Closed form solution for a simple-looking Euclidean geometry problem?

Question

The diagram above is distilled from an engineering problem I'm working on. The goal is to compute $R$ in terms of $r$ and $\theta$, which are known.

The best I have been able to get so far is an equation that combines all $3$ variables, and is therefore numerically solvable for $R$. However it does not seem that it can be rearranged in closed form to isolate $R$. When I posed it to Mathematica, it returned a hideous monstrosity.

I am wondering if there is a more elegant geometrical approach that I'm missing to isolate $R$ in terms of $r$ and $\theta$.

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Here is my work so far:

If we knew $\alpha$, the problem would be trivial. Drop the red perpendicular and observe that: $$ \cos\alpha=\frac{R-r}{R}, $$ which can be rearranged to give: $$ R = \frac{r}{1-\cos\alpha}. $$

We do not know $\alpha$ directly, but we know that $\alpha = \theta-\beta$, and we can compute $\beta$ in terms of $R$ and $r$ from the isosceles triangle: $$ \beta = 2\arcsin\left(\frac{r}{2R}\right) $$

Now we plug our new expression for $\alpha$ into the above expression for $R$ and we get a nasty-looking equation with $R$ on both sides: $$ R = \frac{r}{1-\cos\left(\theta - 2\arcsin\left(\frac{r}{2R}\right) \right)}. $$

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Edit: Numerical Solutions

Here is some data from numerical solutions. I have plotted $R$ vs $r$ with $\theta=\pi/3$ and $R$ vs $\theta$ with $r=1$.

It seems pretty clear that:

• $R$ and $r$ have a directly proportional relationship for fixed $\theta$
• $R$ is inversely proportional to $\theta$ for fixed $R$ (a log-log plot looks straight)

Answer 1

Probably there isn't a nice closed form.

On substituting $z = \frac{r}{2R}$, we get the final expression: $ \cos (\theta - 2\arcsin(z)) = 1 - 2z $

A query on WolframAlpha, solve cos(x - 2arcsin(z)) = 1 - 2z for z, generates an ugly closed form for $z$ .

Answer 2

I got that,

$0=R^2\sin(\frac{\theta}{2})^2-R(\frac{\sqrt{2r}}{2})^2+2\sqrt{R}\cos(\frac{\theta}{2})(\frac{\sqrt{2r}}{2})(\frac{r}{2})-(\frac{r}{2})^2$

This is a quartic polynmial from which one could in theory get the solution in a closed form by looking at the roots, I hope you could see if you agree with it.

First you need to notice that $\sin(\frac{\alpha}{2})=\frac{\sqrt{2Rr}}{2R},\ \sin(\frac{\beta}{2})=\frac{r}{2R}$, I add a new image to see if it helps visualize it, from there the rest is just calculations.

$\cos(\frac{\theta}{2})=\cos(\frac{\alpha+\beta}{2})$

$\cos(\frac{\theta}{2})=-\sin(\frac{\alpha}{2})\sin(\frac{\beta}{2})+\cos(\frac{\alpha}{2})\cos(\frac{\beta}{2})$

$\cos(\frac{\theta}{2})=-\sin(\frac{\alpha}{2})\sin(\frac{\beta}{2})+\sqrt{\cos(\frac{\alpha}{2})^2\cos(\frac{\beta}{2})^2}$

$\cos(\frac{\theta}{2})=-\sin(\frac{\alpha}{2})\sin(\frac{\beta}{2})+\sqrt{(1-\sin(\frac{\alpha}{2})^2)(1-\sin(\frac{\beta}{2})^2)}$

$\cos(\frac{\theta}{2})=(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})+\sqrt{(1-(\frac{\sqrt{2Rr}}{2R})^2)(1-(\frac{r}{2R})^2)}$

$(\cos(\frac{\theta}{2})-(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R}))^2=(1-(\frac{\sqrt{2Rr}}{2R})^2)(1-(\frac{r}{2R})^2)$

$\cos(\frac{\theta}{2})^2-2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})+(\frac{\sqrt{2Rr}}{2R})^2(\frac{r}{2R})^2=1+(\frac{\sqrt{2Rr}}{2R})^2(\frac{r}{2R})^2-(\frac{\sqrt{2Rr}}{2R})^2-(\frac{r}{2R})^2$

$\cos(\frac{\theta}{2})^2-2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})=1-(\frac{\sqrt{2Rr}}{2R})^2-(\frac{r}{2R})^2$

$0=1-\cos(\frac{\theta}{2})^2-(\frac{\sqrt{2Rr}}{2R})^2-(\frac{r}{2R})^2+2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})$

$0=\sin(\frac{\theta}{2})^2-(\frac{\sqrt{2Rr}}{2R})^2-(\frac{r}{2R})^2+2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})$

$0=R^2\sin(\frac{\theta}{2})^2-(\frac{\sqrt{2Rr}}{2})^2-(\frac{r}{2})^2+2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2})(\frac{r}{2})$

$0=R^2\sin(\frac{\theta}{2})^2-R(\frac{\sqrt{2r}}{2})^2+2\sqrt{R}\cos(\frac{\theta}{2})(\frac{\sqrt{2r}}{2})(\frac{r}{2})-(\frac{r}{2})^2$

Answer 3

Except if you want to play with the quartic equation in $\sqrt R$, numerical methods will work nicely.

With regard to your second plot, if instead of plotting $R(\theta)$, you plot $\frac 1{R(\theta)}$, you will notice a much more "pleasant" curve.

Using Saswat Padhi's answer consider that you look for the zero of function $$f(z)=2 \sin ^{-1}(z)+\cos ^{-1}(1-2 z)-\theta$$ $$f'(z)=\frac{2}{\sqrt{1-z^2}}+\frac{1}{\sqrt{z-z^2}}$$ which would be used for Newton method.

In order to save iterations, consider the following table $$\left( \begin{array}{cc} \theta & z \\ 0.00000 & 0.00 \\ 0.55107 & 0.05 \\ 0.84384 & 0.10 \\ 1.09654 & 0.15 \\ 1.33001 & 0.20 \\ 1.55256 & 0.25 \\ 1.76866 & 0.30 \\ 1.98125 & 0.35 \\ 2.19247 & 0.40 \\ 2.40416 & 0.45 \\ 2.61799 & 0.50 \\ 2.83569 & 0.55 \\ 3.05916 & 0.60 \\ 3.29066 & 0.65 \\ 3.53311 & 0.70 \\ 3.79052 & 0.75 \\ 4.06889 & 0.80 \\ 4.37816 & 0.85 \\ 4.73763 & 0.90 \\ 5.19704 & 0.95 \\ 6.28319 & 1.00 \end{array} \right)$$ So, for a given value of $\theta$, a simple table lookup (with or without some interpolation) gives you an estimate of $z$ which will be your initial estimate.

Trying for $\theta=1$, use $z_0=0.15$ and the iterates will be $$\left( \begin{array}{cc} n & z_n \\ 0 & 0.150000 \\ 1 & 0.129986 \\ 2 & 0.130308 \end{array} \right)$$ Trying for $\theta=2$, use $z_0=0.35$ and the iterates will be $$\left( \begin{array}{cc} n & z_n \\ 0 & 0.350000 \\ 1 & 0.354432 \\ 2 & 0.354433 \end{array} \right)$$ Trying for $\theta=3$, use $z_0=0.60$ and the iterates will be $$\left( \begin{array}{cc} n & z_n \\ 0 & 0.600000 \\ 1 & 0.586974 \\ 2 & 0.586915 \end{array} \right)$$

This seems to be quite fast.