Higher stall speed on banking

Nathan Parker on Feb 02, 2012

When you bank, you lose a portion of the vertical component of lift and the stability of the aircraft will cause it to nose over and accelerate until the vertical component of lift is once again equal to the weight of the airplane. Your AoA has not changed, because you have not changed the elevator position, and your elevator controls your AoA. So you will end up at a higher airspeed, but the same AoA.

If you want to restablish the original airspeed, you will have to increase your AoA by pulling back on the yoke. Since every airspeed in a bank will occur at a higher AoA, there is some new higher airspeed where the AoA is equal to the critical AoA.

You can calculate the stalling speed in a bank by using the formula, Vstall = sqrt(1/cos(bank angle))*Vs0.

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Jim Foley on Feb 02, 2012

Download the Pilots Handbook of Aeronautical Knowledge from the FAAs website. Chapter 4, the bottom of page 4-19 is all about the aerodynamic forces in turns.

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Chris on Feb 03, 2012

I have always known it to be because of the following.
In a turn a component of the lift force is tilted to the side. Because of this you require a higher AOA to maintain altitude even at the higher speed to account for this loss of the vertical component. If you do not maintain altitude, your stall speed will not increase.
Stall speed increases by the square root of the load factor. So a plane with a stall speed of 40kts in a 60* turn at 2g, the speed will increase by sqrt(2)*40. Which is 1.41*40=56.4kts.
Remember that the stall speed is not an absolute, it will change depending on the situation.

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Nathan Parker on Feb 03, 2012

“If you do not maintain altitude, your stall speed will not increase.”

This is not true at all. Any stabilized climbing or descending turn will experience the same increase in stall speed.

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Brian on Feb 03, 2012

“Stall speed increases by the square root of the load factor.”

It may be easier for you to think of it in this manner. If you feel heavier in the airplane, your stall speed has increased. If you feel lighter, your stall speed has decreased. If you feel like you feel right now reading this post then you are in 1G. Your stall speed will be as published.

You can fly 1G descending turns. Or you can fly a 60 degree steady descending turn at ~2G. In the later condition, your stall speed is higher. In other words, your butt will give you a far better understanding of when your stall speed is higher. If you feel fat, it’s higher. If you feel normal, it’s normal.

If you want a better understanding of the why a steady turn (keep in mind steady conditions can occur in a climb, level, or descent) has an increased load factor, and thus an increased stall speed, then I suggest reading the link Jim posted. The pictures along with the description will likely prove far more useful than just reading explanations. Of course, feel free to clarify that reading with the folks here. I’m sure any of us would be glad to assist you.

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BC on Feb 05, 2012

Wings stall at a particular angle of attack. That’s all the wing knows. That’s all you need to know. However, since we all don’t have AOA gauges in our planes, we need to somehow equate AOA to speed since we do usually have an airspeed indicator available.
AOA is proportional to airspeed for a given weight and configuration (level unaccelerated flight). Usually, “stall speeds” (I hate that term because you can stall at any speed, or avoid a stall at any speed) are referenced to straight and level 1G flight at max weight.

If the wing stalls at 12 deg AOA, then it will always stall at 12 deg (assuming no configuration/flap changes). Speed doesn’t matter and neither does bank angle, weight or G’s. It stalls at 12 degrees. If you don’t pull to 12 deg, it won’t stall. Simple. If you don’t want to stall, don’t pull on the pole (stick). Remember that FAA written question/answer: “a plane can stall at any speed and at any attitude”, or something to that effect?

Lets say your F-152 (OK, Cessna) stalls at 50 KIAS. That means you’re at 12 degrees AOA (in this example). Fly at 55 KIAS, and you’ll be fine at around 11 deg AOA. Add 1000 lbs of weight, add some back stick to increase the AOA to provide more lift to support the extra weight (in 1G level flight). If that gets you to 12 deg, then you’re stalled at 55 KIAS. Same AOA, though. Better fly at 60 KIAS.

Now, bank it to 60 degrees of bank and add back stick to maintain altitude. Now you’re at double the weight you were when straight and level. The stick is back because you need the additional AOA to create the lift to support the weight of 2 Cessnas (your “lift vector” is canted 60 degrees , but the vertical component of lift is still enough to support the weight of your Cessna because of the increased AOA. Look at that vector diagram someone mentioned).

Now, if this causes the AOA to go over 12 deg, then you’ll be stalled. The bank problem is the same as the weight problem. The plane doesn’t know the difference because it really never knows it’s own attitude. It only knows AOA and how much you’re demanding (pulling on the stick). Well, it also knows speed as kinetic energy available. Fighter pilots know this as “more speed means more G available”.

Basically, increasing lift (due to increasing weight OR increasing bank) at a contant speed means increasing AOA. You can also increase lift at a contant AOA by increasing speed.

In straight and level unaccelerated flight (really, meaning 1G or constant G flight), the AOA will decrease the faster you go (to create the same amount of lift, usually 1x the weight of your plane).

So, putting all these pieces of the picture together, we (finally) come back to the original question of why the stall speed increases. When talking stall speed it means we’re talking about a given (fixed) AOA. If we’re requiring more lift (due to more weight, or more G’s), then it will require more speed. So, while the AOA at stall remains constant, the speed increases.

By the way, all this also applies to all the “V-speeds” (Vx, Vy, max range, etc). They’re all at their particular fixed AOA’s, but the speeds vary with weight just like the stall speed. That’s why airliners calculate takeoff and landing speeds because they’re always different at different weights (but same AOA). That’s why most fighters use AOA for landing: because it’s always the same, even at different weights (they’ll also calculate a speed, as a backup reference, since it’s nice to know energy available. But AOA is usually primary when, say, landing on a carrier).

I suppose I could edit this post and make it much shorter. How about I just appologize for a long post. Hope this helps.

Capt BC

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Young Ho Kim on Feb 05, 2012

With all kind replies above, still not clear.
Is below correct then?

When banked, your load factor increase, need
more lift to maintain altitude. You would need higher speed or higher angle of attack.

When turn, let’s say base to final, you banked,
If you let the plane descend as you normally wish during landing, you are not really increasing AOA,
Your stall speed would be same.

I feel like I am wrong.

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Nathan Parker on Feb 05, 2012

The idea that you need more lift to counter the load factor is very, very incorrect. This is like saying that in a car, you need more power in order to counteract the force that pushing you back into your seat. In reality, the load factor is caused by an increase in lift. In fact, load factor is defined to be Lift/Weight, which should make it clear that a load factor cannot exist without an increase in lift.

When you roll into a 60 degree banked turn, there is no immediate load factor increase because lift has not increased. However, as the aircraft accelerates, the lift will slowly increase until you have twice the lift you had before at the same AoA. Or, you could immediately pull back on the yoke to increase the AoA and avoid the increase in airspeed. In that case, the load factor would immediately increase

Furthermore, it’s misleading to say that you increase the AoA to maintain altitude. Whether you can maintain altitude or not will depend on the drag at the new AoA or the new airspeed. In either case, you’ll have a net increase in drag at the original airspeed and will have a steady descent. Depending on which side of the drag curve you’re on, you might be able to achieve level flight by further increasing your AoA to decrease your airspeed and drag, or you’ll have to increase thrust.

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Brian on Feb 05, 2012

“When banked, your load factor increase…You would need higher speed or higher angle of attack.”

Reading what Nathan posted, do you see how you’re both right and wrong. You’re second half is right, lift would need to increase by either speed (descending) or angle of attack (pulling back).

However, the first part of it is incorrect. Just banking doesn’t increase the load factor. Increasing lift increases the load factor.

–

“When turn…If you let the plane descend as you normally wish during landing, you are not really increasing AOA”

But you see you must increase angle of attack to increase lift; assuming you maintain airspeed. You must increase lift when you bank; either by acceleration or with angle of attack. Try this out in the airplane at a safe altitude with your instructor by setting up for a normal descent using the following:

Power – 1500 RPM
Airspeed – 70 knots
Attitude – Level flight
Flaps – Full

From trimmed level flight at 70 knots, bank to 60 degrees. After you do that keep yourself at 70 knots. You’ll discover that you must pull back to do this, you must increase lift in a bank; either by speed or angle of attack.

*Note: In extreme climbs or descents (greater than 15 degrees) this concept is slightly more complex. But speaking in terms of normal climbs and descents you can expect to increase load factor in a turn.

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BC on Feb 05, 2012

Actually, it’s not incorrect at all. The forces are in balance and it really doesn’t matter which one I list first. I’ve also made several assumptions to keep it simple (like no speed change in the turn, which means adding power to offset drag). Yes, one can roll into a bank and not pull, but that’s not the object of the question. The new lift vector needs to oppose the sum of the weight vector and the centrifugal force vector. Basically, lift countering the load factor. I tried to keep it simple and not use all those terms. But you are correct that load factor is caused by lift (I never said it wasn’t).
Nathan, it appears from your post that you’re familiar with the material. I must then assume you’d be familiar with my assumptions, including steady state situations. No need to pick apart a post by introducing items that obviously don’t apply to the question that was posed.

