When I increase the angle of attack from 0 to 20 the stall begins at 18 deg.

When I decrease the angle of attack from 20 to 0 the stall ends at 13 deg.

Why the difference and what value (13 or 18 deg) should I use as the stall angle?

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    $\begingroup$ From pilot's point of view, the AoA at which the stall begins is the one that matters. During stall recovery procedure the AoA is reduced dramatically, so the difference you mention matters very little. Academically speaking things might be different. $\endgroup$
    – Jpe61
    Commented Sep 19, 2020 at 3:22
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    $\begingroup$ @Jpe61 what if a pilot fully stalls his airplane? Surely then it would matter? $\endgroup$ Commented Sep 19, 2020 at 9:17
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    $\begingroup$ I don't see how. At AoA above 20 degrees, the wing will be fully stalled, below that it is not stalled. If the case is such that lift will be recovered at 13 degrees, then that's how far you have to "push" the plane (or wings) $\endgroup$
    – Jpe61
    Commented Sep 19, 2020 at 15:54
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    $\begingroup$ That is called hysteresis. Quite normal and grows with the speed of your AoA change. $\endgroup$ Commented Sep 20, 2020 at 6:53

1 Answer 1


This is called stall hysteresis. You have two different situations and the flow reacts differently in each of them.

When increasing the AOA

The flow is attached to the wing and the boundary layer is resisting the adverse pressure gradient as much as possible. At some point the flow detach from your profile and you have stalled let's say at 18°. At this point a huge recirculation bubble appears on the succion side of the airfoil.

Decreasing AOA / recovery from the stall

This recirculation bubble represents an area where the average flow speed is close to 0. Looking from the outside, for the free stream flow coming at your wing it looks like you have a new airfoil profile which is made from the old airfoil and the recirculation bubble. Most of the time this recirculation zone extend far beyond the trailing edge of the actual profile, reducing thus the overall aspect ratio (thickness/chord) of the profile. Slimmer profiles have less tolerance to stall and lower stall AOA thus explaining why your recovery only comes when the AOA goes below 13°.

Once the flow is reattached, you are back to the first configuration and you can go back up to the initial stall angle 18°.

Your stall angle is thus 18° but if you are in an airplane it means you'll have to lower the nose below 13° of AOA to recover.

There's an excelent thesis on this subject here from which this picture is taken showing the hysteresis phenomenon. enter image description here

  • $\begingroup$ "slimmer profile"? $\endgroup$ Commented Sep 19, 2020 at 9:19
  • $\begingroup$ Extending the recirculation beyond the TE would reduce aspect ratio. You might make it clearer that it also increases thickness to chord ratio. You answer is very good in that it explains that the recirculation bubble must be broken to reattach flow. Reattachment may be time, as well as angle dependent. I would try a slower (and faster) recovery (in the wind tunnel) to see if this affects the reattachment angle. (keeping in mind, in real life, once the plane starts to sink, AOA goes up, recovery should be ASAP). The recirc bubble has energy, which may account for its resistance. $\endgroup$ Commented Sep 19, 2020 at 10:01
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    $\begingroup$ Not really, as you can see the difference in this case is only 5°, you still have a 13° nose up attitude when the profile recover. Wherase in real life the instant reaction you have to a stall is to release some pressure on the stick and let the nose drop, 5 degree is quite fast to reach. $\endgroup$
    – MaximEck
    Commented Sep 19, 2020 at 11:29
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    $\begingroup$ Furthermore don't forget that stalling doesn't mean instant 0 lift, but rather that your glide ratio drops because of extreme increase in drag and somewhat small decrease on lift depending on the profile. But the airplane doesn't just fall of the sky. You just need to manage your airspeed to ensure your not making the situation worse. And the recovery is quite smooth for a symmetric stall. $\endgroup$
    – MaximEck
    Commented Sep 19, 2020 at 11:35
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    $\begingroup$ Thanks a lot @MaximEck ! You scratched my itchy one. $\endgroup$
    – 3XBULL
    Commented Sep 21, 2020 at 3:07

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