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Old Nov 13, 2012, 08:04 AM
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I go back to my original question:

How do we "know that the flow around the wing does include downward mass acceleration equal to the lift"?

If the wing is the only thing pushing on the air, then that's a fair statement. In most cases, the wing is not the only thing pushing on the air. You can come up with contrived ways in which an airplane might fly in a way that it avoids ground interaction, but in the simplest case of a lifting wing in steady flight, wing interaction with the ground is inevitable. If the ground exerts any additional upward force on the air, then the flow around the wing does not include downward mass acceleration equal to the lift (Newton's Second Law says that it can't).

Bottom line is that if you assume equality between lift and the rate of vertical momentum transfer, you are starting with something that is not generally true.
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Old Nov 13, 2012, 08:42 AM
greg
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Originally Posted by ShoeDLG View Post
If the ground exerts any additional upward force on the air, then the flow around the wing does not include downward mass acceleration equal to the lift (Newton's Second Law says that it can't).
do a horizontal wing (airplane) and a vertical wing (sail) behave differently?
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Old Nov 13, 2012, 09:21 AM
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Originally Posted by ShoeDLG View Post
I go back to my original question:

How do we "know that the flow around the wing does include downward mass acceleration equal to the lift"?

If the wing is the only thing pushing on the air, then that's a fair statement. In most cases, the wing is not the only thing pushing on the air. You can come up with contrived ways in which an airplane might fly in a way that it avoids ground interaction, but in the simplest case of a lifting wing in steady flight, wing interaction with the ground is inevitable. If the ground exerts any additional upward force on the air, then the flow around the wing does not include downward mass acceleration equal to the lift (Newton's Second Law says that it can't).

Bottom line is that if you assume equality between lift and the rate of vertical momentum transfer, you are starting with something that is not generally true.
This next request may benefit you point of view.
Calculate how much air needs to be Accelerated etc to hold an Aircraft in a lift state.
If the amount is more than is evident then your point of view must be true.

For my point of View I can only reefer you to ..<..

ciurpita. I get where your Going.
Same question between a wing upside down and right ways up.
For example an aircraft is going 90deg (Straight up) holds a steady climb then pulls back on the stick to execute a 1G half loop back to straight down. If the earth is pushing the air up and the Aircraft is Pushing the Air up then where does it go ?

I hope we can all Gravitate towards a similar view soon
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Old Nov 13, 2012, 09:35 AM
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Crossplot
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Originally Posted by ShoeDLG View Post
How do we know this?
Because Every action has an Equal and Opposite reaction, and an object at rest or in motion will stay that way unless acted upon by an Extra force.

The tricky part is how some wings work to first allow/force the Air to Accelerate up before it is then Allowed to accelerate back down.

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Crossplot That is one way of looking at it. I see Cl as just a coefficient to trade the lift function with q and S as the variables.

In the F=Ma that lifts is from the accelerations the take place in the flow as it turns around the wing. So we are talking about the path of the flow as it passes the wing . The question being , is there any reason to believe that there is a downward component inherent in the basic (?) flow aft of theTE? The predominant downwash that we see is created by the remnants of the creation of circulation.
yes, but the motion will quickly change to a vortex and the energy imparted to that air is Dissipated to heat through friction.

http://www.rcgroups.com/forums/showthread.php?t=1539175
Some good and Bad Information there ?
I have re read the Question. And , I think I was wrong.
The air aft of the TE may not contain a volume of air in a downward vector with a KE = Weight of aircraft.
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Old Nov 13, 2012, 10:08 AM
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do a horizontal wing (airplane) and a vertical wing (sail) behave differently?
I don't think so, not fundamentally. I think the horizontal momentum transfer to the air in the case of a sail producing a steady side force would be entirely analogous to the vertical momentum transferred to the air in the case of a wing producing steady lift IF you could somehow take the ground out of the picture.

When a wing is in ground effect, the ground pushes up on the air. Newton's Second Law tells you the air's rate of vertical momentum change must be equal to the magnitude of the lift force minus the magnitude of the upward force exerted by the ground.

