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How do wings work?

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2003 Phys. Educ. 38 497

(http://iopscience.iop.org/0031-9120/38/6/001)

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SPECIAL FEATURE: FLIGHTwww.iop.org/journals/physed

How do wings work?Holger Babinsky

Department of Engineering, University of Cambridge, Cambridge CB2 1PZ, UK

E-mail: hb@eng.cam.ac.uk

AbstractThe popular explanation of lift is common, quick, sounds logical and givesthe correct answer, yet also introduces misconceptions, uses a nonsensicalphysical argument and misleadingly invokes Bernoullis equation. A simpleanalysis of pressure gradients and the curvature of streamlines is presentedhere to give a more correct explanation of lift.

M This article features online multimedia enhancements

The science behind aeronautics continues tofascinate and many students are attracted toengineering as a result of an early interest inaircraft. The most commonly asked question ishow a wing can produce lift. Unfortunately themost widely used explanation of lift is wrong in anumber of key points. Not only is this confusingfor students, but in the worst case it can leadto a fundamental misunderstanding of some ofthe most important aerodynamic principles. Inthis article I will demonstrate why the popularexplanation for lift is wrong and then propose analternative explanation.

The popular explanationFigure 1 shows a typical aerofoilthe cross-sectional shape of a wingimmersed in a flowwhere the streamlines have been visualized withsmoke particles. At the front is the stagnation point(S), which is the location where the oncoming flowdivides into that moving above and that movingbelow the wing. The argument revolves aroundthe observation that the distance from this pointS to the trailing edge (T) is greater along theupper surface than along the lower surface. If itis assumed that two neighbouring fluid particleswhich split at S should meet again at T thenthis requires that the average velocity on the uppersurface is greater than that on the lower surface.

Now Bernoullis equation is quoted, whichstates that larger velocities imply lower pressures

ST

Figure 1. Streamlines around an aerofoil sectionvisualized with smoke.

Blow along upper surface of paper

Paper

Figure 2. Paper lifts when air is blown along its uppersurface.

and thus a net upwards pressure force is generated.Bernoullis equation is often demonstrated byblowing over a piece of paper held between bothhands as demonstrated in figure 2. As air is blownalong the upper surface of the sheet of paper it rises

0031-9120/03/060497+07$30.00 2003 IOP Publishing Ltd P H Y S I C S E D U C A T I O N 38 (6) 497

H Babinsky

Mast

Sail

Figure 3. Flow along the cross section of a sail.

and, it is said, this is because the average velocityon the upper surface is greater (caused by blowing)than on the lower surface (where the air is moreor less at rest). According to Bernoullis equationthis should mean that the pressure must be lowerabove the paper, causing lift.

The above explanation is extremely wide-spread. It can be found in many textbooks and,to my knowledge, it is also used in the RAFsinstruction manuals. The problem is that, whileit does contain a grain of truth, it is incorrect in anumber of key places.

Whats wrong with the popularexplanation?The distance argumentWhile the aerofoil of figure 1 does indeed exhibit agreater distance between S and T along the uppersurface, this is not a necessary condition for liftproduction. For example, consider a sail that isnothing but a vertical wing (generating side-forceto propel a yacht). Figure 3 shows the crosssection of a sail schematically and it is obviousthat the distance between the stagnation point andthe trailing edge is more or less the same on bothsides. This becomes exactly true in the absence ofa mastand clearly the presence of the mast is ofno consequence in the generation of lift. Thus,the generation of lift does not require differentdistances around the upper and lower surfaces.

(a) (b) (c)Figure 4. Smoke particles flowing along a lifting aerofoil section.M An MPEG movie of this figure is available from stacks.iop.org/physed/38/497

The equal time argumentIt is often asked why fluid particles should meetup again at the trailing edge. Or, in other words,why should two particles on either side of the wingtake the same time to travel from S to T? There isno obvious explanation and real-life observationsprove that this is wrong. Figure 4 shows a selectionof still frames from a video recording of a smoke-wind-tunnel experiment (the video clip is availablein the online journal). Here, smoke particlesare injected simultaneously upstream of a liftingaerofoil section, generating a line of smoke as seenin figure 4(a). This line of smoke moves withthe flow, dividing into particles travelling aboveand below the aerofoil. By the time the smokehas passed the aerofoil (figure 4(c)) the particlesmoving along the upper surface are clearly aheadof those travelling along the lower surface. Theydo not meet up at the trailing edge. When liftis generated, fluid particles travelling along theupper surface reach the trailing edge before thosetravelling along the lower surface1.

