# Hanging cable (catenary)

1. May 16, 2008

### joel_f

hi~
i need to determine the tension a in the equation of the hanging cable y=acosh(x-h)/a+k with a known length and known location of end points. i figured out how to determine the horizontal translation h but i need a in order to do it and to have the complete equation.

this is for a computer simulation where the weight of the cable is not important. this is why i condensed tension in the formula down to a. if the weight is a requirement then it can be set, in which case i would need to find the horizontal force acting on the cable.

is it possible to find the tension with the given information?

2. May 25, 2008

### benorin

The length of the cable L is given by $$y=2a\mbox{sinh}\left(\frac{d}{2a}\right)$$, where d is the distance between the end points of the cable (I assumed they were at equal heights based on your equation). You can get a numerical approximation of a from this equation after substituting in the known values of d and L.

3. May 26, 2008

### joel_f

would this work for end points of unequal height? am i using the wrong equation for the catenary of this type?

4. May 27, 2008

### benorin

Check this out.

5. Oct 29, 2010

### seuren83

Hello benorin,

I hope you will still get this. I very much liked your extensive answer of the problem.

What if you would replace the force acting on the cable from being gravity to drag. In other words what if the force is not equally distributed. Like a fish on a fishingrod swimming around the fisherman, or a cable with increasing density towards one end.

Can this be solved?

Regards,
Seuren