How do airplanes fly with heavy weight and air resistance?

[russ_watters]

I should have expected that! But I am sure you could design a rocket based on firing ball bearings out of the back which would not rely on aerodynamics to provide the ejecta with momentum.

True. That's basically what an ion engine does.

 

[Drakkith]

This thread is over my head. You need to tell me to get lost again.



We've gone from airfoils, to flat plates, to planing hull boats, to hydrofoil boats, to dirigibles, and now to rockets, and I'm afraid this thread will next delve into how Bumblebees fly.

Please Drak, you are my only hope...

Om...I need you turn around and leave the thread immediately. Exits are located behind you under the flashing sign with the siren going off. There will be a cabby waiting for you outside who will take you to the airport. There's a private jet waiting on the runway that will take you to a tropical island with no internet. You shall remain there until this thread is done.

 

[sophiecentaur]

A gain and a loss are the same thing with opposite signs. They involve exactly the same change to the equation (addition of an energy change term). If the energy change is a friction loss or a pump, the same new equation is used (in the form KE1+PE1=KE2+PE2+E). You can even go a step further and apply it to a heating or cooling coil (where there is now a change in volume of air caused by the change in energy).
Well, that's that overly pedantic thing again. If you derive a new equation from Bernoulli's equation, Bernoulli still helped you -- you needed Bernoulli to get the new equation! Huh? Since you can't generate the new equation without using Bernoulli's original equation and everything about it is the same except the one additional term, it most certainly does help!

I have a feeling that the quantity at issue with flight and lift is more Momentum than KE. That would tie in with the idea of efficiency and deflecting as much air as possible and at a slow velocity. On the runway, the plane needs no energy to maintain its lift because the tarmac goes nowhere.

 

[russ_watters]

I have a feeling that the quantity at issue with flight and lift is more Momentum than KE. That would tie in with the idea of efficiency and deflecting as much air as possible and at a slow velocity. On the runway, the plane needs no energy to maintain its lift because the tarmac goes nowhere.

Huh? You're not suggesting if you shut off the plane's engine but don't apply the brakes it would keep rolling forever (or would only stop due to friction in the wheels), are you? An input of energy is absolutely required to maintain the velocity and therefore lift!

Regardless, this is part of that point noted before: the one-or-the-other idea is what is most wrong here. The fact that you can use Newton's 3rd law doesn't mean you can't also use Bernoulli's principle. Which you choose depends on the information available and personal preference.

 

[sophiecentaur]

Huh? You're not suggesting if you shut off the plane's engine but don't apply the brakes it would keep rolling forever (or would only stop due to friction in the wheels), are you? An input of energy is absolutely required to maintain the velocity and therefore lift!

Regardless, this is part of that point noted before: the one-or-the-other idea is what is most wrong here. The fact that you can use Newton's 3rd law doesn't mean you can't also use Bernoulli's principle. Which you choose depends on the information available and personal preference.

I wouldn't disagree and it's not an either/or. But the lift is surely due to momentum change and the support from the runway is not due to any friction. Better wheel bearings and tyres are the
equivalent of an efficient wing with a good lift / drag ratio. The more lift that can be generated without turbulence (just air deflection), the more efficient the wing.

 

[russ_watters]

I'm not following then how you can say the plane needs no energy to maintain its lift. What the wings do doesn't fundamentally change when the ground gets in the way.

 

[sophiecentaur]

I'm not following then how you can say the plane needs no energy to maintain its lift then. What the wings do doesn't fundamentally change when the ground gets in the way.

I never said that. I said that a certain rate of momentum transfer is required (i.e. lift force). This, of course - as with any aircraft - involves the expenditure of energy. The energy needed will depend upon aerodynamic design, the momentum transfer just depends upon the weight. Connecting those two together is the Efficiency of how well you can deflect enough air at the appropriate velocity.

AS for what happens on the ground, the wings do nothing when stationary and gradually 'do' more as the plane speeds up. But the lift, when near the ground can also be caused by pressure rise, which is a less lossy source of lift. Ground effect is good value for low speed air travel, I have heard.

 

[boneh3ad]

Really? A plane can fly in a purely turbulent state? Perhaps I don't know what turbulence is then.

Yes, it absolutely can and they often do. Based on this statement I suspect you don't actually understand turbulence. Many people think about the airline pilot coming on over the intercom and talking about buckling your seat belts because they are heading into a turbulent patch of air. That is a very narrow scope of what turbulence is, though. Really, it is just chaotic motion of air characterized be increasingly small eddies in a flow with some energy dissipation mechanism. A simple experiment to show laminar and turbulent flow is to light a match and blow it out and watch the smoke rise. The smoke is initially smooth as it first comes off the match head and then starts to twist and turn and mix itself around until you can't see it anymore. That is an example of the transition to turbulent flow. You can see a similar application when you just barely crack your faucet on and see the smooth stream of water coming out and then open it some more and watch it turn "bubbly".

It is usually discussed in the context of a fluid boundary layer. In the case of fastening your seat belts, it is usually often the large turbulent structures associated with the atmospheric boundary layer that cause the plane to buffet. These large eddies are much larger in scale than the wings of the plane though, so the plane will fly just fine through them as long as the pilot is competent.

When talking about flow over the wing, however, far more relevant to its function is the turbulence in the wing boundary layer, which on most modern planes, is turbulent nearly 100% of the time.

Exactly. I wish people would see that this is very relevant and stop trying to adopt one stance or another.
There's a great example of the relevance of this when you compare a displacement boat with a planing boat of the same weight. They are both being supported by the water but the planing boat produces a much smaller wash because the amount of water it needs to displace is spread out over many boat lengths.

I have a feeling that the quantity at issue with flight and lift is more Momentum than KE. That would tie in with the idea of efficiency and deflecting as much air as possible and at a slow velocity. On the runway, the plane needs no energy to maintain its lift because the tarmac goes nowhere.

