visualisation of the lift distribution over a wing

In article ,
Beryl wrote:

snip

No downwash, no lift. No go learn something.
Let's learn here. From you. Is that 90 degree turn *exactly* the same as
a 180 degree turn that directs incoming air back in the opposite
direction?

Read this:

"To determine [the angle represented by a greek letter in the original
text], we observe that no downwash is generated when the wing generates
no lift."

I'm not disagreeing with that. I'll rephrase it, and say no circulation
is generated. It is not even relevant.

http://www.aoe.vt.edu/~cwoolsey/Cour...al/Aerodynamic
Properties.pdf

Read it over and over again until you get it.

Get what? It's about wings and geometry. Find something about air moving
through air.

http://www.onemetre.net/Design/Downwash/Downwash.htm

"The theory of downwash starts by noting that you only get downwash when
you have lift.* No lift, no downwash."

http://amasci.com/wing/airfoil.html

' The "Newton" explanation is wrong because downwash occurs BEHIND the
wing, where it can have no effects? Downwash can't generate a lifting
force? INCORRECT.

Wrong, and silly as well! The above statement caught fire on the
sci.physics newsgroup. Think for a moment: the exhaust from a rocket
or a jet engine occurs BEHIND the engine. Does this mean that
action/reaction does not apply to jets and rockets? Of course not.
It's true that the exhaust stream doesn't directly push on the inner
surface of a rocket engine. The lifting force in rockets is caused
by acceleration of mass, and within the exhaust plume the mass
is no longer accelerating. In rocket engines, the lifting force
appears in the same place that the exhaust is given high velocity:
where gases interact inside the engine.

And with aircraft, the lifting force appears in the same place that
the exhaust (the downwash) is given high downwards velocity. If a
wing encounters some unmoving air, and the wing then throws the air
downwards, the velocity of the air has been changed, and the wing
will experience an upwards reaction force. At the same time, a
downwash- flow is created. To calculate the lifting force of a
rocket engine, we can look exclusively at the exhaust velocity and
mass, but this doesn't mean that the rocket exhaust creates lift.
It just means that the rocket exhaust is directly proportional to
lift (since the exhaust velocity and the lifting force have a
common origin.) The same is true with airplane wings and downwash.
To have lift at high altitudes, we MUST have downwash, and if we
double the downwash, we double the lifting force. But downwash
doesn't cause lift, instead the wing's interaction with the air
both creates a lifting force and gives the air a downwards velocity
(by F=MA, don't you know!)'

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:

Pressure waves can reach the ground, without the air in the column
descending to the ground.
I never said that the particular molecules that the aircraft touches
are the ones that have to reach the ground.
You said "The net flow is downward until it hits the ground and the
momentum is transfer to the earth."

And it is: the *net* flow.

The molecules that "reach" the ground are the ones that
were *already there* at ground level.

Hmmmm...It's not too out of line to think of air particles
at ambient temps as moving at the speed of sound.
Its those air "molecules" that carry sound waves longitudinally.
Air particles don't stay anywhere, in a manner of speaking...

Brian W

In article ,
brian whatcott wrote:

Alan Baker wrote:

Pressure waves can reach the ground, without the air in the column
descending to the ground.
I never said that the particular molecules that the aircraft touches
are the ones that have to reach the ground.
You said "The net flow is downward until it hits the ground and the
momentum is transfer to the earth."

And it is: the *net* flow.

The molecules that "reach" the ground are the ones that
were *already there* at ground level.

Hmmmm...It's not too out of line to think of air particles
at ambient temps as moving at the speed of sound.

Actually, they move at considerably faster than the speed of sound...

Its those air "molecules" that carry sound waves longitudinally.
Air particles don't stay anywhere, in a manner of speaking...

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:
Jim Logajan wrote:
3) Therefore if, say, the downwash is 1 kg/s at any given instant due
to the wing, somewhere else in the fluid there must be an upwash at
that same instant of 1 kg/s. Agree or disagree?

