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In 3D rendering (or geometry for that matter), in the rasterization algorithm, when you project the vertices of a triangle onto the screen and then find if a pixel overlaps the 2D triangle, you often need to find the depth or the z-coordinate of the triangle that the pixel overlaps. Generally, the method consists of computing the barycentric coordinates of the pixel in the 2D "projected" image of the triangle, and then use these coordinates to interpolate the triangle original vertices z-coordinates (before the vertices got projected).

Now it's written in all text books that you can't interpolate the vertices coordinates of the vertices directly but that you need to do this instead:

(sorry can't get Latex to work?)

1/z = w0 * 1/v0.z + w1 * 1/v1.z + w2 * 1/v2.z

Where w0, w1, and w2 are the barycentric coordinates of the "pixel" on the triangle.

Now, what I am looking after, are two things:

  • what would be the formal proof to show that interpolating z doesn't work?
  • what would be the formal proof to show that 1/z does the right thing?

To show this is not home work ;-) and that I have made some work on my own, I have found the following explanation for question 2.

Basically a triangle can be defined by a plane equation. Thus you can write:

Ax + By + Cz = D.

Then you isolate z to get z = (D - Ax - By)/C

Then you divide this formula by z as you would with a perspective divide and if you develop, regroup, etc. you get:

1/z = C/D + A/Dx/z + B/Dy/z.

Then we name C'=C/D B'=B/D and A'=A/D you get:

1/z = A'x/z + B'y/z + C'

It says that x/z and y/z are just the coordinates of the points on the triangles once projected on the screen and that the equation on the right is an "affine" function therefore 1/z is a linear function???

That doesn't seem like a demonstration to me? Or maybe it's the right idea, but can't really say how you can tell by just looking at the equation that this is an affine function. If you multiply all the terms you just get:

A'x + B'y + C'z = 1.

Which is just basically our original equations (just need to replace A' B' and C' with the proper term).

genpfault
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user18490
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  • Unfortunately [LaTeX is not enabled on StackOverflow](http://meta.stackexchange.com/questions/30559/latex-on-stack-overflow). – Aaron D Mar 05 '15 at 17:01
  • Consider the case of subdividing your triangle. Note that the center of the original triangle does not project to the center of the projected corners of the original triangle. – Ben Apr 07 '15 at 22:13

1 Answers1

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Not sure what you are trying to ask here, but if you look at:

1/z = A'x/z + B'y/z + C'

and rewrite it as:

1/z = A'u + B'v + C'

where (u,v) are screen coordinates of the triangle after perspective projection, you can see that the depth (z) of a point on the triangle is not linearly related to (u,v) but 1/depth is and that is what the textbooks are trying to teach you.

morpheus
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