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An airfoil section in a wind tunnel has many static taps on its upper and lower surface. These static ports can only read static pressure which acts perpendicularly to the local airfoil surface.

In place A are static ports that read a pressure value of relative -100 Pa . This pressure acts perpendicularly to the airfoil surface, so it is not perpendicular x co-ordinate. Do we need to convert only this vertical component when we put this value in a diagram when we draw the pressure distribution? The pressure in place A has a vertical component (lift) and also a horizontal component (thrust) .... So how would the integral "know" in which direction the pressure acts?

Can you please explain with an example from start (pressure measurement) to end (calculated forces) how this procedure looks like? (Are the experimentally measured pressures at static ports given in relative or absolute pressure?)

enter image description here

Angle theta is different for every place at airfoil surface, so it cant be one theta from leading edge to trailing edge. Where is here function that describe airfoil top and bottom contours? That function will give us local surface inclination

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  • $\begingroup$ The answer to your next question should cover this nicely. $\endgroup$ Dec 11, 2021 at 16:17
  • $\begingroup$ @PeterKämpf Now I understand this simple task where pressures act perpendicularly to x co-ordinate. I have trouble to understand "real pressure" that acts perpendicularly to airfoil conture, because this pressure is at angle with respect to x co-ordinate. $\endgroup$
    – Jurgen M
    Dec 11, 2021 at 16:34

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You're correct, pressure is a scalar entity, while force is a vector and has both a magnitude and a direction. The pressure ports are a measure of total pressure magnitude only, direction information is not contained in them which is the basis of the question.

For directional information, we need to include:

  • the geometry of the wing profile that was measured;
  • the Angle of Attack during the measurement.

Pressure acting upon a surface creates a force, which is always perpendicular to the surface, also mentioned in the second linked youtube video. So all we need to know is the inclination of the local area at the pressure port. Which we do know, because the profile shape is defined, for instance the NACA 4-digit series (from wikipedia):

enter image description here

Equation for a cambered 4-digit NACA airfoil[edit]

Plot of a NACA 2412 foil. The camber line is shown in red, and the thickness – or the symmetrical airfoil 0012 – is shown in purple. The simplest asymmetric foils are the NACA 4-digit series foils, which use the same formula as that used to generate the 00xx symmetric foils, but with the line of mean camber bent. The formula used to calculate the mean camber line is[4]

\begin{cases}{\dfrac {m}{p^{2}}}\left(2px-x^{2}\right),&0\leq x\leq p,\\{\dfrac {m}{(1-p)^{2}}}\left((1-2p)+2px-x^{2}\right),&p\leq x\leq 1,\end{cases}

I've drawn one force vector F, resulting from the local pressure acting perpendicularly on the surface. There will of course be hedgehog of force vectors all pointing in a different direction - to make sense of it all, they are decomposed into a lift component and a drag component

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  • $\begingroup$ To calculate lift and drag we need pressure distribution that static taps will gives us and f(x) function of upper and lower airfoil contours? So it is not true that with only pressure distribution we can calculate forces like someone mentioned..I know something is missing because pressure is not perpendicular to x co-ordinate when find lift or paralel to x axis when find drag.. $\endgroup$
    – Jurgen M
    Dec 12, 2021 at 11:12
  • $\begingroup$ No, not perpendicular to the x/y-coordinate, perpendicular to the surface. Lift and drag are vertical/horizontal components of the force vectors. $\endgroup$
    – Koyovis
    Dec 12, 2021 at 11:53
  • $\begingroup$ Why we need AoA if we know function of airfoil surface, then AoA is irrelevant? $\endgroup$
    – Jurgen M
    Dec 12, 2021 at 12:18
  • $\begingroup$ It depends how we want to decompose the vectors, lift and drag are defined relative to free air stream. $\endgroup$
    – Koyovis
    Dec 12, 2021 at 12:50
  • $\begingroup$ Is Su in my second video, f(x) function of upper airfoil surface? $\endgroup$
    – Jurgen M
    Dec 14, 2021 at 21:04
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So how would the integral "know" in which direction the pressure acts?

The integral can only integrate in one coordinate direction, so this will tell it what to do. If you integrate over the X coordinate, you will integrate the full pressure over the X projection of the airfoil surface. This will automatically give the lift component only.

If you integrate over the Y coordinate, you will only get the drag contribution. Note that a horizontal stretch of the airfoil has no pressure drag contribution.

Of course, as you correctly observe, the pressure at point A acts on an inclined part of the contour, so it contributes both a lift and a drag component.

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  • $\begingroup$ Angle theta is different for every place at airfoil surface, so it cant be one theta from LE to TE. Can you clarify that? youtube.com/watch?v=tr13vsa0d4s&ab_channel=AerospaceGuy $\endgroup$
    – Jurgen M
    Dec 12, 2021 at 8:44
  • $\begingroup$ @JurgenM If by theta you mean the inclination (better use $\frac{\delta y}{\delta x}$; or explain how you define $\theta$), of course it is different over chord. But you do not need it, all you need is either $\delta x$ or $\delta y$ for pressure integration. $\endgroup$ Dec 12, 2021 at 15:40

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