Timeline for Minimum number of distinct triangles for tesselating the sphere
Current License: CC BY-SA 4.0
21 events
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| Jan 20 at 9:08 | comment | added | Gro-Tsen | Somewhat analogous question to your $k=1$ case, with some partial answers. | |
| S Nov 27, 2020 at 21:05 | history | bounty ended | CommunityBot | ||
| S Nov 27, 2020 at 21:05 | history | notice removed | CommunityBot | ||
| Nov 24, 2020 at 15:28 | history | edited | Arthur B | CC BY-SA 4.0 |
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| Nov 24, 2020 at 15:25 | comment | added | Arthur B | Ah, Henry Segerman just made a similar point on Twitter. twitter.com/henryseg/status/1331254177952133126?s=19. What I had in mind was in fact stronger than convexity (I think) the vertices should lie on the sphere. | |
| Nov 24, 2020 at 15:16 | comment | added | Adam P. Goucher | Do you demand that the polyhedron be convex? Otherwise, there's a really boring $k = 1$ solution (for the weaker notion of convergence): rasterize the sphere by approximating it as the union of many tiny cubes, and then erect a shallow square-based pyramid on every square face that results. | |
| Nov 24, 2020 at 12:14 | comment | added | user44143 | 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 | |
| Nov 24, 2020 at 10:56 | comment | added | Arthur B | These are flat triangles, not spherical triangles, so the set of triangles used doesn't necessarily determine the radius of the sphere they can approximate. | |
| Nov 24, 2020 at 10:50 | comment | added | Arthur B | @Gro-Tsen I don't get that counter argument. The set of triangles from which you build may be finite but we can use as many copies of each kind as you want. | |
| Nov 24, 2020 at 10:12 | comment | added | Gro-Tsen | @MattF. The stronger question you ask (“Is there a finite set of triangles from which you can […] approximate spheres arbitrarily well?”) has a negative answer, because a given finite set of spherical triangles can only be arranged in finitely many ways to tile the (finite-area!) sphere. | |
| S Nov 19, 2020 at 19:11 | history | bounty started | Arthur B | ||
| S Nov 19, 2020 at 19:11 | history | notice added | Arthur B | Draw attention | |
| Nov 15, 2020 at 8:24 | comment | added | user44143 | Got it. I like the questions, and seeing no obvious constructions I’d expect negative answers, but I have no ideas on how to prove that. | |
| Nov 15, 2020 at 6:32 | comment | added | Arthur B | 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. | |
| Nov 15, 2020 at 1:28 | comment | added | user44143 | You could also phrase the first question as: Is there a finite set of triangles from which you can construct polyhedra that approximate spheres arbitrarily well? | |
| Nov 14, 2020 at 16:16 | history | edited | Arthur B | CC BY-SA 4.0 |
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| Nov 14, 2020 at 15:14 | comment | added | Arthur B | Not with the fewest triangles, with the fewest distinct triangles. An isocahedron for instance has exactly 1 distinct triangle. | |
| Nov 14, 2020 at 13:17 | comment | added | DCM | 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? | |
| Nov 14, 2020 at 13:07 | comment | added | DCM | 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?"). | |
| Nov 14, 2020 at 11:31 | history | edited | Arthur B | CC BY-SA 4.0 |
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| Nov 14, 2020 at 11:24 | history | asked | Arthur B | CC BY-SA 4.0 |