catenary

A 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 (http://planetmath.org/HyperbolicFunctions) function in a suitable coordinate system.

Let’s derive the equation  $y=y(x)$  of this curve, called the , in its plane with $x$-axis horizontal and $y$-axis vertical.  We denote the 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,

 $\tan\varphi\;=\;\frac{dy}{dx}\;=\;y^{\prime}.$

In the point, a certain tension $T$ of the wire acts in the direction of the value $a$.  Hence we may write

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

whence the vertical of $T$ is

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

and its differential (http://planetmath.org/Differential)

 $d(T\sin{\varphi})\;=\;a\,d\tan{\varphi}\;=\;a\,dy^{\prime}.$

But this differential is the amount of the supporting   $\sigma\sqrt{1+(y^{\prime}(x))^{2}}\,dx$ (see the arc length).  Thus we obtain the differential equation

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

which allows the separation of variables:

 $\int dx\;=\;\frac{a}{\sigma}\int\frac{dy^{\prime}}{\sqrt{1\!+\!y^{\prime 2}}}$

This may be solved by using the substitution (http://planetmath.org/SubstitutionForIntegration)

 $y^{\prime}\;:=\;\sinh{t},\qquad dy^{\prime}\;=\;\cosh{t}\,dt,\qquad\sqrt{1\!+% \!y^{\prime 2}}\;=\;\cosh{t}$

giving

 $x\;=\;\frac{a}{\sigma}t+x_{0},$

i.e.

 $y^{\prime}\;=\;\frac{dy}{dx}\;=\;\sinh\frac{\sigma(x\!-\!x_{0})}{a}.$

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 equation

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

of the catenary.

Some of catenary

• $\tan\varphi=\sinh\frac{x}{a},\quad\sin\varphi=\tanh\frac{x}{a}$  (cf. the Gudermannian)

• 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 (http://planetmath.org/ExponentialFunction) reflecting the vertical rays.

• If a parabola rolls on a straight line, the focus draws a catenary.

• The involute (a.k.a. the evolvent) of the catenary is the tractrix.

 Title catenary Canonical name Catenary Date of creation 2014-10-26 21:25:30 Last modified on 2014-10-26 21:25:30 Owner pahio (2872) Last modified by pahio (2872) Numerical id 29 Author pahio (2872) Entry type Derivation Classification msc 53B25 Classification msc 51N05 Synonym chain curve Related topic EquationOfCatenaryViaCalculusOfVariations Related topic LeastSurfaceOfRevolution Related topic HyperbolicFunctions Related topic Tractrix Related topic EqualArcLengthAndArea Defines catenary