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Suppose I have a gearbox to convert a high speed, low torque motor into a low speed, high torque output. Every time I add another non-ideal gear, there's an additional power loss due to friction and such.

My question is: Does each gear contribute to power loss equally? My thought process is that the gears at the slow end of the box is rotating much slower, and it's easier to overcome the friction due to the increased torque. Am I thinking of this wrong?

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  • $\begingroup$ What about including lubrication? $\endgroup$
    – Solar Mike
    Dec 10, 2023 at 15:37

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I am a user of, rather than a designer of gearboxes. So take what I say with an appropriately-sized grain of salt.

There are a huge number of exceptions to this statement, but generally, until viscous effects need to be taken into account, each stage tends to have the same proportional torque loss (called efficiency).

I'm working out the math in my head, but when the above statement about efficiency is true, the power loss in each stage should be roughly the same.

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The slower moving gears will have higher torques, so normal force*speed will be the same, so roughly the power loss will be equal in each stage (about 2% per stage as a starting point). This incidentally is why an epicyclic is not a catastrophe, efficiency wise, the many gear pairs are only reacting a small proportion of the torque.

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