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One way to tessellate a 3D sphere is by iterated subdivision of an icosahedron. I am wondering whether this method gives a homogeneous surface density of vertices. To the eye, it seems to do so, and logic indicates that too (each face has the same are in the icosahedron, and faces in each subdivision are created of equal area), but is there some bias that I'm not thinking of?

Otherwise, what tessellation method can yield a constant density of vertices?

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Do you mean the two-dimensional round sphere, as embedded in three-space? Another question: what is the definition of "surface density"? Thanks! – Sam Nead Nov 11 '10 at 13:49
My question is: If I look at the vertices of the tessellation as a set of points on the 3D sphere, how are these points distributed on the surface of the sphere? – F'x Nov 11 '10 at 14:52
Maybe of interest: "Recursive Zonal Equal Area Sphere Partitioning Toolbox", – Hans Lundmark Nov 11 '10 at 15:14
For information about symmetric tilings of the sphere you can look at the book of Conway, Burgiel, and Goodman-Strauss, The Symmetries of Things, A.K. Peters, 2008. – Joseph Malkevitch Nov 11 '10 at 16:05
up vote 14 down vote accepted

There are several possible ways of defining "density on a sphere", each one giving somewhat different results.

Alas, most of them have some "maximum number of vertices" that give exactly equal density. Above that maximum number, further tessellation can at best approximate constant density. (That approximation is more than adequate for many purposes).

"Unfortunately, it is a well-known group theoretical result that there are no completely regular point distributions on the sphere for N > 20." -- Max Tegmark

Equal density as equal areas of the triangles formed by the vertices: You can tessellate a sphere to give a geodesic sphere such that every triangle has exactly equal area, to any desired resolution, using any equal-area projection such as the Snyder equal area projection. (A few people use geodesic grids based on this principle).

Equal density as congruent triangles formed by the vertices: When a person builds a geodesic dome out of panels, it would be super-convenient if every panel were identically the same size and shape. Alas, the maximum "size" is the 120 identical faces of the hexakis icosahedron (aka disdyakis triacontahedron). Any convex solid with more than 120 faces must necessarily have 2 or more kinds of faces.

Equal density as minimum-energy configurations of charged particles: Min-Energy Configurations of Electrons On A Sphere. You can put any integer number of repelling particles on a sphere, and calculate some minimum-energy configuration.

Equal density as equal distance from every vertex to the N nearby vertices: When a person builds a geodesic dome out of struts, it would be super-convenient if all the struts were the same length. Most "naive" methods of dividing the large triangles of the icosahedron into smaller triangles generates lots of different edge lengths; but there are ways to "tweak" the tessellation subdivision in order to minimize the number of different lengths of edges. (Fewer unique lengths requires fewer jigs in manufacturing and fewer spares needed to replace any damaged strut). Alas, the maximum "size" of a strictly convex polyhedron made entirely of equilateral triangles (convex deltahedron) is the 30 edges of the icosahedron. (You could try to make the pentakis dodecahedron out of 60 equilateral triangles, giving 90 equal-length edges, but then it would be slightly concave). Any strictly convex solid made of triangles with more than 30 edges must necessarily have 2 or more lengths of edges.

A few more notes on sphere approximation.

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I've got my answer from one of the links ( given by lhf, in the last paragraph of page 9 and first of page 10:

What's more, they are not asymptotically uniformly distributed. To see this, it is enough to observe that after the first step, the points are centered at 80 spherical triangles which do not all have the same area (the projection process in- creases the areas of the "middle triangles" more than the rest). As the further steps in the process yield the same number of points in each of the 80 triangles, asymptotic uniform distribution cannot hold

Thanks for the link!

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Note that there is another way to tesselate from the icosahedron, which is what I thought you were suggesting. Rather than partition each equilateral triangle into four, one could find the centroid of each triangle and project that out to the sphere. Then all triangles are congruent. But they are no longer equilateral, and subsequent steps distort considerably. – Joseph O'Rourke Nov 11 '10 at 15:23

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