View Full Version : Clip Space coordinate question

azcoder

10-31-2007, 10:37 AM

In a vertex shader, after transforming by the MVP matrix - eg.

gl_Position = gl_ProjectionMatrix * cameraModelViewMatrix * gl_ModelViewMatrix * gl_Vertex;

(I have the camera transform separate from gl_ModelViewMatrix)

The values of gl_Position are mapped to a unit cube, correct? e.g. -1 < x,y,z < 1

I am trying to implement the effect of draping a texture over everything - basically like a projective texture or shadow map.

I thought I could use the x and y coords of

gl_lightPosition = lightProjectionMatrix * lightModelViewMatrix * gl_ModelViewMatrix * gl_Vertex

as the s,t indices into the texture.

But it seems to wrap the texture.

Any feedback/advice is appreciated...

azcoder

10-31-2007, 12:49 PM

Ok, I almost got it.

I need to divide by w, e.g.

s = x/w

t = y/w

Why do I need to do this - can somemone help me understand "homogenous clips space"?

sqrt[-1]

10-31-2007, 04:20 PM

In a vertex shader, after transforming by the MVP matrix - eg.

gl_Position = gl_ProjectionMatrix * cameraModelViewMatrix * gl_ModelViewMatrix * gl_Vertex;

(I have the camera transform separate from gl_ModelViewMatrix)

The values of gl_Position are mapped to a unit cube, correct? e.g. -1 < x,y,z < 1

Wrong, they are mapped from -w < x,y,z < w .

What it sounds like you need to do is supply a w component to your texture coordinates and do a texProj lookup in the fragment shader.

Humus

11-01-2007, 02:43 PM

Why do I need to do this - can somemone help me understand "homogenous clips space"?

That's the perspective division, which is neccesary, well, because we need perspective correction. The value output from the vertex shader is (x, y, z, w). Then you divide by w and get the normalized coordinates (x / w, y / w, z / w). x / w and y / w will be in [-1 .. 1], as will z / w which is used in the depth test.

as will z / w which is used in the depth test.

More precisely the depth test uses window space Z which is

(z/w)*(f-n)/2 + (n+f)/2

where n and f are the parameters do glDepthRange. One thing that D3D does better.

Relic

11-05-2007, 02:58 AM

Unrelated, but a performance tip you should keep in mind:

Your simple line:

gl_Position = gl_ProjectionMatrix * cameraModelViewMatrix * gl_ModelViewMatrix * gl_Vertex;

will do two matrix * matrix and one matrix * vector multiplications which takes 16 + 16 + 4 instructions. Bad.

If you bracket the code so that it resolves to only matrix * vector multiplications like this:

gl_Position = gl_ProjectionMatrix * (cameraModelViewMatrix * (gl_ModelViewMatrix * gl_Vertex));

it will only take 4 + 4 + 4 instructions!

Saved 24 instructions in a single GLSL line. Nice, eh?

And no, a compiler can not do that automatically because the multiplication operator associativity is from left to right.

Better yet, calculate (gl_ProjectionMatrix * cameraModelViewMatrix * gl_ModelViewMatrix) on the CPU and pass the combined matrix as uniform to the shader.

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