polygon scan conversion

Polygon Scan Conversion

Instructor: Paul Merrell

2

Rasterization

• Rasterization takes shapes like triangles and determines which pixels to fill.

3

Filling Polygons

First approach:

1. Polygon Scan-Conversion• Rasterize a polygon scan line by scan line,

determining which pixels to fill on each line.

4

Filling Polygons

Second Approach:

2. Polygon Fill• Select a pixel inside the polygon. Grow outward

until the whole polygon is filled.

5

Why Polygons?

• You can approximate practically any surface if you have enough polygons.

• Graphics hardware is optimized for polygons (and especially triangles)

6

Coherence

Scan-line

Edge

Span

7


Intersection Points

Other points in the span

8


• Process each scan line1. Find the intersections of the scan line with all

polygon edges.

2. Sort the intersections by x coordinate.

3. Fill in pixels between pairs of intersections using an odd-parity rule.

- Set parity even initially.- Each intersection flips the

parity.- Draw when parity is odd.

9

Polygon Scan-Conversion

• Process for scan converting polygons• Process one polygon at a time.• Store information about every polygon edge.• Compute spans for each scan line.• Draw the pixels between the spans.

• This can be optimized using an table that stores information about each edge.

• Perform scan conversion one scan line at a time.

10

Computing Intersections

• For each scan line, we need to know if it intersects the polygon edges.

• It is expensive to compute a complete line-line intersection computation for each scan line.

• After computing the intersection between a scan line and an edge, we can use that information in the next scan line.

11

Scan Line Intersection

Currentscan line

yi+1

Polygon Edges

Intersection points from

previous scan line

Intersection pointsneeded

Previousscan line

yi

12

Scan Line IntersectionUse edge coherence to incrementally update the

x intersections.

We know

Each new scan line is 1 greater in y

We need to compute x for a given scan line,

min@max@

minmax ,yy xx

yymbmxy

11 ii yy

m

byx

(x@ymax,ymax)

(x@ymin,ymin)

13

Scan Line Intersection

So,

and

then

This is a more efficient way to compute xi+1.

m

byx

m

byx i

ii

i

11 ,

11 ii yy

mx

mm

by

m

byx i

iii

1111

14

Edge Tables

• We will use two different edge tables:• Active Edge Table (AET)

• Stores all edges that intersect the current scan line.

• Global Edge Table (GET)• Stores all polygon edges.• Used to update the AET.

15

Active Edge Table

• Table contains one entry per edge intersected by the current scan line.

• At each new scan line:• Compute new intersections for all edges using the

formula.• Add any new edges intersected.• Remove any edges no longer intersected.

• To efficiently update the AET, we will can use a global edge table (GET)

mxx ii

11

16

Global Edge Table Example

GET

0

1

2

3

4

5

6

7

8

Indexed by

scan line

AB EA

(ymax, x@ymin, 1/m)

0 1 2 3 4 5 6 7 8

0

1

2

3

4

5

6

7

8

A

B

C

D

E

Place entries into the GET

based on the ymin values.

8, 6, -3/2

6, 3, 3

8, 0, ¾

5, 2, ¼ 4, 2, -2/3

AB = (2, 1), (3, 5)BC = (3, 5), (6, 6)CD = (6, 6), (3, 8)DE = (3, 8), (0, 4)EA = (0, 4), (2, 1)

DE

CD

BC

17

Active Edge Table Example

• The active edge table stores information about the edges that intersect the current scan line.

• Entries are• The ymax value of the edge

• The x value where the edge intersects the current scan line.

• The x increment value (1/m)

• Entries are sorted by x values.

18


AETScan Line 3

1. The ymax value for that edge2. The x value where the scan line intersects

the edge.3. The x increment value (1/m)

0 1 2 3 4 5 6 7 8

0

1

2

3

4

5

6

7

8

A

B

C

D

E

AB = (2, 1), (3, 5)BC = (3, 5), (6, 6)CD = (6, 6), (3, 8)DE = (3, 8), (0, 4)EA = (0, 4), (2, 1)

AB

4, 2/3, -2/3 5, 5/2, 1/4

EA

(ymax, x, 1/m)

19


AETScan Line 4



Scan Line 4

ymax = 4 for EA, so it gets removed from the AET.

New x value for AB is 5/2 + 1/4 = 11/4

In the GET, edge DE is stored in bucket 4,

so it gets added to the AET.

0 1 2 3 4 5 6 7 8

0

1

2

3

4

5

6

7

8

A

B

C

D

E

AB = (2, 1), (3, 5)BC = (3, 5), (6, 6)CD = (6, 6), (3, 8)DE = (3, 8), (0, 4)EA = (0, 4), (2, 1)

AB

4, 2/3, -2/3 5, 5/2, 1/4

EA

(ymax, x, 1/m)

8, 0, 3/4

DE

5, 11/4, 1/4

20


AETScan Line 5



0 1 2 3 4 5 6 7 8

0

1

2

3

4

5

6

7

8

A

B

C

D

E

AB = (2, 1), (3, 5)BC = (3, 5), (6, 6)CD = (6, 6), (3, 8)DE = (3, 8), (0, 4)EA = (0, 4), (2, 1)

ABDE

(ymax, x, 1/m)

8, 0, 3/4 5, 11/4, 1/48, 3/4, 3/4

Increment x = 0 + 3/4 Remove AB,since ymax = 5.

Add BC since it is in the GET at ymin = 5.

6, 3, 3

BC

21


Scan Line 6

AB = (2, 1), (3, 5)BC = (3, 5), (6, 6)CD = (6, 6), (3, 8)DE = (3, 8), (0, 4)EA = (0, 4), (2, 1)

D

0 1 2 3 4 5 6 7 8

0

1

2

3

4

5

6

7

8

A

B

C

E

Clipping

23

Line Clipping

What happens when one or both endpoints of a line segment are not inside the specified drawing area?

DrawingArea

24

Line Clipping

• Strategies for clipping:a) Check (in inner loop) if each point is inside

Works, but slow

b) Find intersection of line with boundary Correct

if (x ≥ xmin and x ≤ xmax and y ≥ ymin and y ≤ ymax) drawPoint(x,y,c);

Clip line tointersection

25

Line Clipping: Possible Configurations

1. Both endpoints are inside the region (line AB)• No clipping necessary

2. One endpoint in, oneout (line CD)

• Clip at intersection point

3. Both endpoints outsidethe region:

a. No intersection (lines EF, GH)

b. Line intersects the region (line IJ)• Clip line at both intersection points

A

B

C

D

F

E

I

J

G

H

26

Line Clipping: Cohen-Sutherland

• Basic algorithm:• Accept lines that have both

endpoints inside the region.

F

E

Trivially reject

A

BTrivially accept

H

C

D

I

J

G

Clip andretest

Clip the remaining lines at a region boundary and repeat the previous steps on the clipped line segments.

Reject lines that have both endpoints less than xmin or ymin or greater than xmax or ymax.

27

Cohen-Sutherland: Accept/Reject Tests

• Assign a 4-bit code to each endpoint c0, c1 based on its position:

• 1st bit (1000): if y > ymax

• 2nd bit (0100): if y < ymin

• 3rd bit (0010): if x > xmax

• 4th bit (0001): if x < xmin

• Test using bitwise functionsif c0 | c1 = 0000

accept (draw)else if c0 & c1 0000

reject (don’t draw)else clip and retest

01000101 0110

10001001 1010

0001 00100000

28

• Accept/reject/redo all based on bit-wise Boolean ops.

Cohen-Sutherland Accept/Reject

01000101 0110

10001001 1010

0001 00100000

29

Polygon Clipping

What about polygons?

