Dec 20, 2012, 05:38 AM
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Quote:
 Originally Posted by ciurpita how would the answer be different for a vertically mounted rotor (propeller) operating horizontally?
The presence of the ground would still influence the path of a fluid element. I would expect the influence of the ground to be far less for the case where the thrust is parallel to the ground.

The only reason we're talking about helicopters in ground effect is Tim Green asked: "How close to zero can the rate of momentum transfer be, when a 1000lb copter is hovering?"

I responded that: "in the limiting case of ground effect (rotor height very small compared to rotor diameter), the rate of momentum transfer goes to zero."

I'm not exactly sure where the discussion is going, but through a series of follow-on questions, I believe we have agreed:

1. The dynamics of an isolated bowling ball provide an inadequate analogy for describing the dynamics of subsonic fluid flow. The path that a fluid element follows can affect the pressure on the surface of a wing even if that fluid element never makes direct contact with the wing.

2. The momentum of an object that is composed of a collection of elements that are free to move relative to each other (like the air) can be zero even if some or all of the elements are in motion.

I have suggested that the acceleration experienced by a fluid element in the vicinity of a helicopter operating in ground effect is influenced by both the rotor and the ground. Not sure if we agree on this or not.
Last edited by ShoeDLG; Dec 20, 2012 at 05:48 AM.
 Dec 20, 2012, 06:08 AM greg somerset, nj Joined Feb 2005 377 Posts i'm just trying to make sense of this. Sorry if this is tangential to your helicopter analysis. Is this a matter of perspective? Based on your description of "net momentum transfer", of course it's zero. Holes (vacuum) aren't being created in the space above the ground when aircraft move through it. but is it true that no net momentum transfer means no downwash and this means no organized movement of air due to lift? isn't it true that on a smaller scale, < wing chord/10, the air is displaced by the airfoil, first traveling upwards and then downwards on the top surface? Isn't this organized movement of air which means there is a local transfer of momentum in the region close to the airfoil surface? But perhaps more specific to issue being discussed, isn't there is also organized air movement and momentum transfer upward just in front (chord < 10) of the airfoil and something similar behind the airfoil? can you relate what you've been saying on this scale, chord < 10? (Of course it's influenced by the air > chord/10 distance from the airfoil). my understanding is if there is roughly equal organized upward movement of air in front of the airfoil and downward movement behind it, there is no net momentum change on a scale of 2* chord. But the airfoil has affected the air near it (< chord/10). Isn't there some local momentum change < chord/10? i'm not looking for specific answers to my questions, but hoping they give you a sense of my confusion.
Dec 20, 2012, 08:56 AM
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Quote:
 Originally Posted by ShoeDLG The temptation is to respond with the obvious answer: that the air is accelerated downward by the rotor blades. In many ways this answer is correct, but it isn’t complete. If you consider a fluid element near the rotor, the acceleration it experiences is determined by the sum of the forces acting on it (in accordance with Newton’s Second Law). For a rotor operating in ground effect, the forces acting on a fluid element are influenced not only by the presence of the spinning rotor, but by the presence of the ground as well. The path that any given fluid element follows is obviously different depending on whether the rotor is operating near the ground or not. This means that the accelerations experienced by the air are different. Your question is: “What's accelerating the air downward?” The answer is: both the rotor and the ground are responsible for the acceleration experienced by the air.
The ground? Get Real. It can only slow the air down - it cannot accelerate it - the ground doesn't move - not in my world.

To answer my own question correctly - what's accelerating the air downward? THE WING.

Then according to Newton's Third - the wing is reacting with a lifting force equal to the mass of the air times the acceleration of the air.

How do we know the air's accelerated? That air wasn't moving before the wing got to it. It's moving after the wing goes through it though.
How do we know the air has mass? Every thing has mass.
How do we know the wing was the object that moved the air? Nothing else there but the wing. Has to be the wing.
How do we know the wing will react with a lifting force to the action of moving all that air down? Newton's third law - for every action, there's a reaction.

