MODULE 1 : GREAT CIRCLE SAILING 1. Definition For navigation purposes, the definition of a great circle is a circle on the surface of the Earth, the plane of which passes through the Earth's centre. The great circle track is the shortest distance between any two places on the surface of the Earth.

latitude) than either of the two places, origin or destination. High latitudes are often in peril because of bad weather and icing. A safe idea of a veteran sailor is to set a latitude limit for the long voyage plan. This policy is called composite great circle route planning, done with waypoints. To plot a great circle, a gnomonic projection is used. This converts a curved line on the earth's surface to a straight one on the chart. If one were to plot a great To sail a great circle track, the navigator needs to adjust the ship's course continuously because the intersections of the great circle track with the meridians changes constantly (different angles). The great circle track is a curve when plotted on a Mercator Chart. It is unrealistic to attempt to sail on an exact great circle route. Navigators usually divide a great circle track between the initial position and the destination into many much 5

smaller segments (way points) of about one to two day's sailing time (obviously depending on the craft and conditions) and make course adjustments each day at the same time, usually at high noon. The total distance is therefore the sum of the distances of those segments. A potential problem of the great circle track, though is the shortest route between two locations, it also usually tracks closer to the pole (or to higher circle track on a mercator projection, it would consist of a curved line and would be very tiresome to plot.

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General remark: When the position of departure and position of arrival have different latitudes, the sign of the largest value of latitude indicates the elevated pole of the spherical triangle to be used. 2. Distance In the figure below, the distance is the arc AB (d):

In astro navigation we had the formulae to obtain the zenithal distance. When we replace the elements by the ones for great circle sailing, we obtain: cos Mo = cos cola x cos colb + sincola x sincolb x cos ∆g = sin la x sin lb + cos la x cos lb x cos ∆g Mo is expressed in degrees and minutes. They have to be converted to minutes in order to get the distance in miles (1 minute arc = 1 nm). We consider the rhumbline distance (M) and compare with Mo resulting in a gain of distance (always!) ∆M. Taking into consideration possible 7

high latitudes to cross and prevailing weather conditions, the option to sail rhumbline instead of great circle sailing depends on the results. A solution in between is composite sail (see later). 3. Course at start and end of voyage We rather prevent taking calculated elements in new formulae, so we have to find a formulae not taking into account the calculated Mo. The Neper method is a perfect tool for this:

tg (A+B)/2 = cos (NB – NA)/2 / cos (NB + NA)/2 x cotg N/2 tg (A – B) /2 = sin (NB – NA)/2 / sin (NB + NA)/2 x cotg N/2 NB – NA = colb – cola = 90° - lb – (90° - la) = la – lb = ∆l (NB-NA)/2 = ∆l/2 NB + NA = colb + cola = 90° - lb + 90° - la = 180° - (la + lb) (NB + NA)/2 = 90° - (la + lb)/2 = 90° - lm The angle N = ∆g

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In such the formulae become: tg (A+B)/2 = cos ∆l/2 x cosec lm x cotg ∆g/2 tg (A-B)/2 = sin ∆l/2 x sec lm x cotg ∆g/2 The angles A and B are obtained by adding and subtracting the results. Angles A and B will lead to the courses Ro (at departure) and Ro’ (at arrival). Courses are always omni-directionally expressed. A second possibility of calculating the angles A and B is by using the cotangens rule giving access in using the ABC tables of the Norie’s Nautical Tables. Another method is the use of the HO 249 tables ,Volumes 2 and 3,( Sight Reduction Tables). Pub n°249 and the use of its interpolation tables are readily adaptable to solutions of great circle sailing problems. By entering the tables with latitude of departure as “latitude”, latitude of destination as “declination” and difference of longitude as “LHA”, the tabular altitude and azimuth angle may be extracted and converted to distance and course. To find the course at destination, we reverse and start with latitude of destination as “latitude” and latitude of departure as “declination”. 4. Vertex and equatorial intersections The vertex is the highest latitude of a great circle defined by the position of departure and the one of arrival in navigation. In such, the great circle has two vertex, both having the same value but different sign (N/S) and are symmetrical (difference of longitude between the vertexes is 180°). The great circle intersects the equator in two locations called equatorial intersections. Both points have a latitude of 0° and are also symmetrical (diff. Longitude is 180°). The difference in longitude between a vertex and an equatorial intersection is 90°.

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V1

V1 K2

K2 K1

K1 V2

V2 90°

90°

90°

So, when the coordinates of one vertex are known, the coordinates of the other vertex and both equatorial intersections are known. The vertex being the highest latitude of the great circle, the course is 90° or 270° which means we can use the rectangular spherical triangle.

Pn

V

A

PnA = cola , PnV = colv, VPnA = ∆g between A and V 10

What are we looking for? • latitude and longitude of V • distance AV We know the angle V = 90°, so: a. latitude vertex (lv) sin colv sin cola ---------- = ---------sin  sin 90° cos lv -------sin Â

cos la = --------1

cos lv = sin  x cos la As we used calculated elements (A and B), there will be no problem when we can verify the output (see also c.). We can do so by solving the other rectangular spherical triangle PnVB: cos lv = sin B x cos lb Both solutions of lv must be the same. If so, we are also sure that A and B are correct. b. longitude vertex (gv) cos ∆ gav = cotg (90° - colv) x cotg cola = cotg lv x tg la

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c. distances between vertex and position of departure (Moav) / position of arrival (Movb) cos cola = cos colv x cos Moav sin la = sin lv x cos Moav cos Moav = sin la x cosec lv cos Mobv = sin lb x cosec lv As we used a calculated element (lv), there will be no problem when we can verify the output by: Moav + Mobv = Mo To find the equatorial intersection coordinates gk1 and gk2 (latitude is 0°), we simply add/subtract 90° to the longitude of the vertex. The course in the vertex is 90° or 270°. The course in the equatorial intersection (K) is equal to the latitude of the vertex. V

lv

V’

K 90°

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5. Waypoints The waypoints can be chosen by following means : • • • •

same latitude interval (e.g. every 10°) same longitude interval same distance interval same course interval (e.g. every 5°)

The most common used is same longitude interval, with an interval of 5° to 10°.

