Does stall angle of attack in inverted flight change due to the camber of an asymmetric airfoil?

What is the stall angle in an inverted flight configuration? How does it relate to the stall angle in normal flight?

As visible in the image below, in upright flight the lower pressure is on the extrados of the wing. In inverted flight the lower pressure is on the intrados.

(source)

The air stream separation at the stall will occur on a side with different characteristics. One could expect the stall will happen differently.

Equivalent representation, where the gravity is inverted and the wing orientation remains in the same direction. Flying inverted implies flying at a negative angle of attack.

The values taken into account are the airflow direction and the chord line, reflected in the angle of attack value.

As commented, while the two previous images suggest a wing in horizontal flight, the situation can be extrapolated to any stable linear trajectory with a negative AoA.

The horizontal plan or the pitch angle are not necessary to determine the stall angle (though they influence the stall speed).

• Think of it this way. The wing doesn't know it's upside down. All it knows it that the airflow is of a certain speed, from a certain angle. The diagrams you've shown are for different pitch angles. – Simon Sep 17 '15 at 19:11
• @Federico No it's not. Please show me in the lift equations where wing orientation is taken into account. The wing behaves identically for a given speed and a given AoA. – Simon Sep 17 '15 at 19:37
• @Federico Nope, you are confusing pitch with AoA. Aircraft flying straight and level. AoA x degrees. It will generate a force upwards of y newtons. Now turn it upside down and keep the AoA the same. It will now generate a force of y newtons downwards. The force does not change. In order to maintain straight and level, you will need to adjust pitch, such that the A0A is adjusted and lift again = gravity. – Simon Sep 17 '15 at 20:02
• @Federico In the second picture, the pitch is different, and therefore the AoA. It is not a point for debate that for a given AoA and speed, the wing behaves in exactly the same way. I don't get why you are arguing that this is not true. The wing does not know it's upside down. – Simon Sep 17 '15 at 20:07
• – Peter Kämpf Sep 17 '15 at 21:09

3 Answers

Short answer: Asymmetric airfoils have different positive and negative stall angles, the largest absolute value of the two depends on factors like nose shape and camber. With positive camber (normal and utility aircraft), the negative stall angle can be the largest (in absolute values) but the maximum negative lift available before stall will be smaller than for the positive stall. Larger Reynolds numbers push the stall further away in both directions

It depends on the airfoil. With symmetric airfoils, the stall angle is the same for positive and negative stalls. Positively cambered airfoils (the sort mostly used) have their negative stall at a smaller absolute value of the lift coefficient compared to their positive stall, but the stall angle can well be at a higher absolute value.

Below you see a polar plot for a supercritical airfoil which I used for this answer. The positive stall angle of attack is 8°, while the negative one is around -10°.

Polar plot of the R2A airfoil at Mach 0.6 (own work)

The stall angle depends on details of the nose contour and the camber: Positive camber means that the zero lift angle is shifted to negative values, so there is some bias to negative values in the polar. However, if the lower part of the nose has very high curvature, it will create a high suction peak which leads to flow separation just aft of the nose already at a small negative angle of attack.

One extreme case would be the Göttingen 417a airfoil. Airfoiltools unfortunately plots only a range of Reynolds numbers suitable for model airplane enthusiasts, but the plot below should get the point across. The positive stall angle is 12° at the highest Reynolds number, while the negative stall angle is only around -8°.

Gö 417 lift over angle of attack. The lowest Reynolds number (blue line) is 50,000 and the highest is 1,000,000 (olive green line). Note that all curves are XFOIL predictions - real-world data might look differently.

• Since an AOA of 180 degrees is hitting the trailing edge, is 90 degrees from straight above and 270 degrees from straight below, or are they both 90 degrees? – Lnafziger Sep 20 '15 at 15:07
• @Lnafziger: Yes, but 90° is from straight below and 270° is from straight above. The airfoil makes a full rotation around the Y-axis (spanwise axis) for one 360° rotation in its angle of attack (or you rotate the flow direction and keep the airfoil fixed, as you like). – Peter Kämpf Sep 20 '15 at 16:28
• Okay, so what I was trying to say before is that the AOA is not the same for an upright versus an inverted airfoil (in un-disturbed air) because there is a 180'degree difference. For the same AOA, it makes no difference whether the airfoil is upright or inverted, – Lnafziger Sep 20 '15 at 18:07
• @Lnafziger: The simple formula AoA = pitch angle - flight path angle doesn't work in inverted flight. The coordinate system for pitch will not rotate, that for the airfoil's AoA will, however, so the pitch angle is nearly the same small positive value but the AoA is a small negative value in straight inverted flight. Increase pitch, and the AoA will become more negative (there is an additional cosine involved which is 1 in upright and -1 in inverted flight). – Peter Kämpf Sep 20 '15 at 18:49
• @Sean: No, the angle of attack is not relevant. Look at the lift coefficient instead: While upright the max. c$_L$ is 1.3, inverted it is only -0.75. So the inverted stall speed is 31% higher than the upright stall speed. Post-stall behavior is roughly similar, but this being an ISES calculation, the real behavior might be different. – Peter Kämpf Oct 17 '18 at 11:53

