Catenary

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 bridges

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.

Anchoring of marine objects

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

The equation of a catenary in Cartesian coordinates has the form

$${\displaystyle y=a\cosh \left(\frac{x}{a}\right)=\frac{a}{2}\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right),}$$ 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 $${\displaystyle \tan \phi =\frac{s}{a},}$$ where ${\displaystyle \phi }$ is the tangential angle and s the arc length.

Differentiating gives $${\displaystyle \frac{d\phi }{ds}=\frac{{\cos }^2\phi }{a},}$$ and eliminating ${\displaystyle \phi }$ gives the Cesàro equation $${\displaystyle \kappa =\frac{a}{s^2+a^2},}$$ where ${\displaystyle \kappa }$ is the curvature.

The radius of curvature is then $${\displaystyle \rho =a{\sec }^2\phi ,}$$ 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.

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

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

revolved about the ${\displaystyle y}$-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

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

where u is a unit tangent vector.

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

$${\displaystyle T\cos \phi =T_0}$$ and $${\displaystyle T\sin \phi =ws,}$$

and dividing these gives

$${\displaystyle \frac{dy}{dx}=\tan \phi =\frac{ws}{T_0}.}$$

It is convenient to write

$${\displaystyle a=\frac{T_0}{w}}$$

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

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

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 ${\displaystyle dy/dx=s/a}$, 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 ${\displaystyle s_0=0}$ and ${\displaystyle (x,y)=(x_0,y_0)}$.

First, invoke the formula for arc length to get $${\displaystyle \frac{ds}{dx}=\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=\sqrt{1+{\left(\frac{s}{a}\right)}^2},}$$ then separate variables to obtain $${\displaystyle \frac{ds}{\sqrt{1+(s/a{)}^2}}=dx.}$$

A reasonably straightforward approach to integrate this is to use hyperbolic substitution, which gives $${\displaystyle a{\sinh }^{-1}\frac{s}{a}+x_0=x}$$ (where ${\displaystyle x_0}$ is a constant of integration), and hence $${\displaystyle \frac{s}{a}=\sinh \frac{x-x_0}{a}.}$$

But $s/a=dy/dx$, so $${\displaystyle \frac{dy}{dx}=\sinh \frac{x-x_0}{a},}$$ which integrates as $${\displaystyle y=a\cosh \frac{x-x_0}{a}+\delta }$$ (with ${\displaystyle \delta =y_0-a}$ 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 $x_0=0=\delta$ to simplify the result to $${\displaystyle y=a\cosh \frac{x}{a}.}$$

For completeness, the ${\displaystyle y\leftrightarrow s}$ relation can be derived by solving each of the ${\displaystyle x\leftrightarrow y}$ and ${\displaystyle x\leftrightarrow s}$ relations for ${\displaystyle x/a}$, giving: $${\displaystyle {\cosh }^{-1}\frac{y-\delta }{a}=\frac{x-x_0}{a}={\sinh }^{-1}\frac{s}{a},}$$ so $${\displaystyle y-\delta =a\cosh \left({\sinh }^{-1}\frac{s}{a}\right),}$$ which can be rewritten as $${\displaystyle y-\delta =a\sqrt{1+{\left(\frac{s}{a}\right)}^2}=\sqrt{a^2+s^2}.}$$

Alternative derivation

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

$${\displaystyle s=a\tan \phi }$$

it follows that

$${\displaystyle \frac{dx}{d\phi }=\frac{dx}{ds}\frac{ds}{d\phi }=\cos \phi \cdot a{\sec }^2\phi =a\sec \phi }$$ and $${\displaystyle \frac{dy}{d\phi }=\frac{dy}{ds}\frac{ds}{d\phi }=\sin \phi \cdot a{\sec }^2\phi =a\tan \phi \sec \phi .}$$

Integrating gives,

$${\displaystyle x=a\ln (\sec \phi +\tan \phi )+\alpha }$$ and $${\displaystyle y=a\sec \phi +\beta .}$$

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

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

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

Determining parameters

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 ⁠${\displaystyle P_1}$⁠ and ⁠${\displaystyle P_2}$⁠ 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

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

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

$${\displaystyle v=a\cosh \left(\frac{x_2}{a}\right)-a\cosh \left(\frac{x_1}{a}\right).}$$

and the length of the curve from P₁ to P₂ is

$${\displaystyle L=a\sinh \left(\frac{x_2}{a}\right)-a\sinh \left(\frac{x_1}{a}\right).}$$

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

$${\displaystyle L^2-v^2=2a^2\left(\cosh \left(\frac{x_2-x_1}{a}\right)-1\right)=4a^2{\sinh }^2\left(\frac{H}{2a}\right),}$$ so $${\displaystyle \frac{1}{H}\sqrt{L^2-v^2}=\frac{2a}{H}\sinh \left(\frac{H}{2a}\right).}$$

