Spherical Normals

In short: Yes. Yes.

Elaboration:
1) Yes, the normalized position of a sphere is it's normal.
2) Eh, it probably wouldn't matter. Theoretically, you're doing less processing and therefore your floats should lose less precision. There probably wouldn't be a noticeable difference in image quality.

If you know the true value of a normal, you should use that value rather than deriving a value from the shape of the mesh. After all, the mesh is only an approximation of the true shape.

Plus you can probably do it faster. For a unit sphere, the position IS the normal (apart from translating by the position of the sphere's centre). If you know the radius then you can avoid nast sqrts and averaging etc...

Quote:

Original post by d000hg Plus you can probably do it faster. For a unit sphere, the position IS the normal (apart from translating by the position of the sphere's centre). If you know the radius then you can avoid nast sqrts and averaging etc...

Just to clarify. d000hg is simply stating that instead of doing:

Vector3 normal = currentPointOnSphere - sphere.origin;
normal.Normalize();

You could do:

Vector3 normal = currentPointOnSphere - sphere.origin;
normal.Scale( 1 / sphere.radius );

The second method would be less computationally expensive.

And it works because the radius is the length of the calculated vector normal (before scales or normalizes). Hope this helps.

Allright Thanks.