November 2012 - mathfun@mailman.xmission.com

Re: [math-fun] [EXTERNAL] Optimal Dome
by Henry Baker 12 Nov '12

12 Nov '12

FYI -- The Gateway Arch, as does any hanging chain, has a horizontal force component which has to be handled by the foundations of the arch. The concept of using hanging chains with weights attached goes back to Hooke, at least, and for some strange reason this concept of the equivalence of chains and arches through this inversion is called "funicular" by structural engineers. So if you want to know more, include "funicular" in your Google searches. Osserman, Robert. "Mathematics of the Gateway Arch". Notices of the AMS 57, 2, Feb. 2010, 220-229. http://www.ams.org/notices/201002/rtx100200220p.pdf Also, Osserman, Robert. "How the Gateway Arch Got Its Shape": www.msri.org/people/staff/osserman/papers/NNJ_v12n2_Osserman_pp167-189.pdf At 08:38 AM 11/12/2012, Cordwell, William R wrote: >Isn't there a shape that has purely compressional tension? The St. Louis arch? It does not have a vertical tangent, as you note.

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Re: [math-fun] Dome
by Warren Smith 12 Nov '12

12 Nov '12

>From: "Cordwell, William R" <wrcordw(a)sandia.gov> >Isn't there a shape that has purely compressional tension? --"purely compressional tension"?? >The St. Louis arch? It does not have a vertical tangent, as you note. --(thin) arches have only one direction of interest; domes being membranes have two. Hence there is no issue for arches with multi-dimensional stress. For membrane and bulk elasticity, there is such an issue. If you want to control both kinds of stress, you need two functions -- thickness and shape -- to control it with. Heyman's solution (his EQ9) has no vertical tangent, which I presume is related to him getting zero tension. But actually, I guess maybe you could allow a vertical tangent if it occurs exactly at dome base assuming dome base anchored. (If dome base free to slide,then there would be tension in the base-hoop.) Heyman has no base. Heyman's solution goes to infinite height rapidly which is ridiculous. The true solution, I claim, with zero tension imposed, should have a finite maximum allowable height; or if infinite height then this would have to be accomplished by having roughly exponentially-growing thickness (soon violating the thin-membrane assumption) as we go downward.

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Re: [math-fun] Optimal Dome
by Henry Baker 12 Nov '12

12 Nov '12

Hi Warren: Heyman literally "wrote the book": Heyman, J. The Stone Skeleton: Structural Engineering of Masonry Architecture. Cambridge University Press. 1995. Re tension: You are correct. Without some additional horizontal force somewhere, the erf/erg dome can't stand. Chains around the base are a cheap/easy solution to this problem. But the bulk of the dome is only in compression. ---- You might enjoy the "oil lamp" shaped constructions towards the end of the following paper. Curiously, they don't consider 3D constructions. Overhang Mike Paterson, Uri Zwick (Submitted on 12 Oct 2007) How far off the edge of the table can we reach by stacking $n$ identical, homogeneous, frictionless blocks of length 1? A classical solution achieves an overhang of $1/2 H_n$, where $H_n ~ \ln n$ is the $n$th harmonic number. This solution is widely believed to be optimal. We show, however, that it is, in fact, exponentially far from optimality by constructing simple $n$-block stacks that achieve an overhang of $c n^{1/3}$, for some constant $c>0$. http://arxiv.org/abs/0710.2357 At 08:19 AM 11/12/2012, Warren Smith wrote: >So I have a certain amount of skepticism Heyman knows what he is doing.

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[math-fun] Optimal Dome
by Warren Smith 12 Nov '12

