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.
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.