On 21 May at 0630 PDT, (ZD +7), your vessel takes departure at the San Francisco Sea Buoy, LAT 37°45.0'N, LONG 122°41.5'W, enroute to Kobe, LAT 33°52.0'N, LONG 135°00.0'E via great circle. The distance is 4,245 miles, and you estimate that you will average 14.0 knots. What will be your estimated zone time of arrival?

Question

On 21 May at 0630 PDT, (ZD +7), your vessel takes departure at the San Francisco Sea Buoy, LAT 37&deg;45.0'N, LONG 122&deg;41.5'W, enroute to Kobe, LAT 33&deg;52.0'N, LONG 135&deg;00.0'E via great circle. The distance is 4,245 miles, and you estimate that you will average 14.0 knots. What will be your estimated zone time of arrival?

Accepted Answer

Option D: 1342, 3 June. Total distance is 4,245 miles at 14 knots, requiring 303.2 hours (12.63 days). Departing 21 May at 0630, add 12 days and 15 hours. Adjusting for the crossing of the International Date Line (gaining a day) and the zone time difference, the arrival is 1342 on 3 June.