Great Circle Example 1
Home
Flashing Light
Watch Training Workbook
Campus Life
The Joe Show
Deck Skills
Cruise
Navigation
Celestial Nav.
Radar
Meteorology
Ship Structure
Stability
GMDSS
Rules of the Road
Marlinespike
Ship and Cargo Ops.
Shipping Info
FAQs
Shanties and Sea Lore
Terms/Acronyms
World Port/Drink Guide
Forum

You intend to leave 25°00'N, 145°00'E for 36°00'N, 175°00'W the day after tomorrow. What is your initial course and the Great Circle distance of the voyage?

L1: 25°00' N
L2: 36°00'N
λ1: 145°00'E
λ2: 175°00' W

How to Solve:
-First establish what you need to know and what you do already know from the information provided.
-Second, solve for what you don't know. In this case, we do not yet know DLo, so we will solve by adding λ1 to λ2 and subtracting the sum from 360.

-Lastly, now that you only have one unsolved variable, solve for your initial course and distance using the appropriate equations.

Back to GC Sailings

Which Equations To Use:
cos D = sinL1sinL2 + cosL1cosL2cosDLo
tan C = (sin DLo) / ((cosL1tanL2) - (sinL1cosDLo))

What you need:
What you know:
L1
25°00' N
L2
36°00'N
DLo
???

Horizontal Divider 1

Solving

λ1
145°00'E
λ2
175°00' W
DLo
360-(λ12) = 40°(E)

cos D = sinL1sinL2 + cosL1cosL2cosDLo
cos D = (sin 25.0°)(sin 36.0°) + (cos 25.0°)(cos 36.0°)(cos 40.0°)
cos D = 0.810087
D = 35.895° = 2153.7 nm

tan C = (sin DLo) / ((cosL1tanL2) - (sinL1cosDLo))
tan C = (sin 40.0°) / (((cos 25.0°)(tan 36.0°)) - ((sin 25.0°)(cos 40.0°)))
tan C = 1.92033
C = 62.5°
Cn = 062.5°