Why the constant is 5252 and what it is doing there
Power is torque times rotational speed. That is the whole of it: P = Tω, with ω in radians per second. Everything else in this topic is unit bookkeeping, and the mysterious 5252 is the bookkeeping made into a single number.
Work through it. One horsepower is defined as 33,000 foot-pounds-force per minute. A shaft turning at N revolutions per minute sweeps 2πN radians per minute. So a torque of T pound-feet does 2πNT foot-pounds of work per minute, and dividing by 33,000 gives the horsepower: hp = 2πNT/33,000 = NT/5252.113. Rearranged for torque, T = 5252.113 × hp / N. The constant is 33,000/(2π), nothing more.
The metric version is the same identity with different units. One kilowatt is 1,000 newton-metres per second, and N revolutions per minute is 2πN/60 radians per second, so T = 1000 × kW × 60/(2πN) = 9549.297 × kW / N. That constant is 60,000/(2π). If you remember one thing, remember that both constants are just 2π wearing a unit costume.
A useful consequence: since horsepower and torque are equal in magnitude when N = 5252, every dynamometer plot of a machine that reads in horsepower and pound-feet has its power and torque curves crossing at exactly 5,252 rpm. It is not a property of the engine or motor — it is a property of the units.
Which speed to use, and which torque you are asking about
The single biggest source of wrong answers here is using the synchronous speed instead of the nameplate speed. An induction motor never runs at synchronous speed under load; it slips. A four-pole machine on 60 Hz has a synchronous speed of 1,800 rpm and a typical nameplate speed near 1,750 rpm, and that 2.8% difference appears directly in the torque: 10 hp at 1,750 rpm is 30.01 lb-ft, while the same 10 hp at 1,800 rpm is 29.18 lb-ft. Full-load torque means torque at full-load speed.
Then be clear about which torque you want. A motor has several, and they are different numbers on the same machine:
- Full-load torque — rated output power at rated speed. This is what the main result gives you.
- Locked-rotor torque — what the motor produces at zero speed with rated voltage applied. It is not obtainable from the power formula, because mechanical output at zero speed is zero; it is a nameplate or design-letter figure.
- Pull-up torque — the minimum torque developed between zero speed and breakdown. It is the figure that decides whether a high-inertia load actually accelerates.
- Breakdown torque — the maximum the motor can produce before it stalls. Exceed it and speed collapses.
NEMA MG 1 specifies minimum values for locked-rotor, pull-up and breakdown torque by design letter, horsepower and number of poles, which is exactly what the Design A, B, C and D classifications encode. Use the percentage field on this page to convert a data-sheet percentage into real torque, but take the percentage from the machine's own data, not from a rule of thumb.
For an inverter-fed motor the speed is whatever the drive commands, and the constant-torque region below base speed behaves as this formula describes with the power scaled by speed. Above base speed the drive runs in constant-power field weakening, and torque falls as 1/N — which is what the formula says anyway.
Worked example: a 10 hp motor at 1,750 rpm
A 10 hp, four-pole, 460 V motor has a nameplate speed of 1,750 rpm. Find its full-load torque and the torque at 150% of full load.
- Angular velocity. ω = 2π × 1,750 / 60 = 6.283185 × 29.16667 = 183.2596 rad/s.
- Power in watts. 10 hp × 745.6999 = 7,456.999 W.
- Torque in newton-metres. T = P/ω = 7,456.999 / 183.2596 = 40.6907 N·m.
- Torque in pound-feet. 40.6907 / 1.3558179 = 30.0121 lb-ft. Cross-check with the customary constant: 10 × 5,252.113 / 1,750 = 52,521.13 / 1,750 = 30.0121 lb-ft. The two agree.
- At 150% of full load. 30.0121 × 1.50 = 45.0181 lb-ft, or 40.6907 × 1.50 = 61.0361 N·m.
Now the reverse. A torque wrench on the same shaft reads 100 lb-ft while it runs at 1,750 rpm. What power is that? Convert: 100 lb-ft × 1.3558179 = 135.5818 N·m. Then P = Tω = 135.5818 × 183.2596 = 24,845.7 W = 24.846 kW, which is 24,845.7 / 745.6999 = 33.320 hp. Check it with the customary route: 100 × 1,750 / 5,252.113 = 175,000 / 5,252.113 = 33.320 hp. That is well beyond the 10 hp rating, so this machine could not sustain it.
Finally, see what speed does. The same 10 hp at 3,450 rpm gives 10 × 5,252.113 / 3,450 = 15.223 lb-ft, and at 870 rpm it gives 60.369 lb-ft. The 870 rpm machine produces 60.369/15.223 = 3.966 times the torque of the 3,450 rpm one for the same power, which is why an eight-pole motor of a given horsepower is physically far larger: torque, not power, sets the frame size.
How to use the number
Match the motor torque against what the load demands at every speed, not just at rated speed. A centrifugal fan or pump needs torque proportional to the square of speed, so it demands very little at start-up and a standard Design B motor accelerates it easily. A loaded conveyor, a positive-displacement pump or a reciprocating compressor demands close to full torque from zero speed, which is why NEMA Design C exists. A high-inertia load such as a large fan wheel or a centrifuge may need to be checked against pull-up torque and the motor's safe stall time rather than against full-load torque at all.
