catenary: Definition from Answers.com

This article is about the mathematical curve. For other uses, see Catenary (disambiguation).

"Chainette" redirects here. For the wine grape also known as Chainette, see Cinsaut.

In physics and geometry, the catenary is the theoretical shape a hanging chain or cable will assume when supported at its ends and acted on only by its own weight. Its surface of revolution, the catenoid, is a minimal surface and will be the shape of a soap film bounded by two circles. The curve is the graph of the hyperbolic cosine function, which has a U-like shape, similar in appearance to a parabola.

A hanging chain forms a typical catenary.

The silk on a spider's web forming multiple elastic catenaries.

1 History
2 The inverted catenary arch
3 Anchoring of marine vessels
4 Simple suspension bridges
5 Mathematical description
6 Variations
7 See also
8 References
9 External links

History

The word catenary is derived from the Latin word catena, which means "chain". The curve is also called the "alysoid", "funicular", and "chainette". Galileo said that the curve of a chain hanging under gravity would be a parabola, but this was disproved by Joachim Jungius (1587-1657) and published posthumously in 1669.^[1]

In 1691 Leibniz, Christiaan Huygens, and Johann Bernoulli derived the equation in response to a challenge by Jakob Bernoulli. Huygens first used the term 'catenaria' in a letter to Leibniz in 1690, and David Gregory wrote a treatise on the catenary in 1690. However Thomas Jefferson is usually credited with the English word 'catenary'.^[2]

The application of the catenary to the construction of arches is ancient, as described below; the modern rediscovery and statement is due to Robert Hooke, who discovered it in the context of the rebuilding of St Paul's Cathedral,^[3] possibly having seen Huygens' work on the catenary.

In 1671, Hooke announced to the Royal Society that he had solved the problem of the optimal shape of an arch, and in 1675 published an encrypted solution as a Latin anagram^[4] in an appendix to his Description of Helioscopes,^[5] where he wrote that he had found "a true mathematical and mechanical form of all manner of Arches for Building". He did not publish the solution of this anagram^[6] in his lifetime, but in 1705 his executor provided it as Ut pendet continuum flexile, sic stabit contiguum rigidum inversum, meaning "As hangs a flexible cable so, inverted, stand the touching pieces of an arch."

Euler proved in 1744 that the catenary is the curve which, when rotated about the x-axis, gives the surface of minimum surface area (the catenoid) for the given bounding circle.^[7]

The inverted catenary arch

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Arch of Taq-i Kisra in Ctesiphon as seen today

Catenary^[8] arches under the roof of Gaudí's Casa Milà, Barcelona, Spain

The catenary is the ideal curve for an arch which supports only its own weight. When the centerline of an arch is made to follow the curve of an up-side-down (ie. inverted) catenary, the arch endures almost pure compression, in which no significant bending moment occurs inside the material. If the arch is made of individual elements (eg., stones) whose contacting surfaces are perpendicular to the curve of the arch, no significant shear forces are present at these contacting surfaces. (Shear stress is still present inside each stone, as it resists the compressive force along the shear sliding plane.) The thrust (including the weight) of the arch at its two ends is tangent to its centerline.

In antiquity, the curvature of the inverted catenary was intuitively discovered and found to lead to stable arches and vaults. A spectacular example remains in the Taq-i Kisra in Ctesiphon, which was once a great city of Mesopotamia. In ancient Greek and Roman cultures, the less efficient curvature of the circle was more commonly used in arches and vaults. The efficient curvature of inverted catenary was perhaps forgotten in Europe from the fall of Rome to the Middle-Ages and the Renaissance, where it was almost never used, although the pointed arch was perhaps a fortuitous approximation of it.

The Catalan architect Antoni Gaudí made extensive use of catenary shapes in most of his work. In order to find the best curvature for the arches and ribs that he desired to use in the crypt of the Church of Colònia Güell, Gaudí constructed inverted scale models made of numerous threads under tension to represent stones under compression. This technique worked well to solve angled columns, arches, and single-curvature vaults, but could not be used to solve the more complex, double-curvature vaults that he intended to use in the nave of the church of the Sagrada Familia. The idea that Gaudi used thread models to solve the nave of the Sagrada Familia is a common misconception, although it could have been used in the solution of the bell towers.

The Gateway Arch (looking East).

Catenary arch kiln under construction over temporary form

The Gateway Arch in Saint Louis, Missouri, United States follows the form of an inverted catenary. It is 630 feet wide at the base and 630 feet tall.

