I understand that during an ascending or descending turn the angles of attack of the inner wing and outer wing are different.

However, I don't understand why then ascending and turning, the outer wing has greater AOA than the inner wing but during descending and turning, the inner wing has greater AOA than the outer wing.

Clarification: My confusion is that, for example, if the AOA is greater for the outer wing, then the AOA should always be greater for the outer wing regardless the airplane is climbing or descending.

  • $\begingroup$ Do you know what (velocity) components determine the angle of attack? How do these components differ between the inner wing and outer wing? How do these components differ for ascending and descending? $\endgroup$
    – ROIMaison
    Jun 3, 2019 at 8:40
  • $\begingroup$ Really the question should not say that the difference in angle-of-attack is "because they travel at different speeds"-- that implies that the effect should be present in a constant-altitude turn which is not accurate. The question could be improved by deleting the entire first paragraph. $\endgroup$ Jun 4, 2019 at 15:06
  • $\begingroup$ What do you mean exactly by "ascending or descending turn"? $\endgroup$
    – MikeY
    Jun 4, 2019 at 16:47
  • $\begingroup$ @MikeY surely that needs no clarification-- a positive vertical speed or a negative vertical speed. $\endgroup$ Jun 4, 2019 at 16:56
  • $\begingroup$ It's not clear to me that the assertion "..ascending and turning, the outer wing has greater AOA...during descending and turning, the inner wing has greater AOA" is a true statement. $\endgroup$
    – MikeY
    Jun 4, 2019 at 17:19

2 Answers 2


Here's one way-- not the only way-- to look at the dynamics involved--

Imagine an aircraft in a 45-degree-banked constant-altitude turn. Now imagine that the pilot gradually adds more and more power and the aircraft gradually enters a steeper and steeper climb until it is climbing almost straight up. Can you see how to maintain a constant bank angle, the aircraft must actually roll toward the high wingtip? As the climb angle gets steeper, the maneuver gets closer and closer to resembling a vertical rolling climb, with the direction of roll being toward the wingtip that was originally the high wingtip in the constant-altitude turn.

"Fly" through the maneuver with your hand or with a little hand-held model airplane until you understand this.

(Like this-- https://vimeo.com/128025851#t=157s -- video intentionally starts at 2:37)

Now, can you see how a rolling motion always tends to increase the angle-of-attack of the descending wingtip, and to decrease the angle-of-attack of the rising wingtip? As the descending wingtip comes down through the airmass, the local relative wind blows "up from below" compared to the local relative wind closer to the aircraft centerline-- this is an increase in angle-of-attack. Similarly, as the rising wingtip moves up through the airmass, the local relative wind blows "down from above", or at least blows up from below at a shallower angle than the local relative wind closer to the aircraft centerline. This is a decrease in angle-of-attack. These changes in angle-of-attack create an effect known as "roll damping"-- a resistance to rolling. This is why the roll rate doesn't just keep getting faster and faster as long as we hold the ailerons in a deflected position.

So that's the answer to your question. In a constant-banked climbing turn, the aircraft is continually rolling toward the high wingtip, so the high wingtip experiences an increase in angle-of-attack and the low wingtip experiences a decrease in angle-of-attack.

Note that while our thought experiment involved a continual increase in climb angle and climb rate, the basic dynamics we're talking about are present even in a climb at constant angle and rate. In a climbing turn, an aircraft must continually roll toward the high wingtip to hold the bank angle constant. Otherwise the bank angle will increase. It's not a matter of aerodynamics, but rather of three-dimensional geometry.

In a descending turn, everything is the same except that to keep the bank angle constant, the direction of roll must be toward the LOW wingtip, so the low wingtip experiences an increase in angle-of-attack and the high wingtip experiences a decrease in angle-of-attack.

"Fly" through the descending turn case with your hand or a little hand-held model airplane until you understand that the extreme case of a constant-bank descending turn is a vertical rolling spiral, with the direction of roll being toward the wingtip that was originally the low wingtip when the plane was turning with a constant altitude or with a less-than-vertical dive angle.

(Like this-- https://vimeo.com/128025851#t=128s -- video intentionally starts at 2:08)

As a footnote, you can see how aerodynamic "damping" in roll -- the tendency for the roll rate to decrease-- exerts a destabilizing influence in a climbing turn, tending to make the bank angle increase. In a descending turn, aerodynamic "damping" in roll tends to make the bank angle decrease. This is very noticeable in some applications, such as powered hang gliders and trikes.

As another footnote, understand that we are speaking of "ascending" and "descending" relative to the surrounding airmass-- an unpowered glider spiralling up in a thermal updraft is still in a "descending" turn for the purposes of this discussion.

As yet another footnote, we can observe that it is easy to see how the bank angle can change when we pitch up with zero roll rate, in some version of a chandelle or wingover for example. It is harder to see how bank angle can be constant and roll rate can be non-zero in a climbing or diving turn, but it's true, as illustrated in the video links given above. Again, it's fundamentally a matter of 3-dimensional geometry, not aerodynamics.