I wouldn't disagree and it's not an either/or. But the lift is surely due to momentum change and the support from the runway is not due to any friction. Better wheel bearings and tyres are the
equivalent of an efficient wing with a good lift / drag ratio. The more lift that can be generated without turbulence (just air deflection), the more efficient the wing.

So let's talk about energy here for a moment. Think about the force on an airfoil in level flight and the implications for energy. An airfoil moving through the air is going to have two major forces on it: lift and drag. Drag retards the motion, so there is certainly work being done on the airfoil in that case since it is the drag force applied over some distance in the same direction as the drag force. However, lift is orthogonal to the direction of motion, so there is no work being done by the lift. In other words, you can get a complete picture of lift in level flight without considering the energy. It is only when trying to get accurate drag results that energy becomes a factor, and that is the reason why in the early days of fluid mechanics drag was so much more difficult to predict theoretically, and even is still today.

Of course, you can't have lift without drag, so there is still some role of energy there, but you can predict lift without considering it. Momentum is the more important quantity for determining these things since really change in momentum constitutes a force, which is our end goal in the first place.

Agreed. Most of what they say is "wrong" is at worst incomplete, a simplification or not completely applicable. Which of course means that by saying it is wrong, they are oversimplifying! One of the more annoying is their criticism of the applicability of the Venturi tube concept. A wing is basically an inside-out Venturi tube. They criticize the analogy essentially for being an analogy (paraphrase): a wing isn't a Venturi tube because Venturi tubes aren't inside out.

ORLY? Then why is NASA developing an inside-out rocket engine?:

http://en.wikipedia.org/wiki/Aerospike_engine

http://www.nasa.gov/centers/marshall/news/background/facts/aerospike.html

I disagree completely. If one part of a theory is wrong, no matter how right other parts are, then the theory is wrong. The NASA sites do a good job of taking those wrong theories and explaining which parts are correct and then discussing how to make the theory fully correct on the two pages about how to actually explain lift.

And no, an airfoil is not an inverted Venturi tube. Venturi tubes rely on the area change through the duct to accelerate the flow due to conservation of mass. This is not how an airfoil works. In a Venturi tube, there are solid walls which constrict the flow, requiring more velocity in order to pass all the mass coming into the section of the duct where the area is lower. This is not true of an airfoil, which is an open flow that can easily deflect around the airfoil out to infinity if necessary. There is nothing constraining the flow to see it as a smaller area.

With an aerospike engine, you have a couple fundamental differences. First and foremost, you have the fact that this is a compressible flow where the exhaust is coming out of the nozzle supersonically and with a vastly different pressure from the ambient air, so you have what is called a slip line between the exhaust jet and the ambient air. Those work a lot better as a virtual centerline than do random streamlines out at infinity. Even then, this is simplified because the slip line will not be straight and the Venturi effect does not actually hold for supersonic flows anyway, as the relationship between area change and flow rate changes completely.

Second, trying to explain an aerospike engine on a page aimed at the general public in its full physical glory would be fruitless for all but those with a background in compressible gas dynamics. As such, those landing pages are certainly watered down a bit and describe the engines in a way that makes it easy to visualize but leaves out the finer details.

Is the air under the boat supported on skyhooks? It surely rests on the water. Of course the pressure may be low but you can hardly argue it's not there.
We seem to have the big and little endians at work in this thread. "My theory, right or wrong".

Of course not,but what has that got to do with the fact that the supporting force is spread over a bigger area when planing? With a purely displacement boat, it's Archimedes at work and no hydro / aerodynamics. When it's planing, just because there's some fluid flow involved doesn't mean there is not enough force acting on the water to keep the boat from sinking.

I can't imagine another topic that would have people implying that reactionless forces actually exist.

Describing planing in terms of spreading the force out is a bit fishy in my view. Hydrodynamically speaking, it has to do with the boat moving much faster so the momentum it is changing in deflecting the water downward is going to be much greater per contact area of the boat than at a low speed, so that will tend to lift the boat up until it comes to an equilibrium where less area of the boat is in contact with the water.

True, the air pressure absolutely would play some role to hold the boat up while planing. It would almost certainly be orders of magnitude less than the role the water plays, however. The main effect it would have is if the nose comes up too much, allowing the effect of the air to grow enough to generate appreciable lift, and then you see those crashes.

 

[sophiecentaur]

So let's talk about energy here for a moment. Think about the force on an airfoil in level flight and the implications for energy. An airfoil moving through the air is going to have two major forces on it: lift and drag. Drag retards the motion, so there is certainly work being done on the airfoil in that case since it is the drag force applied over some distance in the same direction as the drag force. However, lift is orthogonal to the direction of motion, so there is no work being done by the lift. In other words, you can get a complete picture of lift in level flight without considering the energy. It is only when trying to get accurate drag results that energy becomes a factor, and that is the reason why in the early days of fluid mechanics drag was so much more difficult to predict theoretically, and even is still today.

Just a sample from your long post and I have to reply to this. Just because no work is done ON the plane by the lift force that doesn't mean that no work is done ON the Air that is deflected downward*. You must agree that the lift force must be a reaction against some force and that force is due to the the constant momentum transfer of a finite amount of air moving downwards. This is precisely the same situation as with a rocket on a launch pad, before it starts to lift and with a hovering helicopter. It may be possible to 'ignore' this downward deflection of air in calculating lift from Bernouli but, as we've already discussed, Bernouli is only part of the explanation.

*This is a bit like to old chestnut about why our arm gets tired when holding a heavy book, even when we don't actually raise it. No work on the book doesn't imply no work done at all.

Your comments about planing boats are much along the same lines and, taken at face value, they again imply the existence of a reactionless force, keeping the boat up there. Bottom line is that a force has to come from somewhere.

