Item 0433
OTHER: Flight Dynamics  General  LeadFlap Coupling, for Intermeshing Rotors (k_{pβ}) & PitchLag Coupling (δ_{2}) ? (δ_{4})
See also; OTHER: Flight Dynamics  General  PitchLag Coupling
Objective:
Methods that are Considered Below:
This is not looking too good  Perhaps Kaman did not intend on using the lead flap coupling of the delta3 but the Control System Geometry method would not work with the servoflap control system.
The two sources of rotational acceleration, cyclical Coriolis and flap/lead coupling are out of phase with each other by 45º.
See PROBLEM, below. It appears that what is required is for the blade tip to scribe a figure 8 motion. In other words, the tip advances at the top and the bottom of the flap and it retreats as it passes through the central position of no flap. How can this be accomplished?
This may be the best method.
Robinson uses this however it does not have two rotors working against each other.
Outside Intermeshing Helicopters
Flettner FL282:

From documentation: Blade dampers mounted on the vertical hinge are friction type with steel and brass plates. Load is applied through a rubber disk, pressure being adjusted by turning the damper cap. 
The following are my assumptions regarding leadlag and the FL282 rotor hub.
1/ Each of the 4 blades can leadlag, independent of each other.
2/ The damper resists blade movement in the plane of the rotor disk only.
3/ Greater force is required for the initial movement than the continuing movement, since with sliding surfaces static friction is greater than dynamic friction. This might result in the blade 'wanting' to try and stay in the extreme positions.
4/ There is no provision within the damper for aligning the blade. This must be done by centrifugal force.
K125:
It had two servoflaps; one in front of the blade and one behind the blade. It had a teetering hinge but no delta3. It also had no provision for leadlag.
The rotor was then modified so that the two blades in the same rotor could ledlag, in unison. The leadlag in one rotor was independent from the leadlag in the other rotor. It is impossible from the picture to see if they had incorporated delta3 at this time or not.
The following two rotors have delta3 by Flap Hinge Geometry. This method has a pitchlag coupling. I suspect that Kaman could not use the Control System Geometry, which does not have a pitchlag coupling, because the servoflap control system would not work with it. Could look into this last statement, at some time. The two sources of lead/lag, cyclical Coriolis and pitchlag coupling, are probably the reason for the rotors' lead/lag hinges and dampers etc.

The teetering hinge appears to be at approximately 15º delta3, or perhaps 22º. It appears that; 1/ the 2 blades are able to freely leadlag together a unit. 2/ can leadlag independently of each other but with friction restraint. Note that the lead/lag hinges are offset from the mast. 

