EV Motor Power Calculator

Tractive force and motor power to drive an EV at a chosen speed, road grade and acceleration.
Force & Power

Road Load & Motor Power

F = m·g·Cr·cosθ + ½ρCdA·v² + m·g·sinθ + m·a  •  P = F·v / η
Sedan, 100km/h flat
Sedan, 8% grade
SUV, 120km/h
kg
km/h
%
m/s²
%
Enter values and press Calculate.

The Road-Load Equation

To move a vehicle, the motor must overcome three steady forces plus any acceleration. Rolling resistance (tyres) is m·g·Cr·cosθ; aerodynamic drag is ½·ρ·Cd·A·v² and grows with the square of speed; gradient force for climbing is m·g·sinθ; and acceleration adds m·a. Multiplying the total tractive force by speed gives the wheel power, and dividing by the drivetrain efficiency gives the electrical motor power.

ForceFormula
Rolling resistanceFr = m·g·Cr·cosθ
Aerodynamic dragFd = ½·ρ·Cd·A·v²
Gradient (climb)Fg = m·g·sinθ
AccelerationFa = m·a
Wheel powerP = (Fr+Fd+Fg+Fa) · v

Grade is entered as a percentage (rise/run × 100); the angle is θ = arctan(grade/100). At low speed rolling resistance and grade dominate; at highway speed aerodynamic drag dominates because of the v² term. Uses g = 9.81 m/s² and air density ρ = 1.225 kg/m³.

Real-World Applications & Examples

Worked examples

1. Sedan at 100 km/h, flat. m=1600 kg, Cd=0.28, A=2.3 m², Cr=0.012. v=27.8 m/s. Fr=188 N, Fd=½×1.225×0.28×2.3×27.8²=305 N. P=(188+305)×27.8≈13.7 kW at the wheels.
2. Motor power. At 90% drivetrain efficiency: 13.7/0.9≈15.2 kW (about 20 hp) to cruise — far less than the peak rating.
3. Climbing an 8% grade at 60 km/h. The gradient force m·g·sinθ≈1600×9.81×0.0797=1251 N dominates, pushing wheel power to roughly 28 kW.
4. Highway drag. Doubling speed from 50 to 100 km/h quadruples the drag force — which is why energy use rises steeply on the motorway.
5. Hard acceleration. Adding a=2 m/s² to a 1600 kg car needs an extra m·a=3200 N of force — the reason peak motor power far exceeds cruise power.
6. Consumption estimate. 15.2 kW at 100 km/h means 15200 W / 100 km/h = 152 Wh/km, matching a typical sedan.

Frequently Asked Questions

How do I calculate the power needed to drive an EV?

Add the rolling, aerodynamic, gradient and acceleration forces to get the total tractive force, multiply by speed for wheel power, then divide by drivetrain efficiency: P = (Fr+Fd+Fg+Fa)·v/η.

What is rolling resistance coefficient (Cr)?

Cr is the ratio of rolling drag force to vehicle weight, typically 0.010–0.015 for car tyres on tarmac. Lower values mean less tyre drag and better efficiency.

What is a typical drag coefficient and frontal area?

Modern cars have drag coefficients of about 0.24–0.32 and frontal areas of roughly 2.1–2.8 m². SUVs are higher on both counts, which raises their highway energy use.

Why does aerodynamic drag matter more at high speed?

Drag force grows with the square of speed and drag power with the cube, so at highway speeds aerodynamics dominates the energy budget, while at low speeds rolling resistance and gradient matter more.

How does road grade affect power?

Climbing adds a force m·g·sinθ. Even a modest 8% grade can add over a kilonewton of force for a typical car, often exceeding the combined rolling and drag forces at moderate speed.

How do I convert grade percentage to an angle?

Grade percent is rise over run times 100, so the angle is θ = arctan(grade/100). A 10% grade is about 5.7°. For small grades, sinθ is close to grade/100.

Why is cruise power so much less than motor rating?

Cruising on the flat needs only 10–20 kW, but the motor is rated for acceleration, hill climbing and overtaking, which need brief bursts of 100 kW or more. Peak power is sized for performance, not cruising.

How do I turn power into Wh/km?

Divide the electrical power (in watts) by the speed (in km/h): Wh/km = P(W)/v(km/h). For example, 15 kW at 100 km/h is 150 Wh/km.

What air density should I use?

Standard sea-level air density is about 1.225 kg/m³ at 15 °C. It falls at altitude and in hot weather, slightly reducing aerodynamic drag.

What drivetrain efficiency is realistic?

EV drivetrains (inverter, motor, gearbox) are typically 85–95% efficient. Use around 90% for a first estimate; the motor draws a little more electrical power than the wheels receive.

How much power do I need for a target acceleration?

Add the acceleration force m·a to the road load, then multiply by speed. Because this term is m·a, heavy vehicles need proportionally more force and power to accelerate hard.

Does regenerative braking change these numbers?

The tractive power here is for propulsion. During braking the same forces act in reverse, and regeneration can recover part of the kinetic and gradient energy — see a regenerative braking calculator for that.

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