 

[boneh3ad]

I'm not saying there is no energy expended period to generate lift. Clearly there has to be some energy used, for example, to keep the plane moving. What I am saying is that I don't see where there is any being expended by the air in the generation of lift (or vice versa if you switch the frame of reference). It seems to me that with lift, you have only conservative forces and no work being done on the wing by the lift force. With drag you have non-conservative forces at play and so that is where any expended energy comes into play.

Perhaps that is semantics since you can't have lift without drag so there are definitely energy changes when lift is being generated. My point was meant to be that this energy consideration is concerned with predicting the drag portion of this double-edged sword specifically even though you can't have one without the other.

Regarding the boat, I never said anything about a reactionless force. The boat pushes water down and gets held up as a reaction. That should cover both the action and reaction.

 

[boneh3ad]

I'm going to amend one thing I said here. After thinking a bit further, under the action of viscosity, some of the energy in the boundary layer will be dissipated, particularly if it is turbulent. That will eventually translate to a small but finite loss of momentum compared to the inviscid case in the vertical direction and a much larger effect horizontally, so in essence there is a vertical non-conservative force there and it would allow for an energy dissipation/expenditure associated specifically with the lift portion of the force. It would be small and in many (most?) cases negligible but it would definitely exist. Also, Bernoulli's equation would not cover it without significant modification since it is an inherently viscous phenomenon.

That said, people can and do get very accurate lift prediction from inviscid solvers. The dissipation phenomenon is much stronger in the stream wise direction, which is one of several reasons why you can get a great estimate of lift from inviscid theory but not drag (e.g. in XFOIL).

 

[sophiecentaur]

I'm not saying there is no energy expended period to generate lift. Clearly there has to be some energy used, for example, to keep the plane moving. What I am saying is that I don't see where there is any being expended by the air in the generation of lift (or vice versa if you switch the frame of reference). It seems to me that with lift, you have only conservative forces and no work being done on the wing by the lift force. With drag you have non-conservative forces at play and so that is where any expended energy comes into play.

Perhaps that is semantics since you can't have lift without drag so there are definitely energy changes when lift is being generated. My point was meant to be that this energy consideration is concerned with predicting the drag portion of this double-edged sword specifically even though you can't have one without the other.

Regarding the boat, I never said anything about a reactionless force. The boat pushes water down and gets held up as a reaction. That should cover both the action and reaction.

At first sight, it may seem ok to talk in terms of "frames of reference" regarding 'work done' but is that necessarily valid? Or at least, you have to be more specific about how you define both of the frames. It's a bit hard to define a frame for a distributed mass of air which is moving at different speeds. From the Earth's frame, it is clear that the plane doesn't move vertically but some air does move vertically. From the plane's frame, the Earth doesn't move vertically but some air does. That's as far as I could go. Can you suggest how to do a similar thing with the air?

OK then, I didn't really think you would mean reactionless but which water is pushed down? It is not the same volume of water that corresponds to displacement because only the back wedge of hull is under the water and the boat lifts by over 50% of its submerged depth. The lift force must be coming partly from some air, squashed between the hull and the water and the reaction from water that is being forced downward by the wedge. The prop is low in the water and maintains the nose up, to some extent - plus all power boats are stern heavy with a motor and helmsman at the back. But the fact is that, as soon as the plane starts, the bow wave diminishes (almost vanishes on some boats) and that means less volume of water is actually displaced. So there must be a continual volume water deflected downwards and then left behind over a large area of sea. That was the 'spreading' I referred to; every second, you need to be ejecting / deflecting downwards mass m(of water) at v(down) to provide mv of force to balance Mg where M is boat mass. I reckon m would correspond to some portion of the vertical cross sectional area of submerged part of hull times boat speed (i.e. a sort of shallow virtual trough, left behind.

 

[russ_watters]

boneh3ad, I have a lot more comments, but one basic one: why do you think Bernoulli's implies energy must be expended to generate lift? The point of Bernoulli's is that energy is conserved but can be transferred from one form to another. In this case, from velocity to static pressure.

I think perhaps what you are missing (about Newtons model too) is that after the plane passes, if there is no drag, the air returns to its previous state. It does NOT get pushed to infinity because it IS in a closed container.

 

[Bandarigoda]

I was taught it's according to Bernoullis Principle

 

[cjl]

I'd like to know how you define a portion of lift as coming from an upper or lower surface. I have never run across that statement except here on this forum. To be honest, I find that statement to be incredibly misleading in every logical way I can see it making sense.

It's somewhat misleading, I agree, but I would tend to use it based on the pressure coefficient, and (specifically) whether the integral of the pressure coefficient along the upper or lower surface is of greater magnitude. You can't really look at either surface in isolation, of course, and it is somewhat oversimplifying, but I wouldn't call it completely wrong.

(And yes, I know I'm going back several pages on this reply - I haven't checked this thread since Friday, so I'm still catching up...)

 

[boneh3ad]

At first sight, it may seem ok to talk in terms of "frames of reference" regarding 'work done' but is that necessarily valid? Or at least, you have to be more specific about how you define both of the frames. It's a bit hard to define a frame for a distributed mass of air which is moving at different speeds. From the Earth's frame, it is clear that the plane doesn't move vertically but some air does move vertically. From the plane's frame, the Earth doesn't move vertically but some air does. That's as far as I could go. Can you suggest how to do a similar thing with the air?

Essentially in talking about the frame of the air, it is really just saying the frame of the Earth. Typically, fluid mechanicians treat problems in one of two frames: fixed to the body (in this case a plane or wing) or fixed to the fluid (generally taken to mean the undisturbed free stream, not some individual parcel, and is essentially equivalent to the Earth's frame in this case).

