Measuring Torque of a Stepper Motor Using a Suspended Weight Method

AI Thread Summary
The discussion focuses on measuring the torque of a stepper motor using a suspended weight method, where a weight is attached to the motor via a string. The proposed formula for torque, τ = mgΔL, is debated, with clarifications that the correct approach involves considering the moment arm radius and using τ = mg × r for accurate measurements. Participants suggest alternatives, such as using a pivot for the motor or a spool to increase the radius, which would simplify calculations. The importance of measuring the shaft radius and the angle of rotation is emphasized, as these factors significantly affect torque calculations. Overall, the method can yield a rough estimate of torque, but precision may be limited without proper equipment.
roam
Messages
1,265
Reaction score
12
1. The problem statement, all variables and given/known data

I am trying to experimentally measure the torque of a stepper motor at various velocities. I'm not quite sure how this should be done. Here is a diagram of my proposed method so far:

259uxro.jpg

Basically a weight is to be suspended from the motor by a string, and I will measure by how much the string has moved up after one full revolution. Then I believe the torque must be: ##\tau = m \times g \times \Delta L##. Is this a correct method?

Homework Equations



Torque = Force x Distance

The Attempt at a Solution



I would greatly appreciate it if someone could confirm whether this is a correct method to get a roughly correct measurement of torque (within perhaps ~80% of the manufacturer quotedvalue at a given velocity)?

Do I need to take the shaft diameter into account (or perhaps some other factors)? What is the simplest way of doing this? :confused:
 
Physics news on Phys.org
Torque = Force x Distance
... sort of: the "distance" in question cannot be any old distance - it has to be the length of the moment arm. In your equation ##mg\Delta L## is the work, ##W=Fd##, not the torque.

The equivalent relation for rotation is ##W=\tau\theta## ... where the angle is in radians.
 
  • Like
Likes roam
Another way to measure the torque is to mount the motor on a pivot so the case can rotate (within limits), attach an arm to it with some sort of spring scale on the end.
 
Also in your diagram the Moment Arm - is the radius of the shaft ( from the center to where the string contacts it- so pretty small) -- so you can decrease the needed weight but adding a wheel/spool to the shaft to increase the radius. By choosing a good radius it will make the calculations easier ( e.g. if you want In*Lbs as units- use a 1" radius wheel)
 
  • Like
Likes roam
Thank you Simon, CWatters, Windadct for your responses. I'm still a bit confused.

So, basically, I must measure the radius and use ##\tau= mg \times r##, OR find the angle and use ##\tau = (mg \Delta L)/\theta##?

When the motor turns the string will be wound around the stepper motor shaft, and I must measure the shaft radius with string around it (the amount of string is supposed to change by a minuscule amount whenever I change the speed). I think this would be a very difficult and inaccurate way of measuring ##\tau##...

In the diagram below I have indicated ##\theta##:

i1mpht.jpg


Is this the right idea? This angle can be measured with a protractor. Even if the string is wound several times and we get the same θ, ΔL would be different. Therefore we get different torques each time. So is this a correct method?

P.S. Unfortunately I do not have access to an appropriate spool, or a spring scale at the moment. I have to work with only a ruler and a protractor.
 
Neglecting energy that goes to turn the motor components: $$\tau_{ave} = \frac{mg\Delta L}{2N\pi}$$ would be a reasonable approximation.
Notice that's an average for N entire turns. Picking a big value for N will probably be best.

##\vec \tau = \vec F_g\times \vec r \implies \tau=mgr## ...where r is the radius of the shaft, would be the torque exerted by gravity.
Since the stepper motor accelerates around one turn, it must be providing more torque than that.
 
  • Like
Likes roam
Thank you for the explanation Simon. It's all clear now. :)
 
Back
Top