# 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?

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Does $t$ depend on the position? Is it the tension at some extremity? –  fffred Jul 2 at 20:34
–  Qmechanic Jul 3 at 0:13
fffred: No, the value $t$ is constant along the cable. It is the $x$ (horizontal) component of the tension at any point in the cable. –  user664303 Jul 3 at 8:54
Qmechanic: Thanks. I googled "catenary" before posting, and this lead to what I have already. –  user664303 Jul 3 at 8:59