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Viewing Version
14
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'catenary'
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| Title of object: |
catenary |
| Canonical Name: |
Catenary |
| Type: |
Derivation |
| Created on: |
2005-06-07 08:15:42 |
| Modified on: |
2005-09-15 16:17:16 |
| Classification: |
msc:51N05, msc:53B25 |
| Defines: |
catenary |
| Synonyms: |
catenary=chain curve |
Revision comment (for changes between this and next version):
Preamble:
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\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
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%\usepackage{psfrag}
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\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
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%\usepackage{xypic}
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\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem} |
Content:
A \PMlinkescapetext{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 \PMlinkname{hyperbolic cosine}{HyperbolicFunctions} function in a suitable coordinate system.
Let's derive the equation\, $y = y(x)$\, of this curve, called the {\em catenary}, in its plane with $x$-axis horizontal and $y$-axis vertical.\, We denote the the \PMlinkescapetext{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 \PMlinkescapetext{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 \PMlinkescapetext{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 \PMlinkescapetext{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
$$\sigma\sqrt{1+y'^2}\,dx = a\,dy'$$
or
$$\int dx = \frac{a}{\sigma}\int\frac{dy'}{\sqrt{1+y'^2}}.$$
It may be solved by using the \PMlinkname{substitution}{SubstitutionForIntegration}
$$y' := \sinh{t}, \quad dy' = \cosh{t}\,dt, \quad \sqrt{1+y'^2} = \cosh{t}$$
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
$$y = \frac{a}{\sigma}\cosh\frac{\sigma(x-x_0)}{a}+y_0.$$
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 \PMlinkescapetext{measure units one gets the simple} equation
\begin{align}
y = a\cosh\frac{x}{a}
\end{align}
of the catenary.
\begin{center}
\includegraphics{catenary}
\end{center}
\textbf{Some \PMlinkescapetext{properties} of catenary}
\begin{itemize}
\item $\tan\varphi = \sinh\frac{x}{a}, \quad \sin\varphi = \tanh\frac{x}{a}$
\item The arc length of the catenary (1) from the apex\, $(0,\,a)$\, to the point\, $(x,\,y)$\, is\,\, $a\sinh\frac{x}{a} = \sqrt{y^2-a^2}$.
\item The radius of curvature of the catenary (1) is\, $a\cosh^2\frac{x}{a}$.
\item If a parabola rolls on a straight line, the focus draws a catenary.
\item The involute (or evolvent) of the catenary is the tractrix.
\end{itemize} |
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