The power required to overcome drag is relative to speed cubed. When I'm driving $100 km/h$ my car consumes $~10 $litres/$100km$. At $200 km/h$ the consumption should be $2^3 × 10$ litres/$100km = 80 $litres/$100km$. Obviously, it's a lot less, maybe only double. What is it I don't understand here?
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2$\begingroup$ There are various efficiency factors involved, and efficiency (e.g., engine efficiency) varies with velocity. Without knowing the equations that describe engine efficiency vs. rpm, and various other energy losses vs. speed or rpm, you will not be able to accurately predict fuel consumption. $\endgroup$– David WhiteApr 23, 2017 at 15:18
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2$\begingroup$ Why do you think that drop in km/l should be cubed when drag is cubed??? Do you not get there faster when you're moving faster? $\endgroup$– Hot LicksApr 23, 2017 at 21:00
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3$\begingroup$ Using the speed cubed rule, your per hour consumption should be 8 times more at 200km/h than at 100km/h. That means your per km consumption would be 4 times more. $\endgroup$– Level River StApr 23, 2017 at 21:34
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$\begingroup$ One thing you seem not to be considering is that internal combustion engine efficiency varies considerably with RPM and engine load. You can often find BSFC (Brake Specific Fuel Consumption) for particular cars. For a concrete example, I have a Honda with a VTEC engine and fuel consumption display. The VTEC shift point is about 3000 RPM, and I can see the gauge change about 30% as it shifts in and out. $\endgroup$– jamesqfApr 24, 2017 at 5:51
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2 Answers
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There are two extra things to consider here.
First, in even the absolute simplest case, your car is not just fighting wind resistance (which indeed follows a $F \propto v^2$ law at these velocities) but also various static friction forces, usually following an $F \propto v^0$ law. And as you might imagine some of these forces are dropping based on what gear you're in, as some of the static friction is internal to the engine block. You can also read "constant force" as meaning "constant energy expenditure per unit of distance," which clarifies that something like the pistons compressing air but then that now-hot air being vented out (as it will be) turns out to be a constant force on average.
Now this force that's proportional to the square of velocity might also pick up a horizontal component coming from a cross-breeze of speed $u,$ which looks at first like it doesn't matter (Pythagorean theorem, $|[u,~v]|^2 = u^2 + v^2$) but actually does (because you also have to project it back onto the direction of motion of the car to calculate work, which involves multiplying by $v/\sqrt{u^2 + v^2}$.) So your actual force equation is probably much closer to $$F = F_0 + k v\sqrt{u^2 + v^2},$$ where $k, u$ are probably being approximately constant but potentially $F_0$ might be much lower at 200 km/hr versus 100 km/hr because you have probably shifted to a higher gear. (Realistically the next step in adding accuracy to this model might be writing $k = k_0 + \alpha~u/v$ or so to add the effect that when there is a cross-breeze it takes the drag force over a less-streamlined orientation with respect to the car.)
Second: you are trying to use power $\vec F \cdot \vec v$ to look at fuel consumption per unit distance, but a given amount of fuel probably gives a certain amount of energy and power is an energy expenditure per unit time. Therefore when you want to know fuel consumption per unit distance you need to multiply power by the time it takes per unit distance -- this is the inverse of the velocity. So actually fuel consumption goes like $\vec F \cdot \vec v /|\vec v|$ and your fuel consumption should only scale like: $$F_0 + k v \sqrt{u^2 + v^2}.$$
So in summary, one of these factors of 2 is flat-out wrong for calculating fuel efficiency, the power may go as speed cubed but the energy per unit distance only goes with speed squared in the limit $F_0 = 0,~u = 0.$ The other missing factor of 2 probably comes from the fact that in the lower gear the drag forces $F_0$ and $k v^2$ are approximately comparable, whereas in the higher gear you've reduced $F_0$ considerably by upshifting -- but some components of it probably also come from a slight cross-wind that both acts as a linear drag force and redirects the airflow over a less-aerodynamic profile over the car.
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$\begingroup$ Thanks! So the biggest thing to consider here is fuel consumption per time vs fuel consumption per distance. $\endgroup$– juzzlinApr 23, 2017 at 19:16
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4$\begingroup$ You should also consider that engines have different efficiency at different speeds. You can't always physics your way to the right answer, you have to consider real life engineering too. $\endgroup$ Apr 23, 2017 at 20:12
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2$\begingroup$ @DanielSank Or as I like to say, spherical cars in a vacuum. $\endgroup$– AronApr 24, 2017 at 5:16
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$\begingroup$ @DanielSank Not really at different speeds - it's more about RPM for combustion engines. That's one of the reasons we use transmissions - in an ideal world, no matter your speed, the RPM should be constant. In practice, we tend to use transmissions with a fixed amount of gears, so some speeds are more efficient than others, but it certainly doesn't mean "higher speed, worse efficiency". Electric engines don't really have this problem, which is why diesel-electrics have been so popular in trains - they always run at optimal RPM. But even at poor RPM, drag utterly dominates at high speeds. $\endgroup$– LuaanApr 24, 2017 at 8:43
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It does obey the laws of physics. It is a complicated system that it does not the drag force or the fuel consumption is not always proportional to $v^3$ for all values of $v$.
The general equation for drag is given by:
$$F_{drag} = C_1v + C_2v^2 + C_3v^3 + \ldots$$
The drag force depends on the constants $C_1, C_2, \ldots$ and $v$.
For very small velocities, the lower order terms are more significant compared to the higher order terms. For large velocities, the higher powers become more significant.
The amount of fuel you need is a function of not just of the drag force, but also of other external forces such as friction. The amount of fuel consumed has no decent relationship with the drag force acting. It depends on the time you travelled, how fast you accelerated, etc.
The work done by the drag force is given by:
$$W_{drag} = \int_a^b \vec{F}_{drag}.d\vec{x}$$
If the total energy given by your fuel is $E$, then:
$$E = W_{drag} + K.E_{car} + W_{friction} + W_{other}$$
Each of these terms vary different with velocity. Moreover, the efficiency of the engine changes with the speed of rotation and the gear with which you drove at.
There simply isn't such a cute relation between distance travelled and fuel consumed.
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2$\begingroup$ This same happens also with an electric car and a single gear. But the flaw in my thinking wast to think of fuel consumption per distance instead of fuel consumption per time. CR Drost pointed this out in his answer. $\endgroup$– juzzlinApr 23, 2017 at 19:18