Dec 19, 2012, 05:07 PM
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 Originally Posted by ShoeDLG Just because something is moving doesn't mean that it has momentum. The (linear) momentum of an object is, by definition, equal to the sum of mass times (vector) velocity of each of the elements that make up the object. If some of an object's elements are moving one way and some the other, it is possible for some (or all) of the object's elements to be in motion despite the fact that the object has no net momentum. The air passing through the rotor disk of a helicopter in ground effect cannot flow into the ground or collect below the helicopter. Instead, it is deflected outward and upward. The fact that there is significant upward motion of the air is well illustrated in the attached photo sequence. At any instant in time the air's downward momentum is equal to sum of all the momentum of the downward moving air minus the sum of all the momentum of the upward moving air. The \$64,000 question is: how does the momentum of the downward moving air compare in magnitude to the momentum of the upward moving air? In the limiting case of ground effect, it is easily demonstrated that they are equal in magnitude (no net vertical momentum). To answer your question. In the limiting case of ground effect, the rotors are making the air move, but they aren't changing its net vertical momentum.
Well, if the air's moving, and has no momentum - then you are redefining momemtum using your own physics, and that's a playground I won't play in. Momemtum is mass x velocity - if the air is moving, it has momentum.

F = ma.

The air has mass, and it's being accelerated downward by the rotors. Not up.

Force is an instantaneous thing - and if the rotors are making the air move, then they're reacting to that - and rotors move air down in a hover, so they react with a lifting force - so long as they're moving the air down.

This is Newton's third law. Action and reaction. And it's as true today, as it was when Newton penned it.
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 Dec 19, 2012, 05:10 PM greg somerset, nj Joined Feb 2005 377 Posts under what conditions would it be possible for there be a net transfer of momentum? i think by your definition, there is no net momentum transfer to the air due to the thrust from something as larger as a Saturn-V because ultimately there is an equal movement of air downward and upward
Dec 19, 2012, 05:25 PM
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Quote:
 Originally Posted by Tim Green Well, if the air's moving, and has no momentum - then you are redefining momemtum using your own physics, and that's a playground I won't play in. Momemtum is mass x velocity - if the air is moving, it has momentum.
For an object that consists of multiple elements that can move relative to one another (like the air, for example), the momentum is defined as the sum of the momentum of the constituent elements. Momentum (like force) is a vector quantity with a magnitude and direction. When you consider the momentum of all the elements of an object, it is entirely possible for their sum to be zero (even if the individual elements are in motion).

This isn't my definition for the momentum, it's the one you'll find in any Classical Mechanics textbook.
Last edited by ShoeDLG; Dec 19, 2012 at 05:45 PM. Reason: Grammar correction
Dec 19, 2012, 05:38 PM
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Quote:
 Originally Posted by ciurpita under what conditions would it be possible for there be a net transfer of momentum? i think by your definition, there is no net momentum transfer to the air due to the thrust from something as larger as a Saturn-V because ultimately there is an equal movement of air downward and upward
A rocket is different from a wing (or rotor) in an important way. When propellent is expelled downward from the nozzle of a rocket, it will displace the surrounding air upward just as air moving downward through a rotor disk in ground effect will displace air upward. If the rocket's propellent is significantly more dense than the air it is exhausted into, then the downward motion of the propellent will have significantly more downward momentum than the upward momentum induced in the air.

Momentum exchange is the fundamental mechanism behind rocket thrust. Wing's are different from rockets in subtle but important ways.
 Dec 19, 2012, 05:46 PM greg somerset, nj Joined Feb 2005 377 Posts ok, a rocket isn't the same. bad example. how about a oil tanker propelling itself through water. is the a net transfer of momentum to the water? do you have to define the volume that you sum the particles within? do you understand what i'm getting at?
Dec 19, 2012, 05:49 PM
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Quote:
 Originally Posted by ShoeDLG For an object that consists of multiple elements that can move relative to one another (like the air, for example), the momentum is defined as the sum of the momentum of the constituent elements. Momentum (like force) is a vector quantity with a magnitude and direction. When you consider the momentum of all the elements of an object, it is entirely possible their sum to be zero ( even if the individual elements are in motion). This isn't my definition for the momentum, it's the one you'll find in any Classical Mechanics textbook.
It's not only possible, but guaranteed. However, that doesn't give one the excuse to mangle the obvious - wings accelerate air downward - and they react to that with a lifting force.

