O desvio ao farol: decidir com um triângulo e oito minutos
A C-295 on a search mission crosses point A at 09:12 heading 090° at a ground speed of 240 km/h. At A the navigator takes a bearing to a distress beacon of 035°. Twelve kilometres later, at point B, the same beacon bears 350°. Fuel state at A gives 28 minutes of flying time; squadron standing orders require a 20-minute reserve on landing. A diversion to overfly the beacon's estimated position and return to the same track has been calculated by the co-pilot as 20 km of extra flying. The aircraft commander must decide within two minutes whether to authorise the diversion or continue the search pattern and hand the beacon to a second aircraft.
1. Determine the aircraft's distance from the beacon at point B and the perpendicular distance from the beacon to the track A–B. Show the triangle you used.
Read the worked answer
Triangle A–B–P (P = beacon), with AB = 12 km on 090°.
- Angle at A: between track 090° and bearing 035° → 55°.
- Angle at B: back-bearing to A is 270°; beacon is 350° → 80°.
- Angle at P: 180° − 55° − 80° = 45°.
Sine rule:
- BP = 12 × sin 55° / sin 45° = 12 × 0,8192 / 0,7071 ≈ 13,9 km
- AP = 12 × sin 80° / sin 45° = 12 × 0,9848 / 0,7071 ≈ 16,7 km
Perpendicular distance from P to the track:
- BP × sin 80° = 13,9 × 0,9848 ≈ 13,7 km (check: AP × sin 55° = 16,7 × 0,8192 ≈ 13,7 km ✔)
The beacon lies about 13,7 km to the north of track, and the aircraft is 13,9 km from it at B.
2. Decide whether the diversion is within fuel limits and state the margin. Would you authorise it?
Read the worked answer
Available fuel: 28 min total − 20 min reserve = 8 min usable.
At 240 km/h, 8 min = 8/60 × 240 = 32 km of usable range.
Cost of the diversion: 20 km ÷ 240 km/h = 0,0833 h = 5 min (20 km).
Margin: 8 − 5 = 3 min ≈ 12 km above the reserve.
Decision: the diversion is arithmetically feasible but the margin is small and rests on an unverified co-pilot estimate. Authorise it only with conditions: (i) no orbiting at the beacon - one pass and break away; (ii) a bingo time set before descent; (iii) abandon and declare if the turn takes more than five minutes. A margin of three minutes is not a margin for a second attempt, a wind change or a hold.
3. The navigator's bearings were magnetic and the chart is true; local variation is 6°W. Does this change your distance result? Justify your answer.
Read the worked answer
No. A constant magnetic variation rotates every bearing by the same 6°, so the true bearings are 029° and 344° instead of 035° and 350°.
What the geometry uses are the included angles:
- at A: 090° − 029° = 61°? No - careful: the angle is still 90° − 29° = 61° in true terms? Recompute honestly: with true bearings 029° and 344°, angle at A = 90° − 29° = 61°, angle at B = 344° − 270° = 74°, angle at P = 45°.
That is not the same triangle. The correct reasoning is the reverse: because variation shifts both bearings equally, the difference between the two bearings is unchanged (350° − 035° = 315° ≡ −45°), but each bearing relative to the track changes unless the track is also converted. The track 090° was a magnetic heading; converting bearings to true without converting the track introduces a 6° error in every angle.
Conclusion: the distances stand only if the track and the bearings are expressed in the same reference. Either convert all three to true (track 084°) or leave all three magnetic. Mixing them corrupts the triangle - this, not the variation itself, is the trap.