airplanelift

How Does an Airplane Wing Work? A Primer on Lift

Many people ask how an airplane wing works, and there is any number of answers that are commonly given. You might have even seen vigorous arguments between proponents of competing theories, such as Bernoulli’s principle and Newton’s laws. So who is correct? First, we must discuss what Bernoulli’s equation means.

Bernoulli’s equation

First, what is Bernoulli’s principle? Bernoulli’s principle (or Bernoulli’s equation) is essentially a statement of conservation of energy for a moving fluid that relates the velocity of a fluid to its pressure. Its most common form is

[tex]p + \dfrac{1}{2}\rho v^2 + \rho gz = \mathrm{constant} = p_{\mathrm{t}}.[/tex]

Here, ##p## is the static pressure, ##p_{\mathrm{t}}## is the total pressure, ##\rho## is the density, ##v## is the velocity, ##g## is the acceleration due to gravity, and ##z## is the height. The static pressure essentially represents a form of potential energy per unit volume. The total pressure represents the total energy per unit volume in the system. The term ##^1/_2 \rho v^2## is called the dynamic pressure and is the kinetic energy per unit volume.

This expression is only valid under a set of very limited circumstances, specifically for steady, incompressible, inviscid flow (or along a streamline). There are various other more complicated forms, but this serves to illustrate the physics just fine. Often, this equation is used to compare the pressure and velocity at two different points. In the case of an airfoil where the change in ##z## is very small and can be ignored, this is often written as

[tex]p_1 + \dfrac{1}{2}\rho v_1^2 = p_2 + \dfrac{1}{2} \rho v_2^2.[/tex]

You could rearrange this instead to say

[tex]p_2 = p_1 + \dfrac{1}{2}\rho ( v_1^2 – v_2^2 ).[/tex]

In other words, if ##v_2 > v_1##, then ##p_2 < p_1## and vice versa. There is therefore an inverse relationship between velocity and pressure, which shouldn’t be surprising in light of the energy conservation nature of the equations.

Many of the explanations of lift use this principle to explain how the air moves faster over the top of the airfoil, therefore it has a lower pressure than the bottom, resulting in a net force upward. While this is not, in general, incorrect, it does lend itself to several fallacies. With this in mind, a few incorrect explanations for lift must next be discussed.

Common lift fallacies

Equal transit time

One of the most common myths associated with lift on an airfoil is that of the “equal transit time” theory. In this common explanation, someone may claim that the length of the path over the top of the airfoil is longer, so the air over the top must speed up in order to travel to the end of the airfoil in the same time as the air on the bottom. The problem with this explanation is that nothing forces the air on the top and bottom to traverse the airfoil in the same amount of time. In fact, the air over the top generally travels much faster than that on the bottom and leaves the airfoil well before the air underneath. This fact is illustrated in the following video. For a more detailed breakdown of this issue, see this NASA article.

Venturi effect

Another common explanation for why air might move faster over the top is by comparing the flow over an airfoil to that through a Venturi nozzle. The idea is that you can draw a horizontal line above the airfoil in the free stream representing a streamline, and the resulting shape looks somewhat like a Venturi tube with a constriction in the middle created by the airfoil. Conservation of mass would then imply the air speeds up, resulting in a low pressure over the top.

The first problem here is that this does not correctly predict the velocity observed over the top of an airfoil. An airfoil is not a Venturi tube and there is no magic surface above it that artificially constricts the flow. In fact, if you arbitrarily choose a horizontal streamline above (or below) an airfoil, the answer you get will change depending on which streamline you pick. Further, that answer will only give some average velocity in that section of your “tube” and will likely only be very slightly faster than the free stream, especially if you pick your arbitrary surface far away from the airfoil.

Second, this ignores the effect of the bottom surface, which would naturally constrict the flow below as well using the same logic. This would lead to a lower than ambient pressure on the bottom of the airfoil as well, leading to either negative or positive lift depending on the shape. In reality, it is often (but not always) the case that the pressure below a wing is higher than the ambient pressure. For one example, see the following image, where darker blue means higher pressure.

The Venturi theory could not predict such a case despite the fact that it does occur.

Now, for all that is wrong with the usual Venturi theory, there is one circumstance where it would work. By definition, streamlines in a flow field have no mass flow crossing them. Therefore, if you pick two arbitrary streamlines in any steady flow field, they form what is called a streamtube through which mass is conserved. In this case, the Venturi effect would apply (in the sense of average velocity). So, for a wing, if you happen to know the streamlines (or streamfunction) beforehand, then you can simply select the streamline at the surface of the airfoil and one streamline above it somewhere and it will give you the correct average velocity, and as the upper streamline you choose gets asymptotically closer to the surface, your average velocity will approach the exact velocity near the surface. Of course, all of the streamlines are generally curved, even a great distance away from the airfoil, and this approach requires prior knowledge of the flow field to work. The only way to obtain the required information about the flow field in order to calculate streamlines is to take a different approach.

