Consider sequences of tesselations of the sphere. For instance, one such sequence might start with an icosahedron and proceed by subdividing each triangle face into 4 triangles and projecting the new vertices onto the sphere.

Some sequences of tesselations converge to the sphere in the sense that the maximum distance between the polyhedra and the sphere converges to 0. Some sequences possibly converge more strongly if the density of vertices converges to the uniform distribution over the sphere (this is not the case with the sequence offered as an example above).

  1. Are there convergent sequences of tesselations where the number of distinct triangles involved in each tesselation is bounded for the whole sequence, and if so, what is the lowest bound?

Rephrasing: is there an integer $k$ such that $\forall \epsilon > 0$ there is a polyhedron with triangular faces which approximate a unit sphere within $\epsilon$ and has at most $k$ distinct faces.

  1. If so, do some of these sequences also let the density of vertices approach the uniform distribution?

Clarification: we are looking at the number of distinct triangles. For example, an isocahedron has exactly 1 distinct triangle.

Further clarification: the rephrasing above is not quite accurate. As commenters point out, a simple rasterization of the sphere yields arbitrarily good approximations with just one triangle.

We are looking for polyhedra where the vertices lie on the sphere itself or, at least, convex polyhedra.

  • $\begingroup$ It seems like you're asking several questions here. Am I right to think your main one is along the lines of "for a fixed distance tolerance $\epsilon$, what subdivision strategy approximates the sphere to within tolerance with the fewest triangles?" (possibly with the addendum "if there are several, which one is most uniform?"). $\endgroup$
    – DCM
    Nov 14, 2020 at 13:07
  • $\begingroup$ I think the "icosahedron followed by recursive subdivision" strategy isn't very efficient in this respect - I think it's better to start with the icosahedron then do the subdivision 'in one shot' rather than iteratively. One thing I'm curious about: is your ultimate aim to sample from a uniform distribution on the sphere, or to produce a triangulation? $\endgroup$
    – DCM
    Nov 14, 2020 at 13:17
  • $\begingroup$ Not with the fewest triangles, with the fewest distinct triangles. An isocahedron for instance has exactly 1 distinct triangle. $\endgroup$
    – Arthur B
    Nov 14, 2020 at 15:14
  • 2
    $\begingroup$ That would be a slightly stronger claim. I'm open to the finite set changing as the degree of approximation gets better, so long as the cardinal stays the same. I can imagine situations where a triangle in the set has to become infinitely thin in the limit for instance. $\endgroup$
    – Arthur B
    Nov 15, 2020 at 6:32
  • 1
    $\begingroup$ Here’s a negative answer to my question: If there are only finitely many triangles, then each vertex of the polyhedron has only finitely many possible arrangements. (Eg, we can’t fit more than five equilateral triangles around a single vertex.) For each of those arrangements, there’s a greatest distance from the faces at the vertex to theor circumscribing sphere (or greatest ratio of that distance to the radius of the circumscribing sphere). Thus the finite set of triangles can’t be used to approximate the sphere any better than the least of those greatest distances $\endgroup$
    – Matt F.
    Nov 24, 2020 at 12:14


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