Calculation
P = 2π · n · M / 60
P = M·ω; at fixed speed, double torque means double power.
P = M·ω
Power from Torque and Speed Calculator
P = M·ω; at fixed speed, double torque means double power.
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Result
Select a target and calculate.
Understand the inputs
- Shaft power P — The power transmitted at the shaft – how fast energy flows through the rotation. It follows from torque M and speed n; at the same torque, a higher speed means more power.
- Torque M — The torque transmitted at the shaft. Source: a torque sensor, a motor's nameplate (rated torque), or as the result once P and n are known.
- Rotational speed n — The shaft speed in revolutions per minute (rpm). Source: a tachometer, a motor's nameplate, or a design specification.
Example
47.75 N·m at 1,500 rpm is about 7.5 kW.
Assumptions and limits
Constant torque and 100% shaft-to-load efficiency.
Understand Power from Torque and Speed Calculator
This calculator finds the missing quantity among shaft power P, torque M and rotational speed n from the other two. It shows the direct link between drive sizing, torque and speed, and explains the practical shorthand formula P[kW] ≈ M[N·m]·n[rpm]/9,550.
What does this quantity describe?
The mechanical power of a rotating shaft is the product of torque M and angular velocity ω: P = M·ω. Since rotational speed n is usually given in revolutions per minute, this converts to P = 2π·n·M/60 in watts, or approximately P[kW] = M[N·m]·n[rpm]/9,550.
Shaft power can be pictured like a rower's output: it rises both with the force per stroke (torque) and with the number of strokes per minute (speed). A motor with double the torque at the same speed delivers double the power — as does a motor with double the speed at the same torque.
Formula and variables
P = 2π · n · M / 60
P = 2π · n · M / 60P[kW] ≈ M[N·m] · n[rpm] / 9,550M = P / (2π · n / 60)n = P / (2π · M / 60)
| Symbol / input | Meaning |
|---|---|
| Shaft power P | The power transmitted at the shaft – how fast energy flows through the rotation. It follows from torque M and speed n; at the same torque, a higher speed means more power. |
| Torque M | The torque transmitted at the shaft. Source: a torque sensor, a motor's nameplate (rated torque), or as the result once P and n are known. |
| Rotational speed n | The shaft speed in revolutions per minute (rpm). Source: a tachometer, a motor's nameplate, or a design specification. |
Choose the inputs correctly
Enter two of the three quantities. P is mechanical work per time at the shaft considered, M is torque acting there, and n is the corresponding speed. P requires M and n, M requires P and n, and n requires P and M. All values must refer to the same operating point and shaft; speed is entered in revolutions per minute (rpm), not as angular velocity in rad/s.
How to use the calculator
Select the target quantity and enter the two known values with units. For a quick manual estimate, the shorthand P[kW] ≈ M[N·m]·n[rpm]/9,550 is enough.
Worked example
Given M = 47.75 N·m and n = 1,500 rpm, substitution gives P = 2π · 1,500 · 47.75 / 60 ≈ 7,500 W = 7.5 kW. Using the shorthand: P ≈ 47.75 · 1,500 / 9,550 ≈ 7.5 kW — both methods agree.
Understand the result and units
A shaft power of 7.5 kW at 1,500 rpm is the mechanical power transmitted at that exact operating point. At constant power but lower speed (e.g. after an idealized reduction gearbox), torque increases proportionally. In a real gearbox, output power is lower than input power because of efficiency, so input and output quantities must not be mixed without an efficiency correction.
The SI unit of power is the watt (W), usually kW in drive engineering. Torque is given in N·m, speed in rpm or rad/s. The constant 9,550 comes from 60,000/(2π) and directly converts N·m and rpm into kW.
Typical applications
The formula is used to size electric motors, combustion engines, pumps, fans and gearboxes — determining the required drive power from catalogue torque and speed values, or conversely estimating available torque from a known motor power.
Assumptions, limits and common mistakes
The formula assumes constant torque at the given speed and 100% efficiency between shaft and load. Real drives lose part of the power to bearing, gear and windage friction, so the actually usable output power is below the calculated value.
Common mistake: Do not substitute speed in rpm directly as angular velocity in rad/s into P = M·ω without converting first, using n[rpm]·2π/60 = ω[rad/s]. Also, a torque measured at one speed should not be applied uncritically to other speeds of the same motor, since the torque-speed curve is usually not constant.
Frequently asked questions
Why does the constant 9,550 appear?
It combines the conversion from revolutions per minute to radians per second and from watts to kilowatts: 9,550 ≈ 60,000/(2π), letting P[kW] be calculated directly from M[N·m] and n[rpm].
How are power, torque and speed related in electric motors?
For many motors, rated power is roughly constant over a speed range, so available torque falls as speed rises and vice versa.
Can I size a motor for an application with this formula?
It gives the ideal mechanical power without losses. Real motor sizing additionally needs efficiency, starting torque and the load profile.
Which torque and speed values may I combine?
Only values from the same operating point at the same shaft. Rated torque with no-load speed, or input torque with output speed, does not describe a physically matching power value.
Is the result motor input power or shaft power?
It is mechanical shaft power. A motor's electrical input power is higher because of losses and follows only after accounting for motor efficiency.