Closed form for shape/tension of an elastic cable slung between two points Given the 2D coordinates of two points, $a$ and $b$, between which an elastic cable of known length, $l$, mass per unit length, $m$, and the spring constant, $e$, is slung, I need to compute the shape of the cable, and also the horizontal tension, $t$, in the cable.
So far I have the equations for the x and y coordinates of the cable, parameterized by $p$, which is the distance along the unstretched cable:
$$
f_x(p) = \frac{t}{mg} \sinh^{-1}\left(\frac{mgp}{t}\right) + \frac{tp}{e} + c_x.\\
f_y(p) = \sqrt{\left(\frac{t}{mg}\right)^2 + p^2} + \frac{mgp^2}{2e} + c_y.\\
$$
where $c_x$ and $c_y$ are constants, which leads me to the following triplet of simultaneous equations:
$$
f_x(q) - f_x(r) = a_x - b_x.\\
f_y(q) - f_y(r) = a_y - b_y.\\
|q-r| = l.
$$
in three unknowns, $t$, $q$ and $r$ (given that the constants cancel), where $q$ and $r$ are the values of the parameter $p$ at the points $a$ and $b$ respectively.
How would you compute those unknowns, and can it be done in closed form?
 A: The shape is still a catenary regardless of the degree of elasticity of the cable.
Put a coordinate system on the left support and note the relative coordinates of the right support as $\pmatrix{S & h}$
The equation of the catenary is $$ y(x) = y_C + a \left( \cosh \left( \frac{x-x_C}{a} \right) -1 \right) $$
where $\pmatrix{x_C & y_C}$ is the lowest point on the curve, and $a=\frac{H}{w}$ is the catenary constant, derived from the horizontal tension $H$ and the unit weight $w$ (in force/length units).
To find the curve through the points $\pmatrix{0&0}$ and $\pmatrix{S&h}$ use the following coordinates for the lower point
$$\begin{align}
  x_C & = \frac{S}{2} - a \,\sinh^{-1} \left( \frac{  \frac{h}{a} {\rm e}^{\frac{S}{2 a}} }{ {\rm e}^{\frac{S}{a}}-1 } \right) \\
  y_C & = a \left( 1 - \cosh \left( \frac{x_C}{a} \right) \right)
\end{align} $$
Once the shape is known the rest of the properties can be evaluated
$$ \begin{cases}
  T(x) = H \cosh \left( \frac{x-x_C}{a} \right) & \mbox{Tangential Tension} \\
  V(x) = H \sinh \left( \frac{x-x_C}{a} \right) & \mbox{Vertical Tension} \\
  L(x) = a \sinh \left( \frac{x-x_C}{a} \right) & \mbox{Length of Cable from Lowest Point} \\
  D(x) = \frac{x h}{S} - y(x) & \mbox{Cable Sag from ideal line}
\end{cases}$$
So for example the total length of the cable between supports is 
$$ L = L(S) - L(0) = a \left( \sinh\left( \tfrac{x_C}{a}\right) + \sinh \left( \tfrac{S-x_C}{a} \right) \right) $$
Additionally, the average tension on the cable is found by the integral
$$ P = \tfrac{1}{L} \int \limits_0^S T(x) \sqrt{ 1 + \left( \tfrac{{\rm d}}{{\rm d}x} y(x) \right)^2 } {\rm d}x  = \tfrac{S w}{2} \frac{1+\tfrac{a}{2 S} \left( \sinh\left( \tfrac{2x_C}{a}\right) + \sinh\left(\tfrac{2(S-x_C)}{a}\right)\right)}{\sinh\left( \tfrac{x_C}{a}\right) + \sinh\left( \tfrac{S-x_C}{a}\right)} $$
For an even span, the above average tension is approximated well with $$ P = \frac{2 H + T}{3}$$
Now if the horizontal tension is to be calculated from a) Sag, b) Average Tension or c) Total Length then a numerical method is needed. 
For example if the maximum sag $D_{\rm set}$ is known, then start with an initial guess of $H_{\rm init} =  \tfrac{w \sqrt{S}}{8\,D_{\rm set}}$ and iterate finding $x_C$, and then $D$ and adjusting the tension accordingly until $|D-D_{\rm set}| < {\rm tolerance}$. If you need more details, I propose you ask a new question that references this question.
Below is a screenshot of a catenary shape solver I developed for fun using the above equations.

References:


*

*https://physics.stackexchange.com/a/304484/392

*https://physics.stackexchange.com/a/101787/392
