1649

Item 1649

OTHER: Rotor Concept - Gyro for 2-blade Rotor(s) - Increased Control Authority for a Teetering Rotor:

Objective:

To provide additional cyclic control in rotorcraft that has a 2-blade teetering rotor.

The primary source of the additional control power comes from the inclusion of Offset Teetering Hinges (or alternatively Hub Springs) in the rotorhub.

However, each of these two devices creates a two-per-rev moment (vibration).

Therefore an electrically gyroscope is also included and it is located just below the rotorhub. It consists of a Control Moment Gyroscope that is rotated by a special electric motor.

The Control Moment Gyroscope also creates two-per-rev moments. However, these additional moments are approximately midway between the two-per-rev moments created by the Offset Teetering Hinges (or alternatively Hub Springs).

The four sinusoidal moments per revolution of the rotorhub combine to minimize vibration when off-center cyclic is applied; while at the same time providing greater control of a rotorcraft with a teetering rotor, particularly during a condition of zero-gravity.

In addition, the rotational inertia of this gyroscope allows for a reduction of the blade's tip weight, since it can be used to provide rotation inertia to the teetering rotor.

Drawing: This is a copy of the Electrotor-Plus. The gyro must be modified. Note that flight control rods must go inside the mast.

Specifications:

The Robinson R22 is used as the example.

Helicopter's Main Rotor:

Rotor radius = 151" = 12.58 ft.
Main rotor speed = 510 RPM = 8.5 fps
Tip speed = (12.58 * 2) * (22/7) * 8.5 = 672 fps
Portion of existing tip weight under consideration = 1 lb

Electric Gyro's Rotor:

Rotor radius = 1.0 ft.

Calculations to Determine the Rotational Speed of the Gyroscope:

The following must be checked. Then work on the gyroscopic moments
Polar moment of inertia of mass. [J_M]

Value for disk with mass (m) considered as being concentrated at the radius (r). J_{M =}m * r²

For aerodynamic rotor; J_M = 2 lb * 12.58² = 316.5 ft-lb-sec² [the 2 lbs consists of 1 lb per blade]

For electric rotor; J_M = 2lb * 1² = 2 ft-lb-sec² [the 1 represents the radius of the electric rotor ~ outrunner] in ft-lb-sec²

_{Kinetic Energy - Due to rotation:}

E_KR = (1/2) * J_MO * ω²; [E_KR in ft-lbs], [J_MO ~ moment of inertia of the body about a fixed axis 0, in lbs-ft-sec²]

Kinetic energy of the two tip weights in the aerodynamic rotor = (1/2) * 25.16 * 316.5² = 1,260,167 lbs-ft-sec².

Required Rotational speed of electric rotor = ω= root[E_KR / ((1/2) * J_MO)] = root[1,260,167 / ((1/2) * 2)] = 1,122 radians per second = 10,714 RPM

Phase Angle: 10,714:600 = 17.6:1. The rotorhub has rotated 5º when the gyro has rotated 90º. Therefore the phase angle is 85º.

General Notes:

The electrical power from the engine's alternator may be sufficient to drive the gyro.

An encoder between the mast and the gyro will be required to switch the electric current at the desired timing.

How is the gyro to be turned into a generator so as to back-drive the rotorhub?

This arraignment may not work on a craft with two main rotors. There is the need for lead/lag provision or a CVJ, or a dual universal joint in the hub. ee drawing 1649.dc

For related concept see; Electrotor-Plus

Initially displayed: May 22, 2008 ~ Posted on PPRuNe: May 22, 2008 ~ Posted on Eng-Tips Forum: May 22, 2008 ~ Latest revision; May 22, 2008

The utility invention on this web page and its elaborating linked web pages are publicly disclosed on the Internet to prevent the patenting of this method and system by others. ~ Reference patent law 35 U.S.C. 102 A person shall be entitled to a patent unless - (a) the invention was known ... by others in this country, ..., before the invention thereof by the applicant for patent.