The teetering hinge appears to be at approximately 15º delta3, or perhaps 22º. It appears that; 1/ the 2 blades are able to leadlag together a unit. 2/ can leadlag independently of each other but with some form(s) of restraint. If elastomeric it may provide spring centering effect plus dampening. Note that the lead/lag hinges are offset from the mast. 
Dick DeGraw used standard teetering hinges and found no problems with it. On his helicopter perhaps the leadlag was handled by the oscillating torque (twisting) in the two masts between the 2 rotor disks and/or the Hummingbird was not flown at an extremely fast forward speed.
Dick DeGraw's synchropter employs all three hinges with offset flap hinges for increased control power.
Note: There is a conflict in the above 2 statements regarding provision to handle leadlag.
SynchroLite & Dragonfly
Notes:
It appears certain that there is an alternating leadlag between the two rotors when the tip path planes are not aligned with the mast planes. The tilt of the disks and the cyclical Coriolis effect will cause it. This problem may put excessive loading on the shafts, gears and bearings of the power train. It may cause oscillations at 2P &/or 4P with 2blade rotors and at 6P &/or 6P with 3blade rotors. This leadlag must be taken care of and it may have to be dampened.
On a 2blade rotor, the individual blade may not require leadlag (though there will be some due to Hforce drag) but the complete rotor (hub & 2 blades) will. Theoretically, it can be located at any location between one hub and the other.
I spoke with Marty Hollmann at Aircraft Designs Inc. and he said that there definitely would be some leadlag. The blades will hunt approximately 1 to 2 degrees. He suggested using semiridged teetering rotor hubs.
Additional Information:
To calculate values see; FORM: Rotor  Disk  delta3
Calculated Values: delta3 = 22º; Flap = 10º; Change in Pitch = 4.04º; Lead = 0.35º;
For sketch of hinge, see Sketch B/: OTHER: Flight Dynamics  General  delta3 (pitchflap coupling)
Blade lag angle. (ζ) (zeta). + is opposite of rotor rotation
Increase in a Rotor's RRPM Due to Coning:
To calculate the increase in a blade's rotational velocity when it teeters (flaps) up or down from the plane where the blade's mass is normal to the (virtual) teetering hinge on the mast's axis. I.e. maintaining the kinetic energy when the radius of the blade's mass is reduced
The kinetic energy in a helicopter's rotor is not changed by a change in the coning angle.
Kinetic energy; (E_{KR}) [ftlbs]_{ }= (1/2) * J_{MO} * ω^{ 2}
Polar moment of inertia; (J_{MO}) [slugft^{2}]_{ }= m * r^{2}
ω = angular velocity [radians/second]
m = mass [slugs]
r = radius to center of blade's mass [ft]
Subscript (_{S }) is for the Start values (no cone or smaller cone).
Subscript (_{E }) is for the End values (greater cone).
If the tip path plane is down at the front then (_{S }) will be for 90º and 270º azimuths, and (_{E }) will be for 0º and 180º azimuths.
Since the kinetic energy does not change; E_{KR} = (1/2) * J_{MO} S * ω_{S}^{ 2}_{ }= (1/2) * J_{MO} E * ω_{E}^{ 2}_{ }
Divide both sides by 1/2; J_{MO} S * ω_{S}^{ 2}_{ }= J_{MO} E * ω_{E}^{ 2}_{ }
Polar moment of inertia equals mass time radius squared; J_{MO }= m * r^{2}
Therefore; m_{ S} * r_{S}^{ 2}_{ }* ω_{S}^{ 2}_{ }= m_{ E} * r_{E}^{ 2}_{ }* ω_{E}^{ 2}_{ }
And since the mass does not change; r_{S}^{ 2}_{ }* ω_{S}^{ 2}_{ }= r_{E}^{ 2}_{ }* ω_{E}^{ 2}_{ }
Rearranged; ω_{ E}^{ 2}_{ }= (r_{S}^{ 2}_{ }/ r_{E}^{ 2})* ω_{ S}^{ 2}_{ }
Symbol for the flap angle β (in degrees).
Since the change in the radius is cosine of the Start angle β_{S} minus the cosine of the End angle β_{E}
(r_{S}^{ 2}_{ }/ r_{ E}^{ 2}) = (cosβ_{ S}^{ 2}_{ }/ cosβ_{ E}^{ 2})
Therefore; ω_{ E}^{ 2}_{ }= (cosβ_{ S}^{ 2}_{ }/ cosβ_{ E}^{ 2})* ω_{ S}^{ 2}_{ }
and, if the initial flapping angle is zero (ie. the tip path plane and the mast plane are parallel, then,
ω_{ E}^{ 2}_{ }= (1_{ }/ cosβ_{ E}^{ 2})* ω_{ S}^{ 2}_{ }
and, assigning the initial rotational speed a value of 1, then,
ω_{ E}^{ 2}_{ }= (1_{ }/ cosβ_{ E}^{ 2})* 1_{ }
ω_{ E}^{ 2}_{ }= 1_{ }/ cosβ_{ E}^{ 2}_{ }
Therefore, ω_{ E} is the multiplier of ω_{ S} to obtain the rotational speed of the blade at the increased cone [in radians/second]
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Example, if the cone angle, in the vertical plane of the blade azimuth, (β_{ S}) is 0º, (β_{ E}) is 10º, is 45º
then cosβ_{ E }is 0.9848, is 0.7071
then cosβ_{ E}^{ 2 }is 0.9698, is 0.5000
then ω_{ E}^{ 2}_{ }is 1 / 0.9698 = 1.0311, is 1 / 0.5000 = 2.0000
and ω_{ E }is √1.0311 = 1.0154 * ω_{ S, }√2.0000 = 1.4142 * ω_{ S,}
This means that the rotor's rpm is 01.54% faster when the coning angle has been changed from 0º to 10º. This means that the rotor's rpm would be 41.42% faster if the conning angle was changed from 0º to 45º.
In one revolution, 360º, the rotor will be 1.54% * 360 = 5.544º further advanced than it will be if the conning angle is 0º.
Considering only 90º of rotation, the rotor will have advanced 5.544º / 4 = 1.386º
To then calculate the delta3 (A/ by flap hinge geometry) angle that is required to match the increase in a blade's rotational velocity caused by line 1 above.
The following conclusions have been drawn from the above graphs.
The two sources of rotational acceleration, cyclical Coriolis and flap/lead coupling are out of phase with each other by 45º.
Attempt to Determine the Average Velocity:
Blade pitch [θ] is the driving (oscillating) force,
The relative strength of this force is sin θ. Assume that a theoretical pitch of 90º has a force value of 1.
In the example, the maximum pitch [θ] and flap [β] are both 10º.
This means that the maximum relative force (torque (Q)) is sin(10º), which is 0.1737. This torque will vary from 0.0, to +0.1737, to 0.0 to 0.1737 during a single cycle.
The following table will probably be more accurate if the rows went from 0 to 10 and then the values of 5 and 10 are used

