Measuring shaft torque of a small motor

Question

I would like to determine the mechanical output torque and power of a small motor without using expensive strain guages and similar and have come up with the idea of determining the maximum mass that the motor shaft can lift.

With reference to the image, if I suspend a mass over a pulley and find the maximum mass that the output shaft can lift (and wind the string around its shaft in the process), can I calculate the torque as: T= mgr where m is the mass in Kg, g=9.8 m/s2 and r is the radius of the shaft in metres?

If this is correct then for the setup shown, and assuming the motor could lift no more than 25g, the value of T= 0.025x9.8x0.002= 0.00049 Nm. Then if the shaft is spinning at 1000 rpm, Power= 0.49W

Thanks

Answer 1

This is roughly okay. except you want to make sure there is either no acceleration or if there is it is counted for by just adding the term ma with no g to the torque of the pully,

$T =mgr +mar$.

To make the speed slower and measuring it easier one can choose a 1mm shaft and say 50 gr mass next to a vertical tape measure.

Answer 2

I would suggest mounting the motor on bearings through its axis and then use a torque arm with masses to bring it back to the balance point.

This is how classic dynamometers are designed. Then they use a load cell instead of masses to get an electronic reading...

Answer 3

In this setup is that theoretically, in the steady state everything should work, however the main problem is getting to the steady state. So for example:

• getting to 1000 rpm: if the motor is not able to lift more than 25 g, then it will not be able to accelerate. If the motor is barely able to lift more than 25 gr (lets say 26gr), then it will take a long time to reach 1000 rpm.

• torque rpm dependence: Almost all electric motors have a relationship between torque and rotational velocity. A typical one for DC motors is the following

Figure 1: DC motor torque (source: Robotics SE)

So, the DC motor will accelerate differently at different rpm. So there is a dynamic component.

• length of pulley rope : at 1000 rpm with a 2mm shaft you will have a velocity of about 1 m/s. This is a very high speed for what you are planning, and assuming that your motor takes about 20 seconds to reach 1000 [rpm], then you'd need about 10 [m] or so of wire (which then begs the question how much will the diameter change etc).

one solution.

A solution from the top of my head is the following:

Hook up the motor to a gearbox with a gear ratio of more than 10. The higher the better. Then attach the rope to the shaft. This will have a two fold benefit

1. the speed of the rope will drop (so you won't need too long rope).
2. the torque will be magnified (so you will need to use a higher dead weight). However that is good, because you can easily procure a 2.5 kg load (for a gear ratio 100) and you will have better resolution.

I can provide more details, (but since this is an old post, I don't see too much point).

Answer 4

For future reference,

If you want to measure the shaft torque of a small motor, just couple it with a DC generator (basically a DC motor) and then short circuit the terminals. The power will dissipate as heat, and a resisting torque will be applied to the shaft. If you want to measure it, recall that for any permanent magnet DC motor $$T=K_Ti \tag{1}$$ where $K_T$ is a constant, $T$ is the torque, and $i$ is the current that flows through the generator. You can measure $i$ with an amperemeter and calculate $T$ using Equation 1. You need to find $K_T$ for your generator experimentally.

Furthermore, you can control the load with a potentiometer connected between the generator's terminals. You can even measure the rotation speed with a voltmeter because $$\omega=\frac{V}{K_\text{emf}} \tag{2}$$ where $\omega$ is the rotation speed, $K_\text{emf}$ is a constant, and $V$ is the voltage difference across the generator's terminals.

This is an example of an electric dynamometer.