Great Circle Example 1

Great Circle Example 1

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-(λ1+λ2) = 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°