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CPE 333 : Computer CPE 333 : Computer GraphicsGraphics

มหาวิ�ทยาลัยเทคโนโลัย�มหาวิ�ทยาลัยเทคโนโลัย�พระจอมเกลั�าธนบุ�ร�พระจอมเกลั�าธนบุ�ร�

Dr. Natasha Dejdumrong

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Course DescriptionCourse Description • Line Drawing• Circle Drawing [ Ellipse Drawing ]• 2D and 3D Transformations• Line and Polygon Clipping techniques• 3D Viewing• Geometric Modeling: Representation of Curves and Surfaces, • Solid Modeling• Hidden Surface Removal Algorithms• Illumination and Shading models• Ray tracing

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Text Book and SupplementsText Book and Supplements

• Textbook– Hearn Baker, Computer Graphics with OpenGL,

3rd Ed., Pearson International Edition, 2004.

• Lecture Note• Powerpoints

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ScheduleSchedule

• Classroom: CB40805• Lecture Time: Monday 13:30 – 15:20 [AB]

Tuesday 14:30 – 16:20 [CD]• Lab Time: Monday 14:30 – 16:20 [AB]

Tuesday 16:30 – 17:20 [CD]

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Active LearningActive Learning

• Lecture: 2 Hours• Laboratory: 2 Hours• Teaching Assistant: Pitchaya Jamjun

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MarkingMarking

Midterm Examination 20%

Final Examination 30%

Assignments 40%

Class + Lab Attendance 10%

Total 100%

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Grading PolicyGrading Policy

• 80 or above A• 75-79 B+• 70-74 B• 65-69 C+• 60-64 C• 55-59 D+• 50-54 D• under 50 F

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Chapter 2Chapter 2

Scan Conversion

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OverviewOverview

• Point• Line• Circle• Ellipse

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Raster GraphicsRaster Graphics

A raster graphics image, digital image, or bitmap, is a data file or structure representing a generally rectangular grid of pixels, or points of color, on a computer monitor, paper, or other display device.

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Vector GraphicsVector Graphics

• Vector graphics (also called Geometric Modeling or Object-Oriented Graphics) is the use of geometrical primitives such as points, lines, curves, and polygons, which are all based upon mathematical equations to represent images in computer graphics. It is used by contrast to the term raster graphics, which is the

representation of images as a collection of pixels (dots).

• Raster graphics are distinguished from vector graphics in that vector graphics represent an image through the use of geometric objects such as curves and polygons.

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Raster vs Vector GraphicsRaster vs Vector Graphics

• Raster graphics cannot be scaled to a higher resolution without loss of apparent quality. This is in contrast to vector graphics, which easily scale to the quality of the device on which they are rendered.

• Raster graphics are more practical than vector graphics for photographs and photo-realistic images, while vector graphics are often more practical for typesetting or graphic

design.

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RGBRGB • The color of each pixel is individually defined; images in the

RGB color space, for instance, often consist of colored pixels defined by three bytes—one byte each for red, green and blue.

• Less colorful images require less information per pixel; an image with only black and white pixels requires only a single bit for each pixel.

• A colored raster image (or pixmap) will usually have pixels with between one and eight bits for each of the red, green,

and blue components.

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ResolutionResolution • The quality of a raster image is determined by the total

number of pixels (resolution), and the amount of information in each pixel (often called color depth).

• For example, an image that stores 24 bits of color information per pixel can represent smoother degrees of shading than one that only stores 16 bits per pixel, but not as smooth as one that stores 48 bits.

• Likewise, an image sampled at 640 x 480 pixels (therefore containing 307,200 pixels) will look rough and blocky compared to one sampled at 1280 x 1024 (1,310,720 pixels). Because it takes a large amount of data to store a high-quality image.

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PointPoint

• Real WorldA point has no dimension.

• Practical RepresentationA point has size.