The Bernoulli demonstrationBlowing over a piece of paper does notdemonstrate Bernoullis equation. While it is truethat a curved paper lifts when flow is appliedon one side, this is not because air is moving atdifferent speeds on the two sides. This can easilybe demonstrated by blowing along one side of astraight piece of paper as sketched in figure 5. Inthis case the paper does not experience a forcetowards the side subjected to the faster movingair. The pressure on both sides of the paper is the1 In fact, it can be proved theoretically that if the particles onthe upper surface reach the trailing edge at the same time asthose travelling along the lower surface no lift is produced.

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How do wings work?

Paper (hanging vertically)

Blow air along one side

Figure 5. A straight piece of paper hanging verticallydoesnt move when air is blown along one side.

same, despite the obvious difference in velocity.It is false to make a connection between the flowon the two sides of the paper using Bernoullisequation.

An alternative explanation for liftThe above argument has remained popular becauseit is quick, sounds logical and gives the correctanswer. However, my concern about using thisexplanation is that it introduces misconceptionsabout why aerofoil shapes generate lift, ituses a nonsensical physical argument and itoften includes an erroneous demonstration ofBernoullis equation.

Before we beginsome basic assumptionsThe key to understanding fluid flow around anobject is to examine the forces acting on individualfluid particles and apply Newtons laws of motion.While there are many different types of forcesacting on a fluid particle it is possible to neglectmost of these, such as surface tension and gravity.In fact, for most practical flows the only relevantforces are due to pressure and friction. As a firststep, we can also assume that there are no frictionforces at work either. This is because in mostflows friction is only significant in a very smallregion close to solid surfaces (the boundary layer).Elsewhere, friction forces are negligible.

We shall also assume the flow to be steady.In practice this means that we only considersituations where the overall flowfield does notchange very quickly with time.

With these assumption we can now derive therules governing fluid motion by considering theresultant pressure force acting on an individualfluid particle and applying Newtons second law,which states that force causes acceleration. As we

vStreamline

l

Fluid particle

Figure 6. Fluid particle travelling along a straight line.

shall see, a force acting in the flow direction causesfluid particles to change their speed whereas aforce acting normal to the flow direction causesstreamline curvature (by particle we refer to avery small but finite volume (or element) of thefluid, not individual molecules).

The truth about BernoulliImagine a fluid particle travelling along a straightline (but not at constant velocity) as shownschematically in figure 6. Let the x-directionbe in the direction of motion. If the particle isin a region of varying pressure (a non-vanishingpressure gradient in the x-direction) and if theparticle has a finite size l, then the front of theparticle will be seeing a different pressure fromthe rear. More precisely, if the pressure drops inthe x-direction (dp/dx < 0) the pressure at therear is higher than at the front and the particleexperiences a (positive) net force. Accordingto Newtons second law, this force causes anacceleration and the particles velocity increasesas it moves along the streamline. Conversely, ifthe pressure increases in the direction of the flow,the particle decelerates. This means that if thepressure drops along a streamline, the velocityincreases and vice versa. Bernoullis equationdescribes this mathematically (see the completederivation in the appendix).

However, the fact is often overlooked thatBernoullis equation applies only along a stream-line. There is no explicit relationship between thepressure and velocity of neighbouring streamlines.Sometimes, all streamlines in a flow originatefrom a region where there is uniform velocityand pressure (such as a reservoir or a uniformfree-stream) and in such a case it is possible toapply Bernoullis equation throughout the flow.But in the demonstration of Bernoullis equationshown in figure 2 the air moving along the uppersurface of the paper originates from the mouthof the person performing the experiment and the

November 2003 P H Y S I C S E D U C A T I O N 499

H Babinsky

v Curved streamline

Poutside

Pinside

Pressure force P

outside > Pinside

Figure 7. Fluid particle travelling along a curvedstreamline.

streamlines can be traced right back into thispersons lungs. There is no connection with thestreamlines underneath the paper and Bernoullisequation cannot be applied to compare the pressurein the two regions2. In fact, the pressure in theair blown out of the lungs is equal to that of thesurrounding air (and this is proved when blowingover a straight sheet of paperit doesnt deflecttowards the moving air).

Flow along curved streamlinesNext, examine a particle moving along a curvedstreamline as shown in figure 7. For simplicity wecan assume that the particles speed is constant3.Because the particle is changing direction theremust exist a centripetal force acting normal tothe direction of motion. This force can only begenerated by pressure differences (all other forcesare ignored), which implies that the pressure onone side of the particle is greater than that onthe other. In other words, if a streamline iscurved, there must be a pressure gradient acrossthe streamline, with the pressure increasing in thedirection away from the centre of curvature.

