How does stall depend on angle of attack but not speed?

Question

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?

Answer 1

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.

Answer 2

It's funny you mention a Citabria, because I've actually done exactly what you're talking about 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.

Answer 3

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 static ( = 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.

Answer 4

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 stalled already.

¹_{Taking 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.}

Answer 5

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.

Answer 6

It is unfortunate that more aircraft don’t have AOA indicators. In Navy aircraft, particularly carrier aircraft, it is a primary reference of critical importance.

I was a flight instructor in the Navy, and taught the out of control flight syllabus. Literally a hundred hours “flying” a stalled aircraft. While the behavior when generating very high pitch rates where you would “blow through” the stall AOA could be odd, in general all of the behaviors associated with stall ( buffet, loss of lift) occurred at the exact same AOA. We’d demonstrate this at speeds from 250 kts (6G pull) to 50 kts (vertical climb). We’d control it in beyond stall, losing 15,000 ft while demonstrating what control over the airplane you did have post-stall. Demonstrate flying the aircraft in the vertical at well below what would be stall speed for straight and level, but the plane was still flying since the G load was zero. We’d stall it upside-down at the top of a loop.

Always the same AOA.

Answer 7

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.

Answer 8

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.