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 17:15:09 GMT
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Viewed:
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103 times
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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
(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.
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Message has 1 Reply:
 | | Re: To all my friends!
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| (...) 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 (...) (23 years ago, 7-Nov-01, to lugnet.off-topic.geek)
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Message is in Reply To:
| | To all my friends!
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| 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 (...) (23 years ago, 7-Nov-01, to lugnet.off-topic.geek)
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