# Forces Applied to Bicycle Pedals

1. Nov 26, 2007

### PhilCar

Hi, I’m not an academic and I would like clarification on the following questions. I have searched your forum and the web in general for the answer in order to avoid unnecessarily bothering anyone but was unable to find the answer.

Here are my questions.

Barring minor influences of friction and other minor forces and excluding acceleration and deceleration, for a given bicycle gear setting (front chainring and rear wheel cog) and a given crank arm length, does the pressure applied to a pedal (the force applied with each pedal stroke) vary with cadence if the profile of the road were to remain constant?

For example. Let’s say a rider is using 39 tooth chainring and a 21 tooth cog on the rear wheel and is riding on a constant ratio climb. If they ride at a cadence of 60, there is a certain force applied to each pedal stroke. If they ride the same hill, in the same gear but with a cadence of 90, is the amount of force applied to each stroke the same? I am aware the total amount of power applied would be greater with the higher cadence.

Essentially, aside from differences in the force applied to the pedals during acceleration or deceleration, the force applied to the pedals for a fixed gear ratio with fixed crank lengths and on a fixed ratio climb, for a particular rider and bike combination (implying their overall weight), is a fixed value, regardless of the cadence they are peddling at. Is that statement essentially true or false?

When I refer to the force applied to the pedals I am assuming the cyclist’s technique is the same regardless of their cadence.

I hope I have been clear enough

Cheers,

Phil

2. Nov 27, 2007

### rcgldr

True. If aerodynamic resistance is taken into account the cadence at 90 will require more force to maintain, because of the increase in drag force.

3. Nov 27, 2007

### kokophysics

If you don't consider the friction forces and excluding accelerations, the force is not only the same at any cadence, but is 0. Think about Newton's first law.

4. Nov 27, 2007

### PhilCar

Good point. I guess I need to clarify.

Thanks Kokophysics and thanks Jeff. Something I left unstated in my original post is that I am really only talking about hill climbing and I am not excluding the force of gravity. Therefore the force applied with each pedal stroke is whatever is required, greater than 0, to both counter the force of gravity and provide forward motion.

I'm only concerned with the significant forces at play here and assuming that other relatively minor forces, such as greater wind resistance due to higher speed (when speeds are quite low while climbing steeply), are virtually insignificant. For example, peddling at a higher cadence would likely make the whole pedal stroke (down, back, up and forward) more balanced and therefore more efficient. That is also something I'm intentionally ignoring here.

Under those circumstances, do you now find the statement to be "essentially" true or false?

Cheers,

Phil

5. Nov 27, 2007

### Staff: Mentor

The force will be roughly the same regardless of the speed for low speed hill climbing.

That said, biomechanics may make it more difficult to pedal faster - more force is required to move your legs faster.

6. Nov 27, 2007

### kokophysics

Yes, it would be the same.

The faster you peddle the faster you gain energy (both potential and cinetic) and you gain energy at the same rate you gain velocity. So the faster you peddle, the faster you will ascend and more energy you will win. But this extra energy you win is ONLY provided by the velocity of your peddling which varies, not the force you apply which remains constant.

In other words, as all the power you need to win energy and the power you provide is directly proportional to the velocity, the force remains the same (If you don't change the technique of peddling!!!).

Think that you can be stand up over your bike and peddllig with no hands, so the only force you can provide to the pedals is your own weight and actually you can go faster.

7. Nov 27, 2007

### PhilCar

Thanks again Kokophysics. Your last statement captures the essence of what I was believing was the case in a nutshell. Although it sounds potentially painful so I won't try it! :)

Cheers,

Phil

Last edited: Nov 27, 2007