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What aerodynamic effects actually contribute to producing the lift on an airplane?

I know there's a common belief that lift comes from the Bernoulli effect, where air moving over the wings is at reduced pressure because it's forced to travel further than air flowing under the wings. But I also know that this is wrong, or at best a minor contribution to the actual lift. The thing is, none of the many sources I've seen that discredit the Bernoulli effect explain what's actually going on, so I'm left wondering. Why do airplanes actually fly? Is this something that can be explained or summarized at a level appropriate for someone who isn't trained in fluid dynamics?

(Links to further reading for more detail would also be much appreciated)

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7 Answers 7

up vote 90 down vote accepted

A short summary of the paper mentioned in another answer and another good site.

Basically planes fly because they push enough air downwards and receive an upwards lift thanks to Newton's third law.

They do so in a variety of manners, but the most significant contributions are:

  • The angle of attack of the wings, which uses drag to push the air down. This is typical during take off (think of airplanes going upwards with the nose up) and landing (flaps). This is also how planes fly upside down.
  • The asymmetrical shape of the wings that directs the air passing over them downwards instead of straight behind. This allows planes to fly level to the ground without having a permanent angle on the wings.
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I think a clearer way to state this is to say that the wings push air downwards thus producing lift, and the airfoil shape is simply more efficient that a simpler shape, such as a wing with a rectangular cross-section. There is nothing magical about an airfoil except that it produces the least possible drag for a given amount of lift. –  Colin K Dec 27 '10 at 17:22
How do you not talk about Bernoulli's principle here? Angle of attack has nothing to do with it. They do not push air downwards. This is a massively upvoted and accepted answer on this site? My word. –  Robusto Aug 23 '11 at 1:29
@Robusto: like I said, the airfoil shape produces less drag for a given angle of attack. Angle of attack "has nothing to do with it?" I'm not sure where you got that idea, but AoA is critical to flight dynamics. Whether air is "pushed," pulled, or whatever, it is moved downwards to produce thrust. Newtons third law must be satisfied; a wing is designed to do that as efficiently as possible. –  Colin K Aug 23 '11 at 2:13
@Robusto: I would make a slight correction to Sklivvz's answer. Wings don't just push air downward, they pull it downward. The top surface of the wing is more important than the bottom. If the flow separates from the top surface, the wing stalls. That's what happens at a high-enough angle of attack, and it is exacerbated by anything that makes the surface rough. –  Mike Dunlavey Dec 28 '11 at 13:52
@Sklivvz: Sorry, being a student pilot, and loving physics, this is a favorite topic. Reviewing your answer, I'd point out that aerobatic aircraft have symmetrical airfoils, just so they can fly inverted as well as upright. Positive AofA gives positive lift, and negative negative. Since normal aircraft don't fly inverted, the airfoil is asymmetrical, but that's just an optimization. Aircraft may look like they are level, but the wings are not, they are at an AofA. The slower they go, the higher they must nose-up to get the necessary lift. –  Mike Dunlavey Jan 2 '12 at 21:26

From Stick and Rudder by Wolfgang Langewiesche, page 9, published 1944:

The main fact of all heavier-than-air flight is this: the wing keeps the airplane up by pushing the air down.

It shoves the air down with its bottom surface, and it pulls the air down with its top surface; the latter action is the more important. But the really important thing to understand is that the wing, in whatever fashion, makes the air go down. In exerting a downward force upon the air, the wing receives an upward counterforce--by the same principle, known as Newton's law of action and reaction, which makes a gun recoil as it shoves the bullet out forward; and which makes the nozzle of a fire hose press backward heavily against the fireman as it shoots out a stream of water forward. Air is heavy; sea-level air weights about 2 pounds per cubic yard; thus, as your wings give a downward push to a cubic yard after cubic yard of that heavy stuff, they get upward reactions that are equally hefty.

That's what keeps an airplane up. Newton's law says that, if the wing pushes the air down, the air must push the wing up. It also puts the same thing the other way 'round: if the wing is to hold the airplane up in the fluid, ever-yielding air, it can do so only by pushing the air down. All the fancy physics of Bernoulli's Theorem, all the highbrow math of the circulation theory, all the diagrams showing the airflow on a wing--all that is only an elaboration and more detailed description of just how Newton's law fulfills itself--for instance, the rather interesting but (for the pilot) really quite useless observation that the wing does most of its downwashing work by suction, with its top surface. ...

Thus, if you will forget some of this excessive erudition, a wing becomes much easier to understand; it is in the last analysis nothing but an air deflector. It is an inclined plane, cleverly curved, to be sure, and elaborately streamlined, but still essentially an inclined plane. That's, after all, why that whole fascinating contraption of ours is called an air-plane.

