# Understanding the torque of a motor/gearbox combo

Imagine a typical DC motor / planetary gear combos.

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.

• I carefully read all the (terrific) torque-motor QAs on here, but really coudn't tease out the basic answer, I'm afraid! :O – Fattie Nov 27 '18 at 11:24
• 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. – Jonathan R Swift Nov 27 '18 at 12:20

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!