# Briechistochrone

One of the many sad consequences of the current lockdown, possibly the most unfortunate of all, is that the famous Cooper’s Hill Cheese-Rolling competition will almost certainly not be taking place this year. In the spirit of finding light in the darkness, let’s at least have a look at how we may improve it for next year. With maths!

Now the existing competition is already intense and dangerous enough, but I’ve got a feeling that the competitors would be happy enough to take it up a notch. The cheese already reportedly reaches speeds of up to 70 mph, but could we do better?

Given that the start and ending points of the cheese roll must remain the same (at the top of a hill and the back, very hard, wall of a cottage respectively), all we could do in this optimisation process is change the shape of Cooper’s Hill itself.

Fortunately for us, this problem was solved back in 1696 by such cheese-hungry greats as Newton, Leibniz, and all the Bernoullis, and is known as the brachistochrone problem – what curve would a frictionless bead slide down the quickest? (Assuming fixed start and end positions).

Let’s warm up with the classic solution. Assume the bead starts at $y = 0$, falls a distance $\Delta y = h$ and moves a lateral distance $\Delta x = \ell$ with some velocity $v$:

To calculate the total time taken to travel along the curve, consider the time taken to move along an infinitesimal slice starting at position $(x,y)$:

The distance moved is $\displaystyle{\delta s = \sqrt{\delta x^2 + \delta y^2} = \sqrt{1 + \left(\frac{\delta y}{\delta x}\right)^2}\delta x }$

and the time taken to cover this distance is $\displaystyle{\delta t = \frac{\delta s}{v(x,y)}}$

so the total time to traverse the curve is the integral: $\displaystyle{T = \int \text{d}t = \int \frac{\sqrt{1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2}}{v(x,y)}}\,\text{d}x$

To get the velocity, we turn to physics and note that by conservation of energy at any point on the curve we have $\displaystyle{\frac{1}{2}mv^2 = mgy \quad \Rightarrow \quad v = \sqrt{2gy}}$

where $m$ is the mass of the bead, and $g$ is the acceleration due to gravity.

The problem, then, is to minimise the following integral for all possible $y(x)$: $\displaystyle{\int \mathcal{L}(x, y, y')\,\text{d}x \quad \mathcal{L} = \left(\frac{1 + y'^2}{2gy}\right)^{1/2}}$

where $y'$ is shorthand for the derivative of $y$. This is a classic calculus of variations problem, brought up on this blog many times before, and the solution is given by the Euler-Lagrange equation. (I find it so amazing that you can even solve problems like these, I can’t resist going back to the E-L equation every now and then).

In this case, you simply plug $\mathcal{L}$ into the E-L equation, and end up with $\displaystyle{(1 + y'^2)y = k \qquad \text{(Eq. 1)}}$

for some constant $k$ (to be determined later). This isn’t the easiest equation in the world to solve, but there’s trick – solve for $y$ and $x$ separately in terms of a third variable which parameterises the curve – call it $\theta$, and set $\displaystyle{\frac{\text{d}y}{\text{d}x} = \cot\frac{\theta}{2}} \Rightarrow \text{d}x = \tan\frac{\theta}{2}\text{d}y \qquad \text{(Eq. 2)}$

Substituting (1) into (2) and we get $\displaystyle{y(\theta) = \frac{k}{1 + \cot^2\frac{\theta}{2}}\Rightarrow \text{d}y = k\sin\frac{\theta}{2}\cos\frac{\theta}{2}\text{d}\theta \qquad \text{(Eq. 3)}}$

and then (2) into (3) $\displaystyle{\text{d}x = k\sin^2\frac{\theta}{2}\text{d}\theta \qquad \text{(Eq. 4)}}$

and then upon integration of (3) and (4) we get the famous cycloid solution: $\displaystyle{x(\theta) = k(\theta - \sin\theta)}$ $\displaystyle{z(\theta) = k(1- \cos\theta)}$

