Is a hanging chain a parabola? Comparing catenary and quadratic models

Last winter I hung fairy lights along the fence in our garden, and my younger brother said the drooping wire was “a parabola, like in your maths book”. I agreed with him — until I read that Galileo had thought the same thing in 1638 and was wrong. The curve a hanging chain makes is called a catenary, from the Latin for chain. What surprised me was that almost every picture I found of a catenary looked exactly like a parabola, so I wanted to know how different the two curves really are, and whether the difference is big enough to matter.

The aim of this exploration is to find out how much better a catenary describes a real hanging chain than the best-fitting parabola, and whether the answer depends on how much the chain sags.

I expected the two models to be almost identical for a chain that is pulled nearly straight, because I could not see any difference between the pictures, but I was not sure what would happen for a chain that hangs very low. My plan was to photograph two chains, one shallow and one deep, fit a parabola first because it is the model I already understand, and then learn about the catenary and compare the two fairly, using the same measure of fit for both.

I used a thin metal necklace chain (about 0.5 mm links, so it bends almost freely) and hung it from two drawing pins pushed into a cork board covered with 1 cm squared paper. For chain A the pins were 80 cm apart and the chain hung about 14 cm below them; for chain B I used a longer chain with the pins 70 cm apart, so it hung about 46 cm below them.

I photographed each chain from 2 m away with the camera at the height of the lowest point and parallel to the board, to reduce perspective distortion, and imported the photographs into GeoGebra. I placed the origin at the lowest point of the chain, with the x-axis horizontal, and used the squared paper to set the scale. I then read the height of the chain every 5 cm horizontally. I define:

\(x\) = horizontal distance from the lowest point (cm), \(-40 \le x \le 40\) for chain A and \(-35 \le x \le 35\) for chain B;
\(y\) = height of the chain above its lowest point (cm).

The chain is about 1.5 mm thick in the photographs and I read its centre line, so I estimate each height is accurate to about ±0.2 cm. For this reason I give all measured heights to 1 decimal place. Table 1 shows half of the data for chain B; the full data for both chains (17 and 15 points) is in Appendix 1, and Figure 1 shows both chains.

Both sets of points are symmetric about the y-axis, as I expected from the way the chain was hung. The shallow chain looks like a gentle U. The deep chain has a flatter bottom and much steeper sides, which already made me doubt that one parabola could fit it well.

Because the lowest point is at the origin and the data is symmetric, a parabola through the origin with its vertex there has the form \(y = kx^2\), with only one parameter, \(k\). I wanted the value of \(k\) that makes the parabola as close as possible to all the points, not just the two end points. I measured closeness with the sum of the squared vertical distances (the residuals):

\[ S(k) = \sum_{i=1}^{n} \left( y_i - kx_i^2 \right)^2 . \]

Squaring stops positive and negative residuals from cancelling out. \(S\) is a quadratic function of \(k\), so I could find its minimum with calculus. Differentiating with respect to \(k\) (the \(x_i\) and \(y_i\) are constants):

\[ \frac{dS}{dk} = \sum_{i=1}^{n} 2\left(y_i - kx_i^2\right)\left(-x_i^2\right) = -2\sum x_i^2 y_i + 2k\sum x_i^4 . \]

Setting \(\frac{dS}{dk} = 0\) gives

\[ k = \frac{\sum x_i^2 y_i}{\sum x_i^4} , \]

and since \(\frac{d^2S}{dk^2} = 2\sum x_i^4 > 0\), this is a minimum. This was the first time I had used differentiation on a function whose variable was a parameter rather than \(x\), and it made me understand what my calculator's regression button actually does.

Using a spreadsheet for the sums, chain A gives \(k \approx 0.008533\) with \(S \approx 0.25\text{ cm}^2\), and chain B gives \(k \approx 0.03549\) with \(S \approx 37.13\text{ cm}^2\). One full calculation of the sums is shown in Appendix 2.

For chain A the parabola is excellent: the average squared residual is \(0.25 \div 17 \approx 0.015\text{ cm}^2\), which is smaller than the square of my reading error (\(0.2^2 = 0.04\)). For chain B the average squared residual is about \(2.5\text{ cm}^2\), so a typical point is roughly \(1.6\) cm from the curve — about 8 times my reading error. So the parabola is not good enough for the deep chain, and I needed to find out whether the catenary does better.

My aim was to find how much better a catenary describes a real hanging chain than the best-fitting parabola, and whether this depends on the sag. For a shallow chain (sag about 17% of the span) the two models are indistinguishable with my equipment: both fit to within the reading error. For a deep chain (sag about 66% of the span) the catenary fits about 117 times better by the sum of squared residuals, the parabola has a systematic W-shaped error, and it underestimates the angle at the supports by about 6°.

This explains my original confusion: the pictures I saw were all of fairly shallow chains, where a parabola is an excellent approximation. I now think the right answer to my brother is “it is not a parabola, but for fairy lights you would never notice”.

The main limitation is measurement. Reading heights from a photograph depends on the camera being parallel to the board; a tilt of a few degrees would stretch one side of the chain, and I only checked this by eye. Repeating each photograph three times and averaging would have given me an estimate of the reading error instead of a guess. I also assumed the chain was perfectly flexible and uniform; the clasp of the necklace is heavier than the links, which may explain why chain B's residuals are slightly larger on the right, where the clasp was.

If I extended this exploration, I would like to find the length of each chain from the models and compare it with the length I measured with a tape. That needs the arc length formula, which is not in my course, so I would have to learn it first. I would also like to test whether the parabola becomes the better model when the chain carries a heavy, evenly spread load, as the cables of a suspension bridge do, because I read that this changes the physics.