00:00:00Hg wrote:
> On Fri, 03 Mar 2006 11:57:05 -0800, george wrote:
>
> >> Gravity seems to work to it's own advantage so it's the
> >> ultimate taxing authority in the universe. That really sucks.
> >
> > Air resistance is fricticious
>
> I thought resistance was useless.
>
> At least for Dent and Ford.

I've never sistanced even once

> I asked first.

Ok, vertical momentum of a wing in level flight is zero.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> The hovering spacecraft has zero horizontal and vertical momentum.
> It has weight, directed downwards. The engine accelerates
> mass downward producing an upward force equal in magnitude
> and opposite in direction to the weight of the spacecraft. This
> imparts an acceleration to the spacecraft equal in magnitude and
> opposite in direction from the local acceleration due to gravity.

The flying wing has some horizontal momentum which is secondary here,
and zero vertical momentum.
It also has weight, directed downwards. The wing accelerates
mass downward (mass it finds in the air molecules) producing
an upward force equal in magnitude and opposite in direction to
the weight of the wing (and its presumably attached aircraft. It does
so by finding air in front of it, flinging it downwards and forwards
(which causes the air in front to try to get out of the way by rising).
In the steady state, one can measure high pressure below and low
pressure above, but this is just the macroscopic manifestation of the
greater number of molecular collisions below, and the lesser number of
collisions above. That's what pressure is - we have both agreed on this.

The greater number of collisions below
imparts an acceleration to the aircraft equal in magnitude and
opposite in direction from the local acceleration due to gravity.

Unlike the spacecraft (at least to first order), the wing is actually
supported by the earth, as the pressure below the wing is higher than it
would have been absent the wing's passage, and this higher pressure
(spread out over many square miles) pushes down on the earth with a
force equal to the weight of the aircraft.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> Air resistance is fricticious

Resistance is futile.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

On Fri, 03 Mar 2006 19:05:18 -0500, Morgans wrote:

>
> "00:00:00Hg" > wrote
>>
>> I thought resistance was useless.
>
> Nah, it's "resistance is fruitile."
> <g>

I think I'll have an apple and see if
eating it will reveal the secrets of
gravity as I gaze at the Moon.

"00:00:00Hg" > wrote
>
> I thought resistance was useless.

Nah, it's "resistance is fruitile."
<g>
--
Jim in NC

"00:00:00Hg" > wrote

> I think I'll have an apple and see if
> eating it will reveal the secrets of
> gravity as I gaze at the Moon.

Yeah, but can you tell me the horizontal and vertical components of it's
momentum?

I was thinking apple, but I need two; I'll have a pear, instead. <g>
--
Jim in NC

On Fri, 03 Mar 2006 21:06:30 -0500, Morgans wrote:

>> I think I'll have an apple and see if
>> eating it will reveal the secrets of
>> gravity as I gaze at the Moon.
>
> Yeah, but can you tell me the horizontal and vertical components of it's
> momentum?

Not any more, I'll have to pick another.

Not the Moon... it has no stem.

> You are looking here at the basic question of how does the
> starting vortex form.

No, I'm also looking at how it is maintained.

> You have staked out the position
> that a ground is required for the vortex to form.

No, I've staked out a position that the ground is required for there to
be no net momentum change. The ground is ultimately what the air (given
downward momentum) bounces against, either for real or by proxy.
Granted this is not what provides lift, but it does provide the ultimate
support when the wheels themselves leave the ground.

> Do we agree or disagree that the "wave" i.e.
> starting vortex, however it got started can continue in the
> absence of the ground?

We agree. I do not see however how it can continue in the absence of
energy, and I still maintain that in order to cancel out mv^2/2 of the
wing (which otherwise would be falling), there has to be (locally) an
equal mv^2/2 which the air acquires, and spreads out over the surface of
the earth (where it bounces off, keeping the earth away). Like a
dribbler who supports himself by dribbling, there is lots of momentum
transfer (to the ball, back and forth), which, while it nets to zero,
only does so because of the earth. IF there were no earth, the ball
would never bounce back.

That is not the same as what you seem to think I am maintaining:

> that without a ground an infinite
> wing would require a constant input of infinite energy to
> accelerate the air and give it momentum (and kinetic energy)
> for the uncanceled downwash.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> An object moving through air doesn't cause any significant compression
> (change in volume) of the air until its speed gets close to the speed of
> sound.

Is there not a (slight) pressure increase in front of any object,
especially a blunt one, moving through the air? (if not, what causes
the air to get out of the way, and what causes the breezes as it goes past?)

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> The distinction is that a compressible fluid (commonly called gas)
> undergoes a volume change proportionate to the pressure change

Well, when an object passes through the air, does it not compress the
air in front of it (and rarefy the air behind it)? This is how speakers
work. Those are all pressure changes.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> In open air the volume of air moving around the fan is larger,
> but moving at a lower speed than the air moving through the
> fan so that the momenta of the flow in either direction is equal
> magnitude and opposite in direction to the flow in the other
> direction.

Seems to me "almost equal" would make more sense, otherwise an airplane
propeller would not work. A propeller throws air backwards (alabeit
imperfectly); the airplane moves forwards in response.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

Jose wrote:
> > The hovering spacecraft has zero horizontal and vertical momentum.
> > It has weight, directed downwards. The engine accelerates
> > mass downward producing an upward force equal in magnitude
> > and opposite in direction to the weight of the spacecraft. This
> > imparts an acceleration to the spacecraft equal in magnitude and
> > opposite in direction from the local acceleration due to gravity.
>
> The flying wing has some horizontal momentum which is secondary here,

How much?

> and zero vertical momentum.
> It also has weight, directed downwards. The wing accelerates
> mass downward (mass it finds in the air molecules) producing
> an upward force equal in magnitude and opposite in direction to
> the weight of the wing (and its presumably attached aircraft. It does
> so by finding air in front of it, flinging it downwards and forwards
> (which causes the air in front to try to get out of the way by rising).
> In the steady state, one can measure high pressure below and low
> pressure above, but this is just the macroscopic manifestation of the
> greater number of molecular collisions below, and the lesser number of
> collisions above. That's what pressure is - we have both agreed on this.
>
> The greater number of collisions below
> imparts an acceleration to the aircraft equal in magnitude and
> opposite in direction from the local acceleration due to gravity.

I agree that lift is a force, exerted on the aircraft by the air,
which in steady level flight is equal in magnitude and opposite
in direction to the weight of the aircraft. Energy is 'pumped'
into the air by the plane. There is no need for a net momentum
exchange between the airplane and the air in order for
energy to be exchanged or for forces to be applied.
Indeed, in those last two paragraphs above, you make
no mention of momentum.

BTW, I was wrong to invoke conservation of momentum.
Momentum is conserved in elastic collisions, like the
collision between a cue ball and the eight ball. Momentum
is not conserved in inelastic collisions, like the collision
between a cue ball and a nerf ball.

Roll the airplane into a 90 degree bank. The weight is
now orthogonal to the lift. As teh airplane falls, it
banks even though there is no Earth 'under' the
belly. Why?

--

FF

Jose wrote:
> > In open air the volume of air moving around the fan is larger,
> > but moving at a lower speed than the air moving through the
> > fan so that the momenta of the flow in either direction is equal
> > magnitude and opposite in direction to the flow in the other
> > direction.
>
> Seems to me "almost equal" would make more sense, otherwise an airplane
> propeller would not work. A propeller throws air backwards (alabeit
> imperfectly); the airplane moves forwards in response.
>

For the stationary fan if it were only _almost equal_ then
you would eventually run out of air on one side of the fan.

Air molecules flowing through the propellor cetainly experience
momentum changes. But you can have a net flow of
energy without a net exchange of momentum because
momentum is a vector, energy is a scaler. If the airplane
is in level flight at constant speed it does not NEED to
gain any momentum from the propellor because the
momentum of the airplane is not changing. It needs
force to counter the force of drag.

Consider your example of the person who 'hovers' by
dribbling a basektball. His momentum is zero, the
momentum of the Earth is zero and the momentum
of the ball is constantly changing and reverses twice
each dribble. The dribbler is pumping energy into
the Earth yet there is no net exchange of momentum.

--

FF

>>The flying wing has some horizontal momentum which is secondary here,
> How much?

mv

The air thrown forward (or, if you will, the higher pressure ahead)
tries to reduce that, the engine presumably makes up for it.

> Energy is 'pumped' into the air by the plane.

Yes, and what form does that energy take? I maintain that it takes the
form of a net increase in mv^2/2 over all the air molecules. Since m
doesn't change, and 2 only changes in a pentium, that leaves v to
change. This changes mv, thus momentum.

We agree that there is (microsocopic) momentum transfer at each
collision. We disagree as to whether the net is zero, and I think that
part of that disagreement has to do with just how much of the system we
are looking at.

The wing throws air down. If that causes other air to be squeezed up,
so be it - the wing will grab that air and throw it down again. The air
piles up in front of and below the wing, and ultimately pushes on the
earth. New (undisturbed) air keeps appearing in front of the wing
(where it is pushed up, and then back down). But if, instead of feeding
this system fresh air, we instead feed it the same air, say, by flying
around in circles, there will be a net movement of air. Air will be
sucked from the (infinite amount of) air above, and pushed down into the
(infinite volume of) air below. The next time the wing encounters this
area, there will already be downward movement of air from the first
passage... etc. etc. and so forth.

> Momentum is conserved in elastic collisions

Low speed collisions between air molecules and aluminum sheets are to
first order elastic (although some energy goes into making molecules
wiggle and spin, and I suppose an electron is knocked out every now and
again).

> Roll the airplane into a 90 degree bank. The weight is
> now orthogonal to the lift. As teh airplane falls, it
> banks even though there is no Earth 'under' the
> belly. Why?

I'm not sure I understand the question. But if you put an airplane in a
knife edge and let it dive as it will, and maintain a lift-producing
AOA, the wing will push air in the belly direction, as it pushes itself
against that air in the antibelly direction. Some of that air will
swirl around the wing, but enough of it will dissipate the momentum that
the wing imparted to it over the entire atmosphere, and there will be a
(very) slight breeze blowing in the belly direction.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> For the stationary fan if it were only _almost equal_ then
> you would eventually run out of air on one side of the fan.

