A ship 160m long and 8700 tonne displacement floats at a waterline with Station
AP 1/2 1 2 3 4 5 6 7 71/2 FP
1/2 ordinate 0 2.4 5.0 7.3 7.9 8.0 8.0 7.7 5.5 2.8 0m
While floating at this waterline, the ship develops a list of 10° due to instability.
Calculate the negative metacentric height when the vessel is upright in this condition.
Structured for DG Shipping MEO Class II examination scoring criteria.
1/2 ord | 1/2 ord^3 | SM | Product for second moment |
0 | 0 | 1 | 0 |
2.4 | 13.82 | 4 | 55.28 |
5.0 | 125 | 2 | 250 |
7.3 | 389.02 | 4 | 1556.08 |
7.9 | 493.04 | 2 | 986.08 |
8.0 | 512 | 4 | 2048 |
8.0 | 512 | 2 | 1024 |
7.7 | 456.53 | 4 | 1826.12 |
5.5 | 166.38 | 2 | 332.76 |
2.8 | 21.95 | 4 | 87.8 |
0 | 0 | 1 | 0 |
$$Common \space interval \space (h) \space = \space {{L} \over h} \space = \space {{160} \over 10} \space = \space 16 $$
Second moment of area of waterplane about centreline
$$=\:2\times{{h} \over9}\times\sum I_{CL}$$
$$=\:2\times{{16} \over9}\times8166.12$$
$$=\:29035.1m^4$$
$$BM=\rho\:\times\frac{\sum I_{CL}}{\Delta}$$
$$=\:\frac{1.025\times29035.1}{8700}$$
$$=\:3.421m$$
$$GZ\:=\:\sin\theta\:\left\lbrack GM\:+\:\frac12BM\tan^2\theta\right\rbrack$$
$$at\:angle\:of\:LOLL,\:GZ=0$$
$$0=\sin\theta\left\lbrack GM\:+\:\frac12BM\tan^2\theta\right\rbrack$$
$$GM + {{1} \over 2} BM \times tan^2\theta \space = 0$$
$$GM=\:-\frac12BM\:\times\tan^2\theta$$
$$GM=\:-\frac12\times3.421\times\tan^210$$
$$GM=-0.0531m$$
$$Negative\:metacentric\:height\:GM\:=\:-\:0.0531m$$