 35-121 Room - 50 to 20 25-150 -56 to -10 -70 to 20 25-35 40-80 30-150 - 30 -60 to -20 -60 to -50 25-150 25-150 30-150 437 Measure- ment value (see section VI) Ref 2 2 2 1 1 2 0 2 2 1 1 1 2 2 2 1 2 2 1 1 1 2 0 1 2 1 1 1 2 2 2 0 0 0 1 1 1 2 1 1 2 1 1 1 1 2 2 2 2 0 2 1 0 0 2 1 1 2 0 1 1 1 0 0 0 364 364 179 434 94 428 440 659 660 400 640 401 344 343 344 312 428 02 1 640 640 430 659 661 641 660 400 640 401 344 343 344 175 175 175 640 640 402 660 312 661 600 640 401 640 640 42 7 68 428 260 253 140 401 46 564 563 624 66 445 683 362 362 362 252 564 683 438 Gases Propene Propadiene 2-Methylpropene 1-Butene 1,3-Butadiene Pentene Acetylene RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Solvent Toluene n-Xylene Chloroform Carbon tetrachloride Dichloroethane Acetone Methyl ethyl ketone Diethyl ether Methyl acetate Tetrahydrofuran Dimethylformamide Xleth ylpyrrolidone Baku crude oils 4 Russian crude oils 4 crude oils Diethyl sulfate Aqueous solutions: Solute: KCl. AgNOs 1 N NaCI. L i C 1 , KI, LaCla N-Methylpyrrolidone with 0-60 mole H20 0.5 mole fraction of toluene-heptane, toluene-chloroform, toluene-methyloyclohexane, n-heptane-carbon tetrachlo- ride, acetone-chloroform, acetone-n-hexane Water Methanol Dimethylf ormamide Polyethylene, hydropol, and natural rubber Aqueous solutions: Solute: 0-15% potassium oleate 2 1 detergent solutions 0.5-5% KzCOs Emulsifiers N-hlethylpyrrolidone with 0-60 mole % H20 Polyethylene, hydropol, and natural rubber Water Dichloroethyl ether, ethylene glycol monoethyl ether, fur- fural, nitrobenzene Dinonyl phthalate Triisobutylaluminum 5% aq NaCl Emulsifiers, aq soln Water Water 1% potassium oleate, aq 1-4% Hyamine 1622, aq Paraffin oil Water Methanol Pressure, atm 1-18 1 1 1 1 1 1-18 1 1 3-18 1 1 To 193 1 1 1 50-300 1-300 0-300 50-1300 mm 1 1 0-800 mm 1 400 mm 500-700 mm 1-34 1 1 1 400 mm 500-700 mm 500-700 mm 0-800 mm 1 1 1 73-740 mm 1 1 2-68 1 100-500 mm 1 To 19 1 100-500 mm 1 1 5-39 To 29 1 To 39 1-29 1 To 29 1 3-14 To 39 1 Temp, OC -45 to -25 -60 to -30 -70 to 20 - 30 - 20 - 70 to 20 -45 to -25 - 30 - 70 to 20 -25 to -45 -70 to 20 -70 to 20 25-150 0 to -45 - 70 to 20 - 70 to 20 20-100 20-100 20 0-80 30 13-72 0-45 -60 to -20 25 25 21-104 Room -47 to 20 0 to -45 25 25 25 25 0-45 25 0-70 Room 30-70 0-100 10-50 -5 to 0 38-143 Room 25 25 38-104 Room 25 25 20 1-30 0-30 0-70 20-45 0-30 1-30 0-30 25-60 -76 to -25 -20 to 20 20-45 "Low temp" bleasure- ment value (see section VI) Ref 2 1 1 1 1 1 2 1 1 2 1 1 0 2 1 1 0 0 0 2 2 2 1 1 0 0 2 1 1 2 1 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 2 1 1 1 2 1 0 0 1 0 0 0 2 1 0 0 2 1 0 0 253 362 624 362 362 624 253 362 624 564 624 624 253 562 624 624 682 518 519 608 260 428 565 3 4 3 2 402 403 21 401 622 562 414 4 02 403 403 674 565 414 303 401 175 262 477 303 674 74 401 402 507 493 401 402 507 3i7 387 251 315 168 253 313 59 1 315 533 423 513 253 623 SOLUBILITY OF GASES IN LIQUIDS TABLE 1 ‘ (Continued) 439 Gaaes Solvent Ethanol Methanol, 1-butanol, allyl alcohol, ethylene glycol Diacetone alcohol Acetic acid Methyl acetate Ethyl acetate Ethyl acetoacetate Trimethyl orthoacetate Vinyl acetate Trimethyl orthoformate Diethyl oxalate Triethyl phosphate Ethyl acetate, ethyl formate, methyl acetate, methyl for- mate, isoamyl acetate, isoamyl formate Tetramethyldiamidophosphonyl fluoride, methyl phosphite, methyl orthoacetate, trimethyl meroaptophosphate, methyl borate, methylene diacetate, ethyl perfluorobu- tyrate, ethyl orthoformate n-Hexane n-Hexane, n-octane Cyclohexane Dicyclopentadiene Benzene Toluene m-Xylene g-Xylene Chloroform Dichloroethane Ethyl bromide, ethyl iodide Acetone Butyrolactone Tetrahydrofurfuryl methyl ether, ethylene glycol dimethyl ether, diethylene glycol dimethyl ether, triethylene glycol dimethyl ether, tetraethylene glycol dimethyl ether, diethylene glycol dibutyl ether, diethylene glycol di-t- butyl ether, di-n-hexyl ether Diethyl ether Ethylene glycol monomethyl ether Glyoxol tetramethyl acetal, diethyl oxalate diethyl acetal Acetaldehyde Acetal, methylal., acetaldehyde Dimethylacetamide Dimethylformamide 1-Methylpyrrolidone N-Meth yl-2-pyrrolidinone N-Methylpyrrolidone 1.5-Dimethylpyrrolidone Methylpyrrolidone Pressure, atm 1 1 1 To 11 1 1 To 11 1 To 25 1 1 1 602 mm 1 1 1 3-14 1 721 mm 5-29 To 29 To 39 1 1 3-14 1 1 3-14 1 1 3-14 1 770 mm 1 256-648 mm 1 1 1.4-30 1 1 1 1 1-12 To 25 3.5-14 1 1 1 1 1 1 To 16 1 1 To 25 200-900 mm 3.5-14 0.13-1 1 750 mm 1 To 11 3,5-14 1 Temp, O C -75 to -25 18 25 25 18 “Low temp” 25 - 75 to 45 25-60 25 0 0-40 0 25 25-60 - 10 25 - 25 -20 to 20 3 25 10-40 0-30 20-45 20-60 4 -20 to 20 20-90 -30 to 25 -20 to 20 “Low temp” 20-120 -20 to 20 - 30 25 “Low temp” 19-25 -70 to 40 20-39 3-40 -75 to 45 25-60 -80 to -40 “Low temp’ - 25 0-15 25 25-60 - 20 to 20 -75 to -25 ”Low temp” -75 to 45 25 -75 to 0 - 10 25 25-60 “Low temp” 20-39 15-40 5-40 -20 to 30 -50 to 25 25 20 -35 to 20 15-40 -20 to 30 “Low temp” Measure- ment value (see section VI) 2 0 0 0 0 0 0 1 0 0 2 2 2 0 0 0 1 Ref 423 109 259A 259A 109 623 2598 55 533 259A 632 239 632 2598 533 109 410 1 362 1 513 0 109 0 259A 2 314 0 315 0 253 2 628 0 109 1 513 2 628 1 362 1 513 0 623 2 628 1 513 1 362 0 259A 0 623 2 l98B 0 69 0 387 2 259 1 55 0 533 1 616 0 623 1 362 0 109 0 2598 0 533 1 515 2 423 0 623 1 55 1 410 2 57 0 109 0 259A 0 533 0 623 0 387 0 259A 1 237 1 514 2 562 1 362 1 375 1 155 0 2598 1 514 0 623 440 Gases Propyne 1-Butyne 3-Buten-1-yne RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Solvent Tetramethylurea 2-Methyl-8-ethylpyridine Hexamethylphosphoric triamide Hexamethylphosphorous triamide Aniline, dimethylaniline, nitrobenzene Acetylpyrrolidine, formylpyrrolidine, N-nitrosopyrrolidine, hexamethylphosphoramide, tritetramethylenephosphor- amide Dimethyl sulfoxide Dimethyl sulfoxide, tetramethylene sulfoxide "Solvenon" Dioxane Methylnaphthodioxane, ethylene oxide, 2-methyl-2-me- Tetrahydrofuran thoxy-1,3-dioxane, dioxane (satd with trioxane) Mesityl oxide 1,1,3,3-Tetramethoxypropane Morpholine, N-formylmorpholine 21 solvents Blood, human and animal Lung tissue (blood-free homogenates) Hydrated Fe and AI oxides, soln of egg albumin, gelatin, Liquid NHa Aqueous solutions: Solute: serum, and serum albumin HzSOd, LizS01, LiCl. NaNOs, NaC1, KzSO4, KNOa, KCI, KBr, NH4C1, ZnClz, CdClz, CdBrz, CdIz, MgSO4, ZnSO4, CdSOd, MnSO4, FeSO4 -I- HnSok, NiSO4, Aln(SOd)s, Crz(SO4)s, and Fez(S03a -t H2SO4 Dimethylformamide-water (0-100%), acetonewater (0- Dimethylformamide (with 0-23% HzO) Methanol (with 0.05% Hz0) N-Methylpyrrolidone with 0-60 mole yo Hz0 Potassium methylate (25%) in methanol Dimethylformsmide with 5.4 g NaI/100 g Acetone with 13-30 g of NaI/100 g Dimethylformamide-dioxane (0-100%) Acetone-n-hexane (0.5 mole fraction) Acetone-chloroform, toluene-chloroform (0.5 mole fraction Toluene-dimethylformmide (0.5 mole fraction) Mixtures of CHCL, Hz0, MeOH, or diethylene glycol di- methyl ether with dimethylformamide, dimethyl sulfoxide, or ethylene glycol dimethyl ether loo%), dioxane-water (&loo%) Nonaqueous solutions: mixtures) Water Methanol Aqueous solutions: Solute : 3 w t 70 "&OH Methanol (with 0.05% HnO) Dimethylformamide (with 0-23'% Hz0) Water Methanol 3 w t ' % NaOH, 0.3-10 wt % NHdOH, 20-25 wt ' ?
& NaCl Water Methanol Benzene Toluene p X ylene 2-Chloro-l,3-butadiene Pressure, atm 530-720 mm 3.5-14 1 1 1 1 3.5-14 1 550 mm 1 609 mm 1 583 mm To 29 1-25 1 To 39 To 11 3.5-14 3,5-14 3.5-14 1 1 1 1 1 1 100-400 mm To 1 0-800 mm 1 1 1 1 1 1 1 3-13 To 14 1 1 1 1 1 100-400 mm 1 1 1 1 1 1 1 1 1 1 Temp O C 15-40 25-60 -20 to 30 30-37 25-60 4 25 15 -20 to 30 25 25 20 25 25 25 0-30 0-30 - 75 to 45 20-45 25 -20 to 20 -20 to 30 -20 to 20 37 37 10-40 -50 to -74 0-70 20 25-140 10-78 0-45 23-50 20-39 20-39 20 - 25 - 30 25 0-10 21-104 0-60 Room 0-60 0-60 -70 to 20 25-140 0-60 Room 0-60 0-60 0-60 0-60 2 0 4 0 20-90 20-120 10-50 Measure- ment value (see section VI) Ref 0 0 1 2 0 0 1 1 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 1 2 1 0 1 0 0 1 1 1 1 0 0 0 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 1 259A 533 514 603 533 109 410 575 574 514 410 2598 387 259A 410 259A 315 313 55 253 2598 515 514 516 486 89 567 56 188 387 72 71 565 59 1 387 387 387 362 362 362 517 270 57 1 401 57 1 57 1 71 72 571 40 1 57 1 571 57 1 571 629 629 629 277 SOLUBILITY OF GASES IN LIQUIDS TABLE V (Continued) 441 Gases Diacetylene Carbon tetrafluoride (Freon-14) Chlorotrifluoromethane (Freon-13) Dichlorodifluoromethane (Freon-12) Trichlorofluoromethane (Freon-1 1) Chloroform Dichlorofluoromethane (Freon-21) Chlorodifluoromethane (Freon-22) Trifluoromethane (Freon-23) Dichloromethsne Solvent Aqueous solutions: Dimethylformamide (0-23% in HzO) Methanol (with 0.05% Ha0) 3 wt 70 NHiOH Methanol (with 0.05% Hz0) Dimethylformamide (with 0-23% HzO) Water Isooctane, n-heptane, cyclohexane, benzene Perfluoromethylcyclohexane CCLFCCIFa Water Water Carbitol acetate, carbitol methoxyacetate, 4-methyl-2-pen- tanol acetate Trichlorobenzene Carbitol ethyl ether, dichloroisopropyl ether, y, y’-dichloro- 2,3-Di-~’-ethoxy-~-ethoxydioxane a-Fluoronaphthalene 30 solvents (alcohols, esters, acids, ethers, aldehydes, ke- tones, nitrogen compounds, 2 hydrocarbons, and 2 bro- mides) Dimethyl ether of tetraethylene glycol, diethyl ether o f di- ethylene glycol, ethyl ether o f diethylene glycol acetate n-propyl ether, diethyl ether of ethylene glycol Diethyl ether of tetraethylene glycol Carbitol acetate Dimethyl ether of tetraethylene glycol, diethyl ether of di- ethylene glycol, ethyl ether of diethylene glycol acetate 29 solvents (alcohols, ethers, esters, aldehydes, ketones, ni- trogen compounds, 2 hydrocarbons, and 2 bromides) Ethanol Dimethyl ether of tetraethylene glycol, diethyl ether of di- ethylene glycol, ethyl ether of diethylene glycol acetate Acetone Tetraethylene glycol dimethyl ether Olive oil Dimethyl ether o f tetraethylene glycol, diethyl ether of di- ethylene glycol, ethyl ether of diethylene glycol acetate Diethyl adipate, ethyl laurate, diethyl oxalate Tetraethylene glycol dimethyl ether Triethylene glycol dimethyl ether N,N-Dimethylformamide N,N-Dimethylacetamide Caprylonitrile, succinonitrile, glutaronitrile, adiponitrile, 44 solvents (alcohols, ethers, acids, esters, amines, amides, 78 solvents (esters, ethers. glycols, and others) 82 solvents (esters, glycols, ethers, amines, amides, and sebaconitrile aldehydes, ketones, and oximes) others) Water Dimethyl ether of tetraethylene glycol, diethyl ether of di- ethylene glycol, ethyl ether of diethylene glycol acetate Tetraethylene glycol dimethyl ether N,N-Dimethylformamide Water Diethyl ether of tetraethylene glycol Tetraethylene glycol of dimethyl ether Carbitol acetate, cellosolve acetate Ethyl ether o f diethylene glycol acetate, diethyl ether of di- ethylene glycol, dimethyl ether of tetraethylene glycol Caprylonitrile, benzonitrile, succinonitrile. glutaronitrile, adiponitrile, sebaconitrile Pressure, atm 100-400 mm To 1 1 To 1 100-400 mm To 8 1 1 1 1 To 25 To 8 3.5 3 . 5 3.5 3.5 3.5 3.5 3 . 5 364 mm 364 mm 0.48 0.48 1.0-1.6 mm 0.11 0.7-1.2 mm 1 20-70 mm 0.85 To 1 . 5 0.84 1 To 1 To 7 To 1 To 19 0.85 0.6-1 638 mm 0.85 To 12 5.0 5.5 To 21 To 2 To 2 To 24 181 mm 0.23 181 mm 0.23 0.23 Temp, O C 25-140 20 to -70 0-60 20 to -70 25-140 25-75 7-39 5-36 - 11 to 40 5-35 10-59 25-75 25-75 32 32 32 32 32 32 32 32 32 32 32 25-35 32 25-35 3 20 32 38-107 3 27-90 20-101 38-107 20-101 32 32 32 32 38-107 25-75 10-79 32 3 28-177 38-107 38-107 25-75 32 32 32 32 32 Measure- ment value (see section VI) Ref 1 1 1 1 1 1 2 2 2 2 2 1 1 0 0 0 0 0 2 2 0 0 2 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 0 0 0 1 2 2 2 2 2 2 0 0 2 0 2 2 72 71 57 1 71 72 469 429 15 495 15 62 469 469 677 677 677 677 677 108 678 677 677 078 108 27 078 27 395 383 678 7 395 399 7 7 7 7 110 106 677 678 469 62 078 396 399 7 7 409 077. 395 077 078 110 442 Gaaes Chlorofluoromethane Bromomethane Chloromethane Fluoromethane Iodomethane Chloroethane 1,1,2-Trichloro-1,2,2-tri- fluoroethane (Freon-1 13) 1,2-Dichloro-1,1,2,2-tetra- tluoroethane (Freon-114) l-Chloro-1,1,2,2-tetra- fluoroethane l-Chloro-1,1,2,2,2-penta- fluorethane (Freon-115) 1,1,2,2-Tetrafluoroethane Vinyl chloride Diohlorodifluoroethylene Chlorotrifluoroeth ylene l,2-Epoxyethane (ethylene oxide) RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Solvent 35 solvents (alcohols, ethers, acids, esters, amines, amides, oximes. aldehydes, and ketones) Water Tetraethylene glycol dimethyl ether Dimethyl ether of tetraethylene glycol, diethyl ether of di- ethylene glycol, ethyl ether of diethylene glycol acetate Water DnO Pineapple, mango, and papaya juice Water De0 Ethanol Benzene Carbon tetrachloride Dichloroethane Ethyl ether of diethylene glycol acetate, dimethyl ether of tetraethylene glycol, diethyl ether of diethylene glycol Acetic acid (glacial) Polyethylene 63 solvents (esters, glycols. ethers, substituted benzenes and naphthalenes, and others) Water D P O Water DIO Carbon tetrachloride, dichloromethane Dimethyl ether o f tetraethylene glycol, diethyl ether of di- ethylene glycol, ethyl ether of diethylene glycol acetate 1,1,2,2-Tetrachloroethane, dimethyl ether and diethyl ether of tetraethylene glycol, dimethyl ether of triethylene gly- col, carbitol ethyl ether, carbitol acetate, butyl carbitol acetate diethylene glycol diacetate Dimethyl ether of tetraethylene glycol, diethyl ether of di- ethylene glycol, ethyl ether of diethylene glycol acetate Carbitol acetate Dimethyl ether of tetraethylene glycol, diethyl ether of di- ethylene glycol, ethyl ether of diethylene glycol acetate Tetraethylene glycol dimethyl ether Water Tetraethylene glycol dimethyl ether Methanol Methylcyclohexane Tetrachloroethylene 1,l-Dichloroethane Tetrahydrofuran Dimeth ylformamide Emulsifiers, aq solutions of “26 different solvents” Dimethyl ether of tetraethylene glycol Emulsifiers, aq solutions of Water Dichloroethane 1 Pressure, atm 0.23 1 . 7 1.5-9 1.7 To 950 mm To 1 1 To 1 1 500 mm To 950 mm To 1 To 1 500 mm 500 mm 100-760 mm 500 mm 100-760 mm 2.9 500 mm 1 3 To 950 mm To 1 To 1 To 950 mm To 1 To 1 100-760 mm 0.74 557 mm 0.18 786 mm 1 To 8 To 8 To 5 1 1 1 1 1 1 786 mm 1 20-700 mm 1 0-16.4 Temp, O C 32 10-79 3 35-177 32 0-77 29-40 10-32 29-40 18-32 20 0-7 7 10-59 29-40 29-40 20 20 -10 to 20 20 - 10 to 20 32 20 25 32 0-77 29-40 29-40 0-7 7 29-40 29-40 - 10 to 20 32 32 32 32 32 35-177 25-75 27-86 -10 to 2 0 20-50 -7 to 20 20-50 20-50 20-50 32 5-20 5-20 12-50 100-150 5-20 Measure- ment value (see section VI) Ref 0 2 2 2 2 0 2 2 2 1 2 0 2 2 2 2 2 0 2 0 2 2 1 0 2 2 2 2 2 2 0 2 0 2 0 2 2 1 2 1 2 1 2 2 2 0 0 0 0 0 1 2 0 0 107 62 395 399 678 224 592 198 592 198 390 224 62 592 592 390 390 288 390 288 678 390 414 677 224 592 692 224 592 592 288 678 677 678 677 678 399 469 399 470 382 470 382 382 382 674 53 1 677 674 289 406 100 436 289 Gases Ketene Diethyl ether Carbon monoxide Carbon dioxide SOLUBILITY OF GASES IK LIQUIDS TABLE T' (Continued) Solvent Acetone Lung tissue (blood-free homogenates) Methyl, ethyl, ti-propyl, isopropyl, n-butyl, and isobutyl al- .Ilcohol n-Propyl acetate n-Heptane Hydrocarbon (slack wax, av mol wt 345) Paraffin oil Various hydrocarbons Benzene Toluene 1,2-Dibromoethane n-Perfluoroheptane Aniline, propionitrile, benzyl cyanide, pyridine, nitrobenzene Dimethylformamide Polyethylene, hydropol, natural rubber Carbon disulfide .Smmoniacal cuprous carbonate and cuprous formate solu- tions Hydrated Fe and A 1 oxides; soln of egg albumin, gelatin, serum, and serum albumin cohols IYater \Vater (nomograph) Water (hydrostatic pressure) Ice Methanol Ethanol Ethanol, 95% Alcohol Propanol, 2-propanol, butanol, 2-butanol, 2-methyl-l- propanol 50: 8 0 decanol-dodecanol Cyclotiexanol Oleic acid Ethyl acetate n-Propyl acetate Methyl acetate Ethyl stearate n-Pentane n-Heptane, cyclohexane Oils and kerosenes Paraffin oil Paraffin \-ax Baku crude oils Crude oils, kerosene, gasoline Russian crude oils Pressure, atm 1 1 1 1 1 1-10 (Kp/om) 1 1 1 1 1 1 600-900 mm 1 1 To 4 1 25-500 16-49 25-700 25-700 1 1 1-700 1-20 1 100-700 mm 1 1 16-59 1 5-162 1-102 7-io 3-19 100-760 mm High pressure 1 100-760 mm 1 1 100-760 mm 100 mm 1 100-450 mm 3-15 1 1 1 100-600 mm 1.7-75 1 760 mm 100-760 mm 200-750 mm 258-753 mm 50-300 1 1-300 Temp, "C 37 25-50 20 25 25 106-300 20 25 25 25 25 25 5-40 25 25 10-40 12-40 20-50 0-100 75-120 13-75 25 0-120 10-30 20-75 T o 350 1 1.5-77 37.5 20-30 177-334 20-38 114-348 25-45 25 -20 to - 5 0-75 -29 to -50 25-60 -60 to -26 - 70 to 20 Low temp 2-62 20 Low temp 20-82 20-48 23-62 25-60 -45 t o -25 - 75 to 45 25 25 34-54 38 25 0-20 20-82 20 72 20-100 20-60 20-100 443 Measure- ment value (see section VI) Ref 0 1 2 2 2 2 2 2 0 2 2 2 2 2 1 1 2 0 0 1 1 2 2 2 2 0 0 2 1 0 0 0 2 1 2 2 2 1 2 1 0 1 0 0 1 1 1 0 0 2 0 1 2 1 0 2 1 2 2 1 1 2 2 2 1 1 0 0 0 146 89 215 378 221 216 473 378 9 216 221 221 216 221 237 414 216 24 667 649 673 648 487 428 217 168 147 31 555 214 191 164 450 570 114 165 18 164 668 597 169 172 339 46 633 587 658 625 557 503 378 557 378 3 5 466 533 559 55 221 221 466 476 217 23 378 377 601 682 604 518 444 Gases RUBIN BATTINO AND H. IAWRENCE CLEVER TABLE V (Continued) Solvent Diesel fuels and cracked gas oils Baku and American crude oils Slack waxes Mineral oil Benzene Toluene n-Perfluoroheptane (C4Fp)iN Hexachlorobutadiene Carbon disulfide Methylene chloride Chloroform Carbon tetrachloride 1,2-Dibromoethane Ethylene bromide, propylene chlorobromide Chlorobenzene Acetone Methyl ethyl ketone Butyrokctone Diethyl ether Ethylene glycol monomethyl ether Acetaldehyde Benzylcyanide, propionitrile Aniline Pyridine Dimethylformamide Dimethylacetamide Hexamethylphosphorous triamide Hexamethylphosphoric triamide Methylpyrrolidone Nitrobenzene Tetrahydrofuran Triethyl phosphate Tetramethylurea Polyethylene tetraphthalate Polyethylene, hydropol, and natural rubber Carbon disulfide Butteroil, cottonseed oil, lard Dog fat, human fat, rat-pooled fat Olive oil Whole blood Cerebrospinal fluid Homogenized brain tissue (of cats) Human serum Blood .4mmonia, liquid Chlorine, liquid HCN, liquid Aqueous solutions: 10, 50, 95% alcohol Wine 4% ethanol, beer Aqueous sucrose and citric acid soh Papaw juice Hydrated Fe and A 1 oxides, aqueous egg albumin, gelatin, Aerosol solution (15%) Dimethylformamide (0-23% water) N-Methylpyrrolidone (0-60 mole % water) Monoethanolamine (15%) Mono- and triethanolamine (0.5-5 N ) Mono-, di-, and triethanolamine (3.6-3.9 N) Mono-, di-, and triethanolamine 2.5, 5.0 M H F 0.01 N HC1 AstO-As20s-HBr soh 0.1 N lactic acid serum 9-84 wt 7 0 HzSOd Pressure, atm 3-40 1-10 (Kp/cm') 1 10-95 1 1 1 3-15 1 1 1 1 1 1-20 3-20 1 1 1 1 1 1 1 3-15 740 mm 1 1 1 1 1 1 1 200-900 mm 1-12 1 1 1 1 1 1 1 1 1 1 1 0.8-6.5 1 1 1 100-400 mm 0-800 mm 0.5-6980 mm To 40 7-445 mm 10-760 mm 1 1 1 1 Temp, O C 25-50 40-80 106-300 20-48 30-60 20-35 25 -80 to 0 -45 to -25 25 19-30 4-3 1 - 11 to 55 21 38-54 25 25 25 20-48 25 25 -75 to 45 25-60 -45 to -25 25-60 24 -75 to 45 -75 to -25 25 25 25 -45 to 0 -70 to 20 5 to 40 25-60 25-60 37 -70 to 20 25 -75 to 45 25-60 25-60 25-130 25 25 40-60 25-45 25-45 Room 37.5 37.5 15-38 37.5 -74 to -50 0-2 5 -5 to 10 2-62 a 3 3 25 10-40 25 25-140 0-45 40-140 25-75 50 50 20-30 37.5 20-25 19-76 37.5 Measure- ment value (see section VI) Ref 1 0 2 2 1 2 2 1 2 2 2 2 1 2 2 1 2 2 2 1 2 2 1 0 2 0 2 1 2 2 2 2 2 1 1 0 0 1 1 2 1 0 0 1 1 2 1 2 2 2 1 1 2 0 1 1 0 0 0 0 0 0 0 2 1 1 2 0 0 0 2 1 0 1 1 368 606 473 35 634 217 221 651 559 217 322 322 329 322 79 83 1 221 217 221 35 22 1 221 55 533 559 533 412 55 57 221 221 221 662 625 237 533 533 603 625 221 55 533 533 415 414 217 532 667 667 656 570 570 18 122 56 345 488 503 2 168 379 363 567 217 72 565 283 381 569 568 114 570 49 555 570 SOLUBILITY OF GASES IN LIQUIDS TABLE v (Continued) 445 Gases Carbon oxysulfide, COS Carbon disulfide, CSZ Chlorine monoxide, ClzO Chlorine dioxide, ClOz Boron tri5uoride, BFa Dihorane, BzHa Hydrides: SiH4, GeHd, SnHd, SbHa, H2Se Water vapor Hydrogentsulfide Solvent 0.066 M phosphate buffer 0.5-2.0 M NaCl 0.9% NaCl soln 0.16 M NaCl soln NazSO4-HzSO4 soh KzSO4, KzCOa, KzCOrKHCOa, KOH 10, 20, 30% CaClz soh KC1, NaC1, KNOs, Mg(NOa)r, NaNOz, MgSO4, NazS04 NaOH-NazCOa Glycerol in water, glycerol and sulfuric acid, ethylene gly- col, B,B-dihydroxyethyl ether, tetrahydrofurfuryl alcc- hol, dioxane Binary mixtures: MeOH-acetone, MeOH-CHCls, MeOH- CsHs, acetone-CHCla, acetone-CsHs, CHClrCsHs Methanol Ethanol Paraffin oil Green, spindle, and solar o i l Titanium tetrachloride Water, saline, urine, whole blood, blood plasma, red cells, Methanol Ethanol, acetone Green, spindle, and solar o i l Water Water (nomograph) Aqueous HOC1 Water Carbon tetrachloride 10-76.5% acetic acid soh 9.7-79.170 H2SO4 n-Pentane Benzene, toluene Anisole, phenetole, 2,2’-dichlorodiethyl ether Nitrobenzene Sulfuric acid HF, liquid n-Pentane Diethyl ether Tetrahydrofuran Dimethyl ether of diethylene glycol, di-t-butyl sulfide 2-5.4% NaBH4 in dimethyl ether of diethylene glycol Ethyl Cellosolve, 2,2’-dichlorodiethyl ether, nitrobenzene, bile PFM-4F, paraffin oil PFM-4F, para5n oil Ethyl Cellosolve, 2,2’-dichlorodiethyl ether, nitrobenzene, Hydrocarbon slack wax (av mol wt 345) Triethylenetetramine, tetraethylenepentamine, methylated triethylenetetramine. triacetyltrimethyltriethylenetetram- ine, hexamethylenediamine, N,N-dimethylacetamide, ethylene glycol, glycerol, carbitol, dimethyl ether of tetra- ethylene glycol, triethyl phosphate Water Methanol n-Pentane n-Decane Cyclohexane Spindle, green, and solar oil Benzene Ethyl Cellosolve, chlorex, nitrobenzene, PFMC-4F, paraffin oil Sulfur 15.3% Monoethanolamine in HnO Aq ammonia Aq phenol Aq HC1, ZnClz, and FeClx soln Pressure, atm 1 45-88 1 1 1 25-700 1 0.05-1 , 0 1 760 mm Satn point 1 1 1 5-90 ppm in air Satn point 4.2-8.7 mm 1-88 mm 1-11 10-150 mm 30-125 mm 58-158 mm 40-170 mm 3-7 1 1 1 0-68 0.5-5 1.5-7.7 0-4 5-315 mm 4-300 mm 0.1-35 mm 0.1-35 mm 1-10 (Kp/cm) H20 equil point at 4.5O 0-200 mm To 0.085 Low pressure 15-480 mm 6.8-88 100-800 mm To 48 0.1-35 mm 1 3.0-5.0 1-840 mm 6.8-81 Temp, O C 25-45 172-330 25-45 37.5 25-65 75-120 0-40 18 25 20 -78.5 to 0 20 20 30-100 0-100 37 -78.5 to 0 25-35 30 to 100 3.5 to 20 3.5 0-40 10-45 20 20 20 49-93 23-47 5-170 5-25 5-170 24-98 0-55 0-60 7-50 0-30 30 5-50 5-50 140-300 32.2 0-60 20 -30 to 30 -78.5 to 0 -78.5 to 0 5-170 5-170 10-40 30-100 25-175 5-50 126-444 177-260 40-140 20 20 Measure- ment value (see section VI) Ref 2 2 2 1 0 0 2 2 2 2 2 1 2 2 0 1 1 1 2 0 2 1 1 2 2 2 2 1 2 1 0 1 2 1 1 1 1 1 1 1 2 0 668 165 668 570 556 214 487 391 618 325 333 681 378 378 19 675 409 681 27 19 550 132 508 274 305 305 305 305 88 643 389 78 389 418 405 163 163 671 671 145 146 473 111 1 58 0 226 0 46 1 681 1 680 1 491 1 492 1 615 0 19 0 367 1 145 1 180 1 510 2 283 0 226 0 226 0 290 446 RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE v (Continued) Gases Solvent Hydrogen chloride 24 alcohols 43 solvents (15 alcohols, 8 acids, 20 esters) 25 solvents (19 ethers, 6 glycols) 17 solvents (12 esters, 4 ethers, 1 alcohol) Tetramethox ysilane Tetraethoxysilane Tetrapropoxysilane, ethanol, 2-chloroethanol, ethyl carbon- 19 solvents (14 alcohols, phenol, 4 esters) 35 solvents (15 esters, 8 halides, 7 alcohols, 4 silanes, n- decane) Tetrahydrofuran, tetrahydropyran. diethyl ether, PrrO, BuzO, diethyl sulfide, ethyl nitrate, n-heptane, CClr Dioxane Dibenzyl ether, &@'-dichlorodiethyl ether, o-nitrotoluene, di- Nitrobenzene Anisole n-Butyl, n-pentyl, n-hexyl, n-heptyl, n-octyl, n-nonyl, n- decyl alcohols Ethylene glycol HO(CHz)aOH, HO(CHZ)IOH, HO(CHZ)KOH n-Butane 2,4,4-Trimethyl-l-pentene, 2,4,4-triethyl-Z-pentene n-Heptane n-Hexane n-Decane Cyclohexane Benzene ate, ethyl chloroformate phenyl ether Toluene About 0.05 mole fraction solutions in n-heptane of benzene, toluene, m-xylene, mesitylene, fluorobenzene, ohloroben- zene, hromohemene, iodobenzene, t-butylbenzene, iso- propylbenzene, ethylbenzene, trifluoromethylbenzene About 0.1 mole fraction solutions in toluene of benzene, m- xylene, p-xylene, mesitylene, hemimellitene, pseudocum- ene, chlorobenzene About 0.1 mole fraction solutions in toluene of n-heptane. 1- octene, 2,4.4-trimethyl-l-pentene, 2,4,4-trimethyl-Z-pen- tene, cyclohexene, o-xylene, 1,3,3,4-tetramethylbenzene, 1,2,3,5-tetramethylbenzene, tetrachloroethylene, thiophene Fluorobenzene, chlorobenzene, bromobenzene, iodobenzene Chlorobenzene CHCls, CClr CHCls, CClr, CzH4C1zr CzHzC14 &,%Dichloroethyl ether, anisole Phenetole, n-butyl phenyl ether, diphenyl ether Diamyl ether, diisopropyl ether, dibutyl ether, CzHsOCHz- (ClCH?CH?hO. (CaHs)%O, diisoamyl ether, CHaOCsHs, CsHs, CHaOCHiCsHs, CIHQOCHZCEHK, (ClCHz)zO, (CsHsCH320, (ClCHzCHzCH?)zO, CzHsOCeH, Nitrobenzene m-Nitrotoluene a-Nitrotoluene Thiophene, tetrahydrothiophene, phenyl sulfide, diphenyl sulfide, n-butyl sulfide, di-n-butyl sulfide, isopropyl sul- fide, diisopropyl sulfide AlCla in toluene Hydrogen bromide 2-Chloroethanol 2,2,2-Trichloroethanol n-Pentyl borate Ethanoi n-Butane. n-hexane, AlBra solutions in n-butane and n-hex- ane n-Decane n-Hexane, n-octane, n-decane Benzene. toluene Pressure, atm 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 To 0.424 1 1 4.4-36 4-10 mm 29-71 mm 16-82 1 20-650 mm 100-800 mm 1.5-570 mm 0.2-0.6 6-463 mm l o 174 mm 3.8-6.3 mm 25-215 mm 15-70 mm 2-6 mm 2-6 mm 30-290 mm 23-250 mm 1-60 120-700 mm 1 10-500 mm To 500 mm 1 4-400 mm 14-167 rnm To 500 mm 32-340 mm 1 0.4-0.7 0.5-230 mm 1 1 1 0-2 1 50-700 mm 80-430 mm 25 25 10 10 20-82 -78.5 -78.5 25 0 20-40 10-40 30 25 30-40 -84 to -78.5 25 -78.5 -78.5 -78.5 25 30-40 20-250 Ck25 20-40 20-40 10-30 10 20-40 25 25 25-35 25 0 - 46 Temp, O C 0-18 2-67 0-60 0-5 1 0 1 0 0-63 -78 to 51 -70 to 0 -43 to 0 -15 to 0 0 -61 to 0 -84 to - 4 5 4-39 0-26 0-34 25 5-25 0 25-45 25 Measure- ment value (see section VI) Ref 0 2 2 2 2 2 2 2 2 1 1 1 1 1 2 I 2 2 1 1 1 1 2 2 1 1 1 1 1 2 1 1 1 1 1 1 1 2 0 2 1 2 1 1 1 1 1 2 2 2 2 2 1 1 2 1 1 202 203 212 213 204 204 204 207 208 587 587 587 587 587 251 453 210 210 465 77 77 454 196 650 615 529 453 456 76 75-77 455 77 7.577 77 456 456 586 263 684 457 458 209 456 453 454 458 454 196 76 113 213 213 151 192 196 59 485 SOLUBILITY OF GASES IN LIQUIDS TABLE V (Continued) 447 Gases Hydrogen iodide Ammonia Phosphine Arsine Hydrasoic acid, HN: Nitric oxide, NO Nitrous oxide, NIO Solvent CHCla. CCI4 0- and m-nitrotoluene Thiophene, tetrahydrothiophene, phenyl sulfide, n-butyl sul- fide, di-n-butyl sulfide, isopropyl sulfide, diisopropyl sul- fide, diphenyl sulfide Thiophene, tetrahydrothiophene, isopropyl sulfide, diiso- propyl sulfide, n-butyl sulfide, di-n-butyl sulfide, phenyl sulfide, diphenyl sulfide, n-decane Water (nomograph) Methanol, ethanol Ethanol n-Propyl and isopropyl alcohols Human blood plasma Water Water solutions of NaCI, NaOH, and HiSol Xylene isomers Phenyl chloride Benzene, toluene, CdCla. CzHiClr, kerosene Water Blood, blood serums Aqueous solutions of NaC1, NazS04, NaOH, HzSO4, glycol Monoethanolamine, triethanolamine HCN, methyl nitrile, ethyl nitrile, n-propyl nitrile Tetralin, acetone, CHiCli, CHCla, CCI4, CeHCla, CzH&lz, CH?Clr Water Methyl alcohol Cyclohexane Aqueous solutions of CuSO4, CuCln, MnSO4, Hap&, cOs04, NiSO4, Cun(NHd,Clz, NazSO4 + NaOH, FeSO4, FeCli Aqueous solutions of FeSO4, FeClr Aqueous solutions of FeSO4 Nitrose liquid M’ater %-Heptane, 2,2,4-trimethylpentane, benzene, CCh, CSe Blood Aqueous solutions of bovine serum albumin, bovine hemo- Aqueous solutions of egg albumin, gelatin, serum, serum al- Various homogenized tissues Aqueous solutions of KCl, KNOa, NaCl, NazSO4, Mg(NOs)r, globin, bovine -,-globulin, bovine j3-globulin bumin, hydrated Fe and A 1 oxides MgS04 Xitrogen dioxide, N o t Water Nitroglycerin Nitrosyl chloride, NOCl Cyclohexane Sulfur hexafluoride, SFs Water Sulfur dioxide, SO, Isooctane n-Heptane, cyclohexane, methylcyclohexane, benzene, tolu- +Heptane, isooctane, benzene, CClnFCClzF, CaH&zFs, n-F’erfluoroheptane Carbon disulfide Nitromethane Polyethylene, hydropol, and natural rubber Homogenized lung tissue Water saturated with nitromethane Aqueous solutions of NaCl, LiCl. HC1, KCl, NHICl, BaCla, ene, CCIiFCClFz, SiClr, CClr CCh, cse NazSOi. KI, KBr, KNOa, NMerI, NEtcBr Water Water (nomograph) Pressure, atm 155-665 mm 20-500 mm 1 1 7.5-10.4 mm 1 mm 118-650 mm 325 mm 1 1 1 200 1 200 mm 200 mm 1 1 1 2-229 mm 1 100-800 mm 1 1 100-900 mm 100-800 mm 1 1 1 1 I 1 1 1 1 1 1 1 50-1445 mm To 0.0038 Temp, O C 0-2 5 25 0 0 0-30 25-35 0-3 5 1-42 27-50 24.5 - 15 to 20 0 20 0-26 7-21 21 20 12 14 25 0-50 0-30 10-40 20 10-90 40-95 36 0-40 36-37 37 37.5 37 10-40 37 37 0-40 17-20 20-80 10-40 11-30 0-25 10-30 8-33 25 4-30 15-3 1 0-2 5 2 6 si 25 25 25-115 10-32 Measure- ment value (see section VI) Ref 2 1 2 2 0 1 0 1 1 1 0 0 0 1 1 1 1 1 1 1 2 2 1 0 0 0 0 2 2 2 2 0 1 0 2 1 2 0 0 1 1 1 2 2 2 2 2 1 1 1 1 1 2 1 263 455 196 196 133,137 240 27 241 275 639 639 311 311 311 286 113 286 286 113 113 113 150 498 615 480 480 199 612 278 669 278 241A 122 434 567 17 89 39 1 64 590 615 430 197 322 15 254 322 322 197 414 89 197 430 44 467 129, 130, 134, 135 448 Gases Osmium tetroxide, Os04 Uranium hexafluoride, UFs Methylsilane Methanol Ethanol n-Propyl alcohol Isopropyl alcohol n-Butyl alcohol see-Butyl alcohol &Butyl alcohol Benzene Phenyl chloride Carbon tetrachloride Acetone Nitromethane Methylamine Diethylamine n-Propylamine Isopropylamine n-Butylamine Isobutylamine sec-Butylamine He + CH4 Ar + N t Ar + Nt Ar + N s Ar + CH4 Ar + CHI Ht + O n HI + Nt RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE V (Continued) Solvent Ethanol 1-Heptanol Methyl acetate Ethyl laurate Benzene Solar, spindle, and green oil Acetone Tetraethylene glycol dimethyl ether, N,N-dimethylaceta- mide, N,N-dimethylformamide, 2-octanone, nitrobenzene Dimethyl sulfoxide Titanium tetrachloride Aqueous solutions: ZnSOi NaHSOa NaHSOs NHa NHs Ca(HS0a)r Mg(HSOa), HIS04 Oleum Nitromethane Heavy oils Methyl trichlorosilane Organic Vapors 12 solvents, iccluding amines, amides, nitriles, and glycols Triethylenetetramine, methylated triethylenetetramine, hexamethylenediamine, N,N-dimethylacetamide, ethylene glycol, triethyl phosphate Triethylenetetramine, hexamethylenediamine Triethylenetetramine, hexamethylenediamine Triethylenetetramine, hexamethylenediamine Triethylenetetramine, hexamethylenediamine Triethylenetetramine, hexamethylenediamine Ethanol, acetone Aq 1 N KCI, KI, NMeJ, NEtrRr Ethanol, acetone Ethanol Water Ink and lube oils Olive oil Water, ethylene glycol, glycerol, diethylene glycol Ethylene glycol, glycerol n-Octyl alcohol, ethylene glycol, 1,3-butylene glycol, glyc- erol, diethylene glycol, triethylene glycol Ethylene glycol, diethylene glycol, triethylene glycol Ethylene glycol, di-, tri-, and tetraethylene glycol, hexa- methylenediamine, triethylenetetramine Ethylene glycol Ethylene glycol Mixtures o f Gases Water Sea water Liq NHa Liq NHa Water Liq NHs Water Liq NH8 Pressure, atm 4.7-6.8 mm 2-20 1.4-19 1.2-2.7 1.4-19 1 1 1 0-800 mm 100-750 mm 70-760 mm 729 mm To 0.0038 1 1 0-0.16 1-25 Vap press at 4,5' Vap press at 4. Ko Vap press a t 4.5O Vap press a t 4.5O Vap press at 4.5" Vap press at 4.5O Vap press at 4.5O <2 mm <1 mm 1.0-1.7 mm 2-240 mm 10-140 mm 3-12 mm Vap press at 4.5O Vap press at 4.5O 106 mm 223 mm 24 mm 45 mm 57 mm 200-600 1 25-100 50-75 291-485 36-82 150-800 Temp, O C 2 5 4 5 25-93 25 25-93 26 30-100 25 25-93 0-100 20-100 20-90 15-30 4.5-25 10-25 35 10-32 20-60 25 93 -20 to 50 32 32 32 32 32 32 32 25-35 25 25-35 25-35 15-45 15-35 20 32 32 32 32 32 32 32 25 0-27 0-2 5 25 149 - 10 to 50 Measure- ment value (see section VI) Ref 1 1 1 1 2 0 1 1 0 1 1 2 0 2 1 0 1 1 1 1 1 1 2 2 2 2 2 2 2 2 1 2 1 1 1 0 2 2 2 2 2 2 2 0 2 0 1 0 0 0 0 27 8 238 8 273 19 238 8 574 676 276 348 193 148 394 644 102 467 419 197 357 552 111 111 111 111 111 1 1 1 1 1 1 27 430 27 27 464 464 383 111 111 111 111 111 111 111 441 41 118 116 443 112 683 63 SOLUBILITY OF GASES IN LIQUIDS TABLE V (Continued) 449 Gases H a + N t Ha + CO Hz + COz Hz + COz Hz + K z + CO Hz + 0 2 + CO Nz, 0 2 , COz, CaHs, CaHia Nz + COP Nz + GO2 Nz + COP Nz + CHI Nz + CH4 Nz + CH4 + CzHs + CSHs + C4Hio Nz + COz + CnH8 COi + HzS COz + HzS COZ + CSZ, coz + cos COZ + CHd, COZ + CZH4, COz + CH4 + CaHs, Cot + CzHd + CnHa NzO + 0 2 , Nz0 + ether CHI + C~HIO. CHI + CzHa CsHs + CdHa Natural gas Natural gas Geeem Helium Neon Argon Xenon Hydrogen Nitrogen Oxygen Chlorine Carbon dioxide Solvent Liq NHs Hydrocarbons Water Methanol Diesel fuels Diesel fuels Paraffin oil Kater Water and aq KzSO4, KzCOa, KOH, KzCOa-KHCOs Crude oils Diesel fuels .4q NaCl and CaClz Aq CaCh and NaCl Crude o i l s Water Monethanolamine in water Over piperidine or morpholene-satd kerosene Crude oil Blood Crude oil Paraffin oil Crude oil Water Pressure, atm 100-500 25-300 T o 60 3-40 3-40 25-300 0-300 3-40 50 1-3 0-300 1-3000 mm 0-300 1 High press 180-700 mm 200 1 TABLE VI SOLUBILITY DATA FOR GASES IN MOLTEN SALTS .4ND GLASSES Solvent 3 Na-Ca glasses and 7 alkali glasses Binary and ternary glass melts 2 lithium silicate glasses Pyrex glass Sodium disilicate Pyrex 774 KC1 Gabbrodiabase LiF-BeFs (64:36 mole %) NaF-ZrFc (53:47 mole %) NaF-ZrFa-UFb (50:46:4 mole %) LiF-NaF-KF (46.5:11.5:42.0 mole yo) LiF-BeFz (64:36 mole %) LiF-NaF-KF (46.5:11.5:42.0 mole %) Sodium dirilicate Pyrex 774 LiF-BeFz (64:36 mole %) LiF-NaF-KF (46.5:11.5:42.0 mole %) LiF-BeF, (64:36 mole %) NaF-ZrF, (53:47 mole %), NaF-ZrFrUF4 (50:46:4 mole %) Molten slag Sodium disilicate Pyrex 774 NaOH, KOH NaNOa, KNOa, CsNO: Sodium disilicate Pyrex 774 NaNO8. KNOs. CsNOa NaCl, KC1, MgClz 1 : 1 KCl-NaCI, NaCl-MgCln,~KC1-MgCIz Glass NaCl. KCl NaF-ZrF, (53:47 mole %) NaF-ZrF, (53:47 mole %) Pressure 1 atm 1 atm 10 mm 10 mm 646-698 mm 513, 716 mm 1-2 atm 0.4-2 atm 0.2-2 atrn 1 -2 atrn 1 -2 atm 1 -2 atm 1-2 atm 10 mm 1 0 mm 1-2 atrn 0.5-2 atm 1-2 atrn 1-2 atm 1-2 atm 1 0 mm 1 0 mm 100-800 psia 1 atrn 10 mm 1 0 mm 1 atm 1 atm 1 atm 1 atm I atm Temp, O C -50 to 50 25 - 45 25-50 25-51) 25 50 25-50 4-45 4-45 100 20 40-140 50 37 20 0-70 Temp, O C 1200-1480 1200-1480 1400 25-515 800 1170 900,1300 1300 500-800 600-800 600-800 600-800 500-800 600-800 600-800 800 1170 500-800 600-800 600-800 600-800 600-800 1400- 1 8 0 0 800 1170 410-500 3OC-400 800 1170 300-400 800-1050 565-1050 1100-1315 810-950 Measure- ment value (see section VI) Ref 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 2 1 0 1 0 1 0 0 Measure- ment value (see section VI) 1 0 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 361 9 679 561 368 368 377 679 214 519 368 376 376 519 226 283 236 519 2 4 1 .
