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. image

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 C~d~ 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/m^3^ for air; Wikipedia estimates C~d~ = 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 m^2^, which I’ll revise to 0.7 m^2^ 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

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By Yariv