# What makes air move faster above an airplane wing?

1. Jun 15, 2014

### Maxo

When looking at the shape of an airplane wing from the side, like here I'm wondering what makes the air move faster above the wing. I have some questions about this:

1. It rather looks like there is a longer distance for the air to travel above the wing than below it, but wouldn't a longer travel distance rather imply less speed? Why doesn't it?

2. It also looks like the air must "bend" more when going on top of the wing than below it (more easily seen here: ), wouldn't that also create more air resistance on top of the wing and thereby less speed? Why doesn't it?

3. What is it that makes the air move faster above the wing?

Last edited by a moderator: Apr 19, 2017
2. Jun 15, 2014

### AlephZero

This comes up regularly on PF.Look at the "related discussions" at the bottom of the page and click on the links.

To answer question 1 specifically, since there is a huge amount of nonsense about this on the web: if you look at the animation, the air takes a different amount of time to travel over the top and bottom surfaces of the wing, so trying to find a simple connection between the speed and the shape, or the length of the surface, is irrelevant.

The shape is not very important except for reducing drag. Planes can fly upside down. Even a flat plate will produce plenty of lift, if it is at an angle to the air flow.

3. Jun 16, 2014

### CWatters

I agree with AlephZero. There are many different ways to think about how wings work but many are over simplifications. Including this one...

What happens if you put a finger over the end of a hose pipe to restrict the opening. The water speeds up.

4. Jun 16, 2014

### rcgldr

Lift can be produced with a symmetrical wing or even a flat board. All that is needed is an effective angle of attack to divert the relative flow downwards. Imagine a vertical plane of air, perpendicular to the direction of travel of a wing, and what happens to that air as the wing passes through that imaginary plane. At the bottom of a wing, the air is pushed downward (lift) and a bit forward (drag) by the bottom surface of the wing. At the top of a wing, the upper surface moves forwards and downwards, and the air fills in what would otherwise be a void left behind that moving upper surface. If the effective angle of attack is not excessive, then the air accelerates mostly downward (lift) and a bit forward (drag) to fill in what would otherwise be a void. If the effective angle of attack is excessive, then the air tends to spin (vortices), resulting in less downwards acceleration and more forwards acceleration (stall).

Since the air has momentum, acceleration of air coexists with pressure differentials when a wing produces lift, higher pressure below the wing, lower pressure above the wing. Since air accelerates from higher pressure areas to lower pressure areas, the air ends up moving faster in the lower pressure area above a wing when using the wing as a frame of reference. If using the air as a frame of reference, more of the downwards acceleration occurs above a wing than below a wing, but the maximum speed occurs somewhere behind and below the trailing edge of a wing. So the statement that air flows faster over a wing is based on using the wing as a frame of reference, as opposed to the air itself as a frame of reference.

One of many web sites with a basic decription of how a wing works:

http://www.avweb.com/news/airman/183261-1.html?redirected=1

Last edited: Jun 16, 2014
5. Jun 16, 2014

### CWatters

I can't say that's the best description I've read but I've seen worse.

The subject is much debated on the web. Many people divide into two camps and claim either Newton or Bernoulli has the "real" explanation for how a wing works but it's a false dichotomy. It's both and more.

It's quite interesting to look at plots of lift vs angle of attack. Some non-symmetric sections can produce +ve lift at a -ve angle of attack. This one has a +ve coefficient of lift down to about -4 degrees...

http://www.rcfoamcrafters.com/design_files/naca_airfoil.gif

PS: Called the zero lift angle.

Last edited: Jun 16, 2014
6. Jun 16, 2014

### rcgldr

Newton's third law always holds, the Newton third law pair of forces are the force the wing exerts onto the air, and the force the air exerts onto the wing. Newtons second law, force = mass x acceleration, mostly applies except for ground effect and viscosity issues with interfere with flow. Also standard physics principles mostly apply, such as impulse = force x time = change in momentum of a parcel of air as a wing passes through it.

Using the air as a frame of reference, Bernoulli is somewhat violated by the fact that energy is added to the air by a wing. If the wing is very efficient, such as one with a long wing span, then the amount of energy added versus the lift produced is small, so Bernoulli would mostly hold.

Most cambered airfoils can produce +ve lift at a -ve angle of attack since it's defined by leading edge versus trailing edge, but downwash is still produced, and an alternative is to use effective angle of attack, which is defined to be zero when zero lift is produced.

The main idea is that a wing produces lift by diverting the relative (to the wing) flow downwards.

Getting back to the original questions:

There is a component of acceleration perpendicular to the flow (relative to the wing) to follow that "bend" that results in a reduction in pressure due to the air's momentum preventing it from immediately changing direction. The existance of that low pressure accelerates the air towards the low pressure area from all directions (except air can't flow through the wing), but since the pressure in front of the wing is greater than the pressure behing the wing, there's a net "backwards" acceleration of air over the wing (note that the air was initially pushed forwards a bit in front of the wing in that animation.

Last edited: Jun 16, 2014
7. Jun 16, 2014

### konghouyun

The above surface of the wing have a longer distance for air to flow over.But it takes the same time.So the air flow faster on the above side

8. Jun 16, 2014

### Staff: Mentor

It actually doesn't take the same time -- the two airstreams don't align at the end, the way they started. That's actually one of the common myths about the way wings work.

9. Jun 16, 2014

### rcgldr

This pre-shuttle re-entry prototype glider (a later version had rocket engines and could reach mach 1.6) should dispell any longer path or hump on the top theories. It's an M2-F2 on landing approach with an F-104 Starfighter along side.

10. Jun 17, 2014

### Staff: Mentor

Looks can be deceiving: The top is the longer path -- that's not what the equal transit time misconception gets wrong. In that photo, you can't tell what the angle of attack is, so you don't know where the stagnation point is, so you can't tell which path is longer. Things change when you look at this diagram:

The top looks flat in your picture, but it really isn't -- the down-sloping trailing edge is just hidden behind the tail. And what they define as the 0 aoa line is actually 5 degrees up, geometrically.

Last edited: Jun 17, 2014
11. Jun 17, 2014

### rcgldr

I'll see if I can find the specs again, but I recall that about 2/3rds of the leading portion of the bottom section is a half cone shape that curves back upward and transitions to a flat section that ends at the blunt trailing edge opening in the back where the rocket nozzles were placed in the M2-F3 version. The top half is flat and tapers downwards at around 40% from the leading edge. A good portion of the central part of the bottom surface is a longer path than the top, but not all of the bottom portion.

The main differences from a conventional wing is that the "hump" on the bottom is larger than the "hump" on the top, and the blunt trailing edge needed for the rocket engines. The lift to drag ratio for such a design isn't that great, but that isn't needed for re-entry from space, and the vehicle would need to be able to withstand hypersonic speeds during re-entry, although this was never tested (max speed with rocket engines was mach 1.6).

Here's a newer image I found and rotated 2 degrees to the right to get the ground (note the white "stripe" at the bottom of the image) to appear horizontal, although if the pitcure wasn't taken directly from the side of the aircraft, then the ground apparent orientation can be affected. At higher speed, less angle of attack would be needed. My guess is that the upper front surface was probably angled downwards a bit at high speed (mach .8 or greater).

Here's are links to NASA web pages with images of the M2-F2 and M3-F3:

NASA M2-F2.htm

NASA M2-F3.htm

Last edited: Jun 17, 2014
12. Jun 17, 2014

### Staff: Mentor

In any case, even if you could show that that's an exception, exceptions don't prove rules, they are exceptions.

13. Jun 17, 2014

### rcgldr

The point in this case is an exception that disproves a false premise (the false premise that the larger hump has to be on top, but I'm not sure about the longer path has to be on top, since only a portion of the bottom surface of the M2-F2 involves a longer path.) It might be possible to disprove the longer path premise with a wedge shaped wing (at a specific air speed, it would have a horizontal top and an inclined bottom, equal taper at the trailing edge) although it wouldn't be practical from a lift to drag standpoint.

Last edited: Jun 17, 2014
14. Jun 17, 2014

### A.T.

It doesn't take the same time. The air above the wing reaches the trailing edge before the air below the wing.

15. Jun 17, 2014

### Staff: Mentor

Sorry, but that is a new false premise that you are falling for now. We didn't talk about "larger hump", only longer path. Not the same thing. The problem is that the wing shape is deceiving. While we can't really tell for the M2F2, you can tell for a flat-bottomed wing. The minds-eye draws the chord line geometrically to separate the top and bottom surfaces, but the reality is that the stagnation point moves as the angle of attack changes, which means the top and bottom surfaces aren't separated by the same line all the time. So it only looks like the path is shorter on the bottom: it isn't necessarily true and isn't true under a very negative aoa.

Ie: flip a flat bottomed wing over and it can fly inverted, but only at a high aoa.

16. Jun 17, 2014

### davenn

yup, which is exactly what the animation in the OP also shows

Dave

17. Jun 17, 2014

### rcgldr

I was thinking more along the lines of flipped over and flown backwards (with what is now the trailing edge tapered downwards), which would be similar to the M2-F2, but this is a response to a bit off topic post about the false premise of longer path with equal transit time. I don't know if longer path with unequal transit time is required to produce lift, or at least produce lift efficiently. Previous posts have mentioned that the reduction in pressure above a wing co-exists with faster moving air above a wing, if using the wing as a frame of reference.

18. Jun 17, 2014

### donpacino

Bernoulli's principle accounts for very little of an aircraft's lift. Most of the lift occurs from the angle of attack of the wing.

19. Jun 17, 2014

That is a false dichotomy. Although if you want to get technical about it, Bernoulli's principle can successfully calculate all of the lift on an airfoil given a known velocity field and under the assumption that the flow is not separated.

20. Jun 17, 2014

### mic*

When considering the curiosity of why an aerofoil makes lift, angle of attack should not enter the discussion because it is clear that a non-aerofoil shape will "lift" in the direction of the leading edge as it passes through the air. A child sticking their hand out a car window realises this (not that they should do this).

Bernoulli's principle helps descirbe the way an aerfoil (at 0deg aoa) can make lift, but this is not the wholle story, esp. once considering factors like dihedral, sweep, and even surface finish.

21. Jun 17, 2014

### ModusPwnd

But if the child angles their hand down, the hand goes down. Angle of attack does matter. As the angle changes the force up or down changes.

22. Jun 17, 2014

### rcgldr

Not quite all, since the energy of the air is increased when a wing passes through, Bernoulli is violated somewhat since the Bernoulli principle assumes that total energy remains constant. This issue is more clearly demonstrated in the case of a propeller, where the increase in energy is significant. From this NASA article on propellers:

... at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli'sequation 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 :

propanl.htm

23. Jun 18, 2014

Angle of attack is absolutely critical in determining lift for any shape. In fact, angle of attack is the single most important factor in determining the lift on a shape and has a much greater effect on the lift than do things like thickness or camber. This is doubly true for a supersonic airfoil, where angle of attack is literally the only factor in the lift for a given shape.

Also, the only effect surface finish has on lift is that it has an effect on the laminar-turbulent transition of the wing boundary layer, which can change the effective shape of the wing since turbulent boundary layers are thicker. This is a relatively small effect in most cases, though.

I am sorry but you don't know what you are talking about here. If you have a wing moving through a fluid or with fluid moving over it, if you have a complete picture of the velocity profile, you can calculate lift from Bernoulli's principle to very high accuracy. Not only can you do it, but it is done all the time.

You can't "somewhat" violate Bernoulli's principle. Either the conditions for its use hold (classically inviscid, steady, and incompressible, with few exceptions) or they don't. For the example of the wing, moving through the air at a constant rate, you are basically already maintaining any force or energy balance by assuming the constant rate. In particular, you assume that any thrust generated by a propeller or another engine is exactly countering the drag according to Newton's laws. So then, you have to look at the flow field and look at exactly where the conditions for Bernoulli's equation hold (if they do).

So, for an airfoil moving less than Mach 0.3, the flow is incompressible, so let's stick to that regime. We are already talking about a steady flow, so the remaining question is whether the flow is inviscid or not. You can assume the flow is inviscid and calculate the velocities fairly accurately using a panel method or an Euler code that simply enforces the Kutta condition on the trailing edge (which simulates viscosity without actually acknowledging it exists). In that case, Bernoulli's equation is valid and you can get a good estimate of the drag on a wing from a purely inviscid analysis assuming the wing is not stalled.

Now, you may be saying, "but boneh3ad, air is inherently viscous so that example is not really physically true." True though that may be, as it turns out, you can also handle this easily here. One approach is that you make corrections to the shape of the airfoil in your inviscid analysis to account for the presence of the boundary layer (via displacement thickness) and get a much more accurate estimate of the flow field. In that case your analysis is still inviscid and the flow can still be analyzed with Bernoulli's equation.

The other approach is to actually take viscosity into account and solve the full flow equations (or let nature do it for you). Using one of the many neat properties of fluid flows, you can still get the lift for this situation using Bernoulli's equation despite the fact that the analysis is now viscous in nature. As it turns out, when a viscous fluid flows over a surface, it forms a boundary layer whereby the velocity smoothly goes from zero at the wall to the free stream value. However, this effect is confined to the region very close to the surface, and outside of the boundary layer, the flow behaves exactly like an inviscid flow. It also turns out that nature was kind to us in that the pressure gradient in the wall-normal direction in the boundary layer is effectively zero, meaning the static pressure at the surface is the same as it is in the free stream at the edge of the boundary layer at that point. In other words, you can get your "exact" flow field from a viscous analysis of a wing and simply apply Bernoulli's equation only to the inviscid region of the flow and still come up with the correct lift.

In the inviscid region, the total energy does remain constant, and any energy loss taking place in the viscous region through dissipation is exactly counteracted by assuming a steady flow (and in the real world by the propeller or jet engine). In other words, the conditions for the use of Bernoulli's equation hold. This is also why the same principle does not work for a propeller, since they are, by design, supposed to be dumping much more energy into the flow than the dissipative forces on the propeller itself would require to break even.

Of course, it is still important to understand that Bernoulli's equation in no way explains why the flow moves faster over the upper surface. It is merely a tool that can be used to calculate the lift from a flow field known a priori.

24. Jun 18, 2014

### rcgldr

The increase in overall mechanical energy of the air when a wing passes through the air violates Bernoulli. At some point in the interaction between wing and air, a non-Bernoulli like mechanical interaction with the wing increased the energy of the air (pressure increased without corresponding decrease in speed^2, or speed^2 increased without corresponding decrease in pressure) resulting in a non-zero "exit velocity" (the velocity when the affected air's pressure returns to ambient), and an impulse that travels downwards (and a bit forwards) through the air until it reaches the ground. That impulse is how the weight of the air craft is eventually transmitted to the surface of the earth as part of a closed system (aircraft, air, earth). The pressure at the surface of the earth reflects the weight of the air and any aircraft supported by the air (assuming the aircraft does not a have a vertical component of acceleration).

25. Jun 18, 2014

### CWatters

Nonsense. Many wing sections (particularly those with high camber or flaps) s generate +ve lift at -ve AOA.