
On a ship of 5000 t displacement , a tank is partly full of DO of RD 0.88. If the moment of inertia of the tank about its centreline is 242m^{4} , find the FSC .
Solution:
W = 5000t, RD = 0.88, I = 242m4
We know that :
FSC = (i di /W )
= (242 x 0.88 ) /5000
= 0.042m.

If the tank in Question 1 was partly full of SW instead of DO, find the FSC.
Solution:
RD = 1.025
We know that:
FSC = (i di/W )
= (242 x1.025) /5000
= 0.049m.

On a ship of W 6000 t, KM 7.4m, KG 6.6m, a double bottom tank of i 1200m^{4} is partly full of FW . Find the GM fluid .
Solution:
W = 6000 t & KM = 7.4m,
KG = 6.6m, i = 1200m^{4}, RD = 1.00
GM (solid) = (KM – KG)
= (7.4 – 6.6)
= 0.8m
We know that:
FSC = (i di /W )
= (1200 x 1) /6000
= 0.2m
Fluid GM = GM (solid) – FSC
= (0.8 – 0.2)
= 0.6m

Given the following particulars, Find the GM fluid : W= 8800 t, tank of i = 1166 m4 is partly full of HFO of RD 0.95, KM 10.1m, KG 9.4m .
Solution:
W = 8800 t & i = 1166m^{4},
RD = 0.95,
KM = 10.1 m & KG = 9.0m.
GM (solid) = (KM – KG)
= (10.1 – 9.0)
= 1.1m
We know that:
FSC = (i di /W )
= (1166 x 0.95)/ 8800
= 0.126m
GM fluid = GM (solid) – FSC
= (1.1 – 0.126)
= 0.974m.
the breath of the load water plan of a ship 90m long measured at equal intervals from forward are as follow 0,3.96,8.53,11.58,12.19,12.5,11.58,5.18,3.44, and 0.3m respectively.if the load draft is 5m and the block coefficient is 0.6. find the freshwater allowance and position of the centre of floatation
Sir . There is some mistake in the solution of 20th question of FSC..
Thanks for the easy calculations..