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(...) [x-post and f.u.t. to off-topic.geek since this could get complicated] Well, I have a B.A. in math, but I think this should be a simple enough answer: you can't. The main explanation is that it generally takes three distinct points to (...) (24 years ago, 1-Aug-00, to lugnet.off-topic.fun, 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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| | 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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(...) 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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| | Re: Here's looking at Euclid
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(...) The process depends on whether you want a numerical process or a construction process. I'm going to assume that you want a numerical process and that you know the coordinates of the three points. If you want a construction process to be able (...) (24 years ago, 2-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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In lugnet.off-topic.geek, John Gramley writes: **snip of some rather helpful stuff** How about this: Suppose the two points are vertices of an inscribed octagon whose sides are each of length X. Would that help? Dave! (24 years ago, 2-Aug-00, to lugnet.off-topic.geek)
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(...) Well... yes, I guess... assuming you knew X, though, you could easily calculate the diameter of the circle even without knowing the distance apart the two points were... Then the diameter of the circle would simply be: (...) (24 years ago, 2-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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(...) I have the coordinates of the points, so am I correct in thinking I can calculate X? In any case, thanks to everyone for the help; I have what I need to figure it out! Dave! (24 years ago, 2-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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(...) Nope... again, not unless you know something else (like how many verticies there are in between the two points). You could probably come up with a good guestimate, though, since there would only be 4 possible values for X, all of which you (...) (24 years ago, 2-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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(...) D'oh! I should have specified that the two points are vertices of a single side of the octagon; I'm sure of that much! Am I on the right track here? Assuming that the side of the inscribed octagon is 10 units long, I'm calculating the (...) (24 years ago, 2-Aug-00, to lugnet.off-topic.geek)
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| | Re: Here's looking at Euclid
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(...) I'd say so :) Knowing that they're two verticies side by side helps! By my calculations I get: if length of a single side of the inscribed octagon is 10: Radius of the circle: 13.06562964876 Diameter of the circle: 26.13125929753 Circumference (...) (24 years ago, 3-Aug-00, to lugnet.off-topic.geek)
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