No, the pressure would build up on one side of the fan, and that
pressure would push against the wall and against the other air that is
being pushed by the fan. When the pressure on that side is sufficiently
high, no more (net) air will be able to be smooshed together on that
side, and the air will all be going around.

But a pressure difference will be maintained until the fan is turned off.

> Consider your example of the person who 'hovers' by
> dribbling a basektball. His momentum is zero, the
> momentum of the Earth is zero and the momentum
> of the ball is constantly changing and reverses twice
> each dribble. The dribbler is pumping energy into
> the Earth yet there is no net exchange of momentum.

I agree. Overall, no net change. Microscopically (at each impact)
there is a momentum change. Inbetween dribbles, the earth and the
dribbler experience momentum changes which each dribble then counteracts.

Now look at the same situation with a "basketball transparant" earth,
and an endless supply of basketballs being tossed at the dribbler (who
is backed up against a frictionless wall, so for now we don't need to
consider horizontal forces).

The dribbler keeps on deflecting basketballs downwards, but they don't
bounce back up - they pass through the earth. The dribbler (who
admittedly is no longer really dribbling) is imparting momentum to
basketballs, and once he stops doing that, he will himself experience a
momentum change.

In both cases, as far as the putative dribbler is concerned, he is
throwing basketballs down. He imparts momentum to basketballs, and
really doesn't care what happens to that momentum afterwards.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

In article om>,
wrote:

> Jose wrote:
> > > The hovering spacecraft has zero horizontal and vertical momentum.
> > > It has weight, directed downwards. The engine accelerates
> > > mass downward producing an upward force equal in magnitude
> > > and opposite in direction to the weight of the spacecraft. This
> > > imparts an acceleration to the spacecraft equal in magnitude and
> > > opposite in direction from the local acceleration due to gravity.
> >
> > The flying wing has some horizontal momentum which is secondary here,
>
> How much?
>
> > and zero vertical momentum.
> > It also has weight, directed downwards. The wing accelerates
> > mass downward (mass it finds in the air molecules) producing
> > an upward force equal in magnitude and opposite in direction to
> > the weight of the wing (and its presumably attached aircraft. It does
> > so by finding air in front of it, flinging it downwards and forwards
> > (which causes the air in front to try to get out of the way by rising).
> > In the steady state, one can measure high pressure below and low
> > pressure above, but this is just the macroscopic manifestation of the
> > greater number of molecular collisions below, and the lesser number of
> > collisions above. That's what pressure is - we have both agreed on this.
> >
> > The greater number of collisions below
> > imparts an acceleration to the aircraft equal in magnitude and
> > opposite in direction from the local acceleration due to gravity.
>
> I agree that lift is a force, exerted on the aircraft by the air,
> which in steady level flight is equal in magnitude and opposite
> in direction to the weight of the aircraft. Energy is 'pumped'
> into the air by the plane. There is no need for a net momentum
> exchange between the airplane and the air in order for
> energy to be exchanged or for forces to be applied.
> Indeed, in those last two paragraphs above, you make
> no mention of momentum.

>
> BTW, I was wrong to invoke conservation of momentum.
> Momentum is conserved in elastic collisions, like the
> collision between a cue ball and the eight ball. Momentum
> is not conserved in inelastic collisions, like the collision
> between a cue ball and a nerf ball.

You are incorrect. Momentum is *always* conserved.

>
> Roll the airplane into a 90 degree bank. The weight is
> now orthogonal to the lift. As teh airplane falls, it
> banks even though there is no Earth 'under' the
> belly. Why?

Because the wings are exerting a force on the air and the air
consequently experiences a change in momentum.

The air exerts a force on the wings. In level flight, this force is
countered by an equal and opposite force exerted on the aircraft by the
gravitational attraction of the earth. Without that countering force,
the aircraft would accelerate upward. That's what an unbalanced force
*does*.

But the wings also exert a force on the air (Newton, remember: for every
force there is an equal and opposite, etc., etc.). That force is not
countered by *anything*. Hence, the air is accelerated downward; a
continuous stream of air receives an constant change in momentum.

F = ma; that's the way we normally see it presented. This equation
relates force, mass and acceleration. It assumes a constant force acting
on a constant mass will produce a constant acceleration, and the mass
will start moving faster and faster.

But there is an equally valid presentation of that equation; one which
is more useful for examining what happens with an aircraft moving
through the air:

F = md/t^2; force is equal to mass, times distance, divided by the time
squared. If you keep velocity and time squared together, you get
acceleration of course, but there's no rule that says you have to. In
fact, the rules of equations say exactly the opposite: that an equation
is equally valid regardless of the way you group multiplications and
divisions.

So:

F = m/t * v/t; the force is equal to the rate of mass per unit time,
multiplied by the distance per unit time.

What that says is that if you change the velocity of a given mass flow
(air) by a given velocity, then you will get a given force.

In other words, an aircraft passing through the air will cause a portion
of that air to be disturbed downward. Because the aircraft is moving
forward a constant speed, it imparts a downward velocity to certain mass
of air each unit of time.

The air starts moving downward with a certain velocity.

Once you understand this, you understand why induced drag is less at
hight speeds than low. Go twice as fast, and you encounter twice as much
air in any unit time, and thus only need to impart a velocity to it that
is half as much. But because the kinetic energy involved is proportional
to mass and proportional to the *square* of velocity. Twice as much mass
doubles its contribution to energy lost, but half the velocity
*quarters* its contribution; giving an overall kinetic energy lost to
induced drag of half as much when going twice as fast.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

> Air is pressurized behind the speaker, just as well as the air in front of
> it. That is how bass reflex speakers work.

Yes, but out of phase. Air is pressurized in the direction the speaker
cone is moving. It goes back and forth.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

In article om>,
wrote:

> Jose wrote:
> > > In open air the volume of air moving around the fan is larger,
> > > but moving at a lower speed than the air moving through the
> > > fan so that the momenta of the flow in either direction is equal
> > > magnitude and opposite in direction to the flow in the other
> > > direction.
> >
> > Seems to me "almost equal" would make more sense, otherwise an airplane
> > propeller would not work. A propeller throws air backwards (alabeit
> > imperfectly); the airplane moves forwards in response.
> >
>
> For the stationary fan if it were only _almost equal_ then
> you would eventually run out of air on one side of the fan.
>
> Air molecules flowing through the propellor cetainly experience
> momentum changes. But you can have a net flow of
> energy without a net exchange of momentum because
> momentum is a vector, energy is a scaler. If the airplane
> is in level flight at constant speed it does not NEED to
> gain any momentum from the propellor because the
> momentum of the airplane is not changing. It needs
> force to counter the force of drag.

And both of those forces act on the *air*; hence the air isn't
accelerated forward or backward.

But the forces the aircraft exerts vertically act on two *different*
things. It exerts a downward force on the air and it exerts its upward
force on the *Earth*; hence the force on the air is unbalanced, hence it
must react by moving downward.

>
> Consider your example of the person who 'hovers' by
> dribbling a basektball. His momentum is zero, the
> momentum of the Earth is zero and the momentum
> of the ball is constantly changing and reverses twice
> each dribble. The dribbler is pumping energy into
> the Earth yet there is no net exchange of momentum.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

> F = m/t * v/t; the force is equal to the rate of mass per unit time,
> multiplied by the distance per unit time.

I assume a typo: F = m/t * d/t (since v=d/t)

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

In article >,
Jose > wrote:

> > F = m/t * v/t; the force is equal to the rate of mass per unit time,
> > multiplied by the distance per unit time.
>
> I assume a typo: F = m/t * d/t (since v=d/t)
>
> Jose

You assume correctly. <g>

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

"Jose" > wrote

> Well, when an object passes through the air, does it not compress the
> air in front of it (and rarefy the air behind it)? This is how speakers
> work. Those are all pressure changes.

Air is pressurized behind the speaker, just as well as the air in front of
it. That is how bass reflex speakers work.
--
Jim in NC

"Immanuel Goldstein"
> The Impossibility of Flying Heavy Aircraft Without Training


I would ask Nila Sagadevan to explain the video of Usama Bin Laden gloating
about his accomplishments.


Dallas

"Immanuel Goldstein"
> What hijackers?
> <http://news.bbc.co.uk/1/hi/world/middle_east/1559151.stm>


"Furthermore another article explains that the pilot who lives in Casablanca
was named Walid al-Shri (not Waleed M. al-Shehri) and that much of the BBC
information regarding "alive" hijackers was incorrect according to the same
sources used by BBC."

http://en.wikipedia.org/wiki/Waleed_al-Shehri


Dallas

"cjcampbell"
> Actually, he is not. Not in the US, anyway. There is no one by the name
> of Sagadevan currently holding a pilot certificate of any kind in the
> US

Here he is:
http://www.warpaintofthegods.com/wp/about.cfm



Dallas

Alan Baker wrote:
> In article om>,
> wrote:
>
> > Jose wrote:
> > > > The hovering spacecraft has zero horizontal and vertical momentum.
> > > > It has weight, directed downwards. The engine accelerates
> > > > mass downward producing an upward force equal in magnitude
> > > > and opposite in direction to the weight of the spacecraft. This
> > > > imparts an acceleration to the spacecraft equal in magnitude and
> > > > opposite in direction from the local acceleration due to gravity.
> > >
> > > The flying wing has some horizontal momentum which is secondary here,
> >
> > How much?
> >
> > > and zero vertical momentum.
> > > It also has weight, directed downwards. The wing accelerates
> > > mass downward (mass it finds in the air molecules) producing
> > > an upward force equal in magnitude and opposite in direction to
> > > the weight of the wing (and its presumably attached aircraft. It does
> > > so by finding air in front of it, flinging it downwards and forwards
> > > (which causes the air in front to try to get out of the way by rising).
> > > In the steady state, one can measure high pressure below and low
> > > pressure above, but this is just the macroscopic manifestation of the
> > > greater number of molecular collisions below, and the lesser number of
> > > collisions above. That's what pressure is - we have both agreed on this.
> > >
> > > The greater number of collisions below
> > > imparts an acceleration to the aircraft equal in magnitude and
> > > opposite in direction from the local acceleration due to gravity.
> >
> > I agree that lift is a force, exerted on the aircraft by the air,
> > which in steady level flight is equal in magnitude and opposite
> > in direction to the weight of the aircraft. Energy is 'pumped'
> > into the air by the plane. There is no need for a net momentum
> > exchange between the airplane and the air in order for
> > energy to be exchanged or for forces to be applied.
> > Indeed, in those last two paragraphs above, you make
> > no mention of momentum.
>
> >
> > BTW, I was wrong to invoke conservation of momentum.
> > Momentum is conserved in elastic collisions, like the
> > collision between a cue ball and the eight ball. Momentum
> > is not conserved in inelastic collisions, like the collision
> > between a cue ball and a nerf ball.
>
> You are incorrect. Momentum is *always* conserved.

