Clipping Scribing

8/2/2019 Clipping Scribing

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SUBMISSION BY:- NARESH MEHRA (09010129)

SHYAM SUNDER SINGH (09010149)

10/4/2012

CLIPPING LECTURE SCRIBING |

INDIAN

INSTITUTEOF

TECHNOLOGYCOMPUTERGRAPHICS


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Note:- This pdf contain class notes By Pinaki mitra sir

ON CLIPPING followed by examples of Clipping

2-DIMENSIONAL VIEWING

WORLD CO-ODINATE SYSTEM:Any convenient co-ordinate system

WINDOW: A world co-ordinate area selected for clipping is called window.VIEW PORT: An area on the display to which the window is mapped.


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Window -to- viewport-coordinate transformation:


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We want

(Xv - Xv min) / (Xv max - Xv min) = (Xw Xwmin) / (Xw max Xw min)

(Yv - Yv min) / (Yv max - Yv min) = (Yw Ywmin) / (Yw max Yw min)

Xv = Xv min + (Xw Xw min) . Sx

Yv = Yv min + (Yw Yw min) . Sy

Sx = (Xv max - Xv min) / (Xw max Xw min)

Sy = (Yv max - Yv min) / (Yw max Yw min)

1)We perform the following sequence of operations:Perform a scaling transformation using a fixed point position (Xw min, Yw

min) that scales the window area to the size of viewport.

2) Translate the scaled windows area to the position of the viewport.

Clipping operation

Point clipping

Line clipping

Polygon clipping/area clipping


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Point clipping:

Xw min X Xw max

Yw min Y Yw max

We retain a point P(X,Y) satisfying the above property for display.

line Clipping:

Cohen-sutherland line Clipping:-

Region code:


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X X X X

Bit-4 Bit-3 Bit-2 Bit-1

Above below right left

Bit-1 : is the sign bit of : X- Xw min

Bit-2 : is the sign bit of : Xw max - X

Bit-3 : is the sign bit of : Y -Yw min

Bit-4 : is the sign bit of :Yw max -Y

If the some bit positions of two end points have 1, then this line is reflected

because both are outside the clipping window and in the same region outside

the window.

we perform the logical AND of the region

codes corresponding to P1 and P2. if the result0000.

if the region codes of both P1 & P2 = 0000.

(Trivial acceptance)


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INTERSECTION COMPUTATION:Line segment where P1 =(X1,Y1) & P2=(X2,Y2)

Vertical boundary:

Y = Y1+M(X- X1)

Where X = Xw min or X = Xw max

M = Y2- Y1/ X2 - X1

Horizontal boundary:

X = X1 + (Y - Y1 /M)

Where Y =Yw min or Y =Yw max

CYRUS-BECK & LIANG-BARSKY LINE CLIPPING:Parametric representation of the line for the line .

P(t) = P0 +( P1 P0).t


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Vector : P(t)- PEi

For t =0

P(0) PEi = P0 - PEi

P(1) PEi = P1 - PEi

So Ni . (P(t) PEi ) = 0

Ni . [P0+( P1 P0).t - PEi ] = 0

Ni . [P0 - PEi ] + Ni [P1 P0].t = 0

t = Ni . [P0 - PEi ]/ Ni .[P0 P1].

T can be computed if

1)Ni 2)D 3)Ni: D (h h f d || o )

(0 or multiple intersection)

Dd h do h (.)

So h o


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If moving from P0 to P1 , causes us to cross a particular edge to enter the

edges inside half plane, the intersection is PE(potentially entering), If it causes

to leave the edges inside half plane it is PL(potentially leaving)

Ni. D < 0 => PE (angle >90)

Ni . D > 0 => PL (angle T1 discard the entire segment, otherwise (TE.. TL) determines the

clipped portion of the line segment.

Nicholl-lee Nicholl Line Clipping (only 2-D)


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Line

Case 1:

L,T,R,B possible position for P2.

Case2:


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if P2 lies h o with left & top clipping window

boundaries.

Case3:


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OR

Case 4:P2 lies in region LT if

So 1TR So 12 < So 1TL


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(YT - Y1)/ (XR - X1) < (Y2 -Y1)/ ( X2 - X1) < (YT - Y1)/ (XL - X1)

Trivially if

(YT - Y1) ( X2 - X1) < (XL - X1).(Y2 -Y1)

From parametric equation:

X = X1 + ( X2 - X1) .u

Y = Y1 + ( Y2 - Y1) .u

x- Intersection position on the left window boundary is x=xL with u= (xL-x1/

x2-x1) & y intersection position is

Y = Y1 + ( Y2 - Y1)/ ( X2 - X1) . (XL - X1)

And intersection position with the top boundary with y=yT and u= (xT - y1/ y2-

y1) with

X = X1 + ( X2 - X1)/ ( Y2 - Y1) . (YT - Y1)

POLYGON CLIPPING:

Using line clipping applied to polygonal boundary we obtain a series of

unconnected line segments. We want the output of polygon clipper of be a

sequence of vertices that defines the clipped polygon boundaries.


