Trim or Longitudinal Stability
A ship is floating in salt water at drafts of 6.7 m F and 7.3 m A. MCT 1 cm 250 tonnes m. TPC 10 tonnes. The length of the ship is 120 meters. The center of flotation is amidships; 220 tonnes of cargo is then discharged from a position 24 m forward of the center of flotation. Find the weight of cargo which must now be shifted from 5 m aft of the center of flotation to a position 20 m forward of the center of flotation, to bring the draft aft to 7 meters. Find also the final draft forward.
Bodily rise = W / TPC = 220 / 10 = 22 cm = 0.22 m
Change of trim = (w * d) / MCTC = (220 * 24) / 250 = 21.12 cm by the stern
Change of draft aft = 1/2 * change of trim = 21.12/2 = 10.56 cm = 0.11m
Change of draft fwd = Change of trim aft (As CF is amidships)
Original draft = 7.3m A 6.7m F
Bodily rise = -0.22 m – 0.22m
Change due to trim = + 0.11m – 0.11m
…………………………………………………………………………………
New draft = 7.19 m A 6.37m F
New draft aft = 7.19 m
Aft draft required = 7.0 m
……………………………………….
Reduction required = 0.19 m = 19 cm
Let “w” tonnes of cargo be shifted.
Change of trim = (w * d) / MCTC = 25w / 250 (d = 20 + 5 = 25) by the head
Change of draft aft = 1/2 * change of trim
So, 19 = 1/2 * (25w/250)
w = 380 tonnes of cargo to be shifted to bring aft draft to 7 m.
As CF is amidships, change of draft aft = change of draft fwd
Change of trim = (380 * 25) / 250 = 38 cm = .38 m
So, Final fwd draft = 6.37 m + .38 m = 6.75 m (Ans)