How much upward force does the ground exert on the air? If you consider a patch of the ground with dimensions much smaller than the wing's span, then the upward force exerted by that patch is much smaller than the lift (it goes to zero for a small enough patch). At the other end of the spectrum, if you consider a patch of ground with dimensions many, many times bigger than either the wing's span or altitude, then the upward force exerted by the ground is equal to the lift. It turns out that this is true whether the wing is close to the ground or not.

Bottom line is that if you consider the upward force exerted by a big enough patch of ground, the upward force it exerts is always equal to the lift. This means that regardless of how high you get, you can never really take the ground out of the picture in the case of a steadily lifting wing.

The scientific-sounding way of saying the same thing is that the momentum contained in a slice of a wing's wake is not "Lebesque integrable". The amount of momentum contained in a given slice converges as the size of the slice gets very large compared to the wing's span. However, the value that it converges to depends on the shape of the slice (tall-and-skinny vs short-and-fat).

Does any of this matter? Not really, it's just that if you start out with the premise that the rate of momentum transfer to the air must be equal to the lift force, then any conclusion you draw from that point on is completely suspect (and frequently incorrect).
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Old Nov 13, 2012, 11:30 AM
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This next request may benefit you point of view.
Calculate how much air needs to be Accelerated etc to hold an Aircraft in a lift state.
If the amount is more than is evident then your point of view must be true.
Here's another way of thinking about this... Suppose you had a snapshot of the flow around a wing that would allow you to determine the velocity at any point. If you examined a slice of the wake, could you find enough downward momentum density there to "account for" the lift force? If you were to look just at the vertical region between the wingtips, you'd be able to find more than enough downward momentum density to account for the lift (as long as you made your search region tall enough).

The problem is that you aren't really free to pick and choose where you look for downward momentum density. If you say "the flow around a wing includes downward momentum equal to the lift", implicit in that statement is that you are considering the rate of change of ALL the air influenced by the wing, not just some subset of the air you've chosen to focus on (Newton's Second Law doesn't have a "proximity clause").

OK so what happens if figure out the rate at which the air's vertical momentum is changing when you add up the contributions from ALL the bits of air being influenced? The (not very appealing, but correct) answer is that it depends on how you do the accounting. If you add up all the momentum density in a slice of the wake that extends many spans to the left and right as well as above and below the wing you might think you would get a single answer. However, if you take a slice of the wake that is very big compared to the wingspan in all dimensions, but is overall short and fat, you find that there is no rate of vertical momentum change (the upward and downward contributions cancel exactly). On the other hand, if you take a slice that is tall and skinny (again very large in both dimensions compared to the span), then you find that the rate of momentum change is equal to the lift.

There's no contradiction in this result. If you consider a short-fat region of air containing the wing, this region will have significant area on the "floor and ceiling" for the wing's pressure "footprint" to act. In this case the lift on the wing is balanced by the force exerted on the air by the floor/ceiling. If you consider a very tall-skinny region, there is very little floor/ceiling area for the pressure footprint to act on, and the lift on the wing is balanced by momentum transfer to the air.
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Old Nov 13, 2012, 12:48 PM
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Re: downward aceleration equal to lift.

In the creation of lift by the centipetal accelerations of a turning flow it is shown with simple geometry. The pressure change is the mass inertia resistance (ma) to the accelertion normal to a surface element. Lift is the vertical component of the force of that element. It is equal and opposite to the vertical component of the mass acceleration normal to the surface. (we have an "N' diagram)
From Newton, all forces ballance.

Since the acceleration produces a pressure gradiant, the pressure change is summed accross the entire flowfield. The result is that all of the fluid put in motion by the wing contributes to the lift.
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Old Nov 13, 2012, 01:18 PM
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I have to confess that I have no idea what you mean, but I agree with this part:

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The result is that all of the fluid put in motion by the wing contributes to the lift.
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Old Nov 13, 2012, 02:00 PM
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Next, a report from Prof Irwin Corey------for the uninformed , -the Professor would season his speech with many long and florid, but authentic, words. The professor would then launch into nonsensical observations about anything under the sun, but seldom actually making sense. Changing topics suddenly, he would wander around the stage, pontificating all the while. )
Sounds familiar ----
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Old Nov 13, 2012, 02:06 PM
greg
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Originally Posted by ShoeDLG View Post
I don't think so, not fundamentally. I think the horizontal momentum transfer to the air in the case of a sail producing a steady side force would be entirely analogous to the vertical momentum transferred to the air in the case of a wing producing steady lift IF you could somehow take the ground out of the picture.
"IF"? i don't understand how the ground is in the picture when a plane is rolled 90 deg at 18,000 ft.

