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156 ANTI-ROLL DEVICES FOR SHIPS

Since the stabilizer of the present problem is non-pendulous, the torque L3 = 0, and we may write

ap = L, p, L2 (106)

Since the motor is connected to the gyro-casing by gears of 80 per cent efficiency, and the ratio of the number of teeth on the motor shaft to the number on the gear attached to the gyro-casing is 1 : 100, the torque at the gyro due to the precession motor is

L, = (0.8)(100)L1'

where L,' is the torque at the motor.

The power of the precession motor expressed in horse-power is

2 srnL,'

Ph 33000

whence,

L , = 33 2(3..14)500 14 Pi, = 10.5 P;, at the motor

and

L, [= 80 L,'] = 840 Pit at the gyro

Now we shall find the mean torque L2 acting on the gyro due to the ship's roll. When the mean roll velocity of the ship is iPr, and the gyro-axle is in the equilibrium position, the mean torque acting on the gyro has the value,

L2 = Kswswr

where Ks is the moment of inertia of the gyro-wheel with respect to the spinaxis expressed in slug-feet2 and velocities are expressed in radians per sec. If moment of inertia expressed in pound-feet2 is represented by K,', and spinvelocity when expressed in revolutions per minute is represented by n, then the preceding equation assumes the form:

_ KS 2 an Ks'nwr

L2 32.1 60 w' 307

The mean torque acting on the gyro during the time that the mean velocity of roll is u, and the mean displacement of the gyro-axle from the equilibrium position is ' , has the value

L = K; nw, cos D _ 333350(1310) 7, cos = 1422400 u, cos

307 307

From column 4 of the table (p. 161), the mean roll velocity w, of the ship while the gyro-axle has precessed for 0.25 sec. from the end of a swing is 0.00504 radian per sec.

The mean angular displacement (1) of the gyro-axle from the equilibrium position during any time interval (t' - t) is

'F = 60 - 2' (0,' + ¢2') (107)

where the quantities within the parenthesis represent the values of the instantaneous angular displacement of the gyro-axle at the beginning and end of the time interval. At zero time, ¢,' = 0. At time 0.25 sec., 02 will not be greater than one degree. Suppose that we assume it to be one degree. Whether this assumed value be half the correct value or two times that value

THE ACTIVE TYPE OF GYRO SHIP STABILIZER 157

will make an inappreciable difference in the computed value of the horse-power required to produce the required acceleration of the gyro-axle. Also, as in the subsequent calculations, we shall use a rounded-off number for the horsepower instead of the computed value, it will be safe to assume in the present computation that at time 0.25 sec. 01' = 1°. In this case:

=60°- a(0,'+02')=60-z(~+1)°=59.5° (108)

Hence,

L2 = 1422400 w cos l= 1422400(0.00504)0.507 = 3635lb:ft.

From the assumptions of the problem,

Kp'[= 9 KS'] = 9(333350) lb. ft.2 = 3,000,000lb.-ft 2

and the acceleration of the gyro-axle during the first quarter-second is a = 0.3 radian per sec.

Substituting in (106), these values of ap, L,, L2 and Kp', we have

0.3 - (840 Ph + 3635)32.1

3,000,000

Whence, the power required of the precession motor is Ph = 29 H.P.

In the subsequent computations we shall use 30 H.P.

(d) Values of the Angular Velocity and Displacement of the Gyro-Axle at the End of Each Quarter-Second Interval throughout One Half-Cycle. - Representing the velocity at the beginning and at the end of a time interval t by wo and wp, respectively,

w w /L, + L2 1t = w° +1840 Ph + 1422400 w, cos b t (109)

p[= o + at] = wo + Kp' ) ` Kp J

For the first quarter-second interval t = 0.25 sec., wo = 0 and

uy = 0.00504 rad. per sec.

Hence,

25200 + 1422400(0.00504) cos ~~

WP = 0 + 3,000,000 (0.25)32.1 rod. per see.

We shall now find the mean angular displacement 4' of the gyro-axle from the equilibrium position during this interval. Our method will be to make a guess of the angle moved through during this interval and then check the accuracy of the guess. First we shall test the guess that during this time interval the gyro-axle precesses one degree. In this case the mean displacement from the equilibrium, from (108), would be 59.5° and the angular velocity at the end of the interval would be, from the preceding equation

wp = 0.077 rad. per sec.

An inappreciable error will be made if we assume that during this brief time interval the angular acceleration is constant. In that case, the displacement from the end of a swing would be

0,' + '2 wpt radians

= 0 + ; [0.077(0.25)57.3]° = 0.55°

and the mean angular displacement from the equilibrium position would be

4'=60-;(0+0.55)°=59.72°

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