In essence, the Venturi effect cannot explain or calculate lift except in the specific case where you already know the flow field by other means and can calculate the streamlines, and in that case, since you already know the flow field, doing any sort of analysis based on Venturi is redundant and not useful. For more on the problems with the Venturi theory as it pertains to lift, NASA has another article to browse.

The skipping stone theory

One other common misconception can be described as the skipping stone theory. The general idea is that an airfoil moves over a layer of air in much the same way a stone can move over the surface of a pond and skip rather than sink. While this idea does get one thing correct: namely that lift can be described as a reaction force to deflection of air, it completely ignores the effect of the upper surface, which plays an important role. For more on this theory, see yet another NASA article on the topic.

So where does lift come from?

The simple answer: Newton

This brings us back to our original question: where does lift come from? The simplest answer is that an airfoil (or any lift-generating shape) bends the airflow downward. This is a change in momentum, which, per Newton’s laws, requires a force to be exerted on the air by the airfoil. The equal and opposite force to this is lift.

The practical answer: Bernoulli

However, before you, Newton fans declare total victory, keep in mind that Bernoulli’s equation is also correct when explaining lift. If we start off with the assumption that the air does, in fact, move faster over the top of the airfoil than the bottom, then Bernoulli’s equation does predict a pressure difference and then a net force. This fact is extraordinarily practical because, unlike the downwash produced by an airfoil, it is quite easy to measure the velocity over an airfoil and convert that to pressure to calculate lift or to just measure the pressure directly.

A quick note on viscosity: Some astute observers may point out that real air flows are viscous, so Bernoulli’s equation should not apply. While the former is true, Bernoulli’s equation is still very useful due to a phenomenon called the boundary layer. The effect of viscosity is confined to a very small region very near the surface of an object, and in the direction normal to the surface, the pressure is constant. Therefore, if one uses Bernoulli’s equation to predict the pressure distribution just outside of the boundary layer where viscosity is not important, they are effectively calculating the surface pressure distribution as well, and therefore can calculate the lift.

Why does the air move faster over the top, and for that matter, how do we know how much downwash is produced?

It turns out, these two questions are related, and also why lift is incorrectly explained so often. It also turns out that the answer to this question is extraordinarily complicated. The simplest answer is that inviscid theory predicts one stagnation point at the front where the air impacts the tip of the airfoil, and one at the back at some arbitrary point that would result in no lift and no flow deflection. However, real life is not inviscid. Airfoils take advantage of this fact by incorporating design features that are specially designed to cause the stagnation point on the downstream end of the airfoil to occur at the trailing edge. This is usually done by making the trailing edge effectively sharp, but can also be achieved through other means (see: flatback airfoil). Since that trailing edge stagnation point has now been “artificially” fixed, the airflow must accelerate or decelerate in various places in order to ensure that the conservation laws (conservation of mass, momentum, and energy) are satisfied. This results in a much faster velocity over the top of the airfoil and a deflection of the flow downward.

So, in the end, Newton fans: you are correct! Bernoulli fans: you are also correct (provided you didn’t invoke equal transit time or Venturi)!

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  1. Robert M says:

    Vorticity is so much more productive than primitive variables in explaining and analyzing low Mach number flow, and it fits naturally with the actual physics of a standard airfoil.

    https://en.wikipedia.org/wiki/Kutta–Joukowski_theorem

    For potential flow–no vorticity, or, equivalently, the curl of velocity is everywhere zero–there would be no lift and no drag, so the introduction of vorticity is more than just a pedagogic nicety or a pedantic detail.

    https://physics.stackexchange.com/questions/46131/does-a-wing-in-a-potential-flow-have-lift

    I suspect that the reason that elementary discussions of lift lean so heavily on primitive variables, which are themselves problematical for this problem, is that actually using vorticity to do calculations introduces a fair bit of mathematical baggage.

  2. boneh3ad says:
    lomidrevo

    Many thanks for this article, it helped me a lot to understand the topic!
    However, I think there is missing ρ (density) in the last term of your first equation: gz. (When I compare the units, I miss kg/m[SUP]3[/SUP] in this part of equation)Good catch. I can't believe that's been hanging out like that for a year and a half with no one noticing.

