There’s a lot happening at the start of an Olympic BMX race. Athletes begin at the top of a ramp, which they descend while pedaling and being pulled by gravity. At the end of the ramp, they transition from pointing down to aiming horizontally. You may not think there are many physics problems here, but there are.

### How fast would you go if you didn’t pedal?

One claim about Olympic BMX is that the riders descend the ramp in two seconds at a speed of about 35 mph (15.6 m/s). What if you simply rolled down the incline and let gravity accelerate you? How fast would you be going? Of course, this question depends upon the dimensions of the ramp. An official starting ramp has a height of 8 meters with dimensions something like this (they aren’t completely straight).

Instead of a bike, I’ve placed a frictionless block at the top of the ramp. If I want to determine the speed of this sliding block at the bottom of the ramp, I can start with one of several principles. However, the Work-Energy Principle is the most straightforward approach. This states that the work done on a system is equal to the change in energy.

If I view the block and the Earth as the system, the only external force is the force from the ramp. This force always pushes perpendicular to the direction the block moves such that the total work on the system is zero. That leaves a total change in energy of zero Joules. In this case, there are two types of energy–kinetic energy and gravitational potential energy.

There are two important points about gravitational potential energy:

- The value of
*y*doesn’t really matter. Since the Work-Energy Principle only deals with the change in gravitational potential energy, I only care about the change in*y*. For this situation, I will use the bottom of the ramp as my*y*= 0 meters (but you could put this anywhere). - Again, the change in potential only depends on the change in height. It does not depend on how far the block moves horizontally. This means that the angle of the ramp doesn’t really change the final speed of the block (but only in the case where friction doesn’t matter).

With this in mind, I will call the top of the ramp position 1 and the bottom position 2. The Work-Energy equation becomes:

Since the bikes start from rest, the initial kinetic energy is zero. Also, the final potential energy is zero since I set my *y* value at zero at the bottom. Here I am using *h* as the height of the ramp and the initial y-value. Now, I can solve for the final velocity (the mass cancels) and get:

Using a height of 8 meters and a gravitational constant of 9.8 N/kg, I get a final speed of 12.5 m/s–slower than the 35 mph as stated above. Actually, a real bike would have an even lower speed for two reasons. First, a frictional force would do negative work on the system. Second, bikes have wheels that spin. When a wheel spins, it requires extra energy to make these wheels rotate such that some of the change in gravitational potential energy would be used for rotation instead of translation.

### How much power would it take to start a bike?

Let’s say you have a bike that would reach 10 m/s on its own simply rolling down the ramp. Where does the other 5.6 m/s come from to get up to the starting speed of 35 mph? The athlete. We can fix this by adding another type of energy change in the Work-Energy equation: chemical potential energy. This would be an energy decrease in the person when muscles are used. I can write this as:

Here I am labeling the gravitational potential as *U*_{g} and the chemical potential as *U*_{c}. Putting this all together, I get:

Since the new velocity at the bottom is to be greater than the previous time, the change in chemical potential energy will be negative (which makes sense since the human is using muscles). Using a final speed of 15.6 m/s and a mass of 80 kg (for the rider plus the bike) I get a change in chemical potential energy of 3,462 Joules.

But what about the power? We can define power as the rate that energy changes.

In this case the change in energy is the decrease in chemical potential energy–but what about the time? If I assume a constant acceleration of the bike, then I can calculate the average velocity while on this ramp:

The average velocity also is defined as:

If ?x is the distance down the ramp (the length of the ramp), then I can put this all together to solve for the time interval:

With this and my expression for the change in chemical potential energy, I can calculate the power:

With a ramp length of 20 meters and a final speed of 15.6 m/s, I get an average power of 135 watts. Of course, this is the best-case scenario and also a value for the average power. The actual average power could easily be higher for a number of reasons other than frictional forces. The biggest reason for an increase in power would be the speed. If you have a slightly higher final velocity, this can be a significantly higher kinetic energy (because the velocity is squared). This higher speed would also mean that it takes less time to get to the bottom of the ramp. Put these two factors together and you quickly get crazy high power requirements.

### How many G’s would you pull at the bottom of the ramp?

I drew the ramp with a sharp bottom. Of course, this isn’t how anyone makes an official ramp. The Olympic ramp is curved at the bottom, with a radius of curvature of 10.02 meters (if I’m reading the diagram correct). Why would this circular ending to a ramp cause the bike to accelerate? It has to do with the real definition of acceleration:

In this equation, both the acceleration and the velocities are vectors–this means the direction matters. So, even if you are traveling at a constant speed but change direction you accelerate. This is exactly what happens at the bottom of the ramp:

I’ll skip the derivation for the acceleration due to circular motion (but you can see a more detailed explanation in my ebook – Just Enough Physics). This acceleration would depend on both the radius of the circle as well as the velocity. We call this centripetal acceleration:

Since I already know the velocity (15.6 m/s) and the radius (10.02 m), I can easily calculate the acceleration at the bottom to have a value of 24.3 m/s^{2}. This is an equivalent acceleration of 2.5 G’s–but since we are already at 1 g, you could say this would result in 3.5 G’s (honestly, I’m not sure of the proper G-force convention).

How would you make this acceleration even larger? There are two ways: increase the speed or decrease the radius of curvature. But be careful. If you get an acceleration too large, it’s going to start breaking bikes and maybe even people.