Welcome Guest. Sign in or Signup

20 Answers

Higher stall speed on banking

Asked by: 23542 views Aerodynamics

I was thinking about higher stalling speed in Cessna when banked for turn,

Is below logic correct?

 

when banked, you have less lift as portion of lifting power goes for turning, so the plane sink that change the direction of

relative wind to airfoil in a way that exceed critical AOA even when your attitude is not up.

Is there way to calculate stall speed based on degree of banking and Vso?

Ace Any FAA Written Test!
Actual FAA Questions / Free Lifetime Updates
The best explanations in the business
Fast, efficient study.
Pass Your Checkride With Confidence!
FAA Practical Test prep that reflects actual checkrides.
Any checkride: Airplane, Helicopter, Glider, etc.
Written and maintained by actual pilot examiners and master CFIs.
The World's Most Trusted eLogbook
Be Organized, Current, Professional, and Safe.
Highly customizable - for student pilots through pros.
Free Transition Service for users of other eLogs.
Our sincere thanks to pilots such as yourself who support AskACFI while helping themselves by using the awesome PC, Mac, iPhone/iPad, and Android aviation apps of our sponsors.

20 Answers



  1. 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.
     
     

    +4 Votes Thumb up 5 Votes Thumb down 1 Votes



  2. 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.
     

    0 Votes Thumb up 1 Votes Thumb down 1 Votes



  3. 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.

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  4. 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.

    -1 Votes Thumb up 3 Votes Thumb down 4 Votes



  5. 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.

    +4 Votes Thumb up 4 Votes Thumb down 0 Votes



  6. 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
     
     
     
     
     
     
     

    +13 Votes Thumb up 14 Votes Thumb down 1 Votes



  7. 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.

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  8. 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.

    +2 Votes Thumb up 2 Votes Thumb down 0 Votes



  9. 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.

    +2 Votes Thumb up 2 Votes Thumb down 0 Votes



  10. 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

    +1 Votes Thumb up 1 Votes Thumb down 0 Votes



  11. 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.  

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  12. 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.”  😉

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  13. 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.

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  14. 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.)

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  15. 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.

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  16. 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.

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  17. Derek Schwalenberg on Feb 08, 2012

    0 Votes Thumb up 1 Votes Thumb down 1 Votes



  18. Derek Schwalenberg on Feb 08, 2012

    Image worked in the preview… anyway here is a link to it. http://www.free-online-private-pilot-ground-school.com/images/vg_diagram.gif

    0 Votes Thumb up 1 Votes Thumb down 1 Votes



  19. Jason on Jan 21, 2015

    Young Ho Kim

    I just happened to of stumbled across this thread. As an experienced flight instructor my advice to you would be to consult your “actual” flight instructor at your flight school.

    Guys you must realise that to remain professional as instructors, information must be accurate and explained with the student in mind at all times.

    For a PPL student learning the “basics” of principals of flight for the first time, this thread is not a good start. Its a bloody mess.

    0 Votes Thumb up 0 Votes Thumb down 0 Votes



  20. jmt on Sep 04, 2015

    Keep things simple- I’m assuming that the reason you are asking this is in relation to a landing pattern, which is a very critical phase of flight where being aware of your stall speed and how turning will affect it is very important. I don’t know about everyone else on here but I am trying to do as little math in the cockpit as possible! I like common sense, practical explanations for things as much as possible.

    That being said, a great way to gain real world understanding and appreciation for stall speeds and horizontal vs vertical lift components is practicing turn patterns while maintaining a set altitude and airspeed. A 45 degree turn requires adding a little power and backstick pressure to maintain altitude and airspeed. You will feel the ~2Gs and the “seat of the pants” feel people above have mentioned.

    Now take the same exercise above without the added power and you are practicing approach turn stall recovery. 45 degree AOB, backstick pressure to maintain altitude but NO added power and you will feel the buffeting at a higher airspeed than straight and level. Both of these exercises are designed to teach you how to quickly recognize a dangerous situation PRIOR to hitting your stall speeds. You won’t be using a calculator in the landing pattern (unless something has gone very wrong and you’re doing some fuel calculations).

    Understanding what to look for when you’re flying is more important in my opinion than being able to understand the math behind it. Once you have mastered the skills in the airplane the math and science will all start to make more sense with much less head scratching.

    0 Votes Thumb up 0 Votes Thumb down 0 Votes


Answer Question

Our sincere thanks to all who contribute constructively to this forum in answering flight training questions. If you are a flight instructor or represent a flight school / FBO offering flight instruction, you are welcome to include links to your site and related contact information as it pertains to offering local flight instruction in a specific geographic area. Additionally, direct links to FAA and related official government sources of information are welcome. However we thank you for your understanding that links to other sites or text that may be construed as explicit or implicit advertising of other business, sites, or goods/services are not permitted even if such links nominally are relevant to the question asked.