Solution Q.1.
Dlat = 23o.25S – 20o.4S= 2o.85S = 171’S
Mean Lat = 20o.4S + 1o.425 = 21o.825S
Dlong = 120o.5W – 118o.3W = 2o.2W = 132’W
Dep = d.longxcos(mean lat) = 132 cos(21o.825) = 122’.5E
Tan(course) = Dep / D.Lat = 22’.5 /171’ = 0.716o
0.716-1 = 35o.6 = 36o Therefore, course to steer = S36oW = 216o
Dist = Dep / Sin (course) = 122.5 / Sin(216) = 208’.41
Therefore, distance to new position = 208.41 n.m.
Time to reach SOS position = Dist / Speed
= 208.41 ÷ 25 = 8.3 hrs. = 8 hrs. 18 mins.
ETA = 1230 + 0818 = 2048 GMT
Solution Q.2.
Dist. = 60 n.m.
Dep. = 60.Sin(3o) = 30’ E.
Dlat = 60.Cos(3o) = 51’.96 N.
New Lat = 31o 12’.96 N.
M.lat = 3o 46’.98 N. = 3o.78N.
DLong. = 30.Sec(3o.78) = 34’.92E
New Long = 15o 54’ – 34’.92 = 15o 19’.08W.
New Pos. = 31o 12’.96N. 15o 19’.08W.
Solution Q.3.
Dist. = 2.47 hrs @ 28 knots= 69.16nm
Dep. = 69.16.Sin(148) = 36’.65 E.
Dlat = 69.16.Cos(148) = 58’.65 S.
New Lat = 49o 17’.4 S.
M.lat = 48o 48’.08 S.
DLong. = 36.65.Sec(48.8) = 55’.6E
New Long = 30o 23’.9E
New Pos. = 49o 17’.4 S. 30o 23’.9E
Solution Q.4
Dist. = 24 n.m.
Dep. = 24.Sin(75) = 23’.18 E.
Dlat = 24.Cos(75) = 6’.21 N.
New Lat = 36o 29’.61 N.
M.lat = 36o 26’.5 N.
DLong. = 23.18.Sec(36.44) = 28’.8E
New Long = 08o 46’.6W.
New Pos. = 36o 29’.61 N. 08o 46’.6W.
A fuller explanation of this topic is given in the book ‘Astro Navigation Demystified’.
Home page: www.astronavigationdemystified.com
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