In this video, you see a cyclist testing new aerodynamic wheels from Zipp. Swapping your wheels may seem like a small change, but can make a big difference. From his tests, the rider discovers:

- With conventional wheels, he can ride 20 minutes at an average speed of 41.12 kph with an average power of 379 watts.
- With the Zipp 808 NSW aero wheels he rides 51 minutes at an average speed of 41.13 kph and average power of 344 watts.

Before looking at power and energy, I should go over two small details.

First, how do you measure power? Cyclists can measure power by installing a small computer, called a power meter, that measures the input torque at the pedals or crankshaft and records the rotation angle at timed intervals. If you know the torque and angle, you can calculate the input energy. Dividing this energy by time gives you power.

Second, this isn’t a perfect test of aerodynamics. If you really want to examine the effect of the new wheels, you probably would have to put a bike with a dummy in a wind tunnel. When the reviewer takes his second ride, many things could have changed—wind, body position, amount of sweat on the body—and impacted performance. Let’s assume the only thing that changed was the wheels.

Table of Contents

### Air Drag and Power

What happens when you ride a bike? If you are moving at a constant speed, then the net force on the bike-human system must be zero. In a slightly simplified view, I can draw the following force diagram:

The vertical forces (gravity pulling down and the ground pushing up) don’t really matter here. Just forget about them and pay attention to the horizontal forces. First, let’s look at the air drag. Air acts in complicated ways when an object passes through it. But who cares when we can make a simple model of air drag force? Here’s an expression for the magnitude of this force:

In this model, the air force is proportional to the square of the bike’s speed (*v*). For the other terms, we have:

- ρ is the density of air (around 1.0 kg/m
^{3}). *A*is the cross sectional area of the bike plus the rider (how much of the object interacts with the air).- Finally,
*C*is the drag coefficient. This parameter depends upon the shape of the object. If you change the wheels, it is the value of*C*that should change.

The second horizontal force is the frictional force. An interaction between the road and the tires propels the bike. I know what you’re thinking: Doesn’t the human propel the bike? In a sense, yes. But the reality is sort of complicated. The rider’s power goes through the pedals and chain to the wheel, which turns. But the *force* comes from the tire pushing against the road. So for our energy perspective on this problem let’s just say the human provides the friction force.

Clearly the faster the biker goes, the more human-push will be required. But what about energy? If I have a force pushing in the same direction as the motion of the object, then the work done (by the human) can be calculated as:

In this expression the value of *s* is the distance over which the bike moves. Since the frictional force must be equal in magnitude to the air resistance force, I can write the work as:

Since I really want an expression for the power I can use this work as the change in energy and divide it by some time interval Δt:

The value of *s* divided by Δt is the same as displacement divided by the change in time. This is the definition of average velocity. We get a power expression that doesn’t depend on the distance, just the cube of the velocity.

### The Aero Wheel Data

Got all that? Great. Let’s now apply it to the data from the video. Assuming that the wheel (and the power) was the only thing that changed, what does this say about the wheel? I will call the power required for the normal wheel *P*_{1} and the power for the aero wheel *P*_{2}. The ratio will be:

With the two power values from the video of 379 and 344 watts, this puts the value of the drag coefficient for the aero wheels at 0.908 *C*_{1}. That seems nice. But let’s see how much it would matter for mere mortals.

The cyclist in the video had a speed of about 41 kph (11.39 m/s, or 25.48 mph). What if a more ordinary human rides at 30 kph (8.33 m/s or 18.64 mph)? What kind of power savings would that person see? Granted, I don’t know the drag coefficient nor the cross sectional area. Let me call all of this (along with the density) some value *K*. With his values for speed and power, I get a K (normal wheel) value of 0.256 kg/m.

For a normal human riding at 8.33 m/s (again, 18.64 mph, with the same bike and size as the reviewer), I get a power requirement of 148.0 watts. If I decrease the K value to 0.908*0.256 kg/m = 0.232 kg/m, I get a power requirement of 134.1 watts. At this slower speed, you see a power savings of 13.9 watts compared to the high performance savings of 35 watts.

But what about the total energy used? If you look at the test in the video, the guy rode at 379 watts for 20 minutes. This is a total energy use of 4.5 x 10^{5} Joules (107 food calories). For the longer ride he was at a lower power of 344 watts but for 51 minutes. This is total energy of 1.05 x 10^{6} Joules (251 food calories). I find it interesting that humans aren’t limited by total energy output as much as how quickly they use that energy. But still, he burned just a candy bar worth of energy.

Which brings me back to the original question: Are these wheels worth it? Well, that clearly depends on how much value you put in both speed and in dollars. These wheels, which, according to a quick Google search, cost as much as $3,400, really make a difference only if you routinely achieve high speeds for long periods. If so, these wheels (or wheels like them) may provide the slight edge that brings victory. If not, you might as well save that money for candy bars.