PlanetPhysics/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, $$\tan\varphi = \frac{dy}{dx} = y'.$$ 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 $$T = \frac{a}{\cos\varphi},$$ whence the vertical component of $$T$$ is $$T\sin{\varphi} = a\tan{\varphi}$$ and its differential $$d(T\sin{\varphi}) = a\,d\tan{\varphi} = a\,dy'.$$ 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 $$\begin{matrix} \sigma\sqrt{1+y'^2}\,dx = a\,dy', \end{matrix}$$ which allows the separation of variables: $$\int dx = \frac{a}{\sigma}\int\frac{dy'}{\sqrt{1+y'^2}}$$ This may be solved by using the substitution $$y' := \sinh{t}, \quad dy' = \cosh{t}\,dt, \quad \sqrt{1+y'^2} = \cosh{t}$$ giving $$x = \frac{a}{\sigma}t+x_0,$$ i.e. $$y' = \frac{dy}{dx} = \sinh\frac{\sigma(x-x_0)}{a}.$$ This leads to the final solution $$y = \frac{a}{\sigma}\cosh\frac{\sigma(x-x_0)}{a}+y_0$$ 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 $$\begin{matrix} y = a\cosh\frac{x}{a} \end{matrix}$$ 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 $$x$$-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.