The Impossibility of Flying Heavy Aircraft Without Training

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

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

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

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


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


Yes. The fan continues to do work.

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

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

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


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

But we do presume there is still gravity.


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

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


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


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

--

FF

Jose wrote:
How is momentum conserved when a cue ball hits a nerf ball?

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

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


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

...by the air's eventual collision with the earth.

How is it conserved at the air/airplane collison?

--

FF

Jose wrote:
The flying wing has some horizontal momentum which is secondary here,
How much?

mv

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

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

Yes, and what form does that energy take?

Heat.

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

Yes.

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

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


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

More importantly we disagree on what causes lift.

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

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

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

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

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

--

FF

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.

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

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

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

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

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

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

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

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

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

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

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

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

He uses energy to impart momentum to the basketball

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

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

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

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

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

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

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

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

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

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

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

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

"Dallas" wrote in message
nk.net...

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

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

WOW!! What a fruticake!!

Jose wrote:
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