# Power consumption / energy expenditure to maintain constant velocity while riding a bicycle at $1\times$ velocity vs $2\times$ velocity

Consider the case of someone riding a bicycle on an even horizontal surface at a constant velocity. As they are riding at a constant velocity, there is presumably a constant energy expenditure (which I will quantify by the power) to maintain that velocity.

For the sake of the question, ignore air resistance, but assume friction (between the ground and wheels and within the gears) exists.

Now my question is, what is the relative power consumption of someone riding at velocity $$2v$$ versus someone riding at velocity $$v$$? Do you need to input 4x the power or 2x the power? I have a hard time telling whether it would be 2x or 4x or something else.

## My Attempt

I know power is $$P = Fv$$ where $$F$$ is force applied and $$v$$ is velocity. As the rider turns the pedals, they are applying a torque $$\tau_{1} > 0$$. Assuming the force they apply is constant and the point on the pedal at which they apply it is the same, the torque $$\tau_{1}$$ ought to be constant for a constant velocity ride.

Now at constant velocity, there is some counter-torque $$\tau_{2} < 0$$ due to the friction within the gears of the bike, and presumably the two torques cancel each other out to obtain equilibrium: $$\tau_{1} + \tau_{2} = 0$$.

Now here is where I'm a bit lost: Does the counter-torque $$\tau_{2}$$ depend on the velocity or is it independent of velocity?

1. If $$\tau_{2} = -\beta v$$ ($$\beta$$ is some constant), then the power at 1x velocity is $$P_{1} = \tau_{1}/r \cdot v$$. Given $$\tau_{1} + \tau_{2} = 0$$, we find $$\tau_{1} = \beta v$$ so then $$P_{1} = \beta/r \cdot v^{2}$$. At double the velocity, we find $$P_{2} = \beta/r\cdot 4v^{2}$$. Hence $$P_{2}/P_{1} = 4$$.
2. If $$\tau_{2} = -\beta$$ ($$\beta$$ is some constant), then the power at 1x velocity is $$P_{1} = \tau_{1}/r \cdot v$$. Given $$\tau_{1} + \tau_{2} = 0$$, we find $$\tau_{1} = \beta$$ so then $$P_{1} = \beta/r \cdot v$$. At double the velocity, we find $$P_{2} = \beta/r\cdot 2v$$. Hence $$P_{2}/P_{1} = 2$$.

Now it seems to make sense that you would be using 2x power at 2x velocity, but I also feel as though it would be harder to push on your pedals when you ride at twice the velocity, leading me to think you would be using 4x power at 2x velocity. Which is the correct answer?

Some Notes:

• As @Solomon Slow pointed out, "On a flat road, aerodynamic drag is by far the greatest barrier to a cyclist's speed, accounting for 70 to 90 percent of the resistance felt when pedaling," so the assumption that air drag is negligible in real life scenarios is false. See https://www.exploratorium.edu/cycling/aerodynamics1.html. I suppose introducing air drag would introduce a quadratic drag term, which I left out of my analysis. I do wonder if there is a speed range in which it is negligible nonetheless.
• According to studies into how many calories are burned by cycling, the relationship seems generally to support the heuristic $$P\propto v$$ as opposed to $$P\propto v^{2}$$. See this page, this page, and this calculator.
• Note that there are other factors at play (such as whether the road is sloped or bumpy, how much wind there is, etc.) in real-life that would dispute any simple relationship, but my question is about which approximate relationship is more accurate.
• "As they are riding at a constant velocity, there is presumably a constant energy expenditure (which I will quantify by the power) to maintain that velocity". That is only true because of the need to overcome opposing dissipative forces (mechanical friction, air resistance, etc.) Mar 28 at 21:34
• @BobD True. I appreciate you highlighting this point. In my post, I am including mechanical friction but ignoring air resistance, and my question rests on asking, what is the relationship between that mechanical friction and velocity? Mar 28 at 21:51
• "Ignoring air resistance." At you peril! "On a flat road, aerodynamic drag is by far the greatest barrier to a cyclist's speed, accounting for 70 to 90 percent of the resistance felt when pedaling." exploratorium.edu/cycling/aerodynamics1.html Mar 28 at 22:25
• @SolomonSlow Interesting, I didn't think that was the case. I wonder how that changes things. Mar 28 at 23:52
• @SolomonSlow I am curious if that becomes non-negligible only for high speeds, or if it applies for all speed ranges. In any case, thank you for bringing this to my attention. Mar 29 at 7:35