--- title: "Cable shape and length between two towers" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:cable-geometry-length-between-towers" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-geometry-length-between-towers site: BCT Engineering --- # Cable shape and length between two towers Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2066 Kartik, 6 marks - Video: [Cables - Problem No 1 (supported on the same level & subjected to Uniformly Distributed Load ( UDL )](https://www.youtube.com/watch?v=vEWoLBnX1FQ) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-geometry-length-between-towers - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question Determine the geometry of the shape and calculate the length required for a high tension line between any two towers. Take the span between the two towers and the weight per unit length of the cable to be $\ell$ and $\gamma$ respectively. ## Short answer The cable forms a category y = c[cosh(x/c)-1] where c = H/γ. Its total length between towers of span L is S = 2c sinh(L/(2c)). For small sag, the parabolic approximation is y = γx^2/(2H) and length S ≈ L + 8d^2/(3L), with d the central dip. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does a transmission line between two towers hang as a catenary? A. Because the load is uniform per horizontal metre of span B. Because the tension is constant along the cable C. Because the towers are at different levels D. Because its weight is uniform along the cable length, not along the span Answer: D. A load uniform along the cable length gives tan(psi) = gamma s/H and hence the catenary y = c(cosh(x/c) - 1). (Key point: Why a catenary) 2. In the catenary equations, what is c? A. c = H/gamma, the horizontal tension divided by weight per unit length B. c = d, the sag C. c = gamma l D. c = T_max/H Answer: A. Defining c = H/gamma gives dy/dx = s/c, s = c sinh(x/c) and y = c(cosh(x/c) - 1). (Key point: Parameter c) 3. What is the cable length between towers of span l? A. S = c cosh(l/c) B. S = l + d C. S = 2c sinh(l/2c) D. S = c sinh(l/c) Answer: C. Since s = c sinh(x/c) from the lowest point, doubling at x = l/2 gives S = 2c sinh(l/2c). (Key point: Length of cable) 4. For small sag (d/l <= 1/10), what is the approximation of the cable length and H? A. S = l + 8d^2/(3l); H = gamma l^2/(8d) B. S = l + d; H = gamma l d C. S = l - 8d^2/(3l); H = gamma l/(8d) D. S = 2l; H = gamma l^2/d Answer: A. Treating the load as per horizontal span gives a parabola with H = gamma l^2/(8d) and S about l + 8d^2/(3l). 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