  3. lomidrevo says:

    Many thanks for this article, it helped me a lot to understand the topic!
    However, I think there is missing ρ (density) in the last term of your first equation: gz. (When I compare the units, I miss kg/m[SUP]3[/SUP] in this part of equation)

  4. DoobleD says:

    [QUOTE=”boneh3ad, post: 5321422, member: 268837″]Excellent to hear that. It’s certainly not perfect, though, and I am open to editing it and/or writing a follow-up if more misconceptions arise.[/QUOTE]

    For me, there are two parts I didn’t completely understood :

    – the one circumstance where the venturi effect could actually work (I didn’t see the difference with the other venturi case described above this part)
    – the explanation of why the flow is deflected downward

    However it’s important to say that I am not always comfortable with English language. Neither I did a lot of fluid mechanics. Also I understand that to really understand the downward flow wouldn’t be that easy. That is to say the article might be actually crystal clear for most people, I couldn’t tell.

  5. boneh3ad says:

    [QUOTE=”DoobleD, post: 5321370, member: 562600″]Awesome article. I encountered the NASA pages a while ago and was completely amazed when I discovered that lift is often explained using wrong arguments. I then got completely lost in a ton of different explanations of lift. This article cleared some things.[/QUOTE]

    Excellent to hear that. It’s certainly not perfect, though, and I am open to editing it and/or writing a follow-up if more misconceptions arise.

  6. markuzi says:

    [QUOTE=”John_RB, post: 5311938, member: 555889″]If you get two sheets of A4 (letter) sized paper and hold them parallel with a gap between them and blow into the gap the sheets are sucked together, not blown apart. That’s my practical demo or where lift can come from just by the flow of air[/QUOTE]

    If the venturi effect is applied to the lift on a wing, then an aircraft would never get off the runway because the venturi airflow under the wing would cause the aircraft to be sucked downward to the ground. This is obviously not the case.

    The fact that an aircraft takes off from the runway and flares on a cushion of pressure upon landing is evidence the venturi effect does not cause lift.

    Anyway ….. has anyone actually seen this mythical venturi around a wing.

  7. boneh3ad says:

    [QUOTE=”John_RB, post: 5311938, member: 555889″]If you get two sheets of A4 (letter) sized paper and hold them parallel with a gap between them and blow into the gap the sheets are sucked together, not blown apart. That’s my practical demo or where lift can come from just by the flow of air[/QUOTE]

    The problem with that approach is that you can’t directly use something line Bernoulli’s equation to compare the two streams of air (between the papers and outside of them) because they originate from reservoirs with different total pressures.

    It turns out that human lungs are only typically capable of producing on the order of 2 psi of pressure above atmosphere. It just works out, then, that if you use Bernoulli’s equation on the air in your lungs separately from that outside the papers, that the stream of air between the papers still may end up with a lower static pressure than ambient. If you had an air source with higher pressure blowing between the papers, that wouldn’t necessarily be true and very well could push the pages apart.

  8. John_RB says:

    If you get two sheets of A4 (letter) sized paper and hold them parallel with a gap between them and blow into the gap the sheets are sucked together, not blown apart. That’s my practical demo or where lift can come from just by the flow of air

  9. boneh3ad says:

    [USER=579583]@biggles21[/USER], did you even read the article about which this thread was started? 100+ years of flight and the associated research has provided a whole lot of evidence that supports everything I and other fluid mechanicians have said about this subject.

    You can do the math to calculate the flow around an airfoil and use Bernoulli to get a pressure distribution. You can put the same wing in a wind tunnel or flight with pressure transducers in the surface to measure the pressure distributions and the two match. What’s more, you can put a scale model of a wing in a wind tunnel and measure those pressures and use them to calculate the lift while at the same time measuring the lift directly with a force balance and the two numbers match. How is that not evidence that you can use Bernoulli’s equation and the pressure distribution to calculate lift?

    Like I said, the video you supplied is almost entirely incorrect. It does not seem to understand anything about fluid mechanics. It even cites “magic” as part of the reason Bernoulli is purported to work, which is patently ridiculous. In its own explanation, it purports that lift simply comes from the bottom of an airfoil deflecting air downward, which I already briefly mentioned in this Insight and linked to a NASA article that thoroughly debunks that claim. In the same way, lift on an airplane wing has very little in common with a boat on water (unless that boat is a hydrofoil).

    I don’t know how much other evidence you need. You already have said you don’t believe the 100+ years of textbooks that support my position. Or the 100+ years of wind tunnel tests and flight tests that confirm the mathematics, apparently. Or the 100+ years of aerospace engineers who design airfoils using, at least in early stages, Bernoulli’s equation for part of the job. It sure seems your mind is made up and you have no actual desire to learn here.

  10. boneh3ad says:

    No offense [USER=579583]@biggles21[/USER] but you really shouldn’t try to make definitive statements about things you don’t appear to actually know much about. In fact, almost everything in that video is untrue (most of which was discussed in my Insight or in the linked NASA articles) and whoever made it clearly has no understanding of fluid mechanics.