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SUTHERLAND-HODGEMAN POLYGON CLIPPING:4 possible cases when processing vertices in sequence along the polygonal

boundary:

Out in inin inout

outout

Save v1 ,2 save v2 Save v1 S

none

V1

V

V1

V2

V2

V1

V2

V1

V1 V1

2

3


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{ , 3 } , {, 3 , 4 } , { , 3 , 4 , 5 }, { ,3 ,4 , 5 , 5}

IN SUTHERLAND-HODGEMAN algorithm, there is only one output vertex list ,

the lost vertex is always joined with the 1st vertex.

WEILER ATHERTON POLYGON CLIPPING:For clockwise processing of polygon vertices we use the following:

2

1

56

5

V2

V1

V3

V3

V2


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For an outside-to-inside pair of vertices follow polygonal boundary.

For an inside-to-outside pair of vertices follow the window boundary in a

clockwise direction.

{ v2,2 , v3 , v4 } { 5,5 , v6 }

Examples:-Q1:- Find Region code for given line and clip by Sutherland cohen Algorithm ?

Ans:- According to Sutherland cohen Algorithm

Coordinate of point will be 0101 As it belongs to Left-Below region.

And point B will be 1010 As it lies in Right-Above region.


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:-

Logical AND of A & B is 0000 and none of both is 0000

So it will be partially visible.

As point A & C will be 0001 i.e. nonzero so we can remove it from line segment

and now we have only BC as line segment.

Again B & C = 0000 thats why it also will be partially visible.

By 0-bit = 1 in C we can say that it has left crossed the window. So we remove

CD(left) part from

Line.

Now line segment is BD.

Similarly we can also remove BE and FE segment of line as every point hascoordinate in which

1-bit = 1. I.e. it has crossed the right part of clipping window that why will be

invisible.

And finally we get a Line DF where D & F = 0000 with both are equal to 0000.


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So we can say that Line segment DF will be completely visible as we dont have

any bit equal to

1.

Example 2 :-

Clip below Polygon according to Sutherland-Hodgman algorithm ??

Ans:-

According to Sutherland-Hodgman algorithm we know that

1. Out -> In case:- For V1 -> V2 where V1 is the intersection point, then

we save V1, V2 (I.)

2 .In -> Out case:- For V1 -> V2 where V1 is the intersection point, then

we save V1.

3 .In -> In case :- we save V2.


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4 .Out -> Out case:- We save nothing.

And then we clip by Left -> Top -> Right -> Below.

In given below example

As Line segments from V6 Intersects Window at two points and we get to

points by L1 and L2 from

Case (1) and case 2 respectively.

-When we clip form Left

we get new Vertex list as V2 -> V3 -> V4 -> V5 -> L1 -> L1 -> V1 .

Vertex set after Top Clip :- T2

-> V3

-> V4

-> V5

-> L1

-> L1

-> T1

Vertex set after Right Clip :- R1 -> R2 -> V4 -> V5 -> L1 -> L2 -> T1 -> T2 .

Vertex set after Below Clip :- -> R2 -> B1 -> B2 -> V4 -> V5 -> L1 -> L2 -> T1 -> T2 -

> R1


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And thus we get a New clipped Polygon as below .


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Example 3 :- Weiler-Atherton Algorithm

Find Clipped Polygon for Given Non-convex Polygon and clipping window??

Ans:-

Rules:-

For an outside to inside pair of vertices follow polygonal boundary.

For an inside to Outside pair of vertices follow the window boundary in a

clockwise direction.

In given Below Polygon and window as 0-1 is inside to outside thats why we

include window boundary in Clockwise direction and we get clipped part o-i-j-9.

From 8-9 and 3-4 I.e. outside to inside pair of vertices we get polygonal boundary.

Similarly by Inside to outside clipping of 5-6 We get 5-k-j-4 part.

And we get new Clipped Polygon.


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Example 4: - POINT CLIPPING

Prove that given point will not be visible after clipping???

Ans:-

By condition of Point clipping we know that a point will be inside a clipping

window

If Xmin < = X

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