Extrude shadow volumes to infinity

Question

I want to use the depth-fail algorithm to make shadow volumes and all works great, but i can't figure out how to extrude the shadow volume quads to infinity.

The aim is to create the shadow volume for a triangle that is lit up from a point light. I have red that i first have to change the perspective matrix that it has no far clip pane and set the w coordinate to 0. But what are the x,y,z coordinates then?

An example would be very helpful, but i also want to understand how its done.

Perhaps you can explain why you want to extrude shadow volumes to infinity. — JCooper, Jun 27 '13 at 14:41
Tod avoid bugs in the shadow. It is a method that is often used in the depth-fail algorithm. — user2195441, Jun 27 '13 at 19:47

score 0 · Accepted Answer · answered Jun 28 '13 at 02:10

This link shows an example of a projection matrix. It has the form:

    a   0   b              0
A = 0   d   e              0
    0   0   -(f+n)/(f-n)   -2fn/(f-n)
    0   0   -1             0

f is the far plane and you want f -> infinity.

limit f -> infinity of (f+n)/(f-n) = limit f -> infinity of (1+n/f)/(1-n/f)

and

limit f -> infinity of 2fn/(f-n) = limit f -> infinity of 2n/(1-n/f)

since

f -> infinity => n/f -> 0

your matrix with f -> infinity becomes

    a   0   b    0
B = 0   d   e    0
    0   0   -1   -2n
    0   0   -1   0

if you transform your (x,y,z,w=0) with B you'll get

x' = ax + bz
y' = dy + ez
z' = -z
w' = -z

and the perspective division gives

x' = -ax/z - b
y' = -dy/z - e
z' = 1

While x' and y' are the same as transforming (x,y,z,w=0) with A, z' is now a constant that is always equal to the far plane in normalized device coordinates.

This article shows an efficient implementation of shadow volumes with capping at infinity.

Extrude shadow volumes to infinity

1 Answers1