4 607 377 627 438 Ref 435 546 545 506 444 444 201 301 635 232 232 50 635 232 50 444 444 635 232 50 635 232 451 444 444 589 195 444 444 195 511 511 385 233,346 450 RUBIN BATTINO AND H. LAWRENCE CLEVER T ~ B L E VI (Continued) Gaser Solvent Hydrogen fluoride NaF-ZrFI (53:47, 45:55, 80.5:19.5 mole %) Water vapor Binary and ternary glasses Sodium silicate LiC1-KC1 /50:50, 53:47, 60:40, 69:31 mole yo) Alkali nitrates and perchlorates NaNOa, KNOa, CsNOa CSNO~-B~(NOJ)Z (95.7:4.3 mole %) KNOs-Ba(NOa)t (99:1, 96:4 mole %) KNOrKzCrzO7 (98:2 mole %) NaNOsCa(N0a)z (96:4 mole %) Pressure 0,5-3 atm 0.08-1 atm 3-26 mm To 60 mm 16-21 mm 1 6 -2 1 mm Sulfur trioxide Glass Boron trifluoride LiF-BeFrZrFcThFrUFd (65:28:5: 1: 1 mole %) 1.2-1.9 atm Metal Lithium Sodium Magnesium Aluminum Silicon Chromium Manganese Iron Cobalt Nickel TABLE VI1 SOLUBILITY DATA FOR GASES IN MOLTEN METALS AND ALLOYS (ARRANGED BY IKCREASING ATOMIC NUMBER OF THE METAL) Gas Pressure Nitrogen Oxygen Oxygen Hydrogen Hydrogen Oxygen Hydrogen Nitrogen Nitrogen Hydrogen Nitrogen Oxygen Hydrogen Oxygen Hydrogen Nitrogen 1 atm 1 atm 1 atm 50-800 mm 200-600 mm 1 atm 1 atm 1 atm 1 atrn 1 atm 1 atm 1 atm 1 atm 1 atrn 1 atm 1 atm 1 atm 10-40 mm 20-40 mm 1 atm 1 atm 1 atm To 4 atm 10-30 mm 1 atrn 1 atm Satd FeO slag 1 atm Satd FeO Satd FeO Satd Si02 or MnO Satd FsO H r H z 0 equil H r H z 0 equil Satd Si02 1 atm 1 atm 1 atm 1 atm 1 atm 1 atm Temp, O C 550-800 1250-1750 900-1100 390-480 145-290 300-400 391 332 329 299 900 500-700 Meaaure- ment value (see section VI) 1 0 1 1 1 1 1 1 1 1 0 1 Temp, "C 250-300 250-400 121-204 250-400 130-540 550-775 760 660-1050 700-1000 670-850 700-1000 700-900 1410 1903 1600,1700 1600-1750 1300-1450 1273-1500 1600,1700 1680-2460 1536-1820 1580-1670 1560,1655 1600 Mp to hp 1592 1556-1744 1560,1685 1275-1420 1550-1700 1600 1600 1560 1550-1750 1550-1700 Mp and up 1600 1530-1750 1510-1760 1530-1700 1560-1700 1535-1670 1550-1650 1550-1650 1530-1700 1600,1650 1525-1725 1592-1758 1490-1700 1523-1709 1500-1600 1600 1600,1700 ... Ref 553 546 602 82.83 154 195 195 195 195 195 45 554 Ref 257 256 257 256 448 326 528 158 33 489 462 258 1 636 638 432 433 26 225 638 354 636 664 537 665 86 355 637 370 296 356 535 86 298 538 295 159 242 598 521 536 186 595 598 245 246 184 372 380 534 636 595 636 537 86 638 SOLUBILITY OF GASES IN LIQUIDS Metal Copper Zinc Silver Cadmium Indium Tin Mercury Lead Bismuth Alloys Fe + V, Fe + B Fe + Ni Be + Cr Fe + Xi Ni + Co Fe + Co Steel Fe alloys with Al, B, C, Co, Cu, Ge, P Cu-Zn A1-Cr, A1-Fe, A1-Th, A1-Ti, AI-Sn Cu-Sn Fe-Si Fe-Ti, Fe-Nb. Fe-Ta Mg--41, Mg-Zn AI-Cu, Al-Si Fe with C, Mn and Si Fe-V Mn-Fe, Mn-Si Fe alloys with Cr, Co, Nb, Cu, Mn, Mo, Ni, Fe-Mn Fe alloys with Co, Cu, Ni, C , Rlo, Si, S, and Se Fe-Ni, Fe-Co Fe-Mo-V Fe-Ni, Fe-Mo, Fe-V Fe-V Fe containing P and 0 Fe-Mn, Fe-Si, Fe-Mo Steel Fe-Si Fe-C, Fe-41 Fe-Cr, Fe-V Welding alloys hln-Fe, Mn-Cr, Ni-Cr Fe with small amounts of C, Si, Mn, Cr, Ni Fe-Mn, Fe-Cr, Fe-Ni (entire range of compn) Fe-Cr, Fe-Mn, Fe-Ni Cr-Si Cr-Si Fe-Cr Fe containing S NE-K Fe-S Fe-Ni Fe-Si Fe-Cr Steel Fe-A1 Fe-Si, Fe-Mn, Fe-Si-Mn Fe-Cr, Fe-Ni Si, Ta, Sn, W, and V TABLE VI1 (Continued) Oxygen 1 atm Gas Pressure Hydrogen Sulfur dioxide Hydrogen Kitrogen Krypton Krypton Hydrogen Krypton Krypton Hydrogen Helium Hydrogen Krypton Hydrogen Oxygen Xenon n2 n? Hz H2 H2 Hz Hz H? Ha Hz Ht H2 H2 H2 H2 H? N2 Na h - 2 N2 Na NZ N a N2 ?;a N2 N 2 N2 N? N 2 N? N Z N Z N Z Nz N Z N Z N2 N? 0 2 0 2 O? 0 2 0 2 0 2 0 2 0 2 0 2 0 2 1 atm Satd NiO C 0 4 0 equil 1 atm 5-760 mm 100-760 mm 1-760 mm 1 atm 1 atm 1 atm High press Satd PbO-Si02 Satd PbO-SiOp Air atm Air atm 1 atm 1 atm 1 atm 1 atm 1 atm 1 atm 1 atm 5-760 mm 1 atm 10-40 mm 20-40 mm 1 atm 1 atm 1 atm 1 atm 1 atm 1 atm 1 atm 1 atm 75-570 mrn T o 4 atm Variable 10-30 mm 1 atm 1 atm 1 atm 1 atm 512, 735 mm 1 atm Atm press Variable press Satd FeO Satd Si02 or MnO H r H 2 0 equil Temp, O C 1450-1690 1470-1720 1465-1700 1450-1691 1726-1970°K 945-1100 Mp-1400 11OC-1300 1150,1240 1100-1300 448 1100-1 150 1000-1300 1 100-1 300 1000-1300 20-140 Mp and up 800-1300 516 1000 400-900 1000-1194 350-800 540 1560,1655 1400,1600 1400,1600 1600 1600 1600 1600 1592 1100-1300 1560,1685 760 700-1000 1275-1420 1.580 141 0-1 660 1600 1245-1050 1550-1650 1600 1600-1900 1600-1800 1580-1650 1 600 1560 1600 1550-1750 1550 1550-1600 1600,1700 1530-1750 153O-1750 1600 1600-1750 15.50. 1670 1550,1600 20-176 1550-1600 1470-1720 1600 1650-1760 1600 1550-1650 1550-16RO 1625 Ref 65 182 595 663 65 524 636 306 43 279 190 189 306 258 258 281 280 306 280 281 43 431 306 281 258 126 525 500 30 171 665 304 304 86 86 86 658 63 7 306 33 43 370 296 528 462 356 33 1 25 472 34 535 86 298 298 332 185 538 658 295 159 73 48 638 521 521 521 432 433 350 523 448 87 182 264 125 658 245 246 371 452 RUBIN BATTINO AND H. LAWRENCE CLEVER TABLE VI1 (Continued) .4iloy Fe-Cr, Fe-Ni Fe-Cr, Fe-&Ni Ni-Fe Fe-CrP. Fe-Ni-P Fe-Ti Fe-P Fe-Ni Cu-Xi Gasen Helium Argon Hydrogen Deuterium Nitrogen Oxygen Air Chlorine Methane Ethane Propane Ethylene G r t B 0 2 0 2 O¶ 0 1 0 2 0 2 0 2 0 1 Pressure HrHzO equil HrHzO equil Air atm or satd NiO HrHzO equil HpHzO equil HI-HzO equil HrHzO equil TABLE VI11 PARTIAL MOLAL VOLUMES OF GASES IN LIQUIDS Solvent Water (hydrostatic pressure) Liquid methane Water (hydrostatic pressure) Isooctane, methylcyclohexane, toluene, perfluoromethylcyclo- hexane, carbon tetrachloride, bromoform, carbon disulfide n-Hexane, n-octane, methanol Water Water, methanol Methyl acetate, benzene, acetone, ethyl ether, chlorobenzene, Liquid methane Methane, ethane, ethylene, propane, butane n-Hep tane, n-o ct ane n-Octane, benzene Cyclohexane Benzene, toluene, n-perfluoroheptane Benzene Liquid argon +Heptane, n-octane Benzene, toluene, n-perfluoroheptane Liquid argon Water 'Water (hydrostatic pressure) Water, methanol Methyl acetate, benzene, acetone, ethyl ether, chlorobenzene, CCL n-Hexane Benzene n-Peitluoroheptane Water Water Water, methanol Methyl acetate, benzene, acetone, ethyl ether, chlorobenzene, CCL Water, methanol +Heptane Water Water Water, methanol Methyl acetate, benzene, acetone, ethyl ether, chlorobenzene, n-Hexane, n-heptane, isooctane n-Hexane, n-perfluoroheptane, carbon disulfide Water Water Methyl acetate, benzene, acetone, chlorobenzene, CC14 n-Hexane, n-perfhoroheptane, carbon disulfide Water Water Water, benzene, methanol, tetrahydrofuran Methyl acetate, benzene, acetone, chlorobenzene, CCL CCL CCL Ref 372 47 1450-1691 663 1500-1650 365 1600,1650 380 1500-1650 366 524 1380-1620 596 Temp, O C Temp, O C 25 -183, -167 1-25 25 30 0-50 25 25-100 -183 to -146 -183 to -4 25-50 130-260 20-60 25 25 25 to 50 25 -186 to -153 0 25 0-50 0, 25 30 25 25 0 25 0-50 0, 25 25 38-170 0-50 0, 25 27 25 -186 to -133 0, 25 17-36 38-170 17-30 10-40 0-25 38-170 17-30 25-150 10-40 Pressure, atm Ref 1-102 169 To 160 229 1-102 169 1 282 100 397 324 1 336 1 261 180-220 177 Highpress 37 50-300 To 150 700 1 27-775 50-300 1 1 1-102 1 1 100 27-775 1 1 1-102 1 1 1 1 1 1 1 1 1 1 1 1 1 To 193 1 352 101 337 282,633 340 630 352 282,633 630 359 169 336 261 397 340 220 359 169 336 261 336 219 324 396 336 261 548 220 324 396 261 220 324 396 253 261 SOLUBILITY OF GASES IN LIQUIDS TABLE VIII (Continued) Gases Acetylene Dimethyl ether Methyl chloride Carbon monoxide Carbon dioxide Nitrous oxide Sulfur dioxide Sulfur hexafluoride Carbon t,etrafluoride Solvent Water, benzene, methanol, acetone Water, benzene, methanol, tetrahydrofuran Methyl acetate, benzene, acetone, chlorobenzene, CCl4 21 solvents Methyl acetate, benzene, acetone, chlorobenzene, CC4 Methyl acetate, benzene, acetone, chlorobenzene, CC4, chloro- Water, methanol Methyl acetate, benzene, acetone, ethyl ether, chlorobenzene, cc4 Water Water (hydrostatic pressure) Water, methanol Methyl acetate, benzene, acetone, chlorobenzene, C c 4 Methyl acetate, benzene, acetone, chlorobenzene, cc14 Methyl acetate, benzene, acetone, chlorobenzene, c c 1 4 n-Heptane, isooctane, benzene, CCL, CC12FCCl?F, CeH11C2Fb, n-Heptane, isooctane, benzene, CClr form c s 2 of the experimental values for all the metals except lead for which H was 12 kcal/mole too large. The calculated vibrational entropies agreed within experimental error. VI. SOLUBILITY DATA This section concerns four tables. Table V is ar- ranged according to gas, giving the solvent employed, the temperature and pressure range of the measure- ments, and a reference. In addition, a critical value was assigned to each measurement: 2-the data are quantitative and the precision is probably better than 2%; 1-the data are quantitative but the precision is poorer than 2% ; 0-the data are either just qualita- tive or it was not possible to determine the level of precision of the measurements. These values were arrived at by using the author’s own statements and/or our critical evaluatioii of the method employed. Where there are blanks under the headings of tem- perature and pressure, it was not possible to determine these ranges. For each gas the solvents were sys- tematically and consistently arranged. Reference to nomographs are listed under the gas in this table. Ob- viously no nomograph can be better than the experi- mental data on which it is based. Before using a nomograph the literature should be examined to de- termine whether or not the nomograph has been superseded by more reliable data. Table VI gives references to the solubility of gases in molten salts and glasses and is arranged by gas. Table VI1 gives references to the solubility of gases in molten metals, and alloys. This table is arranged according to increasing atomic weight of the metal with the alloys at the end. No value judgments were made on these solubilities. Table VI11 gives references to the partial molal volume determinations of gases in liquids. This in- Temp, O C 10-40 20-45 10-40 -20 to +20 25 25 0-50 0, 25 0 25 0-50 25 25 10-40 25 27 Pressure, stm 29 To 39 1 3.5-14 1 1 1 1 1 1-102 1 1 1 1 1 1 453 Ref 316 253 261 516 261 261 336 261 359 169 336 261 261 261 254 548 formation is arranged by gas and is included since it is not only interesting in its own right but is valuable for the use and evaluation of theories of gas solubility. ported in part by a PHS Grant No. GM12071 from the National Institutes of Health, Public Health Service. The authors take pleasure in acknowledging the early interest, encouragement, and advice o f Professors P. M. Gross and J. H. Saylor. ACKNOWLEDGMENT.-ThiS investigation Was SUP- VII. REFERENCES (1) Abriksov, N. Kh., Liang, T.-W., and Shashkov, Yu. M., Izv. Akad. Nauk SSSR, Otd. Teckh. Nauk, Met. i Top- livo, 65 (1960); Chem. Abstr., 55, 11050 (1961). (2) Agabal’yants, G. G., Kozenko, E. M., and Merzhanian, A. A., Vinodelie i Vinogradarstzlo SSSR, 14, No. 6, 26 (1954); Chem. 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Each gas has its unique properties and af&ts the chromatographic system in different ways_ The solubility in pure and mixed solvents is explored experimentally and compared with data already in the literature. It is found that the non-linear solubility characteristics in binary solvent systems account for the observed evolution of large quantities of gas when air-saturated solvents are mixed in chromatographs. The de- gassing requirements in one-pump, low-pressure-mixing gradient ar&itectures are compared with those of conventional two-pump, high-pressure-mixing systems. Dissolved oxygen affects detector performance in several ways. It forms a UV light-absorbing complex with many solvents. Changes in oxygen concentration there- fore cause W detector drift. The magnitude of this effect varies markedly with dif- ferent solvents, and is particularly pronounced at wavelengths below 260 nm. Dis- solved oxygen quenches fluorescence of both solvents and solutes. As a consequence, fluorescence detector drift and responsivity depend on oxygen concentrations. Max- imum fluorescence sensitivity can only be achieved with deoxygenated mobile phases. Because of these facts,? analytical precision requires that gas concentrations be carefully controlled. The various control techniques are discussed : heating, boiling, vacuum, ultrasonics and gas sparging. A new method of helium degassing is described which eliminates bubble formation and maintains the level of all other gases at zero concentration. . INTRODU~ON Because air is ubiquitous, the gases which comprise it tend to be found in high-performance liquid chromatography (HPLC) mobile phases. Their presence ac- counts for a variety of effects, many of which interfere with sensitive, precise, trouble- free chromatographic analysis. Although some of these problems have. been under- stood for some time, others have received scant attention. There is little overall awareness of the important role of gases, and that which dues exist is often based on misunderstanding. .- Resent address: Rheodyne Inc., 2809 Tenth Street, BerkeIey, Calif. 94710, U.S.A. 278 S. R. BAKALYAR, M. P. T. BRADLEY, R. HONGANEN The importance of excluding dissolved oxygen from the mobile phase m order to protect labile stationary phases has been mentioned in most standard texts on HpLC1L3. Leitch documented the improved column lifetime and analytical precision resulting from deoxygenated solvents ‘. Exposure of the liquid-liquid partition column (3,3’-oxydipropionitrile) to dissolved oxygen during constant or heavy use severely reduced column life. _ The potential for harm to the mobile phase has also been pointed out. ~utyl ether mobile phase was oxidized during handling and storage in the reservoir, forming peroxides which reacted with the stationary phase and changed its polarity. The susceptibility of other ethers, such as tetrahydrofuran, may also be a probiem’. SnydcP has counciled about the benefits of excluding oxygen in liquid-solid (adsorption) chromatography, since sampk oxidation is often increased by the pres- ence of the adsorbent. A recent example’ in reversed-phase chromatography de- scribed the ease with which aniline and its metabolites were oxidized during chro- mato,orphy, and the improved results with deoxygenated mobile phases compared to mobile phases to which antioxickts had been added. The improvement of detector performance resulting from deoxygenating the mobile phase has also been reported. FOX and Stale9 showed that deoxygenated mob& phases in polycychc aromatic hydrocarbon analysis produced a limit of de- tection for benzo[a]pyrene which was nearly four times as sensitive than when air- saturated mobile phase was used. Different but linear calibration curves were obtained in the presence and absen.ce of dissolved oxygen. Compared to.pyrene, oxygen quench- ing of fluorescence is less important in chrysene but more important in benzo- and dibenzopyrenes. They did not determine how effective or reproducible their deoxy- genation process was- Chamberlain and Marlow’ demonstrated that oxygen dissolved in the mobile phase caused increased noise levels and decreased standing current in the LC electron capture detector they used. Perhaps a more generally encountered problem than any of the above is the occurrence of gas bubbles in the detector1-3. Air dissoked in the carrier at high pres- sure can subsequently form bubbles in the mobile phase as it passes through the detector, causing noise and drift. Gas bubbles also aifect pump performance, but this has not been a serious concern prior to the advent of one-pump, low-pressure- mixing chromatograph architectures10-13. Some of the above reports contain misunderstandings about the solubiiity behavior of air gases. For example, it has been said that the more polar the mobile phase, the greater the tendency to dissolve air ‘s9. Actually the opposite is true. Water is the Ieast hospitable solvent for gases. And the bubble problems are attributed to oxygenxJlJ, whereas nitrogen is as much a’source of difficulty. The purpose of this paper is to report new fkiings on gas hehavior of chro- matographic signifkance, and to describe ways of improving the reliability, sensitivity and analytical precision of HPLC by careful control of dissolved gases. We will pro- ceed by (1) reviewing gas solubillty behavior in pure solvents and binary mixtures, (2) comparing the one-pump, low-pressure-mixing gradient architecture with the two- pump, high-pressure mixing gradient architecture, as to gas solubility probiems, (3) describing methods of controlling dissolved gas concentrations, (4) discussing the “opt-cal properties” of dissolved ,w so as to explain observed UV and fluorescence detector artifacts, including UV detector behavior which has not been reported before DISSOLVED GASES IN HFLC 279 in the chromatographic literature and (5) commenting on other aspects of dissolved gases, including the effect of carbon dioxide on mobile phase pH and therefore on the reproducibility of retention times and peak areas. EXPERIMENTAL CoZunm, soZwnts and samples Staidess-steel columns (250 x 3.1 mm I.D.) packed with IO-pm Spherisorb ODS and 250 x 4.6 mm I.D. stainless-steel columns packed with lo-pm LiChrosorb RP-8 were used. Both are totally porous, bonded reversed-phase packings which are respectively octadecyl (C,,) functionality on spherical silica and octyl (C,) functionality on irregular silica. (Spectra-Physics, Santa Clara, Calif., U.S.A.) Mobile phases were prepared from distilled-in-glass solvents (Burdick & Jack- son, Muskegon, Mich., U.S.A.). Water was prepared by a Mini-Q system, fed by a Milli-RO system, in turn fed by Santa Clara (Calif., U.S.A.) tap water; the Milli- Q had four cartridges, two mixed-bed ion exchangers followed by two activated carbon units (Millipore, Bedford, Mass., U.S.A.). Solvents were degassed as indicated in the , text. Sparging gases were high purity grades, exceeding 99.99 mole% purity_ Air was “breathing quality”. Samples were from Chem Service (West Chester, Pa., U.S.A.) and Aldrich (Milwaukee, Wise., U.S.A.). They were dissoIved in water-methanol and water-acetonitrile mixtures. Control of flow, composition and temperature A Spectra-Physics Model SP 8000 research liquid chromatograph and a Spectra-Physics Model 3500B gradient liquid chromatograph were used. The former employs a single pump, attached to a low-pressure composition forming module (ternary proportioning valve). It has &O. 1 “C column temperature control via a forced air oven. The 3500B system employes a dual reciprocating piston pump for solvent A and an identical but independent one for solvent B. The composition is formed at high pressure in a dynamically stirred chamber. Temperature control was via a water bath. The detectors were Spectra-Physics Model SP 83 10 operated at .254 nm, Model SP 770 variable-wavelength detector and Model SP 970 fluorescence detector. RESULTS AND DISCUSSION Gas solubiZity in pure solvents The solubilities of gases in liquids have long been an area of active interest to chemists. Practical concern has been related to such diverse fields as industrial processes and the composition of artilicial atmospheres. Theoretical concern has been related to the small solubility and the variety of gases available to use as probes for the investigation of liquid and solution structure and properties. A number of ex- cellent review articles exists 15-1. There are several major sources 19-z1 for gas solubility data in pure solvents in addition to these reviews. Solubility data in mixed solvents is of course of the greatest interest to the chromatographer since the use of a pure solvent mobile phase is rare except in exclusion chromatography. Unfortunately, the 280 S. R. BAKALYAR M. P. T. BRADLEY; R- HONGANEN data for such realistic mobile phases is scanty, although a few papers exist=-z which describe the solubility of some air gases-in aqueous alcohol.soluti6n.s. Before proceeding further it may be instructive to review briefly a few of the salient facts about gases. Air comprises 78.08 % N,, 20.95% O,, 0.93 % Ar, 0.03% CO, and less than 0.01% other gases. Gases in a mixture behave essentially independently, the solubility of the individual gases in a mixture of gases being di- rectly proportional to their partial pressures (Dalton’s law). Different gases have dif- ferent solubilities in a given solvent. Different solvents have different solubility prop- erties towards a given gas. The soIubiIity of gases in liquids usually decrease with increasing temperature, but there are numerous exceptions and the correlation. of sohxbility data as a function of temperature is not simple. Gas solubilities in most solvent mixtures, like so many physical phenomena in non-ideal solutions, are not a line+r function of the composition expressed in mole fraction_ Fig. 1 shows the solubiiity of several gases in many solvents using data taken fro& refs. 17 and 18. The solubility is expressed as the moie fraction. A feel for the magnitude of these values can be gained by realking that the carbon dioxide solubility Fig. 1. Gas solubility VS. solvent polarity index. The solubility is expressed as mole fraction x l(r for pure gases in equilibriumwith the solvent at 1 atm and 25 “C. The polarity index is thesnyderparrsm- eter. V, COr; 0, Ar; 6,0=; ~3, Nz; 8, He. of mole fraction 0.01 in benzene is equivalent to about 2.4 ml of carbon dioxide at 1 atm and 25°C per ml of benzene: and the nitrogen soiubility of mole fraction 0.00OO14 in water is equivalent to about 16 ~1 of nitrogen at 1 atm and 25°C per m! of water. The solubility of the various gases in the various solvents thus ranges over a factor of 1000. The solubility is plotted against solvent polarity, expressed as Snyder’s polarity index, Pz6. This solvent characterization parameter is but one of several which have been proposed recently. It can be compared with the Hildebrand solubility parameter @‘So in that values of P’ roughly parallel values of S, and have similar sign&ance. However, whereas 6 is measured for the.pure solvent; and only reflects interactions that exist in the pure solvent, P'. is measured against a variety DLss0LvED GASES IN HPLC 281 of solutes that encompass all possibIe types of interaction: The Snyder parameter therefore may not be the most reasonable one to use if a smooth curve is desired, but it is employed here since it is currently popular in chromatographic literature12. The important point to note is that gases, being non-polar, behave .F one would expect, having increasing solubility as the solvent polarity decreases. It should be point& out that these solubility values represent the amount of gas which willjbe found dissolved in the solvent at equilibrium with one atmosphere of the pure gas over the solvent. So the previously mentioned 2.4 ml.of carbon dioxide in benzene only exists in a pure carbon dioxide atmosphere. Much less is in the solvent exposed to air. Hydrogen is not plotted. It falls midway between the helium and nitrogen curves, except in water, where it is slightly more soluble than nitrogen. The solubility of the rare gases increases with increasing atomic weight- In water the mole fractions for He, Ne, Ar, Kr, Xe and Rn are 0.07, 0.08, 0.25, 0.45, 0.78 and 1.68, respectively. It is clear that water dissoIves the ieast amount of air gases. The common misconception that it is the best solvent may derive from the fact that gas is often seen to come out of solution in water. But this may be because, once degassed, it does not take much gas to redissolve before saturation is reached. The real problem with regard to gas solubility in HPLC, however, comes with mixed solvents, either in isocratic mode where two solvents are mixed by the instrument, or in gradient mode. Gas sohbility in mixed solvents As previously mentioned, there is not much data in the literature on gas sol- ubility in mixed solvents. Most data which exists is for water-alcohol &tures, and we can infer from it what may happen in other binary mixtures of solvents, especially those binary mixtures where the two solvents interact strongly Ben-NaimU studied the solubility of argon in water-methanol systems. A plot of soIubiIity vs. mole fraction of methanol shows the solubiIity generally increasing with increasing methanol content as one would expect. However, it goes through a maximum and then a minimum, especially at low temperature. Cargill and Morrison23 extended the argon study over a wider temperature range, and included water-tert.- butanol systems. The latter exaggerated the peculiar behavior of alcohol binary mix- tures. CargilPJ also studied oxygen and found it to behave like argon. Fig. 2 shows a few of the solubility curves from the Cargill papep. In the paper the solubility was expressed by S,,, defined as the volume of gas in ml, corrected to 273 “K and 1 atm, dissolved by 1 kg of solvent, under a gas pressure of 1 atm. In Fig; 2 it is expressed as mole fraction. Fig. 2 can be contrasted with Fig. 3, which shows the amotint of gas actually present in the solution at various mole fractions, starting with the A and B solvents air saturated. The significant observation is that: when two soIvents, such as water and ethanol, are at equilibrium with the atmosphere (Le., air saturated) and when they are blended to form a mixture (such as in gradient elution), supersaturated con- ditions exist during much of the run, which cause gas to come out of solution until the concentration is at the allowed saturation level. Fig. 4 illustrates what happens in a real chromatographic situation. It is a record of the column inlet pressure during a gradient run of 0 to 100% methanol in water in 5 min. The pumping system is the Spectra-Physics SP 8000 liquid chro- S. R. BAKALYAR, M. P. T. BRADLEY, R HONGANEN f Fig. 2. Oxygen solubility in aqueous alcohol mixtures: moIe fraction oxygen vs. mole fraction alcohol. The solubility is expressed as in Fig. 1. The original data of CargiP was plotted in terms of %,, defmed as the volume of gas in ml, corrected to 0 “C and 1 atm, dissolved by 1 kg of solvent, under a gas pressure of 1 atm, on a logarithmic scaIe. Data reproduced by permission of the publisher. Fig. 3. Oxygen solubiity vs. oxygen concentration. The actual level represents the amount of oxygen in the admixtures, starting with air saturated pure water and air saturated pure ethanol. The saturation level is the-allowed solubility level, the 29.9 “C curve from Fig. 2. Note the non-linear behavior, particularly at the low mole fractions of ethanol. . -OEGASSED- NOT OEGASSEO IAk !isturated I Fig. 4. Bubble formation in pump chambers with undegassed solvents. Cohn%, 250 x 3.1 mm I.D.; packing, IO-pm Spherisorb ODS; solver& O-100% methanol in 5 min; fiow-I&, 5.0 mi/min; pressure, as indic&ed temperature, 25 “C. Recorded on the SP 8000 printer/plotter using the signal from the pre-coIumn pressure transducer. DISSOLVED GASES IN EIPLC .