A ship arrives in port with drafts 6.8m. F and 7.2m A; 500 tonnes of cargo is then discharged from each of 4 holds. The centre of gravity of No. 1 hold is 40m forward of amidships The centre of gravity of No. 2 hold is 25m forward of amidships The centre of gravity of No. 3 hold is 20m aft of amidships The centre of gravity of No. 4 hold is 50m aft of amidships Also 50 tonnes of cargo is loaded in a position whose centre of gravity is 15m aft of amidships, and 135 tonnes of cargo centre of gravity 40m for ward of amidships. TPC 15 tonnes. MCT 1cm 400 tonnes m. The centre of flotation is amidships. Find the final drafts.

A ship arrives in port with drafts 6.8m. F and 7.2m A; 500 tonnes of cargo is then discharged from each of 4 holds.

The centre of gravity of No. 1 hold is 40m forward of amidships

The centre of gravity of No. 2 hold is 25m forward of amidships

The centre of gravity of No. 3 hold is 20m aft of amidships

The centre of gravity of No. 4 hold is 50m aft of amidships

Also 50 tonnes of cargo is loaded in a position whose centre of gravity is 15m aft of amidships, and 135 tonnes of cargo centre of gravity 40m for ward of amidships. TPC 15 tonnes. MCT 1cm 400 tonnes m. The centre of flotation is amidships. Find the final drafts.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 16.13

 

Solution:

Weight	Distance from CF	Moment
-500	-40 (F)	+20000
-500	-25 (F)	+12500
-500	+20 (A)	-10000
-500	+50 (A)	-25000
+50	+15 (A)	+750
+135	-40 (F)	+5400
∑ = -1815		∑ = -7150

 

Change in trim = 7150/400

                             = 17.87 cm by the head

x/t = l/L

=> x/17.87 = ½

=> x = 8.935 cm

 

y/t = L-l/L

=> y/17.8 = ½

=> y = 8.935 cm

Bodily rise = w/TPC = 1815/15 = 121 cm = 1.21 m

New draft Aft = 7.2 – 121 – 0.08935 = 5.901 m

New draft FOR = 6.8 – 1.21 + 0.08935 = 5.67 m