1. ## Something i've always wondered about jetfighters.

I know the principle of an airplane wing is that above the wing, the wind has to go faster around it because of the shape of the wing. Thus creating a higher pressure on the downside of the wing and so it creates lift.

But what happens when a jetfighter flies upside down? Shouldn't it just go down incredibly fast, because you've just created an incredible downforce like a formula 1 car(a spoiler is basically a wing that's upside down).

Schematic:

High airspeed, low pressure
--------------------------------
wing
--------------------------------
Low airspeed, high pressure

How come jet fighters can fly upside down, and not fall down?

I say ninja's are responsible, but any other suggestions(preferably serious ones) are welcome.

2. There are many explanations of lift, but one qutie correct one is that total pressure = static pressure + dynamic pressure. Dynamic pressure = 0.5*density*speed&#178;. Total air pressure is the same above and below the wing, namely the pressure of the atmospheric air at standstill. Air and plane motion are relative when concerning pressures (I know that might sound unbelievable but trust me, it's true), so one can also see the jet as standing still and the air moving. So you've got p_static + 0.5*density*airspeed&#178; = total pressure (constant).

Now, the wing shape slows down the air underneath the wing. That means the dynamic pressure term decreases. As total pressure is constant, that implies an increase in static pressure. Static pressure is the pressure "pushing on the surface of the wing". So underneath the wing, there is an increase of "pushing pressure". Same reasoning, above the wing the shape induces a faster airflow. So the dynamic pressure term increases there, and hence the static pressure decreases. So less pushing force on top of the wing, more below the wing. Result: lift.

Now to arrive at your question: when an aircraft would fly upside down exactly as it should when it flies normally and stable, it would drive itself into the ground. What it does, is press the nose down (as seen from the cockpit) eg up as seen from the ground when it is upside down. Now the inverted wings have a "normal" angle of attack, still generating lift in the healthy upward (as seen from the ground) direction. As the wing is not designed for this, any non-symmetrical airfoil (that is well designed ) will be less efficient in this way.

So the reason why a plane can fly inverted, is that it changes its pitch a bit such that the wing is in / this orientation when flying to the right.

For the same reason, you must pitch up to 10&#176; angle of attack before doing a roll. That way, you compensate for the otherwise downward lift when you're upside down.

I know this could have been much simpler, but this is more or less complete.

For those who want it simple:

"when an aircraft flies upside down, the downward lift would drive it into the ground. Therefore, the inverted aircraft down pushes its nose up as seen from the ground, so the now tilted wings generate an upward instead of inverted lift".

http://www.avsim.com/pages/0803/fs20...tra_05thmb.jpg <<<<--- here you see the subtle nose up (as seen from the ground) pitch of an inverted plane. In airshows, they sometimes doon't do it, but that means they'll lose altitude.

3. I suggest we wait with discussing just *how* the air is decellerated below the wing and even more interesting accelerated on top of it, until the original question is considered answered by frostbyte.

This secondary matter namely has some interesting cause&result matters that I would like to discuss. I can't remember whether I ever had a final answer on them and simply forgot about it, or there is no final answer.

4. Originally Posted by Nicolas
In airshows, they sometimes doon't do it, but that means they'll lose altitude.

Thanks! And btw, I don't think that guy could afford to lose much more altitude...

5. The one in the pic? No indeed . That's why he goes for the pitch up attitude. He does not feel like digging himself in.

airfoils and local airspeed then!

*how come the air is accelerated on top of a wing, and decellerated below the wing?*

Air can flow faster than the total plane airspeed at some parts on top of the wing. My question is whether a decreasing pressure gradient is the driving force behind this acceleration. Because increasing speed also induces a decreasing pressure gradient (chicken, egg...).

My guess is that the airfoil acts like half a nozzle. The expansion of the air in this semi-nozzle makes the pressure drop => decreasing pressure gradient => velocity increases.

Now comes a hurlde: is the decreasing pressure gradient that accelerates the air also the one that generates the upper surface part of the lift, or does the accelerating air induce another pressure effect (ie the reduced static pressure) in order to generate this part of the total lift?

You see it, I have a chicken and egg problem. Any help?

7. You people make me feel so retarted since most of these explinations are totally over my head. Then again, alot of you are real scientist and not just idiots with an AOL connection thinking they have it figured out.

