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enter image description here
(wikimedia.org) A typical curve showing section lift coefficient versus angle of attack for a cambered airfoil.

Based on the graph above, do negative angle-of-attack values create lift? How?

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    $\begingroup$ It's not really clear what you are asking. Is your first question just asking how to read the graph? I'm wondering because the graph clearly shows that there's lift generated with angles of attack between (just over) -5° and +25°. Thus, your first question appears to be about just reading the graph, but your second question, "how?", implies that you are able to read the graph and want an explanation as to why the lift is produced at a negative angle of attack (thus, implying that you don't actually need an answer to your first question). $\endgroup$ – Makyen May 12 '18 at 17:33
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    $\begingroup$ @Makyen His first question asks whether the graph is correct or is an error. $\endgroup$ – Pertinax May 12 '18 at 19:47
  • $\begingroup$ @Pertinax I guess it's possible that's what was intended, but not how I read the question. If it was what was intended, wouldn't it be asked something like "The above graph shows that the airfoil produces lift at some small negative angles-of-attack. Can that really happen? If so, how?" As I read the question, it's asking for people to read the graph and state what the graph says wrt. negative angels-of-attack creating lift. Then, separately, how that happens. $\endgroup$ – Makyen May 12 '18 at 20:08
  • $\begingroup$ You might be interested in my question Can a small GA plane maintain level flight with a significant nose-down attitude? $\endgroup$ – a CVn May 12 '18 at 21:45
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A symmetric airfoil will generate no lift at no angle of attack, and negative lift at a negative angle of attack. However, cambered airfoils are curved such that they will generate lift at small negative angles of attack.

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    $\begingroup$ Actually cambered airfoils generate lift at many angles of attack. At 0 AoA the graph shows how much lift comes only from the camber. It also shows how much negative AoA is needed to overcome the lift from the camber. At -5 AoA the two lift forces are in balance, according to this graph. $\endgroup$ – candied_orange May 12 '18 at 20:16
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The graphic above only shows the positive $C_L$ portion of the complete curve. The line (of course!) continues below the zero $C_L$ line.

Also, the $C_L$ curve for any airfoil, symmetrical or not, must cross the $C_L$ = 0 line at some angle of attack. Whether the airfoil is symmetrical or not simply determines what that zero $C_L$ intersection AOA will be. In fact, for asymmetrical airfoils, it is only a matter of convention as to how to define Angle of Attack. Yes, it is the angle that the Airfoil makes with the relative wind (or the flight path through the air), but how the "Airfoil" is defined can vary. In some cases it is defined as the longest chord line through the airfoil from leading to trailing edge, in some cases it is defined by the bottom of the airfoil, etc.

As to the question "How?", all Aerodynamic forces are generated by the pressure of the air pushing on the surface of the airfoil. At each point on the surface of anything in a fluid, the fluid pushes on the surface, normal (perpendicular) to the surface, with whatever pressure exists at that point. What we call Lift is just an abstraction we create to help visualize and do aerodynamics calculations. It is the component of the sum of all those tiny forces added up (actually, integrated vectorially) which is perpendicular to the flight path of the aircraft. Lift is created because when you incline any airfoil, in any direction, the actual normal pressure at each point on the surface changes*, all over the surface of the airfoil, and the total vector sum of all the forces (on one side versus the other) is no longer balanced.

  • Why does the pressure change from one point to another? When subsonic, fluid flows are incompressible. That means that the density, (and therefore the total pressure they exert), must remain constant. So when a subsonic fluid flows, relative to a surface, since total pressure is constant, (that's the basis of the Bernoulli principle! - See Incompresible flow) and the pressure parallel to the surface (parallel to the local flow), increases, (It must since it's moving!), the pressure normal or perpindicular to the flow decreases correspondingly, to keep the total the same. And it's the normal pressure that's pushing on the surface. Basically, there is only so much energy, and it must be divided, (vectorially) between the parallel flow (dynamic) pressure and the normal pressure.
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  • $\begingroup$ The first couple of paragraphs look perfect to me, but the part at the end where the "parallel pressure" must increase and therefore the "perpendicular pressure" must decrease (to keep the total pressure constant) ... perhaps you should think about that a little more. Or talk with someone who knows something about fluid dynamics. I don't think that part is correct. $\endgroup$ – David K May 13 '18 at 2:22
  • $\begingroup$ Unfortunately, when most folks discuss Bernoulli's principal, and use the word pressure, they are really referring only to the normal pressure or pressure measured across the flow. It is obvious, that if you mount a static port at the nose of the aircraft, pointed forward, it will read higher! It is now a pitot tube, and measures dynamic pressure, not static. When I use the phrase total pressure, I mean the vector sum of the static (or normal) pressure, and the dynamic pressure. This is also referred to as stagnation pressure, and is what must remain constant in incompressible flow. $\endgroup$ – Charles Bretana May 13 '18 at 19:21
  • $\begingroup$ Perhaps you should be more precise in your language, for example say dynamic pressure when you mean dynamic pressure, and stagnation pressure when you mean stagnation pressure. For many people, those are also explained by Bernoulli's principle, but if you call them out by those names at least people will know what you mean. $\endgroup$ – David K May 14 '18 at 2:17
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    $\begingroup$ Air in gas phase is always compressible. The compressibility effects on fluid flow are negligible at low subsonic speeds (Mach<0.3), but saying that the fluid is incompressible is incorrect and confusing. $\endgroup$ – hrobeers Aug 13 at 15:16
  • $\begingroup$ Sorry, Wrong. Subsonic flow is not compressible. If you try to compress it from one side, the gas on the other side just moves out of the way. Indeed, this is the main distinction between subsonic and supersonic flow. The only way to compress subsonic gas is to constrain it in a closed container, which is of course a completely different situation. $\endgroup$ – Charles Bretana Aug 13 at 20:53

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