A ship of 4515 tonnes displacement is upright, and has KG 5.4m and KM 5.8m. It is required to list the ship 2 degrees to starboard and a weight of 15 tonnes is to be shifted transversely for this purpose. Find the distance through which it must be shifted.

A ship of 4515 tonnes displacement is upright, and has KG 5.4m and KM 5.8m. It is required to list the ship 2 degrees to starboard and a weight of 15 tonnes is to be shifted transversely for this purpose. Find the distance through which it must be shifted.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 14.2

 

Solution:

GM = KM – KG = 5.8 – 5.4 = 0.4 m

GG = tanθ * GM = tan2⁰ * 0.4 = 0.014 m

GG₁ = m*d/displacement

=> d = GG₁ * displacement/m

=> d = 0.014*4515/15

=> d = 4.2 m