The Impossibility of Flying Heavy Aircraft Without Training

Jose wrote:
They aren't 'gotten rid of' they are accelerated which causes them to
be spaced farther apart -- thus lowering the pressure.

Accelerating them gets rid of them in the sense I mean, but I suppose I
was sloppy there. In any case, to be accelerated, they need to go
somewhere. The standard explanation is that there is a longer path up
top. The reason there is a longer path is that the air is bent
downwards. If you bend plywood (concave down), the top sheet is
stretched and the bottom sheet is compressed. Same with the air.

There is a longer path along the top because the wing is convex up.


When the air is bent downwards, the air is accelerated downwards. This
causes downwash.

Not until after it passes the high point in the airfoil. Befor it
gets there,
it is accelerated upwards.

Air accelerated downwards by the wing requires (by
Newton) the wing to be accelerated upwards (counteracting in this case
the acceleration due to gravity). It does so in a manner that also fits
Bernoulli's equations.

When the air reaches the trailing edge it is back to where it started.
But in the meantime air above it has begun to flow down. After the
wing has passed the momentum of _that_ downflow carries the air
down past the altitude of the wing. But that is after the wing has
passed. The downflow is -art of what happens as the air in the wake
of the airplane is restored to equilibrium.


The lift is a result of the pressure difference between the lower and
upper surfaces of the wing. The downwash is the result of the momentum
of the air above the rarefied region created by the wing moving
downward.

And the pressure difference is sustained by the wing continually
imparting momentum (indirectly by creating the pressure differential) to
the air above the rarified region.

Regardless, the lift is a result of the pressure differential between
the upper and lower wing surfaces.


The downrushing air starts it s downwash above the wing and does
not pass the wing in the vertical direction until after he wing has
passed.

Matters not. It is another way to look at lift.

No, it is a way of looking at downrushing air that has never
contacted the wing.


[The downrushing air] is not really caused by lift (my mistake),
it is caused by the same phenomenum that causes lift.

Fair enough. What this says is that both ways of looking at it are
valid. Bernoulli is easier to calculate, Newton is easier to conceptualize.


No. That says that the downrushing air and lift are both caused by the
same phenomenum.

--

FF

Not until after it passes the high point in the airfoil. Befor it
gets there, it is accelerated upwards.

Does the air split (top/bottom path) at the same level as it rejoins?
If the air splits at a higher altitude, then the air has to have a net
downward motion to get to the rejoin point.

When the air reaches the trailing edge it is back to where it started.

Is it? I think it's lower than when it started. It certainly is with
any appreciable AOA.

No, it is a way of looking at downrushing air that has never
contacted the wing.

It doesn't matter whether it contacts the wing or not. The contact is
by proxy (by contacting the other molecules of air that contact...the
wing.) You call it pressure. I agree. Pressure is ultimately
newtonian; I think we agree there too.

It's a floor wax. It's a dessert topping.

There is more to that. If this collision occurs in outer space, I
guarantee you that the center of mass will =not= quit moving.
But it will not move in a manner that conserves momentum.

Yes it will. What will not be conserved is macroscopic kinetic energy.

Make the room bigger. Make it an infinite room. At what point does
the fan continue to put momentum into the air mass continuously, and
not just during start up?

At the point when it's an infinite room. The bigger the room, the
longer it takes for a pressure equilibrium to occur. If we deal with
the earth's atmosphere and a propeller, the propeller pushes air back,
which alters the rotational momentum of the earth, in a manner equal
(and opposite) to the amount of rotational (around the earth's center)
momentum the airplane acquires. This could potentially happen until the
earth is spinning godawful fast (faster than the propeller could
handle). In practice we'll get tired of arguing before that.

So, are you saying that in the presence of the Earth there is no
net change in the momentum of the basketballs being thrown by the
dribbler and also no net change in momentum of the air molecules
accelerated by the wing?

Averaged over all basketballs and all air molecules, yes, because the
earth acts as a momentum transfer point.

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

Jose wrote:
Not until after it passes the high point in the airfoil. Befor it
gets there, it is accelerated upwards.

Does the air split (top/bottom path) at the same level as it rejoins?
If the air splits at a higher altitude, then the air has to have a net
downward motion to get to the rejoin point.

Agreed. Clearly this happens with high AOA.


When the air reaches the trailing edge it is back to where it started.

Is it? I think it's lower than when it started. It certainly is with
any appreciable AOA.

Are there any airfoils that produce lift at an AOA at or below zero?


No, it is a way of looking at downrushing air that has never
contacted the wing.

It doesn't matter whether it contacts the wing or not. The contact is
by proxy (by contacting the other molecules of air that contact...the
wing.) You call it pressure. I agree. Pressure is ultimately
newtonian; I think we agree there too.

But the pressure that supplies the lift is in the air below the wing
whereas it is the air above the wing that washes down.You're
attributing lift to the wrong air.

If he air above the wing washed sideways, and not down at all
you'd still get lift as the thumbtack, notecard and soda straw
demonstrates. No downwash from the notecard.




It's a floor wax. It's a dessert topping.

There is more to that. If this collision occurs in outer space, I
guarantee you that the center of mass will =not= quit moving.
But it will not move in a manner that conserves momentum.

Yes it will. What will not be conserved is macroscopic kinetic energy.

Right. My mistake.

...

So, are you saying that in the presence of the Earth there is no
net change in the momentm of the basketballs being thrown by the
dribbler and also no net change in momentum of the air molecules
accelerated by the wing?

Averaged over all basketballs and all air molecules, yes, because the
earth acts as a momentum transfer point.


--


FF

Are there any airfoils that produce lift at an AOA at or below zero?

Depends how AOA is defined (I don't know the precise definition). But
if we define it as the angle between the line from split to rejoin
point, and the direction of travel (or relative wind), then I think not.

But the pressure that supplies the lift is in the air below the wing
whereas it is the air above the wing that washes down. You're
attributing lift to the wrong air.

It doesn't matter. The molecules collide all over the place and
momentum is moved around (but never lost).

If he air above the wing washed sideways, and not down at all
you'd still get lift as the thumbtack, notecard and soda straw
demonstrates. No downwash from the notecard.

Describe the notecard setup better and I will do the experiment, and
then tell you what I think.

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

Jose wrote:
Are there any airfoils that produce lift at an AOA at or below zero?

Depends how AOA is defined (I don't know the precise definition). But
if we define it as the angle between the line from split to rejoin
point, and the direction of travel (or relative wind), then I think not.

ISTR that AOA is defined relative to the chord.


But the pressure that supplies the lift is in the air below the wing
whereas it is the air above the wing that washes down. You're
attributing lift to the wrong air.

It doesn't matter. The molecules collide all over the place and
momentum is moved around (but never lost).

Of course it matters. It is the undisturbed molecules (or minimally
disturbed molecules) under the wing that push up on the wing.

The air flowing over the top of the wing cannot push UP on the
wing.


If he air above the wing washed sideways, and not down at all
you'd still get lift as the thumbtack, notecard and soda straw
demonstrates. No downwash from the notecard.

Describe the notecard setup better and I will do the experiment, and
then tell you what I think.


Take 3 x 5 index card or a similar light small card like a playing
card and push a thumbtack through the center.

Leave the thumbtack in and set the card on a table point up.

Put a soda straw, standing on end over the point of the tack.

Blow through the straw and while blowing lift the straw
straight up.

--

FF