The 1 in 60 rule is not limited to en-route track corrections. It also applies wherever an angle, a distance, and a range form a right-angled (or near-right-angled) triangle. The main examinable applications are:
Glide slope height — finding the correct height at a given range
Rate of descent (ROD) — finding the ROD needed to stay on slope at a given speed
Speed change on a glide slope — adjusting ROD when speed changes
VOR/DME crosstrack — finding distance off an airway centre line
Range from VOR bearing change — finding distance to a VOR by timing a bearing change
📚 Key Approximation
1 NM = 6 080 ft, but 1 NM ≈ 6 000 ft is used for all 1-in-60 calculations. This introduces only ~1% error and is accepted for all exam purposes.
2. Height on a Glide Slope
The same track-error geometry applies in the vertical plane. The angle is the glide slope angle (Z°), the adjacent side is the range, and the opposite side is the height.
Figure 1 — Glide slope geometry — height as a function of angle and range
Derivation (Z = 1°)
Range = 60 ft → Height = 1 ft
Range = 6 000 ft (1 NM) → Height = 100 ft
General rule: Height = 100 × Z × range (NM)
Glide slope height
H (ft) = 100 × Z° × Range (NM)
Equivalently: H = (Z × range_ft) / 60 | where 1 NM = 6 000 ft
Glide slope angle
Height per NM
Common at
2.5°
250 ft/NM
Military fast-jet airfields
3.0°
300 ft/NM
Most civil airports (ILS standard)
3.5°
350 ft/NM
Some civilian airports near terrain
5.5°
550 ft/NM
London City Airport (steep approach)
Worked Examples
Example 1 — 3° slope at 4 NM
H = 100 × 3 × 4 = 1 200 ft
Example 2 — 5.5° slope at 3 NM
H = 100 × 5.5 × 3 = 1 650 ft
3. Rate of Descent (ROD)
The rate of descent required to maintain a glide slope depends on both the slope angle and the ground speed. Think of it this way: in one minute the aircraft travels GS/60 NM, and must descend a height of 100 × Z × (GS/60) feet.
Figure 2 — Rate of descent calculation — range at one minute from touchdown
ROD for a 3° glide slope only
ROD (ft/min) = 5 × Ground Speed (kt)
For other angles: ROD = 5 × GS × (Z / 3)
✎ Why the 5 × Rule works for 3°
At 3°: height per NM = 300 ft. Distance per minute = GS/60 NM.
ROD = 300 × (GS/60) = 5 × GS. The factor of 5 comes directly from 300/60.
Worked Examples
3° slope, GS 120 kt:
ROD = 5 × 120 = 600 ft/min
4° slope, GS 100 kt:
3° ROD = 5 × 100 = 500 ft/min
Scale to 4°: 500 × (4/3) = 667 ft/min (use 670 in practice)
4. Change of Speed on a Glide Slope
Changing speed on a glide slope requires a corresponding change in ROD to stay on the slope.
✈ Speed Change Rules
Decrease speed → Decrease ROD
Increase speed → Increase ROD
Change in ROD for a 3° slope only
ΔROD (ft/min) = 5 × ΔGround Speed (kt)
For other angles: scale by (Z / 3)
Worked Examples
Example 1 — London Heathrow (3°), GS 140 kt
ROD = 5 × 140 = 700 ft/min
Example 2 — London City (5.5°), GS 120 kt
3° ROD = 5 × 120 = 600 ft/min
Scale to 5.5°: 600 × (5.5/3) = 1 100 ft/min
Example 3 — London Heathrow (3°), reduce GS from 140 → 120 kt
ΔSpeed = 20 kt → ΔROD = 5 × 20 = 100 ft/min
Decrease ROD: 700 − 100 = 600 ft/min
Example 4 — London City (5.5°), reduce GS from 120 → 110 kt
3° change: 5 × 10 = 50 ft/min
Scale to 5.5°: 50 × (5.5/3) = 92 ft/min decrease
5. VOR/DME Crosstrack Problems
The 1 in 60 formula applies directly to find how far an aircraft is off an airway centre line when the RMI QDM differs from the published airway QDM.
Distance off = (5 × 48) / 60 = 4 NM right of centre line
Note: RMI < airway QDM → aircraft is RIGHT of centre line
✎ Which side of centre line?
The QDM is the bearing to the VOR. If your QDM is higher than the airway QDM, the VOR is displaced to the right of your nose → you are left of centre line.
If your QDM is lower than the airway QDM, the VOR is displaced left → you are right of centre line.
ℹ Airway Width
Most airways extend 5 NM each side of the centre line (10 NM total). If your distance off is 5 NM or more, you are at or beyond the airway boundary and in potential conflict with ATC clearances.
6. Finding Range from Change of VOR Bearing
If you record the change in QDM (or QDR) from a VOR over a known time and ground speed, you can calculate your range from the VOR using the 1 in 60 rule.
Figure 4 — Finding range from change of VOR bearing
Worked Example
Tracking 090°(M) at 180 kt GS. At 1100 hrs QDM = 002°. At 1105 hrs QDM = 357°.
Bearing change: 002° → 357° = 5° (VOR moved left, so aircraft passed from right side to left side)
Distance flown: 5 min × 180/60 = 15 NM
Using 1 in 60: 5° × Range = 15 × 60 → Range = (15 × 60) / 5 = 180 NM
Range from bearing change
R = (Distance flown × 60) / Bearing change (°)
Range is the perpendicular distance — closest point of approach to the VOR
✎ Practical Notes
The bearing change angle must be small enough for the 1 in 60 approximation to hold (< ~20°)
The midpoint of the time interval should ideally be when the aircraft is abeam the VOR (QDM = 000° or 360°)
This technique works equally with QDM or QDR, as long as you use the change consistently
📚 Chapter 12 Quick Revision
Glide slope height: H = 100 × Z × range(NM)
ROD (3° only): ROD = 5 × GS | other angles: multiply by Z/3
Speed change ROD (3° only): ΔROD = 5 × ΔGS | scale for other angles
VOR/DME crosstrack: DO = (angle off × DME) / 60
VOR range: R = (distance flown × 60) / bearing change
Airway width: 5 NM each side of centre line in most countries
Q1. An aircraft is flying due South. At 1000 hrs, point P bears 267°(T) from the aircraft.
At 1006 hrs, point P bears 275°(T). Ground speed is 120 kt. Estimate the range from point P.
▶ Show answer & workings
Answer: 90 NM
Bearing change: 275° − 267° = 8°
Distance flown in 6 min at 120 kt: 120 × 6/60 = 12 NM
Range: R = (12 × 60) / 8 = 90 NM
Instructor's Note: Source key: 90 NM. Verified.
Q2. You are approaching Innsbruck, Austria on a glide slope of 3.5°. What height (QFE) should you be at 2 NM range?
▶ Show answer & workings
Answer: 700 ft
H = 100 × Z × range = 100 × 3.5 × 2 = 700 ft
Instructor's Note: Source key: 700 ft. Verified.
Q3. You are approaching Rota, Spain, on a glide slope of 2.6°. At what height should you be at 4 NM range?
▶ Show answer & workings
Answer: 1 040 ft
H = 100 × 2.6 × 4 = 1 040 ft
Instructor's Note: Source key: 1040 ft. Verified.
Q4. You are approaching Paris/Charles de Gaulle on a 3° glide slope. What height (QFE) at 2 NM range?
▶ Show answer & workings
Answer: 600 ft
H = 100 × 3 × 2 = 600 ft
Instructor's Note: Source key: 600 ft. Verified.
Q5. Using the glide slope scenarios in Q2–Q4, what rates of descent (ROD) are required to maintain the slope at the given ground speeds?
Instructor's Note: Source key: (a)700, (b)780, (c)750. All verified.
Q6. On approach to London Heathrow runway 27, glide slope 3°, you reduce speed from 150 kt to 120 kt GS. What change in ROD is required to maintain the glide slope?
▶ Show answer & workings
Answer: Decrease ROD by 150 ft/min
Speed change: ΔGS = 150 − 120 = 30 kt
ROD change (3°): ΔROD = 5 × 30 = 150 ft/min
Decrease speed → Decrease ROD by 150 ft/min
Instructor's Note: Source key: decrease ROD by 150 ft/min. Verified.
Q7. You are flying into Gioia Del Colle, Italy, on a glide slope of 2.5°. Due to a hydraulic failure you cannot select flaps; approach speed 220 kt TAS with a 10 kt headwind component.
(a) What ROD is required to maintain the glide slope?
You regain partial hydraulics and select mid-flap, reducing approach speed to 190 kt TAS.
(b) What change in ROD is required?
(c) What is your new ROD?
(b) ROD change on speed reduction: Key point: use TAS change for ΔROD (GS change = TAS change if headwind component unchanged)
ΔTAS = 220 − 190 = 30 kt
3° ΔROD = 5 × 30 = 150 ft/min
Scale to 2.5°: 150 × (2.5/3) = 125 ft/min
Decrease speed → Decrease ROD by 125 ft/min
(c) New ROD:
875 − 125 = 750 ft/min
Instructor's Note: Source key: (a)875, (b)decrease 125, (c)750. All verified. Note: GS used for initial ROD, TAS change used for delta ROD — source uses same values because headwind is constant.
Q8. You are flying an airway with centre line QDM of 137°(M) toward VOR/DME 'A'. Your RMI reads 141°(M) / DME 90 NM.
(a) Are you left or right of centre line?
(b) What is your distance off the airway centre line?
(c) Are you in trouble with ATC? (Airways normally extend 5 NM from centre line.)
▶ Show answer & workings
Answer: (a) Left of centre line | (b) 6 NM | (c) Yes
(a) Side of centre line:
Airway QDM = 137°(M). Your QDM = 141°(M) — higher than centre line QDM.
Higher QDM means VOR is displaced further clockwise from your nose → you are displaced anti-clockwise → LEFT of centre line
(c) ATC concern:
6 NM > 5 NM airway half-width → Yes, you are outside the airway.
You have exceeded the protected airspace boundary and are potentially in conflict with adjacent airways or uncontrolled airspace.
DGCA CPL/ATPL General Navigation Study Notes
Chapter 12 — Other Applications of the 1 in 60 Rule Capt. Pankaj Pahil | www.ghostaviator.com For personal study use only. Ghost Aviator Interactive Colour Edition.
