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I'm a little confused by the angle of attack of an aircraft during climbing flight. Suppose you have an aircraft in level flight which then rotates to some angle x. Just after rotation, I would assume that the angle of attack is equal to x, but after this I get confused. The aircraft continues to climb at the angle x, but does the angle of attack vary with time spent in climbing flight? Or once the rotation is over does the angle of attack go back to whatever it is for steady level flight?

Similarly, if you have an aircraft travelling at steady level flight with net lift = 0N, which then rotates to some angle x where for an angle of attack x the net lift is 50N. Does the aircraft only experience this extra 50N of lift for a brief moment during/after rotation, and following that the net lift acting on the wings is again 0N?

imveryconfused
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Angle of attack is the angle between relative wind, a.k.a. flight path, and the reference datum (chord line in the below image, but might be just given relative to longitudinal axis of the aircraft especially since many aircraft have wing twist, so the chord line varies along span).

Figure 2.2: Pitch + Incidence = Climb + Attack

(figure from How It Flies section 2.4)

Now to you cases

Just after rotation, I would assume that the angle of attack is equal to x, but after this I get confused.

As the pilot pulls on the control column and the elevators deflect up, the aircraft starts to pitch up. As it pitches up, the angle of attack increases. That increases lift, which is proportional to angle of attack (in the non-stalled region).

As the lift increases, the aircraft will start accelerating upwards, which means its flight path angle will increase. The pitch angle still increases at the same rate, but now the flight path angle also does, so their difference—angle of attack—increase will slow down until it stabilises at some value where the flight path angle rate catches up with the pitch rate.

If the pilot then pushes the control column back and neutralizes the elevator deflection, the aircraft will stop pitching up and the angle of attack will decrease again until the flight path rate stops as well.

In fact the angle of attack at which the aircraft will stabilise is proportional to the elevator deflection (for longitudinally stable aircraft), so if the pilot lets the control column return to its trimmed position, the aircraft will assume the original angle of attack again.

I'll refer you to the mentioned How It Flies chapter 2. Angle of Attack Awareness and Angle of Attack Management for further details.

Jan Hudec
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  • 'As the lift increases, the aircraft will start accelerating upwards, which means its flight path angle will increase. This goes against the increase of angle of attack, which stabilises at some value when the flight path angle rate catches up with the pitch rate'. I think this made everything completely clear to me. So the angle of attack will decrease over time after rotation until the AoA is at whatever angle with the flight vector that is required for steady level climb? – imveryconfused Nov 24 '18 at 21:18
  • @imveryconfused, the pitch still increases at the same rate, but the flight path angle now also increases, so their difference—angle of attack—stops increasing. – Jan Hudec Nov 24 '18 at 21:21
  • Thank you so much - this is really helpful. Just as an example, say an aircraft pitches to 5 degrees instantaneously (obviously not possible on many accounts, but just for the sake of argument!) and then stops, the angle of attack at that instant is 5 degrees and the extra lift generated by the wings pushes the aircraft upwards. This changes the flight path of the aircraft, decreasing the AoA over time, until the flight path and the AoA are at some angle apart where the aircraft is now in a steady climb? – imveryconfused Nov 24 '18 at 21:29
  • @imveryconfused, yes, basically back at the same AoA as for level flight, because the lift required during steady climb is a tiny bit less than for level flight (compare vertical flight when weight is completely balanced by engine thrust and lift, which is the horizontal component at that point, is zero) (do read the referenced book, especially the chapters 2 and 4; they do very good job of explaining the physics) – Jan Hudec Nov 24 '18 at 21:33
  • That is really helpful - I can't thank you enough! – imveryconfused Nov 24 '18 at 22:52