The Shape of a Catenary

Version 2

The figure crudely represents a catenary, or the shape of a perfectly flexible, uniform cord that is suspended between two points. The object is to determine both the catenary’s shape and its length. The cord has a linear weight density of *w *units of force per unit of length.

Considering a differential length of cord, the horizontal forces are equal and represent the horizontal force applied by the end points. The vertical forces are gravity, *wdl*, offset by the differential in vertical tension

Total vertical force at some point *x* is found by integration:

.

Because the cord can support no bending moment, the forces acting on the differential at any point must be in line with the cord itself, so that

,

from which

Differentiation yields a hairy differential equation:

Eshbach offers this as the equation of a catenary

but the curve does not pass through the point (0,0). Adding a constant achieves that objective:

Now it must be asked whether Eq. 2 is indeed a solution of Eq. 1. First, find its derivatives

and impose them on Eq. 1:

Evidently Eq. 1 is not only a solution but it defines the relationship between *a* and *H*/*w*.

Now it is only necessary to make Eq. 2 pass through the point (*x*1,*y*1):

which will define *a*. There is no easy solution for *a* except by numerical iteration. When it is known, then the ratio *H*/*w* will be known.

Cord Length

The length of the cord can be found in either of two ways. The first simply uses the familiar expression for the length of any curve:

The second way takes advantage of the fact that the vertical force at each end point must equal half the total weight of the cord, and that total weight is the cord’s length *S* times *w*.

Also, because the cord can support no bending moment, the ratio of forces at each point, including end points, must equal the slope:

From these two equations and the slope previously expressed,

The constant *a* is already known from the solution to Eq. 3, given the end points, ±*x*1 and *y*1.

A Parabola

A parabola can be made to fit between the same points as the catenary and its expression will be

.

This can be plotted as was the catenary, and also its length can be determined the same way:

Then the question is, How much longer the catenary than the parabola? A program has been written to plot both curves and to determine their respective lengths. (This program can be supplied as an application to anyone interested in it. Unfortunately, the application version is 1.7M in size!) In the following table, span is the entire horizontal length of the cord, or 2*x*1, and has been fixed at 100. The height is *y*1.

Two methods were used in creating the table, with different results for the catenaries calculated by different methods. The "exact" method used the equations derived in this discussion and are probably right since there are no approximations except in calculating *a*, which was chosen to satisfy the equality in Eq. 3 to within 1E-12! The "numerical" method was just that: vertical force and *y* were integrated numerically, though in very small steps. Yet the error probably lies within that method simply because of the iteration.

This rationale is somewhat less than satisfying, however, since the plots of the parabola and the "exact" catenary appear so close together that it’s hard to believe the difference in length shown in the table. Yet this is only a subjective observation.

Height (for span=100) |
Catenary length |
Catenary |
Parabola length |

100 |
249.3 |
237.1 |
232.3 |

90 |
229.4 |
218.7 |
214.5 |

80 |
209.8 |
200.7 |
197.0 |

70 |
190.8 |
183.0 |
180.0 |

60 |
172.4 |
165.9 |
163.5 |

50 |
154.9 |
149.6 |
147.9 |

40 |
138.7 |
134.5 |
133.4 |

30 |
124.3 |
121.0 |
120.4 |

20 |
112.6 |
110.1 |
109.8 |

10 |
104.8 |
102.7 |
102.6 |

5 |
102.7 |
100.8 |
100.7 |