computer graphics- scc 342 chapter 2: overview of graphics systems chapter 3: graphics output...

Computer Graphics- SCC 342

Chapter 2: Overview of Graphics SystemsChapter 3: Graphics Output Primitives

Chapter 2: Overview of Graphics Systems

2.1 Video Display Devices

2.2 Raster-Scan Systems

Overview of a graphics system

3

Overview of a graphics system

Input devices include Pointing/locator devices: indicate location on screen

– Mouse/trackball/spaceball

– Data tablet

– Joystick

– Touch pad and touch screen Keyboard device: send character input Choice devices: mouse buttons, function keys

4


5

Cathode-ray-tube (CRT)

CRT basics: The screen output is stored in the frame buffer and

is converted into voltages across the reflection plates

via a digital-to-analog converter (DAG) Light is emitted when electrons hit phosphor But light output from the phosphor decays

exponentially with time, typically in 10 – 60

microseconds

– Thus the screen needs to be redrawn or refreshed

– Refresh rate is typically 60 Hz to avoid flicker (“twinkling”)

– Flicker: when the eye can no longer integrate individual light pulses from a point on screen, e.g., due to low refresh rate

6


Shadow-mask color CRTs Three different colored phosphors (R, G, B) dots are arranged in very small groups (triads) on

coating

We see a mixture of three

colors

Three electron guns (R, G,

B) emit electron beams in a

controlled fashion so that

only phosphors of the

proper colors are excited

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Raster – Scan Display

8


The Frame Buffer Stores per-pixel information

– Depth of a frame buffer: number of bits per pixel

– E.g. for color representation, 1 bit => 2 colors,

8 bits => 256 colors,

24 bits => true color (16 million colors) Color buffer is only one of many buffers, other

information, e.g., depth, can also be used Implemented with special type of memory in

standard PCs or on a graphics card for fast

redisplay Part of standard memory in earlier systems

9


Raster-scan basics: The screen is a rectangular array

of picture elements, or pixels Resolution: determines the

details you can see

number of pixels in an image, e.g.,

1024×768, 1280x1024, 1366 x 768, etc. also in ppi or dpi – pixel or dot per inch

10

Raster-Scan Pattern: Horizontal scan rate: # scan lines per second Interlaced (TV) vs. non-interlaced displays

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Random-Scan Display Also called Vector-scan display Pictures are generated as line drawings

12


Simple raster-graphics system Video controller (display processor) controls

operations of display device.

13

Raster system with fixed portion of main memory reserved for the frame buffer.

14



Video controller

15


Raster – scan display processor

- Digitizes picture definition into a set of pixel values for storage in the frame buffer (scan conversion).

16


Additional functions for Raster – scan display processor

- generating line styles

- display color areas

- transformations

- interface with interactive devices.

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Chapter 3: Graphics Output Primitives

3.5 Line Drawing Algorithms

3.9 Circle- Generating Algorithms

3.13 Pixel Addressing and Object Geometry

3.14 Fill Area Primitives

3.15 Polygon Fill Area

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2D Graphics Primitives Graphics drawing is based on basic geometric

structures called graphics primitives Points Lines Circles Conic Sections


Drawing Points Input specification

Coordinates (x, y) of the point Color Specification

Procedure Write the required color value to the corresponding

position of the frame buffer Color can be specified depending upon the graphics hardware interface

e.g. RGB For a black and white screen whenever a value of 1 is encountered the

graphics system turns the corresponding pixel on

Function Call setPixel (x,y): To turn the pixel (x,y) on getPixel (x,y): To get the pixel value for (x,y)

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Drawing Points Output

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Drawing lines What is a line?