How is momentum conserved when a cue ball hits a nerf ball?

>
> >
> > Roll the airplane into a 90 degree bank. The weight is
> > now orthogonal to the lift. As teh airplane falls, it
> > banks even though there is no Earth 'under' the
> > belly. Why?
>
> Because the wings are exerting a force on the air and the air
> consequently experiences a change in momentum.

Yes, both the airplane and the air experience a net change in
momentum when the aircraft climbs, descends, or banks.

In level flight at constant speed the aircraft has constant horzontal
and zero vertical momentum.

>
> The air exerts a force on the wings. In level flight, this force is
> countered by an equal and opposite force exerted on the aircraft by the
> gravitational attraction of the earth. Without that countering force,
> the aircraft would accelerate upward. That's what an unbalanced force
> *does*.

Yes, no question about weight being balanced by lift.

>
> But the wings also exert a force on the air (Newton, remember: for every
> force there is an equal and opposite, etc., etc.). That force is not
> countered by *anything*. Hence, the air is accelerated downward; a
> continuous stream of air receives an constant change in momentum.

If the air has a net increase in downward momentum, how is
momentum conserved.

>
> F = ma; that's the way we normally see it presented. This equation
> relates force, mass and acceleration. It assumes a constant force acting
> on a constant mass will produce a constant acceleration, and the mass
> will start moving faster and faster.
>
> But there is an equally valid presentation of that equation; one which
> is more useful for examining what happens with an aircraft moving
> through the air:
>
> F = md/t^2; force is equal to mass, times distance, divided by the time
> squared. If you keep velocity and time squared together, you get
> acceleration of course, but there's no rule that says you have to. In
> fact, the rules of equations say exactly the opposite: that an equation
> is equally valid regardless of the way you group multiplications and
> divisions.

>
> So:
>
> F = m/t * v/t; the force is equal to the rate of mass per unit time,
> multiplied by the distance per unit time.
>
> What that says is that if you change the velocity of a given mass flow
> (air) by a given velocity, then you will get a given force.

Yes, Force is the time rate of change of momentum.

>
> In other words, an aircraft passing through the air will cause a portion
> of that air to be disturbed downward. Because the aircraft is moving
> forward a constant speed, it imparts a downward velocity to certain mass
> of air each unit of time.
>
> The air starts moving downward with a certain velocity.

I don't deny that downflow occurs. The pont is that downflow is a
consequence of lift, not the cause of lift, and it is balanced by
upflow, (albeit a more diffuse flow) otherwise the upper atmosphere
would run out of air.

>
> Once you understand this, you understand why induced drag is less at
> hight speeds than low. Go twice as fast, and you encounter twice as much
> air in any unit time, and thus only need to impart a velocity to it that
> is half as much. But because the kinetic energy involved is proportional
> to mass and proportional to the *square* of velocity. Twice as much mass
> doubles its contribution to energy lost, but half the velocity
> *quarters* its contribution; giving an overall kinetic energy lost to
> induced drag of half as much when going twice as fast.
>

Interesting.

--

FF

> How is momentum conserved when a cue ball hits a nerf ball?

The vector sum, before and after, is identical. The vectors themselves
are different (kinetic energy is converted to heat and such) but looking
at both balls, or even looking at a cue ball and a glue ball, the center
of gravity moves with the same velocity before and after.

> If the air has a net increase in downward momentum, how is
> momentum conserved.

....by the air's eventual collision with the earth. Momentum is
similarly conserved when an object merely falls. The momentum gained by
the falling object is cancelled by the momentum acquired by the earth
rising up to meet it. In the case of "mysterious phantom gravity" not
associated with the earth, momentum is not conserved, it disappears into
the phantom gravity. This is one of the reasons why phantom gravity is
not experimentally supported.

If you ignore the earth, you are in the same position.

> I don't deny that downflow occurs. The pont is that downflow is a
> consequence of lift, not the cause of lift, and it is balanced by
> upflow, (albeit a more diffuse flow) otherwise the upper atmosphere
> would run out of air.

If there were no earth for the smooshed-together air to crowd up
against, the upper atmosphere =would= run out of air.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

Jose wrote:
> > Do you agree that the net momentum transfered to the Earth by the
> > air molecules is equal and opposite to the net momentum transferred
> > to the wing by the air molecules?
>
> Yes.
>
> > Do you agree, therefor that there is no net momentum transfered to
> > the air?
>
> Overall, yes. Similarly, there is no net momentum transferred to the
> basketball when it is being used to support a (very fast) dribbler. But
> that is not to say that there is no momentum transfer. The basketball
> certainly moves around. I do agree that the net overall is zero. The
> air does not pile up permanently.

Good. That was my point all along. There is no net momentum
transfered
to the air. There is a net transfer of energy to the air..

>
> > At which ponit the Earth throws the air molecule back up so that the
> > net momemtum transferred to the air molecule is zero (averaged over
> > the entire atmosphere)
>
> Yes.
>
> > [it hits the wing on the way up]
> > Which again transferes an equal and opposite momentum to the
> > molecule which again is transferrred to the Earth leaving no net
> > transfer of momentum to the air.
>
> Yes.
>
> Overall, there is no net (or "permanent") transfer of momentum to the
> air. The air is an intermediary, keeping the wing and the earth apart.
> There is certainly =energy= transfer to the air (mv^2/2), and there is
> a lot of momentum transfer =back=and=forth= with the air, but I will
> agree that the net is zero. The air is sort of a catalyst - ending up
> unchanged as it transfers momentum to the earth and then transfers it
> back from the earth to the wing.

Yes, although we do not yet agree on the details of the mechanism.

>
> So.. after all that, I think we are in agreement - there is no =net=
> (permanent) vertical momentum transfer to the air, but there is locally
> momentum transferred to the air, which carries it to the earth and uses
> it to neutralize the momentum the earth has acquired being attracted to
> the plane, in doing so it acquires momentum in the opposite direction
> and transfers it to the wing, ending the cycle and leavint the air ready
> to act as momentum messenger again.
>

No. Being attracted to something does not cause momentum. There
must be relative motion for momentum.

> It carries momentum messages both ways, they (overall) cancel out, but
> do keep the earth and the wing separated.

No, it is not momentum that keeps the aircraft from falling, it is
lift. The lift is produced by a pressure difference through the
wing.

>
> ===
>
> In addition, the wing is throwing air forwards, due to its AOA and its
> own forward motion. (this acts as drag, counteracted by the engine).
> The air thrown forwards increases the pressure in front of the wing,
> that plus the air thrown down makes the air pressure in front of and
> below the wing higher, causing the air to rise in front of the wing.
> This rising air helps lift the wing; this is the source of induced drag.
> Some of the rising air spills around the wingtips, causing vortices.
> The vortices are not the cause of lift, they are an inescapable side
> effect of lift.
>
> Concur?

No.

--

FF

Jose wrote:
> > For the stationary fan if it were only _almost equal_ then
> > you would eventually run out of air on one side of the fan.
>
> No, the pressure would build up on one side of the fan, and that
> pressure would push against the wall and against the other air that is
> being pushed by the fan. When the pressure on that side is sufficiently
> high, no more (net) air will be able to be smooshed together on that
> side, and the air will all be going around.

If the air is ALL going around then the flow in one direction is equal
to the flow going in the other direction, RIGHT? Not _almost equal_
but _exactly equal_, right?

OK to be clear, by 'flow' I meant rate. While the fan is on there is
a bit more air on one side than the other, but once equilibrium
is achieved the flow rate in one direction equals the flow rate in
the other direction. You have a closed loop. After equilibrium
occurs the fan no longer puts any net momentum into the air
mass. The momenta of the individual air molecules cancel.

>
> But a pressure difference will be maintained until the fan is turned off.
>

Yes. The fan continues to do work.

> > Consider your example of the person who 'hovers' by
> > dribbling a basektball. His momentum is zero, the
> > momentum of the Earth is zero and the momentum
> > of the ball is constantly changing and reverses twice
> > each dribble. The dribbler is pumping energy into
> > the Earth yet there is no net exchange of momentum.
>
> I agree. Overall, no net change. Microscopically (at each impact)
> there is a momentum change. Inbetween dribbles, the earth and the
> dribbler experience momentum changes which each dribble then counteracts.

The collison with the dribbler is inelastic. Energy is conserved,
momentum is not. The dribbler changes the momentum of
the basketbal without changing his momentum. That time
rate of change of the basketball results in a force on the dribbler
that is equal in magnitude and opposite in direction to his weight.

>
> Now look at the same situation with a "basketball transparant" earth,
> and an endless supply of basketballs being tossed at the dribbler (who
> is backed up against a frictionless wall, so for now we don't need to
> consider horizontal forces).

But we do presume there is still gravity.

>
> The dribbler keeps on deflecting basketballs downwards, but they don't
> bounce back up - they pass through the earth. The dribbler (who
> admittedly is no longer really dribbling) is imparting momentum to
> basketballs, and once he stops doing that, he will himself experience a
> momentum change.