12 Nov '12

> From: Henry Baker <hbaker1(a)pipeline.com> > I've always wondered about the shape of the ideal self-supported masonry > dome -- e.g., the Roman Pantheon, Hagia Sophia in Istanbul, St. Peter's > Rome, etc. "Self-supported" here means compression forces only, as masonry > has very little strength in tension or shear. > > Here's the solution, which involves the erf & erg functions, instead of > (exp(x)+exp(-x))/2=cosh(x) (2D catenary). > > z_r = 1/2 r_0 [erg(r/r_0) - sqrt(pi)/2 erf(r/r_0)] > > (equation (9) in [Heyman1998] below) > > Note the obvious analogy to cosh/sinh functions. > > The Roman Pantheon is hemispherical, which is too narrow near the top and > too wide near the bottom. All domes since the Pantheon incorporate one or > more circular iron/steel chains horizontally around the lower portions of > the dome to counteract the spreading forces; --well, now you've blown it. Your stated goal was to get pure compression. But as you just admitted, there is horizontal tension. But in your defense, I claim this is impossible to avoid, at least for a convex dome that eventually has vertical tangent; you unavoidably get compression along vertical "meridians" but tension along "lines of latitude." Jacques Heyman's paper http://rsnr.royalsocietypublishing.org/content/52/1/39.full.pdf you gave derives the "shape of the ideal dome" without, however, DEFINING what it MEANS to be an "ideal dome." So I have a certain amount of skepticism Heyman knows what he is doing. If he knew, then the paper should contain a definition and a theorem? I would propose that the definition should involve uniform stress, which means the meridian-compression stress is everywhere equal to a constant C, and the latitude-tension stress is everywhere equal to a constant T. Assume constant-weight-density material. The solution will involve both specifying the shape (radius vs height curve) of the dome, and also the thickness (assumed thin everywhere, but it varies) as a function of height. There will be a system of 2 differential equations. I claim it is not possible to do it with uniform-thickness; if you tried you would succeed in uniformizing one (say C), but the other would do whatever it wanted, which unless a miracle occurs will not be to be uniform. To uniformize both quantities you need two functions worth of control, i.e. both thickness and shape. Why is uniform-stress <==> optimal shape? Because if nonuniform stress you can remove a little material from a low-stress place and relocate it to a high-stress place, thus getting a better dome (can survive using a weaker-strength material) involving same amount material. But one must admit this argument makes more sense if "stress" is a scalar-valued quantity like in a thin-beam arch (compression only). For a dome, stress is 2-dimensional quantity. [Also there could be a third kind of stress, shear, but due to symmetry the shear will always be zero for a thin dome. For 3D bulk elasticity as opposed to 2D membrane elasticity stress would be 6-dimensional.] So really, I think the real truth is that there is a magic function F(C,T) of the two (or more) kinds of stress, such that the material self-destructs when F>threshold. and really we want to uniformize this FUNCTION, which IS scalar-valued. So if you knew F, then you could derive an optimum shape for a uniform-thickness dome. However, this function is unknown -- the engineers as far as I know have basically never characterized it despite centuries -- and it presumably depends on the material-type. [To see that, consider a liquid "material" which self-destructs if tension>0 but no problem with compression. Obviously this has a different F than a typical solid.] Therefore, to avoid having to worry about that, I'm suggesting uniformizing both C and T individually rather than F(C,T). Notice that this F-issue is not discussed at all by Heyman, again suggesting at a fundamental level he does not know what he is doing. Frankly, I believe this entire area is dominated by people who do not know what they are doing at a fundamental level. [I am not sure what Heyman had in mind, but perhaps it was: uniform compressional stress, uniform thickness, and variable uncontrolled tensional stress? Or perhaps it was: zero tension, and uncontrolled variable compressional stress? Apparently the latter: Heyman in his EQ9 giving his main result, has the property that y, the nega-height, gets infinitely large while x, the radius, grows very slowly. How can this be true while still having uniform stress?? Think it cannot. So I claim Heyman's "optimal" dome, which is infinitely high, would in practice if you built it too high in a constant g-field, self-destruct due to high compressive stress near the base.] Now let me try to (begin to) formulate the math. r(y) = radius as function of height y. q(y) = thickness ditto. The compressional stress at any given y equals the weight above that y (expressible as integral), divided by 2*pi*r*q. This is to be set equal to C to enforce uniform stress. The tensional stress due to weight above a thin band of dome at any given y is proportional to the weight above that y, times the curvature of the meridian-curve there (assume always convex so positive sign), times r/q. There is also a countervailing term arising from the weight within that thin band itself. It is proportional to r*q*density*r'(s) where s is arclength along meridian. [I said "proportional" not "equal" since there will some factors of density, pi, and stuff I'm too lazy to figure out now.] The difference of these has to be set equal to T. You then have a system of 2 integro-differential equations in 2 unknown functions -- which can be converted to pure-differential form by using fundamental theorem of calculus. You then attempt to solve the system, which will probably be impossible in closed form, but let's see. I think Heyman actually did have a cute idea here, which if I understand aright was to set T=0 thus modeling a "liquid" material aka "oversimplified masonry", but he failed to recognize the need for BOTH thickness-curve and shape-curve control to try to uniformize the compressive stress C, therefore obtaining a manifestly-ridiculous result.

