A rope with its two ends held in place forms a curve called a catenary (assuming that the stiffness of the rope is negligible). A catenary takes the shape of the function:
f(x) = a cosh(x/a),
where cosh is the hyperbolic cosine function.

If a 50-foot rope hangs by its ends from two flagpoles, one 50 feet tall and one 40 feet tall, and at its lowest point is 20 feet above the ground, how far apart are the flagpoles?

(In reply to re: Hrmph... by levik)

It's been (mumble)ty-(mumble) years since I've used equations such as this, and I'm going on what "feels right" rather than looking it up, but it seems to me that the formula actually should be [f(x) = a cosh (x/a) + b] where
f(x)= the heitht of the rope at point x
x = the distance from the left pole
R = length of the rope
d = distance between the two poles
a = R/d
h = height of the poles
h(0) = height of the poles such that at point x = d/2 the rope just touchees the ground
b = h - h(0).

This assumes that that the poles are the same height. There would be additional adjustment factors for poles of differing heights.

In this case we are given R = 50; b = 20 and h1 = 40; and h2 = 50. It is enough information to work out d, if we really needed to. Fortunately with these values we don't need to use the formula to do so.

Posted by TomM on 2002-12-24 00:16:29