30

Polygon Clipping: ExampleClip first to ymin and ymax

ymax

ymin

v0

v1v2

v3v4

v5

v6v7

v8

v9

Input vertex list: (v0, v1, v2, v3, v4, v5, v6, v7, v8, v9)

vertex: v0

Inside region: No

Add p0 to output list

Output vertex list:

p0

Line intersect boundary: Yes

p0

31


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v1

inside region: yes

add v1 to output list

Output vertex list:

p0

line intersect boundary: no

, v1

ymax

ymin

p0

32


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v2

inside region: yes

add v2, p1 to output list

Output vertex list:

p0

, v2, p1

line intersect boundary: yes

p1

ymax

ymin

p0 , v1

33


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v3

inside region: no

Output vertex list:

p0

p0, v1, v2, p1


p1

ymax

ymin

34


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v4

inside region: no

add p2 to output list

Output vertex list:

p0

, p2

p1


p2

ymax

ymin

p0, v1, v2, p1

35


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v5

inside region: yes


Output vertex list:

p0

p0, v1, v2, p1, p2

p1

p2


p3

ymax

ymin

, v5, p3

36


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v6

inside region: no

Output vertex list:

p0

p0, v1, v2, p1, p2, v5, p3

p1

p2


p3

ymax

ymin

37


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v7

inside region: no

Output vertex list:

p0

p0, v1, v2, p1, p2, v5, p3

p1

p2


p3

ymax

ymin

38


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v8

inside region: no

add p4 to output list

Output vertex list:

p0

p0, v1, v2, p1, p2, v5, p3

p1

p2

p3


p4

ymax

ymin

, p4

39


v0

v1v2

v3v4

v5

v6v7

v8

v9


vertex: v9

inside region: yes


Output vertex list:

p0

p0, v1, v2, p1, p2, v5, p3, p4

p1

p2

p3p4


p5

ymax

ymin

, v9, p5

40

Polygon Clipping: ExampleThis gives us a new polygon

v1v2

v5

v9

with vertices:

p0

(p0, v1, v2, p1, p2, v5, p3, p4, v9, p5)

p1

p2

p3p4 p5

ymax

ymin

41

Polygon Clipping: Example (cont.)Now clip to xmin and xmax

xmin xmax

Input vertex list: = (p0, v1, v2, p1, p2, v5, p3, p4, v9, p5)

Output vertex list: (p0, p6, p7, v2, p1, p8, p9, p3, p4, v9, p5)

v1v2

v5

v9

p0

p1

p2

p3p4 p5

p6

p7

p8

p9

42

Polygon Clipping: Example (cont.)Now post-process

xmin xmax

v9

v3

Output vertex list: (p0, p6, p7, v2, p1, p8, p9, p3, p4, v9, p5)

Post-process: (p0, p6, p9, p3,) and (p7, v2, p1, p8) and (v4, v9, p5)

p8

p6

v2

p7

p0

p5 p3p4

p9

p1

Transformations

44

General Transformations

• Want to be able to manipulate graphical objects:• Translation: move an object• Rotation: change orientation• Scale: change size• Other transformations: reflection, shear, etc.

• A general 2-D transformation has the form:x´ = fx(x, y)

y´ = fy(x, y)

where x and y are the original coordinates and x´ and y´ are the transformed coordinates.

• In vector form:

y

x

y

x

f

f

y

xpp

p

pp and where

)(

)(

45

General Transformations (cont.)

• General transforms may be non-linear:• Lines do not necessarily map to lines• Every point (along lines, inside shapes, etc.) needs to be transformed• Non-linear Transformations are harder to deal with.• Fortunately, many important transformations are linear.• Non-Linear Example:

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

322

14

22

2

xyy

xyx

x

46

Linear Transformations

• We will use linear transformations:

x´ = fx(x, y) = axx + bxy + cx

y´ = fy(x, y) = ayx + byy + cy

• Linear Transforms can be written using matrix operations:

• Advantages:• Lines transform to lines• Only need to transform vertices.• Computationally efficient• Problem: would like to simplify further get rid of T

TMppp

y

x

yy

xx

c

c

ba

ba

47

Translation

Move object from one place to another:• Forward transform:

x´ = x + tx

y´ = y + ty

or p´ = p + T where

• Inverse transform: p = p´ – T

y

x

t

tT

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

3

4

T

48

Scale

Change an object’s size:• Forward transform:

x´ = sx x

y´ = sy y

or p´ = Sp where

• Inverse transform: p = S-1p´

where

y

x

s

s

0

0 S

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

20

02 S

y

x

s

s

1

11

0

0 S

Why do we get anapparent translation?

