I am looking at buying a DC motor with reasonably high torque characteristics that is essentially going to be for the joint of something like a robotic arm.

I came across several worm drive DC motors that fit the requirements, however I am a bit confused by some terminology. They give the stall torque, stall current but also a hold torque which is 80% of rated load. For a motor that's holding a weight on an arm at some angle, wouldn't the stall torque be the same as the hold torque? My understanding was that stall torque is produced at near zero rotational speed, which is the case for a static system.

Secondly for a worm drive motor, I didn't think that kind of gearbox allowed rotation from the load to the motor by way of the worm drive design. Wouldn't the hold torque of a worm drive motor be the breaking point of the gearbox?

Motor in question: https://www.motiondynamics.com.au/worm-drive-motor-12v-50w-45-65-rpm-5-29nm-torque.html


closed as off-topic by Scott Seidman, PeterJ, Daniel Grillo, Transistor, pipe Aug 23 '16 at 3:39

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  • \$\begingroup\$ No. The hold torque is likely to be the torque a worm gear will hold with the motor unpowered - or at least, driven by no more than its rated current. Stall torque will be what it can exert at its stall current, which is several times its rated current, and often, enough current to destroy the motor in seconds or minutes. \$\endgroup\$ – Brian Drummond Aug 22 '16 at 8:51
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    \$\begingroup\$ I'm voting to close this question as off-topic because it's mechanical engineering \$\endgroup\$ – Scott Seidman Aug 22 '16 at 11:14

Worm drives are unusual gearboxes.

Because of their capacity for high ratio, they can have very low efficiency and still be useful. It's quite difficult to make an efficient worm drive gearbox, you need a low ratio, and usually a multi-start thread on the worm, and often special low friction materials.

If a worm drive has an efficiency of less than 50%, it's impossible to 'back drive' it, that is turn the output shaft and spin the worm. In this case, the 'hold torque', if that's defined as the maximum torque that will be held by the output shaft with the input motor unpowered, would be infinite, or the breaking strength of the gearbox.

As that motor has fairly high power, and specifies a hold torque less than its max output torque (obviously much less than gearbox breakage level), it looks like the efficiency is > 50%, and it can be back driven.

The hold torque is now governed by how much torque it takes to turn the unpowered motor. This will be governed by friction and motor cogging. The first will be poorly defined, depending on lubrication state and bearing smoothness, though the second will be more predictable.

The manufacturer here is giving a torque figure that he expects will always be unable to spin the motor. Given the uncertainties in motor friction, I would not be surprised if you typically need to apply rated torque or even overload torque to get the output shaft to move.

  • \$\begingroup\$ Thanks for the reply, it's a lot clearer now what they mean. Following from this, say I wanted to hold a weight at a fixed position with an active motor - Would I first take away the hold torque and then do my current calculations based on the torque that remains? \$\endgroup\$ – Init33 Aug 22 '16 at 10:01

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