Editing Catenary (section)

===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.