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Excellent explanation! –  Ingo Feb 19 '11 at 10:50

This answer is nothing more than a variation of Sklivv's answer. I simply wish to discuss some quantitative ideas following from Sklivv's answer and discuss what I understand (from an aerospace engineering friend) to be a common conceptual mistake - that the application of "mere surface effects" and "application of Bernoulli's principle" is wrong. These "mere surface effects and Bernoulli's principle" follow from Sklivv's idea as I hope to make clear. Everything in aeroplane physics begins and ends with "aeroplanes thrust air downwards, so the air thrusts aeroplanes up". This answer is written to be understandable to someone like I who know nothing about fluid dynamics - aside from:

  1. The mathematically elegant and thoroughly enjoyable 2D problems tackled with complex variable theory (see Finding Stagnation Points from the complex potential );

  2. That I know that there is Clay Mathematics prize up for grabs for anyone who can prove existence of, or give counterexample against, the existence of smooth, globally well defined solutions to the Navier-Stokes equations;

  3. That aerospace engineering colleagues and friends tell me experimental proof is still queen in this field: most real fluid dynamics involving aeroplane flight heavily leans on phenomenological models tuned by experiment.

I shall answer by taking these points up in turn.

Experiment is Queen

From a particular experimental point of view, there is no mystery why aeroplanes fly. Rather, the better question, in my opinion, is "how do they control the inevitable huge lift forces on them to make the latter stably lift in a constant, vertical direction?"

This experimental view is as follows: think of the Beaufort Scale and other scales used by meteorologists to get across the practical meaning of their wind and other warnings: for example the Fujita Scale for tornadoes and Tropical Cyclone Category Systems, which describe in practical terms the effects of storms of various intensities.

Now I understand flying regulations forbid commercial jetliners from flying at slower than $300\mathrm{km\,h^{-1}}$ before their final approach to the runway. Think about $300\mathrm{km\,h^{-1}}$ airspeed in terms of the scales I have just spoken of: this is an F4 tornado, category 5 cyclone and is well off the 12-class Beaufort scale. Buildings and structures of any shape the size and weight of fully laden aeroplanes are torn up and borne into the sky or utterly torn down and destroyed. There is NO shortage of lift from a $300\mathrm{km\,h^{-1}}$ relative airspeed to hold almost anything the size and weight of fully laden commercial jetliner up: at these airspeeds, almost anything of this size and weight and lighter flies. At least it does so fleetingly: if it isn't designed like an aeroplane, as it moves its attitude changes and so does the direction of ram pressure: it is then likely to be flipped over and dashed catastrophically onto the ground. Put simply: almost everything flies at this airspeed, but only very special things do so stably.

Simple Mathematical Models

We can do a back of the envelope estimation of ram pressure in this case: see my drawing below of a simple aerofoil with significant angle of attack being held stationary in a wind tunnel. I'm going to put some numbers to Sklivvz's description:

Simple Aerofoil

Lets suppose the airflow is deflected through some angle $\theta$ radians to model an aeroplane's attitude (not altitude!) on its last approach to landing or as it takes off, flying at $300\mathrm{km\,h^{-1}}$ airspeed or roughly $80\mathrm{m\,s^{-1}}$. I have drawn it with a steep angle of attack. Air near sea-level atmospheric pressure has a density of about $1.25\mathrm{kg\,m^{-3}}$ (molar volume of $0.0224\mathrm{m^{-3}})$. The change in momentum diagram is shown, whence the change in vertical and horizontal momentum components are (assuming the speed of flow stays roughly constant):

$$\Delta p_v = p_b \sin\theta;\quad\quad\Delta p_h = p_b \,(1-\cos\theta)$$

At the same time, the deflecting wing presents an effective blocking area to the fluid of $\alpha\,A\,\sin\theta$ where $A$ is the wing's actual area and $\alpha$ a scale factor to account for the fact that in the steady state not only fluid right next to the wing is distrubed so that the wing's effective area will be bigger than its actual area. Therefore, the mass of air deflected each second is $\rho\,\alpha\,A\,v\,\sin\theta$ and the lift $L$ and drag $D$ (which force the engines must afford on takeoff) must be:

$$L = \rho\,\alpha\,A\,v^2\,(\sin\theta)^2;\quad\quad D = \rho\,\alpha\,A\,v^2\,(1-\cos\theta)\, \sin\theta$$

If we plug in an angle of attack of 30 degrees, assume $\alpha = 1$ and use $A = 1000\mathrm{m^3}$ (roughly the figure for an Airbus A380 wing area), we get a lifting force $L$ for $\rho = 1.25\mathrm{kg\,m^{-3}}$ and $v = 80\mathrm{m\,s^{-1}}$ of 200 tonne weight. This is rather less than the takeoff weight of a fully laden A380 Airbus (which is 592 tonnes, according to the A380 Wikipedia page) but it is an astonishingly high weight just the same and within the right order of magnitude. As I said, experiment is Queen here. We see that the wing's effective vertical cross section is bigger than the actual wing by a factor of 2 to 3. This is not surprising at steady state, well below speed of sound flow: the fluid bunches up and the disturbance is much bigger than just around the wing's neighbourhood. So, plugging in an $\alpha = 3$ (given the experimental fact that the A380 can lift off at 592 tonnes gross laden weight), we get a drag $D$ of 54 tonne weight (538kN) - about half of the Airbus's full thrust of 1.2MN, so this ties in well with the Airbus's actual specifications, given there must be a comfortable margin to lift the aeroplane out of difficulty when needed.

In these F4 / C5 grade winds (and up to three times faster in normal flight), we see therefore there simply isn't any shortage of lift. The aeronautical engineering problem is more about keeping this plentiful lift stably directed upwards and allowing the aeroplane to hold a steady attitude and keep any torques arising from lift nonuniformity from flipping the plane over.

As the aeroplane picks up speed, the ram pressure calculated above is proportional to the square of the airspeed (see my answer to Drag force at high speeds ), so that at full speed the effect more than accounts for the drop in air density and the shallower angle of attack - we cannot make this downwards ram pressure without overcoming the much greater horizontal hindwards component - drag - so that is important to fly with low angle of attack for good fuel efficiency.

Refining the Mathematical Model

It is important to heed that the above description in terms of momentum difference between incoming air and the downwash begotten by the wing is exactly the same physics as the "more popular" descriptions given in terms of the Bernoulli equation and the integration of pressure around the wing. This is easy to see: the Navier-Stokes equation(See the Wikipedia page for the derivation of the Navier-Stokes equation ), is a very simple application of nothing more than Newton's second and third laws to infinitessimal volumes of fluid, notwithstanding the lack of knowledge about its fundamental mathematical properties (as bespoken by the Clay Mathematics Millenium Prize's unclaimed status: I love the Navier-Stokes equation- such a simple, readily grasped idea so baldly just an embodiment of Newton's laws, yet throwing up profound mysteries that show us scientists how little we yet know about the World). The steady state Navier Stokes equation for a perfect, incompressible fluid is (here $\vec{v}$ is the steady state velocity field and $p$ the scalar pressure field):

$$(\vec{v}\cdot \nabla) \vec{v} = \nabla \left(\frac{|\vec{v}|^2}{2}\right) + \nabla\wedge(\nabla\wedge\vec{v}) = -\nabla p$$

which gives $\nabla\left(p + \frac{|\vec{v}|^2}{2}\right) = 0$ or $p + \frac{|\vec{v}|^2}{2} = const$ for an irrotational flow ($\nabla\wedge\vec{v} = \vec{0}$) when integrated along the integral curve of $\vec{v}$, i.e. a streamline. Or, alternatively, we can argue in a more first principles way in this simple case: the force on a infinitessimal volume is $-\nabla p$ and the acceleration of a particle on the streamline is, by application of the Serret-Frenet formulas (here $s$ is the arc length along the streamline through the particle and $\kappa$ the path's curvature):

$$\mathrm{d}_t (v \hat{\mathbf{t}}) = \mathrm{d}_s v \times \mathrm{d}_t s\, \hat{\mathbf{t}} + v\,\mathrm{d}_s(\hat{\mathbf{t}})\,\mathrm{d}_t s=v\,\mathrm{d}_s v, \hat{\mathbf{t}} - \kappa\,v^2\,\hat{\mathbf{n}}=\mathrm{d}_s \left(\frac{v^2}{2}\right)\, \hat{\mathbf{t}} - \kappa\,v^2\,\hat{\mathbf{n}}$$

whence, on applying $\vec{F} = m \vec{a} \Rightarrow -\nabla p \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z = \rho\,\vec{a}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$, we get:

$$-\nabla p = \rho \left(\mathrm{d}_s \left(\frac{v^2}{2}\right)\, \hat{\mathbf{t}} - \kappa\,v^2\,\hat{\mathbf{n}}\right)$$

which again yields $p + \frac{|\vec{v}|^2}{2} = const$ when integrated along a streamline (here we can see the sideways (normal to streamline) centripetal force $-v^2\,\hat{\mathbf{n}} / R$ given by the wonted $v^2/R$ formula). So we can (and will, below), for example, apply the Theorem of Blasius to calculate lift, and be assured it is no more than a quantification of Sklivv's idea that "aeroplanes thrust air downwards, so the air thrusts aeroplanes up". The pressure difference between the upper and lower surface of a wing exists because the wing is pushing the air down, not a separate phenomenon. Often one hears that the Bernoulli principle applied to wings is wrong: this is not true. There is a fallacy (to be discussed below) as shown by experiment (and, hand-wavingly, by theory) in the wonted demonstration of lift using Bernoulli's principle, but the idea is basically sound, as it must be from its derivation from the Navier-Stokes equation and Newton's laws shown above.

A Joukowsky Aerofoil Calculation and Errors in Wonted Application of Bernoulli's Principle to Wings

We look at a 2D calculation of lift by Bernoulli's principle, or, equivalently, by application of the Theorem of Blasius. The common misconception here is that airflows split at the wing's leading edge and two neighbouring particles will reach the wing's lagging edge at the same time, so that the upper particles must fare the curved surface at higher speeds and therefore the pressure on the upper wing surface is less. Actually, the upper path particles are sped up much more than this explanation implies and reach the wing's lagging edge well before their lower-path-faring neighbours. See this wonderful video from the University of Cambridge, particularly at about 50 seconds in. This fact shows that the circulation $\oint_\Gamma \vec{v}\cdot\mathrm{d}\vec{r}$ around the wing's surface $\Gamma$ is nonzero, a fact which we intuitively expect from simple theory (as shown below) and which is amply confirmed in experiment: see the video, or go to the end of a runway of a large airport on a damp day so that you can let big commercial jetliners fly over you at about 50m height (take your earmuffs). On a damp day, you will see vortices breaking off the wings' outer edges, will see them swirling in the damp air for many seconds in the aeroplane's wake and, if you take you hearing protection off after the aeroplane has pass, you will hear the vortices crackling in the air, sounding a little like waves washing on the beach. This is much more fun than it sounds when your children are badgering you to do such a thing and, from the sights and sounds, I learnt heaps more from doing it that I thought I would. Even though the following calculation has an air of theoretical soundness and "first principles" to it, it is important to understand that it too is an experimental model: the circulation is forced into our description, motivated by the confirmation of the former's existence by experiment. The Kutta-Joukowski Condition (see Wikipedia page for Kutta Condition) as well as the Wikipedia page for the Kutta-Joukowski Theorem is little more than an ad-hoc experimentally motivated fix: it is simply this. When we model the flow with a Joukowski aerofoil (described below), there is a sharp, lagging edge on the wing. This begets a singularity with unphysical, infinite velocties. However, by postulating and choosing the right circulation in the flow, we can put a stagnation point at the lagging edge, thus cancelling the singularity, regularising our solution and also forcing the experimentally observed condition that there is only ever one stagnation point at the wing's leading edge, never elsewhere.

Another way to look at this experimentally motivated condition is well explained in this answer to the Physics SE question Does a wing in a potential flow have lift?. An irrotational, inviscid, incompressible flow cannot alone lift a wing. We add circulation to "fudge" a compensation for this theoretical lack: viscosity is "nature's way of enforcing the Kutta-Joukowsski condition".

So we begin with the complex variable method (see the Wikipedia page for "Potential flow" in the section "Analysis for two-dimensional flow" to study a potential flow i.e. irrotational ($\nabla \wedge = \vec{0}$) velocity field $\vec{v}$ with a potential $\psi$ such that $\vec{v} = -\nabla \psi$ that is also incompressible (continuity equation $\nabla\cdot \vec{v} = \nabla^2 \psi = 0$). See also the Physics SE questions Finding Stagnation Points from the complex potential ).

The main method here is to use the Joukowski transform:

$$\omega(z,\,s_z,\,s_\omega) = \frac{ s_\omega }{2}\left(\frac{z}{ s_z } + \frac{ s_z }{z}\right)$$

to map the potential flow corresponding to a spinning, offset cylinder (see the NASA page “Lift of a Rotating Cylinder" ) into the flow around the image of this cylinder under the Joukowsky transform. The truly weird Flettner Aeroplane actually used spinning cylinders rather than wings to fly successfully. The Joukowsky transform maps the circle $|z| = s_z$ onto the real axis between the points $\omega = \pm s_\omega$ in the $\omega$-plane; this section of the real axis between $\omega = \pm s_\omega $ is then the branch cut for the inverse Joukowski transform. The Joukowsky transform is a two to one mapping, and the branches of the inverse Joukowski transform map the whole $\omega$-Riemann sphere (if we define the stereographic projection so that $|z| = s_\omega $ is the $\omega$-Riemann sphere’s equator) separately to the inside and outside of the circle $|z| = s_z$ in the $z$-plane (which outside and inside can be thought of the Northern and Southern hemispheres of the $z$-Riemann sphere, if the stereographic projection is chosen so that the circle $|z| = s_z$ is the $z$-Riemann sphere’s equator). The $\omega$-Riemann surface is made by slitting two copies of the Riemann sphere along the branch cut and stitching the edges together, to get a double cover of genus nought for the $\omega$-Riemann sphere. For this problem, I define the branch cut as slightly differently from the real axis section between the $\pm s_\omega$, I define it as the path:

$$\operatorname{Im}(\omega) = h \cos\left(\frac{\pi}{2} \operatorname{Re}(\omega)\right)$$

between the two branch points with an adjustable height parameter $h$, for reasons that will become clear.

The radius $r$ of the spinning cylinder radius is chosen so that cylinder surface passes through the point $z=+s_z$, which is the image of one of the branch points in the $\omega$ plane. This achieves the sharp edge that becomes the lagging edge of our aerofoil.

The complex potential for the spinning cylinder is:

$$\Omega(z) = v \,e^{-i\alpha}\,\left(z- \delta\right) + \frac{r^2 \,v\, e^{+i\alpha }}{z- \delta } + i\,a\,\log\left(z - \delta \right)$$

where $\alpha$ is the angle of attack, $\delta = \delta_r + i\,\delta_i$ is the offset and $r$ is the radius of the cylinder steeped in a uniform flow which converges to $v$ metres per second along the positive real axis, as $z\to\infty$. The logarithm and dipole terms put a branch point and pole at the cylinder’s centre, so the flow is perfectly valid outside and on the cylinder. $a$ is the circulation. If we let $\phi$ stand for the angular co-ordinate labelling the cylinder’s edge, there are two stagnation points on the cylinder with angular co-ordinates $\phi_\pm$ where $\mathrm{d}_z \Omega(z) = 0$, i.e. when:

$$e^{i\,(\phi_\pm - \alpha)} = -i\frac{a}{2\,v\,r}\pm\sqrt{1-\left(\frac{a}{2\,v\,r }\right)^2} = \exp\left(-\arcsin\frac{a}{2\,v\,r }\right)$$

Now, we map this flow to the $\omega$ plane and apply the Theorem of Blasius to the image of the offset circle so as to work out the lift on this image. The image can be plotted with the Mathematica command:

$$\small{\mathrm{P[\delta_r\_, \delta_i\_] := \\ ParametricPlot[\{Re[\omega[\delta_r + i \delta_i + \sqrt{(1 - \delta_r)^2 + \delta_i^2} Exp[i \theta]], Im[\omega[\delta_r + i \delta_i + \sqrt{(1 - \delta_r)^2 + \delta_i^2} Exp[i \theta]]\}, \{\theta, 0, 2 \pi\}]}}$$

and the result is drawn below in the $\omega$-plane for $s_z = s_\omega = 1$, $\delta_r = -0.1$, $\delta_i = 0.3$ (i.e. the spinning circle offset so that its centre is at $-0.1+i\,0.2$ and with a radius $r = \sqrt{(1 - \delta_r)^2 + \delta_i^2}$ so that its image passes through the branch point $\omega = +s_\omega = 1$ in the $\omega$-plane:

Joukowski Aerofoil

Now we come to the crucial Kutta-Joukowski postulate, an experimental "fudge". The sharp edge on the aerofoil above would normally map the flow in the $z$-plane so that there were an unphysical infinite velocity at this sharp point. In practice, it is seen in wind tunnel tests that the streamlines stay tangent to the upper surface, and that there is one stagnation point at the wing's leading edge (intuitively the air "crashes" here) and no other stagnation points on either the top of bottom of the wing. Sometimes there is a small region of turbulence around the wing's lagging edge (as in the University of Cambridge video) (i.e. the incompressible potential flow model fails here) or the flow peels smoothly off the lagging edge. The way we achieve effects similar to experiment and "renormalise" our solution is to add the right amount of circulation $a$ to the flow so that one of the stagnation points on the spinning cylinder is mapped to the sharp edge (the branch point at $\omega = +s_\omega$) in the $\omega$-plane: the stagnation thus cancels the otherwise unphysical infinite velocities there and "regularises" our solution. With the radius of the cylinder chosen as $r = \sqrt{(1 - \delta_r)^2 + \delta_i^2}$, it can readily be shown from the equation above for the stagnation point positions that the circulation needed is:

$$a = 2 v\,\delta_i \cos\alpha + 2\,v\,(1-\delta_r) \sin\alpha$$

This then is the wholly experimentally motivated Kutta-Joukowski condition. It is motivated by the knowledge that circulation is observed aroung wings, there is experimentally only one stagnation point on the leading edge of the wing and the fact that the right amount of circulation can reproduce these experimentally seen results.

When this is done, the Blasius theorem lift calculation done around the transformed Joukowski aerofoil in the $\omega$-plane is:

$$\begin{array}{lcl}D_\ell - i\,L_\ell &=& \frac{i\,\rho}{2}\oint_{\Gamma_\omega} (\mathrm{d}_\omega \Omega)^2 \,\mathrm{d} \omega\\ &=& \frac{i\,\rho}{2}\oint_{\Gamma_z} (\mathrm{d}_z \Omega)^2 \frac{1}{\mathrm{d}_z \omega}\,\mathrm{d} z\\ &=& -\pi\,\rho \Sigma[\,\mathrm{residues\,of\,}\,(\mathrm{d}_z \Omega)^2 \frac{1}{\mathrm{d}_z \omega}\,\mathrm{at\,poles\,within\,}\Gamma]\\ &=& -4\,\pi\,i\,\rho\,a\,v\,e^{-i\,\alpha}\end{array}$$

where $\Gamma_\omega$ is the Joukowski aerofoil and $\Gamma_z$ the transformed aerofoil (i.e. the spinning cylinder). So there is no lift without circulation. It's worth stating again:

An irrotational, inviscid, incompressible flow cannot alone lift a wing. We add circulation to "fudge" a compensation for this theoretical lack: viscosity is "nature's way of enforcing the Kutta-Joukowsski condition".

