 | | Re: Here's looking at Euclid —Dave Schuler
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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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| | | | | Re: Here's looking at Euclid —David Eaton
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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 —Dave Schuler
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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 —David Eaton
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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 —Dave Schuler
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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 —David Eaton
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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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