Maximum Cable Tension at the Support (Parabolic)

At the support the cable is doing two things at once. It is pulling horizontally with $H$, the same $H$ it pulls with everywhere. And it is holding up half the load on the span, $wL/2$, because there are two supports and by symmetry they share the weight. The tension in the cable is the resultant of those two, and Pythagoras finishes the job.

Draw the triangle and everything about the page becomes obvious. The horizontal leg is $H$. The vertical leg is $wL/2$. The hypotenuse is the cable tension, and the angle at the base is the angle the cable actually leaves the support at — which is the slope of the parabola there, $4d/L$. That factor of four is worth carrying around: a cable sagging one metre in a twenty metre span leaves its supports at a slope of 0.20, not 0.05. The tangent at the support is four times the average slope of the chord to the low point, because that is what a parabola does.

This is the number the hardware feels. The dead-end clamp, the insulator string, the anchorage, the shackle, the turnbuckle, the factor of safety against the cable's rated breaking strength — all of them are checked against the tension at the support, and none of them against $H$. Confusing the two is the third of the three classic mistakes in this subject, and it is the unconservative one. It survives so long because on a taut cable the two numbers are nearly equal: at a sag ratio of 1 in 20 the difference is a fifth of a percent, well inside anybody's rounding. It stops being nearly equal on a bridge main cable at 1 in 10, where the gap is 2%, and on anything slacker.

The vertical leg has a life of its own, too. $wL/2$ is what the tower or the pier has to carry down, and it is the reason a suspension bridge tower is a compression member sized on the sum of the vertical components from both sides. On a transmission structure it is the vertical load at the crossarm, and it changes with the terrain: a structure at the bottom of a valley carries more than half the span each side, and one on a hilltop can be carrying so little that the insulator string tries to lift — an uplift condition, checked in every line design and remedied with a heavier string or a different sag.

Run the equation backwards and it becomes the design tool. Start with what the cable is allowed to pull to — breaking strength divided by whatever factor of safety the code demands, typically somewhere between 2 and 4 for a permanent line and higher for anything people ride on — and the equation returns the horizontal tension you may string it at. That $H$ then goes to the sag formula, the sag goes to the clearance check, and if the clearance fails, the whole loop runs again with a different span or a different structure height. That is the entire design cycle for a suspended cable, and it is short.