Illustration of a clipping window - raster graphics, Computer Graphics

Illustration of a Clipping window ABCD is placed as follows:

A (100, 10), B (160, 10, C (160, 40), D (100, 40)

By using Sutherland-Cohen clipping algorithm determine the visible portion of the line segments i.e. EF, GH and P1P2. E (50, 0), F (70, 80), G (120, 20), H (140, 80), P1 (120, 5), P2(180, 30).

2065_Illustration of a Clipping window - Raster Graphics.png

Figure: Example of Cohen Sutherland Line Clipping

At first identifying the line P1P2

INPUT: P1(120, 5),   P2(180, 30)

xL = 100,   xR = 160,    yB = 10,    yT = 40

x1  > xL then bit 1 of code -P1 = 0 C1 left = 0

x1  < xR then bit 2 of code -P1 = 0 C1 right = 0

y1 < yB then bit 3 of code -P1 = 1 C1 bottom = 1

 y1  < yT then bit 4 of code -P1 = 0 C1 top = 0

code -P1 = 0100,

x2  > xL then bit 1 of code -P1 = 0 C2 left = 0

x2  > xR  then bit 2 of code -P1 = 1 C2 right = 1

 y2  > y B then bit 3 of code -P1 = 0 C2 bottom = 0

y2  < yT then bit 4 of code -P1 = 0 C2 top = 0

 code -P2 = 0010.

Both code -P1 <> 0 and code -P2 <> 0

then P1P2 not totally visible

code -P1 AND code -P2 = 0000

therefore (code -P1 AND code -P2 = 0)

then line is not fully invisible.

As code -P <> 0

for  i = 1

{

C1 left (= 0) <> 1 then nothing is done. i = i + 1 = 2

}

code -P1 <> 0 and code -P2 <> 0

then P1P2 not totally visible.

code -P1 AND code -P2 = 0000

therefore (code -P1 AND code -P2 = 0)

then line is not fully invisible.

 for   i = 2

     {

C1 right (= 0) <> 1 then nothing is to be done. i = i + 1 = 2 + 1 = 3

}

code -P1 <> 0 and code -P2 <> 0 then P1P2 not totally visible.

code -P1 AND code -P2 = 0000

therefore, (code -P1 AND code -P2 = 0)

then the line is not fully invisible.

 for   i = 3

{

 C1 bottom = 1 then find intersection of P1P2 with bottom edge yB = 10

xB = (180-120)(10-5)/(30-5) + 120

=132

then P1 = (132,10)

 x1  > xL then bit 1 of code -P1 = 0   C1 left = 0

x1  < xR then bit 2 of code -P1 = 0   C1 right = 0

y1  = yB then bit 3 of code -P1 = 0   C1 bottom = 0

y1  < yT then bit 4 of code -P1 = 0   C1 top = 0

code -P1 = 0000

i = i + 1 = 3 + 1 = 4

}

code -P1 <> 0 but code -P2 <> 0

then P1P2 not totally visible.

code -P1 AND code -P2 = 0000

therefore, (code -P1 AND code -P2 = 0)

then line is not fully invisible.

As code -P1 = 0

Swap P1 and P2 along with the respective flags

P1 = (180, 30) P2 = (132, 10) code -P1 = 0010 code -P2 = 0000

C1 left = 0                         C2 left = 0

C1 right = 1                       C2 right = 0

C1 bottom = 0                  C2 bottom = 0

C1 top = 0                         C2 top = 0

Reset i = 1

for i = 1

{

C1 left (= 0) <> 1 then nothing is to be done. i = i + 1 = 1 + 1 = 2

}

code -P1 <> 0, and code -P2 <> 0

then P1P2 is not totally visible.

code -P1 AND code -P2 = 0000

therefore, (code -P1 AND code -P2 = 0)

then line is not fully invisible.

 for i = 2

{

 C1 right   = 1 then find out intersection of P1P2 with right edge xR = 160

yR = (30 - 5)(160 - 120)/(180 - 120) + 5

= 21.667

= 22 then P1 = (160, 22)

 x1  > xL then bit 1 of code -P1 = 0   C1 left = 0

x1  = xR then bit 2 of code -P1 = 0   C1 right = 0

y1  > yB then bit 3 of code -P1 = 0   C1 bottom = 0

y1  < yT then bit 4 of code -P1 = 0   C1 top = 0

 code -P1 = 0000, i = i + 1 = 2 + 1 = 3

}

As both code -P1 = 0 and code -P2 = 0 then the line segment P1P2 is completely visible.