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[math-fun] moRe: EllipticE valuation [Was: EllipticK valuation [Was: Elliptic K π approximation]]
by Bill Gosper 12 Nov '12

12 Nov '12

On Wed, Nov 7, 2012 at 3:05 PM, Bill Gosper <billgosper(a)gmail.com> wrote: The ("complete") first kind elliptic integral K(m) and parameter m are both expressible in 𝜗 constants, and those three+ 𝜗 constants are all expressible in terms of Dedekind η. So Out[840]= EllipticK[(16 η[q]^8 η[q^4]^16)/η[q^2]^24] == (π η[q^2]^10)/(2 η[q]^4 η[q^4]^4) Crude but effective: d/dq and eliminate EllipticK to get EllipticE[(16 \[Eta][q^4]^8)/(\[Eta][q]^8 + 16 \[Eta][q^4]^8)] == -( 1/(\[Eta][q]^8 + 16 \[Eta][q^4]^8))(-((\[Pi] \[Eta][q]^4 \[Eta][q^2]^10)/( 2 \[Eta][ q^4]^4)) - (\[Pi] \[Eta][ q^2]^9 (6 Sqrt[3] q^(3/2) logderiveta[q] \[Eta][q] \[Eta][q^2] \[Eta][q^4] - 30 Sqrt[3] q^(5/2) logderiveta[q^2] \[Eta][q] \[Eta][q^2] \[Eta][q^4] + 24 Sqrt[3] q^(9/2) logderiveta[q^4] \[Eta][q] \[Eta][q^2] \[Eta][ q^4]) (\[Eta][q]^8 + 16 \[Eta][q^4]^8))/(2 q^( 1/3) \[Eta][q]^4 \[Eta][ q^4]^4 (6 Sqrt[3] q^(7/6) logderiveta[q] \[Eta][q] \[Eta][q^4] - 24 Sqrt[3] q^(25/6) logderiveta[q^4] \[Eta][q] \[Eta][q^4])) The remaining problem is those logderivetas, but lo, ((1/(4*x) - ((logderiveta[(1/(E^x))]))/(E^x))/((DedekindEta[((x*I)/(2*Pi))])^4)) is algebraic for (x/π)^2 rational. Unfortunately, (x/π)^2= (i/2-1/√12)^2 (as in the q for k=(-1)^(1/3)) isn't rational, so EllipticE[(-1)^(1/3)] == E^(-(I π/12)) ((2 2^(1/3) π^2)/(3^(3/4) Gamma[1/3]^3) + Gamma[1/3]^3/(4 2^(1/3) 3^(1/4) π)) required Legendre's relation plus numerical empiricism. This, with the K((-1)^(1/3)) result gives closed forms for all 2F1[a+1/2,b+1/2,c;(-1)^(1/3)], a,b,c integers. Likewise for every k for which we find both K(k) and E(k), which are contiguous 2F1[k]. BtW, if you accept MatProd[mat[k],{k,2,a}] as closed form, then I can write 2F1[a+1/2,b+1/2,c;k] in closed form. --rwg

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Re: [math-fun] Fermat's last theorem
by Adam P. Goucher 12 Nov '12

12 Nov '12

Yes, that was conjectured, but Noam Elkies found a counter-example (four fifth powers in the relationship a^5 + b^5 + c^5 = d^5, or something like that). Sincerely, Adam P. Goucher > ----- Original Message ----- > From: David Makin > Sent: 11/12/12 12:29 PM > To: math-fun > Subject: [math-fun] Fermat's last theorem > > Hi, > > Just a quick thought I had the other day, as I understand it Fermat's last theorem (now proved) basically says that for: > > a^p + b^p = c^p > > Then where all variables are integers there are no solutions for a, b and c where p>2. > > My thought was has anyone considered: > > a^p + b^p + c^p = d^p > > or indeed: > > a1^p + a2^p + a3^p + ..... an^p = b^p > > And is it possible that for the case of: > > a^p + b^p + c^p = d^p > > Then there is a solution for a,b,c,d for p=3 but not for p>3 and generally for: > > a1^p + a2^p + a3^p + ..... an^p = b^p > > there's a solution for a1..an and b if p=n but not for p>n ? > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] DC elliptical subway tunnels
by Henry Baker 11 Nov '12