He uses energy to impart momentum to the basketball without
changing his own momentum Energy is conserved, momentum
is not. Work is done. When he stops chucking the basketballs,
gravititational potential energy will be converted to kinetic energy
as he gains momentum by falling. Energy is conserved, momentum
is not. This is in the reference frame of the Earth, of course. In
his reference frame the earth falls toward him and if I am in freefall
next to the dribbler he has no momentum with respect to me.

>
> In both cases, as far as the putative dribbler is concerned, he is
> throwing basketballs down. He imparts momentum to basketballs, and
> really doesn't care what happens to that momentum afterwards.
>

Precisely. He does not need the earth beneath him any more than
an airplane wing needs the Earth beneath it.

--

FF

Jose wrote:
> > How is momentum conserved when a cue ball hits a nerf ball?
>
> The vector sum, before and after, is identical. The vectors themselves
> are different (kinetic energy is converted to heat and such) but looking
> at both balls, or even looking at a cue ball and a glue ball, the center
> of gravity moves with the same velocity before and after.

Perhaps you are not familiar with nerf balls. Nerf balls are foam
rubber. When a cue ball hits a nerf ball (sufficiently large) nerf
ball it stops and the nerf ball just quivers a bit. The center of
mas quits moving. The kinetic energy of the cue ball has been
converted to heat. Energy is conserved, momentum is not.

>
> > If the air has a net increase in downward momentum, how is
> > momentum conserved.
>
> ...by the air's eventual collision with the earth.

How is it conserved at the air/airplane collison?

--

FF

Jose wrote:
> >>The flying wing has some horizontal momentum which is secondary here,
> > How much?
>
> mv
>
> The air thrown forward (or, if you will, the higher pressure ahead)
> tries to reduce that, the engine presumably makes up for it.
>
> > Energy is 'pumped' into the air by the plane.
>
> Yes, and what form does that energy take?

Heat.

> I maintain that it takes the
> form of a net increase in mv^2/2 over all the air molecules.

Yes.

> Since m
> doesn't change, and 2 only changes in a pentium, that leaves v to
> change. This changes mv, thus momentum.

Mass and energy are scalers but velocity is a vector.
You can increase the average velocity of the air molecules
without changing the momentum of the air mass. Indeed,
that is exaclty what happens when you heat air.

>
> We agree that there is (microsocopic) momentum transfer at each
> collision. We disagree as to whether the net is zero, and I think that
> part of that disagreement has to do with just how much of the system we
> are looking at.

More importantly we disagree on what causes lift.

If there is lower pressure on the upper surface of a wing than there
is underneath there will be an upward force on that wing. I think
we agree on this.

You argue that the presssure difference and resulting force
is secondary, lift is actual caused by the reaction of the wing
to the momentum change it induces in the air. But suppose
the wing creates low pressure on the upper surface by throwing
air sideways? You still have a pressure differential and the
resultant force but the only downwash is the air flowing
toward the upper surface of the wing from above to fill in
the rarefied region.

For that matter, consider the common demonstration using a
notecard, thumbtack and a straw. Put the tack through the
middle of a 3x5 index card or something similar. Put a drinking
straw over the thumbtack. Hold the aparatus with the straw
vertical and the notedard down. Blow through the straw and
let go of the notecard. The notecard will be supported by the
Bernouli effect.

The only downwash is through the straw, directed at the notecard,
pushing it down. There is no downwash from the card. The card
does not deflect any air down, it deflects the air sideways.
Yet the card is supported by the pressure differential created
by the Bernouli effect. Horizontal flow accross the upper surface
of the card creates that pressure difference.

Downwash does not cause lift. Downwash is a secondary effect
caused by the same phenomenum that causes lift.

--

FF

>> to neutralize the momentum the earth has acquired being attracted to
>> the plane,
> No. Being attracted to something does not cause momentum. There
> must be relative motion for momentum.

Being attracted to something and having no force resisting the
attraction (which is the case microscopically inbetween collisions)
allows relative motion to occur. That's how things fall down, acquiring
momentum in the process. Of course the earth falls up at the same time,
so depending on whether or not you include the earth, you can argue no
net momentum change.

> No, it is not momentum that keeps the aircraft from falling, it is
> lift. The lift is produced by a pressure difference through the
> wing.

.... which is caused by microscopic collisions, which each transfer
momentum from an air molecule to the wing. This is what pressure is.

"Lift" is a shorthand for this process, the same way raising to a power
is a shorthand for repeated repeated addition.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> After equilibrium
> occurs the fan no longer puts any net momentum into the air
> mass. The momenta of the individual air molecules cancel.

Yes, but only because of the wall, which allows the pressure to build up
on the far side of the fan. Were there no wall (such as for an airplane
propeller), this would not be the case.

> The collison with the dribbler is inelastic. Energy is conserved,
> momentum is not.

Well, only if you treat momentum as a scalar, or deal only with the
momentum of a single particle at a time. If two glueballs collide, (for
simplicity assume they were of equal mass, equal and opposite velocity),
the net (vector) momentum before is zero, but each glueball will have a
finite momentum because it is moving. After the collision, the net
(vector) momentum is zero (the splatball is motionless), and each
glueball component of the splatball is also motionless. The glueballs
have each lost momentum, because they have stopped.

So, while the vector sum of the momenta have not changed, the (scalar)
sum of the absolute values of the momenta have.

Kinetic energy (mv^2/2) is =not= conserved in an inelastic collision,
since v changes, and v^2 is scalar. It is transformed into other forms.
Some of that kinetic energy becomes heat and noise (which is
ultimately molecular kinetic energy), some of it shakes electrons
around, but macroscopic kinetic energy is not conserved for an inelastic
collision.

> He uses energy to impart momentum to the basketball

So, he is "throwing basketballs down". They could just as easily be
very very tiny basketballs; the kind with eight electrons or so.

> Precisely. He does not need the earth beneath him any more than
> an airplane wing needs the Earth beneath it.

No, he doesn't need the earth in order to =stay=up=. But the system
=does= need the earth to satisfy the "no net momentum change in the
basketballs/air" criterion. Absent the earth's surface, there =is= a
net momentum change, whether the basketballs are the size of
basketballs, or the size of air molecules.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> Perhaps you are not familiar with nerf balls. Nerf balls are foam
> rubber. When a cue ball hits a nerf ball (sufficiently large) nerf
> ball it stops and the nerf ball just quivers a bit. The center of
> mas quits moving. The kinetic energy of the cue ball has been
> converted to heat. Energy is conserved, momentum is not.

There is more to that. If this collision occurs in outer space, I
guarantee you that the center of mass will =not= quit moving.

On a pool table, friction with the table is involved, (as is to some
extent rolling moment). The nerf ball starts its quiver in the
direction the cue ball was going. If there is not enough force in the
quiver to break starting friction, then the momentum gets imparted to
the table (and the entire earth, which has no problem absorbing it).

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

>>>If the air has a net increase in downward momentum, how is
>>> momentum conserved.
>> ...by the air's eventual collision with the earth.
> How is it conserved at the air/airplane collison?

(sorry, should have added this to the prevous post)

It is conserved because the wing gets pushed (back) up when the air
molecule gets pushed down.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

"Dallas" > wrote in message
nk.net...
>
> "cjcampbell"
>> Actually, he is not. Not in the US, anyway. There is no one by the name
>> of Sagadevan currently holding a pilot certificate of any kind in the
>> US
>
> Here he is:
> http://www.warpaintofthegods.com/wp/about.cfm
>
WOW!! What a fruticake!!

Jose wrote:
> > Perhaps you are not familiar with nerf balls. Nerf balls are foam
> > rubber. When a cue ball hits a nerf ball (sufficiently large) nerf
> > ball it stops and the nerf ball just quivers a bit. The center of
> > mas quits moving. The kinetic energy of the cue ball has been
> > converted to heat. Energy is conserved, momentum is not.
>
> There is more to that. If this collision occurs in outer space, I
> guarantee you that the center of mass will =not= quit moving.

But it will not move in a manner that conserves momentum.

--

FF

In article m>,
wrote:

> Jose wrote:
> > > Perhaps you are not familiar with nerf balls. Nerf balls are foam
> > > rubber. When a cue ball hits a nerf ball (sufficiently large) nerf
> > > ball it stops and the nerf ball just quivers a bit. The center of
> > > mas quits moving. The kinetic energy of the cue ball has been
> > > converted to heat. Energy is conserved, momentum is not.
> >
> > There is more to that. If this collision occurs in outer space, I
> > guarantee you that the center of mass will =not= quit moving.
>
> But it will not move in a manner that conserves momentum.

You really need to study basic physics...

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

Alan Baker wrote:
> In article m>,
> wrote:
>
> > Jose wrote:
> > > > Perhaps you are not familiar with nerf balls. Nerf balls are foam
> > > > rubber. When a cue ball hits a nerf ball (sufficiently large) nerf
> > > > ball it stops and the nerf ball just quivers a bit. The center of
> > > > mas quits moving. The kinetic energy of the cue ball has been
> > > > converted to heat. Energy is conserved, momentum is not.
> > >
> > > There is more to that. If this collision occurs in outer space, I
> > > guarantee you that the center of mass will =not= quit moving.
> >
> > But it will not move in a manner that conserves momentum.
>
> You really need to study basic physics...
>

Doh!

You'd never guess it but I did get A's in freshman Physics.

Of course I had that bass akwards. Energy can change from
kinetic to gravitatonal potential to heat etc. An elastic collision
is one in which kinetic energy is conserved. In an inelastic collison
some kinetic energy is converted to another form, typically heat.
Momentum is always conserved, as you noted earlier.

--

FF

I'll tell ya what.
I do admire it.

This is the nicest pair of debaters on the planet!

Polite.
Considerate.
Seeem to take time to read, and understand(!) what somebody wrote.
Patiently explaining their own position.
No yellin.
No screamin.
No ugly name callin.

Archie and Jughead ponder quantum pressure fluctuations.

