The Projection Transform looks like this:
$\pccs = \begin{bmatrix}
E & 0 & A & 0 \\
0 & F & B & 0 \\
0 & 0 & C & D \\
0 & 0 & -1 & 0 \\
\end{bmatrix} \; \pvcs$
Let $\pvcs = (x, y, z, 1)$ and let $\pccs = (x', y', z', w')$. Then
$\begin{bmatrix} x' \\ y' \\ z' \\ w' \end{bmatrix}
=
\begin{bmatrix}
E & 0 & A & 0 \\
0 & F & B & 0 \\
0 & 0 & C & D \\
0 & 0 & -1 & 0 \\
\end{bmatrix}
\begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix}$
How $y'$ is Calculated
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:
$\begin{array}{rl}
y' & = F y + B z \\
& = F {-tz \over n} + Bz \\
w' & = -z
\end{array}$
As stated above, the VCS point $({-tz \over n} , z)$ is
transformed to a CCS point for which ${y' \over w'} = 1$.
So
$\begin{array}{rl}
{y' \over w'} & = { {F { -tz \over n} + Bz } \over -z }
& = {t \over n} F - B
& = 1
\end{array}$
So we know that
${t \over n} F - B = 1$.
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
${b \over n} F - B = -1$.
We can then solve the two equations in $F$ and $B$ to get:
$\begin{array}{rl}
F & = {2 n \over t-b} \\
B & = {t+b \over t-b}
\end{array}$
How $x'$ is Calculated
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
$\begin{array}{rl}
E & = {2 n \over r-l} \\
A & = {r+l \over r-l}
\end{array}$
How $z'$ is Calculated
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:
$\begin{array}{rl}
z' & = C z + D \\
w' & = -z
\end{array}$
Substitute $z = -n$ to get one equation. Substitute $z
= -f$ to get another equations. The unknowns are $C$
and $D$:
$\begin{array}{rl}
{z' \over w'} & = {{C (-n) + D} \over n} \\
& = -C + {D \over n} \\
& = -1 \\
\\
{z' \over w'} & = {{C (-f) + D} \over f} \\
& = -C + {D \over f} \\
& = +1
\end{array}$
Solving the equations yields:
$\begin{array}{rl}
C & = {f+n \over n-f} \\
D & = {2fn \over n-f}
\end{array}$
The VCS-to-CCS Transformation Matrix
Bringing everything together:
$\pccs = \begin{bmatrix}
{2n \over r-l} & 0 & {r+l \over r-l} & 0 \\
0 & {2n \over t-b} & {t+b \over t-b} & 0 \\
0 & 0 & {f+n \over n-f} & {2fn \over n-f} \\
0 & 0 & -1 & 0
\end{bmatrix}
\; \pvcs$
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.
Points outside the view frustum must be clipped. The clipping
could be done in almost any coordinate system, but it's best to do
clipping in the CCS because:
- The canonical view volume in the CCS is independent of camera
parameters. That means that clipping in the CCS can be
implemented in hardware which doesn't have to be parameterized ... so
it's fast. There's no point earlier in the sequence of
transformations where clipping can be done in a camera-independent
manner.
- After the perspective division (CCS-to-NDCS) some depth
information is lost, which can result in improper clipping (as
discussed below). So it doesn't make sense to clip after the CCS.
Example of Improper Clipping in the NDCS
In the figure below, the segment $pq$ in the VCS is
transformed to the segment $p'q'$ in the NDCS. If we clip
$p'q'$, the segment will appear to be exiting the far plane of
the canonical view volume. But it should really exit the top
plane!
For example, if $q = [0,0,-100,1]^T$ and $n=1, f=2, r=+1,
l=-1, t=+1, b=-1$ then $q$ is behind the viewpoint (as above)
and the projection matrix is
$\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -3 & -4 \\
0 & 0 & -1 & 0
\end{array}\right]$
Then $q' = P q = [x', y', z', w']^T = [0, 0, -304, -100]^T$
and, upon division by $w'$ we get $[0, 0, 3.04]^T$, which
is ahead of the viewer instead of behind, as the
original $q$ was.
For a better intuition, consider what happens to $q'$ as $q$ is moved
along the $z$ axis of the VCS:
- As $q$ approaches the origin, the segment $pq$
approaches, then passes, the top-front corner of the VCS
view volume. At the same time, the transformed $q'$ in
the NDCS will move right (more distant) along the ${z'
\over w'}$ axis.
- When $q$ touches the origin, the transformed $q'$ in the
NDCS has $w' = 0$ (recall the matrix above, in which $w' =
-z$), and $q'$ lies at infinity on the ${z' \over w'}$
axis. That is, $p'q'$ is parallel to the ${z' \over w'}$
axis.
- As soon as q passes the origin, $q'$ in NDCS has $w' >
0$ and $z' < 0$, so $q'$ lies on the negative ${z' \over
w'}$ axis, and moves in toward the NDCS origin as q
continues inward.
- After $q$ passes the centre of the VCS view volume, $z'$
becomes positive (while $w'$ remains positive), and $q'$
moves to the right of the NDCS origin.
Proper Clipping in the CCS
If the point were in NDCS, we would clip against the six planes
that define the faces of the canonical view volume:
$\begin{array}{rcr}
{x' \over w'} = +1 & \qquad & {x' \over w'} = -1 \\
{y' \over w'} = +1 & & {y' \over w'} = -1 \\
{z' \over w'} = +1 & & {z' \over w'} = -1
\end{array}$
The corresponding planes in the CCS are:
$\begin{array}{rcr}
x' - w' = 0 & \quad & x' + w' = 0 \\
y' - w' = 0 & & y' + w' = 0 \\
z' - w' = 0 & & z' + w' = 0
\end{array}$
Given a segment $p'q'$ in the CCS, we clip against each of
the six planes. Note that these are planes in a 4D
homogeneous space, but clipping works the same in any
dimension, as we'll see below.
A line segment $ab$ has endpoints $a$ and $b$. It can be
written in parametric form as
$\ell(t) = a+t(b-a)$
For $t \in [0,1]$, $\ell(t)$ is between $a$ and $b$. For $t$
outside this interval, $\ell(t)$ is not on the segment
$ab$.
A plane $\pi$ is defined by a normal, $n$, and a distance from
the origin, $d$. All points $x$ on the plane satisfy
the implicit equation
$\pi(x) = n \cdot x - d = 0$
Points above the plane (i.e. on the side of the plane that $n$
points to) will have $\pi(x) > 0$. Points below the plane will
have $\pi(x) < 0$.
Segment $ab$ is clipped by the plane if and only if its
endpoints lie on different sides of the plane. So compute
$\pi(a)$ and $\pi(b)$. Then $ab$ intesects $\pi$ if and only if
$\pi(a)$ and $\pi(b)$ have different signs. (If one or both are
equal to zero, $ab$ is not clipped because it does
not cross the plane.)
If $ab$ is clipped by $\pi$, we need to find the point at which
$ab$ intersects $\pi$, to clip $ab$ at that point.
To find the point of intersection, substitute the parametric
segment equation into the implicit plane equation and solve for
the parameter $t$:
$\begin{array}{l}
0 = \pi( \ell(t) ) \\
0 = \pi( a+t(b-a) ) \\
0 = n \cdot (a+t(b-a)) - d \\
t = { \large d - n \; \cdot \; a \over \large n \; \cdot \; (b-a) }
\end{array}$
Then the intersection point is $\ell(t)$.