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Old Sep 18, 2012, 05:00 PM
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These exercizes are all well and fine but I fail to see their value in a model science forum.
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Old Sep 29, 2012, 01:03 AM
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Sorry to to disrupt the train of thought here but I have a question pertinent to the theme of this thread.

First, we know that the flow around the wing does include downward mass acceleration equal to the lift. There is also a downward momentum aft of the wing that is approximatly equal to lift.

The question is, does anyone have accessto an authoritive source as to downwash beinga true reaction to lift?
A wing generates circulation which establishes the lift then is left behind as downwash.However, the "down" of downwash appears to come only from the mechanics of vortex pairs.
I cannot find anyway to rationalize the "down" of downwash to be a reaction to lift.
(Acknowlrdging Shoes admonition that one needs to understand the entire field)
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Old Nov 03, 2012, 07:27 PM
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Quote:
Originally Posted by Crossplot View Post
Sorry to to disrupt the train of thought here but I have a question pertinent to the theme of this thread.

First, we know that the flow around the wing does include downward mass acceleration equal to the lift. There is also a downward momentum aft of the wing that is approximatly equal to lift.

The question is, does anyone have accessto an authoritive source as to downwash beinga true reaction to lift?
A wing generates circulation which establishes the lift then is left behind as downwash.However, the "down" of downwash appears to come only from the mechanics of vortex pairs.
I cannot find anyway to rationalize the "down" of downwash to be a reaction to lift.
(Acknowlrdging Shoes admonition that one needs to understand the entire field)
The lift formula contains an Acceleration Portion
Lift = Cl 1/2roh V^2 S
Force = MA

The V^2 is the Acceleration part

The CL is how well the sling shot works

1/2 roh is the size or the Shot (So to say) and the reaction (Opposite) to it

S is the # of sling shots or the surface area.

This is the old and still true to date formula for lift that involves reaction.

Though the Cl value is where we see the interesting Pressure functions, as is, CL as a # is almost a meaningless as to what is truly going on.

Sorry No Authority here
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Old Nov 11, 2012, 11:08 PM
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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 predominent downwash that we see is created by the remnants of the creation of circulation.

Have a nice day .
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Old Nov 12, 2012, 04:22 AM
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First, we know that the flow around the wing does include downward mass acceleration equal to the lift.
How do we know this?
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Old Nov 12, 2012, 03:15 PM
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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.

Quote:
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 predominent 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
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Old Nov 12, 2012, 06:42 PM
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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.
So in order for a fluid to exert a force on an object (e.g. for the air to exert lift on a wing), the momentum of the fluid must change (at a rate equal and opposite the exerted force)?
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Old Nov 12, 2012, 06:55 PM
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So in order for a fluid to exert a force on an object (e.g. for the air to exert lift on a wing), the momentum of the fluid must change (at a rate equal and opposite the exerted force)?
That about sums it up. But don't forget the Formula F=MA
If you increase the acceleration rate. You do not need as much mass. So different wing foils take advantage of this to get the same result.

PS
Got a bit lost with the Exerted and exerter. But for the wing to produce lift. The force has to be opposite to gravity, and that happens to be down. So there has to be some air that has had its velocity vector changed to something of a more Downward direction/vector.
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Old Nov 12, 2012, 07:11 PM
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That about sums it up. But don't forget the Formula F=MA
So if the momentum of a fluid must change in order for it to exert a force on an object, then when you're floating on your back in a swimming pool, you must be changing the vertical momentum of the water?
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Old Nov 12, 2012, 07:28 PM
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So if the momentum of a fluid must change in order for it to exert a force on an object, then when you're floating on your back in a swimming pool, you must be changing the vertical momentum of the water?
Not quite. You are displacing the water and causing the total volume to increase. But what you are talking about is similar to a Hot air Balloon or Helium balloon.

http://en.wikipedia.org/wiki/Buoyancy

But if you were made of a material more dense than water you would need to work on the water to stay head above the water. (Like a skinny guy wearing a lead weight belt with out a wetsuit, or other buoyant devices)
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Old Nov 13, 2012, 12:40 AM
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If the following statement is true:

"In order for a fluid to exert a force on an object, the momentum of the fluid must change."

How is it that the water can exert an upward force through buoyancy without its momentum changing?
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Old Nov 13, 2012, 03:00 AM
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Originally Posted by ShoeDLG View Post
If the following statement is true:

"In order for a fluid to exert a force on an object, the momentum of the fluid must change."

