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Old March 1st 06, 11:13 PM posted to rec.aviation.piloting,rec.aviation.homebuilt,rec.aviation.student
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Default lift, wings, and Bernuolli

You seem to think [rising air in front of the wing]
requires the presence of a ground.


No, the rising air is caused by the high pressure in front of the wing,
caused by the wing throwing the air down (and ahead). The ground is not
required for that.

However, the ground =is= required for there to be "no net vertical
motion of air". Were there no ground, the air thrown down would not
bounce back up. The molecules would bounce around the other molecules,
dissipating the motion, but the downward momentum imparted by the wing
to keep it in the air would be equal to the downward momentum the wing
would have acquired had it not been flying. Although dissipated across
the rest of the atmosphere, it would not "disappear".

If air just "smooshed" away wings would not
work to hold us up.


If the wing is motionless, the air =does= just smoosh out of the way,
unlike a runway, which will hold an airplane up. To hold an airplane up
requires a special kind of motion - the kind that has the wing throwing
air down. Landing gear does not (usually!) throw the runway down.

If all this sounds a bit like perpetual motion, it's because
we've left out some details.


Yes.

we're looking at the wing long after it has started
up and everything is in steady state.


Yes. This obscures the "what causes what" question.

Essentially, it causes air
to rise ahead and get pushed down.


I see it as "it throws air down, which causes air in front to rise ahead".

A wing not only keeps the plane away from the earth, it keeps the earth
away from the plane. If you could measure the total forces on the earth
due to everything on top of it (essentially making the earth a giant
bathroom scale), the reading would not change when an airplane takes
off. Even though the plane is not touching the earth, it is throwing
air down at the earth, and that impact registers as weight.


Agreed, but this does not help explain the basics of lift.


It does belie the claim that there is no net downward momentum transfer.
This is only true if the air is allowed to bounce against the earth,
and as you stated earlier, the wing itself doesn't care about the earth.
As far as the wing is concerned, it is throwing air down, dealing with
the side effects (high pressure below and in front) and riding the wave
that it =caused= by throwing the air down in the first place.

If the earth were air-transparant, there would for sure be net downward
momentum of air, equal to the momentum the wing would have acquired had
it been freefalling (which is what flying is preventing).

(from the next post)
The upward momentum comes
from the air molecules. There is no requirement for the
ground. The approaching air molecules below the wing at
higher pressure cause the air molecules ahead to try to
escape. They preferentially escape upwards towards the
approaching low pressure above the wing.


Some upward momentum comes from there. But now that I think further,
the downward momentum imparted by the wing to the air (which then
bounces against the ground) only partly gets transferred back to the
wing. Most of it misses the (relatively) small wing and simply causes a
few more air molecules to escape the earth completely, or at least to
rise higher before gravity reclaims them.

This seems to imply that a wing would not produce lift,
i.e., could not fly, without the ground. That's clearly
incorrect.


Agreed (that a wing could produce lift without the ground). However,
without the ground, there would =not= be no net downward movement of air.

(from the next post)

[The Bernoulli effect] does explain [how the upwash starts]
- by virtue of pressure differentials.


Since pressure is derived from molecular collisions, and the Bernoulli
effect is also ultimately derived from those same collisions, we are
looking at the same thing, but in one case with a shortcut, the other
case on a microscopic level. For those for whom the Bernoulli effect is
a bit mysterious, or at least not obvious, looking at the newtonian
microscopic version is instructive. For those comfortable with
Bernoulli's equations, it provides a quick way to get numeric answers.

Jose
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