OK then, I didn't really think you would mean reactionless but which water is pushed down? It is not the same volume of water that corresponds to displacement because only the back wedge of hull is under the water and the boat lifts by over 50% of its submerged depth. The lift force must be coming partly from some air, squashed between the hull and the water and the reaction from water that is being forced downward by the wedge. The prop is low in the water and maintains the nose up, to some extent - plus all power boats are stern heavy with a motor and helmsman at the back. But the fact is that, as soon as the plane starts, the bow wave diminishes (almost vanishes on some boats) and that means less volume of water is actually displaced. So there must be a continual volume water deflected downwards and then left behind over a large area of sea. That was the 'spreading' I referred to; every second, you need to be ejecting / deflecting downwards mass m(of water) at v(down) to provide mv of force to balance Mg where M is boat mass. I reckon m would correspond to some portion of the vertical cross sectional area of submerged part of hull times boat speed (i.e. a sort of shallow virtual trough, left behind.

The boat could plane (nearly) as well in a vacuum as postulated before (assuming the water doesn't flash evaporate, of course). Consider two cases here. In one, the boat is at rest and displaces the full amount of water as predicted by Archimedes principle since the only force holding it up is the buoyant force. Now, as the boat speeds up, you introduce a second force: the hydrodynamic force of the bottom which is essentially a lift force. As it speeds up, this force gets greater. If the boat stayed equally submerged, it would mean there is a force imbalance somewhere since you would have the full buoyancy combined now with a lift force, so naturally, as you increase speed, the boat will lift out of the water, lowering both the buoyant force and (to a lesser extent) the lift force from the water until the forces are in balance again.

So you are getting lift from the water and buoyancy from the water. The lift generated by the air will tend to be quite small compared to that from the water for most reasonable speeds. Once you get fast enough and the air wets substantially more of the surface than the water I imagine there is a point where the effect of the air approaches the same order of magnitude as the effect of the water, and at that point is probably when you are in danger of the kind of spectacular crashes you see. Also, without the effect of air the limit to how high the boat can come off the water would be lower since there would be no added effect of the air.

That fits neatly into the fact that less water is displaced and you have less of a wake because the boat simply isn't pushing aside as much water and fits neatly into your description of what is happening without requiring the air to have a major effect.

 

[boneh3ad]

boneh3ad, I have a lot more comments, but one basic one: why do you think Bernoulli's implies energy must be expended to generate lift? The point of Bernoulli's is that energy is conserved but can be transferred from one form to another. In this case, from velocity to static pressure.

I think perhaps what you are missing (about Newtons model too) is that after the plane passes, if there is no drag, the air returns to its previous state. It does NOT get pushed to infinity because it IS in a closed container.

I didn't ever say that Bernoulli's equation implies that energy must be expended. In fact, the very fact that you can get a very, very accurate measure of lift straight from Bernoulli's equation implies that little if any energy is dissipated in the generation of lift since the equation assumes none of that occurs in the first place. If energy isn't dissipated, then Bernoulli's equation can handle the situation without any added energy terms (assuming the other assumptions are met, of course).

And I am not missing the fact that without drag the air returns to the previous state. It absolutely is not in a closed container, though, and the air has what is essentially an infinite distance away from the surface in which to relax itself and pass any mass flow requirement through. Using the Venturi idea, you could take literally any free-stream streamline as your "wall" at any distance and have an equally Venturi-looking shape but predict wildly different velocities as a result. If the Venturi analogy worked, then you wouldn't be able to pick an arbitrary streamsurface and get a different answer each time. Consider the following situation:

You have a foil in water and want to look at it in terms of the Venturi theory. Let's say you have incoming flow of 10 m/s, your foil's half-thickness is 0.2 m and you want to arbitrarily choose the streamsurface 1 m above your foil. The incoming mass flow into that control volume is 10000 kg s^-1m^-1 per width of your foil. Divide that by the constricted area over your foil (0.8 m x width) and you come up with 12.5 m/s over that surface. Now do the same thing but arbitrarily choose the surface 2 m above your foil. Now your answer becomes 11.11 m/s. Clearly this doesn't happen in real life, but if you treated it as a half-Venturi, that is exactly what would happen.

It's somewhat misleading, I agree, but I would tend to use it based on the pressure coefficient, and (specifically) whether the integral of the pressure coefficient along the upper or lower surface is of greater magnitude. You can't really look at either surface in isolation, of course, and it is somewhat oversimplifying, but I wouldn't call it completely wrong.

Yeah I eventually justified to myself it looking at the momentum on either side of the slip line emanating from the trailing edge. I think my problem with that statement is simply that far too many people take it to mean that the top side is essentially sucked up harder than the bottom is pushed up, which makes no real sense.

 

[russ_watters]

If the air is pushed to infinity, it is not returning to its initial state and the implications of the two are very different: if it escapes to infinity, it keeps the momentum and energy (thus the plane must lose energy to generate lift) but if it is constrained by its container, both the energy and momentum are returned when it returns to its previous state.

By the way, that's what differentiates wave riding from normal flying: the expending of energy to generate lift. That's why a hydrofoil is much more efficient than a planing hull.

 

[boneh3ad]

No, it isn't pushed to infinity but it has that if it needed it. The point is that the Venturi effect relies on the flow being throttled by some sort of area change. That is not the case for a wing.

 

[rcgldr]

On the runway, the plane needs no energy to maintain its lift because the tarmac goes nowhere.

Since the air can't continue to flow downwards aft of the trailing edge of a wing it ends up flowing forwards and backwards (and outwards at the wing tips). The forward flow that starts under the trailing edge and continues forward under the wing reduces drag, which is partly why ground effect flight takes less power. As mentioned, there's also a pressure increase below the wing, due to the impact of the downwash with the tarmac, which reduces the angle of attack required to keep the aircraft in ground effect, further reducing drag and the power required to maintain ground effect flight.

