Exercise 3 Answers

 Solution Q.1.

Dlat  = 23^o.25S – 20^o.4S= 2^o.85S = 171’S

Mean Lat    = 20^o.4S + 1^o.425  = 21^o.825S

Dlong          = 120^o.5W – 118^o.3W  = 2^o.2W = 132’W

Dep   = d.longxcos(mean lat)  = 132 cos(21^o.825)  = 122’.5E

Tan(course) =   Dep / D.Lat = 22’.5 /171’ = 0.716^o

0.716^-1 = 35^o.6  = 36^o  Therefore, course to steer = S36^oW  = 216^o

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

Return to Exercise 3

Solution Q.2.

Dist.      = 60 n.m.

Dep.         = 60.Sin(3^o)  = 30’ E.

Dlat          = 60.Cos(3^o) = 51’.96 N.

New Lat   = 31^o 12’.96 N.

M.lat        = 3^o 46’.98 N.  = 3^o.78N.

DLong.    = 30.Sec(3^o.78) = 34’.92E

New Long = 15^o 54’ – 34’.92 = 15^o 19’.08W.

New Pos. = 31^o 12’.96N.  15^o 19’.08W.  

Return to Exercise 3

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   = 49^o 17’.4 S.

M.lat        = 48^o 48’.08 S.

DLong.    = 36.65.Sec(48.8) = 55’.6E

New Long = 30^o 23’.9E

New Pos. = 49^o 17’.4 S. 30^o 23’.9E     

Return to Exercise 3

Solution Q.4

Dist.      = 24 n.m.

Dep.         = 24.Sin(75)  = 23’.18 E.

Dlat          = 24.Cos(75) = 6’.21 N.

New Lat   = 36^o 29’.61 N.

M.lat        = 36^o 26’.5 N.

DLong.    = 23.18.Sec(36.44) = 28’.8E

New Long = 08^o 46’.6W.

New Pos. =  36^o 29’.61 N.  08^o 46’.6W.

 Return to Exercise 3

Return to Rhumb line sailing

A fuller explanation of this topic is given in the book ‘Astro Navigation Demystified’.

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