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Given pressure coefficient data for corresponding x/c (position along chord) values on a wing, I am looking to calculate coefficients of drag and lift. Here's what I have: b, MAC, S, sweep angles at LE & c/4, and Cp vs. x/c data for 6 different stations (span position given). Angle of attack is 0 degrees.

I understand that to calculate Cl, I can numerically integrate Cp with respect to x/c from 0 to 1. (ignoring skin friction [which, I know, is not best practice])

To calculate Cd (again ignoring skin friction), I believe I need to perform the same integration but with the Cp values being first multiplied by dz/dx inside the integrand, where z corresponds with the airfoil thickness.

My problem is that I do not have any information on the airfoil thickness. Any thoughts on approaching this? Is this problem possible? If it matters, I'm sourcing my equations from Anderson's Fundamentals of Flight, Ch. 1.

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Without the local inclination of the airfoil surface you will not be able to achieve exact results for drag. You need to integrate just the component orthogonal to the flow direction at infinity for the lift coefficient (because it is defined that way) and parallel to that direction for the pressure part of the drag coefficient.

While it is OK to neglect friction for lift, it does contribute about half of total drag when the flow is attached, so it should not be neglected for calculating total drag.

To illustrate how important the local inclination is, here the c$_{\text p}$ plot of the Eppler 502 at a moderate angle of attack (3°): Pressure distribution over an E502 at 3°

And now the same thing, plotted with vectors. Arrows pointing away from the surface denote suction and vice versa.

Pressure distribution over an E502 at 3° using arrows, courtesy of XFOIL

Since the c$_{\text p}$ plot should use the projected length along the X-axis, this inclination effect is already considered when you integrate the pressure coefficient over chord. Treat chord length as a dimensionless number that runs from 0 to 1. Since the angle of attack is 0°, you need no other correction and the net surface under the c$_{\text p}$ plot is your lift coefficient. Use the results at the six known span positions and interpolate between them.

In principle, the same applies to the drag coefficient, but now you would need a plot which projects the pressure coefficient on a coordinate which runs orthogonal to the flow direction. In inviscid flow, the pressure differences equal out, so it is only the friction-induced changes in the pressure distribution between inviscid and viscous flow which cause drag.

My best advice is to find airfoils with the same pressure distribution and use their drag coefficient values. Make sure you know the Reynolds and Mach numbers, because both have a higher effect on drag than on lift. Some quirks in the pressure distribution can give them away, like the location of the laminar separation bubble on the lower side of the c$_{\text p}$ plot above or the magnitude of the stagnation point pressure coefficient. Regarding the relative thickness of the airfoil: The amount of suction on both sides of an airfoil grows with its relative thickness.

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  • $\begingroup$ Thanks for the detailed answer here, Peter. It sounds like I'm on the right track for getting the approximate lift coefficient, but I may out of luck for Cd unless I can determine the airfoil somehow. I'll have to look into that. I'll add that my AoA=0 comment in my original question was for the sake of simplicity - I actually have data for two separate angles (2 and 4 deg). I'm currently attempting to find the normal and axial force coefficients, then will rotate those w/alpha to find Cd and Cl. Are you aware of any methods to find airfoil shape (or z variation) that could use this data? $\endgroup$ – user26989 Nov 19 '17 at 21:38
  • $\begingroup$ @user26989: There is free software which allows to reverse-engineer (actually, design) an airfoil from a desired pressure distribution. It is called XFOIL and heavily used by model enthusiasts. I used it years ago to create the images in this answer. $\endgroup$ – Peter Kämpf Nov 19 '17 at 22:11
  • $\begingroup$ This looks very interesting, thanks for the link (and all the help here!). $\endgroup$ – user26989 Nov 19 '17 at 22:35

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