visualisation of the lift distribution over a wing

In article ,
Jim Logajan wrote:

Alan Baker wrote:
Jim Logajan wrote:
3) Therefore if, say, the downwash is 1 kg/s at any given instant due
to the wing, somewhere else in the fluid there must be an upwash at
that same instant of 1 kg/s. Agree or disagree?

Agree. At the surface of the Earth.

The qualifier indicates to me that you don't agree with the statement as
written.

You're right. I shouldn't have stated it that way.


Unfortunately, I consider conservation of mass at all points in time in an
incompressible fluid an essential element to understanding the behavior of
downwash. If you don't, then I think any further debate between us is
ended.

Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

That means that there is a constant change of momentum being done on the
air by the aircraft. That means air *must* be moving down (net) after
the aircraft has passed.


(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2) at, say,
20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it maintained
that speed. In the mean time, once the helicopter stopped descending,
conservation of mass in an incompressible fluid seems to require an equal
volume of air to have an upward vector of 20 m/s. So the surface of earth
appears to be irrelevant for over two minutes.)

Nope.

The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

It appears you have never studied fluid dynamics (maybe elementary fluid
statics?) and I doubt that you own any books on the subject.

The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

In other news, 1 + 1 = 2.

That means that there is a constant change of momentum being done on
the air by the aircraft. That means air *must* be moving down (net)
after the aircraft has passed.

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2)
at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it
maintained that speed. In the mean time, once the helicopter stopped
descending, conservation of mass in an incompressible fluid seems to
require an equal volume of air to have an upward vector of 20 m/s. So
the surface of earth appears to be irrelevant for over two minutes.)

Nope.

Dang - I try to use real numbers to establish a baseline example, and you
manage to use a single word to demolish my attempts! Really helpful
mathematical counter-example you produced - not.

The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

I don't know what your problem is - maybe you are thinking this is a
rocket problem where no external fluids are involved and you can't get
your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM.
Whatever the case, you seem to be fixated on applying one conservation
law to one element in the entire system to the exclusion of everything
else.

Best of luck to you.

Jim Logajan wrote:
Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

It appears you have never studied fluid dynamics (maybe elementary fluid
statics?) and I doubt that you own any books on the subject.

The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

In other news, 1 + 1 = 2.

That means that there is a constant change of momentum being done on
the air by the aircraft. That means air *must* be moving down (net)
after the aircraft has passed.

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2)
at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it
maintained that speed. In the mean time, once the helicopter stopped
descending, conservation of mass in an incompressible fluid seems to
require an equal volume of air to have an upward vector of 20 m/s. So
the surface of earth appears to be irrelevant for over two minutes.)
Nope.

Dang - I try to use real numbers to establish a baseline example, and you
manage to use a single word to demolish my attempts! Really helpful
mathematical counter-example you produced - not.

The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

I don't know what your problem is - maybe you are thinking this is a
rocket problem where no external fluids are involved and you can't get
your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM.
Whatever the case, you seem to be fixated on applying one conservation
law to one element in the entire system to the exclusion of everything
else.

Best of luck to you.



Two dimensional Newtonian thinking in a three dimensional non-Newtonian world.

In article ,
cavelamb wrote:

Jim Logajan wrote:
Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

It appears you have never studied fluid dynamics (maybe elementary fluid
statics?) and I doubt that you own any books on the subject.

The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

In other news, 1 + 1 = 2.

That means that there is a constant change of momentum being done on
the air by the aircraft. That means air *must* be moving down (net)
after the aircraft has passed.

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2)
at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it
maintained that speed. In the mean time, once the helicopter stopped
descending, conservation of mass in an incompressible fluid seems to
require an equal volume of air to have an upward vector of 20 m/s. So
the surface of earth appears to be irrelevant for over two minutes.)
Nope.

Dang - I try to use real numbers to establish a baseline example, and you
manage to use a single word to demolish my attempts! Really helpful
mathematical counter-example you produced - not.

The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

I don't know what your problem is - maybe you are thinking this is a
rocket problem where no external fluids are involved and you can't get
your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM.
Whatever the case, you seem to be fixated on applying one conservation
law to one element in the entire system to the exclusion of everything
else.

Best of luck to you.



Two dimensional Newtonian thinking in a three dimensional non-Newtonian world.

LOL

Sorry, caveman....


But conservation of momentum works well enough at the speeds at which
aircraft operate.

And Newton's laws tell us all we need to know.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

In article ,
Jim Logajan wrote:

Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

It appears you have never studied fluid dynamics (maybe elementary fluid
statics?) and I doubt that you own any books on the subject.

Sorry, lad, but conservation of mass is a principle that comes up mostly
in *chemistry*.


The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

In other news, 1 + 1 = 2.

What a pity then that you don't understand it.


That means that there is a constant change of momentum being done on
the air by the aircraft. That means air *must* be moving down (net)
after the aircraft has passed.