In structural engineering a catenary shell is a structural form, usually made of concrete, that follows a catenary curve. The profile for the shell is obtained by using flexible material subjected to gravity, converting it into a rigid formwork for pouring the concrete and then using it as required, usually in an inverted manner.

A kiln, a kind of oven for firing pottery, may be made from firebricks with a body in the shape of a catenary arch, usually nearly as wide as it is high, with the ends closed off with a permanent wall in the back and a temporary wall in the front. The bricks (mortared with fireclay) are stacked upon a temporary form in the shape of an inverted catenary, which is removed upon completion. The form is designed with a simple length of light chain, whose shape is traced onto an end panel of the form, which is inverted for assembly. A particular advantage of this shape is that it does not tend to dismantle itself over repeated heating and cooling cycles — most other forms such as the vertical cylinder must be held together with steel bands.

Anchoring of marine vessels

The catenary form given by gravity is made advantage of in its presence in heavy anchor rodes. An anchor rode (or anchor line) usually consists of chain and/or cable. Anchor rodes are used by ships, oilrigs, docks, and other marine assets which must be anchored to the seabed.

Particularly with larger vessels, the catenary curve given by the weight of the rode presents a lower angle of pull on the anchor or mooring device. This assists the performance of the anchor and raises the level of force it will resist before dragging. With smaller vessels and in shallow water it is less effective.^[9]

The catenary curve in this context is only fully present in the anchoring system when the rode has been lifted clear of the seabed by the vessel's pull, as the seabed obviously affects its shape while it supports the chain or cable. There is also typically a section of rode above the water and thus unaffected by buoyancy, creating a slightly more complicated curve.

Simple suspension bridges

Capilano Suspension Bridge. In simple suspension bridges such as this, where the weight runs parallel to the cables, the cables follow a catenary curve.

Free-hanging chains follow the catenary curve, but suspension bridge chains or cables do not hang freely since they support the weight of the bridge. In most cases the weight of the cable is negligible compared with the weight being supported. When the force exerted is uniform with respect to the length of the chain, as in a simple suspension bridge, the result is a catenary.

Hercilio Luz Bridge, Florianópolis, Brazil. Most suspension bridge cables follow a parabolic, not catenary, curve.

When the force exerted is uniform with respect to horizontal distance, as in a suspension bridge, the result is a parabola, much as Galileo originally claimed.^[10]

When suspension bridges are constructed, the suspension cables initially sag as the catenary curve, before being tied to the deck below, and then gradually assume a parabolic curve as additional connecting cables are tied to connect the main suspension cables with the bridge deck below.

Mathematical description

Equation

Catenaries for different values of the parameter 'a'

The equation (up to translation and rotation) of a catenary in Cartesian coordinates has the form

$y = a \, \cosh \left ({x \over a} \right ) = {a \over 2} \, \left (e^{x/a} + e^{-x/a} \right )$ ,

where $cosh$ is the hyperbolic cosine function. The scaling factor $a$ can be interpreted as the ratio between the horizontal component of the tension on the chain (a constant) and the weight of the chain per unit of length.

The formula for the St. Louis Arch,

$y = -127.7 \; \textrm{ft} \cdot \cosh({x / 127.7 \; \textrm{ft}}) + 757.7 \; \textrm{ft}$ .

is displayed inside.

The Whewell equation for the catenary is

$\tan \varphi = \frac{s}{a}$ .

Differentiating gives

$\frac{d\varphi}{ds} = \frac{\cos^2\varphi}{a}$

and eliminating $\varphi$ gives the Cesàro equation:

$\kappa=\frac{a}{s^2+a^2}$ .

Other properties

A parabola rolled along a straight line traces out a catenary (see roulette) with its focus ^[7].

Square wheels can roll perfectly smoothly if the road has evenly spaced bumps in the shape of a series of inverted catenary curves. The wheels can be any regular polygon save for a triangle, but one must use the correct catenary, corresponding correctly to the shape and dimensions of the wheels.^[11]

A charge in a uniform electric field moves along a catenary (which tends to a parabola if the charge velocity is much less than the speed of light c).^{[citation needed]}

The surface of revolution with fixed radii at either end that has minimum surface area is a catenary revolved about the x-axis.

Derivation

We assume that the path followed by the chain is given parametrically by $\vec{r} = (x, y) = (x(s),\ y(s))$ where $s$ represents arc length and $\vec{r}$ is the position vector. This is the natural parameterization and has the property that $\frac{d\vec{r}}{ds}$ is the unit tangent vector, $\vec{u}$ . The derivation of the curve for an optimal arch is similar except that the forces of tension become forces of compression and everything is inverted.

