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Motor Torque & Horsepower Calculator

Convert between horsepower, kilowatts, torque (lb-ft or N·m), and RPM. Enter any two values and this calculator solves for the rest, with efficiency built in.

Inputs

Results

Torque
0 lb-ft
Horsepower (HP) 0 HP
Kilowatts (kW) 0 kW
Torque (lb-ft) 0 lb-ft
Torque (N·m) 0 N·m
Speed (RPM) 0 RPM
Input Electrical Power —
3-Phase Current (approx) —
Formulas used:
HP = Tlb-ft × RPM ÷ 5252
kW = TN·m × RPM ÷ 9549
Amps (3φ) = kWin × 1000 ÷ (V × √3 × PF)
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Assumes steady-state full-load operation. Add service factor for intermittent peaks and 25% headroom for starting torque.

How Motor Torque and HP Work Together

Horsepower and torque measure two different things. Torque is how hard the shaft twists. HP is how much work the motor can do over time. You cannot have HP without torque, and you cannot use torque at zero RPM to do work. The bridge between them is RPM.

The Key Formulas

In US units: HP = Torque (lb-ft) × RPM ÷ 5252. In metric: kW = Torque (N·m) × RPM ÷ 9549. These two formulas are the same equation, just in different units. One HP equals 0.746 kW. One lb-ft equals 1.356 N·m.

Why 5252?

The number 5252 comes from the definition of one horsepower as 33,000 foot-pounds per minute. Dividing by 2π gets you 5252. It is not magic, just unit math. It also explains why HP and torque curves always cross at exactly 5252 RPM on a dyno chart.

Pro tip: For CNC spindle sizing, look at the torque curve at low RPM. High HP at 10,000 RPM does not help if you are taking heavy cuts at 500 RPM. A motor that makes constant torque from 100 to 3000 RPM is much more versatile for machining.

Motor Efficiency Matters

AC induction motors are 85 to 92 percent efficient at full load. The rest is heat. If you need 10 HP of real shaft power and the motor is 88 percent efficient, the motor actually pulls 11.4 HP worth of electrical power. Premium efficiency motors save energy and run cooler, which extends bearing life.

3-Phase Motor Current

For a 3-phase AC motor, input current (amps) equals kW times 1000 divided by voltage times square root of 3 times power factor. A 5 HP motor (3.73 kW mechanical) at 460 V, 88% PF, 90% efficiency pulls about 5.9 amps per phase at full load. Use this for sizing fuses, contactors, and wire gauge.

Frequently Asked Questions

Torque in pound-feet equals horsepower times 5252 divided by RPM. So a 5 HP motor at 1800 RPM has 14.6 lb-ft of torque. In metric, torque in newton-meters equals 9549 times kilowatts divided by RPM.
Horsepower equals torque in lb-ft times RPM divided by 5252. For example, 10 lb-ft at 3600 RPM equals 6.85 HP. The number 5252 comes from the definition of one horsepower as 33,000 foot-pounds per minute.
One kilowatt equals 1.341 horsepower. So a 5 kW motor is about 6.7 HP. Going the other way, one HP equals 0.746 kW. Motor nameplates often show both values because HP is common in the US and kW is common in Europe and Asia.
On any dyno chart, the HP curve and torque curve in lb-ft always cross exactly at 5252 RPM. This is not a coincidence, it is a math fact from the formula HP = T times RPM divided by 5252. Below 5252 RPM, torque is higher than HP. Above 5252 RPM, HP is higher.
Motor efficiency is the ratio of output mechanical power to input electrical power. Typical AC induction motors are 85 to 92 percent efficient at full load. Premium efficiency motors hit 94 to 96 percent. Below 50 percent load, efficiency drops fast, so size the motor to run near its rated power.
For a 3-phase AC motor, input power in kW equals voltage times amps times power factor times square root of 3, all divided by 1000. A typical 5 HP motor at 460 V pulls about 6.5 amps at full load. Output HP equals input kW times efficiency divided by 0.746.
Power factor is the ratio of real power to apparent power in an AC system. Motors draw reactive power for magnetizing, which lowers the power factor. Typical induction motors have a power factor of 0.80 to 0.90 at full load. Power factor correction capacitors can bring this up to 0.95 or higher.
Cutting torque comes from the material removal rate. For a 0.5 inch diameter endmill cutting 0.1 inch deep in aluminum at 400 SFM, the spindle torque is about 0.6 lb-ft at 3000 RPM, which is 0.35 HP. Steel requires 3 to 5 times more power than aluminum for the same cut.
Torque accelerates the load from zero. HP sustains speed under load. A high-torque low-HP motor starts heavy loads well but cannot keep them spinning fast. A high-HP low-torque motor reaches high speed but may stall under heavy starting load. For CNC spindles, both matter across the RPM range.
Service factor is a multiplier on nameplate HP that tells you short-term overload capacity. A 5 HP motor with SF 1.15 can deliver 5.75 HP for short periods without damage. Use service factor for startup inrush or intermittent peaks, not for continuous duty.

Common Motor Ratings (3-Phase NEMA / IEC)

HP kW Torque @ 1800 RPM Full-Load Amps (460V) Typical Use
0.5 HP0.37 kW1.5 lb-ft (2.0 N·m)1.0 ASmall conveyor, fan
1 HP0.75 kW2.9 lb-ft (4.0 N·m)1.5 ADrill press, pump
3 HP2.24 kW8.8 lb-ft (11.9 N·m)4.2 ABench mill, lathe
5 HP3.73 kW14.6 lb-ft (19.8 N·m)6.8 ABridgeport-class mill
7.5 HP5.6 kW21.9 lb-ft (29.7 N·m)9.6 AVMC, CNC lathe
10 HP7.46 kW29.2 lb-ft (39.6 N·m)12.5 ALarger VMC spindle
15 HP11.2 kW43.8 lb-ft (59.4 N·m)19 AHeavy-duty VMC
25 HP18.6 kW73.0 lb-ft (99 N·m)30 ALarge lathe / mill
50 HP37.3 kW146 lb-ft (198 N·m)60 AProduction machining
100 HP74.6 kW292 lb-ft (396 N·m)118 AIndustrial compressor

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