A uniform, non-elastic, and infinitely flexible cable (or wire, chain, string etc) suspended from two fixed points under the influence of gravity forms a **catenary** - similar to a parabola but not quite!

The cable is assumed to be so thin and flexible that any tension exerted by the cable is parallel to the chain. Furthermore, the chain may be considered a rigid body once it has attained static equilibrium and thus the net force acting on it is zero.

Let **M** be the **mass **of the cable and **L** be its **length**. Let the **linear density** (mass per unit length) be **μ**, so **μ = M/L.**

Consider a section of the cable from its lowest point (vertex) of length **s**. Let the tension at the vertex be **T₁** which acts horizontally and let **T₂** be the tension at the other end making an angle **θ** with the horizontal. The weight of this section is **μsg **where g is the acceleration due to gravity; we use the linear density of the cable since we are not dealing with a point mass.

Resolving horizontally we get,

and resolving vertically we get,

Dividing these equations gives us,

Suppose now we position the vertex A at the origin of an x-y coordinate system and let B have general coordinates (x, y). If **y=f(x)** is the equation of the catenary curve, then tanθ is the gradient of the curve at B, so

Define the constant a as

__Detour into Infinitesimal Calculus__

__Detour into Infinitesimal Calculus__

Suppose we consider the cable section to be very small, say of arc length Δ**s, **then Δ**s** is approximately the hypotenuse of a right-angled triangle:

Thus, by Pythagoras,

So we get

To solve this differential equation, put z=dy/dx which gives,

Hence the equation of the catenary curve is,

I would like to thank the "Physics Ninja" (Dr E.) for his brilliant, detailed video on the derivation of the catenary curve equation.

## Comments