  11. boneh3ad says:

    [QUOTE=”markuzi, post: 5305943, member: 579491″]Go here:

    [URL]https://en.wikipedia.org/wiki/User:Rolo_Tamasi[/URL]

    Lift on a wing is the result of differential radial forces over the wing surface.[/QUOTE]

    That person’s article actually just proposes a different way of calculating the pressures given the velocity field. It is essentially functionally identical to using Bernoulli.

  12. Chestermiller says:

    Part of the issue regarding energy is that, although the forces and the velocities are the same in the two frames of reference (aside from a constant offset equal to the free stream velocity), work is not frame invariant. Fortunately, the net sums of the energy contributions is frame invariant (i.e., the energy balance is satisfied in all frames of reference). The differences in the work are exactly offset by the differences in the kinetic energy.

  13. markuzi says:

    Go here:

    [URL]https://en.wikipedia.org/wiki/User:Rolo_Tamasi[/URL]

    Lift on a wing is the result of differential radial forces over the wing surface.

  14. rcgldr says:

    [QUOTE=”boneh3ad, post: 5305910, member: 268837″]The increase in gravitational potential energy is negligible in a flow like that. It is air[/QUOTE]I meant the decrease in gravitational potential energy of the glider, which is offset by the increase in the energy of the air (since the glider is in a steady descent and not accelerating). It’s a 1500 lb glider descending at 1 mph, 4 hp, 550 ft lbs of energy every second. Sorry I didn’t make it clear I was referring to the potential energy of the glider.

    However, the same issue exists for something simpler, like a parachutist with parachute deployed in a vertical and steady descent. From the ground / air (no wind) frame of reference, the decrease in gravitational potential energy is offset by an increase in energy of the air. From the parachutist’s frame of reference … ? So I’m not sure this was going anywhere.

    … but I see your point, the energy issue shouldn’t matter, it’s the momentum change that corresponds to lift (and drag).

    and thanks for creating the primer.

  15. boneh3ad says:

    The increase in gravitational potential energy is negligible in a flow like that. It is air, so it has very low density and very low change in height. It’s perfectly reasonable to just ignore that effect.

    Otherwise, I’m honestly not sure how I can be any more clear. The flow fields in the two frames are exactly the same only offset by the free stream velocity. Forces such as lift are the direct result of the change of momentum, not energy, and the momentum change is identical in each case.

  16. rcgldr says:

    [QUOTE=”boneh3ad, post: 5305898, member: 268837″]It isn’t really a cause and effect thing. It’s two sides of the same coin.[/QUOTE]The shape of the airfoil, angle of attack, and free stream velocity could be considered as the causes. As you stated, everything else, pressure gradient normal to and in the direction of flow, change in velocity of the affecte air, … , are coexistent.

    [QUOTE=”boneh3ad, post: 5305898, member: 268837″]conservation of energy[/QUOTE]Using the glider model, from the air frame of reference (no wind), the decrease in gravitational potential energy is offset by the increase in the air’s energy (mostly mechanical, some thermal). From the glider’s frame of refernce, it’s not clear to me.

  17. boneh3ad says:

    You have completely more as the quote that you posted earlier then. It says nothing about perpendicular pressure gradients. And it’s really difficult to try and say flow accelerates [I]because[/I] of the pressure gradient instead of vice versus. It isn’t really a cause and effect thing. It’s two sides of the same coin.

    Also, of course the kinetic energy changes. Energy is frame dependent. What is not frame dependent is conservation of energy, so the change in total energy should be the same either way. After all, you’ve removed all of the kinetic energy associated with the free stream velocity both upstream and downstream of the airfoil.

  18. rcgldr says:

    [QUOTE=”boneh3ad, post: 5305881, member: 268837″]That quote has nothing to do with your point. That quote simply says that the air flow bends around the wing and the pressure varies, which corresponds with the velocity bending.[/quote]What I was trying to get at is that the pressure differentials are due to the pressure gradient’s that are perpendicular to the relative flow, coexistant with the curved flow. The idea is that a wing produces lift by curving (diverting) the relative flow. The air travels faster above a wing because the pressure gradient related to curved flow coexists with a low pressure zone just above the wing’s surface.

    [QUOTE=”boneh3ad, post: 5305881, member: 268837″]As to your other paragraph, I’m really having a tough time deciphering your language. … momentum … [/quote]My issue isn’t about the change in momentum, since that is the same in both frames of reference. My issue is about the change in energy of the air by the wing. The change in energy of the air is affected by the frame of reference.

  19. boneh3ad says:

    That quote has nothing to do with your point. That quote simply says that the air flow bends around the wing and the pressure varies, which corresponds with the velocity bending.