- 283 matograph, which has a single pump with two pump chambers each of 4UO-~l displace- ment The flow feedback control has been turned off so that the pulsation (pressure dip) at the be ginning of each pump stroke is evident. This pulsation is a measure of the compliance in the chamber, which in turn is the sum of mechanical elasticity, fluid compressibility and the compr&sibility of any undissolved gas present. The upper curve exhibits the expected rise and fall of pressure- during the gradient, which reflects the viscosity pro& of the gradient mobile phase. The pressure pulses tire uniform, dem- onstrating that there are no undissolved gas bubbles in the chamber. This curve was made using solvents which had been helium “degassed” (see below). The lower curve is similar, but the pressure pulses are irregular. The solvents were not degassed, but were simply equilibrated with air. These irregular pulses result from bubbles of gas entering the pump during the tirst part of the gradient. Later in the gradient, the pressure and solvent composition are such as to redissolve them. In some cases, especially at low operating pressures, the pulses persist because the gas never redissolves. Such be- havior not only causes flow-rate errors, but is deleterious to composition precision. The bubbles in the above example are formed in the low-pressure ternary proportioning valve, which mixes the two pure solvents’ in the proper proportion during the run. Fig. 5 diagrams such a system, and compares it with the conventional two- pump, high-pressure-mixing architecture. It is clear that the environment where the pure solvents are mixed is radically different in the two types. In the conventional architecture, the solvents are mixed at high pressure, where the solubility is much higher. Thus gases do not come out of solution at that point. However, if there is air in the solvents, it will come out of solution when the pressure again reaches one atmosphere, and sometimes sooner if some gas has been picked up at high pressure via a small leak in the system. For this reason, a flow restrictor is often put at the detector outlet, so that only after the detector does the pressure reach a level where gas bubbles form (see the pressure prosle in Fig. 5). GQS control methods There .are two approaches to g& control. One &rives to eliminate ai1 dissolved gases, the other to eliminate or control the concentrations of only certain gases. The former has been the most common approach directed at remedying the various prob- lems described above, ie., gas bubbles, oxidative degradation of samples and phases, and detector artifacts. The approach used most often is vacuum d&gassing’sz, the application of a vacuum to the mobile phase just prior to chromatography. HeatingLs3 and uhrasonic treatment’“2g have also been employed, One of .the gas solubility review9 has a good discussion of degassing methods. It reports that the most com- mon method to degas a solvent in non-chromatographic work is to boil away a por- tion of it under vacuum, a batch binary distillation. The Ramsey-Rayleigh equation for this type of distillation predicts that the evaporation of as little as 0.1% of the solvent should reduce the gas content by several l@lO-fold. But the asstimption of equilibrium is incorrect. In practice W-20 oA of the solvent is evaporated. Other meth- ods include pumping on the frozen solvent or boiling follqwed by sprasng into an evacuated. ff ask. Sparging with a pure gas (bubbling it through the solvent) has most frequently been empIoyed for the elimination of only certain gases’n’. Tbis techiliquk, also re- ferred to as purging or stripping, has been used in gas chromatographic studies“Q’ 284 S. R B_A?CALYAR, M. P. T. BRAD‘LEY, E?A HONGANEN I I I I Fig 5. Comparison of one-pump and two-pump architectun~ The upper system represents a two- pump, high-pressure-mixing system such as the SP 35OOF.L The lower system represents a one-pump, low-pressure-mixing system such as the SP 8000. The pressure drops appear non-linear along the col- umn length only because it is a semi-logarithmic plot. Williams and Milled compared several techniques for purging water: dynamic and static vacuum, with and without manual and ultrasouic agitation; ultrasonic treat- ment alone; and purging with an inert gas. The most effective system tested was inert gas-purging at flow-rates of about 1000 ml/min of 100-ml water samples. This tech- nique removed 95-98 % of the dissolved oxygen in 15-30 set, where the next best te&nique of dynamic vacuum with agitation took l-2 min to remove the same quan- tity of gas. The degassing technique we used was that employed by the SP 8ooO chromato- graph. This has been briefiy described only O~CX’~ and is the subject of a patent application. The method uses helium to sparge all pure solvents (up to three in the ternary-type mobile phase’ control system of the chromatograph). It was found that this is the only gas, with the possible exception of neon, which is capabIe of efimi- nating all the previously mentioned problems. That is, it not only prevents bubble formatiori but eliminates all gases except helium from the tiobile phase. (There is no literature on the helium solubiliq-in binary mixtures.) Our experimental evidence suggests two possible explanations for this. The solubility curve for helium may be non&near, like the. oxygen shown in- Fig. 3. Rut the. amount of gas involved may he so- low, i.e., the absolute -value of the gas volume which is supersaturated may be so smali, that the microbubbles formed do not manifest themselves. Alternatively, the solubilitjr curve may he nearly -liuear, Ieading to only ssnall amollnts of gas which are above the saturation level. Both of these may be operating. .The effectiveness-of this technique for eliminating bubble generation during low-pressure mixing is illustrated in Fig. 6. This plots the volume of air evolved per ml of mobile phase formed by mixing two pure, air-saturated solvents. A Model 74OB pump (dual reciprocating piston, f~dback-controiled type) was used for each solvent. The outputs of the pumps were teed together and the tee exit line directed to a l.O-ml mixing chamber. The position of the inlet and outlet lines, and the shape of the -roof” of the chamber were designed to trap any bubbles formed. After passage of a measured amount of total mobile phase through the chamber at a pre-detertuined solvent composition, the gas bubbles formed were sucked into a precision syringe which was connected to the top of the chamber. Replicate runs were not made, so the detailed shape of the curves is not to be taken as significant. Considerable scatter was suspected. The general shape, however, is no doubt accurate. These curves cor- respond well to the area of supersaturation in Fig. 3. Note that the more similar the two solvents, the less gas was evolved. The hexane-isooctane run produced no bubbles. IQ all cases, no bubbles were observed when the two solvents were helium Fig. 6. Gas evolve& VS. solvent composition. Apparatus and piocedure described in text_ 286 S. R BAKALYAR, M. P. T; BRADLEY,.R.HONGANEN UV absorbance of dissolved gases - -. -. During the above studies of gas solubility and degassing techniques, we noticed that UV detector. baselines sometimes drifted .corrsiderably when degassing was ini- tiated. We have experimentally conkmed that &is is due to the presence of oxygen, as illustrated in Fig. 7. It is a-record of W detector signal at 254 nm and 0.08 AUFS for a l.O-cm cell with methanol flowing. The trace starts with the baseline after the methanol has been sparged with pure oxygen for some time. A stable signal was obtained. Then the sparging gas is changed to air, and- the signal drifts down to a new equilibrated value. Then the gas is switched to helium. Yet another level is found. The ratios of these three signals are: (helium-air-oxygen) 0:0.223:1.00. The ratio of partial pressures, i.e., the theoretical signal levels if due to oxygen concentration, are 0:0.209:1.00. This represents a 6.7 % error from theory. Fig. 7. W detector standii signal vs. oxygen concentration in mobile phase. Solvent, methanol; fiow- rate, 2.0 ml/min; pressure, nominally 1 atm; temperature ambient (approximately 25 “C); detector, Model SP 8200 at 254 nm with a l-cm path cell. The gas is further changed from helium to nitrogen and back to helium. The stable baseline is consistent with the hypothesis that it is the oxygen only which is rewonsible for the signal. Finally, the sparge rate was reduced, allowing air to back dBuse through the yent tube into the solvent bottIe. The upscale drift due to oxygen absorbance is evident. Re-establishment of an adequate sparge rate rapidly brings back baseline stability. The amount of absorbance varies considerably among the common chromato- graphic solvents, as shown in Fig. 8. The. W absorbance at 254 nm was monitored DISSOLVED GASES IN HPLC 287 at a sensitivity of 0.02 AUFS. Note that water exhibits only a very small effect at this wavelength, whereas the tetrahydrofuran baseline changed more than 0.16 AU. The baseline shifts are completely reversible: resumption of air sparging elevates the signal back to the original level. Evaus3”@ and Munck and Scott= demonstrated that dissolved. oxygen gives rise .to such absorption in the far UV in a number of organic solvents. This was-observed for n-hexane, n-heptane, methanol, ethanol, di- &thy1 ether and cycloh&xane. The absorbance was found to be directly proportional to the partial pressure of oxygen above the solution. The absorbance increases at shorter wavelengths. Comparing the absorbance of the amount of oxygen dissolved in the solvent with the absorbance of an equivalent concentration of gaseous oxygen (negligible at the wavelengths examined) leads to a conclusion that- the absorption of the solution must be due to interaction between oxygen and the solvent. I(eS_E H20 _J - / CU,CN - c CH2OH t Fig. 8. W detector standing signal with and without degas&g for various solvents. Solvent, as in- dicated; flow-rate, 15 ml/& (splitter in front of detector providing 5 ml/min detector flow); pressure, 1 atm in solvent reservoir, 2 atm at pump outlet; temperature 23 f 0.5 “C; detector, Model SP 8310 at 254 mn with a O&-cm path cell. Heidt and Ekstrom3s,36 have examined this phenomenon in water. The ab- sorption coefficient, E, was independent of oxygen concentration, increased with de- creasing wavelength, and increased with increases in temperature. They explained the results in terms of the existence of two different hydrates of moleculdt oxygen in water. Pure (deoxygenated) water also was found to absorb UV light and behave in a similar manner, but the magnitude of the effect was very much lessthan the light absorption due to molecular oxygen in ,water. Thus dissolved oxygen a&cts HPLC mobile phases in two ways, one of which is reversible, the other non-reversible. The reversible e&cts have been described above 288 s. R. BAKALYAR, M. P. T. BRADLEY, R. HONGANEN and are due to the absorbance of molecular oxygen itself (small) and to the ab- sorbance of the molecular complex between oxygen and the solvent (large). Removal of oxygen gives rise to a decrease in UV detector baseline signal with this effect. The non-reversible effects are caused by the interaction of oxygen with the solvent t.o form reIativeIy stable chemical species- The cyclic ether tetrahydrofuran (THF) appears to be pamticularly bad in this respect, as it is thought to form a hydroperoxide and a series of unstable peroxides. Fig. 8 has shown that the reversible effect with THF is also large. Fluorescence effects of dissoIved gases The role of oxygen in fluorescence systems is quite complex and has been the subject of debate and experiment for many years. An early publication on the role of oxygen in fluorescence quenching was published by Bowen and Wiiliams3’ who discussed in particular the quenchin, = of aromatic hydrocarbon fluorescence. Later workers38T3g found that the quenching effect of oxygen usually followed the Stem Volmer relationship F,,/F = 1 t K[O,]. Results reported by Parker and Bar- nes3’ for the quenching of the borate-benzoin complex show that at 0.1% (v/v) oxygen in nitrogen in equilibrium with the solvent, ethanol, an 8% error occurs. At 0.8 % oxygen the fluorescence is reduced 43 %, and with air the fluorescence intensity is reduced 94%. The same authors showed that the majority of the effect was re- versible, but that there was a second, slower reaction which was irreversible. Bar and Weinreb= showed that in considering the mechanism of oxygen quenching, the mechanism of excitation is important. If the system is such that the solvent is absorbing the exciting radiation, and the ener.gg is transferred to the solute, quenching occurs by competition between the oxygen and the solute. They report that as the concentration of solute decreases, the quenching effect of oxygen increases. With systems where the solute absorbs the exciting wavelength directly and the sol- vent does not absorb, oxygen quenching is again more efficient at lower concentra- tions but the magnitude of the effect with concentration is not so great. This is postulated to be owing to competition between oxygen quenching and self quenching, which increases considerably with increasing solute concentration. That the effect of quenching is specific to oxygen was confirmed by Furst et aLbo, who examined the fluorescence of a large number of compounds in solutions saturated with oxygen, nitrogen, argon, carbon dioxide, hydrogen and nitrous oxide. All gases except oxygen gave the same fluorescence intensity. Thus, the effect of the other gases is to remove oxygen from the solution without otherwise affecting the fluorescence. The quenching effect varies with compound type, as was briefly mentioned in the introduction. Aromatic hydrocarbons, aliphatic aldehydes and ketones are particularly susceptible to oxygen quenching, whereas substituted aromatics and some heterocyclics are much less susceptible. Clearly, the analyst using HPLC with fluorescence detection must be con- cerned with the variation of the magnitude of the quenching effect among different compounds, and in the non-linear character of this elect. This is particularly true when working at the trace level, where the limits of the detector are being stretched, and the oxygen quenching is at its most efficient. DISSOLVEDGASES IN HPLC 289 UV andjfuorescence eflects in chromatography Fig. 9 shows the effect of dissoIved oxygen on both the detector.baseline and the eluted peak response (height and area) for both UV and fluoresceuCe detectors. The data, made on a dual channel recorder, was co&cted by repetitively injecting the one-peak sampIe every I.8 min using the SP 8CKlO autoinjector. Initially, the mobile phase was helium degassed. At time 0, the helium was replaced by an air sparge. After 16 min, the helium degassing was resumed. The conditions are listed in the figure caption. L A lze sedges SrAeTIIBsPmE -~ETBesPAEGE Fig. 9. UV absorbance detector and fluo rescence detector response to mobile phase oxygen. Column 250 x.4.6 mm I.D.; paSng, l&urn LiChrosorb RP-8; solvent, water-acetonitrZe (25:75); flow-rate, 5.0 ml/min; temperature, 25 “C; detector, Model SP 8310 UV detector at 254 nm with a l&cm path ceU. aad Model SP 770 fluorescence detector at 250 MI exciting wavelength and with 340 ma emission wavelength filter, 1_0,~A full scaIe; sample size, 10~1; sample, naphtbalerte. The UV detector trace exhibits the baseline elevation previously described in association with Figs. 7 and 8. However, the response to the compounds does not change. The fluorescence detector trace .also exhibits a baseline shift. This has not yet been mentioned. The change is in the opposite direction to the W drift. This is because the increased concentration of oxygen quenches the background fluorescence of the mobile phase, whereas in the W detector the oxygen is responsible for in- creased absorbance. The magnitude of this drift is noteworthy. It is about 0.004 PA, or about 40% of the fuIl-scale sensitivity of the detector (0.01 PA full scale). How- ever, the most marked effect of the oxygen is in the sample response, which is only about 20 o/0 of that when the solvent is degassed. _ Fig. 10 also shows these effects, as well as a number of general cha&teristics of the two detectors. Four separate chromatographic runs were made, using both a W and a fluorescence detector on each nm. Thus there are eight chromatograms. 290 S. R BAKAL.YAR, M. P. T. BRADLEY. R. HONGAN!ZN Fig. 10. Characteristics of UV and fluorescence detectors. Cokmn, 250 x 4.6 mm I.D.; packing, lO-+n LiChrosckb RP-8; solvent, water-acetonitrile (40:60) and water-methanol (40:@); flow-rate, 3.0 mI/min; temperature, 25 “C; detector, same as Fig. 9, except UV at 0.01 AUFS aad fk~osescence at 0.5 pA full scale; sample size, 10 ~1; sample, as shown. The upper four chromatograms are run with a mobile phase of 60% acetonitrile in water: The bottom four chromatograms are run under identical conditions, except the mobiIe phase is 60% methanol in water. The four chromatograms on the left were made using helium-degassed mobile phases. The four chromatogmms on the right were made using air sparged (air saturated) mobile phases. A vertical line in- dicates the poik at which the gases were changed. The many aspects of the tiompfex behavior of these two detectors are described below. UV and fluorescent detector response digerences The upper left chromatograms illustrate the well known fact that the two de- tectors respond quite differently to different compounds. Solute 2, nitrobenzene, is not even detected by the fiuorescence detector under thek conditions. These two chromatograms will kerve as a reference against which the other three sets will be compared. . Response d_@erences caused by the solvent The lower left chromatograms are quite different. First, quite obviously be- cause the peaks are more retained. But a close inspection shows that, for some com- pounds, the de&to? response has changed. This is true for both .detectok. The solknt composition of the mobile phase thus has a profomid effect- on response: DISSOLVED GASES IN HPLC 291 BQseiine- and- response changes causeci by dissolved oxygen The upper right .chromatograms, compared to the upper left ones, snow the effects of oxygen. UV detector: baseline change but no peak height or area change. (For a description of the dependence of peak height and area on mobile phase com- position, see ref. 41) Fluorescence: baseline change and peak height and area change. Note that peak 3 has been affected much more than peak 1. Thus an internal standard does not solve the problem. The sensitivity of the fluorescence detector has been reduced-for all peaks, although only slightly in the case of peak 1. The lower right chromatograms, compared to the lower left ones, show sim- ilar but not identical effects of oxygen on the fluorescence detector. Note for example that there is much less change in the relative sizes of peaks 1 and 3, compared to the upper (acetonitrile) chromatograrns. Other eflects of dissolved gases Two major effects of dissolved gases have not yet b&n mentioned : t he changes in refractive index caused by changes in concentrations of various gases, and the changes in pH caused by changes in the concentration of carbon dioxide in unbuffered mobile phases. These effects will be discussed in subsequent papers, but a few com- ments will be made here. The refractive index of the mobile phase is a function of the types and con- centrations of dissolved substances, including gases. The effects of gases are small, but it is likely that, under some circumstances, careful attention. to controlling dis- solved gases will produce more stable refractive index detector baselines, in effect improving sensitivity. The pH of the mobile phase is an important retention variable. Since dissolved carbon dioxide brings the pH 7.0 of pure water down to about 5.5, changing carbon dioxide levels are expected to effect retention times of basic compounds. UV and fluorescence detector response is also a function of pH, since the molar extinction coefficient is often pH dependent. Experiments of the type shovm in Fig. 10 are ex- pected to reveal such effects readily, when appropriate sample types are chosen. C6NCLUSIONS We have demonstrated that gases dissolved in the mobile phase play a complex role in HPLC. Their non-linear solubility behavior in binary mixtures tends to cause the-formation of gas bubbles when solvents are mixed, a process which degrades pump and detector performance. Whereas the gas bubble problem has its solution in the reduction of the con- centration level of all gases, most of the other problems can be dealt with by con--- trolling just the oxygen concentration. Oxygen afkcts the standing signal ‘levels of both UV and fluorescence detectors. It also alfAts the response characteristics of the latter. This is.a complex phenomenon involving solute type, mobile phase solvent composition and oxygen concentration. Much work remains to be-done before we will understand what is happening. In spite of limited understanding of the mech-- anisms, one can at least attempt to hold the oxygen level. constant so as to provide reproducible chromatographic results. The most effective constancy is to take .the oxygen concentration to zero. 292 S. R. BAKALYAR, M. P. T. BRADLEY, R HONGANEN The effect of dissolved gases on the performance of refractive index detectors is yet to be explored, as are the consequences of varying carbon dioxi&.umcentrations on solute retention and on detector response characteristics. The techniques discussed here can be readily applied to such studies. Much remains to be learned about gas solubility, and no doubt the main source of such knowledge wiii be the continued use of gases as probes to study the structure of Iiquids. The increasing use of ternary mixtums in both isocratic and gradient sep- arations provides an even more difiicult theoretical problem In any event, there are a variety of techniques available for controlling gas concentrations and thereby elimi- nating problems which would otherwise exist. The new helium degassing method described appears to be a particularly simple and effective method.. . . NOTE ADDED IN PROOF Subsequent to the submission of this paper, R. W. Cargill sent us a pertinent private communication. His data will be published in J. Chem. GC., Faraday Tram. I (1978) It indicates that the solubility behavior of helium in aqueous alcohol mixtures is very similar (non-Linear) to that shown in Fig. 2 for oxygen, although the magnitude of the solubility is lower. We had postulated two explanations. of why. helium degassing eliminates bubble problems (see the “gas control methods” section). The Cargill data suggests that the low-solubility postulate is correct, and not the linear-behavior postulate. The Cargih paper also makes a significant con- tribution to the understanding of water structure, and will be of interest to those who are working to elucidate retention mechanisms in reversed-phase liquid chro- matography. REFERENCES 1 5. 3. -Kirkland, in J. J. Kirkland (Editor), hhiern Practiie of Liqud Chromatography, Wiley- Interscience, New York, 1971, pp_ 182-184. 2 L. R Snyder and J. J. Kirkhd, Intr&ction to Modern Liquid Chromatography, Wiley, New York, 1974, pp. 95-97 and 273. 3 S. G. Perry, R Amos and P. 1. Brewer, Practid Liqd Chromatography, Plenum, New York, 1972, p_ 174. 4 R E. Leitch, J. Chrunmto~. Sci.. 9 (1971) 531. 5 S. R Bakzdyar, Amer. hb., 10, No. 6 (1978) 43. 6 L. R Snyder, in J. J. Kirkland (Editor), Modern Practice of Liquid Chromatography, Wiley-Inter- science, New York, 1971. pp. 2X-227. 7 L. A. Stemson ard W. J. DeWitte, f. .Chromatogr., 137 (1977) 305. 8 M. A. Fox and S. W. Staky, Am?. Chem., 48 (1976) 992. 9 k T. C~rlain and J. S. Marlow, J. Chronratogr Sci.. 15 (1977) 29. 10 F. W. Karask, ResJDeveIop., 28 (1977) 38. 1 I L. B. Sybrandt and E. F. MOntoya, Amer. Lab., 9, No. 8 (1977) 79. 12 5. R..BzkaIyar R Mcllwritik and E. Roggendosf; J, Chronzmogr., 142 (1977) %53. 13 D. L. Saunders, L Gtromatugr. Sci., 15 (1977) 129. 14 V. E..D&‘Ova. .M. B. Denton and M. F. Burke, Anal. Chem.. 46 (1974) 1363. 15 D. M. Himmelbkm, J,.Ckem._Eng. Da&, 5 (1960) lo 16 R Batt@o and H. L. Clever, Chem. Rev., 66 (1966) 395. 17 E. W&hn Zna R Battino, Ckem_ Rev., 73 (1973) 1. 18 E. Wilhelm, R Battino and R W. Wilcock, CIzem. Rev.. 77 (1977) 219. . DISSOLVED GASES IN HPLC 293 19 A. SeideU, Solubilites of Inorgunic and Metal Organic Compnunds, Van ~osfxand, New York, 3rd ed_. 1940. and supplements. 20 J. H. Hildebrand, J. M. Prausnitz aid R. L. Suxt, Reguhr a& Relcrted solutions, Van Nostrand- Reiiold, New York, 1970, App. 3. 21 H. Stephen and T. Stephen (Editors). Solubiiities of hwrgmic andOrga+c Conzpoum5, Macmihn, New York, 1963. 22 A. Ben-Naim, J. Phys. Chem., 71 (1967) 4002. 23 R.. W. cargill and T. J. Morrison, J. Chem. Sac. Farad. Trarrs. I, 71. (1975) 628. 24 R W. Cargill, J. Ckem. Sot. Fmad. Trcurs. I, 72 (1976) 2296. 25 J. Tokunaga, J. Chem. z&g. Data, 20 (1975) 41. 26 L. R. Snyder, J. Chromfogr., 92 (1974) 223. 27 J. H. Hildebrand and R L. Scott, The Sohbility of Non-dectrolytes, Dover Public&ions, New York, 3rd ed., 1964. 28 J. H. Hi&brand and EL L. Scott, Regular Solrtztixs, Pn&iu~Hall, Wewood CUfs, NJ., 1962. 29 0. A. Kapustina, in L. D. Rozenberg (Editor), Physicazprincipks of Uhr&o& Tedurology, Vol. 1, Plenum, New York, 1973. 30 J. W. Swhnerton, V. J. Liunenbom and C. H. Cheek, Anal. Chem., 34 (lQ62) 483. 31 J. H. Williams and C. D. Miier, Anal. Gem., 34 (1962) 657. 32 D. F. Evans, J.Chem. Sac. (London), (1953) 345. 33 D. F. Evans, J. Chem. Phys., 23 (1955) 1424. 34 A. V. Muck and J. F. Scott, Nature (LxJR&R~, (1956) 587. 35 L. J. Heidt and L. Ekstrom, .l. Amer. Chem. Sac., 79 (1957) 1260. 36 L. J. Heidt and L. Ekstrom, J. Amer. Chem. Sm., 79 (1957) 5587. 37 E. J. Bowen 2nd k H. Wii, Trans. Faraday Sot., 35 (1939) 65. 38 V. Bar and A. We&b, J. Chem. Phys., W 6 (1957) 1412. 39 C. A. Parker and W. J. Barnes, Trmrs. Fmaday Sot_, 82 (1957) 6%. 40 M. 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7170 | https://www.youtube.com/watch?v=izNIep5IyKY | 2.31 Least Number of Coins
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Least Number of Coins
Practice making a specific amount of money with the fewest number of coins.
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Transcript:
let's take a look at the lesson least number of coins it says how do you make 22 cents using the fewest coins well let's look at that number 22. 22 cents all right so we're going to try to use the fewest amount of coins to make 22 cents well i know that a quarter is worth 25 cents a dime is worth 10 cents a nickel is worth 5 cents and a penny is worth 1 cent so 25 cents i can see is already more than 22. so if we tried to use a quarter it would just be too much so we have zero quarters what about dimes could we use dimes to get up to 22 let's see how high we could get with dimes because i know dimes are less than 20 dimes are 10. so how many tens would we need let's see if we have one dime we have 10 cents if we have two dimes we have 20 cents ah so we could have two dimes which is equal to 20 cents and then we need two more cents to get up to 22. and i know that the only coin that is a cent is a penny all right so we need two of those to make two cents so i'm going to put my two up here two pennies so we have twenty plus two is equal to twenty two so we used zero quarters we used two dimes we didn't use any nickels because we wanted to use the fewest amount of coins and then we used two pennies next it asks how do you make 15 cents using the fewest coins so 15 cents all right i know my quarter is 25 my dime is 10 my nickel is 5 and the penny is 1. so would 25 work no 25 is too much so we have zero quarters all right what about 10 cents could we use one of those remember we're starting on the largest side and going down because we want to use the ones that are worth the most so we can use the fewest amount we can use a 10 yes we can use 1 10 because we have 15 cents we could say 10 but if we counted by 10s and tried to use two dimes it would go 10 20 oh 20 would be too much so we just do 10. now can we add a nickel to 10 what is 5 more than 10 let's see 10 11 12 13 14 15. that's exactly what we're looking for we could do one nickel and be right at 15 cents we've only used two coins to make 15 cents did we need any pennies no zero all right zero quarters one dime one nickel and zero pennies this question asks how do you make 77 cents using the fewest coins so i see we have 77 cents we have quarters here that are worth 25 dimes here that are worth 10 nickels here that are worth five and pennies here that are worth one so let's start with our largest one the quarter can we use quarters to make this number yes quarters are less than 77 cents so we know we can use some of those let's count by 25s 25 50 75 100 or in this case one dollar well i know this one is too much so we know we can't have four quarters but i see 75 is less than 77 so we can use three quarters and we will be at 75 cents so if we are at 75 we can count by ones until we get to 77 because i know counting by tens would be too much and counting by fives would be too much let's count by one so we have 75 76 77 so we only need to count by ones twice so we have 75 so one penny would be 76 two pennies would be 77. so we needed two pennies so we had three quarters 75 cents and two more pennies 76 77 so we used zero dimes and zero nickels so it was three quarters which was 75 cents zero dimes zero nickels and two pennies to make it 77 cents next they ask how do you make 88 cents using the fewest coins so we're going to try and make 88 cents well i remember these are worth 25 dimes are worth 10. nickels are worth 5 and pennies are worth 1. let's count by 25s since this number is greater than 25 25 50 75 i know the fourth one would be 100 and that would be too much so i'll stop at 75 so we can use three quarters three quarters would leave us with 75 cents well could i add a dime and still be under this amount let's try let's add 10 so 75 plus 10 is 85 so we can use a dime two because 85 is less than 88. now if we tried to add a nickel to this it would turn into 90. so that would be too much so we can say zero nickels but we do need some pennies how many pennies do we need to get from 85 to 88 let's see 85 and then if we add one it's 86 2 it's 87 and 3 is 88 so it's three pennies so we have three quarters one dime and three pennies so three quarters one dime zero nickels and three pennies next they ask how do you make 80 cents using the fewest coins well i remember that we could count all the way up to 75 with my quarters 75 is still less than 80. so it went 25 50 75 that's three quarters then if i were to add 10 more to 75 that would turn into 85. can we do that or is that too much yeah that's too much we're looking for 80. so there are zero dimes what about if we have 75 cents and we add five more let's see 5 is 5 is 10. it's 80 cents that's exactly what we needed so we just need one nickel and zero pennies so we have three quarters zero dimes one nickel and zero pennies now they want to know how do you make 18 cents using the fewest coins i know my quarter is 25 my dime is 10 my nickel is 5 and a penny is 1. well i know that 25 is already more than 18 so we're going to have zero quarters what about a dime is 10 less than 18 yeah so we can do one dime all right because if we did two dimes it would be 20 cents which would be too much so now can we add a nickel can we add five to our ten yeah because it would turn it into fifteen so let's add one nickel great so now we're at 15 cents if we did another five we'd be at 20 and that would just be too much so let's move on to pennies well let's see how many pennies we need to add if we're at 15 16 17 18. we're at 18 cents how many pennies did we add one two three so it is zero quarters one dime one nickel and three pennies let's take a look at the next problem how do you make 63 cents using the fewest coins i know when i count by 25s it goes 25 50 then 75 but 75 is too much so i know we can have two quarters now let's try counting by tens 50 60 oh 70 would be too much so it's just one dime and then if i added 5 to 60 it would be 65 which would be too much then 60 plus 3 would be 63 so i will add three pennies two quarters one dime zero nickels and three pennies here we have how do you make 92 cents using the fewest coins well i know that 92 cents is a pretty big number so let's start with our quarters which are worth 25 cents so we have 25 50 75 100. well 100 is too much i know we won't have four quarters but we can have three quarters which is 75 cents now if we counted on by 10 more since dimes are our next coin our next problem shows how do you make 92 cents using the fewest coins well quarters are worth 25 dimes are worth 10. nickels are worth five and pennies are worth one i'm going to start with the largest one quarters because 92 is a pretty large number 25 50 75 i know four quarters would be a dollar or 100 cents and we are not at 100 cents over here so i know that three quarters is the most i can use and that is 75 cents now let's try our dimes if we add 10 to 75 it turns into 85. if we added 10 more it would be 95 which would be too much so we know we just need one dime now let's see nickels i know if i add 5 to 85 it will turn into 90. so i can do one nickel i cannot add another nickel because that will be greater than 92 so if i have 90 and i need to get to 92 i just need to add two pennies that is three quarters one dime one nickel and two pennies next we have how do you make 93 cents using the fewest coins well i remember that three quarters is 75 cents one dime is 10 so let's count up by tens 75 85 oh it would be 95 but we're looking for 93 so that would be too high so just one time then i remember 85 plus five is equal to 90 so we can do one nickel and then how many pennies do we need to get from 90 to 93 let's see 90 91 92 93 that's three pennies so we have three quarters one dime one nickel and three pennies last we have how do you make 22 cents using the fewest coins well 22 cents is less than 25 cents so we can't use any quarters so that will be zero but we can try counting by tens 10 20 oh have to stop there so i could do two dimes which is 20 cents and then if i were to try to add 5 it would be 25 which would be too much so we don't need any nickels but how many pennies do we need to get from 20 to 22 yes just two pennies so zero quarters two dimes zero nickels two pennies thanks for your help friends |
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Is the integral of $\frac{1}{x}$ equal to $\ln(x)$ or $\ln(|x|)$?
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$\begingroup$
The inconsistency I see between mathematical subjects is really confusing me. I understand that it isn't possible for $e^x$ to be less than zero for real $x$, which is probably why they say that the integral is $\ln(|x|)$.
Before I ramble on too much, I just want to ask: Is there a set of guidelines to follow to help me choose whether to let the integral of $\frac{1}{x}$ equal to $\ln(x)$ or $\ln(|x|)$?
Thanks, Aralox
integration
logarithms
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edited Jan 11, 2015 at 6:11
Anixx
10.3k11 gold badge3333 silver badges5757 bronze badges
asked Nov 11, 2012 at 3:23
AraloxAralox
22322 silver badges66 bronze badges
$\endgroup$
15
3
$\begingroup$ And no: it is true that $\,e^x>0\,$ for all reals. $\endgroup$
DonAntonio
– DonAntonio
2012-11-11 03:25:49 +00:00
Commented Nov 11, 2012 at 3:25
2
$\begingroup$ Well @Aralox: you can write the argument of the logarithm (to any base) without the absolute value if you make clear, of if you're given, that the argument is positive, otherwise you must use the abs. value. You can drop this requirement if you work with the complex logarithm function, which perhaps you haven't yet studied. $\endgroup$
DonAntonio
– DonAntonio
2012-11-11 03:35:33 +00:00
Commented Nov 11, 2012 at 3:35
1
$\begingroup$ Yes! That's the answer I am looking for! I already understand that the abs brackets can be dropped if x is positive, but I was looking for more. I remember studying the principal logarithm a few months back, and understand that Ln(z) = ln|z| + iArg(z). Could you show me how to integrate 1/x using it? $\endgroup$
Aralox
– Aralox
2012-11-11 03:44:55 +00:00
Commented Nov 11, 2012 at 3:44
1
$\begingroup$ Hmm if I integrate 1, and get x + C, wouldnt f(x) = x + 5 for 1 < x < 4 and x + 15 elsewhere be perfectly valid too? i.e. just saying 'x + C' already implies that you can let C equal to whatever wherever you like? $\endgroup$
Aralox
– Aralox
2012-11-11 22:45:08 +00:00
Commented Nov 11, 2012 at 22:45
1
$\begingroup$ Once again, @RahulNarain: the moment you determine some conditions on the function you already determine the integration constant! I really can't understand what the problem here is...though perhaps I'm missing something, of course. $\endgroup$
DonAntonio
– DonAntonio
2012-11-12 03:26:43 +00:00
Commented Nov 12, 2012 at 3:26
| Show 10 more comments
7 Answers 7
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22
$\begingroup$
Strictly speaking, any function of the form $$F(x) = \begin{cases} \ln x + c_1 & \text{if } x > 0, \ \ln (-x) + c_2 & \text{if } x <0 \end{cases}$$ defined over $\mathbb R \setminus {0}$ is a valid antiderivative. Verify this by differentiating $F(x)$ and getting $F'(x) = 1/x$ back for any $x \ne 0$.
