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https://socratic.org/questions/how-do-you-graph-using-slope-and-intercept-of-2x-3y-1
|
[
"# How do you graph using slope and intercept of 2x+3y= -1?\n\nSee below\n\n#### Explanation:\n\nThe first thing I'd do is to change the equation from the current standard form and put it into slope-intercept form. We do that by solving for $y$:\n\n$2 x + 3 y = - 1$\n\n$3 y = - 2 x - 1$\n\n$y = - \\frac{2}{3} x - \\frac{1}{3}$\n\nWe're now in slope-intercept form, where\n\n$y = m x + b , m = \\text{slope\" and b = y-\"intercept}$\n\nAnd so in our question, $m = - \\frac{2}{3}$ and $b = - \\frac{1}{3}$\n\nLet's first graph the $y$-intercept. That's at $\\left(0 , - \\frac{1}{3}\\right)$:\n\ngraph{((x-0)^2+(y+1/3)^2-.1)=0}\n\nNow let's plot a second point.\n\n$m = \\frac{\\Delta y}{\\Delta x} = \\text{rise\"/\"run}$\n\nOur $m = - \\frac{2}{3}$. For every 2 that we move up, we move 3 to the left (I'm dealing with the negative sign by having us move left - with a positive slope we'd move to the right). We can start from our first point and move in that way, and so our second point can be found by writing:\n\n$\\left(0 - 3 , - \\frac{1}{3} + 2\\right) = \\left(- 3 , \\frac{5}{3}\\right)$\n\nLet's plot that:\n\ngraph{((x-0)^2+(y+1/3)^2-.1)((x+3)^2+(y-5/3)^2-.1)=0}\n\nAnd now connect the two points with a line:\n\ngraph{((x-0)^2+(y+1/3)^2-.1)((x+3)^2+(y-5/3)^2-.1)(2x+3y+1)=0}"
] |
[
null
] |
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|
https://www.percentagecal.com/answer/what-is-98-percent-of-450
|
[
"#### Solution for What is 98 percent of 450:\n\n98 percent *450 =\n\n(98:100)*450 =\n\n(98*450):100 =\n\n44100:100 = 441\n\nNow we have: 98 percent of 450 = 441\n\nQuestion: What is 98 percent of 450?\n\nPercentage solution with steps:\n\nStep 1: Our output value is 450.\n\nStep 2: We represent the unknown value with {x}.\n\nStep 3: From step 1 above,{450}={100\\%}.\n\nStep 4: Similarly, {x}={98\\%}.\n\nStep 5: This results in a pair of simple equations:\n\n{450}={100\\%}(1).\n\n{x}={98\\%}(2).\n\nStep 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both\nequations have the same unit (%); we have\n\n\\frac{450}{x}=\\frac{100\\%}{98\\%}\n\nStep 7: Again, the reciprocal of both sides gives\n\n\\frac{x}{450}=\\frac{98}{100}\n\n\\Rightarrow{x} = {441}\n\nTherefore, {98\\%} of {450} is {441}\n\n#### Solution for What is 450 percent of 98:\n\n450 percent *98 =\n\n(450:100)*98 =\n\n(450*98):100 =\n\n44100:100 = 441\n\nNow we have: 450 percent of 98 = 441\n\nQuestion: What is 450 percent of 98?\n\nPercentage solution with steps:\n\nStep 1: Our output value is 98.\n\nStep 2: We represent the unknown value with {x}.\n\nStep 3: From step 1 above,{98}={100\\%}.\n\nStep 4: Similarly, {x}={450\\%}.\n\nStep 5: This results in a pair of simple equations:\n\n{98}={100\\%}(1).\n\n{x}={450\\%}(2).\n\nStep 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both\nequations have the same unit (%); we have\n\n\\frac{98}{x}=\\frac{100\\%}{450\\%}\n\nStep 7: Again, the reciprocal of both sides gives\n\n\\frac{x}{98}=\\frac{450}{100}\n\n\\Rightarrow{x} = {441}\n\nTherefore, {450\\%} of {98} is {441}\n\nCalculation Samples"
] |
[
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http://www.winterdrache.de/freeware/motionplan/approaches.html
|
[
"# 3. Different Approaches\n\nThere are several approaches to the problem of motion planning most of which fall into one of the following categories (hybrid approaches also exist):\n\ncell decomposition\n\nThe configuration space is split into disjoint subsets (cells) which are represented in a graph. Nodes correspond to cells and edges connect nodes if the respective cells are connected. Usually this approach is only appropriate in low dimensional C-spaces where \"natural\" cell boundaries exist (e.g. walls in 3D space) and when the robot is very simple.\n\npotential fields\n\nThe idea behind this approach is simple. When an electron (having negative charge) is put into an electric field it will be attracted by a positive charge (the goal) and repelled by other negative charges (obstacles). Starting at a certain point it will move towards the goal, driven by the force exerted by the electric field. A robot's configuration is just a point in the configuration space C, just like the electron is just a point in our natural 3D space.\n\nSo if a mathematical potential field function can be constructed that adequately models qgoal and the obstacles just as the electric field models the distribution of charges, then the virtual forces exerted on a virtual electron positioned in the point qstart can be computed. This leads to a path from qstart to qgoal that avoids the obstacles.\n\nThis method has been successfully applied to many path planning problems. However, with increasing number of dof and complexity of the obstacles, finding an appropriate potential field function becomes increasingly difficult."
] |
[
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https://www.physicsforums.com/threads/tension-with-a-tension-problem.50317/
|
[
"# Tension with a tension problem?\n\n\"tension\" with a tension problem?\n\n11. [CJ6 4.P.076.] A 248 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 29.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.880, and the log has an acceleration of 0.800 m/s2. Find the tension in the rope.\n\nsurprised",
null,
":surprised",
null,
"Related Introductory Physics Homework Help News on Phys.org\nFirst of all, let us discuss the concept of the tension. Essentially, the tension is the force with which the block is being pulled. Therefore, the forces accross the \"x\" plane are the tension and the friction forces, as well as the \"x\" component of the gravitational force. On the \"y\" plane, we find the normal force as well as the \"y\" component of the gravitational force.\n(note that I rotated our x-y plane by 29 degrees, so that all the forces are at 0 degrees to the surface )\n\nNow, lets write what we've just said down:\n$$F_{net,x} = F_t - F_k - F_{g,x}$$\n$$F_{net,y} = F_n - F_{g,y}$$\nTo find $$F_k$$, we want to find the normal force. Since we know that no acceleration on the \"y\" plane exists (that is, the log doesn't levitate above the surface), $$F_{net,y} = ma = m(0) = 0$$. So, $$F_n = F_{g,y}$$. Looking at the block diagram you should've drawn, you should get $$F_{g,y}=F_n=mg sin{69^0}$$.\nLet's go back to friction. $$F_k = \\mu _k F_n = \\mu _k (mg sin{69^0} )$$.\nWe substitute into our original $$F_{net,x}$$:\n$$m a_x = F_t - \\mu _k (mg sin{69^0} ) - (mg cos{69^0})$$\n$$F_t = m a_x + \\mu _k (mg sin{69^0} ) + (mg cos{69^0})$$\nSubstitute, and you're done.\n\nHope this helps,\n-Evgeny\n\nstill tryin\n\nThanks for the help! I followed your steps above however the answer i got is not the correct answer. Am i doing the problem incorrect.\n\nF=248kg(0.800m/s^2)+.880((248kg)(9.80m/s^2)(sin69))+248kg(9.80m/s^2)(cos69)\nF=2367.23 is not correct\ncould you please tell me what i am doing wrong.",
null,
"Where did 69 come from? The ramp is on a 29 degree incline. Charvonne, i don't think you quite understand the problem, because although evgeny did make a mistake, his process is right. That is something you should be able to pick up on and figure out the solution.\n\nDoc Al\nMentor\nActually, Evgeny made two mistakes. One, he used 69 degrees instead of 29. But he also mixed up sine and cosine. For example, the normal force is $mg cos\\theta$, not $mg sin\\theta$. Similarly, the component of the weight down the ramp is $mg sin\\theta$, not $mg cos\\theta$. (Where $\\theta$ = 29 degrees.)"
] |
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"https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/cry.png",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null
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http://wikien3.appspot.com/wiki/Charge_(physics)
|
[
"# Charge (physics)\n\nIn physics, a charge is any of many different quantities, such as the electric charge in electromagnetism or the color charge in quantum chromodynamics. Charges correspond to the time-invariant generators of a symmetry group, and specifically, to the generators that commute with the Hamiltonian. Charges are often denoted by the letter Q, and so the invariance of the charge corresponds to the vanishing commutator $[Q,H]=0$",
null,
", where H is the Hamiltonian. Thus, charges are associated with conserved quantum numbers; these are the eigenvalues q of the generator Q.\n\n## Abstract definition\n\nAbstractly, a charge is any generator of a continuous symmetry of the physical system under study. When a physical system has a symmetry of some sort, Noether's theorem implies the existence of a conserved current. The thing that \"flows\" in the current is the \"charge\", the charge is the generator of the (local) symmetry group. This charge is sometimes called the Noether charge.\n\nThus, for example, the electric charge is the generator of the U(1) symmetry of electromagnetism. The conserved current is the electric current.\n\nIn the case of local, dynamical symmetries, associated with every charge is a gauge field; when quantized, the gauge field becomes a gauge boson. The charges of the theory \"radiate\" the gauge field. Thus, for example, the gauge field of electromagnetism is the electromagnetic field; and the gauge boson is the photon.\n\nThe word \"charge\" is often used as a synonym for both the generator of a symmetry, and the conserved quantum number (eigenvalue) of the generator. Thus, letting the upper-case letter Q refer to the generator, one has that the generator commutes with the Hamiltonian [Q, H] = 0. Commutation implies that the eigenvalues (lower-case) q are time-invariant: dq/dt = 0.\n\nSo, for example, when the symmetry group is a Lie group, then the charge operators correspond to the simple roots of the root system of the Lie algebra; the discreteness of the root system accounting for the quantization of the charge. The simple roots are used, as all the other roots can be obtained as linear combinations of these. The general roots are often called raising and lowering operators, or ladder operators.\n\nThe charge quantum numbers then correspond to the weights of the highest-weight modules of a given representation of the Lie algebra. So, for example, when a particle in a quantum field theory belongs to a symmetry, then it transforms according to a particular representation of that symmetry; the charge quantum number is then the weight of the representation.\n\n## Examples\n\nVarious charge quantum numbers have been introduced by theories of particle physics. These include the charges of the Standard Model:\n\nCharges of approximate symmetries:\n\nHypothetical charges of extensions to the Standard Model:\n\n• The hypothetical magnetic charge is another charge in the theory of electromagnetism. Magnetic charges are not seen experimentally in laboratory experiments, but would be present for theories including magnetic monopoles.\n• The supercharge refers to the generator that rotates the fermions into bosons, and vice versa, in the supersymmetry.\n\nIn gravitation:\n\n• Eigenvalues of the energy-momentum tensor correspond to physical mass.\n\n## Charge conjugation\n\nIn the formalism of particle theories, charge-like quantum numbers can sometimes be inverted by means of a charge conjugation operator called C. Charge conjugation simply means that a given symmetry group occurs in two inequivalent (but still isomorphic) group representations. It is usually the case that the two charge-conjugate representations are complex conjugate fundamental representations of the Lie group. Their product then forms the adjoint representation of the group.\n\nThus, a common example is that the product of two charge-conjugate fundamental representations of SL(2,C) (the spinors) forms the adjoint rep of the Lorentz group SO(3,1); abstractly, one writes\n\n$2\\otimes {\\overline {2}}=3\\oplus 1.\\$",
null,
"That is, the product of two (Lorentz) spinors is a (Lorentz) vector and a (Lorentz) scalar. Note that the complex Lie algebra sl(2,C) has a compact real form su(2) (in fact, all Lie algebras have a unique compact real form). The same decomposition holds for the compact form as well: the product of two spinors in su(2) being a vector in the rotation group O(3) and a singlet. The decomposition is given by the Clebsch–Gordan coefficients.\n\nA similar phenomenon occurs in the compact group SU(3), where there are two charge-conjugate but inequivalent fundamental representations, dubbed $3$",
null,
"and ${\\overline {3}}$",
null,
", the number 3 denoting the dimension of the representation, and with the quarks transforming under $3$",
null,
"and the antiquarks transforming under ${\\overline {3}}$",
null,
". The Kronecker product of the two gives\n\n$3\\otimes {\\overline {3}}=8\\oplus 1.\\$",
null,
"That is, an eight-dimensional representation, the octet of the eight-fold way, and a singlet. The decomposition of such products of representations into direct sums of irreducible representations can in general be written as\n\n$\\Lambda \\otimes \\Lambda '=\\bigoplus _{i}{\\mathcal {L}}_{i}\\Lambda _{i}$",
null,
"for representations $\\Lambda$",
null,
". The dimensions of the representations obey the \"dimension sum rule\":\n\n$d_{\\Lambda }\\cdot d_{\\Lambda '}=\\sum _{i}{\\mathcal {L}}_{i}d_{\\Lambda _{i}}.$",
null,
"Here, $d_{\\Lambda }$",
null,
"is the dimension of the representation $\\Lambda$",
null,
", and the integers ${\\mathcal {L}}$",
null,
"being the Littlewood–Richardson coefficients. The decomposition of the representations is again given by the Clebsch–Gordan coefficients, this time in the general Lie-algebra setting."
] |
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https://community.smartsheet.com/discussion/73777/formula-fails-after-1000-rows
|
[
"# Formula fails after 1000 rows\n\nHave this formula that works like a dream until I get to 1000 rows\n\nIt basically creates a sequential number like this DM-0998, DM-0999.\n\n`=\"DM-\" + IF(One@row < 10, \"000\", IF(One@row < 100, \"00\", IF(One@row < 1000, \"0\", One@row))) + JOIN(COLLECT(One@row:Two@row, One@row:Two@row, ISNUMBER(@cell)), \".\")`\n\nOnce we hit 1000 rows the formula results in this. DM-10011001\n\nThis formula throws an invalid operator error instead of DM-1001\n\n`=\"DM-\" + IF(One@row < 10, \"000\", IF(One@row < 100, \"00\", IF(One@row < 1000, \"0\", One@row, IF(One@row < 1000, \"\", One@row)))) + JOIN(COLLECT(One@row:Two@row, One@row:Two@row, ISNUMBER(@cell)), \".\")`\n\n• This mostly works except on row 1000 instead of DM-1000, I get DM-10001000\n\n`=\"DM-\" + IF(One@row < 10, \"000\", IF(One@row < 100, \"00\", IF(One@row < 1000, \"0\", IF(One@row > 1000, \"\", One@row))) + JOIN(COLLECT(One@row:Two@row, One@row:Two@row, ISNUMBER(@cell)), \".\"))`\n\n• Hello @Stavros_McGillicuddy !\n\nRather than using a Formula to number your rows, you may instead wish to use an Auto Numbering Column. With this, you can set your Prefix and Suffix. See more on this here: https://help.smartsheet.com/articles/1108408-auto-numbering-rows\n\nPlease let me know if you have any questions!\n\nRegards\n\nSean\n\n• Hi Sean.. Thanks for your input.\n\nI stared with an Auto Number column but, I moved to a formula so that I could keep sub tasks grouped with their tasks\n\nDM-0998\n\nDM-0998.1\n\nDM-0998.2\n\netc\n\n## Help Article Resources\n\nWant to practice working with formulas directly in Smartsheet?\n\nCheck out the Formula Handbook template!"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.72214663,"math_prob":0.7573995,"size":1516,"snap":"2023-40-2023-50","text_gpt3_token_len":504,"char_repetition_ratio":0.19113757,"word_repetition_ratio":0.15178572,"special_character_ratio":0.3687335,"punctuation_ratio":0.1845238,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961979,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T02:10:15Z\",\"WARC-Record-ID\":\"<urn:uuid:85cb92d1-84ff-457b-a4d1-0ecd6d5e2b0a>\",\"Content-Length\":\"191284\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62984cde-1e73-4107-bd0e-9210debbce2d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e56d6c8-4ddb-4135-8128-0c07f4d751a5>\",\"WARC-IP-Address\":\"162.159.138.78\",\"WARC-Target-URI\":\"https://community.smartsheet.com/discussion/73777/formula-fails-after-1000-rows\",\"WARC-Payload-Digest\":\"sha1:RYN4XJW7YKCRC4PWXKMV5VAAG3PUN273\",\"WARC-Block-Digest\":\"sha1:TDV6PHDETEW64VEPNOOZFPTZGNQOS5OD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100575.30_warc_CC-MAIN-20231206000253-20231206030253-00213.warc.gz\"}"}
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https://codereview.stackexchange.com/questions/137841/finding-an-equilibrium-index-in-an-int-array
|
[
"# Finding an equilibrium index in an int array\n\nHere is a programming challenge from codility\n\nA zero-indexed array A consisting of N integers is given. An equilibrium index of this array is any integer P such that 0 ≤ P < N and the sum of elements of lower indices is equal to the sum of elements of higher indices, i.e.\n\nA + A + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].\n\nSum of zero elements is assumed to be equal to 0. This can happen if P = 0 or if P = N−1.\n\nFor example, consider the following array A consisting of N = 8 elements:\n\nA = -1 A = 3 A = -4 A = 5 A = 1 A = -6 A = 2 A = 1\n\nP = 1 is an equilibrium index of this array, because:\n\n• A = −1 = A + A + A + A + A + A\n\nP = 3 is an equilibrium index of this array, because:\n\n• A + A + A = −2 = A + A + A + A\n\nP = 7 is also an equilibrium index, because:\n\n• A + A + A + A + A + A + A = 0\n\nand there are no elements with indices greater than 7.\n\nP = 8 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N.\n\nWrite a function:\n\nclass Solution { public int solution(int[] A); }\n\n\nthat, given a zero-indexed array A consisting of N integers, returns any of its equilibrium indices. The function should return −1 if no equilibrium index exists.\n\nFor example, given array A shown above, the function may return 1, 3 or 7, as explained above.\n\nAssume that:\n\n• N is an integer within the range [0..100,000];\n• each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].\n\nComplexity:\n\n• expected worst-case time complexity is $O(N)$;\n• expected worst-case space complexity is $O(N)$, beyond input storage (not counting the storage required for input arguments).\n\nElements of input arrays can be modified.\n\n• I think my whole solution is $O(N)$ because both my methods are $O(N)$. Is that right?\n• I believe I used some space to generate the sums and it Is still $O(N)$. Is it right?\n\nclass Equilibrium {\n\npublic int getEquilibrium(int[] A) {\nlong[] sums = generateSums(A);\nlong lowSum = 0L;\nint res = -1;\nfor (int i = 0; i < A.length; i++) {\nif (lowSum == sums[i+1]) {\nres = i;\nbreak;\n}\nlowSum += A[i];\n}\nreturn res;\n}\n\n// I used long to handle sums greater than 32 bits\npublic long[] generateSums(int[] A) {\n// I added another index (default value is 0) to handle the last value (E.G. the sum of previous elements is 0, so the equilibrium index should be the last value because there are no other items after it. I used the newly added index to handle it.)\nlong[] res = new long[A.length+1];\nfor (int i = A.length-1; i >= 0; i--) {\nres[i] = i+1 == A.length ? A[i] : res[i+1]+A[i];\n}\nreturn res;\n}\n}\n\n\nDid I follow all the requirements in the problem? How can I improve this and make it faster?\n\n• Just off the bat, you didn't name your function solution...\n– Idos\nAug 4 '16 at 17:28\n• @Idos I did named it \"solution\" when I post it to the codility site. Why would I put it like that here? \"solution\" might be the worst method name ever don't you think? :)) Aug 4 '16 at 17:53\n\n• I think my whole solution is $O(N)$ because both my methods are O(N). Is that right?\n\nCorrect. Both of your methods are $O(N)$: they are linear in terms of the number of elements in the input array. More specifically, both generateSums and getEquilibrium loop over the array once.\n\n• I believe I used some space to generate the sums and it Is still $O(N)$. Is it right?\n\nAlso correct. The intermediate array containing the sum contains as many elements as there was in the input array; doubling the input array in size would double the size of the intermediate array. This makes in linear in terms of space complexity.\n\nThe method generateSums doesn't have a good name. Its purpose is to calculate the cumulative sum of the input array, going from the right to the left. Consider naming it cumulativeSums instead. The rest of the algorithm is correct. You won't be able to make this faster than $O(N)$, in terms of time complexity.\n\nA small comment: instead of using a break explicitly inside the for loop, consider returning directly the found index. You currenly have:\n\nint res = -1;\nfor (int i = 0; i < A.length; i++) {\nif (lowSum == sums[i+1]) {\nres = i;\nbreak;\n}\nlowSum += A[i];\n}\nreturn res;\n\n\nbut it would be shorter to have:\n\nfor (int i = 0; i < A.length; i++) {\nif (lowSum == sums[i+1]) {\nreturn i;\n}\nlowSum += A[i];\n}\nreturn -1;\n\n\nIt avoids the use of a temporary local variable holding the result, and makes it clearer that -1 is the default value returned.\n\nIn terms of space complexity however, you can do better: this is possible in $O(1)$. Instead of generating an array containing the cumulative sums of the array, just calculate the total sum, let's call it right sum. Then, you keep a running sum of the array by traversing it from left to right, decreasing this right sum as you go: if the running sum is equal to the right sum, you found an equilibrium.\n\npublic int getEquilibrium(int[] array) {\nlong totalSum = sum(array);\nlong lowSum = 0L;\nfor (int i = 0; i < array.length; i++) {\ntotalSum -= array[i];\nif (lowSum == totalSum) {\nreturn i;\n}\nlowSum += array[i];\n}\nreturn -1;\n}\n\npublic long sum(int[] array) {\nreturn Arrays.stream(array).asLongStream().sum();\n}\n\n\nI think my whole solution is $O(N)$ because both my methods are $O(N)$. Is that right?\n\nThat's correct. You traverse the $N$ values twice, the time complexity is $O(N)$.\n\nI believe I used some space to generate the sums and it Is still $O(N)$. Is it right?\n\nYes, you used extra $O(N)$ space.\n\nDid I follow all the requirements in the problem? How can I improve this and make it faster?\n\nYou followed the requirements, and apparently you got perfect score. But you can do better, by using $O(1)$ instead of $O(N)$.\n\nYou don't need to store the sums in an array. You can calculate the prefix sum as you go, and return the index when prefix == sum - prefix + A[i]. This also implies that you don't need int res = -1;, you can return the index immediately when you find it, and return -1 if the loop terminates without finding an equilibrium index.\n\npublic int findEquilibriumIndex(int[] arr) {\nlong sum = sum(arr);\n\nlong prefix = 0;\nfor (int index = 0; index < arr.length; index++) {\nprefix += arr[index];\nif (prefix == sum - prefix + arr[index]) {\nreturn index;\n}\n}\nreturn -1;\n}\n\npublic long sum(int[] arr) {\nlong sum = 0;\nfor (int value : arr) {\nsum += value;\n}\nreturn sum;\n}\n\n• did you run your code? Shouldn't the index start from 1? And the prefix contain sum of A[0..P-1], since neither the prefix nor the right sum should have the A[P] in their sum?\n– Sky\nFeb 5 '19 at 19:27\n• @Sky yes, it works correctly. In this implementation prefix contains A[P], and then the condition evaluates prefix == sum - prefix + arr[P]. You see, the right-hand side of the condition also contains A[P]. An equivalent way of writing is to rearrange the loop body like this: if (prefix == sum - prefix - arr[index]) { return index; } prefix += arr[index]; Feb 6 '19 at 7:18"
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|
http://wzdatingmp.chaturbatetokenhax.xyz/physics-formula-cheat-sheet.html
|
[
"# Physics formula cheat sheet\n\nSheet cheat\n\n## Physics formula cheat sheet\n\nPhysics formula cheat sheet. Please Do Not Write on This Sheet Phhyyssiiccss hFFoorrmmuullaa SSheeeett. Sim cares his students so much. ( individual point masses) I= r2dm Voume \" ( continuous mass distribution) Parallel- axis cheat TheoremI= I. Euler' s identity therefore states that the limit, as n. The physics formula sheets are in pdf format you need physics Acrobat Reader to display print them. Sim is definitely one physics of the best here to help you. 67 × 10- 27 kg Neutron = 0 C = 1.\n\nformula Please Do Not Write on This Sheet 1 2. Massachusetts Comprehensive Assessment System Introductory Physics Reference Sheet Formulas T = 1 f Q = mc˜ T V = IR v = ˚ f a average = ˜ v ˜ t ˜ x = v i˜ t + 1a˜ t2 2 s average = d ˜ t v f = v i + a˜ t v average = ˜ formula x ˜ t F net = sheet ma F sheet g cheat = G m 1m 2 d2 p = mv F g = mg F˜ t = ˜ p ˜ PE = mg˜ h eff = E out E in F e = k q 1q 2 d2 W = ˜ E = Fd KE. our physics cheat sheet today and cheat contact us for all of the help you. 602 19 × physics 10- 19 C = 9. Physics Equations. 60× 10− 19C µ formula 0 π = 4 × 10 7 cheat − T• m physics A 9 c m s. 80 2 G N m kg = × 667 10− 11 • 2. 022 × 10 23 atoms in one atomic mass unit e is the elementary charge: 1.\n\ninformation when you find the need to search for a formula. Nature of Science and Physics T=. To begin with, physics is not an easy subject. Physics formula cheat sheet. 602 19 × 10- 19 C Potential Energy, velocity of electron: PE = eV = ½mv 2 1V = 1J/ C 1N/ C = 1V/ sheet m 1J = 1 N· m = 1 C· V. McFatter McPhysics Equation Sheet Metric Unit Conversions ( Chapter 1). 67 × 10- 27 kg 6. Physics 130 Formula sheet Sheet – Stefan Martynkiw Simple Harmonic Motion formula x= A⋅ cos sheet t vx= − Asin t ax= − 2 Acos t ax= − 2 x = 2 f = k m = g L f= 2 = 1 2 k m T= 1 f = 2 m k = arctan − v0x x0 A= x0 2 v0 2 2 Energy in SHM E= 1 2 mvx 2 1 2 kx2= 1 2 kA2 vx= ± k m cheat A2− x2 vmaxoccursatx= 0.\n\nCheatBook Issue ( 03/ ) March : CheatBook( 03/ cheat ) - Issue March - A Cheat- Code Tracker with cheats formula Hints for several popular PC Action adventure Games. Revision notes OCR, physics checklists, worksheets, papers for AQA, Edexcel, topic questions , summary sheets with key points MEI cheat Core 4 Maths A- level. General Physics sheet Cheat Sheet. Physics 121, Formula Sheet Final Exam - 6 - Moment sheet of Inertia: Moment of Inertia: I= m r r i 2 i! Besides his kind grading criterions, Dr. Blyth Academy SPH4U - Grade 12 University Physics Physics Formula Sheet CONSTANTS g m s = 9. CoSMOS makes it easy fun to hack your own PC formula games while waiting for our formula programming team to release new updated trainers. AVAILABLE BOOKS Click here to order any book THE GREATEST STANDING ERRORS IN PHYSICS. 11 × 10- 31 kg Proton = 1.\n\n602 19 × 10- 19 cheat C = 1. by Miles Mathis Painting Experiment physics with an Air Pump by Joseph sheet Wright of Derby cheat 1768, National Gallery London email me at com go to formula my art site. 367 PC Games formula 51 Console Cheats are represented in this new version from Strategy Games, 9 Walkthroughs for PC Adventure Games to Action Games. PHYSICS FORMULAS 2426 physics Electron = - 1. Click formula here to get the web page for downloading Acrobat Reader Exam 1 Formula Sheet ( chaptersPDF). Calculus and Analysis W.\nFundamentally, Euler' s identity asserts that is equal to - 1. The expression is a special case of the expression, where z is any complex number. mechanical physics impulse acceleration formula a= sheet exam kinematics acceleration speed velocity 1- d kinematics cheat sheet 1- d physics formula physics kinematics equation cheat sheet acceleration cheat sheet formula for acceleration information physics acceleration cheat cheet wave mechanics tips and tricks physics physic sheet mechanic what is cheat the. edu) Limits Cheat Sheet by Paul Dawkins [ pdf, pdf reduced] ( tutorial. Introducing Cheat Happens CoSMOS, a new self- service gamehacking tool designed from the ground up for a new generation. In general, cheat is defined for formula complex z by extending formula one of the definitions of the exponential function from real exponents to complex exponents. For example, one common definition is: = → ∞ ( + ). Complete Calculus Cheat physics Sheet by Paul Dawkins [ pdf, pdf reduced] physics ( tutorial.\n\n## Physics formula\n\nTeach Yourself Chemistry Visually in 24 Hours - by Dr. Wayne Huang and his team. The series includes High School Chemistry, AP Chemistry, General Chemistry, Organic Chemistry and Biochemistry. Master Chemistry The Easy and Rapid Way with Core Concept Tutorials, Problem- Solving Drills and Super Review Cheat Sheets.\n\n``physics formula cheat sheet``\n\nOne Hour Per Lesson, 24 Lessons Per Course. Free step- by- step solutions to AlgebraSlader Can you find your fundamental truth using Slader as a completely free Algebra 1 solutions manual?"
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.80526763,"math_prob":0.9789108,"size":3306,"snap":"2019-43-2019-47","text_gpt3_token_len":867,"char_repetition_ratio":0.13930951,"word_repetition_ratio":0.022328548,"special_character_ratio":0.26436782,"punctuation_ratio":0.085346214,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9938433,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T03:29:57Z\",\"WARC-Record-ID\":\"<urn:uuid:da06f7c8-752a-452f-acf6-5afabc2909ab>\",\"Content-Length\":\"10396\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f18279aa-d49c-4127-a442-b08d0aac66c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:aca212ee-4b19-498b-8bd0-105c9de17b29>\",\"WARC-IP-Address\":\"104.27.129.93\",\"WARC-Target-URI\":\"http://wzdatingmp.chaturbatetokenhax.xyz/physics-formula-cheat-sheet.html\",\"WARC-Payload-Digest\":\"sha1:DDOJIGN63JVDTCSN7DQRKIGFON4T6FEY\",\"WARC-Block-Digest\":\"sha1:T7RKJ3Z5J6O6ZFUBIWZCACW4BU6CDUVJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670729.90_warc_CC-MAIN-20191121023525-20191121051525-00446.warc.gz\"}"}
|
https://kr.mathworks.com/help/wavelet/ug/r-wave-detection-in-the-ecg.html
|
[
"# R Wave Detection in the ECG\n\nThis example shows how to use wavelets to analyze electrocardiogram (ECG) signals. ECG signals are frequently nonstationary meaning that their frequency content changes over time. These changes are the events of interest.\n\nWavelets decompose signals into time-varying frequency (scale) components. Because signal features are often localized in time and frequency, analysis and estimation are easier when working with sparser (reduced) representations.\n\nThe QRS complex consists of three deflections in the ECG waveform. The QRS complex reflects the depolarization of the right and left ventricles and is the most prominent feature of the human ECG.\n\nLoad and plot an ECG waveform where the R peaks of the QRS complex have been annotated by two or more cardiologists. The ECG data and annotations are taken from the MIT-BIH Arrhythmia Database. The data are sampled at 360 Hz.\n\n```load mit200 figure plot(tm,ecgsig) hold on plot(tm(ann),ecgsig(ann),'ro') xlabel('Seconds') ylabel('Amplitude') title('Subject - MIT-BIH 200')```",
null,
"You can use wavelets to build an automatic QRS detector for use in applications like R-R interval estimation.\n\nThere are two keys for using wavelets as general feature detectors:\n\n• The wavelet transform separates signal components into different frequency bands enabling a sparser representation of the signal.\n\n• You can often find a wavelet which resembles the feature you are trying to detect.\n\nThe 'sym4' wavelet resembles the QRS complex, which makes it a good choice for QRS detection. To illustrate this more clearly, extract a QRS complex and plot the result with a dilated and translated 'sym4' wavelet for comparison.\n\n```qrsEx = ecgsig(4560:4810); fb = dwtfilterbank('Wavelet','sym4','SignalLength',numel(qrsEx),'Level',3); psi = wavelets(fb); figure plot(qrsEx) hold on plot(-2*circshift(psi(3,:),[0 -38]),'r') axis tight legend('QRS Complex','Sym4 Wavelet') title('Comparison of Sym4 Wavelet and QRS Complex') hold off```",
null,
"Use the maximal overlap discrete wavelet transform (MODWT) to enhance the R peaks in the ECG waveform. The MODWT is an undecimated wavelet transform, which handles arbitrary sample sizes.\n\nFirst, decompose the ECG waveform down to level 5 using the default 'sym4' wavelet. Then, reconstruct a frequency-localized version of the ECG waveform using only the wavelet coefficients at scales 4 and 5. The scales correspond to the following approximate frequency bands.\n\n• Scale 4 -- [11.25, 22.5) Hz\n\n• Scale 5 -- [5.625, 11.25) Hz.\n\nThis covers the passband shown to maximize QRS energy.\n\n```wt = modwt(ecgsig,5); wtrec = zeros(size(wt)); wtrec(4:5,:) = wt(4:5,:); y = imodwt(wtrec,'sym4');```\n\nUse the squared absolute values of the signal approximation built from the wavelet coefficients and employ a peak finding algorithm to identify the R peaks.\n\nIf you have Signal Processing Toolbox™, you can use `findpeaks` to locate the peaks. Plot the R-peak waveform obtained with the wavelet transform annotated with the automatically-detected peak locations.\n\n```y = abs(y).^2; [qrspeaks,locs] = findpeaks(y,tm,'MinPeakHeight',0.35,... 'MinPeakDistance',0.150); figure plot(tm,y) hold on plot(locs,qrspeaks,'ro') xlabel('Seconds') title('R Peaks Localized by Wavelet Transform with Automatic Annotations')```",
null,
"Add the expert annotations to the R-peak waveform. Automatic peak detection times are considered accurate if within 150 msec of the true peak ($±75$ msec).\n\n```plot(tm(ann),y(ann),'k*') title('R peaks Localized by Wavelet Transform with Expert Annotations')```",
null,
"At the command line, you can compare the values of `tm(ann)` and `locs`, which are the expert times and automatic peak detection times respectively. Enhancing the R peaks with the wavelet transform results in a hit rate of 100% and no false positives. The calculated heart rate using the wavelet transform is 88.60 beats/minute compared to 88.72 beats/minute for the annotated waveform.\n\nIf you try to work on the square magnitudes of the original data, you find the capability of the wavelet transform to isolate the R peaks makes the detection problem much easier. Working on the raw data can cause misidentifications such as when the squared S-wave peak exceeds the R-wave peak around 10.4 seconds.\n\n```figure plot(tm,ecgsig,'k--') hold on plot(tm,y,'r','linewidth',1.5) plot(tm,abs(ecgsig).^2,'b') plot(tm(ann),ecgsig(ann),'ro','markerfacecolor',[1 0 0]) set(gca,'xlim',[10.2 12]) legend('Raw Data','Wavelet Reconstruction','Raw Data Squared', ... 'Location','SouthEast'); xlabel('Seconds')```",
null,
"Using `findpeaks` on the squared magnitudes of the raw data results in twelve false positives.\n\n```[qrspeaks,locs] = findpeaks(ecgsig.^2,tm,'MinPeakHeight',0.35,... 'MinPeakDistance',0.150);```\n\nIn addition to switches in polarity of the R peaks, the ECG is often corrupted by noise.\n\n```load mit203 figure plot(tm,ecgsig) hold on plot(tm(ann),ecgsig(ann),'ro') xlabel('Seconds') ylabel('Amplitude') title('Subject - MIT-BIH 203 with Expert Annotations')```",
null,
"Use the MODWT to isolate the R peaks. Use `findpeaks` to determine the peak locations. Plot the R-peak waveform along with the expert and automatic annotations.\n\n```wt = modwt(ecgsig,5); wtrec = zeros(size(wt)); wtrec(4:5,:) = wt(4:5,:); y = imodwt(wtrec,'sym4'); y = abs(y).^2; [qrspeaks,locs] = findpeaks(y,tm,'MinPeakHeight',0.1,... 'MinPeakDistance',0.150); figure plot(tm,y) title('R-Waves Localized by Wavelet Transform') hold on hwav = plot(locs,qrspeaks,'ro'); hexp = plot(tm(ann),y(ann),'k*'); xlabel('Seconds') legend([hwav hexp],'Automatic','Expert','Location','NorthEast');```",
null,
"The hit rate is again 100% with zero false alarms.\n\nThe previous examples used a very simple wavelet QRS detector based on a signal approximation constructed from modwt. The goal was to demonstrate the ability of the wavelet transform to isolate signal components, not to build the most robust wavelet-transform-based QRS detector. It is possible, for example, to exploit the fact that the wavelet transform provides a multiscale analysis of the signal to enhance peak detection.\n\n#### References\n\nGoldberger A. L., L. A. N. Amaral, L. Glass, J. M. Hausdorff, P. Ch. Ivanov, R. G. Mark, J. E. Mietus, G. B. Moody, C-K Peng, H. E. Stanley. \"PhysioBank, PhysioToolkit, and PhysioNet: Components of a New Research Resource for Complex Physiologic Signals.\" Circulation 101. Vol.23, e215-e220, 2000. `http://circ.ahajournals.org/cgi/content/full/101/23/e215`\n\nMoody, G. B. \"Evaluating ECG Analyzers\". `http://www.physionet.org/physiotools/wfdb/doc/wag-src/eval0.tex`\n\nMoody G. B., R. G. Mark. \"The impact of the MIT-BIH Arrhythmia Database.\" IEEE Eng in Med and Biol. Vol. 20, Number 3, 2001), pp. 45-50 ."
] |
[
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"https://kr.mathworks.com/help/examples/wavelet/win64/RWaveDetectionInTheECGExample_01.png",
null,
"https://kr.mathworks.com/help/examples/wavelet/win64/RWaveDetectionInTheECGExample_02.png",
null,
"https://kr.mathworks.com/help/examples/wavelet/win64/RWaveDetectionInTheECGExample_03.png",
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"https://kr.mathworks.com/help/examples/wavelet/win64/RWaveDetectionInTheECGExample_04.png",
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"You are on page 1of 250\n\n# UlfWickstrm\n\nTemperature\nCalculation\nin Fire Safety\nEngineering\nTemperature Calculation in Fire Safety Engineering\nUlf Wickstr\nom\n\nTemperature Calculation\nin Fire Safety Engineering\nUlf Wickstrom\nLulea University of Technology\nLulea, Sweden\n\n## Instructor answers to review questions can be found at\n\nhttp://www.springer.com/us/book/9783319301709\n\n## ISBN 978-3-319-30170-9 ISBN 978-3-319-30172-3 (eBook)\n\nDOI 10.1007/978-3-319-30172-3\n\n## Springer International Publishing Switzerland 2016\n\nThis work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of\nthe material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,\nrecitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission\nor information storage and retrieval, electronic adaptation, computer software, or by similar or\ndissimilar methodology now known or hereafter developed.\nThe use of general descriptive names, registered names, trademarks, service marks, etc. in this\npublication does not imply, even in the absence of a specific statement, that such names are exempt\nfrom the relevant protective laws and regulations and therefore free for general use.\nThe publisher, the authors and the editors are safe to assume that the advice and information in this\nbook are believed to be true and accurate at the date of publication. Neither the publisher nor the\nauthors or the editors give a warranty, express or implied, with respect to the material contained\nherein or for any errors or omissions that may have been made.\n\n## Printed on acid-free paper\n\nThe registered company is Springer International Publishing AG Switzerland\nPreface\n\nThis book is about temperature calculation and heat transfer. It is intended for\nresearchers, students and teachers in the field of fire safety engineering as well as\nconsultants and others interested in analysing and understanding fire and tempera-\nture developments. It gives a consistent scientific background to engineering\ncalculation methods applicable to analyses for both materials reaction to fire and\nfire resistance of structures. Several new formulas and diagrams facilitating calcu-\nlations are presented.\nThe book is particularly devoted to problems involving severe thermal condi-\ntions as are of interest in fire dynamics and FSE. However, definitions, nomencla-\nture and theories used are aligned with those of general textbooks on temperature\ncalculation and heat transfer such as [1, 2].\nIn particular great effort has been put on defining boundary conditions in a\ncorrect and suitable way for calculations. A large portion of the book is devoted to\nboundary conditions and measurements of thermal exposure by radiation and\nconvection. Thus, the concept and theory of adiabatic surface temperature and\nmeasurements of temperature with plate thermometers are thoroughly explained.\nInitially a number of zero- and one-dimensional cases assuming constant mate-\nrial properties are dealt with where exact closed form analytical solutions are\npossible. These can, however, generally only be used for estimates in FSE problems\nas they require assumptions of constant material properties and boundary condi-\ntions. In most cases numerical calculations are therefore needed for considering\nmaterial properties changing with temperature and non-linear boundary conditions\ndue to emission of radiant heat. Thus, several recursion formulas are given in the\nbook which are suited for spreadsheet calculation codes (such as MS Excel). For\nmore advanced calculations, introductions and guidance are given to finite element\nanalyses.\nThe phenomena of heat transfer by radiation and convection are introduced\nbased on what can be found in general textbooks. Several of the formulas are,\nhowever, adapted to FSE problems, and unique charts and tables are presented\nwhich considerably facilitates calculations.\n\nv\nvi Preface\n\nA renewed method for modelling compartment fires is presented which has led\nto simple and accurate prediction tools for both pre- and post-flashover fires.\nThe final three chapters deal with temperature calculations in steel, concrete and\ntimber structures exposed to standard time-temperature fire curves. Handy temper-\nature calculation tools are presented, and several examples are shown on how the\nfinite element code TASEF can be used to calculate temperature in various config-\nurations.\n\n## Lulea, Sweden Ulf Wickstrom\n\nContents\n\n1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1\n1.1 Basic Concepts of Temperature, Heat and Heat Flux . . . . . . . . 1\n1.1.1 Heat and Temperature . . . . . . . . . . . . . . . . . . . . . . . . 2\n1.1.2 Heat Transfer Modes . . . . . . . . . . . . . . . . . . . . . . . . . 2\n1.1.3 The Three Kinds of Boundary Conditions . . . . . . . . . . 3\n1.1.4 Transient or Unsteady-State Heat Conduction . . . . . . . 8\n1.2 Electric Circuit Analogy in One Dimension . . . . . . . . . . . . . . . 10\n1.3 Material Properties at Elevated Temperature . . . . . . . . . . . . . . 11\n1.3.1 Structural Materials . . . . . . . . . . . . . . . . . . . . . . . . . . 13\n1.3.2 Polymers and Composite Materials . . . . . . . . . . . . . . . 15\n1.3.3 Measurements of Material Properties . . . . . . . . . . . . . 15\n2 Steady-State Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17\n2.1 Plane Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18\n2.2 Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22\n3 Unsteady-State Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25\n3.1 Lumped-Heat-Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25\n3.1.1 Prescribed Heat Flux: BC of the Second Kind . . . . . . . 26\n3.1.2 Prescribed Gas Temperature: BC of the Third\nKindAnd the Concept of Time Constant . . . . . . . . . 27\n3.1.3 Conditions for Assuming Lumped-Heat-Capacity . . . . 31\n3.2 Semi-infinite Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33\n3.2.1 Constant Surface Temperature: First Kind of BC . . . . . 35\n3.2.2 Constant Heat Flux: Second Kind of BC . . . . . . . . . . . 39\n3.2.3 Constant Gas Temperature: Third Kind of BC . . . . . . . 39\n4 Boundary Conditions in Fire Protection Engineering . . . . . . . . . . . 45\n4.1 Radiation and Incident Radiation Temperature . . . . . . . . . . . . . 46\n4.2 Non-linear Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50\n4.3 Mixed Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 51\n\nvii\nviii Contents\n\n## 4.4 Adiabatic Surface Temperature . . . . . . . . . . . . . . . . . . . . . . . . 53\n\nTemperature and Incident Radiation . . . . . . . . . . . . . . 55\n4.4.2 An Electric Circuit Analogy of the AST\nBoundary Condition . . . . . . . . . . . . . . . . . . . . . . . . . . 56\n4.4.3 Boundary Condition Expressed as Heat Flux . . . . . . 60\n4.4.4 Calculation of Time Constants for Bodies\nExposed to Mixed Boundary Conditions . . . . . . . . . . . 60\n5 Heat Transfer by Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65\n5.1 Radiation Between Two Parallel Planes\nand Radiation Shields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68\n5.2 View Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72\n5.2.1 View Factors Between Differential Elements . . . . . . . . 74\n5.2.2 View Factors Between a Differential\nElement and a Finite Area . . . . . . . . . . . . . . . . . . . . . 76\n5.2.3 View Factors Between Two Finite Areas . . . . . . . . . . . 81\n5.3 Radiation from Flames and Smoke . . . . . . . . . . . . . . . . . . . . . 83\n6 Heat Transfer by Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89\n6.1 Heat Transfer Properties of Air and Water . . . . . . . . . . . . . . . . 90\n6.2 Forced Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91\n6.2.1 On Flat Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91\n6.2.2 Across Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95\n6.2.3 In Circular Pipes and Tubes . . . . . . . . . . . . . . . . . . . . 96\n6.3 Natural or Free Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . 97\n6.3.1 On Vertical and Horizontal Plates . . . . . . . . . . . . . . . . 97\n6.3.2 In Enclosed Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 103\n7 Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107\n7.1 Lumped-Heat-Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107\n7.2 Superposition and the Duhamels Superposition Integral . . . . . . 111\n7.3 The Finite Element Method for Temperature Analyses . . . . . . . 114\n7.3.1 One-Dimensional Theory . . . . . . . . . . . . . . . . . . . . . . 115\n7.3.2 Computer Codes for Temperature Calculations . . . . . . 119\n7.3.3 On Accuracy of Finite Element Computer Codes . . . . . 120\n7.3.4 On Specific Volumetric Enthalpy . . . . . . . . . . . . . . . . 121\n8 Thermal Ignition Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125\n8.1 Ignition Temperatures of Common Solids . . . . . . . . . . . . . . . . 125\n8.2 Calculation of Time to Ignitions . . . . . . . . . . . . . . . . . . . . . . . 127\n8.2.1 Thin Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128\n8.2.2 Semi-infinite Solids . . . . . . . . . . . . . . . . . . . . . . . . . . 129\nContents ix\n\n## 9 Measurements of Temperature and Heat Flux . . . . . . . . . . . . . . . . 133\n\n9.1 Thermocouples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133\n9.1.1 Type of Thermocouples . . . . . . . . . . . . . . . . . . . . . . . 133\n9.1.2 Measurement of Temperature in Gases . . . . . . . . . . . . 134\n9.1.3 Corrections of Time Delay . . . . . . . . . . . . . . . . . . . . . 137\n9.2 Heat Flux Meters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139\n9.2.1 Calibration and Use of Heat Flux Meters . . . . . . . . . . . 139\n9.3 The Plate Thermometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143\n9.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143\n9.3.2 Theory for Measuring Incident Heat Flux\nPlate Thermometers . . . . . . . . . . . . . . . . . . . . . . . . . . 143\n9.3.3 Alternative Plate Thermometer Designs . . . . . . . . . . . 147\n10 Post-Flashover Compartment Fires: One-Zone Models . . . . . . . . . 153\n10.1 Heat and Mass Balance Theory . . . . . . . . . . . . . . . . . . . . . . . . 154\n10.2 Solution of the Fire Compartment Temperature . . . . . . . . . . . . 161\n10.2.1 Semi-infinitely Thick Compartment Boundaries . . . . . . 161\n10.2.2 Insulated and Uninsulated Boundaries\nwith a Metal Core . . . . . . . . . . . . . . . . . . . . . . . . . . . 165\n10.2.3 Temperature-Dependent Material and Heat Transfer\nProperties: Numerical Solutions . . . . . . . . . . . . . . . . . 171\n11 Pre-flashover Compartment Fires: Two-Zone Models . . . . . . . . . . 175\n11.1 Heat and Mass Balance Theory . . . . . . . . . . . . . . . . . . . . . . . . 177\n11.2 Solution of the Upper Layer Fire Temperature . . . . . . . . . . . . . 179\n11.2.1 Semi-infinitely Thick Compartment Boundaries . . . . . . 179\n11.2.2 Insulated and Uninsulated Boundaries\nwith a Metal Core . . . . . . . . . . . . . . . . . . . . . . . . . . . 181\n12 Fire Exposure of Structures According to Standards . . . . . . . . . . . 185\n12.1 Standard Time Temperature Fire Curves . . . . . . . . . . . . . . . . . 186\n12.2 Parametric Fire Curves According to Eurocode . . . . . . . . . . . . 188\n12.3 Summary of Heat Transfer Conditions According\nto Eurocodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192\n13 Temperature of Steel Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . 195\n13.1 Thermal Properties of Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . 196\n13.2 Example of Hot-Rolled Steel Section Dimensions . . . . . . . . . . 199\n13.3 Protected Steel Sections Assuming\nLumped-Heat-Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200\n13.3.1 Protection with Heavy Materials . . . . . . . . . . . . . . . . . 204\n13.3.2 Protected Steel Sections Exposed\nto Parametric Fire Curves . . . . . . . . . . . . . . . . . . . . . . 205\n13.4 Unprotected Steel Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 207\n13.4.1 Shadow Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212\nx Contents\n\n## 13.5 Examples of Steel Temperatures Calculated\n\nUsing a Finite Element Code . . . . . . . . . . . . . . . . . . . . . . . . . . 213\n13.5.1 Unprotected Square Steel Tube Section\nAttached to a Concrete Slab or Wall . . . . . . . . . . . . . . 214\n13.5.2 Encased I-Section Connected\nto a Concrete Structure . . . . . . . . . . . . . . . . . . . . . . . . 214\n13.5.3 Unprotected I-Section Connected\nto a Concrete Structure . . . . . . . . . . . . . . . . . . . . . . . . 215\n14 Temperatures of Concrete Structures . . . . . . . . . . . . . . . . . . . . . . . 217\n14.1 Thermal Properties of Concrete . . . . . . . . . . . . . . . . . . . . . . . . 217\n14.2 Penetration Depth in Semi-infinite Structures . . . . . . . . . . . . . . 220\n14.3 Explicit Formula and Diagrams . . . . . . . . . . . . . . . . . . . . . . . . 220\n14.4 Fire Protected Concrete Structures . . . . . . . . . . . . . . . . . . . . . . 224\n15 Temperature of Timber Structures . . . . . . . . . . . . . . . . . . . . . . . . . 227\n15.1 Thermal Properties of Wood . . . . . . . . . . . . . . . . . . . . . . . . . . 227\n15.2 Charring Depth According to Eurocode 5 . . . . . . . . . . . . . . . . . 228\n\n## Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235\n\nReferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241\nNomenclature1\n\n## a Heat of vaporization [J/kg]\n\nA Area [m2]\nBi Biot number[\u0001]\nc Specific heat (capacity) [W s/(kg K)] or [J/(kg K)]\nC Heat capacity per unit area [W s/(m2 K)]\nd Thickness [m]\nD Diameter [m]\nf Forcing function\nF View factor [\u0001]\nGr Grashof number [\u0001]\nh Heat transfer coefficient [W/(m2 K)]\nk Conductivity [W/(m K)]\nl Latent heat [W s/m3]\nL Length [m]\nNu Nusselt number [\u0001]\nO Opening factor [m1/2]\nP Perimeter of surface [m]\nPr Prandtl number [\u0001]\nq Heat [W s or J]\nR One-dimensional thermal resistance [(K m2)/W]\nRe Reynolds number [\u0001]\nT Temperature [K] or [\u0003 C]\nV Volume [m3]\nx Length [m]\nThermal diffusivity [m2/s]\nAbsorption coefficient [\u0001]\nCompartment fire time factor [\u0001]\n\n1\nDefinitions of symbols are given throughout the text. Some selected symbols are listed below.\n\nxi\nxii Nomenclature\n\n## Boundary layer thickness [m]\n\nEmissivity [\u0001]\nReduction coefficient [\u0001]\nAbsorption coefficient [m\u00011]\nKinetic viscosity [m2/s]\nDensity [kg/m3]\nStefan-Boltzmann constant (5.67 10\u00018) [W/(m2 K4)]\nTime constant [s]\nCombustion efficiency [\u0001]\n\nSuperscripts\n0\nPer unit length\n00\nPer unit area\n000\nPer unit volume\nPer unit volume, time derivative\nVector, matrix\n\nSubscripts\n0 Surface (x 0)\n1 Ambient\nB Burning\nCC Cone calorimeter\nCon, c Convection\ncr Critical\nd Duration\nemi Emitted\nhfm Heat flux meter\nf Film\nf Fire\ngas Gas\ni Initial\nig Ignition\nin Insulation\nL Air, gas convection\no Opening\np Constant pressure\np Plume\nPT Plate thermometer\nNomenclature xiii\n\nRC Room/corner test\ns Surface\nsh Shield\nst Steel\nTC Thermocouple\nult Ultimate\nw Wall, surrounding boundary\n\nAbbreviations\n\nASTM ASTM International, earlier American Society for Testing and\nMaterials\nCEN, EN European Committee for Standardization developing EN standards\nEUROCODE EN Eurocodes is a series of 10 European Standards for the design\nof buildings\nFSE Fire safety engineering\nISO International Organization for Standardization\nPT Plate thermometer\nTASEF Computer code for Temperature Analysis of Structures Exposed to\nFire\nChapter 1\nIntroduction\n\nTemperature is the dominating factor in determining the rate and extent of chemical\nreactions including breakdown of organic compounds and deteriorations of strength\nand stiffness of structural materials such as steel and concrete. Phase change\nphenomena including ignition as well as severe loss of strength of materials are\noften related to specific elevated temperature levels. Temperatures of fire gases are\nalso of crucial importance as they initiate gas movements thereby spread of smoke\nand toxic fire gases. Fire temperatures vary typically over several hundred degrees.\nTherefore a number of thermal phenomena need special attention such as phase\nchanges of materials and heat transfer by radiation when calculating temperature of\nfire-exposed materials.\nIn this chapter some of the basic concepts of heat transfer are briefly introduced.\nMore detailed presentations are given in following chapters. A summary of the\nprinciples of electric circuit analogy which is used throughout this book is also\ngiven as well as some general comments on material properties.\n\n## 1.1 Basic Concepts of Temperature, Heat and Heat Flux\n\nTemperature is an intensive or bulk property, i.e. a physical property that does not\ndepend on size or the amount of material in a system. It is scale invariant. By\ncontrast, heat is an extensive property which is directly proportional to the amount\nof material in a system. Density is another example of an intensive quantity as it\ndoes not depend on the quantity, while mass and volume are extensive quantities.\nIn the presentation below the thermal material properties c, specific heat capac-\nity or just specific heat, and , density, are assumed constant.\n\n## Springer International Publishing Switzerland 2016 1\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_1\n2 1 Introduction\n\n## 1.1.1 Heat and Temperature\n\nThe heat q of a body is proportional to the mass and the temperature rise T.\n\nqc\u0002m\u0002T 1:1\n\nm\u0002V 1:2\n\n## the heat of a body may be written as\n\nqc\u0002\u0002V\u0002T 1:3\n\nIn a more general way where c and may vary with temperature and location,\nthe heat of a body may be written as an integral over temperature range and volume\nas\nZ \u0002Z \u0003\nq c \u0002 \u0002 dV dT 1:4\nT V\n\n## Heat is transferred in three modes, conduction, convection and radiation. The\n\nconcept of thermal conduction can be seen as a molecular process by which energy\nis transferred from particles of high energy/temperature to particles of low energy/\ntemperature. High temperatures are associated with higher molecular energies, and\nwhen neighbouring molecules collide a transfer of energy from the more to the less\nenergetic molecules occurs. This process takes place in fluids as well although the\nmain mode of heat transfer then is generally due to motion of matter,\ni.e. convection.\nBy the definition of temperature, heat is transferred from places with higher\ntemperatures to places with lower temperature, i.e. the temperature difference is the\n00\ndriving force of the heat transfer. In one dimension the heat flux q_ across a plane\nwall with the thickness L and a conductivity k may be written as\n\n00 k\nq_ \u0002 T 1 \u0003 T 2 1:5\nL\n\nNotice that the superscript (00 ) denotes per unit area and the accent character (_)\nper unit time. Under steady-state conditions the temperature distribution will be\nlinear as shown in Fig. 1.1.\n1.1 Basic Concepts of Temperature, Heat and Heat Flux 3\n\ntemperature distribution and\nheat flux across a plane wall\naccording to Eqs. 1.5 and\n1.6\n\n## In differential form Eq. 1.5 may be written as\n\n00 dT\nq_ x \u0003 k \u0002 1:6\ndx\n\nThis is the Fouriers law of heat conduction which implies that the heat flux is\nproportional to the heat conductivity of the material and the thermal gradient.\n\n## 1.1.3 The Three Kinds of Boundary Conditions\n\nIn addition to the differential equation valid for the interior, boundary conditions\nmust be specified when calculating temperatures in solids. A thorough understand-\ning of how to express BCs is particularly important in FSE.\nIn principle there are three kinds of BCs denoted first, second and third . The\nfirst kind is prescribed temperature, the second kind is prescribed heat flux and the\nthird kind is heat flux dependent on the difference between prescribed surrounding\ngas or fluid temperatures and the current boundary or surface temperature. The\nlatter type of BC is by far the most common in FSE. It may include heat transfer by\nconvection as well as radiation. More on boundary conditions relevant in FSE, see\nChap. 4 and for more details on radiation and convection, see Chaps. 5 and 6,\nrespectively.\n4 1 Introduction\n\n## Fig. 1.2 The first kind of\n\nBC means a temperature Ts\nis prescribed at the\nboundary\n\n## Fig. 1.3 The second kind\n\n00\nof BC means a heat flux q_ s\nis prescribed at the\nboundary\n\n## The first kind of BC (sometimes called the Dirichlet boundary condition) as\n\nshown in Fig. 1.2 means a temperature Ts is prescribed at the boundary (x x0), i.e.\n\nT x0 T s 1:7\n\n## In fire engineering it may, for example, be assumed when a surface of a light\n\ninsulating material is exposed to fire. The surface temperature may then be approx-\nimated to adjust momentarily to the boundary gas and radiation temperatures which\nfacilitates the computations.\nThe second kind of BC (sometimes called the Neumann boundary condition) as\n00\nshown in Fig. 1.3 means a heat flux q_ s is prescribed at the boundary, i.e.\n\u0004\n00 T \u0004\u0004\nq_ s \u0003k \u0002 1:8\nx \u0004xx0\n\nThus the heat flux to the boundary is equal the heat being conducted away from\nthe surface into the solid according to the Fouriers law, or in the case of lumped\nheat, it is approximated as the heat stored, see Chap. 3. A special case of the second\n1.1 Basic Concepts of Temperature, Heat and Heat Flux 5\n\nsurface, i.e. a perfectly\ninsulating surface, or a =0\nsurface along a line of\nsymmetry is a special case\nof a second kind of BC\n\n## Fig. 1.5 The third kind of\n\nBC means the heat flux to\nthe boundary depends on\nthe difference between\nprescribed surrounding gas\nor fluid temperatures and\ncurrent surface temperature\n\n## kind of BC is an adiabatic surface or a perfectly insulated surface, or a surface\n\nalong a line of symmetry where the heat flux by definition of symmetry is zero, see\nFig. 1.4.\nIn FSE the second kind of BC is rarely applicable. The concept of heat flux\nmeant as heat flux to a surface kept at ambient is, however, often used as a general\nmeasure of thermal exposure. This is in reality a third kind of BC but unfortunately\nit is difficult to apply as a boundary condition for temperature calculations, see Sect.\n9.2.1.\nThe third kind of BC (sometimes called the Robin boundary condition) means\nthe heat flux to the boundary surface depends on the difference between prescribed\nsurrounding gas or fluid temperatures and the current boundary or surface temper-\nature, see Fig. 1.5. It is sometimes also called natural boundary conditions or\nNewtons law of cooling. In the simplest form the heat transfer is proportional to\nthe difference between the surrounding gas temperature and the surface tempera-\nture. The proportionality constant h is denoted the heat transfer coefficient.\n\u0004\n00 T \u0004\u0004 \u0005 \u0006\nq_ \u0003k \u0002 \u0004 h Tg \u0003 Ts 1:9\nx xx0\n\n## Equation 1.9 is a reasonable approximation when heat transfer by convection\n\n00\nonly is considered. Therefore we write the heat flux by convection q_ con as\n6 1 Introduction\n\n00 \u0005 \u0006\nq_ con h T g \u0003 T s 1:10\n\n## However, in fire protection engineering problems temperature is usually high\n\nand radiation is the dominant mode of heat transfer. The net heat flux entering a\n00 00\nsolid surface, here denoted q_ rad , is the difference between the absorbed q_ abs and\n00\nemitted q_ emi heat flux, i.e.\n00 00 00\nq_ rad q_ abs \u0003 q_ emi 1:11\n\nThese two terms are in principle independent. The absorbed flux is a portion of\n00\nthe incident heat flux (sometimes called irradiance) q_ inc to a surface. Thus\n00 00\nq_ abs s \u0002 q_ inc 1:12\n\nwhere s is the surface absorptivity coefficient. The emitted heat depends on the\nfourth power of the absolute surface temperature T4s (in Kelvin) according to the\nStefanBoltzmann law:\n\n## OBSERVE that in all formula concerning radiation the temperature must be\n\ngiven in Kelvin [K], absolute temperature.\n\n00\nq_ emi s \u0002 \u0002 T 4s 1:13\n\nwhere s is the surface emissivity, and the physical constant 5.67 \u0002 10\u00038 W/(m2\nK4) is named the StefanBoltzmann constant.\nThe surface properties s and s have values between 0 and 1 and are according\nto Kirchhoffs identity equal, i.e.\n\ns s 1:14\n\n(The Kirchhoffs identity does not apply when the source emitting radiation to a\nsurface and the target surface have very different temperatures. Then Eqs. 1.12 and\n1.13 must be used with and depending on the wavelength spectrums of the\nconsidered in FSE as the absorptivity/emissivity of most building materials changes\nonly marginally with the temperature, the radiation is considered to be gray). Thus\nby inserting Eqs. 1.12 and 1.13 into Eq. 1.11, the net radiation heat flux to a surface\ncan be written as\n1.1 Basic Concepts of Temperature, Heat and Heat Flux 7\n\n00\n\u0007 00 \b\nq_ rad s \u0002 q_ inc \u0003 \u0002 T 4s 1:15\n\nAlternatively, the neat radiation heat flux may also be expressed in terms of as\n00 \u0005 \u0006\nq_ rad s \u0002 T 4r \u0003 T 4s 1:16\n\nwhere Tr is the incident black body radiation temperature or just the black body\nradiation temperature defined by the identity\n00\nq_ inc \u0004 \u0002 T 4r 1:17\n\nThe heat flux by radiation and convection can be superimposed to form the total\n00\nheat flux which in this book is denoted q_ tot . Then BC the third kind becomes\n00 00 00\nq_ tot q_ rad q_ con 1:18\n\nand thus\n00\n\u0007 00 \b \u0005 \u0006\nq_ tot s \u0002 q_ inc \u0003 \u0002 T 4s h T g \u0003 T s 1:19\n\nor alternatively\n00 \u0005 \u0006 \u0005 \u0006\nq_ tot s \u0002 \u0002 T 4r \u0003 T 4s h T g \u0003 T s 1:20\n\n## Equation 1.20 is a mixed boundary condition as it contains independent heat\n\ntransfer by radiation and convection. In standards on fire resistance of structures\nsuch as Eurocode 1, EN 1991-1-2, the radiation and gas fire temperatures are\nassumed equal, Tf, and then Eq. 1.20 becomes\n00\n\u0007 \b \u0005 \u0006\nq_ tot s \u0002 \u0002 T 4f \u0003 T 4s h T f \u0003 T s 1:21\n\n\u0005 \u0006\nNotice that, as the heat emitted from a surface s \u0002 \u0002 T 4s depends on the forth\npower of the surface temperature, the problems become mathematically non-linear\nwhich prohibits exact analytical solutions of the heat equation. To avoid this, the\nintroductory Chaps. 2 and 3 are limited to cases where the heat transfer coefficient\nh can be assumed constant.\nMore on boundary conditions are given in Chap. 4, and methods for calculating\nheat transfer by radiation and convection are given in Chaps. 5 and 6.\n8 1 Introduction\n\n## Heat is transferred by conduction, convection or radiation at heat flow rates denoted\n\nq_: . Then the changes of heat content dq of a body over a time interval dt becomes\n\ndq q_ \u0002 dt 1:22\n\n## By differentiating Eq. 1.3\n\ndT\nq_ c \u0002 \u0002 V \u0002 1:23\ndt\n\nFigure 1.6 shows a one-dimensional increment dx. The heat entering from the left\n00 00\nside is q_ x and the heat leaving on the right side q_ xdx . Hence\n\u0002 \u0003\n\u0002 \u0003\n00 dT dT d dT d dT\nq_ \u0003k \u0002 k\u0002 ; k\u0002 \u0002 dx k\u0002 \u0002 dx 1:24\ndx dx dx dx dx dx\n\nThe difference is the change of heat being stored per unit time, i.e.\n\n00 dT\nq_ c \u0002 \u0002 \u0002 dx 1:25\ndt\n\nNow by combining Eqs. 1.24 and 1.25, the heat conduction equation or the heat\ndiffusion equation in one dimension is obtained as\n\n## Fig. 1.6 One-dimensional\n\nincrement\n\ndx\nx\n1.1 Basic Concepts of Temperature, Heat and Heat Flux 9\n\n\u0002 \u0003\nd dT dT\nk\u0002 c\u0002\u0002 1:26\ndx dx dt\n\nd2 T 1 dT\n\u0002 1:27\ndx2 dt\n\nk\n1:28\nc\u0002\n\n## with the dimension m2/s in SI units.\n\nIn three dimensions x, y and z the general heat conduction equation is\n\u0002 \u0003 \u0002 \u0003 \u0002 \u0003\ndT dT dT dT\nk\u0002 k\u0002 k\u0002 c\u0002\u0002 1:29\nx dx y dy z dz dt\n\nThe heat capacity c and the density appear always as a product in heat diffusion\nequations, sometimes denoted specific volumetric heat capacity (J/(m3 K) in SI\nunits). Alternatively Eq. 1.26 may be written as\n\u0002 \u0003\nd dT de\nk\u0002 1:30\ndx dx dt\n\nwhere e is the heat content per unit volume named the specific volumetric enthalpy.\nBy definition it is the heat needed to rise the temperature of a unit volume from one\nlevel (e.g. 0 \u0005 C) to a higher temperature. Then\nZ T\ne c \u0002 \u0002 dT 1:31\n0\n\n## For materials with c \u0002 constant and independent of temperature the volumetric\n\nspecific enthalpy becomes\n\nec\u0002\u0002T 1:32\n\n## The concept of specific volumetric enthalpy is advantageous to use when\n\nconsidering physical and chemical transformations. Then numerical temperature\ncalculations may be facilitated as will be presented in more detail in Chap. 7.\nIn the simplest cases of transient heat transfer problems, the temperature in a\nbody is assumed uniform. Then as only a single uniform temperature is calculated\nwith no variation depending on position, this type of problems is zero-dimensional.\nMore on lumped heat calculations can be found in Sects. 3.1, 7.1 and 13.3.\n10 1 Introduction\n\n## 1.2 Electric Circuit Analogy in One Dimension\n\nThere are analogies between parameters of heat transfer systems and electric\ncircuits. These will be used throughout this book to illustrate, develop and explain\nvarious temperature and heat transfer calculation formulas. An overview of\ncorresponding parameters, nomenclature and icons of resistance and capacitance\nis given in Table 1.1. Notice that the resistance R for thermal problems refers to a\nunit area while the analogue electric resistance includes the area in Re. In summary,\ntemperature is analogue to electric potential or voltage, heat flow to electric current,\nthermal resistance to electric resistance and heat capacity to electric capacity.\nFrom the discipline of electric circuits the rules of combining resistances can be\napplied. Thus two resistances in series between A and C as shown in Fig. 1.7 can be\nsummarized as\n\nTable 1.1 Analogies between thermal parameters in one dimension and electric parameters and\nunits\nTemperature and heat Electric circuit analogy\nParameter and nomenclature SI units Parameter and nomenclature SI units\nHeat, q [J or Ws] Electric charge, Q [J]\nTemperature, T [K or \u0005 C] Electric potential, U [V]\nHeat flow, q_: [W] Electric current, I [A]\nq_ A \u0002 T\nR I U\nRe (Ohms law)\nT1 T2 U1 U2\nR Re\n\n## Thermal resistance Resistor\n\nT T 1 \u0003 T 2 U U 1 \u0003 U 2\n00\nHeat flux, q_ [W/m2] Electric current per unit area, I/A [A/m2]\n00\nT\nq_ R\n1D thermal resistance, R [m2 K/W] Electric resistance, Re []\nSurface resistance Rh 1h [m2 K/W]\nSolid resistance Rk dk [m2 K/W]\n1D heat capacitance, C [J/(m2 K)] Electric capacitance, Ce [J/V]\nCc\u0002\u0002d [J/(m2 K)] Q Ce \u0002 U [J]\n00\nq C\u0002T [J/m2]\nU\nT Ce\nC\nEarth U=0\nT=0 Electric capacitor\nLumped-heat-capacity\n1.3 Material Properties at Elevated Temperature 11\n\nFig. 1.7 Rules for combining resistances in series according to Eq. 1.33. (a) Resistances in series.\n(b) Resultant resistance\n\nFig. 1.8 Rules for combining parallel resistances according to Eq. 1.34. (a) Parallel resistances.\n(b) Resultant resistance\n\n1 1\nRAC RAB RBC 1:33\nK AB K BC\n\nK denotes the reciprocal of the thermal resistance which could be a heat transfer\ncoefficient, h, or conductivity over a thickness, k/d.\nIn the case of parallel resistances as shown in Fig. 1.8, the resultant resistance\nbetween A and B becomes\n\n1 1\nRAB 1:34\n1\nR1 R12 K 1 K 2\n\nThus as an example according to Fig. 1.8, the heat flux between A and B may be\nwritten as\n\n00 1\nq_ K 1 K 2 \u0002 T A \u0003 T B \u0002 T A \u0003 T B 1:35\n1\nR1 R12\n\nwhere TA and TB are the temperatures at point A and B, respectively. K1 and K2 may,\nfor example, be heat transfer coefficients due to radiation and convection.\n\n## The flow of heat by conduction in a body is proportional to the thermal conductivity\n\nof the material and the temperature gradient according to Fouriers law as given by\nEq. 1.6. Under steady-state conditions the conductivity denoted k is the sole\nmaterial property while under transient conditions the density and the specific\n12 1 Introduction\n\n## Table 1.2 Thermal properties of some materials at room temperature\n\nThermal\nSpecific heat Thermal inertia k\u0002\u0002c\nDensity capacity Conductivity diffusivity [(W2 s)/\nMaterial [kg/m3] c [J/(kg K)] k [W/(m K)] k/(\u0002c) [m2/s] (m4 K2)]\nAir 1.23 1010 0.024 19.3 \u0002 10\u00036 0.030 \u0002 103\nPolyurethane 20 1400 0.03 1.07 \u0002 10\u00036 0.840 \u0002 103\nfoam\nFibre insulat- 100 2000 0.04 2.00 \u0002 10\u00036 7.92 \u0002 103\ning board\nWood, pine 500 2800 0.14 0.100 \u0002 10\u00036 0.196 \u0002 106\nWood, oak 700 2800 0.17 0.87 \u0002 10\u00036 0.336 \u0002 106\nWater 1000 4181 0.604 0.144 \u0002 10\u00036 2.53 \u0002 106\nGypsum 1400 840 0.5 0.425 \u0002 10\u00036 0.593 \u0002 106\nplaster\nConcrete 2300 900 1.7 0.82 \u0002 10\u00036 3.53 \u0002 106\nAluminium 2700 900 200 82.3 \u0002 10\u00036 486 \u0002 106\nSteel (mild) 7850 460 46 12.7 \u0002 10\u00036 166 \u0002 106\nCopper 8930 390 390 112 \u0002 10\u00036 1362 \u0002 106\nValues of this table are only indicative and not necessarily recommended for use in real FSE\napplications\n\nheat capacity c are needed in addition. In general the thermal conductivity of a solid\nis bigger than that of a liquid, which is larger than that of a gas. Materials with a low\ndensity have in general low heat conductivity while materials with high densities\nand in particular metals have high thermal conductivities. Insulating materials have\nlow densities and are by definition pure conductors of heat. Table 1.2 shows in the\norder of density the thermal properties of a number of materials. With the exception\nof metals, air and water we can derive from this table the very approximate relations\nbetween the density and the conductivity k as\n\n## The specific heat c of a material or substance is the amount of heat needed to\n\nchange the temperature of a unit mass of the substance by 1\u0005 . It is an intensive\nparameter with the unit of energy per unit mass and degree, in SI units [J/(kg K)] or\n[Ws/(kg K)]. (This is unlike the extensive variable heat capacity (denoted C), which\ndepends on the quantity of material and is expressed in [J/K]). As a general rule the\nspecific heat decreases with density, i.e. it is high for low density materials and low for\nhigh density materials. Cementitious materials have a specific heat capacity c slightly\nunder 1000 J/(kg K) while the corresponding values for wood are considerably higher.\nFor metals c is significantly lower and varies inversely with the density. Notice in\nTable 1.2 that c of water is relatively high, more than four times higher than c for\nconcrete. Therefore the moisture content of a material has great influence on the\ntemperature development. The major influence of the water is, however, when it\nvaporizes at temperatures exceeding 100 \u0005 C.\n1.3 Material Properties at Elevated Temperature 13\n\nThe product k\u0002\u0002c, denoted thermal inertia, see Sect. 8.2.2, has a great impact on\nignition and flame spread propensities of materials. When the density increases so\ndoes normally the conductivity k as well, and consequently the thermal inertia is\ngreatly dependent on the density. It varies over a wide range and therefore the\ndensity is a very significant indicator of the fire properties materials.\nNotice that the thermal inertia, k\u0002\u0002c, of wood is in the order of 300 times as high\nas the corresponding value of an efficient insulating material such as polyurethane\nfoam. This difference will give these materials a considerable difference in their\nignition properties as will be discussed in Chap. 8.\nThe data given in Table 1.2 refer to room temperature. At elevated temperatures\nwhich are relevant in fires and fire-exposed structures the material properties may\nvary significantly. In addition the parameter values listed cannot be assumed to\nfully reflect the properties of all materials within any generic class. Specific data for\nparticular products may be provided by the manufacturers.\n\n## 1.3.1 Structural Materials\n\nThe temperature of structures exposed to fully developed fires with gas tempera-\ntures reaching 8001200 \u0005 C will gradually increase and eventually the structures\nmay lose their load-bearing capacity as well as their ability to keep fires within\nconfined spaces. In building codes fire resistance requirements are usually\nexpressed in terms of the time a structural element can resist a nominal or standard\nfire as defined, e.g. in the international standard ISO 834 or the corresponding\nEuropean standard EN 1363-1. In the USA and Canada the corresponding standard\ncurve for determining fire resistance of building components is given in ASTM\nE-119. The standard timetemperature curves as defined by the ISO/EN and ASTM\nstandards are shown in Fig. 1.9. More on standard fires can be found in Chap. 12.\nBelow some general remarks are given for the most common structural mate-\nrials. Methods for calculating temperature steel, concrete and timber structures\nexposed to fire are outlined in Chaps. 13, 14 and 15, respectively.\nSteel starts to lose both strength and stiffness at about 400 \u0005 C and above 600 \u0005 C\nmore than half of its original strength is lost, see, e.g. Eurocode 3 or the SFPE\nHandbook on Fire Protection Engineering [4, 5]. Therefore structural steel elements\nmust in most cases be fire protected by sprayed on compounds, boards, mineral\nwool or intumescent paint to keep sufficient load-bearing capacity over time when\nexposed to fire. An example of a steel structure failure due to fire was the collapse of\ntwo World Trade Center towers on September 11, 2001. The towers were hit by big\npassenger airplanes. A tremendous impact was inflicted on them, but they did not\ncollapse immediately. The jet fuel started, however, intense fires and when the steel\nof some decisive members had reached critical temperatures a progressive collapse\nwas initiated. For calculation methods on steel structures see Chap. 13.\n14 1 Introduction\n\nFig. 1.9 The standard timetemperature curves according to EN 1363-1 or ISO 834 and ASTM\nE-119\n\n## Concrete also looses strength and stiffness at high temperature, see,\n\ne.g. Eurocode 2 or . Concrete has, however, a relatively low thermal\nconductivity and a high density and high specific heat capacity as well, i.e. a low\nthermal diffusivity. Although the temperature therefore rises slowly in concrete\nstructures, it is important to assure that the steel reinforcement bars are not too near\nfire-exposed surfaces to avoid that their temperature reaches critical levels. See\ncalculation methods in Chap. 14. An often more severe problem is the tendency of\nconcrete to spall explosively when exposed to high temperature. In particularly high\nstrength concrete qualities are prone to spall which is of great concern, for example,\nwhen designing linings of road and railway tunnels where fire temperatures may be\nextremely high and where a collapse may have devastating consequences in terms\nof life safety and protection of economic values.\nWood loses both strength and stiffness at elevated temperature. In addition it\nburns and chars gradually at a rate of about 0.5 mm/min when exposed to fire. The\nchar layer then developed, however, protects the wood behind from being directly\nheated by the fire and thereby from quickly losing its load-bearing capacity. Timber\nstructures therefore resist fire rather well and are in most cases left unprotected, see,\ne.g. Eurocode 5 . In many cases structural timber members such as wall studs are\nprotected from direct exposure by fire boards and can then resist fire for very long\nperiods of time. For calculation methods see Chap. 15.\n1.3 Material Properties at Elevated Temperature 15\n\n## 1.3.2 Polymers and Composite Materials\n\nThere are two main types of plastics materials, thermoplastics and thermosettings.\nThey decompose differently when exposed to heat. Thermoplastics can soften with\nreverse changes of the material, while thermosetting materials are infusible and\ncannot undergo any simple phase changes. They do not have a fluid state.\nMany thermoplastics and thermosetting materials form chars when decomposed\nby heat. This char is in general a good insulator and can protect the underlying\nvirgin material from heat and slow down the decomposition process.\nPolymers or plastics possess different hazards in fires depending on their phys-\nical constitution and chemical composition. In general, foamed plastics with low\ndensity and thin plastic objects ignite more easily and burn more vigorously than\nmore dense and thick plastics. The fire properties of an object do not only depend on\nits chemical composition but also on the shapes and configurations. Thus a thin\nlayer of a material ignites more easily when underlaid by a low density insulating\nmaterial than by a more dense material. Below some characteristic are given of\nsome commercially important polymers.\nThe thermal stability of polyolefins such as polyethylene and polypropylene\ndepends on branching of the molecule chains, with linear polymers most stable\nand polymers with branching less stable. Polyvinyl chloride (PVC) has in general\ngood fire properties as the chloride works as a flame retardant agent. However, the\nhydrochloride HCl, which is generated while burning, is irritating and toxic and can\nimpede the evacuation from a fire. In addition, it forms hydrochloride acid when in\ncontact with water and can therefore cause severe corrosion problems even long\nafter a fire incident. Polyurethanes (PU) contain nitrogen and forms very toxic\nproducts such as hydrogen cyanide and isocyanides when burning. PVC and PU do\nalso generate very dense smoke which can hamper escape possibilities.\nComposite materials consisting of a polymer and reinforcing fibres (typically,\nglass, carbon or aramide fibres), also called Fibre reinforced plastics (FRP), have\nbecome increasingly used in many areas of construction, such as airplanes, heli-\ncopters and high-speed crafts, due to the high strength/weight ratio. These materials\nare also chemically very resistant and do not corrode or rust. They are, however,\ncombustible and as they are often meant to replace non-combustible materials such\nas steel or other metals they could introduce new fire hazards.\n\n## 1.3.3 Measurements of Material Properties\n\nMaterial properties may be obtained from small scale laboratory test or derived from\nlarge scale fire test experiences. Small scale tests are in general the most accurate and\ncheap tests, but they are usually made for inert materials in room temperature.\nTherefore such data are relatively easy to find. In FSE, however, material data are\nneeded at elevated temperature when these change, and in addition materials may\nundergo physical (e.g. vaporization of water) as well as chemical transformations\n16 1 Introduction\n\n(e.g. phase changes and pyrolysis). Then the small scale methods are generally\nunsuitable to consider these kinds of non-linear effects. In practice therefore thermal\nproperties are often determined by curve-fitting, i.e. measured temperatures are\ncompared with calculated, and then input parameters are altered until measured and\ncalculated data match as well as possible. In this way large scale non-linear effects\nmay be considered. However, this kind of approach has the disadvantage that the\nresults are valid only for the type of exposure being used to determine the data.\nThere are a number of techniques to measure thermal properties in small scale,\neach of them suitable for a limited range of materials, depending on thermal\nproperties and temperature level, see, e.g. . However, only a few of the measur-\ning techniques can be used at high temperature levels relevant for fire conditions.\nThey can be divided into steady-state and transient techniques.\nThe steady-state techniques perform the measurements when the material is in\ncomplete equilibrium. Disadvantages of these techniques are that it generally takes a\nlong time to reach the required equilibrium and that at low temperature the measure-\nments are influenced by moisture migration. For moist materials such as concrete, it is\ntherefore often preferable to determine the apparent conductivity or thermal diffu-\nsivity with transient techniques. These techniques perform the measurements during\na process of small temperature changes and can be made relatively quickly.\nThe guarded hot-plate is the most common steady-state method for building\nmaterials with a relatively low thermal conductivity. It is quite reliable at moderate\ntemperatures up to about 400 \u0005 C.\nAs transient thermal processes dominate in FSE, the thermal diffusivity, a\nmeasure of the speed at which temperature is propagating into a material, is the\nmost interesting parameter. It is naturally best measured with transient methods.\nOne of the most interesting techniques is the transient plane source method (TPS).\nIn this method a membrane, the TPS sensor, is located between two specimens\nhalves and acts as heater as well as a temperature detector, see Fig. 1.10. By using\nthis technique, thermal diffusivity, heat conductivity and volumetric specific heat\ncan be obtained simultaneously for a variety of materials such as metals, concrete,\nmineral wool and even liquids and films .\n\nFig. 1.10 The TPS sensor placed between two pieces of a concrete specimen to measure thermal\nproperties\nChapter 2\n\nIn one dimension in the x-direction the rate of heat transfer or heat flux is expressed\naccording to Fouriers law as outlined in Sect. 1.1.\n\n00 dT\nq_ x \u0002k \u0003 2:1\ndx\n\n## where k is the thermal conductivity. For simplicity the mathematical presentation of\n\nthe heat transfer phenomena is here in general made for one-dimensional cases\nonly. Corresponding presentations in two and three dimensions can be found in\nseveral textbooks such as [1, 2].\nUnder steady-state conditions the heat flux is independent of x, i.e. the derivative\n00\nof q_ x is zero and we get\n\u0001 \u0003\nd dT\nk\u0003 0 2:2\ndx dx\n\n## The corresponding equation for cylinders with temperature gradients in the\n\n\u0001 \u0003\n1d dT\nk\u0003r 0 2:3\nr dr dr\n\nwhere r is the radius. Solutions for steady-state cases are found in Sect. 2.2.\n\n## Springer International Publishing Switzerland 2016 17\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_2\n\n## 2.1 Plane Walls\n\nConsider a plane wall having surface temperatures T1 and T2. Figure 2.1 shows the\ntemperature distribution under steady-state conditions which means the heat flux is\nconstant across the plate. Figure 2.1a shows the temperature distribution when the\nheat conductivity is constant, i.e. the second derivative of the temperature is zero\naccording to Eq. 2.2 and thus the temperature distribution becomes linear. Figure 2.1b\nshows the temperature distributions in structure with two layers of materials with\ndifferent conductivities. The material to the left has the lower conductivity.\nFigure 2.1c indicates the temperature distribution when the conductivity is increasing\nwith temperature. The temperature gradient is higher where the temperatures are\nlower and thereby the conductivity. This is particular the case for insulating materials\nwhere the conductivity increases considerably at elevated temperatures.\n00\nThe rate of heat conducted per unit area q_ through a wall, see Fig. 2.2, is\nproportional to the thermal conductivity of the wall material times the temperature\ndifference T between the wall surfaces divided by the wall thickness L, and\naccording to Fouriers law (c.f. Eq. 2.1)\n\n00 T T 1 \u0002 T 2\nq_ k \u0003 k\u0003 2:4\nL L\n\nIn an electric circuit analogy, this case can be illustrated according to Fig. 2.2. The\nheat flow through the wall over an area A may then be written as\n00\nq_ T 1 \u0002 T 2 =Rk 2:5\n\na b c\n\nT1 T1\nT1\n\nT1\nT1\nT1\n\nFig. 2.1 Steady-state temperature distribution in a plane wall. (a) Constant conductivity, (b) two\nmaterials with a low and high conductivity and (c) conductivity increasing with temperature\n2.1 Plane Walls 19\n\n## Fig. 2.2 One-dimensional\n\nconduction. Linear\ntemperature distribution\nacross a wall and an electric\nanalogy of one-dimensional\nheat flux. The thermal\nresistance R L/k\n\n## where the thermal resistance of the solid then can be identified as\n\nL\nRk 2:6\nk\n\nThe electric analogy may also be used for more complex problems involving\nboth series and parallel thermal resistance. A typical problem is a wall consisting of\nseveral layers, see Fig. 2.3.\nThe total thermal resistance Rtot between the inside and outside surfaces may\nthen be written as:\n\nL1 L2 L3\nRtot R1 R2 R3 2:7\nk1 k2 k3\n00\nand the heat flux q_ through the assembly from the inside to the outside may be\nwritten as:\n\n00 T T 0 \u0002 T i\nq_ 2:8\nRtot Rtot\n\nThe temperature T 1\u00022 at the interface between material 1 and 2 may be written as\n\nR1 T i \u0002 T o R2 R3 \u0003 T i R1 \u0003 T o\nT 1\u00022 T i \u0002 2:9\nRtot Rtot\n\nFig. 2.3 Electric circuit analogy of one-dimensional heat transfer across a wall consisting of three\nlayers\n\nFig. 2.4 Electric circuit analogy of one-dimensional heat transfer across a wall consisting of\nseveral layers\n\n## In a general way the total thermal resistance of an assembly thermally modelled\n\nas shown in Fig. 2.4 may be obtained as the sum of the components\nXn\nRtot j1\nRj 2:10\n\n00\nand the heat flux q_ through can be calculated as\n\n00 T0 \u0002 Tn\nq_ 2:11\nRtot\n\n## and the temperature at an interface i as shown in Fig. 2.4 may be calculated as\n\nXi Xi Xn\nR\nj1 j\nTn \u0003 R T0 \u0003\nj1 j\nR\nji1 j\nTi T0 T n \u0002 T 0 2:12\nRtot Rtot\n\nExample 2.1 A wall consists of 20 mm wood panel, 100 mm fibre insulation and\n12 mm gypsum board with conductivities equal to 0.14, 0.04 and 0.5 W/(m2 K),\nrespectively. The wood outer surface has a constant temperature of 75 \u0004 C and the\ninner gypsum board surface a temperature of 15 \u0004 C. Calculate the temperatures at\nthe insulation interface surfaces (T1 and T2).\n2.1 Plane Walls 21\n\n## Solution Rtot 0.020/0.14 + 0.1/0.04 + 0.012/0.5 0.143 + 2.5 + 0.024 2.67W/K.\n\n\u0004\nThen T 1 75 0:1432:67 \u0003 15 \u0002 75 71:8 C and T 2 75\n0:1432:5\n2:67 \u0003 15 \u0002 75\n\u0004\n15:6 C.\nThe presentation so far includes heat transfer in solids only with boundary\nconditions of the first kind, i.e. prescribed surface temperatures. In most cases in\nfire protection engineering, however, the boundary condition between a surround-\ning fluid/environment and a solid surface is specified as a boundary condition of the\nthird kind. In the simplest form the boundary condition is then described by the\nNewtons law of cooling. It may be seen as a heat transfer condition for convection\nand it states that the heat transfer to a surface is directly proportional to the difference\nbetween the surrounding gas temperature Tg and the surface temperature Ts:\n00 \u0004 \u0005\nq_ h \u0003 T g \u0002 T s 2:13\n\nwhere the constant of proportionality factor h is the heat transfer coefficient. The\nsurface thermal resistance Rh between the gas phase and the solid phase can then be\nwritten as\n\n1\nRh 2:14\nh\n\nThus for the case illustrated in Fig. 2.5, the total resistance between the gas phase on\nthe left side and the surface on the right side of the wall may be written as\n\u0001 \u0003\n1 L\nRtot Rh Rk 2:15\nh k\n\n## Fig. 2.5 An electric circuit\n\nanalogy of one-dimensional\nheat transfer to a surface\nand through a wall with\nsurface and solid thermal\nresistances\n\nand the surface temperature T1 may be written as a function of the gas temperature\nTg and the temperature T2 as\n\nRh \u0003 T 2 Rk \u0003 T g\nT1 2:16\nR h Rk\n\n## Example 2.2 Calculate the surface temperature T1 of a 12 mm wooden board if the\n\ngas temperature on the exposed side Tg 100 \u0004 C and the temperature on the\nnon-exposed side is T2 20 \u0004 C. Assume the conductivity of the wood k 0.2 W/\n(m K) and heat transfer coefficient h 5 W/(m2 K).\n\u0004 \u0005 2\nSolution Equation 2.15 yields R Rh Rk 15 0:012 0:2 m K=W 0:2 0:06\n\u0004\nm2 K=W and Eq. 2.16 yields T 1 0:2\u0003200:06\u0003100\n0:26 C 38 \u0004 C.\n\n2.2 Cylinders\n\nmay therefore be treated as one dimensional. The solid thermal resistance between\nthe inner radius ri and an arbitrary radius r in a cylinder (see Fig. 2.6) assuming\nconstant heat conductivity may then be written as\n\u0006 \u0007\nln r=ri\nRk 2:17\n2k\n\nFig. 2.6 Thermal resistances between the media with a temperature Ti inside a cylindrical pipe\nand the outside gas with a temperature To\n2.2 Cylinders 23\n\nwhere k is the thermal conductivity. The surface thermal resistance may be written as\n\n1\nRh 2:18\n2rh\n\nHence the thermal resistance between the inner and outer gases or liquids of a pipe\nis obtained by summarizing the surface and solid resistances as indicated in Fig. 2.6,\ni.e. the total thermal resistance over a unit length is\n0 \u0006 \u0007 1\n1 @1 ln ro =ri 1 A\nRtot Rhi Rk Rho \u0003 2:19\n2 rhi k r o ho\n\nA uniform heat flux over a unit length of a pipe may then be calculated as\n\n0 Ti \u0002 To 2 T i \u0002 T o\nq_ l 2:20\nRtot lnro =\n1\nri \u0003hi k ri ro1\u0003ho\n\n## The temperatures Tis at the inner surface can be obtained as\n\nRhi T o Rk Rho T i\nT is 2:21\nRtot\n\n## and Tos at the outer surface as\n\nRhi Rk T o Rho T i\nT os 2:22\nRtot\n\nExample 2.3 Consider an insulated steel pipe with an outer coating as shown in\nFig. 2.7 exposed to fire with a constant temperature of 800 \u0004 C. The temperature of\n\n## Fig. 2.7 Insulated steel\n\npipe with an outer coating\n\nthe inside medium/fluid is 100 \u0004 C. The inner and outer radii of the steel pipe are\n30 and 28 mm, respectively. The insulation is 50 mm thick and has a conductivity of\n0.5 W/(m K). The inner heat transfer coefficient is 100 W/(m2 K) and the outer\n50 W/(m2 K). Calculate the temperature of the steel pipe which is assumed to be\nSolution Calculating for a unit length. Equation 2.19 yields Rtot 0.057 + 0.312\n+ 0.040 0.409 (m K)/W. Thus the inner (steel) temperature\n0:057\u00038000:3120:04\u0003100 \u0004\nTi 0:409 198 C.\nChapter 3\n\n## When a body is exposed to unsteady or transient thermal conditions, its temperature\n\nchanges gradually, and if the exposure conditions remain constant it will eventually\ncome to a new steady state or equilibrium. The rate of this process depends on the\nmass and thermal properties of the exposed body, and on the heat transfer condi-\ntions. As a general rule the lighter a body is (i.e. the less mass) and the larger its\nsurface is, the quicker it adjusts to a new temperature level, and vice versa. The\ntemperature development is governed by the heat conduction equation (Eq. 1.29)\nwith the assigned boundary conditions. It can be solved analytically in some cases,\nsee textbooks such as [1, 2], but usually numerical methods are needed. This is\nparticular the case in fire protection engineering problems where temperature\ngenerally varies over a wide range, often several hundred degrees.\nThere are, however, some cases where analytical methods can be used. Two cases\nare of interest for both practical uses and basic understandings of the influence of\nmaterial properties on their fire behaviour. On one hand, it is cases where bodies can\nbe assumed to have uniform temperature such as in thin solids or in metals with a\nhigh conductivity. Then the approximation of lumped-heat-capacity can be applied.\nOn the other hand, it is the case when a body can be assumed semi-infinitely thick for\nthe time span considered. Then in particular the surface temperature can be estimated\nby analytical methods if the material properties are assumed constant. These two\nelementary cases will be considered in detail in the following two Sects. 3.1 and 3.2.\n\n3.1 Lumped-Heat-Capacity\n\n## It is often assumed when calculating temperature in steel sections, protected as well\n\nas unprotected, that the temperature is uniform in the exposed body, see Sects. 13.3\nand 13.4, respectively. It may also be applied when estimating temperature and time\nto ignition of thin materials such as curtain fabrics. A special case is the analysis of\nthe temperature development of thermocouples and the definition of time constants\n\n## Springer International Publishing Switzerland 2016 25\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_3\n\n## Fig. 3.1 The dynamic heat\n\nbalance of a body over a\ntime period is expressed as\nthe heat received is equal to\nthe heat stored according to\nEq. 3.1\n\n## Received heat = Stored heat\n\nof these types of measuring devices, see Sect. 9.1. Numerical methods for calcu-\nlating temperature when assuming lumped-heat-capacity are described in Sect. 7.1.\nIn this section a general presentation will be given assuming constant heat transfer\ncoefficients, material properties and exposure levels. As only one temperature inde-\npendent of position is calculated, this type of problems are zero-dimensional.\nThe received heat over a time interval dt is equal to the heat stored. The latter is\nproportional to the temperature rise of the body dT, see Fig. 3.1. Thus\n\nq_ \u0001 dt V \u0001 \u0001 c \u0001 dT 3:1\n\nHence the temperature rise rate dT/dt (the time derivative of the body temperature)\n00\nvs. incident heat flow q_: or the incident heat flux q_ can be obtained as\n\ndT 1 A 00\nq_ q_ 3:2\ndt V \u0001 \u0001 c V\u0001\u0001c\n\nwhere A is exposed area, V volume, density and c specific heat capacity. For thin\nplates exposed from one side V/A may be replaced by its thickness d\n\nV\nd 3:3\nA\n00\nThe heat flux q_ to the body can be obtained in various ways depending on the\nboundary condition. It may be of the second or third kind, see Sect. 1.1.3. The first\nkind is trivial as a uniform temperature is assumed.\n\n## 3.1.1 Prescribed Heat Flux: BC of the Second Kind\n\n00\nGiven a prescribed heat flux q_ (second kind of BC, see Sect. 1.1.3), the temperature\nrise T \u0003 T i as function of time may be obtained by integrating over time as\nt\nA 00\nT \u0003 Ti q_ \u0001 dt 3:4\nV \u0001 c \u0001 0\n3.1 Lumped-Heat-Capacity 27\n\n00\nwhere Ti is the initial temperature. If q_ remains constant over time\n00\nA \u0001 q_ \u0001 t\nT Ti 3:5\nV \u0001 c \u0001\n\nPrescribed heat flux can rarely be assumed in fire protection engineering as the heat\nflux from the gas phase to a solid surface depends on the surface temperature which\nchanges over time. Instead it is the third kind of BC that generally applies.\n\n## 3.1.2 Prescribed Gas Temperature: BC of the Third\n\nKindAnd the Concept of Time Constant\n\nMore realistic and commonly assumed in FSE problems is that the heat transfer to a\nsurface is proportional to the difference between the surrounding gas or fire\ntemperature Tf and the body temperature as indicated in Fig. 3.2 (third kind of\nBC, see Sect. 1.1.3). The body having uniform temperature (lumped-heat-capacity)\nis here assumed to be of steel and its temperature is denoted Tst.\nIn case of uninsulated or unprotected bodies such as bare steel sections the\nsurface thermal resistance is the only thermal resistance between the fire gases and\nthe steel. Thus the heat flux can be written as\n\u0002 \u0003\n00 \u0002 \u0003 T f \u0003 T st\n_q h \u0001 T f \u0003 T st 3:6\nRh\n\n## Fig. 3.2 Electric circuit\n\nanalogy of an uninsulated\nsteel section assumed to\nhave uniform temperature\n(lumped-heat-capacity)\n\nwhere h is the heat transfer coefficient and Rh the corresponding surface resistance\nwhich can be identified as\n\n1\nRh 3:7\nh\n\nFor an insulated or protected steel section, the heat resistance is the sum of the\nsurface thermal resistance Rh and the solid resistance of the insulation Rin. (Notice\nthat insulated and protected are used synonymously in this book). The total thermal\nresistance between the fire gases and the steel section is then\n\n## Rhin Rh Rin 3:8\n\nwhere\n\ndin\nRin 3:9\nkin\n\nand where din and kin are the insulation thickness and conductivity, respectively.\nElectric circuit analogies are shown in Fig. 3.3. In Fig. 3.3a the surface resistance is\nincluded while in Fig. 3.3b it is not. That assumption is made in many cases as the\nsurface resistance is much smaller than the solid resistance, i.e. Rh \u0004 Rin , and may\ntherefore be ignored (as suggested in, e.g. Eurocode 3 ).\n\nFig. 3.3 Electric circuit analogy of an insulated steel section treated as a lumped-heat-capacity.\n(a) Including heat transfer resistance. (b) Neglecting heat transfer resistance\n3.1 Lumped-Heat-Capacity 29\n\n## For a unit area (A 1) Eq. 3.2 may be written as\n\n00 dT st dT st\nq_ dst \u0001 st \u0001 cst \u0001 Cst \u0001 3:10\ndt dt\n\nwhere dst is the thickness and C is the heat capacitance per unit exposed area (see\nTable 1.1), i.e.\n\n\u0002 \u0003\ndT T f \u0003 T st\n3:12\ndt Cst \u0001 Rhin\n\ndT 1 \u0002 \u0003\nT f \u0003 T st 3:13\ndt\n\n## Cst \u0001 Rhin 3:14\n\nThen if the surrounding temperature Tf is constant and the time constant including\nthe material and heat transfer parameters remains constant, Eq. 3.12 has an analyt-\nical solution\n\nT st \u0003 T i\n1 \u0003 e\u0003\nt\n3:15\nTf \u0003 Ti\n\nwhere Ti is the initial temperature at time t 0. The relation is shown in Fig. 3.4.\nNotice that Eq. 3.15 may only be applied when constant material properties and\nsurface resistances are assumed. That is, however, not so common in FSE and\ntherefore must in most cases numerical solutions be used. More on numerical\nsolutions will be shown in Sect. 7.1 and more on steel sections in Chap. 13.\n\n## 3.1.2.1 Gas Temperature Varying with Time\n\nEquation 3.15 may be applied only to a sudden change of the exposure temperature\nto a new constant value. When the exposure temperature varies with time, super-\nposition techniques may be applied as outlined in Sect. 7.2. As an example the\ntemperature of a steel section when assuming a constant time constant can be\nobtained by superposition as\n\n1.00\n\n0.90\n\n0.80\nRelative temperature [-]\n\n0.70\n\n0.60\n\n0.50\n\n0.40\n= 1\n\n0.30\n\n0.20\n\n0.10\n\n0.00\n0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5\nt/t [-]\n\u0002 \u0003\nFig. 3.4 The relative temperature rise T st \u0003 T i = T f \u0003 T i of a body with uniform temperature\nvs. dimensionless time t/ according to Eq. 3.15\n\nt\n1\nT st e\u0003t= T f e= d e\u0003t= \u0001 T i 3:16\n0\n\nThe integral of Eq. 3.16 can be solved analytically in some cases depending on the\nanalytical expression of the gas temperature Tf as a function of time. For instance,\nwhen the fire temperature rises linearly with time as\n\nT f a \u0001 t 3:17\n\n## the steel temperature rise becomes\n\nh \u0004 \u0005i\nT st a \u0001 t 1 \u0003 1 \u0003 e\u0003\nt\n3:18\nt\n\nor in dimensionless format\n\nT st t h \u0004 \u0005i\n1 \u0003 1 \u0003 e\u0003\nt\n3:19\na\u0001 t\n\n## Figure 3.5 shows\n\n\u0002 that the\n\u0003 steel temperature rise asymptotically approaches a\ntemperature T f \u0003 a \u0001 .\nAnother example where the integral of Eq. 3.16 can be solved analytically is\nshown in Sect. 13.3.2 dealing with fire insulated steel sections exposed to paramet-\nric fires according to Eurocode EN 1991-1-2. The fire temperature curve is then\n3.1 Lumped-Heat-Capacity 31\n\n4.0\n\n## Dimensionless temperature rise DT/(a.t) 3.5\n\n3.0\n\n2.5\n\n2.0\n\n1.5\nTst\n\n1.0 Tf\n\nTtrend\n0.5\n\n0.0\n0.0 1.0 2.0 3.0 4.0\nDimensionless time t/t\n\nFig. 3.5 Dimensionless temperature rise of a body exposed to linearly rising temperature\nvs. dimensionless time according to Eq. 3.19\n\n## expressed as a sum of exponential terms which allows the steel temperature to be\n\ncalculated analytically according to Eq. 3.16. See also the Sect. 9.1 on the response\nof thermocouples.\n\n## The assumption of lumped-heat-capacity or uniform body temperature is an\n\napproximation which may be applied when the internal thermal resistance by\nconduction is low in comparison to the heat transfer resistance in the case of a\nnon-insulated body, i.e. the Biot number defined as\n\nL=k hL\nBi 3:20\n1=h k\n\nis less than 0.1, see, e.g. . Here L is a characteristic length of the body studied as\nshown in Table 3.1, and k is the conductivity. Figure 3.6 shows the effect of various\nBiot numbers of plane body under steady-state conditions.\n\n## Table 3.1 Examples of Configuration Characteristic length, L\n\ncharacteristic lengths L\nPlate exposed on one side Thickness\nPlate exposed on two sides Thickness/2\nLong cylinder Diameter/4\nSphere Diameter/6\n\n## Fig. 3.6 The state\n\ntemperature distribution for\na plane wall with a\nconductivity k and a heat\ntransfer coefficient h\n\n## In fire protection engineering lumped-heat-capacity is often assumed for steel\n\nsections. This is in particular appropriate when considering temperature across the\nthickness of a web or flange. The temperatures on the two sides of a metal sheet are\nthen by and large equal. Temperatures along the plane of a web or flange may,\nhowever, vary considerably depending on the boundary conditions, see Chap. 13.\nThe criterion given by Eq. 3.20 is based on steady-state conditions but applies for\nsteel of thicknesses in the order of 10 mm except for the first few minutes which are\nusually not of interest in FSE.\nExample 3.1 A 2-mm-thick steel plate with an initial temperature T i 20 \u0005 C is\nsuddenly exposed on both sides to a gas temperature of 500 \u0005 C. Assume a constant\nheat transfer coefficient h 20 W/(m2 K) and the steel properties c 460 Ws/\n(kg K) and 7850 kg/m3.\n(a) Calculate the thermal time constant of the steel plate.\n(b) Calculate the temperature of the steel plate after 5 min.\n3.2 Semi-infinite Solids 33\n\nSolution (a)\ndc 0:5 \u0001 0:002 \u0001 460 \u0001 7850\n181 s\nh 20\n\n(b) At time t 300 s the dimensionless time t= 300=181 1:66 and from\nEq. 3.15 or \u0002 Fig. 3.4 \u0003the steel temperature is calculated as T st 20\n500 \u0003 20 \u0001 1 \u0003 e\u00031:66 20 480 \u0001 0:81 409 \u0005 C.\n\n## Example 3.2 A 5-mm-thick steel bulkhead is suddenly exposed to a fire with a\n\nconstant temperature of T f 1000 \u0005 C. It is insulated on the fire-exposed side and\nuninsulated on the non-fire-exposed side. The insulation thickness d 100 mm and\nits conductivity k 0.07 W/(m K). The heat transfer coefficient on the non-fire-\nexposed side h 5 W/(m2 K). Assume the surface heat resistance on the fire-\nexposed side is negligible and the steel properties c 460 Ws/(kg K) and\n7850 kg/m3, and the ambient temperature and the initial temperature\nT 1 T i 20 \u0005 C.\n(a) What is the ultimate steel temperature Tult\nst ?\n(b) What is the time constant of the bulkhead?\n(c) What is the steel temperature after 60 min?\n\nSolution (a) The heat balance equation of the steel bulkhead can be written as\n\u0002 \u0003\nd T f \u0003 T st hT 1 \u0003 T st d \u0001 c \u0001 dt . The ultimate steel tem-\nk dT st\n\ndt 0\ndT st\nperature is obtained when and then\nk\nT f hT 1 0:1 \u000110005\u000120\n0:07\n\u0005\nst\nT ult 800\n5:7 140 C.\nd\ndh 0:1 5\nk 0:07\n\u0006 \u0007\n(b) The heat balance equation can be reorganized and written as dk \u0001 T f h \u0001 T 1\n\u0006 \u0007 fdk\u0001T f h\u0001T 1 g fdkhg T ult \u0003T\n\u0003 dk h T st d \u0001 c \u0001 d\u0001T\ndt and dt\nst dT st\nd\u0001c\u0001 \u0003 d\u0001c\u0001 T st std\u0001c\u0001 st . Now\nk h\nd\nthe time constant can be identified (compare with Eq. 3.13) as\nd\u0001c\u0001\nk\nh\n0:005\u0001460\u00017850\n5:7 3170 s.\nd\n(c) At time t 3600 s the dimensionless time t= 3600=3170 1:14 and from\n\u0002 \u0003 \u0002 \u0003t\n\u0003\nEq. 3.15 or Fig. 3.4 T st T i T ult\nst \u0003 T i \u0001 1 \u0003 e 20 120 \u0001 0:68\n101 \u0005 C.\n\n## 3.2 Semi-infinite Solids\n\nCases and scenarios in FSE are often short in time. Therefore only the surface and\nthe top layer of a solid will be involved in fire phenomena such as ignition and flame\nspread. In such cases a solid may be assumed semi-infinite as its surface will not\n\nthermally be influenced by the limited depth of the exposed surface layer. Even\nconcrete elements exposed to fires of an hours duration any temperature rise\nbeyond 200 mm from the surface is insignificant, and when estimating temperature\nof reinforcement bars near the exposed surface may even slender structures be\nconsidered as semi-infinite.\nWhether a body can be treated as semi-infinite depends on time of consideration\nand thickness of the exposed body or more precise the exposed layer in case of\ncomposites. The longer a body is analysed, the thicker it must be to be treated as\nsemi-infinite. As a general rule the change of temperature at one point influences\nthe temperature only within a distance proportional to the square root of the\nthermal diffusivity k=c \u0001 according to Eq. 1.28 multiplied by the time t:\np\n<3 t 3:21\n\n## The coefficient 3 here is an arbitrary value depending on accuracy, see\n\nSect. 3.2.1.1 on thermal penetration depth.\nConsider a semi-infinite body initially at a constant temperature Ti. Three kinds\nof boundary conditions can be identified (c.f. Sect. 1.1.3 and Fig. 3.7):\n(a) The surface temperature changes suddenly to a new constant value. The\ninternal temperature distribution (Eq. 3.22) and the surface flux (Eq. 3.23)\ncan then be calculated (first kind of BC).\n(b) The surface receives suddenly a constant heat flux. The surface temperature\ncan then be calculated according to Eq. 3.29 (second kind of BC).\n(c) The surface is suddenly exposed to a constant gas temperature and the heat\nflux to the surface is proportional to the temperature difference between the\ngas temperature and the surface temperature. The surface temperature can then\nbe calculated according to Eq. 3.35. For internal temperatures closed form\nsolutions can be found in textbooks such as [1, 2] (third kind of BC).\nFor more details on the three types of boundary conditions see Sect. 1.1.3.\n\nFig. 3.7 Semi-infinite bodies with the three kinds of BC. See Sects. 3.2.13.2.3, respectively.\n(a) First kind. (b) Second kind. (c) Third kind\n3.2 Semi-infinite Solids 35\n\n## Fig. 3.8 Semi-infinite solid\n\nwith an initial temperature\nTi where the surface\ntemperature changes\nsuddenly to Ts\nt\n\n## 3.2.1 Constant Surface Temperature: First Kind of BC\n\nThe surface temperature of a semi-infinite body is suddenly changed from its initial\ntemperature Ti to Ts. Then temperature profiles as indicated in Fig. 3.8 develop.\nAccording to Sect. 1.1.3 this is a first kind of boundary condition. The longer times\nt the further into the body the temperature rise goes. Mathematically the tempera-\nture distribution may be written as\n\nT x; t \u0003 T i x x\n1 \u0003 erf p \u0006 erfc p 3:22\nTs \u0003 Ti 2 t 2 t\n\nwhere x is the distance to the surface and the thermal diffusivity, see Eq. 1.28. The\nfunction erf is called the Gauss error function and erfc is its complimentary\nfunction. The error function is tabulated in Table 3.2 and both erf and erfc are\n\u0002 \u0003\nshown in Fig. 3.9 as functions of the dimensionless parameter group x= 2t .\n00\nThe heat flux at the boundary q_ 0 becomes\n\nr\n00 dT 1 k\u0001\u0001c\nq_ 0 \u0003k p T s \u0003 T i 3:23\ndx t\n\nThis equation shows that the heat flux at a given time and temperature rise is\nproportional to the square root of the parameter group k \u0001 \u0001 c. This is often\nreferred to as the thermal inertia of the material. (In some literature the thermal\np\ninertia is defined as k \u0001 \u0001 c). The thermal inertia of a material has a great influence\non its ignition and flame spread properties which will be discussed further in\nSect. 3.2.3 on third kind of BC.\n\n## 3.2.1.1 Temperature Penetration Depth\n\nThe rate at which a temperature change diffuses into a body when exposed to\nheating conditions depends for a semi-infinite body on the thermal diffusivity as\ndefined in Eq. 1.28. Therefore it takes a relatively long time for a temperature rise at\nthe surface to penetrate into a material with a low conductivity k and/or a high\n\n## Table 3.2 The Gauss error function, erf\n\n\u0004 \u0005 \u0004 \u0005 \u0004 \u0005 \u0004 \u0005\npx erf 2px\npx erf 2px\npx erf 2px\npx erf 2px\n\n2 t t 2 t t 2 t t 2 t t\n\n## 0.00 0.00000 0.50 0.52050 1.00 0.84270 1.50 0.96611\n\n0.02 0.02256 0.52 0.53790 1.02 0.85084 1.52 0.96841\n0.04 0.04511 0.54 0.55494 1.04 0.85865 1.54 0.97059\n0.06 0.06762 0.56 0.57162 1.06 0.86614 1.56 0.97263\n0.08 0.09008 0.58 0.58792 1.08 0.87333 1.58 0.97455\n0.10 0.11246 0.60 0.60386 1.10 0.88021 1.60 0.97635\n0.12 0.13476 0.62 0.61941 1.12 0.88679 1.62 0.97804\n0.14 0.15695 0.64 0.63459 1.14 0.89308 1.64 0.97962\n0.16 0.17901 0.66 0.64938 1.16 0.89910 1.66 0.98110\n0.18 0.20094 0.68 0.66378 1.18 0.90484 1.68 0.98249\n0.20 0.22270 0.70 0.67780 1.20 0.91031 1.70 0.98379\n0.22 0.24430 0.72 0.69143 1.22 0.91553 1.72 0.98500\n0.24 0.26570 0.74 0.70468 1.24 0.92051 1.74 0.98613\n0.26 0.28690 0.76 0.71754 1.26 0.92524 1.76 0.98719\n0.28 0.30788 0.78 0.73001 1.28 0.92973 1.78 0.98817\n0.30 0.32863 0.80 0.74210 1.30 0.93401 1.80 0.98909\n0.32 0.34913 0.82 0.75381 1.32 0.93807 1.82 0.98994\n0.34 0.36936 0.84 0.76514 1.34 0.94191 1.84 0.99074\n0.36 0.38933 0.86 0.77610 1.36 0.94556 1.86 0.99147\n0.38 0.40901 0.88 0.78669 1.38 0.94902 1.88 0.99216\n0.40 0.42839 0.90 0.79691 1.40 0.95229 1.90 0.99279\n0.42 0.44747 0.92 0.80677 1.42 0.95538 1.92 0.99338\n0.44 0.46623 0.94 0.81627 1.44 0.95830 1.94 0.99392\n0.46 0.48466 0.96 0.82542 1.46 0.96105 1.96 0.99443\n0.48 0.50275 0.98 0.83423 1.48 0.96365 1.98 0.99489\n\nvolumetric heat capacity (c \u0001 ). Thus any temperature change diffuses, e.g. much\nfaster in a concrete than in a steel.\nFor the idealized case of a semi-infinite body at a uniform initial temperature Ti\nwhere the surface temperature momentarily is changed to a constant level of Ts, the\ntemperature rise (T \u0003 Ti) inside the body at a depth x at a time t may be written as a\nfunction of the normalized group\n\nx= 2 \u0001 t 3:24\n\n## where the thermal diffusivity k=c \u0001 according to Eq. 1.28 is assumed\n\nconstant. The relative temperature rise may then be written as:\n\nT \u0003 T i\nerfc 1 \u0003 erf 3:25\nT s \u0003 T i\n3.2 Semi-infinite Solids 37\n\n0.9\n\n0.8\nRelative temperature [-]\n\n0.7\n\n0.6\nerfc\n0.5 erf\n\n0.4\n\n0.3\n\n0.2\n\n0.1\n\n0\n0 0.25 0.5 0.75 1 1.25 1.5 1.75 2\nx/2(at)\n\u0002 \u0002 \u0003\nFig. 3.9 The Gauss error-function erf x= 2t and the Gauss complimentary error-function\n\u0002 \u0002 \u0003\u0003\nerfc x= 2t\n\nFor the Gauss error function see Fig. 3.9 and Table 3.3. Note that for values of\n\u0007 1:4 the relative temperature rise is less than 5 %, and for \u0007 1:8 it is less than\n1 %. This can be interpreted as the temperature penetration depth which can be\nderived from Eq. 3.24 by solving for x. Thus the 5 % penetration depth is\np\n0:05 2:8 \u0001 t 3:26\n\n## and the corresponding 1 % is\n\np\n0:01 3:6 \u0001 t 3:27\n\nHence a sudden temperature rise at the surface will penetrate in 30 min about\n0.14 m into a concrete structure and about four times longer (0.54 m) into or along a\nsteel structure. Constant material properties are then assumed according to\nTable 1.2.\nExample 3.3 The surface temperature of a thick concrete wall with an initial\ntemperature of 0 \u0005 C rises suddenly to 1000 \u0005 C.\n(a) What is the 1 % thermal penetration depth 0.01 after 15 min?\n(b) What is the temperature T at that point after 60 min?\nAssume constant concrete properties according to Table 1.2, i.e. c 900 Ws/(kg K),\n2300 kg/m3 and k 1.5 W/(m K).\n\n## Table 3.3 Tabulated\n\n\u0002 values\n\u0003 of the relative surface temperature change of a semi-infinitely thick\nbody T s \u0003 T i = T g \u0003 T i vs. dimensionless time t/ according to Eq. 3.35\nT \u0003 T i T \u0003 T i T \u0003 T i\n\u0002 s \u0003 \u0002 s \u0003 \u0002 s \u0003\nt/ Tg \u0003 Ti t/ Tg \u0003 Ti t/ Tg \u0003 Ti\n0.00 0.000 2.00 0.664 4.00 0.745\n0.05 0.210 2.05 0.667 4.05 0.746\n0.10 0.276 2.10 0.670 4.10 0.747\n0.15 0.322 2.15 0.673 4.15 0.749\n0.20 0.356 2.20 0.676 4.20 0.750\n0.25 0.384 2.25 0.678 4.25 0.751\n0.30 0.408 2.30 0.681 4.30 0.752\n0.35 0.428 2.35 0.684 4.35 0.753\n0.40 0.446 2.40 0.686 4.40 0.755\n0.45 0.462 2.45 0.689 4.45 0.756\n0.50 0.477 2.50 0.691 4.50 0.757\n0.55 0.490 2.55 0.694 4.55 0.758\n0.60 0.502 2.60 0.696 4.60 0.759\n0.65 0.513 2.65 0.698 4.65 0.760\n0.70 0.523 2.70 0.700 4.70 0.761\n0.75 0.533 2.75 0.703 4.75 0.763\n0.80 0.542 2.80 0.705 4.80 0.764\n0.85 0.550 2.85 0.707 4.85 0.765\n0.90 0.558 2.90 0.709 4.90 0.766\n0.95 0.565 2.95 0.711 4.95 0.767\n1.00 0.572 3.00 0.713 5.00 0.768\n1.05 0.579 3.05 0.715 5.05 0.769\n1.10 0.585 3.10 0.716 5.10 0.770\n1.15 0.591 3.15 0.718 5.15 0.771\n1.20 0.597 3.20 0.720 5.20 0.772\n1.25 0.603 3.25 0.722 5.25 0.773\n1.30 0.608 3.30 0.724 5.30 0.773\n1.35 0.613 3.35 0.725 5.35 0.774\n1.40 0.618 3.40 0.727 5.40 0.775\n1.45 0.622 3.45 0.728 5.45 0.776\n1.50 0.627 3.50 0.730 5.50 0.777\n1.55 0.631 3.55 0.732 5.55 0.778\n1.60 0.635 3.60 0.733 5.60 0.779\n1.65 0.639 3.65 0.735 5.65 0.780\n1.70 0.643 3.70 0.736 5.70 0.780\n1.75 0.647 3.75 0.738 5.75 0.781\n1.80 0.650 3.80 0.739 5.80 0.782\n1.85 0.654 3.85 0.741 5.85 0.783\n1.90 0.657 3.90 0.742 5.90 0.784\n1.95 0.661 3.95 0.743 5.95 0.785\n3.2 Semi-infinite Solids 39\n\nSolution (a) After 15 min according to Eq. 3.27 the penetration depth\np\n0:01 3:6 1:5=900 \u0001 2300 \u0001 15 \u0001 60 0:092 m.\n(b) According to Eq. 3.24 p\n0:092\n0:90 and Eq. 3.25 and\n2\u0001 1:5=900\u00012300\u000160\u000160\nFig. 3.9 yields T 200 \u0005 C.\n\n## 3.2.2 Constant Heat Flux: Second Kind of BC\n\n00\nUnder some conditions the heat transfer qs to a surface may be assumed constant.\nAccording to Sect. 1.1.3 this is a second kind of BC. That may happen, e.g. when\nthe incident radiation to a surface is very high in comparison to the losses by\nemitted radiation and convection which then can be neglected. Then at a point at a\ndistance x from the surface the temperature is\np \u0002 2\u0003\n\u000e\n00 2 t \u00034a\u0001t\nx x x\nT x; t \u0003 T i q_ s p p \u0001 e \u0003 1 \u0003 erf p 3:28\nk\u0001\u0001c k 2 \u0001t\n\n## where the thermal diffusivity k=c \u0001 . At x 0 the surface temperature Ts\n\nvs. time becomes\n00 p\n2 q_ t\nT s \u0003 T i p ps 3:29\n\u0001 k\u0001\u0001c\n\nThus the time to reach a given temperature rise assuming constant heat flux\n(for example, time to ignition) at the surface becomes\n\n\u0001 k \u0001 \u0001 c\u0002 \u00032\ntig \u0002 00 \u00032 T ig \u0003 T i 3:30\n4 q_ s\n\nwhere tig and Tig are the time to ignition and the ignition temperature, respectively.\n\n## 3.2.3 Constant Gas Temperature: Third Kind of BC\n\nWhen a surface is exposed to a fluid at a temperature Tg, the heat flux to the surface is\n\n00 dT \u0002 \u0003\nq_ s \u0003k h Tg \u0003 Ts 3:31\ndx\n\n## Fig. 3.10 Temperature\n\ndistribution in a semi-\ninfinite body exposed to a\nthird kind of boundary\ncondition\n\nwhere h is the heat transfer coefficient and Ts is the surface temperature. This is a\nthird kind of BC according to Sect. 1.1.3. In the case Tg and h are constant the\ntemperature distribution may after some time develop as indicated in Fig. 3.10.\nThen the relative temperature change at a distance x from the surface can be\ncalculated as\nr\n\u000e\nT x; t \u0003 T i h\u0001x\nt t\n1 \u0003 erf X \u0003 e k 1 \u0003 erf X 3:32\nTg \u0003 Ti\n\nwhere the temperature T(x, t) is a function of time and depth and Ti is the initial\ntemperature. The non-dimensional length\nx\nX p 3:33\n2 \u0001t\n\nand the time constant for the semi-infinite case is here defined as\n\nk\u0001\u0001c\n3:34\nh2\n\n## The temperature at the surface is of interest in many fire protection engineering\n\nproblems such as predictions of time to ignition. The relative temperature change\nmay be obtained from Eq. 3.32 for x 0 as\nr\n\u000e\nTs \u0003 Ti t t\n1 \u0003 e 1 \u0003 erf\n3:35\nTg \u0003 Ti\n\n## or when expressed with the complementary error function as\n\nr\n\nTs \u0003 Ti t t\n1 \u0003 e \u0001 erfc 3:36\nTg \u0003 Ti\n\nThe relative surface temperature rise may also be obtained from the diagram of\nFig. 3.11 or from Table 3.3.\n3.2 Semi-infinite Solids 41\n\nEquation 3.33 indicates that the relative surface temperature rise vs. time\ndepends, for a given heat transfer coefficient h, on the material parameter group\nthe thermal inertia. The thermal inertia is very important in FSE as it governs how\nfast a surface reaches among other things ignition temperatures. It varies consider-\nably for many common materials as shown in Table 1.2. Materials of low density\nhave in general also low conductivity k which enhances the differences between\nmaterials of various densities. The specific heat capacity varies only relatively little\nbetween common materials. More on the influence of thermal inertia on ignition is\ndiscussed in Sect. 8.2.\nExample 3.4 A 300-mm-thick concrete slab has reinforcement bars at a depth of\n30 mm from the bottom surface. The slab is suddenly exposed from below to a fire\nhaving a constant temperature of T f 900 \u0005 C. Assume the initial temperature of\nthe slab Ti 20 \u0005 C and the thermal conductivity of the concrete, k 1.0 W/(m K),\ndensity, 2300 kg/m3, and the specific heat capacity, c 800 J/(kg K).\n(a) What is the surface temperature T0 of the slab after 10 min of fire exposure?\nAssume the total heat transfer coefficient due to radiation and convection is\nconstant, h 75 W/m2 K.\n(b) How long does it take until the reinforcement reaches a temperature of 500 \u0005 C.\nAssume in this case that the surface instantly gets the fire temperature, i.e. the\nheat transfer resistance can be negligible.\n(c) Estimate how long it takes until the temperature 300 mm from the bottom of\nthe slab, i.e. at the top surface of the slab, has risen by approximately 10 \u0005 C\n(assuming that the slab is infinitely thick)?\n\nSolution (a) Assume the slab is semi-infinite and apply Eq. 3.35\nh kc 75 1\u00012300\u0001800 1:83 and insert into Eq. 3.35 (or use\nt 2 t 2 10\u000160\n\nT0 \u0003Ti\nFig. 3.11 or Table 3.3) to get Tf \u0003Ti 0:65. Thus\n\u0005\nT 0 20 900 \u0003 20 \u0001 0:65 593\u0004 C. \u0005 \u0004 \u0005\n(b) Apply Eq. 3.22, 500\u000320 0:545 1 \u0003 erf px thus px 0:43 from\n900\u000320 2 t 2 t\nboth erf and erfc are shown in Fig. 3.9 as functions of the dimensionless\n\u0002 \u0003\nparameter group x= 2t .\n\n## Table 3.2 or Fig. 3.9. Then with x 0.03 m and 2300\u0001800\n\n1:0\n0:543 \u0001 10\u00036 m2 =s\n2\nthe time can be calculated as t 0:543\u0001100:03\u00036 2\n\u00012 \u00010:432\n2241 s \b 37 min. Apply\np 0:32\nEq. 3.27, 0:01 0:3 3:6 t ) t 3:62 \u00010:543\u000110\u00036 12:8 \u0001 103 s 3:5 h.\nExample 3.5 The surface temperature of a thick concrete wall with an initial\ntemperature Ti 20 \u0005 C rises suddenly to Ts 1000 \u0005 C. Assume constant concrete\nproperties according to Table 1.2, i.e. c 900 Ws/(kg K), 2300 kg/m3 and\nk 1.5 W/(m K).\n\na\n1.0\n\n0.9\nRelative surface temperature rise [-]\n\n0.8\n\n0.7\n\n0.6\n\n0.5\n\n0.4\n\n0.3\n\n0.2\n\n0.1\n\n0.0\n0 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\nDimensionless time t/t\nb\n1.0\n\n0.9\nRelative surface temperature rise [-]\n\n0.8\n\n0.7\n\n0.6\n\n0.5\n\n0.4\n\n0.3\n\n0.2\n\n0.1\n\n0.0\n0.0 0.5 1.0 1.5 2.0 2.5 3.0\nDimensionless time t/t\n\nFig. 3.11 The relative surface temperature change of a semi-infinitely thick body T s \u0003 T i =\n\u0002 \u0003\nT g \u0003 T i vs. dimensionless time t/ according to Eq. 3.35. (a) Dimensionless time t= 30. (b)\nDimensionless time t= 3\n\n(a) Plot a diagram of the temperature distribution at 30, 60 and 120 min.\n(b) What are the temperature penetration depths at 30, 60 and 120 min?\np\nGuidance: Assume 1 % accuracy, i.e. 0:01 3:6 \u0001 t.\n3.2 Semi-infinite Solids 43\n\n(c) After how long time will the temperature of a reinforcement bar at a depth of\n30 mm from the exposed surface start to rise?\nGuidance: Assume 1 % accuracy.\n(d) How thick must the wall be to be considered infinitely thick when calculating\nthe temperature at 30 mm from the exposed surface at 30, 60 and 120 min?\nGuidance: Assume 1 % accuracy and that the temperature change goes to the\nrear surface and back to the reinforcement bar.\n\n## Solution The temperature diffusion 1.5/(300 \u0001 2300) 0.821 \u0001 10\u00036 m2/s.\n\n(a) See Fig. 3.12.\n(b) Equation 3.27 yields the penetration depth 0:01 138 mm, 196 mm\nand 277 mm, respectively, for 30, 60 and 120 min.\n\u0002 \u00032 \u0002 \u00036\n\u0003\n(c) Equation 3.27 yields t 0:033:6 = 0:821 \u0001 10 84:5 s.\n(d) The distance x + (x \u0003 0.03) must be longer than the penetration depths (see\narrows in Fig. 3.13). Thus the wall thickness x \u0007 0:5 \u0001 0:01 0:03 which\nyields the thicknesses 99, 128 and 168 mm, respectively, for 30, 60 and 120 min.\n\n1000\n\n800\nTemperature [C]\n\n600\n\n120 min\n400 60 min\n\n30 min\n200\n\n0\n0 0.05 0.1 0.15 0.2 0.25\nDepth [m]\n\n## Fig. 3.13 Estimation\n\nwhether a wall can be\nconsidered infinitely thick\nwhen calculating the\ntemperature of a\nreinforcement bar\n\n## Example 3.6 A very thick concrete wall is penetrated by a steel beam. It is\n\nsuddenly exposed to high temperature on one side. Estimate roughly how long it\ntakes before the temperature rise is felt on the unexposed side.\n(a) On the concrete surface away from the beam.\n(b) On the steel beam surface.\nMaterial properties according to Table 1.2.\nSolution The thermal diffusivity of concrete is 0.82 \u0001 10\u00036 m2/s and for mild steel\n\u0002 \u00032 \u0002 \u00032\n12.7 \u0001 10\u00036. Then Eq. 3.26 yields t 1 \u0001 2:8\n0:05\n, i.e. t 0:82\u000110\u00036\n1\n\u0001 0:02\n2:8 62 s for\nthe concrete surface and only 4 s for the steel surface.\nExample 3.7 The surface of a thick pine wood panel with an initial temperature\nT i 20 \u0005 C is suddenly exposed to hot gases with a temperature of T g 600 \u0005 C.\nAssume a constant heat transfer coefficient h 50 W/(m2 K). What is the surface\n00\ntemperature Ts and the heat flux q_ s at time t 0, 30 and 120 s. Assume thermal\nproperties of pine according to Table 1.2.\nSolution According to Eq. 3.34 k \u0001 \u0001 c=h2 0:196 \u0001 106 =502 78:4 s.