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 = y_{A} - y_{B} 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 s_{A} and s_{B}, respectively, then the
total arc length S = s_{A} + s_{B} 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 T_{0} 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 e^{α} and terms are rearranged, the result
is a quadratic equation for e^{α}:

_{}

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 e^{α} (the larger) and e^{β} (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 e^{β}) is less than 1, in which case β is negative. This
means that point B is to the right of O, and l_{B} and s_{B}
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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