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)
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
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.
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