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An airplane needs to move forward to generate lift, and because energy isn't created from nothing, all the kinetic energy of lift comes in the form of drag, where air (air resistance) turns forward motion into upward motion.

Some of the drag on an airplane is therefore "productive" in that it produces lift, and some is "unproductive" in that it takes forward motion from the plane without providing lift.

What percentage of an airplane's forward kinetic energy loss due to drag is productive as opposed to unproductive?

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    $\begingroup$ It appears you are asking what percent of drag is "induced drag", the word for drag due to the creation of lift. Expect to see some graphs in the answers-- . May not be correct to equate drag with kinetic energy loss unless you want to specify a particular thought experiment like thrust is zero and pilot is manipulating controls to maintain altitude as airspeed bleeds off. So you could improve question by deleting reference to KE if you can get to it before an answer is offered. $\endgroup$
    – quiet flyer
    Nov 7, 2019 at 17:47
  • $\begingroup$ Is there any energy lost to lift? It costs a boat nothing to maintain altitude. Maybe once wings are going a certain speed, it costs them nothing to maintain altitude. Would a different wing shape (with no lift) actually have less drag? $\endgroup$
    – Fattie
    Nov 7, 2019 at 18:24
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    $\begingroup$ There is both induced drag, and parasitic drag. See sections 1.2.5, 2.13.8 and 4.5 here in this great e-book "See How It Flies" av8n.com/how $\endgroup$
    – CrossRoads
    Nov 7, 2019 at 18:48
  • $\begingroup$ @Fattie Yes, wing with no lift costs less drag. If a wing can maintain lift without drag penalty, then it's a perpetual motion machine. $\endgroup$
    – JZYL
    Nov 7, 2019 at 20:04
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    $\begingroup$ Can we please find a better "duplicate" question than that? This question is absolutely conceptual, while that question & its answer are entirely math & formulas. I'd almost vote to close as too broad, but a good "here's why it depends" answer could really work. Surely we have one of those somewhere? In a form accessible to someone who's asking this type of question? $\endgroup$
    – Ralph J
    Nov 7, 2019 at 20:56

4 Answers 4

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In the most simple model for subsonic aerodynamics, drag is split into two components:

  1. Zero-lift drag, that is all the drag created when the airplane produces no net lift. This kind of drag has again two components: Friction and pressure drag, that is the aerodynamic drag parallel and perpendicular to the local surface. This drag would dominate in a vertical dive or a zero-g parabola.
  2. Drag created due to lift. Since that was explained mathematically first by using the Biot-Savart law for electromagnetic induction, this is called induced drag. The simplest explanation is: Lift is created by bending the oncoming air slightly downwards, and the reaction force is perpendicular to the mean angle of that airstream. Induced drag is the force component parallel to the initial direction of motion of the air relative to the airplane, and lift is the perpendicular component of that force. Thus, induced drag is lift times half the tangent of the bending angle.

While zero-lift drag increases with dynamic pressure, i.e. with the square of airspeed times density, induced drag decreases with dynamic pressure. Like this: Drag components over speed Drag components over speed for a typical glider (own work). The nonlinearity at the lowest speed is due to flow separation when the lift-creating limits of the aircraft are approached. The physics for large aircraft are the same, only the numbers will be larger.

Due to the dependency on the square of airspeed, the sum of both components has a minimum when they are of equal magnitude. However, given enough thrust, a motorized aircraft can sustain level flight at the far right end of that diagram when lift-dependent drag almost vanishes.

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When flying at the airspeed that yields the maximum L/D ratio, which is also the airspeed that yields the lowest total drag force, 50% of the total drag is "induced drag", i.e. drag due to the creation of lift. At higher airspeeds, a lower % of the total drag is "induced drag". At lower airspeeds, a higher % of the total drag is "induced drag".

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  • $\begingroup$ Ps as an aside, the airspeed for minimum drag referenced in my answer, is NOT the airspeed for minimum drag coefficient. $\endgroup$
    – quiet flyer
    Nov 7, 2019 at 23:57
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No energy is 'lost' to lift. All the energy delivered by the engine is spent, directly or indirectly, in accelerating air downwards in order to produce lift...

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  • $\begingroup$ no, only half the energy. The rest is used to heat the boundary layer. $\endgroup$
    – Peter Kämpf
    Nov 9, 2019 at 18:50
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About 6%

You're looking for the "Lift to Drag" ratio - see here for some examples of different things: https://en.wikipedia.org/wiki/Lift-to-drag_ratio

A paraglider is a 10:1 glide ratio.

A 747 at cruise is 17:1 - thus - 6% of the power keeps it up, and the rest, 94%, offsets the drag of flying at mach 0.85.

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    $\begingroup$ No way. When a high-performance sailplane is cruising at the max L/D ratio, 50% of the drag is induced drag. PS this could be the basis of an answer. $\endgroup$
    – quiet flyer
    Nov 7, 2019 at 23:06
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    $\begingroup$ I don't think this is what was asked. When you have L/D = 17/1, this means you just need 6% of thrust with respect to lift (and weight). The total 'energy supplied' is commensurate with this amount. Now the question is, how much of these 6% is 'useful' (needed for lift) and how much just a 'waste'. $\endgroup$
    – Zeus
    Nov 7, 2019 at 23:13
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    $\begingroup$ That's not what lift-to-drag ratio mean. Lift-to-drag ratio 17 means that it only needs thrust equal 1/17 (~6%) of lift (which is equal to weight) for stable horizontal flight. $\endgroup$
    – Jan Hudec
    Nov 8, 2019 at 6:10

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