Catenary

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.

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 (UK: /kəˈtiːnəri/ kə-TEE-nər-ee, US: /ˈkætənɛri/ KAT-ən-err-ee) 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.

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, or, particularly in the materials sciences, an example of a funicular. Rope statics describes catenaries in a classic statics problem involving a hanging rope.

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. 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. The symmetric modes consisting of two evanescent waves would form a catenary shape.

History

File:GaudiCatenaryModel.jpg

Antoni Gaudí's catenary model at Casa Milà

The word "catenary" is derived from the Latin word catēna, which means "chain". The English word "catenary" is usually attributed to Thomas Jefferson, who wrote in a letter to Thomas Paine on the construction of an arch for a bridge:

I have lately received from Italy a treatise on the equilibrium of arches, by the Abbé Mascheroni. It appears to be a very scientific 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.

It is often said 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°. 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.

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. Some much older arches approximate catenaries, an example of which is the Arch of Taq-i Kisra in Ctesiphon.

File:Analogy between an arch and a hanging chain and comparison to the dome of St Peter's Cathedral in Rome.png

Analogy between an arch and a hanging chain and comparison to the dome of Saint Peter's Basilica in Rome (Giovanni Poleni, 1748)

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 in an appendix to his Description of Helioscopes, 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 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; their solutions were published in the Acta Eruditorum for June 1691. David Gregory wrote a treatise on the catenary in 1697 in which he provided an incorrect derivation of the correct differential equation.

Leonhard 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 circles. Nicolas Fuss gave equations describing the equilibrium of a chain under any force in 1796.

Inverted catenary arch

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.

The Gateway Arch in St. Louis, Missouri, United States, is sometimes said to be an (inverted) catenary, but this is incorrect. It is close to a more general curve called a flattened catenary, with equation y = A cosh(Bx), which is a catenary if AB = 1. 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.

• Catenary[28] arches under the roof of Gaudí's Casa Milà, Barcelona, Spain.
  Catenary arches under the roof of Gaudí's Casa Milà, Barcelona, Spain.
• The Sheffield Winter Garden is enclosed by a series of catenary arches.[29]
  The Sheffield Winter Garden is enclosed by a series of catenary arches.
• The Gateway Arch (St. Louis, Missouri) is a flattened catenary.
  The Gateway Arch (St. Louis, Missouri) is a flattened catenary.
• Catenary arch kiln under construction over temporary form
  Catenary arch kiln under construction over temporary form

Catenary bridges

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. The same is true of a simple suspension bridge or "catenary bridge," where the roadway follows the cable.

A stressed ribbon bridge is a more sophisticated structure with the same catenary shape.

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.

File:Catenary-Comparison.svg

Comparison of a catenary arch (cyan) and a parabolic arch (pink) with the same span and sag. An ideal suspension bridge cable is represented by the catenary curve. The catenary and parabola equations, plotted between -1 and 1, are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = \text{cosh }(x)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = x^2 (\text{cosh }(1) - 1) + 1} , respectively.

Anchoring of marine objects

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.

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.

Mathematical description

Equation

File:Catenary-pm.svg

Catenaries for different values of a

The equation of a catenary in Cartesian coordinates has the form

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = a \cosh \left(\frac{x}{a}\right) = \frac{a}{2}\left(e^\frac{x}{a} + e^{-\frac{x}{a}}\right),} where cosh is the hyperbolic cosine function, and where a is the distance of the lowest point above the x axis. All catenary curves are similar to each other, since changing the parameter a is equivalent to a uniform scaling of the curve.

The Whewell equation for the catenary is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan \varphi = \frac{s}{a},} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} is the tangential angle and s the arc length.

Differentiating gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\varphi}{ds} = \frac{\cos^2\varphi}{a},} and eliminating Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} gives the Cesàro equation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{a}{s^2+a^2},} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa} is the curvature.

The radius of curvature is then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho = a \sec^2 \varphi,} which is the length of the normal between the curve and the x-axis.

Relation to other curves

When a parabola is rolled along a straight line, the roulette curve traced by its focus is a catenary. The envelope of the directrix of the parabola is also a catenary. 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.

File:Square wheels on Catenary road.gif

Square wheels riding smoothly over inverted Catenaries

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.

Geometrical properties

Over any horizontal interval, the ratio of the area under the catenary to its length equals a, 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 x-axis.

Science

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 c).

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = a \cosh^{-1}\left(\frac{x}{a}\right) + b}

revolved about the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} -axis.

Analysis

Model of chains and arches

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. 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. An underlying principle is that the chain may be considered a rigid body once it has attained equilibrium. 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 r = (x, y) = (x(s), y(s)) where s represents arc length and r is the position vector. This is the natural parameterization and has the property that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\mathbf{r}}{ds}=\mathbf{u}}

where u is a unit tangent vector.

File:CatenaryForceDiagram.svg

Diagram of forces acting on a segment of a catenary from c to r. The forces are the tension T₀ at c, the tension T at r, and the weight of the chain (0, −ws). 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. Let c be the lowest point on the chain, called the vertex of the catenary. The slope dy/dx of the curve is zero at c since it is a minimum point. Assume r is to the right of c, whereas the opposite side will be implied by symmetry. The forces acting on the section of the chain from c to r are the tension of the chain at c, the tension of the chain at r, and the weight of the chain. The tension at c is tangent to the curve at c and is therefore horizontal without any vertical component and it pulls the section to the left so it may be written (−T₀, 0) where T₀ is the magnitude of the force. The tension at r is parallel to the curve at r and pulls the section to the right. The tension at r can be split into two components so it may be written Tu = (T cos φ, T sin φ), where T is the magnitude of the force and φ is the angle between the curve at r and the x-axis (see tangential angle). Finally, the weight of the chain is represented by (0, −ws) where w is the weight per unit length and s is the length of the segment of chain between c and r.

The chain is in equilibrium so the sum of three forces is 0, therefore

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T \cos \varphi = T_0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T \sin \varphi = ws\,,}

and dividing these gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}=\tan \varphi = \frac{ws}{T_0}\,.}

It is convenient to write

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a = \frac{T_0}{w}}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}=\frac{s}{a}}

is an equation defining the curve.

The horizontal component of the tension, T cos φ = T₀ is constant and the vertical component of the tension, T sin φ = ws is proportional to the length of chain between r and the vertex.

Derivation of equations for the curve

The differential equation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dy/dx = s/a} , given above, can be solved to produce equations for the curve. We will solve the equation using the boundary condition that the vertex is positioned at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_0=0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,y)=(x_0,y_0)} .

First, invoke the formula for arc length to get Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{ds}{dx} = \sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\left(\frac{s}{a}\right)^2}\,,} then separate variables to obtain Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{ds}{\sqrt{1+(s/a)^2}} = dx\,.}

A reasonably straightforward approach to integrate this is to use hyperbolic substitution, which gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \sinh^{-1}\frac{s}{a} + x_0 = x} (where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0} is a constant of integration), and hence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{s}{a} = \sinh\frac{x-x_0}{a}\,.}

But Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle s/a = dy/dx} , so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx} = \sinh\frac{x-x_0}{a}\,,} which integrates as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = a \cosh\frac{x-x_0}{a} + \delta} (with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x_0=0=\delta} to simplify the result to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = a \cosh\frac{x}{a}.}

For completeness, the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y \leftrightarrow s} relation can be derived by solving each of the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \leftrightarrow y} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \leftrightarrow s} relations for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x/a} , giving: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cosh^{-1}\frac{y-\delta}{a} = \frac{x-x_0}{a} = \sinh^{-1}\frac{s}{a}\,,} so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y-\delta = a\cosh\left(\sinh^{-1}\frac{s}{a}\right)\,,} which can be rewritten as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y-\delta = a\sqrt{1+\left(\frac{s}{a}\right)^2} = \sqrt{a^2 + s^2}\,.}

Alternative derivation

The differential equation can be solved using a different approach. From

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s = a \tan \varphi}

it follows that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dx}{d\varphi} = \frac{dx}{ds}\frac{ds}{d\varphi}=\cos \varphi \cdot a \sec^2 \varphi= a \sec \varphi} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{d\varphi} = \frac{dy}{ds}\frac{ds}{d\varphi}=\sin \varphi \cdot a \sec^2 \varphi= a \tan \varphi \sec \varphi\,.}

Integrating gives,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = a \ln(\sec \varphi + \tan \varphi) + \alpha} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = a \sec \varphi + \beta\,.}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sec \varphi + \tan \varphi = e^\frac{x}{a}\,,} and taking the reciprocal of both sides Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sec \varphi - \tan \varphi = e^{-\frac{x}{a}}\,.}

Adding and subtracting the last two equations then gives the solution Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = a \sec \varphi = a \cosh\left(\frac{x}{a}\right)\,,} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s = a \tan \varphi = a \sinh\left(\frac{x}{a}\right)\,.}

Determining parameters

File:Catenary-tension.svg

Three catenaries through the same two points, for different lengths of the catenary.

