**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

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