# Bicycle Transmission Physics

Bikes are pretty simple machines. Completely enjoyable without special mathematical treatment. But if you’re an obsessive engineering-type geek like me, you’ll feel better when you have a little math to go with all that exercise.

This is a real simplified run through power transmission in a bicycle as I understand it. The numbers check out as do the units of measure. Somebody call bullshit if you see it.

### Definitions

First, some definitions. These are just to clarify the physics meaning of these words.

Power – The rate at which work is done. If you bike on a flat road at constant speed, then your power output was constant for the whole trip.

Work – Power integrated over time. Power multiplied by time if power was constant

Force – A force acts on a body to produce an acceleration.

Torque – A measure of a force’s tendency to produce rotation. Think about a wrench. If you used a wrench that was 1 foot long and applied 25 lbs of force on it, then the torque on the nut is 25 ft*lbs. Double the distance, double the torque.

### Assumptions

Let’s assume that a random dude is driving a bike down the road. The dude weighs 200 lbs and he is driving an old-school ten speed. Let’s also assume that the losses of the bicycle mechanics are small and ignorable (called negligible in the business).

We’ll ignore the wind. Suffice to say that wind resistance increases the power required to move the bike. Quantifying this value is difficult because it depends on a lot of factors. We’ll also ignore rolling resistance. Just know that narrower, higher PSI tires give less rolling resistance.

### Pedals and Crank

Force is applied to pedals. If random dude is standing on the pedal and the pedal is parallel to the flat road, then he’s exerting 200 lbs of force downward on the pedal. Really, you cannot apply your full load into the pedals at all times and the average force will be less, but we’re simplifying. Also, if the pedal is not parallel, then only a trig ratio of the force on the pedal is producing rotation.

The crank is fastened to the bike chainring. The cranks job is to transform the force into a torque. The torque is then applied to the chainring which pulls the chain (no duh).

The torque on the crank is 111.5 ft-lbs if the crank is 170mm or 6.7 inches

200lbs * (6.67 inches/12 inches/ft) = 111.5 ft*lbs

### Chain

To calculate the force and torque delivered by the chain, you have to know these three things.

1. That power is equal everywhere in the system and it is calculated by rotational velocity multiplied by torque.
2. By using the same chain on the front and back gear, we know that the ratio of front and rear gear rotational velocity is the same as the ratio of teeth on the gears.
3. The ratio of torque is inversely proportional to the gear ratio. It has to be to make the power equal in the front and back even though the speed of the front and back gears is different.

### Freewheel

The torque applied to the rear gear is torque on the front gear times the inverse of the ratio of front and rear gears. Assuming a 34 tooth rear gear:

111.5 ft*lbs * 34 teeth/42 teeth = 90.26 ft*lbs

The torque delivered to the rear gear is about 81% of the front chainring while rotational velocity of the freewheel is 1/81% = 123%.

Speed vs. torque is the trade off that is constantly made by switching gears. Remember the power delivered to the back wheel is exactly the same as the power delivered to the pedal, just with a different composition of torque and speed.

### Rear Tire

The torque on the rear gear and rear tire is identical. We know this because the speed of the wheel and rear gear is identical as is the power. The force applied to the road (propelling the bike forward) assuming a 29 inch tire is the chainring torque divided by the radius of the rim.

90.26 ft*lbs / 29 inches/12 inches/ft = 37.35 lbs

This lbs measurement is really a measure of force, not mass. It means that the force applied to the road by the back tire is equivalent to the force of gravity on 37.35 lbs.

### Switching Gears

Let’s run the calx again with a 9 tooth rear gear. The torque on the front doesn’t change, so we’ll leave it at 111.5 ft*lbs. The rear torque is changed to

111.5 ft*lbs * 9 teeth/42 teeth = 23.9 ft*lbs

Leaving the rear tire 9.9 lbs of force to apply to the road.

23.9 ft*lbs / 29 inches/12 inches/ft = 9.9 lbs

### Acceleration

You can see the big difference between a low gear and high gear. The power delivered to the front is constant in both cases. Yet 3.77 times more force is delivered to the rear wheel by the lower gear over the top gear (not coincidentally, this is the ratio of the rear gears 34/9 while keeping the front gear constant). How does it translate into movement? I thought you’d never ask.

Surely you’ve heard that F=ma. Force = mass * acceleration. The equation can be rearranged so that

a = Force/mass

To calculate the bike acceleration, we need to know that the earth’s gravity causes falling objects to accelerate at a standard rate = 32.17 ft/second/second. Then apply this to our force calculation.

a = 37.35 lbs * (32.17 ft/s/s) / 220lbs = 5.46 ft/s/s

A pretty abstract measure, but you can make sense of it by multiplying by a number of seconds. The number means that in order to get up to 5.46 ft/s you need 1 second of travel time. To get the bike up to 6 mph (8.8 ft/s) you need (8.8 ft/s) / (5.46 ft/s/s) = 1.6 seconds

Same measurements with the higher gearing

a = 9.9 lbs * (32.17 ft/s/s) / 220lbs = 1.45 ft/s/s

8.8 ft/s / 1.45 ft/s/s = 6.06 seconds

I know that you already knew how to shift gears and that low gear was for low speed and high gear was for high speed. You already knew that it was harder to accelerate in high gear, but now you know exactly by how much.

## 2 thoughts on “Bicycle Transmission Physics”

1. Damien says:

Hi,

A good article, I enjoyed reading it. Could you explain this further for me please – “Also, if the pedal is not parallel, then only a trig ratio of the force on the pedal is producing rotation”

Thanks

2. Richard Vanderplas says:

Thanks Rob:
Showing how torque and rotation rate progresses through the system has helped me a lot. One thing looks wrong to me. Shouldn’t the wheel radius be 14.5 inches for a 29 inch wheel? Also I would show the calculation: mass = weight / gravity = 220 / 32.17 = 6.84, then show: a = F / m = 37.35 / 6.84 = 5.46. ( or, with radius fix: 74.7 / 6,84 = 10.92 ft/s/s )
Richard