June 2020 - mathfun@mailman.xmission.com

[math-fun] Solar panel calculations
by Henry Baker 29 Jun '20

29 Jun '20

I'm considering putting a solar PV system on my roof, and I thought that I should be able to calculate most of the numbers myself from first principles. I have searched for so-called online 'calculators', but found that most of them are aimed at idiots. Here's what I expected to find: A formula giving the exact sun position throughout the year, given latitude and longitude, as well as a factor which handles the distance from the sun w.r.t. time. Now this only provides the *input* energy. I now need to overlay this with average *weather* data on a per-hour basis to get the expected net energy. Finally, I need to *integrate* these data over an entire year to get the values for my own house. There may be such an online calculator, but I haven't been able to find it yet. It may be possible to factor the weather info out of the integration, so that I could 'correct' my location sunlight & weather data with my own lat/lon azimuth (angle of my house) and roof angle, but I'd need to see the equations to find out if this factoring is possible. I presume that someone on this list has already done these calculations for their own home?

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[math-fun] Maximum number of empty tetrahedra ...
by Dan Asimov 29 Jun '20

29 Jun '20

Given n points {x_1, ..., x_n} on the surface of a 2-sphere S^2, let's call any 3 of them "empty" if they lie on the curve of some circle C ⊂ S^2 that's smaller than a great circle and there is no other point x_j on or inside C. Given n, what is the maximum number of empty triples possible? I can more or less prove that for n = 4, 6, 12 the answer comes from the vertices of a regular tetrahedron, octahedron, icosahedron, giving 4, 8, 20 triples, respectively. Likewise for higher dimensions. Next case: Given n points {x_1, ..., x_n} on the surface of a 3-sphere S^3, call any 4 of them "empty" if they lie on the surface of some 2-sphere S ⊂ S^3 that's smaller than a great sphere and there is no other point x_j on or inside S. I'd certainly guess that the maximum number of empty quadruples for the values n = 5, 8, 10, 120 are 5, 16, 30, 600, respectively. The arrangements would be the vertices of the following: the 5-cell (4-simplex), the 16-cell (a.k.a. 4-orthoplex), the 5-cell and the antipodal 5-cell, the 600-cell, respectively. (Small perturbations of all these arrangements give the same numbers as well.) But what about the in-between values, and asymptotics? —Dan

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[math-fun] socially most distant sudokus
by Veit Elser 29 Jun '20

29 Jun '20

Does anyone know what is the most socially distant set of clues in sudoku (taxicab distance, unless you prefer something else) and can you point me to examples? -Veit

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Re: [math-fun] Solar panel calculations
by Henry Baker 29 Jun '20

29 Jun '20

Hi Thomas: Is there any way to extract the intermediate results of the detailed calculations on Google 'sunroof' ? Otherwise, this web site is completely useless to me, as I want to examine some sensitivities to various parameters. At 08:59 AM 6/29/2020, Thomas Colthurst wrote: >On Mon, Jun 29, 2020 at 11:31 AM Henry Baker <hbaker1(a)pipeline.com> wrote: >> Yes, I've been to that 'sunroof' web site, but it doesn't give me >> any details, so I haven't the slightest idea what calculations >> it is actually performing. >> >> Furthermore, it doesn't know the pitch of my roof, unless it >> is doing some clever image-matching from different satellite >> angles. > >Well, the imagery isn't always from satellites, but yes, it builds a 3d model >of every house. (We didn't build this part, it was already in Google Maps >-- >you can check out the 3d model of your house by going to the satellite >view and then clicking the "3d" button.) > >-Thomas C > >> At 08:16 AM 6/29/2020, Thomas Colthurst wrote: >> >I did this! Well, I and a small team. google.com/sunroof >> > >> >It even does the integral over every pixel of your roof, so >> >as to take into account shading from nearby trees and houses (as >> >well as non-flat roofs, horizon effects, chimneys and other roof >> >obstructions, ...). >> > >> >You might also like NREL's PVWatts calculator <https://pvwatts.nrel.gov/> >, >> >which is less >> >accurate (because it doesn't do the full ray-tracing over the roof >> >thing) but provides more intermediate information. >> > >> >-Thomas C >> > >> >On Mon, Jun 29, 2020 at 10:52 AM Henry Baker <hbaker1(a)pipeline.com> wrote: >> >> I'm considering putting a solar PV system on my roof, and I >> >> thought that I should be able to calculate most of the numbers >> >> myself from first principles. I have searched for so-called >> >> online 'calculators', but found that most of them are aimed at >> >> idiots. >> >> >> >> Here's what I expected to find: >> >> >> >> A formula giving the exact sun position throughout the year, >> >> given latitude and longitude, as well as a factor which >> >> handles the distance from the sun w.r.t. time. >> >> >> >> Now this only provides the *input* energy. >> >> >> >> I now need to overlay this with average *weather* data >> >> on a per-hour basis to get the expected net energy. >> >> >> >> Finally, I need to *integrate* these data over an >> >> entire year to get the values for my own house. >> >> >> >> There may be such an online calculator, but I haven't >> >> been able to find it yet. >> >> >> >> It may be possible to factor the weather info out >> >> of the integration, so that I could 'correct' my >> >> location sunlight & weather data with my own lat/lon >> >> azimuth (angle of my house) and roof angle, but I'd >> >> need to see the equations to find out if this factoring >> >> is possible. >> >> >> >> I presume that someone on this list has already done >> >> these calculations for their own home?

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Re: [math-fun] Number of ways to pick 4 points from an m X n grid?
by Keith F. Lynch 28 Jun '20

28 Jun '20

Neil Sloane <njasloane(a)gmail.com> wrote: > Keith, Very interesting! Thanks. > Sylvester looked at this question, see > https://mathworld.wolfram.com/SylvestersFour-PointProblem.html Sigh. After all that work, I discover that someone did it better the century before last, working with a quill pen by candle-light. At least that lets me check my answers by looking "in the back of the book." > For a square, his theorem is that the prob. of getting a convex > quadrilateral is 25/36 , which is exactly the value you found! The same to the six places I gave, correctly figuring that that's where my luck would run out. I calculated 0.694444085 and posted 0.694444. The correct answer is 0.694444444... with the 4s repeating forever. In the circular case, I calculated 0.7044826905 and posted 0.704483. The correct answer is 0.704479881... so I was off by 3 in the last digit I gave. It should have been 0.704480. Sigh. For five points, I calculated 0.3402739547 and posted 0.340274. The correct answer is 0.3402777777777... with the 7s repeating forever. So I should have posted 0.340278. I was off by 4 in my last digit. For the circular five-point case I calculated 0.3561808623 and posted 0.356181. That page doesn't give the solution. Assuming the solution is in the same form as the circular four-point case, 1 - (i / (j pi^2)), where i and j are integers, (35 and 12 in the 4-point case), if I try 1 - (305 / (48 pi^2)), I get 0.356188312... which differs from my answer by 7 in the sixth digit. Does anyone know if that's the correct formula? But for 1 - (i / (j pi^2)) to be the correct formula for the 3-point case, i would have to be 0. There are no instances of 0,35,305 in OEIS. There are too many instances of 0,35, or even 0,0,35, to look at; is there a way to specify that the sequence must *begin* with 0,35 or 0,0,35? Or to specify that the sequence must be monotonic? Thanks. That page also doesn't address the probabilities of a convex quadrilateral in the 5-point case, for any shape of boundary. At least it agrees that I was right that squares and rectangles would give the same answer, and that (although I didn't say it here) circles and ellipses would, and that equilateral and isosceles triangles would. But apparently *all* triangular boundaries have the same answer? That's not obvious to me.

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Re: [math-fun] Number of ways to pick 4 points from an m X n grid?
by Dan Asimov 28 Jun '20

28 Jun '20

One method I did not see listed in that linked MathWorld article is to choose the points from a standard normal distribution in the plane, i.e., with density given by p(x,y) = (2π)^(-1) exp(-(x^2 + y^2)/2). This is of course circularly symmetric, but it's not at all clear to me that four points picked from this distribution will have the same chance of their convex hull being a quadrilateral as four points picked from the uniform distribution on a disk. Now I'm really curious: What *is* the probability that the convex hull of four points chosen from the standard normal distribution in R^2 will be a quadrilateral??? —Dan ----- Keith, Very interesting! I will have to study your message carefully, but I just wanted to mention something Ed Pegg pointed out. Sylvester looked at this question, see https://mathworld.wolfram.com/SylvestersFour-PointProblem.html For a square, his theorem is that the prob. of getting a convex quadrilateral is 25/36 , which is exactly the value you found! -----

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Re: [math-fun] Number of ways to pick 4 points from an m X n grid?
by Keith F. Lynch 28 Jun '20

28 Jun '20

Neil Sloane <njasloane(a)gmail.com> wrote: > If we take an m X n rectangular grid of points, and choose four of > them (in binomial(m*n,4) ways), there are four possibilities: > (i) the 4 points are collinear, > (ii) 3 points are collinear and one point is not on that line, > (iii) 3 points form a triangle of positive area and the other point is in > the interior of that triangle, > (iv) the 4 points form a convex quadrilateral of nonzero area. > . . . > Which is more likely, (iii) or (iv)? This inspired me to look at the limiting case where m and n are large. Your i and ii disappear. The values of iii and iv don't depend on the aspect ratio of the rectangle. I'm not smart enough to find a closed-form solution. (Is one known?) I generated 64 billion (pseudo-)random sets of four points in a square, using Unix's "random()" function, which generates 31-bit integers. Since that function repeats after about 2^35 (about 34 billion), after every 2 billion sets (16 billion random numbers), I exclusive-ored every random() with a different row of an order-32 Hadamard matrix (with the -1s turned into 0s, and the first column (which was all 1s) stripped off). Every one of the 32 rows differs in exactly 16 bit positions from every other of the 32 rows. I was amused, when I went searching for Hadamard matrices, that I found a table of them on your website. So, how do I determine, given the x and y coordinates of four points, whether they form a convex quadrilateral or not? After some thought, I decided to check, for every one of the six line segments between two of the four points, whether it intersects the unique other segment which has no endpoints in common with it. If they intersect, it's a convex quadrilateral. If they don't, it's a triangle with a point inside. That's 3 checks, not 6, since if I know whether A intersects B, I don't need to check whether B intersects A. I get triangles 0.305556 of the time, and convex quadrilaterals 0.694444 of the time. (The last digit may be off by 1.) In 193 of the 64 billion cases, one of the lines was vertical, and my program divided by zero. To speed things up, I didn't bother to special-case that, or even to figure out what the program would do there, since whatever it did would be lost in the noise, changing at worst the 9th digit of a 6 digit number. I also checked for parallel lines, the other possible source for division by zero. But there were never any parallel lines. But wait, there's more! If the four points happened to all be within the circle inscribed in the square, I also used them to find the solution for a circle instead of a rectangle. This time I got a triangle 0.295517 of the time, and a convex quadrilateral 0.704483 of the time. (Again, the last digit may be off by 1.) This was from 24352319646 circles, compared to an expected (pi/4)^4 * 64 billion = 24352272758 circles. Those numbers agree to six digits, or two parts in a million, which gives me confidence in the random number generator. But wait, there's more! That was so much fun, I decided to go one better. *Five* points. There are three non-infinitesimal possibilities: A triangle with two points in it, a convex quadrilateral with one point in it, or a convex pentagon. With that many points per set, the program was generating random numbers like they were going out of style. So I used a new row of the Hadamard matrix every billion sets, for a total of just 32 billion sets. Nevertheless, with the additional computation per set, the program took longer to run than the 4-point program. Again, I counted intersections between line segments which didn't share any endpoint. This time there were 15, rather than 3, such tests to be made per set. If there's 1 intersection, it's a triangle with two points in it. If there are 3 intersections, it's a convex quadrilateral with one point in it. It there are 5 intersections, it's a convex pentagon. There is never any other number of intersections. This time there were 165 vertical lines. Still no parallel lines. And here are the results: Triangle: 0.104167 Quadrilateral: 0.555559 Pentagon: 0.340274 That's within a rectangular boundary. Here are the results for a circular boundary: Triangle: 0.094992 Quadrilateral: 0.548827 Pentagon: 0.356181 That's with 9563216295 circles, vs. an expected (pi/4)^5 * 32 billion = 9563115149 circles, the same to within about 10 parts in a million. Higher polygons? The number of pairs of segments which share no endpoints seems to be given by A050534, with an offset. So, 45 for six points, 105 for seven, etc. What's the limiting case? For N points, where N is large, just how unlikely is it that they comprise a convex N-gon? And what's the expected order of the bounding polygon? Does it scale with the square root of N? With the natural log of N? Has anyone explored this stuff? Thanks.

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Re: [math-fun] Triangular Puzzle
by Walter Trump 27 Jun '20

27 Jun '20

Eric, I like your triangular puzzle and wrote a program that improves existing solutions by swapping digits. A good approach is to put the digits "1" in the longest diagonal and digits "2" in the next one. A lower bound of the sum is exactly 6000. My best solution so far has a sum of 6073. To improve this solution the integers 50, 60 and 70 should be replaced by smaller integers. It is not possible to replace one of the numbers 111, 222, 333, ..., 999. 555, 666, 888, 444, 999, 70, 333, 222, 111 44, 777, 66, 8, 55, 39, 20, 1 77, 9, 3, 6, 40, 25, 18 99, 5, 7, 30, 26, 14 88, 4, 50, 27, 13 60, 38, 24, 15 33, 2, 19 22, 10 11 Sum = 6073 Walter Trump

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[math-fun] A Chess Prime Tour
by Éric Angelini 26 Jun '20

26 Jun '20

Hello Math-Fun Any legal movement of a chess piece can be coded by a 4-digit integer with the traditional convention of square labeling: two digits for the starting square, two digits for the arrival square (1181 would be a Rook movement, for instance, going from a1 to h1). Question #1 What is the finite set of all such legal chess movements/integers? I guess we would (lexicographically) start the collection with: 1) Legal movements starting on the square 11: 1112, 1113, 1114, 1115, 1116, 1117, 1118, 1121, 1122, 1123, 1131, 1132, 1133, 1141, 1144, 1151, 1155, 1161, 1166, 1171, 1177, 1181, 1188; 2) Legal movements starting on the square 12: 1211, 1213, 1214, 1215, 1216, 1217, 1218, 1221, 1222, 1223, 1224, 1231, 1232, 1233, 1234, 1242, 1245, 1252, 1256, 1262, 1267, 1272, 1278, 1282; 3) Legal movements starting on the square 13: 1311, 1312, 1314, etc. (...) 64) Legal movements starting on the square 88: 8811, 8818, 8822, 8828, 8833, 8838, 8844, 8848, 8855, 8858, 8866, 8867, 8868, 8876, 8877, 8878, 8881, 8882, 8883, 8884, 8885, 8886, 8887. Question #2 How many integers of the above set are prime numbers? (those would be called "prime movements") Question #3 What could be the longest "prime tour" made only by distinct prime movements? (This tour would start and stop on the same chessboard's square). Here is an example of a 6-stage such Tour, made by prime numbers taken from the above subset of "prime movements": 1117–1753–5351–5153–5323–2311. Best, É. [With arrows and colors here, on my personal weblog (after the French chess problems part) --cplease rebuild the antispam link by joining the three hereunder segments] http://cinquantesignes. blogspot.com/2020/06/echecs- au-carre.html

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[math-fun] Farkas' lemma --- affine / partial versions?
by Andres Valloud 25 Jun '20

25 Jun '20

Hi, are there any affine counterparts to Farkas' lemma, or the splitting hyperplane theorem? Are there partial results? Andres.

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