The usual formula, $\ln\lvert x\rvert + c$, is what you get when you pick $c_1 = c_2 = c$. Alternatively, as @Joe says, when you're only considering positive $x$, you can just write $\ln x + c$.
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edited Nov 11, 2012 at 3:42
answered Nov 11, 2012 at 3:35
user856user856
$\endgroup$
3
$\begingroup$ Bravo. Because the domain of $1/x$ consists of two connected components, the constant of integration must be allowed to be different on the two. Not to do so ignores the message and content of the theorem that tells us what the antiderivatives of a continuous function defined on an interval may be. $\endgroup$
Lubin
– Lubin
2012-11-13 03:57:02 +00:00
Commented Nov 13, 2012 at 3:57
$\begingroup$ and if you pick $c_2= c_1 + i \pi$ then you get the answer $\log z$ with the conventional branch choice of the complex logarithm. $\endgroup$
Fabian
– Fabian
2014-07-07 08:19:59 +00:00
Commented Jul 7, 2014 at 8:19
$\begingroup$ Also, if someone can explain what DonAntonio was trying to tell me in the comments on the question, I'd much appreciate it... It's hard to communicate on the internet. $\endgroup$
user856
– user856
2015-01-11 07:13:28 +00:00
Commented Jan 11, 2015 at 7:13
Add a comment |
5
$\begingroup$
In general, it is safe to always write:
$$\int \frac{1}{u} \ du = \ln \left| u \right| +C $$ where $C$ is some constant.
However, if the function is always positive $\forall x \in \Bbb R\setminus {0}$ (assuming that's what you're integrating over), you can drop the absolute value.
Addendum of Complex Logarithm
I think this will be helpful since you wanted to know how to integrate with the complex log. Also, this should be a good read. I'd start around page 74 for what you're looking for.
I feel it might help to note that if we say $\log z$ is any logarithm along some branch $B$, then $(\log z)' = \dfrac{1}{z} \forall z$ not on $B.$ However, no matter how we define the complex logarithm, there will always be some branch that is not holomorphic.
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edited Nov 11, 2012 at 6:00
answered Nov 11, 2012 at 3:30
JoeJoe
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$\endgroup$
3
$\begingroup$ Thanks Joe! that almost immediately answered my question about how it its multivalued, and what exactly the 'principal value' means. Also, I had a read of the wikipedia article a few times, but I don't have a good grasp of what a branch is. Could you give it to me in simple terms? $\endgroup$
Aralox
– Aralox
2012-11-11 04:54:59 +00:00
Commented Nov 11, 2012 at 4:54
$\begingroup$ A branch is some portion of the range of a mutlivalued function in which the function is single-valued. A branch cut is a curve in $\Bbb C$ where a multifunction (a multivalued function) is discontinuous. For learning more about branch cuts, take a look at this question: math.stackexchange.com/questions/37764/… $\endgroup$
Joe
– Joe
2012-11-11 05:35:47 +00:00
Commented Nov 11, 2012 at 5:35
$\begingroup$ Thanks again. Would I be right in saying that by limiting the Arg(z) in "Ln(z) = ln|z| + Arg(z)i" to (-pi, pi], that would be a branch? $\endgroup$
Aralox
– Aralox
2012-11-11 13:04:27 +00:00
Commented Nov 11, 2012 at 13:04
Add a comment |
3
$\begingroup$
Equal to $\ln(|x|)$+c, in which $c$ is constant.
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answered Nov 11, 2012 at 3:26
user48941user48941
$\endgroup$
3
1
$\begingroup$ @DonAntonio: I'd say that $\ln x+C$ makes little sense. Why choose the same constant for $x<0$ and $x>0$? (See Rahul's and countinghaus' answer.) $\endgroup$
Hendrik Vogt
– Hendrik Vogt
2012-11-11 07:09:24 +00:00
Commented Nov 11, 2012 at 7:09
2
$\begingroup$ I've already addressed this question...and I find very weird that you keep on asking me the same under other people's answers. $\endgroup$
DonAntonio
– DonAntonio
2012-11-12 00:59:25 +00:00
Commented Nov 12, 2012 at 0:59
$\begingroup$ This is incomplete, if not incorrect. See Rahul's answer. $\endgroup$
Pedro
– Pedro ♦
2015-01-11 08:04:06 +00:00
Commented Jan 11, 2015 at 8:04
Add a comment |
3
$\begingroup$
The correct answer is$\int \frac1x dx= \ln |x| +C$
The absolute value is sometimes omitted in ODE problems. As for guidelines I would say analyze the problem and see if values of x will be out of the domain when solved.
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answered Nov 11, 2012 at 3:42
jmurray1241jmurray1241
9566 bronze badges
$\endgroup$
3
$\begingroup$ @DonAntonio: I'd say that $\ln x+C$ makes little sense. Why choose the same constant for $x<0$ and $x>0$? (See Rahul's and countinghaus' answer.) $\endgroup$
Hendrik Vogt
– Hendrik Vogt
2012-11-11 07:08:47 +00:00
Commented Nov 11, 2012 at 7:08
$\begingroup$ I've already addressed this question. $\endgroup$
DonAntonio
– DonAntonio
2012-11-12 00:58:41 +00:00
Commented Nov 12, 2012 at 0:58
$\begingroup$ This is incomplete, if not incorrect. See Rahul's answer. $\endgroup$
Pedro
– Pedro ♦
2015-01-11 08:04:08 +00:00
Commented Jan 11, 2015 at 8:04
Add a comment |
2
$\begingroup$
I think the introduction of complex variables here is unnecessary. The function $1/x$ is a perfectly well-defined function on $\mathbb{R} \setminus {0}$ and we can ask if it has an antiderivative on that set. On the positive real axis, $\log(x)$ is an antiderivative. On the negative real axis, $\log(-x)$ is an antiderivative (and $\log(x)$ doesn't make sense). This is confusing, so we write $\log(|x|)$ so that we don't have to remember. Note that $\log(|x|) + C$ is not good notation, since the constant can be different on the negative and positive real axes.
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answered Nov 11, 2012 at 5:57
user29743user29743
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$\begingroup$
I remember studying the principal logarithm a few months back, and understand that Ln(z) = ln|z| + iArg(z). Could you show me how to integrate 1/x using it?
Right, and this is why the "most" correct answer is that $\ln(x)$ would be the antiderivative, considered as a function over the complex numbers (still not defined at zero). So that (as an example) one could use $\ln(-x)=\ln(x)+i\pi$ in the manner of Rahul's answer (with $c_2=i\pi$, which Euler wonderfully treats as a constant to ignore in some of his papers).
This has the problem that it's a multivalued function, but has the advantage that this enables one to choose a consistent branch to integrate over all sorts of interesting contours.
For instance, then the integral of $1/x$ over the unit circle (in $\mathbb{C}$) has to follow from $\ln(1)=0$ to $\ln(1)=2\pi i$ giving $2\pi i-0 = 2\pi i$, where I've here really followed the function up the Riemann surface, as it were - obviously not a function any more. Not how we usually think of the FTC, granted! But that really is the value of the contour integral.
See for one CAS' solution to this issue.
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edited Nov 13, 2012 at 3:27
answered Nov 11, 2012 at 4:28
kcrismankcrisman
2,3052020 silver badges3535 bronze badges
$\endgroup$
5
$\begingroup$ Thank you, that makes so much sense. So in the end I can leave off the abs brackets if I state that my constant is in C rather than R. Could you please give me some examples of its multivalued nature? It's confusing me a little. Unfortunately I don't quite understand the curve integral/Riemann surface you described (although it sounds quite interesting, and I believe I am going to cover it in one of my future units). I have studied how to integrate these things using the sum of residues however. What does FTC stand for? $\endgroup$
Aralox
– Aralox
2012-11-11 04:50:43 +00:00
Commented Nov 11, 2012 at 4:50
1
$\begingroup$ @Aralox and kcrisman: I'm sorry to say this, but I think that this answer doesn't make so much sense. There's no $\ln x$ for negative $x$ in Rahul's answer, and $\ln(-x)=\ln x+i\pi$ is true only for certain branches of the complex logarithm. Moreover, $\ln x$ isn't the antiderivative, but only an antiderivative (cf. Rahul's answer). Finally, if I only look at the question, then taking about complex number only introduces unnecessary confusion here. I'd go with Rahul's answer, Aralox. $\endgroup$
Hendrik Vogt
– Hendrik Vogt
2012-11-11 07:15:01 +00:00
Commented Nov 11, 2012 at 7:15
$\begingroup$ I can see how there should rightly be separate constants for the positive and negative real axis, and that this assumes that they are the same (like a few of the existing answers below). However kcrisman's answer gave me the complex viewpoint that I was looking for. If somehow I could combine his, Rahul's and Joe's answers together, it would be perfect. $\endgroup$
Aralox
– Aralox
2012-11-11 13:11:15 +00:00
Commented Nov 11, 2012 at 13:11
$\begingroup$ Actually I'm rethinking this again - the constant does take into account the pos/neg x axis if it is complex $\endgroup$
Aralox
– Aralox
2012-11-11 22:47:36 +00:00
Commented Nov 11, 2012 at 22:47
$\begingroup$ I'm not suggesting $\ln(-x)=\ln(x)+i\pi$ all the time, just that in Rahul's answer that's how one could interpret this. But I see how that could have been misinterpreted, so I'll edit the answer - thanks. $\endgroup$
kcrisman
– kcrisman
2012-11-13 03:25:37 +00:00
Commented Nov 13, 2012 at 3:25
Add a comment |
0
$\begingroup$
On the negative axis their difference is a constant. Both are antiderivatives.
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answered Jan 11, 2015 at 7:44
orangeskidorangeskid
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$\endgroup$
3
$\begingroup$ Hmm. As real functions only one is defined on the negative axis. As complex functions one is not holomorphic and thus cannot be an antiderivative. I don't see how to reconcile these facts with your answer. $\endgroup$
Jyrki Lahtonen
– Jyrki Lahtonen
2015-01-11 10:37:56 +00:00
Commented Jan 11, 2015 at 10:37
$\begingroup$ $\log(x)$ on the negatives is the restriction of a holomorphic branch of $\log(z)$ on $\mathbb{C}\backslash{0}$, preferably mapping to $\mathcal{Im} \zeta = \pm \pi$ so it coincides on $(0,\infty)$ with the usual $\log$. For instance, $\log(-3) = \log 3 + i \pi$. The equality $\log(z)' = \frac{1}{z}$ ( for any branch one takes) is valid for complex derivatives so also for real derivatives. $\endgroup$
orangeskid
– orangeskid
2015-01-11 11:01:30 +00:00
Commented Jan 11, 2015 at 11:01
$\begingroup$ I know what you mean. My point was that $\log(|z|)$ is nowhere holomorphic, and it wasn't IMO clear exactly which two functions you refer to, when you say Both are antiderivatives. $\endgroup$
Jyrki Lahtonen
– Jyrki Lahtonen
2015-01-11 14:24:30 +00:00
Commented Jan 11, 2015 at 14:24
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7172 | https://www.doubtnut.com/qna/30689108 | How many monochlorobutanes will be obtained on chlorination of n-butane?
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The correct Answer is:B
To determine how many monochlorobutanes can be obtained from the chlorination of n-butane, we can follow these steps: Step 1: Understand the Structure of n-Butane n-Butane has the molecular formula C4H10 and its structure can be represented as: CH3−CH2−CH2−CH3 Step 2: Identify the Hydrogen Atoms In n-butane, there are: - 3 terminal hydrogen atoms (attached to the two CH3 groups) - 6 hydrogen atoms attached to the three CH2 groups Step 3: Chlorination Process During chlorination, one hydrogen atom is replaced by a chlorine atom (Cl). This can occur at different positions in the n-butane molecule. Step 4: Determine Possible Monochlorobutane Products 1. Chlorination at Terminal Carbons (CH3 groups): - If we replace one of the terminal hydrogen atoms from the CH3 group, we will get: CH2Cl−CH2−CH2−CH3 - This product is called 1-chlorobutane. 2. Chlorination at Internal Carbons (CH2 groups): - If we replace one of the hydrogen atoms from the first CH2 group, we will get: CH3−CHCl−CH2−CH3 - This product is called 2-chlorobutane. Step 5: Identify Unique Products - The chlorination at the terminal CH3 groups gives the same product regardless of which terminal hydrogen is replaced. - The chlorination at the internal CH2 groups also gives the same product regardless of which internal hydrogen is replaced. Conclusion Thus, the two unique monochlorobutane products obtained from the chlorination of n-butane are: 1. 1-chlorobutane 2. 2-chlorobutane Therefore, the total number of monochlorobutanes obtained is 2. Final Answer 2 monochlorobutanes will be obtained on chlorination of n-butane. ---
To determine how many monochlorobutanes can be obtained from the chlorination of n-butane, we can follow these steps:
Step 1: Understand the Structure of n-Butane
n-Butane has the molecular formula C4H10 and its structure can be represented as:
CH3−CH2−CH2−CH3
Step 2: Identify the Hydrogen Atoms
In n-butane, there are:
- 3 terminal hydrogen atoms (attached to the two CH3 groups)
- 6 hydrogen atoms attached to the three CH2 groups
Step 3: Chlorination Process
During chlorination, one hydrogen atom is replaced by a chlorine atom (Cl). This can occur at different positions in the n-butane molecule.
Step 4: Determine Possible Monochlorobutane Products
1. Chlorination at Terminal Carbons (CH3 groups):
- If we replace one of the terminal hydrogen atoms from the CH3 group, we will get:
CH2Cl−CH2−CH2−CH3
- This product is called 1-chlorobutane.
Chlorination at Internal Carbons (CH2 groups):
If we replace one of the hydrogen atoms from the first CH2 group, we will get:
CH3−CHCl−CH2−CH3
This product is called 2-chlorobutane.
Step 5: Identify Unique Products
- The chlorination at the terminal CH3 groups gives the same product regardless of which terminal hydrogen is replaced.
- The chlorination at the internal CH2 groups also gives the same product regardless of which internal hydrogen is replaced.
Conclusion
Thus, the two unique monochlorobutane products obtained from the chlorination of n-butane are:
1. 1-chlorobutane
2. 2-chlorobutane
Therefore, the total number of monochlorobutanes obtained is 2.
Final Answer
2 monochlorobutanes will be obtained on chlorination of n-butane.
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7173 | https://tatoeba.org/en/sentences/show/10261144 | Sabía que no era verdad. - Spanish example sentence - Tatoeba
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Sentence #10261144
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Sentence #10261144 — belongs to Seael
Sabía que no era verdad.
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I knew that it wasn't true.
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I knew it wasn't true.
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Sabía que no era cierto.
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Je savais que ce n'était pas vrai.
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Я знал, что это неправда.
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Sabía que no era verdad.
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7174 | https://www.reactgroup.org/toolbox/understand/antibiotics/how-do-antibiotics-work/ | Skip to content
Automatic translation by Google. Limitations and info.
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Understand – Antibiotics
How do antibiotics work?
Antibiotics disrupt essential processes or structures in the bacterial cell. This either kills the bacterium or slows down bacterial growth. Depending on these effects an antibiotic is said to be bactericidal or bacteriostatic.
Bactericidal and bacteriostatic antibiotics
A bactericidal antibiotic kills the bacteria while the bacteriostatic antibiotics stop bacterial growth without killing them. The human immune system is then needed to clear the infection.
Antibiotic targets in bacteria
There are several classes of antibiotics with different mechanisms of action and bacterial targets. In principal, there are three main antibiotic targets in bacteria:
The cell wall or membranes that surrounds the bacterial cell
The machineries that make the nucleic acids DNA and RNA
The machinery that produce proteins (the ribosome and associated proteins)
These targets are absent or structurally different in human and mammalian cells, which means that antibiotics usually do not harm our cells. However, antibiotics can in some cases have unpleasant side effects. Read more under Why should I care? – Risks for the individual and society.
Narrow-spectrum and broad-spectrum antibiotics
Antibiotics can either have a narrow or broad spectrum of activity. Narrow-spectrum antibiotics are more specific and are only active against certain groups or strains of bacteria. Broad-spectrum antibiotics instead inhibit a wider range of bacteria. Narrow-spectrum antibiotics are generally preferable since the effect on other non-disease causing bacteria is more limited. Unfortunately, broad-spectrum antibiotics are often used since it can be difficult for doctors to diagnose the infectious agent due to a lack of training and/or available diagnostic tools. For more information, see How did we end up here – Use and inappropriate use – In human medicine.
See also these selected resources for more details on different antibiotics and their mechanisms of action.
Selected Resources
| | |
--- |
| Resource | Description |
| Eric’s Medical Lectures: Mechanisms and classification of antibiotics | Video. Narrated lecture about antibiotics and their mechanism of action (24 min, YouTube). |
| How antibiotics work | Video. Short video describing the difference between bacteriostatic and bactericidal antibiotics. Outlines why antibiotics are not effective against viruses (3 min, YouTube). |
| A brief overview of classes of antibiotics | Fact sheet. Short description of different classes of antibiotics and their mode of action. |
| Antimicrobial Resistance Learning Site – Pharmacology | Educational material. Learn about concepts related to medical use of antimicrobials and of resistance. |
More from "Antibiotics"
How do antibiotics work?
{280251:JSKG3QWK};{280251:T4EC8RPX};{280251:NCDNQVDB};{280251:7JG6BKN6} nature default 0 11290 |
7175 | https://stackoverflow.com/questions/14177618/1n-exponentiation-in-c | math - (-1)^n exponentiation in c++ - Stack Overflow
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(-1)^n exponentiation in c++
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Asked 12 years, 8 months ago
Modified12 years, 8 months ago
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Consider a convergent serie in the form:
cpp
sum(((-1)^n)something)
where n is the index of iteration (n goes from 1 to infinity).
If we implement direclty the formula, we have std::pow(-1, n) but is there a more "rapid" algorithmic trick to implement that ?
c++
math
optimization
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edited Jan 6, 2013 at 7:13
Ali
58.7k 31 31 gold badges 177 177 silver badges 274 274 bronze badges
asked Jan 5, 2013 at 23:34
VincentVincent
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1 You can always use a boolean condition if pow is too slow.Qaz –Qaz 2013-01-05 23:35:58 +00:00 Commented Jan 5, 2013 at 23:35
Have you benchmarked? How do you know the compiler doesn't automatically optimize away the exponential?Mahmoud Al-Qudsi –Mahmoud Al-Qudsi 2013-01-05 23:36:11 +00:00 Commented Jan 5, 2013 at 23:36
1 How is that convergent? Won't the sum flicker between -something and 0 as n goes from 1 to infinity?mattjgalloway –mattjgalloway 2013-01-05 23:51:43 +00:00 Commented Jan 5, 2013 at 23:51
@mattjgalloway I’m sure something isn’t a variable here, but just a placeholder for a term that is of no matter for this question.Lumen –Lumen 2013-01-06 00:57:58 +00:00 Commented Jan 6, 2013 at 0:57
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Check whether n is even or odd,
cpp
(n % 2 == 0) ? 1 : -1;
does it. If you want to avoid a branch,
cpp
1 - 2(n & 1)
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answered Jan 5, 2013 at 23:36
Daniel FischerDaniel Fischer
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Mats Petersson
Mats PeterssonOver a year ago
I'm pretty sure (n & 1)?-1:1 is the faster method. I'd worry that % 2 makes it actually divide (if n is not unsigned), and multiplication most likely does lead to true multiplication, which is at least a few cycles on a modern processor. gcc at the very least does a good job at sorting out ternary expressions without branching.
2013-01-05T23:42:50.533Z+00:00
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Daniel Fischer
Daniel FischerOver a year ago
@MatsPetersson If the compiler doesn't replace a % 2 with an & 1 (on a two's complement machine), throw it away, it's garbage. The multiplication should only remain one if bit-shifting isn't faster on the platform in question. But I wouldn't expect the branchless code to be faster either. Should both be pretty much the same.
2013-01-05T23:47:21.447Z+00:00
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Mats Petersson
Mats PeterssonOver a year ago
gcc does a and with one, but adds another 4 instructions to cope with the fact that n may be negative. Make it unsigned [assuming that's valid] and it doesn't. movl %edi, %eax shrl $31, %eax leal (%rdi,%rax), %esi andl $1, %esi subl %eax, %esi (That's probably looking very messy in a comment!)
2013-01-05T23:53:19.913Z+00:00
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Daniel Fischer
Daniel FischerOver a year ago
@MatsPetersson Oh four-letter-word, it does indeed. The one time I heed the "write % 2 if you mean to test parity, the compiler will sort it out" exhortations, I shouldn't have, and stuck to instinct.
2013-01-05T23:57:51.697Z+00:00
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Daniel Fischer
Daniel FischerOver a year ago
@Tinctorius I can't deny that. For replacing the modulo operation on signed integers, of course two's complement is a requirement.
2013-01-06T00:32:20.447Z+00:00
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I'm assuming that sum(((-1)^n)something) is pseudocode, and n is a variable bound by sum.
Let's extend that notation to sum(n <- [0,1,2,3..], ((-1)^n)f(n)). Your best option would probably be to first split this into two sums, that you add together:
cpp
sum(n <- [0,2..], ((-1)^n)f(n)) + sum(n <- [1,3..], ((-1)^n)f(n))
In the first term, n is always even, so (-1)^n will always be +1. Analogously, in the second term, it will always be -1. We can now rewrite this as follows:
cpp
sum(n <- [0,2..], f(n)) + sum(n <- [1,3..], -f(n))
Since every term in the second sum is multiplied by a constant, we can move that constant out of the sum:
cpp
sum(n <- [0,2..], f(n)) - sum(n <- [1,3..], f(n))
Now, let's make sure these sums take the same sequences of indices, and substitute 2m and 2m+1 for n:
cpp
sum(m <- [0,1..], f(2m)) - sum(m <- [0,1..], f(2m+1))
Now we can unite these sums again:
cpp
sum(m <- [0,1..], f(2m) - f(2m+1))
Or, if you want pseudo-C:
cpp
T result = 0;
for(m = 0; m < limit; m+=2) {
result += f(m);
result -= f(m+1);
}
This saves you a multiplication by +1 or -1, as most seem to suggest here. Since your sequence is convergent, taking an extra term should not negatively influence the correctness of the answer.
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edited Jan 6, 2013 at 0:11
answered Jan 6, 2013 at 0:03
user824425 user824425
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Yeah, there is a magic trick: (-1)^n == 1 if and only if n is even, and (-1)^n == -1 if and only if n is odd. Thus:
cpp
int p = (n % 2 == 0) ? 1 : -1;
sum(psomething)
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answered Jan 5, 2013 at 23:38
LumenLumen
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If you are doing this in a loop, you could simply do:
```cpp
x = 1; // Assuming we start on n = 0
for(...) // or while(...)
{
sum += x something;
x = -x;
}
```
This is most likely a lot faster than doing checks on n - of course, it DOES assume that all n values are iterated over, and you are not skipping a few here and there...
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answered Jan 5, 2013 at 23:46
Mats PeterssonMats Petersson
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The term ((-1)^n)something evaluates to -something for odd n, or something for even n:
n & 1 ? -something : something
If something is a constant value, then sum(((-1)^n)something) evaluates to -something when the last value of n is odd, or 0 for an even number of summands:
n & 1 ? -something : 0
In this case, the serie would not be convergent.
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edited Jan 6, 2013 at 0:49
answered Jan 5, 2013 at 23:39
GOTO 0GOTO 0
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Mats Petersson
Mats PeterssonOver a year ago
My interpretatin was that "something" is not a constant.
2013-01-05T23:47:32.743Z+00:00
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GOTO 0
GOTO 0Over a year ago
My interpretation is that average developers are really bad at maths.
2013-01-05T23:49:21.293Z+00:00
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Mats Petersson
Mats PeterssonOver a year ago
The term "convergent series" does however imply that it's not a constant something, as that will just oscilate, as you describe, between 0 and -something - that's not convergence in my math book - but that was more about thirty years ago, so maybe "new math" doesn't have the same interpretation of convergence.
2013-01-05T23:55:53.527Z+00:00
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user824425
user824425Over a year ago
@ft1: It's safe to assume that the OP really meant f(n) by something. Combined with the guarantee that the series is convergent, the function f must either be constantly 0, or not constant at all.
2013-01-06T00:20:12.307Z+00:00
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ypercubeᵀᴹ
ypercubeᵀᴹOver a year ago
Hey @MatsPetersson and ft1, you are both saying the same thing. One is saying that "if something is constant, then the series is not convergent", the other is saying "if the series is convergent, then something is not constant". And (according to my even older math books) these two statements are equivalent.
2013-01-06T00:22:57.09Z+00:00
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7176 | https://kidscarehomehealth.com/selective-mutism/ | Close Search
Blog
Selective Mutism – Symptoms, Causes, and Treatment
By KidsCare Home HealthJuly 28, 2025No Comments
According to the Selective Mutism Association1, selective mutism is “an anxiety disorder characterized by a person’s inability to speak in certain social settings such as at school, work, or in the community.” It can manifest at any age, but it most often begins in children aged two to four years old. However, the disorder may not become fully recognized until the child reaches nursery or preschool. Selective mutism is not a choice or a result of intense shyness. Rather, individuals with selective mutism are literally unable to speak even when remaining silent can result in negative consequences, like shame, social ostracism, or punishment.
What is Selective Mutism?
Selective mutism is a consistent failure to speak in specific social situations despite being able to talk elsewhere, which explains the “selective” aspect of the condition. Classified as an anxiety disorder, this mental health condition prevents children (or adults) from talking in certain situations because of fear or anxiety. There are varying degrees of selective mutism, from not speaking but participating fully in activities and appearing social to refusing to speak or participate in activities altogether.
Who Does Selective Mutism Affect?
Selective mutism affects about 1 in 140 young children.2 It is most prevalent in young children. However, in rare cases, adults and teens can suffer from this anxiety disorder as well. Expressive/receptive language disorders, as well as communication disorders, may increase the risk of selective mutism in children. Research also suggests that bilingual children may be at a higher risk of developing the disorder compared to monolingual speakers. Although very rare, traumatic or stressful events may be related to the development of selective mutism. However, most of the individuals with the disorder have no history of traumatic life events.
Signs and Symptoms of Selective Mutism
Selective mutism symptoms often manifest when children begin interacting with people outside their immediate family and comfort zone. For example, a child who speaks comfortably at home may suddenly become silent in public settings or around unfamiliar individuals. This silence is often a response to intense anxiety, rather than a lack of language skills, shyness, or defiance. In other cases, children may appear frozen with fear or completely shut down when expected to speak, and they are unable to express themselves even when they want to.
Other indicators can include:
Struggling to make eye contact when uncomfortable
Presenting as behaviorally inhibited
Relying on pointing, nodding, writing, and other forms of nonverbal communication to answer when spoken to
Speaking through a trusted individual, e.g., whispering an answer to a question to a parent or friend at school
Children diagnosed with selective mutism tend to demonstrate difficulty effectively participating in school or forming friendships with other students their age due to an inability to speak at school or in public. Also, selective mutism symptoms often only occur in specific settings and/or around particular individuals.
Emotional and Social Impact
As we discussed recently with ADHD and connection, mental health disorders can severely impact a child’s ability to socialize or perform in school and other settings. With selective mutism, this is the case in situations outside an individual’s comfort zone. Those affected struggle with speaking in certain situations because of intense anxiety, rather than a language delay, learning disorder, or autism. This fear of speaking can significantly interfere with daily life, particularly in school or other public settings.
What Causes Selective Mutism?
There is no specific cause for selective mutism. In general, it is understood to be “an anxiety disorder related to shyness, social anxiety, and inhibited temperament in which speaking situations are avoided and this avoidant behavior gets reinforced over time.”1
Research also suggests that the following conditions may contribute to a higher risk of selective mutism development:
A genetic link between children with SM and anxious parents or family members
A decreased threshold of excitability in the amygdala, the area of the brain that receives and processes signals of potential threats
Expressive/receptive language and communication disorders
Being bilingual
Co-occurring Conditions
Is selective mutism a disability? Yes, especially in educational or occupational settings where it significantly interferes with communication and daily functioning. There are correlations between the disorder and other mental health disorders, such as selective mutism and autism spectrum disorder. The relation isn’t causal. Rather, it’s just more likely that children with social mutism also suffer from other conditions, such as:
Social anxiety disorder (most common)
Other anxiety disorders
Phobias
Autism spectrum disorder
Separation anxiety
Post-traumatic stress disorder (PTSD)
How is Selective Mutism Diagnosed?
Selective mutism is a mental health disorder that requires diagnosis by a trained professional using the Diagnostic and Statistical Manual of Mental Disorders, 5th Edition. A selective mutism diagnosis must meet the following criteria:
There is a consistent failure to speak in specific social situations where it’s expected, such as school, despite speaking in other situations.
The silence interferes with educational or occupational achievement or social communication.
The inability to speak in social situations extends beyond at least a month (i.e., it continues after the first month of school).
A failure to speak is not due to a lack of knowledge of, or comfort with, the spoken language required in the social situation.
A communication disorder (e.g., stuttering) or another mental disorder, such as autism, schizophrenia, or another psychotic disorder, does not better explain the behavior.
DSM-5 Diagnostic Criteria
To further explain the selective mutism meaning and help health professionals make a diagnosis, the individual should meet the following criteria:
There is a consistent failure to speak in specific social situations where it’s expected, such as school, despite speaking in other situations.
The silence interferes with educational or occupational achievement or social communication.
The inability to speak in social situations extends beyond at least a month (i.e., it continues after the first month of school).
A failure to speak is not due to a lack of knowledge of, or comfort with, the spoken language required in the social situation.
A communication disorder (e.g., stuttering) or another mental disorder, such as autism, schizophrenia, or another psychotic disorder, does not better explain the behavior.
Professional Evaluation
Selective mutism is often diagnosed by an individual’s primary healthcare provider, such as a pediatrician. To reach a diagnosis, they will ask you about your child’s signs and symptoms and review their medical history. They will also review your child’s speech and language development. Bringing your child’s academic reports and teacher comments to the appointment is often helpful. In some cases, the healthcare provider may also want to observe your child at both home and school, or request a video if an in-person evaluation isn’t possible.
Other medical exams may be conducted, such as a neurological exam or a hearing test, as well as an examination of the child’s ears, lips, tongue, and jaws. Overall, the healthcare provider wants to rule out other medical conditions, such as schizophrenia. They may also seek help from other healthcare providers, like a speech-language pathologist (SLP), psychologist, or psychiatrist. The SLP can assess your child’s ability to understand and use language, and a psychologist can investigate emotional issues that may cause the condition.
Treatment and Therapy for Selective Mutism
Most children can overcome selective mutism with appropriate handling and treatment. As with most mental disorders, early intervention is best, as the later the diagnosis occurs in life, the longer treatment may take. The effectiveness of the treatment will depend on several factors, including the duration of symptoms, whether there are additional communication/learning difficulties or anxieties, and the involvement and cooperation of all parties involved.
Cognitive Behavioral Therapy (CBT)
CBT is generally the first-line treatment for selective mutism, and it focuses on identifying and changing unhelpful thought patterns and behaviors that contribute to anxiety. When treating selective mutism, CBT helps individuals become more aware of their anxious thoughts about speaking and understand how these thoughts influence their feelings and behaviors. Then, through gradual, structured exposure to speaking situations they find challenging, individuals can build confidence and reduce anxiety over time.
Speech-Language Therapy
Speech therapy can help individuals overcome any underlying speech problems, if necessary, to improve communication confidence.
School and Family Interventions
Selective autism treatment is most effective as a coordinated effort among family, caregivers, teachers, and healthcare providers. The three pillars of these interventions include parent training, school accommodations, and positive reinforcement. Parent training equips caregivers with tools to reduce pressure around speaking, manage their anxiety, and reinforce progress at home. School accommodations such as reducing performance pressure and offering a safe space or trusted adult for support are essential to treatment. Positive reinforcement helps build confidence by celebrating small steps toward communication, providing a sense of safety and control.
Medication Options
In severe cases, medications, such as selective-serotonin reuptake inhibitors (SSRIs), may be helpful. However, this is typically only recommended when behavioral or speech therapy is not effective on its own.
Selective Mutism vs Other Conditions
Shyness vs Selective Mutism
Selective mutism should not be confused with shyness, as the former is a mental health disorder and the latter is a personality trait. Being shy involves a tendency to withdraw from people, especially strangers. However, shyness does not interfere with one’s daily functioning at the level, if at all, that a mental health disorder does. Almost everyone is shy to some degree, but it does not result in a physical inability to speak.
Selective Mutism and Autism Spectrum Disorder
Selective mutism and autism spectrum disorder are often co-occurring, as autistic individuals are more likely to have selective mutism. However, the two conditions are different, and experts group them into separate categories.
When to Seek Professional Help
Early intervention is essential for managing and treating mental health disorders. When a child speaks comfortably at home but struggles consistently with speaking in specific settings, such as school, social situations, or around unfamiliar people, consider consulting a mental health professional. Selective mutism can impact academic performance, social development, and self-esteem if left untreated. A qualified mental health provider can help create a tailored treatment plan and provide guidance to families and educators to support the child’s progress.
About KidsCare Home Health
KidsCare Home Health is a pediatric home health agency dedicated to serving children with special needs, including those with mental health disorders. With services conveniently available nationwide, we focus on nursing, speech therapy, physical therapy, and occupational therapy, as well as provide case management for children up to 18 years old.
Learn More
Frequently Asked Questions about Selective Mutism
What causes selective mutism?
There is no specific cause for selective mutism. Research suggests that some conditions may contribute to a higher risk, such as bilingualism, decreased threshold of excitability in the amygdala, or expressive/receptive language and communication disorders.
Is selective mutism a disability?
Yes, it is a mental health disability, as it can severely interfere with one’s ability to navigate specific social settings.
Can children outgrow selective mutism?
Children can overcome selective mutism with the proper treatment, care, and support.
How long does treatment take?
The duration of selective mutism treatment depends on several factors: the child’s age, how long the mutism has been present, the severity of the anxiety, and the consistency of support at home and school.
Can selective mutism be cured?
The word “cure” isn’t typically used in clinical settings, as selective mutism is not a disease. As an anxiety-based condition, selective mutism can be effectively treated. Many children do fully overcome it, especially with early, consistent intervention.
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7177 | https://photonics101.com/radiation-and-antennas/finite-dipole-half-wavelength-full-wavelength-double-wavelength.html | The Finite Dipole Antenna - small, half-wavelength and larger
Published in Electrodynamics / Full Electrodynamics / Radiation and Antennas Tags: Far Field / Directivity / Poynting Vector / Dipole / Radiation pattern
One of the most widely used antennas in telecommunication is the dipole antenna. In this problem, we investigate the radiation properties of such an antenna. We first calculate the electromagnetic fields far away from the antenna. Then, as a special cases we discuss half-wavelength, full-wavelength and larger dipole antennas.
As the name suggests, a dipole antenna consist of two terminals (poles), where the radio frequency current is applied by a transmitter or extracted by a receiver. One of the simplest form of dipole antennas consist of two pieces of straight and thin metallic wire, which are oriented on a common axis.
Generally, each of these wires may have different lengths depending on the application. Below an image of such an antenna is shown. In one special case the lengths of both wires are equal to one forth of the wavelength of the radiating wave. Therefore, the length of the whole antenna is equal to half of the wavelength. Thus, this antenna is called half-wave dipole antenna. For the infinitesimal dipole please regard a corresonding worksheet.
Problem Statement
A symmetric and very thin dipole antenna which works at frequency ω is placed in a homogeneous environment with permitivity and permeability of ε and μ. It can be shown that the antenna has approximately the following current distribution.
where, l = 0 is the current amplitude at the feed point of the antenna, k = 2π/λ, λ is the wavelength of the radiating wave, l is the total length of the antenna, δ(ρ) is delta function, ρ is the radial distance from the z axis in cylindrical coordinate, and ez is the unit vector in z direction.
Find the far field electric and magnetic fields E(r,ω), H(r,ω) generated by the antenna.
Approximate your results for (l\ll\lambda) and compare them to the results of "The Infinitesimal Dipole". In this case, calculate the radiation resistance of the antenna.
Sketch approximately the far filed radiation pattern in principal planes for
Half-wave dipole antenna ((l=\lambda / 2))
Full-wave dipole antenna ((l=\lambda))
Somewhat larger antenna, (l=1.4\lambda) and
(l=1.5\lambda)
Hints
In "The Infinitesimal Dipole" it is shown that for an infinitesimal antenna with current distribution of
[\mathbf{J}\left(\mathbf{r},\omega\right)=J_0 \, \delta (\mathbf{r})\, \mathbf{e}_z]
the vector potential is given by [\mathbf{A}(\mathbf{r},\omega)=\frac{\mu\,J_0}{4\pi}\,\frac{ e^{\mathrm{i} k r}}{r}\,\mathbf{e}_z]
Think of an arbitrary antenna as a collection of many infinitesimal antennas, and try to extend this formula to find the vector potential of the dipole antenna.
Radiation resistance (R_a) is defined as below:
[P_r=\frac{1}{2} I_0^2 R_a]
where, (P_r) is the total radiated power by the antenna and (I_0) is feed point current amplitude.
