Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!iuvax!rutgers!mit-eddie!mit-amt!djs From: djs@mit-amt (David J. Sturman) Newsgroups: comp.graphics Subject: Re: Polygon Representation of a Sphere's Surface Message-ID: <3661@mit-amt> Date: 23 Mar 89 22:37:43 GMT References: <270@ai.etl.army.mil> <2908@kalliope.rice.edu> Reply-To: djs@media-lab.media.mit.edu (David J. Sturman) Organization: MIT Media Lab, Cambridge MA Lines: 36 In article beshers@cs.cs.columbia.edu (Clifford Beshers) writes: >In article <2908@kalliope.rice.edu> foo@titan.rice.edu (Mark Hall) writes: > > In article <270@ai.etl.army.mil> richr@ai.etl.army.mil. (Richard Rosenthal) writes: > >I want to have in 3-D space (x, y, z) a polygon near-representation > >of the surface of a sphere where each polygon is identical > >and regular (if that's the word). > > > >Is an icosahedron the right place to start? > > ... > >Not at all. Start with an icosahedron. You now have a polygonal >approximation to a sphere where each face is a triangle. >Inscribe a triangle inside each face with its vertices at the >midpoints of the edges, dividing each face into four triangular >regions. Project the three new points onto the surface of the >sphere. You now have a a polygonal approximation to a sphere >where each face is a triangle. Inscribe a triangle... > It seems to me that this will give you a polyhedral approximation to the sphere but NOT a regular polyhedron. I suspect that the projection of the new points to the the surface of the sphere will unevenly deform the new triangles, since one vertex of each outer triangle is already on the sphere while all three vertices of the inner triangle are moved by the projection. Thus, the interior triangle will no longer be isomorphic to the outer triangles. I have not worked out the geometry, and am working on intuition at this point. David Sturman MIT Media Lab djs@media-lab.media.mit.edu