"TinCanMan" <Tin### [at] hotmail com> wrote:
> ... . The
> limitation here is that you are restricted to the number of points related
> to the iteration level 'n' (I don't know the formula off the top of my
> head).
If you're really doing "standard" triangulation of a sphere with equidistant
points, then it should be something like s*(4^n) triangles where s is the
number of the starting triangles (its easy if you start with an octaeder)
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