The Impossibility of Flying Heavy Aircraft Without Training

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.
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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

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.