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

[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?

 

[boneh3ad]

You won't get lift on any hill, even this "ideal" type you concocted. You will get a heck of a lot of drag and therefore eventually some erosion, but certainly no lift.

Just as an aside, the only way you wouldn't get turbulence is if the wind was pretty much not blowing. Between Görtler vortices on the concave portion, the separation on the convex portion or just plain old' Tollmien-Schlichting waves over the whole thing, it will be nearly impossible to have laminar flow over such a shape at any reasonable velocity.

 

[RandomGuy88]

Boneh3ad is right, there won't be any lift on a hill because there isn't really another side for there to be a pressure difference. For an airfoil there are two sides, the upper surface has lower pressure then the bottom surface and this pressure difference creates lift. On a hill there really is no lower surface is there?

The reason for having a coordinate system that follows the streamlines is so that you can easily determine the forces that are tangential and perpendicular to the the streamlines. This is how you derive the expression relating the streamline curvature and the pressure gradient.

If you have curved streamlines (which are curved because of the solid body) the pressure decreases towards the center of the curvature. If the streamlines were circular the pressure in the center is lowest.

For a 2D hill you will have two areas where the curvature causes the pressure to decreases as you go away from the surface (curvature is concave up). This is at the bottom of the hill on both sides. (I am ignoring the viscosity, so I am not considering the separated wake region). So at the bottom of the hill the pressure increases relative to the freestream static pressure. As the air goes up the hill there is an inflection point and then the curvature is in the other direction (concave down). So now the lowest pressure is at the surface and the pressure increases as you go up from the surface.

So from the flat ground to the top of the hill the pressure first increases then decreases and is lowest at the peak.

 

[boneh3ad]

Don't forget in this "ideal" hill there is a major curvature discontinuity so there would be major pressure spikes at those locations.

 

[SDEric]

Boneh3ad is right, there won't be any lift on a hill because there isn't really another side for there to be a pressure difference. For an airfoil there are two sides, the upper surface has lower pressure then the bottom surface and this pressure difference creates lift. On a hill there really is no lower surface is there?

The reason for having a coordinate system that follows the streamlines is so that you can easily determine the forces that are tangential and perpendicular to the the streamlines. This is how you derive the expression relating the streamline curvature and the pressure gradient.

If you have curved streamlines (which are curved because of the solid body) the pressure decreases towards the center of the curvature. If the streamlines were circular the pressure in the center is lowest.

For a 2D hill you will have two areas where the curvature causes the pressure to decreases as you go away from the surface (curvature is concave up). This is at the bottom of the hill on both sides. (I am ignoring the viscosity, so I am not considering the separated wake region). So at the bottom of the hill the pressure increases relative to the freestream static pressure. As the air goes up the hill there is an inflection point and then the curvature is in the other direction (concave down). So now the lowest pressure is at the surface and the pressure increases as you go up from the surface.

So from the flat ground to the top of the hill the pressure first increases then decreases and is lowest at the peak.

Understand and appreciate the discussion.

The thing is, you don't need the bottom to have higher pressure for the top to have lower pressure. You can demonstrate this with various flat-bottomed airfoils; with the right (mildly convex) curvature of the bottom, you can have net lift of the bottom being zero. The top still has net negative pressure, or net lift. The bottom of the airfoil isn't necessary to generate lift.

I can really understand why the hill shouldn't have lift; I can also understand why this curvature of the airflow is a minor component to overall airfoil lift, but just like wondering whether a flat-bottomed and front-to-back symmetrical airfoil has lift, I wonder whether a hill has lift.

It's a completely symmetrical situation, but here's an aspect that is does not net to zero: the airflow being turned by the concave curvature at the front of the hill and at the back of the hill is slower than the airflow being turned at the top of the hill. Higher speed, greater acceleration, greater force. Thus the sum of the forces is net upward (and leeward, counting drag).

Are the walls of a venturi sucked toward each other?

 

[boneh3ad]

Why SHOULD a hill have lift? Even if the pressure on the hill is lower than atmospheric, it isn't like it is going to suck the hill upward. That isn't how pressure works. There will still be zero pressure on any imaginary surface of the hill not exposed to the air, and the air will exert a force everywhere it touches normal to the surface. That means a hill, in fact, sees a net DOWNWARD aerodynamic force (though zero net force since it doesn't move).

Pressure does not suck. It only pushes. An object (be it a hill or airfoil or anything else) can never experience a be aerodynamic force towards a fluid without a higher pressure fluid on the other side. It is Newton's law.

 

[olivermsun]

Boneh3ad is right, there won't be any lift on a hill because there isn't really another side for there to be a pressure difference.

If there is a pressure difference relative to "ambient" (usu. something like hydrostatic) then will there will be any confusion in calling the net force due to the pressure anomaly "lift"?

Even if the pressure on the hill is lower than atmospheric, it isn't like it is going to suck the hill upward. That isn't how pressure works. There will still be zero pressure on any imaginary surface of the hill not exposed to the air, and the air will exert a force everywhere it touches normal to the surface.

If it is a physical argument about the phenomenon that will be observed under these conditions, then imagine this:

The hill was in equilibrium with the Earth (or whatever was below it, be it trapped gases or underground water if you insist it must be fluid). If there is a flow speedup above the hill, reducing the downward pressure on the hill, and the Earth is still pushing upward, then the hill will probably relax upward a bit. Do you call this lift? Isostatic adjustment?

 

[boneh3ad]

With no wind, the hill exerted a force of X on the ground below it. This force was a combination of the weight of the hill, W, and the force of the atmosphere pushing on the hill, P1. If the wind were to blow, there will be a pressure drop over the hill compared to the pressure in slower moving wind, so the force exerted by the hill will now be Y, the same weight of the hill, W, plus the force exerted on the now very slightly lower pressure surrounding air, P2. The net force is still downward, still dominated by the weight of the hill, and still not lift. Lift is an upward force, and it is impossible to have lift if the top side of an object is surrounded by air and the bottom side is resting against a solid, such as the Earth under the hill.

 

[olivermsun]

The net force is still downward, still dominated by the weight of the hill, and still not lift. Lift is an upward force, and it is impossible to have lift if the top side of an object is surrounded by air and the bottom side is resting against a solid, such as the Earth under the hill.

I'm not sure this answers my question. For the hill surface to "rebound" when the downward pressure of its own weight is decreased by the pressure drop, there has to be some finite period of time when there was a net upward force on the hill surface, causing (allowing?) the hill to rise.

Suppose we modify the scenario and say that the dynamic pressure drop caused by wind blowing over some bump in the ocean surface causes the bump to rise slightly. Was this caused by lift?

Alternatively, suppose we have an airfoil (which we agree has lift) in level flight. Is the weight of the airfoil (and aircraft) canceling the entire upward force due to the flow, in which case there is no net force at all? Is it still lift?

 

[boneh3ad]

No, it would be caused by just a shrinking of the downward force. There is no situation ever where pressure can suck something towards it. If you really look at a situation where you think that is happening, you will find that somewhere is is actually something pushing the object.

Think of it this way: you grab a bowling ball and then stand on a bathroom scale. You then set the bowling ball down and watch your weight (i.e. the force you exert on the scale) lessen. Did you experience lift? No, you just shed a bi of weight and therefore decreased your downward force.

The same is true of a hill. Wind blows over it and whatever minute pressure drop results just ends up being a drop in the effective weight of the hill (and a very, very, very small drop at that), not any sort of upward force. The effective weight of the hill falls and therefore the ground the hill sits on exerts less upward force on the hill since the sum of forces is zero.

There can't be lift because there is nothing under the hill to push on it that is affected in any way by airspeed. The ground merely reacts to the weight of the hill. The same affect could be achieved by you standing on the hill and then leaving it so it weighs less. Certainly the hill didn't experience lift there, no?

 

[olivermsun]

In the case of an airfoil, which we agree can experience lift, there is a pressure force "pushing" from the bottom and creating the net lift. Clearly, pushing vs. pulling is not the source of disagreement here.

Your point is that in the hill example, upward forces in the (solid) medium are unaffected by the external flow, and hence the net force cannot be called "lift."

How about the "intermediate" example of flow over a bump on the ocean surface? Couldn't it be argued that any upward force due to the shear flow (even if the lower layer is prescribed to have zero flow) is a special case of "lift"?

Just wondering where you draw the line, that's all.