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Hopefully, this is the correct forum to ask this. There's lots of stuff online about final drive ratios, but not much with respect to actual acceleration times, and usually it's just anecdotal, butt-dyno kind of stuff.

I understand that increasing the final drive ratio (FDR) will increase torque at the rear wheel and therefore acceleration in a single gear up to a point (because top speed will decrease in that gear), but automobiles/bikes have multi-gear transmissions.

For example, let's say that my bike can reach 60MPH in 1st gear at redline with a stock FDR of 3:1. If I increase it to 4:1, I can accelerate to redline faster, but only to 45MPH, at which point I have to shift to a taller gear. The shift takes time and the gear that you've shifted to is potentially a lower ratio than first gear was with the original FDR. So yes, you got to 45MPH faster, but now your acceleration from 45 to 60 might be slower since you're in second gear. It seems like increasing the FDR just causes you to have to shift more often to get up to speed.

On the other hand, if you lower FDR, you can stay in lower gears for longer and it also has the added benefit of allowing you to cruise at lower RPM (assuming it's not so low that you stall the engine in parking lots or when starting from dead stop).

Basically, my question is, assuming my bike can reach 120MPH with a 3:1 or a 4:1 FDR, will the 4:1 give me a significantly, demonstrably faster acceleration time to 120MPH? I understand I'm using generalities, but that's exactly what I was wondering... in general, would this be the case? Thank you.

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    However, having a shorter final drive also means the engine is revving higher in cruise, which means worse MPG and quadratically worse mechanical wear. Of course you could fit taller top gears to wash this difference out.
    – Harper - Reinstate Monica
    Aug 16, 2022 at 2:37
  • @Harper-ReinstateMonica, right. I was thinking of changing the sprockets on my bike to give a lower ratio for lower revs at highway speed, but I also don't want to compromise acceleration, hence the question. But also, I've never understood why so many people are forever talking about changing their FDR and how much it improved performance. In each gear, sure, the butt-dyno tells you "more torque". But what about actual overall acceleration time?
    – bertmoog
    Aug 16, 2022 at 12:39
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    it's the same thing in reverse, Lowering ratio compromises acceleration, raising ratio improves acceleration. And those guys do race each other and run 0-60s and quarter miles. So they are watching. But yes, it also means more shift actions in the same quarter-mile, so it depends how well you shift.
    – Harper - Reinstate Monica
    Aug 16, 2022 at 16:40

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In general, you're right. The reason is along the lines of what you're saying, but there's another factor I don't think you've included. One of the things the lower (higher numerical) gearing does for your vehicle is, it allows the engine to get to its torque peak faster. Since the engine can rev quicker, it can accelerate the vehicle faster. As you stated, the top end speed is affected because numerically it cannot go as fast, but that is the trade-off. Mind you, this is all a generalization which equates to the vehicle being able to maintain the same amount of traction.

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  • That's true, getting it out of "the hole" faster would definitely offset shift times. But are you of the understanding that, in general, it would be a wash?
    – bertmoog
    Aug 16, 2022 at 12:33
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    Not at all a wash. As I said, it's a trade off. You trade the quicker start for less top end speed. This is how racers have been doing it for years with regards to 1/8 mile racing versus 1/4 mile racing. You want your engine speed to be just maxing out as you cross the finish line. The sooner you can get to your max engine speed, the quicker you'll go down the 1/8 mile. You have to have a little taller gears for the 1/4 mile so you don't run out of gear.
    – Pᴀᴜʟsᴛᴇʀ2 ♦
    Aug 16, 2022 at 14:17

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