Area of Intersection of Two Circular Segments
Date: 04/20/2007 at 03:45:14 From: Claudio Subject: area of two intersecting circular segments Given a circle of radius r and center c, suppose two intersecting chords AB and CD (intersecting in P) form two circular segments. How do I compute the area of the intersection of the two circular segments? I can easily find out the area of the whole circle and of the two circular segments, but I'm having trouble with the area of the intersection. Here's what I think might work: KNOWN VARIABLES W = center of the circle r = radius of the circle AB = first chord CD = second chord P = point of intersection of the two cords a = height of the circular segment created by AB b = height of the circular segment created by CD --- OUTPUT K = area of the overlapping section --- PB = sqrt( r^2 - (r-a)^2 ) CP = sqrt( r^2 - (r-b)^2 ) chord CB = sqrt( PB^2 + CP^2 ) X = area of triangle BCP = PB*CP/2 h = height of the triangle WCB = sqer( r^2 - (CB/2)^2) Y = area of the triangle WCB = h*CB/2 theta = angle of the circular section created by the arc from C to B = 2*arctan((CB/2)/h) Z = the area of the circular section created by the arc from C to B = pi*r^2*theta/(2pi) K = Z - Y + X Cheers, Claudio
Date: 04/23/2007 at 09:28:36 From: Doctor George Subject: Re: area of two intersecting circular segments Hi Claudio, Thanks for writing to Doctor Math. It looks like your analysis is using the Pythagorean theorem with triangles that are not necessarily right triangles. Let me show you what I came up with, then you can compare your results with mine. Problem Statement ----------------- Consider a circle with center O and radius R. Construct chords AB and CD that intersect at point P such that O is not in the region bounded by segments AP, PD and arc AD. Find the area of that region, assuming that the lengths of CA, AD, and DB are known. Solution -------- To simplify notation define the following. a = angle COA (in radians) b = angle AOD c = angle DOB The solution can be written as a function of R, a, b and c. We can relate the angles to the known segment lengths like this. CA = 2R sin(a/2) AD = 2R sin(b/2) (1) DB = 2R sin(c/2) A known theorem about circles tells us that angle CDA = a/2 angle DAB = c/2 Looking at triangle APD we can say angle APD = pi - (a+c)/2 From the law of sines we can write sin[pi - (a+c)/2] sin(c/2) ----------------- = -------- (2) AD PD The area of triangle APD is AD/2 * [PD * sin(a/2)] If we use equations (1) and (2) to substitute for AD and PD we can write the area of triangle APD as sin(a/2) sin^2(b/2) sin(c/2) R^2 ------------------------------ cos[(a+c)/2] Now we just need to add the area bounded by chord AD and arc AD. This area is R^2 (b - sin(b)) ---------------- 2 Does that make sense? Write again if you need more help. - Doctor George, The Math Forum http://mathforum.org/dr.math/
Date: 04/25/2007 at 02:54:04 From: Claudio Subject: Thank you (area of two intersecting circular segments) Dear Dr. George, Thanks very much for your help. The solution you've found is very helpful to me and I really appreciate your time. Cheers, Claudio
Date: 02/21/2016 at 05:57:14 From: Graeme Subject: the area of intersection of two circular segments The total area of intersection was provided above. But I am rather looking for the area of the intersection of the two segments. I looked at that other solution, but did not find it strictly relevant. It has been 40+ years since I did this sort of stuff. I'm having difficulty just setting up the relevant methodology ...
Date: 02/22/2016 at 11:38:41 From: Doctor Rick Subject: Re: the area of intersection of two circular segments Hi, Graeme. As far as I can see, what Doctor George found there *was* the area of overlap between two segments of the circle. He did not, however, state clearly the order of the points around the circle: they are not in the usual alphabetized A, B, C, D! Here is a figure that I believe represents what Doctor George was describing: - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.