# Composite Great Circle Sailing

## Composite Great Circle Sailing

On 5 March 2005 at 1200 hours local time, a vessel is expected to depart from port Elizabeth (South Africa) to arrive at Melbourne (Australia) with a General average speed of 12.5 knots. The Master intends to follow a composite great circle track with a limiting latitude of 42° S from:

Departure position: 34° 05′ S 026° 00′ E.
Landfall position: 39° 00′ S 143° 50′ East.
The Pilotage distance from port Elizabeth to the departure position is 45 miles and from the landfall position to berth Melbourne is 84 miles. Calculate following:

The total distance of the voyage.
ETA local time at landfall position?

Dlong = 143° 50′ – 026°00′ = 117° 50′

## Composite Great Circle Sailing

△ PAV1 , V1 = 90°

PA = 90° – 34° 05′ = 55° 55′

PV1 = 90° – 42° = 48°

By using Napiers rule,

Sin (90 – P1) = tanPV1.tan(90-PA)

cos P1 = tan 48° . tan 34° 05′

P1 = 41.28°

Again using Napiers rule,

sin AV1 = cos (90- P1) . cos(90 – PA) = sin P1 . cos (90 – 55°05′)

sin AV1 = sin 41.28° . cos 34° 55′

AV1 = 33.12° = 1987.3′

Now, In △ PBV2 , V2 = 90°

PB = 90° – 39°00′ = 51°

PV2 = 90° – 42° = 48°

By using Napiers rule,

sin (90 – P3) = tan (90 – PB) . tan PV2 = tan (90 – 51°) . tan (48°)= tan 39° . tan 48°

cos P3 = 0.899

P3 = 25.93°

Again using Napiers rule,

sin BV2 = cos (90- P3) . cos(90 – PB) = sin P3 . cos (90 – 51°) = sin 25.93° . cos 39°

BV2 = 19.86° = 1191.97′

Now, P2 = dlong – (P1+P3) = 117° 50′ – (25.93° + 41.28°) = 50° 37.4′

Distance (departure) = dlong P2 * cos m’lat = 50° 37.4 * cos 42°

= 2257.2 miles

Total distance = 1987.3′ + 1191.97′ + 2257.2 ‘ + 45 ‘ + 84 ‘ = 5565.4 miles (Ans.)

Distance upto Landfall position = 5565.4 – 84 = 5481.4′

Time required = 5481.4 / speed = 5481.4 / 12.5 = 438.5 hours = 18d 6h 30m

Departure time = 05d 12h 00m (Local time)

Time required = 18d 6h 30m

So, LMT at landfall = 23d 18h 30m

Clock advanced = 9h 00m 00s

So, LMT at landfall (with clock) = 24d 3h 30m (Ans.)