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

I have no trouble figuring that out at all. A gas of a different density
within the balloon causes the net upward force on the balloon exerted by
the air outside the balloon to be greater than the net downward force on
it.


(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2)
at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it
maintained that speed. In the mean time, once the helicopter stopped
descending, conservation of mass in an incompressible fluid seems to
require an equal volume of air to have an upward vector of 20 m/s. So
the surface of earth appears to be irrelevant for over two minutes.)

Nope.

Dang - I try to use real numbers to establish a baseline example, and you
manage to use a single word to demolish my attempts! Really helpful
mathematical counter-example you produced - not.

No math is necessary for this. Look up "qualitative analysis".




The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

I don't know what your problem is - maybe you are thinking this is a
rocket problem where no external fluids are involved and you can't get
your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM.
Whatever the case, you seem to be fixated on applying one conservation
law to one element in the entire system to the exclusion of everything
else.

The law I'm focussed on is the one that counts. It doesn't matter
whether the fluid is expelled from inside or whether it's an external
fluid diverted down by the surfaces of the craft.

In order for there to be a continuous force W equaling the weight of the
craft acting on it, the craft must exert a force -W on the fluid. That
-W means that there is a downward change of momentum in the fluid. Since
the fluid is no accelerated indefinitely, there must be a continuous
flow (mass per unit time M/t) of the fluid accelerated to a velocity V
where the equation looks like:

-W = M/t * V

The velocity of the fluid will be:

V = -W/(M/t)

That is inescapable. If the craft weighs 9800N (newtons), and it moves
100kg of air every second, then the air must be moving downward (net,
now!) at 98 m/s.

I'm sorry if you don't get this, but it is very simple and absolutely
irrefutable.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:
In article ,
Jim Logajan wrote:

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

I have no trouble figuring that out at all. A gas of a different density
within the balloon causes the net upward force on the balloon exerted by
the air outside the balloon to be greater than the net downward force on
it.

Nope! The "gas of a different density" inside does not cause the
differential forces exerted by the air outside. Whether the balloon is
filled with phlogiston or concrete, the difference between upward and
downward forces exerted by the air outside depends on the height of the
balloon. That's what my goofy 10,000 foot thick wing was, that would
take off with no airspeed or power. A balloon.

In article ,
Beryl wrote:

Alan Baker wrote:
In article ,
Jim Logajan wrote:

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

I have no trouble figuring that out at all. A gas of a different density
within the balloon causes the net upward force on the balloon exerted by
the air outside the balloon to be greater than the net downward force on
it.

Nope! The "gas of a different density" inside does not cause the
differential forces exerted by the air outside. Whether the balloon is
filled with phlogiston or concrete, the difference between upward and
downward forces exerted by the air outside depends on the height of the
balloon. That's what my goofy 10,000 foot thick wing was, that would
take off with no airspeed or power. A balloon.

You're wrong.

The "net downward force" includes the force of gravity. Which is
lessened because a gas of lower density..

....I can't believe I have to explain things so basic.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:
In article ,
Beryl wrote:

Alan Baker wrote:
In article ,
Jim Logajan wrote:
You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!
I have no trouble figuring that out at all. A gas of a different density
within the balloon causes the net upward force on the balloon exerted by
the air outside the balloon to be greater than the net downward force on
it.
Nope! The "gas of a different density" inside does not cause the
differential forces exerted by the air outside. Whether the balloon is
filled with phlogiston or concrete, the difference between upward and
downward forces exerted by the air outside depends on the height of the
balloon. That's what my goofy 10,000 foot thick wing was, that would
take off with no airspeed or power. A balloon.

You're wrong.

The "net downward force" includes the force of gravity. Which is
lessened because a gas of lower density..

...I can't believe I have to explain things so basic.

I know, but you have to. Your first attempt didn't appear to include
unsaid gravity. Gas that you described as "within the balloon" is now
treated as "part of" the balloon.

In article ,
Beryl wrote:

Alan Baker wrote:
In article ,
Beryl wrote:

Alan Baker wrote:
In article ,
Jim Logajan wrote:
You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!
I have no trouble figuring that out at all. A gas of a different density
within the balloon causes the net upward force on the balloon exerted by
the air outside the balloon to be greater than the net downward force on
it.
Nope! The "gas of a different density" inside does not cause the
differential forces exerted by the air outside. Whether the balloon is
filled with phlogiston or concrete, the difference between upward and
downward forces exerted by the air outside depends on the height of the
balloon. That's what my goofy 10,000 foot thick wing was, that would
take off with no airspeed or power. A balloon.

You're wrong.

The "net downward force" includes the force of gravity. Which is
lessened because a gas of lower density..

...I can't believe I have to explain things so basic.

I know, but you have to. Your first attempt didn't appear to include
unsaid gravity. Gas that you described as "within the balloon" is now
treated as "part of" the balloon.

