Given points $A$, $B$, $C$ and $O$ and radius $r_1$ and $r_2$, find a function $F(r)$ that will return arc length of radius $r_1$ + $r$ bounded between two line segments $AB$ and $AC$. Point $O$ is the circle center, while points $B$ and $C$ lie on the circumference. Point $A$ can be anywhere in the area enclosed by $O$, $B$ and $C$. Radius $r_1$ is a distance from $O$ to $A$. Possible values of $r$ are $0 \le r \le r_2$

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I want to integrate a function of arc length and radius $r_1$ + $r$ over the area enclosed by $A$, $B$ and $C$.


Given input $r$, the function $F(r)$ will return arc length between points $PAB$ and $PAC$. Arc radius is the radius of circle centered at $O$ (not at $A$!). Example image

  • $\begingroup$ What is r and how does it affect anything? Is A fixed? Is r_1? It seems like you are saying contradictory things and it's unclear what you want. Please identify what is allowed to vary as input, and what is the output. As is is I can not make heads or tails out of it. $\endgroup$
    – fleablood
    May 5, 2017 at 23:23
  • $\begingroup$ My apologies. r is the input to function I want to find, given points A, B, C, and O. The r1 and r2 are distances dependent on these points. All points are fixed. The output of F is the arc length of radius r1+r bounded by lines AB and AC. $\endgroup$
    – PSz
    May 5, 2017 at 23:30
  • $\begingroup$ Just a remark/nitpicking, in general given two endpoints, there are two possible arcs. $\endgroup$
    – N.Bach
    May 6, 2017 at 0:02

1 Answer 1


Important additional assumption: we want to measure the "small" arcs with endpoints $PAB(r)$ and $PAC(r)$, and the half-line originating from $O$ and going through $A$ must intersect those arcs.

If you need something more general, the approach below will fail, but it can be adapted.

The arc length in general is "radius" $\times$ "angular width (in radian)". We already know the radius is $r_1+r$, so we need to evaluate the angle. With my additional assumption on the half-line from $O$ through $A$, it suffices to evaluate the two angles $\widehat{PAB\ O\ A}$ and $\widehat{A\ O\ PAC}$. Below I go over how to obtain $\widehat{PAB\ O\ A}$, the other angle can be obtained in a similar way.

Let $x$ be the distance between $A$ and $PAB$, $\alpha$ the angle $\widehat{OAB}$ and $\theta$ our angle of interest $\widehat{PAB\ O\ A}$. Angle $\alpha$ is a parameter of the problem, $x$ and $\theta$ vary with $r$. We only need to know the cosine of $\alpha$, which can be obtained by the relation $$ (r_1+r_2)^2=AB^2+r_1^2-2ABr_1\cos\alpha $$ Next we want $x$: \begin{align*} &(r_1+r)^2=x^2+r_1^2-2xr_1\cos\alpha=(x-r_1\cos\alpha)^2-r_1^2\cos^2\alpha+r_1^2\\ &\iff\quad (x-r_1\cos\alpha)^2=r_1^2+r^2+2r_1r+r_1^2\cos^2\alpha-r_1^2\\ &\iff\quad x=r_1\cos\alpha\pm\sqrt{r^2+2r_1r+r_1^2\cos^2\alpha} \end{align*} Now because $x\ge 0$ and $\sqrt{r^2+2r_1r+r_1^2\cos^2\alpha}\ge\lvert r_1\cos\alpha\rvert$ we actually have $x=r_1\cos\alpha + \sqrt{r^2+2r_1r+r_1^2\cos^2\alpha}$. Next is $\theta$: $$ x^2=(r_1+r)^2+r_1^2-2(r_1+r)r_1\cos\theta $$ So if you put everything together you easily get $\cos\theta$ with respect to the various parameters. Now the problem is that the identity $\theta=\arccos\big(\cos\theta\big)$ is only true when $0\le\theta\le\pi$. That's where my two additional assumptions come into play, since they guarantee $0\le\theta\le\pi$.

  • $\begingroup$ On a side note, I hope this isn't just to obtain the area of the circle enclosed by $AB$ and $AC$. $\endgroup$
    – N.Bach
    May 6, 2017 at 1:08
  • $\begingroup$ No, I need it for calculating integral over this area. Thanks for the answer, it works. $\endgroup$
    – PSz
    May 6, 2017 at 11:44

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