If I roll a bowling ball, once it leaves my hand, I will no longer react to it, as it rolls on down the alley, or gutter. The ball still has momentum, but I'm not reacting to it any more, even though I gave it that momentum.

Same if I roll a dozen balls simultaneously, and they all crash into each other. It takes energy, for me to make a still bowling ball move - but once I've started that ball moving - it won't affect me. And if I'm on a cart, and I throw those dozen bowling balls - my cart will move in the opposite direction - no matter what happens to those balls after they leave me and my cart behind.

Momentum will be conserved, but me and my cart have already reacted to the action of imparting momentum to those balls. With an opposing reaction - we can call cart response, in this case. Rather than lift.
Dec 19, 2012, 05:57 PM
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Quote:
 Originally Posted by Tim Green The air has mass, and it's being accelerated downward by the rotors. Not up.
Ignoring the air accelerated upward by the rotors doesn't make it go away. Look again at the dust in the photo attached to post #838. Before the helicopter arrived, the dust was resting on the ground. Not until helicopter arrived did it get blown upward into the air. In the presence of the ground, the helicopter's rotors most certainly accelerate some air upward.
Dec 19, 2012, 06:03 PM
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Quote:
 Originally Posted by ShoeDLG Ignoring the air accelerated upward by the rotors doesn't make it go away. Look again at the dust in the photo attached to post #838. Before the helicopter arrived, the dust was resting on the ground. Not until helicopter arrived did it get blown upward into the air. In the presence of the ground, the helicopter's rotors most certainly accelerate some air upward.
You are making my point for me - the dust on the ground was moved by air moving downward toward the ground. How else would it pick up the dust? The air hit the ground, and moved the dust upward. But that has nothing to do with the helicoper's rotors having already reacted to the action of accelerating the air downward (toward the dust) - with a lifting force.
Dec 19, 2012, 06:22 PM
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Quote:
 Originally Posted by Tim Green It's not only possible, but guaranteed. However, that doesn't give one the excuse to mangle the obvious - wings accelerate air downward - and they react to that with a lifting force. If I roll a bowling ball, once it leaves my hand, I will no longer react to it, as it rolls on down the alley, or gutter. The ball still has momentum, but I'm not reacting to it any more, even though I gave it that momentum. Same if I roll a dozen balls simultaneously, and they all crash into each other. It takes energy, for me to make a still bowling ball move - but once I've started that ball moving - it won't affect me. And if I'm on a cart, and I throw those dozen bowling balls - my cart will move in the opposite direction - no matter what happens to those balls after they leave me and my cart behind. Momentum will be conserved, but me and my cart have already reacted to the action of imparting momentum to those balls. With an opposing reaction - we can call cart response, in this case. Rather than lift.
This illustrates a fundamental misunderstanding of the mechanics of subsonic fluid flow. If you push against an isolated bowling ball, you will feel a reaction force and the bowling ball will accelerate away. What happens to the bowling ball after you lose contact with it no longer influences the force you feel. If you were to push against another isolated bowling ball, it wouldn't matter if the previous one crashed into a wall or not. However, if you were to push against, not an isolated bowling ball, but a connected collection of bowling balls, the path of a bowling ball that you are currently in contact with WILL be influenced by whether or not the other bowling balls are in contact with a wall.

The fact that the lift and drag experienced by an airplane in ground effect differ from the lift and drag experienced by the same airplane out of ground effect is clear evidence that the behavior of the air in contact with a wing is influenced by the behavior of the air not in contact with the wing. The bowling ball analogy leads to manifestly incorrect conclusions about the behavior of subsonic fluids.
Dec 19, 2012, 07:07 PM
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Quote:
 Originally Posted by ciurpita do you have to define the volume that you sum the particles within?
Absolutely. You can't make a meaningful statement about the momentum of something if you aren't completely clear about what that something is. If you refer to the amount of momentum change that an object imparts to "the air", then you have to consider ALL of the air influenced by the object. This can often be done by considering a volume surrounding the object and setting all dimensions of that volume to be many times the size of the object.

