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Old Dec 31, 2012, 01:50 PM
Ascended Master
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Originally Posted by ShoeDLG View Post
Interesting. One thing that was very apparent when placing a plate above the spinning prop was that the plate wanted to sag from the prop "pulling" it down (reduced pressure on the lower side of the plate).

The prop doesn't just increase the air pressure on the lower plate, it also significantly reduces the air pressure on the upper plate. Definitely consistent with your result.
.
Propellor theory says the spinning prop accelerates the air ahead of it through the prop disk. Probably why no one puts a large plate in front of a prop.
In "Mad, Mad World", Paul Mantz flew a Cessna C-45 through a billboard... I'd love to have seen what any instruction might have picked about any forces/accelerations on the airplane as went through the sign.
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Old Dec 31, 2012, 01:50 PM
Launch the drones ...
Ashtabula, OH USA
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Originally Posted by ShoeDLG View Post
Your explanation is definitely aligned with some of the NASA K-12 websites. If you are content with that level of understanding, that's great, give the subject no more thought. While I don't want to put words in their mouths, it appears there are other contributors here who understand there is more at play than simple momentum exchange (not to be mistaken for an assertion that momentum exchange isn't involved). I have enjoyed exchanging thoughts with them. I have learned from them, and I hope they have taken something away from what I have tried to offer. Your contributions have amounted to little more than contradiction of any suggestion there is any complexity involved (including ideas attributed some very accomplished aerodynamicists).

If you wish to add something constructive to the discussion, you could:

- Provide a rigorous accounting for all of the momentum that a steadily lifting wing transfers to the air, and show that the rate of transfer is in all cases equal to the lift ("There's lots of rate of momentum change, when that wing hits the air" is not a rigorous accounting).

- Show that the vertical forces acting on the air when a helicopter is hovering in ground effect do not sum to zero (as is required for any net momentum change).

- Explain how "kicking air down" accounts for the variation in the drag and lift curve slope experienced by a wing in ground effect.

I still like Nmaster's response when asked if he could help someone understand how wings generate lift: "Nope.. Fluid dynamics is the hardest problem in classical physics." Is lift generation is a very simple phenomenon embedded within a very hard problem? I suppose that's possible, but I think it's more likely there are some subtle features that aren't captured at the K-12 level.
Let me put this another way - my issue with you, is you deny that wings react with a lifting force to the air they shove down (what NASA says, BTW). So, until you understand this, there's simply no point in going anywhere else with you, regarding wings and why they don't drop to the ground.
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Old Dec 31, 2012, 01:53 PM
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Originally Posted by Sparky Paul View Post
.
Propellor theory says the spinning prop accelerates the air ahead of it through the prop disk. Probably why no one puts a large plate in front of a prop.
In "Mad, Mad World", Paul Mantz flew a Cessna C-45 through a billboard... I'd love to have seen what any instruction might have picked about any forces/accelerations on the airplane as went through the sign.
The plate would cause drag, since the plane is supposed to move. However, engineers have no problem placing fan intakes close to another surface - so long as enough air gets in the sides, that the fan is sucking in the amount of air it was designed for.

You could have a turbo fan, that had an L shaped entrance - the thrust would still have the same directional vector as it would with a straight entrance.
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Old Dec 31, 2012, 02:26 PM
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Let me put this another way - my issue with you, is you deny that wings react with a lifting force to the air they shove down. Until you understand this, there's simply no point in going anywhere else with you, regarding wings and why they don't drop to the ground.
Ignatius,

