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| Author |
Message |
| 8 new of 20 responses total. |
wjw
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response 13 of 20:
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Dec 16 16:34 UTC 2000 |
Well, I cheated. The answer is available on the web. In order to spare
you all the extreme temptation I will not post the exact url, but, I
will tell you that my original estimate (rope = .58 times the diameter
of the pen) is *very* close. And the sloution is not trivial or easy.
nspeed's equation looks vaguely similar to the solution I saw, but I did
not do the math to see if they are equivalent.
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ric
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response 14 of 20:
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Dec 16 18:31 UTC 2000 |
(This math problem, by the way, was posted on a BBS that I used to use back
in the mid 80's - perhaps even before I started m-netting).
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rcurl
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response 15 of 20:
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Dec 16 19:45 UTC 2000 |
I saw that it wasn't hard to set up by using the formulae for areas of
segments and sectors of circles. These give simultaneous transcendental
equations, but my need for the answer wasn't great enough to do the work.
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carson
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response 16 of 20:
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Dec 16 20:31 UTC 2000 |
resp:11 (oops, you're right. that'll teach me.)
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flem
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response 17 of 20:
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Dec 17 19:33 UTC 2000 |
re 15: Right. If I'd had the formula handy for the area of a chord, this
would have been a little simpler. Well, it still would have involved arctan,
or something equivalent. But no calc.
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janc
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response 18 of 20:
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Dec 18 02:58 UTC 2000 |
"simultaneous transcendental equations". Sometimes I love mathematical
language.
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rcurl
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response 19 of 20:
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Dec 18 03:14 UTC 2000 |
Pretty transcendental, eh?
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aruba
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response 20 of 20:
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Dec 18 04:45 UTC 2000 |
The best I can do is to say that if p is the desired ratio, and
u = arccos(2p^2-1), then sin(u) - u*cos(u) = pi/2. Or, to put it another
way,
p*sqrt(1-p^2) - (2p^2-1)arccos(p) = pi/4.
If there's a closed form for the solution, I'm interested to see it.
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