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Author Topic: 8-9 more details on answers
 David Burch posted March 22, 2010 09:59 AM                   Here is more of a solution as i worked it.I am using calculator notation: ie dd.mmmm,-----------start at 48.141, - 123.319 and sail on 070T for 16-10.5 = 5.5 nmi, then turn to 130T for 20.4-16.0 = 4.4 nmi then turn to 070 for 23.5-20.4 = 3.1 nmi. Then i end up at 48.143, -123.148. Then plotting this point on the chart, i get range and bearing back to New Dungeness Lt = 6.45 nmi at 122T, this would be the answer to part A.Then we have had current of 1.1 kt toward 270T from 1330 to 1552 or 2h 22m. this is a total drift of 2.6 nmi toward 270. I then move the point from part A, this much to the west and then remeasure distance back to the light and get 8.77 @ 112TSo the answers are reversed in the text and also i get slightly different values. I will add these new ones to the errata.Again, we will each get slightly different answers when plotting. i computed the DR legs then plotted the points electronically. so the summary here is i gotA 6.45 @ 122 T(6.32 @ 120 T book)B 8.77 @ 112 T(8.68 @ 111 T book) From: Starpath, Seattle, WA

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