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A ship of W 16000t, KM 7.5m, KG 6.0m , TPC 25, is listed 3 degree to port .Her present mean draft is 8.6m and she is to finish loading at 8.8m mean draft . Space is available 5m 0f the centre line, on either side. State how much cargo must be stowed on either side to finish upright
Solution :
W = 16000t,
KM, 7.5m KG = 6.0m,
Initial GM = 1.5 m
TPC = 25,
List = 3 degree P
Present draft = 8.6m
Final draft = 8.8
Sinkage available = (8.8 – 8.6)m
= 0.2m
= 20cm
Now , Cargo to load = (Sinkage x TPC)
= (20 x 25)
= 500 t
We know that :
LM =(W x GM tanθ)
Again , Initial LM = (16000 X 1.5 X Tan 30)
= 1257.78 tm
Let ‘X’ t 0f cargo will load on port side
Cargo to be load on stbd side = (500 – X) t
So, final LM caused due to loading of cargo would be
LM(1) = (Weight x distance)
= (X x 5)
= 5X (P)
LM(2) = (Weight x distance)
= (500 –X) x 5
= (2500 – 5X ) (S)
As per above calculation ,
Initial LM = 1257.78 ( P)
Hence to keep vessel upright ,
LM(S) = LM(P)
(2500 – 5X) = (1257.7 + 5X)
2500 – 5X – 5X = 1257.7
(- 10X) = (1257.7 – 2500)
X = (1242.3/10)
= 124 .3 t
Hence cargo loaded is 124.23 t on port side
So, cargo loaded in stbd is (500 – 124.23)
= 375.7 t
Question no.19. Sub question 3
Kg will change when being shifted from derrick to lower hold 2m above kg so GM will also change
Gm= 0.58 and list will be 1.60degree to starboard
Q.7
Final listing moment is 700(p), not starboard
For case 2; since, both the tanks are slack, therefore, Total FSC = 0.3456m x 2 = 0.6912m.
Good luck
Ex -11 List Ques No. 18, Part 2 have a mistake in total Vertical movements…..it should be 87000 not 87200 which gives us a final answer as follows
List = 3* 51′ or 3.85 degrees
Q.7 listing moment is 700 port not stbd