How does an a/c fly invertedly?

[allwerp]

A/Cs have been to have the capability to fly invertedly.But this contradicts Bernoulli's Principle that emphasizes that lift is the consequence of the difference in pressure between the top and the bottom of the wing.....So how does the a/c fly invertedly?:confused:

[Mad (Flt) Scientist]

No violation of Bernouilli required.

If an aircraft had a symmetric section wing (as do some aerobatic planes, for example) then in order to generate some lift it needs to fly at an angle of attack, which essentially makes the wing non-symmetric relkative to the airflow, even though it is geometrically symmetric.

If the same aircraft is flying 'upside down' and wants to generate the same lift in the opposite direction (which is the "new up") it simply flies with it's wing at the same but opposite angle relative to the wind, so that the wing is essentially turned upside down and still generates the same lift in the same way.

For most aircraft, the wing is shaped - cambered - so as to be most efficient generating lift in the traditional direction. But all that means is that you need to have a greater angle when upside down, and it's not as efficient. It's still possible.

If wings could not generate "lift downwards" relative to the aircraft, outside loops and "negative 'g' " manoeuvres would be impossible. They aren't.

[flynverted]

Very carefully :rolleyes: :cool:

[northwing]

The biggest limit is normally the engine fuel and oil system. Most engines will last about 15 secs with the oil that is already circulating but after that the bearings get starved as the oil in the tank gets lifted off the inlet. Aerobatic display aircraft have a special oil system, and possibly fuel system as well, to ensure continued supply when inverted.

Aerodynamically speaking the wing doesn't need to be cambered. You can fly a flat plate - the problem is that it would have no strength and would bend up, not to mention having a rather sudden stall.

[Astra driver]

Following on from Mad scientists' post, an airfoil does not even need to be cambered in order to generate lift; a flat board will generate lift with sufficient angle of attack. For evidence you only need to look at the wings on those cheap balsa wood gliders that we have all played with at one time or another. The airfoil on these is simply another piece of flat balsa, it is the AOA that generates the lift by creating the exact same bernoulli effect, albeit less efficiently than a cambered airfoil.

[beardy]

I believe that there is a flaw in Bernoulli's theory as the sole explanation of lift from an aircraft wing, the lower pressure of air above the upper surface of a wing assumes that the air is accelarated relative to the air close to the under surface. I haven't yet seen an adequate explanation why the air above the upper surface has to have the same mass flow as that under the wing, after all the molecules having been seperated at the leading adge don't have to reunite at the trailing edge. If it doesn't have the same mass flow, the explanation for accelaration is anadequate.

Could someone enlighten me?

[bookworm]

(The worm checks very carefully under the bridge, fearing the worst but unable to detect the unmistakable scent of troll... ;))

You're correct. The simplest explanation is an empirical one: the airflow over the upper surface is accelerated: it's accelerated when there's camber, or an angle of attack with no camber, or when there's both. You can watch it. You can calculate it. And as a consequence there must, by Bernoulli's theorem, be a pressure difference, which causes lift.

Why is it accelerated? Well, it's much harder to explain why fluid flows in the way it does than to watch it, or calculate it. The best I can do is suggest that you look at the trailing edge.

Because the trailing edge slopes downwards, there must be a downward component of velocity of the air just behind the trailing edge. Air doesn't vanish into nothingness, and the fluid mechanical equivalent of "what goes up must come down" kicks in: the only way the air can go down at the trailing edge is if it completes the circuit by going up at the leading edge, faster over the upper surface, down at the trailing edge, and slower under the lower surface. In other words, there must be a circulation around the wing -- which you can see leaving the wingtip as a vortex.

[Astra driver]

The air on the upper surface accelerates because the air is compressed by the greater surface area which can be created either by camber or AOA or a combination of both. (The air is essentially squezed much like when you squeze the end of a water hose to make the water come out faster.)

