Gyroscope effect on ship:

PART A: Gyroscope effect on ship during steering:

1. Rotor rotates clockwise and ship takes right turn:

1. Rotor rotates clockwise and ship takes left turn:

Note:

Diagram

• Initial angular momentum = $\overrightarrow {OX} \ = \ Iw$

• Final angular momentum = $\overrightarrow {OX'} \ = \ Iw$

$\therefore$ Change of angular momentum = $\overrightarrow {OX} \ – \ \overrightarrow {OX}$

= $\overrightarrow {xx}$

= $\overrightarrow {ox} . \ \delta \ \theta$

= $I.w. \delta \ \theta$

$\therefore$ rate of change angular momentum = $I.w. \frac{\delta \ \theta}{\delta \ t}$

$$= \lim_{\delta \to\ 0} I.w. \frac{\delta \ \theta}{\delta t}$$

= $I. \ w. \ \frac{d \ \theta}{dt} \ = \ I.w.w_p.$

Where, $I =$ M.M.I of rotor

$w =$ Angular speed of rotor.

$Wp =$ angular speed of precession.

1. Rotor rotates anticlockwise and ship takes left turn:

1. Rotor rotates anticlockwise and ship takes right turn.

Part B: Gyroscopic effect on ship during pitching:

1. Rotor rotates anticlockwise and ship pitches up with SHM.

Ship moves or turns towards starboard.

1. Rotor rotates anticlockwise and ship pitches down with SHM.

Ship moves or turns towards portside.

1. Rotor rotates clockwise and ship pitches up.

Ship moves towards port side.

1. Rotor rotates clockwise and ship pitches down.

Ship moves towards star board.

PART C: Gyroscopic effect of ship during rolling:

In case of rolling action, the axis of rotation of rotor and axis of rolling of ship are same. Hence there is no gyroscopic effect of ship during rolling.

NOTE:

1. PART A: Steering:

$C \ = \ I.\ w. \ w_p$

2. PART B: Pitching (with SHM)

Diagram

• $\theta \ = \ \phi \ sin \ w_ot$

{ where, $W_o \ = \ \frac{2 \pi}{Tp}$

• $\frac{d \ \theta}{d \ t} = \phi \ W_o \ cos \ w_ot$

If $cos w_ot = 1$

$\therefore$ $w_{p_{max}} \ = \ W_0$

$C \ = \ I.w.w_{p_{max}}$

$I \ = \ mk^2$

$w_p = \frac{v}{R}$

• $\alpha = \frac{d^2 \ \theta}{dt^2} \ = \ -\phi \ w_o^2 \ sin \ w_ot$

If $sin w_ot = 1$

$\therefore$ $\alpha \ max = - \phi \ w_o^2$

. . . . .(Negative sign indicates retardation)

Numericals:

$m = 6000 \ kg$

$N = 2400 \ rpm$

$\therefore$ $w = \frac{2 \ \pi \ \times \ 2400}{60}$

$w = 251.32 \ rad/sec$

$k = 450 \ mm = 0.450 \ m$

Case 1: Rotor rotates anti clockwise when viewed from bow end and steers left:

$\therefore v = 18 \ knots = 18 \times 1860 = 33480 \ m/hr$

$= 9.3 \ m/s$

$\therefore R = 60 \ m$

$I = mk^2$

$= 6000 \times 0.450^2$

$I = 1215 \ kgm^2$

• $wp = \frac{v}{R} = \frac{9.3}{60} = 0.155 \ rad/sec$

$\therefore$ C = I w. Wp

= 1215 x 251.32 x 0.155

$C_{max} = 47329 \times 10^3 \ N.m$

Case 2: Bow moves down and rotor rotates

$Tp = 18 \ sec, \phi = 7.5° = 7.5 \times \frac{\pi }{180} = 0.1309^c$

$w_o = \frac{2 \pi }{Tp} = \frac{2 \pi}{18} = 0.349$ rad/sec

$w_{p_{max}} = \phi . \ w_0 \ = \ 0.1309 \times 0.349 = 0.0456 \ rad/sec$

$C_{max} \ = \ I. w. \ wp_{max} = 1215 \times 251.32 \times 0.0456$

$C_{max} = 13.92 \times 10^3 \ N.m$

Case 3: There is no gyroscopic effect during rolling.

Case 4:

$\alpha_{max} \ = - \phi . \ w_0^2$

$= - 0.1309 \times 0.349^2$

$\alpha _{max} = -0.01594 \ rad/sec^2$

(Negative sign indicates retardation of ship)