# Great Circles

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Great CirclesBy A.W.Strannigan

Great CirclesThe shortest distance between 2 points is a Great Circle Route. A Great Circle route will require a vessel to continuously alter course and it will take vessels into higher latitudes.A/C A/C A/C

A/C

Great CircleDLong

Pole Function of Final Course

V Initial Course B

Destination Point V = Vertex Maximum Latitude Ships Course 090(T) or 270 (T) A Departure Point

Great Circle Course & Distance CalculationsA Great Circle problem is resolved by spherical trig calculations For a Vessel sailing from position A to position B Distance => Cosine Distance =DLong AB x Cos Lat A x Cos Lat B (+ or ) Sine Lat A x Sine Lat B If Latitude A & B are the same name + If Latitude A & B different names -

Great Circle CalculationsCourses => Courses are calculated using the Sine formulaSine a/Sine A = Sine b/Sine B = Sine c/Sine C Courses can also be calculated using ABC

Example 1A vessel is to Sail from Lat 38 42N Long 25 35 E to Lat 40 43N Long 74 00W

Calculate Great Circle distance, Initial Course & Final CoursePole a B Destination b c C

A Departure Point

Example 1DLong = 25 35E 74 00W 99 35W0 (Prime Meridian)

Westerly Course

Example 1 DistanceCos Dist = Cos DLong. Cos Lat A. Cos Lat B (+ or -) Sine Lat A. Sine Lat B => Cos 99 35. Cos 38 42. Cos 40 43. + Sine 3842. Sine 40 43 => -0.09848 + 0.40786 => 0.30938 => 4318.7

GC Sailing :- Finding Initial & Final Course ABC MethodThe A180

142 18 E 37 42 => 109 35E

180 108 07W 71 53

4106S 142 18E A

p B b P a 4643 S 10807W

Example 2 GC DistanceCos Dist = Cos DLong. Cos Lat A. Cos Lat B (+ or -) Sine Lat A. Sine Lat B = Cos 41 06. Cos 4643. Cos 109 35 + Sine 41 06 Sine 4643 = -0.17317 + 0.47855 = 0.30538 = 72.22 Distance = 4333.1 4106S 142 18E A

p B b P a 4643 S 10807W

Example 2 Initial CourseABC A= Tan Lat A/Tan DLong = Tan 41 06/Tan 109 35 = 0.31035 S B = Tan Lat B/Sine DLong = Tan 46 43/ Sine 109 35 = 1.1270 S C = 1.43735 S Tan Az = 1/C x Cos Lat = 1/ 1.43735 x Cos 41 06 = S 42.7 E Initial Course = 137.3 (T)

Example 2 Final CourseABC A= Tan Lat B/Tan DLong = Tan 46 43/Tan 109 35 = 0.37774 S B = Tan Lat A/Sine DLong = Tan 41 06/ Sine 109 35 = 0.92591 S C = 1.30365 S Tan Az = 1/C x Cos Lat = 1/ 1.30365 x Cos 46 43 = S 48.2 W (Reverse) = N 48.2 E Final Course = 048.2 (T)

Example 2 Steaming TimeGC Distance = 4333.1 Vessels Speed = 14kts Time = distance / speed = 4333.1/14 Steaming Time = 309.507 Hours = 12days 21 hours 30 minutes

Example 3Posn A Lat 18 30S 028 00W Posn B Lat 32 30N 048 00E Calculate the Great Circle Distance from A to B The Initial Course The Final Course

Example 3 Answer DistanceDLong => 28 00W => 48 00E =>76 00EP B Lat 32 30N 048 00E

A Lat 18 30S 028 00W

Cos Dist = Cos DLong. Cos Lat A. Cos Lat B (+ or -) Sine Lat A. Sine Lat B = Cos 76. Cos 18 30. Cos 3230N - Sine 18 30 Sine 3230= 5321

Example 3 Initial CourseA = Tan Lat A/ Tan DLong = 0.08342 N B = Tan Lat B/ Sine DLong = 0.65657 N C = 0.73999 => 1/ 0.73999 x Cos 18 30 = N 55E => 055(T)

Example 3 Final CourseA = Tan Lat B/ Tan DLong = 0.15884 S B = Tan Lat A/ Sine DLong = 0.34484 S C = 0.50368 => 1/ 0.50368 x Cos 32 30 = S 67W => Reverse = N 67 E => 067(T)

Example 4Posn A Lat 5 30N 165 30W Posn B Lat 52 30S 135 00E Calculate The Mercator Distance from A to B The Great Circle Distance from A to B The Initial Course The Final Course

Example 4 Mercator DistanceTan Co = DLong/DMP Cos Co = DLat/Distance Distance = DLat/ Cos Co DLong => 165 30W => 135 00E =>59 30 00W DLat DMP Tan Co = 59 30 00/4023.9 = 3570/4023.9 = 0.88719 = 41.6 Distance = 58/ Cos 41.6 = 3480/0.74804 = 4652

=> 5 30N = 328.27 52 30S = 3695.6 58 00 = 4023.9

Example 4 Answer GC DistanceDLong => 165 30W => 135 00E =>59 30 00WB 52 30S 135 00E P Lat 5 30N 165 30W A

Cos Dist = Cos DLong. Cos Lat A. Cos Lat B (+ or -) Sine Lat A. Sine Lat B = Cos 5 30. Cos 52 30. Cos 5930 - Sine 5 30 Sine 5230= 4597

Example 4 Initial CourseA = Tan Lat A/ Tan DLong = 0.0567 S B = Tan Lat B/ Sine DLong = 1.152 S C = 1.5687 => 1/ 1.5687 x Cos 5 30 = S 32.6W => 212.6(T)

Example 4 Final CourseA = Tan Lat B/ Tan DLong = 0.76766 N B = Tan Lat A/ Sine DLong = 0.11175 N C = 0.87941 => 1/ 0.87941 x Cos 52 30 = N 61.8E => Reverse = S 61.8 W => 241.8(T)

Composite Great CirclesGreat Circle routes may take the ship into higher Latitudes There are number of reasons why a vessel may not want to reach the higher latitudes to be found on trans Ocean great circle tracks.Loadline regulations Dangerous ice Bad weather Limits of crew agreement Navigation aids become unreliable Insurance purposes Cargo Considerations Daylight restrictions for deck maintenance Charter impose restrictions Land or Islands on the route

Composite Great CirclesTherefore a composite great circle may be required The composition being that of Great Circle routes and P Parralel Sailing RouteP

V GC2 B

V GC1 A

Composite Great CircleA Composite Great Circle is resolved by Napier Rules Example A ship is to Sail from Durban 30 00S 31 00E to Melbourne 39 00S 144 00E. Charter party instructions require the ship not to pass the 43 Parallel of Latitude. Ship speed 12kts. Calculate:Longitude when the ship arrives at the limiting latitude Initial Course Total Distance Longitude when the ship leaves the limiting latitude Steaming time

Napiers RulesNapiers rules can be applied to any spherical triangle where there is either a 90 side or a 90 angle. THE RULE STATES THE FOLLOWINGSin of the middle Part = Product of the Cosines of the opposite parts Sin of the Middle Part = Product of the tangents of the adjacent Parts (The parts are simply the other sides and angles within the triangle) To use Napiers rules we need to know at least two parts of the triangle, other than the 90 angle.

Composite GCDraw GC RouteA 30 00S 31 00E

43

39 00S 144 00E

Right Angled Spherical TriangleThis triangle has amounts to 6 Parts A 3 angles