RRPM - Searching for some explanations

JC, In the early days of phonographs, drive was by a “clock” spring. You had to wind them up like a mechanical toy. Speed of the turntable was controlled by a friction brake activated by centrifugal weights. That device is called a governor. Lawn mowers have governors, steam locomotives have governors.

But that’s not important.

With very high incidence of rotorblades, so much of the retreating blade is stalled that any attempt at speed increase spreads stall near retreating tips, making rotor disc angle of attack very sensitive to forward speed. Acts like a governor on a rotary lawn mower.
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A metaphor doesn’t translate well; I’ll try to avoid them in the future.
 
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5-10 seconds from hard shock to the absence of shock on the stops can not be considered a normally fast acceleration. What flapping angle is "stick forward nearly stop" despite the disk axis of the mass center? Is it realy higher to the limit (4.5°) I mentioned?

JC, when you take off in a very short distance you then fly level just above the ground to build up speed.......for a gyro to fly in balance (straight and level) rotor thrust vector must pass through its CofM. normal stick range is ~ + or - 9 Deg.

So if you have the stick almost forward... lets say 6 deg.... the rotor trust vector must still pass from C of M. for a level flight, that means flapping is 6 deg. (some compensation depending engine thrust line etc).
 
Nicolas, You are right, but I do not see your stick near the forward stops, in your pictures. What do you think?

Chuck,
your metaphors often facilities understanding. I just do not understand your thinking here.
Of course, the great AoA rotor opposes the forward speed. but my confusion comes from the acceleration of the rotor despite the very large stall on the retreating blade.
Perhaps, the increased drag and reduced post-stall of the airfoil is too bad in my simplified calculation. These values ​​are prevailing in those moments.
 
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JC this was for Mr. Beaty's Experience ... not mine....

" In more recent times, I’ve flown from a rough grass strip and always pull the gyro into the air as quickly as possible. Don’t hit the flap stops but immediately following liftoff, the stick is nearly on the forward stop and the machine flounders, barely flying, for perhaps the first 5-10 seconds."
 
Yes Nicolas, but your reasoning holds true for your picture: With blade flapping 8 or 6° , the pitch balance of your gyro should be obtained with the stick near the forward stop. This does not seem so.
 
In my picture the stick is more forward than neutral and the gyro still rotates nose up.... so rotor thrust vector is NOT on C of M, but more forward....

then the last picture were stick is brought forward to stop rotating nose up, ( and will be kept there for the next 2-4 seconda.. like Mr Beaty explaned) the rotor thrust vector passes through C of M ( because stick is forward now) .......and flapping angle is around 8 deg.... ( most probably by now it will be down to 5 deg... and the rest 3 deg will be the cyclic input to decrease the rotor AoA... the picture most probably was just taken as i started moving the stick forward)
 
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one thing to remember also is that the rotor ( when unloaded and / or turns slower) will not support 1 G immediately when reloaded...

the rotor disk AoA will change (because of retreading blade stall ) EVEN IF YOU do not move the stick and will support the weight it can support at those RRPM . The gyro will be falling through the air because the rotor supports less than its weight and acceleration (downwards) will be slow since the rotor supports a lot of the gyros weight.

Retreading blade stall will have the rotor disk wanting to pitch up in relation to the flying path.... therefore the forward speed of the gyro will be slowed down, this will be putting the gyro closer to a vertical descend ( or more correctly to a slower flying speed) where flapping will decrease and the rotor will accelerate rpm until eventually can support 1 G.

all of the above will happen if you leave the stick at neutral and the frame does not have any tendencies to upset balance ( like HTL etc).

Helping the rotor recover will ensure for less RRPM and height loss. and it will recover from as low as 70% of normal RRPM.

now what is the minimum limit it can recover???????

i do not know..... but if we want we can find out :)
 
Following this fascinating thread with great interest, I have refrained from comment... until now.

...now what is the minimum limit it can recover???????
i do not know..... but if we want we can find out. :)

I do not know who the 'we' will be Nicolas, but it would appear that you, have gone a long way into finding out, for 'us.' :yo:
 
now what is the minimum limit it can recover???????
but if we want we can find out :)
That's what I'm trying to predict, Nicolas. It is difficult to ask a kamikaze to say, by radio, how is the rpm down before losing control ! Well instrumented, your test take off with slow rpm could be used to adjust the theory.
 
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now what is the minimum limit it can recover
To find the minimum in a numerical model (short of a fully fledged 3D coupled CFD analysis)
you would IMHO have to take into account

- Reynolds number dependent lift and drag data for the actual profile
- An inflow distribution as close to the real thing as you can get to (Peters/He/Quang)
- Lift hysteresis due to blade rotation
- Aeroelasticity in bending and torsion of the blade
 
Juergen, It is not very useful to know the limit to one r.p.m closely. I do not believe that 30% of variation of Reynolds number is fundamental. About r.p.m showed by Nicolas, the prop thrust (nose up) alone explains more than 10% loss of rpm.
However, r.p.m acceleration despite the high flapping angle left me puzzled. My spread sheet shows 4,5° and practice of short take off by Nicolas and Chuck shows >6°. I do not understand why.
The distribution of the airflow and the hysteresis that you mentioned, should reduce the longitudinal flapping, moving on the flapping cross. Not increase.
 