I do love it so...:)

On Sat, 4 Mar 2006 10:14:36 -0700, "Matt Barrow"
> wrote:

>
>"Dallas" > wrote in message
nk.net...
>>
>> "cjcampbell"
>>> Actually, he is not. Not in the US, anyway. There is no one by the name
>>> of Sagadevan currently holding a pilot certificate of any kind in the
>>> US
>>
>> Here he is:
>> http://www.warpaintofthegods.com/wp/about.cfm
>>
>WOW!! What a fruticake!!

Too bad that, or those, mind altering experiences didn't give him a
grasp of rationality and reality.

Roger Halstead (K8RI & ARRL life member)
(N833R, S# CD-2 Worlds oldest Debonair)
www.rogerhalstead.com
>
>

wrote:

> A fluid can transmit force without flow in the conventional sense.
> That is the basis for hydraulics.

I guess it depends on what you mean by "conventional sense."

Nothing can transmit a force without some deflection. Some molecules in
the fluid have to move in order to generate a force and typically
movement of a fluid, however minute, is flow.


Matt

So in Larry Niven's Ringworld, when people swim through the ring of
atmosphere to the next Integral Tree, do they set it rotating ever so
slightly in the opposite direction?

(Livin' in a fantasy world since nineteen-sixty-mumblemumble....)


Alan Baker wrote:


>
> No. It is balanced by the downflow eventually transferring its momentum
> back to the earth.
>

>> But it is also experiencing a constant change in momentum in the
>> vertical direction. That's what a force is: a change in momentum over
>> time.
> No. The airplane is in level flight at constant speed.

I'm with fredfighter here, but it is primarily a semantic argument
having to do with frame of reference. Looking from an accelerated frame
of reference (a standstill in the earth's gravitational field) one gets
one answer. Looking from an unaccelerated frame of reference
(freefall), one gets a different answer.

To reconcile them, it is important to treat =everything= from the same
frame of reference, and make appropriate adjustments.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> It does that
> because there are fewer air molecules transfering momentum
> to it from above, that there are from below. But it does not
> do that via a coherent stream of air.

I guess I should have appended this to the previous response...

This is correct. However, a coherent stream of air is not necessary for
this:

>> The movement of the plane towards the earth is transferred to
>> movement of the air towards the earth, which it does until it eventually
>> transfers its momentum back *to* the earth, leaving the system with the
>> same relative momentum with which it began.

to also be correct. A coherent stream of air is not required, nor is it
what I am proposing.

> A fluid can transmit force without flow in the conventional sense.
> That is the basis for hydraulics.

We are not really talking about "flow in the conventional sense", we are
talking about microscopic collisions. Flow may be involved (as in the
flow that causes upwash upflight) but it needn't be (as in the case of
the microscopic dribbler).

> The downflow observed from the wing initiates above the wing
> and flows down behind the wing after the wing has passed.
> It is not the air that suppors the wing.

Well, the only air that supports the wing is are the molecules that
impact it from below. They not only support the wing, they also fight
against the molecules impacting from above. They win, because there are
more of them. There are more of them because of downflow and the
collisions it causes.

> Well then if the downflow is NOT balanced by upflow why doesn't
> the upper atmosphere run out of air?

Because the wing is not of infinite weight. The upper atmosphere in
fact =is= deprived of air while the airplane is in flight... that air is
squeezed down below the wing, increasing the pressure on the surface of
the earth, in an amount exactly equal to the weight of the airplane
(divided by the area of the earth).

If a sufficient (huge!) number of aircraft took to the air, the upper
atmosphere would become measurably thinner. Maybe we should get a grant
to do this experiment using general aviation aircraft - for the good of
Science and the benefit of GA pilots. :)

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> So in Larry Niven's Ringworld, when people swim through the ring of atmosphere to the next Integral Tree, do they set it rotating ever so slightly in the opposite direction?

Yep.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

Matt Whiting wrote:
> wrote:
>
> > A fluid can transmit force without flow in the conventional sense.
> > That is the basis for hydraulics.
>
> I guess it depends on what you mean by "conventional sense."
>
> Nothing can transmit a force without some deflection. Some molecules in
> the fluid have to move in order to generate a force and typically
> movement of a fluid, however minute, is flow.
>

One does not generally refer to compression as flow.

A fluid can transmit force without molecules flowing from the
point of origin to the point of application.

--

FF

Jose wrote:
> > It does that
> > because there are fewer air molecules transfering momentum
> > to it from above, that there are from below. But it does not
> > do that via a coherent stream of air.
>
> I guess I should have appended this to the previous response...
>
> This is correct. However, a coherent stream of air is not necessary for
> this:
>
> >> The movement of the plane towards the earth is transferred to
> >> movement of the air towards the earth, which it does until it eventually
> >> transfers its momentum back *to* the earth, leaving the system with the
> >> same relative momentum with which it began.
>
> to also be correct. A coherent stream of air is not required, nor is it
> what I am proposing.

I inferred coherent flow from 'downwash'. Some persons, perhpas not
yourself, pointed to disturbances on the surface by low flying aricraft
as evidence of downwash. That sounds like coherent flow.

>
> > A fluid can transmit force without flow in the conventional sense.
> > That is the basis for hydraulics.
>
> We are not really talking about "flow in the conventional sense", we are
> talking about microscopic collisions. Flow may be involved (as in the
> flow that causes upwash upflight) but it needn't be (as in the case of
> the microscopic dribbler).
>

When we are discussing the microscopic transmission of momenta
between air molecules whic is the basis for presure, yes. Is that what

you mean by 'downwash' or downflow, as opposed to something that
involves a flow of mass?

> > The downflow observed from the wing initiates above the wing
> > and flows down behind the wing after the wing has passed.
> > It is not the air that suppors the wing.
>
> Well, the only air that supports the wing is are the molecules that
> impact it from below. They not only support the wing, they also fight
> against the molecules impacting from above. They win, because there are
> more of them. There are more of them because of downflow and the
> collisions it causes.

Then it doesn't matter which way the air above the wing flows. If
the air flows sideways, you still have lift. It doesn't have to flow
down.

>
> > Well then if the downflow is NOT balanced by upflow why doesn't
> > the upper atmosphere run out of air?
>
> Because the wing is not of infinite weight. The upper atmosphere in
> fact =is= deprived of air while the airplane is in flight... that air is
> squeezed down below the wing, increasing the pressure on the surface of
> the earth, in an amount exactly equal to the weight of the airplane
> (divided by the area of the earth).

I think that the downflow dispaces other air which flows up to
replace it--conserving momentum and mass.

--

FF

> I inferred coherent flow from 'downwash'.

That coherent flow is not necessary does not mean that coherent flow
does not exist. My point is that the downwash does not have to be
directly from the wing to the earth. It can be very indirect - in a
multiple collision scenario, the existance of new momentum somewhere
imples the existance of opposte new momentum elsewhere, mediated by
collisions which may or may not be "coherent", however you wish to
define it. Momentum is conserved. Always.

> When we are discussing the microscopic transmission of momenta
> between air molecules whic is the basis for presure, yes. Is that what
> you mean by 'downwash' or downflow, as opposed to something that
> involves a flow of mass?

There is downwash, involving a "coherent" acceleration of mass
downwards. Due to an increase in microscopic collisions below (and a
scarcity of them above), there is an incoherent transfer of momentum
(called pressure) to the surrounding air (and ultimately to the earth).

This leads to a condition described as "low pressure above, high
pressure beneath", or equivalently described as "less momentum
transferred via collisions above, more momentum transferred via
collisions below", which supports the wing, propping it up again and
again as it tries to succumb to gravity. We call this lift.

There are some neat bulk equations which help quantify this, which come
embodied in a concept which is useful for understanding this in some
contexts. However, an equivalent (newtonian) concept is more useful for
understanding in other contexts, and explains a few things that are not
addressed by the B word.

> Then it doesn't matter which way the air above the wing flows. If
> the air flows sideways, you still have lift.

No, at least not directly. If there is less momentum transferred from
above than from below, you have lift. This comes from lower pressure
above and higher pressure below. How you get that is ultmately
Newtonian, not magical. Once Newton has his say, Bernoulli can
reformulate it in a useful bulk form.

Consider a flying saucer, composed solely of two disks with no
appreciable space between them. The one below does not spin, the one
above spins rapidly. Should there be lift? Why? Does it matter if the
top disk is rough or smooth?

>> The upper atmosphere in
>> fact =is= deprived of air while the airplane is in flight... that air is
>> squeezed down below the wing, increasing the pressure on the surface of
>> the earth, in an amount exactly equal to the weight of the airplane
>> (divided by the area of the earth).
>
> I think that the downflow dispaces other air which flows up to
> replace it--conserving momentum and mass.

What happens to the downward momentum of the downflowing air when this
happens? The displaced air, flowing upwards, has acquired upward
momentum - where did that come from? (and so far, conservation of mass
has not been an issue)

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

In article . com>,
wrote:

> Alan Baker wrote:
> > In article . com>,
> > wrote:
> >
>
> ...
>
> > > In level flight at constant speed the aircraft has constant horzontal
> > > and zero vertical momentum.
> >
> > True. But it is also experiencing a constant change in momentum in the
> > vertical direction. That's what a force is: a change in momentum over
> > time.
> >
>
> No. The airplane is in level flight at constant speed. Therefore there
> is no change in momentum. The net force on an aircraft in level
> flight at constant speed is zero.
>
> When there is a constant change in momentum the vertical
> direction the airplane climbs or dives.

Incorrect.

You're going to want to go back to your basic physics texts again. Check
the formal definition of "force".

<http://www.rwc.uc.edu/koehler/biophys/2c.html>

"Just as force is the time derivative of momentum, ..."

