Induction Motor Slip Calculator

Synchronous speed from the supply frequency and the pole count, then the slip, the slip speed and the rotor frequency from the speed on your nameplate — with a check on whether it looks right.

Last updated 22 Sep 2026

Induction motor slip

f and poles → synchronous speed → slip

What do you have?

Supply frequency (Hz)50 Hz in India, the Gulf and Europe; 60 Hz in North America and Saudi Arabia. On an inverter drive, the output frequency at the moment you are interested in.

Number of polesPoles, not pole pairs. A 4-pole motor is the ordinary 1,500 rpm machine at 50 Hz. If the plate gives only a speed, the pole count is whichever synchronous speed sits just above it.

Rotor (full-load) speed, rpmThe speed on the nameplate, which is measured at full load.

Slip (%)

The stator winding makes a magnetic field that turns at the synchronous speed set by the supply frequency and the number of poles. The rotor always turns more slowly; the difference is the slip, and it is what induces rotor current and so produces torque. A rotating field cannot be drawn honestly as a still picture, so the two speeds are written on the machine instead of animated.

3.00%Example

50 Hz, 4 poles, nameplate speed 1,455 rpm

Synchronous speed, slip and rotor frequency

n_s = 120 × f ÷ p s = (n_s − n) ÷ n_s slip speed = n_s − n f_rotor = s × f

n_s: synchronous speed in rpm — the speed of the stator’s rotating field
f: supply frequency in hertz
p: number of poles (not pole pairs); always even
n: the rotor’s actual speed, which is what the nameplate gives
s: slip, as a fraction or a percentage of synchronous speed
120: 60 seconds per minute × 2 poles per pole pair

Worked example

50 Hz, 4 poles, nameplate speed 1,455 rpm
n_s = 120 × 50 ÷ 4 = 1,500 rpm
Slip speed = 1,500 − 1,455 = 45 rpm
s = 45 ÷ 1,500 = 3.00%
Rotor frequency = 3.00% × 50 Hz = 1.50 Hz — which is why a rotor bar problem shows up as a sideband that close to the line frequency
The same motor on a 60 Hz supply would turn at 1,800 × 0.97 = 1,746 rpm at the same slip

Synchronous speeds (rpm)

Computed from 120 × f ÷ poles. The last two columns show roughly where a nameplate speed falls for an ordinary motor — which is how to read the pole count off a plate that does not state it: the synchronous speed is the one just above the speed printed.
Poles	50 Hz	60 Hz	At 3% slip, 50 Hz	At 3% slip, 60 Hz
2	3,000	3,600	2,910	3,492
4	1,500	1,800	1,455	1,746
6	1,000	1,200	970	1,164
8	750	900	728	873
10	600	720	582	698
12	500	600	485	582

Why an induction motor must run slow

Feed a three-phase winding from a three-phase supply and the three currents, 120° apart in time and 120° apart in space, add up to a magnetic field of constant magnitude that rotates. How fast it rotates is fixed by the supply frequency and by how many poles the winding is wound for: the field advances two pole pitches every supply cycle, which is 2f/p revolutions per second, or 120f/p revolutions per minute. A 4-pole winding on 50 Hz gives 1,500 rpm; nothing you do to the motor changes that number except changing the frequency.

The rotor cannot keep up, and must not. The rotor of a cage motor has no connection to anything; its currents are induced by the field sweeping past it. If it ever reached synchronous speed the field would be stationary relative to it, nothing would be induced, no current would flow and there would be no torque at all. So it always runs slower, and the shortfall — the slip — is exactly what produces the torque. The example above is 45 rpm behind, which is 3.00% slip. Load the motor harder and it slows a little more; the slip rises until the torque matches.

Rotor frequency is the slip frequency. The rotor conductors see the field pass at the slip speed, so the current in them alternates at s × f — 1.50 Hz here, not 50 Hz. This is why rotor bars can be thick aluminium castings rather than laminated windings, and it is the basis of current-signature fault detection: a broken bar puts sidebands at twice the slip frequency either side of the line frequency.

Is your figure sensible? A standard cage machine — NEMA design A, B or C — runs under 5% slip at full load. NEMA’s own nominal 60 Hz full-load speeds are 3,550, 1,750 and 1,150 rpm for 2, 4 and 6 poles, which against synchronous speeds of 3,600, 1,800 and 1,200 are 1.39%, 2.78% and 4.17%. Big machines go below 1%. Design D motors are deliberately made with 5–8% or 8–13% slip for loads that need a soft, torque-limited start. A figure far outside all of that is almost always an input error rather than a fault: the commonest is choosing the wrong pole count, because the synchronous speed has to be the one just above the nameplate speed. The page flags it rather than diagnosing the motor.

Slip is also the reason an induction motor’s speed is nearly constant but not quite, and the reason a variable-speed drive changes frequency rather than voltage: at a given slip, speed follows frequency. For torque at the speed you land on, use the motor power, torque and speed calculator; for the current the same motor draws, the motor full-load current calculator.

Frequently asked questions

What is the formula for synchronous speed?

120 × frequency ÷ number of poles, in rpm. At 50 Hz a 2-pole motor is 3,000 rpm, a 4-pole 1,500 and a 6-pole 1,000; at 60 Hz the same motors are 3,600, 1,800 and 1,200.

How do I calculate percentage slip?

Subtract the rotor speed from the synchronous speed and divide by the synchronous speed. For a 4-pole 50 Hz motor with a nameplate speed of 1,455 rpm: (1,500 − 1,455) ÷ 1,500 = 3.00%.

What is a normal slip for an induction motor?

Under 5% at full load for a standard cage motor, and often between 1% and 4%. NEMA’s nominal 60 Hz speeds correspond to 1.4%, 2.8% and 4.2% for 2, 4 and 6 poles. High-slip NEMA design D motors are made to run at 5–13%.

How do I find the number of poles from the nameplate speed?

Take the synchronous speed just above it. A plate reading 1,455 rpm on 50 Hz must be a 4-pole motor, because 1,500 is the nearest synchronous speed above; 2,880 rpm would be 2-pole.

Why does an induction motor need slip at all?

Because the rotor current is induced by relative motion between the rotor and the rotating field. At synchronous speed there is no relative motion, no induced voltage, no rotor current and no torque. Slip is the mechanism, not a defect.

What is the rotor frequency of an induction motor?

The slip times the supply frequency. At 3.00% slip on 50 Hz the rotor currents alternate at 1.50 Hz. At standstill the slip is 100% and the rotor frequency equals the supply frequency, which is why the starting current is so large.

Related calculators

References

National Electrical Manufacturers Association. Electric Motor Terminology and Performance Characteristics: “Synchronous Speed = 120 x Frequency / # Poles”; “% Slip = 1 − (Full Load RPM / No Load RPM) × 100”; synchronous speeds of 3600, 1800, 1200, 900 and 720 rpm at 60 Hz for 2, 4, 6, 8 and 10 poles; NEMA designs A, B, C and D, with design D full-load slip of 5–8% or 8–13% and locked-rotor current about 650% of full load for designs B, C and D.
Chapman S J. Electric Machinery Fundamentals, 5th ed. McGraw-Hill, 2012. Chapter 2, Transformers — the ideal transformer, the turns ratio, impedance transformation and voltage regulation; chapter 6, Induction Motors — synchronous speed, slip, rotor frequency and induced torque.
ABB. Softstarter Handbook, publication 1SFC132060M0201. Direct-on-line starting current “Usually between 6-8 times the rated current, but it can be more than 10 times the rated current”; star-delta: “The resulting current when Y-connected will be 1/3 of the current when delta connected” and the torque “ending up being 33% of the torque available when delta connected”; n = 2 × f × 60 / p; s = (n₁ − n)/n₁; T_n = 9550 × P_r/n_r.
IEC 60034-1:2022 (edition 14.0), Rotating electrical machines — Part 1: Rating and performance, whose Table 22 is the schedule of tolerances on the quantities a machine declares.