The Impossibility of Flying Heavy Aircraft Without Training

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?
....

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.

Ok. (I was sloppy - it doesn't "acquire a downward velocity", it really
"endures a downward acceleration", which depending on the initial
vertical velocity may or may not end up with the plane going downward.)

Right, but don't forget that the downward acceleration is constant
without regard to the velocity of the aircraft.

So we are saying the same thing here.

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?

Yes I do. This is what I call "throwi ng the air down". That downward
momentum will remain with the air (dissipated across many other
molecules as it keeps colliding, but never disappearing) until it is
transferred to the earth, which has been accelerating upwards in the
same fashion.

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?

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


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

Ok.

On a macroscopic level, the vertical component of momentum of the
wing is zero.

Yes.

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?

Only in a nonaccelerated frame. We are dealing with an accelerated
frame. Consider a rocketship hovering over the moon. The (macroscopic)
vertical component of its momentum is zero also. However it has to
continually throw down rocket exhaust to stay there.

Instead, let's consider a wing in level flight.

So, without
looking at the rest of the picture, your conclusion about momentum is
flawed.

In the case of the wing, the momentum is transferred a few times... once
when the wing hits the air molecule (throwing the air down), again when
that molecule hits the earth and bounces back (throwing the earth away
from the wing),

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)

and then again when that air molecule (or its proxy)
hits the wing on the way back 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.


Think about a person sitting on a stool. No momentum transfer (or so it
would seem). But then think about a person supporting himself by
dribbling a basketball. There is a lot of momentum transfer, but no
=net= change. The reason there is no net change is that the basketball
keeps pushing the earth away too.

And there is no net transfer of momentum to the basketball. This is
clear as the average velocity of the basketball is zero, even though
the average speed is non-zero.

Think of the example we had earlier of a piston supported by air
pressure in a cylinder. The momenta transferred by air molecules
to the piston is equal and opposite to the momenta transfered by the
air molecules to the bottom of the cylinder. There is no net transfer
of momentum to the air.

--

FF