The Impossibility of Flying Heavy Aircraft Without Training

Jose wrote:
I think you are refering to Newton's third law, often stated as: "For
every action there is an equal and opposite reaction."

Yes.

For an aircraft in level flight, the upwards acceleration due to lift
is counterbalanced by the downward acceleration due to gravity.
This satisfies Newton's third law.

Yes.

For a wing in level flight, the vertical component of momentum is
zero.

No.

Please show us your arithmetic. Suppose a 1500 lb airplane is
flying horizontally at 120 mph at 5000 feet above MSL. What
are the vertical and horizontal components of the momentum
of that aircraft?


That is, on a microscopic scale, no. The wing is constantly
freefalling, then being bounced back up by impact with air molecules.
Averaged over all the molecules, yes, the net is zero (the wing flies)
but on a microscopic scale, the wing is in constant brownian motion.
This implies momentum transfer, and following the momentum on a
microscopic scale is instructive.


OK, show us your arithmetic.

The wing imparts
as much upward momentum to the air as it does downward momentum.

This is where I disagree. Upward momentum gets imparted, but not
(directly) by the wing. Rather, it is imparted by the ground, mediated
through other air molecules.

The ground is stationary. How does the stationary ground impart
momentum to anything?

Of course this wouldn't happen if the wing
didn't pass through and throw the air down to begin with, but the ground
is what ultimately imparts the upwards momentum.

The pressure differential through the wing, from bottom to top,
integrated
over the wing area, provides an upward force for a wing in level
flight.

That's the shortcut. Where does this pressure differential come from -

Bernouli effect.

that is the question.

The downwash behind the aircraft, which is counterbalanced by a more
diffuse upwash around it, is real but not relevent to the issue of
lift.

I disagree here too. It's important in seeing the entire picture.

Well, yes it is part of the entire picture. Its just not relevent to
the
issue of lift, which is only part of the picture.

--

FF

OK, show us your arithmetic.

First, do you agree that air is made of individual molecules separated
by a lot of space compared to the size of the molecules themselves?
Then do you accept that a wing is in freefall during all the (very brief
but very numerous) time in between molecular collisions? (If not, what
holds it up when it is not in contact with any air molecules?)

If so, then during the time it is in freefall, it acquires a downward
velocity. Small, no doubt, but nonzero. The next molecular impact
pushes it back up. On the average they will sum to a net zero vertical
motion. Is this the arithmetic you want to see?

The ground is stationary. How does the stationary ground impart
momentum to anything?

The ground is not stationary. Like the wing, the ground is jiggling
around in brownian motion. Such motion is greatly overwhelmed in
quantity by other things, but it is nonzero. Gravity pulls the ground
towards the airplane just as strongly as it pulls the airplane towards
the ground. This is the same effect as the one that gives high tides on
the side of the earth that is away from the moon.

Where does this pressure differential come from -

Bernouli effect.

That's the shortcut. Where does the Bernoulli effect come from - on a
molecular level? That's what I'm addressing. The Bernoulli effect is a
shortcut for doing the calculation in bulk (where it makes the most
sense if you want a numerical answer) but it all comes from molecular
collisions.

Its just not relevent to the
issue of lift, which is only part of the picture.

We disagree here. Both explanations are true as far as they go, but it
is important to see just how far they go (or don't go). The Beruoulli
effect does not explain, for example, how the earth ultimately supports
the aircraft, nor how the upwash starts (for example, suppose there were
a vertical column of vacuum separated from the air by a very strong
piece of cellophane. A wing travels through the vacuum and penetrates
this cellophane. The air behind the cellophane does not rise up to meet
the wing - it has no idea there's a wing coming. Once the wing has
entered the air, that rising motion will start, but why?

That's the question to which I am applying my molecular model.

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

Jose wrote:
fredfighter wrote:

Please show us your arithmetic. Suppose a 1500 lb airplane is
flying horizontally at 120 mph at 5000 feet above MSL. What
are the vertical and horizontal components of the momentum
of that aircraft?

That is, on a microscopic scale, no. The wing is constantly
freefalling, then being bounced back up by impact with air molecules.
Averaged over all the molecules, yes, the net is zero (the wing flies)
but on a microscopic scale, the wing is in constant brownian motion.
This implies momentum transfer, and following the momentum on a
microscopic scale is instructive.

OK, show us your arithmetic.

First, do you agree that air is made of individual molecules separated
by a lot of space compared to the size of the molecules themselves?

Yes.

Then do you accept that a wing is in freefall during all the (very brief
but very numerous) time in between molecular collisions? (If not, what
holds it up when it is not in contact with any air molecules?)

Yes.


If so, then during the time it is in freefall, it acquires a downward
velocity. Small, no doubt, but nonzero.

Sometimes it does, sometimes it does not. I'll allow as the vertical
component of velocity decreases during that time, for a positive up
coordinate system and a plane in (macroscopic) level flight.

Do you agree that in each collision momentum is transferred to the
air molecule that is equal and opposite to the momentum transferred
to the wing?

The next molecular impact
pushes it back up. On the average they will sum to a net zero vertical
motion. Is this the arithmetic you want to see?

No, I want you to calculate the horizontal and vertical componenets
of momentum for the example I gave, or any other reasonable example
of a fixed wing airplane in horizontal flight.

...

That's the shortcut. Where does the Bernoulli effect come from - on a
molecular level? That's what I'm addressing. The Bernoulli effect is a
shortcut for doing the calculation in bulk (where it makes the most
sense if you want a numerical answer) but it all comes from molecular
collisions.

I agreed quite some time ago that the theoretical basis for
macroscopic gas laws is to be found in statistical mechanics.

On a macroscopic level, the vertical component of momentum of the
wing is zero. Therefor on a macroscopic level, the sum of the
momenta transferred to the air molecules, integrated over all of
the air molecules must also be zero by Newton's third law.

Right?

For an airplane in straight level flight there is no net momentum
transfer in the vertical direction, between the air and the airplane,
just like there is no net vertical force acting on the airplane.

--

FF