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Gentlemen:

I think it would be much easier to mount a rotor on a truck with some instruments and test its behavior at various airspeeds and angles of attack (and rates of change of angles of attack). This would provide precise data without risking anyone's life or destroying entire gyroplanes.
 
will not support 1 G immediately when reloaded
Nicolas is pointing out something very important here, namely that this a time dependent process. In simulation this is called transient behaviour so one actually would have to carry out a time integration from one state to the other. Nicolas' input, measurements and observations are one very valuable contribution from which the "eggheads" that dabble in theory profit vastly. Thanks goes out to him! (just take care, Nicolas.... let your wife approve and sign your test programs ...;-)


the hysteresis that you mentioned, should reduce the longitudinal flapping
Please let me explain why I think to the contrary. Prouty states:
If the angle of attack on a blade element is increasing rapidly the stall will be delayed until some angle is reached considerably beyond the stall angle
Loading a rotor means that you rapidly increase blade angle of attack so you will make use of the mechanism described by Prouty and thus be able to produce much larger flap angles than in the quasi static case.

it would be much easier to mount a rotor on a truck
Much as I like Dougs idea of the truck mounted test bed I wonder if this would work for the case under consideration. You could easily reproduce all angular changes but the gyro also accellerates linearly which you could hardly reproduce with a truck, especially in the z direction.
 
Doug's idea will be nice to get the data (which will be closer to JC's calculations) and then test the same rotor in the air and see the variation. I believe that in flight we will get referent data because the rotor will not be forced to a certain incidence (like in the truck) but will adopt its own incidence ( by varying the flight path) and loading will not be immediate.

Leigh, Juergen, thanks for the concern... my tests were progressive (5-10 RRPM less each time ) and i stopped when the gyro was warning me that i pushing it with a stick shake ( not very severe).

what we will need is couple of rotors (good enought for anmaned fligh but not good for maned flight) donated.. maybe an old engine.. something that runs for 5-10 minutes.... only.

then someone could test the truck part and i could test the air part... ( don't worry i will not be sitting in it.... i will need to investigate the results :) )

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JC,

I know your spread sheet calculate the AoA of sections of the blades through the disk ( i love that part) . How difficult could it be to have it calculate the DRIVING region volume of the rotor in flapping of say 3 deg and 6 deg ?




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Juergen, If the stall angle is pushed away by the hysteresis, then the flapping angle remains small: The flapping angle increases greatly when the increase of the retreating blade A.o.A does not balance the loss of airspeed.

Nicolas, the volume led to the ground by the rotor is produced by the lift , it does not depend on the state of blade stall. Moreover the lift of the rotor is symmetrical, even if the blade stall is not. Thus, A.o.A determined by my spead sheet is correct, even if the blades are stalled (Perhaps not understand your question)
The most likely explanation between my calculations and your take-off tests, is my drag model "post stalled" is pessimistic
 
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JC what i meant was:....

the rotor blade has some part that is driving autorotation (about a third of the blade in the middle).

so... how much driving rotor area is there when the flapping is 3 deg and how much when the flapping is 6 deg. ( or how much lift force component is driving ( autorotating) the rotor in the 2 flapping examples.

See images. (green area)

Also The airfoil continues to produce lift past the stall A .. is just less than Cl max.. and less .. and less.

Another point is that Drag at the blade part that mu is negative helps with autorotation
 

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Nicolas, Here is what gives the spreadsheet (about) with the same speed forward and the same rotor thrust for 75% and 62% of nominal r.p.m and 80% nominal rotor thrust
Of course, the airfoil continues to produce lift past the stall.
Airfoil drag in the inversed flow produces a negligible torque.
Jean Claude
 
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JC, the second graph is the take of situation on our genesis almost. When we have 200 RRPM we go full throttle allowing the gyro to accelerate ( not full back stick) and take of at ~260 - 270 RRPM and 15-20 m/s ( 33-45 mph). .. and surely the retreading side is not all stalled!!

Question.... on the bottom you have 310 RPM 18 deg AoA... do you mean rotor disk AoA?
what is the rotor diam?
 
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Nicolas, Yes 18° is disk A.o.A calculated with my rotor: 6,3m x 0,18m, nominal 417 r.p.m for lift 2120 N.
 
our gyro is 2450 N lift, 7.3m rotor and 340 RRPM.

i will check your graphs and let you know what i think.
 
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