How limber a mast?

dinoa

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How limber should a gyro mast be? What would be the optimum deflection and spring rate at the rotor bolt to isolate 2/rev vibration for say:

RFD 23' rotor
600 lbs lift
36" mast cantilever from clamped upper support to rotor bolt
Mast rake aligned to thrust vector at 60 mph cruise
Mast material 4130, 6061-T6 or composite round or square
 
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Mast stiffness is an important factor in tuning the inplane resonant frequency of the rotor system; mast restraint lowers the resonant frequency of the rotor. The inplane resonant frequency must be above 1/rev if 2/rev vibration is to be avoided. Rotation of the rotor doubles the frequency of vibration fed to the airframe.

Necessary mast springiness depends upon rotor inplane stiffness; the greater the inplane stiffness of the rotor, the stiffer the mast can be while avoiding resonance. A round mast is superior to a square/rectangular mast from a vibration standpoint.

Another aspect of mast stiffness/strength is that it forms a roll bar, protecting the pilot.

To get an idea of the resonant behavior of a rotor, grasp s stick of welding rod ~1/4 of its length from one end between thumb and forefinger and thump it at its center or at either end. It will vibrate for a considerable length of time, depending how close to a node it is held. That mimics the resonant behavior of a rotor; keeping in mind that rotation raises a rotor’s inplane resonant frequency.

My guess is that a 2 inch diameter x 0.065 inch wall 4130 tube will provide adequate rollover protection and minimal 2/rev vibration.
 
That was a really great response to a subtly tricky question, Chuck. There's nothing like a wealth of experience when dealing with such engineering questions.
 
Chuck, I was hoping you would accept the challenge. Would you venture to take a stab at what the magnitude of the 2/rev force is?
 
I do not think I'm a mathematician, Chuck. Hyperbolic sinus or matricial calculus has always remained far from my reach.
Dino, The answer to this question comes from my Excel spreadsheet that calculates the forces produced by the blades every 15 degrees of rotation.
It tells me that at 60 mph, a 24' x 7" seesaw rotor carrying 600 lbs, produces a spinning drag of 15 lbs.

This assumes that the hub moves forward at a strictly uniform speed.
But it is easy to understand that if this force acts without other hindrance than the rotor's own mass, then it disappears in favor of the circular motion 2 / rev it will create.
So, with a rotor mass 50 lbs, the radius of this moving is about 0.07 inch when the mast top not reacts

I still have doubts on the hypothesis of a amplification by a resonance in plane.
I am trying to quantify the natural frequency of a rotating beam, without achieving it.
NACA TN 3459 can only be used for a cantilever beam, not free-free.
 
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Yes very enlightening when you consider the inertial effects of rotor mass as JC pointed out the answer is far more complex.
 
The answer .07" corresponds to an anecdotal remark I saw on the Forum estimating it was about 1/16 of an inch of movement.
 
Chuck,
After careful consideration, I believe now that Naca TN 3459 allows to conclude on the impossibility of resonance "in plane" of our seesaw rotors.
I am unable to follow his math but a diagonal reading shows to we the natural frequency in rotation (ω[SUB]R[/SUB]) is the natural frequency with no rotation (ω[SUB]NR[/SUB]) added to a term related to the rotation (Ω):
ω[SUB]R[/SUB][SUP]2[/SUP] = ω[SUB]NR[/SUB][SUP]2[/SUP] + KΩ[SUP]2 [/SUP]in which K is called Southwell coefficient.
Since the resonance appears due to the spinning drag, then ω[SUB]R[/SUB] = Ω, and at this very moment we will have Ω[SUP]2 [/SUP]= ω[SUB]NR[/SUB][SUP]2[/SUP] + KΩ[SUP]2[/SUP] or ω[SUB]NR[/SUB][SUP]2[/SUP] = Ω[SUP]2[/SUP] (1-K) Thus, there is no resonance solution if K > 1.

Now, the note shows that K = 1.2 for the first mode of a cantilever beam with stiffness and mass uniformly distributed (see fig. 7) and also shows K is roughly proportional to the non-rotating natural frequency ratio for the other modes (see fig. 18)
Thus, since in first mode free-free the natural frequency is 1.6 times higher than in cantilever mode, we will have K = 1.9
So, there is not resonance solution either. Sans titre3.png
 
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JC, I measured the frequency of the free-free mode of a 22 ft. DW rotor suspended by cords tied to the roof beams and to the rotor’s nodal points, excited by a jug saw clamped to the hub. The jug saw was driven by a variable voltage supply, a Variac and frequency was measured with a digital counter.

The rotor was resonant at ~6 Hz. Resonance was obvious.

I realize that rotation raises the frequency but don’t know how much. Also, mass or spring restraint at the center lowers the frequency.