Lift is orthogonal to the direction of motion, so there is no work being done by the lift.

In order to produce lift, a wing has to have an angle of attack with respect to the direction of travel of the air craft. Most of this could be due to the cambered upper surface of a wing. The point here is that there is a vertical component of distance along the lift producing surfaces of a wing. For all normal wings, the direction of the "exit velocity" is mostly downwards (downwash related to lift) and somewhat forwards (forwards wash related to drag).

The situation is somewhat like a propeller, except that a propeller operates within it's on induced intake wash, and the amount of energy added to the air is much greater. From this Nasa article:

We can apply Bernoulli's equation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed by the engine (propeller) violates an assumption used to derive the equation.


http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

Since a wing does not operate in it's on induced intake wash (like a helicopter in forward flight as opposed to hovering), there's some increase in downwash in addition to the pressure jump that occurs with a propeller.

you can get a great estimate of lift from inviscid theory but not drag (e.g. in XFOIL).

I had the impression that XFOIL uses a simplified form of Navier Stokes equations and goes well beyond Bernoulli. Also, doesn't XFOIL produce fairly accurate "polars" for both lift and drag?

 

[rcgldr]

frame of reference.

The frame of reference affects the amount of energy added to the air. For an ideal or perfect wing, using the wing as a frame of reference, this ideal wing diverts the relative air flow by only changing direction and not speed, so that there is no change in energy of the affected air.

A frame could also be chosen so that the forward velocity of the frame of reference equals the average exit velocity of the wash from a wing (with respect to this frame), and again no energy is added to the air in this frame of reference.

Getting back to the ideal wing, you can calculate lift and induced drag:

define θ as angle of diversion.

s = speed of air with respect to the wing.
m = mass of the affected air

s sin(θ) = downwards component of diverted air
s cos(θ) = forwards component of diverted air

$${lift} = \dot m \ s \ {sin}(\theta)$$
$${induced \ drag} = \dot m \ s \ (1 - {cos}(\theta)) = {lift} \ \frac {(1 - {cos}(\theta))} {{sin}(\theta)}$$

 

[boneh3ad]

XFOIL is a panel method code with corrections for viscosity an compressibility I believe. It's lift predictions are remarkably accurate but it's drag predictions can be off. Like any other tool, you just have to know its limits.

 

[Aero_UoP]

 

[WaaWaa Waa]

I had long been under the impression that I understood the principles of winged flight. We were taught early in school that the lift generated in the airplanes' wings is due to Bernoulli's Principle.

Last week as I was contemplating sustained inverted flight of jet fighters it occurs to me that this should not be possible according to Bernoulli's lift as the lift would be pointing downwards. So I started googling and arrived at the conclusion that my knowledge about fixed wing airplane flight is practically nil!

The following link to NASA are interactive and rather fun to play with:

http://www.grc.nasa.gov/WWW/k-12/airplane/wrong1.html

You can navigate to other pages of this site which contains other interesting simulations.

 

[cjl]

I had long been under the impression that I understood the principles of winged flight. We were taught early in school that the lift generated in the airplanes' wings is due to Bernoulli's Principle.

Well, in some sense, it is - or at least bernoulli's principle can be used to model the pressure differences around a low-speed subsonic airfoil (less than mach 0.3 or so), if you know the velocity distribution around the airfoil. That's the key though - knowing the velocity distribution. The incorrect (and wildly popular) explanation that the air flowing over the top is going faster because the path is longer (and the parcels of air take an equal time to transit over the top and bottom surface) is the problem.

In reality, the velocity difference is caused by the interaction of the airflow with the wing shape, specifically the formation of a stagnation point at the trailing edge, and this leads to the air over the top surface actually taking less time to transit than the air over the bottom. Interestingly, this means that the popular explanation would substantially underestimate the lift created by an airfoil.

 

[sophiecentaur]

Simulations are fun but they don't prove anything. You need to be careful not to 'believe' them.

 

[cjl]

Simulations are fun but they don't prove anything. You need to be careful not to 'believe' them.

They don't necessarily prove anything, but those NASA simulations do show the correct behavior (at least the ones I've played with), and they do a good job of dispelling many common misconceptions.

 

[russ_watters]

I never said that.

Though I didn't put them in quotes, "the plane needs no energy to maintain its lift", were in fact your exact words.

AS for what happens on the ground, the wings do nothing when stationary...

Oh. I had no idea you were talking about a plane sitting stationary on its wheels. I don't see how that could be relevant here. That isn't lift.

I said that a certain rate of momentum transfer is required (i.e. lift force). This, of course - as with any aircraft - involves the expenditure of energy.

That is only true if you assume the air's new momentum is carried to infinity like boneh3ad suggested. If the air returns to where it started (and it has to), it starts and ends with zero momentum and zero change in energy.

A hypothetical infinite wingspan or a real wing spanning all the way across a wind tunnel produces no lift-induced drag because the air returns to where it started and there is therefore no energy change or permanent momentum change.

 

[russ_watters]

More complete discussion of what I was seeing this morning:

So let's talk about energy here for a moment. Think about the force on an airfoil in level flight and the implications for energy. An airfoil moving through the air is going to have two major forces on it: lift and drag. Drag retards the motion, so there is certainly work being done on the airfoil in that case since it is the drag force applied over some distance in the same direction as the drag force. However, lift is orthogonal to the direction of motion, so there is no work being done by the lift. In other words, you can get a complete picture of lift in level flight without considering the energy.

That's all correct, except for the minor quibble in the last sentence: you don't need to utilize Bernoulli, but you could, so is it really complete without looking at both? If you utilize Bernoulli without Newton and calculate lift correctly, is it still "complete"?

In either case, you don't need an expenditure of energy for a conservation of energy statement to be useful.

Momentum is the more important quantity for determining these things since really change in momentum constitutes a force, which is our end goal in the first place.

