The Catenary Curve – On the way to the Calculus of Variations

In the beginning of the last decade of the 17th century, Jacob Bernoulli proposed the problem of the catenary; in 1696, Johann Bernoulli proposed the problem of the brachistochrone; some years later, Daniel Bernoulli would suggest to Euler the right functional to solve the elastica problem. The Bernoulli’s family had a great influence in the development of the Calculus of Variations, which was formalized by Euler in his book of 1744. Here the catenary problem and its solution are described in detail.

Definition and first model.

In 1690, Jacob Bernoulli stated the following problem:

Assume you have a perfectly elastic wire (it can be deformed by the action of some force, but its shape is recovered immediately after the forces are ceased) made of a material with uniform density \lambda.  The two extremities of the wire are attached at the top of two columns of the same height. Moreover, assume that the only forces acting on the wire are tension and gravity. What is the equation described by the wire?

Consider any two points p and q in the wire. The diagram of forces below shows the tension and gravity forces acting on the segment pq.

catenary_1
A first model to the free-hanging chain problem.

The tension forces must cancel horizontally otherwise there will exist an acceleration on the chain. Additionally, the resultant vertical tension force must be equal to the weight of the segment pq. The difficulty lies on the computation of the vertical components of tension forces T_R and T_L. Fortunately, there is an alternative.

Catenary derivation.

In the next model, we are going to assume the chain has uniform density \lambda and that p is the lowest point of the curve. Moreover, assume that p is at height zero. We are going to derive the right part of the curve from p. The complete curve follows by symmetry.

 

catenary_2.png
A second and easy model.

Let q \neq p be any point at the right of p. At q there exists a tension force T_q tangent to the curve that can be decomposed in horizontal and vertical components T_{qx} and T_{qy}. Let \theta to denote the angle T_q makes with the horizontal axis.

T_{qx} = T_q\cdot \cos{\theta} \quad \quad T_{qy} = T_q \cdot \sin{\theta}.

Since the object is in equilibrium:

\begin{array}{ll} \displaystyle T_{qx} &=T_0  \\[0.3in] \displaystyle T_{qy} &= W(pq) = \lambda g s, \end{array}

where s denotes the length of the segment pq. Dividing the second equation by the first, we derive the differential equation that describes the right par of the curve.

\displaystyle \frac{ \lambda g s }{T_0} = \frac{\sin{\theta}}{\cos{\theta}} = \frac{dy}{dx}\\[0.3in] \frac{dy}{dx} = \frac{\lambda g s}{T_0} = \frac{s}{a}, \quad a = \frac{T_0}{\lambda g}. \quad \quad (1)

Recall that

\displaystyle s(u) = \int_{0}^{u}{\sqrt{1 + (dy/dx)^2} dx}. \quad \quad\frac{ds}{dx} (u) = \sqrt{1 + (dy/dx)^2(u)}. \quad \quad (2)

Therefore

\displaystyle \frac{dy/dx}{ds/dx} = \frac{s}{a\sqrt{1 + s^2/a^2}} = \frac{s}{\sqrt{a^2 + s^2}} = \frac{dy}{ds}.

Integrating both sides

y(s) = \sqrt{a^2 + s^2}. \quad \quad (3)

Using (1) and (2)

\begin{array}{ll} & \displaystyle \frac{dx}{ds} = \frac{1}{\sqrt{1 + (s/a)^2}}\\[0.3in] \Rightarrow & \displaystyle x(s) = a \cdot \sinh^{-1}{(\frac{s}{a})}\\[0.3in] \Rightarrow & \displaystyle a \cdot \sinh(x/a) = s. \end{array}

Substituting on (3)

\begin{array}{ll} y^2(x) &= a^2 + a^2 \cdot \sinh^2(x/a) \\[0.15in] & = a^2\cosh^2(x/a) \\[0.15in] \Rightarrow y(x) &= a\cosh(x/a). \end{array}

Leibniz, Huygens and Johann Bernoulli correctly showed that the solution of the problem is the curve above, known as catenary.  As a curiosity, Jacob Bernoulli, Johann’s brother and also the person who poses the problem, spent an year trying to prove that the solution was a parabola. In fact, the parabola is the solution for a particular case of the catenary problem.

Take any segment of the curve such that its horizontal projection equals to c. Further, assume that every segment with horizontal projection c has equal mass. In particular, for every segment with horizontal projection dx, let m be its mass.

\begin{array}{ll} & \displaystyle \frac{dy}{dx} = \frac{mgx}{T_0}\\[0.15in] \Rightarrow & \displaystyle y(x) = \frac{mg}{2T_0} \cdot x^2. \end{array}

In a suspension bridge, the curve between two same height collumns is a parabola. The road is much heavier than the wire itself, and it dominates the weight force. Therefore, one can assume the weight force is equal for every segment of the curve with the same horizontal projection.

 

clifton-suspension-bridge
Clifton Bridge: Source

 

Further References:

[1] – Raph LEVIEN. The elastica: A mathematical history. [Link]
[2] – Patricia Radelet-de GRAVE. The Problem of the Elastica Treated by Jacob Bernoulli and the Further Development of this Study by Leonhard Euler. [Link]
[3] – C. TRUESDELL. The influence of elastica on analysis: The classical heritage. [Link]

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