AOA in descending turns

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MrManH

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MrManH
Hello everyone,

I read that during a descending turn, the high wing (outside the turn) has a smaller AOA than the low wing (inside the turn). I have trouble making sense of that as in order to be higher than the low wing, it would need to generate more lift.

Why wouldn't the high wing have the greater AOA as it does in a level turn or climbing turn?

I understand that in a turn the outside wing has a slightly greater airspeed than the inside wing which does contribute to extra lift, but what about the down-aileron of the outside wing? That increases the AOA for that wing doesn't it?

Also as far as I know, the relative wind still comes from below the chord line so why would the AOA "logic" be reversed in a descent?

This is a three-dimensional topic so I understand that it's hard to explain it in writing or with 2D images. If you have a 3D image that illustrates this topic, that would be great.

Thanks!
 
MrManH said:
Hello everyone,

I read that during a descending turn, the high wing (outside the turn) has a smaller AOA than the low wing (inside the turn). I have trouble making sense of that as in order to be higher than the low wing, it would need to generate more lift.

Why wouldn't the high wing have the greater AOA as it does in a level turn or climbing turn?

I understand that in a turn the outside wing has a slightly greater airspeed than the inside wing which does contribute to extra lift, but what about the down-aileron of the outside wing? That increases the AOA for that wing doesn't it?

Also as far as I know, the relative wind still comes from below the chord line so why would the AOA "logic" be reversed in a descent?

This is a three-dimensional topic so I understand that it's hard to explain it in writing or with 2D images. If you have a 3D image that illustrates this topic, that would be great.

Thanks!
Click to expand...

Have you done spin training yet? Makes much more sense after you play with it in the wild.
 
Not yet, I'll be doing that in the near future though.

In the meantime I'd like to understand this behavior based on the theory.
 
aileron is not down or up in a stable turn. it's actually neutral.
 
The outside wing is moving faster, therefore it must have a lower aoa or it would be producing more lift and you'd roll even farther into the turn. This is true regardless of whether you're climbing, descending, or level.
 
Dana said:
The outside wing is moving faster, therefore it must have a lower aoa or it would be producing more lift and you'd roll even farther into the turn. This is true regardless of whether you're climbing, descending, or level.
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So you're saying that in a stabilized turn, the outside wing always has a lower AOA than the inside wing?

By stabilized, I mean the desired bank angle has been established and the ailerons are now neutral.
 
MrManH said:
So you're saying that in a stabilized turn, the outside wing always has a lower AOA than the inside wing?
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Yes.

This sentence in your original post is incorrect:
MrManH said:
in order to be higher than the low wing, it would need to generate more lift.
Click to expand...
It doesn't have to generate more lift to "be higher" if the bank angle is established / constant (as @Dana already said). In fact there can't be a difference in lift or the airplane would still be rolling.

The higher wing did have to generate more lift momentarily to "get higher" (i.e. to roll into that attitude).
 
Both wings have to generate the same lift to maintain a constant roll (bank) angle. The outside wing is going faster and that tends to increase lift as a function of velocity squared. This is offset a bit by the lower AOA. DIhedrial also factors into the equation. But in the end, you do whatever you have to do with the ailerons to maintain a constant bank (if that is your objective) which is the final balance factor.
 
MrManH said:
So you're saying that in a stabilized turn, the outside wing always has a lower AOA than the inside wing?

By stabilized, I mean the desired bank angle has been established and the ailerons are now neutral.
Click to expand...

Think of it like this. It isn't this simple and there are all kinds of other things involved but it might help put the picture together. In a turn the outside wing is moving faster. Now think about stall speeds. The slower you go the closer you are to stalling. So by that simplistic view of it, the inside(slower) wing stalls first. Speed doesn't cause stalls of course. Angle of attack causes stalls. But the speed at which the critical angle of attack occurs is predictible given certain conditions of flight.
 
tiger said:
Yes.

This sentence in your original post is incorrect:

It doesn't have to generate more lift to "be higher" if the bank angle is established / constant (as @Dana already said). In fact there can't be a difference in lift or the airplane would still be rolling.