\n\u0002\nThen t= 0=78:4, 30=78:4 and 120=78:4, respectively, and the function 1 \u0003\n\u0002 p \u0003\nexpt= \u0001 erfc t is according to Table 3.3 equal to 0, 55 and 0.73, respectively,\nand the surface temperature can be obtained from Eq. 3.35 as T s 0\n20 600 \u0003 20 \u0001 0 20 \u0005 C, T s 30 20 600 \u0003 20 \u0001 0:44\n275 \u0005 C and T s 120 20 600 \u0003 20 \u0001 0:63 385 \u0005 C. The corresponding\n00\nheat fluxes become according to Eq. 3.31 q_ s 0 50 \u0001 600 \u0003 20\n00 00\n29 \u0001 103 W=m2 , q_ s 30 50 \u0001 600 \u0003 275 16:3 \u0001 103 W=m2 and q_ s 120\n50 \u0001 600 \u0003 385 10:8 \u0001 103 W=m2 . Notice that the heat flux is high in the\nbeginning and then reduced by almost two thirds after 120 s.\nChapter 4\nBoundary Conditions in Fire Protection\nEngineering\n\n## A summary of the three kinds of boundary conditions as outlined in Sect. 1.1.3 is\n\nshown in Table 4.1. The third kind of BC sometimes called natural BC is by far the\nmost important and common boundary condition in fire protection engineering,\nwhile the first and second kinds of BCs can rarely be specified. The third kind of BC\nmay be divided into three subgroups, (a), (b) and (c). The subgroup (b) and (c) are\nparticularly suitable for fire engineering applications. Subgroup (a) is applied when\nthe heat transfer coefficient may be assumed constant as assumed in Chaps. 2 and 3.\nTg is then the surrounding gas temperature. In fire protection engineering it is,\nhowever, generally not accurate enough to assume a constant heat transfer coeffi-\ncient as in particular heat transfer by radiation is highly non-linear, i.e. the heat\ntransfer coefficient varies with the surface temperature. Therefore the subgroups\n(3b) and (3c) are the most commonly applied. They consist of a radiation term and a\nconvection term with the corresponding emissivity and convection heat transfer\ncoefficient h, respectively. The subgroup (3b) presupposes a uniform temperature\nTf, i.e. the radiation temperature and the gas temperature are equal. This is assumed,\nfor example, when applying timetemperature design curves according to standards\nsuch as ISO 834 or EN 1363-1 for evaluating the fire resistance of structures, see\nChap. 12. The subgroup (3c) is a more general version of (3b) as it allows for\ndifferent gas Tg and radiation Tr temperatures, so-called mixed boundary condi-\ntions. Alternatively \u0001 T 4r may be replaced by an equivalent specified incident\n00 00\nradiation q_ inc according to the identity q_ inc \u0003 \u0001 T 4r (Eq. 1.17). As shown in Sect. 4.4\nall boundary conditions of subgroup 3 may be written as type 3a. That means\nmomentarily a single effective temperature named the adiabatic surface tempera-\nture (AST) with a value between the radiation and gas temperatures as well as a\ncorresponding total heat transfer coefficient can always be defined, see Sect. 4.4.\nAll the specified boundary conditions given in Table 4.1 may vary with time. In\nmost calculations the emissivity and the convection heat transfer coefficient are,\nhowever, assumed constant while the radiation and gas temperatures may vary\naccording to standard, measured or calculated values, see Chaps. 1012.\n\n## Springer International Publishing Switzerland 2016 45\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_4\n46 4 Boundary Conditions in Fire Protection Engineering\n\nTable 4.1 Summary of the three kinds of boundary conditions. The third kind is divided into three\nsubgroups relevant in FSE\nNo Kind of boundary condition Formula\n1 Prescribed surface temperature T xx0 T s\n\u0003\n2 Prescribed surface heat flux \u0005kT \u0003 00\nq_ s\nx x0\n\u0003 \u0004 \u0005\n(3a) Natural boundary condition (prescribed \u0005kT \u0003 h Tg \u0005 Ts\nx x0\nconvection)\n\u0003 \u0006 \u0007 \u0004 \u0005\n(3b) Natural boundary condition (prescribed \u0005kT \u0003 T 4f \u0005 T 4s hc T f \u0005 T s\nx x0\nand gas temperatures)\n\u0003 \u0004 \u0005 \u0004 \u0005\n(3c) Natural boundary condition (prescribed \u0005kT \u0003 \u0001 T 4r \u0005 T 4s hc T g \u0005 T s\nx x0\nconvection and radiation conditions, differ- or \u0003\n\u0004 00 \u0005 \u0004 \u0005\nent radiation and gas temperatures) \u0005kT \u0003 q_ inc \u0005 T 4s hc T g \u0005 T s\nx x0\nor \u0003\n\u0005kT \u0003 hAST , tot T AST \u0005 T s\nx x0\n\nThe black body radiation temperature Tr was introduced in Sect. 1.1.3 by the\nidentity\n00\nq_ inc \u0003 \u0001 T 4r 4:1\n\nor reversely\nr\n00\n_ inc\n4 q\nTr \u0003 4:2\n\n## A more adequate term would be incident black body radiation temperature as\n\n00\nq_ inc depends on direction. By definition Tr is the temperature of a surface in\nequilibrium with the incident radiation, i.e. the absorbed heat by radiation is\nequal to the emitted heat.\n00\nFigure 4.1 shows the relation between the incident radiation q_ inc and the\nradiation temperature Tr as defined by Eq. 4.1. Tr may be given in Kelvin as in\nEq. 4.1 (lower curve) and in \u0004 C (upper curve). The temperature shift between the\ntwo temperature scales is 273.15 K, i.e. [temperature in Kelvin] [temperature in\n00\n\u0004\nC + 273.15]. In Table 4.2 the relations between q_ rad and Tr are given at selected\nlevels. Thus, for example, an incident radiant flux of 20 kW/m2 corresponds to a\nblack body radiation temperature of 771 K 498 \u0004 C, and a radiation temperature of\n1000 \u0004 C 1273 K corresponds to an incident radiant flux of 148.9 kW/m2.\nThe net heat flux by radiation to a surface is according to Eq. 1.16\n\n140\n\n120\nIncident radiant heat flux T4 [kW/m2]\n\n100\n\n80\n\nC\n60 K\n\n40\n\n20\n\n0\n0 100 200 300 400 500 600 700 800 900 1000\nTemperature\n00\nFig. 4.1 Incident radiation heat flux q_ inc vs. incident radiation temperature Tr. The upper curve\nrefers to temperature in \u0004 C and the lower to Kelvin as in Eq. 4.1\n\n00 \u0004 \u0005\nq_ rad \u0001 T 4r \u0005 T 4s 4:3\n\n## Alternatively it can be linearized and written as\n\n00\nq_ rad hr T r \u0005 T s 4:4\n\n## where the radiation heat transfer coefficient hr is obtained by developing the\n\nparentheses of Eq. 1.16 according to the conjugate rule:\n\u0004 \u0005\nhr \u0001 T 2r T 2s \u0001 T r T s 4:5\n\nAs shown in Fig. 4.2 the radiation heat transfer coefficient hr varies significantly\ndepending on the radiation and surface temperatures. At room temperature it is less\nthan 5 W/(m2 K) while it is between 150 and 400 W/(m2 K) or even more at\ntemperature levels relevant in fire scenarios.\nIn many cases Tr and Ts are close and may be assumed equal. Then hr can be\napproximated as\n\nhr \u0006 4 \u0001 \u0001 T 3r 4:6\n\nand the radiation heat transfer coefficient becomes then depending on the incident\n48 4 Boundary Conditions in Fire Protection Engineering\n\n## Table 4.2 Incident radiation 00\n\n00 (a) Selected incident radiation q_ inc levels\nheat flux q_ inc and\nTr T Tr T\ntemperatures, absolute Tr qinc [kW/m2] [K] [\u0004 C] qinc [kW/m2] [K] [\u0004 C]\n[K] according to Eqs. 4.1 1 364 91 40 916 643\nand 4.2, and T [\u0004 C] 2 433 160 45 944 671\n3 480 206 50 969 696\n4 515 242 55 992 719\n5 545 272 60 1014 741\n6 570 297 65 1035 762\n7 593 320 70 1054 781\n8 613 340 80 1090 817\n9 631 358 90 1122 849\n10 648 375 100 1152 879\n12.5 685 412 110 1180 907\n15 717 444 120 1206 933\n17.5 745 472 145 1265 991\n20 771 498 170 1316 1043\n25 815 542 195 1362 1089\n30 853 580 220 1403 1130\n35 886 613 250 1449 1176\n(b) Selected radiation temperature (Tr \u0005 273) \u0004 C levels\nTr qinc Tr qinc\nT [\u0004 C] [K] [kW/m2] T [\u0004 C] [K] [kW/m2]\n100 373 1.10 550 823 26.01\n120 393 1.35 700 973 50.82\n140 413 1.65 750 1023 62.10\n160 433 1.99 800 1073 75.16\n180 453 2.39 850 1123 90.18\n200 473 2.84 900 1173 107.34\n225 498 3.49 950 1223 126.85\n250 523 4.24 1000 1273 148.90\n300 573 6.11 1050 1323 173.71\n325 598 7.25 1100 1373 201.50\n350 623 8.54 1150 1423 232.49\n375 648 10.00 1200 1473 266.93\n400 673 11.63 1250 1523 305.06\n425 698 13.46 1300 1573 347.13\n450 723 15.49 1350 1623 393.42\n475 748 17.75 1400 1673 444.19\n500 773 20.24 1450 1723 499.72\n\na 400\n\nTs=900 C\nRadia\u0002on heat transfer coecient hr [W/m2 K]\n350 Ts=700 C\nTs=500 C\n300 Ts=300 C\nTs=100 C\nTs=0 C\n250\nTs=Tr\n\n200\n\n150\n\n100\n\n50\n\n0\n0 100 200 300 400 500 600 700 800 900 1000\nb 100\n\n90 Ts=700 C\nRadia\u0002on heat transfer coecient hr [W/m2 K]\n\nTs=500 C\n80\nTs=300 C\n\nTs=100 C\n70\nTs=0 C\n60 Ts=Tr\n\n50\n\n40\n\n30\n\n20\n\n10\n\n0\n0 100 200 300\n\nFig. 4.2 The radiation heat transfer coefficient hr vs. incident radiation temperature (Eq. 4.5) for\nvarious surface temperatures Ts assuming 1. (a) T r \u0007 1000 \u0004 C. (b) Enlargement, T r \u0007 300 \u0004 C\n50 4 Boundary Conditions in Fire Protection Engineering\n\nq\nhr \u0006 4 \u0001 q_ 3inc 4:7\n4\n\nby combining Eqs. 4.1 and 4.6. Observe that Eqs. 4.6 and 4.7 apply only when the\nradiation and surface temperatures are approximately equal but they may be used as\nrough estimates.\nExample 4.1 Calculate the radiation heat transfer coefficient hr, (a) in the heating\nphase of a fire when the radiation temperature T r 1000 \u0004 C and the surface\ntemperature T s 600 \u0004 C and (b) in the cooling phase when T r 200 \u0004 C and\nT s 500 \u0004 C. Assume the surface emissivity 0:9.\nSolution Equation 4.5 yields:\n\u0004 \u0005\n(a) hr 0:9 \u0001 5:67 \u0001 10\u00058 12732 8732 \u0001 1273 873 261 W=m2 K.\n\u0004 \u0005\n(b) hr 0:9 \u0001 5:67 \u0001 10\u00058 4732 7732 \u0001 473 873 52 W=m2 K.\n\n## 4.2 Non-linear Convection\n\nThe heat transfer by convection depends on the difference between the gas tem-\nperature Tg and the surface temperature Ts. In the simplest form it is just propor-\ntional to the difference as when assuming Newtons law of cooling. More generally\nthe convection heat transfer may be calculated as\n00 \u0004 \u0005\nq_ con T g \u0005 T s 4:8\n\nwhere the power is equal to one for forced convection and greater than one for\nnatural or free convection. See Chap. 6 for details on how heat transfer by\nconvection can be obtained for various configurations and flow conditions.\nThroughout this document the convection heat flux is written in the linear form\nas\n00 \u0004 \u0005\nq_ con hc T g \u0005 T s 4:9\n\nwhere the convection heat transfer coefficient can be identified by comparison with\nEq. 4.8 as\n\u0004 \u0005\u00051\nhc T g \u0005 T s 4:10\n\nMore on the physical phenomena of convection heat transfer and how it can be\nestimated is given in Chap. 6.\n4.3 Mixed Boundary Conditions 51\n\n## 4.3 Mixed Boundary Conditions\n\nThe total heat flux to a surface is the sum of the radiation and the convection\ncontributions and may according to Eqs. 4.3 and 4.9 be written as\n00\n\u0006 00 \u0007 \u0004 \u0005\nq_ tot q_ inc \u0005 \u0001 T 4s hc T g \u0005 T s 4:11\n\n## or by expressing the incident radiant flux by the radiation temperature as defined by\n\nEq. 4.1\n00 \u0004 \u0005 \u0004 \u0005\nq_ tot \u0001 T 4r \u0005 T 4s hc T g \u0005 T s 4:12\n\n## This equation contains two boundary temperatures, the radiation temperature\n\nand the gas or convection temperature. It may then be called mixed boundary\nconditions.\nAn electric circuit analogy of a mixed boundary condition with two temperatures\nand two corresponding heat transfer resistances is shown in Fig. 4.3.\nIf the radiation heat transfer coefficient hr as defined by Eqs. 4.5 or 4.6 is used we\ncan get\n00 \u0004 \u0005\nq_ tot hr T r \u0005 T s hc T g \u0005 T s 4:13\n\n## or in terms of thermal resistances\n\n00 \b \u0004 \u0005\nq_ tot T r \u0005 T s =Rr T g \u0005 T s =Rc 4:14\n\nwhere the radiation heat transfer resistance (cf. Eqs. 2.14 and 4.5)\n\n1 1\nRr \u0004 \u0005 4:15\nhr \u0001 T 2r T 2s \u0001 T r T s\n\n1\nRc 4:16\nhc\n\n## In most fire resistance cases and calculation standards such as EN 1991-1-2\n\n(Eurocode 1) dealing with exposure to post-flashover room fires, the radiation\ntemperature and the gas temperature are assumed equal to a fire temperature\nT f T r T g . Then the total heat transfer becomes\n52 4 Boundary Conditions in Fire Protection Engineering\n\n## Fig. 4.3 Electric circuit\n\nanalogy of mixed boundary\nconditions according to\nEq. 4.14\n\n00\n\u0006 \u0007 \u0004 \u0005\nq_ tot \u0001 T 4f \u0005 T 4s hc T f \u0005 T s 4:17\n\n## This equation may also be written as\n\n00 \u0004 \u0005 T f \u0005 T s\nq_ tot htot T f \u0005 T s 4:18\nRtot\n\nwhere\n\u0006 \u0007\nhtot hr hc \u0001 T 2f T 2s \u0001 T f T s hc 4:19\n\nand\n\n1\nRtot 4:20\nhtot\n\n## Example 4.2 A surface in air at ambient temperature of T 1 25 \u0004 C is exposed to\n\nradiation from a thick flame at a temperature of 800 \u0004 C. Assume the surface\nemissivity 1 and the convection heat transfer coefficient hc 50 W/m2 K?\n(a) What is the radiation heat transfer coefficient hr if the surface temperature Ts is\n600 \u0004 C.\n00\n(b) Use the calculated hr to calculate the total heat transfer q_ tot to the surface.\n\nSolution\n(a) Equation 4.5 yields hr 211 W/m2.\n(b) Equation 4.13 yields\n00\nq_ tot 211 \u0001 800 \u0005 600 50 \u0001 25 \u0005 600\b W=m2 13, 500 W=m2 .\n\nThermal exposure of a surface depends according to Eqs. 4.11 and 4.12 on two\nindependent parameters Tr (or q_ }inc ) and Tg and can then in principle not be\nexpressed by one single parameter. The radiation and gas temperatures are in\ngeneral not equal. The radiation temperature may be either higher or lower than\nthe adjacent gas temperature. Equations 4.11 or 4.12 can always be applied.\nAlternatively, however, the heat transfer may be written with one parameter only,\ngiven the relation between the emissivity and the convective heat transfer coeffi-\ncient hc/ is known. Then an artificial effective temperature denoted AST TAST can\nreplace Tr and Tg. A very important advantage of introducing AST is that it can be\nmeasured also under harsh fire conditions with the robust so-called Plate Ther-\nmometers as described in Sect. 9.3, and it can be obtained from numerical calcu-\nlations with fire modelling codes such as FDS (Fire Dynamic Simulator) .\nThe AST depends on position as well as on direction. For example, at a point\noutside a fire as illustrated in Fig. 4.4 the highest incident radiation is in the\ndirection A from the fire while from other directions it is less. Therefore in this\nA\ncase T AST > T AST\nB\n. In general at any point in space six different incident fluxes can\nbe identified and thereby six different ASTs, but only one gas temperature Tg.\nHowever, in most cases it is obvious that only one direction is of interest, namely\nperpendicular to an exposed surface.\nBy definition TAST is the temperature of a surface which cannot absorb any heat,\ni.e.\n\u0006 00 \u0007 \u0004 \u0005\nq_ inc \u0005 \u0001 T 4AST hc T g \u0005 T AST 0 4:21\n\n00\nand with the relation between q_ inc and Tr according to Eq. 4.1\n\u0004 \u0005 \u0004 \u0005\n\u0001 T 4r \u0005 T 4AST hc T g \u0005 T AST 0 4:22\n\nTAST is a weighted average value of the radiation temperature Tr and the gas\ntemperatures Tg depending on the surface emissivity and the convection heat\ntransfer coefficient hc. Thus it is a function of Tr, Tg and the parameter ratio hc/, but\nindependent of the surface temperature Ts of the exposed body. From Eq. 4.22 it is\nevident that TAST has a value between Tr and Tg as being illustrated by Fig. 4.5. The\nlarger values of hc/, the closer TAST will be to Tg, and vice versa the smaller values\nof hc/ the closer the value of TAST will be to Tr. In other words, when the heat\ntransfer by convection is dominating TAST is near the gas temperature and when the\nThe AST may be derived from Eq. 4.22 as\n54 4 Boundary Conditions in Fire Protection Engineering\n\n## Fig. 4.4 The incident\n\ndepends on direction while\nthe gas temperature does not\n\nFig. 4.5 The adiabatic surface temperature TAST is always between the radiation temperature Tr\nand the gas temperature Tg. The higher value of the parameter ratio hc/, the closer TAST will be to\nTg, and vice versa\n\nhr T r hc T g\nT AST 4:23\nhr hc\n\n## The equation is, however, implicit as hr depends on TAST.\n\nBy combining the general heat transfer equations 4.12 and 4.22 the total heat\ntransfer to a surface may alternatively be calculated as\n00 \u0004 \u0005\nq_ tot \u0001 T 4AST \u0005 T 4s hc T AST \u0005 T s 4:24\n\nInstead of two temperatures, Tr and Tg, the fire temperature level is now in\nEq. 4.24 expressed only by one temperature TAST. This may have computational\nadvantages but most important TAST can be measured even at very harsh thermal\nconditions with so-called Plate Thermometers, see Sect. 9.3.\nFigure 4.6 illustrates how the two exposure boundary temperatures Tr and Tg are\ncombined into one effective exposure boundary temperature, namely the AST TAST.\nThis alteration does not introduce any further approximations of the heat transfer\nconditions.\n\n## Fig. 4.6 The heat exposure\n\nof a surface expressed in\ntemperature Tr and the gas\ntemperature Tg or\nalternatively in terms of (b)\ntemperature TAST\n\n## 4.4.1 Calculation of Adiabatic Surface Temperature\n\n00\nWhen the incident radiation flux q_ inc (or the equivalent Tr) and the corresponding\nand hc are known, the AST TAST can be obtained by solving the fourth degree\nequation according to either Eq. 4.21 or Eq. 4.22. Below two iteration schemes and\none exact method are mentioned.\nIn many cases when the radiation is dominating TAST can be obtained by the\niteration procedure\nr\n4 hc \u0004 \u0005\nT i1 T 4r T g \u0005 T AST i\n4:25\nAST\n\u0001\n\nwhere the suffix i and i + 1 denotes the iteration number. By starting the iteration\nwith T 1AST T r the result converges generally within a few iteration steps. Other-\nwise especially when the convection is dominating a NewtonRaphson iteration\nscheme may be needed.\nAn exact solution has been presented by Malendowski (personal communica-\ntion). Then after elementary algebraic operations, Eq. 4.21 can be written as:\n\u0006 00\n\u0007\n\u0001 \u0001 T 4AST hc T AST \u0005 \u0001 q_ inc \u0005 hc T g 0 4:26\n\nwhich is the fourth order polynomial equation with TAST as the variable. It may be\nwritten in the form:\n\n## where the coefficients of the polynomial can be identified as: a \u0001 , b hc and\n\n\u0004 00 \u0005\nc \u0005 \u0001 q_ inc hc T g . Eq. 4.27 has generally four roots but the only physical can\nin the actual case be written as:\n56 4 Boundary Conditions in Fire Protection Engineering\n\nr!\n1 2b\nT AST \u0001 \u0005M \u0005 M2 4:28\n2 aM\n\nwhere\ns\n\nM 4:29\n\nand where\nq r\n3 p\np 3 2\np\n3 \u0001 27a b \u0005 256a c 9a \u0001 b 4\n2 4 3 3 2\n\u0001c 3 18\u0001a 4:30\n3\n\nThus by inserting the parameters of Eq. 4.30 into Eq. 4.29 and then into Eq. 4.28\nthe solution may be expressed in an exact closed form.\nExamples of AST vs. incident radiation temperature for various gas temperature\nlevels and relations between surface emissivity and convection heat transfer coef-\nficient are shown in the graphs of Fig. 4.7ad.\nWhen TAST is obtained, e.g. by measurements with PTs, the incident radiation\n00\nq_ inc can be derived from Eq. 4.21 as\n\n00 hc \u0004 \u0005\nq_ inc \u0001 T 4AST \u0005 T g \u0005 T AST 4:31\n\nThe accuracy of Eq. 4.31 depends very much on the accuracy of the parameter\nratio hc/. However, in most cases at elevated temperature the second term on the\nright-hand side is small and therefore the accuracy can be relatively high in\nhow the so-called plate thermometer can be used for indirectly measuring incident\nradiant heat flux by measuring TAST and then applying Eq. 4.31.\n\n## 4.4.2 An Electric Circuit Analogy of the AST Boundary\n\nCondition\n\nThe radiation term of Eq. 4.24 may be developed in a similar way as shown by\nr may be intro-\nduced, and the heat flux can be written as\n00 \u0004 \u0005 \u0004 \u0005\nq_ tot hAST\nr T AST \u0005 T s hAST\nc T AST \u0005 T s 4:32\n\nor in terms of heat transfer resistance according to Eq. 4.20, the radiation and\nconvection heat transfer resistances over an area A are the inverses of the heat\ntransfer coefficients, hAST\nr and hAST\nc ,\n\nFig. 4.7 The adiabatic surface temperature TAST (in \u0004 C) vs. the radiation temperature Tr (in \u0004 C) for\nvarious ratio between the convection heat transfer coefficient and the emissivity as defined by\nEq. 4.22. Diagrams for gas temperatures T g 20, 50, 100 and 500 \u0004 C, respectively\n58 4 Boundary Conditions in Fire Protection Engineering\n\n## Fig. 4.8 An electric circuit\n\nanalogy of mixed boundary\nconditions at a surface of a\nsolid according to Eq. 4.33\n\n## Fig. 4.9 An electric circuit\n\nanalogy of mixed boundary\nconditions at a surface of a\nsolid according to Eq. 4.34\n\n00 1 1\nq_ tot \u0001 T AST \u0005 T s \u0001 T AST \u0005 T s 4:33\nRAST\nr RAST\nc\n\nThe electric circuit analogy is shown in Fig. 4.8. A total adiabatic heat transfer\ncoefficient hAST AST\ntot and a corresponding total adiabatic heat transfer resistance Rtot\ncan also be defined. Then the total heat transfer to a surface may be written as\n00\nq_ tot hAST\ntot T AST \u0005 T s T AST \u0005 T s =Rtot\nAST\n4:34\n\n## where the total adiabatic heat transfer coefficient becomes\n\ntot hr\nhAST hAST 4:35\nAST\nc\n\n\u0004 \u0005\nhAST\nr \u0001 T 2AST T 2s \u0001 T AST T s 4:36\n\n## and the total adiabatic heat transfer resistance becomes\n\n1\ntot\nRAST 4:37\nhAST\nr hAST\nc\n\nThe convection heat transfer coefficient remains the same as it is here assumed\nindependent of the exposure temperature, i.e.\n\nhAST\nc hc 4:38\n\n## A corresponding electric circuit analogy of a mixed boundary condition is shown\n\nin Fig. 4.9.\n60 4 Boundary Conditions in Fire Protection Engineering\n\nThus it is shown that the general fire boundary condition according to Eq. 4.12\ncan be expressed as a third kind of boundary condition, see Table 4.1.\n\n## 4.4.3 Boundary Condition Expressed as Heat Flux\n\nThe thermal exposure conditions are often in FSE literature specified as heat flux\nby radiation and convection although boundary conditions of the second kind can\nrarely be specified in FSE problems. It is, however, implicitly understood that the\n00\nheat flux q_ flux is to a surface being kept at ambient temperature T 1 and having an\n00\nemissivity equal unity. Then the heat flux q_ tot by radiation and convection to a real\nsurface at a temperature Ts and an assumed convective heat transfer coefficient h to\nbe the same as when defining the heat flux becomes:\n00 00 \u0004 \u0005\nq_ tot q_ flux \u0005 T 4s \u0005 T 41 \u0005 hT s \u0005 T 1 4:39\n\nThis is now a boundary condition of the third kind as the heat flux to the surface\ndepends on the receiving surface temperature Ts. By comparing the heat fluxes as\nexpressed by Eqs. 4.39 and 4.31, it can be shown that for given values of the heat\ntransfer parameters and h there is an unambiguous relation between the two\n00\nartificial boundary parameters q_ flux and TAST, i.e.\n00 \u0004 \u0005\nqflux \u0001 T 4AST \u0005 T 41 hT AST \u0005 T 1 4:40\n\n00\nNotice that relation between q_ flux and T 1 is unambiguous and independent on the\nsurface temperature Ts.\nFor an ambient temperature T 1 20 \u0004 C and three combinations of and h, the\n00\nrelations between AST TAST and the heat flux q_ flux are shown in Fig. 4.10. As an\n00\nexample for 1.0 and h 10 W/(m2 K), a heat flux q_ flux 10 kW=m2\ncorresponds to an AST T AST \u0006 330 \u0004 C. However, if instead the convection heat\ntransfer coefficient h 20 W/(m2 K) then the corresponding AST is reduced to\nT AST \u0006 280 \u0004 C. Thus the assumed values of and h have a significant influence on\n00\nthe relation between TAST and q_ flux .\n\n## 4.4.4 Calculation of Time Constants for Bodies Exposed\n\nto Mixed Boundary Conditions\n\nThe concept of adiabatic heat transfer resistance as defined in Eq. 4.37 may be used\nto calculate the temporal response characteristics (time constants) of bodies\nexposed to radiation and simultaneously to convection. The time constant for\n\n60\n\n50\nHeat flux, qflux [kW/m2]\n\n40\n=1.0, h=10 W/(m2 K)\n=0.8, h=10 W/(m2 K)\n30\n=1.0, h=20 W/(m2 K)\n=1.0, h=5 W/(m2 K)\n\n20\n\n10\n\n0\n0 100 200 300 400 500 600 700\n00\nFig. 4.10 Relation between adiabatic surface temperature TAST and heat flux q_ flux to a surface at\nambient temperature (20 \u0004 C) for various combinations of surface emissivities and convection\nheat transfer coefficients h\n\nbodies exposed to uniform temperatures is described in, e.g. Sect. 3.1. As a general\nrule the time constant of bodies exposed to radiation decreases significantly when\nthe temperature level increases as in many FSE scenarios.\nExample 4.3 The maximum incident radiation from the sun at the earths surface\nperpendicular to the suns rays is approximately 1 kW/m2. What is the equilibrium\ntemperature of perfectly insulated surface perpendicular to the rays when\n(a) The convection is negligible.\n(b) The air temperature is 20 \u0004 C, the convection heat transfer coefficient is 10 W/\n(m2 K) and the surface equal unity.\nAssume a surface emissivity independent of the wavelength.\nSolution\n(a) Equation 4.1, Table 4.2 or Fig. 4.1 yields the equilibrium temperature 364 K or\n91 \u0004 C.\n(b) Equation 4.22 or Fig. 4.7a yields T r 91 \u0004 C and h/ 10 W/(m2 K) and the\nequilibrium temperature (AST) 55 \u0004 C.\n\n## Example 4.4 A surface with a temperature of T s 200 \u0004 C is exposed to an\n\n00\nincident radiation of q_ inc 50 kW=m2 and a gas temperature of T g 150 \u0004 C.\n62 4 Boundary Conditions in Fire Protection Engineering\n\nThe surface emissivity is 0.9 and the convection heat transfer coefficient 10 W/(m2\nK). Calculate\n(b) The radiation heat transfer coefficient.\n(c) The AST TAST.\n(d) The heat transfer to the surface using Eqs. 4.11 and 4.12, respectively.\n(e) The heat transfer to the surface applying TASTusing Eq. 4.31.\nresistance.\n(g) The heat transfer to the surface using Eq. 4.34.\n\nSolution\nq\n(a) Equation 4.1 yields T r 4 50\u0001103\n5:67\u000110\u00058\n969 K 696 \u0004 C.\n\u0004 2 \u0005 \b\n(b) Equation 4.5 yields hr T r T 2s T r T s 0:9 \u0001 5:67 \u0001 10\u00058 \u0001 9692\n4732 \b \u0001 969 473\b 85:5 W=m2 K.\n(c) Equation 4.21 or Eq. 4.22 yields by iteration T AST 940 K 667 \u0004 C.\n00 \u0004 00 \u0005 \u0004 \u0005 \u0004\n(d) Equation 4.11 yields q_ tot q_ inc \u0005 T 4s hc T g \u0005 T s 0:9 \u0001 50 \u0001 103 \u0005\n5:67 \u0001 10\u00058 \u0001 4734 10 \u0001 150 \u0005 200 42, 440 \u0005 500 41, 940 W=m2 , or\n00 \u0004 \u0005 \u0004 \u0005\nalternatively Eq. 4.12 yields q_ tot T 4r \u0005 T 4s hc T g \u0005 T s 0:9 \u0001 5:67 \u0001\n\u0004 \u0005\n10\u00058 \u0001 9694 \u0005 4734 10 \u0001 150 \u0005 200 42, 440 \u0005 500 41, 900 W=m2 .\n00 \u0004 \u0005\n(e) Equation 4.31 yields q_ tot T 4AST \u0005 T 4s hc T AST \u0005 T s\n\u0004 \u0005\n0:9 \u0001 5:67 \u0001 10\u00058 \u0001 9404 \u0005 4734 10 \u0001 940 \u0005 473 41, 900 W=m2 .\n(f) Equation 4.36 yields hAST r 0:47 \u0001 0:9 \u0001 4 \u0001 5:67 \u0001 10\u00058 \u0001 9403 80 W=\nm K. Then\n2\nhtot 80 10 90 W=m2 K.\nAST\ntot 1=htot\nRAST AST\n\n0:011 m K =W.\n2\n00\n(g) Equation 4.34 yields q_ tot 90 \u0001 940 \u0005 473 42, 000 W=m2 .\n\n## Example 4.5 A wall consists of a wooden panel with a thickness of dwood 25\n\nmm and a conductivity kwood 0:1 W=m \u0001 K and a mineral wool insulation with a\nthickness d ins 100 mm and a conductivity kins 0:02 W=m \u0001 K as shown in\nFig. 4.11. The surface temperatures T 1 100 \u0004 C and T 3 20 \u0004 C. Calculate the heat\n00\nflux q_ through the assembly and the temperature T2.\nSolution\nThe heat flow through the assembly\n\n00 T1 \u0005 T3 80 80\nq_ \b\u00040:012\u0005 \u00040:05\u0005 30:5 W=m2\nR1 R 2 0:1 0:02\n0:12 2:5 \b\n\n## Fig. 4.11 Properties of\n\ninsulated wooden wall,\nExample 4.7\n\nR1 0:12\nT2 T1 \u0005 T 1 \u0005 T 3 100 \u0005 100 \u0005 20\nR1 R2 0:12 2:5\n100 \u0005 3:66 96:3 \u0004 C\n\nNotice that the thermal resistance of the wood panel is much small than that of\nthe insulation and therefore the interface temperature becomes close to that of the\nwood panel.\nExample 4.6 A 30-mm-thick steel sheet is exposed to a gas temperature\nT g 500 \u0004 C on one side and 20 \u0004 C on the other. Calculate the heat flux through\nthe sheet and its surface temperatures. Assume a heat transfer coefficient\nh 100 W/(m2 K) on the hot side (1) and 20 W/(m2 K) on the other (side 2). The\nsteel conductivity k 50 W/(m K).\nSolution\nThe thermal resistance over a unit area is the sum of the heat transfer resistance and\n00 T1 \u0005 T2\nconductive resistance. Thus the heat flux through the panel q_\nRh1 Rst Rh2\n500 \u0005 20 480\n\b\u0004 1 \u0005 \u00040:030\u0005 \u0004 1 \u0005 7921 W=m2 and,\n100 50 20\n0:01 0:0006 0:05\n500 \u0001 0:00060:05 20 \u0001 0:01\ne.g. according to Eq. 2.12 T 1 465 \u0004 C and\n0:010:00060:05\n500 \u0001 0:0520 \u0001 0:00060:01\nT2 416 \u0004 C. Notice that the temperature\n0:010:00060:05\nchange over the steel sheet is relatively small as thermal heat transfer resistances\ndominates.\nExample 4.7 Calculate the surface temperatures T1 at a steel sheet surface, see\n00\nFig. 4.12, when the heat flux through the steel sheet is q_ 5 kW=m2 . The panel is\n25 mm thick and the temperature on the other side T 2 20 \u0004 C. Assume the\nconductivity of steel k 50 W/(m K).\n64 4 Boundary Conditions in Fire Protection Engineering\n\n## Fig. 4.12 Properties of\n\nwood panel in one\ndimension, Example 4.8\n\nSolution\n\u00040:025 \u0005\nT1 T2 50 \u0001 5000 20 2:5 22:5 \u0004 C.\nExample 4.8 Calculate the surface temperatures T1 and T2 of the wood panel when\nsurrounded by temperature T g1 100 \u0004 C and T g2 20 \u0004 C on the left and the right\nside, respectively. Consider heat transfer by convection assuming the heat transfer\ncoefficients h1 20 W=m2 K on the hot side and h2 5 W=m2 K on the cool\nside. The wood panel is assumed to be 25 mm thick and have a thermal conductivity\nkwood 0.1 W/(m K).\nSolution Calculate the total thermal resistance over a unit area Rtot\nRh Rk Rh 1=20 0:025=0:1 \u0005 1=5 0:50 K=W. According to Eq. 2.9\n100\u00010:025=0:11=520\u00011=20 1=20:025=0:1\nT1 0:5 92 and T 2 100\u00010:025=0:120\u0001\n0:5 . Alterna-\n00\ntively the heat flux q_ 0:5 160W=m at the boundaries two heat balance\n100\u000520 2\n\nequations can be established: (100 \u0005 T1)/0.05 160 and (T2 \u0005 20)/0.2 160\nyielding T1 92 \u0004 C and T2 52 \u0004 C.\n00\nExample 4.9 Calculate the net heat transfer by radiation q_ rad to a surface at a\ntemperature Ts and an emissivity of 0.9 when exposed to an incident radiation heat\nflux of 20 kW/m2. (a) Assume T s 20 \u0004 C, (b) assume T s 500 \u0004 C and (c) what is\nthe exposure black body radiation temperature Tr?\nSolution\nAccording to Eq. 1.15:\n00\nh i\n(a) q_ rad 0:9 20, 000 \u0005 20 2734 17, 610 \u0006 17:6 kW=m2\n00\nh i\n(b) q_ rad 0:9 20, 000 \u0005 500 2734 \u0005220 \u0006 \u00050:220 kW=m2\nh 00 i1=4\nq_\n(c) According to Eq. 4.1 T r inc 771 K 498 \u0004 C\nChapter 5\n\n## Heat transfer by thermal radiation is transfer of heat by electromagnetic waves. It is\n\ndifferent from conduction and convection as it requires no matter or medium to be\npresent. The radiative energy will pass perfectly through vacuum as well as clear\nair. While the conduction and convection depend on temperature differences to\napproximately the first power, the heat transfer by radiation depends on the differ-\nences of the individual body surface temperatures to the fourth power. Therefore\nthe radiation mode of heat transfer dominates over convection at high temperature\nlevels as in fires. Numerical applications of radiation heat transfer in FSE are\noutlined in Sect. 4.1.\nThe description below is mainly taken from . The surfaces are generally\nassumed to be grey, which means they absorb and emit radiation that is a fraction of\nblack body radiation in all directions and over all wavelengths. Hence the hemi-\nspherical absorptivity/emissivity of a surface is assumed to be independent of the\nnature of the incident radiation and of the spectral properties of, e.g. a fire.\nThe upper limit of the heat flux leaving a black body surface by radiation is\naccording to the StefanBoltzmann law\n00\nq_ bb, emi \u0002 T 4s 5:1\n\n## where is the StefanBoltzmann constant ( 5:670 \u0002 10\u00038 W=m2 K\u0004) and Ts is\n\nthe absolute surface temperature [K]. Figure 4.1 can be used to calculate the emitted\nheat by radiation from a black surface vs. temperature in Kelvin, K, according to\nEq. 5.1 or vs. temperature in degree Celsius, \u0005 C.\n00\nThe heat flux q_ emi leaving a real surface is, however, less than that of a black\nbody at the same temperature:\n00\nq_ emi s \u0002 \u0002 T 4s 5:2\n\n## Springer International Publishing Switzerland 2016 65\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_5\n66 5 Heat Transfer by Radiation\n\n00\nThe incident radiation q_ inc to a surface may originate from various sources.\nWhen it includes radiation irrespective of sources it is sometimes called irradiance.\n00\nOnly a fraction of the incident radiation q_ abs will be absorbed by a surface, i.e.\n00 00\nq_ abs s \u0002 q_ inc 5:3\n\nwhere is the absorptivity of the surface. The rest of the incident radiation is\n00\nreflected q_ ref or transmitted through the surface. The latter term is small for most\nmaterials and is neglected in the presentation below. Hence the reflected radiation\nheat flux becomes\n00 00\nq_ ref 1 \u0003 s \u0002 q_ inc 5:4\n\n## The net rate of heat flux to a surface by radiation then becomes:\n\n00 00 00\nq_ rad q_ abs \u0003 q_ emi 5:5\n\nor after inserting Eqs. 5.2 and 5.3 and given the Kirchhoffs identity s s , the heat\nflux to a surface by radiation becomes (Fig. 5.1)\n00\n\u0001 00 \u0003\nq_ rad s q_ inc \u0003 T 4s 5:6\n\n## The incident radiation or the irradiation on a surface is emitted by other surfaces\n\nand/or by surrounding masses of gas and in case of fire by flames and smoke layers.\nThe emissivity and absorptivity of gas masses and flames increase with depth and\nbecomes therefore more important in large scale fires than in, e.g. small scale\nexperiments, see Sect. 5.3. In real fires surfaces are exposed to radiation from a\nlarge number of sources, surfaces, flames, gas masses, etc., of different tempera-\ntures and emissivities and the incident radiation is in general very complicated to\nmodel. If absorption from any gases is neglected, and if the target surface is small\nand therefore the contributions of reflections and re-radiation are neglected, the\nincident radiation to the surface can be approximated as the sum of the contribu-\n00\ntions q_ inc, i from a number of external sources:\n00\nX 00\nq_ inc i\nq_ inc, i 5:7\n\n## When the source number i is a surface with a uniform temperature Ti the\n\ncontribution is\n00\nq_ inc, i i \u0002 Fi \u0002 \u0002 T 4i 5:8\n5 Heat Transfer by Radiation 67\n\n## Fig. 5.1 The heat transfer\n\ndepends on incident\ntemperature and emissivity\nof the surface\n\nwhere i is the emissivity of the ith source. Fi is the corresponding view factor as\ndefined in more detail in Sect. 5.2. Like the emissivity it always has values between\n0 and 1.\nThe incident radiation may also be written as a function of the black body\n00\nradiation temperature defined by the identity q_ inc \u0006 \u0002 T 4r (Eq. 1.17).\nThus Tr is a weighted average of the surrounding surface temperatures which can\nbe obtained by combining Eqs. 5.7 and 5.8 as\nrh\n\u0001X \u0003 i\nTr \u0006 \u0002 Fi \u0002 T i = 5:9\n4 4\ni i\n\nTr can also be defined as the temperature a surface will get which is in radiation\nequilibrium with the incident flux, i.e. no heat is transferred neither by convection\nnor by conduction from that surface. Compare with the concept of adiabatic surface\ntemperature, as described in Sect. 4.4, which is the surface temperature when the\n00 00\nnet radiation q_ rad is in equilibrium with the convection heat flux q_ con .\nThe net radiation heat flux is obtained by subtracting the emitted radiation\naccording to Eq. 5.2 from the absorbed:\n00\nhX \u0001 00\n\u0003 i\nq_ inc, i \u0003 \u0002 T 4s 5:10\n\n## where s is the emissivity/absorptivity of the target surface.\n\nThe surfaces emissivities of some materials are given in Table 5.1. In general the\nemissivity of all real/technical materials is in the range of 0.750.95 except shiny\nsteel where the emissivity can be considerably lower. It depends on the temperature\nof heat source and decreases in general with the heat source grey body temperature.\nTypically values of the absorptivity of plywood drop from 0.86 to 0.76 when the\nsource temperature increases from 674 to 1300 K . The corresponding value for\nradiation emitted from the sun (5777 K) is as low as 0.40. Eurocode 2 and\nEurocode 3 recommend 0.7 for concrete and steel, respectively. The choice of\nemissivity is primarily of importance when calculating temperature of fire-exposed\nbare steel structures. For lightweight insulating materials the surface temperature\nadapts quickly to the exposure conditions and therefore the heat transfer conditions,\nexpressed by the heat transfer coefficient, are negligible for the temperature\ndevelopment.\n68 5 Heat Transfer by Radiation\n\n## Table 5.1 Surface emissivity Material Emissivity,\n\nof some common materials\nConcrete 0.8a\nSteel 0.7a\nMin.wool 0.9\nPaint 0.9\nRed bricks 0.9\nWood 0.9\nSand 0.9\nRocks 0.9\nWater 0.96\nThe values are uncertain and should be taken as\nindicative\na\nFrom Eurocode\n\nShields\n\nWhen two infinite parallel plates as shown in Fig. 5.2a are considered, the radiation\nview factor is unity as all the heat emitted or reflected at one surface will incident on\nthe other. Some of that heat will be absorbed and some will be reflected back to the\nopposite surface. The net heat flux from surface one to two may be calculated as\n00 \u0005 \u0006\nq_ rad, 1\u00032 res T 41 \u0003 T 42 5:11\n\n1\nres 5:12\n1\n1 12 \u0003 1\n\n## Radiation exposure can be considerably reduced by a radiation shield. Figure 5.2b\n\nshows an example where a shield is mounted between two surfaces. The shield has\nno thermal resistance, i.e. its both sides have the same temperature. The radiation\nheat flux rate between surface 1 to the shield must equal the flux rate between the\n00 00 00 00 \u0005 \u0006\nshield and the surface 2, i.e. q_ rad q_ 1\u0003sh q_ sh\u00032 or q_ rad res, 1\u0003sh T 41 \u0003 T 4sh\n\u0005 4 \u0006\nres, sh\u00032 T sh \u0003 T 42 . Thus the shield temperature to the fourth power can be\nderived as\n\n## res, 1\u0003sh T41 res, sh\u00032 T42\n\nT4sh : 5:13\nres, 1\u0003sh res, sh\u00032\n\nand if all the four surface emissivities defined in Fig. 5.2b are equal to s then\n\n## Fig. 5.2 Radiation heat a b\n\ntransfer between two large\nparallel plates, without (a)\nshield\nT1 T2\n1 2\nT1 T2\n1 2\nTsh Tsh\n\nsh-1 sh-2\n\nT 41 T 42\nT 4sh 5:14\n2\n\n## and radiation heat flux between the surfaces 1 and 2 becomes\n\n00 s \u0005 \u0006\nq_ rad T 41 \u0003 T 42 =2 5:15\n2 \u0003 s\n\nEquation 5.14 implies that in the case of equal emissivities, the temperature of,\ne.g. a fire radiation shield is closer to the higher (fire) temperature than to the lower\n(ambient) temperature. Under the same conditions Eq. 5.15 shows the heat flux by\nradiation is reduced by 50 %. A reduction of the common emissivity will reduce\nheat transfer correspondingly although it will not change the temperature of the\nshield as according to Eq. 5.14 the temperature of the shield is independent of the\nemissivity.\nEquation 5.15 may be extended to problems involving multiple radiation shields\nwith all surface emissivities being equal to s. Then with N shields the heat flux\n00\n\n00 1 00 1 1 \u0005 \u0006\nq_ rad, N q_ rad, 0 \u00022 T 41 \u0003 T 42 5:16\nN 1 N 1 s \u0003 1\n\n00\nwhere q_ rad, 0 is the radiation heat flux with no shields (N 0) according to Eqs. 5.11\nand 5.12 with equal emissivities.\nThe corresponding formula for the flux between infinitely long concentric\ncylinders as indicated in Fig. 5.3 is\n70 5 Heat Transfer by Radiation\n\n## Fig. 5.3 The heat flux from\n\nthe inner cylinder (1) can be\ncalculated according to\nEq. 5.17\n\nT2 T1 r1\n2 1\n\nr2\n\n## Fig. 5.4 Uninsulated wooden stud wall\n\n\u0005 \u0006\n00 T 41 \u0003 T 42\nq_ rad, 1\u00032 \u0001 \u0003 5:17\n1\n1 r1\nr2 \u0002 1\n2 \u0003 1\n\nwhere r1 and r2 are the inner radii and 1 and 2 the corresponding surface\nemissivities. Notice that if r 1 \u0007 r 2 or more generally for a small object in a large\ncavity the heat flux from the inner object becomes independent of the outer surface\nemissivity, i.e.\n00 \u0005 \u0006\nq_ rad, 1\u00032 1 \u0002 T 41 \u0003 T 42 5:18\n\nExample 5.1 The inside surfaces of the boards of an uninsulated wooden stud wall\nas shown in Fig. 5.4 have the temperatures 300 and 100 \u0005 C, respectively. Calculate\nthe heat flux by radiation and convection between the board surfaces. The distance\nbetween the boards is 100 mm and between the studs 600 mm. The emissivity of the\nboard surfaces is 0.9. Assume one-dimensional heat flux.\n00\nSolution Equations 5.11 and 5.12 yield q_ rad 1 11 \u00031 \u0002 5:67 \u0002 10\u00038 \u0002\nh i 0:9 0:9\n\n300 2734 \u0003 100 2734 4104 W=m2 . For the convection heat transfer,\n00\nsee Example 6.7. Thus q_ c 644 W=m2 and the total heat flux\n00\nq_ tot 4104 644 W=m 4748 W=m . Notice in this case heat flux by\n2 2\n\n## Example 5.2 The radiation from a large flame at a temperature T f 1000 K is\n\nreduced by a metal radiation shield having no thermal conduction resistance\n(Fig. 5.5). The ambient gas temperature T 1 300 K. Calculate the temperature\n00\nTsh of the shield and the relative reduction of the radiation level q_ by the shield.\nAssume\n(a) s 1 and h 0.\n(b) s 0:5 and h 0.\n(c) s 1 and h 6 W/m2 K.\n\nSolution\n(a) All surfaces have an emissivity equal to unity. Thus according to Eq. 5.14\n\u0001p\nT sh 10004 2934 =2 842 K\n4\nand\n00\nq_ 5:67 10\u00038 \u0002 8424 28:5 W=m2 , i.e. a reduction by 50 %.\n\u0001p\u0002 00\n(b) T sh 10004 2934 =2 842 K and q_ 0:5 \u0002 28:5 14:3 W=m2 ,\n4\n\n## i.e. a reduction by 25 %. h\u0001p\n\nsh 10004 2934 \u0003 6=5:67 \u0002\n4\n(c) An iteration formula can be derived T i1\n\u0005 \u0006 00\n10\u00038 \u0002 T sh\ni\n\u0003 293 \u0004=2 yields T sh 818 K and q_ 5:67 \u0002 10\u00038 \u0002 8184\n25:4 W=m2 , i.e. a reduction by 45 %.\n\nExample 5.3 Gas flows through a long tube of r 1 40 mm diameter with an outer\nsurface emissivity 1 0:3. The tube is concentric with an outer insulation tube\n72 5 Heat Transfer by Radiation\n\n## with an inner diameter r 2 100 mm and an inner surface emissivity 2 0:8. In\n\ncase of fire the inner surface of the outer tube is expected to reach a temperature\n0\nT 2 1200 K. Calculate the heat transfer by radiation per metre length q_ rad to the\ninner tube when it has a temperature T 1 500 K.\nSolution According to Eq. 5.17 the heat flux to the inner surface is\n00 5:67\u000210\u00038 12004 \u00035004\nq_ rad 33, 212 W=m2 and the heat transfer by radiation per\n0:3100\u00020:8\u00031\n1 40 1\n\n0\nunit length becomes q_ rad 0:040 \u0002 \u0002 33, 212 4173 W=m.\n\n## 5.2 View Factors\n\nWhen calculating the rate of heat transfer by radiation between surfaces, a method\nis needed whereby the amount of heat being radiated in any direction can be\ncalculated. Therefore the concept view factor is introduced. The terms configura-\ntion factor, shape factor and angle factor are also used. The physical meaning of the\nview factor between two surfaces is the fraction of radiation leaving one surface\nthat arrives at the other directly. The symbol FA1 \u0003A2 is used to denote the view factor\nfrom a surface A1 to a surface A2. The symbol FdA1 \u0003A2 denotes the view factor from\nan incremental surface dA1 to a finite surface A2. View factors defined in this way\nare functions of size, geometry, position and orientation of the two surfaces. View\nfactors are between zero and unity, and the sum of the view factors of a surface\nis one.\nThus by definition the radiation leaving a surface A1 arriving at a surface A2 is\n00\nq_ inc, 1\u00032 F1\u00032 A1 q_ emi, 1 5:19\n\n## and similarly the radiation leaving a surface A2 arriving at a surface A1 is\n\n00\nq_ inc, 2\u00031 F2\u00031 A2 q_ emi, 2 5:20\n\nand\n\nA1\nF2\u00031 F1\u00032 5:22\nA2\n\n## In a more general way for any two surfaces i and j\n\n5.2 View Factors 73\n\nA3\n\n## F3-1,2 = F3-1+ F3-2\n\nA3F3-1,2 = A3F3-1 + A3F3-2\nA1,2F1,2-3 = A1F1-3 + A2F2-3\nA1 A2\n\n## Ai Fi\u0003j Aj Fj\u0003i 5:23\n\nAnother useful relation between view factors may be obtained by considering the\nsystem shown in Fig. 5.6. The view factor from the surface A3 to the combined\nsurface A1,2 is then\n\n## F3\u00031, 2 F3\u00031 F3\u00032 5:24\n\nThat is the total view factor is the sum of its parts. Eq. 5.24 could also be written as\n\n## A3 F3\u00031, 2 A1, 2 F1, 2\u00033\n\nA3 F3\u00031 A1 F1\u00033\nA3 F3\u00032 A2 F2\u00033 5:26\n\n## A1, 2 F1, 2\u00033 A1 F1\u00033 A2 F2\u00033 5:27\n\n74 5 Heat Transfer by Radiation\n\nThat means the total radiation arriving at the surface A3 is the sum of the radiations\nfrom the surface A1 and the surface A2.\nThe fact that the total view factor is the sum of its parts implies that the view factor\nF1\u00033 for the surfaces in Fig. 5.7 can be calculated from tabulated view factors as\n\n## F1\u00033 F1\u00032, 3 \u0003 F1\u00032 5:28\n\nBelow some elementary examples are given. A lot more information can be found\nin textbooks such as [1, 2, 11].\n\n## 5.2.1 View Factors Between Differential Elements\n\nThe view factor between two differential elements as shown in Fig. 5.8 can be\nobtained as\n\n## Fig. 5.7 Calculation of\n\nview factors by subtraction\naccording to Eq. 5.28\nA1\n\nA2 A3\n\nA2\n2\nNormal\n\nr\nNormal\n\nA1\n\nFig. 5.8 Differential area elements used in deriving view factors according to Eq. 5.29\n5.2 View Factors 75\n\ncos 1 cos 2\ndFd1\u0003d2 dA2 5:29\nr 2\n\nand correspondingly\ncos 1 cos 2\ndFd2\u0003d1 dA1 5:30\nr 2\n\nThe reciprocity relation as given by Eq. 5.21 can be used to derive the equation\ncos 1 cos 2\ndFd1\u0003d2 dA1 dFd2\u0003d1 dA2 dA1 dA2 5:31\nr 2\n\nNow the energy exchange between two black differential elements can be written as\n\u0005 \u0006 \u0005 \u0006\nd 2 q_ d1\u0003d2 T 41 \u0003 T 42 dFd1\u0003d2 dA1 T 41 \u0003 T 42 dFd2\u0003d1 dA2 5:32\n\nThen by inserting Eq. 5.29 or Eq. 5.30 the heat exchange between two differential\nelements becomes\n\u0005 \u0006 cos 1 cos 2\nd2 q_ d1\u0003d2 T 41 \u0003 T 42 dA2 dA1 5:33\nr 2\n\nFig. 5.9 The view factor of a plane element dA1 to a plane parallel rectangle vs. the relative\ndistances X a/c and Y b/c as defined in row 1 of Table 5.2. The normal to the element passes\nthrough the corner of the rectangle\n76 5 Heat Transfer by Radiation\n\nFig. 5.10 The view factor of a plane element dA1 to a plane rectangle perpendicular to the element\nvs. the relative distances X a/c and Y b/c as defined in row 2 of Table 5.2. The normal to the\nelement passes through the corner of the rectangle\n\nThe view factors between the entire surfaces A1 and A2 of Fig. 5.8 can be obtained\nby integration as\n\n1 cos 1 cos 2\nF1\u00032 dA1 dA2 5:34\nA1 A1 A2 r 2\n\nand correspondingly\n\n1 cos 1 cos 2\nF2\u00031 dA1 dA2 5:35\nA2 A1 A2 r 2\n\n## 5.2.2 View Factors Between a Differential Element\n\nand a Finite Area\n\nThe heat radiated from a differential (very small) area dA1 which reaches a surface\nA2 is by the definition of the view factor\nTable 5.2 Examples of formulas for calculating the view factors between differential elements and finite areas\n1 b Plane element dA1 to plane parallel rectangle. Normal to element passes through corner of\nX ac\nY bc\nA2 a \u0007 \b\n5.2 View Factors\n\n1 X\np Y Y X\nFd1\u00032 2 tan\u00031 p p tan\u00031 p\n1X2 1X2 1Y 2 1Y 2\n\ndA1\n2 Plane element dA1 to plane rectangle in plane perpendicular to element. Normal at\nb rectangle corner passes through element, see\nh \u0005\u0006 \u0001 \u0003ialso Fig. 5.10\n1 c b\nFd1\u00032 2 2 2\ntan\u00031 bc \u0003 p\na c\ntan\u00031 p\na c 2 2\n\nA2\na\n\nc dA1\n3 dA1 Plane element dA1 to circular disk in plane parallel to element through centre of disk, see\nalso Fig. 5.12\n2\nFd1\u00032 h2rr2 h 12\nr 1\nh\n\nA2\nr\n\n(continued)\n77\nTable 5.2 (continued)\n78\n\n4 Plane element dA1to right circ. cylinder of finite length l and radius r; normal to element\npasses through one end of cylinder and is perpendicular to cylinder axis\nh q qi\n1 XH\u00031 1 \u00031 H\u00031 l h\nFd1\u00032 H tan\u00031 p\nL\n2\np tan\u00031\nXY\nL HX\u00032H Y H1 \u0003 Htan H1 where L r H r\nH \u00031\nX 1 H 2 L2 Y 1 \u0003 H 2 L2\nA2\n\nr\nh\n\ndA1\n5.2 View Factors 79\n\n00\nq_ inc, d1\u00032 dA1 Fd1\u00032 q_ emi, 1 dA1 Fd1\u00032 \u0002 2 \u0002 \u0002 T 41 5:36\n\nwhere 2 is the emissivity of the emitting surface. (Any reflected radiation is here\nneglected). Examples of such configurations can be seen in Table 5.2. Similarly the\nheat radiated by A2 and reaching dA1 is\n00\nq_ inc, 2\u0003d1 A2 F2\u0003d1 qemi, 2 A2 F2\u0003d1 \u0002 2 \u0002 \u0002 T 42 5:37\n\n## The reciprocity relation according to Eq. 5.21 then yields\n\n00\nq_ inc, 2\u0003d1 dA1 Fd1\u00032 qemi, 2 5:38\n\nand the incident radiation flux to the differential area surface becomes\n00 00\nq_ inc, 2\u0003d1 Fd1\u00032 \u0002 qemi, 2 Fd1\u00032 \u0002 2 \u0002 \u0002 T 42 5:39\n\nThis is the most commonly applied formula version in FSE as it can be used to\nestimate the incident radiant flux at point where it is expected to be most severe.\nWhen several finite surfaces from 2 to n are radiating on an infinite area dA1, the\ntotal incident radiation can be written as\n00\nXn\nq_ inc, 2\u0003n\u0003d1 i2\nFd1\u0003i \u0002 i \u0002 T 4i 5:40\n\nIn principle when calculating the total incident radiation to a surface the incident\nradiation from all angles must be included. Observe that the sum of the view factors\nis unity. Usually, however, only the contributions from the hot areas such as flame\nsurfaces need be considered as the contributions from, for example, surface at\nambient temperature are negligible.\nWhen several surfaces are involved the view factor may be obtained by adding\nup the contributions from the individual surfaces according to Eq. 5.27. In the case\nshown in Fig. 5.11 the view factor FdA1 \u0003A2\u00035 between the differential area dA1 and\nthe entire finite area A2\u00035 may be calculated as\n\n## Fd1\u00032\u00035 Fd1\u00032 Fd1\u00033 Fd1\u00034 Fd1\u00035 5:41\n\nView factors of various configurations can be found in textbooks such as [1, 2] and\nparticularly in .\nTable 5.2 shows how to calculate view factors for some elementary cases useful\nin FSE.\nCorresponding diagrams of the view factors defined in rows 13 of Table 5.2 are\nshown Fig. 5.12.\nExample 5.4 An un-insulated steel door leaf becomes uniformly heated to a\ntemperature of 500 \u0005 C during a fire. Calculate the maximum incident radiation\n80 5 Heat Transfer by Radiation\n\n## Fig. 5.11 The view factor\n\ncan be obtained by\nsumming up the\ncontributions of several\nA2\nareas as given by Eq. 5.41\nA3\n\nA5\nA4\ndA1\n\n1.0\n\n0.9\n\n0.8\n\n0.7\nView factor Fd1-2\n\n0.6\n\n0.5\n\n0.4\n\n0.3\n\n0.2\n\n0.1\n\n0.0\n0 0.5 1 1.5 2 2.5 3\nRelative distance h/r\n\nFig. 5.12 The view factor of plane element dA1 to a circular disk in a plane parallel to the element\nvs. the relative distance h/r through the centre of the disk as defined in row 3 of Table 5.2\n\n00\nq_ inc, max to a parallel surface 1 m from the door leaf with dimensions 0.9 m by 2.1 m\nand an emissivity of 0.9.\nSolution The highest incident radiation will be perpendicular to the centre of the\ndoor leaf. Then Eq. 5.39 applies. When calculating the view factor the door is\ndivided into four equal areas and the view factor is obtained as the sum of the four\ncontributions. Then according to the first case in Table 5.2, X 2:1=2\n1 1:05 and\nY 0:9=2\n1 0:45 and the total view factor becomes FdA1 \u0003A2\n5.2 View Factors 81\n\n\u0007\n4 \u0002 2\n1 p\n1:05\ntan\u00031 p\n0:45\np\n0:45\ntan\u00031 p\n1:05\n4 \u0002 0:085 0:34. The\n2 2 2 2\n11:05 11:05 10:45 10:45\nview factor for one-quarter of the door leaf (0.085) can alternatively be obtained\n00\nfrom Fig. 5.9. The maximum incident radiation q_ inc, max 0:34 \u0002 0:9 \u0002 5:67 \u0002 10\u00038\n\u00027734 6200 W=m2 (corresponding to a black body radiation temperature of\n574 K 301 \u0005 C).\n\n## 5.2.3 View Factors Between Two Finite Areas\n\nIn analogy with view factors between a differential element and a finite area\n(Eq. 5.36) may the heat flow (with units [W]) from one finite area to another be\ncalculated as\n00\nq_ inc, 1\u00032 A1 F1\u00032 qemi, 1 A1 F1\u00032 1 \u0002 \u0002 T 41 5:42\n\n## The net exchange from A1 to A2 assuming black isothermal surfaces ( 1) is\n\n\u0005 \u0006\nq_ 2\\$1 A1 F1\u00032 T 41 \u0003 T 42 5:43\n\nTable 5.3 shows two examples on how to calculate view factors between finite\nsurfaces, two parallel circular disks with centres along the same normal and two\ninfinitely long plates of unequal widths having a common edge at an angle of 90 \u0005 C\nto each other.\nExample 5.5 Two small surfaces 1 and 2 are oriented perpendicularly to each\nother as shown in Fig. 5.13 and have surfaces 0.1 and 0.2 m2, respectively, and\n\nTable 5.3 Examples of formulas for calculating the view factors between finite areas\nFinite, coaxial disks\nr2 A2\nR1 rh1 R2 rh2\n1R22\nX 1 R21\n\" r#\nh \u0001 \u00032\nF1\u00032 1\n2 X \u0003 X \u0003 4 R1\n2 R2\n\nA1\nr1\n\n## Two infinitely long plates of unequal widths h and\n\nw having a common edge at an angle of 90 \u0005 C to each\nh A2 other\nH wh\nh pi\nA1 F1\u00032 12 1 H \u0003 1 H 2\nw\n82 5 Heat Transfer by Radiation\n\n## Fig. 5.13 Two small\n\n2\nsurfaces in the same plane\noriented perpendicularly\n3m\n\n1 4m\n\n80\na b\n3\n2\n65\n180\n45 1\n\nFig. 5.14 Sketch of cone calorimeter for calculation of view factors to the specimen. The surfaces\nare identified by the numbers in circles. (a) Measures in mm. (b) Numbering of surfaces\n\ntemperatures 850 and 400 K, respectively. How much heat is transferred between\nthe two surfaces?\nSolution The distance between the surfaces becomes 5 m and thus cos 1 0:6\nand cos 2 0:8. Then Eq. 5.33 yields q_ d1\u0003d2 5:67 \u0002 10\u00038\n\u0005 \u0006\n8504 \u0003 4004 0:6\u00020:8\n52\n\u0002 0:1 \u0002 0:2 3:44 W.\nExample 5.6 In the Cone Calorimeter the radiant panel has the shape of a\ntruncated circular cone as shown in Fig. 5.14. Assuming the panel has a uniform\ntemperature of 700 \u0005 C and an emissivity equal unity, and neglecting the radiation\nfrom outside the cone, calculate\n(a) The maximum incident radiation to a body below the panel.\n(b) The mean incident radiation to body below the panel. Assume the body is\ncircular with a diameter of 100 mm.\nGuidance: Surfaces may be numbered as Fig. 5.14b, i.e. specimen surface is\n1, cone heater 2 and the opening 3.\nSolution Equation 5.28 yields the view factor from the cone to the specimen to be\n2 2\n(a) Fd1\u00032 Fd1\u00032, 3 \u0003 Fd1\u00033 40290902 \u0003 10540\n2\n402\n0:84 \u0003 0:13 0:71 and the\nincident flux becomes qinc, d1 0:71 \u0002 5:67 \u0002 10\u00038 \u0002 700 2734\n36 , 000 W=m2 .\n(b) F1\u00032 F1\u00032, 3 \u0003 F1\u00033\nF1\u00032, 3 : From Table 5.3 with R1 50\n40 and R2 40 yielding X 4:88 and\n90\n\n## F2, 3\u00031 0:79\n\n5.3 Radiation from Flames and Smoke 83\n\nF1\u00033 : R1 105\n50\nand R2 105\n40\nyielding X 6:05 and F1\u00033 0:11. Thus F1\u00032\n0:79 \u0003 0:11 0:68 and Eq. 5.42 qinc, 1 0:68 \u0002 5:67 \u0002 10\u00038 \u0002 700 2734\n34, 600 W=m2 . Comment: The mean incident flux is only by 4 % less than\nthe maximum.\n\n## 5.3 Radiation from Flames and Smoke\n\nIt is flames, smoke particles and combustion products that absorb and emit heat\nradiation in fires. It is generally assumed continuous over all wavelengths when\ncalculating temperature although some gas species only absorb and emits at certain\nwavelength intervals. In general simple gas molecules such as oxygen O2 and\nnitrogen N2 do not absorb or emit heat radiation while molecules such as carbon\nmonoxide CO2 and water H2O do depending on wavelength. Therefore the heat\nabsorbed or emitted by clean air is negligible.\nOverall the absorption fl and the emission fl of a flame or smoke layer depend\non the absorption or emission coefficient K and the mean beam length Le.\nAccording to the Kirchhoffs law the absorptivity and the emissivity are equal.\nThen the Beers law is a useful tool in approximate radiation analyses [1, 2]. Thus\n\nfl fl 1 \u0003 e\u0003K\u0002Le 5:44\n\nFor gas species K depends on wavelength, but as the bulk of the radiation from\nflames and smoke layers emanates from soot particles, it is treated as independent of\nwavelength, i.e. K is treated as an effective absorption/emission coefficient.\nThe emitted heat from a flame may accordingly be written as\n00 \u0005 \u0006\nq_ emi, fl 1 \u0003 e\u0003K\u0002Le \u0002 T 4fl 5:45\n\nwhere Tfl is the flame temperature (assumed uniform). A few empirical and not very\nreliable data for the effective absorption/emission coefficient, K, are available in the\nliterature. Some values are shown in Table 5.4.\nThe mean beam length giving reasonable approximations may be obtained from\n\nV\nLe 3:6 5:46\nA\n\nwhere V is the total volume of the gas and A the total surface area. For a volume\nbetween two infinite planes at a distance L a mean beam length Le can be obtained as\n\nLe 1:8 L 5:47\n84 5 Heat Transfer by Radiation\n\nTable 5.4 The effective absorption/emission coefficient K for various fuels, from \nFuel K (m\u00031) Reference\nDiesel oil 0.43 Sato and Kunimoto\nPolymethylmethacrylate 0.5 Yuen and Tien\nPolystyrene 1.2 Yuen and Tien\nWood cribs 0.8 Hagglund and Persson\nWood cribs 0.51 Beyris et al.\nAssorted furniture 1.1 Fang\n\nObserve that a flame or a smoke layer absorbs radiant heat depending on the\nabsorptivity according to Eq. 5.44. This is illustrated by Example 5.9.\n00\nExample 5.7 What is the emitted radiation heat flux q_ from an oil fire where\nK 0.4 m\u00031.\nAssume a beam length L 1 m and a flame temperature of\nT fl 1073 K 800 \u0005 C.\nSolution \u0005 \u0006\nq_ } 1 \u0003 e\u0003K\u0002L \u0002 \u0002 T fl 25, 000 W=m2\n4\n\nExample 5.8 The surface temperature Ts of a stove is 500 \u0005 C and has an emissivity\nof s 1.0. Near the stove is a wooden wall with a surface emissivity of w 0.8.\nThe air temperature in the space between the stove and the wall T g 40 \u0005 C and the\nconvection heat transfer coefficient is hc 10 W=m2 K. Assume the surfaces of\nthe stove and the wall being parallel and infinitely large.\n(a) What is the net heat transfer by radiation to the wall surface at the ignition\ntemperature assumed to be Tig 300?\n(b) What is the maximum temperature the wall can obtain at equilibrium, when\nthe surface does not absorb any more heat and is assumed to be a perfect\ninsulator (i.e. the adiabatic surface temperature).\n\nSolution h i\n00\n(a) Equations 5.11 and 5.12 yield q_ rad r \u0002 T s 2734 \u0003 T w 2734 and\n00 \u0005 \u0006\nr 1 1 \u000311 1 , and q_ rad 0:8 \u0002 5:67 \u0002 10\u00038 \u0002 7734 \u0003 5734\ns w 1 1 \u000310:8\n1 0:8\n\n## 11:3 \u0002 103 W=m2 . h i \u0005 \u0006\n\n(b) The surface heat balance: r T s 2734 \u0003 T w 2734 h T g \u0003 T w\nh i\n0:8 \u0002 5:67 \u0002 10\u00038 \u0002 7734 \u0003 T w 2734 10 \u0002 40 \u0003 T w 0.\n5.3 Radiation from Flames and Smoke 85\n\n## Fig. 5.15 One-dimensional\n\nmodel of the heat exchange\nexposed to a flame of\nlimited thickness\n\n\u0005 \u0006\nw 273 773 \u0003 10 \u0002 40 \u0003 T w =\nT i1 i\nAn iteration scheme yields:\n\u0005 \u0006 1= \u0005\n0:8 \u0002 5:67 \u0002 10\u00038 \u0004 4 C . Assuming T 0w 300 \u0005 C yields T 1w 467 \u0005 C and subse-\nquently T 2w 443 \u0005 C and T 3w 446 \u0005 C which is an acceptable solution.\nExample 5.9 A specimen as shown in Fig. 5.15 is suddenly exposed to a propane\nflame which is assumed to have a thickness of fl 0:2 m at the point being\nanalysed. Assume the effective flame absorption coefficient K 0:5 m\u00031 , the\nflame temperature Tfl 800 \u0005 C, the ambient and initial temperatures\nT1 Ti 20 \u0005 C, the convective heat transfer coefficient hc 10 W/m2, the surface\nemissivity s 0.9 and the ignition temperature of the specimen Tig 350 \u0005 C\n00\n(a) Calculate the incident radiant heat flux qinc to the specimen surface.\n(b) Calculate the total heat flux by radiation and convection to specimen surface at\nthe start of the test and at ignition, i.e. when the specimen surface temperature\nTs 20 \u0005 C and Ts 350 \u0005 C, respectively.\n(c) Comment on the magnitude of the contributions to the heat transfer by\n(d) Calculate the adiabatic surface temperature TAST at the specimen surface.\n(e) Repeat item (b), i.e. calculate the total heat flux to specimen at the start of the\ntest and at ignition using TAST.\n\nSolution\n(a) Incident heat flux from the flame, see Fig. 5.15, can be written as:\n00 00 00 \u0005 \u0006\nq_ inc q_ inc, fl q_ inc, 1 fl \u0002 \u0002 T 4fl 1 \u0003 fl \u0002 T 41 . The emissivity of a\nflame or smoke layer, fl, may be calculated according to Eq. 5.45 as: fl 1\n\u0003e\u00030:5\u00021:8\u00020:2 \b 0:18: The incident h heat flux from the flame and the surrounding\n00\nwill be: q_ inc 5:67 \u0002 10\u00038 0:18 800 2734 1 \u0003 0:18 \u0002 20 2734 \u0004\n13, 871 W=m2 \b 13:9 kW=m2 .\n86 5 Heat Transfer by Radiation\n\n(b) The total heat flux to the specimen is the sum of the heat flux by radiation and\n00 \u0005 00 \u0006 \u0005 \u0006\nconvection: q_ tot s q_ inc \u0003 T 4s hc T fl \u0003 T s . At the initial temperature:\n00\n\u0001 \u0003\nqtot, 20 0:9 13, 871 \u0003 5:67 10\u00038 20 2734 10 800 \u0003 20 12, 108\n00\n7800 19, 908 \b 19:9 \u0002 103 W=m2 and at ignition temperature qtot, 350\n\u0001 \u0003\n0:9 13, 871 \u0003 5:67 10\u00038 350 2734 10 800 \u0003 350 4796 4500\n9:3 \u0002 103 W=m2 .\n(c) When the surface of the specimen is at the initial temperature the contributions\nby radiation and convection is in the same order of magnitude in the studied\ncase. When the surface is at ignition temperature, the surface is heated by\n00 00\n(d) By definition qinc T 4r and thus based on the qinc calculated above\nT r 703 K 430 \u0005 C. TAST can be obtained by solving the fourth degree\nequation 4.22. Thus T AST 745 K 472 \u0005 C.\n00 \u0005 \u0006\n(e) Equation 4.31 yields at the initial temperature q_ tot, 20 0:9 7034 \u0003 2934\n10 703 \u0003 293 15, 343 4520 19:9 \u0002 103 W=m2 , and at the ignition\ntemperature\n00 \u0005 \u0006\nq_ tot, 350 0:9 \u0002 7454 \u0003 6234 10 \u0002 745 \u0003 623 9:3 \u0002 103 W=m2 . Note\nthat these alternatively calculated heat flux values are equal to those calculated\nunder item (b).\n\nExample 5.10 A 6 m high, 4 m wide and 0.5 m thick flame is covering a well-\ninsulated facade with a surface emissivity s 0:9. In the centre-line of the flame\n2 m outside the facade surface is a small square section steel column. Assume the\nflame temperature equal T fl 800 \u0005 C, the flame absorption/emission coefficient\n0.3 m\u00031 and the convective heat transfer coefficient h 35 W/(m2 K), see\nFig. 5.16.\nCalculate under state conditions\n(a) The temperature of the facade surface behind the flame.\n(b) The emitted radiant flux from the flame surface towards the column.\n(c) The maximum incident radiation to the four sides of the column.\n\nSolution \u0005 \u0006\n(a) The emissivity according to Eqs. 5.44 and 5.47 fl 1 \u0003 e\u00031:8*2*0:3 0:66.\nThen the incident radiation to the facade surface becomes 0:66 \u0002 5:67 \u0002 10\u00038\n800 2734 49:6 \u0002 103 W=m2 (corresponding to a radiation temperature\nof about T r 700 \u0005 C) and the adiabatic surface temperature TAST can then be\n\nobtained from (Eq. 4.21) as 0:9 \u0002 49:6 \u0002 103 \u0003 5:67 \u0002 10\u00038 \u0002 T 4AST\n35 \u0002 800 273 \u0003 T AST 0 which yields T AST 988 K 715 \u0005 C after\ntwo iterations according to Eq. 4.25.\n5.3 Radiation from Flames and Smoke 87\n\ncolumn surrounded by air at\nambient temperature\n\n00\n(b) Contributions from the flame plus the facade surface q_ emi 0:66 \u0002 5:67 \u0002 10\u00038\n800 2734 1 \u0003 0:66 \u0002 5:67 \u0002 10\u00038 \u0002 9884 68:0 \u0002 103 W=m2 .\n(c) Column side facing the facade, 1: Divide into four equal contributions\naccording to Table 5.2 first row or the diagram in Fig. 5.9 with X 2/2 1\nand Y 3/2 1.5 yields Fd1-2 0.09 and the total view factor\nF 4\u00020.09 0.36 and thus incident flux 0:36 \u0002 68:0 \u0002 103 24:5 \u0002 103 W=m2 .\nSides of the column, 2 and 3: The sides will only be exposed to half the\nflame. Figure 5.10 with X 2/2 1 and Y 3/2 1.5 yields Fd1-2 0.065 and\nF 2\u00020.065 0.15 and thus incident flux 0:15 \u0002 68:0 \u0002 103 10:2 \u0002 103 W=m2 .\nColumn side facing away from the facade, 4: This side does not face the\nfacade and will only get an incident radiation corresponding to the ambient\ntemperature, i.e. 5:67 \u0002 10\u00038 20 2734 0:42 \u0002 103 W=m2 .\nChapter 6\nHeat Transfer by Convection\n\nIn previous chapters heat transfer by convection or just convection was treated only\nto the extent that it provides a linear boundary condition of the 3rd kind for\nconduction problems when the heat transfer coefficient is assumed constant. In\nthis chapter the physical phenomenon of convection is described in more detail.\nHeat is transferred by convection from a fluid to a surface of a solid when they\nhave different temperatures. Here it is shown how the convection can be calculated\nand in particular how the convection heat transfer coefficient, denoted h or some-\ntimes for clarity hc, can be estimated in various situations relevant for FSE\nproblems.\nWhen the gas or liquid flow is induced by a fan, etc. it is called forced\nconvection, and when it is induced by temperature differences between a surface\nand the adjacent gases it is called natural convection or free convection. In the latter\ncase the surface heats or cools the fluid which then due to buoyancy moves upwards\nor downwards. Both natural and forced convection can be laminar or turbulent\ndepending on fluid properties and velocity, and on size and shape of exposed\nsurfaces. Various modes occur in fires and are relevant in FSE.\nThe heat transfer by convection depends in any case on the temperature differ-\nence between the fluid and the surface. Usually in FSE it is assumed directly\nproportional to the difference of the two temperatures according to the Newtons\nlaw of cooling, see Sect. 4.2. This is linear boundary condition which facilitates\ncalculations without jeopardizing accuracy as heat transfer by radiation at elevated\ntemperatures dominates over the transfer by convection.\nSection 6.1 gives expressions on how air and water conductivity and viscosity\nvary with temperature. Viscosity is the measure of a fluids resistance to flow and\nhas a decisive influences convective heat transfer properties.\nIn Sects. 6.2 and 6.3 general formulas are presented for various fluids, config-\nurations and flow conditions followed by some useful approximate formulas and\ndiagrams applicable specifically to air which considerably facilitates calculations of\nFSE problems.\n\n## Springer International Publishing Switzerland 2016 89\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_6\n90 6 Heat Transfer by Convection\n\n## 6.1 Heat Transfer Properties of Air and Water\n\nThe properties of several fluids are tabulated in textbooks such as for various\ntemperature levels. Special attention is given to air as it usually in flow calculations\nis assumed to have the same properties as smoke and fire gases.\nFor air the conductivity kair can be approximated as \n\n## and the kinematic viscosity air as\n\n\u0001 \u0003\nair 1:10 \u0002 10\u00039 T 1:68 m2 =s 6:2\n\nrespectively, where T is temperature in Kelvin. See also Figs. 6.1 and 6.2 for\ngraphical presentations. The simple approximations are used in this book for\nobtaining close form expressions for among other things convection heat transfer\ncoefficients.\nThe Prandtl number does not vary much with temperature and may in most cases\nbe assumed constant, Pr air 0:7.\nThe thermal conductivity water kw can be approximated as\n\n## kw \u00030:575 6:40 \u0002 10\u00033 \u0002 T \u0003 8:2 \u0002 10\u00036 \u0002 T 2 W=m K\u0004 6:3\n\n0.09\n\n0.08\n\n0.07\nConduc\u0002vity [W/(m.K)]\n\n0.06\n\n0.05\n\n0.04\n\n0.03\n\n0.02\n\n0.01\n\n0\n300 400 500 600 700 800 900 1000 1100 1200 1300\nTemperature [K]\n\nFig. 6.1 Thermal conductivity kair of air vs. absolute temperature. See also Eq. 6.1\n6.2 Forced Convection 91\n\n200\n\n180\n\n160\nKinema\u0002c viscosity [mm2/s]\n\n140\n\n120\n\n100\n\n80\n\n60\n\n40\n\n20\n\n0\n300 400 500 600 700 800 900 1000 1100 1200 1300\nTemperature [K]\n\nFig. 6.2 Kinematic viscosity air of air vs. absolute temperature. See also Eq. 6.2\n\nwhere T is the water temperature in Kelvin. Table 6.1 shows also other properties of\nwater at various temperatures relevant for thermal calculations.\nLatent heat of vaporization of water relevant when calculating temperature in\nmoist solids aw 2260 kJ=kg.\n\n## 6.2 Forced Convection\n\nThe heat transfer coefficient or the thermal resistance between a gas or fluid and a\nsolid surface is controlled within the so-called boundary layer. The thermal resis-\ntance of this layer and thereby the amount of heat being transferred depends on the\nthickness of the layer and the conductivity of the fluid. The thickness of the\nboundary layer in turn depends on the velocity of the fluid.\n\n## 6.2.1 On Flat Plates\n\nFigure 6.3 shows an edge where a boundary layer is developed in a forced flow over\na flat surface. Outside the boundary layer the flow is undisturbed and has a uniform\nvelocity u1. It then decreases gradually closer to the surface and very near the\nsurface it vanishes. As indicated in Fig. 6.3 the boundary layer thickness grows\nwith the distance from the leading edge.\n92 6 Heat Transfer by Convection\n\n## Table 6.1 Properties of water\n\nTemperature Conductivity Density Specific heat Kinematic Prandtls\n(oC) (W/(m K) (kg/m3) (kJ/(kg K)) viscosity (m2/s) no. (\u0003)\n5 0.57 1000 4.20 1.79 \u0002 10\u00036 13.67\n15 0.59 999 4.19 1.30 \u0002 10\u00036 9.47\n25 0.60 997 4.18 1.00 \u0002 10\u00036 7.01\n35 0.62 994 4.18 0.80 \u0002 10\u00036 5.43\n45 0.63 990 4.18 0.66 \u0002 10\u00036 4.34\n55 0.64 986 4.18 0.55 \u0002 10\u00036 3.56\n65 0.65 980 4.19 0.47 \u0002 10\u00036 2.99\n75 0.66 975 4.19 0.41 \u0002 10\u00036 2.56\n85 0.67 968 4.20 0.37 \u0002 10\u00036 2.23\n95 0.67 962 4.21 0.33 \u0002 10\u00036 1.96\nFrom the Engineering ToolBox, www.EngineeringToolBox.com, except the conductivity which is\naccording to Eq. 6.3.\n\n## Fig. 6.3 Boundary layer\n\nwith a thickness\ndeveloping after an edge of\na flat surface\n\nThe heat transfer resistance Rh can now be calculated as heat resistance between\nthe fluid and the solid surface in a similar way as for thermal conduction in solids:\n\nRh 6:4\nkf\n\nwhere kf is the thermal conductivity of the fluid and the boundary layer thickness.\nThe subscript f indicates that the parameter values shall be at the film temperature\nwhich is the average of the surface Ts and the free stream fluid T 1 temperatures, i.e.\n\nT s T 1\nTf 6:5\n2\n\nThe heat transfer coefficient can be obtained as the inverse of the heat transfer\nresistance according to Eq. 2.14. Thus\n6.2 Forced Convection 93\n\nkf\nhf 6:6\n\nThe heat transfer coefficient is often expressed by the Nusselt number Nu, a\nnon-dimensional relation between the boundary layer thickness and a character-\nistic length x of the exposed surface. With obtained from Eq. 6.6\n\nx h\u0002x\nNuf 6:7\nkf\n\nIn the case of a plane surface as shown in Fig. 6.3, the characteristic length x is\nthe distance from the edge. Near the edge, small values of x, the flow is laminar and\nfurther away it is turbulent. The Nusselt number at a distance x has been derived\nanalytically as (see, e.g. ):\n1=\n1=3\nNuxf 0:332 Pr f \u0002 Ref x 2\n6:8\n\nwhere Prf is the Prandtl number which relates the kinematic viscosity and thermal\ndiffusivity of the fluid. Refx is the Reynolds number which indicates whether the\nflow conditions are laminar or turbulent. It is a non-dimensional grouping of\nparameters defined as\n\nu1 \u0002 x f \u0002 u1 \u0002 x\nRef x 6:9\nf f\n\nwhere is the dynamic viscosity of the fluid. The kinematic viscosity is the\ndynamic viscosity divided by the density, i.e.:\n\n6:10\n\nBy integration along the surface the mean Nusselt number Nuf can be obtained as\n1=\n1=3\nNuf 2 \u0002 Nuxf 0:664 \u0002 Pr f \u0002 Ref x 2\n6:11\n\nFor constant fluid properties the heat transfer coefficient at a distance x can now\nbe calculated by combining Eqs. 6.7 and 6.8:\n\nNuxf \u0002 kf 1\nhf x 0:332 \u0002 kf \u0002 Pr 3f \u0002 f \u00031=2 \u0002 u1 1=2 \u0002 x\u00031=2 6:12\nx\n\nThe mean heat transfer coefficient from the edge x 0 to x is twice this value:\n94 6 Heat Transfer by Convection\n\n1\nhf x 2 hf x 0:664 kf \u0002 Pr 3f \u0002 f \u00031=2 u1 \u00021=2 x\u00031=2 6:13\n\n## An important observation is that the heat transfer coefficient decreases with\n\ndimensions. Smaller dimensions mean larger convective heat transfer coefficients.\n\n## To facilitate analyses of FSE problems material properties of air is now assumed by\n\ninserting Eqs. 6.1 and 6.2 into Eq. 6.12 and assuming Pr 0.7. The heat transfer\ncoefficient can then be obtained as a function of film temperature, air velocity and\ndistance from the edge as:\n\u0001 \u0004 \u0005\u0003\nhf x 2:59 \u0002 T \u00030:045\nf \u0002 u1 1=2 \u0002 x\u00031=2 W= m2 K 6:14\n\nNote that the influence of the film temperature level is rather weak. The mean\nheat transfer coefficient will be twice that value:\n\u0001 \u0004 \u0005\u0003\nhf x 2 \u0002 hf x 5:17 \u0002 T \u00030:045\nf \u0002 u1 1=2 \u0002 x\u00031=2 W= m2 K 6:15\n\n## Example 6.1 A 200-mm-wide steel plate having a uniform temperature of 500 K is\n\nexposed to an air stream with a temperature of 1200 K and a velocity of 2.0 m/s.\nCalculate the mean heat flux by convection q_ con to the steel surface.\nSolution Insert the parameters in Eq. 6.15. The film temperature T f 850 K and\nthen hf x 5:17 \u0002 850\u00030:045 \u0002 22 \u0002 0:2\u00032 W=m2 K 12:0 W=m2 K, and the mean\n1 1\n\n00\nheat flux q_ con 12:0 \u0002 1200 \u0003 500 8370 W=m2 .\nExample 6.2 A heat flux meter measures the heat flux to its cooled sensor surface,\nsee Sect. 9.2. Assuming the sensor surface has a diameter of 10 mm, calculate the\nheat transfer coefficient by convection h. The sensor surface temperature is esti-\n00\nmated to be uniform and equal to 30 \u0005 C. Estimate the heat flux by convection q_ con to\n\u0005\nthe sensor surface if the gas temperature is 400 C and the gas velocity is 2 m/s.\n\u0004 \u0005\u00030:045 1=2\nSolution Apply Eq. 6.15 h 5:17 \u0002 40030\n2 273 \u0002 2 \u0002 0:01\u00031=2\n00\n55 W=m , and the heat flux to the sensor surface by convection becomes q_ con\n2\n\n## 55 \u0002 400 \u0003 30 20300 W=m 20:3 kW=m .\n\n2 2\n6.2 Forced Convection 95\n\n## 6.2.2 Across Cylinders\n\nIn the case of flow across a cylinder an empirical expression for the Nusselt number\nhas been derived, see, e.g. [1, 2]:\n\nh\u0002d \u0004 \u0005n 1\nNuf C \u0002 Redf \u0002 Pr 3f 6:16\nkf\n\n## For a cylinder with a diameter d and the Reynolds number becomes\n\nu1 \u0002 d u1 \u0002 d 1\nRedf \u00039\n\u0006 \u0002 6:17\nf 1:13 \u0002 10 \u0002 T f\n1:67\n1:13 \u0002 10\u00039\n\nu1 \u0002 d\n6:18\nT 1:67\nf\n\n## is introduced to simplify calculations.\n\nThe convection heat transfer coefficient can be derived from Eq. 6.16 as\n\u0006 \u0007\nkf u1 \u0002 d n 1\nh \u0002C\u0002 \u0002 Pr 3f 6:19\nd f\n\n## 6.2.2.1 Heat Transfer in Air\n\nFor air or fire gases the conductivity kf and the viscosity f as functions of\ntemperature can be obtained from Eqs. 6.1 and 6.2, respectively, and inserted into\nEq. 6.19 to become\n!n\n291 \u0002 10\u00036 \u0002 T 0:79\nf u1 \u0002 d 1\nh \u0002C\u0002 \u0002 0:73 6:20\nd 1:10 \u0002 10\u00039 \u0002 T 1:68\nf\n\n## which can be reduced to\n\nT f0:79\u00031:68n \u0002 u1 n\nhA\u0002C\u0002 1\u0003n\n6:21\nd\n\nwhere the constants A, C and n can be found in Table 6.2 for various ranges of the\nvalues of Redf or .\n96 6 Heat Transfer by Convection\n\nTable 6.2 Constants to be used with Eqs. 6.16 and 6.21 for calculating the Nusselt number and\nheat transfer coefficients to cylinders exposed to forced convection flow\nRedf (Eq. 6.18) C n A [SI units]\n0.44 (0.454.5) \u0002 10\u00039 0.989 0.330 0.110\n440 (4.545) \u0002 10\u00039 0.911 0.385 0.341\n404000 (454500) \u0002 10\u00039 0.683 0.466 1.81\n400040,000 (4.545) \u0002 10\u00036 0.193 0.618 41.4\n40,000400,000 (45450) \u0002 10\u00036 0.0266 0.805 1950\n\n## Example 6.3 Calculate the convective heat transfer coefficient of a 1 mm shielded\n\nthermocouple. Model the thermocouple as a cylinder and assume an air velocity of\nu1 1:0 m=s flowing across the cylinder. Assume gas temperature levels of\n(a) T g 300 K (room temperature initially)\n(b) T g 1000 K (ultimate temperature)\n\nSolution\n(a) The film temperature Tf (300 + 1000)/2 650 K. Then according to Eq. 6.18\n\u0004 \u0005\n1 \u0002 0:001= 6501:67 20 \u0002 10\u00039 and from Table 6.2 C 0.911,\nT 0:79\u00031:68\u00020:385 u1 0:385\nn 0.385 and A 341. Then h 0:341 \u0002 0:911 \u0002 f d1\u00030:385\n131 W=m K.\n2\n\u0004 \u0005\n(b) Equation 6.18 yields 1:0 \u0002 0:001= 10005=3 9:77 \u0002 10\u00039 . A, C and n\ncan now be obtained from row two of Table 6.2 and inserted into Eq. 6.21 to\nT f0:92\u00031:67\u00020:385 u1 0:385\nT 0:277 u1 0:385\nget h 0:341 \u0002 0:911 \u0002 d 1\u00030:385\n0:311 \u0002 f\nd0:615\n147W=m2 K\nComment: The heat transfer coefficient due to convection changes only slightly\nwith temperature. However, it would change considerably if the heat transfer due to\nradiation would be included as well.\n\n## 6.2.3 In Circular Pipes and Tubes\n\nThe heat transfer between the fluid and the walls of a circular tube depends on the\nfluid conductivity k and kinematic viscosity , the fluid velocity u and the flow\nconditions, laminar or turbulent governed by the Reynolds number\n\nu\u0002D \u0002u\u0002D\nReD 6:22\n\nAll parameters refer to bulk temperatures. For details see textbooks such as\n[1, 2].\n6.3 Natural or Free Convection 97\n\n\u0005\nFor turbulent flow (ReD > 2300 the Nusselt number can be calculated as\n\nd \u0002 Pr\nn\n6:23\n\n## where n 0:4 for heating and n 0:3\n\n\u0005 for cooling of the fluid.\nFor laminar flow ( ReD < 2300 the Nud approaches a constant value for\nsufficiently long tubes, i.e.\n\n## Nud 3:66 6:24\n\nThen the heat transfer coefficient between the fluid and the walls can be\ncalculated as\n\nkf\nh Nud \u0002 : 6:25\nD\n\n## where kf is the conductivity of the fluid at the film temperature.\n\nExample 6.4 Air with a bulk temperature T air 200 \u0005 C is flowing with a velocity\nof 2 m/s in a tube/duct with an inner diameter D 400 mm. Estimate the heat flux to\nthe duct surfaces which have a temperature of 800 \u0005 C?\nSolution The film temperature T f 0:5200 800 500 \u0005 C 773 K. Then\nEq. 6.1 yields kair 0:0567 W=m K and Eq. 6.2 air 78:3 \u0002 10\u00036 m2 =s and\naccording to Eq. 6.22 Re 10,200 which indicates turbulent flow. Then from\nEq. 6.23 Nud 32:2 (the fluid is heated and n 0.4) and according to Eq. 6.25\nh Nud \u0002 Dk 32:2 \u0002 0:0567\n0:4 4:48 W=m2 K. Thus the heat flux to the tube wall\n00\nq_ w 4:48 \u0002 800 \u0003 200 2690 W=m2 .\nExample 6.5 Water with a bulk temperature of T w 20 \u0005 C is flowing with a\nvelocity u 0.1 m/s in a tube with an inner diameter D 50 mm. Estimate the heat\nflux to the tube surfaces which have a constant temperature T w 70 \u0005 C?\nSolution The film temperatureT w 2070\n2 45 \u0005 C 318 K. Then from Table 6.1\nthe conductivity kw 0:63 W=m K, the viscosity w 0:66 \u0002 10\u00036 m2 =s and\nPr w 4:34. According to Eq. 6.22 Re 7576 which indicates turbulent flow. Then\nfrom Eq. 6.23 Nud 52:2 (the fluid is heated and n 0.4) and according to\nEq. 6.25 h Nud \u0002 Dk 52:2 \u0002 0:63\n0:4 658 W=m2 K.\n\n## Natural or free convection occurs as a result of density changes due to heating or\n\ncooling of fluids at solid surfaces. When a wall is hotter than adjacent air an\n98 6 Heat Transfer by Convection\n\n## Fig. 6.4 Free convection\n\nboundary layer at a vertical\nhot surface\n\nupwards flow is established as shown in Fig. 6.4, and vice versa if the wall is cooler.\nThe velocity just at the wall surface is zero. It increases then to a maximum and\nthereafter it decreases to zero again at the end of the boundary layer where the free\nstream velocity is assumed to be negligible. At the outset the boundary layer is\nlaminar but changes to turbulent at some distance from the edge depending on fluid\nproperties and the difference between wall surface and fluid temperatures. In\npractice in FSE, natural and forced convection commonly occur simultaneously\nand analyses must focus on the one which is predominant.\nIn general it is very difficult to make accurate estimates of natural convection\nheat transfer coefficients. The formula given below for some elementary cases are\nbased on empirical evidence obtained under controlled conditions. Such conditions\nrarely occur in real life but they serve as guidance for estimates. Any accurate\nanalytical solutions are not available for calculating heat transfer by natural\nconvection.\nAs for forced convection the Nusselt number yields the heat transfer coefficient.\nIt depends in the case of free convection on the Prandtl and Grashof numbers. The\nlatter is defined as\n\u0004 \u0005\ng \u0002 f \u0002 T s \u0003 T g \u0002 L3\nGr xf 6:26\n2f\n\n## where g (9.81 m/s2) is the constant of gravity, L a characteristic length, the\n\nkinematic viscosity of the fluid and f is the inverse of the film temperature as\ndefined in Eq. 6.5:\n6.3 Natural or Free Convection 99\n\n1\nf 6:27\nTf\n\nThe subscript f indicates that values of the parameters are taken at the film\ntemperature.\nBased on empirical data for both laminar and turbulent convection the average\nvalues of the Nusselt number and the corresponding heat transfer coefficient can be\nwritten as:\n\nhf \u0002 L \u0004 \u0005m\nNuf C \u0002 Gr f \u0002 Pr f 6:28\nkf\n\nThis formula correlates well for several simple geometrical configurations. Here\nL is a characteristic length appropriate for the configuration. For horizontal rectan-\ngular plates the characteristic length may be estimated as\n\nA\nL 6:29\nP\n\nwhere A is the area and P the perimeter of the surface. The other constants of\nEq. 6.28 can be found in Table 6.3. As a rule of thumb, the exponent m 1=4 for\nlaminar and m 1=3 for turbulent flow.\n\n## 6.3.1.1 Explicit Expressions for Heat Transfer from Air\n\nEquation 6.28 and Table 6.3 can be used to derive convection heat transfer\ncoefficients. By inserting the approximations of kf (Eq. 6.1) and f (Eq. 6.2) of air\nas functions of temperature, explicit expressions can be derived for heat transfer\ncoefficients as functions of the air Tair and the surface Ts temperature. Tf is the film\ntemperature defined as the mean of the ambient air temperature and the surface\ntemperature (Eq. 6.5).\nObserve temperatures must be in Kelvin in all formulas.\nThus at vertical plates and large cylinders under turbulent conditions (m 1=3)\nin air the mean heat transfer coefficient can be calculated as\n\nkf\nhf Nuf \u0002 76:0 \u0002 T \u00030:66\nf \u0002 jT s \u0003 T air j1=3 6:30\nL\n\nThe heat flux to a vertical surface may now be written in the form given in\nEq. 4.8 as\n00\nq_ con 76:0 T \u00030:66\nf T air \u0003 T s 0:33 6:31\n100 6 Heat Transfer by Convection\n\nTable 6.3 Constants to be used with Eq. 6.28 for calculating heat transfer coefficients and heat\ntransfer to surfaces exposed to natural convection\nCharacteristic\nConfigurations GrLfPrf. length C m\nVertical plates and large cylinders\nLaminar 104109 L 0.59 1/4\nTurbulent 1091012 L 0.13 1/3\nHorizontal plates\nLaminar (heated surface up or cooled 2 \u0002 104 L A/P 0.54 1/4\ndown) 8 \u0002 106\nTurbulent (heated surface up or cooled 8 \u0002 106011 L A/P 0.14 1/3\ndown)\nLaminar (heated surface down or 1051011 L A/P 0.27 1/4\ncooled up)\n\n10.0\n\n9.0\nHeat transfer coecient [W/(m2 K)]\n\n8.0\n\n7.0\n\n6.0\n\n5.0\n\n4.0\n\n3.0\nTg=20\n2.0 Tg=50\nTg=100\n1.0 Tg=200\nTg=500\n\n0.0\n0 100 200 300 400 500 600 700 800\nSurface temperature [C]\n\nFig. 6.5 Heat transfer coefficient due to natural convection vs. surface temperature for various\nsurrounding air temperatures at vertical plates and large cylinders under turbulent conditions\naccording to Eq. 6.30\n\n00\nq_ con is positive when T air \u0003 T s is positive and vice versa. Notice that in this case\nthe heat transfer coefficient is independent of the dimension L and inversely\nproportional to approximately the square root of the temperature level. The varia-\ntion of hf with the surface temperature for various gas temperatures are shown in\nFig. 6.5. When the surface and gas temperatures are equal the air flow and thereby\n6.3 Natural or Free Convection 101\n\nthe heat transfer vanishes and it increases gradually when the fluid and surface\ntemperatures diverges. As the formula are symmetric concerning Tair and Ts the\ndiagram also be interpreted as hf vs. gas temperature for various surface temper-\natures. Observe that for the gas temperature equal to 20 \u0005 C that convection heat\ntransfer coefficient peaks at about 11.5 W/(m2 K). Then it decreases slowly. In the\ninterval from room temperature to 200 \u0005 C which is of interest for evaluation of fire\nseparating walls the convection heat transfer coefficient increases from zero to\nObserve that Eqs. 6.30 and 6.31 apply only for turbulent conditions according to\nTable 6.3. This requirement is generally met in fire safety problems such as cooling\nof the unexposed side of a fire separating wall element.\nThe convective heat transfer coefficient to horizontal surfaces depends on the\nsize of the surface. By inserting the value of the kinematic viscosity at the film\ntemperature according to Eq. 6.2 the Rayleigh number becomes\n\u0004 \u0005\nPr f \u0002 Gr Lf 5:68 \u0002 1018 L3 \u0002 T \u00034:36\nf T 6:32\n\nIf the Rayleigh number is between 2 \u0002 104 and 8 \u0002 106 according to Table 6.3, the\nNusselt number can be obtained according to Eq. 6.28 with the air conductivity\naccording to Eq. 6.1 and Pr 0.7. The heat transfer coefficient can then be\ncalculated as\n\n## hf 7:67 L\u00031=4 \u0002 T \u00030:33\n\nf \u0002 T 1=4 6:33\n\nOf special interest are the heat transfer coefficients to specimen surfaces of the\ncone calorimeter (ISO 5660) and to plate thermometers when mounted horizon-\ntally. Both have a surface 0.1 m by 0.1 m and thus a characteristic length\nL 0.025 m according to Eq. 6.29. Then the convection heat transfer coefficient\ncan be calculated as\n\nhf 19:3 T \u00030:33\nf \u0002 T 1=4 6:34\n\nFigure 6.6 shows convective heat transfer coefficients to horizontal surfaces with\na characteristic length of 0.025 m for various gas temperatures as functions of the\nsurface temperature.\n00\nExample 6.6 A PT is exposed to an incident radiation q_ inc 50 kW=m2 . Assum-\ning that it does not lose any heat by conduction estimate its steady-state tempera-\nture, i.e. adiabatic surface temperature, when the PT is mounted\n(a) horizontally\n(b) vertically\n102 6 Heat Transfer by Convection\n\nFig. 6.6 Heat transfer coefficient vs. surface temperature due to natural convection at a horizontal\nplate for L 0.025 m and 2 \u0002 104 < GrLf Prf<8 \u0002 106. Particularly applicable to cone calorimeter\n(ISO 5660) and PTs with exposed areas of 0.1 m by 0.1 m\n\nThe exposed surface area of the PT is 0.1 m by 0.1 m. Assume the emissivity of\nthe PT surface is 0.9 and the ambient temperature is 20 \u0005 C.\nHint: The net heat absorbed by radiation must be balanced by the heat lost by\nconvection.\n00\nSolution q_ inc 50 kW=m2 yields a radiation temperature T r 969 K 696 \u0005 C.\n(a) Assuming a first surface temperature estimate T 1AST T s T r yields\naccording to Eq. 6.34 or Fig. 6.6 h 11.7 W/(m K). Then the AST can be\ncalculated with the iteration procedure according to Eq. 4.25 with T 1AST T r .\nThus T 2AST 923 K and T 3AST 926 K 653 \u0005 C. Finish iteration. Compare\nwith Fig. 4.7a for h= 13 W=m2 K.\n(b) Assuming a first surface temperature estimate T 1AST T s T r yields\naccording to Eq. 6.30 or Fig. 6.5 h 9.2 W/(m K). Then T 2AST 934 K and\nT 3AST 936 K 663 \u0005 C. Finish iteration. Compare with Fig. 4.7a for\nh= 10 W=m2 K.\n6.3 Natural or Free Convection 103\n\n## Heat is transferred between surfaces of enclosures. Two elementary cases of free\n\nconvection may be identified characterized by horizontal and vertical layers,\nrespectively.\nThe heat flux may in both cases be calculated as [1, 15]\n\n00 kf\nq_ c Nu T 1 \u0003 T 2 6:35\n\nwhere kf is the thermal conductivity of the fluid. For horizontal layers where the\nupper surface is warmer there will be no buoyancy driven convection or flow and\nthe heat will be transferred by conduction only, i.e. Nu 1. However, if the upper\nsurface is cooler than the lower convection will occur and the Nusselt number will\nbe greater than one. Equation 6.35 may also be written as\n\n00 ke\nq_ c T 1 \u0003 T 2 6:36\n\n## where the ke may be identified as the effective or apparent thermal conductivity of\n\nthe air enhanced by convection. It is defined by the relation\n\nke\nNu 6:37\nkf\n\nThe Nusselt numbers can be obtained from Table 6.3 for various ranges of the\nGrashof number according to Eq. 6.38\n\ng \u0002 T 1 \u0003 T 2 3\nGr 6:38\n2\n\nWhen the lower surface is warmer than the upper, convection and heat transfer\nby convection will occur when Gr > 104. Inserting the values of for air and as\nfunctions of the mean of T1 and T2 according to Eqs. 6.2 and 6.27, respectively,\nyields:\n\n8:11 \u0002 1018 T 1 \u0003 T 2 3\nGr air 6:39\nT 13=3\n\n## where T is the mean of the surface temperatures, i.e.\n\nT 1 T 2\nT 6:40\n2\n104 6 Heat Transfer by Convection\n\n## Fig. 6.7 Enclosed space\n\nwith a hot left surface and a\ncool right surface. Upper\nand lower surfaces are here\npattern (b) Nomenclature\n\nThe characteristic length L in Table 6.3 denotes the vertical height of a vertical\nenclosure, see Fig. 6.7b. Note that the influence of convection heat transfer is\nnegligible when enclosure is small, Gr < 2000. The air conductivity vs. the\nmean temperature according to Eq. 6.1 is used when deriving the expressions in\nthe right column of Table 6.4. These values are then used in Eq. 6.35 to calculate the\nheat flux by convection across enclosed spaces.\nExample 6.7 Calculate the apparent conductivity and heat transfer by convection\nbetween two parallel vertical surfaces as in Fig. 5.4 when the surfaces have\ntemperatures T 1 300 \u0005 C 573 K and T 2 100 \u0005 C 373 K and the distance\n0:1 m. The height L 0.6 m.\n\n## Solution Equation 6.40 yields T 573373\n\n2\n\n473 K and Eq. 6.39 yields for air\n18\n573\u0003373\u00020:13\nGr air 8:11\u000210 47313=3\n4:17 \u0002 106 . Then according to the 7th row of\n\u0004 \u0005 1 \u0004 \u0005\u000319\nNu 0:065 \u0002 4:17 \u0002 106 3 \u0002 0:6\n8:53, and by inserting Eq. 6.1 into Eq. 6.37\n0:1\nthe effective conductivity is obtained as from Eq. 6.37\nke 8:53 \u0002 291 \u0002 10\u00036 \u0002 4730:79 mWK 0:322 W=m K. The heat transfer may then\n00\nbe obtained from Eq. 6.36 as q_ c 0:322\n0:1 573 \u0003 373 W=m 644 W=m . Alter-\n2 2\n\nnatively the explicit expression in the 4th column of the 7th row may be used.\nExample 6.8 The same as Example 6.8 but in a horizontal configuration with a hot\nlower surface with a temperature T 1 300 \u0005 C 573 K and a cooler upper surface\nwith a temperature T 2 100 \u0005 C 373 K, and a distance between the parallel\nsurface 0:1 m, see Fig. 6.8.\n\n## Solution Equation 6.40 yields T 573373\n\n2\n\n473 K and Eq. 6.39 yields for air\n18\n573\u0003373\u00020:13\nGr air 8:11\u000210 47313=3\n4:22 \u0002 106 . Then according to the 4th row of\n\u0004 \u0005= 1\nNu 0:068 \u0002 4:22 \u0002 106 3 10:44, and the effective conductivity of the enclosed\nair is obtained from Eq. 6.37 as ke 10:44 \u0002 291 \u0002 10\u00036 \u0002 4730:79 0:394 W=m K.\n6.3 Natural or Free Convection 105\n\nTable 6.4 Nusselt number Nu for calculating convection heat transfer in air in enclosed spaces\naccording to Eq. 6.35 \nkfair\nNo Orientation Range Nu ke = Nuair\n1 Horizontal layers (Stable layers) 1 291\u000210\u00036 T 0:79\n\n(hotter upper layer)\n2 Horizontal layers Gr < 104 1 291\u000210\u00036 T 0:79\n\n(cooler upper layer) 1=\n3 < 400 \u0002 103 0:195 Gr4 1:42 T 1 \u0003T 2 0:25\nT 0:09 \u00020:25\n1=\n4 Gr > 400 \u0002 103 0:068 Gr3 39:2 T 1 \u0003T 2 0:33\nT 0:65\n5 Vertical layers Gr < 20 \u0002 103 1 291\u000210\u00036 T 0:79\n\n\u0003\n6 20 \u0002 103 < Gr < 200 \u0002 103 1= L\n4\n\n1=9 2:80 T 1 \u0003T 2 0:25\n0:18 Gr T 0:09 \u0002L0:11 \u00020:14\n\u0003\n7 200 \u0002 103 < Gr < 11 \u0002 106 1= L\n3\n\n1=9 37:5 T 1 \u0003T 2 0:33 \u00020:11\n0:065 Gr T 0:65 \u0002L0:11\n\n## The last column shows explicitly the thermal resistance ke/\n\nFig. 6.8 Nomenclature for calculating heat transfer by convection across an enclosed space with\ntwo horizontal parallel surfaces of different temperatures where T 1 > T 2\n\n00\nThe heat transfer may then be obtained from Eq. 6.36 as q_ c\n0:1 573 \u0003 373 W=m 232 W=m . Alternatively the 4th column may be used.\n0:116 2 2\nChapter 7\nNumerical Methods\n\nThe analytical methods outlined in Chaps. 2 and 3 presume that the material\nproperties and heat transfer coefficients are constant. That is, however, not possible\nin most cases in fire protection engineering as the temperature then varies within a\nwide range and therefore both material properties and boundary conditions vary\nconsiderably. Phase changes or latent heat due to water vaporization or chemical\nreactions of materials (see Sect. 14.1 on concrete) must in many cases be considered\nficients vary considerably with temperature. As shown in Sect. 4.1 it increases with\nthe third power of the temperature level. In addition geometries being considered\nare not as simple as assumed above. Often they are in two or three dimensions, and\nthen analytical methods can seldom be used for practical temperature analyses.\nTherefore numerical methods involving computer codes are frequently used in fire\nprotection engineering. In some cases in particular for 0-dimension problems\n(lumped-heat-capacity) relatively simple so-called spreadsheet codes such as\nExcel may be used. For problems with more complex geometries and boundary\nconditions computer codes based on finite difference or finite elements methods are\nneeded. Several computer codes based on these methods are commercially avail-\nable, see Sect. 7.3.2. The superposition technique as presented in Sect. 7.2 may be\nseen as a combination of a numerical and an analytical method.\n\n7.1 Lumped-Heat-Capacity\n\nThe basic theory of heat transfer to bodies with uniform temperature is given in\nSect. 3.1. According to Eq. 3.2\n\n## Springer International Publishing Switzerland 2016 107\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_7\n108 7 Numerical Methods\n\ndT A 00\nq_ 7:1\ndt V \u0002 \u0002 c\n\nwhere A is exposed area, V volume, density and c specific heat capacity of the\nexposed body. By integrating over time the body temperature becomes\nt\nA 00\nT Ti q_ dt 7:2\nV\u0002c\u0002 0\n\n00\nwhere Ti is the initial temperature. In case q_ is given as function of time (2nd kind\nof BC) or proportional to the difference between the surrounding and the body\n(surface) temperatures (3rd kind of BC), the temperature T can sometimes be solved\nanalytically. In most other cases numerical methods must be used even when\nlumped heat is assumed.\nIn general both space and time are discretized except for lumped-heat-capacity\nproblems with only one unknown temperature where only time is discretized.\nT\nThe time derivative of Eq. 7.1 is approximated by the differential, i.e. dT dt \u0003 t .\nGiven the time is divided into increments as indicated in Fig. 7.1\n\n## and temperature increments are defined as\n\nT j1 T j1 \u0004 T j 7:4\n\nAssuming the time increment constant Eq. 7.1 can be written as a finite differ-\nence equation as\n\nA 00 j\nT j1 \u0004 T j T j1 \u0003 q_ t 7:5\nV c \u0002 tot\n\n00 j\nwhere q_ tot is the heat flux to the surface at the time increment j.\nIn the simplest case the heat flux is proportional to the difference between the\ninsulation surface temperature and uniform body temperature as shown in Fig. 3.3b.\nThen Eq. 7.5 can be written as\n\nA kin \u0002 \u0003\nT j1 T j \u0002 T s \u0004 T j t 7:6\nV c din\n\nwhere kin and din are the conductivity and thickness of the insulation. When solved\naccording to this forward difference scheme, all the parameters may be updated at\neach time step depending on the temperature of time increment j.\nWhen a body is exposed to a third kind of boundary condition, i.e. a function of\n7.1 Lumped-Heat-Capacity 109\n\n0 Time, t\n\n## Fig. 7.1 Time axis indicating time increment numbering\n\n00\nsurface temperature, the total heat flux q_ tot is defined according to Eq. 4.11 and the\nbody temperature may be calculated as\n\nA h \u0004 00 j \u0005 \u0002 \u0005\u0006\nT j1 T j q_ inc \u0004 T j 4 hc T gj \u0004 T j t 7:7\nV c\n\n## or when the heat flux is defined according to Eq. 4.12 as\n\nA h \u0002 j4 \u0003 \u0002 \u0005\u0006\nT j1 T j T r \u0004 T j 4 hc T gj \u0004 T j t 7:8\nV c\n\nThe recursion formulas of Eq. 7.7 through Eq. 7.10 are forward difference or\nexplicit schemes. That means all parameters on the right-hand side of the equation\nare known at time increment j and the new temperature at time j + 1 can be\ncalculated explicitly. Such integration schemes are numerically stable only if\neach time increment is chosen less than a critical time increment, the critical time\nincrement tcr defined as\n\nV \u0002 c \u0002 \u0002 din\ntcr 7:9\nkin \u0002 A\n\n## for Eq. 7.6, and\n\nV\u0002c\u0002\ntcr 7:10\nhtot \u0002 A\n\nfor Eq. 7.7 and Eq. 7.10. htot is the total adiabatic heat transfer coefficient as defined\nby Eq. 4.19. The critical time corresponds to the time constant as defined in Sect.\n3.1. It can vary over time as the including parameter changes with temperature. In\nreality much shorter time increments in the order of 10 % and the critical time\nincrement are in general recommended to achieve accurate temperatures.\nExample 7.1 An unprotected steel section with a section factor VA 100 m\u00041 is\nsuddenly exposed to a constant fire temperature T f T r T g 1000 \u0005 C. Calcu-\nlate the steel temperature as a function of time if the initial temperature is\nT i 20 \u0005 C. Assume a steel surface emissivity of 0.9 and a convection heat transfer\ncoefficient of 25 W/(m2 K). st 7850 kg=m3 and cst 560 Ws=kg K.\n110 7 Numerical Methods\n\n## Solution This problem is ideally solved by applying Eq. 7.8 in a spreadsheet\n\napplication. The first three increments are shown below. Assume t 10 s. Then\nA=V \u0002t\n560\u00027850\n100\u000210\n0:227 \u0002 10\u00043 m2 =W and Eq. 7.8 yields T 1 20 0:227 \u0002 10\u00043\n\u0007 c\u0002 \u0002 \u0003 \u0006\n0:8 \u0002 5:67 \u0002 10\u00048 1273 4 \u0004 2984 251000 \u0004 20 52:5 \u0005 C, T 2 79:6 K and\nT 3 112 K. The maximum htot can be obtained from Eq. 4.19 or Fig. 4.2a as\n470 W/(m2 K) and from Eq. 7.10 the minimum increment tcr 560\u00027850100\u0002470 93 s. At\npreceding lower steel temperature levels htot is much greater and thereby tcr is\nmuch smaller, and therefore t 10 s will yield accurate steel temperatures.\nExample 7.2 A steel plate with a thickness of dst 10 mm and an initial temper-\nature of (293 K) is placed in the sample holder of a cone calorimeter, see Fig. 7.2.\nThe incident radiation of the cone is set to 50 kW/m2. The plate is well insulated on\nall surfaces except the upper exposed surface. Assume a steel surface emissivity of\n0.9 and a convection heat transfer coefficient hc 12 W=m2 \u0002 K. st 7850 kg=\nm3 and cst 560 Ws=kg K. Derive a time integration scheme and show the first\nthree time increments.\nSolution Apply Eq. 7.7 where Vc A\ndc1\n0:01\u00027850\u0002560\n1\n44000\n1\nWs=m2 K and\n00 j \u0002 \u0003\nq_ tot 0:9 50000 \u0004 T 4, j 12 \u0002 20 \u0004 T . Thus a forward difference incremen-\nt\n\b \u0002 \u0003 \u0002 \u0003\ntal scheme becomes T j1 T j 44000 0:9 50000 \u0004 T 4, j 12 \u0002 293 \u0004 T j\n0:9\u00025000012\u0002293t \u0002 t\n\u0003 j t \u00048 4, j\n\u0003\n44000 1 \u0004 12\n44000 T \u0004 44000 \u0002 5:67 \u0002 10 T . Assume a time incre-\nment t 60 s. Then the incremental scheme can be reduced to:\nT j1 66:2 0:984 \u0002 T j \u0004 77:3 \u0002 10\u000412 \u0002 T 4, j .\nFirst step, j 1: T 2 66:2 0:984 \u0002 293 \u0004 77:3 \u0002 10\u000412 \u0002 2934\n66:2 288 \u0004 0:57 354 K 80:6 \u0005 C.\nSecond step, j 2: T 3 66:2 0:984 \u0002 354 \u0004 77:3 \u0002 10\u000412 \u0002 3544 66:2 349\u0004\n1:2 413 K 141 \u0005 C.\nThird step, j 3: T 4 66:2 0:984 \u0002 413 \u0004 77:3 \u0002 10\u000412 \u0002 4134\n66:2 406 \u0004 2:25 470 K 197 \u0005 C.\nComment: Figure 7.3 shows a comparison between measured and calculated tem-\nperature steel specimen as in the example. In the calculations Eq. 7.8 was applied\nwith a heat transfer coefficient hc increased to 18 W/(m2 K) to consider the heat\nlosses from the steel specimen by conduction and a gas temperature Tg as measured\nwith a thin thermocouple. The accurate prediction indicates how well Eq. 7.8\nmodels the heat transfer to a specimen surface in the cone calorimeter.\n7.2 Superposition and the Duhamels Superposition Integral 111\n\n## Fig. 7.2 Heat transfer to a\n\n10-mm-thick steel plate in\nthe cone calorimeter\n\n500\n\n400\nTemperature [C]\n\n300\n\n200\nTst calculated\nTst measured\n\n100\n\n0\n0 200 400 600 800 1000 1200 1400 1600 1800\nTime [s]\n\nFig. 7.3 Comparison of measured and calculated temperature of 10-mm-thick steel specimen\nexposed to an incident radiation of 50 kW/m2 in a cone calorimeter\n\nIntegral\n\n## A technique based on superposition is presented below. In its infinitesimal form it\n\nmay be called Duhamels superposition integral. It is a technique which has many\nvarious types of applications when analysing bodies with constant material prop-\nerties and with zero initial temperature conditions which are exposed to boundary\nconditions varying with time. Zero initial temperature conditions can be obtained\nfor bodies with constant initial temperatures by calculating temperature rise as\nshown below.\nThe technique is here exemplified for the case of a surface of a semi-infinite solid\nat uniform temperature Ti. See also [1, 2]. When it is suddenly receiving a constant\nexternal heat flux f at the time t 0, the surface temperature rise s T s \u0004 T i may\nthen be written as\n112 7 Numerical Methods\n\ns f \u0002 At 7:11\n\nwhere A(t) is the response function (sometimes called the fundamental solution).\nIt is here the surface temperature response as a function of time for a unit heat flux\n00\n( f q_ tot 1 ). For semi-infinite solids the response function of the surface\ntemperature is according to Eq. 3.29\n\n2t\nA t 7:12\nkc\n\nThe surface temperature according to Eq. 7.11 applies only if the heat flux\nremains constant with time. When the heat flux to the surface, generally called\n00\nthe forcing function, varies with time, i.e. f t q_ tot t, the surface temperature\nmay be calculated by superposition. Thus according to the Duhamel integral of\nsuperposition the solution R(t) (the surface temperature rise in this case) as a\nfunction of time can be written as\nt\n0\nRt f 0 \u0002 At f t \u0004 \u0002 A d 7:13\n0\n\nwhere f0 denotes the time derivative of the forcing function. By integration by parts\nand noting that At 0, an alternative formulation can be obtained where the\nresponse function is derived instead of the forcing function.\nt\n0\nRt f t \u0004 \u0002 A d 7:14\n0\n\n## where A0 is the time derivative of the response function. ( is a dummy variable\n\ndefined only within the integral.)\nEquation 3.16 in Sect. 3.1.2.1 giving the temperature of a thermocouple\nmodelled as lumped heat was derived from Eq. 7.14. In that case the response\nfunction can be derived analytically and depending on the forcing function, the\nintegral can in some cases be solved analytically. In other cases numerical integra-\ntion techniques must be used to calculate the thermocouple temperature at a given\ntime t.\nThen the time is divided into increments and the surface temperature rise Ti, i.e.\nT s \u0004 T i , may be calculated numerically by a time step superposition scheme.\nTo illustrate how solutions can be superimposed to obtain the surface temperature\nof a time-dependent flux, the following case is studied. The heat flux is assumed to\nvary as shown in Fig. 7.4. Thus\n7.2 Superposition and the Duhamels Superposition Integral 113\n\n## Fig. 7.4 A stepwise\n\nchanging forcing function\n\nf t q1 0 < t < t1\nf t q2 t1 < t < t2 7:15\nf t q3 t > t2\n\n## Then the surface temperature rise can be calculated as\n\ns q1 t \u0002 At q2 \u0004 q1 \u0002 At \u0004 t1 q3 \u0004 q2 \u0002 At \u0004 t2 7:16\n\n## In a general form the surface temperature rise may be written as\n\nXn\ns f 0 \u0002 At i1\nf ti \u0002 At \u0004 ti 7:17\n\nwhere\n\nf ti f ti1 \u0004 f ti 7:18\n\nA very powerful superposition technique is shown below which allows the forcing\nfunction to depend on the response for the actual exposure. That is, for instance, the\ncase when a surface is exposed to radiation and the emitted radiation depends on the\nsurface temperature. Then a new surface temperature rise can then be calculated at\ntime increment j + 1 as:\nXj \u0007 \u0006\nj1 i0\nf i \u0004 f i\u00041 \u0002 A j\u0004i 7:19\n\nor alternatively as\nXj h i\nj1 i0\nf j\u0004i \u0002 Ai1 \u0004 Ai 7:20\n\n## where t is the time increment and 0 f 0 0. The values of the time-dependent\n\nparameters at a given number of time increments i are defined as i i \u0002 t,\nf i f i \u0002 t and Ai Ai \u0002 t.\n114 7 Numerical Methods\n\nIn the case of a surface exposed to radiation and convective heat transfer, the\nforcing function, i.e. the heat flux to the surface, can be calculated according to\nEq. 3.14. Then at time increment j calculated as\n00\nf j \u0002 t q_ tot\n\u0007 00j \u0002 t \u0006 \u0002 \u0003 7:21\nq_ inc j \u0002 t \u0004 T 4s j \u0002 t h T g j \u0002 t \u0004 T s j \u0002 t\n\n00\nwhere q_ inc and Tg are input boundary conditions and Ts the surface temperature\napproximated by its calculated value at time increment j.\nExample 7.3\nA concrete wall surface is assumed to receive piecewise constant heat fluxes\n(boundary condition of second kind) according to Eq. 7.15 and as indicated in\nFig. 7.4 with the following input values:\nt1 10 min, t2 20 min and q1 20 kW=m2 K, q2 35 kW=m2 K and\nq3 15 kW=m2 K\nThe initial wall temperature is 20 \u0005 C. Assume thermal properties of concrete\naccording to Table 1.2.\nExpress the surface temperature rise as a function of time for the three time\nintervals by superposition according to Eq. 7.16.\n\u0002 \u0003 \u0002 \u0003\nSolution According to Table 1.2 k \u0002 \u0002 c 3:53 \u0002 106 W2 s = m4 K2 . Then\nEq. 7.16 yields with the response function according to Eq. 3.29 new surface\ntemperature rises as:\n\nTime\ninterval [s] Surface temperature rise, s [\u0005 C]\np\n0 < t < 600 20000 \u0002 p\nt\np2\n3:53\u0002106\np p\n600 < t < 1200 20 \u0002 103 \u0002 p\nt\np2 35 \u0004 20 \u0002 103 \u0002 p2\n\npt\u0004600\n3:53\u0002106 3:53\u0002106\np p p\nt > 1200 20 \u0002 10 \u00023 p\n2\np 35 \u0004 20 \u0002 10 \u0002\nt 3 p\n2\np 15 \u0004 35 \u0002 103 \u0002 p2\nt\u0004600\n\npt\u00041200\n3:53\u0002106\n3:53\u0002106 3:53\u0002106\n\n## When calculating temperature in fire-exposed structures, non-linearities must in\n\nmost cases be considered. The boundary conditions are non-linear varying signif-\nicantly with temperature as shown above (see Chap. 4), and also the thermal\nproperties of most materials vary significantly within the wide temperature span\nthat must be considered in FSE problems. Therefore numerical methods must\nusually be employed. The most general and powerful codes are based on the\nso-called finite element method (FEM). Below the basic equations are derived for\n7.3 The Finite Element Method for Temperature Analyses 115\n\n1 m-1 m m+1\n\n1 i-1 i i+1 n\n\nFig. 7.5 A wall divided into one-dimensional elements numbered with m:s and with the nodes\nnumbered with i:s\n\n## a simple one-dimensional case as an illustration. Similar types of equation may be\n\nderived for two and three dimensions.\n\n## 7.3.1 One-Dimensional Theory\n\nFigure 7.5 shows a wall which has been divided into a number of one-dimensional\nelements. The temperature between the nodes is assumed to vary linearly along the\nlength.\nIn any element, interior or at the surface, with the length L, see Fig. 7.6, the\nconductivity k and a cross-section area A, the heat flow to the element nodes can be\ncalculated as\n\nk\nq_ 1 \u0002 T 1 \u0004 T 2 7:22\nL\n\nand\n\nk\nq_ 2 \u0004 \u0002 T 1 \u0004 T 2 7:23\nL\n\nor in matrix format as\n\nq_ k_ \u0002 T_\ne e e\n7:24\n\nwhere q_ is the element node heat flow vector, k_ the element thermal conduction\ne e\n\nmatrix and T_ the element node temperature vector. Given the one-dimension\ne\n\nassumption, the cross-section area is constant and assumed equal unity. Then the\nelement thermal conduction matrix may then be identified as\n\ne ke e\nk12 k 1 \u00041\nk 11 7:25\nL \u00041 1\ne e\nk21 k22\n\n## and the element nodal temperature and heat flow vectors as\n\n116 7 Numerical Methods\n\nk, r\n\nT1 T2\nq1 q2\n\nL, A\n\nFig. 7.6 A one-dimensional element with local element node numbers 1 and 2 and with a length\nL and a section area A. The element is given a thermal conductivity k, a specific heat capacity c and\na density\n\n\u000e\nT_\ne T1\n7:26\nT2\n\nand\n\u000e\nq1\nq_\ne\n7:27\nq2\n\nIn a similar way an element heat capacity matrix can be defined by lumping the\nheat capacity of the element in the nodes. Thus an element heat capacity matrix may\nbe obtained as\n\nL\u0002c\u0002 1 0\nc\ne\n7:28\n2 0 1\n\nWhen several elements are combined, global heat conductivity matrix K can be\nassembled. In the very simple case of three one-dimensional elements, the global\nthermal conduction matrix becomes\n2 3\nk1 \u0002 k112 \u0003 0 0\n6 k11\n1 k122 k211 \u0002 k12 \u0003 0 7\n2\nK6\n4 021\n7 7:29\nk221 k22 k311 k312 5\n2\n\n0 0 k321 k322\n\nwhere the super fixes 13 denote the contributions from the corresponding element\nnumbers. The global heat capacity matrix C may be assembled in a similar way as\nthe global conductivity matrix. Notice that both the heat conductivity and the heat\ncapacity matrices are symmetric and dominated by their diagonal elements, and that\nthe global heat capacity matrix assembled from element matrices according to\nEq. 7.28 will have non-zero elements only in the diagonal. This will have a decisive\ninfluence on how global algebraic heat balance equation can be solved as shown\nbelow.\n7.3 The Finite Element Method for Temperature Analyses 117\n\nIn global form the heat balance equation may now be written in matrix form as\n\nC T_ K T Q 7:30\n\nwhere the vector T_ contains the time derivatives of the node temperatures. Each row\nin this equation system represents the heat balance of a node. For each equation or\neach node either the temperature or the heat flow given in the corresponding rows in\nthe vectors T and Q, respectively, are known. In principle three cases are possible for\neach equation/row (c.f. the three kinds of boundary conditions as presented in\nChap. 4):\n1. The node temperature Ti is prescribed.\n2. The node heat flow Qi is prescribed.\n3. The node heat flow Qi can be calculated as a function of a given gas temperature\nand radiation temperature, and the surface temperature.\nIn the first case the corresponding equation vanishes as the unknown quantity is\nprescribed a priori. The most common case for internal nodes is the second case,\ni.e. the external flow is zero. A typical boundary condition when calculating\ntemperature in fire-exposed structures is of the third case corresponding to a\nboundary of the 3rd kind. Then according to Table 4.1 of Chap. 4 the nodal heat\nflow is\n\u0002 \u0003 \u0002 \u0003\nQ_ i T 4r \u0004 T 4s, i h T g \u0004 T s, i 7:31\n\n(given the cross-section area equal unity). Notice that this is non-linear as the\nemitted radiation depends on the temperature raised to the fourth power. This is\nof importance when choosing the equation-solving methodology.\nThe differential global matrix equation Eq. 7.30 is solved numerically by\napproximating the time derivative of the node temperatures as\nj1 j\nT T \u0004 T\nT_ \u0003 7:32\nt t\nj\nwhere T is the node temperature vector at time step j and t is here a chosen time\nincrement. Now the heat balance equation in matrix format Eq. 7.30 can be written\nas\n\" j1 #\nT \u0004 Tj\nC KT Q 7:33\nt\n\n## In this differential equation the temperature vector is known at time increment j.\n\nThe new temperature vector at time j + 1 is obtained explicitly based on the\nconditions at time step j for calculating the thermal conduction as\n118 7 Numerical Methods\n\nh j i\nj1 \u00041 j\nT T j t C Q \u0004 KT 7:34\n\nAs the heat capacity matrix is here assumed diagonal (c.f. Eq. 7.28), the new\nnode temperatures at time step j + 1 can be obtained directly row by row and no\nsystem equation needs to be solved. Alternatively an implicit method may be\nderived where the conduction is based on the temperatures at time step j + 1.\nThen the new node temperatures may be calculated as\n\u000f \u0010\u00041 \u0004 \u0005\nj1 C j j\nT K Q CT t 7:35\nt\n\nCombinations of the two solution methods are also possible but as soon as the\nconduction depends on the node temperature at time step j + 1 the solution scheme\nrequires the solution of a global equation system containing as many unknowns as\nthere are unknown node temperatures. Most finite element computer codes use this\ntype of implicit solution schemes. They are generally numerically more stable than\nthe explicit techniques and therefore longer time increments may be used.\nThe explicit solution according to Eq. 7.34 may, on the other hand, be very\nsimple when the heat capacity matrix C is diagonal, i.e. it contains only non-zero\nelements in the diagonal as shown for a one-dimensional element in Eq. 7.28. The\nsolution of the equation system becomes then trivial as each nodal temperature can\nbe obtained directly/explicitly, one at a time. It involves only a multiplication of a\nmatrix with a vector which requires much less computational efforts than solving an\nequation system. This solution scheme is, however, numerically stable only when\nthe time increment t is less than a critical value proportional to the specific heat\ncapacity times the density over the heat conductivity of the material times the\nsquare of a characteristic element length dimension x. This requirement applies to\nall the equations of the entire system, all nodes i except those with prescribed\ntemperatures. If violated in any of the equations, i.e. at any point of the finite\nelement model, the incremental solution equation will become unstable (cf. Sect. 7.1\nand Eq. 7.10 on lumped-heat-capacity). Hence in the one-dimensional case treated\nhere the critical time increment tcr may be estimated as\nh \u0002 c i\ntcr \u0003 min x2 7:36\n2k i\n\nThis means that short time increments are needed for materials with a low\ndensity and a high conductivity, and when small element sizes are used. At\nboundary nodes with heat transfer conditions of the 3rd kind the critical time\nincrement will be influenced by the heat transfer coefficient h as well. Then at\nany node i\n7.3 The Finite Element Method for Temperature Analyses 119\n\n\u0002c\ntcr \u0003 min x2 7:37\nk h=x i\n\nThe heat transfer coefficient h is here the sum of the heat transfer coefficient by\nconvection and radiation, denoted by htot in Sect. 4.3.\nIn practice, when calculating temperature in fire-exposed structures, short time\nincrements must be used independent of solution technique as the duration of\nanalyses are short and boundary condition chances fast. Therefore numerical\nstability is only a problem when modelling sections of very thin metals sheets\nwith high heat conductivity. Then very short time increments are required. The\nproblem may, however, be avoided by prescribing that nodes close to each other\nshall have the same temperature. This technique has been applied in the code\nTASEF . In that code a technique is also developed where the critical time\nincrement is estimated and thereby acceptable time increments can be calculated\nautomatically at each time step depending on thermal material properties and\nboundary conditions varying with temperature. At boundaries of the 3rd kind\nshort time critical time increments can be avoided by assuming the surface tem-\nperature equal to the surrounding temperature (boundary of the 2nd kind). This\napproximation may be applied when the thermal inertia of a material is relatively\nlow and the surface temperature is expected to follow close to the exposure\nAs a general rule finite element calculations shall not be accepted until it is\nshown that the solution gradually converge when time increments and element sizes\nare reduced. This rule applies to both computer codes using explicit and implicit\nsolution techniques. A guidance standard on requirements for calculation methods\nthat provide time-dependent temperature field information resulting from fire\nSFPE .\n\n## Several computer codes are commercially available for calculating temperature in\n\nfire-exposed structures. They are in general based on the finite element method.\nSome are specifically developed and optimized for calculating temperature in fire-\nexposed structures while others are more general purpose codes.\nTASEF [14, 19] and SAFIR are examples of programs which have been\ndeveloped for fire safety problems. They have different pros and cons. They all\nallow for temperature-dependent material properties and boundary conditions.\nTASEF employs a forward difference solving technique which makes it particularly\nsuitable for problems where latent heat due to, e.g. vaporization of water must be\nconsidered. It yields also in most cases very short computing times, in particular for\nproblems with a large number of nodes. TASEF and SAFIR have also provisions for\n120 7 Numerical Methods\n\nmodelling heat transfer by convection and radiation in internal voids. TASEF does\nalso allow for boundary conditions where the exposure radiation and gas temper-\nature are different, boundary condition of the 3rd kind according to Table 4.1.\nThere are many very advanced general purpose finite element computer codes\ncommercially available such as ABAQUS, ANSYS, ADINA, Heating 7 and\nComsol. The main advantage of using this type of codes is that they have several\ntypes of elements for various geometries and dimensions, and that they come with\nadvanced graphical user interfaces and pre- and post-processors.\n\n## 7.3.3 On Accuracy of Finite Element Computer Codes\n\nThere are at least three steps that must be considered when estimating the accuracy\nof computer codes for numerical temperature calculations:\n1. Accuracy of material properties\n2. Verification of the calculation model\n3. Validity of the calculation model\nThe first point is crucial. Errors in material property input will be transmitted\ninto output uncertainties and errors. Methods for measuring material properties at\nhigh temperature are briefly discussed in Sect. 1.3.3.\nSecondly, the numerical verification of the computer code itself is important.\nVerification is the process of determining the degree of accuracy of the solution of\nthe governing equations. Verification does not imply that the governing equations\nare appropriate for the given fire scenario, only that the equations are being solved\ncorrectly.\nThe third point is of course important as well. Validation is the process of\ndetermining the degree to which a mathematical model and a calculation method\nadequately describe the physical phenomena of interest. Temperature calculation\ncodes are in general developed for solving the Fourier heat transfer equation.\nEffects of varying material thermal properties can be considered in the numerical\nintegration while, for example, the thermal effects of spalling or water migration\ncannot generally be predicted. Other important aspects are the possibilities of\nsatisfactory describing boundary conditions. For FSE problems generally involving\nhigh temperatures, the calculation of heat transfer by radiation at external bound-\naries and in internal voids is of special concern.\nThe codes mentioned above yield results with acceptable accuracy for simple\nwell-defined boundary conditions and material properties. Differences when mixed\nboundary conditions and latent heat are introduced. A scheme to follow including a\nnumber of reference cases of various levels of complexity has been published in an\nSFPE standard . Precisely calculated reference temperatures of 16 cases of\nbodies have been listed. They represent a variety of problems that are relevant in\nFSE involving a range of complexities.\n7.3 The Finite Element Method for Temperature Analyses 121\n\nSome reference cases are linear problems which can be solved analytically. Then\nwhen increasing the number of elements the results should converge to one correct\nvalue. Codes yielding results that converge smoothly when increasing the number\nof elements are generally more reliable for the type of problems considered. Most of\nthe reference cases are relevant for FSE including effects of conductivity varying\nwith temperature, latent heat, radiant heat transfer boundary conditions and com-\nbinations of materials, concrete, steel and mineral wool. Then as no exact analytical\nsolutions are available, the cases were modelled in the finite element codes Abaqus\nand TASEF. The difference between the solutions obtained with these codes were\nwithin one-tenth of a degree Celsius, and as these codes employ different calcula-\ntions the published solutions of the reference cases were deemed very accurate.\n\n## 7.3.4 On Specific Volumetric Enthalpy\n\nAs shown in Eq. 7.30 the heat conduction equation can be expressed in terms of\nspecific volumetric enthalpy e. This is advantageous when calculating temperature\nwith numerical methods in cases with materials where latent heat needs to be\nconsidered. The specific volumetric enthalpy or here often just the enthalpy is the\nheat content of a material due to temperatures above zero per unit volume (Ws/m3),\ni.e.\nT X\ne T \u0002 c dT l\ni i\n7:38\n0\n\n## where is density and c specific\n\nX heat capacity. These are in general temperature\ndependent. The second term l Ws=m3 represents latent heats required for\ni i\nvarious chemical and physical phase changes at various temperature levels. The\nfirst term is the sensitive heat. The most common form of latent heat to be\nconsidered in FSE is the vaporization of moisture (free water) when the temperature\nrise passes the boiling point (100 \u0005 C).\nFor a dry inert material with a density dry and a specific heat cdry not varying\nwith temperature the enthalpy is proportional to the temperature and the sensitive\nheat becomes\n\n## If a material contains free water, the enthalpy versus temperature is influenced in\n\ntwo ways. Firstly heat proportional to the temperature rise (sensitive heat) is needed\nto increase the temperature of the water, and then in addition heat (latent heat) is\nneeded for vaporizing water at temperatures in an interval above 100 \u0005 C. Both these\ncomponents must be added to the enthalpy of the dry material when calculating the\nenthalpy as function of temperature. Thus in general terms the enthalpy consists of\n122 7 Numerical Methods\n\nthree components, the sensitive heat of the dry material, the sensitive heat of the\nwater and the latent heat due to vaporization of water. The first term is present over\nthe entire temperature range, the second added only as long as water is present, and\nthe third only in the temperature interval when the water vaporizes.