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Re: [math-fun] A modest proposal for carbon sequestration in space
by Henry Baker 03 Aug '06

03 Aug '06

True, but I'm not saying that you don't have to pump. Furthermore, in order to "sequester" the CO2, we only have to get it into orbit, so we don't have to send it any further than geosynchronous orbit. The pressure in the pipe goes down exponentially with height, and the molecules at the top are going extremely fast. If they bounce off the inside of the pipe without losing energy (not a good assumption), then they should come flying out the end. The failure mode is probably that the temperature gets so hot above the "atmosphere", that it melts the carbon tube. Even if the tube doesn't melt, it may radiate away an enormous amount of the energy which we would rather be used to propel the molecules _up_ the tube. So the pressure & density at high altitudes inside the tube would be much higher than the pressure & density outside the tube, so my assumptions about the density inside & outside would no longer hold. So, ideally, we might rather have a true "nanotube" in which the CO2 molecule travels as if in a waveguide, rather than ballistically. Then we would need an enormous number of them in parallel to carry a substantial amount of gas. Launching the molecules into such a waveguide might then be problematic. BTW, I understand that we actually _do_ lose H2 into space all the time, but probably not at an appreciable rate. At 12:00 PM 8/1/2006, Eugene Salamin wrote: >--- Henry Baker <hbaker1(a)pipeline.com> wrote: > >> Eugene: >> >> My intuition is that CO2 is only modestly heavier than N2+O2, so you >> only have to lift the _difference_ between the weight of the CO2 and >> the atmospheric pressure due to N2+O2. > >Then it should take no energy to dispose of the atmosphere into space. > >Gene

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[math-fun] Bay Area field trip to computer history museum?
by Jason Holt 03 Aug '06

03 Aug '06

Any math-funners in the bay area want to take an hour or two to visit the computer history museum in Mountain View? http://www.computerhistory.org/ It's open Wednesday, Friday & Sunday: 1:00 to 4:00pm and Saturday: 11:00am to 5:00pm. They're currently running a theme of computer chess. -J

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[math-fun] P.S. Finite simple group lyrics
by Mike Stay 03 Aug '06

03 Aug '06

http://www.math.northwestern.edu/~matt/kleinfour/lyrics/finite.html -- Mike Stay metaweta(a)gmail.com http://math.ucr.edu/~mike

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[math-fun] Finite simple group of order two
by Mike Stay 03 Aug '06

03 Aug '06

http://www.youtube.com/watch?v=UTby_e4-Rhg&search=finite%20simple%20group -- Mike Stay metaweta(a)gmail.com http://math.ucr.edu/~mike

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Re: [math-fun] Diffusing number sequence
by Daniel Asimov 03 Aug '06

03 Aug '06

Bill C. writes: << Suppose that one has four nonnegative integers in a row, A B C D. Create a subsequent row by abs(A-B) abs(B-C) abs(C-D) abs(D-A). Repeating this process will eventually produce 0 0 0 0, the end state. The object is to create a "long" chain of rows ending in 0 0 0 0. I can "backup" chains to create arbitrarily long chains, but the process seems a bit unwieldy. Is there any formulaic way of specifying an initial row that will produce a chain of length n? >> Not quite what you asked, but see this interesting paper that appeared in the Monthly last year: http://www.trinity.edu/vadim/difboxfinal.pdf which shows that when you take certain reasonable equivalence classes of the (cyclically ordered) four numbers, surprising things happen. --Dan

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Re: [math-fun] Apostol's Riemann song for your iPod
by David Gale 03 Aug '06

03 Aug '06

I'm quite excited after all these years to be able to solve a problem which is at least related to mathematics. The tune to which Apostol wrote the Riemannian lyrics is "Sweet Betsy from Pike" "Oh do you remember sweet Betsy from Pike, Crossed the wide border with her lover Ike With a [something] and [something etc......] Tweedle dum da daa di [something, with necessary number of syllables] Next Question? dg At 07:13 PM 8/1/2006, you wrote: >Henry, > >Thanks for the link > > < http://olimu.com/Riemann/Song.htm > > >to John Derbyshire singing Tom Apostol's song, "Where are >the Zeroes of Zeta(s) ?" Quite a nice little ditty. > >The tune was so familiar, but I couldn't place it at first. > >SO: Trivia question: What familiar old song uses the same tune? > >--Dan > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] poly (regular) polyhedra; nasty floats
by N. J. A. Sloane 03 Aug '06

03 Aug '06

Let me draw your attention to this old paper of mine! %title Four Icosahedra Can Meet at a Point %author H. Rossat %author N. J. A. Sloane %journal Geometriae Dedicata %volume 27 %pages 219-222 %date 1988 %Z Item 141 %type article Neil

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Re: [math-fun] poly (regular) polyhedra; nasty floats
by Schroeppel, Richard 03 Aug '06

03 Aug '06

Mike Stay asks > I haven't heard of polypolyhedra before. Is this what you're talking about? > http://www.langorigami.com/science/polypolyhedra/polypolyhedra.php4 No. AFAIK, I made up the word independently, but the site above is using a different meaning. My meaning is that you have a bunch of (same-sized) regular solids, and you glue them together, matching up faces. Much like polyominos (squares) and polycubes. I have some counts for polypolygons (polypentagons &c) at http://crd.lbl.gov/~dhbailey/expmath/workshop2004/schroeppel-talk.pdf starting on page 23. The polygons don't tile the plane (exc. 3,4,6), so the polypolygons don't make good packing puzzles. Dan suggests using spherical & hyperbolic spaces. The 2D (surface of a sphere) case is finite, so I'd expect the same in 3D by analogy. I've always been curious about the hyperbolic cases, but never worked out the arithmetic. It's necessary to do exact symbolic arithmetic for this stuff: There are messy cases when things just touch exactly, and floating point isn't able to resolve if they are slightly separated, abut, or overlap. The planar case can have polygons that meet at bare vertices, or meet vertex-touches-edge, or edges meet that are offset. I imagine the 3D case is worse, with the only compensation being that there are only five solids to consider (and cubes are easy). Polypolygons (except for 3,4,6) have a dense set of possible center locations, but only one or two possible orientations. The 3D situation looks as if both center positions and polyhedron orientations are dense (except cubes). The flat-space 2D and 3D cases only require algebraic numbers. The spherical spaces can be tackled by embedding in the next larger Rn. Dan (or anyone): Does hyperbolic geometry require doing logs & exponential nasties, or is there a model of the space where the coordinates for polygons etc. can be done with algebraic numbers? I spoke too fast when I said "no four icosahedra vertices can fit around a point". My arithmetic shows a face-face angle of 138.2 degrees, and the vertex occludes a solid angle of (5 x 138.2 - 540) = 151 degrees excess, so 720/151 = 4.7... could fit around a point, based solely on angular-area considerations. I think that the icosahedron case, which requires big spherical regular pentagons, still cannot fit 4 onto a sphere because of wasted space when the pentagons are close together, but area considerations alone don't prove it. Incidentally, I get the following numbers (based only on angular area) for the maximum number of regular polyhedra around a point: tetrahedron 22 cube 8 octahedron 9 dodecahedron 4 icosahedron 4 (really 3 ?) The cube and dodecahedron cases are easily achievable, but I wonder about the others. Surely the recreational math people have looked at this? --------- Speaking of nasty floating point problems: The ill-condition of the matrix { 1/Fib(i+j) } makes it difficult to compute the inverse using limited-precision floating point. If we change the problem to {1/sqrt(Fib(i+j))} or even {sqrt(Fib(i+j)}, we wreck the rational arithmetic approach, and get a hard problem. Maybe for the next Oxford bloatnum problem list. Rich

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Re: [math-fun] poly (regular) polyhedra
by Daniel Asimov 03 Aug '06

03 Aug '06

Rich writes: << I've been working out counts of polypolyhedra by hand. (regular polyhedra, left-right mirror pairs count as only 1) volume 1 2 3 4 5 polytetrahedra 1 1 1 3 7 polycubes 1 1 2 7 ... polyoctahedra 1 1 3 11 polydodecahedra 1 1 3 16 polyicosahedra 1 1 4 33 I find the visualization is hard, and diagrams only help a little. Notes: 3 dodecahedra will fit around an edge, and 4 will fit around a vertex. 3 icosahedra will not fit around an edge, nor 4 around a vertex. 3 icosahedra will fit around a vertex. >> This might also be fun to try in the "other two geometries", S^3 and H^3, where a regular polyhedron's max # per vertex (or per edge) will depend on its size (and the question would presumably be for joining congruent regular polylhedra). H^3 as you may know is especially flexible, in that a polyhedron's size can be chosen so that *any* preassigned number (greater than the Euclidean case) will fit around an edge -- and any preassigned number at all will fit about a vertex. --Dan

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[math-fun] poly (regular) polyhedra
by Schroeppel, Richard 03 Aug '06

03 Aug '06

I've been working out counts of polypolyhedra by hand. (regular polyhedra, left-right mirror pairs count as only 1) volume 1 2 3 4 5 polytetrahedra 1 1 1 3 7 polycubes 1 1 2 7 ... polyoctahedra 1 1 3 11 polydodecahedra 1 1 3 16 polyicosahedra 1 1 4 33 I find the visualization is hard, and diagrams only help a little. Notes: 3 dodecahedra will fit around an edge, and 4 will fit around a vertex. 3 icosahedra will not fit around an edge, nor 4 around a vertex. 3 icosahedra will fit around a vertex. Rich

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