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 Assistant Professor Mechanical Department 1

RANDOM SCAN SYSTEM 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 Of Engineering

RASTER SCAN SYSTEM 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 figure below: Mechanical Engineering Department – School Of Engineering

REPRESENTATION OF LINE AND CIRCLE Mechanical Engineering Department – School Of Engineering

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

DDA ALGORITHM ΔY = ΔX ΔX = ΔY YK+1 = YK + ΔY YK+1 = YK + ΔY CASE 1 = Xn − X1 ΔX = +1 if Xn > X1 ΔX = -1 if Xn < X1 Yn − Y1 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 ΔY = ΔX Xn –X1 ΔX = Fn –F1 ΔY Xn –s1 YK+1 = YK + ΔY YK+1 = YK + ΔY XK+1 = XK + ΔX XK+1 = XK + ΔX Mechanical Engineering Department – School Of Engineering

EXAMPLE EXAMPLE 1: Generate a straight line connecting two points (1, 2) and (8, 6) using DDA algorithm. Estimation of increments Xn – X1 = 8 – 1 = 7 Yn – Y1 = 6 – 2 = 4 X1 = 1 Xn = 8 Y1 = 2 Yn = 6 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 Of Engineering

EXAMPLE cont.. POINT 1 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 POINT 3 X3 = X2+ ΔX = 2.5 + 1 = 3.5 Y3 = Y2+ ΔY = 3.071 + 0.571 = 3.642 POINT 4 X4 = X3+ ΔX = 3.5 + 1 = 4.5 Y4 = Y3+ ΔY = 3.642 + 0.571 = 4.213 POINT 5 X5 = X4+ ΔX = 4.5 + 1 = 5.5 Y5 = Y4+ ΔY = 4.213 + 0.571 = 4.784 POINT 6 X6 = X5+ ΔX = 5.5 + 1 = 6.5 Y6 = Y5+ ΔY = 4.784 + 0.571 = 5.355 POINT 7 X7 = X6+ ΔX = 6.5 + 1 = 7.5 Y7 = Y6+ ΔY = 5.355 + 0.571 = 5.926 POINT 8 X8 = X7+ ΔX = 7.5 + 1 = 8.5 Y8 = Y7+ ΔY = 5.926 + 0.571 = 6.497 (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 Of Engineering

BRESENHAM’S ALGORITHM 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 parameter P1 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 Of Engineering

EXAMPLE Example 2: Generate a straight line connecting two end points (21, 11) and (26, 15) using Bresenham’s algorithm. 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) Point 1: (x1, y1) = (21, 11) P1 = 2Yc – Xc = 8 – 5 = 3 Point 2: 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 Of Engineering 11

EXAMPLE cont.. Point 3: X3 = x2 + Δx = 22 + 1 = 23 Y3 = Y2 + Δy = 12 + 1 = 13 (x3, y3) = (23, 13) P3 = P2 + 2Yc – 2Xc = 1 + 8 – 10 = -1 (P < 0) Point 4: X4 = x3 + Δx = 23 + 1 = 24 Y4 = Y3 = 13 P4 = P3 + 2Yc = -1 + 8 = 7 Point 5: X5 = x4 + Δx = 24 + 1 = 25 Y5 = Y4 + Δy = 13 + 1 = 14 (x4, y4) = (24, 13) (P > 0) (x4, y4) = (25, 14) P5 = P4 + 2Yc – 2Xc = 7 + 8 – 10 = 5 (P > 0) Point 6: 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 Of Engineering 12