Solve power using linear or rotational motion. Choose units for work, speed, force, and time. Export results to share calculations with your team quickly.
| Mode | Inputs | Result (W) | Result (kW) |
|---|---|---|---|
| Work ÷ Time | W = 5000 J, t = 10 s | 500 | 0.5 |
| Force × Velocity | F = 200 N, v = 3 m/s, θ = 0° | 600 | 0.6 |
| Torque × Angular speed | τ = 50 N·m, ω = 1500 rpm | 7854 | 7.854 |
Mechanical power describes how fast energy is transferred by motion. In machines, it links performance to real limits like motor size, fuel use, and heat. For example, 1 mechanical horsepower is about 745.7 W, so a 2 hp tool can deliver roughly 1.49 kW before losses.
This calculator supports three industry-standard models: P = W/t for energy over time, P = F·v·cos(θ) for linear motion, and P = τ·ω for rotating shafts. These forms are consistent because they all represent energy rate, expressed in watts.
Inputs can arrive in mixed units, so conversions are essential. Work may be entered as J, kJ, MJ, Wh, kWh, or ft·lbf. Speed can be m/s, km/h, mph, or ft/s. Torque can be N·m, kN·m, or lbf·ft. The result table converts to W, kW, MW, mechanical hp, and metric PS.
Use the work method when you already know energy transfer. If 5,000 J is done in 10 s, the ideal power is 500 W. This mode is common for lifting tests, lab experiments, and energy audits where time is measured directly and the work estimate is reliable.
The force method is sensitive to direction. Only the component of force along the motion does work, handled by cos(θ). With θ = 0°, power is maximized; at θ = 90°, power becomes 0 W. Negative power can occur if θ is greater than 90°, indicating braking or energy absorption.
For motors and turbines, torque and angular speed are often measured. If τ = 50 N·m at 1500 rpm, ω ≈ 157.08 rad/s and ideal power is about 7.85 kW. This directly supports sizing of gearboxes, couplings, and variable-speed drives.
Real systems lose energy to friction, heat, and electrical losses. Enter an efficiency to estimate delivered output: Pdelivered = Pideal × η. Typical ranges are 70–95% for many mechanical transmissions, but check manufacturer data when precision is needed.
Use the detail table to verify conversions before trusting the final number. Sanity checks help: doubling force doubles power, doubling speed doubles power, and doubling rpm doubles rotational power. If results look too large, confirm time units and angle sign.
Ideal power comes directly from the selected equation. Delivered power multiplies ideal power by your efficiency setting. Use delivered power when you want a realistic output estimate for a machine or drivetrain.
Enter the angle between the force direction and the velocity direction. If you push exactly along the motion, use 0°. If the force is perpendicular to motion, use 90°.
Yes. In the force method, angles greater than 90° make cos(θ) negative, representing braking or energy absorption. In rotating systems, negative torque relative to rotation can also indicate regenerative or resisting power.
It uses ω = rpm × 2π/60. This conversion is required because torque power uses P = τ·ω with ω in rad/s for watts.
If you have manufacturer data, use that value. Otherwise, choose a reasonable estimate based on the system: well-lubricated gear trains and belts are often high, while older or heavily loaded systems may be lower.
No. Mechanical horsepower is about 745.7 W, while metric PS is about 735.5 W. The calculator shows both to avoid confusion when comparing equipment ratings.
Use Work ÷ Time when energy transfer is known, Force × Velocity for linear motion with direction, and Torque × Angular speed for motors and shafts. Choose the mode that matches your measurements.
Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.