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Everyone says that the angle of attack is what determines a stall, not the speed. I understand the theory and understand that it is separation of the airflow that matters for stalling.

However, I don’t understand in a practical sense. Let’s say you’re in a Citabria going at 100 knots. If you pull up extremely fast, you can get a high angle of attack, beyond what you’d need to stall at 60 knots, yet you wouldn’t stall straight away. If you stayed at that angle of attack, you’d quickly slow, then stall. But if I’m right that you wouldn’t stall straight away, then it seems like the angle of attack is not the only thing that matters.

What am I missing? What’s wrong in my argument?

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If only the angle of attack would have been what was determining stall, I don't think the Apollo program would have been especially successful :) – Speldosa Mar 31 '14 at 0:32
@Speldosa AoA is the only thing which determines a stall. Though I'm not sure how that relates to the Apollo program. – Bret Copeland Mar 31 '14 at 0:46
@Speldosa: pitch angle is not the same as the angle of attack, I don't think. – Qantas 94 Heavy Mar 31 '14 at 0:51
This would be a wonderful question to answer with an animated gif. – Steve V. Mar 31 '14 at 3:01
Why did Apollo need any lift at all, exactly? – cpast Apr 9 '14 at 4:13
up vote 27 down vote accepted

I believe you are confusing the wing angle of attack with the pitch of the aircraft. Aircraft moving at a slow, near-stall speed, despite pointing the nose up, will still be traveling more or less horizontally. Their VSI instrument will read near zero. Whereas, if you take an aircraft moving quickly and pull the nose up to the same angle, the aircraft will, obviously, climb rapidly.

Why does this matter? The angle of attack is defined based on the wing's motion through the relative wind. The wing's orientation relative to the ground isn't involved in the definition in any way. When the aircraft as a whole is climbing, the relative wind is coming down from above. As a result the angle of attack is reduced, compared to what it would be if the plane were not climbing.

Just to show some quick numbers, suppose you took an aircraft moving at 100 kts in still air and pulled the nose up so that you are now climbing at 3,000 FPM (most aircraft will lose speed doing this, but the math is valid until the airplane slows down). $1knot\approx100FPM$, so you'll now have an upward vector of 30 knots. Your 100 kt airspeed is now moving up at an angle. A little trigonometry:

$$\sin(x)=\frac{30}{100}$$ $$x=17.46°$$

So, your angle of attack is 17.46 degrees farther away from stalling when climbing at 3000FPM than it would be if your aircraft had the same pitch but was in level flight.

However, few aircraft have the engine power to sustain a climb at this rate. The aircraft will bleed off speed, and as the speed bleeds off, the aircraft will slow, the climb rate will decrease, the aircraft's velocity will become closer to horizontal, and, eventually, the aircraft will stall if the pitch is held constant.

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When you say "angle of attack is 17.46 degrees farther away from stalling" I think that is a bit confusing. You could clarify that that aircraft is now pitched upwards 17.46 degrees, but the angle of attack does not increase by that much. In fact, AOA would decrease, right? – fooot Mar 31 '14 at 16:04
super helpful answer, thanks a lot! clarified my understanding and makes sense. – Peter Mar 31 '14 at 17:28
I agree with @fooot, your explanation itself seems to confuse AoA and pitch in the fourth paragraph. I think what you're trying to say is that pitch has increased, but angle of attack has not, and therefore the pitch angle which can be achieved without stalling has also increased. You should also point out that, in your example, all of the climbing force is generated by increased thrust, whereas most aircraft climb using a combination of thrust and lift, and additional lift requires a higher angle of attack, or a higher airspeed (those are the two ways to generate more lift). – Bret Copeland Mar 31 '14 at 17:28
Unfortunately, my first comment got mangled. What I meant to say in it was this: The explanation is right but it's possible that it could be phrased differently. I wanted to contrast the climb scenario with the level slow flight scenario by showing that, for a given pitch, AoA is reduced in a climb. – fluffysheap Mar 31 '14 at 19:15
Finally, lift is actually not increased in a climb. It increases only momentarily, to enter the climb, but in a sustained climb, it's the same. If you climb without changing speed, the only effect changing your AOA will be the slight influence of the lift vector no longer pointing straight up, but angled slightly aft; but part of the thrust vector will also be pointing up, so the magnitude (or even direction!) of this change will depend on the exact aircraft and parameters of the climb. A handy way to think of it: airspeed * AoA = G-force. – fluffysheap Mar 31 '14 at 19:20

It's funny you mention a Citabria, because I've actually done exactly what you're talking about that in exactly that airplane. Not that it really matters, because this will apply in any airplane.

