Olin Lathrop wrote:

> Gerhard Fiedler wrote:
>> because we assumed previously that the plane flies at constant
>> height, so the distance between plane and Earth is constant, hence
>> no vertical speed, hence no vertical momentum.
>=20
> We loosly said "constant height". =20

"We" did not "loosly" say "constant height" -- when I wrote "constant
height", I actually meant "constant height". This apparently wasn't true
for you... (Back to the issue about messages that lead to
misinterpretations? Maybe another time.)

> That was not strictly true.  It is true for any ordinary human
> observation and for any practical use of flying a airplane, but it is
> not strictly true.  However, you're trying to split hairs so the
> definition of "constant height" becomes important.=20
>=20
> Let's say the atmosphere around the earth is uniform and perfectly
> still before the plane flies thru it. =20

Ok.

> The plane is flying at a steady speed, and such that it is neither
> climbing or falling from it's point of view of flying thru the air.=20

This is not what I was saying... I meant "constant height" as in
"constant distance from Earth", as I wrote a few times.

> This is ever so slightly different from not climbing of falling from
> the point of view with respect to the ground.  The difference would
> be probably not be measureable even with today's most advanced
> instruments, yet it is there.=20

I explained very clearly before what I mean with "constant height", and
I think my definition of "constant height" is in alignment with the
typical understanding of it:

>> ... we assumed a plane with a constant speed at constant height (that
>> is, at constant distance from the Earth). "Constant distance" means
>> "constant vertical speed of zero" which means "no vertical
>> acceleration" and also "no vertical momentum".

To simplify things, I assumed so. Your idea was to look at this from a
"macro" point of view. (I assume you don't have the thread's history in
front of you, and I don't know how good your memory is, but I can quote
you if you don't remember and are unable to look up the thread's
history.) So I followed your way to look at this.

In a "macro" point of view, there are no individual collisions with
individual air molecules, there is just an amorphous mass of air that
may move or stand still, and a plane wing that moves /at constant speed
and height/ through this mass.


> For the plane to fly "level" thru the air (perpendicular to the
> gravity vector), it is actually getting a tiny amount closer to the
> earth.  For it to fly at level altitude with respect to the earth, it
> is actually going down a tiny amount with respect to the air
> (slightly down from perpendicular to the gravity vector).=20

When I say "at constant height", I didn't mean to imply an angle of
attack or anything, just "at constant height". It was assumed that the
AOA is adjusted so that at the given speed and with the given wing
profile and air density and ..., the plane flies at constant height.
What's so difficult to understand with "constant height"?

> So in the end we're left with the three players, the plane, the earth,
> and the air.  When the plane is flying perfectly perpendicular to the
> gravity vector, it is imparting downward momemtum to the air near it
> and upward momentum to the earth.  Since the earth is so massive, the
> speed from the momentum imparted by the plane is very very small.=20

You again have not defined your reference system. How can you even think
of applying Newton's rules without defining a reference system? How do
you measure speed if you don't define a reference system? You seem to
look at this with a pre-Newtonian understanding that assumed an absolute
frame of reference, sort of an "ether". I don't want to get into
discussions around this, so let's stick with Newton -- but in order to
talk in Newton's terms, you /have/ to define a reference system. Which
you never do, so pretty much everything you say is not valid, or at best
unintelligible, from a Newtonian point of view. There is no (Newtonian)
meaning of "speed" or "momentum" without a reference system.=20

The normal reference system for things happening on Earth is the Earth,
often with a few assumptions (we talked about this). You rejected the
Earth as reference system, for reasons that are not quite clear, but you
never defined what else you're using as reference system when you talk
about "speed" and "momentum". They need a reference system to make any
sense at all. If you just started to do clean physics, and began with
defining your reference system, you'd realize that some things are not
exactly the way you tried to put them while avoiding to define a
reference system.

> I think we're going in circles and not making any headway in this
> argument. I've got things to do and I think I've explained things as
> well as I can. Either someone else can jump in here that may have a
> different way to illuminate the issue, or you can look up some
> aerodynamics texts if you want to be convinced that a plane leaves a
> net downdraft.  I have run out of ways to explain it, and patience to
> argue it.=20

I understand... :)  But I thought you could, too. I may be wrong...

Gerhard
--=20
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