Thread: Catenary

Equation of a catenary is
y = a * cosh(x/a) + C

I am told that two telephone poles are 10 m apart and that the poles are 6 m tall. The minimum clearance (distance of the cable to the ground) is 4 m. Question asks how long can the cable be.

So. I know that once I find the equation for the catenary the cable makes, I get the integral and use the arc-length formula. Unfortunately, I don't know how to get the 'a' in the formula.

Here's my attempt so far.
C doesn't really matter. The lowest point at x = 0 is a, since cosh(0) = 1. That means the point at x = -5 and x = 5 is a+2. So I get an equation:
a+2 = a * cosh(5 / a)

But I'm not sure how to solve it.
a * arccosh[(a+2) / a] = 5

That's as far as I got, and I'm rather sure I'm not doing it right. The assignment is focused on linearization and the like (Newton's method, fixed-point iteration, etc.).

Since the poles are 6 meters high and the clearance is 4 m, then the sag is 2 m.

We can use Newton's method to find the length.

The sag is given by

b is the distance from the origin to one of the poles. 5 m.



Let

Then,





If

Then,

Try an initial guess for u. Then, once it converges, 'a' can be found by just subbing it into

The length can be found by plugging 'a' into the length formula:



We can derive the length by: