Definition of Departure
Departure is the distance between two meridians measured along a specified parallel of latitude, expressed in nautical miles.
This may sound simple, but the key insight is that the same change of longitude does not represent the same east-west distance at every latitude. The closer you are to the poles, the shorter the E-W distance for a given change of longitude — because the meridians converge.
Figure 1 — Two meridians separated by the same change of longitude — the east-west distance between them varies with latitude
Figure 2 — Effect of latitude: the same ch.long gives shorter departure as latitude increases toward the poles
Think of it this way: 360 degrees of longitude span the full circumference of the Earth. At the Equator, the circumference is about 21 600 NM. At 60°N the circumference of the parallel is only about 10 800 NM — exactly half. One degree of longitude at 60°N is therefore half of one degree of longitude at the Equator.
Why Departure Varies with Latitude
The key observation is about the two extreme cases:
Figure 3 — At the Equator the meridians are at their maximum separation — departure is greatest here
Figure 4 — At the poles meridians converge to a single point — departure is zero regardless of longitude difference
- At the Equator (lat = 0°): The parallels of latitude are great circles. One minute of arc of longitude = one minute of arc of a great circle = 1 NM. Departure is maximum here.
- At the Poles (lat = 90°): All meridians meet at a single point. Departure = 0 regardless of how far apart the meridians are in longitude.
- At intermediate latitudes: Departure is proportional to cos(latitude). The cosine function has exactly the right behaviour: cos(0°) = 1.0 (full departure at Equator) and cos(90°) = 0 (zero at Poles).
Why cosine, not sine?
Because departure depends on the radius of the parallel, and the radius of any parallel at latitude φ = R × cos(φ), where R is the Earth's radius. The circumference at any latitude is 2π × R × cos(φ) — it shrinks by the cosine factor as you move toward the poles.
The Departure Formula
Derivation
At the Equator, 1 minute of longitude = 1 NM. At latitude φ, the parallel has a circumference equal to cos(φ) of the Equatorial circumference. Therefore:
departure = ch.long (in minutes) × cos(latitude)
Units to Remember
| Item | Unit | Conversion |
|---|---|---|
| Change of Longitude | Minutes of arc | degrees × 60 |
| Departure | Nautical Miles | — |
| Latitude | Degrees | used directly in cos( ) |
| Kilometres to NM | ÷ 1.852 | 1 NM = 1.852 km |
| NM to km | × 1.852 | — |
Figure 5 — Two meridians joined by a parallel of latitude — the east-west measurement is departure
Figure 6 — Departure sketch for the 52°N example: ch.long = 20° = 1200′; departure = 1200 × cos(52°) = 739 NM
The 52°N Example
Worked Calculation
Change of longitude = 20° = 20 × 60 = 1200 minutes of longitude
Latitude = 52°N; cos(52°) = 0.6157
Departure = 1200 × 0.6157 = 738.8 NM
Three Rearrangements
The one formula has three uses depending on what is unknown:
| Unknown | Formula |
|---|---|
| Departure | Dep = ch.long(min) × cos(lat) |
| Change of Longitude | ch.long(min) = Dep ÷ cos(lat) |
| Latitude | cos(lat) = Dep ÷ ch.long(min) → lat = arccos(…) |
Types of Departure Questions
Exam questions on departure fall into two categories:
Type 1: Variations on the Basic Formula
These questions give you two of the three variables (ch.long, lat, departure) and ask you to find the third. The steps are always:
- Convert all units (km to NM, degrees to minutes if needed)
- Identify which variable is unknown
- Rearrange the formula and solve
- Convert the answer back if needed (minutes of longitude to degrees + minutes)
Type 1, Example A — Find New Position
Aircraft at 60°00′N 005°22′W flies 165 km due East. New position?
89.1 = ch.long × cos(60°) = ch.long × 0.5
ch.long = 89.1 / 0.5 = 178.2 minutes = 2°58.2′E
Type 1, Example B — Find Latitude
In which latitude does a difference of 44°11′ of longitude = 2 000 NM departure?
lat = arccos(0.7544) = 41°N or S
Type 2: Departure at One Latitude → Find It at Another
When a flight goes south then east then north (or vice versa), the departure flown is at a different latitude from the starting point. You need to find the departure at the original latitude that corresponds to the same change of longitude.
