Stall speed in climbing turn
 

Can anyone shed some light on how there is an increase in the stalling speed in a climbing turn. I cannot see how there can be an increase unless there is an increase in wing loading and I cannot see how wing loading is increased if there is no back pressure on the CC, am I missing something here?

The turn implies an increase in loading.

G

Compared to level flight there must be an increase in back pressure on the CC (assuming it's trimmed for level flight). If you look at the forces in a turn you'll see that lift has to increase to maintain level flight. This increase in the lift force and no change in the weight means that there will be an increase in the load factor, ie: you will be experiencing more than 1G. Stall speed = stall speed @ 1G x sq.root(load factor).

Now in a climbing turn, you'll have different angle of attack on the wings. In a climbing turn, the outside wing has a higher angle of attack which translates to a higher stall speed. You will get a wing drop because the outside wing will stall first.

I should also point out that you can't determine if wing loading has increased by seeing if the back pressure on the CC has increased. I could trim out that pressure and have the airplane fly in a climbing turn without touching the controls.

Thank you both. Still puzzling over that. Taking the opposite example. If I fly along in the cruise and make a 60 degree banked turn without applying any back pressure and let go there will not be a 41 % increase in stalling speed until I apply back pressure to turn the spiral into a level turn-do you agree?

Go back to basic training, think of the lift vectors, and how it tilts in a turn. To keep the same vertical component you need to increase the total lift perpendicular to the aerofoil because it is now banked. The lift is increased by increasing the angle of attack. Take it to extremes and keep banking and increasing the angle of attack to maintain the vertical component and eventually you will exceed the critical angle and stall the aerofoil at the same speed.

IF you do not increase the angle of attack you will have less vertical lift and descend as you said, and the stall speed will not change.

Remember stall is a function of angle of attack.

Does it?

My understanding is that a change in stall speed is affected by a change in weight (and therefore G loading) with all other things (configuration) being equal. If you have a wing at a higher angle of attack, it might be closer to stalling in terms of slowing down faster, and perhaps likely to stall before the other wing is things are not symetrical, but not at a higher speed.

Correct me if I'm wrong, but the stall speed for a 1.5 G wings level pull up, during which we will presume that both wings stalled at the same time, will be the same speed as a stall in a turn (climbing dscending not a factor) at 1.5G. Just in the 1.5G turn there is a possiblity of the outer wing stallling before the inner - yet at the same speed as the pullup.

If the wings do not stall together, the first wing to stall, will define the stall as having happened, and the speed can be observed.

Happily, my experiences testing for 23.203 Accelerated turning stalls, in single and multi types up to Cheyenne, King Air and Twin Otter have never shown me a wing drop of any concern for the 30 degree bank turn requirement. Slight increase in speeds, but only attributed to the spightly increased G.

Pull what... you're correct. If you banked the airplane and let go, essentially the stall speed would remain the same. But if you were now to pull back and stall the airplane (after having put it in a 60 degree bank and let the nose drop) you'd notice that you'd stall at a higher speed. That's just because you had to increase the G loading to slow it down, which is essentially directly correlated to the stall speed.

Pilot DAR...

For what ever reason, the angle of attack is higher because the wing/airplane requires more lift at the speed that it's at. If it's at a higher weight or in a turn or pulling out from a dive, it will require a higher amount of lift and therefore a higher angle of attack. If, for whatever reason, the angle of attack is higher for a given condition, the stall speed will be higher than the 1G stall speed.

Say the critical AOA is 16 and current AOA is 8. Assuming that the Cl is linear up until the critical AOA I'll say that each degree of AOA is equal to 5 knots, meaning that if I want to balance the lift and decrease the AOA by 1, I need to increase speed by 5. In the lift equation, V is squared but I'm going to ignore that for now as it doesn't change the outcome of this thought experiment. So, if I add weight, for the same speed I need to increase the angle of attack. Now let's say the current AOA is at 12. I only need to slow down 20 knots (4 degrees AOA) to get to the stall now. At the original weight and AOA of 8, I needed to slow down 40 knots (8 degrees AOA) to get to the stall. In a turn, it's the same. I need to increase lift to stay in a level turn so instead of increasing speed, I increase angle of attack and that in turn will increase my stall speed.

