Why does a large drive sprocket require more power to accelerate at the same rate as a smaller drive sprocket if the overall gear ratio is the same (with all other components such as the chain the same between systems)?

As an example, we could compare 11/33 to 13/39 (same 1/3 ratio). Why does the 13 tooth sprocket need more power to accelerate at the same rate as the 11 tooth setup?

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Can we agree that the top sprocket will need more torque to accelerate an atv at the same rate as the same atv with the bottom sprocket? And assuming the overall gear ratio is the same.

The moment of inertia differences of the sprockets are insignificant. Assume the moments of inertia are equal.


Let's assume the sprocket is a disk, then its moment of inetria is given by:

$$ I = \frac{mr^2}{2} $$

where $m$ is the mass of the sprocket and $r$ is it's radius. At some angular velocity $\omega$ the kinetic energy of the disk is:

$$ E = \frac{1}{2} I \omega^2 = \frac{mr^2 \omega^2}{4} $$

So the energy required to accelerate the disk to some angular frequency $\omega$ is proportional to its radius squared, and therefore the power required to maintain a constant angular acceleration is proportional to $r^2$. If we consider just the sprocket, i.e. power is consumed nowhere else, then the 13 tooth sprocket will take about 40% more power than the 11 tooth sprocket for the same angular acceleration.

But, assuming this is related to your previous question, the power used in acelerating the sprocket is likely to be a tiny fraction of the power used to accelerate the whole vehicle. If you post the weight of the sprockets (all four of them) and the vehicle we can work out that fraction.

  • $\begingroup$ The weight of the larger sprocket = the weight of the smaller sprocket. We can even assume that the sprockets are machined such that the angular acceleration (without a chain) would be the same per unit of power applied. Yet there is still a difference in acceleration. Why is that? $\endgroup$ – Randy Aug 19 '13 at 6:18
  • $\begingroup$ @udiboy If the weight on earth is the same, wouldn't the mass be the same? Also, if the angular acceleration is the same, then the mass isn't important. The only thing that changes here is the diameter of the sprockets and the mass of the chain (longer chain for taller sprockets). $\endgroup$ – Randy Aug 19 '13 at 9:28
  • $\begingroup$ @Randy, I'm really really sorry... I left that comment in a hurry, and so I didn't complete it. What I meant to say was "weight(mass in fact) and moment of inertia are two very different concepts". Sorry again. $\endgroup$ – udiboy1209 Aug 19 '13 at 11:02
  • $\begingroup$ @Randy: the moment of inertia depends on the size as well as the weight. The larger sprocket would have to weight 40% less than the smaller to have the same moment of inertia. $\endgroup$ – John Rennie Aug 19 '13 at 14:01
  • $\begingroup$ What if the mass of the larger was concentrated near the hub while the smaller had its mass concentrated at the perimeter? $\endgroup$ – Randy Aug 19 '13 at 14:11

This doesn't take into account that with increased sprocket size you increase the leverage of the pull of the sprocket. To say it takes 40% more hp to go from a 11t to a 13t is ludacris. If you go 11/33 to a 13/39 on a motorcycle, this will act totally different. I'm not sure if you drag raced bikes, rode flat track, or motocross, but when you take into the account the weight of the bike the size of the wheels, the three links added weight is almost zeroed out.


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