49

Scale

• Properties of the scale:• The scale is performed relative to the origin.• A scale factor greater than one enlarges the

objects and moves it away from the origin.• A scale factor less than one shrinks the object and

moves it towards the origin.

• Usually this isn’t what we want• Generally we would like to scale about the object’s

center - not the coordinate origin.

50

Scale can accomplish the following transforms:• Uniform scale: sx = sy

• Preserves angles, but not lengths• Non-uniform scale: sx sy

• Doesn’t preserve angles or lengths

• Reflection about the

x-axis: sx = 1, sy = –1

y-axis: sx = –1, sy = 1

line y = –x: sx = –1, sy = –1• Reflection preserves angles

and lengths• What is the inverse matrix?• A reflection matrix is its own

inverse.

Scale: Transforms

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

20

02 S

20

05.0 S

y = –x

10

01 S

10

01 S

10

01 S

51

Rotation

Given a point p on the plane, how do we rotate that point about the origin?

y

xp

22 yxr

x

yarctan

)sin(

)cos(

r

r

y

xp

r

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

p = (x, y)x = r cos()

y = r sin()

)sin(

)cos(

ry

rx

1. Convert point p to polar coordinates:

52

Rotation (cont)

Now, let’s rotate that point by about the origin

)sin(

)cos(

r

r

y

xp

r

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

p = (x, y)

)cos()sin()cos()sin(

)sin()sin()cos()cos(

rr

rr

y

xp

y

x

yx

yx

)cos()sin(

)sin()cos(

)cos()sin(

)sin()cos(p

p´= (x’, y’)

r

4. Substitute x and y back in:

3. Apply the sum of angles formula

2. To rotate degrees, simply add to :

)sin(

)cos(

ry

rx

53

Rotation (cont)

So, given a point p that we wish to rotate about the origin, we simply multiply p by the rotation matrix

)cos()sin(

)sin()cos(

54

Apply the rotation to each vertex of the object• Forward transform:

x´ = x cos() – y sin()y´ = x sin() + y cos()

or p´ = Rp where

• Inverse rotation: p = R-1p´

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

Rotating an Object

)θcos()θsin(

)θsin()θcos( R

T

1

)θcos()θsin(

)θsin()θcos(

)θcos()θsin(

)θsin()θcos(

R

R

= 45°

55

Rotations• Right-hand rule: Stick your thumb

towards the z-axis. Your fingers rotate in the positive direction as they curl.

• If is positive, the rotation is counterclockwise about the origin.

r

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

p = (x, y)

p´= (x’, y’)

r

r

x

y

0 1 2 3 4 5 6 701

2

3

4

5

6

7

p = (x, y)

p´= (x’, y’)

• If is negative, the rotation is clockwise about the origin.

56

• How can we do more complex transformations?• Scale an object in an arbitrary direction (other than in x and/or

y)• Rotate an object about a point (other than the origin)• Etc.

• Composition: combine basic transforms to achieve a more complex one

• Example: Scale by ½ at a 30° angle• Rotate clockwise 30° ( = – / 6): p´ = RTp• Scale in x by ½: p´´ = Sp´• Rotate counter-clockwise 30° ( = / 6): p´´´ = Rp´´• Final result: p´´´ = Rp´´ = RSp´ = RSRTp

Composition

Recall that a negative rotation matrix is the transpose of its corresponding positive rotation matrix.

57

Composition: Scale at Arbitrary Angle

Scale object at an arbitrary angle :

1. Rotate clockwise by : p´ = RTp

2. Scale about origin: p´ = SRTp

3. Rotate back (counter-clockwiseby ): p´ = RSRTp

10

0 xs

S

)θcos()θsin(

)θsin()θcos( TR

)θcos()θsin(

)θsin()θcos( R

x

y

0 1 2 3 401

2

3

4

5

2/31/2-

1/22/3

)6/cos()6/sin(

)6/sin()6/cos( TR

= 30°

x

y

0 1 2 3 401

2

3

4

5

x

y

0 1 2 3 401

2

3

4

5

= 30°

10

05.0 S

58

• Multiple transformations: Multiply each point (i.e., vertex) in an object by each basic transformation

p´ = RSRTp

• Requires multiple matrix multiplies per vertex

• Lots of extra computation if transforming thousands (or millions) of points.

