The Impossibility of Flying Heavy Aircraft Without Training

[Alan Baker]

In article >,
"00:00:00Hg" > wrote:

> On Fri, 03 Mar 2006 08:00:45 -0800, fredfighter wrote:
>
> >
> > Jose wrote:
> >> > Suppose we have a 1500 lb airplane in level flight at 120 mph.
> >> > What are its horizontal and vertical components of momentum?
>
> Zero at equalibrium.

Incorrect. It has considerable horizontal momentum and no vertical
momentum.

>
> >>
> >> Suppose we have a 1500 lb rocketship hovering over the moon on its
> >> rocket exhaust. What are its horizontal and vertical componnts of
> >> momentum?
>
> Again, zero.

Correct, but how can you possibly get zero for this answer and think
that an aircraft moving horizontally at 120 mph has no horizontal
momentum...

<snip>

--
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."

[00:00:00Hg]

On Fri, 03 Mar 2006 17:56:39 +0000, Alan Baker wrote:

>> >> > Suppose we have a 1500 lb airplane in level flight at 120 mph.
>> >> > What are its horizontal and vertical components of momentum?
>>
>> Zero at equalibrium.
>
> Incorrect. It has considerable horizontal momentum and no vertical
> momentum.

I thought the focus was forces.

The 'aircraft' with respect to the media, air.

If the ground is at play with respect to horizontal 'componet'
is it in the form of 'gravity' (a force) or geologial energy
of some sort expressed as the force of location?

This is hightly interesting. Never thought of it before.

[[email protected]]

00:00:00Hg wrote:
> On Fri, 03 Mar 2006 17:56:39 +0000, Alan Baker wrote:
>
> >> >> > Suppose we have a 1500 lb airplane in level flight at 120 mph.
> >> >> > What are its horizontal and vertical components of momentum?
> >>
> >> Zero at equalibrium.
> >
> > Incorrect. It has considerable horizontal momentum and no vertical
> > momentum.

Whereas for the hovering spacecraft both components are zero.

>
> I thought the focus was forces.

It should be.

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.

Now of course weight is a convenient fiction. There is really no such
thing as gravitational force, what we model as a force acting at a
distance is in reality the distortion of spacetime in the presence of
mass. Perhaps other forces are similarly ficticious.

But how sure can we be that mass and velocity are any less ficticious
than force?

--

FF

[00:00:00Hg]

On Fri, 03 Mar 2006 10:55:59 -0800, fredfighter wrote:

> Now of course weight is a convenient fiction.

Can I allocate excess fat that same definition?

> There is really no such
> thing as gravitational force, what we model as a force acting at a
> distance is in reality the distortion of spacetime in the presence of
> mass. Perhaps other forces are similarly ficticious.

I hope not the Air Force.

So you want to bring general and special relativity into
the frey `eh? Newton ain't good enough for you, huh? Ok.

Gimme your Lorentz transformations for -Mach 1 to +Mach 1
at the transition point. I wanna see how time and gravity
are related to mass transactions. The speed of sound must
be a nodal harmonic of the speed of light. I wanna see it
too. Gimme gimme...

>
> But how sure can we be that mass and velocity are any less ficticious
> than force?

Gravity seems to work to it's own advantage so it's the
ultimate taxing authority in the universe. That really sucks.

[00:00:00Hg]

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.

[00:00:00Hg]

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?

[george]

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

[Jose]

> 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.

[Jose]

> 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.

[Jose]

> 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.

[00:00:00Hg]

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.

[Morgans]

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

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

[Morgans]

"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

[00:00:00Hg]

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.

[Jose]

> 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.

[Jose]

> 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.

[Jose]

> 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.

[Jose]

> 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]

>>What is the net momentum change when the airplane falls to the ground?
>
> The vertical compenent first rises from zero to Vt * M where Vt is the
> terminal velocity of the falling aircraft and M is the mass of the
> falling
> aircraft. Then the vertical component of momentum RAPIDLY drops
> to zero again after the aircraft contacts the ground.

Well, actually, only sorta. The momentum of the airplane is equal to
the momentum of the earth, except in sign. Net is zero. The center of
mass of the earth/airplane does not move.

Leave the earth out of it and just look at the aircraft, and you are
correct. And to keep an airplane up, in view of this acceleration, an
opposite acceleration needs to be applied. Air must be thrown down with
sufficient (net) force to counteract gravity's attempt to accelerate the
wing downwards. All this air piles up against the earth (just a little
bit, but enough that, when multplied over the earth's surface it adds up
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.

[[email protected]]

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

[[email protected]]

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

[Jose]

>>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.

[[email protected]]

Jose wrote:
> >>What is the net momentum change when the airplane falls to the ground?
> >
> > The vertical compenent first rises from zero to Vt * M where Vt is the
> > terminal velocity of the falling aircraft and M is the mass of the
> > falling
> > aircraft. Then the vertical component of momentum RAPIDLY drops
> > to zero again after the aircraft contacts the ground.
>
> Well, actually, only sorta. The momentum of the airplane is equal to
> the momentum of the earth, except in sign. Net is zero. The center of
> mass of the earth/airplane does not move.
>
> Leave the earth out of it and just look at the aircraft, and you are
> correct. And to keep an airplane up, in view of this acceleration, an
> opposite acceleration needs to be applied. Air must be thrown down with
> sufficient (net) force to counteract gravity's attempt to accelerate the
> wing downwards.

No. You can also generate an upward force on an airplane by
creating low pressure over the upper surface of the wing while
the pressure below the wing remains at ambient. I dunno if
there are any airfoils that leave the air below the wing exactly
the same as ambient, but if there were, it would fly. There is
no NEED to throw anything downward.

--

FF

[Jose]

> 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.

[Jose]

> You can also generate an upward force on an airplane by
> creating low pressure over the upper surface of the wing while
> the pressure below the wing remains at ambient. I dunno if
> there are any airfoils that leave the air below the wing exactly
> the same as ambient, but if there were, it would fly. There is
> no NEED to throw anything downward.

I suppose a wing that gobbled up air molecules from the top of the wing
and beamed them into outer space would do the trick. Another way would
be to supercool the top surface, and let the general gas law reduce the
pressure above. But doing either one, air above the air above the wing
would rush down, as the air below the wing pushes the wing up into that
same space. The two will collide, or the wing will have passed by then.
In the latter case, downward momentum has been imparted to the air
above the air above the wing, which gets dissipated as I argued for
conventional wings.

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

[Alan Baker]

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."

[Jose]

> 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.

[Alan Baker]

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."

[Jose]

> 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.

[Alan Baker]

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."

[Morgans]

"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

[Dallas]

"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

[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

[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

[[email protected]]

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

[[email protected]]

Jose wrote:
> > You can also generate an upward force on an airplane by
> > creating low pressure over the upper surface of the wing while
> > the pressure below the wing remains at ambient. I dunno if
> > there are any airfoils that leave the air below the wing exactly
> > the same as ambient, but if there were, it would fly. There is
> > no NEED to throw anything downward.
>
> I suppose a wing that gobbled up air molecules from the top of the wing
> and beamed them into outer space would do the trick.

Or blew them out the rear for thrust.

> Another way would
> be to supercool the top surface, and let the general gas law reduce the
> pressure above. But doing either one, air above the air above the wing
> would rush down, as the air below the wing pushes the wing up into that
> same space. The two will collide, or the wing will have passed by then.
> In the latter case, downward momentum has been imparted to the air
> above the air above the wing, which gets dissipated as I argued for
> conventional wings.
>

Yes and that is what a conventional wing does. It creates lower
pressure above the wing so that the ambient or near ambient
pressure below the wing pushes up on the wing creating lift.

The air from above that low pressure region begins moving down
into that region but doesn't get there until after the wing has
passed. Downwash occurs, as you describe in the paragraph
above. It is a consequence of lift, not the cause. In fact the
energy put into the air by the downwash phenomenum is wasted.
A more efficient wing will produce less downwash than a less efficient
one, for the same lift.

--

FF

not the cause.

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

[Jose]

> 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]

> Downwash occurs, as you describe in the paragraph
> above. It is a consequence of lift, not the cause.