11 Nov '12

I had occasion to ride the DC metro this past week, and noticed that many of the tunnels -- or at least the subway station tunnels -- appear to be elliptical in cross-section. The DC subway tunnels are particularly deep, at least compared with NY & Boston. I seem to recall that one goal was to provide some cover in case of nuclear attack. In any case, these tunnels are most likely under the water table, so that simply keeping their elliptical shape requires some effort. In the subway stations, the opposing train tracks are closer together than the foci of the ellipse. Indeed, the foci appear to be right in the middle of the passenger boarding area. This means that the "whispering gallery" effect provides cross-talk between the passengers waiting on either side of the subway tunnel, and means that the noise level is much higher than it would otherwise be. While the coffering of the elliptical surface does cut down on some of the noise, it is still nearly impossible to hear any announcements from the public speaker system. I guess there is no hope of reducing the noise level without completely covering the elliptical surface with sound deadening materials. On the plus side, I did notice that when 2 DC trains passed on another on opposite tracks, I didn't get the "high air pressure"/"low air pressure" sequence that I've noticed in the Boston subway.

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[math-fun] 3D catenary uses erf & erg
by Henry Baker 11 Nov '12

11 Nov '12

I've always wondered about the shape of the ideal self-supported masonry dome -- e.g., the Roman Pantheon, Hagia Sophia in Istanbul, St. Peter's Rome, etc. "Self-supported" here means compression forces only, as masonry has very little strength in tension or shear. Here's the solution, which involves the erf & erg functions, instead of (exp(x)+exp(-x))/2=cosh(x) (2D catenary). z_r = 1/2 r_0 [erg(r/r_0) - sqrt(pi)/2 erf(r/r_0)] (equation (9) in [Heyman1998] below) Note the obvious analogy to cosh/sinh functions. The Roman Pantheon is hemispherical, which is too narrow near the top and too wide near the bottom. All domes since the Pantheon incorporate one or more circular iron/steel chains horizontally around the lower portions of the dome to counteract the spreading forces; the Pantheon avoids the need for these chains with an absolutely massive circular drum structure which acts like a continuous circular buttress. Hagia Sophia was originally built without chains, but after several collapses, chains were finally added to stabilize the dome. The Pantheon is a living example of Archimedes' Hat Box Theorem: the right cylindrical drum encloses a perfect sphere touching the center of the Pantheon floor and the hemispherical interior of the dome. From the outside, the dome looks a lot more like a cone truncated at the eye; the space between the cone and the hemispherical internal surface is more-or-less filled with material, but with coffering. As the paper below demonstrates, Hooke's guess at the appropriate curve (z_r ~ r^3) (used in Wren's St. Paul's Cathedral, London) is the first term in the Taylor series for the correct erg/erf function. The most interesting feature of this equation is the fact that at r=0 the radius of curvature is infinite (i.e., a plane), unlike a spherical done, which has a finite radius of curvature. But an infinite radius of curvature implies that the compressive forces at r=0 are also infinite (in order to keep the load in the very center from falling down). But the Romans guessed/intuited this: the Pantheon cleverly cuts out the center & leaves the famous "eye" hole; Wren's St. Paul's also includes a hole in the center. Heyman, Jacques. "Hooke's Cubico-Parabolical Coniod". Notes Rec. R. Soc. Lond. 52(1), 39-50 (1998). http://rsnr.royalsocietypublishing.org/content/52/1/39.full.pdf

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[math-fun] generalization of an ellipse to 3 foci ???
by Henry Baker 11 Nov '12

11 Nov '12

What would be the appropriate generalization of an ellipse to more than 2 foci ?

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[math-fun] mental calculation; trifocal ellipses
by rcs＠xmission.com 11 Nov '12

11 Nov '12

A recent "World Cup" contest for mental calculation--- The events go up to cube roots. http://www.recordholders.org/en/events/worldcup/2012/ Willem Bouman, mental factoring of 4 & 5 digit numbers: http://www.recordholders.org/downloads/worldcup/factorization-tutorial.pdf --- Henry Baker asks > What would be the appropriate generalization of an ellipse to more > than 2 foci ? Obvious candidates for trifocals: Assume the three foci are a triangle, probably scalene. a) The string & pencil construction. This will give a curve with six sections of ellipses. b) The locus of points whose sum-of-distances to the foci is constant. This extends naturally to three dimensions. c) Combining A & B in a natural way in 3D: the constraint for our pencil is that the tetrahedron from the pencil to the focal triangle has a constant area -- as if it were a soap film, perhaps. Rich

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