The role of the foam was to show that if something held up the helicopter, and in so doing experienced no net vertical force, that thing would experience no vertical acceleration. This is patently obvious when that thing happens to be a solid (e.g. a foam block), and apparently quite elusive when that thing happens to be a fluid (e.g. the air). You are correctect that the size of the foam is of no consequence.

And by "fluid" I mean a substance where stresses are proportional to strain rates, not necessarily a "liquid".

I do understand fluid also fluidics - having done control systems using same laws of physics as you may refer to .
In this case there are no moving parts and the control is either digital or analogue-depending on desired results..
Anyway
Merry Christmas and Happy New Year to all .
and Season's Greetings to the politically correct crowd.

ShoeDLG --

Would you be interested in pursueing r/2 off thread?

butcharts2@aol.com

Dear Fellow RC-Groupers:

The PIPE Here...whenever I see the "issue" of "what causes a plane to have lift while in flight" come up, I just refer people to get a good look at pages 7 through, say about page 11 of the famous Wolfgang Langewiesche-authored book "Stick and Rudder: An Explanation of the Art of Flying", in more-or-less continuous print for nearly SEVENTY years as of 2012.

Page 7 of the book, in the chapter "How a Wing is Flown", has the very aptly named section entitled "Forget Bernoulli's Theorem", a VERY appropriate title for those of us, who like yours truly, has had numerous migraines manifest themselves whenever someone mentions that filthy eight letter word "calculus" at any time.

As that OH-so-true maxim "a picture is worth a thousand words"...or PERHAPS a few dozen integral calculus (there's that FILTHY word again!) symbols...tells us, the pair of illustrations on page eight of the book really "tells us like it is". That pair of illustrations possesses the following captions, depicting a light Cessna 195 style aircraft from a nose-on view about to fly through something like a fog bank near the ground, and then flying through it. The first caption reads:

"The airplane keeps itself up by beating the air down. If one could lay a smoke cloud into an airplane's path..."

then with the second image showing the Cessna 195-like aircraft having flown THROUGH the cloud...

"..the airplane, having flown through it, would squash it down."

My late RC flight instructor (left us sixteen years ago this month) Sammy Frey told me a generation ago that Langewiesche's book was the best he had ever read for ANY RCer to start to understand "why a model aircraft behaves the way it does" when we fly our birds...and all this time later, whenever I pick up my own personal copy to read and re-read whenever I have a moment to do it, I cannot help but agree with Sammy's assertion from so long ago.

That's just my take on it...I simply wish that calculus (again with the filthy word!) was NOT so scary for my brain, and that I had managed to get some help learning how to understand it when I tried to go for an engineering degree in the early 1980s !!

Yours Sincerely and Happy 2013,

The PIPE....!!

Originally Posted by The PIPE

Dear Fellow RC-Groupers:

The PIPE Here...whenever I see the "issue" of "what causes a plane to have lift while in flight" come up, I just refer people to get a good look at pages 7 through, say about page 11 of the famous Wolfgang Langewiesche-authored book "Stick and Rudder: An Explanation of the Art of Flying", in more-or-less continuous print for nearly SEVENTY years as of 2012.

Page 7 of the book, in the chapter "How a Wing is Flown", has the very aptly named section entitled "Forget Bernoulli's Theorem", a VERY appropriate title for those of us, who like yours truly, has had numerous migraines manifest themselves whenever someone mentions that filthy eight letter word "calculus" at any time.

As that OH-so-true maxim "a picture is worth a thousand words"...or PERHAPS a few dozen integral calculus (there's that FILTHY word again!) symbols...tells us, the pair of illustrations on page eight of the book really "tells us like it is". That pair of illustrations possesses the following captions, depicting a light Cessna 195 style aircraft from a nose-on view about to fly through something like a fog bank near the ground, and then flying through it. The first caption reads:

"The airplane keeps itself up by beating the air down. If one could lay a smoke cloud into an airplane's path..."

then with the second image showing the Cessna 195-like aircraft having flown THROUGH the cloud...

"..the airplane, having flown through it, would squash it down."

