If a device is rotating at a steady rate or is stationary, the torques applied to it must add up to zero -- any torque applied at one point must be taken off at another. So, with a bicycle crankset, the torque applied at the pedals is equal and opposite that taken off by the chain, neglecting the small amount lost in friction.
Let's look at a bicycle drivetrain starting with the cyclist's feet. Torque is conveyed from a pedal through the crank -- and from the left crank, then also through the bottom-bracket spindle -- to the chainwheel. Generally, the cyclist's rising leg applies a light torque opposite that of the descending leg. The chain, at the chainwheel, produces a torque equal and opposite the sum of the torques applied at the pedals.
Let's put some numbers to this We assume that the left leg is descending in mid-stroke, applying a force of 100 pounds directly downward on the left pedal, while the rising right leg is applying a reverse force of 10 pounds to the right pedal. Crank length is usually given in millimeters, but we're using English measurement here, as it's more familiar to most English-speaking readers. We'll assume 0.56 foot (170 mm) cranks.
The torque on the bottom-bracket spindle is 56 pound-feet: the 100-pound force at the pedal, times the 0.56 foot length of the crank. The torque at the chainwheel is slightly less, 50.4 pound feet, after we subtract the -5.6 pound-foot torque from the right pedal.
Now, let's assume a 50-tooth chainwheel. This has a radius of about 4 inches, or 0.33 feet. We can now calculate the chain tension:
50.4 pound-feet/0.33 feet = 153 pounds.
We'll assume a 20-tooth sprocket, with a radius of 1.6 inches -- 0.13 foot at the rear wheel. The 153-pound chain tension produces a 20.2 pound-foot torque at this sprocket:
153 pounds * 0.13 foot = 20.2 pound-feet.
(We could also simply note the ratio of tooth counts of the chainwheel and sprocket, 50/20, which would give us the same torque ratio.)
We'll also assume a 28-622 rear tire, with a nominal diameter of 26.76 inches and radius, 13.38 inches, or 1.11 feet.
We can now calculate the drive force where the rubber meets the road:
20.2 pound-feet / 1.11 feet = 18.2 pounds.
The ratio of the force at the pedals to that at the road is 90 : 18.2, or 4.95.
In these calculations, we neglect forces which would not contribute to torque: pedal force not in the direction of rotation, and the weight on the rear wheel. We also neglect friction, which reduces the drive force by a few percent. Numbers are rounded -- close, but not exact.
The ratio of the force at the pedals to drive force at the road is the
gain ratio, which can be calculated more simply as the ratio of road speed to pedal speed, like this:
.(1.11 feet /.56 feet) * 50 teeth/20 teeth = 4.95
This calculation is simple and elegant, but our torque calculations give us some additional useful results:
- the torque on the bottom-bracket axle
- the effect of the reverse force from the cyclist's rising leg,
- and the tension on the chain.