Today: Viewport Coordinates Polygon Scan Conversion...

Lecture 8: 1 October 2002

Today:Viewport CoordinatesPolygon Scan Conversion & FillingSubtleties of Floating Point vs. IntegerShadow RulesHidden Surface EliminationSpatial Coherence

Thursday:Assignment 2 (* scene exhibition)Mysteries of homogeneous coordinates explainedAssignment 4 (Polygon fill with ivscan) posted

Friday:Assignment 3 (Polygon wireframe rendering) due 5pm

MIT 6.837 Computer Graphics Tuesday, 1 October 2002 (L8)

Framebuffer and Discrete Coordinates

Usually “RGB” or “24-bit” or “full color” framebuffer2D array of xres × yres pixels (picture elements)

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RGB colordepth...

depth

yres

xres

At each pixel can be stored:Color (3 × 8 bit RGB = 24 bits, sometimes 36)Depth (b = 24 or b = 32 bits: values 0..2b − 1)Other attributes: α, stencil, window ID, etc.

High-end machine may store > 200 bits per pixel!


Discretizing the Normalized View Volume

NDC parallelepiped is x = ±1, y = ±1, z = ±1Map this parallelepiped into 3D “device coordinates”

Normalized Device Coordinates 3D Screen Space

x1 x2

y1

y2

xres

yres

0 xres−1 0

yres −1

zres

0

2b −1

S

(−1, −1, −1)

(1, 1, 1)

y

x

z

x, y interval:range of values depends on framebuffer resolutiontypically specified as a rectangular “viewport”:

(x1, y1) to (x2, y2)z interval:

range of values depends on precision of (depth buffer)b bits: unsigned integer in range [0..2b − 1]


Viewport Transformation (first attempt)

Pixel centers at integer coordinates; pixel (0, 0) at lower leftEasy: map x ∈ [−1.. + 1] to xS ∈ [x1..x2],and map y ∈ [−1.. + 1] to yS ∈ [y1..y2]

x1 x2

y1

y2

xres

yres

0 xres−1 0

yres −1

Then:xS = x1 + x+1

2 (x2 − x1)

yS = y1 + y+12 (y2 − y1)

... Problem ?


Viewport Transformation (incorrect)

This transform maps NDC extreme points to pixel centers

xS = x1 +x + 1

2(x2 − x1); yS = y1 +

y + 1

2(y2 − y1)

Example:

xS(−1) = x1 +−1 + 1

2(x2 − x1) = x1; yS(−1) = y1 +

−1 + 1

2(y2 − y1) = y1

x1 x2

y1

y2

Pixels have non-zero extent,which we must take into account!


Viewport Transformation (corrected)

Accounting for pixel extents means mapping:x ∈ [−1.. + 1] to xS ∈ [x1 − 1

2..x2 + 12]

y ∈ [−1.. + 1] to yS ∈ [y1 − 12..y2 + 1

2]

0

h−1

0 w−1

x1 x2

y1

y2

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( x1 − 1/2, y1− 1/2 )

( x2 + 1/2, y2 + 1/2 )

xS = x1 − 0.5 +x + 1

2(x2 − x1 + 1)

yS = y1 − 0.5 +y + 1

2(y2 − y1 + 1)

Note: this is a floating point coordinate system!Depth (z ∈ [−1.. + 1]) is mapped linearly as:

zS =z + 1

2(2b − 1)


Viewport Transformation

Matrix formulation

0

h−1

0 w−1

x1 x2

y1

y2

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( x1 − 1/2, y1− 1/2 )

( x2 + 1/2, y2 + 1/2 )

S =

w2 0 0 (x1 + w−1

2 )

0 h2 0 (y1 + h−1

2 )0 0 r

2r2

0 0 0 1

where w = x2 − x1 + 1, h = y2 − y1 + 1, r = 2b − 1S takes NDC coordinates to (floating point) 3D Screen SpaceRounding to integer is deferred until framebuffer assignment


Subtlety of Viewport Mapping

What happens if aspect ratio of view frustum, viewport differ?

eye

projection plane

(left, bottom, −near)

−z

xy

z = −near

right−handed; view is along −z axis

(right, top, −near)

(right * f/n, top * f/n, −far)

far clip planez = −far

defined by 6 parameters: left, right, bottom, top, near, far

NDC

viewport 1

viewport 2


Preserving Aspect Ratio

Viewport “aspect ratio” is ratio of largest to smallest dimensionStrategies for preserving aspect ratio:

1. Assume field of view equal in x, yScale NDC into smaller of viewport dimensions

NDC

viewport 1

viewport 2w > h

w < h

2. Derive x, y fields of view from viewport width, height

w

h

How ?


