Catenary

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File:Kette Kettenkurve Catenary 2008 PD.JPG
This chain, whose ends hang from two points, forms a catenary.
File:SpiderCatenary.jpg
The silk on this spider web forms multiple elastic catenaries.

In physics and geometry, a catenary (Template:IPAc-en Template:Respell, Template:IPAc-en Template:Respell) is the curve that an idealized hanging chain or cable assumes under its own weight when supported only at its ends in a uniform gravitational field.

The catenary curve has a U-like shape, superficially similar in appearance to a parabola, which it is not.

The curve appears in the design of certain types of arches and as a cross section of the catenoid—the shape assumed by a soap film bounded by two parallel circular rings.

The catenary is also called the alysoid, chainette,<ref name="MathWorld">MathWorld</ref> or, particularly in the materials sciences, an example of a funicular.<ref>e.g.: Template:Cite book</ref> Rope statics describes catenaries in a classic statics problem involving a hanging rope.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

Mathematically, the catenary curve is the graph of the hyperbolic cosine function. The surface of revolution of the catenary curve, the catenoid, is a minimal surface, specifically a minimal surface of revolution. A hanging chain will assume a shape of least potential energy which is a catenary.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> Galileo Galilei in 1638 discussed the catenary in the book Two New Sciences recognizing that it was different from a parabola. The mathematical properties of the catenary curve were studied by Robert Hooke in the 1670s, and its equation was derived by Leibniz, Huygens and Johann Bernoulli in 1691.

Catenaries and related curves are used in architecture and engineering (e.g., in the design of bridges and arches so that forces do not result in bending moments). In the offshore oil and gas industry, "catenary" refers to a steel catenary riser, a pipeline suspended between a production platform and the seabed that adopts an approximate catenary shape. In the rail industry it refers to the overhead wiring that transfers power to trains. (This often supports a contact wire, in which case it does not follow a true catenary curve.)

In optics and electromagnetics, the hyperbolic cosine and sine functions are basic solutions to Maxwell's equations.<ref>Template:Cite book</ref> The symmetric modes consisting of two evanescent waves would form a catenary shape.<ref>Template:Cite journal</ref><ref>Template:Cite journal</ref><ref>Template:Cite journal</ref>

HistoryEdit

The word "catenary" is derived from the Latin word catēna, which means "chain". The English word "catenary" is usually attributed to Thomas Jefferson,<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref><ref>Template:Cite book</ref> who wrote in a letter to Thomas Paine on the construction of an arch for a bridge:

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I have lately received from Italy a treatise on the equilibrium of arches, by the Abbé Mascheroni. It appears to be a very scientifical work. I have not yet had time to engage in it; but I find that the conclusions of his demonstrations are, that every part of the catenary is in perfect equilibrium.<ref>Template:Cite book</ref>{{#if:|{{#if:|}}

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It is often said<ref name="Lockwood124"/> that Galileo thought the curve of a hanging chain was parabolic. However, in his Two New Sciences (1638), Galileo wrote that a hanging cord is only an approximate parabola, correctly observing that this approximation improves in accuracy as the curvature gets smaller and is almost exact when the elevation is less than 45°.<ref>Template:Cite book</ref> The fact that the curve followed by a chain is not a parabola was proven by Joachim Jungius (1587–1657); this result was published posthumously in 1669.<ref name="Lockwood124">Lockwood p. 124</ref>

The application of the catenary to the construction of arches is attributed to Robert Hooke, whose "true mathematical and mechanical form" in the context of the rebuilding of St Paul's Cathedral alluded to a catenary.<ref>Template:Cite journal</ref> Some much older arches approximate catenaries, an example of which is the Arch of Taq-i Kisra in Ctesiphon.<ref>Template:Cite book</ref>

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<ref>cf. the anagram for Hooke's law, which appeared in the next paragraph.</ref> in an appendix to his Description of Helioscopes,<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> 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 to this anagram<ref>The original anagram was abcccddeeeeefggiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux: the letters of the Latin phrase, alphabetized.</ref> 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."