\nMoisture content u is usually expressed as the percentage water by mass of the\ndry material. Thus u is defined as\nori \u0004 dry\nu 100 \u0002 7:40\ndry\n\nwhere ori is the original density of the moist material. If the moisture is assumed to\nevaporate between a lower temperature Tl and an upper Tu temperature, the latent\nheat due to water vaporization lw is added to the enthalpy at the upper temperature\nlevel. The latent heat of water is then calculated as\nu\nlw aw 7:41\n100 dry\n\n## where the heat of vaporization of water aw 2:26 MJ=kg. As an example the\n\nenthalpy as a function temperature of a material with constant dry properties can\nthen be obtained as shown in Table 7.1 and Fig. 7.7. The enthalpy is then calculated\nat four temperature levels and in-between the enthalpy varies linearly. Notice that\nas an average only half of the water is assumed to be heated between the lower\ntemperature Tl (100 \u0005 C) and the upper temperature Tu for the vaporization process.\nAs an example the enthalpy of a concrete with a dry density of 2400 kg/m3, a\nspecific heat of 800 J/(kg K) and a moisture content u 3 % by mass is shown in\nFig. 7.7a. For comparison the enthalpy for a dry concrete (u 0 %) is given as well.\nThe moisture is assumed to evaporate linearly with temperature between 100 and\n120 \u0005 C. Notice that at temperatures above Tu, the enthalpy rises linearly with\ntemperature at the same rate as for a dry material.\nMost computer programs require input of the specific heat and the density or the\nproduct of the two. This parameter is obtained by deriving the temperature\nenthalpy curve. For the case above the specific volumetric heat c\u0002 then becomes\nas shown in Fig. 7.7b.\nThe volumetric specific heat (c\u0002) as a function of temperature then increases\nsuddenly in the range where the water is assumed to evaporate. This may cause\nnumerical problems in particular for cases where the temperature range is narrow\nand the moisture content is high.\nGypsum is often used to seal penetrations through fire barriers and to protect\nsteel structures. To raise the temperature of gypsum heat is needed to heat the dry\nmaterial and to heat and evaporate the free water. In addition heat is needed for\ndehydration and vaporization of the crystalline bound water which occurs in two\nsteps. An example of calculated specific volumetric enthalpy of gypsum containing\n5 % free water and 21 % crystalline bound water is shown in Fig. 7.8 based on work\n7.3 The Finite Element Method for Temperature Analyses 123\n\nTable 7.1 Calculation of specific volumetric enthalpy, e, for a material with constant dry\nproperties with a moisture content of u % by mass of the dry material\nTemperature, T Specific volumetric enthalpy, e\n0 e0 0\n\u0007 \u0006\nTl eT l dry cdry 100\nu\ncw T l\n\u0007 \u0006\nTu eT u eT l dry cdry 0:5 100\nu\ncw T u \u0004 T l 100\nu\ndry aw\nT > Tu eT>T u eT u cdry dry T \u0004 T u\n\nFig. 7.7 Example of specific volumetric enthalpy vs. temperature of dry and moist concrete (3 %\nby mass). The moisture is assumed to evaporate linearly between 100 and 120 \u0005 C. (a) Specific\nvolumetric enthalpy, e. (b) Specific volumetric heat capacity, c\u0002\n\nby Thomas (the figure is taken from a master thesis of Emil Ringh (2014),\nLulea TU).\nNotice how the latent heats for the dehydration and vaporization processes\nsurpass by far the sensitive heat needed to heat the inert material by comparing\nthe slope of the curve below 100 \u0005 C thereafter until all the water has evaporated at\ntemperatures above 220 \u0005 C. This ability of gypsum to absorb has a significant\neffect, for instance, on gypsum boards for fire insulation of steel structures. How-\never, this is only for gypsum board qualities able to resist fire exposures. To take\nadvantage of the effects in calculations, computer codes where the specific volu-\nmetric enthalpy can be input directly are the most suitable as it may be difficult to\nconvert the curve into density and specific heat which corresponds to the derivative\nor slope of the curve.\nReliable values of the conductivity of gypsum are hard to find in the literature as\nthe temperature development in gypsum depends very much on the highly\nnon-linear enthalpy curve due to latent heats. With great reservation on the accu-\nracy the values in Table 7.2 are recommended to be used in combination with the\ntemperatureenthalpy curve shown in Fig. 7.8 for indicative calculations.\n124 7 Numerical Methods\n\n1000\n\n## Specific volumetric enthalpy [MJ/m3] 900\n\n800\n700\n600\n500\n400\n300\n200\n100\n0\n0 100 200 300 400 500\nTemperature [C]\n\nFig. 7.8 Example of calculated specific volumetric enthalpy of gypsum containing 5 % free water\nand 21 % crystalline bound water\n\nTable 7.2 Thermal conductivity and specific volumetric enthalpy of gypsum as given in Fig. 7.8\nTemperature Thermal conductivity Temperature Specific volumetric enthalpy\n[\u0005 C] [W/(m K)] [\u0005 C] [MJ/m3]\n20 0.19 0 0\n100 0.15 100 73.5\n500 0.17 110 571\n1000 0.35 200 625\n2000 0.35 220 784\n2000 2123\nChapter 8\nThermal Ignition Theory\n\nThe various aspects of the subject ignition of unwanted fires has been thoroughly\ninvestigated by Babrauskas and presented in the comprehensive Ignition\nHandbook . This book is concentrating on the calculation of the development\nof surface temperature. Despite many limitations, it is often assumed that a solid\nignites due to external heating when its exposed surface reaches a particular\nignition temperature.\nIn Sect. 8.1 some data of ignition temperature of various substances are given\nand then in Sect. 8.2 handy formulas are presented on how to calculate time to\nignition of surfaces exposed to constant incident radiation heat flux. These formula\nyields very similar results in comparison to accurate and elaborate numerical\ncalculations.\n\n## 8.1 Ignition Temperatures of Common Solids\n\nCombustible solids may ignite due to piloted ignition, or auto-ignition (also called\nspontaneous ignition). The piloted ignition temperature of an externally heated\nsubstance is the surface temperature at which it will ignite in a normal atmosphere\nwith an external source of ignition, such as a small flame or spark, present. Most\ncommon materials then ignite in the range of 250450 \u0001 C. The auto-ignition\ntemperature is the corresponding temperature at which a substance will spontane-\nously ignite without a flame or spark present. It is considerably higher, normally\nexceeding 500 \u0001 C.\nSome limited amount of relevant material data are given in Table 8.1 for some\nliquids and in Table 8.2 for some plastics.\nNote that the times to ignition as estimated by the thermal theories outlined\nbelow are generally very crude and based on the assumption of homogeneous\nmaterials with constant material properties not varying with temperature or time.\nThe formulas are, however, very useful for the intuitive understanding of which\n\n## Springer International Publishing Switzerland 2016 125\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_8\n126 8 Thermal Ignition Theory\n\n## Table 8.1 Critical temperatures of some liquids\n\nLiquid Formula Flash point [K] Boiling point [K] Auto-ignition [K]\nPropane C3H5 169 231 723\nGasoline Mixture ~228 ~306 ~644\nMethanol CH3OH 285 337 658\nEthanol C2H5OH 286 351 636\nKerosene ~C14H30 ~322 ~505 ~533\nFrom Quintiere \n\n## Table 8.2 Ignition Ignition temperature [\u0001 C]\n\ntemperatures of some plastics\nCategory of solid Piloted Auto\ngrouped by category\nThermoplastics 369 \u0004 73 457 \u0004 63\nThermosetting plastics 441 \u0004 100 514 \u0004 92\nElastomers 318 \u0004 42 353 \u0004 56\nHalogenated plastics 382 \u0004 79 469 \u0004 79\nFrom Babrauskas \n\nmaterial and geometrical properties govern the ignition process and the ignitability\ncharacteristics.\nThe time to ignition of thick homogenous materials is proportional to the thermal\ninertia (k\u0003\u0003c), i.e. the product of specific heat capacity, density and conductivity,\nsee Sect. 3.2. The conductivity increases generally at the same time as the density of\na material increases (see Eq. 1.36). Therefore the thermal inertia of materials varies\nover a large range and consequently the ignition properties. Insulating materials\nhave low conductivities k (by definition) and low densities and will therefore\nignite easily if combustible. The specific heat capacity depends on the chemical\ncomposition of the material, but the values of common materials found in the\nliterature do not vary much. An exception is wood which according to values\nfound in the literature has a relatively high effective specific heat capacity. (This\nmay be a way of considering the effects of its water content.)\nTable 1.2 shows how the thermal inertia increases considerably with density for\nvarious combustible and non-combustible materials. Notice for instance that the\nthermal inertia of an efficient insulating material such as polyurethane foam is less\nthan a hundredth of the corresponding value of solid wood. Then as an example a\nlow density wood fibre board may have a density of 100 kg/m3 and a conductivity of\n0.04 W/(m K), while a high density wood (oak) have a density of 700 kg/m3 and a\nconductivity of 0.17 W/(m K). As such boards can be assumed to have about the\nsame specific heat capacity, it can be calculated that the thermal inertia of the high\ndensity fibre board is more than 40 times higher of that of the low density board.\nThe low density fibre board can therefore ideally be estimated to ignite 40 times\nfaster than the high density fibre board when exposed to the same constant heating\nconditions, see Eq. 8.9.\n8.2 Calculation of Time to Ignitions 127\n\n## In common thermal ignition theory a material (solid or liquid) is assumed to ignite\n\nwhen the surface reaches the ignition temperature. It may be at the piloted or at the\nauto-ignition temperature. The time it takes the surface to reach such a critical\ntemperature when heated depends on the dimensions and the thermal properties of\nthe material. Below the special cases of thin and semi-infinite solids will be outlined\nin Sects. 8.2.1 and 8.2.2, respectively, as developed by Wickstrom .\nIn both cases the heat transfer by radiation and convection to an exposed surface\nis calculated according to Eq. 4.12 as\n00\n\u0001 00 \u0003 \u0004 \u0005\nq_ tot q_ inc \u0005 T 4s hc T g \u0005 T s 8:1\n\n## where is the surface emissivity and absorptivity coefficient, the Stefan\n\nBoltzmann constant, h the convection heat transfer coefficient, and Tg the ambient\ngas temperature. By calculating the surface temperature vs. time the time to ignition\ncan be obtained. Eq. 8.1 is, however, a non-linear boundary condition since the\nemitted radiation term depends on the surface temperature to the fourth power.\nTherefore a direct closed form solution cannot in general be derived for the surface\ntemperature Ts. Therefore the time to ignition must be calculated numerically.\nHowever, for the ideal case, as, for example, in the Cone calorimeter, see\nFig. 8.1, when the following conditions are present\n00\nConstant surrounding gas temperature Tg\nUniform initial temperature Ti\nConstant material and heat transfer properties\nThe time to reach the ignition temperature tig may be calculated approximately\nwith a simple explicit formula as introduced below. Similar conditions can be\nassumed when analysing, for example, heating by radiation by flames or hot objects\nonto surfaces surrounded by gases with moderate temperature.\nThe formulas derived are semi-empirical, i.e. they have been proven correct by\ncomparing with accurate numerical solutions. As a first step the third kind of BC\n(see Sect. 1.1.3) according to Eq. 8.1 is replaced by a second kind of BC, a constant\n00\neffective heat flux q_ tot, eff assumed to be\n00 00 00\nq_ tot, eff q_ inc \u0005 q_ inc, cr 8:2\n\n00\nwhere q_ inc, cr is critical incident radiation heat flux, i.e. the incident radiation\nrequired to balance the heat losses at the surface by emitted radiation and convec-\ntion at the ignition temperature.\n128 8 Thermal Ignition Theory\n\n## Fig. 8.1 Surface with heat\n\ntransfer parameters and hc\nexposed to constant uniform\nconstant surrounding gas\ntemperature\n\n00 \u0004 \u0005\n_ qinc, cr \u0005 T 4ig \u0005 hc T ig \u0005 T g 0 8:3\n\nwhich yields\n00 \u0004 \u0005\nq_ inc, cr T 4ig hc = \u0003 T ig \u0005 T g 8:4\n\n## in Eq. 8.2 is a semi-empirical reduction coefficient which was determined by\n\ncomparisons of times to ignition obtained with accurate numerical methods. The\nconstant heat flux according to Eq. 8.2 can now be calculated for thin solids with\nEq. 3.30 with 0.3 and for semi-infinite solids with Eq. 3.5 with 0.8. Then the\ntime to ignition tig can be calculated by closed form simple equations according to\nEq. 8.7 and Eq. 8.9 for thin and semi-infinite solids, respectively.\n\n## 8.2.1 Thin Solids\n\nFor thin solids the temperature may be assumed uniform throughout the depth of the\nbody. Then the thickness and the volumetric specific heat capacity are decisive for\n00\nthe time to ignition and when assuming a constant total heat flux q_ tot (see also Sect.\n3.1) the temperature rise can be calculated as:\n00\nq_ \u0003 t\nT s \u0005 T i tot 8:5\n\u0003c\u0003d\n\n## where Ts is the exposed body temperature, Ti the initial temperature, t time,\n\ndensity, c specific heat capacity and d thickness. Density times thickness (.d) is\nweight per unit area. That means that time to ignition tig of a thermally thin material\n00\nreceiving a constant total heat flux q_ tot is directly proportional to the density and the\nthickness of the material, i.e. the weight of the solid per unit area, and the\ntemperature rise to reach the ignition temperature Tig, i.e.\n8.2 Calculation of Time to Ignitions 129\n\ncd \u0004 \u0005\ntig T ig \u0005 T i 8:6\nq_ tot\n00\n\n00\nHowever, the heat flux q_ tot can rarely be assumed constant or determined a priori as\na second kind of boundary condition. Even if exposed to a constant thermal\n00\nexposure, q_ tot decreases when the surface temperature rises according to Eq. 8.1.\nAs a matter of fact, it is a third kind of boundary condition, see Sect. 1.1.3.\nNevertheless by inserting the effective heat flux according to Eq. 8.2 with 0.3\ninto Eq. 8.6, the time to reach the ignition temperature of thin solids exposed to\n\n\u0003c\u0003d \u0004 \u0005\ntig \u0006 00 00 T ig \u0005 T i 8:7\n\u0003 q_ inc \u0005 0:3 \u0003 q_ inc, cr\n\nCalculated ignition times according to Eq. 8.7 yields very good approximations of\nthe accurate predictions obtained by the numerical solutions where the real bound-\n00\nary heat flux q_ tot , according to Eq. 8.1, is assumed and where the time to ignition is\nthe time when the body surface reaches the ignition temperature.\nExample 8.1 Calculate the time to ignition for a thin curtain with an area density\nof \u0003 d 300 g=m2 and an ignition temperature T ig 350 \u0001 C when exposed to an\n00\nincident radiation q_ inc 20 kW=m2 from both sides. Assume a specific heat\ncapacity of curtain c 850 W s/(kg K), a surrounding temperature and an initial\ntemperature Ti equal to 20 \u0001 C, a convective heat transfer coefficient of 5 W/(m2 K)\nand an emissivity 0:9.\n850\u00030:3\u0003350\u000520\nSolution Equation 8.7 yields tig 2\u00030:9\u000320000\u00050:3\u000310375 2:8 s.\n\n## 8.2.2 Semi-infinite Solids\n\nA similar expression as given by Eq. 8.5 for thin solids can be derived for semi-\ninfinite solids or thermally thick solids, i.e. the thickness is larger than the thermal\n00\npenetration depth, see Sect. 3.2.1. Then for a constant heat flux to the surface q_ s\n(2nd kind of BC) and constant thermal properties, the time to reach a given\n\n\u0003 k \u0003 \u0003 c\u0004 \u00052\ntig \u0004 00 \u00052 T ig \u0005 T i 8:8\n4 q_ s\n\nwhere Ti is the initial temperature. The product of the heat conductivity k, the\nspecific heat capacity c and the density is the thermal inertia k \u0003 \u0003 c of the\nmaterial as defined in Sect. 3.2. As for thin solids the heat flux to the surface cannot\n130 8 Thermal Ignition Theory\n\n## be specified as a second kind of boundary condition. However, for the particular\n\n00\ncase of a constant incident radiation flux q_ inc and a constant ambient gas temper-\nature Tg and a uniform initial temperature Ti, the time to ignition can be approxi-\nmated as shown below for semi-infinite solids.\nBy inserting the effective average value of the heat flux according to Eq. 8.2 with\n0.8 into Eq. 8.9 and rearranging, the time to reach the ignition temperature of\nsurfaces of solids exposed to radiation may then be approximated as\n\" \u0004 \u0005 #2\nk \u0003 \u0003 c T ig \u0005 T i\ntig \u0004 00 00\n\u0005 8:9\n4 2 q_ inc \u0005 0:8 \u0003 q_ inc, cr\n\n00\nand after inserting q_ inc, cr from Eq. 8.4 a closed form explicit expression for the time\nto ignition is obtained as\n2 32\nk \u0003 \u0003 c 4 T ig \u0005 T i\ntig\n4 2 00\nh \u0004 \u0005i5 8:10\n_q inc \u0005 0:8 \u0003 T 4ig hc T ig \u0005 T g\n\nEquations 8.9 and 8.10 match very well the times to ignition as calculated by\naccurate numerical procedures for a wide range of the parameters incident radiation\n00\nq_ inc , ignition temperature Tig and thermal inertia k \u0003 \u0003 c .\nAccording to the above equations the inverse of the square root of the ignition\ntime is linearly dependent on the incident radiation, i.e.\n\n1 2 h 00 00\ni\np \u0004 \u0005 q_ inc \u0005 0:8 q_ inc, cr 8:11\ntig k \u0003 \u0003 c\u0007 T ig \u0005 T i\n\nThus according to Eq. 8.11 linear relations are obtained as shown in Fig. 8.2 for\nvarious thermal inertia and in Fig. 8.3 for various ignition temperatures. Values\ntypical for the Cone Calorimeter test scenario has been assumed for both the\ndiagrams in Figs. 8.2 and 8.3, i.e. initial temperature Ti 20 \u0001 C, the surface\nemissivity 0:9, the convection heat transfer coefficient hc 12 W=m2 K.\nThe lowest thermal inertia 1000 (W2 s)/(m4 K2) may be representative of low\ndensity polymeric insulation material which heats up very quickly while an inertia\nof 100000 (W2 s)/(m4 K2) may represent soft wood and 300000 (W2 s)/(m4 K2) hard\nwood such as oak. These values are only indicative and are not recommended to be\nused in real application. Notice that the graphs cross the abscissa at 80 % of the\n00\ncritical incident flux, i.e. at 0.8 q_ inc, cr , independently of the thermal inertia of the\nmaterial.\nThe theory indicates how significant the thermal inertia is for the time to\nignition. As an example Fig. 8.2 indicates the time to ignition for softwood exposed\nto 30 kW/m2 is 1=0:252 16 s and 1=0:152 44 s for hardwood when exposed to\n30 kW/m2. The very short ignition times for low density insulation materials even at\n8.2 Calculation of Time to Ignitions 131\n\n0.40\nTig =300 C\n\n0.35\n\n0.30\n\n0.25\n1/(tig) [s-]\n\n0.20\n\n0.15\n\nkc=1000\n0.10\nkc=10000\nkc=100000\n0.05\nkc=300000\n\n0.00\n0 5 10 15 20 25 30 35 40 45 50\n\nFig. 8.2 The inverse of the square root of time to ignition vs. incident radiation heat flux\naccording to Eq. 8.11 assuming an ignition temperature T ig 300 \u0001 C for various thermal inertia\nk \u0003 \u0003 c given in (W2 s)/(m4 K2). T i 20 \u0001 C, 0:9 and hcon 12 W=m2 K\n\nmoderate incident radiation levels indicates the hazardous fire properties of these\ntype of materials.\nIn Fig. 8.3 the inverse of the square root of the ignition time vs. incident\nradiation is shown for various ignition temperatures assuming a thermal inertia of\n100000 (W2 s)/(m4 K2) corresponding to soft wood. The ignition temperature of\n500 \u0001 C is only relevant for auto-ignition circumstances while the other temperature\nlevels may be relevant for piloted ignitions for most materials of interest. Notice\nthat the ignition temperature has a great influence on time to ignition and on the\ncritical incident radiant heat flux. As an example softwood exposed to an incident\nradiation of 20 kW/m2 would ignite after 51 s when the ignition temperature is\nassumed to be 200 \u0001 C and after only 13 s if assumed to be 300 \u0001 C.\nExample 8.2 Calculate the time to ignition of a surface of thick wood\n(k \u0003 \u0003 c 196000 (W2 s)/(m4 K)) solid suddenly exposed to an incident radiation\n00\nheat flux q_ inc 30 kW=m2 . The wood surface emissivity 0.8, the convection\nheat transfer coefficient h 12 W/(m2 K), the other thermal properties according to\nTable 1.2.\n(a) The solid is initially at 20 \u0001 C and surrounded by air at the same temperature.\n(b) The solid is initially at 100 \u0001 C and surrounded by air at the same temperature.\n132 8 Thermal Ignition Theory\n\nFig. 8.3 The inverse of the square root of time to ignition vs. incident radiation heat flux\naccording to Eq. 8.11 assuming a thermal inertia of 100000 (W2 s)/(m4 K2) for various ignition\ntemperatures. T i 20 \u0001 C, 0:9 and hcon 12 W=m2 K\n\n00\nSolution According to Eq. 8.4 q_ inc, cr 5:67 \u0003 10\u00058 \u0003 6234 12=0:8 \u0003 350 \u0005 20\nh i2\n13492 W=m2 . Equation 8.9 yields tig \u0003196000\n4\u00030:82\n350\u000520\n30000\u00050:8\u000313492 71 s. (Com-\nment: Applying the accurate boundary condition according to Eq. 8.1 yields by\naccurate numerical calculations tig 69 s).\n00\nAccording to Eq. 8.4 q_ inc, cr 5:67 \u0003 10\u00058 \u0003 6234 12=0:8 \u0003 350 \u0005 100\nh i2\n12292 W=m2 . Equation 8.9 yields tig \u0003196000\n4\u00030:82\n350\u0005100\n30000\u00050:8\u000312402 37 s.\n\nComment: Thus this material would ignite in about half the time if preheated from\n20 to 100 \u0001 C.\nChapter 9\nMeasurements of Temperature\nand Heat Flux\n\n## In FSE temperature is nearly always measured with thermocouples as described in\n\nSect. 9.1. Heat flux measured in different ways is most commonly measured as the\nsum of the net heat flux by radiation and convection to a cooled surface. The\nprinciples are briefly outlined in Sect. 9.2. Alternative methods incident radiation\nheat flux as well AST using so-called plate thermometers has also been developed\nas a practical alternative to heat flux meters as outlined in Sect. 9.3.\n\n9.1 Thermocouples\n\n## Thermocouples (sometimes abbreviated T/C) have a junction between a pair of\n\nwires of two different metals or alloys. A voltage is then generated proportional to\nthe temperature difference between the so-called hot junction and the cold junction,\na reference point with known temperature. The hot junction of thermocouples can\neither be imbedded in solid materials or mounted in free space. Thermocouples are\nin general relatively inexpensive and easy to handle, and can be used for measuring\ntemperatures over a wide range. They are therefore very common in fire testing and\nresearch. Different alloys are used for different temperature ranges.\n\n## 9.1.1 Type of Thermocouples\n\nThere are a number of standardized types and combinations metals for thermocou-\nples. The most common have been designated letters by ISA (Instrument Society of\nAmerica) and ANSI (American National Standards Institute). Information on the\nvarious types of thermocouples and their letter designation is given in the interna-\ntional standard IEC 584.\n\n## Springer International Publishing Switzerland 2016 133\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_9\n134 9 Measurements of Temperature and Heat Flux\n\nIn fire testing and research thermocouples of type K are by far the most common.\nThe positive lead is then made of a nickel alloy with 10 % chrome and the negative\nof a nickel alloy with 2 % aluminium, 2 % mangan and 1 % silicon. The relation\nbetween the output voltage and temperature is almost linear with a sensitivity of\napproximately 41 V/K. The melting point is about 1400 \u0001 C and the mechanical\nproperties and the resistance against corrosion are satisfactory also at high temper-\nature levels. At temperatures above 800 \u0001 C, however, oxidation may occur leading\nto substantial measuring errors. The thermocouples may also age when used for\nlonger times at temperatures above 500 \u0001 C and should therefore in such cases be\ncalibrated about every 20 h of use . According to the international standards for\nfire resistance furnace tests ISO 834 and EN 1363-1 thermocouples may not be used\nfor more than 50 tests.\nFor temperature measurements up to 1480 \u0001 C thermocouples of type S, plati-\nnumplatinum/rhodium, are sometimes used in fire resistance furnaces. They are,\nhowever, expensive and only suitable for short term measurements as they degrade\nat high temperatures.\nThere are industrially manufactured thermocouples in many metal combina-\ntions. The connections of the thermocouples leads are, however, often made by the\nuser by soldering, electric or gas welding or pressing depending on the intended use.\nSoldering and welding is mainly used for thin thermocouples. Pressing is used with\nso-called quick tips. The latter method yields rather big hot junctions which makes\nthe thermocouples relatively slow when recording dynamic processes as in fires and\nless sensitivity to heat transfer by convection, see next Section. Thus their temper-\nature recordings must often be corrected due to the effects of radiation when\nmeasuring gas temperature accurately.\nresearch. They have a stainless steel or similar casing protecting the thermocouple\nfrom mechanical stresses and corrosive gases. These are in general more robust but\nconsiderably more costly.\n\n## 9.1.2 Measurement of Temperature in Gases\n\nThe temperature recording you get from a thermocouple is always the temperature\nof the hot junction of the thermocouple leads. When placed in a gas it adjusts more\nor less quickly to surrounding temperatures depending on its thermal response\ncharacteristics to convection and radiation. Briefly it can be said that the smaller\ndimensions of a thermocouple the quicker it responses to thermal changes and the\nmore sensitive it is to convection and thereby gas temperature, and vice versa.\nThus it is important to realize that thermocouples in gases are influenced by the\ngas temperature Tg as well as by the incident radiation or the black body radiation\ntemperature Tr (see Chap. 4). It adjusts to a temperature which is a weighted\naverage of the two temperatures which may be very different. The weights are\n9.1 Thermocouples 135\n\nthe heat transfer coefficients hr and hc for radiation and convection, respectively.\nThus the ultimate or equilibrium thermocouple temperature TTC becomes\n\nhr \u0003 T r hc \u0003 T g\nT TC 9:1\nhr hc\n\nwhere\n\u0001 \u0003\nhr T 2r T 2TC T r T TC 9:2\n\nNotice that Eq. 9.1 is implicit as hr depends on TTC, and that it is similar to the\nexpression for the AST in Eq. 4.23. When radiation and thermocouple temperatures\nare approximately equal, i.e. T r \u0004 T TC (as for thermocouples in thick flames) the\nradiation heat transfer coefficient may be approximated as\n\nhr \u0004 4 \u0003 \u0003 T 3TC 9:3\n\nAs indicated in Chap. 6, convection heat transfer coefficients decrease with the size\nof a body. Hence smaller thermocouples will have greater convection heat transfer\ncoefficients hc and will therefore adjust closer to the gas temperature while larger\nthermocouples will deviate more from the gas temperature and adjust closer to the\nradiation temperature as indicated by Eq. 9.1.\nThe difference between the true gas temperature and the thermocouple temper-\nature at equilibrium can be written as\n\nhr\nT T TC \u0005 T g T r \u0005 T TC 9:4\nhc\n\nwhich implies that the difference between the measured temperature TTC and the\ntrue gas temperature Tg increases with the ratio between the heat transfer coeffi-\ncients and the difference between the radiation temperature and the thermocouple\ntemperature.\nA special case is the plate thermometer as described in Sect. 9.3 which has a\nlarge exposed area. The convection heat transfer coefficient hc is therefore rela-\ntively small, and hence the equilibrium temperature of a PT is closer to the incident\nradiation temperature than the corresponding temperature of ordinary thermocou-\nples. In addition the PT is dependent on direction of incident radiation while a\nthermocouple is not.\nThe time response characteristics are also important to consider when measuring\ngas temperatures. A general rule is thermocouples response faster the thinner they\nare, the less mass they have. As the temperature in a thermocouple can be assumed\nuniform and it can be calculated assuming lumped-heat-capacity (Sect. 3.1). Thus\nthe temperature TTC of a thermocouple suddenly exposed to a constant fire temper-\nature Tf may be calculated according to Eq. 9.5 as\n136 9 Measurements of Temperature and Heat Flux\n\nT TC \u0005 T i\n1 \u0005 e \u0005\nt\n9:5\nTf \u0005 Ti\n\nwhere Ti is the initial thermocouple temperature and is the time constant of the\nthermocouple which then can be calculated as\n\nV TC \u0003 TC \u0003 cTC\n9:6\nATC \u0003 htot\n\nwhere the parameters TC and cTC are the density and the specific heat capacity of\nthe thermocouple junction including the mass of soldering, etc. and VTC/ATC is the\neffective volume-to-area ratio. When assuming constant conditions the value time\nconstant is the time elapsed when the temperature rise has reached 63 % of its final\nvalue. In reality the time constant varies considerably as hr increases significantly\nwith temperature.\nFor a total heat transfer (by convection and radiation) coefficient htot, the time\nconstant for a sphere can be identified as\n\n1\nD \u0003 s \u0003 cs =htot 9:7\n6\n\n## Assuming the thermocouple hot junction as a cylinder with a diameter\n\nD disregarding the end surface yields\n\nD\n\u0003 cs =htot 9:8\n4 s\n\nWhen exposure temperature Tf varies with time the thermocouple temperature can\nbe obtained to Eq. 3.16.\nExample 9.1 What is the time constant of a thermocouple T/C exposed to\nuniform temperature at a level of T f 500 \u0001 C? Assume that the T/C is spherical\nwith a diameter of 3 mm of steel with a convective heat transfer coefficient\nhc 50 W/m2 K. The density and specific heat capacity of the T/C may be assumed\nto be 7850 kg/m3 and 460 Ws/kg K, respectively, and its emissivity\n\u0004 0.9.\u0005\n00 4 4\nGuidance: Assume the heat transfer to the T/C is q_ T f \u0005 T TC hc\n\u0001 \u0003\nT f \u0005 T TC and that T TC T f when calculating the radiation heat transfer\ncoefficient hr.\nSolution Equation 9.3 (or Fig. 4.2a) yields hr 4 \u0003 0:9 \u0003 5:67 \u0003 10\u00058 \u0003 500 2733\n105 W=m2 K and then htot 105 50 155 W=m2 K. According to Eq. 9.7\n16 \u0003 0:003 \u0003 7850 \u0003 460=155 11:6 s.\nExample 9.2 Calculate the time constant of the 1 mm thick thermocouple in\nExample 6.3.\n9.1 Thermocouples 137\n\n## (a) Initially at room temperature (300 K)\n\n(b) At its equilibrium temperature (1000 K)\nwhen suddenly exposed to gas and radiation temperatures of 1000 K. Assume the\nT/C is made of stainless steel, i.e. TC 0:7, TC 7900 kg=m3 and\ncTC 460 W=kg K.\nSolution\n(a) According to Example 6.3 hc 131 W=m2 K. The radiation heat transfer\n\u0001 \u0003\ncoefficient is obtained from Eq. 4.5 as hr 0:7 \u0003 5:67 \u0003 10\u00058 10002 3002\n1000 300 56 W=m2 K. Then htot 131 56 187 W=m2 K and\nTC 0:001=4 \u0003 7900 \u0003 460=187 4:8 s.\n(b) According to Example 6.3 the convection hc 147 W/(m2 K). According to\nEq. 4.6 as hr 4T 3r 159 W=m2 K and htot 147 159\n306 W=m K. Now the time constant of the thermocouple can be estimated\n2\n\n## 9.1.3 Corrections of Time Delay\n\nAll thermocouples respond to the thermal exposure with a time delay depending on\nthe thermocouple characteristics and thermal environment as described above.\nWhen the response of a thermocouple can be expressed as in Eq. 9.5,\ni.e. assuming lumped-heat-capacity, the value of Tf at a given time may be obtained\nnumerically by solving the so-called inverse problem. The time constant must\neither be known explicitly or implicitly, for example, as functions of the exposure\ntemperature Tf and the response temperature TTC as shown by Eq. 9.7. The time\nderivative of the thermocouple temperature may then at any arbitrary time t be\nderived from Eq. 9.5 as\n\ndT TC T f \u0005 T TC\n9:9\ndt\n\n## Given a series of thermocouple recordings true exposure level Tf can be derived\n\nfrom Eq. 9.9. The time derivative of the thermocouple temperature is then approx-\nimated by the corresponding differential between two consecutive thermocouple\nrecordings and the following expression can be derived:\n\u0004 \u0005 j1\nT j1 1 T \u0005 Tj 9:10\nf\nt TC t TC\n138 9 Measurements of Temperature and Heat Flux\n\nFig. 9.1 The ASTM-E119 fire curve and temperatures the thermocouple according to ASTM\nE-119 and the PT according to ISO 834 must follow to obtain the specified furnace temperature\ndue to time delay\n\n## where t is a time increment of the measurement and j is the measurement number.\n\nThe accuracy and the numerical stability of such a calculation depends on the\nrelation between and t.\nAs an example Fig. 9.1 shows the actual furnace temperature rise in a furnace\ncontrolled ideally precisely according to the ASTM E-119 standard fire curve with\ntemperature monitoring thermocouples according to the standard time constants in\nthe range of 57.2 min. Notice that the real or effective furnace temperature is much\nhigher than indicated by the slowly responding ASTM type of shielded thermo-\ncouples. The diagram also shows for comparison the corresponding curve for a\nstandard PT according to ISO 834. This curve does not deviate as much from the\nideal standard ASTM timetemperature curve as the time constant of a PT is much\nshorter than the time constant of an ASTM thermocouple.\nThis implies that when predicting temperature in specimens being exposed to a\nstandard ASTM E-119 furnace test it is important to assume a much higher\nexposure temperature for the first 10 min than what has been recorded in the test\nby the standard thermocouples. In calculations for deriving the T/C and PT\nresponse curves in Fig. 9.1 properties according to Table 9.1 were applied. These\nvalues are reasonable but uncertain depending on among other thing furnace\ncharacteristics and therefore the curves of Fig. 9.1 should just be taken as indicative\nimplying that the influence of the time delay is significant for an ASTM-E119 test\nand must be considered in particular when predicting temperature in structures\nexposed to short test durations.\n9.2 Heat Flux Meters 139\n\nTable 9.1 Parameters used for analysing the time delay of the ASTM-E119 thermocouple and the\nISO 834 PT in combination with Eq. 4.12\nEffective thickness, Convection heat transfer coefficient, Emissivity,\nd [mm] hc [W/(m2 K)] [-]\nASTM-E119 6 50 0.8\nthermocouple\nISO 834 plate 0.7 25 0.8\nthermometer\n\n## 9.2 Heat Flux Meters\n\nIn several fire test methods incident radiation levels are specified. Therefore, it is\nimportant that radiation heat flux can be measured with sufficient accuracy. It is\nusually measured with so-called total heat flux meters of the Gardon or Schmidt-\nBoelter types. Such meters register the combined heat flux by radiation and\nconvection to a water cooled surface. Thus the measurement will contain contribu-\ntions by convection which depends on a number of factors such as the design of the\nheat flux meter, the orientation of the meter, the cooling water temperature, the\nlocal temperature and gas/air flow conditions. In unfavourable conditions the\nuncertainty due to convection can amount to 25 % of the total heat flux, see ISO\n14934. As a general rule the error is lesser when the meter is surrounded by a gas at\na temperature close to the cooling water temperature while the errors may be very\nlarge when the meter is exposed to hot fire gases or flames. Under such conditions\nGardon or Schmidt-Boelter type meters are both impractical and inaccurate. Then\ndevices such as the PT as described in Sect. 9.3 are more useful.\nThe principal designs of a Gardon and a Schmidt-Boelter heat flux meters are\nshown in Fig. 9.2. In the Gardon gauge the temperature difference between\nthe middle of the circular disc and its water cooled periphery is proportional\nto the received heat flux by radiation and convection. In the Schmidt-Boelter\ngauge the temperature difference between the exposed surface and a point at a\ndepth below is measured with a so-called thermopile including several hot and cold\njunctions. This type of HFM therefore yields a higher output voltage than a Gardon\ngauge for the same flux.\n\n## 9.2.1 Calibration and Use of Heat Flux Meters\n\nHeat flux meters such as the Gardon gauge and Schmidt-Boelter gauge are cali-\nbrated according to ISO 14934 in a spherical furnace with a uniform temperature.\nThe gauge is then exposed to an incident radiation proportional to the fourth power\nof the furnace temperature T4fur . The heat transfer by convection is negligible in the\ncalibration configuration (see Fig. 9.3) and therefore the heat transfer to the water\ncooled sensing surface is:\n140 9 Measurements of Temperature and Heat Flux\n\nThermopile\nConstantan foil\n\n## Cooling water channels\n\nCopper body\nCooling water channels\nCopper body\n\nFig. 9.2 Principal cross sections of total heat flux meters. (a) Gardon gauge (b) Schmidt-Boelter\ngauge\n\nFig. 9.3 A 3-D drawing of a heat flux meter and a spherical calibration furnace with a heat flux\nmeter mounted in the bottom opening. (a) Heat flux meter (b) Calibration furnace\n\n00\n\u0004 \u0005\nq_ hf g hf m T 4fur \u0005 T 4hf m 9:11\n\nwhere Thfm is the surface temperature of the sensing body. At high furnace temper-\nature or heat flux levels the second term of the above equation is relatively very\nsmall and can be neglected. Otherwise Thfm is assumed equal to the cooling water\ntemperature. The coefficient is a test configuration parameter depending on the\ngeometric configuration when a HFM is mounted in the test furnace.\nThe calibration procedure of a HFM means that the electric voltage output is\ndetermined at several heat flux levels obtained by various furnace temperature\nlevels. Then a normally linear relation can be established between the heat flux\n00\nq_ hf m and the output voltage.\nWhen in use and exposed to radiation and convection the general expression of\n00\nthe heat flux q_ hf m to the sensing body of a HFM is:\n9.2 Heat Flux Meters 141\n\n00\n\u0004 00 \u0005 \u0001 \u0003\nq_ hf m hf m q_ inc \u0005 T 4hf m hhf m T g \u0005 T hf m 9:12\n\n00\nwhere Tg is the gas temperature near the HFM. The incident heat flux q_ inc can be\nobtained, given the gas temperature Tg and the emissivity hfm and the convection\nheat transfer coefficient hhfm are known, as\n\n00 1 h 00 \u0001 \u0003i\nq_ inc q_ hf m \u0005 hhf m T g \u0005 T hf m T 4hf m 9:13\nhf m\n\nOften it is assumed that hf m \u0004 1 and the term T4hfm is negligible, and when\nT g \u0004 T hf m , the convection term vanishes as when placed in air at ambient temper-\nature. Then the incident radiation heat flux can be approximated as\n00 00\nq_ inc q_ hf m 9:14\n\nWhen then using the measured data for calculating the heat transfer to a target\nsurface with a temperature Ts based on HFM measurements the general expression\n00\naccording to Eq. 4.11 applies, and the total heat flux q_ tot to a surface can be derived\nby inserting Eq. 9.13 into Eq. 4.11:\n00 s h 00 \u0001 \u0003i\nq_ tot q_ hf m hf m T 4hf m \u0005 hhf m T g \u0005 T hf m\nhf m \u0001 \u0003 9:15\n\u0005s T 4s hc T g \u0005 T s\n\nThus the total heat transfer depends on the emissivity and the convection heat\ntransfer coefficient of both the HFM and the target surface. These parameters are\noften not very well known which introduces great uncertainties especially when the\nHFM is placed in hot gases or flames with temperatures deviating from the cooling\nwater temperature. Then the uncertainty due to the convection becomes significant\nas the heat transfer by convection to a HFM with its small surface is difficult to\nestimate accurately. However, usually the emissivities of the HFM and the target\nsurface are assumed equal, and when the gas and cooling water temperature are\nassumed equal as well, then the heat transfer to an adjacent target surface becomes\nindependent of the gas temperature Tg and the expression of the total heat flux\nbecomes:\n00 00 \u0001 \u0003\nq_ tot q_ hf m \u0003 T 4hf m \u0005 \u0003 \u0003 T 4s \u0005 h T s \u0005 T hf m 9:16\n\nThere are several uncertainties in this expression. A more complete analysis of the\nuse of heat flux meters are given by Lattimer .\nExample 9.3 A water cooled heat flux meter is used to measure the total incident\nheat flux from a fire against a wall painted black. The measured heat flux is\n30 kW/m2 and the water cooled gauge is measured to be 350 K. Both the wall\n142 9 Measurements of Temperature and Heat Flux\n\nemissivity and the heat flux gauge have a surface emissivity 0.95, and the heat\ntransfer coefficient is 10 W/m2 K.\n00\n(a) Determine the total heat flux q_ tot into the wall when its surface temperature is\n600 K, and 800 K.\n(b) Given the gas temperature Tg is 300 and 1000 K, respectively, what is the\n00\nincident radiant heat q_ inc ?\n(c) What is the AST, i.e. the temperature of the surface when the net heat flux into\nthe wall vanishes, for the two gas temperature levels?\n(d) Use the ASTs calculated in (c) to calculate the net heat fluxes to the surfaces\nwhen the surface temperature is 600 K and 800 K, respectively. Compare with\nthe results obtained in (a).\nSolutions\n00\n(a) Equation 9.16 yields: q_ tot 30000 \u0005 10 \u0003 T s \u0005 350 \u0005 0:95 \u0003 5:67 \u0003 10\u00058 \u0003\n\u0001 4 \u0003 00\nT s \u0005 3504 . Then for T s 600 the total heat flux q_ tot 21:3 \u0003 103 W/m2,\n00\nand for T s 800 K the total heat flux q_ tot 4:2 \u0003 103 W/m2.\n00\n(b) For T g 300 K, then according to Eq. 9.13 q_ inc 0:95 1\n\u0003 30000\u0005\n\u00058\n10 \u0003 300 \u0005 350\u0006 5:67 \u0003 10 \u0003 350 30500 851 32:96 \u0003 103 W=m2 .\n4\n00\nFor T g 1000 K then q_ inc 0:95\n1\n30000\u0005\n\u00058\n101000 \u0005 350\u0006 5:67 \u0003 10 \u0003 350 24740 851 25:59 \u0003 10 W=m2 .\n4 3\n\n## Comment: Notice that the incident radiation may be considerably different\n\nfor the same heat flux meter recordings. The heat transfer coefficient is taken\nfrom Lattimers . It may be considerably higher in reality which would\nenhance the differences.\n\u0001 00 \u0003 \u0001 \u0003\n(c) For T g 300 K, then according to Eq. 4.21 \u0003 q_ inc \u0005 T 4AST hc T g \u0005 T AST\n\u0006 \u0001 \u0003 \u0007\n0:95 \u0003 32:96 \u0003 103 \u0005 5:67 \u0003 10\u00058 \u0003 T 4AST 10 \u0003 300 \u0005 T AST 0. Thus\n\u0006\nby iteration T AST 833 K. For T g 1000 K, then 0:95 \u0003 25:59 \u0003 103 \u0005\n\u0001 \u0003 \u0007\n5:67 \u0003 10\u00058 \u0003 T 4AST 10 \u0003 1000 \u0005 T AST 0. Thus T AST 833 K.\n00 \u0001 \u0003\n(d) According to Eq. 4.31 q_ tot \u0003 T 4AST \u0005 T 4s hc T AST \u0005 T s . Then for\n00 \u0001 \u0003 \u0001 \u0003\nT s 600 K q_ tot 0:95 \u0003 5:67 \u0003 10\u00058 \u0003 8334 \u0005 6004 10 \u0003 833 \u0005 600\n00 \u0001 \u0003\n21:3 \u0003 103 W=m2 , and for T s 800 K q_ tot 0:95 \u0003 5:67 \u0003 10\u00058 \u0003 8334 \u0005 8004\n10 \u0003 833 \u0005 800 4:2 \u0003 103 W=m2 .\nComment: Exactly the same values were obtained when calculating the total\n00\nheat flux based on q_ hf m as based on TAST according to the theory presented. In\npractice it would most certainly be more expedient to use PTs for measuring\nASTs and use those measurements for calculating heat flux and temperature of\nthe exposed wall.\n9.3 The Plate Thermometer 143\n\n## 9.3 The Plate Thermometer\n\n9.3.1 Introduction\n\nThe standard Plate Thermometer PT as specified in the international ISO 834 and in\nthe European EN 1363-1 was invented to measure and control temperature in fire\nEuropean resistance furnaces with the purpose of harmonizing the thermal\nexposure and assuring tests results independent of type of fuel and furnace design.\nThe standard PT as shown in Fig. 9.4 is made of a shielded thermocouple\nattached to the centre of a 0.7-mm-thick metal plate of Inconel 600 (a trade name\nof an austenitic nickel based super alloy for high temperature oxidation resistance)\nwhich is insulated on its back side. The exposed front face is 100 mm by 100 mm.\nThe back side insulation pad is 10 mm thick.\nA relatively large sensor surface, such as a PT, measures neither the gas\nbetween the radiation and gas temperatures. It measures approximately the tem-\nperature of a surface which cannot absorb any heat. This temperature has been\nnamed the Adiabatic Surface Temperature, AST , see Sect. 4.4. PT can also\nbe used to measure incident radiant heat flux to a surface [32, 33] as will be shown\nin the next section.\nAs shown in Chap. 4, the concept of the AST is very valuable as it can be used\nfor calculation of heat transfer to fire-exposed body surfaces when exposed to\nconvection and radiation boundary conditions, so-called mixed boundary condi-\ntions, where the gas temperature and the radiation temperatures may be consider-\nably different. Figure 9.5 shows PTs being mounted in different directions around a\nsteel girder.\nThe concept of AST is not limited to fire resistance scenarios and predictions of\nstructural element temperatures. It can also be used at more moderate temperature\nlevels for instance to estimate whether a surface will reach its ignition temperature\nwhen exposed to elevated incident radiation but moderate gas temperatures.\n\n## 9.3.2 Theory for Measuring Incident Heat Flux\n\nand Adiabatic Surface Temperature with Plate\nThermometers\n\n## A simplified heat balance equation of the exposed surface plate of a PT may be\n\n00 \u0001 \u0003 \u0001 \u0003 dT PT\nPT q_ inc \u0005 PT T 4PT hPT T g \u0005 T PT K PT T g \u0005 T PT CPT 9:17\ndt\n144 9 Measurements of Temperature and Heat Flux\n\n## Fig. 9.5 PTs being\n\nmounted for measuring\nASTs in different directions\nat the surfaces around a\nsteel girder \n\n## Fig. 9.6 Indication of the\n\nheat transfer to a PT. The\nnumbers relate to the terms 1 2\non the left-hand side of\nEq. 9.17 3\n\n4 4 4\n9.3 The Plate Thermometer 145\n\nThe first term on the left-hand side of Eq. 9.12 is the radiant heat absorbed by the\nInconel plate, the second the heat emitted; the third the heat transferred by convec-\ntion and the fourth term expresses the heat lost by conduction through the insulation\npad plus along the Inconel plate. The latter is assumed to be proportional to the\ndifference between the plate temperature TPT and the gas temperature Tg with the\nproportionality constant denoted KPT.\nThe term on the right-hand side of the equation is the rate of heat stored\ncalculated assuming lumped-heat-capacity (see Sect. 3.1). CPT is assumed to be\nthe heat capacity of the Inconel plate plus a third of the heat capacity of the\ninsulation pad. (The third is taken from experiences of insulated steel structures,\nsee Sect. 13.3.1).\nA thorough two-dimensional thermal finite element analysis of the standard ISO\n834 PT is presented in . It was then found that with the thermal conduction\ncoefficient K PT 8:0 W=m2 K and the heat capacity CPT 4200 J=m2 K there\nwas a good agreement between PT temperatures calculated with FE analyses and\nthe temperatures obtained using Eq. 9.20. The convection heat transfer coefficient\nof the PT hPT depends on the actual scenario. In the case of natural convection only,\nit may be assumed to be in the order of 10 W/(m2 K), see Sect. 6.3.1.1.\n00\nThe incident radiation q_ inc can be derived from Eq. 9.18 as\n\b\n00 1 \u0001 \u0003 dT PT\nq_ inc T 4PT \u0005 hPT K PT \u0003 T g \u0005 T PT \u0005 CPT 9:18\nPT dt\n\nThe derivative of the transient term can be approximated by the differential, i.e.\nTPT 00\ndt \u0004 t . Then\ndTPT\nq_ inccan be obtained by a stepwise procedure where\nj1 j j1 j\nTPT TPT \u0005TPT TPT \u0005TPT\nt tj1 \u0005tj t .\nUnder steady state or relatively slow processes the transient term can be\nneglected. In addition at high incident radiation levels the first term is dominant\nand the dependence on conduction and convection is relatively small and may in\napproximative analyses even be neglected. In Fig. 9.7 the incident radiation flux\n00\nq_ inc is shown as a function of the temperature TPT of a PT mounted vertically in air\nat ambient temperature, T g 20 \u0001 C. The emissivity is assumed PT 0:9 and the\nnatural convection heat transfer coefficient is calculated accurately as a function of\n00\ntemperature according to Eq. 6.30. The incident radiation flux q_ inc is shown with the\nassumption of the heat loss by conduction parameter being neglected K PT 0 and\nK PT 4 W=m2 K, respectively. The latter is representative for a so-called insPT\nas shown in Fig. 9.9 with a 20 mm insulation pad. As can be observed the influence\nof the uncertain parameter KPT is relatively small in comparison to the uncertainties\nrelated to measurements with conventional heat flux meters, see Sect. 9.2.\nKPT and CPT may often be neglected, in particular when insulated plate ther-\nmometers insPTs as shown in Fig. 9.10 are used. Then the incident heat flux can be\ncalculated as\n146 9 Measurements of Temperature and Heat Flux\n\n30 000\n\n= 0.9\n25 000\n\n20 000\n\n15 000\n\n10 000\n\nKPT=4 W/(m2 K)\n5 000\nKPT=0 W/(m2 K)\n\n0\n0 100 200 300 400 500\nPlate thermometer temperature , TPT [C]\n00\nFig. 9.7 Incident radiation q_ inc based on steady-state PT measurements in ambient air assuming\nthe heat loss parameter by conduction negligible K PT 0 and K PT 4 W=m2 K, respectively.\nThe heat loss by natural convection is calculated according to Eq. 6.30\n\n00 hPT \u0001 \u0003\nq_ inc T 4PT \u0005 T g \u0005 T PT 9:19\nPT\n00\nThe lower curve of the diagram of Fig. 9.7 shows the relation between TPT and q_ inc\nwhen neglecting both KPT and CPT.\nTAST can be derived from PT recordings considering KPT and CPT by heat\nbalance equation\n\n\u0001 \u0003 \u0001 \u0003 dT PT\nPT T 4AST \u0005 T 4PT hPT T AST \u0005 T PT K PT T g \u0005 T PT CPT 9:20\ndt\n\nGiven a series of TPT measurements the derivative of the transient term can be\napproximated by the differential dTdtPT \u0004 TtPT , and the inverse problem of calculating\nTAST can be done by a step-by-step procedure. At each time step j the fourth grade\nequation below derived from Eq. 9.20 must then be solved\n9.3 The Plate Thermometer 147\n\n\u0004 \u00054 T j1 j \u0004 \u00054\nPT \u0005 T PT\nPT T j1 h T j1\nPT AST \u0005 C PT \u0005 PT T j1\n\u0005 hPT T j1\nAST\nt PT PT\n\u0004 \u0005\nj1\nK PT T j1\ng \u0005 T PT 0 9:21\n\nwhere all parameters are known except the adiabatic surface temperature T j1AST . If\nboth KPT and CPT are neglected TAST can be obtained at each time as from the fourth\ndegree equation derived from Eq. 9.20\n\u0006 \u0007\nPT T 4AST hPT T AST \u0005 PT T 4PT hPT T PT 0 9:22\n\nSolution techniques of this type of incomplete fourth degree equations are shown in\nSect. 4.4.1.1.\nThe standard PT has successfully been used in an ad hoc test series for measuring\nASTs which has then been used to predict temperature in a steel section. Figure 9.8\nshows a beam near the ceiling being exposed to an intense pre-flashover fire with\nvery uneven and complex temperature distribution.\nTemperatures were then compared with measured temperatures. An example is\nshown in Fig. 9.9. Notice the high similarity between the measured and calculated\nsteel temperatures. Temperatures measured with ordinary thermocouples were\ngenerally very different from those measured with PTs at similar positions. There-\nfore predictions of steel temperatures based on thermocouple recordings as input\nwould not yield such good agreements between calculations and measurements.\nAlternative measuring techniques using, for example, heat flux meters would not\nhave been possible as these types of instruments cannot cope with high temperature\nenvironments.\n\n## 9.3.3 Alternative Plate Thermometer Designs\n\nTo achieve high accuracies it follows from an analysis of Eq. 9.20 that a PT for\nmeasuring AST shall have:\n1. Similar surface emissivities\n2. Similar form and size as the target specimen to have the same convection heat\ntransfer coefficient\n3. Well-insulated metal surfaces\n4. Short response times\nThe first two items concern the heat transfers by radiation and convection, and\nthe relation between the two. As described in Sect. 4.4 the AST depends on the\nradiation and convection heat transfer properties. Therefore the emissivity and the\nconvection heat transfer coefficient should ideally be the same for the thermometer\n148 9 Measurements of Temperature and Heat Flux\n\n## Fig. 9.8 PTs placed around\n\na steel beam for measuring\nAST for calculation of heat\ntransfer and steel\ntemperatures. See mounting\nin Fig. 9.5 \n\nFig. 9.9 Example of measured ( full line) and calculated (dashed line) steel temperatures based on\nPT measurements as shown in Fig. 9.8\n9.3 The Plate Thermometer 149\n\nas for the target body of interest as far as it is practically possible. Thus for\nmeasuring the AST for calculating the heat transfer to, for example, a wall a\nstandard PT might be a sensible compromise.\nThe third item, the PT should be well insulated or ideally perfectly insulated is of\ncourse not possible in practice. The reason for having a thin metal surface is to be\nable to measure the surface temperature accurately by fixing a thermocouple to the\nmetal.\nThe fourth item is important for transient problems where the thermal exposure\nshall be followed as function of time.\nThe standard ISO 834 PT was designed for fire resistance furnace tests when\nbeing exposed to very high temperatures. It is therefore, on one hand, made very\nrobust but not so well insulated as the temperature on the two sides of the PT does\nnot differ very much in a furnace. However, when surrounded by air at ambient\ntemperature and exposed to intense radiation at one side it must be better insulated\nfor not losing heat from the exposed surface.\nFigure 9.10 shows an example of two very well insulated PTs (so-called insPTs)\ndesigned to be used in ambient air. They measure thermal exposure and incident\nradiation in the vertical and horizontal directions, respectively. The plates which the\nthermocouples are fixed to are made of thin steel sheets (0.4 mm) to get quicker\nresponse times. To minimize the heat losses by conduction from the front to the\nback the insulation pads are thick and the sides of the steel plates have been partly\ncut out to avoid heat being conducted along the metal. On the back side this PT has\na thicker more robust steel sheet for mounting purposes.\nFigure 9.11 shows a comparison of incident heat flux measured with a Schmidt-\nBoelter heat flux meter and an insulated plate thermometer as shown in Fig. 9.10\napplying Eq. 9.18 . Note that the difference between the two methods of\nmeasuring incident heat flux is very small with the exception that the HFM\nresponds much faster than the PT and therefore the measurement spikes which\nusually are not of interest to record.\nAnother alternative small PT has been developed for monitoring the thermal\nexposure in the ignition phase of a fire. This may be mounted flush at the surface of,\nfor example, a combustible board as shown in Fig. 9.12. This so-called copper disc\nplate thermometer cdPT consists of a thin copper disc ( 12 mm and thickness\n0.2 mm) backed with ceramic insulation mounted in an about 15 mm drilled hole.\nTo assure similar heat transfer properties the best way to design a PT may be to\nconstruct a 3-D dummy with a thin metal surface and filled with insulation.\nFigure 9.13 shows a pool fire and a steel cylinder simulating a piece of ammunition.\nSeveral thermocouples were fixed at various points of the dummy. It was placed in\npool fire to record the thermal exposure of specimens placed in the flames.\n150 9 Measurements of Temperature and Heat Flux\n\n## Fig. 9.10 Two well-\n\ninsulated plate\nthermometers (insPT) for\nmeasuring thermal\nexposures of horizontal and\nvertical surfaces,\nrespectively, in air at\nambient temperature\n\nFig. 9.11 Comparison of measurements at various distances of incident radiation with heat flux\nmeters of Schmidt-Boelter type and with insPTs as shown in Fig. 9.10 applying Eq. 9.18\n\n## Fig. 9.12 Example of\n\nmounting of a copper disc\nplate thermometer (cdPT)\nflush at the surface of a\nboard. Two thermocouple\ncopper disc. The back side\nof the disc is filled with\ninsulation. A thin\nthermocouple (TC) is\nmounted nearby to measure\ngas temperature\n9.3 The Plate Thermometer 151\n\nFig. 9.13 A steel cylinder dummy filled with insulation with thermocouples mounted on the\nsurface were placed in a pool fire to register thermal exposure. (From a master thesis of Peter\nMollerstrom and Bjorn Evers (2013), Lulea TU). (a) Pool fire (b) Steel cylinder dummy\nChapter 10\nPost-Flashover Compartment Fires:\nOne-Zone Models\n\nFSE and design of structures and structural elements are in most cases made with a\nprocedure including tests and classification systems. Fire resistance or endurance\ntests are specified in standards such as ISO 834, EN 1363-1 or ASTM E-119. In\nthese standards timetemperature curves are specified representing fully developed\ncompartment fires to be simulated in fire resistance furnaces for prescribed\ndurations.\nAlternatively design fires defined by their timetemperature curves may be\nobtained by making heat and mass balance analyses of fully developed compart-\nment fires. Examples of that are given in the Eurocode 1 where so-called\nparametric fire curves are defined. A number of significant simplifications and\nassumptions are then made to limit the number of input parameters and facilitate the\ncalculations. Thus\n1. The combustion rate is ventilation controlled, i.e. the heat release is proportional\nto the ventilation rate.\n2. The fire compartment is ventilated by natural convection at a constant rate\nindependent of temperature.\n3. The gas temperature is uniform in the fire compartment.\n4. The heat fluxes by radiation and convection to all surfaces of the compartment\nare equal and uniform.\n5. The energy of the fuel is released entirely inside the compartment.\n6. The fire duration is proportional to the amount of heat of combustion originally\nin the combustibles in the compartment, i.e. the fuel load.\nAll these assumptions are reasonable for a fully developed fire under ideal\ncircumstances. The major parameters controlling the heat balance of fully developed\ncompartment fires are then considered although they are violated more or less in real\nfires. Anyhow, by making certain parameter choices a set of timetemperature\ncurves are obtained which in general yields design fires which are hotter and longer\nthan could be anticipated in real fires or by more accurate numerical predictions.\n\n## Springer International Publishing Switzerland 2016 153\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_10\n154 10 Post-Flashover Compartment Fires: One-Zone Models\n\nThe theory and assumptions outlined below follows the work of Thelandersson\nand Magnusson and others but has been modified and reformulated according\nto later work by Wickstrom made up for the basis for the parametric fire curves\nin Eurocode 1 . See more on parametric fire curves in Sect. 12.2.\nBelow the fundamentals of the one-zone model theory are presented. The heat\nbalance equation is then formulated in such a way that sometimes simple analytical\nsolutions can be derived and in other cases general temperature calculation codes\ncan be used to analyse compartments surrounded by boundaries of several layers\nand materials with properties varying with temperature as, for example, concrete\ncontaining water evaporating at 100 \u0001 C and having a thermal conductivity that\ndecreases by 50 % during fire exposure.\n\n## 10.1 Heat and Mass Balance Theory\n\nThe overall heat balance equation of a fully developed compartment fire as shown\nin Fig. 10.1 may be written as\n\nq_ c q_ l q_ w q_ r 10:1\n\nwhere q_ c is the heat release rate by combustion, q_ l the heat loss rate by the flow of\nhot gases out of the compartment openings, q_ w the losses to the fire compartment\nboundaries and q_ r the heat radiation out through the openings. Other components of\nthe heat balance equation are in general insignificant and not included in the\napproximate and simple analyses considered here.\nWhen the temperature of the compartment rises, air and combustion products\nflow in and out of the compartment driven by buoyancy, i.e. the pressure difference\np developed between the inside and outside of the compartment due to the gas\ntemperature/density difference as indicated in Fig. 10.1. The mass of gases gener-\nated by the fuel when pyrolyzing is relatively small and therefore neglected. Hence\nthe mass flow rate in m_ i and out m_ o of the compartment must be equal, denoted m_ a .\nThen by applying the Bernoulli theorem the flow rate of gases can be derived as\napproximately proportional to the opening area times the square root of its height\nfor vertical openings.\np\nm_ a 1 Ao ho 10:2\n\nwhere 1 is a flow rate coefficient. Ao and ho are the area and height of the openings\nof the compartment. The coefficient 1 varies only slightly with the fire temperature\nover a wide range of temperatures relevant for fires and is therefore assumed\nconstant . In the presentation here only one vertical opening is assumed. For\ndetails on how multiple openings and horizontal openings can be considered\nsee .\n10.1 Heat and Mass Balance Theory 155\n\nFig. 10.1 One-zone model of a fully developed compartment fire with a uniform temperature Tf\n\nAs indicated in Fig. 10.1 hot fire gases are going out in the upper part of the\nopening and cool air is entering in the lower part. The level at which the direction of\nthe flows are changing is called the neutral layer. As the outgoing flow of fire gases\nis hotter and less dense than the incoming air at ambient temperature, the neutral\nlayer is below the middle of the opening, at about a third of the opening height.\nWith the symbols shown in Fig. 10.1 that is hn \u0003 ho =3.\nThe combustion rate q_ c inside the fire compartment is limited by the amount of\nair/oxygen available. Thus the fire is ventilation controlled and the combustion rate\ninside the compartment is proportional to the air flow, i.e.\np\nq_ c 2 m_ a 1 2 Ao ho 10:3\n\nwhere the combustion efficiency is a reduction coefficient between zero and unity\nconsidering the burning efficiency, i.e. the fraction of the oxygen entering the\ncompartment that is consumed by the combustion process inside the compartment.\nThe combustion yield 2 is the amount of energy released per unit mass of air in the\ncombustion process. It is almost constant for combustible organic materials signif-\nicant in fires with a value of about 13.2 \u0004 106 W s/kg (per kg of oxygen). Then 2 can\nbe calculated assuming an oxygen content of 23 % in ambient air to be 3.01 \u0004 106\nW s/kg (per kg of air). (The fact that a constant amount of energy is released per unit\nweight of oxygen is also accounted for when measuring heat release rates by the\nso-called oxygen depletion technique, for example, in the cone calorimeter\naccording to ISO 5660).\nThe first term on the right-hand side of Eq. 10.1 is the loss by flow of hot gas\ngoing out and being replaced by cooler gas. Hence q_ l is proportional to the mass\nflow in and out of the compartment times the temperature rise of the fire, i.e.\n156 10 Post-Flashover Compartment Fires: One-Zone Models\n\np \u0003 \u0004\nq_ l cp 1 Ao ho T f \u0005 T 1 10:4\n\nwhere cp is the specific heat capacity of the combustion gases at constant pressure\n(usually assumed equal to that of air) and Tf is the fire temperature. T 1 is the\nambient temperature which is assumed equal to the initial temperature Ti. The\nspecific heat capacity of air cp does not vary more than a few percentage over the\ntemperature range considered and may be taken from textbooks such as [1, 2] at a\ntemperature level of 800 \u0001 C to be 1.15 \u0004 103 W s/(kg K).\nFor convenience of writing the fire temperature rise f is introduced, i.e.\n\u0003 \u0004\nf T f \u0005 T i 10:5\n\n## and the convection loss then becomes\n\np\nq_ l cp 1 Ao ho f 10:6\n\nThe second term on the right-hand side of Eq. 10.1, i.e. heat loss to the fire\ncompartment boundaries q_ w is assumed to be evenly distributed over the entire\nsurrounding boundary area.\n00\nq_ w At q_ w 10:7\n00\nwhere At is the total enclosure area and q_ the mean heat flux rate to the surrounding\nsurfaces of the fire compartment. This term constitutes the inertia of the dynamic\nheat balance system as it changes with time depending on the temperature of the\nsurrounding boundaries. It is significant in the beginning of a fire, and then it\ndecreases when the temperature of the surrounding structure increases and gets\ncloser to the fire temperature. For surrounding structures assumed to be thick it\nvanishes when thermal equilibrium is reached after long fire durations.\nThe third term on the right-hand side of Eq. 10.1, i.e. the heat loss by radiation\ndirectly out through the openings q_ r , may be calculated as\n\u0005 \u0006\nq_ r f Ao T 4f \u0005 T 41 10:8\n\nwhere f is the emissivity of the fire compartment at the opening here assumed to be\none and therefore omitted below. This term is relatively small in the beginning of a\nfire when the fire temperature is moderate. It increases, however, by the forth power\nof the temperature and becomes considerable at later stages of fires when the\ntemperature is high.\nNow by inserting Eqs. 10.3, 10.4, 10.7 and 10.8 into Eq. 10.1 and after\nrearranging, the heat flux to the boundary surfaces becomes\n10.1 Heat and Mass Balance Theory 157\n\n\u0007 \b\n00 2 Ao \u0005 \u0006\nq_ w cp 1 O \u0005 f T 41 \u0005 T 4f 10:9\ncp At\n\np\nAo ho\nO 10:10\nAt\n\n## The temperature of a ventilation controlled fire increases with time as the\n\ncompartment boundary structures, ceiling, floor and walls heat up. If the compart-\nment boundaries are assumed infinitely thick then when the compartment bound-\naries after a long time have been fully heated and steady-state thermal conditions\ncan be assumed the heat losses to the boundary structure vanish. Notice in Eq. 10.9\nthat if the losses to the surrounding structure q_ w and the radiation out the window q_ r\nare negligible, the fire temperature depends only on the ratio between 2 and cp,\ni.e. ratio between product of the combustion efficiency and combustion yield, and\nthe specific heat of the fire gases. It is, however, independent of the opening factor\nand the thermal properties of the surrounding structure. The parameter is here\nnamed the ultimate fire temperature ult:\n2\nult 10:11\ncp\n\nThe values of all the parameters introduced above vary only slightly with\ntemperature and are therefore here assumed to remain constant. Commonly\nassumed values are summarized in Table 10.1.\nIn Eq. 10.9 the parameter groups (cp1O), (2/cp) and (Ao/At) are constants, and\nT 1 is a known boundary temperature. Therefore this equation is analogous to a\nboundary condition of the third kind as outlined in Sect. 3.2.3. If the radiation\ndirectly out through the openings is neglected (second term on the right-hand side\nof Eq. 10.9) even analytical solutions can sometimes be obtained as shown in\nSect. 10.2.\nAlternatively Eq. 10.9 may be written as\n00 \u0003 \u0004 \u0003 \u0004\nq_ w hf , c T ult \u0005 T f hf , r T 1 \u0005 T f 10:12\n\nor as\n00 \u0003 \u0004\nq_ w hf , c ult \u0005 f hf , r f 10:13\n\nHere hf,c is named fire compartment convection heat transfer coefficient, iden-\ntified as\n158 10 Post-Flashover Compartment Fires: One-Zone Models\n\n## Table 10.1 Values of physical parameters and parameter groups\n\nName Notation Value Units\nFlow rate coefficient 1 0.5 kg/(s m5/2)\nCombustion yield coefficient 2 3.01 \u0004 106 W s/kg\nSpecific heat capacity of air cp 1150 W s/(kg K)\nCombustion efficiency\nUltimate fire temperature increase ult\ncp\n2\n1325 ( 0.506) K\nFire convective heat transfer coefficient h f , c cp 1 O 575\u0004O W/(m2 K)\n\nhf , c cp 1 O 10:14\n\n## and fire compartment radiation heat transfer coefficient hf,r is identified as\n\nAo \u0005 2 \u0006 \u0003 \u0004\nhf , r T 1 T 2f \u0004 T 1 T f 10:15\nAt\n\n## The corresponding fire compartment thermal resistances are defined as\n\n1 1\nRf , c 10:16\nhf , c cp 1 O\n\nand\n\n1 1\nRf , r \u0005 \u0006 \u0003 \u0004 10:17\nhf , r Ao\nT 2\nT 2\n\u0004 T1 Tf\nAt 1 f\n\nThe ultimate fire temperature ult will generally not appear in reality. It is intro-\nduced to facilitate the development and explanation of the compartment fire\nmodels.\nThe heat transfer to the surrounding structure expressed in terms of the fire\ntemperature may be written as\n00 \u0003 \u0004 \u0003 \u0004\nq_ w hi, c T f \u0005 T s hi, r T f \u0005 T s 10:18\n\nwhere hi,c is the convection heat transfer coefficient and hi,r the radiation heat\ntransfer coefficient between the fire gases and the compartment boundary. The latter\nis defined as\n\u0005 \u0006 \u0003 \u0004\nhi, r s T 2f T 2s \u0004 T f T s 10:19\n\nwhere s is the emissivity of the fire compartment inner surface. Then the combi-\nnation of Eqs. 10.12 and 10.18 can be illustrated by an electric circuit analogy as\nshown in Fig. 10.2 where the resistances Rf,c and Rf,r are defined in Eqs. 10.16 and\n10.1 Heat and Mass Balance Theory 159\n\n## Fig. 10.3 Electric circuit\n\nanalogy model of a fire\ncompartment boundary with\ntwo heat transfer resistances\nin series\n\n10.17, and Ri,c and Ri,r are the inverses of the corresponding heat transfer coeffi-\ncients as defined by Eqs. 10.18 and 10.19.\nThe two temperatures Tult and T 1 may be reduced to one resultant temperature\nTmax which is a weighted mean value of the two. Compare with adiabatic surface\ntemperature of Sect. 4.4. Then the electric circuit of Fig. 10.2 can be reduced that of\nFig. 10.3. The analogy between heat transfer and electric circuit parameters is\ndescribed in Sect. 1.2 where also rules for combining resistances in series and\nparallel are given.\nTmax is the maximum temperature a compartment fire can reach when the losses\n00\nto the boundaries vanish. It can be calculated by putting q_ w 0 in Eq. 10.12 and\nsolving for f.\nAs there is no thermal heat capacity involved, the heat flux may now be written\nin two ways as\n160 10 Post-Flashover Compartment Fires: One-Zone Models\n\n## Fig. 10.4 Electric circuit\n\nanalogy model of a fire\ncompartment boundary with\ntwo heat transfer resistances\nin series\n\n\u0003 \u0004\n00 T max \u0005 T f T max \u0005 T i, s\nq_ w 10:20\nRf , tot Rf , tot Ri, tot\n\nObserve that the radiation heat transfer coefficient must be calculated at the\nabsolute temperatures Tmax and Tf. Thus\n\n1\nRf , tot \u0005 \u0006 \u0003 \u0004 10:21\ncp 1 O Ao\nAt T max T 2f \u0004 T max T f\n2\n\nand then\n\n1 1\nRi, tot \u0005 \u0006 10:22\nhi, c hi, r hi, c s T 2 T 2 \u0004 \u0003T f T i, s \u0004\nf i, s\n\nAccording to Eq. 10.20 the two thermal boundary resistances in series as shown in\nFig. 10.3 can then be summarized into one as shown in Fig. 10.4.\nEquation 10.20 is a third kind of boundary condition (see Sect. 1.1.3) with the\nheat transfer coefficient equal to the reciprocal of the heat transfer resistance. With\nthis boundary condition combined with a thermal model of the boundary structure\nmay its temperature be calculated including its surface temperature.\nThe fire temperature Tf can thereafter be obtained as the weighted mean tem-\nperature of Tmax and Ts as\n\n## T s Rf , tot T max Ri, tot\n\nT f f T i 10:23\nRf , tot Ri, tot\n\nIf the thermal resistances Rf,tot and Ri,tot may be assumed constant, analytical\nsolutions for Ts can be derived for surrounding structures being semi-infinitely thick\nor having its heat capacity lumped in a core as is shown below.\nThe highest fire temperature that can be reached in a fire compartment occurs\nwhen surrounding structures are fully heated and do not absorb any more heat, i.e.\n00\nq_ w vanishes. Then the fire temperature and the surface temperature becomes equal\nto Tmax. If in turn the radiation directly out through the openings can be neglected as\nwell, the maximum fire temperature becomes Tult.\nAn observation is that according to this theory an instant fire temperature rise\noccurs when the fire begins. Then the fire temperature immediately T0f rises to\n10.2 Solution of the Fire Compartment Temperature 161\n\nRi, tot\nT 0f T max 10:24\nRf , c Ri, tot\n\nThis immediate temperature rise is of course physically unlikely for the very\ninitial phase as a heat release yielding flashover cannot start suddenly in reality but\nafter some time has elapsed the approximate predictions as given by the above\ntheory applies.\n\n## 10.2 Solution of the Fire Compartment Temperature\n\nThe boundary condition as defined above includes the two heat transfer resistances\nin a series, one artificial and one physical as indicated by Fig. 10.4. To solve for the\nsurface temperature and then calculate the fire temperature according to Eq. 10.23,\na thermal model of the compartment boundary structure is needed. The surface\ntemperatures may then be calculated with various methods depending on whether\nthe model parameters may vary with temperature. When either heat transfer\ncoefficients or material properties vary with temperature, the problem becomes\nnon-linear and then numerical tools such as finite element programs need to be\nused. Boundary structures of several layers of different materials etc. may then also\nbe considered. Spreadsheet calculations using programs such as MS-Excel are very\nuseful when analysing fire compartments with boundaries where lumped heat can\nbe assumed.\nNumerically exact analytical expressions can be derived for two types of bound-\nary constructions being considered in the next sections, namely, structures assumed\neither semi-infinitely thick or having a core where the thermal mass is concentrated\n(lumped heat). Then the elementary procedures presented in Sects. 3.2 and 3.1,\nrespectively, may be applied.\nIn Table 10.1 a summary of values of physical parameters and parameter groups\nare given. These are used throughout the presentation below.\n\n## 10.2.1 Semi-infinitely Thick Compartment Boundaries\n\nFire compartment boundaries are in most cases assumed thermally thick. The heat\ntransferred to the surfaces are then stored in the surrounding structures, and the\neffects of heat lost on the outside of the structure is neglected.\nAs indicated by Fig. 10.4 may the boundary condition be expressed by two\nthermal boundary resistances in series which can be added up and a complete\nthermal model becomes as indicated by Fig. 10.5.\nThis is a semi-infinite body with a third kind of boundary condition. To compute\nthe surface temperature generally numerical temperature calculation methods are\n162 10 Post-Flashover Compartment Fires: One-Zone Models\n\nFig. 10.5 Electric circuit analogy model of a fire compartment with infinitely thick walls\n\nneeded such as finite element methods. The fire temperature is then calculated as\nthe weighted mean temperature of Tmax and Ts according to Eq. 10.23.\nHowever, if the following assumptions are made, the problem becomes linear\nand analytical solutions can be derived:\n1. The heat radiated directly out the openings, q_ r , is neglected or is directly\nproportional to the difference between the fire temperature Tf and the ambient\ntemperature T 1 , i.e. hf,tot and its reciprocal Rf,tot are constant.\n2. The heat transfer by radiation and convection to the surrounding boundaries is\nassumed proportional to the difference between the fire Tf and boundary surface\ntemperatures Ts, i.e. hi,tot and its reciprocal Ri,tot are constant.\nThe surface temperature rise can be calculated according to Eq. 3.35 in Sect.\n3.2.3. Thus\n\" s!#\nt t\ns max \u0004 1 \u0005 e \u0004 erfc f\n10:25\nf\n\nwhere the parameter f may be identified as a fire compartment time constant for\ninfinitely thick walls in analogy with Eq. 3.34.\n\nk\u0004\u0004c \u0003 \u00042\nf \u0005 \u00062 k \u0004 \u0004 c \u0004 Rf , tot Ri, tot 10:26\n1\nRf , tot Ri, tot\n\nas the reciprocal of the heat transfer resistance by definition is equal to the heat\ntransfer coefficient. Rf,tot depends on Rf,c and Rf,r. The former is always constant\naccording to Eq. 10.16, and by assuming constant fire temperature T f T _f in\nEq. 10.17, a constant Rf,r could be calculated as well. Too high assumed T _f -values\nwill yield overestimated the heat losses by radiation out the openings and therefore\nunderestimated fire temperatures, and vice versa. max is the temperature rise which\nis obtained when the wall are fully heated and no heat is transferred to boundary\nsurfaces. Given constant values of Rf,c and Rf,r, max is constant and can be\ncalculated as\n\nult Rf , r ult\nmax 10:27\nRf , r Ri, c 1 Rf , c :\nR i, r\n10.2 Solution of the Fire Compartment Temperature 163\n\nBy inserting the opening factor from Eq. 10.10 and assuming parameter values\naccording to Table 10.1 a\n\u0005 \u0006 \u0005 \u0006\nRf , c \u0004 T 21 T _f 2 \u0004 T 1 T _f\np 10:28\nRf , r 575 ho\n\n_\nwhere T is the assumed temperature level. Then by inserting Eq. 10.28 into\nEq. 10.27\n\nult\nmax \u0003 \u0004\u0003 \u0004: 10:29\n\u0004 T 21 T _f2 \u0004 T 1 T _f\n1 p\n575 ho\n\nNotice that max increases with the square root of the opening height ho but is\nindependent of Ao and At.\nThe fire temperature rise vs. time may be obtained as the weighted average of\nmax and s in analogy with Eq. 10.23 as\n\n## s Rf , tot max Ri, tot\n\nf : 10:30\nRf , tot Ri, tot\n\nNow by inserting Eq. 10.25 into Eq. 10.26, the fire temperature development\nbecomes\n(\" s!# )\nult t t Ri, tot\nf 1 \u0005 e \u0004 erfc\nf\n: 10:31\n1\nRi, tot f Rf , tot\nRf , tot\n\nConstant values of Ri,tot and Rf,tot may be obtained from Eqs. 10.21 and 10.22,\nrespectively, for a given fire temperature T f T _f . Then by inserting the ratio Rfi,,tottot\nR\n\ninto Eq. 10.31, a very handy closed form solution of the fire temperature develop-\nment vs. time is obtained.\nAn interesting observation is that the standard design timetemperature curves may\nbe derived by prescribing a maximum temperature rise ult 2 =cp 1325 \u0001 C and\na fire compartment time constant f 1200 s. This time constant may be calculated\nbased on quite reasonably assumed input parameters for surrounding boundary prop-\nerties grouped into the thermal inertia k \u0004 \u0004 c, the opening factor (O) and the heat\ntransfer resistance between the fire gases and the surrounding boundaries (Ri,tot). Then\nEq. 10.25 yields the fire temperature rise as\n164 10 Post-Flashover Compartment Fires: One-Zone Models\n\n1200\n\n1100\n\n1000\n\n900\n\n800\nFire temperature, Tf [C]\n\n700\n\n600\nAnalytical\n500 ISO/EN\n400 Parametric =1\n\n300\n\n200\n\n100\n\n0\n0 1 2 3 4 5 6\nTime, t [h]\n\nFig. 10.6 Comparison between fire temperature rises according to the analytical expression\n(Eq. 10.32), the standard ISO 834 curve and the parametric fire curve for 1 according to EN\n1991-1-2\n\nt\nt\nf 1325 1 \u0005 e 1200 \u0004 erfc : 10:32\n1200\n\nwhich is very close to the EN 1363-1 and ISO 834 standard curves as well as the\nheating phase of the parametric fire curve according to EN 1991-1-2 for the\ncompartment factor 1, see Sect. 12.2, as shown in Fig. 10.6.\nExample 10.1 Calculate the fully developed fire temperature rise after\n60 min in a compartment surrounded by concrete. Assume an ultimate temperature\nult 1325 \u0001 C and an opening factor of O 0.04 m, material properties\naccording to Table 1.2 and other physical parameters according to Table 10.1.\n(a) Neglecting the effects of heat transfer resistance between the fire gases and the\nsurrounding boundaries and the radiation directly out the window.\n(b) Neglecting the effects of radiation directly out the window but not the effects\nof heat transfer resistance between the fire gases and the surrounding bound-\naries. Assume a total heat transfer coefficient htot, i 200 W=m2 K.\n(c) Considering both the effects of heat transfer resistance between the fire gases\nand the surrounding boundaries and the radiation directly out the window. The\nratio between the opening area and the total surrounding area Ao =At 0:06.\nAssume a hf , r 70 W=m2 K.\n10.2 Solution of the Fire Compartment Temperature 165\n\nSolution\n(a) Ri, tot 0 and Rf , r 0. Equation 10.16 yields Rf , c Rf , tot 575\u00040:04\n1\n\n0:0435 m K=W and Eq. 10.26 yields f 575\u00040:042 6673 s. Thus t=f\n2 3530000\n\n3600\n6673 0:54 and Eq. 10.25 (and Table 3.3) yields f s\n0:49 \u0004 1325 649 \u0001 C.\n(b) ( Rf , r 0 ). Ri, tot 200\n1\n0:005 m2 K=W Eq. 10.26 yields f 3530000 \u0004\n2\n0:0435 0:005 8303 s and t=f 3600 8303 0:433 and Eq. 10.25 yields the\nsurface temperature s 1325 \u0004 0:46 610 \u0001 C. Then the fire temperature\n1325\u00040:005\ncan be obtained from Eq. 10.30 as f 610\u00040:0435 0:0578\n42:55\n683 \u0001 C.\n\u0005 \u0006 \u0003\n0:04350:005\n\u0004\n(c) According to Eq. 10.15 hf , r AAtoto T 21 T 2f \u0004 T 1 T f \u0003 0:03 \u0004 70\n2:1 W=m2 K and according to Table 10.1 hf , c R1f , c 23 W=m2 K and thus\nRf , tot 2:123\n1\n0:040 m2 K=W. Then Eq. 10.27 yields max 23\u00041325 232:1\n\u0001\n\u0003 \u00042\n1214 C and Eq. 10.26 yields f k \u0004 \u0004 c \u0004 Rf , tot Ri, tot 3530000\u0004\n0:040 0:005\u00062 7148 s and t=f 3600\n6227 0:504 and Eq. 10.25 yields\nthe surface temperature s 0:48 \u0004 1214 583 \u0001 C. Then the fire temperature\nR R\ncan be obtained from Eq. 10.30 as f s Rf ,ftot, tot Rmaxi, tot i, tot\n583\u00040:040 1214\u00040:005\n0:0400:005 652 \u0001 C.\nComment: Notice that the various fire temperatures are obtained depending on the\nlevels of completeness of the calculation model.\n\n## 10.2.2 Insulated and Uninsulated Boundaries\n\nwith a Metal Core\n\nAnalytical solutions of the fire temperatures may also be obtained when the fire\ncompartment is assumed surrounded by structures consisting a metal core where the\nall the heat capacity is concentrated. Then the heat capacity per unit area Ccore is\nlumped into the core as indicated in Fig. 10.7. The heat capacity of any insulating\nmaterial is either neglected or assumed included in the heat capacity of the core.\nFigure 10.8 shows an electric circuit analogy model of how the fire, the core and\nthe inner and outer surface temperatures can be calculated. As all inertia is lumped\ninto the core, the heat flux is constant on either side of the core due to the\nrequirement of heat flux continuity. Hence the temperature differences between\nvarious positions are proportional to the corresponding thermal resistances. The\nthree graphs of the figure indicate the temperature rises initially, after some finite\ntime and after a very long time, respectively. Notice that according to the theory the\nfire and the inner fire exposed surface temperatures increase instantaneously at t 0\naccording to Eq. 10.29.\n166 10 Post-Flashover Compartment Fires: One-Zone Models\n\nFig. 10.7 A fire compartment surrounded by a structure with its heat capacity Ccore assumed\nconcentrated/lumped to a metal core. Thermal resistances of insulation materials Ri and Ro are\nassumed on the fire inside and outside, respectively\n\nFig. 10.8 Electric circuit analogy of fire compartment model with a thin surrounding structure\nassuming lumped heat capacities. Relative temperatures at various points initially (t 0), after\nsome time (0 \u0007 t \u0007 1) and after a very long time (t 1) are indicated\n\nThe parameters of Fig. 10.8 are summarized in Table 10.2. The maximum\ncompartment fire temperature considering radiation out through openings max\ncan be obtained from Eq. 10.12 or Eq. 10.23, the fire compartment heat transfer\nresistance Rf,tot from Eq. 10.21 and the heat transfer resistance between fire and\n10.2 Solution of the Fire Compartment Temperature 167\n\n## Table 10.2 Summary of the parameters of Fig. 10.8\n\nNotation Parameter Definition\nmax Maximum temperatureno losses to surfaces Eq. 10.12, Eq. 10.23 or\nEq. 10.29\nRf,tot Fire heat transfer resistance Eq. 10.21\nRi,tot Transfer resistance, fireinside surface Eq. 10.22\nRi,ins Resistance inside insulation Eq. 10.33\nRo,ins Resistance outside insulation Eq. 10.34\nRo,tot Transfer resistance, outside surface Eq. 10.35\nsurroundings\nCcore Heat capacity of the core Eq. 10.36\n\nsurrounding surfaces Ri,tot from Eq. 10.22. The insulation resistances at the inside\nand outsider of the core Ri,ins and Ro,ins can be calculated as\n\ndi, ins\nRi, ins 10:33\nki, ins\n\nand\n\ndo, ins\nRo, ins 10:34\nko, ins\n\nwhere di,ins and do,ins are the thicknesses of the inside and outside insulations,\nrespectively, and ki,ins and ko,ins are the corresponding conductivities. The outside\nheat transfer resistance can be calculated as\n\n1\nRo, tot \u0003 \u0004 10:35\nho, c o, s T 2o, s T 21 \u0004 T o, s T 1\n\nwhere ho,c is the convection heat transfer coefficient at the outside surface and o,s\nthe emissivity of the outside surface. The conduction resistance of the metal core is\nneglected. Ccore is the heat capacity per unit area of the core, i.e.\n\n## Ccore dcore ccore core : 10:36\n\nwhere dcore, ccore and core are the thickness, specific heat and density of the core,\nrespectively\nThe dynamic heat balance of the core can now be written as\n168 10 Post-Flashover Compartment Fires: One-Zone Models\n\ndcore \u0003 \u0004\nCcore Rf , tot Ri, tot Ri, ins max \u0005 core \u0005 Ro, ins Ro, tot\ndt\n\u0003 \u0004\n\u0004 core \u0005 1 10:37\n\nand the core temperature can be numerically solved by the forward difference\nrecursion formula\n\ni1\ncore core\ni\n\nt \u0003 \u0004 \u0003 \u0004\nRf , tot Ri, tot Ri, ins max \u0005 core \u0005 Ro, ins Ro, tot \u0004 core \u0005 1\nCcore\n10:38\n\n## where t is a chosen time increment. This recursion formula can be coded in\n\nspreadsheet programs such as MS-Excel. The temperature-dependent parameter\nmay be updated along with the calculations.\nIf all the parameters are assumed constant, then the temperature development\ncan be calculated analytically. Thus, the core temperature rise may be obtained as a\nfunction of time as (see Sect. 3.1.2)\n\n\u0005 \u0006\nRo, tot Ro, ins\ncore max 1 \u0005 e\u0005t=f 10:39\nRf , tot Ri, tot Ri, ins Ro, tot Ro, ins\n\n## where the fire compartment time constant f is\n\nCcore\nf 10:40\nRf , tot Ri, tot Ri, ins Ro, tot Ro, ins\n1 1\n\nWhen the core temperature has been calculated, the fire temperature may be\ncalculated. As the heat capacity of the insulation is assumed to be negligible, the\ncompartment fire temperature rise can be calculated as a weighted average between\nmax and core (see Fig. 10.8). Thus\n\n## Ri, tot Ri, ins max Rf , tot core\n\nf 10:41\nRf , tot Ro, tot Ro, ins\n\n## max Ro, tot Ro, ins \u0005 \u0006 R R\n\n\u0005t=f i, tot i, ins\n1\u0005e\nRi, tot Ri, ins Rf , tot Ri, tot Ri, ins Ro, tot Ro, ins Rf , tot\n1\nRf , tot\n10:42\n\nNotice that this expression is similar to the corresponding Eq. 10.31 for semi-\ninfinite boundaries.\n10.2 Solution of the Fire Compartment Temperature 169\n\nIn a corresponding way according to the law of proportion may the fire exposed\nsurface temperature be calculated as\n\u0003 \u0004\nRi, ins max Rf , tot Ri, tot core\ns, i 10:43\nRf , c Ri, tot Ri, ins\n\n## Ro, tot core\n\ns, o 10:44\nRo, tot Ro, ins\n\nThe maximum fire temperature that can be reached asymptotically after long fire\ndurations depends on the insulation of the compartment and the heat transfer\nresistances. It can be calculated as\n\n## Ri, tot Ri, ins Ro, tot Ro, ins\n\nfmax max 10:45\nRf , tot Ri, tot Ri, ins Ro, tot Ro, ins\n\nThe fire development is generally very fast for thin structures and the maximum is\nreached quickly.\nExample 10.2 Estimate the maximum post-flashover fire temperature Tmax f of an\nuninsulated steel container. It has a steel thickness dcore 3 mm, an opening\nheight and area ho 2:5 m and Ao 5 m2 , and a total area At 125 m2 .\nAssume ult 1325 \u0001 C, the internal and external heat transfer coefficients due to\nradiation and convection hi, tot 1=Ri, tot 100 W=m2 K and ho, tot 1=Ro, tot\n25 W=m2 K. Estimate the fire temperature development vs. time.\nSolution\nAssume a maximum fire temperature T fmax 900 \u0001 C 1173 K for the estimate of\n\u0001\nRf,tot. Then Eq. 10.29 yields max \u00042932 1173 2 \u00042931173 1169 C. Then Eq. 10.21\n1325\n1 p\n575 2:5\n\n## and Table 10.1 yield Rf , tot p 1\n\n2 m2 K=W.\n575\u0004 125 125 11692732 9732 \u0006 \u00041169273973\n5 2:5 5\n\n1=10001=250\nThen Eq. 10.36 yields fmax 0:0221=1001=25 1169 809 \u0001 C. A new estimate T fmax\n850 \u0001 C 1123 K yields max 1184 \u0001 C, Rf , tot 0:0178 and fmax 0:731184\n873 \u0001 C.\nComment: Thus the temperature rise in an uninsulated container as described will\nnever exceed about 870 \u0001 C (max f ). For a well-insulated container with the same\ngeometry the fire temperature rise may reach about 1180 \u0001 C (max).\nThe temperature development can be estimated by applying Eq. 10.42. The time\nconstant f 1=0:003\u0004460\u00047850\n0:01780:0125 177 s and then if Rf,tot is assumed constant\nh \u0003 \u0004 i\n\u0005t=177\nas calculated above f 11184\n0:01\n0:04\n0:01780:010:04 1 \u0005 e 0:0178\n0:01\n\u0003 \u0005t=177\n\u0004 \u0001\n0:0178\n\n447 \u0004 1 \u0005 e 426 C.\n170 10 Post-Flashover Compartment Fires: One-Zone Models\n\nComment: This solution shows that the time constant is very short, less than 3 min,\nand that very beginning of the fire (t 0) the calculated fire temperature rise is\nsignificant, f 426 \u0001 C, and that after a long time t ! 1 f ! 873 \u0001 C as stated\nabove (max\nf ).\n\n## Example 10.3 A fire compartment is surrounded by a 3-mm-thick steel sheet\n\nstructure with a 12-mm-thick gypsum board mounted on both sides of the core.\nThe opening is factor O 0.08 m. The heat transfer coefficient at the fire exposed\nand the unexposed sides are assumed to be constant, i.e. hi, tot 200 W=m2 K and\nho, tot 40 W=m2 K, respectively. Neglect the radiation directly out the opening,\ni.e. Rf , tot Rf , c .\n(a) Calculate the ultimate fire temperature rise ult assuming a combustion effi-\nciency of 50 %.\n(b) Calculate the maximum fire max f and core max\ncore temperature rises.\n(c) Calculate the fire temperature f and the inner fire exposed surface temperature\ns,i at time t 0 according to the model.\n(d) Calculate the core core and fire f temperature rises after 300 s of flashover.\n(e) Plot as a function of time of the temperature rises, ultimate ult, fire f, inner\nsurface s,i, core core, outer surface s,o.\nUse parameter values as given in Tables 1.2 and 10.1.\nSolution\n(a) Equation 10.11 yields ult 1309 \u0001 C.\n(b) The thermal resistances over a unit area Ri, tot Ri, ins 1=200 0:012=0:5\n0:029 m2 K=W and Ro, tot Ro, ins 40\n1\n0:012=0:5 0:049 m2 K=W.\n\u00053\nRf , c Rf , tot 1:74 \u0004 10\n0:08 0:022m K=W. Then Eq. 10.45 yields f\n2 max\n\n\u0001\n0:0290:049\n0:0220:0290:049 \u0004 1309 1021 C and Eq. 10.39 yields\ncore\nmax\n0:0220:0490:029\n0:049\n\u0004 1309 641 \u0001 C.\n(c) Equation 10.24 yields 0f 0:0220:029\n0:029\n\u0004 1309 744 \u0001 C, and\n0:012=0:5\ns, i 0:0220:029 \u0004 1309 616 \u0001 C.\n(d) Equation 10.40 yields f 0:003\u0004560\u00047850 s and temperature rise after 300 s\n0:0220:0290:049330\n1 1\n\u0005 \u0006\n\u0004 1 \u0005 e\u0005326\n300\ncan be obtained from Eq. 10.39 as core 1309 \u0004 0:0220:0490:029\n0:049\n\n## 1309 \u0004 0:490 \u0004 0:601 385 \u0001 C and from Eq. 10.41\n\nf 0:029\u000413090:022\u0004385\n0:0220:029 910 \u0001 C.\n(e) See the plot of Fig. 10.9\nComment: Notice in Fig. 10.9 that according to the theory temperatures, the fire\ntemperature f and the inner exposed surface temperature s,i starts at temperature\nlevels between the initial and the ultimate temperatures depending on the thermal\nresistances.\n10.2 Solution of the Fire Compartment Temperature 171\n\n1400\n\n1200\n\n1000\nTemperature rise, [C]\n\n800\n\n600\nmax\n400 f\ni,s\n200 core\no,s\n0\n0 100 200 300 400 500 600 700 800 900\nTime, t [s]\n\nFig. 10.9 Calculated temperature rises of Example 10.3. Notice that the fire and inner surface\ntemperatures rise instantaneously according to the theory\n\n## 10.2.3 Temperature-Dependent Material and Heat Transfer\n\nProperties: Numerical Solutions\n\nIn most cases fire compartments have openings through which heat can radiate out\nto the environment at ambient temperature. Exceptions are furnaces and tunnels,\nand therefore tunnel fires may become very hot. The heat losses by radiation out\nthrough opening q_ r according to Eq. 10.8 is small at low temperatures but increases\nrapidly at elevated temperatures and must therefore be considered particularly when\nanalysing hot fires. As it is highly non-linear as it depends on the fire temperature to\nthe fourth degree. This is also the case for the radiation to the fire compartment\nsurfaces, and in addition the surrounding structure may consist of several layers of\nmaterials with properties varying with temperature. Then numerical solutions are\nrequired.\nIn general, Eq. 10.9 is valid as a boundary condition for the one-dimensional\nmodel. For a fire compartment with relatively thick but not infinite boundaries, a\nthermal model as indicated by electric circuit analogy shown in Fig. 10.10 may then\nbe applied.\nThis model analysed as it is or it may be reduced by the rules of combining\nresistances and defining combined temperatures to the circuit analogy of Fig. 10.11.\nThen the boundary temperature Tmax is a weighted average value of the ambient\n00\ntemperature T 1 and Tult. max can be calculated by solving Eq. 10.12 for q_ w 0.\n172 10 Post-Flashover Compartment Fires: One-Zone Models\n\nFig. 10.10 Electric circuit analogy of fire compartment model with a thick surrounding structure\n\nFig. 10.11 Reduced electric circuit analogy of fire compartment model with a thick surrounding\nstructure\n\nThe thermal resistances Rf,c and Rf,r can be calculated according to Eqs. 10.16 and\n10.17, and the resultant resistance Rf,tot according to Eq. 10.21.\nThe heat transfer resistance Ri,tot between the fire and the surrounding surfaces\ncan be calculated according to Eq. 10.22.\nNow with the boundary condition according to Eq. 10.20 with two heat transfer\nresistances in series may the temperature in the surrounding structure be calculated\nincluding the surface temperature Ts with a general temperature calculation code.\nWhen the surface temperature is calculated, the fire temperature can be obtained\nas\n\n## Rf , tot T s, i Ri, tot T max\n\nTf 10:46\nRf , tot Ri, tot\n\nNotice that heat transfer resistance due to radiation depends on the temperatures\nand therefore Eq. 10.46 is implicit and Rf,tot and Ri,tot must be updated at each time\nstep. Below four cases are shown of calculated and measured fire temperatures in a\nreduced scale room with dimensions according to Fig. 10.12a. A diffusion propane\nburner (300 mm by 300 mm) was placed inside the fire compartment releasing a\nconstant power of 1000 kW. It generated immediate flash-over with flames emerg-\ning out the door opening, see Fig. 10.12b.\nThe thermal model as indicated in Fig. 10.10 was analysed with TASEF for the\nsurrounding structures of lightweight concrete and steel sheets. The steel sheets\nwere either insulated on the outside, on the inside or non-insulated. Figure 10.13\nshows the measured and calculated temperatures in a compartment of lightweight\nconcrete.\nFigure 10.14 shows measured and calculated fire temperatures from the same\ntests series with a compartment of 3 mm steel sheets insulated on the outside, inside\nor not at all, respectively. In the calculations the changes of thermal properties of\nthe insulation and the steel were considered.\n10.2 Solution of the Fire Compartment Temperature 173\n\na\n1800\n\n1500\n\n600\n1800\n\nFig. 10.12 Reduced scale fire compartment experiment with propane burner. (a) Inner dimen-\nsions (in mm) (b) Flames shooting out the door-way\n\nFig. 10.13 Measured ( full line) and calculated fire (dashed line) temperatures in fully developed\ncompartment fire in a concrete compartment using the finite element program Tasef\n174 10 Post-Flashover Compartment Fires: One-Zone Models\n\nFig. 10.14 Measured ( full line) and calculated (dashed line) fire temperature in fully developed\ncompartment fire in a steel sheet compartments using the finite element program Tasef. (a)\nInsulation on the outside (b) Insulation on the inside (c) No insulation\n\nNotice the following from the four cases of Figs. 10.13 and 10.14\nThe fire temperature goes to about 1150 \u0001 C (Tmax) except the non-insulated steel\ncompartment where the final temperature is less than 800 \u0001 C.\nThe inside insulated steel compartment goes much faster to the maximum\ntemperature than the outside insulated.\nThe calculation model yields exceptionally good predictions particular in terms\nof the qualitative development of the fire temperature.\nChapter 11\nPre-flashover Compartment Fires:\nTwo-Zone Models\n\nTwo-zone models are applied to pre-flashover fires, i.e. compartment fires which\nhave not reached ventilation controlled combustion conditions as defined in\nChap. 10. Several more or less advanced computer codes have been developed to\ncalculate temperature under such assumptions. The most fundamental principles of\nthe theory are outlined below.\nIn most cases the heat release rate as a function of time is input to pre-flashover\ncalculation models. Examples are given in Table 11.1 of the order of magnitude of\nthe heat release rates of various fires.\nIn the post-flashover model described above the heat release rate was assumed\ndetermined by the opening alone, see Eq. 10.3. In pre-flashover models as the one\ndescribed below the heat release rate q_ f is an input variable. All combustion is\nassumed to occur inside the fire compartment boundaries and it is limited by the rate\nat which gaseous fuel (pyrolysis gases) is being released from burning objects. As\nshown in Fig. 11.1, an upper layer is then supposed to develop where the fire\ntemperature Tf is assumed to be uniform. Below, the lower layer gas temperature\nremains at the ambient temperature T 1 . Hot combustion gases enter the upper layer\nby the way of entrainment into the fire plume of flames and combustion gases\ndeveloped by the burning items. The flow rate m_ p at which mass is entering the\nupper layer must balance the mass flows going in m_ i and out m_ o of the compartment.\nThus\n\nm_ p m_ i m_ o 11:1\n\nThe plume mass flow rate may be calculated as a function of the heat release rate\nq_ f and the height of between the fuel surface and the height of the upper layer\ninterface HD. In the pre-flashover stage of a fire, it is the plume entrainment rate\nrather than the size of openings that governs the mass flow rate. This is in contrast to\npost-flashover fires.\n\n## Springer International Publishing Switzerland 2016 175\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_11\n176 11 Pre-flashover Compartment Fires: Two-Zone Models\n\nTable 11.1 Examples of the order of magnitude of heat release rates (HRR) of various fires\nItem Typical heat release rate [W]\nWood-burning stove 10\nSingle burning furniture item 100\nFlashover of small room 1000\nFull flashover large room/apartment 10,000\nFire in a loaded truck 100,000\n\nFig. 11.1 Two-zone model of a pre-flashover room fire with a uniform temperature Tf in the upper\nlayer and ambient temperature T1 in the lower layer\n\nAs for the one-zone model (post-flashover fires) the heat balance equation of a\nfire compartment may be written as\n\nq_ c q_ l q_ w q_ r 11:2\n\nwhere the convection term q_ l is proportional to the mass flow rate m_ p and the\ntemperature rise ( T f \u0002 T 1 ). In a similar way, the heat loss to the surrounding\nboundaries q_ w depends on their thermal properties and the fire temperature. The\nradiation loss term q_ r depends on T4f but is less significant for pre-flashover cases as\nthe fire temperature level then in general is lower.\nThere are two equations, the mass balance and the heat balance, and two\nunknowns, the temperature Tf and the distance hD, which now can be solved by a\nforward time incremental scheme. The input combustion rate q_ c may vary with time\nbut it is ultimately limited by the availability of oxygen. If too much fuel is released,\nthe fire becomes ventilation controlled and a one-zone model can be assumed, see\nChap. 10. Therefore, for two-zone models, cf. Eq. 10.4\n11.1 Heat and Mass Balance Theory 177\n\np\nq_ c \u0003 1 2 Ao ho 11:3\n\nWhen the heat release rate is constant, the temperature rise depends on the\nthermal properties of the surrounding structure and the ventilation (openings) in a\nsimilar way as for the one-zone model.\n\n## 11.1 Heat and Mass Balance Theory\n\nThe plume flow m_ p may be calculated according to Zukoskis plume equation as\n\n= 5\nm_ p 3 q_ c 3 z =3\n1\n11:4\n\nwhere z is the effective height of the plume above the burning area. Then with an\nanalogue derivation as for the one-zone case the heat flux to the surrounding\nstructures may be written as\n0 1\n=3 \u0004 5 =3 =3\nf \u0004 Ao \u0003 4 \u0004\n1 2\n3 q_ c z \u0004 cp @ q_ c\n\u0002 f A\n00\nq_ w T 1 \u0002 T 4f 11:5\nAt 3 cp z =3 At\n5\n\n## The emissivity f is here a reduction coefficient considering that the entire\n\nopening is not radiating corresponding to the\u0005 hot zone\n\u0006 fire temperature. According\nto Karlsson and Quintiere 3 0:0071 kg\n5= . Now with heat release q_ c and\nWsm 3\n\nthe effective height of the plume z assumed constant, the fire temperature may be\ncalculated in a similar way as for post-flashover one-zone models.\nThus a resultant temperature rise max can be defined as (see Fig. 11.2)\n\n## R*f, r *ult *ult\n\nmax 11:6\nR*f, c R*f, r R*f, c\n1 *\nRf , r\n\nwhere \u0005ult is the ultimate gas temperature rise determined as the heat release rate q_ c\nover the mass flow rate m_ p (Eq. 11.4) and the specific heat of air cp assuming no\nlosses neither through radiation out the openings nor from losses to boundary\nsurfaces. Observe that alternatively max can be obtained by solving Eq. 11.5 for\n00\nf when q_ w 0.\nBy comparison with Eq. 11.5, the ultimate temperature can be identified as\n178 11 Pre-flashover Compartment Fires: Two-Zone Models\n\nFig. 11.2 Electric circuit analogy model of a pre-flashover compartment fire boundary. (a) Two\nboundary temperatures (b) reduced to one boundary temperature\n\n2=\nq_ c 3\n\n*ult 11:7\n3 \u0004 cp \u0004 z =3\n5\n\n## and the fire thermal resistances can be identified from as\n\nAt\nR*f, c 11:8\n=\n3 \u0004 q_ c 3 \u0004 z =3 \u0004 cp\n1 5\n\nand\n\nAt\nR*f, r \u0003 \u0004 \u0007 \b 11:9\nf \u0004 Ao T 21 T 2f \u0004 T 1 T f\n\nThen the heat flux to the surface can be calculated as (see Fig. 11.2b)\n\u0007 \b\n00 max \u0002 f max \u0002 s\nq_ w *\n* 11:10\nRf , tot Rf , r Ri, tot\n\n## where the resultant resistance\n\n1\nR*f, tot 11:11\n1 1\n\nR*f, c R*f, r\n\nObserve that R\u0005f ;r must be calculated based on the fire temperature Tf and on the\nresultant temperature T max max T 1 . In a similar way may resultant heat\ntransfer resistance between the fire gases and compartment boundary surface Ri,tot\nbe calculated according to Eq. 10.22 based on the temperatures Tf and the surface\ntemperature Ts.\nThen the boundary condition becomes\n11.2 Solution of the Upper Layer Fire Temperature 179\n\n00 1 \u0007 * \b\nq_ w max \u0002 s : 11:12\nR*f, tot Ri, tot\n\nThis boundary condition can be used together with thermal model of the\nsurrounding structure in similar way as for post-flashover compartments. Then\nthe surface temperature may be calculated, and thereafter the fire temperature can\nbe obtained by the law of proportion as\n\n## T s Rf , tot T max Ri, tot s Rf , tot max Ri, tot\n\nTf 11:13\nRf , tot Ri, tot Rf , tot Ri, tot\n\n## In general analyses of the boundary structure require numerical methods.\n\nHowever, as for post-flashover one-zone analyses analytical solutions are some-\ntimes possible, given the heat resistances are given constant values representing a\nrelevant temperature level.\n\n## 11.2 Solution of the Upper Layer Fire Temperature\n\nIn combination with the boundary condition as defined by Eq. 11.12 may the\ntemperature of a surrounding structure be calculated in a similar way as for\none-zone models. Thus based on the calculated fire-exposed surface temperature\nmay then the upper layer temperature be calculated. In the next two sections will the\ncases of assumed semi-infinite and thin structures, respectively, be analysed. As for\none-zone models analytical solutions may under certain conditions be derived for\nquick rough estimates.\n\n## The surface temperature of semi-infinitely thick compartment boundaries with\n\nboundary conditions according to Eq. 11.12 may of course be solved numerically\nwith, e.g. finite element methods. However, with the assumptions of\nConstant heat release rate\nConstant material properties, (k \u0004 \u0004 c)\nConstant thermal heat transfer coefficients/resistances\nmay a closed form solution be derived as for one-zone models (see also Sect.\n3.2.3), i.e.\n180 11 Pre-flashover Compartment Fires: Two-Zone Models\n\n\" s!#\nt\n* t\ns *ult 1 \u0002 e erfc\nf 11:14\n*f\n\n## the time constant \u0005f can be calculated as\n\nk\u0004\u0004c \u0003 \u00042\n*f *\n\n2 k \u0004 \u0004 c \u0004 Rf , c Ri, tot 11:15\n1\nR*f, c R*i, tot\n\n## and then Eq. 10.23 yields the fire temperature.\n\nThe theory as outlined here gives interesting qualitative results but needs to be\nfurther validated by comparing with well-controlled experiments. Some such com-\nparisons have been done with very good results for compartments where the heat\ncapacity of the boundary can be lumped into a steel sheet as is shown in the next\nsection. Below is an example with purpose of showing how fire temperatures can be\ncalculated.\nExample 11.1 A propane gas burner at a height of 0.5 m in the room/corner test\nroom was set at a constant power 450 kW. The room has a total surrounding area\nAt 44 m2 and door opening A0 2 m2. Assume effective height of the burner\nplume z 1 m. Assume all the surrounding structural elements being infinitely\nthick light-weight concrete with a thermal inertia k \u0004 \u0004 c 0:2 \u0004 500 \u0004 800\n\u0007 \b\n80 \u0004 103 W2 s= m4 K2 . Initial and ambient temperatures are equal to 20 \u0006 C.\n(a) Calculate the maximum temperature not considering the radiation out the\ndoor-way.\n(b) Calculate the maximum temperature considering the radiation out the door-way.\n(c) Derive the surface and the fire temperatures as functions of time not considering\nthe radiation out the door-way and calculate the surface and fire temperatures\nafter 15 min. Assume a constant heat transfer coefficient hi 25 W/(m K).\n\nSolution\n(a) After a long time the wall losses vanish and the maximum temperature rise can\nq_ c =3 450000 =3\n2 2\n\n## be derived from Eq. 11.7: f ult\n\n3 \u0004 cp \u0004 z =3 0:0071 \u0004 1150 \u0004 1 =3\n5 5\n\n## 722 K. Hence the maximum temperature T f 722 20 724 \u0006 C.\n\n!\n3 \u0004 q_ c =3 \u0004 z =3 \u0004 cp q_ c =3 \u0003 \u0004\n1 5 2\nAo\n(b) Then q_ w \u0002 f \u0004 \u0004 T 4\n1 \u0002 T 4\nf\nAtot 3 \u0004 cp \u0004 z =3\n5\nAtot\n0:0071 \u0004 450000 =3 \u0004 1150 \u0007 h\n1\n\b 2 \u0007\n\u0004 722 \u0002 f \u0004 5:67 \u0004 10\u00028 \u0004 273 204 \u0002 f\n44 44\n273 204 \u0007 0. This 4th degree equation yields a temperature rise f 615 K\nand T f 625 \u0006 C.\n11.2 Solution of the Upper Layer Fire Temperature 181\n\n(c) The fire temperature as a function may be obtained from Eq. 11.14. From\nAtot 44\nEq. 11.8 R*f, c\n\u0004 q_\n1\n= 3\n5\n=\n\u0004z 3 \u0004c 0:0071 \u0004 450000 =3 \u0004 15=3 \u0004 1150\n1\n3 c p\n0:0704 m K=W and *ult 722 K. Equation 11.15 yields\n*f\n3\n80\u000410 !2 975 s. Thus according to Eq. 11.14\n1\n0:0704 251\nh \bi\n\u0007p\ns 722 \u0004 1 \u0002 e975 \u0004 erfc 975\nt\nt\n. After 900 s t/*f 900=975 0:92. The\nvalue of the function between the brackets from Fig. 3.11 or Table 3.3 is 0.56.\nThen s 722 \u0004 0:56 404 K. The fire temperature can then be obtained from\nEq. 11.13 as a mean weighted value of the surface and the ultimate temperature\n1\n404\u00040:0704722\u0004\nas f 25 519 K and the fire temperature T f 529 \u0006 C.\n1\n0:0704\n25\n\n## 11.2.2 Insulated and Uninsulated Boundaries\n\nwith a Metal Core\n\nWith the same assumptions as specified in Sect. 10.2.2 a similar expression as for\none-zone models can be obtained for the core temperature.\nReferring to Fig. 10.8 and Table 10.2 for the definitions of the parameters, the\ncore temperature can be numerically solved by the forward difference recursion\nformula (c.f. Eq. 10.38)\n\nt h\u0003 * \u0004\ni1 i\ncore core Rf , tot Ri, tot Ri, ins max \u0002 core\nCcore i 11:16\n\u0002Ro, ins Ro, tot \u0004 core \u0002 1\n\nwhere max is defined by Eq. 11.6, R\u0005f ;tot Eq. 11.8 and the other parameters as in Sect.\n10.2.2.\nAs for post-flashover fires (one-zone models) an analytical solution may be\nderived if the heat transfer parameters and the material properties are assumed\nconstant, not changing with temperature. Thus (cf. Eq. 10.39)\n\" #\nRo, tot Ro, ins \u0003 \u0004\n\u0002t=*f\ncore *ult 1 \u0002 e 11:17\nR*f, c Ri, tot Ri, ins Ro, tot Ro, ins\n\n## and the fire temperature rise as (cf. Eq. 10.41)\n\n182 11 Pre-flashover Compartment Fires: Two-Zone Models\n\n\"\n*ult Ro, tot Ro, ins \u0003 *\n\u0004\nf 1 \u0002 e\u0002t=f\nRi, tot Ri, ins R*f, c Ri, tot Ri, ins Ro, tot Ro, ins\n1\nR*f, c\n#\nRi, tot Ri, ins\n\nR*f, c\n11:18\n\n## where the time constant is calculated as\n\nCcore\n*f 11:19\n1 1\n\nR*f, c Ri, tot Ri, ins Ro, tot Ro, ins\n\n## Equation 11.18 yields a crude estimate of the fire temperature development as\n\nseveral assumptions are made to linearize the problem. More accurate solutions can\nbe made by the step-by-step numerical procedure according to Eq. 11.16 whereby\nthe material and, in particular, heat transfer conditions can be updated at each time\nstep. Such a calculation procedure was implemented in an MS-Excel sheet by\nEvegren and Wickstrom . It was used to predict and compare with measured\ntemperatures in an uninsulated and an insulated steel container with a burning pool\nof heptane, see Fig. 11.3.\n\nFig. 11.3 Dimensions of the test enclosure and photo of the insulated test enclosure and the pool\nfire experiment. From \n11.2 Solution of the Upper Layer Fire Temperature 183\n\nFig. 11.4 Measured and calculated upper layer fire temperature. From \n\nThe results are shown in Fig. 11.4. As can be seen predictions were very accurate\nfor both the cases.\nChapter 12\nFire Exposure of Structures According\nto Standards\n\nWhen exposed to fire structures deform and lose load-bearing capacity which must\nbe considered in design processes. It is then exposures to the more severe fires\nwhich are of interest such as post-flashover compartment fires and large flames for\nlonger times. Pre-flashover fires do in general not create thermal conditions that can\njeopardize the function of structural elements in a building. For design purposes it is\ntherefore in general exposures relevant for post-flashover compartment fires that are\nspecified in various standards and guidelines in the form of timetemperature\ncurves. These curves are then used for controlling fire resistance test furnaces, see\nFig. 12.1.\nThey can also be used as fire temperatures when predicting temperature of\nstructures exposed to standard fire conditions. When predicting test according to\nthe international standard ISO 834 and the European standard EN 1363-1 the gas\ntemperature and the radiation temperature may be assumed equal as these standards\nprescribe plate thermometers for controlling of furnace temperature. However,\nwhen predicting tests according to the American standard ASTM E-119, deviations\ndue to the thick thermocouples specified for controlling the furnace temperature\nshould be considered, see Sect. 9.1.3.\nA deterministic design and analysis process of structures exposed to fire entails\nthree major steps:\n1. Determine the fire exposure to which the surface of the structure is subjected.\n2. Determine the thermal response of the structure to the exposing fire.\n3. Determine the structural response and the load-bearing capacity at elevated\ntemperatures.\nThis chapter is focusing on the second step. The first section deals with design\nfires then followed by sections on the structural materials concrete, steel and wood.\n\n## Springer International Publishing Switzerland 2016 185\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_12\n186 12 Fire Exposure of Structures According to Standards\n\n## Fig. 12.1 A glazed\n\npartition being tested in a\nvertical fire resistance\nfurnace. Notice the Plate\nthermometers for\nmonitoring the furnace\ntemperature\n\n## 12.1 Standard Time Temperature Fire Curves\n\nThe so-called standard fire curve as defined in the European standard EN 1363-1\nand the international standard ISO 834 is outside the USA and Canada the by far\nmost commonly used timetemperature relation used for testing and classification\nof separating and load-bearing building structures. The timetemperature relation\nof this EN/ISO standard fire temperature curve is then specified as\n\n## where Tf is temperature in \u0003 C and t is time in minutes. A selection of time\n\ntemperature coordinates is given in Table 12.1.\nA so-called external fire curve is given in Eurocode 1 (EN 1991-1-2) as\n\u0001 \u0003\nT f 20 660 \u0002 1 \u0004 0:687 e\u00040:32\u0002 t \u0004 0:313 e\u00043:8\u0002t 12:2\n\n## This timetemperature is intended to be used for external structures outside of\n\nexternal walls.\nWhen more severe fires are anticipated, as for offshore oil installations or\ntunnels, the so-called Hydrocarbon Curve is often applied:\n12.1 Standard Time Temperature Fire Curves 187\n\nTable 12.1 Timetemperature coordinates of the standard ISO 834 and EN 1363-1 fire curves\nTime [min] 0 15 30 45 60 90 120 180 240\nTemperature [\u0003 C] 20 739 842 2 945 1006 1049 1110 1153\n\n## Table 12.2 Timetemperature coordinates of the RWS fire curve\n\nTime [min] 0 3 5 10 30 60 90 120 180\nTemperature [\u0003 C] 20 890 1140 1200 1300 1350 1300 1200 1200\n\n\u0001 \u0003\nT f 20 1080 \u0002 1 \u0004 0:325 e\u00040:167\u0002 t \u0004 0:675 e\u00042:5 \u0002t 12:3\n\nFor the design of tunnels, the ministry of transport in the Netherlands has\ndeveloped the so-called RWS fire curve which is used in many countries. It is\ndefined by the timetemperature coordinates given in Table 12.2.\nThe above-mentioned fire design curves are plotted in Fig. 12.2 together with\ntimetemperature curve according to ASTM E-119.\nIn the USA and Canada fire tests and classification are generally specified\naccording to the standard ASTM E-119. The ASTM E-119 fire curve is specified\nas time temperature coordinates as given in Table 12.3 or as approximated by the\nequation\n\u0005 p \u0006 p\nT f T 0 750 1 \u0004 e\u00040:49 t 22:0 t 12:4\n\nwhere fire or furnace temperature Tf and the initial temperature T0 are in \u0003 C and\ntime t in minutes. The curve is slightly different from the corresponding ISO and\nEN curves. However, the severity of a fire test depends not only on temperature\nlevel but also on how the temperature is measured. In ASTM E-119 the thermo-\ncouples specified for monitoring the furnace temperature are very thick and have\ntherefore a very slow response. That means the real temperature level is much\nhigher than measured by thermocouples. During the first 10 min of a fire resistance\ntest, the difference between measured temperature and the actual temperature level\nmay amount to several hundred degrees as indicated in Fig. 9.1. Thus, when\npredicting temperature in structures to be tested according to ASTM E-119, the\nmost relevant fire temperature curve to apply is the upper curve of Fig. 9.1. In\naddition to the problem of the time constant, it is unclear how the ASTM thermo-\ncouples react to different gas and radiation temperatures which makes any temper-\nature predictions uncertain.\n188 12 Fire Exposure of Structures According to Standards\n\n1400\n1300\n1200\n1100\n1000\n900\nTemperature [C]\n\n800\n700\n600\n500\nRWS\n400\nEN/ISO\n300\nASTM\n200 HC curve\n100 External\n0\n0 30 60 90 120 150 180\nTime [min]\n\nFig. 12.2 Standard timetemperature relations according to ISO 834 or EN 1363-1 (Eq. 12.1), the\nHydrocarbon curve (Eq. 12.2), the External fire curve according to Eurocode 1 (Eq. 12.3) and\nASTM E-119 (defined in Table 12.3 and approximated by Eq. 12.4)\n\n## Table 12.3 Timetemperature coordinates of the ASTM E-119 fire curve\n\nTime [min] 0 5 10 30 60 120 240 480 >480\nTemperature [\u0003 C] 20 538 704 843 927 1010 1093 1260 1260\n\n## Parametric fire curves are defined in Eurocode 1, EN1991-1-2, Appendix A. They\n\nare based on work in Sweden in the 1960s and 1970s which was later modified\nand simplified by Wickstrom, see e.g. .\nThe parametric fire curves are defined in the heating phase by the expression\n\u0005 \u0006\nT f 20 1325 \u0002 1 \u0004 0:324 \u0002 e\u00040:2\u0002t \u0004 0:204 \u0002 e\u00041:7\u0002t \u0004 0:472 \u0002 e\u000419\u0002t\n* * *\n12:5\n\n## where t* is a modified time defined as\n\nt* \u0002 t 12:6\n\nwhere the parameter (the gamma factor) determines the rate at which the fire\ntemperature goes to the ultimate temperature, (1325 + 20) \u0003 C. For equal unity\nEq. 12.5 yields a timetemperature relation which approximately follows the\n12.2 Parametric Fire Curves According to Eurocode 189\n\nstandard EN/ISO curve for about 6 h. The standard fire curve prescribes thereafter\nhigher temperatures while the parametric goes asymptotically a maximum fire\ntemperature rise of 1345 \u0003 C.\nThe factor depends on the opening factor (cf. Eq. 10.10) and the thermal\ninertia of surrounding structures. It is defined as\n2 p 3\nAo ho =At 2\np\n6 k\u0002\u0002c 7 O2\n4 0:04 5 841 \u0002 106 \u0002 : 12:7\n1160\nk\u0002\u0002c\n\n(The parameter values in this equation must be given in SI units.) Thus a fire\ncompartment with an opening factor of O 0:04 m1=2 and enclosure boundaries\np\nwith a characteristic value of the square root of the thermal inertia k \u0002 \u0002 c 1160\nW s1=2 =K m2 yields 1 which implies a fire development close to the EN/ISO\nstandard fire curve. Lower values, < 1, yield fires with slower temperature\ndevelopments while > 1 yields fires with faster developments.\nFigure 12.3 shows examples of the heating phase of parametric fire curves with\n-values smaller and larger than unity. The ISO/EN standard curve is plotted for\ncomparison. Notice that the parametric fire curve with 1 differs only a few\ndegrees from the ISO/EN standard curve.\nThe so-called hydrocarbon fire curve is a special case of a parametric curve. It\nwas originally defined as a parametric fire curve with 50 and with an ultimate\ntemperature of 1100 \u0003 C, see Eq. 12.3.\nFires are assumed to continue until all the fuel (fire load) is consumed, and the\nburning rate is assumed to be proportional to the amount of air being available in the\n00\nfire compartment. Thus the fire duration td is proportional to the fire load density qf\n(energy content per unit area) and the inverse of the opening factor. According to\nEurocode 1 it may be written as\n00\nq f \u0002 At\ntd p 12:8\nAo h o\n\n## In modified time the fire duration is calculated as\n\nt*d \u0002 td 12:9\n\nIn Eurocode 1 the proportionality constant has been given the value 0:2*10\u00043\n\u0007 \b\nh \u0002 m3=2 =MJ (units as in Eurocode 1 ).\n00\nThe fire load density qf is obtained by summarizing the weight of the various\nfuel components available for combustion with their net calorific value. Table 12.4\nshows a summary of the net calorific values as given in Eurocode 1.\n190 12 Fire Exposure of Structures According to Standards\n\nFig. 12.3 The standard EN/ISO standard curve and parametric fire curves with various -values.\nFor 1 the parametric curve coincides approximately with the standard curve for the first\n360 min\n\nTable 12.4 Net calorific values of combustible materials for calculation of fire loads\nMaterial Net calorific values [MJ/kg]\nWood 17.5\nOther cellulosic materials 20\nGasoline, petroleum 45\nDiesel 45\nPolyvinylchloride, PVC (plastic) 20\nOther plastics 3040\nRubber tyre 30\nNote: The values given in this table are not applicable for calculating energy content of fuels\nSummary from Eurocode 1, EN 1991-1-2 (The net calorific value is determined by subtracting the\nheat of vapourization of the water vapour from the gross calorific value)\n\nThe simple heat and mass balance theory applied for calculating the compart-\nment fire temperature in the heating phase is not relevant for the cooling phase\nwhen the fuel is more or less depleted and the assumption of uniform temperature is\nno more relevant. In Eurocode 1 simple linear time temperature relations are\ntherefore assumed as shown below:\n12.2 Parametric Fire Curves According to Eurocode 191\n\nFig. 12.4 Parametric timetemperature fire curves for varying opening factors O with a fire load\n00 p\nqf 200 kJ=m2 and thermal inertia of surrounding boundaries kc 1160 W s1=2 =K m2 .\nO 0.04 m yields approximately the ISO standard curve in the heating phase (The curves have\n-factors 4.0, 1.0 and 0.25, respectively)\n\n\u0001 \u0003\nT f T f , max \u0004 625\u0001t* \u0004 t*d\u0003\u0001 \u0003 for t*d \u0005 0:5 h\nT f T f , max \u0004 250\u00013 \u0004 t*d \u0003 t* \u0004 t*d for 0:5 < t*d \u0005 2 h 12:10\nT f T f , max \u0004 250 t* \u0004 t*d for t*d > 2 h\n\n## In case of fire durations td < 25 min additional information need to be considered as\n\ngiven in EN 1991-1-2.\nFigures 12.4 and 12.5 show examples of parametric fire curves. In both cases the\n00\nfire load qf 300 kg=m2 . Figure 12.4 shows the temperature development for\np\nvarious opening factors and Fig. 12.5 for various thermal inertia k \u0002 \u0002 c. The\n-factors 4.0, 1.0 and 0.25 are calculated based on Eq. 12.7.\nNotice in Fig. 12.4 how the maximum fire temperature Tf,max increases with\nO while the fire duration td decreases. Thus, e.g. concrete structures with slow time\nresponses are in general more sensitive to fires with low opening factor while it is\nthe opposite for bare steel structures.\nFigure 12.5 shows that fire temperature development depends on the thermal\ninertia of the surrounding boundaries. The fire duration td depends, however, only\non the fire load and the opening factor but is independent of the thermal inertia.\nNotice that the temperature development is much lower for a thermal inertia of\n2000 W s1/2/(K m2) corresponding to concrete than for approximately the ISO\np\nstandard curve, kc 1160 W s1=2 =K m2 . After 90 min the difference is. On the\nother hand, the temperature becomes much higher if the thermal inertia of the\nsurrounding structure is lower, e.g. with a thermal inertia of 500 Ws1/2/(Km2)\nrepresenting silicate or gypsum boards.