• Analytic GeometryA point P = (x, y) [For 2 Dimension]

A point P = (x, y, z) [For 3 Dimension]

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Raster GraphicsRaster Graphics

• Raster Graphics• Pixels

Example

Point(4.3, 5.3) = Pixel(4, 5)

Point(3.8, 4.8) = Pixel(4, 5)

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))(),((),( yxPixelyxPoint IntInt

Straight LineStraight Line

(x2,y2)

(x1,y1)

x

y

y = m x +b

x

ym

y

x

b

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Equation : y = m x + b

b : y-intercept m : dy/dx

F(x, y) : Ax + By +C = 0

F(x, y) : dy x - dx y + b dx = 0

Therefore, A = dy, B = -dx, C = b dx

Straight LineStraight Line

bxdx

dyy

U(xi+1, yi+1)

M

D (xi, yi )

(xi+1, yi )

Bresenham’s Line AlgorithmBresenham’s Line Algorithm

U(xi+1, yi+1)

M

D (xi, yi )

(xi+1, yi )

U(xi+1, yi+1)

M

D (xi, yi )

(xi+1, yi )

Midpoint Line AlgorithmMidpoint Line Algorithm

Suppose that point (xi, yi) has been plotted.

We move xi to xi + 1.

The problem is how to select between two pixels,

U (xi + 1, yi + 1) and D (xi + 1, yi ).

For this purpose, we consider the middle pixel

M (xi + 1, yi + 1/2).

d = F(M) = a (xi + 1) + b( yi + 1/2) + c

IF d > 0 , choose U

d < 0 , choose D

d = 0 , choose either U or D, so choose U.

• When D is chosen, M is incremented one step in the x direction. So

dnew = F (xi +2, yi + 1/2)

= a (xi + 2) + b (yi + 1/2) + c

while

dold = F (xi + 1, yi + 1/2) = a (xi + 1) + b (yi + 1/2) + c

So the increment in d (denoted dD) is

dD = dnew - dold = a = dy

• When U (xi + 1, yi + 1) is chosen, M is incremented one step in both directions:

dnew = F (xi +2, yi + 3/2)

= a (xi + 2) + b( yi + 3/2) + c

= dold + a + b

So the increment in d (denoted dU ) is

dU = a + b = dy - dx

• First, we have the point (x1, y1). So M (x1 + 1, y1 + 1/2) and

F(M) = a (x1 + 1) + b (y1 + 1/2) + c

= F (x1, y1 ) + a + b/2

Since F (x1 , y1) = 0, we have

d = d1 = dy - dx/2

In order to avoid a division by 2, we use 2d1 instead.

Afterward, 2d is used. So, with d used in place of 2d, we have

• First set d1 = 2dy - dx

• IF di >= 0 THEN

xi+1 = xi + 1, yi+1 = yi + 1 and di+1 = di + 2 (dy-dx)

• IF di < 0 THEN

xi+1 = xi + 1, yi+1 = yi and di+1 = di + 2dy

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Midpoint Line Algorithm [Scan-convert the line between (x1, y1) and (x2, y2)]

dx = x2 - x1;

dy = y2- y1;

d = 2 * dy - dx; /* initial value of d */

dD = 2 * dy; /* increment used to move D */

dU = 2 * (dy-dx); /* increment used to move U */

x = x1;

y = y1;

Plot(x,y); /* the first pixel */

While (x < x2)

if d < 0 then

d = d + dD; / * choose D */

x = x + 1;

else

d = d + dU; /* choose U */

x = x + 1;

y = y + 1;

endif

Plot(x,y); /* the selected pixel closest to the line */

EndWhile

Midpoint Line AlgorithmMidpoint Line Algorithm

Since for the points, x < y, consequently the algorithm can apply.

Here dy = 11 - 8 = 3, dx = 9-5 = 4

• First d1 = 2dy - dx = 6 - 4 = 2 > 0 Select U

So the new point is (6, 9) and

d2 = d1 + 2 (dy - dx) = 2 + 2(-1) = 0 Select U

The chosen pixel is (7, 10) and

d3 = d2 + 2 (dy - dx) = 0 +2(-1) = -2 < 0 Select D

The chosen pixel is (8, 10) then

d4 = d3 + 2dy = - 1 + 6 = 5 > 0 Select U

The chosen pixel is (9, 11).

