I'm familiar with the equation for the sag of a cable:

$$x = \frac{wx(L-x)}{2T}$$

Where w = the weight per length, T = the tension and L = the length of the span between supports. I'm wondering if there is a similar equation for the sag of a cable that has two different weights. For instance, if two cables with different cross sectional areas were tied or spliced together.

I'm just hoping to be pointed in the right direction, if there is one. I'm hoping there is, but more realistically I expect this can't be done statically and I will need to use some sort of FEM package.

Many thanks.

  • 1
    $\begingroup$ Does this answer your question? Catenary curved string with differing linear densities - Linear density distribution $\endgroup$
    – Kyle Kanos
    Feb 12 at 13:57
  • $\begingroup$ @KyleKanos: The OP's question is a bit simpler than the full catenary equation, I think; they appear to be in the limit where the stretching of the string is negligible (which is why their solution for the shape is a quadratic polynomial rather than a hyperbolic cosine.) Still, the techniques on your answer there should apply. $\endgroup$ Feb 12 at 16:01
  • $\begingroup$ @MichaelSeifert There is no stretching in the catenary solution. $\endgroup$
    – Buzz
    Feb 12 at 21:50
  • 1
    $\begingroup$ @Buzz: You're of course correct; I don't know why I said that. What I meant was that the OP's solution assumes the limit where the deviations from horizontal are "small" ($y' \ll 1$) — which is not the same thing as "no stretching." Mea culpa. $\endgroup$ Feb 12 at 21:57

1 Answer 1


This can be solved exactly by noting that (in this limit) the displacement of the cable satisfies the differential equation $T y'' = - w y$ at all points (look at the forces in an infinitesimal element of the rope to see this.) This generalizes in a straightforward way to the case where $w(x)$ is a function of the position $x$.

In the case where you have cables of two different weights, you would write down the general solution to the above ODE for each part of the cable separately, subject to the boundary conditions at either end and the additional requirement that $y$ and $y'$ are continuous at the location of the splice. This then leads to a system of four equations (ends of the cable and continuity at the splice) and four unknowns (two free parameters for each region) that can be solved.

  • $\begingroup$ I suspect that in general the derivatives won't necessarily match at the join. $\endgroup$
    – PM 2Ring
    Feb 12 at 14:19
  • $\begingroup$ @PM2Ring: If the derivatives don't match at the join then there will be a net force from the tension on either side of the join point. This leads to a net force on an object with zero mass which is inconsistent. You'd get an difference between the derivatives if, for example, the mass density of the string included a point mass somewhere (modeled as a delta function.) This might happen, for example, if you tied the strings together with a big old knot; but for a smooth splice, where the mass density doesn't diverge, the derivatives will match. $\endgroup$ Feb 12 at 15:58
  • $\begingroup$ Ok, that makes sense. But when I try to picture a light thread connected to a rope, I see a kink at the join. I need to do the experiment... $\endgroup$
    – PM 2Ring
    Feb 12 at 17:37
  • 2
    $\begingroup$ @PM2Ring: If the rope is stiff and resists bending, and the tension on the thread isn't strong enough to overcome it, you can get a noticeable kink at the join. But the differential equation assumes stiffness is negligible. $\endgroup$ Feb 12 at 19:39
  • $\begingroup$ I should know this stuff. ;) FWIW, I made a plotter using that last equation in my answer. $\endgroup$
    – PM 2Ring
    Feb 13 at 2:28

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