Now we substitute the Kutta-Joukowski condition to get:

$$D_\ell + i\,L_\ell = 8\,\pi\,i\,\rho\,v^2\,\left(\delta_i\,\cos\alpha + (1-\delta_r)\,\sin\alpha\right) \frac{s_z^2}{s_\omega} e^{+i\alpha}$$

We now need to scale the velocities so that the relative airspeeds are equal in the $\omega$- and $z$-planes.

The above is the force per unit length (in direction normal to the page) on the wing and its direction is the direction in the $\omega$-plane. We have:

$$\lim\limits_{\omega\to\infty} \left(\mathrm{d}_\omega \Omega(\omega(z))\right) = \lim\limits_{z\to\infty} \left(\mathrm{d}_z\Omega(\omega(z))\right) \lim\limits_{\omega\to\infty} \left(\mathrm{d}_\omega z\right) = 2 \,e^{-i\alpha} v \frac{s_z}{s_\omega}$$

so we need $s_\omega = 2$ and $s_z = 1$, then $\delta$ will be a dimensionless parameter defining the offset of the $z$-plane cylinder as a fraction of its radius. But now the $\omega$-plane planform width of the wing is 4 units. Moreover, the above calculation yields the force per unit length (normal to the 2D flow). So we divide the result for $s_\omega = 2$ and $s_z = 1$ by 4 and then scale up by the total wing area to get the total force on the wing. Furthermore, we need to we rotate the flow in the sketch below so that the incoming flow is horizontal (i.e. in the direction of the aeroplane's relative air velocity) in the $\omega$-total force on the wing above becomes:

$$D + i\,L = \pi\,i\,\rho\,v^2\,A\,\left(\delta_i\,\cos\alpha + (1-\delta_r)\,\sin\alpha\right)$$

We witness the d'Alembert paradox: the perfect flow cannot model the drag. Now let's put some numbers in. If we put $\delta = 0$, then the wing is simply the straight branch cut between $\omega = \pm 1$, so we have a version of the calculation I began with but now refined to take account of the full flow pattern. With $\alpha = 0.3$ (a little less than 20 degrees), $\rho = 1.25\mathrm{kg\,m^{-3}}$, $v=80\mathrm{m\,s^{-1}}$ and $A = 850\mathrm{m^2}$, we get $L=643\mathrm{tonne}$, pretty near to the Airbus's fully laden takeoff weight. If we chose the parameters $\delta_i = 0.2$, $\delta_r =-0.1$ to give a wing shape that does not seem too fanciful for a jetliner wing with the lagging edge flaps wound fully out for takeoff and landing (see the plot below) we get about 1200 tonnes lift for our $300\mathrm{km\,h^{-1}}$ airspeed. Clearly this is optimistic and the overreckonning arises from the assumption of equal effectiveness of whole wingspan, whereas the tips will clearly not be well modelled by 2D flow. Not all of the wings will work as modelled, thus the $A$ in this formula is somewhat less than the planform area. What the flow model does show (see below), however, is that the effective vertical cross section presented to the incoming air is much greater than the tilted area $A \,\sin\theta$ assumed in the very simple model at the beginning of my answer. At steady state, a considerable cross section of air both above and below the vertical cross section is bent downwards and contributes to the effect "aeroplanes thrust air downwards, so the air thrusts aeroplanes up" described in Sklivv's answer.

Now, to plot the complete transformed flow in the $\omega$-plane, we must use the inverse Joukowski transform. To do this successfully, one must use the right branches of the inverse transform in the right co-ordinate patches. For Mathematica, which puts the branch cut for the square root function along the negative real axis (the namespace std::sqrt in Microsoft Visual C++ puts it along the positive real axis), we define the following chart functions, which are particular branches of the inverse transform:

$$\zeta_1(\omega) = \frac{s_z}{s_\omega}\left(\omega- i \sqrt{\omega-s_\omega}\,\sqrt{-\left(\omega+s_\omega\right)}\right)$$ $$\zeta_2(\omega) = \frac{s_z}{s_\omega}\left(\omega+ i \sqrt{\omega-s_\omega}\,\sqrt{-\left(\omega+s_\omega\right)}\right)$$ $$\zeta_3(\omega) = \frac{s_z}{s_\omega}\left(\omega- \sqrt{\omega^2-s_\omega^2}\right)$$ $$\zeta_4(\omega) = \frac{s_z}{s_\omega}\left(\omega+ \sqrt{\omega^2-s_\omega^2}\right)$$

and then the following Mathematica commands will plot the full flow:

$$\small{\mathrm{\Omega[z\_,\,\delta\_,\,v\_,\,r\_,\,a\_,\,\alpha\_,\,s\_]:= v\,e^{-i\,\alpha}\left(\frac{z}{s}-\delta\right) + \frac{r^2\,v\,e^{i\,\alpha}}{\frac{z}{s}-\delta} + i\,a\,Log\left[\frac{z}{s}-\delta\right]}}$$ $$\small{\mathrm{G[z\_,\,\delta_r\_,\,\delta_i\_,\,\alpha\_]:=\Omega\left[z,\,\delta_r+i\,\delta_i,\,1,\,\sqrt{(1-\delta_r)^2 + \delta_i^2},2\,\delta_i Cos[\alpha] + 2\,(1-\delta_r)\,Sin[\alpha],\,\alpha,\,1\right]}}$$

$$\small{\mathrm{S[\delta_r\_, \delta_i\_, \alpha\_, h\_, c\_] := \\ Show[ContourPlot[ Im[If[(Abs[x] < 1 ) \wedge (y > 0) \wedge (y < h\, Cos[\pi x/2]), G[\zeta_1[x + i y], \delta_r, \delta_i, \alpha]], If[x < 0, G[\zeta_3[x + i y], \delta_r, \delta_i, \alpha]], G[\zeta_4[x + i y], \delta_r, \delta_i, \alpha]]]]], \{x, -2, 2\}, \{y, -2, 2\}, Contours \to c, MaxRecursion\to 2, PlotPoints \to 300, AspectRatio \to 1], P[\delta_r, \delta_i, \{Black, Thick\}]]}}$$

where $\mathrm{P}[]$ is the parametric plot command above used to plot the aerofoil. The above use of the branch functions works for $\delta_r < 0$: other branches are needed for correct results when $\delta_r > 0$. The parameter $h$ bends the branch cut so that it bows upwards and stays inside the aerofoil, thus allowing the branches of the inverse Joukowsky transform to plot the mapped cylinder flow properly. Drawn below is the outcome from the command $\mathrm{S[-0.1, 0.2, 0.2, 0.2, 100]}$, i.e. the flow around the wing for an angle of attack of 0.2 radians, the circle offset parameters of $-0.1 + 0.2\,i$, a bow in the branch cut so that $h=0.2$. Witness the branch cut inside the aerofoil below and also how far from the wing's surface its effect stretches. The effective vertical component of the wing's area that is presented to the flow is clearly much greater than the actual vertical component of the wing's area, so the factor of 2 to 3 scaling in the A380 Airbus lift as reckoned by the simple fluid deflexion calculation seems highly plausible and unsurprising.

Joukowski Aerofoil Flow

Lastly, to come the full circle, here is an animation to be found on web pages "Irrotational plane flows of an inviscid fluid" at the University of Genoa's environmental engineering department; see http://www.diam.unige.it/~irro/. The animation shows the progress of fluid particles for the Joukowski aerofoil flow, illustrates the assertion that the flow above the wing traverses the wing much more quickly than the flow underneath and lastly, shows very well the main thesis that "aeroplanes thrust air downwards".

Joukowsku Aerofoil Animation

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@DImension10AbhimanyuPS I once did a fluid dynamics course when I was very young and it is the kind of thing that drives a physicist / mathematician mad. The "theory" is all rules of thumb and its a mad mishmash of cookbook physics and mathematical abuse. This of course is owing to the mathematical complexity - the existence of the Clay mathematics prize shows how much we really know about deep fluid dynamics (although numerical models are getting very good). I decided very early on that the only rigorous knowledge in this field is experiment, so I insist on explanations in these terms. –  WetSavannaAnimal aka Rod Vance Sep 18 '13 at 7:40
@DImension10AbhimanyuPS, why does every other question have to be edited by you? It looks suspiciously like you're doing it for the rep. I don't see the point in editing a question which has been inactive for a long long time, just to add one silly "faq" tag –  Pranav Hosangadi Sep 23 '13 at 5:56
@PranavHosangadi No I don't do it for rep (you don't get rep for editing posts once you have > 2000 rep) and retagging is useful, and "faq" is not silly. see also the meta post meta.physics.stackexchange.com/questions/4653/… –  Dimensio1n0 Sep 23 '13 at 7:54
If editing answers brings the question to the top of math.se, then more traffic sees your answer and you will likely get more upvotes. So editing posts may garner rep, indirectly. –  ErikE Jun 11 '14 at 20:41

Since you asked for an explanation appropriate to an non-specialized audience, maybe this will do: "A Physical Description of Flight; Revisited" by David Anderson & Scott Eberhardt. It is a revision of the earlier "A Physical Description of Flight" (HTML version).

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Really great paper. –  Justin L. Nov 6 '10 at 7:34
+1 This looks like a pretty handy explanation. –  Noldorin Nov 7 '10 at 23:07
A block-quote or more extensive description would be more helpful than just the link. –  YatharthROCK Jun 29 '14 at 9:30

Without going into the excellent and detailed mechanics explaining reaction lift that others have provided for this answer, I just want to say that contrary to popular belief/high school physics textbooks, airplanes do not fly solely on account of Bernoulli's principle. According to Walter Lewin's excellent "For the Love of Physics":

"Bernoulli's principle accounts for 20% of an airplane's lift, the rest is provided by reaction lift."

