This article is under construction.
- Completed: demonstration of using differential calculus to find the equation of the catenary curve.
- Completed: demonstration that identifying the point of minimum potential energy is identical to identifying force equilibrium.
- Still to do: numerical analysis implementation of finding the shape of the catenary curve, using the same implementation structure as applied in diagrams 4, 5 and 6 presented in the article with discussion of Hamilton's stationary action.
Calculating the shape of the Catenary
Statics and dynamics
Statics is a branch of mechanics in the following sense: if motion would be prevented altogether then nothing can change. Sufficient opportunity to move is an absolute necessity in order to have statics at all.
Generally a case is referred to as a case in statics when the following applies:
- the setup is such that motion of the system results in dissipation of energy, with a finite amount of (potential) energy available.
- The nature of the setup is such that as it dissipates energy the system proceeds to a point where there is no longer any opportunity to dissipate energy.
Example: a marble rolling around in roughly circular motion in a bowl. The kinetic energy of the rolling motion is drained by friction. As the marble loses kinetic energy the force of gravity pulls the marble down the slope. The conversion of potential energy to kinetic energy partially offsets the loss of kinetic energy to friction. The radius of circular motion keeps shrinking. When the marble arrives at the bottom of the bowl all the available potential energy has been converted. The marble sitting at the bottom of the bowl is the end state of the system because there is no longer any opportunity to dissipate energy.
The end state of the system is a state of force equilibrium. It is only when the resultant of the forces that are acting in the system reaches zero that there is no longer opportunity to dissipate energy.
In statics it isn't necessary to examine how the system will proceed to the end state. In statics it is sufficient to probe whether or not there is a state of force equilbrium.
In the diagram the curving part of the dashed line represents the hanging chain. The vertical lines left and right represent additional chain length, the weight of the additional chain length provides tensioning force.
In the diagram the force that the hanging chain exerts at the cusp is decomposed in two components. The magnitude of the vertical component is equal to the weight of the length of chain that is being suspended in between the cusps. The magnitude of the horizontal component follows from the angle of the chain (at the cusp).
The two piles of chain left and right represent that surplus chain just piles up. As a consequence the tension force is constant; only the length of chain from the cusp to the pile contributes to tensioning. I will refer to this tension as the provided tension.
The total tension force exerted by the catenary (at the cusp) is the resultant force of the horizontal and vertical component. I will refer to that tension as the cusp tension. The cusp tension and the provided tension are acting in opposition to each other; I will refer to the resultant effect as non-equilibrium.
With the checkbox 'Non-equilibrium' checked: as you move the slider to the left and to the right the vertical bar next to the left chain pile represents whether or not the cusp tension is in equilibrium with the provided tension. As displayed in the diagram The length of the non-equilibrium bar is not in 1-on-1 proportion with the component forces. For clarity the length of the non-equilibrium vertical bar is doubled.
When the non-equilibrium points upward: that is a state where the provided tension force is not enough, and when released the chain will sag. When the non-equilibrium points downward there is a surplus of provided tension force, and the suspended chain will be pulled upward.
The number displayed above the right side cusp is the length of the chain as a function of the value of the slider.
Shape of the Catenary
The shape of the catenary curve can be found by setting up a differential equation.
Since the shape is symmetric it is sufficient to evaluate from the midpoint to the cusp.
|TH||The horizontal component of the tension|
|λ||The weight per unit of length|
|L||the length of the chain from the midpoint to the x-coordinate.|
The weight that has to be supported at coordinate x is given by multiplying the length L with the weight per unit of length: λL
The differential equation setup here uses the following property: since gravity is acting in the vertical direction the horizontal component of the tension is constant from cusp to cusp. It is not a function of the x-coordinate; it can be treated as a constant.
In preparation: from midpoint to cusp the slope of the curve increases; the length of chain per unit of x-coordinate increases accordingly. (1.1) gives an expression for dL/dx, which will later be used.
At every point, from the mid point to the cusp, the slope of the curve is the ratio of horizontal tension component and vertical tension component:
(1.1) gives an expression in terms of the derivative of L with respect to x, so in order to make use of (1.1) we need to rearrange (1.2) such that we can take the derivative of L with respect to x.
TH and λ are constants, so we can treat λ/TH as a constant.
Taking the derivative:
Combining (1.4) and (1.1) allows us to eliminate the quantity dL, arriving at a differential equation that is is purely in terms of the cartesian coordinates x and y:
Given the form of the right hand side of (1.5) it is clear that if the differential equation has a solution at all it must involve the hyperbolic cosine; the sinh and cosh are related according to the equation: sinh²+1=cosh²
(1.6) incorporates the ratio TH/λ such that it satisfies (1.5)
Coordinate system of the diagram
The coordinate system is chosen such that the two cusps are located at x=-1 and x=1 respectively. The mass of chain is chosen such that the mass per unit of length is one unit of mass. The provided tension is set up such that at the equilibrium position the horizontal component of the tension in the catenary comes out as 1 unit of force. That is why when the slider is in the 1.00 position the line that represents the horizontal component of the tension force is 1 unit long; it represents 1 unit of force.
Statics, general nature of variational approach
Before discussing the proces of using calculus of variations to find the shape of the catenary I first want to discuss the nature of variational approach in general. In this stage the catenary curve is used for the purpose of illustration.
A mass m, subject to a gravitational acceleration g, is subject to a force mg. When that mass descends over a height h the amount of gravitational potential energy that is released is mgh (As mentioned earlier, to keep the numbers simple the mass per unit of length of chain is chosen such that the gravitational force per unit of length is one unit of force.
Potential energy is obtained by taking the integral of the exerted force with respect to position coordinate.
As we know, integration and differentiation are inverse operations: when you differentiate potential energy with respect to the position coordinate you recover the exerted force.
The value of the slider is used as variational parameter. I will refer to that value as p. In the graphlet the shape of the chain as a function of parameter p is implemented as follows:
More specifically: to make the cusps come out at y=0 the height of the cusps as a function of the variational parameter is subtracted.
Graphlet 2: the right side sub-panel displays the rate of change of the potential as a function of the value of the variational parameter. I will abbreviate 'rate of change of potentential as function of p' to: potential-change-rate
The graph for the potential-change-rate of the catenary section is overall below the graph for the potential-change-rate of the tensioning section.
The tensioning section and the catenary section are in counter-motion; when the catenary section descends the tensioning section moves upwards. In this diagram the potential-change-rate of the tensioning section has been mirrored in the x-axis. The mirroring of the graph allows direct comparison.
Graphlet 3 displays the difference between the two change-rates. For clarity: in the right side sub-panel the vertical scaling has been increased by a factor of 10
The line with long dashes displays the difference between the two change-rates.
The line with the short dashes displays the integral. This integral is the resultant potential as a function of the variational parameter.
Derivative of the potential energy
Let's say you have decided to find the equilibrium position by sweeping out variation, and evaluating the resultant potential energy. You are looking for the point of minimum potential energy.
You obtain an expression for the potential energy as a function of the variational parameter.
In mathematics, to find the point where a function is minimal you take the derivative, and you find the point where the derivative is zero.
Applying that in statics: you start with an expression for the potential, and you take the derivative with respect to the spatial coordinate. Tnat means you recover the force.
That is why differential approach and variational approach are equivalent: in both cases the evaluation is evaluation of the (resultant) force that is being exerted.
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Last time this page was modified: April 30 2022