Consequently, the visible portion of input line P1P2 is P'1P'2 where, P1 = (160, 22) and

P2 = (132, 10).

For the line EF

1)      The endpoint codes are allocated code:

code - E → 0101

code - F → 1001

2) Flags are allocated for the two endpoints:

Eleft = 1 (as x coordinate of E is less than xL)

Eright = 0,  Etop = 0 and Ebottom = 1

As the same,

Fleft = 1,  Fright = 0,  Ftop = 1 and Fbottom = 0

3) Because codes of E and F are both not equivalent to zero the line is not wholly visible.

4) Logical intersection of codes of E and F is not equivalent to zero. Consequently, we may avoid EF line and declare it as wholly invisible.

Identifying the line GH:

a) The endpoint codes are allocated:

code - G → 0000 and

code - H → 1000

b)   Flags are allocated for the two endpoints:

Gleft = 0,  Gright = 0,  Gtop = 0 and Gbottom = 0.

As the same,

Hleft = 0,  Hright = 0,  Htop = 1 and  Hbottom = 0.

c) Because, codes of G and H are both not equivalent to zero according to the line is not totally visible.

d)   Logical intersection of codes of G and H is equivalent to zero consequently we cannot specify it as completely invisible.

f)   Because, code - G = 0, Swap G and H with their flags and set i = 1

Implying   Gleft = 0,  Gright = 0,  Gtop = 1 and  Gbottom = 0.

Hleft = 0,  Hright = 0,  Htop = 0 and  Hbottom = 0.

The same as G → 1000 and H → 0000

6) Because, code - G <> 0 then

for i = 1,

{since Gleft = 0

i = i + 1 = 2

go to 3

}

The conditions 3 and 4 don't hold and so we can't declare line GH as completely visible or invisible.

for i = 2, {since Gright = 0

i = i + 1 = 3

go to 3

}

The conditions 3 and 4 don't hold and so we can't declare line GH as completely visible or invisible.

for i = 3, {since Gbottom = 0

i = i + 1 = 4

go to 3

}

The conditions 3 and 4 don't hold and so we can't declare line GH as completely visible or invisible.

for i = 4, {since Gtop = 1

Intersection along with top edge, as P(x, y) is found as given below:

Any of line passing via the points G, H and a point P(x, y) is given via y - 20 = {(180 - 20) /(140 - 120)}(x - 120) or, y - 20 = 3x - 360 or, y - 30 = -340

Because, the y coordinate of every point on line CD is 40, consequently we put y = 40 for the point of intersection P(x, y) of line GH along with edge CD.

40 - 3x = -340 or, - 3x = - 380

Or else x = 380/3 = 126.66 ≈ 127

Consequently, the point of intersection is P (127, 40). We allocate code to it.

Because, the point lays on edge of the rectangle hence the code allocated to it is 0000. Here, we allocate G = (127, 40); i = 4 + 1 = 5. State 3 and 4 are again checked. Because, codes G and H are both are equivalent to 0, hence, the line among H(120, 20) and G(127, 40) is wholly visible.

Posted Date: 4/3/2013 3:09:42 AM | Location : United States







Related Discussions:- Illustration of a clipping window - raster graphics, Assignment Help, Ask Question on Illustration of a clipping window - raster graphics, Get Answer, Expert's Help, Illustration of a clipping window - raster graphics Discussions

Write discussion on Illustration of a clipping window - raster graphics
Your posts are moderated
Related Questions
Given two triangles P along with vertices as P1(100,100,50), P2(50,50,50), P3(150,50,50) and q along with vertices as Q1(40,80,60), q2(70,70,50), Q3( 10,75,70), determine that tria

Modelling and Rendering a surface Common practice in modelling and rendering a surface is to approximate it by a polygonal surface.  By a polygonal surface we mean a surface whi

Polygon Rendering and Ray Tracing Methods In previous section we had discussed various methods for visible-surface detection, although in order to produce visibility the prese

Basic Ray Tracing Algorithm - Polygon Rendering The Hidden-surface removal is the most complete and most versatile method for display of objects in a realistic fashion. The co


What is uniform rational splines

Image Capture Formats: Video cameras appear in two various image capture formats: progressive and interlaced scan. Interlaced Scan It is a technique of enhancing the

what is staircase appearance

Question: (a) Describe in details the meaning of the following terms often available in drawing tools: (i) Welding objects; (ii) Trimming objects; (iii) Intersecting obje

Advantages of JPEG Images Huge compression ratios mean sooner download speeds. JPEG produces outstanding results for main photographs and complicated images. JPEG s