Is a 24-hour nonstop flight fuel-efficient?

Before checking sources

Matt's first pass

My intuition is that the energy cost of maintaining airspeed and staying aloft during the flight will be much greater than the energy cost of takeoff, so I wanted to test that first.

I assumed an Airbus airplane weighs somewhere between a large elephant, about 10^4 kg, and an adult blue whale, about 10^5 kg. I guessed the midpoint, 5 x 10^4 kg. Then I added the weight of about 240 passengers at 90 kg each, an additional 2.2 x 10^4 kg, so I used a mass without fuel of about 7.2 x 10^4 kg.

The average speed for 23,000 km in 24 hours and 24 minutes is about:

distance ~= 2.3 x 10^7 m
time ~= 8.8 x 10^4 s

speed ~= distance / time
      ~= (2.3 x 10^7 m) / (8.8 x 10^4 s)
      ~= 2.5 x 10^2 m/s

To figure out takeoff energy, I set it as approximately equal to the kinetic energy of an Airbus moving at 250 m/s:

takeoff KE ~= 1/2 x mass x speed^2
           ~= 1/2 x (7.2 x 10^4 kg) x (2.5 x 10^2 m/s)^2

I wrote that down as about 6.3 x 10^4 J, though that arithmetic is one of the things to check later.

For the energy cost of staying aloft, I assumed wind resistance would be negligible compared with the energy cost of overcoming gravity. Looking at slices of 1 second, the plane should drop by about 10 m per second, so we need to overcome that gravitational energy each second.

energy per second ~= m x g x h
                  ~= (7.2 x 10^4 kg) x (10 m/s2) x (10 m)
                  ~= 7.2 x 10^6 J/s

whole-trip energy ~= (7.2 x 10^6 J/s) x (8.8 x 10^4 s)
                  ~= 6.3 x 10^11 J

From this, I thought the cost of maintaining altitude was about seven orders of magnitude greater than the energy cost of takeoff. Intuitively, more stops and carrying less fuel weight should mean much lower energy expenditure.

If overcoming gravity during flight is the main energy expenditure, and jets are about 50% efficient, I divided by about 9 x 10^7 J/L and solved for approximate fuel volume per kg of flying mass. I got about 0.10 L per kg.

fuel for 7 x 10^4 kg airplane ~= 7,000 L

fuel for carrying that fuel ~= 700 L

fuel for carrying that fuel ~= 70 L

fuel for carrying that fuel ~= 7 L

whole trip fuel ~= 7,777 L

If instead there was a stop at the halfway point, I estimated about 3,500 L of fuel for each half-trip, plus fuel to carry that fuel, for a full-trip total near 7,367 L. In that model, stopping midway would save about 410 L of fuel.

Post-check reflection

Matt's reflection

My airplane mass estimate was about one-third of what it probably should have been, maybe a bit lower. That was one of the less significant errors.

I also did not have the right mental model for what sort of energy expenditure I needed to calculate: drag was the big issue rather than overcoming gravity directly. I was not familiar with using the L/D ratio to estimate cruise work, so I was way low on energy usage. I also did not do a great job of figuring out the cost to take off and underestimated both. In the end, I was a few orders of magnitude low on the fuel required for the trip.

My intuitions were still correct about whether a direct flight was more efficient than a partial trip: it is probably more efficient to stop and refuel. But I expect the draw to the direct flight is the time savings. We are happy to spend more and consume more fuel if it saves us half a day.