Force Applied To Pedal on Bike

[ollie_craw]

Homework Statement

I think that the equation for the force someone exerts on their bike pedals or the force forward is stated below. can anyone confirm this is true?

Homework Equations

Fapplied by rider = (weight of rider) (Radius Crank /Radius Tire) (Gear Ratio)

The Attempt at a Solution

 

[andrewkirk]

It's broadly correct. There are some nuances though.

1. Gear Ratio in the above formula, for a derailleur bike, needs to be the number of teeth on chainring divided by number on rear sprocket. In cycling, ratios are often quoted as 'rollout distances', which is the distance traveled for one turn of the cranks. Such a ratio should not be used in the above formula since it already contains the tyre radius, as well as being multiplied by $2\pi$.

2. The force applied by the rider is unlikely to be their full weight, because some weight may be borne by the seat and the other pedal.

 

[ollie_craw]

It's broadly correct. There are some nuances though.

1. Gear Ratio in the above formula, for a derailleur bike, needs to be the number of teeth on chainring divided by number on rear sprocket. In cycling,
ratios are often quoted as 'rollout distances', which is the distance traveled for one turn of the cranks. Such a ratio should not be used in the above formula since it already contains the tyre radius, as
well as being multiplied by $2\pi$.


2. The force applied by the rider is unlikely to be their full weight, because some weight may be
borne by the seat and the other pedal.

You're incorrect on the gear ratios part. It needs to be used in the equation and it is the rear sprocket divided by the front chainring

 

[ollie_craw]

It's broadly correct. There are some nuances though.

1. Gear Ratio in the above formula, for a derailleur bike, needs to be the number of teeth on chainring divided by number on rear sprocket. In cycling,
ratios are often quoted as 'rollout distances', which is the distance traveled for one turn of the cranks. Such a ratio should not be used in the above
formula since it already contains the tyre radius, as
well as being multiplied by $2\pi$.


2. The force applied by the rider is unlikely to be their full weight, because some weight may be
borne by the seat and the other pedal.

Just realized I'm wrong. What equations would I need to have acting against the bike? For the force of gravity would it just be mass times g times sin of the angle?

 

[CWatters]

Your approach isn't necessarily wrong. It depends what you are trying to calculate?

Its not easy to calculate all the forces acting on a bike unless you have data on rolling resistance and drag.

Yes the force due to gravity is mgsin(angle of slope).

 

[ollie_craw]

Your approach isn't necessarily wrong. It depends what you are trying to calculate?

Its not easy to calculate all the forces acting on a bike unless you have data on rolling resistance and
drag.

Yes the force due to gravity is mgsin(angle of slope).

Would the force applied also be times the sin of the angle if you're going uphill?

 

[CWatters]

That's the angle I assumed you were referring to earlier.

Can I suggest you make a diagram?

 

[Nidum]

https://www.physicsforums.com/threads/biking-up-a-hill-in-different-gears.872222/#post-5476399

http://www.real-world-physics-problems.com/bicycle-physics.html

 

[andrewkirk]

You're incorrect on the gear ratios part. It needs to be used in the equation and it is the rear sprocket divided by the front chainring

Yes, I hadn't noticed that you were multiplying by the gear ratio rather than dividing by it. Gear ratios on bikes are usually quoted as a ratio of distance rather than ratio of force, so one would define gear ratio as teeth on chainring divided by teeth on rear sprocket, which is the reciprocal of the mechanical advantage.

I don't know if this is still the practice, but when I was racing in the seventies one referred to the gear ratio in a typical top gear (back then, when one had only 2x5 = 10 speeds) of 52 x 13 as 108 inches, which was 52 teeth (on chainring) divided by 14 teeth (on rear sprocket) times 27 inches (diameter of wheel). This then had to be multiplied by pi to get the rollout distance. In Europe the practice was to quote the rollout distance, so it included the pi factor and was expressed in metres.

The angle of the hill won't make any difference to the ratio of circumferential force of foot on pedal to force of tyre on road. That ratio is completely determined by the geometry and gear train. However the angle of the hill may affect the rider's position on the bike, which may change how much of their weight is applied to the pedal. For instance on a very steep slope, one tends to stand, so that more of one's weight goes onto the pedal. There's also the issue of the angle between the force applied to the pedal and the circumferential direction, as it's only the component of the former in the latter direction that transmits to the rear wheel. That angle will vary with the position of the crank, but there will be one spot in each revolution at which the two directions align. That's the position of maximum power and component manufacturers have sometimes constructed elliptical chainrings that vary the gear ratio as the crank rotates, in order to even out the delivery of power. Pro riders sometimes use them in time trials but they've never really caught on in a big way.

The angle of the hill will however determine what proportion of the force of tyre on road goes towards lift and what goes towards acceleration and countering drag. If the rider is going slowly so that drag is small, the steeper the hill, the smaller the proportion of push goes towards acceleration. That's why when one is climbing a steep hill slowly, one can be pushing hard on the pedals but not accelerating.

 

[haruspex]

the angle of the hill may affect the rider's position on the bike, which may change how much of their weight is applied to the pedal.

No, the rider can always apply full weight to the pedal, regardless of the hill. But it is not just a matte of weight. The rider can pull upwards on the handlebars to increase the downward force on the pedals.

Without cleats, full torque is only achieved when the pedals are in the horizontal position. To get uphill, the cyclist has to achieve sufficient torque averaged over a full turn of the pedals. The easiest way to figure this out may be in terms of energy. How much work can the cyclist do by transferring weight from pedal to pedal in one rotation of the crankshaft?

With cleats, torque is also increased by pulling upwards with the lower foot. When the pedals are in the vertical position, some torque may still be achievable by pushing forwards with the upper foot and backwards with the lower. However, that implies a torque about a vertical axis, which the rider needs to derive from somewhere else. The handlebars might supply some of this, but that's limited.