# Computer graphics Scan conversion of Line, circle & ellipse.

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14-Dec-2015

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Computer graphics Scan conversion of Line, circle & ellipse Slide 2 Scan conversion is the process of converting basic, low level objects in to their corresponding pixel map representations. This is often an approximation to the object, since the frame buffer is a discrete grid. Slide 3 Direct or floating or simple algorithm Digital Differential Analyzer algorithm (DDA) Bresenham's algorithm or midpoint algorithm General Bresenham's algorithm Line Drawing algorithms Description: Given the specification for a straight line (parameters). find the collection of addressable pixels which most closely approximates this line (mathematical equations). Draw the scan conversion of lines (frame buffer map). Goals Straight lines should appear straight. Lines should start and end accurately. Lines should be drawn as rapidly as possible. matching endpoints with connecting lines. Lines should have constant brightness. Slide 4 1. A straight line may be defined by two endpoints and an equation If the two endpoints used to specify a line are (X1,Y1) and (X2,Y2),then the equation of the line is used to describe the X, Y coordinates of all the pointes that lie between these two endpoints. x1 y1 x2 Y2 Description: Direct method dy > dx negativeVIII dy < dx negativeVII positiveVI dy > dx positiveV negativeIV dy < dx negativeIII positiveII dy > dx positiveI ratio Slope (m) Octant Octant covering in 2D space Dx = x2 - x1 2. The line equation depend on octant which contain line Dy = y2 - y1 Slide 5 Y= M * X + B Where: Where: M is the slope of the line. B is the Y intercept. B= Y M * X Note : Note : The slope between any point (X1,Y1) on the line and (X2,Y2) M = DY/ DX 3. The equation of straight line in octants Dx > Dy x go from x1 to x2 with values (+1) or (-1) EX1:- draw line between(0,1) and ( 5,4) Sol:- M= 4 -1 / 5-0 = 3/5 = 0.6, B=1, y= (0.6)*X+1 x go from x1 to x2 with integer values (+1) in each step x go from x1 to x2 with integer values (+1) in each step 0 1 2 3 4 5 0 1 2 3 4 x012345y11.62.22.83.44 draw(0,1)(1,2)(2,2)(3,3)(4,3)(5,4) Slide 6 X= Y B / M 4. The equation of straight line in octants Dy > Dx y go from y1 to y2with values (+1) or (-1) y go from y1 to y2 with values (+1) or (-1) Ex: draw line between(1,2) and ( 3,6) sol:- m= 2, B= 0 XYdraw12(1,2) 1.53(2.3) 24(2,4) 2.55(3,4) 36(3,6) 1 2 3 4 5 1 2 3 4 5 6 Slide 7 Ex: draw line between(2,-2) and ( 6,-6) Sol:- m= -4 / 4 = -1, B= -2 (-1*2) = 0 m= -4 / 4 = -1, B= -2 (-1*2) = 0 XYdraw 2-2(2,-2) 3-3(3,-3) 4-4(4,-4) 5-5(5,-5) 6-6(6,-6) 2 3 4 5 6 -2 -3 -4 -5 -6 Slide 8 features: The algorithm performs a floating-point multiplication for every step in x. This method therefore requires an enormous number of floating-point multiplications, and is therefore expensive. Round functions are needed not general for all octants Can get gaps in the line. example: y = 10.x + 2 x=1, y=12; x=2, y=22. Slide 9 Algorithm for octant with x1Dy), slope( positive, negative)) Slide 10 Digital Differential Analyzer (DDA) Digital Differential Analyzer (DDA) 1. We calculate the length of the line in the X direction ( number of pointes) by the equation Abs(y2-y1) 2. calculate the length of the line in the Y direction ( number of pointes) by the Equation Abs(x2-x1) 3. The Length estimates is equal to the larger of the above two equations. Where ABS is a function takes the positive of the arguments. length=max( Abs(y2-y1), Abs(x2-x1) Description: Slide 11 4. The increment steps ( dX and dY ) are used to increment the X and Y coordinates for the next pointes to be plotted. Dx= (x2-x1) / length Dy= (y2-y1) / length 5. each point will calculate as equation 7. first point we add 0.5 x = x + Dx y = y + Dy x = x1 + 0.5 x = x1 + 0.5 y = y1 + 0.5 y = y1 + 0.5 6. Integer function works as follow( round down) Ex.Integer (8.5) = 8 Ex. Integer (8.5) = 8 Integer (-8.5) = -9 Integer (-8.5) = -9 Slide 12 Ex1: Consider the line from (-8,-4) to (0,0)Use DDA to scan conversion of line. Sol 1 : X1=-8 ; Y1=-4 ; X2=0 ; Y2=0 ; Length= 8 dX= 1; dY=0.5 ; X=-7.5 ; Y=-3.5 IXYdraw -7.5-3.5(-8,-4) 1-6.5-3(-7,-3) 2-5.5-2.5(-6,-3) 3-4.5-2(-5,-2) 4-3.5-1.5(-4,-2) 5-2.5(-3,-1) 6-1.5-0.5(-2,-1) 7-0.50(-1,0) 80.50.5(0,0) -8 -7 -6 -5 -4 -3 -2 -1 0 -4 -3 -2 -1 0 Slide 13 Start Length=ABS (X2-X1) If length < ABS (Y2-Y1) Then Length=ABS (Y2-Y1) dX = (X2 - X1) / Length dY = (Y2 - Y1) / Length X=X1 + 0.5 Y=Y1 + 0.5 Plot (Integer (X), Integer (Y)) For I=1 to Length Begin X=X + dX Y=Y + dY Plot (Integer (X), Integer (Y)) End Finish Algorithm DDA Slide 14 Features of DDA 1- The algorithm is orientation dependent 2- The end point accuracy deteriorates 3- The algorithm suffer from the fact that it must be performed using floating point arithmetic HW1:Consider the line from (0,0) to (-8,-4) HW1: Consider the line from (0,0) to (-8,-4) evaluate the DDA algorithm evaluate the DDA algorithm