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Old Dec 30, 2012, 04:15 PM
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Originally Posted by ciurpita View Post
Wheli = Fair = d(m.v)/dt = ma

causing the air to accelerate downward and increase its velocity

Fscale = Fair = d(m.v)/dt = ma

causing the air to de-accelerate and decrease its downward velocity
This may seem like semantics, but neither of the above two expressions for the air's rate of momentum change are correct. Newton's Second Law says that an object's rate of momentum change is equal to the unbalanced force acting on it.

The only correct way to write Newton's Second Law is:

Sum (F_air) = W_helicopter - F_scale = d(mv)/dt = 0

By separating the forces and applying Newton's Second Law twice, you are implying that the air gains downward momentum at a rate equal to the helicopter's weight and then loses downward momentum at a rate equal to the reading on the scale.

The fact that it is incorrect to apply Newton's Second Law in "pieces" becomes readily apparent if the helicopter is supported not by the air but by a foam block. If you were to write two separate equations for this case you would get:

W_helicopter = F_block = d(mv)/dt = ma_block

suggesting the weight of the helicopter increases the block's downward velocity

F_scale = F_block = d(mv)/dt = ma_block

suggesting the force from the scale decreases the block's downward velocity

Obviously the downward velocity of the block never changes.

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.

Imagine that the helicopter isn't pushing down on the air, but is instead pushing down on a stream of independent BB's. In this case, you could draw a box around the helicopter, and for the helicopter to be in a steady hover, the BB's would have to transport downward momentum out of the box at a rate equal to the helicopter's weight. In other words, the helicopter would be increasing the downward momentum of the stream of BB's at a rate equal to its weight. The scale would obviously remove the downward momentum at the same rate.

This is only true for the stream of BB's because they are independent. The path (and velocity) of a BB leaving the helicopter is not influenced in any way by the presence of the ground. Each BB is first influenced only by the helicopter and then influenced only by the ground. The performance of a helicopter using independent BB's to hover would not depend in any way on its height above the scale.

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.


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Originally Posted by ciurpita View Post
ok, the net result is no change in momentum, but what happens to the air between rotor and scale compared to the same mass of air without a helicopter? what is the velocity of the air above the rotor, halfway between the rotor and scale, and at the scale?
The answers to those questions depend on the height of the rotor above the scale!
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Old Dec 30, 2012, 06:34 PM
Launch the drones ...
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Quote:
Originally Posted by ShoeDLG View Post
-In order for the helicopter to be in a steady hover, the upward force exerted by the air on the helicopter must be equal to the helicopter's weight.

-Newton's Third Law tells you that the force exerted by the helicopter on the air is equal and opposite the force exerted by the air on the helicopter. Therefore the helicopter exerts a downward force on the air equal to its weight.

-The reading on the scale tells you that the air is exerting a downward force on it equal to the helicopter's weight.

-Newton's Third Law tells you that the force exerted by the air on the scale is equal and opposite the force exerted by the scale on the air. Therefore the scale exerts an upward force on the air equal to the helicopter's weight.

-The net vertical force acting on the air is the vector sum of the downward force exerted on it by the helicopter (equal in magnitude to the helicopter's weight) and the upward force exerted on it by the scale (also equal in magnitude to the helicopter's weight). The vertical forces acting on the air sum to zero.

-Newton's Second Law tells you that an object's rate of momentum change is equal to the net force acting on it. The net vertical force acting on the air is zero, therefore it's rate of vertical momentum change is also zero.
There's lots of rate of momentum change, when that wing hits the air - but this matters not a bit, to you, somehow.
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Old Dec 30, 2012, 07:17 PM
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Originally Posted by ShoeDLG View Post
Sparky,

What if you were to measure the force on the scale in the case where you place a large plate just above the rotor of the single-rotor helicopter? I predict that you would see a noticeable drop in the force measured by the scale.

I suspect that you would have to fix the helicopter in place for this because a stable hover would be next to impossible to achieve. I predict that the lift generated by the rotor would go up even as the reading on the scale went down.
.
Kinda needed more hands than I have.
I put the scale on an armchair, and photographed a Shaung Ma 9104 just above the scale, with and without a glass plate above the rotor, I got about the same numbers for the downwash.
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Old Dec 31, 2012, 12:21 AM
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Originally Posted by Tim Green View Post
There's lots of rate of momentum change, when that wing hits the air - but this matters not a bit, to you, somehow.
You asked for clarification and I offered a straightforward accounting of the forces acting on the air. Newton's Laws indicate very clearly that the air's net rate of vertical momentum change must be zero. If you believe Newton's Laws, there's no getting around that.

