I'm writing a program that calculates the intersection of a ray with a sphere S that is transformed by a 4x4 matrix M. The ray's parametric equation is as follows (o = origin vector, d = direction vector): r(t) = o + dt. To find the intersection with the sphere, I will apply the inverse of M (M') on the ray and calculate the intersection of M'r(t) with S. M'r(t) = M'o + M'dt

Let r1(t) = [0,0,0] + t[0,0,-1]

Here's where I'm confused: If M translates 2 units in the negative z plane, then then r2(2) = M'o + M'dt = [0,0,2] + t[0,0,1] will result in a ray that points in the opposite direction of S.

The way I see it I should apply translation transformations only to the rays origin but my lecture notes state otherwise. What am I not understanding? Lecture note exerpt:

For a pixel $p(x,y)$, we form a parametric equation for the ray as: $$ \vec{R}(t)=\vec{e}+\vec{c}t $$ where $\vec{e}=(0,0,-d)$ and $\vec{c}=(x,y,0)-\vec{e}$. The vector $(x,y,d)$ is thus in the direction of the ray. If we create and instance of a generic object and place it where we cant it in the scene, then there exits a transformation matrix $M$ that performs this operation: $$ \vec{q}=M\vec{q}' $$ where $\vec q$ is the generic object transformed as we desire it to appear in the scene. At this point, we need to find out if the ray for a given pixel intersects this object. An efficient way of performing this computation is to keep the object in its generic form $\vec q'$ and apply the inverse transformation $M^{-1}$ to the ray $\vec R(t)$ instead. The transformed ray can thus be written in the following way: $$ \vec R(t)M^{-1}=M^{-1}(\vec e+\vec ct)=M^{-1}\vec e+M^{-1}\vec ct $$ Conveniently, the values for $t$ at intersections are the same in both spaces.


Translating a vector doesn't make sense, because a vector doesn't have a position; only translation of points makes sense.

The usual way to handle this in computer graphics is as follows: when you "lift" 3D points and vectors into 4D, you use $w=1$ as the 4th coordinate of a point, but you use $w=0$ as the 4th component of a vector. Then, when you multiply a vector by a $4 \times 4$ matrix, you end up multiplying the "translation" elements of the matrix by zero, so no translation gets applied to the vector.

  • $\begingroup$ I knew I was missing something simple. Thanks for answering; this program is due tomorrow $\endgroup$ – Mark McKenna Dec 5 '13 at 19:12

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