Stability – I : Chapter 11

Ship’s wt	KG	VM	d	LM
8500t	8.3m	70550tm	389.23tm
(-) 200t	4m	(-) 800tm	5S	1000 P
(-)300t	5m	( -)1500tm	2P	600 S
(+)100t	2m	(+) 200tm	4S	400 S
200t( Shift)	2m	(+) 400tm	3P	600P

Final W = 8100t                                  Final VM = 68850tm                               FLM=289.23tm

We know that:  
Final GM = (Final VM )/(Final W)
Final KG = (68850 /8100)
= 8.5m

Final GM = (KM – KG)
= (9.3 – 8.5)
=      0.8m

We know that:
tanθ = Final LM/(W x GM)
=   289.23/(8100 x 0.8)
= 2degree 33.3minute

14. A ship of 15000t displacement, KG 8.7m KM 9.5m is listed 10deg to port. The following cargo work was carried out:

• 500t loaded KG 8.0m, 5m stbd of CL .
• 300t discharged, KG 4.0m, 4 port of CL.

Find the quantity of SW ballast that must be transferred transversely to bring the vessel up right , the tank centres the being  12m apart.

NOTE : Since vessel is required upright , it is not necessary to calculate the final KG or GM unless specifically asked).

Solution:

Displacement (W) = 15000t,
KG = 8.7m, KM = 9.5m & list = 10 degree (P)
Final W = (15000 + 500) – 300  = 15200 t
Initial LM = 15000 x 0.8 x tan 10 = 2115.9 tm (P)

We know that:
Listing moment caused
LM(1) = (500 x 5)
= 2500 tm (S)

LM (2) = (300 x 4)
= 1200 tm (S)

Total listing moment = (2500 + 1200) tm
= 3700 tm (S)

According to question and calculation done above ,
Initial listing moment = 1584.1 tm (S)

So, maintain vessel upright same LM must be created on (P) side.
Again, we know that  LM = w x d
1584.1 tm   = (w x 12)
= (1584.1/12)
= 132 t

Hence , 132t SW should transferred to (P) side to keel the vessel upright.