The motion of the rotor is certainly adding downward momentum to some of the air, and that local momentum exchange does matter. However, despite this local momentum exchange, the rotor never changes the net vertical momentum of the air. For every element of air displaced downward by a rotor in ground effect, another element of must be displaced upward by the same amount. This corresponding displacement doesn't happen at some time in the future, it happens simultaneously. If this were not the case, the air would collect below the rotor in the same way the BB's collect on the scale in the example in post #901. The air doesn't collect below the rotor. Instead it circulates outward and then upward once outboard of the rotor.

Suppose you neglected the air going upward outboard of the rotor and focused only on the down-going column of air right below the rotor. Is the rate at which the rotor adds downward momentum to that column of air equal to the weight of the helicopter? Suppose we carefully measured it and found that it was. If we changed the altitude of the helicopter, we would change the velocity of the air at any point below the rotor (note that this is not the case for a stream of independent BB's). The change in velocity means that we would find a different value for the rate of momentum change at every altitude. This is problematic if you are suggesting the rate of downward momentum change is always equal to the weight of the helicopter.
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Old Dec 31, 2012, 12:39 AM
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Sparky,

That looks like an exceptionally dangerous setup. Thanks for doing that!

Would be interesting to see how the reading changes (if at all) as the glass got closer and closer to the rotor.
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Old Dec 31, 2012, 01:37 AM
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Originally Posted by ShoeDLG View Post
Sparky,

That looks like an exceptionally dangerous setup. Thanks for doing that!

Would be interesting to see how the reading changes (if at all) as the glass got closer and closer to the rotor.
.
It gets real noisy as the flybar hits the glass.
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Old Dec 31, 2012, 07:35 AM
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Sparky,

I was inspired by your effort and decided to give it a try. Rather than try to hold a helicopter in place, I fixed a prop vertically to blow down on a scale covered by a large plastic plate. I put another plate over the top of this and supported it so that I could adjust its height above the rotor. The pitch of the prop is far from ideal for this setup, but it's what I had on hand.

Prop diameter: 9.5"
Prop height above lower plate: fixed at 3.5"
Prop height below upper plate: adjusted from 1" to 5"
Scale reading with rotor stopped: 0-10 g
Scale reading with rotor turning and no upper plate: 1,000 - 1,020 g

The first two attachments below are crude pictures of the setup. The next shows the definition of the parameters I used. The last attachment shows the results.

On the results graph:
-The triangles show the ratio of F_scale (the force measured by the scale with the upper plate at a given height) to F_no-plate (the force measured by the scale with the upper plate removed).
-The squares show the ratio of T_rotor (the upward rotor thrust with the upper plate at a given height) to T_no-plate (the rotor thrust with the upper plate removed).

The force measured by the scale drops off very rapidly as the upper plate is moved closer to the prop from above. With 1" of separation, the force measured by the scale dropped to 115 g (just over 10% of what it showed with no plate).

Although my method for measuring the thrust of the rotor was crude, it's good enough to show that the rotor's thrust isn't changing by much.

This setup could certainly be improved in a number of ways, but the variation of the force measured by the scale was very repeatable.

EDIT: I came up with a much improved method for measuring the thrust of the rotor (accurate to about 5% of the total thrust). This method showed no discernable difference in the thrust due to the presence of the upper plate (at all upper plate heights).
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Old Dec 31, 2012, 10:40 AM
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Quote:
Originally Posted by ShoeDLG View Post
You asked for clarification and I offered a straightforward accounting of the forces acting on the air. Newton's Laws indicate very clearly that the air's net rate of vertical momentum change must be zero. If you believe Newton's Laws, there's no getting around that.

The motion of the rotor is certainly adding downward momentum to some of the air, and that local momentum exchange does matter. However, despite this local momentum exchange, the rotor never changes the net vertical momentum of the air. For every element of air displaced downward by a rotor in ground effect, another element of must be displaced upward by the same amount. This corresponding displacement doesn't happen at some time in the future, it happens simultaneously. If this were not the case, the air would collect below the rotor in the same way the BB's collect on the scale in the example in post #901. The air doesn't collect below the rotor. Instead it circulates outward and then upward once outboard of the rotor.

