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Stall Speed Increase during Bank Q

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Well-known member
Nov 25, 2001
OK, gang. I admit it...I'm not worthy to be in this forum. I have too many "basic" questions ruminating around my cranium. (For example, how fast does gravity travel? How do traffic lights know when I am approaching? Why does Bobbysamd continue to support the Broncos?)

The big aviation question I am left with is this: Why does the stall speed of a wing/aircraft increase during a banked turn? We all know it does, but I'm not confident I can explain the physics of the situation.

First, an assumption: I surmise that this increase in stall speed is only valid for level flight conditions, so I rule out descending banked turns as part of the "scenario."

Next, the lift component issue. To continue level flight, the overall lift from the wing must be increased to compensate for that loss of lift component due to the bank angle of the aircraft. That vector analysis alone, however, does not address why the stall speed increases--it only says that one would reach the critical angle of attack in a bank. The aircraft does not care what it's orientation is--that is why any aircraft can stall in any attitude.

My guess is that the answer lies in the real life application of the total lift equation, e.g. part of the l = rho v2 gonkulation. Is load factor a player in this matter?

Can anyone educate me in simple, fighter pilot terms? Be sure to use your hands.

Your first assumption is correct, it's all based on level flight, and your discussion of the lift vector is also right on.

In order to maintain level flight we need the vertical component of the lift vector to be the same in all cases, regardless of the airplane's exact angle of bank or attitude. When the aircraft is in a bank, achieving the same vertical component of lift requires a greater total lift vector. To increase the total lift we require an increase in angle of attack. So if we imagine two aircraft, in level flight, one wings-level and the other turning, at the same airspeed the turning aircraft will be carrying a higher angle of attack than the wings-level aircraft.

As such, the turning aircraft has less of a margin between it's present AOA and it's critical AOA. That critical AOA signifies the wing working it's hardest. If both aircraft began decellerating, both would increase their AOA to maintain altitude. The turning aircraft hits its critcal AOA first (though still above book Vs) and stalls, while the wings level aircraft can slow down a few more knots before it too reaches critical AOA (at Vs) and departs.

So, the short version: Bank angle makes the wing work harder to hold up the same amount of airplane, thus that wing will run out of oomph sooner (i.e., faster) than if it were wings level.

Clear as mud? And I'm supposed to be able to convey this to primary students...
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Bank Angle and Broncos

Nice explanation, Cardinal, especially the portion about increasing AOA. The FTH has a good diagram of the vectors. Wolfgang Langwische, in "Stick and Rudder," explained it well in his book.

Eagleflip, I've been a Broncos fan since 1960. While I realize this is not the Pro Football Message Board, suffice it to say they ran into bad luck with McCaffrey getting hurt the first game, Terrell being in and out of the lineup, other key injuries, and assorted shenanigans, e.g. the Eddie Kennison business. They'll be back.
Fighter pilot Terms..


If you pull hard on the stick in any attitude to the point that the airplane starts to shake..Why is it shaking?

The wing is designed to operate effeciently within a certain load range..That load range is based on breaking the plane limits and desired performance..

As you go out to the edge of the load limit based on performance..And you cant change the shape of the wing..The shaking is caused by less than effecient airflow..The disturbed airflow due to high load will cause the wing to quit working sooner than it would otherwise with smoother airflow..

Remember best cornering speed?

Plowing versus cutting..

I can do this in Rotor Head terms as well..But it will cost you..

Thanks for the explanation, all.

Now, Bobbysamd...You've got to remember that I don't actually follow pro sports at all...I figured of all people to throw a softball at, you'd be one of the few to positively flame on me...

So...is there a general rule of thumb for determining the stall speed in any given bank? I'd think that it would be somewhat related to the cosine of the bank angle, but the cosine of 60 degrees (for example) is .50...I don't think stall speed increases by half at that bank angle, so that's not the answer.

I somewhat remember a Peter Garrison article in Flying Mag. on the subject--anyone else remember what he said?

Stall Speed and Denver Donkos

Somehow, I recall that you multiply the stall speed by load factor to determine stall speed in a bank. I may be wrong about that. Anyway, as bank increases load factor increases.

I guess I swung and missed on your softball. I always liked football better than baseball anyway. I didn't think I was flaming, just trying to rationalize a disappointing season. :( :)

Fly safe, and try some stalls in medium to steep turns accelerated stalls next time you're up and observe the airspeed. Go Broncs!
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There is a chart plotting bank angle against load factor against stall speed increase. I think its in the old Flight Training Handbook, Ch 17, but I don't have one and they're getting hard to find. The Gleim CFI book has it reprinted.

You are on the right track, load factor is the inverse cosine of the bank angle. ( 1/Cos60 ) Modest bank angles, as you well know, generate very small increase in load factor, but at 60 degrees it starts to get serious

45 degrees = 1.4 Gs, 1.20 x Vs
60 degrees = 2.0 Gs. 1.40 x Vs
75 degrees = 4.0 Gs, 2.00 x Vs

Much beyond that and the the values go off the chart. I think Garrison has written on this subject several times, but not very recently. He is an aerodynamic and literary god among men, IMHO.
Another one...

Here's a another stall speed/load factor related question, one of my favorites for students working on CFI. Here is the scenario, I take a regular C-172 or PA-28, go up and fly the aircraft so that the airspeed falls below the white arc. The stall warn didn't sound, and I didn't stall, no buffet or drop. How did I do it?
A Couple Ways

1. You are operating at less than max gross weight since the marked stall speeds are for max gross. (less weight = less AOA for a given airspeed at a constant altitude)

2. You forgot the pitot cover :)


I was thinking at first it was a CAS problem, but I'll vote for slip. After all, in a slip, the relative wind isn't packing the pitot properly but kind of blowing across it.

PS - Pitot cover is good. :D
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Slip is a good answer, however, who wants to try a slip below Vso. Better be good on that spin recovery. :)

Ok, here's a question. What happens to the aircraft's stall speed as the load factor goes less than 1?
Vs would decrease. The airplane wieghs less, and thus needs less AOA to keep it's weight airborne at a given speed which translates to a lower Vs. I'm not sure if that's physically accurate, but it works in my head.

Now, how does altitude affect Vs?
Just wanted to bring this back to the top because I'm kind of curious of what he has to say his answer is.....