You're so self-satisfied with your insight that you haven't bothered to try to understand what I've written (or what Lissaman, or Kroo, or Drela have written on this subject). I have never suggested (nor have they) that wings don't experience a reaction force from pushing down on the the air. We differ on what happens to the air's momentum as a result of the force exerted on it by a lifting wing. Newton's Second Law does not require the air to experience a change in momentum equal to the lift force as you state. In the case of a wing in ground effect, the air's rate of vertical momentum change is easily demonstrated to be less than the lift.
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Old Dec 31, 2012, 04:00 PM
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Ignatius,
. In the case of a wing in ground effect, the air's rate of vertical momentum change is easily demonstrated to be less than the lift.
In this case momentum gets to the ground, air pressure works the same way like a landing gear to keep the aircraft aloft. In free atmosphere there is no high pressure cushion to add lift, because there is no wall to contain it, instead of beeing pressurized the air accellerates -> momentum.
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Old Dec 31, 2012, 04:19 PM
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Idle mind, work in progress...
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Old Dec 31, 2012, 06:00 PM
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Ignatius,
I have never suggested (nor have they) that wings don't experience a reaction force from pushing down on the the air.
Well, my stars. I don't believe what I'm reading. What an amazing world, we live in - will wonders never cease? You've come over to NASA's side.

My work is done - time to play now. So, what do you want to know about the air's momentum - and under what circumstances? Cmon - let's rumble.
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Old Dec 31, 2012, 06:18 PM
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Poking a sharp stick in my eye would be preferable to more of your buffoonery. Congratulations on learning to read (although the big boy pants don't appear imminent). NOTAC.
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Old Jan 01, 2013, 08:03 AM
greg
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Newton's Second Law does not say that the air must acquire downward momentum at a rate equal to the weight of the helicopter any more than it says the foam block must acquire downward momentum at a rate equal to the weight of the helicopter. It is POSSIBLE that the air does gain momentum at a rate equal to the helicopter's weight, only to have it later removed by the scale, but Newton's Second Law certainly doesn't require this to be the case.
...

In the case of a helicopter using the air to hover, the acceleration experienced by an element of the air is at every instant influenced by both the rotor and the ground. This is borne out by the well established fact that the hover performance of a helicopter depends very strongly on the height of the rotor above the ground.
sounds like you're saying you need to consider the air, at least in steady state, as monolithic mass, like, as you've said, a block of foam.

but while considering the flow of air over an airfoil, isn't the air analyzed as small air parcels that follow streamlines and influenced by adjacent parcels?
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Old Jan 01, 2013, 09:53 AM
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Poking a sharp stick in my eye would be preferable to more of your buffoonery..
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Old Jan 01, 2013, 10:45 AM
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Greg,

No, that's not what I'm suggesting.

Suppose you have a rigid object with equal and opposite forces (F_1 and F_2) acting on it: F_1 = - F_2

It is obvious in the case of a rigid object that it experiences no acceleration (change in its momentum) if the two forces acting on it are equal and opposite. This is consistent with Newton's Second Law:

Sum(F_object) = F_1 + F_2 = d(p_object)/dt = 0,

where p_object is the object's (vector) momentum: p_object = Sum(m_i*v_i).

The product m_i*v_i represents the mass times (vector) velocity of an element of the object. The sum is carried out over all of the elements that make up the object.

Given that the rigid object experiences no change in its momentum (no acceleration), it's misleading at best to suggest:

F_1 = d(p_object)/dt

and

F_2 = d(p_object)/dt.

The object does not accelerate in one direction due to F_1 and accelerate an equal amount in the opposite direction (decelerate) due to F_2.

Suppose instead of a rigid object, you have an object made up of a collection of elements that are free to move relative to one another. If this object also has equal and opposite forces acting on it, then Newton's Second Law applies in exactly the same way:

Sum(F_object) = F_1 + F_2 = d(p_object)/dt = 0.

In the case where the object is not rigid, does it accelerate in one direction due to F_1 and accelerate in the opposite direction an equal amount due to F_2?

Because the object is not rigid, this is now possible. Take the case where the object is made up of only two elements: m_1 and m_2. Suppose F_1 acts only on m_1 and F_2 acts only on m_2. If m_1 and m_2 do not exert any force on each other, then it is correct to write:

F_1 = d(p_1)/dt = m_1*d(v_1)/dt = m_1*a_1

and

F_2 = d(p_2)/dt = m_2*d(v_2/dt = m_2*a_2.

