Originally Posted by PeterO_UK

Something else that occured to me last night... On a head with delta hinges, when the head and spindle are tilted to the right with respect to the rotor disk, as well as the "twist" cyclic applied to the blades in the fore and aft positions, there will also be cyclic applied in the side-to-side positions because the blades will appear to have flapped up and down with respect to the head.

However this second cyclic input would seem to add to the gyroscopic effects of applying the force directly to the head and in the case of right tilt will make the rotors tilt nose down. This nose down tilt in the fore positon will again appear like flapping and will increase the pitch , which is also what the "twist" cyclic pitch is trying to do in that position. Now, the question to ask is "Is this effect is significant ?" bearing in mind it doesn't occure on heads without delta hinges and they do still fly OK.

PeterO_UK

not for small delta 3 angles. As you have discovered delta 3 produces an out of phase response. Too much delta 3 creates an unstable situation. From what I've read delta 3 creates a lot of arguments in the full sized aircraft. This out of phase response is problematic for me.

Hello Mickey.........

Okay, on a teetering-head with some ideal frictionless bearing for the teetering, the spindle cannot influence the rotors plane of rotations. The steering works with cyclic pitch only.

Ba adding simply some piece of foam to the teetering-head, the spindle will influence the rotors plane of rotation.

The significant differences caused by simply squeezing some foam in the teetering head are caused by the influence the spindle makes on the plane of rotation. The spindles input on the cyclicity remains the same.
Okay, the foam might also influence the way the blades teeter, but if I spin up my handheld rotor very fast and take it out of the airstream, its still turning a few seconds without teetering and I can still feel the difference if the foam is inside or not. And if I add tip weights to the rotor, I can feel that the influence of the foam is reduced and the difference between free and foamed are smaller.

On the flex-plate-heads you have to deal with the flexibility: To soft you don´t have much control, to hard you have to much control. I think to change the thickness of a well-working flex-plate could turn this gyro to an unflyable one.

And if it depends on the flexibility, it also depends on the inertia of the rotor-disc and its gyroscopic stability, because they work together and moderate the control inputs.

Bye
Mike

Originally Posted by Hephaistos

Jochen, I didn’t understand perfectly what you have done in the last step of your changed document.

Mike,

I referenced a lift-slope curve generator I had already taken out of an earlier document. Now it's safe, I hope.

Jochen

Originally Posted by Hephaistos

Hello Mickey.........

Okay, on a teetering-head with some ideal frictionless bearing for the teetering, the spindle cannot influence the rotors plane of rotations. The steering works with cyclic pitch only.

Agreement. This has been my point all along. Even the three bladers and flex plates steer the same way because ....

Originally Posted by Hephaistos

Ba adding simply some piece of foam to the teetering-head, the spindle will influence the rotors plane of rotation.

... Because this influence is out of phase, right left spindle will make nose up/down rotor response, so it is not helpful in steering. Rather the foam allows the rotor to put force on the spindle providing more authoritative control.
If you make a non-aerodynamic test rig, just a wire with weights on the end for a rotor and spin it with a drill motor you will see that you can't force it to tilt directly, instead you have to tilt 90 degrees ahead. It's the same as the spinning bicycle wheel.
Draw a simple diagram of a rotor with a foam or stiff plate. Show where the tilting spindle applies a force to the blade around the rotation, show the blade movement 90 degrees later. Do the same with the cyclic input from the spindle, show the additional force due to cyclic, and the resultant rise 90 degrees later. You will immediately be able to see that the cyclic input from the spindle is moving the rotor the direction you desire but the force input from the spindle is making the rotor move at right angles to the spindle.
Now notice that with a flex plate or foam that when the blade does move as a result of cyclic that the blade through the flex plate applies torque to the fuselage to make it move the right direction and with a teetering rotor it does not, it just makes the lift force move off to the side and that makes the fuselage drag along in the right direction (this explains the relative softness of a teetering rotor response).
A single diagram will explain all....

Hello............

Mickey, I played a little with my handheld-rotor-toy again. I fixed some really heavy tip weights (50 grams each) to it and spinned it by hand, say 60 turns per minute.

EDIT START

60 RPM are completely wrong . I estimated about 6 turns per second and, at 6 o´clock in the morning right after waking up and drinking the first coffee, I multiplied with ten instead of sixty . So spinning the rotor-toy by hand makes 360 turns per minute.

EDIT END


Running in calm air I think the aerodynamic effects are quite small. With free teetering the rotor keeps its plane of rotation perfectly.
With the foam inside, when I move the spindle, I first hear the foam work (its some strange sound you only hear if its completely quiet in the room, it comes from the foam moving relative to the mechanics if the bleade teeters), then I see the out-of-phase response (its size depends on how fast I move the spindle). The out-of-phase response is combined with some already changed plane of rotation towards the direction i moved the spindle. Then the out of phase response decreases and the plane of rotation (from now on PofR) is orthogonal to the spindle again.

So its different steps:

1) moving the spindle
2) initial change of PofR according to spindle movement caused by foam
3) out-of phase-response caused by 2) changes PofR in a second direction (precession)
4) the spindle stops moving
5) rotor is teetering strongly (depends on how fast the spindle was moved)
6) out-of phase-response decreases because the input of moving the spindle is gone and the foam is absorbing its energy
7) rotor has reached its new PofR and teetering is stopped


Jochen, I had a closer look at the Drake-formula following rate again. The more I look at it, the more I get the feeling that there has been made some simplification. I was ment to explain why a model-helicopter needs a flybar and that worked. Perhaps Hillers original formula was different.
You calculated the torque of the blade and the effect of the torque to its angular acceleration and got quite close to Drakes formula. What i miss is the resistance of the rotor disc against changes of its PofR. This effect is there, like on the gyroscopes for building the artificial horizon. I had a look at the formulas for gyroscopes, but I still work to get it inside my head.
http://www.wmi.badw.de/teaching/Lect...ik_I_Kap_2.pdf
This is a link to a GERMAN university script explaining this things.

Bye
Mike

P.S. I attached a picture of the rotor-toy when editing for those who havent seen it.

Mickey,

from the formula for the angular velocity ω = T / I *t (torque divided by moment of inertia times time) I've derived formulas for the angular velocity of a one-bladed main rotor as well as a one-bladed control rotor. The significant difference between my and Drake's formulas is that I'm missing a quotient of 2 (integer number) both times. Could this '2' be the - up to now missing - number of blades of the rotor?

Jochen

Mike, interesting test, glad you are so energetic.

Originally Posted by Hephaistos

RUnning in calm air I think the aerodynamic effects are quite small.

I'm not certain that they are insignificant. You won't know till you try this with a non aerodynamic device. Also 60 RPM is fairly slow, at some point in the reduction in RPM all the resonant/precession effects start to fall apart and you can directly manipulate the rotor.

Hello......

Mickey, If I move the spindle of the free teetering rotor it keeps its plane of rotation without any changes. So the aerodynamic influence is small enough to be ignored.

EDIT START
Mickey doubted the existence of precession at a RPM of 60, a wrong value I initialyy posted .
I made a mistake when calculating the RPM, the real RPM is 360 and not 60 (still estimated, but reality is much closer to 360 than to 60 )
EDIT END

360 RPM is enough to get strong reactions of precession, because I put this heavy tip-weights to it. If I fix the teetering bearing and move the spindle, I have to be shure to have a good grip so the rotor will not fall down.
Perhaps I get a stroboscopic light from my cousin to make a photographic examination with several flashes and a long exposal time at the camera.
But 360 RPM are so slow that I can see the movements if I use neon-lights (they flash with 50 Hz if you have some non-electronic-ones)
I am shure that the result will be the same if I replace the blade with a welding wire, but for making some photo it would be better to do so to eliminate all reason for doubt.

Jochen, If you talk about the factor 2 you miss it one time in this, and the other time in the opposite direction like I do?

And considering a movement in all three axis, would it not be neccesery to consider the moment of inertia in all three axis?
The Drake-formula describes something aerodynamic happening in two rotatative axis, and for this its really simple, so I am a little in doubt

Let XXX be the length of blade axis
Let YYY be the chord axis
Let ZZZ be the spindle axis

For two blades the complete INERTIA:

I Rotor XXX = 2 * I Blade XXX
I Rotor YYY = 2 * I Blade YYY
I Rotor ZZZ = 2 * I Blade ZZZ

And I YYY can be simplified and set equal with I ZZZ

And for three blades we get the following

I Rotor XXX >>3 * I Blade XXX
I Rotor YYY >> 3 * I Blade YYY
I Rotor ZZZ = 3 * I Blade ZZZ

this results from the angle of 120 degrees, the masses concerning the XXX and YYY are not aligned anymore, and if the is distance beetwen, we get higher values.