• Composition: matrix multiply is associative• Multiply matrices first to create a single matrix• Multiply each point by a single matrix

p´ = RSRTp = M p

• Much more efficient

Composition and Efficiency

59

Rotation about an Arbitrary Point

• Our rotations (so far) have been about the origin

• This rotation “moves” the entire object as it rotates

• Usually when we want to rotate an object, we want to rotate it about its center (or some other fixed point)

• How can we do this?• Answer: Apply a sequence of transformations

60

Composition: Rotation About a Point

Rotate object about an arbitrary point T:

1. Translate T to origin: p´ = p – T

2. Rotate about origin: p´ = R(p – T)

3. Translate back: p´ = R(p – T) + T

)θcos()θsin(

)θsin()θcos( R

= 45°

y

x

t

tT x

y

0 1 2 3 401

2

3

4

5

x

y

0 1 2 3 401

2

3

4

5

2

2

T

x

y

0 1 2 3 401

2

3

4

5

61

Rotation about an Arbitrary Point

• Standard format for rotating about an arbitrary point:

• Translate point of rotation to the origin• Rotate about the origin• Translate the point back to its original position

• Compute the composite transformation that will accomplish this, then apply that transformation to all vertices in the object.

62

Homogeneous Coordinates

• Problem:• Rotation, scale, and shear multiply a matrix with p: p´ = Mp • Translation adds a vector to p: p´ = p + T • Would like to treat all transformations the same

• Optimize the hardware• Compose transformations

• Solution: homogeneous coordinates• Increase point’s dimensionality add a third coordinate w:

• Two homogeneous points (p1 and p2) specifythe same 2-D Cartesian point if:

p1 = cp2 for some real-valued scalar number c

w

y

x

p

63

Homogeneous Coordinates (cont.)

• With homogeneous coordinates, the x-y plane is a two-dimensional sub-space in 3-D

• Although homogeneous points have three coordinates, they correlate to positions on a 2-D plane

• Can use any 2-D plane that doesn’tinclude the origin

• For simplicity, choose the planew = 1 [x y 1]T = [X Y 1]T

x

y

w

0

0

0

1

0

0

w

wy

wx

y

x

y

x

2

2

2

1

11

/

/

Y

X

wy

wx

w

y

x

Homogeneous point

2-D Cartesiancoordinates

64

Homogeneous Coordinates

• For 2D transformations, the points are now 3-vectors

• And the transformation matrices are 3x3 matrices

w

y

x

p

333231

232221

131211

ttt

ttt

ttt

T

65

Homogeneous Coordinates: Translation

• What does the translation matrix look like?• To translate a point p to point p’, we need

with

and

100

10

01

,

1

y

x

t

t

Ty

x

p

' Tpp

11

'

'

' y

x

ty

tx

y

x

p

66

Homogeneous Coordinates: Scale

• What does the scale matrix look like?• To scale a point p to point p’, we need

with

and

,

1

y

x

p

' Spp

11

'

'

' yS

xS

y

x

p y

x

100

00

00

y

x

S

S

S

67

Homogeneous Coordinates: Rotation

• Rotation has a similar derivation. Remember the 2D rotation matrix:

• When we move to homogeneous coordinates, we have

• We apply this matrix to each vertex of a polygon to rotate the polygon as a whole

100

0cossin

0sincos

R

θcosθsin

θsinθcos R

68


Now let’s try that rotation about an arbitrary point T:1. Translate to origin: p´ = T1p

2. Rotate about origin: p´ = RT1p

3. Translate back: p´ = T2RT1p

100

0)54cos()54sin(

0)54sin()54cos(

R

= 45°

100

210

201

T

x

y

0 1 2 3 401

2

3

4

5

x

y

0 1 2 3 401

2

3

4

5

x

y

0 1 2 3 401

2

3

4

5

100

210

201

T

100

210

201

T

69


• So

• Where M is the composite transformation matrix

Mp

pRTTp

12

100

828.707.707.

