Download presentation

Presentation is loading. Please wait.

Published byLayne Templeton Modified over 3 years ago

1
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Clipping Concepts, Algorithms for line clipping 1 of 16 Clipping - 10/16/12

2
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Line Clipping in 2D 2 of 16 Clipping - 10/16/12

3
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Parametric form for line segment Line is in clip rectangle if parametric variables t line and s edge both in [0,1] at intersection point between line and edge of clip rectangle Slow, must intersect lines with all edges Parametric Line Formulation For Clipping 3 of 16 Clipping - 10/16/12

4
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Clip Rectangle Cohen-Sutherland Line Clipping in 2D 4 of 16 Clipping - 10/16/12

5
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Very similar to 2D Divide volume into 27 regions (Picture a Rubik’s cube) 6-bit outcode records results of 6 bounds tests 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 OC 0 = 0 and OC 1 = 0 can be trivially accepted Lines lying entirely in a volume outside of a plane can be trivially rejected: OC 0 AND OC 1 0 (i.e., they share an “outside” bit) Cohen-Sutherland Line Clipping in 3D 5 of 16 Clipping - 10/16/12 Back plane 000000 (in front) 100000 (behind) Front plane 010000 (in front) 000000 (behind) Bottom plane 000000 (above) 000100 (below) Left plane 000001 (to left of) 000000 (to right of) Top plane 001000 (above) 000000 (below) Right plane 000000 (to left of) 000010 (to right of)

6
CS123 | INTRODUCTION TO COMPUTER GRAPHICS 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 one or 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 line’s two outcodes differ in the corresponding bit, the line has one vertex on each side of the edge, thus crosses pick an order for checking edges: top – bottom – right – left compute the intersection point the clip edge fixes either x or y can substitute into the line equation iterate for the newly shortened line, “extra” clips may happen (e.g., E-I at H) Cohen-Sutherland Algorithm (1/3) Clip rectangle D C B A E F G H I 6 of 16 Clipping - 10/16/12

7
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Cohen-Sutherland Algorithm (2/3) 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 if TOP then x = x0 + (x1 – x0) * (ymax – y0) / (y1 – y0); y = ymax; else if BOTTOM then x = x0 + (x1 – x0) * (ymin – y0) / (y1 – y0); y = ymin; else if RIGHT then y = y0 + (y1 – y0) * (xmax – x0) / (x1 – x0); x = xmax; else if LEFT then y = y0 + (y1 – y0) * (xmin – x0) / (x1 – x0); x = xmin; if (x0, y0 is the outer point) then x0 = x; y0 = y; ComputeOutCode(x0, y0, outcode0) else x1 = x; y1 = y; ComputeOutCode(x1, y1, outcode1) until done 7 of 16 Clipping - 10/16/12

8
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Similar algorithm for using 3D outcodes to clip against canonical parallel view volume: Cohen-Sutherland Algorithm (3/3) 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 if TOP then x = x0 + (x1 – x0) * (ymax – y0) / (y1 – y0); z = z0 + (z1 – z0) * (ymax – y0) / (y1 – y0); y = ymax; else if BOTTOM then x = x0 + (x1 – x0) * (ymin – y0) / (y1 – y0); z = z0 + (z1 – z0) * (ymin – y0) / (y1 – y0); y = ymin; else if RIGHT then y = y0 + (y1 – y0) * (xmax – x0) / (x1 – x0); z = z0 + (z1 – z0) * (xmax – x0) / (x1 – x0); x = xmax; else if LEFT then y = y0 + (y1 – y0) * (xmin – x0) / (x1 – x0); z = z0 + (z1 – z0) * (xmin – x0) / (x1 – x0); x = xmin; else if NEAR then x = x0 + (x1 – x0) * (zmax – z0) / (z1 – z0); y = y0 + (y1 – y0) * (zmax – z0) / (z1 – z0); z = zmax; else if FAR then x = x0 + (x1 – x0) * (zmin – z0) / (z1 – z0); y = y0 + (y1 – y0) * (zmin – z0) / (z1 – z0); z = zmin; if (x0, y0, z0 is the outer point) then x0 = x; y0 = y; z0 = z; ComputeOutCode(x0, y0, z0, outcode0) else x1 = x; y1 = y; z1 = z; ComputeOutCode(x1, y1, z1, outcode1) until done 8 of 16 Clipping - 10/16/12

