Determine the free surface correction for a fuel oil tank 30 ft. long by 40 ft. wide by 15 ft. deep, with a free surface constant of 3794. The vessel is displacing 7,000 tons in saltwater.
A. 0.35 foot
B. 0.54 foot
C. 0.65 foot
D. 1.38 feet
A. 0.35 foot
B. 0.54 foot
C. 0.65 foot
D. 1.38 feet
This is a very simple, but very tricky questions.
ReplyDeleteFor those who have my book, STABILITY AND TRIM FOR THE SHIP'S OFFICER, 4th Edition, you can turn to pages 169 to 171. The title of this section is FREE SURFACE CONSTANTS. The symbol for Free Surface Constant is a small italics "i". Please note a lot of the license questions have been based on information in my book. not just in the USA but around the World.
In this case if you sort of studied but did not really understand free surface constants you might try to calculate it, however it has been given to you as 3,794 ft-tons (the units of which are not given in the question). So to get the answer all you need to do is divide the Free Surface Constant by the Displacement in Long tons and the correction will be in feet. This is how it is done manually in a vessel's Stability Booklet or Manual. The correction is than either added to KG or subtracted from GM which will produce GM Available, symbol for which is GMo.
FS Correction = GGo = FS Constant/Displacement = 3,794 ft-LT/ 7,000 LT = 0.542 feet
So the best answer is B.
Now lets see what is wrong with this question. Using the information given we should be able to calculate the Free Surface Constant with the following formula:
Free Surface Constant = i = rlb^3/12x35 = rlb^3/420
Assuming the density of the FO is about 0.9 and the vessel is in Sea Water density 1.025 the value of r = 0.9 / 1.025 = 0.8780.
Evaluating we find i = (0.8780)(30)(40)(40)(40)/420 = 4,013.71 ft-tons
Dividing this value by the Displacement we get 4023.72/7000 = 0.5748 ft which is not a choice given. So the body of the question does not "ring true" with the facts given.
The experienced mariner would have probable just divided the Free Surface Constant given by the Displacement and got the correct answer and moved on to the next question. On the other hand a lot of time can be wasted with this built into the body of the question and not the choices offered.
Do you agree or disagree?
Bravo is the quick and dirty answer.
ReplyDeleteI agree with the fact that without the "actual fuel specific gravity" the question gives you a lot of "extra information' aka distraction to consider.
working the formula backwards 3794/((30*40*40*40)/420)=0.8299375 for the Sp.G.....