Pressure times area is also force. Does that make one "more important" than the other? No. Just two ways of calculating the same thing.

I disagree completely. If one part of a theory is wrong, no matter how right other parts are, then the theory is wrong.

No parts of the theory are wrong, they are just sometimes misstated/misused. Its like when the Newton's 3rd Law method is misstated/misused; that doesn't mean the whole concept of applying Newton's 3rd Law is wrong, it just means people don't understand it. For example, the "equal transit time" misconception is not actually part of the Bernoulli/Venturi explanation of lift, it is an extension/implication that people mistakenly generate from it. Similarly:

In a Venturi tube, there are solid walls which constrict the flow, requiring more velocity in order to pass all the mass coming into the section of the duct where the area is lower. This is not true of an airfoil, which is an open flow that can easily deflect around the airfoil out to infinity if necessary. There is nothing constraining the flow to see it as a smaller area.

This is one of the wrong implications people who get too hung up on Newton's 3rd law take away from it and one of the reasons why it is a bad idea to focus on it and ignore Bernoulli. The Newton's 3rd law explanation deals only with the momentum change of the air mass at the time it is imparted and doesn't say anything at all about what happens after the wing passes - to anything; the mass, volume, presssure, etc. Concluding that the air is unconstrained and keeps that new-found momentum forever/out to infinity is very, very wrong. As I said this morning, if that were true, airplanes flying around the world would gradually be increasing the pressure of the lower atmosphere and decreasing the pressure of the upper atmosphere, in addition to hurling air out of the atmosphere and off into space. That is, of course, silly, right? The air returns to the state it started in because it is constrained. Don't let the fact that it can travel a long way without bouncing back trick you into thinking that that's "infinite".

If we assume no drag, after the wing passes, that air's momentum downard is bounced-back by the surrounding air and the air eventually returns to its original state. The air near the wing is constrained by the air far away from the wing. (by the way, this is what I thought sophie was getting at with the airplane on a runway thought experiment -- that the air can't move away from the wing and that changes something fundamentally.)

This wrong implication doesn't make the use of Newton's 3rd law to lift completely wrong (or wrong at all, when applied correctly) just as the wrong equal transit time implication doesn't make the Bernoulli/Venturi tack wrong.

But that's why I like the Bernoulli/Venturi analysis better: it is a more complete picture of what is happening to the air: it doesn't stop at the wing.

The air is like a bunch of spring-mass systems lined-up next to each other, initially at rest. If you hit one with a hammer every second, you impart some momentum to them. You can then analyze what happens by using the momentum change to calculate force. But you can also use the kinetic energy imparted to calculate force. the fact that one method works doesn't tell us the other doesn't work. At the same time, the fact that you can ignore the spring when using the momentum method should not make you think that the spring isn't there. That's the error being made when drawing conclusions from the simplifying assumption the Newton/momentum method for lift.

Second, trying to explain an aerospike engine on a page aimed at the general public in its full physical glory would be fruitless for all but those with a background in compressible gas dynamics. As such, those landing pages are certainly watered down a bit and describe the engines in a way that makes it easy to visualize but leaves out the finer details.

No doubt, it involves a different region of flow, but that doesn't change the fact that the other side is constrained by the atmosphere.

 

[russ_watters]

No, it isn't pushed to infinity but it has that if it needed it.

You can't have both at the same time. I agree that it isn't pushed to infinity. So why not? Answer: it is constrained not to by the air around it.

The point is that the Venturi effect relies on the flow being throttled by some sort of area change. That is not the case for a wing.

It is indeed the case for the wing because the air around the wing constrains the air and throttles it!

In either case, the nuts and bolts of why the speed of the air increases over the wing isn't what makes the Venturi effect work so well in describing lift: it is the fact that the velocity change can be exactly translated into lift via Bernoulli's equation that makes it fit so well. The Venturi effect is just a simplified demonstration of Bernoulli's principle. Just because it isn't throttled in exactly the same way as in a Venturi tube, that doesn't mean it isn't being throttled or that the Bernoulli effect doesn't apply.

Try this thought experiment: Take a two-dimensional venturi tube (not circular in cross section) and pull the two sides apart while maintaining the same freestream velocity. The velocity profile will rapidly change, then change more and more slowly. Question: after the sides get very far apart, how does the velocity profile along the walls continue to change and why? Does it:
1. Continue to change in proportion to the area change, approaching equal velocity along the entire wall?
2. Drop to a certain minimum velocity change and stay at that new velocity profile, since the other side is too far away to continue interacting with it?

In any case, a couple of days ago you said:

As CWatters alludes to, this causes a downright comical debate between two warring factions who, in essence, are both correct anyway.

Why have you hardened your position so much since then?

 

[sophiecentaur]

This is all too much to answer at this late hour but I can only say that, to maintain a mass at a given height, no work needs to be done (a book on a shelf demonstrates this). This is what I meant when I wrote that a plane needs no energy to have lift (in princiiple). Providing the force, in the absence of a shelf, requires energy because it is necessary to push air downwards constantly. Doesn't a helicopter do just this? You can certainly feel and see the downwash of a hovering helicopter. Where is the difference in principle? I don't see why Newton 3 is being ignored - just because what happens to the deflected air is a bit nebulous. It is quite reasonable that the complicated way the air flows over the wing is easier to analyse if this downward wash is ignored and (apparently) the Bernouli calculations give a reasonable answer but they don't (according to that NASA link, I think) give the whole story.
The idea of momentum being "bounced back" is just an arm waving argument. What does it bounce against, if not some more air lower down - which will recoil, conserving momentum down there as well? If you look at the air that remains behind an aircraft that has just passed, there is a pair of vortices (horizontal axes) which consist of air going down where they touch and up at the outside. The KE of this, eventually disperses. Nothing "bounces back up"

 

[russ_watters]

This is the part that is wrong:

Providing the force, in the absence of a shelf, requires energy because it is necessary to push air downwards constantly.

