Catenary

 ( Roulettes ) 

Another roulette, formed by rolling a line on a catenary, is another line. This implies that can roll perfectly smoothly on a road made of a series of bumps in the shape of an inverted catenary curve. The wheels can be any regular polygon except a triangle, but the catenary must have parameters corresponding to the shape and dimensions of the wheels.

The surface of revolution with fixed radii at either end that has minimum surface area is a catenary

$$y\; =\; a\; \backslash cosh^\{-1\}\backslash left(\backslash frac\{x\}\{a\}\backslash right)\; +\; b$$

revolved about the $y$-axis.

Let the path followed by the chain be given parametrically by where represents arc length and is the position vector. This is the natural parameterization and has the property that

$$\backslash frac\{d\backslash mathbf\{r\}\}\{ds\}=\backslash mathbf\{u\}$$

where is a unit tangent vector.

A differential equation for the curve may be derived as follows.Following Routh Art. 443 p. 316 Let be the lowest point on the chain, called the vertex of the catenary.Routh Art. 443 p. 317 The slope of the curve is zero at since it is a minimum point. Assume is to the right of since the other case is implied by symmetry. The forces acting on the section of the chain from to are the tension of the chain at , the tension of the chain at , and the weight of the chain. The tension at is tangent to the curve at and is therefore horizontal without any vertical component and it pulls the section to the left so it may be written where is the magnitude of the force. The tension at is parallel to the curve at and pulls the section to the right. The tension at can be split into two components so it may be written , where is the magnitude of the force and is the angle between the curve at and the -axis (see tangential angle). Finally, the weight of the chain is represented by where is the weight per unit length and is the length of the segment of chain between and .

The chain is in equilibrium so the sum of three forces is , therefore

$$T\; \backslash cos\; \backslash varphi\; =\; T\_0$$ and $$T\; \backslash sin\; \backslash varphi\; =\; ws\backslash ,,$$

and dividing these gives

$$\backslash frac\{dy\}\{dx\}=\backslash tan\; \backslash varphi\; =\; \backslash frac\{ws\}\{T\_0\}\backslash ,.$$

It is convenient to write

$$a\; =\; \backslash frac\{T\_0\}\{w\}$$

which is the length of chain whose weight is equal in magnitude to the tension at .Whewell p. 67 Then

$$\backslash frac\{dy\}\{dx\}=\backslash frac\{s\}\{a\}$$

is an equation defining the curve.

The horizontal component of the tension, is constant and the vertical component of the tension, is proportional to the length of chain between and the vertex.Routh Art 443, p. 318

 A minor variation of the derivation presented here
 can be found on page 107 of Maurer.
 A different (though ultimately mathematically equivalent) derivation,
 which does not make use of hyperbolic function notation,
 can be found in Routh
 (Article 443, starting in particular at page 317).
     

First, invoke the formula for arc length to get $$\backslash frac\{ds\}\{dx\}$$

 = \sqrt{1+\left(\frac{dy}{dx}\right)^2}
 = \sqrt{1+\left(\frac{s}{a}\right)^2}\,,
     

   = dx\,.
     

A reasonably straightforward approach to integrate this is to use hyperbolic substitution, which gives $$a\; \backslash sinh^\{-1\}\backslash frac\{s\}\{a\}\; +\; x\_0\; =\; x$$ (where $x\_0$ is a constant of integration), and hence $$\backslash frac\{s\}\{a\}\; =\; \backslash sinh\backslash frac\{x-x\_0\}\{a\}\backslash ,.$$

But $s/a\; =\; dy/dx$, so $$\backslash frac\{dy\}\{dx\}\; =\; \backslash sinh\backslash frac\{x-x\_0\}\{a\}\backslash ,,$$ which integrates as $$y\; =\; a\; \backslash cosh\backslash frac\{x-x\_0\}\{a\}\; +\; \backslash delta$$ (with $\backslash delta=y\_0-a$ 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 $x\_0=0=\backslash delta$ to simplify the result to $$y\; =\; a\; \backslash cosh\backslash frac\{x\}\{a\}.$$

