Length of a catenary in terms of sag and span?

August 3, 2008 6:43 AM

I am building a zip line. The supports are 70m apart and one is 7m higher than the other. I want the lowest point of the zip line to be 2m lower than the lower support. How long is the wire?

Obviously the curve is a catenary but everything online is expressed in terms of the wire tension and mass per unit length, which I want to eliminate from consideration.

To be a bit clearer about the parameters, assume one support is at (0,0) and the second is at (70, -7). I want the lowest point of the zip line to be at y= -9. I'm trying to calculate the arc length along the catenary from one support to the other.

Alternatively, assuming the supports are at (0,0) and (0,L) and that the sag is S, is there a function f(L,S) which gives the arc length of the catenary in terms of the span L and sag S?

Thanks!

posted by unSane to science & nature (17 answers total)

In reality, you can't neglect the tension and weight of the wire. This document will help you calculate the length only based on the assumption that the wire doesn't stretch and is completely flexible. Of course under tension, all wires will stretch.

A parabolic curve will approximate a catenary. Maybe try that formula instead.
posted by JJ86 at 7:16 AM on August 3, 2008

Are you specifying the lowest point for the bare wire? Of course, when loaded with a human the wire shape will probably be more like a triangle rather than a catenary, and will be lower than the bare-wire height.
posted by schrodycat at 7:46 AM on August 3, 2008 [1 favorite]

Seconding schrodycat. Under the weight of a human the line will be pulled into a reasonably straight line, given the much lower weight of the wire itself (at a guess, less than 10kg).

So you're really dealing with two triangles those length depends on where the rider is along the wire. The sum of the two hypotenuses(?) is the total length of the wire. The lowest point is going to occur pretty close to the lower support, although I haven't calculated it.

Elasticity in the cable is likely to be a factor anyway, as is the eventual stretching of the cable after being used for a while.
posted by le morte de bea arthur at 8:39 AM on August 3, 2008

if you assume there's no stretching and that the wire forms two straight lines (support - rider - support) then the rider (or rather, the point on the wire at which the rider is attached) traces out a portion of an ellipse whose foci are the points where the wire is attached to the support. so you would need to find the ellipse whose shape matches your requirements. this is non trivial, at least for me (probably easiest to do with a computer program that repeatedly improves an initial guess).

in practice you want to make the system adjustable in some way, so that you can tweak it when you first build it, and also later to take up any slack.
posted by not sure this is a good idea at 9:01 AM on August 3, 2008 [1 favorite]

(you can see what i mean about the ellipse here - run that little demo)
posted by not sure this is a good idea at 9:03 AM on August 3, 2008

Once you've got your wire (<73.5 m, plus allowance for fittings) be sure your arrangement allows for later tightening. My zip line used to stretch and had to be shortened. I used nylon/poly rope and of course your steel line won't stretch nearly as much, but don't make it unchangeable.
posted by airplain at 9:25 AM on August 3, 2008

If you don't leave a way to adjust the length easily, you'll have to invent one next year when the wire has stretched out. Worry about that instead of "measure twice, cut once."
posted by fantabulous timewaster at 9:55 AM on August 3, 2008

Yes, it's a catenary. I haven't checked the algebra myself, but I think you can use eqn. 4 on the Mathworld entry to solve for the arc length. Use equations (1)-(2) to solve for a and determine the value of t at the end of your line first.

Of course, if you're actually constructing this line in real life you'll want to pay attention to the comments above too!
posted by caek at 10:40 AM on August 3, 2008

As a professional in this field, here are a few things to think about:

Cable length on zip lines is usually not determined before hand, and is a function of the lowest drape being where the participant exits the zip-line, either onto a platform or a ladder, with the drape itself being the brake which slows thing down. The length of a two meter drape from the end support is dependent on where the drape is located

If you are using cable clamps to secure the end it is pretty easy to pull the cable through a slightly tightened clamp using a come-along until you find your exact length.

There should be minimal cable stretch but leave enough cable on the far end for adjustments as needed.

If this is just hypothetical, forget all the above.
posted by Xurando at 12:25 PM on August 3, 2008 [1 favorite]

I agree that the ellipse method is going to be the more useful given a load on the line. And it shouldn't be too hard to get the solution. It will help to rotate your coordinate system to one where the supports are both on the horizontal axis.

As for your original question regarding catenaries, however, you shouldn't worry about the mass per unit length. This is a gravitational problem, and the answer simply can't depend on the mass. Feel free to pick units where the mass of your cable is 1. Now write down your equation for the catenary, y(x). You should have 3 unknowns in the equation. Find the point xc where dy/dx=0. Then your 3 constraints are: y(0)=0, y(L)=-7, y(xc)=-9. Use those to solve for your unknowns. Now to get the length of the cable, you want to (probably numerically) evaluate the integral Int(Sqrt(1+(dy/dx)^2), {x, 0, L}). The same basic idea applies to the ellipse thing.

But as others have pointed out, the practical method probably doesn't require any of that.
posted by dsword at 1:32 PM on August 3, 2008

dsword mentioned: As for your original question regarding catenaries, however, you shouldn't worry about the mass per unit length. This is a gravitational problem, and the answer simply can't depend on the mass.

Well F=mg, so yeah mass is required if you want to solve the gravity aspect.
posted by JJ86 at 6:27 PM on August 3, 2008

i think what he was trying to say is that it's only relative mass - its distribution - that is important here (the physics is the same no matter what the value of g, so you have a free factor to play with, which means the absolute value of the mass is irrelevant). all that's necessary is to assume that the cable is uniform.

but the lack of clarity was matched only by his optimism at the maths involved in solving for the correct ellipse :o)
posted by not sure this is a good idea at 6:39 PM on August 3, 2008

Thanks for the answers. The reason for the question was the slightly unusual design of my zip line. It is supported by two 16' tall 6x6 posts, leaning away from the zip line, with the tension supplied by guy wires behind the poles. The zip line itself is constant length but the poles can be pulled further back by the guy lines, decreasing the sag and increasing the tension. We used a come along attached to an independent anchor (my tractor) to dial in the tension, then locked it in using the guy lines. I'm going to be adding turnbuckles to the guy lines at one end.

The design turns out to work very well and is easy to adjust, athough raising the poles was a non-trivial task.
posted by unSane at 7:24 PM on August 3, 2008

By the way, the triangle approximation works very well. However the mass of the person riding the line has a large effect on the sag... steel wire doesn't stretch much under load but the geometry means that even a very small stretch translates to a considerable amount of sag. So my five year old son rides above my head, but when I get on it, my ass drags on the grass!
posted by unSane at 7:30 PM on August 3, 2008 [1 favorite]

JJ86:

Well F=mg, so yeah mass is required if you want to solve the gravity aspect.

F=ma also. How fast does a 5 kg bowling ball fall near the surface of the Earth? The OP is looking for a length, not a mass times a length, not a mass divided by a length or any other quantity involving a mass. So, in working through the problem, anywhere a mass appears, it must be divided by another quantity with dimension of mass. But there's only one mass involved in the problem: the mass of the cable. Now, if the cable has a non-uniform distribution, then there will be a geometrical factor associated with the distribution, but the total mass of the cable will still not come into the answer. Likewise, if you have a person riding the cable, you will have a quantity involving something like the ratio of the mass of the person to the mass of the cable.
posted by dsword at 8:26 PM on August 3, 2008

dsword: You only need the mass to be able to calculate the uniformly distributed force of the cable's weight. There is no need to divide by any other masses.

:Basic Engineering:
posted by JJ86 at 6:03 AM on August 4, 2008

I'm sorry, but you are very clearly mistaken. I suggest you work through the problem of finding the equation of a catenary going through the points (0,0), (70,-7), and having a minimum at y=-9. You will find that there is at most one such curve. The length of this curve clearly does not depend on any mass scale--no mention of mass is even necessary in formulating the problem! If you don't believe this is actually the same problem, then by all means go through the process of applying the variational principle to minimize the potential for a wire of mass M, demand that it pass through these same points by adjusting the tension and the boundary conditions, and finally calculate the arc length to see that the answer is the same and is independent of M.

:Basic Calculus:

On the other hand, if you want to know the tension on the supports ( [T]=[F]=kg * m / s^2 ), then yes, you need to know the mass of the cable. Likewise, if you allow the cable to stretch (say, according to Hooke's Law), then you will need to know the mass. But if only interested in the arc length of a curve of a specified shape going through specified points, forget about the mass. Call it 1 and be done with it.
posted by dsword at 11:00 AM on August 4, 2008