a good pdf on clipping

8/11/2019 A good pdf on Clipping

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CS123 | INTRODUCTION TO COMPUTER GRAPHICS

Andries van Dam

ClippingConcepts, Algorithms for

line clipping

Clipping - 10/16/12


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Andries van Dam

Clipping endpoints

If


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Andries van Dam

Clip Rect Divide plane into 9 regions

Compute the sign bit of 4 comparisons between a vertex and anedge

; ; ;

point lies inside only if all four sign bits are 0, otherwise exceeds edge

4 bit outcode records results of four bounds tests: First bit: above top edge

Second bit: below bottom edge

Third bit: to the right of right edge

Fourth bit: to the left of left edge Compute outcodes for both vertices of each line (denoted OC0

and OC1)

Lines with OC0 = 0 (i.e., 0000) andOC1 = 0 can be triviallyaccepted.

Lines lying entirely in a half plane outside an edge can be triviallyrejectedif (OC0ANDOC1) 0 (i.e., they share an outside bit)

Cohen-Sutherland Line Clipping in 2D

Clipping - 10/16/12


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Andries van Dam

Very similar to 2D

Divide volume into 27 regions (Picture aRubiks cube)

6-bit outcode records results of 6 boundstests First bit: behind back plane

Second bit: in front of front plane

Third bit: above top plane

Fourth bit: below bottom plane

Fifth bit: to the right of right plane

Sixth bit: to the left of left plane

Again, lines with OC0= 0 and OC1= 0 canbe trivially accepted

Lines lying entirely in a volume outside of

a plane can be trivially rejected: OC0AND

OC10 (i.e., they share an outside bit)

Cohen-Sutherland Line Clipping in 3D

Clipping - 10/16/12

Back plane

000000 (in front)

100000 (behind)

0

0

0

0

00

Top plane

001000 (above)

000000 (below)

Right plane

000000 (to left of)

000010 (to right of)


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Andries van Dam

If we can neither trivially accept/reject (T/A, T/R),divide and conquer

Subdivide line into two segments; then T/A or T/R oneor both segments:

use a clip edge to cut line

use outcodes to choose the edges that are crossed

for a given clip edge, if a lines two outcodes differ in thecorresponding bit, the line has one vertex on each side of theedge, thus crosses

pick an order for checking edges: top bottom right left

compute the intersection point

the clip edge fixes eitherxory

can substitute into the line equation

iterate for the newly shortened line, extra clips mayhappen (e.g., E-I at H)

Cohen-Sutherland Algorithm (1/3)

D

C B

A

E

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Andries van Dam

=0 + ( 0)and =0 + 1

( 0)

Algorithm:


ComputeOutCode(x0, y0, outcode0);

ComputeOutCode(x1, y1, outcode1);

repeat

check for trivial reject or trivial accept

pick the point that is outside the clip rectangle

ifTOP then

x = x0 + (x1 x0) * (ymax y0) / (y1 y0);

y = ymax;

else ifBOTTOMthen

x = x0 + (x1 x0) * (ymin y0) / (y1 y0);

y = ymin;

else ifRIGHTthen

y = y0 + (y1 y0) * (xmax x0)

x = xmax;

else ifLEFTthen

y = y0 + (y1 y0) * (xmin x0)

x = xmin;

if (x0, y0 is the outer point) then

x0 = x; y0 = y; ComputeOutCod

else

x1 = x; y1 = y; ComputeOutCod

until done

Clipping - 10/16/12


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Andries van Dam

Similar algorithm for using 3D outcodes to clip against canonical parallel vi


xmin = ymin = -1; xmax = ymax = 1;

zmin = -1; zmax = 0;

ComputeOutCode(x0, y0, z0, outcode0);

ComputeOutCode(x1, y1, z1, outcode1);

repeat

check for trivial reject or trivial accept

pick the point that is outside the clip rectangle

ifTOP then

x = x0 + (x1 x0) * (ymax y0) / (y1 y0);

z = z0 + (z1 z0) * (ymax y0) / (y1 y0);

y = ymax;else ifBOTTOMthen

x = x0 + (x1 x0) * (ymin y0) / (y1 y0);

z = z0 + (z1 z0) * (ymin y0) / (y1 y0);

y = ymin;

else ifRIGHTthen

y = y0 + (y1 y0) * (xmax x0) / (x1 x0);

z = z0 + (z1 z0) * (xmax x0) / (x1 x0);

x = xmax;

else ifLEFTthen

y = y0 + (y1 y0) * (xmin x0) / (x1

z = z0 + (z1 z0) * (xmin x0) / (x1 x = xmin;

else if NEAR then

x = x0 + (x1 x0) * (zmax z0) / (z1

y = y0 + (y1 y0) * (zmax z0) / (z1

z = zmax;

else if FAR then

x = x0 + (x1 x0) * (zmin z0) / (z1

y = y0 + (y1 y0) * (zmin z0) / (z1

z = zmin;

if (x0, y0, z0 is the outer point) then

x0 = x; y0 = y; z0 = z;

