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Old 8th Apr 2022, 21:47
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danandrews
 
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Originally Posted by Theholdingpoint
You can't get it, because the "usual" lift formula is just a highly semplified equation, which has nothing to do with the real one.

Just think about the rest of the expression:
  • v: velocity of the undisturbed air. Not really useful when you realize that the air hitting your plane is not "undisturbed"
  • 1/2: coefficient coming from the previous integral of v
  • rho: mean density of still air. Again, not really a factor, especially if M>0.3
  • S: top projection of the wing area. Not counting wing tickness (big factor), real wing area (another big factor), wing profile (huge factor), curvature (...), fuselage area (...), tailplane (...), engine nacelles and so on...
Lift is just the normal component of the total areodynamic force acting on an airplane, which is given by the 3D integral of the stresses acting on the entire machine.
So that formula has the following meaning: airplane A performance (in terms of lift) is equal to the performance of a model airplane flying at v, in a pocket of air with constant density rho, having a wing area S, multiplied by a factor called CL.
Airplane B, flying in the same air, with the same wing area but with a different wing profile (just an example of an influencing factor) will have a different value of CL.
This is useful in aircraft design, as it's a quick way to compare the performance of different airplanes at the same environmental conditions.

We could then say that CL is just a number used to sum up a lot of different variables (AoA, wing profile, Reynolds number, downwash, compressibility, local windspeeds, surfaces, etc), nothing more than that.
Well done with your critical thinking, keep up the good work!

Tinstaafl static pressure doesn't generate lift. Temperature is an influencing factor, but it's not really a useful form of energy for lift generation (we could even say that it's a negative contribution, as higher T-->lower rho, less lift, less thrust).
Theholdingpoint Thank you I appreciate this reply, took me a while to get it at first. My incorrect assumption I believe was thinking that the rest of the equation was representative of all of the energy that could be extracted from the air. And that the co-efficient was simply the proportion that was extracted.

I think I see your point. When the relationship between some factor (wing size, velocity etc) to lift created, is simply observed, or rather, easily derived, it makes sense to put that into the equation. Nice simple relationships. The multitude of other contributing factors to the final value of lift are too complicated to measure and/or express simply. Thus they take the form of a dimensionless co-efficient? Am I getting that right?

I must say this has troubled me since I started my flying training many moons ago. So thank you!

And apologies for the late reply to all who took the time to answer my question in this thread.
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