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Re: thrust on airplanes.



Hi Folks --

It is important to recognize the distinction between
-- the data, and
-- an interpretation placed upon the data.

At 05:31 PM 4/13/01 -0500, brian whatcott wrote:
...
<http://www.aero.ufl.edu/~issmo/mav/MSU/MSU.htm>

OK, good, there is some data there. Figure 8 in particular.

(07:38 PM 4/12/01 -0500)

W = F x V

...the elementary definition of work shows that there is a
fairly constant thrust X airspeed product for constant horsepower.

1) Please, let's not use the word "work" to refer to something that is
properly called "power".

During the slow take off phase, this means there is much more thrust
available, so that for a given mass flow through the prop disk, there is
more airflow acceleration than in faster phases of flight.

2) The data does _not_ support that interpretation.

==============================================

(07:20 AM 4/13/01 -0500)

... an initially increasing
thrust (until the prop unstalls) and thereafter decreasing thrust ...

3) The data does _not_ support that interpretation, either.

In particular, there is no reason to believe that these propellers are
limited by blade stall in any regime.

At 05:31 PM 4/13/01 -0500, brian whatcott continued:

It's a little surprising at first sight that the engine thrust is
maximal at zero airspeed.

4) I don't see why this should be surprising.

But this is deceiving.
A model airplane prop has such high solidity that it never works at
truly zero airspeed.

5) In experiments of this sort, the sensible and conventional rule is that
"the airspeed" refers to the airspeed in the far field. So the point
labelled "zero airspeed" in the aforementioned Figure 8 is truly zero
airspeed. This makes sense, because thrust times airspeed is the power
delivered to the airframe, which is obviously zero in the static case.

6) There is nothing deceiving about this. These model props do not behave
significantly differently from full-scale airplane props.

7) The fall-off in thrust at higher airspeeds does not support the
"constant power" interpretation, either. These are fixed-pitch propellers.
At some high airspeed you will overrun the geometric pitch, whereupon the
thrust will pass through zero and become negative. Any adherence to the
constant-power law is fortuitous.

==================================

The fact that the propeller develops more-or-less constant thrust at low
airspeeds depends neither on blade stall nor on "solidity" or lack thereof.
a) It's obvious that we can build a propeller that is unstalled in the
static case, just by giving it a modest geometric pitch.
b) It's a fact that all practical propellers are remarkably efficient, and
can be modelled to a good approximation by an "actuator disk" that just
takes air over a certain area A and throws it backwards with a velocity V.
A four-bladed prop on a Cessna 152 would _not_ be twice as effective as a
two-bladed prop.

It's also obvious that an actuator disk (or anything else) in the static
situation will develop a bounded amount of thrust and will impart zero
power (F*V) to the airframe. We really don't need to know any details of
the propeller operation to know this. F*V=0 when V=0. Searching for a
"local" explanation in terms of blade stall or "solidity" is pointless.

===============

For real airplanes, there comes a point during the takeoff roll where the
"constant thrust" approximation should be phased out; the thrust will
decrease according to the constant-power law or otherwise. This crossover
in particular, and takeoff performance in general, is notoriously difficult
to calculate. One can draw parallels to other notoriously difficult
problems, such as the transition to/from hover in a helicopter.

Handwaving and guessing are unlikely to shed much light on these problems.

If anybody would like to have a serious discussion, please start by reading
the "Takeoff" chapter in von Mises _Theory of Flight_ and the "Hover"
section in Stepniewski & Keys _Rotary-Wing Aerodynamics_.