Is the equation $A = \frac{d^2}{8}\left(\alpha-\sin\alpha\right)$ solvable for $\alpha$?

Question

I have been given this equation. $A$ and $d$ are known, and I want to solve to $\alpha$. $$A=\frac{\pi d^2}{4}\cdot\frac{\alpha}{2 \pi}-\frac{d\,\sin\left(\frac{\alpha}{2}\right)\left[\frac{d}{2}-\frac{d}{2}\left(1-\cos\left(\frac{\alpha}{2}\right)\right)\right]}{2}$$ Until now, I've tried my best to land at this: $$A=\frac{\pi\,d^2}{4}\cdot\frac{\alpha}{2 \pi}-d^2\sin\left(\frac{\alpha}{2}\right)\left[-\frac{1-4\cos\left(\frac{\alpha}{2}\right)}{8}\right]$$ Now I don't know how to proceed or if this can even be solved.

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Edit by @Blue. As @AdrianKeister shows, the equation should actually reduce to

$$ A = \frac{d^2}{8}\left(\;\alpha - \sin \alpha\;\right)$$

I've used this form to provide a more-informative title for this question.

Answer 1

Given that $\alpha$ appears both inside and outside the trig functions, I'm nearly certain you'll have to resort to a numerical solution. You might find this little article I wrote helpful: https://www.winemantech.com/blog/the-numerical-analysis-of-finding-the-height-of-a-circular-segment.

By the way, I'm not sure I trust your algebra there:

$$A=\frac{\pi d^2}{4}\cdot\frac{\alpha}{2 \pi}-\frac{d\,\sin\left(\frac{\alpha}{2}\right)\left[\frac{d}{2}-\frac{d}{2}\left(1-\cos\left(\frac{\alpha}{2}\right)\right)\right]}{2}$$ $$A=\frac{\alpha d^2}{8}-\frac{d^2\,\sin\left(\frac{\alpha}{2}\right)\left[\frac{1}{2}-\frac{1}{2}+\frac12\cos\left(\frac{\alpha}{2}\right)\right]}{2}$$ $$A=\frac{\alpha d^2}{8}-\frac{d^2\,\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)}{4}$$ $$A=\frac{\alpha d^2}{8}-\frac{d^2\,\sin(\alpha)}{8}$$ $$A=\frac{d^2}{8}\left[\alpha-\sin(\alpha)\right].$$ Now you can see my blog post is very close to what you need.

Answer 2

As Adrian Keister already answered, you need to solve the equation $$\beta=\alpha-\sin(\alpha)\qquad \text{where}\qquad \beta=\frac{8A}{d^2}$$ and a numerical method (Newton method will be the simplest) will be required.

However, approximations could be done. For example, if $\beta$ is small, Taylor series built around $\alpha=0$ would give $$\beta=\frac{\alpha ^3}{6}-\frac{\alpha ^5}{120}+\frac{\alpha ^7}{5040}-\frac{\alpha ^9}{362880}+O\left(\alpha ^{11}\right)$$ and using series reversion $$\alpha=\sqrt[3]{6} \beta ^{1/3}+\frac{ 1}{10}\beta+\frac{3\ 3^{2/3} }{350 \sqrt[3]{2}}\beta ^{5/3}+\frac{\sqrt[3]{3} }{350\ 2^{2/3}}\beta ^{7/3}+O\left(\beta ^3\right)$$

For illustration purposes, $$\left( \begin{array}{ccc} \beta & \text{approximation} & \text{exact} \\ 0.00 & 0.00000 & 0.00000 \\ 0.25 & 1.17122 & 1.17123 \\ 0.50 & 1.49722 & 1.49730 \\ 0.75 & 1.73605 & 1.73633 \\ 1.00 & 1.93387 & 1.93456 \\ 1.25 & 2.10733 & 2.10876 \\ 1.50 & 2.26458 & 2.26717 \\ 1.75 & 2.41030 & 2.41461 \\ 2.00 & 2.54744 & 2.55420 \\ 2.25 & 2.67799 & 2.68810 \\ 2.50 & 2.80340 & 2.81799 \\ 2.75 & 2.92471 & 2.94517 \\ 3.00 & 3.04274 & 3.07077 \end{array} \right)$$ which seems to be quite reasonable.

Edit

We can make the approximation looking nicer using $\beta=\frac{\gamma ^3}{6}$ (or $\gamma=\sqrt[3]{6\beta}$) and get $$\alpha=\gamma +\frac{\gamma ^3}{60}+\frac{\gamma ^5}{1400}+\frac{\gamma ^7}{25200}+\cdots$$

We could even make the approximation better continuing the expansion and transforming the result as a Padé approximant to get $$\alpha=\gamma\, \frac{1-\frac{1493 }{21120}\gamma^2+\frac{167 }{268800}\gamma^4} { 1-\frac{123 }{1408}\gamma^2+\frac{403 }{295680}\gamma^4}$$ leading to $$\left( \begin{array}{ccc} \beta & \text{approximation} & \text{exact} \\ 0.00 & 0.00000 & 0.00000 \\ 0.25 & 1.17123 & 1.17123 \\ 0.50 & 1.49730 & 1.49730 \\ 0.75 & 1.73633 & 1.73633 \\ 1.00 & 1.93456 & 1.93456 \\ 1.25 & 2.10874 & 2.10876 \\ 1.50 & 2.26713 & 2.26717 \\ 1.75 & 2.41452 & 2.41461 \\ 2.00 & 2.55403 & 2.55420 \\ 2.25 & 2.68780 & 2.68810 \\ 2.50 & 2.81747 & 2.81799 \\ 2.75 & 2.94431 & 2.94517 \\ 3.00 & 3.06939 & 3.07077 \end{array} \right)$$