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I'm less interested in the empirical answer to this question and more interested in the theoretical physical dynamics that govern this.

I understand that the rider applies some force to the pedals (which could also be thought of as a torque given the length of the crank arm),but the answer isn't simply F=ma because there are gearing ratios involved. In short, I'd like to know the process of calculations by which one could approach this problem and determine the maximum acceleration of a rider.

I suspect some assumptions will be necessary:

  • Max force applied to pedals = rider weight (this is probably not quite correct if using clipless pedals but probably fairly close?)
  • Optimal gearing ratio is available
  • Let's assume a relatively instantaneous problem regime, i.e. a 2-second standing start or something like that

If more information or assumptions are needed, please advise. I'm also open to migrating this to say SE physics if that's a more appropriate place for it.

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    One significant factor is geometry. Accelerating too swiftly can produce a "wheelie" where the front tire rises off the ground, and, in the most extreme case, where the bike flips over. Commented Feb 26, 2021 at 0:25
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    @DanielRHicks with the gearing and posture one would certainly be using for optimal & sustained acceleration, it'll be difficult to produce enough torque to meaningfully raise the front wheel
    – Paul H
    Commented Feb 26, 2021 at 0:42
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    would this be a better fit on the physics SE?
    – GageMartin
    Commented Feb 26, 2021 at 1:08
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    @GageMartin seems reasonable. One of the remits of Bicycles.SE is "practical, real world problems" and this question seems to be purely theory with both assumptions and disregard of practical considerations.
    – Criggie
    Commented Feb 26, 2021 at 2:22
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    Max force applied to pedals is much higher than the rider weight and it has more to do with pushing on the handlebars than with clips or clipless pedals although they do play a role. Commented Feb 26, 2021 at 8:36

3 Answers 3

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The power-drag equation is well-understood if not widely-known, and is discussed in this bicycles.SE answer. In particular, the drag equation shows the component for changes in kinetic energy, viz., acceleration. You're asking about a standing start, so initial speed is low and we can ignore the aerodynamic drag component. In addition, let's assume the rider and bicycle is on a flat surface, so we can also ignore the potential energy component. Finally, although there is a rolling and mechanical resistance component to drag, it is typically small so let's ignore it here.

What's left is the kinetic energy component. F = m a is the correct equation to use, but you have to convert F at the pedal to F available at the rear wheel. Then a = F / m. There are a series of lever arms between the pedal and the point at which the rear wheel touches the ground involving the length of the crank arm, the gear ratio, and the radius of the rear wheel. Let's say the total lever arm "gain ratio" is c. Then F Newtons of force at the pedal translate to F / c Newtons at the rear wheel. If the total mass of bike and rider is m, then the acceleration of the bike and rider is F / (m * c). Martin et al. (2006) uses this technique to model sprint performance, with very good results.

Maximum pedal force can actually be greater than body weight, if you can hold onto the handlebars. In fact, maximum pedal force approaches 1RM ("1 rep max") force from a leg press. A world-class track sprinter has a 1RM of roughly 4x body mass; more commonly, most of us can produce pedal forces equivalent to at least twice body mass. So, depending on the gain ratio calculated above, maximum acceleration is likely to be somewhere in the ballpark of 0.2g to 0.4g, or between roughly 2 and 4 m/s^2.

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    One thing which confuses me: If I can squat or leg press twice my own body weight, wouldn’t that mean that I can accelerate at 2g, which is obviously not the case?
    – Michael
    Commented Feb 26, 2021 at 8:39
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    @Michael g is close to 9.8 m/sec^2 in the direction of the center of the Earth. We're translating that into acceleration orthogonal to that, so do the calc in Newtons and deflate by what I called 'c' to get force available at the rear wheel (which will vary by crank length, gear ratio, and wheel radius). Then divide that by mass. There is a small amount of rolling resistance that dissipates some of the available force, but it's relatively small since Crr is small.
    – R. Chung
    Commented Feb 26, 2021 at 15:54
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    I was curious about tire slippage under max acceleration, but an experiment with dynamic friction coefficients looks like you'd be fine. Static friction from a standing start would be even higher. Commented Feb 26, 2021 at 20:56
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enter image description here

This image depicts a SWAG at the line between the center of gravity of cyclist + bike and the center of force propelling the bike. (Actually, the center of force would probably be the point of contact for the rear tire, but I'd already drawn this picture when that occurred to me.)

Force will be divided between that used to propel the bike forward and that used to rotate the bike+rider upward. The upward force would be proportional to the sine of the angle relative to ground, which I make out to be around 60 degrees in picture, yielding a sine value of around 0.87. The forward force would be proportional to the cosine of the angle -- about 0.5.

To keep the front tire on the ground it's necessary to keep the upward force below 1g.

So, if the upward force needs to stay below 1, the forward force needs to stay below (1/0.87) times 0.5, or about 0.57.

So the maximum practical acceleration would be about 0.57 times G, or around 5.8 m/s/s.

I suspect that bikes designed for "drag racing" have a longer rear triangle and a more prone rider position to reduce depicted the angle (to maybe 30-45 degrees) in order to make higher acceleration practical.

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    This reasoning doesn't really apply for human-powered bicycles, because the available power is much, much less than what would be needed for lifting the front wheel.
    – ojs
    Commented Feb 28, 2021 at 13:25
  • @ojs - So no one's ever done a wheelie?? Commented Feb 28, 2021 at 18:16
  • Yes, the trick is to shift weight back and pull on the bars. You don't even need to pedal to lift the front wheel that way.
    – ojs
    Commented Feb 28, 2021 at 18:44
  • @ojs - Runners have been clocked with an acceleration nearing 10 m/s/s. Commented Feb 28, 2021 at 19:02
  • Yes. Sprinters start from a crouched position, have a starting block to push against with both legs, don't maintain that acceleration for long and most importantly, don't have gears. I guess with low enough gear one could pull off the unwanted wheelie, but by the time they have switched to higher gear the competition who started with higher gear will be far away.
    – ojs
    Commented Feb 28, 2021 at 19:52
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Assuming that the bike is built for the purpose of achieving maximum possible instantaneous acceleration, there is exactly one limiting factor: The grip of the rear tire on the road.

Force on the pedals is not a limiting factor because you allow for optimal gearing. In theory, you could start with a vanishingly small gear from resting, and thus put an arbitrarily high torque on the rear wheel.

Geometry is not a limiting factor because you can build the bike to have the rear wheel in just the right place, such that the entire weight is on the rear wheel during the acceleration.

But the material of the rear tire and the road surface determine how much horizontal force can be transmitted with a given weight.

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