I am having a bit of confusing as to why people say planes gets more efficient as they fly higher. Because on a typical long haul flight the Fuel Flow of the aircraft will obviously be much more that when it was lighter as it was heavy but my main question is since aircraft FF decrease with weight but at the same altitude isn't it more efficient to keep flying at the same altitude instead of climbing further? Here below are some charts on the B737 FF getting lesser with the altitude as it gets lighter. Notice how the FF goes up as you climb higher even though its lighter. Looking at the charts I suppose isn't it more efficient to keep flying at the same altitudes? Can anyone help me clarify this? Thanks

B737 FF Charts

  • $\begingroup$ If you stay in the gray boxes, every time the aircraft gets lighter your fuel flow decreases, even though you may have to fly higher to stay inside the gray. For example, from 60000 kg at 37000 feet, if you reduce the weight to 50000 you have to climb to 38000, but the fuel flow decreases from 1135 to 1055. At a given weight, of course, there is an upper limit to how high you can fly, and being too close to that limit isn't good. So obviously "fly higher to save fuel" is a gross oversimplification. $\endgroup$
    – David K
    Commented Jul 23, 2020 at 13:07

1 Answer 1


When airplane mass, altitude, N1 (engine spool speed) and Mach number change together, it is very hard to come to a meaningful conclusion what is better for fuel flow. However, you see that the grey boxes tend to show the smallest fuel flow and they sit at the upper altitudes, so altitude seems to help. Going above 90% N1, however, seems to increase fuel flow again.

With increasing altitude, the flight Mach number needs to be increased to keep indicated airspeed constant. I expect that the chart lists trim points at or near the optimum polar point for cruise, at least that is what the almost constant indicated airspeed for each mass suggests. With increasing Mach, drag goes up but so does the distance covered per time, so there is again an optimum where the transsonic drag increase is still tolerable and outweighed by the higher speed over ground.

Therefore, higher flight speed is one reason why airliners like to climb to the tropopause. But there is more: The thermodynamic efficiency of a heat machine grows with the ratio of temperatures in its thermodynamic cycle. Since the lowest temperature is determined by outside air, and outside air temperature drops with altitude while the top temperature is limited by the engine's materials, flying higher improves the thermodynamic efficiency of the engines.

  • $\begingroup$ So if in general optimum doesn't necessarily mean better fuel flow only? It involves many factors like these? Looking from the charts, I am assuming now I am taking off with the weight of 70 tons and I will cruise at FL330 and after say I burned of some fuel and I am at 65 tons now and I want to step climb to FL350, the improvement of fuel flow from 330 to 350 looks very small. And the n1 has increased to 87 from 85. Is this how a step climb is done? $\endgroup$
    – Joe Wie
    Commented Jul 22, 2020 at 19:25
  • 3
    $\begingroup$ @JoeWie: The optimum is defined in a very practical way: To quickly get a defined payload from A to B for the least amount of fuel. Quickly, because faster aircraft can fly more trips per day. This can be expanded: To fly a defined payload from A to B with the smallest possible fleet and the least amount of fuel. For step climbs please look here $\endgroup$ Commented Jul 22, 2020 at 20:11
  • $\begingroup$ Doesn't also propulsive efficiency of turbines increase with velocity due to the pressure recovery? That is another reason to fly high where the true airspeed is higher. $\endgroup$
    – Jan Hudec
    Commented Jul 27, 2020 at 17:10
  • $\begingroup$ @JanHudec: No, efficiency gains are from lower ambient temperature. Pressure recovery increases mass flow and thrust, but the fuel flow has to match the air mass flow and efficiency is not improved. $\endgroup$ Commented Jul 27, 2020 at 20:23
  • $\begingroup$ @PeterKämpf, Hm, I mixed in the pressure recovery incorrectly. Actually I meant due to the lower difference between intake and exhaust velocity, since the exhaust velocity remains fairly constant. $\endgroup$
    – Jan Hudec
    Commented Jul 27, 2020 at 20:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .