This is an ECQB-PPL question specifically asking what happens before the stall. I do understand that after the critical angle has been reached, the airfoil is no longer generating lift (so obviously the lift decreases) and parasitic drag increases dramatically. However, it is not clear to me why this is true when approaching the stall condition.

The magnitude of the lift vector is described by the following formula and $C_{lift}(\alpha)$ still increases (albeit slowly) when approaching the stall condition $\alpha_s$:

$$ L = \frac{1}{2} C_{lift}(\alpha) \rho V^2 S $$

Or is it a purely geometrical consideration, and what is meant by this question is that as the lift vector is assumed to be perpendicular to the airfoil chord average local relative airflow, its vertical projection (that should counteract weight) decreases, and its horizontal projection adds to the drag?

Just to highlight the reason why I'm confused: my understanding of the definition of "stall" is that $C_L$ has flattened out and maximum lift has been reached. Therefore, I thought that the lift would still (slowly) increase, and not decrease when approaching the stall (but not having reached it yet).

This is assuming that the airplane is otherwise in a stable level flight.

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    $\begingroup$ Who ever said that lift decreases while approaching a stall? That sounds incorrect to me. $\endgroup$ Jan 9 at 11:20
  • $\begingroup$ This is an exam question from the current ECQB-PPL 21.2 set (E51-200511-00068). Specifically, it asks "How do lift and drag change when approaching a stall condition?" and the accepted answer is "Lift decreases and drag increases". $\endgroup$ Jan 9 at 11:53
  • $\begingroup$ Anyways in my opinion approaching it means when you reach. $\endgroup$
    – albiremo
    Jan 9 at 11:59
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    $\begingroup$ Stall point is defined as the peak in the $C_L$-$\alpha$ curve. So when approaching stall, the rate of increase reduces, bit $C_L$ still increases. $\endgroup$
    – Koyovis
    Jan 9 at 14:38
  • $\begingroup$ The question should be why does coefficient of lift decrease when approaching stall. Once example is a high g turn or loop, where the lift increases with g force acceleration while approaching but not yet in a stalled state. If the angle of attack is excessive, typically one wing stalls before the other, resulting in a snap roll. $\endgroup$
    – rcgldr
    Jan 10 at 2:00

4 Answers 4


After the stall is reached the airfoil still generates lift, but less than before the stall. Plus drag has greatly increased due to the separated boundary layer on top of the wing, which causes pressure drag.

enter image description hereImage source

Figure above shows why the boundary layer separates: the airflow close to the skin is slowed down by friction, and does not have enough kinetic energy to overcome the pressure gradient along the upper aft side. Static pressure in the Wake region of the picture is quite low, causing the pressure drag.

Loss of lift is also because of the separated airflow, the flow aft of the separation point is lost for lift generation. Lift loss is not due to wing tilt: the lift vector is perpendicular to the airspeed vector, not to the wing chord.

enter image description hereImage source @ Re=10$^6$

Above process takes place gradually: $C_L$ increases linearly with angle of attack until the separation point is located at the wing aft tip (about $\alpha$ = 12° in the plot), then gradually diminishes with AoA until it peaks out at $\alpha$ = 16°. At this max $C_L$ peak there is quite some separated airflow at the top wing area already.

Looking at the example above, the point at $\alpha$ = 12° would be the point of stall onset, where the lift coefficient starts to reduce and turbulence may be detected but lift still increases with AoA. The point at $\alpha$ = 16° is the stall point: further increase of AoA results in a reduction of lift.

  • $\begingroup$ I see, so it's really the magnitude of the lift vector that decreases, and the main contributing factor when approaching the stall condition is $S$, being strongly reduced by growing separation, whereas slightly increasing $C_L$ and slightly decreasing $V^2$ do not play an important role here, correct? $\endgroup$ Jan 9 at 9:09
  • $\begingroup$ It is the magnitude indeed. Approaching stall, as in no flow separation yet, means the airfoil is not yet stalled. It is the flow separation that causes non-linear changes. S does not change, it is the projected wing area. $\endgroup$
    – Koyovis
    Jan 9 at 9:16
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    $\begingroup$ Sorry, unfortunately, I still don't understand your answer :-( Then why does the magnitude of the lift vector decrease before reaching the stall (as in the original question)? $\endgroup$ Jan 9 at 9:23
  • $\begingroup$ Have edited the answer. Magnitude of the lift vector peaks at the point of stall, by definition. $\endgroup$
    – Koyovis
    Jan 9 at 14:39

A simple and arguably easy to understand answer is that the wing does not produce separate forces called Lift and Drag. It produces force at every point of the wing *, pointed in a direction normal (perpendicular) to the surface of the wing at each point. Lift is just the sum of the the components of all these forces that are normal or perpendicular to the flight path of the airplane (it's direction of flight through the air), and Drag is just the components of all these forces that are parallel to the flight path. (See image in @Koyovis answer above ... All those little graphics attached to the upper surface of the wing exists everywhere - at every single point on the surface of the entire airframe), and every single one of them produces a Lift component, and a Drag component.

So, as an airfoil approaches a stall condition, it's Angle of Attack (AOA), is increasing, i.e., the angle of the wing with respect to the flight path is getting steeper, so every one of those actual real forces acting at each point of the airfoil is tilting more and more aft (to the rear). Therefore the component of those forces which is perpendicular to the flight path (LIFT) is decreasing, and the component parallel to the flight path (DRAG) is increasing.

  • Aerodynamic forces are just the static pressure of the air pushing on the surface of any airfoil (or, in fact, any shape) moving through the air. These forces are caused by the collisions of air molecules hitting the surface and rebounding. In these collisions, momentum is exchanged, and the aerodynamic force is just the result of the change in momentum. The formula is just F=ma.

This formula is actually a shortcut for F=dP/dt. Where P is the momentum (Mass x Velocity or mV), so dP is the change in momentum, and dP/dt is the rate of change of the Momentum. It is the same as F=ma because a is the rate of change of Velocity (dV/dt) so F=ma is the same as F=m x dV/dt which is the same as F = d(mV)/dt, which is the same as F=dP/dt since P = mV

  • $\begingroup$ Thank you everybody for your answers and sorry for the delay, I was really busy with the exams! It's hard to choose the "right" answer, but I guess this answer explains it best in the spirit of ECQB and confirms my suspicion, that they basically meant that (assuming that the airflow is horizontal), the vertical component decreases and the horizontal component increases as the lift vector rotates (even if the magnitude of "lift" itself still increases until $\alpha_s$)... $\endgroup$ Jan 27 at 10:40
  • $\begingroup$ @Yuri, I might add the trivial correction to your parenthetical comment about the flow being horizontal. It doesn't matter whether it is horizontal, Instead of making that assumption, all your concepts and statements are equally true if you think about things from the frame of reference of the aircraft, and it's flight path through the air (opposite the relative wind), and not of the ground. The Lift is then the component of the total force perpendicular to the flight path, and the drag is the component parallel with the flight path. $\endgroup$ Jan 27 at 16:44
  • $\begingroup$ Of course, you are right - I was just using terms like "horizontal" and "vertical" as the question pertained to straight and level flight, and it's easy to imagine and explain. Thank you for the elaboration! $\endgroup$ Jan 28 at 17:18
  • $\begingroup$ Yes, sir, right on! $\endgroup$ Jan 29 at 1:54

The answer start by understanding how are aerodynamic forces generated. The aerodynamic forces are a result of the pressure distribution around the airfoil and of the shear forces due to the viscosity (look at the wikipedia definition, without entering too much into details).

enter image description here

Starting from this, we can have a look to a simple case (NACA0012) in which we have a symmetric airfoil at 3 degrees. The flow field result to be "attached" to the airfoil and we have the maximum efficiency, because we do not have "ricirculation zones" (mathematically we can pose the Hypothesis of Potential flow and say we have no zone with vorticity different from 0, if you want to know more have a look at Helmoltz' theorems). The more we increase the angle the more we happen to have an adverse gradient in the back of the upper face of the airfoil. At a certain point that adverse gradient brings the boundary layer to separate and the flow being no more attached (have a look at this to understand the mechanism). This bring to a different pressure distribution as you can see from the picture of Thwaites I attach to you here.

enter image description here

As you can see the resulting of the pressure distribution (the area under the line) is less prominent. This can be easily described with the Bernoulli theorem. You can imagine that the fluid vane is limited by the de-attached flow and hence you have an increase of pressure with respect to the attached flow. So finally this is the physics behind the stall and the degradation of the aerodynamic performances (abrupt decrease in lift and increase in drag, due to the form drag). Regarding this exists different type of stalls. To know a bit more, I suggest you to read this NASA report.

The fact that the lift decreases before the stall depends on the type of stall that you have. On the ECQB I think that they mean with "stall" not the degradation of aerodynamic coefficient stall but the case in which the flow is completely separated. There are stalls in which a laminar separation bubble is placed on the upper faces, but you have a re-attachement due to turbulent transition. You can see an example in the figure below (leading edge short bubble).

enter image description here

Here you have a decrease of the lift before the complete separation of the flow, hence it would give an answer to the ECQB question.

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    $\begingroup$ Thank you, but it seems that like @Koyovis you are explaining what happens after (or at) the stall (abrupt decrease in lift and increase in drag), but my question is why in ECQB they claim that lift decreases before reaching the stall. $\endgroup$ Jan 9 at 11:27
  • $\begingroup$ Edited the answer to try to address the specific question. $\endgroup$
    – albiremo
    Jan 9 at 11:43

At airfoiltools.com you can examine many relationships of lift and drag without worrying about the "right" answer on a test.

In one particular course of study, "approaching stall" may be the warning buffet of airflow separation from the trailing edge of the wing. With a magnifying glass, you may see lift begin to decrease here as drag continues to increase. Maybe in a range of a few tenths of a degree AoA, before the wing fully stalls.

But, as the "polar graphs" indicate, and as a pilot, the exact AoA this happens is trivial, and different airfoils can have vastly different behaviors near stall. Is is best to use V$^2$ for adequate lift when flying and stay away from excessive AoA.

A possible theoretical explanation for a certain wing may be within the 3D realm of examination, that being the wing tip vortices begin to curl back into the top of the wingtip more at higher AoA, thereby reducing total lift just before stall.

One might speak to their professor regarding that subject.


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