Another way to 'visualize' this is to study how turbulent the airflow is over the wing. Obviously, you'd need a windtunnel of some sort to do this. But I could tell you when the wing was going to stall without knowing the AOA. The benefit here is that I could still tell you when the wing will stall when it's covered in a bit of ice or in a lot of ice. Essentially, the AOA is just a crude measurement of what the air is doing over the wing. Just like ice or other factors will change the critical AOA and current AOA, different G loading will change the stall speed and such. They're all related.

Check out this: Roll-Wise Torque Budget [Ch. 9 of See How It Flies]

As far as I know, that's the regulation for flight testing aircraft for the stall but that doesn't mean that both wings have stalled.

Here is the military regulation:

“The stall speed (equivalent airspeed) at 1 g normal to the flight path is the highest of the following:
1. The speed for steady straight flight at CLmax (the first local maximum of lift coefficient versus α which occurs as CL is increased from zero).
2. The speed at which uncommanded pitching, rolling, or yawing occurs.
3. The speed at which intolerable buffet or structural vibration is encountered."
MIL-STD-1797A

I'm still thinking about this, but I am not sure I agree/understand...

Yes, for a given angle of bank, which corresponds to a given G load, there will be a stall speed increase. Ignoring changes in configuration (flaps, slats, ice) the angle of attack at which that airfoil stalls is proportional to the lift it must create (weight X G), but nothing else. If for a given condition of flight, the pilot increases the angle of attack, the pilot takes that wing closer to stalling, but the speed at which the stall occurs is not changing because the angle of attack is changing. If the change of angle of attack (with no other configuration changes) resulted in a change of the stall speed, the angle of attack indicator would not really be a useful indicator of the approach to the stall.

If stall speed changed relative to angle of attack, it would be necessary for flight manuals to contain stall speed charts for angle of attack in addition to stall speeds for angle of bank. Then every aircraft would have to be equipped with an angle of attack indicator so the pilot could apply that chart.

The section of airfoil at any spanwise point along the wing (and in co-ordinated flight) does not know if it is in level or banked flight. It just knows it is reacting a given load (weight X G). It knows its angle of attack, and how close it is getting to its Cl max, but it will wait to get to that Cl max before it stalls. Increasing angle of attack takes it closer to that Cl max and the stall. but does not change Cl max, so the stall speed is not being changed.

When you "go ballistic" over the top of a push over, it is possible that you are not reducing the angle of attack as the pitch attitude of the aircraft (relative to earth) reduces. You have, however dramatically reduced the stall speed, as you have reduced G, and the demand for lift.

I'm not asserting my expertise on the foregoing, but it is what I have understood from training from several sources...

Confusing isnt it?

If you fly a conventional light aircraft along hands off trimmed for S & L flight and select full power it will go into a climbing turn- how then is any loading applied if you do not touch the CC, so how does the stalling speed increase?

The wing (not the engine) lifts the aircraft by creating a lift vector opposite to weight (vertically down). With any angle of bank, the actual lift vector has to be greater than weight, since there is a horizontal component to add. Therefore the lift force at the same airspeed is greater, so alpha must increase. If you are already at max alpha, then you stall, alternatively you increase airspeed to increase lift at the same alpha.

QED.

But that doesnt prove how the stallling speed increases in a climbing turn! For the stalling speed to increase in flight without reconfiguartion there has to be a increase in weight either from ice or loading. Loading can only be produced by pilot input in most normal situations I can think of. Ive just given you one example of the aircraft climbing without any CC input at all-so how did stalling speed increase?

Pilot DAR...

Correct. Just to clarify that it's a given angle of bank in a level turn that will correspond to a given G load.

