This thread will show you how to calculate the power required at the wheels to travel at a particular speed in your car on flat ground with no breeze. If you find this article difficult to understand you may wish to read

Power (kW) and Energy (kWh). You can use this eqution and its results to work out your range, size your battery pack, size your motor, work out your top speed or a whole range of things.

**1. The POWER requirements of your car at a particular speed is:**
Power in Watts = ((Mass in kg) (9.8m/sē) (Velocity in m/s) (Rolling Resistance)) + ((0.6465) (Coefficient of Drag) (Area in mē) (Velocity^3))
The 9.8m/s is acceleration due to gravity, the 0.6465 is 0.5 times the density of air in kg/m^3. If you enter the required numbers it will give you your power consumption in watts. Once you have worked that out you can use it to work out the following.

**2. The ENERGY required to maintain the speed in Step 1. for a certain period of time is given by:**
Energy in Watt-Hours = (Power in Watts)(Time in Hours)

For example the power required to travel 60mph (converted to m/s) might be 20kW, driving for ninety minutes would mean you use (20 000)(1.5) = 30kWh of energy. Assuming no efficiency losses (in the drive train or batteries) you would get 90 miles of range with a 30kwh pack. You could use this value for

Sizing your Battery Pack, since the energy stored in your battery pack will be its voltage times its amp-hour rating but please allow for your drive train inefficiency and the characteristics of your batteries.

**3. The EFFICIENCY of you vehicle at the speed from Step 1. is given by:**
Efficiency in Wh/mile = ((Power in Watts)(1 hour))/(velocity in mph)

*
So if it takes 20kW to travel 60mph then you will get (20 000W)(1hr)/60 = 333Wh/mile. Please note that is the efficiency after the wheel, please take into account the energy losses of the electrical and mechanical drive train in order to better reflect your real world efficiency.

A great read on automotive power:

http://wps.aw.com/wps/media/objects/...cs/topic02.pdf
* This equation is incorrect. "Efficiency in Wh/mile = ((Power in Watts)(1 hour))/(velocity in mph)" It should read " Efficiency in Wh/mile = (Power in Watts)/(speed in mph)". Of course the Power in Watts is that which is required for that particular speed. The term "speed" should be used because velocity typically infers a vector quantity.