A ship of 5000 tonnes displacement has KG 4.2m and KM 4.5m, and is listed 5 degrees to port. Assuming that the KM remains constant, find the final list if 80 tonnes of bunkers are loaded in No. 2 starboard tank whose centre of gravity is 1 metre above the keel and 4 metres out from the centreline.

 A ship of 5000 tonnes displacement has KG 4.2m and KM 4.5m, and is listed 5 degrees to port. Assuming that the KM remains constant, find the final list if 80 tonnes of bunkers are loaded in No. 2 starboard tank whose centre of gravity is 1 metre above the keel and 4 metres out from the centreline.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 14.1

Solution:

GM = KM – KG = 4.5 – 4.2 = 0.3 m

Weight (ton)	KG (m)	TCG (m)	Moment (keel)	Moment(ton.m)
5000	4.2	0	21000	0
+80	1	4	80	320
∑ = 5080			∑ = 21080	∑ = 320

 

New KG = moment/weight = 21080/5080 = 4.1496 m

Final GM = KM – KG = 4.5 – 4.149 = 0.35 m

As ship is already listed 5⁰ to port side-

GG₁ = moment/mass

=> moment = GM*tanθ * mass

=> moment = 0.3*tan5 * 5000

                     = 131.23 ton-m

Final moment = 320 – 131.23 = 188.767 ton-m

Final GG₁ = moment/mass

                 = 188.767/5080

                 = 0.037 m

 

   Tanθ = GG₁/GM

=> θ = tan-1 (0.037/0.35)

=> θ = 6⁰3` to starboard