Quote:Original post by glSmurf
ok ...this was abit harder then I first thought. I can't even compute if the two ellipsoids intersect

It is more than a bit harder than you think it is. The lengthy answer is given in this paper:

Wang, Wenping. Jiaye Wang. Myung-Soo Kim. "An Algebraic Condition for the Separation of Two Ellipsoids." Computer Aided Geometric Design, vol. 18, no. 6, pp. 531-539, July 2001.

You can find the paper online here:

http://3map.snu.ac.kr/mskim/ftp/ellipsoid.pdf

The short answer, though, is: give up on ellipsoids!

I always thought it could be done rather easily but never tested the method I'm thinking about. So your expertise might save me some time :). The way I see it:

- Take the shortest distance between the two ellipsoid up vectors.
- The two points on both up vectors can be (should already be) expressed as a parametric value along their respective axis kn € [0;1].
- The radius of the cross section along each up vectors can be computed using cos(kn) * uradius.
- And from there overlap is easy to tell by comparing distances...

Am I overlooking something?

Another thing I was thinking about is that it should be possible to use SAT with perfect sphere and alikes. A sphere is a convex with an infinity of planes and edges so provided that you pick the correct ones SAT should work perfectly. This could extend to ellipsoid as well.
My 20 yen.

oooo...I haven't looked yet but I think geometrictools.com might have something...as it certainly had something strange about basis vectors...

Also, I remember from an geometrictools.com ellipsoid-to-triangle collision paper I read, peroxide.dk I think, they simplified the problem by making a matrix to transform the ellipsoid into a sphere then applying the matrix to all of the triangles.

Perhaps you could do that to make a sphere-to-ellipsoid collision?

EDIT: put in clickies and wanted to report that I believe geometric tools 'only' has intersection for 2D ellipsoids, but maybe you could project your 3D ellipsoids into the three planes? I not sure that would work though :S

I am going to try this later on today, because I think it's doable, and not that hard due to symmetry. If my method works, it will undoubtedly reduce to the same math in the paper that Christer referred to.

So, you have two arbitrarily oriented ellipsoids A and B, with radii A.xyz and B.xyz, and you want to find their contact point & normal.

You want to put B in a space such that A becomes a unit sphere. If B and A are axis aligned, this is easily done by :

B.x /= A.x; B.y /= A.y; B.z /= A.z, and similarly for the center of B.

If A and B are not axis aligned, then you can't represent the ellipse's radius by 3 scalars, but by 3 vectors.

If you rotate these vectors into A's axis aligned space, then scale down the resulting vectors by A.xyz, rotate the ellipsoid center into A's space and scale down, you will now have arbitrary ellipsoid B vs unit sphere A.

Now we can rotate around the unit sphere A with no problem, so we can rotate the ellipsoid B so that its radius vectors become
axis aligned, and of the form :

b.x = < * 0 0 >
b.y = < 0 * 0 >
b.z = < 0 0 * >


Now we have an axis-aligned ellipsoid vs a sphere. We could add the sphere's radius of 1 to the ellipsoid radii and simply test the origin WRT the ellipse.

So it boils down to taking advantage of symmetry to reduce the problem to something conceptually simpler.

Once you've calculated your point in A's sphere space, you'd need to go through the inverse of whatever transforms you've made up to this point, and then plug the point into your Ellipsoid->GetNormal( point ) function to get the normal at that point.