 | | Re: Here's looking at Euclid
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(...) if (...) Ok, I was more bored than I thought (assuming degrees, not radians): F(x,y)=90*pi*(sqrt((...2)^2))/2)/ (sqrt((x1-x)^2+(y1-y)^2))) SO, using the formula L = (all that garbage) and the equasion I got before: (...) you can solve for both (...) (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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(...) note (...) I think he means if you know *both* the distance on the circumfrence *and* the straight line. Or maybe he just missed that. But if you know both, then yes, you can. -Shiri (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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(...) You sure about that? Let's say I have a basketball and a baseball. I use a bit of string to make two dots one inch apart on both of them. Are the two balls now the same size? (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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(...) Hmm... That was my first instinct reaction, however, the thought occurred that perhaps what is known about the two points is their distance apart as measured along the circumfrence of the circle? (assuming closest distance, but furthest would (...) (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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(...) Yeah, I was afraid of that. Believe it or not, almost immediately after I posted I was sitting at a circular table with a can of Coke, and I realized that the same 1 1/2 inch that defines the diameter of the can only describes a tiny portion (...) (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
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