Principal planes are defined as below:
Principal E-plane which includes the direction of maximum radiation and the electric field.
Principal H-plane which includes the direction of maximum radiation and the magnetic field.
Okay, let us move on to find the solution for the finite dipole antenna.
1. Far Field Determination
In this part we first generalize our findings in "The Infinitesimal Dipole" to an arbitrary current distribution. In other words, we want to find the vector potential (\mathbf{A}(\mathbf{r},\omega)) for an arbitrary current distribution (\mathbf{J}\left(\mathbf{r'},\omega\right)). And, to do that we use the results of "The Infinitesimal Dipole".
The linearity of Maxwell equations and constitutive relations allow us to use the superposition principle. It means that we can decompose the current distribution into arbitrarily number of parts, then calculate the vector potential for each part, and finally add all vector potentials which is equal to the vector potential of the original current distribution.
Based on this fact, the current distribution can be decomposed into infinite infinitesimal current pieces. One of these tiny pieces of current which is located at position (\mathbf{r'}) can be written as (\mathbf{J}\left(\mathbf{r'},\omega\right)dV' \, \delta (\mathbf{r-r'})), where (dV') is a volume element. Therefore, according to "The Infinitesimal Dipole" the contribution of this infinitesimal current to the total vector potential is (\frac{\mu\,\mathbf{J}\left(\mathbf{r'},\omega\right)dV'}{4\pi}\,\frac{ e^{\mathrm{i} k |\mathbf{r'-r}|}}{|\mathbf{r'-r}|}), where (|\mathbf{r'-r}|) indicates the distance between the tiny part of current and the observation point. Now, to obtain the total vector potential we need just to add up the contributions of all infinitesimal currents, that is: \begin{equation} \Rightarrow\mathbf{A}(\mathbf{r},\omega)=\int_{V'}\frac{\mu\,\mathbf{J}\left(\mathbf{r'},\omega\right)}{4\pi}\,\frac{ e^{\mathrm{i} k |\mathbf{r'-r}|}}{|\mathbf{r'-r}|} dV' \tag{1}\label{AJ} \end{equation}
In this equation, (V') denotes the volume of the current distribution. Note that equation [\ref{AJ}] is a general formula that can be applied to any current distribution.
Now, one might use equation [\ref{AJ}] to find (\mathbf{A}(\mathbf{r},\omega)), then (\mathbf{E}(\mathbf{r},\omega)) and (\mathbf{H}(\mathbf{r},\omega)), and finally approximate them to find far fields. As you might also noticed, there are some drawbacks in this procedure. First of all, most of the time it is not possible to solve equation [\ref{AJ}] exactly. Second, we do not need to know the exact form of vector potential to find far fields. It is sufficient to know the vector potential for far zone to find far fields. Therefore, it is wise to approximate equation [\ref{AJ}] for far zone.
In reality the distance between antennas are far larger than the dimensions of the antennas themselves, i.e. (\forall \mathbf{r'} \in V': r\gg r'). Therefore, one can use this fact to simplify equation [\ref{AJ}]. To do that we can approximate (|\mathbf{r'-r}|) in the denominator of the kernel of the integral in equation [\ref{AJ}] by the first term of its Taylor expansion. \begin{equation} \Rightarrow\frac{1}{|\mathbf{r'-r}|}\simeq \frac{1}{r}\quad\quad, r\gg r' \tag{2}\label{Ap1} \end{equation}
However, we cannot use this approximation for the exponent; because the exponential function which pure imaginary argument is a periodic function which varies much faster than the denominator. Therefore, we should use a better approximation for that. Thus, we approximate the exponent by the first two terms of its Taylor expansion: \begin{eqnarray} |\mathbf{r'-r}|&=&\sqrt{r'^2+r^2-2\,r'\,r\,\cos(\theta)}\nonumber\&=&r-r'\cos(\theta)+\frac{r'^2\sin^2(\theta)}{2r}+...\tag{3}\label{Apr}\&\simeq & r-r'\cos(\theta)\nonumber\&=& r-\,\mathbf{r'}.\,\mathbf{e}_r\nonumber \end{eqnarray} \begin{equation} \Rightarrow e^{\mathrm{i} k |\mathbf{r'-r}|} \simeq e^{\mathrm{i} k(r-\,\mathbf{r'}.\,\mathbf{e}_r)}\quad\quad, r\gg r' \tag{4}\label{Ap2} \end{equation} where (\theta) is the angle between (\mathbf{r'}) and (\mathbf{r}), and (\mathbf{e}_r) is the unit vector in radial direction.
Now, by substituting equation [\ref{Ap1}] and [\ref{Ap2}] into equation [\ref{AJ}] we obtain: [\mathbf{A}(\mathbf{r},\omega)\simeq\int_{V'}\frac{\mu\,\mathbf{J}\left(\mathbf{r'},\omega\right)}{4\pi}\frac{ e^{-\mathrm{i} k\, (r-\mathbf{r'}.\,\mathbf{e}_r)}}{r} dV'] \begin{equation} \Rightarrow\mathbf{A_{ff}}(\mathbf{r},\omega)=\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}\int_{V'}\mathbf{J}\left(\mathbf{r'},\omega\right) e^{-\mathrm{i} k\, \mathbf{r'}.\,\mathbf{e}_r}dV' \tag{5}\label{Aff} \end{equation} Equation [\ref{Aff}] gives us the far zone vector potential. And, we will use it as the fundamental formula in almost all problems related to antennas.
Now, we need to know the conditions under which our approximation is valid which defines the far field zone. To obtain equation [\ref{Aff}] from equation [\ref{AJ}] we used two approximations: one in the exponent and one in the denominator of the kernel of the integral. The impact of our approximation for the denominator is negligible in comparison to the approximation of the exponent; because, (e^{\mathrm{i} k |\mathbf{r'-r}|}) varies much faster than (\frac{1}{|\mathbf{r'-r}|}) for (r\gg r'). Therefore, we only consider the error introduced to (\mathbf{A}(\mathbf{r},\omega)) by approximating the exponent.
It has been shown by numerical and experimental tests that when the maximum error in the exponent is smaller than (\frac{\pi}{8}), the approximation for vector potential is acceptable and (\mathbf{A_{ff}}(\mathbf{r},\omega)) gives good results. According to equation [\ref{Apr}], the most significant neglected term in approximating the exponent is (k\frac{r'^2\sin^2(\theta)}{2r}). Therefore the maximum error is equal to (k\frac{r'^2_{max}}{2r}) (when (\theta=\frac{\pi}{2}) and (r') is maximum). Thus: [k\frac{r'^2_{max}}{2r} < \frac{\pi}{8}\; \Rightarrow\frac{2\pi}{\lambda}\frac{r'^2_{max}}{2r} < \frac{\pi}{8} \Rightarrow \frac{8 r'^2_{max}}{\lambda} < r\ , \mathrm{so} ]
\begin{equation} \frac{2L^2}{\lambda} < r \tag{6}\label{ff} \end{equation} in which (L=2r_{amx}) is the largest dimension of the antenna. Equation [\ref{ff}] determines the far zone for an antenna with largest dimension of (L). Therefore, equation [\ref{Aff}] is valid as far as equation [\ref{ff}] is satisfied, which is the case for most practical applications.
Equation [\ref{Aff}] can be interpreted as follows. (\mathbf{A_{ff}}(\mathbf{r},\omega)) can be expressed as the product of a spherical wave (\left(\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}\right)) and a directional weighting function (\left(\mathbf{T}(\theta,\phi)=\int_{V'}\mathbf{J}\left(\mathbf{r'},\omega\right) e^{-\mathrm{i} k\, \mathbf{r'}.\,\mathbf{e}_r}dV'\right)). Note that the dependency of (\mathbf{T}) on (\theta) and (\phi) originates from the dependency of (\mathbf{e}_r) to (\theta) and (\phi) (actually (\mathbf{e}_r(\theta,\phi))). We actually expected this result.
Because as the waves leave the radiating source, they approach the spherical form regardless of the shape of the radiator. However, the shape of the radiator determines the strength of the spherical wave in each direction in space which is embedded in the weighting function (\mathbf{T}(\theta,\phi)).
Considering the above decomposition, one can obtain (\mathbf{H}(\mathbf{r},\omega)) as shown below. \begin{eqnarray} \mathbf{H}\left(\mathbf{r},\omega\right)&=&\frac{1}{\mu} \nabla\times\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)=\frac{1}{\mu} \nabla\times\left(\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}\mathbf{T}(\theta,\phi)\right)\tag{7}\label{HA}\&=&\frac{1}{4\pi}\left[ \nabla\left(\frac{e^{\mathrm{i} k r}}{r}\right)\times \mathbf{T}(\theta,\phi)+\frac{e^{\mathrm{i} k r}}{r}\,\nabla\times\mathbf{T}(\theta,\phi)\right]\nonumber\&=&\frac{e^{\mathrm{i} k r}}{4\pi r}\left[\left(\mathrm{i}k-\frac{1}{r}\right)\mathbf{e}_r\times \mathbf{T}(\theta,\phi)\right.\nonumber\&&\quad\quad +\frac{1}{r\sin(\theta)}\left(\frac{\partial}{\partial\theta}\left(T_\phi(\theta,\phi)\sin(\theta)\right)-\frac{\partial}{\partial \phi}T_\theta(\theta,\phi)\right)\mathbf{e}_r\nonumber\&&\quad\quad+\left. \frac{1}{r\sin(\theta)}\frac{\partial}{\partial \phi}T_r(\theta,\phi)\mathbf{e}_\theta-\frac{1}{r}\frac{\partial}{\partial \theta}T_r(\theta,\phi)\mathbf{e}_\phi\right]\nonumber \end{eqnarray} which is a little bit complicated. However, as far as (k\gg\frac{1}{r}) (or equivalently (r\gg\lambda)), the above equation can be approximated as below. \begin{eqnarray} \mathbf{H}\left(\mathbf{r},\omega\right)&\simeq &\frac{e^{\mathrm{i} k r}}{4\pi r}\,\mathrm{i}k\mathbf{e}_r\times \mathbf{T}(\theta,\phi)\nonumber\&=&\frac{1}{\mu}\mathrm{i}k\mathbf{e}_r\times\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)\tag{8}\label{HAff} \end{eqnarray}
If you compare equation [\ref{HAff}] with equation [\ref{HA}], it seems that what we have done is just the substitution of (\nabla) with (\mathrm{i}k\mathbf{e}_r). Actually, it can be shown that when (r\gg\lambda), all field quantities can be approximated by using this substitution (you can check it for (\mathbf{E}\left(\mathbf{r},\omega\right))).
Using this substitution, one can approximate (\mathbf{E}\left(\mathbf{r},\omega\right)) for (r\gg\lambda) as below. \begin{eqnarray} \mathbf{E}\left(\mathbf{r},\omega\right)&=& \mathrm{i} \omega\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)-\frac{\nabla\big(\nabla\cdot\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)\big)}{\mathrm{i}\omega\mu\,\varepsilon}\nonumber\&\simeq &\mathrm{i}\omega\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)-\frac{\mathrm{i}k\mathbf{e}_r\big(\mathrm{i}k\mathbf{e}_r\cdot\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)\big)}{\mathrm{i}\omega\mu\,\varepsilon}\nonumber\&=&\mathrm{i}\omega\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)-\mathrm{i}\omega\mathbf{e}_r\left(\mathbf{e}_r\,.\,\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)\right)\nonumber\&=&\mathrm{i}\omega\left(A^\theta_\text{ff}\left(\mathbf{r},\omega\right)\mathbf{e}_\theta+A^\phi_\text{ff}\left(\mathbf{r},\omega\right)\mathbf{e}_\phi\right)\nonumber\&=&\mathrm{i}\omega\mathbf{A^t_{ff}}\left(\mathbf{r},\omega\right)\tag{9} \end{eqnarray} where by (\mathbf{A^t_{ff}}\left(\mathbf{r},\omega\right)) we mean the transverse component of (\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)).
Let us now summarize what we have done so far.
Firstly, we approximated the vector potential for far zone (\frac{2L^2}{\lambda} < r): \begin{equation} \mathbf{A_{ff}}(\mathbf{r},\omega)=\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}\int_{V'}\mathbf{J}\left(\mathbf{r'},\omega\right) e^{-\mathrm{i} k\, \mathbf{r'}.\,\mathbf{e}_r}dV'\tag{10}\label{Af} \end{equation}
Secondly, we approximated the fields for (r\gg\lambda): \begin{equation} \mathbf{H_{ff}}\left(\mathbf{r},\omega\right)=\frac{1}{\mu}\mathrm{i}k\mathbf{e}_r\times\mathbf{A_{ff}}\left(\mathbf{r},\omega\right)\tag{11}\label{Hf} \end{equation} \begin{equation} \mathbf{E_{ff}}\left(\mathbf{r},\omega\right)=\mathrm{i}\omega\mathbf{A^t_{ff}}\left(\mathbf{r},\omega\right)\tag{12}\label{Ef} \end{equation} These three equations (equation [\ref{Af}], [\ref{Hf}], and [\ref{Ef}]) govern the far zone of an arbitrary antenna. From now on, we consider everything at far zone unless explicitly stated. Therefore, we neglect the prefix "far field" and subscript "ff" for field quantities; for example instead of far field vector potential, we simply say vector potential and write (\mathbf{A}\left(\mathbf{r},\omega\right)).
Now we have every thing at hand. Therefore, let us calculate the vector potential for the dipole given in this problem by substituting the current distribution in equation [\ref{Af}]. \begin{eqnarray} \Rightarrow\mathbf{A}(\mathbf{r},\omega)&=&\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}\int_{-\frac{l}{2}}^{\frac{l}{2}}\int_{0}^{\infty}\int_{0}^{2\pi}\left[I_0 \, \sin\left(k(\frac{l}{2}-|z'|)\right) \,\frac{\delta(\rho')}{2\pi\rho'}\, \mathbf{e}_z\right] e^{-\mathrm{i} k\, \mathbf{r'}.\,\mathbf{e}_r} \rho' d\phi' d\rho' dz'\nonumber\&=& \frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}\int_{-\frac{l}{2}}^{\frac{l}{2}}\left[I_0 \, \sin\left(k(\frac{l}{2}-|z'|)\right)\, \mathbf{e}_z\right] e^{-\mathrm{i} k\,z'\cos(\theta)} dz'\nonumber \end{eqnarray}
It is easy but lengthy to calculate the integral. Below we calculate it step by step. First we expand the exponential into sin and cos functions. The integral over the imaginary part is zero since the function is odd. Thus, only the cos function remains. \begin{eqnarray} \Rightarrow\mathbf{A}(\mathbf{r},\omega)&=&\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}I_0\,\mathbf{e}_z\int_{-\frac{l}{2}}^{\frac{l}{2}}\sin\left(k(\frac{l}{2}-|z'|)\right)\big[\cos\left(\mathrm{i} k\,z'\cos(\theta)\right)-\mathrm{i}\sin\left(\mathrm{i} k\,z'\cos(\theta)\right)\big] dz'\nonumber\&=&2\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}I_0\,\mathbf{e}_z\int_{0}^{\frac{l}{2}}\sin\left(k(\frac{l}{2}-z')\right)\cos\left(\mathrm{i} k\,z'\cos(\theta)\right) dz'\nonumber \end{eqnarray} Then we use "(\sin(x)\cos(y)=\frac{1}{2}\left[\sin(x+y)+sin(x-y)\right])" to calculate the integral. \begin{eqnarray} \Rightarrow\mathbf{A}(\mathbf{r},\omega)&=&\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}I_0\,\mathbf{e}_z\int_{0}^{\frac{l}{2}}\left[\sin\left(k(\frac{l}{2}-z'+z'\cos(\theta))\right)+\sin\left(k(\frac{l}{2}-z'-z'\cos(\theta))\right) dz'\right]\nonumber\&=&\frac{\mu\,e^{\mathrm{i} k r}}{4\pi r}I_0\,\mathbf{e}_z \left[\frac{\cos(k\frac{l}{2}\cos(\theta))-\cos(k\frac{l}{2})}{k(1-\cos(\theta))}+\frac{\cos(k\frac{l}{2}\cos(\theta))-\cos(k\frac{l}{2})}{k(1+\cos(\theta))}\right]\nonumber\&=&\frac{\mu I_0}{2\pi}\frac{e^{\mathrm{i} k r}}{r}\,\frac{\cos(k\frac{l}{2}\cos(\theta))-\cos(k\frac{l}{2})}{k\sin^2(\theta)}\,\mathbf{e}_z\tag{13} \end{eqnarray}
Now, we can calculate (\mathbf{H}(\mathbf{r},\omega)) and (\mathbf{E}(\mathbf{r},\omega)) using equation [\ref{Hf}] and [\ref{Ef}]. \begin{equation} \Rightarrow\mathbf{H}(\mathbf{r},\omega)=\frac{-\mathrm{i} I_0}{2\pi}\frac{e^{\mathrm{i} k r}}{r}\,\frac{\cos(k\frac{l}{2}\cos(\theta))-\cos(k\frac{l}{2})}{\sin(\theta)}\,\mathbf{e}_\phi\tag{14}\label{Hl} \end{equation} \begin{equation} \Rightarrow\mathbf{E}(\mathbf{r},\omega)=\frac{-\mathrm{i} I_0}{2\pi}\sqrt{\frac{\mu}{\varepsilon}}\frac{e^{\mathrm{i} k r}}{r}\,\frac{\cos(k\frac{l}{2}\cos(\theta))-\cos(k\frac{l}{2})}{\sin(\theta)}\,\mathbf{e}_\theta\tag{15}\label{El} \end{equation} As can be seen, in far zone the fields are tangential and locally resemble plane waves.
2. Calculation of the Radiation Resistance - Small Dipole
One special case is when the length of the dipole antennas is much smaller that the wavelength ((l\ll \lambda)). Under this condition the current distribution on the antennas can be approximated as below. [ \mathbf{J}\left(\mathbf{r},\omega\right) = \left{ \begin{array}{l l} I_0\left(k(\frac{l}{2}-|z|)\right) \,\frac{\delta(\rho)}{2\pi\rho}\, \mathbf{e}_z & \quad |z| < \frac{l}{2}\ 0 & \quad |z|\geq\frac{l}{2} \end{array} \right.] where we approximated (\sin(x)) by (x). As can be seen, the current attains its maximum value at the center of the antenna and linearly decrease to zero as it approaches the edges. That is why this current distribution is called "triangular". This is a very good approximation for current distribution on short dipoles.
To calculate the radiated electric and magnetic fields in this case, one can again use equation [\ref{Af}] to find (\mathbf{A}\left(\mathbf{r},\omega\right)), and then equation [\ref{Hf}] and [\ref{Ef}] to find (\mathbf{H}\left(\mathbf{r},\omega\right)) and (\mathbf{E}\left(\mathbf{r},\omega\right)). Instead, we can just approximate equation [\ref{Hl}] and [\ref{El}] using (\cos(x)\simeq1-\frac{x^2}{2}). \begin{eqnarray} \Rightarrow\mathbf{H}(\mathbf{r},\omega)&=&\frac{-\mathrm{i} I_0}{2\pi}\frac{e^{\mathrm{i} k r}}{r}\,\frac{1-\frac{\left(k\frac{l}{2}\cos(\theta)\right)^2}{2}-1+\frac{\left(k\frac{l}{2}\right)^2}{2}}{\sin(\theta)}\,\mathbf{e}_\phi\nonumber\&=&\frac{-\mathrm{i} k I_{tot}}{4\pi}\frac{e^{\mathrm{i} k r}}{r}\, \sin(\theta)\,\mathbf{e}_\phi\tag{16}\label{Htra} \end{eqnarray} \begin{eqnarray} \Rightarrow\mathbf{E}(\mathbf{r},\omega)&=&\frac{-\mathrm{i} I_0}{2\pi}\sqrt{\frac{\mu}{\varepsilon}}\frac{e^{\mathrm{i} k r}}{r}\,\frac{1-\frac{\left(k\frac{l}{2}\cos(\theta)\right)^2}{2}-1+\frac{\left(k\frac{l}{2}\right)^2}{2}}{\sin(\theta)}\,\mathbf{e}_\theta\nonumber\&=&\frac{-\mathrm{i} k I_{tot}}{4\pi}\sqrt{\frac{\mu}{\varepsilon}}\frac{e^{\mathrm{i} k r}}{r}\, \sin(\theta)\,\mathbf{e}_\theta\tag{17}\label{Etra} \end{eqnarray} where, [I_{tot}=\int_\frac{-l}{2}^\frac{l}{2}I_0\left(k(\frac{l}{2}-|z'|)\right)dz'=I_0 k \frac{l^2}{4}]
By comparing these fields to those of an infinitesimal antenna (see again "The Infinitesimal Dipole"), it can be seen that they are the same except that the amplitude of the fields is different which is obviously due to different sizes of antennas.
Now let us calculate the radiation resistance of this antenna. The radiation resistance of an antennas is a resistance which can dissipate the same amount of power as the antenna radiate, if it is connected to the same source. In other words, if we neglect the ohmic loss in an antenna, the radiation resistance is exactly the real part of the input impedance of the antenna. The input impedance of an antenna is very important since it should match the input impedance of the transmitter (or receiver) in order to maximize the power transfer. It is not always easy to calculate the imaginary part of the input impedance since it depends of the near field behavior of the antenna.
However, the real part (radiation resistance) is easy to calculate, since it is connected to the radiated power which can be calculated using far fields. The radiation resistance is related to the transmitted power as follows, \begin{equation} P_r=\frac{1}{2} I_0^2 R_a\tag{18}\label{Ra} \end{equation} where, the left hand side is the total radiated power by antenna, and the right hand side is the average power dissipated by a resistor when it is connected to an alternating current source with an amplitude of (I_0) (which is also equal to the feeding current of the antenna).
Therefore, we first calculate the Poynting vector using far fields given in equation [\ref{Etra}] and [\ref{Htra}]. \begin{eqnarray} \mathbf{S}\left(\mathbf{r},\omega\right)&=&\frac{1}{2}\,\text{Real}\Big{\mathbf{E}\left(\mathbf{r},\omega\right)\times \mathbf{H^}\left(\mathbf{r},\omega\right)\Big} \nonumber\&=&\frac{ \mu^\frac{5}{2} \varepsilon^\frac{3}{2}I_0^2\omega^4 l^4}{512 \pi ^2}\frac{1}{r^2}\,sin^2(\theta)\mathbf{e}_r\tag{19}\label{Sff} \end{eqnarray} Then we calculate the radiated power by integrating the Poynting vector over the solid angle of a sphere. \begin{eqnarray} P_r&=&\int_0^{2\pi}\int_0^\pi\mathbf{S}\left(\mathbf{r},\omega\right)\,.\,\mathbf{e}_r\,r^2\sin(\theta)d\theta d\phi\nonumber\&=&\frac{ \mu^\frac{5}{2} \varepsilon^\frac{3}{2}I_0^2\omega^4 l^4}{256 \pi}\int_0^\pi\sin^3(\theta)d\theta\nonumber\&=&\frac{ \mu^\frac{5}{2} \varepsilon^\frac{3}{2}I_0^2\omega^4 l^4}{192 \pi}\tag{20}\label{Pr} \end{eqnarray} Now using equation [\ref{Ra}], radiation resistance can be obtained. \begin{equation} R_r=\frac{2P_r}{I_0^2}=\frac{ \mu^\frac{5}{2} \varepsilon^\frac{3}{2}\omega^4 l^4}{96 \pi} \tag{21} \end{equation} Note that this formula is valid as far as (l\ll \lambda).
As an example, the radiation resistance for a simple radio antenna with length of (30^\text{cm}), working at (100^\text{MHz}), and placed in air ((\Rightarrow l=\frac{\lambda}{10}\ll\lambda)) is approximately (195^{m\Omega}), which is extremely small. This resistance it much smaller than the (50^\Omega) output impedance of standard transmitters (or input impedance of standard receivers), and as a result a matching network is needed to convert (195^{m\Omega}) to (50^{\Omega}), which of course adds some complexity to our telecommunication system. That is why short dipoles are not good radiators. One way to increase the radiation resistance is to increase the length of the dipole. That is one reason why half-wave dipoles are more popular that short dipoles.
3. Important Special Cases: small dipole-, half-wavelength- and larger antennas
In this section we would like to see what will happen to the radiation pattern of a dipole antenna as its length is increased. To do so, we consider four dipoles with different lengths.
Using equation [\ref{El}] and [\ref{Hl}], one can calculate the Poynting vector for a general dipole as below.
[\begin{eqnarray} \mathbf{S}\left(\mathbf{r},\omega\right)&=&\frac{1}{2}\,\text{Real}\Big{\mathbf{E}\left(\mathbf{r},\omega\right)\times \mathbf{H^}\left(\mathbf{r},\omega\right)\Big} \nonumber\&=&\frac{I^2_0}{4\pi^2}\sqrt{\frac{\mu}{\varepsilon}}\frac{1}{r^2}\,\frac{\left[\cos(k\frac{l}{2}\cos(\theta))-\cos(k\frac{l}{2})\right]^2}{\sin^2(\theta)}\,\mathbf{e}_r\tag{22}\label{Sl} \end{eqnarray}]
For each case we just need to substitute the length of the antenna in equation [\ref{Sl}]. Therefore we obtain:
[\begin{equation} \mathbf{S}\left(\mathbf{r},\omega\right) =\mathbf{e}_r \, \frac{I^2_0}{4\pi^2}\sqrt{\frac{\mu}{\varepsilon}}\frac{1}{r^2}\,\left{ \begin{matrix} \frac{\cos^2(\frac{\pi}{2}\cos(\theta))}{\sin^2(\theta)}\quad\quad\quad\quad &,(l=\frac{\lambda}{2})\ \frac{\left[\cos(\pi\cos(\theta))+1\right]^2}{\sin^2(\theta)}\quad\quad &,(l=\lambda)\ \frac{\left[\cos(1.4 \pi\cos(\theta))-\cos(1.4\pi)\right]^2}{\sin^2(\theta)}\quad\quad &, (l=1.4 \lambda)\ \frac{\cos^2(\frac{3\pi}{2}\cos(\theta))}{\sin^2(\theta)}\quad\quad &, (l=1.5\lambda) \end{matrix} \right. \end{equation}]
As can be seen it is a little bit difficult to plot the radiation pattern by hand (which is almost always the case). However, there are simple ways to approximately sketch the radiation patterns to get a feeling about them. First of all, note that the radiation pattern is rotationally symmetric about the z-axis ((\phi)-independent). Secondly, find the directions in which the radiation pattern is zero (nulls of the antenna). Thirdly, between two consecutive nulls there is a local radiation maximum (which is called a lobe). Fourth, find the maximum direction of radiation, and plot approximately the radiation pattern in principal planes.
As an example, we do this procedure for the third dipole antenna (l=1.4\lambda) as below. \begin{equation} S_r^n\left(\mathbf{r},\omega\right)=\frac{\left[\cos(1.4 \pi\cos(\theta))-\cos(1.4\pi)\right]^2}{\sin^2(\theta)}\quad\quad,(l=1.4\lambda) \tag{23}\label{S1.4} \end{equation} Here, (S_r^n\left(\mathbf{r},\omega\right)) means the normalized radial component of the Poynting vector.
The nulls of the radiation pattern can be easily found. [S_r^n\left(\mathbf{r},\omega\right)=0\rightarrow\cos(1.4 \pi\cos(\theta))-\cos(1.4\pi)=0] [\rightarrow \theta=\arccos {1,-1+\frac{2}{1.4},1-\frac{2}{1.4},-1}] [\Rightarrow Nulls:\quad \theta_\text{n}\simeq{0,64.6^\circ,115.4^\circ,180^\circ}] From the above information it can be inferred that the radiation pattern has three lobes: one in ((0,64.6^\circ)), one in ((64.6^\circ ,115.4^\circ)), and the last one in ((115.4^\circ ,180^\circ)). The radiation maxima are also located approximately between the nulls, i.e. at (\theta_m\simeq{32.3^\circ ,90^\circ ,147.7^\circ}).
Now we need to determine the principal planes. By substituting the values of (\theta_m) in equation [\ref{S1.4}], it appears that the maximum radiation occurs at (\theta =90^\circ). Since the antenna is omni-directional we are free to choose (\phi) for the direction of maximum radiation. Therefore, we choose (\mathbf{e}_r(\theta=90^\circ,\phi=0)) as the maximum direction of radiation. From equation [\ref{El}] we also know the electric field is directed along (\mathbf{e}_\theta(\theta=\pi,\phi=0)=-\mathbf{e}_z). Therefore the principal E-plane is x-z plane. In a similar way the principal H-plane can be determined as x-y plane.
Considering all of the above facts, the radiation pattern for dipole antenna with (l=1.4\lambda) can be sketched as shown in Fig. 1 and 2.
The radiation in three dimensions is also shown in Fig. 3 below.
Using a computer program such as Mathematica or MATLAB, one can easily plot the exact radiation pattern. The radiation patterns for all other dipoles are plotted in Fig. 4 (due to rotational symmetry, the patterns are plotted only in E-plane).
Note that each pattern is normalized to its own maximum. For comparison the radiation pattern of an infinitesimal dipole is also plotted.
As can be seen, by increasing the length of the antenna the main lobe of the radiation pattern gets narrower and as a result directivity increases. However, as the length of the dipole becomes larger than the wavelength, two side lobes are added to the radiation pattern (as can be seen in figure (1) for (l=1.4 \lambda)).
As the length of the dipole is increased further, the side lobes become stronger (see Fig. 4 and compare the patterns for (l=1.4 \lambda) and (l=1.5 \lambda)). These side lobes decrease the directivity of the dipole.
That was quite a bit of work, wasn't it?!
In essence the far field of the finite dipole antenna is influenced by the separation of the charges. For some special cases we have seen how much.
We strongly encourage you to implement some of the formulas yourself and verify our radiation pattern! You will learn a lot about antenna theory by adapting the graphs and incorporating the mathematical formulas.
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7178 | https://www.youtube.com/watch?v=Yl1aOD3CZbM | Statistics - Find the weighted mean
MySecretMathTutor
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This video covers how to find the weighted mean for a set of data. Remember that each data point is multiplied by a given weight, and then divided by the total weight. for more videos visit
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Transcript:
in this video we're going to look at how to find a weighted mean now this is slightly different than a normal mean in that a certain weight gets worked into all of our data points to be a little bit more specific here's what a weighted mean is in a weighted mean we look for a mean of each value when it is multiplied by some sort of weight and we can use the following formula to help us out in this formula you can see each of these little XIs is a data point and these W are the weights so all of them get multiplied by a weight now we're dividing by the sum of all the weights that way we get our mean to see exactly how this works let's look at an example so according to a certain teacher they like to grade their class on a waiting system this means that every category in their class has a different uh weight associated with it so their homework grade weight is is 25 quiz grade weight is 30 test weight is 10 and final exam is 35 what the weights tell us is that the final exam counts for quite a bit and and so do the quizzes but looks like since the weight on the test is very small that doesn't count for a whole lot so using these weights uh suppose we have a student who got an 88 on their homework a 71 on their quiz 97 on their test and a 90 on their final exam we're going to to find their average score or their weighted mean according to the teacher's waiting scale so let's go ahead and do that we need to take each of the students scores these are like our data points and multiply them by their respective weight let's see they got a 20 five weight for the homework category so 25 88 we got a weight of 30 on 71 weight of 10 on 97 and a weight of 35 on 90 so every single data point is multiplied by a weight now we need to add up all of our weights to see what we need to divide by so 25 and 30 is 55 65 and 35 so we need to divide by 100 when we're done let's grab our calculators and quickly add this up all right so on the top top 8,450 and all of this is being divided by 100 so it shows that our students average score is an 84 and a half now this kind of makes sense in the context of the students actual scores because you can see that their quiz grade is quite low and the quizzes have a pretty significant weight on them so even though they're doing really good in their test and their final exam uh it's not helping out as much as it needs to because the test is only worth uh a weight of 10 so remember when looking for the weighted mean take your data points and multiply them by their weight and divide by the total weight |
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Base Area of a Triangular Prism – Definition, Formulas, Examples
Home » Math Vocabulary » Base Area of a Triangular Prism – Definition, Formulas, Examples
What Is the Base Area of a Triangular Prism?
Base Area of a Triangular Prism Formulas
How to Calculate the Base Area of a Triangular Prism
Solved Examples on Base Area of a Triangular Prism
Practice Problems on Base Area of a Triangular Prism
Frequently Asked Questions about Base Area of a Triangular Prism
What Is the Base Area of a Triangular Prism?
The area of the triangle located at the base of a triangular prism is referred to as the base area of a triangular prism. It is measured in square units.
A triangular prism is a polyhedron. It has three rectangular sides (lateral faces) and two triangular faces (one is the base and the other is the top). The two triangular faces are parallel and congruent. The base area of a triangular prism is the area of the base triangle.
Take a look at the net of a triangular prism, which clearly shows the three rectangular lateral faces and two triangular bases.
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Base Area of a Triangular Prism Formulas
To find the base area of a triangular prism, we must know different formulas for finding the area of a triangle. Let’s discuss some important formulas that can be helpful in finding the base area of a triangular prism.
Base Area of a Triangular Prism Formula When the Lengths of Sides of the Triangle Are Given
If the length of the sides of a triangular prism’s base is known, then its base area can be determined by using Heron’s formula.
Base area of a triangular prism = Area of the base triangle = A
where
a, b, and c are the lengths of the sides of the base triangle, and
s = semiperimeter .
Base Area of a Triangular Prism Formula When the Base and Height of the Triangle Are Given
If the length of the base triangle and its height is known, then the base area of a triangular prism can be calculated as follows:
Base area of a triangular prism = Area of the triangular base
If the base “b” and height “h” of the base triangle is given, then the base area of the triangular prism .
Let’s summarize!
| | |
--- |
| Triangular Prism Base Area Formulas | |
| What do we know about the base triangle? | Formula |
| Equilateral Triangle | |
| Base = b, Height = h | |
| Side-lengths: a, b, c, and semiperimeter s | |
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How to Calculate the Base Area of a Triangular Prism
We can determine the base area of a triangular prism by finding the area of the base triangle. Depending on the known parameters and type of triangle in the base, various formulas can be used to determine the area of the base of a triangular prism.
Let’s see how to calculate the base area of a triangular prism.
Equilateral Triangle
If the base triangle is an equilateral triangle and each side length is “a,” then the base area of the triangular prism is given by
Triangle with base b and height h
If the base “b” and height “h” of the base triangle are given, then the base area of the triangular prism can be calculated as
.
Facts about the Base Area of a Triangular Prism
A triangular prism has two triangular bases and three rectangular sides (lateral faces).
The triangular prism is said to be semiregular if the bases are equilateral triangles and the other faces are squares rather than rectangles.
The total surface area of a triangular prism is the sum of the areas of the two triangular bases and the lateral surface area.
Conclusion
In this article, we have learned about the base area of a triangular prism for the different types of triangular bases. We discussed important formulas for finding the base area of a triangular prism. Let’s use these formulas to solve a few examples and practice problems.
Solved Examples on Base Area of a Triangular Prism
What is the base area of a triangular prism if the right triangle at its base has the base = 3 units and height = 4 units?
Solution:
The base of the given triangular prism is a right triangle whose base is 3 units and height 4 units.
Area of triangle with base b and height
Base area of triangular prism
Base area of triangular prism = 6 square units
Find the base area of the triangular prism whose sides are 3 inches, 4 inches, and 5 inches respectively.
Solution:
a = 3 inches, b = 4 inches and c = 5 inches
Using the Heron’s formula, we write
Area of triangle
Semiperimeter inches
Substituting the values of s, a, b, and c in the above formula, we get
Area of triangle
Area of triangle square inches
If the height of the triangular base of a triangular prism is 5 inches and the base length is 12 inches, find the base area of the triangular prism.
Solution:
h = 5 inches and b = 12 inches
Area of the triangle when height and base side are given
∴ Area of triangle square inches
Base area = 30 square inches
If the base area of a triangular prism is and the base length is 9 inches, then find the height of the triangular base.
Solution:
The base area of a triangular prism=180 inch2 and b=9 inches
The base area of a triangular prism
inches
Hence, the height of the triangular bases is 40 inches.
If the base of a triangular prism is an equilateral triangle and the perimeter of the base is 120 feet, find the base area of the prism.
Solution:
Let the side of equilateral triangle
Perimeter feet
feet
Area of triangle
Area of triangle square feet
The base sides of a triangular prism are in the ratio 7 : 5 : 10 and its perimeter is 22 units. Find the base area of the triangular prism.
Solution:
Let the sides be 7x, 5x, and 10x.
Perimeter
Therefore, we get the values
a = 7 , b = 5, c = 10
inches
∴ Area of triangle
∴ Area of triangle
∴ Area of triangle square units
Practice Problems on Base Area of a Triangular Prism
Base Area of a Triangular Prism - Definition, Formulas, Examples
Attend this quiz & Test your knowledge.
1
A triangular prism has ______ triangular faces.
1
2
3
4
CorrectIncorrect
Correct answer is: 2
A triangular prism has two triangular faces (one at the top and the other at the bottom).
2
The base area of the triangular prism whose base-edges are x units, y units, and z units is ______.
square units
square units
square units
CorrectIncorrect
Correct answer is:
Area of the triangular prism whose triangular base has edges x units, y units, and z units is given by Heron's formula.
square units
3
If the base-edge of the triangular base of the triangular prism is 24 inches and its base area is 60 square units, then the height of the triangle is _______.