Yes, 1G stall speed is directly related to 1G weight. As you increase your G loading, the stall speed will change. Vs2=Vs1 x sq.rt(load factor) --- Vs1=Vs x sq.rt(current weight/max gross weight)

Vs1 will be the new stall speed for the reduced weight. Plug that into the Vs2 equation and that will give you the stall speed at a particular load factor.

If you're doing this all at 1G, then that's correct, ie: you're increasing your angle of attack by slowing down. The stall speed won't change because you're still at 1G. If you increase your G loading, your stall speed will increase according to the Vs2 equation above.

In level flight at 1G that's correct. But in a turn that's not correct. Just imagine a highly maneuverable fighter, for example, in level flight at 300 kts with a stall speed of 100 kts. If the pilot yanked back on the stick, the aircraft would pitch up abruptly and would be able to stall at that very high speed. He would feel a high G loading and that G loading would be what causes the stall speed to increase. All these equations go back to the Lift equation.

Exactly! In the push-overs, you will have a lower stall speed than published and, like you pointed out, it's because of reduced load factor (G). The same relationship still happens when you have positive load factor - the stall speed will now increase.

It seems like you understand that G directly affects stall speed but don't see how angle of attack plays into it. Going back to the lift equation -- Lift = 0.5 rho Vsquared S Cl

That can be rewritten for our purposes as -- Weight = Vsquared AoA

So, at a higher G, I essentially 'weight' more and need proportionally more lift. To balance the equation I can either increase velocity or increase the AoA. If I keep the same AoA and increase the speed, I'll increase the lift to balance the weight. If I keep the same speed and increase the AoA, I'll increase the lift to balance the weight. I could do any combination of either and come out with an increase in lift to balance the weight. In a steep turn you have to add power because to maintain level, you're increasing your angle of attack. That increases drag which must be offset by thrust or you will slow down.

If I wanted to accelerate in level flight, I add power to accelerate but I decrease the AoA so that I don't get any net increase in upwards force.

italia458, your statement "I increase angle of attack and that in turn will increase my stall speed" is very confusing because it is too general and wrong most of the time. When you increase your angle of attack by slowing down in level flight, stall speed stays just the same.

As you also pointed out, stall speed is function of the square root of the load factor. You know all that, it is just the way you write it that makes it look more complicated than it really is.

Because you are no longer trimmed. The excess power is converted into climb with your airspeed more or less constant - assuming no phugoid. If you let the aircraft bank one side or the other, some of that excess power will be converted into higher speed. Higher speed without changing trim or touching the controls means the aircraft wants to pull up, or pull up in a turn. Thus stall speed increases.

This actually hurts to read most of these explanations. It's embarrassing. Have we all passed basic aerodynamics or not?

Genghis explained it best.

Thanks for the detailed expeanation Italia, but I'm just not getting the part about increased AoA incresing stall speed. It looks like we agree on all the rest.

Though I have done high speed pull ups in a C150 Aerobat (just 'cause I could) and made it "stall" at an unusually high speed, I don't really think that counts, or qualifies as a true explanation of the relationship between AoA and stall speed. Aside from possibly ripping it off, any airfoil can be taken to an extreme AoA at any speed, and "stall", but doing that does not conform to the accepted flight test technique for determining stall speeds, so I don't accept it as relevent.

I maintain that were AoA a factor in stall speeds, flight manual data and airspeed indicator markings, and flight training would be very much more complicated than they are.

When stalling planes, which I do regularly, I always consider contaminated airfoils, high lift devices, G loading, and weight. AoA is just something which is a biproduct of the event.

Though I see the formula you present, I admit to not quite seeing how it applies in real world flying.

If AoA were a factor in a change in stall speed, how would an AoA indicator be of any use for aircraft attitude control to prevent unintended stalls? There you are refering to the AoA indicator as you carefully fly the aircraft in the approach to stall regime, and the stall speed is changing because of the increasing AoA, and suddenly moves in to meet you before you reach the critical AoA? I think not.

