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# Curling: The sport of mathematicians

Tomorrow marks the beginning of events at the $2022$ Beijing Winter Olympics, and boy am I excited. This means that over the course of the next few weeks I, like many others, will become an armchair expert and superfan of some of the coolest sports in existence (pun entirely intended).

One sport I always find myself become enthralled by is the curling which I attribute to the combination of my Scottish blood, and how scientific curlers must be. It’s some of this science that I’d like to share with you today.

For those that don’t know, curling is a sport played on ice where players take turns to push stones along the ice aiming to get the stone as close to a target as possible. Once one player has “thrown” the stone, the other players can sweep the ice to help achieve this. So how does the all work?

Let’s start at the very beginning: when the “lead” (the player who throws the stone) throws the stone. There are really three things that the lead can control.

First is the turn or the curl of the stone. This is the angular velocity, which is what ends up causing the trajectory of the stone to curve. I’m sure you’re now wanting to know how this works. You’re not alone. So do physicists. So maybe by the next Winter Olympics, I’ll have an answer for you there, but for now I’m going to assume that the stones are thrown with no curl. Don’t worry though – there is still some interesting mathematics to be done!

The second thing that the lead can control is the line. This is the direction that the stone would travel were it not for the turn.

The third and final factor is the velocity of the stone. Rather confusingly, the technical curling term for this is the weight, but for the sake of clarity I’m going to refer to it as the velocity.

Now for some equations! Let’s let the velocity of the stone is ${v}_{0}$. Remember, we’re assuming that there is no curl to the stone so we can ignore that. The mass of a curling stone is $19.1$kg.

Therefore using the equation ${K}_{e} = \frac{1}{2}m{v}^{2}$, where ${K}_{e}$ is the kinetic energy, we get that the kinetic energy of the stone when it’s released from the lead players hand is:

$9.55 \times {{v}_{0}}^{2}$

It’s fairly intuitive that the stone will keep moving until it no longer has any kinetic energy (and by substituting $v=0$ into our kinetic energy equation, we can see this). So what’s going to cause the stone to lose energy? Well this is going to have to be a force of some kind which brings us to our second equation $W = F \times d$. This states that the work done ($W$) is the product of the force doing the work ($F$), and the distance($d$).

And what is this mystic force? A pat on the back if you’ve already worked this one out. It’s friction! Of course, there are other factors that will cause the stone to lose energy, but friction is by far the greatest one.

Our equation for friction (${F}_{r}$) is ${F}_{r} = \mu N$, where $N$ is the normal reaction force. In the case of our stone, this is equal to the weight of the stone (i.e. the mass times the gravitational field strength $g \ approx 9.8$). We’ll get to what $\mu$ is later, so for now will leave it as is.

Therefore ${F}_{r} \approx \mu \times 19.1 \times 9.8 \approx 187.2 \mu$. We can put this into our earlier equation now, to get that the work done by friction (i.e. the energy lost due to friction) is $\approx 187.2 \mu d$.

Let’s think about when the stone stops moving. We’ve already said that this is when the stone has no kinetic energy. If we assume (as we are doing) that all energy loss is due to friction then the total work done by friction must equal the initial kinetic energy. This means we get the following equation:

$9.55 \times {{v}_{0}}^{2} \approx 187.2 \mu d$

We can, however, make this a tad simpler. See on both sides of this equation, we use the fact that the mass of the stone is 19.1. Suppose we hadn’t done that. Then the equation would look like this:

$\frac{1}{2} m \times {{v}_{0}}^{2} \approx 9.8 m \mu d$

So we can cancel the $m$s to get:

$\frac{1}{2} \times {{v}_{0}}^{2} \approx 9.8 \mu d$

A bit of rearranging and we get an equation for the distance over which friction has acted. A little bit of thought tells us that this is the displacement of the stone (i.e. how far it has travelled)

$d \approx \frac{{{v}_{0}}^{2}}{19.6 \mu}$

Wonderful! We have an equation for the distance the stone travels, something that’s very useful to know when curling. What’s more, it depends on ${v}_{0}$ which is something that the lead player can control! Ideal! (I would like to point out here that it is definitely not something I personally could control, which is why I’m very impressed by curlers!)

This leaves two main questions though: what is $\mu$, and why the sweeping? I’ve been watching a lot of Death in Paradise recently, and if I’ve learned one this, it’s that whenever you’re left with two mysteries, they’re probably connected. This is no different.

$\mu$ is the coefficient of friction, and is non-negative. It is a measure of how much friction there is between two surfaces. The higher the value of $\mu$, the more friction there is between the two surfaces and two perfectly smooth surfaces have $\mu = 0$.

$\mu$ depends on both surfaces and in the case of curling, this is the bottom of the stone, and the ice. The surface of the stone is fairly fixed and for the purposes of today’s discussion, uninteresting. But the ice on the other hand is a whole different ball game (or stone game I suppose). One lesser known fact about curling ice is that it is not smooth. Two layers of water droplets are sprayed on the ice to create a “pebbled” effect. This means $\mu$ is a positive value. And this is where the sweeping comes in.

By sweeping, players smooth out the surface of the ice, thus reducing the value of $\mu$. Going back to our equation from earlier ($d \approx \frac{{{v}_{0}}^{2}}{19.6 \mu}$), this means that the stone travels further, and hopefully closer to the target.

While curling, players are constantly assessing the smoothness of the ice, the velocity of the stone, and the distance they want it to travel. And that’s ignoring the curl, or the tactics of hitting other stones out of the way. And that’s pretty cool in my book.