And for a formula working with two rotative axis, you need two Inertias. For the two bladed rotor, the are equal.
But for the three-bladed they are not equal. So a formula working with only one Inertia can NEVER describe a three bladed rotor.

Thats enough for now, I am going to pick up my work again!

Bye
Mike

Mike,

I think you can only do this sort of calculation for a single bladed rotor. A two-bladed teetering head rotor - turning, but not moving - will give you double the lift, but will not tilt the rotor plane in any direction, as the second blade prevents that. And since I know from my own experience that there's a very noticable difference between two- and three-bladed rotors, I was exploring the possibility that the addinitional 2 might be there to take the number of blades into acoount. Something analog to integrating a sinus curve over a whole period to get the area between the waveform and the baseline. The result is zero, but if you integrate only over half a period nd double the value, you get the correct result.

Jochen

Hello.........

Mickey, I made a mistake with the RPM. I edited the posts with the wrong RPM. I tried to keep the history by surrounding the edits with EDIT START - EDIT END, I think its okay like this. Your doubts about existence of precession hopefully fall away now.

Jochen, calculating a single-bladed rotor is shurely not the thing we dream of. The values you get just show you the direction you go, but they dont help you to decide what will work and what doesn´t. The more I try to find something useful to help us out of this situation, the more I doubt that there could be some more or less complicated formula. A useful formula you put the parameters of your construction inside and you get some answer if its controllable or not.

Thinking about the twist of the blade while flying points in the same way, and I think nobody will doubt that the twist is important. Perhaps the only way to crack this problem is to have some FSM-motion-simulation. For this, you calculate the forces and moments on the system at a certain time, then you use this to calculate the position at the next timestep, and then you start again. If the timesteps are small enough, you get the right results.

The TERVAMAKI-Simulation for real autogyro works the same way, perhaps its time to have something like this for model-autogyro. I can use the software we have at work for this things, my boss is always happy if somebody has fun with the programs we use at work and therefrom gets knowledge he can sale to our customers.

But I think without the help and input from you all I cannot do this.

Bye
Mike

I had a bit of a realisation this afternoon when I was telling a fellow modeler at work about this discussion....

Anyway, I suddenly realised what Micky has been saying about servos moving the hub of direct control heads.

Because the shaft and hub are only moved through a few degrees, the torque supplied from the servos only needs to overcome a fraction of the centrifugal force from the blades.

See the attached diagram

The black line represents the hub tilted through 7 degrees (exeagerated).
The red line is the 12.5N force that Micky calculated for a typical blade.
Resolving the red line normal to the hub gives a 3.4 Ncm torque at the hub bearing.
The yellow line represents the 500g airframe hanging below the rotor.
If the servos moved the airframe 7 degrees to one side, it would produce a 9.1 Ncm torque at the hub bearing. (I forgot to put on the diagram that the CofG is 15cm below the hub bearing).

Clearly it is easier for the servos to move the hub through 7 degrees than to move the airframe through 7 degrees.

PeterO_UK

UPDATE: It's not quite as clear cut as I thought... Firstly my blades are about twice as heavy as Micky used in the calculation, so the 12.5N should be more like 25N. Secondly I forgot about the other two blades, which will combine to exert the same torque from the other side.
So the torque to rotate the head 7degrees is more like 13.6Ncm. I also got the airframe weight wrong, so the other torque should be 10 Ncm.

If the two figures were equal, I think it mean that if the servos moved to put a 7 degree angle between the rotor hub and the airframe, then the hub would move 3.5 degrees one way and the airframe would move 3.5 degrees the other way.

PeterO_UK

Hello………

Peter, I am sorry that I have to tell you that the things are not as easy as you imagine. This problem has been discussed for several times, so I decided to make a little picture (attached) and try to explain it from my point of view.

The first important thing is to choose the right point of view. You have to fix one point of the system for further examination. If you want to decide if the rotor or the airframe will move, this point can neither be the rotor nor the airframe.
Theoretically you can make the examination from every point of the system, but for easier calculation it’s commonly used to take one point with importance to the system. I decided to fix the head-joint for my explanation, this gives the rotor and the airframe the possibility to move, and the torques calculated for the head-joint are the torques that are needed from the control-servos to keep the head-joint in this position (without respect of eventually used levers for angle-reduction and torque-amplification).

For a fixed-wing you can fix the CofG in your imagination without making significant errors, if you move the rudder there will by some reaction of the airframe to the opposite direction (only from the masses inertia), but the inertia of the rudder is very small in comparison to the inertia of the airframe, so you can ignore it.

For an autogyro the things are different. If the rotor is not turning, and you move the servos, for example when checking their function, you can see some little reaction of the airframe, because the mass of the rotor is bigger in relation to the mass of the airframe in comparison to the rudder of the fixed-wing.

Getting the autogyro to air, the things get different again. The rotor starts to turn, and as we all know, the faster its turning, the more it resists to changes in its plane of rotation. Or, in other words, its inertia in comparison to the inertia of the airframe becomes really important.

For a practical experiment I suggest to do the following: Take place on a swivel chair and spread your arms to the side. Lift your feet so they don’t touch the floor. Now imagine that your head is the head-joint, your right hand is the airframe and your left hand the rotor. If you move only one hand fore- or backwards, the swivel-chair will start to turn. Only if you move both hands together, you can prevent the chair from turning. As your hands have the same inertia, you have to move them with the same speed. Now take something heavy to one hand and try again to move your hands fore and aft without making the chair turn. The hand with the weight now makes smaller movements and the other bigger ones. That’s how my examination of the autogyro works.

Now have a look at the drawing I attached. You see the rotors lift and drag and the vector-addition to get the result. The result is translated to the head joint I choose for point of view.
When the result at its origin at the rotors middle is not aligned with the head-joint, you create a torque also shown in the drawing. With some clever construction you can minimize this torque, but that’s something else.
For the airframe with the prop it’s the same.
For flying at level, straight ahead with constant speed, all forces and torques have to sum up at zero. That’s the condition of stability for the system.
Now we move the control-servo a little. It will apply a torque to the rotor and to the airframe.

And now we go for the important part: This torque is exactly the same for the rotor and the airframe, because every action is always producing a reaction of exactly the same size but with opposite direction.

How far the airframe and the rotor will move depends only on their individual moments of inertia, because they have to take the same torque.

So the airframe and the rotor will both move, causing reactions like precession, cyclic pitch, misalignment of the CofG etc, and a few seconds later the system is balanced again and we have a new stable system with a different direction of flight.

Until now, I explained this with assuming a non-flexing, non-teetering, completely rigid rotor-hub. If you can accept my conclusion that there is a torque applied to the rotor-spindle, the next step would be the examination how the different types of rotors react to the applied torques. As my experimentation with the rotor-toy has shown, there are significant differences, and these differences have significant influence on the reaction to the applied torque from the servo.

Bye
Mike

Originally Posted by Hephaistos

Hello………

Peter, I am sorry that I have to tell you that the things are not as easy as you imagine.

I've not read much of the rest of your long post because it seems you've totally missed the point of what I was describing.

My diagram only concerns the forces on the hub of a hinged head. It only concerns the forces acting on the hub when the servos move.

Now take something heavy to one hand and try again to move your hands fore and aft without making the chair turn. The hand with the weight now makes smaller movements and the other bigger ones. That’s how my examination of the autogyro works.

I was right then ! That's pretty much the same point I was making......

PeterO_UK

Hello Peter........

The effect you have described in your post is pointing in the right direction. As you see, we get the same result with different ways from our examination concerning the torque.
Having the same torque on both moving parts is not a result, its a basic rule of static systems like this. To have an idea which movement this torque will cause, its necesssery to have a look at their moments of inertia.
I am sorry that my post made you angry. I just intended to pick up your idea an extend it a little.

Bye
Mike

I'm working on an interactive graphical representation of this stuff from which I hope to be able to make a short animation, but I'm going flying now so it won't be finished until tomorrow maybe.

PeterO_UK