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
Explain the merits and demerits of Penetration techniques. The merits and demerits of the Penetration techniques are as follows:     It is an inexpensive method.     It h

Steps uses in the Cohen Sutherland Line Clipping Algorithm are: Figure: Steps for Cohen Sutherland Line Clipping STEP 1: Input:  x L , x R , y T , y B



What do you mean by emissive and non-emissive displays? EMISSIVE: The emissive display changes electrical energy into light energy. The plasma panels, thin film ele

1. Compare Bresenham line generation with Digital Differential Analyzer line generation. Ans.   Bresenham line generation algorithm is better than Digital Differential Analyze

PIXEL PAINT: The pixel paint file format permits a document to be opened in the pixel paint and pixel paint professional graphics application. Such format permits you to identify

computer animation Note : This is to be noticed that computer animation can also be produced by changing camera parameters as its position, orientation and focal length, as w

The image you have been given for contour extraction is shown in Figure 1. The method for constructing the search space is shown in Figure 2. It is generated from two initial discr

Define Octrees?  Hierarchical tree structures called octrees, are used to show solid objects in some graphics systems. Medical imaging and other applications that needs display