A ship leaves port with drafts 7.6m F and 7.9m A; 400 tonnes of bunkers are burned from a space whose centre of gravity is 15m forward of the centre of flotation, which is amidships. TPC 20 tonnes. MCT 1cm 300 tonnes m. Find the minimum amount of water which must be run into the forepeak tank (centre of gravity 60m forward of the centre of flotation) in order to bring the draft aft to the maximum of 7.7m. Find also the final draft forward.

A ship leaves port with drafts 7.6m F and 7.9m A; 400 tonnes of bunkers are burned from a space whose centre of gravity is 15m forward of the centre of flotation, which is amidships. TPC 20 tonnes. MCT 1cm 300 tonnes m. Find the minimum amount of water which must be run into the forepeak tank (centre of gravity 60m forward of the centre of flotation) in order to bring the draft aft to the maximum of 7.7m. Find also the final draft forward.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 16.20

 

Solution:

For 400 tons of bunkers burning-

Bodily rise = w/TPC = 400/20 = 20 cm = 0.2 m

Change in trim = w*d/MCTC = 400*15/300 = 20 cm = 0.2 m

 

x/t = l/L

=> x/0.2 = ½

=> x = 1/10 m

 

y/t = L-l/L

=> y/0.2 = ½

=> y = 1/10 m

 

New draft AFT = 7.9 – 0.2 + 1/10 = 7.8 m

New draft FOR = 7.6 – 0.2 – 1/10 = 7.3 m

Let,

‘w’ tons of water must be run into the forepeak tank.

Bodily sinkage = w/TPC = w/20 cm = w/2000 m

New draft AFT = 7.8 + w/2000

Required draft = 7.7 m

Change in draft = 0.1 + w/2000

Change in trim = w*d/MCTC

                         = 60w/300

                         = w/5

x/t = l/L

=> x/(w/5) = ½

=> x = w/10 cm

=> x = w/1000 m

 

Now,

0.1 + w/2000 = w/1000

=> 0.1 = w/2000

=> w = 200 tons

Bodily sinkage = 200/20 = 10 cm = 0.1 m

Change in trim = w*d/MCTC

                         = 200*60/300

                         = 40 cm

                         = 0.4 m

y/t = L-l/L

=> y/0.4 = ½

=> y = 1/5 m

Final draft = 7.3 + 0.1 + 1/5 = 7.6 m