September 2014 - mathfun@mailman.xmission.com

[math-fun] mile-high (or more) towers?
by Bill Gosper 21 Sep '14

21 Sep '14

WDS>Suppose we wanted to create an extremely high tower (think of it as a very long vertical rod) by combining the facts that (a) compressive strength of materials is large (b) "buckling" instabilities can be prevented by active control. Another idea is an extremely large-span arch (if a constant-diameter rod is bent into the shape of a "cycloid" "catenary"?--hgb curve, then the stresses in this arch are purely compressive). (Of course, it'd be a bummer if you had a power or software failure... or an airplane hit it...) [lots&lots of interesting analysis]<WDS John McCarthy once computed that an affordable column of tungsten (I forget his dimensions, something like a hundred meters by a few inches) would be a crust penetrator. Hans Moravec designed towers many miles high rigidified by a circulating stream of superconducting "niobium cannonballs". You could also build very large "vacuum balloons" thus stiffened. --rwg

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Re: [math-fun] mile-high (or more) towers?
by Henry Baker 21 Sep '14

21 Sep '14

"cycloid" -> "catenary" https://en.wikipedia.org/wiki/Sagrada_Fam%C3%ADlia At 01:19 PM 9/21/2014, Warren D Smith wrote: >Another idea is an extremely large-span arch (if a constant-diameter rod is bent into >the shape of a "cycloid" curve, then the stresses in this arch are purely compressive).

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[math-fun] mile-high (or more) towers?
by Warren D Smith 21 Sep '14

21 Sep '14

Suppose we wanted to create an extremely high tower (think of it as a very long vertical rod) by combining the facts that (a) compressive strength of materials is large (b) "buckling" instabilities can be prevented by active control. Another idea is an extremely large-span arch (if a constant-diameter rod is bent into the shape of a "cycloid" curve, then the stresses in this arch are purely compressive). (Of course, it'd be a bummer if you had a power or software failure... or an airplane hit it...) In fact back when I was an undergrad at MIT, some engineering prof was assuring us gullible students this was a revolutionary idea that would completely change structural engineering, and he was procuring big grants to do it. These are fine fantasies; now let's think a little about reality. (The results of my analysis will suggest said professor was completely full of shit, albeit apparently a successful con-man.) For a vertical tower, you could keep the stabilization turned on as you built it. However for an arch, I'm not seeing how to do that. So it seems as though an actively-stabilized arch would be pretty much impossible to build. (You could build with reinforcement which later was removed, but not if we are doing this for an enormous arch beyond the size attainable without active stabilization...) Percy Bridgeman, in his investigations of high pressures, was able to reach up to 5 GPa pressures using high strength steels and 10 GPa pressures (100,000 Kgforce per square cm) using tungsten-carbide/cobalt (6 wt% Co) composite material. Later, 300 GPa pressures (comparable to those at Earth's core) were achieved using "anvils" made of single crystal diamond and 600 GPa using "2 stage" diamond anvils with the second (very small) stage made of "nanocrystalline poly-crystal diamond." Unfortunately, all of these apparatus eventually fail, one reason being "creep." Even at only 0.5 GPa, today's commercial steel high pressure equipment says you have to replace the "Bridgeman seal" every 2 years. In view of this, I doubt it is possible to maintain non-hydrostatic compressive stresses of more than about 300 MPa (equals 30000 kgforce/cm^2) in any large structure. That stress level would be sufficient to support a vertical steel rod, of uniform diameter, 38 km high. The shrinkage of steel at this stress level is about 1.5 parts per thousand. The infrastructure (sensors and actuators) required to perform the active control would add additional weight -- so in view of that plus the desire for some safety margin, perhaps the upper limit really is only about 20 km. In contrast, the highest tower yet built is 830 meters in Dubai. This "Burj Khalifa" is a non-guyed skyscraper containing residences and hotels. Its cross-section is a 3-pointed star at base about 170 meters wide. (Some other recent tall buildings have parts with separate towers linked by girder/walkways and with wind-turbines attached to those girders which can provide perhaps up to 20% of building's electrical needs.) It is mainly made of steel-reinforced concrete and appears to be a technical success but economic failure. Stresses caused by the blowing of the wind cause the top to sway about 2 meters horizontally, and there is a swinging "tuned mass damper" inside the tower near the top which is something of a semi-active control to prevent resonant swaying. As far as I understand this device is not controlled by any intelligent mechanism, but it is an intentionally designed dynamic non-rigid mechanism which I suppose could be considered as an "analog computer." Also the exterior of the building is intentionally shaped in funny ways that computer simulations of wind flow think make it unlikely to get into a destructive resonance with large wind vortices. Previously the highest were radio masts made of metal periodic "lattices" (triangles) with guy wires and dampers. Specifically the previous record holder was the "Warsaw radio mast": 646 meters high, had equilateral triangle base 5 meters on each side, and weighed about 420 metric tons, of which the guy wires were about 80 tons. It was completed in the early 1970s but collapsed in 1991. The immediate cause of the collapse was a maintenance error, but it suffered from metal fatigue caused by wind stresses, and from difficulty trying to keep it painted. This tower had 3 vertical steel tubes linked by latticework. Each of the 3 main tubes was 245mm in diameter with wall thickness varying from 8 to 34 mm. Thus the cross-sectional area of the tower near the base was 3 * 245 * pi * 34 mm^2 = 78.5 cm^2 for a compressive stress of 420000 kgforce / 78.5 cm^2 = 5350 kgforce/cm^2 = 53 MPa. So... first of all, passive prevention of buckling, via guy wires, only cost about 20% of the Warsaw tower weight. If you also count all the latticework, then maybe we are up to 50%. If active control cost more weight than that, what would be the point? Well, I suppose there are circumstances where it is infeasible to guy a structure... and the mass of the guy wires presumably grows quadratically or faster with tower height while the mass of an active control system might be hoped to grow only linearly... In a speed-V wind, assuming drag coefficient of 1, the total torque near bottom on a height-H vertical tube of radius R, is about H^2*R*V^2 kg*sec^2/m^5. Note the H^2, quadratic, growth term, indicating that linear-growth hope is eventually dead meat. If V=200 km/hr = 55.556 meter/sec and H=650 meter and R=0.245 meter, we find torque = 320 million newton*meters. If this is 10^9 in all (3 tubes) then the force on one tube that needs to be applied to counter this torque is about 2*10^8 newtons which is 2*10^7 kgforce which is 20000 metric tons of force. This far exceeds the weight of the tower supported by that tube at only 420/3=140 metric tons. So actually, maximum wind-caused bending stresses far exceed the gravitational stresses, hence the guy wires are very important. Active control could be done by measuring torque (bending stress) on the rod, and applying compensating torques. In this way, the bending stresses everywhere could permanently be kept below some constant. Glued-on silicon and metal-foil strain gauges could be used for this, but you need temperature compensation and compensation for stresses caused by the solidification of the glue. The correct tower shape is not necessarily exactly vertical if, e.g. one side gets hotter. For example if one of the three Warsaw tubes got 5C hotter than the other two tubes, it would expand 65 parts per million which would be 4.2 cm of extra tower height. This would bend the tower with a radius of curvature of about 77 km, causing the top of the tower to move 2.7 meters horizontally away from vertical. I believe that the highest winds also were expected to cause about that same amount of bend. Thus the correct shape is not "vertical"; it is "the one which keeps all bending stresses bounded." With the 3-tube Warsaw-type design, torques could be applied by pulling cables between the 3 tubes. If there were no wind then I would say this is feasible. However, counteracting the torques from wind seems to be to require far too large torques to be feasibly applied in this way. In short removing the guy wires from the Warsaw mast and using active control instead, would be asinine. SUMMARY: I'm not seeing advantages for active stabilization of structures the size of the largest yet built -- up to 800 meters -- and in fact the whole idea seems to be ridiculous in view of wind loads. But if we were building on the moon (vacuum, no wind) then it might be the best way to build huge towers.