A set of connected points satisfying y = mx+c

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Δx

Δy

c

(x1,y1)

(x2,y2)

2 1

2 1

1 1

y y ym

x x x

c y mx


In graphics, usually we are given the end points (xa,ya) and (xb,yb) of the line for drawing

A simple algorithm Calculate the slope

Start with (xa,ya) k=1 and set xk+1=xk+1

Then find Put a pixel at (xk+1 , round(yk+1))

Continue till the other xb end point is reached

|m|<1 ensures that the points will be continuous

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1 1

b a

b a

k k k k

y y ym

x x x

y m x

y y m x x

(xa,ya)

(xb,yb)

1k ky y m

The digital differential analyzer (DDA) line drawing algorithm


For |m|>1 Gaps appear

Solution Use

Start with (xa,ya) and set yk+1=yk+1

Put a pixel at (round(xk+1) , yk+1)

Continue until the end point y2 is reached

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(xa,ya)

(xb,yb)

11

k kk k

y yx x

m

1

1k kx x

m


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(xa,ya)


DDA Characteristics Efficiency

Better than direct line drawing y=mx+c Eliminates Multiplication

But uses rounding off which is an expensive operation

Uses floating point operations Round off errors can cause the line to drift away

from the true line segment Especially for long lines Can be a problem for systems with low precision

numeric representation

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Round off Error Accumulation for DDA

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Drift caused by round-off error accumulation

Intended Line


The Bresenham Line Algorithm The Bresenham algorithm is

another incremental scan conversion algorithm

The big advantage of this algorithm is that it uses only integer calculations

Jack Bresenham worked for 27 years at IBM before entering academia. Bresenham developed his famous algorithms at IBM in the early 1960s

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Bresenham Algorithm: Main Idea Move across the x axis in unit intervals and at each

step choose between two different y coordinates to select the point closest to original line

2 3 4 5

2

4

3

5

(xk, yk)

(xk+1, yk)

(xk+1, yk+1)For example, from position (2, 3) we have to choose between (3, 3) and (3,

4)

We would like the point that is closer to the original line

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At sample position xk+1 the vertical separations from the mathematical line are labelled dupper and dlower

The y coordinate on the mathematical line at xk+1 is:

bxmy k )1(

y

yk

yk+1

xk+1

dlower

dupper

Our assumption here is that the line has a positive

slope less than one 30

Deriving The Bresenham Line Algorithm


So, dupper and dlower are given as follows:

and:

We can use these to make a simple decision about which pixel is closer to the mathematical line

klower yyd

kk ybxm )1(

yyd kupper )1(

bxmy kk )1(1

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Deriving The Bresenham Line Algorithm (cont…)


This simple decision is based on the difference between the two pixel positions:

Let’s substitute m with ∆y/∆x where ∆x and ∆y are the differences between the end-points:

122)1(2 byxmdd kkupperlower

)122)1(2()(

byxx

yxddx kkupperlower

)12(222 bxyyxxy kk

cyxxy kk 22Constant terms collected in ‘c’

Since in our case Δx is positive therefore this term has the same sign

as dlower-dupper

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So, a decision parameter pk for the kth step along a line is given by:

The sign of the decision parameter pk is the same as that of dlower – dupper

If pk is negative, then we choose the lower pixel, otherwise we choose the upper pixel

cyxxy

ddxp

kk

upperlowerk

22

)(

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Remember coordinate changes occur along the x axis in unit steps so we can do everything with integer calculations

At step k+1 the decision parameter is given as:

Subtracting pk from this we get:


cyxxyp kkk 111 22

)(2)(2 111 kkkkkk yyxxxypp

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But, xk+1 is the same as xk+1 so:

where yk+1 - yk is either 0 or 1 depending on the sign of pk

The first decision parameter p0 is evaluated at (x0, y0) is given as:


1 12 2 ( )k k k kp p y x y y

0 2p y x

0 0 0

0 0

0 0

0 0

2 2

2 2 2 (2 1)

2 2 2 (2 1)

2 2 2 (2 1)

2 2 2 (2 1)

2 2 2 2 (2 1)

2

k k k

k k

p y x x y c

y x x y y x b

p y x x y y x b

y x x m x b y x b

yy x x x b y x b

x

y x y x x b y x b

y x

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BRESENHAM’S LINE DRAWING ALGORITHM (for |m| < 1.0)

1. Input the two line end-points, storing the left end-point in (x0, y0)

2. Plot the point (x0, y0)

3. Find:

4. At each xk along the line, starting at k = 0, perform the following test. If pk < 0, the next point to plot is (xk+1, yk) and:

Otherwise, the next point to plot is (xk+1, yk+1) and:

5. Repeat step 4 (Δx – 1) times

xyp 20

ypp kk 21

xypp kk 221

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Bresenham Example Let’s have a go at this Let’s plot the line from (20, 10) to (30, 18) First off calculate all of the constants:

Δx: 10 Δy: 8 2Δy: 16 2Δy - 2Δx: -4

Calculate the initial decision parameter p0:

p0 = 2Δy – Δx = 6

37

17

16

15

14

13

12

11

10

18

292726252423222120 28 30

k pk (xk+1,yk+1)

0

1

2

3

4

5

6

7

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Bresenham Example


Generalization of the Bresenham Line Drawing Algorithm

For lines with slopes greater than 1 Inter-change the roles of the x and y directions

Take a unit step in y direction and calculate x values nearest to the line path

For lines with negative slopes Procedure is generally similar except that now one

coordinate decreases as the other increases Vertical, Horizontal and Diagonal lines

Can be drawn as special cases without processing by the line plotting method

Interchange of the starting and ending points can be checked

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Characteristics of Bresenham Line Drawing Algorithm

Uses only integer calculations More efficient Paves the way for making more complex curves

based upon a similar logic Take the screen pixel closest to the original curve

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Properties of circles: A Circle is the set of points that are all at a given

distance r (called radius) from a center point (xc,yc)

The equation for a circle is:

The circle is a frequently used component in pictures and graphics

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2 2 2c cx x y y r


A circle can be drawn by using the following equation by moving in unit steps from xc-r to xc+r and calculating the corresponding y value at each position

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22c cy y r x x


Problems with this approach Computations Spacing between points is not uniform

Reason Absolute value of the slope exceeds 1

Solution-1 Switch the roles of x and y when the absolute

value of slope exceeds 1 Increment y by one unit and calculate

corresponding x

Still computationally complex

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Problems… Solution-2

Use parametric polar form of the circle

Start with initial angle equal to zero and increment the angle to get and then plot the next point

The increment in the angle should be small (1/r) to ensure a continuous boundary or if a large step is used then the points can be connected by straight lines to approximate a circular boundary

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cos

sin

c

c

x x r

y y r


Reducing computational complexity Using 8-way symmetry

However, both the algorithms described earlier involve complex floating point calculations

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The Midpoint Circle Drawing Algorithm The midpoint circle algorithm has been

developed and patented by J. Bresenham Characteristics of MPCA

Incremental in nature Uses a decision parameter to find the pixel

closest to the circumference of the original circle

Uses (almost all) integer operations Uses 8-way symmetry to reduce computations

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Main Idea Assume we are drawing the

2nd octant of the circle Assume we have just plotted

point (xk,yk)

We have two choices for the next point (xk+1,yk) and (xk+1,yk-1)

This decision is made by calculating whether (xk+1,yk-1/2) is outside the circle or inside it

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Main Idea This can be done by using the circle function

based on the equation of the circle

Thus our decision parameter is

48

0,

( , ) 0,

0,circf x y

boundary circle theinside is ),( if yx

boundary circle on the is ),( if yx

boundary circle theoutside is ),( if yx

222 )21()1(

)21,1(

ryx

yxfp

kk

kkcirck


Main Idea We decide yk+1 on the basis of pk

pk+1 is given by,

49

1

if 0

1k k

kk

y py

y else

1 1 1

2 2 21

22 21

1( 1, )211 1 ( )2

11 1 2 1 2

k circ k k

k k

k k k

p f x y

x y r

x x y r

- 1


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2 2 2

22 21 1

22 21 1

2 2 2

22

1

2 22 2

1 1

2 21 1

1( 1) ( )2

11 1 2 1 2

11 1 2 1 2

1( 1) ( )2

1 12 1 1 ( )2 2

1 12 1 1 4 4

2 1 1

k k k

k k k k

k k k k k

k k

k k k

k k k k k

k k k k k

p x y r

p x x y r

p p x x y r

x r y

x y y

x y y y y

x y y y y

2 2k+1 k k+1 k k+1 k kp = p + y - y - y - y + 2 x +1 +1


The initial decision parameter can be obtained by evaluating the circle function at (x0,y0-1/2)=(0,r-1/2)

