Ship’s wt |
FKG |
VM |
LM |
| 12000 | 7.25 | 87000 | 0 |
| 200 | 10 | 2000 | 1400 S |
Final W= 12200 t Final VM = 89200 FLM 1400 S
We know that:
Final KG = (Final VM/Final W)
Final KG = (89200/12200)
= 7.311m
Again, Final GM = (KM –KG)
= (9.0 – 7.311)
= 1.689m
tanθ = 1400/(12200 x 1.689)
= 3 degree 53.2min
-
A ship of W 10000t, KM 7.3m , KG 6.8m is listed 5degre to port . A heavy lift weighing 100t, lying 6m to port 0f the centre line and KG 10.0 m, is to be shifted to the lower hold KG 2.0m on the centre line of the ship , by the ship’s jumbo derrick whose head is 28m above the keel. Find
- The list as soon as the derrick takes the load .
- The list when the derrick swings to the load to the centre line.
- The list after the shifting is over.
Solution:
Displacement (W) = 10000t,
KM = 7.3m &
KG = 6.8m ,
List = 5 degree(P)
w = 100t, d = 6m to port of the CL KG = 10m & KG = 2.0m
Derrick head = 28m .
We can calculate :
Initial LM = (W x GM x tanθ)
= (10000 x 0.5 x tan50)
= 437.44 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