Fluid mechanics, why does the air flow faster over the wing?

[SDEric]

Could one of you physics gurus answer this one. There was an answer to this question posted early in this thread, but I'd like to make sure it reflects the consensus opinion.

If you had an airfoil that was half a cylinder, with the flat bottom at a zero angle of attack (that is, parallel to the direction of flow), would lift be generated?

It wouldn't even have to be an entire half a cylinder; any section of a cylinder would work for the thought experiment. The point being, if you had an airfoil that was symmetrical front to back, and at a zero angle of attack, would lift be generated?

The NASA program "foilsim" says lift would be generated. If this is true, I would love an explanation.

I understand the mechanics of air being deflected down by both the top and the bottom of a wing, and the whole idea of "turning" air flow. I am curious whether the Bernoulli effect alone creates lift.

 

[MaxManus]

My post is a year old, and I don't remember the equations but yes the there will be generated lift with zero angle of attach. But I am not sure of your last question. If you agree on the bernoulli effect, then why not agree in an ideal example where you haven't taken the mass of the into the problem?

 

[rcgldr]

One of the links from one of my earlier posts is broken now. There's an archived version of it, but most of the images are gone:

archive_/~weltner/Mis6/mis6.html

As far as a true half cylinder goes, you might be able to generate the 2d coordiantes of a half cylinder based air foil and use XFOIL to calculate the polars (lift and drag versus AOA, Re, ...).

http://web.mit.edu/drela/Public/web/xfoil

A true half cylinder would generate a lot of drag. If the air foil was made of a thin slice of the upper part of a cylinder, a flat bottom and a circular arc of a few degrees on the top, it might work in theory, but actual examples appear to need a non-zero angle of attack to produce lift. A frisbee is similar to this, except the bottom surface is hollow so is similar in shape to the upper surface, and from what I recall, a frisbee needs an angle of attack to generate lift. An aerobie is a flying ring that combines a special triangular outer spoiler rim with an airfoil section similar to a thin slice of a cylinder for the inner shape, but it still needs some non-zero angle of attack to produce lift. The spoiler rim is designed to keep the center of lift near the center of the ring over a reasonable range of angle of attacks to eliminate pitch torque which would result in a roll due to gyroscopic precession.

http://aerobie.com/about/ringscientificpaper.htm

 

[SDEric]

Thanks! I can easily see how a slice of a cylinder would create lift if it were at a non zero angle of attack. The angled bottom would deflect air down, and the curved top would turn the upper air stream down. The net effect would absolutely be to push air down. Probably the optimum angle of attack would be with the initial portion of the upper surface tangential to the airflow, but that's just a guess.

I just cannot see how lift would be generated at a zero angle of attack. Sure, the air following the rear portion of the curve would be turned down and would lift the back portion of the cylinder section. But the exact opposite effect would take place at the front of the cylinder section. It seems to me that the net effect would be to only apply torque to the cylinder section.

NASA's Foilsim program seems to say that a cylinder section would have lift at a zero angle of attack. But that is a simulator, and may or may not be completely correct.

Here's the "Misinterpretations of Bernoulli's Law" paper you were attempting to link to above: http://plato.if.usp.br/2-2007/fep0111d/Bernoulli.pdf

 

[AlephZero]

Bernouilli's equation is just Newton's laws of motion applied to a fluid. It tells you how the changes of fluid velocity around the body are related to the changes in pressure over the surface of the body, but it doesn't tell you anything about why the velociity and pressure distributions are the way they are.

The thing that "causes" the lift is the fluid viscosity and the effect it has on the boundary layer of the flow. Bernouilli's equation doesn't included any viscosity terms, so it can't possibly tell you about that. In fact it there was no viscosity, there would be no lift and drag forces on a body of any shape, at any angle of attack (and there would also be no boundary layer, and no turbulence in the flow).

I had a quick look at the NASA foilsim web pages. There was some theory there but I couldn't find a complete explanation of what the program does. But from what it did say, you are quite right to question whether the results would be correct for something that doesn't "look like a normal aerofoil". Many real-world computer methods in fluid dynamics only work well in particular situations. Competely "general purpose" computational fluid dynamics software may take too long to run.

It is true the half-cylinder is symmetrical front-to-back, but the airflow pattern around it is not symmetrical, because of the air viscosity. The boundary layer becomes thicker as the air flows over the body, and at some point it will probably separate from the surface.

The idea of "angle of attack" is not obvious for something like a half cylinder. If the flat surface was parallel to the far-field airflow, the stagnation point that defines the "leading edge position" would not be at the corner between the flat and curved surfaces, it would be some point along the curved top surface. In that sense, the airflow has a non-zero angle of attack. Alternatively, if you bisected the 90-degree angle between the curved and flat surfaces, you could argue that the angle of attack was actually -45 degrees.

 

[SDEric]

Bernouilli's equation is just Newton's laws of motion applied to a fluid. It tells you how the changes of fluid velocity around the body are related to the changes in pressure over the surface of the body, but it doesn't tell you anything about why the velociity and pressure distributions are the way they are.

The thing that "causes" the lift is the fluid viscosity and the effect it has on the boundary layer of the flow. Bernouilli's equation doesn't included any viscosity terms, so it can't possibly tell you about that. In fact it there was no viscosity, there would be no lift and drag forces on a body of any shape, at any angle of attack (and there would also be no boundary layer, and no turbulence in the flow).

I had a quick look at the NASA foilsim web pages. There was some theory there but I couldn't find a complete explanation of what the program does. But from what it did say, you are quite right to question whether the results would be correct for something that doesn't "look like a normal aerofoil". Many real-world computer methods in fluid dynamics only work well in particular situations. Competely "general purpose" computational fluid dynamics software may take too long to run.

It is true the half-cylinder is symmetrical front-to-back, but the airflow pattern around it is not symmetrical, because of the air viscosity. The boundary layer becomes thicker as the air flows over the body, and at some point it will probably separate from the surface.

The idea of "angle of attack" is not obvious for something like a half cylinder. If the flat surface was parallel to the far-field airflow, the stagnation point that defines the "leading edge position" would not be at the corner between the flat and curved surfaces, it would be some point along the curved top surface. In that sense, the airflow has a non-zero angle of attack. Alternatively, if you bisected the 90-degree angle between the curved and flat surfaces, you could argue that the angle of attack was actually -45 degrees.

Thanks for the informative post. (Aside: So an airfoil would not generate lift in liquid helium?!?)

I chose a half cylinder for the thought experiment, but really it is a general question: Is lift generated by an airfoil that is (1) symmetrical front to back, and (2) at a zero angle of attack?

In a similar question, does wind blowing across a plain "lift" a small hill it encounters, assuming that the wind speed and hill size and shape are such as to create no turbulence? I have seen discussion that claims that there is higher pressure on the windward and leeward sides of the hill, and lower pressure on the peak. If true, this would help explain why sand dunes exist, or at least why waves are formed when air blows across water. But is the net effect on the hill upward?

 

[rcgldr]

Thanks for the informative post. (Aside: So an airfoil would not generate lift in liquid helium?

I'm not sure if it's impossible to deflect a flow for even a near inviscid fluid if the flow speed is fast enough that momentum prevents the pressures from equalizing. If you dropped a very light (hollow) sphere in a very tall column of liquid helium would it accelerate the same as it would in a vacuum? To eliminate the issue of pressure increasing with depth, consider accelerating a charged hollow sphere with an electrical field through a long container filled with liquid helium, free of gravitational effects, and compare this to the sphere accelerating in a vacuum.

Is lift generated by an airfoil that is (1) symmetrical front to back, and (2) at a zero angle of attack?

Symmetrical means the same shape above and below. Such a shape should not generate lift at zero angle of attack (assuming that it's isn't a transitory state with angle of attack changing over time).

In a similar question, does wind blowing across a plain "lift" a small hill it encounters ... not turbulent ... is the net effect on the hill upward?

Not sure, laminar (non-turbulent) flow is rare except at low speeds. For an air foil with a flat bottom and a curved top and zero angle of attack (the flat bottom horizontal), if the peak of the curve is near the leading edge, you get positive lift, if it's near the trailing edge you get negative lift (and a lot of drag, it would be similar to an inverted wedge). At somepoint in between, there would be zero lift, but I don't know if this corresponds to having the peak exactly at the middle.

 

[RandomGuy88]

If you dropped a very light (hollow) sphere in a very tall column of liquid helium would it accelerate the same as it would in a vacuum?

No. An object that is accelerated in any fluid (even an inviscid fluid) will experience drag due to the "added mass" effect. Basically if the object is accelerating in a fluid it must also be accelerating the fluid itself. So if you apply a given force F to an object of mass m it will not accelerate at a rate of F/m because you must also consider the mass of the fluid that is accelerated.

So the sphere dropped into a column of liquid helium would a lower acceleration than a sphere dropped in a vacuum.

In addition to this there is also buoyancy to consider.

 

[daqddyo1]

MaxManus - Are you still here?
The air moves faster over the rounded top suface of the wing simply because it has to travel further than the air that travels below the wing. When the air at the front (leaing edge) of the wing separates into two parts, those parts must meet at the trailing edge of the wing at the same time. Think about what would happen if this were not the case.

I have played with a perfect airfoil in a wind tunnel. You get no lift if the bottom edge of the wing is parallel to the airflow. As soon as you present an angle of attack (tilt the wing up a bit), lift occurs. There is an optimum angle of attack for maximum lift. This depends in part to the relative windspeed. With an angle of attack and sufficient airspeed, the air passing over the top of the wing behaves as if it were trying to separate away from the upper surface causing a reduction in pressure, resulting in lift. As well, below the wing, the air srikes that surface at a slight relative angle which also contribute to the lift.

 

[RandomGuy88]

MaxManus - Are you still here?
The air moves faster over the rounded top suface of the wing simply because it has to travel further than the air that travels below the wing. When the air at the front (leaing edge) of the wing separates into two parts, those parts must meet at the trailing edge of the wing at the same time. Think about what would happen if this were not the case.

NOOOOOOOOOOOO! That is absolutely INCORRECT! There is NO reason that the two parts of air that split at the leading edge must meet again at the trailing edge.

Lift is created because the streamlines must curve due to the body. A pressure gradient is then created due to this streamline curvature. Think of a particle of air traveling in a circle. There is a centrifugal force on this particle and this centrifugal force is balanced by the pressure. So for air moving in a circle the pressure at the center of the circle is lower then the edge of the circle. So if you think air flowing over the surface of an airfoil, it curves and on the upper surface the centrifugal force is trying to push the air particle away from the surface so the pressure at the surface decreases in order to balance this centrifugal force. On the lower surface the same mechanism causes the pressure to be higher at the surface and this pressure difference causes lift.

 

[daqddyo1]

One is always in danger when referring to centrifugal force as an action. It is a reaction force and is exerted BY a particle forced to move in a curved path and NOT on that particle. The air moving in the curved path over the top of the wing has some upward momentum that it didn't have before the wing hit it. This means there is less air density close to the top of the wing.
At sufficient airspeed, this increases the effective pressure difference to provide the lift force provided by the wing.


If the air passing over the top of the wing continually takes more (or less) to time pass over the wing, then somewhere there is going to be a great buildup of of air.

P.S. I'm going to bed now.

 

[SDEric]

If the air foil was made of a thin slice of the upper part of a cylinder, a flat bottom and a circular arc of a few degrees on the top, it might work in theory, but actual examples appear to need a non-zero angle of attack to produce lift. A frisbee is similar to this, except the bottom surface is hollow so is similar in shape to the upper surface, and from what I recall, a frisbee needs an angle of attack to generate lift.

According to one UC Davis M.S. thesis, a frisbee generates some lift at a zero angle of attack. See figure 2-3 in http://morleyfielddgc.files.wordpress.com/2009/04/hummelthesis.pdf.

I'd love to have an intuitive understanding of this lift.

According to Foilsim, there is lift on a flat-bottomed, ellipse topped shape even when the flat bottom is parallel to the airflow. It says there is positive pressure on the leading curve and on the trailing curve, and negative pressure at the top. I understand the first part; that seems obvious, but I don't get why the top would have lower pressure than ambient and I don't get why the trailing edge would have higher pressure. If anyone can help me understand this, I would be most grateful.

 

[rcgldr]

For an air foil with a flat bottom and a curved top and zero angle of attack (the flat bottom horizontal), if the peak of the curve is near the leading edge, you get positive lift, if it's near the trailing edge you get negative lift (and a lot of drag, it would be similar to an inverted wedge). At somepoint in between, there would be zero lift, but I don't know if this corresponds to having the peak exactly at the middle.

According to one UC Davis M.S. thesis, a frisbee generates some lift at a zero angle of attack. ... I'd love to have an intuitive understanding of this lift.

According to Foilsim, there is lift on a flat-bottomed, ellipse topped shape even when the flat bottom is parallel to the airflow. It says there is positive pressure on the leading curve and on the trailing curve, and negative pressure at the top.

From my quoted post, there is a flat bottom, smoothly curved top airfoil with zero lift at zero angle of attack, but the peak is probably a bit aft of the middle of the chord length. The frisbee article doesn't mention the wind tunnel air speed versus the speed of a typical throw. If the speed in the wind tunnel was signficantly slower, then tubulent effects would be reduced. Also a frisbee doesn't have a flat bottom, the bottom has a similar shape to the top, but the effects may be similar to a flat bottom, ellipse top airfoil shape.

Because of the convex curvature of relative flow, there's a low pressure area near and a bit aft of the peak. There may be places on such an airfoil where pressure above is greater than pressure below, but apparently there is sufficient area where the pressure above is less than the pressure below to result in a net upwards force (lift). Turbulent flow could reduce or eliminate this lift.

As far as an explanation for this lift, it's because there is net downwash aft of the airfoil. The airfoil's presence in a relative horizontal flow causes upwash on the leading edge, then curves this flow back downwards again. If a downwash is created, the air foil passes by the affected air before it stops such a downwash. If the total effect of the upwash, bending of flow, and then downwash results in a net downwash of air as the air foil passes through, then lift is created.

 

[boneh3ad]

The air moving in the curved path over the top of the wing has some upward momentum that it didn't have before the wing hit it. This means there is less air density close to the top of the wing.
At sufficient airspeed, this increases the effective pressure difference to provide the lift force provided by the wing.

It means no such thing. The air need not change density for lift to be generated. In fact, if it had to, the Wright brothers would never have left the ground and birds wouldn't be able to glide. In a flow below Mach 0.3, density is effectively constant. Above Mach 0.3, density plays very little role until you approach sonic conditions.

If the air passing over the top of the wing continually takes more (or less) to time pass over the wing, then somewhere there is going to be a great buildup of of air.

No there isn't. In reality, the air moving over the top of an airfoil passes over the wing much faster than the air passing over the bottom surface. This occurs because the effect of viscosity ensures that the rear stagnation point must be located at the sharp trailing edge (a concept known as the Kutta condition). The top and bottom streams of air leave the trailing edge parallel to one another but at different velocities. This tends to keep the flow turning for a little distance past the trailing edge, contributing to the downwash.

 

[SDEric]

From my quoted post, there is a flat bottom, smoothly curved top airfoil with zero lift at zero angle of attack, but the peak is probably a bit aft of the middle of the chord length. The frisbee article doesn't mention the wind tunnel air speed versus the speed of a typical throw. If the speed in the wind tunnel was signficantly slower, then tubulent effects would be reduced. Also a frisbee doesn't have a flat bottom, the bottom has a similar shape to the top, but the effects may be similar to a flat bottom, ellipse top airfoil shape.

Because of the convex curvature of relative flow, there's a low pressure area near and a bit aft of the peak. There may be places on such an airfoil where pressure above is greater than pressure below, but apparently there is sufficient area where the pressure above is less than the pressure below to result in a net upwards force (lift). Turbulent flow could reduce or eliminate this lift.

Thanks. I think the general question is, for frisbees, hills, and semi-circular cross section airfoils, how does an object, symmetrical front to rear, and oriented parallel to the air flow, generate lift? It just begs for the (incorrect) longer path explanation.

As far as an explanation for this lift, it's because there is net downwash aft of the airfoil. The airfoil's presence in a relative horizontal flow causes upwash on the leading edge, then curves this flow back downwards again. If a downwash is created, the air foil passes by the affected air before it stops such a downwash. If the total effect of the upwash, bending of flow, and then downwash results in a net downwash of air as the air foil passes through, then lift is created.

The problem I see with this explanation is that the airflow turned down at the back seems like it is exactly offset by the airflow turned up at the front.

But that view is wrong. The airflow is turned up at the front, and by the time it reaches the middle it is back to moving parallel to the direction of travel. So it has been turned up and then turned back to parallel. Then it turns down, and by the time it leaves the back, it is turned down and is left turned down. So you are exactly right. All general discussion of airfoils focuses on the more complicated situation in which the airfoil has a shape not symmetrical front to rear, and angled to the airflow. What happened to the physics principles of deconstruction? No wonder everyone is frustrated by these explanations. Even that stupid NASA website boils down to "it's too complicated to explain clearly".

I recall Feynman complaining that no one had worked out the physics of such an obvious phenomenon as wind generated waves.

 

[DrDu]

I think it should be possible to get some symmetric airfoil with the Joukowski transformation.
You can try it out here (Java does not work for me):

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

 

[olivermsun]

Thanks. I think the general question is, for frisbees, hills, and semi-circular cross section airfoils, how does an object, symmetrical front to rear, and oriented parallel to the air flow, generate lift? It just begs for the (incorrect) longer path explanation.

Hills are easy -- the flow only goes around one side of the hill.

Are you sure that frisbees and semicircular airfoils generate lift when oriented parallel to the flow?

I recall Feynman complaining that no one had worked out the physics of such an obvious phenomenon as wind generated waves.

It's a lot more complicated because the waves (which have a whole set of dynamics of their own) exert a feedback on the wind stress.

 

[SDEric]

Hills are easy -- the flow only goes around one side of the hill.

Ah! That is fallacy #1. The air flowing over the top of the wing has NOTHING to do with the air flowing under the bottom! Assuming that the two flows are somehow connected leads one to the "equal transit time" fallacy.

Are you sure that frisbees and semicircular airfoils generate lift when oriented parallel to the flow?

I am not sure. The simulations, and supposedly some wind tunnel data, say there is lift, but I have a hard time developing an intuition why it exists.

It's a lot more complicated because the waves (which have a whole set of dynamics of their own) exert a feedback on the wind stress.

Sure, but again, make it simple. A small, smooth wave (that is, not breaking), a wind speed low enough to not cause turbulence but faster than the propagation speed of the waves. For some reason the wave builds. I can understand why the wave moves (same as drag on the wing) but why does the wave build? I think it is because pressure is higher at the base of the wave and less at the top, much like how the curved top wing has some lift.

 

[olivermsun]

Ah! That is fallacy #1. The air flowing over the top of the wing has NOTHING to do with the air flowing under the bottom! Assuming that the two flows are somehow connected leads one to the "equal transit time" fallacy.

You yourself agreed in an earlier post that the net momentum change of the flow over the wing is important (the flow remains "downturned" after it passes by the wing). My point is that this condition is impossible for a symmetric hill (the flow cannot leave with net downward momentum since the ground is in the way). This places strong constraints on the net lift produced by the hill.

I am not sure. The simulations, and supposedly some wind tunnel data, say there is lift, but I have a hard time developing an intuition why it exists.

Could you link to some simulation results? The diagrams I was able to find of lift generated by a Frisbee all show an angle of attack, but my search was by no means exhaustive.

I think it is because pressure is higher at the base of the wave and less at the top, much like how the curved top wing has some lift.

I guess here it may be helpful to point out that pressure differences exist over airfoils, whether or not a net lift is generated.

 

[RandomGuy88]

For the flow over a 2D airfoil the downwash exactly equals the upwash and there is no net downward momentum added to the flow. The force on the airfoil is a result of the difference in pressure between the upper and lower surface. This pressure difference is due to streamline curvature.

Before you argue that there can be no force without a net downwash consider this example.

Imagine you have an open loop wind tunnel. So the air is pulled in from the atmosphere at one end and exhausted at the other end. Because the flow is accelerated from rest to whatever speed you want in the test section the pressure drops considerably. As a result, the pressure in the test section is less than atmospheric pressure. So if you have a door that gives you access to the test section and this door opens out it will be very difficult to open this door because of the pressure difference between the low pressure in the tunnel and the atmospheric pressure outside. The door is getting sucked into the tunnel. This is analogous to the lift of a 2D airfoil because there is a force but there is no net downwash. The force is only a result of the pressure difference.

Half a cylinder with the bottom face parallel to the flow generates lift because the pressure decreases over the rounded surface. This pressure decreases because of streamline curvature. The same is true of a hill and a frisbee. A frisbee generates lift a zero degrees angle of attack. Think about it like this, when you throw a frisbee horizontally it doesn't drop as fast as if you just dropped it straight down right? The lift is less than the weight so it still falls but just not as quickly.

 

[SDEric]

You yourself agreed in an earlier post that the net momentum change of the flow over the wing is important (the flow remains "downturned" after it passes by the wing). My point is that this condition is impossible for a symmetric hill (the flow cannot leave with net downward momentum since the ground is in the way). This places strong constraints on the net lift produced by the hill.

The hill has an element not present in the airfoil: flat ground surrounding it. The air flowing over the hill is deflected yet again after it passes the hill, so afterward it is again flowing parallel to the ground.

The flat ground in front of the hill similarly participates in deflecting the air up. The upward deflection of the air starts before the hill.

I just thought of something hilarious: If you put a hill in the flat part of an airplane wing, would the lift increase? I know you wouldn't likely do that because the penalty in drag would not be worth it, but it's an entertaining thought experiment.

Could you link to some simulation results? The diagrams I was able to find of lift generated by a Frisbee all show an angle of attack, but my search was by no means exhaustive.

NASA has on the web "Foilsim": http://www.grc.nasa.gov/WWW/k-12/airplane/foil3.html
It does not do a frisbee, but it does various shapes of similar cross section. All convex top shapes with less convex (through to concave) bottom shapes have lift when parallel to the flow.

Here are some supposed experimental results (though it is just a line drawn on a piece of paper): Figure 2-3 in http://morleyfielddgc.files.wordpres...mmelthesis.pdf

I guess here it may be helpful to point out that pressure differences exist over airfoils, whether or not a net lift is generated.

It is easy to understand that there are local pressure differences over the top of a frisbee, but it is hard for me to understand why the sum of those pressure differences is less than zero.

 

[boneh3ad]

Sure, but again, make it simple. A small, smooth wave (that is, not breaking), a wind speed low enough to not cause turbulence but faster than the propagation speed of the waves. For some reason the wave builds. I can understand why the wave moves (same as drag on the wing) but why does the wave build? I think it is because pressure is higher at the base of the wave and less at the top, much like how the curved top wing has some lift.

Not really. Ocean waves are decently well-understood these days. They are just one example of what is called the Kelvin-Helmhotz instability which arises, among other situations, when there is a velocity difference across the interface of two fluids. The behavior can be modeled reasonably well and has been studied quite a bit. it is the same physical mechanism for many of the patterns on Saturn and Jupiter and various cloud formations in the atmosphere.

 

[olivermsun]

Not really. Ocean waves are decently well-understood these days. They are just one example of what is called the Kelvin-Helmhotz instability which arises, among other situations, when there is a velocity difference across the interface of two fluids.

I'm pretty sure that ocean surface waves are not universally examples of K-H instability!

 

[boneh3ad]

I'm pretty sure that ocean surface waves are not universally examples of K-H instability!

Well there are, of course, the cases such as tsunamis or other seismically generated waves and any large scale wave associated with tides. However, most of the ones that you see of relatively short spatial scale are wind-generated, and are examples of the Kelvin-Helmholtz instability.

 

[olivermsun]

Before you argue that there can be no force without a net downwash consider this example.

Imagine you have an open loop wind tunnel...This is analogous to the lift of a 2D airfoil because there is a force but there is no net downwash. The force is only a result of the pressure difference.

Unless I'm misunderstanding the situation, the net pressure forces on the wind tunnel are zero and hence there's nothing to balance in a 3rd law sense.

I agree with you about the hill however. What I said earlier was incorrect. This flow should behave just like half the flow over a cylinder, so it's entirely consistent that a net lift can be generated on the upper half of the cylinder (the hill) with no net momentum change to the air (since it's balanced by the "virtual" lower half).

 

[DrDu]

Yesterday I also found this simulator,
http://www.grc.nasa.gov/WWW/K-12/airplane/map.html
which allows you to simulate a curved plate.
Even for an tilt angle 0 and a symmetric plate it produces a lift force.
I believe this example was first solved by Kutta analytically.

 

[olivermsun]

Neat! Thanks for the links to the simulations.

The reason for my comments earlier is that I've tended to describe lift over a cambered foil as being qualitatively due to the "effective" angle of attack, with respect to a symmetric airfoil, which is incorporated into the foil shape (difference between zero-lift and chord lines)

However, the curved plate example convinces me that the above is not a good qualitative description of lift generated over some commonly encountered shapes.

 

[RandomGuy88]

Here is a short paper discussing how streamline curvature is responsible for lift.

http://ocw.mit.edu/courses/aeronaut...namics-fall-2005/study-materials/4liftgen.pdf

 

[SDEric]

I am afraid I am not smart enough to understand even the beginning of that paper. I don't understand the significance of a coordinate system that rotates and follows the streamlines.

But I think I finally understand the pressure difference as a result of curving the airflow. For the airflow to turn over a convex surface, the surface as to "pull" the air to make it follow the curve, to turn away from the ambient air. Hence the upper surface has lower pressure than the ambient air.

For a concave surface, the surface has to "push" the air to make it turn toward the ambient. Hence the concave surface has higher pressure than the ambient.

This is confusing when one attempts to think about a hill. If the hill rises straight out of the ground, there is a point of infinite concave curvature at the base, followed by moderate convex curvature. To keep the thought experiment clear, one has to consider a moderate convex hill surrounded by moderate concave curvature.

In the area in which the curvature is concave, there is positive pressure. Once at the inflection point is past, the pressure becomes negative. After the air flows over the top of the hill and passes the inflection point on the other side, the pressure becomes positive again.

I think it is important when constructing these thought experiments to remove the discontinuities. I think the airflow, thanks to its viscosity, smooths out the discontinuities, but the discontinuities are hard for me to handle intuitively.

 

[SDEric]

I think I devised a thought experiment that completely isolates the effect being discussed.

Say you have a hill that rises up out of a flat plain. The beginning of the hill has radius of curvature R. It smoothly rises up through π/2 radian, from the plain; 1/4 of a circle; then the surface of the hill is perpendicular to the plain. Then the hill curves, again radius R but this time convex, in the other direction, until it is 1/4 of a circle (π/2 radian) from the plain again. Then the concave curvature of radius R begins again, until the curve reaches the plain again.

Total height of the hill is 2R.

The hill has infinite width to either side. The plain extends to infinity in front of and in back of the hill.

The wind is blowing, parallel to the plain, speed v. Air speed is low enough to create no turbulence. Air density is ρ.

What is the lift on the hill per meter of width?