visualisation of the lift distribution over a wing

In article ,
Jim Logajan wrote:

Alan Baker wrote:
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*.

Please don't patronize when you've never studied a subject. A moment's
research would have prevented you from posting something that incredibly
ignorant. You might want to look up the "axioms of fluid dynamics" before
you further add to your public embarrassment.

I'm not the least embarrassed, because I know what "axiom" actually
means.


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.

Shrug - your insults are lame and tiresome, but I will admit you're
patronization is irritating. I do make mistakes about things I was
taught, but I should point out to you that I did well enough in college
physics to earn an undergraduate degree in the subject. Unlike you, I
have had to solve a **** load of problems involving conservation of
momentum to prove I understood the basics - including conservation of
momentum in quantum mechanical systems. So far as I know, you HAVEN'T had
to prove your mastery of the subject with ANYONE.

Conservation of linear momentum is another of the axioms of fluid
dynamics, lad.


(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".

Well that probably explains your problem - you don't know how to set up
the math properly, so you have no way to validate whether your
"qualitative analysis" is correct.

I do know how to set up the math properly. I've known since I was about
17.


Ironically, all your posts contain the same violation of conservation of
momentum - and yet you keep pointing to that concept as vindication.

Nope.


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.

You can't "focus" on one conservation law because the number of
constraining equations has to equal the number of variables. Doing so
simply leads to an infinite number of bogus results.

Actually, you're wrong.


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.

Sigh. This is a case where a little knowledge is a dangerous thing. It
would take a book to explain the problem with your conceptual view of
fluid dynamics. I don't have that sort of patience.

Fluid dynamics isn't even an issue here. Force and momentum are.

The aircraft experiences a force from the air and therefore MUSR impart
a force on the air.

Force IS change of momentum with respect to time.


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

Of course if you had read my earlier post you'd see I'd ALREADY USED THAT
EQUATION. But you obviously aren't familiar with the conventions used in
fluid dynamics, so you probably had no clue what my "m_dot" meant or how
I got the figures I did.

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.

You math is correct and no one has denied there is a downwash (why you
think otherwise continues to baffle me) yet your "net, now" comment
violates conservation of momentum.

It doesn't. You're simply wrong. The momentum of the system as a whole
remains constant, but the air gets some downward momentum, which is only
netted out when it finally gets transferred to the earth (which has some
upward momentum from the plane's gravity pulling up on it).


Here's why:

If we choose a reference frame so that at T=0 everything in the system is
stationary with respect to that frame, we set the net momentum of the
system to 0. Then, so long as the system remains closed, at all other
times the conservation of momentum must yield 0.

We cannot choose a reference frame where everything is stationary with
respect to it. The aircraft is moving with respect to the air.


But according to your "qualitative analysis" the net vertical momentum
P_net_z increases with time T, like so:

P_net_z(T) = (100 kg/s)*T*V

Only if you ignore the upward momentum of the earth. The earth is part
of the system, too.


That's because the earth and the airplane maintain zero vertical
momentums (P_earth_z(T) = 0, P_plane_z(T) = 0,) and there appears to be
nothing in your conceptual view of the situation to correct that
violation of conservation of momentum.

Only because you're ignoring the upward momentum of the earth.


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

It appears to salve your ego to ascribe assertions to me that I never
made and then tell the world that those falsehoods prove I don't "get
it." Probably because you've grown so much hubris and so little humility.

LOL

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