The Latitude Ratio Formula
When you know departure at latitude B and need to find departure at a different latitude A (where both A and B share the same meridian spacing), use:
This formula saves a step of working out the change of longitude explicitly. You simply divide the departure at the known latitude by the cosine of the known latitude, then multiply by the cosine of the target latitude.
Figure 7 — Departure problem with multiple legs — example 3 diagram showing tracks and the key is that the N-S legs cancel, leaving only the E-W leg as the net answer
Ratio Formula Example — Example 3
Start G (40°S) → H: 180° 240 NM → J: 270° 240 NM → K: 000° 240 NM
Find RL bearing and distance K from G.
The 240 NM west from H is at 44°S.
The 240 NM north takes J (44°S) back to K (40°S).
So K is at the same latitude as G (40°S) but directly west.
Dep at 40°S = Dep(44°S) × cos(40°)/cos(44°) = 240 × 0.7660/0.7193 = 255.6 NM
Inspection of Answers — A Time-Saving Technique
For MCQ examinations, it is often not necessary to calculate the exact answer. By understanding the direction of the effect, you can eliminate impossible options and identify the correct one without arithmetic.
Example — The North/South/East/West Box
Aircraft at 27°00′N 170°00′W travels 3000 km S, 3000 km E, 3000 km N, 3000 km W. Final position?
General Rule for the "Box" Problem
S → E → N → W starting in the Northern Hemisphere: net displacement is west
(eastward leg at lower lat covers less longitude than westward leg at higher lat).
N → E → S → W: net displacement is east.
In the Southern Hemisphere, the directions reverse.
Practice Questions
Questions 1–10 require calculation; Q11–15 are multiple-choice. All solutions are shown.
Q1A flight is to be made along the parallel of latitude from A (48°00′N 04°00′W) to B (48°00′N 02°27′E). Calculate the distance.
Departure = 387 × cos(48°) = 387 × 0.6691 = 259 NM
Q2An aircraft flies 1000 NM along a rhumb line track of 090°(T) from C (36°00′N 174°45′E) to D. What is the longitude of D?
1236 min = 20°36′ of longitude eastward from 174°45′E.
174°45′E + 20°36′E = 195°21′E = 360° − 195°21′ = 164°39′W
Q3Starting from E (50°N): E→F 000°(T) 300 NM; F→G 090°(T) 300 NM; G→H 180°(T) 300 NM. What is the RL bearing and distance of H from E?
F to G = 300 NM east at 55°N: ch.long = 300/cos55° = 300/0.5736 = 523 min = 8°43′.
G to H = 300 NM south → lat of H = 50°N (same as E).
E and H are both at 50°N, H is 8°43′ east of E.
Departure at 50°N = 523 × cos50° = 523 × 0.6428 = 336 NM.
Bearing of H from E = 090°(T) (due east). Distance = 336 NM.
Q4What is the track and distance along the parallel of 80°S from 176°15′W to 179°45′E?
Departure = 240 × cos(80°) = 240 × 0.1736 = 41.7 NM. Track = 270°(T).
Q5In which latitude is a difference of longitude of 44°10′ equivalent to a departure of 2295 NM?
cos(lat) = 2295 / 2650 = 0.8660
lat = arccos(0.866) = 30°N or S (sin 30° = 0.5; cos 30° = 0.866 ✓)
Q6Aircraft departs J (36°00′S 130°14′E) at 0946 UTC, flying RL track 270° at FL100, temp 0°C, Mach 0.81, wind component 35 kt tail. Position at 1004 UTC?
TAS = 0.81 × 643 = 521 kts (use 520 kts).
GS = 520 + 35 = 555 kts.
Time = 1004 − 0946 = 18 min. Distance = 555 × 18/60 = 166.5 NM.
Heading west at 36°S: ch.long(min) = 166.5 / cos(36°) = 166.5 / 0.809 = 205.8 min ≈ 206 min = 3°26′.
New longitude = 130°14′E − 3°26′ = 126°48′E.
Position: 36°00′S 126°48′E.
Q7(a) At which latitude is the departure in NM between two points equal to: (i) their difference in longitude in minutes? (ii) half their difference in longitude in minutes? (b) At 52°N: (i) departure between 136°16′W and 43°44′E? (ii) shortest distance between those same points?
(a)(ii) For Dep = ½ ch.long: cos(lat) = 0.5 → lat = 60°N or S.
(b)(i) 136°16′W to 43°44′E = 180° = 10 800 minutes. Dep = 10 800 × cos(52°) = 10 800 × 0.6157 = 6 653 NM.
