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Showing posts from April, 2018

Daily Question for April 28, 2018

A ship is inclined by moving a weight of 30 tons a distance of 30 ft. from the centerline. A 28-foot pendulum shows a deflection of 12 inches. Displacement including weight moved is 4,000 tons. KM is 27.64 feet. What is the KG? A. 21.34 feet     B. 22.06 feet     C. 22.76 feet     D. 23.21 feet

Daily Question for April 27, 2018

Your vessel's draft is 16'-00" fwd. and 18'-00" aft. The MT1 is 500 ft-tons. How many tons of water must be shifted from the after peak to the forepeak, a distance of 250 feet, to bring her to an even draft forward andaft? A. 52 tons    B. 50 tons    C. 48 tons    D. 24 tons

Daily Question for April 26 2018

You are reading the draft marks. The top 2 inches of the 9 forward is visible above the water level, and the water level is four inches below the 10 aft. What is the mean draft? A. 9'-10"    B. 9'-06"   C. 9'-04"   D.  9'-02"

Daily Question for April 25, 2018

A vessel aground may have negative GM since the __________. A. decrease in KM is equal to the loss of draft B. virtual rise of G is directly proportional to the remaining draft C. displacement lost acts at the point where the ship is aground D. lost buoyancy method is used to calculate KM, and KB is reduced

Daily Question for April 22, 2018

A cargo of 100 tons is to be loaded on deck 20 feet from the ship's centerline. The ship's displacement including the 100 tons of cargo will be 10,000 tons and the GM two feet. The list of the vessel after loading this cargo will be __________. A. 5.4°   B.  5.7°   C.  5.9°   D.  6.1°

Daily Question for April 20. 2018

Your ship of 12,000 tons displacement has a center of gravity of 21.5 feet above the keel. You run aground and estimate the weight aground is 2500 tons. The virtual rise in the center of gravity is __________. A. 1.26 feet   B. 3.80 feet   C. 4.80 feet   D. 5.66 feet

Daily Question for April 19, 2018

A vessel aground may have negative GM since the __________. A. decrease in KM is equal to the loss of draft B. virtual rise of G is directly proportional to the remaining draft C. lost buoyancy method is used to calculate KM, and KB is reduced D. displacement lost acts at the point where the ship is aground

Daily Question for April 18, 2018

You are on a supply run to an offshore drilling rig. On board is the cargo listed. What is the height above the main deck of the center of gravity of the cargo? I. Intermediate drill casing - 10 lengths each 16 inches in diameter. Each length weighs 1.7 long tons. The center of gravity above the main deck of the casing stow is 1.8 feet. II. Crated machine parts and assorted pipe fittings - 6 crates stowed two high. Each crate is 4'L X 3.5'W X 3'H. Each crate weighs 840 lbs. III. 10 each - 55 gallon drums of lube oil stowed on end. Each drum weighs 462 pounds, is 26 inches in diameter and 32 inches high. IV. Dry stores - 12 containers stowed two high. Each container weighs 0.9 long ton and measures 6'L X 4'W X 3'H. A. 1.20 feet   B. 1.64 feet   C. 2.26 feet   D. 3.00 feet

Daily Question for April 17, 2018

Your vessel's draft is 24'-06" forward and aft. The MT1 of your vessel is 1000 ft-tons. How many tons of cargo must be loaded in number 4 hold, which is 100 feet abaft the tipping center, if she is to have a 2 foot drag? A. 120 tons B. 240 tons C. 300 tons D. 480 tons

Daily Question for April 15, 2018

Reserve buoyancy is __________. A. also called GM B. the void portion of the ship below the waterline which is enclosed and watertight C. affected by the number of transverse watertight bulkheads D. the watertight portion of a vessel above the waterline

Daily question for April 14, 2018

A vessel's LCG is determined by __________. A. dividing the total longitudinal moment summations by displacement B. dividing the total vertical moment summations by displacement C. multiplying the MT1 by the longitudinal moments D. subtracting LCF from LCB

DAily Question for April 10, 2018

On a vessel of 15,000 tons displacement, compute the reduction in metacentric height due to free surface in a hold having free water in the tank tops. The hold is 50 feet long and 60 feet wide. The reduction in metacentric height is __________. A. 1.54 feet B.  1.59 feet C.  1.63 feet D. 1.71 feet

Daily Question for April 8, 2018

A tank 36 ft. by 36 ft. by 6 ft. is filled with water to a depth of 5 ft. If a bulkhead is placed in the center of the tank running fore-and-aft along the 36-foot axis, how will the value of the moment of inertia of the free surface be affected? A. The moment of inertia would remain unchanged. B. The moment of inertia would be 1/4 its original value. C. The moment of inertia would be 1/2 the original value. D. None of the above

Daily Question for April 5, 2018

Your vessel has a midships engine room and the cargo is concentrated in the end holds. The vessel is __________. A. sagging with tensile stress on main deck B. sagging with compressive stress on main deck C. hogging with tensile stress on main deck D. hogging with compressive stress on main deck

Daily Question for April 1, 2018

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