It is now possible to derive two equations which together define the shape of the curve and the tension of the chain at each point. This is done by a careful inspection of the various forces acting on a small segment of the chain and using the fact that these forces must be in balance if the chain is in static equilibrium.

First, let $\vec{T}=\vec{T}(s)$ be the force of tension as a function of $s$ . 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, $\vec{T}$ must be parallel to the chain. In other words,

$(1)\qquad\vec{T} = T \vec{u}$

where $T$ is the magnitude of $\vec{T}$ , a positive scalar function of $s$ .

Second, let $\vec{G}=\vec{G}(s)$ be the external force per unit length acting on a small segment of a chain as a function of $s$ . The forces acting on the segment of the chain between $s$ and $s + Δ s$ are the force of tension $\vec{T}(s+\Delta s)$ at one end of the segment, the nearly opposite force $-\vec{T}(s)$ at the other end, and the external force acting on the segment which is approximately $\vec{G}\Delta s$ . These forces must balance so

$\vec{T}(s+\Delta s)-\vec{T}(s)+\vec{G}\Delta s \approx \vec{0}$ .

Divide by $Δ s$ and take the limit as $\Delta s \to 0$ to obtain

$(2)\qquad\frac{d\vec{T}}{ds} + \vec{G} = \vec{0}$ .

Note that, up till now, no assumptions have been made regarding the force $\vec{G}$ , so equations (1) and (2) can be used as the starting point in the analysis of a flexible chain acting under any external force. The next step is to put in the specific expression for $\vec{G}$ and solve the resulting equations.

In this case, $\vec{G} = (0, -\lambda g)$ where the chain has constant mass per unit length $λ$ and the only external force acting on the chain is that of a uniform gravitational field $\vec{g} = (0, -g)$ . So we have

$\frac{d\vec{T}}{ds} = (0, \lambda g)$ .

Integrating we get,

$\vec{T} = (c, \lambda g s + d)$ .

Note that at the minimum the curve is horizontal and $\vec{T} = T \vec{u} = T (1, 0) = (T, 0) = (c, \lambda g s + d)$ . So $c$ is the tension of the chain at its lowest point this point occurs at $s = - d / λ g$ . The point from which $s$ is measured is arbitrary, so pick this point to be the minimum, giving $d = 0$ . The equation becomes

$\vec{T} = (c, \lambda g s)$ .

Note that the horizontal component of the tension is a constant.

From here, we can continue the derivation in two ways.

Alternative 1

If $\varphi$ is the tangential angle of the curve then $\vec{T}$ is parallel to $(1, \tan \varphi)$ so

$\tan \varphi = \frac{\lambda g s}{c}$ .

Write $a = \frac{c}{\lambda g}$ to combine constants and obtain the Whewell equation for the curve,

$\tan \varphi = \frac{s}{a}$ .

In general, parametric equations can be obtained from a Whewell equation by integrating:

$x = \int \cos \varphi \, ds$

$y = \int \sin \varphi \, ds$

To find these integrals, make the substitution $\tan{\varphi} = \sinh u$ (or $\varphi = \mbox{gd} u$ where $gd$ is the Gudermannian function).

Then $s = a \ \sinh u$ and

$x = \int \cos {\varphi} ds = \int \mbox{sech} u \, a \cosh u du = a \int du = au + \alpha$

$y = \int \sin {\varphi} ds = \int \tanh u \, a \cosh u du = a \int \sinh u du = a \cosh u + \beta$ .

We can eliminate u to obtain

$y = a \cosh \frac{x-\alpha}{a} + \beta$

where $α$ and $β$ are constants to be determined, along with $a$ , by the boundary conditions of the problem. Usually these conditions include two points from which the chain is being suspended and the length of the chain.

Alternative 2

From

$\vec{T} = T \vec{u} = T \left(\frac{dx}{ds}, \frac{dy}{ds}\right) = (c, \lambda g s)$ ,

$\frac{dy}{dx} = \frac{dy}{ds}\left/\frac{dx}{ds}\right. = \frac{\lambda g s}{c} = \frac{s}{a}$ ,

where $a = \frac{c}{\lambda g},$ same as before. Then

$\frac{dx}{ds} =1\left/\frac{ds}{dx}\right. = \frac{1}{\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}} = \frac{a}{\sqrt{a^2+s^2}}$

and

$\frac{dy}{ds} = \frac{dy}{dx}\cdot\frac{dx}{ds} = \frac{s}{\sqrt{a^2+s^2}}$ .

