Nicolas Alvarez wrote:

>> Any clearer? ;-)
> 
> Thanks; shows I'm still too noob for this kind of things.
> 
> 
> 
> (that's a no)

OK, well because I'm feeling bored, let me try again...



Light travels in essentially straight lines. If we assume a ray starts 
at position S and travels parallel to the vector D, we can write

   R(t) = Dt + S

That is, the vector D scaled by the scalar t, plus the vector S.

A sphere is the set of all points within a specific distance of a 
certain point. So if C is the center of our sphere, and r is the radius 
of the sphere, then the sphere consists of all points less than r units 
from the point C. We're interested in the *surface* of the sphere, which 
is the set of all points *exactly* r units away from C. We can write that as

   |P - C| = r

Solving this equation for any vector P gives us the surface coordinates. 
But that's not what we're trying to do. We want to know whether any 
point in space lines on the surface of the sphere *and* on a given ray. 
In other words, solve simultaneous equations. Non-linear ones!

But we'll get to that in a second. First, notice that the dot product of 
a vector with itself is equal to the length of that vector squared. In 
oter words,

   vlength(V) = sqrt(V # V)

That means we can write

   sqrt((P - C) # (P - C)) = r

Alternatively, we can avoid the expensive square root operation by

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

Finally, with a tiny bit of rearranging, we have

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

(If the thing on the left *equals* the thing on the right, then 
subtracting one from the other must logically result in zero.)

I'm going to be fanciful and write the above in shorthand:

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

It's a slight abuse of notation, but it's less typing.

Now if we replace our unknown point P with our ray equation R(t) we get

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

If I put X = S - C, we get

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

Opening the brackets as per the standard binomial theorum, we have

   D^2 t^2 + 2 D X t + X^2 - r^2 = 0

By which of course I actually mean

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

This matches the form of the standard quadratic

   a t^2 + b t + c = 0

if we set

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

We can now apply the usual formula

   t = (-b +/- sqrt(b^2 + 4ac)) / 2a

Assuming that the square root part (the "determinant") is strictly 
positive, we obtain two purely-real solutions. These are the values for 
t where the ray "enters" and "leaves" the sphere. If the determinant is 
negative, that basically means the ray doesn't touch the sphere at all.

Assuming we *do* have an intersection, just calculate

   Dt + S

for the two values of t to obtain the actual points of intersection.



Is *that* any clearer? ;-)

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

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