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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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(...) 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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(...) 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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(...) 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 (...) 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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| | Re: Here's looking at Euclid
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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: (...) (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
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