Subject:
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Re: Here's looking at Euclid
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Tue, 1 Aug 2000 19:17:51 GMT
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Viewed:
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209 times
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Anyway, I'm assuming you know BOTH the distance between the two points (D),
AND the distance along the circumference (L)... I just did it again to be
totally geeky, without assuming knowledge of x1,y1 and x2,y2:
In the equasion:
pi*(sqrt((D^2/2)-2*(D/sqrt(2))*x))*(sin^-1(D/(2*sqrt((D^2/2)-2*(D/sqrt(2))
*x))))=90*L
solve for x... then:
Diameter = 2*sqrt((D^2/2)-2*(D/sqrt(2))*x)
DaveE
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Message is in Reply To:
 | | Re: Here's looking at Euclid
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| (...) note (...) Well, I'm assuming that we know the coordinates of the two points-- in which case, you'll get slightly different results on the baseball and the basketball... Using the 1 inch string on the basketball will create two points which (...) (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
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