For the case of an oil tanker propeller (that isn't in proximity to a sea wall), there will certainly be net momentum exchange. It's important to realize that the wake structure of a propeller or rotor is very different from that of a wing. The unique wake structure of a wing has some interesting implications. The link in post #842 shows how you can arrive at different values for the rate of momentum transfer depending on the shape of the volume that surrounds the wing. This is why I have been suggesting that the rate at which a wing transfers momentum to the air "depends on how the air is bounded". If you go waaaay back to post #191 in this thread, you'll find a pretty interesting article written for the American Institute of Aeronautics and Astronautics by the late Peter Lissaman (his work is definitely worth a Google search).

He captures my thoughts pretty well in the article:

"In wind tunnels, for example, all the cross stream forces are represented by pressure perturbations on the boundaries - there can't be any downwash, that's what the walls are there to prevent! Interestingly, all aircraft operating on any planet (that covers most cases of interest!) operate in ground effect in the sense that their wake is long compared to their height from the ground plane, and so produce no net downwash momentum in an infinitely large volume. It's a good thing we ave already shown that downwash is not required for lift!"

So rockets are different from propellers and rotors, and propellers and rotors are different from (steadily) lifting wings.
Dec 19, 2012, 07:16 PM
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Quote:
 Originally Posted by ShoeDLG This illustrates a fundamental misunderstanding of the mechanics of subsonic fluid flow. If you push against an isolated bowling ball, you will feel a reaction force and the bowling ball will accelerate away. What happens to the bowling ball after you lose contact with it no longer influences the force you feel. If you were to push against another isolated bowling ball, it wouldn't matter if the previous one crashed into a wall or not. However, if you were to push against, not an isolated bowling ball, but a connected collection of bowling balls, the path of a bowling ball that you are currently in contact with WILL be influenced by whether or not the other bowling balls are in contact with a wall. The fact that the lift and drag experienced by an airplane in ground effect differ from the lift and drag experienced by the same airplane out of ground effect is clear evidence that the behavior of the air in contact with a wing is influenced by the behavior of the air not in contact with the wing. The bowling ball analogy leads to manifestly incorrect conclusions about the behavior of subsonic fluids.
I used bowling balls to demonstrate that your net momentum transfer theory cannot be used to dismiss Newton's third law of action and reaction when wings acclerate air downward. Net momentum is conserved with bowling balls - but the cart still moves backward in reaction to throwing the balls.

That's all I used the bowling ball analogy for. Not as a model for air, but as a model to disprove your theory that since net momentum transfer = 0, a cart for instance, cannot react when a bowling ball is ejected from it.

So, back to my original point - net momentum transfer = zero cannot be used to say that wing's don't react with lift, to the action of accelerating air downward.

Doesn't matter if it's a fluid - it has mass, and it can be accelerated. What's accelerating the air downward? The wing or rotor. What's reacting to the air it accelerates downward? The wing or rotor.
Dec 19, 2012, 07:40 PM
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Quote:
 Originally Posted by Tim Green So, back to my original point - net momentum transfer = zero cannot be used to say that wing's don't react with lift, to the action of accelerating air downward.
So I think we've established that, according to the definition of linear momentum, it is possible for elements of an object to be in motion, but for the object to have no net momentum.

Your question was: If the Rotors aren't transferring momentum to the air - then what's making the air move?

Again, net momentum transfer is not required for the air to move.
Dec 19, 2012, 07:51 PM
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Quote:
 Originally Posted by ShoeDLG So I think we've established that, according to the definition of linear momentum, it is possible for elements of an object to be in motion, but for the object to have no net momentum. Your question was: If the Rotors aren't transferring momentum to the air - then what's making the air move? Again, net momentum transfer is not required for the air to move.
What's accelerating the air downward?
Dec 20, 2012, 04:59 AM
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Quote:
 Originally Posted by Tim Green What's accelerating the air downward?
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.
Dec 20, 2012, 05:21 AM
greg
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Quote:
 Originally Posted by ShoeDLG The answer is: both the rotor and the ground are responsible for the acceleration experienced by the air.
how would the answer be different for a vertically mounted rotor (propeller) operating horizontally?

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