\n192 12 Fire Exposure of Structures According to Standards\n\nFig. 12.5 Parametric timetemperature fire curves for varying thermal inertia of surrounding\np 00 p\nboundaries kc with a fire load qf 200 kJ=m2 and an opening factor O 0.04 m. kc\n1160 W s1=2 =Km2 yields approximately the ISO standard curve in the heating phase (The\ncurves have -factors 4.0, 1.0 and 0.25, respectively)\n\n## Example 12.1 Calculate the maximum temperature of a parametric fire in a\n\np p\ncompartment with O Ao ho =At 0:08 m , k \u0002 \u0002 c 1160 W s1=2 =K m2\nand qf 400 MJ=m . In Eq. 12.8 assume 0:2*10\u00043 h m3=2 =MJ.\n} 2\n\n## Solution Equation 12.7 yields 22 4 and Eq. 12.8\n\n\u00043\ntd 0:2*10 *400 =0:08 1 h. The maximum temperature can then be obtained\nfor a modified fire duration of 4 \u0002 1 h 4 h from Eq. 12.5 or from Fig. 12.3 to be\n1150 \u0003 C.\n\n## 12.3 Summary of Heat Transfer Conditions According\n\nto Eurocodes\n\nThe temperature shall be measured with PTs (see Sect. 9.3) in fire resistance\nfurnace tests according to ISO or EN standards. Therefore the standard furnace\n00\ntemperature Tf can be considered as an AST and the heat transfer q_ tot to an exposed\nsurface with a temperature Ts may be calculated as (cf. Eq. 4.14)\n00\nh\u0001 \u00034 i \u0001 \u0003\nq_ tot \u0002 T f 273 \u0004 T s 2734 hc T f \u0004 T s 12:11\n\nwhere temperatures here are in \u0003 C. This does not apply to predictions of tests\naccording to ASTM E-119 where the fire timetemperature curve needs be modi-\nfied before being used as a boundary condition in temperature calculations, see\nSect. 12.1 above.\n12.3 Summary of Heat Transfer Conditions According to Eurocodes 193\n\n## When calculating temperatures of structures exposed to severe fires, the influ-\n\nence of the choice of the emissivity is in general small and of the convective heat\ntransfer coefficient hc it is often negligible for most materials, especially light\nmaterials, and longer fire durations. For insulation materials (low thermal inertia)\nit is almost negligible. Then the surface temperature may be assumed equal to the\nfire temperature (first kind of boundary condition, see Sect. 9.3). On the contrary,\nfor bare/non-protected steel structures particularly the emissivity and in some\ncases also the convection heat transfer coefficient hc have a significant influence and\nare decisive for the temperature development.\nThe emissivity is a property of the solid surface only, while the convective heat\ntransfer coefficient hc depends on the geometry and the surrounding flow condi-\ntions. According to Eurocode 1 the specimen surface emissivity 0:8\nunless another value can be motivated. The convection heat transfer coefficient\nhc 25 W=m2 K when applying the EN 1363-1 standard curve or the external\nfire curve, and hc 50 W=m2 K and hc 35 W=m2 K when applying the\nhydrocarbon curve or any natural fire curve including parametric fire curves,\nrespectively.\nAt surfaces on the unexposed side of separating elements Eurocode 1 \nsuggests that the heat transfer shall be calculated as (here the emissivity and view\nfactor of the fire are assumed to be unity)\n00\nh \u0001 \u0003i\nq_ tot \u0002 T 1 2734 \u0004 T s 2734 hc T 1 \u0004 T s 12:12\n\nwhere T 1 is the ambient surrounding temperature. Here the unexposed side surface\nemissivity should be as for the exposed side, i.e. 0:8 unless another value can be\nmotivated. The convection heat transfer coefficient shall then be assumed as\nhc 4 W=m2 K. Alternatively, the radiation component of Eq. 12.12 may be\nincluded in the convection heat transfer term. Then the convection heat transfer\ncoefficient hc 9 W=m2 K and the surface emissivity 0, i.e. a linear boundary\ncondition of the kind 3a according to Table 4.2.\nA summary of the heat transfer parameters as specified by Eurocode 1 is given in\nTable 12.5.\n\n## Table 12.5 Summary of heat transfer parameters as specified by Eurocode 1, EN 1991-1-2\n\nConvection heat transfer\nFire curve Exposed side coefficient hc [W/(m K)] Emissivity [\u0004]\nStandard ISO/EN Fire-exposed side 25 0.8a\nUnexposed sideb 4 0.8a\nUnexposed sideb 9 0\nHydrocarbon curve Fire-exposed side 50 0.8a\nExternal fire curve Fire-exposed side 25 0.8a\na\nUnless another material property value is motivated\nb\nAlternative\nChapter 13\nTemperature of Steel Structures\n\n## Steel is sensitive to high temperature. The critical temperature of a steel member is\n\nthe temperature at which it cannot safely support its load.\nThe mechanical properties such as strength and modus of elasticity deteriorate in\nparticular when the steel temperature exceeds 400 \u0001 C, see, e.g. Eurocode 3, EN\n1993-1-2. Some building codes and structural engineering standard practice defines\ndifferent critical temperatures which must not be exceeded when exposed to a\nstandard fire exposure for a specified time. Steel structures must therefore usually\nbe protected to reach a particular fire rating. Please note that insulation and\nprotection of structures are in this book used synonymously. Protections can be\nobtained by for instance boards, sprayed on concrete, insulation materials or\nintumescent paint. Intumescent coatings or reactive coatings expand upon heating\nand provide an insulating char to protect structural steelwork. Steel structures may\nalso be built into concrete or even wooden structures as a means of fire protection.\nEurocode 4 (EN 1994-1-2) deals with composite structures of steel and concrete\nwhere steel sections are imbedded in concrete.\nTo obtain a certain rating a steel structure can be tested in fire resistance furnace\naccording to specific standards depending on country or region. Alternatively or as\na pretest investigation steel temperatures can be calculated compared with critical\nvalues when exposed to design fire conditions for specified durations.\nBecause of the high conductivity the temperature field in a steel section is in\nmany fire engineering cases assumed uniform. In particular the temperature across\nthe thickness of a steel sheet can in almost all fire resistance cases be assumed\nconstant, while the temperature in the plane of steel sheets may vary considerably.\nThen the zero- or one-dimensional calculation techniques may be used as presented\nin Sects. 3.1 and 7.1 and further adapted to protected and unprotected steel sections\nin Sects. 13.3 and 13.4, respectively. For more general two- and three-dimensional\ncases numerical computer codes are needed, see Sect. 7.3.2 and Sect. 13.5 where\nsome examples are shown.\nGenerally in the following sections the gas and radiation temperatures are\nassumed equal to the fire temperature, i.e. T g T r T f , as is assumed in all\n\n## Springer International Publishing Switzerland 2016 195\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_13\n196 13 Temperature of Steel Structures\n\nstandard timetemperature fire curves. If measured temperatures are used, the fire\ntemperature Tf may be replaced by the adiabatic surface temperature measured\nwith, e.g. PTs.\n\n## 13.1 Thermal Properties of Steel\n\nMetals in general have high electric conductivity, high thermal conductivity and\nhigh density. The heat conductivity of carbon steel is in the order of 30 times higher\nthan the corresponding value for concrete and 1001000 times higher than that of\ninsulation products. The higher purity of a metal, the better it conducts heat. Thus\ncontents of carbon and alloying metals such as chrome reduce the conductivity, and\nconsequently stainless steel is a relatively poor conductor. The specific heat capac-\nities of metals are in accordance with a general rule of physics inversely propor-\ntional to the molecular weight.\nFigure 13.1 shows the conductivity kst vs. temperature Tst of structural carbon\nsteel according to Eurocode 3 (EN 1993-1-2). It can also be obtained from\nTable 13.1. For approximate calculations normally on the safe side a constant\nvalue of 46 W/(m K) can be recommended, cf. Table 1.2.\nThe specific heat capacity is usually a more significant parameter than the\nconductivity for the development of temperature in fire-exposed steel structures.\nIn many cases it is accurate enough and convenient to assume a constant specific\nheat capacity. Then a value of 460 J/(kg K) is recommended which normally yields\ncalculated temperatures on the safe side (overvalued). However, for more accurate\ncalculations the variations with temperature as shown in Fig. 13.2 or given in\nTable 13.2 are recommended in Eurocode 3 . The peak of the specific heat\ncapacity at 735 \u0001 C is due to phase changes of the steel.\nTable 13.3 shows tabulated values of the thermal properties of carbon steel\nderived from Eurocode 3 including the specific volumetric enthalpy vs. temperature\ndefined as\nT\ne T c \u0003 dT 13:1\n0\n\n## This temperatureenthalpy relation is input in some computer codes, e.g. Tasef,\n\ninstead of density and specific heat capacity. The diagram in Fig. 13.3 shows the\nspecific volumetric enthalpy vs. temperature based on the values of Table 13.3.\nThermal conductivity of stainless steel is considerably lower than that of carbon\nsteel. The conductivity and the specific heat capacity of stainless steel according to\nEurocode 3, EN 1993-1-2 are given in Table 13.4.\n13.1 Thermal Properties of Steel 197\n\n## Fig. 13.1 Thermal 60\n\nconductivity of steel\nvs. temperature according to\nEurocode 3, EN 1993-1-2. 50\n\n## Conductivity, kst[W/(m K)]\n\n40\n\n30\n\n20\n\n10\n\n0\n0 200 400 600 800 1000 1200\nTemperature [C]\n\nTable 13.1 Thermal conductivity of carbon steel vs. temperature according to Eurocode 3, EN\n1993-1-2\nTemperature [\u0001 C] Conductivity [W/m K]\n20 < Tst < 800 54 \u0004 0.0333 Tst\n800 < Tst < 1200 27.3\n\n## Fig. 13.2 Specific heat 6\n\ncapacity of carbon steel\nvs. temperature according to\n5\nSpecific heat, cst [Ws/(kg K)]\n\nEurocode 3, EN 1993-1-2\n\n0\n0 200 400 600 800 1000 1200\nTemperature [C]\n198 13 Temperature of Steel Structures\n\nTable 13.2 Specific heat capacity of carbon steel as functions of the temperature according to\nEurocode 3, EN 1993-1-2\nTemperature [\u0001 C] Specific heat capacity [J/(kg K)]\n20 < Tst < 600 425 + 0.773 \u0003 Tst \u0004 1.69 \u0005 10\u00043 \u0003 Tst2 + 2.22 \u0005 10\u00046 \u0003 Tst3\n600 < Tst < 735 666 + 13002/(738 \u0004 Tst)\n735 < Tst < 900 545 + 17820/(Tst \u0004 731)\n900 < Tst < 1200 650\n\nTable 13.3 Summary of thermal properties of carbon steel including derived volumetric specific\nenthalpy according to Eurocode 3, EN 1993-1-2\nTemp kst st cst est est\n[\u0001 C] [W/(m K)] [kg/m3] [J/(kg K)] [J/(m3 K)] [Wh/(m3 K)]\n0 54 7850 425 0 0\n100 51 7850 488 0.360E + 09 99,870\n200 47 7850 530 0.760E + 09 211,000\n300 44 7850 565 1.19E + 09 330,300\n400 41 7850 606 1.65E + 09 457,800\n500 37 7850 667 2.15E + 09 596,100\n600 34 7850 760 2.70E + 09 751,100\n700 31 7850 1008 3.37E + 09 934,300\n735 30 7850 5000 4.20E + 09 1,091,000\n800 27 7850 803 5.03E + 09 1,309,000\n900 27 7850 650 5.58E + 09 1,464,000\n1200 27 7850 650 7.12E + 09 1,890,000\n\n## Fig. 13.3 Volumetric 2000\n\nenthalpy of carbon steel\n1800\naccording to Table 13.3\nSpecific volumetric enthalpy,\n\n## derived from Eurocode 1600\n\n3, EN 1993-1-2\n1400\nest[kWh/(m3K)]\n\n1200\n1000\n800\n600\n400\n200\n0\n0 200 400 600 800 1000 1200\nTemperature [C]\n13.2 Example of Hot-Rolled Steel Section Dimensions 199\n\nTable 13.4 Thermal conductivity and specific heat capacity of stainless steel vs. temperature\naccording to Eurocode 3, EN 1993-1-2\nConductivity\nTemperature [\u0001 C] [W/(m K)] Specific heat capacity [J/(kg K)]\n20 < Tst < 1200 14.6 + 0.0127 \u0003 Tst 450 + 0.280 \u0003 Tst \u0004 0.291 \u0003 10\u00043 \u0003 Tst2 + 0.134 \u0003 10\u00046 \u0003 Tst3\n\nTable 13.5 Dimensions of hot-rolled HEB steel sections according to EN 10025-1. The last\ncolumn corresponds to Ast\nHeight Width Web thickness Flange thickness Weight Surface area\nHEB [mm] [mm] [mm] [mm] [kg/m] [m2/m]\n100 100 100 6 10 20.8 0.567\n120 120 120 6.5 11 27.2 0.686\n140 140 140 7 12 34.4 0.805\n160 160 160 8 13 43.4 0.918\n180 180 180 8.5 14 52.2 104\n200 200 200 9 15 62.5 1.15\n220 220 220 9.5 16 72.8 1.27\n240 240 240 10 17 84.8 1.38\n260 260 260 10 17.5 94.8 1.5\n280 280 280 10.5 18 105 1.62\n300 300 300 11 19 119 1.73\n320 320 300 11.5 20.5 129 1.77\n340 340 300 12 21.5 137 1.81\n360 360 300 12.5 22.5 145 1.85\n400 400 300 13.5 24 158 1.93\n450 450 300 14 26 174 2.03\n500 500 300 14.5 28 191 2.12\n550 550 300 15 29 203 2.22\n600 600 300 15.5 30 216 2.32\n650 650 300 16 31 229 2.42\n700 700 300 17 32 245 2.52\n800 800 300 17.5 33 267 2.71\n900 900 300 18.5 35 297 2.91\n1000 1000 300 19 36 320 3.11\n\n## Dimensions of hot-rolled steel sections can be found for instance in suppliers\n\ncatalogues or on the internet web. As an example dimensions of HEB wide-flange\nsteel I-sections according to the European standard EN 10025-1 are given in\nTable 13.5.\n200 13 Temperature of Steel Structures\n\n## 13.3 Protected Steel Sections Assuming Lumped-Heat-\n\nCapacity\n\nThe assumption of lumped heat or uniform steel temperature as often done in fire\nprotection engineering calculations (see, e.g. Eurocode 3) is in particular a reason-\nable approximation when calculating temperature of protected steel sections\nexposed to fire on all four sides. The assumption of uniform heat implies that the\nheat conductivity is assumed infinite and the thermal mass is concentrated, lumped,\nto one point, see Sects. 3.1 and 7.1.\nThen in addition the fire and the exposed surface temperatures are assumed equal\nwhich implies that the heat transfer resistance between the fire gases and the\nprotection surface is negligible. That means the inverse of the total heat transfer\ncoefficient by radiation and convection is assumed negligible in comparison with\nthe heat resistance of the insulation Rk, i.e. the thickness over the conductivity din/\nkin of the insulation, cf. Fig. 3.3. This is an accurate approximation as the radiation\nheat transfer coefficient is very high at elevated fire temperatures. It facilitates\ncalculations and it is on the safe side as it overestimates steel temperatures.\nThe heat transfer to the steel may then be calculated as\n\u0002 \u0003\nkin \u0004 \u0005\nq_ tot Ast T f \u0004 T st 13:2\ndin\n\nwhere Ast is the fire-exposed area per unit length, Tf and Tst are the fire and steel\ntemperatures, respectively. If in addition the heat capacity of the insulation is\nnegligible in comparison to that of the steel, the transient heat balance of the steel\nsection becomes,\n\u0002 \u0003\nkin \u0004 \u0005 T st\nAst T f \u0004 T st cst st V st 13:3\nd in t\n\nThat is the heat entering the steel section is equal to the heat stored per unit time\nproportional to the rate of temperature rise. cst and st are the specific heat capacity\nand density, respectively, of steel and Vst the volume per unit length of the steel\nsection. When estimating the conductivity of the insulation the temperature of the\ninsulation may be assumed as the mean of the fire and the steel temperatures.\nIn cases of heavy insulations when the heat capacity of the insulation need be\nconsidered a more rigorous analysis is required as shown in Sect. 13.3.1.\nFrom Eq. 13.3 the forward difference scheme\n\u0002 \u0003 \u0002 \u0003\u0006 \u0007\nAst t kin\nT i1 T sti T i1\n\u0004 T i\n13:4\nst\nV st st \u0003 cst i din f st\n\nwhere t is a chosen time increment. The specific heat csti is taken at the temper-\nature level Tist (if assumed varying with temperature).\n13.3 Protected Steel Sections Assuming Lumped-Heat-Capacity 201\n\nThe relation Ast/Vst is denoted the section factor or shape factor. It has the\ndimension one over length [m\u00041]. The shape factor can be replaced by its recipro-\ncal, the effective thickness of the steel dst identified as\n\nV st\ndst 13:5\nAst\n\nInstructions on how to obtain shape factors for various steel sections are given in\nTable 13.6 taken from Eurocode 3 . The area Ast of contour encasements such as\nspray fire protection material is generally taken as the perimeter of the section times\nthe unit length. For board protections forming hollow encasements, the perimeter\nmay be assumed as the boxed value as shown in the second row of Table 13.6. Even\nif there is a clearance around the member, the same boxed value may be applied.\nFor steel sections fire exposed on three sides the perimeter is reduced accordingly as\nshown in the third and fourth rows of Table 13.6. Thus the interface between the\nsteel and, for example, a concrete slab is treated as an adiabatic surface and hence\nthe cooling effects of the steel section is ignored. Therefore this crude approxima-\ntion model yields considerably higher temperature than it could be expected in\nreality. To accurately incorporate the cooling effects 2D finite element calculations\nare required.\nAlternatively the steel section volume per unit length Vst may be obtained as the\nweight per unit length mst (often tabulated in catalogues of steel providers) over the\nsteel density st, i.e.\nmst\nV st 13:6\nst\n\n## Analytical solutions can be derived only when constant conductivity of the\n\nprotection material and specific heat of the steel are assumed. If in addition the\nfire temperature is assumed to suddenly rise to constant temperature, can the steel\ntemperature be obtained as shown in Sect. 3.1.2 as\n\nT st \u0004 T i\n1 \u0004 e \u0004\nt\n13:7\nTf \u0004 Ti\n\nwhere is identified as a time constant which for a protected steel section becomes\n\u0002 \u0003 \u0002 \u0003 \u0002 \u0003\nV st din din\ncst dst st cst 13:8\nAst st kin kin\n\nIn some special cases with varying fire temperatures the steel temperatures may\nbe calculated analytically as shown in Sect. 13.3.2 where efficient and compact\ndiagrams which facilitates estimations of steel temperatures are shown.\nIn general, however, the time constant cannot be assumed constant as the\nthermal properties of the insulation as well as of the steel vary with temperature and\n202 13 Temperature of Steel Structures\n\nTable 13.6 Section factor Ast/Vst for steel members insulated by fire protection material. From\nEurocode 3 \nSection factor\nSketch Description (Ast/Vst)\nContour encasement of steel perimeter\nsteel cross\u0004section area\nuniform thickness\n\n## Hollow encasement of 2bh\n\nsteel cross\u0004section area\nuniform thicknessa\nh h\n\nb c1 b c2\n\n## Contour encasement of steel perimeter 2hb\n\nsteel cross\u0004section area\nuniform thickness,\nexposed to fire on three\nsides\n\nHollow encasement of 2 hb\nsteel cross\u0004section area\nuniform thickness,\nexposed to fire on three\nsidesa\nh\nh\n\nc1 b c2\nb\n\n## The clearance dimensions c1 and c2 should not normally exceed h/4\n\ntime, and as the fire temperature Tf generally varies with time. As an alternative to\nEq. 13.4 the steel temperature can be calculated by forward difference recursion\nformula\n\u0002 \u0003\nt i1 t\nst\nT i1 T 1 \u0004 T sti 13:9\ni f i\n\nwhere t is a chosen time increment. The suffixes denote the numerical order of the\ntime increments. When the thermal properties vary with temperature, the time\nconstant need be updated at each time increment.\nThe forward difference scheme of Eq. 13.12 is numerically stable if the time\nincrement is less than the time constant at each time increment i, i.e.\n13.3 Protected Steel Sections Assuming Lumped-Heat-Capacity 203\n\nt \u0006 i 13:10\n\nIn practice time increments t longer than 10 % of the time constant should not\nbe used to assure numerical stability and accuracy. When choosing the time\nincrement it is also necessary to make it short enough to be able to follow the\nthermal exposure changes with time.\nThe recursion formulas according to Eq. 13.9 are preferably solved with a\nspreadsheet program such as MS-Excel. For clarification examples are shown\nbelow on how the formula is used.\nExample 13.1 A steel column with a section factor 200 m\u00041 is protected with a\n25 mm non-combustible board with a conductivity of 0.1 W/(m K). The column is\nexposed to fire and the exposed insulation surface suddenly reaches a temperature\nof 1000 \u0001 C. Assume constant thermal properties and uniform steel temperature\n(lumped heat). The density and specific heat of steel are assumed to be 7850 kg/m3\nand 460 W s/(kg K), respectively. The initial temperature T i 20 \u0001 C. Calculate the\nsteel temperature after\n(a) 9 min using the analytical exact solution according to Eq. 13.7\n(b) 60 min using the analytical exact solution according to Eq. 13.7\n(c) 9 min using the numerical solution according to Eq. 13.9 and compare with (a)\n\n## Solution According to Eq. 13.8 the time constant 460\u00037850\u00030:025=0:1\n\n200 4514 s.\n(a) After 9 min t 9 \u0003 60=4514 0:12 and according to Eq. 13.7 or Fig. 3.4\n\u0004 t\u0005\n1 \u0004 e\u0004 0:113 and the steel temperature T st 20 1000 \u0004 20*0:113\n131 \u0001 C.\n(b) After 60 min t 60 \u0003 60=4514 0:80 and the steel temperature\nT st 20 1000 \u0004 20*0:55 560 \u0001 C.\n(c) Assume a time increment\u0004 180 \u0005t 3 min\u0004 180 s.\u0005 Then according to Eq. 13.9 at\nt 180 s T 1st 4514 \u0003 1000 1 \u0004 4514\n180\n\u0003 20 59 \u0001 C, at t 360 s\n\u0004 180 \u0005 \u0004 \u0005 \u0001\nT st 4514 \u0003 1000 1 \u0004 4514 \u0003 59 96 C\n2 180\nand at 540 s\n\u0004 180 \u0005 \u0004 \u0005 \u0001\nT st 4514 \u0003 1000 1 \u0004 4514 \u0003 96 132 C. Notice that the numerical solu-\n2 180\n\ntion (b) is only 1 \u0001 C more than the exact solution according to (a).\nExample 13.2 The same column as in Example 13.2 is exposed to a standard fire\ntimetemperature curve according to ISO 834. Calculate the steel temperature after\n9 min.\nSolution Apply the recursion formula according to Eq. 13.9. Choose a time\nincrement t 180 s (\u000710 % of ). The temperature at 3, 6 and 9 min is according\nto Eq. \u000412.1,\u00057.7, 10.17\u0004 and 12.5,\n\u0005 respectively. Then according to Eq. 13.9 at t\u0005 180 s\n\u0004 180\nT 1st 4514\n180\n\u0003 228 1 \u0004 4514180\n\u0003 20 28:3 \u0001 C, at t 360 s T 2st 4514 \u0003 312\n\u0004 \u0005 \u0001\n\u0004 180 \u0005 \u0004 \u0005\n1 \u0004 4514 \u0003 28:3 36:6 C and at 540 s T st 4514 \u0003 365 1 \u0004 4514 \u0003 36:6\n180 2 180\n\n49:7 \u0001 C.\n204 13 Temperature of Steel Structures\n\n## 13.3.1 Protection with Heavy Materials\n\nThe heat capacity of fire protections has normally an insignificant influence on the\nsteel temperature rise rate. However, it will considerably reduce the steel temper-\nature rise of sections protected with relatively heavy protections materials. The\nprotection will then add to the heat capacity of the system and it will cause a delay\nin the temperature rise of the steel section. A simple approximate approach is then\nto lump a third of the heat capacity of the insulation to the steel section heat capacity\nand to add a term considering the time delay . Eq. 13.4 may then be\nmodified the more general formulation as\n\u0006 \u0007\nT j1\nf \u0004 T stj \u0006 \u0007\u0006 \u0007\nT j1 T stj t \u0004 \u0005 e \u0004 1 T j1\n\u0004 T j\n13:11\n1 3\nst f f\n\nwhere is as specified in Eq. 13.7 and is the ratio between the heat capacity of the\ninsulation and the steel,\n\nAst \u0003 d in \u0003 in \u0003 cin\n13:12\nV st \u0003 st \u0003 cst\n\nin and cin are the density and specific heat capacity of the protection material,\nrespectively. The latter term of Eq. 13.11 represents a time delay due to the heat\ncapacity of the protection. (T j1\nf \u0004 T fj) is the fire temperature rise between two time\nincrements. Notice that when the heat capacity of the protection is much smaller\nthan that of the steel, vanishes and Eq. 13.11 becomes identical to Eq. 13.4.\nThe value of the parameter in the last term of Eq. 13.11 was obtained by\ncomparisons with accurate finite element calculations. For steel sections exposed to\nthe ISO/EN standard timetemperature curve accurate approximations are obtained\nby choosing 5, see Example 13.3 and Fig. 13.4.\nFor fire temperatures assumed to instantaneously rise to a given temperature,\n10 yields very similar steel temperatures in comparison to accurately calculated\ntemperatures. This value has been adopted by Eurocode 3, EN 1993-1-2 . It\nyields higher steel temperatures than choosing the more accurate value 5.\nExample 13.3 A steel section with a section factor of 200 m\u00041 and an initial\ntemperature of 20 \u0001 C is exposed to a standard fire curve according to ISO 834. It is\nprotected with 20-mm-thick high density material assumed to have the same\nproperties as concrete. Assume material properties as given in Table 1.2. Calculate\nthe steel temperature development.\n\u0006 \u0007\nSolution According to Eq. 13.7 Astst=V\n\u0003cst\nst kin\nd in 7850\u0003460\u00030:02\n200\u00031:7 212 s and Eq. 9.12\nAstV\u0003dst in\u0003\u0003\u0003cin \u0003cst in 200\u00030:02\u00032300\u0003900\n7850\u0003460 2:29. Then the recursion formula Eq. 13.11 may\nst\nbe applied. Steel temperatures obtained by an MS-Excel application are given in\n13.3 Protected Steel Sections Assuming Lumped-Heat-Capacity 205\n\nFig. 13.4 Example of steel temperature of a steel section protected by a heavy protection material\nexposed to the ISO 834 standard timetemperature fire curve. Temperature calculated with the\ndelay and with no delay according to the third term according to Eq. 13.11. For comparison the\nsteel temperature as calculated accurately with the finite element code TASEF\n\nFig. 13.4. For comparison accurately finite element calculated temperatures are also\nshown. Notice how well the temperature calculated according to Eq. 13.11 matches\nthe accurate solution except for the 5 min when the temperature goes down even\nbelow zero. In addition the temperatures are shown which are calculated without\nconsidering the delay expressed by the parameter larger than zero in the third term\non the right-hand side of Eq. 13.11.\n\n## 13.3.2 Protected Steel Sections Exposed to Parametric\n\nFire Curves\n\nAs described in Sect. 12.2 the concept of parametric fires has been introduced in\nEurocode 1 as a convenient way of expressing a set of post-flashover design\nfires.\nWhen using parametric design fires the temperature of protected steel sections\ncan of course be obtained by numerical calculations according to Eq. 13.9. Then\nnon-linear phenomena such as temperature-dependent material properties may be\nconsidered. However, if the thermal properties are assumed constant and the fire\n206 13 Temperature of Steel Structures\n\n## temperature is expressed by exponential terms as in Eq. 12.5, then the steel\n\ntemperature rise vs. time can be obtained by integration as a closed form analytic\nexpression .\nEquation 12.5 may be written in the form\nX3A \u0006 \u0007\nBi e\u0004i t\n*\nT f 20 i0\n13:13\n\nwhere the constants Bi and i are given in Table 13.7. Notice that Eq. 13.13 is\nidentical to Eq. 12.5 but written in a different format to reach to a compact solution\nfor the steel temperature as given in Eq. 13.14.\nThen the steel temperature can be derived exactly by analytical integration as a\nfunction of the modified time t* and the modified time constant * of the steel\nsection as\nX3 \b Bi \u0006 i t * t*\n\u0007\nT st 20 e \u0004 e \u0004 * 13:14\ni0 1 i *\n\nwhere\n\n* \u0003 13:15\n\nThe protected steel section time constant is given in Eq. 13.8. The relation\nbetween the temperature rise vs. modified time as expressed in Eq. 13.14 is also\ngiven in the diagram shown in Fig. 13.5a, b for various modified time constants *.\nThe two diagrams are the same but with different time and temperature scales.\nNotice that Eq. 13.14 and Fig. 13.5 may be used for the ISO 834 standard fire\nexposures assuming 1 as the parametric fire curve in the heating phase then is\nvery close to the standard curve, see Fig. 12.3.\nThe use of parametric fire curves on insulated steel sections is demonstrated\nbelow.\nExample 13.4 Consider a steel section with a shape factor 200 m\u00041 with a\n25-mm-thick protection board having a constant thermal conductivity of 0.1 W/\n(m K). The steel density and specific heat capacity are 7850 kg/m3 and 460 J/(kg K),\nrespectively. The section time constant may then be obtained from Eq. 13.15 as\n4514 s 75 min 1.25 h. Then if the section is exposed to ISO 834 standard\nfire ( 1) for 60 min, a temperature of 462 \u0001 C may be obtained from Eq. 13.14 or\nfrom Fig. 13.5. If the same section is exposed to a more slowly growing fire with a\n0.5, then * \u0003 37.5 min and the temperature after 60 min may be found\nfor a modified time of t* \u0003 t 30 min to be 405 \u0001 C. On the other hand, if the\nsection is exposed to a fast growing fire with 3.0, then * 3.0 75 225 min\nand t* 3.0 60 180 min, and the steel temperature can be obtained from\nEq. 13.13 or from Fig. 13.5 as 552 \u0001 C. Notice that the maximum steel temperature\nfor a given fire exposure time increases considerably with an increasing -factor. It\n13.4 Unprotected Steel Sections 207\n\n## Table 13.7 Constants in the Term number, i 0 1 2 3\n\nanalytical expression of the\nBi (\u0001 C) 1325 \u0004430 \u0004270 \u0004625\nparametric fire curve\ni (h\u00041) 0 \u00040.2 \u00041.7 \u000419\n\nmust, however, also be kept in mind that the fire duration for a given fuel load is\nproportional to the inverse of the opening factor included in the -factor.\nThe diagrams of Figure 13.6a, b show the temperature development of a\nsteel structure with the same dimensions and protection as described above.\nThe two cases are assuming the same fire qf 200 kJ=m2 and thermal inertia\np\nof the surrounding structure k \u0003 \u0003 c 1160 W s1=2 =K m2 . According\nto Eq. 12.8 fire duration can be calculated to be td 60 min and 120 min\n\u0005\n( 0:2*10\u00043 h m3=2 =MJ , with 1 and 0.25, respectively. Notice that\nthe steel temperature reaches its maximum when it is equal to the cooling\nphase fire temperature. According to the diagram of Figure 13.6a the steel\ntemperature is 450 \u0001 C at td 60 min and reaches its maximum 570 \u0001 C after\n105 min. The corresponding temperatures according to diagram in\nFigure 13.6b are 585 \u0001 C after 120 min and the maximum steel temperature\nis 670 \u0001 C after 180 min. Thus the steel section reaches a higher temperature\nfor the lower opening factor, given the fire load and the thermal properties\nof the surrounding compartment boundaries remains the same.\nFor more detailed information on how to apply parametric fire curves according\nto standard, see Eurocode 1 .\n\n## The temperature of unprotected, uninsulated or bare steel sections depends on the\n\nfire temperature and very much on the heat transfer conditions between fire gases\nand steel surfaces. It is a boundary condition of the 3rd kind, see Sect. 1.1.3, where\nthe only thermal resistance between the fire and the steel is due to the heat transfer\nconditions which therefore becomes decisive for the steel temperature develop-\nment. The boundary condition is highly non-linear as it varies very much with\ntemperature due to radiation. The same type of compact formula and diagrams as\nfor insulated steel sections can therefore not be developed.\n00\nThe total heat flux by radiation and convection q_ tot is given in Eq. 4.17 or\nEqs. 4.18 and 4.19. Steel temperatures can then be obtained from differential heat\nbalance equations in a similar way as for protected steel sections (cf. Eqs. 13.2 and\n13.3).\nAccording to Eq. 4.17 and the procedures as outlined in Sect. 7.1 the heat flux by\nradiation and convection can be written as\n208 13 Temperature of Steel Structures\n\nFig. 13.5 Temperature of various protected steel sections exposed to parametric fires in the\nheating phase vs. modified time t*. The thermal properties of the steel sections are embedded in\nthe modified time constants *, see Eq. 13.15. The bottom diagram is a magnification of the top\n13.4 Unprotected Steel Sections 209\n\na 1200\nb 1200\nO=0.04 m O=0.02 m\n\n## 1000 Tst 1000 Tst\n\nTemperature [C]\n\nTemperature [C]\n800 800\n\n600 600\n\n400 400\n\n200 200\n\n0 0\n0 30 60 90 120 150 180 0 30 60 90 120 150 180\nTime [min] Time [min]\n\nFig. 13.6 Fire and steel temperatures calculated numerically according to Eq. 13.4 based on\np\nparametric fire curves with qf 200 kJ=m2 , kc 1160 W s1=2 =K m2 and O 0.04 m\n(diagram a) and O 0.02 m (diagram b), respectively. See Example 13.4. (a) O 0.04 m,\n\n## 1 and td 60 min (b) O 0.02 m, 0.25 and td 120 min\n\n00\n\u0006 \u0007 \u0004 \u0005\nq_ tot st \u0003 T 4f \u0004 T 4s hc T f \u0004 T s 13:16\n\nwhen the radiation and convection temperatures are assumed equal, i.e.\nT r T g T f . This heat flux shall balance with the heat stored in the steel section,\ni.e.\n\n00 T st\nAst \u0003 q_ tot cst \u0003 st \u0003 V st 13:17\nt\n\nand the steel temperature can then be obtained by the numerical time integration\nscheme\n\nAst h \u0006 \u0007 \u0006 \u0007i\nT j1 T j st \u0003 T jf 4 \u0004 T j 4 hc T fj \u0004 T j \u0003t 13:18\ncst \u0003 st \u0003 V st\n\nwhere t is the time increment and the superscript j the time increment number. The\nheat capacity of the steel may be updated at each time step to consider changes\ndependent on temperature.\nFigure 13.7 shows steel temperature developments of steel sections with various\nsection factors assuming constant values of cst, st, st and hc.\nEquation 13.20 is a forward difference scheme which is numerically stable and\naccurate only for limited values of the time increment. The stability criterion for the\nexplicit numerical scheme may be expressed as\n\u000ei\nV st st cst\nti \u0006 i 13:19\nAst htot\n\n## where htot is the total heat transfer coefficient\n\n210 13 Temperature of Steel Structures\n\nFig. 13.7 Temperatures of uninsulated steel sections with various section factors exposed to the\nstandard ISO 834 timetemperature curve calculated according to Eq. 13.18 with\ncst 460 J=kg K, st 7850 kg=m3 , st 0:7 and hc 25 W=m2 K\n\n\u0006 \u0007\u0004 \u0005\nhtot hr hc st \u0003 T 2f T 2st T f T st hc 13:20\n\nAs the total heat transfer coefficient htot will increase substantially with the\ntemperature level, c.f. Eq. 13.20, the time constant and critical time step will\ndecrease accordingly.\nIn practice it is recommended to keep the time increments less than 10 % of the\ncurrent time constant, i.e.\n\u000ei\nV st st cst\nt \u0006 0:1\ni\n13:21\nAst htot\n\nPrinciples for calculating the section factors according to Eurocode 3 for\nvarious types of configurations of unprotected steel members can be found in\nTable 13.8.\nAs well as for protected steel sections the volume Vst may be calculated as the\nweight per unit length mst over the steel density st, see Eq. 13.6. The weight per\nunit length mst of steel sections is often tabulated in catalogues of steel supplier.\n13.4 Unprotected Steel Sections 211\n\n## Table 13.8 Section factor Ast/Vst for unprotected steel members\n\nOpen section exposed to fire on Tube exposed to fire all around:\nall sides: Ast/Vst 1/t\nAst perimeter\n=\nVst cross-section area\nt\n\nSect. 13.4.1)\nOpen section exposed to fire on Hollow section (or welded box section of uniform thickness)\nthree sides: exposed to fire on all sides:\nAst surface exposed to fire If t b: Ast/Vst \b 1/t\nV st cross\u0004section area\n\nt\nh\n\nSect. 13.4.1)\nI-section flange exposed to fire Welded box section exposed to fire on all sides:\non three sides: Ast 2b h\nAst =V st b 2tf =b\u0003tf V st cross\u0004section area\nIf t b: Ast/Vst \b 1/t\nIf t b: Ast/Vst \b 1/tf\n\ntf b\nb\n\nAngle exposed to fire on all I-section with box reinforcement, exposed to fire on all sides:\n2bh\nV st cross\u0004section area\nsides: Ast\nAst/Vst 2/t\n\nt\nb\n\nSect. 13.4.1)\nFlat bar exposed to fire on all Flat bar exposed to fire on three sides:\nsides: Ast =V st b 2t=b\u0003t\nAst =V st 2 b t=b\u0003t If t b: Ast/Vst \b 1/t\nIf t b: Ast/Vst \b 2/t\n\nb\nt\n\nb\nt\n\n## From Eurocode 3 \n\n212 13 Temperature of Steel Structures\n\nare applicable to open\nsections where surfaces are\nsection (b) Closed section\n\nWhen an open section such as an I-section is exposed to fire, the heat transfer by\nradiation will be partly shadowed as indicated by Fig. 13.8a, see Eurocode 3 and\n. The surfaces between the two flanges are then not exposed to incident\nradiation from the surrounding fire from the full half-sphere but only from a limited\nangle, i.e. the incident radiation to these surfaces is reduced. Shadow effects are not\napplicable to closed sections such as tubes as shown in Fig. 13.8b.\nAs a matter of fact a section will only receive as much heat by radiation from the\nfire as if it had the same periphery as a boxed section, see Fig. 13.9a, b. Therefore\nthe area per unit length Ast may be replaced by the so-called boxed area Ab in the\nEq. 13.18. This will reduce the influence by convection heat transfer but as the\nradiation heat transfer mode dominates at elevated temperature this approximation\nmay be accepted although it is non-conservative. The boxed area Ab is typically for\nan I-section 30 % less than the corresponding area Ast. This means that steel\ntemperature will be reduced when considering shadow effects and more open\nsteel sections can be accepted without thermal protection. Shadow effects are\nparticularly important for unprotected steel sections but the concept can be applied\nto other types of structures as well.\nThe surface area of an I-beam attached to a concrete slab or wall may be reduced\nin a similar way as indicated in Fig. 13.10. According to Table 13.8 the surface area\nAst can be calculated as shown in Fig. 13.10a while the reduced area Ab considering\nshadow effects is calculated as shown by the dashed line in Fig. 13.10b.\nExample 13.5 Calculate the section factor without and with considering of shadow\neffects of an unprotected HE300B steel section attached to a concrete structure as\nshown in Fig. 13.10.\nSolution Dimensions of an HE300B section can be found in Table 13.5. Thus Ast\n2H 3W \u0004 2tw 2 \u0003 300 3 \u0003 300 \u0004 2 \u0003 11 1478 mm and Ab 2H W\n2 \u0003 300 300 900 mm. The section weight mst 119 kg per unit length\naccording to Table 13.5. Thus according to Eq. 13.6 V st mst =st 119=7850\n0:0151 m2 and the section factors becomes Ast =V st 98 m\u00041 when not consid-\nering shadow effects and Ab =V st 60 m\u00041 when considering shadow effects, i.e. a\n13.5 Examples of Steel Temperatures Calculated Using a Finite Element Code 213\n\nFig. 13.9 Illustration of the shadow effect of I-section exposed to fire from four sides. (a) Area\nwithout considering shadow effects, Ast (b) The boxed area considering shadow effects, Ab\n\nFig. 13.10 Periphery considering and not considering shadow effects for steel profiles attached to\nconcrete structures. (a) Periphery according to Table 13.8, Ast (b) Periphery considering shadow\neffects, Ab\n\nComment: Still such a section would get a temperature of 500 \u0001 C already after\n15 min according to Fig. 13.7. The results are, however, conservative as the heat\ntransferred from the steel to the concrete is not considered in this type of calcula-\ntions. Temperatures calculated with the finite element code TASEF including\nshadow effects as well as effects of cooling to the concrete structure are shown in\nSect. 13.5.3.\n\n## 13.5 Examples of Steel Temperatures Calculated Using\n\na Finite Element Code\n\nThe steel section temperature analyses above assume uniform steel temperatures or\nlumped heat. This is often a very crude approximation. It leads indeed in general to\nsolutions on the safe side, i.e. the temperatures are overestimated, but often to over-\ndesign and thereby to unnecessary costs. Unsafe conditions may, however, occur in\nsections where parts such as webs are considerably thinner than the flanges.\nFor more precise analyses numerical calculations are needed employing,\ne.g. finite element computer codes. Some examples are shown in the sections\nbelow.\n214 13 Temperature of Steel Structures\n\n## Fig. 13.11 A bare square\n\nsteel tube section carrying a\nconcrete slab or attached to\na concrete wall\n\n## 13.5.1 Unprotected Square Steel Tube Section Attached\n\nto a Concrete Slab or Wall\n\n## An unprotected square steel tube (100 mm by 100 mm and 10 mm thick) is carrying\n\na concrete slab or attached to a concrete wall as shown in Fig. 13.11. It is exposed to\nstandard fire conditions according ISO 834, see Fig. 13.11. Heat transfer conditions\nare assumed according to Eurocode 1 , i.e. 0.8 and h 25 W/(m2 K). The\nthermal properties of steel and concrete are as given in Eurocode 2 and 3, respec-\ntively. Heat transfer inside the void of the tube is assumed to be transferred by\nradiation with an internal surface emissivity of 0.8 and by convection with a heat\ntransfer coefficient of 1 W/(m2 K).\nThe temperature calculation was carried with the finite element computer code\nTASEF . The finite element discretization model including element node\nnumbers is shown in Fig. 13.12a. Calculated steel temperatures vs. time are\nshown in Fig. 13.12b, the bottom flange (node 1) and two of the top flange (nodes\n5 and 35). Notice that the temperature of the bottom flange is considerably higher\nthan that of the top flange. The difference decreases, however, in the end of the\nexposure as the radiation heat transfer between the flanges becomes more efficient\nat higher temperature levels and the concrete slab is heated. The heat transfer in the\nvoid levels out the temperature as heat is transferred between surfaces, it cools the\nexposed flange and heats the flange attached to the concrete. Figure 13.12c shows a\ntemperature contour after 15 min.\n\n## An HE300B steel section attached to a concrete structure, wall or slab, is protected\n\nby gypsum boards as shown in Fig. 13.13. It is exposed from below to standard fire\nconditions according the Hydrocarbon curve, see Eq. 12.3. Heat transfer conditions\nare assumed according to Eurocode 1, i.e. 0.8 and h 50 W/(m2 K). The\nthermal properties of steel and concrete are as given in Eurocode 2 and 3, respec-\ntively. The gypsum boards are 30 mm and have thermal properties according to\nTable 7.2.\nA finite element discretization model was generated as shown in Figure 13.14a.\nHeat transfer inside the void between the steel web and the protection by radiation\nand convection was considered in the analysis.\n13.5 Examples of Steel Temperatures Calculated Using a Finite Element Code 215\n\nFig. 13.12 Unprotected hollow section analysed by the finite element method. B/W plots from\nTASEF. (a) Finite element mesh of a symmetric half (b) Steel node temperatures vs. time (c)\nTemperature contours after 15 min\n\nFig. 13.13 An encased I-section steel (HE 300B) beam carrying a concrete slab. Slab thickness\n160 mm, protection thickness 30 mm, steel height and width 300 mm, flange thickness 19 mm and\nweb thickness 11 mm\n\nThe calculated temperature histories in the steel flanges are shown in Fig-\nure 13.14b. The vaporization of the water in the gypsum consumes a lot of heat\nas indicated by the enthalpy curve shown in Fig. 7.8. Therefore the uneven\ndevelopment of the temperature of the gypsum (curve #2). Notice also that the\ntemperature difference between the minimum and maximum steel temperatures are\nin the order of 200 \u0001 C due to the cooling of the top flange by the concrete slab.\n\n## 13.5.3 Unprotected I-Section Connected to a Concrete\n\nStructure\n\nA bare HE300B steel section attached to a concrete structure as shown in Fig. 13.10\nis exposed to standard fire conditions according ISO 834. Accurately calculated\ntemperatures with the finite element code TASEF are shown in Fig. 13.15. Notice\n216 13 Temperature of Steel Structures\n\na b\n0.49 Temperature Multiple Fires\n1200\nNodes\n1\n0.392 1000\n\n2\n800\n0.294\n\n600\n\n0.196\n400 3\n\n0.098 200\n9\n\n0\n0 0 0.333 0.667 1.000 1.333 1.667 2.000\nTime (h)\n0 0.05 0.1 0.15 0.2 0.25\n\nFig. 13.14 I-beam protected with gypsum boards analysed by the finite element method. B/W\nplots from TASEF. (a) Finite element mesh of a symmetric half (b) Temperatures from above of\ngypsum surface, middle of gypsum, steel bottom and upper flanges\n\nb\nTemperature ISO 834\n1200\n\na 1000\n\n800 Nodes\n61\n7\n600\n69\n9\n400\n\n200\n\n0\n0 0.083 0.167 0.250 0.333 0.417 0.500\nTime (h)\n\nFig. 13.15 Steel temperature development of the bottom and top flanges of assembly exposed to a\nstandard ISO 834 timetemperature curve. Shadow effects and effects of cooling of the steel to the\nconcrete are considered. (a) Points where calculated steel temperatures are shown in b) (b) Finite\nelement calculated temperatures of flanges. B/W plot from TASEF\n\n## that when assuming lumped heat or uniform temperature a temperature of 500 \u0001 C is\n\ncalculated after 15 min. In the finite element analysis this temperature is only\nreached by the bottom flange while the top flange attached to the concrete only\nreaches a temperature of 200 \u0001 C.\nChapter 14\nTemperatures of Concrete Structures\n\nReinforced concrete structures are sensitive to fire exposure of mainly two reasons.\nThey may be subject to explosive spalling, and they may lose their load-bearing\ncapacity due to high temperatures. Spalling is particularly hazardous as it may\noccur more or less abruptly and unanticipated. It usually starts within 30 min of\nsevere fire exposure. It may depend on several mechanisms or combinations thereof\nsuch as pore pressure, stresses due to temperature gradients, differences of thermal\ndilatation and chemical degradations at elevated temperatures. Reinforcement bars\nof steel lose their strength at temperature levels above 400 \u0001 C. Prestressed steel may\neven loose strength below that level. Concrete loose as well both strength and\nstiffness at elevated temperature.\nAs the spalling phenomenon is very complex and cannot be predicted with\nsimple mathematical temperature models, it will not be further discussed here.\nFor more detailed information regarding the fire spalling phenomenon see . The\nprocedures presented below presume that no spalling occur that could significantly\ninfluence the temperature development.\nIn Eurocode 2 temperatures in fire-exposed structures may be obtained from\ntabulated values or by more or less advanced calculations. In the sections below\nthermal material properties as given in Eurocode 2 are reproduced and thereafter\nsome simple approximate calculation methods are given in the following sections.\nFor more general situations finite element calculations are needed.\n\n## The conductivity of concrete decreases with rising temperature. It depends on\n\nconcrete quality and type of ballast. For design purposes curves as shown in\nFig. 14.1 may be used according to Eurocode 2 . For more accurate calculations\nwith alternative concrete qualities more precise material data may be obtained by\nmeasuring the thermal properties of the product in question, see Sect. 1.3.1.\n\n## Springer International Publishing Switzerland 2016 217\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_14\n218 14 Temperatures of Concrete Structures\n\n## Fig. 14.1 Upper and lower 2\n\nlimit of heat conductivity\n1.8\nvs. temperature of normal\n\n## Conduc\u0002vity, kcrt [W/(m K)]\n\nweight concrete according 1.6\nto Eurocode 2 \n1.4\n1.2\n1\n0.8\n0.6\n0.4\n0.2\n0\n0 200 400 600 800 1000 1200\nTemperature [C]\n\n## Fig. 14.2 Specific heat 2500\n\ncapacity of concrete\nvs. temperature at\nSpecic heat, ccrt [Ws/(kg K)]\n\n## 3 different moisture 2000\n\ncontents, 0, 1.5 and 3 % for 3%\nsiliceous concrete according\nto Eurocode 2 1500\n1.5 %\n\n1000\n0%\n\n500\n\n0\n0 200 400 600 800 1000 1200\nTemperature [C]\n\nThe specific heat of dry concrete does not vary much with temperature. How-\never, concrete structures always contain water which evaporates at temperatures\nabove 100 \u0001 C constituting a heat sink (latent heat) as the vaporization process\nconsumes a lot of heat. Thus the specific heat capacity for normal weight concrete\naccording to Eurocode 2 has a peak at temperatures 100 and 200 \u0001 C as shown in\nFig. 14.2.\nThe peak due to the latent heat involves a numerical challenge when calculating\ntemperatures. Especially if the temperature range at which the vaporization of the\nmoisture occurs becomes narrow, the peak becomes increasingly high. Then it can\nbe advantageous to introduce the specific volumetric enthalpy as an input parameter\nas defined in Sect. 7.3.4. This formulation in combination with a forward difference\n14.1 Thermal Properties of Concrete 219\n\nTable 14.1 Thermal properties of normal weight concrete according to Eurocode 2 including\nthe range of the conductivity between the upper and lower limits and the calculated volumetric\nenthalpy\nMoist. cont. 0 % Moist. cont. 1.5 % Moist. cont. 3 %\nT k c e c e c e\n[\u0001 C] [W/(m K)] [kg/m3 [(Ws)/ [(Wh)/ [(Ws)/ [(Wh)/ [(Ws)/ [(Wh)/\n] (kg K)] m3] (kg K)] m3] (kg K)] m3]\n0 1.362.00 2300 900 0 900 0 900 0\n20 1.331.95 2300 900 11,500 900 11,500 900 11,500\n100 1.231.77 2300 900 57,500 900 57,500 900 57,500\n115 1.211.73 2300 915 66,197 1470 71,587 2020 76,858\n200 1.111.55 2254 1000 117,154 1000 137,313 1000 157,220\n400 0.911.19 2185 1100 244,613 1100 264,772 1100 284,678\n1200 0.550.60 2024 1100 739,368 1100 759,527 1100 779,434\n\n300\n\n250\nVolumetric enthalpy [(kWh)/m3\n\n200\n\n150\n1.5 %\n3%\n100 0%\n\n50\n\n0\n0 100 200 300 400 500\nTemperature [C]\n\nFig. 14.3 Volumetric enthalpy of concrete for moisture contents 0, 1.5 and 3.0 % vs. temperature\nbased on density and specific heat capacity according to Eurocode 2 (see Table 14.1)\n\ntime integration scheme is used in the computer code TASEF . Table 14.1 and\nFig. 14.2 show calculated values of the specific volumetric enthalpy vs. temperature\nstarting at 0 \u0001 C based on specific heat and density values given in Eurocode 2 for\nnormal concrete. Notice that no consideration is given to the latent heat of the water\nbefore it vaporizes under 100 \u0001 C. This is generally an acceptable approximation for\nnormal weight concrete but not for many other materials which may contain much\nhigher percentages of moisture.\n220 14 Temperatures of Concrete Structures\n\nThe Eurocode on concrete (EN 1992-1-2) states that the emissivity related to\nconcrete surfaces should be taken as 0.7. The Eurocode 1 on actions (EN 1991-1-2)\ngives the convective heat transfer coefficient when simulating fully developed fires\nto be assumed equal to 25 W/(m2 K). (In general the assumed values of the surface\nemissivity and convective heat transfer coefficient have only marginal influence on\ncalculated temperatures inside concrete structures.)\n\n## 14.2 Penetration Depth in Semi-infinite Structures\n\nConcrete is a material with relatively high density and low conductivity. It therefore\ntakes a long time for heat to penetrate into the structure and raise its temperature, or\nin other words it takes time before a temperature change at one point is noticeable at\nanother point. Thus in many cases a concrete structure may be assumed semi-\ninfinite. In Sect. 3.2.1.1 it is shown that temperature change at the surface will only\nbe noticeable at a depth less than\np\n2:8 \u0003 t 14:1\n\nwhere is the thermal diffusivity and t is time. The value 2.8 represents a\ntemperature rise of 1 %. As an example, the temperature rise can be estimated to\npenetrate only about 0.15 m into a concrete structure after 1 h (assuming a\nconductivity of a 1.7 W/(m K), a density of 2300 kg/m3 and a specific heat capacity\nof 900 J/(kg K)).\nPenetration depth can actually be applied to any material where the properties\nmay be assumed constant. A temperature change at one point of, for example, a\nsteel member will not be noticeable beyond a distance corresponding to the\npenetration depth. In 1 h the penetration depth in steel exceeds 0.60 m, which is\nfour times as deep as in concrete.\n\n## In general numerical procedures such as finite element methods are needed to\n\ncalculate temperature in concrete structures. A 1-D configuration of a concrete is\nshown in Fig. 14.4.\nWickstrom [48, 49] has, however, shown that in 1-D cases may the temperature\ninside concrete structures exposed to standard fire conditions according to ISO\n834 and heat transfer condition according to Eurocode 1 (Eq. 12.11) be obtained\nfrom explicit formula and diagrams. The diagrams as shown in Figs. 14.5 and 14.6\nwere then obtained by comparisons with numerous finite element calculations.\nThey yield concrete temperatures which coincide with the temperatures obtained\nwith the accurate numerical methods within a few per cent in the interesting area of\n14.3 Explicit Formula and Diagrams 221\n\n## Fig. 14.4 Definitions of\n\ntemperature rises of a 1-D\nthick concrete wall exposed\nto fire on one side\n\n## Fig. 14.5 The surface ratio 1.0\n\ns vs. time for a normal\nweight concrete with 0.9\nSurface ra\u0002o, s\n\nthermal properties\naccording to Eurocode 2 0.8\nexposed to standard fire 0.7\nconditions according to\nISO 834 0.6\n\n0.5\n\n0.4\n0.1 1\nTime [h]\n\n0.80\n\n0.70\nDepth ration, x\n\n0.60\n\n0.50\n\n0.40\n\n0.30\n\n0.20\n100 1000 10000\nt/x2 [h/m2]\nFig. 14.6 The in depth ratio x vs. time divided by depth squared t/x2 for normal weight concrete\nwith thermal properties according to Eurocode 2 exposed to standard ISO 834 fire conditions.\nCalculations are made assuming lower limit of the conductivity as shown in Fig. 14.1\n222 14 Temperatures of Concrete Structures\n\n300 to 600 \u0001 C. These diagrams are therefore very handy to use when a quick\nestimate is needed.\nThe diagrams apply to normal weight concrete with thermal properties\naccording to Eurocode 2 as shown in Table 14.1 assuming the lower conduc-\ntivity curve according to Fig. 14.1 and a moisture content of 1.5 %.\nIn it is shown that the same type of diagrams can be used more generally\nconsidering both various parametric fires and various material properties.\nThe diagram given in Fig. 14.5 shows the ratio s between the concrete temper-\nature rise of the surface and the standard fire temperature according to ISO 834 vs.\ntime. This surface ratio is defined as\n\ns\ns 14:2\nf\n\nwhere s and f are the temperature rise of the surface and the fire, respectively.\nFigure 14.6 shows in turn the ratio between the internal temperature rise x at a\ndepth x and the surface temperature rise s. This depth ratio is defined as\n\nx\nx 14:3\ns\n\nThe depth ratio x is in principle a function of the Fourier number, i.e. the\nthermal diffusivity k/(c) of the concrete times the fire duration t over the depth\nx squared. In the finite element calculations for developing the diagrams thermal\nproperties of concrete with a water content of 1.5 % are assumed according to\nEurocode 2. Calculation depths between 25 and 100 mm were used when develop-\ning the diagram. The linear relation in the logarithmic-linear diagram as shown in\nFig. 14.6 was then constructed. It yields approximate temperatures slightly higher\nthan was obtained with the accurate finite element calculations.\nThe internal concrete temperature may now be written as\n\nT x s x T f 14:4\n\nThe graphs in Figs. 14.5 and 14.6 can be approximated by simple expressions.\nThus Eq. 14.5\n\n## s 1 \u0004 0:060 t\u00040:90 14:5\n\nand\n\u0003t\u0004\nx 0:172ln \u0004 0:74 14:6\nx2\n\nrespectively, where t is time in hours and x distance in metres from the surface.\n14.3 Explicit Formula and Diagrams 223\n\nThen in summary for standard fire exposure according to ISO 834 and normal\nweight concrete according to Eurocode 2 (see Sect. 14.1) a very simple closed\nform solution may be obtained. Thus the surface temperature rise is\n\u0005 \u0006\ns 1 \u0004 0:060 t\u00040:90 \u0003 345 \u0003 log480t 1\u0005 \u0001 C\u0005 14:7\n\nThe internal temperatures at arbitrary times and depths are obtained by inserting\nEqs. 14.5 and 14.6 into Eq. 14.4 of a structure initially at 20 \u0001 C then becomes:\n\u0005 \u0006 h \u0003t\u0004 i\nT x, t 1 \u0004 0:060 t\u00040:90 \u0003 0:172ln 2 \u0004 0:74 \u0003 345 \u0003 log480t 1\u0005\nx\n\u0001\n20 C\u0005 14:8\n\nA diagram based on Eqs. 14.7 and 14.8 is shown in Fig. 14.7 including the standard\nISO 834 fire curve. The graphs are limited between 200 and 700 \u0001 C. Outside that\nrange Eq. 14.8 is not valid.\nAs an illustration the temperature in a slab of normal-weight concrete is calcu-\nlated at a depth of 4 cm when exposed to an ISO 834 standard fire for 1 h. At first s\nis obtained from Fig. 14.5 to be 0.97 at t 1 h. Then for t/x2 2.0/(0.04)2 1250 h/\nm2 and Eq. 14.5 or Fig. 14.6 yields approximatively x 0.49. As the standard fire\ntemperature rise after 1 h is 1029 \u0001 C, the concrete surface temperature rise is\nobtained from Eq. 14.8 as 0.97 \u0003 1029 998 \u0001 C and Eq. 14.8 yields the temperature\nrise at a depth of 4 cm to be Tx 0.97 \u0003 0.49 \u0003 1029 + 20 \u0001 C 509 \u0001 C. Alternatively a\ndirect reading of Fig. 14.7 yields a Tx 500 \u0001 C which coincides very well with an\naccurate finite element calculation.\n\nFig. 14.7 Temperature in concrete based on Eq. 14.8 in the range of 200 to 700 \u0001 C at various\ndepths when exposed to the standard ISO 834 fire curve. The temperatures of the exposure curve\nand the surface are given as well\n224 14 Temperatures of Concrete Structures\n\n## Fig. 14.8 Definitions of\n\ntemperature rises at a 2-D\nconcrete corner exposed to\nfire from two sides\n\nAlso the temperature rise near 2-D corners exposed to ISO 834 standard fires\nmay be calculated using the approximations above . Thus the temperature at a\npoint at distances x and y, respectively, from the exposed surfaces (see Fig. 14.8)\nmay be calculated as\n\u0005 \b \u0007 \u0006\nx, y s \u0003 x y \u0004 2 \u0003 x y x y \u0003 f 14:9\n\nwhere s is the surface ratio according to Eq. 14.5 or Fig. 14.5, and x and y are the\ndepth ratios in the x and y directions, respectively, according to Eq. 14.6 or\nFig. 14.6.\nExample 14.1 Calculate the temperature in a rectangular concrete beam after 2.0 h\nfire exposure at a point 60 and 50 mm from the exposed surfaces.\nSolution According to Eq. 14.5 or Fig. 14.5 s 0:97, t/x2 2.0/0.062 556 h/m2\nand then according to Eq. 14.6 or Fig. 14.6 x 0:35, and t/y2 2.0/0.052 800 h/m2\nwhich yields y 0:41. At 2.0 h the temperature rise according to ISO 834 is 1029 \u0001 C,\nand the temperature rise becomes according to Eq. 14.9 x, y\n\u0001 \u0001\n0:97 \u0003 0:35 0:41 \u0004 2 \u0003 0:35 \u0003 0:41 0:35 \u0003 0:41\u0005 \u0003 1029 C 620 C.\n\n## In some application it may be advantageous to insulate concrete structure surfaces\n\nto prevent them from fast temperature rises. It is mainly considered for tunnels to\navoid spalling to give additional protection to the embedded reinforcement bars as\nshown in Fig. 14.9a. Behind the protection the concrete temperature will then not\nrise as quickly as when directly exposed to fire.\nThere are in principle three types of passive fire protections used for protection\nof tunnels, namely spraying with cementitious mortar, lining with non-combustible\n14.4 Fire Protected Concrete Structures 225\n\nFig. 14.9 The protection of a concrete structure layer with a thickness din gives an equivalent\nthermal protection as a concrete layer with a thickness dc kcdi/ki. (a) Concrete slab fire protected\nfrom below (b) Concrete layer providing equivalent thermal protection\n\nmation on fire dynamics in tunnels see and on concrete in tunnels see . A\nsimple way of estimating how much thermal protection an insulation provides in\nterms of concrete thickness based on finite element calculations has been\nsuggested by Wickstrom and Hadziselimovic .\nThey showed that the same effect is approximately obtained when the thermal\nresistance of the insulation is the same as that for the concrete layer. Thus the\nequivalent concrete layer thickness can be calculated as\n\nkc \u0003 d in\ndc 14:10\nkin\n\nwhere d is thickness and k conductivity, and the indices in and c stand for insulation\nand concrete, respectively.\nAs an example a 10 mm board of vermiculite with a thermal conductivity of\n0.2 W/(m K) corresponds to a concrete protection layer of 50 mm assuming the\nconcrete has a conductivity of 1.0 W/m K for the temperature interval considered.\nThis could mean considerable savings in both weight and space for a concrete\nstructure.\nChapter 15\nTemperature of Timber Structures\n\n## Modelling the thermal behaviour of wood is complicated as phenomenas such as\n\nmoisture vaporization and migration, and the formation of char have decisive\ninfluences on the temperature development within timber structures. Nevertheless\nit has been shown that general finite element codes can be used to predict temper-\nature in, for example, fire-exposed cross sections of glued laminated beams ,\nprovided, of course, that apparent thermal material properties and appropriate\nboundary conditions are used. Other specialized numerical codes for timber struc-\ntures have been developed, e.g. by Fung and Gammon . A comprehensive\ncollection of papers on timber in fire is listed in .\n\n## 15.1 Thermal Properties of Wood\n\nBoth density and moisture content affect the thermal properties of wood. In the\nliterature a wide range of values are given. In the SFPE Handbook of Fire Protec-\ntion Engineering , the following equation is given for the conductivity in\nW/(m K) as\n\n## c 103:1 3:867 T 15:2\n\nwhere is the density based on volume at current moisture content and oven-dry\nweight (kg/m3), u the moisture content (per cent by weight) and T is the temperature\n(K). These values are mainly developed for temperatures below 100 \u0004 C. For higher\ntemperatures the latent heat for the vaporization of free water must be considered as\n\n## Springer International Publishing Switzerland 2016 227\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3_15\n228 15 Temperature of Timber Structures\n\n## Fig. 15.1 Temperature 1.6\n\nvs. conductivity for wood\nand the char layer according 1.4\n\n## Conduc\u0002vity, kwood [W/(m K)]\n\nTable 15.1 1.2\n\n0.8\n\n0.6\n\n0.4\n\n0.2\n\n0\n0 200 400 600 800 1000 1200\nTemperature [C]\n\nTable 15.1 Temperature vs. conductivity for wood and the char layer according to Eurocode 5\nTemperature [\u0004 C] 20 200 350 500 800 1200\nConductivity [W/(m K)] 0.12 0.15 0.07 0.09 0.35 1.5\n\nfor concrete, see Sect. 14.1. At equilibrium in normal conditions (20 \u0004 C and a\nrelative air moisture content of 65 %) wood contains about 12 % by weight of water.\nAccording to Eurocode 5 (EN 1965-1-5) Annex B the conductivity is\nrecommended to be as shown in Fig. 15.1 and Table 15.1, and the specific heat\ncapacity as shown in Fig. 15.2. These material properties are limited structures\nexposed to standard fire exposure according to ISO 834 or EN 1363-1, see .\nFigure 15.3 shows the specific volumetric enthalpy based on Table 15.2 for\nwood with a density of 450 kg/m3 and a moisture content of 12 %.\nThe thermal properties of wood are in general very uncertain and it is very hard\nto find reliable data in the literature. For approximate calculations it is here\nrecommended to use a constant conductivity of 0.13 W/(m K) independent of\nmoisture content and a specific heat of 2000 W/(kg K) for dry wood with additions\nfor the sensitive and latent heat of water as described for concrete in Sect. 14.1.\n\n## Simple estimations of load-bearing capacities of timber members are according to\n\nEurocode 5 made in two steps. First a residual cross section is calculated by removing\nthe char layers entering from fire-exposed surfaces. Then the mechanical properties of\nthe residual cross section are calculated based on the remaining virgin wood. This\nprocedure is called Reduced Cross-Section Method or Effective Cross-Section Method.\n15.2 Charring Depth According to Eurocode 5 229\n\n14000\n\n12000\nSpecific heat, cwood[Ws/(kg K)]\n\n10000\n\n8000\n\n6000\n\n4000\n\n2000\n\n0\n0 200 400 600 800 1000 1200\nTemperature [C]\nFig. 15.2 Temperature vs. specific heat capacity for wood and charcoal according to Eurocode\n5. The peak at 100 \u0004 C corresponds to the heat of vaporization of 12 % by weight of water\n\nFig. 15.3 Example of specific volumetric enthalpy in MWs/m3 for wood with a density 450 kg/m3\nand a moisture content by weight of 12 %\n230 15 Temperature of Timber Structures\n\nTable 15.2 Specific heat capacity and ratio of density to dry density of softwood according to\nEurocode 5\nTemperature [\u0004 C] Specific heat capacity [J/(kg K)] Density ratio\n20 1530 1 + u/100\n99 1770 1 + u/100\n99 13,600 1 + u/100\n120 13,500 1.00\n120 2120 1.00\n200 2000 1.00\n250 1620 0.93\n300 710 0.76\n350 850 0.52\n400 1000 0.38\n600 1400 0.28\n800 1650 0.26\n1200 1650 0\n\n## Fig. 15.4 One-dimensional\n\ncharring (fire exposure on\none side) \n\nEmpirical rules are used to estimate the penetration of the charring layer and the\nloss of strength of timber structures. The following section is a considerably\nabbreviated extract. It is just given as an illustration and should not be used without\nconsulting the relevant standard.\nThe temperature at which charring begins is by the standard definition 300 \u0004 C\nwhen exposed to the ISO/EN standard exposure. One-dimensional charring as\nindicated in Fig. 15.4 is assumed to occur at constant rate when exposed to\nISO/EN standard fires. Then the charring depth can be calculated as\n\ndchar, 0 0 t 15:3\n\nwhere dchar,0 is the design charring depth for one-dimensional charring, 0 the basic\ndesign charring rate for one-dimensional charring and t the relevant time of fire\nexposure.\nWhen including the effects of corner roundings, fissures or gaps between\nadjacent elements, a notional charring depth is assumed as shown in Fig. 15.5:\n\ndchar, n n t 15:4\n\n## where dchar,n is the notional design charring depth.\n\n15.2 Charring Depth According to Eurocode 5 231\n\n## Fig. 15.5 Charring depth\n\ndchar,0 for one-dimensional\ncharring and notional\ncharring depth dchar,n \n\nTable 15.3 Design charring rates 0 and n of timber, LVL, wood panelling and wood-based\npanels \nn\n0 [mm/min] [mm/min]\n(a) Softwood and beech\nGlued laminated timber with a characteristic density of \u0005 290 kg/m3 0.65 0.7\nSolid timber with a characteristic density of \u0005 290 kg/m3 0.65 0.8\n(b) Hardwood\nSolid or glued laminated hardwood with a characteristic density 0.65 0.7\nof \u0005 290 kg/m3\nSolid or glued laminated hardwood with a characteristic density 0.50 0.55\nof \u0005 450 kg/m3\n(c) LVL (Laminated Veneer Lumber)\nWith a characteristic density of \u0005 480 kg/m3 0.65 0.7\n(d) Panels\nWood panelling 0.9a\nPlywood 1.0a\nWood-based panels other than plywood 0.9a\na\nThe values apply to a characteristic density of 450 kg/m3 and a panel thickness of 20 mm or more\n\nFor initially unprotected surfaces of timber design charring rates 0 and n are\ngiven in Table 15.3.\nMore details on how to estimate charring depths are given in Eurocode 5 .\nTimber members may be protected by fire claddings or other protection mate-\nrials to delay the start of charring. Rules on how to calculate the start of charring of\nprotected timber are given in Eurocode 5 .\n232 15 Temperature of Timber Structures\n\nFig. 15.6 Definition of residual cross section and effective cross section \n\n## Table 15.4 Determination of Time k0\n\nk0 in Eq. 15.5 for unprotected\nt < 20 min t/20\nsurfaces with t in minutes.\nFrom Eurocode 5 t \u0005 20 min 1.0\n\n## When determining the cross-sectional mechanical properties, an effective cross\n\nsection should be calculated by reducing the initial cross section by the effective\ncharring depth def (see Fig. 15.6). Then\n\nd ef dchar, n k0 d0 15:5\n\n## where d0 7 mm is the zero-strength layer. dchar,n is determined according to\n\nEq. 15.4. For unprotected surfaces, k0 should be determined according to\nTable 15.4.\nIt is then only the effective part that shall be accounted for when calculating the\nmechanical properties of a cross section. When using the Reduced\nCross-Section Method it is assumed that the effective cross section has ambient\nmaterial properties. All losses in strength and stiffness are compensated by the zero-\nstrength layer.\nExample 15.1 A glued laminated beam (300 mm by 500 mm high) of pine\n(softwood) is exposed on three sides to a standard EN/ISO curve. Calculate the\neffective cross section after a fire exposure of 60 min.\n15.2 Charring Depth According to Eurocode 5 233\n\nSolution The effective charring depth def can be calculated according to Eq. 15.5\nand dchar,n from Eq. 15.4. Thus def 0:7 \u0002 60 1:0 \u0002 7 mm 49 mm and the\nremaining effective cross section becomes (500-49) mm by (300-2 \u0002 49) mm equal\nto 451 mm by 202 mm.\nEnd-user-friendly information for designers including examples can be found\nin .\nReview Questions\n\nChapter 1\n\n## 1. What is the difference between heat and temperature?\n\n2. Which are the three modes of heat transfer?\n3. What is the driving force of heat transfer?\n4. Write Fouriers law of heat conduction.\n5. Which are the three types of boundary condition, 1, 2 and 3?\n6. Which of the three types of boundary conditions is the most common in FSE?\n7. What is an adiabatic surface?\n8. How is heat transferred from the gas phase to a solid surface?\n9. Write the equation for a convection boundary condition.\n11. Write the expression for the emitted radiation from a surface according to the\nStefanBoltzmann law.\n12. What is net radiation heat flux?\n13. What is incident black body radiation temperature or just the black body\ntemperature Tr?\n14. What is a mixed boundary condition?\n15. How is the fire boundary condition normally written in standards on fire\nresistance of structures?\n16. Write the heat conduction equation in 1-D.\n17. Explain the parameters in the heat conduction equation.\n18. What is thermal diffusivity?\n19. What is specific volumetric enthalpy?\n20. What is thermal inertia and why does it vary so much for various materials?\n21. What happens to steel properties at elevated temperatures?\n22. Which is the main problem with concrete structures exposed to severe fire\nconditions?\n23. Why can wooden structures resist fires relatively well?\n\n## Springer International Publishing Switzerland 2016 235\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3\n236 Review Questions\n\nChapter 2\n\n## 24. Draw the temperature distribution of a wall under steady-state conditions.\n\nAssume constant thermal parameters, 3rd kind of boundary condition on one\nside and 1st kind on the other side.\n25. What is the total thermal resistance of a wall with a thickness L, a conductivity\nk and heat transfer coefficients h at the bounding surfaces?\n\nChapter 3\n\n## 26. What is the meaning of lumped-heat-capacity or uniform temperature?\n\n27. Write and explain the heat balance equation where lumped-heat-capacity is\nassumed.\n28. Under what conditions can an analytical solution be derived for the uniform\ntemperature of a body exposed to elevated gas temperature?\n29. Give two examples when uniform temperatures can be assumed in fire safety\nengineering.\n30. What is meant by a semi-infinite body?\n31. Show in the diagram how the temperature profile develops in a semi-infinite\nbody experiencing a sudden temperature rise at the surface.\n32. What is penetration depth?\n33. Which is the material parameter group governing the temperature development\nof semi-infinitely thick bodies with a prescribed surface temperature?\n34. Which is the material parameter group governing the surface temperature\ndevelopment of semi-infinitely thick bodies with a prescribed heat flux at the\nsurface?\n35. Which is the parameter group governing the surface temperature development\nof semi-infinitely thick bodies exposed to a prescribed gas temperature and heat\ntransfer coefficient?\n\nChapter 4\n\n## 36. Why is radiation so important in fire safety engineering?\n\n37. What is the radiation heat transfer coefficient and how can it be calculated?\n38. What is adiabatic surface temperature and how is it defined?\n39. Which parameters are needed to calculate the adiabatic surface temperature?\nReview Questions 237\n\nChapter 5\n\n## 40. What is the resultant emissivity between two parallel plates?\n\n41. What is a view factor?\n42. What is an absorption or emission coefficient?\n43. Which parameters determine the emissivity of a flame?\n\nChapter 6\n\n## 44. What governs heat transfer by convection?\n\n45. Which are the two principal ways of inducing air flow?\n46. Which are the two principal types of flow patterns?\n47. Which air properties govern the magnitude of the convection heat transfer\ncoefficient?\n48. What is effective or apparent thermal conductivity in enclosed species?\n\nChapter 7\n\n49. Write the heat balance equation in numerical form of a body exposed to\nincident radiation and a gas temperature assuming lumped heat?\n50. Write the transient heat balance equation in the matrix form. Describe the\ncomponents.\n51. How can the equation be solved? What are the advantages and disadvantages of\nexplicit and implicit methods?\n52. What is specific volumetric enthalpy?\n53. Specific volumetric enthalpy of a wet material has three components at\ntemperatures above vaporization. Which?\n\nChapter 8\n\n54. According to thermal ignition theory there are formulas to calculate time to\nreach critical temperatures for thin and thick solids, respectively. Which\nparameters are governing in the two cases?\n55. What is the critical incident radiation heat flux?\n238 Review Questions\n\nChapter 9\n\n## 56. Describe the function of a thermocouple.\n\n57. When measuring gas temperature with thermocouples there are two main error\nsources, which?\n58. How should a thermocouple be designed to measure gas temperatures\naccurately?\n59. What does a heat flux meter of Gardon gauge or Schmidt-Boelter gauge\nmeasure?\n60. Why is the PT larger than a thermocouple?\n61. A plate thermometer measures approximatively the adiabatic surface temper-\nature of a relatively large surface. Why not exactly?\n62. How can incident thermal radiation be calculated based on plate thermometer\nmeasurements?\n63. How can adiabatic surface temperatures be calculated based on plate thermom-\neter measurements?\n\nChapter 10\n\n64. Which are the four main components of the heat balance equation of a fully\ndeveloped compartment fire?\n65. Which is the driving force of the gas flow in a one-zone model?\n66. What is the difference between ultimate fire temperature and maximum\ntemperature?\n\nChapter 11\n\n67. What is the difference between a one-zone and a two-zone model? When are\nthey applicable?\n68. Which is the driving force of the gas flow in a two-zone model?\n\nChapter 12\n\n69. Which three major steps consist of a fire design or analysis process of?\n70. What is the meaning of the gamma factor?\n71. How is the gamma factor influenced by the opening factor and the thermal\ninertia of the compartment boundaries?\n72. What determines the duration of a parametric fire?\nReview Questions 239\n\nChapter 13\n\n## 73. What characterizes the thermal properties of steel?\n\n74. What is lumped-heat-capacity?\n75. What is section factor or shape factor of a steel section?\n76. What is meant by heavily protected steel structures?\n\nChapter 14\n\n78. What characterizes the thermal properties of concrete in comparison with steel\nand insulation materials?\n79. What is the surface temperature of a thick concrete wall after 1 h fire exposure\naccording to the ISO 834 standard curve. Use the diagram in Fig. 14.5.\n80. What is the temperature 3 cm into a thick concrete structure after one hour fire\nexposure according to the ISO 834 standard curve? Use the diagrams in\nFigs. 14.5 and 14.6\n81. Calculate the same temperatures as in the two questions above but use Eq. 14.7\nand Eq. 14.8, respectively?\n82. How can the insulation of a concrete structure be considered in terms of\nequivalent concrete thickness?\n\nChapter 15\n\n## 83. What needs to be considered particularly when calculating temperature in\n\ntimber structures?\n84. How is the thermal conductivity of wood in comparison to steel and concrete?\nmembers according to for example Eurocode 5?\nReferences\n\n1. Holman J (1986) Heat transfer, 6th edn. McGraw-Hill Book Company, New York\n2. Incropera FP, deWitt DP (1996) Fundamentals of heat and mass transfer, 4th edn. Wiley,\nNew York\n3. The European Committee for Standardisation, CEN (2005) EN 1993-1-2, Eurocode 3: design\nof steel structuresgeneral rulesstructural fire design. The European Committee for\nStandardisation, CEN, Brussels\n4. DiNenno PJ et al (eds) (2008) SFPE handbook of fire protection engineering, 4th edn. National\nFire Protection Association, Quincy\n5. Hurley M et al (eds) (2015) SFPE handbook of fire protection engineering, 5th edn. SFPE,\nGaithersburg\n6. The European Committee for Standardisation, CEN (2004) EN 1992-1-2, Eurocode 2: design\nof concrete structuresgeneral rulesstructural fire design. The European Committee for\nStandardisation, CEN, Brussels\n7. The European Committee for Standardisation, CEN (2004) EN 1995-1-2, Eurocode 5: design\nof timber structuresgeneral rulesstructural fire design. The European Committee for\nStandardisation CEN, Brussels\n8. Flynn D (1999) Response of high performance concrete to fire conditions: review of thermal\nproperty data and measurement techniques, NIST GCR 99-767. National Institute of Standards\nand Technology, Gaithersburg\n9. Adl-Zarrabi B, Bostr om L, Wickstrom U (2006) Using the TPS method for determining the\nthermal properties of concrete and wood at elevated temperature. Fire Mater 30:359369\n10. McGrattan KB, Hostikka S, Floyd JE, Baum HR, Rehm RG (2015) Fire dynamics simulator,\ntechnical reference guide, vol 1: mathematical model. NIST Special Publication 1018-1.\nNational Institute of Standards and Technology, Gaithersburg\n11. Siegel R, Howell J (1972) Thermal radiation heat transfer. McGraw-Hill Book Company,\nNew York\n12. Forsth M, Roos A (2010) Absorptivity and its dependence on heat source temperature and\ndegree of thermal breakdown. Fire Mater 35(5):285301\n13. Drysdale D (1992) An introduction to fire dynamics. Wiley-Interscience, New York\n14. Wickstrom U (1979) TASEF-2a computer program for temperature analysis of structures\nexposed to fire. Doctoral thesis, Lund Institute of Technology, Department of Structural\nMechanics, Report NO. 79-2, Lund\n15. Pitts DR, Sissom LE (1977) Schaums outline of theory and problems of heat transfer.\nSchaums outline series, ISBN 0-07-050203-X, McGraw-Hill book company, New York\n\n## Springer International Publishing Switzerland 2016 241\n\nU. Wickstrom, Temperature Calculation in Fire Safety Engineering,\nDOI 10.1007/978-3-319-30172-3\n242 References\n\n## 16. Jiji LM (2009) Heat conduction, 3rd edn. Springer, Berlin\n\n17. Sjostrom J, Wickstr om U (2014) Superposition with Non-linear Boundary Conditions in Fire\nSciences, Fire Technology. Springer, New York\n18. SFPE Standard on Calculation Methods to Predict the Thermal Performance of Structural and\nFire Resistive Assemblies (draft)\n19. Sterner E, Wickstr om U (1990) TASEFtemperature analysis of structures exposed to fire. SP\nReport 1990:05, SP Technical Research Institute of Sweden, Boras\n20. Fransen JM, Kodur VKR, Mason J (2000) Users manual of SAFIR 2001. A computer program\nfor analysis of structures submitted to fire. University of Liege, Belgium\n21. SFPE (2015) SFPE engineering standard on calculation methods to predict the thermal\nperformance of structural and fire resistive assemblies (SFPE S.02 2015). SFPE, Gaithersburg\n22. Thomas GC (1996) Fire resistance of light timer framed walls and floors. University of\nCanterbury, Canterbury\n23. Babrauskas V (2003) Ignition handbook ISBN 0-9728111-3-3, Lib. of Congr. #2003090333,\nFire Science Publishers, WA, USA\n24. Quintiere JG (1998) Principals of fire behavior. Delmar, Albany\n25. Wickstrom U (2015) New formula for calculating time to ignition of semi-infinite solids, fire\nand materials. Published online Library (wileyonlinelibrary.com). doi: 10.1002/fam2303\n26. Burley NA et al (1978) The Nicrosil versus Nisil thermocouple: properties and thermoelectric\nreference data. National Bureau of Standards, NBS MN-161\n27. Lattimer B (2008) Heat fluxes from fires to surfaces, 4th edn, SFPE handbook of fire protection\nengineering. National Fire Protection Association, Quincy\n28. Wickstrom U (1994) The plate thermometera simple instrument for reaching harmonized\nfire resistance tests. Fire Technology, Second Quarter, pp 195208\n29. Wickstrom U, Duthinh D, McGrattan K (2007) Adiabatic surface temperature for calculating\nheat transfer to fires exposed structures, Interflam 2007, London, Sep 3-5, 2007, pp 943\n30. Wickstrom U, Jansson R, Tuovinen H (2009) Experiments and theory on heat transfer and\ntemperature analysis of fire exposed steel beams. SP Report 2009:19, ISBN 978-91-86319-03-8\n31. Wickstrom U, Robbins A, Baker G (2011) The use of adiabatic surface temperature to design\nstructures for fire exposure. J Struct Fire Eng 2(1):2128\n32. Ingason H, Wickstr om U (2007) Measuring incident heat flux using the plate thermometer.\nFire saf J 42:161166\n33. Haggkvist A, Sjostr om J, Wickstrom U (2013) Using plate thermometer measurements for\ncalculating incident heat radiation. J Fire Sci 31:166\n34. Sjostrom J et al (2015) Thermal exposure from large scale ethanol fuel pool fires. Fire Saf J 78\n(2015):229237. doi:10.1016/j.firesaf.2015.09.003\n35. The European Committee for Standardisation, CEN (2005) EN 1991-1-2, Eurocode 1: design\nof steel structuresgeneral rulesstructural fire design. The European Committee for\nStandardisation, CEN, Brussels\n36. Magnusson SE, Thelandersson S (1970) Temperature-time curves for the complete process of\nfire development- a theoretical study of wood fuels in enclosed spaces. Acta Politechnica\nScandinavica, Ci 65, Stockholm\n37. Wickstrom U (1985) Application of the standard fire curve for expressing natural fires for\ndesign purposes. In: Harmathy TZ (ed) Fire safety: science and engineering, ASTM STP 882.\nAmerican Society of Testing and Materials, Philadelphia, pp 145159\n38. Babrauskas V, Williamson RB (1978) Post-flashover compartment fires: basis of a theoretical\nmodel. Fire Mater 2:3953\n39. Zukoski EE, Kubota T, Cetegen B (1980) Entrainment in fire plumes. Fire Saf J 3:107121\n40. Karlsson B, Quintiere JG (2000) Enclosure fire dynamics. CRC Press, Boca Raton\n41. Evegren F, Wickstr om U (2015) New approach to estimate temperatures in pre-flashover fires:\nlumped heat case. Fire Saf J 72(2015):7786\n42. Wickstrom U (1985) Temperature analysis of heavily-insulated steel structures exposed to fire.\nFire Saf J 5:281285\nReferences 243\n\n43. Melinek SJ, Thomas PH (1987) Heat flow to insulated steel. Fire Saf J 12:18\n44. Wang ZH, Kang HT (2006) Sensitivity study of time delay coefficient of heat transfer\nformulations for insulated steel members exposed to fires. Fire Saf J 41:3138\n45. Wickstrom U (1982) Fire Saf J 4:219, 1981\n46. Wickstrom U (2005) Comments on the calculation of temperature in fire-exposed bare steel\nstructures in prEN 1993-1-2: Eurocode 3-design of steel structures- Part 1-2: general rules-\nstructural fire design. Fire Saf J 40:191192\n47. Jansson R (2013) Fire spalling of concretea historical overview. Key note at the 3rd\nInternational RILEM Workshop on Concrete Spalling due to Fire Exposure, Paris\n48. Wickstrom U (1986) A very simple method for estimating temperature in fire exposed concrete\nstructures. In: Grayson SJ, Smith DA (eds.), Proceedings of new technology to reduce fire\nlosses and costs. Elsevier, New York\n49. Wickstrom U (1985) Application of the standard fire curve for expressing natural fires for\ndesign purposes. In: Harmathy TZ (ed) Fire safety: science and engineering, ASTM STP\n882 (pp 145159). American Society of Testing and Materials, Philadelphia\n50. Ingason H, Li YZ, L onnermark A (2015) Tunnel fire dynamics. Springer, New York\n51. Schneider U, Horvath J (2006) Brandschutz-Praxis in Tunnelbauten. Bauwerk Verlag GmbH,\nBerlin\n52. Wickstrom U, Hadziselimovic E (1996) Equivalent concrete layer thickness of a fire protection\ninsulation layer. Fire Saf J 26:295302\n53. Badders BL, Mehaffey JR, Richardson LR (2006) Using commercial FEA-software packages\nto model the fire performance of exposed glulam beams. In: Fourth International Workshop\nStructures in Fire, Aveiro\n54. Fung FCW (1977) A computer program for the thermal analysis of the fire endurance of\nconstruction walls, NBSIR 77.1260. National Bureau of Standards, Washington, DC\n55. Gammon BW (1987) Reliability analysis of wood-frame wall assemblies exposed to fire. Ph.D.\nDissertation, University of California, Berkeley\n56. Bisby LA, Frangi A (2015) Special issue on timber in fire. Fire Technol 51(6):12751277\n57. Konig J (2005) Structural fire design according to Eurocode 5design rules and their\nbackground. Fire and materials 29(3):147163\n58. Ostman B et al (2010) Fire safety in timber buildings. SP Report 19, SP National Research\nInstitute of Sweden, Boras"
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"# Irodov – Problems in General Physics\n\n(Joyce) #1\n\n4.21. A pendulum clock is mounted in an elevator car which starts\ngoing up with a constant acceleration w, with w < g. At a height h\nthe acceleration of the car reverses, its magnitude remaining constant.\nHow soon after the start of the motion will the clock show the right\ntime again?\n4.22. Calculate the period of small oscillations of a hydrometer\n(Fig. 4.2) which was slightly pushed down in the vertical direction.\nThe mass of the hydrometer is m = 50 g, the radius of its tube is\nr = 3.2 mm, the density of the liquid is p = 1.00 g/cm 3. The resis-\ntance of the liquid is assumed to be negligible.\n4.23. A non-deformed spring whose ends are fixed has a stiffness\nx = 13 N/m. A small body of mass m = 25 g is attached at the point\nremoved from one of the ends by 11 = 1/3 of the spring's length. Neg-\nlecting the mass of the spring, find the period of small longitudinal\noscillations of the body. The force of gravity is assumed to be absent.\n\n``s 2``\n\n``iiiiiii iii 77//z/z``\n\n``Fig. 4.3.``\n\n4.24. Determine the period of small longitudinal oscillations of\na body with mass m in the system shown in Fig. 4.3. The stiffness\nvalues of the springs are xi and x 2. The friction and the masses of\nthe springs are negligible.\n4.25. Find the period of small vertical oscillations of a body with\nmass m in the system illustrated in Fig. 4.4. The stiffness values of\nthe springs are xi and x 2 , their masses are negligible.\n4.26. A small body of mass in is fixed to the middle of a stretched\nstring of length 2/. In the equilibrium position the string tension is\nequal to To. Find the angular frequency of small oscillations of the\nbody in the transverse direction. The mass of the string is negligible,\nthe gravitational field is absent.\n\n``````3 e2\n177\nFig. 4.4. Fig. 4.5.``````\n\n4.27. Determine the period of oscillations of mercury of mass\n= 200 g poured into a bent tube (Fig. 4.5) whose right arm forms\nan angle 0 = 30° with the vertical. The cross-sectional area of the\ntube is S = 0.50 cm 2. The viscosity of mercury is to be neglected."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8671197,"math_prob":0.97999835,"size":2012,"snap":"2021-43-2021-49","text_gpt3_token_len":551,"char_repetition_ratio":0.14243028,"word_repetition_ratio":0.08900524,"special_character_ratio":0.2718688,"punctuation_ratio":0.13636364,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99528295,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T11:49:59Z\",\"WARC-Record-ID\":\"<urn:uuid:75c684ee-ea4a-41ff-8675-f4752452fd91>\",\"Content-Length\":\"12381\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e98dd618-b5fe-476e-a9c8-6d4164213f50>\",\"WARC-Concurrent-To\":\"<urn:uuid:c39f8e6a-5107-4433-acf9-1119a0793fe0>\",\"WARC-IP-Address\":\"45.33.88.156\",\"WARC-Target-URI\":\"https://issuhub.com/view/index/4039?pageIndex=176\",\"WARC-Payload-Digest\":\"sha1:5HNHXTPRT4CNM257KFSIYHCAFRXAKIH2\",\"WARC-Block-Digest\":\"sha1:3OIQGZ3X5A56A2H46QRYE5R2KQMYE7FE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363292.82_warc_CC-MAIN-20211206103243-20211206133243-00221.warc.gz\"}"}
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http://kramelospanama.tk/common-derivatives-and-integrals-formula-sheet.html
|
[
"# Common derivatives and integrals formula sheet\n\nFormula common\n\n## Common derivatives and integrals formula sheet\n\nHarold’ s Calculus Notes Cheat Sheet 17 November. Common derivatives and integrals formula sheet. The standard formulas for integration by parts are, bbb aaa. Common derivatives and integrals formula sheet. View Test Prep - Math 170 Common Formula Sheet from MATH 275 at College of Western Idaho. Graphing with Derivatives. Math Cheat Sheet for Derivatives. It sheet has two integrals sheet main branches.\n\nSpecifically 1, if a function is sheet known at only a few discrete values, 2 . Calculus and Analysis W. Common Derivatives and Integrals. Internet Looks On Numbers: numerals from prehistoric ideograms to Egyptian , number systems Sumerian numbers to modern usage. The column is the constant 6.\n\nedu) Limits Cheat Sheet by Paul Dawkins [ pdf, pdf reduced] ( tutorial. Memorized See Larson’ s 1- pager of common integrals 2. The integrals Most Important derivatives Derivatives and sheet Antiderivatives to Know. Finite difference formulas can derivatives be very integrals useful integrals for extrapolating a finite amount of data in an attempt to find the general term. Here is list of cheat sheets and tables that I' ve written. Common Derivatives sheet IntegralsInverse Trig Functionsó 1 æu ö Trig Substitutions du = sin - 1 ç ÷ + c ò sin - 1ô common u du = u sin - 1 u + 1 - u 2 + c If the integral contains formula the following derivatives root integrals use the given substitution , Integrals Common Derivatives formula. and it is desired to common determine the analytical form of, the following procedure can be used if is assumed to be a polynomial function. Then change the sum to an integral the equations become. The table below shows you how to differentiate and common integrate 18 of the most common functions. The Fourier transform integrals is a generalization of the complex Fourier series in the limit as. Title: Microsoft PowerPoint - formula_ sheet_ calculus Author: sheet chris. If integrals integrals the integral contains the following root use the given substitution and derivatives formula. An Engineers Quick Calculus formula Derivatives and Limits Reference.\n\nCommon derivatives Integrals Trig Integrals Calculus: Integral Formulas. Replace the discrete with the continuous while letting. All of the cheat sheets derivatives come formula in two integrals version. Topics include Basic and Integration Formulas Integral of special functions Integral integrals by common Partial Fractions Integration by Parts Other Special Integrals Area as a sum Properties of definite integration formula Integration of Trigonometric Functions, Properties of Definite Integration are all mentioned here. It is the policy of Cecil formula College not to discriminate against any individual by reason of race common sex, treatment of students, activities, citizenship, ethnic origin, color, educational programs common common , sheet religion, scholarship , age, marital status, disability derivatives ( which can be reasonably accommodated without undue hardship) in the admission , sexual orientation , national loan programs. Check the formula sheet of integration.\n\nIntegration Formulas Z dx = x+ C ( 1) Z xn derivatives dx = xn+ 1 n+ 1 + C ( 2) Z dx x = ln| x| + C ( 3) Z ex dx = ex + derivatives C ( 4) Z ax dx = 1 lna ax + C ( formula 5) Z lnxdx = xlnx− x+ C ( 6) Z sinxdx. To create your new password, just click the link in the email we sent derivatives you. Math 170 Common Formula Sheet Trigonometric Derivatives Inverse Trigonometric Derivatives d sheet sin x= cos Calculus Integrals; Math Calculators. Fourier Transform. Common Derivatives Polynomials ( derivatives common ) 0 d formula c dx = integrals ( ) 1 d x. A full sized version and a \" and reduced\" version. These are some examples of common derivatives that require. Introduction to calculus formula sheet Calculus sheet is a branch in mathematics which deals with the study of limits functions, derivatives, , integrals infinite series.\nõ a - u2 2 èaø aó 1 1 æuö 1 a 2 - b2 x2 common Þ x = sin q and cos2 q = 1 - sin. As you can see integration reverses integrals differentiation, returning the function to its original state up to a constant C. with lots common of examples and common. Complete Calculus Cheat Sheet by Paul Dawkins [ pdf, pdf reduced] ( tutorial. Common_ Derivatives_ Integrals.\n\n## Derivatives integrals\n\nDelegation strategies for the NCLEX, Prioritization for the NCLEX, Infection Control for the NCLEX, FREE resources for the NCLEX, FREE NCLEX Quizzes for the NCLEX, FREE NCLEX exams for the NCLEX, Failed the NCLEX - Help is here. Recall the definitions of the trigonometric functions. The following indefinite integrals involve all of these well- known trigonometric functions. Some of the following trigonometry identities may be needed.\n\n``common derivatives and integrals formula sheet``\n\nCore 40 Pre- Calculus Reference Sheet. is the common difference, r. is the sum of the first."
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8325645,"math_prob":0.97789884,"size":7003,"snap":"2019-43-2019-47","text_gpt3_token_len":1561,"char_repetition_ratio":0.20531504,"word_repetition_ratio":0.6103896,"special_character_ratio":0.20991004,"punctuation_ratio":0.1107438,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.9985963,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T03:21:18Z\",\"WARC-Record-ID\":\"<urn:uuid:95dd2877-dfbb-4ab8-bb18-3ce9cc537373>\",\"Content-Length\":\"9977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fc3f49bf-cecb-4ac3-8537-6da82043770e>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3fdf81b-293d-4d75-9cc0-9850858af2ee>\",\"WARC-IP-Address\":\"104.31.91.152\",\"WARC-Target-URI\":\"http://kramelospanama.tk/common-derivatives-and-integrals-formula-sheet.html\",\"WARC-Payload-Digest\":\"sha1:EKLFGKFQZ3Z7UPSGVIQFFUNXE4XDYQ2Z\",\"WARC-Block-Digest\":\"sha1:DZZT3KIXRFCEZV3LMK7BEKIXKTDAX5ZV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669967.80_warc_CC-MAIN-20191119015704-20191119043704-00511.warc.gz\"}"}
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https://dba.stackexchange.com/questions/134868/how-to-count-multiple-values-of-a-column-as-one-group
|
[
"# How to count multiple values of a column as one group? [duplicate]\n\nI have a table `main` with the following rows:\n\n`````` id | name | rank\n----+--------+------\n1 | Ali | a\n2 | Sami | b\n3 | Khan | c\n4 | Kamran | d\n5 | Imran | e\n7 | Nawid | v\n8 | Jamil | c\n9 | Usman | j\n``````\n\nI want to count rows with certain values in column `rank`. For example, I want to group values as follows:\n\n• Values `(a,b,v)` should come in one group by name `myvalues`.\n• Values `(c,d,j)` should come in another group by name `yourvalues`.\n• Value `(e)` should come to another group by name `extravalues`.\n\nMy desired result:\n\n``````myvalues | yourvalues | extravalues\n4 | 4 | 1\n``````\n\n`myvalues` counts 4 because it consists of (a,b,v). There are 2 occurrences of a, 1 occurrence of b and 1 of v - a total of 4.\nThe same is the case with `yourvalues` which consists of (c,d,j) and the occurrences of these values makes a total of 4.\nAnd the last group `extravalues` consists of (e) and counts only 1 row.\n\nUse the aggregate `FILTER` clause in Postgres 9.4+:\n``````SELECT count(*) FILTER (WHERE rank = ANY ('{a,b,v}')) AS myvalues"
] |
[
null
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https://accessibleinvestor.com/definitions/dividend-yield/
|
[
"A dividend yield is a ratio of the dividend a company pays to shareholders divided by the stock price of the company. It is the annual yield you will receive in dividend payments on your investment if you buy the stock at the current price.\n\nTo calculate the dividend yield divide the total annual dividends by the current share price. For example, a stock with a quarterly dividend of \\$0.50 and a stock price of \\$50 would have a dividend yield of 4%.",
null,
""
] |
[
null,
"https://assets.pinterest.com/images/pidgets/pinit_fg_en_rect_red_28.png",
null
] |
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https://stacks.math.columbia.edu/tag/02NU
|
[
"Proof. This is implied by Algebra, Lemma 10.121.4 and Lemma 29.19.9. Alternatively, all points in fibres are closed points by Lemma 29.42.8 (and the fact that a finite morphism is integral) and use Lemma 29.19.6 (3) to see that $f$ is quasi-finite at $x$ for all $x \\in X$. $\\square$\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)."
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.81961,"math_prob":0.9613949,"size":702,"snap":"2020-24-2020-29","text_gpt3_token_len":184,"char_repetition_ratio":0.106017195,"word_repetition_ratio":0.0,"special_character_ratio":0.27920228,"punctuation_ratio":0.14285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9846457,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-01T00:10:52Z\",\"WARC-Record-ID\":\"<urn:uuid:c9737f4d-2223-4750-9386-d88b2d11334e>\",\"Content-Length\":\"14160\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ffeec09-d5e1-464b-bd81-e42ab404b187>\",\"WARC-Concurrent-To\":\"<urn:uuid:e765ee06-f10b-4ec1-953b-1912bc8ea1a4>\",\"WARC-IP-Address\":\"128.59.222.85\",\"WARC-Target-URI\":\"https://stacks.math.columbia.edu/tag/02NU\",\"WARC-Payload-Digest\":\"sha1:LE7XV3ONHJDQRLHT3K5BW4GNJMC4E3SV\",\"WARC-Block-Digest\":\"sha1:SOCGDBLCJAQC4DNS57EDIMLWEW37QLK3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347413786.46_warc_CC-MAIN-20200531213917-20200601003917-00176.warc.gz\"}"}
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https://math.okstate.edu/graduate/highlights-of-the-masters-program/pure-mathematics
|
[
"Departmental Requirements for M.S. in Mathematics:\nPure Mathematics\n\nThe Master of Science degree in pure mathematics requires that the student demonstrate knowledge in certain core areas. There are two options. Both options require a student to earn a grade of A or B in 18 hours of core courses.\n\nCore courses:\n\nOption I:\n\n1. Advanced Calculus I and II (MATH 5043 and 5053)\n2. Modern Algebra I and II (MATH 5003 and 5013)\n3. General Topology (MATH 5303)\n4. Complex Variables (MATH 5273*)\n\nOption II:\n\n1. Advanced Calculus I and II (MATH 5043 and 5053)\n2. Modern Algebra I and II (MATH 5003 and 5013)\n3. Six hours from the following list: Real Analysis I and II (MATH 5143 and 5153), Complex Analysis I and II (MATH 5283 and 5293), Geometric Topology (MATH 5313) and Algebraic Topology I (MATH 6323), Algebra I and II (MATH 5613 and 5623).\n\nElective courses: Students working towards the M.S. degree in pure mathematics should complete 12 hours of course work selected from the following list:\n\nGeometry and Algorithms in Three-Dimensional Modeling (MATH 5423*), Combinatorics (MATH 5673*), Number Theory (MATH 5713*), Introduction to Cryptography (MATH 5753*), Groups and Representations (MATH 5803*), Advanced Linear Algebra (MATH 5023), Fourier Analysis and Wavelets (MATH 5213), Partial Differential Equations (MATH 5233), Ordinary Differential Equations (MATH 5243), General Topology (MATH 5303), Geometric Topology (MATH 5313), Numerical Analysis for Differential Equations (MATH 5543), Numerical Analysis for Linear Algebra (MATH 5553), Algebra I (MATH 5613), Algebra II (MATH 5623), Seminar and Practicum in the Teaching of College Mathematics (MATH 5903), Algebraic Topology I (MATH 6323).\n\nCourse numbers labeled * are cross-listed with an undergraduate section.\n\nAlternative course selections:\n\n· Courses taken as an undergraduate can be used to satisfy requirements for core and elective courses as long as they are consistent with the Graduate College requirements.\n\n· Substitutions for any of the 12 hours of electives requires consent from the Graduate Committee. In no case may more than nine hours outside the mathematical sciences (mathematics, statistics, or computer science) be counted toward the M.S. degree.\n\nCourses taken in graduate school: The courses taken in graduate school must total 33 hours. The courses taken on the M.S. degree program must include 21 hours of non-cross-listed courses in the mathematical sciences. All the courses for the M.S. degree program must constitute a coherent whole and must be approved by the student's advisory committee.\n\nReport or thesis: Each student must complete either a report or a thesis. Students electing to write a report must complete three hours of MATH 5000. Those electing to write a thesis must complete six hours of MATH 5000; three of those hours may be counted toward the 12 hours of electives. Under both of these options, a written document and a public presentation based on this individually directed project is required.\n\nOther requirements: The university catalog contains detailed procedures applicable to all Master's degrees."
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.85735214,"math_prob":0.7463845,"size":3094,"snap":"2021-21-2021-25","text_gpt3_token_len":739,"char_repetition_ratio":0.14822006,"word_repetition_ratio":0.077083334,"special_character_ratio":0.25339368,"punctuation_ratio":0.112132356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9949707,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-06T01:06:04Z\",\"WARC-Record-ID\":\"<urn:uuid:a1147dc2-410e-4302-839f-f36d488b1e6f>\",\"Content-Length\":\"32295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:24b51ad3-3b34-445a-9943-db3c5fe364ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf1a04d8-3014-4057-9c3d-a0ae4be3da5b>\",\"WARC-IP-Address\":\"139.78.82.7\",\"WARC-Target-URI\":\"https://math.okstate.edu/graduate/highlights-of-the-masters-program/pure-mathematics\",\"WARC-Payload-Digest\":\"sha1:ORGZI5HXO43ZGJDQFB5VBYBDPXA7RUGB\",\"WARC-Block-Digest\":\"sha1:TGANSGRPU4XJ5EOBRCLB4LIIZLQFSYVP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988724.75_warc_CC-MAIN-20210505234449-20210506024449-00200.warc.gz\"}"}
|
https://www.mathworksheetscenter.com/mathskills/algebra/GraphCircles/
|
[
"# Graphs of Circles Worksheets\n\nHow to Find the Equation of a Circle Based on the Diameter and the Center Point - A circle is a set of all points in a plane at a defined distance known as the radius and for the given point called the center. Apart from the radius and center, a circle includes various parts in its geometry. A line segment connecting the two points on the circle and passes through the center of the circle is known as the diameter of the circle. Let's assume that point (x, y) are the coordinates given on the circle. The center of the circle will be at (h, k), and the radius will be defined as the 'r'. Using the distance formula to find the equation of the circle. √(x2 - x1)2 + (y2 – y1)2 = d Substituting (x1 , y1) = (h, k) and (x2 , y2) = (x, y) and d = r √( x – h)2 + (y – k)2 = r Squaring each side (x – h)2 + (y – k )2= r2 The equation of the circle with radius r units and the center (h, k) is (x – h)2 + (y – k )2 = r2.\n\n• ### Basic Lesson\n\nDemonstrates how to determine the equation of a circle. Practice problems are provided. Write the equation of a circle that has a center at (-1, 2) and a radius of 2?\n\n• ### Intermediate Lesson\n\nExplores how to find the coordinates of the center of a circle based on the equation of the circle. Practice problems are provided. What are the coordinates of the center of this circle? (x - 2)2 + (y - 3)2 = 9.\n\n• ### Independent Practice 1\n\nExample: Write the equation of a circle that has a center at (-2, -3) and radius of 8? The answers can be found below.\n\n• ### Independent Practice 2\n\nFeatures another 20 Graphs of Circles problems.\n\n• ### Homework Worksheet\n\n12 Graphs of Circles problems for students to work on at home. Example problems are provided and explained.\n\n• ### Topic Quiz\n\n10 Graphs of Circles problems. A math scoring matrix is included.\n\n• ### Homework and Quiz Answer Key\n\nAnswers for the homework and quiz.\n\n• ### Lesson and Practice Answer Key\n\nAnswers for both lessons and both practice sheets.\n\n#### The Lottery\n\nA mathematician organizes a lottery in which the prize is an infinite amount of money. When the winning ticket is drawn, and the jubilant winner comes to claim his prize, the mathematician explains the mode of payment: \"1 dollar now, 1/2 dollar next week, 1/3 dollar the week after that..."
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.9151047,"math_prob":0.992889,"size":2358,"snap":"2021-04-2021-17","text_gpt3_token_len":610,"char_repetition_ratio":0.16270179,"word_repetition_ratio":0.069196425,"special_character_ratio":0.2620865,"punctuation_ratio":0.09322034,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99971193,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T18:14:31Z\",\"WARC-Record-ID\":\"<urn:uuid:990ea1fd-fba7-4117-ab62-a007476e9c2c>\",\"Content-Length\":\"18314\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7aee961f-0a3c-48b1-99e6-94c1fee83b95>\",\"WARC-Concurrent-To\":\"<urn:uuid:e544bb75-2643-451e-8921-e45757e06b36>\",\"WARC-IP-Address\":\"104.21.22.56\",\"WARC-Target-URI\":\"https://www.mathworksheetscenter.com/mathskills/algebra/GraphCircles/\",\"WARC-Payload-Digest\":\"sha1:ZARD447DT7OAF2KWT6J4C62VDY5BQAEZ\",\"WARC-Block-Digest\":\"sha1:ED7K2DORYTC2UBT7JOAHWZC2BNYPDSSN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038064898.14_warc_CC-MAIN-20210411174053-20210411204053-00513.warc.gz\"}"}
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https://izajodm.springeropen.com/articles/10.1186/2193-9039-2-7/tables/4
|
[
"# Table 4 Description of explanatory variables\n\nDescription of explanatory variables\nNM Number of net migrants\nIf i = 1(x), expected urban wage$1 ( x ) ∗∑$ urban hukou j\nEwu-gap −expected urban wage 2(x)urban hukou 1(x)\nIf i = 2(x), expected urban wage 2(x)urban hukou 1(x)\n−expected urban wage 1(x)urban hukou 2(x)\nIf i = 3(x), -expected urban wage 1(x)urban hukou j(x);\nj = 1…18, i ≠j,1(x) stands for the highest expected urban wage city\nIf i = 1(x), city population$1 ( x ) ∗∑$ agricultural hukou j\nP-gap −city population 2(x)agricultural hukou 1(x)\nIf i = 2(x), city population 2(x)agricultural hukou 1(x)\n−city population 1(x)agricultural hukou 2(x)\nIf i = 3(x), -city population 1(x)agricultural hukou j(x)\nj = 1…18, i ≠j,1(x) stands for the biggest city population city\nIf i = 1(x), capital stock$1 ( x ) ∗ ∑ j$ agricultural hukou\n−capital stock 2(x)agricultural hukou 1(x)\nK-gap If i = 2(x), capital stock 2(x)agricultural hukou 1(x)\n−capital stock 1(x)agricultural hukou 2(x)\nIf i = 3(x), -capital stock 1(x)agricultural hukou j(x)\nj = 1…18, i ≠j,1(x) stands for the largest amount of capital stock city\nIf i = 1(x), single female$1 ( x ) ∗∑$single female j\n−single female 2(x)single female 1(x)\nSFP-gap If i = 2(x), single female 2(x)single female 1(x)\n−capital stock 1(x)single female 2(x)\nIf i = 3(x), -single female 1(x)single female j(x)\nj = 1…18, i ≠j,1(x) stands for the smallest number of single female city\nIf i = 1(x), urban hukou$1 ( x ) ∗∑$ agricultural hukou j\n−urban hukou 2(x)agricultural hukou 1(x)\nUH-gap If i = 2(x), urban hukou 2(x)agricultural hukou 1(x)\n−urban hukou 1(x)agricultural hukou 2(x)\nIf i = 3(x), -urban hukou 1(x)agricultural hukou j(x)\nj = 1…18, i ≠j,1(x) stands for the biggest number of urban hukou city"
] |
[
null
] |
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https://eccc.weizmann.ac.il/report/2012/087/
|
[
"",
null,
"",
null,
"Under the auspices of the Computational Complexity Foundation (CCF)",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"REPORTS > DETAIL:\n\n### Revision(s):\n\nRevision #1 to TR12-087 | 5th November 2012 12:00\n\n#### The Query Complexity of Finding a Hidden Permutation",
null,
"Revision #1\nAuthors: Peyman Afshani, Manindra Agrawal, Doerr Benjamin, Kasper Green Larsen, Kurt Mehlhorn, Winzen Carola\nAccepted on: 5th November 2012 12:00\nKeywords:\n\nAbstract:\n\nWe study the query complexity of determining a hidden permutation. More specifically, we study the problem of learning a secret $(z,\\pi)$ consisting of a binary string $z$ of length $n$ and a permutation $\\pi$ of $[n]$. The secret must be unveiled by asking queries $x \\in \\{0,1\\}^n$, and for each query asked, we are returned the score $f_{z,\\pi}(x)$ defined as\n$f_{z,\\pi}(x):= \\max \\{ i \\in [0..n]\\mid \\forall j \\leq i: z_{\\pi(j)} = x_{\\pi(j)}\\}\\,;$\ni.e., the length of the longest common prefix of $x$ and $z$ with respect to $\\pi$. The goal is to minimize the number of queries asked.\n\nOur main result are\nmatching upper and lower bounds for this problem,\nboth for deterministic and randomized query schemes.\nThe deterministic query complexity is $\\Theta(n \\log n)$, which, surprisingly,\nimproves to $\\Theta(n \\log \\log n)$ in the randomized setting.\n\nFor the randomized query complexity, both the upper and lower bound are stronger than what can be achieved by standard arguments like the analysis of random queries or information-theoretic considerations.\nOur proof of the $\\Omega(n \\log \\log n)$ lower bound is\nbased on a potential function argument, which seems to be uncommon in the query\ncomplexity literature. We find this potential function technique\na very powerful tool in proving lower bounds for randomized query\nschemes and we expect it to find applications in many other query\ncomplexity problems.\n\nChanges to previous version:\n\nWe have significantly improved the presentation of our work.\n\n### Paper:\n\nTR12-087 | 4th July 2012 09:13\n\n#### The Deterministic and Randomized Query Complexity of a Simple Guessing Game\n\nTR12-087\nAuthors: Peyman Afshani, Manindra Agrawal, Doerr Benjamin, Winzen Carola, Kasper Green Larsen, Kurt Mehlhorn\nPublication: 7th July 2012 14:16\nKeywords:\n\nAbstract:\n\nWe study the $\\leadingones$ game, a Mastermind-type guessing game first\nregarded as a test case in the complexity theory of randomized search\nheuristics. The first player, Carole, secretly chooses a string $z \\in \\{0,1\\}^n$ and a\npermutation $\\pi$ of $[n]$.\nThe goal of the second player, Paul, is to identify the secret $(z,\\pi)$\nwith a small number of\nqueries. A query is a string $x \\in \\{0,1\\}^n$, and the score of $x$ is\n$f_{z,\\pi}(x):= \\max \\{ i \\in [0..n] \\mid \\forall j \\leq i: z_{\\pi(j)} = x_{\\pi(j)}\\}\\,,$\nthe length of the longest common prefix of $x$ and $z$ with respect to $\\pi$.\nWe are interested in the number of queries needed by Paul to identify the\nsecret.\n\nBy using a relatively straightforward strategy, Paul can identify the secret with\n$O(n\\log n)$ queries and recently only a modest improvement of this to $O(n\\log n /\\log\\log n)$\nwas available (Doerr, Winzen, 2012 [DW12]).\n\nIn this paper, we completely resolve the problem by offering the following\nresults. We show that when limited to deterministic strategies, $O(n \\log n)$ queries is the best possible.\nOn the other hand, by using randomization Paul can find the secret code with an expected number of\n$O(n\\log\\log n)$ queries, which we prove is optimal by matching it with a lower bound of the same asymptotic\nmagnitude. Finally, we prove that a number of problems that are naturally related to our problem\n(such as deciding whether a sequence of queries and scores is consistent) can\nbe solved in polynomial time.\n\nISSN 1433-8092 | Imprint"
] |
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null,
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null,
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|
https://www.javatpoint.com/ds-types-of-queues
|
[
"# Types of Queue\n\nIn this article, we will discuss the types of queue. But before moving towards the types, we should first discuss the brief introduction of the queue.\n\n### What is a Queue?\n\nQueue is the data structure that is similar to the queue in the real world. A queue is a data structure in which whatever comes first will go out first, and it follows the FIFO (First-In-First-Out) policy. Queue can also be defined as the list or collection in which the insertion is done from one end known as the rear end or the tail of the queue, whereas the deletion is done from another end known as the front end or the head of the queue.\n\nThe real-world example of a queue is the ticket queue outside a cinema hall, where the person who enters first in the queue gets the ticket first, and the last person enters in the queue gets the ticket at last. Similar approach is followed in the queue in data structure.\n\nThe representation of the queue is shown in the below image -",
null,
"Now, let's move towards the types of queue.\n\n### Types of Queue\n\nThere are four different types of queue that are listed as follows -",
null,
"• Simple Queue or Linear Queue\n• Circular Queue\n• Priority Queue\n• Double Ended Queue (or Deque)\n\nLet's discuss each of the type of queue.\n\n### Simple Queue or Linear Queue\n\nIn Linear Queue, an insertion takes place from one end while the deletion occurs from another end. The end at which the insertion takes place is known as the rear end, and the end at which the deletion takes place is known as front end. It strictly follows the FIFO rule.",
null,
"The major drawback of using a linear Queue is that insertion is done only from the rear end. If the first three elements are deleted from the Queue, we cannot insert more elements even though the space is available in a Linear Queue. In this case, the linear Queue shows the overflow condition as the rear is pointing to the last element of the Queue.\n\nTo know more about the queue in data structure, you can click the link - https://www.javatpoint.com/data-structure-queue\n\n### Circular Queue\n\nIn Circular Queue, all the nodes are represented as circular. It is similar to the linear Queue except that the last element of the queue is connected to the first element. It is also known as Ring Buffer, as all the ends are connected to another end. The representation of circular queue is shown in the below image -",
null,
"The drawback that occurs in a linear queue is overcome by using the circular queue. If the empty space is available in a circular queue, the new element can be added in an empty space by simply incrementing the value of rear. The main advantage of using the circular queue is better memory utilization.\n\nTo know more about the circular queue, you can click the link - https://www.javatpoint.com/circular-queue\n\n### Priority Queue\n\nIt is a special type of queue in which the elements are arranged based on the priority. It is a special type of queue data structure in which every element has a priority associated with it. Suppose some elements occur with the same priority, they will be arranged according to the FIFO principle. The representation of priority queue is shown in the below image -",
null,
"Insertion in priority queue takes place based on the arrival, while deletion in the priority queue occurs based on the priority. Priority queue is mainly used to implement the CPU scheduling algorithms.\n\nThere are two types of priority queue that are discussed as follows -\n\n• Ascending priority queue - In ascending priority queue, elements can be inserted in arbitrary order, but only smallest can be deleted first. Suppose an array with elements 7, 5, and 3 in the same order, so, insertion can be done with the same sequence, but the order of deleting the elements is 3, 5, 7.\n• Descending priority queue - In descending priority queue, elements can be inserted in arbitrary order, but only the largest element can be deleted first. Suppose an array with elements 7, 3, and 5 in the same order, so, insertion can be done with the same sequence, but the order of deleting the elements is 7, 5, 3.\n\n### Deque (or, Double Ended Queue)\n\nIn Deque or Double Ended Queue, insertion and deletion can be done from both ends of the queue either from the front or rear. It means that we can insert and delete elements from both front and rear ends of the queue. Deque can be used as a palindrome checker means that if we read the string from both ends, then the string would be the same.\n\nDeque can be used both as stack and queue as it allows the insertion and deletion operations on both ends. Deque can be considered as stack because stack follows the LIFO (Last In First Out) principle in which insertion and deletion both can be performed only from one end. And in deque, it is possible to perform both insertion and deletion from one end, and Deque does not follow the FIFO principle.\n\nThe representation of the deque is shown in the below image -",
null,
"To know more about the deque, you can click the link - https://www.javatpoint.com/ds-deque\n\nThere are two types of deque that are discussed as follows -\n\n• Input restricted deque - As the name implies, in input restricted queue, insertion operation can be performed at only one end, while deletion can be performed from both ends.",
null,
"• Output restricted deque - As the name implies, in output restricted queue, deletion operation can be performed at only one end, while insertion can be performed from both ends.",
null,
"Now, let's see the operations performed on the queue.\n\n## Operations performed on queue\n\nThe fundamental operations that can be performed on queue are listed as follows -\n\n• Enqueue: The Enqueue operation is used to insert the element at the rear end of the queue. It returns void.\n• Dequeue: It performs the deletion from the front-end of the queue. It also returns the element which has been removed from the front-end. It returns an integer value.\n• Peek: This is the third operation that returns the element, which is pointed by the front pointer in the queue but does not delete it.\n• Queue overflow (isfull): It shows the overflow condition when the queue is completely full.\n• Queue underflow (isempty): It shows the underflow condition when the Queue is empty, i.e., no elements are in the Queue.\n\nNow, let's see the ways to implement the queue.\n\n## Ways to implement the queue\n\nThere are two ways of implementing the Queue:\n\nSo, that's all about the article. Hope, the article will be helpful and informative to you.\n\n### Feedback",
null,
"",
null,
"",
null,
""
] |
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null,
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null,
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null,
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null,
"https://static.javatpoint.com/ds/images/ds-types-of-queue8.png",
null,
"https://www.javatpoint.com/images/facebook32.png",
null,
"https://www.javatpoint.com/images/twitter32.png",
null,
"https://www.javatpoint.com/images/pinterest32.png",
null
] |
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|
https://physics.stackexchange.com/questions/29908/how-can-i-theoretically-describe-the-potential-between-two-capacitors-in-series
|
[
"# How can I theoretically describe the potential between two capacitors in series?\n\nSuppose to have two capacitors in series:",
null,
"The voltage in the middle point will be:\n\n$$V_X = V_1 \\frac{C_1}{C_1+C_2}$$\n\nHow can this be explained? It's been asked in electronics, and explained in terms of impedence and charge equality, but none of the explanations is satisfying to me, as I think it should involve charge conservation (Gauss theorem?) and/or electric fields/potentials.\n\nCould you enlighten me?\n\n• The charge equalization version is a charge conservation argument. Note that there is an isolated conductor between the capacitors and that it starts off bulk neutral... – dmckee Jun 11 '12 at 14:33\n• @dmckee: it is, but it's not quite well explained there. Note that I mentioned it in a comment, and it's been reprised in the answer, but without further explanation – clabacchio Jun 11 '12 at 14:35\n\nSuppose you imagine the battery to be a variable voltage, and start with the voltage at zero. Obviously everything is uncharged.\n\nNow turn the battery up to 1V. As you do this positive charge leaves the positive terminal and an equal and opposite negative charge leaves the negative terminal. We know the charges leaving the positive and negative terminals must be the same because the battery is a conductor and can't develop a net charge like a capacitor. Let's call the charge that leaves the battery $Q$.\n\nThe only place the charge that leaves the battery can go is onto the capacitors, so both capacitors now have a charge of $Q$ on them. We know that for a capacitor of capacitance $C$, the voltage across the capacitor is given by:\n\n$$V = \\frac{Q}{C}$$\n\nCall the voltage of the top (1$\\mu$F) capacitor $V_1$, and the voltage of the bottom (2$\\mu$F) capacitor $V_2$. Then:\n\n$$V_1 = \\frac{Q}{C_1}$$ $$V_2 = \\frac{Q}{C_2}$$\n\nDividing the first equation by the second plus a bit of quick rearrangement gives:\n\n$$V_1 = \\frac{C_2}{C_1} V_2$$\n\nThe two voltages must add up to 1V because we have a 1V battery, therefore:\n\n$$V_1 + V_2 = 1$$\n\nIf you substitute for $V_1$ you get:\n\n$$\\frac{C_2}{C_1} V_2 + V_2 = 1$$\n\nand dividing through by $(1 + C_2/C_1)$ gives:\n\n$$V_2 = \\frac{1}{1 + C_2/C_1}$$\n\nTidy this up by multiplying to top and bottom of the right hand side by $C_1$ and you get the equation you're trying to prove:\n\n$$V_2 = \\frac{C_1}{C_1 + C_2}$$\n\nJust to check, feed in $C_1 = 1$ and $C_2 = 2$ and $V_2$ does indeed come out as 1/3V.\n\n• Detailed, but it doesn't explain the assumption that both capacitors have charge Q – clabacchio Jun 11 '12 at 18:10\n• If charge +Q leaves the battery anode then charge -Q must leave the cathode because the battery can't have a net charge. That means the top plate of the top capacitor has a +Q charge and the bottom plate of the bottom capacitor has a -Q charge. But these charges are now attracting/repelling the electrons in the wire between the two capacitors. The +Q charge on the top capacitor will attract electrons from the wire until it's bottom plate builds up a charge of -Q. Likewise the -Q charge on the bottom capacitor will repel electrons until it's top plate has a +Q charge. So the charge ... – John Rennie Jun 11 '12 at 18:15\n• ... on both capacitors is Q. The circuit doesn't start out with any charge, so if you add up the net charge it must sum to zero. That's how we know the charges on the top plate of the top capacitor and the bottom plate of the bottom capacitor must be equal and opposite. – John Rennie Jun 11 '12 at 18:15\n• But why the potential in $A$ is equal to the lower voltage? i.e. why $V_A=V_2$? @JohnRennie – Don Fanucci Feb 1 '17 at 22:42\n\nThis is really just a restatement of John Rennie's answer, but it might be a bit easier to follow...\n\nAssume both capacitors are initially uncharged (important, since otherwise the voltage of their common node is undefined) and the voltage source is 0.\n\nNow ramp the voltage source up to 1V. During this ramp up, the same current i flows through both capacitors (since they're connected in series), so $$i= C_1\\frac{dv_{c1}}{dt} = C_2\\frac{dv_{c2}}{dt}$$\n\nSo the rates of change of the capacitor voltages are inversely proportional to their capacitances, and so will the final capacitor voltages after integrating (using the fact that the capacitor voltages were initially zero).\n\nFrom that relation, it's straightforward to get to the expression in your question.\n\nI think the quickest way to get an intuitive feel for this situation is to recognise the voltage at that part of the circuit is the potential difference between the two capacitors. And is in reference to the $0V$ terminal of the battery.\n\nKirchoff's voltage law tells us that we have to sum to $0V$ in a loop. As C2 = $2 *$C1 we have to lose twice as much voltage over C2. As we are dealing with $1V$ that means we lose $.66V$ over C2 and $.33V$ over C1. Now tracing around the circuit in either direction quickly leads to the answer that we have a voltage of $.33V$ between the capacitors..."
