Author

Topic: 89 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 1610.5 = 5.5 nmi, then turn to 130T for 20.416.0 = 4.4 nmi then turn to 070 for 23.520.4 = 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 @ 112T
So 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 got
A 6.45 @ 122 T (6.32 @ 120 T book)
B 8.77 @ 112 T (8.68 @ 111 T book)
From: Starpath, Seattle, WA


