Derivation of the Equation

A heavy cable suspended horizontally between two points takes the shape of a catenary.


  catenary[x_,a_] := a Cosh[x/a]

We can use an animation to determine how the parameter, a, effects the catenary.


  Table[
  	Plot[catenary[x,a], {x,-50,50},PlotRange -> {0,120}],
  {a,10,100,10}];

So a determines the amount of sag in the cable.

Suppose a cable is to be suspended between two telephone utility poles that 100 meters apart, as in the illustration below. If the cable will be allowed to sag 10 meters in the middle, how long does the cable need to be?

Mathematica note: This illustration was created by plotting the catenary in Mathematica , using the Copy and Paste commands to paste the catenary into SuperPaint, drawing the rest of the illustration in SuperPaint, and Pasting the illustration back into Mathematica . It is in PICT format, so it is probably not portable to DOS or NeXT computers. PostScript graphics are portable between platforms. However, after several unsuccessful attempts to import this illustration as a PostScript graphic, I gave up. If you want to try this, you might ask the woman in the illustration for help. She seems to have her act together.

We first need to find the value of a to give a sag of 10 meters. From the illustration, you can see that we need to find a so that

catenary[50,a] - catenary[0,a] == 10

Up to An Example: The Length of a Suspended Cable