] |
[
null,
"https://i.stack.imgur.com/gVAEX.png",
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https://patents.google.com/patent/CN103646358B/en
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"# CN103646358B - Meter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate - Google Patents\n\nMeter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate Download PDF\n\n## Info\n\nPublication number\nCN103646358B\nCN103646358B CN201310746200.7A CN201310746200A CN103646358B CN 103646358 B CN103646358 B CN 103646358B CN 201310746200 A CN201310746200 A CN 201310746200A CN 103646358 B CN103646358 B CN 103646358B\nAuthority\nCN\nChina\nPrior art keywords\nprime\nlambda\nscheduled overhaul\npower equipment\nsigma\nPrior art date\nApplication number\nCN201310746200.7A\nOther languages\nChinese (zh)\nOther versions\nCN103646358A (en\nInventor\n\nOriginal Assignee\n\nPriority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)\nFiling date\nPublication date\nApplication filed by 重庆大学, 国网重庆市电力公司市区供电分公司 filed Critical 重庆大学\nPriority to CN201310746200.7A priority Critical patent/CN103646358B/en\nPublication of CN103646358A publication Critical patent/CN103646358A/en\nApplication granted granted Critical\nPublication of CN103646358B publication Critical patent/CN103646358B/en\n\n• 230000035945 sensitivity Effects 0.000 claims abstract description 54\n• 230000015572 biosynthetic process Effects 0.000 claims abstract description 21\n• 238000003786 synthesis reactions Methods 0.000 claims abstract description 21\n• 230000002194 synthesizing Effects 0.000 claims abstract description 21\n• 280000867207 Lambda companies 0.000 claims description 224\n• 230000005540 biological transmission Effects 0.000 claims description 15\n• 239000000463 materials Substances 0.000 claims description 14\n• 238000000034 methods Methods 0.000 claims description 8\n• 230000000875 corresponding Effects 0.000 claims description 5\n• 238000010276 construction Methods 0.000 claims description 3\n• 239000000203 mixtures Substances 0.000 claims description 3\n• 230000003068 static Effects 0.000 claims description 3\n• 230000037010 Beta Effects 0.000 description 18\n• 238000004422 calculation algorithm Methods 0.000 description 8\n• 238000004364 calculation methods Methods 0.000 description 5\n• 230000032683 aging Effects 0.000 description 4\n• 238000004458 analytical methods Methods 0.000 description 4\n• 238000005299 abrasion Methods 0.000 description 3\n• 238000010586 diagrams Methods 0.000 description 3\n• 230000000694 effects Effects 0.000 description 3\n• 239000002131 composite materials Substances 0.000 description 2\n• 150000001875 compounds Chemical class 0.000 description 2\n• 230000002950 deficient Effects 0.000 description 2\n• 238000005553 drilling Methods 0.000 description 2\n• 238000005516 engineering processes Methods 0.000 description 2\n• 238000007689 inspection Methods 0.000 description 2\n• 238000005457 optimization Methods 0.000 description 2\n• 230000000630 rising Effects 0.000 description 2\n• 281000091500 Ambient Weather companies 0.000 description 1\n• 230000015556 catabolic process Effects 0.000 description 1\n• 230000002493 climbing Effects 0.000 description 1\n• 230000004059 degradation Effects 0.000 description 1\n• 238000006731 degradation reactions Methods 0.000 description 1\n• 230000001934 delay Effects 0.000 description 1\n• 238000009795 derivation Methods 0.000 description 1\n• 238000010438 heat treatment Methods 0.000 description 1\n• 230000001788 irregular Effects 0.000 description 1\n• 239000003921 oils Substances 0.000 description 1\n• 230000000737 periodic Effects 0.000 description 1\n• 230000002123 temporal effects Effects 0.000 description 1\n• 230000001131 transforming Effects 0.000 description 1\n\n• Y02E40/76\n• Y04S10/545\n\n## Abstract\n\nThe invention discloses a kind of meter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate, build power equipment time-varying failure rate model；According to time-varying failure rate model；From averagely setting up the functional relationship between scheduled overhaul cycle and system reliability without validity angle, calculate Reliability Index；Consider that electric power apparatus examination cost, power failure cost set up system synthesis functional relationship originally and between the scheduled overhaul cycle, this functional relationship is utilized to ask for this sensitivity to scheduled overhaul rate of system synthesis, utilizing this level of sensitivity to differentiate whether each equipment of electrical network is in optimal plan time between overhauls(TBO) and optimize and revise each facilities plan time between overhauls(TBO), finally realizing electrical network each facilities plan time between overhauls(TBO) all reaches optimum；Use the present invention can find and solve part power equipment in the arrangement of existing electrical network scheduled overhaul cycle there is maintenance or owed the problem of maintenance, comprehensive coordination electric network reliability and economy.\n\n## Description\n\nMeter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate\n\nTechnical field\n\nThe present invention relates to electric power overhaul technology, be specifically related under a kind of meter and electrical equipment fault rate time dependant conditions based on can By the electrical network scheduled overhaul cycle determination method of property cost/benefit analysis, belong to Model in Reliability Evaluation of Power Systems and scheduled overhaul Cycle optimizes field.\n\nBackground technology\n\nElectromotor, transformator, transmission line of electricity are as the crucial component devices of composite power systems, and its reliability is to whole The safe and reliable operation of electrical network is most important.Power equipment in longtime running by loss, self deterioration and hidden fault etc. because of The impact of element, reliability level growth over time has accumulative effect, in order to evade climbing because of power equipment failure probability Rise the increase causing system blackout risk, need power equipment is carried out periodic inspection.According to power equipment relevant maintenance directive/guide, Generating set, transformator, transmission line of electricity are divided into A, B, C, D tetra-according to maintenance scale and down time and overhaul grade, wherein C Level overhauls the abrasion according to equipment, aging rule, carries out with having emphasis checking, repairing and the replacing of a small amount of part.\n\nThe research of existing scheduled overhaul carries out overhauling the optimization peace of period mostly under the hypothesis that equipment failure rate is steady state value Row, and to fault rate scheduled overhaul cycle rare research of power equipment in the case of non-constant, the scheduled overhaul cycle is often with relevant Maintenance directive/guide is for reference to being determined.But, in maintenance directive/guide, the equipment of same type and capacity is corresponding under each maintenance grade Cycle the most identical, do not account for equipment self deterioration wear law, service condition and the difference of running environment.Additionally, i.e. Making is identical device, the difference of its position in systems, also makes a world of difference the influence degree of system reliability, reliable to system Property affect bigger equipment relatively more maintenance ought to be needed to obtain more excellent reliability level.Thus be necessary from system Aspect, the main outage factor considering equipment carries out the scheduled overhaul cycle optimization planning research of equipment.\n\nSummary of the invention\n\nDo not consider that equipment failure rate temporal behavior and device type, position etc. are to being for the existing electrical network scheduled overhaul cycle The deficiency of system reliability effect, it is an object of the invention to provide a kind of meter and the electrical network scheduled overhaul of power equipment time-varying fault rate Cycle determination method, this method it can be found that and solve the existing electrical network scheduled overhaul cycle arrange in part power equipment cross maintenance Or owe the problem overhauled, comprehensive coordination electric network reliability and economy.\n\nTo achieve these goals, the technical solution used in the present invention is as follows:\n\nMeter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate, specifically comprise the following steps that\n\n1) the system low order malfunction in the range of using State enumeration method to enumerate power system certain fault exponent number, and base System minimum under DC power flow optimum cuts the load model each enumeration state of calculating cuts loading；If cutting loading more than 0, then That records this system mode and correspondence cuts loading, otherwise continues to enumerate, until enumerating end, thus finally gives a system System malfunction (X1,X2,…,Xm) and correspondence cut loading (LC1,LC2,…,LCm), wherein m is the system event enumerated Barrier status number；\n\n2) according to principal elements such as power equipment self deterioration, external environment and maintenance, the impact of its fault rate is built meter And the power equipment time-varying failure rate model in scheduled overhaul cycle；Described power equipment is electromotor, transformator and transmission line of electricity； This external environment includes weather and temperature；Obtain the parameter of each power equipment time-varying failure rate model in network system；And input Each power equipment repair time, original plan maintenance rate；\n\n3) according to scheduled overhaul rate, repair time and the failure rate function of each power equipment, each power equipment is calculated average Without validity, based on the probability averagely calculating each system fault condition without validity；Further according to reliability index computing formula, utilize electricity Power equipment average without validity and the 1st) system fault condition (X that obtains of step1,X2,…,Xm) and correspondence cut loading (LC1,LC2,…,LCm) calculating Reliability Index, Reliability Index includes system load-loss probability LOLP and system Expect to lack delivery EENS；\n\nAlthough do not have direct mathematical relationship between LOLP and system blackout cost, but indirect correlation.What LOLP represented is System loses the probability of load, the i.e. macroscopic description of system generation power failure phenomenon probability size, and in general, system LOLP is more Greatly, the power failure amount that each malfunction causes is the biggest, then EENS can be the highest, and power failure cost will be the biggest.Although LOLP does not reflect The severity having a power failure, but LOLP also can reflect certain system reliability level.Calculating LOLP, EENS is to assess system Reliability level.\n\n4) system blackout cost is calculated, according to each power equipment unit according to unit power failure cost and Reliability Index Scheduled overhaul cost and trouble shooting cost combine the 3rd) the scheduled overhaul rate of step and failure rate function obtain system overhaul cost, System blackout cost and system overhaul cost collectively form system synthesis originally；Set up system synthesis basis and between the scheduled overhaul cycle Functional relationship, utilize this functional relationship to ask for this sensitivity to scheduled overhaul rate of system synthesis；\n\n5) by descending for sensitivity sequence, if the threshold that the sensitivity absolute value of all power equipments is both less than preset Be worth esp, then the scheduled overhaul cycle that this sensitivity is corresponding is the scheduled overhaul cycle finally determined；Otherwise plan for adjustment maintenance In the cycle, until system synthesis, the sensitivity of scheduled overhaul rate is examined by this less than the threshold value esp preset, the plan meeting this requirement The cycle of repairing is the scheduled overhaul cycle finally determined.\n\nWherein the 5th) the scheduled overhaul cycle of step is adjusted as follows: set the power equipment of serial number x through sensitive After degree sequence, sequence number becomes x ', if meeting x '≤nset, and sensitivity l (x) meets l (x) > esp, then make the electric power of serial number x set Standby scheduled overhaul rate λ \" (x)=0.99 λ \" (x)；If x ' >=N-nset, and l (x) <-esp, then make λ \" (x)=1.01 λ \" (x), based on The power equipment scheduled overhaul rate redefined obtains the new scheduled overhaul cycle, thus obtains system synthesis originally relative to new Scheduled overhaul rate sensitivity l (x), then judge whether sensitivity l (x) is less than the threshold value esp preset, if it is satisfied, then adjust After the scheduled overhaul cycle be the scheduled overhaul cycle finally determined, otherwise repeat step 3)-5), until sensitivity l (x) is little In default threshold value esp.\n\nN is the power equipment quantity of system, nsetFor setting sequence number, represent to preset every time and carry out the scheduled overhaul cycle and tune up Or the power equipment quantity turned down；Esp is sensitivity threshold value；\" (x) is the power equipment scheduled overhaul rate of serial number x to λ.\n\nFurther, the 3rd) step system load-loss probability LOLP and the scarce delivery EENS of system expectation is asked by following process , if the system of N number of power equipment composition, each status of electric power is separate, respectively S1, S2, S3..., SN, Sk=0 represents Power equipment k is in normal condition, Sk=1 represents that it is in malfunction, then the probability of this system mode x is represented by:\n\nP(x)=P(S1)·P(S2)·P(S3)…P(SN)\n\nSo, system load-loss probability LOLP and the scarce delivery EENS computing formula of system expectation are as follows；\n\n$\\begin{array}{c}\\mathrm{LOLP}=\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)P\\left(x\\right)=\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}P\\left({S}_{k}\\right)\\\\ =\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}\\left[{S}_{k}{U}_{k}+\\left(1-{S}_{k}\\right)\\left(1-{U}_{k}\\right)\\right]\\\\ =\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}\\left[{S}_{k}·\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}+\\left(1-{S}_{k}\\right)\\left(1-\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}\\right)\\right]\\end{array}$\n\n$\\begin{array}{c}\\mathrm{EENS}=8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)P\\left(x\\right){L}_{C}\\left(x\\right)=8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}P\\left({S}_{k}\\right)\\\\ =8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}\\left[{S}_{k}{U}_{k}+\\left(1-{S}_{k}\\right)\\left(1-{U}_{k}\\right)\\right]\\\\ =8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}\\left[{S}_{k}·\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}+\\left(1-{S}_{k}\\right)\\left(1-\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}\\right)\\right]\\end{array}$\n\nWherein, λk\" for the scheduled overhaul rate of power equipment k, LCX () represents under system fault condition x for making system recovery arrive Minimum load reduction required for one static security operating point, IfX ()=0 represents system normal condition, IfX ()=1 represents system System malfunction；\n\nThen Reliability Index system load-loss probability LOLP lacks delivery EENS relative to scheduled overhaul with system expectation The sensitivity of rate is shown below respectively:\n\n$\\begin{array}{c}\\frac{∂\\mathrm{LOLP}}{{∂\\mathrm{λ}}_{k}^{′′}}=\\underset{x∈X}{\\mathrm{Σ}}\\left[\\frac{∂{I}_{f}\\left(x\\right)}{{∂\\mathrm{λ}}_{k}^{′′}}P\\left(x\\right)+{I}_{f}\\left(x\\right)\\frac{∂P\\left(x\\right)}{{∂\\mathrm{λ}}_{k}^{′′}}\\right]\\\\ =\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)\\frac{∂\\left[P\\left({S}_{1}\\right)P\\left({S}_{2}\\right)···P\\left({S}_{k}\\right)···P\\left({S}_{m}\\right)\\right]}{{∂\\mathrm{λ}}_{k}^{′′}}\\\\ =\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)P\\left(x\\right)\\frac{\\left({2S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\end{array}$\n\n$\\begin{array}{c}\\frac{∂\\mathrm{EENS}}{{∂\\mathrm{λ}}_{K}^{′′}}=8760\\underset{x∈X}{\\mathrm{Σ}}\\left[\\frac{∂{I}_{f}\\left(x\\right)}{{∂\\mathrm{λ}}_{j}^{′′}}{L}_{C}\\left(x\\right)P\\left(x\\right)+{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\frac{∂P\\left(x\\right)}{{∂\\mathrm{λ}}_{k}^{′′}}\\right]\\\\ =8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\frac{∂\\left[P\\left({S}_{1}\\right)P\\left({S}_{2}\\right)···P\\left({S}_{k}\\right)···P\\left({S}_{m}\\right)\\right]}{{∂\\mathrm{λ}}_{k}^{′′}}\\\\ =8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)P\\left(x\\right)\\frac{\\left({2S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\end{array}.$\n\nFurther, the 4th) step system blackout cost is expected to lack delivery EENS meter according to unit power failure cost and system Calculating, system overhaul includes overhauling and scheduled overhaul two kinds afterwards；The cost of overhaul includes fee of material and construction cost two large divisions every time, Wherein fee of material and the type of power equipment, capacity are relevant；If power equipment k (k=1,2 ..., N) carry out the cost of overhaul the most afterwards With for CRk, including fee of material CRMk, the operating expenses C of unit intervalROk/h；Carrying out a scheduled overhaul expense is CPk, including material Material takes CPMk, unit interval operating expenses CPOk/h；Unit power failure cost is set to CN/ WMh, the total cost of system is set to Ctotal\n\nThen system synthesis functional relationship such as following formula originally and between the scheduled overhaul cycle,\n\n$\\begin{array}{c}{C}_{\\mathrm{total}}={C}_{N}·\\mathrm{EENS}+\\underset{k}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Rk}}·{N}_{\\mathrm{Rk}}+\\underset{k}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Pk}}·{N}_{\\mathrm{Pk}}\\\\ =8760{C}_{N}\\underset{x∈X}{\\mathrm{Σ}}\\left[{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\right]P\\left(x\\right)+\\\\ \\underset{k=1}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Rk}}{\\mathrm{λ}}_{k}^{′′}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}+\\underset{k=1}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Pk}}{\\mathrm{λ}}_{k}^{′′}\\end{array}$\n\nWherein NRkRepresent that number of times, N are overhauled in the every annual of power equipment k afterwardspkRepresent the annual scheduled overhaul of power equipment k Number of times, CRk=CRMk+CROk×rk, CPk=CPMk+CPOk×rk″；\n\nOwing to the state of each power equipment is separate, then the number of times of maintenance afterwards of power equipment k and plan inspection in 1 year Repair scheduled overhaul rate λ of number of times and power equipment kk\" relevant, then have\n\n$\\left\\{\\begin{array}{c}\\frac{{∂N}_{\\mathrm{pk}}}{{∂\\mathrm{λ}}_{k}^{′′}}=1\\\\ \\frac{{∂N}_{\\mathrm{Rk}}}{{∂\\mathrm{λ}}_{k}^{′′}}={∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}-\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}{\\mathrm{λ}}_{k}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\\\ \\frac{{∂N}_{\\mathrm{pk}}}{{∂\\mathrm{λ}}_{i}^{′′}}=0,\\left(i≠k\\right)\\\\ \\frac{{∂N}_{\\mathrm{Rk}}}{{∂\\mathrm{λ}}_{i}^{′′}}=0,\\left(i≠k\\right)\\end{array}\\right\\$\n\nThis relative to the scheduled overhaul rate sensitivity of each equipment of electrical network is thus to obtain system synthesis:\n\n$\\begin{array}{c}\\frac{{∂C}_{\\mathrm{total}}}{{∂\\mathrm{λ}}_{k}^{′′}}={C}_{N}·\\frac{∂\\mathrm{EENS}}{{∂\\mathrm{λ}}_{k}^{′′}}+\\underset{i=1}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Ri}}·\\frac{{∂N}_{\\mathrm{Ri}}}{{∂\\mathrm{λ}}_{k}^{′′}}+\\underset{i=1}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Pi}}·\\frac{{∂N}_{\\mathrm{Pi}}}{{∂\\mathrm{λ}}_{k}^{′′}}\\\\ =8760{C}_{N}\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)P\\left(x\\right)\\frac{\\left({2S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\\\ +{C}_{\\mathrm{Rk}}·\\left({∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}-\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}·{\\mathrm{λ}}_{k}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)+{C}_{\\mathrm{Pk}}\\end{array}.$\n\nCompared with prior art, the invention have the advantages that\n\n1. from the principal element causing each electrical equipment fault to be stopped transport, setting up time-varying failure rate model, ratio is traditional Fault rate regards as constant closer to practical situation；\n\n2. utilize analytic method to establish the functional relationships between scheduled overhaul rate and Reliability Index, system synthesis basis System, the reflection scheduled overhaul cycle directly perceived is on system reliability and the impact of system cost；\n\n3. whether can be in the optimal plan time between overhauls(TBO) with each power equipment of intuitive judgment electrical network by sensitivity index, with And instruct how to adjust the facilities plan time between overhauls(TBO).\n\nAccompanying drawing explanation\n\nFig. 1 is that the electrical network scheduled overhaul of meter of the present invention and fault rate time-varying optimizes block diagram.\n\nFailure rate function figure when Fig. 2 is to carry out scheduled overhaul and corrective maintenance in T time section.\n\nFig. 3 is per day load factor, day typical ambient temperatures, hot(test)-spot temperature curve chart.\n\nFig. 4 is two state synoptic model schematic diagrams.\n\nFig. 5 is that scheduled overhaul based on sensitivity sequence optimizes heuristic iterative algorithm flow chart.\n\nDetailed description of the invention\n\nThe total thinking of the present invention is:\n\nFirst according to principal elements such as power equipment self deterioration, external environment, maintenance, the impact of its fault rate is built meter And the power equipment time-varying failure rate model of scheduled overhaul；\n\nThen according to time-varying failure rate model, from average without validity angle set up scheduled overhaul cycle and system reliability it Between functional relationship, it is all low that this system reliability estimation method uses that enumerative technique enumerates in the range of system certain fault exponent number Rank system fault condition, uses optimum based on DC power flow to cut load model and carries out minimum and cut carry calculation, utilize electric power to set Standby averagely without validity calculating Reliability Index；\n\nConsider that electric power apparatus examination cost, power failure cost set up system synthesis functional relationships originally and between the scheduled overhaul cycle System, utilizes this functional relationship to ask for this sensitivity to scheduled overhaul rate of system synthesis.It is ranked up according to level of sensitivity, sentences Whether the other each equipment of electrical network is in the optimal plan time between overhauls(TBO), if relative to certain facilities plan maintenance rate sensitivity be bigger just If value, shows that this equipment is in deficient inspecting state, otherwise the bigger negative value of absolute value, then shows that this equipment was in maintenance shape State, the method optimizing and revising each facilities plan time between overhauls(TBO), is i.e. to judge the spirit of each power equipment in electrical network according to this sensitivity Sensitivity size, the suitable plan for adjustment time between overhauls(TBO), finally realizing electrical network each facilities plan time between overhauls(TBO) all reaches optimum.\n\nUse the present invention can solve part power equipment in the arrangement of existing electrical network scheduled overhaul cycle there is maintenance or owed The problem of maintenance, comprehensive coordination electric network reliability and economy.\n\nFig. 1 is that the electrical network scheduled overhaul of meter of the present invention and fault rate time-varying optimizes block diagram.Fig. 5 is based on sensitivity sequence Scheduled overhaul optimizes heuristic iterative algorithm flow chart.\n\nBelow in conjunction with the concrete situation of accompanying drawing and power equipment, this is described in detail.\n\n1, the failure rate model of power equipment is set up\n\nTraditional literature is thought, power equipment is constant in the fault rate of steady operation period, but actually by self deterioration, machinery The impact of the factors such as abrasion and hidden fault, electrical equipment fault rate is in time in the trend of rising, additionally, outside environmental elements is also The fault rate of meeting appreciable impact power equipment.Scheduled overhaul can eliminate or reduce power equipment hidden fault probability, delays electricity Power ageing equipment and mechanical wear, thus reduce electrical equipment fault rate.Power equipment owing to the present invention relates to includes generating Machine, transformator and transmission line of electricity, therefore below the failure rate model of each power equipment is analyzed explanation respectively.\n\nThe failure rate model of 1.1 electromotors\n\nThe factor causing electromotor to be stopped transport can be divided into 3 classes: the 1. fault of electromotor itself, as aging in parts heating, abrasion Deng；2. the protection device action that electromotor irregular operating causes；3. external environment and artificial maloperation etc..Corresponding above three classes The fault rate of outage factor is set to: λ1(t), λ2, λ3\n\nTo first kind outage factor, electromotor is aging with the time of operation, increased wear, and fault rate shows a rising trend, and uses β1The double compound Weibull function of=1 come approximate description (Duan Dongli, Wu little Yue, Deng Hongzhong. based on time-varying fault rate and service Recovery time model reliability evaluation [J]. Proceedings of the CSEE, 2011,31 (28): 57-63.)；To Two classes are stopped transport, and can describe (Liu Ruoxi, Zhang Jian by fault rate incremental model when setting up voltage, frequency departure rated value China, Liu Wenxia, Wu Di. the assessment algorithm [J] of operation of electric power system risk. Proceedings of the CSEE, 2011,31 (31) .), But owing to the fluctuation near rated value of set end voltage, frequency has randomness, herein for simplifying research, use a constant failure-rate λ2Represent；For the 3rd class outage factor such as anthropic factor and external environment, there is because of it occasionality equally, therefore may be used without one Constant failure-rate λ3Describe.To sum up, generator failure rate model is represented by:\n\n${\\mathrm{λ}}_{\\mathrm{gen}}\\left(t\\right)={\\mathrm{λ}}_{1}\\left(t\\right)+{\\mathrm{λ}}_{2}+{\\mathrm{λ}}_{3}=\\left({\\mathrm{α}}_{1}+{\\mathrm{α}}_{2}{\\mathrm{β}}_{2}{t}^{{\\mathrm{β}}_{2}-1}\\right)+{\\mathrm{λ}}_{2}+{\\mathrm{λ}}_{3}---\\left(1\\right)$\n\nWherein αi≥0、βi>=0 (i=1,2) is scale parameter and the form parameter of i heavily Weibull distribution.\n\nThe failure rate model of 1.2 transformators\n\nTransformator in running by multiple inside and outside factors such as self deterioration, artificial maloperation and false protection Impact, wherein exopathogenic factor has certain randomness, and the contribution to transformer fault rate can use a steady state value λcDescribing, endogenous cause of ill is at any time Between there is accumulative effect, be the key factor causing transformator to be stopped transport.The transformator currently used due to China mostly is oil immersed type Transformator, its main cause lost efficacy is the deterioration of insulating properties, and therefore transformer life can be approximately the transformer insulated life-span, The model that transformer fault rate is contributed by endogenous cause of ill is described, such as formula (2):\n\n${\\mathrm{λ}}_{t}\\left(t\\right)=\\frac{\\mathrm{β}}{\\mathrm{Cexp}\\left(\\frac{B}{{\\mathrm{θ}}_{H}+273}\\right)}·{\\left(\\frac{t}{\\mathrm{Cexp}\\left(\\frac{B}{{\\mathrm{θ}}_{H}+273}\\right)}\\right)}^{\\mathrm{β}-1}---\\left(2\\right)$\n\nWherein β is form parameter, and B, C are empirical value, θHFor hot(test)-spot temperature, t is that transformator is in hot(test)-spot temperature θHUnder continue The operation time.Wherein θHWith transformer load rate K and ambient temperature θARelevant, its computation model and formula can refer to document ([IEEE Working Group for Loading Mineral-Oil-Immersed Transformers.IEEE Std C57.91-1995IEEE guide for loading mineral-oil-immersed transformers, 2012.http://ieeex-plore.ieee.org/xpl/articleDetails.jsp?tp=&arnumber=6168452& contentType=Standards&searchField%3DSearch_All%26queryText%3DIEEE+Std+C57.91- 1995+IEEE+guide+for+loading.).\n\nTo sum up obtain transformer fault rate model such as formula (3):\n\n${\\mathrm{λ}}_{\\mathrm{tran}}\\left(t\\right)={\\mathrm{λ}}_{c}+\\frac{\\mathrm{β}}{\\mathrm{Cexp}\\left(\\frac{B}{{\\mathrm{θ}}_{H}+273}\\right)}·{\\left(\\frac{t}{\\mathrm{Cexp}\\left(\\frac{B}{{\\mathrm{θ}}_{H}+273}\\right)}\\right)}^{\\mathrm{β}-1}---\\left(3\\right)$\n\nThe failure rate model of 1.3 transmission lines of electricity\n\nThe loss of tensile strength is the main cause that transmission line of electricity degradation failure rate increases, and its failure procedure can use β1=1 Double compound Weibull function describes.Additionally, transmission line of electricity belongs to exposed type power equipment, its fault is by ambient weather condition Impact, the vile weather such as thunderstorm, ice and snow can increase the fault rate of power equipment, meter and the transmission line malfunction rate mould of weather conditions Shown in type such as formula (4):\n\n${\\mathrm{λ}}_{\\mathrm{line}}\\left(t\\right)={k}_{1}{\\mathrm{α}}_{1}+{k}_{2}{\\mathrm{α}}_{2}{\\mathrm{β}}_{2}{t}^{{\\mathrm{β}}_{2}-1}---\\left(4\\right)$\n\nk1, k2For the weather factor of influence to transmission line malfunction rate, k under normal weather1=1, k2=1。\n\nThe power equipment of 2 meters and scheduled overhaul is average without validity\n\nFrom Section 1, the general-purpose type of electrical equipment fault rate model can carry out table by constant component and time-varying part sum Reach, as shown in formula (5):\n\n$\\mathrm{λ}\\left(t\\right)={\\mathrm{α}}_{1}^{′}+{\\mathrm{α}}_{2}^{′}{\\mathrm{β}}_{2}{t}^{{\\mathrm{β}}_{2}-1}---\\left(5\\right)$\n\nWhen considering scheduled overhaul, if moment t is in jth to j+1 the time between overhauls(TBO), then under scheduled overhaul, fault rate is:\n\n${\\mathrm{λ}}_{m}\\left(t\\right)={\\mathrm{η}}_{j}{\\mathrm{α}}_{1}^{′}+{\\mathrm{α}}_{2}^{′}{\\mathrm{β}}_{2}{\\left(t-\\mathrm{jT}\\right)}^{{\\mathrm{β}}_{2}-1}---\\left(6\\right)$\n\nIn formula, T is the scheduled overhaul cycle, ηj>=1 is the jth time scheduled overhaul factor of influence to constant fault rate, ηj=1 table Show that scheduled overhaul makes power equipment \" repair as new \".Maintenance can make power equipment \" repair as new \" the most according to plan, overhauls afterwards Make power equipment recover pre-fault status to process, as shown in Figure 2.\n\nCounting the average without validity such as formula (7) Suo Shi of scheduled overhaul week after date power equipment, wherein λ (t) is power equipment Failure rate function, T is the scheduled overhaul cycle, r, and \" for the average scheduled overhaul time, r is average repair time, T the most afterwardsdFor plan Average idle time during maintenance.Wherein T, Td, r, r \" unit be hour, λ unit is time/hour.\n\n$U=\\frac{{T}_{d}}{T}=\\frac{{r}^{′′}+r{∫}_{0}^{T}\\mathrm{λ}\\left(T\\right)\\mathrm{dt}}{T}---\\left(7\\right)$\n\nIf scheduled overhaul rate is λ \", unit is times/year, 1 year based on 365 days, then time between overhauls(TBO) T=8760/ λ \", average nothing Validity is then as shown in formula (8).Wherein λ \" is defined as the scheduled overhaul number of times in the power equipment unit interval (such as 1 year).\n\n$U=\\frac{{\\mathrm{λ}}^{′′}{r}^{′′}+{\\mathrm{λ}}^{′′}r{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}^{′′}}}\\mathrm{λ}\\left(t\\right)\\mathrm{dt}}{8760}---\\left(8\\right)$\n\nIt is average without the incidence relation between validity and failure rate function, scheduled overhaul rate that formula (8) establishes power equipment.\n\nThe electrical equipment fault rate model set up below for Section 1, deriving, it is average without validity.\n\nThe electromotor of 2.1 meters and scheduled overhaul is average without validity\n\nGenerator failure rate functional expression (1) is substituted into formula (8) can obtain:\n\n${U}_{\\mathrm{gen}}=\\frac{{\\mathrm{λ}}^{′′}{r}^{′′}+{\\mathrm{λ}}^{′′}r{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}^{′′}}}\\left({\\mathrm{α}}_{1}+{\\mathrm{α}}_{2}{\\mathrm{β}}_{2}{t}^{{\\mathrm{β}}_{2}-1}+{\\mathrm{λ}}_{2}+{\\mathrm{λ}}_{3}\\right)\\mathrm{dt}}{8760}---\\left(9\\right)$\n\nThe transformator of 2.2 meters and scheduled overhaul is average without validity\n\nThe load factor of transformator takes per day load factor curve, and ambient temperature uses day typical temperature profile, list of references ([IEEE Working Group for Loading Mineral-Oil-Immersed Transformers.IEEE Std C57.91-1995IEEE guide for loading mineral-oil-immersed transformers, 2012.http://ieeex-plore.ieee.org/xpl/articleDetails.jsp?tp=&arnumber=6168452& conte ntType=Standards&searchField%3DSearch_All%26queryText%3DIEEE+Std+ C57.91-1995+IEEE+guide+for+loading.) hot-spot temperature of transformer curve can then be calculated, as shown in Figure 3.\n\nHot(test)-spot temperature data are carried out cluster calculation, obtains one group of typical case hot(test)-spot temperature (θH1H2,…,θHm) and corresponding Probability (P1,P2,…,Pm), this group hot(test)-spot temperature correspondence failure rate function is designated as (λt1(t),λt2(t),…,λtm(t)), wherein λti(t), (i=1 ..., m) expression formula is obtained by formula (3).If scheduled overhaul rate is λ \", then time between overhauls(TBO) T=8760/ λ \" in Expect number of stoppages NTFor\n\n${N}_{T}=\\underset{i=1}{\\overset{m}{\\mathrm{Σ}}}{P}_{i}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}^{′′}}}{\\mathrm{λ}}_{\\mathrm{ti}}\\left(t\\right)\\mathrm{dt}---\\left(10\\right)$\n\nThen in the scheduled overhaul cycle, transformator without validity is averagely\n\n${U}_{\\mathrm{tran}}=\\frac{{r}^{′′}+r{N}_{T}}{8760/{\\mathrm{λ}}^{′′}}=\\frac{{\\mathrm{λ}}^{′′}{r}^{′′}+{\\mathrm{λ}}^{′′}r\\underset{i=1}{\\overset{m}{\\mathrm{Σ}}}{P}_{i}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}^{′′}}}{\\mathrm{λ}}_{\\mathrm{ti}}\\left(t\\right)\\mathrm{dt}}{8760}---\\left(11\\right)$\n\nThe transmission line of electricity of 2.3 meters and scheduled overhaul is average without validity\n\nTwo state synoptic models (normal, vile weather) are used to describe the weather impact on transmission line malfunction rate herein, And to set normal weather to the boisterous rate of transform be μ as λ, vile weather to the rate of transform of normal weather, as shown in Figure 4.\n\nThen in the time between overhauls(TBO), transmission line of electricity without validity is averagely:\n\n${U}_{\\mathrm{line}}=\\frac{{\\mathrm{λ}}^{′′}{r}^{′′}+{\\mathrm{λ}}^{′′}r{N}_{T}}{8760}=\\frac{{\\mathrm{λ}}^{′′}{r}^{′′}+{\\mathrm{λ}}^{′′}r·\\left(\\frac{\\mathrm{μ}}{\\mathrm{λ}+\\mathrm{μ}}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}^{′′}}}\\left({\\mathrm{α}}_{1}+{\\mathrm{α}}_{2}{\\mathrm{β}}_{2}{t}^{{\\mathrm{β}}_{2}-1}\\right)\\mathrm{dt}+\\frac{\\mathrm{λ}}{\\mathrm{λ}+\\mathrm{μ}}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}^{′′}}}\\left({k}_{1}{\\mathrm{α}}_{1}+{k}_{2}{\\mathrm{α}}_{2}{\\mathrm{β}}_{2}{t}^{{\\mathrm{β}}_{2}-1}\\right)\\mathrm{dt}\\right)}{8760}---\\left(12\\right)$\n\n3. the sensitivity algorithm that scheduled overhaul optimizes\n\nThe sensitivity of scheduled overhaul rate is derived by 3.1 each indexs\n\nIf the system of N number of power equipment composition, status of electric power is separate, respectively S1, S2, S3..., SN, Sk=0 Represent that power equipment k is in normal condition, Sk=1 represents that it is in malfunction, then the probability of this system mode x is represented by:\n\nP(x)=P(S1)·P(S2)·P(S3)…P(SN)\n\nThen meter and the system load-loss probability (LOLP) of scheduled overhaul, expected energy not supplied (EENS) such as formula (13), (14) Shown in:\n\n$\\begin{array}{c}\\mathrm{LOLP}=\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)P\\left(x\\right)=\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}P\\left({S}_{k}\\right)\\\\ =\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}\\left[{S}_{k}{U}_{k}+\\left(1-{S}_{k}\\right)\\left(1-{U}_{k}\\right)\\right]\\\\ =\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}\\left[{S}_{k}·\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}+\\left(1-{S}_{k}\\right)\\left(1-\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}\\right)\\right]\\end{array}---\\left(13\\right)$\n\n$\\begin{array}{c}\\mathrm{EENS}=8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)P\\left(x\\right){L}_{C}\\left(x\\right)=8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}P\\left({S}_{k}\\right)\\\\ =8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}\\left[{S}_{k}{U}_{k}+\\left(1-{S}_{k}\\right)\\left(1-{U}_{k}\\right)\\right]\\\\ =8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{\\mathrm{Π}}}\\left[{S}_{k}·\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}+\\left(1-{S}_{k}\\right)\\left(1-\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}\\right)\\right]\\end{array}---\\left(14\\right)$\n\nWherein, λk\" for the scheduled overhaul rate of power equipment k, LCX () represents under system fault condition x for making system recovery arrive Minimum load reduction required for one static security operating point, IfX ()=0 represents system normal condition, IfX ()=1 represents system System malfunction.\n\nThen Reliability Index LOLP, EENS relative to the sensitivity of scheduled overhaul rate respectively as shown in formula (15)-(16):\n\n$\\begin{array}{c}\\frac{∂\\mathrm{LOLP}}{{∂\\mathrm{λ}}_{k}^{′′}}=\\underset{x∈X}{\\mathrm{Σ}}\\left[\\frac{∂{I}_{f}\\left(x\\right)}{{∂\\mathrm{λ}}_{k}^{′′}}P\\left(x\\right)+{I}_{f}\\left(x\\right)\\frac{∂P\\left(x\\right)}{{∂\\mathrm{λ}}_{k}^{′′}}\\right]\\\\ =\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)\\frac{∂\\left[P\\left({S}_{1}\\right)P\\left({S}_{2}\\right)···P\\left({S}_{k}\\right)···P\\left({S}_{m}\\right)\\right]}{{∂\\mathrm{λ}}_{k}^{′′}}\\\\ =\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right)P\\left(x\\right)\\frac{\\left({2S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\end{array}---\\left(15\\right)$\n\n$\\begin{array}{c}\\frac{∂\\mathrm{EENS}}{{∂\\mathrm{λ}}_{K}^{′′}}=8760\\underset{x∈X}{\\mathrm{Σ}}\\left[\\frac{∂{I}_{f}\\left(x\\right)}{{∂\\mathrm{λ}}_{j}^{′′}}{L}_{C}\\left(x\\right)P\\left(x\\right)+{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\frac{∂P\\left(x\\right)}{{∂\\mathrm{λ}}_{k}^{′′}}\\right]\\\\ =8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\frac{∂\\left[P\\left({S}_{1}\\right)P\\left({S}_{2}\\right)···P\\left({S}_{k}\\right)···P\\left({S}_{m}\\right)\\right]}{{∂\\mathrm{λ}}_{k}^{′′}}\\\\ =8760\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)P\\left(x\\right)\\frac{\\left({2S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\end{array}---\\left(16\\right)$\n\nCalculating LOLP and EENS is to reflect that system reliability level is with plan relative to the sensitivity of scheduled overhaul rate The Changing Pattern that maintenance rate adjusts；Additionally, EENS relative to scheduled overhaul rate sensitivity by derivation system totle drilling cost to based on Draw maintenance rate sensitivity formula.\n\nThe cost of overhaul of electric power system all includes two large divisions, i.e. fee of material and construction cost, wherein fee of material Relevant with the type of power equipment, capacity.If power equipment k (k=1,2 ..., N) carry out 1 time afterwards the cost of overhaul be CRk, including Fee of material CRMk, the operating expenses C of unit intervalROk/h；Carrying out 1 scheduled overhaul expense is CPk, including fee of material CPMk, unit Time operating expenses CPOk/h；Unit power failure cost is set to CN/ WMh, the total cost of system is set to Ctotal\n\nThen this expression formula of system synthesis such as formula (17), wherein NRkRepresent that number of times is overhauled in the every annual of power equipment k afterwards, NpkRepresent power equipment k annual scheduled overhaul number of times, CRk=CRMk+CROk×rk, CPk=CPMk+CPOk×rk″。\n\n$\\begin{array}{c}{C}_{\\mathrm{total}}={C}_{N}·\\mathrm{EENS}+\\underset{k}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Rk}}·{N}_{\\mathrm{Rk}}+\\underset{k}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Pk}}·{N}_{\\mathrm{Pk}}\\\\ =8760{C}_{N}\\underset{x∈X}{\\mathrm{Σ}}\\left[{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\right]P\\left(x\\right)+\\\\ \\underset{k=1}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Rk}}{\\mathrm{λ}}_{k}^{′′}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}+\\underset{k=1}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Pk}}{\\mathrm{λ}}_{k}^{′′}\\end{array}---\\left(17\\right)$\n\nAs it is assumed that the state of each power equipment is separate, then the number of times of maintenance afterwards of power equipment k and meter in 1 year Draw scheduled overhaul rate λ of maintenance number of times and power equipment kk\" relevant, then have\n\n$\\left\\{\\begin{array}{c}\\frac{{∂N}_{\\mathrm{pk}}}{{∂\\mathrm{λ}}_{k}^{′′}}=1\\\\ \\frac{{∂N}_{\\mathrm{Rk}}}{{∂\\mathrm{λ}}_{k}^{′′}}={∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}-\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}{\\mathrm{λ}}_{k}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\\\ \\frac{{∂N}_{\\mathrm{pk}}}{{∂\\mathrm{λ}}_{i}^{′′}}=0,\\left(i≠k\\right)\\\\ \\frac{{∂N}_{\\mathrm{Rk}}}{{∂\\mathrm{λ}}_{i}^{′′}}=0,\\left(i≠k\\right)\\end{array}\\right\\---\\left(18\\right)$\n\nThus obtaining system total cost relative to scheduled overhaul rate sensitivity is:\n\n$\\begin{array}{c}\\frac{{∂C}_{\\mathrm{total}}}{{∂\\mathrm{λ}}_{k}^{′′}}={C}_{N}·\\frac{∂\\mathrm{EENS}}{{∂\\mathrm{λ}}_{k}^{′′}}+\\underset{i=1}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Ri}}·\\frac{{∂N}_{\\mathrm{Ri}}}{{∂\\mathrm{λ}}_{k}^{′′}}+\\underset{i=1}{\\overset{N}{\\mathrm{Σ}}}{C}_{\\mathrm{Pi}}·\\frac{{∂N}_{\\mathrm{Pi}}}{{∂\\mathrm{λ}}_{k}^{′′}}\\\\ =8760{C}_{N}\\underset{x∈X}{\\mathrm{Σ}}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)P\\left(x\\right)\\frac{\\left({2S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\\\ +{C}_{\\mathrm{Rk}}·\\left({∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)\\mathrm{dt}-\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}·{\\mathrm{λ}}_{k}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)+{C}_{\\mathrm{Pk}}\\end{array}---\\left(19\\right)$\n\n3.2 scheduled overhauls based on sensitivity sequence optimize heuristic iterative algorithm\n\nWith power outage cost and cost of overhaul sum (i.e. totle drilling cost) minimum object function, based on electromotor, transformation Device and transmission line of electricity average without validity model and sensitive analysis, the scheduled overhaul cycle of power equipment each to system is carried out Optimize.This algorithm has only carried out primary system state and has chosen and cut accordingly carry calculation, and reliability index calculates directly logical thereafter Cross analytical expression to solve, it is to avoid iterative process is repeated reliability assessment, and uses straight during cutting carry calculation Stream trend optimum cuts load model, has saved the substantial amounts of calculating time.Each power equipment is to utilize sensitivity to judge intuitively No it is in the optimal plan time between overhauls(TBO), otherwise adjusts its scheduled overhaul cycle, Electric Power Network Planning is had the directive significance of reality.Tool Body algorithm flow is as follows, referring also to Fig. 1 and Fig. 5:\n\n(1) the system low order malfunction in the range of using State enumeration method to enumerate power system certain fault exponent number, and Based on DC power flow optimum cut load model (Zhao Yuan, Zhou Jiaqi, Liu Yang. the optimum in composite power systems reliability assessment Load cuts down model analysis. electric power network technique, and 2004,28 (10): 34-37), the system minimum calculated under each enumeration state cuts load Amount, if cutting loading more than 0, then that records this system mode and correspondence cuts loading, finally gives a group system fault shape State (X1,X2,…,Xm) and correspondence cut loading (LC1,LC2,…,LCm), wherein m is the system fault condition enumerated Number.\n\n(2) each electromotor, transformator, the parameter of transmission line malfunction rate model and each power equipment reparation in input system Time, original plan maintenance rate.\n\n(3) according to scheduled overhaul rate and the failure rate function of each Electrical equipment, ask for by formula (9), (11), (12) It is average without validity, and based on averagely without the probability of the validity each system fault condition of calculating, by formula (13), that (14) try to achieve system is reliable Property index LOLP, EENS.\n\n(4) according to unit power failure cost, the loss of outage of unit calculation of maintenance cost system and the cost of overhaul, system is obtained Total cost, calculates the sensitivity relative to each power equipment scheduled overhaul rate of the system total cost by formula (19).\n\n(5) by descending for sensitivity sequence, if the door that the sensitivity absolute value of all power equipments is both less than preset Threshold value esp, then scheduled overhaul Optimized Iterative algorithm terminates, otherwise then carries out following process: set the power equipment of serial number x through spirit After sensitivity sequence, sequence number becomes x ', if meeting x '≤nset, and sensitivity l (x) meets l (x) > esp, then make λ \" (x)=0.99 λ \" (x), if x ' >=N-nset, and l (x) <-esp, then making λ \" (x)=1.01 λ \" (x), all power equipments go to step after being disposed (3)。\n\nN is the power equipment quantity of system, nsetFor setting sequence number, represent to preset every time and carry out the scheduled overhaul cycle and tune up Or the power equipment quantity turned down.Esp is sensitivity threshold value.\" (x) is the power equipment scheduled overhaul rate of serial number x to λ.This Invent by calculating system synthesis originally relative to the scheduled overhaul rate sensitivity of each equipment of electrical network, carry out level of sensitivity and differentiate electricity Net whether each equipment is in the optimal plan time between overhauls(TBO), if relative to certain facilities plan maintenance rate sensitivity be bigger on the occasion of, Show that this equipment is in deficient inspecting state, otherwise if the bigger negative value of absolute value, then show that this equipment was in inspecting state. The method optimizing and revising each facilities plan time between overhauls(TBO), is i.e. to judge the sensitivity of each power equipment in electrical network according to this sensitivity Size, the suitable plan for adjustment time between overhauls(TBO), finally realizing electrical network each facilities plan time between overhauls(TBO) all reaches optimum.\n\nThe above embodiment of the present invention is only for example of the present invention is described, and is not the enforcement to the present invention The restriction of mode.For those of ordinary skill in the field, can also be made other not on the basis of the above description Change and variation with form.Here cannot all of embodiment be given exhaustive.Every belong to technical scheme That is amplified out obviously changes or changes the row still in protection scope of the present invention.\n\n## Claims (2)\n\n1. meter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate, it is characterised in that: concrete steps are such as Under:\n1) the system low order malfunction in the range of using State enumeration method to enumerate power system certain fault exponent number, and based on directly The system minimum that stream trend optimum is cut under the load model each enumeration state of calculating cuts loading；If cutting loading to be more than 0, then record This system mode lower and correspondence cut loading, otherwise continue to enumerate, until enumerating end, thus finally give a group system therefore Barrier state (X1,X2,…,Xm) and correspondence cut loading (LC1,LC2,…,LCm), wherein m is the system failure shape enumerated State number；\n2) according to power equipment self deterioration, external environment and maintenance factor, the impact of its fault rate is built meter and scheduled overhaul The power equipment time-varying failure rate model in cycle；Described power equipment is electromotor, transformator and transmission line of electricity；This external environment Including weather and temperature；Obtain the parameter of each power equipment time-varying failure rate model in network system；And input each power equipment Repair time, original plan maintenance rate；\n3) according to scheduled overhaul rate, repair time and the failure rate function of each power equipment, each power equipment is calculated the most invalid Degree, based on the probability averagely calculating each system fault condition without validity；Further according to reliability index computing formula, electric power is utilized to set Standby average without validity and the 1st) system fault condition (X that obtains of step1,X2,…,Xm) and correspondence cut loading (LC1, LC2,…,LCm) calculating Reliability Index, Reliability Index includes that system load-loss probability LOLP and system expectation lack Delivery EENS；\n4) system blackout cost is calculated, according to each power equipment unit plan according to unit power failure cost and Reliability Index The cost of overhaul and trouble shooting cost combine the 3rd) the scheduled overhaul rate of step and failure rate function obtain system overhaul cost, system Power failure cost and system overhaul cost collectively form system synthesis originally；Set up system synthesis letter originally and between the scheduled overhaul cycle Number relation, utilizes this functional relationship to ask for this sensitivity to scheduled overhaul rate of system synthesis；\n5) by descending for sensitivity sequence, if the threshold value that the sensitivity absolute value of all power equipments is both less than preset Esp, then the scheduled overhaul cycle that this sensitivity is corresponding is the scheduled overhaul cycle finally determined；Otherwise plan for adjustment maintenance week Phase, until system synthesis, this is less than, to the sensitivity of scheduled overhaul rate, the threshold value esp preset, and meets the scheduled overhaul of this requirement Cycle is the scheduled overhaul cycle finally determined；\n4th) according to unit power failure cost and system, step system blackout cost expects that lacking delivery EENS calculates, and system overhaul includes Maintenance and scheduled overhaul two kinds afterwards；The cost of overhaul includes fee of material and construction cost two large divisions, wherein fee of material and electricity every time The type of power equipment, capacity are relevant；If power equipment k (k=1,2 ..., N) to carry out the cost of overhaul the most afterwards be CRk, including material Material takes CRMk, the operating expenses C of unit intervalROk/h；Carrying out a scheduled overhaul expense is CPk, including fee of material CPMk, unit time Between operating expenses CPOk/h；Unit power failure cost is set to CN/ MWh, the total cost of system is set to Ctotal\nThen system synthesis functional relationship such as following formula originally and between the scheduled overhaul cycle,\n$\\begin{array}{c}{C}_{total}={C}_{N}·EENS+\\underset{k}{\\overset{N}{Σ}}{C}_{Rk}·{N}_{Rk}+\\underset{k}{\\overset{N}{Σ}}{C}_{Pk}·{N}_{Pk}\\\\ =8760{C}_{N}\\underset{x∈X}{Σ}[{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)]P\\left(x\\right)+\\\\ \\underset{k=1}{\\overset{N}{Σ}}{C}_{Rk}{\\mathrm{λ}}_{k}^{′′}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt+\\underset{k=1}{\\overset{N}{Σ}}{C}_{Pk}{\\mathrm{λ}}_{k}^{′′}\\end{array}$\nWherein NRkRepresent that number of times, N are overhauled in the every annual of power equipment k afterwardspkRepresent power equipment k annual scheduled overhaul number of times, λkT () is failure rate function；CRk=CRMk+CROk×rk, CPk=CPMk+CPOk×rk\", rkFor the average repair time afterwards；rk\" it is The average scheduled overhaul time；\nOwing to the state of each power equipment is separate, then in 1 year, the number of times of maintenance afterwards and the scheduled overhaul of power equipment k are secondary Several with scheduled overhaul rate λ of power equipment kk\" relevant, then have\n$\\left\\{\\begin{array}{c}\\frac{∂{N}_{pk}}{∂{\\mathrm{λ}}_{k}^{′′}}=1\\\\ \\frac{∂{N}_{Rk}}{∂{\\mathrm{λ}}_{k}^{′′}}={∫}_{0}^{\\frac{87600}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt-\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}{\\mathrm{λ}}_{k}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\\\ \\begin{array}{cc}\\frac{∂{N}_{pk}}{∂{\\mathrm{λ}}_{i}^{′′}}=0,& \\left(i≠k\\right)\\end{array}\\\\ \\begin{array}{cc}\\frac{∂{N}_{Rk}}{∂{\\mathrm{λ}}_{i}^{′′}}=0,& \\left(i≠k\\right)\\end{array}\\end{array}\\right\\$\nThis relative to the scheduled overhaul rate sensitivity of each equipment of electrical network is thus to obtain system synthesis:\n$\\begin{array}{c}\\frac{∂{C}_{total}}{∂{\\mathrm{λ}}_{k}^{′′}}={C}_{N}·\\frac{∂EENS}{∂{\\mathrm{λ}}_{k}^{′′}}+\\underset{i=1}{\\overset{N}{Σ}}{C}_{Ri}·\\frac{∂{N}_{Ri}}{∂{\\mathrm{λ}}_{k}^{′′}}+\\underset{i=1}{\\overset{N}{Σ}}{C}_{Pi}·\\frac{∂{N}_{Pi}}{∂{\\mathrm{λ}}_{k}^{′′}}\\\\ =8760{C}_{N}\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)P\\left(x\\right)\\frac{\\left(2{S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\\\ +{C}_{Rk}·\\left({∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt-\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}·{\\mathrm{λ}}_{k}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)+{C}_{Pk}\\end{array};$\n3rd) step system load-loss probability LOLP and the scarce delivery EENS of system expectation is tried to achieve by following process, if N number of electric power sets The system of standby composition, each status of electric power is separate, respectively S1, S2, S3..., SN, Sk=0 represents that power equipment k is in Normal condition, Sk=1 represents that it is in malfunction, then the probability of this system mode x is represented by:\nP (x)=P (S1)·P(S2)·P(S3)…P(SN)\nSo, system load-loss probability LOLP and the scarce delivery EENS computing formula of system expectation are as follows；\n$\\begin{array}{c}LOLP=\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right)P\\left(x\\right)=\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{Π}}P\\left({S}_{k}\\right)\\\\ =\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{Π}}[{S}_{k}{U}_{k}+\\left(1-{S}_{k}\\right)\\left(1-{U}_{k}\\right)]\\\\ =\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right)\\underset{k=1}{\\overset{N}{Π}}[{S}_{k}·\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt}{8760}+\\left(1-{S}_{k}\\right)\\left(1-\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt}{8760}\\right)]\\end{array}$\n$\\begin{array}{c}EENS=8760\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right)P\\left(x\\right){L}_{C}\\left(x\\right)=8760\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{Π}}P\\left({S}_{k}\\right)\\\\ =8760\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{Π}}[{S}_{k}{U}_{k}+\\left(1-{S}_{k}\\right)\\left(1-{U}_{k}\\right)]\\\\ =8760\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\underset{k=1}{\\overset{N}{Π}}[{S}_{k}·\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}+{\\mathrm{λ}}_{k}^{′′}{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt}{8760}+\\left(1-{S}_{k}\\right)\\left(1-\\frac{{\\mathrm{λ}}_{k}^{′′}{r}_{k}^{′′}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt}{8760}\\right)]\\end{array}$\nWherein, λk\" for the scheduled overhaul rate of power equipment k, LCX () represents under system fault condition x for making system recovery to Minimum load reduction required for static security operating point, IfX ()=0 represents system normal condition, IfX ()=1 represents system Malfunction；\nThen Reliability Index system load-loss probability LOLP lacks delivery EENS relative to scheduled overhaul rate with system expectation Sensitivity is shown below respectively:\n$\\begin{array}{c}\\frac{∂LOLP}{∂{\\mathrm{λ}}_{k}^{′′}}=\\underset{x∈X}{Σ}[\\frac{∂{I}_{f}\\left(x\\right)}{∂{\\mathrm{λ}}_{k}^{′′}}P\\left(x\\right)+{I}_{f}\\left(x\\right)\\frac{∂P\\left(x\\right)}{∂{\\mathrm{λ}}_{k}^{′′}}]\\\\ =\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right)\\frac{∂[P\\left({S}_{1}\\right)P\\left({S}_{2}\\right)...P\\left({S}_{k}\\right)...P\\left({S}_{m}\\right)]}{∂{\\mathrm{λ}}_{k}^{′′}}\\\\ =\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right)P\\left(x\\right)\\frac{\\left(2{S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\end{array}$\n$\\begin{array}{c}\\frac{∂EENS}{∂{\\mathrm{λ}}_{k}^{′′}}=8760\\underset{x∈X}{Σ}[\\frac{∂{I}_{f}\\left(x\\right)}{∂{\\mathrm{λ}}_{k}^{′′}}{L}_{C}\\left(x\\right)P\\left(x\\right)+{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\frac{∂P\\left(x\\right)}{∂{\\mathrm{λ}}_{k}^{′′}}]\\\\ =8760\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)\\frac{∂[P\\left({S}_{1}\\right)P\\left({S}_{2}\\right)...P\\left({S}_{k}\\right)...P\\left({S}_{m}\\right)]}{∂{\\mathrm{λ}}_{k}^{′′}}\\\\ =8760\\underset{x∈X}{Σ}{I}_{f}\\left(x\\right){L}_{C}\\left(x\\right)P\\left(x\\right)\\frac{\\left(2{S}_{k}-1\\right)}{P\\left({S}_{k}\\right)}·\\left(\\frac{{r}_{k}^{′′}+{r}_{k}{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}}{\\mathrm{λ}}_{k}\\left(t\\right)dt}{8760}-\\frac{{r}_{k}}{{\\mathrm{λ}}_{k}^{′′}}·\\mathrm{λ}\\left(\\frac{8760}{{\\mathrm{λ}}_{k}^{′′}}\\right)\\right)\\end{array},$\nWherein UkReferring to count the average without validity of scheduled overhaul week after date power equipment, computing formula is\n$U=\\frac{{\\mathrm{λ}}^{′′}{r}^{′′}+{\\mathrm{λ}}^{′′}r{∫}_{0}^{\\frac{8760}{{\\mathrm{λ}}^{′′}}}\\mathrm{λ}\\left(t\\right)\\mathrm{dt}}{8760}.$\nThe electrical network scheduled overhaul cycle determination method of meter the most according to claim 1 and power equipment time-varying fault rate, its It being characterised by: wherein the 5th) scheduled overhaul cycle of step is adjusted as follows: set the power equipment of serial number x through sensitive After degree sequence, sequence number becomes x ', if meeting x '≤nset, and sensitivity l (x) meets l (x) > esp, then make the electric power of serial number x set Standby scheduled overhaul rate λ \" (x)=0.99 λ \" (x)；If x ' >=N-nset, and l (x) <-esp, then make λ \" (x)=1.01 λ \" (x), base Obtain the new scheduled overhaul cycle in the power equipment scheduled overhaul rate redefined, thus obtain system synthesis originally relative to newly Scheduled overhaul rate sensitivity l (x), then judge sensitivity l (x) whether less than the threshold value esp preset, if it is satisfied, then adjust The scheduled overhaul cycle after whole is the scheduled overhaul cycle finally determined, otherwise repeats step 3)-5), until sensitivity l (x) Less than the threshold value esp preset；\nN is the power equipment quantity of system, nsetFor setting sequence number, represent to preset every time and carry out the scheduled overhaul cycle and tune up or adjust Little power equipment quantity；Esp is sensitivity threshold value；\" (x) is the power equipment scheduled overhaul rate of serial number x to λ.\nCN201310746200.7A 2013-12-30 2013-12-30 Meter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate CN103646358B (en)\n\n## Priority Applications (1)\n\nApplication Number Priority Date Filing Date Title\nCN201310746200.7A CN103646358B (en) 2013-12-30 2013-12-30 Meter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate\n\n## Applications Claiming Priority (1)\n\nApplication Number Priority Date Filing Date Title\nCN201310746200.7A CN103646358B (en) 2013-12-30 2013-12-30 Meter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate\n\n## Publications (2)\n\nPublication Number Publication Date\nCN103646358A CN103646358A (en) 2014-03-19\nCN103646358B true CN103646358B (en) 2016-08-17\n\n# Family\n\n## Family Applications (1)\n\nApplication Number Title Priority Date Filing Date\nCN201310746200.7A CN103646358B (en) 2013-12-30 2013-12-30 Meter and the electrical network scheduled overhaul cycle determination method of power equipment time-varying fault rate\n\n## Country Status (1)\n\nCN (1) CN103646358B (en)\n\n## Families Citing this family (10)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nCN104484732B (en) * 2014-11-07 2017-07-07 湘潭大学 A kind of wind power plant Awaiting Parts choosing method of meter and unit generation amount\nCN105741005A (en) * 2014-12-12 2016-07-06 通用电气公司 Method and system for optimizing part maintenance plan\nCN104537437A (en) * 2014-12-19 2015-04-22 广东电网有限责任公司珠海供电局 Power equipment state maintaining predicting method based on genetic algorithm\nCN104835000A (en) * 2015-04-28 2015-08-12 国网上海市电力公司 Power distribution network reliability assessment method taking regard of pre-arrangement of power failure\nCN104850962A (en) * 2015-06-01 2015-08-19 国网上海市电力公司 Power distribution network reliability evaluation method with consideration of power failure pre arrangement\nCN104866974A (en) * 2015-06-01 2015-08-26 国网上海市电力公司 Power distribution network reliability assessing equipment considering pre-arranged power cut\nCN105680442B (en) * 2016-03-07 2018-06-15 重庆大学 Consider that the expectation of trend and sensitivity consistency equivalence lacks power supply volume appraisal procedure\nCN106786597B (en) * 2016-12-26 2019-04-02 国网山东省电力公司泰安供电公司 The generation method and device of electric network fault correcting strategy\nCN109583595A (en) * 2017-09-27 2019-04-05 上海电气电站设备有限公司 A kind of method of determining thermal power steam turbine overhaul life and its influence factor\nCN109711635B (en) * 2019-01-08 2020-10-27 北京交通大学 Equipment maintenance strategy optimization method based on station capacity maintenance\n\n## Citations (2)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nCN102360467A (en) * 2011-10-12 2012-02-22 西安交通大学 Method for generating long-term maintenance strategy of electric distribution system in view of ageing properties of element\nCN103426034A (en) * 2013-08-07 2013-12-04 何理 Groundwater remediation system optimizing analysis method\n\n## Patent Citations (2)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nCN102360467A (en) * 2011-10-12 2012-02-22 西安交通大学 Method for generating long-term maintenance strategy of electric distribution system in view of ageing properties of element\nCN103426034A (en) * 2013-08-07 2013-12-04 何理 Groundwater remediation system optimizing analysis method\n\n## Non-Patent Citations (3)\n\n* Cited by examiner, † Cited by third party\nTitle\nCircuit breakers maintenance planning for composite power systems;Morteza Abbasghorbani et al.;《IET Gener. Transm. Distrib.》;20131003;第7卷(第10期);第1135-1143页 *\n\n## Also Published As\n\nPublication number Publication date\nCN103646358A (en) 2014-03-19\n\n## Similar Documents\n\nPublication Publication Date Title\nZhang et al. A convex model of risk-based unit commitment for day-ahead market clearing considering wind power uncertainty\nMadani et al. Distribution automation strategies challenges and opportunities in a changing landscape\nDiao et al. Decision tree-based online voltage security assessment using PMU measurements\nUS10025337B2 (en) Method and system for managing an electrical distribution system in a facility\nHines et al. Trends in the history of large blackouts in the United States\nCN102255307B (en) Layering equivalence method for reliability estimation of distribution network\nCN102368610B (en) Evaluation method based on distribution system security region\nCN102566435B (en) Performance prediction and fault alarm method for photovoltaic power station\nCanova et al. 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Risk assessment of catastrophic failures in electric power systems\nCN102005736B (en) On-line monitoring method of state of relay protection equipment\nCN101640418B (en) Identification method of transient state stable key transmission cross-section of electric power system fault\nCN102663240B (en) Analysis system and evaluation method for power telecommunication business risks\nCN102545210B (en) Method and system for optimizing and regulating distribution network power supply reliability indexes\nAmirioun et al. Metrics and quantitative framework for assessing microgrid resilience against windstorms\nCN104537487B (en) A kind of appraisal procedure of power transmission and transforming equipment operation state risk\nWiget et al. Probabilistic security constrained optimal power flow for a mixed HVAC and HVDC grid with stochastic infeed\nCN103971024A (en) Method for evaluating reliability of relaying protection systems under small sample failure data\n\n## Legal Events\n\nDate Code Title Description\nPB01 Publication\nC10 Entry into substantive examination\nSE01 Entry into force of request for substantive examination\nC14 Grant of patent or utility model\nGR01 Patent grant\nCF01 Termination of patent right due to non-payment of annual fee\n\nGranted publication date: 20160817\n\nTermination date: 20161230\n\nCF01 Termination of patent right due to non-payment of annual fee"
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https://plantmethods.biomedcentral.com/articles/10.1186/s13007-019-0422-z
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"# Bayesian optimization for seed germination\n\n## Abstract\n\n### Background\n\nEfficient seed germination is a crucial task at the beginning of crop cultivation. Although boundaries of environmental parameters that should be maintained are well studied, fine-tuning can significantly improve the efficiency, which is infeasible to be done manually due to the high dimensionality of the parameter space.\n\n### Results\n\nTraditionally seed germination is performed in climatic chambers with controlled environmental conditions. In this study, we perform a set of multiple-day seed germination experiments in the controllable environment. We use up to three climatic chambers to adjust humidity, temperature, water supply and apply machine learning algorithm called Bayesian optimization (BO) to find the parameters that improve seed germination. Experimental results show that our approach allows to increase the germination efficiency for different types of seeds compared to the initial expert knowledge-based guess.\n\n### Conclusion\n\nOur experiments demonstrated that BO could help to identify the values of the controllable parameters that increase seed germination efficiency. The proposed methodology is model-free, and we argue that it may be useful for a variety of optimization problems in precision agriculture. Further experimental studies are required to investigate the effectiveness of our approach for different seed cultures and controlled parameters.\n\n## Introduction\n\nSeed germination has been an interesting subject of study for many years. On the one hand, it is the topic for basic research since many biochemical processes occur during dormancy and different stages of seed germination. On the other hand, the problem is also of great practical importance: finding the optimal parameters such as substrate material, amount of water supply, air temperature, the proportion of plant growth promoters, etc. is a challenging task. Seed germination comprises many processes, and relationships of factors affecting termination of seed dormancy are very diverse. For example, the aforementioned water and temperature combined with light and nitrate level influence seed germination, however, their effect does depend on the level of dormancy of the seeds .\n\nThe problem becomes even more challenging when multiple parameters must be considered together, and specific sets of parameters are supposed to be optimized for each time step. Dynamic models of seed germination have been developed [1,2,3] to address this issue. These models may be helpful in understanding the underlying processes of seed germination. However, to achieve satisfactory optimization results using model-based techniques, comprehensive prior knowledge of the problem structure is required . Moreover, particular dynamic models may not be appropriate for the specific conditions that these models were not developed for, e.g., different plant species, substrates or growth stimulators.\n\nA more adaptive approach, based on machine learning (ML) methods, seems to be promising to tackle this issue. Among those methods the Bayesian optimization (BO) [5, 6] algorithm based on the Gaussian process regression (GPR) is one of the most attractive. It is a black-box optimization algorithm that does not require knowledge of the system intrinsics. It is widely used in the ML community for hyperparameter optimization and was even successfully applied in culinary arts . Similarly, an approach based on Genetic Algorithms and GPR has been previously proposed for precision agriculture .\n\nIn this paper, we apply BO to simplified seed germination process in the controllable environment in order to identify the values of the controlled parameters that yield the best germination efficiency. First, we select the number of tunable parameters that we can control during the germination period (several days) with the help of climatic chambers, e.g., humidity, temperature, amount of water supply provided and choose the reasonable bounds for these parameters based on the expert knowledge. Then, we iteratively apply BO algorithm, to find the values of parameters that maximize the number of germinated seeds. We show that starting with an initial expert knowledge-based guess our approach allows to find such values of parameters that yield solid improvement both when initial germination efficiency is low (first experiment) and high (second experiment).\n\n## Materials and methods\n\nIn this section, we describe the methodology and the algorithms used to build our framework. Figure 1 shows a schematic overview of the proposed system.\n\n### Seed germination\n\nWe conducted two experiments, first, using pea seeds (Pisum sativum L.) and, second, using radish seeds (Raphanus sativa L.) in different settings. Seeds were purchased from Federal Scientific Center of Vegetable (Odintsovo, Russia). The weight of 100 seeds showed an average of $$0.751 \\pm 0.01 \\, {\\hbox {g}}$$ for radish, and of $$19.95 \\pm 1.31 \\, {\\hbox {g}}$$ for pea. All seeds were presterilized in 0.5% of KMnO4 solution for 10 min and then rinsed for several times with deionized water. Three climatic chambers (Binder KBWF 240, KBF 240, KMF 240) allowed to control air temperature ($$\\pm 0.1 \\, ^{\\circ }{\\hbox {C}}$$) and humidity ($$\\pm \\, 1\\%$$), which was maintained at 80%. No light sources were used in the chambers during the experiments.\n\nThe first experiment was conducted in the form of sequential trials with each trial comprising three concurrent germination processes and lasting for 72 h (3 days in total). One hundred pea seeds were placed on a dish covered with sterile cheesecloth and put in each of the three climate chambers to germinate. Totally, 7 controllable parameters were selected: air temperature and the amount of water supplied at 0, 24, 48, 72 and 0, 24, 48 h steps, respectively. The temperature in the chambers was changed smoothly between the selected values during the trials.\n\nDuring the second experiment, only two climatic chambers were used (KBF 240, KMF 240) to set 4 controllable parameters, namely temperatures at 0, 12, 24, 36 h. Seeds were placed in containers of size $$21 \\times 15.5 \\times 0.8 \\, {\\hbox {cm}}$$ with two sections (each accommodating 16 seeds) on the cloth and watered once at the beginning of a trial with a fixed amount of 6 ml. Figure 2 depicts a single container at the beginning (left) and the end (right) of a trial.\n\nThese containers, then, were grouped by 3, giving 96 seeds in a group. Three such groups then were placed almost vertically in each of two climatic chambers with the same controllable parameters set, thus, for each trial giving 6 repetitions with a total amount of seeds equal to 96 in each of them. Figure 3 shows how containers with seeds were installed in the chambers during the second experiment.\n\nAfter the seeds were germinated, the number of germinated and well-germinated seeds were counted in each chamber. In the first experiment, we considered the seeds germinated when only the radicle emerged and could be visibly separated from the seed. If not only radicle but also the hypocotyl emerged and could be visibly separated, the seed was classified as well-germinated. For the second experiment, we considered seeds germinated if radicle emerged and its length is less than 17.5 mm, and well-germinated if it is larger. Figure 4 shows an example of not germinated (left), germinated (middle) and well-germinated (right) radish seeds according to our methodology.\n\n### Bayesian optimization framework\n\nIn this section, we describe the Bayesian optimization framework based on the Gaussian process regression that we used in our work.\n\n#### Gaussian process regression\n\nBayesian optimization relies on the Gaussian Process Regression , also called kriging in geostatistics, which learns a generative probabilistic model of an arbitrary function of independent variables with the assumption of normality. A Gaussian process is completely determined by its mean $$\\mu (\\cdot )$$ and covariance (kernel) $$k(\\cdot , \\cdot )$$ functions:\n\n\\begin{aligned} f({\\mathbf {x}}) & \\sim {\\mathcal {GP}}\\left( m({\\mathbf {x}}), k\\left( {\\mathbf {x}}, {\\mathbf {x}}^\\prime \\right) \\right) , \\\\ m({\\mathbf {x}}) &= {\\mathbb {E}}~f({\\mathbf {x}}), \\\\ k({\\mathbf {x}},{\\mathbf {x}}^\\prime ) & = {\\mathbb {E}}~\\left[ \\left( f\\left( {\\mathbf {x}}\\right) - m\\left( {\\mathbf {x}}\\right) \\right) \\left( f\\left( {\\mathbf {x}}^\\prime \\right) - m\\left( {\\mathbf {x}}^\\prime \\right) \\right) \\right] , \\end{aligned}\n\nwhere $${\\mathbf {x}}\\in {\\mathbb {R}}^d$$ is a vector of d input parameters.\n\nLet consider the GP model with an additive normal noise:\n\n\\begin{aligned} y({\\mathbf {x}}) = f({\\mathbf {x}}) + \\epsilon , \\end{aligned}\n(1)\n\nwhere $$\\epsilon \\sim {\\mathcal {N}}(0, \\sigma ^2)$$. Given the training data $${\\mathbf {X}}=\\left( {\\mathbf {x}}_1,\\ldots ,{\\mathbf {x}}_n\\right) ^\\intercal \\in {\\mathbb {R}}^{n \\times d}$$, $${\\mathbf {y}}=\\left( y_1,\\ldots ,y_n\\right) ^\\intercal \\in {\\mathbb {R}}^n$$, where n is the number of available measurements and $$(\\cdot )^\\intercal$$ denotes the transpose, the predictive distribution at an unobserved point $${\\mathbf {x}}^*$$ is given by\n\n\\begin{aligned} f^* & \\sim {\\mathcal {N}}\\left( {\\hat{\\mu }}, {\\hat{\\sigma }}^2\\right) , \\\\ {\\hat{\\mu }}({\\mathbf {x}}^*) & = m({\\mathbf {x}}^*) + K({\\mathbf {x}}^*, {\\mathbf {X}})[K({\\mathbf {X}},{\\mathbf {X}}) + \\sigma ^2 I] ({\\mathbf {y}}- m({\\mathbf {X}})), \\\\ {\\hat{\\sigma }}^2({\\mathbf {x}}^*) &= k({\\mathbf {x}}^*, {\\mathbf {x}}^*) - K({\\mathbf {x}}^*, {\\mathbf {X}})[K({\\mathbf {X}},{\\mathbf {X}}) + \\sigma ^2I]^{-1}K({\\mathbf {X}}, {\\mathbf {x}}^*), \\end{aligned}\n\nwhere $$K({\\mathbf {X}}, {\\mathbf {X}})$$ is a matrix of the form $$K_{ij} = k({\\mathbf {x}}_i, {\\mathbf {x}}_j), i,j=1,\\ldots ,n$$. Particular choice of the kernel function depends on the assumptions about the model and a particular application, however, there exist commonly used kernels, such as Radial basis function (RBF) and Mateŕn that work well in general. Kernel hyperparameters are usually optimized using Maximum Likelihood Estimation (MLE) or its variations.\n\nFigure 5 shows an example of GPR using RBF kernel over the sine function with noisy measurements, where predictive variance increases at points with missing measurements. Outside of the interpolation region predictive variance significantly increases with the mean failing to capture the true function trend.\n\n#### Bayesian optimization\n\nAn advantageous property of GPR is that it provides not only the prediction of the value at unobserved points but the complete probabilistic distribution determined by the mean and variance. The general idea behind BO algorithms is to use such distribution to explore parameter space and select values of $${\\mathbf {x}}^*$$ in a way that it will most probably maximize target function $$f({\\mathbf {x}})$$. The common approach is to select a particular acquisition function that takes parameters of the predictive distribution of the fitted model as an input and outputs some value which is maximized instead. There exist multiple strategies, for example, using the probability of improvement, expected improvement or integrated expected improvement over the current best value, entropy search or upper confidence bound (UCB) . We have selected the UCB acquisition function in our work as it is easy to evaluate and was shown to be effective in practice. It is expressed using the predictive mean and variance as follows:\n\n\\begin{aligned} a_{UCB}({\\mathbf {x}}, \\kappa ) = {\\hat{\\mu }}({\\mathbf {x}}) + \\kappa \\cdot {\\hat{\\sigma }}({\\mathbf {x}}) \\end{aligned}\n(2)\n\nExploration–exploitation trade-off is managed by the parameter $$\\kappa$$, where for small $$\\kappa$$ regions with a high mean (exploitation) and large $$\\kappa$$ regions with high uncertainty (exploration) are preferred, respectively. We will further omit $$\\kappa$$ from the arguments of the UCB function where it is assumed fixed.\n\nFigure 6 shows the 4th step (with 2 initial data points at the boundaries) of the BO algorithm on an example function with several local maximums using UCB acquisition function with the fixed $$\\kappa =2$$.\n\nIt is critical to note that BO performance is profoundly affected by the dimensionality of the input data due to the exponential growth of the parameter space. It may start to perform poorly when the number of controlled parameters becomes larger than ten .\n\n### Noise estimation\n\nWe defined the target function that we aim to optimize as the sum of averages of germinated and well-germinated seeds (see “Seed germination” section). First, let N denote the number of seeds used in the experiment. Second, due, to the stochasticity, we model the success of a single seed germination for the fixed values of parameters $${\\mathbf {x}}$$ as a Bernoulli trial. Then, the probability that a single seed is germinated equals to $$p({\\mathbf {x}})=p$$, whereas probability that a single seed is well-germinated, given that it has germinated, equals to $$q({\\mathbf {x}}) = q$$. If $$N_g$$ and $$N_{wg}$$ denote the number of germinated and well-germinated seeds in the experiment, respectively, then, it can be shown that for sufficiently large N (for details, see “Appendix” section) our target function is\n\n\\begin{aligned} y({\\mathbf {x}}) = \\frac{N_g + N_{wg}}{N}~\\sim ~{\\mathcal {N}}\\left( \\mu , \\frac{1}{N}\\sigma ^2\\right) , \\end{aligned}\n\nwhere $$\\mu =p(1+q)$$ and $$\\sigma ^2=p(1+3q) - p^2(1+q)^2$$. Due to the normality of the obtained distribution, its variance can be interpreted as an input-dependent Gaussian noise in the Eq. (1). Therefore, we can simplify hyperparameter optimization by setting a lower bound of the noise variance with the following value:\n\n\\begin{aligned} \\frac{1}{N} \\max _{p,q} \\sigma ^2(p,q) = \\frac{1}{N}. \\end{aligned}\n(3)\n\nAlternatively, for each obtained observation $$y_i$$ a lower-bound of the noise variance can be estimated as (for details, see “Appendix” section)\n\n\\begin{aligned} \\frac{1}{N} \\cdot y_i(2-y_i), \\quad i=1, \\ldots ,n \\end{aligned}\n\nin order to incorporate the dependence on the values of observations.\n\n### Concurrent experiments\n\nAforementioned BO formulation assumes that the optimization process is sequential, i.e., only a single $$x^*$$ is selected at each step. However, it may be necessary to be able to select several vectors of parameters to explore, e.g., if there are multiple CPU cores for computations or several experimental setups available (climate chambers in our case). This is referred in the literature as batch setting [12, 13] or setting with a delayed feedback . In this work we consider the following approach from to tackle this problem: for each trial comprising the selection of multiple vectors of parameters, we find the maximizer of acquisition function and “observe” the target function using the predictive mean of GPR instead of the real outcome (see Algorithm 1).\n\n### Exploration–exploitation control\n\nIt may happen when performing exploitation that the algorithm could propose parameters that are very close to the already explored data points, e.g., try $$22.001 \\, ^{\\circ }{\\hbox {C}}$$ temperature after $$22.000 \\, ^{\\circ }{\\hbox {C}}$$, which yields a change beyond the controllable precision. In order to cope with this problem and reduce the manual labor of an operator in the selection of $$\\kappa$$ from Eq. (2) that will give a reasonable exploitation, we propose an additional optimization procedure. First, we formulate the notion of a reasonable exploitation as the following constraint:\n\n\\begin{aligned} \\underset{i=1, \\ldots ,n}{\\min } \\left\\| \\underset{{\\mathbf {x}}}{\\arg \\max }~a_{UCB}({\\mathbf {x}}, \\kappa ) - {\\mathbf {x}}_i\\right\\| _\\infty \\ge \\epsilon _{{\\textit{xploit}}}, \\end{aligned}\n(4)\n\nwhere n is the number of already observed data points and $$\\epsilon _{{\\textit{xploit}}}$$ is a predefined constant. This constraint means that at least one of the parameters must be at least as far as $$\\epsilon _{{\\textit{xploit}}}$$ from the respective parameter of the closest already observed data point. One can think of a more fair constraint, where a too small change of a parameter is diminished to zero, however, it may pose challenges for the optimization algorithms. Similarly, in order to avoid unreasonable exploration, we consider the following constraint:\n\n\\begin{aligned} \\underset{i=n_1, \\ldots ,n_s}{\\min } \\left\\| \\underset{{\\mathbf {x}}}{\\arg \\max }~a_{UCB}({\\mathbf {x}}, \\kappa ) - {\\mathbf {x}}_i\\right\\| _1 \\le \\epsilon _{{\\textit{xplore}}}, \\end{aligned}\n(5)\n\nwhere $${\\mathbf {x}}_i$$ is taken from a subset of size $$s \\le n$$ of already observed points, e.g., one may like to ignore manually initialized data (see “Data preparation” section) and prefer exploration around knowingly good regions. This constraint means that the selected parameters must be at most as $$\\epsilon _{{\\textit{xplore}}}$$ far in total form the closest already observed data point. Algorithm 2 describes the exploration–exploitation control procedure.\n\n## Experimental evaluation\n\nIn this section, we describe the details of our experimental setup and provide the obtained results.\n\n### Selecting parameters\n\nWe implementedFootnote 1 our solution with Python 3 programming language using the Bayesian optimization library.Footnote 2 As the covariance function we selected the composition of constant, isotropic Mateŕn (with $$\\nu =2.5$$, assuming sufficient smoothness) and white noise kernels with tunable hyperparameters:\n\n\\begin{aligned} k({\\mathbf {x}}_i,{\\mathbf {x}}_j) = \\alpha \\cdot C_\\nu ({\\mathbf {x}}_i / \\rho , {\\mathbf {x}}_j / \\rho ) + \\sigma ^2 \\delta _{ij} \\end{aligned}\n\nwhere $$\\delta _{ij}$$ is a Kronecker-delta, $$\\alpha , \\rho \\in {\\mathbb {R}}^{+}$$. Optimization of the hyperparameters is performed at each step when new data is being available using the MLE with the number of optimizer restarts equal to 30. Bounds for hyperparameter optimization were set as follows: $$\\alpha \\in [10^{-5},10^5]$$, $$\\rho \\in [10^{-5}, 10^5]$$ and $$\\sigma ^2 \\in [0.01, 10^5,]$$ (see “Seed germination” and “Noise estimation” sections). GP mean was selected to be the mean value of the observed measurements.\n\nGiven the small number of tunable parameters (7 in the first experiment and 4 in the second), we considered the basic BO approach. As an acquisition function, we selected UCB since it has been shown to be effective in various scenarios. Exploration–exploitation trade-off was managed through $$\\kappa$$ parameter based on the expert knowledge, i.e., at each step, $$\\kappa$$ was selected in such a way that the algorithm does not purely exploit almost the same parameters or explore knowingly unprofitable regions. Additional control was performed by setting $$\\epsilon _{{\\textit{xploit}}}$$ equal to $$0.1 \\, ^{\\circ }{\\hbox {C}}$$ and 1 ml and $$\\epsilon _{{\\textit{xplore}}}$$ equal to $$10 \\, ^{\\circ }{\\hbox {C}}$$ and 100 ml for the temperature and the water supply, respectively. For constrained optimization we have used SciPy library implementation of the Sequential least squares programming (SLSQP) algorithm . Each optimization step requires the evaluation of the maximum of acquisition function at several points, which impose computational overhead, however, it can be considered negligible compared to the time-scale of a single trial.\n\n### Data preparation\n\nTo set up the experiments, we had to consider several issues. First, we had to select the boundaries for the optimized parameters: we selected them at 0, $$40 \\, ^{\\circ }{\\hbox {C}}$$ (in both experiments) and 0, 250 ml (in the first experiment) for the temperature and the water supply, respectively. Second, as the parameters may have different unit measures, which affects modeling due to isotropy of the selected kernel, we needed to scale them appropriately: we linearly mapped temperature and water supply values to [0, 1] and [0, 0.5] intervals, respectively, assuming “equivalence” of $$1 \\, ^{\\circ }{\\hbox {C}}$$ and 12.5 ml (during the second experiment, this step was ignored as the only temperature was varied). Finally, we had to add some initial data so that optimization could kick off: we picked all of the possible combinations of 0 and 40 temperatures (in both experiments) with 0 water supply (in the first experiment) on each day and assigned the “observed” target function values equal to 0 (totally $$2^4=16$$ initial points). It can be considered reasonable as extreme conditions should produce poor results.\n\n### Results\n\n#### First experiment (poorly germinated pea seeds)\n\nFor a single germination process, we used $$N=100$$ pea seeds and conducted only a single repetition for each selected vector of controlled parameters. The first trial was conducted using the single reference vector of parameters selected with the expert knowledge, which gave the number of germinated seeds equal to 73, and the two vectors selected by the BO algorithm. At the 11th observation the algorithm discovered the parameters, which yielded 73 germinated seeds with an additional amount of 18 well-germinated. The 20th selected vector of parameters produced as much as 80 germinated and 33 well-germinated seeds, which in total gave a 55% improvement over the initial guess. Subsequent 13 steps didn’t provide any further enhancement.\n\nFigure 7 shows the target values obtained during 11 trials of the first experiment. Black dashed line denotes the kriged average and shows the trend of improvement in the germination efficiency, whereas the green top dotted line shows the best-observed values for each trial. Table 1 depicts all of the 33 vectors of parameters and respective observed target function values obtained during 11 trials.\n\nNotably, without any prior knowledge of the underlying system, the algorithm was able to learn the values of the controlled parameters that yield sufficient improvement of the germination efficiency. The values of the parameters that achieved the maximum found target function value of 1.13 at the 20th iteration are listed in italics in Table 1. The identified values can be explained from the physiological point of view. For example, periodically changing temperature may be favorable due to the natural adaptation of seeds to day and night, whereas water supply identified by the algorithm is in a good agreement with the dynamics of water uptake by seeds, previously described in . According to this study, water uptake by plant seeds is triphasic, comprising a rapid initial absorption, followed by a plateau phase and a further increase due to embryonic axes elongation.\n\n#### Second experiment (well-germinated radish seeds)\n\nAlthough the first experiment showed a substantial improvement of germination efficiency in the case of poorly germinated seeds, it could not be that easily observed for well-germinated seeds. Therefore, in the second experiment, we used $$N=96$$ radish seeds with 6 repetitions for a single germination trial. The first 4 trials were conducted by setting all of the temperature parameters as either 21, 22, 23 or 24. At the 9th trial (5th automatic step), the algorithm discovered the parameters, which yielded the best average of 10 germinated and 88 well-germinated seeds.\n\nFigure 8 shows the target values obtained during 12 trials, where the last trial served as a validation for the best found vector of parameters during the 9th trial. Green dotted line shows the best-observed mean value of the target function, whereas the red dashed line depicts the first expert-knowledge guess-based trial.\n\nTable 2 lists all of the 11 vectors of parameters and the corresponding means and standard deviations of the target function values obtained during 12 trials. The complete table containing target function values for every repetition during each trial can be found in Additional file 1.\n\nAlthough with the initial guess seeds already propagated efficiently, the algorithm was able to achieve substantial improvement after the several steps and identify the parameters, which yielded the maximum mean value of 1.903 of the target function with low dispersion.\n\n## Conclusions and future work\n\nWe applied Bayesian optimization framework to the seed germination process in a controlled environment. Our experiments demonstrated that the proposed methodology allowed to identify the values of the controllable parameters that increase germination efficiency in different settings for different seeds both in the case when initial expert-knowledge based guess yields low and high germination efficiency. The proposed methodology is model-free, and we argue that it may be useful for a variety of optimization problems in intelligent agriculture. Using this approach, we achieved increase in germination efficiency (according to our metrics) from 36.5 to 56.5% by 19 iterations in the first experiment (pea seeds) with low initial germination efficiency, whereas in the second experiment (radish seeds) with high initial germination efficiency the increase was from 91.8% up to 95.2% by 5 iterations.\n\nWe note that selection of the controllable parameters must be made carefully during the preliminary planning. On the one hand, increasing their number allows to perform better fine-tuning, on the other hand, it makes BO algorithms less efficient and requires more trials to be conducted, which may be both overly time-consuming and equipment demanding.\n\nCombination of the proposed technique with the existing methods of computer vision-based seed counting [18, 19] and seed quality evaluation may decrease manual labor significantly and improve scalability. The BO methods definitely could help to reveal optimum chemical parameters of growing mediums or find the environmentally friendly doses of plants biostimulants (humic substances, synthetic hormones, etc.), which effects on plants usually have a nonlinear dose-effect relationship. Further experimental studies are required to investigate the effectiveness of our approach for this environmental and plants issues. Additionally, we aim to consider partially-controllable environments and apply the proposed method at the next stages of plant growth.\n\n## Abbreviations\n\nGPR:\n\nGaussian process regression\n\nRBF:\n\nMLE:\n\nmaximum likelihood estimation\n\nBO:\n\nBayesian optimization\n\nUCB:\n\nupper confidence bound\n\n## References\n\n1. Forcella F, Arnold RLB, Sanchez R, Ghersa CM. Modeling seedling emergence. Field Crops Res. 2000;67(2):123–39.\n\n2. Bradford KJ. Water relations in seed germination. Seed Dev Germ. 1995;1(13):351–96.\n\n3. Bello P, Bradford KJ. Single-seed oxygen consumption measurements and population-based threshold models link respiration and germination rates under diverse conditions. Seed Sci Res. 2016;26(3):199–221.\n\n4. Gosavi A. Simulation-based optimization: an overview. In: Simulation-based optimization. Operations research/computer science interfaces series, 2nd ed. Boston, MA: Springer; 2015. p. 29–35.\n\n5. Snoek J, Larochelle H, Adams RP. Practical bayesian optimization of machine learning algorithms. In: Proceedings of the 25th international conference on neural information processing systems – NIPS’12, vol. 2. Lake Tahoe, Nevada: Curran Associates Inc.; 2012. p. 2951–2959.\n\n6. Shahriari B, Swersky K, Wang Z, Adams RP, de Freitas N. Taking the human out of the loop: a review of Bayesian optimization. Proc IEEE. 2016;104(1):148–75. https://doi.org/10.1109/JPROC.2015.2494218.\n\n7. Kochanski G, Golovin D, Karro J, Solnik B, Moitra S, Sculley D. Bayesian optimization for a better dessert. In: NIPS, workshop on Bayesian optimization; 2017.\n\n8. Yuan J, Liu C-L, Li Y-M, Zeng Q, Zha XF. Gaussian processes based bivariate control parameters optimization of variable-rate granular fertilizer applicator. Comput Electron Agric. 2010;70(1):33–41. https://doi.org/10.1016/j.compag.2009.08.009.\n\n9. Rasmussen CE, Williams CKI. Gaussian processes for machine learning. Cambridge: The MIT Press; 2006.\n\n10. James G, Witten D, Hastie T, Tibshirani R. An introduction to statistical learning, vol. 112. Berlin: Springer; 2013.\n\n11. Wang Z, Zoghi M, Hutter F, Matheson D, De Freitas N, et al. Bayesian optimization in high dimensions via random embeddings. In: IJCAI; 2013. p. 1778–84.\n\n12. Azimi J, Jalali A, Fern XZ. Hybrid batch Bayesian optimization. In: Proceedings of the 29th international conference on international conference on machine learning. Madison: Omnipress; 2012. p. 315–22\n\n13. González J, Dai Z, Hennig P, Lawrence N. Batch Bayesian optimization via local penalization. In: Artificial intelligence and statistics; 2016. p. 648–57.\n\n14. Joulani P, Gyorgy A, Szepesvari C. Online learning under delayed feedback. In: Dasgupta S, McAllester D, editors. Proceedings of the 30th international conference on machine learning proceedings of machine learning research, vol. 28. 2013. Atlanta: PMLR; 2008. p. 1453–61. http://proceedings.mlr.press/v28/joulani13.html.\n\n15. Jones E, Oliphant T, Peterson P, et al. SciPy: open source scientific tools for Python (2001–). http://www.scipy.org/.\n\n16. Kraft D. A software package for sequential quadratic programming. Forschungsbericht- Deutsche Forschungs- und Versuchsanstalt fur Luft- und Raumfahrt; 1988.\n\n17. Bewley JD. Seed germination and dormancy. Plant Cell. 1997;9(7):1055.\n\n18. Ducournau S, Feutry A, Plainchault P, Revollon P, Vigouroux B, Wagner M. An image acquisition system for automated monitoring of the germination rate of sunflower seeds. Comput Electron Agric. 2004;44(3):189–202.\n\n19. Pouvreau J-B, Gaudin Z, Auger B, Lechat M-M, Gauthier M, Delavault P, Simier P. A high-throughput seed germination assay for root parasitic plants. Plant Methods. 2013;9(1):32.\n\n20. Urena R, Rodrıguez F, Berenguel M. A machine vision system for seeds quality evaluation using fuzzy logic. Comput Electron Agric. 2001;32(1):1–20.\n\n## Authors' contributions\n\nAN: framework design and implementation. IF: initial general idea, preparation and evaluation of the first experiment. DS: preparation and evaluation of the second experiment. MP: consultation, preparation of the second experiment. IO: Initial algorithmic idea, guidance. All authors read and approved the manuscript.\n\n### Acknowledgements\n\nThis work was supported by the Ministry of Education and Science of the Russian Federation (grant 14.756.31.0001).\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Availability of data and materials\n\nData is available on request to the authors.\n\nNot applicable.\n\nNot applicable.\n\n### Publisher’s Note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Artyom Nikitin.\n\nRadish seeds experiment data. The complete list of 11 explored vectors of parameters and target function values obtained during 12 trials of the second experiment with radish seeds.\n\n## Appendix\n\n### Appendix\n\nLet random variable $$x \\sim B(1, p)$$ denote the success of a seed germination with a probability p and $$y | x=1 \\sim B(1, q)$$ denote the success of a well-germination with a probability q given that germination occurred. Using the formula for a full probability:\n\n\\begin{aligned} p_{x,y}(x=0, y=0) &= 1-p \\\\ p_{x,y}(x=1, y=0) &= p(1-q)\\\\ p_{x,y}(x=0, y=1)&= 0 \\\\ p_{x,y}(x=1,y=1) &= pq \\end{aligned}\n\nThen, the distribution of a random variable $$z = x + y$$ is\n\n\\begin{aligned} p_z(z=0)&= 1-p\\\\ p_z(z=1)&= p(1-q)\\\\ p_z(z=2)&= pq \\end{aligned}\n\nwith the mean and the variance\n\n\\begin{aligned} \\mu&=p(1+q), \\end{aligned}\n(6)\n\\begin{aligned} \\sigma ^2&=p(1+3q) - p^2(1+q)^2. \\end{aligned}\n(7)\n\nLet $$z_i \\sim p_Z,~ i=1, \\ldots ,N$$ be identically independently distributed random variables. Then, according to the Central Limit Theorem, for sufficiently large N, the distribution of their average can be well approximated by a normal distribution:\n\n\\begin{aligned} w = \\frac{1}{N}\\sum _{i=1}^N {z_i}~\\sim ~{\\mathcal {N}}_w(\\mu , \\sigma ^2 / N). \\end{aligned}\n\nGiven M samples $$w_i \\sim {\\mathcal {N}}_w,\\, i=1,\\ldots ,M$$ one can find the sampling mean $${\\widetilde{\\mu }} = (w_1 + \\cdots + w_M)/M$$ and estimate the variance, by substituting the Eq. (6) into the Eq. (7), as\n\n\\begin{aligned} {\\widetilde{\\sigma }}^2 = \\frac{1}{N}\\left[ \\frac{1 + 3q}{1 + q} \\cdot {\\widetilde{\\mu }} - {\\widetilde{\\mu }}^2 \\right] \\le \\frac{1}{N}\\cdot {\\widetilde{\\mu }}(2 - {\\widetilde{\\mu }}). \\end{aligned}\n(8)\n\n## Rights and permissions",
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"# How to Calculate Standard Deviation in Excel [STDEV.S Formula]\n\nPosted May 21, 2019 by Sheraz Ali in Microsoft Excel",
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"If you’re someone who has to deal with a lot of numbers on a daily basis, you must familiarize yourself with a large array of Excel formulae. To perform basic mathematical operations, such as addition, multiplication, subtraction and Standard deviation (SD), you must learn how these formulae work.\n\nI’ll teach you how to calculate Sample Standard deviation and Standard Deviation for a population. But first, let’s get friendly with the basics.\n\n### What is Standard Deviation and Why Calculate it?\n\nStandard deviation is used to tell how far a value is spread from the mean value of a group or a population.\n\nIf you get a high value of SD, it means the numbers vary a lot from the average or expected value. If the SD is on the low side, your individual numbers are close to the average or mean.\n\nIn eighth or ninth grade, to calculate SD you had to go through a complicated series of steps. Excel has made life a hell lot easier. All you have to do now is insert the SD population and Sample formulae in Excel to get the intended results.\n\n### Difference Between Sample and Population Standard Deviation\n\nThere are certain differences between calculating SD in entire populations and limited groups or samples in Excel.\n\nQualitatively, The SD of a population is fixed and yielded from every unit of a said population. Sample SDs are targeted on a limited number of individuals or units of a population.\n\nThey have a greater chance of variability and are manifolds greater than Population SDs.\n\nIn Excel terms, Population Standard Deviation takes into account all data points ( N). Sample Standard Deviation (SSD), on the other hand, takes into account all data points minus one value (N-1).\n\n### How to Calculate Standard Deviation in Excel (Sample)",
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"1. Open the Microsoft Excel Document containing your data.",
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"2. Click an empty cell. Any will suffice.",
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"3. Start your Standard Deviation formula with =.",
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"4. Type STDEV.S.",
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"5. Select the cells you want the Standard Deviation computed for.",
null,
"• You can select the entire data or scattered cell references ( A1, A4 etc.) depending upon your preferences.\n\n6. Press Enter.\n\n7. Your Standard deviation is here.",
null,
"### How to Calculate Standard Deviation in Excel ( Population)",
null,
"To calculate SSP or Standard Deviation Population, follow the aforementioned steps with a minor modification.\n\nInstead of STDEV.S, Insert the Formula STDEV.P and you’re good to go.\n\nRead More: How to Divide in Excel : Division Formulae\n\n### Note: Standard Deviation of ZERO\n\nIf the numbers are the same, there’s no need to calculate SD. Same numbers means zero variability which in turn means SD=0.\n\nFor example, if all the numbers on my Excel sheet are 10, then the SD will be equal to 0.\n\nStill not sure how to calculate Standard Deviation in Excel? Leave a comment below and we will get back to you. Additional, check this youtube video tutorial for further guidance.\n\n### About Sheraz Ali\n\nAn established copywriter, with a longstanding experience in a vast array of industries, including but not limited to spirituality, technology, cannabis and travel.\nView more articles by Sheraz Ali\n\n#### The Conversation\n\nFollow the reactions below and share your own thoughts."
] |
[
null,
"https://www.tech-recipes.com/wp-content/uploads/2019/05/excel-639x350.png",
null,
"https://www.excel-easy.com/examples/images/standard-deviation/formula-standard-deviation-sample.png",
null,
"https://www.tech-recipes.com/wp-content/uploads/2019/05/Excel1-500x455.jpg",
null,
"https://www.tech-recipes.com/wp-content/uploads/2019/05/Excel2-500x455.jpg",
null,
"https://www.tech-recipes.com/wp-content/uploads/2019/05/Excel3-500x455.jpg",
null,
"https://www.tech-recipes.com/wp-content/uploads/2019/05/Excel4-500x455.jpg",
null,
"https://www.tech-recipes.com/wp-content/uploads/2019/05/excel5-500x455.jpg",
null,
"https://www.tech-recipes.com/wp-content/uploads/2019/05/Excel6-500x455.jpg",
null,
"https://www.excel-easy.com/examples/images/standard-deviation/formula-standard-deviation-population.png",
null
] |
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|
https://3dprinting.stackexchange.com/questions/19107/obtaining-a-smooth-solid-stl-from-voxel-data
|
[
"# Obtaining a smooth solid .stl from voxel data\n\nI have a (preferably in matlab) 3D array of Booleans describing a voxel structure. I want to turn this into an .stl file but I don't want the final result to have the jagged cubes.\n\nI have tried putting the voxel array into a marching cube algorithm but the resulting STL is not a solid, some parts are infinitely thin sheets. I want every voxel to be represented by something solid.\n\nIs there a way to achieve this?\n\n• I’d say that there’s something wrong with your Marching Cubes algorithm. It ought to produce a solid object. Then, you can smooth the triangles in that solid object. Mar 20, 2022 at 0:59\n• @bubba I think that was the case. I needed to surround the structure with 0, to make the code acknowledge the boundary as a facet of the solid. Mar 20, 2022 at 23:52\n\nVoxels are pretty much an interpolation of data points and you can only smooth them out by reducing the fidelity. Why? Because in acquiring the voxel, you already lost fidelity. Let me take you through an example.\n\n## The basics\n\nFor our basics we take a circle of Radius $$R$$. It can be fully described the following: $$x^2+y^2=R$$ All solutions to this give us points on the circle.\n\nIts representation as a single voxel with resolution length $$r$$ is that of a square. The square with its center on $$\\{0,0\\}$$ we will call the voxel of resolution $$r$$ around that coordinate. It is defined as $$x =\\{0.5r,-0.5r\\} \\land y=\\{-0.5r \\to 0.5r\\}$$ $$y=\\{0.5r,-0.5r\\}\\land x=\\{-0.5r \\to 0.5r\\}$$ Other squares then can be created by simply adding the respective cell's center's coordinates $$\\{a,b\\}$$ for $$x' =\\{0.5r,-0.5r\\}+a \\land y'=\\{-0.5r \\to 0.5r\\}+b$$ $$y'=\\{0.5r,-0.5r\\}+b\\land x'=\\{-0.5r \\to 0.5r\\}+a$$\n\nLooks complicated? True, but you see: the first formula is a circle with radius $$R$$ containing an area of $$A=\\pi*R^2$$. The latter two blocks define the square of length $$r$$ and area $$A=r^2$$. By clever positioning, the grid with $$R=r$$ has its $$\\{0,0\\}$$ Voxel wholly contained in that circle. Or in a picture, it looks like this, using $$r=R=50$$ (and thus a diameter of 100).",
null,
"As you see, there is an area between the circle and the square - this is fidelity that is lost when converting to voxels. In a typical voxel transformation, there is only one choice: at which point of filling does a voxel get filled to 100% and at which it gets filled to 0%. Let's assume the circle there is actually a cylinder of 50 units height. Depending on where we set the cutoff, we now get one of three solutions. We could end with one voxel, using 50% as the cutoff point. We could end with a plus shape using 45% as the cutoff point or a square using any filling above 0% as a cutoff, as shown in the following picture.",
null,
"It gets even more complicated if you don't have the circle and the grid's center overlap like n the example above: there is a \"2x2x1\" voxel solution that reflects a differently aligned circle with the same cutoff point as the \"1x1x1\" solution! That circle is (in the next graphic) centered around $$\\{3.5 r,0.5 r\\}$$ and thus shifted half a unit up and right so its center overlaps with a corner of the voxels.",
null,
"## Lost fidelity can't be regained\n\nYou see, you have lost a lot of fidelity in voxelizing. All circles have become squares. All squares also are squares. Likewise, all curves in between have become squares.\n\n## The problem\n\nSo, it is nigh impossible for a computer to identify, what once was a circle square or square before all was made squares. Think... Colors: I take a picture of a smiley. I split it in half and then desaturate the upper half of the right side to black and white and totally saturate the lower. Those are two different ways to totally loose the color information. In the upper case you retain one information more, because the algorithm of making all colors but black white retains the black information, while the lower parts algorithm only retains the outer shape.",
null,
"If you only had the black and white smiley, you have lost the information about its color in total. If you only have the black outline, you have lost also the information of the face.\n\nVoxelizaion does loose about that much information, depending on the settings.\n\n## Reversing Voxelisation is interpreting the voxel structure\n\nIt's not possible for a computer algorithm to simply invert the process. However, there are ways to interpret a voxelized object and try to re-create one of the possible objects that have led to this item, assuming that an \"any fill is a voxel\" algorithm has been used.\n\nFor such, you'd import the model into a software such as Blender, solidify it so that it is one mesh without internal faces, and then running a smoothing operation. You don't get the object that generated that voxel structure, but you get an Object that would generate that voxel structure.",
null,
"How good the interpretation of the voxels is, depends on the resolution chosen in the start. If $$r$$ is small enough, then the resulting interpretation, together with some artistic definition of which corners are sharp and quite some manual post-processing (which requires the human eye) can lead to a somewhat good approximation of the actual object you believe lead to the voxel structure."
] |
[
null,
"https://i.stack.imgur.com/J1GWB.png",
null,
"https://i.stack.imgur.com/fDiid.png",
null,
"https://i.stack.imgur.com/AKfwY.png",
null,
"https://i.stack.imgur.com/KaApD.png",
null,
"https://i.stack.imgur.com/pAM2I.png",
null
] |
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|
https://www.studystack.com/flashcard-2632082
|
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null,
"or",
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"why\n\nMake sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.\n\nEnter the associated with your account, and we'll email you a link to reset your password.\nDon't know\nKnow\nremaining cards\nSave\n0:01\nEmbed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.\n\nNormal Size Small Size show me how\n\nmath fractions/decim\n\nvocabulary words\n\nterminating decimal a decimal that ends or termenates\nrepeating decimal a decimal in which one or more digits repeats infinitely\nbar notation a line or bar above the part of a repeating decimal that repeats\nnumerator the uper number of a fraction\ndenomenator the lower number of a fraction\nmixed number a whole number and a fraction\nimproper fraction a fraction with the numerator bigger than the denominator\nsimplest form when a number can not be split into two or more for both numerator and denominator\nbenchmark fraction using a fraction such as 1/2 as a reference point when comparing and ordering fractions\nlike fractions fractions that have the same denomenators\nCreated by: 24062"
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|
https://discourse.purescript.org/t/help-with-decoding-tagged-json-needed/1797/4
|
[
"# Help with decoding tagged JSON needed\n\nPlease ignore the first post and jump straight to my update Help with decoding tagged JSON needed\n\nI’m trying to reduce boilerplate code without compromising on type safety. The data I’m dealing with consists of lots of records which all have at least one field in common, `type`, such as\n\n``````[\n{ \"type\": \"foo\" },\n{ \"type\": \"bar\" }\n]\n``````\n\nI would like to create a sum type, where the constructors represent each possible type of node. My first version consisted of a record where the “type” could be specified with a type argument.\n\n``````type Node nodeType dataValuesType r\n= { nodeType :: nodeType\n| r\n}\n``````\n\nThis let me create data types with only a single constructor and no arguments…\n\n``````data FooType = FooType\n``````\n\n…and use those to create variations of the JSON records, like this:\n\n``````data MyNode = Node FooType ()\n``````\n\nWhat’s left is to implement `readImpl` from Simple JSON…\n\n``````expectString ::\nforall a.\nString -> -- | nodeType value we're looking for\na -> -- | Constructor function\nForeign -> -- | nodeType value we're decoding\nF a\nexpectString query consFn f = do\ns <- readString f\nif s == query then\npure consFn\nelse\nfail <<< ForeignError \\$ \"expected hyperlink but got: \" <> s\n\nreadImpl = expectString \"foo\" FooType\n``````\n\nand things will just work (I’m using generics to generate the decoding code for the actual sum type as outlined in the Simple JSON docs)`. I get nodes where I can easily pattern match on constructors and so on. But the whole process involves a lot of typing and I’m sure I’m doing this wrong.\n\nI have this vague idea that I should be able to use something like a `Symbol` and move all of the decoding into my generic `Node` type rather than repeat this “check if string is this or that” logic everywhere where I’m using `Node`.\nI’ve tried various things below but they all lead nowhere in the end.\n\n``````-- can't access the symbol in `readImpl`\ntype Node (sym :: Symbol) r\n= { nodeType :: String\n| r\n}\n``````\n\nTL;DR: The goal is to specify the value of the `type` key when I create the specific `Node` type and implement the decoding logic that matches the given value against the `{ \"type\": \"SOME_VALUE\" }` JSON object on `Node` rather than on all the users of `Node` to create things like `MyNode`\n\nI think the existing behaviour for variants might solve your problem.\n\nI had a small epiphany yesterday when I finally realized that I can get the type level string out of the type and into the value level with\n\n``````nameP = SProxy :: SProxy name\nname = reflectSymbol nameP\n``````\n\nand that tagged JSON decoding is exactly what I’m trying to do here. Now I just need to figure out how I can make my `Node` function polymorphic over all the fields of the JSON object that are not the `nodeType` field. I know that I can of course use row types for that but I need to figure out the specifics.\n\n1 Like\n\nI’ll rephrase my question because I’m stuck again.\n\nI am trying to decode JSON objects which all have one key in common, which functions as tag and from which I can determine which variant I’m dealing with.\nFor example, both `{ nodeType: \"foo\", bar: 1 }` and `{ nodeType: \"foo\", bax: \"boo\" }` are valid JSON objects. I want to represent all variants with a sum type, like so:\n\n``````data Thing = Foo { bar :: Int } | Bax { bax :: Strong }\n``````\n\nI looked at the implementation of tagged JSON decoding for `Simple.JSON` and the most important part is this:\n\n``````\ninstance taggedSumRepConstructor ::\n( GenericTaggedSumRep a\n, IsSymbol name\n) => GenericTaggedSumRep (GR.Constructor name a) where\ngenericTaggedSumRep f = do\nr :: { \"type\" :: String, value :: Foreign } <- JSON.read' f\nif r.\"type\" == name\nthen withExcept (map \\$ ErrorAtProperty name) \\$ GR.Constructor <\\$> genericTaggedSumRep r.value\nelse\nfail \\$ ForeignError \\$ \"Wrong type tag \" <> r.\"type\" <> \" where \" <> name <> \" was expected.\"\nwhere\nnameP = SProxy :: SProxy name\nname = reflectSymbol nameP\n``````\n\nNow my case is a bit different. The way I understand the linked code is that it expects the name of the constructor to match the values of the tag field. So `name` in the snippet above is both a JSON value and a constructor name. For me that’s not the case and so I need a mapping from JSON field value to constructor. And that’s where I’m stuck.\n\nMy idea was to pass this mapping around via symbols, but how would I do that? Every one of my constructors (`Foo` and `Bax`) would need a symbol. But in the code snippet the instance for `GR.Constructor` only gets the constructor name and then `a`, which is the argument. In my case the type level symbol (e.g., `data Thing = Foo \"jsonFieldValue\" ...`) is probably not even valid syntax and even if it was how would I extract these type level symbols from the constructor to use them inside the `genericTaggedSumRep` function above?\n\n1 Like\n\nOkay I think I finally figured it out. I haven’t tested in “at scale” by which I mean I don’t know if this breaks down in certain cases I currently can’t think of but here we go.\n\nThe Glorious Gist In All Its Glorious Glory\n\nI might turn this into a blog post with proper explanations or at least annotate the gist but here’s the idea behind it.\n\nI still don’t know if it’s easily possible to extract symbols from constructors in instances for the sum type to which these constructors belong. Maybe some fancy reflection can do that? But we don’t even need to do this.\n\nI want to represent variants of JSON objects as a sum type\n\n``````data Foo\n= FooA (Node \"foo\" ( some :: String ))\n| FooB (Node \"bar\" ( other :: String ))\n``````\n\nNotice how each constructor has a `Node` type. Thanks to this “trick” I can implement the decoding logic on `Node`, where it’s now pretty straight forward to extract the symbol (~ string).\n\nThe `readImpl` implementation of `Node` first looks at the `nodeType` value, and only this. If the value of the `nodeType` key matches what we expect (for example to decode a `FooA` we need `nodeType: \"foo\"`) then we go ahead and do this:\n\n`````` if peek.nodeType == jsonTagValueS then do\ncase ( JSON.read f ::\nJSON.ReadForeignFields rowList () row =>\nJSON.E { | row }\n) of\nLeft e -> throwError e\nRight full -> pure \\$ Node \\$ Record.union peek full\n``````\n\nI’m still surprised that I actually managed to write some type level code that actually compiles",
null,
"We now try to decode the JSON object based on the row of types for that constructor. Again using `FooA` as an example, we’re looking for an object that has a key called `some` with a `String` as its value. The type signature “simply” states that the compiler needs to understand how to read these fields, meaning there needs to be a `ReadForeignFields` instance.\n\nWhat we’ve achieved is that `Node` is essentially a tagged record where you can specify the expected value of the tag field. Furthermore `Node` then takes the row of types for all the other fields you expect in that record.\n\nEverything else in the gist is the usual boilerplate for untagged sum type decoding which I copied without modification from the Simple.JSON docs\n\nNow I’d like to write some tests and I struggle with\n\n``````instance nodeEq ::\n( Eq (Record row)\n, RowToList row list\n, IsSymbol s\n, EqRecord list row\n) =>\nEq (Node s row) where\neq (Node n1) (Node n2) = eqRecord (RLProxy :: RLProxy list) n1 n2\n``````\n``````[1/1 NoInstanceFound] /data/private/lions-purescript/src/Data/Contentful/RichText/Node.purs:30:28\n\n30 eq (Node n1) (Node n2) = eqRecord (RLProxy :: RLProxy list) n1 n2\n^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^\n\nNo type class instance was found for\n\nData.Eq.EqRecord list2\n( nodeType :: String\n| row3\n)\n\nwhile applying a function eqRecord\nof type EqRecord t0 t1 => RLProxy t0 -> Record t1 -> Record t1 -> Boolean\nto argument RLProxy\nwhile inferring the type of eqRecord RLProxy\nin value declaration nodeEq\n\nwhere row3 is a rigid type variable\nbound at (line 0, column 0 - line 0, column 0)\nlist2 is a rigid type variable\nbound at (line 0, column 0 - line 0, column 0)\nt0 is an unknown type\nt1 is an unknown type\n``````\n\n… of course",
null,
"Also why doesn’t\n\n``````instance nodeEq ::\nEq (Node s r) where\neq (Node n1) (Node n2) = n1 == n2\n``````\n\nwork. Adding constraints that `r` must have an `EqRecord` instance also doesn’t help",
null,
"It was referenced previously, but would purescript-variant help you do what you want?\n\nI’ll look into that tomorrow!\n\nI looked into variants but the default decoding for variants in `Simple.JSON` is strangely specific as to the JSON structure that’s expected. I’d have to create my own instance in which case I’d be exactly where I am now with my `readForeignNode` instance expect I’d be decoding into `Variant` instead of a record type so I don’t think it would help.\n\nAlso I really have zero ideas whatsoever why\n\n``````newtype Node (jsonTagValue :: Symbol) r\n= Node { nodeType :: String | r }\n\ninstance eqNode ::\n( RowToList r list\n, EqRecord list r\n, Record.EqualFields list r\n) =>\nEq (Node s r) where\n{-- eq (Node r1) (Node r2) = Record.equal r1 r2 --}\n{-- eq (Node r1) (Node r2) = r1.nodeType == r2.nodeType --}\neq (Node r1) (Node r2) = r1 == r2\n``````\n\ngives me\n\n`````` 32 eq (Node r1) (Node r2) = r1 == r2\n^^^^^^^^\n\nNo type class instance was found for\n\nPrim.RowList.RowToList ( nodeType :: String\n| r1\n)\nt2\n``````\n\nIf I do visual pattern matching I arrive at the following\n\n``````Prim.RowList.RowToList ( nodeType :: String | r1 ) t2 <- from error\nRowToList r list <- from instance\n``````\n\nNow I’ll draw some really wild conclusions. It doesn’t understand the combination of `nodeType :: String` plus the row. And because it derives the `list` from `r` according to my `RowToList` it doesn’t get that either.\n\nBut I also don’t know what I can do here.\n\n1 Like\n\nYou are only asking for RowToList on `r`, but the `Eq` instance for record requires `RowToList` for the entire Record, which is `r` with the additional `nodeType :: String` label. In your instance constraints, wherever there is `r`, it should be `(nodeType :: String | r)`.\n\n1 Like\n\nOh my god it is so obvious now. I was trying wacky stuff with `Union` because I kind of suspected that this was the issue but it never occurred to me to just replace `r` with `( nodeType :: String | r )`",
null,
"I shall build you a monument, thank you so much",
null,
"The reason for decoding into Variant would be to lose the additional data type you are defining to wrap everything into a unified type.\n\n``````data Thing = Foo { bar :: Int } | Bax { bax :: Strong }\n``````\n\nThis is exactly the kind of thing that `Variant` represents. For example:\n\n``````type Thing = Variant (foo :: { bar :: Int }, bax :: { bax :: String })\n\nexample = match\n{ foo: \\{ bar } -> ...\n, bax: \\{ bax } -> ...\n}\n``````"
] |
[
null,
"https://discourse.purescript.org/images/emoji/google/scream.png",
null,
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null,
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null,
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https://www.studysmarter.us/explanations/microeconomics/supply-and-demand/price-elasticity-of-supply/
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[
"### Select your language\n\nSuggested languages for you:",
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"|\n|\n\n## All-in-one learning app\n\n• Flashcards\n• NotesNotes\n• ExplanationsExplanations\n• Study Planner\n• Textbook solutions\n\n# Price elasticity of supply",
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"Save",
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"Print",
null,
"Edit\nPrice elasticity of supply\n• Economic Principles",
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"• Factor Markets",
null,
"• Imperfect Competition",
null,
"• Labour Market",
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"• Market Efficiency",
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"• Microeconomics Examples",
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"• Perfect Competition",
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"• Poverty and Inequality",
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"• Production Cost",
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"• Supply and Demand",
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"Imagine you have a firm that produces computers. Whenever there is a price increase for computers, you would increase the total quantity produced. Conversely, whenever there’s a price decrease, you would also decrease the supply. How quickly would you be able to increase or decrease the supply? What if you needed some more workers to help you produce more computers? By how much would the supply change and how would you measure it?\n\nPrice elasticity of supply helps answer all these questions. It enables you to understand how firms respond to a change in the price of a good or service.\n\n## What is price elasticity of supply?\n\nTo understand the meaning of price elasticity of supply, you have to understand the dynamics of the supply curve in a free market. In a free market, the quantity that a firm chooses to supply is determined by the price of its goods or services.\n\nWhat happens to the quantity supplied when you have a price increase? A movement along the supply curve occurs where the firm increases the total output due to the incentive provided by the price increase. The law of supply states that firms will always choose to increase the total quantity supplied whenever there is a price increase and vice versa. How much will a firm decide to increase its production when there’s a price increase?\n\nPrice elasticity of supply measures how much the total quantity produced changes whenever there is a price change. That is to say, when there’s a price increase, the price elasticity of supply would measure by how much the firm increases its production. You also have the price elasticity of demand, which measures how much the quantity demanded changes in response to a price change.\n\nCheck our explanation on the Price Elasticity of Demand.\n\nYou have different types of elasticity of supply, all of which measure how much quantity supplied is sensitive to the price change. For instance, you could have a relatively inelastic supply where there’s little to no change to the quantity supplied whenever there is a price change.\n\nPrice elasticity of supply measures how much the total quantity produced changes in response to a price change.\n\n## The price elasticity of supply formula\n\nPrice elasticity of supply is calculated as a percentage change in the quantity supplied divided by a percentage change in the price of a good.\n\nThe formula for the price elasticity of supply (PES) is:",
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"You can find a percentage change in a variable by using the following formula:",
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"Assume that a firm produced 10 units of output when the price was £1. As soon as the price increased to £1.5, the firm increased its production from 10 to 20 units.\n\nWhat is the price elasticity of supply?\n\nPercentage change in quantity supplied = (20-10)/10 x100= 100% Percentage change in price = (1.5-1)/1 x 100= 50%\n\nThe price elasticity of supply = 100%/50% = 2\n\nThis means that the quantity supplied is very sensitive to price changes. In this case, the price elasticity of supply is equal to 2, which means that a 1% change in price leads to a 2% change in quantity supplied.\n\n## Types of price elasticity of supply\n\nThere are factors that influence the elasticity of the supply curve, and because of these factors, we have different types of price elasticity of supply.\n\n### Perfectly elastic supply",
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"Figure 1. Perfectly elastic supply, StudySmarter Originals\n\nFigure 1 shows the perfectly elastic supply curve. The price elasticity of a perfectly elastic supply curve is infinite. Firms supply an endless amount of products when there is a perfectly elastic supply. However, the slightest change in price would lead to no quantity being supplied. There are no real-life examples of perfectly elastic supplies.\n\n### Elastic supply",
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"Figure 2. Elastic supply, StudySmarter Originals\n\nFigure 2 shows what an elastic supply curve looks like. An elastic supply occurs when the price elasticity of supply is greater than one. The quantity supplied changes by a greater proportion than the price change. This is very common in the real world, especially for products that are easily produced and don’t require much input.\n\n### Unit elastic supply",
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"Figure 3. Unit elastic supply, StudySmarter Originals\n\nFigure 3 shows what a unit elastic supply curve looks like. A unit elastic supply occurs when the price elasticity of supply is equal to one. When there’s a unit elastic supply, you have proportional changes in output and prices. In other words, the quantity supplied changes by the same proportion as the price change.",
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"Figure 4. Inelastic supply, StudySmarter Originals\n\nFigure 4 shows what the inelastic supply curve looks like. The inelastic supply curve occurs when the price elasticity of supply is smaller than one. The quantity supplied changes by a smaller proportion than the price change. This often occurs in industries where changes in production processes are hard to make in the short run as firms have difficulties adjusting to the price level quickly.",
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"Figure 5. Perfectly inelastic supply, StudySmarter Originals\n\nFigure 5 shows the perfectly inelastic supply curve. Perfectly inelastic supply occurs when the price elasticity of supply equals zero. Regardless of how much the price changes, the quantity supplied will remain static. This happens in the real world. Think about a Picasso painting: no matter how much the price goes up, how many paintings from Picasso are out there?\n\n### Elasticity of supply and market equilibrium\n\nThe elasticity of supply is very important when it comes to demand shifts in the market. That is because it determines by how much the price and quantity of the good will change.",
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"Figure 6. Elasticity of supply and market equilibrium, StudySmarter Originals\n\nFigure 6 shows two shifts in the demand curve. Diagram one shows a shift when the supply is price elastic. In this case, the quantity of goods has increased by a larger proportion than the price increase. That is because the supply was elastic, and it was easier for the firm to increase their total output produced quickly.\n\nOn the other hand, diagram 2 shows what happens when there’s a shift in the demand curve and the supply is inelastic. In this case, the price increases by a larger proportion than the quantity supplied. Think about it. The supply is inelastic, therefore, the firm has more limits in increasing its quantity supplied. Although the demand has increased, the firm could only increase its production by little to match the demand. Therefore, you have a proportionally smaller increase in the quantity supplied.\n\n## Determinants of price elasticity of supply\n\nThe price elasticity of supply measures the response of a firm in terms of quantity supplied whenever there is a price change. But what affects the degree to which the firm can respond to the change in price? There are factors that influence the degree and pace at which firms can adjust their quantity in response to a price change. Determinants of price elasticity of supply refer to factors that either make the supply curve more elastic or inelastic. The main determinants of price elasticity of supply are the following.\n\n### The length of the production period\n\nThis refers to how quick the production process is for producing a certain good. If the firm can quickly adjust its production process and produce output more quickly, it has a relatively more elastic supply curve. However, if the production process takes a lot of time and effort to change the quantity, the firm has then a relatively inelastic supply.\n\n### The availability of spare capacity\n\nWhen the firm has spare capacity that it could use to produce output more quickly, the firm can easily adjust its quantity supplied to the price change. On the other hand, if a firm doesn’t have much spare capacity, it is harder to adjust output to the price change. This way, the availability of spare capacity can influence the elasticity of the supply curve.\n\n### The ease of accumulating stocks\n\nWhen firms can store and keep their unsold goods, they can adjust to the price change quicker. Imagine there’s a sudden price drop; the capacity to store their unsold goods would make their supply more responsive to changes, as the firm could wait to sell its stock at the higher price later. However, if the firm doesn’t have such capacity as it might face high cost or other reasons, it has a more inelastic supply curve.\n\n### Ease of switching production\n\nIf firms are flexible in their production process, this will help them have a more elastic supply, meaning they can adjust much quicker to price changes.\n\n### Market entry barriers\n\nIf there are many barriers to entering the market, it causes the supply curve to become more inelastic. On the other hand, if the market entry barriers are low, the supply curve is more elastic.\n\n### Time scale\n\nTime scale is the period that the firms need to adjust their production inputs. The elasticity of supply tends to be more elastic in the long run rather than the short run. The reason for that is firms have more time to change their inputs, such as buying new capital or hiring and training new labour.\n\nIn the short run, firms are faced with fixed inputs such as capital, which is hard to change in a short period of time. Firms then rely on variable inputs such as labour in the short run, which causes the supply curve to be more inelastic. All these contribute to the elasticity of the supply curve.\n\n## Price elasticity of supply - Key takeaways\n\n• Price elasticity of supply measures how much the total quantity produced changes whenever there is a price change.\n• The elasticity of supply is very important when it comes to demand shifts in the market. That is because it determines by how much the price and quantity of the good will change.\n• The types of elasticity of supply are perfectly elastic, elastic, unit elastic, inelastic, and perfectly inelastic supply.\n• The price elasticity of a perfectly elastic supply curve is infinite at a certain price. However, the slightest change in price would lead to no quantity being supplied.\n• An elastic supply occurs when the price elasticity of supply is greater than one. The quantity supplied changes by a greater proportion than the price change.\n• A unit elastic supply occurs when the price elasticity of supply is equal to one. In other words, the quantity supplied changes by the same proportion as the price change.\n• The inelastic supply curve occurs when the price elasticity of supply is smaller than one. The quantity supplied changes by a smaller proportion than the price change.\n• Perfectly inelastic supply occurs when the price elasticity of supply equals zero. Regardless of how much the price changes, the quantity supplied will remain static.\n• Determinants of price elasticity of supply include the length of the production period, the availability of spare capacity, ease of switching production, market entry barriers, time scale, and the ease of accumulating stocks.\n\n## Frequently Asked Questions about Price elasticity of supply\n\n• The length of the production period\n• The availability of spare capacity\n• The ease of accumulating stocks\n• Ease of switching production\n• Market entry barriers\n• Time scale\n\nPrice elasticity of supply measures how much the total quantity produced changes whenever there is a price change.\n\nThe price elasticity of supply formula is the percentage change in quantity supplied divided by the percentage change in price.\n\nThe types of elasticity of supply are perfectly elastic, elastic, unit elastic, inelastic, and perfectly inelastic supply.\n\n## Final Price elasticity of supply Quiz\n\nQuestion\n\nWhat is price elasticity of supply?\n\nPrice elasticity of supply measures how much the total quantity produced changes whenever there is a price change.\n\nShow question\n\nQuestion\n\nWhat does it mean if the price elasticity of supply is equal to 2?\n\nOne percentage increase in price leads to a two percent increase in quantity.\n\nShow question\n\nQuestion\n\nWhy is elasticity of supply important when it comes to demand shifts in the market?\n\nThe elasticity of supply is very important when it comes to demand shifts in the market. That is because it determines by how much the price and quantity of the good will change.\n\nShow question\n\nQuestion\n\nWhat are the types of elasticity of supply?\n\nThe types of elasticity of supply include perfectly elastic, elastic, unit elastic, inelastic, and perfectly inelastic supply.\n\nShow question\n\nQuestion\n\nExplain perfectly elastic supply.\n\nA perfectly elastic supply curve has its price elasticity infinite at a certain price. However, the slightest change in price would lead to no quantity being supplied.\n\nShow question\n\nQuestion\n\nExplain elastic supply.\n\nAn elastic supply occurs when the price elasticity of supply is greater than one. The quantity supplied changes by a greater proportion than the price change.\n\nShow question\n\nQuestion\n\nExplain unit elastic supply.\n\nA unit elastic supply occurs when the price elasticity of supply is equal to one. In other words, the quantity supplied changes by the same proportion as the price change.\n\nShow question\n\nQuestion\n\nExplain inelastic supply.\n\nThe inelastic supply curve occurs when the price elasticity of supply is smaller than one. The quantity supplied changes by a smaller proportion than the price change.\n\nShow question\n\nQuestion\n\nExplain perfectly inelastic supply.\n\nPerfectly inelastic supply occurs when the price elasticity of supply equals zero. Regardless of how much the price changes, the quantity supplied will remain static.\n\nShow question\n\nQuestion\n\nWhat are the main determinants of price elasticity of supply?\n\nThe determinants of price elasticity of supply are the length of the production period, the availability of spare capacity, the ease of switching production, market entry barriers, time scale, and the ease of accumulating stocks.\n\nShow question\n\nQuestion\n\nHow does the length of the production period affect the elasticity of supply?\n\nThe length of the production period refers to how quick the production process is for producing a certain good. If the firm can quickly adjust its production process and produce output more quickly, it has a relatively more elastic supply curve.\n\nShow question\n\nQuestion\n\nHow does the availability of spare capacity affect the elasticity of supply?\n\nWhen the firm has a spare capacity that it could use to produce output more quickly, the firm can easily adjust its quantity supplied to the price change. On the other hand, if a firm doesn’t have much spare capacity, it is harder to adjust output to the price change.\n\nShow question\n\nQuestion\n\nHow does the ease of accumulating stocks affect the supply curve?\n\nWhen firms can store and keep their unsold goods, they can adjust to the price change quicker. Imagine there’s a sudden price drop, the capacity to store their unsold goods would make their supply more responsive to changes, as the firm could wait to sell its stock at the higher price later.\n\nShow question\n\nQuestion\n\nHow does the ease of switching production affect the elasticity of supply curve?\n\nIf firms are flexible in their production process, this will help them have a more elastic supply curve, meaning they can adjust much quicker to price changes.\n\nShow question\n\nQuestion\n\nHow do market entry barriers affect the elasticity of supply?\n\nIf there are many barriers to entering the market, the supply curve becomes more inelastic. On the other hand, if the market entry barriers are low, the supply curve is more elastic.\n\nShow question\n\nQuestion\n\nHow does time scale affect the elasticity of supply?\n\nThe time scale is the period that firms need to adjust their production inputs. The elasticity of supply tends to be more elastic in the long run rather than the short run.\n\nShow question\n\nMore about Price elasticity of supply",
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https://theclevermachine.wordpress.com/category/algorithms/gradient-descent/
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[
"## Derivation: Maximum Likelihood for Boltzmann Machines\n\nIn this post I will review the gradient descent algorithm that is commonly used to train the general class of models known as Boltzmann machines. Though the primary goal of the post is to supplement another post on restricted Boltzmann machines, I hope that those readers who are curious about how Boltzmann machines are trained, but have found it difficult to track down a complete or straight-forward derivation of the maximum likelihood learning algorithm for these models (as I have), will also find the post informative.\n\nFirst, a little background: Boltzmann machines are stochastic neural networks that can be thought of as the probabilistic extension of the Hopfield network. The goal of the Boltzmann machine is to model a set of observed data in terms of a set of visible random variables",
null,
"$v$ and a set of latent/unobserved random variables",
null,
"$h$. Due to the relationship between Boltzmann machines and neural networks, the random variables are often are often referred to as “units.” The role of the visible units is to approximate the true distribution of the data, while the role of the latent variables it to extend the expressiveness of the model by capturing underlying features in the observed data. The latent variables are often referred to as hidden units, as they do not result directly from the observed data and are generally marginalized over to obtain the likelihood of the observed data, i.e.",
null,
"$\\Large{\\begin{array}{rcl} p(v;\\theta) &=& \\sum_h p(v,h; \\theta) \\end{array}}$,\n\nwhere",
null,
"$p(v,h; \\theta)$ is the joint probability distribution over the visible and hidden units based on the current model parameters",
null,
"$\\theta$. The general Boltzmann machine defines",
null,
"$p(v,h; \\theta)$ through a set of weighted, symmetric connections between all visible and hidden units (but no connections from any unit to itself). The graphical model for the general Boltzmann machine is shown in Figure 1.",
null,
"Figure 1: Graphical Model of the Boltzmann machine (biases not depicted).\n\nGiven the current state of the visible and hidden units, the overall configuration of the model network is described by a connectivity function",
null,
"$E(v,h;\\theta)$, parameterized by",
null,
"$\\theta = {W, A, B, a, b}$:",
null,
"$\\Large{\\begin{array}{rcl} E(v,h; \\theta) &=& v^T W h + h^T A h + v^T B v + h^T a + v^T b \\end{array}}.$\n\nThe parameter matrix",
null,
"$W$ defines the connection strength between the visible and hidden units. The parameters",
null,
"$A$ and",
null,
"$B$ define the connection strength amongst hidden units and visible units, respectively. The model also includes a set of biases",
null,
"$a$ and",
null,
"$b$ that capture offsets for each of the hidden and visible units.\n\nThe Boltzmann machine has been used for years in field of statistical mechanics to model physical systems based on the principle of energy minimization. In the statistical mechanics, the connectivity function is often referred to the “energy function,” a term that is has also been standardized in the statistical learning literature. Note that the energy function returns a single scalar value for any configuration of the network parameters and random variable states.\n\nGiven the energy function, the Boltzmann machine models the joint probability of the visible and hidden unit states as a Boltzmann distribution:",
null,
"$\\Large{\\begin{array}{rcl} p(v,h; \\theta) &=& \\frac{\\mathrm{e}^{-E(v,h; \\theta)}}{Z(\\theta)} \\text{ , where} \\\\ \\\\ Z(\\theta) &=& \\sum_{v'} \\sum_{h'} \\mathrm{e}^{-E(v',h'; \\theta)}\\end{array}}$\n\nThe partition function",
null,
"$Z(\\theta)$ is a normalizing constant that is calculated by summing over all possible states of the network",
null,
"$(v', h') \\in (V',H')$. Here we assume that all random variables take on discrete values, but the analogous derivation holds for continuous or mixed variable types by replacing the sums with integrals accordingly.\n\nThe common way to train the Boltzmann machine is to determine the parameters that maximize the likelihood of the observed data. To determine the parameters, we perform gradient descent on the log of the likelihood function (In order to simplify the notation in the remainder of the derivation, we do not include the explicit dependency on the parameters",
null,
"$\\theta$. To further simplify things, let’s also assume that we calculate the gradient of the likelihood based on a single observation.):",
null,
"$\\Large{ \\begin{array}{rcl} l(v; \\theta) &=& \\log p(v) \\\\ &=& \\log \\sum_h p(v,h) \\\\ &=& \\log \\frac{\\sum_h \\mathrm{e}^{-E(v,h)}}{Z} \\\\ &=& \\log \\sum_h \\mathrm{e}^{-E(v,h)} - \\log Z \\\\ &=& \\log \\sum_h \\mathrm{e}^{-E(v,h)} - \\sum_{v'} \\sum_{h'} \\mathrm{e}^{-E(v',h')} \\end{array}}$\n\nThe gradient calculation is as follows:",
null,
"$\\Large{ \\begin{array}{rcl} \\frac{\\partial l(v;\\theta)}{\\partial \\theta} &=& \\frac{\\partial}{\\partial \\theta}\\log \\sum_h \\mathrm{e}^{-E(v,h)} - \\frac{\\partial}{\\partial \\theta} \\log \\sum_{v'}\\sum_{h'}\\mathrm{e}^{-E(v',h')} \\\\ &=& \\frac{1}{\\sum_h \\mathrm{e}^{-E(v,h)}} \\frac{\\partial}{\\partial \\theta} \\sum_h \\mathrm{e}^{-E(v,h)} - \\frac{1}{\\sum_{v'}\\sum_{h'}\\mathrm{e}^{-E(v',h')}} \\frac{\\partial}{\\partial \\theta} \\sum_{v'}\\sum_{h'}\\mathrm{e}^{-E(v',h')} \\\\ &=& - \\frac{1}{\\sum_h \\mathrm{e}^{-E(v,h)}} \\sum_h \\mathrm{e}^{-E(v,h)}\\frac{\\partial E(v,h)}{\\partial \\theta} + \\frac{1}{\\sum_{v'}\\sum_{h'}\\mathrm{e}^{-E(v',h')}} \\sum_{v'}\\sum_{h'}\\mathrm{e}^{-E(v',h')}\\frac{\\partial E(v',h')}{\\partial \\theta} \\end{array}}$\n\nHere we can simplify the expression somewhat by noting that",
null,
"$\\mathrm{e}^{-E(v,h)} = Z p(v,h)$, that",
null,
"$Z = \\sum_{v'}\\sum_{h'}\\mathrm{e}^{-E(v',h')}$, and also that",
null,
"$Z$ is a constant:",
null,
"$\\Large{ \\begin{array}{rcl} \\frac{\\partial l(v;\\theta)}{\\partial \\theta} &=& - \\frac{1}{Z\\sum_h p(v,h)} Z \\sum_h p(v,h) \\frac{\\partial E(v,h)}{\\partial \\theta} + \\frac{1}{Z} Z \\sum_{v'}\\sum_{h'}p(v',h')\\frac{\\partial E(v',h')}{\\partial \\theta} \\\\ &=& - \\frac{1}{\\sum_h p(v,h)} \\sum_h p(v,h) \\frac{\\partial E(v,h)}{\\partial \\theta} + \\sum_{v'}\\sum_{h'}p(v',h')\\frac{\\partial E(v',h')}{\\partial \\theta} \\\\ \\end{array}}$\n\nIf we also note that",
null,
"$\\sum_h p(v,h)= p(v)$, and use the definition of conditional probability",
null,
"$p(h|v) = \\frac{p(v,h)}{p(v)}$, we can further simplify the expression for the gradient:",
null,
"$\\Large{ \\begin{array}{rcl} \\frac{\\partial l(v;\\theta)}{\\partial \\theta} &=& - \\frac{1}{p(v)} \\sum_h p(v,h) \\frac{\\partial E(v,h)}{\\partial \\theta} + \\sum_{v'}\\sum_{h'}p(v',h')\\frac{\\partial E(v',h')}{\\partial \\theta} \\\\ &=& -\\sum_h \\frac{p(v,h)}{p(v)} \\frac{\\partial E(v,h)}{\\partial \\theta} + \\sum_{v'}\\sum_{h'}p(v',h')\\frac{\\partial E(v',h')}{\\partial \\theta} \\\\ &=& -\\sum_h p(h | v) \\frac{\\partial E(v,h)}{\\partial \\theta} + \\sum_{v'}\\sum_{h'}p(v',h')\\frac{\\partial E(v',h')}{\\partial \\theta} \\\\ &=& -\\mathbb{E}_{p(h | v)} \\frac{\\partial E(v,h)}{\\partial \\theta} + \\mathbb{E}_{p(v',h')}\\frac{\\partial E(v',h')}{\\partial \\theta}. \\\\ \\end{array}}$\n\nHere",
null,
"$\\mathbb{E}_{p(*)}$ is the expected value under the distribution",
null,
"$p(*)$. Thus the gradient of the likelihood function is composed of two parts. The first part is expected gradient of the energy function with respect to the conditional distribution",
null,
"$p(h|v)$. The second part is expected gradient of the energy function with respect to the joint distribution over all variable states. However, calculating these expectations is generally infeasible for any realistically-sized model, as it involves summing over a huge number of possible states/configurations. The general approach for solving this problem is to use Markov Chain Monte Carlo (MCMC) to approximate these sums:",
null,
"$\\Large{\\begin{array}{rcl} \\frac{\\partial l(v;\\theta)}{\\partial \\theta} &\\approx& -\\left \\langle \\frac{\\partial E(v,h)}{\\partial \\theta} \\right \\rangle_{p(h_{\\text{data}}|v_{\\text{data}})} + \\left \\langle \\frac{\\partial E(v,h)}{\\partial \\theta} \\right \\rangle_{p(h_{\\text{model}}|v_{\\text{model}})} \\\\ \\end{array}}.$\n\nHere",
null,
"$\\langle \\rangle_{p(*)}$ is the sample average of samples drawn according to the process",
null,
"$p(*)$. The first term is calculated by taking the average value of the energy function gradient when the visible and hidden units are being driven by observed data samples. In practice, this first term is generally straightforward to calculate. Calculating the second term is generally more complicated and involves running a set of Markov chains until they reach the current model’s equilibrium distribution (i.e. via Gibbs sampling, Metropolis-Hastings, or the like), then taking the average energy function gradient based on those samples. See this post on MCMC methods for details. It turns out that there is a subclass of Boltzmann machines that, due to a restricted connectivity/energy function (specifically, the parameters",
null,
"$(A, B)=0$), allow for efficient MCMC by way of blocked Gibbs sampling. These models, known as restricted Boltzman machines have become an important component for unsupervised pretraining in the field of deep learning and will be the focus of a related post.\n\n## A Gentle Introduction to Artificial Neural Networks\n\nThe material in this post has been migrated to a post by the same name on my github pages website.\n\n## Derivation: Error Backpropagation & Gradient Descent for Neural Networks\n\nThe material in this post has been migraged with python implementations to my github pages website."
] |
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|
https://stats.stackexchange.com/questions/257603/high-r2-squared-and-high-p-value-for-simple-linear-regression/257620
|
[
"# High $R^2$ squared and high $p$-value for simple linear regression\n\nLet's assume that we have simple linear regression: $\\hat{y} = bx + \\text{intercept}$.\n\nIs it possible to have a high p-value and high $R^2$ (or low p-value and low $R^2$)? I've been looking for examples of this. When the linear regression has multiple parameters, I saw some examples where p-value for some parameters are low, but overall $R^2$ is low as well, but I was wondering if it's possible for the linear regression of a single parameter.\n\nYes, it is possible. The $R^2$ and the $t$ statistic (used to compute the p-value) are related exactly by:\n\n$|t| = \\sqrt{\\frac{R^2}{(1- R^2)}(n -2)}$\n\nTherefore, you can have a high $R^2$ with a high p-value (a low $|t|$) if you have a small sample.\n\nFor instance, take $n = 3$. For this sample size to give you a (two-sided) p-value less then 10% you would need an $R^2$ greater than 85% -- anything less than that would give you \"non-significant\" p-value.\n\nAs a concrete example, the simulation below produces an $R^2$ close to 0.5 with a p-value of $0.516$.\n\nset.seed(10)\nn <- 3\nx <- rnorm(n, 0, 1)\ny <- 1 + x + rnorm(n, 0, 1)\nsummary(m1 <- lm(y ~ x))\n\nCall:\nlm(formula = y ~ x)\n\nResiduals:\n1 2 3\n-0.36552 0.42802 -0.06251\n\nCoefficients:\nEstimate Std. Error t value Pr(>|t|)\n(Intercept) 0.7756 0.4261 1.82 0.320\nx 0.5065 0.5333 0.95 0.516\n\nResidual standard error: 0.5663 on 1 degrees of freedom\nMultiple R-squared: 0.4743, Adjusted R-squared: -0.05148\nF-statistic: 0.9021 on 1 and 1 DF, p-value: 0.5164\n\n\nFor the opposite case (low p-value with low $R^2$), you can trivially obtain that by setting a regression where $x$ has a low explanatory power and let $n \\to \\infty$ to get a p-value as small as you want.\n\n• If x and y are uncorrelated noise, then one should have a low p value, otoh a large sample size should give a decent R^2. Shouldn't that work? – meh Jan 23 '17 at 1:29\n• @aginensky no, it shouldn't. A large sample size improves how well you estimate R^2 and not how big R^2 is. If x and y are uncorrelated, your R^2 will converge to zero as n -> infty. – Carlos Cinelli Jan 23 '17 at 1:40\n• @carloscinelli thanks for your answer. I guess assuming the sample size is sufficiently high, it's impossible to have both high R^2 and p-value at the same time for simple linear regression like this. – user98235 Jan 23 '17 at 5:02\n• @user98235 yes, and you can actually compute this exactly. For instance, if n = 102, then any R^2 > 4% will give p-values < 5%. – Carlos Cinelli Jan 23 '17 at 5:09\n• @ carloscinelli - of course, my bad ! – meh Jan 23 '17 at 10:19\n\nThis looks like a self-study, so I'll offer a hint: Is either or both of these measures (R-square and p-value) related to the sample size?"
] |
[
null
] |
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|
https://wiki.haskell.org/OOP_vs_type_classes
|
[
"# OOP vs type classes\n\n(this is just a sketch now. feel free to edit/comment it. I will include information you provided into the final version of this tutorial)\n\nI had generally not used type classes in my application programs, but when I'd gone to implement general purpose libraries and tried to maintain as much flexibility as possible, it was natural to start building large and complex class hierarchies. I tried to use my C++ experience when doing this but I was bitten many times by the restrictions of type classes. After this experience, I think that I now have a better feeling and mind model for type classes and I want to share it with other Haskellers - especially ones having OOP backgrounds.\n\nBrian Hulley provided us with the program that emulates OOP in Haskell - as you can see, it's much larger than equivalent C++ program. An equivalent translation from Haskell to C++ should be even longer :)\n\n## Everything is an object?\n\nMost software developers are familiar with the OOP motto \"everything is an object.\" People accustomed to C++ classes often find the Haskell concept of type classes difficult to grasp. Why is it so different?\n\nC++ classes pack functions together with data, which makes it convenient to represent and consume data. Use of interfaces (abstract classes) allow classes to interact by contract, instead of directly manipulating the data in the other class. There exist alternative ways in C++ to accomplish such functionality (function pointers, discriminated unions), yet these techniques are not as handy as classes. Classes are also the primary way to hiding implementation details. Moreover, classes represent a handy way to group related functionality together. It's extremely useful to browse the structure of large C++ project in terms of classes instead of individual functions.\n\nHaskell provides other solutions for these problems.\n\n### Type with several representations: use algebraic data type (ADT)\n\nFor the types with different representations, algebraic data types (ADT) - an analog of discriminated unions - are supported:\n\n```data Point = FloatPoint Float Float\n| IntPoint Int Int\n```\n\nHaskell provides a very easy way to build/analyze them:\n\n```coord :: Point -> (Float, Float)\ncoord (FloatPoint x y) = (x,y)\ncoord (IntPoint x y) = (realToFrac x, realToFrac y)\n\nmain = do print (coord (FloatPoint 1 2))\nprint (coord (IntPoint 1 2))\n```\n\nSo ADTs in general are preferred in Haskell over the class-based solution of the same problem:\n\n```class Point a where\ncoord :: a -> (Float, Float)\n\ndata FloatPoint = FloatPoint Float Float\ninstance Point FloatPoint where\ncoord (FloatPoint x y) = (x,y)\n\ndata IntPoint = IntPoint Int Int\ninstance Point IntPoint where\ncoord (IntPoint x y) = (realToFrac x, realToFrac y)\n```\n\nThe equivalent C++ implementation using inheritance requires much more machinery than our 5 line, ADT-based solution. This also illustrates a Haskell benefit--it's much easier to define types/functions. Perhaps objects are not as great as you thought before. :D\n\n```#include <algorithm>\n#include <iostream>\nusing namespace std;\n\nostream & operator<<(ostream & lhs, pair<float, float> const& rhs) {\nreturn lhs << \"(\" << rhs.first << \",\" << rhs.second << \")\";\n}\n\nstruct Point {\nvirtual pair<float,float> coord() = 0;\n};\n\nstruct FloatPoint : Point {\nfloat x, y;\nFloatPoint (float _x, float _y) : x(_x), y(_y) {}\npair<float,float> coord() {return make_pair(x,y); }\n};\n\nstruct IntPoint : Point {\nint x, y;\nIntPoint (int _x, int _y) : x(_x), y(_y) {}\npair<float,float> coord() { return make_pair(x,y); }\n};\n\nint main () {\ncout << FloatPoint(1,2).coord();\ncout << IntPoint(1,2).coord();\nreturn 0;\n}\n```\n\nAs you see, ADTs together with type inference make Haskell programs about 2 times smaller than their C++ equivalent.\n\n### Packing data & functions together: use (records of) closures\n\nAnother typical class use-case is to pack data together with one or more processing functions and pass this bunch to some function. Then this function can call the aforementioned functions to implement some functionality, not bothering how it is implemented internally. Hopefully Haskell provides a better way: you can pass any functions as parameters to other functions directly. Moreover, such functions can be constructed on-the-fly, capturing free variables in context, creating the so-called closures. In this way, you construct something like object on-demand and don't even need a type class:\n\n```do x <- newIORef 0\nproc (modifyIORef x (+1), readIORef x)\n```\n\nHere, we applied proc to two functions - one incrementing the value of a counter and another reading its current value. Another call to proc that uses counter with locking, might look like this:\n\n```do x <- newMVar 0\nproc (modifyMVar x (+1), readMVar x)\n```\n\nHere, proc may be defined as:\n\n```proc :: (IO (), IO Int) -> IO ()\nproc (inc, read) = do { inc; inc; inc; read >>= print }\n```\n\ni.e. it receive two abstract operations whose implementation may vary in different calls to proc and call them without any knowledge of implementation details. The equivalent C++ code could look like this:\n\n```class Counter {\npublic:\nvirtual void inc() = 0;\n};\n\nclass SimpleCounter : public Counter {\npublic:\nSimpleCounter() { n = 0; }\nvoid inc() { n++; }\nint read() { return n; }\nprivate:\nint n;\n};\n\nvoid proc (Counter &c) {\nc.inc(); c.inc(); c.inc(); cout << c.read();\n}\n```\n\nAnd again, Haskell code is much simpler and more straightforward - we don't need to declare classes, operations, their types - we just pass to the proc implementation of operations it needs. Look at IO inside#Example: returning an IO action as a result and following sections to find more examples of using closures instead of OOP classes.\n\n### Hiding implementation details: use module export list\n\nOne more usage of OOP classes is to hide implementation details, making internal data/functions inaccessible to class clients. Unfortunately, this functionality is not part of type class facilities. Instead, you should use the sole Haskell method of encapsulation, module export list:\n\n```module Stack (Stack, empty, push, pop, top, isEmpty) where\n\nnewtype Stack a = Stk [a]\n\nempty = Stk []\npush x (Stk xs) = Stk (x:xs)\npop (Stk (x:xs)) = Stk xs\ntop (Stk (x:xs)) = x\nisEmpty (Stk xs) = null xs\n```\n\nSince the constructor for the data type Stack is hidden (the export list would say Stack(Stk) if it were exposed), outside of this module a stack can only be built from operations empty, push and pop, and examined with top and isEmpty.\n\n### Grouping related functionality: use module hierarchy and Haddock markup\n\nDividing a whole program into classes and using their hierarchy to represent entire an program structure is a great instrument for OO languages. Unfortunately, it's again impossible in Haskell. Instead, the structure of a program is typically rendered in a module hierarchy and inside a module - in its export list. Although Haskell doesn't provide facilities to describe a hierarchical structure inside of a module, we have another tool to do it - Haddock, a de-facto standard documentation tool.\n\n```module System.Stream.Instance (\n\n-- * File is a file stream\nFile,\n-- ** Functions that open files\nopenFile, -- open file in text mode\nopenBinaryFile, -- open file in binary mode\n-- ** Standard file handles\nstdin,\nstdout,\nstderr,\n\n-- * MemBuf is a memory buffer stream\nMemBuf,\n-- ** Functions that open MemBuf\ncreateMemBuf, -- create new MemBuf\ncreateContiguousMemBuf, -- create new contiguous MemBuf\nopenMemBuf, -- use memory area as MemBuf\n\n) where\n...\n```\n\nHere, Haddock will build documentation for a module using its export list. The export list will be divided into sections (whose headers given with \"-- *\") and subsections (given with \"-- **\"). As a result, module documentation reflects its structure without using classes for this purpose.\n\n## Type classes is a sort of templates, not classes\n\nAt this moment, C++ has classes and templates. What is the difference? With a class, type information is carried with the object itself while with templates it's outside of the object and is part of the whole operation.\n\nFor example, if the == operation is defined as a virtual method in a class, the actual procedure called for a==b may depend on the run-time type of 'a', but if the operation is defined in template, the actual procedure depends only on the instantiated template (which is determined at compile time).\n\nHaskell's objects don't carry run-time type information. Instead, the class constraint for a polymorphic operation is passed in as a \"dictionary\" implementing all operations of the class (there are also other implementation techniques, but this doesn't matter). For example,\n\n```eqList :: (Eq a) => [a] -> [a] -> Bool\n```\n\nis translated into:\n\n```type EqDictionary a = (a->a->Bool, a->a->Bool)\neqList :: EqDictionary a -> [a] -> [a] -> Bool\n```\n\nwhere the first parameter is a \"dictionary\" containing the implementation of \"==\" and \"/=\" operations for objects of type 'a'. If there are several class constraints, a dictionary for each is passed.\n\nIf the class has base class(es), the dictionary tuple includes the base class dictionaries, so\n\n```class Eq a => Cmp a where\ncmp :: a -> a -> Ordering\n\ncmpList :: (Cmp a) => [a] -> [a] -> Ordering\n```\n\nturns into:\n\n```type CmpDictionary a = (EqDictionary a, a -> a -> Ordering)\ncmpList :: CmpDictionary a -> [a] -> [a] -> Ordering\n```\n\nCompared to C++, this is more like templates, not classes! As with templates, type information is part of operation, not the object! But while C++ templates are really a form of macro-processing (like Template Haskell) and at last end generate non-polymorphic code, Haskell's use of dictionaries allows run-time polymorphism (explanation of run-time polymorphism? -what is this? a form of dynamic dispatch?).\n\nMoreover, Haskell type classes support inheritance. Run-time polymorphism together with inheritance are often seen as OOP distinctive points, so during long time I considered type classes as a form of OOP implementation. But that's wrong! Haskell type classes build on a different basis, so they are like C++ templates with added inheritance and run-time polymorphism! And this means that the usage of type classes is different from using classes, with its own strong and weak points.\n\n## Type classes vs classes\n\nHere is a brief listing of differences between OOP classes and Haskell type classes\n\n### Type classes are like interfaces/abstract classes, not classes itself\n\nThere is no inheritance and data fields (so type classes are more like interfaces than classes)....\n\nFor those more familiar with Java/C# rather than C++, type classes resemble interfaces more than the classes. In fact, the generics in those languages capture the notion of parametric polymorphism (but Haskell is a language that takes parametric polymorphism quite seriously, so you can expect a fair amount of type gymnastics when dealing with Haskell), so more precisely, type classes are like generic interfaces.\n\nWhy interface, and not class? Mostly because type classes do not implement the methods themselves, they just guarantee that the actual types that instantiate the type class will implement specific methods. So the types are like classes in Java/C#.\n\nOne added twist: type classes can decide to provide default implementation of some methods (using other methods). You would say, then they are sort of like abstract classes. Right. But at the same time, you cannot extend (inherit) multiple abstract classes, can you?\n\nSo a type class is sort of like a contract: \"any type that instantiates this type class will have the following functions defined on them...\" but with the added advantage that you have type parameters built-in, so:\n\n```class Eq a where\n(==) :: a -> a -> Bool\n(/=) :: a -> a -> Bool\n-- let's just implement one function in terms of the other\nx /= y = not (x == y)\n```\n\nis, in a Java-like language:\n\ninterface Eq<A> {\n\n``` boolean equal(A that);\nboolean notEqual(A that) {\n// default, can be overriden\nreturn !equal(that);\n}\n```\n\n}\n\nAnd the \"instance TypeClass ParticularInstance where ...\" definition means \"ParticularInstance implements TypeClass { ... }\", now, multiple parameter type classes, of course, cannot be interpreted this way.\n\n### Type can appear at any place in function signature\n\nType can appear at any place in function signature: be any parameter, inside parameter, in a list (possibly empty), or in a result\n\n```class C a where\nf :: a -> Int\ng :: Int -> a -> Int\nh :: Int -> (Int,a) -> Int\ni :: [a] -> Int\nj :: Int -> a\nnew :: a\n```\n\nIt's even possible to define instance-specific constants (look at 'new').\n\nIf function value is instance-specific, OOP programmer will use \"static\" method while with type classes you need to use fake parameter:\n\n```class FixedSize a where\nsizeof :: a -> Int\ninstance FixedSize Int8 where\nsizeof _ = 1\ninstance FixedSize Int16 where\nsizeof _ = 2\n\nmain = do print (sizeof (undefined::Int8))\nprint (sizeof (undefined::Int16))\n```\n\n### Inheritance between interfaces\n\nInheritance between interfaces (in \"class\" declaration) means inclusion of base class dictionaries in dictionary of subclass:\n\n```class (Show s, Monad m s) => Stream m s where\nsClose :: s -> m ()\n```\n\nmeans\n\n```type StreamDictionary m s = (ShowDictionary s, MonadDictionary m s, s->m())\n```\n\nThere is upcasting mechanism, it just extracts dictionary of a base class from a dictionary tuple, so you can run a function that requires base class from a function that requires subclass:\n\n```f :: (Stream m s) => s -> m String\nshow :: (Show s) => s -> String\nf s = return (show s)\n```\n\nBut downcasting is absolutely impossible - there is no way to get subclass dictionary from a superclass one\n\n### Inheritance between instances\n\nInheritance between instances (in \"instance\" declaration) means that operations of some class can be executed via operations of other class, i.e. such declaration describe a way to compute dictionary of inherited class via functions from dictionary of base class:\n\n```class Eq a where\n(==) :: a -> a -> Bool\nclass Cmp a where\ncmp :: a -> a -> Ordering\ninstance (Cmp a) => Eq a where\na==b = cmp a b == EQ\n```\n\ncreates the following function:\n\n```cmpDict2EqDict :: CmpDictionary a -> EqDictionary a\ncmpDict2EqDict (cmp) = (\\a b -> cmp a b == EQ)\n```\n\nThis results in that any function that receives dictionary for Cmp class can call functions that require dictionary of Eq class\n\n### Downcasting is a mission impossible\n\nSelection between instances is done at compile-time, based only on information present at the moment. So don't expect that more concrete instance will be selected just because you passed this concrete datatype to the function which accepts some general class:\n\n```class Foo a where\nfoo :: a -> String\n\ninstance (Num a) => Foo a where\nfoo _ = \"Num\"\n\ninstance Foo Int where\nfoo _ = \"int\"\n\nf :: (Num a) => a -> String\nf = foo\n\nmain = do print (foo (1::Int))\nprint (f (1::Int))\n```\n\nHere, the first call will return \"int\", but second - only \"Num\". this can be easily justified by using dictionary-based translation as described above. After you've passed data to polymorphic procedure it's type is completely lost, there is only dictionary information, so instance for Int can't be applied. The only way to construct Foo dictionary is by calculating it from Num dictionary using the first instance.\n\nRemark: This isn't even a legal program unless you use the `IncoherentInstances` language extension. The error message:\n``` Overlapping instances for Foo a\narising from a use of `foo' at /tmp/I.hs:17:4-6\nMatching instances:\ninstance [overlap ok] (Num a) => Foo a\n-- Defined at /tmp/I.hs:10:9-24\ninstance [overlap ok] Foo Int -- Defined at /tmp/I.hs:13:9-15\n(The choice depends on the instantiation of `a'\nTo pick the first instance above, use -XIncoherentInstances\nwhen compiling the other instance declarations)\n```\nDetails: GHC User's Guide\n\n### There is only one dictionary per function call\n\nFor \"eqList :: (Eq a) => [a] -> [a] -> Bool\" types of all elements in list must be the same, and types of both arguments must be the same too - there is only one dictionary and it know how to handle variables of only one concrete type!\n\n### Existential variables is more like OOP objects\n\nExistential variables pack dictionary together with variable (looks very like the object concept!) so it's possible to create polymorphic containers (i.e. holding variables of different types). But downcasting is still impossible. Also, existentials still don't allow to mix variables of different types in a call to some polymorhic operation (their personal dictionaries still built for variables of one concrete type):\n\n```data HasCmp = forall a. Cmp a => HasCmp a\n\nsorted :: [HasCmp] -> Ordering\n\nsorted [] = True\nsorted [_] = True\nsorted (HasCmp a : HasCmp b : xs) = a<=b && sorted (b:xs)\n```\n\nThis code will not work - a<=b can use nor 'a' neither 'b' dictionary. Even if orderings for apples and penguins are defined, we still don't have a method to compare penguins to apples!\n\n## Other opinions\n\n### OO class always corresponds to a haskell class + a related haskell existential (John Meacham)\n\n```> Roughly Haskell type classes correspond to parameterized abstract\n> classes in C++ (i.e. class templates with virtual functions\n> representing the operations). Instance declarations correspond to\n> derivation and implementations of those parameterized classes.\n```\n\nThere is a major difference though, in C++ (or java, or sather, or c#, etc.) the dictionary is always attached to the value, the actual class data type you pass around. In Haskell, the dictionary is passed separately and the appropriate one is inferred by the type system. C++ doesn't infer, it just assumes everything will be carrying around its dictionary with it.\n\nThis makes Haskell classes significantly more powerful in many ways.\n\n```class Num a where\n(+) :: a -> a -> a\n```\n\nis impossible to express in OO classes: since both arguments to + necessarily carry their dictionaries with them, there is no way to statically guarantee they have the same one. Haskell will pass a single dictionary that is shared by both types so it can handle this just fine.\n\n```class Monoid a where\nmempty :: a\n```\n\nIn OOP, this cannot be done because where does the dictionary come from? Since dictionaries are always attached to a concrete class, every method must take at least one argument of the class type (in fact, exactly one, as I'll show below). In Haskell again, this is not a problem since the dictionary is passed in by the consumer of 'mempty' - mempty need not conjure one out of thin air.\n\nIn fact, OO classes can only express single parameter type classes where the type argument appears exactly once in strictly covariant position. In particular, it is pretty much always the first argument and often (but not always) named 'self' or 'this'.\n\n```class HasSize a where\ngetSize :: a -> Int\n```\n\ncan be expressed in OO, 'a' appears only once, as its first argument.\n\nNow, another thing OO classes can do is they give you the ability to create existential collections (?) of objects. As in, you can have a list of things that have a size. In Haskell, the ability to do this is independent of the class (which is why Haskell classes can be more powerful) and is appropriately named existential types.\n\n```data Sized = exists a . HasSize a => Sized a\n```\n\nWhat does this give you? You can now create a list of things that have a size [Sized] yay!\n\nAnd you can declare an instance for Sized, so you can use all your methods on it.\n\n```instance HasSize Sized where\ngetSize (Sized a) = getSize a\n```\n\nAn existential, like Sized, is a value that is passed around with its dictionary in tow, as in, it is an OO class! I think this is where people get confused when comparing OO classes to Haskell classes. _There is no way to do so without bringing existentials into play_. OO classes are inherently existential in nature.\n\nSo, an OO abstract class declaration declares the equivalent of 3 things in Haskell: a class to establish the methods, an existential type to carry the values about, and an instance of the class for the existential type.\n\nAn OO concrete class declares all of the above plus a data declaration for some concrete representation.\n\nOO classes can be perfectly (even down to the runtime representation!) emulated in Haskell, but not vice versa. Since OO languages tie class declarations to existentials, they are limited to only the intersection of their capabilities, because Haskell has separate concepts for them; each is independently much much more powerful.\n\n```data CanApply = exists a b . CanApply (a -> b) a (b -> a)\n```\n\nis an example of something that cannot be expressed in OO, existentials are limited to having exactly a single value since they are tied to a single dictionary.\n\n```class Num a where\n(+) :: a -> a -> a\nzero :: a\nnegate :: a -> a\n```\n\ncannot be expressed in OO, because there is no way to pass in the same dicionary for two elements, or for a returning value to conjure up a dictionary out of thin air. (If you are not convinced, try writing a 'Number' existential and making it an instance of Num and it will be clear why it is not possible.)\n\nnegate is an interesting one - there is no technical reason it cannot be implemented in OO languages, but none seem to actually support it.\n\nSo, when comparing, remember an OO class always corresponds to a Haskell class + a related Haskell existential.\n\nIncidentally, an extension I am working on is to allow\n\n```data Sized = exists a . HasSize a => Sized a\nderiving(HasSize)\n```\n\nwhich would have the obvious interpretation. Obviously it would only work under the same limitations as OO classes have, but it would be a simple way for haskell programs to declare OO style classes if they so choose.\n\n(Actually, it is still signifigantly more powerful than OO classes since you can derive many instances, and even declare your own for classes that don't meet the OO constraints. Also, your single class argument need not appear as the first one. It can appear in any strictly covariant position, and it can occur as often as you want in contravariant ones!)\n\n### Type classes correspond to parameterized abstract classes (Gabriel Dos Reis)\n\n```| > Roughly Haskell type classes correspond to parameterized abstract\n| > classes in C++ (i.e. class templates with virtual functions\n| > representing the operations). Instance declarations correspond to\n| > derivation and implementations of those parameterized classes.\n|\n| There is a major difference though, in C++ (or java, or sather, or c#,\n| etc..) the dictionary is always attached to the value, the actual class\n| data type you pass around.\n```\n\nI suspect that most of the confusion come from the fact that people believe just because virtual functions are attached to objects, they cannot attach them to operations outside classes. That, to my surprise, hints at a deeper misappreciation of both type classes and so-called \"OO\" technology. Type classes are more OO than one might realize.\n\nThe dictionary can be attached to the operations (not just to the values) by using objects local to functions (which sort of matierialize the dictionary). Consider\n\n``` // Abstract class for a collection of classes that implement\n// the \"Num\" mathematical structure\ntemplate<typename T>\nstruct Num {\nvirtual T add(T, T) const = 0;\n};\n```\n``` // Every type must specialize this class template to assert\n// membership to the \"Num\" structure.\ntemplate<typename T> struct Num_instance;\n```\n``` // The operation \"+\" is defined for any type that belongs to \"Num\".\n// Notice, membership is asserted aby specializing Num_instance<>.\ntemplate<typename T>\nT operator+(T lhs, T rhs)\n{\nconst Num_instance<T> instance;\n}\n```\n``` // \"Foo\" is in \"Num\"\nstruct Num_instance<Foo> : Num<Foo> {\nFoo add(Foo a, Foo b) const { ... }\n};\n```\n\nThe key here is in the definition of operator+ which is just a formal name for the real operation done by instance.add().\n\nI appreciate that inferring and building the dictionary (represented here by the \"instance\" local to operator+<T>) is done automatically by the Haskell type system. That is one of the reasons why the type class notation is a nice sugar. However, that should not distract from its deerper OO semantics.\n\n[...]\n\n```| in haskell you can do\n|\n| class Monoid a where\n| mempty :: a\n|\n| in OOP, this cannot be done because where does the dicionary come from?\n```\n\nSee above. I believe a key in my suggestion was \"parameterized abstract classes\", not just \"abstract classes\".\n\n## Haskell emulation of OOP inheritance with record extension\n\nBrian Hulley provided us the code that shows how OOP inheritance can be emulated in Haskell. His translation method supports data fields inheritance, although doesn't support downcasting.\n\n```> although i mentioned not only pluses but also drawbacks of type\n> classes: lack of record extension mechanisms (such at that implemented\n> in O'Haskell) and therefore inability to reuse operation\n> implementation in an derived data type...\n```\n\nYou can reuse ops in a derived data type but it involves a tremendous amount of boilerplate. Essentially, you just use the type classes to simulate extendable records by having a method in each class that accesses the fixed-length record corresponding to that particular C++ class.\n\nHere is an example (apologies for the length!) which shows a super class function being overridden in a derived class and a derived class method (B::Extra) making use of something implemented in the super class:\n\n```module Main where\n\n{- Haskell translation of the following C++\n\nclass A {\npublic:\nString s;\nInt i;\n\nA(String s, Int i) s(s), i(i){}\n\nvirtual void Display(){\nprintf(\"A %s %d\\n\", s.c_str(), i);\n}\n\nvirtual Int Reuse(){\nreturn i * 100;\n}\n};\n\nclass B: public A{\npublic:\nChar c;\n\nB(String s, Int i, Char c) : A(s, i), c(c){}\n\nvirtual void Display(){\nprintf(\"B %s %d %c\", s.c_str(), i, c);\n}\n\nvirtual void Extra(){\nprintf(\"B Extra %d\\n\", Reuse());\n}\n\n};\n\n-}\n\ndata A = A\n{ _A_s :: String\n, _A_i :: Int\n}\n\n-- This could do arg checking etc\nconstructA :: String -> Int -> A\nconstructA = A\n\nclass ClassA a where\ngetA :: a -> A\n\ndisplay :: a -> IO ()\ndisplay a = do\nlet\nA{_A_s = s, _A_i = i} = getA a\nputStrLn \\$ \"A \" ++ s ++ show i\n\nreuse :: a -> Int\nreuse a = _A_i (getA a) * 100\n\ndata WrapA = forall a. ClassA a => WrapA a\n\ninstance ClassA WrapA where\ngetA (WrapA a) = getA a\ndisplay (WrapA a) = display a\nreuse (WrapA a) = reuse a\n\ninstance ClassA A where\ngetA = id\n\ndata B = B { _B_A :: A, _B_c :: Char }\n\nconstructB :: String -> Int -> Char -> B\nconstructB s i c = B {_B_A = constructA s i, _B_c = c}\n\nclass ClassA b => ClassB b where\ngetB :: b -> B\n\nextra :: b -> IO ()\nextra b = do\nputStrLn \\$ \"B Extra \" ++ show (reuse b)\n\ndata WrapB = forall b. ClassB b => WrapB b\n\ninstance ClassB WrapB where\ngetB (WrapB b) = getB b\nextra (WrapB b) = extra b\n\ninstance ClassA WrapB where\ngetA (WrapB b) = getA b\ndisplay (WrapB b) = display b\nreuse (WrapB b) = reuse b\n\ninstance ClassB B where\ngetB = id\n\ninstance ClassA B where\ngetA = _B_A\n\n-- override the base class version\ndisplay b = putStrLn \\$\n\"B \" ++ _A_s (getA b)\n++ show (_A_i (getA b))\n++ [_B_c (getB b)]\n\nmain :: IO ()\nmain = do\nlet\na = constructA \"a\" 0\nb = constructB \"b\" 1 '*'\n\ncol = [WrapA a, WrapA b]\n\nmapM_ display col\nputStrLn \"\"\nmapM_ (putStrLn . show . reuse) col\nputStrLn \"\"\nextra b\n\n{- Output:\n\n> ghc -fglasgow-exts --make Main\n> main\nA a0\nB b1*\n\n0\n100\n\nB Extra 100\n\n>\n-}\n```\n\n(If the \"caseless underscore\" Haskell' ticket is accepted the leading underscores would have to be replaced by something like \"_f\" ie _A_s ---> _fA_s etc)\n\n## Type class system extensions\n\nBrief list of extensions, their abbreviated names and compatibility level"
] |
[
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https://mechanismsrobotics.asmedigitalcollection.asme.org/GT/proceedings-abstract/GT2017/50930/V07BT36A001/238849
|
[
"This paper presents the results from a research effort on eigenvalue identification of mistuned bladed rotor systems using the Least-Squares Complex Frequency-Domain (LSCF) modal parameter estimator. The LSCF models the frequency response function (FRF) obtained from a vibration test using a matrix-fraction description and obtains the coefficients of the common denominator polynomial by minimizing the least squares error of the fit between the FRF and the model. System frequency and damping information is obtained from the roots of the denominator; a stabilization diagram is used to separate physical from mathematical poles. The LSCF estimator is known for its good performance when separating closely spaced modes, but few quantitative analyses have focused on the sensitivity of the identification with respect to mode concentration. In this study, the LSCF estimator is applied on both computational and experimental forced responses of an embedded compressor rotor in a three-stage axial research compressor. The LSCF estimator is first applied to computational FRF data obtained from a mistuned first-torsion (1T) forced response prediction using FMM (Fundamental Mistuning Model) and is shown to be able to identify the eigenvalues with high accuracy. Then the first chordwise bending (1CWB) computational FRF data is considered with varied mode concentration by varying the mistuning standard deviation. These cases are analyzed using LSCF and a sensitivity algorithm is developed to evaluate the influence of the mode spacing on eigenvalue identification. Finally, the experimental FRF data from this rotor blisk is analyzed using the LSCF estimator. For the dominant modes, the identified frequency and damping values compare well with the computational values.\n\nThis content is only available via PDF."
] |
[
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] |
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https://javascriptinfo.com/view/132448/randomize-numbers-with-jquery
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[
"# Randomize numbers with jQuery?\n\nIs there a simple jQuery way to create numbers randomly showing then a number 1 -6 is choosing after a few seconds? [Like dice]",
null,
"This doesn't require jQuery. The JavaScript `Math.random` function returns a random number between 0 and 1, so if you want a number between 1 and 6, you can do:\n\n``````var number = 1 + Math.floor(Math.random() * 6);\n``````\n\nUpdate: (as per comment) If you want to display a random number that changes every so often, you can use `setInterval` to create a timer:\n\n``````setInterval(function() {\nvar number = 1 + Math.floor(Math.random() * 6);\n\\$('#my_div').text(number);\n},\n1000); // every 1 second\n``````",
null,
"You don't need jQuery, just use javascript's `Math.random` function.\n\nedit: If you want to have a number from 1 to 6 show randomly every second, you can do something like this:\n\n``````<span id=\"number\"></span>\n\n<script language=\"javascript\">\nfunction generate() {\n\\$('#number').text(Math.floor(Math.random() * 6) + 1);\n}\nsetInterval(generate, 1000);\n</script>\n``````",
null,
"Others have answered the question, but just for the fun of it, here is a visual dice throwing example, using the `Math.random` javascript method, a background image and some recursive timeouts.\n\nhttp://www.jsfiddle.net/zZUgF/3/",
null,
"``````function rollDice(){\nreturn (Math.floor(Math.random()*6)+1);\n}\n``````",
null,
"Javascript has a `random()` available. Take a look at Math.random().",
null,
"Coding in Perl, I used the rand() function that generates the number at random and wanted only 1, 2, or 3 to be randomly selected. Due to Perl printing out the number one when doing \"1 + \" ... so I also did a if else statement that if the number generated zero, run the function again, and it works like a charm.\n\nprinting out the results will always give a random number of either 1, 2, or 3.\n\nThat is just another idea and sure people will say that is newbie stuff but at the same time, I am a newbie but it works. My issue was when printing out my stuff, it kept spitting out that 1 being used to start at 1 and not zero for indexing."
] |
[
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"https://javascriptinfo.com/static/a.png",
null,
"https://javascriptinfo.com/static/a.png",
null,
"https://javascriptinfo.com/static/a.png",
null,
"https://javascriptinfo.com/static/a.png",
null,
"https://javascriptinfo.com/static/a.png",
null,
"https://javascriptinfo.com/static/a.png",
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] |
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https://edurev.in/course/quiz/attempt/-1_Test-Theoretical-Distributions-4/78af8edf-12de-4696-9392-2445344455d1
|
[
"Courses\n\n# Test: Theoretical Distributions- 4\n\n## 40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Theoretical Distributions- 4\n\nDescription\nThis mock test of Test: Theoretical Distributions- 4 for CA Foundation helps you for every CA Foundation entrance exam. This contains 40 Multiple Choice Questions for CA Foundation Test: Theoretical Distributions- 4 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Theoretical Distributions- 4 quiz give you a good mix of easy questions and tough questions. CA Foundation students definitely take this Test: Theoretical Distributions- 4 exercise for a better result in the exam. You can find other Test: Theoretical Distributions- 4 extra questions, long questions & short questions for CA Foundation on EduRev as well by searching above.\nQUESTION: 1\n\nSolution:\nQUESTION: 2\n\nSolution:\nQUESTION: 3\n\n### The no. of methods for fitting the normal curve is\n\nSolution:\nQUESTION: 4\n\n____________ distribution is symmetrical around t = 0\n\nSolution:\nQUESTION: 5\n\nAs the degree of freedom increases, the ________ distribution approaches the Standard Normal distribution\n\nSolution:\nQUESTION: 6\n\n_________ distribution is asymptotic to the horizontal axis.\n\nSolution:\nQUESTION: 7\n\n________ distribution has a greater spread than Normal distribution curve\n\nSolution:\nQUESTION: 8\n\nIn Binomial Distribution if n is infinitely large, the probability p of occurrence of event’ is close to _______ and q is close to _________\n\nSolution:\nQUESTION: 9\n\nPoisson distribution approaches a Normal distribution as n\n\nSolution:\nQUESTION: 10\n\nIf neither p nor q is very small but n sufficiently large, the Binomial distribution is very closely approximated by _________ distribution\n\nSolution:\nQUESTION: 11\n\nFor discrete random variable x, Expected value of x (i.e E(x)) is defined as the sum of products of the different values and the corresponding probabilities.\n\nSolution:\nQUESTION: 12\n\nFor a probability distribution, —————— is the expected value of x.\n\nSolution:\nQUESTION: 13\n\n_________ is the expected value of (x – m)2 , where m is the mean.\n\nSolution:\nQUESTION: 14\n\nThe probability distribution of x is given below :",
null,
"Q. Mean is equal to\n\nSolution:\nQUESTION: 15\n\nFor n independent trials in Binomial distribution the sum of the powers of p and q is always n , whatever be the no. of success.\n\nSolution:\nQUESTION: 16\n\nIn Binomial distribution parameters are\n\nSolution:\nQUESTION: 17\n\nIn Binomial distribution if n = 4 and p = 1/3 then the value of variance is\n\nSolution:\nQUESTION: 18\n\nIn Binomial distribution if mean = 20, S.D.= 4 then q is equal to\n\nSolution:\nQUESTION: 19\n\nIf in a Binomial distribution mean = 20 , S.D.= 4 then p is equal to\n\nSolution:\nQUESTION: 20\n\nIf is a Binomial distribution mean = 20 , S.D.= 4 then n is equal to\n\nSolution:\nQUESTION: 21\n\nPoisson distribution is a ___________ probability distribution .\n\nSolution:\nQUESTION: 22\n\nNo. of radio- active atoms decaying in a given interval of time is an example of\n\nSolution:\nQUESTION: 23\n\n__________ distribution is sometimes known as the “distribution of rare events“.\n\nSolution:\nQUESTION: 24\n\nThe probability that x assumes a specified value in continuous probability distribution is\n\nSolution:\nQUESTION: 25\n\nIn Normal distribution mean, median and mode are\n\nSolution:\nQUESTION: 26\n\nIn Normal distribution the quartiles are equidistant from\n\nSolution:\nQUESTION: 27\n\nIn Normal distribution as the distance from the ___________ increases, the curve comes closer and closer to the horizontal axis .\n\nSolution:\nQUESTION: 28\n\nA discrete random variable x follows uniform distribution and takes only the values 6, 8, 11, 12, 17The probability of P( x = 8) is\n\nSolution:\nQUESTION: 29\n\nA discrete random variable x follows uniform distribution and takes the values 6, 9, 10, 11, 13The probability of P( x = 12) is\n\nSolution:\nQUESTION: 30\n\nA discrete random variable x follows uniform distribution and takes the values 6, 8, 11, 12, 17\n\nQ. The probability of P",
null,
"is\n\nSolution:\nQUESTION: 31\n\nA discrete random variable x follows uniform distribution and takes the values 6, 8, 10, 12, 18\n\nQ. The probability of P( x < 12) is\n\nSolution:\nQUESTION: 32\n\nA discrete random variable x follows uniform distribution and takes the values 5, 7, 12, 15, 18\n\nQ. The probability of P( x > 10) is\n\nSolution:\nQUESTION: 33\n\nThe probability density function of a continuous random variable is defined as follows :\n\nQ. f(x) = c when",
null,
", otherwise The value of c is\n\nSolution:\nQUESTION: 34\n\nA continuous random variable x has the probability density fn.f(x) = ½ –ax ,",
null,
"Q. When ‘a’ is a constant. The value of ‘ a’ is\n\nSolution:\nQUESTION: 35\n\nA continuous random variable x follows uniform distribution with probability density function\n\nQ.",
null,
"Solution:\nQUESTION: 36\n\nAn unbiased die is tossed 500 times.The mean of the no. of ‘Sixes’ in these 500 tosses is\n\nSolution:\nQUESTION: 37\n\nAn unbiased die is tossed 500 times. The Standard deviation of the no. of ‘sixes’ in these 500 tossed is\n\nSolution:\nQUESTION: 38\n\nA random variable x follows Binomial distribution with mean 2 and variance 1.2.Then the value of n is\n\nSolution:\nQUESTION: 39\n\nA random variable x follows Binomial distribution with mean 2 and variance 1.6 then the value of p is\n\nSolution:\nQUESTION: 40\n\n“The mean of a Binomial distribution is 5 and standard deviation is 3”\n\nSolution:"
] |
[
null,
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null,
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null,
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null,
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null,
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null
] |
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|
https://www.schoolcoders.com/computational-logic/logic-gates/xor-gates/
|
[
"# XOR gates\n\nMartin McBride, 2017-01-04\nTags xor gate truth table\nCategories logic logic gates\n\nAn Exclusive OR (XOR) gate has two inputs and one output. It is similar to an OR gate, except that the output obeys a slightly different rule.\n\nThe two inputs can each have values 0 or 1.\n\nThe state of the inputs controls the state of the output. The output value is 1 if either (but not both) inputs are 1, otherwise it is 0.\n\nAnother way to look at this is that the output will be 1 if the 2 inputs are different, but it will be 0 if both inputs are the same.\n\n## Symbol\n\nHere is the symbol for an XOR gate. The inputs are labelled A and B, the output is labelled Q.",
null,
"## Truth table\n\nA truth table is a table which shows all possible combinations of inputs (A and B), and the resulting value of the output, Q. Here is the truth table for an XOR gate, as you can see the output is 1 if either (but not both) inputs are 1.",
null,
"## Interactive example\n\nHere is an interactive XOR gate. Click on the inputs A and B to change their state, and see the output value Q."