How is it that the water can exert an upward force through buoyancy without its momentum changing?
It has to do with the topic of lift/force, we are talking about lift of objects that are Heavier (Denser) than the Medium they are flying (With wings) in.
Or so I thought
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Old Nov 13, 2012, 04:15 AM
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In the case of a lifting wing, you invoked Newton's Laws to suggest that the lift force exerted on the wing must be accompanied a change in the air's momentum.

In the case of a body floating in water, you seem OK with the idea that no change in the water's momentum is required for the water to provide an upward force.

Newton's Second Law says that an object's rate of momentum change is proportional to the unbalanced force acting on it.

In the bouyancy case, its obvious that the water in a pool doesn't experience an unbalanced force when you climb into the pool (the bottom of the pool will push up on the water with the same force that your weight pushes down).

You can write Newton's Second Law as it applies to a lifting wing:

Air's rate of vertical momentum change = Downard force exerted by the wing on the air - Upward force exerted by the ground on the air

Only in the case where the upward force exerted by the ground on the air is zero can you conclude that "the flow around the wing does include downward mass acceleration equal to the lift".

Neither people floating in pools nor wings are exempt from Newton's Laws. The same connection between forces and momentum exchange that exists in one case must exist in the other.
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Old Nov 13, 2012, 07:16 AM
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Only in the case where the upward force exerted by the ground on the air is zero can you conclude that "the flow around the wing does include downward mass acceleration equal to the lift".
why is the above true?

while the pressure regions above and below the wing produce an upward force on the wing surface, don't they also exert a force on the air above and below the pressure regions. If the air displaces, then it is accelerated downward. But where would it displace too?

the air below the wing is blocked by air below it, and a vacuum would be created by the displacement of the air above the wing. The only place i can see it going is forward -- any air displaced below the wing movs forward and upward to fill in the vacuum left by the air displaced above the wing.

if true, then there would be little net movement of air if the air displaces (equal movement down and up). I don't understand why there wouldn't be displacement, but isn't that what the 2d theory suggests?

If so, why is there no displacement when there is a force applied to the air?

greg
looking for understanding
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Old Nov 13, 2012, 07:20 AM
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Originally Posted by ShoeDLG View Post
In the case of a lifting wing, you invoked Newton's Laws to suggest that the lift force exerted on the wing must be accompanied a change in the air's momentum.
Quote:
YES.
By the nature of the Wing it works that way (Unlike an Air Balloon)
In the case of a body floating in water, you seem OK with the idea that no change in the water's momentum is required for the water to provide an upward force.
Quote:
Not talking about Buoyancy, without refreshing and further education on it on my behalf.
.

Newton's Second Law says that an object's rate of momentum change is proportional to the unbalanced force acting on it.

In the bouyancy case, its obvious that the water in a pool doesn't experience an unbalanced force when you climb into the pool (the bottom of the pool will push up on the water with the same force that your weight pushes down).
Quote:
OK, I will talk a little.
That's a strange way of looking at it. But I agree that there is more Pressure on the pools bottom.
You can write Newton's Second Law as it applies to a lifting wing:

Air's rate of vertical momentum change = Downard force exerted by the wing on the air - Upward force exerted by the ground on the air
Quote:
I can see that. No wait. Is that a minus ?
Only in the case where the upward force exerted by the ground on the air is zero can you conclude that "the flow around the wing does include downward mass acceleration equal to the lift".
Quote:
Not quite true, Its due to the Ground that there is Air Pressure of such a value, And because of the Ground, the Air will dissipate its energy into vortices and heat more quickly.
Neither people floating in pools nor wings are exempt from Newton's Laws. The same connection between forces and momentum exchange that exists in one case must exist in the other.
Can you blow a candle out from a distance of 15ft by Just blowing through a straw?

Its very hard to do (I doubt any one could), but to suggest that because the candle did not go out you did not blow any air in that direction is Presumptuous. The energy was dissipated

If you fly out of ground effect you are above the reaction time of air and ground and wing, and are needing more work to be done to the air than when you are in ground effect.

An aircraft is flying in a very tight turn (Bank Angle of close to 90deg).
Q/ Which way is the force vector that Caused it to change its Velocity vector.
A/ It's near Horizontal.
The horizon is not needed to be a wall or anything similar to what you are suggesting with the Ground and Lift.
It's the local air that was affected.
Of cause, This air being accelerated near Horizontally had energy added to it, but, it will dissipate and finally turn to heat before circling the earth. (may be even before getting close to the Horizon ?)

Bryce.
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