Timeline for How to make sure polygon is drawn flat in 3D?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:55 | history | edited | CommunityBot | replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/ | |
| Apr 2, 2016 at 1:46 | history | edited | J. M.'s missing motivation | CC BY-SA 3.0 | deleted 19 characters in body |
| Aug 2, 2015 at 3:45 | history | edited | J. M.'s missing motivation | CC BY-SA 3.0 | just checked, it's also in an earlier version |
| Aug 2, 2015 at 0:08 | history | edited | Michael E2 | CC BY-SA 3.0 | Bug information |
| Aug 2, 2015 at 0:07 | comment | added | Michael E2 | I see the same bug in V9.0.1. | |
| Jun 5, 2015 at 19:33 | history | edited | Michael E2 | edited tags | |
| Jun 4, 2015 at 5:40 | history | tweeted | twitter.com/#!/StackMma/status/606334659593682944 | ||
| Jun 3, 2015 at 17:50 | comment | added | BlacKow | @Guesswhoitis. The shift doesn't seem to work because all calculations are exact, so shifting a face back and forth gives the same values for vertex coordinates. The problem occurs during rendering. | |
| Jun 3, 2015 at 17:48 | vote | accept | BlacKow | ||
| Jun 3, 2015 at 3:14 | comment | added | J. M.'s missing motivation | Why, did you already try applying your pentagram[] to a tilted but shifted pentagon? | |
| Jun 3, 2015 at 3:04 | comment | added | BlacKow | But the face will still be tilted (not parallel to coordinate plane), what's the point of shifting it? | |
| Jun 3, 2015 at 2:57 | comment | added | J. M.'s missing motivation | Now that I think about it, it might be a bit easier to just compute the barycenter of a face, shift the face so that the barycenter is at the origin, perform your pentagram transformation, and then shift back to the original position. (For the barycenter, use averagepoints[], which is also in the link I previously gave. | |
| Jun 3, 2015 at 2:56 | comment | added | BlacKow | Calculating normal is straightforward in my case: connecting center of dodecahedron with center of pentagon will give normal vector. | |
| Jun 3, 2015 at 2:53 | comment | added | BlacKow | You are basically suggesting to make a pentagram in 2D and replicate it with appropriate transformations. But the resulting polygons will be all independent, they won't share vertices as they do now. So it will be workaround for particular case for visualization purpose. I'm trying to figure out how to force mathematica to draw my set of points correctly. Also all coordinates can be calculated with arbitrary precision. How can I specify tolerance for a point to belong to a plane? | |
| Jun 2, 2015 at 23:10 | answer | added | Michael E2 | timeline score: 9 | |
| Jun 2, 2015 at 22:19 | comment | added | J. M.'s missing motivation | As for computing the normal of a face, use the newellNormals[] routine from here. | |
| Jun 2, 2015 at 22:13 | comment | added | J. M.'s missing motivation | A sketch: find the normal to the plane of a face, use that to rotate the face so that the face is parallel to a coordinate plane (one of the components becomes zero), work out your transformation there, and then rotate back. | |
| Jun 2, 2015 at 22:05 | history | asked | BlacKow | CC BY-SA 3.0 |