How do airplanes fly with heavy weight and air resistance?

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Airplanes achieve flight by generating lift through the interaction of their wings with airflow, which can be explained by both pressure differences and the downward deflection of air. The wings create lift by being tilted slightly upward, causing air to move faster over the top and creating lower pressure, while the force of wind hitting the bottom of the wing also contributes to lift. The engines, or fans, produce thrust by expelling air backward, which propels the airplane forward, overcoming air resistance. The tail stabilizes the aircraft and aids in steering, ensuring balance among various forces acting on the plane. Understanding these principles clarifies how airplanes can fly despite their weight and air resistance.
  • #61
OmCheeto said:
hmmm... Where's Drakkith when I need him?

You rang?
 
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  • #62
OmCheeto said:
Thanks! Mystery solved.

Boy those Glenn people are blowhards. And kind of hypocritical too. If one clicks on the "For kids" link, you'll find the following explanation:

Which sounds like the Bernoulli explanation, which they claim is wrong in "Incorrect Theory #1".

Typical government organization. First they lie to you, then they point their finger at you for for spreading their disinformation.

:mad:
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?:
The spike forms one side of a virtual bell, with the other side being formed by the outside air...
http://en.wikipedia.org/wiki/Aerospike_engine
With the aerospike, the ramp serves as the inner wall of the virtual bell nozzle, while atmospheric pressure serves as the "invisible" outer wall. The combustion gasses race along the inner wall (the ramp) and the outer wall (atmospheric pressure) to produce thrust.
http://www.nasa.gov/centers/marshall/news/background/facts/aerospike.html
 
  • #63
sophiecentaur said:
The only heavier than air machine that doesn't rely on Bernouili is a rocket.
Oh, I think I can find application of Bernoulli's principle in a rocket...
 
  • #64
Drakkith said:
You rang?

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

Drakkith said:
It's over your head, Om. Go away.

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... :cry:
 
  • #65
Buckleymanor said:
One of the boats is supported by the water.The planing boat is supported by both the water and air.Donald Campbell and his ill fated record attempt comes to mindhttp://www.youtube.com/watch?v=D6a--7mYpbQ
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".
 
  • #66
sophiecentaur said:
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".
Well you could remove it, the air that is.
Would an electric driven propellar boat placed in a vacuum still plane as well as one that is not.
Would the wash be the same?
 
  • #67
Buckleymanor said:
Well you could remove it, the air that is.
Would an electric driven propellar boat placed in a vacuum still plane as well as one that is not.
Would the wash be the same?

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.
 
  • #68
russ_watters said:
Oh, I think I can find application of Bernoulli's principle in a rocket...

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. :wink:

Now, can you suggest a helium balloon design that involves reaction (please don't go into the kinetic theory of gases for the answer)?
 
  • #69
rcgldr said:
One issue with Bernoulli is that it doesn't deal with energy added to the air.

russ_watters said:
Not in its basic form, but there is no reason why you couldn't put it in there.
http://webpages.eng.wayne.edu/~ah8818/Webpage1.htm
I consider the common statement that Bernoulli's equation doesn't/can't deal with energy loss/gain to be a major misconception about it.
That web page is about losses of energy (conversion of mechanical energy into heat due to friction). The issue I have with Bernoulli and how a wing generates lift is the increase in energy (wrt air). If you modify Bernoulli to take into account the increase in energy, I don't consider that Bernoulli anymore, since Bernoulli doesn't help in calculating the increase in energy. I think the reason the energy increase is often ignored is that it's usually small compared to the change in momentum for most wings.
 
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  • #70
rcgldr said:
That web page is about losses of energy (conversion of mechanical energy into heat due to friction). The issue I have with Bernoulli and how a wing generates lift is the increase in energy (wrt air).
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). The impact to the equation is basically the same as adding or deleting the head term in the equation.

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).
If you modify Bernoulli to take into account the increase in energy, I don't consider that Bernoulli anymore...
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!
...since Bernoulli doesn't help in calculating the increase in energy.
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!
 
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  • #71
sophiecentaur said:
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. :wink:
True. That's basically what an ion engine does.
 
  • #72
OmCheeto said:
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... :cry:

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.
 
  • #73
russ_watters said:
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.
 
  • #74
sophiecentaur said:
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.
 
  • #75
russ_watters said:
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.
 
  • #76
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.
 
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  • #77
russ_watters said:
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.
 
  • #78
OmCheeto said:
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.

sophiecentaur said:
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.

sophiecentaur said:
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.

sophiecentaur said:
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.

russ_watters said:
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.

sophiecentaur said:
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".

sophiecentaur said:
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.
 
  • #79
boneh3ad said:
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.
 
  • #80
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.
 
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  • #81
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).
 
  • #82
boneh3ad said:
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.
 
  • #83
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.
 
  • #84
I was taught it's according to Bernoullis Principle
 
  • #85
boneh3ad said:
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...)
 
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  • #86
sophiecentaur said:
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).

sophiecentaur said:
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.
 
  • #87
russ_watters said:
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.

cjl said:
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.
 
  • #88
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.
 
  • #89
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.
 
  • #90
sophiecentaur said:
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.

boneh3ad said:
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.

boneh3ad said:
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?
 
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