How much of the energy of a car is required for overcoming air resistance?

How much of the energy of a car is required for overcoming air resistance? For example, assume a car going 65 mph has a mileage of 25 gpm. If all air resistance was eliminated, approximately what would be the gain in mileage? I am trying to get a general idea of how much air resistance contributes to the forces opposing the forward motion of a car.

• fluid-dynamics
• efficient-energy-use

$\begingroup$ Since I don't have exact numbers, all I can say is almost all the opposing force is due to air resistance. It's proportional to the square of speed. But the engine itself is not super efficient and a lot of energy is lost in the engine itself. So there's a distinction between "total losses" and "opposing forces". $\endgroup$

$\begingroup$ @tpg2114 I am including internal friction and other inefficiencies in the engine as "opposing forces". $\endgroup$

$\begingroup$ Well, they don't oppose the motion of the car though. A car sitting perfectly still will have losses, despite having no motion or forces to oppose. $\endgroup$

4 Answers 4

This started as a comment but I will expand it as an answer since it warrants it.

When you look at something like miles per gallon, you are looking at all of the total losses in the car. This is air drag, friction with the ground, and losses in the engine itself. To put it into perspective, the average efficiency of an internal combustion engine is only around 20%. So there's a lot of energy lost just in the engine.

But to answer your direct question, almost all of the opposing force on the vehicle is air drag. Friction with the ground is very small. The rolling resistance is proportional to the weight of the vehicle and for car tires the coefficient is small, we'll round and say 0.01.

For air drag, it depends on the drag coefficient and surface area of the car. This coefficient is orders of magnitude larger than the rolling resistance. But drag also depends on the speed squared. So doubling your speed causes a force 4 times larger. This is a completely dominant effect once the speed is big enough to be interesting (a few mph).

To put some equations to it, $F_f = c_f W$ is the rolling resistance force with $c_f$ the coefficient of friction and $W$ the weight of the car. We will assume that the weight is proportional to the volume. $F_d = 1/2 \rho c_d A U^2$ is the air drag force with $\rho$ the air density, $c_d$ the drag coefficient, $A$ the surface area and $U$ the speed.

If $W = \rho_c g V$ where $\rho_c$ is an approximate density of the car. We can approximate a car as a rectangle with height $h$, width $d$ and length $l$. This means the volume of the car is $V = hdl$ and the frontal surface area is $A = dh$ giving $A = V/l$.

With that out of the way, let's take the ratio of the forces:

That looks kind of confusing, so let's just look at the terms in it. If the car is roughly 2 meters wide and 5 meters long (not uncommon numbers), then we can guess that $l \approx 5$. I'm just guessing here, but I'll say $\rho_c \approx 100$ (let's say 1500 kg and using the dimensions for our cube). The density of air is around 1. So this gives $\rho/\rho_c \approx 0.01$. We know that $c_d/c_f \approx 50$ also. This gives us something that looks like:

$$ \frac \approx 50\times0.01\times U^2/5/10 \approx 0.1\times U^2$$

which turns out to be a really simple:

which means that once you get over roughly $11~\text$ for speed (roughly 21 mph), the air drag begins to take over as dominant. Obviously this is all approximate, but it should make it pretty clear that once a car is moving, air drag dominates.

I should note that the surface are approximation here is much larger than the real surface area. However, it doesn't change the results by all that much. The point is, at any speed you are likely to encounter outside of a parking lot, the air drag will contribute the majority of the opposition.