So the equation for the area of the segment of a circle, given that you know the radius and height of the segment is:

$$\text {area} = (r^2)(\cos^{-1}\frac{r-h}{r})-(r-h)(\sqrt {(2rh-h^2)}$$

where $r$ = radius of the circle and $h$ = height of the given segment

Is there any way to make h the subject of this equation?

  • $\begingroup$ Use mathjax please $\endgroup$ – QuIcKmAtHs Jan 24 '18 at 14:11
  • $\begingroup$ What do you mean height of segment? Or do you mean a sphere $\endgroup$ – QuIcKmAtHs Jan 24 '18 at 14:15
  • $\begingroup$ As in the distance between the chord and the outside of the circle $\endgroup$ – PQuinn95 Jan 24 '18 at 14:18
  • $\begingroup$ Why is that the height? It is very unclear. Please post an image if possible $\endgroup$ – QuIcKmAtHs Jan 24 '18 at 14:19
  • $\begingroup$ I believe it is also called a sagitta if that helps? $\endgroup$ – PQuinn95 Jan 24 '18 at 14:20

As said in comments and in the linked post, you face the transcendental equation in $h$ $$\text{area}=r^2 \cos ^{-1}\left(\frac{r-h}{r}\right)-(r-h) \sqrt{2 h r-h^2}$$ and numerical methods are required.

To make the problem simpler, let $h=r x$ and $a=\frac{\text{area}}{r^2}$ which means that we look for the zero of function $$f(x)= \cos ^{-1}(1-x)+(x-1)\sqrt{(2-x) x}-a$$ $$f'(x)=2 \sqrt{(2-x) x}$$ Using Newton method, the iterates will be given by $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ but the problem is to find a good estimate of the starting value $x_0$.

For small $x$, this can be done using a Taylor expansion $$\cos ^{-1}(1-x)+(x-1)\sqrt{(2-x) x}=\frac{4\sqrt{2}}{3} x^{3/2}+O\left(x^{5/2}\right)$$ giving as an estimate $$x_0=\frac 12 \left(\frac{3a}{2}\right)^{2/3}\tag 1$$ It can be made better building a simple Padé approximant $$\cos ^{-1}(1-x)+(x-1)\sqrt{(2-x) x}\approx \frac{4 \sqrt{2}}3\frac{ x^{3/2}}{ 1+\frac{3 }{20}x}$$ Let $x=y^2$ to reduce the problem to the solution of the cubic equation in $y$ $$80 \sqrt{2} y^3-9 a y^2-60 a=0$$ Its only real solution is given by $$y=\frac{3 a}{80 \sqrt{2}}\left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{128000}{9 a^2}\right)\right) \right)\tag 2$$ For illustration purposes, let us try using $$a=0.5\implies y\approx 0.655985\implies x\approx 0.430316$$ (notice that $(1)$ would give $x_0\approx 0.412741 $).

Now, Newton method will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.4303160220 \\ 1 & 0.4324621306 \\ 2 & 0.4324601954 \end{array} \right)$$ which is the solution for ten significant figures.


Take care : the above formula does not work well if $a$ is close to $\frac \pi 2$ (that is to say for $x$ close to $1$). In such a case, using a very truncated Taylor expansion will give as an estimate $$x_0=1+\frac{a}{2}-\frac{\pi }{4}$$ Let us try using $$a=1.5\implies x=\frac{7-\pi }{4}\approx 0.964602$$ Newton method will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.9646018366 \\ 1 & 0.9645944381 \end{array} \right)$$ which is the solution for ten significant figures.


If you want an estimate which is more than reasonable for all $a$'s, you could use $$x_0=\left(\frac{2a}{\pi }\right)^{3/4} $$ which, for the working examples would give $0.423777$ and $0.966003$ which looks to be pretty good.

We can still improve the quality of the estimate using for all $a$'s $$x_0=\left(\frac{2a}{\pi }\right)^{f(a)} $$ where $$f(a)=\frac{373}{540}+\frac{89 }{482}a-\frac{659 }{1363}a^2+\frac{1146 }{1229}a^3-\frac{872 }{891}a^4+\frac{625 }{1202}a^5-\frac{118 }{1083}a^6$$ where all parameters are very highly significant.

For the working examples, this would give $0.432536$ and $0.964677$.

  • $\begingroup$ Since, in order to compute $x_0$ of my very last approximation, you will need logarithms, you could change the exponent from $\frac 3 4$ to $\frac{328}{445}$. Used for the worked examples, this would give $0.430092$ and $0.966579$. I am ready to bet that, for any $0 \leq a \leq \frac \pi 2$, Newton method would converge in a couple of iterations for six significant figures. $\endgroup$ – Claude Leibovici Jan 25 '18 at 11:11

No, there is none. The $\cos$ function (and thus its inverse) is non algebraic. If you must have $h$ defined explicitly, then you must be content with some approximation of $\arccos$.


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