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In a typical airliner, what fraction of fuel/power/energy is used for providing lift and overcoming lift-related drag

Do the density of the air and the speed through it influence this ratio or do they affect lift and drag equally? If they matter, assume typical cruise altitude and speed.

To illustrate, if you took such a plane and:

  • attached its engines directly to the fuselage
  • removed all wings and ailerons (essentially only keeping the fuselage)
  • let the fuselage run along the flightpath on frictionless rails

how much more efficient would it be? Nothing else changes, so it's still in level "flight" at the same altitude and speed.

Optionally, in what ballpark is the amount of extra fuel consumed to initially climb to that altitude, minus any amount saved when descending to land?
Is it significant enough that, for instance, the length of a flight significantly affects the average fuel consumed per kilometer traveled horizontally?

I'd be happy with any answer that fits the description in the first sentence, but if you want details, I'll just pick:

  • Boeing 737 MAX 7
  • Altitude FL390
  • Speed 0.789 Mach
  • Payload 20'000 kg
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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Federico
    Commented Sep 16, 2022 at 20:55

2 Answers 2

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EDIT: I've found the polar for the B747 :-) Relevant calculation appended at the end of the answer.

Very little since aircraft are extremely optimised.

Engine's fuel consumption is proportional to the generated thrust, so the question becomes how much thrust is needed when the lift is null. At least for level flight, thrust equals drag and therefore this can be rephrased as: how much is the drag coefficient when the lift coefficient is zero, i.e. how much is $C_D$ @ $C_{L=0}$.

So: FC $\propto$ T = D $\rightarrow$ $C_D$ @ $C_{L=0}$ (FC = fuel consumption)

To answer this, the drag polar of the aircraft must be known. Online i could find only the following polars from this source:

F-16 polar

Let's take for example the plot with diamonds for the F-16, since I like this plane :-)

The $C_D$ @ $C_{L=0}$ is some 0.022 while at "load factor=1.0", i.e at level flight, it is 0.025. This corresponds to a 12% reduction in drag (and thrust) when the lift is null. This 12% represents the share of drag which is correlated with lift.

How does this reduction of drag/thrust affect the fuel consumption? For the Mach number and the height given in the polar (M=0.9 at 30,000ft), the fuel consumption of a typical afterburning turbofan is more or less linearly proportional to the thrust and therefore our flying-with-no-lift F-16 would need some 12% less fuel.

So, the fraction of fuel consumed for keeping a plane in the air is just this 12% of the total, very little indeed!

Here is the polar of a B747-100 taken from the legendary Dr. Jan Roskam, Airplane Design Part VI, DARcorporation:

747 polar

So, following the same procedure as before, $C_D$ @ $C_{L=0}$ is 0.019 at Mach 0.86. For the cruise condition I used some halfway values i.e. height of 10km, Mach as before and mass of 250,000kg. This gives a $C_L=0.38$ and a relevant $C_D=0.023$ which corresponds to a -17% in drag i.e. thrust.

For a typical high-bypass turbofan at that height and Mach, the fuel consumption doesn't change linearly like for the afterburning turbofan of the F-16 but a bit less than linearly. So the 17% reduction in drag/thrust translates into a 14% reduction in fuel consumption.

Then again: very little indeed.

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  • $\begingroup$ Thanks, I have to say that's a lot less than I expected. Of course that data is with (now redundant) wings still attached, but I presume removing those wouldn't put the results in another ballpark. $\endgroup$
    – Bart van Heukelom
    Commented Sep 14, 2022 at 22:48
  • $\begingroup$ I'm curious though, how did they find those numbers in your source? Specifically, how/when does a real plane, with the wings attached and moving through the air, have a lift coefficient of 0? If I understand it correctly, it can be increased by using e.g. flaps. But the wings are fixed. I guess by compensating the lift with ailerons, or transiently by flying with the roof tilted towards the incoming airstream a bit, using momentum. But surely those greatly affect the drag as well, and at different rates too? I guess my understanding isn't quite complete yet. $\endgroup$
    – Bart van Heukelom
    Commented Sep 14, 2022 at 22:49
  • $\begingroup$ @BartvanHeukelom: Note that this is at Mach 0.9. Going that fast creates a lot of drag. If you want to save fuel, go a lot slower, especially if you're getting lift for free. (Like a train, where the wheels keep it off the ground with minimal rolling resistance regardless of speed.) You're talking about a plane which has these big bulky wings sticking out into the air, creating drag for no benefit (because you're ignoring lift requirements), not a bullet train, so you want to go slower to reduce total energy lost to drag over the distance you have to travel. $\endgroup$
    – Peter Cordes
    Commented Sep 15, 2022 at 1:21
  • $\begingroup$ It’s not the drag at Cl=0. There is no Cl. So you can choose the alpha that has minimum drag, not the alpha that gives Cl=0. $\endgroup$
    – Jim
    Commented Sep 15, 2022 at 5:50
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    $\begingroup$ @BartvanHeukelom: anyway very interesting question 🖖 $\endgroup$
    – sophit
    Commented Sep 15, 2022 at 6:16
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Airplanes not only have to hold themselves up, they also have to overcome drag from speed. But to give you some idea, ton miles per gallon is used in the transportation industry.

Trains are at least twice as efficient than tractor-trailers, who are around 4x more efficient than a 747, which (because it can fly high) is around 2x better than the best 4 engine piston prop era aircraft.

The selling point of aircraft is speed. Ground transportation is slower, but more efficient for one trip.

The aircraft, with its speed, can make 10 trips for each one a train or truck (forget about a barge unless it's your yearly harvest) can make.

In the end, the analysis goes to the bottom line profit for each trip × the number of trips made per given amount of time. A FedEx jet stuffed with overnight mail will probably be a huge money maker.

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    $\begingroup$ I would also add container ships, which are the cheapest way to move goods around the world $\endgroup$
    – sophit
    Commented Sep 14, 2022 at 19:31
  • $\begingroup$ Oh yeah for sure, land or water based transport must be many times more efficient using that measure, or it wouldn't be that popular. But that efficiency is affected by many variables, even when ignoring things like the cost of requisite infrastructure. (1) $\endgroup$
    – Bart van Heukelom
    Commented Sep 14, 2022 at 22:59
  • $\begingroup$ I'm trying to break the comparison down into the separate components a bit. I figured speed would be a very significant factor considering the exponential losses of drag, but that's suddenly irrelevant if you compare the plane against e.g. some kind of future maglev train thing that travels just as fast, through similar air (by whatever means). So I wondered what the next big difference would be there, and guessed maintaining lift would be a good candidate. (2) $\endgroup$
    – Bart van Heukelom
    Commented Sep 14, 2022 at 22:59
  • $\begingroup$ @BartvanHeukelom: don't forget density: jetliners normally fly where density is more or less ⅓ the one at sea level. So in your hypothetical aircraft generating lift for free, drag reduces of the same factor $\endgroup$
    – sophit
    Commented Sep 15, 2022 at 6:10

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