A ship 120m long floats in salt water at drafts 5.8m F and 6.6m A. TPC 15 tonnes. MCT 1cm 300 tonnes m. Centre of flotation is amidships. What is the minimum amount of water ballast required to be taken into the fore peak tank (centre of gravity 60m forward of the centre of flotation) to reduce the draft aft to 6.5 metres? Find also the final draft forward.

A ship 120m long floats in salt water at drafts 5.8m F and 6.6m A. TPC 15 tonnes. MCT 1cm 300 tonnes m. Centre of flotation is amidships. What is the minimum amount of water ballast required to be taken into the fore peak tank (centre of gravity 60m forward of the centre of flotation) to reduce the draft aft to 6.5 metres? Find also the final draft forward.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 16.19

 

Solution:

Bodily sinkage = w/TPC = w/15

 

New draft AFT = 6.6 m + w/15 cm

Required draft = 6.5 m

Change in draft = 0.1 + w/1500

Change in trim = w*d/MCTC

                         = w*60/300

                         = 6w/30                     

                         = w/5

x/t = l/L

=> x/(w/5) = 60/120

=> x = w/10 cm

=> x = w/1000 m

 

Now,

0.1 + w/1500 = w/1000

=> 0.1 = w/3000

=> w = 300 tons

 

Bodily sinkage = 300/15 = 20 cm = 0.2 m

Change in trim = w*d/MCTC

                         = 300*60/300

                         = 60 cm

                         = 0.6 m

y/t= L-l/L

=> y/0.6 = 120-60/120

=> y = 3/10 m

New draft FOR = 5.8 + 0.2 + 3/10 = 6.3 m