Editing Catenary (section)

==Analysis==

===Model of chains and arches===
In the [[mathematical model]] the chain (or cord, cable, rope, string, etc.) is idealized by assuming that it is so thin that it can be regarded as a [[curve]] and that it is so flexible any force of [[Tension (physics)|tension]] exerted by the chain is parallel to the chain.<ref>[[#Routh|Routh]] Art. 442, p. 316</ref> The analysis of the curve for an optimal arch is similar except that the forces of tension become forces of [[Compression (physics)|compression]] and everything is inverted.<ref>{{cite book| last=Church| first=Irving Porter| title=Mechanics of Engineering| url=https://archive.org/details/mechanicsengine06churgoog| year=1890| publisher=Wiley| page=[https://archive.org/details/mechanicsengine06churgoog/page/n410 387]}}</ref>
An underlying principle is that the chain may be considered a rigid body once it has attained equilibrium.<ref>[[#Whewell|Whewell]] p. 65</ref> Equations which define the shape of the curve and the tension of the chain at each point may be derived by a careful inspection of the various forces acting on a segment using the fact that these forces must be in balance if the chain is in [[static equilibrium]].

Let the path followed by the chain be given [[parametric equations|parametrically]] by {{math|1='''r''' = (''x'', ''y'') = (''x''(''s''), ''y''(''s''))}} where {{mvar|s}} represents [[arc length]] and {{math|'''r'''}} is the [[position vector]]. This is the [[Unit speed parametrization|natural parameterization]] and has the property that

<math display=block>\frac{d\mathbf{r}}{ds}=\mathbf{u}</math>

where {{math|'''u'''}} is a [[unit tangent vector]].

[[File:CatenaryForceDiagram.svg|thumb|Diagram of forces acting on a segment of a catenary from {{math|'''c'''}} to {{math|'''r'''}}. The forces are the tension {{math|'''T'''<sub>0</sub>}} at {{math|'''c'''}}, the tension {{math|'''T'''}} at {{math|'''r'''}}, and the weight of the chain {{math|(0, −''ws'')}}. Since the chain is at rest the sum of these forces must be zero.]]
A [[differential equation]] for the curve may be derived as follows.<ref>Following [[#Routh|Routh]] Art. 443 p. 316</ref> Let {{math|'''c'''}} be the lowest point on the chain, called the vertex of the catenary.<ref>[[#Routh|Routh]] Art. 443 p. 317</ref> The slope {{math|{{sfrac|''dy''|''dx''}}}} of the curve is zero at {{math|'''c'''}} since it is a minimum point. Assume {{math|'''r'''}} is to the right of {{math|'''c'''}} since the other case is implied by symmetry. The forces acting on the section of the chain from {{math|'''c'''}} to {{math|'''r'''}} are the tension of the chain at {{math|'''c'''}}, the tension of the chain at {{math|'''r'''}}, and the weight of the chain. The tension at {{math|'''c'''}} is tangent to the curve at {{math|'''c'''}} and is therefore horizontal without any vertical component and it pulls the section to the left so it may be written {{math|(−''T''<sub>0</sub>, 0)}} where {{math|''T''<sub>0</sub>}} is the magnitude of the force. The tension at {{math|'''r'''}} is parallel to the curve at {{math|'''r'''}} and pulls the section to the right. The tension at {{math|'''r'''}} can be split into two components so it may be written {{math|1=''T'''''u''' = (''T'' cos ''φ'', ''T'' sin ''φ'')}}, where {{mvar|T}} is the magnitude of the force and {{mvar|φ}} is the angle between the curve at {{math|'''r'''}} and the {{mvar|x}}-axis (see [[tangential angle]]). Finally, the weight of the chain is represented by {{math|(0, −''ws'')}} where {{mvar|w}} is the weight per unit length and {{mvar|s}} is the length of the segment of chain between {{math|'''c'''}} and {{math|'''r'''}}.