We perhaps disagree merely on the idea of which "causes" the other. How
do you figure that lift causes downwash? Lift doesn't happen unless the
air above is rarified. The air above is not rarified until some of the
molecules are gotten rid of somehow. The process of getting rid of
those molecules is just a newtonian process (which lends itself to
certain bulk equations).

They are different ways of looking at the same thing, depending on which
aspect you want to concentrate on.

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

[[email protected]]

Jose wrote:
> > Downwash occurs, as you describe in the paragraph
> > above. It is a consequence of lift, not the cause.
>
> We perhaps disagree merely on the idea of which "causes" the other. How
> do you figure that lift causes downwash? Lift doesn't happen unless the
> air above is rarified. The air above is not rarified until some of the
> molecules are gotten rid of somehow.

No.

> The process of getting rid of
> those molecules is just a newtonian process (which lends itself to
> certain bulk equations).

They aren't 'gotten rid of' they are accelerated which causes them to
be spaced farther apart -- thus lowering the pressure.

>
> They are different ways of looking at the same thing, depending on which
> aspect you want to concentrate on.
>

The lift is a result of the pressure difference between the lower and
upper surfaces of the wing. The downwash is the result of the momentum
of the air above the rarefied region created by the wing moving
downward.

The downrushing air starts it s downwash above the wing and does
not pass the wing in the vertical direction until after he wing has
passed.
It does not contribute to lift.

It is not really caused by lift (my mistake), it is caused by the same
phenomenum that causes lift.

--

FF

[[email protected]]

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

[[email protected]]

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

[[email protected]]

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

[[email protected]]

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

[Jose]

> They aren't 'gotten rid of' they are accelerated which causes them to
> be spaced farther apart -- thus lowering the pressure.

Accelerating them gets rid of them in the sense I mean, but I suppose I
was sloppy there. In any case, to be accelerated, they need to go
somewhere. The standard explanation is that there is a longer path up
top. The reason there is a longer path is that the air is bent
downwards. If you bend plywood (concave down), the top sheet is
stretched and the bottom sheet is compressed. Same with the air.

When the air is bent downwards, the air is accelerated downwards. This
causes downwash. Air accelerated downwards by the wing requires (by
Newton) the wing to be accelerated upwards (counteracting in this case
the acceleration due to gravity). It does so in a manner that also fits
Bernoulli's equations.

> The lift is a result of the pressure difference between the lower and
> upper surfaces of the wing. The downwash is the result of the momentum
> of the air above the rarefied region created by the wing moving
> downward.

And the pressure difference is sustained by the wing continually
imparting momentum (indirectly by creating the pressure differential) to
the air above the rarified region.

> The downrushing air starts it s downwash above the wing and does
> not pass the wing in the vertical direction until after he wing has
> passed.

Matters not. It is another way to look at lift.

> [The downrushing air] is not really caused by lift (my mistake),
> it is caused by the same phenomenum that causes lift.

Fair enough. What this says is that both ways of looking at it are
valid. Bernoulli is easier to calculate, Newton is easier to conceptualize.

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

[Jose]

>> 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.

[Jose]

> 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.

[Jose]

> 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.

[Jose]

>>>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.

[Matt Barrow]

"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!!

[[email protected]]

Jose wrote:
> > They aren't 'gotten rid of' they are accelerated which causes them to
> > be spaced farther apart -- thus lowering the pressure.
>
> Accelerating them gets rid of them in the sense I mean, but I suppose I
> was sloppy there. In any case, to be accelerated, they need to go
> somewhere. The standard explanation is that there is a longer path up
> top. The reason there is a longer path is that the air is bent
> downwards. If you bend plywood (concave down), the top sheet is
> stretched and the bottom sheet is compressed. Same with the air.

There is a longer path along the top because the wing is convex up.

>
> When the air is bent downwards, the air is accelerated downwards. This
> causes downwash.

Not until after it passes the high point in the airfoil. Befor it
gets there,
it is accelerated upwards.

> Air accelerated downwards by the wing requires (by
> Newton) the wing to be accelerated upwards (counteracting in this case
> the acceleration due to gravity). It does so in a manner that also fits
> Bernoulli's equations.

When the air reaches the trailing edge it is back to where it started.
But in the meantime air above it has begun to flow down. After the
wing has passed the momentum of _that_ downflow carries the air
down past the altitude of the wing. But that is after the wing has
passed. The downflow is -art of what happens as the air in the wake
of the airplane is restored to equilibrium.

>
> > The lift is a result of the pressure difference between the lower and
> > upper surfaces of the wing. The downwash is the result of the momentum
> > of the air above the rarefied region created by the wing moving
> > downward.
>
> And the pressure difference is sustained by the wing continually
> imparting momentum (indirectly by creating the pressure differential) to
> the air above the rarified region.

Regardless, the lift is a result of the pressure differential between
the upper and lower wing surfaces.

>
> > The downrushing air starts it s downwash above the wing and does
> > not pass the wing in the vertical direction until after he wing has
> > passed.
>
> Matters not. It is another way to look at lift.

No, it is a way of looking at downrushing air that has never
contacted the wing.

>
> > [The downrushing air] is not really caused by lift (my mistake),
> > it is caused by the same phenomenum that causes lift.
>
> Fair enough. What this says is that both ways of looking at it are
> valid. Bernoulli is easier to calculate, Newton is easier to conceptualize.
>

No. That says that the downrushing air and lift are both caused by the
same phenomenum.

--

FF

[[email protected]]

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

[Jose]

> Not until after it passes the high point in the airfoil. Befor it
> gets there, it is accelerated upwards.

Does the air split (top/bottom path) at the same level as it rejoins?
If the air splits at a higher altitude, then the air has to have a net
downward motion to get to the rejoin point.

> When the air reaches the trailing edge it is back to where it started.

Is it? I think it's lower than when it started. It certainly is with
any appreciable AOA.

> No, it is a way of looking at downrushing air that has never
> contacted the wing.

It doesn't matter whether it contacts the wing or not. The contact is
by proxy (by contacting the other molecules of air that contact...the
wing.) You call it pressure. I agree. Pressure is ultimately
newtonian; I think we agree there too.

It's a floor wax. It's a dessert topping.

>>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.

Yes it will. What will not be conserved is macroscopic kinetic energy.

> Make the room bigger. Make it an infinite room. At what point does
> the fan continue to put momentum into the air mass continuously, and
> not just during start up?

At the point when it's an infinite room. The bigger the room, the
longer it takes for a pressure equilibrium to occur. If we deal with
the earth's atmosphere and a propeller, the propeller pushes air back,
which alters the rotational momentum of the earth, in a manner equal
(and opposite) to the amount of rotational (around the earth's center)
momentum the airplane acquires. This could potentially happen until the
earth is spinning godawful fast (faster than the propeller could
handle). In practice we'll get tired of arguing before that. :)

> So, are you saying that in the presence of the Earth there is no
> net change in the momentum of the basketballs being thrown by the
> dribbler and also no net change in momentum of the air molecules
> accelerated by the wing?

Averaged over all basketballs and all air molecules, yes, because the
earth acts as a momentum transfer point.

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

[Alan Baker]

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."

[[email protected]]

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

[[email protected]]

Jose wrote:
> > Not until after it passes the high point in the airfoil. Befor it
> > gets there, it is accelerated upwards.
>
> Does the air split (top/bottom path) at the same level as it rejoins?
> If the air splits at a higher altitude, then the air has to have a net
> downward motion to get to the rejoin point.

Agreed. Clearly this happens with high AOA.

>
> > When the air reaches the trailing edge it is back to where it started.
>
> Is it? I think it's lower than when it started. It certainly is with
> any appreciable AOA.

Are there any airfoils that produce lift at an AOA at or below zero?

>
> > No, it is a way of looking at downrushing air that has never
> > contacted the wing.
>
> It doesn't matter whether it contacts the wing or not. The contact is
> by proxy (by contacting the other molecules of air that contact...the
> wing.) You call it pressure. I agree. Pressure is ultimately
> newtonian; I think we agree there too.

