A ship of 8500 tonnes displacement has TPC 10 tonnes, MCT 1cm 100 tonnes m and the centre of flotation is amidships. She is completing loading under coal tips. Nos. 2 and 3 holds are full, but space is available in No. 1 hold (centre of gravity 50m forward of amidships), and in No. 4 hold (centre of gravity 45m aft of amidships). The present drafts are 6.5m F and 7m A, and the load draft is 7.1m. Find how much cargo is to be loaded in each of the end holds so as to put the ship down to the load draft and complete loading on an even keel.

 A ship of 8500 tonnes displacement has TPC 10 tonnes, MCT 1cm 100 tonnes m and the centre of flotation is amidships. She is completing loading under coal tips. Nos. 2 and 3 holds are full, but space is available in No. 1 hold (centre of gravity 50m forward of amidships), and in No. 4 hold (centre of gravity 45m aft of amidships). The present drafts are 6.5m F and 7m A, and the load draft is 7.1m. Find how much cargo is to be loaded in each of the end holds so as to put the ship down to the load draft and complete loading on an even keel.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 16.1

Solution:

TPC = W/(mass*d)

=> W = TPC * mass*d

=> W = 10*(710 – (650+700)/2) cm

=> W = 350 tons

 

Change in draft AFT = 7.1 – 7 = 0.1 m

Change in draft FOR = 7.1 – 6.5 = 0.6 m

Trim = 0.6 – 0.1 = 0.5 m by the head

Let, w ton was loaded in No-1

and (350-w) ton was loaded in No-4

Weight	Distance from CF	Moment
W	-50 (F)	-50w
350-w	+45 (A)	15750-45w
∑ = 350		∑ = 15750-95w

 

Change in trim = trimming moment/MCTC

=> -50 cm = 15750-95w/100

=> -5000 = 15750 – 95w

=> w = 218.42 tons

Load at No-1 hold = 218.42 tons

Load at No-4 hold = 350-218.42 = 131.58 tons