Stability – I : Chapter 10

Ship’s weight	KG	VM
5000t	7.0m	35000tm
(+) 400 (load)	1.315m	(+) 526tm

 Final W = 5400t                                                                                      Final VM=    35526tm

We know that:
Final KG = (Final VM) / (Final W)
Final KG = (35526 / 5400)
= 6.578m

GM Solid   = (KM – KG)
= (7.2 – 6.578)
= 0.622m

We know that:

FSC = (i di/ W)
= (LB³ x di)/ (12 x W )
= (16 x 10³ x 0.95)/(12 x 5400)
= 0.234m

GM (fluid )= GM (solid) – FSC
= (0.622 – 0.234)
= 0.388m.

9. A vessel has two deep tanks port and starboard, each 12m long , 5m wide and 8m deep . The port side is full of SW while the starboard side is empty. W = 9840 t, KM = 8.5m KG = 8.0m . Calculate the GM if fluid if SW is transferred from P to S until each tank has equal quantity of ballast .

Solution:

Volume of tank = (L x B x D )
= (12m x 5m x 8m)

W = 9840t , KM = 8.5m and  KG = 8.0m

We can calculate Mass :
Mass of the tank = (Volume x Density)
= (12 x 5 x 8 x 1.025)
= 492t

According to question, 1/2 of water transferred to stbd
= (492/2)
= 246t

KG, of the port side tank after shifting
= (4 + 2)m
= 6m

Now,   Stbd side tank = (d/2)
= (4/2)m
= 2m

Again, d = ( KG of port  tank  – KG of stbd tank)
=  (6 – 2)m
= 4m