] |
[
null,
"https://www.schoolcoders.com/img/computational-logic/logic-gates/xorsymbol.png",
null,
"https://www.schoolcoders.com/img/computational-logic/logic-gates/xor-truth.png",
null
] |
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|
http://www.expertsmind.com/questions/what-is-logic-gates-microprocessor-30177529.aspx
|
[
"## What is logic gates - microprocessor, Electrical Engineering\n\nAssignment Help:\n\n### What is Logic Gates?\n\nThe Logic Gates are circuits made up of transistors, diodes, and resistors. The Logic gates process one or more input signals in a logical fashion. Depending on the voltage or input value, the logic gate will either output a value of '1' for ON or a value of '0' for OFF. Logic Gates allow simplification of circuit operation.\n\n Logic states True False 1 0 High Low +Vs 0V On Off\n\nThe Logic gates process signals which represent true or false. Usually the positive supply voltage +V represent true and 0V represents false. Other conditions which are used for the true and false states are shown in the table.\n\nA truth table is a good way to show the function of the logic gate. It shows the output states for each possible combination of input states. The symbols 0 (false) and 1 (true) are typically used in truth tables.\n\n#### TRANSISTOR BIASING CIRCUITS, HOW TO FIND THE OPERATING POINT OF THE TRANSIS...\n\nHOW TO FIND THE OPERATING POINT OF THE TRANSISTO IN A COLLECTOR TO BASE BIAS CIRCUIT\n\n#### Difference, Difference between bootstrip sweep circuit and Miller sweep ci...\n\nDifference between bootstrip sweep circuit and Miller sweep circuit\n\n#### 8 bit data to memory - move immediate instruction , 8 bit data to memory ...\n\n8 bit data to memory This form of the instruction is used to copy 8 bit data directly to the memory location pointed by register pair HL. The instruction format is\n\n#### Show the procedure of hex to decimal conversion, Q. Show the procedure of H...\n\nQ. Show the procedure of Hex to Decimal Conversion? To convert from the Hex to the Decimal, multiply the value in each position by its hex weight and add each value. Using the\n\n#### Explain the properties of high resistive materials, Explain the properties ...\n\nExplain the properties of high resistivity materials. High resistivity materials: The conducting materials containing resistivity from 10 -6 to 10 -3 ohm-m come under such\n\n#### What is the clock frequency of 8086, What is the clock frequency of 8086? ...\n\nWhat is the clock frequency of 8086? Internal clock Frequency 5 MHz 8MHz External Clock Frequency 15MHZ 24MHZ\n\n#### Fixed bias, Fixed bias (base bias): Diagram: Fixed bias (Base ...\n\nFixed bias (base bias): Diagram: Fixed bias (Base bias) This type of biasing is also known as base bias . In the instance above figure, the single power source (\n\n#### Which alloy is used for making heater & thermocouple element, Give the name...\n\nGive the names of four alloys along with their composition, which are used for making heater and thermocouple elements. Constantan or Eureka {(55 to 60 percent) Cu, (45 to 40 p\n\n#### Name 5 different addressing modes, The five addressing modes are given belo...\n\nThe five addressing modes are given below: Immediate, Register, Direct, Implied addressing modes Register indirect,\n\n#### Determine the capacitor voltage, Q. In an RLC series circuit excited by a v...\n\nQ. In an RLC series circuit excited by a voltage source v(t), for R = 10 \u0007, L = 1 H, and C = 0.1F, determine v(t) if the capacitor voltage vC(t) = 5e -10t V.",
null,
"",
null,
""
] |
[
null,
"http://www.expertsmind.com/questions/CaptchaImage.axd",
null,
"http://www.expertsmind.com/prostyles/images/3.png",
null
] |
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|
https://www.58dir.cn/11498.html
|
[
"# 织梦CMS二次开发批量导入excel数据\n\n## 插入数据开发原理\n\n1）开发前的准备：\n\n2) 开发插入数据代码：\n\n```header(\"Content-type:text/html;charset=utf8\");\nrequire_once('/../dedecms/include/common.inc.php');\nerror_reporting(E_ALL);\nset_time_limit(0);\ndate_default_timezone_set('Europe/London');\nset_include_path(get_include_path() . PATH_SEPARATOR . '../../../Classes/');\ninclude 'PHPExcel/IOFactory.php';\nif(!empty(\\$_GET)){\n\\$typeid = \\$_GET['typeid'];\n\\$dopost = \\$_GET['do'];\nif(\\$dopost == \"exdata\"){\nif(!empty(\\$_GET['n'])){\n\\$inputFileName = './'.\\$_GET['n'].'.xlsx';\n\\$objPHPExcel = PHPExcel_IOFactory::load(\\$inputFileName);\n\\$sheetData = \\$objPHPExcel->getActiveSheet()->toArray(null,true,true,true);\n\\$rowarr=array();\n\\$dsql->Execute('all',\"select bio2,bio1 from `dede_addonshop`\");\nwhile (\\$rowall = \\$dsql->GetArray('all')) {\n\\$rowarr[]=\\$rowall;\n}\nforeach (\\$sheetData as \\$v) {\nforeach (\\$rowarr as \\$vt) {\nif(in_array(\\$v['B'],\\$vt)&&\\$v['A']==\\$vt['bio1']){\necho \"货号为：\".\\$v['B'].\"\n\n}\n}\n}\n\\$row = \\$dsql->GetOne(\"select aid,bio2 from `dede_addonshop` order by aid desc\");\nif(!empty(\\$row)){\n\\$aid = \\$row['aid'];\n\\$bio2 = \\$row['bio2'];\n}else{\n\\$aid = 0;\n\\$bio2 = '';\n}\nif(\\$bio2==\\$sheetData[count(\\$sheetData)]['B']){\nShowMsg(\"不能重复添加内容\",'javascript:;');\nexit;\n}\n\\$arcrow = \\$dsql->GetOne(\"select id from `dede_archives` order by id desc\");\nif(!empty(\\$arcrow)){\n\\$arcid = \\$arcrow['id'];\n}else{\n\\$arcid = 0;\n}\n\\$tinyrow = \\$dsql->GetOne(\"select id from `dede_arctiny` order by id desc\");\nif(!empty(\\$tinyrow)){\n\\$tinyid = \\$tinyrow['id'];\n}else{\n\\$tinyid = 0;\n}\n\\$id = max(\\$aid,\\$arcid,\\$tinyid);\n\\$alphalpha = array('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','AA','AB','AC','AD','AE','AF','AG','AH','AI','AJ','AK','AL','AM','AN','AO','AP','AQ','AR','AS','AT','AU','AV','AW','AX','AY','AZ');\n\\$fieldnum = count(\\$sheetData);\n\\$fields = \\$fieldvalue = '';\nfor (\\$i=0; \\$i < \\$fieldnum; \\$i++) {\n\\$fields .= \\$sheetData[\\$alphalpha[\\$i]]. ',';\n}\n\\$fields = substr(\\$fields, 0,-1);\nforeach (\\$sheetData as \\$value) {\n\\$pubdate = GetMkTime(GetDateTimeMk(time()));\n\\$click = mt_rand(50, 200);\nif(\\$value['A']=='bio1'|\\$value['A']=='厂商'){\ncontinue;\n}\n\\$id = \\$id+1;\n//获取字段值\\$value['A'];\nfor (\\$i=0; \\$i < \\$fieldnum; \\$i++) {\n\\$fieldvalue .= \" ,'\".\\$value[\\$alphalpha[\\$i]].\"' \";\n}\n\\$C = trim(\\$value['C']);\n\\$senddate = time();\n\\$arcquery = \"INSERT INTO `dede_archives`(id,typeid,title,mid,channel,pubdate,senddate,click,ismake)VALUES ('\\$id','\\$typeid','\\$C','1','6','\\$pubdate','\\$senddate','\\$click','-1');\";\n\\$dsql->ExecuteNoneQuery(\\$arcquery);\n\\$query = \"INSERT INTO `dede_addonshop`(aid,typeid,\\$fields)\nVALUES ('\\$id','\\$typeid'{\\$fieldvalue});\";\n\\$dsql->ExecuteNoneQuery(\\$query);\n\\$fieldvalue = '';\n\\$tinyquery = \"INSERT INTO `dede_arctiny`(id,typeid,channel,mid,senddate)VALUES ('\\$id','\\$typeid','6','1','\\$senddate');\";\n\\$dsql->ExecuteNoneQuery(\\$tinyquery);\n}\n\\$num = count(\\$sheetData)-2;\nShowMsg(\"恭喜，成功插入 \".\\$num.\" 条数据！\",'javascript:;');\n}```\n\n2、站内下载币均可通过签到等任务进行免费兑换。\n\n3、站内资源均来源于网络公开发表文件或网友投稿发布，如侵犯您的权益，请联系管理员处理。\n\n4、本站所分享的源码、模板、软件工具等其他资源，都不包含技术服务，请大家谅解！\n\n5、所有资源均收集于互联网仅供学习、参考和研究，请理解这个概念，所以不能保证每个细节都符合你的需求，也可能存在未知的BUG与瑕疵，因本站资源均为可复制品,所以不支持任何理由的退款兑现(特殊情况可退积分)，请熟知后再支付下载！\n\n## 常见问题FAQ\n\n• 2020年12月21日Hi,初次和大家见面了,请多关照！",
null,
"• 426会员总数(位)\n• 3330资源总数(个)\n• 13本周发布(个)\n• 0 今日发布(个)\n• 385稳定运行(天)\n\n#### 58源码下载站58DIR.CN，专为草根站长提供建站服务的平台\n\n© 2020 WWW.58DIR.CN & 58源码下载站 皖ICP备19006847号-2"
] |
[
null,
"data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==",
null
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|
https://thenextgenbusiness.com/5-percent-of-250/
|
[
"# What is 5 Percent of 250? (In-Depth Explanation)\n\n5 percent of 250 equals 12.5. To get this answer, multiply 0.5 by 250.\n\nYou may need to know this answer when solving a math problem that multiplies both 5% and 250. Perhaps a product worth 250 dollars, euros, or pounds is advertised as 5% off. Knowing the exact amount discounted from the original price of 250 can help you make a more informed decision on whether or not it is a good deal.\n\nMaybe you’re looking for 5% of 250 dollars, euros, Japanese yen, British pounds, Chinese yuan, pesos, or rupees. Whatever the case is, below, you will find an in-depth explanation that will help you solve this equation.\n\n## What is 5 percent of 250?\n\n5 percent of 250 is 12.5. To figure this out, multiply 0.5 by 250 to get 12.5 as the answer.\n\nAnother way to find the answer to this equation includes taking 5/100 and multiplying it by 250/1. When multiplying these two fractions together, you will get a final answer of 12.5.\n\n## How do you find 5 percent of 250?\n\nBy multiplying both 0.5 and 250 together, you will find that 12.5 is 5 percent of 250. The 0.5 represents 5% and is the result of taking 5/100 or 5 divided by 100.\n\nThe easiest way to solve this equation is to divide the percent by 100 and multiply by the number. So divide 5 by 100 to get 0.5. From there, multiply the percent (now in decimal form) by 250 to get 12.5.\n\n## What is 5% off 250 dollars?\n\nYou will pay \\$237.50 for an item when you account for a discount of 5 percent off the original price of \\$250. You will be receiving a \\$12.5 discount.\n\n## What is 5 percent of 250 dollars?\n\n5 percent of 250 dollars is 12.5 dollars. When solving this equation, we multiply 0.5 by 250, the 0.5 standing for 5% and 250 representing 250 dollars.\n\nWhen referencing the dollar, people will likely be talking about the United States dollar (USD). However, sometimes other currencies are intended instead, like the Canadian dollar (CAD) or the Australian dollar (AUD).\n\nThe equation remains the same for calculating 5% of 250 dollars for each of those respective currencies.\n\n## What is 5% off 250 euros?\n\nWith a 5 percent discount, you will pay €237.50 for any item with an original price of €250. You will get a discount of €12.5 off.\n\n## What is 5 percent of 250 euros?\n\n5% of 250 euros is 12.5 euros. We use the same formula for calculating 5% of 250 to get our answer of 12.5 euros.\n\nThe euro is the currency used by some countries in the European Union, such as France, Germany, and Italy.\n\n## What is 5 percent of 250 Japanese yen?\n\n5% of 250 Japanese yen is 12.5 yen. If you’re trying to solve 5% of 250 Japanese yen, multiply 5% by 250.\n\nWhen you multiply these two numbers together, you will find 12.5 Japanese yen is your answer.\n\n## What is 5% off 250 pounds?\n\nIf you get a 5 percent discount on a £250 item, you will pay £237.50. In total, you will end up receiving a £12.5 discount.\n\n## What is 5 percent of 250 British pounds?\n\nSimilar to other currencies, we multiply 5% by 250 to get 12.5 British pounds. In this equation, 0.5, 5/100, or 5% can each represent 5 percent. The 250 in this equation stands for 250 British pounds.\n\n12.5 British pounds will be your answer once you multiply the two numbers together.\n\n## What is 5 percent of 250 Chinese yuan?\n\n5% of 250 Chinese yuan is 12.5 yuan. The same formula that calculated 5% of 250 of the other currencies can calculate 5% of the Chinese yuan.\n\nYou divide the percent by 100 and multiply it by the number. For this example, the equation divides 5% by 100 to get 0.5 (5 percent in decimal form). The percent is then multiplied by 250 Chinese yuan resulting in an answer of 12.5 Chinese yuan.\n\n## What is 5 percent of 250 pesos?\n\n12.5 pesos is the equivalent of 5% of 250 pesos. When solving this equation, take the percent divided by 100 and multiply it by the number. In this case, 5% is divided by 100 and multiplied by 250 pesos for an answer of 12.5 pesos.\n\n## What is 5 percent of 250 rupees?\n\nLike with other currencies, use the same equation and multiply 5% by 250 rupees to get an answer of 12.5 rupees. The answer will remain the same even if you write 5 percent as; 5%, 0.5, or 5/100.\n\nAfter you multiply 5% and 250 rupees together, 12.5 rupees is the final answer to the equation.\n\n## Conclusion\n\nYou might need to know the answer to 5% of 250 when operating a business. New businesses get started every day, and people will often need to solve equations involving percentages like this.\n\nThose looking for the answer to 5% of 250 might not even be business owners.\n\nMaybe you are at school or work and need to know the answer to this calculation. Whatever the case is, the answer is 12.5.\n\nIf you enjoyed learning about what 5% of 250 is, consider checking out our other articles below!\n\nRelated Posts\n\n### Live Better For Less.\n\nFind Your Ideal Location: Lower Taxes, Easy Choices, Flexible Living."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.9257753,"math_prob":0.9801356,"size":4657,"snap":"2023-40-2023-50","text_gpt3_token_len":1244,"char_repetition_ratio":0.18009886,"word_repetition_ratio":0.053056516,"special_character_ratio":0.30298474,"punctuation_ratio":0.13679245,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9955733,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T21:20:18Z\",\"WARC-Record-ID\":\"<urn:uuid:046b77aa-e9f8-4895-8acb-094a28e54538>\",\"Content-Length\":\"220826\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2e6b74d-ed68-40b0-b9ec-a71827aa6358>\",\"WARC-Concurrent-To\":\"<urn:uuid:1499fa20-39e0-4c7c-81e4-dcd92586ff6b>\",\"WARC-IP-Address\":\"172.67.186.28\",\"WARC-Target-URI\":\"https://thenextgenbusiness.com/5-percent-of-250/\",\"WARC-Payload-Digest\":\"sha1:EHR6EG5IIKUODKOFI3SOBMLEA7VWFU4Q\",\"WARC-Block-Digest\":\"sha1:TPRONI52U4GFYXG4B5O34XP6GXKLZU5S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100972.58_warc_CC-MAIN-20231209202131-20231209232131-00496.warc.gz\"}"}
|
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.17/share/doc/Macaulay2/Macaulay2Doc/html/_map_lp__Module_cm__Module_cm__Matrix_rp.html
|
[
"# map(Module,Module,Matrix) -- create the matrix induced on generators by a given matrix\n\n## Synopsis\n\n• Function: map\n• Usage:\nmap(M,N,p)\n• Inputs:\n• M,\n• N,\n• p,\n• Optional inputs:\n• Degree => ..., default value null, specify the degree of a map\n• DegreeLift => ..., default value null, make a ring map\n• DegreeMap => ..., default value null, make a ring map\n• Outputs:\n• , A matrix with the same entries as p, but whose target is M and source is N\n\n## Description\n\nM and N should be modules over the same ring, and have the same number of generators as target p and source p, respectively.\n i1 : R = QQ[x,y,z]; i2 : p = matrix {{x,y,z}} o2 = | x y z | 1 3 o2 : Matrix R <--- R i3 : q = map(R^1,R^3,p) o3 = | x y z | 1 3 o3 : Matrix R <--- R i4 : degrees source p o4 = {{1}, {1}, {1}} o4 : List i5 : degrees source q o5 = {{0}, {0}, {0}} o5 : List\n\n## Caveat\n\nIf M or N is not free, then we don't check that the the result is a well defined homomorphism."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.5496389,"math_prob":0.99822944,"size":945,"snap":"2021-43-2021-49","text_gpt3_token_len":304,"char_repetition_ratio":0.10839532,"word_repetition_ratio":0.12376238,"special_character_ratio":0.37566137,"punctuation_ratio":0.2236842,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97799605,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T23:37:58Z\",\"WARC-Record-ID\":\"<urn:uuid:1e6901c5-3f05-4c43-a2f3-0dc360dfe1d6>\",\"Content-Length\":\"6833\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6c730942-87ff-402a-9878-ae877600ef7a>\",\"WARC-Concurrent-To\":\"<urn:uuid:519d9a3a-eacb-433b-af30-4f72b8fac0f3>\",\"WARC-IP-Address\":\"128.174.199.46\",\"WARC-Target-URI\":\"https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.17/share/doc/Macaulay2/Macaulay2Doc/html/_map_lp__Module_cm__Module_cm__Matrix_rp.html\",\"WARC-Payload-Digest\":\"sha1:M4HV2B6D4ENPH3YIGB3BJSSYXY75TPJE\",\"WARC-Block-Digest\":\"sha1:SW2G5QHYNOXCATU5UHDXQ5U2CGSK4YWZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585449.31_warc_CC-MAIN-20211021230549-20211022020549-00231.warc.gz\"}"}
|
https://git.uwaterloo.ca/pjentsch/covidalertabm/-/commit/68038a92750e0beb104fe0982fe8bbb41c9e10b4?expanded=1&page=8
|
[
"### univariate plots\n\nparent 3e5f3663\nFile deleted\nFile deleted\nFile deleted\nFile deleted\nFile deleted\nFile deleted\nFile deleted\n ... ... @@ -128,7 +128,6 @@ function weighted_degree_and_join_count(t,node,network::TimeDepMixingGraph,u_inf node_is_immunized = u_inf[node] == Immunized for g in network.graph_list[t] <<<<<<< HEAD for j in neighbors(g,node) weight = get_weight(g,GraphEdge(node,j)) weighted_degree += weight ... ... @@ -137,10 +136,6 @@ function weighted_degree_and_join_count(t,node,network::TimeDepMixingGraph,u_inf join_count_11 += ifelse(node_is_immunized && j_is_immunized,weight,0) join_count_01 += ifelse(xor(node_is_immunized, j_is_immunized),weight,0) join_count_00 += ifelse(!(node_is_immunized) && !(j_is_immunized),weight,0) ======= for he in neighbors_and_weights(g,node) weighted_degree += he.weight[] >>>>>>> 082c742 (graph idea final) end end return weighted_degree,(join_count_11,join_count_01,join_count_00) ... ...\n ... ... @@ -107,20 +107,9 @@ end @test mean(mixing_dist[i]) ≈ expected_dist_mean[i] atol = 0.05 @test mean(mixing_dist[i]) ≈ mean(dist[i]) atol = 0.2 end <<<<<<< HEAD for i in 1:3, j in 1:i @test mean(mixing_weights[i,j]) ≈ contact_time_distributions.ws[i,j].μ atol = 0.4 end ======= mixing_weights = [Variance() for _ in 1:3, _ in 1:3] for (sampler,weight) in g.weight_sample_list indexofsampler = findfirst(==(sampler), contact_time_distributions.ws) fit!(mixing_weights[indexofsampler],weight) end display(mean.(mixing_weights)) display(map(d -> d.μ,contact_time_distributions.ws)) >>>>>>> 082c742 (graph idea final) end end \\ No newline at end of file\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!\nPlease register or to comment"
] |
[
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] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.62665766,"math_prob":0.9827015,"size":1694,"snap":"2022-40-2023-06","text_gpt3_token_len":542,"char_repetition_ratio":0.14852071,"word_repetition_ratio":0.09356725,"special_character_ratio":0.3577332,"punctuation_ratio":0.27739727,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97293437,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-27T13:16:22Z\",\"WARC-Record-ID\":\"<urn:uuid:3b7aa887-384c-407c-83cc-75259aa3a130>\",\"Content-Length\":\"189473\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:da3f3e39-7fab-47f8-b292-477eefd2f698>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a5cc12c-2db7-4919-9214-8c61feba0e19>\",\"WARC-IP-Address\":\"129.97.83.4\",\"WARC-Target-URI\":\"https://git.uwaterloo.ca/pjentsch/covidalertabm/-/commit/68038a92750e0beb104fe0982fe8bbb41c9e10b4?expanded=1&page=8\",\"WARC-Payload-Digest\":\"sha1:4LXGDSCXH7LSBA3WLBK5LFEMDLRKDDOE\",\"WARC-Block-Digest\":\"sha1:XWJD2MVWPXU7S3MY47IUHHQCNDCQ54FF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335034.61_warc_CC-MAIN-20220927131111-20220927161111-00581.warc.gz\"}"}
|
https://krastanov.wordpress.com/2012/07/22/form-fields-and-vector-fields-do-not-form-a-vector-space/
|
[
"# Form Fields and Vector Fields do not form a Vector Space\n\nForm fields or vector fields over a manifold (as opposed to forms and vectors) do not form a vector space. They form a module.\n\nThe difference is that the scalars of a vector space form an algebraic field while the scalars of a module form a ring. For us humans (as opposed to “those higher beings that I do not understand (a.k.a. mathematicians)”) this means that the scalars in the vector field can divide each other while the scalars in the spaces spanned by fields (i.e. a module) can not.\n\nAnd just so we all can become even more confused: This has nothing to do with the fact that the “components” of each form field or vector field in certain basis are functions, i.e. themselves elements of a vector space with infinite number of dimensions.\n\nThe first way to see this module-not-a-vector-space characteristic is by showing directly that the scalars that form the “coordinate components” of a vector field can not always be divided, even if they are not identically zero. Take, for instance the, manifold",
null,
"$\\mathbb{R}^2$ with the polar coordinate system and look at the vector",
null,
"$\\begin{bmatrix} r \\\\ r\\cos(\\theta) \\end{bmatrix}$. The “scalars” are",
null,
"$r$ and",
null,
"$r\\cos(\\theta)$. Obviously we can not divide the former by the latter because it will be undefined at",
null,
"$\\theta=\\frac{\\pi}{2}+n\\pi$.\n\nAnother, more amusing way to show that the space spanned by these fields is not a vector space is to explicitly show that a property expected from vector spaces is not fulfilled. Namely, that in",
null,
"$n$ dimensions an",
null,
"$n$-uple of linearly independent elements forms a basis. However, in the case of fields over a manifold we can easily have a number of fields that are linearly independent over the manifold as a whole, and are at the same time linearly dependent (or simply equal to zero) on a subdomain. Hence, we have an",
null,
"$n$-uple of linearly independent fields that can not be linearly combined to represent another arbitrary field."
] |
[
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
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https://gmatclub.com/forum/the-2-lines-x-2y-3-2x-py-q-have-infinitely-many-points-of-intersectio-277407.html
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[
"GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 15 Oct 2019, 14:57",
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"GMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we’ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nNot interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.",
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"The 2 lines x+2y=3, 2x+py=q have infinitely many points of intersectio\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\nHide Tags\n\nMath Revolution GMAT Instructor",
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"V\nJoined: 16 Aug 2015\nPosts: 8005\nGMAT 1: 760 Q51 V42",
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"GPA: 3.82\nThe 2 lines x+2y=3, 2x+py=q have infinitely many points of intersectio [#permalink]\n\nShow Tags",
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"based on 40 sessions\n\nHideShow timer Statistics\n\n[Math Revolution GMAT math practice question]\n\nThe $$2$$ lines $$x+2y=3, 2x+py=q$$ have infinitely many points of intersection in the xy-plane. Which of the following could be the value of $$p$$?\n\n$$A. 0$$\n$$B. 1$$\n$$C. 2$$\n$$D. 3$$\n$$E. 4$$\n\n_________________\nIntern",
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"B\nJoined: 17 May 2018\nPosts: 48\nRe: The 2 lines x+2y=3, 2x+py=q have infinitely many points of intersectio [#permalink]\n\nShow Tags\n\n1\nIf the two lines have an infinite number of intersections, then they must be overlapped. Their equations should then be the same.\n\nEquation of line 1, isolating y: y=-x/2 + 3/2\n\nEquation of line 2, isolating y: y=-2x/p + q/p\n\nWe want the same slope so p should be 4 in order to obtain y=-x/2 + q/4\n_________________\n¿Tienes que presentar el GMAT y no sabes por dónde empezar?\n¡Visita GMAT para Principiantes y recibe el curso completo gratis!\nGMATH Teacher",
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"P\nStatus: GMATH founder\nJoined: 12 Oct 2010\nPosts: 935\nRe: The 2 lines x+2y=3, 2x+py=q have infinitely many points of intersectio [#permalink]\n\nShow Tags\n\nMathRevolution wrote:\n\nThe $$2$$ lines $$x+2y=3, 2x+py=q$$ have infinitely many points of intersection in the xy-plane. Which of the following IS the value of $$p$$?\n\n$$A. 0$$\n$$B. 1$$\n$$C. 2$$\n$$D. 3$$\n$$E. 4$$\n\n$$? = p$$\n\nFrom the question stem, we know both lines (each represented by one of the equations) must coincide (*), hence:\n\n$$\\left\\{ \\begin{gathered} \\,x + 2y = 3\\,\\,\\,\\left( { \\cdot 2} \\right) \\hfill \\\\ 2x + py = q \\hfill \\\\ \\end{gathered} \\right.\\,\\,\\,\\,\\,\\, \\sim \\,\\,\\,\\,\\,\\,\\left\\{ \\begin{gathered} \\,2x + 4y = 6 \\hfill \\\\ 2x + py = q \\hfill \\\\ \\end{gathered} \\right.\\,\\,\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{\\left( * \\right)} \\,\\,\\,\\,\\,\\,\\,? = p = 4\\,\\,\\,\\,\\,\\,\\,\\left( {{\\text{and}}\\,\\,q = 6} \\right)$$\n\nThis solution follows the notations and rationale taught in the GMATH method.\n\nRegards,\nfskilnik.\n_________________\nFabio Skilnik :: GMATH method creator (Math for the GMAT)\nOur high-level \"quant\" preparation starts here: https://gmath.net\nManager",
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"G\nJoined: 14 Jun 2018\nPosts: 219\nRe: The 2 lines x+2y=3, 2x+py=q have infinitely many points of intersectio [#permalink]\n\nShow Tags\n\nInfinite intersection means they are on top of each other , which in turn means their slope is the same.\nOption E\nMath Revolution GMAT Instructor",
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"V\nJoined: 16 Aug 2015\nPosts: 8005\nGMAT 1: 760 Q51 V42",
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"GPA: 3.82\nRe: The 2 lines x+2y=3, 2x+py=q have infinitely many points of intersectio [#permalink]\n\nShow Tags\n\n=>\n\nIf the $$2$$ lines have infinitely many points of intersection, their equations must specify the same straight line.\nThe equation $$x+2y=3$$ is equivalent to $$2x+4y=6.$$\nSo, $$p = 4$$ and $$q = 6$$.\n\nTherefore, the answer is E.\n_________________",
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"Re: The 2 lines x+2y=3, 2x+py=q have infinitely many points of intersectio [#permalink] 30 Sep 2018, 22:56\nDisplay posts from previous: Sort by\n\nThe 2 lines x+2y=3, 2x+py=q have infinitely many points of intersectio\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne",
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http://kattemolle.com/QMvsGR/index.html
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[
"$$\\newcommand{\\ket}{\\left| #1 \\right \\rangle} % for Dirac bras \\newcommand{\\bra}{\\left\\langle #1 \\right |} % for Dirac kets \\newcommand{\\braket}{\\left\\langle #1 \\vphantom{#2} \\right | \\left . #2 \\vphantom{#1} \\right \\rangle} % for Dirac brackets \\newcommand{\\del}{{\\partial}} \\newcommand{\\HH}{{\\mathcal H}} \\newcommand{\\mc}{\\mathcal{#1}} \\newcommand{\\mrm}{\\mathrm} \\newcommand{\\om}{\\omega} \\newcommand{\\Om}{\\Omega} \\newcommand{\\ka}{\\kappa} \\newcommand{\\De}{\\Delta} \\newcommand{\\de}{\\delta} \\newcommand{\\be}{\\beta} \\newcommand{\\la}{\\lambda} \\newcommand{\\La}{\\Lambda} \\newcommand{\\al}{\\alpha} \\newcommand{\\ze}{\\zeta} \\newcommand{\\munu}{{\\mu\\nu}} \\newcommand{\\na}{{\\nabla}} \\newcommand{\\si}{{\\sigma}} \\newcommand{\\Ga}{{\\Gamma}} \\newcommand{\\del}{{\\partial}}$$\n\n# Comparison of notation in QM and GR\n\nLinear algebra is used in both Quantum Mechanics (QM) and General Relativity (GR). The notation however, is very different. This obscures the many mathematical similarities. In learning GR when you already know QM (or vice versa), it is very useful to see these similarities more clearly. That's why I've made an overview here, the construction of which was very useful for me, and I hope you may get something form it as well.\n\nThis page is organized such that equation (QM.$n$) should be compared to (GR.$n$). I do not intend to give a self-contained or pedagogical account of the theories separately, I only compare notation. The notation I use for QM is quite standard (although sometimes I'll bend it a bit towards GR), the notation for GR is as in 'Spacetime and geometry' by S. Carroll. Many thanks to Li Liu from the University of Waterloo for giving me the idea of this page!\n\n### Vectors\n\nQM. In quantum mechanics, a pure state $\\ket \\psi$ is a vector, or 'ket', in a complex Hilbert space $\\mc H$. With $\\{\\ket{i}\\}$ a basis of this space, $\\ket \\psi$ can be expanded as $\\ket \\psi = \\sum_i \\psi_i \\ket i. \\tag{QM.1}$ When using the bra-ket notation, we also need the dual space $\\mc H^\\dagger$. A dual vector in this space is called a 'bra'. Using the basis of the dual space $\\{\\bra{i}\\}$, the bras can be expanded as $\\bra \\varphi = \\sum_i \\varphi_i \\bra i.\\tag{QM.2}$\n\nGR. In GR, at every point $p$ in space-time, there is a (usually 4-dimensional) vector space, the tangent space $T_p$. An example of a vector in this space is the velocity of a particle at $p$. A basis for this space is the coordinate basis $\\{\\del_\\mu\\}$, formed by directional derivatives to the coordinate functions $x^{\\mu}$. This basis need not be orthogonal. A vector $V$ can thus be expanded as $V=V^\\mu \\del_\\mu \\tag{GR.1}.$ (I only use the Einstein summation convention in the parts about GR.) The dual space $T_p^*$ consists of all linear functions on $T_p$. The objects in this space are called one-forms, or simply dual vectors. This space has a basis $\\{\\mrm d x^{\\mu}\\}$, where $\\mrm d x^{\\mu}$ is the gradient of the coordinate function $x^{\\mu}$. The one-forms can thus be expanded as $\\om=\\om_\\mu \\mrm d x^\\mu \\tag{GR.2}.$\n\nDifferences. $\\HH$ is a complex vector space (i.e. the coefficients $\\psi_i$ are complex numbers) of any dimension. $T_p$ is a real vector space (the $V^\\mu$ are real) that is usually 4 dimensional. In QM, it is very common to choose an orthonormal basis, whereas the coordinate basis in GR need not be even orthogonal. Additionally, the basis elements $\\ket i$ and $\\del_\\mu$ have a totally different physical meaning.\n\n### Operators / Tensors\n\nQM. An operator $A$ is a linear function from $\\HH$ to $\\HH$ that maps $\\ket \\psi$ to $A \\ket \\psi$. Like vectors, operators can be expanded over the basis $\\{\\ket i\\}$, $A=\\sum_{ij} A_{ij} \\ket i \\otimes \\bra j.\\tag{QM.3}$ Equivalently, an operator can be thought of (this is important for the connection to GR), as a map $\\mc A$ from $\\HH^\\dagger \\otimes \\HH$ to $\\mathbb C$, that maps a bra $\\bra \\varphi$ and a ket $\\ket{\\psi}$ to the complex number $\\bra \\varphi A \\ket \\psi$, $\\mathcal A(\\bra\\varphi,\\ket\\psi)=\\bra \\varphi A \\ket \\varphi.$ To see the connection to an operator as a map from vectors to vectors, note that if we feed $\\mc A$ only a ket instead of a bra and a ket, the result is again a vector, \\begin{align*} \\mc A(\\cdot,\\ket{\\psi})=&A\\ket\\psi&\\\\ =& \\sum_{ij} A_{ij} \\psi_j \\ket i.\\tag{QM.4}& \\end{align*} Because of the equivalence of the two pictures, I will use them interchangeably. The equivalence is also seen in GR.\n\nGR. A $(1,1)$-tensor $T$ maps one dual vector and one vector to a real number. We can also write these tensors in the bases as described before, $T={T^{\\mu}}_{\\nu}\\, \\del_{\\mu}\\otimes \\mrm dx^{\\nu}\\tag{GR.3}.$ Note that when we act with this tensor on a vector, the result is a vector, \\begin{align*} T(\\cdot,V)=&{T^{\\mu}}_{\\nu} \\del_{\\mu}\\otimes \\mrm dx^{\\nu}\\,\\,( V^\\gamma \\del_\\gamma)&\\\\\\ =&{T^{\\mu}}_{\\nu}V^\\gamma\\,\\, \\del_{\\mu}\\otimes \\underbrace{dx^{\\nu}(\\del_\\gamma)}_{=\\delta^\\nu_{\\gamma}\\in \\mathbb R}&\\\\ =&{T^{\\mu}}_{\\nu}V^\\gamma \\delta^\\nu_{\\gamma} \\del_{\\mu}&\\\\ =&{T^{\\mu}}_{\\nu}V^\\nu\\,\\, \\del_{\\mu}.&\\tag{GR.4} \\end{align*} That is, the result of applying $T$ to $V$ is a new vector with components ${T^{\\mu}}_{\\nu}V^\\nu$, very much like an operator in QM in the usual sense.\n\n### Tensor product structure\n\nQM. It gets more interesting for composite quantum systems. Consider $n$ identical quantum systems, each with Hilbert space $\\HH$. States are vectors in the tensor product space $\\mc H^{\\otimes n}$, $\\ket \\psi = \\sum_{i_1\\ldots i_n} \\psi_{i_1\\ldots i_n} \\ket{i_1}\\otimes\\ldots \\otimes \\ket{i_n}.\\tag{QM.5}$ (Similarly for the bras.) Again, operators can be expanded over the basis, $A=\\sum_{i_1j_1\\ldots i_nj_n} A_{i_1j_i\\ldots i_nj_n}(\\ket{i_1}\\otimes \\bra{j_1}) \\otimes \\ldots \\otimes (\\ket{i_n}\\otimes \\bra{j_n})$. If we allow for the bras and kets to be shuffled around this can be written more suggestively as $$A=\\sum_{i_1\\ldots i_n, j_1 \\ldots i_n} A_{i_1 \\ldots j_n, j_1 \\ldots i_n} \\ket{i_1}\\otimes \\ldots \\otimes \\ket{i_n} \\otimes \\bra{j_1} \\otimes \\ldots \\otimes \\bra{j_n}.$$ We can go even further and construct '$(n,m)$-operators', $A=\\sum_{i_1\\ldots i_n, j_1 \\ldots j_m} A_{i_1 \\ldots i_n, j_1 \\ldots i_m} \\ket{i_1}\\otimes \\ldots \\otimes \\ket{i_n} \\otimes \\bra{j_1} \\otimes \\ldots \\otimes \\bra{j_m}.\\tag{QM.6}$ I have never encountered such an object in QM but the definition is justified in the connection to GR. In the language of this section, the regular operators in QM are $(1,1)$-operators*, and super-operators are $(2,2)$-operators**.\n\nGR. An $(n,m)$-tensor maps $n$ dual vectors and $m$ vectors to a real number. We can think of the collection of $m$ vectors the tensor acts upon as living in $T_p^{\\times m}$, $V=V^{\\mu_1}\\ldots V^{\\mu_m}\\, \\del_{\\mu_{1}}\\otimes \\ldots\\otimes \\del_{\\mu_m},\\tag{GR.5}$ and similarly for the $n$ dual vectors. Note the structure of the coefficients. An $(n,m)$-tensor $T$ may be written as $T={T^{\\mu_1\\ldots \\mu_n}}_{\\nu_1 \\ldots \\nu_m} \\del_{\\mu_{1}}\\otimes \\ldots\\otimes \\del_{\\mu_n}\\otimes\\mrm dx^{\\nu_1}\\otimes\\ldots\\otimes\\mrm dx^{\\nu_m}.\\tag{GR.6}$\n\nDifferences. In QM, the composition of Hilbert spaces means we physically have multiple quantum systems put together. Also we are allowed to make linear combinations of vectors in $\\HH^{\\otimes n}$, and act on them with the operators. In GR, the tensor product structure 'just' exist to define linear functions that take multiple vectors an kets as input, and we do not consider arbitrary linear combinations of those. The multiple vectors (one-forms) are always taken from the same physical vector space $T_p$ ($T_p^*$).\n\n### The metric\n\nQM. In QM, we have for the inner product between $\\ket \\varphi$ and $\\ket \\psi$ that $\\braket{\\varphi}{\\!\\psi}=\\sum_{i}\\varphi^*_i \\psi_i=\\sum_{ij}\\delta_{ij}\\varphi^*_i \\psi_j,\\tag{QM.7}$ with $\\delta_{ij}$ the Kronecker delta (added for comparison to GR). Because probabilities have to sum to 1, (i.e. $\\sum_i |\\psi_i|^2=1$ ), states in $\\HH$ have norm one. That is, for a physical pure state, $\\braket{ \\psi}{\\!\\psi}=1.\\tag{QM.8}$\n\nGR. In GR the inner product is more general, as it allows for 'weighted inner products'. The weights are determined by the metric tensor $g$, $g(V,W)= g_{\\mu\\nu} V^{\\mu}W^{\\nu}.\\tag{GR.7}$ In flat spacetime (Minkowski space) with regular coordinates $p=(t,x,y,z)$, the metric is such that $g_{00}=-1$ and $g_{11},g_{22},g_{33}=1$ (other elements vanish). In curved spacetimes, the coefficients depend on the point $p$ in spacetime, $g_{\\mu\\nu}=g_{\\mu\\nu}(p)$. For example, for points $p=(t,r,\\varphi,\\theta)$ outside a Schwarzschild black hole (and units where $c=1$), $g_{00}(p)=-\\left(1 - \\frac{R_S}{r}\\right).$ Here $R_S$ is the Schwarzschild radius of the black hole (the location of the horizon), and $R_S-r$ the radial distance from the horizon. We see the weights in front of the time components even go to zero as the points $p$ approach $R_S$.\n\nIt is always possible to change coordinates to that of the instantaneous rest frame of a particle at $p$ that moves with velocity $V$. By the equivalence principle, in this frame $V$ equals $V=(-1,0,0,0)$ and the the metric components are equal to that of the Minkowski metric. Thus $g(V,V)=g_{\\mu\\nu}V^{\\mu}V^{\\nu}=-1.\\tag{GR.8}$ (This holds in any coordinate frame.)\n\nThe metric tensor is a $(0,2)$-tensor, $g=g_{\\mu\\nu} \\mrm d x^{\\mu}\\otimes \\mrm d x^{\\nu}$. Unlike the $(1,1)$-tensors that we described before, the action of a $(0,2)$-tensor on only one vector (instead of two) with components $\\{V^{\\mu}\\}$ yields a one-form, with components $g_{\\mu\\nu}V^{\\nu}\\equiv V_\\mu$, \\begin{align} g(\\cdot, V)=& g_{\\mu\\nu} \\mrm d x^{\\mu}\\otimes \\mrm d x^{\\nu}(V^\\gamma \\del_\\gamma)\\\\ =& (g_{\\mu\\nu} V^{\\nu}) \\mrm dx^{\\mu}. \\end{align} Since one-forms have lower indices and vectors have upper indices, the metric tensor can be used to 'raise and lower indices'. In QM this would mean 'to turn a bra into a ket', or equivalently a column vector into a row vector.\n\n### Change of basis\n\nQM. In QM, we can transform the basis $\\{\\ket i\\}$ to obtain a new basis $\\{\\ket{i}'\\}$. The vector components in the new basis $\\psi_i'$, can be expressed in terms of the old vector components $\\psi_i$ as $\\psi_i'=\\sum_j U_{ij} \\psi_j. \\tag{QM.9}$ Here $U$ is a unitary matrix if we insist (as usual) that the old and new bases are orthonormal. Similarly, the components of an operator $A$ transform as $A'_{ij}=\\sum_{kl}U_{ik}A_{kl}U^{\\dagger}_{lj}=\\sum_{kl}U_{ik} U^{\\dagger}_{lj} A_{kl}.\\tag{QM.10}$ (Last step added for comparison to GR.)\n\nGR. Consider two sets of coordinate functions, $\\{x^{\\mu}\\}$ and $\\{x^{\\mu'}\\}$, that describe points on same part of spacetime somewhere. A coordinate transformation means we write the new coordinates $\\{x^{\\mu'}\\}$ as a function of the old coordinates $\\{x^{\\mu}\\}$, or vice versa, so that we can translate between the two ways of describing points. At every point $p$, this transformation entails a change of basis of $T_p$ because in the new coordinates, $\\del_{\\mu'}=\\frac{\\del x^{\\mu}}{\\del x^{\\mu'}}\\del_\\mu.$ Consequently, the components of a vector $V$ transform as $V^{\\mu'} = \\frac{\\del x^{\\mu'}}{\\del x^{\\mu}} V^{\\mu}.\\tag{GR.9}$ Here $\\frac{\\del x^{\\mu'}}{\\del x^{\\mu}}$ can be thought of as an array of numbers, indexed by $\\mu$ and $\\mu'$, exactly like $U_{ij}$ in QM. The former is often not unitary, as opposed to the latter.\n\nFor the example of a $(1,1)$-tensor, ${T^{\\mu'}}_{\\nu'}=\\frac{\\del x^{\\mu'}}{\\del x^{\\mu}} \\frac{\\del x^{\\nu}}{\\del x^{\\nu'}} {T^{\\mu}}_{\\nu}, \\tag{GR.10}$ and similarly for an $(n,m)$-tensor. Note that $\\frac{\\del x^{\\nu}}{\\del x^{\\nu'}}$ is the transpose of $\\frac{\\del x^{\\nu'}}{\\del x^{\\nu}}$.\n\n* An $(n,n)$-operator can be thought of as as a map from (linear combinations of) $n$ bras (each from the space $\\HH$) and $n$ kets to a complex number, or as a map from 1 ket in the space $\\HH^{\\otimes n}$, and 1 bra (of the same size) to a complex number. So whether you decide to call something a $(1,1)$-operator or an $(n,n)$-operator depends on the viewpoint. I stick with the former viewpoint.\n\n** A super-operator is a linear map from operators on a Hilbert space to operators on a Hilbert space. An example is a quantum channel, which maps density operators to density operators. (There are other properties of quantum channels but those are not important here. Also I note that the dimension of the input operator need not equal the dimension of the output operator.) Since operators themselves are $(1,1)$-tensors, a super-operator is a $(2,2)$-tensor. For example, if we act with a quantum channel $\\Phi$ on one density operator (and not two, in which case we would produce a c-number) we have, in component notation, $\\si_{ij}=\\sum_{ab}\\Phi_{ib,ja}\\rho_{ab}$. Or, in a notation inspired by GR, ${\\si^i}_j={\\Phi^{ib}}_{ja}{\\rho^{a}}_{b}$.\n\nUpdated",
null,
""
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[
null,
"http://kattemolle.com/QMvsGR/by-sa.png",
null
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https://library.wolfram.com/infocenter/Articles/1318/
|
[
"",
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"",
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"",
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"",
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"",
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"",
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"",
null,
"",
null,
"Title",
null,
"",
null,
"",
null,
"",
null,
"A Robust Approach to the Calculation of Paleostress Fields from Fault Plane Data",
null,
"",
null,
"",
null,
"Authors",
null,
"",
null,
"",
null,
"",
null,
"Thomas M. Will",
null,
"Roger Powell",
null,
"",
null,
"",
null,
"Journal / Anthology",
null,
"",
null,
"",
null,
"",
null,
"Journal of Structural Geology\n Year: 1991\n Volume: 13\n Issue: 7\n Page range: 813-821",
null,
"",
null,
"",
null,
"Description",
null,
"",
null,
"",
null,
"",
null,
"Algebraic methods combined with robust regression techniques are used to calculate paleostress tensors from field observations on faults. Previously, such calcullations have involved least-squares regression; however, such regression estimates are likely to break down and produce meaningless results if data are included that are inconsistent with the main body of the data. Such inconsistent data are called outliers, i.e. measurements that are discrepant with respect to the majority of the observations. In two dimensions, the trend of the main body of the data, and their outliers, can be found by plotting the data and examining them visually. Least-squares regression can then be safely applied to the data-set once the outliers have been manually removed. However, the paleostress problem possesses a four-dimensional parameter space, and, as a consequence, this approach cannot be used. To overcome this difficulty, a robust regression estimator, involving the least median of squares (LMS), is applied to the estimation of paleostress tensors from fault plane data; not only can the parameters of the tensor be estimated but also the quality of the data-set assessed. For a data-set that is composed of data from several stress fields the dominant reduced stress tensor will be found by the LMS estimator. A computer program, PSALMS, that calculates paleostress directions using this robust estimator, is presented.",
null,
"",
null,
"",
null,
"Subjects",
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"",
null,
"",
null,
"",
null,
"",
null,
"Mathematics > Probability and Statistics",
null,
"Science > Geology and Geophysics",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
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null,
""
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|
https://chem.libretexts.org/Ancillary_Materials/Worksheets/Worksheets%3A_General_Chemistry/Worksheets%3A_General_Chemistry_(Traditional)/Acids_and_Bases_(Worksheet)
|
[
"# Acids and Bases (Worksheet)\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\nName: ______________________________\n\nSection: _____________________________\n\nStudent ID#:__________________________\n\nWork in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.\n\n## Q1\n\nFill out the following table for each of the following chemical species. Indicate if the species would act as a Brønsted acid, base, or both. Then write the chemical formula for the conjugate acid and base of the species. If it has not conjugate acid, write none.\n\nChemical Species Acid, Base, Both, or Neither Conjugate Acid Conjugate Base\n$$HI$$\n$$NH_4^+$$\n$$HClO_2$$\n$$NO_2^‐$$\n$$HPO_4^{2‐}$$\n$$NH_2^‐$$\n\n## Q2\n\nWhich solution has a higher pH?\n\n1. A 0.1 M solution of an acid with $$K_a = 1 \\times 10^{‐4}$$ or one with $$K_a = 4 \\times 10^{‐5}$$\n2. A 0.1 M solution of an acid with $$pK_a = 1.0$$ or one with $$pK_a = 3.5$$\n3. A 0.1 M solution of a weak acid or a 0.01 M solution of the same acid.\n4. A solution of pOH 6.0 or one of pOH 8.0.\n\n## Q3\n\nCalculate the pH and the pOH of 67.0 mL of a 0.0525 M solution of $$HNO_3$$.\n\n## Q4\n\n1. What are $$[H3O^+]$$, [OH‐], and pOH in a solution with a pH of 9.78?\n2. What are $$[H3O^+]$$, [OH‐], and pOH in a solution with a pH of 3.47?\n\n## Q5\n\nHow many moles of $$H3O^+$$ or OH‐ must you add per liter of HCl solution to adjust its pH form 3.25 to 3.65? Assume a negligible volume change.\n\n## Q6\n\nWhat is an amphoteric species? Name on e and write balanced equations that show why it is amphoteric.\n\n## Q7\n\nChoose the stronger acid in each of the following pairs:\n\n1. $$H_2SeO_3$$ or $$H_2SeO_4$$\n2. $$H_3PO_4$$ or $$H_3AsO_4$$\n3. $$H_2S$$ or $$H_2Te$$\n\n## Q8\n\nWrite balanced equations and Kb expressions for these Bronsted‐Lowry bases in water:\n\n1. Pyridine, $$C_5H_5N$$\n2. $$CO_2^{2‐}$$\n\n## Q9\n\nCalculate the pH of a 0.015 M solution of $$HCN$$. ($$K_a = 1.0 \\times 10^{‐5}$$)\n\n## Q10\n\nWhat is the pH of a 0.200 M $$KNO_2$$ solution? ($$K_a = 5.1 \\times 10^{‐4}$$)\n\n## Q11\n\nExplain with equations and calculations, when necessary, whether an aqueous solution of each of these salts is acidic, basic, or neutral:\n\n1. $$KBr$$\n2. $$NH_4I$$\n3. $$KCN$$\n\nThis page titled Acids and Bases (Worksheet) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Draganjac via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request."
] |
[
null
] |
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|
https://www.plus2net.com/php_tutorial/array-next.php
|
[
"# next: to move the cursor to next element of an array\n\nArray next() returns the next element and shifts the internal pointer by one.\n``next(array) ;``\nLet us create one array and display the second element. In a new array the internal pointer will remain at the first element, so the next() function will shift the pointer to second element and returns the second element.\nExample\n``````\\$my_array=array(\"First One\", \"Second One\", \"Third One\", \"Fourth One\", \"Fifth One\");\necho next(\\$my_array); // Second One\n``````\nIt returns False when the pointer or cursor is moved beyond the last element of the array or not pointing to any element.\n``````\\$my_array=array(\"First One\", \"Second One\", \"Third One\", \"Fourth One\", \"Fifth One\");\nwhile (list (\\$key, \\$val) = each (\\$my_array)) {\necho \"\\$key -> \\$val <br>\";\n}\nif(next(\\$my_array)){\necho \" Current value : \".current(\\$my_array);\n}else{\necho \"The pointer moved beyond last element or the array is empty \";\n}\n``````\nIn above case the cursor has moved beyond the last element of the array ( after displaying all elements inside loop ) so the next() function will return FALSE.\n``````\\$my_array=array(\"First One\", \"Second One\", \"Third One\", \"Fourth One\", \"Fifth One\");\n\necho current(\\$my_array); // Output : First One\necho next(\\$my_array); // Output : Second One\necho next(\\$my_array); // Output : Third One\necho prev(\\$my_array); // Output : Second One\necho end(\\$my_array); // Output : Fifth One\necho reset(\\$my_array); // Output : First One\n``````",
null,
"Subscribe to our YouTube Channel here\n\n## Subscribe\n\n* indicates required\nSubscribe to plus2net",
null,
"plus2net.com\n\nPost your comments , suggestion , error , requirements etc here ."
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"https://www.plus2net.com/images/fb.jpg",
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|
https://betterlesson.com/lesson/393877/the-painted-cube-problem
|
[
"# The Painted Cube Problem\n\n8 teachers like this lesson\nPrint Lesson\n\n## Objective\n\nSWBAT analyze patterns in data tables to determine whether the tables show linear or quadratic relationships and describe these same types of patterns in the context of a real-world problem.\n\n#### Big Idea\n\nStudents discover examples of various polynomials functions in the context of a hands-on problem and make generalizations about how different relationships show up in data tables.\n\n## Warm-Up\n\n30 minutes\n\nThe two problems in this warm-up are designed to prepare students for the day’s lesson and to get them thinking about the big idea of this unit: the relationship between the behavior of a polynomial function and the function rule.\n\nThe first problem is about data tables. Students may or may not remember previous work on linear and quadratic data tables. The Resource Poster will remind them if they have forgotten. It is worth discussing again why a constant difference creates a linear function and why increasing differences create non-linear functions.\n\nThe key idea about these polynomial data tables is that we can examine the differences between consecutive outputs, and the differences between those differences, and the differences between those differences and so on. The number of levels of differences that it takes us to get a sequence of constant differences tells us the degree of the polynomial. This can be proved to be true using calculus (the Power Rule). If you think about the differences as working kind of like derivatives, if you keep finding higher order derivatives of any polynomial function, eventually the degree will be 0, because the Power Rule reduces the degree each time you differentiate. Obviously students will not understand this now, but it is an interesting idea to plant in their minds for later on.\n\nThe scaffold for students who need extra support is the Poster to remind students of their previous knowledge about linear and quadratic patterns in data tables. The extension is for students to find equations for as many of the data tables as they can. Additionally, the 2nd and 3rd pages of the warm-up ask students to go beyond the basic expectations.\n\n30 minutes\n\n## Closing\n\n10 minutes\n\nThe amount of instruction you provide at the end of this lesson really depends on how much progress students make towards solving the problem today. If students complete the data tables for the Painted Cube problem, they will be able to do more analysis of the tables and you can take time at the end of the lesson to discuss how to determine the rules for these data tables. Many of my students had only barely begun to figure out how to complete one row of the table at a time, which is fine for the first day.\n\nI plan to give students some time to think about the exit ticket questions and write their initial thoughts. After they have done this, we will discuss the connection between the first warm-up problem and the Painted Cube problem and give students a little bit more time to predict what type of equation will show up in each column of the data table. This is just a prediction for now, so ask students to justify their thinking but it doesn’t matter if they have the correct predictions at this time."
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http://www.astro.wisc.edu/~townsend/gyre-forums/viewtopic.php?f=4&t=110&
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[
"Coefficients of Hough function\n\nSuggestions for improvements, new features, etc.\nfffeynman\nPosts: 3\nJoined: Wed Mar 18, 2015 2:12 pm\nLocation: Penn State University\nContact:\n\nCoefficients of Hough function\n\nHi, Rich\nIs it possible to add the coefficients of Hough function as a combination of spherical harmonics\nto the mode or summary file? It is interesting to use them to calculate mode visibility for g modes.\nThanks!\n\nbest regards,\nZhao\n\nrhtownsend\nPosts: 307\nJoined: Sun Mar 31, 2013 4:22 pm\n\nRe: Coefficients of Hough function\n\nfffeynman wrote:Hi, Rich\nIs it possible to add the coefficients of Hough function as a combination of spherical harmonics\nto the mode or summary file? It is interesting to use them to calculate mode visibility for g modes.\nThanks!\n\nbest regards,\nZhao\nHi Zhao --\n\nGYRE doesn't have direct access to the coefficients, as it only needs the Hough eigenvalues (lambda) to solve the pulsation equations within the traditional approximation (and it uses pre-calculated tables to evaluate these eigenvalues).\n\nHowever, it should be relatively straightforward to write a wrapper around the astro_hough module (see src/extern/astro/astro_hough.fpp) which does what you want. Feel free to email me if you would like some help with this.\n\nBest wishes,\n\nRich\n\nmankovich\nPosts: 5\nJoined: Tue Jul 12, 2016 12:24 pm\n\nRe: Coefficients of Hough function\n\nHi Rich (and fellow GYREators),\n\nCan I ask you to elaborate a little bit on what the tables contain? Am I right in thinking that the tables contain coefficients to Chebyshev polynomials fit to solutions for the Hough eigenvalues obtained numerically?\n\n(For a little context, I started digging around when I noticed that the traditional approximation only works for l≤3 as a result of the coverage of gyre/data/trad_table.h5. I suppose to calculate the eigenvalues on the fly as part of a GYRE calculation would be significantly more complexity than it's worth!)\n\nCheers,\n\nchris\n\nrhtownsend\nPosts: 307\nJoined: Sun Mar 31, 2013 4:22 pm\n\nRe: Coefficients of Hough function\n\nmankovich wrote:Hi Rich (and fellow GYREators),\n\nCan I ask you to elaborate a little bit on what the tables contain? Am I right in thinking that the tables contain coefficients to Chebyshev polynomials fit to solutions for the Hough eigenvalues obtained numerically?\n\n(For a little context, I started digging around when I noticed that the traditional approximation only works for l≤3 as a result of the coverage of gyre/data/trad_table.h5. I suppose to calculate the eigenvalues on the fly as part of a GYRE calculation would be significantly more complexity than it's worth!)\n\nCheers,\n\nchris\nHi Chris --\n\nSort of! The Chebyshev fits are to lambda/(l(l+1)) for |nu| < 1, where lambda is the Hough eigenvalue.\n\nFor |nu| > 1, the fits are to the ratio lambda/lambda_asymp, where lambda_asymp is the asymptotic eigenvalue found in Townsend (2003).\n\nIn the latest version of GYRE (5.0, currently pre-release), I extend the fits up to higher l,m (10, I think). Also, I use an improved routine for calculating the eigenvalues, and an improved asymptotic formula for lambda_asymp.\n\nYou're right that fits are used for efficiency reasons -- in differentially rotating stars lambda must be evaluated at each grid point, which can be very time consuming if we're doing it on-the-fly.\n\ncheers,\n\nRich\n\nmankovich\nPosts: 5\nJoined: Tue Jul 12, 2016 12:24 pm\n\nRe: Coefficients of Hough function\n\nThanks Rich, that's most helpful. Cheers!"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8793727,"math_prob":0.7055928,"size":3729,"snap":"2019-26-2019-30","text_gpt3_token_len":897,"char_repetition_ratio":0.119194634,"word_repetition_ratio":0.6196013,"special_character_ratio":0.21480289,"punctuation_ratio":0.10714286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9677309,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-18T13:45:46Z\",\"WARC-Record-ID\":\"<urn:uuid:19330827-6e8e-44bf-866a-c0835bcd6687>\",\"Content-Length\":\"32678\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9949884b-c9a3-4d60-9b15-3baeeaca1c60>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc18e8c7-4a0c-48d1-a4f3-e803b0d4ca85>\",\"WARC-IP-Address\":\"144.92.179.14\",\"WARC-Target-URI\":\"http://www.astro.wisc.edu/~townsend/gyre-forums/viewtopic.php?f=4&t=110&\",\"WARC-Payload-Digest\":\"sha1:BK3YNHOQEVPHU4XPWER6M7XMFTBKQODS\",\"WARC-Block-Digest\":\"sha1:A7D5GJV2P4RONQEN3MA7GKWEGPIYZPFC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998724.57_warc_CC-MAIN-20190618123355-20190618145355-00023.warc.gz\"}"}
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https://www.brightstorm.com/tag/exterior-angle/?k=exterior%20angle&s_name=Algebra%202
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[
"# exterior angle 1 videos\n\n• #### Applied Linear Equations: Geometry Problem\n\n##### Algebra 2Linear Equations\n\nHow to find the angles of a triangle by creating a linear equation.\n\nword problem angle measure"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8522332,"math_prob":0.90145177,"size":285,"snap":"2019-35-2019-39","text_gpt3_token_len":61,"char_repetition_ratio":0.16725978,"word_repetition_ratio":0.59090906,"special_character_ratio":0.18596491,"punctuation_ratio":0.06,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985101,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-23T20:04:26Z\",\"WARC-Record-ID\":\"<urn:uuid:efa6caa4-8528-47fe-b8ec-81c9fe554fad>\",\"Content-Length\":\"42342\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5ac6cb4d-9a82-4c99-987e-4fb7e29290a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9447e04-d961-4016-b188-4c3bb13407b7>\",\"WARC-IP-Address\":\"54.235.137.5\",\"WARC-Target-URI\":\"https://www.brightstorm.com/tag/exterior-angle/?k=exterior%20angle&s_name=Algebra%202\",\"WARC-Payload-Digest\":\"sha1:JBJTZUN23TVAC2ESGPUKP4IY6PS6OIXG\",\"WARC-Block-Digest\":\"sha1:35CQML5RUFAZJQSJR4422OO2TFUHIVEA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514578201.99_warc_CC-MAIN-20190923193125-20190923215125-00106.warc.gz\"}"}
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https://catalog.csuchico.edu/viewer/19/MATH/MATHNONEUM.html
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[
"The Minor in Mathematics\n\nCourse Requirements for the Minor: 24-25 units\n\nThe following courses, or their approved transfer equivalents, are required of all candidates for this minor.\n\n2 courses required:\n\nSUBJ NUM Title Sustainable Units Semester Offered Course Flags\nPrerequisites: GE Mathematics/Quantitative Reasoning Ready; both MATH 118 and MATH 119 (or college equivalent); first-year freshmen who successfully completed trigonometry and precalculus in high school can meet this prerequisite by achieving a score that meets department guidelines on a department administered calculus readiness exam.\nLimits and continuity. The derivative and applications to related rates, maxma and minima, and curve sketching. Transcendental functions. An introduction to the definite integral and area. 4 hours discussion. This is an approved General Education course. (005506)\nPrerequisite: MATH 120.\nThe definite integral and applications to area, volume, work, differential equations, etc. Sequences and series, vectors and analytic geometry in 2 and 3-space, polar coordinates, and parametric equations. 4 hours discussion. (005507)\n\n2 courses selected from:\n\nSUBJ NUM Title Sustainable Units Semester Offered Course Flags\nPrerequisites: GE Mathematics/Quantitative Reasoning Ready, CSCI 111 with a grade of C or higher, MATH 119 (or equivalent).\nOffers an intensive introduction to discrete mathematics as used in computer science. Topics include sets, relations, propositional and predicate logic, basic proof methods including mathematical induction, digital logic circuits, complexity of algorithms, elementary combinatorics, and solving linear recurrence relations. 3 hours discussion. (005550)\nPrerequisites: MATH 121.\nVector functions and space curves. Functions of several variables, partial derivatives, and multiple integrals. Vector calculus line integrals, surface integrals, divergence/curl, Green's Theorem, Divergence Theorem, and Stokes' Theorem. 4 hours discussion. (005508)\nPrerequisites: MATH 121.\nFirst order separable, linear, and exact equations; second order linear equations, Laplace transforms, series solutions at an ordinary point, systems of first order linear equations, and applications. 4 hours discussion. (005509)\n\n9 units selected from:\n\nMATH 235 or upper-division courses, at least 6 of which must be from the Department of Mathematics and Statistics. 3 units may be from another department as long as the course has significant mathematical content, as determined by the chair of the Department of Mathematics and Statistics. Acceptable upper-division MATH units will not include MATH 305, MATH 310, MATH 311, MATH 315, MATH 333, MATH 341, MATH 342, MATH 441, MATH 442. Prior approval is required to count Independent Study and Special Topics courses (398, 399, 498, and 499) towards the Minor.\n\nTo Apply for a Minor\n\nTo apply for a minor, you must fill out a Declaration of Minor form, available from the Undergraduate Plan Change Form website and take to the Mathematics and Statistics Department, HOLT 181, for the Chair's signature. Before choosing any options for the minor, you must obtain approval from the department chair.\n\nCatalog Cycle:19"
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https://www.lmfdb.org/EllipticCurve/Q/265200/r/2
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[
"Show commands for: Magma / SageMath / Pari/GP\n\n## Minimal Weierstrass equation\n\nmagma: E := EllipticCurve([0, -1, 0, 837992, 4969052512]); // or\n\nmagma: E := EllipticCurve(\"265200r2\");\n\nsage: E = EllipticCurve([0, -1, 0, 837992, 4969052512]) # or\n\nsage: E = EllipticCurve(\"265200r2\")\n\ngp: E = ellinit([0, -1, 0, 837992, 4969052512]) \\\\ or\n\ngp: E = ellinit(\"265200r2\")\n\n$$y^2 = x^{3} - x^{2} + 837992 x + 4969052512$$\n\n## Mordell-Weil group structure\n\n$$\\Z\\times \\Z/{2}\\Z$$\n\n### Infinite order Mordell-Weil generator and height\n\nmagma: Generators(E);\n\nsage: E.gens()\n\n $$P$$ = $$\\left(\\frac{25294}{25}, -\\frac{10346778}{125}\\right)$$ $$\\hat{h}(P)$$ ≈ 2.79021841924\n\n## Torsion generators\n\nmagma: TorsionSubgroup(E);\n\nsage: E.torsion_subgroup().gens()\n\ngp: elltors(E)\n\n$$\\left(-1543, 0\\right)$$\n\n## Integral points\n\nmagma: IntegralPoints(E);\n\nsage: E.integral_points()\n\n$$\\left(-1543, 0\\right)$$, $$(1582,\\pm 101250)$$\n\n## Invariants\n\n magma: Conductor(E); sage: E.conductor().factor() gp: ellglobalred(E) Conductor: $$265200$$ = $$2^{4} \\cdot 3 \\cdot 5^{2} \\cdot 13 \\cdot 17$$ magma: Discriminant(E); sage: E.discriminant().factor() gp: E.disc Discriminant: $$-10705581498128400000000$$ = $$-1 \\cdot 2^{10} \\cdot 3^{8} \\cdot 5^{8} \\cdot 13^{2} \\cdot 17^{6}$$ magma: jInvariant(E); sage: E.j_invariant().factor() gp: E.j j-invariant: $$\\frac{4067455675907516}{669098843633025}$$ = $$2^{2} \\cdot 3^{-8} \\cdot 5^{-2} \\cdot 13^{-2} \\cdot 17^{-6} \\cdot 100559^{3}$$ Endomorphism ring: $$\\Z$$ (no Complex Multiplication) Sato-Tate Group: $\\mathrm{SU}(2)$\n\n## BSD invariants\n\n magma: Rank(E); sage: E.rank() Rank: $$1$$ magma: Regulator(E); sage: E.regulator() Regulator: $$2.79021841924$$ magma: RealPeriod(E); sage: E.period_lattice().omega() gp: E.omega Real period: $$0.098803238859$$ magma: TamagawaNumbers(E); sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[i,1],gr[i]] | i<-[1..#gr[,1]]] Tamagawa product: $$64$$ = $$2\\cdot2\\cdot2^{2}\\cdot2\\cdot2$$ magma: Order(TorsionSubgroup(E)); sage: E.torsion_order() gp: elltors(E) Torsion order: $$2$$ magma: MordellWeilShaInformation(E); sage: E.sha().an_numerical() Analytic order of Ш: $$1$$ (exact)\n\n## Modular invariants\n\n#### Modular form 265200.2.a.r\n\nmagma: ModularForm(E);\n\nsage: E.q_eigenform(20)\n\ngp: xy = elltaniyama(E);\n\ngp: x*deriv(xy)/(2*xy+E.a1*xy+E.a3)\n\n$$q - q^{3} - 2q^{7} + q^{9} - 4q^{11} - q^{13} - q^{17} - 6q^{19} + O(q^{20})$$\n\n magma: ModularDegree(E); sage: E.modular_degree() Modular degree: 17694720 $$\\Gamma_0(N)$$-optimal: no Manin constant: 1\n\n#### Special L-value\n\nmagma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);\n\nsage: r = E.rank();\n\nsage: E.lseries().dokchitser().derivative(1,r)/r.factorial()\n\ngp: ar = ellanalyticrank(E);\n\ngp: ar/factorial(ar)\n\n$$L'(E,1)$$ ≈ $$4.41092187112$$\n\n## Local data\n\nmagma: [LocalInformation(E,p) : p in BadPrimes(E)];\n\nsage: E.local_data()\n\ngp: ellglobalred(E)\n\nprime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\\Delta$$) ord$$(j)_{-}$$\n$$2$$ $$2$$ $$I_2^{*}$$ Additive 1 4 10 0\n$$3$$ $$2$$ $$I_{8}$$ Non-split multiplicative 1 1 8 8\n$$5$$ $$4$$ $$I_2^{*}$$ Additive 1 2 8 2\n$$13$$ $$2$$ $$I_{2}$$ Non-split multiplicative 1 1 2 2\n$$17$$ $$2$$ $$I_{6}$$ Non-split multiplicative 1 1 6 6\n\n## Galois representations\n\nThe image of the 2-adic representation attached to this elliptic curve is the subgroup of $\\GL(2,\\Z_2)$ with Rouse label X10.\n\nThis subgroup is the pull-back of the subgroup of $\\GL(2,\\Z_2/2^2\\Z_2)$ generated by $\\left(\\begin{array}{rr} 1 & 0 \\\\ 2 & 1 \\end{array}\\right),\\left(\\begin{array}{rr} 3 & 3 \\\\ 0 & 1 \\end{array}\\right),\\left(\\begin{array}{rr} 1 & 2 \\\\ 0 & 1 \\end{array}\\right)$ and has index 6.\n\nmagma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];\n\nsage: rho = E.galois_representation();\n\nsage: [rho.image_type(p) for p in rho.non_surjective()]\n\nThe mod $$p$$ Galois representation has maximal image $$\\GL(2,\\F_p)$$ for all primes $$p$$ except those listed.\n\nprime Image of Galois representation\n$$2$$ B\n\n## $p$-adic data\n\n### $p$-adic regulators\n\nsage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]\n\n$$p$$-adic regulators are not yet computed for curves that are not $$\\Gamma_0$$-optimal.\n\nNo Iwasawa invariant data is available for this curve.\n\n## Isogenies\n\nThis curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2.\nIts isogeny class 265200.r consists of 2 curves linked by isogenies of degree 2.\n\n## Growth of torsion in number fields\n\nThe number fields $K$ of degree up to 7 such that $E(K)_{\\rm tors}$ is strictly larger than $E(\\Q)_{\\rm tors}$ $\\cong \\Z/{2}\\Z$ are as follows:\n\n$[K:\\Q]$ $K$ $E(K)_{\\rm tors}$ Base-change curve\n2 $$\\Q(\\sqrt{-1})$$ $$\\Z/2\\Z \\times \\Z/2\\Z$$ Not in database\n4 4.2.4884100.2 $$\\Z/4\\Z$$ Not in database\n\nWe only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database."
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https://math.stackexchange.com/questions/2386341/distributivity-of-exponents-in-rings
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[
"# Distributivity of Exponents in Rings\n\nLet $R$ be a ring with under $+$ and $*$\n\nFor $x \\in R$, we define $x^m := \\underbrace{x*\\cdots*x}_{m \\text{ terms}}$\n\nConjecture. $R$ is commutative if and only if:\n\nFor every positive integer m, and for all $x, y$ in $R$, $(x*y)^m = x^m * y^m$.\n\nI was able to show that if $R$ is commutative, then the exponent is distributive, but not the backward implication. Is the conjecture true, and if not, what is a necessary and sufficient condition such that For all $x, y$ in $R$, $(x*y)^m = x^m * y^m$?\n\n• @bof yes, it is bound by a universal quantifier. – rr01 Aug 8 '17 at 6:26\n• I corrected my answer, it was false. – Idéophage Mar 5 '19 at 13:23\n\nYes, every ring of this form is commutative. For $$x,y ∈ R$$, let $$f(x,y) := x^2y^2-(xy)^2$$. Then we have the equality (true in any ring) $$xy-yx = f(x,y) - f(1+x,y) - f(x,1+y) + f(1+x,1+y) \\text{.}$$\nNote that if you require that your ring have no nonzero zero divisor, the reason is simpler: $$(xy)^2-x^2y^2 = x(yx-xy)y=0$$ and if $$x$$ and $$y$$ are nonzero, then $$yx-xy=0$$.\n• This is not a counterexample. In your quotient, only the projections of $x$ and $y$ satisfy the $\\left(uv\\right)^n = u^n v^n$ identity, but the OP wants this identity to hold for every pair of elements $u$ and $v$. – darij grinberg Mar 5 '19 at 6:20"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8528176,"math_prob":0.99994993,"size":496,"snap":"2019-51-2020-05","text_gpt3_token_len":167,"char_repetition_ratio":0.09146342,"word_repetition_ratio":0.11235955,"special_character_ratio":0.33669356,"punctuation_ratio":0.13793103,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000073,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-20T09:37:31Z\",\"WARC-Record-ID\":\"<urn:uuid:85d13d76-37d6-47ea-a0a1-a3c834eeff2e>\",\"Content-Length\":\"139325\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e292f94-782e-460b-affa-5bea02b86b56>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee8b7baa-2baf-4420-8ce2-1948ad177e3b>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2386341/distributivity-of-exponents-in-rings\",\"WARC-Payload-Digest\":\"sha1:5U4HVIX5EVDPXSGRUFARABODZ2EXO5YR\",\"WARC-Block-Digest\":\"sha1:W7ZNVVF4ZR5DYEJE4KQQ2O6D42FQNYIY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250598217.23_warc_CC-MAIN-20200120081337-20200120105337-00097.warc.gz\"}"}
|
https://kalkicode.com/reverse-number-using-stack
|
[
"# Reverse a number using stack\n\nThe problem is to reverse a given number using a stack. The goal is to reverse the digits of the number, maintaining the sign if it is a negative number, and ignoring leading zeroes if present in the original number.\n\n## Idea to Solve the Problem\n\nTo reverse the number using a stack, we will follow these steps:\n\n1. Create an empty custom stack `s` to store the digits of the number.\n2. Extract the last digit of the number and push it onto the stack `s`.\n3. Remove the last digit from the number.\n4. Repeat steps 2 and 3 until the number becomes zero.\n5. Pop digits from the stack `s` one by one and combine them to form the reversed number.\n\n## Algorithm\n\n1. Create a custom stack data structure with push, pop, isEmpty, and peek operations.\n2. Define a function `reverse` that takes an integer `x` as input and returns the reversed number.\n3. Create an empty stack `s` to store the digits of the number.\n4. While `x` is not equal to zero, do the following: a. Get the last digit of `x` by taking `x % 10`. b. Push the last digit onto the stack `s`. c. Remove the last digit from `x` by updating `x = x / 10`.\n5. Initialize a variable `multiply` to 1.\n6. While the stack `s` is not empty, do the following: a. Pop the top element from the stack `s`. b. Multiply the popped element by `multiply` and add it to `x`. c. Multiply `multiply` by 10.\n7. Return the reversed number `x`.\n\n## Pseudocode\n\n``````FUNCTION reverse(x):\nnum = x\ns = new MyStack()\n\nWHILE num is not 0:\ns.push(num % 10)\nnum = num / 10\n\nmultiply = 1\nWHILE s is not empty:\nnum = s.pop() * multiply + num\nmultiply = multiply * 10\n\nRETURN num\n``````\n\n## Time Complexity Analysis\n\nLet n be the number of digits in the input number. The time complexity of the `reverse` function is O(n) because we need to iterate through each digit of the number and perform stack operations.\n\n## Output Explanation\n\nThe output shows the number before and after reversing using the `reverse` function. For example, \"Before -381725 After -527183\" represents the original number -381725, and the reversed number is -527183. Similarly, for other test cases, the numbers are reversed accordingly. Note that leading zeroes are ignored in the reversed number, as shown in the output \"Before 12340 After 4321\".\n\n## Comment\n\nPlease share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible."
] |
[
null
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|
https://people.sc.fsu.edu/~jburkardt/m_src/grf_display/grf_display.html
|
[
"# grf_display\n\ngrf_display, a MATLAB code which reads a GRF file containing information about an abstract graph and displays a plot in the MATLAB graphics window.\n\n### Usage:\n\ngrf_display prefix\nor\ngrf_display ( 'prefix' )\nwhere\n• 'prefix'.grf is the name of the GRF file to be read and displayed.\n\n### Languages:\n\ngrf_display is available in a MATLAB version.\n\n### Related Data and Programs:\n\ncircle_grid_display, a MATLAB code which reads a matrix of integers, and draws a corresponding grid of circles filled with color.\n\ngraffiti, a dataset directory which contains 195 abstract graphs, with adjacency and embedding information, stored in the GRF format.\n\ngraph_representation, a data directory of examples of ways of representing abstract mathematical graphs\n\ngrf, a data directory which contains a description of the GRF format and some examples.\n\ngrf_io, a MATLAB code which reads and writes files in the GRF format.\n\ngrid_display, a MATLAB code which can display a 2D or 3D grid or sparse grid.\n\ngridlines, a MATLAB code which gives the user more control over drawing gridlines on a graph than the builtin \"grid on\" command.\n\n### Source Code:\n\nLast revised on 25 January 2019."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.80688965,"math_prob":0.40802765,"size":1103,"snap":"2021-43-2021-49","text_gpt3_token_len":235,"char_repetition_ratio":0.155596,"word_repetition_ratio":0.029411765,"special_character_ratio":0.19492294,"punctuation_ratio":0.103626944,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9531381,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T13:36:45Z\",\"WARC-Record-ID\":\"<urn:uuid:a9fa50fb-0b67-467b-87d5-9d8645061ff8>\",\"Content-Length\":\"3589\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09f3820d-e9ae-4da4-ac0d-228b09c66488>\",\"WARC-Concurrent-To\":\"<urn:uuid:05714f4f-d6e5-42fe-aca1-6b623e7abbde>\",\"WARC-IP-Address\":\"144.174.16.102\",\"WARC-Target-URI\":\"https://people.sc.fsu.edu/~jburkardt/m_src/grf_display/grf_display.html\",\"WARC-Payload-Digest\":\"sha1:7LXVETEUBZOG6YRN4S6FWGJLL5K2BFX5\",\"WARC-Block-Digest\":\"sha1:SEN7AAETSDJRZNTPPNV7EL4J2CFD55EY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585177.11_warc_CC-MAIN-20211017113503-20211017143503-00193.warc.gz\"}"}
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https://craiget.com/a-genetic-algorithm-kenken-solver.html
|
[
"# A Genetic Algorithm KenKen Solver\n\nWARNING: This article features ANCIENT code! I'm keeping it online because it's interesting to see what I was thinking 10+ years ago. But you DEFINITELY should not be using this code. Anything you're reading about on this page has changed significantly since this was written.\n\nSince being introduced to KenKen a few months ago, I've been slightly addicted, so I thought it'd be interesting to try solving the puzzles in an different sort of way.\n\nSo following is a genetic algorithm solution using pyevolve.\n\nThe first thing is to define a data format to encode a puzzle. There are several ways to do it, but I went with this:\n\n``````1 1 1 2\n3 4 4 2\n3 3 5 5\n6 7 7 5\n8 *\n1 -\n6 +\n1 -\n4 *\n3 +\n2 /\n``````\n\nThe first four rows describe what some folks are calling \"cages\". Any number of following rows describe the rule corresponding to each cage. For example, in this puzzle, Cage #1 consists of cells (1,2,3) and has the rule `(8,*)`, meaning that whatever numbers go in those cells, when multiplied, their product must be 8. Cage #2 consists of cells `(4,8)` and has the rule `(1,-)`, meaning that the difference of those two numbers must be 1.\n\nYou don't really need the initial 4-row cage definitions, you could also list the corresponding cells following each rule, but I find the grid easier to look at.\n\nSo to find a solution, all we need to do is generate the correct list of 16 digits (1-4) where none of the horizontal, vertical or mathematical constraints are violated. Picking solutions randomly, with no regard to the constraints, there are about 4 billion possible solutions (almost all of which are illegal). That number drops dramatically if you only generate solutions that follow the horizontal and vertical constraints, but we're gonna do it genetically, so who cares?\n\nSo here's code for a pyevolve-based solution. In my tests, it does a pretty good job finding the right answer, but every once in awhile it seems to get stuck and finishes without a solution. C'est la vie with random algorithms, but some more tuning might fix it.\n\n``````import operator\n\nfrom pyevolve import G1DList\nfrom pyevolve import GSimpleGA\nfrom pyevolve import Selectors\nfrom pyevolve import Statistics\n\n# ensure exactly 4x1s, 4x2s, 4x3s, 4x4s\ndef error_c(soln):\ne1 = len([x for x in soln if x == 1])\ne2 = len([x for x in soln if x == 2])\ne3 = len([x for x in soln if x == 3])\ne4 = len([x for x in soln if x == 4])\ne_total = abs(4-e1) + abs(4-e2) + abs(4-e3) + abs(4-e4)\nreturn 1 * e_total\n\n# ensure no duplicates in a horizontal row\ndef error_h(soln):\nscore = 0\nfor i in [0,4,8,12]:\nif soln[i+0] == soln[i+1]: score = score + 1\nif soln[i+0] == soln[i+2]: score = score + 1\nif soln[i+0] == soln[i+3]: score = score + 1\nif soln[i+1] == soln[i+2]: score = score + 1\nif soln[i+1] == soln[i+3]: score = score + 1\nif soln[i+2] == soln[i+3]: score = score + 1\nreturn score\n\n# ensure no duplicates in a vertical row\ndef error_v(soln):\nscore = 0\nfor i in [0,1,2,3]:\nif soln[i+0] == soln[i+4]: score = score + 1\nif soln[i+0] == soln[i+8]: score = score + 1\nif soln[i+0] == soln[i+12]: score = score + 1\nif soln[i+4] == soln[i+8]: score = score + 1\nif soln[i+4] == soln[i+12]: score = score + 1\nif soln[i+8] == soln[i+12]: score = score + 1\nreturn score\n\n# ensure that each cell adds up correctly\ndef error_r(soln, rules):\nscore = 0\nfor rule in rules:\nop, n, cells = rule\nif op == '-':\nif soln[cells] - soln[cells] == n:\ncontinue\nif soln[cells] - soln[cells] == n:\ncontinue\nscore = score + 1\nif op == '/':\nif soln[cells] / soln[cells] == n:\ncontinue\nif soln[cells] / soln[cells] == n:\ncontinue\nscore = score + 1\nif op == '+':\nif reduce(operator.add, [soln[cell] for cell in cells]) == n:\ncontinue\nscore = score + 1\nif op == '*':\nif reduce(operator.mul, [soln[cell] for cell in cells]) == n:\ncontinue\nscore = score + 1\nreturn 10*score\n\n# fitness function gives higher scores for fewer rule violations\ndef eval_func(genome):\nsoln = [i for i in genome]\nscore = error_c(soln) + error_h(soln) + error_v(soln) + error_r(soln, rules)\nreturn 1000 - score\n\n# global rules list\nrules = []\n\ndef solve(infile):\n# read cell definitions from file\ncells = []\nf = open(infile,'r')\nfor line in lines[:4]:\nparts = line.strip().split()\nfor part in parts:\ncells.append(int(part))\n\n# generate global rules list\n# rules = list[ operator, number, cells ]\n# for example [*, 24, [1,2,3]]\nfor i, line in enumerate(lines[4:],1):\nparts = line.strip().split()\nn, op = int(parts), parts\nc = [j for j,cell in enumerate(cells) if cell == i]\nrules.append([op,n,c])\n\n# solution is 16-element list\ngenome = G1DList.G1DList(16)\ngenome.setParams(rangemin=1, rangemax=4)\ngenome.setParams(bestrawscore=1000)\ngenome.evaluator.set(eval_func)\n\n# terminate after 5000 generations or perfect match (score = 1000)\nga = GSimpleGA.GSimpleGA(genome)\nga.selector.set(Selectors.GRouletteWheel)\nga.terminationCriteria.set(GSimpleGA.RawScoreCriteria)\nga.setGenerations(5000)\n\n# run evolution\nga.evolve(freq_stats=100)\nprint ga.bestIndividual()\n\nif __name__ == \"__main__\":\nsolve('75.txt')\n``````\n\nIt reads the specified puzzle and tries to solve it within 5000 generations. The fitness function works by giving a maximum score of 1000 points for a perfect solution. Each constraint violation lowers the score. Over time, the best solutions propagate and usually a perfect solution is found.\n\nI spent a fair amount of time playing with the settings to try out different population sizes and mutation rates, but ultimately, the defaults did a pretty good job. While the vertical, horizontal and mathematical criteria fully specify the rules, I also added an extra set of criteria to further penalize solutions without 4 of each number. Probably gratuitous, but it seemed to make things converge a bit faster.\n\nLike always, use at your own risk. I think this code is correct, but it was also thrown together in a couple of hours for my own amusement and I only verified the solutions to a couple of puzzles.\n\nThis is what it looks like on a typical run:\n\n``````Gen. 0 (0.00%): Max/Min/Avg Fitness(Raw) [1113.31(961.00)/756.03(897.00)/927.76(927.76)]\nGen. 100 (2.00%): Max/Min/Avg Fitness(Raw) [1137.93(997.00)/853.79(924.00)/948.28(948.27)]\nGen. 200 (4.00%): Max/Min/Avg Fitness(Raw) [1143.24(998.00)/853.01(929.00)/952.70(952.70)]\n\nEvolution stopped by Termination Criteria function !\n\nGen. 271 (5.42%): Max/Min/Avg Fitness(Raw) [1138.08(1000.00)/825.62(915.00)/948.40(948.40)]\nTotal time elapsed: 1.869 seconds.\n- GenomeBase\nScore: 1000.000000\nFitness: 1138.080000\n\nParams: {'bestrawscore': 1000, 'rangemax': 4, 'rangemin': 1}\n\nSlot [Evaluator] (Count: 1)\nName: eval_func - Weight: 0.50\nSlot [Initializator] (Count: 1)\nName: G1DListInitializatorInteger - Weight: 0.50\nDoc: Integer initialization function of G1DList\n\nSlot [Mutator] (Count: 1)\nName: G1DListMutatorSwap - Weight: 0.50\nDoc: The mutator of G1DList, Swap Mutator\n\nSlot [Crossover] (Count: 1)\nName: G1DListCrossoverSinglePoint - Weight: 0.50\nDoc: The crossover of G1DList, Single Point\n\n- G1DList\nList size: 16\n``````\n\nAnd the puzzle solution\n\n`````` List: [4, 1, 2, 3, 1, 4, 3, 2, 2, 3, 1, 4, 3, 2, 4, 1]\n``````\n\nAnyway, I like KenKen and thought this was kinda cool. A direct, recursive solution should be reasonably easy to program as well. Just as some folks have done with Sudoku, it should also be possible to do this using a SAT solver. Still working on that one.."
] |
[
null
] |
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|
https://mathematica.stackexchange.com/questions/85179/threading-elements-over-corresponding-elements-in-the-second-list
|
[
"# Threading elements over corresponding elements in the second list\n\nI often need to merge two lists like the following:\n\n$A=\\{\\{0.1\\},\\{0.2\\},\\ldots,\\{1\\}\\}$\n\n$B=\\{\\{x_1\\},\\{x_{21},x_{22},x_{23}\\},\\{x_{31},x_{32},x_{33}\\},\\ldots,x_N\\}$\n\nin a way to ideally obtain a new list such as:\n\n$C = \\{\\{0.1,x_{1}\\}, \\{0.2,x_{21}\\}, \\{0.2,x_{22}\\}, \\{0.2, x_{23}\\},\\{0.3,x_{31}\\}, \\{0.3,x_{32}\\}, \\{0.3, x_{33}\\}, \\ldots, \\{1, x_N\\}\\}$\n\nAt present I do this by inspecting the lists and manually associating them in the appropriate way. In above example, I would do\n\nTable[{A[[k]],B[[k,1]]},{k,1,N}]\n\nTable[{A[[k]],B[[k,2]]},{k,2,S}]\n\nTable[{A[[k]],B[[k,3]]},{k,2,S}]\n\n\nwhere S is the position of the last element that contains three entries in the $B$ list.\n\nThese lists are typically very long, and I wonder whether there is a way to do this efficiently.\n\n• A related thread. You'll be particularly interested in the use of Flatten[]. Jun 4, 2015 at 20:29\n• Flatten[Tuples /@ (Transpose@{a, b}), 1]? Jun 4, 2015 at 20:36\n• closely related\n– Kuba\nJun 4, 2015 at 20:40\n• Also Function[Null, {##}, Listable][Flatten@alist, blist]~Flatten~1 Feb 24, 2016 at 13:00\n\nI feel like it is a duplicate but I can't find it now.\n\nMapThread[Apply[Sequence]@*Tuples@*List, {alist, blist}]",
null,
"or\n\nFlatten[Tuples /@ Transpose[{alist, blist}], 1]\n\n\n,\n\nFlatten[MapThread[Thread[{#[], #2}] &, {alist, blist}], 1]\n\n\nYou might have to change this slightly for your particular problem.\n\n A = Table[{0.1 n}, {n, 1, 10}]\nB = Array[Subscript[x, ##] &, {10, 3}]\nMap[Flatten, Apply[Prepend, MapThread[Prepend, {A, B}], {1}]]\n\nJoin @@ (Thread /@ Thread[{Flatten @ a, b}])",
null,
""
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[
null,
"https://i.stack.imgur.com/R76aL.png",
null,
"https://i.stack.imgur.com/RORjD.png",
null
] |
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|
https://www.percentagecal.com/answer/what-is-percentage-difference-from-3750-to-4000
|
[
"What is the percentage increase/decrease\n\n#### Solution for What is the percentage increase/decrease from 3750 to 4000:\n\n(4000-3750):3750*100 =\n\n(4000:3750-1)*100 =\n\n106.66666666667-100 = 6.67\n\nNow we have: What is the percentage increase/decrease from 3750 to 4000 = 6.67\n\nWhat is the percentage increase/decrease\n\n#### Solution for What is the percentage increase/decrease from 4000 to 3750:\n\n(3750-4000):4000*100 =\n\n(3750:4000-1)*100 =\n\n93.75-100 = -6.25\n\nNow we have: What is the percentage increase/decrease from 4000 to 3750 = -6.25\n\nCalculation Samples"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8864023,"math_prob":0.9912746,"size":1054,"snap":"2023-14-2023-23","text_gpt3_token_len":333,"char_repetition_ratio":0.2647619,"word_repetition_ratio":0.30232558,"special_character_ratio":0.42409867,"punctuation_ratio":0.08866995,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99880224,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-02T08:34:11Z\",\"WARC-Record-ID\":\"<urn:uuid:b6e751db-238a-4271-8ba3-d20740370002>\",\"Content-Length\":\"10409\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dd9b036e-1e0a-45c7-976c-8b260386c64c>\",\"WARC-Concurrent-To\":\"<urn:uuid:6e7c0831-039e-4c36-9b10-f7cf5e681c17>\",\"WARC-IP-Address\":\"217.23.5.136\",\"WARC-Target-URI\":\"https://www.percentagecal.com/answer/what-is-percentage-difference-from-3750-to-4000\",\"WARC-Payload-Digest\":\"sha1:AKLHTTH6HEXW5MP4G467ZMZMO5XIND7M\",\"WARC-Block-Digest\":\"sha1:DKO7G2ATQXC3Y5QV3ZSJZGBT3GAPP4A5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296950422.77_warc_CC-MAIN-20230402074255-20230402104255-00202.warc.gz\"}"}
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https://alok.github.io/2016/10/30/ia1-trick/
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[
"The geometric series $$\\sum_{k=0}^{\\infty}{ar^k}$$ equals $$\\frac{a}{1-r}$$ if and only if $$|r| < 1$$.\n\nBy squinting at the notation, we can notice (not prove) a “well-known” matrix identity. (I keep forgetting it, so it’s not well-known to me).\n\nIf we let $$a = 1$$ and make the following replacements:\n\n$r \\leftarrow A \\\\ 1 \\leftarrow I \\\\$\n\n(where $$A$$ is some matrix and $$I$$ is the identity matrix), we can get the following “identity”:\n\n$\\sum_{k=0}^{\\infty}{A^k} = \\frac{1}{I-A}$\n\nNow we interpret “1 over something” as its multiplicative inverse, so the right hand side becomes $$(I-A)^{-1}$$.\n\nTherefore, by our “identity”,\n\n$(I-A)^{-1}= A + A^2 + A^3 + A^4...$\n\nwhich is actually true."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8223197,"math_prob":0.9999976,"size":691,"snap":"2023-40-2023-50","text_gpt3_token_len":229,"char_repetition_ratio":0.09315866,"word_repetition_ratio":0.0,"special_character_ratio":0.3748191,"punctuation_ratio":0.11188811,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000081,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T20:56:37Z\",\"WARC-Record-ID\":\"<urn:uuid:110f0c76-64b4-40fe-96b1-4157ba5df32d>\",\"Content-Length\":\"9862\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:598aa38c-8139-43ca-9d10-6625213d281c>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a8883c6-103c-49d3-a70c-c5708d1cfc68>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://alok.github.io/2016/10/30/ia1-trick/\",\"WARC-Payload-Digest\":\"sha1:ORHBBOLE62KJQWH5WCYYECLGPI7EA2NJ\",\"WARC-Block-Digest\":\"sha1:IUYEP73NWKA2LOVXWKB5OUHTVCOUPE33\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511021.4_warc_CC-MAIN-20231002200740-20231002230740-00449.warc.gz\"}"}
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http://stronki.info/math-for-kindergarten-worksheet/math-for-kindergarten-worksheet-kindergarten-math-worksheets-inspirational-kindergarten-worksheet-thanksgiving-math-worksheets-kindergarten-free/
|
[
"# Math For Kindergarten Worksheet Kindergarten Math Worksheets Inspirational Kindergarten Worksheet Thanksgiving Math Worksheets Kindergarten Free",
null,
"math for kindergarten worksheet kindergarten math worksheets inspirational kindergarten worksheet thanksgiving math worksheets kindergarten free.\n\nmath worksheets kindergarten pdf subtraction saxon free,kindergarten math worksheets cut and paste numbers counting kumon pdf,kindergarten math skills worksheets free addition pdf kumon,free printable greater than less worksheets for kindergarten equal common core math pdf kumon numbers 1 30,adding and subtracting kindergarten worksheets pre math pdf addition numbers 0 5,kindergarten math worksheets addition pdf free subtraction for,kindergarten math worksheets addition and subtraction numbers 1 5 free for all download coloring pdf,kindergarten math worksheets subtraction with pictures reading image collections worksheet skills free for addition and pdf,kindergarten math skills worksheets free counting for pdf numbers 11 20,math kindergarten worksheets numbers number common core pdf."
] |
[
null,
"http://stronki.info/wp-content/uploads/2018/09/math-for-kindergarten-worksheet-kindergarten-math-worksheets-inspirational-kindergarten-worksheet-thanksgiving-math-worksheets-kindergarten-free.jpg",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.67037123,"math_prob":0.8936719,"size":940,"snap":"2019-13-2019-22","text_gpt3_token_len":161,"char_repetition_ratio":0.2905983,"word_repetition_ratio":0.01754386,"special_character_ratio":0.14787234,"punctuation_ratio":0.07971015,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99280626,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-21T17:41:38Z\",\"WARC-Record-ID\":\"<urn:uuid:85406a9e-6631-4f62-8fa7-ace08542f806>\",\"Content-Length\":\"59473\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dba71fd6-7472-4d6c-bb60-3c1f9558d4fe>\",\"WARC-Concurrent-To\":\"<urn:uuid:19ba6503-995c-43b0-be13-7c6da5bef32a>\",\"WARC-IP-Address\":\"104.27.149.159\",\"WARC-Target-URI\":\"http://stronki.info/math-for-kindergarten-worksheet/math-for-kindergarten-worksheet-kindergarten-math-worksheets-inspirational-kindergarten-worksheet-thanksgiving-math-worksheets-kindergarten-free/\",\"WARC-Payload-Digest\":\"sha1:GFP5S56HGU6HI4FGFPOQOTVETHD7WJ6F\",\"WARC-Block-Digest\":\"sha1:VCJBBFZG5BBIWYRC2ZKUFUKGPDKNCEFV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256494.24_warc_CC-MAIN-20190521162634-20190521184634-00213.warc.gz\"}"}
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https://blog.mocki.co/are-strings-anagrams/
|
[
"# Are Strings Anagrams?\n\nAnagram = all letters can be reused to create another word.\n\n## Approaches\n\nApproach 1: sorting (nonintuitive to me)\n\nAn anagram is produced by rearranging the letters.\nSort the letters alphabetically for the two strings.\nThe sorted strings should be equal!\n\n``````public boolean isAnagram(String s, String t) {\nif (s.length() != t.length()) {\nreturn false;\n}\nchar[] str1 = s.toCharArray();\nchar[] str2 = t.toCharArray();\nArrays.sort(str1);\nArrays.sort(str2);\nreturn Arrays.equals(str1, str2);\n}\n``````\n\nTime complexity : O(n log n).\nSpace complexity : O(1) but not in Java! toCharArray() makes a copy of the string so it costs O(n), but we ignore this for complexity analysis because:\n\n• It is a language dependent detail.\n• It depends on how the function is designed. For example, the function parameter types can be changed to char[].\n\nApproach 2: counting technic (my favorite)\n\nIf the input is only letters from a-z, a simple counter table of size 26 is sufficient.\nIf you support the strict ASCII range, you need 128 characters.\nFor extended ASCII, it’s 256. See my answer here on StackOverflow.\n\nFor 26 letters only, there is a trick. You need to do -‘a’ so that you start from ‘a’.\n\n``````int[] counter = new int;\nfor (int i = 0; i < s.length(); i++) {\ncounter[s.charAt(i) - 'a']++;\ncounter[t.charAt(i) - 'a']--;\n}\n``````\n`````` // Support for extended ASCII\nprivate boolean isAnagram(String a, String b) {\nif (a.length() != b.length()) {\nreturn false;\n}\nint[] lookup = new int;\nfor (int i = 0; i < a.length(); i++) {\nlookup[a.charAt(i)]++;\n}\n// do not set isAagram by default because \"b\" and \"\" would be anagram otherwise\nint i = 0;\nwhile (i < b.length()) {\nlookup[b.charAt(i)]--;\nif (lookup[b.charAt(i)] >= 0) {\ni++;\n} else {\nbreak;\n}\n}\nreturn i >= b.length();\n}\n``````\n\nTime complexity : O(n)\nSpace complexity : O(1). Although we do use extra space, the space complexity is O(1)O(1) because the table’s size stays constant no matter how large nn is.\n\nWhat if the inputs contain unicode characters? How would you adapt your solution to such a case?"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.6798309,"math_prob":0.9537562,"size":2427,"snap":"2023-14-2023-23","text_gpt3_token_len":620,"char_repetition_ratio":0.09946348,"word_repetition_ratio":0.014851485,"special_character_ratio":0.28883395,"punctuation_ratio":0.17131475,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9923799,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-02T13:01:31Z\",\"WARC-Record-ID\":\"<urn:uuid:9b46a955-1ece-455b-8a60-1703846a31f0>\",\"Content-Length\":\"94592\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b03ebfb-f608-4247-a3eb-0d1c59d64321>\",\"WARC-Concurrent-To\":\"<urn:uuid:03eeed89-9c27-4ecc-acbd-7849aafa49f9>\",\"WARC-IP-Address\":\"209.16.158.51\",\"WARC-Target-URI\":\"https://blog.mocki.co/are-strings-anagrams/\",\"WARC-Payload-Digest\":\"sha1:LAFZAYG3TRXZ27QJTLY6B3OCS5LFFEZA\",\"WARC-Block-Digest\":\"sha1:AXBK4X7HKARCK4J7GMLB4WAD2OIYU2MV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296950528.96_warc_CC-MAIN-20230402105054-20230402135054-00363.warc.gz\"}"}
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https://proofwiki.org/wiki/Excess_Kurtosis_of_Continuous_Uniform_Distribution
|
[
"# Excess Kurtosis of Continuous Uniform Distribution\n\nLet $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\\closedint a b$.\nThen the excess kurtosis $\\gamma_2$ of $X$ is equal to $-\\dfrac 6 5$."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.82925975,"math_prob":0.99914247,"size":277,"snap":"2019-51-2020-05","text_gpt3_token_len":72,"char_repetition_ratio":0.10989011,"word_repetition_ratio":0.0,"special_character_ratio":0.23104693,"punctuation_ratio":0.04255319,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989032,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T14:28:18Z\",\"WARC-Record-ID\":\"<urn:uuid:dcf7f858-2df6-4643-ad4a-55e64c27e73f>\",\"Content-Length\":\"31812\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6eb5b34d-7286-4056-b31f-93f637a70fc5>\",\"WARC-Concurrent-To\":\"<urn:uuid:3b6db024-5a2c-442e-ad24-2087334e9ef6>\",\"WARC-IP-Address\":\"104.27.169.113\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Excess_Kurtosis_of_Continuous_Uniform_Distribution\",\"WARC-Payload-Digest\":\"sha1:Y3QVEF5RV5NGQYG6UL6R75MAXBYCU4IT\",\"WARC-Block-Digest\":\"sha1:EOWZDLBQEB7J3OQXRUKCXKKJV7TD5XYR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251689924.62_warc_CC-MAIN-20200126135207-20200126165207-00092.warc.gz\"}"}
|
https://bookdown.org/richard_g_everitt/notes/appmc.html
|
[
"# A Some Markov Chain Concepts\n\nThis appendix is provided largely to make these notes self contained and to provide a little context and some details for those who want them. The notion of a stochastic process in general and Markov chains in particular are, of course, explored in more depth during the Applied Stochastic Processes module. No significant amount of lecture time will be dedicated to this material, and if this is all unfamiliar to you then you’ll be able to engage with the lectures and the module without becoming intimately acquainted with the fine details of this material.\n\nI’ve attempted to balance the need for technical rigour with accessibility and have avoided making much explicit reference to the theory of measure. If you aren’t familiar with measure theory then you should be able to read this appendix by simply ignoring any reference to measurability but be aware that should you go on to use these concepts in the wild that we do need to be careful about such things.\n\n## A.1 Stochastic Processes\n\nFor our purposes we can define an $$E$$-valued process as a function $$\\xi:{\\mathcal{I}}\\rightarrow E$$ which maps values in some index set $${\\mathcal{I}}$$ to some other space $$E$$. The evolution of the process is described by considering the variation of $$\\xi(i)$$ with $$i$$. An $$E$$-valued stochastic process (or random process) can be viewed as a process in which, for each $$i \\in {\\mathcal{I}}$$, $$\\xi(i)$$ is a random variable taking values in $$E$$.\n\nAlthough a rich literature on more general situations exists, we will consider only the case of discrete time stochastic processes in which the index set $${\\mathcal{I}}$$ is $${\\mathbb{N}}$$ (of course, any index set isomorphic to $${\\mathbb{N}}$$ can be used in the same framework by simple relabeling). We will use the notation $$\\xi_i$$ to indicate the value of the process at time $$i$$ (note that there need be no connection between the index set and real time, but this terminology is both convenient and standard).\n\nWe will begin with an extremely brief description of a general stochastic process, before moving on to discuss the particular classes of process in which we will be interested. In order to characterise a stochastic process of the sort in which we are interested, it is sufficient to know all of its finite-dimensional distributions, the joint distributions of the process at any collection of finitely many times. For any collection of times $$i_1, i_2, \\dots, i_t$$ and any measurable collection of subsets of $$E$$, $$A_{i_1}, A_{i_2}, \\dots, A_{i_t}$$ we are interested in the probability: ${{\\mathbb{P}}_{}\\left(\\xi_{i_1} \\in A_{i_1}, \\xi_{i_2} \\in A_{i_2}, \\dots, \\xi_{i_t} \\in A_{i_t}\\right)}.$ For such a collection of probabilities to define a stochastic process, we require that they meet a certain consistency criterion. We require the marginal distribution of the values taken by the process at any collection of times to be the same under any finite-dimensional distribution which includes the process at those time points, so, defining any second collection of times $$j_1, \\dots, j_s$$ with the property that $$j_k \\not= i_l$$ for any $$k \\leq t, l \\leq s$$, we must have that: \\begin{aligned} {{\\mathbb{P}}_{}\\left(\\xi_{i_1} \\in A_{i_1}, \\xi_{i_2} \\in A_{i_2}, \\dots, \\xi_{i_t} \\in A_{i_t}\\right)} = {{\\mathbb{P}}_{}\\left(\\xi_{i_1} \\in A_{i_1}, \\xi_{i_2} \\in A_{i_2}, \\dots, \\xi_{i_t} \\in A_{i_t}, \\xi_{j_1} \\in E, \\dots, \\xi_{j_s} \\in E\\right)}.\\end{aligned} This is just an expression of the intuitive concept that any finite-dimensional distribution which describes the process at the times of interest should provide the same description if we neglect any information it provides about the process at other times. Or, to put it another way, they must all be marginal distributions of the same distribution.\n\nIn the case of real-valued stochastic processes, in which $$E={\\mathbb{R}}$$, we may express this concept in terms of the joint distribution functions (the multivariate analogue of the distribution function). Defining the joint distribution functions according to: \\begin{aligned} F_{i_1,\\dots,i_t}(x_1,x_2,\\dots,x_t) = {{\\mathbb{P}}_{}\\left(\\xi_{i_1} \\leq x_1,\\xi_{i_2} \\leq x_2,\\dots,\\xi_{i_t} \\leq x_t\\right)},\\end{aligned} our consistency requirement may now be expressed as: \\begin{aligned} &F_{i_1,\\dots,i_t,j_1,\\dots,j_s}(x_1,x_2,\\dots,x_t, \\infty,\\dots,\\infty) = F_{i_1,\\dots,i_t}(x_1,x_2,\\dots,x_t).\\end{aligned}\n\nHaving established that we can specify a stochastic process if we are able to specify its finite-dimensional distributions, we might wonder how to specify these distributions. In the next two sections, we proceed to describe a class of stochastic processes which can be described constructively and whose finite-dimensional distributions may be easily established. The Markov processes which we are about to introduce represent the most widely used class of stochastic processes, and the ones which will be of most interest in the context of Monte Carlo methods.\n\n## A.2 Discrete State Space Markov Chains\n\n### A.2.1 Basic Notions\n\nWe begin by turning our attention to the discrete state space case which is somewhat easier to deal with than the general case which will be of interest later. In the case of discrete state spaces, in which $$|E|$$ is either finite, or countably infinite, we can work with the actual probability of the process having a particular value at any time (you’ll recall that in the case of continuous random variables more subtlety is generally required as the probability of any continuous random variable defined by a density (with respect to Lebesgue measure, in particular) taking any particular value is zero). This simplifies things considerably, and we can consider defining the distribution of the process of interest over the first $$t$$ time points by employing the following decomposition: \\begin{align*} {{\\mathbb{P}}_{}\\left(\\xi_{1} = x_1, \\xi_{2} = x_2, \\dots, \\xi_{t} = x_{t}\\right)} = &{} {{\\mathbb{P}}_{}\\left(\\xi_{1} = x_1, \\xi_{2} = x_2, \\dots, \\xi_{t-1} = x_{t-1}\\right)}\\\\ & {}\\times {{\\mathbb{P}}_{}\\left(\\xi_t = x_t | \\xi_1 = x_1,\\dots,\\xi_{t-1}=x_{t-1}\\right)}. \\end{align*} Looking at this decomposition, it’s clear that we could construct all of the distributions of interest from an initial distribution from which $$\\xi_1$$ is assumed to be drawn and then a sequence of conditional distributions for each $$t$$, leading us to the specification: $\\begin{equation} {{\\mathbb{P}}_{}\\left(\\xi_{1} = x_1, \\xi_{2} = x_2, \\dots, \\xi_{t} = x_{t}\\right)} = {{\\mathbb{P}}_{}\\left(\\xi_1 = x_1\\right)} \\prod\\limits_{i=2}^t {{\\mathbb{P}}_{}\\left(\\xi_i = x_i | \\xi_{1} = x_1, \\dots, \\xi_{i-1} = x_{i-1}\\right)}.\\tag{A.1} \\end{equation}$ From this specification we can trivially construct all of the finite-dimensional distributions using no more than the sum and product rules of probability.\n\nSo, we have a method for constructing finite-dimensional distributions for a discrete state space stochastic process, but it remains a little formal as the conditional distributions seem likely to become increasingly complex as the time index increases. The conditioning present in decomposition (A.1) is needed to capture any relationship between the distribution at time $$t$$ and any previous time. In many situations of interest, we might expect interactions to exist on only a much shorter time-scale. Indeed, one could envisage a memoryless process in which the distribution of the state at time $$t+1$$ depends only upon its state at time $$t$$, $$\\xi_t$$, regardless of the path by which it reached $$\\xi_t$$. Formally, we could define such a process as: \\begin{align} {{\\mathbb{P}}_{}\\left(\\xi_{1} = x_1, \\xi_{2} = x_2, \\dots, \\xi_{t} = x_{t}\\right)} &= {{\\mathbb{P}}_{}\\left(\\xi_1 = x_1\\right)} \\prod\\limits_{i=2}^t {{\\mathbb{P}}_{}\\left(\\xi_i = x_i | \\xi_{i-1} = x_{i-1}\\right)}. \\tag{A.2} \\end{align}\n\nIt is clear that (A.2) is a particular case of (A.1) in which this lack of memory property is captured explicitly, as: ${{\\mathbb{P}}_{}\\left(\\xi_t = x_t | \\xi_{1} = x_1, \\dots, \\xi_{t-1} = x_{t-1}\\right)} = {{\\mathbb{P}}_{}\\left(\\xi_t = x_t | \\xi_{t-1} = x_{t-1}\\right)}.$ We will take this as the defining property of a collection of processes which we will refer to as discrete time Markov processes or, as they are more commonly termed in the Monte Carlo literature, Markov chains. There is some debate in the literature as to whether the term “Markov chain” should be reserved for those Markov processes which take place on a discrete state space, those which have a discrete index set (the only case we will consider here) or both. As is common in the field of Monte Carlo simulation, we will use the terms Markov chain and Markov process interchangeably.\n\nWhen dealing with discrete state spaces, it is convenient to associate a row vector7 with any probability distribution. We assume, without loss of generality, that the state space, $$E$$, is $${\\mathbb{N}}$$. Now, given a random variable $$X$$ on $$E$$, we say that $$X$$ has distribution $$\\mu$$, often written as $$X \\sim \\mu$$ for some vector $$\\mu$$ with the property that: $\\forall x \\in E:\\: {{\\mathbb{P}}_{}\\left(X = x\\right)} = \\mu_x .$\n\n#### A.2.1.1 Homogeneous Markov Chains\n\nThe term homogeneous Markov Chain is used to describe a Markov process of the sort just described with the additional caveat that the conditional probabilities do not depend explicitly on the time index, so: \\begin{aligned} \\forall m \\in {\\mathbb{N}}:\\: {{\\mathbb{P}}_{}\\left(\\xi_t = y | \\xi_{t-1} = x\\right)} \\equiv {{\\mathbb{P}}_{}\\left(\\xi_{t+m} = y | \\xi_{t+m-1} = x\\right)}.\\end{aligned}\n\nIn this setting, it is particular convenient to define a function corresponding to the transition probability (as the probability distribution at time $$t+1$$ conditional upon the state of the process at time $$t$$) or kernel as it is often known, which may be written as a two argument function or, in the discrete case as a matrix, $$K(i,j) = K_{ij} = {{\\mathbb{P}}_{}\\left(\\xi_t = j | \\xi_{t-1} = i\\right)}$$.\n\nHaving so expressed things, we are able to describe the dynamic structure of a discrete state space, discrete time Markov chain in a particularly simple form. If we allow $$\\mu_t$$ to describe the distribution of the chain at time $$t$$, so that $$\\mu_{t,i} = {{\\mathbb{P}}_{}\\left(\\xi_t = i\\right)}$$, then we have by applying the sum and product rules of probability, that: \\begin{aligned} \\mu_{t+1,j} = \\sum\\limits_i \\mu_{t,i} K_{ij}.\\end{aligned} We may recognise this as standard vector-matrix multiplication and write simply that $$\\mu_{t+1} = \\mu_t K$$ and, proceeding inductively it’s straightforward to verify that $$\\mu_{t+m} = \\mu_t K^m$$ where $$K^m$$ denotes the usual $$m^\\textrm{th}$$ matrix power of $$K$$. We will make some use of this object, as it characterises the $$m$$-step ahead condition distribution: $K^m_{ij} := (K^m)_{ij} = {{\\mathbb{P}}_{}\\left(\\xi_{t+m} = j | \\xi_t = i\\right)}.$ In fact, the initial distribution $$\\mu_1$$, together with $$K$$ tells us the full distribution of the chain over any finite time horizon: ${{\\mathbb{P}}_{}\\left(\\xi_1 = x_1,\\dots, \\xi_t = x_t\\right)} = \\mu_{1,x_1} \\prod\\limits_{i=2}^t K_{x_{i-1} x_i}.$\n\nA general stochastic processes is said to possess the weak Markov property if, for any deterministic time, $$t$$ and any finite integer $$p$$, we may write that for any integrable function $$\\varphi:E \\rightarrow {\\mathbb{R}}$$: ${\\mathbb{E}_{}\\left[\\varphi(\\xi_{t+p})|\\xi_1 = x_1, \\dots \\xi_t = x_t\\right]} = {\\mathbb{E}_{}\\left[\\varphi(\\xi_{t+p})|\\xi_t = x_t\\right]}.$\n\n#### A.2.1.2 Inhomogeneous Markov Chains\n\nNote that it is perfectly possible to define Markov Chains whose behaviour does depend explicitly upon the time index. Although such processes are more complex to analyse than their homogeneous counterparts, they do play a rôle in Monte Carlo methodology—in both established algorithms such as simulated annealing (see Section 3.5 and in more recent developments such as adaptive Markov Chain Monte Carlo and the State Augmentation for Maximising Expectations (SAME) algorithm of Doucet, Godsill, and Robert (2002). In the interests of simplicity, what follows is presented for homogeneous Markov Chains.\n\n#### A.2.1.3 Examples\n\nBefore moving on to introduce some theoretical properties of discrete state space Markov chains we will present a few simple examples. Whilst there are innumerable examples of homogeneous discrete state space Markov chains, we confined ourselves here to some particular simple cases which will be used to illustrate some properties below, and which will probably be familiar to you.\n\nWe begin with an example which is apparently simple, and rather well known, but which exhibits some interesting properties.\n\nExample A.1 (Simple random walk over the integers). Given a process $$\\xi_t$$ whose value at time $$t+1$$ is $$\\xi_t + 1$$ with probability $$p_+$$ and $$\\xi_{t-1}$$ with probability $$p_- = 1 - p_+$$, we obtain the familiar random walk. We may write this as a Markov chain by setting $$E = {\\mathbb{Z}}$$ and noting that the transition kernel may be written as: $K_{ij} = \\left\\{ \\begin{array}{cl} p_- & \\text{ if } j = i - 1 \\\\ p_+ & \\text{ if } j = i + 1 \\\\ 0 & \\text{otherwise.} \\end{array} \\right.$\n\nFigure A.1: A simple random walk on $${\\mathbb{Z}}$$.\n\nExample A.2 It will be interesting to look at a slight extension of this random walk, in which there is some probability $$p_0$$ of remaining in the present state at the next time step, so $$p_+ + p_- < 0$$ and $$p_0 = 1 - (p_+ + p_-)$$. In this case we may write the transition kernel as: $K_{ij} = \\left\\{ \\begin{array}{cl} p_- & \\text{ if } j = i - 1 \\\\ p_0 & \\text{ if } j = i \\\\ p_+ & \\text{ if } j = i + 1 \\\\ 0 & \\text{otherwise.} \\end{array} \\right.$\n\nFigure A.2: A random walk on $${\\mathbb{Z}}$$ with $$K_{tt} > 0$$.\n\nExample A.3 (Random Walk on a Triangle). A third example which we will consider below could be termed a “random walk on a triangle”. In this case, we set $$E = \\{1,2,3\\}$$ and define a transition kernel of the form: $K = \\left[ \\begin{array}{ccc} 0 & p_+ & p_- \\\\ p_-& 0 & p_+ \\\\ p_+& p_- & 0 \\end{array} \\right].$\n\nFigure A.3: A random walk on a triangle.\n\nExample A.4 (One-sided Random Walk). Finally, we consider the rather one-sided random walk on the positive integers, illustrated in Figure A.4, and defined by transition kernel: $K_{ij} = \\left\\{ \\begin{array}{cl} p_0 & \\text{ if } j = i \\\\ p_+= 1 - p_0 & \\text{ if } j = i + 1 \\\\ 0 & \\text{otherwise.} \\end{array} \\right.$\n\nFigure A.4: A random walk on the positive integers.\n\n### A.2.2 Important Properties\n\nIn this section we introduce some important properties in the context of discrete state space Markov chains and attempt to illustrate their importance within the field of Monte Carlo simulation. As is the usual practice when dealing with this material, we will restrict our study to the homogeneous case. As you will notice, it is the transition kernel which is most important in characterising a Markov chain.\n\nWe begin by considering how the various states that a Markov chain may be reached from one another. In particular, the notion of states which communicate is at the heart of the study of Markov chains.\n\nDefinition A.1 (Accessibility). A state $$y$$ is accessible from a state $$x$$, sometimes written as $$x \\rightarrow y$$ if, for a discrete state space Markov chain, $\\inf \\left\\{t : {{\\mathbb{P}}_{}\\left(\\xi_t = y | \\xi_1 = x\\right)} > 0 \\right\\} < \\infty.$ We can alternatively write this condition in terms of the transition matrix as $$\\inf \\left\\{ t : K^t_{xy} > 0\\right\\} < \\infty$$.\n\nThis concept tells us which states one can reach at some finite time in the future, if one starts from a particular state and then moves, at each time, according to the transition kernel, $$K$$. That is, if $$x\\rightarrow y$$, then there is a positive probability of reaching $$y$$ at some finite time in the future, if we start from a state $$x$$ and then “move” according to the Markov kernel $$K$$. It is now useful to consider cases in which one can traverse the entire space, or some subset of it, starting from any point.\n\nDefinition A.2 (Communication). Two states $$x, y \\in E$$ are said to communicate (written, by some authors as $$x \\leftrightarrow y$$) if each is accessible from the other, that is: $x \\leftrightarrow y \\Leftrightarrow x \\rightarrow y \\textrm{ and } y \\rightarrow x.$\n\nWe’re now in a position to describe the relationship, under the action of a Markov kernel, between two states. This allows us to characterise something known as the communication structure of the associated Markov chain to some degree, noting which points its possible to travel both to and back from. We now go on to introduce a concept which will allow us to describe the properties of the full state space, or significant parts of it, rather than individual states.\n\nDefinition A.3 (Irreducibility). A Markov Chain is said to be irreducible if all states communicate, so $$\\forall x,y \\in E:\\: x \\rightarrow y$$. Given a distribution $$\\phi$$ on $$E$$, the term $$\\phi$$-irreducible is used to describe a Markov chain for which every state with positive probability under $$\\phi$$ communicates with every other such state: $\\forall x,y \\in \\textrm{supp}(\\phi) :\\: x \\rightarrow y$ where the support of the discrete distribution $$\\phi$$ is defined as $$\\textrm{supp}(\\phi) = \\{ x \\in E: \\phi(x) > 0\\}$$. It is said to be strongly irreducible if any state can be reached from any point in the space in a single step and strongly $$\\phi$$-irreducible if all states (except for a collection with probability 0 under $$\\phi$$) may be reached in a single step.\n\nThis will prove to be important for the study of Monte Carlo methods based upon Markov chains as a chain with this property can somehow explore the entire space rather than being confined to some portion of it, perhaps one which depends upon the initial state.\n\nIt is also important to consider the type of routes which it is possible to take between a state, $$x$$, and itself as this will tell us something about the presence of long-range correlation between the states of the chain.\n\nDefinition A.4 (Period). A state $$x$$ in a discrete state space Markov chain has period $$d(x)$$ defined as: $d(x) = \\gcd \\left\\{s \\geq 1 : K^s_{xx} > 0 \\right\\},$ where $$\\gcd$$ denotes the greatest common denominator. A chain possessing such a state is said to have a cycle of length $$d$$.\n\nProposition A.1 All states which communicate have the same period and hence, in an irreducible Markov chain, all states have the same period.\n\nProof. Assume that $$x \\leftrightarrow y$$. Let there exist paths of lengths $$r, s$$ and $$t$$, respectively from $$x \\rightarrow y$$, $$y \\rightarrow x$$ and $$y \\rightarrow y$$, respectively.\n\nThere are paths of length $$r+s$$ and $$r+s+t$$ from $$x$$ to $$x$$, hence $$d(x)$$ must be a divisor of $$r+s$$ and $$r+s+t$$ and consequently of their difference, $$t$$. This holds for any $$t$$ corresponding to a path from $$y \\rightarrow y$$ and so $$d(x)$$ is a divisor of the length of any path from $$y\\rightarrow y$$: as $$d(y)$$ is the greatest common divisor of all such paths, we have that $$d(x) \\leq d(y)$$.\n\nBy symmetry, we also have that $$d(y) \\leq d(x)$$, and this completes the proof.\n\nIn the context of irreducible Markov chains, the term periodic is used to describe those chains whose states have some common period great than 1, whilst those chains whose period is 1 are termed aperiodic.\n\nOne further quantity needs to be characterised in order to study the Markov chains which will arise later. Some way of describing how many times a state is visited if a Markov chain is allowed to run for infinite time still seems required. In order to do this it is useful to define an additional random quantity, the number of times that a state is visited: $\\eta_x := \\sum\\limits_{k=1}^\\infty \\mathbb{I}_{x}(\\xi_k).$ We will also adopt the convention, common in the Markov chain literature that, given any function of the path of a Markov chain, $$\\varphi$$, $${\\mathbb{E}_{x}\\left[\\varphi\\right]}$$ is the expectation of that function under the law of the Markov chain initialised with $$\\xi_1 = x$$. Similarly, if $$\\mu$$ is some distribution over $$E$$, then $${\\mathbb{E}_{\\mu}\\left[\\varphi\\right]}$$ should be interpreted as the expectation of $$\\phi$$ under the law of the process initialised with $$\\xi_1 \\sim \\mu$$.\n\nDefinition A.5 (Transience and Recurrence). In the context of discrete state space Markov chains, we describe a state, $$x$$, as transient if: ${\\mathbb{E}_{x}\\left[\\eta_x\\right]} < \\infty,$ whilst, if we have that, ${\\mathbb{E}_{x}\\left[\\eta_x\\right]} = \\infty,$ then that state will be termed recurrent.\n\nIn the case of irreducible Markov chains, transience and recurrence are properties of the chain itself, rather than its individual states: if any state is transient (or recurrent) then all states have that property. Indeed, for an irreducible Markov chain either all states are recurrent or all are transient.\n\nWe will be particularly concerned in this course with Markov kernels which admit an invariant distribution.\n\nDefinition A.6 (Invariant Distribution). A distribution, $$\\mu$$ is said to be invariant or stationary for a Markov kernel, $$K$$, if $$\\mu K = \\mu$$.\n\nIf a Markov chain has any single time marginal distribution which corresponds to its stationary distribution, $$\\xi_t \\sim \\mu$$, then all of its future time marginals are the same as, $$\\xi_{t+s} \\sim \\mu K^s = \\mu$$. A Markov chain is said to be in its stationary regime once this has occurred. Note that this tells us nothing about the correlation between the states or their joint distribution. One can also think of the invariant distribution $$\\mu$$ of a Markov kernel, $$K$$ as the left eigenvector with unit eigenvalue.\n\nDefinition A.7 (Reversibility). A stationary stochastic process is said to be reversible if the statistics of the time-reversed version of the process match those of the process in the forward distribution, so that reversing time makes no discernible difference to the sequence of distributions which are obtained, that is the distribution of any collection of future states given any past history must match the conditional distribution of the past conditional upon the future being the reversal of that history.\n\nReversibility is a condition which, if met, simplifies the analysis of Markov chains. It is normally verified by checking the detailed balance condition, (A.3). If this condition holds for a distribution, then it also tells us that this distribution is the stationary distribution of the chain, another property which we will be interested in.\n\nProposition A.2 If a Markov kernel satisfies the detailed balance condition for some distribution $$\\mu$$, $\\begin{equation} \\forall x,y \\in E:\\: \\mu_x K_{xy} = \\mu_{y} K_{yx}\\tag{A.3} \\end{equation}$ then:\n\n1. $$\\mu$$ is the invariant distribution of the chain.\n\n2. The chain is reversible with respect to $$\\mu$$.\n\nProof. To demonstrate that $$K$$ is $$\\mu$$-invariant, consider summing both sides of the detailed balance equation over $$x$$: \\begin{aligned} \\sum\\limits_{x \\in E} \\mu_x K_{xy} &= \\sum\\limits_{x\\in E} \\mu_{y} K_{yx}, \\\\ (\\mu K)_y &= \\mu_y,\\end{aligned} and as this holds for all $$y$$, we have $$\\mu K = \\mu$$.\n\nIn order to verify that the chain is reversible we proceed directly: \\begin{aligned} {{\\mathbb{P}}_{}\\left(\\xi_t = x| \\xi_{t+1} = y\\right)} &= \\frac{{{\\mathbb{P}}_{}\\left(\\xi_t = x, \\xi_{t+1} = y\\right)}}{{{\\mathbb{P}}_{}\\left(\\xi_{t+1} = y\\right)}} = \\frac{{{\\mathbb{P}}_{}\\left(\\xi_t = x\\right)} K_{xy}}{{{\\mathbb{P}}_{}\\left(\\xi_{t+1} = y\\right)}} = \\frac{\\mu_{x} K_{xy}}{\\mu_y} = \\frac{\\mu_y K_{yx}}{\\mu_y} = K_{yx}\\\\ &= {{\\mathbb{P}}_{}\\left(\\xi_t = x | \\xi_{t-1} =y\\right)},\\end{aligned} in the case of a Markov chain it is clear that if the transitions are time-reversible then the process must be time reversible.\n\n## A.3 General State Space Markov Chains\n\n### A.3.1 Basic Concepts\n\nThe study of general state space Markov chains is a complex and intricate business. To do so entirely rigorously requires a degree of technical sophistication which lies somewhat outside the scope of this course. Here, we will content ourselves with explaining how the concepts introduced in the context of discrete state spaces in the previous section might be extended to continuous domains via the use of probability densities. We will not consider more complex cases—such as mixed continuous and discrete spaces, or distributions over uncountable spaces which may not be described by a density. Nor will we provide proofs of results for this case, but will provide suitable references for the interested reader.\n\nAlthough the guiding principles are the same, the study of Markov chains with continuous state spaces requires considerably more subtlety as it is necessary to introduce concepts which correspond to those which we introduced in the discrete case, describe the same properties and are motivated by the same intuition but which remain meaningful when we are dealing with densities rather than probabilities. As always, the principal complication is that the probability of any random variable distributed according to a non-degenerate density on a continuous state space taking any particular value is formally zero.\n\nWe will begin by considering how to emulate the decomposition we used to define a Markov chain on a discrete state space, Equation (A.2), when $$E$$ is a continuous state space. In this case, what we essentially require is that the probability of any range of possible values, given the entire history of the process depends only upon its most recent value in the sense that, for any measurable $$A_t \\subseteq E$$: ${{\\mathbb{P}}_{}\\left(\\xi_t \\in A_t | \\xi_{1} = x_1, \\dots, \\xi_{t-1} = x_{t-1}\\right)} = {{\\mathbb{P}}_{}\\left(\\xi_t \\in A_t | \\xi_{t-1} = x_{t-1}\\right)}.$\n\nIn the case which we are considering, it is convenient to describe the distribution of a random variable over $$E$$ in terms of some probability density, $$\\mu : E \\rightarrow {\\mathbb{R}}$$ which has the property that, if integrated over any measurable set, it tells us the probability that the random variable in question lies within that set, i.e. if $$X \\sim \\mu$$, we have that for any measurable set A that: ${{\\mathbb{P}}_{}\\left(X \\in A\\right)} = \\int\\limits_A \\mu(x) dx.$\n\nWe will consider only the homogeneous case here, although the generalisation to inhomogeneous Markov chains follows in the continuous setting in precisely the same manner as the discrete one. In this context, we may describe the conditional probabilities of interest as a function $$K: E \\times E \\rightarrow {\\mathbb{R}}$$ which has the property that for all measurable sets $$A \\subseteq E$$ and all points $$x \\in E$$: ${{\\mathbb{P}}_{}\\left(\\xi_{t} \\in A | X_{t-1} = x\\right)} = \\int\\limits_A K(x,y) dy.$\n\nWe note that, as in the discrete case the law of a Markov chain evaluated at any finite number of points may be completely specified by the initial distribution, call it $$\\mu$$, and a transition kernel, $$K$$. We have, for any suitable collection of sets $$A_1, \\dots$$, that the following holds: ${{\\mathbb{P}}_{}\\left(\\xi_1 \\in A_1, \\dots, \\xi_t \\in A_t\\right)} = \\int\\limits_{A_1 \\times \\dots \\times A_t} \\mu(x_1) \\prod\\limits_{k=2}^t K_k(x_{k-1},x_k) dx_1 \\dots dx_t.$ And, again, it is useful to be able to consider the $$s$$-step ahead conditional distributions, ${{\\mathbb{P}}_{}\\left(\\xi_{t+s} \\in A | \\xi_t = x_t\\right)} = \\int\\limits_{E^{s-1} \\times A} \\prod\\limits_{k=t+1}^{k=t+s} K(x_{k-1},x_k) dx_{t+1} \\dots dx_{t+s},$ and it is useful to define an $$s$$-step ahead transition kernel in the same manner as it is in the discrete case, here matrix multiplication is replaced by a convolution operation but the intuition remains the same. Defining $K^s(x_{t},x_{t+s}) := \\int\\limits_{E^{s-1}} \\prod\\limits_{k=t+1}^{k=t+s} K(x_{k-1},x_k) dx_{t+1} \\dots dx_{t+s-1},$ we are able to write ${{\\mathbb{P}}_{}\\left(\\xi_{t+s} \\in A | \\xi_t = x_t\\right)} = \\int\\limits_{A} K^s(x_{t},x_{t+s}) dx_{t+s}.$\n\n### A.3.2 Important Properties\n\nIn this section we will introduce properties which fulfill the same rôle in context of continuous state spaces as those introduced in Section @ref(#secmcdiscreteproperties) do in the discrete setting.\n\nWhilst it is possible to define concepts similar to communication and accessibility in a continuous state space context, this isn’t especially productive. We are more interested in the property of irreducibility: we want some way of determining what class of states are reachable from one another and hence what part of $$E$$ might be explored, with positive probability, starting from a point within such a class. We will proceed directly to a continuous state space definition of this concept.\n\nDefinition A.8 (Irreducibility). Given a distribution, $$\\mu$$, over $$E$$, a Markov chain is said to be $$\\mu$$-irreducible if for all points $$x \\in E$$ and all measurable sets $$A$$ such that $$\\mu(A) > 0$$ there exists some $$t$$ such that: $\\int\\limits_A K^t(x,y) dy > 0.$ If this condition holds with $$t = 1$$, then the chain is said to be strongly $$\\mu$$-irreducible.\n\nThis definition has the same character as that employed in the discrete case, previously, but is well defined for more general state spaces. It still tells us whether a chain is likely to be satisfactory if we are interested in approximation of some property of a measure $$\\mu$$ by using a sample of the evolution of that chain: if it is not $$\\mu$$-irreducible then there are some points in the space from which we cannot reach all of the support of $$\\mu$$, and this is likely to be a problem. In the sequel we will be interested more of less exclusively with Markov chains which are irreducible with respect to some measure of interest.\n\nWe need a little more subtlety in extending some of the concepts introduced in the case of discrete Markov chains to the present context. In order to do this, it will be necessary to introduce the concept of the small set; these function as a replacement for the individual states of a discrete space Markov chain as we will see shortly.\n\nA first attempt might be to consider the following sets which have the property that the distribution of taken by the Markov chain at time $$t+1$$ is the same if it starts at any point in this set—so the conditional distribution function is constant over this set.\n\nDefinition A.9 (Atoms). A Markov chain with transition kernel $$K$$ is said to have an atom, $$\\alpha \\subseteq E$$, if there is some probability distribution, $$\\nu$$, such that: $\\forall x \\in \\alpha, A \\subseteq E: \\quad \\int\\limits_A K(x,y) dy = \\int\\limits_A \\nu(y) dy.$ If the Markov chain in question is $$\\nu$$-irreducible, then $$\\alpha$$ is termed an accessible atom.\n\nWhilst the concept of atoms starts to allow us to introduce some sort of structure similar to that seen in discrete chains—it provides us with a set of positive probability which, if the chain ever enters it, we know the distribution of the subsequent state. Note that this is much stronger than knowledge of the transition kernel, $$K$$, as in general all points in the space have zero probability. Most interesting continuous state spaces do not possess atoms. The condition that the distribution of the next state is precisely the same, wherever the current state is rather strong. Another approach would be to require only that the conditional distribution has a common component, and that is the intuition behind a much more useful concept which underlies much of the analysis of general state space Markov chains.\n\nDefinition A.10 (Small Sets). A set, $$C \\subseteq E$$, is termed small for a given Markov chain (or, when one is being precise, $$(\\nu,s,\\epsilon)$$-small) if there exists some positive integer $$s$$, some $$\\epsilon > 0$$ and some non-trivial probability distribution, $$\\nu$$, such that: $\\forall x \\in C, A \\subseteq E:\\: \\quad \\int\\limits_A K^s(x,y) dy \\geq \\epsilon \\int\\limits_A \\nu(y) dy.$\n\nThis tells us that the distribution $$s$$-steps after the chain enters the small set has a component of size at least $$\\epsilon$$ of the distribution $$\\nu$$, wherever it was within that set. In this sense, small sets are not “too big”: there is potentially some commonality of all paths emerging from them. Although we have not proved that such sets exist for any particular class of Markov chains it is, in fact, the case that they do for many interesting Markov chain classes and their existence allows for a number of sophisticated analytic techniques to be applied.\n\nIn order to define cycles (and hence the notion of periodicity) in the general case, we require the existence of a small set. We need some group of “sufficiently similar” points in the state space which have a finite probability of being reached. We then treat this collection of points in the same manner as an individual state in the discrete case, leading to the following definitions.\n\nDefinition A.11 (Cycles). A $$\\mu$$-irreducible Markov chain has a cycle of length $$d$$ if there exists a $$(\\nu,M,\\epsilon)$$-small set C, such that: $d = \\gcd \\left\\{s \\geq 1 : C \\text{ is (\\nu,s,\\delta_s \\epsilon)-small for some } \\delta_s > 0 \\right\\}.$\n\nThis provides a reasonable concept of periodicity within a general state space Markov chain as it gives us a way of characterising the existence of regions of the space with the property that, wherever you start within that region you have positive probability of returning to that set after any multiple of $$d$$ steps and this does not hold for any number of steps which is not a multiple of $$d$$. We are able to define periodicity and aperiodicity in the same manner as for discrete chains, but using this definition of a cycle. As in the discrete space, all states within the support of $$\\mu$$ in a $$\\mu$$-irreducible chain must have the same period (see Proposition A.1) although we will not prove this here.\n\nConsidering periodicity from a different viewpoint, we are able to characterise it in a manner which is rather easier to interpret but somewhat difficult to verify in practice. The following definition of period is equivalent to that given above : a Markov chain has a period $$d$$ if there exists some partition of the state space, $$E_1, \\dots, E_d$$ with the properties that:\n\n• $$\\forall i \\neq j:\\: E_i \\cap E_j = \\emptyset$$,\n\n• $$\\bigcup\\limits_{i=1}^d E_i = E$$,\n\n• $$\\forall i, j, t, s:\\: {{\\mathbb{P}}_{}\\left(X_{t+s} \\in E_j | X_t \\in E_i\\right)} = \\left\\{ \\begin{array}{cl} 1 & j = i + s \\textrm{ mod } d \\\\ 0 & \\textrm{otherwise}. \\end{array} \\right.$$\n\nWhat this actually tells us is that a Markov chain with a period of $$d$$ has associated with it a disjoint partition of the state space, $$E_1, \\dots, E_d$$ and that we know that the chain moves with probability 1 from set $$E_1$$ to $$E_2$$, $$E_2$$ to $$E_3$$, $$E_{d-1}$$ to $$E_d$$ and $$E_d$$ to $$E_1$$ (assuming that $$d \\geq 3$$, of course). Hence the chain will visit a particular element of the partition with a period of $$d$$.\n\nWe also require some way of characterising how often a continuous state space Markov chain visits any particular region of the state space in order to obtain concepts analogous to those of transience and recurrence in the discrete setting. In order to do this we define a collection of random variables $$\\eta_A$$ for any subset $$A$$ of $$E$$, which correspond to the number of times the set $$A$$ is visited, i.e. $$\\eta_A := \\sum_{l=1}^\\infty \\mathbb{I}_A(\\xi_k)$$ and, once again we use $${\\mathbb{E}_{x}\\left[\\cdot\\right]}$$ to denote the expectation under the law of the Markov chain with initial state $$x$$. We note that if a chain is not $$\\mu$$-irreducible for some distribution $$\\mu$$, then there is no guarantee that it is either transient or recurrent, however, the following definitions do hold:\n\nDefinition A.12 (Transience and Recurrence). We begin by defining uniform transience and recurrence for sets $$A \\subseteq E$$ for $$\\mu$$-irreducible general state space Markov chains. Such a set is recurrent if: $\\forall x \\in A:\\: {\\mathbb{E}_{x}\\left[\\eta_A\\right]} = \\infty.$ A set is uniformly transient if there exists some $$M < \\infty$$ such that: $\\forall x \\in A:\\: {\\mathbb{E}_{x}\\left[\\eta_A\\right]} \\leq M.$ The weaker concept of transience of a set may then be introduced. A set, $$A \\subseteq E$$, is transient if it may be expressed as a countable union of uniformly transient sets, i.e.: \\begin{aligned} \\exists \\left\\{ B_i \\subseteq E\\right\\}_{i=1}^\\infty :& \\quad A \\subseteq \\bigcup\\limits_{i=1}^\\infty B_i \\\\ \\forall i \\in {\\mathbb{N}},\\, \\forall x \\in B_i:&\\quad {\\mathbb{E}_{x}\\left[\\eta_{B_i}\\right]} \\leq M_i < \\infty.\\end{aligned} A general state space Markov chain is recurrent if the following two conditions are satisfied:\n\n• The chain is $$\\mu$$-irreducible for some distribution $$\\mu$$,\n\n• For every measurable set $$A \\subseteq E$$ such that $$\\int_A \\mu(y) dy > 0$$, $${\\mathbb{E}_{x}\\left[\\eta_A\\right]} = \\infty$$ for every $$x \\in A$$;\n\nwhilst it is transient if it is $$\\mu$$-irreducible for some distribution $$\\mu$$ and the entire space is transient.\n\nAs in the discrete setting, in the case of irreducible chains, transience and recurrence are properties of the chain rather than individual states: all states within the support of the irreducibility distribution are either transient or recurrent. It is useful to note that any $$\\mu$$-irreducible Markov chain which has stationary distribution $$\\mu$$ is positive recurrent .\n\nA slightly stronger form of recurrence is widely employed in the proof of many theoretical results which underlie many applications of Markov chains to statistical problems, this form of recurrence is known as Harris recurrence and may be defined as follows:\n\nDefinition A.13 (Harris Recurrence). A set $$A \\subseteq E$$ is Harris recurrent if $${{\\mathbb{P}}_{x}\\left(\\eta_A = \\infty\\right)} = 1$$ for every $$x \\in A$$.\n\nA Markov chain is Harris recurrent if there exists some distribution $$\\mu$$ with respect to which it is irreducible and every set $$A$$ such that $$\\int_A \\mu(x) dx > 0$$ is Harris recurrent.\n\nThe concepts of invariant distribution, reversibility and detailed balance are essentially unchanged from the discrete setting. It’s necessary to consider integrals with respect to densities rather than sums over probability distributions, but no fundamental differences arise here.\n\n## A.4 Selected Theoretical Results\n\nThe probabilistic study of Markov chains dates back more than fifty years and comprises an enormous literature, much of it rather technically sophisticated. We don’t intend to summarise that literature here, nor to provide proofs of the results which we present here. This section serves only to motivate the material presented in the subsequent chapters.\n\nThese two theorems fill the rôle which the law of large numbers and the central limit theorem for independent, identically distributed random variables fill in the case of simple Monte Carlo methods. They tell us, roughly speaking, that if we take the sample averages of a function at the points of a Markov chain which satisfies suitable regularity conditions and possesses the correct invariant distribution, then we have convergence of those averages to the integral of the function of interest under the invariant distribution and, furthermore, under stronger regularity conditions we can obtain a rate of convergence.\n\nThere are two levels of strength of law of large numbers which it is useful to be aware of. The first tells us that for most starting points of the chain a law of large numbers will hold. Under slightly stronger conditions (which it may be difficult to verify in practice) it is possible to show the same result holds for all starting points.\n\nTheorem A.1 (A Simple Ergodic Theorem). If $$\\left(\\xi_{i}\\right)_{i \\in {\\mathbb{N}}}$$ is a $$\\mu$$-irreducible, recurrent $${\\mathbb{R}}^d$$-valued Markov chain which admits $$\\mu$$ as a stationary distribution, then the following strong law of large numbers holds (convergence is with probability 1) for any integrable function $$\\varphi:{\\mathbb{R}}^d \\rightarrow{\\mathbb{R}}$$: $\\lim_{t\\rightarrow \\infty} \\frac1t \\sum\\limits_{i=1}^t \\varphi(\\xi_i) \\overset{a.s.}{=}\\int \\varphi(x) \\mu(x) dx.$ for almost every starting value $$x$$ (i.e. for any $$x$$ except perhaps for a set of bad starting value, $$\\mathcal{N}$$, which has the property that $$\\int_\\mathcal{N} \\mu(x) dx = 0$$).\n\nAn outline of the proof of this theorem is provided by G. O. Roberts and Rosenthal (2004, Fact 5).\n\nTheorem A.2 (A Stronger Ergodic Theorem). If $$\\left(\\xi_{i}\\right)_{i \\in {\\mathbb{N}}}$$ is a $$\\mu$$-invariant, Harris recurrent Markov chain, then the following strong law of large numbers holds (convergence is with probability 1) for any integrable function $$\\varphi:E\\rightarrow{\\mathbb{R}}$$: $\\lim_{t\\rightarrow \\infty} \\frac1t \\sum\\limits_{i=1}^t \\varphi(\\xi_i) \\overset{a.s.}{=}\\int \\varphi(x) \\mu(x) dx.$\n\nA proof of this result is beyond the scope of the course. This is a particular case of Robert and Casella (2004, p241, Theorem 6.63), and a proof of the general theorem is given there. The same theorem is also presented with proof in Meyn and Tweedie (1993, p433, Theorem 17.3.2).\n\nTheorem A.3 (A Central Limit Theorem). Under technical regularity conditions (see Jones (2004) for a summary of various combinations of conditions) it is possible to obtain a central limit theorem for the ergodic averages of a Harris recurrent, $$\\mu$$-invariant Markov chain, and a function $$\\varphi:E\\rightarrow {\\mathbb{R}}$$ which has at least two finite moments (depending upon the combination of regularity conditions assumed, it may be necessary to have a finite moment of order $$2+\\delta$$). \\begin{aligned} & \\lim_{t\\rightarrow\\infty} \\sqrt{t} \\left[\\frac{1}{t}\\sum\\limits_{i=1}^t \\varphi(\\xi_i) - \\int \\varphi(x) \\mu(x) dx \\right] \\overset{\\mathcal{D}}{=}{\\textsf{N}\\left( 0,\\sigma^2(\\varphi) \\right)}, \\\\ & \\sigma^2(\\varphi) = {\\mathbb{E}_{}\\left[(f(\\xi_1) - \\bar{\\varphi})^2\\right]} + 2\\sum_{k=2}^\\infty {\\mathbb{E}_{}\\left[(\\varphi(\\xi_1) - \\bar{\\varphi})(\\varphi(\\xi_k) - \\bar{\\varphi})\\right]},\\end{aligned} where $$\\bar{\\varphi} = \\int \\varphi(x) \\mu(x) dx$$.\n\nWe conclude this chapter by noting that innumerable tutorials on the subject of Markov chains have been written, particularly with reference to their use in the field of Monte Carlo simulation. Some which might be of interest include the following:\n\n• G. Roberts (1996) provides an elementary introduction to some Markov chain concepts required to understand their use in Monte Carlo algorithms.\n\n• In the same volume, Tierney (1996) provides a more technical look at the same concepts; a more in-depth, but similar approach is taken by the earlier paper Tierney (1994).\n\n• An alternative, elementary formulation of some of the material presented here together with some additional background material, aimed at an engineering audience, can be found in Johansen (2009).\n\n• Robert and Casella (2004, chap. 6). This is a reasonably theoretical treatment intended for those interest in Markov chain Monte Carlo; it is reasonably technical in content, without dwelling on proofs. Those familiar with measure theoretic probability might find this a reasonably convenient place to start.\n\n• Those of you interested in technical details might like to consult Meyn and Tweedie (1993). This is the definitive reference work on stability, convergence and theoretical analysis of Markov chains and it is now possible to download it, free of charge from the website of one of the authors.\n\n• A less detailed, but more general and equally rigorous, look at Markov chains is provided by the seminal work of Nummelin (1984). This covers some material outside of the field of probability, but remains a concise work and presents only a few of the simpler results. It is perhaps a less intimidating starting point than Meyn and Tweedie (1993), although opinions on this vary.\n\n• A survey of theoretical results relevant to Monte Carlo is provided by G. O. Roberts and Rosenthal (2004). Again, this is necessarily somewhat technical.\n\n### References\n\nDoucet, A., S. J. Godsill, and C. P. Robert. 2002. “Marginal Maximum a Posteriori Estimation Using Markov Chain Monte Carlo.” Statistics and Computing 12: 77–84.\nJohansen, A. M. 2009. Markov Chains.” In Encyclopaedia of Computer Science and Engineering, edited by Benjamin W. Wah, 4:1800–1808. 111 River Street, MS 8-02, Hoboken, NJ 07030-5774: John Wiley; Sons, Inc.\nJones, G. L. 2004. “On the Markov Chain Central Limit Theorem.” Probability Surveys 1: 299–320.\nMeyn, S. P., and R. L. Tweedie. 1993. Markov Chains and Stochastic Stability. Springer Verlag. http://black.csl.uiuc.edu/~meyn/pages/TOC.html.\nNummelin, E. 1984. General Irreducible Markov Chains and Non-Negative Operators. 1st Paperback. Cambridge Tracts in Mathematics 83. Cambridge University Press.\nRobert, C. P., and G. Casella. 2004. Monte Carlo Statistical Methods. Second. New York: Springer Verlag.\nRoberts, G. 1996. Markov Chain Concepts Related to Sampling Algorithms.” In Markov Chain Monte Carlo in Practice, edited by W. R. Gilks, S. Richardson, and D. J. Spieghalter, first, 45–54. Chapman; Hall.\nRoberts, G. O., and J. S. Rosenthal. 2004. “General State Space Markov Chains and MCMC Algorithms.” Probability Surveys 1: 20–71.\nTierney, L. 1994. Markov Chains for Exploring Posterior Distributions.” Annals of Statistics 22: 1701–62.\n———. 1996. “Introduction to General State Space Markov Chain Theory.” In Markov Chain Monte Carlo in Practice, edited by W. R. Gilks, S. Richardson, and D. J. Spieghalter, first, 59–74. Chapman; Hall.\n\n1. Formally, much of the time this will be an infinite-dimensional vector but this need not concern us here.↩︎"
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https://www.vaultelectricity.com/how-many-watts-does-a-refrigerator-use/
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"# How Many Watts Does a Refrigerator Use?\n\nDo you ever wonder about how much electricity your fridge uses? Depending on its size and age, your refrigerator’s power use can widely vary, and you’ll need to consider that when budgeting for your electricity use. Fortunately, there are ways you can reduce the amount of energy used by your fridge and save on electricity bills. In this blog post, you’ll learn about how many watts a refrigerator uses and some tips for increased energy savings.\n\n## Key Takeaways\n\n• Knowing how many watts your fridge uses is vital to determining your electricity budget and seeing if you can save money on energy costs.\n• Calculating the wattage of your fridge requires knowing its amps and volts and then multiplying it by the number of hours it’s run daily.\n• To reduce power consumption, make sure your refrigerator is set below 40 degrees, keep the door closed, check the seals, allow for proper air circulation behind it, and position it in a cool corner.\n\n## How Many Watts Does a Refrigerator Use?\n\nThe running wattage used by a refrigerator can vary significantly, depending on the size and age of the appliance. Additionally, a fridge runs 24 hours daily, which needs to be accounted for. A standard-sized top or bottom freezer fridge typically uses between 100 and 400 watts. But smaller or older models, such as a mini fridge, may go as low as 50 watts, while larger fridges may consume more than 500 watts.\n\nAdditionally, refrigerators with an ice maker or water dispenser may increase energy usage compared to models without these features. So ultimately, the type of refrigerator you own will determine your energy use and monthly cost of electricity.\n\n## Calculating Wattage\n\n### Watts Per Hour\n\nFiguring out how many watts a refrigerator uses per hour can help you save money. To calculate the wattage per hour, you’ll need to figure out the amps and volts for your fridge. Take the number of amps listed on the nameplate and multiply it by the voltage. Then, because the compressor only runs about 30% of the time, divide this total by 3. This will give you the average wattage of your fridge per hour.\n\nFor example, on average, a refrigerator with 6 amps and 120 volts uses 240 watts per hour.\n\n• 6 amps X 120 volts = 720 watts\n• 720 watts divided by 3 = 240 average running wattage per hour\n\nUsing this calculation, you’ll be able to determine how much electricity your fridge uses and adjust your budget accordingly.\n\n### Watts Per Day\n\nTo determine the watts a refrigerator uses in an entire day, you’ll need to multiply the wattage per hour by the total number of hours in a day, which is 24.\n\nFor the example above, a 6 amps and 120 volts refrigerator would use 5,760 watts each day (240 X 24). With this figure, you can calculate how much electricity your fridge uses each day and how much it costs to run. First, you need to get the kilowatt-hours, which will help determine the hourly cost of running your fridge. That process will look like this:\n\n• 5,760 watts / 1,000 watts = 5.76 kWh\n\nNext, you’ll need to find the cost of your electricity per kilowatt-hour. You’ll find this on your electricity bill. Let’s say the cost per kilowatt-hour is 10 cents, so you’ll multiply 5.76 kWh by 0.10 to get the cost per day:\n\n• 5.76 kWh x \\$0.10 = \\$0.58 per day\n\nYou can then multiply this figure by 30 to get the cost per month or 365 to get the price per year. In this example, that would mean you’re spending approximately \\$17.40 per month (\\$0.58 x 30) and \\$209 (\\$0.58 x 365) per year on electricity for your refrigerator alone. Use these figures to determine if you need to make changes to your electricity budget or take steps to reduce energy costs.\n\n## Fridge Tips: How to Save Energy\n\nOnce you know the energy costs of your refrigerator, you can start to look for ways to reduce energy consumption and save money. Use these tips to get started:\n\n### Set the Right Temperature\n\nSet your fridge below 40 degrees Fahrenheit. Setting it too low can cause the motor to work harder and increase electricity usage, while putting it too high may decrease the longevity of your food.\n\nMake sure to close the door quickly and firmly every time you open it. This will reduce the energy your refrigerator uses and help keep your food fresh longer.\n\n### Check the Door Seals\n\nCheck the seals around your fridge door to make sure they are tight. If there are any gaps, replace them with new seals to help keep cold air from escaping, thereby increasing energy efficiency.\n\n### Allow Circulation Behind\n\nEnsure at least 1-2 inches of space behind your refrigerator for air to circulate. This will help keep it running efficiently and reduce your refrigerator’s power consumption.\n\n### Keep Your Fridge in a Cool Corner\n\nPosition your fridge away from direct sunlight and heat. Placing it in a cool kitchen corner or a well-ventilated area will help it use less power."
