Timeline for Most regular way to triangulate $\mathbb{R}^3$?
Current License: CC BY-SA 3.0
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| Mar 27, 2014 at 5:52 | history | edited | user137794 | CC BY-SA 3.0 |
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| Mar 26, 2014 at 22:36 | comment | added | Wlodek Kuperberg | Though I understand you do not require that all tetrahedra of the triangulation be congruent, your question seems closely related to the still unsolved problem of classifying the tetrahedra that tile $\mathbb{R}^3$. For a survey of known results, see Egon Schulte's article arxiv.org/pdf/1005.3836.pdf | |
| Mar 26, 2014 at 16:17 | comment | added | user126154 | I would check what happens by using close-packing of spheres and taking dual tessellations. en.wikipedia.org/wiki/Close-packing_of_equal_spheres | |
| Mar 26, 2014 at 15:41 | comment | added | Mirko | An estimate (rather trivial) would be the solid angle at the corner of a pyramid with a square base, with height 1/2 the side of the base: Six of these form a cube. (This ought not be the best.) We need to further split the pyramid into two tetrahedra (making the solid angle even smaller). | |
| Mar 26, 2014 at 14:19 | review | First posts | |||
| Mar 26, 2014 at 14:19 | |||||
| Mar 26, 2014 at 14:04 | history | asked | user137794 | CC BY-SA 3.0 |