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sprocket question

A

Anonymous-0

Well-known Member
we have 9 teeth on the "drive" shaft and 17 teeth on the "driven" shaft. Would like to keep the driven shaft going the same speed, but would like to make the drive shaft have more teeth,(just think it would work better if there were more teeth on the sprocket) like maybe a dozen or more. Anybody know if there is an easy formula to figure this out? If I put a dozen on the drive, how many need to go on the driven?
 
You need to keep the same ratio of teeth. 9/17= .5294. If you want to increase the drive sprocket by 33% 9*1.333=12 then you also need to increase the driven sprocket by 33%. 17*1.333=22.6 teeth. a 22 tooth sprocket will give you a ratio of 12/22=.545 (driven spocket would turn slightly faster) and 23 would give you 12/23=.521 (driven slightly slower). Since 17 is prime, if you want the EXACT ratio you would have to go up to an 18 on the drive and 34 on the driven.
 
You presently have a 1.8 to 1 ratio so any change to input rato would have to be made to output to maintain same speed.
 
Just a suggestion, but I would go with ratio that is a 'hunting tooth' situation. Then the teeth wear evenly. So at least one sprocket will end up with an odd number of teeth.
Also, with larger sprockets, you will increase the chain velocity and you have to be careful as higher speeds sometimes require special lubrication.
 
Seems like an odd number of links in the chain with two even number teethed gears would accomplish the same thing. Even if the chain and the gears were even, the teeth should still "hunt" as long as #teethgear1 x #teethgear2 does NOT equal #links in the chain.
 
Don't most smaller chains run every pin in the sprocket anyway? I know on something like a cat, you want odd and even numbers but something like a motorcycle doesn't matter.
 
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