In 1691, Gottfried Leibniz, Christiaan Huygens, and Johann Bernoulli derived the equation in response to a challenge by Jakob Bernoulli;<ref name="Lockwood124"/> their solutions were published in the Acta Eruditorum for June 1691.<ref>Template:Citation</ref><ref name="calladine" >Template:Citation</ref> David Gregory wrote a treatise on the catenary in 1697<ref name="Lockwood124"/><ref>Template:Citation</ref> in which he provided an incorrect derivation of the correct differential equation.<ref name="calladine" />

Leonhard Euler proved in 1744 that the catenary is the curve which, when rotated about the Template:Mvar-axis, gives the surface of minimum surface area (the catenoid) for the given bounding circles.<ref name="MathWorld"/> Nicolas Fuss gave equations describing the equilibrium of a chain under any force in 1796.<ref>Routh Art. 455, footnote</ref>

Inverted catenary archEdit

Catenary arches are often used in the construction of kilns. To create the desired curve, the shape of a hanging chain of the desired dimensions is transferred to a form which is then used as a guide for the placement of bricks or other building material.<ref>Template:Cite book</ref><ref> Template:Cite book</ref>

The Gateway Arch in St. Louis, Missouri, United States is sometimes said to be an (inverted) catenary, but this is incorrect.<ref>Template:Citation</ref> It is close to a more general curve called a flattened catenary, with equation Template:Math, which is a catenary if Template:Math. While a catenary is the ideal shape for a freestanding arch of constant thickness, the Gateway Arch is narrower near the top. According to the U.S. National Historic Landmark nomination for the arch, it is a "weighted catenary" instead. Its shape corresponds to the shape that a weighted chain, having lighter links in the middle, would form.<ref>Template:Cite journal</ref><ref name="nrhpinv2">Template:Citation and Template:NHLS url Template:Small</ref>

Catenary bridgesEdit

File:Soderskar-bridge.jpg
Simple suspension bridges are essentially thickened cables, and follow a catenary curve.
File:Puentedelabarra(below).jpg
Stressed ribbon bridges, like the Leonel Viera Bridge in Maldonado, Uruguay, also follow a catenary curve, with cables embedded in a rigid deck.

In free-hanging chains, the force exerted is uniform with respect to length of the chain, and so the chain follows the catenary curve.<ref>Template:Cite book</ref> The same is true of a simple suspension bridge or "catenary bridge," where the roadway follows the cable.<ref>Template:Cite book</ref><ref>Template:Cite book</ref>

A stressed ribbon bridge is a more sophisticated structure with the same catenary shape.<ref>Template:Cite book</ref><ref>Template:Cite journal</ref>

However, in a suspension bridge with a suspended roadway, the chains or cables support the weight of the bridge, and so do not hang freely. In most cases the roadway is flat, so when the weight of the cable is negligible compared with the weight being supported, the force exerted is uniform with respect to horizontal distance, and the result is a parabola, as discussed below (although the term "catenary" is often still used, in an informal sense). If the cable is heavy then the resulting curve is between a catenary and a parabola.<ref name="Lockwood122">Lockwood p. 122</ref><ref> {{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

File:Comparison catenary parabola.svg
Comparison of a catenary arch (black dotted curve) and a parabolic arch (red solid curve) with the same span and sag. The catenary represents the profile of a simple suspension bridge, or the cable of a suspended-deck suspension bridge on which its deck and hangers have negligible weight compared to its cable. The parabola represents the profile of the cable of a suspended-deck suspension bridge on which its cable and hangers have negligible weight compared to its deck. The profile of the cable of a real suspension bridge with the same span and sag lies between the two curves. The catenary and parabola equations are respectively, <math>y = \text{cosh } x </math> and <math>y = x ^ 2 [(\text{cosh }1) - 1] + 1</math>

Anchoring of marine objectsEdit

File:Catenary.PNG
A heavy anchor chain forms a catenary, with a low angle of pull on the anchor.

The catenary produced by gravity provides an advantage to heavy anchor rodes. An anchor rode (or anchor line) usually consists of chain or cable or both. Anchor rodes are used by ships, oil rigs, docks, floating wind turbines, and other marine equipment which must be anchored to the seabed.

When the rope is slack, the catenary curve presents a lower angle of pull on the anchor or mooring device than would be the case if it were nearly straight. This enhances the performance of the anchor and raises the level of force it will resist before dragging. To maintain the catenary shape in the presence of wind, a heavy chain is needed, so that only larger ships in deeper water can rely on this effect. Smaller boats also rely on catenary to maintain maximum holding power.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

Cable ferries and chain boats present a special case of marine vehicles moving although moored by the two catenaries each of one or more cables (wire ropes or chains) passing through the vehicle and moved along by motorized sheaves. The catenaries can be evaluated graphically.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

Mathematical descriptionEdit

EquationEdit

File:Catenary-pm.svg
Catenaries for different values of Template:Mvar

The equation of a catenary in Cartesian coordinates has the form<ref name="Lockwood122"/>

<math display=block>y = a \cosh \left(\frac{x}{a}\right) = \frac{a}{2}\left(e^\frac{x}{a} + e^{-\frac{x}{a}}\right),</math> where Template:Math is the hyperbolic cosine function, and where Template:Mvar is the distance of the lowest point above the x axis.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> All catenary curves are similar to each other, since changing the parameter Template:Mvar is equivalent to a uniform scaling of the curve.

The Whewell equation for the catenary is<ref name="Lockwood122"/> <math display=block>\tan \varphi = \frac{s}{a},</math> where <math>\varphi</math> is the tangential angle and Template:Mvar the arc length.

Differentiating gives <math display=block>\frac{d\varphi}{ds} = \frac{\cos^2\varphi}{a},</math> and eliminating <math>\varphi</math> gives the Cesàro equation<ref>MathWorld, eq. 7</ref> <math display=block>\kappa=\frac{a}{s^2+a^2},</math> where <math>\kappa</math> is the curvature.

The radius of curvature is then <math display=block>\rho = a \sec^2 \varphi,</math> which is the length of the normal between the curve and the Template:Mvar-axis.<ref>Routh Art. 444</ref>

Relation to other curvesEdit

When a parabola is rolled along a straight line, the roulette curve traced by its focus is a catenary.<ref name="Yates 13"/> The envelope of the directrix of the parabola is also a catenary.<ref>Yates p. 80</ref> The involute from the vertex, that is the roulette traced by a point starting at the vertex when a line is rolled on a catenary, is the tractrix.<ref name="Yates 13"/>

Another roulette, formed by rolling a line on a catenary, is another line. This implies that square wheels 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.<ref>Template:Cite journal </ref>

Geometrical propertiesEdit

Over any horizontal interval, the ratio of the area under the catenary to its length equals Template:Mvar, independent of the interval selected. The catenary is the only plane curve other than a horizontal line with this property. Also, the geometric centroid of the area under a stretch of catenary is the midpoint of the perpendicular segment connecting the centroid of the curve itself and the Template:Mvar-axis.<ref>Template:Cite journal</ref>

ScienceEdit

A moving charge in a uniform electric field travels along a catenary (which tends to a parabola if the charge velocity is much less than the speed of light Template:Mvar).<ref>Template:Cite book</ref>

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

<math display="block">y = a \cosh^{-1}\left(\frac{x}{a}\right) + b</math>

revolved about the <math>y</math>-axis.<ref name="Yates 13">Template:Cite book</ref>

AnalysisEdit

Model of chains and archesEdit

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 exerted by the chain is parallel to the chain.<ref>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 and everything is inverted.<ref>Template:Cite book</ref> An underlying principle is that the chain may be considered a rigid body once it has attained equilibrium.<ref>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 parametrically by Template:Math where Template:Mvar represents arc length and Template:Math is the position vector. This is the natural parameterization and has the property that

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

where Template:Math is a unit tangent vector.

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

The chain is in equilibrium so the sum of three forces is Template:Math, 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 Template:Math.<ref>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, Template:Math is constant and the vertical component of the tension, Template:Math is proportional to the length of chain between Template:Math and the vertex.<ref>Routh Art 443, p. 318</ref>

Derivation of equations for the curveEdit

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.
 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).

</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 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 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 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 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 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 derivationEdit

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 Template:Mvar and Template:Mvar-axes can be shifted so Template:Mvar and Template:Mvar 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 parametersEdit

File:Catenary-tension.svg
Three catenaries through the same two points, depending on the horizontal force Template:Mvar.

In general the parameter Template:Mvar 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 Template:Math is to the left of Template:Math and let Template:Mvar be the horizontal and Template:Mvar be the vertical distance from Template:Math to Template:Math. Translate the axes so that the vertex of the catenary lies on the Template:Mvar-axis and its height Template:Mvar 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 Template:Math and Template:Math be Template:Math and Template:Math 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 Template:Math to Template:Math is

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

When Template:Math 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 Template:Mvar and must be solved numerically. Since <math>\sinh(x)/x</math> is strictly monotonic on <math>x > 0</math>,<ref>See Routh art. 447</ref> there is at most one solution with Template:Math and so there is at most one position of equilibrium.

However, if both ends of the curve (Template:Math and Template:Math) are at the same level (Template:Math), it can be shown that<ref>Archived at GhostarchiveTemplate:Cbignore and the Wayback MachineTemplate:Cbignore: {{#invoke:citation/CS1|citation |CitationClass=web }}Template:Cbignore</ref> <math display=block>a = \frac {\frac14 L^2-h^2} {2h}\,</math> where L is the total length of the curve between Template:Math and Template:Math and Template:Mvar is the sag (vertical distance between Template:Math, Template:Math 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 Template:Math and Template:Math which are located at the same level (Template:Math).

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

Tension relationsEdit

There is a simple relationship between the tension in the cable at a point and its Template:Mvar- and/or Template:Mvar- 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, 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 Art 443, p. 318</ref> <math display=block>T = wy = wa \cosh\frac{x}{a}\,.</math>

Variational formulationEdit

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 Template:Mvar 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 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.

Generalizations with vertical forceEdit

Nonuniform chainsEdit

If the density of the chain is variable then the analysis above can be adapted to produce equations for the curve given the density, or given the curve to find the density.<ref>Following Routh Art. 450</ref>

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

<math display=block>\int_\mathbf{c}^\mathbf{r} w\, ds\,,</math>

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

<math display=block>T \cos \varphi = T_0</math> and <math display=block>T \sin \varphi = \int_\mathbf{c}^\mathbf{r} w\, ds\,,</math> and therefore <math display=block>\frac{dy}{dx}=\tan \varphi = \frac{1}{T_0} \int_\mathbf{c}^\mathbf{r} w\, ds\,.</math>

Differentiation then gives

<math display=block>w=T_0 \frac{d}{ds}\frac{dy}{dx} = \frac{T_0 \dfrac{d^2y}{dx^2}}{\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}}\,.</math>

In terms of Template:Mvar and the radius of curvature Template:Mvar this becomes

<math display=block>w= \frac{T_0}{\rho \cos^2 \varphi}\,.</math>

Suspension bridge curveEdit

File:Golden Gate Bridge, SF.jpg
Golden Gate Bridge. Most suspension bridge cables follow a parabolic, not a catenary curve, because the roadway is much heavier than the cable.

A similar analysis can be done to find the curve followed by the cable supporting a suspension bridge with a horizontal roadway.<ref>Following Routh Art. 452</ref> If the weight of the roadway per unit length is Template:Mvar and the weight of the cable and the wire supporting the bridge is negligible in comparison, then the weight on the cable (see the figure in Catenary#Model of chains and arches) from Template:Math to Template:Math is Template:Mvar where Template:Mvar is the horizontal distance between Template:Math and Template:Math. Proceeding as before gives the differential equation

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

This is solved by simple integration to get

<math display=block>y=\frac{w}{2T_0}x^2 + \beta</math>

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.<ref>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.</ref>

Catenary of equal strengthEdit

In a catenary of equal strength, the 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 per unit length, the weight, Template:Mvar, per unit length of the chain can be written Template:Mvar, where Template:Mvar is constant, and the analysis for nonuniform chains can be applied.<ref>Following Routh Art. 453</ref>

In this case the equations for tension are

<math display=block>\begin{align} T \cos \varphi &= T_0\,,\\ T \sin \varphi &= \frac{1}{c}\int T\, ds\,. \end{align}</math>

Combining gives

<math display=block>c \tan \varphi = \int \sec \varphi\, ds</math>

and by differentiation

<math display=block>c = \rho \cos \varphi</math>

where Template:Mvar is the radius of curvature.

