You may be surprised to find out that an airplane can stall at any attitude if the critical angle of attack is exceeded, but it cannot stall at 0 G.

This quote found in the April 2014 issue of Flying magazine, but it seems counterintuitive to me.

Why can't an airplane stall at 0G?

  • $\begingroup$ do you mean in space. zero G or when the plane is dropping in a rate that the force felt by the plane is Zero G? $\endgroup$
    – user16299
    Aug 7, 2016 at 13:45
  • $\begingroup$ @george - Either, but in space there wouldn't be an atmosphere anyway. $\endgroup$
    – Steve V.
    Aug 7, 2016 at 23:56
  • $\begingroup$ To get 0g and stall at the same time, you basically need drag and thrust to cancel each other or both equal to 0. The relative angle of wing and engine are installed makes it impossible for drag and thrust to cancel each other at high angle of attack. I guess it's possible on a missile whose wing can rotate above critical AoA? $\endgroup$ May 14, 2017 at 0:53

5 Answers 5


It's probably more accurate to say "you cannot maintain 0 G in a stall". Even a stalled airfoil generates some lift -- the stall just means that the lift coefficient drops below the best the airfoil can do, not that it becomes zero. Since the stalled wing produces lift, and this lift is generally not in a direction where it can be canceled out by thrust, it will pull the aircraft away from the freefall trajectory.

And even that isn't completely true in the theoretical extreme. If we imagine an aircraft with fully variable angle of incidence (never mind the engineering problems in building such a thing), we can imagine rotating the wings up to around 90° angle of attack. Then, by symmetry, they will not produce any lift, yet it will be fully stalled. There will be lots and lots of drag, but sufficiently powerful thrust can offset that and make the net motion of the aircraft follow a zero G / freefall trajectory.

  • 1
    $\begingroup$ +1, this is actually a good way of interpreting the "wrong" statement. $\endgroup$ Mar 30, 2014 at 14:29
  • 2
    $\begingroup$ Well, the practical use of this is that you push 0G and then you can retract flaps/slats/whatever immediately and keep 0G until you've reached best-glide speed. So while "you cannot maintain 0G in a stall" is more accurate, it is also less useful. $\endgroup$
    – Jan Hudec
    Mar 30, 2014 at 19:51
  • $\begingroup$ Experimental tilt-wing aircraft have been built that can do exactly what you describe. $\endgroup$ Sep 29, 2015 at 3:58

At 0 G the plane does not need to generate any lift, therefore there is no critical angle past which the airfoil cannot generate the lift required.

In addition, a sustained 0 G maneuver is a parabolic arc, which means that in theory your flight path should keep your angle of attack at zero the entire time (in a theoretical, symmetrical airfoiled aicraft). You can't stall because your AoA is constantly 0 degrees and cannot exceed the critical angle.

  • 5
    $\begingroup$ The critical angle of attack is the point where the airflow separates, not where the airfoil no longer generates lift. Why you shouldn't be able to have a stalled airfoil in 0G seems wrong to me, it would be very hard to maintain 0G though as you'd effectively be a big speed-brake, not what you want to be (a ballistic arc, as you say, is a lot more effective). The key here is "sustained 0G". $\endgroup$
    – falstro
    Mar 30, 2014 at 9:51
  • $\begingroup$ The more general case (not requiring a symmetrical airfoil) is to point out that you can't stall because the wing is at the zero-lift angle-of-attack throughout the 0G maneuver. At least, to a first approximation, i.e. assuming that the thrust line is exactly parallel to the instantaneous direction of the flight path. $\endgroup$ Apr 16, 2023 at 14:32

Look at it backwards.

  • Why are you at zero G? Because your downward acceleration is equal to the acceleration due to gravity.
  • Why are you accelerating downwards so fast? Because you're not creating any lift to counteract gravity.
  • Why are you not creating any lift? Because you are at zero degrees angle of attack!

And that's why you can't stall at zero G. You're always also at zero degrees angle of attack (you can get pedantic, depending how you measure it) when you're at zero G. And stalling is always caused by high angles of attack. And if that doesn't make 100% sense, you should look more into stalls before you dig too deep into how G loading affects stall speed.

EDIT: Many people here seem to be replacing the world "airplane" in the question with the word "wing". 0g flight involves the entire airplane, not just the wing. Sure, you could take a Cessna, strap highly controllable rocket engines to the top of it and drop it from another airplane. Then, use the rocket engines to counteract drag and give downward acceleration equal to the force of gravity. That would give you a stalled wing and 0g.

But that's not really an airplane anymore, and it doesn't really answer the question.

To stall, the wing must exceed the critical angle of attack. That means it must have an angle of attack above zero. That means lift is being created, which means an upward force is acting upon the airplane. That means you are no longer at 0g.

The point isn't that once an airplane is at 0g it becomes unstallable. The point is you have to leave 0g to stall it.

It's also an example of how airspeed is irrelevant to stalling. (Sounds crazy, since we talk about stalling speed!) Stalling isn't about the speed or amount of air flowing over the wings, it's about the angle, path and flow. A wing on the same airplane can be not stalled at 5mph, or stalled at 100mph. It just depends on the angle.

  • 1
    $\begingroup$ Consider a biplane, with one wing at high incidence and the other at the opposite, low incidence. Both are stalled, but each cancels out the lift of the other. Voilá, a stalled airplane at 0 g. Easier than using rockets, too. $\endgroup$ May 2, 2014 at 21:43

Critical angle of attack or stall angle of attack is by definition the maximum angle at which the stream over the airfoil is still attached. It is also the angle of maximum lift. Beyond this angle, the stream over the airfoil becomes unattached, leading to a sudden decrease in lift.

enter image description here

Now, I think the article is somehow misleading. I think what they mean by stall is the airplane lift not being able to compensate weight. Since the weight in 0G is 0, then obviously it won't "stall" in this definition.

A harder question is whether the airfoil can stall in 0G according to the actual definition. Well, if the airplane is free falling in a straight line with its wings parallel to the ground, the airfoil is stalling, but in reality it will likely enter spiral dive, the airfoil will produce lift (which produces the centripetal force of the spiral) and won't be 0G anymore. An airplane is so that it will naturally adapt to the relative air stream, so it's hard to maintain 0G. You do this following a ballistic trajectory, on which at the top portion of the trajectory you get a short free fall feeling, where the airfoil is not producing lift (zero angle of attack).

source: http://en.wikipedia.org/wiki/Low-g_condition http://en.wikipedia.org/wiki/Angle_of_attack#Critical_angle_of_attack


Aircraft is flying in the air by 1 G force that resist the gravity of earth and keep the aircraft stay airborne. So if we lose this lifting force that is generated by the differential pressure between the upper and lower side of the wings, in this case the weight force, it will overcomes the lift and will pull that weight of aircraft to the downward.

So the zero G means the weight will be a little bit more than of lift, And eventually the zero G will account as weightlessness but it will be moving toward the centre of gravity (the earth). Why? because we are still flying at the atmosphere environments.

  • 1
    $\begingroup$ I am an enthusiast, not a pilot nor an engineer. I'm sorry, but I don't follow your explanation. $\endgroup$
    – CGCampbell
    Mar 20, 2015 at 11:32
  • 4
    $\begingroup$ this does not answer the question $\endgroup$
    – Federico
    Mar 20, 2015 at 14:11

You must log in to answer this question.

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