You are basically comparing a situation with positive $\alpha$ (above) with a situation with negative $\alpha$ (below).

You could obtain the same situation by pitching down: see how in the second picture your flow direction arrives from above the chord line (in the reference frame of the wing).

If your airfoil would have been symmetrical, the critical positive and negative $\alpha$ would have the same absolute value, only opposite sign, but you show a cambered airfoil.

I currently do not have my aerodynamics book at hand, but google helps us:

As you can see in the image, adding camber to an airfoil shifts its $C_{L\alpha}$ line towards negative values. This is desirable because in this way you can have lift even when the angle of attack is 0 (and with no or little drag increase). Another consequence it that the maximum positive $\alpha$ will be smaller than the uncambered case and the negative one will be even more negative (but with some limitations, the Runge-Kutta condition at the trailing edge will affect the shape of the negative $C_{L\alpha}$ curve)

• I understand the role played by the camber to maintain the AOA near 0 (and minimal drag) in cruise. What I don't understand is whether this camber will affect the stall angle in inverted flight. – mins Sep 17 '15 at 20:44
• @mins inverted flight or negative AoA is the same thing. – Federico Sep 17 '15 at 20:49
• The way to prove it would be to continue to plot through to -25 degrees AOA. The symmetrical plot would be a mirror image. The cambered would not. BTW, cambering the wing, as airliners do, vastly increases lift generated, giving airliners a speed envelope few aircraft enjoy. Figure 43.6 may be an oversimplification for comparative purposes. – Robert DiGiovanni Oct 19 '18 at 7:05

Notice inverted flight is very rare in birds, which have heavily cambered wings similar to the GO 417.

It is important to remember that inverting an asymmetric wing, in simple terms, is reversing all that is good about the lift generation "right side up". The result is a loss of lift at that speed and angle of attack. Further more, there is a very good chance the stall angle of attack of the inverted wing is lower.

In the case of a heavily undercambered wing, inversion would likely be a disaster. Wind tunnel/smoke would show heavy turbulence over the top and very little lift from pressure underneath the wing. Notice the two great virtues of thin undercambered wings have been reversed.

On the other extreme, fully symmetrical wings show little difference when inverted and are popular in aerobatic aircraft.

The flat bottom will behave better than the undercambered, but when inverted will generate less lift at a given AOA and exhibit flat plate stalling characteristics (stalling at a lower AOA).

Amazing how much interest the movie is generating. We'll have to give Denzel a symmetric wing, and Sully a good set of floats.

• "Further more, there is a very good chance the stall angle of attack of the inverted wing is lower.": The question is about the change in the stall AoA, and you don't seem to know the answer. – mins Oct 19 '18 at 10:48
• Well, what do you think? For the flat bottom (see picture) inverted, the stall angle will probably be less. Give me a wind tunnel, I'll give you proof. – Robert DiGiovanni Oct 19 '18 at 12:56
• The point is I don't see what does your imprecise post add to the very well documented selected answer from a knowledgeable expert. – mins Oct 19 '18 at 15:47
• 1. Haven't heard your "knowledgeable" answer yet. 2. If you think my answer is incorrect, please say why. 3. My answer is based on NACA short films. 4. Please don't waste comment space on personal opinions. 5. If you think wings should be flat on top and rounded on the bottom, we have wind tunnels for that. 6. As a retired researcher, my standard of "knowing" is validation by testing. As I have not tested that particular wing, I say "seems likely". If you disagree, fine. I will be happy to consider your point of view – Robert DiGiovanni Oct 19 '18 at 15:49
• "Haven't heard your "knowledgeable" answer yet": Not mine, this one. Documented with the checked values -10° vs +8° in contradiction with your assumption "the stall angle of attack of the inverted wing is lower". You could now tell me -10° is lower than +8, indeed. – mins Oct 19 '18 at 16:13