But the pressure that supplies the lift is in the air below the wing
whereas it is the air above the wing that washes down.You're
attributing lift to the wrong air.

If he air above the wing washed sideways, and not down at all
you'd still get lift as the thumbtack, notecard and soda straw
demonstrates. No downwash from the notecard.



>
> It's a floor wax. It's a dessert topping.
>
> >>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.
>
> Yes it will. What will not be conserved is macroscopic kinetic energy.

Right. My mistake.

> ...
>
> > So, are you saying that in the presence of the Earth there is no
> > net change in the momentm of the basketballs being thrown by the
> > dribbler and also no net change in momentum of the air molecules
> > accelerated by the wing?
>
> Averaged over all basketballs and all air molecules, yes, because the
> earth acts as a momentum transfer point.
>

--


FF

[Richard Lamb]

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...:)

[Jose]

> Are there any airfoils that produce lift at an AOA at or below zero?

Depends how AOA is defined (I don't know the precise definition). But
if we define it as the angle between the line from split to rejoin
point, and the direction of travel (or relative wind), then I think not.

> But the pressure that supplies the lift is in the air below the wing
> whereas it is the air above the wing that washes down. You're
> attributing lift to the wrong air.

It doesn't matter. The molecules collide all over the place and
momentum is moved around (but never lost).

> If he air above the wing washed sideways, and not down at all
> you'd still get lift as the thumbtack, notecard and soda straw
> demonstrates. No downwash from the notecard.

Describe the notecard setup better and I will do the experiment, and
then tell you what I think.

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

[Roger]

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
>
>

[[email protected]]

Jose wrote:
> > Are there any airfoils that produce lift at an AOA at or below zero?
>
> Depends how AOA is defined (I don't know the precise definition). But
> if we define it as the angle between the line from split to rejoin
> point, and the direction of travel (or relative wind), then I think not.

ISTR that AOA is defined relative to the chord.

>
> > But the pressure that supplies the lift is in the air below the wing
> > whereas it is the air above the wing that washes down. You're
> > attributing lift to the wrong air.
>
> It doesn't matter. The molecules collide all over the place and
> momentum is moved around (but never lost).

Of course it matters. It is the undisturbed molecules (or minimally
disturbed molecules) under the wing that push up on the wing.

The air flowing over the top of the wing cannot push UP on the
wing.

>
> > If he air above the wing washed sideways, and not down at all
> > you'd still get lift as the thumbtack, notecard and soda straw
> > demonstrates. No downwash from the notecard.
>
> Describe the notecard setup better and I will do the experiment, and
> then tell you what I think.
>

Take 3 x 5 index card or a similar light small card like a playing
card and push a thumbtack through the center.

Leave the thumbtack in and set the card on a table point up.

Put a soda straw, standing on end over the point of the tack.

Blow through the straw and while blowing lift the straw
straight up.

--

FF

[Matt Whiting]

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

[Stella Starr]

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.
>

[Jose]

>> 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.

[Jose]

> 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.

[Jose]

> 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.

[[email protected]]

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

[[email protected]]

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

[Jose]

> 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.

[Alan Baker]

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."

[Alan Baker]

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."

[Richard Lamb]

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

[Alan Baker]

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."

[Richard Lamb]

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?

[Morgans]

> 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>

[Jose]

> 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.

[[email protected]]

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

[Jose]

> 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.

[Jose]

> 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.

[Matt Whiting]

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

[David CL Francis]

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

[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

[David CL Francis]

On Fri, 3 Mar 2006 at 21:10:20 in message
. com>,
wrote:

>No. You can also generate an upward force on an airplane by
>creating low pressure over the upper surface of the wing while
>the pressure below the wing remains at ambient. I dunno if
>there are any airfoils that leave the air below the wing exactly
>the same as ambient, but if there were, it would fly. There is
>no NEED to throw anything downward.

Wrong there is a need. The pressure change you postulate would itself
have an effect on the air and cause it to change direction.

Helicopters are the easiest things to consider. They expend an awful lot
of power just to hang there. Where does all that energy go? It goes into
changing the momentum of a large amount of air downwards. The rate of
change of momentum then being equal to the weight of the helicopter.
--
David CL Francis

[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

[David CL Francis]

On Fri, 3 Mar 2006 at 04:46:18 in message
. com>,
wrote:

>The momentum of an airplane in level flight at
>constant speed is constant. Conservation of
>momentum REQUIRES that there is no net
>change in themomentum of the air. There is
>momentum exchanged between the airplane
>and the air. But there is no NET momentum
>change in the air.

You are perhaps neglecting the fact that in momentum exchanges energy is
generally lost.

Your statement is true in a way. If you consider a lump of air striking
something and being defected then it is analogous to the impact between
two billiard balls. If they strike at an angle then momentum is
conserved but it is redistributed and energy is generally lost.

The point you are making makes no difference to the outcome. The energy
lost has to be exactly balanced by the energy input to the system by the
engines or by potential energy. The energy is what is needed to
continually add to the downward momentum imparted. The local change in
momentum is provided by the aircraft descending exchanging potential
energy for kinetic or by having a source of power.
--
David CL Francis

[David CL Francis]

On Sat, 4 Mar 2006 at 05:38:33 in message
. com>,
wrote:

>A more efficient wing will produce less downwash than a less efficient
>one, for the same lift.

Yes but it still has to provide the exact same amount of rate of change
of momentum. It tends to move a bigger mass of air slower but at the
same momentum change.
--
David CL Francis

[[email protected]]

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

[Richard Lamb]

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]

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....

[Jose]

> 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.

[Alan Baker]

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."

[[email protected]]

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."

[[email protected]]

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

[[email protected]]

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

[[email protected]]

David CL Francis wrote:
> On Sat, 4 Mar 2006 at 05:38:33 in message
> . com>,
> wrote:
>
> >A more efficient wing will produce less downwash than a less efficient
> >one, for the same lift.
>
> Yes but it still has to provide the exact same amount of rate of change
> of momentum. It tends to move a bigger mass of air slower but at the
> same momentum change.

An infinite wing has no induced downwash.

How is it supported?

--

FF

[[email protected]]

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

[Richard Lamb]

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

[Jose]

> 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.

[Jose]

> 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.

[Jose]

> An infinite wing has no induced downwash.
>
> How is it supported?

It has downwash. There is upflow in front of the wing, and downwash
behind the wing. There is more downwash than upflow, this counteracts
mg of the wing.

There is no vortex at the wingtips (no wingtips) but there is a vortex
in the direction of flight.

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

[00:00:00Hg]

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.

[Jose]

> 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.

[Jose]

> 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.

[Jose]

> 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.

[Jose]

> I think you'll agree:
>
> 1) It is useful to have a specific term for each variable that needs
> to be considered.
>
> 2) It is condusing to have two or more words for the same variable.

This is true. I think that in this case we should use more basic terms
(such as "local downflow velocity") for the velocity of the downflow
near the wing.

>> The upper
>> atmosphere =is= (slightly) depleted by the flight.
>
> And then returns to ambient pressure, at a slightly higher
> termperature, after the wing has passed, right?

Wrong. It does not return to ambient pressure (regardless of
temperature) until the airplane lands.

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

[Jose]

> After the airplane lands how does the air return to ambient
> pressure, at a higher temperature, (or at any rate at a not
> lower temperature) without any upward flow?

After the airplane lands, the downflow from the wings ends. The
downflow from the wings is what keeps the air above rarified, and the
air below squished. Once that is removed, the pressure below can
relieve itself by having the air flow upwards a sufficient amount.

AT THAT POINT, the air has returned to its normal pressure distribution,
(albeit slightly warmer). But so long as the airplane is flying, it has
to be supported by the air, and the pressure below the wing is greater
than (and the pressure above the wing is less than) it would have been
absent the continuous downflow induced by the wing in flight.

Similarly, for a fan in a closed room, the air pressure on one side of
the room will remain higher than the pressure on the other side, until
the fan is turned off. Then the air will spring back.

And similarly, when you sit on a chair, it deforms slightly. When you
get up, it springs back (unless you broke it!)

