The areas of a ship’s water-planes commencing from the load water-plane and spaced at equidistant intervals down to the inner bottom, are: 2500, 2000, 1850, 1550, 1250, 900 and 800sq. m, respectively. Below the inner bottom is an appendage 1 metre deep which has a mean area of 650sq. m. The load draft is 7 metres. Find the load displacement in salt water, the Fresh Water Allowance, and the height of the centre of buoyancy above the keel.

The areas of a ship’s water-planes commencing from the load water-plane and spaced at equidistant intervals down to the inner bottom, are:

2500, 2000, 1850, 1550, 1250, 900 and 800sq. m, respectively.

 Below the inner bottom is an appendage 1 metre deep which has a mean area of 650sq. m. The load draft is 7 metres. Find the load displacement in salt water, the Fresh Water Allowance, and the height of the centre of buoyancy above the keel.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 10.7

 

Solution:

Aw	SM	f(∇)	Levers(x)	f(m)
2500	1	2500	6	15000
2000	4	8000	5	40000
1850	2	3700	4	14800
1550	4	6200	3	18600
1250	2	2500	2	5000
900	4	3600	1	3600
800	1	800	0	0
		∑ = 27300		∑ = 97000

 

Volume = h/3 * ∑f(∇)

               = 1/3 * 27300

               = 9100 m³

Total volume = 9100 + 650 = 9750 m³

Displacement = 9750*1.025 = 9993.75 tons

TPC = A_w*1.025/100

        = 2500*1.025/100

        = 25.625 ton/cm

FWA = ∆/4*TPC

          = 9993.75/4*25.625

          = 97.5 mm

COG from inner bottom = ∑f(m) / ∑f(∇) * h

                                             = 97000/27300 * 1

                                             = 3.55 m

COG from keel = 3.55 + 1 = 4.55 m

KB = V₁x₁ + V₂x₂/V₁+V₂

      = 9100*4.55 + 650*0.5 / 9100+650

      = 4.3 m above the keel