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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 242m4 , 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.
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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.
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On a ship of W 6000 t, KM 7.4m, KG 6.6m, a double bottom tank of i 1200m4 is partly full of FW . Find the GM fluid .
Solution:
W = 6000 t & KM = 7.4m,
KG = 6.6m, i = 1200m4, 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
In question #2 doesn’t it need to convert into 0.050??
Where is the solution for que 18&19 ?
yes, question number 20, solid gm is 0.9, subtract fsc and get fluid gm and then apply RM=WGMSinx
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
Very helpful to solve questions
Sir . There is some mistake in the solution of 20th question of FSC..
Thanks for the easy calculations..