    As to your other paragraph, I’m really having a tough time deciphering your language. The bottom line is that if you think from the frame of the wing and moving air, the wing deflects air downward. In oother words, the wake is going to have a downward component to its velocity. If you change the reference frame by subtracting the free stream velocity, that doesn’t affect the downward velocity at all. The downward momentum change remains the same in both frames, ergo so does the lift.

    If you want to talk drag, the wake coming off of an airfoil is going to be slightly slower than the free stream in the wing’s frame, so when you subtract off the free stream velocity, that results in a slight forward velocity, which you also frequently cite.

  20. rcgldr says:

    [QUOTE=”boneh3ad, post: 5305856, member: 268837″]You would essentially just subtract off the free stream velocity of the plane.[/QUOTE]If I do that, than what was 0 velocity from the air’s frame of reference becomes a negative velocity from the plane’s frame of reference.

    From the planes frame of refrence, the slight decrease in speed due to drag is multiplied by the relative flow: 1/2 m (v – Δv)^2 – 1/2 m v^2 ~= – m v Δv. So from the plane’s frame of reference, it would seem the loss in energy due to drag would be greater than the increase in energy due to downwash, which I assume is a fraction of the relative flow speed. From a wing frame of reference, I don’t see how diversion of flow could increase the energy of the air. It’s easier to see this if only drag is considered, like a bus. From the air’s frame of reference, the energy of the air is increased by the bus, but from the bus frame of reference (bus moving with no wind, or bus not moving with an oncoming wind), the relative flow is slowed down, decreasing the energy of the air. Maybe I’m missing something here.

    I did find this old quote from Mark Drela related to my point about curved flow

    [i]Subject: Re: How does a wing produce lift
    Date: Wed, 17 Feb 1999 14:19:57 GMT
    Newsgroups: rec.models.rc.air

    In article <7a9l31$23u$1@pale-rider.INS.CWRU.Edu> …

    Professor Mark Drela, MIT Aero and Astro wrote:

    “Bending the free airstream causes pressure gradients” on the wingsurface
    (wing at an AOA) This wing deflection of the airflow also causes vertical
    momentum on the inert airmass (downwash).[/i]

  21. boneh3ad says:

    You would essentially just subtract off the free stream velocity of the plane (which would be nearly horizontal but not quite in that case). You would end up with the flow field you often like to cite where the air will be left with a slightly downward Calicut. In your example, about 9 knots. Subtracting the “horizontal” velocity doesn’t change that “downward” component.

  22. rcgldr says:

    [QUOTE=”boneh3ad, post: 5305815, member: 268837″]Where are those numbers from and what do they mean? [/QUOTE]I thought it was a Cessna, but the numbers are off (too low). These are old notes. Doing a web search I found an article that mentions Cessna, downwash, and mass flow, but that’s not where I got my old (over 15 , maybe 20 years old) notes from.

    [i]If we estimate that the average vertical component of the downwash of a Cessna 172 traveling at 110 knots to be about 9 knots, then to generate the needed 2,300 lbs of lift the wing pumps a whopping 2.5 ton/sec of air! In fact, as will be discussed later, this estimate may be as much as a factor of two too low. [/i]

    [URL]http://www.allstar.fiu.edu/aero/airflylvl3.htm[/URL]

    It’s easier to consider the case of a high end glider like the Nimbus with a 60:1 glide ratio. At 60 mph, descent rate is 1 mph, average weight 1500 lbs. The decrease in (the gliders) gravitational potential energy, and increase in total energy of the air translates into 4 hp. Most of the energy added to the air would be mechanical, some of it thermal. This is a better example to show energy is added to the air by the glider, from the air’s frame of reference. How would I convert this data to the glider’s (wing) frame of reference?

    [QUOTE=”Ironmanlet, post: 5305797, member: 579130″]amount of time in general.[/QUOTE]From the wing’s frame of reference, the air is slowed down due to drag.

  23. boneh3ad says:

    Your s=d/t claim only works if you assume t is constant. What I’m saying is that is not necessarily true. It is [I]not[/I] true if the object is producing lift. The wing certainly does change the speed but there’s no reason that the time of a particle moving from front to back has to be the same as the time of a particle moving the same distance in the free stream. That is not the mechanism by which the air is accelerated. Now, it may work out that a symmetric airfoil at zero angle of attack exhibits this property, but it’s not a general rule.

    Now that I think a little more, except for the small region near the airfoil where viscosity has sapped some of the energy, it’s quite possible that a symmetric airfoil at zero angle of attack would exhibit an equal (or very close to it) time for a particle to move from one edge to the other near the airfoil as compared to the same distance in the free stream. That’s just not a general rule.

    Just take a look at this video. In the beginning part where they just release the small puffs of smoke, it’s clear that the regions near the airfoil are traveling the same distances are regions in the undisturbed air in dramatically different amounts of time. That’s what I’m trying to illustrate.