>
> ...
> > >
> > > Yes, no question about weight being balanced by lift.
> > >
> > > >
> > > > But the wings also exert a force on the air (Newton, remember: for every
> > > > force there is an equal and opposite, etc., etc.). That force is not
> > > > countered by *anything*. Hence, the air is accelerated downward; a
> > > > continuous stream of air receives an constant change in momentum.
> > >
> > > If the air has a net increase in downward momentum, how is
> > > momentum conserved.
> >
> > By the increased upward momentum of the earth.
> >
> > The earth pulls the plane downward, and the plane pulls the earth
> > upward. The movement of the plane towards the earth is transferred to
> > movement of the air towards the earth, which it does until it eventually
> > transfers its momentum back *to* the earth, leaving the system with the
> > same relative momentum with which it began.
> >
>
> It is the coilumn of air under the wing that supports the weight
> of the plane. It does that, on the statistical mechanical level
> via a series of minute momentum exchanges. It does that
> because there are fewer air molecules transfering momentum
> to it from above, that there are from below. But it does not
> do that via a coherent stream of air.
>
> A fluid can transmit force without flow in the conventional sense.
> That is the basis for hydraulics.
>
> The downflow observed from the wing initiates above the wing
> and flows down behind the wing after the wing has passed.
> It is not the air that suppors the wing.
>
> ...
>
> > > I don't deny that downflow occurs. The point is that downflow is a
> > > consequence of lift, not the cause of lift, and it is balanced by
> > > upflow, (albeit a more diffuse flow) otherwise the upper atmosphere
> > > would run out of air.
> >
> > No. It is balanced by the downflow eventually transferring its momentum
> > back to the earth.
>
> Well then if the downflow is NOT balanced by upflow why doesn't
> the upper atmosphere run out of air?

Because the air contacts the earth and *stops* moving downward.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

In article <c2OOf.597091$084.130237@attbi_s22>,
Stella Starr > wrote:

> So in Larry Niven's Ringworld, when people swim through the ring of
> atmosphere to the next Integral Tree, do they set it rotating ever so
> slightly in the opposite direction?

Yes. If they were to keep swimming in the same direction.

>
> (Livin' in a fantasy world since nineteen-sixty-mumblemumble....)
>
>
> Alan Baker wrote:
>
>
> >
> > No. It is balanced by the downflow eventually transferring its momentum
> > back to the earth.
> >

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

This has all been very interesting, but there is a basic assumption
that seems to be glossed over.

I was thinking, that to really get anything out of all this,
shut the engine off!

Everybody has been _assumin'_ straight and level flight.
I suppose that's ok for academic discussion, but for learnin'
aerodynamics, let's just assume the engine quit and take it
from there.

L / D

Just a thought...


Richard

In article et>,
Richard Lamb > wrote:

> This has all been very interesting, but there is a basic assumption
> that seems to be glossed over.
>
> I was thinking, that to really get anything out of all this,
> shut the engine off!
>
> Everybody has been _assumin'_ straight and level flight.
> I suppose that's ok for academic discussion, but for learnin'
> aerodynamics, let's just assume the engine quit and take it
> from there.
>
> L / D
>
> Just a thought...
>
>
> Richard

It doesn't really make any difference.

In a constant glide, the aircraft now does have momentum with respect to
the earth, but it is *still* incurring the same forces.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

Alan Baker wrote:

> In article et>,
> Richard Lamb > wrote:
>
>
>>This has all been very interesting, but there is a basic assumption
>>that seems to be glossed over.
>>
>>I was thinking, that to really get anything out of all this,
>> shut the engine off!
>>
>>Everybody has been _assumin'_ straight and level flight.
>>I suppose that's ok for academic discussion, but for learnin'
>>aerodynamics, let's just assume the engine quit and take it
>>from there.
>>
>> L / D
>>
>>Just a thought...
>>
>>
>>Richard
>
>
> It doesn't really make any difference.
>
> In a constant glide, the aircraft now does have momentum with respect to
> the earth, but it is *still* incurring the same forces.
>
Right.

But now we might actually get something from the discussion.

Like how much power is actually required for S&L?

Effects of speed on glide angle?

And, what happens when you get a wee bit too slow?

Or, if that's too boring...

What happens to the boundary layer?
Is it ticklish?

And what about those long and short bubbles?

> wrote

> > Because the wing is not of infinite weight. The upper atmosphere in
> > fact =is= deprived of air while the airplane is in flight... that air is
> > squeezed down below the wing, increasing the pressure on the surface of
> > the earth, in an amount exactly equal to the weight of the airplane
> > (divided by the area of the earth).
>
> I think that the downflow dispaces other air which flows up to
> replace it--conserving momentum and mass.

I think I will create a new award. I'm not sure what the prize or trophy
will be yet.

I'm calling it "Rec.Aviation Geek of the Decade", or perhaps of "The
Century."

I am in total awe and amazement, that you and Jose have tied for this award,
based on how long you two have kept this amazingly boring subject alive. I
just CAN'T believe it !!!

Now, continue on, or not.

Please, use your restraint, and common sense. Use the "or not." <g>
--
Jim in NC
(mostly, using his right to use the "ignore thread" button! <g>

> let's just assume the engine quit and take it
> from there.
>
> L / D

What is L? What is D?

That's the fundamental question being discussed. The engine (or the
earth's gravity) merely supplies the force. But once you introduce the
idea of gliding, you also need to address the things that gliders
address - ridge lift, thermals, messy stuff like that, which are all
ways of getting free energy from the sun.

Calm air, flat ambient earth, no engine, the airplane will descend.

Now explain to me how autogyros work. :)

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

Morgans wrote:
>
> ...
>
> Now, continue on, or not.
>
> Please, use your restraint, and common sense. Use the "or not." <g>
>

Well I'm really hoping that Jose tries the card thumbtack soda straw
thing.

--

FF

> I am in total awe and amazement, that you and Jose have tied for this award,
> based on how long you two have kept this amazingly boring subject alive.

Great discoveries are often made in the seventh decimal place. :)

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> Well I'm really hoping that Jose tries the card thumbtack soda straw
> thing.

Actually, I did try it and it didn't "work" (that is, the card didn't
float, which is what I think you expect to happen). I'm probably doing
it wrong so I'll keep at it. When I get it to work, I'll report what
happened and why (in newtonian terms) I think it did.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

wrote:
> Matt Whiting wrote:
>
wrote:
>>
>>
>>>A fluid can transmit force without flow in the conventional sense.
>>>That is the basis for hydraulics.
>>
>>I guess it depends on what you mean by "conventional sense."
>>
>>Nothing can transmit a force without some deflection. Some molecules in
>>the fluid have to move in order to generate a force and typically
>>movement of a fluid, however minute, is flow.
>>
>
>
> One does not generally refer to compression as flow.
>
> A fluid can transmit force without molecules flowing from the
> point of origin to the point of application.

True, but most hydraulic systems of any usefulness require the fluid to
flow to some degree. I hydraulic system that simply statically supports
a load isn't a very interesting system and could easily be done without
using a fluid and for much less money. :-)

Matt

On Fri, 3 Mar 2006 at 15:21:11 in message
>, "00:00:00Hg"
> wrote:
>On Fri, 03 Mar 2006 01:27:46 +0000, David CL Francis wrote:
>
>> Above
>> Mach one the air does not detect the approaching aircraft! :-)
>
> If it did, what would happen?

The whole point is that disturbances in the air are propagated at or
near the velocity of sound. It follows that at supersonic speeds
nothing happens to the air until it reaches the supersonic aircraft, or
vice versa.
--
David CL Francis

On Fri, 3 Mar 2006 at 02:27:02 in message
et>, Richard Lamb
> wrote:
>I hate to be a spoil sport (or dullard?), but...
>
>the (stationary) air does WHAT (as the wing passes by)???

The nature of things is such that the situation does not change if you
change the frame of reference. It is normal in doing calculations to
start with a frame of reference based on the aircraft. If you follow the
aircraft then the air is going past it.

The presence of the wing changes the air flowing past the aircraft in
the same way as if you consider the aircraft passing through the air.
The 'stationary' air as you call it has its local velocity and direction
changed by the aircraft.
--
David CL Francis

On Fri, 3 Mar 2006 at 05:30:06 in message
. com>,
wrote:

>Newton had three laws of motion, you're ignoring the first.
>Is there a net change inmomentum of the fan? If not,
>how can there be a net change of momentum of the air?
>
I am ignoring nothing. The above statement is wrong. You agree below
that energy is put into the air. In the case of a fan that energy goes
into increasing the velocity of the air. The rate of change of momentum
(mass flow times velocity increase) produces forces that increase the
momentum of the air. Energy changes momentum. Momentum destroyed turns
back into energy.

This argument is hung up on the idea that the air returns to a steady
state eventually - which it does! But not quite back to where it was
because of losses Nevertheless energy is lost and replaced by the
engines of the aircraft.

>There no question that energy is put into the air. There is
>no net change in momentum, of the air. otherwise all the
>air would pile up on one side of the fan and there would
>be a vacuum on the inlet side. Air moving through the
>fan in one direction is offset by air moving around the fan
>in the other direction.
>
The air slows down and looses energy and momentum far away from the
aircraft - so what? Any small drop in pressure at the fan also reaches
back and develops flow some way in front of the fan. For lift purposes
it does not matter much. The air may or may not make its way back to the
inlet again, some of it will.

>In open air the volume of air moving around the fan is larger,
>but moving at a lower speed than the air moving through the
>fan so that the momenta of the flow in either direction is equal
>magnitude and opposite in direction to the flow in the other
>direction.

Except for losses that occur due to friction and eddies that float away
to dissipate themselves elsewhere. But the answer must be still be so
what? Momentum is not conserved because energy has been added. Are you
saying that a land vehicle with a horizontal fan to drive it along rails
will not accelerate and move? Will the vehicle not build up momentum
because of this?
--
David CL Francis

Alan Baker wrote:
> In article . com>,
> wrote:

....

>
> >
> >
> > Well then if the downflow is NOT balanced by upflow why doesn't
> > the upper atmosphere run out of air?
>
> Because the air contacts the earth and *stops* moving downward.
>

Could you define downflow?