The higher wing did have to generate more lift momentarily to "get higher" (i.e. to roll into that attitude).
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Thanks for clarifying and I agree that my original sentence was incorrect.

From a mathematical demonstration, I can see how the high wing AOA would have to be lower at a greater airspeed in order to match the amount of lift produced by the slower wing. I wish I could also demonstrate this visually but I can't visualize the relative wind and the chord line of both wings at different AOA while the airplane is in a turn. EDIT: I think the spiraling staircase analogy will suffice for this.

I can no longer find the link that claimed that in a climbing turn, the high wing would be the one with the greatest AOA. I guess we've ruled that as an incorrect statement.

This leads me to my next question. Why would the high wing drop first in a slip? The FAA says this: "If the airplane is slipping toward the inside of the turn at the time the stall occurs, it tends to roll rapidly toward the outside of the turn as the nose pitches down because the outside wing stalls before the inside wing."

Would that behavior change based on whether it's a side-slip (straight flight path) or slipping turn?
 
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Grab a copy of "Stick and Rudder" by W. Langewiesche at your earliest convenience. He explains all of this in great clarity. Available on Amazon. Best $20 you'll spend in flight training.
 
I think I just answered my own question regarding the high wing stalling in a slip.

In order to maintain the bank angle while slipped, the high wing must keep its aileron down to counter the opposing yawing moment. The high wing is therefore maintained at a higher AOA and would be the first to stall.

Do you agree?
 
Chip Sylverne said:
Grab a copy of "Stick and Rudder" by W. Langewiesche at your earliest convenience. He explains all of this in great clarity. Available on Amazon. Best $20 you'll spend in flight training.
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This is great advice. I would follow it. My advice is to follow Chip's advice.
 
I recently saw a very good video of someone explaining how a skid in an airplane is a useless dangerous thing, to be avoided, but a slip is not, and has uses and is not dangerous (other than be careful about descent and the ground and all) and during his explanation he had a model airplane in his hands and pointed out that your eyes, using a model like that, could be used AS the relative wind airflow direction.
I too was having problems figuring out what AOA was doing in certain maneuvers, but his point is really good because of you take the model and point it directly at yourself, your eyes are the airflow, so now put it in a banking situation, level with your eyes...you see what the relative wind is.
Does that make sense? It might be ridiculously obvious but for me it was a huge help. If in a maneuver from level with eye position to the model, if you have to look up its a higher angle of attack.
 
MrManH said:
I think I just answered my own question regarding the high wing stalling in a slip.

In order to maintain the bank angle while slipped, the high wing must keep its aileron down to counter the opposing yawing moment. The high wing is therefore maintained at a higher AOA and would be the first to stall.

Do you agree?
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I'm not an expert and the advice above to read "Stick and Rudder" is great. But I think you are focusing too much on ailerons. While they certainly can have an effect(especially if you move them the wrong way in or near stall), in a steady state turn(bank rather) situation they are mostly neutral(air-frame dependent).

Most of the time it's the relative airspeed of the wing that dictates its relative AOA(to the other wing)
 
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Chip Sylverne said:
Grab a copy of "Stick and Rudder" by W. Langewiesche at your earliest convenience. He explains all of this in great clarity. Available on Amazon. Best $20 you'll spend in flight training.
Click to expand...

Eric Stoltz said:
This is great advice. I would follow it. My advice is to follow Chip's advice.
Click to expand...

LongRoadBob said:
I recently saw a very good video ...and during his explanation he had a model airplane in his hands and pointed out that your eyes, using a model like that, could be used AS the relative wind airflow direction.
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Toggle full screen and turn display to portrait:

https://archive.org/details/StickAndRudderAnExplanationOfTheArtOfFlying/page/n131

Dihedral.JPG

Although used to explain something else his illustration answers your questions about angle of attack in a slip and, if you imagine the plane in a bank rather than level flight, in a turn. The effect of dihedral increases the angle of attack on the advanced or inside wing. But despite that fact, the other wing stalls first! You can see, though, the fuselage interferes with airflow on the lagging wing which is the outside wing in a turn with respect to relative wind.
 