Example. Scan-convert the line between (5, 8) and (9, 11).

ExampleExample

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Bresenham’s Line Algorithm can work only with the line segment with slope m from 0 to 1. (45o)

y

x

RemarksRemarks

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Bresenham’s Line Algorithm can work only with the line segment with slope m from 0 to 1. (45o)

RemarksRemarks

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Bresenham’s Line Algorithm can work only with the line segment with slope m from 0 to 1. (45o)

RemarksRemarks

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x2+ y2 = r2

F(x,y) = x2 + y2 - r2

F(x,y) = (x - xc)2+ (y - yc)2 - r2

x = r cos(theta)

y = r sin(theta)

CircleCircle

r y

x

y

x

P x r x ( , )2 2 y

r cos

r sin

P=(r cos , r sin )

x

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F(x,y) = (x - xc)2+ (y - yc)2 - r2

IF F(x,y) = 0 :

Point (x,y) is on the circle.

IF F(x,y) < 0 :

Point (x,y) is inside the circle.

IF F(x,y) > 0 :

Point (x,y) is outside the circle.

F(x,y) < 0

F(x,y) > 0

CircleCircle

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Midpoint Circle AlgorithmMidpoint Circle Algorithm

-y 45

+x

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P1 = (x, y) P5 = (-y, -x)

P2 = (y, x) P6 = (-y, -x)

P3 = (-y, x) P7 = (y, -x)

P4 = (-x, y) P8 = (x, - y)

(2, 8) (y, x)

(-2, 8) (-y, x)

y

(8, 2) (x, y)

x

(x, -y) (8, -2)

(y, -x) (2, -8)

(-y, -x) (-2, -8)

(-x, -y) (-8, -2)

(-8, 2) (-x, y)

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Eight-Way SymmetryEight-Way Symmetry

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d = F(M) = F(xi, yi) = xi2+ yi

2 - r2

IF d < 0, choose U

d > 0, choose D

d = 0, choose either U or D,

so choose U.

Midpoint AlgorithmMidpoint Algorithm

P(xi, yi) U(xi + 1, yi )

D(xi +1, yi - 1)

M

(b)

Let dold = F(xi+1, yi - 1/2)

= (xi + 1)2 + (yi - 1/2)2 - r2

If dold < 0, then U (xi+1, yi) is chosen and the next midpoint will be one increment over x.

Thus

dnew = F(xi+2, yi - 1/2)

= dold + 2xi + 3

The increment in d is

dU = dnew - dold = 2xi + 3

IF dold >= 0, M is outside the circle and D is chosen.

The new midpoint will be one increment over x and one increment down in y:

dnew = F (xi + 2, yi - 3/2)

= dold + 2xi - 2yi + 5

The increment in d is therefore

dD = dnew - dold = 2(xi - yi ) + 5

Initial point : (0, r).

The next midpoint lies at (1, r- 1/2) and so

F (1, r- 1/2) = 1 + (r- 1/2)2 - r2 = 5/4 - r

To avoid the fractional initialization of d,

we take h = d - 1/4.

So the initials value of h is 1-r and the comparison d < 0 becomes h < -1/4.

However, since h starts out with an integer value and is incremented with integer values ( dU and dD),

we can change the comparison to h < 0.

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h = 1-r; /* initialization */

x = 0;

y = r;

Plot(x,y);

While y > x

if h < 0 then /* Select U */

dU = 2*x + 3;

h = h + dU;

x = x + 1;

else /* Select D */

dD = 2*(x-y) + 5;

h = h + dD;

x = x + 1;

y = y - 1;

endif

Plot(x,y);

EndWhile

Midpoint Circle AlgorithmMidpoint Circle Algorithm