I have assembled a benchtop torque measurement stand for evaluating electric motor and controller efficiency.

The motor is 49kV , 5kW brushless outrunner.

bldc torque stand

Input power reading is taken from motor controller (VESC6, FOC). Current is limited to 140A, Input voltage is 48V.

$$RPM=480$$ $$P_{input}=5800W$$ $$Q_{calculated}=\frac{P_{input}*9.549}{RPM} = 117 Nm$$ $$Q_{assumed}=0.8*Q_{calculated}=94Nm$$

The torque stand consists of two load-cells holding the motor stator. Each cell was calibrated and tested for 5g error and is rated for 10kg load. After the motor is installed, the weight reading is offset to 0.

The rotor of the motor (shaft) is supported by the bearing and so the axial weights are compensated.

The averaged mass readings from the load-cells at peak input power are: $$m_{topcell}=4328g$$ $$m_{botcell}=-4244g$$

I assume the torque arm: $$L_{arm}=0.065m$$ and Force measured: $$F_{cell}=\frac{(m_{topcell}-m_{botcell})}{1000}G=\frac{8572}{1000}*9.81=84N$$

And the output torque:$$ Q_{real}=F_{cell}L_{arm}=84*0.065=5.47Nm$$

Clearly, $5.47 N$ isn't even close. So either mass or torque arm is wrong.


So, the question is - what is my torque arm?

I have revised the approach and made $1.35m$ lever. The mass measured with digital scale was $1520g$, which is equal to $20.13Nm$. If I plug this into another torque equation: $$Q=8.3\frac{I_A}{K_V}$$ I get $$Q=8.3\frac{140}{49}=23.7Nm$$

So, perhaps the true question I should be asking is
Why cannot I apply classic torque ($Q_{calculated}$) formula to an electric motor?


1 Answer 1


Whre are you measuring your torque arm from and to?

The rotational centre to the point of applied load resistance would be my suggestion, but that does not seem to be what you have at 6.5cm...

The simplest I used was a long torque arm with masses to hang on it to return it to its static position, worked fine...

  • $\begingroup$ I ran extra experiment and yes - what a surprising result. See "Edit" section. As for 6.5cm - this is a length to the strain-gauge on the load cell. However, I can see that the centre of a friction point is at 3.5cm (in between two bolts). So, the question is - how do I define a friction point? As that seems to impact my results by mere +-200%. $\endgroup$ Commented Jul 12, 2019 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.