Cable shape and length between two towers

Data: a transmission line hangs between two towers at the same level, span $\ell$, with weight $\gamma$ per unit length of cable (so the load is uniform along the cable length, not along the span). Let the sag (dip) at mid-span be $d$.

Shape: catenary

Take the origin at the lowest point $O$. For the portion of the cable of length $s$ from $O$ to a point $P(x, y)$, the forces are the horizontal tension $H$ at $O$, the weight $\gamma s$ and the tension $T$ at $P$:

$$\tan\psi = \frac{dy}{dx} = \frac{\gamma s}{H}$$

Let $c = H/\gamma$. Then $\dfrac{dy}{dx} = \dfrac{s}{c}$ and $ds = \sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx$, which gives $s = c\sinh\dfrac{x}{c}$. Integrating,

$$y = c\left(\cosh\frac{x}{c} - 1\right)$$

This is the equation of a catenary.

Sag and horizontal tension

At the tower, $x = \ell/2$:

$$d = c\left(\cosh\frac{\ell}{2c} - 1\right), \qquad c = \frac{H}{\gamma}$$

Length of the cable between the towers

$$S = 2c\,\sinh\frac{\ell}{2c}$$

Maximum tension at the towers: $T_{max} = \gamma\,(c + d) = H + \gamma d$.

Parabolic approximation (small sag, $d/\ell \le 1/10$)

When the sag is small, the weight per unit length can be taken as acting per unit horizontal span ($w = \gamma$). The cable is then a parabola:

$$y = \frac{\gamma x^2}{2H}, \qquad H = \frac{\gamma \ell^2}{8d}$$ $$S = \ell\left[1 + \frac{8}{3}\left(\frac{d}{\ell}\right)^2 - \frac{32}{5}\left(\frac{d}{\ell}\right)^4 + \dots\right]$$

so, to a good approximation, $S \approx \ell + \dfrac{8d^2}{3\ell}$, and the tension at the tower is $T = \sqrt{H^2 + (\gamma\ell/2)^2}$.

Example: $\ell = 300$ m, $d = 6$ m: $S \approx 300 + \dfrac{8\times 36}{900} = 300.32$ m.