God I love this place

8. Actually, the bernoulli effect accounts for very little of the lift generated by an airplane - the vast majority of the lift is generated for 2 reasons. First, the wing is slightly angled up, even in level flight (a couple degrees, no more). Most airplanes have the wings angled relative to the airplane for this. This pushes air down. Newton's third law comes into effect - plane pushes air down, air pushes plane up. The second reason is that the air flowing along the top surface of the wing is going down at the rear of the wing (look at the shape). This again causes the plane to go up from newton's third law. Air goes down, plane goes up. In the case of fighters and stunt aircraft, they use a symmetrical airfoil - the top and bottom surfaces have the same shape (some use a semi-symmetrical, where they are slightly different shapes to increase efficiency while cruising in level flight). They get basically all of the lift from angle of attack. A good demonstration of this is when you stick your hand out the window of a car (at speed). If you hold it level, you feel drag, but no lift. Angle the front of your hand up, and you get lift. Angle it down, and you get lift (but towards the ground). This is how a fighter or stuntplane gets basically all of its lift.

9. You should not feel like an idiot at all. I wondered myself as well. It's because the pilot only needs a small pitch up attitude in order to get healthy lift at inverted mode that the answer isn't obvious. But you did identify the problem with flying upside down, so you took a first large step in understanding! Many people don't even realize that when you fly upside down like you would in normal attitude, your wings would drive you into the ground (I'm ignoring some side effects of flying inverted on lift generation here ).

And if you feel an idiot because I know more about aircraft: I should know more about it, as I'm an aerospace engineer. I'm sure there are subjects on which you know more.

I see we're both from Belgium btw .

10. Originally Posted by cjl
Actually, the bernoulli effect accounts for very little of the lift generated by an airplane - the vast majority of the lift is generated for 2 reasons. First, the wing is slightly angled up, even in level flight (a couple degrees, no more). Most airplanes have the wings angled relative to the airplane for this. This pushes air down. Newton's third law comes into effect - plane pushes air down, air pushes plane up. The second reason is that the air flowing along the top surface of the wing is going down at the rear of the wing (look at the shape). This again causes the plane to go up from newton's third law. Air goes down, plane goes up. In the case of fighters and stunt aircraft, they use a symmetrical airfoil - the top and bottom surfaces have the same shape (some use a semi-symmetrical, where they are slightly different shapes to increase efficiency while cruising in level flight). They get basically all of the lift from angle of attack. A good demonstration of this is when you stick your hand out the window of a car (at speed). If you hold it level, you feel drag, but no lift. Angle the front of your hand up, and you get lift. Angle it down, and you get lift (but towards the ground). This is how a fighter or stuntplane gets basically all of its lift.
Newton's third law is often used to explain lift, but it is incomplete (and unclear) at worst and an alternative for the pressure explanation at best.

Bernouilli's law is fundamental for lift generation. It does not contribute for a small part; the exchange between static and dynamic pressure is what generates lift. (well, under the assumptions of Bernouilli's law, such as incompressible flow)

btw many wings indeed are placed at an angle of attack of about 3&#176; onto the fuselage, but remember that curved airfoils can generate lift at zero degrees angle of attack.

11. Yes, and in large part due to the air travelling downward at the rear of the wing. I'm not saying the bernoulli effect is irrelevant - it is not completely irrelevant, but it is not the primary cause of lift on most wings. The velocity difference in air over the top and the bottom of the wing is present, but not as large as many people imply, and certainly not large enough to generate all the lift that most planes need to get off the ground.

12. In my opinion, that is completely wrong.

Unless I have remembered it totally wrong, I was learned that the Bernouilli effect explains the result of the difference in airspeed above and below the wing, and that this result (the pressure difference) is what generates the lift. So the Bernouilli effect is (within its assumptions of validity) the FULL lift force.

Newton's third law can be used as an alternative explanation, but as gas dynamics are in my opinion unclear for Newton's third law (you don't really see the gas as mass particles interacting with the wing) I prefer the pressure explanation. So only when explained in its totallity, Newton's 3rd law explains lift, but then still it does nothing more than giving another way of expressing Bernouilli's law. Bernouilli's law expresses the results the conservation of energy principle expressed in pressures for fluids. Obviously, an object (wing) in a pressure gradient will feel a force (lift). Newton's third law is the rough way to explain this force, Bernouilli's principle uses the particulars of fluids (ie pressure) to explain this. But I believe it is clear that both are based on the conservation of energy and describe exactly the same, namely the generation of the full lift force.

I'm talking in general here, leaving out some details filtered by assumptions. So "full" is not REALLY "full", but full enough for all practical purposes and the spirit of this post.

13. FrostByte, think of it this way: when you're next on the interstate, open the window and stick your hand straight out a bit (hold your hand tightly outward, like the proverbial karate chop). Rotate your wrist to tilt your thumb upward, then downward. Feel the force of the wind on your hand? That's lift. The angle of your hand is what aviators call "angle of attack".

The Bernoulli effect counts (and engineers use it to design efficient wings), but angle of attack counts a lot more, especially when doing aerobatics, with helicopters, and with propellers.

So, if you're flying upside down, how do you counteract Bernoulli? Simply by altering the angle of attack to compensate, and pushing the nose of your plane away from the ground a bit.

This is probably oversimplified somewhat, but it should get you thinking in the right direction.

14. The Bernoulli effect counts (and engineers use it to design efficient wings), but angle of attack counts a lot more, especially when doing aerobatics, with helicopters, and with propellers.
And that is wrong too in my opinion. What happens when changing the angle of attack is also explained fully by the Bernouilli effect! (well, up to the point of stall at least ).

Lift is NOT built up of a bit of angle of attack, a bit of Newton's third law and a bit of Bernouilli.

Lift force is the net force generated by the interaction of streaming air and the wing. This lift force can be explained through the theory of Newton's third law, or alternatively through the theory of Bernouilli.

I hope it is clear now. If anyone can correct me with references, I'm open to that.

One more thing about Bernouilli's effect: what is dictating for the lift force is the static pressure term of Bernouilli's law. So if you want to get an idea of the lift force, you can try to imagine the air pressure on each part of the wing. The difference gives the direction of lift "downstream". I'm just saying this to give a way to estimate lift intuitively on more complex shapes.

15. The Bernoulli explanation and the change in air momentum explanation are equivalent, but incomplete. Neither explain the key element, which is why the air leaves the trailing edge moving downwards. The "air hits the lower surface wing and bounces down" thing only works at very low air densities where the mean free path of the air molecules approaches the wing dimensions.

If you try to calculate a normal density low-speed airflow round an aerofoil ignoring viscosity, you'll find that the flow on the lower surface will go around the trailing edge to meet the upper surface flow and leave the wing at a point on the upper surface forward of the trailing edge, with no lift or drag generated. To get a reasonable approximation to the real flow, you have to artificially add a restraint, known as the Kutta condition, that the flow leaves at the trailing edge.

The Kutta condition is an effect of viscosity. Viscosity slows the air closest to the aerofoil surface, producing what is known as a boundary layer. The air in the boundary layer has less energy and if the external flow is slowing down the boundary layer can can run out of kinetic energy and come to a halt. If this happens the external flow no longer follows the direction of the surface and is said to separate. Aft of the separation point the boundary layer air is moving in a direction opposite to the external flow. Depending on the shape of the surface, the flow can remain separated or re-attach to give a "bubble" of separated flow. The acceleration then deceleration round a sharp corner is practically guaranteed to produce separation and this is what causes the flow condition at the sharp trailing edge of an aerofoil.

Stalling happens when separation starts upstream of the trailing edge, typically at high incidences where the flow on the upper surface experiences a long region aft of the leading edge where it is slowing down.

16. ## Re: Something i've always wondered about jetfighters.

It's even more interesting when two guys are holding a gate!

17. The Bernoulli explanation and the change in air momentum explanation are equivalent, but incomplete. Neither explain the key element, which is why the air leaves the trailing edge moving downwards. The "air hits the lower surface wing and bounces down" thing only works at very low air densities where the mean free path of the air molecules approaches the wing dimensions.
It is true that momentum and Bernouilli are equivalent and ignore viscosity. What is unclear from your explanation in my opinion, is that the "air hits lower surface wing and bounces down" theory, also known as the "skipping stone" theory, is extremely incomplete, much worse than Bernouilli and momentum. That is because this theory -unlike Bernouilli and momentum- ignores the upper surface of the wing.

If you try to calculate a normal density low-speed airflow round an aerofoil ignoring viscosity, you'll find that the flow on the lower surface will go around the trailing edge to meet the upper surface flow and leave the wing at a point on the upper surface forward of the trailing edge, with no lift or drag generated. To get a reasonable approximation to the real flow, you have to artificially add a restraint, known as the Kutta condition, that the flow leaves at the trailing edge.
I think that you are mixing things up a bit. Viscosity indeed is needed to predict separation. Using Bernouilli (or momentum) Without viscosity, a ball in an airflow would have no drag. the famous paradox of aerodynamics .

