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Re: Gear Ratio Math?


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Posted by Alberta Mike on January 19, 2002 at 18:18:05 from (209.47.89.126):

In Reply to: Gear Ratio Math? posted by K-Mo on January 19, 2002 at 06:06:48:

I'm cloudy on how the 3 gears in each "set" work together. I was visualizing the 3 gears in each set working together so that the first one drives the reverse idler (which changes the rotation direction). Then the third gear is on the same shaft as the reverse idler gear which would keep the third one turning in "reverse". I'm a little cloudy on my transmission layout as to how the gears are related to each other. Is this setup I describe here the way the three gears are tied into each other? Are there other gears before or after the ones mentioned?

Then, I didn't really say the math in the original post was incorrect (although I'm not sure how he got those numbers to be honest), I just looked at it in a different way (although I should have said for example gearing up 1.57 X's instead of 157% which was not really a correct way of stating it). Now, an idler will change direction yes, but the number of teeth have to somehow come into the overall ratio of the system. Just because it's changing direction doesn't nulifiy the gearing ratio that exists any more than it would for any two gears working together (directly or by chain linkage, etc).

And yes, I think the driving:driven ratio is usually the way ratios are presented but I didn't do that. For example, if you had a 10 tooth gear driving a 50 tooth gear, you'd have a 1:5 "low" gearing ratio. But at the same time, you could say it is a 5X's gear reduction. Both are really the same thing and that's what I did.

Anyways, I'm still not sure how these 3 gears in each "set" work together physically so maybe someone could straighten me out on that part and I can head scratch some more.


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