Timeline for Finding the Complete Silhouette Area Formula Given a Convex Polyhedron
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 22, 2024 at 2:35 | history | edited | cvgmt | edited tags | |
| Oct 18, 2024 at 14:53 | answer | added | cvgmt | timeline score: 2 | |
| Oct 18, 2024 at 4:23 | history | edited | user64494 | CC BY-SA 4.0 | edited title |
| Oct 17, 2024 at 19:56 | history | edited | Romogi | CC BY-SA 4.0 | Eliminated a paragraph. |
| Oct 12, 2023 at 15:27 | comment | added | Romogi | @GregHurst Here is the partial cuboid formula. It is not general, because it only works when $\gamma = 0$. It would be great if all of the angles can be between $[-2\pi, 2\pi]$ which would mean the general solution is a piecewise function. And it would be great to be able to do it for any convex polyhedron. I feel like there would be a way for Mathematica to solve this fairly easily instead of focusing so much on the Mathematics behind it. | |
| Oct 12, 2023 at 15:24 | comment | added | Romogi | @GregHurst You gave a great comment! While, yes your solution finds the area after a particular rotation there is a drawback. It is numeric, so it doesn't give a general formula for finding the exact area after a known roll-pitch-yaw. It isn't computationally optimal and it doesn't help to find interesting mathematical relationships. | |
| Oct 12, 2023 at 0:38 | comment | added | Greg Hurst | I'm not sure what you're asking. But if you want the area of the xy-projection, you could find the area of the convex hull of the projected vertices. Or you could project the mesh itself and take 0.5 it's area since a convex polyhedron has an 'up' side and a 'down' side. SeedRandom[1]; reg = RandomPolyhedron[{"ConvexHull", 20}]; convex hull area: Area[ConvexHullMesh[reg[[1, All, 1 ;; 2]]]] (* 0.477824 *) projection area: 0.5Area[MeshRegion[reg[[1, All, 1 ;; 2]], Polygon[reg[[2]]]]] (* 0.477824 *) | |
| Oct 11, 2023 at 21:21 | history | edited | Romogi | CC BY-SA 4.0 | Simplified title because roll, pitch, and yaw is implied in the question. |
| Oct 11, 2023 at 21:10 | history | asked | Romogi | CC BY-SA 4.0 |