The integrals of the right hand sides of these equations can be found using standard techniques giving

$x = a\ \operatorname{arcsinh}(s/a) + \alpha,\ y = \sqrt{a^2+s^2} + \beta$ .

Again, $α$ and $β$ are constants to be determined, along with $a$ , by the boundary conditions of the problem.

Variations

Elastic Catenary

In an elastic catenary, the cable replaced by a spring and is no longer assumed to be of fixed density, but is allowed to stretch in accordance with Hooke's Law. In this case, the mass per unit length is no longer constant but can be given as

$\lambda = \frac{\lambda_0}{1+\epsilon T}$

where $λ 0$ is the mass per unit length for the chain in its relaxed state and $ε$ is the spring constant. As in the earlier derivation,

$\frac{d\vec{T}}{ds} = (0, \lambda g)$ .

So the horizontal component of $\vec{T}$ , $T \cos \varphi$ is a constant c. Putting this into the equation for density produces

$\lambda = \frac{\lambda_0}{1+ c\epsilon \sec \varphi}$ .

Then the equation for the vertical component of $\vec{T}$ is

$\frac{d}{ds}(T \sin \varphi) = c\frac{d}{ds} (\tan \varphi) = \frac{g\lambda_0}{1+ c\epsilon \sec \varphi}$ ,

or, combining constants,

$\frac{d}{ds} (\tan \varphi) = \frac{1}{a+ b\sec \varphi}$ .

Using the substitution $\tan{\varphi} = \sinh u$ gives

$\frac{d}{ds} (\sinh u) = \cosh u \frac{du}{ds} = \frac{1}{a+ b\cosh u}$

$\cosh u (a+ b\cosh u) = \frac{ds}{du}$ .

Parametric equations can be obtained by integrating:

$x = \int \cos \varphi \, ds = \int \mbox{sech} u \cosh u (a+ b\cosh u)du = \int (a+ b\cosh u)du= au + b\sinh u + \alpha$ ,

$y = \int \sin \varphi \, ds = \int \tanh u \cosh u (a+ b\cosh u)du = \int \sinh u(a+ b\cosh u)du = a\cosh u + \tfrac{b}{2}\sinh^2 u + \beta$ .

When b = 0, corresponding to a completely inelastic cable, this is simply the catenary. When a = 0, corresponding to the case there the cable essentially has length 0 in its relaxed state, similar to a Slinky, this is a parabola. When a and b are both >0 then the curve is intermediate between a catenary and a parabola.

Equal resistance catenary

In an equal resistance catenary, cable is strengthened according to the magnitude of the tension at each point, so its resistance to breaking is constant along its length. Assuming that the strength of the cable is proportional to its density, the mass per unit length can be given as

λ = λ r T

where $λ r$ is the mass per unit length per unit of tension force required for the chain to resist breaking. As in the earlier derivation,

$\frac{d\vec{T}}{ds} = (0, \lambda g)$ .

So the horizontal component of $\vec{T}$ , $T \cos \varphi$ is a constant c. Putting this into the equation for density produces

$\lambda = \lambda_r c \sec \varphi$ .

Then the equation for the vertical component of $\vec{T}$ is

$\frac{d}{ds}(T \sin \varphi) = c\frac{d}{ds} (\tan \varphi) = g \lambda_r c \sec \varphi$ ,

or, combining constants,

$\frac{d}{ds} (\tan \varphi) = \frac{1}{a} \sec \varphi$

$\frac{d}{ds}\left(\frac{dy}{dx}\right) = \frac{1}{a} \frac{ds}{dx}$ .

Multiplying both sides by $d s / d x$ gives

$\frac{d^2y}{dx^2} = \frac{1}{a} \left(\frac{ds}{dx}\right)^2 = \frac{1}{a} \left[1+\left(\frac{dy}{dx}\right)^2\right]$ .

This can be reduced to a differential equation of degree one using separation of variables to obtain

$\arctan \frac{dy}{dx} = \frac{x}{a}-\alpha$

$\frac{dy}{dx} = \tan\left(\frac{x}{a}-\alpha\right)$ .

Another integration produces

$y = -a\ln \cos\left(\frac{x}{a}-\alpha\right) + \beta$ .

