Ship’s wt |
KG |
VM |
| 4950 t | 4.85m | 24007.5 tm |
| (+)50t | 1.25m | (+ )62.5 tm |
Final W = 5000 t Final VM = 24070 tm
We know that:
Final KG = (Final VM)/( Final W)
Final KG = (24070/5000)
= 4.814m
Final GM = (KM –KG)
= (5.79 – 4.814) m
= 0.976m.
Tanθ = LM/(Weight x GM)
= 200/ (5000 x 0.976)
θ = 2.34 degree
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A weight of 100t is discharge from a position 2.45m above the Keel and 6m to port of the centre line of a ship of w 10000t, KM 8.25m, KG 7.45m. Find the list
Solution:
Weight discharged (w) = 100t, & KG = 2.45m
Transverse distance = 6m to part from CL
W = 10000 t,
KM = 8.25 m,
KG = 7.45m
We know that:
LM caused = (weight x distance)
= (100 x 6) tm
= 600 tm (P)
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