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Of course aerodynamics factors into this question, and the faster you are moving the more air you have to push out of your way, the more energy you use. But would the difference be only a small percentage change or would it be a lot more than that. Essentially would a certain unit of energy be able to move a certain unit of mass a relatively fixed distance, or will that distance be reduced with increases in speed.

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3 Answers 3

The drag on a moving object is given by the approximate expression:

$$ F = \frac{1}{2} \rho v^2 C_d A $$

For my Ford Focus $C_d$ = 0.32 and $A$ = 2.12 m$^2$. The density of air at STP is about 1.2 kg/m$^3$, so at the UK motorway cruising speed of 70 mph (31.3 m/sec) the drag is:

$$ F = \frac{1}{2} \times 1.2 \times 31.3^2 \times 0.32 \times 2.12 = 399 N$$

The power expended in overcoming this drag is force times velocity:

$$ P = F v = 399 \times 31.3 = 12.5kW $$

If we do the same calculation for 50 mph (22.35 m/sec) we get P = 4.5kW. The batteries in the EV1 electric car hold from 16 to 26kWh depending on which battery pack was used. Assume my Focus has a battery with the higher capacity then at 70 mph we get a range of 146 miles and at 50 mph we get 289 miles.

In practice I don't think the difference is as great as that. With my (petrol fuelled) Ford Focus the fuel consumption at 50 is about two thirds of that at 70, and the calculation above suggests I should see about a halving of fuel consumption. The difference is presumably down to frictional and viscous losses in the engine and transmission.

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"With my (petrol fuelled) Ford Focus [...]" Internal combustion vehicles have a speed and gear-train dependent engine efficiency factor which really wrecks merry havoc with this kind of BoTE work. My father-in-law really does see the difference between 45 and 65 (MPH) in his Prius, and often chooses the slower back roads when he isn't in a rush because of this. –  dmckee Feb 27 '13 at 20:36
    
You got 12.5kW and 4.5kW, isn't that be suggesting a 3x spread instead of a 2x spread? –  Dan Neely Feb 27 '13 at 21:09
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@DanNeely: remember that at 70 mph the power may be 3x higher but you're moving 40% further in each second. The energy per unit distance works out a factor of 2 different. –  John Rennie Feb 28 '13 at 6:43

Air resistance goes up with the square of your speed. Double the speed, quadruple the air resistance. At low speeds, most of your power is used to overcome friction with the ground and in your mechanical systems. Those losses go up linearly. Somewhere around 10-60MPH (depending on the vehicle) the energy required to defeat air resistance becomes higher, and continues to grow faster, than the energy required to defeat friction.

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Sorry for the duplicate response Sparr; I was already far in the response when I noticed there was another. –  joshphysics Feb 27 '13 at 18:14

Well there are at least three factors that affect the range:

  1. Energy dissipated as heat through friction inside of the engine/car (there are a lot of moving parts in there that waste a lot of energy).

  2. Air resistance (as you point out).

  3. Acceleration. When a car accelerates, the engine does work.

The air resistance (drag) that a car experiences is proportional to the square of its instantaneous speed at high speeds, and it is proportional to the speed itself at low speeds. In either case, the we see that at low speeds, where air resistance becomes negligible, we have to worry most about frictional energy losses, while at high speeds, we have to worry most about drag.

Having said all of this, suppose we are driving at low speeds in a very aerodynamic vehicle, then our range will increase significantly if we don't start and stop constantly. This happens because starting and stopping requires acceleration, and acceleration requires a large amount of energy to be expended because the engine needs to do work to propel the car forward. So to be most efficient, you should generally drive at a relatively low speed without stopping.

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