Projection matrix point to sphere surface

I need to project a 3D object onto a sphere's surface (uhm.. like casting a shadow).

AFAIR this should be possible with a projection matrix.

If the "shadow receiver" was a plane, then m开发者_StackOverflow社区y projection matrix would be a 3D to 2D-plane projection, but my receiver in this case is a 3D spherical surface.

So given sphere1(centerpoint,radius),sphere2(othercenter,otherradius) and an eyepoint how can I compute a matrix that projects all points from sphere2 onto sphere1 (like casting a shadow).

Do you mean that given a vertex v you want the following projection:

v'= centerpoint + (v - centerpoint) * (radius / |v - centerpoint|)

This is not possible with a projection matrix. You could easily do it in a shader though.

Matrixes are commonly used to represent linear operations, like projection onto a plane. In your case, the resulting vertices aren't deduced from input using a linear function, so this projection is not possible using a matrix.

If the sphere1 is sphere((0,0,0),1), that is, the sphere of radius 1 centered at the origin, then you're in effect asking for a way to convert any location (x,y,z) in 3D to a corresponding location (x', y', z') on the unit sphere. This is equivalent to vector renormalization: (x',y',z') = (x,y,z)/sqrt(x^2+y^2+z^2).

If sphere1 is not the unit sphere, but is say sphere((a,b,c),R) you can do mostly the same thing: (x',y',z') = R*(x-a,y-b,z-c) / sqrt((x-a)^2+(y-b)^2+(z-c)^2) + (a,b,c). This is equivalent to changing coordinates so the first sphere is the unit sphere, solving the problem, then changing coordinates back.

As people have pointed out, these functions are nonlinear, so the projection cannot be called a "matrix." But if you prefer for some reason to start with a projection matrix, you could project first from 3D to a plane, then from a plane to the sphere. I'm not sure if that would be any better though.

Finally, let me point out that linear maps don't produce division-by-zero errors, but if you look closely at the formulas above, you'll see that this map can. Geometrically, that's because it's hard to project the center point of a sphere to its boundary.