So in this case, the object experiences no net change in its momentum, but part of the object (m_1) experiences a rate of change of momentum equal to one of the forces acting on it (F_1), while another part of the object (m_2) experiences a rate of change of momentum equal to the other force acting on it (F_2).

Does this describe the behavior of the air in the case of a helicopter in ground effect? Does the force exerted on the air by the rotor (call it F_1) change the momentum of one portion of the air (call it m_1) at a rate equal to F_1, while the force exerted on the air by the ground (call it F_2) changes the momentum of another portion of the air (call it m_2) at a rate equal to F_2 (equal and opposite F_1)?

Because the air is not rigid, this is possible. However, because adjacent elements of the air can exert forces on one another, the air is different from an object made up of two independent elements. Suppose m_1 exerts a force on m_2 equal to F_12. Newton's Third Law says that m_2 exerts a force on m_1 equal to -F_12. If the rotor only exerts a force on m_1 and the ground only exerts a force on m_2, then we can write Newton's Second Law as:

F_1 + F_12 = d(p_1)/dt = m_1*d(v_1)/dt = m_1*a_1

and

F_2 - F_12 = d(p_2)/dt = m_2*d(v_2)/dt = m_2*a_2.

This shows that only in the special case where you can separate the air into two portions such that:

1. The two portions exert no net force on one another (F_12 = 0)

2. The rotor directly pushes only on one portion and the ground directly pushes only on the other

can you say that the air experiences a downward acceleration equal to the force exerted on it by the rotor and an upward acceleration equal to the force exerted on it by the ground. This analysis considers the air to be a collection of particles that are free to move relative to each other, not a rigid monolithic mass.
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Old Jan 01, 2013, 11:46 AM
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...
can you say that the air experiences a downward acceleration equal to the force exerted on it by the rotor and an upward acceleration equal to the force exerted on it by the ground. This analysis considers the air to be a collection of particles that are free to move relative to each other, not a rigid monolithic mass.
.
Well, kinda... The ground prevents the air blown down from continuing down, and it flows out sideways accelerated in a different direction... but no longer affects the rotor until the unaffected air around it pushes on it... as seen in the heli reloading at the lake image.
The downwash from an airplane has no effect on the airplane once it's off the surfaces of the plane.
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Old Jan 01, 2013, 11:56 AM
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.The downwash from an airplane has no effect on the airplane once it's off the surfaces of the plane.
Then why does the performance of an airplane change when the airplane is in ground effect?
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Old Jan 01, 2013, 12:30 PM
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Then why does the performance of an airplane change when the airplane is in ground effect?
One theory is that the ground affects normal flow patterns, so that there's less upwash on the wings when they are close to the ground. This means that less downwash is needed to maintain lift, thus reducing AOA and drag - so a low winged plane near the ground has extra lift. (During normal flight, some downwash, is counteracted by the upwash - meaning wings will sometimes make more downwash than the plane's weight requires, to maintain lift - depends on the AOA and design of the wing - just how much downwash is counteracted by the upwash)

But I would like to see the behaviour of the airstreams for wings close to the ground, before I took a stand, one way or another.
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Old Jan 01, 2013, 12:47 PM
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Then why does the performance of an airplane change when the airplane is in ground effect?
.
The air has to go somewhere. The downwash has a higher pressure than the ambient air that isn't beneath the plane. When the plane is within a wing span of the ground, the higher air pressure "floats" the plane because it can't get away from the plane quickly enough.. The dust being blown off the runway when the Shuttle lands shows this very well. SR-71s are said to benefit from this also when they land.
VTOL planes are typically tested over an open grid pit which takes the air blown down and deflects it from feeding back to the airframe/motor. VTOLs are frequently plagued by ingesting the hot air circulating back up around and back through the motor... Like the downwash on the heli, but it's hotter and less dense.
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