2 3D Viewing Process  3D viewing process MC Model Space Model Transformation Model Transformation WC World Space Viewing Transformation Viewing Transformation VC View Space Viewport Transformation Viewport Transformation DC Display Space Normalization & Clipping Normalization & Clipping NC Normalize Space Projection Transformation Projection Transformation PC Projection Space

Simplified View The Data Flow: 3D Polygons (+Colors, Lights, Normals, Texture Coordinates…etc.)  2D Polygons  2D Pixels (I.e., Output Images) Transform (& Lighting) Rasterization

Attributes  Varying Transform (& Lighting) Rasterization vertex attributes: Color, Normal, Tex_coord, …etc. vertex attributes: Color, Normal, Tex_coord, …etc. vertex shader vertex shader fragment shader fragment shader varying variables: interpolated from vertices to internal pixels varying variables: interpolated from vertices to internal pixels

Key Questions Q1: What screen pixels are covered by the triangles? Q2: How to interpolate the attributes?

Rasterization

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 5e] Section : –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)

Using Edge Equations Find the edge equations for P1-P2, P2-P3, P1-P3 For each pixel, test which sides of the edges it is at. Q: Do we need to test every pixel on the screen? P1 P2 P3 ege defined by Ax+By+c=0

Example P1 P2 P3 ege defined by Ax+By+C=0

Example Now we have P1-P3 defined by 2x+y-4=0 To test if a point (u,v) is inside the triangle, we calculate 2(u)+(v)-4. For example, (0,0) leads to 2(0)+(0)-4=-4 So the negative side is inside. Use P2 to decide which side is inside. Negate A,B,C if you prefer the positive side P1 P2 P3 Ax+By+C=0

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)

13 Attribute Derivation Color Interpolation (255, 0, 0) (0, 255, 0)(0, 0, 255) (255, 0, 0) (0, 0, 0) Red (0, 0, 0) (0, 255, 0) Green (0, 0, 255) (0, 0, 0) Blue

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 P1=(x1,y1,z1) P3=(x3,y3,z3) P2=(x2,y2,z2)

Interpolation in World Space vs Screen Space But, remember that we are interpolating on screen coordinates (x’, y’): P’2 P’1 P2 P1 P3 x’ y’

Let P’ 1 =(x’ 1, y’ 1 ); P’ 2 =(x’ 2, y’ 2 ) and P’ 3 =(x’ 3, y’ 3 )= (1-s)(x’ 1, y’ 1 ) + s(x’ 2, y’ 2 ) 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)

Interpolation OpenGL uses interpolation to find proper texels from specified texture coordinates

Perspectively Correct Interpolation The previous slide is not “perspectively correct” Consider the interpolation along the vertical (Y) direction. Should the checkers spaced equally? Use the s to t conversion shown earlier, then you get the perspectively correct interpolation.

Appendix

Derivation of s and t Two end points P 1 =(x 1, y 1, z 1 ) and P 2 =(x 2, y 2, z 2 ). Let P 3 =(1-t)P 1 +(t)P 2 After projection, P 1, P 2, P 3 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 ).

(x’ 1, y’ 1 ), (x’ 2, y’ 2 ), (x’ 3, y’ 3 ) are obtained from P 1, P 2, P 3 by:

Since We have:

When P 3 is projected to the screen, we get (x’ 3, y’ 3 ) by dividing by w, so: But remember that (x’ 3, y’ 3 )=(1-s)(x’ 1, y’ 1 ) + s(x’ 2, y’ 2 ) Looking at x coordinate, we have

We may rewrite s in terms of t, w 1, w 2, x’ 1, and x’ 2. In fact, or conversely Surprisingly, x’ 1 and x’ 2 disappear.