Agree. At the surface of the Earth.

The qualifier indicates to me that you don't agree with the statement as
written.

Unfortunately, I consider conservation of mass at all points in time in an
incompressible fluid an essential element to understanding the behavior of
downwash. If you don't, then I think any further debate between us is
ended.

(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2) at, say,
20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it maintained
that speed. In the mean time, once the helicopter stopped descending,
conservation of mass in an incompressible fluid seems to require an equal
volume of air to have an upward vector of 20 m/s. So the surface of earth
appears to be irrelevant for over two minutes.)

Alan Baker wrote:
...
First of all, the downward motion of the vortex clearly carries right
out the bottom of the frame.
Are you impaired? The airplane is approaching the camera. The camera is
looking up at the airplane. The bottom of the frame contains the distant
background. Objects farther than the airplane appear lower in the frame.
If the camera was above the approaching airplane and looking down at
it, distant objects would appear higher in the frame than the airplane.

None of which refutes what I said.

Oh, it was simply interesting to you that the vortex goes off into the
distance, right out the bottom of the picture.

In article ,
Jim Logajan wrote:

Alan Baker wrote:
Jim Logajan wrote:
3) Therefore if, say, the downwash is 1 kg/s at any given instant due
to the wing, somewhere else in the fluid there must be an upwash at
that same instant of 1 kg/s. Agree or disagree?

Agree. At the surface of the Earth.

The qualifier indicates to me that you don't agree with the statement as
written.

You're right. I shouldn't have stated it that way.


Unfortunately, I consider conservation of mass at all points in time in an
incompressible fluid an essential element to understanding the behavior of
downwash. If you don't, then I think any further debate between us is
ended.

Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

That means that there is a constant change of momentum being done on the
air by the aircraft. That means air *must* be moving down (net) after
the aircraft has passed.


(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2) at, say,
20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it maintained
that speed. In the mean time, once the helicopter stopped descending,
conservation of mass in an incompressible fluid seems to require an equal
volume of air to have an upward vector of 20 m/s. So the surface of earth
appears to be irrelevant for over two minutes.)

Nope.

The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

In article ,
Beryl wrote:

Alan Baker wrote:
...
First of all, the downward motion of the vortex clearly carries right
out the bottom of the frame.
Are you impaired? The airplane is approaching the camera. The camera is
looking up at the airplane. The bottom of the frame contains the distant
background. Objects farther than the airplane appear lower in the frame.
If the camera was above the approaching airplane and looking down at
it, distant objects would appear higher in the frame than the airplane.

None of which refutes what I said.

Oh, it was simply interesting to you that the vortex goes off into the
distance, right out the bottom of the picture.

No. That shows that the air continues to move downward far below the
small portion of the vortex which is moving up.

The net movement of the air after the plane's passing must be downward,
because the plane is exerting a force on the air.

Force is a change of momentum with respect to time.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg

Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

It appears you have never studied fluid dynamics (maybe elementary fluid
statics?) and I doubt that you own any books on the subject.

The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

In other news, 1 + 1 = 2.

That means that there is a constant change of momentum being done on
the air by the aircraft. That means air *must* be moving down (net)
after the aircraft has passed.

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2)
at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it
maintained that speed. In the mean time, once the helicopter stopped
descending, conservation of mass in an incompressible fluid seems to
require an equal volume of air to have an upward vector of 20 m/s. So
the surface of earth appears to be irrelevant for over two minutes.)

Nope.

Dang - I try to use real numbers to establish a baseline example, and you
manage to use a single word to demolish my attempts! Really helpful
mathematical counter-example you produced - not.

The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

I don't know what your problem is - maybe you are thinking this is a
rocket problem where no external fluids are involved and you can't get
your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM.
Whatever the case, you seem to be fixated on applying one conservation
law to one element in the entire system to the exclusion of everything
else.

