Catenary

y = (a/2)(exp(x/a)+exp(-x/a))

or, equivalently,

y = a cosh(x/a)

where "cosh" denotes the hyperbolic cosine. As mentioned in the Webster, this is the curve taken by a homogenous, flexible and inextensible chain hanged by its two extremities. Galileo thought this curve would be a parabola; in 1690, Jakob Bernouilli challenged his colleagues to find the exact mathematical expression, and in 1691, Leibniz, Huygens and Johann Bernouilli all found independently the answer to this question. The catenary curve is the solution to the following differential equation:

y'' = (m/c) sqrt(1+y'²)

which describes the condition for static equilibrium, m being the mass density of the chain and c the tension along the horizontal direction. Follow "A hanging string forms a hyperbolic cosine curve" for a complete demonstration... It is interesting to note that when a parabola is rolling on a line, the path followed by the focus is a catenary.

Primary source: www.2dcurves.com

The style of overhead wire used by many electric railways is often referred to as catenary, because it contains a roughly catenary curve. The wire that takes this form is called the messenger wire and it is loosely hung between widely spaced suspension points. From this messenger wire hang drop wires of varying lengths connect to the contact wire which is as near smoothly horizontal as can be achieved. This is known as catenary suspension and is utilised because it allows the contact wire to be kept relatively straight and horizontal while permitting quite wide spacing of supports.

In some applications, another wire, the auxilary wire, is above and parallel to the contact wire, attached to it at regular and close intervals. The purpose of this wire is that it contains more copper than the others, allowing for better electrical conductivity. The added copper makes it softer, which would not be desirable in either the messenger wire or contact wire, for reasons of strength and wear resistance; the addition of an extra wire which is neither load bearing nor touched by the locomotives' pantographs enables better conductivity without compromise.

It has become common to refer to the whole arrangement as 'catenary', even though only the messenger wire describes a catenary curve; moreover, lazy usage has seen the term 'catenary' used to describe all overhead wire systems, even those that use no catenary curve at all!

The tension is also kept up by way of large counterweights, usually hung from one of the supports, to ensure the wire won't droop (pantograph height is a precise thing). When this is done, the catenary is divided into sections. One wire will join the existing wire from an addition support, and the two will run parallel. After passing another support or two, the first wire will then run off to one side to either terminate or attach to the counterweight, while the second is now the power supply wire. This ensures uninterupted power for the train (otherwise you'd have that subway syndrome where the lights go off in the car as it passes over a switch where there's no third rail). This sometimes is how section breaks are isolated, in addition to being a convenient way to keep the wires up!

Edit: I've now discovered, in the UK at least, that the side-to-side movement of the wire is 50cm

Cat"e*na*ry (?), Cat`e*na"ri*an (?), a. [L. catenarius, fr. catena a chain. See Chain.]

Relating to a chain; like a chain; as, a catenary curve.

© Webster 1913.

Cat"e*na*ry, n.; pl. Catenaries (). Geol.

The curve formed by a rope or chain of uniform density and perfect flexibility, hanging freely between two points of suspension, not in the same vertical line.

© Webster 1913.

q = (mass per unit length)*(gravitational acceleration).

In many problems you can assume that the tension is constant over the length of the string, but when the string's own mass is in consideration in the problem you can no longer do this, as we'll see. So we'll say that, over the length dL, the tension on the string changes from T to T+dT. We'll also say that the angle the string makes with the horizontal goes from A to A+dA.

Now since this segment of string is immobile (i.e. it has no acceleration), all forces on it must cancel each other. We'll balance the forces of tension and gravity in the horizontal, then the vertical direction.

In the horizontal direction the only forces we have are the tensions on each end. Tension, by definition, is a force pointing in the same direction as the string does. The horizontal component of the tension therefore involves the cosine of the string's angle with the horizontal. The force equation is this:

F_horiz = 0 = -T cos(A) + (T+dT) cos(A+dA)

The addition rule for the cosine is

cos(x+y) = cos(x) cos(y) - sin(x) sin(y)

so

cos(A+dA) = cos(A) cos(dA) - sin(A) sin(dA) = cos(A) - dA sin(A)

Here we took the first-order approximations for trig functions of small angles: the cosine of a small angle equals one and the sine of a small angle equals the angle itself. Now back in our force equation we have

0 = -T cos(A) + T cos(A) + dT cos(A) - T dA sin(A) - dT dA sin(A)

The first two terms cancel and the last term, with the product of two infinitesimals, is negligible compared to the other remaining terms, leaving us with

dT cos(A) = T dA sin(A)

(Incidentally, this is where things would fall apart if we took T to be constant over the whole string:
we'd have only 0 = T dA sin(A), leaving us puzzled as to which of those factors is supposed to always be zero.)
Separating the two variables,

dT/T = tan(A) dA

Integrating both sides

ln(T) = -ln(abs(cos(A))) + C

T = C_T sec(A)

In the last step we took the exponential of both sides, cancelling the logarithms and redefined the integration constant so that it'd be proportional to the tension. This constant basically just depends on how the string is hung.

Now in the vertical direction, we have gravity as well as tension:

F_horiz = 0 = -q dL - T sin(A) + (T+dT) sin(A+dA)

We'll again expand according to the addition rule for the sine:

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

Using the first-order approximations again, this gives

0 = -q dL - T sin(A) + T sin(A) + dT sin(A) + T dA cos(A) + dT dA cos(A)

Cancelling and removing the negligible double-infinitesimal term we get

q dL = dT sin(A) + T dA cos(A)

Now we'll substitute in the solution we obtained for the tension. Differentiating, we have

dT = C_T tan(A) sec(A) dA

so

q dL = C_T tan²(A) dA + C_T dA = C_T dA (tan²(A) + 1)

so

q dL = C_T sec²(A) dA

At this point we could solve this differential equation and obtain the inclination angle as a function of length along the string. But to find the shape of the actual curve that the string forms, it's better to substitute Cartesian coordinates. A and dL are both definable in terms of differentials of Cartesian coordinates:

A = tan^-1(dY/dX)

so

dA = cos²(tan^-1(dY/dX)) d(dY/dX)

(look it up in a table of derivatives if you don't believe me... ) and

dL = sqrt((dX)² + (dY)²) = sqrt(1 + (dY/dX)²) dX

We'll define the slope dY/dX as a new variable S for notational convenience and take X as the independent variable. Our new force equation is this:

q sqrt(1+S²) dX = C_T sec²(tan^-1(S)) cos²(tan^-1(S)) dS

The cosines and secants nicely cancel, leaving

(q/C_T) dX = dS/sqrt(1+S²)

The nicest way to integrate the right half of this equation is to substitute a hyperbolic sine for S. The hyperbolic sine and cosine have the properties that each one is the derivative of the other and

cosh²(a) - sinh²(a) = 1

so

sqrt(1+sinh²(a)) = cosh(a)

So if we substitute S = sinh(a) we get

dS/sqrt(1+S²) = cosh(a) da/cosh(a) = da

This leaves the equation, after integration and resubstitution, as a fairly simple

(q/C_T) X = sinh^-1(dY/dX) + C_X

sinh((q/C_T) X - C_X) dX = dY

As noted before, the hyperbolic cosine and sine are each other's derivatives, so the integral is simple and leaves us with an expression for the curve the string forms in Cartesian coordinates:

Y = (C_T/q) cosh((q/C_T) X - C_X) + C_Y