Solving $R\space \sinh\frac{D}{R}=k$ for $R$ Does a solution exist for $R$ in this equation?  I can't seem to solve it either analytically or numerically.
$$R\space \sinh\frac{D}{R}=k$$
 A: $$R\space \sinh\frac{D}{R}=k$$
$$\text{Let}\qquad x=\frac{D}{R}\qquad;\qquad R=\frac{D}{x}\qquad;\qquad C=\frac{k}{D}$$
$$\boxed{\sinh(x)=C\:x}$$
This equation is well known. For example see a discussion about the existence of root(s) in : Solving $\sinh x = kx$
The root cannot be expressed with a finite number of standard functions.
On a formal view-point (of no interest in practice) an analytical form of solution is :
$$R=\frac{D}{f^{-1}(k/D)}$$
where $f^{-1}$ is the inverse function of $f(x)=\frac{\sinh(x)}{x}$ . But this function $f^{-1}$ is not standard.
Of course, numerical solving is possible thanks to iterative methods for numerical solving of non-linear equations (For example Newton-Raphson method https://mathworld.wolfram.com/NewtonsMethod.html). 
A: Starting from @JJacquelin's answer, you are looking for the zero of function
$$f(x)=\frac{\sinh(x)}x-C$$ The algebraic expression varies very fast which is never very good for numerical methods.
Consider instead looking for the zero of function (assuming $C>0$)
$$g(x)=\log \left(\frac{\sinh (x)}{x}\right)-\log(C)$$ which is much smoother and then better conditioned for the solver.
For sure, we need a starting point. A simple estimate could be
$$x_0=2.35 \big[\log(C)\big]^{3/4}$$  For testing, let me try for $C=123456789$. The iterates of Newton method will be
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 21.07424431 \\
 1 & 22.43727831 \\
 2 & 22.43517918 \\
 3 & 22.43517917
\end{array}
\right)$$
