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 - 16:18:22 CEST
This archive was generated by Yannick on Sat Dec 14 2019 - 19:13:46 CET