This is a transcendental equation in a and must be solved numerically. Since ${\displaystyle \sinh (x)/x}$ is strictly monotonic on ${\displaystyle x>0}$, 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 $${\displaystyle a=\frac{\frac{1}{4}{L}^2-h^2}{2h}}$$ 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 $${\displaystyle L=2a\sinh \frac{H}{2a}}$$ and $${\displaystyle H=2a\mathrm{arcosh}\frac{h+a}{a}}$$ 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: $${\displaystyle (T\cos \phi {)}^2+(T\sin \phi {)}^2=T_0^2+(ws{)}^2}$$ which (recalling that ${\displaystyle T_0=wa}$) can be rewritten as $${\displaystyle \begin{array}{cc}{T}^2({\cos }^2\phi +{\sin }^2\phi )& =(wa{)}^2+(ws{)}^2\\ T^2& =w^2(a^2+s^2)\\ T& =w\sqrt{a^2+s^2}.\end{array}}$$ But, as shown above, ${\displaystyle y=\sqrt{a^2+s^2}}$ (assuming that ${\displaystyle y_0=a}$), so we get the simple relations $${\displaystyle T=wy=wa\cosh \frac{x}{a}.}$$

Variational formulation

Consider a chain of length ${\displaystyle L}$ suspended from two points of equal height and at distance ${\displaystyle D}$. The curve has to minimize its potential energy $${\displaystyle U={\int }_0^{D}wy\sqrt{1+y^{\prime 2}}dx}$$ (where w is the weight per unit length) and is subject to the constraint $${\displaystyle {\int }_0^D\sqrt{1+y^{\prime 2}}dx=L.}$$

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

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

$${\displaystyle {\int }_{\mathbf{c}}^{\mathbf{r}}wds,}$$

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

$${\displaystyle T\cos \phi =T_0}$$ and $${\displaystyle T\sin \phi ={\int }_{\mathbf{c}}^{\mathbf{r}}wds,}$$ and therefore $${\displaystyle \frac{dy}{dx}=\tan \phi =\frac{1}{T_0}{\int }_{\mathbf{c}}^{\mathbf{r}}wds.}$$

Differentiation then gives

$${\displaystyle w=T_0\frac{d}{ds}\frac{dy}{dx}=\frac{T_0{\displaystyle \frac{d^{2}y}{d{x}^2}}}{\sqrt{1+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}}.}$$

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

$${\displaystyle w=\frac{T_0}{\rho {\cos }^2\phi }.}$$

Suspension bridge curve

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

$${\displaystyle \frac{dy}{dx}=\tan \phi =\frac{w}{T_0}x.}$$

This is solved by simple integration to get

$${\displaystyle y=\frac{w}{2T_0}{x}^2+\beta }$$

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

$${\displaystyle \begin{array}{cc}T\cos \phi & =T_0,\\ T\sin \phi & =\frac{1}{c}\int Tds.\end{array}}$$

Combining gives

$${\displaystyle c\tan \phi =\int \sec \phi ds}$$

and by differentiation

$${\displaystyle c=\rho \cos \phi }$$

where ρ is the radius of curvature.

The solution to this is

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

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

$${\displaystyle x=c\phi ,s=\ln \left(\tan \left(\frac{\pi +2\phi }{4}\right)\right).}$$

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

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

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 $${\displaystyle \frac{ds}{dp}=1+\frac{T}{E}.}$$ 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

$${\displaystyle T\cos \phi =T_0,}$$ and $${\displaystyle T\sin \phi =w_{0}p,}$$

from which

$${\displaystyle \frac{dy}{dx}=\tan \phi =\frac{w_{0}p}{T_0},T=\sqrt{T_0^2+w_0^2{p}^2},}$$

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 $${\displaystyle a=\frac{T_0}{w_0}}$$ so $${\displaystyle \frac{dy}{dx}=\tan \phi =\frac{p}{a}\text{and}T=\frac{T_0}{a}\sqrt{a^2+p^2}.}$$

Then $${\displaystyle \begin{array}{cc}\frac{dx}{ds}& =\cos \phi =\frac{T_0}{T}\\ \frac{dy}{ds}& =\sin \phi =\frac{w_{0}p}{T},\end{array}}$$ from which $${\displaystyle \begin{array}{rlrlrl}\frac{dx}{dp}& =\frac{T_0}{T}\frac{ds}{dp}& & =T_0\left(\frac{1}{T}+\frac{1}{E}\right)& & =\frac{a}{\sqrt{a^2+p^2}}+\frac{T_0}{E}\\ \frac{dy}{dp}& =\frac{w_{0}p}{T}\frac{ds}{dp}& & =\frac{T_{0}p}{a}\left(\frac{1}{T}+\frac{1}{E}\right)& & =\frac{p}{\sqrt{a^2+p^2}}+\frac{T_{0}p}{Ea}.\end{array}}$$

Integrating gives the parametric equations

$${\displaystyle \begin{array}{cc}x& =a\mathrm{arsinh}\left(\frac{p}{a}\right)+\frac{T_0}{E}p+\alpha ,\\ y& =\sqrt{a^2+p^2}+\frac{T_0}{2Ea}{p}^2+\beta .\end{array}}$$

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

$${\displaystyle \begin{array}{cc}x& =a\mathrm{arsinh}\left(\frac{p}{a}\right)+\frac{T_0}{E}p,\\ y& =\sqrt{a^2+p^2}+\frac{T_0}{2Ea}{p}^2\end{array}}$$

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,

$${\displaystyle \mathbf{T}=T\mathbf{u},}$$

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

$${\displaystyle \mathbf{T}(s+\mathrm{\Delta }s)-\mathbf{T}(s)+\mathbf{G}\mathrm{\Delta }s\approx \mathbf{0}.}$$

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

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

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.