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
distribution is somewhat skewed with higher "point density" about the
original vertices.

Mersenne twister can probably produce a random enough (as in "uniform")
distribution but I'd rather do it exactly if possible.  I need to do it once
to generate an input table for an application, so time and algorithm
efficiency is irrelevant.

The exact number of points does not matter as long as it is large enough.


Best regards,
Ilia

"Chris Johnson" <chris(at)chris-j(dot)co(dot)uk> wrote in message
news:40946490$1@news.povray.org...
> If you can cope with a number close to 60000, start with an octahedron (or
> possibly an icosahedron), then subdivide each triangle into four (by
drawing
> a new triangle with its vertices the midpoints of the edges of the
original
> triangle), then rescale the position of these midpoints onto the sphere
> surface. This subdivision is then repeated.
>
> If you need exactly 60,000 points, there are ways that I can see of doing
> it, but they're not very nice.
>
> -Chris
>
>

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