L D
Ship in
Q 10 10000 – 8.954 89540 – 2027. 4
NO1 DB – 200 1.15 – 230 (-) 399
and so on .
Note – 5: When dealing with several tank’s in a ship calculation such as this the change of KG of tank due to change in its sounding may be ignored. This is the actual practice at sea .
FOR EXAMPLE: When half of the HFO (i.e. 100t from NO3 C is pumped out , the moment of the discharge weight about keel would 100X 0.6 = 60 tm .
Solution:
Ship’s wt |
KG |
VM |
FSM |
10000 | 8.954 | 89540tm | 2087.4 |
(- ) 200 (dish) | 1.15 | (-) 230tm | (- )399.0 |
(- ) 100 (dish) | 0.60 | (-)60 tm | ( +)1140.0 |
(- ) 200(dish) | 0.70 | (- )140 tm | (-) 600.0 |
Final W = 9500 Final VM=89110tm Final FSM= 2228.4tn
We know that:
Final KG = (Final VM/ Final W)
Final KG = (89110/9500)
Final KG = 9.38m
Again, FSC = (FSM /W)
= (2228.4 /9500)
= 0.234m
GM Solid = (KM –KG)
= (9.9 – 9.38)
= 0.52m
GM fluid = GM (solid) – FSC
= (0.52 – 0.234)
= 0.286m.
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..