A parabola and a catenary can look very similar but are not the same. The graph of
y = x²
is a parabola and the graph of
y = cosh(x) = (ex + e–x)/2
is a catenary. You’ve probably seen parabolas in a math class; you’ve seen a catenary if you’ve seen the St. Louis arch.
Depending on the range and scale, parabolas and catenaries can be too similar to distinguish visually, though over a wide range enough range the exponential growth of the catenary becomes apparent.
For example, for x between -1 and 1, it’s possible to scale a parabola to match a catenary so well that the graphs practically overlap. The blue curve is a catenary and the orange curve is a parabola.
The graph above looks orange because the latter essentially overwrites the former. The relative error in approximating the catenary by the parabola is about 0.6%.
But when x ranges over -10 to 10, the best parabola fit is not good at all. The catenary is much flatter in the middle and much steeper in the sides. On this wider scale the hyperbolic cosine function is essentially e|x|.
Here’s an intermediate case, -3 < x < 3, where the parabola fits the catenary pretty well, though one can easily see that the curves are not the same.
Now for some details. How are we defining “best” when we say best fit, and how do we calculate the parameters for this fit?
I’m using a least-squares fit, minimizing the L² norm of the error, over the interval [M, M]. That is, I’m approximating
c + kx²
and finding c and k that minimize the integral
The optimal values of c and k vary with M. As M increases, c decreases and k increases.
It works out that the optimal value of c is
and the optimal value of k is
Here’s a log-scale plot of the L² norm of the error, the square root of the integral above, for the optimal parameters as a function of M.
7 thoughts on “Best approximation of a catenary by a parabola”
I’m confused by this, specifically by your second graph above. I thought a catenary was the shape adopted by a wire hanging from its endpoints, but I find it hard to believe a wire would ever take on the shape shown in that graph, unless the middle part were lying on the ground. Am I making an easy-to-explain error?
@Isaac: Think of a very, very long wire.
The vertical scale in the middle graph represents 3 orders of magnitude more distance than the vertical scale in the other graphs. If the horizontal and vertical scales were equal, as they roughly are in the other two graphs, you’d see a more gradual bend.
Of course! Thanks very much, I should have seen that. If we take the units as metres, the wire is hanging from 7km up, and is hanging so close to vertical nobody can tell the difference except right near the bottom.
This is neat.
Can you show an example fitting a hyperbola?
Nice post, as usual. Is your least squares method converging to the Chebyshev’s Polynomial in that interval?
Hi! Could you please explain the working of how you got the optimal values of c and k? Thank you !
It’s been a while since I wrote this, but I probably used one of Mathematica’s Minimize functions.