Gearing question

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Old 05-08-2016, 05:24 PM
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Question Gearing question

Hi,

I have been trying to get my project to put a 49cc pocket bike engine on a razor pocket mod started, I have to order the sprockets and need some help about the gearing and confirming that I am right and it will work.

So the scooter originally has a 11 tooth sprocket on the motor and a 80 tooth sprocket on the rear wheel, the motor spins at a max rpm of about 2,750rpm and that makes the scooter go 15mph with that gearing, so it stands to reason that if the engine was spinning the same gear that was on the motor at twice the rpm it would go twice the speed (at least once it gets up to speed and the clutch isn't slipping that much), do you think this is right and it will work that way? Also. The engine rotates counter clockwise the the wheel sprocket is on the left side when looking at the scooter from the front, so it would spin the wrong way if the engine sprocket was on the left side and connected to the wheel sprocket directly so I have to mount the engine the other way around and run a shaft to the other side, the left side of the shaft will have the motor sprocket on it and the right side will have a 11 tooth sprocket, the motor can either have a 7 tooth sprocket or a 10 tooth sprocket on it, I was thinking the 10 tooth sprocket would be better since that would be a 1.1 gear ratio and my theory is that it would spin the gear on the other side at about the same rpm and with the same torque as the engine as if the engine was just directly connected to the wheel sprocket. Sorry for the novel but I don't know how else to explain it. Do you think this is right and will work that way? Please let me know what you think and I am open to any other designs you may have for the shaft to transfer the power to the other side. I'm not sure of the exact max rpm of the engine but I've heard somewhere around 6,000rpm to 12,000rpm, the scooter is already geared for torque and has a 7.27 gear ratio so I am thinking it should have enough torque without adding any on the other side. Thanks for any replies
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