Here is an add I saw claiming that CSX freight trains can move 1 ton of material 500 miles on 1 gallon of diesel fuel. Wow, that is something I have to look into. Here is the site of CSX with the info I will use. From that site:

CSX claims that EPA claims that a for every ton-mile, a truck emits about 3 times more nitrogen oxide and particulates than a locomotive does.

CSX can move 1 ton of freight 500 miles with 1 gallon of diesel (on average).

Their locomotives (that seems like an archaic term) uses auxiliary power for idling times so that the diesel engine can be shut down.

Using throttle optimization based on train load and location and stuff.

They appear to have other initiatives to be a greener company. All this seems nice. However, I still think the 1 gallon thing is a little far fetched. So, let me do a basic calculation. Let me estimate the frictional force per ton of freight over this 500 mile trip. I will assume a straight track with no hills (this is my spherical cow assumption). Also, it seems like a fair estimate to say that one car can carry about 100 tons of freight. So, here is a diagram for one of those cars.

Suppose I move this railroad car 500 miles. How much work would have to be done by the engine? I called this force F pull , but maybe it should have been labeled "engine". Anyway, if the train is moving at a constant speed, then F pull = F friction (just talking magnitudes here). So, I can write:

Now, I assume that this car has a weight of 100 tons then it should use 100 gallons of diesel. According to Wikipedia, diesel has an energy density of 37.3 MJ/L (141 MJ/gal). As for efficiency, this D.O.E. paper lists the efficiency at 45%. So, how much work went into the car over this 500 mile trip?

If I use a value of 500 miles (8 x 105 meters), I can get a value for the frictional force.

The usual model for the frictional force (even though this is rolling friction or something like that) says that:

Where F N is the force the ground pushes up on the car and μ is the coefficient of friction. In this model, μ depends on the two types of surfaces interacting. Since the car is in equilibrium in the vertical direction, this means that F N = mg. For a 100 ton car, the mass would be 9.07 x 104 kg. Putting this in, I get the following for the coefficient of friction.

That seems pretty small. Oh, I know this isn't real friction but still seems small. Well, maybe it isn't too crazy. This site lists some values for the coefficient of friction for ice. For ice skates on ice, it seems the value could be as low as 0.005.

But what about air resistance? This is one thing the train can do well. What happens as you add more and more train cars to the total train? The total drag due to the air resistance might increase, but not that much for each additional car. The primary contributors to the drag are likely (just guessing here) the first and last car. So, the drag force on a train of 5 cars and 10 cars will probably be quite similar.

What about the claim from CSX? ——————————

I am going to go ahead and put a "plausible" tag on this claim. Seems like it could be true.