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Imagine a typical DC motor / planetary gear combos.

enter image description here

Is my thinking correct:

Say you have a 100 kg mass on a frictionless surface. The gearbox output shaft has a 0.10m arm with which you push the mass. (ie, linearly a few inches). You want to accelerate the mass at 0.1 m/s2.

Does that mean, it would take a force of 10 N (ie 100 x 0.1), and hence a torque of 1 Nm (10 x 0.1)?

If that's correct, and say after the gearbox the spindle indeed delivers 1 N. It's a 6000 rpm motor and a 100:1 gearbox.

In fact, would that mean the motor itself indeed delivers "$1 \text{Nm}$" .. it "doesn't matter" about the gearbox?

Or, if the output spindle is indeed delivering "$1 \text{Nm}$", would that mean the motor itself to purchase would be rated 1/100 = $10 \text{mNm}$ ?

{ie, 10 milliNewton-meter ... 0.01 Newton-meter!}

The figure seems way too low, but I'm likely mistaken.

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  • $\begingroup$ I carefully read all the (terrific) torque-motor QAs on here, but really coudn't tease out the basic answer, I'm afraid! :O $\endgroup$ – Fattie Nov 27 '18 at 11:24
  • $\begingroup$ You seem to be missing the 'm' off the torques in the latter half of your question - I've added these in bold in my quotes below, but you may wish to edit the question for correctness. Try using LaTeX to improve readability of numbers/units - I use the format $12345\text{ Units}$ , inputted as"dollar 12345 \text{ Units} dollar", but you can experiment with your own style. $\endgroup$ – Jonathan R Swift Nov 27 '18 at 12:20
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There are a lot of questions in here, I'll try to go through one by one:

Say you have a 100 kg mass on a frictionless surface. The gearbox output shaft has a 0.10m arm with which you push the mass. (ie, linearly a few inches). You want to accelerate the mass at 0.1 m/s2.

Does that mean, it would take a force of 10 N (ie 100 x 0.1), and hence a torque of 1 Nm (10 x 0.1)?

$F=m*a$, so yes, it would take a force of $10\text{N}$. and $T=F*L$, so yes, that corresponds to $1\text{Nm}$ in this case. So far, so good.

If that's correct [it is!], and say after the gearbox the spindle indeed delivers 1 Nm. It's a 6000 rpm motor and a 100:1 gearbox.

In fact, would that mean the motor itself indeed delivers "1 Nm" .. it "doesn't matter" about the gearbox?

No. It does matter about the gearbox. In this example, you have a $6000\text{rpm}$ motor, which corresponds to an output shaft speed of $60\text{rpm}$, and $1\text{Nm}$ If you change the gearbox, then these two output values will vary accordingly. i.e. you could have $30\text{rpm}$, and $2\text{Nm}$, or $120\text{rpm}$, and $0.5\text{Nm}$. Theoretically

Or if the output spindle is indeed delivering "1 N", would that mean the motor in question to purchase would be rated 1/100 = 10 mNm ?

Yes, this is correct. The actual motor, before the gearbox, would need to produce about $10\text{mNm}$

The figure seems way too low, but I'm likely mistaken.

Yes, you're mistaken, it's not that low. Here's the first datasheet that I found for a generic $12\text{V}$, $6000\text{rpm}$ motor, which lists the torque at maximum efficiency as $16.5\text{mNm}$.

This brings us onto an important point - Motor efficiency and maximum power output etc. changes as the (motor) spindle speed varies - you need to make sure that your design case is set appropriately to operate primarily near the maximum efficiency point where possible. That's the subject of another question/answer, though. You may be able to find existing questions on this topic, too - I've certainly answered at least one myself!

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