I have assembled a benchtop torque measurement stand for evaluating electric motor and controller efficiency.
The motor is 49kV , 5kW brushless outrunner.
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.
$$L_{arm\_assumed}=\frac{Q_{assumed}}{F_{cell}}=\frac{94Nm}{84N}=1.11m$$
So, the question is - what is my torque arm?
EDIT
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?
