In the previous post, we discovered that the kinetic energy of a drifting continent is of the same general magnitude as that of a moving bicycle and its rider – 1500 joules would be a typical figure.

I went on to calculate that, whereas it takes me only about 10 seconds to get my bike up to full speed, it would take me hundreds of years to get the continent up to its tiny full speed were I to put my shoulder against it and push (assuming that it was perfectly free to move). How can this be, when the amount of energy that I’m giving each of these objects is the same?

The problem is that when I push the continent, I am, effectively, in the wrong gear.

On a bike with gears, you’ve got a range of choices about how you power it: you can ride in a high gear, pedalling slowly but pushing hard on the pedals, or ride in a low gear, pedalling more quickly but pushing less hard on the pedals. There’s a simple tradeoff: if you want to pedal half as fast, you’ve got to push twice as hard for the same effect.

But there’s a limit to how hard you can push on the pedals, which means that if you move up too far up through the gears, there comes a point where you can no longer make up for the decreased pedalling rate by pushing harder on the pedals, and the power that you can supply to the bicycle falls.

Anyone who’s tried to accelerate a bicycle when they are in too high a gear will have experienced this problem, and it’s what I experience when I try to push the continent directly. Because the top speed of the continent is extremely low (about the speed of a growing fingernail), I’m necessarily pushing it very slowly as I accelerate it. This means that to give it energy at the rate that I want to (1500 joules in 10 seconds, like the bike) I would have to push it impossibly hard – the force needed is about the same as the weight of a 300-metre cube of solid rock.

Is there a way that we can put me into a lower gear, so that I can push with a force that suits me, over a longer distance, and still apply the very high force over a short distance to the continent?

Yes. Just as we’ve all used a screwdriver as a lever to get the lid of a tin of paint off, so I could use a lever to move the continent. Similarly to the bike gears, the lever allows me to exchange pushing hard over a small distance with pushing less hard over a longer distance. To do the job, the lever would need to be long enough to allow me to push, with all my might, through a distance of about 2.5 metres, with the short arm of the lever pushing the continent. We’d need an imaginary immoveable place for me to stand, and we could use the edge of the neighbouring continent as the pivot (just as we use the rim of a paint tin as the pivot). The catch is the length of the lever: if the short arm was 1 metre long, the long arm would be about 1.5 million kilometres long.

Simon Gage of Edinburgh International Science Festival suggested a more compact arrangement: a bicycle with an extremely low gear ratio, with the front wheel immobilised on the neighbouring continent (assumed immoveable), and the back wheel resting on the continent we’re trying to accelerate. A transmission giving 17 successive 4:1 speed reductions would do the job nicely. Ten seconds of hard pedalling would get the continent up to full speed. To me on the saddle, it shouldn’t feel any different to accelerating my bike away from the lights.

*A wee caveat*. This is a thought experiment, and we’ve swept some fairly significant engineering issues under the carpet. The rearmost parts of the power train would be moving at speeds that are literally geological, so in reality it would take me years of pedalling to take all of the slack and stretch out of the system. These parts would also be transmitting mountainous forces, and so they’d need to be supernaturally strong. There will be frictional losses. And then there’s the issue of transmitting a gigantic force to the continent through the contact of a bike tyre on the ground.

### The calculations

What force is required to accelerate the Eurasian plate to top speed in 10 seconds?

The top speed of the plate is 3.2 × 10^{-10} ms^{-1}. If I accelerate it uniformly, its average speed will be half of this, and so in the 10 seconds over which I hope to accelerate it, it will travel 1.6 × 10^{-9} m.

Now *W = fd*

where *W* is the work that I do on the plate (ie the kinetic energy that I give it), *f* is the force that I apply to it, and *d* is the distance through which I push the plate. Rearranging gives us

*f = W/d*

We know *W* from the previous post (it’s 1500 joules) and we’ve just calculated *d*. Thus *f* works out at about 9.4 × 10^{11} newtons.

For comparison, a 300-metre cube of rock of density 2700 kg m^{-3} will have a weight of (300 m)^{3} × 2700 kg m^{-3} × 9.81 m s^{-2} = 7 × 10^{11} newtons roughly.

#### The lever

When a lever is used to amplify a force, the ratio of the lengths of the arms of the lever needs to be the same as the ratio of the two forces. Suppose that I can push with a force equal to my own body weight, about 600 newtons. If I’m to use a lever to amplify my push of 600 N to a force of 9.4 × 10^{11} N, the ratio of the lengths of the arms needs to be (9.4 × 10^{11})/600, or roughly 1.5 × 10^{9}. So if the short arm of the lever is 1 metre long, the long arm needs to be about 1.5 × 10^{9} metres long, which is 1.5 million kilometres. For comparison, the Moon is about 400,000 kilometres away.

To do 1500 joules of work with a force of 600 N, I’d need to push over a distance of 2.5 metres (because 600 × 2.5 = 1500).

#### The bicycle gearing

I estimated that it takes me 15 pedal revolutions to get my bike up to full speed. Knowing the length of the pedal cranks, I know the total distance that I have pushed the pedals through, and I know how much work I have done on the bicycle – 1500 joules. (I’m ignoring energy losses here, because they are small at low speeds on a bike and the calculation is highly approximate anyway). Using *work done = force × distance*, this gives an average force on the pedals of about 94 newtons.

The 17 stages of 4:1 reduction mean that the back wheel is rotating 417 = 1.7 × 10^{10} times slower than I’m pedalling. The pedalling force is amplified in the same ratio, to give a force on the teeth of the rearmost gear of 1.6 × 10^{12} newtons. We now have to allow for the fact that the radius of the rear wheel is about twice the length of the pedal crank. This roughly halves the force available at the rim of the rear wheel, giving a force of about 8 × 10^{11} newtons, which is close to what we need.