This relationship (derived mathematically inthe appendix) between pressure fields and flowcurvature is very useful for the understanding offluid dynamics (although it doesnt have a name).Together with Bernoullis equation, it describesthe relationship between the pressure field and theflow velocity field. A good demonstration of this2 One might argue that the air blown over the top surfacedoes eventually come to rest in the room and at that stagea connecting streamline might be drawn towards the lowersurface. However, in this case the flow is clearly affected byfrictionthis is what brings the flow to restand Bernoullisequation only describes frictionless flows.3 This in turn implies, according to Bernoullis equation, thatthe pressure along the streamline is constant (dp/dx = 0).

Concentricstreamlinesin a vortex(or tornado)

Pressure dropstowards centre

Figure 8. Pressure gradient across streamlines in avortex.

Figure 9. Streamlines around a lifting curved plateaerofoil.

is a tornado (or any vortex, such as that seen ina bathtub). As sketched in figure 8, an idealizedvortex consists of concentric circles of streamlines.The above relationship implies that there is apressure gradient across these streamlines, withthe pressure dropping as we approach the core.This explains why there are such low pressures inthe centre of vortices (and why tornados suckobjects into the sky). In a real, three-dimensional,tornado, the streamlines are not circles but spiralswhich originate from somewhere far away wherethe air is at rest and at atmospheric pressure.Applying Bernoullis equation to each of thesestreamlines shows that the velocities increase thecloser we get to the vortex core (which is what weobserve in nature)4.4 At the very centre, the assumption of frictionless flow is nolonger justified (because streamlines in opposite directions getvery close to each other, creating a strong shear flow) and theabove arguments no longer hold (the eye of the storm has lowflow velocities and low pressure).

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Figure 10. Simulated streamlines around thin andthick aerofoils.

Lift on aerofoilsNow we can return to the original problem.Figure 9 shows a schematic sketch of thestreamlines around the simplest lifting aerofoila curved plate. Far away the air is undisturbedby the presence of the wing, the pressure isatmospheric (= patm) and the streamlines arestraight and horizontal. Now consider movingalong a line from point A towards the surface,keeping on a path that is always perpendicular tothe local streamline direction. Starting at A wenote that the streamlines are straight and paralleland therefore there is no pressure gradient in thedirection of the dashed line. However, closerto the aerofoil streamlines become increasinglycurved and there must now be a pressure gradientacross the streamlines. From the direction ofcurvature we note that the pressure drops as wemove downwards. By the time we reach theaerofoil surface at B the pressure is noticeablylower than that at A (pB < patm). In the sameway we can imagine moving from C to D. Again,as we approach the aerofoil streamlines exhibitmore and more curvature but this time the pressureincreases towards the surface. At D the pressureis therefore greater than that at C (pD > patm).Hence pB < pD and this generates a resultant

(a)

(b)

(c)Figure 11. Streamlines around a symmetrical aerofoilat various angles of attack. (a) Positive angle of attack:lift points up. (b) Zero angle of attack: no lift.(c) Negative angle of attack: lift points down.

pressure force on the aerofoil, acting upwards, i.e.lift.

From the above we learn that any shapethat introduces curvature into the flowfield cangenerate lift. Aerofoils work because the flowfollows the local surface curvature on the upperand lower surfaces. It is not necessary to considerfrictional forces to explain lift, however; it is onlydue to the action of friction that streamlines takeup the pattern we would intuitively expect, sostrictly speaking lift would not be possible withoutfriction.

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(a) low angle of attack (b) high angle of attack (c) stalled flowFigure 12. Streamlines around an aerofoil at increasing angle of attack.

Some observations resulting from thisexplanationFollowing this line of argument it is possible tomake some interesting observations. For example,consider the difference between the streamlinesover a thin and a thick aerofoil as shownschematically in figure 10 (determined from acomputer simulation). Despite the difference inthickness, both have similar flow patterns abovethe upper surface. However, there is considerabledifference in the flow underneath. On the thinaerofoil the amount of flow curvature below thewing is comparable to that above it and we mightconclude that the overpressure on the underside isjust as large as the suction on the upper surfacethe two sides contribute almost equally to thelift. In the case of the thick aerofoil, however,there are regions of different senses of curvaturebelow the lower surface. This suggests that therewill be areas with suction as well as areas withoverpressure. In this case the lower surface doesnot contribute much resultant force and we canconclude that thin aerofoils are better at generatinglift. This is generally true, and birds tend to havethin curved wings. Aircraft do not, because of thestructural difficulties of making thin wings, andalso because the volume contained in the wing isuseful, e.g. for fuel storage.

