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Thunder Tiger Raptors 30-90 - Imperio > Raptor feathering shaft
 
 
Virtual1
Senior Heliman
Location: Waterloo, Iowa - USA

I just got done replacing my feathering shaft... for some unknown reason one of those aluminum collars froze up on the shaft and would not move. (i.e. not even when persuaded with a hammer!) Dunno what was up with that.

Anyway, while messing with that, I was pondering the head design, and it occurred to me that both blades sort of float on the head, in the dampeners. There is a very small amount of ability for the feathering shaft to move laterally, (even with the bolts tightened fully onto each end of the shaft) which would seem to have the potential to create an imbalance in the head if one side was farther out than the other.

Has anyone else noticed this? Does it matter?

Oh, something else I learned today, feathering shafts are 200% easier to change out if you grease them... those dampeners get quite a death-grip on the shafts otherwise, ya darn near have to pound 'em out.
04-28-2002 Over year old.
HOMEPAGE  
 
 
Waldo
Veteran
Location: Middle of the corn patch (Iowa)

You'll notice that they fly 100% better with a greased spindle and dampners, not greasing causes vibes, wobbles and shakes. Hey did you get any flying in Friday? I went out all afternoon burned up a gallon.

[b]Bill[/b]
04-28-2002 Over year old.
 
 
Virtual1
Senior Heliman
Location: Waterloo, Iowa - USA

I thought about it, but I was dead tired and had to work that night, so I slept all day. Bummer too, looks like it may have been the only good flying day we're going to get anytime soon. Wind advisories and thunderstorms are really getting me down.
04-28-2002 Over year old.
HOMEPAGE  
 
 
pwood
Key Veteran
Location: Dubai - UAE

Hi, I thought exactly the same thing about the pottential offset causing a vibration. I had a "heated" discution with some of the guys here at runryder but was still not convinced so went back to my old text books and found it the chapter "motion in a circle". The force outwards actually decreases with an increasing radius. (with the same mass and rotational speed) Still not convinced, I took my well greased R30 to the field and pushed the spindle to one extreme. Flew it arround with lots of vibration for 2 min. and then the vibration went away and it had re-centred itself. Science works !

Paul
05-01-2002 Over year old.
 
 
Greg McFadden
Key Veteran
Location: Spokane Valley, WA

Actually, that is incorrect. the equation for force normal to a rotating mass is as follows (All of this comes from Engineering Mechanics : Dynamics by Hibbeler and is corroberated by Physics with modern physics by Serway and further by Modern Physics by Kenneth Krane)


F=(m*V^2)/r

so if you keep V constant then increasing radius does indeed reduce the force, however V is the linear velocity, and is NOT held constant if you have a constant head speed, which is what I try to have.

Now if you remember back to physics or dynamics.

V=w*r where w is the angular velocity in Radians per second and r is the radius in meters.

So if you plug that into the prior equation and simplify a bit, you get

F=m*(w^2)*r

So now if you hold headspeed constant (constant angular velocity, w) then the force will increase for a increasing radius.

Now that small of a delta in the radius will cause vibrations but it is such a small difference that other factors can greatly influence the results such as a resonance situation in the head or any one of a number of other factors. (man, I could have so much fun plugging one of these heads into a modern computational fluid dynamics and stress analysis sim.. anyone have about 200,000$ they would like to donate for that? ) Try a more extreem case and see if it really goes away. or better yet don't because it would be dangerous.

The silence often, of pure innocence persuades, when speaking fails
05-01-2002 Over year old.
 
 
Sillyness
Senior Heliman
Location: Little Rock AR

Numbers

An interesting note:

If you convert all the units to compatible ones (i.e., linear velocity in m/s, blade weight in kg, radius of center of mass of the blade in m), you come up with this:

For a 600mm ( .6m) blade, weighing 130g (.13kg) and a CG at approx 400mm (.4m) estimated, and a head speed of 1800rpm (30rps), you get a force exerted on the blade, grips, bearings, etc... of:

1850Newtons!!!

For those not familiar with newtons, if you divide them by 9.8, that will give you the kilograms subjected to earth's gravity it would feel like(Newton is a force, while kilogram in a mass... 1 kilogram exerts 9.8 Newtons under the influence of the earth's gravity).

Anywho, in simple terms, that would feel like aprox 188 kg, or 415 lbs!!!

If anyone reads this and wants to double check my math, I don't really believe it myself... unreal!!!
05-06-2002 Over year old.
 
 
Doug
Elite Veteran
Location: Naples Florida....

Any (unrestrained) rotating assembly will rotate about it's center of mass. If the rotor is balanced it won't shake no matter how much lateral "play" exists
05-06-2002 Over year old.
 
 
Greg McFadden
Key Veteran
Location: Spokane Valley, WA

However, these heads are not unrestrained. Once it is offset the center of rotation is offset from the center of mass.

The silence often, of pure innocence persuades, when speaking fails
05-06-2002 Over year old.
 
 
Doug
Elite Veteran
Location: Naples Florida....

For all practical porpoises they are unrestrained. The total centripetal force is thousands of times the (mast/heli) rocking inertia force at this scale.
05-06-2002 Over year old.
 
 
Greg McFadden
Key Veteran
Location: Spokane Valley, WA

so what you are saying is that if I pit a longer feathering spindle in and offset my feathering spindle by an inch or more from CG, it will center itself again?

The silence often, of pure innocence persuades, when speaking fails
05-06-2002 Over year old.
 
 
Doug
Elite Veteran
Location: Naples Florida....

As odd as that sounds, yes the system will spin about it's center of mass.
05-06-2002 Over year old.
 
 
Greg McFadden
Key Veteran
Location: Spokane Valley, WA

Hate to break it to you but I just posed this question to one of my physics professors. He and I agreed that it will not center itself based off of your assumptions. I also tried a little experiment with a bit of 1/8" stainless rod and a broken tailboom. Took the end of the boom with the holes not the slots (after cutting off the bent portion) and off centered the rod and spun it. low and behold the rod did not center itself it instead flung itself across the room. The question I posed was as follows. We then chatted about a heli-head and other heli stuff.

You have a rod of uniform mass per unit length (simulates two blades with a feathering spindle between them). The CM is of course at the center of the rod. You place stoppers a distance X from the Center of Mass of the rod (CM) then you have a frictionless sleeve that can slide anywhere between the two stoppers. (that simulates the head)

so we have a problem roughly like this, if I can do it justice in text:
{edit}
the forum keeps erasing the spaces and screwin up the picture so the spaces have been replaced by the letter "a" remove the a's and draw it and it will be right {edit}

aaaaaa|aaaaa---|
-----------|------------|-----------
aaaaaa|aaaaa---|
aaaaaaaaaaaa|

The two short horizontal pieces are the sleeve and the vert. piece that attaches to them is the drive shaft (otherwise known as the center of rotation) if that is a distance y from the CM of the rod, and we torque the head so that a constant angular velocity of 1000rpm is attained. This converts to an angular velocity of 104.7 radians per second. The linear speed then of the CM is w*r which gives you 104.7*y. if the rod has a mass M and you look at the force normal to the direction of motion (the force we are interested in) then you get F=Mv^2/r which simplifies to F=M*10966*y. Now this force will be required to hold the rod from flying off to the left and will be applied by the sleve on the right stopper. There is NO WAY given the explanations given previously for this to center itself. If I remove the stoppers the rod flies off. it does not center itself.

BTW, as we also discussed, only a free body will rotate about its center of mass of its own acord. if I take the two blades connected to the feathering spindle and throw them in a vaccume they will rotate about their CM. this is not a free system, it is a driven system.

The silence often, of pure innocence persuades, when speaking fails
05-07-2002 Over year old.
 
 
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Thunder Tiger Raptors 30-90 - Imperio > Raptor feathering shaft
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