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
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