Area x velocity = constant

In this case the area is reduced by the camber so therefore the velocity has to increase.

Static pressure + dynamic pressure = constant

We have now increased the dynamic pressure of the air so therefore the static pressure on the upper surface is reduced, creating a net pressure difference between the lower and upper surface. Net result is a lifting action.

The wing tip vortices are caused by the higher pressure under the wing trying to equalize with the lower pressure on top of the wing as it passes through the air.

[beardy]

Bookworm,
Thanks, I understand that there is an accelaration of air over the upper surface of a plane. Bernoulli's theorem applies to constant mass flow and is normally illustrated in a venturi which is a closed system. A free wing is not a closed system and need not necessarily have the same constant mass flow above and below the wing.
If I understand your explanation of the vortex theory then there must be an equal and opposite force to the weight of the lifted stucture under the wing. Is this why I am buffeted by a helicopter in the hover in that a mass of air equivalent to the weight of the helicopter is accelarated by 1g by the rotating wing to oppose the acceelaration of earth's gravity? If so how why don't I feel this under an aircraft, I would have expected a 230 tonne downforce shortly after rotation of a large jet would be quite noticeable in the area just off the runway?

Astra Driver
I am not sure which greater area you are refering to and how it compresses air, your equation would seem to indicate expansion (if you are refering to wing area). Bernoulli's theorem is only valid for an incompressible fluid in closed system. The air is compressible and a wing exists in an open system, it is not one half of a venturi.

[bookworm]

Thanks, I understand that there is an accelaration of air over the upper surface of a plane. Bernoulli's theorem applies to constant mass flow and is normally illustrated in a venturi which is a closed system. A free wing is not a closed system and need not necessarily have the same constant mass flow above and below the wing.

Bernoulli's theorem also applies along any streamline, and everywhere in irrotational flow (which this sort of flow is, within the fluid). A closed system is not required.

If so how why don't I feel this under an aircraft, I would have expected a 230 tonne downforce shortly after rotation of a large jet would be quite noticeable in the area just off the runway?

The force on the ground is there. It's spread over a considerable surface area though.

[beardy]

I may not have used the phrase correctly, but I understand a closed system as being one from which there is no transfer of mass nor energy. I believe that this is a requirement of Bernoulli's theorem. The venturi is a classic closed system, to achieve the effect of a closed system in free air you would have to specify the boundary beyond which the energy/mass exchange is negligible.

If there is a force on the ground, is there also a rise in local air pressure, if so does this imply a rise in pressure under the wing which would also contribute to the lift. Presumably, if the force on the ground is equivalent to the weight of the aircraft there would be no need for a reduction in pressure over the wing c.f. the hovercraft.

[bookworm]

I believe that this is a requirement of Bernoulli's theorem

Not really. Bernoulli's theorem deals with the variation of pressure and velocity at different points within a fluid. It's not really a system level thing, but a characteristic of the Euler fluid. It's certainly not necessary to demonstrate a system as closed with respect to mass flow to apply Bernoulli's theorem -- that would be somewhat difficult in that its prime purpose is to relate the properties of the fluid at different points on a streamline.

If there is a force on the ground, is there also a rise in local air pressure, if so does this imply a rise in pressure under the wing which would also contribute to the lift. Presumably, if the force on the ground is equivalent to the weight of the aircraft there would be no need for a reduction in pressure over the wing c.f. the hovercraft.

This all depends on where you draw the arbitrary boundary around your "system". I think that, in a sense, you're double-counting.

The air is merely the medium through which the ground supports the aircraft. The ground applies a force to the air. The air applies a force to the aircraft.

In the same way, when the aircraft is parked, the ground applies a force to the tyre. The tyre applies a force to the aircraft.

But that doesn't mean that there are no stresses in the tyre, does it? To apply the upward force to the aircraft, the tyre must deform around the wheel, creating more pressure on the lower part of the wheel than on the upper. In the same way, the air "deforms" around the wing, causing the pressure difference we're discussing. The fact that the ground applies a force to the tyre, or the air, doesn't change that.

[beardy]

derivation of Bernouli's theorem (http://astron.berkeley.edu/~jrg/ay202/node86.html)
Indicates that the presumption is of constant energy, towhit there can be no addition or subtraction of mass/energy, i.e. a closed system. In terms of lift since what is under discussion is a change of pressure for Bernoulli's theorem (which assumes conservation of energy) to apply there must be no mass/energy added or subtracted to explain the change of pressure due solely to the change of velocity of the particles of air.

Be that as it may, my original question was why, in what is an open system (i.e. free air) does the air have to accelerate? this is of course the basis of using Bernoulli's theorem to explain lift. There is a detectable acceleration and a detectable pressure drop so the conservation of energy implies that there is an acceleration, there are alternative methods of conserving energy, altering the mass, temperature or entropy of the system are all possibilities. What role does viscosity play?

Bernoulli's theorem would lead me to expect a windsock to collapse in on itself. It doesn't, but then the sock itself is not in a closed system where energy is conseved, there is the area outside the sock to be taken in to consideration.

Accepting that there is a decrease in pressure over the wing I have difficulty in accepting the hydraulic explanation of an increase in ground level pressure under a wing that is producing lift. There are notable exceptions, Wing In Ground Effect (WIGE) springs to mind.

[bookworm]

Indicates that the presumption is of constant energy, towhit there can be no addition or subtraction of mass/energy, i.e. a closed system.

No, a closed fluid element, not a closed system. There's a significant difference when the element is moving around the system!

Accepting that there is a decrease in pressure over the wing I have difficulty in accepting the hydraulic explanation of an increase in ground level pressure under a wing that is producing lift. There are notable exceptions, Wing In Ground Effect (WIGE) springs to mind.

The WIGE concentrates the force in a smaller area. If you imagine moving the wing continuously to a greater height, the pressure on the ground becomes less, but it acts over a much greater area. It doesn't have to be over very much area before the pressure gets lost in the noise -- but the integral, the force, is the same.

By Newton's 3rd Law, the wing applies a force to the atmosphere. If an opposing force on the atmosphere is not provided by something, the entire atmosphere would accelerate downwards. Given that the global atmosphere is not a still fishtank but a turmoil of large scale motions in which a few million tons is neither here nor there, that's not an unreasonable model. But what happens at the boundaries of the model, whether it be miles away from the aircraft or hundreds of feet away, makes little difference to the pressures at the wing surface.

[RomeoTangoFoxtrotMike]

I've found www.av8n.com/how/htm/airfoils.html (http://www.av8n.com/how/htm/airfoils.html) to a very helpful explanation.

HTH,

/RTFM

[beardy]

Thank you so much for the link, it makes it all clear. Apart from the windsock!

bookworm,

Thanks for your effort I now believe you about surface pressure, I am sorry I should not have doubted. I had forgotten the question of a freighter full of budgies (does it weigh less if they are roosting or flying?) The point about closed system or elements of a system does not appear to be very important now.

[hobie]

Someone wrote this but I can't remember who ....

"Here is a list of things you need in an airplane intended for upside-down flight:

You need super-duper seatbelts to keep the pilot from flopping around.

You need to make sure the airframe is strong enough to withstand extra stress, including stress in new directions.

You need to make sure that the fuel, engine oil, and battery acid stay where they are supposed to be.

You will notice that changing the cross-sectional shape of the wing is not on this list. Any ordinary wing flies just fine inverted. Even a wing that is flat on one side and curved on the other flies just fine inverted. It may look a bit peculiar, but it works."


:ok:

[bookworm]

Thanks for your effort I now believe you about surface pressure, I am sorry I should not have doubted.

Oh but you should have doubted... That's what this is all about. :) Thanks for the thought-provoking discussion.