airplanelift

How Does an Airplane Wing Work: a Primer on Lift

[Total: 4    Average: 5/5]

Many people ask how an airplane wing works, and there are 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 air flow 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.

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 air flow 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)!

Click For Forum Comments

PhD - Fluid mechanics, nonlinear dynamical systems, hydrodynamic stability. Swimming, grilling, beer, movies, watching most sports, playing a few, and a dash of video games
52 replies
Newer Comments »
  1. skrat
    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

  2. biggles21
    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.

  3. biggles21
    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.

  4. DoobleD
    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.

  5. boneh3ad
    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.

  6. boneh3ad
    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]Introduction to Flight[/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.

  7. rcgldr
    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.

  8. boneh3ad
    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 at 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.

  9. rcgldr
    rcgldr says:

    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.

    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.

  10. Chestermiller
    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

  11. Ironmanlet
    Ironmanlet 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”

    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.

    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)

  12. Ironmanlet
    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.

  13. rcgldr
    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:

    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.

    [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.

    radio control helicopters

    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.

  14. Ironmanlet
    Ironmanlet says:

    relative flow is diverted with no change in speed, so no change in energy, and no work performed by the wing onto the air.

    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

  15. Ironmanlet
    Ironmanlet says:

    relative flow is diverted with no change in speed, so no change in energy, and no work performed by the wing onto the air.

    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

Newer Comments »

Leave a Reply

Want to join the discussion?
Feel free to contribute!

Leave a Reply