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DGCA Air Navigation · 83 questions

1 in 60 Rule & Bearings: DGCA Air Navigation Questions and Answers

83 practice questions on 1 in 60 Rule & Bearings with answers and explanations, plus the key concepts and formulas, from the TrueHeading question bank.

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About this topic

Use this page to revise 1 in 60 Rule & Bearings for the DGCA Air Navigation paper. Below are the key ideas first, then 12 practice questions with answers and explanations, picked from the 83 questions in the TrueHeading bank for this topic.

In short: At 60 NM from the point of departure, 1° of track error gives 1 NM of lateral displacement. Distances scale in proportion.

Open the full set in the student zone to answer every question, track your accuracy and take timed mock tests.

Key concepts: 1 in 60 Rule & Bearings

6 ideas to know cold.

State the 1 in 60 rule.

At 60 NM from the point of departure, 1° of track error gives 1 NM of lateral displacement. Distances scale in proportion.

Track error∘=off-track distancedistance flown×60\text{Track error}^\circ=\frac{\text{off-track distance}}{\text{distance flown}}\times 60
1°1 NM off60 NM along track1 in 60 rule1° of error ≈ 1 NM off per 60 NM flowntrack error° = (off-track ÷ distance flown) × 60

How do you regain track using 1 in 60?

Add the track error and the closing angle.

Closing angle=off-trackdistance to go×60\text{Closing angle}=\frac{\text{off-track}}{\text{distance to go}}\times 60
  • To rejoin track in the same distance you have already flown, alter heading by twice the track error
  • Turn the correct way: towards the track

QDM, QDR, QUJ and QTE?

The Q-codes for bearings:

  • QDM: magnetic bearing to the station
  • QDR: magnetic bearing from the station
  • QUJ: true bearing to the station
  • QTE: true bearing from the station

Tip: QDM and QDR differ by 180°, as do QUJ and QTE.

Time to a station from a bearing change?

Fly at a steady speed on a track passing abeam, note the time for a given bearing change.

Time to station (min)=60×time between bearings (min)change of bearing (∘)\text{Time to station (min)}=\frac{60\times\text{time between bearings (min)}}{\text{change of bearing (}^\circ)}
  • Valid when the bearing change is measured at the wingtip or on the beam
  • Distance = ground speed × time

State the 1-in-60 rule.

At 60 NM from the start, an error of 1 NM equals 1° of track error.

Error∘=off-track NM×60distance gone NM\text{Error}^\circ=\dfrac{\text{off-track NM}\times60}{\text{distance gone NM}}
  • Track error (°) = off-track distance × 60 / distance gone
  • Closing angle (°) = off-track distance × 60 / distance to go
  • Heading change to regain track = track error + closing angle

How do you work out the descent angle and rate of descent with the 1-in-60 rule?

Descent angle (°) = height to lose / (distance in NM × 100). Rate of descent ≈ groundspeed × 5 for 3°.

ROD≈GS×5 (ft/min, for 3∘)\text{ROD}\approx GS\times 5\ (\text{ft/min, for }3^\circ)
  • 3° path: 300 ft per NM
  • Rate of descent (ft/min) ≈ GS (kt) × 5 (more precisely × 5.3 for 3°)
  • To lose 9 000 ft at 3°: 30 NM

Practice questions: 1 in 60 Rule & Bearings

Tap “Show answer” after you have tried each one.

Q1After flying 25 NM of a 100 NM leg, an aircraft is 6 NM off the intended track. Using the 1-in-60 rule, the heading change needed to arrive at the destination is approximately:

  1. 19°
  2. 13°
  3. 11°
  4. 23°
Show answer

Answer: A. 19°

Track error 6/25×60 = 14.4°; closing angle 6/75×60 = 4.8°; total ≈ 19°.

Q2After flying 60 NM of a 120 NM leg, an aircraft is 5 NM off the intended track. Using the 1-in-60 rule, the heading change needed to arrive at the destination is approximately:

  1. 16°
  2. 20°
  3. 10°
  4. 7°
Show answer

Answer: C. 10°

Track error 5/60×60 = 5.0°; closing angle 5/60×60 = 5.0°; total ≈ 10°.

Q3After flying 50 NM of a 100 NM leg, an aircraft is 3 NM off the intended track. Using the 1-in-60 rule, the heading change needed to arrive at the destination is approximately:

  1. 17°
  2. 7°
  3. 1°
  4. 10°
Show answer

Answer: B. 7°

Track error 3/50×60 = 3.6°; closing angle 3/50×60 = 3.6°; total ≈ 7°.

Q4After flying 20 NM of a 300 NM leg, an aircraft is 10 NM off the intended track. Using the 1-in-60 rule, the heading change needed to arrive at the destination is approximately:

  1. 35°
  2. 38°
  3. 36°
  4. 32°
Show answer

Answer: D. 32°

Track error 10/20×60 = 30.0°; closing angle 10/280×60 = 2.1°; total ≈ 32°.

Q5An aircraft is 8 NM off track after flying 75 NM from the departure point. The track error is approximately:

  1. 10°
  2. 6°
  3. 16°
  4. 14°
Show answer

Answer: B. 6°

8 ÷ 75 × 60 = 6.4°.

Q6An aircraft is 6 NM off track after flying 50 NM from the departure point. The track error is approximately:

  1. 7°
  2. 10°
  3. 17°
  4. 3°
Show answer

Answer: A. 7°

6 ÷ 50 × 60 = 7.2°.

Q7After flying 100 nm an aircraft is 4 nm off the planned track. The track error is approximately:

  1. 2.4°
  2. 1.2°
  3. 1500.0°
  4. 4.8°
Show answer

Answer: A. 2.4°

TE = 60 × off-track ÷ distance gone = 60 × 4 ÷ 100 = 2.4°.

Q8After flying 120 nm an aircraft is 8 nm off the planned track. The track error is approximately:

  1. 8.0°
  2. 4.0°
  3. 900.0°
  4. 2.0°
Show answer

Answer: B. 4.0°

TE = 60 × off-track ÷ distance gone = 60 × 8 ÷ 120 = 4.0°.

Q9An aircraft is 6 nm off track after 100 nm, with 110 nm still to go. What total heading change is needed to reach the destination (track error plus closing angle)?

  1. 3.3°
  2. 3.6°
  3. 0.3°
  4. 6.9°
Show answer

Answer: D. 6.9°

TE = 60×6/100 = 3.6°. CA = 60×6/110 = 3.3°. Total = 6.9°.

Q10An aircraft is 5 nm off track after 40 nm, with 80 nm still to go. What total heading change is needed to reach the destination (track error plus closing angle)?

  1. 11.2°
  2. 7.5°
  3. 3.8°
  4. 22.5°
Show answer

Answer: A. 11.2°

TE = 60×5/40 = 7.5°. CA = 60×5/80 = 3.8°. Total = 11.2°.

Q11An aircraft has flown 90 nm and is 4 nm off its planned track. To regain the track after a further 100 nm, the closing angle (relative to the original track) is about:

  1. 3°
  2. 2°
  3. 5°
  4. 1°
Show answer

Answer: B. 2°

Closing angle = 60 × off-track ÷ distance to regain = 60 × 4 ÷ 100 ≈ 2.4°.

Q12A constant 3° track error is maintained for 80 nm. How far off track will the aircraft be?

  1. 4.0 nm
  2. 2.2 nm
  3. 2.0 nm
  4. 8.0 nm
Show answer

Answer: A. 4.0 nm

Off-track distance = TE × distance ÷ 60 = 3 × 80 ÷ 60 = 4.0 nm.

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