The derivation of the catenary can be split up in two parts. The first part is the general derivation of the symmetric catenary cable

and the second part is more complicated derivation of the asymmetric catenary cable. In order to understand the derivation of the

asymmetric catenary it is best to first try to understand to derivation of the equation for the symmetric catenary. On this page the

more complicated derivation of the asymmetric catenary cable is explained.

The asymmetric catenary cable is for instance the cable of a gondola hanging between two support towers, an electrical cable or

a mooring line hanging down from a vessel and going into the anchor point with an inclination.

 

 

Asymmetric Catenary Cable

 

Consider the cable, hanging under its own weight (with weight per unit length μ), depicted in the following figure:

Using:

 

  (gravity is often included in μ)

 

it is assumed that point A is above point B, so that the height difference d = yA - yB  is positive, and is given by:

 

 

                                              (1)

 

If we now define:

 

 

 

with

 

then:

                                                                                      (2)

 

If the arc lengths of the right-hand and left-hand segments are sA and sB, respectively, then the total arc length S = sA + sB is given by:

 

                           

                                                                                       (3)

using:

 

 

If we square both sides of Eqs (2) and (3) and subtract the former from the latter, then:

 

 

When using the trigonometric rules for sinh and cosh:

 

and using:

                  

this results in:

 

or

                                             (4)

or

                                                                                        (5)

 

    This equation can be solved by the solver function in Excel. The solver function finds values for T0 and λ with μ and L given. Note that when d = 0 the equation is the same as in the symmetric case.

If β = 2λ - α is now substituted into Eq. (3) and the sinh functions are expressed in terms of exponentials, the resulting equation is:

 

                                             (6)

 

and if both sides are multiplied by eα and terms are rearranged, the result is a quadratic equation for eα:

 

                           

or

                           

or

                           

or

                                                                                      (7)

 

with              

 

Note that if the substitution α = 2λ - β were made instead, the resulting equation for β would have exactly the same form, so that the two roots of Equation (7) are eα (the larger) and eβ (the smaller), namely:

 

                                                        (8)

 

These results can be checked to make sure that β + α = 2λ

It may turn out that the smaller root (that is eβ) is less than 1, in which case β is negative. This means that point B is to the right of O, and lB and sB are treated as negative quantities, since there is no physical cable between O and B.

 

The equation for the shape the cable and the tension in the cable can be used for any given combination of two end points of a catenary and the cable length. For a symmetric catenary the vertical distance d between the end points is simply zero.

 

Besides the general explanation of the catenary cable this site also contains “did you know that?”s about catenaries. For the interested

readers who want to know just a bit more than average, click this link: Did you know that???

 

 

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