Synchronous Speed from Frequency and Poles

Also known as motor RPM from frequency · 1800 RPM motor

Ns=2fpN_{s} = \frac{2f}{p}

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

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A pair of poles takes one full electrical cycle to sweep the field once around, so the field turns f/(p/2) = 2f/p revolutions per second. Multiply by 60 and you get the version every electrician knows: Ns = 120f/p in rpm. At 60 Hz that gives 3600, 1800, 1200 and 900 rpm for 2, 4, 6 and 8 poles; at 50 Hz, 3000, 1500, 1000 and 750. Enter f in hertz and read Ns in rpm and the two forms agree exactly.

Pole count is fixed by the winding, so before variable-frequency drives the only way to change a motor's speed was to change machines or use pole-changing windings. A VFD attacks the f instead, which is why a 4-pole motor on a 30 Hz drive turns near 900 rpm. Two traps: p counts poles, not pole pairs, and the actual shaft speed is always a percent or two below Ns because of slip — a true synchronous machine is a different animal with a DC-excited or permanent-magnet rotor.

Synchronous Speed from Frequency and Poles
Ns=2fpN_{s} = \frac{2f}{p}
Where
  • NsN_{s}= Synchronous speed
  • ff= Supply frequency
  • pp= Number of poles
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