HobbyKing.com New Products Flash Sale
Reply
Thread Tools
This thread is privately moderated by mnowell129, who may elect to delete unwanted replies.
Old Jan 19, 2009, 04:35 PM
Registered User
Joined Nov 2004
3,327 Posts
Quote:
Originally Posted by JochenK
Mike,

I'm not sure what the factor a really means. John Drake had an original factor a of 5.73 which he then reduced to 3.75 and I don't know how he got this factor. That's one thing, the other one's that the formula is for helicopter blades. From what I've seen, these blades can't move 'vertically' as our gyro blades do around the flapping hinges, and that may influence the absolute values calculated. For relative calculations the formula seems good enough.

Jochen
"a" is the lift curve slope, about 2*pi for infiinite wings, but reduced by aspect ratio and some other factors. 3.75 is a reasonable lift curve slope with these size blades and aspect ratios.
mnowell129 is offline Find More Posts by mnowell129
Reply With Quote
Sign up now
to remove ads between posts
Old Jan 19, 2009, 04:47 PM
Registered User
Joined Nov 2004
3,327 Posts
Quote:
Originally Posted by Hephaistos
I get something like a = 5.7, for the rotor of the RPG with a Reynolds number of about 120000, I get a = 6.9.
This can't be, for an infinite wing the theoretical value is 2*pi = 6.28.

Quote:
Originally Posted by Hephaistos
The value for a, which describes the steepness of the dragpolar at a certain point, will be different for every angle of attack you extract it.
It's the steepness of the lift curve slope with the angle of attack in radians, not the drag polar. The lift curve slope is basically linear up to near stall, so a single value is pretty constant through any region of interest we would be concerned with.
mnowell129 is offline Find More Posts by mnowell129
Reply With Quote
Old Jan 20, 2009, 01:00 AM
Tannenzäpfle-Terminator
Hephaistos's Avatar
Baden-Baden, Germany
Joined Dec 2008
436 Posts
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 control-inputs, and thus lowers the following-rate and makes the model controlable.
How much the inputs are smoothened, thus depends on the head-construction: GF-flex-plate: more direct input, teetering-head: more smoothened and delayed, teetering-head with foam: something adjustable in between.
And not to forget: the tip weights also increase the gyroscopic stability of the rotor!

Bye
Mike
Hephaistos is offline Find More Posts by Hephaistos
Reply With Quote
Old Jan 20, 2009, 03:31 AM
Registered User
JochenK's Avatar
Joined Jan 2005
1,286 Posts
Quote:
Originally Posted by Hephaistos
I get something like a = 5.7, for the rotor of the RPG with a Reynolds number of about 120000, I get a = 6.9.
Mickey,

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 three-bladers 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?
JochenK is offline Find More Posts by JochenK
Last edited by JochenK; Jan 20, 2009 at 04:51 AM.
Reply With Quote
Old Jan 20, 2009, 05:58 AM
Registered User
Joined Nov 2004
3,327 Posts
Quote:
Originally Posted by Hephaistos
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.
Please explain how tilting spindles are not cyclic pitch like a swashplate.

Quote:
Originally Posted by Hephaistos
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 control-inputs, and thus lowers the following-rate and makes the model controlable.
How do you account for the 90 degree phase lag in a gyroscope? Please explain how weight shift can work with precession in play? Also please explain how you can shift the CG of a teetering head gyro?
Quote:
Originally Posted by Hephaistos
And not to forget: the tip weights also increase the gyroscopic stability of the rotor!
Please explain how this provides stability.
mnowell129 is offline Find More Posts by mnowell129
Reply With Quote
Old Jan 20, 2009, 06:02 AM
Registered User
Joined Nov 2004
3,327 Posts
Quote:
Originally Posted by JochenK
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.
agreed. Something that bugs me in the FR formula is the result is in rad/sec/rad lag.
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.
mnowell129 is offline Find More Posts by mnowell129
Reply With Quote
Old Jan 20, 2009, 06:10 AM
Tannenzäpfle-Terminator
Hephaistos's Avatar
Baden-Baden, Germany
Joined Dec 2008
436 Posts
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 following-rate 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 two-bladed 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 control-input 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 rotor-hub. Because of its flexibility, the control-input 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 rotor-hub (you misalign the CofG with the rotor-spindle, 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 = Following-rate 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 following-rate FA that is in the good range……
Bye
Mike
Hephaistos is offline Find More Posts by Hephaistos
Reply With Quote
Old Jan 20, 2009, 06:19 AM
Registered User
JochenK's Avatar
Joined Jan 2005
1,286 Posts
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 two-bladed versions with nearly identical fuses. The calculated following rates are roughly the same for the two- and three-bladed rotors, but in reality there's a big difference.

Jochen
JochenK is offline Find More Posts by JochenK
Reply With Quote
Old Jan 20, 2009, 06:47 AM
Registered User
Joined Nov 2004
3,327 Posts
Quote:
Originally Posted by JochenK
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 two-bladed versions with nearly identical fuses. The calculated following rates are roughly the same for the two- and three-bladed rotors, but in reality there's a big difference.

Jochen
I think what is missing is that the following rate per blade for amount of cyclic input is the same. But, the three bladed head opposes the cyclic input at the servos allowing only small amounts of cyclic input, so even though the blade follows quickly the amount of overall input is at a low rate. The two bladed rotor does not provide any feedback to the servos and immediately follows the entire commanded input. I think that if you had a three bladed head with very small flapping offsets you would have a much higher overall rotor response. Because in the three bladed heads, the flap offset is usually pretty large, this provides a lot of control input resistance. I don't know how to mathematically characterize this.
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.
mnowell129 is offline Find More Posts by mnowell129
Reply With Quote
Old Jan 20, 2009, 07:19 AM
Registered User
PeterO_UK's Avatar
Joined Aug 2006
1,461 Posts
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
PeterO_UK is offline Find More Posts by PeterO_UK
Reply With Quote
Old Jan 20, 2009, 07:52 AM
Registered User
Joined Nov 2004
3,327 Posts
Quote:
Originally Posted by PeterO_UK
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
I wasn't referring specifically to Jochen's designs, just in general with the flex plate designs. My guess is that Jochen's designs are more affected by control feedback and other factors. The two bladed design has very little resistance to control input. I still think this is a big factor.
mnowell129 is offline Find More Posts by mnowell129
Reply With Quote
Old Jan 20, 2009, 08:31 AM
Registered User
JochenK's Avatar
Joined Jan 2005
1,286 Posts
Mickey,

I think the secret of the following rate of multi-bladed rotors may be hidden in the moment of inertia. Drake calculates the moment of inertia for a single blade. For a two-blader you get double that moment and for a three-blader it's trice as much, and the following rate gets reduced accordingly.

Jochen
JochenK is offline Find More Posts by JochenK
Reply With Quote
Old Jan 20, 2009, 09:38 AM
Registered User
PeterO_UK's Avatar
Joined Aug 2006
1,461 Posts
Quote:
Originally Posted by mnowell129
I wasn't referring specifically to Jochen's designs, just in general with the flex plate designs. My guess is that Jochen's designs are more affected by control feedback and other factors. The two bladed design has very little resistance to control input. I still think this is a big factor.
When I get the chance I'll try and mount my flycam such that it can see the servos and the horizon. Then we'll be able to see how the model actually reacts to roll input.


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
PeterO_UK is offline Find More Posts by PeterO_UK
Reply With Quote
Old Jan 20, 2009, 09:50 AM
Registered User
Joined Nov 2004
3,327 Posts
Quote:
Originally Posted by PeterO_UK
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.
There are two gaping holes in this argument that no one has yet to explain.
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.
mnowell129 is offline Find More Posts by mnowell129
Reply With Quote
Old Jan 20, 2009, 09:51 AM
Registered User
JochenK's Avatar
Joined Jan 2005
1,286 Posts
Folks,

I've had another idea concerning the following rate. Attached.

Jochen

Edit: I've added another thought to the document.
JochenK is offline Find More Posts by JochenK
Last edited by JochenK; Jan 21, 2009 at 07:11 AM.
Reply With Quote
Reply


Thread Tools