Pipeline Transformations: Summary

Prepend S to existing matrix string:

xSySzS

= S P E MO

xOyOzO1

MO Modeling transformations (Object to World)(One matrix per object)

E Re-express World coordinates in Eye coord. sys.Translate, then Rotate (E = RT)

P Perspective transformation into NDCDevice-independent (produces canonical coords.)Projection from 4- to 3-vector happens after P

S Transformation into screen coordinatesDevice-dependent


Scan-Conversion Preliminaries

3D Polygon Scan ConversionInterpolation of Color, Depth

Conversion of continuous primitiveto its discrete pixel coverage

Color (depth, etc.) associated with verticesinterpolate across polygon interior

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f(A,B)

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f(A,C)

A

B

C

Step along left, right triangle edgesinterpolate color (depth, etc.) at pixels


Preliminaries: Linear Interpolation

Consider interpolation along line segment

P(α) = p + α ( q − p ) = (1 − α) p + α q q

p

Coefficients non-negative, sum to 1 ⇒ interpolationyields a convex combination of input samples

Can interpolate geometry this way:PointsVectors (e.g., Normals)

And other attributes as well:ColorsTransparencyTexture coordinates...

But, must take care to work in correct coordinate system


Gouraud Interpolation

Depth, color associated with polygon verticesDepth: z coordinate, from perspective transformColor: radiometry (shading), from lighting model

We will use nested interpolation algorithm:Interpolate polygon edges to find span endpoints

Interpolate span interiors to find pixel values

��IA

IB

IC

ys

xs

yA

yB

yC

y IP IAB IAC

xAB xP xAC

Terminology: Phong Shading, Gouraud Interpolation


Scan Converting One Triangle

Position (x, y) and intensity I given at each vertex A, B, C:

��IA

IB

IC

ys

xs

yA

yB

yC

y IP IAB IAC

xAB xP xAC

∆y, ∆I

∆x

FillPoly (xA, yA, IA, xB, yB, IB, xC, yC, IC) {∆XAB = xB−xA

yB−yA; ∆XAC = xC−xA

yC−yA

∆IAB = IB−IAyB−yA

; ∆IAC = IC−IAyC−yA

xAB = xAC = xA ; IAB = IAC = IA

For y ∈ yA..yB

FillSpan(y, xAB, IAB, xAC, IAC)

xAB += ∆XAB

xAC += ∆XAC

IAB += ∆IAB

IAC += ∆IAC

For y ∈ yB..yC

...

}


Pseudo code for Span Filling

XL, IL XR, IR

��

IA

IB

IC

ys

xs

yA

yB

yC

y IP XL, IL

xAB xP xAC

∆y, ∆I

∆x

XR, IR

FillSpan(y, xL, IL, xR, IR) {∆I = IR−IL

xR−xLI = IL + ∆I · (�xL� − xL)x = �xL�while ( x ≤ xR )

Raster.setPixel(x, I)x = x + 1I += ∆I

}


Recap: High-Level Strategy

1) Shade vertices2) Transform, project vertices into screen space3) Interpolate geometry, color along polygon boundaries4) Interpolate color in polygon interior

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f(A,B)

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f(A,C)

A

B

C


Pitfalls of Gouraud Interpolation

Consider a local light source illuminating a large planar areaWhat causes the “diffuse highlight” at left? (Hint: does -L vary?)

Why is highlight not produced at right?What’s the difference between the two figures?


Improving Gouraud Interpolation

Introduce more samples!

Or, call shader at every pixelWe’ll do this in Assignments 5 & 6 (ivcast and ivray)You can try it in ivscan if you wish


Screen Space Color Interpolation

Is color interpolation correct under perspective?


Screen Space Color Interpolation

No! Example: apparent motion of “gray point” or “halfway point”:

Causes phenomenon known as color “swimming” under rotationInterpolation of eye-space shading parameters in screen space:

I1

I2

(A+B)/2

A B

not ( I1+I2 )/2!

How can this be fixed? (Two ways.)


Hidden-Surface Elimination

So far, have scan-converted only one polygonIf polygons overlap, must eliminate hidden portions

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FillSpan() bottomed out to setPixel(x, I)Now, must arbitrate among competing spans

Several methods (trade reduced memory for increased time):Full-frame z bufferSingle-span z buffer – intermediate choiceNo z buffer – maintain ASL (Active Span List)


Visibility resolution with full-frame Z-buffer

Keep an array of depth values, one per pixel

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At start of frame, all depths initialized to “far away” (z = zmax)Now, scan convert each polygon independentlyWrite color value conditionally at each pixel:

If z < zstored thenupdate stored colorzstored = z

Advantages?Simple; Running time is linear (why?)


Z-buffer Memory Requirements

What are framebuffer memory requirements?1024 × 1280 = 11

4 MpixelsR, G, B, α channels 8 bits / pixel = 4 bytes / pixelDepth (z) values 32 bits / pixel = 4 bytes / pixelDouble-buffering doubles color storage requirementTotal: [5Mb (RGBα) + 5Mb (z)] ×2 = 20 Mb!

First framebuffer (512 × 512) cost $80,000 (late 1960’s)Even today, fast full-precision z-buffer & logic are expensiveAre there reasons not to use a full-frame z-buffer?


Z-buffering

Reasons not to use a full-frame z bufferOutput medium:

High-resolution anti-aliased film outputExample: 3000 × 3000 × 256 samples per pixel = 9Gb of depth memory

EfficiencyScene may have large number of invisible elementsWe’ll revisit this topic in Visibility lectures

Today, we’ll assume no full-frame z-bufferClassical techniques, lots of literatureIntroduction to visible-surface algorithmsLater in term, we’ll study “object-space visibility” algorithms


Scan-line Hidden Surface Algorithms

We’ll work with just one raster of framebuffer memory (color, z)“Sweep” horizontal scanline upward over scene (from raster 0 to h − 1):

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For now, assume polygons are convex, with no horizontal edgesExactly 0 or 2 edges at each scanlineEach polygon has in/out flag (used later)

Scan-line algorithm maintains two invariants:1. Data structure contains portion of scene intersected by

current scanline, ordered by x intercept of edges2. Renderer outputs all pixels on current scanline before proceeding


Initialization

Precompute: Edge Table (ET), one entry per scan line

0

23

1 BA BC

4567

ymax xint

∆x

color, ∆col

11 126 6

depth, ∆depth

next

.

.

.

.

.

.

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0123

4567

9

1011

8

0 1 2 3 4 5 6 7 9 10 118 12

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A C

B

D

F

E

12

13 14 15

Polygon ABC

poly

ED

112

.

.

.

DF

8

101112

9

121

.

.

.

AC

Polygon DEF

1114

.

.

.

EF

Each entry is a linked list of EdgeRecs, sorted by xint, containing:float yend: y of top edge endpointfloat xint, ∆x: current x intersection, delta wrt yfloat colcurr, ∆col: current color, delta wrt yfloat zcurr, ∆z: current depth, delta wrt yPointer to polygon from which edge came (for matching left, right edges)next: pointer to next record, or NULL

At start of frame, insert all edges into ET... How long does precomputation take in total?


Initialization (cont.)

Input: list of polygons, each specified as ordered list of edges

0

23

1 BA BC

4567

ymax xint

∆x

color, ∆col

11 126 6

depth, ∆depth

next

.

.

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.

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4567

9

1011

8

0 1 2 3 4 5 6 7 9 10 118 12

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A C

B

D

F

E

12

13 14 15

Polygon ABC

poly

ED

112

.

.

.

DF

8

101112

9

121

.

.

.

AC

Polygon DEF

1114

.

.

.

EF

Think of Edge Table as a bucketed list of “events”:ET is static data structureEvent: start/finish of an edge (i.e., vertex)

occurs within interval [y, y + 1)Within each bucket, events are sorted by x coordinate


Active Edge List

Active edge list (AEL) is a dynamic data structureIt is initially empty (scan line beneath bottom of viewport)

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A C

B

D

F

E

y < −0.5

y1

AEL at y1: BA DF BC EF

ymax xint

∆x

color, ∆col

4 12 13 52.. 6.. 7.. 11..

depth, ∆depth

next...

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.

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.

.

.

.

.

.

.

.

.

y > h−0.5

y sweep

It is incrementally maintained to storeall edges intersecting scanline, ordered by x

It is empty at termination (scan line above top of viewport)(This is a useful sanity check for your implementation)


When Does AEL Change State?

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A C

B

D

F

E

y + 1 y

When a vertex is encounteredI.e., when an edge begins or endsAll such events pre-stored in Edge Table!

How do y pre-sorting, x pre-sorting improve efficiency?When two edges change order along a scanline

I.e., when edges cross each other!How to detect this efficiently?


To be continued Thursday...


Today: Viewport Coordinates Polygon Scan Conversion...

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