Sure there are side shows, lots of complex math needed, in order to actually design a wing. But the lifting mechanism is straight down simple - the wing reacts to the air it causes to move downward per Newton's third law.
Last edited by Tim Green; Dec 20, 2012 at 09:06 AM.
Dec 20, 2012, 11:16 AM
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Quote:
 Originally Posted by Tim Green The ground? Get Real. It can only slow the air down - it cannot accelerate it - the ground doesn't move - not in my world. To answer my own question correctly - what's accelerating the air downward? THE WING. Then according to Newton's Third - the wing is reacting with a lifting force equal to the mass of the air times the acceleration of the air. How do we know the air's accelerated? That air wasn't moving before the wing got to it. It's moving after the wing goes through it though. How do we know the air has mass? Every thing has mass. How do we know the wing was the object that moved the air? Nothing else there but the wing. Has to be the wing. How do we know the wing will react with a lifting force to the action of moving all that air down? Newton's third law - for every action, there's a reaction. Sure there are side shows, lots of complex math needed, in order to actually design a wing. But the lifting mechanism is straight down simple - the wing reacts to the air it causes to move downward per Newton's third law.
I 'll be darned - - I foolishly believed lift occurred because there was a pressure difference .
pressure higher on bottom than top- seems to want to equalize - for some strange reason and as long as pressure difference is sufficient - you got lift. Complex, math required - In some cases - in most nope
Dec 20, 2012, 11:52 AM
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Quote:
 Originally Posted by Tim Green The ground? Get Real. It can only slow the air down - it cannot accelerate it - the ground doesn't move - not in my world.
"Acceleration" in the Newtonian sense refers to a change in (vector) velocity. If the ground is slowing the air down, it is accelerating it (and there is a force required to create the change velocity).

No one said anything about the ground moving. If you don't think the stationary ground can accelerate the air, then I recommend you jump off a tall building and tell us if you experience a reaction force as you come to a stop.

Quote:
 Originally Posted by Tim Green Sure there are side shows, lots of complex math needed, in order to actually design a wing. But the lifting mechanism is straight down simple - the wing reacts to the air it causes to move downward per Newton's third law.
I will continue point out that the mechanism you describe is generally incorrect. Your assertion that a wing must change the air's vertical momentum is nothing more than an expression of faith. Until you actually compare the momentum in a wing's upwash to momentum to its downwash you are speculating about net momentum transfer. Simply stating that there is more momentum in the downwash than in the upwash does not make it so.
Dec 20, 2012, 12:28 PM
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Quote:
 Originally Posted by richard hanson I 'll be darned - - I foolishly believed lift occurred because there was a pressure difference . pressure higher on bottom than top- seems to want to equalize - for some strange reason and as long as pressure difference is sufficient - you got lift. Complex, math required - In some cases - in most nope
Feeling a bit sarcastic today, are we? Anyway, F=MA is the math you need to understand first, before trying anything more complex.
Dec 20, 2012, 12:39 PM
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Quote:
 Originally Posted by ShoeDLG "Acceleration" in the Newtonian sense refers to a change in (vector) velocity. If the ground is slowing the air down, it is accelerating it (and there is a force required to create the change velocity). No one said anything about the ground moving. If you don't think the stationary ground can accelerate the air, then I recommend you jump off a tall building and tell us if you experience a reaction force as you come to a stop. I will continue point out that the mechanism you describe is generally incorrect. Your assertion that a wing must change the air's vertical momentum is nothing more than an expression of faith. Until you actually compare the momentum in a wing's upwash to momentum to its downwash you are speculating about net momentum transfer. Simply stating that there is more momentum in the downwash than in the upwash does not make it so.
The ground absorbs energy from the air - it doesn't add energy. Only moving things can add energy to the air. Nor is the ground, part of the plane.

Anyway, it's a moot point - cause you seem to be denying that wings move air (accelerate the air downward) ... just so you can deny that the wings react to the air they move with a lifting force.

