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I've seen somewhat conflicting information on when planes stall. I've seen references to "stall speed," apparently a speed below which the airplane will stall, but stalling also seems to be caused by the angle of attack. The tag wiki for says:

A stall is an aerodynamic condition wherein the angle of attack of a wing increases beyond the "critical angle of attack", causing the wing to cease generating lift. It's important to note that stalls can happen at any airspeed and in any attitude; the only cause of a stall is exceeding the critical angle of attack.

Under what circumstances do airplanes stall? How are stall speed and critical angle of attack related? Is stall speed the airspeed at which the critical angle of attack goes below 0°, causing a stall in level flight? Or is stall speed even a real thing?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Ralph J
    Dec 11, 2022 at 23:15

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Stall at subsonic speeds depends ONLY on the AoA; you can (for real) fly an aircraft at half it's stall speed without having it stalled. For that, you will have to fly in NOT MORE THAN a quarter-G flight (since you now have 1/4th the dynamic pressure). So long as the critical AoA has not been exceeded, the aircraft will not stall.

But turns out, planes must fly in a 1G flight most of the time. And if all other values are held constant (mass, configuration, CG position etc.), with engines producing minimum/negligible thrust, in a 1G flight, then the speed (IAS) at which this critical AoA will be achieved can be determined and consistently reproduced. This is the stall speed, and an aircraft physically cannot fly below this speed - in a 1G flight.

In the lift equation, hold the following three variables constant: lift, because in 1G flight, we need a constant lift for a constant weight. Lift coefficient, hold it at its maximum value (Cl-max), since this is the point beyond which stall occurs. And surface area, since we're only using one particular aerofoil. With these three constant values put in, the remaining variable - dynamic pressure, is determined; there is only one value for dynamic pressure (IAS) which now fits in this equation. This IAS is the stall speed. If an aircraft, originally flying at this speed slows down any further, the pilot will instinctively pull back on the control column to obtain a higher Cl, which they need in order for the aforementioned constant lift to be maintained. Unfortunately, the Cl is already at its maximum value, and pitching up further will only reduce the Cl, since the aircraft will have stalled.

The aircraft will physically not be able to support a 1G flight below this speed - the 1G stall speed, but that does not mean that the aircraft also cannot support a less than 1G flight.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Ralph J
    Dec 11, 2022 at 7:09
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Stall is defined as the point where the lift curve slope over angle of attack becomes flat or the lift coefficient peaks. It is caused by progressive flow separation on the suction side and happens generally when the airplane is too slow to create the desired lift. In a tight turn this happens at a higher speed than in level flight, and in a parabola flight the airplane will never slow down enough to stall.

In rare circumstances you also can stall an airplane by flying too fast.

The stall angle of attack is not constant but varies with speed, altitude, Mach number and the rate of angle of attack increase, as discussed here and here. Since the lift curve slope of a wing does not change with the pitch rate, a high pitch rate will indeed increase the stall AoA by up to 50%.

Speed and altitude effects are expressed in the Reynolds number, and this will also shift the AoA up by several degrees when increased from, say, 200,000 to 5,000,000. The influence of the Mach number will really manifest itself above Mach 0.5, but if the leading edge radius is small, it can already make a difference of several degrees between incompressible conditions and Mach 0.3. Stall at higher Mach numbers is more complex, because before lift drops, the wing will experience increasing buffeting, which by itself will limit the operational AoA.

Next, pitch angle and angle of attack are not the same, but they differ by the flight path angle and the wind angle of attack, which is nonzero if you fly in an up- or downdraft. This is discussed in detail here.

If your engine or airspeed allows, you can fly a full loop without ever stalling the airplane.

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Stall is a complex phenomenon that happen when the airflow on the upper surface of a wing cannot follow the wing's curvature anymore. At that point there is a:

  • drop in lift;
  • rise in drag;
  • and a bigger nose-down tendency.

