In the last log I talked about the basic calculations behind triangle rasterization. But we ended up with a few interpolations (and therefore, divisions) we needed to do every scanline. These are costly in both time and resources, so I want to avoid doing those, or at least try to do them only once per triangle.

One of the main reasons why triangles are used to render 3D graphics is their mathematical unambiguity. For every three points in space,there is exactly one triangle. What's more, a triangle is always flat, unlike quadrangles which can be folded along a diagonal. We can extend this flat surface to get a plane. In essence, the surface of the triangle is a small cut-out of that infinite plane.

There is another way to define a plane: using two vectors and a base point. And no matter where you are on the plane, those two vectors remain the same. So why not use those vectors to define the surface of our triangle?

We're actually already using three of those vectors in a slightly different form: our gradients. The combination of m1, m2 or m3 with the corresponding gradients for the z-axis or colour components gives us the vectors along the edges of the triangle. To make calculation a bit simpler, we would like one the vectors to point along the scanline, so we can calculate the horizontal gradient from that (our n-values before).

I ended up using the following values:

The m1, m2 and m3 values remain the same, as they are used to define the edges of the triangle.

For mz, mr, mg and mb, the vertical gradients for the z-axis and each colour component, I will use the values along the same edge as m1, the edge that goes from the top-most to bottom-most vertex, so we can keep using them for the entire triangle as we descend along that edge.

The horizontal gradients nz, nr, ng and nb will require a bit more work. I will calculate the vector on the widest part of the triangle (along a scanline), the height of the middle vertex (x2, y2). With some mathematical acrobatics I ended up with the following formula for each component c:

nc = (c2-(c3-c1)*Q-c1)/(x2-(x3-x1)*Q-x1)

Where Q is

Q = (y2-y1)/(y3-y1)

Now we only have to calculate the gradients, both vertical and horizontal, one per triangle.

I'll put this in a block called 'PreCalc', for pre-calculation. In the block diagram I've also split the drawing of each scanline into two: CalcLine, which calculates the values along the edges, and DrawLine, which fills the pixels in between.