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Get a motor’s torque from its nameplate, a shaft’s power from measured torque, or the speed for a given power and torque. Add the reduction to the machine and see the torque reaching the shaft, what the reducer must handle with a service factor and the peak at starting. Units convert on their own: hp, kW, lbf·ft, lbf·in and N·m.
Pre-sizing and estimating tool: results do not replace design, calculations or reports signed by a licensed professional, nor verification against the applicable codes and standards.
Constant-power curve (hyperbola): the same motor can give high torque slowly or low torque fast.
How to read: the horizontal axis is on a log scale (each 1-2-5-10 mark multiplies speed). Every point on the curve has the same power; moving left (less speed) makes torque rise. The "motor" point is the input shaft; the "output" point is the machine shaft after the reduction: lower speed, higher torque, almost on the same curve (the small gap is the drive loss).
What the machine shaft, coupling and reducer must withstand.
How to read: the top bar is the continuous working torque; the middle one, that torque × service factor, the minimum rated torque of the reducer or coupling to select; the bottom one, the short peak at starting, which shaft, key and reducer must survive without breaking.
Standard ratings and rated torque at typical full-load speed (about 1,750 rpm at 60 Hz and 1,450 rpm at 50 Hz; each nameplate varies from 1,700 to 1,780 and 1,400 to 1,480 rpm).
How to read: find the motor rating and read the shaft torque at your supply frequency. The highlighted row is the closest to your case. 2-pole motors have half this torque; 6-pole about 1.5 times; 8-pole twice.
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Divide power by angular speed: T = P ÷ ω. In practice, T (lbf·ft) = 5252 × hp ÷ rpm. In the example, 10 hp at 1,770 rpm gives 5252 × 10 ÷ 1,770 ≈ 29.7 lbf·ft (40.2 N·m). With kW and N·m, use 9549 × kW ÷ rpm.
There is no direct conversion: hp is power and lbf·ft is torque. You need the speed. With it, lbf·ft = 5252 × hp ÷ rpm. The same 10 hp motor gives 29.7 lbf·ft at 1,770 rpm, but a 10 hp 6-pole motor (1,175 rpm) gives about 44.7 lbf·ft.
Multiply input torque by the ratio and efficiency: T₂ = T × i × η. In the example, 29.7 lbf·ft × 20 × 0.94 ≈ 558 lbf·ft at 88.5 rpm. Power barely changes (it only loses the 6% efficiency); the reducer trades speed for torque.
A multiplier covering shocks, frequent starts and hours of use (AGMA load classes). The selected gearbox must have a rated torque equal to or above the working torque × service factor. In the example, 558 lbf·ft × 1.25 ≈ 697 lbf·ft: choose a unit rated at least that.
Here torque is that of a rotating shaft transmitting power (motor, gearbox, machine). Bolt tightening torque is the wrench torque that stretches a bolt to preload a joint; it depends on diameter, grade and friction, not on power. For that, use the bolt torque calculator.
Because they turn slower for the same power. A 10 hp 2-pole motor (3,500 rpm) gives about 15 lbf·ft; 4-pole (1,770 rpm), 29.7 lbf·ft; 6-pole (1,175 rpm), 44.7 lbf·ft. It is also larger and more expensive, so a 4-pole motor with a gearbox is often the better choice.
It can, if the gearbox is at its limit. On an across-the-line start an induction motor gives 150% to 250% of rated torque for a few seconds, and that peak reaches the machine shaft multiplied by the ratio: about 1,116 lbf·ft in the example. A soft starter or VFD limits the peak.
Important notice: Indus Metrics tools produce estimates and pre-sizing results from the data you enter and from published engineering methods. They do not replace design, calculations or reports signed by a licensed professional engineer (PE), nor verification against the applicable codes, standards and regulations, such as OSHA rules and the ASME codes. Material and fuel properties vary: check manufacturer data or your own process analyses before deciding. By using this site you agree to the Terms of Use.
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