rasterisation - university of exeterempslocal.ex.ac.uk/people/staff/pjk205/ecm2410/slides07.pdf ·...

ECM2410: Graphics and Animation

Rasterisation

Part I: Simple Lines

Rasterisation 1/16

Rendering a scene

User space Device space

Compose Render

Affine transformation

Transform

Compose from primitives (lines, ellipses, etc) in user space.

Transform affine transformations:lines→ lines;ellipses→ ellipses

Render primitives on device.Rasterisation Line Rasterisation 2/16

Rendering a scene

Compose Render

Transform

Rendering a scene

Compose Render

Transform

Rendering a scene

Compose Render

Transform

Rendering a scene

Compose Render

Transform

Rasterisation

Vector displaysPen plottersTektronix cathode ray tubes

Raster displaysMonitorsTelevisionsLaser, ink-jet, dot matrix printersLiquid crystal displaysArrays of LEDs

Rasterisation is the conversion of ideal, mathematicalgraphics primitives onto raster displays.

Drawing linesDrawing circles and ellipsesFilling areas

Hearn & Baker, Chapter 2 for displays.Rasterisation Line Rasterisation 3/16

Rasterisation

Framebuffer

Raster display is a 2D array of picture elements: pixelsFramebuffer: in-memory image of the display.

Screen coordinates of pixelsare centered at the integers.

Pixels are individually set orcleared (random access).Sequential readout byphysical device.

Graphical objects arecontinuous.Framebuffer is discrete.

Rasterisation / Framebuffer Line Rasterisation 4/16

Framebuffer

Rasterisation

Rasterisationis the conversion of ideal, mathematical graphicsprimitives onto raster displays.

Efficiencywe seek efficient algorithms for computing thediscrete approximation to continuous description.

Scan Conversionrasterisation line by line of the raster display.

Rasterisation

Line Segments

Problem: Draw a line from (Xstart ,Ystart) to (Xend ,Yend )Simplified: Draw a line from (0, 0) to (Xend ,Yend )

(Xend > 0 and Yend > 0)

(19,4)

Grid marks pixel centres

Example: Line from (0, 0) to (Xend ,Yend )= (19, 4)

Lines Line Rasterisation 6/16

Line Segments

(19,4)

Line Segments

(19,4)

Line Segments

(19,4)

Line Segments

(19,4)

Line Segments

(19,4)

Line Segments

(19,4)

Line Segments

(19,4)

Which pixels to illuminate?

Illuminate every pixel the line intersects

Illuminate the vertically closest pixel

Lines / Which Pixel? Line Rasterisation 7/16

Gradient

Gradient of line

y = m x + c

m =Yend − Ystart

Xend − Xstart

Simplified line from (0, 0) to (Xend ,Yend ):

y = m x , m =Yend

Lines / Gradient Line Rasterisation 8/16

Gradient

Gradient of line

y = m x + c

m =Yend − Ystart

Xend − Xstart

y = m x , m =Yend

Gradient

Gradient of line

y = m x + c

m =Yend − Ystart

Xend − Xstart

y = m x , m =Yend

Gradient

Gradient of line

y = m x + c

m =Yend − Ystart

Xend − Xstart

y = m x , m =Yend

Gradient

Gradient of line

y = m x + c

m =Yend − Ystart

Xend − Xstart

y = m x , m =Yend

Gradient

Gradient of line

y = m x + c

m =Yend − Ystart

Xend − Xstart

y = m x , m =Yend

Gradient

Gradient of line

y = m x + c

m =Yend − Ystart

Xend − Xstart

y = m x , m =Yend

Naïve Midpoint Algorithm

Illuminate pixel vertically closest to the line

4/198/19 16/19

double m = Yend / Xend;int x = 0;double y = 0.0;

while ( x <= Xend ) {y = m * x;iy = round(y);setpixel(x, iy);x++;

DDA Digital Differential Analyser. Hearn & Baker, p86.

Floating point operations and round() are computationallyexpensive.

Lines / Naïve Algorithm Line Rasterisation 9/16

4/198/19 16/19

Remove round()

Line coherence: As x 7→ x + 1 then y 7→ y + m

4/1916/19

At each step: move right one pixeland up by 4

midpoint indicates pixel centreabove illuminated pixel.

When y crosses verticalmidpoint between two centresmove up and compute newmidpoint target.

double m = Yend / Xend;int x = 0;int iy = 0;double y = 0.0;double midpoint = 0.5;

while (x <= Xend) {setpixel(x, iy);x++;y += m;

if (y > midpoint) {iy++;midpoint += 1.0;

Lines / Eliminating Floats Line Rasterisation 10/16

Remove round()

4/1916/19

Remove round()

4/1916/19

Remove round()

4/1916/19

Remove round()

4/1916/19

Remove round()

4/1916/19

Remove round()

4/1916/19

Integers only

m =Yend

Xendy = m x =

Yend xXend

Multiply m, y and midpoint bytwice total change in x ;(i.e. by 2Xend )

m = 2 Yend y = 2 Yendx

19 19 57midpoint19 57

int m = 2 * Yend;int x = 0;int iy = 0;int y = 0;int midpoint = Xend;

if (y > midpoint) {iy++;midpoint += 2 * Xend;

Integers only

m =Yend

Xendy = m x =

Yend xXend

19 19 57midpoint19 57

Integers only

m =Yend

Xendy = m x =

Yend xXend

19 19 57midpoint19 57

Integers only

m =Yend

Xendy = m x =

Yend xXend

19 19 57midpoint19 57

Integers only

m =Yend

Xendy = m x =

Yend xXend

19 19 57midpoint19 57

Integers only

m =Yend

Xendy = m x =

Yend xXend

19 19 57midpoint19 57

Integers only

m =Yend

Xendy = m x =

Yend xXend

19 19 57midpoint19 57

Bresenham’s algorithm

Eliminate midpoint by adjusting y tostart at the midpoint and decreasingto 0.

3 −5 => 332511

Hearn & Baker, page 88

int m = 2 * Yend;int x = 0;int iy = 0;int y = Xend;

while (x <= Xend) {setpixel(x, iy);x++;y -= m;