All examples for the hanging cable problems that I've seen involve something like calculating the unknown distance between poles for a cable of a known length and known sag.
I've been trying to figure out how to re-arrange the equations to determine an unknown sag from a known cable length and known distance apart.
For example, what is the sag for a cable 3m long between two points 2m apart.
This was one of the sources I've looked at THE HANGING CABLE PROBLEM FOR PRACTICAL APPLICATIONS as well as the Wikipedia entry for Catenaries. The first demonstrates how to re-write the width in terms of the sag and length.
This is for a personal project where I'm hanging Christmas lights where I've now ended up down a rabbit hole and out of my depth mathematically speaking.
Best Answer
Rewrite the equation $$a \sinh \left(\frac{x}{a}\right)=y$$ as $$\frac{\sinh(t)}t=k \qquad \text{with}\qquad t=\frac{x}{a}\qquad \text{and}\qquad k=\frac{y}{x}$$
Using Taylor series $$\frac{\sinh(t)}t=1+\frac{t^2}{6}+\frac{t^4}{120}+\frac{t^6}{5040}+O\left(t^8\right)$$ Use series reversion $$t=u-\frac{u^3}{40}+\frac{107 u^5}{67200}+O\left(u^7\right)\qquad \text{with}\qquad u= \sqrt{6(k-1)}$$
For your case where $k=y=\frac 32$, $x=1$ $$t\sim \frac{21041 \sqrt{3}}{22400} \implies a=\frac 1 t = 0.614640$$