    [MEDIA=youtube]6UlsArvbTeo[/MEDIA]

    [USER=17595]@rcgldr[/USER] Where are those numbers from and what do they mean? The idea of a mass flow doesn’t even make sense without some area through which it is traveling. A single velocity doesn’t make sense either, as the wake behind a wing will have a much more complicated profile. Do have a diagram to explain what you mean here, because I’m not following.

  24. rcgldr says:

    [quote]work done … frame of reference[/quote]Air frame of reference – From some old notes: Cessna at cruise speed, downwash speed ~= 4.5 m/s, mass flow ~= 2500 kg/s, work done per second ~= 25,300 joules (25,300 watts ~= 34 horsepower). How do I get this from the wing’s frame of reference?

  25. Ironmanlet says:

    I am not talking about the difference in time between the top of the chorde and the bottom. I am simply talking about the amount of time in general. The relationship to the top of the airfoil with the bottom of the airfoil is irrelevant. Without relation of one side of the airfoil to the other. Even if it is symmetrical, the speed of the air has to change due to the increase of distance. I am not an expert. To be honest, I am an idiot. But this I know. You can not move air at a greater distance in the same time at same speed.

    S=D/T

  26. boneh3ad says:

    I’ve got to think about this for a second. At nonzero pitch, the flow definitely does not take the same time to get from front to back as the case of no airfoil. After all, it doesn’t take the same time to get from front to back between the top and bottom in that case so it is impossible for it to be the same as the no airfoil case. At zero angle of attack I’m not as certain. I can test the idea out a bit later to show one way or the other.

  27. Ironmanlet says:

    so if an airfoil (greater distance) moves through the air, the air molecules on either side of the airfoil take a different amount of time to travel that greater distance than if there was not an airfoil (greater distance) passing through the air?

    That would either one, make a vacuum in our atmosphere or two, make the air occupy the same space as the airfoil.

  28. boneh3ad says:

    The two frames of reference in question here are related by a simply Galilean transformation and are otherwise functionally identical (already stated by [USER=345636]@Chestermiller[/USER] while I was typing this). The work done in each frame of reference is the same. The lift calculated is the same.

    As for why Bernoulli’s equation applies here, consider that Bernoulli’s equation is an energy conservation statement at its heart, where total energy in the flow is represented by total pressure. If it applies along a streamline, then if every streamline originates in a region with the same total pressure, then it applies everywhere that hasn’t been acted on by a non-conservative process (i.e. where flow energy has been conserved). Since Bernoulli’s equation generally only applies in an inviscid flow, it is therefore valid everywhere here.

    I’ll also remind everyone that we are considering lift here. If you want to calculate an accurate value for drag (which in general cannot be done) you cannot neglect viscosity.

    [QUOTE=”Ironmanlet, post: 5305751, member: 579130″]Even at flat pitch for a symmetrical airfoil, the air is moving a greater distance at the same time.[/QUOTE]

    Nothing requires any sort of “same time” constraint here.

  29. Ironmanlet says:

    [QUOTE=”Chestermiller, post: 5305756, member: 345636″]You are aware that the physics is invariant to the choice of the inertial frame of reference of the observer, correct? So as long as the wing is not accelerating relative to the ground, the analysis of the flow by an observer traveling along with the velocity of the wing yields the same results as the analysis of the flow by an stationary observer, aside from a constant velocity difference.[/QUOTE]

    I love you

  30. Chestermiller says:

    [QUOTE=”rcgldr, post: 5305742, member: 17595″]For me one of the confusing issues is the effect of the frame of reference.[/QUOTE]
    You are aware that the physics is invariant to the choice of the inertial frame of reference of the observer, correct? So as long as the wing is not accelerating relative to the ground, the analysis of the flow by an observer traveling along with the velocity of the wing yields the same results as the analysis of the flow by an stationary observer, aside from a constant velocity difference.

  31. Ironmanlet says:

    [QUOTE=”rcgldr, post: 5305742, member: 17595″]relative flow is diverted with no change in speed, so no change in energy, and no work performed by the wing onto the air.[/QUOTE]

    there has to be a change in speed. Even at flat pitch for a symmetrical airfoil, the air is moving a greater distance at the same time. Seeing as how Speed is a function of time and distance, if distance increases but time remains constant, the speed must change. It does not matter that the speed above and below the airfoil is changing at the same rate, it is still changing. If the speed of air did not change going over an airfoil even at flat pitch on a symmetrical airfoil, while the distance increases and the time remains constant, there would be a vacuum on the trailing edge of the airfoil. And no one here will suggest that is happening.

    Even if it is not a perfect airfoil, you can’t just move something through air (at any speed) and have it result in no work is being performed.