The chain is in equilibrium so the sum of three forces is {{math|'''0'''}}, therefore

<math display=block>T \cos \varphi = T_0</math>
and
<math display=block>T \sin \varphi = ws\,,</math>

and dividing these gives

<math display=block>\frac{dy}{dx}=\tan \varphi = \frac{ws}{T_0}\,.</math>

It is convenient to write

<math display=block>a = \frac{T_0}{w}</math>

which is the length of chain whose weight is equal in magnitude to the tension at {{math|'''c'''}}.<ref>[[#Whewell|Whewell]] p. 67</ref>  Then

<math display=block>\frac{dy}{dx}=\frac{s}{a}</math>

is an equation defining the curve.

The horizontal component of the tension, {{math|1=''T'' cos ''φ'' = ''T''<sub>0</sub>}} is constant and the vertical component of the tension, {{math|1=''T'' sin ''φ'' = ''ws''}} is proportional to the length of chain between {{math|'''r'''}} and the vertex.<ref>[[#Routh|Routh]] Art 443, p. 318</ref>

===Derivation of equations for the curve===
The differential equation <math>dy/dx = s/a</math>, given above, can be solved
to produce equations for the curve.
<ref>
  A minor variation of the derivation presented here
  can be found on page 107 of [[#Maurer|Maurer]].
  A different (though ultimately mathematically equivalent) derivation,
  which does not make use of hyperbolic function notation,
  can be found in [[#Routh|Routh]]
  (Article 443, starting in particular at page 317).
</ref>
We will solve the equation using the boundary condition that
the vertex is positioned at <math>s_0=0</math> and <math>(x,y)=(x_0,y_0)</math>.

First, invoke the formula for
[[Arc length#Finding arc lengths by integration|arc length]]
to get
<math display=block>\frac{ds}{dx}
  = \sqrt{1+\left(\frac{dy}{dx}\right)^2}
  = \sqrt{1+\left(\frac{s}{a}\right)^2}\,,</math>
then [[Separation of variables#Ordinary differential equations (ODE)|separate variables]]
to obtain
<math display=block>\frac{ds}{\sqrt{1+(s/a)^2}}
    = dx\,.</math>

A reasonably straightforward approach to integrate this is to use
[[Trigonometric substitution#Hyperbolic substitution|hyperbolic substitution]],
which gives
<math display=block>a \sinh^{-1}\frac{s}{a} + x_0 = x</math>
(where <math>x_0</math> is a [[constant of integration]]),
and hence
<math display=block>\frac{s}{a} = \sinh\frac{x-x_0}{a}\,.</math>

But <math display=inline>s/a = dy/dx</math>, so
<math display=block>\frac{dy}{dx} = \sinh\frac{x-x_0}{a}\,,</math>
which [[Hyperbolic functions#Standard integrals|integrates as]]
<math display=block>y = a \cosh\frac{x-x_0}{a} + \delta</math>
(with <math>\delta=y_0-a</math> being the constant of integration satisfying the boundary condition).

Since the primary interest here is simply the shape of the curve,
the placement of the coordinate axes are arbitrary;
so make the convenient choice of <math display=inline>x_0=0=\delta</math>
to simplify the result to
<math display=block>y = a \cosh\frac{x}{a}.</math>

For completeness, the <math>y \leftrightarrow s</math> relation can be derived by
solving each of the <math>x \leftrightarrow y</math> and <math>x \leftrightarrow s</math> relations for <math>x/a</math>, giving:
<math display=block>\cosh^{-1}\frac{y-\delta}{a} = \frac{x-x_0}{a} = \sinh^{-1}\frac{s}{a}\,,</math>
so
<math display=block>y-\delta = a\cosh\left(\sinh^{-1}\frac{s}{a}\right)\,,</math>
which [[Inverse hyperbolic function#Composition of hyperbolic and inverse hyperbolic functions|can be rewritten]] as
<math display=block>y-\delta = a\sqrt{1+\left(\frac{s}{a}\right)^2} = \sqrt{a^2 + s^2}\,.</math>

===Alternative derivation===
The differential equation can be solved using a different approach.<ref>Following Lamb p. 342</ref> From

