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
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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
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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.