Of course gas within the balloon is part of the ballon. Nothing I said
ever implied otherwise.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:
Jim Logajan wrote:
Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in
this, you're missing out from the start. It's conservation of
*momentum* that's in play here.

It appears you have never studied fluid dynamics (maybe elementary
fluid statics?) and I doubt that you own any books on the subject.

Sorry, lad, but conservation of mass is a principle that comes up
mostly in *chemistry*.

Please don't patronize when you've never studied a subject. A moment's
research would have prevented you from posting something that incredibly
ignorant. You might want to look up the "axioms of fluid dynamics" before
you further add to your public embarrassment.

The aircraft has a force exerted on it equal to its weight. That
means that the aircraft must be exerting a force on the air in the
opposite direction.

In other news, 1 + 1 = 2.

What a pity then that you don't understand it.

Shrug - your insults are lame and tiresome, but I will admit you're
patronization is irritating. I do make mistakes about things I was
taught, but I should point out to you that I did well enough in college
physics to earn an undergraduate degree in the subject. Unlike you, I
have had to solve a **** load of problems involving conservation of
momentum to prove I understood the basics - including conservation of
momentum in quantum mechanical systems. So far as I know, you HAVEN'T had
to prove your mastery of the subject with ANYONE.

(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped
from a balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's
engine immediately started. After a small drop it levels out and
maintains a downwash of air moving through its 6 m diameter disk
(A ~= 28 m^2) at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe
= Fg.)

It would take ~150 s for that downwash to reach the ground if it
maintained that speed. In the mean time, once the helicopter
stopped descending, conservation of mass in an incompressible
fluid seems to require an equal volume of air to have an upward
vector of 20 m/s. So the surface of earth appears to be irrelevant
for over two minutes.)

Nope.

Dang - I try to use real numbers to establish a baseline example, and
you manage to use a single word to demolish my attempts! Really
helpful mathematical counter-example you produced - not.

No math is necessary for this. Look up "qualitative analysis".

Well that probably explains your problem - you don't know how to set up
the math properly, so you have no way to validate whether your
"qualitative analysis" is correct.

Ironically, all your posts contain the same violation of conservation of
momentum - and yet you keep pointing to that concept as vindication.

I don't know what your problem is - maybe you are thinking this is a
rocket problem where no external fluids are involved and you can't
get your mind around the fact that THIS ISN'T A BLOODY ROCKET
PROBLEM. Whatever the case, you seem to be fixated on applying one
conservation law to one element in the entire system to the exclusion
of everything else.

The law I'm focussed on is the one that counts. It doesn't matter
whether the fluid is expelled from inside or whether it's an external
fluid diverted down by the surfaces of the craft.

You can't "focus" on one conservation law because the number of
constraining equations has to equal the number of variables. Doing so
simply leads to an infinite number of bogus results.

In order for there to be a continuous force W equaling the weight of
the craft acting on it, the craft must exert a force -W on the fluid.
That -W means that there is a downward change of momentum in the
fluid.

Sigh. This is a case where a little knowledge is a dangerous thing. It
would take a book to explain the problem with your conceptual view of
fluid dynamics. I don't have that sort of patience.

Since the fluid is no accelerated indefinitely, there must be a
continuous flow (mass per unit time M/t) of the fluid accelerated to a
velocity V where the equation looks like:

-W = M/t * V

Of course if you had read my earlier post you'd see I'd ALREADY USED THAT
EQUATION. But you obviously aren't familiar with the conventions used in
fluid dynamics, so you probably had no clue what my "m_dot" meant or how
I got the figures I did.

The velocity of the fluid will be:

V = -W/(M/t)
That is inescapable. If the craft weighs 9800N (newtons), and it moves
100kg of air every second, then the air must be moving downward (net,
now!) at 98 m/s.

You math is correct and no one has denied there is a downwash (why you
think otherwise continues to baffle me) yet your "net, now" comment
violates conservation of momentum.

Here's why:

If we choose a reference frame so that at T=0 everything in the system is
stationary with respect to that frame, we set the net momentum of the
system to 0. Then, so long as the system remains closed, at all other
times the conservation of momentum must yield 0.

But according to your "qualitative analysis" the net vertical momentum
P_net_z increases with time T, like so:

P_net_z(T) = (100 kg/s)*T*V

That's because the earth and the airplane maintain zero vertical
momentums (P_earth_z(T) = 0, P_plane_z(T) = 0,) and there appears to be
nothing in your conceptual view of the situation to correct that
violation of conservation of momentum.

I'm sorry if you don't get this, but it is very simple and absolutely
irrefutable.

It appears to salve your ego to ascribe assertions to me that I never
made and then tell the world that those falsehoods prove I don't "get
it." Probably because you've grown so much hubris and so little humility.