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Vessel in same condition as in question 10, loads 1000t cargo in NO2 LH KG 4m ; 2000t cargo NO4 LH KG 5m . Find the final GM fluid, given that the final KM is 10.0m .
Solution:
Ship’s wt |
KG |
VM |
10000t | 8.954m | 89540 tm |
(+ ) 1000 | 4m | (+) 4000 tm |
(+) 2000 | 5m | (+)10000 tm |
Final W = 13000 Final VM = 103540tm
We know that:
Final KG = (Final VM/ Final W)
Final KG = (103540 / 13000 )
= 7.964m
KM = 10m ( given)
Again, GM solid = ( KM – KG)
= (10 – 7.964) m
= 2.036m
Now, FSC =( FSM/ W)
= (2087.4 /13000)
= 0.1605m
GM fluid = GM (solid) – FSC
= (2.036 – 0.1605)
= 1.875m.
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..