Rotational Speed Calculator
Calculate rotational speed in RPM (revolutions per minute), RPS (Hz), radians per second (rad/s), degrees per second, and rotational period (T). Solve from counted revolutions and time, linear cutting/surface speed and diameter, or AC electric motor poles and frequency.
Step-by-Step Calculation Breakdown
- Find Rotational Frequency in RPS (Hz): f = Revolutions / Time_sec = 350.00 rev / 12.00 s = 29.1667 RPS (Hz)
- Convert RPS to Revolutions per Minute (RPM): RPM = f × 60 = 29.1667 × 60 = 1,750.00 RPM
- Compute Angular Velocity (ω) & Period (T): ω = 2π × f = 183.26 rad/s (10,500.0 °/s) | T = 1 / f = 34.29 ms/rev
- Calculate Shaft Torque (@ 10.0 HP): Torque = (5,252.11 × 10.0 HP) / 1,750.00 RPM = 30.01 lb-ft (40.69 N·m)
Rotational Speed Formulas & Engineering Conversions
1. Rotational Speed from Revolutions & Time
RPM = (N_rev / t_sec) × 60 | f (Hz) = N_rev / t_sec
Dividing total completed revolutions by elapsed seconds gives rotational frequency in hertz (RPS). Multiplying by 60 converts RPS to RPM.
2. Angular Velocity (ω) & Rotational Period (T)
ω (rad/s) = (2π × RPM) / 60 | T (sec) = 60 / RPM
One full turn equals 2π radians (360°). The rotational period T is the exact time required to complete a single revolution.
3. Spindle / Shaft RPM from Surface Speed (SFM)
RPM = (12 × SFM) / (π × D_in) ≈ (3.8197 × SFM) / D_in
Used in CNC milling, lathe turning, and pulley drives to determine the required spindle RPM for a desired cutting speed (Surface Feet Per Minute) and tool diameter.
4. AC Motor Synchronous & Operating Speed
N_sync = (120 × f_Hz) / Poles | RPM = N_sync × (1 − Slip)
Determines the magnetic field speed of an AC motor from line frequency (60 Hz or 50 Hz) and stator pole count, minus induction rotor slip.
Illustrative Example
Scenario (Illustrative): An optical tachometer counts 350 revolutions of a 10 HP drive shaft over 12 seconds. Find the shaft's rotational speed in RPM, RPS (Hz), rad/s, time per revolution, and output torque.
- Revolutions per Second (RPS / Hz):
350 rev / 12 s = 29.17 RPS (Hz). - Revolutions per Minute (RPM):
29.1667 RPS × 60 s/min = 1,750.00 RPM. - Angular Velocity (ω):
2π × 29.1667 Hz = 183.26 rad/s (10,500.0 deg/s). - Period per Revolution (T):
1 / 29.1667 Hz = 0.0343 seconds (34.29 ms per turn). - Shaft Torque at 10 HP:
(5,252.11 × 10 HP) / 1,750 RPM = 30.01 lb-ft (40.69 N·m).
How to Use This Calculator
- Select Your Input Method: Pick Revolutions / Angle & Time when you know turn count and duration, Surface Speed & Diameter to compute spindle/wheel RPM from SFM, FPM, MPH, or m/s, or AC Motor Poles & Slip for electric motor speed.
- Enter Your Measurements: Provide the known values and select your preferred units. Optionally enter a reference radius or shaft horsepower/kilowatts to compute tangential speed and shaft torque.
- Review Multi-Unit Outputs: Read RPM, RPS (Hz), rad/s, deg/s, RPH, period per revolution (ms/rev and s/rev), and shaft torque (lb-ft and N·m).
- Copy, Print, or Reset: Use Copy Result to copy the complete breakdown, Print / Save PDF to save a clean sheet, or Reset Defaults to restore starting inputs.
Frequently Asked Questions
Are rotational speed (RPM) and angular velocity (rad/s) the same thing?
2π radians, ω = (2π × RPM) / 60 ≈ 0.10472 × RPM.
How do machinists calculate spindle rotational speed (RPM) from SFM?
RPM = (12 × SFM) / (π × D_in). In shop shorthand, 12 / π ≈ 3.82, so machinists often estimate RPM ≈ (3.82 × SFM) / D_in.
Why does a 60 Hz 4-pole AC motor run at ~1,750 RPM instead of 1,800 RPM?
N_s = (120 × 60 Hz) / 4 poles = 1,800 RPM. An induction motor rotor must spin slightly slower than the magnetic field so that magnetic flux lines cut the rotor bars and induce current and torque. That speed difference is called slip (typically 1.5% to 3.5% at full load, or 50 RPM on a 1,800 RPM motor, resulting in 1,750 RPM).
How does rotational speed affect shaft torque at a constant horsepower?
HP = Torque_lb-ft × RPM / 5,252.11). For a fixed horsepower rating, shaft torque is inversely proportional to RPM—reducing rotational speed by half through a 2:1 gearbox doubles the available output torque.