In my earlier days, my job was to fly newly STOL kitted Cessnas, for the purpose of confirming the correct set up of the stall warning vane (it must be repositioned, when the airfoil cuff is installed). That vane operates based simply upon AoA and stagnation point on the leading edge. If varying or increasing AoA changed the unacellerated stall speed, my task would have been impossible....

During recent flight testing in a Piper Cheyenne II, which required my repeated approach to the stall, I was using the AoA to get close, without getting too far into it. With some practice, I found flying well into the red of the AoA no problem. Indeed, the aircraft had a much more forgiving stall than I expected (based upon the urban myths of that type). That flight test was cut short with a landing gear problem (when the left main seemed to get stuck up during configuration change for test). After half an hour of no luck, I used the descent to get a flyby to do a few very steep turns to build up some G and try to pull it down. The Cheyenne does not seem to care for that kind of flying. It did not bite me, but it sure nibbled! I stopped doing that!

Could you offer some more explicit real world examples of civil aircraft for which the AoA is a stated factor in stall speed determination, beyond the agreed effect of G on stall speed? How is this documented in the flight manual and pilot training material?

I'm trying to see what you're saying, but I'm not there yet....

in agreement with Pilot_DAR here

For a given configuration etc. a wing will stall at a critical AoA. That critical AoA is NOT a function of my current AoA - if my stall AoA is 10 degrees, it doesn't matter if I'm at 1 or 9 right now - 10 is still the critical AoA.

Indeed, even under 'g', if we ignore wing deformation, the wing still stalls at the same critical AoA - and generates the same lift coefficient as it does so. The increased stall speed under 'g' is how the load factor is compensated for, since the CL can't change.

I *can* think of some esoteric ways to have current AOA affect apparent stall speed, but they depend on some quite convoluted systems architecture and in most cases are artificial, and would in fact not be seen if you did a 'true' stall.

I think two of my sentences are the ones that are causing the confusion. When taken in the context I was explaining, I think they make perfect sense but I accept that I might not have explained it well to someone else. I'll try to explain them better below.

This quote is talking about why, in a climbing turn, the outside wing stalls first. If you look at the link I included, it shows that in a climbing turn, the outside wing is at a higher angle of attack. Assuming that both wings will stall at the same angle of attack, if I now slow down in this condition (outer wing having a higher angle of attack), the outer wing will stall first as it will reach the critical angle of attack first - assuming that I don't change anything else. This is what I meant when I said the outer wing will have a higher stalling speed in this case.

I think we all agree that in a level turn you need to increase the lift, which causes an increase in load factor (and an increase in stall speed) to maintain your altitude. So now in the banked condition, I need to find a way to increase the lift.

Lift = 0.5 rho Vsquared S Cl

I can simplify it since I'll assume that the density stays the same and that the surface area of the wing doesn't change. It then becomes:

Lift = Vsquared Cl

To increase lift I need to either increase the speed or increase the Cl or a combination thereof. Cl essentially translates to AoA. So I either increase my AoA or I increase my speed to get the extra lift. Assuming I do not want to increase my speed (like I stated in the quote), I need to increase my AoA. Increasing my AoA in this situation puts the aircraft into a load factor higher than 1. Since the stall speed is directly related to load factor, I have now increased the stall speed in this scenario. I think the confusion came from me not summarizing that this increase in AoA, increases the load factor (because I'm maintaining the same speed which results in more lift than weight now) and, therefore, increases the stall speed.

I was only talking about the airplane in a turned condition with no increase in speed. In that case, I believe it's perfectly correct to say that the stall speed increase is due directly to the AoA increase - that is the root cause in this case. In a 1G condition - ie: slowing down in level flight - this is not the case as the 1G condition dictates the stall speed will remain the same.

If it's still not clear, let me know.

