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Слайды и текст к этой презентации:
№1 слайд
Содержание слайда: Drawing Triangles
CS 445/645
Introduction to Computer Graphics
David Luebke, Spring 2003
№2 слайд
Содержание слайда: Admin
Homework 1 graded, will post this afternoon
№3 слайд
Содержание слайда: Rasterizing Polygons
In interactive graphics, polygons rule the world
Two main reasons:
Lowest common denominator for surfaces
Can represent any surface with arbitrary accuracy
Splines, mathematical functions, volumetric isosurfaces…
Mathematical simplicity lends itself to simple, regular rendering algorithms
Like those we’re about to discuss…
Such algorithms embed well in hardware
№4 слайд
Содержание слайда: Rasterizing Polygons
Triangle is the minimal unit of a polygon
All polygons can be broken up into triangles
Convex, concave, complex
Triangles are guaranteed to be:
Planar
Convex
What exactly does it mean to be convex?
№5 слайд
Содержание слайда: Convex Shapes
A two-dimensional shape is convex if and only if every line segment connecting two points on the boundary is entirely contained.
№6 слайд
Содержание слайда: Convex Shapes
Why do we want convex shapes for rasterization?
One good answer: because any scan line is guaranteed to contain at most one segment or span of a triangle
Another answer coming up later
Note: Can also use an algorithm which handles concave polygons. It is more complex than what we’ll present here!
№7 слайд
Содержание слайда: Decomposing Polys Into Tris
Any convex polygon can be trivially decomposed into triangles
Draw it
Any concave or complex polygon can be decomposed into triangles, too
Non-trivial!
№8 слайд
Содержание слайда: Rasterizing Triangles
Interactive graphics hardware commonly uses edge walking or edge equation techniques for rasterizing triangles
Two techniques we won’t talk about much:
Recursive subdivision of primitive into micropolygons (REYES, Renderman)
Recursive subdivision of screen (Warnock)
№9 слайд
Содержание слайда: Recursive Triangle Subdivision
№10 слайд
Содержание слайда: Recursive Screen Subdivision
№11 слайд
Содержание слайда: Edge Walking
Basic idea:
Draw edges vertically
Fill in horizontal spans for each scanline
Interpolate colors down edges
At each scanline, interpolate
edge colors across span
№12 слайд
Содержание слайда: Edge Walking: Notes
Order vertices in x and y
3 cases: break left, break right, no break
Walk down left and right edges
Fill each span
Until breakpoint or bottom vertex is reached
Advantage: can be made very fast
Disadvantages:
Lots of finicky special cases
Tough to get right
Need to pay attention to fractional offsets
№13 слайд
Содержание слайда: Edge Walking: Notes
Fractional offsets:
Be careful when interpolating color values!
Also: beware gaps between adjacent edges
№14 слайд
Содержание слайда: Edge Equations
An edge equation is simply the equation of the line containing that edge
Q: What is the equation of a 2D line?
A: Ax + By + C = 0
Q: Given a point (x,y), what does plugging x & y into this equation tell us?
A: Whether the point is:
On the line: Ax + By + C = 0
“Above” the line: Ax + By + C > 0
“Below” the line: Ax + By + C < 0
№15 слайд
Содержание слайда: Edge Equations
Edge equations thus define two half-spaces:
№16 слайд
Содержание слайда: Edge Equations
And a triangle can be defined as the intersection of three positive half-spaces:
№17 слайд
Содержание слайда: Edge Equations
So…simply turn on those pixels for which all edge equations evaluate to > 0:
№18 слайд
Содержание слайда: Using Edge Equations
An aside: How do you suppose edge equations are implemented in hardware?
How would you implement an edge-equation rasterizer in software?
Which pixels do you consider?
How do you compute the edge equations?
How do you orient them correctly?
№19 слайд
Содержание слайда: Using Edge Equations
Which pixels: compute min,max bounding box
Edge equations: compute from vertices
Orientation: ensure area is positive (why?)
№20 слайд
Содержание слайда: Computing a Bounding Box
Easy to do
Surprising number of speed hacks possible
See McMillan’s Java code for an example
№21 слайд
Содержание слайда: Computing Edge Equations
Want to calculate A, B, C for each edge from (xi, yi) and (xj, yj)
Treat it as a linear system:
Ax1 + By1 + C = 0
Ax2 + By2 + C = 0
Notice: two equations, three unknowns
Does this make sense? What can we solve?
