williams-et-al-1983-apple-ii-computer-graphics

CHAPTER 11-GRAPHS AND CHARTS 13 1After you have run the program a few times with your own data, we cango through it and hit the highlights.7¥' DEF FN X (T) =8Y, DEF FN Y(T) =8Y, * COS (T)7Y, * SIN (T)The fu nctions defi ned here are used to calculate the X and Y-coordinatesof points on the circle. The values 80 and 70 define the radius of the circle(two radii?!?). The dots on the screen are slightly higher than they are wide,so if you let both numbers be 80 (as theoretically they should), the pielooks more like an egg.12¥' READ NUM%13¥' DIM DTA( NUM%)14Y, FOR . I = 1 TO NUM%15¥' READ DTA(I)16Y, TTL = TTL + DTA( I)17¥' NEXT INUM% is the number of regions, and the data values are read into the arrayDTA. Since a total is required to calculate the relative size of each "slice,"we accumulate the data values in TTL.24¥' FOR T = Y, TO 1. 57 STEP • Y,125¥' X = FN X(T) :Y = FN Y(T)26Y, HPLOT X + 14Y,, Y + ' 8y,27¥' HPLOT -X + 14Y,, Y + 8Y,28¥' HPLOT X + 14¥', -Y + 8Y,29¥' HPLOT -X + 14Y,, -Y + 8Y,3¥'¥' NEXT TThe circle is drawn using a little simple trigonometry. (There is really nosuch thing as simple trigonometry!) You will notice here (and also later inthe program) that we place the center at (140,80) by adding 140 to each Xand 80 to each Y as we plot.The real question is, why does T go from 0 to 1.57? The SIN and COSfunctions we use measure a circle in radians instead of degrees. There areroughly 6.28 radians in a fu ll circle (3.14 in a semicircle), and we use thesymmetry of the circle so that we only need to calculate the points for onequarterof the circle. The result of all this is that we go one-quarter of theway to 6.28, and one-quarter of 6.28 is the elusive 1.57.36Y, FOR I = 1 TO NUM%37¥' ACC = ACC + DTA(I)38¥' R = (ACC / TTL) * 6.2839¥' HPLOT 14Y,, 8Y, TO FN X (R) + 14Y,,FN Y(R) + 8¥'4Y,Y, NEXT I41Y, END