Bernoulli's Principle ... and Lift
motivated by e-mail from Bob W ( who keeps me busy :^)

There's this common explanation of what provides the lift, for an airplane.
It goes like this:
 The top surface of a typical aircraft wing is "longer", front-to-back, than the bottom surface. The air which goes OVER the wing must then travel farther than the air which goes BELOW the wing ... as in Fig. 1. That means the upper flow of air is faster and, according to Bernoulli's Principle, the pressure is less above the wing. That decreased pressure acts like a partial vacuum (above the wing) ... and that's what provides the lift that keeps the plane in the air. Figure 1

>Bernoulli's Principle?
Yes. I think it was Daniel.
Although it's a mathematical statement, it says that the greater the speed of a fluid (and that includes air), the less the pressure.
You open the door of a plane while it's in flight and people get sucked out.
You open the window of your car while driving and the pipe smoke gets sucked out.
>You smoke while driving?
Well, sometimes ... but I keep the window open, relying on Bernoulli!

Well, I've always accepted that common explanation, but ...
>But now you're going to tell me that Bernoulli was wrong.
No, of course not. I'm going to tell you that using Bernoulli's Principle to explain lift ... that explanation is wrong.
>And you just learned this?
Well ... uh, yes. The point is this:

• The air, after leaving the wing, has a DOWNWARD velocity. (It didn't ... before the wing hit it!)
• It's the wing that caused this downward velocity ... and that took some force, on the air.
• According to Newton's Third Law of Motion, every action has an equal and opposite reaction.
• You push DOWN on something ... it pushes back UP.
• It's this UP push (the "reaction" on the wing due to its pushing the air down) ... this UP push provides the lift.
 >The air has a downward velocity? Are you sure of that? After leaving the wing? It's a "downwash". It leaves the back of the wing. So ... you don't want to follow a plane too closely >And you knew all this? Well ... uh, I read it here where there's this really neat picture of a plane flying through fog and pushing down the fog after it passes. >That's the trailing downdraft? The downwash. Yes. >That plane looks like a Bombardier Challenger. Don't you have Bombardier stock? Don't remind me.I also got e-mail from Carl P who notes that, in a helicopter: "As each rotor passes over head, the downward force ... hits the vehicle with a very hard thud" "The fraction of time that each rotor is over the fuselage, its lift is cancelled out". Figure 2
>So you think it's that downdraft that provides the lift?
Downwash! Me? Think? I just regurgitate what others say. For example, Dr. Max. W. Munk, Ph.D, Dr. Eng. is a world class aerodynamicist.
In "The Principles of Aerodynamics" he says:
"An airplane would never be pressed upward by the air unless it first pressed the air down. In airplane flight, air must be deflected downward continuously. Fresh, resting, peaceful air is continually waked up from its slumber and set in motion down toward the ground. The air thus disturbed resists that motion, thereby pressing the airplane upward."

>So how much lift do you get?
Aah, that's another of Newton's Laws ... the second, which we can use like so (a quick-and-dirty calculation):
 The (average) force required to impart a velocity Δv metres/second to a mass M kilograms in a time Δt seconds is   F = M Δv /Δt   Newtons. (That's the second law.) Suppose that the mass of air which is given a downward velocity of Δv by the wing is M (something like Fig. 3). Before the plane arrives, it was at rest. After the plane passes, it's moving ... with a downward component Δv. This mass experiences the presence of the plane for the time Δt, namely the time for the plane to travel a distance ΔL. If the plane is travelling horizontally at the speed V, then V = ΔL / Δt so that Δt = ΔL / V That makes F = M Δv /Δt   = M V Δv / ΔL Figure 3
>I have a question!
Shoot.
>Is this exact ... or what?
Didn't I say "quick-and-dirty"? It's mostly dirty, but it'll give us some idea of the lift provided by that downdraft.
>Downwash!
Uh, yes. Anyway, continuing:

• The Mass M is some volume (call it Vol  m3) multiplied by the density of air (call it ρ  kg/m3): M = ρ Vol  kg.
• The volume of air moved is some horizontal area (A  m2) times some vertical distance: Vol = A h m3
• So the Mass is M = ρ A h kg
• Our force is then: F = M V Δv / ΔL = ρ A h V Δv / ΔL
>Another question!
Yes?
>Do we know all these numbers?
Well, to be consistent with our "quick-and-dirty" methodology, we'll assume:
• The value of ΔL is just the front-to-back length of the wing.
• Further, we'll assume that A is the area of the wing: A = wingspan*ΔL
• That gives F = ρ A h V Δv / ΔL = (wingspan) ρ h V Δv
• Finally we'll assume that the downward velocity imparted to the mass M is proportional to the plane's speed: Δv = k V
• Then F = k (wingspan) ρ h V2
 >Question! Yes, what is it? >That "k" hides all the dirty stuff, eh? You can choose it to get a correct answer, eh? Okay, following that link, above, we take the velocity of the Mass M in the direction of the angle of attack of the wing. We'll call it θ, as in Fig. 4 The DOWNWARD component of this velocity is then Δv = V sin(θ) so k = sin(θ). So [1]                 F = (wingspan) sin(θ) ρ h V2 Figure 4
>Do you believe it?
Well it has the correct dimensions, namely kg*m/sec2 which is Newtons.

The interesting thing is that, if you know the force required to keep the plane aloft (that's the lift) then this magic quick-and-dirty formula [1] gives an estimate of "h", hence of how much air needs to be moved !
However, we can try it out, like so (using numbers similar to those in the link, for a Cessna 172):

 wingspan = 10 m sin(θ) = 0.1 about 5.7 degrees ρ = 1 kg/m3 h = 5 m (pure invention) V = 100 m/s (or 360 km/hr) F = 47,900 Newtons (about 4900 kg-force)
... which is about 5 times the weight of the Cessna.

 Wingspan = metres Angle of attack = degrees Density of fluid = kg/m3 Height of mass moved = metres Speed of plane = km/hour Lift = kilogram-force ... quick-and-dirty

>That h value is strange. You won't find it at the Boeing website, eh? You invent it and ...
Yes, but we'd expect h to depend (in some complicated way!) upon the front-to-back dimension of the wing.
If we increase the wing area we'd expect to move more air.
We'd expect h, hence the lift, to increase. Somehow we should incorporate the wing area into our equation ... not just the wingspan.
Okay, so notice that in the quick-and-dirty formula [1], namely F = (wingspan) sin(θ) ρ h V2, the product (wingspan) h has the dimensions of area: metres2.
Let's just call it A. We then get:
[2]                 F = A sin(θ) ρ V2
But sin(θ) is dimensionless (being a ratio of like quantities, like the sides of a right triangle), so A ρ V2 incorporates ALL the dimensions of lift.
So we can rewrite [2] like so:
[3]                 F = CL A (1/2)ρV2
where CL is a dimensionless quantity called the Coefficient of Lift.

>Dimensionless?
It'd have the same value in any system of units: metres/kilograms/seconds or miles/pounds/hours or lightyears/...
>Yeah, yeah ... I get it.
Good. So now we take A as the surface area of the wing and have, finally:
 Lift = CL A (1/2)ρV2 where CL = Coefficient of Lift   a dimensionless quantity depending upon the Angle Of Attack (AOA), shape of wing etc. A = area of wing   in metres2 ρ = density of air   in kilograms/metres3 V = speed of aircraft   in metres/second

Taken from this site