
In a vessel of 8800 tonnes displacement and KG 6.2m, 200 tonnes of cargo was loaded in the lower hold 1.7m above the keel. Find the final KG.
Solution:
Given:
Weight = 8800t,
Original KG =6.2m,
Cargo loaded = 200t,
KG = 1.7m
Ship’s weight 
KG 
VM 
8800 t  6.2m  54560tm 
(+) 200 load  1.7m  Final VM 54900 tm 
Final W – 9000 t Final VM 54900 tm
So, we can calculate
Final KG = (Final VM/ Final W)
Final KG = (54900 / 9000)
= 6.1m
Here ‘VM’ stands for vertical moments.

600 tonnes of cargo were discharged from a vessel from the upper 11m above the keel. In the original KG displacement were 6m and 12600 tonnes, calculate the KG.
Solution:
Given:
Weight =12600 t
Original KG =6m
Cargo discharged = 600 t,
Cargo discharge 11 m above the keel
Ship’s weight 
KG 
VM 
12,600t  6m  75,600tm 
( ) 600t ( discharge)  11m  () 6600tm 
Final Weight =12000 t Final VM =69000tm
We know that :
Final KG = (Final VM / Final Weight)
Final KG = ( 69,000 / 12000)
= 5.75m

In a vessel of 9900 tonnes displacement and KG 4m, a heavy lift of 100 tonnes is loaded on the UD (KG 15m ). Find the final KG.
Solution:
Ship’s weight 
KG 
VM 
12,600t  4m  39,600tm 
(+)100 t load  15m  ( +)1500tm 
Final W 10000t Final VM =41000tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = ( 41000 / 10000 )
= 4.11m.

500 tonnes of cargo of discharge from the lower the lower hold (KG 3m ) of a vessel whose displacement and KG before discharging were 11500 tonnes and 6.3m . Find the final KG.
Solution:
Given:
Weight = 11500t
Initial KG = 6.3m
Cargo discharged = 500t, KG =3m
Ship’s weight 
KG 
VM 
11500t  4m  39,600tm 
(+)100 t load  15m  ( +)1500tm 
Final W =11000t Final VM =70950tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (70,950/11000)
=6.45m

500 tonnes of cargo was shifted 15 meters vertically downwards in a vessel of 10000 tonnes displacement. Find the effect it has on the KG of the vessel and whether KG increases or decreases.
Solution:
Given:
Weight =10000t,
Cargo shifted downward =500t,
We know that:
GG1() = (w x d)/( total weight)
= (500 x 15)/ 10000
=0.75m
Since, The weight is shifted vertically downward, the KG of ship will also go down parallel to the KG of the weight
Hence, KG of the ship decrease by 0.75m

In a vessel of 9000 tonnes displacement , KG 10.5m, 300 tonnes of cargo was shifted from the LH (KG 2.5m)to the UDC (KG 11.5m). Find the resultant KG of teh vessel.
Solution :
Weight =9000 t
Original KG = 10.5m ,
Cargo shifted = 300 t
Vertical Distance of cargo shifted
= (height of UD – height of LH)
= (11.5 2.5)
= 9m upwards.
Ship’s weight 
KG 
VM 
9000t  10.5m  94500tm 
(+)300t  9m  (+)2700tm 
Final W = 9000 Final VM =200t
We know that :
Final KG = (Final VM / Final Weight)
Final KG = (97200 / 9,000)
= 10.8m.

In a vessel of 9009 tones displacement, KG 8.7m, how many tonnes of cargo can be loaded on the upper deck (KG 15m ) so that the final KG would be comes 9m.
Solution:
Given:
Weight = 9009 t,
Original KG =8.7m,
Let cargo loaded be X tonnes
Final KG =9m
Ship’s weight 
KG 
VM 
9009 t  8.7m  94500tm 
(+) X t (load)  15m  (+) (15 X) tm 
Final W = (9009 + X) t Final VM =(78378.3 + 15X) tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (78378.3 +15X ) /(9009 + X)
9.0 = (78378.3 + 15X)/ (9009+X)
9 x (9009+X) = (78378.3 + 15X)
(81081 + 9X) = (15X + 78378.3)
Hence, 6X = 78378 .3 – 81081
6X = 2702.7t
X = 450.45 t

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.