The Projection Matrix

From the transformation matrix above, $y' = F y + B z$. Consider a 2D slice at $x = 0$ through the view volume (shown on the left below) and the corresponding 2D slice at ${x' \over w'} = 0$ through the canonical view volume (shown on the right below):

Points on the top line of the view volume satisfy ${y \over -z} = {t \over n}$. That is, there points are of the form $({-tz \over n} , z)$. These points correspond to points on the top line of the canonical view volume that satisfy ${y' \over w'} = 1$.

Substitute a point $({-tz \over n} , z)$ into the $y'$ and $w'$ lines of the transformation matrix:

As stated above, the VCS point $({-tz \over n} , z)$ is transformed to a CCS point for which ${y' \over w'} = 1$. So

So we know that

By a similar argument (using VCS points $( {-b z \over n}, z )$ on the bottom line which map to CCS points for which ${y' \over w'} = -1$) we can show that

We can then solve the two equations in $F$ and $B$ to get:

This is done analagously to $y'$.

Given $\ell$ and $r$ which are the left and right limits in the $x$ direction of the view volume, we apply the same method as for $y'$ and get

VCS points on the near plane ( $z = -n$ ) map to CCS points on the line ${z' \over w'} = -1$, and VCS points on the far plane ( $z = -f$ ) map to CCS points on the line ${z' \over w'} = +1$.

From the transformation matrix:

Substitute $z = -n$ to get one equation. Substitute $z = -f$ to get another equations. The unknowns are $C$ and $D$:

Solving the equations yields:

Bringing everything together:

Other VCS-to-CCS transformation matrices exist. In particular, orthographic projection is sometimes used.

See functions frustum(), perspective(), and ortho() in 00-intro/src/linalg.cpp in openGL.zip.