Timeline for Hexagon collision detection for fast moving objects?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 26, 2014 at 12:22 | comment | added | sam hocevar | @LuisAlves yes and no… all the logic works, but you’ll have to replace vB-vA with g(t)-f(t) where f and g are A and B’s positions over time. Since this is no longer a straight line, you’ll have to solve a box — parametric curve intersection problem. | |
| Oct 25, 2014 at 19:42 | comment | added | Luis Alves | Is it possible to use this solution for accelarated movement? | |
| May 29, 2013 at 13:25 | history | bounty awarded | API-Beast | ||
| May 29, 2013 at 13:25 | vote | accept | API-Beast | ||
| S May 24, 2013 at 20:28 | history | edited | sam hocevar | CC BY-SA 3.0 | fix diagrams |
| S May 24, 2013 at 20:28 | history | suggested | jrsala | CC BY-SA 3.0 | added helpful stuff I hope? |
| May 24, 2013 at 20:25 | comment | added | sam hocevar | Mmmh, okay, I guess I'll just call them C and D instead of being so confusing. | |
| May 24, 2013 at 20:19 | review | Suggested edits | |||
| May 24, 2013 at 20:28 | |||||
| May 24, 2013 at 19:56 | comment | added | jrsala | +1. This is very elegant. However, it's A + (-B) you must compute, not A+B. As you say, it's "Minkowski subtraction", not addition. It just so happens that when B is a rectangle and the origin of the space is it center, then -B = +B. Also, the point is not "B-B": if you computed the difference of all b and b' vectors in B, you wouldn't get a point, you would get 2*B ;) | |
| May 24, 2013 at 18:46 | comment | added | sam hocevar | @Mr.Beast I did my homework and came up with a simpler and hopefully understandable solution. | |
| May 24, 2013 at 18:45 | history | edited | sam hocevar | CC BY-SA 3.0 | rework solution |
| May 24, 2013 at 14:39 | comment | added | API-Beast | @SamHocevar If you could revise the answer that would be great. | |
| May 23, 2013 at 9:09 | comment | added | sam hocevar | @madshogo You're right. I assumed that the timestep was small enough compared to object sizes that this wouldn't be a problem, but it's certainly not very robust in the general case. I'll look into fixing it. | |
| May 22, 2013 at 17:58 | comment | added | jrsala | -1: This method does not in any way allow you to be sure collision happens. It only allows you to be sure it doesn't happen, in the case where the segment and the extruded volume do not intersect. But it is entirely possible that they intersect and yet that there is no collision. What's wrong is the "Now you can use [...] simple segment-segment intersection to decide where the collision occurred" part. | |
| May 22, 2013 at 17:39 | comment | added | API-Beast | Looks like a rather interesting approach, however, I don't grasp it 100% yet, what happens when the object is really small and is moving between the two lines? i.imgur.com/hRolvAF.png | |
| May 22, 2013 at 16:38 | history | answered | sam hocevar | CC BY-SA 3.0 |