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For example if the first gear has 32 teeth and the second 64, if the third also has 32 teeth would the output of the third retain the mechanical advantage created by the first 32 tooth gear powering the second?

32:64:32 or 1:2:1

Thanks for your time and help!

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if you have them in line like the following picture then NO.

enter image description here

The first and the last gear will have the same speed and the same torque (actually slightly less due to losses).

if you wanted to retain a mechanical advantage you'd need 3 shafts and at least 4 gears like in the following image.

enter image description here

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    $\begingroup$ You might say that the gear chain you have in first picture just cares about the teethcount of the gears in the ends irrespectively how many gears there are the gears in middle are only losses. Its a common misconception for laymen. $\endgroup$ – joojaa Oct 29 at 20:02
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    $\begingroup$ Why would that second image help? Gear 4 is even smaller than Gear 2, so wouldn't you wind up with something that has even less RPM than the input gear? $\endgroup$ – nick012000 Oct 30 at 6:05
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    $\begingroup$ @nick012000 I updated the image. you were right about the previous image, it was ambiguous whether it offered or not mechanical advantage. $\endgroup$ – NMech Oct 30 at 10:04
  • $\begingroup$ As an aside, would one really choose small ratios of teeth numbers or try to do something like 64/31 instead of 64/32? $\endgroup$ – Peter - Reinstate Monica Oct 30 at 12:05
  • $\begingroup$ Are you asking whether its good practice about the teeth having 1 as the lowest common denominator? $\endgroup$ – NMech Oct 30 at 12:31
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No. We can show this with math.

The first gear pair has a ratio of 1:2, and the second pair has a ratio of 2:1. So, discounting losses, for every turn of the input shaft, the second shaft will make 1/2 of a turn. For the second pair, for every turn of the second shaft, the third shaft will turn twice.

We can then multiply these together to see how many turns the third shaft will make compared to the first shaft: 1/2 * 2 = 1. Therefore, for every turn of the first shaft, the third shaft will turn once, and there is no mechanical advantage for the overall system.

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