Suppose you neglected the air going upward outboard of the rotor and focused only on the down-going column of air right below the rotor. Is the rate at which the rotor adds downward momentum to that column of air equal to the weight of the helicopter? Suppose we carefully measured it and found that it was. If we changed the altitude of the helicopter, we would change the velocity of the air at any point below the rotor (note that this is not the case for a stream of independent BB's). The change in velocity means that we would find a different value for the rate of momentum change at every altitude. This is problematic if you are suggesting the rate of downward momentum change is always equal to the weight of the helicopter.
Nice story, but as scientists, we must stick to the facts. Rotors kick air down, not up. The force added to that air is measurable with a scale. In a hover, the force matches the weight of the chopper. Therefore, there's only one conclusion to draw - NASA draws the right conclusion - why can't you?.

Lift is a reaction to shoving air downwards. It even adds up - the weight of the chopper hovering over a scale to the weight shown on the scale from the choppers air hitting it. They'll pretty well match, depending on the skill of the pilot doing the hovering.

Put yourself in NASA's shoes for once, and see lift not as some mysterious thing requiring wierd incantations about net momentum not changing, but as something simple enough, anyone with a fan in their hands can understand it. Cause it is, that simple.

Now, trying to design a wing requires math - but that's the same for missiles and rockets, where the lift is more obvious then with planes (but not choppers). Choppers have an obvious similarity with missiles and rockets, in that they all visually shove something down, in order to lift. But,.a chopper's rotors are it's wings, and you have to use that, to get your head around wings - they also react to shoving air down, with lift.
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Last edited by Tim Green; Dec 31, 2012 at 11:00 AM. Reason: removed some sarcasm
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Old Dec 31, 2012, 12:03 PM
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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.
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Old Dec 31, 2012, 12:35 PM
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Helicopter downwash.... note the extent before it curls up.
This can be "seen" flying an r/c heli close to any object on the ground.
In the two heli image, that flat 1/4" plate will stop the heli from crossing it if the heli approaches slowly.
This behavior is seen any time the heli is approaching anything taller than the ground.. a reluctance to advance above the block without some forward stick.
The rotor downwash ahead of the heli feeds back to the rotor, while the downwash directly below just keeps going down.. if even for only a 1/4".
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Old Dec 31, 2012, 01:45 PM
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Did some fan-into-sail tests a while back. I put the motor pylon from one of my planes on a 3-wheeled cart, with a large foam plate at one end.
With the prop blowing at the plate, the cart would reluctantly move forward, the direction a plane would move with that setup.
Reversing the rotation with the prop blowing away from the plate, the cart would move rapidly in the opposite direction to the first configuration.
Not something many propellor people test for.
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Old Dec 31, 2012, 02:00 PM
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Originally Posted by ciurpita View Post
what's the definition of net?
In the context of a wing generating lift, the "net" rate at which the wing imparts vertical momentum to the air is the vector sum of the mass times acceleration of every element of air being accelerated by the wing.

How much air is being accelerated by a steadily lifting, subsonic wing at any moment? All of the air "connected" to the wing by other air (I'm sure there's a topographically correct way to say this). As you move away from the wing, the magnitude of the accelerations obviously die away. However, there is usually a lot more mass of air that is far from the wing than close to the wing. While the velocities induced in the air far from the wing may be very small, the momentum induced is not necessarily small. If you choose to neglect the induced momentum of the air far from a lifting wing, you have to be very careful or you will get the wrong value.

What does acceleration of the air far from the wing have to do with the lift? When you invoke Newton's Second Law to connect the lift force on the wing to air's rate of momentum change, there is no "proximity clause" that says momentum changes close to the wing are more important than momentum changes far from the wing.
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Old Dec 31, 2012, 02:08 PM
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Originally Posted by Sparky Paul View Post
Not something many propellor people test for.
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.
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Old Dec 31, 2012, 02:28 PM
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Quote:
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.
Just proves what everyone else knows, but you lot still argue, ..... that wings suck them selves up.
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Old Dec 31, 2012, 02:44 PM
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Originally Posted by Sparky Paul View Post
Did some fan-into-sail tests a while back. I put the motor pylon from one of my planes on a 3-wheeled cart, with a large foam plate at one end.
With the prop blowing at the plate, the cart would reluctantly move forward, the direction a plane would move with that setup.
Reversing the rotation with the prop blowing away from the plate, the cart would move rapidly in the opposite direction to the first configuration.
Not something many propellor people test for.
Newton would have expected nothing less.
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