(these solutions assume that the bead starts at position $(0,0)$). The final task, before we can visualise this curve, is to add the boundary conditions. We know that the bead starts at the origin, and so at $\theta = 0$, but at what $\theta = \theta^*$ does it end? To calculate this we need to solve the coupled nonlinear equations for $\theta^*, k$: $\displaystyle{\ell = k(\theta^* - \sin\theta^*)}$ $\displaystyle{h = k(1- \cos\theta^*)}$

At this point we must turn to a computer. Here are some solutions for $h = 1$ and varying $\ell$:

You can see that for $\ell/h \ll 1$, the solution is monotonic – $y$ decreases as $x$ increases (in fact $y(x) \propto -x^{2/3}$).

This is true for $\ell/h < \pi/2$ – at which point the end of the curve has zero slope. Beyond this it’s preferable for the bead to dip below the end of the curve – even though the total distance travelled is longer, the average speed is higher.

Now this is all very interesting, but at Cooper’s Hill we’re not dealing with beads on a line, we’ve got 4 kilograms of weapons-grade Double Gloucester to worry about. The question is then: what implications are there for the brachistochrone from considering a cylinder rather than a bead?

The first thing I thought about was the rotational kinetic energy of the cheese (surely a painful consideration for the contestants). If we say that the cheese is rotating at a rate $\omega$ with radius $r$, and that it does not slip against the ground, so that $v = \omega r$, then the energy conservation equation becomes $\displaystyle{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3}{4}mv^2 = mgy}$

where $I = mr^2/2$ is the moment of inertia of a cylinder. You can think of this effect as changing the strength of $g$. As $g$ doesn’t appear in the cycloid solution above, then adding the rotational kinetic energy won’t change the shape of the optimal curve (though it can change to total time to move across it).

One thing which will have an effect though is the displacement of the centre of mass of the cheese – while the cylinder touches the curve at $(x,y)$, the CoM is not at the same position:

If the tangent to the curve is $\displaystyle{\mathbf{t} = \left(\begin{array}{l}\text{d}x\\\text{d}y\end{array}\right)}$

then the unit normal from the curve to the CoM is $\displaystyle{\hat{\mathbf{n}} = \frac{1}{\sqrt{1 + y'^2}}\left(\begin{array}{l}y'\\-1\end{array}\right)}$

and the energy equation becomes the uglier $\displaystyle{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3}{4}mv^2 = mg\left(y - \frac{1}{\sqrt{1 + y'^2}}\right)}$

At this point my tolerance for algebra finally snapped, and I turned to the very helpful paper ‘Brachistochrone for a rolling cylinder‘, which frankly I should have done much sooner.

Translating their Eq. 2.10 into the same notation as my Eq 1., we have: $\displaystyle{(1 + y'^2)y - r\sqrt{1 + y'^2} = k \qquad \text{(Eq. 5)}}$

which you’ll notice turns into Eq. 1 when $r \rightarrow 0$. It turns out you can solve Eq. 5 using the same trick as before, presenting the solutions $\displaystyle{x(\theta) = k(\theta - \sin\theta) - r\cos\frac{\theta}{2}+r}$ $\displaystyle{z(\theta) = k(1- \cos\theta) + r\sin\frac{\theta}{2}}$

which again, recover those above when $r \rightarrow 0$. Great!

Plotting this new solution then for a cylinder of radius 0.1 we see:

…very little difference. As you might expect because the CoM is shifted higher, the cylinder solution ends up dipping a little faster than the point solution, but it pretty similar.

If you make the cylinder much larger the difference is more apparent:

but at this point the cheese is the size of the hill, and so the whole thing becomes a bit pointless.

Having a point is decidedly not the point here, or ever, so I’ll nevertheless declare this investigation a cheesy success. I keenly await the call from the headquarters atop Cooper’s Hill – together we can roll that cheese more optimally than it’s ever been rolled before.