Other than that, I think we're all in agreement.

italia458 you obviously have some difficulties with equations and aerodynamics, I don't think this is the right place to address those at that level. Maybe you should read a good book on this topic, I recommend Darrol Stinton.

In a turn, it is not the outside wing that has the higher stall speed, it is not the inside wing either. They both have the same stall speed, otherwise the designer of the aircraft needs to be shot. You will also find that in a turn, it is the inside wing that will stall first because it is flying at a slightly lower speed and higher angle of attack - just the opposite of what you are saying.

On your statement that increasing angle of attack increases stall speed: it is the load factor that increases stall speed, it does not matter how you increase the load factor. Your are stating a general problem via one of its particularities - so that requires further explanations and it is a lot of trouble for nothing.

If you want a good understanding of stall, focus on angle of attack. The way you express stall margin as a speed that is above or below the one of your airplane is extremely confusing and impractical.

Machdiamond... I don't think I have difficulty with equations and aerodynamics like you stated. Can you point out somethings that I said wrong?

That was my quote and I was talking about a climbing turn. That's the key difference.

Roll-Wise Torque Budget [Ch. 9 of See How It Flies]

Maybe to yourself it doesn't matter what increases the load factor but there is nothing wrong with stating that the increase in load factor was due to the increase in AoA. I don't believe anything I've said is wrong. If you were to read what I said and follow along with my examples I was giving, it's pretty clear to me. I think we're all past the fact that load factor directly affects the stall speed, according to the equation that I provided!

No I was not! It was a very specific example and I clearly noted that.

I have a good understanding of stall. This is the Flight Testing forum - I think talking about things other than the standard stuff you would get told in a beginner pilot course on aerodynamics would be appropriate. It seems that me going one step beyond the load factor bit and identifying that, in this case, the increase in AoA is what caused that increase in stall speed was too much.

@italia458

Sorry but you are thoroughly "wrapped around the axle".

Increasing angle of attack DOES NOT increase the stall speed. Increasing load factor does - everyone agrees with that - but you are confusing a coincidental factor with a CAUSAL factor.

This can be seen by considering that I can increase angle of attack independent of load factor - a simple slow-down does that - and the stall speed doesn't change, even though the AoA changes throughout the manoeuvre - the time history of AoA in no way affects the final stall speed for a suitably slow manoeuvre.

Arguing that because load factor affects Vs, and AoA can be associated to load factor, then AoA affects stall is misguided logic. I could as well try claiming that bank angle affects stall speed, because bank angle can be related to load factor under certain circumstances. That would be just as wrong.

AoA does not have a direct effect, or a causal effect, on stall speed. Never has, never will.

italia,

You are very wrong about the outside wing having a higher angle of attack.

The extreme example of this is a helicopter blade.

The tip is always at a dramatically lower AoA than the root.

:O

I defined a condition and for that condition I specified that angle of attack was changing the stall speed. There is nothing wrong with that statement.

It's exactly the same as saying bank angle increases stall speed - under the condition of level flight, that is 100% correct.

I understand this perfectly well - I would suggest you re-read what I said.

Really? Well you should write to John Denker and tell him that he's wrong.

I would just suggest you look into the helicopter blade thing again. I would guess that you're actually talking about 'washout' and not a climbing turn in an airplane. The propeller blades on an airplane also have washout.

There are moments when the explanation of a problem is getting much more complicated than it deserves.

Deneker, who so far as I know tends to get his maths right, shows that in a climbing turn [ reference ] the outside wing is at a slightly higher angle of attack. Therefore if the aeroplane is stalled in a climbing turn, the outside wing will stall slightly ahead of the inside wing (which may not have done).

This makes intuitive sense, if only because in my experience most aeroplanes stalled in a turn tend to roll either towards, or through, wings level.


Now, the speed the aeroplane registers as the stall speed is from a single point on the airframe.

So, at a given reference speed, in a single system (the aeroplane) the outside wing stalls before the inside wing. In a certain light then, the outside and higher AoA wing has a lower stalling speed (as it has stalled) than the inside and thus lower AoA wing, which may not have done.

