Question

A ship sails for 5 hours from position A: (27° 38' N, 112° 45' W) at a speed of 16 knots on a course of 295°T.
Determine the course angle.
How far does the ship travel at the end of 5 hours?
Find the departure, DEP, for the final position.
Find the difference in latitude , DLAT ( to the nearest degree and minute) for the final position.
Determine the latitude of the final position B.

Answer 1

To determine the course angle, we subtract 180° from the course of 295°:

Course angle = 295° - 180° = 115°

To find the distance traveled by the ship, we multiply the speed of 16 knots by the duration of 5 hours:

Distance = 16 knots * 5 hours = 80 nautical miles

To find the departure (DEP) for the final position, we use the formula:

DEP = Distance * sin(course angle)

DEP = 80 nautical miles * sin(115°)

Using a calculator, we find that sin(115°) ≈ 0.9063, so

DEP ≈ 80 nautical miles * 0.9063 ≈ 72.504 nautical miles

To find the difference in latitude (DLAT) for the final position, we use the formula:

DLAT = DEP * sin(course)

DLAT = 72.504 nautical miles * sin(115°)

Again, using a calculator, we find that sin(115°) ≈ 0.9063, so

DLAT ≈ 72.504 nautical miles * 0.9063 ≈ 65.8116 nautical miles

To find the latitude of the final position B, we subtract the DLAT from the latitude of position A:

Latitude of final position = 27° 38' N - 65.8116 nautical miles

To convert nautical miles to degrees and minutes, we use the approximation that 1 nautical mile is roughly equal to 1 minute of latitude. Therefore:

65.8116 nautical miles ≈ 65.8116 minutes

So, the latitude of the final position B is approximately:

27° 38' N - 65.8116 minutes ≈ 27° 38' - 65' 49" = 27° 38' - 65' = 26° 33' N.