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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 equation 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 to maintain a particular speed is:
Power = (power to overcome rolling drag) + (power to overcome aerodynamic drag)
so
Power in Watts = ((Mass in kg) (9.8m/s²) (Velocity in m/s) (Rolling Resistance)) + (0.5)(1.293 kg/m³) (Coefficient of Drag) (Area in m²) (Velocity^3))

The 9.8m/s is acceleration due to gravity, 0.5 is a constant from the derivation of the formula, the 0.6465 is the density of air in kg/m^3 (a different air density could be substituted if desired for different conditions). The first velocity is the speed that the vehicle is moving on the road; the second velocity is the relative speed of airflow over the vehicle, which is equal to the road speed of the vehicle in the absence of wind. 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 60 mph (converted to m/s) might be 20kW, driving for ninety minutes would mean you use (20 000)(1.5) = 30 kWh 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 20 kW to travel 60 mph then you will get (20 000 W)(1hr)/(60 miles/hr) = 333 Wh/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/877/898586/topics/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.
 

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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
 

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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
 

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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.
 

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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.
 

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Apologies, I wanted to ask a question based on the Power requirement formula - I was unsure which whether the Velocity^3 mentioned in the formula was referring to the velocity at the particular speed the EV was travelling
 

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I wanted to ask a question based on the Power requirement formula - I was unsure which whether the Velocity^3 mentioned in the formula was referring to the velocity at the particular speed the EV was travelling
I have clarified the original post, adding this explanation of the "velocity" of the first formula:
The first velocity is the speed that the vehicle is moving on the road; the second velocity is the relative speed of airflow over the vehicle, which is equal to the road speed of the vehicle in the absence of wind.
So, yes, it is the speed at which the EV is travelling.

You may have been confused by post #6, which introduced incorrect interpretations of the terms of the formula, and was not answered.
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-

...
3)Velocity in m/s=Velocity of vehicle that is in initially (Like 0m/s).What is it ? CONFUSING
...
8)Velocity^3=Velocity of vehicle ?Final Velocity? CONFUSING
No, these are not the correct velocities. The formula being discussed calculates the power to overcome rolling and aerodynamic drag at a constant speed. The "velocity" values are not initial and final velocities during acceleration; they are both the speed at which the EV is travelling.

To clarify this, I added:
Power = (power to overcome rolling drag) + (power to overcome aerodynamic drag)
... and the explanation that this is the power required to maintain a particular speed, not to accelerate or to climb a grade.
 

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... and the explanation that this is the power required to maintain a particular speed, not to accelerate or to climb a grade.
Well it does mean that since the drag is only there when you reach that point that you will get at that point.
Its not that you would need more power to come there. Offcourse all conditions must be equal as in the calculation.
Also i'm not saying that you will get there quick...especially the last couple of kph/mph will take ages because the delta in forces gets smaller, the acceleration will slink fast...
 
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