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As Mr. John D. Anderson states, the aerodynamic forces on a body are due to only two sources: pressure distribution over the body and shear stress distribution.

Shear stress is acting tangential to the surface, so to some amount, friction is also contributing to the lift of a body, not only to drag. But what I most of the time see, is that one is looking at the pressure distribution and integrating this over the surface to get the resulting force.

Do we neglect the part of friction lift because it is so little? But for drag we account for it?

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    $\begingroup$ You may add a link to the state of J.Anderson. Moreover, here are some related questions on this website and other resources to help you understand lift generation. $\endgroup$
    – Manu H
    Commented Apr 17, 2020 at 6:58
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    – FreeMan
    Commented Apr 20, 2020 at 13:59

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First, let me illustrate how I understand your question. Please comment accordingly if I got you wrong.

Below you see the friction drag computed by XFOIL for an airfoil at 3° angle of attack.

E502mod at 3° AoA, friction plot

Friction drag over chord for an E502mod airfoil at 3° AoA. Blue: Top surface, Red: Bottom surface. Source: XFOIL 6.97.

And here is a plot of the pressure distribution at that point:

Same airfoil, now with arrows to indicate pressure

Where the longest arrow pointing at the airfoil contour points to is the stagnation point. So the large blue peak in the first plot is exactly in the region of the nose where its curvature is strongest and the surface is mostly vertical. Integrating all the friction drag over the height of the airfoil should result in a small upwards component from that high local friction around the nose. Lift, from friction, in other words.

Did I understand that correctly?

I've never seen that component of lift being mentioned separately, and even XFOIL does not include it in its calculations (just checked the FORTRAN source I have lying around). You assume correctly that this component is indeed negligibly small, it is about two orders of magnitude smaller than the pressure forces.

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  • $\begingroup$ When airfoil is at a non-zero AOA, more shear forces will translate into lift component. Doesn't xfoil compute far-field lift and drag instead of surface integration? $\endgroup$
    – JZYL
    Commented Apr 17, 2020 at 19:09
  • $\begingroup$ From web.mit.edu/aeroutil_v1.0/xfoil_doc.txt; The lift and moment coefficients CL, CM, are calculated by direct surface pressure integration: / _ / CL = L/q = | Cp dx CM = M/q = | -Cp [(x-xref) dx + (y-yref) dy] / / _ where x = x cos(a) + y sin(a) ; a = angle of attack _ y = y cos(a) - x sin(a) $\endgroup$
    – user46017
    Commented Apr 21, 2020 at 3:13
  • $\begingroup$ @m2as3registeredservicesohmone: You are right. In XFOIL.f, subroutine CLCALC, the lift coefficient is summed up in the line CL = CL + DX* AG (plop over N panels). No additional summand using DY, so tangential forces are neglected. Since they are two magnitudes lower than pressure, this makes sense. $\endgroup$
    – Peter Kämpf
    Commented Apr 21, 2020 at 22:45
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Let's assume that all of the shear stresses have been converted to lift. Shear stresses manifest in the form of skin friction drag which, for a fully turbulent boundary layer (producing more shear stresses than a laminar one) over a flat plate, has an approximate solution given by (Ref. Anderson, Fundamentals of Aerodynamics):

$$C_f=\frac{0.074}{Re_c^{1/5}}$$

Let's assume a Reynolds number ($Re_c$) of 0.1 million and 2 sides, this gives a drag coefficient of 0.015. If you assume a lift slope of $2\pi$, this translates to an error on attack of attack of only 0.14 deg, even if all the shear stresses have been converted to lift. Needless to say, shear stress is a negligible error component for lift prediction.

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