A BASIC program for the circle Message #8 Posted by Tom Sherman on 26 Nov 2002, 5:57 p.m., in response to message #7 by Mike
Mike,
Don't play around with it too much! The expression is a curiosity, but totally impractical. Ordinarily, if we know any one of the three features of a circle  diameter, area, circumference  we can directly calculate the other two. That is, if we have one we can get two. But with my expression put in the solver, we need two to get one! Not much of a bargain.
Much better would be a simple BASIC program. The following is written for an HP71B:
10 DESTROY ALL @ FIX 2 @ DELAY 2
20 DISP "DIAM=1,AREA=2,CIRC=3"
30 DELAY 8
40 INPUT "START WITH 1,2,or 3?";N
50 IF N>1 THEN 100
60 INPUT "DIAMETER =";D
70 DISP "AREA =";PI*D*D/4
80 DISP "CIRCUMFERENCE =";PI*D
90 GOTO 200
100 IF N>2 THEN 160
110 INPUT "AREA =";A
120 D=SQR(4*A/PI)
130 DISP "DIAMETER =";D
140 DISP "CIRCUMFERENCE =";PI*D
150 GOTO 200
160 INPUT "CIRCUMFERENCE =";C
170 D=C/PI
180 DISP "DIAMETER =";D
190 DISP "AREA =";PI*D*D/4
200 END
Tom
