catenary | Physics Library

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 $T$

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

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

-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

. 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

$\tan\varphi = \sinh\frac{x}{a}, \quad \sin\varphi = \tanh\frac{x}{a}$
The arc length of the catenary (2) from the apex $(0,\,a)$ to the point $(x,\,y)$ is $a\sinh\frac{x}{a} = \sqrt{y^2-a^2}$ .
The radius of curvature of the catenary (2) is $a\cosh^2\frac{x}{a}$ , which is the same as length of the normal line of the catenary between the curve and the -axis.
The catenary is the catacaustic of the exponential curve reflecting the vertical rays.
If a parabola rolls on a straight line, the focus draws a catenary.
The involute (or evolvent) of the catenary is the tractrix.

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