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Does any of this matter? Not really, it's just that if you start out with the premise that the rate of momentum transfer to the air must be equal to the lift force, then any conclusion you draw from that point on is completely suspect (and frequently incorrect).
I hope i'm correct is believing that the pressure regions above and below the wing exert a force on the wing. I also believe they exert a force on the surrounding air, as well. If the air is free to move, why wouldn't that force accelerate it?

can you please explain why there is no vertical airflow immediately above or below the pressure regions adjacent to to the wing? Why would there be flow near the wing tips but not the wing root (i realize it can't flow laterally thru the wing, but why not longitudinally)?

i'm not arguing that there must be flow or no flow. But if there is no flow, no displacement of the air, why not?
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Old Nov 13, 2012, 03:26 PM
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Based purely on an appeal to Newton's Second Law, the air's instantaneous rate of change of "sideways" momentum in response to a wing at 90 degrees of bank at 18,000 feet must be equal and opposite the lift force acting on the wing. Assuming there's no vertical terrain up there ;-).

That said, you can't (correctly) jump to the conclusion that there is always equality between the lift and the air's rate of momentum change. In the case of steady flight above the earth (which represents a substantial fraction of all flight), the wing is pushing the air one way and the ground is pushing it the other way. It's the difference in how hard they push that will determine the air's rate of momentum change.

I'm not suggesting the wing can't change the air's net momentum under certain conditions, I'm simply saying that there are many cases where direct equality does not exist between the lift force on the wing and the air's net rate of momentum change.
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Old Nov 13, 2012, 04:27 PM
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As the air flows around a lifting wing, there is a balance at every point between the pressure differences and the local accelerations. The pressure gradients and accelerations are inextricably connected, and I think it's easy to get into trouble by suggesting that either one "causes" the other.

A lifting wing acts much like a turning vane, imparting downward momentum to the air passing above and below it. The pressure difference between the top and bottom of the wing couldn't exist without this flow turning (unless you get REALLY close to the ground). It's very tempting to end the story there and conclude that the lift force on the wing is due to the momentum exchange associated with this turning. There's a potential problem with that conclusion in the case of a 3D wing. It ignores the fact that the air passing outboard of the wingtips gets deflected upward (and a lot more air passes outboard of the wingtips than between them). To say that the lift acting on the wing is a product of momentum exchange, the lift must equal the air's total rate of momentum change (accounting for both upward and downward changes). A couple hundred posts ago I posted a .pdf file where I went through the accounting exercise of comparing the upward and downward exchange rates for a wing in steady flight. When you go through this exercise, you discover that the relationship between the lift and the momentum exchange isn't nearly as straightforward as it is often made out to be here.
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Old Nov 13, 2012, 05:24 PM
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I have to confess that I have no idea what you mean, but I agree with this part:
I do have a problem writing to others without knowledge of thier understanding of the areas that I am addressing .

What portion of the responders here are aquainted with the accelerations involved in turning flow? When flow turns the velocity vectors are changing. This involves very serious acceleration. This is the same acceleration derived from centrapetal acceleration (normal) of the flow passing the curve, v^2/R Times rho gives a pressure gradient dp/dR. The important momentum change here is the change of the flow velocity vector.

As previous, the vertical component of the normal mass acceleration is exactly(?) equal to the vertical contribution to lift from any surface element.

I don't care much for momentum. Momentum is a result and tells me very little of cause.

Somebody let me know if there is a better way to tell the story.
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Old Nov 15, 2012, 06:21 AM
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http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html
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Old Nov 15, 2012, 07:24 AM
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All very esoteric - but how it relates to model science -simply escapes me.
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