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

[Jose]

> When you say a)"rarified" and b)"squished" do you mean...

Ultimately I mean lower pressure and higher pressure. Specifically I am
talking about the extra pressure that is distributed across the entire
earth's surface while the airplane is flying. Two things are being
disscussed here - what keeps the airplane up, and what is ultimately
supporting it; I was addressing the latter.

> By "flow upwards a sufficient amount" I presume you are
> thinking about some sort of density variation changing the
> volume of air below the wing that is later released?

Yes, though I am not talking just about the air immediately below the
wing, but of all the air that is pressing against the earth. The thing
that prevented this (net) upward flow while the plane is flying is the
continued downflow from the wing. Once that stops, the air can spring back.

> This concept of "springing back" implies both pressure and
> density changes. While it's true that gas does change
> volume and density when pressure is applied, when studying
> the phenomenon of lift at subsonic speeds, we usually ignore
> the density changes.

Yes, and that is a good approximation for some analyses. It does leave
something out though, and sometimes the thing that is left out is the
answer being sought. When I jump up, I push the earth down. This can
usually be ignored, but it is necessary to complete the analysis of all
the forces and their conservation. On a larger scale (the moon orbiting
the earth) it becomes significant.

> If this is a discussion about lift ( I apologize if it's not
> :-) and not just a pure discussion of the physics of a
> compressible gas, it's not clear to me why you would want to
> consider compressible effects. It's pretty universally
> agreed we can ignore them.

The contention is that there is =no= net downflow. That contention is
not true (although it's "pretty close"). The difference between "no"
and "almost no" is what holds the airplane up.

> But of course, it doesn't have to compress (at least not in
> theory) and when you sit on a rock, the extent to which it
> does compress is so small we can easily ignore it.

Well, actually it does have to compress. That's the source of the
force. Even a rock compresses. We can ignore it for most practical
purposes, but not when you are asking where the force comes from. And
that is the question being addressed.

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

[David CL Francis]

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

[Jose]

> I don't think leaving out density changes leaves out very
> much that's relevant to lift, but I have to admit to not
> being sure exactly what question this discussion is trying
> to answer, so I'll hold off any comment.

The question was whether Bernoulli (did I get it right?) supersedes
Newton. I maintain that, while Bernoulli's equations are very useful,
they obscure something Newtonian about the source of the force. If you
have lower pressure above and higher pressure below, you get lift. But
you can't get to that condition without throwing air down. Consider a
"wing" with a flat bottom. Define the JAOA to be the angle the =bottom=
makes with the airflow. In this case the JAOA is zero. The top of the
wing is an arc. The air has to go a longer way around the top of the
arc, so the conventional Bernoulli argument would be that there is lower
pressure above for this reason alone.

I don't think that kind of wing, in that configuration, would generate
any lift. If it does, there will be downwash. Increase the JAOA and
you certainly get downwash, and you will also get lift. The two will
balance.

The argument of "no net downwash" has to do with whether the air comes
back up. It does not (completely) until the airplane lands.

I will grant that, once the plane is flying over the earth, there will
be no net accumulation of air below - the attempt at accumulating the
air will be counteracted by the increased pressure (which is also
causing the upwash ahead). But enough air will already have been
accumulated below (and will remain accumulated below until the end of
the flight) so that that increased pressure will support the aircraft.

In the (silly) configuration where there is no earth, and no gravity,
there will be no pressure accumulation. The high pressure below will,
in addition to pushing air ahead up, will continue to push air below
down. This flow will dissipate, but not disappear. The aircraft cannot
be supported by the earth (like sitting on a rock), so it has to be
supported by downward thrust (downflow). As far as the wing is
concerned, this is what happens anyway. The wing sees rising air,
flings it down, and keeps going.

> Taken in its entirety, I'd say the [no net downflow idea] is false.
> Pressure differences hold the plane up. Force holds the plane up.
> Neither *requires* any net downflow.

Yes. But to get the plane flying there is some net downflow; enough to
increase the global pressure by (weight of airplane)/(area of earth).
When the plane lands, there is a net upflow to release this pressure.

Globally, because of the earth's surface, there is no net downflow
during steady state flight. (I believe that's the point you want me to
concede - I do concede that point). However, this is because of the
earth's surface. Locally, the wing is changing the vertical velocity of
the air it encounters. Locally, the wing is throwing air down. This
has the consequence (which the wing doesn't care about) that air rises
up to meet it, because the air density is mostly unchanged. But, that
tiny density change caused by the tiny pressure change is what
ultimately enforces the "no net downwash" because of the earth's surface.

> So if I understand this, you are saying that if air were
> incompressible, there would be no lift?

No. If air were truely incompressible, there would be no downflow at
all; the entire earth would be pushed away just as it does when I jump.

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

[Jose]

> 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.

[[email protected]]

Jose wrote:
> > I don't think leaving out density changes leaves out very
> > much that's relevant to lift, but I have to admit to not
> > being sure exactly what question this discussion is trying
> > to answer, so I'll hold off any comment.
>
> The question was whether Bernoulli (did I get it right?) supersedes
> Newton.

Maybe I mised something becuase I did not see that particular question
posed.


I maintain that, while Bernoulli's equations are very useful,
> they obscure something Newtonian about the source of the force. If you
> have lower pressure above and higher pressure below, you get lift. But
> you can't get to that condition without throwing air down.

No, you can throw it horizontally.

> ...
> Globally, because of the earth's surface, there is no net downflow
> during steady state flight. (I believe that's the point you want me to
> concede - I do concede that point).

Dunno about him, but that has been my point.
....
>
> > So if I understand this, you are saying that if air were
> > incompressible, there would be no lift?
>
> No. If air were truely incompressible, there would be no downflow at
> all; the entire earth would be pushed away just as it does when I jump.
>

If the Earth is pushed away, wouldn't that stretch out
the air molecules between the plane and the Earth decreasing
the pressure below the wing?

--

FF

[Jose]

> No, you can throw it horizontally.

.... where it throws the air in the way down to get it out of the way.
Why down? Because the wing is going up, and below is where the (new)
room is.

> If the Earth is pushed away, wouldn't that stretch out
> the air molecules between the plane and the Earth decreasing
> the pressure below the wing?

Not if the air is truly incompressible. There would be more air
molecules between the wing and the earth because the wing is up there
and not down here.

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

[Jose]

[snip]

I think we are pretty much in full agreement, and were merely using
different words, or stressing concepts that don't matter for the image
being discussed by the other.

When an infinite (or 2D) wing starts its takeoff roll, is there not a
vortex generated? Is not some air thrown down at the outset? (Granted
the vortex isn't at the wingtips and has a different axis, but air goes
down behind the wing and up ahead of it, which wasn't the case before
the takeoff roll).

Steady state, the wing keeps throwing air down but it keeps coming back
up (in front of it). I suppose I am using the term "throwing air down"
a bit more sloppily than you are taking it - I do mean it to be the same
as "imparting a downward acceleration". However, in all the pictures
I've seen, air does leave the wing with a downward velocity with a wing
in level flight. I just find the image to be more useful to me than the
"high pressure low pressure" because it better explains where the
pressure comes from.

> It still sounds like you are just saying that since some air
> is somehow held down so it can later "come back up" you are
> simply saying that there is some density increase for higher
> pressure air. While true to a tiny extent, this is
> irrelevant to the whole issue.

I'm saying it just to answer the (valid) question of "where the air
goes". Just enough of it gets scrunched up to hold the airplane up.
PV=nRT. In this case the V is big (and not really well defined), and
nRT is effectively constant.

> Even if the gravity does not
> affect the fluid, an underwater wing moved through the water
> at an AOA will apply a force to the fluid and cause it to
> move down. The longer you do this, the longer that force
> will accelerate the ball of fluid down.

Exactly. And this is the reason I hold to "the wing throws the air
down, the earth gets in the way".

> You don't need any compression to have pressure. Pressure
> is sufficient for lift. A perfectly incompressible fluid
> still has pressure.