In reality, when you come down to aerofoil level, stall speed is a pointless concept - it only has any validity when you apply it to a whole aeroplane, not to a bit of aerofoil. At the level of an aerofoil: stalling alpha is a useful concept, as is Reynolds number, but stall speed is not.

Between the two wings in a turning aeroplane, the tiny difference between speeds will make a miniscule difference in Re and have absolutely no relevance, since Re only really has appreciable effects with orders of magnitude, not the small percentage change that applies here.

G

Thanks for the explanation Genghis!

Yes, No and Maybe
 

Italia,

For a lot of flight dynamics debates, the tricky bit is asking the right question rather than having the right answers! And I'm not sure that "Stall speed in a climbing turn" is really helping us out here, because it doesn't let us separate out the variables properly.

Once we talk about a climbing-, rather than a level-, turn we open up a world of possibilities. Any amount of g, any amount of lift, and any helical climb angle (just to screw up Mr Deneker's maths).

So, to have a sensible answer, we really need to only change one variable at a time. Are you asking about changing angle of bank at 1 g, or holding a given angle of bank and varying the normal acceleration?

Andy

PS For what it's worth, the Deneker explanation doesn't help much if you're considering a constant angle of bank turn. If the outer wing is traveling faster and at a higher angle of attack, then it would be producing more lift than the inner wing. This would give a rolling moment if you were not balancing this out with some out-turn aileron. This aileron is effectively reducing the angle of attack of the outside wing (excuse the horribly imprecise terminology).

PPS In essence, if your outer wing is travelling faster but you're not rolling, its at a lower angle of attack!

Andy... I agree the question being asked wasn't 'ideal' for all aspects of aerodynamics. But I do believe it was a valid question with a valid answer. In the first paragraph of my first post in this thread, I stated clearly that the stall speed was a function of the load factor and provided the equation for this. Further explanation was to expand on this and give a more detailed look at the condition of a climbing turn.

Agreed. There's also the fact that, in any climb, thrust usually has a downward component which reduces the requirement of lift from the wings. For example, if you were to go from a level flight condition at 150 kts into a climb while remaining at the same speed, the angle of attack would decrease. But I don't believe that these factors that affect climbs and/or turns would affect the comparison of angle of attack between the two sides of the wing.

I don't believe that's quite true. The only part I believe that's in contention is when the ailerons are deflected to prevent the airplane from increasing the angle of bank. When you deflect the ailerons, you're essentially changing the deflection angle (downwash) of the airflow. Generally, positive camber is used on wings so that the deflection angle of the airflow is downward (with reference to the chord line) which, in turn, lifts the wing. When you deflect the aileron on the outside wing upwards, that deflects a short band of airflow (the air flowing over the ailerons) over the wing at a slightly less downward angle. You could consider it upwards to make it easier to visualize but it's not necessarily upwards relative to the chord line. This band of airflow that's now being directed at an upwards angle, relative to the rest of the wing, is what balances the lift on the wings and ensures that the airplane does not increase its bank in the turn. The opposite happens on the inside wing where the aileron is deflected down. Since angle of attack is defined as the angle between the chord line and the relative airflow, the angle of attack is not changing. You're just changing the upwash and downwash. This video shows this: Airfoil with movable flap in smoke tunnel - YouTube

You've essentially just changed the shape of the wing which changes the coefficient of lift of the wing (similar to extending flaps) - increasing lift for the inside wing and decreasing lift for the outside wing in a turn, while remaining at the same angle of attack. It's the change in deflection of airflow at the rear of the airfoil that is ensuring you're not increasing bank in a turn.

Like Genghis said - the outside wing is at a higher angle of attack than the inside and that's why when you stall you generally will roll towards wings level or through depending on the airplane.

EDIT: I redid this post and changed a few things. Should be more clear/correct.