Wow.
 Dec 20, 2012, 12:56 PM Registered User Germany, BW, Stuttgart Joined Mar 2012 1,205 Posts Energy and momentum are two fundamentally different quantities. The ground doesn't need to change the air's energy in order to change its momentum
 Dec 20, 2012, 01:12 PM Registered User Joined Dec 2011 62 Posts [QUOTE=ShoeDLG;23559575]. I think we are in complete agreement about how the radial pressure gradient at any point in the fluid is related to the acceleration normal to the flow streamlines: (1) dr = -(r/v)dv or equivalently: dv = (-v/r)dr (2) dp = density *( v^2 / r) * dr = density *( v^2 / r) * (-r/v)*dv (3)dp = - density * v * dv QUOTE] (823) Thanks again for the imput, However, there is a little more work to be done! As stated, your equations fit the min pressure point on the circular cylinder where dv/v = .5 When we move to the eliptical cylinder where the max velocity = v=V(1+h/l), dv/v falls as h/l gets smaller. The same problem arises when we move away from min pressure point on all cylinders. Either the .5 is a constant, based apon its being defined by the circular cylinder being a dipol or a correction needs to be added for different h/l. The momentum of air passing the min pressure point is the same for equal h but the vertical profle changes with h/l.
Dec 20, 2012, 01:46 PM
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Quote:
 Originally Posted by ShoeDLG The bottom line remains that you can't determine the pressure at a point on the surface of a cylinder or airfoil purely from the boundary conditions there. You have to know the conditions over the entire fluid boundary,:.
One has to keep things in context. For the hundreds of web pages and the thousands of visiters, 98% are only looking for principles, they are not going to try to solve any four element equations.

To the best of my knowledge no one that completely understands has produced a truly accurate description of lift for the "common" man. They have produced papers of "chicken scratches" for each other. When I started this what was out there was terrible.

As for conditions at a point on the surface, if you have the local velcity it was established from the source/ sink displacement pattern that shapes the entire field. It also includes the overall effects of circulation. That should be close enough for principles.
Dec 20, 2012, 05:01 PM
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Quote:
 Originally Posted by Tim Green Feeling a bit sarcastic today, are we? Anyway, F=MA is the math you need to understand first, before trying anything more complex.
Sarcastic -
not really
prove my staement wrong.
 Dec 21, 2012, 01:37 AM Registered User Germany, BW, Stuttgart Joined Mar 2012 1,205 Posts A little knowledge is a dangerous thing. We “know” F=ma, so if I push on something it will accelerate. Example, I lay a piece of plywood on the floor and stand on top of it. If the plywood has mass m and I have weight W then the piece of plywood accelerates downward at a rate a = W/m. Same thing applies to a helicopter hovering in ground effect. If the helicopter has weight W and the air has mass m, then the air accelerates downward at a rate a = W/m.
 Dec 21, 2012, 07:10 AM Registered User United States, UT, Salt Lake City Joined Oct 2007 9,921 Posts If I paint a picture of a piece of plywood on the floor and stand on it - Does the picture accelerate? Just wondering - Merry Christmas! Last edited by richard hanson; Dec 21, 2012 at 08:01 AM.
Dec 21, 2012, 12:54 PM
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Quote:
 Originally Posted by ShoeDLG Energy and momentum are two fundamentally different quantities. The ground doesn't need to change the air's energy in order to change its momentum
We're working on wing's reaction to the air - not the ground's.
Dec 21, 2012, 12:58 PM
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Quote:
 Originally Posted by ShoeDLG A little knowledge is a dangerous thing. We “know” F=ma, so if I push on something it will accelerate. Example, I lay a piece of plywood on the floor and stand on top of it. If the plywood has mass m and I have weight W then the piece of plywood accelerates downward at a rate a = W/m. Same thing applies to a helicopter hovering in ground effect. If the helicopter has weight W and the air has mass m, then the air accelerates downward at a rate a = W/m.
Great - so therefore, the chopper will react with an equal and opposing force called lift.

Do you agree with the above statement?

If not - you clearly don't understand physics, particularly Newton's third law.