Goal: solve for A & B in terms of C
№22 слайд
Содержание слайда: Computing Edge Equations
Set up the linear system:
Multiply both sides
by matrix inverse:
Let C = x0 y1 - x1 y0 for convenience
Then A = y0 - y1 and B = x1 - x0
№23 слайд
Содержание слайда: Computing Edge Equations:
Numerical Issues
Calculating C = x0 y1 - x1 y0 involves some numerical precision issues
When is it bad to subtract two floating-point numbers?
A: When they are of similar magnitude
Example: 1.234x104 - 1.233x104 = 1.000x101
We lose most of the significant digits in result
In general, (x0,y0) and (x1,y1) (corner vertices of a triangle) are fairly close, so we have a problem
№24 слайд
Содержание слайда: Computing Edge Equations:
Numerical Issues
We can avoid the problem in this case by using our definitions of A and B:
A = y0 - y1 B = x1 - x0 C = x0 y1 - x1 y0
Thus:
C = -Ax0 - By0 or C = -Ax1 - By1
Why is this better?
Which should we choose?
Trick question! Average the two to avoid bias:
C = -[A(x0+x1) + B(y0+y1)] / 2
№25 слайд
Содержание слайда: Edge Equations
So…we can find edge equation from two verts.
Given three corners C0, C1, C0 of a triangle, what are our three edges?
How do we make sure the half-spaces defined by the edge equations all share the same sign on the interior of the triangle?
A: Be consistent (Ex: [C0 C1], [C1 C2], [C2 C0])
How do we make sure that sign is positive?
A: Test, and flip if needed (A= -A, B= -B, C= -C)
№26 слайд
Содержание слайда: Edge Equations: Code
Basic structure of code:
Setup: compute edge equations, bounding box
(Outer loop) For each scanline in bounding box...
(Inner loop) …check each pixel on scanline, evaluating edge equations and drawing the pixel if all three are positive
№27 слайд
Содержание слайда: Optimize This!
findBoundingBox(&xmin, &xmax, &ymin, &ymax);
setupEdges (&a0,&b0,&c0,&a1,&b1,&c1,&a2,&b2,&c2);
/* Optimize this: */
for (int y = yMin; y <= yMax; y++) {
for (int x = xMin; x <= xMax; x++) {
float e0 = a0*x + b0*y + c0;
float e1 = a1*x + b1*y + c1;
float e2 = a2*x + b2*y + c2;
if (e0 > 0 && e1 > 0 && e2 > 0) setPixel(x,y);
}
}
№28 слайд
Содержание слайда: Edge Equations: Speed Hacks
Some speed hacks for the inner loop:
int xflag = 0;
for (int x = xMin; x <= xMax; x++) {
if (e0|e1|e2 > 0) {
setPixel(x,y);
xflag++;
} else if (xflag != 0) break;
e0 += a0; e1 += a1; e2 += a2;
}
Incremental update of edge equation values
(think DDA)
Early termination (why does this work?)
Faster test of equation values
№29 слайд
Содержание слайда: Edge Equations:
Interpolating Color
Given colors (and later, other parameters) at the vertices, how to interpolate across?
Idea: triangles are planar in any space:
This is the “redness”
parameter space
Note:plane follows form
z = Ax + By + C
Look familiar?
№30 слайд
Содержание слайда: Edge Equations:
Interpolating Color
Given redness at the 3 vertices, set up the linear system of equations:
The solution works out to:
№31 слайд
Содержание слайда: Edge Equations:
Interpolating Color
Notice that the columns in the matrix are exactly the coefficients of the edge equations!
So the setup cost per parameter is basically a matrix multiply
Per-pixel cost (the inner loop) cost equates to tracking another edge equation value
№32 слайд
Содержание слайда: Triangle Rasterization Issues
Exactly which pixels should be lit?
A: Those pixels inside the triangle edges
What about pixels exactly on the edge? (Ex.)
Draw them: order of triangles matters (it shouldn’t)
Don’t draw them: gaps possible between triangles
We need a consistent (if arbitrary) rule
Example: draw pixels on left or top edge, but not on right or bottom edge
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