In your question, you said that you understand angle of attack is what causes the stall. But I'm not sure that you understand that given the same wing, it's always the same angle. I say that because of this:

you can get a high angle of attack, beyond what you’d need to stall at 60 knots,

The angle of attack you need to stall remains the same, regardless of speed. Perhaps things are different in the supersonic realm, but this is good enough for Citabrias.

You're right that if you were cruising at 100kts and suddenly pulled back on the stick, you would slow down before stalling. But that's not what causes the stall. The stall is caused by high angle of attack, and that's caused by elevator position.

Stick position is the single best predictor of when an airplane will stall, and nobody talks much about it. I can also say your example isn't 100% accurate, because I've actually done it. If you cruise at 100kts, then slam the stick back as hard as you can, you'll stall with a minimal loss of speed beforehand. And if you wanted, you could have a higher entry speed than 100kts and stall at 100kts. Eventually, you get into structural issues caused by the excessive g-loading.

Stalling isn't just caused by angle of attack, it's always caused by the same angle of attack. I hope this answers your question.

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Well, for the example correct answer is the part where you say "If you cruise at 100kts, then slam the stick back as hard as you can, you'll stall with a minimal loss of speed beforehand.". – Jan Hudec Mar 31 '14 at 7:15
If I take the 60 knots in the question, stall loading at 100 knots would be just 2.67G. But internet search gives me just 44 knots and that gives me 5.17G at 100 knots, or just a little over the structural limit. You'll probably bleed off the 2 knots to get within limit in the entry to the manoeuvre. – Jan Hudec Mar 31 '14 at 7:19
Yeah, the 44kts for Vs is correct. It's also important to remember that changes dramatically with weight. And the Citabria 7ECA has at least two gross weights, depending on the wing struts. – Jungroth Mar 31 '14 at 7:59
Consider the zoom maneuver. At any point in the zoom (easier as the aircraft slows), merely centering the stick will stall the airplane. – rbp Nov 27 '15 at 1:05

But if I’m right that you wouldn’t stall straight away

You will stall right away. You won't pitch down immediately though.

Immediately when you exceed 2.67G1, the aircraft will start to buffet and kick back somewhat as pulling more on the yoke no longer causes increase in lift and the pitch up rate and acceleration will stop increasing. But the pitch will not stop increasing. The wings are still generating some lift, just less than before stall. So you'll continue climbing until you run out of kinetic energy (which you will faster than usual because drag is increased in stall) and decelerate below the speed at which the stalled wings can't produce enough lift to balance the weight. At that point your speed will still be higher than the 60 knots, because at 60 knots the wings can balance the weight when not stalled, but in this case they are staled already.

1Taking 100 kts cruise and 60 kts vs. Internet search gives me just 44 knots for vs and that would mean 5.17G for stall at 100 knots while the certified limit is 5G, so you shouldn't be doing it at 100 knots, only up to 98.

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The foot note and the "practical sense" of the original question reminds me of "weather penetration speed": A certain speed at which the aircraft will stall before reaching it's structural limits, should it be hit by excessive air loads. – radarbob Apr 2 '14 at 2:46
@radarbob: I don't think that would be the reason. The main reason for "weather penetration speed" is to have sufficient margin from Vne and Mne. Especially exceeding the later is dangerous, because exceeding Mne also causes loss of lift. – Jan Hudec Apr 2 '14 at 4:51

Imagine you throw an airplane with the bottoms of the wings facing forward at 500 knots. The aircraft is going very fast, but rest assured it is stalled. It could recover very, very easily because it has so much kinetic energy and there as a huge amount of airflow, but it is not producing lift the way itshould.

