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  • If the pressure goes always from high to low, we should have a compression, due the lower pressure of the air stream over the upper surface, and an expansion in the airstream in the lower surface.

  • So What causes the "suction" on the upper surface of an airfoil and the "pushing" in the lower one?

  • I undestand why there is a region of high pressure and a low pressure around the airfoil. But I am a little bit confused about the effects of the external pressure (Free stream) on the airfoil stream.

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    $\begingroup$ Welcome to aviation.SE! We have a lot of questions on this site about lift. This one and this one might be helpful places to start (and check the related questions list for each of them). If those don't help then maybe you could add some more detail to your question about what you find confusing? $\endgroup$
    – Pondlife
    Jan 4 at 17:59
  • $\begingroup$ Lift is not just about suction and pushing. Thats just a VERY simplified way of explaining to laymen how lift works. A good place to start to get an idea of some of the mechanism involved is here: youtube.com/watch?v=PF22LM8AbII $\endgroup$
    – Invariant
    Jan 18 at 15:24
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The key to the lift generation is the curvature of the streamlines. Forget the popular, but unfortunately wrong explanations based on the non-equidistant particle paths.

The simplest example will be if you place a plate with an angle of attack in the subsonic flow: the streamlines will get curved on the both plate sides as the flow turns. Therefore, even a flat plate will generate a lift once its angle of attack is different from zero. As the streamlines are turning on the upper surface, the pressure will get reduced from the free-stream to the airfoil surface. In other words, moving away from the airfoil, the pressure will increase - in the direction of the radius of curvature of the streamlines. Important to note is, that this pressure drop is normal to the streamlines. This change in pressure is simply an effect of turning the flow, i.e. changing its direction. Mathematically, the pressure gradient normal to the streamline, dp/dn, can be expressed as $$ \frac{dp}{dn} = \rho \times \frac{V^2}{R} , $$ where R is the radius of curvature the streamline. Derivation of this equation can be obtained if writing the momentum equations in the natural coordinates.

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  • $\begingroup$ Would you mind to explain the symbols in your equation? That will certainly make it easier to understand. Otherwise a fine answer. +1 $\endgroup$ Jan 18 at 21:29
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I’m not sure where to pitch this but I’ll have a go... There are a few ways to look at it - perhaps simplest is that the top surface is longer the an the straight-line distance across the chord and so the airflow must speed up as it is effectively stretched. If the air were stationary then the surrounding air would move in to equalise the pressure, but it’s a dynamic situation and so the surrounding air will accelerate inwards and so much of the pressure differential remains.

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    $\begingroup$ Sounds like a classic "equal transit time fallacy" to me. The bottom of the wing also has a longer than straight chord line distance. $\endgroup$
    – Sanchises
    Jan 18 at 9:29
  • $\begingroup$ True, for an undercambered wing it’s useful to consider the acceleration of the air as it moves downward to follow the surfaces of the wing. Still, the notion holds that the surrounding air takes time to accelerate in to equalise pressure and so this only works because it’s a dynamic situation. $\endgroup$
    – Frog
    Jan 18 at 18:46

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