DGCA CPL / ATPL • General Navigation

Chapter 10
The 1 in 60 Rule

DGCA CPL/ATPL Study Notes • Interactive Colour Edition
Track Error Angle • TMG • Glide Slope • Closing Angle
Compiled by Capt. Pankaj Pahil
www.ghostaviator.com

Contents

  1. 1. The 1 in 60 Rule — Concept & Application
  2. 2. Measuring Angles in Flight
  3. 3. Geometry of the 1 in 60 Rule — The Radian Derivation
  4. 4. The Tangent Explanation
  5. 5. Expanding or Contracting the Triangle
  6. 6. The Formula & Worked Applications
  7. Practice Questions & Detailed Answers
  8. Master Reference Tables

1. The 1 in 60 Rule — Concept & Application

✈ The Core Idea

For any angle up to about 20°, if the adjacent side of a right-angled triangle is 60 units long, then the length of the opposite side (in the same units) is numerically equal to the angle in degrees.

In flight: if you have travelled 60 NM along-track and you are N NM off-track, your track error angle is approximately N degrees.

Figure 1 — The 1 in 60 Rule — a right-angled triangle with an adjacent side of 60 units: the angle z° equals the opposite side in the same units
Figure 1 — The 1 in 60 Rule — a right-angled triangle with an adjacent side of 60 units: the angle z° equals the opposite side in the same units

This is an enormously useful tool for in-flight calculations, removing the need for a protractor. The rule works because:

2. Measuring Angles in Flight

Suppose you are flying on a planned track of 100°(T). After 60 NM from your last on-track fix, you identify a ground feature that is 4 NM right of track.

✎ Applying the rule directly:

Adjacent = 60 NM, Opposite = 4 NM → Track Error Angle = 4° right
Your Track Made Good (TMG) = 100° + 4° = 104°(T)

No protractor needed — simply read off the angle from the numbers. The next chapter covers what to do once you have the error angle.

3. Geometry of the 1 in 60 Rule — The Radian Derivation

Why does this rule work? The key lies in the definition of a radian.

Figure 2 — A circle of 1 metre radius — travelling 1 metre along the circumference subtends angle θ at the centre
Figure 2 — A circle of 1 metre radius — travelling 1 metre along the circumference subtends angle θ at the centre
Figure 3 — The arc of 1 metre on a unit circle subtends θ = 57.3° at the centre (1 radian)
Figure 3 — The arc of 1 metre on a unit circle subtends θ = 57.3° at the centre (1 radian)

The Radian Calculation

Step-by-step derivation

Circle circumference: 2πr = 2 × 3.142 × 1 = 6.284 m
Full circle = 360° corresponds to 6.284 m of arc
1 metre of arc corresponds to θ, where: θ/1 = 360°/6.284 → θ = 57.3°
Scaling: This ratio holds for any circle radius — 57.3° always corresponds to an arc = radius
Rounding: 57.3 → 60 (introduces ~5% error — acceptable for flight calculations)

📚 Why 60 and not 57.3?

Strictly, this should be the "1 in 57.3 rule". But 57.3 is awkward for mental arithmetic at high workload in the cockpit. Rounding to 60 introduces only ~5% error, which is perfectly acceptable for navigation.

4. The Tangent Explanation

An alternative derivation shows why the rule works for a flat right-angled triangle (rather than an arc).

Figure 4 — Right-angled triangle with angle z — adjacent, opposite, and hypotenuse
Figure 4 — Right-angled triangle with angle z — adjacent, opposite, and hypotenuse
Figure 5 — Tangent table: values of tan z for angles 1° to 20°
Figure 5 — Tangent table: values of tan z for angles 1° to 20°
Figure 6 — Expanded table: 60 × tan z compared with z — showing the close numerical correlation
Figure 6 — Expanded table: 60 × tan z compared with z — showing the close numerical correlation

The Tangent Table

Angle z10°15°20°
tan z0.0170.0350.0870.1760.2680.364
60 × tan z1.022.105.2210.5616.0821.84

Up to about 10°, the correlation between angle z and 60 × tan z is excellent (within 1%). Above 15°–20° the error grows, but in practice you should rarely need the rule for angles that large.

