USCG 2260-SA-4
STEPS
SOLUTION:
Displacement =
18000 tons + 50.0 tons weight
= 18050 tons
50 tons Suspended weight
75 feet Off centerline
112 feet Boom height
56 feet Stowed above baseline
3.5º Angle of Inclination
GM =
Weight x Distance
Displacement x Tan Ø
Where: Weight = Weight moved measured in long tons
Distance = Distance the weight is moved from the original CG in fee
Tan Ø = Tangent of the angle of inclination or deflection
Displacement = Vessel's displacement measured long ton
GM =
50 tons x 75 feet
=
3750.0 ft-tons
18050 tons x Tan (3.5º)
1103.98 tons
GM =
3.40'
Red = table values, etc.
Green = ANSWER
Blue = information provided.
Black = calculated values.
LEGEND:
You are making a heavy lift with the jumbo boom. Your vessel displaces 18,000 T. The 50-ton
weight is on the pier, and its center is 75 feet to starboard of the centerline. The head of the
boom is 112 feet above the base line, and the center of gravity of the lift when stowed on deck
will be 56 feet above the base line. As the jumbo boom takes the strain, the ship lists 3.5°.
What is the GM when the cargo is stowed?
56 feet
112 feet
75 feet
CL
50 tons
Baseline
Principles of Stability
Stability and Trim Calculations - Correction for List and Off-Center Weight
2260-sa-4
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