>
> > I have done that, but the distribution is not even as
> > the four triangles resulting from division of each
> > original one are not equilateral.  The final distrib-
> > ution is somewhat skewed with higher "point density"
> > about the original vertices.
>
> This will not be the case if you start with an icosahedron
>

Unfortunately, it is the case.  Just to prove it, take a look at a triangle
with the following apical positions A=1 0 0 , B=0 1 0, C=0 0 1.
Let's divide sides AB and AC in half and normalize the resulting points.
AB/2 = D = {cos(45), cos(45), 0}  and  AC/2 = E = {cos(45), 0, cos(45)}.
The resulting triangle  ADC  is not equiltateral :
AD = AE = sqrt (pow(1-cos(45),2)+pow(cos(45),2)) = 0.76
DE = sqrt(0+ 1/2 + 1/2) = 1.

It's the same story with icosahedron.



> > The exact number of points does not matter as long as
> > it is large enough.
>
> 60k is simple to generate exactly. I don't have the vocabulary to explain
> how, but I can generate this array in a few minutes and email it to you if
> you would like.

If you could send me your algorithm I would most certainly appreciate it.  I
want to be able to generate any number of points to model the sphere, I am
not sure what number will satisfy my task but the more the better.

Ilia

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