This picture (source) nicely depicts this detachment of the airflow:

 wing stalling

Stall depends (almost) only on the angle of attack of the wing. The easiest way to visualize it is by looking at the plot of lift coefficient $C_l$, which has typically a trend like the one underlined in blue here under:

 lift coefficient

The top of that curve is the maximum lift coefficient $C_{l_{max}}$ which can be extracted from that wing before its stalling; in this case $C_{l_{max}}$ is some 1.6 reached at a AoA of some 16°. Beyond that value the wing stalls as signalised by the sudden drop of $C_l$. Note how $C_{l_{max}}$ depends on Reynolds number $R$ and surface finishing (roughness).

By definition, lift is calculated as:

$L=½ \rho V² S C_l$

where $\rho$ is air density, $S$ wing surface and $C_l$ the lift coefficient as just seen.

Now, being $C_{l_{max}}$ known, from this lift equation we can simply calculate the speed $V$ at which the wing stall:

$V_{stall}=\sqrt{ \frac{L}{½\rho S C_{l_{max}} } }$

For obvious safety reasons $V_{stall}$ should be as low as possible i.e. the denominator as high a possible. This can be achieved using a bigger wing surface $S$ and/or increasing $C_{l_{max}}$ via high-lift devices like flaps and slats. Flying high (lower density) increases stall speed (which is more dangerous).


Bonus material

As visible in the plot, $C_{l_{max}}$ depends on the Reynolds number and on the surface finishing. In particular if the the Reynolds number (i.e. speed) is smaller, $C_{l_{max}}$ becomes also smaller. A dirty wing is also going to have a lower $C_{l_{max}}$, 1.2 instead of 1.6 in our example. This implies that a jetliner in landing phase after having had a rough day is going to have a slightly lower $C_{l_{max}}$ i.e. higher stall speed.

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  • $\begingroup$ That's a good explanation. Is this V-stall a TAS? Because stall IAS doesn't change with density; also, if it is TAS, I believe that the density in the denominator should be relative density and not absolute density. $\endgroup$ Dec 9, 2022 at 0:18
  • $\begingroup$ @AdityaSharma V is airflow's speed as seen by the wing. $\endgroup$
    – sophit
    Dec 9, 2022 at 7:25
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    $\begingroup$ Just verified, the V here is the TAS. Also, the density here should be relative density (absolute density divided by ISA-MSL density). Thanks $\endgroup$ Dec 9, 2022 at 8:19
  • $\begingroup$ @AdityaSharma ok $\endgroup$
    – sophit
    Dec 9, 2022 at 8:20
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    $\begingroup$ @AdityaSharma: 🖖 $\endgroup$
    – sophit
    Jan 6, 2023 at 22:16
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stalling occurs when a wing exceeds a critical Angle of Attack

That is the stock answer, the right answer, passes the written test.

But as a pilot, operating an aircraft, reading your POH, it says stall speed.

"references to "stall speed", apparently a speed below which a plane will stall"

How is this related to Angle of Attack, and why is stalling speed higher in a turn?

the real reason for stalling is that load force is more than the amount of lift force the wing can create under a given set of conditions. The pilot exceeds the critical Angle of Attack by pulling the stick back too far in an effort to create adequate lift.

For any given aircraft, the given set of conditions are:

Lift = Load (Mass × G force) = Density × Wing Area × Coefficient × V$^2$

Defining "Stall speed" based on Load covers all bases of flight.

Stall is defined in terms of speed for many GA aircraft because other variables are more or less constant and Angle of Attack is at its lift producing maximum

In level flight, at 1 G, the stall speed is, for example: 50 knots (flaps up).

Here, we may ask what is the "coefficient of lift"? This term is based on Angle of Attack and Airfoil type. When you lower your flaps, airfoil type changes, enabling increased lift without increasing aircraft AoA$^1$ or airspeed$^2$. Now, one can go slower before stalling, for example: stall speed 45 knots (flaps down).

Now for turns: stall speed 60 degree turn.

Load factor is 2 G. Coefficient of lift, density, and wing area are constant, velocity must be greater. You can read it from your POH, or calculate it as: V$^2$ = 2 × 50$^2$

V stall = 1.41 × 50 knots or 71 knots

Now for cases where Load is less than 1 G as previously discussed in current answers.