2707.707.

100

210

201

100

2707.707.

2707.707.

100

210

201

100

0)54cos()54sin(

0)54sin()54cos(

100

210

201

12RTTM

70

3D Transformations

111000

100

010

001

z

y

x

z

y

x

tz

ty

tx

z

y

x

t

t

t

Tpp

111000

000

000

000

zs

ys

xs

z

y

x

s

s

s

z

y

x

z

y

x

Spp

Mpp

11000

z

y

x

dcba

dcba

dcba

zzzz

yyyy

xxxx

General Case:

Translation: Scale:

71

3D Rotation

Rotate about x-axis Rotate about y-axis

Rotate about z-axis

1000

0100

00cossin

00sincos

zR

1000

0cos0sin

0010

0sin0cos

yR

1000

0cossin0

0sincos0

0001

xR

These rotations can be combined to rotate about an arbitrary axis.

Viewing

73

Graphics PipelineModel Space

World Space

Eye/CameraSpace

Screen Space

Model Transformations

Viewing Transformation

Projection & WindowTransformation

74

Viewing Transformations• Projection: take a point from m dimensions to n

dimensions where n < m• There are essentially two types of viewing transforms:

• Orthographic: parallel projection• Points project directly onto the view plane• In eye/camera space (after viewing

transformation): drop z

• Perspective: convergent projection• Points project through the origin

onto the view plane• In eye/camera space (after viewing

transformation): divide by z

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Projection Environment

x

y

z• The view plane is in the xy plane and passes through the origin

• The view direction is parallel to the z axis

• The eye or camera position is on the +z axis, a distance d from the origin.

• We will use a right-handed view system

d

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Parallel Projection

x

z

(x, y, z)

• So z’ = 0, x’ = x

• Looking down the y axis:

(x’, y’, z’)

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Parallel Projection• Thus, for parallel, orthographic projections,

x’ = x, y’ = y, z’ = 0

• So, to perform a parallel projection on an object, we can use matrix multiplication

Mpp '

1000

0000

0010

0001

M

What is M?i.e., we simply drop the z coordinate

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• Points project through the focal point (e.g., eyepoint) onto the view plane:

• Projection lines converge

Perspective Projection

x

y

z

Virtual Image Plane

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Perspective Projection Computation

zd

x

d

x

'

Looking down the y axis:

x

x’

d

d-z

By similar triangles:

zd

xdx

'

d

zdx

x

'

d

zx

x

1

'

view plane

eye

p = (x, y, z)

p’ = (x’, y’, z’)x

z

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zd

y

d

y

'

Looking down the x axis:

yy’

d

d-z

By similar triangles:

zd

ydy

'

d

zdy

y

'

d

zy

y

1

'

view planeeye

p = (x, y, z)

p’ = (x’, y’, z’)

y

z

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• So, we have

• what is z’?

• can we put this into matrix form?

d

zy

y

1

'

d

zx

x

1

'

0,

1,

1)',','( have weso ,0'

d

zy

d

zx

zyxz

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• So, the matrix that will give us the correct perspective is:

1

1000

0000

001

10

0001

1

z

y

x

d

z

d

z

Mbad

This works - what is the problem with it?

Answer: The entries in the matrix are point dependent!i.e., every point will have to have a different matrix

83

Perspective Projection Computation• Solution: use homogeneous points (remember them?)

• Our Cartesian point is:

• A homogeneous point that is equivalent to our desired Cartesian point is:

• can we come up with a matrix that gives us what we need, but is point independent?

d

zyx 10

d

z

d

zy

d

zx

1

0,

1,

1

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• We want:

11

0 z

y

x

ponm

lkji

hgfe

dcba

y

x

dz

0,0,0,1 so, , dcbadczbyaxx

0,0,1,0 so, , hgfehgzfyexy

0,0,0,0 so, ,0 lkjilkzjyix

1,1

,0,0 so, ,1 pd

onmpoznymxd

z

so,

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• So, the new matrix we get is:

11

00

0000

0010

0001

d

M per

This gives us correct results, and is point independent

polygon scan conversion

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