I believe my 1/2 Torque assumption is wrong. Mea culpa.
That means two rotors will unwind the motor at double the rate of the single rotor.
So, it's an even bigger advantage.
Sorry to all for the stupid mistake.
The key problem remains, however, that one cannot determine how much vertical friction is keeping the helicopter aloft.
The argument I heard used was that if the helicopter were really "stuck", then it wouldn't fall down when the torque from the motor is gone. But this is a weak argument.
The examples I provided though were not hypothetical but I observed personally as a event assistant captain.
-Kang
--- In Indoor_Construction_at_yahoogroups.com, "Jeff" <janderson@...> wrote:
>
> Wow, ask a question you get an answer. Still working on understanding
> it all, but see questions interspersed. I'm using the rich text and
> highlighting in Red italics. Not sure how this will work for everyone's
> reader.
>
> Jeff Anderson
> Livonia, MI
>
>
> --- In Indoor_Construction_at_yahoogroups.com, "Yuan Kang Lee"
> <ykleetx_at_> wrote:
> >
> > Let me try to provide a quick explanation why the overall rotor RPM
> slows down. And I'll also try to explain why this "loophole" is almost
> non-enforceable because it is impossible to determine at an event if the
> helicopter is holding itself aloft or has the help of vertical friction
> due to the rotor touching the ceiling/obstruction and having the rotor
> stopped.
> >
> > I'm going to make gross simplifications -- please bear with me.
> >
> > Assume the rotors are equal pitch, equal area.
> >
> > in normal operations, a rotor will rotate at an RPM (f)so that the
> overall drag on that rotor is equal to the torque supplied by the motor.
> Since the drag is proportional to RPM^2, the RPM will be proportional to
> the square root of the Torque:
>
> First question, in drag are you also including lift? Seems needed, but
> not sure it changes the relationship.
>
> >
> > RPM is proportional to sqrt (Torque).
> >
> > For the specific case of a helicopter with two rotors, each rotor sees
> 1/2 torque, we have
> >
>
> OK, there you lost me. Why does each rotor see only half the torque?
> Half the power I get, but not half the torque. Both rotors are tied to
> opposite ends of the motor. I should have equal and opposite torque,
> static or dynamic the same depending only on the winds in a given motor.
> Since the rest of the argument depends on this, I'll save the rest of my
> comments on your theoretical explanation till I get this part.
>
> >
> > RPM of one rotor is proportional to sqrt (0.5 x Torque)
> >
> > Overall unwinding of the motor has 2 x RPM, hence the overall
> unwinding rate is
> >
> > rubber unwinding rate of two rotors is proportional to 2 x sqrt (.5 x
> Torque)
> >
> > Now consider the case when one of the rotors is stopped, and all
> torque is supplied say to the bottom rotor.
> >
> > RPM of one rotor is porportional to sqrt (Torque)
> >
> > Hence only one rotor is turning, the motor unwinding rate is the same:
> >
> > rubbber unwinding rate of one rotor is proportional to sqrt ( Torque )
> >
> >
> > So, how do you compare the two rates. It is easy to show that the
> proportional factor is the same constant in both cases, and are
> functions of AoA, prop area, and air density. Hence the two motor
> unwinding rates are comparable:
> >
> > unwinding rate of two rotors / unwinding rate of one rotor =
> >
> > 2 * sqrt (0.5 * T) / sqrt (T) = sqrt (2) = 1.4
> >
> > That is, the two rotors unwind at a rate 40% more than a single rotor.
> >
> >
> > So, the overall unwinding advantage is real. Not by 100 % but by 40%.
> >
> > =======
> <SNIP>
> I was going to comment on your examples on enforceability but realized I
> was getting into opinion, hypothetical examples and rule interpretation
> so I'm not sure of the value added.
>
Received on Wed Mar 30 2011 - 16:34:46 CEST
This archive was generated by Yannick on Sat Dec 14 2019 - 19:13:46 CET