Stall speed in climbing turn
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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?
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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?
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.
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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.
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.
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higher angle of attack which translates to a higher stall speed
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.
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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
Pilot DAR...
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.
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.
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.
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
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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...
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.
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...
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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?
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?
Last edited by Pull what; 11th May 2012 at 13:27.
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.
QED.
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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?
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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.
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.
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 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.
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.
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.
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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.
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.
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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?
This actually hurts to read most of these explanations. It's embarrassing. Have we all passed basic aerodynamics or not?
Genghis explained it best.
Genghis explained it best.
Last edited by Runaway Gun; 11th May 2012 at 18:10.
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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....
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....
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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.
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.
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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.
In a climbing turn, the outside wing has a higher angle of attack which translates to a higher stall speed.
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.
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.
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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.
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.
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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.
This quote is talking about why, in a climbing turn, the outside wing stalls first.
Roll-Wise Torque Budget [Ch. 9 of See How It Flies]
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...
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.