Pressure is sufficient for lift. but a perfectly incompressible fluid
is an ideal that does not exist. The electrons get squished a little
harder; that supplies the force. Granted this is not important in the
calculation of what pressure does, and the distances involved are less
than miniscule, but it is important conceptually in seeing "just how
things work" on a basic level. "It goes =somewhere=."

I think we are basically in agreement with the physics, just disagree as
to what parts of the physics are important, because we are evaluating
importance differently.

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

[David CL Francis]

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

[David CL Francis]

On Thu, 9 Mar 2006 at 16:56:45 in message
>, Jose
> wrote:

>Pressure is sufficient for lift. but a perfectly incompressible fluid
>is an ideal that does not exist. The electrons get squished a little
>harder; that supplies the force. Granted this is not important in the
>calculation of what pressure does, and the distances involved are less
>than miniscule, but it is important conceptually in seeing "just how
>things work" on a basic level. "It goes =somewhere=."

No metal that is incompressible exists either. The point is lost about
the fundamentals of Bernoulli' theorem. Compressibility is not a part of
the simple theory and it applies just as much to water in pipes.

If water flows through a pipe and comes to a place where the pipe
narrows smoothly then the water speeds up as the mass flow through a
pipe is constant along its length. The pressure then drops as some of
the pressure energy is transformed into kinetic energy. Many measurement
devices depend on this as do carburettors. The keels and rudders of
boats generate lift in just the same way as wings.
--
David CL Francis

[Bush]

It was a challenge back in the early 1980's being thrust from a Cessna
402 and taking off, climbing, holding, and approaching in a 747 and
walking away from the landing however that's how we did it back then
it really flies like a big 172, you just need to be patient in the
turns, it takes time for the ailerons and roll spoilers to react
(ailerons, youv'e seen pictures) otherwise you get some flutter.

Bush


On Wed, 22 Feb 2006 06:15:59 -0600, Immanuel Goldstein
> wrote:

