Stability 1 – Chapter 6 Centre of Gravity
- A heavy lift derrick , whose head is 20m above the keel , is to shift a locomotive weighing 300 tonnes from the UD (KG 8m ) to the LH (KG 2m). If the displacement and initial KG of the vessel were 12000 tonnes and 7.6m, find the KG of the vessels
- When the derrick has taken the weight off the UD and
- After shifting is over.
Solution:
Given:
Weight = 12000t,
Initial KG =7.6m,
Cargo shifted = 300t,
Change of KG = (8 – 2)
= 6m downwards
Derrick Height = 20m,
KG of the Weight = 8m
(since the weight is on the UD)
Case – 1
| Ship’s weight | KG | VM |
| 12000 t | 7.6m | 91200tm |
| 300t (shift) | 12m | (+) 3600 tm |
Final W =12000 t Final VM =94800tm
We know that:
Final KG = (Final VM / Final Weight)
FKG = (94800 / 12000)
= 7.9m
Case – 2
| Ship’s weight | KG | VM |
| 12000 t | 7.6m | 91200tm |
| 300t (shift) | 6m | (-)1800 tm |
Final W =12000 Final VM = 89400tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (89400 / 12000)
= 7.45m.
- On a vessel of 4,950 tonnes displacement, KG 9.2m, the ship’s jumbo derrick is to be load a weight of 50 tonnes from the wharf , on to the UD( KG 8m) . if the head of the derrick is 25m above the keel , calculate the KG of the vessel
- when the weight is hanging by the derrick on the centre line but 2m above the UD , and
- after loading.
Solution:
Given:-
Weight = 4950t,
Initial KG = 9.2m
Cargo loaded = 50t ,
UD KG =8m
Height of the derrick = 25m
Case – 1
| Ship’s weight | KG | VM |
| 4950 t | 9.2m | 45540 tm |
| (+) 50t (load) | 25m | (+) 1250 tm |
Final W = 5000t FVM = 46790tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (46790 / 5000)
= 9.358m
Case – 2
| Ship’s weight | KG | VM |
| 4950 t | 9.2m | 45540 tm |
| (+) 50t (load) | 8m | (+) 400 tm |
Final W = 5000t Final VM = 45940tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (45940 / 5000)
= 9.188m
- A ship’s derrick, whose head is 22m above the keel, is used to be discharge a weight of 20 tonnes (KG 5m), lying on the centre line. If the vessel’s displacement and KG before discharging were 6000 tonnes and 8m. Calculate the KG
- As soon as the derrick lifts the weight and
- After discharging .
Solution:-
Given:-
Weight = 6000t,
Original KG = 8m,
Cargo discharged = 20t,
KG of cargo discharged = 5m
Height of the derrick = 22m
Case – 1
| Ship’s weight | KG | VM |
| 6000t | 8m | 48 000 tm |
| 20t | 17m | (+) 340 tm |
Final W = 6000 Final VM 48340 tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (48340 / 6,000)
= 8.056m
Case – 2
| Ship’s weight | KG | VM |
| 6000t | 8m | 48 000 tm |
| (-)20t (discharge) | 5m | (-) 100 tm |
Final W =5980t Final VM = 47900 tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (47900/5980)
= 8.010m.