The Catenary

(notes edited by Roberto Bigoni)


catenaria

A thin, inextensible, perfectly uniform and flexible rope, suspended between the points A and B, in condition of equilibrium in the constant Earth field g, assumes the shape done by a flat curve said Catenary.

In order to deduce analytically the equation of such curve, we can choose a cartesian orthogonal reference system having origin in the middle point of AB, horizontal axis of abscissas and vertical axis of ordinates.

Let P be a point on arc VP. There are two forces wich act on the punctiform rope element in position P, whose length be dl and mass be dm:

The condition of equilibrium of the point P is due to the fact that the sum of the tangential components of t and p is balanced by the reaction due to inextensibility of the rope, while the components of p and t perpendicular to the tangent balance themselves: therefore their modules are equal.

If we name θ the angle the geometrical tangent to the point P forms with the axis of the abscissas, we shall have

Eqn1.gif

Eqn2.gif

Let be y(x) the equation of the curve that describes the disposition of the rope in the chosen system of reference and (x ; y) the cartesian coordinates of the generic P point in such system: tan θ is given by the derivative of the y(x):

Eqn3.gif

The weight p is due to the weight of the segment VP of the rope.

Let v be the abscissa of the vertex, λ the constant linear density of the rope and g the module of the gravitational acceleration: the module of the weight p is given by

Eqn4.gif

and therefore

Eqn5.gif

Deriving both the terms of (5)

Eqn6.gif

and therefore

Eqn7.gif

If we introduce the constant k:

Eqn8.gif

the differential equation admits the solution

Eqn9.gif

The symbol cosh denotes the hyperbolic cosine.

If we know the coordinates a and b of B (and therefore the coordinates of A, symmetrical of B with respect to the origin O) and the length L of the rope (wich obviously must be greater of AB), we have

Eqn10.gif

Eqn11.gif

From (10) we get

Eqn12.gif

and from (11)

Eqn13.gif

Dividing term by term the (12) and (13) we have

Eqn14.gif

and we can express the function (9) in the following form

Eqn15.gif

From (14) we can also deduce

Eqn16.gif

Replacing (16) in (12) we obtain

Eqn17.gif

Given the values of a, b and L , the equation (17) allows to approximate the value of k (different from the banal solution 0).

Since y(a)=b, from (15) we have

Eqn18.gif

Annulling the derivative of y(x) in (15)

Eqn19.gif

we can get the abscissa v of vertex V

Eqn20.gif

The ordinate of V turns out therefore

Eqn21.gif

The modulus T of the tension, done by the sum of p and t, acting on the point P(x,y) of the rope can be calculated using the Pythagoras's theorem. From (8), (3) e (9) we have

Eqn23.gif

therefore

Eqn24.gif

Eqn25.gif

Since the value of g is expressed as 9.81 m/s2, the following Javascript application requires measures in metric units.

Parameters of the equation Eqn22.gif
input
abscissa a
ordinate b
length
density
graph
k
h
β
vertex: x
vertex: y
punto
x
ordinate
tension

 


last revision: May 2015