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Calculating Circle Radius

Date: 01/29/2001 at 14:32:11
From: Lyle Galloway
Subject: Calculating circle radius from arc and chord lengths

Hi -

circle, knowing only the length of an arc on that circle and the 
chord length of that same arc. It is relatively easy to determine 
this if I know the rise, or bulge, of the arc (distance from the 
midpoint of the chord to the midpoint of the arc), but I can't seem 
to come up with a simple formula involving just arc and chord 

This is similar to an answer in your archives, 

  Calculating the Radius from a Chord   

but the rise is known in that case. 

Thanks for any suggestions.

Date: 01/29/2001 at 17:11:49
From: Doctor Rob
Subject: Re: Calculating circle radius from arc and chord lengths

Thanks for writing to Ask Dr. Math, Lyle.

You have a segment of a circle, bounded by an arc of the circle and
the chord subtending it. Let the length of the arc be s, the length
of the chord be c, and the radius of the circle be r.

For a diagram, see the Ask Dr. Math FAQ on the segment of a circle:   

Then solve 

       c/s = sin(x)/x,

for x (in radians), which must be done numerically (see below).

       r = s/(2*x).

Here 2*x is the measure in radians of the central angle subtending
the arc.

For Newton's Method, see

  Area of a Segment from Arc and Chord Length   

There is also an infinite series that can solve sin(x)/x = k,
0 < x <= Pi. Let y = 1 - k. Then 0 < y < 1, and the following infinite
series converges in this range: 

   x = sqrt(6*y)*(1 + 3*y/20 + 321*y^2/5600 + 3197*y^3/112000 +
             445617*y^4/27596800 + 1766784699*y^5/179379200000 + ...).

This converges rapidly for large values of k = 1 - y.

Another series is 

   x = Pi/(1+k) - (Pi^3*k^3/6)(1 - 4*k + [10+9*Pi^2/20]*k^2 -
         [20+16*Pi^2/5]*k^3 + ...),
     = Pi/(1+k) - (Pi^3*k^3/[6*(1+k)^4]) - (3*Pi^5*k^5/40)*(1 -
         64*k/9 + ...).

This converges rapidly for small values of k.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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