Calculate circle segment as proportion of total circle area How can I find the height of the arced portion of a circular segment ('h' in this link) when I only know the radius of the circle and the area of the segment.
The link above gives me an equation for calculating the area of the segment. So I can set this to fraction of the total circle, 'f'.
$$f\pi r^2 = r^2 cos^{-1} \left(\frac{r-h}{r}\right)-(r-h)\sqrt{2rh-h^{2}} $$
I would then need to solve this for 'h', which I'm struggling to do.
I have also tried using the angle instead, which gives a simpler equation. I set the area to the proportion of the total area again.
$$f\pi r^2 = \left(\frac{\theta\pi}{360} - \frac{\sin(\theta)}{2}\right)r^{2}$$
But I have not managed to solve this for $\theta$ either. Ultimately, I want to be able to mark out segments on a circle that correspond with specific proportions of the circle, and need to know where to place the markers. In this particular case the circle has a radius of 23.5mm and the segment's area is $\frac{3}{8}$ of the total circle area.
 A: I think you're overcomplicating it.
Since the ratio $f$ of sector area to circle area always equals the ratio of $θ$ to 2π radians, we have:
$θ = 2\pi f$ in radians or $360^\circ\cdot f$ in degrees.
If we bisect $θ$, we have two right triangles, the hypotenuse being $R$, and the height, $r$. We want to find $r$ since $h + r = R$. We know that $\cos(θ/2) = r/R$ by definition, so:
$r = R\cos(θ/2)$.
Therefore:
$h = R - r$,
$= R - R\cos(θ/2)$,
$= R\,( 1- \cos(\pi f) )$ in radians,
or $R\,(1-\cos(180^\circ\cdot f)$ in degrees.
For $f = \frac{3}{8}$ and $R = 23.5\,\text{mm}$, $h = 14.5\,\text{mm}$.
Please do follow up in the comments if needing further clarification.
A: $\begin{array}\\
f\pi r^2 
&= r^2 \cos^{-1} \left(\frac{r-h}{r}\right)-(r-h)\sqrt{2rh-h^{2}}\\
&= r^2 \cos^{-1} \left(1-h/r\right)-r^2(1-h/r)\sqrt{2h/r-(h/r)^{2}}\\
&= r^2\left( \cos^{-1} \left(1-h/r\right)-(1-h/r)\sqrt{2h/r-(h/r)^{2}}\right)\\
f\pi
&=  \cos^{-1} \left(1-h/r\right)-(1-h/r)\sqrt{2h/r-(h/r)^{2}}\\
&=  \cos^{-1} \left(1-h/r\right)-(1-h/r)\sqrt{1-1+2h/r-(h/r)^{2}}\\
&=  \cos^{-1} \left(1-h/r\right)-(1-h/r)\sqrt{1-(1-h/r)^2}\\
&=  \cos^{-1}(u)-u\sqrt{1-u^2}
\qquad u=1-h/r\\
&=  \cos^{-1}(\cos(v))-\cos(v)\sqrt{1-\cos^2(v)}
\qquad u=\cos(v)\\
&=  v-\cos(v)\sin(v)\\
&=  v-\sin(2v)/2\\
&=  (2v-\sin(2v))/2\\
2f\pi
&=  2v-\sin(2v)\\
\end{array}
$
Solve this for $v$
in terms of $f$
(has to be done numerically),
then get
$u=\cos(v),
h=r(1-u)
$.
For small $v$,
$2v-\sin(2v)
\approx 2v-(2v-4v^3/3)
=4v^3/3
$
so,
for small $f$,
$\begin{array}\\
v 
&\approx (3f\pi/2)^{1/3}\\
u
&=\cos(v)\\
&\approx 1-v^2/2\\
&\approx 1-(3f\pi/2)^{2/3}/2\\
h
&\approx r(3f\pi/2)^{2/3}/2\\
\end{array}
$
