1142

Item 1142

DESIGN: SynchroLite ~ Rotor - Disk - Lift Distribution re: Vibration w/ 2 Blades

Objective:

To see if an Offset Teetering Hinge, or a Hub Spring, on intermeshing rotors with two blades each will cause excessive vibration during horizontal flight.

A side note: The use of an offset teetering hinge or hub spring should reduce the amount of teeter required. This should reduce the Corollas effect and therefore the vibration caused by it. Not quite so fast. The hub spring will cause the blade to bend in the out of plane direction and this will probably cause some cyclical Corollas.

Conclusion:

NO: It will not cause a longitudinal vibration during longitudinal flight (forward cyclic stick).

YES. It will cause a 2P lateral vibration, during longitudinal flight (forward cyclic stick).

During lateral flight, the tip of the disk will be less because the moment to roll the fuselage will be less.

For solution to this problem (good luck) see [Dumb Thoughts] at the bottom of this page.

Drawing:

Notes:

The tipping is up at the bottom of the sketch. [azimuth (ψ) 0º]

Upward forces give positive values (increased thrust) in the lower right quadrant of the combined rotor disks.

For simplicity, the moments are based on a force of +1lb, at a rotor radius of 100" and azimuth (ψ) 0º (@ bottom of sketch).

The considered force is only that which is contributed by the offset hinge or hub spring.

Port Rotor Moment

Starboard Rotor Moment

Combined Moment

Longitudinal Moment:

	ψ	Longitudinal arm x Force	ψ	Longitudinal arm x Force	Sum of moments
A	0	cos(0)²=1.00	270	cos(90)²=0.00	(1+0)100 =100*
B	15	cos(15)²=0.93	285	cos(75)²=0.07	(.93+.07)100=100*
C	30	cos(30)²=0.75	300	cos(60)²=0.25	(.75+.25)100=100*
D	45	cos(45)²=0.50	315	cos(45)²=0.50	(.50+.50)100=100*

Lateral Moment:

	ψ	Lateral arm x Force	ψ	Lateral arm x Force	Sum of moments
A	0	((sin(0)100)-12)cos(0)=-12	270 (90)	((sin(90)100)+12)cos(90)=0	-12+0=-12 ⁽¹⁾
B	15	((sin(15)100)-12)cos(15)=+13	285 (75)	((sin(75)100)+12)cos(75)=+28	+13+28=+41
C	30	((sin(30)100)-12)cos(30)=+33	300 (60)	((sin(60)100)+12)cos(60)=+43	+33+43=+76
D	45	((sin(45)100)-12)cos(45)=+42	315 (45)	((sin(45)100)+12)cos(45)=+58	+42+58=100 ⁽¹⁾

Direction of rotation: Inside Advancing, (breaststroke). 0º ψ at the bottom of the page.

The large black letters are at the location of the combined port and starboard rotor moments. A,B,C and D are at 15º intervals. E, F and G etc. will move back toward the longitudinal axis and for the next 90º move into the aft port quadrant.

The squared in two 'Longitudinal' columns the % of the longitudinal load of 1 x the % of the longitudinal distance to 100

The '100' in the 'Combined Moment - Longitudinal' is the rotor radius

The RRPM is roughly 600 and the occurrence will be 2P about the X-axis. This is 20 cycles per second, which is near the most objectionable frequency.

The vibration will be most noticeable when the cyclic stick is a extreme positions, such as abrupt maneuvers and during fast forward flight

Perhaps the frame can be tuned to minimize this. See xxxx.html

This is the same value, actually slightly more with position A, as the longitudinal one. But, the lateral resistance to roll is less than that of pitch.

Related Pages:

Dumb Thoughts:

Ref; OTHER: Mechanical - General - Spring-Weight-Damper System

Dynamic Antiresonant Vibration Isolator (DAVI) [Information on DAVI ~ RW p.233]. I think that a problem with this method is that the tuning bar mass would have to be unacceptably large.

Simple Passive Isolator with Active Damper. [Information on Simple Passive Isolator ~ RW p.6 & 231]. Could the damper be quickly adjusted by a PC and sensors on mast and frame, which are sampling, vibration, frequency, amplitude and azimuth, plus RRPM. Something must maintain the force between 2P forces from the hub spring or bi-teetering, and, it cannot be something that has a large mass or long moment arm. Could it be something aerodynamic?

Last Revised: June 19, 2009