For completeness, the $y\; \backslash leftrightarrow\; s$ relation can be derived by solving each of the $x\; \backslash leftrightarrow\; y$ and $x\; \backslash leftrightarrow\; s$ relations for $x/a$, giving: $$\backslash cosh^\{-1\}\backslash frac\{y-\backslash delta\}\{a\}\; =\; \backslash frac\{x-x\_0\}\{a\}\; =\; \backslash sinh^\{-1\}\backslash frac\{s\}\{a\}\backslash ,,$$ so $$y-\backslash delta\; =\; a\backslash cosh\backslash left(\backslash sinh^\{-1\}\backslash frac\{s\}\{a\}\backslash right)\backslash ,,$$ which can be rewritten as $$y-\backslash delta\; =\; a\backslash sqrt\{1+\backslash left(\backslash frac\{s\}\{a\}\backslash right)^2\}\; =\; \backslash sqrt\{a^2\; +\; s^2\}\backslash ,.$$

$$s\; =\; a\; \backslash tan\; \backslash varphi$$

it follows that

$$\backslash frac\{dx\}\{d\backslash varphi\}\; =\; \backslash frac\{dx\}\{ds\}\backslash frac\{ds\}\{d\backslash varphi\}=\backslash cos\; \backslash varphi\; \backslash cdot\; a\; \backslash sec^2\; \backslash varphi=\; a\; \backslash sec\; \backslash varphi$$ and $$\backslash frac\{dy\}\{d\backslash varphi\}\; =\; \backslash frac\{dy\}\{ds\}\backslash frac\{ds\}\{d\backslash varphi\}=\backslash sin\; \backslash varphi\; \backslash cdot\; a\; \backslash sec^2\; \backslash varphi=\; a\; \backslash tan\; \backslash varphi\; \backslash sec\; \backslash varphi\backslash ,.$$

Integrating gives,

$$x\; =\; a\; \backslash ln(\backslash sec\; \backslash varphi\; +\; \backslash tan\; \backslash varphi)\; +\; \backslash alpha$$ and $$y\; =\; a\; \backslash sec\; \backslash varphi\; +\; \backslash beta\backslash ,.$$

As before, the and -axes can be shifted so and can be taken to be 0. Then

$$\backslash sec\; \backslash varphi\; +\; \backslash tan\; \backslash varphi\; =\; e^\backslash frac\{x\}\{a\}\backslash ,,$$ and taking the reciprocal of both sides $$\backslash sec\; \backslash varphi\; -\; \backslash tan\; \backslash varphi\; =\; e^\{-\backslash frac\{x\}\{a\}\}\backslash ,.$$

Adding and subtracting the last two equations then gives the solution $$y\; =\; a\; \backslash sec\; \backslash varphi\; =\; a\; \backslash cosh\backslash left(\backslash frac\{x\}\{a\}\backslash right)\backslash ,,$$ and $$s\; =\; a\; \backslash tan\; \backslash varphi\; =\; a\; \backslash sinh\backslash left(\backslash frac\{x\}\{a\}\backslash right)\backslash ,.$$

Relabel if necessary so that is to the left of and let be the horizontal and be the vertical distance from to . Translate the axes so that the vertex of the catenary lies on the -axis and its height is adjusted so the catenary satisfies the standard equation of the curve

$$y\; =\; a\; \backslash cosh\backslash left(\backslash frac\{x\}\{a\}\backslash right)$$

and let the coordinates of and be and respectively. The curve passes through these points, so the difference of height is

$$v\; =\; a\; \backslash cosh\backslash left(\backslash frac\{x\_2\}\{a\}\backslash right)\; -\; a\; \backslash cosh\backslash left(\backslash frac\{x\_1\}\{a\}\backslash right)\backslash ,.$$

and the length of the curve from to is

$$L\; =\; a\; \backslash sinh\backslash left(\backslash frac\{x\_2\}\{a\}\backslash right)\; -\; a\; \backslash sinh\backslash left(\backslash frac\{x\_1\}\{a\}\backslash right)\backslash ,.$$

When is expanded using these expressions the result is

$$L^2-v^2=2a^2\backslash left(\backslash cosh\backslash left(\backslash frac\{x\_2-x\_1\}\{a\}\backslash right)-1\backslash right)=4a^2\backslash sinh^2\backslash left(\backslash frac\{H\}\{2a\}\backslash right)\backslash ,,$$ so $$\backslash frac\; 1H\; \backslash sqrt\{L^2-v^2\}=\backslash frac\{2a\}H\; \backslash sinh\backslash left(\backslash frac\{H\}\{2a\}\backslash right)\backslash ,.$$

This is a transcendental equation in and must be solved numerically. Since $\backslash sinh(x)/x$ is strictly monotonic on $x\; >\; 0$,See Routh art. 447 there is at most one solution with and so there is at most one position of equilibrium.

