On 13 October your 0515 zone time (ZT) fix gives you a position of LAT 26°53.0'N, LONG 90°05.0'W. Your vessel is on course 068°T, and your speed is 7.8 knots. Local apparent noon (LAN) occurs at 1145 zone time, at which time a meridian altitude of the Sun's lower limb is observed. The observed altitude (Ho) for this sight is 54°51.5'. What is the calculated latitude at LAN?
• Meridian Passage (LAN) Latitude: Relationship between calculated latitude, Sun’s declination, and meridian altitude (Ho). • Run between DR positions: Advancing your 0515 DR position to the time of LAN using course and speed. • Declination from Nautical Almanac: Using date (13 October) to determine the Sun’s declination and applying it correctly for meridian altitude problems.
• How do you advance your 0515 DR position to 1145 using the given course and speed? What is your new DR latitude at LAN time? • For a lower-limb meridian altitude of the Sun, what is the basic formula that connects latitude, declination, and meridian altitude? How does it differ if you are in the same hemisphere as the Sun vs. opposite hemisphere? • On 13 October, is the Sun’s declination north or south? Based on that, should you add or subtract declination in your latitude computation?
• Confirm the time interval between 0515 ZT and 1145 ZT and correctly convert it to hours and tenths of an hour for the run calculation. • Verify the distance run using ( \text{Distance} = \text{Speed} \times \text{Time} ) and resolve it into a change of latitude along the given course (068°T). • Double-check that you are using the correct meridian altitude formula for the Sun’s lower limb and the correct sign (+/–) for the Sun’s declination on 13 October before choosing an answer.
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