View Full Version : oblique frustum

DarkWIng

04-02-2004, 05:56 AM

I've been trying to implement nVidia's "oblique frustum culling" demo's technique in my engine for clipping at waterplane, but without luck. Mostly becouse I don't understand the math bedind it.

I want to create a function that takes old projection matrix, old modelview matrix (from demo looks like I need it) and some clip plane and outputs new projection matrix.

Demo looks like it's building a modification(shear?) matrix using old inverted&trnsponded MVP. I have no idea why/how this works. Can someone give me some info about it. Or some links that explain this.

DarkWIng

04-05-2004, 11:25 PM

Can anyone give me a little help with this. Like someone who wrote the demo or implemented it?

I also would like to know something about it. I hear d some rumors about this kind of clipping for months and I'm courious.

Eric Lengyel

05-07-2004, 12:05 AM

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'.

DarkWIng

05-15-2004, 01:47 AM

I just now noticed reply. Thank you Eric, that was most helpfull.

Eric Lengyel

05-15-2004, 08:17 PM

I'm actually doing some more research on this technique at the moment. It turns out that the far plane is completely hosed, and your z-buffer precision is cut roughly in half. These effects can be minimized with a little more work -- I'll make another post once I've nailed everything down myself. The above technique should work pretty well in most real-world cases.

Eric Lengyel

05-20-2004, 02:05 PM

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

Thank you very much for this. I still have to take a in-depth look at it since I'm pretty full of stuff to do but I appreciated your contribution a lot.

Thank you Eric!

Eric Lengyel

06-02-2004, 11:23 PM

Sample code for this technique is now posted on terathon.com (http://www.terathon.com/) .

Hi Eric, i've noticed the code you published in gpg5 and the code in theraton.com are different. In particular the line (in theraton.com)

Vector4D c = clipPlane * (2.0F / Dot(clipPlane, q));

is

Vector4D c = clipPlane * (-2.0F / Dot(clipPlane, q));

in GPG5, anyway none of them work for me, i don't know exactly why yet, but keep on looking for the error

Ok, i am particulary idiot, 2 hours after posting that i discovered the error, now seems oblique frustum works :)

At least it helped me to detect several caveats in my algorithsm and my way of thinking :)

Toni

P.S I still don't know why that -2 is different in GPG tho

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