Calculating Circle RadiusDate: 01/29/2001 at 14:32:11 From: Lyle Galloway Subject: Calculating circle radius from arc and chord lengths Hi - I am trying to find a formula that will give me the radius of a 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 lengths. This is similar to an answer in your archives, Calculating the Radius from a Chord http://mathforum.org/dr.math/problems/auer8.18.98.html 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: http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment Then solve c/s = sin(x)/x, for x (in radians), which must be done numerically (see below). THERE IS NO FORMULA TO DO THIS ALGEBRAICALLY. Then 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 http://mathforum.org/dr.math/problems/castricone.11.27.00.html 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 http://mathforum.org/dr.math/ |
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