--

FF

David CL Francis wrote:
> On Fri, 3 Mar 2006 at 02:27:02 in message
> et>, Richard Lamb
> > wrote:
>
>> I hate to be a spoil sport (or dullard?), but...
>>
>> the (stationary) air does WHAT (as the wing passes by)???
>
>
> The nature of things is such that the situation does not change if you
> change the frame of reference. It is normal in doing calculations to
> start with a frame of reference based on the aircraft. If you follow the
> aircraft then the air is going past it.
>
> The presence of the wing changes the air flowing past the aircraft in
> the same way as if you consider the aircraft passing through the air.
> The 'stationary' air as you call it has its local velocity and direction
> changed by the aircraft.

Yeahbut...

A handy frame of reference is - handy.

But it can be very misleading.....

Richard Lamb wrote:
> David CL Francis wrote:
>
>> On Fri, 3 Mar 2006 at 02:27:02 in message
>> et>, Richard Lamb
>> > wrote:
>>
>>> I hate to be a spoil sport (or dullard?), but...
>>>
>>> the (stationary) air does WHAT (as the wing passes by)???
>>
>>
>>
>> The nature of things is such that the situation does not change if you
>> change the frame of reference. It is normal in doing calculations to
>> start with a frame of reference based on the aircraft. If you follow
>> the aircraft then the air is going past it.
>>
>> The presence of the wing changes the air flowing past the aircraft in
>> the same way as if you consider the aircraft passing through the air.
>> The 'stationary' air as you call it has its local velocity and
>> direction changed by the aircraft.
>
>
> Yeahbut...
>
> A handy frame of reference is - handy.
>
> But it can be very misleading.....
>
>
For instance?

If the air is moving, we expect a lower pressure. Nod to Bernoulli.

But the air would also be moving along the bottom side of the wing also?

And what would that do to the pressure under the wing?

And if the pressure under the wing is below ambient....

> Momentum is not conserved because energy has been added. Are you saying that a land vehicle with a horizontal fan to drive it along rails will not accelerate and move? Will the vehicle not build up momentum because of this?

Momentum is always conserved. If you see momentum disappearing, you are
not looking at the whole system. In the case of the land vehicle
propelled by a fan, the air blown back acquires momentum in one
direction, exactly balanced by the momentum that the vehicle acquires,
plus the (rotational) momentum (due to wheel friction) that the earth
acquires.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

In article . com>,
wrote:

> Alan Baker wrote:
> > In article . com>,
> > wrote:
>
> ...
>
> >
> > >
> > >
> > > Well then if the downflow is NOT balanced by upflow why doesn't
> > > the upper atmosphere run out of air?
> >
> > Because the air contacts the earth and *stops* moving downward.
> >
>
> Could you define downflow?

Sure.

The aircraft passes through and air moves downward. As it moves its
motion is dissipated into more and more air moving less and less, but
eventually the momentum that was transferred to it is transferred back
to the earth.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

Alan Baker wrote:
> In article . com>,
> wrote:
>
> > Alan Baker wrote:
> > > In article . com>,
> > > wrote:
> >
> > ...
> >
> > >
> > > >
> > > >
> > > > Well then if the downflow is NOT balanced by upflow why doesn't
> > > > the upper atmosphere run out of air?
> > >
> > > Because the air contacts the earth and *stops* moving downward.
> > >
> >
> > Could you define downflow?
>
> Sure.
>
> The aircraft passes through and air moves downward. As it moves its
> motion is dissipated into more and more air moving less and less, but
> eventually the momentum that was transferred to it is transferred back
> to the earth.
>
> --
> Alan Baker
> Vancouver, British Columbia
> "If you raise the ceiling 4 feet, move the fireplace from that wall
> to that wall, you'll still only get the full stereophonic effect
> if you sit in the bottom of that cupboard."

Jose wrote:
> > Well I'm really hoping that Jose tries the card thumbtack soda straw
> > thing.
>
> Actually, I did try it and it didn't "work" (that is, the card didn't
> float, which is what I think you expect to happen). I'm probably doing
> it wrong so I'll keep at it. When I get it to work, I'll report what
> happened and why (in newtonian terms) I think it did.
>

Well after reading that I went and tried it myself and blew the card
off the end of the straw so I must be doing it wrong too!

I've known of this trick from childhood, (yes, I realize that some
of you are thinking that could mean I first learned of it a few days
ago) so by now I can't remember exactly how or even if I did it
myself. Memory is like that.

Could we reduce the crossposting? I think one newsgroup is more
than sufficient. You chose, and I'll follow.

--

FF

David CL Francis wrote:
> On Fri, 3 Mar 2006 at 05:30:06 in message
> . com>,
> wrote:
>
> >Newton had three laws of motion, you're ignoring the first.
> >Is there a net change inmomentum of the fan? If not,
> >how can there be a net change of momentum of the air?
> >
> I am ignoring nothing. The above statement is wrong. You agree below
> that energy is put into the air. In the case of a fan that energy goes
> into increasing the velocity of the air. The rate of change of momentum
> (mass flow times velocity increase) produces forces that increase the
> momentum of the air. Energy changes momentum. Momentum destroyed turns
> back into energy.

Well I'm sorry to see that I an not the only one who was confused
on this issue. In Newtonian dynamics, energy is always conserved,
mass is always conserved, and momentum is always conserved.

When the momentum of a body changes, then energy is converted
from one form to another, but the momentum of a body can only
change by being transferred to another body. Momentum,
like energy and mass, is never destroyed.

I'm not fond of 'unit analysis' but consider that the units of energy
and the units of momentum when reduced to fundamental units,
are different. Conversion between the two is impossible.

>
> This argument is hung up on the idea that the air returns to a steady
> state eventually - which it does! But not quite back to where it was
> because of losses Nevertheless energy is lost and replaced by the
> engines of the aircraft.

Yes, the airplane puts energy into the air. But in the closed system
that consists of the airplane and the atmosphere, or the fan and
the air in the room, momentum is conserved, just as mass and
energy are.

>
> > ... Air moving through the
> >fan in one direction is offset by air moving around the fan
> >in the other direction.
> >
> The air slows down and looses energy and momentum far away from the
> aircraft - so what? Any small drop in pressure at the fan also reaches
> back and develops flow some way in front of the fan. For lift purposes
> it does not matter much. The air may or may not make its way back to the
> inlet again, some of it will.

If only some of it does, then mass is not conserved. ALL of it,
or rather an equivalent amount of displaced mass makes it
back to the inlet of the fan. In order to make it back, it has
velocity. for a rather slow fan in a rather small room the velocity
through the fan may be ten times the average velocity of the
air moving around the fan in the opposite direction. That's
OK, but conservation of momentum requires that ten times
the mass be moving in that opposite direction at one tenth the
'fan' velocity and a moments consideration should convince you
that this also conserves mass.

In te case of the aircraft, the fan is moving through the air so that
when the air (or rather an equivalent displaced mass of air) returns
to the inlet, the inlet has move on.

>
> >In open air the volume of air moving around the fan is larger,
> >but moving at a lower speed than the air moving through the
> >fan so that the momenta of the flow in either direction is equal
> >magnitude and opposite in direction to the flow in the other
> >direction.
>
> Except for losses that occur due to friction and eddies that float away
> to dissipate themselves elsewhere.

No. The turbulance dissipates energy, (that is to say it converts to
heat) not momentum.

Momentum is always conserved.

--

FF

wrote:
> Alan Baker wrote:
> > In article . com>,
> > wrote:
> >
> > > Alan Baker wrote:
> > > > In article . com>,
> > > > wrote:
> > >
> > > ...
> > >
> > > >
> > > > >
> > > > >
> > > > > Well then if the downflow is NOT balanced by upflow why doesn't
> > > > > the upper atmosphere run out of air?
> > > >
> > > > Because the air contacts the earth and *stops* moving downward.
> > > >
> > >
> > > Could you define downflow?
> >
> > Sure.
> >
> > The aircraft passes through and air moves downward. As it moves its
> > motion is dissipated into more and more air moving less and less, but
> > eventually the momentum that was transferred to it is transferred back
> > to the earth.
> >

I was hoping for a mathematical defintion, rather than a description
of the process. That would minimize my opportunity to draw an
incorrect inference. In this regaerd, a mathematical definiton
would be best.

I infer from your description the definition: "downflow is a flow
of air from the airplane toward the ground". That removes a
potential abiguity, whether downflow was a flow of momentum
through the air, (like a pressure wave) or a flow of mass.
Is that how you define downflow, as a flow of air molecules
(with mass) toward the ground?

Can you state a mathematical definition of downflow?

Fred Thomas' in _Fundamentals of Sailplane Design_ defines
the freestream velocity, then states the relationship between
that, the local velocity near the airfoil and the induced downwash
as a vector sum and goes on to show how this produces an
effective angle of attack less than the geometric angle of attack.
But he does not present a separate mathematical defintion of
induced downwash or the local velocity of the air near the wing
so the vector sum above does not serve (within the context of
his discussion) to define either term.

But it is clear that the induced downwash is a velocity, not a
a massflow. Yes, it is mass that has that velocity but the
parameter _induced downwash_ is a velocity.

So, can we agree to the definition of downflow as a flow
of air toward the ground and define the induced downwash
as the velocity of that air near the wing?

Meanwhile:

Earlier I wrote:

The point is that downflow is a consequence of,
not the cause of lift, and it is balanced by
upflow, (albeit a more diffuse flow) otherwise
the upper atmosphere would run out of air.
[and later corrected this to: downflow is a
consequence of the same phenomenum that
produces lift, not the cause of lift]

You replied:

No. It is balanced by the downflow eventually
transferring its momentum back to the earth.

So I asked:

Well then if the downflow is NOT balanced by
upflow why doesn't the upper atmosphere run
out of air?

Your response:

Because the air contacts the earth and *stops*
moving downward.

Earlier you corrected me regarding conservation of
momentum. Now, consider conservation of mass.

That the downflow stops upon contacting the Earth does
NOT explain why the upper atmosphere is not depleted
of air.

Plainly if air flows to the Earth and *stops* there as
you wrote, it has displaced other air which flowed
up to replace it, right?