Aside from aerobatics, and an Aviation Stack Exchange favourite: lift is less than weight in a climb , we have the emergency descent. Yes, that's right, lift is also less than weight in a descent (as a mirror image of a climb) because now the drag component (rather than a thrust component) is helping resist gravity. This means Angle of Attack will be less than that required for level flight. Emergency descent turns are also performed at a higher airspeed, which additionally makes them safer.

$^1$ lowering flaps does increase local AoA at the wing. This is one reason that flaps are placed next to the fuselage, effectively increasing the "washout" of the wing tips. This ensures the warning buffet will start near the fuselage. The net effect is a lowering of stall speed (an increase in maximum lift for a given airspeed).

$^2$ drag also increases. Throttle must be added to maintain airspeed (or aircraft must lower its pitch to the horizon).

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  • $\begingroup$ That seems right; just wanted to point out, load is not weight × G-force. Weight itself is a force, and with G-force I suppose you mean normal acceleration. Load is mass × normal acceleration. Also on that note, this might seem unrelated, but how does climbing or descending affect the rate of turn for a given bank angle? Because apparently climb/descent reduces the load factor. $\endgroup$ Dec 9, 2022 at 22:25
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    $\begingroup$ @AdityaSharma my guess is that because the required lift is less, the planes trim speed is reduced, the rate of turn is less. Also to be considered is the flight path, which is no longer a horizontal circle, but a helix. Taken to extreme, the plane increasingly rolls about the point rather than pitching and yawing about it as in a level turn. Good future question for ASE (I rarely dive or climb steeper than 10 degrees). $\endgroup$ Dec 10, 2022 at 0:01
  • $\begingroup$ Thank you very much for the input; your statement sounds quite reasonable, although regarding trim speed, yes it'll reduce and that will have an effect, but let's assume it constant, because that was the case I'm interested in. Also worth noting is that it's the horizontal component of lift that provides the required centripetal force for a turn, and in our case, that lift has reduced. I still don't understand this fully, and if this has already not been answered, I'd love to ask this question myself. Good day! $\endgroup$ Dec 10, 2022 at 0:19
  • $\begingroup$ @AdityaSharma neither do I, but thankfully it's not "the slower plane with the same lift has the greater rate of turn". And the helix path is a greater distance to travel. From experience, any type of complex manuever is best done with excess speed. We calculated 71 knots for a 60 degree turn stall (when 1 G stall was 50 knots), be we do them at 90 knots to be safe. But I will look forward to your next post, and good day also. $\endgroup$ Dec 10, 2022 at 0:58
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A stall disrupts the laminar flow of wind across the top wing surface, which is what's producing lift.

Because of that I have my doubts that it is ONLY dependent on AoA. In fact, I'm pretty sure you could induce a stall without changing the AoA at all using a turbulator on the leading edge (a la original KasperWing) to create eddies across the top of the wing.

Change my mind. :-)

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Ralph J
    Dec 9, 2022 at 14:56
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    $\begingroup$ "A stall disrupts the laminar flow". Being the flow laminar is not necessary. Indeed if the boundary layer is turbolent it might even delay stall. $\endgroup$
    – sophit
    Dec 9, 2022 at 15:24
  • $\begingroup$ Oh sure, you could establish a tight boundary layer that was chaotic. But that would only establish the laminar flow above the boundary layer, right? Also, a high speed stall doesn't involve the AoA only. In fact, it isn't the AoA that causes the stall. It's the rapid change of the AoA that triggers the turbulence over the top of the wing. So, I am still not convinced that "only AoA" causes a stall. I think that's roughly correct but not quite precise. $\endgroup$
    – rakehell
    Dec 9, 2022 at 22:11
  • $\begingroup$ I think the truth is a stall doesn't have to be triggered by a high angle of attack. It's the laminar air flow detachment that causes it. So anything that detaches the airflow can cause a stall. $\endgroup$
    – rakehell
    Dec 10, 2022 at 20:37
  • $\begingroup$ Ralph, another comment has 5 replies. How come you moved mine to chat after only one reply? $\endgroup$
    – rakehell
    Dec 10, 2022 at 20:38

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