Download ppt - Triangle Scan Conversion. 2 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Rasterization Rasterization (scan conversion) –Determine which

Triangle Scan Conversion

2 Angel: Interactive Computer Graphics 5E © Addison-Wesley

2009

Rasterization• Rasterization (scan conversion)

– Determine which pixels that are inside primitive specified by a set of vertices

– Produces a set of fragments– Fragments have a location (pixel location) and other

attributes such color and texture coordinates that are determined by interpolating values at vertices

• Pixel colors determined later using color, texture, and other vertex properties

Triangle Area Filling Algorithms

• Why do we care about triangles?

• Edge Equations

• Edge Walking

Do something easier!

• Instead of polygons, let’s do something easy!

• TRIANGLES! Why? 1) All polygons can be broken into triangles

2) Easy to specify

3) Always convex

4) Going to 3D is MUCH easier

Polygons can be broken down

Triangulate - Dividing a polygon into triangles.

Is it always possible? Why?

Any object can be broken down into polygons

Specifying a model

• For polygons, we had to worry about connectivity AND vertices.

• How would you specify a triangle? (What is the minimum you need to draw one?)– Only vertices

(x1,y1) (x2,y2) (x3,y3)

– No ambiguity– Line equations

A1x1+B1y1+C1=0 A2x2+B2y2+C2=0 A3x3+B3y3+C3=0

Triangles are always convex

• What is a convex shape?

An object is convex if and only if any line segment connecting two points on its boundary is contained entirely within the object or one of its boundaries. Think about scan lines again!

Scan Converting a Triangle

• Recap what we are trying to do

• Two main ways to rasterize a triangle– Edge Equations

• A1x1+B1y1+C1=0

• A2x2+B2y2+C2=0

• A3x3+B3y3+C3=0

– Edge Walking

Types of Triangles

What determines the spans? Can you think of an easy way to compute spans?

What is the special vertex here?

Edge Walking

• 1. Sort vertices in y and then x

• 2. Determine the middle vertex

• 3. Walk down edges from P0

• 4. Compute spans

P0

P1

P2

Edge Walking Pros and Cons

Pros• Fast• Easy to implement in

hardware

Cons• Special Cases• Interpolation can be

tricky

Color Interpolating

P0

P1

P2

(?, ?, ?)

(?, ?, ?)

Edge Equations

• A1x1+B1y1+C1=0

• A2x2+B2y2+C2=0

• A3x3+B3y3+C3=0

• How do you go from:

x1, y1 - x2, y2 toA1x1+B1y1+C1?

P0

P1P2

Given 2 points, compute A,B,C

0

0

11

00

CByAx

CByAx

0

0

11

00

C

C

B

A

yx

yx

C

C

B

A

yx

yx

11

00

1

1

11

00 CB

A

yx

yx

01

01

0110 xx

yy

yxyx

C

B

A

C = x0y1 – x1y0

A = y0 – y1

B = x1 – x0

Edge Equations• What does the edge

equation mean?

• A1x1+B1y1+C1=0

• Pt1[2,1], Pt2[6,11]

• A=-10, B=4, C=16

• What is the value of the equation for the: – gray part

– yellow part

– the boundary line

• What happens when we reverse P0 and P1?

P0

P1

Combining all edge equations

1) Determine edge equations for all three edges

2) Find out if we should reverse the edges

3) Create a bounding box

4) Test all pixels in the bounding box whether they too reside on the same side

P2

P1

P0

Edge Equations: Interpolating Color

• Given redness at the 3 vertices, set up the linear system of equations:

• The solution works out to:

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 (which is?)– A: 1 add

Pros and Cons of Edge Equations

• Pros• If you have the right

hardware (PixelPlanes) then it is very fast

• Fast tests• Easy to interpolate

colors

• Cons• Can be expensive if

you don’t have hardware

• 50% efficient

Recap

P0

P1

P2

P1

P2

P0


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Scan Conversion of Line Segments

• Start with line segment in window coordinates with integer values for endpoints

• Assume implementation has a write_pixel function

y = mx + h


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DDA Algorithm• Digital Differential Analyzer

– DDA was a mechanical device for numerical solution of differential equations

– Line y=mx+ h satisfies differential equation where m = y/x = y2-y1/x2-x1

• Along scan line x = 1 at each iteration of the loop

For(x=x1; x<=x2,ix++) { y+=m; write_pixel(x, round(y), line_color)}


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Problem

• DDA = for each x plot pixel at closest y– Problems for steep lines


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Using Symmetry

• Use for 1 m 0

• For m > 1, swap role of x and y– For each y, plot closest x


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Bresenham’s Algorithm

• DDA requires one floating point addition per step

• We can eliminate all fp through Bresenham’s algorithm

• Consider only 1 m 0

– Other cases by symmetry• Assume pixel centers are at half integers

• If we start at a pixel that has been written, there are only two candidates for the next pixel to be written into the frame buffer


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Candidate Pixels

1 m 0

last pixel

candidates


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Decision Variable

-

d = x(b-a)

d is an integerd > 0 use upper pixeld < 0 use lower pixel


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Incremental Form• More efficient if we look at dk, the value of

the decision variable at x = k

dk+1= dk –2y, if dk <0dk+1= dk –2(y- x), otherwise

•For each x, we need do only an integer addition and a test•Single instruction on graphics chips