Item 1444

**DESIGN: Electrotor-Plus ~ ****MRGA**** - ****Gyro**** - General**

Overview:

- The rotor's speed will be about 666 RPM. Therefore a 2P vibration caused by a hub spring or an offset teetering hinges will result in a vibration frequency of 22.2 cycles per second.
- From the graph on this page OTHER: Aerodynamics - Vibration - Rotor Induced - Overview it can be seen that a velocity of 1" per/sec would result in a constant discomfort.
- The Gyro also causes a vibration frequency of 44.4 cycles per second. However, the velocity will now be reduced to (sin(45º) x 2) -1 = 0.41" and this removes the vibration from the zone of constant discomfort.
- In other words;

- The hub spring or offset teetering hinges provides additional cyclic control; at the expense of adding vibration.
- The gyro reduces this additional by 1.41 - 1 = 59%.

Description:

- The horizontal axis on the sketch below is the azimuth of Blade A about the rotorcraft, where 0º
*ψ*is aft. - The vertical axis is the teetering angle of the pair of rotor blades. It is also the moment that is generated by the pair of blades and transmitted directly to the mast. It is also the moment from the gyro which is transmitted to the mast up to 90º in advance of the blade that is inducing this moment.
- These moments [
*J*] are in addition to the control force at the top of the mast that is created by a basic teetered rotor disk._{M} - The rotor is teetered down at the front of the rotorcraft (180º azimuth). [
*β*~ positive is upward]

- Numbers on above drawing relate to Blade A:
- 1/ Maximum upward flap. Maximum blade generated upward moment (by offset or spring) directly to the mast. No moment from blade to gyro.
- 2/ Maximum downward moment from blade to gyro. No moment from blade to mast.
- 3/ Maximum downward moment from gyro to mast.
- 4/ Maximum downward flap. Maximum blade generated downward moment (by offset or spring) directly to the mast. No moment from blade to gyro.

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I think that the offset and the hub spring will assist the blade's out-of-plane moment as the blades move toward 0º teeter (flap) and resist the blade's out-of-plane moment as the blades move away from 0º teeter.

- 90º phase angle between hub spring moment and gyro moment
- I think that the moment from the gyro may be less than 90º ahead of the moment from the rotor offset or hub-spring. They are both rotating about the vertical axis. If the ratio between the motor's RPM and the rotorhub's RPM is 10:1 then the rotorhub will have rotated 90º/10 = 9º when the motor has rotated the 90º. This may likely mean that the gyro induced moment will be 99º behind the preceeding blade induced moment and 81º ahead of the following blade induced moment.
- Increasing the motor's speed and the reduction ratio will reduce this phasing problem.
- Perhaps the inclusion of a very stiff spring and a damper between the gyro induced moment and the mast will delay its moment sufficiently and thereby crease the desired 90º phasing. This spring and damper may also improve the control moment gyroscopic performance by allowing more of the moment to be transferred. Per only a prototype test will tell.
- Also see the lowering of the mass of the motor's rotor; below.
- Obtaining a constant moment throughout the 360º of rotation.
- If the hub spring moment and the gyro moment are sine waves then the total moment will vary around the mast. It will be highest where the sine waves intersect.
- If a linear increase and decrease in both moments can be produced then the total moment should remain constant. On a hub spring perhaps this can be done by the construction of the elastomeric spring, by giving it a triangular shape and compressing one edge toward the plane of the other two edges.
*See sketch in binder.*

- The phase angle between the input torque to the gyro and the out put torque from the gyro may be something less than 90º because the input torque (rotorhead) is rotating about the same axis as that of the gyro's wheel. That said, perhaps it may be exactly 90º since the amount of the input torque to the gyro is varying in a sinusoidal manner as it rotates about this axis.
- Should it be discovered that the phase angle is definitely less than 90º, there is the ability to still center the torque from the gyro to the mast 90º ahead of the torque from the hub spring to the mast by angling the moment rods away from vertical thereby having their bottom end ahead of there top end.
- July 31, 2008 ~ The thinking is to lower the mass of the motor's rotor, and of course the motor's stator. This lowering will give better clearance between the rotorhub and the blades of the other rotor or allow for reduced obliquity and/or stagger. It will also put the center of the mass below the center of the universal joint and this might be beneficial to phasing it may/should cause the gyro induced moment to start earlier (and end earlier), there by moving the phase angle closer to 90º, hopefully.

**Spring - About Y-axis: ***See last note.*

- Can Torque be transfered from about X-axis to about Y-axis if Y-axis has a very large polar moment of inertia?
- The Moment of Inertia of the motor's rotor [J
_{M}XY] is 1,000th that of the craft's Moment of Inertia [J_{M}M]. Will springs in Force Rods help with gyroscopic precession? - Will this be required to assist with torque transfer from X-axis to Y-axis.
- The spring cannot be on the rods. It must be between the gyro and the craft. In addition, for the spring to be between the qyro and the craft then the mast (rotating tube) must not tilt because if it does it will change the pitch of both blades.
- A/ Idea; What if the knuckle joint, which is out side the Hooke's joint, was replaced by a Hooke's joint? It would be spring loaded between the Y-axis and a bearing on the mast thereby allowing the gyro to precess?
- B/ Consider replacing the pair of bearings on the Knuckle joint with special (elastomeric?) bearings. These special bearings will allow the desired rotation about the Y-axis, PLUS they will allow spring resistant rocking about the X-axis.
- These springs will change the azimuth of the moment imparted to the mast by the gyro. This is because it is intended that this gyro generated moment vary with the rate of tilting whereas a spring would associate the amount of moment with the force (compression) of the spring. Ie. The 90º advance of the gyro-induced moment ahead of the rotorhub-induced moment will probably be much less than 90º.
- See Damper - In X-axis Moment Rods.

Damper - In X-axis Moment Rods:

- Place a damper in the moment arm(s) (not a spring). This will allow the
*Δβ*to differ from the_{YOKE}*Δβ*. In addition, it will only exert a moment when the arm(s) is moving. I.e. when the rotor-blades are in the process of teetering._{GYRO}**Unfortunately, this means that the Yoke(s) will experience cyclical Coriolis but the motor will not.**Can the motors be electronically accelerated and decelerated at ((600 RRPM /60) *2) = 20 cycles per second? - The amount of damping
*+Δβ*and*-Δβ*must be equal in both directions. The motor will probably require a centering spring, to rectify any discrepancy with the damping. - Note that currently only 1549.html and 1548_B.html utilize control rods. !547.html and 1550.html may not work.

Cyclical Coriolis Due to the Hooke's Joint:

- If the teetering angle of the rotorhub and the motor are not equal at all angles the velocity ratio about the R-axis and Z-axis will differ. This will result in an acceleration and deceleration of the motor at a rate of twice the rotorhubs RRPM. See; [Source ~ RW p.155]. Perhaps the motor-controller can adjust for this, however it may add to complexity and an additional potential failure point.
- Can the angular momentum of the motor be reduced and thereby give the rotorhub and the gyro equal angles?
*(Δβ*)._{YOKE}= Δβ_{GYRO}

Lower Hooke's Joint on Rotorhub:

- This joint is handling a significant torque as it returns the gyro to the level plane.

Control Moment Gyroscope:

- http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19670029734_1967029734.pdf Have printed pages 1 to 20 and 93 to 106 ~ on control moment gyroscope.
- Excellent information site on gyroscopes (including an online calculator of values). http://www.gyroscopes.org/
- A Geometric Study of Single Gimbal Control Moment Gyros

Direct Torque Control Moment Gyroscope:

- GB 886305
- US 5,386,738
- US 5,476,018

Related Pages:

- OTHER: Flight Dynamics - General - Precession - Gyroscopic and Aerodynamic
- OTHER: ~ Rotor Concepts - Gyro for 2-blade Rotor(s)
- http://www.gressaero.com/files/infosheet.pdf

**Found this question and answer on ****www.Gyroscopes.org****. It is the same as I was asking on Eng-Tips:**

**Subject: **What happens if the gyro cannot precess?

**Question: **If you apply a couple to the gyroscope but stop it from precessing what happens?

For example I hang a weight on the gyroscope, but it cannot precess about a vertical axis because of a stop.

Does it move downwards instead of precessing, if so how fast?

The applied couple with horizontal axis should generate angular momentum parallel to this axis. How

does this show up if the gyro cannot precess? Is there another couple which cancels it? If so where?

**Answer:** A gyro is a flywheel with freedom to move at right angles to the spin axis. Remove this freedom to move and it is no longer a gyro.

Where Precession occurs it is always at right angles to the applied torque.

In the case you have given, there are two torques applied at the same time.

There is the torque due to gravity and the torque from the stop.

Torque is a vector quantity and therefore the sum of these two forces will produce a resultant torque. The Gyroscope will precess at right angles to and at a speed determined by the vector sum of the applied torques.

<< How does this show up if the gyro cannot precess? >>

If a gyro is restrained so that it cannot precess it ceases to be a gyro.

However any angular motion of a flywheel IS precession, if a gyro is forced into angular motion, in whatever direction, then the gyroscopic couple accompanying it is at a right angle to that motion.

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Now to consider the ramifications. In my requirement the 'torque due to gravity' is the flapping of the blades and the 'stop' is the large moment of inertia of the total rotorcraft.

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Last Revised: January 26, 2009