The Impossibility of Flying Heavy Aircraft Without Training

So how do these equations relate to our two-dimensional airfoil? Look
again at
the Clark Y and notice that an airfoil is a curved shape. While the bottom
is
relatively flat, the top surface is thicker and more curved. Thus, when
air
passes over an airfoil, that flow over the top is squeezed into a smaller
area
than that airflow passing the lower surface. The Continuity equation tells
us
that a flow squeezed into a smaller area must go faster, and the Bernoulli
equation tells us that when a flow moves faster, it creates a lower
pressure.


I don't quite understand the "squeezed into a smaller area". I Understood
that the flow over the top surface had to travel further (thus faster) over
the longer curved distance to get from the leading edge to the back of the
airfoil. I am just a lay person and do not even play an aeronautical
engineer on TV so I may be totally mistaken.

The "squeezed into a smaller area" part comes from the classic example
of the effect in a venturi. If a (compressible) fluid flows from a fat
tube into a thin tube and back into a fat tube, it is being "squeezed
into a smaller area" when it's in the thin tube. The pressure in the
thin tube is lower.

As for the wing, a lot is left out of the explanation. Not all things
are equal, and you need to take that into account. For example,
although the path over the top is longer, at the end, the air is not put
back the way it was prior to passage. The air molecules are moving
downards. This is required by the way the trailing edge of the wing is
angled. It didn't start out that way, therefore force must be applied
to the molecules to make this happen. This can only come from the wing,
and that's what holds the wing up.

Symmetric airfoils generate lift too if they are at the proper angle of
attack. Thin symmetric airfoils generate lift, but the path over the
top and bottom is then nearly equal in length.

Hollow airfoils (think just the top surface of the wing, with the bottom
surface and some of the leading edge removed) will also hold a plane
up, and the path over the top and bottom is identical. What is
different (before and after) in all cases is that the air has acquired a
downward velocity, and this has to be balanced by an upward force
applied by the air to the wing.

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

If a (compressible) fluid flows from a fat tube into a thin tube and
back into a fat tube, it is being "squeezed into a smaller area" when
it's in the thin tube.

The Bernoulli equation only applies to *incompressible* flow.

This is required by the way the trailing edge of the wing is angled.


What about reflexed airfoils?

The Bernoulli equation only applies to *incompressible* flow.

Right. But the Bernoulli principle applies to compressible and
incompressible flow. It is the source of the venturi effect, which
occurs in air, among other places.

What about reflexed airfoils?

I am not familiar with them. The trailing edge is depicted as pointing
up; does it do so when the wing is at a lifting angle of attack? Is
there an airflow separation near the trailing edge? If air is not being
deflected downwards somewhere, we're back to the lifting fairies.

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

But the Bernoulli principle applies to compressible and
incompressible flow.

True. But your original text suggested that the fluid *had* to be
compressible.

The trailing edge is depicted as pointing up; does it do so when the
wing is at a lifting angle of attack?

Good question. I don't know. Anyway, remember that there is no
overall downward movement of the air unless there exist wingtip
vortices. This suggests that the important downward momentum of air
happens *after* the air leaves the trailing edge. I'm not certain
that what direction the trailing edge is actually pointing is
critical. I'm skeptical that if you have an airfoil generating
positive lift, just tilting the trailing edge upwards slightly is
going to kill that lift.

Also, although theory (Kutta Condition) says that the air flow leaves
the trailing edge smoothly, my understanding is that in real life, the
rear stagnation point will be somewhat on the top surface of the
airfoil anyway.

But your original text suggested that the fluid *had* to be
compressible.

Yes, it did suggest that. It shouldn't have. My clumsiness.

Anyway, remember that there is no
overall downward movement of the air unless there exist wingtip
vortices.

An infinitely wide wing has no wingtips. You suggest it could not
provide lift. I've read the "wingtip vortices provide lift" papers, I'm
not convinced that the correlation implies a causation in that
direction. I see it as: the wing causes downwash which provides lift
(action-reaction) and =that= creates vortices. The higher pressure air
underneath the wing has to go somewhere - around (the wingtip) and up
makes sense to me, and that is a vortex.

What happens in the two-dimensional case?

I'm skeptical that if you have an airfoil generating
positive lift, just tilting the trailing edge upwards slightly is
going to kill that lift.

I agree with you. The downward movement of air is being generated over
the entire wing chord, and has some depth to it too. I suspect (without
solving any equations) that tilting the trailing edge upwards slightly
(that's what ailerons do, sort of) does reduce lift, but as the airflow
right underneath the airfoil gets deflected upwards, the air further
away (below) does not, leaving a small lower pressure region just below
the upward pointing trailing edge. The rest of the mass of air below
this small low pressure region continues downwards through its momentum.

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

An infinitely wide wing has no wingtips. You suggest it could not
provide lift.

Heck no. It just does so with no *net* downward momentum of air. In
other words, the upward momentum ahead of the wing is equal to the
downward momentum at the rear of the wing.

I've read the "wingtip vortices provide lift" papers,

I'm not proposing that.

the wing causes downwash which provides lift (action-reaction) and
=that= creates vortices.

Infinite wings have no downwash, yet provide lift. By *definition*,
downwash is caused by wingtip vortices.

The higher pressure air underneath the wing has to go somewhere -
around (the wingtip) and up makes sense to me, and that is a vortex.

Except when you don't have wingtips. ;-)

Heck no. It just does so with no *net* downward momentum of air. In
other words, the upward momentum ahead of the wing is equal to the
downward momentum at the rear of the wing.

What upward momentum? The air starts out still, and infinitely wide
wing comes flying through it, when all is done air has to have been
deflected downwards, at the very least to satisfy Newton.

Behind the wing, air is moving downwards, there is higher pressure below
the wing's path, and lower pressure above the wing's path. The air will
recover, as the pressure equalizes (eventually pushing down on the
earth's surface and up into the lower pressure region)

Infinite wings have no downwash, yet provide lift. By *definition*,
downwash is caused by wingtip vortices.

Well, then an infinite wing ought to have something that is not called
downwash, but keeps Newton happy. Call it downflow, call it pressure
bounce.. but it's something.

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

"Greg Esres" wrote in message
oups.com...
An infinitely wide wing has no wingtips. You suggest it could not
provide lift.

Heck no. It just does so with no *net* downward momentum of air. In
other words, the upward momentum ahead of the wing is equal to the
downward momentum at the rear of the wing.


Disagree. There is a *net* downward momentum of air. Otherwise there is no
lift. However the downward momentum for any finite section of the infinite
wing is infinitesimal. Note, however, that even though in Calculus 100 we
assume that an infinitesimal is approximately equal to zero and can be
ignored, it is only approximately to zero and only very very near zero.

While zero times any number is still zero, almost zero times infinity is
NOT zero. Ergo, the downwash is not zero either.



I've read the "wingtip vortices provide lift" papers,

I'm not proposing that.

the wing causes downwash which provides lift (action-reaction) and
=that= creates vortices.

Infinite wings have no downwash, yet provide lift. By *definition*,
downwash is caused by wingtip vortices.

No. wingtip vortices are caused by downwash. Infinite wings don't have
wingtip vortices because they don't have ends, not because they don't have
downwash.

Highflyer
Highflight Aviation Services
Pinckneyville Airport ( PJY )