For what it's worth, this helps explain the change in airflow pattern at the leading edge of the airfoil in the YouTube video I posted, when the flap is deflected down. https://www.box.com/s/18a7421d629f1346c56b

Mad Scientist....

I understand what you're meaning but I don't agree with exactly what you said. This picture is a performance chart for a BE55. It shows the relationship between bank angle and stall speed. Performance charts usually list the conditions that the chart will be accurate but it doesn't even mention that the airplane needs to be in a level turn for this relationship to be accurate. Disregarding what I think of that, I believe they must have thought that it was unnecessary to include that condition considering that the relationship between stall speed and angle of bank is so well know and I'm pretty sure that everyone to pass a Private pilot flight test knows that 60 degrees of bank = 2G which has a corresponding increase in stall speed. It's a condition that is normally assumed when talking about angle of bank and stall speed.

http://i.imgur.com/wZY2c.jpg

I see nothing wrong with describing the relationship between AoA and stall speed as long as I have specified the conditions in which that relationship is true - which I did.

Italia458

The chart in your link is almost certainly referring to load factors and the consequent increase in stall speed IN A SUSTAINED LEVEL TURN. It doesn't say so on the placard but I suppose it assumes that anyone sitting in the pilots' seat knows this. If you are saying that the chart implies anything more than that you are wrong. Likewise, Mad Flt Scientist is, as usual, right

Dick

Dick... I'm not saying the chart implies anything more than that!

I think you should read my post again and the quote I used from Mad Scientist. Mad Scientist said you couldn't say bank angle affects stall speed and I say you can, if you specify the conditions, and I included the picture to help prove my point. You should read the thread again to understand why the 'bank angle affects stall speed' statement came up.

I agree, Mad Scientist is quite knowledgeable and is usually correct. But I am allowed to disagree with someone and I've stated why I disagree in this case.

Italia,

It's a comprehension thing. You said "It shows the relationship between bank angle and stall speed". But it doesn't, and that statement is wrong. There is no general relationship between bank angle and stall speed.

Dick

Italia, my apologies.

You are quite correct about the helicopter blade. (I just went and drew the diagrams!):O


However, I think Andy is quite correct about the angle of attack having to reduce to offset the higher airspeed.

You can call it what you like, ie downwash etc, but at the end of the day lift is all about AoA and airspeed, and if the airspeed has increased then the AoA has to reduce to stop a roll.

Yes that is by movement of aileron in a conventional aircraft thus you can talk about deflection angle etc.

You could, however design an aircraft that could rotate the wings independently about their wing spar rather than change their shape.
Such a wing would still require you to have to rotate them to a position giving a lower AoA for the angle of bank to be stable.


No matter how you go about it, in a stable climbing turn to the left, there is a steady rate of roll to the right, which suggests a higher lift on the inside of the turn, which points to a higher AoA not lower. In a stable descending turn to the left, there is a steady roll to the left.

Lift is simply AoA and airspeed. Downwash etc are just confusions.

Genghis

"This makes intuitive sense, if only because in my experience most aeroplanes stalled in a turn tend to roll either towards, or through, wings level."

There are many other possibilities for this.

For example, as soon as the aircraft stalls in a turn, it is no longer balancing the forces of "G" against gravity to leave you in balanced flight. ie you have gone from a balanced turn, to flying in a straight line with one wing down thus there is a sideslip towards the inner wing potentially giving it greater lift.

Or maybe the differential ailerons/rudder required to maintain a stable turn effect things?

Going on what John Farley has just emailed me(agreeing with Genghs) the clue is in the fact that no aircraft can turn without a turning force.

So I would say then we are still talking about an increase in loading because there must be a centre seeking force which in turn must increase the loading.

In relation to a full power climbing turn stall in a SE prop A/C, surely there must be a RAF difference due to slipstream over the roots but is that equal over both roots in a balanced climbing turn?

It distresses me occasionally that I no longer go to any of the same meetings at JF, the arguments we could have were wonderful. If he's agreeing with me now, I must have finally learned something.