⚠ Validity Limit

The 1 in 60 rule is acceptably accurate for angles up to ~20°. Above this, the tangent relationship becomes significantly non-linear. For track errors larger than 20°, use the nav computer or trigonometry.

5. Expanding or Contracting the Triangle

Fixes don't always occur at exactly 60 NM intervals. The 1 in 60 rule can be adapted for any along-track distance using the principle of similar triangles.

Figure 7 — 8 NM off track in 60 NM — track error = 8°
Figure 7 — 8 NM off track in 60 NM — track error = 8°
Figure 8 — Same 8° track error at 30 NM along-track gives only 4 NM off track — similar triangles
Figure 8 — Same 8° track error at 30 NM along-track gives only 4 NM off track — similar triangles

✎ Scale the Cross-Track Distance

For a given track error angle, the cross-track error scales proportionally with along-track distance:

6. The Formula & Worked Applications

Track Error Angle
z = (Distance Off / Distance Gone) × 60
Or rearranged: Distance Off = (z × Distance Gone) / 60

Applications

Use caseKnownFindFormula
Track error angleDistance off, distance goneAngle zz = (off/gone) × 60
Cross-track distanceAngle z, distance goneDistance offoff = (z × gone) / 60
Along-track distanceAngle z, distance offDistance gonegone = (off × 60) / z
Glide slope heightGS angle, rangeHeightH = (angle × range_ft) / 60
Glide slope rangeGS angle, heightRangerange_ft = (H × 60) / angle

✈ TMG vs Track Error

Track Made Good (TMG) = planned track ± track error angle
If aircraft is left of track: TMG = planned track − error (track is less than planned)
If aircraft is right of track: TMG = planned track + error (track is more than planned)

✎ Glide Slope Worked Example

GS angle 3°, range 2 NM (1 NM = 6 000 ft):
Range = 2 × 6 000 = 12 000 ft
Height = (3 × 12 000) / 60 = 600 ft

GS angle 2.5°, height 1 000 ft QFE (1 NM = 6 000 ft):
Range (ft) = (1 000 × 60) / 2.5 = 24 000 ft
Range (NM) = 24 000 / 6 000 = 4 NM

📚 Quick Revision — Chapter 10

Practice Questions & Detailed Answers

12 calculation questions • Track error, TMG, glide slope, closing angle • Full worked solutions

Q1. You are flying from A to B. Position: 60 NM outbound from A, 7 NM left of track. Track error angle?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 7° Left

z = (7 / 60) × 60 = left. At exactly 60 NM, the formula simplifies directly — the off-track distance IS the angle.

Instructor's Note: Source key: 7°L. Verified.
Q2. Flying C to D. Position: 120 NM outbound, 8 NM right of track. Track error angle?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 4° Right

z = (8 / 120) × 60 = right.
At 120 NM, halve the off-track distance to get the angle: 8 ÷ 2 = 4°.

Instructor's Note: Source key: 4°R. Verified.
Q3. Flying E to F. Position: 90 NM outbound, 6 NM right of track. Track error angle?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 4° Right

z = (6 / 90) × 60 = 360 / 90 = right.
At 90 NM, multiply off-track by 2/3: 6 × (60/90) = 4°.

Instructor's Note: Source key: 4°R. Verified.
Q4. Flying G to H. Position: 30 NM outbound, 4 NM left of track. Track error angle?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 8° Left

z = (4 / 30) × 60 = 240 / 30 = left.
At 30 NM (half of 60), double the off-track: 4 × 2 = 8°.

Instructor's Note: Source key: 8°L. Verified.
Q5. Flying J to K, planned track 045°(T). Position: 80 NM outbound, 4 NM left of track. Track Made Good?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 042°(T)

Step 1 — Track error angle: z = (4 / 80) × 60 = 3°
Step 2 — Apply: Aircraft is left of track, so TMG < planned.
TMG = 045° − 3° = 042°(T)

Instructor's Note: Source key: 042°(T). Verified.
Q6. Flying L to M, planned track 220°(T). Position: 45 NM outbound, 3 NM right of track. Track Made Good?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 224°(T)

