April 2007 - mathfun@mailman.xmission.com

Re: [math-fun] mechanical engineering sanity check
by Eugene Salamin 27 Apr '07

27 Apr '07

Let rho = density, g = gravity, sigma[i,j] = stress tensor, f[i] = body force per unit volume, F[i] = surface force per unit area, n[i] = unit surface normal, k = length scale factor. The stress obeys the following two equations (plus one more homogeneous equation that I do not need here). sum(diff(sigma[i j], x[j]), ,j) = f[i] within the volume, sum(sigma[i j] n[j], j) = F[i] on the bounding surface. Suppose the system must look the same under length scale change. Then sigma[i j] must be invariant. The body force is gravity, so f = rho g. But the LHS scales as 1/k. Hence rho g must be scaled as 1/k. The LHS side of the second equation is independent of k. On the RHS, the reaction force scales as rho g k^3 and the contact area scales as k^2, so F scales as rho g k, and this is then independent of k. So your final sentence is correct. Gene ----- Original Message ---- From: R. William Gosper <rwg(a)osots.com> To: math-fun(a)mailman.xmission.com Sent: Tuesday, April 24, 2007 11:26:01 AM Subject: [math-fun] mechanical engineering sanity check I have a 1/k scale model of a weight on a tripod (http://gosper.org/legbox3.jpg) Alan Adler tells me beam strength scales with the width and the square of the thickness, so k^3. Likewise the load, but there is also a factor of k snapping moment. So for large k, the legs would need to thicken like k^(4/3). (DaVinci's analysis of elephants, giraffes, tree-trunks, ...?) But my k is only about 2.5, and the model is fairly robust, and I want to see how the full-scale would load without the extra leg thickening. So I just need to scale the *density* by k, right? I.e., on a planet with 2g, all the animals could be half scale. --rwg _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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Re: [math-fun] mechanical engineering sanity check
by Phil Carmody 25 Apr '07

25 Apr '07

From: "R. William Gosper" <rwg(a)osots.com> > > I have a 1/k scale model of a weight on a tripod > (http://gosper.org/legbox3.jpg) Alan Adler tells me beam strength > scales with the width and the square of the thickness, so k^3. What happened to the cube in d = 4Fl^3/(Yab^3) d = displacement F = force l = length of the beam a = width of beam b = thickness of beam Y = YoungÂs modulus Or have I misunderstood the question? Phil () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani] __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] Re: Path Question
by James Propp 25 Apr '07

25 Apr '07

I mentioned the Ehrenfeucht-Mycielski sequence and gave a reference, but I realize that few of you will take the trouble to look at the PDF unless I do a bit of advertising. Here's a passage from a different paper "On the Ehrenfeucht-Mycielski Balance Conjecture" by Kieffer and Szpankowski (available from the web at www.cs.purdue.edu/homes/spa/papers/em.pdf) "In the paper Ehrenfeucht and Mycielski (1992), an interesting binary sequence was defined, since termed the EM-sequence, which seems to possess pseudorandomness properties. The EM-sequence is sequence A038219 in the encyclopedia Sloane (2007), and is generated via an algorithm described in Sloane (2007) as follows: "The sequence starts 0,1,0 and continues according to the following rule: find the longest sequence at the end that has occurred at least once previously. If there are more than one previous occurrences select the last one. The next digit of the sequence is the opposite of the one following the previous occurrence." For example, the first 50 terms of the EM-sequence are 01001101011100010000111101100101001001110100011000. Despite the simplicity of this algorithm, not very much is known about the asymptotics of the EMsequence. It is natural to conjecture that the EM-sequence behaves as a typical sequence generated by a binary IID process. In particular, we would expect that the averages of the initial segments of the EM-sequence converge to 1/2; this is called the balance conjecture." The balance conjecture remains open, as far as I know. Jim

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[math-fun] mechanical engineering sanity check
by R. William Gosper 25 Apr '07

25 Apr '07

I have a 1/k scale model of a weight on a tripod (http://gosper.org/legbox3.jpg) Alan Adler tells me beam strength scales with the width and the square of the thickness, so k^3. Likewise the load, but there is also a factor of k snapping moment. So for large k, the legs would need to thicken like k^(4/3). (DaVinci's analysis of elephants, giraffes, tree-trunks, ...?) But my k is only about 2.5, and the model is fairly robust, and I want to see how the full-scale would load without the extra leg thickening. So I just need to scale the *density* by k, right? I.e., on a planet with 2g, all the animals could be half scale. --rwg

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[math-fun] Re: Path Question
by James Propp 23 Apr '07

23 Apr '07

Sorry to take so long responding! Michael Greenwald writes: >Second, I am wondering about the intuition behind the error >metric that was outlined in James Propp's email. If I understand >it correctly, the idea is to find a construction for an infinite >string S in which the "distributional error" [I don't know any >term (let alone a *better* term) for the error computation used] >over substrings of *all* sizes n < N decreases asymptotically >fastest (I think someone said that log(N)/N is optimal, no?) with >the length of the string prefix. I probably said it wrong, so let me try to say it a different way. The thing I want is an infinite sequence S of 0's and 1's with the following countably many properties: 1) The number of times "0" occurs in the first N bits of S goes to infinity like N/2 + O(log N) 2) The number of times "1" occurs in the first N bits of S goes to infinity like N/2 + O(log N) 3) The number of times "00" occurs in the first N bits of S goes to infinity like N/4 + O(log N) 4) The number of times "01" occurs in the first N bits of S goes to infinity like N/4 + O(log N) etc. (with one property for each finite string of 0's and 1's). So the quantification is that for each n, and for each bit-string of length n, there exists a constant C such that for all N > 1, the number of times the bit-string of length n occurs in the first N bits of S differs from (N-k+1)/2^n by less than C times log N. Note that the constant C may depend on the particular bit-string of length n one is looking at. Note also that I've used N > 1 rather than N > N_0 (for some N_0); this cuts down on the number of quantified variables with no real change in meaning (the only price you pay is that you might have to use a bigger C). >So I'm wondering whether >there's any intuition about why we might care about all length >substrings equally, even substrings pretty close to N in length, >rather than focusing on strings that are small compared to N. I don't see any reason to think that a sequence of the kind I want will have good properties for the initial N terms vis-a-vis substrings of length n when n is a lot bigger than log N. Steve Witham writes: >Wouldn't it be easier to look at the minimum frequency instead of >the distribution of frequencies? I realize it's not the same >question, but as long as you're hunting in the dark. I agree: for fixed n and N, one could look at the sequence s of length n that occurs the least number of times in one's sequence S of length N, and try to make this quantity as large as possible by choosing S cleverly. One candidate for S that deserves more study is the very pretty Ehrenfeucht-Mycielski sequence: for an introduction see Klaus Sutner's paper www.cs.cmu.edu/~sutner/papers/em-sequence.pdf . It's mildly scandalous that we still don't know whether 0's and 1's occur in the E-M sequence with asymptotic frequency 1/2 (let alone finite subsequences of length greater than one!). Jim

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[math-fun] Re: Path Question
by James Propp 23 Apr '07

23 Apr '07

Dan writes: >A folk theorem holds that the vector defined by > > V_n := (phi,phi^2,...,phi^n), > >where p is the golden ratio, can't be beaten, at least in the limit. I don't understand this. Even if we just look at the first three components of the vector, we see that [m phi^3] (which I'll remind everyone means the fractional part of m phi^3, under Dan's notation) is equal to [m phi^2] plus [m phi], for all m, so the set of multiples of (phi,phi^2,phi^3) mod (1,1,1) stays in the part of the torus (x,y,z) with x+y=z mod 1, and hence cannot be dense in T^3. Or am I missing something? Leaving that quibble aside, I like Dan's geometric point of view. Even if (phi,phi^2,...,phi^n) is the wrong vector to use, some vectors are going to be okay. And Dan's concatenation method is bound to give something decent even if it doesn't give N/log N. Thanks, Dan! Jim

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CORRECTION Re: [math-fun] Re: Path Question
by Daniel Asimov 23 Apr '07

23 Apr '07

CORRECTION: << (*) floor{2*x_1}, . . ., floor{2*x_n} >> should correctly read: << (*) floor(2*{x_1}), . . ., floor(2*{x_n}) >> Sorry about that. --Dan

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Re: [math-fun] Re: Path Question
by Daniel Asimov 23 Apr '07

23 Apr '07

Jim asks: << Can anyone find an explicit construction of an infinite sequence S of 0's and 1's with the property that for any finite bit-string s, the relative frequency with which s occurs in the first N bits of S converges to (1/2)^(length of s) with error that falls like (log N)/N? >> I have an idea for a strategy that may result in a solution. In the cubical torus T^n = R^n/Z^n one may wish to find a vector v in R^n such that stepping along T^n in equal steps each equal to [v] (= the image of v in T^n by the quotient map R^n -> T^n) gets you dense in T^n as fast as possible. I.e., define R_v(n) := the radius of the largest open ball in T^n lying in the complement of {[v], [2v], . . ., [nv]}. We want to find v such that R_v(n) decreases as fast as possible. A folk theorem holds that the vector defined by V_n := (phi,phi^2,...,phi^n), where p is the golden ratio, can't be beaten, at least in the limit. (This is based on the non-folk theorem that stepping along the circumference-1 circle T = R/Z in steps of equal length gets you dense in T as fast as possible by using steps of length phi.) THE PRESENT PROBLEM may be thought of as similar to a series of such torus problems on tori T^n, where a point [x_1, . . .,x_n] of T^n is mapped to the 0-1 sequence (*) floor{2*x_1}, . . ., floor{2*x_n} where {y} denotes the fractional part of y. The idea being, to get a 0-1 sequence with all the right probabilities having the least error, PERHAPS it will work to let n = 1, 2, 3, . . . for L(n) consecutive steps, and stepping through T^n as in the above paragraph, and *concatenating* all the length-n 0-1 sequences obtained from each step via (*). I suspect L(n) should be something of the order of L(n) = 2^n. The step itself, while stepping through T^n, should perhaps be the vector V_n as above. QUESTION: What if this whole thing were collapsed down to the 1-dimensional version of just stepping through T = R/Z with a constant steplength = phi, and writing down the sequence e_n = floor(2*n*phi) --Dan

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[math-fun] Re: Path Question
by James Propp 23 Apr '07

23 Apr '07

I wrote: >(with one property for each finite string of 0's and 1's). So >the quantification is that for each n, and for each bit-string of >length n, there exists a constant C such that for all N > 1, >the number of times the bit-string of length n occurs in the >first N bits of S differs from (N-k+1)/2^n by less than C times >log N. For "N-k+1" read "N-n+1". Jim

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Re: [math-fun] The Plumber's Nightmare Graph
by Daniel Asimov 23 Apr '07

23 Apr '07

Ed wrote: << I found this paper about the Klein graph quite nice. http://www.liga.ens.fr/~deza/withFowler/AdditionKlein.pdf They call it the Plumber's Nightmare Graph, which is an interesting name. >> In case anyone is interested in some context for this: This is the edge graph (the 1-skeleton) of the so-called "regular map" of the 3-holed torus by 24 heptagons, 3 per vertex. A regular map on a surface has a combinatorial automorphism group that's transitive on flags: A flag is any triple given by (v,e,f) for v = vertex lying in e = edge lying in f = face. There are 7x24/3 = 56 vertices, each lying in 3 edges, each lying in 2 faces, giving a total of 56x3x2 = 336 flags, and by transitivity 336 is also the order of the automorphism group. What's special about this particular 3-holed torus (as a Riemann surface) is that for any compact Riemann surface of genus >= 2, its group of conformal transformations G is finite and must satify Hurwitz's bound #(G) <= 84(g-1). There is a *unique* Riemann surface M_3 of genus 3 (up to conformal equivalence) that reaches this bound of 84x2 = 168 conformal equivalences, and since no Riemann surface of genus 2 meets Hurwitz's bound, this is the "first" Riemann surface that achieves this maximal symmetry. The group Aut(M_3) of order 168 is the unique simple group of order 168, mysteriously isomorphic to both PSL(3,Z_2) and PSL(2,Z_7). Is is also a subgroup of the Lie group PSL(2,C). If anti-conformal equivalences are thrown in as well, one gets the same group G of order 336 above. The Riemann surface M_3 has many (real analytic) metric realizations, among them one of constant negative curvature (unique if the curvature is fixed, at say -1). This can be constructed from 24 regular heptagons in the hyperbolic plane, each of the unique size that allows exactly 3 of them to fit about any vertex. But it has an even more beautiful metric realization M_K given by one homogeneous complex polynomial defined on the complex projective plane CP^2: the polynomial was discovered by Klein and is X Y^3 + Y Z^3 + Z X^3 = 0 (where CP^2 is defined as the quotient of C^3 - {0} by (X Y Z) ~ (cX cY cZ) for all nonzero c in C). M_K does not have constant curvature, but it is a minimal surface in CP^2. --Dan

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