Best of luck to you.

Jim Logajan wrote:
Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

It appears you have never studied fluid dynamics (maybe elementary fluid
statics?) and I doubt that you own any books on the subject.

The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

In other news, 1 + 1 = 2.

That means that there is a constant change of momentum being done on
the air by the aircraft. That means air *must* be moving down (net)
after the aircraft has passed.

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2)
at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it
maintained that speed. In the mean time, once the helicopter stopped
descending, conservation of mass in an incompressible fluid seems to
require an equal volume of air to have an upward vector of 20 m/s. So
the surface of earth appears to be irrelevant for over two minutes.)
Nope.

Dang - I try to use real numbers to establish a baseline example, and you
manage to use a single word to demolish my attempts! Really helpful
mathematical counter-example you produced - not.

The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

I don't know what your problem is - maybe you are thinking this is a
rocket problem where no external fluids are involved and you can't get
your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM.
Whatever the case, you seem to be fixated on applying one conservation
law to one element in the entire system to the exclusion of everything
else.

Best of luck to you.



Two dimensional Newtonian thinking in a three dimensional non-Newtonian world.

In article ,
Jim Logajan wrote:

Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in this,
you're missing out from the start. It's conservation of *momentum*
that's in play here.

It appears you have never studied fluid dynamics (maybe elementary fluid
statics?) and I doubt that you own any books on the subject.

Sorry, lad, but conservation of mass is a principle that comes up mostly
in *chemistry*.


The aircraft has a force exerted on it equal to its weight. That means
that the aircraft must be exerting a force on the air in the opposite
direction.

In other news, 1 + 1 = 2.

What a pity then that you don't understand it.


That means that there is a constant change of momentum being done on
the air by the aircraft. That means air *must* be moving down (net)
after the aircraft has passed.

You must have a devil of a time figuring out what keeps balloons afloat,
what with no handy downward moving air!

I have no trouble figuring that out at all. A gas of a different density
within the balloon causes the net upward force on the balloon exerted by
the air outside the balloon to be greater than the net downward force on
it.


(Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a
balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine
immediately started. After a small drop it levels out and maintains a
downwash of air moving through its 6 m diameter disk (A ~= 28 m^2)
at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.)

It would take ~150 s for that downwash to reach the ground if it
maintained that speed. In the mean time, once the helicopter stopped
descending, conservation of mass in an incompressible fluid seems to
require an equal volume of air to have an upward vector of 20 m/s. So
the surface of earth appears to be irrelevant for over two minutes.)

Nope.

Dang - I try to use real numbers to establish a baseline example, and you
manage to use a single word to demolish my attempts! Really helpful
mathematical counter-example you produced - not.

No math is necessary for this. Look up "qualitative analysis".




The conservation of momentum says that there cannot be an equal amount
of air moving upward at an equal speed.

I don't know what your problem is - maybe you are thinking this is a
rocket problem where no external fluids are involved and you can't get
your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM.
Whatever the case, you seem to be fixated on applying one conservation
law to one element in the entire system to the exclusion of everything
else.

The law I'm focussed on is the one that counts. It doesn't matter
whether the fluid is expelled from inside or whether it's an external
fluid diverted down by the surfaces of the craft.

In order for there to be a continuous force W equaling the weight of the
craft acting on it, the craft must exert a force -W on the fluid. That
-W means that there is a downward change of momentum in the fluid. Since
the fluid is no accelerated indefinitely, there must be a continuous
flow (mass per unit time M/t) of the fluid accelerated to a velocity V
where the equation looks like:

-W = M/t * V

The velocity of the fluid will be:

V = -W/(M/t)

That is inescapable. If the craft weighs 9800N (newtons), and it moves
100kg of air every second, then the air must be moving downward (net,
now!) at 98 m/s.

I'm sorry if you don't get this, but it is very simple and absolutely
irrefutable.

--
Alan Baker
Vancouver, British Columbia
http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg