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Co-ordinates of a dodecahedron

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badhorsey | 14:09 Mon 30th Jan 2012 | Science
7 Answers
Hi,

I'm trying to plot a dodecahedron on an ancient 3D program (VU-3D on the ZX Spectrum, in fact).

What I need is a set of co-ordinates for the vertices, either absolute or relative... it doesn't matter which way up it is!

Thanks!
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Question Author
Thanks, but I know what a dodecahedron is - I just need the co-ordinates of the corners to plot one in three-space.

For example, if I were plotting a 4 x 4 x 4 cube, it would be the following:

0, 0, 0
0, 4, 0
0, 4, 4
0, 0, 4
4, 0, 0
4, 4, 0
4, 4, 4
4, 0, 4

- So I'm looking for the same co-ordinate set for an example dodecahedron.

Size / scale / inversion doesn't matter, I can transpose these.
Hi badhorsey
This link DOES give you the coordinates. There are 20 points in all. Scroll down and you will see them.
The first set of numbers in brackets has 8 combinations, the other three have 4 combinations each giving the total of 8+(3 x 4)=20
Here is an extract from the link given by Howard:

The following Cartesian coordinates define the vertices of a dodecahedron centered at the origin:[1]
(±1, ±1, ±1) [8 combinations]
(0, ±1/φ, ±φ) [4 combinations]
(±1/φ, ±φ, 0) [4 combinations]
(±φ, 0, ±1/φ) [4 combinations]

where φ = (1 + √5) / 2 is the golden ratio (also written τ) ≈ 1.618. The edge length is 2 / φ = √5 – 1. The containing sphere has a radius of √3.

The comments in square brackets are mine.
Question Author
Hi, thanks for that - I saw all of that - but I don't have the mathematical skills to understand it!
OK, here goes:
(1,1,1)
(1,1,-1)
(1,-1,1)
(1,-1,-1)
(-1,1,1)
(-1,1,-1)
(-1,-1,1)
(-1,-1,-1)
(0,0.856,1.168)
(0,0.856,-1.168)
(0,-0.856,1.168)
(0,-0.856,-1.168)
(0.856,1.168,0)
(0.856,-1.168,0)
(-0.856,1.168,0)
(-0.856,-1.168,0)
(0.856,0,1.168)
(0.856,0,-1.168)
(-0.856,0,1.168)
(-0.856,0,-1.168)
Question Author
Thank you!

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