This is a problem in "Differential Equations: A Modern Approach with Wavelets" by Krantz:

If the length of any small portion of an elastic cable of uniform density is proportional to the tension in it, then show that it assumes the shape of a parabola when hanging under its own weight.

After not knowing how to approach the problem, I decided to research it and I ended up finding about the elastic catenary. And another source says that it is a non-algebraic curve, so definitely not a parabola.

Is the problem wrong? or is it that the problem is not that of the elastic catenary?

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    $\begingroup$ Cross-posted in Mathematics SE as Help checking if this variation of the catenary problem makes the curve become a parabola or if it is actually not correct. Please note that SE policy is that cross-posting is strongly discouraged : meta.stackexchange.com/q/64068. Choose the site you think will give you the best answers and delete on other sites before answers start accumulating on multiple sites. $\endgroup$ Commented Nov 15, 2020 at 19:10
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    $\begingroup$ You are correct. The problem set by Krantz is wrong. A parabola results if the loading remains uniform horizontally, as for a suspension bridge with relatively light cables. Your 2nd source shows the transition from catenary to parabola but does not explain what physical approximation could result in a parabola. Uniform pressure on light elastic circular membranes is used to create parabolic mirrors. $\endgroup$ Commented Nov 15, 2020 at 19:20

2 Answers 2


For a chain, the shape approximates a catenary, but not exactly because each link is rigid and cannot have continuous curvature.

The defining property of a catenary is that for any segment the total vertical load must balance with the weight of the segment.

$${\rm d}V = w \,{\rm d}s = w \sqrt{1 + (y')^2} {\rm d}x$$

Here $V$ is the vertical load, $w$ is the weight per length, ${\rm d}s$ is a small segment length, and $y(x)$ is the shape. Also $H$ is the fixed horizontal component of tension.

Since the tension must be tangent to the curve $y' = V/H$ or

$$ y'' = \frac{1}{H} \frac{{\rm d}}{{\rm d}x} V = \frac{w}{H} \sqrt{1+(y')^2}$$

with solution

$$ y' = \sinh \left( \tfrac{x-x_0}{a} \right) $$ where $a=H/w$.

The shape is found by integrating the above as

$$ y = y_0 + a \left( \cosh \left( \tfrac{x-x_0}{a} \right) - 1 \right) $$

Now consider a span $S$ and a cable hanging about two equal height points along the span.

This makes the mid-point sag equal to $$ D = a \left( \cosh \left( \frac{S}{2 a} \right) - 1 \right) $$

The length of the cable is

$$ L = 2 a \sinh\left( \frac{S}{2 a} \right) $$

From the tangency constraint, the vertical tension is $V = H y'$ or

$$ V = a w \sinh\left( \frac{S}{2 a} \right) $$

This vertical tension is exactly half the total weight of the cable $$V = \tfrac{1}{2} w L$$, or each tower supports half the cable.

If the shape wasn't a catenary, but a parabola, the above fact would not be true.

For a parabola, the vertical load is slightly less than the total weight of the cable.

$$ V_{\rm para} =\left( 1 - \tfrac{S^2}{6 a^2} \right) \left( \tfrac{1}{2} w L \right)$$

Now the total support of the towers is less than the weight of the cable, which means the parabola cannot be the final resting shape of the cable.

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    $\begingroup$ Does this solution include the stretching of the chain? $\endgroup$ Commented Sep 5, 2021 at 10:10
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    $\begingroup$ Yes, I forgot to mention that the entire chain shares the horizontal tension component $H$. $\endgroup$ Commented Sep 5, 2021 at 16:12
  • $\begingroup$ Is there any stretching at all in this solution? I mean the solution doesn't involve any elastic coefficients characterizing the cable. $\endgroup$ Commented Oct 7, 2021 at 16:12
  • $\begingroup$ @user8736288 The elastic properties of the cable do not affect the general shape (catenary). Beyond the shape analysis above, once you go into the sag and tension analysis then the elasticity is important. There are two main effects to consider, initial stress-strain, and creep over time. PS. You can look into this application I made that solves the catenary problem for multiple spans. $\endgroup$ Commented Oct 7, 2021 at 17:03
  • $\begingroup$ If elastic properties are not included in the analysis in the first place, it doesn't really come as a surprise that they do not affect the general solution obtained! My intuition was that parts of the cable with larger tension should undergo larger strains and the lineic mass should also vary accordingly, which would further complexify the analysis. These effects must be negligible in the end but I'm pointing at this since Krantz seems to rely on Hooke's law to derive its mysterious parabola shape. $\endgroup$ Commented Oct 7, 2021 at 20:07

For a cable hanging vertically from a support, it is true for any small portion: $S \sigma(y) + \rho S \Delta y = S \sigma(y+\Delta y)$, where $S$ is the section and $\rho$ the density. So, in the limit when $\Delta y$ goes to zero: $$\frac{\partial \sigma}{\partial y} = \rho$$

For a linear elastic cable: $$\sigma = E \epsilon_y \implies \frac{\partial \epsilon_y}{\partial y} = \frac{\rho}{E}$$

The strain is the derivative of the displacement in relation to the length: $$\epsilon_y = \frac{\partial u_{y}}{\partial y}$$

What leads to the equation:

$$\frac{\partial^2 u_{y}}{\partial y^2} = \frac{\rho}{E}$$

So, the new elongated coordinate y' has a quadratic relation to the old y.

On the other hand, the Poisson coeficient is a relation between the side strain and the longitudinal strain. If the initial cable diameter is $D$, the new diameter as a function of the original length of the cable is:

$D'(y) = D(1 - \epsilon_x) = D(1 - \nu \epsilon_y) $

That expression is linear with $y$, due to the definition of $\epsilon_y$

The relation $D'(y')$ must be then quadratic (parabolic), with the max. ( = initial) diameter at the lower end of the cable and minimal at the top (where it is hanging from) where the stress, strain and Poisson contraction is maximum.

In other words, the cilindrical shape of the cable turns into a parabolic shape.


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