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I was wanting exactly the same thing. I found these coords in John
VanSickle's rock generator and extracted them to a PoV array as follows:
#declare qA=(sqrt(5)+1)/4;
#declare qB=0.5/sqrt(qA*qA+1/4);
#declare qA=qA/sqrt(qA*qA+1/4);
#declare P = array[12] {
< qA, qB,0>,
<-qA, qB,0>,
< qA,-qB,0>,
<-qA,-qB,0>,
<0, qA, qB>,
<0,-qA, qB>,
<0, qA,-qB>,
<0,-qA,-qB>,
< qB,0, qA>,
< qB,0,-qA>,
<-qB,0, qA>,
<-qB,0,-qA>
}
Hope this helps,
Mike Andrews.
Mick Hazelgrove wrote:
> Help
>
> How can I calculate the 3D positions of the vertices of an Icosahedron
>
> Thanks in advance
>
> Mick Hazelgrove
Post a reply to this message
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com> wrote:
: Just wondering.... How did you do that? The precision is
: impressive.
Those vertices can be calculated analytically (ie. you can write an exact
equation describing a vertex of the icosahedron). Then you can just calculate
a numerical value with any precision you want (and your calculator supports).
--
- Warp. -