The parameter a is the minimal y coordinate of the points of the catenary. In practical applications, neither this parameter, nor the axes where the catenary has its standard form are known a priori.

These data can be determined from the position of two given points Template:Tmath and Template:Tmath and the length L of the catenary between these points as follows:

Relabel if necessary so that P₁ is to the left of P₂ and let H be the horizontal and v be the vertical distance from P₁ to P₂. Translate the axes so that the vertex of the catenary lies on the y-axis and its height a is adjusted so the catenary satisfies the standard equation of the curve

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = a \cosh\left(\frac{x}{a}\right)}

and let the coordinates of P₁ and P₂ be (x₁, y₁) and (x₂, y₂) respectively. The curve passes through these points, so the difference of height is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v = a \cosh\left(\frac{x_2}{a}\right) - a \cosh\left(\frac{x_1}{a}\right)\,.}

and the length of the curve from P₁ to P₂ is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L = a \sinh\left(\frac{x_2}{a}\right) - a \sinh\left(\frac{x_1}{a}\right)\,.}

When L² − v² is expanded using these expressions the result is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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)\,,} so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac 1H \sqrt{L^2-v^2}=\frac{2a}H \sinh\left(\frac{H}{2a}\right)\,.}

This is a transcendental equation in a and must be solved numerically. Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sinh(x)/x} is strictly monotonic on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x > 0} , there is at most one solution with a > 0 and so there is at most one position of equilibrium.

However, if both ends of the curve (P₁ and P₂) are at the same level (y₁ = y₂), it can be shown that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a = \frac {\frac14 L^2-h^2} {2h}\,} where L is the total length of the curve between P₁ and P₂ and h is the sag (vertical distance between P₁, P₂ and the vertex of the curve).

It can also be shown that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L = 2a \sinh \frac {H} {2a}\,} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = 2a \operatorname {arcosh} \frac {h+a} {a}\,} where H is the horizontal distance between P₁ and P₂ which are located at the same level (H = x₂ − x₁).

The horizontal traction force at P₁ and P₂ is T₀ = wa, where w is the weight per unit length of the chain or cable.

Tension relations

There is a simple relationship between the tension in the cable at a point and its x- and/or y- coordinate. Begin by combining the squares of the vector components of the tension: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (T\cos\varphi)^2 + (T\sin\varphi)^2 = T_0^2 + (ws)^2} which (recalling that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_0=wa} ) can be rewritten as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}} But, as shown above, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = \sqrt{a^2 + s^2}} (assuming that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_0=a} ), so we get the simple relations Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = wy = wa \cosh\frac{x}{a}\,.}

Variational formulation

Consider a chain of length Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} suspended from two points of equal height and at distance Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D} . The curve has to minimize its potential energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U = \int_0^D w y\sqrt{1+y'^2} dx } (where w is the weight per unit length) and is subject to the constraint Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^D \sqrt{1+y'^2} dx = L\,.}

The modified Lagrangian is therefore Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L} = (w y - \lambda )\sqrt{1+y'^2}} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda } is the Lagrange multiplier to be determined. As the independent variable Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} does not appear in the Lagrangian, we can use the Beltrami identity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}-y' \frac{\partial \mathcal{L} }{\partial y'} = C } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} is an integration constant, in order to obtain a first integral Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{(w y - \lambda )}{\sqrt{1+y'^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.

Generalizations with vertical force

Nonuniform chains

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.

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_\mathbf{c}^\mathbf{r} w\, ds\,,}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T \cos \varphi = T_0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T \sin \varphi = \int_\mathbf{c}^\mathbf{r} w\, ds\,,} and therefore Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}=\tan \varphi = \frac{1}{T_0} \int_\mathbf{c}^\mathbf{r} w\, ds\,.}

Differentiation then gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}}\,.}

In terms of φ and the radius of curvature ρ this becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w= \frac{T_0}{\rho \cos^2 \varphi}\,.}

Suspension bridge curve

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. If the weight of the roadway per unit length is w 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 c to r is wx where x is the horizontal distance between c and r. Proceeding as before gives the differential equation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}=\tan \varphi = \frac{w}{T_0}x\,. }

This is solved by simple integration to get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=\frac{w}{2T_0}x^2 + \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.

