Worked: arrival position from course and distance
1The lesson in plain words
Given a start, a course and a distance, find d.lat and departure, then the arrival latitude, the mean latitude, d.long and the arrival longitude.
The method
Turn the course into its quadrant angle. d.lat = d × cos C and dep = d × sin C, named from the quadrant. Apply d.lat to the start latitude: that gives the arrival latitude. Mean latitude = halfway between the two. d.long = dep ÷ cos (mean lat). Apply it to the start longitude. Spanish exams call this the situación de estima de llegada. Both examples are notional arithmetic exercises; their starts and straight tracks must not be used as navigable routes.
Example 3, the data
From 43°20.0′N 008°30.0′W, near A Coruña, you estimate a true course over the ground of 315° for 48.0 M. 315° is N 45° W. d.lat = 48.0 × cos 45° = 33.9′ N. Departure = 48.0 × sin 45° = 33.9 M W.
Example 3, the answer
Arrival latitude = 43°20.0′ + 33.94′ = 43°53.9′N. Mean latitude ≈ 43°37.0′N. Using unrounded values, d.long = dep ÷ cos(mean lat) = 46.88′ W, giving 009°16.9′W. Notice that 33.9 M of departure became 46.9′ of longitude: at this latitude a minute of longitude is only about 0.72 M.
Example 4, the data
From 28°05.0′N 015°25.0′W, near Las Palmas, you estimate a true course over the ground of 240° for 300 M. 240° is S 60° W. d.lat = 300 × cos 60° = 150.0′ S = 2°30.0′ S. Departure = 300 × sin 60° = 259.8 M W.
Example 4, the answer
Arrival latitude = 28°05.0′ − 2°30.0′ = 25°35.0′N. Mean latitude = 26°50.0′N. d.long = 259.8 ÷ cos 26°50′ = 259.8 ÷ 0.8923 = 291.2′ W = 4°51.2′ W. Arrival longitude = 015°25.0′ + 4°51.2′ = 020°16.2′W. The Las Palmas exercise track crosses Gran Canaria. These estimated ground inputs give an EP, not a fix. For a DR, use the inputs defined in the DR/EP unit.
2Key data
- Order
- d.lat and dep → arrival lat → mean lat → d.long → arrival long
- Example 3
- 315°, 48 M from 43°20′N 008°30′W → 43°53.9′N 009°16.9′W
- Example 4 working
- d.lat 150.0′ S, dep 259.8 M W, mean lat 26°50′N
- Example 4
- 240°, 300 M from 28°05′N 015°25′W → 25°35.0′N 020°16.2′W
- Spanish term
- Situación de estima de llegada
3Key points
- 1
Find the arrival latitude first: you need it for the mean latitude.
- 2
With non-zero departure and mean latitude away from 0°, d.long in minutes exceeds departure in miles.
- 3
Name d.lat and departure from the quadrant of the course.
- 4
This calculated position is an estimate: confirm it with a fix.
4Frequent mistakes
Wrong: Add the departure in miles straight onto the longitude.
Right: Convert first: d.long = dep ÷ cos(mean latitude). At mean latitude 43°37′N, departure 33.94 M gives about 46.9′ of longitude.
Wrong: Use the start latitude instead of the mean latitude for a long run.
Right: Find the arrival latitude, then use the latitude halfway between start and arrival.
Wrong: Apply a west d.long to an east longitude by adding.
Right: Going west from an east longitude, the longitude decreases; past 0°, it becomes west.
5Practice case & exam keys
From 37°10.0′N 000°20.0′E, you estimate a ground course of 220° for 50 M. A crew member calls the result a fix at 36°31.7′N 000°20.2′E. Accept it?
Show the answer
No. It is an EP, and the longitude is west. 220° is S 40° W. d.lat = 50 cos 40° ≈ 38.3′ S; dep = 50 sin 40° ≈ 32.1 M W. Arrival latitude ≈ 36°31.7′N; mean latitude ≈ 36°50.8′N. Using unrounded values, d.long = 40.16′ W. From 000°20.0′E, this crosses Greenwich to 000°20.2′W. Confirm the estimate with a fix.
Exam keys
- d.lat = d cos C; dep = d sin C; named from the course's quadrant.
- Arrival latitude first, then mean latitude, then d.long = dep ÷ cos (mean lat).
- Watch the names when crossing the equator or the Greenwich meridian.
- Worked: 315°, 48 M → 43°53.9′N 009°16.9′W; 240°, 300 M → 25°35.0′N 020°16.2′W.
Finish this unit
When you've read it, match the terms, then answer three questions. Get them all right to finish the unit; opening the page alone doesn't count.
My notes for this unit
No notes