Road Vehicle Power Demand

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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

Acceleration

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

Gradients

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.
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12 thoughts on “Road Vehicle Power Demand”
  2. 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?
     Reply
    • 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
       Reply
  4. 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.
     Reply
    • 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.
       Reply
  5. Hey Nigel, do you have this page published already? I did the calculations as well and got a little bit of difference in energy / distance
     Reply
    • Hello Andre, the calculations are exactly as per the equations. Maybe I should publish the calculation in full and then we can discuss? Best regards, Nigel
       Reply
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