

Joined Nov 2004
3,914 Posts

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Joined Nov 2004
3,914 Posts

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Hello..........
Mickey, thanks for explaining the factor a. Now I got it. I just made a mistake with the words I used, I wrote dragpolar and thought about the lift curve.............. So there is a difference if I use this formula on a helicopter, where I directly manipulate the pitch via the swashplate or on a autogyro, where I manipulate the blades pitch indirectly with moving the CofG below of the rotor. If the rotor is running stable, like the wheels of a bicycle, it will take some time until the different CofG has made the effect on the rotor. So the gyroscopic stability of the rotor smoothens and delays my controlinputs, and thus lowers the followingrate and makes the model controlable. How much the inputs are smoothened, thus depends on the headconstruction: GFflexplate: more direct input, teeteringhead: more smoothened and delayed, teeteringhead with foam: something adjustable in between. And not to forget: the tip weights also increase the gyroscopic stability of the rotor! Bye Mike 



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I made that mistake, not Mike. Never having heard of the lift curve slope steepness before and not knowing that it's a function of the aspect ratio of the wing/blade, I had a look at the polars of the SG 6042, took a δ CL and a δ α reading around an angle of incidence of 2° and processed the values. I'll change my post accordingly. With the new values of a from the calculator you mentioned, I calculated the follwing rates of my three very tame threebladers and got values that were in the same region as the main rotor's following rate in Drake's example, far to high. It seems to me that the following rate of the whole system is a good deal slower than the following rate of the rotor alone. Jochen P.S.: Mike, can you do another copy/paste, please? 



Joined Nov 2004
3,914 Posts

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Joined Nov 2004
3,914 Posts

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I'd like to see Hiller's original paper. Also Hiller helicopters reportedly had very sluggish controls. I think Hiller was conservative on the numbers. 




Hello…….
Now we have the correct value for the factor a, but the following rate still is too high. As you all fly happily with your gyros, the followingrate has to be tamed somewhere. There are to ways we know exactly how the rotor is tamed: FLYBARLESS: Following rate tamed with some gyros and a microcontroller working with fast servos. Wasn’t there a guy from Australia on the forum who builds a flybarless twobladed gyro who tamed the following rate with a gyro? He was warned by Mickey, but he got this gyro to fly, with the known difficulties of course……… FLYBAR: We know this works with success. You give some input with the control servos, but the controlinput is tamed by the flybar, the flybar also stabilizes the rotor. As we know the rotors of all the other gyros are tamed somehow, I just think about where this could be done, and had the Idea that this happens at the rotorhub. Because of its flexibility, the controlinput doesn´t act immediately on the pitch of the blade, they are smoothened a little and so the rotor is tamed……. The grade of taming is varied by different parameters: *** GS = Gyroscopic stability of the rotor (the thing that makes you ride a bicycle) *** FL = Flexibility of the rotorhub (you misalign the CofG with the rotorspindle, but if you have a very soft hub, it takes some time until the spindle and the CofG are aligned again) *** AD = Aerodynamic damping of the airframe *** MD = Mechanical damping due to the mass of the airframe and its specific moment of inertia (all parts concentrated versus large distances between them) *** IT = Induced torque by misaligning (the more torque, the more input on the rotor) *** FH = Followingrate for Helicopter calculated with the formula by John Drake So a formula for the following rate of an autogyro FA could look like this: FA = ( IT * FH ) / ( GS * FL * AD * MD ) Okay. That’s not THE formula; it just shows where I would put the values: FH and IT make high following rates, GS, FL, AD and MD tame it. Perhaps there is somebody out there who knows a lot more about maths like I do who will try it…. Perhaps this formula could help to understand a little more about the bigger models and the small foamies; I think they have found different solution how to influence the parameters above to get a solution to the formula above which results in a total followingrate FA that is in the good range…… Bye Mike 



Mickey,
what I'm mainly missing is a factor that takes the number of blades into account. I've got both the Minimum and Micromum in three. and twobladed versions with nearly identical fuses. The calculated following rates are roughly the same for the two and threebladed rotors, but in reality there's a big difference. Jochen 


Joined Nov 2004
3,914 Posts

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A second factor is the following rate calculation assumes a free hinge at the blade end. The two bladed, certainly the teetering, likely has a much freer hinge line with respect to the three bladed. This provides additional damping in the three bladed that doesn't show up in the two blade. Also note that a flapping offset of 0 is assumed in the following rate calculations and that the two blader has likely a much smaller effective hinge offset, probably right at the hub itself, where the three bladers real offset is somewhere out close to the blade. I suspect that if you made a three bladed head with three independent tabs for flex plate, you would get higher control response. Finally for the same diameter the three bladed head can turn slower, reducing the following rate. My guesses in summary: Some combination of control feedback reduction plus end damping considerations. 




Micky,
maybe I've missed the point (in which case please ignore this), but if you compare Jocken's 2 and 3 bladed head designs the hinge lines are the same in both cases. http://www.rcgroups.com/forums/attac...mentid=1471302 http://www.rcgroups.com/forums/attac...mentid=1606407 Interesting discussion though..... PeterO_UK 


Joined Nov 2004
3,914 Posts

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Mickey,
I think the secret of the following rate of multibladed rotors may be hidden in the moment of inertia. Drake calculates the moment of inertia for a single blade. For a twoblader you get double that moment and for a threeblader it's trice as much, and the following rate gets reduced accordingly. Jochen 



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My "gut feeling" is that the horizon will initially roll the wrong way entering a turn as the airfame is moving under the rotor disk and it will be the consequential offset in the CofG that applies the force to the rotor disk. PeterO_UK 



Joined Nov 2004
3,914 Posts

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First hole: If the rotor is a gyroscope and you shift the CG, say to the right, applying a force to the rotor, then the rotor gyroscopically precesses NOSE DOWN (assume CCW rotation), not to the right. Second hole: The teetering rotor offers no resistance to the spindle movement. How then does one move the fuselage? So you move the spindle, so what? the rotor offers no resistance to this movement. So what does the fuselage push against to move itself for weight shift. If the fuselage moves at all, I believe what happens is 1) Fuse moves to the right. 2) Weight of fuse pulls it back vertical. 3) The now tilted spindle applies cyclic pitch to the rotor. 4) The rotor now moves to the new position with cyclic pitch. 




Folks,
I've had another idea concerning the following rate. Attached. Jochen Edit: I've added another thought to the document. 