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download to your computer. Click on subject below for details.
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Sample solution for Vertical Center of
Gravity (VCG) problem
You have 600 tons of below deck tonnage.
There is no liquid mud aboard. If you have 150 tons of cargo above deck
with a VCG above the deck of 2.8 feet, what is the maximum allowed VCG
of the remainder of the deck cargo that is permitted? (See illustration
D036DG, stability letter for M.V. Hudson)
A. 1.96 feet Incorrect.
B. 2.25 feet Incorrect.
C. 3.20 feet Correct.
D. 3.55 feet Incorrect.
Hudson stability letter and Loading Diagram
click to view D036DG
found in the stem of the question:
a. No liquid mud onboard
b. 600 tons of below deck tonnage.
Taking these two variables you would
consult the Loading Diagram, found on the third page in Diagram D036DG,
to find the MAXIMUM amount of cargo allowed above deck.
On the horizontal grid find 600 tons of
below deck tonnage and run vertically until you intersect with the line
labeled "WITHOUT LIQUID MUD". From this point run horizontally to find
the "ABOVE DECK CARGO", which is
300 Long Tons.
Looking at the first page of
Diagram D036DG in paragraph #3 you will find that "the height above the
main deck of the center of gravity of the deck cargo shall not exceed
the value shown on the LOADING DIAGRAM
To find VCG of Remaining Deck Cargo :
VCG = Vert. Moment / Total Weight
3.0 feet = (420 + 150X) ft-tons / 300
150X ft-tons = 480 Tons
X = 3.2 feet
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