From: ksbrown@seanet.com (Kevin Brown) Newsgroups: sci.math.research Subject: Re: How to pixelize sphere? Date: Wed, 29 Jul 1998 00:48:23 GMT On Mon, 27 Jul 1998 Fred S Long wrote: > At one time I looked into this sort of problem as a way of producing > a grid on the earth's surface for numerical prediction of weather. > It occured to me that one might imagine confining electrons to the > surface of the sphere and letting the repulsive forces between them > force them into a stable configuration. Then, as suggested, form > the Voroni tesselation. Never reached the point of a simulating > the idea. The general problem of determining the minimal (local and global) potential energy configurations of n points on a sphere has been extensively studied for various definitions of the potential. With V(r) equal to 1/r^12 - 2/r^6 this sometimes called the Lennard-Jones Problem, and with V(r) = 1/r (the case you mentioned) it's called the Thomson problem (although I'm not sure if this was named after Sir Joseph John Thomson, Sir George Paget Thomson, or even William Thomson, Lord Kelvin). People have also studied other criteria for "uniformly" distributing points and regions on a sphere, including equal area partitioning. A survey of this topic was presented in a recent article in Mathematical Intelligencer (Winter 97 issue). Also, if your browser supports Java applets, you can examine some solutions of the Thomson problem for n up to 32 on the web at http://www.seanet.com/~ksbrown/kmath005.htm Notice that although there are unique stable solutions for most values of n in this range, it is believed that the number of local minima actually increases exponentially in the long run. (The smallest n to have more than one stable solution is n=16.) However, the exact number of minima is only known for the smaller values of n. Of all these solutions, the most appealing in some ways is the case n=4!=24, which is a polyhedron whose surface includes 24/4 perfect squares, 24/3 perfect triangles, and 24/2 additional "bi-gons", arranged so that each vertex is identical. In addition, the verticies have a "handedness", so there are really two distinct versions of this solid. ============================================================================== From: Jeff Erickson Newsgroups: sci.math.research Subject: Re: How to pixelize sphere? Date: Wed, 29 Jul 1998 15:28:18 -0400 Paul Hughett wrote: > > Tom Chou (tc208@damtp.cam.ac.uk) wrote: > > : Does anyone have any pointers for dividing up > : the surface of a sphere into as many identical > : polygons as possible? I need somewhere between 10^4-10^6 > : (quite a bit more than 20!) > : surface polygons with as few defects as possible. > > If you insist on identical (congruent) polygons, I suspect > that 20 is the best you can do. Actually, you can get any even number of congruent triangles with a regular bipyramid. Put n equally spaced points on the equator, add the north and south poles, take the convex hull, et voila! 2n congruent triangles! Even if you exclude bipyramids, you can get 120 congruent triangles. Start with a regular icosahedron. Barycentrically subdivide each facet into six congruent 30-60-90 triangles and project the result out onto the sphere. Equivalently, project the vertices, edge midpoints, and facet centroids of the icosohedron onto its circumsphere and take the convex hull. You get the same result if you start with a regular dodecahedron and split each facet into 10 congurent triangles, or if you start with a rhombic triacontahedraon and split each facet into 4 congruent triangles. The last construction gives the polyhedron its name: "disdyakis triacontahedron". For a picture, see: http://www.astro.virginia.edu/~eww6n/math/DisdyakisTriacontahedron.html 120 is the largest number of congruent triangles you can get without using a bipyramid. (I realize this is probably not what the original questioner wanted, but he did say "congruent", not "regular"!) --- Jeff Erickson Department of Computer Science jeffe@cs.uiuc.edu University of Illinois http://sal.cs.uiuc.edu/~jeffe Urbana-Champaign