Lecture Note
University
University of California San DiegoCourse
CSE167 | Computer GraphicsPages
4
Academic year
2023
Jake Tomson
Views
0
Initial decision parameter P = (xx + 1) 2 + (yk 1/2) - 92 = (0+1) + 9-1/2)2 - r2 (2) 2 - r 2 1+92 - 2r X = I r 1/4 = 1 1 1 1 - 2 4 = 5 - r 4 The initial decision parameter is obtained by evaluating the circle function starting at position ( no / yo) = (0,9) If the radius r is specified as an integer we simply round Po to 1-92 ie Po = 1-r Ques Given a circle radius r= 10 Determine positions along the circle octant in the first quadrant from n = 0 to x=y n= 10- Po= 1-19 9 L 0 P, P + 2n K ++ 9 + 2XI +1 1= 6x40
Date P2=Pk + 2n +1 K+1 = -6 +2 x 2 + 1 x Pk -6+5 O -9 = (1,10) = -1.20 I -6 (2,10) 2 - I (3,10) P3=Pk+2n +1 3 6 (4,9) k+1 4 -3 (5,9) = 1 + 2 x 3 +1 8 15 (6,8) -1+6+1 6 5 (7,7) = +6 >0 P4 Pk+2x +1- P6 Pv+ 2x +1 - K+1 2yk+1 K+1 2nkti = 2n,+2 = 14 20 41-16 2ykth = 2yk-2 117-21-16 2F4570 2x = (2 x4) + 2 OK = K+1 = 10 NK:4 - 2ykt 2yk-2 YK=95 2yrti = (2x9) 2 = 16-2 =530 X 6/- = S**It P4 6 + 10717-20 = 16 - 20+1 his = 16-19 05 d- =-3
Ques Given a circle radius r=8 octant determine in positions along the circle to the first quadrant from x=0 n.y. ( n zy) r=8 Po - r = 1-8-2-20 PeaPEt 2n 71 k+1 x Pk 2-742x1+1 2-7+3 0 -7 (1,8) -4120 11 -4 (2.8) 2 1 (3,7) P2 Pk+2x 3 -6 (4,7) k+1 $ + (5,6) -14-2x2+1 5 +2 ( 5,5) - 4+4+1 05-2-4+5 1 1 70 = 01 P3 = Px+2x +1 - 2y K+1 k+1 PXS = 1+6+271-14*2 1 - 8 - 1 - 16 0-11-1 z 10-16 - 66
P5 = Pkt 2n +1-2ykth k+1 = 3+ (2x5+2)+1- 2 X = 3+10+2+1-12 - - = IN(3 - 14 = Q Z ZO = 1313 +1-3-12-2 = 16 -14 2-20-00
Initial Decision Parameter
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