Talk:PlanetPhysics/Catenary

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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 \htmladdnormallink{graph}{http://planetphysics.us/encyclopedia/Bijective.html} of the hyperbolic cosine \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} in a suitable coordinate \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html}.

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 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 \htmladdnormallink{force}{http://planetphysics.us/encyclopedia/Thrust.html} acting on an infinitesimal portion of the wire having the projection $dx$ on the $x$-axis.\, Because of the \htmladdnormallink{equilibrium}{http://planetphysics.us/encyclopedia/InertialSystemOfCoordinates.html}, this force must be equal the weight\, $\sigma\sqrt{1+(y'(x))^2}\,dx$ (see the arc length).\, Thus we obtain the \htmladdnormallink{differential equation}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} \begin{align} \sigma\sqrt{1+y'^2}\,dx = a\,dy', \end{align} which allows the \htmladdnormallink{separation of variables}{http://planetphysics.us/encyclopedia/SeparationOfVariables.html}: $$\int dx = \frac{a}{\sigma}\int\frac{dy'}{\sqrt{1+y'^2}}$$ 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 \htmladdnormallink{position}{http://planetphysics.us/encyclopedia/Position.html} 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{align} y = a\cosh\frac{x}{a} \end{align} of the catenary.

\textbf{Some 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 (2) 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 (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. \item The catenary is the \htmladdnormallink{catacaustic}{http://planetphysics.us/encyclopedia/Catacaustic.html} of the exponential curve reflecting the vertical rays. \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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