This thread is privately moderated by mnowell129, who may elect to delete unwanted replies.
Jan 19, 2009, 04:35 PM
Mickey from Orlando. Really.
Joined Nov 2004
4,257 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.
Jan 19, 2009, 04:47 PM
Mickey from Orlando. Really.
Joined Nov 2004
4,257 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.
 Jan 20, 2009, 01:00 AM Tannenzäpfle-Terminator 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
Jan 20, 2009, 03:31 AM
Registered User
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?
Last edited by JochenK; Jan 20, 2009 at 04:51 AM.
Jan 20, 2009, 05:58 AM
Mickey from Orlando. Really.
Joined Nov 2004
4,257 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.
Jan 20, 2009, 06:02 AM
Mickey from Orlando. Really.
Joined Nov 2004
4,257 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.
 Jan 20, 2009, 06:19 AM Registered User 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
Jan 20, 2009, 06:47 AM
Mickey from Orlando. Really.
Joined Nov 2004
4,257 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.
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.
 Jan 20, 2009, 07:19 AM Registered User Joined Aug 2006 1,464 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
Jan 20, 2009, 07:52 AM
Mickey from Orlando. Really.
Joined Nov 2004
4,257 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.
 Jan 20, 2009, 08:31 AM Registered User 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
Jan 20, 2009, 09:38 AM
Registered User
Joined Aug 2006
1,464 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
Jan 20, 2009, 09:50 AM
Mickey from Orlando. Really.
Joined Nov 2004
4,257 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.
Jan 20, 2009, 09:51 AM
Registered User
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.