(b)(ii) Shortest distance from 52°N 136°W to 52°N 044°E goes over the North Pole. Route = (90°−52°)×2 = 76° of latitude change × 60 NM/° = 4 560 NM.
Q8Aircraft leaves L (37°S) flying L→M 270°(T) at 310 kt for 80 min; M→N 180°(T) at 280 kt; N→P 090°(T) at 250 kt. P is due south of L. N reached 93.5 min after leaving N. Calculate distance and time M to N.
ch.long = 413.3/cos(37°) = 413.3/0.7986 = 517 min = 8°37′ west.
P is due south of L, so N→P must cover the same 8°37′ going east.
N to P: 93.5 min at 250 kt = 389.6 NM (east).
ch.long for N→P: 517 min → lat of N/P: cos(lat) = 389.6/517 = 0.7537 → lat ≈ 41°S.
Distance M to N = difference in latitude × 60: 41°S − 37°S = 4° = 240 NM.
Time M to N = 240/280 = 0.857 hr = 51.25 min.
Q9Aircraft Q (GS 301 kt) flies parallel 46°N through 10° of longitude. Aircraft R (GS 364 kt) takes the same time. At what latitude does R fly?
Time Q = 417/301 = 1.385 hr = 83 min.
R in 83 min at 364 kts covers 364 × 83/60 = 503.5 NM.
cos(lat_R) = 503.5/600 = 0.8392 → lat = arccos(0.8392) = 33°N or S.
Q10S (36°N 10°E, GS 470 kt, RL 090°) and T (30°N 10°E, GS 150 kt, RL 090°) depart 1522Z. S turns onto 180° at 13°E. Which reaches 30°N 13°E first, and when? Where is the other aircraft at that time?
Dep S leg 1 = 180 × cos(36°) = 145.6 NM at 470 kt = 18.5 min.
Then 36°N→30°N = 6° = 360 NM at 460 kt = 47.0 min. Total S = 65.5 min.
T (30°N, 150 kt): 10°E→13°E at 30°N = 180 × cos(30°) = 155.9 NM at 150 kt = 62.0 min.
T arrives first (62.0 min vs 65.5 min — 3.5 min earlier → 1522 + 62 min = 1624Z).
When T arrives, S is 65.5−62.0 = 3.5 min short. At 460 kt: 3.5×460/60 = 26.8 NM ≈ 27 NM ≈ 27 min lat.
S position: 30°27′N 13°E.
Q11Aircraft starts 0410S 17822W, heads true north 2950 NM, then 90° left for 314 km RL. Final position?
90° left from north heading = west. 314 km ÷ 1.852 = 169.5 NM.
ch.long = 169.5/cos(45°) = 169.5/0.7071 = 239.7 min = 3°59.7′ ≈ 4° west.
17822W + 4° = 17°82+240=178+60+22+60? → 17822W + 0400 = 18222W = 360°−182°22′ = 177°38′E.
Answer: 45°00′N 177°38′E → d (4500N 17738E).
Q12Aircraft at 50°N 006°E: 300 NM South, 300 NM East, 300 NM North, 300 NM West. Final position relative to start?
N leg: back to 50°N. W leg at 50°N: ch.long = 300/cos50° = 467 min = 7°47′ W.
Net: went 7°4′ E then 7°47′ W → net = 0°43′ W. Final position is west of start.
Answer: c
Q13Aircraft departs 04°00′N 170°00′W: 600 NM south, 600 NM east, 600 NM north, 600 NM west. Final position?
N: 06°S → 04°N. W at 04°N: ch.long = 600/cos4° = 600/0.9976 = 601.4 min = 10°1.4′ W.
Net: 10°3.3′ E − 10°1.4′ W = 1.9′ E. Start 170°00′W − 1.9′ = 169°58.1′W.
Answer: c
Q14Aircraft flying eastwards at 60°N at 240 kts goes once round Earth. At what speed must another fly eastwards along the Equator to take the same time?
Time = 10 800 / 240 = 45 hours.
Equatorial circumference = 360 × 60 = 21 600 NM.
Required speed = 21 600 / 45 = 480 kts. Answer: c
Q15Position 58°33′N 174°00′W. Fly exactly 6 NM eastwards. New position?
ch.long = 6 / 0.5217 = 11.5 min of longitude.
Going east from 174°00′W: 174°00′ − 11.5′ = 173°48.5′W.
Answer: d — 58°33′N 173°48.5′W