>The Impossibility of Flying Heavy Aircraft Without Training
>
>Nila Sagadevan | February 21 2006
>
>Nila Sagadevan is an aeronautical engineer and a qualified pilot of heavy aircraft.
>
>[...]
>
>What follows is an attempt to bury this myth once and for all, because I’ve
>heard this ludicrous explanation bandied about, ad nauseum, on the Internet and
>the TV networks—invariably by people who know nothing substantive about flight
>simulators, flying, or even airplanes.
>
>A common misconception non-pilots have about simulators is how “easy” it is to
>operate them. They are indeed relatively easy to operate if the objective is to
>make a few lazy turns and frolic about in the “open sky”. But if the intent is
>to execute any kind of a maneuver with even the least bit of precision, the task
>immediately becomes quite daunting. And if the aim is to navigate to a specific
>geographic location hundreds of miles away while flying at over 500 MPH, 30,000
>feet above the ground the challenges become virtually impossible for an
>untrained pilot.
>
>And this, precisely, is what the four hijacker pilots who could not fly a Cessna
>around an airport are alleged to have accomplished in multi-ton, high-speed
>commercial jets on 9/11.
>
>For a person not conversant with the practical complexities of pilotage, a
>modern flight simulator could present a terribly confusing and disorienting
>experience. These complex training devices are not even remotely similar to the
>video games one sees in amusement arcades, or even the software versions
>available for home computers.
>
>In order to operate a modern flight simulator with any level of skill, one has
>to not only be a decent pilot to begin with, but also a skilled instrument-rated
>one to boot — and be thoroughly familiar with the actual aircraft type the
>simulator represents, since the cockpit layouts vary between aircraft.
>
>The only flight domains where an arcade/PC-type game would even begin to
>approach the degree of visual realism of a modern professional flight simulator
>would be during the take-off and landing phases. During these phases, of course,
>one clearly sees the bright runway lights stretched out ahead, and even
>peripherally sees images of buildings, etc. moving past. Take-offs—even
>landings, to a certain degree—are relatively “easy”, because the pilot has
>visual reference cues that exist “outside” the cockpit.
>
>But once you’ve rotated, climbed out, and reached cruising altitude in a
>simulator (or real airplane), and find yourself en route to some distant
>destination (using sophisticated electronic navigation techniques), the
>situation changes drastically: the pilot loses virtually all external visual
>reference cues. S/he is left entirely at the mercy of an array of complex flight
>and navigation instruments to provide situational cues (altitude, heading,
>speed, attitude, etc.)
>
>In the case of a Boeing 757 or 767, the pilot would be faced with an EFIS
>(Electronic Flight Instrumentation System) panel comprised of six large
>multi-mode LCDs interspersed with clusters of assorted “hard” instruments. These
>displays process the raw aircraft system and flight data into an integrated
>picture of the aircraft situation, position and progress, not only in horizontal
>and vertical dimensions, but also with regard to time and speed as well. When
>flying “blind”, I.e., with no ground reference cues, it takes a highly skilled
>pilot to interpret, and then apply, this data intelligently. If one cannot
>translate this information quickly, precisely and accurately (and it takes an
>instrument-rated pilot to do so), one would have ZERO SITUATIONAL AWARENESS.
>I.e., the pilot wouldn’t have a clue where s/he was in relation to the earth.
>Flight under such conditions is referred to as “IFR”, or Instrument Flight Rules.
>
>And IFR Rule #1: Never take your eyes off your instruments, because that’s all
>you have!
>
>The corollary to Rule #1: If you can’t read the instruments in a quick, smooth,
>disciplined, scan, you’re as good as dead. Accident records from around the
>world are replete with reports of any number of good pilots — I.e., professional
>instrument-rated pilots — who ‘bought the farm’ because they screwed up while
>flying in IFR conditions.
>
>Let me place this in the context of the 9/11 hijacker-pilots. These men were
>repeatedly deemed incompetent to solo a simple Cessna-172 — an elementary
>exercise that involves flying this little trainer once around the patch on a
>sunny day. A student’s first solo flight involves a simple circuit: take-off,
>followed by four gentle left turns ending with a landing back on the runway.
>This is as basic as flying can possibly get.
>
>Not one of the hijackers was deemed fit to perform this most elementary exercise
>by himself.
>
>In fact, here’s what their flight instructors had to say about the aptitude of
>these budding aviators:
>
>Mohammed Atta: "His attention span was zero."
>
>Khalid Al-Mihdhar: "We didn't kick him out, but he didn't live up to our standards."
>
>Marwan Al-Shehhi: “He was dropped because of his limited English and
>incompetence at the controls.”
>
>Salem Al-Hazmi: "We advised him to quit after two lessons.”
>
>Hani Hanjour: "His English was horrible, and his mechanical skills were even
>worse. It was like he had hardly even ever driven a car. I’m still to this day
>amazed that he could have flown into the Pentagon. He could not fly at all.”
>
>Now let’s take a look at American Airlines Flight 77. Passenger/hijacker Hani
>Hanjour rises from his seat midway through the flight, viciously fights his way
>into the cockpit with his cohorts, overpowers Captain Charles F. Burlingame and
>First Officer David Charlebois, and somehow manages to toss them out of the
>cockpit (for starters, very difficult to achieve in a cramped environment
>without inadvertently impacting the yoke and thereby disengaging the autopilot).
>One would correctly presume that this would present considerable difficulties to
>a little guy with a box cutter—Burlingame was a tough, burly, ex-Vietnam F4
>fighter jock who had flown over 100 combat missions. Every pilot who knows him
>says that rather than politely hand over the controls, Burlingame would have
>instantly rolled the plane on its back so that Hanjour would have broken his
>neck when he hit the floor. But let’s ignore this almost natural reaction
>expected of a fighter pilot and proceed with this charade.
>
>Nonetheless, imagine that Hanjour overpowers the flight deck crew, removes them
>from the cockpit and takes his position in the captain’s seat. Although weather
>reports state this was not the case, let’s say Hanjour was lucky enough to
>experience a perfect CAVU day (Ceiling And Visibility Unlimited). If Hanjour
>looked straight ahead through the windshield, or off to his left at the ground,
>at best he would see, 35,000 feet -- 7 miles -- below him, a murky
>brownish-grey-green landscape, virtually devoid of surface detail, while the
>aircraft he was now piloting was moving along, almost imperceptibly and in eerie
>silence, at around 500 MPH (about 750 feet every second).
>
>In a real-world scenario (and given the reported weather conditions that day),
>he would likely have seen clouds below him completely obscuring the ground he
>was traversing. With this kind of “situational non-awareness”, Hanjour might as
>well have been flying over Argentina, Russia, or Japan—he wouldn’t have had a
>clue as to where, precisely, he was.
>
>After a few seconds (at 750 ft/sec), Hanjour would figure out there’s little
>point in looking outside—there’s nothing there to give him any real visual cues.
>For a man who had previously wrestled with little Cessnas, following freeways
>and railroad tracks (and always in the comforting presence of an instructor),
>this would have been a strange, eerily unsettling environment indeed.
>
>Seeing nothing outside, Mr. Hanjour would be forced to divert his attention to
>his instrument panel, where he’d be faced with a bewildering array of
>instruments. He would then have to very quickly interpret his heading, ground
>track, altitude, and airspeed information on the displays before he could even
>figure out where in the world he was, much less where the Pentagon was located
>in relation to his position!
>
>After all, before he can crash into a target, he has to first find the target.
>
>It is very difficult to explain this scenario, of an utter lack of ground
>reference, to non-pilots; but let it suffice to say that for these incompetent
>hijacker non-pilots to even consider grappling with such a daunting task would
>have been utterly overwhelming. They wouldn’t have known where to begin.
>
>But, for the sake of discussion let’s stretch things beyond all plausibility and
>say that Hanjour—whose flight instructor claimed “couldn’t fly at all”—somehow
>managed to figure out their exact position on the American landscape in relation
>to their intended target as they traversed the earth at a speed five times
>faster than they had ever flown by themselves before.
>
>Once he had determined exactly where he was, he would need to figure out where
>the Pentagon was located in relation to his rapidly-changing position. He would
>then need to plot a course to his target (one he cannot see with his
>eyes—remember, our ace is flying solely on instruments).
>
>In order to perform this bit of electronic navigation, he would have to be very
>familiar with IFR procedures. None of these chaps even knew what a navigational
>chart looked like, much less how to how to plug information into flight
>management computers (FMC) and engage LNAV (lateral navigation automated mode).
>If one is to believe the official story, all of this was supposedly accomplished
>by raw student pilots while flying blind at 500 MPH over unfamiliar (and
>practically invisible) terrain, using complex methodologies and employing
>sophisticated instruments.
>
>To get around this little problem, the official storyline suggests these men
>manually flew their aircraft to their respective targets (NB: This still
>wouldn’t relieve them of the burden of navigation). But let’s assume Hanjour
>disengaged the autopilot and auto-throttle and hand-flew the aircraft to its
>intended—and invisible—target on instruments alone until such time as he could
>get a visual fix. This would have necessitated him to fly back across West
>Virginia and Virginia to Washington DC. (This portion of Flight 77’s flight path
>cannot be corroborated by any radar evidence that exists, because the aircraft
>is said to have suddenly disappeared from radar screens over Ohio, but let’s not
>mull over that little point.)
>
>According to FAA radar controllers, “Flight 77” then suddenly pops up over
>Washington DC and executes an incredibly precise diving turn at a rate of 360
>degrees/minute while descending at 3,500 ft/min, at the end of which “Hanjour”
>allegedly levels out at ground level. Oh, I almost forgot: He also had the
>presence of mind to turn off the transponder in the middle of this incredibly
>difficult maneuver (one of his instructors later commented the hapless fellow
>couldn’t have spelt the word if his life depended on it).
>
>The maneuver was in fact so precisely executed that the air traffic controllers
>at Dulles refused to believe the blip on their screen was a commercial airliner.
>Danielle O’Brian, one of the air traffic controllers at Dulles who reported
>seeing the aircraft at 9:25 said, “The speed, the maneuverability, the way that
>he turned, we all thought in the radar room, all of us experienced air traffic
>controllers, that that was a military plane.”
>
>And then, all of a sudden we have magic. Voila! Hanjour finds the Pentagon
>sitting squarely in his sights right before him.
>
>But even that wasn’t good enough for this fanatic Muslim kamikaze pilot. You
>see, he found that his “missile” was heading towards one of the most densely
>populated wings of the Pentagon—and one occupied by top military brass,
>including the Secretary of Defense, Rumsfeld. Presumably in order to save these
>men’s lives, he then executes a sweeping 270-degree turn and approaches the
>building from the opposite direction and aligns himself with the only wing of
>the Pentagon that was virtually uninhabited due to extensive renovations that
>were underway (there were some 120 civilians construction workers in that wing
>who were killed; their work included blast-proofing the outside wall of that wing).
>
>I shan’t get into the aerodynamic impossibility of flying a large commercial
>jetliner 20 feet above the ground at over 400 MPH. A discussion on ground effect
>energy, tip vortex compression, downwash sheet reaction, wake turbulence, and
>jetblast effects are beyond the scope of this article (the 100,000-lb jetblast
>alone would have blown whole semi-trucks off the roads.)
>
>Let it suffice to say that it is physically impossible to fly a 200,000-lb
>airliner 20 feet above the ground at 400 MPH.
>
>The author, a pilot and aeronautical engineer, challenges any pilot in the world
>to do so in any large high-speed aircraft that has a relatively low wing-loading
>(such as a commercial jet). I.e., to fly the craft at 400 MPH, 20 feet above
>ground in a flat trajectory over a distance of one mile.
>
>Why the stipulation of 20 feet and a mile? There were several street light poles
>located up to a mile away from the Pentagon that were snapped-off by the
>incoming aircraft; this suggests a low, flat trajectory during the final
>pre-impact approach phase. Further, it is known that the craft impacted the
>Pentagon’s ground floor. For purposes of reference: If a 757 were placed on the
>ground on its engine nacelles (I.e., gear retracted as in flight profile), its
>nose would be almost 20 above the ground! Ergo, for the aircraft to impact the
>ground floor of the Pentagon, Hanjour would have needed to have flown in with
>the engines buried 10-feet deep in the Pentagon lawn. Some pilot.
>
>At any rate, why is such ultra-low-level flight aerodynamically impossible?
>Because the reactive force of the hugely powerful downwash sheet, coupled with
>the compressibility effects of the tip vortices, simply will not allow the
>aircraft to get any lower to the ground than approximately one half the distance
>of its wingspan—until speed is drastically reduced, which, of course, is what
>happens during normal landings.
>
>In other words, if this were a Boeing 757 as reported, the plane could not have
>been flown below about 60 feet above ground at 400 MPH. (Such a maneuver is
>entirely within the performance envelope of aircraft with high wing-loadings,
>such as ground-attack fighters, the B1-B bomber, and Cruise missiles—and the
>Global Hawk.)
>
>The very same navigational challenges mentioned above would have faced the
>pilots who flew the two 767s into the Twin Towers, in that they, too, would have
>had to have first found their targets. Again, these chaps, too, miraculously
>found themselves spot on course. And again, their “final approach” maneuvers at
>over 500 MPH are simply far too incredible to have been executed by pilots who
>could not solo basic training aircraft.
>
>Conclusion
>The writers of the official storyline expect us to believe, that once the flight
>deck crews had been overpowered, and the hijackers “took control” of the various
>aircraft, their intended targets suddenly popped up in their windshields as they
>would have in some arcade game, and all that these fellows would have had to do
>was simply aim their airplanes at the buildings and fly into them. Most people
>who have been exposed only to the official storyline have never been on the
>flight deck of an airliner at altitude and looked at the outside world; if they
>had, they’d realize the absurdity of this kind of reasoning.
>
>In reality, a clueless non-pilot would encounter almost insurmountable
>difficulties in attempting to navigate and fly a 200,000-lb airliner into a
>building located on the ground, 7 miles below and hundreds of miles away and out
>of sight, and in an unknown direction, while flying at over 500 MPH — and all
>this under extremely stressful circumstances.
>
>Complete text:
><http://physics911.net/sagadevan.htm>

