A ship of length 120 m displaces 11750 t when floating in sea water of density 1025 kg/m3. The Centre of gravity is 2m above centre of buoyancy and water plane area is defined by following equidistant half breadth given below.
Station AP 1 2. 3 4 5 6 7 FP
Half breadth (m) 3.3 6.8 7.6 8.1 8.1 8.0 6.6 2.8 0
Calculate each of the following
1. The area of water plane
2. The position of centroid of water plane from midships.
3.The second moment of area of water plane about a transverse axis through the centroid.
4.The moment to change trim one centimetre.
Station AP 1 2. 3 4 5 6 7 FP
Half breadth (m) 3.3 6.8 7.6 8.1 8.1 8.0 6.6 2.8 0
Calculate each of the following
1. The area of water plane
2. The position of centroid of water plane from midships.
3.The second moment of area of water plane about a transverse axis through the centroid.
4.The moment to change trim one centimetre.
in the calculation of summation of c the value will be 576.4 m4.
ReplyDeleteThanks for the info.
DeleteCould you explain the use of h*lever for calculation of products for I and II moments. It is different from other problems you have solved have calculation of CoB and CoF is done.
ReplyDeleteHere two stations have same 1/2 ordinate so we take lever from centre.
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