Induction Motor Slip

Also known as rotor slip · slip percentage

s=100(NsNr)Nss = \frac{100 \, (N_{s} - N_{r})}{N_{s}}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

An induction motor's rotor must lag the rotating field — if it ever caught up, the field would stop sweeping past the rotor bars, no voltage would be induced, and no torque would exist. That lag is slip. A 4-pole motor on 60 Hz has a synchronous speed of 1800 rpm; a nameplate reading of 1750 rpm means s = 100 × 50/1800 ≈ 2.8%, entirely typical for a healthy machine at full load.

Slip is a free load indicator: it is very nearly proportional to torque, so a tachometer reading tells you how hard the motor is working without a single meter lead. At no load slip falls under 1%; at breakdown torque it may hit 20%. Rising slip on a familiar machine means added mechanical load, low voltage, or a broken rotor bar. Note that this solver treats rpm as a frequency, so answers appear in hertz unless you pick rpm from the table.

Induction Motor Slip
s=100(NsNr)Nss = \frac{100 \, (N_{s} - N_{r})}{N_{s}}
Where
  • ss= Slip
  • NsN_{s}= Synchronous speed
  • NrN_{r}= Rotor speed