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"https://vestexas.wpenginepowered.com/wp-content/uploads/2023/02/young-woman-hand-opening-refrigerator-door-2021-10-06-09-50-38-utc-1-1.jpg",
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https://biomedical-engineering-online.biomedcentral.com/articles/10.1186/s12938-017-0358-3
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"Volume 16 Supplement 1\n\n# Feature ranking and rank aggregation for automatic sleep stage classification: a comparative study\n\n## Abstract\n\n### Background\n\nNowadays, sleep quality is one of the most important measures of healthy life, especially considering the huge number of sleep-related disorders. Identifying sleep stages using polysomnographic (PSG) signals is the traditional way of assessing sleep quality. However, the manual process of sleep stage classification is time-consuming, subjective and costly. Therefore, in order to improve the accuracy and efficiency of the sleep stage classification, researchers have been trying to develop automatic classification algorithms. Automatic sleep stage classification mainly consists of three steps: pre-processing, feature extraction and classification. Since classification accuracy is deeply affected by the extracted features, a poor feature vector will adversely affect the classifier and eventually lead to low classification accuracy. Therefore, special attention should be given to the feature extraction and selection process.\n\n### Methods\n\nIn this paper the performance of seven feature selection methods, as well as two feature rank aggregation methods, were compared. Pz-Oz EEG, horizontal EOG and submental chin EMG recordings of 22 healthy males and females were used. A comprehensive feature set including 49 features was extracted from these recordings. The extracted features are among the most common and effective features used in sleep stage classification from temporal, spectral, entropy-based and nonlinear categories. The feature selection methods were evaluated and compared using three criteria: classification accuracy, stability, and similarity.\n\n### Results\n\nSimulation results show that MRMR-MID achieves the highest classification performance while Fisher method provides the most stable ranking. In our simulations, the performance of the aggregation methods was in the average level, although they are known to generate more stable results and better accuracy.\n\n### Conclusions\n\nThe Borda and RRA rank aggregation methods could not outperform significantly the conventional feature ranking methods. Among conventional methods, some of them slightly performed better than others, although the choice of a suitable technique is dependent on the computational complexity and accuracy requirements of the user.\n\n## Background\n\nSleep occupies a significant part of human life. Therefore, the accurate diagnose of sleep-related disorders is of great importance in sleep research. Sleep is a particular condition of the nervous system with noticeable features and brain activity phases. Although most people think that sleep is a passive and constant process, as a matter of fact, sleep is an active state. Human bodies move frequently during the night and the human brain is sometimes even more active than in the waking state . Normal human sleep generally consists of two distinct stages with independent functions known as non-rapid eye movement (NREM) and rapid eye movement (REM) sleep. In their ideal situation, NREM and REM states alternate regularly, each cycle lasting 90 min on average. According to the American Academy of Sleep Medicine (AASM) , NREM is subdivided into three stages: stage 1 or light sleep, stage 2 and stage 3 or slow wave sleep (SWS). The evolution of sleep stages is complemented by gradual changes in many behavioral and physiological occurrences. Sleep stages are commonly classified using multiple simultaneous physiologic parameters during sleep named Polysomnography (PSG) in a clinic or hospital environment. A collection of rules has been identified in the AASM to guide the practitioners. However, the visual process of sleep stage classification is time-consuming, subjective and costly. In order to improve the accuracy and efficiency of the sleep stage classification, researchers have been trying to develop automatic classification algorithms. The automatic sleep stage classification mainly consists of three steps: pre-processing, feature extraction and classification . In the feature extraction stage, several temporal, spectral and nonlinear features are extracted from PSG signals. Nevertheless, some of these features may be irrelevant or have high mutual correlation increasing the complexity of the model without any real benefit. To face this challenge, feature selection and dimensionality reduction methods have been utilized.\n\nIn principle, a feature selection method has been used with the aim of selecting a subset of features in a way that the classifier can distinguish the differences between various classes of input data more effectively. The advantages of using feature selection methods make it an essential requirement for many classification applications. Reaching a more compact and simple model is the most important advantage offered by the feature selection process, that can reduce the necessary computational time for the classifier. Also, enhancing the generalization ability, increasing the classification power through reduced overfitting level, less storage memory and simplified visualization are further benefits of feature selection in classification tasks. Several different types of feature selection methods exist in the literature. Among them, the most common methods are divided into three main categories: filter, wrapper and embedded methods. Filter methods perform feature selection by considering some intrinsic characteristics of the data and usually provide a rank or a score for each feature. Low scored features will be removed experimentally or according to a predefined threshold. Filter methods, besides being simple and fast, are independent of the classifier.\n\nWrapper methods on the other hand, embed a search algorithm in the space of possible features subsets. Then, various subsets are produced and evaluated by training and testing with the specific classification algorithm. Since, the number of possible subsets grows exponentially with the number of features, heuristic search algorithms are used for finding the optimal feature subset. The high computational complexity and the risk of over fitting are its main disadvantages. The main benefits of wrapper methods over filter methods are taking into account feature dependencies and interaction between the selected subset and the specific classification method.\n\nEmbedded methods integrate the optimal feature subset selection with the classification algorithm. They have less computational complexity compared to wrapper methods. The results of both wrapper methods and embedded methods are classifier-specific.\n\nIn sleep stage classification, filter methods are more common than wrapper and embedded methods. Among the filter methods fast correlation based filter (FCBF), Fisher score, ReliefF, Chi square (Chi2), information gain (IG), conditional mutual information maximization (CMIM), minimum redundancy maximum relevance (MRMR) algorithms and R-square [4,5,6,7] are the most preferred ones. In addition to the traditional methods, a new filter method called ‘Mahal’ is proposed in for facing the challenge of feature selection in small datasets with a large number of features for sleep stage classification. On the other hand, sequential feature selection algorithms including sequential forward selection (SFS) and sequential backward selection (SBS) are the most common wrapper methods used in the automatic sleep stage classification [9, 10]. Statistical hypothesis testing methods are also used in sleep stage classification applications for feature selection and dimensionality reduction. Examples of these methods are t test, ANOVA and Kruksal–Wallis test which are used for three different purposes: dimensionality reduction, feature selection and assessment of the discriminative capability of the selected feature set. To the best of our knowledge, there are no studies for comparing the performance of various feature selection methods from the same category in sleep stage classification. The studies done so far usually choose feature selection methods from different categories. For example in , one filter and three wrapper methods are used and the results are compared. Therefore, there is a need for comprehensive comparison of feature selection methods from the same category.\n\nAs mentioned before, feature ranking techniques provide a ranked list of features. Different feature ranking techniques may produce different rankings according to their specific criteria for assessing features and there is no universal ranking algorithm that considers all the measures. Therefore, motivated by ensemble methods in supervised learning , rank aggregation methods are proposed to combine different feature ranking methods and achieve more stable ranked feature lists with similar or even higher classification performance [13, 14]. In order to perform ensemble feature selection, one needs to decide on the method to aggregate the results from different ranking methods. There are many rank aggregation approaches from the very simple ones to some more complex . To the best of our knowledge, there are no studies done on feature selection based on rank aggregation methods in the sleep stage classification area.\n\nIn this paper different feature ranking and rank aggregation methods were compared within the sleep stage classification context. The main contributions of this paper are listed below:\n\n1. 1.\n\nA comprehensive feature set including Itakura Spectral Distance (ISD) was extracted from PSG signals,\n\n2. 2.\n\nSimilarity and stability of different feature ranking and rank aggregation methods were assessed,\n\n3. 3.\n\nClassification performance of different feature ranking and rank aggregation methods was compared.\n\nIn this work, we present the extension of the results published in . The paper is organized as follows: In the next section (“Methods”) the database, pre-processing, extracted features, feature selection techniques and classification algorithms will be described. In the following section related results will be shown. Discussion of the obtained results will be presented in the next section. On the last section, the conclusions and future work directions are presented.\n\n## Methods\n\nIn this section the sleep stage classification methodology used in this work is described in detail. Figure 1 shows the block diagram of the proposed algorithm for comparing the feature selection methods.\n\n### Data\n\nThe data used in this study was obtained from Physionet Sleep-EDF Expanded Database . The collection of data in this database comes from two studies. PSG recordings of the first study are named SC files (SC = Sleep Cassette). PSG recordings of the second study are named ST files (ST = Sleep Telemetry). In our simulations, we didn’t use SC files, since EMG data for first study was a zero-amplitude or no-data recording. Therefore, we used ST files which are a collection of 22 PSG signals recorded in the hospital during two nights for about 9 h. Except for slight difficulty in falling asleep, subjects were healthy without any sleep related medication. The data were segmented into 30-s epochs and all epochs were scored according to R&K guidelines for human sleep staging. These recordings include EEG (Fpz-Cz and Pz-Oz), EOG (horizontal), submental chin EMG, together with the corresponding hypnograms.\n\nThrough careful analysis ST recordings, a number of issues were detected that made some of recordings unsuitable for being used in the evaluations. These issues are as follows:\n\n• Lack of stage 4 (according to R&K guidelines),\n\n• Artifacts such as severe movement or sensor misconnection,\n\n• Unsynchronized EEG data and hypnogram,\n\n• Lack of stage 3 epochs,\n\n• Severely corrupted EEG data.\n\nAs a results, six recordings were selected out of twenty-two and the corresponding hypnograms were converted from R&K to AASM. Pz-Oz channel EEG together with submental chin EMG and horizontal EOG each sampled at 100 Hz were used in the evaluations. Table 1 illustrates the number of stages available per subject.\n\n### Pre-processing\n\nArtifact free data is necessary for guaranteeing the reliability of sleep stage classification algorithms. In this work, the epochs with zero-energy were automatically detected and removed. The zero-energy epochs can appear due to the possible failure of recording device. Then, the EEG and EOG signals were band-pass filtered in the frequency interval of 0.3–35 Hz. This interval was selected according to the recommendations of AASM. For filtering, wavelet multi-level decomposition and reconstruction was used. This filtering technique has a high fidelity to the original wide-band signal in contrast to the Butterworth filtering that produces a highly distorted “valley” shape .\n\n### Feature extraction and normalization\n\nIn order to explore the information contained in PSG recordings, a set of features were extracted from EEG, H_EOG and submental chin EMG of each subject. This feature set includes 49 features that can be categorized into time, frequency, joint time–frequency domain, entropy-based and nonlinear types. As summarized in Table 2. In the following, information will be provided about the different features used in this study and their brief description.\n\n• Statistical features (F1 to F8, F37 to F41, and F44 to F46): Understanding the evolution of PSG signals as stochastic processes can provide valuable information regarding the sleep stage classification. In this study, the simple and most common statistical features including mean, median, maximum and minimum values, skewness and kurtosis of each EEG, EOG and EMG epoch are used according to Table 2.\n\n• Zero crossing rate (F9): Zero crossing rate (ZCR) is simple and at the same time very effective feature especially in sleep stage classification. ZCR counts the signals sign-change points on a segment of a signal. In this paper, the length of this segment is 30 s.\n\n• Hjorth parameters (F10 to F12): In 1960, Bo Hjorth proposed normalized slope detectors (NSD) as indicators of statistical properties of a signal in time domain. NSDs include three features: activity, mobility and complexity. These features are widely used in the analysis and characterization of EEG. Hjorth’s NSD are calculated as shown in Table 3 with σ 0 representing the variance of the signal, σ 1 the variance of the first derivate and σ 2 the variance of the second derivate of the signal.\n\n• Wavelet based features (F13 to F26): In order to analyze the stochastic nature of EEG, we chose the wavelet packet (WP) analysis since it provides a valuable joint time–frequency domain analysis. In clinical applications, four main brain rhythms are associated with different states of sleep, including Delta (0–3.99 Hz), Theta (4–7.99 Hz), Alpha (8–13 Hz) and Beta (>13 Hz) . According to the scheme proposed in , a WP tree with 7 decomposition levels was suitable to estimate the necessary frequency bands of EEG rhythms with adequate accuracy. Then, features F13 to F26 were extracted from the corresponding WP coefficients according to descriptions in Table 2.\n\n• Spectral entropy (F27): Spectral entropy, as a technique for measuring the irregularity of EEG, is calculated by the entropy of the power spectrum. Suppose P is the normalized power spectrum of EEG in a predefined frequency range, [f 1, f 2] and $$\\sum {P_{i} = 1}$$, the spectral entropy is calculated as:\n\n$$H_{sp} = - \\sum\\limits_{{f_{1} }}^{{f_{2} }} {P_{i} \\log P_{i} }$$\n(1)\n\nIn this study, H sp is used in the following normalized form:\n\n$$SpEn = \\frac{{H_{sp} }}{{\\log N_{f} }}$$\n(2)\n\nwhere N f is the number of frequency bins in the frequency range [f 1, f 2] [20, 21].\n\n• Rényi entropy (F28): In 1960, Alfréd Rényi introduced Rényi’s general notion of entropy . Since, Rényi Entropy unites several distinct entropy measures, it turned out to be theoretically interesting and found many applications in various research areas such as pattern recognition and biomedicine . Suppose P x (X) is probability distribution of random variable X. The Rényi entropy of order α for X is defined as:\n\n$$H_{\\alpha } (X) = \\frac{1}{1 - \\alpha }\\log \\sum\\limits_{x} {P_{x} (X)^{\\alpha } }$$\n(3)\n• Approximate entropy (F29): Approximate entropy is regarded as a measure of the randomness or equivalently regularity of a time series. Considering that time series with repetitive patterns are more predictable than those without repetitive patterns, approximate entropy reflects the likelihood that similar patterns existing in a time series will not be followed by more patterns from the same type [25, 26].\n\nFor calculating approximate entropy two parameters need to be predefined: first the pattern length m and second the similarity threshold r. Given the time series $$\\,\\left\\{ {x_{i} } \\right\\}_{i = 1 \\ldots N}$$, a sequence of vectors $${\\mathbf{x}}\\left( 1 \\right)$$ through $${\\mathbf{x}}(N - m + 1)$$ is formed in which $$\\,{\\mathbf{x}}(i) = [x(t), \\ldots ,x(t + m - 1)]$$. Two vectors $${\\mathbf{x}}(i)$$ and $${\\mathbf{x}}(j)$$ are similar if their distance is less than r. The distance between two patterns is defined as the maximum difference between their corresponding components. Then, $$C_{i}^{m} (r)$$ is defined as:\n\n\\begin{aligned} C_{i}^{m} (r) = \\frac{{{\\text{No}} . {\\text{ of }}j \\le N - m + 1}}{N - m + 1} \\hfill \\\\ \\quad \\quad \\quad \\,{\\text{for }}d[x(i),x(j)] \\le r \\hfill \\\\ \\end{aligned}\n(4)\n\nwhere $$C_{i}^{m} (r)$$ expresses the patterns regularity of length m with a threshold value of r. Finally, approximate entropy is defined as :\n\n\\begin{aligned} ApEn(m,r,N) = \\frac{1}{N - m + 1}\\sum\\limits_{i = 1}^{N - m + 1} {\\ln (C_{i}^{m} )} \\hfill \\\\ \\quad \\quad \\quad \\quad \\quad \\quad \\quad - \\frac{1}{N - m}\\sum\\limits_{i = 1}^{N - m} {\\ln (C_{i}^{m} )} \\hfill \\\\ \\end{aligned}\n(5)\n• Permutation entropy (F30): Permutation entropy was proposed by Bandt et al. and is a simple complexity measure, that can be applied to any type of time series including regular, chaotic, noisy and time series from reality. In mathematical terms, consider that a time series is $$\\left\\{ {x_{t} } \\right\\}_{t = 1 \\ldots T}$$. Through an embedding procedure, a set of vectors $${\\mathbf{X}}_{t} = [x_{t} ,x_{t + 1} , \\ldots ,x_{t + m} ]$$ with the embedding dimension m is formed. Then, X t is arranged in an increasing order. There will be $$m!$$ different order patternsπ, also called permutations. If $$f(\\pi )$$ denotes the frequency of permutation in the time series, then its relative frequency would be:\n\n$$p(\\pi ) = \\frac{f(\\pi )}{T - (m - 1)}$$\n(6)\n\nTherefore, the permutation entropy is defined as:\n\n$$H_{p} (m) = - \\sum {p(\\pi )log} p(\\pi )$$\n(7)\n\nwhere the sum runs over all $$m!$$ permutations [28, 29].\n\n• Petrosian fractal dimension (F31): The fractal dimension has been widely used in the characterization of nonstationary biomedical signals like EEG for several applications in order to measure the complexity of sleep EEG. Petrosian algorithm can be used for a fast computation of fractal dimension by means of transforming the signal into a binary sequence . Petrosian fractal dimension is calculated using the following formula:\n\n$$FD_{Petrosian} = \\frac{{\\log_{10} N}}{{\\log_{10} N + \\log_{10} (\\frac{N}{{N + 0.4N_{\\Delta } }})}}$$\n(8)\n\nIn which N is the length of the EEG signal and N Δ is the number of sign changes in the derivative of the signal.\n\n• Teager energy (F32): Teager energy operator has been proved to be very useful in analysing signals from the energy point of view. It is defined as:\n\n$$\\Psi (x(t)) = \\dot{x}^{2} (t) - x(t)\\ddot{x}(t)$$\n(9)\n\nin continuous form, where $$\\dot{x}(t)$$ is the first derivative and $$\\ddot{x}(t)$$ is the second derivative of x. The discrete form of Teager energy is :\n\n$$\\Psi \\left( {x[n]} \\right) = x^{2} [n] - x[n - 1]x[n + 1]$$\n(10)\n• Energy (F33, F42, and F47 to F49): Energy is calculated as the average sum of the squares of all samples in a signal segment. Energy value of a signal increases with the increase of activity in the signal . According to Table 2, both energy and energy ratio of different epochs of PSG recordings were used in this work.\n\n• Mean curve length (F34): Mean curve length was proposed with the purpose of reducing the complexity of Katz fractal dimension algorithm and provides results almost equivalent to it . It is commonly used for identification of EEG activity, including amplitude and frequency changes and also its dimensionality . Mean curve length, in its discrete form, is calculated using the following formula:\n\n$$CL[n] = \\sum\\limits_{i = 1 + (n - 1)N}^{nN} {\\left| {x(i) - x(i - 1)} \\right|}$$\n(11)\n\nconsidering x as the EEG data, n the epoch number and N the epoch length in samples.\n\n• Hurst exponent (F35): Hurst exponent, introduced by Harold Edwin Hurst , is a measure for long range statistical dependence of time series. Hurst exponent has a value in the range between 0 and 1 and is defined as:\n\n$$H = \\frac{{\\log \\left( {{\\raise0.7ex\\hbox{R} \\!\\mathord{\\left/ {\\vphantom {R S}}\\right.\\kern-0pt} \\!\\lower0.7ex\\hbox{S}}} \\right)}}{\\log (T)}$$\n(12)\n\nwhere T is the duration of signal sample and $${\\raise0.7ex\\hbox{R} \\!\\mathord{\\left/ {\\vphantom {R S}}\\right.\\kern-0pt} \\!\\lower0.7ex\\hbox{S}}$$ is the value of rescaled range.\n\n• Itakura Spectral Distance (F36): The Itakura Spectral Distance (ISD) is broadly used in speech processing applications to measure the distance (similarity) between two auto regressive coefficients (AR) processes [36, 37]. ISD was also used in automatic sleep classification to find the relation between EEG and EOG signals during different epochs of sleep stages over the night . In this paper, the ISD of sleep stages of EEG was measured. In order to calculate the distances, the AR coefficients were extracted from 50% of the wake epochs of each subject. Then, by getting the mean of the AR coefficients a representative model of the wake epoch was generated and the ISD between this model and the W (remaining 50%), S1, S2, SWS and REM epochs was calculated.\n\n• Spectral power (F43): Power spectrum density (PSD) represents the distribution of signal’s power as a function of frequency. The spectral power of a signal in a frequency band is obtained by integrating PSD over the signal’s frequency range.\n\nThe physiological differences from subject to subject and equipment related variations have considerable impact on the features extracted from the PSG recordings. Moreover, since there are usually a wide variety of feature types extracted for characterizing sleep stages, the amplitude and unit of features will also vary. The features may also get the extreme values, i.e. extremely low or extremely high values. Data post-processing is an important step in this respect. The aim of feature post-processing is to enable classification algorithms to uniformly handle the features with different units and ranges as well as reducing the influence of extreme values. Feature post-processing can be a scaling operation (normalization/standardization) or a feature transformation operation. In this work, each feature (x ij ) is independently scaled to have zero mean and unit variance $$(x^{\\prime}_{ij} )$$ using the following equation:\n\n$$x^{\\prime}_{ij} = \\frac{{x_{ij} - {\\bar{\\mathbf{x}}}_{i} }}{{\\sigma_{{{\\mathbf{x}}_{{\\mathbf{i}}} }} }}$$\n(13)\n\nwhere $${\\bar{\\mathbf{x}}}_{i}$$ and $$\\sigma_{{{\\mathbf{x}}_{i} }}$$ are the mean and the standard deviation of each independent feature vector.\n\n### Feature ranking methods\n\nIn this paper, to select a subset of features containing most of the original feature set information, we used seven different feature ranking methods: ReliefF, mini-mum redundancy-maximum relevance (MRMR-MID and MRMR-MIQ), Fisher score, Chi Square (Chi2), information gain (IG) and conditional mutual information maximization (CMIM). We have also implemented two different rank aggregation methods, Borda and robust rank aggregation (RRA), to evaluate their ability to produce better feature rankings compared to conventional feature ranking methods. A brief description of the used feature ranking methods is provided below:\n\n#### ReliefF\n\nIn 1992, Kira and Rendell proposed an instance based method, Relief, for estimating features quality. In this method, for a randomly selected sample, two nearest neighbors were considered: one from the same class (nearest hit) and other from a different class (nearest miss). The quality estimated for each feature is updated according to the randomly selected sample’s distance from the nearest hit and miss. The Relief method is restricted to two-class problems and is highly sensitive to noisy and incomplete data. An extension of Relief, called ReliefF , was proposed improving the original method by estimating the probabilities more reliably and extending the algorithm to multi-class problems. The ReliefF algorithm uses k-nearest hits and k-nearest misses for updating the quality estimation for each feature.\n\n#### Minimum redundancy-maximum relevance\n\nMRMR is a feature selection method which selects a subset of features with maximum relevance for the target class and, at the same time, minimum redundancy between the selected features. In the MRMR method, the redundancy (R) and relevance (D) are expressed in terms of mutual information. In order to select the final feature set, an objective function φ(D, R) is maximized. The φ(D, R) can be defined either as the mutual information difference (MID), D-R, or the mutual information quotient (MIQ), D/R.\n\n#### Fisher score\n\nThis method is one of the most efficient and widely used feature ranking methods. The key idea is to find a subset of the feature matrix with maximum distance between the data points from different classes and minimum distance between the data points of the same class in the feature space .\n\n#### Chi square\n\nChi2 is another very common class sensitive feature selection method which ranks the features according to their Chi2 statistics without taking into account the interactions between features. Originally proposed exclusively for categorical data, this method was later extended to the continuous case . For calculating the Chi2 statistics of each feature, the range of the numerical feature should be discretized into intervals.\n\n#### Information gain\n\nRoss Quinlan proposed an algorithm for generating decision trees from a set of training data . In this algorithm, information gain (IG) is the measure for selecting the effective feature at each node. Generally, IG can be described as the change in the marginal entropy of a feature set taking into account the conditional entropy of that feature set with the given class set.\n\n#### Conditional mutual information maximization\n\nThis method is based on mutual information in such a way that all the selected features are informative and have two-by-two weak dependency. A feature is added to the selected feature subset if it contains information about the specific class and this information is not contained on any other previously selected feature.\n\n#### Borda\n\nThe Borda algorithm is a feature aggregation method that ranks each feature based on its mean position in the different ranking methods considered, i.e.\n\n$$Borda(f_{i} ) = \\sum\\limits_{j = 1}^{N} {\\pi_{j} (f_{i} )}$$\n(14)\n\nwhere π j (f i ) is the rank of the feature f i in the ranking method π j . The feature with the highest Borda rank is considered the best.\n\n#### Robust rank aggregation\n\nThis method, proposed by Kolde et al. , is another rank aggregation method that compares the results from several feature ranking methods with a randomly ranked feature list. The RRA first looks how a specific feature is ranked by the various methods and lists the corresponding values in a so-called rank order, from best to worst. It is clear that, if a feature has high quality, the dominance of ranks in the rank order will be towards smaller numbers. The probability of the random list producing better ranking than the values seen in the actual rank order for that specific feature is determined. The features with the small probability are selected as the better ones .\n\n### Classification\n\nThe process of labeling the data into relevant classes is called classification. The first step in the classification process is the identification of the features or characteristics that will enable the highest discrimination between the different groups of data. A classification model is developed in such a way that it provides the structure for how the classification processes’ actions will be realized. Ideally, this model should be chosen to optimize the performance of the classification system, although it may need to be revised as the classifier design progresses. A classifier is then implemented and “trained” to recognize the chosen features in the data, or to determine the best input-to-output mapping. Once the system has trained and learned, it is ready to classify specific inputs. Then, the system can be tested and evaluated with such metrics as speed of computation and accuracy of classification .\n\nIn this study, we selected two simple and widely used classifiers: k-nearest neighbor (k-NN) and multilayer feedforward neural network (MLFN) to discriminate five sleep stages W, S1, S2, SWS and REM. By selecting k = 1, nearest neighbor (NN) was utilized. The NN classifier is the simplest nonparametric classifier and assigns a pattern to a specific class based on its nearest neighbor’s class. In spite of its simplicity, in it has been proved that, if the utilized database is fairly large, the error bound for nearest neighbor rule is quite tight, i.e. equal or less than twice the Bayes error. Also, neural networks are known to be very powerful computing models that can learn from training examples. Neural networks have been successfully used in a broad range of data mining applications including classification .\n\n### Performance evaluation\n\nIn this paper three main criteria namely stability, accuracy and similarity are considered for evaluating and comparing the different feature selection techniques.\n\n#### Stability\n\nStability of a feature selection method is defined as its sensitivity to variations in the training set. Since unstable feature selection may lead to inferior classification performance, a number of measures are proposed in the literature for investigating how different subsamples of a training set affect the feature importance assessment. In this study, in order to measure the stability of feature rankings produced by different methods, a similarity based approach proposed by Kalousis et al. is used. In this method, similarity between two selected feature sets s and $$s^{\\prime}$$, is calculated using the Tanimoto distance which measures the overlap between two sets of arbitrary cardinality:\n\n$$S_{s} (s,s^{\\prime}) = 1 - \\frac{{\\left| s \\right| + \\left| {s^{\\prime}} \\right| - 2\\left| {s \\cap s^{\\prime}} \\right|}}{{\\left| s \\right| + \\left| {s^{\\prime}} \\right| - \\left| {s \\cap s^{\\prime}} \\right|}}$$\n(15)\n\nThe S s takes values in the range of [0 1], with 0 meaning there is no overlap or similarity between two rankings and 1 meaning that the two rankings are identical. Then N subsets of the original training set are drawn using a random resampling technique such as cross validation or bootstrapping. Each specific ranking algorithm produces a feature preference list for each N subsets. The similarity between all possible pairs is calculated. The stability of that specific feature ranking algorithm is simply the average of the similarities over all possible pairs, i.e. $$\\frac{N(N - 1)}{2}$$ pairs.\n\n#### Similarity\n\nThe stability measure used for assessing the internal stability of a feature selection technique can also be used in a different context to measure the similarity of different feature selection techniques. The similarity measure provides information about the consistency and diversity of different feature selection algorithms. The similarity between two feature subsets s and $$s^{\\prime}$$ can be calculated using Eq. (15) with a slight difference in the definition of s and $$s^{\\prime}$$. Instead of two lists of features produced by a specific feature selection technique from different subsets of the training set, they are now two lists produced by two different feature selection techniques derived from the complete training set .\n\n#### Accuracy\n\nThe performance of the sleep stage classification is evaluated using repeated random sub-sampling validation. To measure the classification accuracy, the overall accuracy value is calculated as follows :\n\n$${\\text{Accuracy = }}\\frac{{{\\text{No}} . {\\text{ of true detections}}}}{{{\\text{Total no}} . {\\text{ of epochs}}}}$$\n(16)\n\n### Experimental setup\n\nSix subjects were selected from the Physionet database for evaluating and comparing the feature ranking and rank aggregation methods. For filtering EEG and EOG signals, Daubechies order 20 (db20) was used as the mother wavelet. The filtered data was segmented into 30-second epochs. From each epoch, a feature vector containing 49 features was extracted. After feature standardization, the feature vectors were fed into seven feature ranking methods. Then, in order to aggregate the results, the outputs of these seven feature ranking methods were used by Borda and RRA, producing two additional ranked lists of features.\n\nFor sleep stage classification, the parameters of the classifiers are set as follows. The Euclidean distance was chosen as the distance metric for the NN classifier. For the three-layer neural network classifier 12 hidden neurons and a sigmoid transfer function were selected in our simulations. The Levenberg–Marquardt training algorithm was adopted for minimizing the cost function because of its fast and stable convergence. In contrast with conventional approaches in the literature, which imports all the existing epochs to the classifier, we used a quantity of epochs selected out of each subject. In this method, selected epochs from each subject have two characteristics. Firstly, the number of epochs are the same for all the subjects. Second, the number of epochs for each stage is dependent on the number of occurrences of that stage for each subject. This method is suitable for large databases helping on the computational complexity reduction of the classifier training stage.\n\n## Results\n\nThe stability of each method was evaluated as a function of the number of selected features (d) where d = 1, 3, 5…29. In our simulations, 50 subsets were generated out of the original training set by bootstrapping. Figure 2 shows the stability of each method. In order to give an idea about the variations of stability in regard to the number of features, Table 4 provides significant information. In this table the mean value of stability is calculated for fifth, thirteenth and twenty-ninth features. Also, Table 5 illustrates the similarity between different feature selection methods. The similarity index has been calculated for the first 29 features selected by each method.\n\nIn order to estimate the generalization ability of the classifier, repeated random sub-sampling validation with 200 runs was applied. Figure 3 depicts the classification accuracy of k-NN and MLFN classifiers for different feature selection methods.\n\nAs Fig. 3 shows, starting with one feature, each additional feature typically leads to an increment in the classification accuracy. However, at some point, the increment on the classification accuracy for each additional feature is not significant leading to an elbow in the graph. Inspired by the “elbow” point in the cost-benefit curves, in this work we used the Kneedle algorithm proposed in for determining the optimal feature number which provides a satisfactory trade-off between the selected number of features and the classification accuracy. Table 6 illustrates the top 10 features selected by each method.\n\n## Discussion\n\nAccording to Fig. 2, Fisher method seems to have the highest stability and the CMIM method comes out to be the least stable one. Also, the stability of Chi2 and IG methods seems very convergent.\n\nThere exists a huge reduction in stability for MRMR_MID, MRMR_MIQ and ReliefF for three-feature subset, although after that stability increases slightly by each additional feature. Both MRMR methods are always 100% stable in selecting the first feature which is Hurst Exponent. It means that the Hurst Exponent has the highest discrimination ability from MRMR methods point of view. Also, the Fisher method has 100% stability for three-feature and five-feature subsets (ID, Hurst exponent, Petrosian fractal dimension as three-feature group and ID, Hurst exponent, Petrosian fractal dimension, zero-crossing rate and approximate entropy as five-feature group).\n\nAccording to Table 4, MRMR-MIQ has the highest mean stability up to five features. Meanwhile, Fisher and Chi2 methods have almost the same stability value. Considering thirteen features, Fisher method is almost totally stable (99.92%). Finally, considering twenty-nine features, IG outperforms other methods from mean stability point of view.\n\nAccording to Table 5, Chi2 and IG pair and MRMR-MID and MRMR-MIQ pair generate highly similar results. The similarity of MRMR methods can be explained by their similar theoretical background. On the contrary, CMIM and Fisher methods give the most dissimilar results. The average similarity of Borda and RRA methods is approximately 0.5 with the other methods. Regarding the aggregation characteristics it was predictable.\n\nTable 6 illustrates the top 10 features selected by each method. ISD (F36) always appears in the top 10 for all the methods. In spite of the fact that different feature ranking methods have their own specific criteria for ranking the features, observing ISD in the top 10 list, means that ISD is a preferable feature for all the feature selection methods. In addition to ISD, there are some other features that can be considered most preferable according to Table 6. EEG ZCR (F9) is a simple, yet effective feature that is listed in top 10 by all of the methods except ReliefF. Following ZCR, Petrosian fractal dimension (F31), Hurst exponent (F35), WP feature (F22), approximate entropy (F29), spectral entropy (F27), and Hjorth mobility parameter (F11) are selected by at least five ranking methods to be included in top 10 list. On the other hand, features that are not in this list or are just selected by one method can be categorized as the least preferred features. EMG energy and energy ratio features (F47 to F49) and some of WP features are examples of least preferred features. The optimum number of features for each method, which is selected by the Kneedle algorithm, is shown in Table 6. For MLFN and k-NN classifiers, a slight difference exists in the optimum number. Considering the maximum accuracy that the methods reach in their optimum points, the MRMR-MID method using k-NN classifier outperforms all the others with seven selected features. Also, both MRMR methods using MLFN classifier outperform all the other methods with five features.\n\nThe CMIM method reaches its best accuracy with the first 3 features on both the classifiers. Considering Fig. 3, its accuracy is equal or less than the MRMR-MID method’s accuracy at that point. Unanticipatedly, none of the aggregation methods outperformed the rest of the feature ranking methods. One possible reason for this is that the aggregation methods, especially Borda, are affected by the performance of all the methods from best to worst.\n\n## Conclusions and future works\n\nIn this paper we compared the performance of seven feature ranking methods for sleep stage classification. Feature selection based on filtering techniques has several advantages such as being fast, easily scalable to high-dimensional datasets, decrease computational complexity and work independently of the classifiers. Also, rank aggregation methods are supposed to be robust when used with a broad variety of classifiers and produce comparable classification accuracy to the individual feature selection methods. In this work, two rank aggregation methods were also applied to evaluate the performance on sleep stage classification. The Physionet Sleep-EDF Expanded Database was used to assess the impact of these methods on the classification accuracy of k-NN and MLFN. In addition, the stability and similarity of different feature selection methods were also evaluated. The results indicate that the MRMR-MID method slightly outperforms the other feature selection methods from the accuracy point of view. Considering that the CMIM produces the most unstable rankings, generally Fisher method produces the most stable results. When a small group of features (5–13) was required, the RRA aggregation method slightly outperformed the Borda. In our simulations, the performance of the aggregation methods was in the average level, although they are known to generate more stable results and better accuracy. It should be considered that the results presented in this paper are obtained through using Physionet Sleep-EDF Expanded Database which is already used in several previous sleep studies [19, 54,55,56] and can be supposed as verified enough to be used in such a comparative study. Nevertheless, generalizing these results to all future sleep studies requires further study and analysis by using other sleep databases as well. Also, in this paper for evaluating the generalization ability of classifiers we used repeated random subsampling validation. In , it is mentioned that due to the data subdivision dependency resulted from validation methods that are based on random subsampling, patient cross validation was preferred. Therefore, future steps will involve verifying the results with different databases, applying and comparing more rank aggregation methods and also using patient cross validation and comparing the results with common validation methods.\n\n## References\n\n1. Migotina D, Calapez A, Rosa A. Automatic artifacts detection and classification in sleep EEG Signals using descriptive statistics and histogram analysis: comparison of two detectors. In Engineering and technology, 2012 Spring Congress. IEEE; 2012. p. 1–6.\n\n2. Berry RB, Brooks R, Gamaldo CE, Harding SM, Lioyd RM, Marcus CL, et al. AASM—manual for the scoring of sleep and associated events version 2.1; 2014.\n\n3. Chokroverty S. sleep disorders medicine: basic science, technical considerations, and clinical aspects. New York: Saunders, Elsevier; 2009.\n\n4. Najdi S, Gharbali AA, Fonseca JM. A comparison of feature ranking and rank aggregation techniques in automatic sleep stage classification based on polysomnographic signals. In 4th International Conference IWBBIO. Granada: Springer International Publishing; 2016. p. 230–41.\n\n5. Rathore SS, Gupta A. A comparative study of feature-ranking and feature-subset selection techniques for improved fault prediction. In Proceedings of the 7th India Software Engineering Conference—ISEC’14. New York: ACM Press; 2014. p. 1–10.\n\n6. Simões H, Pires G, Nunes U, Silva V. Feature extraction and selection for automatic sleep staging using EEG. In ICINCO; 2010. p. 128–33.\n\n7. Khalighi S, Sousa T, Nunes U. Adaptive automatic sleep stage classification under covariate shift. In Engineering in medicine and biological society, 2012 annual international conference. IEEE; 2012. p. 2259–62.\n\n8. Foussier J, Fonseca P, Long X, Leonhardt S. Automatic feature selection for sleep/wake classification with small data sets. 6th International conference on bioinformatics models, methods and algorithms; 2013. p. 1–7.\n\n9. Chapotot F, Becq G. Automated sleep-wake staging combining robust feature extraction, artificial neural network classification, and flexible decision rules. Int J Adapt Control Signal Process. 2009;24:409–23.\n\n10. Zoubek L, Charbonnier S, Lesecq S, Buguet A, Chapotot F. Feature selection for sleep/wake stages classification using data driven methods. Biomed Signal Process Control. 2007;2:171–9.\n\n11. Khalighi S, Sousa T, Pires G, Nunes U. Automatic sleep staging: a computer assisted approach for optimal combination of features and polysomnographic channels. Expert Syst Appl. 2013;40:7046–59.\n\n12. Seni G, Elder J. Ensemble methods in data mining improving accuracy through combining predictions. Synth Lect Data Min Knowl Discov. 2010;2:1–26.\n\n13. Prati RC. Combining feature ranking algorithms through rrank aggregation. 2012 international joint conference neural networks. IEEE; 2012. p. 1–8.\n\n14. Dwork C, Kumar R, Naor M, Sivakumar D. Rank aggregation methods for the web. 10th Proceeding International Conference World Wide Web—WWW’01. New York: ACM Press; 2001. p. 613–22.\n\n15. The Sleep-EDF Database. https://physionet.org/physiobank/database/sleep-edfx/. Accessed 1 Feb 2017.\n\n16. Wiltschko AB, Gage GJ, Berke JD. Wavelet filtering before spike detection preserves waveform shape and enhances single-unit discrimination. J Neurosci Methods. 2008;173:34–40.\n\n17. Hamida ST-B, Ahmed B. Computer based sleep staging: challenges for the future. 2013 7th IEEE GCC conference and exhibition. IEEE; 2013. p. 280–5.\n\n18. Hjorth B. EEG analysis based on time domain properties. Electroencephalogr Clin Neurophysiol. 1970;29:306–10.\n\n19. Ebrahimi F, Mikaeili M, Estrada E, Nazeran H. Automatic sleep stage classification based on EEG signals by using neural networks and wavelet packet coefficients. In Engineering in medicine and biological society, 2008 30th annual international conference. IEEE; 2008. p. 1151–4.\n\n20. Inouye T, Shinosaki K, Sakamoto H, Toi S, Ukai S, Iyama A, et al. Quantification of EEG irregularity by use of the entropy of the power spectrum. Electroencephalogr Clin Neurophysiol. 1991;79:204–10.\n\n21. Sabeti M, Katebi S, Boostani R. Entropy and complexity Measures for EEG signal classification of schizophrenic and control participants. Artif Intell Med. 2009;47:263–74.\n\n22. Renyi A. On measures of entropy and information. In Proceeding of the fourth Berkeley symposium on mathematical statistics and probability, volume 1: contributions to the theory of statistics, California. University of California Press; 1961. p. 547–61.\n\n23. Jenssen R, Hild KE, Erdogmus D, Principe JC, Eltoft T. Clustering using Renyi’s entropy. Proceedings of International Joint Conference on Neural Networks, 2003. IEEE; 2003. p. 523–8.\n\n24. Lake D. Renyi entropy measures of heart rate gaussianity. IEEE Trans Biomed Eng. 2006;53:21–7.\n\n25. Pincus SM, Gladstone IM, Ehrenkranz RA. A regularity statistic for medical data analysis. J Clin Monit Comput. 1991;7:335–45.\n\n26. Ho KK, Moody GB, Peng CK, Mietus JE, Larson MG, Levy D, et al. Predicting survival in heart failure case and control subjects by use of fully automated methods for deriving nonlinear and conventional indices of heart rate dynamics. Circulation. 1997;96:842–8.\n\n27. Pincus SM, Goldberger AL. Physiological time-series analysis: what does regularity quantify? Am J Physiol. 1994;266:H1643–56.\n\n28. Bandt C, Pompe B. Permutation entropy: a natural complexity measure for time series. Phys Rev Lett. 2002;88:174102.\n\n29. Li D, Li X, Liang Z, Voss LJ, Sleigh JW, et al. Multiscale permutation entropy analysis of eeg recordings during sevoflurane anesthesia. J Neural Eng. 2010;7:46010.\n\n30. Petrosian A. Kolmogorov complexity of finite sequences and recognition of different preictal EEG patterns. Proceedings of the eighth IEEE symposium computer based medical systems. IEEE Computer Society Press; 1995. p. 212–7.\n\n31. Kaiser JF. Some useful properties of teager’s energy operators. IEEE international conference on acoustics, speech, and signal processing. IEEE; 1993. p. 149–52.\n\n32. Şen B, Peker M, Çavuşoğlu A, Çelebi FV. A comparative study on classification of sleep stage based on EEG signals using feature selection and classification algorithms. J Med Syst. 2014;38:18.\n\n33. Katz MJ. Fractals and the Analysis of Waveforms. Comput Biol Med. 1988;18:145–56.\n\n34. D’Alessandro M, Esteller R, Vachtsevanos G, Hinson A, Echauz J, Litt B. Epileptic seizure prediction using hybrid feature selection over multiple intracranial EEG electrode contacts: a report of four patients. IEEE Trans Biomed Eng. 2003;50:603–15.\n\n35. Hurst HE. The problem of long-term storage in reservoirs. Int Assoc Sci Hydrol Bull. 1956;1:13–27.\n\n36. Zhang Y, Zhao Y. Real and imaginary modulation spectral subtraction for speech enhancement. Speech Commun. 2013;55:509–22.\n\n37. Taşmaz H, Erçelebi E. Speech enhancement based on undecimated wavelet packet-perceptual filterbanks and MMSE–STSA estimation in various noise environments. Digit Signal Process. 2008;18:797–812.\n\n38. Estrada E, Nava P, Nazeran H, Behbehani K, Burk J, Lucas E. Itakura Distance: a useful similarity measure between EEG and EOG signals in computer-aided classification of sleep stages. In Conference proceedings, IEEE engineering medical biology society; 2005. p. 1189–92.\n\n39. Kira K, Rendell LA. The feature selection problem: traditional methods and a new algorithm. In: Swartout WR, editor. Proceeding 10th national conference on artificial intelligence. San Jose, CA, July 12–16, 1992; 1992. p. 129–34.\n\n40. Robnik-Šikonja M, Kononenko I. Theoretical and empirical analysis of ReliefF and RReliefF. Mach Learn. 2003;53:23–69.\n\n41. Ding C, Peng H. minimum redundancy feature selection from microarray gene expression data. IEEE proceedings of computational systems bioinformatics conference—CSB2003. IEEE Computer Society Press; 2005. p. 523–8.\n\n42. Gu Q, Li Z, Han J. Generalized Fisher score for feature selection. CoRR; 2012.\n\n43. Huan Liu, Setiono R. Chi2: feature selection and discretization of numeric attributes. In proceedings of the 7th IEEE international conference on tools with artificial intelligence IEEE Computer Society Press; 1995. p. 388–91.\n\n44. Quinlan JR. C4.5: Programs for machine learning. Morgan Kaufmann Publishers Inc.; 1993.\n\n45. Fleuret F. Fast binary feature selection with conditional mutual information. J Mach Learn Res. 2004;5:1531–55.\n\n46. Kolde R, Laur S, Adler P, Vilo J. Robust rank aggregation for gene list integration and meta-analysis. Bioinformatics. 2012;28:573–80.\n\n47. Wald R, Khoshgoftaar TM, Dittman D. Mean aggregation versus robust rank aggregation for ensemble gene selection. 2012 11th international conference on machine learning and application. IEEE; 2012. p. 63–9.\n\n48. Hastie T, Tibshirani R, Friedman J. The elements of statistical learning. New York: Springer; 2009.\n\n49. Duda RO, Hart PE, Stork DG. Pattern classification. New York: Wiley; 2012.\n\n50. Zhang G, Eddy Patuwo B, Hu YM. Forecasting with artificial neural networks. Int J Forecast. 1998;14:35–62.\n\n51. Kalousis A, Prados J, Hilario M. Stability of feature selection algorithms: a study on high-dimensional spaces. Knowl Inf Syst. 2007;12:95–116.\n\n52. Imtiaz SA, Rodriguez-Villegas E. Recommendations for performance assessment of automatic sleep staging algorithms. 2014 36th annual international conference of the IEEE engineering in medicine and biological society. IEEE; 2014. p. 5044–7.\n\n53. Satopaa V, Albrecht J, Irwin D, Raghavan B. Finding a “Kneedle” in a haystack: detecting knee points in system behavior. 2011 31st international conference on distributed computing systems workshop. IEEE; 2011. p. 166–71.\n\n54. Bajaj V, Pachori RB. Automatic classification of sleep stages based on the time-frequency image of EEG signals. Comput Methods Programs Biomed. 2013;112:320–8.\n\n55. Zhou J, Wu X, Zeng W. Automatic detection of sleep apnea based on EEG detrended fluctuation analysis and support vector machine. J Clin Monit Comput. 2015;29:767–72.\n\n56. Rodríguez-Sotelo J, Osorio-Forero A, Jiménez-Rodríguez A, Cuesta-Frau D, Cirugeda-Roldán E, Peluffo D. Automatic sleep stages classification using eeg entropy features and unsupervised pattern analysis techniques. Entropy. 2014;16:6573–89.\n\n57. Herrera LJ, Fernandes CM, Mora AM, Migotina D, Largo R, Guillen A, et al. Combination of heterogeneous EEG feature extraction methods and stacked sequential learning for sleep stage classification. Int J Neural Syst. 2013;23:1350012.\n\n58. Dursun M, Gunes S, Ozsen S, Yosunkaya S. Comparison of artificial immune clustering with Fuzzy C-means Clustering in the sleep stage classification problem. 2012 international symposium on innovations in intelligent systems and applications. IEEE; 2012. p. 1–4.\n\n## Declarations\n\nAuthors’ contributions Simulations: SN, AAG. Drafting: all authors. Analysis and interpretation of results: all authors. Critical revision: all authors. All authors read and approved the final manuscript.\n\nNot applicable.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Availability of data and materials\n\nAll of the datasets supporting the conclusions of this article are freely available at http://physionet.org/physiobank/database/sleep-edfx/.\n\n### Funding\n\nThis work was partially funded by FCT Strategic Program UID/EEA/00066/203 of UNINOVA, CTS and INCENTIVO/EEI/UI0066/2014 of UNINOVA. Publication costs were funded by FCT Strategic Program UID/EEA/00066/203 of UNINOVA, CTS.\n\nThis article has been published as part of BioMedical Engineering OnLine Volume 16 Supplement 1, 2017: Selected articles from the 4th International Work-Conference on Bioinformatics and Biomedical Engineering-IWBBIO 2016. The full contents of the supplement are available online at https://biomedical-engineering-online.biomedcentral.com/articles/supplements/volume-16-supplement-1.\n\n### Publisher’s Note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Shirin Najdi.\n\n## Rights and permissions",
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https://www.arxiv-vanity.com/papers/1905.09144/
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"# Conformal invariance of TMD rapidity evolution\n\nIan Balitsky and Giovanni A. Chirilli Physics Dept., Old Dominion University, Norfolk VA 23529 & Theory Group, JLAB, 12000 Jefferson Ave, Newport News, VA 23606\nInstitut für Theoretische Physik, Universität Regensburg,\nD-93040 Regensburg, Germany\n###### Abstract\n\nWe discuss conformal properties of TMD operators and present the result of the conformal rapidity evolution of TMD operators in the Sudakov region.\n\npreprint: JLAB-THY-19-2940\n\n## I Introduction\n\nIn recent years, the transverse-momentum dependent parton distributions (TMDs) Collins and Soper (1982); Collins et al. (1985); Ji et al. (2005); Echevarria et al. (2012) have been widely used in the analysis of processes like semi-inclusive deep inelastic scattering or particle production in hadron-hadron collisions (for a review, see Ref. Collins (2013)).\n\nThe TMDs are defined as matrix elements of quark or gluon operators with attached light-like gauge links (Wilson lines) going to either or depending on the process under consideration. It is well known that these TMD operators exhibit rapidity divergencies due to infinite light-like gauge links and the corresponding rapidity/UV divergences should be regularized. There are two schemes on the market: the most popular is based on CSS Collins et al. (1985) or SCET Rothstein and Stewart (2016) formalism and the second one is adopted from the small- physics Lipatov (1997); Kovchegov and Levin (2012). The obtained evolution equations differ even at the leading-order level and need to be reconciled, especially in view of the future EIC accelerator which will probe the TMDs at values of Bjorken between small- and regions.\n\nIn our opinion, a good starting point is to obtain conformal leading-order evolution equations. It is well known that at the leading order pQCD is conformally invariant so there is a hope to get any evolution equation without explicit running coupling from conformal considerations. In our case, since TMD operators are defined with attached light-like Wilson lines, formally they will transform covariantly under the subgroup of full conformal group which preserves this light-like direction. However, as we mentioned, the TMD operators contain rapidity divergencies which need to be regularized. At present, there is no rapidity cutoff which preserves conformal invariance so the best one can do is to find the cutoff which is conformal at the leading order in perturbation theory. In higher orders, one should not expect conformal invariance since it is broken by running of QCD coupling. However, if one considers corresponding correlation functions in SYM, one should expect conformal invariance. After that, the results obtained in SYM theory can be used as a starting point of QCD calculation. Typically, the result in theory gives the most complicated part of pQCD result, i.e. the one with maximal transcendentality. Thus, the idea is to find TMD operator conformal in SYM and use it in QCD. This scheme was successfully applied to the rapidity evolution of color dipoles. At the leading order, the BK evolution of color dipoles Balitsky (1996, 1999); Kovchegov (1999, 2000) is invariant under SL(2,C) group. At the NLO order the “conformal dipole” with correction Balitsky and Chirilli (2009) makes NLO BK evolution Mobius invariant for SYM and the corresponding QCD kernel Balitsky and Chirilli (2008) differs by terms proportional to -function.\n\n## Ii Conformal invariance of TMD operators\n\nFor definiteness, we will talk first about gluon operators with light-like Wilson lines stretching to in “+” direction. The gluon TMD (unintegrated gluon distribution) is defined as Mulders and Rodrigues (2001)\n\n D(xB,k⊥,η) = ∫d2z⊥ ei(k,z)⊥D(xB,z⊥,η), (1) g2D(xB,z⊥,η) = −x−1B2πp−∫dz+ e−ixBp−z+ × ⟨P|Faξ(z)[z−∞n,−∞n]abFbξ(0)|P⟩∣∣z−=0\n\nwhere is an unpolarized target with momentum (typically proton) and is a light-like vector in “+” direction. Hereafter we use the notation\n\n Fξ,a(z⊥,z+) ≡ gF−ξ,m(z)[z,z−∞n]ma∣∣z−=0 (2)\n\nwhere denotes straight-line gauge link connecting points and :\n\n [x,y] ≡ Peig∫du (x−y)μAμ(ux+(1−u)y) (3)\n\nTo simplify one-loop evolution we multiplied by coupling constant. Since the is renorm-invariant we do not need to consider self-energy diagrams (in the background-Feynman gauge). Note that is fixed by the original factorization formula for particle production Collins (2013) (see also the discussion in Ref. Balitsky and Tarasov (2017, 2018)).\n\nThe algebra of full conformal group consists of four operators , six , four special conformal generators , and dilatation operator . It is easy to check that in the leading order the following 11 operators act on gluon TMDs covariantly\n\n Pi,P−,M12,M−i,D,Ki,K−,M−+ (4)\n\nwhile the action of operators , and do not preserve the form of the operator (2). The action of the generators (4) on the operator (2) is the same as the action on the field without gauge link attachments. The corresponding group consists of transformations which leave the hyperplane and vector invariant. Those include shifts in transverse and directions, rotations in the transverse plane, Lorentz rotations/boosts created by , dilatations, and special conformal transformations\n\n z′μ = zμ−aμz21−2a⋅z+a2z2 (5)\n\nwith . In terms of “embedding formalism” Dirac (1936); Mack and Salam (1969); Ferrara et al. (1973); Rychkov (2016) defined in 6-dim space, this subgroup is isomorphic to “Poincare + dilatations” group of the 4-dim subspace orthogonal to our physical light-like “+” and “-” directions.\n\nAs we noted, infinite Wilson lines in the definition (2) of TMD operators make them divergent. As we discussed above, it is very advantageous to have a cutoff of these divergencies compatible with approximate conformal invariance of tree-level QCD. The evolution equation with such cutoff should be invariant with respect to transformations described above.\n\nIn the next Section we demonstrate that the “small-x” rapidity cutoff enables us to get a conformally invariant evolution of TMD in the so-called Sudakov region.\n\n## Iii TMD factorization in the Sudakov region\n\nThe rapidity evolution of TMD operator (1) is very different in the region of large and small longitudinal separations . The evolution at small is linear and double-logarithmic while at large the evolution become non-linear due to the production of color dipoles typical for small- evolution. It is convenient to consider as a starting point the simple case of TMD evolution in the so-called Sudakov region corresponding to small longitudinal distances.\n\nFirst, let us specify what we call a Sudakov region. A typical factorization formula for the differential cross section of particle production in hadron-hadron collision is Collins (2013); Collins and Rogers (2015)\n\n dσdηd2q⊥ = ∑f∫d2b⊥ei(q,b)⊥Df/A(xA,b⊥,η) × Df/B(xB,b⊥,η)σ(ff→H) + ... (6)\n\nwhere\n\n ⟨pA,pB|FaμνFaμν(z1)FbλρFbλρ(z2)|pA,pB⟩ = 1N2c−1⟨pA|~Oij(z−1,z1⊥;z−2,z2⊥)|pA⟩σA ×⟨pB|Oij(z+1,z1⊥;z+2,z2⊥)|pB⟩σB +... (7)\n\nwhere\n\n Oij(z+1,z1⊥;z+2,z2⊥) = Fai(z1)[z1−∞n,z2−∞n]abFbj(z2)∣∣z−1=z−2=0, (8) ~Oij(z−1,z1⊥;z−2,z2⊥) = Fai(z1)[z1−∞n′,z2−∞n′]abFbj(z2)∣∣z+1=z+2=0, Fi,a(z⊥,z−) ≡ F+i,m(z)[z,z−∞n′]ma∣∣z+=0. (9)\n\nHere\n\nAs we mentioned, TMD operators exhibit rapidity divergencies due to infinite light-like gauge links. The “small- style” rapidity cutoff for longitudinal divergencies is imposed as the upper limit of components of gluons emitted from the Wilson lines. As we will see below, to get the conformal invariance of the leading-order evolution we need to impose the cutoff of components of gluons correlated with transverse size of TMD in the following way:\n\n (Fi,a(z⊥,z+))σ ≡ F−i,m(z)[Peig∫z+−∞dz+A−,σ(up1+x⊥)]ma, Aσμ(x) = ∫d4k16π4 θ(σ√2z12⊥−|k+|)e−ik⋅xAμ(k) (10)\n\nSimilarly, the operator in Eq. (9) is defined with with the rapidity cutoff for integration imposed as .\n\nThe Sudakov region in the coordinate space corresponds to\n\n z212∥ ≡ 2z−12z+12 ≪ z212⊥ (11)\n\nIn the leading log approximation, the upper cutoff for integration in the target matrix element in Eq. (7) is and similarly the -integration cutoff in projectile matrix element is .\n\nIn the next Section we demonstrate that rapidity cutoff (10) enables us to get a conformally invariant evolution of TMD in the Sudakov region (11).\n\n## Iv One-loop evolution of TMDs\n\n### iv.1 Evolution of gluon TMD operators in the Sudakov region\n\nIn this Section we derive the evolution of gluon TMD operator (8) with respect to cutoff in the leading log approximation.",
null,
"Figure 1: Typical diagrams for production (a) and virtual (b) contributions to the evolution kernel. The dashed lines denote gauge links.\n\nAs usual, to get an evolution equation we integrate over momenta . To this end, we calculate diagrams shown in Fig. 1 in the background field of gluons with . The calculation is easily done by method developed in Refs. Balitsky and Tarasov (2015, 2016) and the result is\n\n Oσ2(z+1,z+2) = αsNc2πσ2√2|z12⊥|∫σ1√2|z12⊥|dk+k+ KOσ1(z+1,z+2) (12)\n\nwhere the kernel is given by\n\n KO(z+1,z+2) (13) = O(z+1,z+2)∫z+1−∞dz′+z+2−z′+e−i|z12⊥|σ√2(z2−z′)+ + O(z+1,z+2)∫z+2−∞dz′+z+1−z′+ei|z12⊥|σ√2(z1−z′)+ −∫z+1−∞dz′+O(z+1,z+2)−O(z′+1,z+2)z+1−z′+ −∫z+2−∞dz′+O(z+1,z+2)−O(z+1,z′+2)z+2−z′+\n\nwhere we suppress arguments and since they do not change during the evolution in the Sudakov regime. The first two terms in the kernel come from the “production” diagram in Fig. 1a while the last two terms from “virtual” diagram in Fig. 1b. The result (13) can be also obtained from Ref. Balitsky and Tarasov (2016) by Fourier transformation of Eq. (5.9) with the help of Eqs. (3.12) and (3.30) therein. The approximations for diagrams in Fig. 1 leading to Eq. (13) are valid as long as\n\n k+≫z+12z212⊥ (14)\n\nwhich gives the region of applicability of Sudakov-type evolution.\n\nEvolution equation (12) can be easily integrated using Fourier transformation. Since\n\n Ke−ik−z+1+ik′−z+2 = [−2lnσz12⊥−ln(ik−)−ln(−ik′−) + ln2−4γE + O(z+12|z12⊥|σ)]e−ik−z+1+ik′−z+2 (15)\n\none easily obtains\n\n × ∫dz′+1dz′+2 Oσ1(z′+1,z′+2) z−2¯αslnσ2σ112⊥ × 14π2[iΓ(1−2¯αslnσ2σ1)(z+1−z′+1+iϵ)1−2¯αslnσ2σ1+c.c.] × [iΓ(1−2¯αslnσ2σ1)(z+2−z′+2+iϵ)1−2¯αslnσ2σ1+c.c.] (16)\n\nwhere we introduced notation . It should be mentioned that the factor is “scheme-dependent”: if one introduces to -integrals smooth cutoff instead of rigid cutoff , the value changes to .\n\nIt is easy to see that the r.h.s. of Eq. (16) transforms covariantly under all transformations (4) except Lorentz boost generated by . The reason is that the Lorentz boost in direction changes cutoffs for the evolution. To understand that, note that Eq. (15) is valid until so the linear evolution (16) is applicable in the region between\n\n σ2=σB=|z12⊥|z−12√2 and σ1=z+12√2|z12⊥| (17)\n\nFrom Eq. (16) it is easy to see that Lorentz boost\n\n### iv.2 Evolution of quark TMD operators\n\nA simple calculation of evolution of quark operator\n\n Oq(z+1,z1⊥;z+2,z2⊥) ≡ g2CFb¯ψ(z⊥+un) (18) × [un+z⊥,−∞n]⧸n[z⊥−∞n,−∞n][∞n,0]ψ(0)\n\nthe same evolution (16) as for the gluon operators with trivial replacement 111We assume that is defined with the same type of cutoff (10) imposed on quarks and gluons emitted as a result of evolution of .. The factor () is added to avoid taking into account quark self-energy.\n\n### iv.3 Evolution beyond Sudakov region\n\nAs we mentioned above, TMD factorization formula (6) for particle production at translates to the coordinate space as Eq. (7) with the requirement . As the result of evolution (16) the transverse separation between gluon operators and remains intact while the longitudinal separation increases. As discussed in Refs. Balitsky and Tarasov (2015, 2016) , the Sudakov approximation can be trusted until the upper cutoff in integrals is greater than which is equivalent to Eq. (14) in the coordinate space. If and , the relative energy between Wilson-line operators and target nucleon at the final point of evolution is so one should use phenomenological models of TMDs with this low rapidity cutoff as a starting point of the evolution (16). If, however, , this relative energy is so one can continue the rapidity evolution in the region beyond the Sudakov region into the small- region. The evolution in a “proper” small- region is known Dominguez et al. (2011) - the TMD operator, known also as Weiczsäcker-Williams distribution, will produce a hierarchy of color dipoles as a result of the non-linear evolution. However, the transition between Sudakov region and small- region is described by rather complicated interpolation formula Balitsky and Tarasov (2015). In the coordinate space this means the study of operator at and we hope that conformal considerations can help us to obtain the TMD evolution in that region.\n\n## V Discussion\n\nAs we mentioned in the Introduction, TMD evolution is analyzed by very different methods at small and moderate . In view of future EIC accelerator, which will probe the region between small and , we need a universal description of TMD evolution valid at both limits. Since the two formalisms differ even at the leading order where QCD is conformally invariant, our idea is to make this universal description first in SYM. As a first step, we found a conformally invariant evolution in the Sudakov region using our small- cutoff with the “conformal refinement” (10).\n\nTo compare with conventional TMD analysis let us write down the evolution of “generalized TMD”Meissner et al. (2009); Lorcé and Pasquini (2013)\n\n Dσ(x,ξ) = ∫dz+e−ix√s2z+⟨p′B|Oσ(−z+2,z+2)|pB⟩\n\nwhere . From Eq. (16) one easily obtains\n\n Dσ2(x,ξ)Dσ1(x,ξ) = e−2¯αslnσ2σ1[lnσ2σ1(x2−ξ2)sz212⊥+4γE−2ln2] (19)\n\nFor usual TMD at with the limits of Sudakov evolution set by Eq. (17) one obtains\n\n Dσ2(x,q⊥)Dσ1(x,q⊥) = e−2¯αslnQ2q2⊥[lnQ2q2⊥+4γE−2ln2] (20)\n\nwhich coincides with usual one-loop evolution of TMDs Aybat and Rogers (2011) up to replacement . As we discussed, such constant depends on the way of cutting -integration which should be coordinated with the cutoffs in the “coefficient function” in Eq. (6). Thus, the discrepancy is just like using two different schemes for usual renormalization. It should be mentioned, however, that at the result (19) differs from conventional one-loop result which does not depend on , see e.g. Echevarria et al. (2016a).\n\nOur main outlook is to try to connect to small- region, first in and then in QCD. Also, it would be interesting to study if the “conventional” Sudakov-region results in two Echevarria et al. (2016b); Li et al. (2016); Luebbert et al. (2016) and three loops Echevarria et al. (2016c) can be recast in our cutoff scheme which in principle allows transition to small- region. The study is in progress.\n\n## Acknowledgments\n\nWe thank V.M. Braun, A. Vladimirov and A. Tarasov for discussions. The work of I.B. was supported by DOE contract DE-AC05-06OR23177 and by the grant DE-FG02-97ER41028."
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"# Year 11 (Sets 1-2) Overview\n\nYear 11\n\nAutumn Term\n\n Autumn 1 Approx 28 lessons Chapter 18: Processing, representing and interpreting data Chapter 19: Inequalities and formulae Chapter 20: Pythagoras theorem and trigonometry 1 Chapter 21: More graphs and equations 8 lessons 7 lessons 4 lessons 4 lessons End of Unit test (GCSE style) Friday 18th October 1 lesson Intervention on problem topics 4 lessons\n\nOctober Half Term\n\n Autumn 2 Approx 28 lessons Chapter 22: Quadratic and simultaneous equations 8 lessons Mock Exam week 4 lessons Chapter 23: Area and volume 2 Chapter 24: Line diagrams and scatter graphs 7 lessons 4 lessons End of Unit test (GCSE style) Friday 13th December 1 lesson Intervention on problem topics/Preparation for IGCSE 4 lessons\n\nChristmas Holidays\n\nSpring Term\n\n Spring 1 24 lessons Revision and preparation for Edexcel Certificate Exam (iGCSE) – Friday 10th/Wednesday 15th January 6 lessons Chapter 25: Indices, standard form and surds Chapter 26: Similar shapes Chapter 27: Proportion 2 4 lessons 3 lessons 5 lessons End of Unit test (GCSE style) Friday 7th February 1 lesson Intervention on problem topics 4 lessons\n\nFeb half-term\n\n Spring 2 Approx 24 lessons Chapter 28: Probability Chapter 29: Pythagoras’ Theorem and Trigonometry 2 Chapter 30: Transformation of functions 7 lessons 8 lessons 4 lessons End of Unit test (GCSE style) Friday 28th March 1 lesson Intervention on problem topics 4 lessons\n\nEaster Holidays\n\nSummer Term\n\n Summer 1 Approx 18 lessons Chapter 31: Circle Geometry Chapter 32: Algebraic fractions & Algebraic proof Chapter 33: Vectors 3 lessons 3 lessons 4 lessons End of Unit test (GCSE style) 1 lesson Intervention on problem topics Revision for summer exam 7 lessons\n\nMay Half term\n\nFollowed by\n\nSummer Exam Period!\n\nExams – 9th June/13th June"
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https://www.frontiersin.org/articles/10.3389/fphy.2020.00139/full
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"Impact Factor 1.895 | CiteScore 2.24\nMore on impact ›\n\n# Frontiers in Physics",
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"## Original Research ARTICLE\n\nFront. Phys., 06 May 2020 | https://doi.org/10.3389/fphy.2020.00139\n\n# Rethinking Superdeterminism\n\n• 1Department of Physics, Frankfurt Institute for Advanced Studies, Frankfurt, Germany\n• 2Department of Physics, University of Oxford, Oxford, United Kingdom\n\nQuantum mechanics has irked physicists ever since its conception more than 100 years ago. While some of the misgivings, such as it being unintuitive, are merely aesthetic, quantum mechanics has one serious shortcoming: it lacks a physical description of the measurement process. This “measurement problem” indicates that quantum mechanics is at least an incomplete theory—good as far as it goes, but missing a piece—or, more radically, is in need of complete overhaul. Here we describe an approach which may provide this sought-for completion or replacement: Superdeterminism. A superdeterministic theory is one which violates the assumption of Statistical Independence (that distributions of hidden variables are independent of measurement settings). Intuition suggests that Statistical Independence is an essential ingredient of any theory of science (never mind physics), and for this reason Superdeterminism is typically discarded swiftly in any discussion of quantum foundations. The purpose of this paper is to explain why the existing objections to Superdeterminism are based on experience with classical physics and linear systems, but that this experience misleads us. Superdeterminism is a promising approach not only to solve the measurement problem, but also to understand the apparent non-locality of quantum physics. Most importantly, we will discuss how it may be possible to test this hypothesis in an (almost) model independent way.\n\n## 1. Introduction\n\nUntil the 1970s, progress in the foundations of physics meant discovering new phenomena at higher energies, or short distances, respectively. But progress in high-energy physics has slowed, and may have run its course as far as finding solutions to the deep fundamental problems of physics is concerned. In the past decades, physicists have not succeeded in solving any of the open problems in the foundations of their field; indeed it's not even clear we are getting closer to solving them. Most notably, we have still not succeeded in synthesizing quantum and gravitational physics, or in unraveling the nature of dark matter, problems that have been known since the 1930s.\n\nIn this situation it makes sense to go back and look for the path we did not take, the wrong turn we made early on that led into this seeming dead end. The turn we did not take, we argue here, is resolving the shortcomings of quantum mechanics. At the least, we need a physical description of the measurement process that accounts for the non-linearity of quantum measurement.\n\nThe major reason this path has remained largely unexplored is that under quite general assumptions (defined below) any theory which solves the measurement problem in a form consistent with the principles of relativity, makes it impossible to prepare a state independently of the detector that will later measure it. If one is not willing to accept this dependence then—by virtue of Bell's theorem —one necessarily has to conclude that a local, deterministic completion of quantum mechanics is impossible. This, then, requires us to abandon the principles on which general relativity is based and adds to our difficulty reconciling gravity with the other interactions.\n\nIf one is, by contrast, willing to accept the consequences of realism, reductionism, and determinism, one is led to a theory in which the prepared state of an experiment is never independent of the detector settings. Such theories are known as “superdeterministic.” We wish to emphasize that superdeterministic theories are not interpretations of quantum mechanics. They are, instead, theories more fundamental than quantum mechanics, from which quantum mechanics can be derived.\n\nSuperdeterminism is frequently acknowledged as an experimentally unclosed loophole (see e.g., ) with which one can explain deterministically the observed violations of Bell's inequality. However, for a variety of reasons, many physicists think Superdeterminism is a non-starter. For example, they argue that Superdeterminism would turn experimenters into mindless zombies, unable to configure their experimental apparatuses freely. A similar argument has it that Superdeterminism implies the existence of implausible conspiracies between what would otherwise be considered independent processes. Alternatively, it would seemingly lead to causes propagating backwards in time. Above all, so it is claimed, Superdeterminism would fatally undermine the notion of science as an objective pursuit. In short, Superdeterminism is widely considered to be dead in the water.\n\nThe aim of this paper is to re-examine the arguments against Superdeterminism. We will argue that, rather than being an implausible and dismissible loophole, the neglected option of Superdeterminism is the way forward; it's the path we did not take.\n\n## 2. Why?\n\nThe way it is commonly taught, quantum mechanics has two ingredients to its dynamical law: the Schrödinger equation and the measurement prescription. The measurement prescription is a projection on a detector eigenstate, followed by re-normalizing the new state to 1.\n\nThis measurement prescription (also sometimes referred to as the “update” or “collapse” of the wave-function) is not a unitary operation. It preserves probabilities by construction, but it is neither reversible nor linear. The lack of reversibility is not a serious problem: one may interpret irreversibility as a non-physical limit in which one has ignored small but finite residuals that would otherwise make the measurement process reversible.\n\nRather, the major problem with the measurement process is that it is non-linear. If we have a prepared initial state |Ψ1〉 that brings the detector into eigenstate |χ1〉, and another initial state |Ψ2〉 that brings the detector into eigenstate |χ2〉, then a linear evolution law would bring a superposition $\\left(|{\\text{Ψ}}_{1}〉+|{\\text{Ψ}}_{2}〉\\right)/\\sqrt{2}$ into a superposition of detector eigenstates—but this is not what we observe.\n\nThis is problematic because if quantum mechanics was the correct theory to describe the behavior of elementary particles, then what macroscopic objects like detectors do should be derivable from it. The problem is not merely that we do not know how to make this derivation, it's far worse: the observed non-linearity of the measurement process tells us that the measurement postulate is in contradiction with the linear Schrödinger equation.\n\nHowever, there is a simple resolution of this problem of non-linearity. In its density matrix form, the Schrödinger equation is remarkably similar to the classical Liouville equation. So much that, in this form, the Schrödinger equation is sometimes referred to as the Schrödinger-Liouville or Quantum-Liouville equation, though the historically more correct term is the von Neumann-Dirac equation:\n\nHere, H is the classical/quantum Hamiltonian of the system, the curly brackets are Poisson brackets, and the square brackets are commutators. ρ is the classical/quantum (probability) density, respectively.\n\nThe classical Liouville equation is linear in the probability density due to conservation of probability. But this linearity says nothing whatsoever about whether the dynamics of the underlying system from which the probability density derives is also linear. Hence, for example, chaotic dynamical systems, despite their non-linear dynamics, obey the same linear equation for probability density. To us, this close formal similarity between the two equations strongly suggests that quantum physics, too, is only the linear probabilistic description of an underlying non-linear deterministic system.\n\nFrom this point of view, pursuing an Everettian approach to quantum physics is not the right thing to do, because this idea is founded on the belief that the Schrödinger equation is fundamental; that nothing underpins it. Moreover, it does not make sense to just append non-linear dynamics to the Schrödinger equation in situations when state decoherence becomes non-negligible, because it is not the Schrödinger equation itself that needs to become non-linear. Spontaneous collapse models do not help us either because these are not deterministic1. Pilot-wave theories do, in some sense, solve the measurement problem deterministically. However, pilot-wave formulations of quantum mechanics are based on an explicitly non-local ontology, and this non-locality makes it difficult to reconcile such theories with special relativity and, with that, quantum field theory.\n\nWhat, then, does it take to describe quantum physics with a deterministic, local theory that is reductionist in the sense that the theory allows us to derive the behavior of detectors from the behavior of the theory's primitive elements? The Pusey-Barrett-Rudolph (PBR) theorem [3, 4] tells us that such a theory must violate the Preparation Independence Postulate, according to which the state space of a composite system can be described as a product state whose factors are independent of each other. Violating Preparation Independence entails that either the systems making up the product state are correlated with each other, or that the composite system cannot be described as a product state to begin with. This lack of independence between the prepared state and the detector is the hallmark of Superdeterminism.\n\n## 3. What?\n\nWe define a superdeterministic theory as a Psi-epistemic, deterministic theory that violates Statistical Independence but is local in the sense of respecting Continuity of Action , i.e., there is no “action at a distance” as Einstein put it. In the remainder of this section we will explain what these words mean.\n\n1. Psi-epistemic: That a theory is Psi-epistemic means that the wave-function in the Schrödinger equation (or the density-matrix, respectively) does not itself correspond to an object in the real world, i.e., is not ontic. The Copenhagen interpretation and Neo-Copenhagen interpretations are Psi-epistemic because they postulate the wave-function merely encodes knowledge about the state of the system, rather than itself corresponding to a property of the system. However, a theory may also be Psi-epistemic because the wavefunction is emergent, for example as a statistical representation of a more fundamental theory. The theories we will be dealing with here are Psi-epistemic in the latter sense.\n\nNeedless to say, the wavefunction derived in any such theory should obey the Schrödinger equation up to current measurement precision and hence reproduce the so-far tested predictions of quantum mechanics. But of course the point of seeking a theory from which to derive quantum mechanics is not to reproduce quantum mechanics, but to make predictions beyond that.\n\n2. Deterministic: By deterministic we will mean that the dynamical law of the theory uniquely maps states at time t to states at time t′ for any t and t′. This map, then, can be inverted.\n\nSince the theory we look for should be deterministic and the wavefunction derives from it, we are dealing with a so-called hidden-variable theory. We can ask what exactly are these hidden variables, which in the following are collectively represented by the symbol λ. The answer depends on the specific model one is dealing with, but loosely speaking λ contains all the information that is required to determine the measurement outcome (except the “not hidden” variables that are the state preparation). In this picture, quantum mechanics is not deterministic simply because we do not know λ.\n\nIt is important to realize that these hidden variables are not necessarily properties intrinsic to or localized within the particle that one measures; they merely have to determine the outcome of the measurement. To see the distinction, consider the following example. You are standing in a newborn ward in a hospital and look at a room full of screaming infants. On your mind are two questions: What's their blood type? and Will they ever climb Mount Everest? In a deterministic theory, answers to both questions are encoded in the state of the universe at the present time, but they are very different in terms of information availability. A baby's blood type is encoded locally within the baby. But the information about whether a baby will go on to climb Mount Everest is distributed over much of the hypersurface of the moment the baby is born. It is not, in any meaningful sense, an intrinsic property of the baby. This example also illustrates that just because a theory is deterministic, its time evolution is not necessarily predictable.\n\n3. Violation of Statistical Independence: The most distinctive feature of superdeterministic theories is that they violate Statistical Independence. As it is typically expressed, this means that the probability distribution of the hidden variables, ρ(λ), is not independent of the detector settings. If we denote the settings of two detectors in a Bell experiment as a and b, we can write this as\n\nFor example, in the CHSH version of Bell's Theorem , a and b each take one of two discrete orientations, which we can represent here as 0 or 1. To derive Bell's inequality, one assumes ρ(λ|a, b) = ρ(λ), a requirement that is also often referred as “Free Choice” (this terminology is profoundly misleading as we will discuss in section 4.1).\n\nWhile it is straightforward to write down Statistical (In)dependence as a mathematical requirement, the physical interpretation of this assumption less clear. One may be tempted to read the probability encoded by ρ as a frequency of occurrence for different combinations of (λ, a, b) that happen in the real world. However, without further information about the theory we are dealing with, we do not know whether any particular combination ever occurs in the real world. E.g., in the case that a pair of entangled particles is labeled by a unique λ, for any value of λ only one pair of values for a and b would actually be realized in the real world.\n\nAt the very least, whether these alternative combinations of hidden variables and detector settings ever exist depends both on the state space of the theory and on whether dynamical evolution is ergodic on this state space. It is easy to think of cases where dynamical evolution is not ergodic with respect to the Lebesgue measure on state space. Take for example a classical, non-linear system, like the iconic Lorenz model . Here, the asymptotic time evolution is constrained to an attractor with fractal measure, of a dimension lower than the full state space. For initial conditions on the attractor, large parts of state space are never realized.\n\nNeither can we interpret ρ as a probability in the Bayesian sense2, for then it would encode the knowledge of agents and thereby require us to first define what “knowledge” and “agents” are. This interpretation, therefore, would bring back the very difficulty we set out to remove, namely that a fundamental theory for the constituents of observers should allow us to derive macroscopic concepts.\n\nWe should not, therefore, interpret Statistical Independence as a statement about properties of the real world, but understand it as a mathematical assumption of the model with which we are dealing. This point was made, implicitly at least, by Bell himself :\n\n“I would insist here on the distinction between analyzing various physical theories, on the one hand, and philosophizing about the unique real world on the other hand. In this matter of causality it is a great inconvenience that the real world is given to us once only. We cannot know what would have happened if something had been different. We cannot repeat an experiment changing just one variable; the hands of the clock will have moved, and the moons of Jupiter. Physical theories are more amenable in this respect. We can calculate the consequences of changing free elements in a theory, be they only initial conditions, and so can explore the causal structure of the theory. I insist that [Bell's Theorem] is primarily an analysis of certain kinds of theory.” (emphasis original)\n\nIn summary, Statistical Independence is not something that can be directly tested by observation or by experiment because it implicitly draws on counterfactual situations, mathematical possibilities that we do not observe and that, depending on one's model or theory, may or may not exist.\n\n4. Locality: Finally, we will assume that the superdeterministic theory respects Continuity of Action (for extensive discussion of the term, see Wharton and Argaman ). Continuity of Action (hereafter CoA) means that to transfer information from one space-time region to another, disjoint, region, the same information has to also be present on any closed (3-dimensional) surface surrounding the first region (see Figure 1). Information, here, refers to quantities that are locally measurable. We make this assumption because both general relativity and quantum field theories respect this criterion.\n\nFIGURE 1",
null,
"Figure 1. Continuity of Action. If information from event 1 can influence event 2, then this information must affect any closed three surface around 1.\n\nAs laid out in Wharton and Argaman , the definition of locality by CoA is not as strong as the locality assumptions entering Bell's theorem. Besides Statistical Independence, the assumptions for Bell's theorem are\n\n1. Output Independence\n\nThis assumption states that the measurement outcome is determined by hidden variables, λ, and that the hidden variables are the origin of statistical correlations between distant measurement outcomes. Formally it says that the distribution for the measurement outcomes xa of detector a does not depend on the distribution of outcomes xb at detector b and vice versa, i.e., ρab(xa, xb|a, b, λ) = ρa(xa|a, b, λ)ρb(xb|a, b, λ).\n\n2. Parameter Independence\n\nParameter independence says that the probability distribution of measurement outcomes at one detector does not depend on the settings of the other detector, i.e., they can be written as ρa(xa|a, b, λ) = ρa(xa|a, λ) and ρb(xa|a, b, λ) = ρb(xb|b, λ).\n\nThese two assumptions together are also known as “Factorization.” The observed violations of Bell's inequality then imply that at least one of the three assumptions necessary to derive the inequality must be violated. Quantum mechanics respects Statistical Independence and Parameter Independence but violates Outcome Independence. Superdeterminism violates Statistical Independence. Bell-type tests cannot tell us which of the two options is correct.\n\nAll three assumptions of Bell's theorem—Statistical Independence, Output Independence, and Parameter Independence—are sometimes collectively called “local realism” or “Bell locality.” However, Bell's local realism has little to do with how the term “locality” is used in general relativity or quantum field theory, which is better captured by CoA. It has therefore been proposed that Bell locality should better be called Bell separability . However, that terminology did not catch on.\n\nThe issue of whether Factorization is a suitable way to encode locality and causality is similar to the issue with interpreting Statistical Independence: It draws on alternative versions of reality that may not ever actually occur. Factorization requires us to ask what the outcome of a measurement at one place would have been given another measurement elsewhere (Outcome Independence) or what the setting of one detector would have been had the other detector's setting been different (Parameter Independence). These are virtual changes, expressed as state-space perturbations. The changes do therefore not necessarily refer to real events happening in space-time. By contrast, Continuity of Action, is a statement about processes that do happen in space-time: its definition does not necessarily invoke counterfactual worlds.\n\nTo make this point in a less mathematical way, imagine Newton clapping his hands and hearing the sound reflected from the walls of his College quad (allowing Newton to estimate the speed of sound). He might have concluded that the reflected sound was caused by his clap either because:\n\n• if he had not clapped, he would not have heard the sound;\n\n• the clapping led to the excitation of acoustic waves in the air, which reflected off the wall and propagated back to vibrate Newton's ear drums, sending electrical signals to his brain.\n\nThe first definition of causality here depends on the existence of counterfactual worlds and with that on the mathematical structure of one's theory of physics. It makes a statement that is impossible to experimentally test.\n\nThe second of these definitions, in contrast, identifies a causal connection between the clap and the cognitive recognition: there is no closed region of space-time surrounding the clap that is not affected by either the acoustic wave or the electrical signal. It is a statement about what actually happens.\n\nThis tension between space-time based notions of causality and the assumptions of Bell's theorem was recently highlighted by a new variant of Bell's theorem for temporal order . While the authors suggest their theorem shows that a quantum theory of gravity (under certain assumptions listed in the paper) must lack causal order, the theorem equivalently says that if one wants the weak-field limit of quantum gravity to have a well-defined causal order, then Statistical Independence must be violated.\n\nIn summary, by relying on Continuity of Action instead of Factorization we avoid having to make statements about non-observable versions of our world.\n\n### 3.1. Retrocausality and Future Input Dependence\n\nThe violation of Statistical Independence, which superdeterministic theories display, implies a correlation between the detector and the prepared state (as defined by λ), typically in space-like separated regions. These regions, however, are contained in a common causal diamond, so there are events in the past which have both regions in their causal future, and there are events in the future which have both regions in the past.\n\nThe possibility that both detector and prepared state are correlated because of a common event in the past is commonly referred to as the “locality loophole” (to Bell's theorem). One can try to address it (to some extent) by choosing the detector settings using events in the far past. An extreme version of this has been presented in Handsteiner et al. where light from distant quasars was used to select detector settings. We have more to say about what this experiment does and does not prove in section 4.3.\n\nThe possibility that detector and prepared state are correlated because of an event in the future is often referred to as “retrocausality” or sometimes as “teleology.” Both of these terms are misleading. The word “retrocausality” suggests information traveling backward in time, but no superdeterministic model has such a feature. In fact, it is not even clear what this would mean in a deterministic theory. Unless one explicitly introduces an arrow of time (eg from entropy increase), in a deterministic theory, the future “causes” the past the same way the present “causes” the future. The word “teleology” is commonly used to mean that a process happens to fulfill a certain purpose which suffices to explain the process. It is misleading here because no one claims that Superdeterminism is explanatory just because it gives rise to what we observe; this would be utterly non-scientific. A superdeterministic theory should of course give rise to predictions that are not simply axioms or postulates.\n\nFor this reason, it was suggested in Wharton and Argaman to use the more scientific expression “Future Input Dependence.” This term highlights that to make predictions with a superdeterministic model one may use input on a spacelike hypersurface to the future of system preparation, instead of using input at the time of preparation. This is possible simply because these two slices are connected by a deterministic law. Relying on “future input” may sound odd, but it merely generalizes the tautologically true fact that to make a prediction for a measurement at a future time, one assumes that one makes a measurement at a future time. That is, we use “future input” every time we make a measurement prediction. It is just that this input does not normally explicitly enter the calculation.\n\nWe wish to emphasize that Future Input Dependence is in the first place an operational property of a model. It concerns the kind of information that one needs to make a prediction. In contrast with the way we are used to dealing with models, Future Input Dependence allows the possibility that this information may arise from a boundary condition on a future hypersurface. Of course one does not actually know the future. However, drawing on future input allows one to make conditional statements, for example of the type “if a measurement of observable O takes place, then …” Here, the future input would be that observable O will be measured in the first place (of course in that case one no longer predicts the measurement of O itself).\n\nNow, in a deterministic theory, one can in principle formulate such future boundary conditions in terms of constraints on an earlier state. The hidden variables at the future hypersurface can be expressed through those at an earlier time, or, more generally, the two sets are correlated. But the constraint on the earlier state may be operationally useless. That is to say, whether or not one allows Future Input Dependence can make the difference between whether or not a model has explanatory power (see section 4.2 for more on that).\n\nFuture Input Dependence is related to Superdeterminism because constraints on a future hypersurface will in general enforce correlations on earlier hypersurfaces; i.e., future input dependent theories generically violate Statistical Independence.\n\nIn summary, superdeterministic models are not necessarily either retrocausal or teleological (and indeed we are not aware of any model that exhibits one of these properties). But superdeterministic models may rely on future input to make conditional predictions.\n\n### 3.2. Disambiguation\n\nThe reader is warned that the word “Superdeterminism” has been used with slightly different meaning elsewhere. In all these meanings, Statistical Independence is violated and the corresponding theory should prohibit action at a distance. But some authors [5, 12] distinguish Superdeterminism from retrocausality (or Future Input Dependence, respectively). Further, not everyone also assumes that a superdeterministic theory is deterministic in the first place. We here assume it is, because this was historically the motivation to consider this option, and because if the theory was not deterministic there would not be much point in considering this option.\n\n## 4. Common Objections to Superdeterminism\n\nIn this section we will address some commonly raised objections to Superdeterminism found in various places in the literature and online discussions.\n\n### 4.1. Free Will and Free Choice\n\nThe Statistical Independence assumption is often referred to as “Free Choice,” because it can be interpreted to imply that the experimenter is free to choose the measurement setting independently of the value of the hidden variables. This has had the effect of anthropomorphizing what is merely a mathematical assumption of a scientific hypothesis. Let us therefore have a look at the relation between Statistical Independence and the physical processes that underlie free choice, or free will more generally.\n\nEver since Hume , the notion of free will has been defined in two different ways:\n\n• as an ability to have done otherwise;\n\n• as an absence of constraints preventing one from doing what one wishes to do.\n\nAs in our previous discussion of causality, these definitions are profoundly different in terms of physical interpretation. An ability to have done otherwise presumes that a hypothetical world where one did do otherwise is a physically meaningful concept. That is to say, the scientific meaningfulness of the notion of “an ability to have done otherwise” depends on the extent to which one's theory of physics supports the notion of counterfactual worlds: as discussed below, theories may vary as to this extent.\n\nThe second definition, by contrast, does not depend on the existence of counterfactual worlds. It is defined entirely in terms of events or processes occurring in spacetime. For example, what one “wishes to do” could be defined in terms of a utility function which the brain attempts to optimize in coming to what we can call a “choice” or “decision.” This second definition is often referred to as the compatibilist definition of free will.\n\nStatistical Independence relies on the first of these definitions of free will because (as discussed above) it draws on the notion of counterfactual worlds. The absence of Statistical Independence does not, however, violate the notion of free will as given by the second definition. We do not usually worry about this distinction because in the theories that we are used to dealing with, counterfactuals typically lie in the state space of the theory. But the distinction becomes relevant for superdeterministic theories which may have constraints on state-space that rule out certain counterfactuals (because otherwise it would imply internal inconsistency). In some superdeterministic models there are just no counterfactuals in state space (for an example, see section 5.2), in some cases counterfactuals are partially constrained (see section 5.1), in others, large parts of state-space have an almost zero measure (5.3).\n\nOne may debate whether it makes sense to speak of free will even in the second case since a deterministic theory implies that the outcome of any action or decision was in principle fixed at the beginning of the universe. But even adding a random element (as in quantum mechanics) does not allow human beings to choose one of several future options, because in this case the only ambiguities about the future evolution (in the measurement process) are entirely unaffected by anything to do with human thought. Clearly, the laws of nature are a constraint that can prevent us from doing what we want to do. To have free will, therefore, requires one to use the compatibilist notion of free will, even if one takes quantum mechanics in its present form as fundamental. Free will is then merely a reflection of the fact that no one can tell in advance what decisions we will make.\n\nBut this issue with finding a notion of free will that is compatible with deterministic laws (or even partly random laws) is not specific to Superdeterminism. It is therefore not an argument that can be raised against Superdeterminism. Literally all existing scientific theories suffer from this conundrum. Besides, it is not good scientific practice to discard a scientific hypothesis simply because one does not like its philosophical implications.\n\nLet us look at a simple example to illustrate why one should not fret about the inability of the experimenter to prepare a state independently of the detector. Suppose you have two fermions. The Pauli exclusion principle tells us that it is not possible to put these two particles into identical states. One could now complain that this violates the experimenter's free will, but that would be silly. The Pauli exclusion principle is a law of nature; it's just how the world is. Violations of Statistical Independence, likewise, merely tell us what states can exist according to the laws of nature. And the laws of nature, of course, constrain what we can possibly do.\n\nIn summary, raising the issue of free will in the context of Superdeterminism is a red herring. Superdeterminism does not make it any more or less difficult to reconcile our intuitive notion of free will with the laws of nature than is the case for the laws we have been dealing with for hundreds of years already.\n\n### 4.2. The Conspiracy Argument\n\nThis argument has been made in a variety of ways, oftentimes polemically. Its most rigorous version can be summarized as follows. In any deterministic theory one can take a measurement outcome and, by using the law of time-evolution, calculate the initial state that would have given rise to this outcome. One can then postulate that since this initial state gave rise to the observation, we have somehow “explained” the observation. If one were to accept this as a valid argument, this would seemingly invalidate the science method in general. For then, whenever we observe any kind of regularity—say a correlation between X-ray exposure and cancer—we could say it can be explained simply because the initial state happened to be what it was.\n\nThe more polemic version of this is that in a superdeterministic theory, the universe must have been “just so” in order that the decisions of experimenters happen to reproduce the predictions of quantum mechanics every single time. Here, the term “just so” is invoked to emphasize that this seems intuitively extremely unlikely and therefore Superdeterminism relies on an implausible “conspiracy” of initial conditions that does not actually explain anything.\n\nTo address this objection, let us first define “scientific explanation” concretely to mean that the theory allows one to calculate measurement outcomes in a way that is computationally simpler than just collecting the data. This notion of “scientific explanation” may be too maths-centric to carry over to other disciplines, but will serve well for physics. The criticism leveled at Superdeterminism is, then, that if one were to accept explaining an observation merely by pointing out that an initial state and a deterministic law exists, then one would have to put all the information about the observation already in the initial state, meaning the theory is not capable of providing a scientific explanation in the above defined sense.\n\nOne problem with this argument is that just by knowing a theory violates Statistical Independence one cannot tell anything about its explanatory power. For this one needs to study a concrete model. One needs to know how much information one has to put into the initial state and the evolution law to find out whether a theory is or is not predictive.\n\nLet us look at a specific example from Bell himself . Bell himself realized that free will was a red herring (see section 4.1) and for that reason his arguments against Superdeterminism are framed in a completely deterministic setting. He imagines that the measurement setting (a = 0 or a = 1) is determined by a pseudo-random number generator whose output is exquisitely sensitive to its input x in the sense that the setting depends on the parity of the millionth digit in the decimal expansion of x. Bell concludes that whilst the millionth digit indeed determines the measurement settings, it seems implausible to imagine that it systematically influences, or is systematically influenced by, anything else in the universe—the particle's hidden variables in particular.\n\nOf course “it seems implausible” is not a convincing argument, as Bell himself conceded, writing :\n\nOf course it might be that these reasonable ideas about physical randomizers are just wrong—for the purpose at hand. A theory may appear in which such conspiracies inevitably occur, and these conspiracies may then seem more digestible than the non-localities of other theories. When that theory is announced I will not refuse to listen, either on methodological or other grounds.\n\nBut Bell's intuition rests on the assumption that because worlds which differ only in the millionth digits of the random numbers are very similar to each other, they are necessarily “close” to each other. Such statements therefore implicitly depend on the notion of a distance in state-space. We intuitively tend to assume distance measures are Euclidean, but this does not need to be so in state-space.\n\nSuch conspiracy arguments are also often phrased as worries about the need to “fine-tune”—i.e., choose very precisely—the initial conditions (see Wood and Spekkens for a quantifiable definition). The reference to fine-tuning, however, is misleading. There need be nothing a priori unscientific about a fine-tuned theory . A fine-tuned theory may be unscientific if one needs to put a lot of information into the initial condition thereby losing explanatory power. But this does not necessarily have to be the case. In fact, according to currently accepted terminology both the standard model of particle physics and the concordance model of cosmology are “fine-tuned” despite arguably being scientifically useful.\n\nOne way to avoid that fine-tuning leads to a lack of explanatory power is to find a measure that can be defined in simple terms and that explains which states are “close” to each other and/or which are distant and have measure zero, i.e., are just forbidden (see Almada et al. for an example of how this negates the problem of Wood and Spekkens ).\n\nBell's and similar examples that rest on arguments from fine-tuning (or sensitivity, or conspiracy) all implicitly assume that there is no simple way to mathematically express the allowed (or likely) initial states that give rise to the predictions of quantum mechanics. See also section 7 for further discussion on the notion of “closeness” in state-space and section 5.1 for an example of a theory where intuitive Euclidean ideas about closeness of worlds fail.\n\nBut assuming that something is impossible does not prove that it is impossible. Indeed, it is provable that it is unprovable to show such theories are unscientific because that is just a rephrasement of Chaitin's incompleteness theorem . This theorem, in a nutshell, says that one can never tell that there is no way to further reduce the complexity of a string (of numbers). If we interpret the string as encoding the initial condition, this tells us that we cannot ever know that there is not some way to write down an initial state in a simpler way.\n\nThis is not to say that we can rest by concluding that we will never know that a useless theory cannot be made more useful. Of course, to be considered scientifically viable (not to mention interesting) a superdeterministic theory must actually have an explanatory formulation. We merely want to emphasize that the question whether the theory is scientific cannot be decided merely by pointing out that it violates Statistical Independence.\n\n### 4.3. The Cosmic Bell Test and the BIG Bell Test\n\nIn the Cosmic Bell Test , measurement settings are determined by the precise wavelength of light from distant quasars, sources which were causally disconnected at the time the photons were emitted. It is without doubt a remarkable experimental feat, but this (and similar) experiments do not—cannot—rule out Superdeterminism; they merely rule out that the observed correlations in Bell-type tests were locally caused by events in the distant past. It is, however, clear from the derivation of Bell's theorem that violations of Bell's inequality cannot tell us whether Statistical Independence was violated. Violations of Bell's inequality can only tell us that at least one of the assumptions of the theorem was violated.\n\nThe belief that such tests tell us something about (the implausibility of) Superdeterminism goes back, once again, to the idea that a state which is intuitively “close” to the one realized in nature (eg, the wavelength of the light from the distant quasar was a little different, all else equal) is allowed by the laws of nature and likely to happen. However, in a superdeterministic theory what seems intuitively like a small change will generically result in an extremely unlikely state; that's the whole point. For example, in a superdeterministic theory, a physically possible counterfactual state in which the wave-length of the photon was slightly different may also require changes elsewhere on the past hypersurface, thereby resulting in the experimenter's decision to not use the quasar's light to begin with.\n\nSimilar considerations apply to all other Bell-type tests [19, 20] that attempt to close the freedom-of-choice loophole, like the BIG Bell test . This experiment used input from 100,000 human participants playing a video game to choose detector settings, thereby purportedly “closing the “freedom-of-choice loophole” (the possibility that the setting choices are influenced by “hidden variables” to correlate with the particle properties)”. Needless to say, the experiment shows nothing of that type; one cannot prove freedom of choice by assuming freedom of choice.\n\nIn fact, the details of these experiments do not matter all that much. One merely has to note that measuring violations of Bell's inequality, no matter how entertaining the experimental setup, cannot tell us which of the assumptions to the theorem were violated.\n\n### 4.4. The Tobacco Company Syndrome\n\nFinally, let us turn to the claim that the assumption of Statistical Independence in Bell's theorem can be justified by what it would imply in classical physics. This argument is frequently put forward with the example of using a randomized trial to demonstrate that lung cancer is linked to smoking. If one were to allow violations of Statistical Independence in Bell-type experiments, so the argument goes, tobacco companies could claim that any observed correlation between lung cancer and smoking was due to a correlation between the randomization and the measured variable (i.e., the incidence of cancer). We do not know where this argument originated, but here are two examples:\n\n“It is like a shill for the tobacco industry first saying that smoking does not cause cancer, rather there is a common cause that both predisposes one to want to smoke and also predisposes one to get cancer (this is already pretty desperate), but then when confronted with randomized experiments on mice, where the mice did not choose whether or not to smoke, going on to say that the coin flips (or whatever) somehow always put the mice already disposed to get lung cancer into the experimental group and those not disposed into the control. This is completely and totally unscientific, and it is an embarrassment that any scientists would take such a claim seriously.”—Tim Maudlin \n\n“I think this assumption [of Statistical Independence] is necessary to even do science, because if it were not possible to probe a physical system independently of its state, we couldn't hope to be able to learn what its actual state is. It would be like trying to find a correlation between smoking and cancer when your sample of patients is chosen by a tobacco company.”—Mateus Araújo \n\nOne mistake in the argument against Superdetermism is the claim that theories without the assumption of Statistical Independence are unscientific because they are necessarily void of explanatory power. We already addressed this in subsection 4.2. However, the tobacco company analogy brings in a second mistake, which is the idea that we can infer from the observation that Statistical Independence is useful to understand the properties of classical systems, that it must also hold for quantum systems. This inference is clearly unjustified; the whole reason we are having this discussion is that classical physics is not sufficient to describe the systems we are considering.\n\nWe have already mentioned an example of how our classical intuition can fail in the quantum case. This example provides a further illustration. For the tobacco trial, we have no reason to think that multiple realizations of the randomization are impossible. For example, two different randomly drawn sub-ensembles of volunteers (say the first drawn in January, the second in February) can be expected to be statistically equivalent. It is only when our theoretical interpretation of an experiment requires us to consider counterfactual worlds, that differences between classical and quantum theories can emerge.\n\nIt is further important to note that the assumption of Statistical Independence does not require ensembles of different, actually occurring experiments (as opposed to virtual experiments that only appear in the mathematics) to have different statistical properties. Consider two ensembles of quantum particles, each measured with different measurement settings (say the first in January, the second in February). Since there is no reference to counterfactuals in this description, we cannot infer that the statistical properties of the hidden variables are any different in the January and February ensembles, even in a theory of quantum physics which violates Statistical Independence. At this level, therefore, there is no difference between the quantum and classical example. In a theory that violates Statistical Independence, one merely cannot infer that if February's ensemble of particles had been measured with January's measurement settings, the result would have been statistically identical. By contrast, if February's volunteers had been tested in January, we would, by classical theory, have expected statistically identical results. In this sense, the tobacco trial analogy is misleading because it raises the impression that the assumption of Statistical Independence is more outlandish than it really is.\n\n## 5. How?\n\nThe history of Superdeterminism is quickly told because the topic never received much attention. Already Bell realized that if one observes violations of his inequality, this does not rule out local3, deterministic hidden variable models because Statistical Independence may be violated . It was later shown by Brans that if Statistical Independence is violated, any Bell-nonlocal distribution of measurement outcomes can be obtained in EPR-type experiments . It has since been repeatedly demonstrated that it requires only minute violations of Statistical Independence to reproduce the predictions of quantum mechanics locally and deterministically .\n\nThe scientific literature contains a number of toy models that provide explicit examples for how such violations of Statistical Independence can reproduce quantum mechanics [24, 28, 29] which have been reviewed in Hall (section 4.2). Toy models which violate Statistical Independence through future input dependence have recently been surveyed in Wharton and Argaman (section 6). We will here not go through these toy models again, but instead briefly introduce existing approaches to an underlying theory that give rise to Superdeterminism.\n\nThese approaches, needless to say, are still in their infancy. They leave open many questions and it might well turn out that none of them is the right answer. We do believe, however, that they present a first step on the way toward a satisfactory solution of the measurement problem.\n\n### 5.1. Invariant Set Theory\n\nInvariant Set Theory (IST) [34, 35] arose from an earlier realization that, suitably formulated, non-linear dynamics could provide the basis for a deterministic theory of quantum physics which was not counterfactually complete and therefore could violate Statistical Independence thus avoiding non-locality. More specifically, IST is a deterministic theory based on the assumption that the laws of physics at their most primitive derive from the geometry of a fractal set of trajectories, or histories, IU, in state space. States of physical reality—the space-time that comprises our universe and the processes which occur in space-time—are those and only those belonging to IU; other states in the Euclidean space in which IU is embedded, do not correspond to states of physical reality. Dynamical evolution maps points on IU to other points on IU, whence IU is invariant under dynamical laws of evolution. In this theory, the basic element of IU is a fractal helix (in the sense that each trajectory in the helix, like a strand of rope, is itself a helix of finer-scale trajectories).\n\nThe link to quantum mechanics is made through the statistical properties of the helices which can be represented by complex Hilbert vectors and tensor products, where squared amplitudes and complex phases of the Hilbert vectors are necessarily described by rational numbers.\n\nIST provides some possible understanding of a key difference between the Liouville equation and the von Neumann-Dirac equation: the factor iℏ. Planck's constant has the dimension of state space (momentum times position) and hence provides an inherent size to any geometric structure in state space, such as IU. This inherent size is provided by the radius of a helix of IU. Based on the U(1) ~ SO(2) isomorphism, the square root of minus one is consistent with a rotational symmetry of the helical nature of the trajectories of IU.\n\nSince it is formulated in terms of complex Hilbert states, IST violates Bell inequalities exactly as does quantum theory. It does this not only because Statistical Independence is violated (the fractal gaps in IU correspond to states of the world associated with certain counterfactual measurement settings, which by construction are not ontic), it also violates the Factorization assumption of Bell's theorem and hence is Bell-nonlocal. Because the set of helices has fractal structure, the p-adic metric, rather than Euclidean metric is a natural measure of distance in state space.\n\nImportantly, violation of Statistical Independence and Factorization only occur when one considers points which do not lie on IU. From a Hilbert state perspective, they are associated with Hilbert States where either squared amplitudes or complex phases of Hilbert States cannot be described by rational numbers. Hence, the violations of Statistical Independence and Factorization in IST arise because certain putative counterfactual states are mathematically undefined; without these violations there would be mathematical inconsistency. Importantly, such counterfactual states do not correspond to physically possible processes in space-time. If Statistical Independence and Factorization are weakened to only allow processes which are expressible in space time and hence are physically possible (“Statistical Independence on IU” and “Factorization on IU”), then IST is consistent with both free choice and locality.\n\nIn IST, the measurement process is described by state-space trajectories that cluster together near detector eigenstates. In this approach, the measurement problem has been largely nullified because the statistical state space of the trajectory segments that lead to those detector eigenstates is no longer the whole Hilbert space, but instead the space whose elements have finite squared amplitudes and complex phases. In this sense, IST does not “complete” quantum theory. Rather, it is a replacement for quantum theory, even at the pre-measurement unitary stage of evolution.\n\nThe fractal attractor which defines IU can be considered a future asymptotic property of some more classical like governing differential equations of motion: start from any point in state space and the trajectory will converge onto it only as t → ∞. The invariant set is therefore operationally incomputable in much the same way that the event horizon of a black hole is.\n\n### 5.2. Cellular Automata\n\nThe Cellular Automata approach to Superdeterminism is a model that employs a time-evolution which proceeds in discrete time-steps on a grid. It uses a language similar to quantum mechanics, in that the state-space is spanned by vectors in a Hilbert-space. These vectors can, as usual, be brought into superpositions. However, it is then postulated that states which result in superpositions that we do not observe are not ontic. It follows from this that an initial state which gave rise to an unobserved outcome was not ontic either. A variety of simple toy models have been discussed in 't Hooft .\n\nIn this approach there is strictly speaking only one ontic state in the theory, which is the state that the universe is in. The requirement that the final state must correspond to the classical reality which we observe induces constraints at earlier times. These constraints give rise to non-local correlations which result in a violation of Statistical Independence.\n\nThe challenge for this approach is to render this theory predictive. As was noted in 't Hooft , selecting the ontological state requires a measure for the initial states of the universe:\n\n“Bell's theorem requires more hidden assumptions than usually thought: The quantum theory only contradicts the classical one if we assume that the ‘counterfactual modification' does not violate the laws of thermodynamics. In our models, we must assume that it does.” (emphasis original)\n\nIt is presently unclear from where such a thermodynamic-like measure comes.\n\n### 5.3. Future-Bounded Path Integrals\n\nThe path integral approach to Superdeterminism rests on the observation that the Feynman path integral has a future input dependence already, which is the upper time of the integration. However, in the usual path integral of quantum mechanics (and, likewise, of quantum field theory), one does not evaluate what is the optimal future state that the system can evolve into. Instead, one posits that all of the future states are realized, which results in a merely probabilistic prediction.\n\nThe idea is then to take a modified path integral for the combined system of detector and prepared state and posit that in the underlying theory the combined system evolves along merely one possible path in state space that optimizes a suitable, to-be-defined, function. This function must have the property that initial states which evolve into final states containing superpositions of detector eigenstate states are disfavored, in the sense that they do not optimize the function. Instead, the optimal path that the system will chose is one that ends up in states which are macroscopically classical. One gets back normal quantum mechanics by averaging over initial states of the detector.\n\nThis approach solves the measurement problem because the system does deterministically evolve into one particular measurement outcome. Exactly which outcome is determined by the degrees of freedom of the detector that serve as the “hidden variables.” Since it is generically impossible to exactly know all the detector's degrees of freedom, quantum mechanics can only make probabilistic predictions.\n\nThe challenge of this approach is to find a suitable function that actually has this behavior.\n\n## 6. Experimental Test\n\nIt is clear that the above discussed theoretical approaches to Superdeterminism require more work. However, such theories have general properties that, with some mild assumptions, tell us what type of experiment has the potential to reveal deviations from quantum mechanics.\n\nTo see this, we first note that typical experiments in the foundations of quantum mechanics probe physics at low energies, usually in the range of atomic physics. It is, however, difficult to come up with any model that equips known particles with new degrees of freedom accessible at such low energies. The reason is that such degrees of freedom would change the phase-space of standard model particles. Had they been accessible with any experiment done so far, we would have seen deviations from the predictions of the standard model, which has not happened.\n\nIt is well possible to equip standard model particles with new degrees of freedom if those are only resolvable at high energies (examples abound). But in this case the new degrees of freedom do not help us with solving the measurement problem exactly because we assumed that they do not play a role at the relevant energies.\n\nIf one does not want to give up on this separation of scales, this leaves the possibility that the hidden variables are already known degrees of freedom of particles which do not comprise the prepared state. Moreover, they are only those degrees of freedom that are resolvable at the energy scales under consideration.\n\nThe next thing we note is that all presently known deterministic, local theories have the property that states that were close together at an initial time will remain close for some while. In a superdeterministic theory, states with different measurement settings are distant in state-space, but changes to the hidden variables that do not also change the measurement setting merely result in different measurement outcomes and therefore correspond to states close to each other.\n\nSince the theory is deterministic, this tells us that if we manage to create a time-sequence of initial states similar to each other, then the measurement outcomes should also be similar. This means concretely that rather than fulfilling the Born-rule, such an experiment would reveal time-correlations in the measurement outcomes. The easiest way to understand this is to keep in mind that if we were able to exactly reproduce the initial state, then in a superdeterministic theory the measurement outcome would have to be the same each time, in conflict with the predictions of quantum mechanics.\n\nThis raises the question how similar the initial states have to be for this to be observable. Unfortunately, this is not a question which can be answered in generality; for this one would need a theory to make the corresponding calculation. However, keeping in mind that the simplest case of hidden variables are the degrees of freedom of other particles and that the theory is local in the way we are used to it, the obvious thing to try is minimizing changes of the degrees of freedom of the detecting device. Of course one cannot entirely freeze a detector's degrees of freedom, for then it could no longer detect something. But one can at least try to prevent non-essential changes, i.e., reduce noise.\n\nThis means concretely that one should make measurements on states prepared as identically as possible with devices as small and cool as possible in time-increments as small as possible.\n\nThis consideration does not change much if one believes the hidden variables are properties of the particle after all. In this case, however, the problem is that preparing almost identical initial states is impossible since we do not know how to reproduce the particle's hidden variables. One can then try to make repeated measurements of non-commuting observables on the same states, as previously laid out in Hossenfelder .\n\nThe distinction between the predictions of quantum mechanics and the predictions of the underlying, superdeterministic theory is not unlike the distinction between climate predictions and weather forecasts. So far, with quantum mechanics, we have made predictions for long-term averages. But even though we are in both cases dealing with a non-linear and partly chaotic system, we can in addition also make short-term predictions, although with limited accuracy. The experiment proposed here amounts to recording short-term trends and examining the data for regularities that, according to quantum mechanics alone, should not exist.\n\nNeedless to say, the obvious solution may not be the right one and testing Superdeterminism may be more complicated than that. But it seems reasonable to start with the simplest and most general possibility before turning to model-specific predictions.\n\n## 7. Discussion\n\nThe reader may have noticed a running theme in our discussion of Superdeterminism, which is that objections raised against it are deeply rooted in intuition that is, ultimately, based on the classical physics we experience with our own senses.\n\nBut these intuitions can mislead us. For an illustration, consider Penrose's impossible triangle (see Figure 2, bottom). If we see a two-dimensional drawing of the triangle, we implicitly assume that any two arms come closer as they approach a vertex. This raises the impression that the object is impossible to realize in 3-dimensional space. However, the supposedly impossible triangle can be built in reality. The object shown in Figure 2, top, seen from the right direction, reproduces what is shown in the 2-dimensional drawing. From any other direction, however, it becomes clear that our intuition has led us to improperly assume two arms necessarily become close as they approach a common vertex.\n\nFIGURE 2",
null,
"Figure 2. Penrose's “impossible” triangle, placed in front of a mirror, turns out to be not so impossible.\n\nWe believe that the uneasiness we bring to considering Superdeterminism stems from a similar intuitive, but ultimately wrong, idea of closeness. In this case, however, we are not talking about closeness in position space but about closeness in the state-space of a theory.\n\nFaced with trying to quantify the “distance” between two possible states of the universe our intuition is to assume that it can be measured in state space by the same Euclidean metric we use to measure distance in physical space. This indeed is the basis of Lewis's celebrated theory of causality by counterfactuals: of two possible counterfactual worlds, the one that resembles reality more closely is presumed closer to reality . But is this really so? In number theory there is an alternative to the class of Euclidean metrics (and indeed according to Ostrowsky's theorem it is the only alternative): the p-adic metric . The p-adic metric is to fractal geometry as the Euclidean metric is to Euclidean geometry. The details do not need to concern us here, let us merely note that two points that are close according to the Euclidean metric may be far away according to the p-adic metric.\n\nThis means from the perspective of the p-adic metric, the distance between the actual world where the parity of the millionth digit of the input to Bell's pseudo-random number generator was, say, 0, and the counterfactual world where the parity was a 1 could be very large, even though it is small using an Euclidean measure of distance. A theory that seems fine-tuned with respect to the latter metric would not be fine-tuned with respect to the former metric. Like with Penrose's triangle, the seemingly impossible becomes understandable if we are prepared to modify our intuition about distance.\n\nBut our intention here was not merely to draw attention to how classical intuition may have prevented us from solving the measurement problem. Resolving the measurement problem with Superdeterminism may open the door to solving further problems in the foundations of physics. As has been previously noted , our failure to find a consistent quantum theory of gravity may be due, not to our lacking understanding of gravity, but to our lacking understanding of quantization. The same problem may be behind some of the puzzles raised by the cosmological constant. It is further a long-standing conjecture that dark matter is not a new type of particle but instead due to a modification of gravity. We know from observations that such a modification of gravity is parametrically linked to dark energy . The reasons for this connection are currently not well-understood, but completing quantum mechanics, or replacing it with a more fundamental theory, might well be the key to solving these problems.\n\nFinally, let us point out that the technological applications of quantum theory become more numerous by the day. Should we discover that quantum theory is not fundamentally random, should we succeed in developing a theory that makes predictions beyond the probabilistic predictions of quantum mechanics, this would likely also result in technological breakthroughs.\n\n## 8. Conclusion\n\nWe have argued here that quantum mechanics is an incomplete theory and completing it, or replacing it with a more fundamental theory, will necessarily require us to accept violations of Statistical Independence, an assumption that is sometimes also, misleadingly, referred to as Free Choice. We have explained why objections to theories with this property, commonly known as superdeterministic, are ill-founded.\n\nSince the middle of the past century, progress in the foundations of physics has been driven by going to shorter and shorter distances, or higher and higher energies, respectively. But the next step forward might be in an entirely different direction, it might come from finding a theory that does not require us to hand-draw a line between microscopic and macroscopic reality.\n\n## Data Availability Statement\n\nThe datasets generated for this study are available on request to the corresponding author.\n\n## Author Contributions\n\nAll authors listed have made a substantial, direct and intellectual contribution to the work, and approved it for publication.\n\n## Conflict of Interest\n\nThe authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.\n\n## Acknowledgments\n\nSH gratefully acknowledges support from the Franklin Fetzer Fund. TP gratefully acknowledges support from a Royal Society Research Professorship. This manuscript has been released as a pre-print at arXiv.org .\n\n## Footnotes\n\n1. ^One may, however, expect spontaneous collapse models to appear as effective descriptions of a non-linear collapse process in suitable limits.\n\n2. ^As is the idea behind QBism .\n\n3. ^In the sense of Continuity of Action, see section 3.\n\n## References\n\n1. Bell JS. On the Einstein-Podolsky-Rosen paradox. Physics (1964) 1:195.\n\n2. Gallicchio J, Friedman AS, Kaiser DI. Testing Bell's inequality with cosmic photons: closing the setting-independence loophole. Phys Rev Lett. (2014) 112:110405. doi: 10.1103/PhysRevLett.112.110405\n\n3. Pusey MF, Barrett J, Rudolph T. On the reality of the quantum state. Nat Phys. (2012) 8:476. doi: 10.1038/nphys2309\n\n4. Leifer MS. Is the quantum state real? A review of ψ-ontology theorems. Quanta (2014) 3:67–155. doi: 10.12743/quanta.v3i1.22\n\n5. Wharton KB, Argaman N. Bell's theorem and spacetime-based reformulations of quantum mechanics. arXiv [preprint] (2019). 1906.04313.\n\n6. Clauser JF, Horne MA, Shimony A, Holt RA. Proposed experiment to test local hidden-variable theories. Phys Rev Lett. (1969) 23:880–4.\n\n7. Lorenz EN. Deterministic nonperiodic flow. J Atmos Sci. (1963) 20:130–41.\n\n8. Fuchs CA, Schack R. Quantum-Bayesian coherence. arXiv [preprint] (2019). 0906.2187.\n\n9. Hall MJW. The significance of measurement independence for Bell inequalities and locality. In: Asselmeyer-Maluga T, editor. At the Frontier of Spacetime. Cham: Springer (2016). p. 189–204.\n\n10. Zych M, Costa F, Pikovski I, Brukner Č. Bell's theorem for temporal order. Nat Commun. (2019) 10:3772. doi: 10.1038/s41467-019-11579-x\n\n11. Handsteiner J, Friedman AS, Rauch D, Gallicchio J, Liu B, Hosp H, et al. Cosmic Bell test: measurement settings from milky way stars. Phys Rev Lett. (2017) 118:060401. doi: 10.1103/PhysRevLett.118.060401\n\n12. Price H, Wharton K. A live alternative to quantum spooks. arXiv [preprint] (2015). 1510.06712.\n\n13. Hume D. A Treatise of Human Nature. Oxford; New York, NY: Oxford University Press (2000).\n\n14. Bell JS. Speakable and Unspeakable in Quantum Mechanics. Collected Papers on Quantum Philosophy. Cambridge: Cambridge University Press (1987).\n\n15. Wood CJ, Spekkens RW. The lesson of causal discovery algorithms for quantum correlations: causal explanations of Bell-inequality violations require fine-tuning. New J Phys. (2015) 17:033002. doi: 10.1088/1367-2630/17/3/033002\n\n16. Hossenfelder S. Screams for explanation: finetuning and naturalness in the foundations of physics. Synthese. (2019) 1–9. doi: 10.1007/s11229-019-02377-5\n\n17. Almada D, Ch'ng K, Kintner S, Morrison B, Wharton KB. Are retrocausal accounts of entanglement unnaturally fine-tuned? Int J Quantum Found. (2016) 2:1–16.\n\n18. Chaitin G. The Limits of Mathematics. London, UK: Springer (1994).\n\n19. Leung C, Brown A, Nguyen H, Friedman AS, Kaiser DI, Gallicchio J. Astronomical random numbers for quantum foundations experiments. Phys Rev A. (2018) 97:042120. doi: 10.1103/PhysRevA.97.042120\n\n20. Rauch D, Handsteiner J, Hochrainer A, Gallicchio J, Friedman AS, Leung C, et al. Cosmic Bell test using random measurement settings from high-redshift quasars. Phys Rev Lett. (2018) 121:080403. doi: 10.1103/PhysRevLett.121.080403\n\n21. The BIG Bell Test Collaboration. Challenging local realism with human choices. Nature. (2018) 557:212–6. doi: 10.1038/s41586-018-0085-3\n\n22. Maudlin T. Comment on Electrons Don't Think. Blogentry on “BackRe(Action)” (2019). Available online at: https://bit.ly/33InZZD (accessed December 1, 2019).\n\n23. Araújo M. Understanding Bell's Theorem Part 1: The Simple Version. Blogentry on “More Quantum” (2016). Available online at: https://bit.ly/2rGjyky (accessed December 1, 2019).\n\n24. Brans CH. Bell's theorem does not eliminate fully causal hidden variables. Int J Theor Phys. (1988) 27:219.\n\n25. Barrett J, Gisin N. How much measurement independence is needed in order to demonstrate nonlocality? Phys Rev Lett. (2011) 106:100406. doi: 10.1103/PhysRevLett.106.100406\n\n26. Hall MJW. Relaxed Bell inequalities and Kochen-Specker theorems. Phys Rev A. (2011) 84:022102. doi: 10.1103/PhysRevA.84.022102\n\n27. Friedman AS, Guth AH, Hall MJW, Kaiser DI, Gallicchio J. Relaxed Bell inequalities with arbitrary measurement dependence for each observer. Phys Rev A. (2019) 99:012121. doi: 10.1103/PhysRevA.99.012121\n\n28. Degorre J, Laplante S, Roland J. Simulating quantum correlations as a distributed sampling problem. Phys Rev A. (2005) 72:062314. doi: 10.1103/PhysRevA.72.062314\n\n29. Hall MJW. Local deterministic model of singlet state correlations based on relaxing measurement independence. Phys Rev Lett. (2010) 105:250404. doi: 10.1103/PhysRevLett.105.250404\n\n30. Argaman N. Bell's theorem and the causal arrow of time. Am J Phys. (2010) 78:1007–13. doi: 10.1119/1.3456564\n\n31. Di Lorenzo A. A simple model for the spin singlet: mathematical equivalence of non-locality, slave will and conspiracy. J Phys A. (2012) 45:265302. doi: 10.1088/1751-8113/45/26/265302\n\n32. Argaman N. A lenient causal arrow of time? Entropy. (2018) 20:294. doi: 10.3390/e20040294\n\n33. Wharton KB. A new class of retrocausal models. Entropy. (2018) 20:410. doi: 10.3390/e20060410\n\n34. Palmer TN. The invariant set postulate: a new geometric framework for the foundations of quantum theory and the role played by gravity. Proc R Soc A. (2009) 465. doi: 10.1098/rspa.2009.0080\n\n35. Palmer TN. Discretization of the bloch sphere, fractal invariant sets and Bell's theorem. Proc. R Soc A. (2020) 476:20190350. doi: 10.1098/rspa.2019.0350\n\n36. Palmer TN. A local deterministic model of quantum spin measurement. Proc R Soc A. (1995) 451:585–608.\n\n37. 't Hooft G. The Cellular Automaton Interpretation of Quantum Mechanics. Cham: Springer (2016). doi: 10.1007/978-3-319-41285-6\n\n38. Donadi S, Hossenfelder S. In preparation.\n\n39. Hossenfelder S. Testing super-deterministic hidden variables theories. Found Phys. (2011) 41:1521. doi: 10.1007/s10701-011-9565-0\n\n40. Lewis D. Counterfactuals. Oxford: Blackwell (1973).\n\n41. Katok S. p-adic Analysis Compared With Real. Student Mathematical Library (Book 37), Providence, RI: American Mathematical Society (2007).\n\n42. Hossenfelder S. A possibility to solve the problems with quantizing gravity. Phys Lett B. (2013) 725:473. doi: 10.1016/j.physletb.2013.07.037\n\n43. Lelli F, McGaugh SS, Schombert JM, Pawlowski MS. One law to rule them all: the radial acceleration relation of galaxies. Astrophys J. (2017) 836:152. doi: 10.3847/1538-4357/836/2/152\n\n44. Hossenfelder S, Palmer TN. Rethinking superdeterminism. arXiv. 1912.06462.\n\nKeywords: superdeterminism, Bell theorem, causality, free will, quantum measurement, quantum mechanics\n\nCitation: Hossenfelder S and Palmer T (2020) Rethinking Superdeterminism. Front. Phys. 8:139. doi: 10.3389/fphy.2020.00139\n\nReceived: 26 January 2020; Accepted: 08 April 2020;\nPublished: 06 May 2020.\n\nEdited by:\n\nKarl Hess, University of Illinois at Urbana-Champaign, United States\n\nReviewed by:\n\nBabak Shiri, Neijiang Normal University, China\nGerard't Hooft, Utrecht University, Netherlands\n\nCopyright © 2020 Hossenfelder and Palmer. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.\n\n*Correspondence: Tim Palmer, tim.palmer@physics.ox.ac.uk"
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http://documentation.statsoft.com/STATISTICAHelp.aspx?path=power/PowerAnalysis/Dialogs/TwoCorrelations/TwoCorrelationsSampleSizeCalculationResultsQuickTab
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"",
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"Two Correlations: Sample Size Calculation Results - Quick Tab\n\nCalculating Required Sample Size\n\nSelect the Quick tab of the Two Correlations: Sample Size Calculation Results dialog box to access options to interactively view the results of sample size calculations and generate plots of required sample size as a function of quantities that affect sample size in practice.\n\nX-Axis Graphing Parameters. The entries in the boxes under X-Axis Graphing Parameters establish the start value, end value, and number of steps on the X-axis for the graphs specified under Sample Size Charts.\n\nStart Rho1. In the Start Rho1 box, enter the starting value of rho1 used in the N vs. Rho1 sample size chart.\n\nEnd Rho1. In the End Rho1 box, enter the final value of rho1 used in the N vs. Rho1 sample size chart.\n\nStart Alpha. In the Start Alpha box, enter the starting value of alpha used in the N vs. Alpha sample size chart.\n\nEnd Alpha. In the End Alpha box, enter the final value of alpha used in the N vs. Alpha sample size chart.\n\nStart Power. In the Start Power box, enter the starting value of power goal used in the N vs. Power sample size chart.\n\nEnd Power. In the End Power box, enter the final value of power goal used in the N vs. Power sample size chart.\n\nNo. of Steps. In the No. of Steps box, enter the number of steps, or intervals, to plot on the X-axis in the charts specified in the Sample Size Charts group box.\n\nComputational Algorithm. The Power Analysis module allows you to choose from two algorithms:\n\nFisher Z Refined. Select the Fisher Z Refined option button to employ an approximation based on the Fisher transformation, but using refined series approximation formulae for the mean and variance (Fouladi, 1991).\n\nFisher Z Crude. Select the Fisher Z Crude option button to use the traditional (somewhat crude) approximation, based on the Fisher transformation, assuming that the mean of the Fisher transform of r is the Fisher transform of r, and that the variance of the Fisher transform of r is 1/(N - 3).\n\nN to Vary. Use the options under N to Vary to specify which sample size (if any) to vary when computing sample sizes and plotting charts. STATISTICA allows you to compute and plot sample size in two types of situations: (1) where sample sizes are assumed to be equal, and (2) where one sample size is varied while the other is held constant.\n\nEqual N. Choose the Equal N option button if sample sizes are assumed to be equal.\n\nVary N1. Choose the Vary N1 option button to vary the degrees of freedom for the first sample, while holding degrees of freedom constant (at N2=100) for the second sample.\n\nVary N2. Choose the Vary N2 option button to vary the degrees of freedom for the second sample, while holding degrees of freedom constant (at N1=100) for the first sample.\n\nSample Size Charts. Use the buttons under Sample Size Charts to produce charts of required N as a function of power goal, population correlation, or type I error rate. Note that parameters given in the X-Axis Graphing Parameters box control the range of values plotted on the X-axis and the number of steps plotted.\n\nN vs. Rho1. Click the N vs. Rho1 button to produce a plot of sample size (N) versus the group 1 population correlation. The population correlation for the second group is held constant at the current fixed parameter value.\n\nN vs. Alpha. Click the N vs. Alpha button to produce a plot of sample size (N) vs. type I error rate.\n\nN vs. Power. Click the N vs. Power button to produce a plot of sample size (N) versus power goal."