    Two particles of matter can not occupy the same place at the same time. Air is made of particles and so are the wings. Work is being performed by the wing.

    Edit: just saw your reply to my RC heli post. So it has a governor to keep the rotor at a certain speed? like most modern helis. okay

  32. Ironmanlet says:

    [QUOTE=”rcgldr, post: 5305742, member: 17595″]relative flow is diverted with no change in speed, so no change in energy, and no work performed by the wing onto the air.[/QUOTE]

    there has to be a change in speed. Even at flat pitch for a symmetrical airfoil, the air is moving a greater distance at the same time. Seeing as how Speed is a function of time and distance, if distance increases but time remains constant, the speed must change. It does not matter that the speed above and below the airfoil is changing at the same rate, it is still changing. If the speed of air did not change going over an airfoil even at flat pitch on a symmetrical airfoil, while the distance increases and the time remains constant, there would be a vacuum on the trailing edge of the airfoil. And no one here will suggest that is happening.

    Even if it is not a perfect airfoil, you can’t just move something through air (at any speed) and have it result in no work is being performed.

    Two particles of matter can not occupy the same place at the same time. Air is made of particles and so are the wings. Work is being performed by the wing.

    Edit: just saw your reply to my RC heli post. So it has a governor to keep the rotor at a certain speed? like most modern helis. okay

  33. rcgldr says:

    For me one of the confusing issues is the effect of the frame of reference. From the wings perspective, consider the case of an ideal wing where the relative flow is diverted with no change in speed, so no change in energy, and no work performed by the wing onto the air. From the air’s perspective, the affected air ends up with a non zero “exit velocity” (it’s velocity when it’s pressure returns to ambient), so work is done. From this perspective, as the air crosses the plane swept by the wing, work is performned on the air, mostly in the form of a pressure jump from below ambient to above ambient. Bernoulli applies to the air as it approaches that plane from above and departs from that plane below, but Bernoulli is violated as the air crosses the plane because that is where the wing performs work on the air. It’s similar to situation of a propeller as noted in this NASA article:

    [i]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[/i] (propeller) [i] violates an assumption used to derive the equation.[/i]

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

    I always thought that the argument against Bernoulli was based on the work done using the air’s frame of reference, not the validity of using Bernoulli’s equation to calculate pressures given a velocity field using the wing as a frame of reference.

    [quote]radio control helicopters[/quote]The aerobatic ones generally have a pitch angle range around +/- 12 degrees. There’s a throttle versus pitch angle curve programmed into the transmitter to keep the rotor speed near constant in spite of the torque load due to higher pitch angles.

  34. Ironmanlet says:

    at 1:37 this dude literally takes off upside down and at 2:06 he simply hovers upside for a large amount of time. I am no genius, but in my mind, there is no way he can do that without negative pitch in the blades. Even if it is not a result of just bernoulli but also newton, there has to be negative pitch.

    [MEDIA=youtube]l5FqYiZb_5s[/MEDIA]

  35. Ironmanlet says:

    [QUOTE=”Greg Bernhardt, post: 5305189, member: 1″]A comment from a Google+ PF follower “If the Bernoulli effect were the primary cause of lift, planes would not be able to fly upside-down”[/QUOTE]
    [QUOTE=”boneh3ad, post: 5305301, member: 268837”]Well I’d request said commenter supply proof of that claim because there is none; it’s not true. Bernoulli’s equation certainly does not preclude an airplane from flying upside down provided a sufficient amount of angle of attack. It will just come with a larger amount of drag.[/QUOTE]

    Agreed with Bone here. You can easily engineer an airfoil to exceed below flat pitch into negative pitch. I have never used a remote control helicopter, but I am sure most of them fly upside down easily by being able to do this.

    (the pitch angle of my rotor blades on my actual helicopter does not go below flat pitch, the collective just does not go down that far, lol)

  36. Chestermiller says:

    Hi rcgldr:

    I think what boneh3ad is alluding to (bh, correct me if I’m wrong), the potential flow equations (or Navier Stokes equations, if you want to be more sophisticated) you use to calculate the velocity distribution (if you are doing a model, rather than obtaining field velocity data) already take into account the accelerations you are referring to. So, provided you correctly solved the fluid dynamics equations, your results for the velocity distribution should be consistent with the Bernoulli equation, and the corresponding pressure distribution from Bernoulli equation should correctly predict the lift.

    Chet

  37. rcgldr says:

    [QUOTE=”boneh3ad, post: 5305651, member: 268837″]If you already know the velocity field and you know the ambient pressure outside of the region of influence of the airfoil, you can calculate the total pressure and everything else you need to calculate lift from Bernoulli’s equation provided there is no separation and you already have another means of calculating the velocity field.[/QUOTE]The issue that I recall was that the calculation of the velocity field had to consider the effects of the flow perpendicular to the surface of a wing, not a post correction done after the velocity field was calculated.

