The catenary curve is the shape of a chain hanging between two equal-height poles under the influence of gravity. But the derivation of the (hyperbolic cosine) curve equation from the physics traditionally assumes a uniform gravitational field. Suppose instead one uses the non-uniform gravitational field that diminishes with distance from the center of the Earth. (Perhaps this would be relevant for a very long chain that sags significantly.) Does this lead to an interesting curve, known in some closed form? Or just to a differential equation that can only be solved numerically?

I ask this primarily out of curiosity, so please interpret in that spirit!

Well, the next physically interesting situation is the skipping rope equation, where the gravitational field is proportional to the distance from a fixed axis. Clebsch, 1860. You may find a lot of material googoling "skipping rope equation" ("courbe de la corde à sauter" or "Springseilkurve" should you read French or German).

You might look at Robert Osserman's article The mathematics of the gateway arch, which discusses many generalizations of the catenary problem. I'm not sure that it does exactly what you want, however.

  • Great reference, great paper--Thanks! He explores versions of a "weighted catenary," where the chain's density varies along its length. He does not seem to mention the natural condition of using the inverse-square law of gravity. But it seems likely that one could select a density variation to simulate the effect of true gravity... – Joseph O'Rourke Jun 25 '10 at 17:16

Have a look at

J.Denzler, A.M.Hinz, Catenaria Vera---the True Catenary, Exposition. Math. 17 (1999), 117--142. (MR 2000e:49006, ZB 943.49001).

—A.M.Hinz (Maribor, Slovenia)

(Image from the cited paper added by J.O'Rourke:)

This started out as a comment, but it wouldn't fit in the box:

If you are interested in deviation from the catenary caused by variation in the strength of gravity, you should be almost as interested in deviation caused by variation in its direction. Imagine a suspension bridge with a span of length s and towers of height h. These towers will be further apart at the top, by a factor of 1 in R/h, where R is the radius of the Earth. (For a concrete example, the Akashi Kaikyō Bridge, the longest in the world, has s=1991m and h=283m, giving a factor of about 1 in 22000, or 9cm.) The difference in the force of gravity between top and bottom is R2/(R+h)2, or 1 in R/2h.

So the directional effect is half as strong as the height-induced effect. Whether they affect the shape of the curve in the same proportion is another question.

  • Yes, you are absolutely correct! Surely some classical scholar worked out the chain curve under true gravity, both in magnitude and direction... – Joseph O'Rourke Jun 27 '10 at 1:34

The skipping rope equation is known as the troposkien. It is well known to anyone who designs Darrieus style vertical axis wind turbines. It is to the best of my knowledge solved only iteratively.

  • See also Pietro Majer's answer on the skipping-rope equation. – Joseph O'Rourke May 11 '12 at 12:13

If one uses the functions $f_d(t) = (\cos (d t))^{\frac 1d}$ (which were implicitly introduced by Colin Maclaurin) one can compute the catenaries for central gravitational fields which are proportional to a power of the distance to the Sun. These are curves of the form $rf(\theta)=1$ where $f$ is a function of the above type (or a dilation and rotation thereof).

Interestingly, one can also get catenaries for parallel forces from the same functions. In fact, parametrised curves of the form $(F(t),f(t))$ where $f$ is one of the above functions and $F$ is a primitive, give the catenaries for gravitational fields which are proportional to powers of the distance to the $X$-axis. These curves, the Maclaurin catenaries, are new.

Not relevant to the question but worth mentioning is that if one rotates them about the $X$-axis, they provide examples of surfaces for which all six parameters in the fundamental forms ($E$, $F$, $G$, $L$, $M$, and $N$) are proportional to powers of the above distance.

Details in arXiv 1102.1579

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.