The flow calculations using Bernoulli assume inviscid flow. In order to include viscous effects but stay within the inviscid calculation model, the Kutta condition is applied. Basically the Kutta condition is the extra equation needed to make the viscous problem fully defined in the inviscid scheme (single solution). Using the Kutta condition, separation effects can be calculated, so suddenly our ball does generate drag. And separation will be calculated for wings at (relatively) high angle of attack, giving a subtle change in lift compared to the 'no separation' simple Bernouilli approach, but mainly a much more precise drag calculation.

Without the Kutta condition, a symmetrical wing (wrt the freestream flow, so no angle of attack) will generate no drag. And no lift, which is logical for a symmetric airfoil at zero angle of attack. In reality, it will generate indeed no lift, but it will generate drag due to the viscous effects. HOWEVER, using only inviscid Bernoulli calculations, one can calculate lift and drag! Your claim that no lift or drag are calculated is wrong. What is true, is that this method makes a small error in lift and a large error in drag, due to not taking separation effects into account. The Bernouilli theory on airfoils does predict lift (and drag) through its predicted pressure difference on top and bottom (and front versus back). For laminar flow, Bernouilli can predict lift very accurately.

Bernouilli without the Kutta Condition does not predict the downwash, but -and this is very important- for laminar flow, the effects of the flow on the wing itself are predicted by bernouilli, so lift is predicted. I have difficulties in formulating this. So Bernouilli does not predict what the air will do behind the wing, but it does predict what the effect of it is on the wing, and hence on the lift. All this excluding separation effects though.

********
There is a lengthy article on the internet talking about "Bernouilli lift" and "Reaction Lift" as sources of lift for wings. In my opinion, the pressure increase due to the stagnation of deflecting air in what the author calls "reaction lift" is pefectly explained by Bernouilli's formula, so I do not agree that both are different. The author attributes Bernouilli lift only to wing shape and not angle of attack, which is incorrect in my opinion as changing the angle of attack can be seen as changing the wing shape wrt the flow, again an argument why both are the same. Many sources distinguish Bernouilli lift and and Reaction LIft, but as I just argued I think this distinction is not necessary. Obviously it is interesting to know the lift of an airfoil at zero angle of attack and its change due to angle of attack, but again I argue that the whole Cl/alpha curve can be explained by Bernouilli lift (without viscosities of course), without having to separate between Bernouilli lift and Reaction Lift.

btw the author also uses the wrong explanation of travel length (as if 2 air molecules are married and therefore HAVE to meet up together at the trailing edge...)
*********

That said, I agree with you gwiz that you need viscosity to explain lift and certainly drag in a more complete fashion than the less complete Bernouilly or equivalent momentum explanations.

18. Originally Posted by Nicolas
I think that you are mixing things up a bit. Viscosity indeed is needed to predict separation. Using Bernouilli (or momentum) Without viscosity, a ball in an airflow would have no drag. the famous paradox of aerodynamics .

The flow calculations using Bernoulli assume inviscid flow. In order to include viscous effects but stay within the inviscid calculation model, the Kutta condition is applied. Basically the Kutta condition is the extra equation needed to make the viscous problem fully defined in the inviscid scheme (single solution). Using the Kutta condition, separation effects can be calculated, so suddenly our ball does generate drag. And separation will be calculated for wings at (relatively) high angle of attack, giving a subtle change in lift compared to the 'no separation' simple Bernouilli approach, but mainly a much more precise drag calculation.
On the contrary, I think you are confusing the Kutta condition with general separation.

For inviscid flow around a ball, you can calculate a lifting situation by specifying a point for the flow to leave the aft side of the ball that is off the axis of the flow direction. This approximates the well-known sports phenomenon that spin makes a ball swerve. The Kutta condition applies the same technique to an aerofoil, and specifies that the flow leaves at the sharp trailing edge.

In viscous flow a ball generates drag even if it is not spinning, because the flow on all sides separates upstream of where it would in inviscid flow, there is no longer a common point where all the flow leaves.

So we have the following possibilities:

True inviscid flow, no lift, no drag.

Inviscid flow with leaving point specified: lift, no drag.

Inviscid flow with separation point upstream of trailing edge specified: lift and drag

Viscous flow on symmetric shape: drag, no lift.

Viscous flow on asymmetric shape: lift and drag.

Viscous flow with separation upstream of trailing edge: more drag, less lift.

Of course, at transonic and supersonic speeds you can also get drag in inviscid flow, but that's another matter.

19. Indeed, let's stick to low speed subsonic here .

The problem I have with your list:

Simple Bernouilli models fall under inviscid flow with no leaving point specified, right? Because for a leaving point, you need another condition. But simple Bernouilli predicts lift as far as I remember! (due to the pressure profile it predicts). Or am I forgetting that in those cases a leaving point is specified? I thought that in Anderson's Introduction to Flight, lift calculations were made already before going into the Kutta condition.

