Rasterization

May 14, 2007

Triangles Only• We will discuss the rasterization of

triangles only.

• Why?– Polygon can be decomposed into triangles.– A triangle is always convex.– Results in algorithms that are more

hardware friendly.

Being Hardware Friendly• [Angel 4e] Section 7.11.3 or • [Angel 3e] Section 8.11.6:

– Intersect scan lines with polygon edges.

– Sort the intersections, first by scan lines, then by order of x on each scan line.

– It works for polygons in general, not just in triangles.

– O(n log n) complexity feasible in software implementation only (i.e., not hardware friendly)

Color and Z

• Now we know which pixels must be drawn. The next step is to find their colors and Z’s.

• Gouraud shading: linear interpolation of the vertex colors.

• Isn’t it straightforward?– Interpolate along the edges. (Y direction)– Then interpolate along the span. (X

direction)

Interpolation in World Space vs Screen Space

• p1=(x1, y1, z1, c1); p2=(x2, y2, z2, c2); p3=(x3, y3, z3, c3) in world space

• If (x3, y3) = (1-t)(x1, y1) + t(x2, y2) then z3=(1-t)z1+t z2; c3=(1-t)c1+t c2

• But, remember that we are interpolating on screen coordinates (x’, y’):

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• Does s=t? If not, should we compute z3 and c3 by s or t?

• Express s in t (or vice versa), we get something like:

• So, if we interpolate z on screen space, we get the z of “some other point on the line”

• This is OK for Z’s, but may be a problem for texture coordinates (topic of another lecture)

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Appendix

Derivation of s and t

• Two end points P1=(x1, y1, z1) and P2=(x2, y2, z2). Let P3=(1-t)P1+(t)P2

• After projection, P1, P2, P3 are projected to (x’1, y’1), (x’2, y’2), (x’3, y’3) in screen coordinates. Let (x’3, y’3)=(1-s)(x’1, y’1) + s(x’2, y’2).

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