The Impossibility of Flying Heavy Aircraft Without Training

In article ,
Jose wrote:

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

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

Jose

You assume correctly. g

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

"Jose" wrote

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

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

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


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


Dallas

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


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

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


Dallas

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

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



Dallas

Alan Baker wrote:
In article . 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.

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



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

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

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

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


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

Yes, no question about weight being balanced by lift.


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

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


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

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

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


So:

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

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

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


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

The air starts moving downward with a certain velocity.

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


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


Interesting.

--

FF

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

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

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

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

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

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

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

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

Jose wrote:
Do you agree that the net momentum transfered to the Earth by the
air molecules is equal and opposite to the net momentum transferred
to the wing by the air molecules?

Yes.

Do you agree, therefor that there is no net momentum transfered to
the air?

Overall, yes. Similarly, there is no net momentum transferred to the
basketball when it is being used to support a (very fast) dribbler. But
that is not to say that there is no momentum transfer. The basketball
certainly moves around. I do agree that the net overall is zero. The
air does not pile up permanently.

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


At which ponit the Earth throws the air molecule back up so that the
net momemtum transferred to the air molecule is zero (averaged over
the entire atmosphere)

Yes.

[it hits the wing on the way up]
Which again transferes an equal and opposite momentum to the
molecule which again is transferrred to the Earth leaving no net
transfer of momentum to the air.

Yes.

Overall, there is no net (or "permanent") transfer of momentum to the
air. The air is an intermediary, keeping the wing and the earth apart.
There is certainly =energy= transfer to the air (mv^2/2), and there is
a lot of momentum transfer =back=and=forth= with the air, but I will
agree that the net is zero. The air is sort of a catalyst - ending up
unchanged as it transfers momentum to the earth and then transfers it
back from the earth to the wing.

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


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


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

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

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


===

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

Concur?

No.

--

FF

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

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

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

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


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


Yes. The fan continues to do work.

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

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

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


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

But we do presume there is still gravity.


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

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


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


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

--

FF

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

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

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


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

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

How is it conserved at the air/airplane collison?

--

FF