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.
Best Answer
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.