Power from Force and Velocity Calculator

Definition of power and the formula P = F·v

Power P is the rate of energy transfer, that is the work done per unit time: P = dW/dt. If the work of a force F over a displacement ds is dW = F·ds·cos θ, where θ is the angle between the force and the displacement, then dividing by dt gives P = F·v·cos θ, because ds/dt = v. For a force directed along the motion cos θ = 1 and the formula reduces to P = F·v.

This formula describes instantaneous power, the value at a given moment, not an average. When force and speed are constant the two coincide; when they vary in time, the average power is the total work divided by the time. From P = F·v it follows at once that with a limited engine power, a higher speed means less available force — hence the gearbox, which trades power for more force at lower speed.

Historical note: Watt, horsepower, and the watt

James Watt, selling improved steam engines in the 1780s, needed a measure to compare them with horses working mill wheels. From measurements at a brewery he adopted that a horse lifts 33,000 pounds one foot in one minute: thus horsepower was born, today 745.7 W. The continental metric horsepower (PS, ch, KM) was later defined as 75 kgf·m/s, that is 735.49875 W; the two units differ by 1.4 %.

The metric unit of power was named after Watt at the 1882 meeting of the British Association for the Advancement of Science, and in 1889 the second International Electrical Congress adopted it. The watt entered the SI as a derived unit: 1 W = 1 J/s = 1 kg·m²/s³. In the United States engine power is still quoted in horsepower, while European Union directives since 2010 require kilowatts, with PS allowed only as a supplementary value.

Derivation from the definition of work

Step 1. The work of a constant force F over a displacement s along its direction is W = F·s. Step 2. The average power over time t is P_avg = W/t = F·s/t. Step 3. For uniform motion s/t is the speed v, so P = F·v. In the limit t → 0 the same calculation gives the instantaneous power P = F·v(t) for variable motion as well.

Step 4. For F = 600 N and v = 25 m/s: P = 600·25 = 15,000 W = 15 kW; in horsepower 15,000/745.7 = 20.1 hp, in metric horsepower 15,000/735.5 = 20.4 PS. Step 5. Work in one hour at constant power: W = P·t = 15,000 W·3,600 s = 5.4·10⁷ J = 54 MJ = 15 kWh. The same result follows from W = F·s with the distance s = v·t = 25·3,600 = 90 km (56 mi): 600 N·90,000 m = 5.4·10⁷ J.

Assumptions: force along the motion and instantaneous power

The formula P = F·v assumes that the entered force is the component along the velocity. Engine thrust, air drag, or the tension in a tow rope usually meet this condition; gravity on a level road or the centripetal force in circular motion are perpendicular to the velocity and contribute no power. When the force acts at an angle θ, the value F·cos θ should be entered in the “Force” field.

The second assumption concerns the instant. The calculator reports the power at a given moment; for a trip with varying speed and force one must integrate F·v over time or divide the total work by the time. Third: the power is the mechanical power at the wheel, hook, or piston. The engine power quoted by a manufacturer is larger by the drivetrain losses (usually 10–15 %), and the power drawn from the grid by an electric motor is larger by the losses in windings and iron.

Power on the road: where the kilowatts come from

Air drag grows with the square of speed: F_d = ½·C_d·ρ·A·v², so the power to overcome it grows with the cube of speed. For a passenger car with C_d·A ≈ 0.7 m² at ρ = 1.2 kg/m³ and 25 m/s the drag is about 260 N, and the rolling resistance (coefficient 0.01, mass 1,500 kg) about 150 N; with a margin for road irregularities that gives the order of 600 N and 15 kW — as in the example on this page. At 50 m/s (112 mph) the drag alone grows to 1,050 N and the power to more than 60 kW.

A bicycle on a level road at 30 km/h (19 mph) meets a total resistance of about 20 N, which gives 166 W — the power a trained amateur sustains for an hour. Uphill on a 10 % grade with a mass of 80 kg the weight component of 78 N is added, so even at 2 m/s one needs 200 N·2 m/s = 400 W. A 40-ton truck on the highway needs about 15 kN of tractive force and 375 kW, the power of a typical 500 hp engine.

Units and conversions: W, kW, hp, PS, and kWh

Conversion factors used by the calculator: 1 kW = 1,000 W, 1 MW = 10⁶ W, 1 hp = 550 ft·lbf/s = 745.6999 W, 1 PS = 735.49875 W (exactly, from the definition 75 kgf·m/s at g_n = 9.80665 m/s²). In US units the force is given in pounds-force (1 lbf = 4.4482 N) and the speed in ft/s or mph; the product in lbf·ft/s is divided by 550 to obtain hp. Work in one hour: 1 kWh = 3.6 MJ.

Orders of magnitude for orientation: a resting human gives off about 80–100 W of heat, a professional cyclist sustains 400 W for an hour, a working horse — despite the name of the unit — about 750 W over a working day and up to several kilowatts in a short burst, a passenger car engine 80–200 hp, an electric locomotive 4–6 MW, a jet engine of an airliner in cruise 10–20 MW of thrust power, a power plant unit 500–1,000 MW.

Sources

The definition, derivation, and historical note come from general physics textbooks and metrology sources: D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, chapter on work and kinetic energy (power); H. D. Young, R. A. Freedman, University Physics, chapter “Work and Kinetic Energy,” section “Power”; R. P. Feynman, R. B. Leighton, M. Sands, The Feynman Lectures on Physics, vol. I, chapter 13 (Work and Potential Energy); BIPM, The International System of Units (SI), 9th edition (the watt as a derived unit); NIST SP 811 (conversion factors for hp, lbf); SAE J1349 (engine power rating); H. W. Dickinson, James Watt: Craftsman and Engineer, Cambridge University Press, 1936. This page is for educational purposes.