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I was riding my 14-speed road bicycle and something interesting occurred to me. When in the lower gears, a steady pedaling of ~60 rpm can only get 8-9 mph, however, in the highest of gears, that same cadence in the higher gears might be 20-25 mph.

Of course some of that energy goes into deforming the road, making noise, my body radiating heat, dissipation due to air resistance, you name it. But probably most of it goes into my motion. And since I have the same cadence in each case, I should get roughly the same power output.

So how does this fit into the law of energy conservation?

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  • $\begingroup$ "Since I have the same cadence in each case, I should get roughly the same power output" - Power depends on two factors: velocity and force. Are you pushing harder on the pedals to get the same cadence at higher gears? $\endgroup$ – probably_someone Jul 3 '18 at 1:20
  • $\begingroup$ Slightly, yes, but for my mass about ~80 kg that amounts to about one calorie (4350 J). Seems like that cannot be compensated by the extra force on the pedals. $\endgroup$ – zhutchens1 Jul 3 '18 at 1:24
  • $\begingroup$ How are you doing that calculation? Calories and J are units of energy, rather than force. $\endgroup$ – probably_someone Jul 3 '18 at 1:26
  • $\begingroup$ Yeah, I know that. For the two cases $\Delta T = 0.5 m (v_f^2 - v_i^2)$ for a difference in energy of ~1 calorie for the two states. For $F\Delta r$ work I feel the extra force for the higher gear is so minimal that it doesn't account for that energy difference. $\endgroup$ – zhutchens1 Jul 3 '18 at 1:35
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    $\begingroup$ Ah, here's where the confusion lies. The power you expend on a bicycle to maintain speed is not equivalent to your kinetic energy per second. It's equivalent to the energy dissipation rate from things like axle friction, air resistance, and deformation of the tires. The magnitude of the dissipation rate is not necessarily strongly related to your kinetic energy. After all, according to Newton's First Law, to maintain a certain speed (and thus a certain kinetic energy) you only need to make sure the net force on you is zero. The force on the pedals is just balancing out dissipation. $\endgroup$ – probably_someone Jul 3 '18 at 1:42
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The power you expend on a bicycle to maintain speed is not equivalent to your kinetic energy per second. It's equivalent to the energy dissipation rate from things like axle friction, air resistance, and deformation of the tires. The magnitude of the dissipation rate is not necessarily strongly related to your kinetic energy. After all, according to Newton's First Law, to maintain a certain speed (and thus a certain kinetic energy) you only need to make sure the net force on you is zero. The force on the pedals is just balancing out dissipation.

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For the same gradient road, at the same speed, at two different gear settings, power required (energy input per unit time) is the same. The gear friction will not change much with modern efficient bicycles (Ref). Why does power remain unchanged?

Power equals Force x velocity:

$P = Fv\tag1$

or in the case of rotation:

$P = \tau \omega\tag2$

Where $\omega$ is rotational speed on the axis you are pedaling on. As cadence increases, i.e. pedaling side sprocket is smaller, the $\omega$ on the pedaling side increases, but required torque $\tau$ decreases proportionally, therefore power remains the same.

If, however your speed increases, air resistance is bound to be a factor, so going faster will require torque $\tau$ to decrease slower than rotational speed increase, thus increasing power input. This is because torque $\tau$ is related to required forward force by:

$\Delta\tau = r_{w}g_r\Delta F_f/r_s \tag3$

where $r_w$ is the radius of the driven wheel, $g_r$ is the gear ratio between the driven wheel sprocket and the pedal sprocket radius at which one is pedaling, and $F_f$ is the force required to move the bicycle forward against your mass, parts friction and air resistance, $r_s$ is the radius from the pedal to the centre of the driving sprocket. Power will therefore increase at the rate that torque fails to keep up with pedal sprocket rotational speed increase.

If you cycle uphill, but maintain the same speed as on a flat road, you will have to expend more power, why? You are gaining gravitational potential energy. For the same bicycle speed and same cadence, in addition to the power required to move the bicycle across a flat road, you are lifting your body and the bicycle to higher ground, so you will have to use more power. This will be manifest in the higher torque input into the pedals in equation $2$ above.

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Having a fixed cadence does not mean having a fixed power output. The power output depends on both the cadence and the force you apply to the pedals. In a high gear and a fixed cadence, you will have to push harder against the pedals to maintain your speed, while in a low gear you won't have to push hard. This is why it "feels easy" to ride in a low gear.

Of course, I second the points made by probably_someone in their answer as well.

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It is actually all about momentum and energy transfer. In your bicycle you will find a spiky disc kind of thing which the chain rotates when you pedal. If you see a similar thing is there on the back wheel. But when you change the gear you actually change the radius of that disc. And note that maximum energy transference is required (to produce same velocity or momentum) when any two things or objects are identical. See picture.enter image description here

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    $\begingroup$ Can you crop the picture a bit? $\endgroup$ – user191954 Jul 3 '18 at 3:57

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