ComputeOutCode(x0, y0, z0, outcode

else

x1 = x; y1 = y; z1 = z;

ComputeOutCode(x1, y1, z1, outcode

until done

Clipping - 10/16/12


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Andries van Dam

Dont round and then scan

convert, because the line will

have the wrong slope: calculate

decision variable based on pixel

chosen on left edge (remember:

= + )

Horizontal edge problem:

Clipping/rounding produces pixelA; to get pixel B, round up x of the

intersection of line with

= and pick pixel

above:

Scan Conversion after Clipping

B A

x = xmin

y = ymin

y = ymin 1

y = ymin 1/2

(, + )

(, ( + ))

=

Clipping - 10/16/12


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Andries van Dam

Sutherland-Hodgman Polygon Clipping

The 2D Sutherland-Hodgman algorithm generalizes to higher dime

We can use it to clip polygons to the 3D view volume one plane at a

Section 36.5 in textbook


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Andries van Dam

Use parametric line formulation:

Determine where line intersects the

infinite line formed by each clip

rectangle edge

solve for t multiple times dependingon the number of clip edges crossed

decide which of these intersections

actually occur on the rectangle

For : use any point on edge Ei

Cyrus-Beck/Liang-Barsky Parametric Line Clippi

=0 + 1 0

Clipping - 10/16/12


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Andries van Dam

Now solve for the value of tat the intersection of P0P1with the edge Ei:

First, substitute for P(t):

Next, group terms and distribute dot product:

Let Dbe the vector from P0toP1=(P1P0), andsolve for t:

note that this gives a valid value of tonly if thedenominator of the expression is nonzero.

For this to be true, it must be the case that: Ni0 (that is, the normal should not be 0; this could

occur only as a mistake)

D0 (that is, P1P0)

Ni D0 (edge Eiand line Dare not parallel; if theyare, no intersection).

The algorithm checks these conditions.

Cyrus-Beck/Liang-Barsky Parametric Line Clippi

[0 + 1

0 +

= [

Clipping - 10/16/12


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Andries van Dam

Eliminate s outside [0,1] on the line

Which remaining s produce interior

intersections? Cant just take the innermost values!

Move from P0to P1; for a given edge,just before crossing:

if Ni D < 0 Potentially Entering (PE);

if Ni D > 0

Potentially Leaving (PL) Pick inner PE/PL pair: for with

max , for with min , and > 0, < 1

If


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Andries van Dam

Cyrus-Beck/Liang-Barsky Line Clipping AlgorithmPre-calculate Niand select PEi

for each edge;

foreach line segment to be clipped

ifP1= P0 then line is degenerate so clip as a point;

elsebegin

tE= 0; tL= 1;

foreach candidate intersection with a clip edge

ifNi D 0 then{Ignore edges parallel to line}

begin

calculate t; {of line and clip edge intersection}

use sign of Ni

D to categorize as PE or PL;if PE thentE= max(tE,t);

ifPL thentL= min(tL,t);

end

iftE> tLthen return nil

else return P(tE) and P(tL) as true clip intersections

end

Clipping - 10/16/12


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Andries van Dam

D = P1 P0= (x1 x0, y1 y0)

Leave PEi

as an arbitrary point on clip edge: its a free variable and d

Parametric Line Clipping for Upright Clip Rectan

Calculations for Parametric Line Clipping Algorithm

(x0-x,y0- ymax)(x, ymax)(0,1)top: y = ymax

(x0-x,y0- ymin)(x, ymin)(0,-1)bottom: y = ymin

(x0- xmax,y0-y)(xmax,y)(1,0)right: x = xmax

(x0- xmin,y0-y)(xmin, y)(-1,0)left: x = xmin

P0-PEiPEi

Normal NiClip EdgeiD

iN

PPi

N

t

=0

(

)01

(

min0(

xx

xx

)01(

ma0(

xx

xx

)01

(

mi0(

yy

yy

)01

(

ma0(

yy

yy

Clipping - 10/16/12


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Andries van Dam

Examine t:

Numerator is just the directed distance to an edge; sign co

to OC

Denominator is just the horizontal or vertical projection o

dxor dy; sign determines PE or PL for a given edge

Ratio is constant of proportionality: how far over from P

intersection is relative to dxor dy

Parametric Line Clipping for Upright Clip Rectan

Clipping - 10/16/12

a good pdf on clipping

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