## Wednesday, August 11, 2010

### Teminal Velocity

The impetus for this post lies with three facts. First, I like to bike to work. Second, Cornell sits on a hill. And finally, I'm not very brave.

As a result of all of these, along with Ithaca's less-than-optimal road maintenance, my semi-daily rides home tend to produce a lot of wear on my brakes as I cruise downhill at what appears to me to be very high speeds. I began to ponder just how high this speed really is, and if I could reduce my use of the brakes or if I'm going to end up using them anyway at the bottom of the hill.

Figure 1: An inclined plane

So, I asked myself, what do I remember about bikes going down the hill? Well, I remember the good old inclined plane (figure 1), and I remember that air resistance is proportional to velocity, so that the equation of motion is given by
$ma = mg\sin\theta - \alpha v.$

I had no idea what α was, though. My first stop in considering it was naturally Wikipedia. A quick search came up with the formula
$m a = mg\sin\theta - \frac{1}{2}\rho A C_d v^2$
where ρ is the density of air, A the projected area of the body and Cd the drag coefficient

The first thing to notice here is that I was wrong - drag in a fluid acts like the velocity squared, and not the velocity. Second, we can easily determine terminal velocity out of this formula - it's the speed at which the sum of the forces equals to zero, or
$v_t = \sqrt{\frac{2mg\sin\theta}{\rho A C_d}}.$

We can throw in some numbers into that. ρ = 1.2 kg/m3 for air; Wikipedia estimates Cd = 0.9 for a cyclist. For the mass, we need to add up mine (~75 kg), the bike's (15-20 kg) and my bag's (let's say 5 kg). We come to about 100 kg, give or take 5%. A is a little harder to estimate, but height times width gives me an initial guess of 0.62 m2, which I'll revise to 0.7 m2 to account for the bike, flailing arms and fashionable helmet, up to about 10% accuracy.

We're left with sinθ, which varies by road, but in general we expect the terminal velocity to look like
$v_t \approx \left(50 \pm 3 \rm{m/s}\right) \sqrt{\sin\theta}.$

This appears not-unreasonable. For an 8% grade like we have down University avenue this yields about 50 km/h and for a 13% grade like we have down Buffalo street this will bring us up to a respectable 65 km/h. Both, incidentally, are faster than I'm willing to go down a badly maintained, not entire straight road.

So we have some numbers, and I begin to feel justified about pressing those breaks often, but all of this is really an introduction for the next post, in which I go against all my theorist instincts and take some data in the field. Stay tuned.

1. That's - sigh - about 30 mph and 40 mph, respectively, in crazy units