You are making two assumptions, one explicit and one implicit. That the cross-section area is constant (it’s not, it depends on the angle of attack, which is smaller at higher speeds) and that the air density is constant (it’s not, supersonic planes fly at much higher altitudes, where the air is thinner).
The ideal gas law applies, at least nearly enough. So PV = nRT. By saying the density is equal between the top and bottom, you are also saying the pressure is equal. The air around the wing is having it's momentum changed, not it's pressure. At least, at sub mach speeds.
> It's not as simple as v^2, you have to consider the compressibility effects (wave drag where the pressure wave moves along with the plane, like a bow wave on a ship).
I think it is that simple. Those effects are the reason the drag coefficient is changing, you don't need to account for that twice.
> Basically all aerospace engineering undergrads get taught early on that airfoil drag is just zero in 2D incompressible and inviscid flows, [...]
Don't you need some adicional hypothesis, like that all the path integrals of the wind field are zero? Most of the easy part of the analysis of the lift in a wing assumes that air is inviscid (except for the initial generation of the vortex around the wing). [And if the speed is much smaller than the speed of sound, you can approximate that the fluid is incompressible.]
> Additionally, when plane speeds up, more drag is produced, slowing it down.
Wait, is it speeding up here or slowing down? Slowing down means deceleration, speeding up is acceleration, and it can't be doing both at the same time.
> When the plane slows down, it produces less drag, allowing to to pick up more speed.
Same deal?
I think what he's getting at is that drag increases with the square of speed, but it's a very confusing way of explaining it.
> Is there a nice way to derive this?
“It has been found both experimentally and theoretically that, if the aerodynamic force is applied at a location 1/4 chord back from the leading edge on most low speed airfoils, the magnitude of the aerodynamic moment remains nearly constant with angle of attack. Engineers call the location where the aerodynamic moment remains constant the aerodynamic center (ac) of the airfoil” [1].
As you go faster, it goes from quarter chord to half. For a rectangle, the chord length is equal to the airfoil length.
Why quarter? It comes from thin-airfoil theory [2].
> As for CO2, the plane flying slower is more fuel-efficient.
It's counter-intuitive, but an aircraft can be more efficient at higher airspeeds. The reason for this is that, unlike a ground vehicle, an aircraft in level flight must expend energy generating lift to balance its weight. This loss of energy shows up as lift-induced drag [1]. As airspeed increases, the losses to lift-induced drag decrease, while the losses to aerodynamic drag increase. Where the point of minimum total drag lies depends on several factors including weight and altitude. See [2], which contains a chart for a generic aircraft which is 5% less efficient at Mach 0.72 than it is at Mach 0.80.
> Why would pressure (Bernoulli out of Euler) propagate supersonically while momentum (Newton) does so subsonically?
I'm sorry, I didn't understand the question.
But in supersonic flight, with a flat plate, you don't have any rotation in the game, as illustrated here [0]. And yet you will be producing a lot of lift.
No, it really is the pressure alone. And viscous drag, if you want to be pedantic. Those are the only forces at play, the rest is only a side effect of those forces.
> If the turning of the gas was the necessary mechanism for lift, planes in supersonic flight would fall out of the sky
Where are you getting this? Why would pressure (Bernoulli) propagate supersonically while momentum (Newton) does so subsonically?
> Instead of relying on an airfoil shape for lift, you could fly by sucking air from the top of your wing and dumping out the back of your plane
Wings are bigger than engines. That’s the leverage you get with a lifting body: you move more molecules than your thruster alone.
The correct answer here is unintuitive. But the very wrong answer is pressure alone. (As the article we’re commenting on clearly shows with its brilliant flat-cardboard example. You don’t need camber to have a lifting body, just angle of attack.)
>The thing I never found satisfying was this notion that the air over the top moves faster because it has further to go
On the one hand I agree that it is a stupid way to phrase it. On the other hand if the air doesn't "make it" then there is nothing where the wing just was aka a vacuum. The low pressure area that forms above the wing sucks the air along making it faster. Why doesn't all the air rush to fill the low pressure area? Well for air below the wing there is a wing in the way, air above the air flowing over the wing does rush down to fill the void providing lift, air behind the wing does as well creating some drag.
Same for angle of attack it deflects the air that would normally be above and behind the wing down ,making a low pressure area form above the wing which the air speeds to fill.
> Yes, the air above the wing needs to travel a longer distance with the typical section used in wings, which means that it goes faster than the air below the wing.
Both of these sub-clauses are true, but the "which means" connecting them aren't. There's no law of physics saying a fluid that has a longer path ahead of it speeds up in anticipation.
> With the pressure argument, you have not explained why the air is faster over the top surface of the wing, you have just stated that it is.
That's true. However ...
> ... the circulation generated by the wings ...
You haven't explained why there is "circulation generated by the wings", you have just stated that there is.
Note, if you take the full flow over the wings and subtract the vector field that is the undisturbed average flow then you are left with a circulation. The two are effectively equivalent, so "explaining" by saying that there is "a circulation" is basically saying that the air flows faster over the top and slower underneath.
It's a non-explanation.
I do have an explanation as to why the air flows faster over the top, but this discussion is too small to contain it.
PS: I agree that all the modelling and equations work very well, and from that point of view its "well understood". However, the discussion here shows that for many people, even those who are technically capable, it is not "well understood".
> The air pressure is proportional to either v or v² depending on the speed and atmosphere.
1. I think you mean "air resistance". Yes?
2. If so, then no, air resistance transitions from (linear) Stokes drag at low velocities to (aptly named) quadratic drag at higher velocities, as a function of velocity, but not a function of air pressure. So not "either v or v^2", but a combination of the two factors.
Nit: aerodynamic drag force is modeled as quadratic (squared) with speed. The power required is force times velocity, which means the power is cubic with speed, but the force is only quadratic.
With regard to footnote 7: As someone who has played a very small part in correcting some misconceptions about how airfoils generate lift, I find that I now have to apply some corrections the other way: there is necessarily a pressure difference across an airfoil if it is generating lift, and Bernoulli's equation does correctly describe the relationship between that difference in pressure and the equally real difference in local airflow speeds (up to airspeeds where compressibility is an issue, as Bernoulli's equation is only strictly correct for incompressible flows.) What was most wrong with the old explanations was the use of a fallacious 'equal transit time' argument in an attempt to explain the difference in speeds.
You are making two assumptions, one explicit and one implicit. That the cross-section area is constant (it’s not, it depends on the angle of attack, which is smaller at higher speeds) and that the air density is constant (it’s not, supersonic planes fly at much higher altitudes, where the air is thinner).
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