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INTERACTIVE COMPUTER GRAPHICS

INTERACTIVE COMPUTER GRAPHICS. Presented by PUDI SATYANARAYANA MURTHY M.Tech , Assistant Professor DEPARTMENT OF MECHANICAL ENGINEERING VISAKHA INSTITUTE OF ENGINEERING & TECHNOLOGY. CHAPTER 2.2: INTERACTIVE COMPUTER GRAPHICS. Prof. RATNADEEPSINH M. JADEJA

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INTERACTIVE COMPUTER GRAPHICS

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  1. INTERACTIVE COMPUTER GRAPHICS Presented by PUDI SATYANARAYANA MURTHY M.Tech, Assistant Professor DEPARTMENT OF MECHANICAL ENGINEERING VISAKHA INSTITUTE OF ENGINEERING & TECHNOLOGY

  2. CHAPTER 2.2: INTERACTIVE COMPUTERGRAPHICS Prof. RATNADEEPSINH M.JADEJA Assistant Professor MechanicalDepartment 1

  3. RANDOM SCANSYSTEM In Random Scan System, an electron beam is directed to only those parts of the screen where a picture is to be drawn. The picture is drawn one line at a time, so also called vector displays or stroke writing displays. After drawing the picture the system cycles back to the first line and design all the lines of the picture 30 to 60 time each second. Mechanical Engineering Department – School OfEngineering

  4. It is the most common type of graphics monitor based on television technology. In a raster scan system, the electron beam is swept across the screen, one row at a time from top to bottom. When electron beam moves across each row the beam intensity is turned ON and OFF to create a pattern of illuminated spots. Picture definition is stored in a memory called frame buffer which holds the set of intensity values, which are then retrieved from the frame buffer and pointed on the screen one row at a time as shown in figurebelow: RASTER SCANSYSTEM Mechanical Engineering Department – School OfEngineering

  5. REPRESENTATION OF LINE ANDCIRCLE Mechanical Engineering Department – School OfEngineering

  6. DDAALGORITHM Y dF =n= ΔY = Fn –F1 =CONSTANT ΔX Xn–s1 dX Y = MX +C (Xn,Yn) Xn–X1 ΔX= ΔY Fn–F1 ΔY= ΔX YK+1 Fn–F1 Xn–s1 ΔY YK ΔX YK+1 = YK +ΔY (X1,Y1) Similarly X XK XK+1 XK+1 = XK +ΔX Mechanical Engineering Department – School OfEngineering

  7. Yn −Y1 CASE 1 = Xn −X1 ΔX =+1 if Xn >X1 ΔX =-1 if Xn <X1 DDAALGORITHM CASE 2 = Yn −Y1 ΔY=+1 if Yn > Y1 ΔY=-1 if Yn <Y1 • Xn −X1 Cal cul at∶e Cal cul at∶e Fn–F1 Xn–X1 ΔY= ΔX ΔY ΔX= Fn–F1 Xn–s1 YK+1 = YK +ΔY YK+1 = YK +ΔY XK+1 = XK +ΔX XK+1 = XK +ΔX Mechanical Engineering Department – School OfEngineering

  8. EXAMPLE 1: Generate a straight line connecting two points (1, 2) and (8, 6) using DDAalgorithm. EXAMPLE Estimation ofincrements Xn – X1 = 8 – 1 =7 Yn – Y1 = 6 – 2 =4 Y1 =2 Yn =6 X1 =1 Xn =8 Xn−X1 > and Xn >X1 So, ΔX =1 Yn −Y1 and 4 7 Fn–F1 ΔY= ΔX= =0.571 Xn–s1 Mechanical Engineering Department – School OfEngineering