<math display=block>s = a \tan \varphi</math>

it follows that

<math display=block>\frac{dx}{d\varphi} = \frac{dx}{ds}\frac{ds}{d\varphi}=\cos \varphi \cdot a \sec^2 \varphi= a \sec \varphi</math>
and
<math display=block>\frac{dy}{d\varphi} = \frac{dy}{ds}\frac{ds}{d\varphi}=\sin \varphi \cdot a \sec^2 \varphi= a \tan \varphi \sec \varphi\,.</math>

Integrating gives,

<math display=block>x = a \ln(\sec \varphi + \tan \varphi) + \alpha</math>
and
<math display=block>y = a \sec \varphi + \beta\,.</math>

As before, the {{mvar|x}} and {{mvar|y}}-axes can be shifted so {{mvar|α}} and {{mvar|β}} can be taken to be 0. Then

<math display=block>\sec \varphi + \tan \varphi = e^\frac{x}{a}\,,</math>
and taking the reciprocal of both sides
<math display=block>\sec \varphi - \tan \varphi = e^{-\frac{x}{a}}\,.</math>

Adding and subtracting the last two equations then gives the solution
<math display=block>y = a \sec \varphi = a \cosh\left(\frac{x}{a}\right)\,,</math>
and
<math display=block>s = a \tan \varphi = a \sinh\left(\frac{x}{a}\right)\,.</math>

===Determining parameters===
[[Image:Catenary-tension.svg|350px|thumb|Three catenaries through the same two points, depending on the horizontal force {{mvar|T<sub>H</sub>}}.]]
In general the parameter {{mvar|a}} is the position of the axis. The equation can be determined in this case as follows:<ref>Following Todhunter Art. 186</ref>

Relabel if necessary so that {{math|''P''<sub>1</sub>}} is to the left of {{math|''P''<sub>2</sub>}} and let {{mvar|H}} be the horizontal and {{mvar|v}} be the vertical distance from {{math|''P''<sub>1</sub>}} to {{math|''P''<sub>2</sub>}}. [[Translation (geometry)|Translate]] the axes so that the vertex of the catenary lies on the {{mvar|y}}-axis and its height {{mvar|a}} is adjusted so the catenary satisfies the standard equation of the curve
 
<math display=block>y = a \cosh\left(\frac{x}{a}\right)</math>

and let the coordinates of {{math|''P''<sub>1</sub>}} and {{math|''P''<sub>2</sub>}} be {{math|(''x''<sub>1</sub>, ''y''<sub>1</sub>)}} and {{math|(''x''<sub>2</sub>, ''y''<sub>2</sub>)}} respectively. The curve passes through these points, so the difference of height is

<math display=block>v = a \cosh\left(\frac{x_2}{a}\right) - a \cosh\left(\frac{x_1}{a}\right)\,.</math>

and the length of the curve from {{math|''P''<sub>1</sub>}} to {{math|''P''<sub>2</sub>}} is

<math display="block">L = a \sinh\left(\frac{x_2}{a}\right) - a \sinh\left(\frac{x_1}{a}\right)\,.</math>

When {{math|''L''<sup>2</sup> − ''v''<sup>2</sup>}} is expanded using these expressions the result is

<math display="block">L^2-v^2=2a^2\left(\cosh\left(\frac{x_2-x_1}{a}\right)-1\right)=4a^2\sinh^2\left(\frac{H}{2a}\right)\,,</math>
so
<math display="block">\frac 1H \sqrt{L^2-v^2}=\frac{2a}H \sinh\left(\frac{H}{2a}\right)\,.</math>

This is a transcendental equation in {{mvar|a}} and must be solved [[Numerical analysis|numerically]]. Since <math>\sinh(x)/x</math> is strictly monotonic on <math>x > 0</math>,<ref>See [[#Routh|Routh]] art. 447</ref> there is at most one solution with {{math|''a'' > 0}} and so there is at most one position of equilibrium.

