If the power output of the traction motor is not known and only the voltage and current draw are known, an estimate of the motor power output can be calculated from the following formula:
P = V * I * efficiency
where:
V = motor voltage
I = motor current at normal max load (NOT stall current)
efficiency = 0.85 (typical) [may vary from 0.75~0.9 depending on motor]
for example a 24V 150W motor is rated at 7.15A on the name plate, this gives 24 * 7.15 * 0.85 = 146W, which is a close approximation to the rated value.
Example 1: For a locomotive using a 350W 2500RPM motor using 85mm dia wheels, with a reduction ratio of 4.25:1 and having this drive on each of 2 bogies gives a tractive effort of 27.3 kg.
Example 2: For a locomotive using a 450W motor with an attached gearbox giving 420RPM output, and having 90mm dia wheels, with a reduction ratio of 0.9:1 from motor to axle (i.e. a step up) and using this drive on each of 2 bogies gives a tractive effort of 41.7 kg. Note the actual motor RPM is irrelevant, it is only the output RPM of the gearbox that matters, giving an effective motor speed of 420RPM to use in calculations.
Note the accuracy of these calculations is not necessarily exact, and figures calculated here may vary from results published elswhere. They are calculated here from the formulas above to 3 or 4 significant digits only.
Disclaimer: Results to be used at your own risk. Calculator provided for your personal use only.