In closing, here is a link to a diagram from John S. Dencker's "See How it Flies Website" that illustrates how a rolling motion creates a difference in angle-of-attack between the two wingtips. The diagram is not dealing specifically with the constant-bank case, and also it is really aimed at explaining "adverse yaw" which is a separate issue, but it still may be helpful-- https://www.av8n.com/how/htm/yaw.html#sec-adverse-yaw

  • $\begingroup$ With a descending turn, don't you actually mean a negative G turn (with the lift forces pointing downwards)? I don't see how the AoA can otherwise be higher at the low wingtip. $\endgroup$
    – hrobeers
    Jun 4, 2019 at 15:53
  • 1
    $\begingroup$ @hrobeers-- certainly not. I'm sorry, I'm not understanding your point. My answer explains why the AoA is higher at the low wingtip than the high wingtip-- it's due to the non-zero roll rate, even though bank angle is constant. $\endgroup$ Jun 4, 2019 at 16:46
  • $\begingroup$ @ quiet flyer-- ok I see, thanks for clarification. $\endgroup$
    – hrobeers
    Jun 4, 2019 at 18:14
  • $\begingroup$ @quietflyer thank you for your answer. forgive my ignorance but I think you explained it from a practical perspective but I was looking for a more theoretical answer. You seem to show that the airplane rolls one way so AOA is changing. But I thought it should be that AOA changes (for some reason I dont understand) so the airplane rolls one way? $\endgroup$
    – nouveau
    Jun 5, 2019 at 15:37
  • $\begingroup$ Obviously the video would have been made slightly differently, starting with the climbing case, if it had been made specifically for this answer $\endgroup$ Jun 6, 2019 at 11:42

For the moment let the AOB of the aircraft be zero, assuming it is actually in a flat turn (just ruddering it's way around). Stealing this answer from Reddit, best description I have seen (my bold emphasis below), Props to grizzleeadam


Think of a spiral staircase with an inner handrail and outer handrail. Both handrails must ascend the same vertical distance, however, the outer rail travels a longer distance over a more sweeping path, as it has a larger radius. This means the inner rail is shorter, and reaches the same vertical height in less distance travelled - it must have a steeper angle.

Now try to imagine the rails as the relative wind of a climbing airplane. If the inner wing laid flat across the inner handrail, due to the steeper angle of the inner rail, the outer wing would have a gap at the leading edge - this translates to a higher AOA on the outside wing in a climbing turn.

For a descending airplane, this time the outer wing must lay flat against the outer rail, and the steeper slope of the inner rail would cause a gap at the inner leading edge - a higher AOA of the inner wing in a descending turn

Here's a picture, with the railings unrolled. The inside wing follows the inside rail on a steeper path. Let the wing lay flat on it. The outside wing follows the outside rail on a less steep path. The gap between the wing leading edge and the outside rail in the increase in AOA that it sees as it follows the rail in its spiral upward.

enter image description here

For a descent, the roles reverse, with the outside wing now having a lower AOA in the inside.

To accommodate a bank angle in understanding the problem, just shift the inside railing down.


Another picture. This time, I have a body fixed reference frame with the X axis out the nose, the Y axis out the left wing, and Z axis up (I know, should be flipped, but just looks better this way). The body fixed frame is embedded in an inertial frame with about 45 degrees of roll (picture the left wing going out the Y axis) and a little bit of pitch up. This is an aircraft in a climbing turn. The rotation vector in the inertial frame is straight up (dashed line). This rotation must be the vector sums of rotations about the X, Y, and Z axes. The amount of those rotations is just the projection of the rotation vector onto those axes (the thick black arrows). Because the nose is pitched up, there is projection onto the X axis, therefore there is roll.

enter image description here

  • $\begingroup$ Yes that is another valid way to look at the situation. The rails represent the direction of the local relative wind at each wingtip. It appears we could add the bank angle into the picture simply by lowering the height of the entire inside rail by some amount. Once we do this, the roll rate can no longer be zero. It's interesting how the same problem can be approached in several different ways. Is it possible that neither answer fully considers ALL the relevant effects at play? $\endgroup$ Jun 4, 2019 at 17:54
  • $\begingroup$ I think that if we correctly match the bank angle to the turn rate, so we aren't skidding around, we end up w/ a situation where it is equally valid to say that the change is angle-of-attack is caused by the roll rate or by the "spiral staircase" effect-- the difference in direction of local relative wind between the two wingtips. You can't have one without the other. $\endgroup$ Jun 4, 2019 at 18:01
  • $\begingroup$ See my update on bank angle. I agree with you on the roll rate. If you take a model airplane and give it some bank and pitch, and then rotate it through a vertical axis (climbing or descending turn) you do get some roll, pitch, and yaw rate. $\endgroup$
    – MikeY
    Jun 4, 2019 at 18:18
  • 1
    $\begingroup$ @Mike Y for any relative wind, the gap is either at the LEADING edge of the inside wing or the TRAILING edge of the outside wing. Just grabbed a ruler and tried it. $\endgroup$ Jun 5, 2019 at 0:52
  • $\begingroup$ @MikeY thanks for the answer. From your quotation, I still struggle to understand that "due to the steeper angle of the inner rail, the outer wing would have a gap at the leading edge - this translates to a higher AOA on the outside wing in a climbing turn." I cannot visualize what gap this is. Do you mind elaborating a bit on this? Thanks again. $\endgroup$
    – nouveau
    Jun 5, 2019 at 15:32

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