Out of a circular piece of paper, you wish to form a cone cup, so you cut out a circle wedge (with its extreme at the circle center) and join the resulting straight sides, forming a conical cup.
What size should the wedge be, to maximize the capacity of the cone?
(In reply to
re: Extension to this problem... -- numerical solution by Charlie)
The individual volume figures on my previous post were limited in accuracy due to the specification of the starting and increment values with single precision constants. The total was unaffected so the solution is still the same. However, the correct numbers in the final table are:
116.640 0.104000632107 0.352639958810 0.456640590917
116.641 0.104002310809 0.352638280138 0.456640590947
116.642 0.104003989522 0.352636601448 0.456640590970
116.643 0.104005668245 0.352634922741 0.456640590986
116.644 0.104007346979 0.352633244017 0.456640590996
116.645 0.104009025724 0.352631565275 0.456640591000
116.646 0.104010704480 0.352629886516 0.456640590996
116.647 0.104012383246 0.352628207740 0.456640590986
116.648 0.104014062023 0.352626528946 0.456640590969
116.649 0.104015740811 0.352624850135 0.456640590946
As mentioned, the answer, 116.645 is still the same, with the same total volume.
The #'s on the constants in the program make them double-precision:
DEFDBL A-Z
pi = ATN(1) * 4
dr = pi / 180
c = 2 * pi
FOR angle = 116.64# TO 116.65# STEP .001#
c1 = c * angle / 360: c2 = c - c1
r1 = c1 / (2 * pi): r2 = c2 / (2 * pi)
h1 = SQR(1 - r1 ^ 2): h2 = SQR(1 - r2 ^ 2)
v1 = pi * h1 * r1 * r1 / 3: v2 = pi * h2 * r2 * r2 / 3
PRINT USING "###.### ##.############ ##.############ ##.############"; angle; v1; v2; v1 + v2
NEXT angle
Without the #'s the angles were near 116.640, 116.641, etc., but not to the full 15-decimal (double) precision.
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Posted by Charlie
on 2005-04-10 14:29:47 |
re(2): Extension to this problem... -- numerical solution