However, if both ends of the curve ( and ) are at the same level (), it can be shown thatArchived at Ghostarchive and the target="_blank" rel="nofollow"> Wayback Machine: $$a\; =\; \backslash frac\; \{\backslash frac14\; L^2-h^2\}\; \{2h\}\backslash ,$$ where L is the total length of the curve between and and is the sag (vertical distance between , and the vertex of the curve).

It can also be shown that $$L\; =\; 2a\; \backslash sinh\; \backslash frac\; \{H\}\; \{2a\}\backslash ,$$ and $$H\; =\; 2a\; \backslash operatorname\; \{arcosh\}\; \backslash frac\; \{h+a\}\; \{a\}\backslash ,$$ where H is the horizontal distance between and which are located at the same level ().

The horizontal traction force at and is , where is the weight per unit length of the chain or cable.

The modified Lagrangian is therefore $$\backslash mathcal\{L\}\; =\; (w\; y\; -\; \backslash lambda\; )\backslash sqrt\{1+y\text{'}^2\}$$ where $\backslash lambda$ is the Lagrange multiplier to be determined. As the independent variable $x$ does not appear in the Lagrangian, we can use the Beltrami identity $$\backslash mathcal\{L\}-y\text{'}\; \backslash frac\{\backslash partial\; \backslash mathcal\{L\}\; \}\{\backslash partial\; y\text{'}\}\; =\; C$$ where $C$ is an integration constant, in order to obtain a first integral $$\backslash frac\{(w\; y\; -\; \backslash lambda\; )\}\{\backslash sqrt\{1+y\text{'}^2\}\}\; =\; -C$$

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.

Let denote the weight per unit length of the chain, then the weight of the chain has magnitude

$$\backslash int\_\backslash mathbf\{c\}^\backslash mathbf\{r\}\; w\backslash ,\; ds\backslash ,,$$

where the limits of integration are and . Balancing forces as in the uniform chain produces

$$T\; \backslash cos\; \backslash varphi\; =\; T\_0$$ and $$T\; \backslash sin\; \backslash varphi\; =\; \backslash int\_\backslash mathbf\{c\}^\backslash mathbf\{r\}\; w\backslash ,\; ds\backslash ,,$$ and therefore $$\backslash frac\{dy\}\{dx\}=\backslash tan\; \backslash varphi\; =\; \backslash frac\{1\}\{T\_0\}\; \backslash int\_\backslash mathbf\{c\}^\backslash mathbf\{r\}\; w\backslash ,\; ds\backslash ,.$$

Differentiation then gives

$$w=T\_0\; \backslash frac\{d\}\{ds\}\backslash frac\{dy\}\{dx\}\; =\; \backslash frac\{T\_0\; \backslash dfrac\{d^2y\}\{dx^2\}\}\{\backslash sqrt\{1+\backslash left(\backslash dfrac\{dy\}\{dx\}\backslash right)^2\}\}\backslash ,.$$

In terms of and the radius of curvature this becomes

$$w=\; \backslash frac\{T\_0\}\{\backslash rho\; \backslash cos^2\; \backslash varphi\}\backslash ,.$$

$$\backslash frac\{dy\}\{dx\}=\backslash tan\; \backslash varphi\; =\; \backslash frac\{w\}\{T\_0\}x\backslash ,.$$

This is solved by simple integration to get

$$y=\backslash frac\{w\}\{2T\_0\}x^2\; +\; \backslash beta$$

and so the cable follows a parabola. If the weight of the cable and supporting wires is not negligible then the analysis is more complex.Ira Freeman investigated the case where only the cable and roadway are significant, see the External links section. Routh gives the case where only the supporting wires have significant weight as an exercise.

In this case the equations for tension are

Combining gives

$$c\; \backslash tan\; \backslash varphi\; =\; \backslash int\; \backslash sec\; \backslash varphi\backslash ,\; ds$$

and by differentiation

$$c\; =\; \backslash rho\; \backslash cos\; \backslash varphi$$

where is the radius of curvature.

The solution to this is

$$y\; =\; c\; \backslash ln\backslash left(\backslash sec\backslash left(\backslash frac\{x\}\{c\}\backslash right)\backslash right)\backslash ,.$$

In this case, the curve has vertical asymptotes and this limits the span to . Other relations are

$$x\; =\; c\backslash varphi\backslash ,,\backslash quad\; s\; =\; \backslash ln\backslash left(\backslash tan\backslash left(\backslash frac\{\backslash pi+2\backslash varphi\}\{4\}\backslash right)\backslash right)\backslash ,.$$

The curve was studied 1826 by Davies Gilbert and, apparently independently, by Gaspard-Gustave Coriolis in 1836.

