A ship of 7200 tonne displacement has KG 5.20 m, KB 3.12 m and KM 5.35 m. 300 tonne of fuel at Kg 0.6 m are now used. Ignoring free surface effect and assuming the KM remains constant, calculate the angle to which the vessel will heel.

A ship of 7200 tonne displacement has KG 5.20 m, KB 3.12 m and KM 5.35 m. 300 tonne of fuel at Kg 0.6 m are now used. Ignoring free surface effect and assuming the KM remains constant, calculate the angle to which the vessel will heel.

Reference: REED'S NAVAL ARCHITECTURE FOR MARINE ENGINEERS (4^th edition), Exercise – 5.17

 

Solution:

New KG = 7200*5.2 – 300*0.6/ 7200-300

               = 5.40 m

GM = KM – KG = 5.35 – 5.4 = -0.05 m

BM KM – KB = 5.35 – 3.12 = 2.23 m

Tanθ = √((-2GM)/BM)

=> Tanθ = √((-2*(-0.05))/2.23)

=> θ = 11⁰ 57`