Doesn't a helicopter do just this?

Yes, which is why helicopters are much, much less efficient than airplanes.

The idea of momentum being "bounced back" is just an arm waving argument. What does it bounce against, if not some more air lower down - which will recoil, conserving momentum down there as well? If you look at the air that remains behind an aircraft that has just passed, there is a pair of vortices (horizontal axes) which consist of air going down where they touch and up at the outside. The KE of this, eventually disperses. Nothing "bounces back up"

Flip the issue over: If the vortices constantly carry air downwards, why hasn't the distribution of air in the atmosphere permanently changed due to a hundred years of airplane flight and millions of years of birds?

Also (not as important): the vortices actually are a result of drag, not lift. If you span a wing across a wind tunnel, there are no voritces and no lift-induced drag.

 

[rcgldr]

... the air's momentum is carried to infinity

until something stops it. Mechanical energy can be converted into heat, but forces and impulses don't just vanish (they may spread out, but they don't vanish).

Consider a closed system consisting of a sealed container filled with air. The weight of the air is exerted onto the container via a pressure differential, higher at the bottom, lower at the top so that the downforce exerted onto the container by the air equals the weight of the air. Next, add a small aircraft model inside the container, at rest on the bottom of the container. The total weight of the closed system is the sum of the weight of the container, the air inside the container, and the model. Next the model is flying in circles within the container, with no vertical component of acceleration. Again the total weight of the closed system remains the same. The downforce exerted onto the container due to pressure differential now equals the sum of the weight of the air and the small model.

The earth, the atomosphere, and any object supported by the atmoshpere can also be considered a closed system. The force that gravity exerts on an aircraft is transmitted via a continuous impulse generated due to lift through the atmosphere back to the ground where that force is opposed as part of an extended Newton third law pair.

 

[boneh3ad]

More complete discussion of what I was seeing this morning:
That's all correct, except for the minor quibble in the last sentence: you don't need to utilize Bernoulli, but you could, so is it really complete without looking at both? If you utilize Bernoulli without Newton and calculate lift correctly, is it still "complete"?

In either case, you don't need an expenditure of energy for a conservation of energy statement to be useful.

I agree. I have at no point indicated that one measure (Bernoulli or Newton) is any more correct than the other and that it depends on the situation which is more useful. My comments have largely be directed at other statements stating that energy is necessarily expended.

Pressure times area is also force. Does that make one "more important" than the other? No. Just two ways of calculating the same thing. No parts of the theory are wrong, they are just sometimes misstated/misused. Its like when the Newton's 3rd Law method is misstated/misused; that doesn't mean the whole concept of applying Newton's 3rd Law is wrong, it just means people don't understand it.

Right, and along that same line, at no point have I said that using Bernoulli's equation is wrong. What I said is that the way the NASA article explains it in the "incorrect theory" page includes both Bernoulli (correct) and the equal transit time idea (incorrect) and it properly notes which are correct and says that the combination of the two is an incorrect theory. On its own, Bernoulli's equation is not a theory or explanation for the origin of lift, or at least not a complete one.

Concluding that the air is unconstrained and keeps that new-found momentum forever/out to infinity is very, very wrong. As I said this morning, if that were true, airplanes flying around the world would gradually be increasing the pressure of the lower atmosphere and decreasing the pressure of the upper atmosphere, in addition to hurling air out of the atmosphere and off into space. That is, of course, silly, right? The air returns to the state it started in because it is constrained. Don't let the fact that it can travel a long way without bouncing back trick you into thinking that that's "infinite".

You can't have both at the same time. I agree that it isn't pushed to infinity. So why not? Answer: it is constrained not to by the air around it.
It is indeed the case for the wing because the air around the wing constrains the air and throttles it!

In an ideal flow, it would continue out to infinity. The air does not actually do this in real life because real life is not potential flow. It is the action of viscosity that will tend to bring the wake of the plane back toward the conditions of the "undisturbed" free stream. However, even with viscosity the effect of the airfoil is felt to fairly large distances away from the surface because it is unconstrained. This is why, in a wind tunnel, you have to pay close attention to the size of the model, otherwise you end up producing an actual Venturi-like effect instead of the normal flow over the wing.

If we assume no drag, after the wing passes, that air's momentum downard is bounced-back by the surrounding air and the air eventually returns to its original state. The air near the wing is constrained by the air far away from the wing. (by the way, this is what I thought sophie was getting at with the airplane on a runway thought experiment -- that the air can't move away from the wing and that changes something fundamentally.)

If you have no drag, you have no lift and no deflection of streamlines.

This wrong implication doesn't make the use of Newton's 3rd law to lift completely wrong (or wrong at all, when applied correctly) just as the wrong equal transit time implication doesn't make the Bernoulli/Venturi tack wrong.

But that's why I like the Bernoulli/Venturi analysis better: it is a more complete picture of what is happening to the air: it doesn't stop at the wing.

I never said that it made the Bernoulli approach wrong. I said it made the Bernoulli combined with Venturi/equal transit time approach wrong. The equal transit time and Venturi approach are both wrong on their own merits. Bernoulli is completely valid in getting very accurate estimates of lift (particularly if you take into account the displacement thickness), but not so much drag.

The air is like a bunch of spring-mass systems lined-up next to each other, initially at rest. If you hit one with a hammer every second, you impart some momentum to them. You can then analyze what happens by using the momentum change to calculate force. But you can also use the kinetic energy imparted to calculate force. the fact that one method works doesn't tell us the other doesn't work. At the same time, the fact that you can ignore the spring when using the momentum method should not make you think that the spring isn't there. That's the error being made when drawing conclusions from the simplifying assumption the Newton/momentum method for lift.
No doubt, it involves a different region of flow, but that doesn't change the fact that the other side is constrained by the atmosphere.

Whether or not this is true depends on the velocity of the plane traveling through the air (or air traveling over the plane, pick your favorite). Below Mach 0.3, air is incompressible and you won't have any of this spring effect.

In either case, the nuts and bolts of why the speed of the air increases over the wing isn't what makes the Venturi effect work so well in describing lift:

It doesn't do well at all at describing lift. See my previous example.

it is the fact that the velocity change can be exactly translated into lift via Bernoulli's equation that makes it fit so well.

Provided, of course, that you have the correct velocity distribution over the airfoil, which you cannot get from the Venturi effect.

The Venturi effect is just a simplified demonstration of Bernoulli's principle. Just because it isn't throttled in exactly the same way as in a Venturi tube, that doesn't mean it isn't being throttled or that the Bernoulli effect doesn't apply.

Oh, Bernoulli's principle applies to the Venturi effect just as well as it does to a wing. That doesn't mean that the Venturi effect applies to the wing. The phenomena are totally different. The Venturi effect is an inviscid phenomenon based on mass flow considerations. The reason the flow over a wing is faster has to do with the shape of the wing (particularly the trailing edge) and viscosity.

Try this thought experiment: Take a two-dimensional venturi tube (not circular in cross section) and pull the two sides apart while maintaining the same freestream velocity. The velocity profile will rapidly change, then change more and more slowly. Question: after the sides get very far apart, how does the velocity profile along the walls continue to change and why? Does it:
1. Continue to change in proportion to the area change, approaching equal velocity along the entire wall?
2. Drop to a certain minimum velocity change and stay at that new velocity profile, since the other side is too far away to continue interacting with it?

As you do this, the velocity over the constricted portion of the "tube" would approach the value of the inlet velocity asymptotically. Eventually, you wouldn't even be able to measure the Venturi effect over what has now become a bump in the wall because the area change is so infinitesimally small compared to the overall area. At that point, the variations in flow over that bump would be dominated by the effect of the shape of the bump itself.

In any case, a couple of days ago you said: Why have you hardened your position so much since then?

I haven't. Both the Bernoulli and Newton approaches are correct. Bernoulli in particular, however, requires you to know why the velocity over the wing is faster than under it, and that is not correctly described by the Venturi effect.

This is all too much to answer at this late hour but I can only say that, to maintain a mass at a given height, no work needs to be done (a book on a shelf demonstrates this). This is what I meant when I wrote that a plane needs no energy to have lift (in princiiple). Providing the force, in the absence of a shelf, requires energy because it is necessary to push air downwards constantly.

Yes, but consider holding that book up. The shelf is not expending energy. The only reason your arm is expending energy is because your arm is not a rigid structure and must use energy to hold itself rigid. Essentially, your muscles are dissipating energy in order to stay rigid. It is a dissipative phenomenon. A plane, on the other hand, requires the dissipative action drag in order to generate lift. However, this action primarily acts on the horizontal motion, so given the flow field around the wing, it is perfectly reasonable to get very accurate lift estimates from assuming it to be a non-dissipative system. Drag is simply more difficult because you must take that dissipation into account to predict it accurately.

The idea of momentum being "bounced back" is just an arm waving argument. What does it bounce against, if not some more air lower down - which will recoil, conserving momentum down there as well? If you look at the air that remains behind an aircraft that has just passed, there is a pair of vortices (horizontal axes) which consist of air going down where they touch and up at the outside. The KE of this, eventually disperses. Nothing "bounces back up"

This is a very good point, and builds on my previous one. The two wingtip vortices from a typically commercial airliner will remain below the plane for miles. They do not bounce back up, nor do they dissipate until miles behind the plane. Even then, the only reason they dissipate is because of viscosity. Otherwise they would persist all the way back to the airport (presuming you somehow locked that rear stagnation point without the action of viscosity of course).

Flip the issue over: If the vortices constantly carry air downwards, why hasn't the distribution of air in the atmosphere permanently changed due to a hundred years of airplane flight and millions of years of birds?

Viscosity is dissipative, and any energy we add to it does nor persist forever. It will always bring it back toward its equilibrium state.

Also (not as important): the vortices actually are a result of drag, not lift. If you span a wing across a wind tunnel, there are no voritces and no lift-induced drag.

No, they are a result of lift, but are also related to drag. The wingtip vortices form because the wing has to end somewhere. You can look at it a few ways. One, with low pressure on the top and high pressure on the bottom of the wing as necessary for lift, the air on the bottom near the tip will tend to move from high to low pressure, meaning up and around the wing tip, generating a vortex. If the wing generates lift, it will generate this vortex.

The other way to look at it is that in the frame of reference of the stationary wing, the flow is faster over the top than the bottom. If you subtract out the mean flow, you end up with what looks like a vortex traveling around the wing. This vortex, as a result of what is called the Kutta condition, is effectively superimposed over the flow of the wing, and on a finite-span wing, it will get carried away from the wing tip rearward by the free stream. These vortices, if you look at them head on, will come from each wingtip and are counter-rotating in such a way that the flow between them is downward. They are thus intimately related to lift.

In a wind tunnel, these vortices don't form because you are probably looking at studies over 2D wings that span the whole tunnel, so there is no reason for them to form. If you have a large wind tunnel with a scale model of a plane, including the 3D, finite-span wing, you would absolutely have the vortices.

A cool picture of wingtip vortices:

 

[Traz 0]

So, I'm getting a bit lost here, but, I have an additional question. In flight school we were taught that lift is perpendicular to the chord line drawn from the leading edge of the wing to the trailing edge. Which makes sense under the Bournoulli model we were also taught. Is this still considered true? And, if not, how was this missed by aerospace engineers, who must surely rely on such calculations?