The solution to this is

<math display=block>y = c \ln\left(\sec\left(\frac{x}{c}\right)\right)\,.</math>

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

<math display=block>x = c\varphi\,,\quad s = \ln\left(\tan\left(\frac{\pi+2\varphi}{4}\right)\right)\,.</math>

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".<ref>Template:Cite journal </ref>

Elastic catenaryEdit

In an elastic catenary, the chain is replaced by a spring which can stretch in response to tension. The spring is assumed to stretch in accordance with Hooke's law. Specifically, if Template:Math is the natural length of a section of spring, then the length of the spring with tension Template:Mvar applied has length

<math display=block>s=\left(1+\frac{T}{E}\right)p\,,</math>

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

The equations for tension of the spring are

<math display=block>T \cos \varphi = T_0\,,</math> and <math display=block>T \sin \varphi = w_0 p\,,</math>

from which

<math display=block>\frac{dy}{dx}=\tan \varphi = \frac{w_0 p}{T_0}\,,\quad T=\sqrt{T_0^2+w_0^2 p^2}\,,</math>

where Template:Mvar is the natural length of the segment from Template:Math to Template:Math and Template:Math is the weight per unit length of the spring with no tension. Write <math display=block>a = \frac{T_0}{w_0}</math> so <math display=block>\frac{dy}{dx}=\tan \varphi = \frac{p}{a} \quad\text{and}\quad T=\frac{T_0}{a}\sqrt{a^2+p^2}\,.</math>

Then <math display=block>\begin{align} \frac{dx}{ds} &= \cos \varphi = \frac{T_0}{T} \\[6pt] \frac{dy}{ds} &= \sin \varphi = \frac{w_0 p}{T}\,, \end{align}</math> from which <math display=block>\begin{alignat}{3} \frac{dx}{dp} &= \frac{T_0}{T}\frac{ds}{dp} &&= T_0\left(\frac{1}{T}+\frac{1}{E}\right) &&= \frac{a}{\sqrt{a^2+p^2}}+\frac{T_0}{E} \\[6pt] \frac{dy}{dp} &= \frac{w_0 p}{T}\frac{ds}{dp} &&= \frac{T_0p}{a}\left(\frac{1}{T}+\frac{1}{E}\right) &&= \frac{p}{\sqrt{a^2+p^2}}+\frac{T_0p}{Ea}\,. \end{alignat}</math>

Integrating gives the parametric equations

<math display=block>\begin{align} x&=a\operatorname{arsinh}\left(\frac{p}{a}\right)+\frac{T_0}{E}p + \alpha\,, \\[6pt] y&=\sqrt{a^2+p^2}+\frac{T_0}{2Ea}p^2+\beta\,. \end{align}</math>

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

<math display=block>\begin{align} x&=a\operatorname{arsinh}\left(\frac{p}{a}\right)+\frac{T_0}{E}p\,, \\[6pt] y&=\sqrt{a^2+p^2}+\frac{T_0}{2Ea}p^2 \end{align}</math>

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

Other generalizationsEdit

Chain under a general forceEdit

With no assumptions being made regarding the force Template:Math acting on the chain, the following analysis can be made.<ref>Follows Routh Art. 455</ref>

First, let Template:Math be the force of tension as a function of Template:Mvar. 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, Template:Math must be parallel to the chain. In other words,

<math display=block>\mathbf{T} = T \mathbf{u}\,,</math>

where Template:Mvar is the magnitude of Template:Math and Template:Math is the unit tangent vector.

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

<math display=block>\mathbf{T}(s+\Delta s)-\mathbf{T}(s)+\mathbf{G}\Delta s \approx \mathbf{0}\,.</math>

Divide by Template:Math and take the limit as Template:Math to obtain

<math display=block>\frac{d\mathbf{T}}{ds} + \mathbf{G} = \mathbf{0}\,.</math>

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, Template:Math where the chain has weight Template:Mvar per unit length.

See alsoEdit

NotesEdit

Template:Reflist

BibliographyEdit

Further readingEdit

External linksEdit

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