Catenary

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For the railroad term see Overhead lines

For its use in ring theory, see Catenary ring.

"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 catenary.

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

[edit] 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 claimed 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]

[edit] The inverted catenary arch

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.

Arch of Taq-i Kisra in Ctesiphon as seen today

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.

Catenary arches under the roof of Gaudí's Casa Milà, Barcelona, Spain

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).

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.

Catenary arch kiln under construction over temporary form

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.

[edit] Suspension bridges

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

Free-hanging chains follow the curve of the hyperbolic function above, but suspension bridge chains or cables, which are tied to the bridge deck at uniform intervals, instead follow a parabolic curve, much as Galileo originally claimed.^[8]

When the force exerted is uniform with respect to horizontal distance, as in a suspension bridge, the result is a parabola. 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.

When suspension bridges are constructed, the suspension cables initially sag as the catenaric function, 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.

[edit] Mathematical description

[edit] 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

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

From this the Cesàro equation can be derived by differentiation:

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

[edit] Derivation

Let $\vec{G}$ be the external force per unit length acting on a small segment of a chain, and let $\vec{T}$ be the force of tension as a function of $s$ ,

the distance along the chain from some fixed point on it. 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 $\vec{G}\Delta s$ . The chain is assumed to be in static equilibrium so these forces must balance, therefore

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

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

$\frac{d\vec{T}}{ds} + \vec{G} = \vec{0}$ .

The flexibility of the chain implies that $\vec{T}$ is parallel to the curve of the chain, so we may assume $\vec{T} = T \frac{d\vec{r}}{ds}$ where $\frac{d\vec{r}} {ds}$ is the unit vector tangent to the curve and $T$ is the magnitude of $\vec{T}$ and is a positive scalar function of $s$ . $\vec{r} = \begin{pmatrix}x \\y\end{pmatrix}$ is the position vector of a point on the chain and is a function of $s$ .

We assume 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 $\vec{g} = \begin{pmatrix}0\\-g\end{pmatrix}$ , so $\vec{G} = \begin{pmatrix}0,\\-\lambda g\end{pmatrix}$ and

$\frac{d\vec{T}}{ds} = \begin{pmatrix}0\\ \lambda g\end{pmatrix}$ .

Integrating we get,

$\left( T \frac{d \vec{r}}{ds} \right)_s - \left( T \frac{d \vec{r}}{ds} \right)_m = \begin{pmatrix} 0 \\ \lambda g \left( s-s_m \right)\end{pmatrix}$

where the subscripts s and m are the limits of integration which are an arbitrary point on the curve and the point at which the curve is at a minimum height respectively. But at the minimum the curve is horizontal and $\left(\frac{\vec{r}}{ds}\right)_m = \begin{pmatrix} 1\\0\end{pmatrix}$ so,

$\vec{T} = \begin{pmatrix}T_m\\ \lambda g \left(s-s_m\right)\end{pmatrix}$ .

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

[edit] Continue alternative 1

If $\varphi$ is the tangential angle of the curve then $\vec{T}$ is parallel to $\begin{pmatrix}1\\ \tan \varphi\end{pmatrix}$ so

$\tan \varphi = \frac{\lambda g \left(s - s_m\right)}{T_m}$ .

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

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

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.

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.

[edit] Continue alternative 2

The vector equation above can be broken up into two scalar parts:

$\begin{array}{lcl} T \frac{dx}{ds} = T_m \\\\ T \frac{dy}{ds} = \lambda g \left( s-s_m \right) \end{array}$

Note that the first equation indicates that the horizontal component of the tension is the same everywhere along the chain. We can solve the first equation for $T = T_m \frac{ds}{dx}$ and use this to eliminate $T$ in the second:

$\frac{dy}{ds} = \frac{\lambda g}{T_m} \left(s-s_m\right) \frac{dx}{ds}$

Again setting $a=\frac{T_m}{\lambda g}$ and defining the the dimensionless parameter σ = (s-s_m)/a reduces this equation to:

$\frac{dy}{ds} = \sigma \frac{dx}{ds}.$

Noting that,

$1 =\left| \frac{d\vec{r}}{ds} \right|^2 = \left( \frac{dx}{ds} \right)^2 + \left( \frac{dy}{ds} \right)^2 = \left(1+\sigma^2\right) \left( \frac{dx}{ds} \right)^2$

we find,

$\frac{dx}{ds} = \frac{1}{\sqrt{1 + \sigma^2}} \qquad \frac{dy}{ds} = \frac{\sigma}{\sqrt{1+\sigma^2}}$

