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Deck A nswers 1. A. Water temperature Incorrect answer. A standard thermometer is used to measure the jacket water temperature. B. Incorrect answer. A standard pressure gauge is used for this measurement. Water pressure C. Exhaust temperature D. Air box pressure 2. Correct answer. Reference: Kates and Luck; Diesel and High Compression Gas Engines. A pyrometer is generally considered as a unit for measuring high-temperatures that would be encountered in the exhaust system. Incorrect answer. Customarily, a manometer is utilized to measure air box pressure. Incorrect answer. The regulation requires that the system stop the siphoning of liquid through the pump including within the vessel itself. Stopping the Áow at the deck manifold would not stop the internal transfer on most piping conÀgurations. B. Prevent the oil from leaving the shore facility Incorrect answer. The question is inquiring as to the requirements of the emergency pump control on a tank vessel, not the shore facility. C. Prevent the oil from siphoning through the pump Correct answer. Reference: 33 CFR 155.780. "If an emergency pump control is used, it must stop the Áow of oil or hazardous material if the oil or hazardous material could siphon through the stopped pump." D. 3. A. Stop the Áow of oil at the main deck manifold None of the above Incorrect answer A. 6324.2 nm, 310.3°T Incorrect answer 6345.3 nm, 301.7°T Incorrect answer C. 6398.0 nm, 298.3°T Incorrect answer D. 6445.2 nm, 240.3°T Correct. Reference: Bowditch; The American Practical Navigator Reference: Plant; Formula for the Mariner. B. The problem can be solved utilizing the following formulas: Cos Distance = (Cos Lat1 × Cos Lat2 × Cos Dlo) ± (Sin Lat1 × Sin Lat2) Cos Initial Course = (Sin Lat2 – (Cos Distance × Sin Lat1)) ÷ (Sin Distance × Cos Lat1) Cos Distance = (Cos Lat1 × Cos Lat2 × Cos Dlo) ± (Sin Lat1 × Sin Lat2) Dlo = (180° – 151°12.7'E) + (180° – 122°27.8'W) = 86.3250° (Cos Lat1 × Cos Lat2 × Cos Dlo) = .042060826 (Sin Lat1 × Sin Lat2) = .341441441 Cos Distance = (Cos Lat1 × Cos Lat2 × Cos Dlo) ± (Sin Lat1 × Sin Lat2) Subtract when crossing the equator Cos Distance = (.0420608260) – (.341441441) Cos Distance = (-0.29938) Distance = 107.4204 × 60° = 6445.2264 nm Cos Initial Course = (Sin Lat2 – (Cos Distance × Sin Lat1)) ÷ (Sin Distance × Cos Lat1) Lat2 is negative when crossing the equator (Sin Lat2 – (Cos Distance × Sin Lat1)) = (-.373731964) (Sin Distance × Cos Lat1) = .753998648 Cos Initial Course = (-.373731964) ÷ (.753998648) Cos Initial Course = 0.4957 Initial Course = N 119.7137° W Initial Course = (360°-119.7137°) Initial Course = 240.2862° 4. A. She must show a stern light Incorrect. A vessel under oars may show the lights prescribed for a sailing vessel, and if she did, a stern light would be included. B. Correct answer. Reference: Inland and International Rule 25d(ii): "A vessel under oars may exhibit the lights prescribed in this rule for sailing vessels, but if she does not, she shall have ready at hand an electric torch or lighted lantern showing a white light which shall be exhibited in sufÀcient time to prevent collision." She is allowed to show the same lights as a sailing vessel C. She must show a Àxed wall-round white light Incorrect. This is not a requirement of a vessel under oars. D. She must show a day-shape of a black cone Incorrect. There is no day shape requirement for a vessel under oars. 70 Proceedings Winter 2012 | Spring 2013 www.uscg.mil/proceedings