The Impossibility of Flying Heavy Aircraft Without Training

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

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


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

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

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

--

FF

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

mv

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

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

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

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

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

Momentum is conserved in elastic collisions

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

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

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

Jose
--
Money: what you need when you run out of brains.
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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

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.

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.

In article . com,
wrote:

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

The flying wing has some horizontal momentum which is secondary here,

How much?

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

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

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


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

You are incorrect. Momentum is *always* conserved.


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

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

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

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

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

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

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

So:

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

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

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

The air starts moving downward with a certain velocity.

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

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

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

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

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

In article . com,
wrote:

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

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


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

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

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

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


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

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

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

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

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

In article ,
Jose wrote:

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

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

Jose

You assume correctly. g

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