--

FF

T o d d P a t t i s t wrote:

> David CL Francis > wrote:
>
>
>>The nature of things is such that ....
>
>
> I've been following along (more or less :-) and chose
> David's post to jump in again, since, from experience I have
> great trust in any analysis by David.
>
> This thread, however, seems to wander all over the place.
> It looks like one participant will make one set of
> assumptions, then another will assume something different.
>
> I see the following discussions going on:
>
> 1) A pure thread related purely to lift and Bernoulli. In
> this thread, the subject matter is maximally simplified by
> a) using the standard Bernoulli assumptions, inviscid flow,
> incompressible, subsonic, etc., b) ignoring parasitic drag
> c) using 2-D flow (or equivalently infinite wing)
> assumptions and looking at steady state conditions. This
> gets to the heart of upwash and downwash.
>
> 2) The same as 1) above, but looking at 3-D flow. Now we
> have induced drag and the wing/fan produces a net motion of
> the air as it passes through. Much of this discussion seems
> focused on issues relating to closed systems (rooms, earth
> with ground, etc.) and what happens to the air, how big a
> system should be looked at, etc.
>
> 3) The same as 2) above, but with viscosity added so that
> the air ultimately stops its motion and heats up due to
> viscous losses.
>
> Quite honestly, for most of the posts here, I can't figure
> out what assumptions lie behind the comments.

Todd,
Your #2 and #3 is where I wanted to go with thus mess,
Thankee.

The down wash, being transferred from air near the wing to air
far away from the wing...

Air is quite springy stuff.

The energy transfer is spread over an increasing area (or volume)
and quickly reduced in magnitude - to the point where it is no
longer detectable (without invoking Steven Hawking).

For all practical purposes, that would seem to indicate that the
"down flow" would not reach the ground before being dissipated into
the larger air mass. (not arguing against the eventual contact with
the entire surface of the planet. But that doesn't help us understand
basic aerodynamics!)

Only when the wing is close to the ground is the down wash detectable
because it hasn't had time (or room) to be absorbed/dissipated.

Now, while the above is obviously not true in the molecular sense,
it may help us understand the practical parts better.

Also, add #4?

While we have been concentrating on the pressure side (bottom) of the
wing, it's the upper surface that has the greater influence here.

The only way I see of increasing pressure on the bottom surface is to
increase speed (or density?).

But the top side is where the pressure is reduced.
And there are a lot of factors that effect that part.

Thickness of the camber line is a big one.
Deeper camber tends to cause a lower pressure on top - hence more lift
for a given surface and speed.

This is most often accomplished by deploying flaps.
True they go down into the stream on the bottom side - and probably do
to some degree - invoke some impact lift (pressure) on the bottom.

But the curvature of the airfoil has increased also - and the camber
line has deepened - and the apparent angle of attack has increased.

These factors further decrease the lowered pressure field on top of the
wing - WAY more than any useful increase in pressure below it.

Lastly (for now), if we are indeed pressing down on the air below the
wing, we are also Pulling Down on the air above it...

The air below presses against the earth. As I've said before, that one is
so obvious (that we stop looking?).

But I think the low pressure field Above the wing is also pulling down
on the atmosphere above it.

While air pressure decreases with altitude me may think that the field
above the wing dissipates quicker. Maybe true, BUT - the pressure field
Above the wing is of much higher magnitude - so maybe not.

Well, so much for my silly idea.
I don't know how to analyze that one mathematically.

I'd really like to see what the bigger brains can make of it.

Richard

> Could we reduce the crossposting? I think one newsgroup is more
> than sufficient. You chose, and I'll follow.

It's being posted to piloting, homebuilt, and student. We could easily
lose homebuilt. Should we lose student?

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> But in the closed system
> that consists of the airplane and the atmosphere, or the fan and
> the air in the room, momentum is conserved, just as mass and
> energy are.

Except that that's not a closed system. You need the earth to close the
system. No earth but phantom gravity, and to conserve momentum, the air
will continue to downflow, which is what would happen.

> If only some of it does, then mass is not conserved. ALL of it,
> or rather an equivalent amount of displaced mass makes it
> back to the inlet of the fan.

No, mass can be conserved by having some of it pile up. This is in fact
what happens. The pressure on one side of the room goes up. Guess why.

> In te case of the aircraft, the fan is moving through the air so that
> when the air (or rather an equivalent displaced mass of air) returns
> to the inlet, the inlet has move on.

In the case of the aircraft (propeller), the air does not return to the
inlet. It keeps on being blown back, since there is no wall to stop it.
The momentum stays with the air and the earth (which starts to spin a
little one way) until the airplane lands, and pushes the earth the other
way. Momentum is always conserved.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

On Mon, 06 Mar 2006 17:56:42 -0800, fredfighter wrote:

> Could you define downflow?

It happens to geese as they exceede the speed of down.

> So, can we agree to the definition of downflow as a flow
> of air toward the ground and define the induced downwash
> as the velocity of that air near the wing?

I don't think this is a useful definition. Downflow and downwash are
the actual movement of something, not merely the velocity of that movement.

> Plainly if air flows to the Earth and *stops* there as
> you wrote, it has displaced other air which flowed
> up to replace it, right?

Only if pressure is constant. (at constant temperature). However,
pressure does not remain constant. The pressure below the wing (and
thus against the earth) increases due to the extra molecules that have
been thrown down. Those molecules came from above the wing. The upper
atmosphere =is= (slightly) depleted by the flight.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> I'd really like to see what the bigger brains can make of it.

I think you did fine. I will take issue with:

> The air below presses against the earth. As I've said before, that one is
> so obvious (that we stop looking?).

The air does press on the earth, and this is "where the momentum goes",
which is a big question in one of the poster's minds. No earth, nothing
to press against, and the momentum just keeps on going down. It is
(thus) not true that there is no local momentum transfer. That is one
of the points I was making. There is of course no global momentum
transfer once all parts of the closed system are taken into account.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

> Could we reduce the crossposting? I think one newsgroup is more
> than sufficient. You chose, and I'll follow.

Absent protest, from just after "now" on, I'll reply and post this
thread only to r.a.piloting.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

On Tue, 7 Mar 2006 at 05:06:54 in message
>, Jose
> wrote:

>Momentum is always conserved. If you see momentum disappearing, you
>are not looking at the whole system. In the case of the land vehicle
>propelled by a fan, the air blown back acquires momentum in one
>direction, exactly balanced by the momentum that the vehicle acquires,
>plus the (rotational) momentum (due to wheel friction) that the earth
>acquires.

From a Physics book:

A jet of water merges from a hose pipe of a cross sectional area 5x
10^-3 m^2 and strikes a wall at right angles. Calculate the force on the
wall assuming the water is brought to rest and does not rebound.
(Density of water = 1.0 x 10^3 Kg m^-3)

After explaining the simple calculation which gives a force of 45N it
goes on to say; " in practice the horizontal momentum of the water is
seldom completely destroyed and so the answer is only approximate."

~~~~~~~~~~
Any changes to the entire earth as a result are insignificant. Closed
systems are adequate for most practical purposes.

All these arguments about the 'total system' are irrelevant to
considering the kind of problem we have here. As in many problems you do
not need to include the whole universe to get practical and accurate
answers.

In the same way including discussions about molecular velocities beating
on the sides of the aircraft is a mere distraction. At normal altitudes
these effects are negligible compared to the consideration of the air as
an incompressible fluid.

Some aircraft can maintain a 7g turn banked at the appropriate angle.
(81.8 degrees approximately). What happens to that 7 times pressure on
the earth now? The same calculations will give accurate figures of lift
only slightly affected by the small difference of speed between the two
wings and the circular path. The earth hardly comes into it for
accuracy except that it is gravity that is being balanced.

--
David CL Francis

> A jet of water merges from a hose pipe of a cross sectional area 5x 10^-3 m^2 and strikes a wall at right angles. Calculate the force on the wall assuming the water is brought to rest and does not rebound. (Density of water = 1.0 x 10^3 Kg m^-3)
>
> After explaining the simple calculation which gives a force of 45N it goes on to say; " in practice the horizontal momentum of the water is seldom completely destroyed and so the answer is only approximate."

Is this a US book? This is why the US lags in science.

The original question is ok (after all, in physics we use cylindrical
cows, frictionless surfaces, and point masses). But the comment at the
end is very misleading. The momentum is never "destroyed". It is
actually transferred to the wall, and thus to the earth. What they are
probably trying to say is that there is usually some rebound of the
water, and it sprays all over the place rather than becoming embedded
like machine gun bullets in sand... which would have been a better example.

> Any changes to the entire earth as a result are insignificant.

Depends whether you are trying to understand the fundamental physics or
just trying to calculate an answer.

Jose
--
Money: what you need when you run out of brains.
for Email, make the obvious change in the address.

On Wed, 8 Mar 2006 at 23:12:19 in message
>, Jose
> wrote:

>The original question is ok (after all, in physics we use cylindrical
>cows, frictionless surfaces, and point masses). But the comment at the
>end is very misleading. The momentum is never "destroyed". It is
>actually transferred to the wall, and thus to the earth. What they are
>probably trying to say is that there is usually some rebound of the
>water, and it sprays all over the place rather than becoming embedded
>like machine gun bullets in sand... which would have been a better example.
>
I have got to be quick here - busy the next couple of days!

Their statement may be a bit sloppy but the fact is that because the
wall is very rigid and firmly fixed the ground an accurate calculation
of the force can be made by assuming the momentum is destroyed. The
effect on the earth is so small that it is minuscule compared to the
practical result.

>> Any changes to the entire earth as a result are insignificant.
>
>Depends whether you are trying to understand the fundamental physics or
>just trying to calculate an answer.

I thought the basic principles have been stated and stated! It is
whether or not we have a clear understanding of lift or not that seems
to be going around in circles. What happens to the earth a long way away
from the aircraft is, for all practical purposes insignificant and not
worth worrying about. It certainly does not prove that down wash has
nothing to do with lift for example as has been stated.
--
David CL Francis