Visibility Determination AKA, hidden surface elimination - PowerPoint PPT Presentation

Visibility Determination • AKA, hidden surface elimination Visibility Algorithms Roger Crawfis CIS 781 This set of slides reference slides used at Ohio State for instruction by Prof. Machiraju and Prof. Han-Wei Shen. Hidden Lines Hidden Lines Removed

Topics Hidden Surfaces Removed � Backface Culling � Hidden Object Removal: Painters Algorithm � Z-buffer � Spanning Scanline � Warnock � Atherton-Weiler � List Priority, NNA � BSP Tree � Taxonomy Where Are We ? Back-face Culling � Canonical view volume (3D image space) � Clipping done � division by w � z > 0 � Problems ? y y clipped line clipped line x x 1 � Conservative algorithm 1 near far z z 0 1 � Real job of visibility never solved image plane near far

Back-face Culling Back-face Culling • If a surface’s normal is pointing in the same general • Only handles faces oriented direction as our eye, then this is a back face away from the viewer: • The test is quite simple: if N * V > 0 then we reject the surface – Closed objects – Near clipping plane does not intersect the objects • Gives complete solution for a single convex • If test is in eye- space, then if polyhedron. N z > 0 reject. • Still need to sort, but we have reduced the number of primitives to sort. Painters Algorithm Point sorting vs Polygon Sorting � Sort objects in depth order � Draw all from Back-to-Front (far-to-near) • What does it mean to sort two line � Simply overwrite the existing pixels. segments? � Is it so simple? – Zmin? at z = 22, at z = 18, at z = 10, – Zmax? – Slope? – Length? X z Y

3D Cycles Form of the Input � How do we deal with cycles? Object types: what kind of objects does it handle? � How do we deal with intersections? � convex vs. non-convex � How do we sort objects that overlap in Z? � polygons vs. everything else - smooth curves, non- continuous surfaces, volumetric data Z Form of the output Object Space Algorithms � Volume testing – Weiler-Atherton, etc. Precision: image/object space? � input: convex polygons + infinite eye pt � output: visible portions of wireframe edges � Object Space � Image Space � Geometry in, geometry out � Geometry in, image out � Independent of image resolution � Visibility only at pixel centers � Followed by scan conversion

Image-space algorithms Conservative Visibility Algorithms � Traditional Scan Conversion and Z-buffering � Viewport clipping � Hierarchical Scan Conversion and Z-buffering � Back-face culling � input: any plane-sweepable/plane-boundable � Warnock's screen-space subdivision objects � preprocessing: none � output: a discrete image of the exact visible set Z-buffer: Scanline Z-buffer I. for each polygon do � Z-buffer is a 2D array that stores a depth value for each pixel. for each pixel (x,y) in the polygon’s projection do z := -(D+A*x+B*y)/C; � InitScreen : DrawZpixel(x, y, z, polygon’s color); for i := 0 to N do for j := 1 to N do II. for each scan-line y do Screen[i][j] := BACKGROUND_COLOR ; Zbuffer[i][j] := ∞ ; for each “in range” polygon projection do for each pair (x 1 , x 2 ) of X-intersections do � DrawZpixel (x, y, z, color) for x := x 1 to x 2 do if (z <= Zbuffer[x][y]) then z := -(D+A*x+B*y)/C; DrawZpixel(x, y, z, polygon’s color); Screen[x][y] := color; Zbuffer[x][y] := z; If we know z x,y at (x,y) then: z x+1,y = z x,y - A/C

Incremental Scanline Incremental Scanline (contd.) + + + = Ax By Cz D 0 � All that was about increment for pixels on each scanline. − + + ( Ax By D ) = ≠ z , C 0 � How about across scanlines for a given pixel ? C On a scan line Y = j , a constant � Assumption: next scanline is within polygon Thus depth of pixel at ( x 1 = x + ∆ x , j ) − + + − + + ( Ax Bj D ) ( Ax Bj D ) − + + + + ( Ax By D ) ( Ax By D ) − = + z z 1 − = + z z 1 1 C C 1 C C − A ( x x ) − A ( y y ) − = z z 1 − = z z 1 1 C 1 C A B = − ∆ , since ∆ x = 1, z z ( ) x = − ∆ , since ∆ y = 1, z z ( ) y 1 C 1 C A B = − z z = − z z 1 C 1 C Z-buffer - Example Non-Planar Polygons ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ P 3 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ − ( y y ) Z-buffer = + − z z ( z z ) 1 s ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ a 1 4 1 − ( y y ) 1 4 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ − ( y y ) ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ = + − z z ( z z ) 1 s b 1 2 1 − ( y y ) ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ 1 2 P 4 P 2 − ( x x ) = + − z z ( z z ) a p p a b a − ( x x ) y s z a z p z b a b Screen P 1 Bilinear Interpolation of Depth Values