LongRoadBob said:
I recently saw a very good video of someone explaining how a skid in an airplane is a useless dangerous thing, to be avoided, but a slip is not, and has uses and is not dangerous (other than be careful about descent and the ground and all) and during his explanation he had a model airplane in his hands and pointed out that your eyes, using a model like that, could be used AS the relative wind airflow direction.
I too was having problems figuring out what AOA was doing in certain maneuvers, but his point is really good because of you take the model and point it directly at yourself, your eyes are the airflow, so now put it in a banking situation, level with your eyes...you see what the relative wind is.
Does that make sense? It might be ridiculously obvious but for me it was a huge help. If in a maneuver from level with eye position to the model, if you have to look up its a higher angle of attack.
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Was this the one? I think it helps answer the OP question.

 
Go do some glider flying to really see the effect of this. The long wingspan and slow speed (thus small turning circle) really accentuate the tendency of the plane to overbank into the turn. That small circle and long wingspan generate a really big airpseed difference across the two wings. As a result, you find yourself flying with a lot of opposite aileron in a 45 degree bank when thermalling.
 
^^ last 2.

Ok, I was a bit too generic in my "neutral" comment. Yes, there are planes and bank angles that require aileron input to stay in constant bank. Often opposite(also opposite to OP) to the turn. However, for the purpose of OP question, my statement was mostly accurate. Aileron isn't what's deciding the AOA in his OP.
 
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genna said:
^^ last 2.

Ok, I was a bit too generic in my "neutral" comment. Yes, there are planes and bank angles that require aileron input to stay in constant bank. Often opposite(also opposite to OP) to the turn. However, for the purpose of OP question, my statement was mostly accurate. Aileron isn't what's deciding the AOA in his OP.
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Actually it is, the effective AOA of the outboard wing is lower, because the aileron is deflected upward. Likewise the inboard with is higher because the down aileron. This is why skidding turns can have the tendency to quickly develop into a spin. That higher AOA in the inside wing causes that wing to stall first. But the OP was a bit confused thinking that the outside wing has 'down' aileron... its actually the opposite in a steady state turn.
 
George Mohr said:
Actually it is, the effective AOA of the outboard wing is lower, because the aileron is deflected upward. Likewise the inboard with is higher because the down aileron. This is why skidding turns can have the tendency to quickly develop into a spin. That higher AOA in the inside wing causes that wing to stall first. But the OP was a bit confused thinking that the outside wing has 'down' aileron... its actually the opposite in a steady state turn.
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Isn't this statement relatively aircraft specific based on its lateral stability and bank angle? I'll have to pay closer attention next time I get into a 30 degree bank in the 172RG to see if I'm actually using much opposite aileron to hold that bank. I'm guessing your use of the word "opposite" includes neutral aileron?
 
George Mohr said:
Actually it is, the effective AOA of the outboard wing is lower, because the aileron is deflected upward. Likewise the inboard with is higher because the down aileron. This is why skidding turns can have the tendency to quickly develop into a spin. That higher AOA in the inside wing causes that wing to stall first. But the OP was a bit confused thinking that the outside wing has 'down' aileron... its actually the opposite in a steady state turn.
Click to expand...

Look at OP: "..I understand that in a turn the outside wing has a slightly greater airspeed than the inside wing which does contribute to extra lift, but what about the down-aileron of the outside wing? That increases the AOA for that wing doesn't it?.."

What he is saying is that he still has down aileron on the outside wing. That's not a condition that you would normally encounter in constant bank and coordinated turn(his question was about coordinated turn)

Yes, opposite aileron would lower AOA of the outside wing. But that's a function of counteracting of the higher lift caused by higher aifrlow(among other things) on that wing.

So I agree with your statements. What I was trying to convey is that original post's premise is flawed.


EDIT: I should read your post more carefully. You say the same thing i say at the end
 
MrManH said:
Isn't this statement relatively aircraft specific based on its lateral stability and bank angle? I'll have to pay closer attention next time I get into a 30 degree bank in the 172RG to see if I'm actually using much opposite aileron to hold that bank. I'm guessing your use of the word "opposite" includes neutral aileron?
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You probably won't notice it in a 30 deg bank, but you will in a steep turn(45+)
 
genna said:
You probably won't notice it in a 30 deg bank, but you will in a steep turn(45+)
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I definitely agree for steep turns.

I think we're all in agreement here, I should have set some boundaries in my example by specifying a "normal" bank angle to avoid the topic of overbanking tendencies!

Thanks everyone that answered in this thread, I'm satisfied with the answers. I'll also be ordering the book.
 
George Mohr said:
Actually it is, the effective AOA of the outboard wing is lower, because the aileron is deflected upward. Likewise the inboard with is higher because the down aileron. This is why skidding turns can have the tendency to quickly develop into a spin. That higher AOA in the inside wing causes that wing to stall first.
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Whelp, I'm here to tell you that you can neutralize the ailerons and still have the apparent lower angle of attack wing stall first.

Let's say you want to test this out in a bank to the right, here's how to set it up. Start off very near stall speed in a bank to the left. Again, to the LEFT. Hold the ailerons neutral, but use right rudder (not a bunch) and start easing back to the stall with the yoke so as to attain the break after the plane has passed through level flight and is banked to the right. In this configuration you'll be skidding toward the high side, the left wing, but the low wing will stall first. Dihedral effect at this point has the left wing at a higher AoA and vice versa. So, the low AoA wing stalls first without any confounding factor due to aileron usage.
 
dtuuri said:
Whelp, I'm here to tell you that you can neutralize the ailerons and still have the apparent lower angle of attack wing stall first.

Let's say you want to test this out in a bank to the right, here's how to set it up. Start off very near stall speed in a bank to the left. Again, to the LEFT. Hold the ailerons neutral, but use right rudder (not a bunch) and start easing back to the stall with the yoke so as to attain the break after the plane has passed through level flight and is banked to the right. In this configuration you'll be skidding toward the high side, the left wing, but the low wing will stall first. Dihedral effect at this point has the left wing at a higher AoA and vice versa. So, the low AoA wing stalls first without any confounding factor due to aileron usage.
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Huh?
 
Salty said:
Huh?
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Your airplane doesn't have ailerons that go down only up, if I understood your explanation of the still shots in your crosswind landing video thread, right? In that case, you don't need to hold the ailerons neutral. Just do a stall while slipping as in the picture on page 126 of Stick and Rudder. The lagging wing at the lower AoA (due to dihedral effect) will stall first. I think because the fuselage acts as a huge stall strip and disturbs the airflow.
 
Sam D said:
Was this the one? I think it helps answer the OP question.

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At 10:20 that guy says that if you crank the wheel left the left aileron goes down and creates a high angle of attack.

Um, nope... sorry but how did he ever get through his “controls free and CORRECT” pre-takeoff check?
 
+1 for Stick and Rudder.

From BoldMethod, "The answer is really simple - don't use rudder to tighten a turn."
https://www.boldmethod.com/learn-to-fly/aerodynamics/slip-skid-stall/

Consider Emergency Maneuver Training to not only learn the theory behind skids, spins, etc., but
to make responding to spins (or incipient spins) automatic. It's also a blast once you begin to get comfortable getting in and out of spins.
 
MrManH said:
I read that during a descending turn, the high wing (outside the turn) has a smaller AOA than the low wing (inside the turn). I have trouble making sense of that as in order to be higher than the low wing, it would need to generate more lift.
Click to expand...

In a steady state turn, both wings would be generating the same amount of lift. If the higher wing were generating more lift, the bank angle would be increasing.

genna said:
aileron is not down or up in a stable turn. it's actually neutral.
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Only in theory. ;)

MrManH said:
I think we're all in agreement here, I should have set some boundaries in my example by specifying a "normal" bank angle to avoid the topic of overbanking tendencies!
Click to expand...

Ah, but if you think about why there is such a thing as overbanking tendencies, it's very relevant here.
 
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