Solution Naval Numerical 30

Numerical 30: A ship 130m long displaces 14000 tonne when floating at draughts of 7.5m forward and 8.10m aft. GML –125m, TPC – 18,LCF-3m aft of midships. Calculate the final draughts when a mass of 180 tonne lying 40m aft of midships is removed from the ship.

Solution: 

Given:

L =130 m

△ = 14000 t

$d_f$ = 7.5 m

$d_a$ = 8.1 m

$GM_L$ = 125 m

TPC = 18 t/cm

LCF = 3 m (aft pf midship)

to find, final draughts when a mass of 180 t lying 40m aft of midships is removed.

 

Bodily rise = $\frac{mass}{TPC}$ = $\frac{180}{18}$ = 10 cms or 0.1 m

MCT1cm = $\frac{GM_L\times \Delta}{100L}$

 = $\frac{125\times 14000}{100\times 130}$ = 134.6 t-m

change in trim = $\frac{mass\times (position\, of\, mass\, from\, midship) }{MCT1cm}$

= $\frac{180\times (40-3)}{134.6}$

= 49.48cm

Change forward = + $\frac{change\, in\, trim }{L}\left ( \frac{L}{2}+ distance\, of\, mass\, from\, midship \right )$ 

change forward = + $\frac{49.48 }{130}\left ( \frac{130}{2}+ 3 \right )$ =+25.88 cm

Change Aft = - $\frac{change\, in\, trim }{L}\left ( \frac{L}{2} - distance\, of\, mass\, from\, midship \right )$ 

Change Aft = - $\frac{49.48 }{130}\left ( \frac{130}{2} - 3 \right )$ =+23.60 cm 

New draught fwd = Old draught fwd -Bodily rise + change fwd 

= 7.5 + 0.1 + 0.2588 = 7.6588m

New draught aft = Old draught aft -Bodily rise - change aft

= 8.1 - 0.1 - 0.236 = 7.764m