Can you help me out on that?

Edit: I think that in calculating Bernouilli's equation over the surface of the wing, you implicitly apply the Kutta condition. At least the effect is the same: the streamline is assumed to leave at the trailing edge. That would explain calculating lift with Bernouilli already before learning about Kutta itself. Assuming pressure perpendicular to the local surface, the resulting pressure vector pattern does predict a drag on aerodynamic wing profiles that way. It may be clear that this drag calculation is far from complete .

**********
I was confusing the Kutta Condition with separation indeed.

20. By specifying a leaving point, you are specifying a circulation about your aerofoil. In its simplest form, you have a uniform flow perturbed by a vortex with its axis normal to the flow. The lift is proportional to this circulation.

If you do not specify a leaving point, the leaving point that appears is that corresponding to zero circulation and hence no lift. For an asymmetic shape, this means that the pressure forces given by Bernoulli cancel out and there is no net force.

In inviscid flow with a Kutta condition, the low pressure on the forward-facing leading edge of the airfoil cancels the drag due to the upper/lower surface pressure difference.

21. In inviscid flow with a Kutta condition, the low pressure on the leading edge of the airfoil cancels the drag due to the upper/lower surface pressure difference.
I do not understand this. Why a low pressure on the leading edge? There's a stagnation point overthere, so I would assume a high pressure.

Furthermore how do you arrive at a horizontal (=parallel to the freestream flow) drag force due to a vertical (upper/lower) pressure difference?
************

If you do not specify a leaving point, the leaving point that appears is that corresponding to zero circulation and hence no lift. For an asymmetic shape, this means that the pressure forces given by Bernoulli cancel out and there is no net force.
But the way I explained it, by intrinically assuming the streamline follows the wing surface, is that indeed assuming the Kutta condition?

22. Yes, having flow following the surface means it can only separate at the sharp trailing edge, so you've got a Kutta condition.

For the drag force, I was thinking just of lift on an aerofoil at incidence where the upper/lower surface pressure difference normal to the aerofoil has an aftwards component. Camber makes things more complicated.

Lift = Normal force x cos(incidence)

Drag = Normal force x sin(incidence)

The forward stagnation point is only at the leading edge at zero incidence. As incidence increases, the stagnation point moves round to the lower surface. The "upper surface" flow initially goes backwards along the lower surface from the stagnation point then round the leading edge to the upper surface.

23. OK now I understand that sentence; I thought you were talking about zero incidence and had mistyped you or something like that .

Thanks for helping me out on the Kutta condition thing, now I know how come one can calculate lift using Bernouilli without consciously using Kutta: it's implied by equating the surface to a streamline.

It never hurts to be refreshed on things, thanks .

I also hope it is clear now to others that momentum and Bernouilli explanations are equivalent.

Well, unless you split into "reaction lift" and "bernouilli lift" but I've already given my opinion on splitting equivalent things...

25. Originally Posted by Nicolas
I also hope it is clear now to others that momentum and Bernouilli explanations are equivalent.

Well, unless you split into "reaction lift" and "bernouilli lift" but I've already given my opinion on splitting equivalent things...
Quite, you can split lift into an incidence-produced component and a zero-incidence/camber component if you like, but in both cases you can measure the lift by either integrating the pressure distribution (Bernoulli) or by measuring the change in momentun of the air (reaction).

26. Originally Posted by captain swoop
If I was trying to explain it to you in person, I'd be drawing lots of little diagrams, but in a forum it's too much effort to do this.

27. Quite, you can split lift into an incidence-produced component and a zero-incidence/camber component if you like, but in both cases you can measure the lift by either integrating the pressure distribution (Bernoulli) or by measuring the change in momentun of the air (reaction).
That's exactly my stance on the subject .

28. Originally Posted by captain swoop
You can correct errors and make new ones in the same post. And by the time those are corrected, there's two things to be (re)learned .

And completely in the spirit of this board, a website has been debunked along the way .

Lovely board when used properly.

29. The OP deals with leveled (though inverted) flight. How do pilots keep altitude when the bank angle is 90 deg? By compensating with the rudder, having it act as an elevator? I do it with my models and it works fine.

30. Originally Posted by Argos
The OP deals with leveled (though inverted) flight. How do pilots keep altitude when the bank angle is 90 deg? By compensating with the rudder, having it act as an elevator? I do it with my models and it works fine.
If your speed's high enough, you may be able to get enough lift from the fuselage and fin, otherwise my advice is not to stay there too long.

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