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