51r

rr

rfp circ

45

)21(1

)21,1(

22

0

2 21 1 1

1

1

1

1

- - -

if 0

1

2 1 2 1 i

2

f 0

1

1

2

1

k kk

k

k k k

k k k k k k

kk

k

k

k

y py

y else

p x y p

p x el

p p y y y y x

pse

If the redius is an integer then 5/4

can be rounded off to 1 to ensure

integer operations

- 1

The Mid-Point Circle Algorithm…

1. Input radius r and circle centre (xc, yc), then set the coordinates for the first point on the circumference of a circle centred on the origin as:

2. Calculate the initial value of the decision parameter as:

3. Starting with k = 0 at each position xk, perform the following test. If pk < 0, the next point along the circle centred on (0, 0) is (xk+1, yk) and:

),0(),( 00 ryx

rp 45

0

12 11 kkk xpp

52

The Mid-Point Circle Algorithm…

Otherwise the next point along the circle is (xk+1, yk-1) and:

4. Determine symmetry points in the other seven octants

5. Move each calculated pixel position (x, y) onto the circular path centred at (xc, yc) to plot the coordinate values:

6. Repeat steps 3 to 5 until x >= y

111 212 kkkk yxpp

cxxx cyyy

53


Mid-Point Circle Algorithm Example To see the mid-point circle algorithm in action

lets use it to draw a circle centred at (0,0) with radius 10

54

Mid-Point Circle Algorithm Example (cont…)

9

76543210

8

976543210 8 10

10 k pk (xk+1,yk+1) 2xk+1 2yk+1

0

1

2

3

4

5

655

Mid-Point Circle Algorithm Exercise

Use the mid-point circle algorithm to draw the circle centred at (0,0) with radius 15

56


For displaying objects, their mathematical points are converted into pixel coordinates.

For preserving object geometry, display objects to correspond to dimensions given by mathematical points.

Screen grid coordinstes

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Raster-scan algorthims refrenced the center of screen pixel position

Anothe refrencing method: pixel boundary

Area occubied by pixel at (x,y) = area of unit square with diagonally oppsite corners at (x,y) and (x+1,y+1)

Advatages:- Avoids half-integer pixel boundries- Facilitats precies object representation- Simplifies processing of many scan-

conversion steps.

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3.14 Fill Area Primitives

Most graphics routins require the fill area to specified as polygon

- their boundries are described by linear equations.

- most surface can be approximated by polygons (called surface tessellation)

59


Polygons: plane figure specified by three or more vertices and connected by edges

Has all vertices within plane- no edge crossing Polygon classification:

- Convex: interior angle = or less than 180 degrees

- Concave: interior angle greater than 180 degrees

Implementations for filling algorithms are complicated for concave polygons

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convexconcave


Identifying concave polygons

- extension of some edges will intersect other edges

- take a look at the polygon vertex positions relative to the extension line of any edge: some vertices are on one side and some are on the other

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- Split concave polygons into a set of convex polygons

- Using edge vectors and edge cross products

- Use vertex position relative to edge extension line.

62

Splitting concave polygons


63

Splitting concave polygons Assume all polygons in xy plane Ek = Vk+1 – Vk

Calculate cross product of successive edge vectors

- if at least z component of one vector is negative, then concave polygon.

- no successive vertices are collinear. Split along the line of the first edge vector in the

cross product pair.



64



65


Splitting concave polygons Another method is a rotational method

66


67

Splitting convex polygons into triangle (tessellation)

- define any sequence of three consecutive vertices to a triangle.

- delete middle vertex from the original list

- continue forming triangle

- stop when only three vertices are remaining.


Inside-Outside test Also called odd-even test

68

P

Q


Winding-Number test Count counterclockwise encirclements of point

69

P

Q


Polygon tables objects in a scene are described as sets of

polygon surface facets Object description includes:

- Coordinate information (geometry for the polygon facets)

- Surface parameters(color, transparency, and light-reflection properties)

Object information are organized in tables.

70


Polygon tables

71


Polygon tables Important for check for consistency and

completeness:

1. each pixel is an end point for at least two edges.

2. every edge is part of at least one polygon

3. every polygon is closed

4. each polygon has at least one shared edge

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computer graphics- scc 342 chapter 2: overview of graphics systems chapter 3: graphics output...

Documents

randomscan display

rasterscan pattern

vectorscan display pictures

scan lines

raster system

tube crt slide

frame buffer scan conversion

input devices