The Impossibility of Flying Heavy Aircraft Without Training

On Wed, 1 Mar 2006 at 22:43:58 in message
.com,
" wrote:

It discussed the Bernuolli theory of flight- and (if I recall) quite
conclusively proved that one of the _fundamental_ assumptions of the
Bernuolli theory- that air that travels path over the top of the wing
is flowing appreciably faster than air that flows over the bottom- is
simply incorrect in a compressible fluid....

There is not really a Bernoulli theory of lift. Bernoulli's theory
shows the relationship between the velocity and pressure of fluid flow
when energy is not added or removed and the flow is subsonic. It is a
very simple theory which is correct for much of the time. It quite
accurately, at lower speeds, represents the velocity and pressures
between streamlines.

The air does flow faster over the top than the bottom and for the lower
subsonic region air behaves very closely to being incompressible.
Generally pressure changes are transmitted at the velocity of sound.

At high subsonic and of course at supersonic speed the effect of
compressibility cannot be ignored.

Shock waves form, first on places like the top surface of the wing where
the air first reaches the velocity of sound. As the speed rises they
become bigger and move towards the leading and trailing edges. Above
Mach one the air does not detect the approaching aircraft! :-)

I have just read a few more messages in this thread and discussing lift
in this general way without maths and without using at least simple
physics and slowly developing the methods is almost futile.

What's it matter about lift as long as the aircraft fly? !!!!!
--
David CL Francis

David CL Francis wrote:

The air does flow faster over the top than the bottom and for the lower
subsonic region air behaves very closely to being incompressible.
Generally pressure changes are transmitted at the velocity of sound.


I hate to be a spoil sport (or dullard?), but...

the (stationary) air does WHAT (as the wing passes by)???


))

David CL Francis wrote:
On Tue, 28 Feb 2006 at 07:42:32 in message
. com,
wrote:

For a fan in open air the momentum of the air moving through the
fan is equal and opposite to the momentum of the air moving around
the fan to replace the air removed from the front of the fan. There is
no net momentum change in the air. For ducted flow that returns the
air to the front fo the fan, the net momentum is also zero. Net flow
and
net momentum through any closed loop is zero--else the 'loop'
is not 'closed'.

Followjng a wing in level flight, the downward momentum of the
air in the downwash is equal and opposite to the upward momentum
of the air to either side that moves up to replace the air that washes
down. There is no net momentum change in the air.

This whole discussion is becoming weird. The power required to drive a
fan goes somewhere. There is and must be a net increase in the air
velocity across the fan so there is a change of momentum from the air
entering to the air leaving. The power input results in a momentum
change. This principle applies to fans, helicopters, wings and other
things - even rowing! The air that leaves the driven fan, jet engine,
ducted fan, wing is the result of the thrust or vice versa, how ever you
like to think of it.


The momentum of an airplane in level flight at
constant speed is constant. Conservation of
momentum REQUIRES that there is no net
change in themomentum of the air. There is
momentum exchanged between the airplane
and the air. But there is no NET momentum
change in the air.

However, that speeded up air dissipates itself in the atmosphere
gradually giving up energy to the surrounding air as it all eventually
steadies down again. What does that do? Well in the ultimate I guess it
raises the temperature of the atmosphere slightly!

Yes. In closed loop ducted flow it can raise the temperature
of the air a lot.


An airscrew does much the same. It captures air from a tube somewhat
larger that its diameter, speeds it up and it goes out the back faster
then it came in. if you had a closed circuit like a wind tunnel it still
requires power to accelerate the air to the required speed. Less than an
open system though a carefully shaped return duct will slow the air down
again ready to be accelerated again. In that cases the losses have to be
made up.

What is a wing but a kind of linear fan?

There is a net transfer of energy from the airplane to the air.

There is not a net transfer of momentum from the airplane to the air.

Momentum is a vector, energy is a scaler.

--

FF

Jose wrote:
Most aerodynamic equations dealing with low subsonic speeds treat air as an
incompressible fluid because compressibility doesn't have a significant
effect until you approach sonic speeds.

Isn't compressiblity what causes pressure changes (absent temperature
changes)?


No.

Compressible fluids (commonly called liquids) also experience
pressure changes. THis is used advantageously for hydraulic
power.

The distinction is that a compressible fluid (commonly called gas)
undergoes a volume change proportionate to the pressure change,
while the volume of an incompressible fluid changes little with
pressure. Compressible fluids obey Charles' law,
(or is it Boyle's law?):

P1 * V2 = P2 * V1

--

FF

In article .com, george
says...

Isn't a Bernouli what an Arab wears?

under his theorum :-)

Is that near his axiom?
Sorry I couldn't help myself.

Chuck S

The momentum of an airplane in level flight at
constant speed is constant. Conservation of
momentum REQUIRES that there is no net
change in themomentum of the air.

What is the net momentum change when the airplane falls to the ground?

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

Suppose we have a 1500 lb airplane in level flight at 120 mph.
What are its horizontal and vertical components of momentum?

Suppose we have a 1500 lb rocketship hovering over the moon on its
rocket exhaust. What are its horizontal and vertical componnts of momentum?

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

Jose wrote:
Suppose we have a 1500 lb airplane in level flight at 120 mph.
What are its horizontal and vertical components of momentum?

Suppose we have a 1500 lb rocketship hovering over the moon on its
rocket exhaust. What are its horizontal and vertical componnts of momentum?


I asked first.

--

FF

Jose wrote:
The momentum of an airplane in level flight at
constant speed is constant. Conservation of
momentum REQUIRES that there is no net
change in the momentum of the air.

What is the net momentum change when the airplane falls to the ground?


The vertical compenent first rises from zero to Vt * M where Vt is the
terminal velocity of the falling aircraft and M is the mass of the
falling
aircraft. Then the vertical component of momentum RAPIDLY drops
to zero again after the aircraft contacts the ground.

--

FF

On Fri, 03 Mar 2006 08:00:45 -0800, fredfighter wrote:


Jose wrote:
Suppose we have a 1500 lb airplane in level flight at 120 mph.
What are its horizontal and vertical components of momentum?

Zero at equalibrium.


Suppose we have a 1500 lb rocketship hovering over the moon on its
rocket exhaust. What are its horizontal and vertical componnts of momentum?

Again, zero.



I asked first.

Same answer, different criteria. Net energy for a given mass will be
the same whether the craft is flying, rocketing, or in orbit.

F=MA

But, only in the instance of 'flying' does Bernuolli apply.