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[math-fun] Hallowe'en spacefill
by Bill Gosper 21 Sep '14

21 Sep '14

Simply sampling Mandelbrot's Kochfill at triple frequency makes a nice spiderweb. gosper.org/halloween.png --rwg Re Cambridge U. Press, Finch's Mathematical Constants is certainly a marvel.

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[math-fun] spigot algorithms
by Henry Baker 21 Sep '14

21 Sep '14

Is there a list of spigot algorithms somewhere?

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[math-fun] Seeking Yaglom and Yaglom help
by James Propp 19 Sep '14

19 Sep '14

Can anyone with access to a copy of Yaglom and Yaglom's two-volume work "Challenging Problems with Elementary Solutions" give me a reference for their proof of the fact that, given a sequence of length 2n+1 consisting of n terms equal to -1 and n+1 terms equal to +1, there's exactly one integer k between 0 and 2n such that cyclically shifting the original sequence k positions to the left yields a sequence in which all non-empty partial sums are positive? I know a nice picture proof of this, a version of which appears in Yaglom and Yaglom: Draw a lattice path of length 2n+1 consisting of diagonal steps in the (+1,+1) direction and (+1,-1) direction (northeast and southeast if you prefer), with northeast steps corresponding to +1's in the sequence and southeast steps corresponding to -1's in the sequence. Now repeat this finite lattice path infinitely often in both directions, obtaining a periodic infinite path whose slope is 1/(2n+1). Draw a line of this slope above the lattice path, and start lowering it until it just touches the lattice path. We need to show that the contact points are 2n+1 units apart in the horizontal direction. But this follows from the fact that n and n+1 are relatively prime. I'd like to refer to the Yaglom and Yaglom problem and solution in an article I'm writing. I don't have a copy of Yaglom and Yaglom, but I bet many of you do! (I think Y&Y use rightward and eastward steps, rather than northeast and southeast steps. But these memories are thirty years old.) Thanks, Jim Propp

