catenary

A chain or a homogeneous flexible thin wire takes a form resembling an arc of a parabola when suspended at its ends. The arc is not from a parabola but from the graph of the hyperbolic cosine function in a suitable coordinate system.

Let's derive the equation $ y = y(x)$ of this curve, called the catenary, in its plane with $ x$ -axis horizontal and $ y$ -axis vertical. We denote the line density of the weight of the wire by $ \sigma$ .

In any point $ (x,\,y)$ of the wire, the tangent line of the curve forms an angle $ \varphi$ with the positive direction of $ x$ -axis. Then,

$\displaystyle \tan\varphi = \frac{dy}{dx} = y'.$

In the point, a certain tension $ T$ of the wire acts in the direction of the tangent; it has the horizontal component $ T\cos\varphi$ which has apparently a constant value $ a$ . Hence we may write

$\displaystyle T = \frac{a}{\cos\varphi},$

whence the vertical component of $ T$ is

$\displaystyle T\sin{\varphi} = a\tan{\varphi}$

and its differential

$\displaystyle d(T\sin{\varphi}) = a\,d\tan{\varphi} = a\,dy'.$

But this differential is the amount of the supporting force acting on an infinitesimal portion of the wire having the projection $ dx$ on the $ x$ -axis. Because of the equilibrium, this force must be equal the weight $ \sigma\sqrt{1+(y'(x))^2}\,dx$ (see the arc length). Thus we obtain the differential equation

$\displaystyle \sigma\sqrt{1+y'^2}\,dx = a\,dy',$ (1)

which allows the separation of variables:

$\displaystyle \int dx = \frac{a}{\sigma}\int\frac{dy'}{\sqrt{1+y'^2}}$

This may be solved by using the substitution

$\displaystyle y' := \sinh{t}, \quad dy' = \cosh{t}\,dt, \quad \sqrt{1+y'^2} = \cosh{t}$

giving

$\displaystyle x = \frac{a}{\sigma}t+x_0,$

i.e.

$\displaystyle y' = \frac{dy}{dx} = \sinh\frac{\sigma(x-x_0)}{a}.$

This leads to the final solution

$\displaystyle y = \frac{a}{\sigma}\cosh\frac{\sigma(x-x_0)}{a}+y_0$

of the equation (1). We have denoted the constants of integration by $ x_0$ and $ y_0$ . They determine the position of the catenary in regard to the coordinate axes. By a suitable choice of the axes and the measure units one gets the simple equation

$\displaystyle y = a\cosh\frac{x}{a}$ (2)

of the catenary.

Some properties of catenary



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