  38. boneh3ad says:

    I am pretty sure we have been over this in some thread somewhere, but there are no corrections of any sort needed due to acceleration perpendicular to streamlines. That really doesn’t have an effect of any sort in this context. There is no flow normal to a streamline by definition, and no flow normal to the wing surface [I]at[/I] the surface because that would violate mass continuity. The only acceleration in the case of curved streamlines is centripetal acceleration due to the curvature. Since Bernoulli’s equation cares only about the speed of a flow (the velocity is squared), if you already know the velocity field and you know the ambient pressure outside of the region of influence of the airfoil, you can calculate the total pressure and everything else you need to calculate lift from Bernoulli’s equation provided there is no separation and you already have another means of calculating the velocity field.

  39. rcgldr says:

    When using Bernoulli to calculate pressure by calculating the relative speeds over (or under) tiny panel sections of the surface of a wing, I seem to recall that the pressure gradient related to curved flow required some adjustments to deal with acceleration perpendicular to streamline flow. Something to do with the macro scale flow having a component perpendicular to the surface of a wing, and it’s interaction with the relative flow parallel to the wing surface, as the perpendicular flow has to be diverted into the diretion of the wing’s surface, except for the downwards flow behind the trailing edge. Maybe this only applied to a specific type of model used to generate polars for a wing.

    I’m wondering if there’s a relatively simple explanation of the effects of the component of flow perpendicular to the surface of a wing.

  40. boneh3ad says:

    What level of material interests you?

    For example, the go-to text for introducing undergraduate engineers to the basic principles of flight is [url=http://www.amazon.com/Introduction-Flight-John-Anderson/dp/0078027675/ref=sr_1_2?ie=UTF8&qid=1449005674&sr=8-2&keywords=introduction+to+flight][I]Introduction to Flight[/I][/url] by John Anderson. It’s pretty expensive but the basics haven’t changed so you can certainly look into an older edition or check it out at a library if you have access.

  41. boneh3ad says:

    Well I’d request said commenter supply proof of that claim because there is none; it’s not true. Bernoulli’s equation certainly does not preclude an airplane from flying upside down provided a sufficient amount of angle of attack. It will just come with a larger amount of drag.

  42. Greg Bernhardt says:

    A comment from a Google+ PF follower “If the Bernoulli effect were the primary cause of lift, planes would not be able to fly upside-down”

  43. Greg Bernhardt says:

    [QUOTE=”boneh3ad, post: 5304350, member: 268837″]I should have titled it “Lift or: How I Learned to Stop Worrying and Love Bernoulli [I]and[/I] Newton.” That’s a real missed opportunity.[/QUOTE]
    Ripe for the sequel :)

  44. DoobleD says:

    Awesome article. I encountered the NASA pages a while ago and was completely amazed when I discovered that lift is often explained using wrong arguments. I then got completely lost in a ton of different explanations of lift. This article cleared some things.

  45. biggles21 says:

    Well, if you're a scientist then back up what you say with evidence, logic and reason.  I've spoken with the FAA, CAA; EASA; Boeing and Airbus, as well as a few aerodynamic professors; and none have been able to provide any evidence that contradicts what I say. None. Zero.  If I'm wrong this should be easy. I find you rather offensive and rude. Someone who resorts to personal attacks rather than scientific debate, indicates that they've nothing to defend their position – so have already lost the debate. So you should stop being rude.

  46. biggles21 says:

    Hang on. You lot are still arguing over how planes fly and what causes lift? After 100 years of data on aviation? Really? This should be a no-brainer.  Planes fly based on the same principles that bird stay in the air. see: " How planes stay airborne – Summary" on youtube https://www.youtube.com/watch?v=8yw81t7N6vY&feature=youtu.be It's obvious that Bernoulli's principles don't apply to aviation, as aviation doesn't actually use any of his mathematical formula or equations in the calculation of lift.   Also, Newton can only DESCRIBE some of the forces involved in lift, it cannot EXPLAIN lift. Flying text books list the critical factors that affect lift (speed, wing area, air density, ….) ; none  of these factors include the low air pressure on the top of the wing. Yeah I know, the same text book will contradict itself and say that lift is caused by low air pressure on the wing.

  47. skrat says:

    I have finished my first Bologna in Physics and am currently working on a masters degree. My idea is to someday study the very details of fluid dynamics and use my knowledge in flying industry. So, perhaps not the very basic literature, however PHD articles are probably (at the moment) too advanced for me. Something in between would be perfect.I will for sure check that Introduction to Flight you suggested! Thank you! If you remember any more, don't hesitate!cheers

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