[0,7,5] [6,7,5] [0,6,7] 5 5 5 5 5 5 5 7 5 5 5 5 5 5 5 5 5 5 5 6 7 5 6 7 5 5 5 5 4 5 6 7 5 5 5 3 4 5 6 7 5 5 2 3 4 5 6 7 5 [0,1,5] [0,1,2] [5,1,7] 5 5 5 5 5 5 5 ∞ 5 5 5 5 5 5 5 ∞ 5 5 5 5 5 5 ∞ ∞ 5 5 5 5 5 5 ∞ ∞ 5 5 5 5 5 ∞ ∞ ∞ 5 5 5 5 5 ∞ ∞ ∞ 5 5 5 5 ∞ ∞ ∞ ∞ 5 5 5 5 ∞ ∞ ∞ ∞ 5 5 5 ∞ ∞ ∞ ∞ ∞ 4 5 5 7 ∞ ∞ ∞ ∞ 5 5 ∞ ∞ ∞ ∞ ∞ ∞ 3 4 5 6 7 ∞ ∞ ∞ 5 ∞ ∞ ∞ ∞ ∞ ∞ ∞ 2 3 4 5 6 7 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ Non Trivial Example ? Example Rectangle: P1(10,5,10), P2(10,25,10), P3(25,25,10), P4(25,5,10) Triangle: P5(15,15,15), P6(25,25,5), P7(30,10,5) Frame Buffer: Background 0, Rectangle 1, Triangle 2 Z-buffer: 32x32x4 bit planes

Z-Buffer Advantages Z-Buffer Disadvantages � Does not do transparency easily � Simple and easy to implement � Aliasing occurs! Since not all depth questions can be resolved � Amenable to scan-line algorithms � Anti-aliasing solutions non-trivial � Can easily resolve visibility cycles � Shadows are not easy � Handles intersecting polygons � Higher order illumination is hard in general Spanning Scan-Line Spanning Scan Line Algorithm Can we do better than scan-line Z-buffer ? • Use no z-buffer � Scan-line z-buffer does not exploit • Each scan line is subdivided into several "spans" � Scan-line coherency across multiple scan-lines • Determine which polygon the current span belongs to � Or span-coherence ! • Shade the span using the current polygon’s color � Depth coherency • Exploit "span coherence" : • For each span, only one visibility � How do you deal with this – scan-conversion algorithm and a little test needs to be done more data structure – Assuming no intersecting polygons.

Spanning Scan Line Algorithm Spanning Scan Line Algorithm • A scan line is subdivided into a sequence of spans • Each span can be "inside" or "outside" polygon areas – "outside“: no pixels need to be drawn (background color) – "inside“: can be inside one or multiple polygons • If a span is inside one polygon, the pixels in the span will be drawn with the color of that polygon • If a span is inside more than one polygon, then we need to compare the z values of those polygons at the scan line edge intersection point to determine the color of the pixel When there are multiple Determine a span is inside or polygons outside (single polygon) • When a scan line intersects an edge of a • Each polygon will have its own in/out flag polygon • There can be more than one polygon having – for a 1 st time, the span becomes "inside" of the the in/out flags to be "in" at a given instance polygon from that intersection point on – for a 2 nd time, the span becomes "outside“ of the • We want to keep track of how many polygons polygon from that point on the scan line is currently in • Use a "in/out" flag for each polygon to keep • If there is more than one polygon "in", we track of the current state need to perform z value comparison to • Initially, the in/out flag is set to be "outside" determine the color of the scan line span (value = 0 for example). Invert the tag for “inside”.

Many Polygons ! Z value comparison ET x y max ∆ x poly-ID • When the scan line intersects an edge, leaving the top-most polygon, we use the color of the remaining polygon if there is now only 1 polygon "in". poly-ID A,B,C,D color in/out flag PT • If there is still more than one polygon with an "in" flag, we need to perform z comparison, but only when the � Use a PT entry for each polygon scan line leaves a non-obscured polygon. � When polygon is considered, Flag is true � Multiple polygons can have their flags set to true � Use IPL as active In-Polygon List ! Spanning Scan-Line: Example Example 1 Y AET IPL BG a I x 0 , ba , bc, x N BG, BG+S, BG IV II x 0 , ba , bc, 32, 13, x N BG, BG+S, BG, BG+T, BG S T III x 0 , ba ba , 32, ca, 13, x x N BG, BG+S, BG+S+T, BG+T, BG III x 0 , , 32, ca, 13, BG, BG+S, BG+S+T, BG+T, BG N 2 III IV x 0 , ba , ac, 12, 13, x N BG, BG+S, BG, BG+T, BG c 1 BG a II 3 IV S I T b 2 III c X 0 X N II 3 I Think of ScanPlanes to understand ! b