The Impossibility of Flying Heavy Aircraft Without Training

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

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.

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 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

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.

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

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."

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."

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