Computer Graphics. Brian Wyvill. Clipping. cpsc/enel P 1 H K AB AC DE FG HI JK

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1 Comuter Grahic 1001 A 1000 F G E C B D 0101 H K I J P2 P3 Outide Inide P4 P1 Cliing by AB AC DE FG HI JK AND Inide outide Polygon edge i outut cli boundary Inide outide Inide outide i i outut no outut 2nd outut inide outide Brian Wyvill cc/enel P 1

2 Cliing Line Cohen and Sutherland Cliing Algorithm Cliing i erformed on line whoe endoint lie outide a articular cliing region. (A window manager window for examle.) It i oible to do thi whilt can converting the rimitive. In general, determine the bit code according to the oition of the endoint w.r.t the window. Left of Window: Bit 0 et x < xmin Right of Window: Bit 1 et x > xmax Below WIndow: Bit 2 et y < ymin Above Window: Bit 3 et y > ymax AB AC DE FG HI JK A 1000 F 1010 A 0001 C 0101 H G E B D The world ace can be lit into region. An outcode i calculated according to the region that the endoint occuy. K I J AND Non zero are trivial reject Both zero are trivial accet Other line are further ubdivided. Line AB i clied againt the to edge and then a new outcode caclulated for A, the line A B i then clied again againt the left edge. cc/enel P 2

3 Cohen and Sutherland continued x1,y1 x1,y1 x0,y0 y=yt Note Problem with vertical line if loe i ued. P0=x0,y0 x=xl x0,y0 y0=m * x0 + c y1=m * x1 + c m = (y1 y0)/(x1 x0) c = (y1 m * x1) y = m * xl+ (y1 m * x1) x = xl Parametric Line P(t) = P0 + (P1 P0)*t 0<=t<=1 x=x0+t*(x1 x0) y=y0+t*(y1 y0) e.g. Left edge t = (xl x0)/(x1 x0) y = y0+((xl x0)/(x1 x0))*(y1 y0) e.g To edge t = (yr y0)/(y1 y0) x = x0+((yr y0)/(y1 y0))*(x1 x0) Bai for Cyru Beck cc/enel P 3 P1=x1,y1

4 Cyru Beck (Liang Barky) Cliing Four value of t calculated correonding to each edge. Decide which are valid. P(t)=P0+(P1 P0)*t B: N. (P(t) A) < 0 C: N. (P(t) A) = 0 D: N. (P(t) A) > 0 We want to find P(T)=C : N. (P(t) A) = 0 N. or (P0+(P1 P0)*t A)=0 N. (P0 A) + N. (P1 P0)*t =0 t=n. (P0 A)/( N. (P1 P0)) N P0 t=0 D A C P1 t=1 t will be valid if N. (P1 P0) i non zero B Point A i choen checkfor: N = 0 oo error P1 = P0 degenerate line N. (P1 P0)=0 P0 P1 Parallel to cli edge cc/enel P 4

5 Cyru Beck (Liang Barky) Cliing continued QE t=1 QL E.g. 1 t=n. (P0 A)/( N. (P1 P0)) t=0 For each line egment four value of t are found one for each cliing edge. A i choen to be a vertex of the cli rectangle. N will be different for each edge. firt check that 0<=t<=1 (ele reject that value of t) direction of egment i from t=0 oibly croing at QE to t=1 cro at QL. QL QE t=0 t=1 E.g. 2 te>tl reject Claify the t value by inecting ign: N. (P1 P0) <0 > QE (angle > 90) correonding to te QL E.g. 3 N. (P1 P0) >0 > QL (angle < 90) correonding to tl In E.g. 3 chooe larget te and mallet tl QE QE QL t=1 t=0 cc/enel P 5

6 Efficiency N. (P1 P0) Let D = (P1 P0) e.g. Nx = 1 Ny = 0 N A P1 N. D = (Nx*Dx+Ny*Dy) = Dx P0 In general N. (P1 P0) reduce to +/ Dx or Dy the ign determine QL or QE t=n. (P0 A)/( N. (P1 P0)) For the above examle of N: (P0x Ax) / Dx In general it i the ditance to the edge along X or Y divided by Dx or Dy The C&S algorithm would comare P0x and Ax to comute the outcode. Several comarion would be done for line croing many boundary. Note that the ign have not been cancelled a the ign of numerator and denominator are ued by the algorithm. F&VD P120 cc/enel P 6

7 Polygon Cliing Sutherland Hodgman algorithm. Cli arbitrary olygon againt rectangular cli boundary. Strategy Divide and Conquor: Cli each olygon edge againt a cli edge and re enter algorithm cliing next cli edge with clied olygon. cc/enel P 7

8 Polygon Cliing continued Inide Polygon edge i outut cae 1 cae 2 cae 3 cae 4 outide cli boundary Inide outide Inide outide i i i outut no outut i i firt outut i 2nd outut inide outide v0 v1 v2 v3 v4 v5... vn v0 v1 v2 v3 v4 v5... vn Cli all line in the olygon againt a cli edge, write out new array of vertice cc/enel P 8

9 Polygon Cliing Examle cae 1 cae 2 cae 3 cae 4 Inide outide Inide outide Inide outide i i firt outut Polygon edge i outut cli boundary i i i outut no outut i 2nd outut inide cc/enel P 9 outide A each new vertex i outut the clier call itelf with the new vertex. (Re entrant). The olygon become ucceively clied to the boundary. Thi ieline ha been imlemented in hardware.

10 Inide Outide Tet P2 A general tet for inide outide : ince a x b = b x a let V1 = P1P2 x P1P3 let V2 = P1P2 x P1P4 tet ign of V1 and V2 P3 V1=(x2 x1)*(y3 y1) (y2 y1)*(x3 x1) Outide Inide P4 P1 In the above examle x2=x1 and y2>y1 and x3<x1 V1 = (+ve)*( ve) > 0 P3 lie to the left of P1P2 if P1P2 x P1P3 > 0 cc/enel P 1 0