My late RC flight instructor (left us sixteen years ago this month) Sammy Frey told me a generation ago that Langewiesche's book was the best he had ever read for ANY RCer to start to understand "why a model aircraft behaves the way it does" when we fly our birds...and all this time later, whenever I pick up my own personal copy to read and re-read whenever I have a moment to do it, I cannot help but agree with Sammy's assertion from so long ago.

That's just my take on it...I simply wish that calculus (again with the filthy word!) was NOT so scary for my brain, and that I had managed to get some help learning how to understand it when I tried to go for an engineering degree in the early 1980s !!

Yours Sincerely and Happy 2013,

The PIPE....!!

YES - same as NASA is saying - action, and reaction - Lift is a reaction to accelerating a lot of air downward (the action).

Originally Posted by TangoKilo

Is it not the fact that force equals the rate of change of momentum?

F=d(m.v)/dt

The rate of change of momentum of the air being caused to move by the rotor blades must be equal to the helicopters weight.

Yes indeed. And you can hover a model copter, weighing 1lb, over a scale large enough so that all the copters air hits it, that scale will read roughly 1lb, just from the air hitting it.

Originally Posted by Tim Green

And you can hover a model copter, weighing 1lb, over a scale large enough so that all the copters air hits it, that scale will read roughly 1lb, just from the air hitting it.

And if you understand Newton's Second and Third Laws, you will conclude from this that there is no net vertical force acting on the air, meaning that its rate of change of vertical momentum is zero.

You would be doing yourself a disservice to make up your mind on this subject before trying understand the analysis done by late Peter Lissaman (you can read about some of his work here).

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

In his presentation titled "The Meaning of Lift" he thoughtfully examines the relationship between lift and momentum. His conclusion boils down to:

"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!"

Although his approach is lighthearted it does include Calculus. If the calculus in his analysis is not accessible to you, it may still be worth recognizing that the relationship between lift and momentum exchange may be more subtle that it might appear.

Originally Posted by ShoeDLG

And if you understand Newton's Second and Third Laws, you will conclude from this that there is no net vertical force acting on the air, meaning that its rate of change of vertical momentum is zero.

What on earth, are you on about? Please clairify.

Originally Posted by ShoeDLG

You would be doing yourself a disservice to make up your mind on this subject before trying understand the analysis done by late Peter Lissaman (you can read about some of his work here).

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

In his presentation titled "The Meaning of Lift" he thoughtfully examines the relationship between lift and momentum. His conclusion boils down to:

"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!"

Although his approach is lighthearted it does include Calculus. If the calculus in his analysis is not accessible to you, it may still be worth recognizing that the relationship between lift and momentum exchange may be more subtle that it might appear.

There's no net downwash? Really?

If you place a scale above a hovering chopper - scale reads zero, or close to it.
If you place a scale below a hovering chopper - scale registers the weight of the chopper.

If your man was right, fans would blow air both directions - and they don't, not in my world.

-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.

Heli-hovering.... Bernoulli and Newton..
Bernoulli works all the time, but can be thwarted.

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.

Originally Posted by ShoeDLG

-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.

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

causing the air to accelerate downward and increase its velocity

Originally Posted by ShoeDLG

-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.

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

causing the air to de-accelerate and decrease its downward velocity

Originally Posted by ShoeDLG

-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.

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?

what's the definition of net?

Originally Posted by ShoeDLG

Bernoulli equation without having to resort to a mysterious and unsupportable r/2 “integration factor". .

For anyone that was following pressure change from normal acceleration:
Shoe was right in saying that the pressure sum accumulated accross the field from normal acceleration (turning flow) Requires a proper math solution.

However, for practical understanding, we can resort to old fashoned graphing.
At the surface we have the curve r and the velocity relative to it providing a dP/dr pressure gradient at the surface. For the gradients of r's outboard we can use v = V(1+a^2/r^2).
If we plot the pressure gradients, density*v^2/r, against r , the area under the plot curve will be the local pressure difference from P(total).

If we count squares we find that it is equal to the surface gradient times 1/2 the surface r. This leaves us uderstanding the source of lift while able to contiue the use of the Bernoulli equation