Stability – I : Chapter 11

15. A bulk is carrier presently of 12250t, KM 9.8m , KG 9.0m has a list of 6dee to starboard .she then load 1250 t of ore (KG 8m , 2m to stbd of centre line ) and discharges 250t of ore (KG 2m 5m , from star board of centre line). 160t of SW ballast is then transferred from the stbd shoulder tank to the port DB tank vertically downwards by 9m and transversely  by 10m ) Find the final list assuming that they are no slack tanks given the final KM is 9.6m .

Solution:

Ship’s wt	KG	VM	D	LM
12250t	9.0m	110250		1116.57(S)
(+) 1250t	8m	(+) 10000	2S	2500(S)
(-) 250t	2m	(-) 500	5S	1250(P)
160t (Shift)	9m	(-) 1440	10P	1600(P)

Final W = 13250t                                     Final VM = 118310 tm                 FLM =766.57(S)

We know that :

Final KG = (Final VM) /(Final W)
= (118310/13250)
= 8.929 m

Again,

Initial LM = ( W x GM tanθ)
= (12250 x 0.8 x tan 6.5⁰)
= 1116.57 (S)

GM = ( KM – KG)
= (9.6 – 8.929)
= 0.671m

tanθ = ( Final LM)/( W x GM)
= (766.57)/( 13250 x 0.671)
= 4.93 degree(S)