Recently, it was shown that this type of catenary could act as a building block of electromagnetic metasurface and was known as "catenary of equal phase gradient".

$$s=\backslash left(1+\backslash frac\{T\}\{E\}\backslash right)p\backslash ,,$$

where is a constant equal to , where is the stiffness of the spring.Routh Art. 489 In the catenary the value of is variable, but ratio remains valid at a local level, soRouth Art. 494 $$\backslash frac\{ds\}\{dp\}=1+\backslash frac\{T\}\{E\}\backslash ,.$$ The curve followed by an elastic spring can now be derived following a similar method as for the inelastic spring.Following Routh Art. 500

The equations for tension of the spring are

$$T\; \backslash cos\; \backslash varphi\; =\; T\_0\backslash ,,$$ and $$T\; \backslash sin\; \backslash varphi\; =\; w\_0\; p\backslash ,,$$

from which

$$\backslash frac\{dy\}\{dx\}=\backslash tan\; \backslash varphi\; =\; \backslash frac\{w\_0\; p\}\{T\_0\}\backslash ,,\backslash quad\; T=\backslash sqrt\{T\_0^2+w\_0^2\; p^2\}\backslash ,,$$

where is the natural length of the segment from to and is the weight per unit length of the spring with no tension. Write $$a\; =\; \backslash frac\{T\_0\}\{w\_0\}$$ so $$\backslash frac\{dy\}\{dx\}=\backslash tan\; \backslash varphi\; =\; \backslash frac\{p\}\{a\}\; \backslash quad\backslash text\{and\}\backslash quad\; T=\backslash frac\{T\_0\}\{a\}\backslash sqrt\{a^2+p^2\}\backslash ,.$$

Then from which

Integrating gives the parametric equations

Again, the and -axes can be shifted so and can be taken to be 0. So

are parametric equations for the curve. At the rigid limit where is large, the shape of the curve reduces to that of a non-elastic chain.

First, let be the force of tension as a function of . The chain is flexible so it can only exert a force parallel to itself. Since tension is defined as the force that the chain exerts on itself, must be parallel to the chain. In other words,

$$\backslash mathbf\{T\}\; =\; T\; \backslash mathbf\{u\}\backslash ,,$$

where is the magnitude of and is the unit tangent vector.

Second, let be the external force per unit length acting on a small segment of a chain as a function of . The forces acting on the segment of the chain between and are the force of tension at one end of the segment, the nearly opposite force at the other end, and the external force acting on the segment which is approximately . These forces must balance so

$$\backslash mathbf\{T\}(s+\backslash Delta\; s)-\backslash mathbf\{T\}(s)+\backslash mathbf\{G\}\backslash Delta\; s\; \backslash approx\; \backslash mathbf\{0\}\backslash ,.$$

Divide by and take the limit as to obtain

$$\backslash frac\{d\backslash mathbf\{T\}\}\{ds\}\; +\; \backslash mathbf\{G\}\; =\; \backslash mathbf\{0\}\backslash ,.$$

These equations can be used as the starting point in the analysis of a flexible chain acting under any external force. In the case of the standard catenary, where the chain has weight per unit length.

• Catenary arch
• Chain fountain or self-siphoning beads
• Funicular curve
• Overhead catenary – power lines suspended over rail or tram vehicles
• Roulette (curve) – an elliptic/hyperbolic catenary
• Troposkein – the shape of a spun rope
• Weighted catenary

• Frank Swetz (1995). 9780883857038, MAA. . ISBN 9780883857038

• Catenary curve calculator
• Catenary at The Geometry Center
• "Catenary" at Visual Dictionary of Special Plane Curves
• The Catenary - Chains, Arches, and Soap Films.
• Cable Sag Error Calculator – Calculates the deviation from a straight line of a catenary curve and provides derivation of the calculator and references.
• Dynamic as well as static cetenary curve equations derived – The equations governing the shape (static case) as well as dynamics (dynamic case) of a centenary is derived. Solution to the equations discussed.
• The straight line, the catenary, the brachistochrone, the circle, and Fermat Unified approach to some geodesics.
• Ira Freeman "A General Form of the Suspension Bridge Catenary" Bulletin of the AMS