Departure

The cosine of latitude converts angular longitude spacing into the east-west distance along that latitude.

Set up the calculation

1. Find the change of longitude and choose the shorter east or west direction unless the route states a particular direction.
2. Convert the whole change of longitude to minutes of arc.
3. Use the latitude of the parallel on which the distance is measured.
4. Multiply the longitude minutes by cosine latitude.
5. Attach east or west direction to the movement when a new position is required.

Worked example: 20 degrees at 52 degrees

Two meridians differ by 20 degrees at latitude 52 degrees. Convert 20 degrees to 1,200 minutes. Departure = 1,200 × cos 52 degrees = 738.8 NM. The distance is about 739 NM along the parallel.

A quick latitude table

Worked example from Indian longitudes

At 20 degrees north, the longitude difference between 073 degrees east and 083 degrees east is 10 degrees, or 600 minutes. Departure = 600 × cos 20 degrees = 563.8 NM east. The India-based positions change the setting, not the method.

If departure is known, divide by cosine latitude to recover change of longitude. Then apply the direction carefully to the starting longitude.

Types of departure questions

Worked example: distance to a new longitude

An aircraft at 60 degrees north, 005 degrees 22 minutes west flies 165 km due east. Using 1 NM = 1.852 km, the distance is about 89 NM. Change of longitude = 89 ÷ cos 60 degrees = 178 minutes = 2 degrees 58 minutes. Moving east from 005 degrees 22 minutes west gives 002 degrees 24 minutes west.

Crossing 180 degrees

Longitude wraps at 180 degrees. If an eastward change carries the calculation beyond 180 degrees east, subtract the excess from 180 degrees and continue in west longitude. For example, from 176 degrees 36 minutes east, an eastward change of 19 degrees 04 minutes reaches 164 degrees 20 minutes west.

Sign discipline

Compute the magnitude first. Then decide whether the destination is east or west of the start. With like-named longitudes, subtract the smaller value from the larger. With unlike names, add them, then use the smaller angle if the total exceeds 180 degrees.

The same formula can identify the latitude of a parallel or transfer a known departure to another latitude.

Worked example: find the parallel

A change of longitude of 44 degrees 11 minutes is 2,651 minutes. If its departure is 2,000 NM, cos latitude = 2,000 ÷ 2,651 = 0.7544. The inverse cosine gives about 41 degrees. Cosine has the same positive value in both hemispheres, so the answer can be 41 degrees north or 41 degrees south unless other information fixes the hemisphere.

Given departure at one latitude, find it at another

A 240 NM east-west leg at 44 degrees south spans a fixed change of longitude. Find the distance along 40 degrees south for the same meridians: departure at 40 = 240 × cos 40 degrees ÷ cos 44 degrees = 255.6 NM. The lower-latitude parallel is larger, so the result must exceed 240 NM.

Inspection of the answers

Before calculating, check direction and scale. If latitude increases for the same longitude interval, departure must decrease. If an answer at 60 degrees is larger than the equatorial value, reject it. If a computed cosine ratio is greater than 1, the given data or the algebra is wrong.

Departure is also the east-west component of a short route. Combined with change of latitude, it supports plane-sailing and mid-latitude calculations.

Resolve an oblique track

An aircraft flies 120 NM on track 060 degrees true. North-south component = 120 × cos 60 degrees = 60 NM north. Departure = 120 × sin 60 degrees = 103.9 NM east. For a short leg centred on 25 degrees north, change of longitude = 103.9 ÷ cos 25 degrees = 114.6 minutes, or 1 degree 54.6 minutes east.

The mid-latitude approximation

A short oblique leg runs from 20 degrees north to 24 degrees north with 3 degrees of longitude change. Mean latitude is 22 degrees. Departure ≈ 180 × cos 22 degrees = 166.9 NM. Change of latitude is 4 degrees, or 240 NM. The straight plane-sailing distance is √(240² + 166.9²) = 292.3 NM.

Rectangular routes do not normally close

Equal east-west distances flown on different latitudes produce different changes of longitude. If an aircraft flies east on a lower latitude and west by the same distance on a higher latitude, the higher-latitude leg covers more longitude. It finishes west of the starting meridian after returning to the starting latitude.

For example, consider 300 NM east at 45 degrees north and 300 NM west at 50 degrees north. The eastward longitude change is 300 ÷ cos 45 degrees = 424.3 minutes. The westward change is 300 ÷ cos 50 degrees = 466.7 minutes. The net displacement is 42.4 minutes west.

Oxford's inspection example uses four equal 3,000 km legs, south, east, north and west, starting at 27 degrees north, 170 degrees west. The aircraft returns to 27 degrees north but finishes farther west, at about 173 degrees 18 minutes west, because the westbound leg is flown on the smaller higher-latitude parallel.

The safest exam method is always the same: convert longitude to minutes, use the cosine of the correct latitude, then perform a direction and scale check.

Source-worked calculations

Oxford calculation patterns

A complete exam workflow

1. Sketch the parallel and mark east or west before using the calculator.
2. Convert degrees and minutes of longitude to one minute value.
3. Select actual latitude for a parallel, or mean latitude for a short oblique leg.
4. Multiply for departure; divide for longitude; take inverse cosine for latitude.
5. Return minutes to degrees and minutes, wrap correctly at 180 degrees, and check whether poleward distance is smaller.