Pitch in deg. 
Relative Torque^{(1)} 
Relative Acceleration ???^{ (2)} 

0.5 
0.0087 
0.0087 

1.5 
0.0262 
0.0349 

2.5 
0.0436 
0.0785 

3.5 
0.0610 
0.1395 

4.5 
0.0785 
0.2180 

5.5 
0.0958 
0.3138 

6.5 
0.1132 
0.4270 

7.5 
0.1305 
0.5575 

8.5 
0.1478 
0.7053 

9.5 
0.1650 
0.8703 

Sum 
0.8703 


Average 
0.0870 

(1) This is just the sine of the angle
(2) This is just a running sum of the relative Torques.
The mean lead advancement in 90º of rotation is (0.2180 /0.8703) * 1.386º = 0.3472º
Rotary Motion w/ Constant Acceleration: (from Machinery Handbook p.151)
Angle of rotation (θ) (Δψ ?) 90º * 0.01745 = 1.5705 [radians]
Initial angular velocity (ω_{O}) () 360º * 0.01745 = 6.282 [radians per second]
Final angular velocity (ω_{F}) () 6.282 * 1.0154 = 6.3825 [radians per second]
Angular acceleration (α) () [radians, per second, per second] α = (ω_{F } ω_{O}^{2}) / 2θ

α = (ω_{F } ω_{O}^{2}) / 2θ 

α = (6.3825  6.282^{2}) / ( 2 * 1.5705) 

α =250.6925 / 3.141 

α = 79.8130 
Time (t) [in seconds] t = 2θ / (ω_{F }+_{ }ω_{O})

t = 2θ / (ω_{F }+_{ }ω_{O}) 

t = ( 2 * 1.5705) / (6.3825 + 6.282) 

t = 3.141 / 12.6645 

t = 
From the above graph, it looks like the delta3 angle should be 21.8885º
If the calculations are incorrect, consider having delta3 by flap hinge geometry at 40º and delta3 by control system geometry at  25º for at 40º flap  lead/lag and a 15º flap  pitch
~ or ~
Perhaps excessive lead will cause other problems with the rotor, particularly because it is the blade tip that is leading, not the whole blade or the whole rotor. ~ or ~ Perhaps Kaman could not have a hub with combined delta3s because his pitch mechanism was inside the hub and it operates a flap not a pitch horn. Note that the Kaman's lead/lag hinges, like others, causes the top to lead.
15° Offset Teetering Hinge: (ref. PitchTeeter Coupling  delta3)
Using flap hinge geometry (type A).
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I believe that operational activity is as follows;
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Calculation Method 1 and Calculation Method 2 give 2 different values. 1 is twice 2. One or both have a problem or misunderstanding. See: OTHER: Flight Dynamics  General  delta3 (pitchflap coupling) for the same concern.
Calculation Method 1:
It looks like a change of 15 to 16 degrees in the teetering hinge will cause the blades to advance the correct amount to handle leadlag. This action also causes a decrease in the input pitch angle.
In the following, a 10° of flap and a 15° teetering angle will advance the blade 0.8015" but cause a 2.68° decrease in pitch angle.
See the lower portion of the drawing for elaboration;
It may be more accurate to base the above calculations on inertia OTHER: Flight Dynamics  General  Rotor Inertia instead of radius, because speed is to a power
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See the upper portion of the drawing.
The elevation etc., in respect to the teetering axis, of the connection between the horizontal pitch horn and the vertical pitch link may offer the opportunity to vary the resultant actions.
Additional Sources of Vibration, which delta3 cannot remove:
2P Vibration Hforce Drag:
Mismatched Radius of Gyration between Blades on Same Rotor.
Drag hinges may be required to handle the above two problems. Perhaps the vertical pivot between the blade grip and the blade root can be given a lead/lag spring & damper.
Calculation Method 2: 
Current Program in Access Database: FORM: Rotor  Disk  delta3
X = RightAngle  delta3
B = Cos(X)
D = Sin(X)
E = D
F = Cos(Beta) * E
Y = Atn(F / B)
zeta = X  Y
In the above, a 10° of teeter (flap) beta and a 15° delta3 will advance the blade 0.221 degrees
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There's a discrepancy between the above methods 1 & 2.
Also, I do not know if the delta3 angle should be 15.5 degrees, 45 degrees or some other value. Note that the Kaman appears to be about 15º
For more information on delta3 see: [OTHER: Flight Dynamics  General  PitchFlap Coupling (delta3)]
For variation of RRPM caused by cyclical Coriolis / Knuckle joint effect see: DESIGN: Rotor  Disk  Semirigid Teetering
To run calculations see: FORM: Rotor  Disk  delta3
Method 2/ CircularElliptical Flexible Synchro Gears
Have a rod connected to the teetering rotor hub and passing down the hollow mast into the gearbox. (The rods could go down the outside as is standard.) The other rotor hub and mast will have its own rod also. Both rods will be connected to a device in the gearbox. Movement of either of the rods up or down from its mid position will cause the device to change both Synchro ring gears from a circular shape to a slightly elliptical shape, perhaps by rocking on constant velocity joints located in the center of both gears. The Synchro ring gear of associated with the moved rod will be compressed slightly in the same vertical plane as its associated rotor blades. The other masts Synchro ring gear will be slightly elongated in the vertical plane of its rotor mast. Note: The center to center distance of the Synchro ring gears never varies.
Calculation of Rotational Speed Increase
Using the computer's calculator, the cosine of an angle (in degrees) gives the ratio of the length of the adjacent side / the length of the hypotenuse. Consider the hypotenuse as the rotor disk radius and the angle is the degree of flap. The percentage increase in speed of the flapped rotor bladed = 1/(inverse cosine of the flap; in degrees)
Hypothetical extreme example: If the cosine of 60 degrees (i.e. flap of 60 degrees) is 0.5 then in this case the speed of the blade will be twice what it is 1/4 of a revolution later. I THINK. Conservation of momentum?
A flap of 2 degrees (cosine 2 = 0.99939) will there for result in a speed increase of 1/0.99939 = 0.06 %.
A flap of 5 degrees (cosine 5 = 0.9962) will there for result in a speed increase of 1/0.9962 = 0.381 %.
A flap of 10 degrees (cosine 10 = 0.9848) will there for result in a speed increase of 1/0.9848 = 1.54 %.
If flap is 5 degrees and diameter of Synchro ring gear is 8" then the probable change in one gear is (.381/(100*2))*8" = +).0152" and the other 0.0152", approximately 1/64".
The flapping is twice per rotor revolution. If the rotor rpm is 600 then the flapping will be 1200 times per second, which is 20 times per second.
Method 3/ Elastomeric coupling in Rotor Hub
Consider Lord Dynaflex LCR Series coupling (have hard copy) or L type jaw coupling.
Method 4/ Synchro Gears Inline
Locate 4 identical angular spur gears in a line. The outside 2 are mounted on the masts. The inside 2 are located in bearings on eccentric cams. These cams are offset 90° to each other and they will shift these gears from sidetoside between the outer gears, 20 times per second. This should allow the 2 rotors to leadlag. This idea will have to be though through to make sure pitch radiuses between adjacent gears remain constant etc.
Method 5/ Replace the Teetering Hinges with Constant Velocity Hinges
This should remove any oscillations due to cyclical Coriolis.
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