Vertex

10°

10°

10°

10°

10°

10°

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When inserting waypoints, we always start for calculations from the vertex. In such we keep rectangular spherical triangles using simplified formulae. Pn

Vertex WP3 WP2 A

WP1

To determine the latitude, longitude and distances, we will use each time a new rectangular spherical triangle. E.g., for WP3 we use the triangle Pn-V-WP3, for WP2 the triangle Pn-V-WP2, … The waypoints on the other side of the vertex (to position of arrival when vertex lies in between position of departure and arrival) are symmetrical to the ones between A and vertex. They have symmetrically the same latitude and ∆g. For each triangle we obtain the distance (Mo) to the vertex: Mowp3v, Mowp2v, …. To obtain the distances between the waypoints: Mowp1v – Mowp2v = Mowp1wp2,… The sum of all those partial distances and Moawp1 must be equal to Moav (see distance position departure to vertex). What do we have to calculate? • latitude of waypoints • distance waypoint to vertex • course at each waypoint 14

Considering triangle PnVWP3, we see: Angle V = 90°, colv and ∆g (e.g. 10°) is known. Cos ∆g = cotg (90° -colv) x cotg colwp3 = cotg lv x tg lwp3 (latitude WP3) : tg l wp3 = cos ∆g x tg lv cos colwp3 = cos colv x cos Mowp3v sin lwp3 = sin lv x cos Mowp3v (distance WP3V) : cos Mowp3v = sin lwp3 x cosec lv sin Awp3 sin V (90°) ------------ = ---------------sin colv sin colwp3 (course at waypoint): sin Awp3 = cos lv x sec lwp3

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For calculations, a table can be used: Item ∆g cos ∆g tg lv tg lwp lwp sec lwp cos lv sin Awp Awp sin lwp cosec lv cos Mowpv Mowpv

WP3 10° 0,98481

WP2 20° 0,93969

WP1 30° 0,86602

6. Composite sail The method of composite sailing involves placing two vertexes on the limiting latitude that will give two great-circle paths, one from the point of departure, the other from the great-circle destination. In this case, the composite course is in three sections, the starting great-circle path, the East or West path along the limiting latitude, and a second (different) great-circle path to the destination point. The limiting latitude is chosen by the navigator taking into account: - the latitude of the original great circle vertex - weather conditions in higher latitudes - obstacles such as land masses The total distance of composite sail is less than rhumbline sailing but more than the great circle sailing. In such it is a compromise in respect with the prevailing situation in higher latitudes.

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Pn

Max. latitude V

V’ B

A

Composite sailing: Original great circle: V is the vertex of the first great circle defined by A and V, V’ is the second vertex defined by the second great circle (B and V’). The formulae to define the course of departure and arrival in previous chapter can be used. To define the vertex V: Consider the rectangular spherical triangle APnV: Cos lv = cos lA x sin  Cos ∆g = cotg lv x tg lA Cos MoAV = sin la x cosec lv 17

In rectangular spherical triangle BPnV’, replace A by B and lA by lB. To obtain the rhumbline distance between V and V’ (parallel sailing): M = ∆e = ∆g x cos lmax ∆g = gv – gv’ 7. Traverse distance to an obstacle It is important to know the shortest distance to an obstacle (island, coral reef, wreck,…) and also to know the time you reach this point. In such we need the great circle distance to the obstacle and the great circle distance till the traverse position. By means of spherical triangulation, both requests can be calculated. Method: a. calculate the orthodromic start course of the original track b. calculate the orthodromic course of the track from departure to the obstacle c. make the difference between both angles a and b d. calculate the distance from departure to the obstacle e. determine the traverse distance with the elements: right angle, c and d f. calculate the distance from departure to the position of traverse distance by using the cosinus formulae using the two other sides of the rectangular triangle

obstacle

departure

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8. Internet sites http://www.csgnetwork.com/marinegrcircalc.html Designation

Value

Meridional Parts

Required Data Entry Departure Position Co-Ordinates Latitude

Degrees

Minutes

North

Longitude

Degrees

Minutes

West

Destination Position Co-Ordinates Latitude

Degrees

Minutes

North

Longitude

Degrees

Minutes

West

Calculated Results Initial Course

Degrees True

Rounded

Distance

Nautical Miles

Rounded

Distance

Statute Miles

Rounded

Distance

Kilometers

Rounded

Required Waypoint Entry Way Point Longitude

Degrees

Minutes

West

Calculated Waypoint Results

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Way Point Latitude

??

Degrees

??

Minutes

??

N/S

Version 1.4.5 http://www.islandwaterworld.com/greatcircle.htm Great Circle Sailing Below is a form using JavaScript that solves the Great Circle Sailing problem : Origin (Initial Position) Latitude :

degrees

minutes

North

Longitude :

degrees

minutes

East

Destination (Final Position) Latitude :

degrees

minutes

North

Longitude :

degrees

minutes

East

http://www.wcrl.ars.usda.gov/cec/java/lat-long.htm http://www.csgnetwork.com/marinegrcircalc.html

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