How to calculate per mile cost

of an Electric Vehicle

The calculation of the cost of running an electric vehicle depends on how the vehicle is driven, and the cost of electric power. What I am attempting to do on this page is show the cost of the various forces acting against the vehicle.

Any vehicle moving on the ground and through air has resistance losses. Energy is needed to overcome rolling resistance that is include the flexing of the tires and the friction within the drivetrain. The rolling resistance of a typical radial tire is about .01, meaning that the force needed to push that tire is about 1% of the weight put on the tire. Bias-ply require about four times as much energy to push them, for a given weight, and special low resistance tires made for electric vehicles may require half the energy. Note that rolling resistance is measured at maximum inflation pressure and that decreasing tire pressure will increase rolling resistance. [The tires have to flex more.]

Air resistance is the major loss above about 35 mph. Air resistance depends on frontal area of the vehicle, the corefficent of drag, altitude and temperature. To simplify this discussion, sea level and 68 degrees Fahrenheit (20 degrees Celsius) is assumed. For frontal area in square feet and speed in miles per hour, this leads to an air resistance coefficient of 0.459. Air resistance increases with frontal area and the square of speed.

With this information, a short basic program [shown below] can be used to calculate the cost of running on level ground at a steady speed as a function of vehicle weight and speed. Please go see Air Resistance Calculations

For an example, a 3000 lb car with .01 rolling resistance, 20 ft2 frontal area, a 0.46 coefficent of drag, running level at steady speed:

at 25 mph uses 3.26 hp and costs  3.2 cents/mile,
at 35 mph uses 6.25 hp and costs  4.4 cents/mile,
at 45 mph uses 10.9 hp and costs  6.0 cents/mile,
at 55 mph uses 17.8 hp and costs  7.9 cents/mile,
at 65 mph uses 27.3 hp and costs 10.3 cents/mile,

assuming $0.20 / KW-Hr, 85% motor & 70% battery efficiency.

To this steady state cost must be added the cost of going up a hill or increasing speed.

The energy needed to drive a 3000 lb vehicle up 1000 ft. is 3000000 ft-lb which is equal to 1.13 KW-Hr. If you multiply by 14/10 (battery eff) and 0.85 (motor eff) you need 1.86 KW-Hr from the wall. At $0.20 per KW-HR, this is a cost of about $0.37

To accellerate a 3000 lb vehicle from 0 to 60 mph (88 fps) is the same as lifting the vehicle 120.4 feet. In other words, With 1.0 g accelleration, 2.74 seconds is needed to go from 0 to 60 mph in a distance of 120.4 ft. Therefore 120.4 * 3000 = 361194 ft-lb or .136 KW-Hr Multiply by 14/10 (battery eff) and 0.85 (motor eff) = .162 KW-Hr. At $0.20 per KW-HR, this is a cost of $0.0324.

In short, the cost of running the vehicle goes up with weight (linear) and the cost of overcoming air resistance goes up with the square of speed.