[John Doe]

Hmmm one has to wonder just where he is going with this story?!
So either the hijackers were better pilots than this story would imply or,
they had help from someone else!
anyway that is the way "I" look at it. We may never know the "true" story of
9/11/01




> On Wed, 22 Feb 2006 06:15:59 -0600, Immanuel Goldstein
> > wrote:
>
> >The Impossibility of Flying Heavy Aircraft Without Training
> >
> >Nila Sagadevan | February 21 2006
> >
> >Nila Sagadevan is an aeronautical engineer and a qualified pilot of heavy
aircraft.
> >
> >[...]
> >
> >What follows is an attempt to bury this myth once and for all, because
I've
> >heard this ludicrous explanation bandied about, ad nauseum, on the
Internet and
> >the TV networks-invariably by people who know nothing substantive about
flight
> >simulators, flying, or even airplanes.
> >
> >A common misconception non-pilots have about simulators is how "easy" it
is to
> >operate them. They are indeed relatively easy to operate if the objective
is to
> >make a few lazy turns and frolic about in the "open sky". But if the
intent is
> >to execute any kind of a maneuver with even the least bit of precision,
the task
> >immediately becomes quite daunting. And if the aim is to navigate to a
specific
> >geographic location hundreds of miles away while flying at over 500 MPH,
30,000
> >feet above the ground the challenges become virtually impossible for an
> >untrained pilot.
> >
> >And this, precisely, is what the four hijacker pilots who could not fly a
Cessna
> >around an airport are alleged to have accomplished in multi-ton,
high-speed
> >commercial jets on 9/11.
> >
> >For a person not conversant with the practical complexities of pilotage,
a
> >modern flight simulator could present a terribly confusing and
disorienting
> >experience. These complex training devices are not even remotely similar
to the
> >video games one sees in amusement arcades, or even the software versions
> >available for home computers.
> >
> >In order to operate a modern flight simulator with any level of skill,
one has
> >to not only be a decent pilot to begin with, but also a skilled
instrument-rated
> >one to boot - and be thoroughly familiar with the actual aircraft type
the
> >simulator represents, since the cockpit layouts vary between aircraft.
> >
> >The only flight domains where an arcade/PC-type game would even begin to
> >approach the degree of visual realism of a modern professional flight
simulator
> >would be during the take-off and landing phases. During these phases, of
course,
> >one clearly sees the bright runway lights stretched out ahead, and even
> >peripherally sees images of buildings, etc. moving past. Take-offs-even
> >landings, to a certain degree-are relatively "easy", because the pilot
has
> >visual reference cues that exist "outside" the cockpit.
> >
> >But once you've rotated, climbed out, and reached cruising altitude in a
> >simulator (or real airplane), and find yourself en route to some distant
> >destination (using sophisticated electronic navigation techniques), the
> >situation changes drastically: the pilot loses virtually all external
visual
> >reference cues. S/he is left entirely at the mercy of an array of complex
flight
> >and navigation instruments to provide situational cues (altitude,
heading,
> >speed, attitude, etc.)
> >
> >In the case of a Boeing 757 or 767, the pilot would be faced with an EFIS
> >(Electronic Flight Instrumentation System) panel comprised of six large
> >multi-mode LCDs interspersed with clusters of assorted "hard"
instruments. These
> >displays process the raw aircraft system and flight data into an
integrated
> >picture of the aircraft situation, position and progress, not only in
horizontal
> >and vertical dimensions, but also with regard to time and speed as well.
When
> >flying "blind", I.e., with no ground reference cues, it takes a highly
skilled
> >pilot to interpret, and then apply, this data intelligently. If one
cannot
> >translate this information quickly, precisely and accurately (and it
takes an
> >instrument-rated pilot to do so), one would have ZERO SITUATIONAL
AWARENESS.
> >I.e., the pilot wouldn't have a clue where s/he was in relation to the
earth.
> >Flight under such conditions is referred to as "IFR", or Instrument
Flight Rules.
> >
> >And IFR Rule #1: Never take your eyes off your instruments, because
that's all
> >you have!
> >
> >The corollary to Rule #1: If you can't read the instruments in a quick,
smooth,
> >disciplined, scan, you're as good as dead. Accident records from around
the
> >world are replete with reports of any number of good pilots - I.e.,
professional
> >instrument-rated pilots - who 'bought the farm' because they screwed up
while
> >flying in IFR conditions.
> >
> >Let me place this in the context of the 9/11 hijacker-pilots. These men
were
> >repeatedly deemed incompetent to solo a simple Cessna-172 - an elementary
> >exercise that involves flying this little trainer once around the patch
on a
> >sunny day. A student's first solo flight involves a simple circuit:
take-off,
> >followed by four gentle left turns ending with a landing back on the
runway.
> >This is as basic as flying can possibly get.
> >
> >Not one of the hijackers was deemed fit to perform this most elementary
exercise
> >by himself.
> >
> >In fact, here's what their flight instructors had to say about the
aptitude of
> >these budding aviators:
> >
> >Mohammed Atta: "His attention span was zero."
> >
> >Khalid Al-Mihdhar: "We didn't kick him out, but he didn't live up to our
standards."
> >
> >Marwan Al-Shehhi: "He was dropped because of his limited English and
> >incompetence at the controls."
> >
> >Salem Al-Hazmi: "We advised him to quit after two lessons."
> >
> >Hani Hanjour: "His English was horrible, and his mechanical skills were
even
> >worse. It was like he had hardly even ever driven a car. I'm still to
this day
> >amazed that he could have flown into the Pentagon. He could not fly at
all."
> >
> >Now let's take a look at American Airlines Flight 77. Passenger/hijacker
Hani
> >Hanjour rises from his seat midway through the flight, viciously fights
his way
> >into the cockpit with his cohorts, overpowers Captain Charles F.
Burlingame and
> >First Officer David Charlebois, and somehow manages to toss them out of
the
> >cockpit (for starters, very difficult to achieve in a cramped environment
> >without inadvertently impacting the yoke and thereby disengaging the
autopilot).
> >One would correctly presume that this would present considerable
difficulties to
> >a little guy with a box cutter-Burlingame was a tough, burly, ex-Vietnam
F4
> >fighter jock who had flown over 100 combat missions. Every pilot who
knows him
> >says that rather than politely hand over the controls, Burlingame would
have
> >instantly rolled the plane on its back so that Hanjour would have broken
his
> >neck when he hit the floor. But let's ignore this almost natural reaction
> >expected of a fighter pilot and proceed with this charade.
> >
> >Nonetheless, imagine that Hanjour overpowers the flight deck crew,
removes them
> >from the cockpit and takes his position in the captain's seat. Although
weather
> >reports state this was not the case, let's say Hanjour was lucky enough
to
> >experience a perfect CAVU day (Ceiling And Visibility Unlimited). If
Hanjour
> >looked straight ahead through the windshield, or off to his left at the
ground,
> >at best he would see, 35,000 feet -- 7 miles -- below him, a murky
> >brownish-grey-green landscape, virtually devoid of surface detail, while
the
> >aircraft he was now piloting was moving along, almost imperceptibly and
in eerie
> >silence, at around 500 MPH (about 750 feet every second).
> >
> >In a real-world scenario (and given the reported weather conditions that
day),
> >he would likely have seen clouds below him completely obscuring the
ground he
> >was traversing. With this kind of "situational non-awareness", Hanjour
might as
> >well have been flying over Argentina, Russia, or Japan-he wouldn't have
had a
> >clue as to where, precisely, he was.
> >
> >After a few seconds (at 750 ft/sec), Hanjour would figure out there's
little
> >point in looking outside-there's nothing there to give him any real
visual cues.
> >For a man who had previously wrestled with little Cessnas, following
freeways
> >and railroad tracks (and always in the comforting presence of an
instructor),
> >this would have been a strange, eerily unsettling environment indeed.
> >
> >Seeing nothing outside, Mr. Hanjour would be forced to divert his
attention to
> >his instrument panel, where he'd be faced with a bewildering array of
> >instruments. He would then have to very quickly interpret his heading,
ground
> >track, altitude, and airspeed information on the displays before he could
even
