# OOW SQA June 2005 Stability and Operations Solved

Section A:

1. (a) Define each of the following terms:

(i) Center of Gravity:
Is the point in a body through which the weight of the body is considered to act vertically downward.

The position of center of gravity of a ship and it’s contents is denoted by ‘G’ and it’s height above the keel (K) by ‘KG’

(ii) Centre of Buoyancy:
Is the point in a body through which the weight of the body is considered to act vertically downward.

The position of center of gravity of a ship and it’s contents is denoted by ‘G’ and it’s height above the keel (K) by ‘KG’

(iii) Initial Metacentric Height (GM):
It is the distance measured along the centerline between G and M in the initial upright position.

1. (b) Sketch a transverse labelled diagram illustrating a heeled vessel in stable equilibrium: 1 (c):  A vessel is initially displacing 8,300t. KG 7.64m; KM 8.92m (constant)

A 110t weight is to be shifted from a position on the centreline Kg 2.4m to a position Kg 11 .2m on the centreline using the vessel’s own derrick. The derrick head is 37m above the keel.

Calculate the vessel’s GM for EACH of the following:
(i) When the weight is lifted clear of its initial position;  (5) To Calculate the Vertical Shift of G
GGv    =    w x sv    =    110 x (37.0 – 2.4)    =    0.459 m
∆                      8 300

To Calculate the New GM: (GM = KM-KG)
KM                    8.920 m

KG                –   7.640 m
GMOLD              1.280 m
GGv            –   0.459 m
GMNEW                  0.821 m

The GM after the lift is 0.82 m

(ii) When the weight has been shifted;  (5) To Calculate the Vertical Shift of G
GGv    =    w x sv    =    110 x (11.20 – 2.40)    =    0.117 m

∆                        8 300

To Calculate the Final GM
GMOLD              1.280 m

GGv             –   0.117 m
GMNEW                  1.163 m

The GM after the shift is 1.16 m

(iii) When the weight has been discharged ashore.  (5) To Calculate the Vertical Shift of G
GGv    =    w x sv    =    110 x (7.64 – 2.4)    =    0.070 m

∆ – w                8 300 – 110
To Calculate the Final GM
GMOLD              1.280 m

GGv           –      0.070 m
GMNEW                  1.210 m

The GM after the shift is 1.21 m

2. (a) (i) State Archimedes principle: (3)
When a body is wholly or partially immersed in a liquid, it experiences an upthrust (apparent loss of weight – termed Buoyancy force (BF)), equal to the mass of liquid displaced.

1) The volume of liquid displaced by a floating body is equal to the underwater volume.
2) The weight of liquid displaced by a floating body is equal to the weight of the body.
3) A floating body experiences an upthrust equal to the weight of water displaced, which is equal to the weight of the body.

2. (a) (ii)        Define TPC. (2) :
The TPC for any given draught is the weight which must be loaded or discharged to change the ship’s mean draught by one centimeter.
TPC = WPA x ρ
100

2. (b) A vessel is loading in SW of RD 1.025. Waterline Length 89m; Breadth 22m; Cw 0.83. Calculate the TPC. (4)

TPC    =          Aw  x ρ
100
=        (L x B x Cw) x ρ
100
=         (89 x 22 x 0.83) x 1.025
100
TPC    =          16.658
Ans: The TPC is 16.66

2. (c) Datasheet Q2 refers to a vessel with initial draughts of forward 4.6m, aft 4.8m in FW. The vessel then loads 3 830t of cargo. Find EACH of the following:

(i)        the initial displacement;  (3)

draughtf 4.600 m + draughta   4.800 m  / 2
= Mean Draught   4.700 m in FW

From Hydrostatic Table;
Ans: The Initial Displacement is 9 190t

(ii)        the final displacement;  (1)
To Calculate the New Displcement
Δ OLD                       9 190.0 t

Cargo Loaded     –   3 830.0 t
Δ NEW                    13 020.0 t

(iii)       the final mean draught in FW.  (7)

To Calculate the New Draught Scroll to Top