作者：未知 来源：月光软件站 加入时间：2005-2-28　月光软件站

Let P = (Px, Py, Pz, Pw) be the eye-space plane to which you would like to clip.
This will replace your ordinary near plane and should be facing away from the
camera, so Pz < 0.
 
Let Q be an eye-space point.  If P dot Q = 0, then Q lies on the plane P.
 
What we want to do is modify the projection matrix M so that points Q for
which P dot Q = 0 get transformed in such a way that the transformed z-coordinate
is the negation of the transformed w-coordinate.  This corresponds to
the projected z-coordinate that you would ordinarily get for a point lying
on the near plane.  After division by the w-coordinate, you get -1.
 
The projection matrix M transforms points from eye space into homogeneous clip
space.  To obtain the clip-space plane P', we need to multiply P by the inverse
transpose of M.  (This is because planes are covariant vectors.)  The clip-space
plane P' is given by
 
       -1 T
P' = (M  )  P
 
Let Q' be the projection of the point Q, given by Q' = MQ.  If Q lies on the plane P,
then P' dot Q' = 0, but we want it to be -Q'w so that it corresponds to the near plane.
So we force it by subtracting 1 from the w-coordinate of P'.  Before doing this,
however, we want to divide P' by its z-coordinate as to avoid scaling the z
direction in clip space.  Doing all of this gives us the following plane P":
 
       P'x    P'y       P'w
P" = (-----, -----, 1, ----- - 1)
       P'z    P'z       P'z
 
We integrate this into the projection matrix M by constructing a new projection
matrix M' as follows.
 
     [  1    0    0    0  ]
     [                    ]
     [  0    1    0    0  ]
M' = [                    ] M
     [ P"x  P"y   1   P"w ]
     [                    ]
     [  0    0    0    1  ]
 
The matrices M and M' only differ in the entries of the 3rd row, so we could just
calculate replacements for 3rd-row entries of M as follows:
 
M'(3,1) = P" dot M_1
M'(3,2) = P" dot M_2
M'(3,3) = P" dot M_3
M'(3,4) = P" dot M_4
 
where M_i means the i-th column of M and M'(i,j) means the (i,j)-th entry of M'.

Okay -- here's the short, short version of the optimal implementation of the oblique frustum.
 
Let C = camera-space clipping plane. Assume the camera is on the negative side of the plane: C_w < 0.
 
Let M be the original projection matrix; M must be invertible, but otherwise we don't care what
it is. We'll modify the third row of M so that the near plane coincides with the arbitrary
clipping plane C.  We aren't allowed to modify the fourth row of M because doing so would screw
up the perspective-correct vertex attribute interpolation. (The fourth row usually just
moves the z-coordinate of a camera-space point into the w-coordinate of a clip-space point.)
 
Given a projection matrix M, the near plane is always M_4 + M_3, and the far plane is always
M_4 - M_3. (M_i means the i-th row of M.)  To force the near plane to coincide with the
clipping plane C, we must have M_3 = aC - M_4, where a is some positive scale factor that we
can adjust.
 
Why adjust a?  Because now our far plane F has been moved to F = 2*M_4 - aC, which is not
generally parallel to C.  F intersects C on the x-y plane, so it's really not in a good
position.  The best we can do is minimize the size of the view frustum by choosing the
constant a so that F contains the corner Q of the view frustum opposite the near plane C.
This point Q is given by
 
Q = M^-1 * (sgn(C_x), sgn(C_y), 1, 1),
 
where M^-1 is the inverse of the projection matrix. (The points (+/-1, +/-1, 1, 1) are the
four corners of the view frustum on the far plane in clip space.)  The scale factor a is now
given by
 
a = (2*M_4 dot Q) / (C dot Q).
 
Since M_4 is usually (0, 0, -1, 0), this can be simplified a little.
 
All we have to do to the original projection matrix M is replace the third row with aC - M_4.
 
-- Eric Lengyel

相关软件：