Is there a wider question that you have coming out of this?

Yes. I got a result that a truck used less energy than a car pound for pound for a given mileage even assuming the engine efficiency was the same. This seemed puzzling to me because if anything I'd expect it to be the opposite (truck is less aerodynamic). I was wondering whether this has any basis in the physical world or if it was just an artifact of calculation imprecision. One possibility that occurred to me is air resistance is a major factor at highway speeds and I figured a big factor in air resistance is going to be frontal area, which will scale up more slowly than a vehicle's mass and volume (a big truck looks long in comparison to its width and height compared to a car, and all else being equal volume and hence freight capacity scales by the cube while area scales by the square).

I think I found newbielink:http://physics.ucsd.edu/do-the-math/2011/07/100-mpg-on-gasoline/

[nonactive], it gives a formula for air resistance:

1/2CdpAD(v^2)

Cd = drag coefficient

p = 1.3 kg/m^3 (density of air at sea level)

A = cross sectional area of the vehicle

D = distance travelled

v = velocity

They say Cd for a Prius is .25, for a sedan is .3, and for a SUV or pickup truck .5-.6. I'm not sure what the proper units are for the rest but I'll use meters. I'll approximate the car's frontal area as 1.5X2=3m and the truck's as 3X3=9m, so the truck has 3X the frontal area of the car, and give the car a Cd of .25 and the truck a Cd of .6. The rest of the variables I'll keep constant (65 mph, 65 miles travelled, converted to meters and m/s).

Car = 1/2(.25)(1.3)(3)(104,000)(29^2) = 32,799,000

Truck = 1/2(.6)(1.3)(9)(104,000)(29^2) = 306,998,640

I get 9.36 times the air resistance for the truck as for the car.

I'm not sure I'm doing this right (especially as I think newbielink:http://mb-soft.com/public2/car.html

[nonactive] suggests I should drop the 1/2 from the equation - and I get more consistent results without it), but generally playing around with the formula I get in the ballpark of 4-10 times the air resistance for increasing frontal area by a couple of times and increasing Cd by a factor of 2 or so (which I think would probably model the comparison between a car and a large truck fairly well).

Now according to newbielink:http://www.consumerreports.org/cro/cars/tires-auto-parts/car-maintenance/fuel-economy-save-money-on-gas/overview/index.htm

[nonactive] at highway speeds air drag can account for half or more of energy consumption. And the truck I was using in my calculation was 36 tons while the car was 1.5 tons. So this fits with the at first puzzling result I got, that a truck used less energy pound for pound for the same mileage; it's 24 times heavier but only gets maybe 5-10 times the air resistance because of its less aerodynamic frontal profile and bigger frontal area. So it looks like there's an economy of scale here with big vehicles: air resistance scales with frontal area, which scales up slower than volume and mass assuming a relatively long vehicle (like a truck with a trailer).

Does this sound reasonable to you guys?