5 units
8 units
4 units
3 units
CorrectIncorrect
Correct answer is: 5 units
Area of the triangle when height and base side are given
units
4
If the sides of the triangular base of a prism are 10 units, 12 units, and 6 units, then the base area is equal to ___________.
square units
square units
square units
square units
CorrectIncorrect
Correct answer is: square units
Sides of the triangular base of a prism are 10 units, 12 units, and 6 units.
Here,
∴ Base areas of triangular prism
square units
5
If the base of a triangular prism is an equilateral triangle with a side length of 16 inches. Then the base area of the prism is _______.
CorrectIncorrect
Correct answer is:
If the base triangle is equilateral and each side length is “a,” then the base area of the triangular prism
Frequently Asked Questions about Base Area of a Triangular Prism
What is the number of triangles and rectangles in a triangular prism?
What do you mean by the scalene triangular base of a prism?
What do you mean by the base area of a triangular prism?
What is the base area of a triangular prism if the base is an equilateral triangle?
How do you find the base area of a triangular prism? |
7180 | https://missdcoxblog.wordpress.com/2020/07/04/differentiation-and-multiple-choice-questions/ | missdcoxblog
My views on Teaching & Education
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Differentiation and multiple choice questions
Using multiple choice questions (MCQs) with students has become more popular, probably due to online systems that can provide pre-made questions or that they mark themselves; both saving teachers time.
However, if you wish to write your own, research suggests that you follow some simple guidelines. I will be looking at these and suggesting how you can differentiate using multiple choice quizzes by showing how you can increase complexity. This might be appropriate when considering the different year groups or students that you teach or the purpose of the quiz. If you want it to be quick and ‘easy’, choose the first types of question.
The key for differentiation using MCQs is to try increase challenge for students in the subject content rather than making the questions more complex in terms of procedure or literacy.
Research suggests:
The following suggestions are roughly in order of increasing difficulty:
Simple question, one word answers
Sticking to tier 2 vocabulary keeps things simple and with one ‘word’ answers these can be quick and easy to answer for students.
This swaps ‘rules’ for ‘commandments’ which is a slightly more complex way to check vocabulary understanding
Students need to know what ‘Decalogue’ means and then know what the answer is
Simplify: Make the distractors very obviously wrong. This may be appropriate in some educational settings, for some students.
Extend: Adding in tier 3 vocabulary means you are also testing knowledge of keywords which requires a ‘double step’; understanding the keyword and knowing the correct answer.
Extend: Answers that are more than one word
2. Identify the incorrect statement
These are more challenging as they require students to carefully read all the answers rather than scan for the answer that they know is correct.
3. The best answer
This is more complicated because it requires students to read carefully to identify the subtle differences between the answers and know which is 100% correct rather than partially correct or is the overall ‘best’ answer. These are complex as the distractors can be correct but less likely or important.
4. Extended explanations
Using extended explanations are appropriate in some subjects and are especially useful for A level where more complex, (literacy based) reasoning is needed.
Extend: Use more complex subject language (tier 3) in the question and answers.
Strategies for differentiating all the above
5. Multiple correct answers
This is probably the easiest way of increasing challenge in content without increasing the literacy demands. Students need to select multiple correct answers.
Extend: Don’t tell students how many of the answers are correct. This makes it significantly more challenging.
6. Give the option ‘I don’t know yet’
One of the key issues with MCQs for assessing knowledge is that students can guess and they always have a 1 in 3/4 chance of getting the answer right. If you really don’t want them to guess so you have true picture of what they do/don’t know, adding ‘I don’t know yet’ can help. However they need to be trained to use it! It can go against the whole ‘try your best’ mantra that students may have had drilled into them as they are admitting they don’t know when they could actually guess and might get it right. We spend time going through this with them.
7. Carefully crafted distractors
This is where knowledge of potential misconceptions is really useful. There are certain topics/questions with students where there are often common misconceptions. Using these misconceptions as your distractors makes the question more challenging because instinctively students might go for the misconception answer/s.
Further reading
View at Medium.com
Blake Harvard is the KING of MCQs and he has several blogs on them including different ways of using them for diagnosis:
View at Medium.com
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Ratios and Proportional Relationships Error Analysis
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Error analysis is a great way to get your students engaging in higher level thinking! My students love being the teacher and finding other students' errors. In this product, you will get 12 different error analysis problems all related to ratios and proportional relationships. In each problem, the work of two different students will be shown. One student has the correct answer while the other student does not. This activity requires more rigor than traditional error analysis because students not only have to find an error, but they also have to figure out which student is correct!
Specific topics in this product include:
• Identifying Proportional Relationships
• Constant of Proportionality
• Writing an Equation for a Proportional Table
• Unit Rates with Complex Fractions
• Explaining the Meaning of an Ordered Pair
• Multi-Step Percent Word Problems
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These error analysis activities can be used in a variety of ways! In my classroom, I have used them as:
• Bellwork/Opening Problems throughout my unit
• Tickets out the Door
• Stations - students find the error at each station
• Small Group Instruction
Check out some of my other products related to ratios and proportions:
Unit Rates with Complex Fractions Lesson Plan
Percent Proportion vs. Percent Equation
Constant of Proportionality Pyramid Puzzle
Constant of Proportionality Jigsaw
Proportional vs. Non-Proportional Relationships Mazes
Proportional Relationships Bundle
Constant of Proportionality - Odd Man Out
Proportional Relationships Tournament Challenge
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Ratios and Proportional Relationships Error Analysis
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This was good for an Error Analysis activity, which I try to do a few times a week with my students.
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Error analysis activities are awesome for student collaboration, discussion and critical thinking. I love that these pages require students to share advice with the incorrect student. Thank you for creating and sharing!
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CCSS 7.RP.A.1
Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction ½/¼ miles per hour, equivalently 2 miles per hour.
CCSS 7.RP.A.2
Recognize and represent proportional relationships between quantities.
CCSS 7.RP.A.2a
Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin.
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Pair of Straight Lines
The equation ax 2+ 2hxy + by 2+ 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b doesn’t variables simultaneously.
Let a ≠ 0.
Now, the above equation becomes
a 2 x 2+ 2ax (hy + g) = aby 2– 2afy – ac
on completing the square on the left side, we get,
a 2 x 2+ 2ax (hy + g) = y 2(h 2– ab) + 2y (gh – af) + g 2– ac.
i.e. (ax + hy + g) =+√y 2(h 2–ab)+2y(gh–af)g 2–ac
We cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is,
(gh – af)2= (h 2– ab) (g 2– ac)
i.e. g 2 h 2– 2afgh + a 2 f 2= g 2 h 2– abg 2– abg 2– ach 2+ a2bc
cancelling and diving by a, we have the required condition
abc + 2fgh – af 2– af 2– bg 2– ch 2= 0
Illustration:
What is the point of intersection of two straight lines given by general equation ax 2+ 2hxy + by 2+ 2gx + 2fy + c = 0?
Solution:
The general solution is
ax 2+ 2hxy + by 2+ 2gx + 2fy + c = 0 …… (1)
Let (α, ß) be the point of intersection we consider line paralleled transformation.
x = x’ +α, y = y’ +ß
From (1) we have
a(x’ + α)2+ 2h(x’ +α) (y’ +ß) + b(y’ +ß)2+ 2g(x’ +α) + 2f(y’ + ß) + c = 0
⇒ ax’2+ 2hx’y’ + by’2+ a α 2+ 2h αß+ b ß 2+ 2g α+ 2fß + 2x’(a α+ h ß+ g) + 2y’ + 2y’ (h α+ bß + f) = 0
⇒ ax’2+ 2hx’y’ + by’2+ 2x’(a α+ hß + g) + 2g’ + 2y’ (h α+ b ß+ f) = 0
Which must be in the form
ax'2+ 2hx’y’ + by’ = 0
This cannot be possible unless
aα+ hß + g = 0
h α+ b ß+ f = 0
Solving
α/hf–bg = ß/hg–af = 1/ab–h 2
α= hf–bg/ab–b 2,ß= hg–af/ab–h 2
Illustration:
Represent lines y = 2x and y = 3x by a homogeneous equation of second degree
Solution:
(y – 2x) (y – 3x) = 0
Or 6x 2– 5xy + y 2= 0
Illustration:
Represent lines parallel to y = 2x and y = 3x by a second degree equation
Solution:
(y – 2x – c 1) (y – 3x – c 2) (where c 1 and c 2 are constants)
= 6x 2– 5xy + y 2+ (3c 1+ 2c 2) x + (– c 1– c 2) y + c 1 c 2= 0
Note:
1.Homogeneous part is same as for the equation of above illustration. Therefore, the homogeneous part of a general second degree equation determines the slope of the lines i.e. lines parallel to ax 2+ 2hxy + by 2+ c = 0 and through the origin are represented by the equation ax 2+ 2hxy + by 2= 0
2.The equation ax 2+ 2hxy + by 2+ 2fy + c = 0 represents a pair of parallel straight lines if h/a = b/h = f/g or bg 2= af 2
The distance between them is given by 2√g 2–ac/a(a+b) or √f 2–bc/b(a+b)
Illustration:
Does the second degree equation x 2+ 3xy + 2y 2– x – 4y – 6 = 0 represents a pair of lines. If yes, find their point of intersection.
Solution:
We observe that
a = 1, h =3/2, b = 2, g = –1/2, f = 2, c = – 6
abc + 2fgh – af 2– bg 2– ch 2= – 12 + 3 – 4 – 1/2 + 27/2 = 0
Therefore the given second-degree equation represents a pair of lines, x 2+ 3xy + 2y 2– x – 4y – 6 = (x + 2y + 2) (x + y – 3).
Consider the equations formed by first two rows of.
i.e. ax + hy + g = 0 and hx + by + f = 0
i.e. x +3/2 y – 1/2 and 3/2 x + 2y – 2 = 0
Solving these, we get the required point of intersection.
i.e. 2x + 3y – 1 = 0
3x – 4y – 4 = 0
Solving the above equation, we get x = 8, y = –5.
Note :
(2x + 3y – 1)(3x + 4y – 4)≠ x 2+ 3xy + 2y 2– x – 4y – 6.
To read more,Buy study materials ofStraight Linescomprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematicshere.
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| | | | | | | | | | | | | | --- --- --- --- --- --- | | x^2 | x^{\msquare} | \log_{\msquare} | \sqrt{\square} | \nthroot[\msquare]{\square} | \le | \ge | \frac{\msquare}{\msquare} | \cdot | \div | x^{\circ} | \pi | | \left(\square\right)^{'} | \frac{d}{dx} | \frac{\partial}{\partial x} | \int | \int_{\msquare}^{\msquare} | \lim | \sum | \infty | \theta | (f\:\circ\:g) | f(x) | |
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--- --- ---
| - \twostack{▭}{▭} | \lt | 7 | 8 | 9 | \div | AC |
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| ▭\:\longdivision{▭} | \right) | . | 0 | = | + | y |
\mathrm{partial\:fractions} \mathrm{substitution} \mathrm{long\:division} \mathrm{trigonometric\:substitution} \mathrm{by\:parts}
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Introduction to Integration
What is an Integration?
Integration is the union of elements to create a whole. Integral calculus allows us to find a function whose differential is provided, so integrating is the inverse of differentiating. It defines and computes the area of a region constrained by the graph of a function. Integration developed historically from the process of exhaustion, in which inscribing polygons approximated the area of a curved form.
We distinguish integration into two forms: definite and indefinite integrals. Fundamental instruments in calculus, differentiation and integration have extensive use in mathematics and physics. Leibniz created the ideas of integration. Let us investigate integration, its features, and some of its effective approaches.
Integration - An Inverse Process of Differentiation
Integration is the opposite of differentiation basically. Integration helps us to determine the original function of a derivative if provided one.
If $\frac{d(F(x))}{dx}=f(x)$, then $\int f(x)dx=F(x)+C$ . This is known as indefinite integrals.
For Example
Suppose $f(x)=x^3$
The derivative of f(x) is $f'(x)=3x^2$
The antiderivative of $3x^2$ is $x^3$
So the derivative of any constant is zero and anti-derivative of any expression will contain arbitrary constant denoted by C that is $ \int 3x^2dx=x^3+C$
Therefore, if $\frac{dy}{dx}=f(x)$, then we can write it as y = $\int f(x)dx=F(x)+C$ where:
$\int f(x)dx$ will represent the complete class of integral.
C is an arbitrary constant.
x is the variable of equation.
The symbol $\int$ denotes the integral.
f(x) is the integrand.
Rules of Integration
Sum and Difference Rules:
$\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx$
$\int [f(x) - g(x)] dx = \int f(x) dx - \int g(x) dx$
For example: $\int (x^2 + 3x) dx = \int x^2 dx + \int 3x dx$
$= \frac{x^3}{3} + \frac{3x^2}{2} + C$
Power Rule:
$ \int x^ndx= \frac {x^{(n+1)}}{n+1} +C $
Please note here n$\neq$-1
For example: $\int x^5dx=\frac{x^6}{6}+C$
Exponential Rules:
$ \int e^xdx=e^x+C$
$ \int a^xdx= \frac{a^x}{ln(a)}+C$
$ \int ln(x)dx=xln(x)-x+C$
Constant Multiplication Rule:
$ \int adx=ax+C $
Reciprocal Rule:
$ \int \frac{1}{x}dx=ln|x|+C $
Properties of Integration
Properties of indefinite
$\int [f(x) \pm g(x)]dx= \int f(x)dx \pm \int g(x)dx $
$ \int kf(x)dx = k \int f(x)dx $ (here k is the constant)
$ \int f(x)dx= \int g(x)dx $ if $ \int [f(x)-g(x)]dx=0 $
By collabrating these properties, we derive: $\int \sum k _nf_n(x)dx=\sum k_n\int f_n(x)dx $
What are some uses for an integral calculator and what is it exactly?
One may find the integrals of functions by use of an integral calculator—a mathematical tool. Solving complex integration problems in a quick and exact way is the main application for this instrument in the domains of education, engineering, and physics. It can manage definite as well as indefinite integrals.
Implementations of an integral calculator:
Two examples of solving definite and indefinite integrals include computing the area under a curve or finding the antiderivative.
Double-checking hand computations helps one verify integration answers.
"Handling complex functions" is the capacity to combine activities that are challenging for manual handling.
Applications in Physics and Engineering: Applied to derive motion equations, work done, and areas under curves by use of these programs.
Improving learning means giving students help understanding integration techniques and their uses.
For example using an integration calculator we can find:
$\int (3x^2-2x+1)dx =3\frac{x^{2+1}}{2+1}-2 \frac{x^{1+1}}{1+1}+x+C$ $=3\frac{x^{3}}{3}-2 \frac{x^{2}}{2}+x+C$ $=x^3-x^2+x+C$
Methods of Integration
The simplest basic search might not always be enough to find an essential. We use several techniques for integration to help to simplify functions into normal forms. The main strategies are listed here:
1. Integration by Decomposition
In this method we need to breakthe function into basic parts: $$\int \frac{x^2-x+1}{x^3}dx$$ Expanding: $$\int (\frac{x^2}{x^3}-\frac{x}{x^3}+\frac{1}{x^3})dx $$ Applying the basic rules: $ log |x|+ \frac {1}{x}-\frac{1}{2x^2}+C$
2. Integration by Substitution
For simplyfing the integral change the variables: $$ \int sin(mx)dx $$ Let mx=t, so $dx=dt/m$ Therefore, $ \frac{1}{m} \int sin t dt= -\frac{1}{m}cos t +C $
3. Integration using Partial Fractions
We use this function for rational functions: $ \int \frac{1}{(x+1)(x+2)}dx=\frac{A}{x+1}+ \frac{B}{x+2} $ so now we can integrate them seperately and solve for the value of A and B.
4. Integration by Parts
Derived from the product rule of differentiation: $$ \int udv = uv - \int vdu $$
Applications of Integration
Integration has great use in domains like physics, engineering, and economics. Of the greatest significance are:
One may get areas by use of the area under curves.
The volume computation is mostly about figuring the solid of rotation volume.
Applications of physics abound in motion problems, work done, and the center of mass.
Calculation of probability density functions makes use of statistics and probability.
This extensive reference to integration addresses its basic ideas, guidelines, and methods, which provide the foundation for more complex uses of calculus.
Frequently Asked Questions (FAQ)
What is the use of integration in real life?
Integrations is used in various fields such as engineering to determine the shape and size of strcutures. In Physics to find the centre of gravity. In the field of graphical representation to build three-dimensional models.
What is the best integral calculator?
Symbolab is the best integral calculator solving indefinite integrals, definite integrals, improper integrals, double integrals, triple integrals, multiple integrals, antiderivatives, and more.
What does to integrate mean?
Integration is a way to sum up parts to find the whole. It is used to find the area under a curve by slicing it to small rectangles and summing up thier areas.
Why users love our Integral Calculator
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7185 | https://mindyourdecisions.com/blog/2022/05/10/a-challenging-geometry-problem-from-india-three-pairwise-tangent-circles-and-a-circumscribing-circle/ | Skip to content
Mind Your Decisions
Math Videos, Math Puzzles, Game Theory. By Presh Talwalkar
A Challenging Geometry Problem From India – Three Pairwise Tangent Circles And A Circumscribing Circle
If you buy from a link in this post, I may earn a commission. This does not affect the price you pay. As an Amazon Associate I earn from qualifying purchases. Learn more.
Posted May 10, 2022 By Presh Talwalkar. Read about me, or email me.
I thank Aditya for the suggestion! A version of this problem was given to 14-17 year old students in India.
Three congruent circles are pairwise tangent and each has a radius equal to 2. A circle circumscribes the three circles. Calculate the total area shaded in blue. The blue region is comprised of two parts. One region is the three circular sectors of the small circles enclosed by the line segments connecting the three small circle’s centers. The other region is outside the three small circles and bound by the large circumscribing circle (exclude the area in between the three small circles).
As usual, watch the video for a solution.
A Challenging Geometry Problem From India – Three Pairwise Tangent Circles And A Surrounding Circle
Or keep reading. . .
"All will be well if you use your mind for your decisions, and mind only your decisions." It costs thousands of dollars to run a website and your support matters. If you like the posts and videos, please consider a monthly pledge on Patreon. You may also consider a one-time donation to support my work.
. . . . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To A Challenging Geometry Problem From India
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
I will present one way to calculate the answer. First calculate (I) = the area of the large circle and subtract the areas of the three smaller circles. Then subtract the area (II) in between the three smaller circles, which is equal to the area of the equilateral triangle between the circle’s centers minus the area of the three circular sectors. Finally add the area (III) of the three circular sectors.
Suppose the large circle has a radius equal to R. Each small circle has a radius equal to 2. Subtracting the area of 3 small circles from the large circle gives the expression for the area of region (I):
(I)= πR2 – 3π(2)2 = πR2 – 12π
We will calculate R below, after we calculate the area of region (II).
Each small circle has a radius equal to 2. The triangle formed by connecting the three small circle’s center has three sides equal to two radii with length 2 + 2 = 4. Thus the triangle is an equilateral triangle with s = 4. Furthermore each circular sector will have a central angle equal to 60 degrees, so three circular sectors will be equivalent to a semicircle with a central angle equal to 180 degrees. Subtracting the area of a semicircle with radius r = 2 from an equilateral triangle with side length 4 gives the area of region (II):
(II)= s2(√3)/4 – πr2/2 = 42(√3)/4 – π(22)/2 = 4(√3) – 2π
(III)πr2/2 = 2π
The blue regions have a total area equal to the difference of (I) and (II), which is:
(I) – (II) + (III)= πR2 – 12π – (4(√3) – 2π) + 2π = πR2 – 8π – 4(√3)
Radius of the circumscribing circle
Connect the centers of the 3 small circles to form an equilateral triangle.
If two circles are tangent, then their centers and the tangent point are collinear. Thus the circumscribing circle’s center is collinear with the lower left circle’s center and its tangent point. This line also bisects the angle of the equilateral triangle. The large circle’s center is also equidistant from the 3 smaller circle’s centers, so it is along the perpendicular bisector of each side of the equilateral triangle. Construct one such perpendicular bisector, leading to a 30-60-90 right triangle. The longer leg is equal to the radius of the small circle, 2, so the shorter leg will be 2/√3 and the hypotenuse will be 4/√3. The large circle’s radius is then 4/√3 plus one small circle’s radius length, so it is 2 + 4/√3.
We can now solve the problem:
(I) – (II) + (III)= πR2 – 8π – 4(√3) = π(4 + 16/√3 + 16/3) – 8π – 4(√3) = π(16/√3 + 4/3) – 4(√3) ≈ 26.281
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7186 | https://www.youtube.com/watch?v=OaKCYDZPVDo | Solving Polynomial INEQUALITIES | jensenmath.ca |
JensenMath
236000 subscribers
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Description
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Posted: 24 Feb 2020
Learn to solve polynomial inequalities by graphing and also by using a factor table (sign chart). This example will walk you through the 3 steps you will use:
1) Rearrange all terms to one side
2) Fully factor
3) Graph or make a factor table
Find additional resources at jensenmath.ca here:
Link to desmos graph used in example:
12 comments
Transcript:
Intro this is a tutorial on solving polynomial inequality so in this video I'm going to take you through an example where we figure out when is this polynomial less than or equal to negative 16 and it's an inequality specifically because we have this inequality symbol here the question doesn't say when is it equal it says when is it less than or equal to so there's a couple strategies we can use when solving polynomial inequalities one we could solve by graphing I know that we could solve it using something called a factor table or sign chart I'll take you through both methods so before we begin if you haven't subscribed to the channel yet make sure you do that right now so that you can easily access tutorials on you know any topic you're going to learn in high school math so here we go [Music] so step one whenever solving a Steps polynomial inequality is moving all the terms to one side so that it says it's less than or greater than zero so I'm going to start by moving this negative 16 to the other side and we can just follow a normal a liberal rules been rearranging an equation as long as you remember that if you ever multiply or divide by a negative number you have to breathe you have to reverse the inequality symbol but in this case I'm just adding 16 to both sides so I don't have to reverse the inequality it stays just like this let me fix that symbol a little bit less than or equal to zero so the reason why we do this is because now we're interested in when is this function less than or equal to zero and that's gonna have the same set of answers as when this function was less than or equal to negative 16 but this one's a lot easier to work with because if you think about it logically now we're just interested in while this function you know this functions degree three negative leading coefficient so I know it goes from two to four and it's degree three so it could have two turning points in between with a y-intercept of 16 so it looks something like this we're interested in when is it less than or equal to zero so when is it below the x-axis so I'd say in this interval and this interval now this graph isn't perfectly accurate but you can see why having it set less than or equal to zero is useful because then we're just looking for when are the Y values of the function less than or equal to zero for what range of X values is this true so in order to be able to graph this perfectly accurate accurately we're gonna have to find what are these x-intercepts here where the function switches from being positive to negative or vice versa so if we want to find the x-intercepts you know by now that we need to get this into factored form so that's why step two is factoring the polynomial so when factoring step one should be always to check for a common factor so I noticed that I could common factor out negative two from all of the terms x cubed plus 3x squared minus 6x minus eight less than or equal to serial and now I'm interested in factoring what I have inside here further right I want to get into fully factored form so can find the x-intercepts so it's four terms so I would try grouping but you'll notice grouping won't work with this we wouldn't get a common binomial so when that happens you're going to have to test for zeroes and do your synthetic or long and division to get it into factored form so I'm going to want to try and factor this further I'm just gonna call this f of X to keep my work organized now we need to try and guess and check integers that could make this zero and the only integers that could work so I call these possible zeros our factors of eight so plus or minus one to four or eight those are the set of integers that could make this become zero we just have to test until we find one that works if we plugged 1 in for X you know it wouldn't equal zero but if we collect negative 1 in for X I'll get negative 1 plus 3 plus 6 minus 8 so that's 2 plus 6 is 8 minus 8 is 0 so I just figured out that F at negative 1 equals 0 so that tells me that X plus 1 is a factor of ethics so I can off to the side here I can divide this polynomial here by X plus 1 and I'll use synthetic division I think that's a little bit quicker so I figured negative one was the zero the coefficients of the function of 1 3 negative 6 negative 8 do this since that division quickly negative 1 times 1 is negative 1 3 plus negative 1 is 2 negative 1 times 2 is negative 2 add get negative a multiply get 8 and there's my remainder of 0 so what I have here is my remainder my X to the 0 term my X to the 1 term and my X to the 2 term that are in my quotient so I can now factor this 2 negative 2x plus 1 times my quotient x squared plus 2x minus 8 with no remainder less than or equal to 0 now that it's in factored form oh I can factor this one further right this is a degree 2 polynomial so it's a quadratic I can factor that further if I can find numbers that multiply to negative 8 and add to 2 and those numbers do exist they are 4 and negative 2 so there we go now that's full factored form right each factor is degree one and I have this constant factor in front I could divide that over but I don't have to reverse the inequality so I'm just gonna leave it just to keep this as straightforward as possible so I know this function has x-intercepts at x equals negative 1 negative 4 and 2 right each of those values would me cheat each of those values would make this whole product become 0 so this tells me when it's equal to 0 by only 1 is it less than or equal to 0 so we're going to use those x-intercepts as the dividing points either in our graph or in a factor table so I'll show you both methods so our graph we know there's x-intercepts I'll just roughly plot them at negative 4 negative 1 and 2 and the function remember this is the factored form version of it's the factored form version of this function so how do y intercept of 16 so let me just roughly plot that as well I mean the y intercept isn't really useful for this but just to keep the graph as accurate as possible and remember that function up there where he talked about it it was a negative leading coefficient and an odd degree so we know that goes from Q 2 to Q 4 so I know it's going to go from here to here and if you look at each of the factors each of the factors was order 1 so I know it's gonna go straight throughout each of the x-intercepts so the function is going to look something like this so that's a very rough sketch of what the function would look like and now we're interested in explaining when is this function less than or equal to zero so for what X values do the Y values of the function go to zero or below so we would be interested in this part of the function and also this part of the function in those two intervals between X values of negative 4 negative 1 and between 2 and infinity the Y values of the function would be negative so that's going to be the answer to our inequality so now we can read our answer in two different ways we could write it in bracket notation so it's between negative four and negative ones we want to describe our interval as saying the inequality is true when we have an x value there is an element of numbers between negative 4 and negative 1 including negative 4 and negative 1 that's why I have the square brackets or if we have an x value between 2 and infinity including 2 round bracket at infinity always a round bracket at infinity or we could write Graphs this as an inequality we could say between negative 4 and negative 1 including those two that's why there's the equal sign under or X is greater than or equal to 2 so hopefully understand why this is the answer right these are the X values that make the original inequality true these are the input values so if I input a value into the original inequality that is within this interval or this interval it'll make the inequality true so let me give you a better graphical representation so this is just a very rough graph let me show you what the desmos graph looks like so here's the same function graphed on desmos and you'll notice while I'm moving the slider here just take a look at the point so the point has the x value and the y value so look between negative 4 and negative 1 and I'm talking about X values between negative 4 and negative 1 look at what the Y values are the Y values are all negative Y values are all negative between negative 4 negative 1 but what about when we go to X values between negative 1 & 2 well the Y values are all positive there right the functions above the x-axis and then what about for X values bigger than 2 well all the Y values are now negative so the original question wants to know when is this function less than or equal to 0 so when are the Y values negative for what X values are the Y values negative for X values between negative 4 and 1 and also for X values bigger than 2 and now notice we included negative 4 negative 1 and 2 in our answer because the funk the original question wants to know not only when is it less than 0 it also wants to include the times when it's equal to 0 so when our final answer that's why we put square brackets in our bracket notation or when we wrote it as inequalities we put lines underneath so now let me show you how we could get the exact same answers using another method called the factor table or sign chart and how does that work well at the of it we put negative infinity on the left we put infinity on the far right and between it what we do is we divide it up into intervals based on where the x-intercepts are well we figured out the x-intercepts were at negative 4 negative 1 & 2 so negative 4 negative 1 & 2 and basically what we're going to do instead of actually graphing it and looking at the graph we're just going to choose test points and plug those test points into the original equation and figure out are the Y values positive or negative without actually having to look at the graph so we're going to have to choose a test point within each interval so I'm going to choose some test points in each of these intervals now so between negative infinity and negative 4 I'll choose negative 5 but you know negative 10 would have worked as well and between negative 4 and negative 1 I'm going to choose negative 2 between negative 1 & 2 I'm going to choose 0 in between 2 and infinity I'm going to choose 3 so these are all X values that we are going to input into the function and test whether the function is positive or negative and when we test it in the function we're gonna do it in a bit of an organized way we're going to test it in each of the factors separately instead of plugging it into the entire original function and then evaluating it it's actually easier if we plug it into each factor separately let me show you what I mean by that so we put each of the factors down the side here so negative 2 was a factor of our function X plus 4 was a factor X plus 1 was a factor and X minus 2 is a factor and now what we do is we test each of these numbers into each of these factors and then just record whether our result would be positive or negative so well negative 2 is always negative I could just complete that real quickly but when negative 5 is plugged into X plus 4 I would get negative 5 plus 4 which is negative 1 when negative 5 is plugged into X plus 1 I would get negative 5 plus 1 which is negative 4 and negative 5 minus 2 is negative 7 so notice what we have figured out here is that each of the factors in this factored form equation here would become negative so I would have a negative factor times a negative factor times a negative factor times a negative factor an even amount of negative factors gives us an overall result of a positive function meaning that it is bigger than zero at that point so this is no part of our answer notice the function in that interval we just tested in the negative infinity negative for the function is positive that this part of the function all is positive Y values well how about between negative four and negative one half with this part what are all these Y values so we test negative two into the function and it makes this factor positive it makes this factor negative it makes this factor negative an odd number of negative factors results in an overall function that is negative and notice yes all the Y values are negative there how about zero positive positive negative we have two negative factors which results in a positive function and look the Y values in that section are all positive tests three positive positive positive so we have one negative factor which means the original function has negative Y values in that section so we just use the X intercepts as the dividing points and then choose test values in each interval plug them into the factors of the original function and figure out the sign of the original function and notice how this table reveals the final answer to us the original question said when is it less than or equal to zero so when are the Y values negative and these two intervals so between negative four and negative one and between two and infinity so if we look at our final answer that's exactly what we had here between negative 4 negative 1 and 2 to infinity and we included those values because there is an equal sign under the inequality in the original question all right so that's the end of this tutorial hopefully it helped make sure you go to Jensen math dot CA if you haven't been there yet to get any supporting materials and some extra practice questions and also make sure you subscribe to the channel |
7187 | https://www.mathletetraining.com/math_wiki/thales-of-miletus/ | Thales of Miletus – Mathlete Training Centre
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Thales of Miletus
Monday, 04/01/2021
Thales of Miletus was a Greek mathematician, astronomer and pre-Socratic philosopher from Miletus in Ionia, Asia Minor. He was one of the Seven Sages of Greece.
Thales used geometry to calculate the heights of pyramids and the distance of ships from the shore. He is the first person known to have used deductive reasoning applied to geometry, by deriving four corollaries to Thales’ theorem.
Thales’s theorem
Thales’s theorem states that if A, B, and C are distinct points on a circle where the line AC is a diameter, then the angle ∠ABC is a right angle.
Intercept theorem
The Intercept theorem provides the ratios between the line segments created when two parallel lines are intercepted by two intersecting lines.
It is sometimes called also “Thales’ Theorem”.
It states that if m and n are parallel lines, and XC and XB are two intersecting lines that intercept m and n, then the following relationship between the lengths of the line segments is true:
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HomeMathematicsSimilarity Transformations
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Similarity Transformations
Difficulty Level: Basic | Created by: CK-12
Last Modified: Aug 07, 2013
Learning Objectives
Draw a dilation of a given figure.
Plot an image when given the center of dilation and scale factor.
Determine if one figure is the dilation of another.
Review Queue
Are the two quadrilaterals similar? How do you know?
What is the scale factor from \begin{align}XYZW\end{align} to \begin{align}CDAB\end{align}? Leave as a fraction.
Quadrilateral \begin{align}EFGH\end{align} has vertices \begin{align}E(-4, -2), F(2, 8), G(6, 2)\end{align} and \begin{align}H(0, -4)\end{align}. Quadrilateral \begin{align}LMNO\end{align} has vertices \begin{align}L(-2, -1), M(1, 4), N(3, 1),\end{align} and \begin{align}O(0, -2)\end{align}. Determine if the two quadrilaterals are similar. Explain your reasoning.
Know What? One practical application of dilations is perspective drawings. These drawings use a vanishing point (the point where the road meets the horizon) to trick the eye into thinking the picture is three-dimensional. The picture to the right is a one-point perspective and is typically used to draw streets, train tracks, rivers or anything else that is linear.
There are also two-point perspective drawings, which are very often used to draw a street corner or a scale drawing of a building.
Both of these drawing are simple representations of one and two perspective drawings. Your task for this Know What? is to draw your own perspective drawing with either one or two vanishing points and at least 5 objects. Each object should have detail (windows, doors, sign, stairs, etc.)
Dilations
A dilation makes a figure larger or smaller, but has the same shape as the original. In other words, the dilation is similar to the original.
Transformation: An operation that moves, flips, or changes a figure to create a new figure. Transformations that preserve size are rigid and ones that do not are non-rigid.
Dilation: A non-rigid transformation that preserves shape but not size.
All dilations have a center and a scale factor. The center is the point of reference for the dilation (like the vanishing point in a perspective drawing) and scale factor tells us how much the figure stretches or shrinks. A scale factor is typically labeled \begin{align}k\end{align} and is always greater than zero. Also, if the original figure is labeled \begin{align}\triangle ABC\end{align}, for example, the dilation would be \begin{align}\triangle A'B'C'\end{align}. The \begin{align}'\end{align} indicates that it is a copy. This tic mark is said “prime,” so \begin{align}A'\end{align} is read “A prime.” A second dilation would be \begin{align}A''\end{align}, read “A double-prime.”
Example 1: The center of dilation is \begin{align}P\end{align} and the scale factor is 3. Find \begin{align}Q'\end{align}.
Solution: If the scale factor is 3 and \begin{align}Q\end{align} is 6 units away from \begin{align}P\end{align}, then \begin{align}Q'\end{align} is going to be \begin{align}6 \times 3 = 18\end{align} units away from \begin{align}P\end{align}. Because we are only dilating apoint, the dilation will be collinear with the original and center.
Example 2: Using the picture above, change the scale factor to \begin{align}\frac{1}{3}\end{align}. Find \begin{align}Q''\end{align}.
Solution: Now the scale factor is \begin{align}\frac{1}{3}\end{align}, so \begin{align}Q''\end{align} is going to be \begin{align}\frac{1}{3}\end{align} the distance away from \begin{align}P\end{align} as \begin{align}Q\end{align} is. In other words, \begin{align}Q''\end{align} is going to be \begin{align}6 \times \frac{1}{3} = 2\end{align} units away from \begin{align}P\end{align}. \begin{align}Q''\end{align} will also be collinear with \begin{align}Q\end{align} and center.
Example 3: \begin{align}KLMN\end{align} is a rectangle with length 12 and width 8. If the center of dilation is \begin{align}K\end{align} with a scale factor of 2, draw \begin{align}K'L'M'N'\end{align}.
Solution: If \begin{align}K\end{align} is the center of dilation, then \begin{align}K\end{align} and \begin{align}K'\end{align} will be the same point. From there, \begin{align}L'\end{align} will be 8 units above \begin{align}L\end{align} and \begin{align}N'\end{align} will be 12 units to the right of \begin{align}N\end{align}.
Example 4: Find the perimeters of \begin{align}KLMN\end{align} and \begin{align}K'L'M'N'\end{align}. Compare this to the scale factor.
Solution: The perimeter of \begin{align}KLMN = 12 + 8 + 12 + 8 = 40\end{align}. The perimeter of \begin{align}K'L'M'N' = 24 + 16 + 24 + 16 = 80\end{align}. The ratio of the perimeters is 80:40 or 2:1, which is the same as the scale factor.
Example 5: \begin{align}\triangle ABC\end{align} is a dilation of \begin{align}\triangle DEF\end{align}. If \begin{align}P\end{align} is the center of dilation, what is the scale factor?
Solution: Because \begin{align}\triangle ABC\end{align} is a dilation of \begin{align}\triangle DEF\end{align}, we know that the triangles are similar. Therefore the scale factor is the ratio of the sides. Since \begin{align}\triangle ABC\end{align} is smaller than the original, \begin{align}\triangle DEF\end{align}, the scale factor is going to be a fraction less than one, \begin{align}\frac{12}{20}=\frac{3}{5}\end{align}.
If \begin{align}\triangle DEF\end{align} was the dilated image, the scale factor would have been \begin{align}\frac{5}{3}\end{align}.
If the dilated image is smaller than the original, then the scale factor is \begin{align}0<k<1\end{align}.
If the dilated image is larger than the original, then the scale factor is \begin{align}k>1\end{align}.
Dilations in the Coordinate Plane
In this text, the center of dilation will always be the origin, unless otherwise stated.
Example 6: Determine the coordinates of \begin{align}\triangle ABC\end{align} and \begin{align}\triangle A'B'C'\end{align} and find the scale factor.
Solution: The coordinates of \begin{align}\triangle ABC\end{align} are \begin{align}A(2, 1), B(5, 1)\end{align} and \begin{align}C(3, 6)\end{align}. The coordinates of \begin{align}\triangle A'B'C'\end{align} are \begin{align}A'(6, 3), B'(15, 3)\end{align} and \begin{align}C'(9, 18)\end{align}. By looking at the corresponding coordinates, each is three times the original. That means \begin{align}k = 3\end{align}.
