Subject:
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Re: To all my friends!
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 7 Nov 2001 20:25:03 GMT
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Viewed:
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109 times
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In lugnet.off-topic.geek, Larry Pieniazek writes:
> In lugnet.off-topic.geek, Dave Schuler writes:
> > The way I show my gratitude when I receive informational help from people is
> > by shamelessly exploiting their knowledge again, so here goes:
> >
> > If I know the lengths of two chords of a circle, both of which are smaller
> > than the diameter and which are parallel to each other at a known distance,
> > how do I compute the radius of the circle?
> > I've looked on various geometry sites on the web, but I can't find a
> > particular means to make this calculation.
>
> I am just guessing, but don't 4 points uniquely (over) determine a circle,
> because any 3 noncollinear points actually uniquely determine a circle? If
> you can calculate the radius from any 3 non collinearpoints you can do it
> from this info as you can determine 3 points of a normalised circle from
> this info.
>
> (Say the chord lengths are C1 and C2, and the distance D...)
>
> Put C1 on the X axis and C2 above it
>
> the circle defined by these chords has points at
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> (C1/2,0)
> (-C1/2,0) from the first chord
>
> and
>
> (C2/2,D)
> (-C2/2,D) from the second chord
>
> Pick any three and if oyu have a radius from 3 points thingie you're there.
r = sqrt[(A^2/8D + D/2 - B^2/8D)^2 + B^2/4]
where:
A = chord 1 length
B = chord 2 length
D = perpindicular distance between A and B
or you could search on cyclic quadrilateral and solve for r using that mess,
simplifying for the quadrilateral = trapezoid.
-Rob.
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Message has 1 Reply:
Message is in Reply To:
 | | Re: To all my friends!
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| (...) I am just guessing, but don't 4 points uniquely (over) determine a circle, because any 3 noncollinear points actually uniquely determine a circle? If you can calculate the radius from any 3 non collinearpoints you can do it from this info as (...) (23 years ago, 7-Nov-01, to lugnet.off-topic.geek)
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