Md. Tawhid Hassan

  A magic square in math is a square array of numbers in which each row, column, and diagonal of numbers have the same sum.  It's a hand written magic box of 31 rows and columns. Sum of each row & column is = 14,911

The areas of a ship’s water-planes, commencing from the load water-plane and spaced 1 metre apart, are as follows: 800, 760, 700, 600, 450 and 10sq. m, respectively. Midway between the lowest two water-planes the area is 180sq. m. Find the load displacement in salt water, and the height of the centre of buoyancy above the keel. Reference: Ship Stability for Masters & Mates (6 th edition), Exercise – 10.12 Solution: Area SM f( ∇) Levers(x) f(m) 800 1 800 5 4000 760 4 3040 4 12160 700 2 1400 3 4200 600 4 2400 2 4800 450 1.5 675 1 675 180 2 360 0.5 180 10 0.5 5 0 0   ...

The following table gives the area of a ship’s water-plane at various drafts:   Draft (m)     6           7           8   Area (sq. m)     700 760      800 Find the volume of displacement and approximate mean TPC between the drafts of 7 and 8m. Reference: Ship Stability for Masters & Mates (6 th edition), Exercise – 10.11   Solution: Volume = h/12 * (5y 1 + 8y 2 – y 3 )               = 1/12 * (5*800 + 8*760 – 700)               = 781.67 m 3 TPC = A w *1.025/100         = 781.67*1.025/100         = 8.01 ton/cm

The transverse horizontal ordinates of a ship’s amidships section commencing from the load waterline and spaced at 1 metre intervals are as follows: 16.30, 16.30, 16.30, 16.00, 15.50, 14.30 and 11.30m, respectively.   Below the lowest ordinate there is an appendage of 8.5sq m. Find the area of the transverse section. Reference: Ship Stability for Masters & Mates (6 th edition), Exercise – 10.10   Solution: ord SM f(A) 16.3 1 16.3 16.3 4 65.2 16.3 2 32.6 16 4 64 15.5 2 31 14.3 4 57.2 11.3 1 11.3     ∑ = 277.6   A w = h/3 * ∑ f(A)        = 1/3 * 277.6         = 92.533 m 2 Waterplane area = 92.533 + ...

Three consecutive ordinates in a ship’s water-plane area are: 6.3, 3.35 and 0.75m, respectively.   The common interval is 6m. Find the area contained between the last two ordinates. Reference: Ship Stability for Masters & Mates (6 th edition), Exercise – 10.9   Solution: Area = h/12 * (5y 1 + 8y 2 – y 3 )           = 6/12 * (5*0.75 + 8*3.35 – 6.3)           = 12.125 m 2

A ship’s water-plane is 80 metres long. The breadths commencing from forward are as follows: 0, 3.05, 7.1, 9.4, 10.2, 10.36, 10.3, 10.0, 8.84, 5.75 and 0m, respectively.   The space between the first three and the last three ordinates is half of that between the other ordinates. Calculate the area of the water-plane, and the position of the centre of flotation. Reference: Ship Stability for Masters & Mates (6 th edition), Exercise – 10.8   Solution:   Breadth SM f(A) Levers(x) f(m) FOR 0 0.5 0 -3.5 0   3.05 2 6.1 -3 -18.3   7.1 1.5 10.65 -2.5 -26.625   9.4 4 37.6 -2 -75.2   10.2 2 20.4 -1 -20.4 ...