Use torque, not power, when sizing shafts, keys, couplings and gearboxes. Every mechanical component in the drivetrain is stressed by torque; power is only relevant through the speed it happens to be delivered at. A 10 hp gearbox output at 87 rpm carries 603.7 lb-ft, twenty times what the same power carries at the motor shaft, and that ratio is exactly the gear ratio.
Watch what happens when a motor is oversized. Torque at any given load is set by the load, not by the motor, so a 20 hp motor driving a 6 hp load produces 6 hp worth of torque and runs at poor power factor and reduced efficiency doing it. The motor efficiency calculator quantifies that penalty from a set of clamp-meter readings.
Finally, remember the direction of the derivation. Torque from power is a full-load rated quantity; power from torque is an instantaneous measurement. If you convert a measured breakaway torque into horsepower, you get a number that the motor may deliver for a few seconds and could never sustain, and reading it as a power rating is a real mistake. For converting between power units on their own, use the horsepower to kilowatts calculator.
Torque per unit of power at standard motor speeds
| Poles | Synchronous rpm (60 Hz) | Typical full-load rpm | lb-ft per hp at full load | N·m per kW at full load |
|---|---|---|---|---|
| 2 | 3,600 | 3,450 | 1.5223 | 2.7679 |
| 4 | 1,800 | 1,750 | 3.0012 | 5.4568 |
| 6 | 1,200 | 1,160 | 4.5277 | 8.2321 |
| 8 | 900 | 870 | 6.0369 | 10.9762 |
| 10 | 720 | 690 | 7.6117 | 13.8396 |
| 12 | 600 | 575 | 9.1341 | 16.6075 |
The full-load column uses the representative nameplate speeds shown, not the synchronous speeds — always use your own nameplate figure. On a 50 Hz supply the synchronous speeds are 3,000, 1,500, 1,000 and 750 rpm for 2, 4, 6 and 8 poles, and every torque figure rises by 60/50 = 1.2 for the same power.
Mistakes and limits
- Using synchronous speed for full-load torque. An induction motor slips. A four-pole machine's 1,750 rpm nameplate speed gives 2.9% more torque than its 1,800 rpm synchronous speed does for the same power.
- Confusing input and output power. The nameplate horsepower is mechanical output at the shaft. Input power is higher by the reciprocal of efficiency, and using input power here overstates torque.
- Expecting locked-rotor torque from this formula. Mechanical output at zero speed is zero, so P = Tω cannot produce starting torque. That figure comes from the motor's design letter and data sheet.
- Reading a peak torque as a power rating. Converting a measured breakaway or shock torque into horsepower gives a number the machine may deliver briefly and cannot sustain.
- Mixing pound-feet with foot-pounds of energy. They have the same dimensions and different meanings. Torque is lb-ft; work is ft-lbf. Keeping them separate avoids a whole class of errors.
- Forgetting the gearbox. Torque multiplies by the reduction ratio and speed divides by it, less the gearbox efficiency. Size the low-speed shaft for the multiplied torque.
- Assuming a 50 Hz motor behaves like a 60 Hz one. The same machine on 50 Hz runs 5/6 of the speed, so for the same power it produces 20% more torque, and its magnetic circuit sees a higher volts-per-hertz unless the voltage is reduced in proportion.
Key terms
- Slip
- The difference between synchronous and actual speed, expressed as a fraction of synchronous speed. It is what makes an induction motor produce torque; at zero slip it produces none.
- Design letter
- The NEMA MG 1 classification (A, B, C, D) that sets minimum locked-rotor, pull-up and breakdown torques and maximum locked-rotor current. Design B is the general-purpose default.
- Breakdown torque
- The maximum torque a motor develops at rated voltage and frequency without an abrupt drop in speed. Loading beyond it stalls the machine.
- Service factor
- A multiplier on nameplate horsepower that the motor may deliver continuously at rated voltage and frequency without exceeding its temperature rise limit. A 1.15 service factor motor is not a motor with 15% more torque available indefinitely at rated temperature.
Where torque sits in a motor calculation
Torque is the mechanical half of a motor problem; current is the electrical half. They are linked through voltage, power factor and efficiency, and each side answers different questions. Torque decides the frame size, the shaft, the coupling and whether the load will start. Current decides the conductor size, the overload relay setting and the branch-circuit protection — the motor full-load amps calculator and the motor overload and breaker sizing calculator cover that path under NEC Article 430.
The two meet at starting. Locked-rotor torque and locked-rotor current are properties of the same design, and you generally cannot raise one without raising the other. That is why reduced-voltage starting reduces starting torque quadratically: torque goes as the square of applied voltage, so starting at 65% voltage gives 42% of the locked-rotor torque. A soft starter that solves an inrush problem can create an acceleration problem, and checking the load's demand against the reduced torque is the calculation that catches it.
Variable-frequency drives change the shape of the problem rather than the physics. Below base speed a drive holds volts per hertz constant, so flux and therefore available torque stay roughly constant while power scales with speed. Above base speed the voltage cannot rise further, flux falls, and available torque falls approximately as 1/N with power roughly constant. Both regimes obey P = Tω; what changes is which of the two is held fixed.
For the driven machine, remember that fans and centrifugal pumps follow affinity laws: flow with speed, torque with speed squared, power with speed cubed. Reducing such a load's speed by 20% cuts its power demand by nearly half, which is the whole economic case for variable-speed drives on fans and pumps and is a far larger effect than any efficiency improvement in the motor itself.