Question

(a) Explain the term volumetric heeling moments. (6)

(b) A ship 85 m long displaces 8100 tonne when floating in sea water at draughts of 5.25 m forward and 5.55 m aft. TPC 9.0, GMl 96 m, LCF 2 m aft of midships. It is decided to introduce water ballast to completely submerge the propeller and a draught aft of 5.85 m is required. A ballast tank 33 m aft of midships is available. Find the least amount of water required and the final draught forward. (10)

Answer 1

$$Let\:m=mass\:of\:ballast\:required$$

$$MCT1_{\operatorname{\mathrm{cm}}}=\frac{\Delta\times GM_{L}}{100L}$$

$$=\frac{8100\times96}{100\times85}$$

$$=91.48ton.m$$

$$Trimming\:moment=m\times d$$

$$=m\left(33-2\right)$$

$$=31m$$

$$Change\:of\:trim=\frac{T\times M}{MCT_{1\operatorname{\mathrm{cm}}}}$$

$$=\frac{31\times m}{91.48}cm\:by\:stern$$

$$Bodily\:Sinkage=\frac{m}{TPC}$$

$$=\frac{m}{9.0}$$

$$Draft\:Aft\:d_{A^{}1}=d_{A}+\frac{m}{TPC}+\frac{t}{L}\left(\frac{L}{2}-x\right)$$$$5.85=5.55+\frac{m}{9\times100}+\frac{31m}{91.48\times100\times85}\left(\frac{85}{2}-2\right)$$

$$5.85-5.55=2.7258\times10^{-3}m$$

$$m=110\:tonnes$$

$$Draft\:Fwd\:d_{F1^{}}=d_{F}+\frac{m}{TPC}-\frac{t}{L}\left(\frac{L}{2}+x\right)$$

$$=5.25+\frac{110}{9\times100}-\frac{0.372}{85}\left(\frac{85}{2}+2\right)$$

$$=5.25+0.1222-0.1951$$

$$d_{F1^{}}=5.177m$$

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