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 17:56:16 GMT
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Viewed:
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201 times
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In lugnet.off-topic.fun, John Gramley writes:
> In lugnet.off-topic.fun, Dave Schuler writes:
> > Since I haven't had geometry in a zillion years, I'm ill-equipped to solve a
> > little dilemma I've run across, and I thought a few of the more math-minded
> > among us might be able to help. Here goes:
> >
> > Given two points on a circle, can one compute the diameter of the circle if
> > the distance between the two points is known? How would one do so?
> > In the name of mercy, please keep it simple in accordance with my mathless
> > brain, but I appreciate any help you folks can give!
>
> [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
> determine a unique circle. You can get away with two distinct points only if
> you know something else about them, like that they're endpoints of the
> diameter.
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 work as well) Hmm... I think in that case, it is possible...
Let's say you've got x1,y1 and x2,y2, and you know the distance between the
two points along the circumference is L. The smallest circle possible is if
the center of the circle is at (x1+x2)/2,(y1+y2)/2. Further, we know that both
points are equidistant from the center of the circle, so the center of the
circle MUST be on the line:
y=((x1-x2)/(y1-y2))x+y1+y2+((x1^2+x2^2)/(y1-y2))
(Forgive me if I made an error in algebra)
Next, construct a function F(x,y) that takes a point (the center of the
circle) and calculates the partial circumference between the given two points
(assume closest distance again). I'm bored, but I'm not bored enough to figure
that one out without either a geometry book or a calc book (Not really sure
I'd NEED a calc book... maybe to solve it... dunno). Anyway, it's possible to
find that equasion. Next, knowing that the output of F(x,y) = L, and knowing
that x,y must lie on the aformentioned line, you can solve for the center of
the circle (the required x,y into F to get L), let's say x0,y0. And now the
diameter's just:
2*sqrt((x1-x0)^2+(y1-y0)^2)
Anyway, sorry to bore the non-math types, but assuming that you mean distance
along the circumference, yes, you can. (But if you don't know if it's the
minimal or maximal distance between the points, you may get two answers-- note
above I assumed minimal distance)
That make any sense whatsoever?
DaveE
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Message has 2 Replies:
 | | 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)
| | | 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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Message is in Reply To:
| | Re: Here's looking at Euclid
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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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