Catenary of equal strength

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, w, per unit length of the chain can be written T/c, where c is constant, and the analysis for nonuniform chains can be applied.

In this case the equations for tension are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} T \cos \varphi &= T_0\,,\\ T \sin \varphi &= \frac{1}{c}\int T\, ds\,. \end{align}}

Combining gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c \tan \varphi = \int \sec \varphi\, ds}

and by differentiation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c = \rho \cos \varphi}

where ρ is the radius of curvature.

The solution to this is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = c \ln\left(\sec\left(\frac{x}{c}\right)\right)\,.}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = c\varphi\,,\quad s = \ln\left(\tan\left(\frac{\pi+2\varphi}{4}\right)\right)\,.}

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".

Elastic catenary

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 p is the natural length of a section of spring, then the length of the spring with tension T applied has length

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s=\left(1+\frac{T}{E}\right)p\,,}

where E is a constant equal to kp, where k is the stiffness of the spring. In the catenary the value of T is variable, but ratio remains valid at a local level, so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{ds}{dp}=1+\frac{T}{E}\,.} The curve followed by an elastic spring can now be derived following a similar method as for the inelastic spring.

The equations for tension of the spring are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T \cos \varphi = T_0\,,} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T \sin \varphi = w_0 p\,,}

from which

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}=\tan \varphi = \frac{w_0 p}{T_0}\,,\quad T=\sqrt{T_0^2+w_0^2 p^2}\,,}

where p is the natural length of the segment from c to r and w₀ is the weight per unit length of the spring with no tension. Write Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a = \frac{T_0}{w_0}} so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}=\tan \varphi = \frac{p}{a} \quad\text{and}\quad T=\frac{T_0}{a}\sqrt{a^2+p^2}\,.}

Then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{dx}{ds} &= \cos \varphi = \frac{T_0}{T} \\[6pt] \frac{dy}{ds} &= \sin \varphi = \frac{w_0 p}{T}\,, \end{align}} from which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Integrating gives the parametric equations

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Again, the x and y-axes can be shifted so α and β can be taken to be 0. So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

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

Other generalizations

Chain under a general force

With no assumptions being made regarding the force G acting on the chain, the following analysis can be made.

First, let T = 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, T must be parallel to the chain. In other words,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T} = T \mathbf{u}\,,}

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

Second, let G = 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 + Δs are the force of tension T(s + Δs) at one end of the segment, the nearly opposite force −T(s) at the other end, and the external force acting on the segment which is approximately GΔs. These forces must balance so

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}(s+\Delta s)-\mathbf{T}(s)+\mathbf{G}\Delta s \approx \mathbf{0}\,.}

Divide by Δs and take the limit as Δs → 0 to obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\mathbf{T}}{ds} + \mathbf{G} = \mathbf{0}\,.}

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, G = (0, −w) where the chain has weight w per unit length.

See also

• 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

Bibliography

• Lockwood, E.H. (1961). "Chapter 13: The Tractrix and Catenary". A Book of Curves. Cambridge.
• Salmon, George (1879). Higher Plane Curves. Hodges, Foster and Figgis. pp. 287–289.
• Routh, Edward John (1891). "Chapter X: On Strings". A Treatise on Analytical Statics. University Press.
• Maurer, Edward Rose (1914). "Art. 26 Catenary Cable". Technical Mechanics. J. Wiley & Sons.
• Lamb, Sir Horace (1897). "Art. 134 Transcendental Curves; Catenary, Tractrix". An Elementary Course of Infinitesimal Calculus. University Press.
• Todhunter, Isaac (1858). "XI Flexible Strings. Inextensible, XII Flexible Strings. Extensible". A Treatise on Analytical Statics. Macmillan.
• Whewell, William (1833). "Chapter V: The Equilibrium of a Flexible Body". Analytical Statics. J. & J.J. Deighton. p. 65.
• Template:Mathworld

Further reading

• Swetz, Frank (1995). Learn from the Masters. MAA. pp. 128–9. ISBN 978-0-88385-703-8.
• Venturoli, Giuseppe (1822). "Chapter XXIII: On the Catenary". Elements of the Theory of Mechanics. Trans. Daniel Cresswell. J. Nicholson & Son.

External links

• O'Connor, John J.; Robertson, Edmund F., "Catenary", MacTutor History of Mathematics archive, University of St Andrews
• Template:PlanetMath
• 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