Since i don’t have time to go into great detail on this subject, and you and others with a higher level of knowledge want all the details, here’s a quick reference that will please everyone, including those who want a quick “picture worth a thousand words” and rocket scientists alike (the vector analysis diagram explains it all):

http://en.wikipedia.org/wiki/Load_factor_(aeronautics)

Capt BC

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Wes Beard on Feb 05, 2012

Young Ho Kim,

Keep in mind most of the aviation books out there don’t discuss load factor in any other state than level flight. The airplane is not climbing or descending at all during the turn. In this one example, load factor is directly related to the angle of bank using the formula Load Factor = 1 / cos (bank angle). To maintain enough “vertical” lift to counteract weight you will need to either increase your true airspeed or pull back on the yoke and increase the AoA.

If you base to final question, you are correct that the stall speed would be the same as you banked and descended. The key to remember is that your pitch attitude during a stall is going to be lower than in a level flight condition.

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Nathan Parker on Feb 05, 2012

BC:

Forces are not in balance when the analysis is conducted from an inertial frame of reference, as is usually the case. During a turn, there is a constant acceleration towards the center of the circle. The horizontal component of lift is not balanced by any other force; indeed, if it were, no turn would occur. Centrifugal force is a “pseudo force” and doesn’t belong in any vector diagram for an aircraft turn, except perhaps to dismiss it.

I do think a reader of your post would gather that load factor existed independently of the lift vector and that would cripple any further understanding of this subject. My view is that it is never pedagogically defensible to create an obstacle to future learning just to make things simple. Better to use the language that our textbooks often do: “That’s beyond the scope of this course.”

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BC on Feb 06, 2012

I also have a degree in physics, but I agree with you that our quest for the perfect diagram/explanation is far beyond the scope of this topic. Sometimes things have to be simplified for an explanation that can be understood by students, even though there are some technical “deviations” . And I know you have a background in physics because no one else would know that “centrifugal” force is really a misnomer. Good on you.

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Brian on Feb 06, 2012

Wes, “If you base to final question, you are correct that the stall speed would be the same as you banked and descended. ”

If I’m reading you right, you’re claiming that a base to final bank would have the same stall speed as level flight so long as we descend. This is incorrect. However, it could be made to be true if your descent angle were steep enough that the vertical component of drag, combined with the vertical component 1G wing lift, were enough to be the opposing weight.

Keep in mind, though, that this is highly unlikely in typical descents with typical banks. Further, it would only be exactly true at one exact bank angle with a corresponding exact descent angle. Change either, and the other would have to change as well to keep your statement true. Clear as mud yes?

Adding pitch angle into this discussion only makes it unecessarily complicated because now we’re talking 3D vector diagrams instead of 2D. This is why I made a point to note the importance of treating anything less than 15 degrees of pitch as a “normal flight” condition. Meaning, treat turn forces as if you’re in level flight so long as pitch angle < 15. I.E. For all normal flight this applies: Load Factor = 1 / cos (bank angle) … (Source of this statement: Aerodynamics for Naval Aviators.)

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Nathan Parker on Feb 06, 2012

“If I’m reading you right, you’re claiming that a base to final bank would have the same stall speed as level flight so long as we descend. This is incorrect. ”

Yes, and this is what gripes me about the language “increase AoA to maintain altitude”, because it implicitly suggests that if you don’t try to maintain altitude, you don’t increase the stall speed. Sloppy language on the part of the instructor can absolutely destroy knowledge in the head of the student. I know that people groan and roll their eyes when I quibble about the words used to describe something, but I can point over and over again to misunderstood concepts that are dervied from those incorrect words.

BTW, I calculate that you need a descent angle of 60 degrees to make a 60 degree banked turn a 1 g manuever.

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Brian on Feb 06, 2012

“This is why I made a point to note the importance of treating anything less than 15 degrees of pitch as a ‘normal flight’ condition.”

Instead of just arbitrarily claiming the above, let me show you. I’m sorry for the math, but it’s necessary to see the truth behind the statement. If you’re not comfortable with math, fear not, simply look at the bold text and read the conclusion section.

– Necessary Formulas –
Level Flight Load Factor = 1 / Cos (Bank Angle)
Load Factor Corrected for Pitch = [Cos (Pitch Angle) * Load Factor in Pounds] / Aircraft Weight
– The Example –
Condition: 2500 pound airplane — 30 degree bank — 15 degree descent.

Level Flight Load Factor = 1 / Cos (30) –> 1.15

Since we need load factor in pounds –> 1.15 * 2500 = 2887
Load Factor Corrected for Pitch = [Cos (15) * 2887] / 2500 –> 1.12
–Conclusion–
The difference between a level 30 degree bank load factor (1.15) and 30 degree bank while in a 15 degree descent (1.12) is only 0.03. Sure, the descent load factor is less than the level flight condition. But to say load factor is non existant in any descent, well that’s incorrect. Further, with an error of a mere 2.6 percent we conclude that, with pitch attitudes of 15 degrees or less, the level flight load factor formula is sufficiently accurate.

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