Timeline for Test if two curves intersect before finding roots
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24, 2010 at 12:51 | comment | added | Alexander Tsepkov | Well, the reason is that I already implemented a solver using implicitization, but would like to save time by only calling the quartic root finder when there actually are roots to find, preferably on the interval t=[0,1]. | |
| Aug 23, 2010 at 23:03 | history | edited | Gerry Myerson | CC BY-SA 2.5 |
corrected spelling
|
| Aug 23, 2010 at 22:07 | comment | added | J. M. isn't a mathematician | Why isn't Bezier clipping (e.g. dx.doi.org/10.1016/0010-4485(90)90039-F ) suitable for your purposes? | |
| Aug 23, 2010 at 19:29 | answer | added | Thierry Zell | timeline score: 1 | |
| Aug 23, 2010 at 19:06 | comment | added | Alexander Tsepkov | yes, they're are actually quadratic bezier curves presented in parametric form: x(t)=A.X(t-1)^2+B.X(t-1)t+C.Xt^2, y(t)=A.Y(t-1)^2+B.Y(t-1)t+C.Y^2 | |
| Aug 23, 2010 at 19:05 | history | edited | Alexander Tsepkov |
Changed tags from intersection-theory
|
|
| Aug 23, 2010 at 18:38 | comment | added | damiano | From your question it seems that you have two parametric curves in the plane and you want to find their intersection points, am I right? Are the parametric equations polynomials? | |
| Aug 23, 2010 at 17:57 | comment | added | José Figueroa-O'Farrill | I'm afraid that the tag is not appropriate, as a simple detour via wikipedia would have told you. | |
| Aug 23, 2010 at 17:48 | history | asked | Alexander Tsepkov | CC BY-SA 2.5 |