However, if both ends of the curve ({{math|''P''<sub>1</sub>}} and {{math|''P''<sub>2</sub>}}) are at the same level ({{math|''y''<sub>1</sub> {{=}} ''y''<sub>2</sub>}}), it can be shown that<ref>Archived at [https://ghostarchive.org/varchive/youtube/20211205/T-gUVEs51-c Ghostarchive]{{cbignore}} and the [https://web.archive.org/web/20201028222841/https://www.youtube.com/watch?v=T-gUVEs51-c Wayback Machine]{{cbignore}}: {{cite web| url = https://www.youtube.com/watch?v=T-gUVEs51-c| title = Chaînette - partie 3 : longueur | website=[[YouTube]]| date = 6 January 2015 }}{{cbignore}}</ref>
<math display=block>a = \frac {\frac14 L^2-h^2} {2h}\,</math>
where L is the total length of the curve between {{math|''P''<sub>1</sub>}} and {{math|''P''<sub>2</sub>}} and {{mvar|h}} is the sag (vertical distance between {{math|''P''<sub>1</sub>}}, {{math|''P''<sub>2</sub>}} and the vertex of the curve).

It can also be shown that
<math display=block>L = 2a \sinh \frac {H} {2a}\,</math>
and
<math display=block>H = 2a \operatorname {arcosh} \frac {h+a} {a}\,</math>
where H is the horizontal distance between {{math|''P''<sub>1</sub>}} and {{math|''P''<sub>2</sub>}} which are located at the same level  ({{math|''H'' {{=}} ''x''<sub>2</sub> − ''x''<sub>1</sub>}}).

The horizontal traction force at {{math|''P''<sub>1</sub>}} and {{math|''P''<sub>2</sub>}} is {{math|''T<sub>0</sub>'' {{=}} ''wa''}}, where {{mvar|w}} is the weight per unit length of the chain or cable.

===Tension relations===
There is a simple relationship between the tension in the cable at a point and its {{mvar|x}}- and/or {{mvar|y}}- coordinate. Begin by combining the squares of the vector components of the tension:
<math display=block>(T\cos\varphi)^2 + (T\sin\varphi)^2 = T_0^2 + (ws)^2</math>
which (recalling that <math>T_0=wa</math>) can be rewritten as
<math display=block>\begin{align}
T^2(\cos^2\varphi + \sin^2\varphi) &= (wa)^2 + (ws)^2 \\[6pt]
T^2 &= w^2 (a^2 + s^2) \\[6pt]
T &= w\sqrt{a^2+s^2} \,.
\end{align}</math>
But, [[#Derivation of equations for the curve|as shown above]],
<math>y = \sqrt{a^2 + s^2}</math> (assuming that <math>y_0=a</math>), so we get the simple relations<ref>[[#Routh|Routh]] Art 443, p. 318</ref>
<math display=block>T = wy = wa \cosh\frac{x}{a}\,.</math>

=== Variational formulation ===
Consider a chain of length <math>L</math> suspended from two points of equal height and at distance <math>D</math>. The curve has to minimize its potential energy
<math display=block> U = \int_0^D w y\sqrt{1+y'^2} dx </math>
(where {{mvar|w}} is the weight per unit length) and is subject to the constraint 
<math display=block> \int_0^D \sqrt{1+y'^2} dx = L\,.</math>

The modified [[Calculus of variations|Lagrangian]] is therefore 
<math display=block> \mathcal{L} = (w y - \lambda )\sqrt{1+y'^2}</math>
where <math>\lambda </math> is the [[Lagrange multiplier]] to be determined. As the independent variable <math>x</math> does not appear in the Lagrangian, we can use the [[Beltrami identity]] 
<math display=block> \mathcal{L}-y' \frac{\partial \mathcal{L} }{\partial y'} = C </math>
where <math>C</math> is an integration constant, in order to obtain a first integral
<math display=block>\frac{(w y - \lambda )}{\sqrt{1+y'^2}} = -C</math>

This is an ordinary first order differential equation that can be solved by the method of [[separation of variables]]. Its solution is the usual hyperbolic cosine where the parameters are obtained from the constraints.