Road Vehicle Power Demand

Let’s look at the road vehicle power demand for the simple steady state condition. We will look at the tractive effort and power required:

  • Aerodynamic forces
  • Rolling resistance
  • Acceleration
  • Gradients
  • Hotel loads

Aerodynamic Forces

The aerodynamic forces on the vehicle are proportional to velocity squared and hence become dominant at high speed. FA is the frontal area and Cd is the drag coefficient. ρair is the density of air and we assume 1.225kg m-3

Rolling Resistance

In the simplest terms the rolling resistance is a fixed term dependent on the vehicle mass, gravity and the tyre coefficient of rolling resistance (Rr).

Typical values for car tyres are an Rr ~ 0.006 to 0.01


If we accelerate the car very quickly we need to apply a force to do so.


Drive the vehicle up a slope and the additional force is dependent on the mass, gravity and the angle of the slope.

Total Traction Force

The 5% grade at 65mph (105km/h), at GVW, assuming a tyre rolling resistance coefficient of 0.0044, a motor+inverter efficiency of 94% (assuming this peak has been set for the highway cruise) and a gearbox efficiency of 0.97 give a requirement of 698kW of electrical power.

This post has been built based on the sponsorship of: h.e.l group and LiTech Power

Hotel Loads

These are all of the loads on the car such as: 12V electrical system load and HV loads such as air conditioning or heating.

The preset values are those for the estimation of the Tesla Semi. Change the gradient to 0% and you will see the energy use at 105km/h on the flat calculated as just below 1.1kWh/km.

Many questions about the Energy per Unit Distance Requirement from the Battery, this is a simple calculation taking the output power from the battery in Watts and dividing by the Vehicle Speed [km/h].

16 thoughts on “Road Vehicle Power Demand”

  1. How is the output at motor shaft calculated? for a speed of 132km/h and an acceleration of 1.67m/s, the output shaft power is coming as 214kW something. I have not entered the shaft dia (for motor). Is there a default shaft dia already considered in the formula?

    • All of the forces acting on the vehicle are added up, this gives total tractive effort required at the Tyre contact with the road.

      This is converted into a power: work is force times displacement (Work = Force x Distance), and velocity is displacement over time (Velocity = distance / time), power equals force times velocity: Vehicle Power = Force x Velocity

      The motor has to drive through a gearbox, here I have used a single number for the efficiency of that gearbox. Motor Power = Vehicle Power x Gearbox Efficiency

  2. Gearbox efficiency can be significant in terms of marginal gains (or losses) and is by no means a constant (even down to differential cooling impacts depending on location). Even without such complications, model shown does illustrate how the hard work on optimising battery can be undone, at least the trade-off costs are more tangible for an EV. Keep up the good work Nigel.

    • Completely agree, the gearbox efficiency is a complex system in itself and the single number does not do it justice. As you say, the intention is to use this for the high level directional calculations.

    • Hi Vishal, explanation added to the bottom of the page: Many questions about the Energy per Unit Distance Requirement from the Battery, this is a simple calculation taking the output power from the battery [Watts] and dividing by the Vehicle Speed [km/h].

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