> >figure out where in the world he was, much less where the Pentagon was
located
> >in relation to his position!
> >
> >After all, before he can crash into a target, he has to first find the
target.
> >
> >It is very difficult to explain this scenario, of an utter lack of ground
> >reference, to non-pilots; but let it suffice to say that for these
incompetent
> >hijacker non-pilots to even consider grappling with such a daunting task
would
> >have been utterly overwhelming. They wouldn't have known where to begin.
> >
> >But, for the sake of discussion let's stretch things beyond all
plausibility and
> >say that Hanjour-whose flight instructor claimed "couldn't fly at
all"-somehow
> >managed to figure out their exact position on the American landscape in
relation
> >to their intended target as they traversed the earth at a speed five
times
> >faster than they had ever flown by themselves before.
> >
> >Once he had determined exactly where he was, he would need to figure out
where
> >the Pentagon was located in relation to his rapidly-changing position. He
would
> >then need to plot a course to his target (one he cannot see with his
> >eyes-remember, our ace is flying solely on instruments).
> >
> >In order to perform this bit of electronic navigation, he would have to
be very
> >familiar with IFR procedures. None of these chaps even knew what a
navigational
> >chart looked like, much less how to how to plug information into flight
> >management computers (FMC) and engage LNAV (lateral navigation automated
mode).
> >If one is to believe the official story, all of this was supposedly
accomplished
> >by raw student pilots while flying blind at 500 MPH over unfamiliar (and
> >practically invisible) terrain, using complex methodologies and employing
> >sophisticated instruments.
> >
> >To get around this little problem, the official storyline suggests these
men
> >manually flew their aircraft to their respective targets (NB: This still
> >wouldn't relieve them of the burden of navigation). But let's assume
Hanjour
> >disengaged the autopilot and auto-throttle and hand-flew the aircraft to
its
> >intended-and invisible-target on instruments alone until such time as he
could
> >get a visual fix. This would have necessitated him to fly back across
West
> >Virginia and Virginia to Washington DC. (This portion of Flight 77's
flight path
> >cannot be corroborated by any radar evidence that exists, because the
aircraft
> >is said to have suddenly disappeared from radar screens over Ohio, but
let's not
> >mull over that little point.)
> >
> >According to FAA radar controllers, "Flight 77" then suddenly pops up
over
> >Washington DC and executes an incredibly precise diving turn at a rate of
360
> >degrees/minute while descending at 3,500 ft/min, at the end of which
"Hanjour"
> >allegedly levels out at ground level. Oh, I almost forgot: He also had
the
> >presence of mind to turn off the transponder in the middle of this
incredibly
> >difficult maneuver (one of his instructors later commented the hapless
fellow
> >couldn't have spelt the word if his life depended on it).
> >
> >The maneuver was in fact so precisely executed that the air traffic
controllers
> >at Dulles refused to believe the blip on their screen was a commercial
airliner.
> >Danielle O'Brian, one of the air traffic controllers at Dulles who
reported
> >seeing the aircraft at 9:25 said, "The speed, the maneuverability, the
way that
> >he turned, we all thought in the radar room, all of us experienced air
traffic
> >controllers, that that was a military plane."
> >
> >And then, all of a sudden we have magic. Voila! Hanjour finds the
Pentagon
> >sitting squarely in his sights right before him.
> >
> >But even that wasn't good enough for this fanatic Muslim kamikaze pilot.
You
> >see, he found that his "missile" was heading towards one of the most
densely
> >populated wings of the Pentagon-and one occupied by top military brass,
> >including the Secretary of Defense, Rumsfeld. Presumably in order to save
these
> >men's lives, he then executes a sweeping 270-degree turn and approaches
the
> >building from the opposite direction and aligns himself with the only
wing of
> >the Pentagon that was virtually uninhabited due to extensive renovations
that
> >were underway (there were some 120 civilians construction workers in that
wing
> >who were killed; their work included blast-proofing the outside wall of
that wing).
> >
> >I shan't get into the aerodynamic impossibility of flying a large
commercial
> >jetliner 20 feet above the ground at over 400 MPH. A discussion on ground
effect
> >energy, tip vortex compression, downwash sheet reaction, wake turbulence,
and
> >jetblast effects are beyond the scope of this article (the 100,000-lb
jetblast
> >alone would have blown whole semi-trucks off the roads.)
> >
> >Let it suffice to say that it is physically impossible to fly a
200,000-lb
> >airliner 20 feet above the ground at 400 MPH.
> >
> >The author, a pilot and aeronautical engineer, challenges any pilot in
the world
> >to do so in any large high-speed aircraft that has a relatively low
wing-loading
> >(such as a commercial jet). I.e., to fly the craft at 400 MPH, 20 feet
above
> >ground in a flat trajectory over a distance of one mile.
> >
> >Why the stipulation of 20 feet and a mile? There were several street
light poles
> >located up to a mile away from the Pentagon that were snapped-off by the
> >incoming aircraft; this suggests a low, flat trajectory during the final
> >pre-impact approach phase. Further, it is known that the craft impacted
the
> >Pentagon's ground floor. For purposes of reference: If a 757 were placed
on the
> >ground on its engine nacelles (I.e., gear retracted as in flight
profile), its
> >nose would be almost 20 above the ground! Ergo, for the aircraft to
impact the
> >ground floor of the Pentagon, Hanjour would have needed to have flown in
with
> >the engines buried 10-feet deep in the Pentagon lawn. Some pilot.
> >
> >At any rate, why is such ultra-low-level flight aerodynamically
impossible?
> >Because the reactive force of the hugely powerful downwash sheet, coupled
with
> >the compressibility effects of the tip vortices, simply will not allow
the
> >aircraft to get any lower to the ground than approximately one half the
distance
> >of its wingspan-until speed is drastically reduced, which, of course, is
what
> >happens during normal landings.
> >
> >In other words, if this were a Boeing 757 as reported, the plane could
not have
> >been flown below about 60 feet above ground at 400 MPH. (Such a maneuver
is
> >entirely within the performance envelope of aircraft with high
wing-loadings,
> >such as ground-attack fighters, the B1-B bomber, and Cruise missiles-and
the
> >Global Hawk.)
> >
> >The very same navigational challenges mentioned above would have faced
the
> >pilots who flew the two 767s into the Twin Towers, in that they, too,
would have
> >had to have first found their targets. Again, these chaps, too,
miraculously
> >found themselves spot on course. And again, their "final approach" maneuv
ers at
> >over 500 MPH are simply far too incredible to have been executed by
pilots who
> >could not solo basic training aircraft.
> >
> >Conclusion
> >The writers of the official storyline expect us to believe, that once the
flight
> >deck crews had been overpowered, and the hijackers "took control" of the
various
> >aircraft, their intended targets suddenly popped up in their windshields
as they
> >would have in some arcade game, and all that these fellows would have had
to do
> >was simply aim their airplanes at the buildings and fly into them. Most
people
> >who have been exposed only to the official storyline have never been on
the
> >flight deck of an airliner at altitude and looked at the outside world;
if they
> >had, they'd realize the absurdity of this kind of reasoning.
> >
> >In reality, a clueless non-pilot would encounter almost insurmountable
> >difficulties in attempting to navigate and fly a 200,000-lb airliner into
a
> >building located on the ground, 7 miles below and hundreds of miles away
and out
> >of sight, and in an unknown direction, while flying at over 500 MPH - and
all
> >this under extremely stressful circumstances.
> >
> >Complete text:
> ><http://physics911.net/sagadevan.htm>
>

[Orval Fairbairn]

In article <M2%Rf.145850$H%4.115608@pd7tw2no>,
"John Doe" > wrote:

> Hmmm one has to wonder just where he is going with this story?!
> So either the hijackers were better pilots than this story would imply or,
> they had help from someone else!
> anyway that is the way "I" look at it. We may never know the "true" story of
> 9/11/01

*OR* -- it didn't take as much skill as the "9/11 Truth Movement" would
have us believe. I go with this one, talking from 46 years of pilot
experience.