Proceedings magazine is a communication tool for the Coast Guard's Marine Safety & Security Council. Each quarterly magazine focuses on a specific theme of interest to the marine industry.
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66 Proceedings Spring 2014 www.uscg.mil/proceedings A nswers Deck 1. A. Water temperature Incorrect answer. A standard thermometer is used to measure the jacket water temperature. B. Water pressure Incorrect answer. A standard pressure gauge is used for this measurement. C. Exhaust temperature Correct answer. A pyrometer is generally considered as a unit for measuring high tempera- tures that would be encountered in the exhaust system. D. Air box pressure Incorrect answer. Customarily a manometer is utilized to measure air box pressure. 2. A. Stop the fow of oil at the main deck manifold Incorrect answer. The regulation requires that the system stop the siphoning of liquid through the pump including within the vessel itself. Stopping the fow at the deck manifold would not stop the internal transfer on most piping confgurations. B. Prevent the oil from leaving the shore facility Incorrect answer. The question refers 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 fow of oil or hazardous material if the oil or hazardous material could siphon through the stopped pump." D. None of the above Incorrect answer. 3. A. 6324.2 miles, 310.3°T Incorrect answer. B. 6345.3 miles, 301.7°T Incorrect answer. C. 6398.0 miles, 298.3°T Incorrect answer. D. 6445.2 miles, 240.3°T Correct answer. The problem can be solved utilizing the following formulas: Cos Distance=(Cos Lat 1 × Cos Lat 2 × Cos Dlo) ± (Sin Lat 1 × Sin Lat 2 ) Cos Initial Course= (Sin Lat 2 – (Cos Distance × Sin Lat 1 )) ÷ (Sin Distance × Cos Lat 1 ) Cos Distance=(Cos Lat 1 × Cos Lat 2 × Cos Dlo) ± (Sin Lat 1 × Sin Lat 2 ) Dlo = (180° – 151°12.7 'E) + (180° – 122°27.8 'W) = 86.3250° (Cos Lat 1 × Cos Lat 2 × Cos Dlo) = .042060826 (Sin Lat 1 × Sin Lat 2 ) = .341441441 Cos Distance=(Cos Lat 1 × Cos Lat 2 × Cos Dlo) ± (Sin Lat 1 × Sin Lat 2 ) Subtract when crossing the equator Cos Distance = (.0420608260) – (.341441441) Cos Distance = (-0.29938) Distance = 107.4204 x 60° = 6445.2264 nm Cos Initial Course= (Sin Lat 2 – (Cos Distance × Sin Lat 1 )) ÷ (Sin Distance × Cos Lat 1 ) Lat 2 is negative when crossing the equator (Sin Lat 2 – (Cos Distance × Sin Lat 1 )) = (-.373731964) (Sin Distance × Cos Lat 1 ) = .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 answer. A vessel under oars may show the lights prescribed for a sailing vessel, and if she did, a stern light would be included. B. She is allowed to show the same lights as a sail- ing vessel. 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 exhib- ited in suffcient time to prevent collision." C. She must show a fxed all-round white light. Incorrect answer. This is not a requirement of a vessel under oars. D. She must show a day- shape of a black cone. Incorrect answer. There is no day shape requirement for a vessel under oars. Spring2014_FINAL.indd 66 3/21/14 11:14 AM