Walter Lewin also poses an insightful question if planes really fly due to the equal transit theory and Bernoulli's principle (they do not!). enter image description here

"...then how do planes fly upside down?"

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The problem is that the faulty theory has the name "Bernoulli" attached to it. The true Bernoulli principle, and reaction explanation, are the same. –  Mike Dunlavey Sep 20 '13 at 12:47
@MikeDunlavey agreed: see the comments in my answer about the Navier-Stokes equation. Also, from a aerospace engineering friend, the problem is the "equal transit time" assumption whereas experimentally the upper transit time is of the order of one half the lower (as shown by either the simple potential flow 2D model or in the University of Cambridge video, particularly at about 50 seconds in), and since Bernoulli pressure drop is proportional to $v^2$ this makes for a huge difference between the equal transit time theory and reality. –  WetSavannaAnimal aka Rod Vance Sep 20 '13 at 13:02
The question is indeed an insightful one so +1, but when flying upside down the angle of attack needs to be adjusted so that there is a higher speed over the uppermost (formerly underside) of the wing. Bernoulli's principle still works applied to the realistic flow models, i.e. with much shorter transit time on the uppermost (formerly underside) of the wing. –  WetSavannaAnimal aka Rod Vance Sep 20 '13 at 13:14
That is true. The variable angle of attack on the wings on a fixed wing aircraft when the aircraft is level would require the wings themselves to rotate. Of course, on most airliners, only the elevator control surfaces can adjust their angle with the planes fuselage, while the wings are always fixed in position. I guess that's why we don't see too many demonstrations of airliners flying upside down! –  shortstheory Sep 23 '13 at 6:29
@shortstheory: Airliners are quite capable of negative-G flight (but not for long, because of oil sumps etc.) In fact they have to be strong enough to handle multiple Gs up or down. On a fixed wing aircraft the angle of attack is changed by pitching the nose up or down. The purpose of the elevator is to control the angle of attack of the main wing by pitching the entire aircraft up or down. Notice, next time you fly, how when the aircraft slows down for landing, it pitches up, because at the slower speed more angle of attack is needed. –  Mike Dunlavey Nov 19 '13 at 20:00

Consider the velocity field of the particles in the air mass in a 2D projection of the X(forward) and Z(up) axes. For each particle, Integrate over area and time, to derive the center of air-mass momentum (p) before and after the passage of the airplane : dp/dt. (On a very calm morning, with no wind or turbulence, the center of air-mass and its momentum is stationary in Z(assume level un-accelerated flight), and equal to the True Airspeed in X pointing in the aft -X direction. Integrate over the area and you will find that the center and momentum of the particle and vector field has changed, with passage of the plane. This center of air mass and center of momentum will move forward(+X) and downward(-Z) relative to its original state. The equal and opposite momentum change with time dp/dt of the airplane is a force. We might label the -X component "drag" and the +Z component "lift"(careful: the airplane coordinate system is different from the stationary airmass). This is a dissipative system, so don't wait too long after the plane passes to record the vector field. We can observe this process in contrails on clear days when the high altitude air is cold and relatively moist. Sadly since we mostly view them from below with a projection along the Z, we miss the downward component of the momentum field. You can see this as a test pilot, flying as chase wing-man, in formation (projection in the Y-Z plane from behind or X-Z from the side). Expand this model to 3D to include lateral or Y axis flow and effects! I suggest this "p-dot"(dp/dt) of momentum-change explanation is better, than "pushing" or "pulling" the air downwards, because the later may confuse position and momentum in the view of the reader. This is also the first term(LHS) in the beautiful Euler-LaGrange equation, which would lead to an even more elegant analysis of this question!

As a new user, I will need to figure out how to attach the appropriate Figures and Equations to this post...-thanks

Note: The drag equation is really the ideal gas law, except density replaces m/V.

P/rho = R T :

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Please Use MaThJaX. –  Dimensio1n0 Nov 16 '13 at 3:15

Fluid interactions with solid bodies depend on the fluid properties and the geometry of the object. In the case of an aeroplane, we have air as our fluid and an aerofoil geometry. The aerofoil geometry is designed on purpose to force fluid under it preferentially to above it. This results in a pressure difference, which then leads to a buoyancy force that accelerates the wing according to Newton's second law (lift). Bernoulli's law is relevant for calculating the fluid problem.

So, to achieve flight, all you need is some well-designed aerofoils and some way of imparting an initial velocity. To keep flying you need to keep your speed high and to keep flying stably you need a well-designed aircraft with the centre of mass, centre of thrust and centre of lift being in the same position.

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protected by Qmechanic Oct 31 '13 at 21:51

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