Ok, amazingly, travelling at 150 km/hr uses 125% more fuel than travelling at 100 km/hr. But, you’ll arrive and turn off your engine 20 minutes sooner, so won’t that negate to a certain extent the extra fuel used? I mean, you’re using 100% less fuel for 20 minutes aren’t you?

As I said, I’m not very bright, maybe I’ve got it all wrong… :-O

So: at 150km/hour (41.7 m/s) the aerodynamic drag is (0.41/2)*(1.16)*(41.7^2)*(3.15)=1302 N (say 1300 Newtons). The other external force to be overcome by converting the chemical potential energy of gasoline to mechanical energy in the cylinders is tire rolling resistance. At typical highway speeds, as best I can determine, this number is about 0.015 times vehicle weight. I’ll assume it increases linearly with speed and that 150 km/hr is 1.5 times typical highway speed, so I’ll estimate 1.5*1.5=2.25 times vehicle weight. My LR3 typically weighs about 26000 N so this force is something like 585 N so the total external force to be overcome is about 1885 N. Now, force times speed is power, so I need the burning of fossil fuels to give me energy at the rate of 1885 N *41.7 m/s or 78,600 watts. This is equivalent to 105 horsepower. Whew.

The same calculations for the speed of 100 km/hr (27.8 m/s) yield an aerodynamic drag force of 579 N and a rolling resistance of 390 N for a total of 969 N. At 27.8 m/s this requires 26,900 watts or 36 horsepower.

All this is very interesting, but since I’ve assumed that rolling resistance and all manner of dissipative forces in the car (pumping of fluids, driveline friction, etc.) are linear functions of speed as is distance covered per second, the ratio of fuel consumption rates at those two speeds will depend only on aerodynamic drag, which increases as a power function of speed. In my LR3 therefore, I should use )(1300-589)/579)*100%=124.5% more fuel at 150 km/hour. I’m more than willing to call it 125% . And since the only variable in the aerodynamic drag calculation specific to my vehicle is Cd and it is a linear factor, the same ratio will hold for any vehicle.

So, with your permission, we’ll agree on 125% more fuel (or 225% as much fuel) and a time savings of 20 minutes for a 100 km trip.

Why do I burn more fuel the faster I go?

Basically, higher speed involves a higher force related to air resistance. As speed goes up, the force required to push a vehicle through air rises significantly.

The power required to overcome the aerodynamic drag is given by:

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 80 km/h may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 160 km/h requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice faster. Since power is the rate of doing work, four times a work in half the time requires eight times the power.

If the image insertion didn’t work above, you can check out the formula at
http://www.clean.ns.ca/default.asp?id=190&pagesize=1&sfield=content.id&search=179&mn=1.21.52.160

It also has extra information about our program!

I have to question your figures, considering that going 100 kilometers at 100 kilometers per hour takes an hour, going the same distance at 150 kilometers per hour takes 40 minutes. I’m thinking that’s a savings of 20 minutes, which makes sense – going 50% faster should reduce your time by 33% (I majored in mathematics).

I’m not sure that you would use 150% more fuel. That would be a reduction in m.p.g. of 60% (the same as an increase in liters per kilometer of 150%). No figures I have determined from my experience or calculation indicate a decrease of that magnitude in fuel economy, even for a vehicle with very poor aerodynamics (the type that would be most affected).

Nothing personal, but when such fundamental errors are made, it calls into question the accuracy of your other points as well.

I calculated here (purely using estimations): http://hamiltonianfunction.blogspot.com/2006/04/ween-us-from-imported-oil-maybe-not.html that elimination of drive through windows would save the U.S. about 350,000 barrels of oil per year, or about 0.007% of our imports. But I stopped using them despite this very small number.

The time factor is not insignificant. See http://hamiltonianfunction.blogspot.com/2006/04/use-of-time.html for some numbers.

I guess the undertone of the article was meant to be kind of sarcastic; I think that many of these things really aren’t a great inconvenience in the grand scheme of things, however, we tend to make a bigger deal out of them than what we really should.

I suppose I didn’t lay on the sarcasm as thickly as I had intended.