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?
• Transverse GM (metacentric height) and how it relates to list angle when a weight is shifted off centerline • Using GM = (w × d) / (Δ × tan θ) when a weight is moved transversely on or off the ship • Effect of lifting a weight from the pier to the ship: change in KG and how that affects final GM when the weight is stowed at a different height
• First, treat the initial lift: what is the horizontal shift of the ship’s center of gravity when the 50-ton weight 75 ft off center is taken by the ship? Use that to relate the observed 3.5° list to the initial GM during the lift. • Next, adjust the vertical center of gravity for moving the 50-ton weight from the pier (not in Δ) to its stowed position at 56 ft above the base line. How does this change KG and therefore GM compared to the value during the lift? • Check that you are using the correct displacement (18,000 T) and weight units consistently, and that you convert the list angle to radians or use tan θ correctly in your GM formula.
• Be sure you are using tan(3.5°), not 3.5, in the GM formula. • Confirm whether the 50-ton weight is initially part of the ship’s displacement when it is still on the pier (this affects how you set up Δ in each step). • After finding the GM during the lift, carefully compute the change in KG when the weight goes from being suspended at 112 ft to being stowed at 56 ft, and then recompute the final GM from KM − KG.
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