Step 1 — Track error: z = (3 / 45) × 60 = 4°
Step 2 — Apply: Aircraft is right of track, so TMG > planned.
TMG = 220° + 4° = 224°(T)

Instructor's Note: Source key: 224°(T). Verified.
Q7. Flying N to P, planned track 315°(T). Position: 40 NM outbound, 6 NM left of track. Track Made Good?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 306°(T)

Step 1 — Track error: z = (6 / 40) × 60 = 9°
Step 2 — Apply: Aircraft is left of track, so TMG < planned.
TMG = 315° − 9° = 306°(T)

Instructor's Note: Source key: 306°(T). Verified.
Q8. A surveyor stands 660 metres from a mast and measures an elevation angle of 4° to the top. What is the height of the mast?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 44 metres

Distance off: H = (z × distance) / 60 = (4 × 660) / 60 = 2 640 / 60 = 44 m

The surveyor's 660 m horizontal distance is the 'adjacent' and the mast height is the 'opposite'.

Instructor's Note: Source key: 44 m. Verified.
Q9. Instrument approach, glide slope 3.00°, exactly 2 NM from touchdown. What height should you be passing? (Use 1 NM = 6 000 ft)
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 600 ft

Step 1 — Convert range to feet: 2 NM × 6 000 = 12 000 ft
Step 2 — Height: H = (3 × 12 000) / 60 = 36 000 / 60 = 600 ft

Instructor's Note: Source key: 600 ft. Verified.
Q10. Instrument approach, glide slope 2.5°, correctly on slope, passing 1 000 ft QFE. Range from touchdown? (1 NM = 6 000 ft)
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 4 NM

Step 1 — Find range in feet: range_ft = (H × 60) / angle = (1 000 × 60) / 2.5 = 60 000 / 2.5 = 24 000 ft
Step 2 — Convert to NM: 24 000 / 6 000 = 4 NM

Instructor's Note: Source key: 4 NM. Verified.
Q11. Flying Q to R, planned track 125°(T). Position: 40 NM from R, 2 NM left of track. What track must you fly to arrive overhead R?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 128°(T)

Closing angle: z = (2 / 40) × 60 = 3°
Direction: Aircraft is left of track. To close onto R, turn right (increase track).
Track to R = 125° + 3° = 128°(T)

Note: When flying toward the destination with a closing angle, the correction adds to or subtracts from the planned track to point at R — not at a parallel offset.

Instructor's Note: Source key: 128°(T). Verified.
Q12. Flying S to T, planned track 272°(T). Position: 50 NM from T, 5 NM right of track. What track must you fly to arrive overhead T?
Anchor: §6 — Formula
▶ Show answer & workings
Answer: 266°(T)

Closing angle: z = (5 / 50) × 60 = 6°
Direction: Aircraft is right of track. To close onto T, turn left (decrease track).
Track to T = 272° − 6° = 266°(T)

Instructor's Note: Source key: 266°(T). Verified.

Master Reference Tables — Chapter 10

The 1 in 60 Formulae

FindFormulaNotes
Track error anglez = (off / gone) × 60z in degrees
Cross-track distanceoff = (z × gone) / 60Units consistent
Along-track distancegone = (off × 60) / z
Glide slope height (ft)H = (angle × range_ft) / 601 NM = 6 000 ft for calcs
Glide slope range (ft)range_ft = (H × 60) / angleThen ÷ 6 000 for NM

Key Facts

ParameterValue
Exact value (radians)57.3°
Practical approximation60° (~5% error)
Valid angle rangeup to ~20°
Left of track → TMGplanned − error (TMG less than planned)
Right of track → TMGplanned + error (TMG more than planned)
Closing angle (left)planned + closing angle
Closing angle (right)planned − closing angle

Answer Key — No ⚑ Flags

Q123456789101112
Ans 7°L4°R4°R8°L 042°T224°T306°T 44 m600 ft4 NM 128°T266°T

DGCA CPL/ATPL General Navigation Study Notes
Chapter 10 — The 1 in 60 Rule
Capt. Pankaj Pahil  |  www.ghostaviator.com
For personal study use only. Ghost Aviator Interactive Colour Edition.
Capt. Pankaj Pahil
www.ghostaviator.com