From: Robin Chapman Newsgroups: sci.math Subject: Re: Coordinates for great rhombicubeoctahedron, icosahedron?? Date: Sun, 05 Jul 1998 08:36:49 GMT In article <359CFAFD.242B8E4C@showme.missouri.edu>, Len Barbour wrote: > > I need to construct computer-generated diagrams of both a great > rhombicubeoctahedron and an icosahedron. > > Could someone please send me a list of cartesian coordinates for the > vertices of each of these solids? > > I know that there are several versions of these solids (e.g. stellar, > truncated, etc.) and this is not exactly my usual field of expertise. > The icosahedron I need consists of 20 equilateral triangles (12 > vertices) as the faces. The great rhombicubeoctahedron has 12 squares, > 8 hexagons and 6 octagons (48 vertices). > > I should be able to supply the coordinates to the software (POV-Ray) as > a series of triangles in order to construct the solid shapes. For the icosahedron take (0, 1, (1 + sqrt(5))/2) and the 11 other points obtained by cyclic permuations of the coordinates and by replacing any coordinate by its negative. For the other shape take (1, 1 + sqrt(2), 1 + 2sqrt(2)) and the 47 other points obtained by arbitrary permutations of the coordinates and by replacing any coordinate by its negative. In each case the figure will have a side-length of 2. Robin Chapman -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/rg_mkgrp.xp Create Your Own Free Member Forum ============================================================================== From: Robin Chapman Newsgroups: sci.math Subject: Re: Cartesian Polyhedron Descriptions Date: Tue, 24 Nov 1998 10:46:26 GMT In article <73cvgn$t75$1@news.indigo.ie>, "JPCostello" wrote: > I have been looking for somewhere on the web where I may find the regular > pythogorean polyhedra and its subgroups described in three dimentional > Carthesian notation (Vertices x,y,z)). > > The tetrahedron and octahedron are simple because they incribe themselves > into the cube so easily, But I cant seem to get a grip on the dodec or icos. > I know that the mirror each other so one set is as good as another. Every > time I try to section off one or the smaller I end up with 6 sided faces, > which lends the thought that the truncuated dodecahedron may hold the > anwser. For the vertices of the icosahedron take (+-1, +-t, 0), (0, +-1, +-t) and (+-t, 0, +-1) where all possibilities of signs are taken and t = (1 + sqrt(5))/2 is the golden ratio. For the dodecahedron take its vertices to be the centres of the faces of this icosahedron. Robin Chapman + "They did not have proper SCHOOL OF MATHEMATICal Sciences - palms at home in Exeter." University of Exeter, EX4 4QE, UK + rjc@maths.exeter.ac.uk - Peter Carey, http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda, chapter 20 -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own ============================================================================== From: ken@straton.demon.co.uk (Ken Starks) Newsgroups: sci.math Subject: Re: Carthesian Polyhedron Descriptions Date: Tue, 24 Nov 1998 11:43:53 GMT "JPCostello" wrote: > Can any one explain how to, or help with co-ordinate sets. Three Golden Rectangles in the three coordinate planes make an icosohedron. Let phi = (1 - sqrt(5))/2 plane z = 0 : x = +/- phi y = +/- 1 plane y = 0 : z = +/- phi x = +/- 1 plane x = 0 : y = +/- phi z = +/- 1 I can send you a graphic if you wish. Ken, __O _-\<,_ (_)/ (_) Virtuale Saluton.