CISC/CMPE 454 - Clipping

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 $\ell(t)$.

Clipping in the CCS

Example of Improper Clipping in the NDCS

Proper Clipping in the CCS

Clipping a Segment Against a Plane