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r.p.m. are used to control the precessional speed of the gyro-axle. The full power of the motor together with the torque due to the ship's roll are used to accelerate the precessional velocity for 1.25 sec. after the start of precession. During the first quarter second, the acceleration is 0.3 rad. per sec. per sec. Assume that there is a lag of 0.5 sec. between the inception of a roll and the inception of precession. The precessional velocity attained in about 1.25 see. after the inception of precession is maintained constant by the use of a motor and brakes for such a length of time that a constant deceleration, then developed, will cause the gyro-axle to come to rest 120° from one end of a swing in the 6.5 sec. of one-half period.

The precession motor is connected to the gyro-casing by gears having diameters in the ratio of 1 to 100. Assume that the efficiency of the gears is 80 per cent, and that the moment of inertia of the gyro-wheel, casing, and gear with respect to the precession axis is nine times that of the gyro-wheel with respect to the spin-axle.

Compute: (a) The moment of inertia and the mass which the gyro-wheel must have if it consists of a uniform disk 8 ft. in diameter; (b) values of the roll velocity of the ship at quarter-second intervals throughout one-half cycle of roll where the maximum amplitude of roll is 2° from the vertical; (c) the horse-power of the motor which, making 500 r.p.m., will produce a mean acceleration of the gyro-axle of 0.3 rad. per see. per sec. during the first quartersecond of precession; (d) values of the precessional velocity and angular displacement of the gyro-axle at the end of each quarter-second interval throughout one half-cycle; (e) the time at which the final deceleration of the precessional velocity of the gyro-axle should begin; (f) the gyroscopic torque opposing roll at the end of each half-second during one half-cycle of roll.

(g) Construct a table with the following data arranged in consecutive columns: tr, time of roll reckoned from the end of an oscillation; tp, time of precession reckoned from the end of an oscillation; wr, instantaneous roll velocity; wr, mean roll velocity during the preceding time interval; wp, instantaneous velocity; 0', angular displacement of the gyro-axle from end of oscillation; ot, displacement of gyro-axle from equilibrium position; 4,, mean angular displacement of gyro-axle during preceding time-interval; Lg, gyroscopic torque opposing roll.

(h) Plot on the same time-axis, curves coordinating

wr and t,, 't and t,, Lg and t,

Solution. (a) The Required Moment of Inertia of the Gyro-Wheel. - From the equation of U. S. Admiral D. W. Taylor, the required moment of inertia of gyro-wheel and axle, with respect to the spin-axis, expressed in pound-feet2, is

Ks = 12250DHT n

Substituting in this equation the data of the problem:

_1225(5°)(2200 tons) (2.5ft.)(13sec.)KS   33000


8   r.p.m.


The mass of a uniform disk 8 ft. in diameter that will have the above moment of inertia, from (22):

2K5   2(333350lb.-ft.2) = 41670 lb.   (101)

MP r]-   16ft.z

(b) Velocities of Ship Roll at Instants One Quarter-Second Apart throughout a Roll of Amplitude Two Degrees and Period Thirteen Seconds. - In the case of a body vibrating with simple harmonic motion of rotation of period T, the angular velocity at time t after leaving the end of an oscillation

(41), is

2 at

wt = We stn y -

where we represents the velocity when traversing the equilibrium position. If the amplitude of vibration measured from the equilibrium position be 0 radians or ¢°,

we = 2,± = 2 "    radians per sec.   (102)

T 57.3 T] 9.12 T



wt = 912° T sin 1360° t T radians per sec.   (103)

We shall assume that during one roll of a ship, the motion is simple harmonic. Then, if the stabilizer keeps the ship within 2 degrees of the vertical, the velocity of roll at time t, of the ship considered in this problem is

w"   (9.1) (13) sin 360 tr = 0.017 sin (27.7 tr) radians per sec.   (104) 13 The mean roll velocity during an interval while the instantaneous roll velocity changes at a uniform rate from w; to w;' is

wr = w, -l- 2 (wr" - wr')   (105)

Values of the instantaneous roll velocities at the end of half-second intervals throughout a half-cycle of roll are given in column 3 of the table (p. 161). Values of the mean roll velocities during each half-second interval are given in column 4.

(c) The Horse-Power of the Motor Which, Making 500 R.P.M., will Produce a Mean Acceleration of the Gyro-Axle of 0.3 Radian per Second per Second During the First Quarter-Second of Precession. - The mean angular acceleration during any time interval


a=K, P

where Lp represents the mean torque acting upon the precessing system with respect to the axis of the gudgeons about which the system precesses, and Kp' represents the moment of inertia of the same system with respect to the same axis. The total torque is the sum of the torque, L1, due to the precession motor, the torque, L2, due to the ship's roll and the torque, L,, due to the center of mass of the system being not on the axis of precession.

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