Nicolas Alvarez wrote:

> As far as I know, the way POV-Ray does it (and probably most others) is 
> by creating a transformation matrix of all the individual 
> transformations, then inversing it, and using it to transform *rays* 
> before doing the intersection calculations.

I believe that is correct. (Transforming the object equation would be 
much harder...)

> But don't trust me on this. I still haven't managed to solve the 
> ray-sphere intersection equation...

Suppose you have a sphere with center C [a vector] and radius r [a 
scalar]. The equation for this is

   (P - C)^2 - r^2 = 0

(Where squaring a vector *really* means taking the dot product of the 
vector with itself.) Here P is your unknown - but it's a vector. If we 
replace that with the ray equation

   Ray(t) = Dt + S

then we obtain

   (Dt + S - C)^2 - r^2 = 0

If we set V = S - C then we have

   (Dt + V)^2 - r^2 = 0

and expanding we find

   D^2t^2 + 2VDt + V^2 - r^2 = 0

Writing "#" to explicitly mean "dot product", what I really mean is

   (D # D) t^2 + 2 (V # D) t + (V # V) - r^2 = 0

Notice this is a quadratic in one unknown - t, a scalar. In fact, what 
we have is

   a t^2 + b t + c = 0

where

   a = D # D
   b = 2 (V # D)
   c = (V # V) - r^2

Solving this yields a value for t which satisfies the equation. If you 
substitute this t back into the original ray equation, it gives you a 
point in space for the intersection.

Notice that if you compute multiple intersections, the one with the 
lowest value for t is guaranteed to be the "front" one [assuming that S 
is the "start point" of the ray. Oh, and if t is negative, the 
intersection is "behind" the camera or whatever, and you should ignore 
that intersection...]

Any clearer? ;-)

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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