9
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Scan Conversion after Clipping BA x = x min y = y min y = y min – 1 y = y min – 1/2 Clip rectangle 9 of 16 Clipping - 10/16/12

10
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Sutherland-Hodgman Polygon Clipping 10 of 16 The 2D Sutherland-Hodgman algorithm generalizes to higher dimensions We can use it to clip polygons to the 3D view volume one plane at a time Section 36.5 in textbook

11
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Cyrus-Beck/Liang-Barsky Parametric Line Clipping (1/3) 11 of 16 Clipping - 10/16/12

12
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Now solve for the value of t at the intersection of P 0 P 1 with the edge E i : First, substitute for P(t): Next, group terms and distribute dot product: Let D be the vector from P 0 to P 1 = (P 1 – P 0 ), and solve for t: note that this gives a valid value of t only if the denominator of the expression is nonzero. For this to be true, it must be the case that: N i 0 (that is, the normal should not be 0; this could occur only as a mistake) D 0 (that is, P 1 P 0 ) N i D 0 (edge E i and line D are not parallel; if they are, no intersection). The algorithm checks these conditions. Cyrus-Beck/Liang-Barsky Parametric Line Clipping (2/3) 12 of 16 Clipping - 10/16/12

13
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Cyrus-Beck/Liang-Barsky Parametric Line Clipping (3/3) 13 of 16 Clipping - 10/16/12

14
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Cyrus-Beck/Liang-Barsky Line Clipping Algorithm Pre-calculate N i and select P E i for each edge; for each line segment to be clipped if P 1 = P 0 then line is degenerate so clip as a point; else begin t E = 0; t L = 1; for each candidate intersection with a clip edge if Ni D 0 then {Ignore edges parallel to line} begin calculate t; {of line and clip edge intersection} use sign of N i D to categorize as PE or PL; if PE then t E = max(t E,t); if PL then t L = min(t L,t); end if t E > t L then return nil else return P(t E ) and P(t L ) as true clip intersections end 14 of 16 Clipping - 10/16/12

15
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © D = P 1 – P 0 = (x 1 – x 0, y 1 – y 0 ) Leave P E i as an arbitrary point on clip edge: it’s a free variable and drops out Parametric Line Clipping for Upright Clip Rectangle (1/2) Calculations for Parametric Line Clipping Algorithm (x 0 -x,y 0 - y max )(x, y max )(0,1)top: y = y max (x 0 -x,y 0 - y min )(x, y min )(0,-1)bottom: y = y min (x 0 - x max,y 0 -y)(x max,y)(1,0)right: x = x max (x 0 - x min,y 0 -y)(x min, y)(-1,0)left: x = x min P 0 -P E i PEiPEi Normal N i Clip Edge i 15 of 16 Clipping - 10/16/12

16
CS123 | INTRODUCTION TO COMPUTER GRAPHICS Andries van Dam © Examine t: Numerator is just the directed distance to an edge; sign corresponds to OC Denominator is just the horizontal or vertical projection of the line, dx or dy; sign determines PE or PL for a given edge Ratio is constant of proportionality: “how far over” from P 0 to P 1 intersection is relative to dx or dy Parametric Line Clipping for Upright Clip Rectangle (2/2) 16 of 16 Clipping - 10/16/12

17
Embedded Font (Hidden slide, to ensure that all characters are embedded for future editing) !@#$%^&*()_+- =~`,./;’[]{}:”>?< ABCDEFGHIJKLMNOPQRSTUVWXYZ a b c d e f g h i j k l m n o p q r s t u v w x y z 1234567890 ABCDEFGHIJKLMNOPQRSTUVWXYZ a b c d e f g h i j k l m n o p q r s t u v w x y z 1234567890 ABCDEFGHIJKLMNOPQRSTUVWXYZ a b c d e f g h i j k l m n o p q r s t u v w x y z 1234567890 17

Similar presentations

OK

CS 325 Introduction to Computer Graphics 03 / 08 / 2010 Instructor: Michael Eckmann.

CS 325 Introduction to Computer Graphics 03 / 08 / 2010 Instructor: Michael Eckmann.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on nuclear family and joint family band Ppt on site selection of thermal power plant Ppt on impact of social networking sites Ppt on review of literature examples Probability for kids ppt on batteries Download ppt on transportation in human beings A ppt on football Ppt on digital bike locking system Ppt on disk formatting freeware Ppt on 10 sikh gurus sikhism