0682

Item 0682

OTHER: Flight Dynamics - General - Pitch-Lag Coupling, (δ₂) ? (δ₄)

Notes:

Another source of pitch-flap coupling is the mean lag angle (ζ₀), due to the rotor torque. [Source ~ HT p.239]

This expression [not put on this web page] gives the lag damping required for stability, or alternatively the maximum pitch-lag allowed. For a given rotor, the stability decreases as the coning angle (β₀) increases with thrust. Positive pitch-lag coupling (lag back, pitch down) is destabilizing for articulated rotors. [Source ~ HT p.658]

Information and equations for lag can be found in - Equilibrium about the drag hinge [Source ~ AH p.143].

"It has been general practice to locate the drag pivot outboard of the flapping pivot. This relationship involves a relatively wide variation in the application of the loads to the flapping pivot bearings inasmuch as the lagging position of the blade during power operation may be 15 to 25-degrees or even more, behind the radial position which the blade assumes during autorotation operation." [Source ~ MDD p.21].

"The forward blade stop should permit the blade to move to a position at least 3-degrees and preferably 5-degrees ahead of the autorotational position which is the true radial with the blade c.g. passing through the center of rotation. The lag stop should be located 3 to 5-degrees behind the worst operating position of the blade." [Source ~ MDD p.23].

For a technical and mathematical consideration of Pitch-Lag Coupling see: Rotary Wing Structural Dynamics and Aeroelasticity, page 315- 323. In fact see the whole chapter "Instabilities of Rotors'.

Leishman mentions this in 'The Jump' Takeoff' section of his article 'Development of the Autogyro; A technical Perspective'. It mentions that Cierva used this in his modified C-30.

This is related to the Torque ~ Pitch on this page. OTHER: Flight Dynamics - General - Cross-Coupling

Also consider the alternative of piezoelectric pitch control. I understand that some German company is using this method on propellers.

Consider the "lean" of the hinge X-axis and/or y-axis.

Increase in a Rotor's RRPM Due to Coning:

The following is from OTHER: Flight Dynamics - General - Lead-Flap Coupling, for Intermeshing Rotors (kpβ) & Pitch-Lag Coupling

The following may be revised for Pitch-Lag. For Electrotor-SloMo - coaxial

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) [ft-lbs]= (1/2) * J_MO * ω²

Polar moment of inertia; (J_MO) [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; E_KR = (1/2) * J_MO S * ω_S²= (1/2) * J_MO E * ω_E²

Divide both sides by 1/2; J_MO S * ω_S²= J_MO E * ω_E²

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

Therefore; m_S * r_S²* ω_S²= m_E * r_E²* ω_E²

And since the mass does not change; r_S²* ω_S²= r_E²* ω_E²

Rearranged; ω_E²= (r_S²/ r_E²)* ω_S²

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_S²/ r_E²) = (cosβ_S²/ cosβ_E²)

Therefore; ω_E²= (cosβ_S²/ cosβ_E²)* ω_S²

and, if the initial flapping angle is zero (ie. the tip path plane and the mast plane are parallel, then,

ω_E²= (1/ cosβ_E²)* ω_S²

and, assigning the initial rotational speed a value of 1, then,

ω_E²= (1/ cosβ_E²)* 1

ω_E²= 1/ cosβ_E²

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_Eis 0.9848, is 0.7071

then cosβ_E²is 0.9698, is 0.5000

then ω_E²is 1 / 0.9698 = 1.0311, is 1 / 0.5000 = 2.0000

and ω_Eis √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

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_O²) / 2θ

	α = (ω_F- ω_O²) / 2θ
	α = (6.3825 - 6.282²) / ( 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. Pitch-Teeter Coupling - delta3)

Using flap hinge geometry (type A).

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I believe that operational activity is as follows;

Initially, the mast plane and the tip path plane are parallel, and are both rotating at Ω. [the rotational speed of the rotor].

Then the tip path plane is tilted down at the front and up at the back.

The rotational speed of the tip path plane when the blades are at azimuth 0º and 180º will now be Ω+x.

The rotational speed of the tip path plane when the blades are at azimuth 90º and 270º will now be Ω-x.

Therefor, as the blades rotate, the average rotational velocity of the tip path plane will remain Ω and the constant rotational velocity of the mast plane will remain Ω, as well.

The angle of the 'delta3 by flap hinge geometry' must therefor be set so that its the in-plane component will match and absorb the lead and the lag caused by the cyclical Coriolis.

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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;

Assume that at 'mean thrust' (normal steady state load) the teetering hinge and the CGs of all the blades are in a common plane.

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.

The following must be reviewed.

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

Increase in rotational speed @ maximum flap of 10° = (102"-100.45") / 100.45" = 1.543% [from lower portion of above drawing].

In other words, if the coning were to be changed from 0° to 10°, the rotor would be revolving 1.543% faster.

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.

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.]

This means that the maximum accelerations are at azimuths 45° and 225°. The maximum decelerations are at azimuths 135° and 315°.

The circumference = 22/7 * 2 * R = 3.14286 * 2 * 102" = 641.14284"

The 1/4 circumference is = 160.2857"

At the faster speed the blade tip will have traveled (100.507715 / 100) * 160.2857 = 161.09949".

This is an advancement, for the 1/4 rotation, of 161.09949 - 160.2857 = 0.81379"

Sine of angle (of lead advancement) is 0.81379 / 102 = 0.007978333

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.

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

Other Web Page at this Site on Auto-Collective ~ Torque Pitch

Note: The total weight, which varies according to the weight of the pilot, will probable effect the RRPM-Pitch-Torque balance. Can it be pre-flight adjustable to suit each flight?

OTHER: Miscellaneous - Thoughtless Ideas - Torque/Pitch Collective Rotor Hub

OTHER: Miscellaneous - Thoughtless Idea - Constant Speed Rotor

OTHER: Flight Dynamics - General - Cross-Coupling # Torque-Pitch

OTHER: Flight Dynamics - General - Lead-Flap Coupling, for Intermeshing Rotors & Pitch-Lag Coupling

OTHER: Flight Dynamics - General - Pitch-Lag Coupling

DESIGN: Single-Bladed All Electric Rotor- Rotor Hub - Pitch-Torque Coupling

DESIGN: Electrotor-Simplex ~ Rotor - Hub - Overview of Gimbaled w/ Torque Collective B

DESIGN: Electrotor-Simplex ~ Rotor - Hub - Overview of Gimbaled w/ Torque/Pitch Collective & Teetering Drive:

Last Revised: December 13, 2009