Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random | Template Test |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Talkback

Downloads

Information
[parent] catenary (Application)

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

  • $\tan\varphi = \sinh\frac{x}{a}, \quad \sin\varphi = \tanh\frac{x}{a}$
  • 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.



Anyone with an account can edit this entry. Please help improve it!

"catenary" is owned by pahio.

View style:

Other names:  chain curve

This object's parent.

Cross-references: catacaustic, position, separation of variables, differential equation, equilibrium, system, function, graph
There are 2 references to this object.

This is version 1 of catenary, born on 2009-04-13.
Object id is 639, canonical name is Catenary.
Accessed 551 times total.

Classification:
Physics Classification: 45. (Classical mechanics of discrete systems)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:

No messages.

Testing some escape charachters for html category with a generator has an injective cogenerator" now escape ” with "