Terminology.  The following terms are used in this exercise:  Rhumb Line, Departure, D.Long, D.Lat, Mean Lat,      Mid. Lat.To find the meaning of these terms, click here:   

Short distance sailing is a term which is applied to sailing along a rhumb-line for distances less than 600 nautical miles.  The following formulae are used extensively in short distance sailing:

To Calculate Departure when the course is not known:  dep.= d.long cos(mean lat)

To Calculate Departure when the course is Known:  Dep  = Dist x Sin(course)

To Calculate Distance when departure and course are known:

Dist  =   Dep  / Sin (course)

To Calculate Dlat when the distance and course are known: DLat = Dist x Cos(course)

To Calculate Course to Steer (the rhumb line course between two points):

Tan(course) =   Dep / D.Lat

To calculate Dlong (difference in longitude corresponding to the departure):

DLong.  =  Dep. x Sec(Mean.Lat)   or   Dlong =      Dep  / Cos(Mean.Lat)

 Example.  What is the rhumb line course to steer and the distance to travel from position 40^o.5N, 43^o.0W to position 42^o.25N 41^o.8W? 

Solution:

Dlat      = 42^o.25N – 40^o.5N  = 1^o.75N = 105’

Mean Lat = 40^o 30’N + (105′ / 2) = 40^o 30’N +  52’.5  = 41^o 22’.5N = 41^o.38

Dlong        = 43^o.0W – 41^o.8W  = 1^o.2E = 72’E

Dep           = d.long x cos(mean lat)  = 72 cos(41^o.38)  = 54’.02

Tan(course)  = Dep / D.Lat   =    54.02 / 105  = 0.51

Therefore course =  0.51^-1 = 27^o

Therefore, Course to steer = 027^o 

Dist         =     Dep / Sin (course)  =  54.02 / Sin(27)  = 120’

Therefore, Distance to new position = 120 n.m.

Exercise Questions

 Question 1.  At 1230 GMT, a ship receives an S.O.S. signal from a vessel in position 23^o.25S, 120^o.5W.  Her own D.R. position is 20^o.4S, 118^o.3W. What is the rhumb line course to steer to reach the S.O.S. position and what will be the E.T.A. if the speed is 25 knots?

Question 2.  If a ship starts from position 30^o 21’N, 15^o 54’W and travels on a course of 030^o at 12 knots for 5 hours, what will be its new position?

 Question 3. At 0930, a ship’s position is 48^o 18’.75S, 29^o  28’.30E.  Course is 148^o  and speed is 28 knots.  What will be the ship’s position at 1158?

Question 4.  A yacht’s D.R. position at 0800 is 36^o 23’.4N.  09^o 15’.4W.  The Skipper estimates that the course and speed made good for the next 4 hours was 075^o, 6 knots.  What was the estimated position at 1200?

  ‘Click here for exercise answers’

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

Home page:  www.astronavigationdemystified.com