Towed cables

Instead of gravity, we assume we have a cylindrical cable that is acted on by drag forces due to the movement of some surrounding fluid (e.g. air or water). The velocity relative to the cable is assumed to be a constant $\vec{v} = (0, -v)$ . (Velocity is assumed to be vertical here to preserve similarities with the gravitational case.) To compute the force due to drag, write $\vec{v} = \vec{v}_u + \vec{v}_n$ where $\vec{v}_u$ and $\vec{v}_n$ respectively are the components parallel to and orthogonal to the cable. The cable is assumed to be smooth so the force on the cable due to $\vec{v}_u$ is taken to be negligible. The force acting on the cable, following the Drag equation is

$\vec{G} = -c |\vec{v}_n|^2 \vec{n}$

where $c$ is a constant depending on the density of the fluid, the diameter of the cable, and the Drag coefficient. If $\vec{n} = (-\sin\varphi,\ \cos\varphi)$ denotes the unit normal vector, then

$\vec{v}_n = \vec{v}\cdot\vec{n}\ \vec{n} = -\cos\varphi\ \vec{n}$ .

$\vec{G} = -c \cos^2\varphi\ \vec{n}$ .

From equations (1) and (2) above,

$\frac{d\vec{T}}{ds} = \frac{dT}{ds} \vec{u} + T\frac{d\vec{u}}{ds} = \frac{dT}{ds} \vec{u} + T\frac{d\varphi}{ds}\vec{n} = c \cos^2\varphi\ \vec{n}$ .

Setting the coefficients of $\vec{u}$ and $\vec{n}$ equal produces

$\frac{dT}{ds} = 0,\,T\frac{d\varphi}{ds} = c \cos^2\varphi$ .

So T is a constant in this case and combining constants in the second equation gives

$\frac{d\varphi}{ds} = \frac{\cos^2\varphi}{a}$

which is one of the equations for the catenary given above. This is a case where a different expression for the force acting on the chain/cable produce the same curve but a different expression for tension.

In applications, the force of gravity and additional terms in the force due to drag may be added to the expression for force, yielding equations that must be solved numerically.

References

^ Swetz, Faauvel, Bekken, "Learn from the Masters", 1997, MAA ISBN 0883857030, pp.128-9
^ "Catenary" at Math Words
^ http://links.jstor.org/sici?sici=0035-9149(200105)55%3A2%3C289%3AMAMSTO%3E2.0.CO%3B2-X
^ cf. the anagram for Hooke's law, which appeared in the next paragraph.
^ Arch Design
^ The original anagram was "abcccddeeeeeefggiiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux": the letters of the Latin phrase, alphabetized.
^ ^a ^b http://mathworld.wolfram.com/Catenary.html
^ [1]
^ Chain, Rope, and Catenary - Anchor Systems For Small Boats
^ Paul Kunkel (June 30, 2006). "Hanging With Galileo". Whistler Alley Mathematics. http://whistleralley.com/hanging/hanging.htm. Retrieved March 27, 2009.
^ "Roulette: A Comfortable Ride on an n-gon Bicycle" by Borut Levart, Wolfram Demonstrations Project, 2007.

External links

Wikimedia Commons has media related to: Catenary

Wikisource has the text of the 1911 Encyclopædia Britannica article Catenary.

Weisstein, Eric W., "Catenary" from MathWorld.
"Catenary" at Encyclopédie des Formes Mathématiques Remarquables
"Elastic catenary" at Encyclopédie des Formes Mathématiques Remarquables
"Catenary of equal resistance" at Encyclopédie des Formes Mathématiques Remarquables
"Catenary" at Visual Dictionary of Special Plane Curves
Hanging With Galileo - mathematical derivation of formula for suspended and free-hanging chains; interactive graphical demo of parabolic vs. hyperbolic suspensions.
Catenary Demonstration Experiment - An easy way to demonstrate the Mathematical properties of a cosh using the hanging cable effect. Devised by Jonathan Lansey
Horizontal Conveyor Arrangement - Diagrams of different horizontal conveyor layouts showing options for the catenary section both supported and unsupported
Catenary curve derived - The shape of a catenary is derived, plus examples of a chain hanging between 2 points of unequal height, including C program to calculate the curve.
Cable Sag Error Calculator - Calculates the deviation from a straight line of a catenary curve and provides derivation of the calculator and references.
Hexagonal Geodesic Domes - Catenary Domes, an article about creating catenary domes

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	catenoid (mathematics)
	catenary suspension (engineering)

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catenary

Contents

History

The inverted catenary arch

Anchoring of marine vessels

Simple suspension bridges

Mathematical description

Equation

Other properties

Derivation

Alternative 1

Alternative 2

Variations

Elastic Catenary

Equal resistance catenary

Towed cables

See also

References

External links