A frequent question is how aircraft manage tofly upside down. To demonstrate this, figure 11shows the streamlines over a symmetrical aerofoilat positive, zero and negative angles of attack.Just by judging the degree of flow curvatureabove and below the wing it can be seen thatthis aerofoil produces positive, zero and negativelift respectively. Negative lift (which wouldbe required for flying upside down) is simplya question of the angle of attack at which theaircraft flies. Even non-symmetrical aerofoils can

generate negative lift, but they do require moresevere negative angles of attack (because they stillproduce positive lift at zero angle of attack) whichmakes them less suitable for flying upside down.

This also demonstrates the significance ofangle of attack, as seen again in figure 12. Asthe angle of attack of a wing increases, more flowcurvature is introduced above the wingcompare(b) with (a)and more lift is generated. However,at some point the flow is no longer capable offollowing the sharp curvature near the nose andit separates from the surface. As a result theamount of streamline curvature above the winghas reduced considerably (see figure 12(c)), whichcauses a sharp drop-off in lift force. Unfortunatelythe process of flow detaching from the surfaceoften happens instantaneously when the angle ofattack is increased, making the loss of lift rathersudden and dangerousthis is called stall.

ConclusionIn this article I have attempted to give a hands-onand correct explanation for Bernoullis equation,the relationship between streamline curvature andpressure, and lift. To explain lift it is notnecessary to go through all of the above stepsin the argument. Most students will be happywith the streamline pattern around a lifting wing(figure 1)because it intuitively looks rightand this should be exploited. A short, butcorrect, explanation might start by discussingthe existence of transverse pressure gradients incurved streamlines and applying this knowledge tothe flowfield around an aerofoil in a similar mannerto that shown in figure 9. This should explainwhy pressures on the two sides of an aerofoil aredifferent. There is no need even to introduceBernoullis equation or discuss the rather subtlesignificance of friction.

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AcknowledgmentsI would like to thank my colleagues, RexBritter, Peter Davidson, Will Graham, JohnHarvey, Harriet Holden, Tim Nickels and LenSquire, for many useful discussions and helpfulcomments. Will Graham also provided thecomputer simulations of aerofoil streamlines.

AppendixPressure gradient along streamlinesBernoullisequation

vStreamline

l

x

p p+dp

Surface area AFluid particle

In the above figure the cubic fluid particleexperiences a pressure p behind and a slightlydifferent pressure p + dp in front. This causes anacceleration according to Newtons second law:

F = ma = mdvdt

.

Here the resultant pressure force is F = dp A,which is negative because it points in the xdirection. The mass of the fluid particle can bedetermined from its volume and the fluid density: m = lA.

The magnitude of the pressure changebetween front and back can be determined fromthe pressure gradient in the streamwise direction(dp/dx) and the size of the particle:

dp = l dpdx

.

Combining all of the above gives

l dpdx

A = lA dvdt

which simplifies to

dp = dvdt

dx.

Noting that dx/dt = v, this can be rewritten asdp = v dv.

Now we can integrate between two points (1and 2) along a streamline to relate the pressure

difference between these points to the velocitydifference: 2

1dp =

21

v dv

p2 p1 = (v222

v21

2

)

which can be rearranged as

p1 +

2v21 = p2 +

2v22 .

Because points 1 and 2 are arbitrary locationsalong the streamline, the above equation canbe used to connect any two locations alonga streamlinethis is in essence Bernoullisequation. Note that the streamline need not bestraight.

Pressure gradient across curved streamlines

vCurved streamline

Poutside

Pinside

Pressure forceP

outside > Pinside

h

Surface area A

The centripetal force is F = mv2/R and wecan define pinside = p and poutside = p + dp.Similar to before we note that m = Ah anddp = h(dp/dn), where n is the coordinate in thedirection normal to the streamline (pointing awayfrom the centre of curvature).

Combining all of the above yields

F = A dp = Ahdpdn

= Ahv2

R

which can be simplified to

dpdn

= v2

R

which expresses the pressure gradient acrossstreamlines in terms of the local radius of curvatureR and the flow velocity v. If a streamline isstraight, R and dp/dn = 0. Therefore,there is no pressure gradient across straightstreamlines.

Received 9 September 2003PII: S0031-9120(03)68660-0

November 2003 P H Y S I C S E D U C A T I O N 503