Okay, you have learned … 2D Graphics Pipeline OpenGL drawing Graphics processing consists of many stages: Viewport and World Window setup main() { glViewport(0,0,300,200); Object Modeling Object glMatrixMode(GL_PROJECTION); World Coordinates glLoadIndentity(); Local Coordinates transformation gluOrtho2D(-1,1,-1,1); glBegin(GL_QUADS); glColor3f(1,1,0); glVertex2i(-0.5,-0.5); glVertex2i(+0.5,0); glVertex2i(+0.5,+0.5); (next page) glVertex2i(-0.5,+0.5); glEnd(); } 2D Graphics Pipeline (2) Clipping and Rasterization OpenGL does these for you – no explicit OpenGL functions needed for doing clipping and rasterization Clipping window to Object Applying Object Clipping – Remove objects that are outside the viewport World Coordinates world window subset mapping world window Rasterization (scan conversion) – Convert high Simple 2D Drawing Pipeline level object descriptions to pixel colors in the frame buffer Object Display Rasterization Screen coordinates 1 2D Point Clipping 2D Line Clipping Determine whether a Determine whether a (xmax, ymax) point (x,y) is inside or (xmax, ymax) line is inside, outside, or outside of the world partially inside window? If a line is partially If (xmin <= x <= xmax) inside, we need to and (ymin <= y <= ymax) (xmin, ymin) display the inside (xmin, ymin) segment then the point (x,y) is inside else the point is outside Trivial Accept Case Trivial Reject Case (Xmax, Ymax) Lines that are clearly inside Lines that are clearly outside the world the world window - what are p1 p1 they? window – what are they? p2 Xmin <= P1.x, P2.x <=xmax p1.x, p2.x <= Xmin OR Ymin <=P1.y, P2.y <= Ymax (Xmin, Ymin) p1.x, p2.x >= Xmax OR p2 p1.y, p2.y <= ymin OR p1.y, p2.y >= ymax 2 Non-Trivial Cases Non-trivial case clipping Lines that cannot be trivially Compute the line/window p2 rejected or accepted boundary edges intersection One point inside, one point There will be four outside intersections, but only one or two are on the window p1 Both points are outside, but not “trivially” outside edges These two points are the Need to find the line end points of the desired segments that are inside line segment Rasterization (Scan Conversion) Rasterization Algorithms Convert high-level geometry description A fundamental computer graphics function to pixel colors in the frame buffer Determine the pixels’ colors, illuminations, textures, etc. Implemented by graphics hardware Rasterization algorithms Viewport Lines Rasterization Transformation Circles Triangles Polygons 3 Rasterize Lines Line Drawing Algorithm (1) Why learn this? 8 7 Line: (3,2) -> (9,6) Understand the discrete nature of computer graphics 6 5 ? Write pure device independent graphics 4 programs (Palm graphics) 3 2 Become a graphics system developer 1 0 1 2 3 4 5 6 7 8 9 10 11 12 Line Drawing Algorithm (2) Line Drawing Algorithm (3) Slope-intercept line equation Given the line equation y = mx + b, and end points (x0,y0) (x1, y1) Y = mx + b Walk through the line: starting at (x0,y0) Given two end points (x0,y0), (x1, y1), If we choose the next point in the line as X = x0 + Δx how to compute m and b? Y = ? Y = y0 + Δx* m (x1,y1) (x1,y1) m = (y1-y0) / (x1 – x0) = y0 + Δx* (dy/dx) dy = dy / dx dy (x0,y0) (x0,y0) dx dx b = y1 – m * x1 4 Line Drawing Algorithm (4) Line Drawing Algorithm (5) (x1,y1) X = x0 Y = y0 How about a line like this? Illuminate pixel (x, int(Y)) X = x0 + 1 Y = y0 + 1 * m Can we still increment X by 1 at each Step? Illuminate pixel (x, int(Y)) The answer is No. Why? X = X + 1 Y = Y + 1 * m We don’t get enough samples Illuminate pixel (x, int(Y)) How to fix it ? … Increment Y (x0, y0) Until X == x1 Line Drawing Algorihtm (6) Line Drawing Algorithm (7) X = x0 Y = y0 (x1,y1) Illuminate pixel (x, int(Y)) The above is the simplest line drawing Y = y0 + 1 X = x0 + 1 * 1/m algorithm Illuminate pixel (x, int(Y)) Not very efficient Optimized algorithms such integer DDA Y = Y + 1 X = X + 1 /m and Bresenhan algorithm (section 8.10) Illuminate pixel (x, int(Y)) are typically used … Not the focus of this course (x0,y0) Until Y == y1 5.