I am trying to do a simulation on the start up speed to a system powered by an AC motor.

As part of this I am trying to find a mathematical torque/speed curve that I can use.

I expect a curve that looks something like this.

enter image description here

I have used the equations HERE as reference for the below calculations.

My general process is

  • Determine motor resistance & inductance based on published data for full load conditions
  • Use this to plot a curve of the motors torque speed relationship

However the curve I am getting looks nothing like the published curve at low speeds. Have I done anything wrong, or is there a more accurate model I can use.

enter image description here

I am using a WEG motor as a reference with the following published data

  • Frame: 315H/G
  • Output: 185 kW
  • Frequency: 50 Hz
  • Poles: 6
  • Full load speed: 988
  • Slip: 1.20 %
  • Voltage: 415 V
  • Rated current: 328 A
  • Locked rotor current: 2000 A
  • Locked rotor current (Il/In): 6.1
  • No load current: 119 A
  • Full load torque: 1789 Nm
  • Locked rotor torque: 190 %
  • Breakdown torque: 210 %


The best I can fit based on Charles Cowie's Answer

  • Rth = 0.03424
  • Xth = 1E-9
  • R2 = 0.0020499
  • X = 1E-8

This still looks nothing like the published curve I am going to give up and go with just taking samples from number of motor curves. But I would still like to get a solution to this.

New Attempt


2 Answers 2


There are a few things you have wrong. First, 1789 Nm is the rated ("full load") torque, not the maximum torque. The maximum torque is usually called the "breakdown torque" and according to your data, it is 210% of the full load torque. This puts the maximum torque at 3756.9 Nm. The locked rotor torque also isn't 42.93 Nm, but rather 190% of the full load torque. This puts the locked rotor torque at 3399.1 Nm.

Try fixing those things. Also notice that the graph provided plots the speed, torque, and current as a ratio of the rated values.

  • \$\begingroup\$ Thanks for the catch, I can get a bit better curve but not too much. The values change the magnitude but not the shape. \$\endgroup\$
    – Hugoagogo
    Commented Sep 6, 2015 at 9:43
  • \$\begingroup\$ Your locked rotor torque is showing under 1000 Nm on your new curve. It should be just under 3400 Nm. \$\endgroup\$
    – Eric
    Commented Sep 6, 2015 at 17:30
  • \$\begingroup\$ I could get a better fit at the start of the curve at the cost of the location and height of the peak \$\endgroup\$
    – Hugoagogo
    Commented Sep 6, 2015 at 20:59

I think that the problem may be that you don't have sufficient information about the motor. What is the basis for your calculation of K?

The information presented here seems to follow the methods used in the texts that I have and the presentations of the material that I have seen.

Edit 1 To get good results, you really need good values for all of the equivalent circuit parameters. I found a paper entitled "Estimation of induction motor equivalent circuit parameters from nameplate data" by Keun Lee, Div. of Eng., Colorado Sch. of Mines, Golden, CO, USA; Frank, S. ; Sen, P.K. ; Polese, L.G. ; Alahmad, M. ; Waters, C. I will try to get a copy.

  • \$\begingroup\$ My k is as defined in the linked page \$\endgroup\$
    – Hugoagogo
    Commented Sep 5, 2015 at 21:12
  • \$\begingroup\$ The linked page defines K as the ratio of the rotor voltage to the stator voltage. How is that related to your calculation of K? \$\endgroup\$
    – user80875
    Commented Sep 5, 2015 at 22:42
  • \$\begingroup\$ The page linked in my question defines k as 3/(2pi*ns) where ns is the synchronous rpm \$\endgroup\$
    – Hugoagogo
    Commented Sep 5, 2015 at 23:44
  • \$\begingroup\$ I had a crack with the equations in your link, helped a bit, see question. \$\endgroup\$
    – Hugoagogo
    Commented Sep 6, 2015 at 9:45
  • \$\begingroup\$ If you want to continue pursue the problem here, I will see what more I can add when I have time. \$\endgroup\$
    – user80875
    Commented Sep 6, 2015 at 14:58

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