Defining the dimensionless parameters ξ = (x-x_m)/a and η = (y-y_m)/a and substituting places these equations in a form which is easier to integrate:

$\frac{d\xi}{d\sigma} = \frac{1}{\sqrt{1 + \sigma^2}} \qquad \frac{d\eta}{d\sigma} = \frac{\sigma}{\sqrt{1+\sigma^2}}$

The integrals of these equations are,

$\xi = \sinh^{-1}(\sigma) \qquad \eta = \sqrt{1 + \sigma^2} - 1$ .

Solving the first for $σ = sinh(ξ)$ and substituting into the second equation gives,

$\eta = \sqrt{1 + \sinh^2(\xi)} - 1 = \cosh(\xi) - 1$

And finally transforming back to x and y gives equations for the catenary in the form

$y = y_m + a\left[\cosh\left(\frac{x-x_m}{a}\right) - 1\right] \qquad a=\frac{T_m}{\lambda g}$

and the parametric form

$x = x_m + a\sinh^{-1}\Big(\frac{s-s_m}{a}\Big)$

$y = y_m - a + \sqrt{a^2 + \left(s-s_m\right)^2}.$

Including s_m six constants are needed to specify the cantenary curve and its endpoints. So either three points on the curve or the values of x, y and s at the endpoints will do. If only the values at the endpoints are known these can be substituted into the equations for the catenary to give simultaneous equations for the unknowns. For example, if s=y=0 at one end of the chain the parametric equation for y above gives the relation

$s_m^2=y_m\left(y_m-2a\right)$ .

If s_m is not needed then just the values of the coordinates at the two endpoints and either the sag (y_m) or "tensional length" (a) will suffice to specify the problem. (The tensional length is the equivalent length of chain whose weight is equal to the horizontal tension.)

[edit] 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 ^[9].

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).

[edit] 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 mostly of chain 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^[10].

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.

[edit] 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

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

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 $\vec{T}$ , $\tau cos \varphi$ 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

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

or, combining constants,

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

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

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

or

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

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.

[edit] Towed cables

When a cable is subject to wind or water flows, the drag forces lead to more general shapes, since the forces are not distributed in the same way as the weight. A cable having radius $a$ and specific gravity $σ$ , and towed at speed $v$ in a medium (e.g., air or water) with density $ρ 0$ , will have an $(x, y)$ position described by the following equations ^[11]:

$\frac{{dT}}{{ds}}=-\rho _{0}\left( {\sigma -1}\right) \pi a^{2}g\sin \phi -\rho _{0}v^{2}\pi aC_{T}\cos \phi ;$

$T\frac{{d\phi }}{{ds}}=-\rho _{0}\left( {\sigma -1}\right) \pi a^{2}g\cos \phi +\rho _{0}av^{2}\left[ {C_{D}\sin \phi +\pi C_{N}}\right] \sin \phi ;$

$\frac{{dx}}{{ds}}=\cos \phi ;$

$\frac{{dy}}{{ds}}=-\sin \phi .$

Here $T$ is the tension, $φ$ is the incident angle, $g = 9.81 m / s 2$ , and $s$ is the cable scope. There are three drag coefficients: the normal drag coefficient $C D$ ( $\approx 1.5$ for a smooth cylindrical cable); the tangential drag coefficient $C T$ ( $\approx 0.0025$ ), and $C N$ ( $= 0.75 C T$ ).

The system of equations has four equations and four unknowns: $T$ , $φ$ , $x$ and $y$ , and is typically solved numerically.

[edit] Critical angle tow

Critical angle tow occurs when the incident angle does not change. In practice, critical angle tow is common, and occurs far from significant point forces.

