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This image displays "Pressure vectors and flow over cambered section". As far as I understood fluid dynamics, the static pressure is lower in areas where the fluid flows at a higher speed. When looking at a cambered airfoil, the lower pressure above and the higher pressure under the foil cause an upwards force, i.e., lift. Why are the pressure vectors above the foil larger than those under it? Is it maybe not the static pressure that's displayed but the dynamic pressure?

enter image description here

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  • $\begingroup$ Doesn't this depend strongly on the angle of attack? Intuitively at a high angle of attack the pressure under the wing should be higher. $\endgroup$ Commented Oct 17, 2022 at 19:50
  • $\begingroup$ Obviously, this image shows a low to 0° AOT. Cambered airfoils can create lift at 0° AOT. I guess the question is: Is the depicted pressure force distribution the one for dynamic or static pressure? $\endgroup$
    – Mino
    Commented Oct 17, 2022 at 20:16
  • $\begingroup$ I never liked the terminology to begin with. Pressure is a measurable quantity. It is different when the wing moves from when the wing is stationary relative to the air. Static pressure would be the measured pressure on a stationary wing and dynamic pressure the measured pressure on a moving wing. Since pressure can never be negative, the only thing that can possibly be shown here is the pressure difference between the two, but that, of course, is not an actual physical pressure. Pressure is also not a vector but a scalar, so at most they are showing the pressure force. $\endgroup$ Commented Oct 17, 2022 at 20:30
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    $\begingroup$ First of all, pressure is not a vector physical quantity. Then, try to have a look at my answer below $\endgroup$
    – basics
    Commented Oct 17, 2022 at 22:42
  • $\begingroup$ Notice - the vectors have points on the ends. That's the direction they pull or push. $\endgroup$ Commented Nov 11, 2022 at 15:06

2 Answers 2

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In plots like this, vectors usually represent $-(P(\mathbf{x}_s) - P_{atm}) \mathbf{\hat{n}}(\mathbf{x}_s)$.

Although these vectors has no direct physical meaning, since the stress acting on the airfoil is $-P(\mathbf{x}_s)\mathbf{\hat{n}}(\mathbf{x}_s)$, we can integrate them to get the aerodynamic force acting on the airfoil

$-\displaystyle \oint_S(P(\mathbf{x}_s) - P_{atm}) \mathbf{\hat{n}}(\mathbf{x}_s) = - \oint_S P(\mathbf{x}_s) \mathbf{\hat{n}}(\mathbf{x}_s) = \mathbf{F}^{aero}$,

since $\oint_S P_{atm}\mathbf{\hat{n}}(\mathbf{x}_s) =0$.

Once you master this representation, it's very intuitive since you immediately see the regions of the airfoil where:

  • the pressure is much lower than the atmospheric pressure where suction occurs (typically the regions close to the leading edge on the upper sides) and regions; arrows pointing outwards of the airfoil;
  • the pressure is higher than the atmospheric pressure (typically the small regions around the stagnation point; in your plot on the whole lower side of the airfoil, either because of large angle of incidence or because the plot is very qualitative); arrows pointing inwards.

Note 1. Pressure always pushes on a surface, so suction is referred to the ambient pressure or neighboring regions with higher pressure.

Note 2. You can write the pressure difference as a function of the magnitude of the free-stream velocity $U_\infty$ and local velocity $U(\mathbf{x}_s)$, by means of Bernoulli's theorems. As an example, for a steady, incompressible, irrotational flow you can write

$P(\mathbf{x}_s) - P_\infty = \dfrac{1}{2} \rho U_\infty^2 - \dfrac{1}{2} \rho U^2(\mathbf{x}_s)$

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  • $\begingroup$ I initially guessed −(P(xs)−Patm)n^(xs) , too, but if you look closely then you will notice that the vectors are not even pointing in the surface normal direction. Unless that is just an inaccuracy of the plot, could it be that they also represent the shear force parallel to the surface? Even that doesn't seem to make sense at the front of the wing, though. What am I missing? $\endgroup$ Commented Oct 18, 2022 at 3:34
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    $\begingroup$ I guess low-quality plot $\endgroup$
    – basics
    Commented Oct 18, 2022 at 6:02
  • $\begingroup$ Fair enough. Thanks. $\endgroup$ Commented Oct 18, 2022 at 7:31
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    $\begingroup$ Nice. If you're fone with the answer, please accept it, so that it's solved. A minor detail about what "low-quality" may mean: the picture looks stretched horizontally, s.t. angles are not conserved. If you find the right ratio of the original picture, the original picture is likely to be ok, with vectors orthogonal to the surface $\endgroup$
    – basics
    Commented Oct 18, 2022 at 7:51
  • $\begingroup$ I think that the pressure in the figure is referring to gauge pressure, not absolute. The gauge pressure can certainly be negative. Secondly, for an Euler fluid (non-viscous), the pressure force per unit area is oriented normal to the surface, and can be represented as a vector. $\endgroup$ Commented Oct 18, 2022 at 11:28
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Turns out, this is in fact a display of the pressure COEFFICIENT, so the c_p value and not static or dynamic pressure. It's given by:

enter image description here

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  • $\begingroup$ That is still a scalar and not a vector. From a physical point of view the diagram, as shown, is a horrible way of presenting data. $\endgroup$ Commented Oct 18, 2022 at 20:04

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