  9. POINT1 X1 = X1+ 0.5 = 1 + 0.5 =1.5 Y1 = Y1+ 0.5 = 2 + 0.5 =2.5 POINT 2 X2 = X1+ ΔX = 1.5 + 1 =2.5 Y2 = Y1+ ΔY = 2.5 + 0.571 =3.071 POINT3 X3 = X2+ ΔX = 2.5 + 1 =3.5 Y3 = Y2+ ΔY = 3.071 + 0.571 =3.642 POINT4 X4 = X3+ ΔX = 3.5 + 1 =4.5 Y4 = Y3+ ΔY = 3.642 + 0.571 =4.213 POINT5 X5 = X4+ ΔX = 4.5 + 1 =5.5 Y5 = Y4+ ΔY = 4.213 + 0.571 =4.784 POINT6 X6 = X5+ ΔX = 5.5 + 1 =6.5 Y6 = Y5+ ΔY = 4.784 + 0.571 =5.355 POINT7 X7 = X6+ ΔX = 6.5 + 1 =7.5 Y7 = Y6+ ΔY = 5.355 + 0.571 =5.926 POINT8 X8 = X7+ ΔX = 7.5 + 1 =8.5 Y8 = Y7+ ΔY = 5.926 + 0.571 =6.497 EXAMPLEcont.. (X8, Y8) = (8,6) (X1, Y1) = (1,2) (X2, Y2) = (2,3) (X3, Y3) = (3,3) (X4, Y4) = (4,4) (X5, Y5) = (5,4) (X6, Y6) = (6,5) (X7, Y7) = (7,5) Mechanical Engineering Department – School OfEngineering

  10. BRESENHAM’SALGORITHM Input the two end points (X1, Y1) and (Xn,Yn). Calculate : Xc = Xn – X1 and Yc = Yn –Y1 If Xc > Yc then slop is less then 1 so, ΔX =1 Calculate the starting value of decision parameterP1 as P1 = 2Yc –Xc If Pk < 0 The next point is (Xk+1, Yk) and Pk+1 = Pk +2Yc If Pk > 0 The next point is (Xk+1, Yk+1) and Pk+1 = Pk + 2Yc –2Xc Mechanical Engineering Department – School OfEngineering

  11. Example 2: Generate a straight line connecting two end points (21, 11) and (26, 15) using Bresenham’s algorithm. EXAMPLE Solution: X1 =21 X2 =26 Y1 =11 Y2 =15 Xc = Xn – X1 = 26 – 21 =5 Yc = Yn – Y1 = 15 – 11 =4 (Xc > Yc, so slop is less then 1 so, ΔX =1) Point1: (x1, y1) = (21,11) P1 = 2Yc – Xc = 8 – 5 =3 Point2: X2 = x1 + Δx = 21 + 1 =22 Y2 = Y1 + Δy = 11 + 1 =12 (P >0) (x2, y2) = (22,12) (P >0) P2 = P1 + 2Yc – 2Xc = 3 + 8 – 10 =1 Mechanical Engineering Department – School OfEngineering 11

  12. Point3: X3 = x2 + Δx = 22 + 1 =23 Y3 = Y2 + Δy = 12 + 1 =13 EXAMPLEcont.. (x3, y3) = (23,13) P3 = P2 + 2Yc – 2Xc = 1 + 8 – 10= -1 (P <0) Point4: X4 = x3 + Δx = 23 + 1 =24 (x4, y4) = (24,13) Y4 = Y3 =13 P4 = P3 + 2Yc = -1 + 8 =7 Point5: X5 = x4 + Δx = 24 + 1 =25 Y5 = Y4 + Δy = 13 + 1 =14 (P >0) (x4, y4) = (25,14) P5 = P4 + 2Yc – 2Xc = 7 + 8 – 10= 5 (P >0) Point6: X6 = x5 + Δx = 25 + 1 =26 Y6 = Y5 + Δy = 14 + 1= 15 (x4, y4) = (26,15) P6 = P5 + 2Yc – 2Xc = 5 + 8 – 10= 3 (P >0) Mechanical Engineering Department – School OfEngineering 12

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