Pole Malone

The arclength of the catenary is $Sinh(x)$ , so first we solve for $x_0$ in

$Sinh(x_0)=\dfrac{50}{40}x_0$

which numerically works out to $x_0=1.1827255398...$

Then, $a=\dfrac{40}{x_0}$ , and we have the equation for the catenary that passes through $(-40,30)$ and $(40,30)$

$f_a(x)=\dfrac{a}{2}\left( e^{-\frac{x}{a}} + e^{-\frac{x}{a}} \right) - \dfrac{a}{2} \left( e^{-x_0} + e^{x_0} \right)+ 30$

so that at $x=0$ , the wire is

$a-\dfrac{a}{2} \left( e^{-x_0} + e^{x_0} \right) + 30 = 3.4562490858...$

above the ground