Dang, I should know better than to respond before checking the whole string. Sorry.
Jeff Anderson
Livonia, MI
--- In Indoor_Construction_at_yahoogroups.com, "ray_harlan" <rbharlan@...> wrote:
>
> I must be getting tired. The straight pin stop isn't feasible, of course. It would lock the top rotor to the motorstick, to which the bottom rotor already is locked. Doubt the model would fly.
>
> It is feasible to stop the top rotor with a pair of vertical sticks. They can easily influence the load on the scale through friction against the rotor, so some averaging may be needed. But something should be gleaned from this experiment, at least from the free-top- rotor case.
>
> Ray
>
> --- In Indoor_Construction_at_yahoogroups.com, "ray_harlan" <rbharlan@> wrote:
> >
> > Kang has some salient points regarding the fact that one cannot discern whether or not the bottom rotor is providing lift of at least the total weight of the model and rubber, so one can't tell if the model is actually flying. That is why the AMA rules disqualify stopped-top-rotor flights.
> >
> > The math is a bit off. Torque supplied to each rotor is the full torque of the motor. Think of it this way. If you put a torque meter at each end of the motor, they both would read the same and it would be the full torque of the motor. If you have a motor test rig with a torque meter at one end and a prop at the other, the torque would not change if you suddenly grabbed the prop and stopped it. This says that the torque delivered to the prop doesn't depend on what the prop is doing.
> >
> > What does happen when the model hits the ceiling, is that the angle of attack of the rotors increases dramatically and the props slow down, closer to a stalled condition. The change in lift is probably messy to analyze. If we assume that the model is rising at a constant velocity before hitting the ceiling (lift = drag), when it does hit, the lift may increase dramatically because of the increased angle of attack (but diminished some by the decreased RPM). That this is plausible is borne out by Sarath's experience with her model staying up a longer time yet falling normally. The scenario probably is that there is sufficient excess lift (above weight)even with a stopped prop to keep the model aloft for a while. The stopped prop keeps the motor unwinding at about half the rate of both rotors unwinding, further adding to the time.
> >
> > An easy way to test this hypothesis might be to put a weight, several times heavier than the helicopter plus motor on an electronic scale and tie a string to it and the helicopter. Wind up the motor and put it on the model. Let it "fly" and see what the change in weight is on the scale. Then stop the top rotor by putting a straight pin in the prop hook and let it jamb against the motorstick. Then see what the change in weight is, for the same motor torque. The tether might help to stabilize the model.
> >
> > Ray
> >
> > --- In Indoor_Construction_at_yahoogroups.com, "Yuan Kang Lee" <ykleetx@> 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:
> > >
> > > RPM is proportional to sqrt (Torque).
> > >
> > > For the specific case of a helicopter with two rotors, each rotor sees 1/2 torque, we have
> > >
> > > 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%.
> > >
> > > =======
> > > Now, it is true that with enough torque, one rotor turning has enough lift to hold the helicopter up. But I see two main problems. The biggest problem is that no one can determine without careful experimentation at what point the lift falls short of holding the helicopter aloft. The reason is that in almost all cases, when the top rotor stops due to static friction, there is static friction in the vertical direction that also holds the helicopter up.
> > >
> > > I will give you a couple of examples that I saw personally:
> > >
> > > - the use of tape on the top antenna to stop or slow down the top rotor. There is clearly friction in the vertical direction.
> > >
> > > - having the rotor stopped as it is stuck against a girder, even though the rotor appears to be "loose" and not stuck but the rotor stops turning. Again, it is impossible that some vertical friction is not in place
> > >
> > > - using a "wedge" at the top of the antenna to stick in a crevice. Same. The torque of the motor must put some vertical friction in the crevice.
> > >
> > >
> > > The second problem is this: without using the ceiling / building as a "crutch", the one rotor would not hover in a stable manner. Yes, one rotor might have enough lift in the vertical direction to keep the copter aloft -- but the copter would never fly on the one rotor because it would be unstable.
> > >
> > > All opinions. Please forgive me if I made a mistake.
> > >
> > > Regards,
> > > -Kang
> > >
> > > --- In Indoor_Construction_at_yahoogroups.com, "abcd4321" <sarathjaladi@> wrote:
> > > >
> > > > Ok so we were testing our heli today and we got the ceiling bumper at the top of the prop stuck into a crack in the ceiling. the prop stopped spinning and it was only sustained by the lower prop. I am fairly certain of this because the whole copter moved up and down in the crack and after about 6 minuets it left the crack and flew down in the normal manner, as it would have had it not gotten stuck, and landed with a normal amount of winds left. This Leads me to believe that it is possible stop the rotation of the top rotor and still have a legal flight which lasts much longer than a normal flight. However one must find a way of stopping the rotation which does not support any of the mass of the copter.
> > > >
> > > > --- In Indoor_Construction_at_yahoogroups.com, "Bill Gowen" <wdgowen@> wrote:
> > > > >
> > > > > Note to the moderator - the below is not a rules related post!
> > > > >
> > > > > Without the benefit of any actual knowledge on my part I think you can say that when both rotors are spinning they are getting equal torque from the motor and would therefore have something close to the same RPM. When you stop one of them the torque to the other one does not increase so there shouldn't be any major change in the RPM of that rotor. Therefore it will take something close to twice as long for the motor to unwind.
> > > > >
> > > > > In the process of attempting to design a helicopter myself I experimented with fins replacing the bottom rotor. This configuration wouldn't generate enough lift to fly. My GUESS is that the same would probably apply to most twin rotor copters. I think without the benefit of support by the ceiling they probably wouldn't stay in the air.
> > > > >
> > > > > Notice all the disclaimers in the above opinions.
> > > > >
> > > > >
> > > > > And part of the issue with on-line responses, if you are not there to see it, it could be hard to judge if held up mechanically or aerodynamically. Which, to me, makes all the difference in the world.
> > > > >
> > > > > Its also why I asked the theoretical question. Because if I can't explain what I think I'm seeing 30 ft over my head against bright lights, is my interpretation correct?
> > > > > - Is the longer flight time reasonably explainable from a purely theoretical aerodynamic sense?
> > > > > - If some one could point me to the relevant propellor equations I might be able to puzzle it out with some examples.
> > > > > - Seems like a simple question, but I may be oversimplifying. Which has lower drag?
> > > > > --Two propellors identical but opposite pitch props working together.
> > > > > -- or one of those propellors working by itself
> > > > > -- to produce the same lift (or maybe exceed the minimum lift?).
> > > > > - True for all operating conditions, or just some?
> > > > > - Because if the drag isn't lower, how can the flight be longer?
> > > > > - And is it a small or large effect?
> > > > >
> > > > > Thanks,
> > > > >
> > > > > Jeff Anderson
> > > > > Livonia, MI
> > > > >
> > > >
> > >
> >
>
Received on Fri Apr 01 2011 - 04:03:28 CEST
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