Item 0433

OTHER: Flight Dynamics - General - Lead-Flap Coupling, for Intermeshing Rotors (kpβ) & Pitch-Lead Coupling (δ4)

Objective:

Methods that are Considered Below:

  1. Use flap hinge geometry, as does Kaman

    2. 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 servo-flap 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?

  2. Use circular-elliptical flexible synchro gears
  3. Elastomeric coupling in rotor hub
  4. 4 synchro gears In-line
  5. Replace the teetering hinges with constant velocity hinges.

    6. This may be the best method.

  6. Let the long masts handle the twist.

Robinson uses this however it does not have two rotors working against each other.

Outside Intermeshing Helicopters

Flettner FL-282:

 

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 lead-lag and the FL-282 rotor hub.

1/ Each of the 4 blades can lead-lag, 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.

Kaman:

K-125:

It had two servo-flaps; 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 lead-lag.

The rotor was then modified so that the two blades in the same rotor could led-lag, in unison. The lead-lag in one rotor was independent from the lead-lag 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 pitch-lag coupling. I suspect that Kaman could not use the Control System Geometry, which does not have a pitch-lag coupling, because the servo-flap 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 pitch-lag coupling, are probably the reason for the rotors' lead/lag hinges and dampers etc.

K-225:

 

The teetering hinge appears to be at approximately 15º delta3, or perhaps 22º.

It appears that; 1/ the 2 blades are able to freely lead-lag together a unit. 2/ can lead-lag independently of each other but with friction restraint.

K-Max:

 

The teetering hinge appears to be at approximately 15º delta3, or perhaps 22º.

It appears that; 1/ the 2 blades are able to lead-lag together a unit. 2/ can lead-lag independently of each other but with some form(s) of restraint. If elastomeric it may provide spring centering effect plus dampening.

Dick DeGraw's Hummingbird:

Dick DeGraw used standard teetering hinges and found no problems with it. On his helicopter perhaps the lead-lag 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 lead-lag.

SynchroLite & Dragonfly

Notes:

It appears certain that there is an alternating lead-lag 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 2-blade rotors and at 6P &/or 6P with 3-blade rotors. This lead-lag must be taken care of and it may have to be dampened.

On a 2-blade rotor, the individual blade may not require lead-lag (though there will be some due to H-force 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 lead-lag. The blades will hunt approximately 1 to 2 degrees. He suggested using semi-ridged 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 (pitch-flap 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; (EKR) [ft-lbs] = (1/2) * JMO * ω 2

Polar moment of inertia; (JMO) [slug-ft2] = m * r2

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; EKR = (1/2) * JMO S * ωS 2 = (1/2) * JMO E * ωE 2

Divide both sides by 1/2; JMO S * ωS 2 = JMO E * ωE 2

Polar moment of inertia equals mass time radius squared; JMO = m * r2

Therefore; m S * rS 2 * ωS 2 = m E * rE 2 * ωE 2

And since the mass does not change; rS 2 * ωS 2 = rE 2 * ωE 2

Rearranged; ω E 2 = (rS 2 / rE 2)* ω S 2

Symbol for the flap angle β (in degrees).

Since the change in the radius is cosine of the Start angle βS minus the cosine of the End angle βE

(rS 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

Method 1/ Graph of Blade Motions Resulting from Flapping, delta3 by flap hinge geometry; and delta4??:

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.

PROBLEM:

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 (ωO) () 360º * 0.01745 = 6.282 [radians per second]

Final angular velocity (ωF) () 6.282 * 1.0154 = 6.3825 [radians per second]

Angular acceleration (α) () [radians, per second, per second] α = (ωF - ωO2) / 2θ

 

α = (ωF - ωO2) / 2θ

 

α = (6.3825 - 6.2822) / ( 2 * 1.5705)

 

α =250.6925 / 3.141

 

α = 79.8130

Time (t) [in seconds] t = 2θ / (ωF + ωO)

 

t = 2θ / (ωF + ωO)

 

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. Pitch-Teeter 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 (pitch-flap 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 lead-lag. 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;
    1. Assume that at 'mean thrust' (normal steady state load) the teetering hinge and the CGs of all the blades are in a common plane.
    2. At 'mean thrust' when the tip path plane changes from that of the mast plane, the radius of gyration, at the azimuths of the maximum positive and the maximum negative teeter, decreases by the cosine of the angle between these two planes. I.e. a 10º teeter will subject the radius of gyration to a multiplier of 0.9848, at the azimuths of maximum positive and negative teeter.
    3. The following must be reviewed.
    4. then the radius of gyration multiplier will have a smaller value. This is a change of 0.0152. If the coning angle changes from 3º to 13º then the change is (0.9986 - 0.9744) = 0.0242
    5. Increase in rotational speed @ maximum flap of 10° = (102"-100.45") / 100.45" = 1.543% [from lower portion of above drawing].
    6. In other words, if the coning were to be changed from 0° to 10°, the rotor would be revolving 1.543% faster.
    7. The blade tip advances, both, when it flaps up and when it flaps down from the mast plane. Therefore, if the rotor disk were tipped 10° down at azimuth 270° then the tip speed would be 100% at 0° and 180° azimuths and 101.543% at 90° and 270° azimuths.
    8. Average increase in speed = 1.543 / 2 = .771% and this will take place at 45°, 135°, 225° and 315° azimuth, where the flapping angle is 5°. [This is not the exact way to calculate this, but it is good enough.]
    9. This means that the maximum accelerations are at azimuths 45° and 225°. The maximum decelerations are at azimuths 135° and 315°.
    10. The circumference = 22/7 * 2 * R = 3.14286 * 2 * 102" = 641.14284"
    11. The 1/4 circumference is = 160.2857"
    12. At the faster speed the blade tip will have traveled (100.507715 / 100) * 160.2857 = 161.09949".
    13. This is an advancement, for the 1/4 rotation, of 161.09949 - 160.2857 = 0.81379"
    14. Sine of angle (of lead advancement) is 0.81379 / 102 = 0.007978333
    15. Therefore, the angle (of lead advancement) (δψ) is 0.45712967 degrees.

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 H-force 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 - Pitch-Flap Coupling (delta3)]

For variation of RRPM caused by cyclical Coriolis / Knuckle joint effect see: DESIGN: Rotor - Disk - Semi-rigid Teetering

To run calculations see: FORM: Rotor - Disk - delta3

Method 2/ Circular-Elliptical 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 In-line

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 side-to-side between the outer gears, 20 times per second. This should allow the 2 rotors to lead-lag. 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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Last Revised: June 29, 2007