Timeline for Break polyhedron into tetrahedron
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 19, 2020 at 7:26 | comment | added | Hugh Thomas | @AlecJacobson I don't understand your comment. If $v_0$ lies in the same plane as some triangle, then that triangle lies in a face containing $v_0$, so I do not build the cone over it. | |
| May 9, 2020 at 23:33 | comment | added | Alec Jacobson | If the (convex) input polyhedron contains non-triangular faces, then this algorithm could lead to degenerate (zero volume) tetrahedra. Consider if your chosen vertex lies on the same plane as a non-incident triangle. A fix would be to create the triangulation with triangles fans from the chosen vertex for all incident faces. | |
| Dec 3, 2009 at 15:22 | comment | added | Gabriel Benamy | I saw, but from my understanding of the problem, it's just "list of tetrahedrons". | |
| Dec 3, 2009 at 15:22 | vote | accept | Graviton | ||
| Dec 3, 2009 at 15:11 | comment | added | Hugh Thomas | Not as easy as all that. See the comment to the question by Nurdin Takenov -- if you don't assume convexity, you may have to introduce new vertices. | |
| Dec 3, 2009 at 14:59 | comment | added | Gabriel Benamy | If the initial polyhedron is concave, then it's simple to split it into multiple convex polyhedrons; then just apply the procedure to each faction. | |
| Dec 3, 2009 at 14:49 | comment | added | Mariano Suárez-Álvarez | ...assuming the initial polihedron is convex (which you cannot tell from the combinatorial data alone) | |
| Dec 3, 2009 at 14:36 | history | answered | Hugh Thomas | CC BY-SA 2.5 |