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/s²) (Velocity in m/s) (Rolling Resistance)) + ((0.6465) (Coefficient of Drag) (Area in m²) (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.

[BLSTIC]

It's worth noting that an EF Falcon (a 1550kg Australian Ford family car) consumes 13kw at the wheels to drive at 105km/h. The car has a cd of 0.31, and is normally powered by a 4 litre engine. It gets 7.5l/100km in this state, and is about 18% efficient.

That figure was actually measured on a dyno, it is not a calculated figure.

It's also worth noting that unless you want to spend an hour getting to your cruise speed you need more power. A figure of around 35kw/1000kg has been proven to be adequate for safe highway overtaking and hill climbing

[Franky.EV]

I've manage to put all these Formulas in a oooCalc spreadsheet : https://www.diyelectriccar.com/forums...tor-45278.html.

I also take into account, grade, power to run the ancillaries, gearbox, etc. to get as close of possible of the need of the ev'ers.

All blue text in the spreadsheet are parameters you can modify.

[dralorg]

Hi, please can you tell me what's the name of the Book you've linked?

Thanks

[winzeracer]

Originally Posted by dralorg

Hi, please can you tell me what's the name of the Book you've linked?

Thanks

Looks like Corbin was the one to paste that link you can PM him here
https://www.diyelectriccar.com/forums/member.php?u=10828

Or if you have specific questions on this matter ask them in the Technical Discussions area. Welcome to the forum!

[ashkar_malik]

Kindly Clarify the formula.I am getting Confused.
Power in Watts = ((Mass in kg) (9.8m/s²) (Velocity in m/s) (Rolling Resistance)) + ((0.6465) (Coefficient of Drag) (Area in m²) (Velocity^3))
For Example-

1)Mass in Kg=Mass Of Vehicle
2)9.8m/s2=Gravity Value
3)Velocity in m/s=Velocity of vehicle that is in initially (Like 0m/s).What is it ? CONFUSING
4)Rolling Resistance =>Fr=Rolling coefficient(like-0.03)*(1200Kg)mass of Vehicle*Gravity(9.8m/s)
5)0.6465=Constant value
6)Coefficient of Drag=0.42(Calculated via Formula)
7)Area in m2=Area of Vehicle(EV)
8)Velocity^3=Velocity of vehicle ?Final Velocity? CONFUSING

[brian_]

The main post's power requirement is shown to maintain speed. That's important to know, but some people might think that's all they need to drive the car. As already mentioned, more power is needed to accelerate:

The POWER required to accelerate your car at a particular speed is:
Power in Watts = ((Mass in kg) (acceleration rate in m/s²) (Velocity in m/s)

This is power needed to accelerate, in addition to the power needed to keep the car moving

So, for instance, if your 1000 kg car needs 8 kW to maintain 20 m/s (72 km/h, 45 mph), and you want to be able to get up to 25 m/s (90 km/h, 56 mph) in a few (say 5) seconds, you need the 8 kW to keep it moving plus another 1000 kg * (5 m/s / 5 s) * 20 m/s = 20 kW, for a total of 28 kW.

This illustrates that while power needed to maintain speed is critical for total energy consumption and range, the peak power installed in a car is determined largely by acceleration requirements.

[brian_]

There's another reason that more power is needed than just the amount required to maintain speed: climbing. When the car climbs, it gains potential energy. That must come from the motor.

The POWER required to raise your car up a grade at a particular speed is:
Power in Watts = (Mass in kg) (acceleration of gravity in m/s²) (grade as a fraction) (Velocity in m/s)

This formula is calculating the weight of the car multiplied by the vertical component of the speed; as usual, power is force (in this case the downward force which is weight) multiplied by speed (in this case vertical component of speed).
The acceleration of gravity is 9.8 m/s² (round it out to 10 m/s² for planning purposes)
Grade is normally given as a percentage; for instance, a steep major highway grade might be 8%, so the corresponding value for the formula is 0.08.

This is power needed to climb, in addition to the power needed to keep the car moving.

So, for instance, if your 1000 kg car needs 10 kW to maintain 25 m/s (90 km/h, 56 mph) on flat ground, but you want to maintain that speed up a 4% grade, you need the 10 kW to keep it moving plus another 1000 kg * 9.81 m/s * 0.04 * 25 m/s = 9.8 kW, for a total of 19.8 kW... or about 20 kW.

This illustrates that while power needed to maintain speed is critical for total energy consumption and range, the sustainable power installed in a car is determined largely by grade climbing requirements.