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Sample solution for Vertical Center of Gravity (VCG) problem M.V. Hudson

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.

Solution:

 M.V. Hudson stability letter and Loading Diagram click to view D036DG

1. Important information 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.

2. 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 (3.0 feet)."

To find VCG of Remaining Deck Cargo :

VCG = Vert. Moment / Total Weight

3.0 feet = (420 + 150X) ft-tons / 300 Tons

150X ft-tons = 480 Tons

X = 3.2 feet