Setting $\frac{{d\phi }}{{ds}}=0$ leads to an equation for the critical angle:

$\rho _{0}\left( {\sigma -1}\right) \pi a^{2}g\cos \phi =\rho _{0}av^{2}\left[ {C_{D}\sin \phi +\pi C_{N}}\right] \sin \phi .$

If $π C N < < C D sinφ$ , the formula for the critical angle becomes

$\rho _{0}\left( {\sigma -1}\right) \pi a^{2}g\cos \phi =\rho _{0}av^{2}{ C_{D}\sin }^{2}{\phi ;}$

or

$\left( {\sigma -1}\right) \pi ag\cos \phi =v^{2}{C_{D}\sin }^{2}{\phi =}v^{2} {C_{D}}\left( 1-\cos ^{2}\phi \right) ;$

or

$\cos ^{2}\phi +\frac{\left( {\sigma -1}\right) \pi ag}{v^{2}{C_{D}}}\cos \phi -1=0;$

leading to the rule-of-thumb formula

$\cos \phi =-\frac{\left( {\sigma -1}\right) \pi ag}{2v^{2}{C_{D}}}+\sqrt{1+ 4\frac{\left( {\sigma -1}\right) ^{2}\pi ^{2}a^{2}g^{2}}{v^{4}{C_{D}^{2}}}}.$

The drag coefficients of a faired cable are more complicated, involving loading functions that account for drag variation as a function of incidence angle.

[edit] Other uses of the term

In railway engineering, a catenary structure consists of overhead lines used to deliver electricity to a railway locomotive, multiple unit, railcar, tram or trolleybus through a pantograph or a trolleypole. These structures consist of an upper structural wire in the form of a shallow catenary, short suspender wires, which may or may not contain insulators, and a lower conductive contact wire. By adjusting the tension in various elements the conductive wire is kept parallel to the centerline of the track, reducing the tendency of the pantograph or trolley to bounce or sway, which could cause a disengagement at high speed.
In nonrigid and semi-rigid airships, a catenary curtain is a fabric and cable internal structure used to distribute the weight of the gondola across a large area of the ship's envelope.
In conveyor systems, the catenary is the portion of the chain or belt underneath the conveyor that is traveling back to the start. It is the weight of the catenary that keeps tension in the chain or belt.
For backpacking tents, a catenary curve along the ridgeline reduces the amount of flap and sag in the material^[12]. this is mainly used on tarpaulins and other ultralight tents.

[edit] See also

Roulette (curve) - an elliptic/hyperbolic catenary

[edit] 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
^ Paul Kunkel (June 30, 2006). "Hanging With Galileo". Whistler Alley Mathematics. http://whistleralley.com/hanging/hanging.htm. Retrieved on March 27, 2009.
^ "Roulette: A Comfortable Ride on an n-gon Bicycle" by Borut Levart, Wolfram Demonstrations Project, 2007.
^ Chain, Rope, and Catenary - Anchor Systems For Small Boats
^ A.P. Dowling, "The dynamics of towed flexible cylinders. Part 2. Negatively buoyant elements" (1988). Journal of Fluid Mechanics, 187, 533-571.
^ http://www.backpacking.net/makegear/cat-tarp/index.html

[edit] External links

Wikimedia Commons has media related to: Catenary

Weistein, Eric W., "Catenary" from MathWorld.
"Catenary" at Encyclopédie des Formes Mathématiques Remarquables
"Elastic catenary" 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
Catenary - wolfram.com

[0] Swetz, Faauvel, Bekken, "Learn from the Masters", 1997, MAA ISBN 0883857030, pp.128-9

[1] "Catenary" at Math Words

[2] ttp://links.jstor.org/sici?sici=0035-9149(200105)55%3A2%3C289%3AMAMSTO%3E2.0.CO%3B2-X

[3] . the anagram for Hooke's law, which appeared in the next paragraph.

[4] Arch Design

[5] The original anagram was "abcccddeeeeeefggiiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux": the letters of the Latin phrase, alphabetized.

[Catenary_at_Mathworld-6] ttp://mathworld.wolfram.com/Catenary.html

[7] Paul Kunkel (June 30, 2006). "Hanging With Galileo". Whistler Alley Mathematics. http://whistleralley.com/hanging/hanging.htm. Retrieved on March 27, 2009.

[8] "Roulette: A Comfortable Ride on an n-gon Bicycle" by Borut Levart, Wolfram Demonstrations Project, 2007.

[9] Chain, Rope, and Catenary - Anchor Systems For Small Boats

[10] A.P. Dowling, "The dynamics of towed flexible cylinders. Part 2. Negatively buoyant elements" (1988). Journal of Fluid Mechanics, 187, 533-571.

[11] ttp://www.backpacking.net/makegear/cat-tarp/index.html

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