# How Does an Airplane Wing Work: a Primer on Lift - Comments

• Insights
Gold Member
boneh3ad submitted a new PF Insights post

How Does an Airplane Wing Work: a Primer on Lift Continue reading the Original PF Insights Post.

• S.G. Janssens, tech99 and Greg Bernhardt

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Really nice Insight!

Gold Member
I should have titled it "Lift or: How I Learned to Stop Worrying and Love Bernoulli and Newton." That's a real missed opportunity.

• AlanDRPh and Greg Bernhardt
I should have titled it "Lift or: How I Learned to Stop Worrying and Love Bernoulli and Newton." That's a real missed opportunity.
Ripe for the sequel :)

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"

Gold Member
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.

• Greg Bernhardt
Hi,

Thanks,

MJ

• Greg Bernhardt
Gold Member
What level of material interests you?

For example, the go-to text for introducing undergraduate engineers to the basic principles of flight is Introduction to Flight 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.

• Greg Bernhardt
rcgldr
Homework Helper
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.

Gold Member
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.

rcgldr
Homework Helper
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.

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Chestermiller
Mentor
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

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)

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.

• AlanDRPh
rcgldr
Homework Helper
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.

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

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.

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.

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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

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Chestermiller
Mentor
For me one of the confusing issues is the effect of the frame of reference.
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.

• Greg Bernhardt
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.

I love you

• Greg Bernhardt
Gold Member
The two frames of reference in question here are related by a simply Galilean transformation and are otherwise functionally identical (already stated by @Chestermiller 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.

Even at flat pitch for a symmetrical airfoil, the air is moving a greater distance at the same time.
Nothing requires any sort of "same time" constraint here.

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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.

Gold Member
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.

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

rcgldr
Homework Helper
work done ... frame of reference
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?

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Gold Member
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 not 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.

@rcgldr 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.

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• cjl
rcgldr
Homework Helper
Where are those numbers from and what do they mean?
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.

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.

http://www.allstar.fiu.edu/aero/airflylvl3.htm

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?

amount of time in general.
From the wing's frame of reference, the air is slowed down due to drag.

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