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Well, recovery in this situation might be somewhat complicated by the disintegration of the airframe. :) – reirab Dec 18 '14 at 21:07

The stall Angle of Attack (AoA) is not fixed, but increases with pitch rate and - to a lesser extent - with the Reynolds Number.

When a wing stalls, the boundary layer in the rear part of a wing stops and even reverses it's flow direction, causing separation. For the outer air flow, this looks as if the wing has grown thicker there and has a smaller AoA than before, without separation. This causes the lift loss of a stalled wing. The "history" of the local boundary layer influences this - if it has seen a high acceleration around the nose of the airfoil, it has to perform a steep deceleration over the remainder of the wing. Friction has already reduced the energy of this boundary layer, and the steep deceleration ends in separation further downstream.

If the stall AoA is approached rapidly, the boundary layer on the rear wing has still the characteristics which go with the low AoA that prevailed when that parcel of air flowed around the wing's nose. Therefore, it has more energy left and is less prone to separation. The effect is an increase in the stall AoA with pitch rate, to a point where the total lift of the wing is 50% more than that at the stationary AoA at the same speed. Of course, this is a dynamic stall with a load factor much higher than 1. For more details, I refer you to NACA TN 2525 from 1951. No price for guessing which airplane was used.

On the flip side, the lift drops much more than in a stationary ( = slow pitch rate) stall. A docile stall behavior can now become abrupt! Another consequence of this lift overshoot is the possibility of a hysteresis loop, especially in helicopter, propeller and turbine blades where strong and cyclic changes in AoA are possible. This is called lift flutter and causes high mechanical stresses and vibration. See Sighard Hörner's "Fluid Dynamic Lift", page 4-24 and 25 for more.

The Reynolds number effect is less pronounced, but still gives an increase in stall c$_{l max}$ of 15 - 25% between $Re = 10^6$ and $Re = 5\cdot 10^6$. Details depend on the particular airfoil. Abbott-Doenhoff or the Wortmann catalog have lots of data on this.

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Please bear in mind that by using a ⋅ (\cdot) to imply multiplication can be confusing to a lot of people, since it looks identical to a decimal point. \times is probably a better bet. See also this, which corroborates the ambiguity: When should \cdot be used to indicate multiplication? – Danny Beckett Apr 28 '14 at 23:51
+1 for the talk about pitch rate influence. Is there some relation in pitch rate vs speed where this effect becomes noticeable? – Radu094 Apr 29 '14 at 4:20
@Radu094: Yes, the pitch rate needs to go up linearly with air speed to keep the effect constant. NACA TN 2525 gives the parameter $\frac{c}{v}\cdot \frac{d\alpha}{dt}$ to calculate the maximum lift coefficient (c = wing chord). Values of that parameter up to 0.66 were tested, and the maximum lift coefficient increased linearly over the whole range. For the stationary case the parameter was generally less than 0.05. – Peter Kämpf Apr 29 '14 at 8:02
Peter, there's no need to be so infuriated. Of course I read the link; there are answers, but there are also 2 or 3 comments/paragraphs in answers about how it is ambiguous. I'd've pinged you in chat instead, but you haven't visited it. This was more of a polite recommendation, not me trying to force something upon you. I was merely trying to help you improve clarity. – Danny Beckett Apr 29 '14 at 9:15

To understand this, I like to think of a jet with afterburner pointed nearly vertical. It's not stalled, the angle of attack is nearly 0, because the relative wind is going nearly straight down due to all the thrust. But "straight down" is still going smoothly across the wings, no stall.

Similarly, a Cessna 172 going 10 kts ground speed, pitched level to the horizon, is going to be decending far faster than the 10 kts of ground speed, so the relative wind is hitting the bottom of the wing. Just another way of describing a high angle of attack.

This may only make sense in my head, but it works for me.

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You wouldn't necessarily stall if you applied abrupt elevator at, say, 100kts due to there being enough inertia to ensure that the airflow relative to the chord line (AoA) did not, infact, exceed the critical angle.

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