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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. $\endgroup$ 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

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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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  • $\begingroup$ Interesting to note: proportionally lift induced drag almost vanishes (compared to form drag), but in terms of force units, it may actually increase if wing (and fuselage) AoA is less than optimal L/D AoA at higher airspeeds. $\endgroup$ Nov 16, 2022 at 17:39
  • $\begingroup$ @RobertDiGiovanni That is right, but do not forget that the lift coefficient goes down linearly with dynamic pressure but the induced drag coefficient goes down with the square of the lift coefficient. So absolute induced drag is proportional to the lift coefficient. $\endgroup$ Nov 16, 2022 at 18:58
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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. $\endgroup$ Nov 7, 2019 at 23:57
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By definition, energy is transferred to an aircraft via the application of thrust along a displacement. For the sake of simplicity we consider a displacement unitary and energy is therefore just proportional to thrust.

In level flight, thrust simply equals drag and therefore the question can be rephrased as: how much changes drag when lift goes to zero, i.e. how much changes $C_D$ @ $C_{L=0}$.

Those values are plotted on the polar of the aircraft, like this one for a B747-100 taken from ¹:

747 polar

So, from this polar it's easy to see that $C_D$ @ $C_{L=0}$ is some 0.019 at Mach 0.86.

This value has to be compared with the drag when the lift is actually being produced. Let's consider then a standard cruise condition at height of 10km, Mach as before and mass of 250,000kg (halfway between MTOW and OEW for a B747-100). This cruise condition gives a $C_L=0.38$ and from the polar $C_D=0.023$.

The difference with the previous value is $0.023-0.019=0.004$ which corresponds (at the cruise condition just given) to a drag due to lift of $28kN$. This value multiplied by the displacement (in $m$) is the energy lost to lift we were looking for. For a range of say $8'500km$ that gives an energy lost to lift of some $240GJ$.

Two side notes: obviously during cruise height, speed, Mach and weight change and also the 0.004 that we have calculated changes as well; anyway the order of magnitude for the energy is correct. Even if the energy lost to lift look big, it is actually only 0.004 of the original 0.023 i.e. it corresponds only to the 17% of the whole energy used to keep the jumbo jet in flight at 10km and Mach 0.87.


¹ Dr. Jan Roskam, Airplane Design Part VI, DARcorporation

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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. $\endgroup$ Nov 9, 2019 at 18:50
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Simply BY DEFINITION, no energy can be lost to "Lift". Lift is the portion of all aerodynamic force on the aircraft that lies normal (perpendicular) to the flight path motion of the aircraft. So being perpendicular to the flight path, it can neither accelerate nor decelerate the aircraft.

Drag is the portion of the total aerodynamic force that lies parallel to the direction opf motion. So, simply by definition, it is the drag that slows the aircraft.

But your question (restated as "How much of an airplane's forward energy is lost to Aerodynamic force?" is a good one, and it is dependent on the Angle of Attack, (AOA). The higher the AOA, the more the total Aerodynamic force vector is tilled aft (backwards), from the aircraft direction of motion. This means that as AOA increases, the Drag component of that force (as a percentage of the total) will increase, and the Lift component percentage will decrease.

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The surprise of some engineers about comparing kinetic energy to lift?!?.

Yes, let's look at the formulas, and how we apply them. In steady state flight, we talk about the 4 FORCES of flight. F = kg m/s$^2$ = ma: Lift, Gravity, Thrust, Drag

So why not Kinetic Energy KE = kg m$^2/s^2$ = F × d, or even Power P = kg m$^2/s^3$ = F × d/t?

Because, in describing forces on an aircraft in steady state flight, everything happens in the same distance at the same time, so d/t cancels out, leaving a balance of forces.

However, we can easily determine total drag from the glide ratio.

The energy state of an aircraft is its potential energy mgh + its kinetic energy 1/2mv$^2$. Assuming it maintains constant speed in a glide and flys in a straight line at constant AOA to its landing point, energy consumed is all mgh: kg m$^2$/s$^2$.

There for total drag force × distance = mgh

So if the glide ratio is 10 to 1, total kinetic energy converted from potential energy is around kg m/s$^2$ × distance = kg × gravity (m/s$^2$) × height. Canceling units we have, for a 1000 lb airplane (thrust = drag): 100 lbs of thrust force is required for level flight.

To determine lift to drag ratios, we move to the wind tunnel, and develop data points for the formula Ctotal drag = Cdrag form + Cdrag induced. As seen in Peter Kampf's graph, this will vary depending on airspeed and angle of attack.

A better wing will have a higher lift to total drag ratio, and considering that these are aircraft, a better question may be how much "forward energy" is lost NOT creating lift.

Drag is a product of motion, and apportioned to lifting AND moving forward, nothing is wasted or lost. "Forward energy" is thrust × distance. What one can do next is to begin to work on efficiency, so one may need less to go as high and as far.

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  • $\begingroup$ -1 because I can barely read the math :( Please use Mathjax to make it more understandable. $\endgroup$ Nov 10, 2019 at 3:34
  • $\begingroup$ Revived with Mathjax corrections. $\endgroup$ Nov 16, 2022 at 17:07
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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. $\endgroup$ 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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