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Synchronous Speed and Slip Calculator

The stator's rotating field turns at a speed fixed by supply frequency and pole count. The rotor must lag it to generate torque at all — that lag, as a fraction, is the slip.

Synchronous Speed & Slip — N_s = 120·f/P

Synchronous speed of the rotating field, and (if you enter the rotor speed) the slip and rotor-frequency of an induction motor. Poles must be an even number.

Synchronous speed N_s
Slip s
Slip (percent)
Rotor frequency f_r
  1. 1.Given: f = 50 Hz, P = 4 poles, N_r = 1440 rpm
  2. 2.FormulaN_s = 120 × f ÷ PSubstituteN_s = 120 × 50 ÷ 4ResultN_s = 1500 rpm
  3. 3.Formulas = (N_s − N_r) ÷ N_sSubstitutes = (1500 − 1440) ÷ 1500Results = 0.04 (4 %)
  4. 4.FormulaRotor frequency f_r = s × fSubstitutef_r = 0.04 × 50Resultf_r = 2 Hz
Common trap: The rotor can never reach synchronous speed — if it did, there'd be no relative motion, no induced EMF, and no torque. A healthy induction motor runs at a few percent slip; a slip of 0 means it's a synchronous machine, not an induction one.

The formula

N_s = 120f / P ; s = (N_s − N_r) / N_s ; f_r = s × f

N_s
synchronous speed of the rotating field (rpm)
f
supply frequency (hertz)
P
number of poles (always even)
N_r
actual rotor speed (rpm)
s
slip (fraction or per cent)

Worked example

A 4-pole induction motor on a 50 Hz supply, running at 1440 rpm.

  1. N_s = (120 × 50) / 4 = 1500 rpm
  2. s = (1500 − 1440) / 1500 = 60 / 1500 = 0.04
  3. Rotor frequency: f_r = 0.04 × 50 = 2 Hz

1500 rpm synchronous, 4 % slip, and rotor currents at just 2 Hz.

Where you'll use it

Every induction machine question, and the reason a nameplate says 1440 rpm rather than 1500. An induction motor can never reach synchronous speed — at zero slip there is no relative motion, no induced rotor current and therefore no torque.

The laws behind it

Parts this applies to

Inductor / CoilTransformer (power)Current / Potential Transformer (CT/PT)DC MotorSpeaker / MicrophoneRCCB / RCD / GFCICurrent Sensor (Hall / Shunt / CT)LVDT & Displacement Transducers

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