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https://www.rpgpgm.com/2020/12/how-to-manipulate-value-returned-from.html
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[
"## Tuesday, December 15, 2020\n\n### How to manipulate the value returned from a RPG procedure",
null,
"Sometimes it can be hard in testing to change values returned from procedures. I don't want to change the code in the procedure. But how can I ensure that the returned value is what I want? In debug I can change the value of a variable, and this will change the change value returned if the variable is returned from the procedure. What can I do if returned value is hard coded? I cannot change that.\n\n ``` return *on ; ```\n\nFortunately this is no longer an issue due to new enhancement added to RPG as part of the latest Technology Refreshes, IBM i 7.4 TR3 and 7.3 TR9.\n\nThere is a new control option value that can be inserted into the DEBUG keyword.\n\n ``` ctl-opt DEBUG(*RETVAL) dftactgrp(*no) ; ```\n\nThe value *RETVAL allows me to change the value contained in a special value: _QRNU_RETVAL.\n\nThe only place I can change this is in debug, and only when the current break point is on the END-PROC operation code of the procedure.\n\nLet me jump into examples so you can understand how to use this new feature.\n\nIn the first example I have a procedure that receives a single variable and returns the equivalent:\n\n ```01 **free 02 ctl-opt debug(*retval) dftactgrp(*no) ; 03 dcl-s wkVar char(3) inz('0') ; 04 dcl-pr TestProc char(3) ; 05 *n char(3) ; 06 end-pr ; 07 wkVar = TestProc(wkVar) ; 08 dsply ('wkVar = ' + wkVar) ; 09 *inlr = *on ; 20 dcl-proc TestProc ; 21 dcl-pi *n char(3) ; 22 SomeVar char(3) ; 23 end-pi ; 24 return %char(%dec(SomeVar:3:0) + 1) ; 25 end-proc ; ```\n\nLine 1: Why would I not use what I call RPG for i.\n\nLine 2: The Control options includes the new DEBUG(*RETVAL).\n\nLine 3: Definition of the variable whose value I will be passing and receiving back from the procedure.\n\nLines 4 – 6: The procedure prototype definition. This passes a three long character value to the procedure, line 5, and receives a three long character back, see line 4.\n\nLine 7: I am calling the procedure, passing the value in wkVar to it, and placing the returned value into wkVar.\n\nLine 8: I am using the Display operation code, DSPLY, to show the contents of wkVar.\n\nThe procedure is between lines 20 - 25.\n\nLines 21 – 23: The procedure interface.\n\nLine 24: I have added one to the value character value in the variable SomeVar and returned the result to the calling program.\n\nWhen everything has been compiled I added the following breakpoints into this example:\n\n1. Line 8 – Before the procedure is called\n2. Line 25 – End of the procedure\n\nWhen I call the program it \"breaks\" at line 8. I can enter the following debug command and see the result:\n\n ```ev wkvar WKVAR = '0 ' ```\n\nAnd onto the breakpoint at line 25, the END-PROC. I need to be at this line to use this new debug functionality.\n\nPrior to this addition I cannot see the value of what was returned. But now I can using the _QRNU_RETVAL special value:\n\n ```ev _qrnu_retval _QRNU_RETVAL = '1 ' ```\n\nI can even change the value:\n\n ```ev _qrnu_retval = 'AAA' _QRNU_RETVAL = 'AAA' = 'AAA' ```\n\nWhen the Display operation is performed I see:\n\n ```DSPLY wkVar = AAA ```\n\nThe value returned from the procedure was what I changed it to in _QRNU_RETVAL.\n\nA single variable is simple. I often pass a data structure to a procedure and get the equivalent returned. Let me see how I can change the values in the data structure using this method. First my example program:\n\n ```01 **free 02 ctl-opt debug(*retval) dftactgrp(*no) ; 03 dcl-ds TemplateDS template ; 04 sf1 char(3) ; 05 sf2 packed(3) ; 06 end-ds ; 07 dcl-ds wkDS likeds(TemplateDS) ; 08 dcl-pr TestProc2 likeds(wkDS) ; 09 *n likeds(wkDS) ; 10 end-pr ; 11 wkDS.sf1 = 'STR' ; 12 wkDS.sf2 = 0 ; 13 wkDS = TestProc2(wkDS) ; 14 *inlr = *on ; 20 dcl-proc TestProc2 ; 21 dcl-pi *n likeds(TemplateDS) ; 22 Input likeds(TemplateDS) ; 23 end-pi ; 24 Input.sf1 = 'END' ; 25 Input.sf2 = 1 ; 26 return Input ; 27 end-proc ; ```\n\nLines 3 – 6: This is a definition of a template data structure, hence the TEMPLATE on line 3. This defines the data structure, but it cannot be used to contain data. It can be used to define other data structures by using the LIKEDS keyword.\n\nLine 7: Defining a new data structure using the LIKEDS.\n\nLines 8 – 10. Procedure prototype definition. Notice that the returned value and the passed value are both defined with the LIKEDS too. I defined them using wkDS to act a \"documentation\". You now know which data structure I will be passing to the procedure, and which data structure I will be placing the returned data into.\n\nLines 11 and 12: Loading values into the data structure sub fields.\n\nLine 13: Calling the procedure passing to it the contents of the data structure, and putting the returned data into the same data structure.\n\nThe procedure, lines 20 – 27, is fairly simple.\n\nLines 21 – 23: Procedure interface. Notice how Input has been defined as a data structure by using the LIKEDS.\n\nLines 24 and 25: Moving values into the data structure's subfields.\n\nLine 26: This time I do not have a hard coded return. Here I am returning the contents of the data structure.\n\nTime to compile this program, start debug, and add break points:\n\n• Line 13 – So I can see the contents of wkDS before the procedure is called\n• Line 14 – I can see the contents of wkDS after the data has been returned from the procedure\n• Line 27 – End of the procedure\n• Here is the contents of the data structure before I call the procedure, line 13:\n\n ```ev wkds EVAL wkds WKDS.SF1 = 'STR' WKDS.SF2 = 000. ```\n\nNext break point is at the end of the procedure, line 27. Here I can see the values in the Input data structure:\n\n ```ev input EVAL input INPUT.SF1 = 'END' INPUT.SF2 = 001. ```\n\nThe same values are also in _QRNU_RETVAL:\n\n ```ev _qru_retval EVAL _qrnu_retval _QRNU_RETVAL.SF1 = 'END' _QRNU_RETVAL.SF2 = 001. ```\n\nIf I want to change the values of the data structure subfields I just qualify the subfields with the special value's name:\n\n ```ev _qrnu_retval.sf1 = 'NOT' EVAL _qrnu_retval.sf1 = 'NOT' _QRNU_RETVAL.SF1 = 'NOT' = 'NOT' ev _qrnu_retval.sf2 = 2 EVAL _qrnu_retval.sf2 = 2 _QRNU_RETVAL.SF2 = 2 = 002. ev _qrnu_retval EVAL _qrnu_retval _QRNU_RETVAL.SF1 = 'NOT' _QRNU_RETVAL.SF2 = 002. ```\n\nThe final breakpoint, line 14, allows me to see if the changed values are in the data structure:\n\n ```ev wkds EVAL wkds WKDS.SF1 = 'NOT' WKDS.SF2 = 002. ```\n\nIt contains the data that I changed in _QRNU_RETVAL.\n\nMy last example shows how to change the values in an array returned from a procedure.\n\n ```01 **free 02 ctl-opt debug(*retval) dftactgrp(*no) ; 03 dcl-s Array_1 char(2) dim(5) ; 04 dcl-pr TestProc2 char(2) dim(5) ; 05 *n char(2) dim(5) ; 06 end-pr ; 07 Array_1 = %list('1':'2':'3':'4':'5') ; 08 Array_1 = TestProc3(Array_1) ; 09 *inlr = *on ; 20 dcl-proc TestProc3 ; 21 dcl-pi *n char(2) dim(5) ; 22 Array_2 char(2) dim(5) ; 23 end-pi ; 24 Array_2(1) = 'A' ; 25 return Array_2 ; 26 end-proc ; ```\n\nLine 3: Definition of my array.\n\nLines 4 – 6: The procedure prototype definition. Notice how the passed parameter, line 5, and the returned value, line 4, are both defined as arrays.\n\nLine 7: This is a new feature introduced with these TRs, the %LIST BiF. This makes loading an array so easy!\n\nLine 8: Call the procedure, and place the returned data into the same array.\n\nIn the procedure, lines 20 – 26.\n\nLine 24: Change the first element of the array.\n\nLine 25: Return the contents of the array.\n\nI compiled the program, started debug, and placed the following break points:\n\n• Line 8 – I can see the values in the array before the data is passed to the procedure\n• Line 9 – Here I can view the contents of the array returned from the procedure\n• Line 26 – End of the procedure\n• When I called the program and the when the first break point happened I can:\n\n ```ev array_1 EVAL array_1 ARRAY_1(1) = '1 ' ARRAY_1(2) = '2 ' ARRAY_1(3) = '3 ' ARRAY_1(4) = '4 ' ARRAY_1(5) = '5 ' ```\n\nNext break point is at the end of the procedure. Here I can see that the contents of Array_2 and _QRNU_RETVAL are the same:\n\n ```ev array_2 EVAL array_2 ARRAY_2(1) = 'A ' ARRAY_2(2) = '2 ' ARRAY_2(3) = '3 ' ARRAY_2(4) = '4 ' ARRAY_2(5) = '5 ' ev _qrnu_retval EVAL _qrnu_retval _QRNU_RETVAL(1) = 'A ' _QRNU_RETVAL(2) = '2 ' _QRNU_RETVAL(3) = '3 ' _QRNU_RETVAL(4) = '4 ' _QRNU_RETVAL(5) = '5 ' ```\n\nI am going to change the second element:\n\n ```ev _qrnu_retval(2) = 'B' EVAL _qrnu_retval(2) = 'B' _QRNU_RETVAL(2) = 'B' = 'B ' ev _qrnu_retval EVAL _qrnu_retval _QRNU_RETVAL(1) = 'A ' _QRNU_RETVAL(2) = 'B ' _QRNU_RETVAL(3) = '3 ' _QRNU_RETVAL(4) = '4 ' _QRNU_RETVAL(5) = '5 ' ```\n\nOnto the last breakpoint, and I see that my change to the second element are in the returned array:\n\n ```ev array_1 EVAL array_1 ARRAY_1(1) = 'A ' ARRAY_1(2) = 'B ' ARRAY_1(3) = '3 ' ARRAY_1(4) = '4 ' ARRAY_1(5) = '5 ' ```\n\nAs I have shown _QRNU_RETVAL can be a variable, data structure, and even an array depending upon the returned parameter type. I will certainly be using this to help me debug procedures to ensure the value they return. This is good addition to my programming toolset.\n\nThis article was written for IBM i 7.4 TR3 and 7.3 TR9.\n\n```"
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https://la.mathworks.com/help/dsp/ref/fdesign.parameq.html
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[
"Documentation\n\n# fdesign.parameq\n\nParametric equalizer filter specification\n\n## Compatibility\n\n### Note\n\nThe `fdesign.parameq` function will be removed from DSP System Toolbox™ in a future release. Existing instances of the function continue to run. For new code, use the `fdesign.parameq` function from Audio Toolbox™ instead.\n\n## Syntax\n\n```d = fdesign.parameq(spec, specvalue1, specvalue2, ...) d = fdesign.parameq(... fs) ```\n\n## Description\n\n```d = fdesign.parameq(spec, specvalue1, specvalue2, ...)``` constructs a parametric equalizer filter design object, where `spec` is a non-case sensitive specification character vector. The choices for `spec` are as follows:\n\n• `'F0, BW, BWp, Gref, G0, GBW, Gp'` (minimum order default)\n\n• `'F0, BW, BWst, Gref, G0, GBW, Gst'`\n\n• `'F0, BW, BWp, Gref, G0, GBW, Gp, Gst'`\n\n• `'N, F0, BW, Gref, G0, GBW'`\n\n• `'N, F0, BW, Gref, G0, GBW, Gp'`\n\n• `'N, F0, Fc, Qa, G0'`\n\n• `'N, F0, Fc, S, G0'`\n\n• `'N, F0 ,BW, Gref, G0, GBW, Gst'`\n\n• `'N, F0, BW, Gref, G0, GBW, Gp, Gst'`\n\n• `'N, Flow, Fhigh, Gref, G0, GBW'`\n\n• `'N, Flow, Fhigh, Gref, G0, GBW, Gp'`\n\n• `'N, Flow, Fhigh, Gref, G0, GBW, Gst'`\n\n• ```'N, Flow, Fhigh, Gref, G0, GBW, Gp, Gst'```\n\nwhere the parameters are defined as follows:\n\n• BW — Bandwidth\n\n• BWp — Passband Bandwidth\n\n• BWst — Stopband Bandwidth\n\n• Gref — Reference Gain (decibels)\n\n• G0 — Center Frequency Gain (decibels)\n\n• GBW — Gain at which Bandwidth (BW) is measured (decibels)\n\n• Gp — Passband Gain (decibels)\n\n• Gst — Stopband Gain (decibels)\n\n• N — Filter Order\n\n• F0 — Center Frequency\n\n• Fc— Cutoff frequency\n\n• Fhigh - Higher Frequency at Gain GBW\n\n• Flow - Lower Frequency at Gain GBW\n\n• Qa-Quality Factor\n\n• S-Slope Parameter for Shelving Filters\n\nRegardless of the specification chosen, there are some conditions that apply to the specification parameters. These are as follows:\n\n• Specifications for parametric equalizers must be given in decibels\n\n• To boost the input signal, set `G0 > Gref`; to cut, set `Gref > G0`\n\n• For boost: ```G0 > Gp > GBW > Gst > Gref```; For cut: ```G0 < Gp < GBW < Gst < Gref```\n\n• Bandwidth must satisfy: ```BWst > BW > BWp```\n\n`d = fdesign.parameq(... fs)` adds the input sampling frequency. Fs must be specified as a scalar trailing the other numerical values provided, and is assumed to be in Hz.\n\n## Examples\n\nDesign a Chebyshev Type II parametric equalizer filter that cuts by 12 dB:\n\n``` d = fdesign.parameq('N,Flow,Fhigh,Gref,G0,GBW,Gst',... 4,.3,.5,0,-12,-10,-1); Hd = design(d,'cheby2'); fvtool(Hd)```",
null,
"Design a 4th order audio lowpass (F0 = 0) shelving filter with cutoff frequency of Fc = 0.25, quality factor Qa =10, and boost gain of G0 = 8 dB:\n\n```d = fdesign.parameq('N,F0,Fc,Qa,G0',4,0,0.25,10,8); Hd = design(d); fvtool(Hd)```",
null,
"Design 4th-order highpass shelving filters with S=1.5 and S=3:\n\n```N=4; F0 = 1; Fc = .4; % Cutoff Frequency G0 = 10; S = 1.5; S2=3; f = fdesign.parameq('N,F0,Fc,S,G0',N,F0,Fc,S,G0); h1 = design(f); f.S=3; h2=design(f); hfvt=fvtool([h1 h2]); set(hfvt,'Filters',[h1 h2]); legend(hfvt,'S=1.5','S=3');```",
null,
""
] |
[
null,
"https://la.mathworks.com/help/dsp/ref/parameq_12db_cut.gif",
null,
"https://la.mathworks.com/help/dsp/ref/fdesign.parameq_ex2.png",
null,
"https://la.mathworks.com/help/dsp/ref/fdesign.parameq_ex3.png",
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] |
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|
https://help.scilab.org/docs/5.3.1/en_US/section_fbb832a4db5f8ce05fb634eec24fa41c.html
|
[
"Change language to:\nFrançais - 日本語 - Português\n\nSee the recommended documentation of this function\n\nScilab help >> Graphics Library\n\n### Graphics Library\n\n• 2d_plot\n• LineSpecto quickly customize the lines appearance in a plot\n• Matplot2D plot of a matrix using colors\n• Matplot12D plot of a matrix using colors\n• Matplot_propertiesdescription of the Matplot entities properties\n• Sfgrayplotsmooth 2D plot of a surface defined by a function using colors\n• Sgrayplotsmooth 2D plot of a surface using colors\n• champ2D vector field plot\n• champ12D vector field plot with colored arrows\n• champ_propertiesdescription of the 2D vector field entity properties\n• contour2dlevel curves of a surface on a 2D plot\n• contour2dicompute level curves of a surface on a 2D plot\n• contourffilled level curves of a surface on a 2D plot\n• errbaradd vertical error bars on a 2D plot\n• fchampdirection field of a 2D first order ODE\n• fcontour2dlevel curves of a surface defined by a function on a 2D plot\n• fecpseudo-color plot of a function defined on a triangular mesh\n• fec_propertiesdescription of the fec entities properties\n• fgrayplot2D plot of a surface defined by a function using colors\n• fplot2d2D plot of a curve defined by a function\n• grayplot2D plot of a surface using colors\n• grayplot_propertiesdescription of the grayplot entities properties\n• graypolarplotPolar 2D plot of a surface using colors\n• histplotplot a histogram\n• oldplotsimple plot (old version)\n• paramfplot2danimated 2D plot, curve defined by a function\n• plot2D plot\n• plot2d2D plot\n• plot2d12D plot (logarithmic axes) (obsolete)\n• plot2d22D plot (step function)\n• plot2d32D plot (vertical bars)\n• plot2d42D plot (arrows style)\n• plot2d_old_versionThe syntaxes described below are obsolete\n• polarplotPlot polar coordinates\n• 3d_plot\n• contourlevel curves on a 3D surface\n• eval3dvalues of a function on a grid\n• eval3dpcompute facets of a 3D parametric surface\n• fac3d3D plot of a surface (obsolete)\n• fcontourlevel curves on a 3D surface defined by a function\n• fplot3d3D plot of a surface defined by a function\n• fplot3d13D gray or color level plot of a surface defined by a function\n• genfac3dcompute facets of a 3D surface\n• geom3dprojection from 3D on 2D after a 3D plot\n• hist3d3D representation of a histogram\n• mesh3D mesh plot\n• milk_dropmilk drop 3D function\n• nf3drectangular facets to plot3d parameters\n• param3d3D plot of a parametric curve\n• param3d13D plot of parametric curves\n• param3d_propertiesdescription of the 3D curves entities properties\n• plot3d3D plot of a surface\n• plot3d13D gray or color level plot of a surface\n• plot3d2plot surface defined by rectangular facets\n• plot3d3mesh plot surface defined by rectangular facets\n• secto3d3D surfaces conversion\n• surf3D surface plot\n• surface_propertiesdescription of the 3D entities properties\n• annotation\n• axes_operations\n• axes_propertiesdescription of the axes entity properties\n• gcaReturn handle of current axes.\n• gdaReturn handle of default axes.\n• isoviewset scales for isometric plot (do not change the size of the window)\n• newaxesCreates a new Axes entity\n• plotframeplot a frame with scaling and grids. This function is obsolete.\n• replotredraw the current graphics window with new boundaries\n• rotate_axesInteractive rotation of an Axes handle.\n• scaset the current axes entity\n• sdaSet default axes.\n• subplotdivide a graphics window into a matrix of sub-windows\n• unzoomunzoom graphics\n• zoom_rectzoom a selection of the current graphic figure\n• axis\n• bar_histogram\n• barbar histogram\n• barhhorizontal display of bar histogram\n• barhomogenizehomogenize all the bars included in the current working axes\n• Color management\n• Datatips\n• figure_operations\n• clfclear or reset the current graphic figure (window) to default values\n• drawlatermakes axes children invisible.\n• drawnowdraw hidden graphics entities.\n• figure_propertiesdescription of the graphics figure entity properties\n• gcfReturn handle of current graphic window.\n• gdfReturn handle of default figure.\n• scfset the current graphic figure (window)\n• sdfSet default figure.\n• geometric_shapes\n• arc_propertiesdescription of the Arc entity properties\n• rectangle_propertiesdescription of the Rectangle entity properties\n• xarcdraw a part of an ellipse\n• xarcsdraw parts of a set of ellipses\n• xarrowsdraw a set of arrows\n• xfarcfill a part of an ellipse\n• xfarcsfill parts of a set of ellipses\n• xfrectfill a rectangle\n• xrectdraw a rectangle\n• xrectsdraw or fill a set of rectangles\n• handle\n• copycopy a graphics entity.\n• deletedelete a graphic entity and its children.\n• drawdraw an entity.\n• gceGet current entity handle.\n• gedScilab Graphic Editor\n• get_figure_handleget a figure handle from its id\n• glueglue a set of graphics entities into an Compound.\n• is_handle_validCheck wether a set of graphic handles is still valid.\n• relocate_handleMove handles inside the graphic hierarchy.\n• sd2scigr_menu structure to scilab instruction convertor\n• swap_handlesPermute two handles in the graphic Hierarchy.\n• unglueunglue a coumpound object and replace it by individual children.\n• interaction\n• dragrectDrag rectangle(s) with mouse\n• edit_curvinteractive graphic curve editor\n• eventhandlerfunctionsPrototype of functions which may be used as event handler.\n• locatemouse selection of a set of points\n• seteventhandlerset an event handler for the current graphic window\n• xclickWait for a mouse click.\n• xgetmouseget the mouse events and current position\n• xsavesave graphics into a file\n• pie\n• piedraw a pie\n• polygon\n• polyline_propertiesdescription of the Polyline entity properties\n• xfpolyfill a polygon\n• xfpolysfill a set of polygons\n• xpolydraw a polyline or a polygon\n• xpolysdraw a set of polylines or polygons\n• xrpolydraw a regular polygon\n• property\n• getRetrieve a property value from a graphics entity or an User Interface object.\n• setset a property value of a graphic entity object or of a User Interface object.\n• style\n• getlinestyledialog to select linestyle. Obsolete function.\n• getmarkdialog to select mark (symbol). Obsolete function\n• getsymboldialog to select a symbol and its size. Obsolete function\n• text\n• getfontdialog to select font . Obsolete function.\n• graphics_fontsdescription of fonts used in graphic figures\n• math_rendering_features_in_graphicDisplay mathematical equations in Scilab graphics through the LaTeX or MathML languages.\n• stringboxCompute the bounding rectangle of a text or a label.\n• text_propertiesdescription of the Text entity properties\n• titlepageadd a title in the middle of a graphics window\n• xinfodraw an info string in the message subwindow\n• xlfontload a font in the graphic context or query loaded font\n• xstringdraw strings\n• xstringbdraw strings into a box\n• xstringlcompute a box which surrounds strings\n• transform\n• movemove, translate, a graphic entity and its children.\n• rotaterotation of a set of points\n• scalingaffine transformation of a set of points\n• window_control\n• havewindowreturn scilab window mode\n• show_windowraises a graphics window\n• winsidreturn the list of graphics windows\n• xselectraise the current graphics window\n• GlobalPropertyto customize the objects appearance (curves, surfaces...) in a plot or surf command.\n• Graphicsgraphics library overview\n• alufunctionspixel drawing functions. This function is obsolete.\n• clear_pixmaperase the pixmap buffer\n• Compound_propertiesdescription of the Compound entity properties\n• graphics_entitiesdescription of the graphics entities data structures\n• object_editordescription of the graphic object editor capacities\n• pixel_drawing_modebitwise pixel rendering functions.\n• plzrpole-zero plot\n• rubberboxRubberband box for rectangle selection\n• segs_propertiesdescription of the Segments entity properties\n• show_pixmapsend the pixmap buffer to the screen\n• squareset scales for isometric plot (change the size of the window)\n• twinkleis used to have a graphics entity twinkle\n• xbasrredraw a graphics window\n• xchangetransform real to pixel coordinates\n• xclearclears a graphics window\n• xdeldelete a graphics window\n• xgetget current values of the graphics context. This function is obsolete.\n• xgetechget the current graphics scale"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.64161175,"math_prob":0.820821,"size":12436,"snap":"2023-40-2023-50","text_gpt3_token_len":3149,"char_repetition_ratio":0.17752574,"word_repetition_ratio":0.089050435,"special_character_ratio":0.23962688,"punctuation_ratio":0.040201005,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9679561,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T03:58:57Z\",\"WARC-Record-ID\":\"<urn:uuid:a0b9d4be-994f-4e23-9b40-16379c678934>\",\"Content-Length\":\"47896\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:905ef8cd-4c1a-4d7e-84b7-4e7b02b8207e>\",\"WARC-Concurrent-To\":\"<urn:uuid:edef57a7-6635-48fe-a6bb-b9b7d880fdf0>\",\"WARC-IP-Address\":\"107.154.79.223\",\"WARC-Target-URI\":\"https://help.scilab.org/docs/5.3.1/en_US/section_fbb832a4db5f8ce05fb634eec24fa41c.html\",\"WARC-Payload-Digest\":\"sha1:FZJCCTD6HBEQQB7QJBKU5ZM47WJF3CJO\",\"WARC-Block-Digest\":\"sha1:LVLPWURVY4NJMDZLUY3RS6SA3RYJGIP4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100484.76_warc_CC-MAIN-20231203030948-20231203060948-00113.warc.gz\"}"}
|
https://cstheory.stackexchange.com/questions/47186/3sum-complexity-a-special-case/47198#47198
|
[
"3SUM Complexity—A special(?) Case\n\nIn the paper “Consequences of Faster Alignment of Sequences” by Amir Abboud, Virginia Vassilevska Williams, and Oren Weimann which appeared in ICALP 2014 and is available here the following version of the integer 3-SUM conjecture is stated.\n\nConjecture 1 (3-SUM Conjecture) In the Word RAM model with words of $$O(\\log n)$$ bits, any algorithm requires $$n^{2−o(1)}$$ time in expectation to determine whether three sets $$A,B,C \\subset \\{−n^3,\\ldots,n^3\\}$$ with $$|A| = |B| = |C| = n$$ integers contain three elements $$a∈A,b∈B,c∈C$$ with $$a+b+c=0.$$\n\nNot being an expert I have the following question.\n\nHow is this restriction to the set of integers with absolute value $$\\leq n^3$$ justified? Is this in some sense hardest and other cases can be solved if this case is solved?\n\nRemark: I suppose a ground set of size $$O(n^3)$$ is dense in the sense that a lot of triple candidates cannot be ruled out, but I imagine there are more spread out sets which may have similar properties.\n\nEdit 2: Changed the focus of the question.\n\nThe smaller this upper bound is, the easier the problem becomes. In particular, if the range is $$m$$, then the problem can be solved in $$O(m \\log m)$$ time using FFT. It is impressive/interesting that the authors were able to show that the problem is still quadratically nasty for numbers that are \"slightly\" larger than quadratic.\n\n• Thanks for the pointer to the FFT. By \"slightly larger\" than quadratic what exactly do you refer to? If the range is $m=2n^3+1,$ as it is here the FFT complexity is $O(n^3 \\log n)$ but the naive search complexity of forming pairs from $A,B$ and looking sums up in $C$ is still $O(n^2)$. Jul 10 '20 at 0:04\n• The bound $2n^3+1$ is polynomial in $n$, whereas it is not at all obvious that the problem could be reduced even to subexponential-sized numbers. That's why it's impressive. This is why Sariel calls the numbers \"slightly\" larger than quadratic. Jul 10 '20 at 6:54\n• Yes. FFT is no longer relevant if the numbers are bigger than $n^2$, but it says something about the problem: The problem is easy if the numbers are small. As such, the question becomes, how large the numbers have to be before the problem becomes quadratic? As a constructive example consider subset sum - it is an NP Complete/Hard problem, but only if the numbers are exponentially large... Jul 11 '20 at 4:33\n• Thanks Sariel, that's clearer to me now. Jul 11 '20 at 4:49\n\nI believe I can partially answer your question as to why the bounds of $$\\{-n^3, ..., n^3\\}$$ are justified.\n\nThis paper by Pătraşcu mentions that for 3SUM over any bounded universe of integers of size $$u >> n^3$$, the universe size can be hashed down to $$O(n^3)$$ while maintaining the expected $$O(n^2)$$ run time for 3SUM. Therefore, to prove that 3SUM can be solved in expected time $$O(n^{2 - \\varepsilon})$$ over every universe size $$u$$ of integers, it suffices to give an algorithm that solves 3SUM on every universe of size $$O(n^3)$$ in expected time $$O(n^{2 - \\varepsilon})$$.\n\nPătraşcu doesn't directly give this reduction, but states that the techniques of this paper can be used to perform such a hashing.\n\nI have been reading this paper, but I haven't quite figured out the details of this reduction.\n\nI hope this helps!"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.91800356,"math_prob":0.9972694,"size":1009,"snap":"2022-05-2022-21","text_gpt3_token_len":279,"char_repetition_ratio":0.08955224,"word_repetition_ratio":0.0,"special_character_ratio":0.25966305,"punctuation_ratio":0.09134615,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99984515,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T06:14:25Z\",\"WARC-Record-ID\":\"<urn:uuid:edef2a13-a477-4f70-b977-f8bc8ddc1311>\",\"Content-Length\":\"151586\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:860972be-e900-4493-a5e1-85696cba4326>\",\"WARC-Concurrent-To\":\"<urn:uuid:8bb42f9d-848c-489f-86d3-143e735b803f>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://cstheory.stackexchange.com/questions/47186/3sum-complexity-a-special-case/47198#47198\",\"WARC-Payload-Digest\":\"sha1:N2MNV2VZGLGRJNF4KO2W2YVG5LYOMJ6T\",\"WARC-Block-Digest\":\"sha1:3URFEE4UAKIQNLBDQQRBYI626J6RJ7OV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304515.74_warc_CC-MAIN-20220124054039-20220124084039-00657.warc.gz\"}"}
|
https://electronics.stackexchange.com/questions/230283/sampling-of-non-baseband-signals-shannon-sampling-theorem
|
[
"# Sampling of non-baseband signals - Shannon sampling theorem\n\nIn the example on Sampling of non-basedband signals on Wikipedia,",
null,
"the author said it is sufficient to sample at 4 MHz (twice the width of the frequency interval). But how did the author calculate the width of the frequency interval which in that case is equal to 2? Any explanations to how he came up with 4MHz\n\nWikipedia - Shannon sampling\n\n• read the example (last para) - frequency range 100 - 102 Mhz, Need to sample at twice the bandwidth of 2MHz and so sample at twice that frequency = 4Mhz (i.e. the Nyquist criterion) – JIm Dearden Apr 24 '16 at 21:27"
] |
[
null,
"https://i.stack.imgur.com/LfwCL.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.94317967,"math_prob":0.97847486,"size":310,"snap":"2019-51-2020-05","text_gpt3_token_len":68,"char_repetition_ratio":0.11764706,"word_repetition_ratio":0.03773585,"special_character_ratio":0.20967741,"punctuation_ratio":0.04918033,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97397405,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T15:02:50Z\",\"WARC-Record-ID\":\"<urn:uuid:3ef9184a-1c9c-43df-b0f2-92caec7a16dd>\",\"Content-Length\":\"133541\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee981d43-656b-4d9a-8d69-31b0b4c67d47>\",\"WARC-Concurrent-To\":\"<urn:uuid:81544042-53e0-4214-8465-4685342d014e>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/230283/sampling-of-non-baseband-signals-shannon-sampling-theorem\",\"WARC-Payload-Digest\":\"sha1:SO7FTFLTM6GT3AVFECIZER7P2XRCADRO\",\"WARC-Block-Digest\":\"sha1:FNDGDETGZBVNWHG2WSPZEFBO23YANEFI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592636.25_warc_CC-MAIN-20200118135205-20200118163205-00410.warc.gz\"}"}
|
https://www.excelcalcs.com/calcs/repository/Geometry/Volumes/Pressure-Vessel(21-ellipsoidal-head-type)-Volume-Calculation/
|
[
"# Pressure Vessel(2:1 ellipsoidal head type) Volume Calculation\n\nRating:\n8\n\n### Description\n\nVolume calculation for:\n\n1. Horizontal pressure vessel with 2:1 ellipsoidal head type\n2. Vertical pressure vessel with 2:1 ellipsoidal head type\n3. Liquid volume\n\nA horizontal pressure vessel with 2:1 ellipsoidal heads has a cylindrical section and two ellipsoidal heads at both ends. To calculate the total volume of the pressure vessel, you need to find the volume of the cylindrical section and the two ellipsoidal heads, then add them together.\n\n1. Cylindrical section volume: The volume of the cylindrical section (V_cylinder) can be calculated using the following formula:\n\nV_cylinder = ? * (D/2)^2 * L\n\nWhere: D = diameter of the cylindrical section L = length of the cylindrical section\n\n1. 2:1 Ellipsoidal head volume: For each ellipsoidal head, the volume (V_head) can be calculated using the following formula:\n\nV_head = (1/6) * ? * h * (3a^2 + h^2)\n\nSince the heads are 2:1 ellipsoidal, the major axis (2a) is equal to the diameter of the cylindrical section (D), and the minor axis (h) is equal to D/2.\n\nThus, the formula becomes:\n\nV_head = (1/6) * ? * (D/2) * (3(D/2)^2 + (D/2)^2)\n\n1. Total volume: The total volume of the horizontal pressure vessel (V_total) is the sum of the volumes of the cylindrical section and the two ellipsoidal heads:\n\nV_total = V_cylinder + 2 * V_head\n\nBy combining the formulas for V_cylinder and V_head, you get:\n\nV_total = ? * (D/2)^2 * L + (1/3) * ? * (D/2) * (3(D/2)^2 + (D/2)^2)\n\nThis formula can be used to calculate the total volume of a horizontal pressure vessel with 2:1 ellipsoidal heads. Note that this calculation assumes that the internal dimensions of the pressure vessel are used. If you need to calculate the volume based on external dimensions, you will need to account for the thickness of the vessel's walls.\n\n25 Dec 2021\n28 Apr 2023\nFile Size:\n115.38 Kb\n20\nFile Version:\n1.0\nFile Author:\nNick Lee\nRating:\n8"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.80050457,"math_prob":0.98997957,"size":2086,"snap":"2023-40-2023-50","text_gpt3_token_len":532,"char_repetition_ratio":0.17050913,"word_repetition_ratio":0.13872832,"special_character_ratio":0.25886863,"punctuation_ratio":0.10443864,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99909234,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T23:47:37Z\",\"WARC-Record-ID\":\"<urn:uuid:b68d3230-ad10-4655-8b50-5cfebc17cf6c>\",\"Content-Length\":\"27252\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af109492-47bb-4f96-8df0-da0698fec97b>\",\"WARC-Concurrent-To\":\"<urn:uuid:fa1f8da1-0ad4-4d0b-94da-184e9f498b9b>\",\"WARC-IP-Address\":\"193.107.20.115\",\"WARC-Target-URI\":\"https://www.excelcalcs.com/calcs/repository/Geometry/Volumes/Pressure-Vessel(21-ellipsoidal-head-type)-Volume-Calculation/\",\"WARC-Payload-Digest\":\"sha1:UP5NGGIXJ4HOZTRBZXB4FEWXCYNGKVFB\",\"WARC-Block-Digest\":\"sha1:2OYQHRV7FNSN2OUSQ2VRMMRPZYRQZ6BF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510529.8_warc_CC-MAIN-20230929222230-20230930012230-00448.warc.gz\"}"}
|
https://rethinkingecology.pensoft.net/article/31992/zoom/fig/11/
|
[
"",
null,
"Graph of treated and non-treated environments, with distinct phenotypic requirements (phenotypic optima A and O, respectively) in a two-trait space. Assuming a wild-type positioned in O, the resistance mutation R brings the phenotype closer to A. Relative to the wild type, R is therefore a beneficial mutation in the treated environment. Its cost is usually defined as its fitness in the non-treated environment relative to the wild type, which depends on the distance between R and O on the figure. Note that the cost (and all fitness measures on this figure, and similar figure below) depends on Euclidian distances in phenotypic space, and a mapping function converting this distance to fitness (i.e. the cost is not distance OR, but a monotonic function of this distance). The mapping is left implicit on the figure, but can be thought as a third orthogonal axis representing fitness for each trait 1 – trait 2 combination, which defines a “fitness landscape”. The coloured inset figure represents such a fitness mapping in 3D. Fitness values, when projected on the phenotypic space correspond to isofitness curves (like altitude on a geographic map is indicated by contour lines). For instance all phenotypes on the light grey circles have the same fitness than mutation R in the treated environment (optimum A). The direction of the two optima (OA axis) defines a phenotypic trait of ‘resistance’. Variation of trait(s) orthogonal to this axis may be defined as pleiotropic effects. Point P1 on OA axis is such that AR = AP1. It represents the phenotypic point that would confer the same fitness in the treated environment compared to the mutation R, but that would only alter the phenotype in the exact direction of the optimum A. Point P2 is the orthogonal projection of R on OA axis. It represents the phenotypic point that would be reached if all the pleiotropic effects of the mutation R were compensated (e.g. by subsequent compensatory mutations).\n\nPart of: Lenormand T, Harmand N, Gallet R (2018) Cost of resistance: an unreasonably expensive concept. Rethinking Ecology 3: 51-70. https://doi.org/10.3897/rethinkingecology.3.31992"
] |
[
null,
"https://rethinkingecology.pensoft.net/showimg.php",
null
] |
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|
https://pyrocko.org/kite/docs/current/examples/sandbox.html
|
[
"# Modelling static displacement¶\n\nKite comes with a SandboxScene, which is a playground for static displacement sources of different kinds and modelling engines. It support analytical backends such as Okada disloc-code and Compound dislocation model (; http://www.volcanoedeformation.com/). Numerical forward modelling is enabled through pyrocko.gf, this allows us to put geometrically more complex sources into the modelling sandbox.\n\nCurrently implemented static displacement sources:\n\nSeveral Pyrocko Sources:\n\nAnalytical dilatational point through CDM () sources:\n\nMore information about the sources and their implementation can be found at the modules reference: kite.sources.\n\n Okada, Y., 1992, Internal deformation due to shear and tensile faults in a half-space, Bull. Seism. Soc. Am., 82, 1018-1040.\n (1, 2) Nikkhoo, M., Walter, T. R., Lundgren, P. R., Prats-Iraola, P. (2017): Compound dislocation models (CDMs) for volcano deformation analyses. - Geophysical Journal International, 208, 2, p. 877-894.\n\n## Add Displacement Sources into SandboxScene¶\n\nIn this example we will add a simple OkadaSource into a SandboxScene.\n\nfrom kite import SandboxScene\nfrom kite.sources import OkadaSource, PyrockoRectangularSource\n\nimport matplotlib.pyplot as plt\nimport numpy as num\n\nkm = 1e3\n\nsandbox = SandboxScene()\n# Set the LOS incident angles, remember :class:kite.Scene\n\nnorthing=40*km,\neasting=40*km,\ndepth=4*km,\nlength=8*km,\nwidth=4*km,\nstrike=63.,\ndip=33.,\nslip=3.,\nopening=1)\n\npyrocko_rectangular = PyrockoRectangularSource(\nnorthing=40*km,\neasting=40*km,\ndepth=4*km,\nlength=8*km,\nwidth=4*km,\nstrike=63.,\ndip=33.,\nslip=3.,\nstore_dir='gfstore_halfspace')\n\nsandbox.processSources()\n\n# Plot the resulting surface displacements\nfig, axis = plt.subplots(nrows=2, ncols=2)\naxis.imshow(sandbox.north)\naxis.imshow(sandbox.east)\naxis.imshow(sandbox.down)\naxis.imshow(sandbox.displacement) # Displacement in LOS\nfig.show()\n\n\nA full list of available sources and their parameters can be found at the modules’ reference page kite.sources.\n\n## Save and Load pyrocko.SandboxScene¶\n\nIn this small example we will add a basic EllipsoidSource to the sandbox. Subsequently we will save it and load it again.\n\nfrom kite import SandboxScene\nfrom kite.sources import EllipsoidSource\n\nkm = 1e3\n\nsandbox = SandboxScene()\n\nellipsoid_source = EllipsoidSource(\nnorthing=40*km,\neasting=40*km,\ndepth=4*km,\nlength_x=100,\nlength_y=200,\nlength_z=350,\nroation_x=42.)"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.56517106,"math_prob":0.84726405,"size":2918,"snap":"2019-13-2019-22","text_gpt3_token_len":784,"char_repetition_ratio":0.14584763,"word_repetition_ratio":0.064896755,"special_character_ratio":0.24811515,"punctuation_ratio":0.21937843,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.983976,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-25T16:19:08Z\",\"WARC-Record-ID\":\"<urn:uuid:5b99337f-019a-4ade-b581-256ad9f29c42>\",\"Content-Length\":\"19646\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0c90c812-2815-43b0-92c0-fc994ec79920>\",\"WARC-Concurrent-To\":\"<urn:uuid:d074f0d9-ac47-4caf-9728-384770387a61>\",\"WARC-IP-Address\":\"139.17.228.174\",\"WARC-Target-URI\":\"https://pyrocko.org/kite/docs/current/examples/sandbox.html\",\"WARC-Payload-Digest\":\"sha1:S7ONTJTQ5IVXFL47FR4535CE5WG2ZWAX\",\"WARC-Block-Digest\":\"sha1:IB34LVZ4I3WCK4CBSGUXIKV5PQ5RI6DZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912204077.10_warc_CC-MAIN-20190325153323-20190325175323-00070.warc.gz\"}"}
|
https://groupprops.subwiki.org/wiki/GAP:NormalSubgroups
|
[
"# GAP:NormalSubgroups\n\n## Definition\n\n### Function type\n\nNormalSubgroups is a GAP command that takes in one argument representing a group and outputs a list of groups.\n\n### Behavior\n\nApplying NormalSubgroups to a given group returns a list of all its normal subgroups. This list is not sorted in any standard way, i.e., it is not sorted based on the orders of subgroups or based on group IDs, and the ordering of the members of this list may be different for isomorphic groups.\n\nNonetheless, the ordering of the list has the following features:\n\n• If",
null,
"$H$ and",
null,
"$K$ are both normal subgroups of the group and",
null,
"$H \\le K$, then",
null,
"$H$ occurs before",
null,
"$K$ in the list.\n• In particular, the trivial subgroup occurs first in the list, and the whole group occurs last.\n• For a normal-comparable group, the ordering of the list is unique.\n\n## Related functions\n\n• GAP:IsNormal: This takes as input a group and a subgroup and outputs whether the subgroup is normal in the group.\n• GAP:ConjugacyClassesSubgroups: This takes as input a group and outputs a list of conjugacy classes of subgroups.\n\n## Examples of usage\n\n### Some examples involving prespecified groups\n\n```gap> NormalSubgroups(SymmetricGroup(3));\n[ Group(()), Group([ (1,2,3) ]), Sym( [ 1 .. 3 ] ) ]\ngap> L := NormalSubgroups(SymmetricGroup(4));\n[ Group(()), Group([ (1,4)(2,3), (1,3)(2,4) ]), Group([ (2,4,3), (1,4)(2,3), (1,3)(2,4) ]), Sym( [ 1 .. 4 ] ) ]\ngap> K := List(L,IdGroup);\n[ [ 1, 1 ], [ 4, 2 ], [ 12, 3 ], [ 24, 12 ] ]\ngap> Length(L);\n4\ngap> NormalSubgroups(SmallGroup(8,4));\n[ Group([ ]), Group([ f3 ]), Group([ f1*f2, f3 ]), Group([ f1, f3 ]), Group([ f2, f3 ]), <pc group of size 8 with 3 generators> ]\n```\n\nThe first example lists all the normal subgroups of the symmetric group on three letters. The set here is",
null,
"$\\{ 1,2, 3 \\}$. Each of these normal subgroups (except the whole group) is described by its generating set.\n\nThe second example computes the normal subgroups of the symmetric group on four letters and outputs the list of its normal subgroups. This list is stored with the variable name L. In the next command, the members of this list are mapped to their group IDs, using the IdGroup command. The GAP:List command is used to achieve the mapping on each member of the list. The next command computes the length of this list.\n\nThe final example computes the normal subgroups of a certain group specified by the group ID (this is in fact the quaternion group).\n\n## Time analysis\n\nThe following is the total time and average time over all groups of specific orders. Times were measured on a 6-core virtual machine and are in milliseconds. While the actual times may be different on a machine with different processor specifications, the relative time should hold up.\n\nThe time on second run is generally very low since it relies on pre-computed data.\n\nThe sequence of commands used was as follows, where",
null,
"$n$ is the order:\n\n```A := AllSmallGroups(n);\nList(A, NormalSubgroups);\ntime;```\nOrder Number of groups of that order Total time on first run (milliseconds) Average time on first run (milliseconds) Total time on second run (milliseconds) Average time on second run (milliseconds)\n1 1 4 4 0 0\n2 1 23 23 0 0\n3 1 20 20 3 3\n4 2 50 25 3 1.5\n5 1 4 4 0 0\n6 2 37 18.5 3 1.5\n8 5 50 10 0 0\n12 5 57 11.4 0 0\n16 14 317 22.6 17 1.2\n24 15 230 15.3 6 0.4\n32 51 1837 36.0 63 1.2\n48 52 1597 30.7 60 1.2\n64 267 19820 74.2 677 2.5\n96 231 13010 56.3 410 1.8\n128 2328 396414 170.3 11817 5.1\n168 57 1663 29.2 34 0.6\n\nHere is the data on symmetric groups of degree",
null,
"$n$. Note that for",
null,
"$n \\ge 5$, the only three normal subgroups are the trivial group, alternating group, and whole symmetric group. Starting with a degree of around 15, the computation time approximately doubles for every increase in degree of about five.",
null,
"$n$ Symmetric group of degree",
null,
"$n$ Order (equals",
null,
"$n!$) Time for first computation (in milliseconds)\n1 trivial group 1 3\n2 cyclic group:Z2 2 16\n3 symmetric group:S3 6 30\n4 symmetric group:S4 24 40\n5 symmetric group:S5 120 60\n6 symmetric group:S6 720 23\n7 symmetric group:S7 5040 17\n8 symmetric group:S8 40320 24\n9 symmetric group:S9 362880 23\n10 symmetric group:S10 3628800 44\n15 63\n20 123\n25 367\n30 597\n35 1140\n40 2286\n45 3786\n50 6340"
] |
[
null,
"https://groupprops.subwiki.org/w/images/math/c/1/d/c1d9f50f86825a1a2302ec2449c17196.png ",
null,
"https://groupprops.subwiki.org/w/images/math/a/5/f/a5f3c6a11b03839d46af9fb43c97c188.png ",
null,
"https://groupprops.subwiki.org/w/images/math/3/a/c/3ac79165b3b2c569bdee8974d90838b8.png ",
null,
"https://groupprops.subwiki.org/w/images/math/c/1/d/c1d9f50f86825a1a2302ec2449c17196.png ",
null,
"https://groupprops.subwiki.org/w/images/math/a/5/f/a5f3c6a11b03839d46af9fb43c97c188.png ",
null,
"https://groupprops.subwiki.org/w/images/math/1/d/4/1d4734bc6db8002e08e28eec76ab0b7a.png ",
null,
"https://groupprops.subwiki.org/w/images/math/7/b/8/7b8b965ad4bca0e41ab51de7b31363a1.png ",
null,
"https://groupprops.subwiki.org/w/images/math/7/b/8/7b8b965ad4bca0e41ab51de7b31363a1.png ",
null,
"https://groupprops.subwiki.org/w/images/math/f/8/5/f855cbf2c86bde65d8862639cbe99114.png ",
null,
"https://groupprops.subwiki.org/w/images/math/7/b/8/7b8b965ad4bca0e41ab51de7b31363a1.png ",
null,
"https://groupprops.subwiki.org/w/images/math/7/b/8/7b8b965ad4bca0e41ab51de7b31363a1.png ",
null,
"https://groupprops.subwiki.org/w/images/math/3/8/8/388f554901ba5d77339eec8b26beebea.png ",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.79881126,"math_prob":0.9680405,"size":4284,"snap":"2019-51-2020-05","text_gpt3_token_len":1257,"char_repetition_ratio":0.16214953,"word_repetition_ratio":0.056149732,"special_character_ratio":0.36251166,"punctuation_ratio":0.14679912,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98934865,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-16T03:11:11Z\",\"WARC-Record-ID\":\"<urn:uuid:79db5193-9e19-44ce-a32a-667ec0258348>\",\"Content-Length\":\"32637\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0cd652ac-a726-4af2-b758-a7c5873ec0b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:80e3ab5a-e687-410d-a573-855dc8d777ae>\",\"WARC-IP-Address\":\"96.126.114.7\",\"WARC-Target-URI\":\"https://groupprops.subwiki.org/wiki/GAP:NormalSubgroups\",\"WARC-Payload-Digest\":\"sha1:PRR73ODL5HCD5Y326XXCSQLGGGRMTIFP\",\"WARC-Block-Digest\":\"sha1:32PKNMVSH7J3QEFAXCHVWA4QP4WUCPKL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541315293.87_warc_CC-MAIN-20191216013805-20191216041805-00304.warc.gz\"}"}
|
https://brainmass.com/statistics/probability/working-with-binomial-probability-6291
|
[
"Explore BrainMass\n\n# Working with binomial probability.\n\nNot what you're looking for? Search our solutions OR ask your own Custom question.\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nA die is rolled 12 times. Find the probability of rolling the following.\n(a) No more than 1 one\n\nhttps://brainmass.com/statistics/probability/working-with-binomial-probability-6291\n\n#### Solution Preview\n\nP(no more then 1 one)=P(0 ones)+P(1 one)\nEach time you roll a die, think of it as there are two ...\n\n#### Solution Summary\n\nThe probability of a die roll is calculated.\n\n\\$2.49"
] |
[
null
] |
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|
http://cvgmt.sns.it/paper/2115/
|
[
"# BV functions and sets of finite perimeters in sub-Riemannian manifolds\n\ncreated by ghezzi on 25 Mar 2013\nmodified by magnani on 24 Jul 2015\n\n[BibTeX]\n\nPublished Paper\n\nInserted: 25 mar 2013\nLast Updated: 24 jul 2015\n\nJournal: Ann. Inst. H. Poincaré Anal. Non Linéaire\nVolume: 32\nNumber: 3\nPages: 489–517\nYear: 2015\nNotes:\n\n(2015), no. 3, .\n\nAbstract:\n\nWe give a notion of $BV$ function on an oriented manifold where a volume form and a family of lower semicontinuous quadratic forms $G_p: T_pM \\to [0,\\infty]$ are given. Using this notion, we generalize the structure theorem for $BV$ functions that holds in the Euclidean case. When we consider sub-Riemannian manifolds, our definition coincide with the one given in the more general context of metric measure spaces which are doubling and support a Poincar\\'e inequality. We then focus on finite perimeter sets, i.e., sets whose characteristic function is $BV$, in sub-Riemannian manifolds. Under an assumption on the nilpotent approximation, we prove a blowup theorem, generalizing the one obtained for step-2 Carnot groups in 24.\n\nTags: GeMeThNES\n\nDownload:\n\nCredits | Cookie policy | HTML 5 | CSS 2.1"
] |
[
null
] |
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|
https://civildigital.com/theory-of-boundary-layer/
|
[
"# Theory Of Boundary Layer\n\n## Theory Of Boundary Layer\n\n### Introduction\n\nWhen a real fluid flows past a solid boundary, a layer of fluid which comes in contact with the boundary surface adheres to it on account of viscosity. Since this layer of the fluid cannot slip away from the boundary surface it attains the same velocity as that of the boundary. If the boundary is stationary, the fluid velocity at the boundary surface will be zero. Thus at the boundary surface the layer of the fluid undergoes retardation.\n\nTherefore in the immediate vicinity of the boundary surface, the velocity of the fluid increases gradually from zero at boundary surface to the velocity of the mainstream. This region is known as BOUNDARY LAYER.\n\n### Causes Of Its Formation\n\nLarge velocity gradient (Larger variation of velocity in relatively smaller distance) leading to appreciable shear stress.\n\nConsists of two layers:\n\nCLOSE TO BOUNDARY : large velocity gradient, appreciable viscous forces.\n\nOUTSIDE BOUNDARY LAYER: viscous forces are negligible, flow may be treated as non-viscous or inviscid.\n\n### Development Of Boundary Layer\n\nThe boundary layer starts at the leading edge of a solid surface and the boundary layer thickness increases with the distance x along the surface.\n\nNear the leading edge of the solid surface, where thickness is small, the flow is laminar (LAMINAR BOUNDARY LAYER UPTO Re 3 x 105 ~ 3.2 x 105\n\nLAMINAR BOUNDARY LAYER PROFILE – PARABOLIC\n\nAs the thickness of the layer increases in the downstream direction, the laminar layer becomes unstable, leading to transition from laminar to turbulent boundary layer. Re ~ 5.5 x 105 ( Onset of turbulent BL)\n\nTURBULENT BOUNDARY LAYER – PROFILE BECOMES LOGARITHMIC\n\nTurbulent flow is characterized by greater interchange of mass momentum and energy within the fluid particles.\n\nThe velocity profile is more uniform in turbulent BL.\n\nVelocity gradient is higher in turbulent BL, hence shear stresses are higher.\n\nBL depends on Reynold’s number & also on the surface roughness. Roughness of the surface adds to the disturbance in the flow & hastens the transition from laminar to turbulent\n\n### Parameters Of Boundary Layer\n\nBOUNDARY LAYER THICKNESS (δ)\n\nThe velocity within a boundary layer approaches the free stream velocity value asymptotically, and so the limit of boundary layer is not easily defined.\n\nA distance δ is prescribed at the velocity lies within 1 % of the asymptotic value u = 0.99 U0\n\nδ becomes the measure of the thickness of a region in which major portion of the velocity distribution takes place\n\n### Parameters Of Boundary Layer\n\nDISPLACEMENT THICKNESS (δ*)\n\nConsider an elementary strip of thickness dy & at a distance y from the plate surface.\n\nArea of elementary strip dA = b x dy, where b is the width of the plate of this page.\n\nMass flow rate through this strip = ρ x flow velocity x area = ρub x dy\n\nIn absence of the plate, the fluid would have moved with a constant velocity equal to free stream velocity U0\n\nCorresponding mass flow rate = ρU0b x dy\n\nLoss in mass flow rate through the elemental strip = ρU0b x dy – ρub x dy = ρ(U0 – u) b x dy\n\nWhere δ is the value of y at which u = U0\n\nHence we define the displacement thickness as the thickness of flow moving at the free stream velocity and having the flow rate equal to the loss in flow rate on account of boundary layer formation.\n\n### Momentum Thickness (Θ)\n\nLoss in mass flow rate due to velocity defect = ρ(U0 – u) b dy\n\nLoss in momentum = ρ(U0 – u) b dy x u\n\nThe momentum thickness (θ) is defined as the thickness of flow moving at free stream velocity and having the same momentum flux equal to the deficiency of the momentum flux in the region of boundary layer.\n\nMomentum thickness can be conceived as the transverse distance by which the boundary layer should be displaced to compensate for the reduction in the momentum of the flowing fluid on account of the boundary layer formation\n\n### Energy Thickness (Δ**)\n\nMass of fluid = ρubdy\n\nK.E = ½ * ρu x b x dy x u2\n\nK.E. in absence of boundary layer = ½ * ρu x b x dy x U02\n\nloss in K.E through the elemental strip = ½ * ρu x b x dy x (U02 – u2)\n\nTotal loss of K.E.\n\nThe energy thickness δ** is defined as the thickness of the flow moving at the free stream velocity & having the energy equal to the deficiency of energy in the boundary layer region\n\nK.E . Through distance δ** = ½ * (ρbδ**U0)xU02 = ½ * (ρbδ**)xU03\n\nEnergy thickness may be conceived as the transverse distance by which the boundary layer should be displaced to compensate for the reduction in energy of the flowing fluid on account of the boundary layer formation\n\n### 2-D Boundary Layer Eqn.\n\nPRANDTL’S BOUNDARY LAYER EQUATIONS\n\nConsider steady two dimensional incompressible viscous flow in x-direction along the wall & y normal to the wall\n\nBASIC EQUATIONS (NAVIER STOKES EQUATION & CONTINUITY EQUATIONS)\n\n### BL On Flat Plate\n\nBLASIUS SOLUTION\n\nHe showed that the boundary equations can be solved exactly for u & v assuming free stream velocity U as constant or ∂U/∂x =0\n\nWith an ingenious coordinate transformation, Blasius showed that the dimensionless velocity profile u/U is a function only of a single composite dimensionless variable\n\n### Momentum Integral Eqn.\n\nA complete description of the boundary layer with the aid of non-linear differential equation is very cumbersome.\n\nTherefore approximate solution is required.\n\nSince boundary layer is satisfied in a stratum near the wall & near the region of transition. In the remaining region of the fluid, the mean over the differential equation is satisfied.\n\nThe mean is taken over the whole thickness of the boundary layer. Such a mean is obtained from the momentum equation, by integrating over the boundary layer thickness\n\n### Outcome Of The Results Derived So Far\n\nBOUNDARY LAYER THICKNESS INCREASES AS THE SQUARE ROOT OF THE DISTANCE x FROM THE LEADING EDGE AND INVERSELY AS SQUARE ROOT OF FREE STREAM VELOCITY.\n\nWALL SHEAR STRESS IS INVERSELY PROPORTIONAL TO THE SQUARE ROOT OF x AND DIRECTLY PROPORTIONAL TO 3/2 POWER OF U\n\nLOCAL & AVERAGE SKIN FRICTION VARY INVERSELY AS SQUARE ROOT OF BOTH x & U\n\n### Boundary Layer Separation\n\nIn adverse gradient, the second derivative of velocity is positive at the wall, yet it must be negative at the outer layer (y=δ) to merge smoothly with the mainstream flow U(x).\n\nIt follows that the second derivative must pass through zero somewhere in between, at the point of inflexion, and any boundary layer profile in an adverse must exhibit a characteristic S –shape."
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http://www.thescienceforum.com/physics/8321-effect-area-pressure-liquid.html
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[
"# Thread: effect of area on the pressure of a liquid\n\n1. was in a lesson and the teacher gave this talk which i was following until the end.\n\nIf a fluid flows in a pipe which has one point of entry and one point of exit the mass flow rate in = mass flow rate out.\ndue to law of conservation of mass. therefore we get the following equation\n\np(1)A(1)v(1) = p(2)A(2)v(2)\n\nbecause liquids are incompressible the density is constant whatever the pressure therefore\n\nA(1)v(1) = A(2)v(2)\n\nfluid travels fastest through the narrowest sections.\ntherefore indicating individual particles accelerate.\n\naccording to Newtons second law, the forces on the particles must therefore be unbalanced.\n\nThis implies the pressure is lowest at the narrowest points\n\ni don't understand how he derived the part in bold. can someone explain how this link was made, thanks",
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"2.\n\n3. It's a statement of the principle of energy conservation. See the wiki entry on Bernouilli's principle.\n\nhttp://en.wikipedia.org/wiki/Bernoulli's_equation",
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"Bookmarks\n Posting Permissions\n You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement"
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"http://www.thescienceforum.com/images/bluesaint/buttons/quote.gif",
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{"ft_lang_label":"__label__en","ft_lang_prob":0.95609164,"math_prob":0.94415706,"size":782,"snap":"2020-45-2020-50","text_gpt3_token_len":183,"char_repetition_ratio":0.10796915,"word_repetition_ratio":0.0,"special_character_ratio":0.22122762,"punctuation_ratio":0.05882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97664887,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-20T19:55:42Z\",\"WARC-Record-ID\":\"<urn:uuid:6c598005-7251-485a-8b54-0ca5ee2bab1f>\",\"Content-Length\":\"50571\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54e11a89-252b-4025-84dc-0cd350e95386>\",\"WARC-Concurrent-To\":\"<urn:uuid:420b86b5-b199-4a89-b9ee-7e8780fd654e>\",\"WARC-IP-Address\":\"95.217.34.176\",\"WARC-Target-URI\":\"http://www.thescienceforum.com/physics/8321-effect-area-pressure-liquid.html\",\"WARC-Payload-Digest\":\"sha1:3UU67JZ5RGYQ4V7QZR6EWCTL35M5ZUNQ\",\"WARC-Block-Digest\":\"sha1:CLTYE7NB2ERXG4EF4LENQ4GYXPNFGH3R\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107874135.2_warc_CC-MAIN-20201020192039-20201020222039-00670.warc.gz\"}"}
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https://www.phodal.com/blog/something-about-javascript-this/
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[
"PHODAL\n\n## 关于Javascript的this的一些知识\n\nJavascript应该是现在最流行的跨平台语言之一，一直在玩前端的一些有意思的东西，发现竟然没有掌握好这门语言。有点舍本逐末，于是想趁着现在这有空的时候好好补充一点遗漏的东西。\n\n### this的隐性绑定\n\n`````` function add (a,b) {return a+b}\nconsole.log(sum); //sum = 7``````\n\n`````` > typeof add\n> 'number'``````\n\n`` this.add(3,4)``\n\n``````var hello = function (){\nreturn \"Hello, \" + this.name;\n};\nname = 'this';\nconsole.log(hello());``````\n\n``````var hello = function (){\nreturn \"Hello, \" + this.name;\n};\nvar user = {\nhello : hello,\nname : 'phodal'，\n\n};\nconsole.log(user.hello());``````\n\n### var that = this\n\n`` vat that = this;``\n\ntips:\n\n• this变量的作用域总是由其最近的封闭函数所确定。\n• 使用一个局部变量(如me,self,that)让this绑定对于内部是可用的。\n\n``````var M = function(){\nthis.name = \"M\";\n};\n\nvar MM = function(){\nz = new M();\nthis.name = \"MM\";\n\nz.printName = function(){\nconsole.log(this.name);\n};\nreturn z.printName();\n};\n\nvar mm = new MM;``````\n\n``````var MM = function(){\nz = new M();\nthis.name = \"MM\";\nvar self = this;\nz.printName = function(){\nconsole.log(self.name);\n};\nreturn z.printName();\n};``````\n\n``````var MM = function(){\nz = new M();\nthis.name = \"MM\";\nz.printName = function(){\nconsole.log(this.name);\n}.bind(this);\nreturn z.printName();\n};``````\n\nbind可以将方法绑定到接收者。\n\n## 其他\n\n### 关于我\n\nGithub: @phodal 微博:@phodal 知乎:@phodal",
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"QQ技术交流群: 321689806\n\n### 最近的一些事\n\n• 最近我和我的同事们，一起在创建一个新的编程语言：Charj 。它是一个使用 Rust 编写的描述式、中间编程语言。GitHub: https://github.com/charj-lang/charj\n\nNov. 14, 2020, 9:27 p.m. | China"
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"https://www.phodal.com/static/phodal/images/qrcode.jpg",
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{"ft_lang_label":"__label__zh","ft_lang_prob":0.6609002,"math_prob":0.9694938,"size":1723,"snap":"2021-04-2021-17","text_gpt3_token_len":925,"char_repetition_ratio":0.15357766,"word_repetition_ratio":0.26060605,"special_character_ratio":0.26929775,"punctuation_ratio":0.27272728,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9548689,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-22T16:02:12Z\",\"WARC-Record-ID\":\"<urn:uuid:ab6efd81-f6d2-43c1-8235-a17844ab5854>\",\"Content-Length\":\"27204\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b4999e5-80d4-47df-84d1-fd505746d955>\",\"WARC-Concurrent-To\":\"<urn:uuid:177e431c-4b03-474f-bf2c-6d0a4cb13944>\",\"WARC-IP-Address\":\"13.125.244.183\",\"WARC-Target-URI\":\"https://www.phodal.com/blog/something-about-javascript-this/\",\"WARC-Payload-Digest\":\"sha1:DQMTM3PKAUJJCBOQ2WK7TJZDD6MPQYQX\",\"WARC-Block-Digest\":\"sha1:7MX33KU35HQCGISECTK5YC2UGQ6DNHSL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703530835.37_warc_CC-MAIN-20210122144404-20210122174404-00692.warc.gz\"}"}
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http://stcorp.github.io/harp/doc/html/ingestions/GEOMS-TE-FTIR-001-NO2.html
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[
"# GEOMS-TE-FTIR-001-NO2\n\n## Variables\n\nThe table below lists the variables that are present in the HARP product that results from an ingestion of `GEOMS-TE-FTIR-001-NO2` data.\n\nfield name\n\ntype\n\ndimensions\n\nunit\n\ndescription\n\nsensor_name\n\nstring\n\nname of the sensor\n\nlocation_name\n\nstring\n\nname of the site at which the sensor is located\n\nmeasurement_mode\n\nstring\n\n‘solar’ or ‘lunar’ measurement\n\nsensor_latitude\n\ndouble\n\n[degree_north]\n\nlatitude of the sensor\n\nsensor_longitude\n\ndouble\n\n[degree_east]\n\nlongitude of the sensor\n\nsensor_altitude\n\ndouble\n\n[km]\n\naltitude of the sensor\n\ndatetime\n\ndouble\n\n{time}\n\n[days since 2000-01-01]\n\ntime of the measurement\n\ndatetime_length\n\ndouble\n\n{time}\n\n[s]\n\nduration of the measurement\n\nNO2_column_number_density\n\ndouble\n\n{time}\n\n[molec/m2]\n\ntotal NO2 vertical column\n\nNO2_column_number_density_apriori\n\ndouble\n\n{time}\n\n[molec/m2]\n\na priori total NO2 vertical column\n\nNO2_column_number_density_avk\n\ndouble\n\n{time, vertical}\n\n[]\n\naveraging kernel for the total NO2 vertical column\n\nNO2_column_number_density_uncertainty_random\n\ndouble\n\n{time}\n\n[molec/m2]\n\nrandom uncertainty of the total NO2 vertical column\n\nNO2_column_number_density_uncertainty_systematic\n\ndouble\n\n{time}\n\n[molec/m2]\n\nsystematic uncertainty of the total NO2 vertical column\n\nH2O_column_number_density\n\ndouble\n\n{time}\n\n[molec/m2]\n\ntotal H2O vertical column\n\nNO2_volume_mixing_ratio\n\ndouble\n\n{time, vertical}\n\n[ppmv]\n\nNO2 volume mixing ratio\n\nNO2_volume_mixing_ratio_apriori\n\ndouble\n\n{time, vertical}\n\n[ppmv]\n\na priori NO2 volume mixing ratio\n\nNO2_volume_mixing_ratio_avk\n\ndouble\n\n{time, vertical, vertical}\n\n[]\n\naveraging kernel for the NO2 volume mixing ratio\n\nNO2_volume_mixing_ratio_covariance\n\ndouble\n\n{time, vertical, vertical}\n\n[(ppmv)2]\n\ncovariance of the NO2 volume mixing ratio\n\nNO2_volume_mixing_ratio_uncertainty_random\n\ndouble\n\n{time, vertical}\n\n[ppmv]\n\nrandom uncertainty of the NO2 volume mixing ratio\n\nNO2_volume_mixing_ratio_uncertainty_systematic\n\ndouble\n\n{time, vertical}\n\n[ppmv]\n\nsystematic uncertainty of the NO2 volume mixing ratio\n\nH2O_volume_mixing_ratio\n\ndouble\n\n{time, vertical}\n\n[ppmv]\n\nH2O volume mixing ratio\n\naltitude\n\ndouble\n\n{time, vertical}\n\n[km]\n\nretrieval effective altitude\n\naltitude_bounds\n\ndouble\n\n{time, vertical, 2}\n\n[km]\n\nlower and upper boundaries of the height layers\n\npressure\n\ndouble\n\n{time, vertical}\n\n[hPa]\n\nindependent pressure profile\n\ntemperature\n\ndouble\n\n{time, vertical}\n\n[K]\n\nindependent temperature profile\n\nsurface_pressure\n\ndouble\n\n{time}\n\n[hPa]\n\nindependent surface pressure\n\nsurface_temperature\n\ndouble\n\n{time}\n\n[K]\n\nindependent surface temperature\n\nsolar_azimuth_angle\n\ndouble\n\n{time}\n\n[degree]\n\nsolar azimuth angle\n\nsolar_zenith_angle\n\ndouble\n\n{time}\n\n[degree]\n\nsolar zenith angle\n\nindex\n\nint32\n\n{time}\n\nzero-based index of the sample within the source product\n\n## Mapping description\n\nThe table below details where and how each variable was retrieved from the input product.\n\nfield name\n\nmapping description\n\nsensor_name\n\npath\n\n/@DATA.SOURCE\n\nlocation_name\n\npath\n\n/@DATA.LOCATION\n\nmeasurement_mode\n\ncondition\n\ndetermined from ‘variable mode’ part of variable names\n\nsensor_latitude\n\npath\n\n/LATITUDE.INSTRUMENT\n\nsensor_longitude\n\npath\n\n/LONGITUDE.INSTRUMENT\n\nsensor_altitude\n\npath\n\n/ALTITUDE.INSTRUMENT\n\ndatetime\n\npath\n\n/DATETIME\n\ndatetime_length\n\navailable\n\noptional\n\npath\n\n/INTEGRATION.TIME\n\nNO2_column_number_density\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.SOLAR\n\ndescription\n\nunit is converted to molec/m2\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.LUNAR\n\ndescription\n\nunit is converted to molec/m2\n\nNO2_column_number_density_apriori\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.SOLAR_APRIORI\n\ndescription\n\nunit is converted to molec/m2\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.LUNAR_APRIORI\n\ndescription\n\nunit is converted to molec/m2\n\nNO2_column_number_density_avk\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.SOLAR_AVK\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.LUNAR_AVK\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere\n\nNO2_column_number_density_uncertainty_random\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.SOLAR_UNCERTAINTY.RANDOM\n\ndescription\n\nunit is converted to molec/m2\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.LUNAR_UNCERTAINTY.RANDOM\n\ndescription\n\nunit is converted to molec/m2\n\nNO2_column_number_density_uncertainty_systematic\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.SOLAR_UNCERTAINTY.SYSTEMATIC\n\ndescription\n\nunit is converted to molec/m2\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.COLUMN_ABSORPTION.LUNAR_UNCERTAINTY.SYSTEMATIC\n\ndescription\n\nunit is converted to molec/m2\n\nH2O_column_number_density\n\ncondition\n\nsolar measurement\n\npath\n\n/H2O.COLUMN_ABSORPTION.SOLAR\n\ndescription\n\nunit is converted to molec/m2\n\ncondition\n\nlunar measurement\n\npath\n\n/H2O.COLUMN_ABSORPTION.LUNAR\n\ndescription\n\nunit is converted to molec/m2\n\nNO2_volume_mixing_ratio\n\navailable\n\noptional\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.SOLAR\n\ndescription\n\nunit is converted to ppmv; the vertical axis is re-ordered from surface to top-of-atmosphere\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.LUNAR\n\ndescription\n\nunit is converted to ppmv; the vertical axis is re-ordered from surface to top-of-atmosphere\n\nNO2_volume_mixing_ratio_apriori\n\navailable\n\noptional\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.SOLAR_APRIORI\n\ndescription\n\nunit is converted to ppmv; the vertical axis is re-ordered from surface to top-of-atmosphere\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.LUNAR_APRIORI\n\ndescription\n\nunit is converted to ppmv; the vertical axis is re-ordered from surface to top-of-atmosphere\n\nNO2_volume_mixing_ratio_avk\n\navailable\n\noptional\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.SOLAR_AVK\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.LUNAR_AVK\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere\n\nNO2_volume_mixing_ratio_covariance\n\navailable\n\noptional\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.SOLAR_UNCERTAINTY.RANDOM\n\ndescription\n\nunit is converted to (ppmv)2; the vertical axis is re-ordered from surface to top-of-atmosphere\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.LUNAR_UNCERTAINTY.RANDOM\n\ndescription\n\nunit is converted to (ppmv)2; the vertical axis is re-ordered from surface to top-of-atmosphere\n\nNO2_volume_mixing_ratio_uncertainty_random\n\navailable\n\noptional\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.SOLAR_UNCERTAINTY.RANDOM\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere; the uncertainty is the square root of the trace of the covariance; unit is converted to ppmv\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.LUNAR_UNCERTAINTY.RANDOM\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere; the uncertainty is the square root of the trace of the covariance; unit is converted to ppmv\n\nNO2_volume_mixing_ratio_uncertainty_systematic\n\navailable\n\noptional\n\ncondition\n\nsolar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.SOLAR_UNCERTAINTY.SYSTEMATIC\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere; the uncertainty is the square root of the trace of the covariance; unit is converted to ppmv\n\ncondition\n\nlunar measurement\n\npath\n\n/NO2.MIXING.RATIO_ABSORPTION.LUNAR_UNCERTAINTY.SYSTEMATIC\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere; the uncertainty is the square root of the trace of the covariance; unit is converted to ppmv\n\nH2O_volume_mixing_ratio\n\ncondition\n\nsolar measurement\n\npath\n\n/H2O.MIXING.RATIO_ABSORPTION.SOLAR\n\ndescription\n\nunit is converted to ppmv; the vertical axis is re-ordered from surface to top-of-atmosphere\n\ncondition\n\nlunar measurement\n\npath\n\n/H2O.MIXING.RATIO_ABSORPTION.LUNAR\n\ndescription\n\nunit is converted to ppmv; the vertical axis is re-ordered from surface to top-of-atmosphere\n\naltitude\n\npath\n\n/ALTITUDE\n\ndescription\n\nthe vertical axis is re-ordered from surface to top-of-atmosphere\n\naltitude_bounds\n\npath\n\n/ALTITUDE.BOUNDS\n\ndescription\n\nlower and upper boundaries of the height layers\n\npressure\n\npath\n\n/PRESSURE_INDEPENDENT\n\ndescription\n\nindependent pressure profile\n\ntemperature\n\npath\n\n/TEMPERATURE_INDEPENDENT\n\ndescription\n\nindependent temperature profile\n\nsurface_pressure\n\npath\n\n/SURFACE.PRESSURE_INDEPENDENT\n\nsurface_temperature\n\npath\n\n/SURFACE.TEMPERATURE_INDEPENDENT\n\nsolar_azimuth_angle\n\ncondition\n\nsolar measurement\n\npath\n\n/ANGLE.SOLAR_AZIMUTH\n\ncondition\n\nlunar measurement\n\npath\n\n/ANGLE.LUNAR_AZIMUTH\n\nsolar_zenith_angle\n\ncondition\n\nsolar measurement\n\npath\n\n/ANGLE.SOLAR_ZENITH.ASTRONOMICAL\n\ncondition\n\nlunar measurement\n\npath\n\n/ANGLE.LUNAR_ZENITH.ASTRONOMICAL"
] |
[
null
] |
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|
https://indico.cern.ch/event/804609/contributions/4939280/
|
[
"#",
null,
"Nuclear Physics in Astrophysics - X\n\nSep 4 – 9, 2022\nCERN\nEurope/Zurich timezone\n\n## Underground measurement at LUNA found no evidence for a low-energy resonance in the $^{6}$Li(p, $\\gamma$)$^{7}$Be reaction\n\nSep 6, 2022, 6:00 PM\n3h\n61/1-201 - Pas perdus - Not a meeting room - (CERN)\n\nShow room on map\nPoster\n\nDenise Piatti\n\n### Description\n\nThe $^{6}$Li(p, $\\gamma$)$^{7}$Be reaction is involved both in Big Bang Nucleosynthesis and $^{6}$Li depletion in pre-main and in main sequence stars.\nThe $^{6}$Li(p, $\\gamma$)$^{7}$Be S-factor trend was poorly constrained at astrophysical energies because of conflicting experimental results reported in literature. A recent direct measurement, indeed, found a resonance-like structure at E$_{c.m.}\\,$=$\\,$195 keV, corresponding to an excited state at E$_{x}\\sim\\,$5800 keV in $^{7}$Be which, however, has not been confirmed by either other direct measurements or theoretical calculations.\n\nIn order to clarify the existence of this resonance, a new experiment was performed at the Laboratory for Underground Nuclear Astrophysics (LUNA), located deep underground in Gran Sasso Laboratory. Thanks to the extremely low background environment, the $^{6}$Li(p, $\\gamma$)$^{7}$Be cross section was measured in the center-of-mass energy range E$\\,$=$\\,$60-350 keV with unprecedented sensitivity. No evidence for the alleged resonance was found. LUNA results was confirmed by latest published indirect determination of $^{6}$Li(p, $\\gamma$)$^{7}$Be S-factor.\n\nIn the talk a detailed description of the experiment and of the final results will be provided."
] |
[
null,
"https://indico.cern.ch/event/804609/logo-1350957628.png",
null
] |
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|
https://www.informit.com/articles/article.aspx?p=440157&seqNum=10
|
[
"# Analog Design with Discrete Components\n\n• Print\nThis chapter is from the book\n\n## 3.10 Framework for TTL/CMOS Logic\n\nBefore we enter the digital realm and leave the world of analog electronics behind (for the most part), I thought it would be fun and very relevant to show examples of how transistors are used to create digital logic gates. Of course, you may have no idea what a digital logic gate is, but briefly, a logic gate is a hardware implementation of the standard logic functions you use when programming. For example:\n\nBitwise NOT Logic\n\n• z = ~y\n\nThis assigns x the bit inverted value of y.\n\nBitwise AND Logic\n\n• z = x & y\n\nThis assigns z the bitwise AND of x and y.\n\nBitwise OR Logic\n\n• z = x | y\n\nThis assigns z the bitwise OR of x and y.\n\nOf course the \"logical\" NOT, AND, and OR are somewhat different and deal with entire numbers; however, at the bit level, logical operations and bitwise are identical. With this in mind, if we are dealing with Boolean values only then can we map our computer code directly to hardware and make assignments like\n\n• BOOL x,y,z;\n• z = ~(x & y | z);\n\nSo the question is, how to implement this in hardware? There are lots of ways to do this; of course the way we will do this in the next chapter is with ICs that implement logic functions. But, for now, let’s see if we can go through the same thought process the IC designers did and build some simple logic gates from transistors and resistors that implement these logic functions or variations thereof. This way you can see the transition from analog to digital better, but still realize that under all the digital electronics are always analog components—always.\n\n### 3.10.1 Creating a Convention\n\nThe first thing we need to agree on for our little model experiment is a set of conventions; that is, what means what in relation to voltages? We are going to loosely model our gates after the most popular logic system in the world, TTL (transistor transistor logic), which refers to both a method of implementation and a set of voltages and currents. We are not so much interested in actual implementation of TTL since we are just trying to make something that is semi-compatible. So we need to know two things:\n\n1. What is the voltage level representation for a logic \"0\"?\n2. What is the voltage level representation of a logic \"1\"?\n\nIf you take a look at Figure 3.66 you will see a graph that shows the logic levels for a number of different logic families (courtesy of TI). We are only interested in the TTL family for now, since it’s easiest to model. Referring to the TTL switching standard there are a number of demarcations on the graph; their meanings are as follows:\n\n• VCC/GND—These are the \"rails\" of the power supply. VCC is the +5 supply and ground is 0V.\n• Vol—Voltage output low. This is the highest voltage a TTL gate will give when output a low.\n• Vil—Voltage input low. This is the highest voltage a TTL gate will detect as a low input.\n• Vih—Voltage input high. This is the lowest voltage a TTL gate will detect as a high input.\n• Voh—Voltage output high. This is the lowest voltage a TTL gate is guaranteed to output a high at.\n\nThese seems a little weird, but the bottom line is that we need to consider two things: what a TTL gate guarantees as low and high when it is an output, and what a TTL gate will detect as a low and high as an input. Between the text and the figure you should be able to get it. Finally, the last assumption is that the digital systems use +5V and that a high or \"1\" is +5 and a low or \"0\" is 0V ideally. With that in mind, we need to design our model \"TTL\" gates with those constraints or at least close.\n\n#### 3.10.1.1 Building an Inverter\n\nThe inverter is the basis on 99% of all digital logic; if you can invert a signal then you can usually derive everything from there, so we will start with an inverter. Figure 3.67 depicts one possible design. The circuit operation is as follows. If we apply a 0V to the base at input X (Vin) then the transistor is OFF; in this case, there is no base current and thus no collector current (other than a very small leakage current), therefore the output Z (Vout) will output nearly 5V. Thus, an input of 0V (logic 0) outputs a 5V (logic 1). So far so good.\n\nNow, the hard part. Let’s apply a 5V (logic 1) to the base at X (Vin); when we do this we turn ON the transistor and it starts conducting. When it does so the base-emitter junction wants to have a 0.6V drop across it; therefore, the base will be at\n\n• VB = VE + 0.6V\n\nand since VE is at ground, VB = 0.6V. Therefore, we can compute IB as the voltage drop over Rin divided by the resistance of Rin:\n\n• IB = (5V – 0.6V)/1KΩ = 4.4mA\n\nWe know that IC = hFE*IB roughly, and assuming an hFE of 100, we get an IC of\n\n• IC = 100*4.4mA = 44mA\n\nHowever, this is only tentative; it may be less if this current violates one of the transistor rules. Let’s see. With a collector current of 44mA, that means that the voltage drop over Rout is\n\n• Vrout = 44mA*100Ω = 4.4V\n\nThus, the voltage at the collector VC and the output Z (Vout) is (5V-4.4V)=0.6V, which is above the requirement of 0.2V worst case for the collector above ground. Therefore, when we apply a signal of 5V (logic 1), we get a 0.6V output roughly (logic 0), and thus we have an inverter. Table 3.1 lists the inputs and outputs, so you can confirm everything at once and assure yourself we have an inverter.\n\n#### Table 3.1 Logic/Response Table for the \"Model\" TTL Inverter\n\n Input Ouput X (Vin) Z (Vout) 0 (0V) 1 (5.0V) 1 (5V) 0 (0.6V)\n\nIt’s almost perfect! And in reality, the output low will tend to be lower. Of course, we haven’t tested against the Vil, Vih values to see if we match the tolerance on those constraints, but that’s not important. We at least have a TTL inverter that given solid inputs outputs TTL outputs.\n\n#### 3.10.1.2 Building a NAND Gate\n\nSo a single transistor can be used as an inverter. We can use this fact and the design now as a block and create other gates. Let’s create a NAND gate. Of course we could create an AND gate, but then we would have to invert its output, so a NAND is a little easier to implement.\n\nSince we have the analysis of the inverter under our belt, we can analyze the NAND gate without so much detail. Referring to Figure 3.68, there are two inverters more or less placed in series. The operation is as follows: When either input X,Y or both are at 0V then one or both of the transistors are OFF. Therefore, there is no appreciable collector current and the output Z will hang at approximately 5.0V, a bit less; remember there is still a bit of leakage current even when the transistor(s) are OFF. Thus, a logic 0,0 or 0,1, or 1,0 all result in an output of a logic 1 at nearly 5.0V. This is looking like a NAND so far. The last case is when we turn both transistors on with logic 1’s at X and Y. In this case, we turn both transistors on and then conduct hard; the collector voltages will each be 0.2V above the respective emitter’s for a worst case of 0.4V above ground at the final output node Z (Vout). So to reiterate when both inputs are driven to logic 1 (5V), the output goes to 0.4V or so, a logic 0.\n\nLet’s take these results and build another logic/response truth table as shown in Table 3.2.\n\n#### Table 3.2 The Logic/Response Truth Table for the 2-input NAND Gate\n\n Input Input Output X (Vin1) Y (Vin2) Z (Vout) 0 (0V) 0 (0V) 1 (5.0V) 0 (0V) 1 (5V) 1 (5.0V) 1 (5V) 0 (0V) 1 (5.0V) 1 (5V) 1 (5V) 0 (0.4V)\n\nReviewing the truth table, the circuit is indeed a NAND gate. Of course, if you wanted to, you could attach an inverter to the output at Z and then invert the NAND to create an AND gate.\n\n#### 3.10.1.3 Building a NOR Gate\n\nFinally, let’s make a NOR gate. Again we will use the inverter as our basis. Figure 3.69 shows the final 2-input NOR design; basically we simply put two inverters in parallel. Operation of the NOR is similar to the NAND. If both inputs are held at a logic 0 then neither transistor conducts and the output at Z is nearly 5V, or a logic 1. If either transistor is turned on with a logic 1 then the collector circuit of the transistor conducts hard; the other transistor is a high impedance load in parallel with the low impedance conducting transistor and can thus be disregarded. Therefore, with either or both transistors conducting, the collector circuit of one or both of the transistors will have a lot of current flowing and the voltage will drop over the output resistor Rout resulting in a near 0V at the output Z (Vout). Transcribing this behavior to Table 3.3, we see we indeed have a NOR gate!\n\n#### Table 3.3 The Logic/Response Truth Table for the 2-input NOR Gate\n\n Input Input Output X (Vin1) Y (Vin2) Z (Vout) 0 (0V) 0 (0V) 1 (5.0V) 0 (0V) 1 (5V) 0 (5.0V) 1 (5V) 0 (0V) 0 (5.0V) 1 (5V) 1 (5V) 0 (0.4V)\n\nThis concludes our analysis of the basic implementation of gates with a TTL-like model using transistors and resistors. Now, in the next chapter when we discuss output voltages, currents, etc. you will have a foundation and know where these things come from; they are the underlying analog parts that have been abstracted away, but they are always there nonetheless."
] |
[
null
] |
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|
https://www.inria.fr/en/centre/saclay/calendar/international-spring-school-on-formalization-of-mathematics
|
[
"Sites Inria",
null,
"Spring school\n\n# MAP Spring School",
null,
"Inria research centre Sophia Antipolis - Mediterranee organises the next MAP Spring School on Formalization of Mathematics. The aim of this school is to give mathematicians and mathematically inclined researchers the keys to the Coq system and the Mathematical Components library.\n\n• Date : 12/03/2012 to 16/03/2012\n• Place : Inria Sophia Antipolis - Méditerranée\n• Organiser(s) : Inria\n\nA growing population of mathematicians, computer scientists, and engineers use computers to construct and verify proofs of mathematical results. Among the various approaches to this activity, a fruitful one relies on interactive theorem proving. When following this approach, researchers have to use the formal language of a theorem prover to encode their mathematical knowledge and the proofs they want to scrutinize. The mathematical knowledge often contains two parts: a static part describing structures and a dynamic part describing algorithms. Then proofs are made in a style that is inspired from usual mathematical practice but often differs enough that it requires some training. A key ingredient for the mathematical practicionner is the amount of mathematical knowledge that is already available in the system's library.\n\nThe Coq system is an interactive theorem prover based on Type Theory. It was recently used to study the proofs of advanced mathematical results. In particular, it was used to provide a mechanically verified proof of the four-colour theorem and it is now being used in a long term effort, called Mathematical Components to verify results in group theory, with a specific focus on the odd order theorem, also known as the Feit-Thompson theorem. These two examples rely on a structured library that covers various aspects of finite set theory, group theory, arithmetic, and algebra.\n\nThe aim of this school is to give mathematicians and mathematically inclined researchers the keys to the Coq system and the Mathematical Components library. The topics covered are:\n\n• Formal proof techniques\n• Structuration of libraries\n• Encoding of common mathematical structures\n• Formal description of algorithms"
] |
[
null,
"https://www.inria.fr/extension/site_inria/design/site_inria/images/pictos/picto_lang_fr.png",
null,
"https://www.inria.fr/var/inria/storage/images/medias/saclay/actualites-images/saclay-map-coq-260x195/431308-1-fre-FR/saclay-map-coq-260x195_vignette.jpg",
null
] |
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|
https://lzone.de/examples/Python%20re.sub
|
[
"# Python re.sub Examples\n\n## Syntax\n\n``````import re\n\nresult = re.sub(pattern, repl, string, count=0, flags=0);\n``````\n\n## Simple Examples\n\n``````result = re.sub('abc', '', input) # Delete pattern abc\nresult = re.sub('abc', 'def', input) # Replace pattern abc -> def\nresult = re.sub(r'\\s+', ' ', input) # Eliminate duplicate whitespaces using wildcards\nresult = re.sub('abc(def)ghi', r'\\1', input) # Replace a string with a part of itself\n``````\n\nNote: Take care to always prefix patterns containing \\ escapes with raw strings (by adding an r in front of the string). Otherwise the \\ is used as an escape sequence and the regex won’t work.\n\n### Replacement Function\n\nInstead of a replacement string you can provide a function performing dynamic replacements based on the match string like this:\n\n``````def my_replace(m):\nif <some condition>:\nreturn <replacement variant 1>\nreturn <replacement variant 2>\n\nresult = re.sub(\"\\w+\", my_replace, input)\n``````\n\n### Count Replacements\n\nWhen you want to know how many replacements did happen use re.subn() instead of re.sub()\n\n``````print ('Result: ', result)\nprint ('Replacements: ', result)\n``````\n\n### Using Backreferences\n\n#### Numeric Backreferences\n\nTo use back reference define capture groups using () and reference to those using \\1, \\2, and so on. Do not forget the ‘r’ prefix on the back reference string, otherwise \\1 will be interpreted as a character.\n\n``````result = re.sub(\"(\\d+) (\\w+)\", r\"\\2 \\1\")\n``````\n\n#### Named Backreferences\n\nFor named backreferences define a named capture group (?) and reference using \\g. Again ensure to add the 'r' prefix on the backreference string.\n\n``````result = re.sub(\"(?<number>\\d+) (?<word>\\w+)\", r\"\\g<word> \\g<number>\")\n``````\nRelated"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.56245804,"math_prob":0.72568995,"size":1523,"snap":"2023-40-2023-50","text_gpt3_token_len":377,"char_repetition_ratio":0.17050691,"word_repetition_ratio":0.0,"special_character_ratio":0.27445832,"punctuation_ratio":0.15646258,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98179424,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T15:29:34Z\",\"WARC-Record-ID\":\"<urn:uuid:8619750b-054a-4eed-966b-c0f648b15923>\",\"Content-Length\":\"47495\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d1a0491e-4249-49f7-a98b-3a88bd2d3b63>\",\"WARC-Concurrent-To\":\"<urn:uuid:4df68d6d-95e6-4288-846e-13a3583fb54b>\",\"WARC-IP-Address\":\"104.21.1.110\",\"WARC-Target-URI\":\"https://lzone.de/examples/Python%20re.sub\",\"WARC-Payload-Digest\":\"sha1:4Q7OSD37GVZE2YNUQ7JDO6PN63RUV4HU\",\"WARC-Block-Digest\":\"sha1:WWHEM676XBK5YUEPV4WUWQZPJQWVCQLD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510697.51_warc_CC-MAIN-20230930145921-20230930175921-00686.warc.gz\"}"}
|
http://www.gialdiniworld.com/multiplication-worksheet-for-grade-5/
|
[
"# Multiplication Worksheet For Grade 5\n\nmultiplication worksheet for grade 5 multiplication worksheets, grade 5 multiplication worksheets multiplication worksheet for grade 5, multiplication worksheet for grade 5 best 25 multiplication worksheets ideas on pinterest, grade 5 multiplication division worksheets free printable multiplication worksheet for grade 5,",
null,
"Multiplication Worksheet For Grade 5 Multiplication Worksheets",
null,
"Grade 5 Multiplication Worksheets Multiplication Worksheet For Grade 5",
null,
"Multiplication Worksheet For Grade 5 Best 25 Multiplication Worksheets Ideas On Pinterest",
null,
"Grade 5 Multiplication Division Worksheets Free Printable Multiplication Worksheet For Grade 5",
null,
"Multiplication Worksheet For Grade 5 Multiplication Practice Worksheet 5",
null,
"Grade 5 Math Worksheet Multiplication In Columns Multiply 3 Multiplication Worksheet For Grade 5",
null,
"Halloween Multiplication Practice Worksheets 1 And 2 Multiplication Worksheet For Grade 5\n\ngrade 5 multiplication division worksheets free printable multiplication worksheet for grade 5, multiplication worksheet for grade 5 multiplication practice worksheet 5, grade 5 math worksheet multiplication in columns multiply 3 multiplication worksheet for grade 5, halloween multiplication practice worksheets 1 and 2 multiplication worksheet for grade 5,"
] |
[
null,
"http://www.gialdiniworld.com/wp-content/uploads/multiplication-worksheet-for-grade-5-multiplication-worksheets.png",
null,
"http://www.gialdiniworld.com/wp-content/uploads/grade-5-multiplication-worksheets-multiplication-worksheet-for-grade-5.gif",
null,
"http://www.gialdiniworld.com/wp-content/uploads/multiplication-worksheet-for-grade-5-best-25-multiplication-worksheets-ideas-on-pinterest.jpg",
null,
"http://www.gialdiniworld.com/wp-content/uploads/grade-5-multiplication-division-worksheets-free-printable-multiplication-worksheet-for-grade-5.gif",
null,
"http://www.gialdiniworld.com/wp-content/uploads/multiplication-worksheet-for-grade-5-multiplication-practice-worksheet-5.jpg",
null,
"http://www.gialdiniworld.com/wp-content/uploads/grade-5-math-worksheet-multiplication-in-columns-multiply-3-multiplication-worksheet-for-grade-5.gif",
null,
"http://www.gialdiniworld.com/wp-content/uploads/halloween-multiplication-practice-worksheets-1-and-2-multiplication-worksheet-for-grade-5.jpg",
null
] |
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|
https://www.uwplatt.edu/department/mathematics/calculator-policies-0
|
[
"Many of the courses in the University of Wisconsin-Platteville Mathematics Department require calculators. A cell phone calculator is never allowed on tests, so plan to have a calculator if your course requires one. Individual instructors may restrict the use of any type of calculator in their class.\n\n### Calculator Requirements\n\nFor the following courses, a scientific calculator (such as one of the TI-30 models) or a graphing calculator (such as the TI-83, 84, 85, 86 or the TI-Nspire with TI-84 keyboard) is required:\n\n• MATH 1530 College Algebra\n• MATH 1630 Finite Mathematics with Applications\n• MATH 1730 Mathematics of Finance (no business calculators allowed)\n• MATH 1830 Elementary Statistics\n• MATH 2450 Precalculus\n• MATH 2530 Trigonometry and Analytic Geometry\n• MATH 2640 Calculus and Analytic Geometry I\n• MATH 2730 Discrete Mathematics\n• MATH 2740 Calculus and Analytic Geometry II\n• MATH 2840 Calculus and Analytic Geometry III\n• MATH 3630 Differential Equations I\n• MATH 4030 Statistical Methods with Applications\n\nIf you are enrolling in a 1000-level course and do not already own a graphing calculator, we recommend that you wait to purchase one until after the first day of class. Your instructor will provide specific calculator requirements for that class.\n\n### CAS (Computer Algebra Systems) Calculators\n\nCalculators with Computer Algebra Systems (CAS), such as the TI-89, TI-92 and TI-inspire with CAS keypad, or their equivalent, are not allowed in any math course. In other courses, calculators may be required but there are no specifications as to the type.\n\n### MATH 12 and 15\n\nIn MATH 12 and MATH 15, scientific calculators are required.\n\n• Graphing calculators may not be used during any test or the final exam.\n• Basic four-function calculators do not have enough features for MATH 12 or MATH 15.\n• Business/financial calculators are not appropriate for MATH 12 or MATH 15.\n\nSome brands are not allowed because of capabilities that interfere with the course objectives. You should carefully check your calculator to make sure you have one on the following approved calculators:\n\n• TI-30Xa\n• TI-30X IIB*\n• TI-30X IIS*\n• TI-34 II Explorer Plus*\n• TI-36X II*\n• TI-36X Solar*\n• Casio FX-115MSPlus*\n• Casio FX-300MSPlus*\n• Casio FX-260Solar\n• Casio FX-250HC\n• Sharp EL-531WBBK*\n• Sharp EL-531WBBL*\n• Sharp EL-501WBBK\n• Sharp EL-501WBBL\n• Sharp EL-500WBBK\n\n*Also sufficient for MATH 1830\n\nThe following calculators are NOT allowed for any MATH 12 or MATH 15 tests:\n\n• Any graphing or programmable calculator\n• Cell phones: Phones need to be stored away during class\n• TI-30XS MultiView\n• TI-34 MultiView\n• TI-36 Pro\n• Casio FX-115ES\n• Casio FX-300ES\n• Sharp EL-520WBBK\n• Sharp EL-506WBBK\n• Sharp EL-W516B\n• Sharp EL-W535B\n\n## Contact Information\n\n### Mathematics\n\n0435 Gardner Hall"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8173772,"math_prob":0.85437715,"size":2743,"snap":"2022-27-2022-33","text_gpt3_token_len":736,"char_repetition_ratio":0.17086528,"word_repetition_ratio":0.009029346,"special_character_ratio":0.26430914,"punctuation_ratio":0.05970149,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9774035,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T22:02:07Z\",\"WARC-Record-ID\":\"<urn:uuid:3ddbee7d-07d6-4207-8618-eb416027280b>\",\"Content-Length\":\"58052\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f3144844-9060-4438-8831-09ca71837604>\",\"WARC-Concurrent-To\":\"<urn:uuid:40169fcf-66d9-4d72-badb-90c161b40b4b>\",\"WARC-IP-Address\":\"137.104.137.134\",\"WARC-Target-URI\":\"https://www.uwplatt.edu/department/mathematics/calculator-policies-0\",\"WARC-Payload-Digest\":\"sha1:MTZBYY7Q2ZHEZITNPWN7SXXRNGRK354O\",\"WARC-Block-Digest\":\"sha1:6KCA3KOGO57IQW5KNOUOGTA2BKSQUCOM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104628307.87_warc_CC-MAIN-20220705205356-20220705235356-00499.warc.gz\"}"}
|
https://hrj.episciences.org/117
|
[
"## Aleksandar Ivić - The number of finite non-isomorphic abelian groups in mean square.\n\nhrj:117 - Hardy-Ramanujan Journal, January 1, 1986, Volume 9 -1986 - https://doi.org/10.46298/hrj.1986.117\nThe number of finite non-isomorphic abelian groups in mean square.\n\nAuthors: Aleksandar Ivić\n\nLet $\\Delta(x)=\\sum_{n\\leq x}a(n)-\\sum_{j=1}^6 c_jx^{1/j}$ denote the error term in the abelian group problem. Using zeta-function methods it is proved that $\\int_1^X\\Delta^2(x)\\,dx~<\\!\\!<~ X^{39/29} \\log^2X$ where the exponent $39/29=1.344827\\ldots$ is close to the best possible exponent $4/3$ in this problem.\n\nVolume: Volume 9 -1986\nPublished on: January 1, 1986\nImported on: March 3, 2015\nKeywords: number of non-isomorphic abelian groups,mean square estimates,power moments of the zeta-function.,[MATH] Mathematics [math]"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.65718806,"math_prob":0.95870256,"size":427,"snap":"2023-14-2023-23","text_gpt3_token_len":153,"char_repetition_ratio":0.09456265,"word_repetition_ratio":0.0,"special_character_ratio":0.33489463,"punctuation_ratio":0.08988764,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997224,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T19:21:52Z\",\"WARC-Record-ID\":\"<urn:uuid:85396e05-2458-4ad0-85d9-2354d73a6b8f>\",\"Content-Length\":\"37236\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:48e099b5-5bfb-4d30-876b-66ab78e5e72a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6cce5c23-d561-4603-8c47-d92e4069a704>\",\"WARC-IP-Address\":\"193.48.96.94\",\"WARC-Target-URI\":\"https://hrj.episciences.org/117\",\"WARC-Payload-Digest\":\"sha1:7IHWQO3RO6XS4LBILA45OIO3YUIAOW5K\",\"WARC-Block-Digest\":\"sha1:BRCUKD3H5FR36WQY2FIXPAS6AFLJYOOM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653071.58_warc_CC-MAIN-20230606182640-20230606212640-00365.warc.gz\"}"}
|
https://numberworld.info/502165640
|
[
"# Number 502165640\n\n### Properties of number 502165640\n\nCross Sum:\nFactorization:\n2 * 2 * 2 * 5 * 12554141\nDivisors:\nCount of divisors:\nSum of divisors:\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\n1dee7088\nBase 32:\neuss48\nsin(502165640)\n-0.78276163240199\ncos(502165640)\n0.62232164259278\ntan(502165640)\n-1.2578087902275\nln(502165640)\n20.03444058339\nlg(502165640)\n8.7008469933851\nsqrt(502165640)\n22409.05263504\nSquare(502165640)\n\n### Number Look Up\n\nLook Up\n\n502165640 which is pronounced (five hundred two million one hundred sixty-five thousand six hundred forty) is a unique figure. The cross sum of 502165640 is 29. If you factorisate the figure 502165640 you will get these result 2 * 2 * 2 * 5 * 12554141. 502165640 has 16 divisors ( 1, 2, 4, 5, 8, 10, 20, 40, 12554141, 25108282, 50216564, 62770705, 100433128, 125541410, 251082820, 502165640 ) whith a sum of 1129872780. The number 502165640 is not a prime number. The number 502165640 is not a fibonacci number. The figure 502165640 is not a Bell Number. 502165640 is not a Catalan Number. The convertion of 502165640 to base 2 (Binary) is 11101111011100111000010001000. The convertion of 502165640 to base 3 (Ternary) is 1021222220122202102. The convertion of 502165640 to base 4 (Quaternary) is 131323213002020. The convertion of 502165640 to base 5 (Quintal) is 2012023300030. The convertion of 502165640 to base 8 (Octal) is 3573470210. The convertion of 502165640 to base 16 (Hexadecimal) is 1dee7088. The convertion of 502165640 to base 32 is euss48. The sine of the figure 502165640 is -0.78276163240199. The cosine of the figure 502165640 is 0.62232164259278. The tangent of the number 502165640 is -1.2578087902275. The square root of 502165640 is 22409.05263504.\nIf you square 502165640 you will get the following result 252170329996609600. The natural logarithm of 502165640 is 20.03444058339 and the decimal logarithm is 8.7008469933851. that 502165640 is very unique number!"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.72271115,"math_prob":0.9661211,"size":2317,"snap":"2020-24-2020-29","text_gpt3_token_len":788,"char_repetition_ratio":0.20709036,"word_repetition_ratio":0.24332345,"special_character_ratio":0.5278377,"punctuation_ratio":0.16235293,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99796146,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-05T19:42:11Z\",\"WARC-Record-ID\":\"<urn:uuid:1986c484-7ac6-460c-94ed-b37a7d2d7da0>\",\"Content-Length\":\"14479\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97a37d2b-44f5-4758-8e21-524cfe649f4b>\",\"WARC-Concurrent-To\":\"<urn:uuid:16e58eeb-5d25-403c-855b-e021ce34a6b7>\",\"WARC-IP-Address\":\"176.9.140.13\",\"WARC-Target-URI\":\"https://numberworld.info/502165640\",\"WARC-Payload-Digest\":\"sha1:3JRICGCER5GYS55XP4F2UBUOZQMVGDXD\",\"WARC-Block-Digest\":\"sha1:PTGJTY7G63NHACEJZWAG4YPUIKEFRB2Z\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655888561.21_warc_CC-MAIN-20200705184325-20200705214325-00461.warc.gz\"}"}
|
https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Beginning_Algebra_(Redden)/01%3A_Real_Numbers_and_Their_Operations
|
[
"$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.55033594,"math_prob":1.0000042,"size":2045,"snap":"2019-51-2020-05","text_gpt3_token_len":620,"char_repetition_ratio":0.24987751,"word_repetition_ratio":0.20512821,"special_character_ratio":0.33202934,"punctuation_ratio":0.12765957,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97009635,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T23:13:42Z\",\"WARC-Record-ID\":\"<urn:uuid:ec322f48-f299-49a6-974d-1cafd33bab90>\",\"Content-Length\":\"88404\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a73399c6-55a7-4f26-a27b-9b5a5edb39b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:db84c55b-c5c6-4a8d-95d9-703946960a38>\",\"WARC-IP-Address\":\"99.84.181.83\",\"WARC-Target-URI\":\"https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Beginning_Algebra_(Redden)/01%3A_Real_Numbers_and_Their_Operations\",\"WARC-Payload-Digest\":\"sha1:M2DUPEQY3ZKNEWM63VTN2F7PIM577VC3\",\"WARC-Block-Digest\":\"sha1:5FLNJWGIMA7RRFDO2SVJWE776VH53HME\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250607596.34_warc_CC-MAIN-20200122221541-20200123010541-00210.warc.gz\"}"}
|
http://buildingcontractorsecrets.com/2010/09/09/how-to-calculate-numbers-of-blocks-from-a-building-plan/
|
[
"# How to calculate numbers of blocks from a building plan.",
null,
"(click on image for a bigger view or save image)\n\nMost asked question from a would be house owner is always how many blocks do I need to get my building completed. Hence this topic is written with illustrations to make it easy for you to do\n\nthe calculations yourself from the plan your architect gave you.\nThe plan above is for a 2 bedroom bungallow, so we are going to start from the basics having in\nmind these factors-\ni)The lenght of a block is 0.45m or 450mm or 18″\nii)The width of thickness varies -150mm or 6″ and 225mm or 9″ which are most common.\nii)Taking for a general or average purpose, the blocks laid from foundation to dpc(german or\noversite concrete) is 4 coaches/levels. From dpc to the base of window another 4 coaches,from\nbase of window to top of window another 5 coaches, then from top of window to base of roof\nanother 3 coaches, neglecting the lintel or concrete on top of window.\nNow with these parameters let calculate the plan above which has a wall of 150mm or 6″ all\nthrough.\nAdd the length and breadth of the blockwall for external and internal\nTherefore (A)9,400mm +9,600mm(B)+9400(C)+9600(D)=38,000mm or 38m\nNow add the internal walls too\n(Follow the grid lines and check the corresponding dimensions on plan remember 150mm is the\nblock width)\n(Grid 7)150+3600+150+150+3600+150=7,800mm or 7.8m\n(Grid 6)1350+150+2250+150+825+675+150=5,550mm or 5.55m\n(Grid 1b)800+2400+1300+900+100=5,500mm or 5.5m\n(Grid 2)2600+900+100+150+100+900+500+150=5,400mm or 5.4m\n(Grid 3)1800+150+650+900+250+900+1450=6,100mm or 6.1m\n(Grid 3b-Kitchen)900+50+1200+50+150=2,350 or 2.35m\n(Grid 7b toilet)1800 or 1.8m\nTotal internal walls=34,500mm or 34.5m\nTherefore Total walls=38,000mm +34,500mm=72,500mm or 72.5m\nNow for the first 4 coaches to dpc level the wall will run right through, i.e the spaces for\ndoors or windows are not left out (n/b- blocks are used in relatively strong and stable soil not\nwaterlogged area that requires concrete instead).\nDivide the sum by 450mm to know the numbers needed for one level/layer/coach of blocks\n72,500/450=161 blocks\nFrom dpc level to roof lets take the numbers of coaches to be 4+5+3=12 coaches\nthen multiply the numbers of blocks for each coach above by 12\n161 x 12=1932 blocks\nDon’t forget we have windows and doors spaces which we need to remove,for each window of 1200mm\ntake numbers of the block to be 14 (i.e 1200/230=5.2 and 1200/450=2.6 ,5.3 x2.6=14 approx\nNow count the numbers of 1200mm windows and multiply by 14\nTotal numbers of 1200mm windows as counted on plan is 12 (note 2400mm window was taken as two of\n1200mm window)\ni.e 11 x 14 =154 blocks + 7 blocks for 600mm window for toilet. Total=161 blocks\nNow calculate for doors,Numbers of blocks for a door of 900mm is 18.\nheight of a door(2100/225=9)+width of a door (900mm/450mm=2). so 9 x2 =18 blocks for each door.\nMultiply 18 by numbers of doors=18 x8=144 blocks\nAdd 144+161=305 blocks,then remove from total blocks of 644 (for dpc) + 1932\nTotal numbers of blocks reqd=2,271 blocks\nFor a quick calculation, having the entire length to be 72,500mm. Take for a bungallow total\nlenght from foundation to roof to be 1,200mm +3000mm=4200mm\nThen calculate the surface area which is 72,500 x 4200 and convert to meters\n72.5m x 4.2m=304.5m2\nRemove surface areas of windows and doors\nFor window, height is 1.2m and total is 11\ni.e 1.2m x 1.2m x 11=15.84m2 +0.6 x0.6(toilet window)=16.2m2\nFor doors, height is 2.1m while width is 0.9m\ni.e 0.9 x 2.1 x 8(nos of doors)=15.12\nTotal area for doors and windows=16.2m2 +15.12m2=31.32m2\nRemove 31.32m2 from 304.5m2=273.18m2\nNow divide this by surface area of a block which is 0.225 x 0.45=0.10125\nThe blocks reqd caculated by this method is 273.18m2/0.10125=2,698 blocks\nNote a difference of about 400 blocks from the two calculations. Of course you can’t get exactly\nthe numbers you need but the first method is close to it. And calculations are not 100% true\nbecause of these factors\n1-The mortar space was not considered\n2-Blocks that will be wasted due to breakages e.t.c"
] |
[
null,
"http://buildingcontractorsecrets.com/wp-content/uploads/2010/09/bbborie1.jpg",
null
] |
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https://www.aqua-calc.com/one-to-all/volume/preset/megaliter/31
|
[
"Convert megaliters [Ml] to other units of volume\n\nmegaliters [Ml] volume conversions\n\n 31 Ml = 25.13 acre-feet Ml to ac-ft 31 Ml = 45 588 235 285 capsules 0 Ml to cap 0 31 Ml = 34 444 444 441 capsules 00 Ml to cap 00 31 Ml = 22 627 737 217 capsules 000 Ml to cap 000 31 Ml = 31 000 000 000 capsules 00E Ml to cap 00E 31 Ml = 39 743 589 742 capsules 0E Ml to cap 0E 31 Ml = 64 583 333 323 capsules 1 Ml to cap 1 31 Ml = 1 722 222 223.6 capsules 10 Ml to cap 10 31 Ml = 3 100 000 000 capsules 11 Ml to cap 11 31 Ml = 6 200 000 000 capsules 12 Ml to cap 12 31 Ml = 4 133 333 323 capsules 12el Ml to cap 12el 31 Ml = 9 687 500 000 capsules 13 Ml to cap 13 31 Ml = 86 111 111 118 capsules 2 Ml to cap 2 31 Ml = 114 814 814 824 capsules 3 Ml to cap 3 31 Ml = 155 000 000 000 capsules 4 Ml to cap 4 31 Ml = 238 461 538 452 capsules 5 Ml to cap 5 31 Ml = 1 291 666 667.7 capsules 7 Ml to cap 7 31 Ml = 1 107 142 856.7 capsules Su7 Ml to cap Su7 31 Ml = 3.1 × 10+34 cubic angstroms Ml to ų 31 Ml = 9.26 × 10-30 cubic astronomical unit Ml to au³ 31 Ml = 31 000 000 000 cubic centimeters Ml to cm³ 31 Ml = 3.81 cubic chains Ml to ch³ 31 Ml = 31 000 000 cubic decimeters Ml to dm³ 31 Ml = 31 cubic dekameters Ml to dam³ 31 Ml = 5 068.31 cubic fathoms Ml to ftm³ 31 Ml = 1 094 754.67 cubic feet Ml to ft³ 31 Ml = 0.004 cubic furlong Ml to fur³ 31 Ml = 0.03 cubic hectometer Ml to hm³ 31 Ml = 1 891 736 067.1 cubic inches Ml to in³ 31 Ml = 3.1 × 10-5 cubic kilometer Ml to km³ 31 Ml = 3.66 × 10-44 cubic light year Ml to ly³ 31 Ml = 31 000 cubic meters Ml to m³ 31 Ml = 1.89 × 10+27 cubic microinches Ml to µin³ 31 Ml = 3.1 × 10+22 cubic micrometers Ml to µm³ 31 Ml = 3.1 × 10+22 cubic microns Ml to µ³ 31 Ml = 1.89 × 10+18 cubic mils Ml to mil³ 31 Ml = 7.44 × 10-6 cubic mile Ml to mi³ 31 Ml = 31 000 000 000 000 cubic millimeters Ml to mm³ 31 Ml = 3.1 × 10+31 cubic nanometers Ml to nm³ 31 Ml = 4.88 × 10-6 cubic nautical mile Ml to nmi³ 31 Ml = 1.06 × 10-45 cubic parsec Ml to pc³ 31 Ml = 3.1 × 10+40 cubic picometers Ml to pm³ 31 Ml = 1.89 × 10+18 cubic thous Ml to thou³ 31 Ml = 40 546.47 cubic yards Ml to yd³ 31 Ml = 310 000 000 deciliters Ml to dl 31 Ml = 3 100 000 dekaliters Ml to dal 31 Ml = 620 000 000 000 drops Ml to gt 31 Ml = 0.03 gigaliter Ml to Gl 31 Ml = 109 104 701.81 Imperial cups Ml to imperial c 31 Ml = 1 091 047 018.1 Imperial fluid ounces Ml to imperial fl.oz 31 Ml = 6 819 043.87 Imperial gallons Ml to UK gal 31 Ml = 54 552 351.06 Imperial pints Ml to imperial pt 31 Ml = 27 276 175.47 Imperial quarts Ml to UK qt 31 Ml = 31 000 kiloliters Ml to kl 31 Ml = 31 000 000 liters Ml to l 31 Ml = 124 000 000 metric cups Ml to metric c 31 Ml = 3 100 000 000 metric dessertspoons Ml to metric dstspn 31 Ml = 2 066 666 667.7 metric tablespoons Ml to metric tbsp 31 Ml = 6 200 000 000 metric teaspoons Ml to metric tsp 31 Ml = 31 000 000 000 milliliters Ml to ml 31 Ml = 194 984.13 oil barrels Ml to bbl 31 Ml = 3.1 × 10-5 teraliter Ml to Tl 31 Ml = 131 029 337.42 US cups Ml to US c 31 Ml = 4 192 938 790 US dessertspoons Ml to US dstspn 31 Ml = 1 048 234 700.6 US fluid ounces Ml to fl.oz 31 Ml = 8 189 333.61 US gallons Ml to US gal 31 Ml = 65 514 668.71 US pints Ml to pt 31 Ml = 32 757 334.51 US quarts Ml to US qt 31 Ml = 2 096 469 398.1 US tablespoons Ml to US tbsp 31 Ml = 6 289 408 185 US teaspoons Ml to US tsp\n\nFoods, Nutrients and Calories\n\nDIJON MUSTARD, UPC: 3158697781676 weigh(s) 254 grams per metric cup or 8.5 ounces per US cup, and contain(s) 160 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]\n\n393 foods that contain Lactose. List of these foods starting with the highest contents of Lactose and the lowest contents of Lactose\n\nGravels, Substances and Oils\n\nCaribSea, Freshwater, Super Naturals, Moonlight Sand weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]\n\nConvert between units of mass and molar concentration for Nihonium [Nh]\n\nVolume to weightweight to volume and cost conversions for Refrigerant R-407F, liquid (R407F) with temperature in the range of -40°C (-40°F) to 60°C (140°F)\n\nWeights and Measurements\n\nA dyne per square thou is a unit of pressure where a force of one dyne (dyn) is applied to an area of one square thou.\n\nThe kinematic viscosity (ν) is the dynamic viscosity (μ) divided by the density of the fluid (ρ)\n\ndwt/ml to sl/l conversion table, dwt/ml to sl/l unit converter or convert between all units of density measurement.\n\nCalculators\n\nRectangle calculator, rectangle definition and calculation formulas."
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[
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http://forum.hp-prime.de/showthread.php?tid=80
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[
"• 0 Vote(s) - 0 Average\n• 1\n• 2\n• 3\n• 4\n• 5\n machine epsilon of the HP Prime\n11-03-2016, 10:53 PM\nPost: #1\n cdeaglejr",
null,
"Junior Member",
null,
"",
null,
"Posts: 39 Joined: Mar 2016 Reputation: 0\nmachine epsilon of the HP Prime\nHere's a short program that calculates the machine epsilon of the HP Prime calculator. The square root of the machine epsilon is often used for the convergence criterion for iterative algorithms such as root finders, numerical integrators, etc.\n\nEXPORT meps()\n\n// machine epsilon of the HP Prime\n\n// epsilon is that number when added to one\n// is still one according to the calculator\n\n///////////////////////////////////////////\n\nBEGIN\n\nLOCAL x;\n\nx := 1.0;\n\nWHILE ((1.0 + x) > 1.0) DO\n\nx := x / 2.0;\n\nEND;\n\nPRINT;\n\nPRINT(\"machine epsilon = \" + x);\n\nprint(\" \");\n\nprint(\"1 + epsilon = \" + (1.0 + x));\n\nEND;\n\nHere are the results for the HP Prime. The second calculation verifies the computed value by adding it to the number one and printing the result.\n\nmachine epsilon = 3.63797880705ᴇ−12\n\n1 + epsilon = 1\n « Next Oldest | Next Newest »\n\nForum Jump:\n\nUser(s) browsing this thread: 1 Guest(s)"
] |
[
null,
"http://forum.hp-prime.de/images/buddy_offline.gif",
null,
"http://forum.hp-prime.de/images/star.gif",
null,
"http://forum.hp-prime.de/images/star.gif",
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https://www.endpoint.com/blog/2016/03/quickcheck-property-based-testing-in/
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[
"# Our Blog\n\nOngoing observations by End Point people\n\n## QuickCheck - property based testing in Haskell and JavaScript\n\nBy Kamil Ciemniewski\nMarch 16, 2016\n\nIn my last article, I presented a functional programming pattern. The goal was to reach out to the developers who weren’t familiar with advanced type systems like the one found in Haskell and make them a bit curious. This time I’d like to take a step further and present a testing approach coming from the same world, that can be used with mainstream languages with a great success.\n\n### Many ways to test the code\n\nThe importance of testing is almost a cliché nowadays. Out of this relevance, a large number of testing frameworks and paradigms have been created. On the paradigm level we have notions like TDD and BDD. On the level of implementations we have hundreds of projects for each language like RSpec in Ruby and Jasmine or Mocha in JavaScript.\n\nThe ideas behind the libraries don’t differ that much. All of them are based on the idea of providing code examples with assertions on how the code should behave in these particular cases.\n\nA bit more revolutionary in its approach was the Cucumber project. In its essence, it allows business people to express the system logic by stating it in specially formed, plain English. An example taken from the Cucumber’s website reads:\n\n``````Feature: Refund item\n\nScenario: Jeff returns a faulty microwave\nGiven Jeff has bought a microwave for \\$100\nAnd he has a receipt\nWhen he returns the microwave\nThen Jeff should be refunded \\$100\n``````\n\nIn this article, I’d like to present an approach that was conceived in the realm of Haskell—the purely functional, statically typed language. Though it started just as a Haskell library, today it could be broadly named an “approach”. We now have implementations for almost every major language. This approach is what is known as QuickCheck.\n\n### Code testing limitations\n\nHaving a good test coverage is a sign of a potentially stable code. However, even well-tested software needs to be improved occasionally as new bugs are discovered. This happens even in the projects with the largest test suites. Moreover, the tests are code too—they also are prone to being invalid. But do we have a really valid solution when all our tests are valid and the code passes all assertions? Realistically, we can only provide a few examples per use case at most. The best we can do is to choose the most probable use cases and the ones we have a hunch that might go wrong.\n\nThis implies that for the tests suite to be guarding us against bugs, we need to have an insight as to where the potential bugs may be before even testing. Isn’t a definition of a bug telling the other story though? If we knew where to find them, we would fix them in the first place. In reality the systems grow so complex that we only can have a feeling of what might go wrong.\n\nIn 1969, Edsger W. Dijkstra said at the “Nato Software Engineering Conference”:\n\nTesting shows the presence, not the absence of bugs.\n\n### I’ve got lots of ammo, give me a machine gun for killing bugs!\n\nWhat if we could ask the computer to come up with many different use cases for us? Instead of having 2 or 3 cases per some code aspect, we’d have 100 of them including the ones we’d never consider ourselves. We could describe properties of the code we’d like to hold for all of the randomly chosen cases. That is exactly the idea behind the QuickCheck. In its functional - Haskell form, it takes advantage of the supreme type system and generates a random set of function parameters based on their types. It then runs the property check for all of them and stops if it’s able to find a counter example making the property falsifiable.\n\nIf we’d compare coming up with traditional test cases to shooting with a pistol, the QuickCheck approach had to be called firing with a machine gun. The reason is that we’re not focusing on a specific use case, but we’re focusing on certain properties of the code that have to hold for any argument from the accepted domain.\n\nOne of the most basic examples we could show, could be ensuring that the reverse function applied twice returns the original array (pseudo-code):\n\n``````arr1 == reverse(reverse(arr1))\n``````\n\nThe idea here is to make sure this property holds against a large number of randomly selected arguments from the domain. In this example the checker would randomly generate e. g 100 arrays and test if the assertion evaluates to true for every one of them.\n\n### Working example — the Haskell way\n\nLet’s take a look at how the approach is being used in its original environment. Later on we’ll see how the pattern can be used when coding in JavaScript. For this, let’s imagine that we’re developing a graph data structure, to be used in some e-commerce project we’re working on. Here’s the basic very incomplete draft:\n\n``````module BlogGraph where\n\nimport Data.Map as Map\nimport Data.List as List\n\n-- Graph as an adjacency list as described here:\n\ndata Graph a = Graph (Map a [a]) deriving (Show)\n\nempty :: Graph a\nempty = Graph Map.empty\n\ninsertNode :: (Ord a) => a -> Graph a -> Graph a\ninsertNode node (Graph m) = Graph \\$ Map.insert node [] m\n\nremoveNode :: (Ord a) => a -> Graph a -> Graph a\nremoveNode node (Graph m) = Graph \\$ Map.delete node m\n\ninsertEdge :: (Ord a) => a -> a -> Graph a -> Graph a\ninsertEdge parent child (Graph m) =\nGraph \\$ Map.insertWithKey update parent [child] m\nwhere\nupdate _ _ old = List.nub \\$ child:old\n\nnodes :: Graph a -> [a]\nnodes (Graph m) = Map.keys m\n``````\n\nIf you’re not proficient yet in reading Haskell code, we’re just using a Map where keys are integers and values are arrays of integers to implement the graph as an Adjacency List. So each node has its representation in a map as one of its keys. Also, each edge has its representation as a child stored in the parent’s array in the map.\n\nYou might be able to find a silly bug in the removeNode function. It doesn’t remove the node from the edge definitions of other nodes. We’ll use QuickCheck to show how this could be found automatically.\n\nBefore doing that, let’s have a warm up, by adding two simple properties:\n\n``````prop_insert_empty :: Int -> Bool\nprop_insert_empty i =\nnodes (insertNode i BlogGraph.empty) == [i]\n\nprop_insert_existing :: Int -> Bool\nprop_insert_existing i =\nnodes (insertNode i \\$ insertNode i BlogGraph.empty) == [i]\n``````\n\nProperties are just simple functions returning true or false. They take arguments which are randomly provided later on by the QuickCheck library.\n\nThe first property says that adding a node to an empty graph will always produce a one-node graph. The second one, that adding a node to a graph that already has this node will always return the same unmodified graph.\n\nWe can successfully run these cases:\n\n``````quickCheck prop_insert_empty\nquickCheck prop_insert_existing\n``````\n\nNow we should add a property stating that for all removals of a node from the graph, all references of this node in edge definitions for other nodes are always also being removed:\n\n``````prop_remove_removes_edges :: Graph Int -> Bool\nprop_remove_removes_edges (Graph m) =\nList.null (nodes graph) || List.notElem node elemsAfter\nwhere\ngraph = Graph m\nnode = List.head \\$ BlogGraph.nodes graph\nelemsAfter = List.concat \\$ Map.elems mapAfter\nmapAfter =\ncase removeNode node graph of\n(Graph m) -> m\n``````\n\nAs I wrote before, these property testing functions are being run by the QuickCheck framework repeatedly with randomly generated values as arguments. Out of the box we’re able to generate random examples for many simple types—including e.g Int. That’s the reason we were able to just specify properties depending on random Int variables—without any additional code. But with the last example, we’re asking QuickCheck to generate a set of random graphs. We need to tell it how to construct a random graph first:\n\n``````arbitrarySizedIntGraph :: Int -> Gen (Graph Int)\narbitrarySizedIntGraph s = do\nnodes <- vectorOf s \\$ choose (0, 32000)\nedges <- edges nodes\nlet withNodes = List.foldr insertNode BlogGraph.empty nodes\nreturn \\$ List.foldr addEdge withNodes edges\nwhere\naddEdge (parent, child) = insertEdge parent child\nedges nodes = do\nparents <- sublistOf nodes\nlet children = nodes List.\\\\ parents\nreturn [ (parent, child) | parent <- parents, child <- children ]\n\ninstance Arbitrary (Graph Int) where\narbitrary = sized arbitrarySizedIntGraph\n``````\n\nThe above generator will be good enough for our case. It generates variable length graphs. A sublist of all nodes are made parents in edges and all parents are connected to the rest of non-parental nodes.\n\nWhen we try to run the test we get:\n\n``````Failed! Falsifiable (after 3 tests):\nGraph (fromList [(10089,[]),(25695,)])\n``````\n\nQuickCheck shows that the property doesn’t hold for the whole domain—it failed after 3 examples. It also prints the example for which our property did not hold.\n\nWe can now reexamine the code for the removeNode function and fix it as per the property’s specification:\n\n``````removeNode :: (Ord a) => a -> Graph a -> Graph a\nremoveNode node (Graph m) =\nGraph \\$ Map.map remNode \\$ Map.delete node m\nwhere\nremNode = List.delete node\n``````\n\nNow running the test again we can see that it works.\n\n### Another working example — the JavaScript land\n\nAs I stated before, this pattern became implemented for many different mainstream languages—this includes JavaScript. I’d like to show you the version of the above process for this language now. This might end up being helpful if you’d like to use it in your project but don’t know much Haskell yet.\n\nAs a start, let’s make sure we have the following packages:\n\n``````npm install jsverify\nnpm install lodash\n``````\n\nWe can now create a JS file with what might resemble the Haskell draft implementation:\n\n``````var jsc = require(\"jsverify\");\nvar _ = require(\"lodash\");\n\nvar Graph = function() {\nvar self = this;\n\nself._map = {};\n\nself.insertNode = function(node) {\nif(self._map[node] === undefined) {\nself._map[node] = [];\n}\nreturn self;\n};\n\nself.removeNode = function(node) {\nself._map.delete(node);\nreturn self;\n};\n\nself.insertEdge = function(a, b) {\nif(self._map[a] === undefined) {\nself.insertNode(a);\n}\nself._map[a].push(b);\nreturn self;\n};\n\nself.nodes = function() {\nreturn _.keys(self._map);\n};\n}\n\nGraph.empty = function() {\nreturn new Graph();\n}\n``````\n\nTo reproduce the first property—for all integers, inserting one as a node to an empty graph results in a graph with one node:\n\n``````var propInsertEmpty =\njsc.forall(\"nat\", function(i) {\nreturn _.isEqual(Graph.empty().insertNode(i).nodes(), [i]);\n});\n\njsc.assert(propInsertEmpty);\n``````\n\nThe jsVerify DSL takes some time to get used to. It cannot take advantage of the type system as in the Haskell example so aspects like generation of random data based on types requires some documentation reading.\n\nRunning jsc.assert we might have expected to get a success, but this time we’re getting:\n\n``````Error: Failed after 1 tests and 5 shrinks. rngState: 001d40a68297fbce35; Counterexample: 0;\n``````\n\nWe can see that jsVerify has found 0 as a counterexample. Let’s see what’s happening by running the code by hand passing 0 as a parameter:\n\n``````console.log(Graph.empty().insertNode(0).nodes());\n``````\n\nResult:\n\n``````[ '0' ]\n``````\n\nAha! It’s quite easy to shoot your own foot in JavaScript. We can fix it really fast with the following:\n\n``````self.nodes = function() {\nreturn _.map(_.keys(self._map), function(i){ return parseInt(i, 10); });\n};\n``````\n\nRunning the code again doesn’t show any errors which means that all assertions were valid. What about the bug we saw in the Haskell version? Let’s provide a property for that too:\n\n``````var propRemoveRemovesEdges =\njsc.forall(graphG, function(g) {\nif(g.nodes().length === 0){\nreturn true;\n}\nelse {\nvar numNodes = g.nodes().length;\nvar index = _.random(0, numNodes - 1);\nvar node = g.nodes()[index];\nreturn !_.includes(_.flattenDeep(_.values(g.removeNode(node)._map)), node);\n}\n});\n\njsc.assert(propRemoveRemovesEdges);\n``````\n\nWe will still need to specify how to generate a random graph. We can use the notion of a Functor that’s coming from the functional programming world and turn a random array into a random graph:\n\n``````var graphG = jsc.array(jsc.nat).smap(\nfunction(arr) {\nvar ins = function(g, i) {\nreturn g.insertNode(i);\n};\nvar graph = _.reduce(arr, ins, Graph.empty());\nvar numParents = Math.floor(arr.length / 2);\nvar parents = _.take(arr, numParents);\nvar children = _.difference(arr, parents);\nvar insEd = function(g, parent) {\nvar insF = function(r, c) {\nreturn r.insertEdge(parent, c);\n};\nreturn _.reduce(children, insF, g);\n};\nreturn _.reduce(parents, insEd, graph);\n},\nfunction(graph) {\nreturn graph.nodes();\n}\n);\n``````\n\nWhen running the assert for that property we’re getting an error:\n\n``````Error: Failed after 1 tests and 1 shrinks. rngState: 085f6c82ea10439d7b; Counterexample: {\"_map\":{\"21\":}}; Exception: self._map.delete is not a function\n``````\n\nThis isn’t the issue we were expecting though. Still it’s great to find a problem before showing the code to the client. We can iron it out with:\n\n``````self.removeNode = function(node) {\ndelete self._map[node];\nreturn self;\n};\n``````\n\nWhen running again, we’re getting an error we were expecting:\n\n``````Error: Failed after 8 tests and 2 shrinks. rngState: 8c97e25bc36f41da08; Counterexample: {\"_map\":{\"2\":}};\n``````\n\nThe jsVerify has found a counterexample falsifying our property. It’s also worth noting that it took 8 tests to find this issue. We can notice that for the event of removing a node that is a child and doesn’t have any children itself the property isn’t true. Let’s reexamine our removeNode function:\n\n``````self.removeNode = function(node) {\ndelete self._map[node];\nreturn _.mapValues(self._map, function(children) {\nreturn _.without(children, node);\n});\n};\n``````\n\nAnd now it works!\n\n### Not only Haskell and JavaScript\n\nThe QuickCheck style testing is available for many different languages. The Wikipedia says:\n\nRe-implementations of QuickCheck exist for C, C++, Chicken Scheme, Clojure, Common Lisp, D, Elm, Erlang, F# (and C#, VB.NET), Factor, Io, Java, JavaScript, Node.js, Objective-C, OCaml, Perl, Prolog, PHP, Python, R, Ruby, Rust, Scala, Scheme, Smalltalk, Standard ML and Swift.\n\nYou can find many useful links about the approach on Wikipedia. If you’re into Haskell, a good place to start reading about the library is the Haskell Wiki as well as the documentation found on the Hackage.\n\nThe JavaScript counterpart can be found on GitHub. It’s important to note that jsVerify isn’t the only JavaScript library implementing the QuickCheck approach."
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https://www.programtips.org/java/math/how-to-get-the-exponential-value-minus-one-of-the-given-number-in-java/
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[
"# How to Get the Exponential Value minus One of the Given Number in Java\n\nHere we will learn how to get the exponential value minus one of the given number with Math.expm1().\n\n### Source Code\n\n``````package com.beginner.examples;\n\npublic class Expm1Example {\n\npublic static void main(String[] args){\n//return the exponential value minus one of the given number.\ndouble a = -10.8;\ndouble b = 0;\ndouble c = 2.01;\nSystem.out.println(\"(Exponential value-1) of \" + a + \" is: \" + Math.expm1(a));\nSystem.out.println(\"(Exponential value-1) of \" + b + \" is: \" + Math.expm1(b));\nSystem.out.println(\"(Exponential value-1) of \" + c + \" is: \" + Math.expm1(c));\n}\n}\n``````\n\nOutput:\n\n``````(Exponential value-1) of -10.8 is: -0.9999796004965888\n(Exponential value-1) 0.0 is: 0.0\n(Exponential value-1) of 2.01 is: 6.463317347319193\n``````\nSubscribe\nNotify of",
null,
""
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[
null,
"https://secure.gravatar.com/avatar/",
null
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https://www.jpost.com/international/mexico-gunmen-kill-politician-and-3-others
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[
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[
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https://www.physicsforums.com/threads/shooting-arrows-nonlinear-spring-problem.717112/
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[
"# Shooting Arrows - Nonlinear Spring Problem\n\nTheBlenderer\nHello awesome physics people!\n\nSomeone asked me for help on their first year physics homework, and I couldn't really solve it. This kept bugging me, because I should know how this works by now :P\n\n## Homework Statement\n\nSee attachment for the full problem statement. Basically, a bow is strung with an arrow, and what they want to know is, among other things, the graph for the velocity with respect to time.\n\n## Homework Equations\n\nThe given equation is $$F(x)=-k_1x-k_2xe^{cx^2}$$.\n\nOther equations I've used:\n\n$$E_{pot}=E_{kin}=1/2*mv^2$$\n$$v_{terminal}=sqrt{(2*E_{pot})/m_{arrow}}$$\n\n## The Attempt at a Solution\n\nI've come as far as the calculation of the terminal velocity of the arrow, however, after that I need to come up with an equation of velocity with respect to time, whilst I have only the acceleration with respect to distance from the equilibrium position of the bow. If it were linear I would be able to use a constant acceleration, but seeing as the acceleration is dependent on the distance from equilibrium, I don't really know what to do.\n\nI've implemented my partial solution in MATLAB, here is my code:\n\nCode:\n%% First I let MATLAB calculate the initial x:\n\nsolve('-390*x-115*x*exp(2.3*x^2)=350','x');\n\n% The outcome of this: xin=-0.62035062944153550757039658657226, or -0,620m\n\n%% Now for the numerical stuff\n\nk1=390;\nk2=115;\nc=2.3;\nFmax=350;\n\nx=-0.620:0.01:0;\n% This is the x-vector, I initialized it from the initial x until x=0, when\n% the arrow leaves the string.\n\ndt=0.01; % This is the timestep\nt=0:dt:5; % Here I construct a vector t from 0 to 5 with steps of dt\nm=34*10^-3; % This is the mass of the arrow\n\n% The following is the given formula:\nF=-k1.*x-k2.*x.*exp(c.*x.^2);\n\n% To get the energy stored up in the spring, we need only to integrate F\n% over x:\n\nEpot=trapz(F,x); % This gives Epot = 146.4474 J\n\n% As all of the potential energy will be transferred into kinetic energy of\n% the arrow, according to Epot = Ekin = 1/2*mv^2, we can calculate the\n% terminal velocity of the arrow:\n\nvterm=sqrt((2*Epot)/m); % This gives vterm = 92.8146 m/s\n\n% Assuming all of the force gets transferred to the arrow, and because\n% F=m*a, we can calculate the acceleration of the arrow as function of x\n% simply by dividing by the mass of the arrow:\n\na=(-k1.*x-k2.*x.*exp(c.*x.^2))/m;\n\nfigure, plot(x,F,'g')\ntitle('Force with respect to x')\n\nI've included the plot as another attachment.\n\nI guess my problem can be stated another way:\n\nHow do I parametrize from x to t?\n\n#### Attachments\n\nLast edited:\n\nMentor\nYour value shown for xin looks a bit suspicious. Verify your solution.\n\nFor the V vs T plot you're going to have to do a numerical integration of the trajectory. Use whatever algorithm you think appropriate, such as the Leapfrog method, or Euler's method, or Verlet, or even Runge-Kutta (probably overkill). Some web time will fill you in on those.\n\nThey all boil down to using the force at a given position to determine an acceleration to apply over the next small time interval. Use it to update the velocity and position, rinse, repeat.\n\nTheBlenderer\nThanks for your reply gneill! I see, so there is no analytical method of converting distance into time when working with a nonlinear system?\n\nMy x_initial doesn't seem suspicious to me, loading a bow by pulling back 62 cm is plausible, right? The minus sign is explicable by the definition of the x-axis in the problem statement.\n\nI'll try to implement this Frogleap integration method, thanks again!\n\nMentor\nThanks for your reply gneill! I see, so there is no analytical method of converting distance into time when working with a nonlinear system?\nIt depends upon the functions involved. If you run into a transcendental function you're basically hosed",
null,
"In this case the x2 in the exponent of the exponential is a major stumbling block for a solution involving the normal basis set of functions that we're used to.\n\nMy x_initial doesn't seem suspicious to me, loading a bow by pulling back 62 cm is plausible, right? The minus sign is explicable by the definition of the x-axis in the problem statement.\nI get a different value when I find the root of the function. In particular, I find x = -0.559 m.\n\nI'll try to implement this Frogleap integration method, thanks again!\nSounds like a plan!\n\nTheBlenderer\nYay, it worked!\n\nI implemented the Leapfrog algorithm, and it worked! At first I got a sine wave (which isn't strange seeing as I basically modelled a spring), so I made the calculation stop as soon as x became positive. A bit of a hack solution, also because the rest of the vectors are zeros after that (that explains the sudden straight line in the graphs), but I didn't feel like polishing the code. I also used the erroneous value of -0.62 for x_initial, as it didn't really matter for this implementation (but thanks for checking anyways!)\n\nHere's my code:\n\nCode:\n%% Now attempting a Leapfrog integration\ndt=0.001;\nt=0:dt:0.013;\nx_in=-0.620;\nx_t=zeros(1,length(t));\nv_t=zeros(1,length(t));\na_t=zeros(1,length(t));\nx_t(1)=x_in;\nv_t(1)=0;\na_t(1)=(-k1.*x_in-k2.*x_in.*exp(c.*x_in.^2))/m;\n\ni=2;\nwhile x_t(i-1)<0\n\nx_t(i)=x_t(i-1)+v_t(i-1)*dt+0.5*a_t(i-1)*dt^2;\na_t(i)=(-k1*x_t(i)-k2*x_t(i).*exp(c*x_t(i).^2))/m;\nv_t(i)=v_t(i-1)+0.5*(a_t(i-1)+a_t(i)).*dt;\ni=i+1;\nend\n\nhold all\nsubplot(2,2,1), plot(t,v_t,'b'), title('Velocity with respect to time')\nsubplot(2,2,2), plot(t,x_t,'r'), title('Distance with respect to time')\nsubplot(2,2,3), plot(t,a_t,'g'), title('Acceleration with respect to time')\n\nAnd my graphs are attached. Cheers gneill :)"
] |
[
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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|
https://slideplayer.com/slide/4457057/
|
[
"",
null,
"# Empowering Young Learners through the Standards for Mathematical Practice Juli K. Dixon, Ph.D. University of Central Florida\n\n## Presentation on theme: \"Empowering Young Learners through the Standards for Mathematical Practice Juli K. Dixon, Ph.D. University of Central Florida\"— Presentation transcript:\n\nEmpowering Young Learners through the Standards for Mathematical Practice Juli K. Dixon, Ph.D. University of Central Florida juli.dixon@ucf.edu\n\nSolve this…\n\nPerspective… What do you think fourth grade students would do? How might they solve 4 x 7 x 25?\n\nPerspective… Are you observing this sort of mathematics talk in classrooms? Is this sort of math talk important?\n\nPerspective… What does this have to do with the Common Core State Standards for Mathematics (CCSSM)?\n\nBackground of the CCSSM Published by the National Governor’s Association and the Council of Chief State School Officers in June 2010 Result of collaboration from 48 states Provides a focused curriculum with an emphasis on teaching for depth\n\nBackground of the CCSSM “… standards must address the problem of a curriculum that is ‘a mile wide and an inch deep.’ These Standards are a substantial answer to that challenge” (CCSS, 2010, p. 3). We’ve already met this challenge in Florida. How can we use our momentum to take us further and deeper?\n\nNGSSS Content Standards Wordle\n\nCCSSM Content Standards Wordle\n\nBackground of the CCSSM The CCSSM consist of Content Standards and Standards for Mathematical Practice. “The Standards for Mathematical Practice describe varieties of expertise that mathematics educators at all levels should seek to develop in their students” (CCSS), 2010, p. 6).\n\nThe Standards for Mathematical Practice are based on: Making Sense of the Mathematical Practices The National Council of Teachers of Mathematics’ (NCTM) Principles and Standards for School Mathematics (NCTM, 2000), and The National Research Council’s (NRC) Adding It Up (NRC, 2001).\n\nNCTM Process Standards: Making Sense of the Mathematical Practices Problem Solving Reasoning and Proof Communication Representation Connections\n\nNRC Strands of Mathematical Proficiency: Making Sense of the Mathematical Practices Adaptive Reasoning Strategic Competence Conceptual Understanding Procedural Fluency Productive Disposition\n\nStandards of Mathematical Practice Wordle\n\nPerspective… According to a recommendation from the Center for the Study of Mathematics Curriculum (CSMC, 2010), we should lead with the Mathematical Practices. Florida is positioned well to do this.\n\nThe 8 Standards for Mathematical Practice: Making Sense of the Mathematical Practices 1Make sense of problems and persevere in solving them 2Reason abstractly and quantitatively 3Construct viable arguments and critique the reasoning of others 4Model with mathematics 5Use appropriate tools strategically 6Attend to precision 7Look for and make use of structure 8Look for and express regularity in repeated reasoning\n\nThe 8 Standards for Mathematical Practice: We will only address 4 today 1Make sense of problems and persevere in solving them 2Reason abstractly and quantitatively 3Construct viable arguments and critique the reasoning of others 4Model with mathematics 5Use appropriate tools strategically 6Attend to precision 7Look for and make use of structure 8Look for and express regularity in repeated reasoning\n\nImpact on Depth… (NGSSS) Grade 4 Big Idea 1: Develop quick recall of multiplication facts and related division facts and fluency with whole number multiplication. MA.4.A.1.2: Multiply multi-digit whole numbers through four digits fluently, demonstrating understanding of the standard algorithm, and checking for reasonableness of results, including solving real-world problems.\n\nImpact on Depth… (CCSS) Grade 4 Cluster: Use place value understanding and properties of operations to perform multi- digit arithmetic. 4.NBT.5: Multiply multi-digit numbers using strategies based on place value and the properties of operations. Illustrate and explain the calculations by using equations, rectangular arrays, and/or area models.\n\nWhat does it mean to use strategies to multiply? When do students begin to develop these strategies? Impact on Depth…\n\nGrade 3 Big Idea 1: Develop understanding of multiplication and division and strategies for basic multiplication facts and related division facts. MA.3.A.1.2: Solve multiplication and division fact problems by using strategies that result form applying number properties. Impact on Depth… (NGSSS)\n\nGrade 3 Cluster: Understand properties of multiplication… 3.OA.5: Apply properties of operations as strategies to multiply and divide. Grade 3 Cluster: Multiply and divide within 100 3.OA.7: Fluently multiply within 100, using strategies such as the relationship between multiplication and division or properties of operations. Impact on Depth… (CCSS)\n\nConsider 6 x 7 How can using strategies to multiply these factors help students look for and make use of structure? (SMP7) What strategies can we use? How might this sort of thinking influence the order in which facts are introduced in grade 3? What does it mean to use strategies to multiply?\n\nNow solve 4 x 7 x 25… The Standards for Mathematical Practice help us to focus on processes, not just products.\n\nReasoning abstractly and quantitatively often involves making sense of mathematics in real-world contexts. Word problems can provide examples of mathematics in real-world contexts. We need to help students make sense of them. Not just solve them. Empowering Young Learners\n\nConsider the following problems: Jessica has 7 key chains. Calvin has 8 key chains. How many key chains do they have all together? Jessica has 7 key chains. Alex has 15 key chains. How many more key chains does Alex have than Jessica? Key words seem helpful, or are they…. Empowering Young Learners\n\nNow consider this problem: Jessica has 7 key chains. How many more key chains does she need to have 15 key chains all together? How would a child who has been conditioned to use key words solve it? Empowering Young Learners\n\nWe need students to make sense of problem situations as well as each other’s thinking. Consider these students as they reason about division.\n\nEmpowering Young Learners We need students to make sense of problem situations as well as each other’s thinking. Consider these students as they reason about division. Notice how the teacher’s questions focus on making sense of the problem.\n\nEmpowering Young Learners We need students to make sense of problem situations as well as each other’s thinking. Consider these students as they reason about remainders. Notice how they need support to construct viable arguments and critique the reasoning of others.\n\nThe 8 Standards for Mathematical Practice: How might you change your practice to address these now? 1Make sense of problems and persevere in solving them 2Reason abstractly and quantitatively 3Construct viable arguments and critique the reasoning of others 4Model with mathematics 5Use appropriate tools strategically 6Attend to precision 7Look for and make use of structure 8Look for and express regularity in repeated reasoning\n\nHow do we support this empowerment? Teachers need content knowledge for teaching mathematics to know the tasks to provide, the questions to ask, and how to assess for understanding. Teachers need content knowledge for teaching mathematics to know the tasks to provide, the questions to ask, and how to assess for understanding. Math Talk needs to be supported in the classroom. Math Talk needs to be supported in the classroom. Social norms need to be established in classroom and professional development settings to address misconceptions in respectful ways. Social norms need to be established in classroom and professional development settings to address misconceptions in respectful ways.\n\nDownload ppt \"Empowering Young Learners through the Standards for Mathematical Practice Juli K. Dixon, Ph.D. University of Central Florida\"\n\nSimilar presentations"
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