For both the cantilever and free-free modes, hub stiffness plays a leading role since most inplane deflection occurs at the hub rather than along the blades. This being the case, centrifugal stiffening plays a smaller tole than would be the case for a beam of uniform stiffness.

JC, check this video at 32.30. Mathematician Arthur Young, inventor of the underslung see-saw rotor, believed inplane resonance was the cause of 2/rev vibration.

https://www.youtube.com/watch?v=ZzjhJULj5OY
 
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No Title

Working backwards, solving for Force at .07" deflection, 2 X 0.065" 4130 mast, 36"cantilever gives 25lbs.
Despite all its advantages steel is heavy. A comparison of various masts with identical deflection with 2" outside diameter
wallrelative weightdensity
4130.065"100%7.85
Ti alloy 0.2"173%4.4
T6 6061 0.25"123%2.7
graphlite.095"30%1.6

Graphlite was evaluated because it is a standardized, manufactured product produced to close tolerances. It exhibits superior fatigue properties and can be used in homebuilt fabrications with repeatable results. A link showing its use in sailplane wing spars https://www.ihpa.ie/carbon-dragon/in...ite-carbon-rod . It would seem an ideal material for rotor spars as well.

Drawing shows about $350 worth of graphlite rod nested on a GRP mandrel in a 60" X 2"mast
 

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C. Beaty;n1131849 said:
I realize that rotation raises the frequency but don’t know how much.
The resonance of 6Hz you obtained with your the jig saw is ω[SUB]NR[/SUB]. To know the natural frequency during the rotation, you must apply the formula: ω[SUB]R[/SUB][SUP]2[/SUP] = ω[SUB]NR[/SUB][SUP]2[/SUP] + KΩ[SUP]2[/SUP] with K = 1.9 (I hope it true)
For example, with ω[SUB]NR[/SUB] = 6 Hz, a rotation of 300 rpm changes the natural frequency so that ω[SUB]R[/SUB][SUP]2[/SUP] = 6[SUP]2[/SUP] + (1.9 * 5[SUP]2[/SUP]) Hence ω[SUB]R[/SUB] = 9.1 Hz
Since it the spinning drag shakes the rotor at 5Hz, the vibratory motion is not amplified.
You can check this formula has no solution that gives a natural frequency equal to the frequency of rotation, therefore the amplitude of the motion is never very affected.

For both the cantilever and free-free modes, hub stiffness plays a leading role since most inplane deflection occurs at the hub rather than along the blades. This being the case, centrifugal stiffening plays a smaller role than would be the case for a beam of uniform stiffness.
I had already shown that an articulated beam at the root still has the own frequency equal to twice the frequency of rotation

dinoa;n1131861 said:
Working backwards, solving for Force at .07" deflection, 2 X 0.065" 4130 mast, 36"cantilever gives 25lbs.
Despite all its advantages steel is heavy.
If you are looking for the softer mast, to avoid vibration are transmitted from the rotor to the airframe, then we must reduce his diameter and the elasticity modulus. It seems like you're doing the opposite.

A link showing its use in sailplane wing spars . It would seem an ideal material for rotor spars as well.
A blade has other requirements than a sailplane wing, Dino. Of course, it must be rigid to allow the larger aspect ratio. But we need enough weight, because the centrifugal force decreases the coning
 
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quote "If you are looking for the softer mast, to avoid vibration are transmitted from the rotor to the airframe, then we must reduce his diameter and the elasticity modulus. It seems like you're doing the opposite."

I was trying to duplicate the deflection of a 4130 steel mast that empirically has been found to work well. How soft or limber is the question. A very limber mast can be achieved that meets stress requirements. This is often the issue with CF where you design for deflection not stress as is the case in aircraft structures like a sailplane wing as Marske points out in the link about graphlite.

quote "A blade has other requirements than a sailplane wing, Dino. Of course, it must be rigid to allow the larger aspect ratio. But we need enough weight, because the centrifugal force decreases the coningA blade has other requirements than a sailplane wing, Dino. Of course, it must be rigid to allow the larger aspect ratio. But we need enough weight, because the centrifugal force decreases the coning weights for coning"

I think tip weights could be used to tune coning. Leading edge weight would also be needed for chordwise balance.
 
You are right, Dino, but it is barely more efficient to use the expensive ultra-light materials and then add the depleted uranium. Just much more expensive
 
You are the one that is right. I was under the impression the difference would be significant.
 
Jean Claude
I don't participate in this forum anymore, but I just had to log on to congratulate you for your elegant demonstration that it is probably impossible to run at the in-plane free-free natural frequency.
I say probably because it's difficult to imagine that Arthur Young got it wrong and by stiffening the rotor in-plane he appears to have resolved the problem.
Mike G
 
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