G

Sorry to drag this back to angle of attack, but I have been looking at Denkers document and having a think.

Can somebody please explain to me where I am wrong in thinking that part of his document is flawed.

In his section on climbing and descending turns linked to earlier, he says

"In a level turn both wingtips are moving horizontally. In a climbing turn, both wingtips will be climbing, but they will not make equal angles to the horizon. This is because the climb angle depends on the ratio of the vertical speed to the forward speed. As a result of the different climb angles, we get different angles of attack for the two wingtips. The geometry of the situation is shown infigure 18.6 (in the chapter on spins)"

I have given this a lot of thought, and I don't believe that the angle to the horizon or climb or descent has anything whatsoever to do with angle of attack.

In the referred to figure 18.6, he shows that in a spin or descent the outer wing has reduced angle of attack, and says that in a climb this changes.

I think this is wrong.

I don't think motion relative to gravity/centre of the earth is in anyway pertinent, rather I think that relative airflow is everything.

I don't believe that an AoA sensor on the wing would notice any difference between a climbing or descending turn or indeed level.

If figure 18.6 is true in a descent, if you now point the nose up above the horizon, what has changed? Lift is not relative to the horizon, it is relative to the wing/AoA.

I don't get it.

Somebody please explain slowly.

Because in a turn, the outside wing is going faster. More forward speed gives a lower 'relative airflow' which equals higher AOA.

http://i13.photobucket.com/albums/a2...aker/lift2.jpg

Same applies in reverse in a descent, the outside wing has reduced AOA.

Tourist, I am with you on that one.

Angle of attack only relates to the angle between the trajectory and the body axis. The outside world does not matter at all (downwind turn anyone?).

Where the earth reference system gets involved is in defining the direction of the weight vector, which affects the left and right wing equally.

There are only three things that affect local angle of attack differently on your right or left wing in a turn: yaw rate, roll rate and aileron deflection. Also the propeller slipstream but I'll leave that one aside.

You will exhibit a yaw rate in a turn, which means lower speed for the inside wing thus higher local angle of attack. Roll rate is zero once a turn is established, thus no effect. Aileron deflection in a sustained coordinated turn will be positive or negative depending on the turn rate and the aircraft stability parameters. On most airplanes in shallow turns you will hold a bit of aileron in the direction of the turn so that means your outside wing will have increased local angle of attack. If you slow down or pull g's and stall in that configuration, your outside wing will stall first.

In a steep sustained and coordinated turn, with most airplanes, you will have to hold opposite aileron. This is to compensate the reduced lift of the inside wing which now has a much lower speed. This increases the inside wing angle of attack which is already high to start with, so your inside wing will stall first if you slow down (or pull more g's).

I hope this clears some confusion.

Background Noise.

Thanks for that, I think I get the whole helical motion thing.

If I have it right,

Both wings climb or descend the same amount, but one travels further to do it, thus if both wings have the same mounting angle on the fuselage, one has a higher angle of attack.


This does, however leave us with a different problem.

If the outer wing in a climb has higher airspeed and AoA, and lift is dependent upon AoA and lift, then how come there is no change of angle of bank, ie balanced lift?

I still believe the statement "the outer wing has a higher AoA in a climbing turn" is incorrect.

What it should say is "the outer wing in a climb would have a higher AoA, but to maintain a constant angle of bank in a climbing turn, the pilot differentially adjusts the AoA of the wings using the ailerons to reduce the outer wing lift to match the inner wing"


This, of course is actually slightly simplistic, because in a climbing turn to the left, the aircraft is actually in a continuous roll to the right, so the lift on the right wing must actually be slightly lower. The converse is true in the descent.

Outer wing has the down going aileron so that might be one reason.

I think it's more complicated than that though. Some years back I saw a glider accident where the inner wing stalled first. Possible because the combination of large span and tight skidding turn meant the inner wing was slower and partly blanketed