Again, the center, original point, and dilated point are collinear. Therefore, you can draw a ray from the origin to \begin{align}C', B',\end{align} and \begin{align}A'\end{align} such that the rays pass through \begin{align}C, B,\end{align} and \begin{align}A\end{align}, respectively.
Let’s show that dilations are a similarity transformation (preserves shape). Using the distance formula, we will find the lengths of the sides of both triangles in Example 6 to demonstrate this.
\begin{align}& \underline{\triangle ABC} && \underline{\triangle A'B'C'}\
& AB=\sqrt{(2-5)^2+(1-1)^2}=\sqrt{9}=3 && A'B'=\sqrt{(6-15)^2+(3-3)^2}=\sqrt{81}=9\
& AC=\sqrt{(2-3)^2+(1-6)^2}=\sqrt{26} && A'C'=\sqrt{(6-9)^2+(3-18)^2}=\sqrt{234}=3 \sqrt{26}\
& CB=\sqrt{(3-5)^2+(6-1)^2}=\sqrt{29} && C'B'=\sqrt{(9-15)^2+(18-3)^2}=\sqrt{261}=3 \sqrt{29}\end{align}
From this, we also see that all the sides of \begin{align}\triangle A'B'C'\end{align} are three times larger than \begin{align}\triangle ABC\end{align}. Therefore, a dilation will always produce a similar shape to the original.
In the coordinate plane, we say that \begin{align}A'\end{align} is a “mapping” of \begin{align}A\end{align}. So, if the scale factor is 3, then \begin{align}A(2, 1)\end{align} is mapped to (usually drawn with an arrow) \begin{align}A'(6, 3)\end{align}. The entire mapping of \begin{align}\triangle ABC\end{align} can be written \begin{align}(x,y) \rightarrow (3x, 3y)\end{align} because \begin{align}k = 3\end{align}. For any dilation the mapping will be \begin{align}(x, y) \rightarrow (kx, ky)\end{align}.
Know What? Revisited Answers to this project will vary depending on what you decide to draw. Make sure that you have at least five objects with some sort of detail. If you are having trouble getting started, go to the website:
Review Questions
Given \begin{align}A\end{align} and the scale factor, determine the coordinates of the dilated point, \begin{align}A'\end{align}. You may assume the center of dilation is the origin.
\begin{align}A(3, 9), k = \frac{2}{3}\end{align}
\begin{align}A(-4, 6), k = 2\end{align}
\begin{align}A(9, -13), k = \frac{1}{2}\end{align}
Given \begin{align}A\end{align} and \begin{align}A'\end{align}, find the scale factor. You may assume the center of dilation is the origin.
\begin{align}A(8, 2), A'(12, 3)\end{align}
\begin{align}A(-5, -9), A'(-45, -81)\end{align}
\begin{align}A(22, -7), A(11, -3.5)\end{align}
In the two questions below, you are told the scale factor. Determine the dimensions of the dilation. In each diagram, the black figure is the original and \begin{align}P\end{align} is the center of dilation.
\begin{align}k = 4\end{align}
\begin{align}k = \frac{1}{3}\end{align}
In the two questions below, find the scale factor, given the corresponding sides. In each diagram, the black figure is the original and \begin{align}P\end{align} is the center of dilation.
Find the perimeter of both triangles in #7. What is the ratio of the perimeters?
Writing What happens if \begin{align}k = 1\end{align}?
The origin is the center of dilation. Find the coordinates of the dilation of each figure, given the scale factor.
\begin{align}A(2, 4), B(-3, 7), C(-1, -2); k = 3\end{align}
\begin{align}A(12, 8), B(-4, -16), C(0, 10); k = \frac{3}{4}\end{align}
Multi-Step Problem Questions 15-21 build upon each other.
Plot \begin{align}A(1, 2), B(12, 4), C(10, 10)\end{align}. Connect to form a triangle.
Make the origin the center of dilation. Draw 4 rays from the origin to each point from #15. Then, plot \begin{align}A'(2, 4), B'(24, 8), C'(20, 20)\end{align}. What is the scale factor?
Use \begin{align}k =4\end{align}, to find \begin{align}A''B''C''\end{align}. Plot these points.
What is the scale factor from \begin{align}A'B'C'\end{align} to \begin{align}A''B''C''\end{align}?
Find (\begin{align}O\end{align} is the origin):
\begin{align}OA\end{align}
\begin{align}AA'\end{align}
\begin{align}AA''\end{align}
\begin{align}OA'\end{align}
\begin{align}OA''\end{align}
Find:
\begin{align}AB\end{align}
\begin{align}A'B'\end{align}
\begin{align}A''B''\end{align}
Compare the ratios:
\begin{align}OA:OA'\end{align} and \begin{align}AB:A'B'\end{align}
\begin{align}OA:OA''\end{align} and \begin{align}AB:A''B''\end{align}
Algebra Connection For questions 22-27, use quadrilateral \begin{align}ABCD\end{align} with \begin{align}A(1, 5), B(2, 6), C(3, 3)\end{align} and \begin{align}D(1, 3)\end{align} and its transformation \begin{align}A'B'C'D'\end{align} with \begin{align}A'(-3, 1), B'(0, 4), C'(3, -5)\end{align} and \begin{align}D'(-3, -5)\end{align}.
Plot the two quadrilaterals in the coordinate plane.
Find the equation of \begin{align}\overleftrightarrow{CC'}\end{align}.
Find the equation of \begin{align}\overleftrightarrow{DD'}\end{align}.
Find the intersection of these two lines algebraically or graphically.
What is the significance of this point?
What is the scale factor of the dilation?
Construction We can use a compass and straight edge to construct a dilation as well. Copy the diagram below.
Set your compass to be \begin{align}CG\end{align} and use this setting to mark off a point 3 times as far from \begin{align}C\end{align} as \begin{align}G\end{align} is. Label this point \begin{align}G'\end{align}. Repeat this process for \begin{align}CO\end{align} and \begin{align}CD\end{align} to find \begin{align}O'\end{align} and \begin{align}D'\end{align}.
Connect \begin{align}G', O'\end{align} and \begin{align}D'\end{align} to make \begin{align}\triangle D'O'G'\end{align}. Find the ratios, \begin{align}\frac{D'O'}{DO}, \frac{O'G'}{OG}\end{align} and \begin{align}\frac{G'D'}{GD}\end{align}.
What is the scale factor of this dilation?
Describe how you would dilate the figure by a scale factor of 4.
Describe how you would dilate the figure by a scale factor of \begin{align}\frac{1}{2}\end{align}.
Review Queue Answers
Yes, all the angles are congruent and the corresponding sides are in the same ratio.
\begin{align}\frac{5}{3}\end{align}
Yes, \begin{align}LMNO \sim EFGH\end{align} because \begin{align}LMNO\end{align} is exactly half of \begin{align}EFGH\end{align}.
Notes/Highlights
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Description
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Difficulty Level
Basic
Tags
Problem Solving, ratio, cross products, proportions, ratios, transformations, indirect measurement, similarity, polygons, Extremes, Parallel Lines, Perimeter, cross multiplication, proportion, distance formula, self-similarity, fractals, dilations, proportionality, CK.MAT.ENG.SE.1.Geometry-Basic.7, corresponding angles, dilation, AAA, SAS, SSS, corresponding sides, Sierpinski triangle, transversals, Triangle Proportionality Theorem, proportional segments, AA, scale factor, similar polygons, means, CK.MAT.ENG.SE.2.Geometry.7, (32 more)
Subjects
mathematics
Grades
9, 10
Standards Correlations
-
Concept Nodes
License
CC BY NC
Language
English
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Date Created
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Last Modified
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7189 | https://www.hilotutor.com/archives_celerity.html | Make Your Point > Archived Issues > CELERITY Send Make Your Point issues straight to your inbox. | | | | | | | | | | | | --- --- --- --- --- | | | | --- | | | | | | | connect today's word to others: | Let's say you're a firefighter arriving at a burning house. If you act with celerity, you act with swiftness: you know what to do, so you leap instantly into action. If you act with a___rity, you act with cheerful eagerness: you smile and whistle as you make your way confidently into the house. | | | make your point with... | "CELERITY" "Celerity" comes from the Latin word for "swift," which explains why it looks a little like "accelerate." When you do something with celerity, you do it with speed or swiftness. Part of speech: Uncountable noun. (Like "milk," "rice," and "advice," uncountable nouns are words for stuff that can’t be broken into exact units. You talk about "some milk," "the rice," and "a lot of advice," but you don’t say "a milk," "three rices," or "many advices." Likewise, talk about "the celerity," "this celerity," "its celerity," "such celerity," "no celerity," and so on, but don’t say "one celerity" or "celerities." An exception is that you can say "a celerity" if you're talking about a specific level of speed: "He reacted with a celerity that shocked me.") Other common forms: none How to use it: When "speed," "swiftness," and "rapidity" aren't fancy or formal enough for your context, pick "celerity" instead, but be aware that not that many English speakers know it. Talk about someone doing something with celerity. You might act or decide with celerity, donate or contribute with celerity, improve or advance with celerity, follow or obey with celerity, move or whirl around with celerity, enter or exit with celerity, respond or reply or react with celerity, etc. Or, talk about the celerity of something: the celerity of the deer's reactions, her celerity of reply, the celerity of their victory. Add an adjective, if you want: "she moved with astonishing celerity," "they reacted with surprisingcelerity," "he spoke with great celerity." Although it's usually people who act with celerity, you might also talk about things happening with celerity: "the virus is mutating with celerity," "the canoes slipped with celerity across the quiet lake waters." | | | examples: | Seeing the distressed driver in her stalled car, Chad responded with celerity, helping other bystanders to push it into a safe position. | | | study it now: | Look away from the screen to explain the definitionin your own words. You’ll know you understand what "celerity" means when you can explain it without saying "quickness" or "speediness." | | | try it out: | Fill in the blanks: "With great celerity, (someone) (was doing something)." Example: "With great celerity, citizens were flocking to donate blood to those who had been injured in the attack." | | | before you review: | | Disclaimer: When I write definitions, I use plain language and stick to the words' common, useful applications. If you're interested in authoritative and multiple definitions of words, I encourage you to check a dictionary. Also, because I'm American, I stick to American English when I share words' meanings, usage, and pronunciations; these elements sometimes vary across world Englishes.
| | | | | | | | | | | |
--- --- --- --- ---
| | | | --- | | | | | | | connect today's word to others: | Let's say you're a firefighter arriving at a burning house. If you act with celerity, you act with swiftness: you know what to do, so you leap instantly into action. If you act with a___rity, you act with cheerful eagerness: you smile and whistle as you make your way confidently into the house. | | | make your point with... | "CELERITY" "Celerity" comes from the Latin word for "swift," which explains why it looks a little like "accelerate." When you do something with celerity, you do it with speed or swiftness. Part of speech: Uncountable noun. (Like "milk," "rice," and "advice," uncountable nouns are words for stuff that can’t be broken into exact units. You talk about "some milk," "the rice," and "a lot of advice," but you don’t say "a milk," "three rices," or "many advices." Likewise, talk about "the celerity," "this celerity," "its celerity," "such celerity," "no celerity," and so on, but don’t say "one celerity" or "celerities." An exception is that you can say "a celerity" if you're talking about a specific level of speed: "He reacted with a celerity that shocked me.") Other common forms: none How to use it: When "speed," "swiftness," and "rapidity" aren't fancy or formal enough for your context, pick "celerity" instead, but be aware that not that many English speakers know it. Talk about someone doing something with celerity. You might act or decide with celerity, donate or contribute with celerity, improve or advance with celerity, follow or obey with celerity, move or whirl around with celerity, enter or exit with celerity, respond or reply or react with celerity, etc. Or, talk about the celerity of something: the celerity of the deer's reactions, her celerity of reply, the celerity of their victory. Add an adjective, if you want: "she moved with astonishing celerity," "they reacted with surprisingcelerity," "he spoke with great celerity." Although it's usually people who act with celerity, you might also talk about things happening with celerity: "the virus is mutating with celerity," "the canoes slipped with celerity across the quiet lake waters." | | | examples: | Seeing the distressed driver in her stalled car, Chad responded with celerity, helping other bystanders to push it into a safe position. | | | study it now: | Look away from the screen to explain the definitionin your own words. You’ll know you understand what "celerity" means when you can explain it without saying "quickness" or "speediness." | | | try it out: | Fill in the blanks: "With great celerity, (someone) (was doing something)." Example: "With great celerity, citizens were flocking to donate blood to those who had been injured in the attack." | | | before you review: | |
| | |
--- |
| | |
| |
| connect today's word to others: |
Let's say you're a firefighter arriving at a burning house.
If you act with celerity, you act with swiftness: you know what to do, so you leap instantly into action.
If you act with a___rity, you act with cheerful eagerness: you smile and whistle as you make your way confidently into the house.
| |
| make your point with... |
"CELERITY"
"Celerity" comes from the Latin word for "swift," which explains why it looks a little like "accelerate."
When you do something with celerity, you do it with speed or swiftness.
Part of speech:
Uncountable noun.
(Like "milk," "rice," and "advice," uncountable nouns are words for stuff that can’t be broken into exact units. You talk about "some milk," "the rice," and "a lot of advice," but you don’t say "a milk," "three rices," or "many advices."
Likewise, talk about "the celerity," "this celerity," "its celerity," "such celerity," "no celerity," and so on,
but don’t say "one celerity" or "celerities."
An exception is that you can say "a celerity" if you're talking about a specific level of speed: "He reacted with a celerity that shocked me.")
Other common forms:
none
How to use it:
When "speed," "swiftness," and "rapidity" aren't fancy or formal enough for your context, pick "celerity" instead, but be aware that not that many English speakers know it.
Talk about someone doing something with celerity. You might act or decide with celerity, donate or contribute with celerity, improve or advance with celerity, follow or obey with celerity, move or whirl around with celerity, enter or exit with celerity, respond or reply or react with celerity, etc.
Or, talk about the celerity of something: the celerity of the deer's reactions, her celerity of reply, the celerity of their victory.
Add an adjective, if you want: "she moved with astonishing celerity," "they reacted with surprisingcelerity," "he spoke with great celerity."
Although it's usually people who act with celerity, you might also talk about things happening with celerity: "the virus is mutating with celerity," "the canoes slipped with celerity across the quiet lake waters."
| |
| examples: |
Seeing the distressed driver in her stalled car, Chad responded with celerity, helping other bystanders to push it into a safe position.
| |
| study it now: |
Look away from the screen to explain the definitionin your own words. You’ll know you understand what "celerity" means when you can explain it without saying "quickness" or "speediness."
| |
| try it out: |
Fill in the blanks: "With great celerity, (someone) (was doing something)."
Example: "With great celerity, citizens were flocking to donate blood to those who had been injured in the attack."
| |
| before you review: |
| |
| make your point with... |
"Celerity" comes from the Latin word for "swift," which explains why it looks a little like "accelerate."
When you do something with celerity, you do it with speed or swiftness.
Part of speech:
Uncountable noun.
(Like "milk," "rice," and "advice," uncountable nouns are words for stuff that can’t be broken into exact units. You talk about "some milk," "the rice," and "a lot of advice," but you don’t say "a milk," "three rices," or "many advices."
Likewise, talk about "the celerity," "this celerity," "its celerity," "such celerity," "no celerity," and so on,
but don’t say "one celerity" or "celerities."
An exception is that you can say "a celerity" if you're talking about a specific level of speed: "He reacted with a celerity that shocked me.")
Other common forms:
none
How to use it:
When "speed," "swiftness," and "rapidity" aren't fancy or formal enough for your context, pick "celerity" instead, but be aware that not that many English speakers know it.
Talk about someone doing something with celerity. You might act or decide with celerity, donate or contribute with celerity, improve or advance with celerity, follow or obey with celerity, move or whirl around with celerity, enter or exit with celerity, respond or reply or react with celerity, etc.
Or, talk about the celerity of something: the celerity of the deer's reactions, her celerity of reply, the celerity of their victory.
Add an adjective, if you want: "she moved with astonishing celerity," "they reacted with surprisingcelerity," "he spoke with great celerity."
Although it's usually people who act with celerity, you might also talk about things happening with celerity: "the virus is mutating with celerity," "the canoes slipped with celerity across the quiet lake waters."
| |
| examples: |
Seeing the distressed driver in her stalled car, Chad responded with celerity, helping other bystanders to push it into a safe position.
| |
| study it now: |
Look away from the screen to explain the definitionin your own words. You’ll know you understand what "celerity" means when you can explain it without saying "quickness" or "speediness."
| |
| try it out: |
Fill in the blanks: "With great celerity, (someone) (was doing something)."
Example: "With great celerity, citizens were flocking to donate blood to those who had been injured in the attack."
| |
| before you review: |
| |
| study it now: |
Look away from the screen to explain the definitionin your own words. You’ll know you understand what "celerity" means when you can explain it without saying "quickness" or "speediness."
| |
| try it out: |
Fill in the blanks: "With great celerity, (someone) (was doing something)."
Example: "With great celerity, citizens were flocking to donate blood to those who had been injured in the attack."
| |
| before you review: |
| |
| try it out: |
Fill in the blanks: "With great celerity, (someone) (was doing something)."
Example: "With great celerity, citizens were flocking to donate blood to those who had been injured in the attack."
| |
| before you review: |
| |
| before you review: |
Spend at least 20 seconds occupying your mind with the game below. Then try the review questions. Don’t go straight to the review now—let your working memory empty out first.
| |
| review today's word: |
1. One opposite of CELERITY is
| |
| a final word: |
Make Your Point is crafted with love and brought to you each weekday morning by Liesl Johnson, a reading and writing tutor on a mission to explore, illuminate, and celebrate words.
Subscribe to "Make Your Point" for a daily vocabulary boost.
© Copyright 2017 | All rights reserved. |
7190 | https://www.ck12.org/flexi/cbse-math/factorization-using-identities/how-to-factor-trinomials-when-'a'-is-not-1/ | Flexi answers - How to factor trinomials when 'a' is not 1? | CK-12 Foundation
All Subjects
CBSE Math
Factorization using Identities
Question
How to factor trinomials when 'a' is not 1?
Flexi Says:
We know how to factor quadratic trinomials \begin{align}(ax^2+bx+c)\end{align} where \begin{align}a = 1\end{align} using methods we have previously learned. To factor a quadratic polynomial where \begin{align}a \neq 1\end{align}, we should factor by grouping using the following steps:
Step 1: We find the product \begin{align}ac\end{align}.
Step 2: We look for two numbers that multiply to give \begin{align}ac\end{align} and add to give \begin{align}b\end{align}.
Step 3: We rewrite the middle term using the two numbers we just found.
Step 4: We factor the expression by factoring out the common binomial.
Analogy / Example
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How to remember algebraic identities?Evaluate using suitable identities. 497 \times 505How to do difference of squares?
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7191 | https://people.sc.fsu.edu/~jburkardt/py_src/subset_sum_brute/subset_sum_brute.html | subset_sum_brute
subset_sum_brute
subset_sum_brute, a Python code which uses brute force to solve the subset sum problem, to find a subset of a set of integers which has a given sum.
We are given a collection of (21) weights and a target value (24639098). We seek a combination of the weights which adds up to the target value.
The function subset_sum_brute() simply considers every possible subset of the weights, determines its sum, and compares that to the target value. The first case in which the target value is matched is returned as the solution.
Licensing:
The information on this web page is distributed under the MIT license.
Languages:
subset_sum_brute is available in a C version and a C++ version and a Fortran90 version and a MATLAB version and an Octave version and a Python version.
Related Data and Programs:
closest_pair_brute, a Python code which uses brute force to solve a 2D version of the closest pair problem, which identifies the closest pair of points in a given collection.
knapsack_01_brute, a Python code which uses brute force to solve small versions of the 0/1 knapsack problem;
matrix_chain_brute, a Python code which finds the cost of the most efficient ordering to use when multiplying a sequence of matrices, using brute force.
partition_brute, a Python code which uses a brute force method to find solutions of the partition problem, in which a set of integers must be split into two subsets with equal sum.
python_combinatorics, a Python code which considers a variety of problems in combinatorics involving counting, combinations, permutations, and so on.
satisfy_brute, a Python code which uses brute force to find all assignments of values to a set of logical variables which make a complicated logical statement true.
tsp_brute, a Python code which is given a city-to-city distance table, and solves a (small) traveling salesperson problem (TSP), using brute force.
Reference:
Silvano Martello, Paolo Toth,
Knapsack Problems: Algorithms and Computer Implementations,
Wiley, 1990,
ISBN: 0-471-92420-2,
LC: QA267.7.M37.
Source Code:
subset_sum_brute.py, the source code.
subset_sum_brute.sh, runs all tests.
subset_sum_brute.txt, the output file.
Last revised on 25 October 2022. |
7192 | https://www.scribd.com/document/857528056/Percent-Problems | Percent Problems | PDF
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Percent Problems
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K u t a S o f t w a r e - I n f i n i t e A l g e b r a 1 N a m e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Period_ Date_______
Percent Problems
Solve each problem. Round to the nearest tenth or tenth of a percent.
1) W h a t p e r c e n t o f 2 9 i s 3?2) W h a t p e r c e n t o f 3 3.5 i s 2 1?3) W h a t p e r c e n t o f 5 5 i s 3 4?4) 4 1% o f 7 8 i s w h a t?5) 2 8% o f 6 3 i s w h a t?6) 5 8% o f w h a t i s 6 3.4?7) 1 i s w h a t p e r c e n t o f 5 2.6?8) W h a t p e r c e n t o f 3 8 i s 1 5?9) 4% o f 7 3 i s w h a t?1 0) W h a t i s 1 2% o f 1 7.5?1 1) 7 9% o f 6 7 m i l e s i s w h a t?1 2) W h a t i s 5 9% o f 1 4 m?
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1 3) 1 1 2 m i n u t e s i s 7 6% o f w h a t?1 4) W h a t i s 1 6% o f 4 3 m i n u t e s?1 5) $7 3 i s w h a t p e r c e n t o f $1 2 5?1 6) W h a t i s 9 0% o f 1 3 0 i n c h e s?1 7) W h a t i s 6 8% o f 1 1 8 t o n s?1 8) W h a t p e r c e n t o f 1 8 0.4 m i n u t e s i s 2 5.7 m i n u t e s?1 9) 1 6 i n c h e s i s 3 5% o f w h a t?2 0) 9 0% o f 5 4.4 h o u r s i s w h a t?2 1) 1 4 0 f t i s 9 7% o f w h a t?2 2) W h a t i s 1 7 0% o f 9 7 t o n s?2 3) W h a t i s 1 0 3% o f 1 2 7 t o n s?2 4) 1 0 2 h o u r s i s 9 4% o f w h a t?
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K u t a S o f t w a r e - I n f i n i t e A l g e b r a 1 N a m e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Period_ Date_______
Percent Problems
Solve each problem. Round to the nearest tenth or tenth of a percent.
1) What percent of 29 is 3?10.3%2) What percent of 33.5 is 21?62.7%3) What percent of 55 is 34?61.8%4) 41% of 78 is what?32 5) 28% of 63 is what?17.6 6) 58% of what is 63.4?109.3 7) 1 is w hat percent of 52.6?1.9%8) What percent of 38 is 15?39.5%9) 4% of 73 is what?2.9 10) What is 12% of 17.5?2.1 11) 79% of 67 miles is what?52.9 miles 12) What is 59% of 14 m?8.3 m
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13) 112 minutes is 76% of what?147.4 minutes 14) What is 16% of 43 minutes?6.9 minutes 15) $73 is what percent of $125?58.4%16) What is 90% of 130 inches?117 inches 17) What is 68% of 118 t ons?80.2 tons 18) What percent of 180.4 minutes is 25.7 minutes?14.2%19) 16 inches is 35% of what?45.7 inches 20) 90% of 54.4 hours is what?49 hours 21) 140 ft is 97% of what?144.3 ft 22) What is 170% of 97 tons?164.9 tons 23) What is 103% of 127 tons?130.8 tons 24) 102 hours is 94% of what?108.5 hours
-2-
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7193 | https://fiveable.me/key-terms/honors-pre-calc/sec | Sec - (Honors Pre-Calculus) - Vocab, Definition, Explanations | Fiveable | Fiveable
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key term - Sec
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Definition
The secant function, denoted as 'sec', is a trigonometric function that represents the reciprocal of the cosine function. It is one of the fundamental trigonometric functions used in the study of right triangle trigonometry.
5 Must Know Facts For Your Next Test
The secant function is defined as the reciprocal of the cosine function, meaning $sec(x) = \frac{1}{\cos(x)}$.
The secant function is used to find the length of the hypotenuse of a right triangle, given the length of the adjacent side and the angle.
The secant function is one of the six basic trigonometric functions, along with sine, cosine, tangent, cotangent, and cosecant.
The secant function is useful in various applications, such as surveying, navigation, and engineering, where the length of the hypotenuse is required.
The domain of the secant function is all real numbers except for $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer, as the cosine function is zero at these values.
Review Questions
Explain the relationship between the secant function and the cosine function.
The secant function is the reciprocal of the cosine function, meaning that $sec(x) = \frac{1}{\cos(x)}$. This relationship is fundamental to understanding the secant function, as it means that the secant function can be used to find the length of the hypotenuse of a right triangle when the length of the adjacent side and the angle are known. The secant function is often used in conjunction with the cosine function to solve problems in right triangle trigonometry.
Describe the domain of the secant function and explain the significance of the excluded values.
The domain of the secant function is all real numbers except for $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer. This is because the cosine function is zero at these values, and the secant function is the reciprocal of the cosine function. As a result, the secant function is undefined at these points, as division by zero is not allowed. Understanding the domain of the secant function is crucial in applying it correctly to solve problems in right triangle trigonometry.
Analyze the practical applications of the secant function in fields such as surveying, navigation, and engineering.
The secant function is widely used in various fields, such as surveying, navigation, and engineering, due to its ability to determine the length of the hypotenuse of a right triangle. In surveying, the secant function is used to measure the distance to an inaccessible point by using the length of the adjacent side and the angle. In navigation, the secant function is employed to calculate the distance between two points on the Earth's surface, given the latitude and longitude coordinates. In engineering, the secant function is utilized in the design of structures, such as bridges and buildings, where the length of the hypotenuse is a critical factor in ensuring structural integrity. Understanding the applications of the secant function is essential for students studying right triangle trigonometry.
Related terms
Cosine (cos):The cosine function is a trigonometric function that represents the ratio of the adjacent side to the hypotenuse of a right triangle.
Reciprocal Function:
A reciprocal function is a function that is the inverse of another function, where the input and output values are interchanged.
Right Triangle Trigonometry:The study of trigonometric functions and their relationships within the context of right triangles.
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0 |
7194 | https://www.quora.com/How-do-you-find-the-length-of-a-line-with-two-coordinates | Something went wrong. Wait a moment and try again.
Lines (general)
Distance Rule
Coordinates of Points
Coordinate Plane
Coordinate Systems
Distance in Mathematics
Distance Function
5
How do you find the length of a line with two coordinates?
·
To find the length of a line segment defined by two coordinates (x1,y1) and (x2,y2), you can use the distance formula derived from the Pythagorean theorem. The formula is:
d=√(x2−x1)2+(y2−y1)2
Where:
- d is the length of the line segment,
- (x1,y1) are the coordinates of the first point,
- (x2,y2) are the coordinates of the second point.
If you have two points (1,2) and (4,6), you can find the length as follows:
Calculate the differences:
x2−x1=4−1=3
y2−y1=6−2=4
Plug these values into the formula:
[math]d = \sqrt{(3)^2 + (4)^2} = [/math]
To find the length of a line segment defined by two coordinates (x1,y1) and (x2,y2), you can use the distance formula derived from the Pythagorean theorem. The formula is:
d=√(x2−x1)2+(y2−y1)2
Where:
- d is the length of the line segment,
- (x1,y1) are the coordinates of the first point,
- (x2,y2) are the coordinates of the second point.
Example
If you have two points (1,2) and (4,6), you can find the length as follows:
Calculate the differences:
x2−x1=4−1=3
y2−y1=6−2=4
Plug these values into the formula:
d=√(3)2+(4)2=√9+16=√25=5
So, the length of the line segment is 5 units.
Related questions
How can we find the length of the line segment joining two points on a coordinate plane?
How do you find the length of a coordinate?
How do you find the length of a line between two points?
How do you find the coordinates of a straight line when given some sort of ratio?
How do you find coordinates on a line equation?
John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math
·
Author has 9K answers and 23.4M answer views
·
5y
Draw a horizontal line from one coordinate and a vertical line from the other coordinate until they meet at a common point.
For example, point (3, 7) and point (8, 4):
There are two possible common points (3, 4) or (8, 7)
Let’s use (8, 7)
Draw a horizontal line from (3, 7) to (8, 7). It is five units long.
Draw a vertical line from (8, 4) to (8, 7). It is three units long.
Draw a
Draw a horizontal line from one coordinate and a vertical line from the other coordinate until they meet at a common point.
For example, point (3, 7) and point (8, 4):
There are two possible common points (3, 4) or (8, 7)
Let’s use (8, 7)
Draw a horizontal line from (3, 7) to (8, 7). It is five units long.
Draw a vertical line from (8, 4) to (8, 7). It is three units long.
Draw a line between your two points.
You now have a triangle, 5 on one side and 3 on ...
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5y
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Eric Hawk
Author has 3K answers and 3.9M answer views
·
5y
Do you mean the length of a line segment given the coordinates of its two endpoints? If so:
Distance - Wikipedia
In analytic geometry, the distance between two points of the xy-plane can be found using the distance formula. The distance between (x1, y1) and (x2, y2) is given by:
d=√(Δx)2+(Δy)2=√(x2−x1)2+(y2−y1)2.
Similarly, given points (x1, y1, z1) and (x2, y2, z2) in three-space, the distance between them is:
{\displaystyle d={\sqrt {(\Delta x)^{2}+(\Delta y)^{2}+(\Delta z)^{2}}}={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{
Do you mean the length of a line segment given the coordinates of its two endpoints? If so:
Distance - Wikipedia
In analytic geometry, the distance between two points of the xy-plane can be found using the distance formula. The distance between (x1, y1) and (x2, y2) is given by:
d=√(Δx)2+(Δy)2=√(x2−x1)2+(y2−y1)2.
Similarly, given points (x1, y1, z1) and (x2, y2, z2) in three-space, the distance between them is:
d=√(Δx)2+(Δy)2+(Δz)2=√(x2−x1)2+(y2−y1)2+(z2−z1)2.
These formula are easily derived by constructing a right triangle with a leg on the hypotenuse of another (with the other leg orthogonal to the plane that contains the 1st triangle) and applying the Pythagorean theorem. In the study of complicated geometries, we call this (most common) type of distance Euclidean distance, as it is derived from the Pythagorean theorem, which does not hold in non-Euclidean geometries.
Related questions
How do you find the coordinates of a point on a line?
How do you find the coordinates of a point on a straight line when given two points on that line and the slope and y-intercept?
How do you find the coordinates of a point if you know the coordinates of two other points and the distance between them?
How do you find the coordinates of a point if you are given two lines?
How do you find a line equation (on Earth) using two GPS coordinates (geometry, spherical coordinates, math)?
George Ivey
Former Math Professor at Gallaudet University
·
Author has 23.7K answers and 2.6M answer views
·
2y
Originally Answered: How do you find the length of a line segment when given only two points on it?
·
IF those two points happen to be the segments endpoints then you measure the distance between them. If not it is impossible,
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Prabal Kalia
Student at Indian Institute of Information Technology, Kottayam (IIIT Kottayam) (2018–present)
·
8y
Related
How can I find the distance between two line?
You can only find the distance between two parallel lines, as the rest would intersect at one point or coincide.
first step make both the line equations to the form
l1=ax+by=c1
l2=ax+by=c2
then apply the formula,
Hope it helps!
You can only find the distance between two parallel lines, as the rest would intersect at one point or coincide.
first step make both the line equations to the form
l1=ax+by=c1
l2=ax+by=c2
then apply the formula,
Hope it helps!
Bob Collier
Former EE Designed Specialized Computers for 33 Years.
·
Author has 3.1K answers and 1.6M answer views
·
2y
Related
How do you calculate the distance between two lines using coordinates?
Unless they are parallel, the ‘distance between them’ is meaningless. They will cross at one point and get infinitely far apart -twice.
If you have 2 points on one line and one on the other, w the 2, find ‘m’ in
y = mx + b
for that line.
Then from the point on the other one (c, d), you can construct a perpendicular
d = -c/m + e
that will cross the first line. This will give you a triangle and the separation is one of the sides.
Could be you can do this with just one point on each line [instead of the 2 on one that I posed]. I didn’t bother th check that out. You can if you want to.
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Former Journeyman Wireman IBEW
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Author has 33.3K answers and 5.3M answer views
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1y
Related
What is the length of a line segment if given the coordinates of its two endpoints?
You can use Pythagoras
Construct a right triangle
Using your line as the hypotenuse
You draw one vertical one horizontal
Line
They are parallel with the axis'
You have 2 straight lines on the grid now
Your point’s coordinates tell you that
Bill Scott
Author has 300 answers and 104K answer views
·
2y
Related
How can we find the length of the line segment joining two points on a coordinate plane?
Use the Pythagorean Theorem:
Graph the points at (2,7) and (6,4). We are going to find the length of that line segment.
The formula is √(x2-x1)^2 + (y2-y1)^2
x2-x1 means take the x values of the two points and subtract them. y2-y1 means do the same thing for the y values.
2 - 6 = -4; 7 - 4 = 3.
(-4)^2 = 16. 3^2= 9.
√16 + 9 = √25 = 5
As you can see from the numbers above, I chose points that would make a Pythagorean triple:
One side is 3, the other side is 4 and the hypotenuse is 5.
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Mike McLaury
Former Retired at Computer Consultant
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Author has 210 answers and 88.5K answer views
·
Updated 5y
Related
How do you find the length of a line between two points?
It depends on the situation.
If you are a carpenter, use a tape measure.
Or you may need a surveyer.
If this is a math problem, with the rectangular coordinates of two points given as (a,b) and (c,d), then sqrt((d-b)^2 + (c-a)^2)).
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7195 | https://it.wikipedia.org/wiki/Carbonato_di_potassio | Carbonato di potassio - Wikipedia
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Disambiguazione – "Potassa" rimanda qui. Se stai cercando la località italiana della provincia di Grosseto, vedi Potassa (Gavorrano).
Questa voce sull'argomento sali è solo un abbozzo.
Contribuisci a migliorarla secondo le convenzioni di Wikipedia.
| Carbonato di potassio |
| Carbonato di potassio anidro |
| Nome IUPAC |
| carbonato di potassio |
| Nomi alternativi |
| potassa, triossocarbonato(IV) di dipotassio |
| Caratteristiche generali |
| Formula bruta o molecolare | K 2 CO 3 |
| Massa molecolare (u) | 138,21 g/mol |
| Aspetto | solido bianco |
| Numero CAS | 584-08-7 |
| Numero EINECS | 209-529-3 |
| PubChem | 11430 e 516886 |
| DrugBank | DBDB13977 |
| SMILES | C(=O)([O-])[O-].[K+].[K+] |
| Proprietà chimico-fisiche |
| Densità (g/cm 3, in c.s.) | 2,43 (20 °C) |
| Solubilità in acqua | 1120 g/L (20 °C) |
| Temperatura di fusione | 891 °C (1.164 K) |
| Indicazioni di sicurezza |
| Simboli di rischio chimico |
| |
| attenzione |
| Frasi H | 315 - 319 - 335 |
| Consigli P | 302+352 - 305+351+338 |
| Modifica dati su Wikidata·Manuale |
Il carbonato di potassio (o potassa) è il sale di potassio dell'acido carbonico, di formula K 2 CO 3. A temperatura ambiente si presenta come un solido bianco inodore, molto solubile in acqua, dando una soluzione decisamente basica. È un composto nocivo[senza fonte] e irritante. La potassa viene usata nella produzione artigianale del sapone e del vetro.
Etimologia
[modifica | modifica wikitesto]
Il termine potassa prende il nome dal tedesco 'Pottasche', composto da Pott (= vaso) ed Asche (= cenere).
Venne presumibilmente scoperta nel 1650 da Johann Rudolph Glauber e ricavata per evaporazione del soluto da ceneri, in prevalenza di faggi e querce, in apposite capanne site in prossimità delle vetrerie silvane 'Waldglashütten' delle foreste germaniche, ove veniva già prodotto dal XII secolo il 'vetro silvano', 'Waldglas' anche 'vetro da legna', 'Holzglas' dalle sabbie quarzifere dei Tauri e dal tipico colore verdastro impiegato in finestre e per la realizzazione di semplici contenitori. Venne aggiunta per abbassar la temperatura di rammollimento del nuovo 'vetro di potassa' ('Pottascheglas') e per migliorarne la trasparenza, permettendo d'estenderne la produzione fino al XIX secolo, superando la crisi del legname del XVII secolo.
Storia
[modifica | modifica wikitesto]
Storia Il carbonato di potassio è il componente principale della potassa e della cenere perlifera, più raffinata. Storicamente, la cenere perlifera veniva creata cuocendo la potassa in un forno per rimuovere le impurità. La polvere bianca e fine che rimaneva era la cenere perlifera. Il primo brevetto rilasciato dall'Ufficio Brevetti degli Stati Uniti fu assegnato a Samuel Hopkins il 31 luglio1790 per un metodo migliorato per la produzione di potassa e cenere perlifera.
Note
[modifica | modifica wikitesto]
^scheda del carbonato di potassio su IFA-GESTISArchiviato il 16 ottobre 2019 in Internet Archive.
Altri progetti
[modifica | modifica wikitesto]
Altri progetti
Wikizionario
Wikimedia Commons
Wikizionario contiene il lemma di dizionario «carbonato di potassio»
Wikimedia Commons contiene immagini o altri file sul carbonato di potassio
Collegamenti esterni
[modifica | modifica wikitesto]
(IT,DE,FR) Carbonato di potassio, su hls-dhs-dss.ch, Dizionario storico della Svizzera.
Preparazione del carbonato di potassio dalla cenere di legno, su itchiavari.org.
| mostra V·D·M Sali di potassio |
| Composti inorganici | K 2 O·KO 2·KAl(SO 4)2·KBr·KBrO 3·KCN·KCNO·KCl·KClO 3·KClO 4·KF·KCr(SO 4)2·KH·KHCO 3·KHF 2·KHSO 3·KHSO 4·KI·KIO 3·KH(IO 3)2·KIO 4·KMnO 4·K 2 MnO 4·KNO 2·KNO 3·KOCN·KOH·KSCN·K 2 CO 3·K 2 CrO 4·K 2 Cr 2 O 7·K 2 S·K 2 SO 3·K 2 SO 4·K 2 S 2 O 5·K 2 S 2 O 7·K 2 S 2 O 8·K 2 S 4 O 6·K[Sb(OH)6]·K 2 SiO 3·K 3[Fe(CN)6]·K 4[Fe(CN)6]·KH 2 PO 4·K 2 HPO 4·K 3 PO 4·K 2 TeO 3 |
| Composti organici | Acetato di potassio·Ascorbato di potassio·Biftalato di potassio·Bitartrato di potassio·Citrato di potassio·Formiato di potassio·Ossalato di potassio·Sorbato di potassio·Tartrato di potassio·Tartrato di sodio e potassio |
| Controllo di autorità | Thesaurus BNCF36697·LCCN(EN)sh85105587·GND(DE)4163061-0·BNF(FR)cb12191119d(data)·J9U(EN,HE)987007529543405171·NDL(EN,JA)00572682 |
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7196 | https://math.stackexchange.com/questions/430722/covered-10x10-rectangle-with-l-shapes-trominos | discrete geometry - Covered 10x10 rectangle with L-shapes trominos - Mathematics Stack Exchange
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Covered 10x10 rectangle with L-shapes trominos
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We have given L-shaped trominos and a square of size 10x10. Give a nice proof, that 18 L-trominos is the minimal number with which the square can be covered such that it is impossible to insert one more L-tromino. I give an example with 18 L-trominos.
discrete-geometry
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edited Jun 27, 2013 at 11:13
Raskolnikov
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asked Jun 27, 2013 at 11:02
andrewandrew
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Cool problem, I can't figure it out. You could consider reading the '96 polyominoes paper: citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.17.3375. I tried using some of those ideas but wasn't able to conclude anything.Samuel Reid –Samuel Reid 2014-04-22 20:41:14 +00:00 Commented Apr 22, 2014 at 20:41
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Depends what you think is a "nice proof." Here are some possibilities.
By cases
Suppose that only 17 L-trominoes are placed on the grid; then 51 squares are occupied, and 49 are free.
Consider the grid to consist of 2x2 blocks in 5 rows and 5 columns.
If any one of these 25 blocks contains three free spaces, then we could place a tromino there. Otherwise, every block has at most two free squares, so there are 24 blocks with exactly two free and two occupied squares, and one block with three occupied squares.
Now in each 2x2 block, the free squares are either adjacent, or "checkered". At each corner between four blocks, you can have at most two free squares (otherwise, you could insert an L-tromino there.) Using these rules and enumerating the possible combinations of blocks, you can eventually rule them all out and conclude that an additional tromino can be placed.
The number of cases really seems to proliferate thanks to the single block with three occupied squares, though, so covering all the cases is quite a rabbit hole.
Computationally
Create the graph whose 324 vertices are all the possible positions of an L-tromino on the grid, with edges between any two mutually exclusive placements (L-tromino positions which share a vertex.)
Then compute the independent domination number of the graph.
An independent dominating set is an independent set (so, a set of tromino positions which do not overlap), such that every other vertex is adjacent to at least one (so, no more trominos may be placed); the smallest independent dominating set is what we seek.
Here are 9 lines of Sage code that do that:
N = 10
rcp = cartesian_product([range(N-1), range(N-1), range(4)])
pos = [0, 1, N, N+1]
def posns(r, c, p):
return {rN + c + pos[i] for i in {0,1,2,3}-{p}}
def isedge(rcp0, rcp1):
return rcp0 != rcp1 and bool(posns(rcp0) & posns(rcp1))
G = Graph([rcp, isedge])
print(G.dominating_set(independent=True, value_only=True))
Here, rcp is the set of possible tromino locations, encoded as triples (r,c,p)(r,c,p) for the tromino occupying the 2x2 block with upper left corner in row r r and column c c, and omitting the p p th square (numbered from 0 to 3 in left-to-right reading order.) The function posns returns a set of the three squares occupied by a given tromino, where the squares are numbered from 0 to 99. & is the set intersection operator, and in Python a set is considered True iff it's nonempty.
The answer is 18, as you can see in this SageMathCloud notebook.
(Note: this takes many minutes to run; the cached results should be visible at the notebook. For some reason I always find it necessary to open a SageMathCloud link, go back from a login screen, and open it a second time to get a public notebook to actually display.)
By referencing the OEIS
Check out OEIS A238381.
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edited May 10, 2017 at 1:09
answered May 9, 2017 at 22:20
Nick MatteoNick Matteo
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7197 | https://www.youtube.com/watch?v=81jxfxLg9tg | Introduction to Complex Numbers: Lecture 1 - Oxford Mathematics 1st Year Student Lecture
Oxford Mathematics
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Posted: 1 Dec 2024
To make sure our students, who come from all over the world, are up to speed for the challenges ahead, this lecture recaps much of the work they should have done at high school, providing a short description of some of the basics of complex numbers.
But even in an recap there are subtle differences and pointers to what's to come.
You can watch the subsequent lecture (from a few years ago) here:
You can watch many other student lectures via our main Student Lectures playlist (also check out specific student lectures playlists):
All first and second year lectures are followed by tutorials where students meet their tutor in pairs to go through the lecture and associated problem sheet and to talk and think more about the maths. Third and fourth year lectures are followed by classes.
458 comments
Transcript:
[Music] okay so this is the course on the introduction to complex numbers so here's my notation okay I'm going to use this character to mean the real numbers and I'm going to use this character to mean of the complex numbers okay okay so let's get going so what does something belonging to this set c mean z in c means that Z is equal to a + I B where A and B are real and i^ 2 is minus1 okay this guy here is called the real part of the complex number and we usually write it as re of Z and this real number here is the imaginary part which you usually write as M of Zed okay so let's just look at some simple properties so first of all if we have two complex numbers his one a plus I B and here's another one W is or Omega I can't write the difference so it's always the same C + I D okay so here necessarily by what I've just said A and B and C and D are real numbers okay then what's Zed plus W that's just equal to a + C plus I B + d add complex numbers in a very simple way and you notice that this is also equal to W plus Z because you know that the order in which you add two real numbers together does not matter okay and what's Zed times W and in common with most other mathematicians when I don't put anything between two objects it means I'm multiplying them right I will not put dot I will not put times I will just leave a space leave nothing right so Z W is well a + I B C + I D and now we just expand this in the usual sort of way so this is a c and because i^ 2 is minus one it's minus b d and then we have the cross terms I into and I'm going which one am I'm going to write first a d plus b c okay so that's the product of two complex numbers and you notice also that this is actually you can work out for yourself this is the same as W Z okay okay so what about zed ided w okay so for this we have to note that um if I take W and I multiply it by this guy when I put a minor sign in here okay then this by the definition of the product here is C c² plus d^2 okay because the cross terms cancel and this you'll notice is a real number so these were clearly complex numbers right but this is a real number and so then if we want to do the division of Z by W well this is equal to a + i b divided by C + IID and then we multiply this by one okay I'll put the in bracket so there's just no ambiguity here and we multiply this by one C minus ID over C minus IID okay and now I can multiply the bottom line the denominator and get that this is equal to and I'm not good at these things so AC plus BD I guess and therefore um divided by c^ 2 + d^2 that's the real part and the imaginary part is equal to uh I'll write it this way BC minus a d is that right BC minus a d That's right divided by again the same denominator c^ 2 plus d^2 okay that's d okay and this is true of course provided that that W is not zero in the complex numbers which is obviously in real numbers 0 plus I Z okay so as long as W is not zero then C2 + d s is clearly a positive real number and so we can actually divide okay so that's some simple things and in general this is a fairly useful property I guess in general um Z equals W this is equality of two complex numbers if and only if AAL C the real parts are the same and b equal D the imaginary parts are the same and what this has the consequence is that um you can uh equate real and imaginary Parts okay can everybody see this I'm looking at the people on the background especially the person over there doesn't see it can you see this okay and you can hear me okay okay okay so in this here uh we have made use of the conjugate and what's the conjugate the conjugate is something we write as Zed Bar right and Zed Bar is a minus I B so in other words if Zed is a plus i b z bar just changes the sign of the imaginary part and we note as we've seen above that zed z bar is real okay and and and so and it's also non- negative okay so the real part of Zed is equal to Z plus Z bar/ 2 that cancels off the imaginary parts and the imaginary part of Zed is right you're listening good Zed minus z bar over I 2 I right okay just carry on here and further the following are true for all Zed and W in the complex numbers Zed plus W bar right so this is a z plus W is a complex number so you can take as conjugate and this turns out to be the same as z bar plus W bar that is you can conjugate and add or you can add and conjugate zed w is a complex number you can take its conjugate and this is equal to Zed Bar W bar okay perhaps I put those a little too far apart there we go you can conjugate multiply or you can multiply and conjugate get the same thing and Zed over W is a complex number you can take its conjugate this is z bar divided by W bar when w is not zero okay so you can go and prove these These Are A simple exercise for you to go and prove okay okay uh the need for complex numbers um arises from polinomial okay so for example if I have a polom right so here's a polinomial polom of argument and I'll call the argument X that's all right isn't it is equal to X cubed so this is a cubic polinomial + 2x^2 + 2x + 1 so there's a cubic polinomial right okay so this I claim uh satisfies that uh P if you evaluate at minus one then this is a zero because 1 - 2 + 2 - 1 is zero okay and so that means I can write P of X is equal to x + 1 that's telling me that it vanishes when X is minus one and then I have to find out what's the remaining term there must be an x s to give me an X cubed there must be a one to give me plus one and I think there must be an X to make sure that the other two terms work right okay and so what do we see we see the other roots of this polinomial satisfy that x^2 + x + 1 is zero and we know how to find the roots of a quadratic we can simply apply the the formula and so the roots are well the minus B plus and minus the sare < TK of b ^ 2us 4 a c that's 4 1 1 all over 2 a okay this is equal to minus a half plus or minus I < tk3 over two so the roots are necessarily complex in this case i^2 is minus one as before so this simple example um illustrates uh several things okay that are generally true things that are generally true about any polinomial so I chose a specifically a cub pubic there but this is there are several things any of them will will satisfy okay so um let me write general polom and this time I'll write it in terms of an argument Zed okay this is a0 plus A1 Z Plus A2 z^ 2 and so on plus a n Z to the N okay here n is an integer and it's b b these ajs are complex numbers okay and this is true for J equals not 1 up to n okay so this is the way we usually think of polinomial we'll see them in this course in many many many different forms this is perhaps the the way that you've seen the most simply written down before okay so what are the things that are true the first thing is the the fundamental uh theorem of algebra I am going to quote this for you I am not going to prove it to you you need a lot more mathematical Machinery to be able to prove this and you'll learn that mathematical Machinery in the next few years okay here's the fundamental theorem of algebra and this says that P of Zed is equal to a subn that's this guy times Zed minus and I call these gamma Z minus Gamma 1 time z- gamma 2 time Z minus gamma n okay for n complex numbers gamma J right right I've said it twice here right written in words and written it down here in mathematics uh so that's right and this is for uh this is for um J = 1 up to n in other words every polinomial with a degree n which the highest power is z to the N let's say has n complex Roots um right so that's the fundamental theorem of algebra that's the first result here's the second result I just don't think this has a name so I'm just going to call it a theorem so if these ajs are not complex at all well are complex but they're also real okay if the ajs are real okay all of them then and I'm going to use this character a little bit this character means there exists it's not just me that uses this character you'll see it a great deal many of my colleagues will also use it okay then there exists a uh a number r with 0 less equal R less equal n uh with the following BR true P of Zed is equal to a subn again the same guy as here or the same guy as there um time Z - alpha 1 time Z - Alpha R this is this r that exists okay times q1 of Zed times Qs of Zed okay where r + 2 s is equal to n okay where also um the alpha JS are real right up to R there's R of them and the term QJ of Zed is precisely um a quadratic polinomial z^ 2 plus b J Z Plus CJ so quadratic polinomial with a coefficient one here okay so in this case BJ and CJ are real numbers okay and these things are also equal to zed minus gamma and Zed minus gamma bar so this this is the conjugate this guy okay sorry gamma and c that's right okay and this is all true for J = 1 up to S everybody can see this I'm sorry I put it under the shadow right I'm not trying to hide it this is the conjugate so you can write down any polinomial that has real coefficients it may have a number of real Roots either between zero and N of them and then it has the remaining terms which are also can be written as real polinomial real quadratic polinomial and those quad real quadratic polinomial have two complex Roots which are conjugates of each other and you can see that in the example above right minus a half plus and minus i3 too that's a complex root times it's and its conjugate okay okay so I'm just going to write one more thing down here which is a consequence of this thus if um all the coefficients are in are real okay okay and Gamma being a complex number is a root of P okay then gamma bar is also root a p you can see your disadvantage now can't do anything about this okay there so then we come on to the polar form okay and the polar form is like the following I can draw a picture and this picture has the real axis going out this way and the imaginary axis going up this way that's the usual way it's done maybe on some other planet they do it the other way around but we don't we do it this way around then I can put down here the real part of a complex number here Z is equal to x + i y and this is the real part is the abissor if you like it's the x coordinate and here's the imaginary part over here and this is why is the imaginary part of Zed oh WR this very well okay so that's this number here this is just supposed to be a broken line that indicates that's where X Plus i y is but we know the points in a complex plane you can all rep also represent in polar form so if I draw a straight line between these two things I can write the distance of this point from the origin as R and this angle here as Theta and this would be the polar representation of of a complex number Z = X Plus i y so R is equal to the square root and I shall make emphasize it by saying plus the square root right it's the positive square root it's a distance it cannot be negative it's plus the square < TK of x^2 + y^2 and this is called the modulus this is the modulus of Zed and it's usually written as um this and uh oh this is a very much thinner one and a Theta is the inverse tan of uh X y/x I don't know why I need a bracket here I don't really need a bracket this is the inverse tangent of Y /x okay is the argument okay it's the argument and I sometimes put an e in but I don't think there's any it's the argument of Zed and this is written as well usually different things in different languages but usually this that's what I shall use okay so the first thing to note is that the argument um uh is not defined the argument is not defined for Zed which is is the zero complex number R is perfectly well defined but I think that's the only complex number for which we cannot Define an argument okay so you notice that if Y is positive and X is zero then argument of Zed is just Pi / 2 right and if Y is negative these are real numbers right so I can say this and x equals z then the argument of Zed is 3 Pi / two okay and also uh clearly um Theta Theta + 2 pi Theta minus 2 pi Theta oh theta plus 4 Pi Etc right you see what I'm doing here these all Define okay they all Define uh the same argument that is if you give me an argument of theta plus 18 Pi that's exactly the same argument as the number that has Theta as its polar angle right um uh I should just tell you um if zero is and I think less equal Theta and this is less than 2 pi okay uh then this is the principal argument the principal value the principal value for the argument okay so we can write a polar form I'm sorry standing back I realized that my straight line from the origin to zal X Plus i y looks slightly curved I'm sorry that's a human fallibility you can't draw exactly straight lines it's supposed to be a straight line okay note that um zedar the conjugate of Z said um has modulus R right it's the same as the modulus of Zed it's the same distance from the origin right and argument minus Theta okay that is if you look at the conjugate on this diagram here the conjugate is somewhere down here right and so the angle is minus th okay too much also notice that um you know you can write X is R cosine Theta so this is just trigonometry okay and Y is equal to R sin Theta okay this is just to express this expresses X and Y in terms of R and Theta of course okay okay here's a proposition for you I claim this is true right for all Zed and W in the complex numbers we have um that the modulus of zed w is equal to the modulus of Zed time the modulus of w I'm not claiming that's immediate but tell you it's true the modulus of Zed over W is the modulus of Zed divided by the modulus of w and this is true of course if W is not the zero complex number and the modulus of Zed squared that's just the square of a real number is equal to zed z bar so I close claim that those things are all true I also claim that the argument of ZW is equal to the argument of Zed plus the argument of w okay and this is true of course when Zed isn't zero and W isn't zero because the arguments not defined if either of them is I thought the co of the inverse tent was he's trying to be smart notice he's showing you that he's listened to the lecture this morning the co-domain of the tangent right yeah it shouldn't it be between half it's up to you it's up to you so I can choose and probably on the planet Zog they do choose the co-domain of the T inverse tangent to between 17 pi and 19 Pi it makes no difference at all it's up to you you can either decide to be between 0 and 2 pi that's what I'm doing or you can decide to be minus pi and Pi if you wish to it's not going to you're being formulaic here this is not mathematics it's not going to change the way mathematics works but if that's what you were taught then open your mind and be flexible right some of what you were taught at school is probably sort of true right what I'm telling you here is true and I'm not being arrogant when you start to do analysis you will see that we have to be be very very precise to make sure we can have firm foundations so you may think this lecture is very simple but it's also correct that's your first that should be stage zero that should be what you desire right something it's correct but thank you I do appreciate Interruption time this was not foolish I'm sorry I didn't mean to put you down at all than you all right so the argument of ZW is the argument of Z Plus the argument of w and the argument of Zed over W okay is equal to the argument of Zed oh brackets minus the argument of w okay okay and again I need both Zed and W to be non zero not just because I can't divide by a zero number but because the arguments are not defined these two number are not defined if either of them is zero okay okay so that told you some things but you notice I didn't tell you anything about the um the modulus or argument of Z plus W and here we have to um go into inequalities and this is called the triangle inequality okay okay the triangle in equality tells you that if you have Z + W and you take its modulus then this is no bigger than the modulus of Zed plus the modulus of w so you don't have you like in equalities I know when you're at school you like everything to be this equivalent and so on but this is not true right Z plus W is not equal to zed mod Z plus mod W but it's always smaller that this is called the triangle inequality and you can go and persuade yourselves by drawing pictures of triangles why this might be the name given to this inequality okay um so as I claimed was a proposition um for proofs okay uh let's um I'll leave these for you as an exercise I've told you plenty to to to work these out or uh you can um see some some of these statements um you can see the see the um outline notes see the online notes okay so there's independently written online notes written by my colleague Richard Earl which you can find on the website for this course I put them up okay and you'll find some of the proofs written there but let me just end this lecture with just a few um a few of these proofs of the things that I've said today okay so for example these are just examples of some of the proofs that we would need one of the thing things I claimed was true was that ZW bar was equal to z bar W bar I claimed that on The Middle on the middle um Oh I thought it was on the middle board uh where did where did I claim that somewhere you can see you can see the board I can't see the board from down here that's the problem right anyway I claimed this somewhere um and uh well here's a proof Zed is equal to a + I okay and W oh we don't need W do we actually oh yes we do yeah W is C plus ID sorry yes we do here right okay and so what's z w ZW is equal to a c minus B b d right plus I into a d + BC okay and so if you want to take the conjugate of this guy we change this right we change the sign of the imaginary part and then what's z bar W bar is a minus IB that's z bar time C minus ID that's W bar and that's equal to a c plus b d is that right so why did I get a minus here it's minus oh I squ is minus one there you go good some people are awake things in right AC minus h AC minus BD and then we have um tell me again if I'm wrong I have an minus I a D+ b c oh and that's the same as this right so that's your proof oh and uh I'll prove another result that I quoted on the on the board there let's suppose that I have gamma in C okay and a polinomial p of gamma is equal to zero right so gamma is a a complex root of a polom okay okay then I certainly have that P of gamma bar is equal to a0 this my same polinomial I wrote down that plus A1 z Oh gamma gamma bar plus A2 gamma bar squar plus plus a n time gamma bar to the N okay I'm proving this yeah okay so this is this okay um and so if the ajs are in fact real okay if the ajs are real J equals not one up to n so all of the coefficients are real then this means of course that AJ is equal to AJ bar it's pretty conjugate of a real number as itself right so uh P of P of gamma bar is equal to well instead of a not I can write down a not bar because that's the same number and instead of A1 I can write down A1 bar time gamma bar that guy Plus A2 bar gamma bar s plus plus a n Bar gamma bar to the N okay but now what have I told you about taking conjugates I told you that the conjugate of a product is the product of the conjugates and so I can write this as a z bar plus A1 gamma bar right and here's just a product three items in it but they're just product so this is A2 gamma squ bar okay plus oh plus a n gamma to the N bar this is a product of n plus1 items and each of those products I can between two things I can apply my my rule that the product of con the product of conjugates is the conjugate of the product okay so this equality here is using the properties above okay and this is equal to well hang on a second this is the modulus of a complex number plus the modulus of a complex number and the modulus of a complex number plus the modulus of a complex number is the modulus of the Su on the top there right conjugate conj conjugate sorry conjugate gosh the top result on the board up there Zed plus W bar is z bar plus W bar right so this is equal to a0 plus A1 gamma plus A2 gamma SAR plus plus a n gamma to the n bar right that's just using the result at the top of the board there many times but a finite number of times right and so what's this this is p of gamma bar and that is the conjugate of Z because P of gamma is zero and the conj of 0 is 0 minus i0 and that's the same as 0 plus I Z so that proves this for you for you people over there that proves this result for you right don't sit there tomorrow right I can't do anything about this I can't knock this thing away okay so tomorrow I will carry on with this simple Excursion into complex numbers I suspect that many of you have seen quite a lot of this material before perhaps not presented in exactly the way that I'm presenting it perhaps so I don't I don't know but um um yeah so I see you tomorrow morning okay this is my job right this is my job I'll let you applaud my lectures if and only if when the council worker comes to empty your dust bins you clap them while they do their job okay this is my job so please this [Music] a [Music] |
7198 | https://pilotinstitute.com/pitot-static-system/ | Published Time: 2024-12-24T20:33:07+00:00
The Pitot-Static System: How It Works - Pilot Institute
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The Pitot-Static System: How It Works
By Pilot Institute
Posted on December 24, 2024 - 9 minute read
Table of Contents
[x] Key Takeaways
[x] What Is the Pitot-Static System?
[x] Components of the Pitot-Static System
[x] Operating Principle
[x] How Does the System Give Inputs to the ASI?
[x] How Does the System Give Inputs to the Altimeter?
[x] How Does The System Give Inputs To The VSI?
[x] Blockages In The Pitot-Static System
Blocked Pitot Tube With Clear Drain Hole
Blocked Pitot Tube and Drain Hole
Blocked Static Port
[x] How Do We Fix Failures In Flight?
Fix Pitot Tube Blockages
Fix Static Port Blockages
[x] How To Prevent Blockages On The Ground?
[x] Why Not Have The Pitot Heat On All The Time?
[x] Other Sources Of Errors
Small Errors in the Static Port
Small Errors in the Pitot Tube
[x] How Do We Counter The Small Errors?
[x] Conclusion
“Excuse me, sir, do you know how fast you were going? And how high were you flying?” The answers come from a clever little setup on your airplane: a metal tube sticking into the airstream and a tiny hole on the fuselage.
These two simple features power three of the most important cockpit instruments. This setup is called the pitot-static system.
In this article, we’ll break down how the pitot-static system works, what it measures, and why it’s so important.
Key Takeaways
The pitot-static system measures airspeed, altitude, and vertical speed.
Blockages in the pitot tube or static port can lead to inaccurate or failed instrument readings.
Pilots can use pitot heat or alternate static sources to handle blockages in-flight.
Preflight checks, including removing covers, help prevent blockages caused by dirt, insects, or ice.
What Is the Pitot-Static System?
The pitot-static system measures an aircraft’s airspeed, altitude, and vertical speed using air pressure differences.
The respective instruments for showing these readings are:
The Airspeed Indicator (ASI)
The Altimeter
The Vertical Speed Indicator (VSI)
Components of the Pitot-Static System
Have you ever noticed the little L-shaped component on the front of an airplane? That’s called a pitot tube.
It helps measure how fast the airplane is going by measuring the air pressure. And do you see that little hole on the back? That’s for draining out any moisture that might get inside.
The other important part of the system is a tiny hole on the side of the airplane, called a static port. This one measures the atmospheric pressure without being affected by the aircraft’s movement.
Both the pitot tube and static port are connected to instruments through tiny pipes that carry the air pressure readings.
If you’re in a modern, fancy airplane with glass primary flight displays instead of round dials, then it uses something called an Air Data Computer to show these readings. Cool, right?
Operating Principle
According to Bernoulli’s Theorem, the total pressure on an airplane stays the same. There are two kinds of pressures that an airplane faces.
One is called dynamic pressure, which happens when the airplane moves through the air. The faster it goes, the stronger this pressure gets.
The second kind is static pressure, caused by the weight of the air above everything. If you fly higher, there’s less air above you and so less static pressure.
When you add these two types of pressure together, you get the total pressure on the airplane.
How Does the System Give Inputs to the ASI?
The Airspeed Indicator is the only instrument that uses both the pitot tube and the static port.
As the aircraft moves, air exerts pressure against its motion—this is called dynamic pressure.
The pitot tube, facing the oncoming wind, measures this dynamic pressure along with the static (atmospheric) pressure.
Together, this is called total pressure.
To calculate airspeed accurately, we need to separate the dynamic pressure from the total pressure.
Since static pressure changes with altitude and weather, the static port provides a constant reading of the current atmospheric pressure. This allows the system to subtract it and display the correct airspeed.
The Airspeed Indicator (ASI) uses inputs from both the pitot tube and the static port, separated by a diaphragm.
Static pressure fills the ASI case, while the pitot tube sends total pressure to the diaphragm.
The difference between these pressures creates dynamic pressure, causing the diaphragm to move. This movement drives the ASI needle, showing airspeed.
Some pitot tubes combine static and total pressure internally, sending only dynamic pressure to the ASI for a simpler design.
Remember, the ASI measures airflow relative to the aircraft, so strong wind gusts or turbulence can affect its readings.
How Does the System Give Inputs to the Altimeter?
The atmospheric, or static pressure, varies with altitude and weather. The altimeter uses the input from the static port to determine the altitude.
The altimeter is the only instrument in small aircraft that gives the pilot altitude information.
The altimeter measures altitude by comparing static pressure in its sealed chamber to a fixed pressure inside a diaphragm, set at 29.92 inHg (standard sea-level pressure).
As the aircraft climbs and static pressure decreases, the pressure difference moves the diaphragm, which turns the altimeter’s needles.
Since atmospheric pressure changes with weather, pilots must adjust the altimeter using the local barometric pressure shown in the Kollsman Window. If not adjusted, altitude readings may be inaccurate, risking collisions or terrain hazards.
Temperature also affects the altimeter—colder air makes your true altitude lower than indicated, so pilots should account for this, especially over mountains.
How Does The System Give Inputs To The VSI?
The Vertical Speed Indicator also relies solely on the static pressure input. As the name implies, the VSI shows the rate of the aircraft’s climb or descent.
It shows the rate of climb or descent in feet per minute. When you’re flying at a constant altitude, it should indicate zero if it’s been calibrated properly.
In the VSI’s case, the static pressure enters the instrument through the static pressure line that you can see in the diagram.
This line connects to the inside of the diaphragm. The static line also has a restricted hole that leaks pressure at a calibrated rate to the instrument casing.
Both of the inputs come from the static port, but the inside of the diaphragm has an unrestricted opening, and the VSI case has a metered opening.
When the aircraft climbs or descends, the static pressure will change instantly in the unrestricted opening, but the case pressure will take longer to adjust because of the restricted opening.
This creates differential pressure acting on the diaphragm, which then moves and rotates the connected needle too.
The faster the rate of climb or descent, the greater the differential pressure, and the more the needle will swing.
Blockages In The Pitot-Static System
The pitot tube and static ports are exposed to ice, dirt, moisture, or insects. If they’re blocked, your instruments can give false or no readings. In Instrument Meteorological Conditions (IMC), you must rely on instruments, so knowing how to diagnose these issues is critical.
In 1996, Aeroperu Flight 603 crashed when static ports were taped over, leading to conflicting instrument readings. To avoid pitot-static errors, keep the pitot hole, static port, and drain hole clear.
Blocked Pitot Tube With Clear Drain Hole
If the pitot tube is blocked but the drain hole stays open, air escapes, and the ASI drops to zero as pressures equalize.
Blocked Pitot Tube and Drain Hole
If both are blocked, trapped pressure prevents the ASI from measuring airspeed. At a steady altitude, it will display a constant, unchanging speed.
During a climb, it shows increasing airspeed; during descent, it shows decreasing airspeed, regardless of actual speed. This can mislead you into reducing throttle, risking a stall.
Blocked Static Port
A blocked static port traps pressure, affecting all three instruments. The VSI shows zero, the altimeter freezes, and the ASI gives incorrect readings.
The farther you fly from the blockage altitude, the greater the ASI error—showing higher speeds during descent and lower speeds during climbs
How Do We Fix Failures In Flight?
Fix Pitot Tube Blockages
Icing is the most common in-flight blockage, especially for the pitot tube, which faces airflow and has a narrow opening.
If you see moisture, turn on the pitot heat to activate the heating elements and prevent ice buildup. If ice forms, it takes time to melt, draining through the drain hole.
One tragic case was Air France Flight 447, where iced pitot tubes caused faulty readings, leading to a stall and crash.
Fix Static Port Blockages
Static ports rarely ice over, but if they do, use the alternate static source if your aircraft has one. This switches to cabin air pressure, which isn’t as accurate but keeps your instruments working.
If your aircraft lacks an alternate static source, break the glass on a pitot-static instrument to let cabin air into the system. If you must, sacrifice the Vertical Speed Indicator, as it’s the least critical instrument.
How To Prevent Blockages On The Ground?
When your aircraft is parked on the tarmac, the pitot-static system faces other risks. Windy conditions over time can increase the chances of blockages from dirt and dust.
Small holes like the pitot tube can attract insects, even for nest-building.
Here’s some perspective of how serious this is: The likely cause of the crash of Birgenair Flight 301, a Boeing 757 carrying 189 people, was a wasp nest built inside the pitot tube.
The aircraft had spent at least two days parked with the pitot tubes uncovered, which was enough time for the wasps to block one of the pitot tubes.
To prevent this, pilots or ground crew cover the pitot tubes with a bright “Remove Before Flight” cover.
On your preflight walkaround, always check that these covers are removed. Forgetting one means losing airspeed information until you land.
Static port blockages are even more critical since they affect all three pitot-static instruments. The Aeroperu crash is a strong reminder to inspect static ports during your walkaround too.
Why Not Have The Pitot Heat On All The Time?
In flight, the surrounding air cools the pitot tube, but on the ground, heat builds up if you leave the pitot heat on. The tube can reach several hundred degrees, which can burn you or melt the cover when securing the aircraft.
Always cover the pitot tube after a flight, but wait until it cools down. Leaving the heat on also wears out the heating elements and adds strain to the electrical system.
In small airplanes, the alternator or generator is the primary source of electrical power during flight. In case of alternator failure, the airplane switches to battery power, which can drain quickly in an emergency. Use pitot heat only when flying in visible moisture at temperatures near or below freezing.
Other Sources Of Errors
The pitot-static system also encounters errors that are less drastic but still fairly significant. A small amount of error is acceptable for normal flight operations, but it’s not practical to completely remove it.
Small Errors in the Static Port
The static port is vulnerable to position errors. Aircraft manufacturers try to place the static port in a position on the fuselage where it can measure the atmospheric pressure without any influence from the aircraft’s movement.
In reality, it’s not possible to have zero position error at all angles of attack the aircraft experiences. Still, the static port is placed in a location where the error is as low as possible.
Small Errors in the Pitot Tube
The pitot tube doesn’t have position error since it measures total pressure, but it does experience alignment error.
If you’re not flying straight and level—like in a bank, sideslip, or steep nose-up or nose-down attitude—the airflow to the pitot tube can be blocked by the fuselage or misaligned with the relative wind.
This reduced airflow lowers dynamic pressure, causing the ASI to show incorrect airspeed. Alignment error usually occurs when the relative wind is angled more than 15 degrees to the pitot tube, such as during a high Angle of Attack before a stall.
In these situations, rely on the stall warning horn, not the ASI, to detect a stall.
How Do We Counter The Small Errors?
You can reduce small errors in the pitot-static system by using static ports on both sides of the fuselage. This balances pressure readings during a bank or side wind. Pitot tubes, on the other hand, typically work independently to provide redundancy.
Larger aircraft often have multiple, independent pitot-static systems—one for the captain, one for the First Officer, and sometimes a third for backup—providing redundancy.
During takeoff, the Pilot Monitoring (PM) calls out airspeed, usually at 80 knots, and the Pilot Flying (PF) verifies it matches their ASI.
Even if your general aviation aircraft has only one system, calling out airspeed on takeoff ensures the pitot tube is working and uncovered.
Conclusion
The Airspeed Indicator, Altimeter, and Vertical Speed Indicator are important tools for any pilot. It’s amazing how such a simple system, powered by air pressure, keeps you informed during flight.
But when something goes wrong, it can get tricky fast. Learning all about how the pitot-static system works—and how it can fail—helps you stay ahead of potential issues.
The more you know, the safer you’ll feel on your next flight.
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Jun 19, 2025
Written By Annika De Lathauwer
How can I become a lawyer in the UK with a US degree?
Jun 19, 2025
Written By Annika De Lathauwer
Is it possible to become a lawyer in the UK with a US degree? The short answer is yes.There are also many options depending on which jurisdiction and which type of law you would like to work in. Once you have decided these you can start to follow a certain route to requalify into one of the areas listed below.
Which Jurisdiction?
The first question you must ask yourself is which jurisdiction you would like to work in. There is no one UK law, instead it is divided into three groups: England and Wales, Scotland, and Northern Ireland. These all function slightly differently and have their own separate characteristics.
Coming to the UK with a US law degree?
Join AllAboutLaw’s Digital Campus for free weekly virtual events with leading UK law firms and law schools—everything you need to explore your legal career options.
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England and Wales
Solicitor
The current way for a foreign qualified lawyer to become a solicitor in England and Wales is to sit the Solicitor’s Qualifying Examination (SQE1 and SQE2). Previously the system involved sitting the Qualified Lawyer’s Transfer Scheme (QLTS), however, this has been phased out since September 2021.
A US lawyer must have a university degree or equivalent, pass both parts of the SQE exam, and complete two years of Qualifying Work Experience (QWE).
Depending on your qualifications and work experience, you might be eligible to skip some or all of the SQE exams, as well as have overseas work experience count towards the two year requirements. Learn more about the rules surrounding exemptions here.
Barrister
For a qualified foreign lawyer to become a barrister in England and Wales they must apply to the Bar Standards Board (BSB) to provide evidence of their legal qualification. If you receive full exemption from training, you’ll be able to practise as a barrister once you’ve been admitted to an Inn and Called to the Bar. However, this exemption may be conditional on passing specific Law Conversion Module Assessments or the Bar Transfer Test.
If you receive partial exemption, you’ll need to complete the necessary academic, vocational or work-based training, and potentially certain Law Conversion Module Assessments.
Scotland
Solicitor
In order to become a solicitor in Scotland you will have to secure a certificate of eligibility making sure that you are fit to practise law in Scotland. This will involve filling out an application and paying a fee. You must then pass the Qualified Lawyers Assessment (QLA), which covers 11 subjects.
After this, you must then apply to be admitted by the Roll of Solicitors. You will then receive a practising certificate to be able to work as a Scottish solicitor.
Advocate
To become an advocate (barrister) in Scotland, you will have to complete the Scottish bar exams, and the Diploma in Legal Practice from a university in Scotland. This is a year-long practical course. The final step is ‘devilling’, a nine-month pupillage.
75% of ULaw students passed the January 2025 SQE1 assessments first time
The industry average first attempt pass rate is 60%. Find out how ULaw can help you pass the SQE.
Learn more
Northern Ireland
Solicitor
As an American lawyer wanting to train as a solicitor in Northern Ireland, you will have to meet the following benchmarks. You must have trained as a solicitor, be in possession of a solicitors practising certificate, and have your name on the roll, which is a list of all the solicitors of the Court of Judicature in Northern Ireland.
You should fill out an application and pay to get a practising certificate and requalify as a lawyer in Northern Ireland. Registration must be renewed every year you intend to work in the country.
Barrister
You must contact the Law Society of Northern Ireland to check eligibility and apply for admission, in order to become a barrister in the country.
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