Introduction to Computer Graphics 5. Clipping

Transcription

1 Introduction to Computer Grapics 5. Clipping I-Cen Lin, Assistant Professor National Ciao Tung Univ., Taiwan Textbook: E.Angel, Interactive Computer Grapics, 5 t Ed., Addison Wesley Ref:Hearn and Baker, Computer Grapics, 3rd Ed., Prentice Hall

2 Objectives Introduce basic implementation strategies 2D Clipping Lines Polygons Clipping in 3D

3 Clipping 2D against clipping window 3D against clipping volume Easy for line segments polygons Hard for curves and text Convert to lines or polygons first

4 Clipping 2D Line Segments Brute force approac: compute intersections wit all sides of clipping window Inefficient: one division per intersection H J G B F I A D C E

5 Coen-Suterland Algoritm Idea: eliminate as many cases as possible witout computing intersections Start wit four lines tat determine te sides of te clipping window y = y max x = x min x = x max y = y min

6 Case 1 Case 1: bot endpoints of line segment inside all four lines Draw (accept) line segment as is y = y max x = x min x = x max y = y min

7 Case 2 Case 2: bot endpoints outside all lines and on same side of a line Discard (reject) te line segment y = y max x = x min x = x max y = y min

8 Case 3 One endpoint inside, one outside Must do at least one intersection y = y max x = x min x = x max y = y min

9 Case 4 Bot outside May ave part inside Must do at least one intersection y = y max x = x min x = x max y = y min

10 Defining Outcodes For eac endpoint, define an outcode [b0 b1 b2 b3 ] Outcodes divide space into 9 regions Computation of outcode requires at most 4 subtractions b0 = 1 if y > ymax, 0 oterwise b1 = 1 if y < ymin, 0 oterwise b2 = 1 if x > xmax, 0 oterwise b3 = 1 if x < xmin, 0 oterwise

11 Using Outcodes AB: outcode(a) = outcode(b) = 0 Accept line segment B A 0000

12 Using Outcodes CD: outcode (C) = 0, outcode(d) 0 Compute intersection Location of 1 in outcode(d) determines wic edge to intersect wit Note if tere were a segment from C to a point in a region wit 2 ones in outcode, we migt ave to do two interesections D C

13 Using Outcodes EF: outcode(e) logically ANDed wit outcode(f) (bitwise) 0 Bot outcodes ave a 1 bit in te same place Line segment is outside of corresponding side of clipping window reject 1010 F E 0110

14 Using Outcodes GH and IJ: same outcodes, neiter zero but logical AND yields zero Sorten line segment by intersecting wit one of sides of window Compute outcode of intersection (newendpoint of sortened line segment) Reexecute algoritm 1001 H 1000 J 1010 G 0001 I

15 Efficiency In many applications, te clipping window is small relative to te size of te entire data base Most line segments can be eliminated based on teir outcodes.

16 Polygon Clipping Not as simple as line segment clipping Clipping a line segment yields at most one line segment Clipping a polygon can yield multiple polygons However, clipping a convex polygon can yield at most one oter polygon

17 Tessellation and Convexity One strategy is to replace nonconvex (concave) polygons wit a set of triangular polygons (a tessellation) Also makes fill easier

18 Clipping as a Black Box Consider line segment clipping as a process tat takes in two vertices and produces eiter no vertices or te vertices of a clipped line segment.

19 Pipeline Clipping of Line Segments Clipping against eac side of window is independent of oter sides Can use four independent clippers in a pipeline

20 Pipeline Clipping of Polygons Suterland-Hodgman algoritm Strategy used in SGI Geometry Engine Small increase in latency

21 Coen-Suterland Metod in 3D Use 6-bit outcodes Wen needed, clip line segment against planes

22 Coen-Suterland Metod in 3D 1 1 1, 1 1, 1 p p p z y x Ceck for outcodes: p p p z y x z y x z y x z y x Since SRT Divide,, z y x To avoid unnecessary float division, We can ceck

23 Coen-Suterland Metod in 3D If outcode(a)==outcode(b)==0 Accept te wole line segment. If(outcode(A) and outcode(b))!=0 Reject te line segment. Oter cases Calculate an intersection (according to outcode bits) Ten ceck outcode again Note: use parametric forms x = x a + (x b - x a )u y = y a + (y b - y a )u z = z a + (z b - z a )u = a + ( b - a )u

24 Polygon Clipping in 3D Similar to 2D clipping Bounding box Clipping wit eac clipping plane Etc..

25 Bounding Boxes Rater tan doing clipping on a complex polygon, we can use an axis-aligned bounding box or extent Smallest rectangle aligned wit axes tat encloses te polygon Simple to compute: max and min of x and y