Timeline for Approximating surface normals from signed distance bounds
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 4, 2024 at 14:47 | vote | accept | biquart | ||
| Oct 4, 2024 at 11:11 | answer | added | lightxbulb | timeline score: 3 | |
| Oct 1, 2024 at 14:59 | comment | added | Kevin Reid | Someone (could be you, could be @lightxbulb, could be someone else) should write and post an answer (that can then be marked as accepted, if you like it). Questions should not be left with only comments. | |
| Oct 1, 2024 at 13:20 | comment | added | biquart | Yes, I've read that hit and trial is usually the preferred method to get the step size, when you apply some unusual transformation that gives only a distance bound. I'm sort of a first timer on Stack Exchange. Since I think you've addressed nearly everything (thanks!), what would be the appropriate way to close the question? | |
| Sep 30, 2024 at 16:05 | comment | added | lightxbulb | Actually that's not necessary with the numerical approximation, it would still work for functions that are almost everywhere differentiable (you can see this from the box sdf for example), and I think those constraints can be relaxed further. I would prefer using the analytical gradient most of the time though, as the step size in the numerical approximation is notoriously hard to get right. | |
| Sep 30, 2024 at 14:42 | comment | added | biquart | Yeah, it would also need to be differentiable, at least | |
| Sep 30, 2024 at 10:15 | comment | added | lightxbulb | To be precise you probably need a bit more than continuity. | |
| Sep 30, 2024 at 9:07 | comment | added | biquart | Hmm, I see. This is quite enlightening, thanks for the comment, that makes sense. I was reasoning this way - if f approximates g, then it doesn't necessarily follow that the derivative of f approximates that of g. Which is true of course, but approximation at one point, and approximation over a surface are quite different. | |
| Sep 29, 2024 at 10:45 | comment | added | lightxbulb | My guess is that since $f(p)=0$ must still hold at the points of the surface, and that $f$ is continuous, as you get closer to the boundary the normal must converge to the actual one. | |
| S Sep 28, 2024 at 6:11 | review | First questions | |||
| Sep 28, 2024 at 16:08 | |||||
| S Sep 28, 2024 at 6:11 | history | asked | biquart | CC BY-SA 4.0 |