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[math-fun] triangular window?
by Bill Gosper 18 Sep '14

18 Sep '14

I often use small windows as straightedges to slide around on various displayed graphics, and sometimes wish for slanted edges. Is there a way to hack up a transparent triangle using the alpha channel? E.g., Mavericks offers a crazy variable transparency, variable blur Terminal window. An inconvenient alternative is to rotate screenshots with Mathematica. --rwg

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[math-fun] Craig Gentry, 2014 MacArthur Fellow, for FHE
by Henry Baker 17 Sep '14

17 Sep '14

FYI -- It's so ironic that the crypto.stanford.edu website listed below is considered "untrusted" due to a bad crypto certificate. http://www.macfound.org/fellows/914/ MacArthur Fellows / Meet the Class of 2014 Craig Gentry Computer Scientist Research Scientist, Cryptography Research Group IBM Thomas J. Watson Research Center Yorktown Heights, NY Age: 41 Published September 16, 2014 Craig Gentry is a computer scientist fueling a revolution in cryptography and theoretical computer science through his elegant solutions to some of the disciplineÂs most challenging open problems. In 2009, Gentry published a plausible candidate construction of a fully homomorphic encryption (FHE) scheme, answering a problem posed in 1978 and thought by many to be impossible to resolve. FHE makes it possible to perform arbitrary computations (mathematical operations like sum or product as well as more complicated operations) on encrypted data while it remains encrypted and without needing a secret key, and while keeping the individual pieces of data hidden and secure. For example, a web application could prepare an individualÂs tax return using her encrypted financial information, without seeing any data in the clear. Although FHE is not yet practical for widespread implementation, GentryÂs breakthrough has initiated a surge of new research globally in computer science as it has enormous implications for making the cloud computing environment more secure and compatible with data privacy for individuals. Gentry went on to tackle another longstanding open problem, moving beyond cryptographic bilinear maps (which have been used widely in encryption, cryptographic digital signatures, and key agreement) to multilinear maps. The potential for multilinear maps was described a decade ago, and while applications such as broadcast encryption schemes and unique signature schemes have been discussed, Gentry and his collaborators were the first to publish a plausible candidate multilinear map. In 2013, he and his colleagues leveraged those findings to present the first example of cryptographic software obfuscation, a discovery that is paving the way for the encryption of entire programs while keeping their functionality intact, thereby making it infeasible to reverse engineer the program. Drawing on techniques from mathematics and computer science, Gentry is inspiring not only a flood of practical applications but also opening doors for new intellectual pursuits across the whole of cryptograp hy. Craig Gentry received a B.S. (1995) from Duke University, a J.D. (1998) from Harvard Law School, and a Ph.D. (2009) from Stanford University. Prior to his Ph.D., he worked as an intellectual property lawyer (1998Â2000) and as a senior research engineer for DoCoMo USA Labs (2000Â2005). He is currently a research scientist in the Cryptography Research Group at the IBM Thomas J. Watson Research Center. His scientific research has appeared in proceedings such as STOC (Proceedings of the Annual ACM Symposium on Theory of Computing), FOCS (Proceedings of the Annual IEEE Symposium on Foundations of Computer Science), Crypto, and Eurocrypt, among others. For More Information Website: https://crypto.stanford.edu/craig/easy-fhe.pdf

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[math-fun] Philae landing: equation of motion
by Guy Haworth 17 Sep '14

17 Sep '14

What is the equation of motion of the PHILAE lander, relative to the '2km sphere' comet 67P, assuming: - it is released with a velocity vertically downwards of v1 m/s about the comet at time t(0), x(0) km above the surface - acceleration at the surface is 10-^3 m/s/s If x = k.t^(2/3), then d2x/dt2 = -(2/9)k^3/x^2 ... which may have something to do with it. Incidentally, it sounds like ESA are going to release the Philae lander at a much greater height than originally planned. Maybe this is because they cannot predict and deliver ROSETTA's position as accurately as they expected. Thanks. G

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[math-fun] mod 2^k mod p
by Mike Stay 16 Sep '14

16 Sep '14

Say p is prime and k is floor(lg(p)/2). If r < 2^{2k} < p then write r as (b, a) where b, a are each k-bit numbers. Are there efficient ways to compute f(b, a, b', a') = (b xor b', a xor a') g(b, a, b', a') = (b*b' in GF(2^k), a*a' in GF(2^k)) when they're defined? Or similarly for a prime field q instead of GF(2^k)? -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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