## Rhumb Line Sailing

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 40o.5N, 43o.0W to position 42o.25N 41o.8W?

Solution:

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

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

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

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

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

Therefore course =  0.51-1 = 27o

Therefore, Course to steer = 027o

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 23o.25S, 120o.5W.  Her own D.R. position is 20o.4S, 118o.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 30o 21’N, 1554’W and travels on a course of 030o at 12 knots for 5 hours, what will be its new position?

Question 3. At 0930, a ship’s position is 48o 18’.75S, 29o  28’.30E.  Course is 148o  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 36o 23’.4N.  09o 15’.4W.  The Skipper estimates that the course and speed made good for the next 4 hours was 075o, 6 knots.  What was the estimated position at 1200?

1. Tin Zaw Cho says: