visualisation of the lift distribution over a wing

In article ,
Jim Logajan wrote:

Alan Baker wrote:
In article ,
Beryl wrote:

Alan Baker wrote:

It's like the downwash argument. You can say "IT DOESN'T MATTER",
when people argue that the air behind an aircraft is not deflected
downward, but it *does* matter. Having an accurate understanding of
the physical processes of flight matters.

It isn't really deflected downward, not for long anyway. It's
churning in a torus. Like a smoke ring.

No.

It really *is* deflected downward.

It's also deflected upward. ;-)

Here's why:
Because the airplane and the Earth have zero relative vertical velocity
during straight and level flight, conservation of momentum requires the
net vertical flow of air to also be zero.

Therefore in subsonic flows where the fluid is assumed incompressible, to
the extent any fluid is moving downward, conservation of mass requires an
equal amount of mass must be moving upward (the continuity requirement.)

Hence airplanes must cause air to move in circles.

Nope. Wrong.

The aircraft is experience an force upward the entire time it is in
flight. That force means there must be an equal force acting on the air,
and since the air was not moving vertically (in our idealized case for
this discussion) before the aircraft arrived, the force exerted on it
must mean that it is moving downward afterward it has passed.


The edges of the deflected area churn, and the air that is deflected
ends up getting diffused among all the other air below *it*, but it
really is deflected downward.

Yes some deflection downward occurs. But I don't know that it could be
said to "diffuse" in any sense due to conservation of mass and momentum
requirements.

As the air the plane has forced downward encounters more air, the
momentum is diffused so that a greater and greater mass of air moves
downward at smaller and smaller velocities (net)...

....until it encounters the ground.


And eventually, that downward deflection makes it way until it -- very
diffusely -- impacts upon the surface of the earth. That is the only
thing that finally stops it.

That assertion is not true in general. What appears to happen instead is
that any downward deflection is quickly reversed, leading to what is
known as a shed vortex. Here are some links on the subject:

http://www.grc.nasa.gov/WWW/K-12/airplane/shed.html

Sorry, but the vortex is an edge effect. The net flow is downward.

http://www.grc.nasa.gov/WWW/K-12/airplane/downwash.html

While the deflected flow doesn't need to reach the surface of the earth
for the airplane to stay aloft, an increase in air _pressure_ would
eventually make its way to the surface.

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

Alan Baker wrote:
...

As the air the plane has forced downward encounters more air, the
momentum is diffused so that a greater and greater mass of air moves
downward at smaller and smaller velocities (net)...

...until it encounters the ground.

Yikes! Eventually, ALL of the air will be on the ground!

And eventually, that downward deflection makes it way until it -- very
diffusely -- impacts upon the surface of the earth. That is the only
thing that finally stops it.
That assertion is not true in general. What appears to happen instead is
that any downward deflection is quickly reversed, leading to what is
known as a shed vortex. Here are some links on the subject:

http://www.grc.nasa.gov/WWW/K-12/airplane/shed.html

Sorry, but the vortex is an edge effect. The net flow is downward.

Mankind needs to set up fans IMMEDIATELY to move air back up into the sky!

Alan Baker wrote:
Jim Logajan wrote:
Because the airplane and the Earth have zero relative vertical
velocity during straight and level flight, conservation of momentum
requires the net vertical flow of air to also be zero.

Therefore in subsonic flows where the fluid is assumed
incompressible, to the extent any fluid is moving downward,
conservation of mass requires an equal amount of mass must be moving
upward (the continuity requirement.)

Hence airplanes must cause air to move in circles.

Nope. Wrong.

The aircraft is experience an force upward the entire time it is in
flight. That force means there must be an equal force acting on the
air, and since the air was not moving vertically (in our idealized
case for this discussion) before the aircraft arrived, the force
exerted on it must mean that it is moving downward afterward it has
passed.

Nothing you wrote in your paragraph contradicts anything in my
paragraphs. So I'm at a loss therefore as to what specific statements you
claim are wrong. So how about I break it down into smaller claims and you
tell me which of these statements you agree with and which you disagree
with:

1) Conservation of momentum requires that at all times during flight that
net vertical momentum of the total system must be zero. Agree or
disagree?

2) We can treat air as incompressible, so conservation of mass means the
net vertical mass flow of the system during level flight must be zero.
Agree or disagree?

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?

4) Because upwash mass rate equals downwash mass rate, at some point the
downward flow reverses direction and becomes the upwash. Agree or
disagree?

Yes some deflection downward occurs. But I don't know that it could
be said to "diffuse" in any sense due to conservation of mass and
momentum requirements.

As the air the plane has forced downward encounters more air, the
momentum is diffused so that a greater and greater mass of air moves
downward at smaller and smaller velocities (net)...

...until it encounters the ground.

Keep in mind that balloons need no downwash to stay aloft. Yet we know
from conservation laws that the *static* pressure on the surface of the
earth must be increase due to their presence. Nothing you've written
rules out the possibility that the *dynamic* pressure of the downwash
translates into a *static* pressure increase well before the downwash
reaches the surface of the earth. The physics of the situation do not
seem to rule out that a priori.

In article ,
Jim Logajan wrote:

Alan Baker wrote:
Jim Logajan wrote:
Because the airplane and the Earth have zero relative vertical
velocity during straight and level flight, conservation of momentum
requires the net vertical flow of air to also be zero.

Therefore in subsonic flows where the fluid is assumed
incompressible, to the extent any fluid is moving downward,
conservation of mass requires an equal amount of mass must be moving
upward (the continuity requirement.)

Hence airplanes must cause air to move in circles.

Nope. Wrong.

The aircraft is experience an force upward the entire time it is in
flight. That force means there must be an equal force acting on the
air, and since the air was not moving vertically (in our idealized
case for this discussion) before the aircraft arrived, the force
exerted on it must mean that it is moving downward afterward it has
passed.

Nothing you wrote in your paragraph contradicts anything in my
paragraphs. So I'm at a loss therefore as to what specific statements you
claim are wrong. So how about I break it down into smaller claims and you
tell me which of these statements you agree with and which you disagree
with:

1) Conservation of momentum requires that at all times during flight that
net vertical momentum of the total system must be zero. Agree or
disagree?

2) We can treat air as incompressible, so conservation of mass means the
net vertical mass flow of the system during level flight must be zero.
Agree or disagree?

Agree, but that system needs to include the Earth itself, doesn't it?


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.

4) Because upwash mass rate equals downwash mass rate, at some point the
downward flow reverses direction and becomes the upwash. Agree or
disagree?

Agree. At the surface of the Earth.


Yes some deflection downward occurs. But I don't know that it could
be said to "diffuse" in any sense due to conservation of mass and
momentum requirements.

As the air the plane has forced downward encounters more air, the
momentum is diffused so that a greater and greater mass of air moves
downward at smaller and smaller velocities (net)...

...until it encounters the ground.

Keep in mind that balloons need no downwash to stay aloft. Yet we know
from conservation laws that the *static* pressure on the surface of the
earth must be increase due to their presence. Nothing you've written
rules out the possibility that the *dynamic* pressure of the downwash
translates into a *static* pressure increase well before the downwash
reaches the surface of the earth. The physics of the situation do not
seem to rule out that a priori.

Conservation of momentum does.

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

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.

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.

(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.)

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 ,
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