# difficult .... list question ??

Discussion in 'Stability' started by abdelrhman shah, Dec 19, 2014.

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### abdelrhman shahJunior Member

aship of 7500 tonnes diplacment has KM=8.6 m , KG=7.8 ,and 20m beam .
A quantity of deck cargo is lost from starboard side (KG=12m ,and centre of gravity is 6m in form the rail ) . if the resulting list is( 3 degrees and 20 minutes) to port ...find how much deck cargo was lost.

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### abdelrhman shahJunior Member

aship of 7500 tonnes diplacment has KM=8.6 m , KG=7.8 ,and 20m beam .
A quantity of deck cargo is lost from starboard side (KG=12m ,and centre of gravity is 6m in form the rail ) . if the resulting list is( 3 degrees and 20 minutes) to port ...find how much deck cargo was lost.

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### TANSLSenior Member

The answer is not easy because, losing cargo, the ship displaces less and metacenter position is no longer the initial, ie KM varies.
Heel is 3 degrees, which means that the amount of charge lost is large.

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### abdelrhman shahJunior Member

???
????????????

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### abdelrhman shahJunior Member

ofcourse it is not easy ..... i got a headache from thinking ...

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### TANSLSenior Member

What I really mean is that with the information you give us is impossible to solve the problem, if you don´t do certain assumptions that are not correct.

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### philSweetSenior Member

Sometimes it is best to get a feel for the problem's magnitude. Perhaps you just assume 75 tons is lost, 1%, and see what would happen. Then you can better judge what additional info is needed or what assumptions may be appropriate to make.

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### abdelrhman shahJunior Member

I dont think so......because this question in( ship stability for master and mates) and already has a final answer ..... (91.9 tons)

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### TANSLSenior Member

As I said before, if you assume that the displacement does not change, the position of metacenter does not change, all of which is not correct, the problem is solvable. If you do not do certain assumptions, there is no solution.
But I do not understand why you pose a question you already know the solution.
By the way, ask your teacher if what I say is not true.
Cheers.

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### philSweetSenior Member

If you assume a 75 ton loss, then the BM increases by about 1% (you have to figure KB out from the info provided, I'm thinking 3 is a decent guess), the VCG decreases to about (780-12)/99 , and there is a 4 cm lateral offset to the CG. Extrapolating from the calculated list this generates, I get 98 tons. If you don't correct for either VCG or BM, you get 87 tons. If you correct for VCG and not BM, you get 92 tons. The calculation can be repeated using the extrapolated tonnage for a better result.

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### TANSLSenior Member

At some moment in the calculation process is necessary, I wonder, use the displacemente value. But that value, which initially is 7500 tons, no longer is know as the ship missed some unknown load.
Would be more accurate, knowing the curve of GZ values for a displacement of 7500 tons, calculate the GZ value for the angle of 3 degrees 20 minutes. This value gives the heeling moment which allows us to calculate the weight that is lost.
Assuming, of course, that the displacement of 7500 tons is after losing weight.

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### viking northVINLAND

I think this has been one of several enquirers from the middle east area, must be exam time

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### TANSLSenior Member

What I do not understand is why students do not raise doubts to their teachers. Maybe because they do not trust them?

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### viking northVINLAND

We assume it's students -- maybe it's the teacher--

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### TANSLSenior Member

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