GEN 14: 1 in 60 Rule – LMS | INSIGHT AVIATION

Results

📄 Notes ⚡ Cheat Sheet

Q1. A pilot receives the following signals from a VOR DME station: radial 180 deg +/- 1 deg, distance = 200 NM. What is the approximate error?

(+/- 1 NM)

(+/- 3.5 NM)

(+/- 2 NM)

(+/- 7 NM)

(+/- 3.5 NM) –

Calculate using the 1 in 60 rule formula: Error = (Distance * Angle) / 60.

Error = (200 * 1) / 60 = 3.33 NM.

The closest option provided is 3.5 NM.

Q2. A ground feature was observed on a relative bearing of 315 deg and five minutes later on a relative bearing of 280 deg. The aircraft heading was 165(M) variation 25W drift 10 Right and GS 360 kt. When the relative bearing was 270 deg, the distance and true bearing of the aircraft from the feature was:

30 NM and 050 deg

40 NM and 110 deg

40 NM and 290 deg

30 NM and 060 deg

30 NM and 050 deg –

True Heading = 165(M) – 25(Var) = 140(T).

Bearing Calculation: At Relative Bearing 270 (which is 90 deg Left), True Bearing = Heading – 90. TB = 140 – 90 = 050 deg.

Distance Calculation: The problem implies the standard “doubling the angle” scenario where the time is measured between 45 degrees off the nose (315 RB) and the beam (270 RB). Speed 360 kts for 5 mins = 30 NM. In a 45-90 triangle, distance flown equals distance to the station abeam.

Q3. The distance between two waypoints is 200 NM, To calculate compass heading, the pilot used 2 deg E magnetic variation instead of 2 deg W. Assuming that the forecast W/V applied, what will the off track distance be at the second waypoint?

14 NM

21 NM

0 NM

7 NM

14 NM –

Determine total angular error: The difference between 2E and 2W is 4 degrees.

Apply 1 in 60 rule: Off Track Distance = (Angle * Distance) / 60.

Calculation: (4 * 200) / 60 = 13.33 NM.

14 NM is the closest valid option.

Q4. Given: Distance ‘A’ to ‘B’ is 100 NM, Fix obtained 40 NM along and 6 NM to the left of course. What heading alteration must be made to reach ‘B’?

18 deg Right

6 deg Right

9 deg Right

15 deg Right

15 deg Right –

Track Error (TE) = (60 * 6) / 40 = 9 degrees.

Closing Angle (CA) = (60 * 6) / (100 – 40) = 6 degrees.

Total Correction = TE + CA = 15 degrees.

Direction: Aircraft is Left of course, so turn Right.

Q5. A ground feature was observed on a relative bearing of 315 deg and 3 MIN later on a relative bearing of 270 deg The W/V is calm; aircraft GS 180 kt. What is the minimum distance between the aircraft and the ground feature?

3 NM

12 NM

9 NM

6 NM

9 NM –

This is a standard isosceles triangle problem. The bearing changes from 315 (45 deg left) to 270 (90 deg left/beam).

Distance flown = Distance to station at the beam.

Calculation: 180 kts * (3/60 hours) = 9 NM.

Q6. Given: Distance A to B = 120 NM, After 30 NM aircraft is 3 NM to the left of course. What heading alteration should be made in order to arrive at point ‘B’?

8 deg left

4 deg right

8 deg right

6 deg right

8 deg right –

Track Error = (60 * 3) / 30 = 6 degrees.

Closing Angle = (60 * 3) / (120 – 30) = (180 / 90) = 2 degrees.

Total Correction = 8 degrees.

Direction: Left of course requires a Right turn.

Q7. Given: Distance ‘A’ to ‘B’ is 90 NM, Fix obtained 60 NM along and 4 NM to the right of course. What heading alteration must be made to reach ‘B’?

16 deg Left

4 deg Left

8 deg Left

12 deg Left

12 deg Left –

Track Error = (60 * 4) / 60 = 4 degrees.

Closing Angle = (60 * 4) / (90 – 60) = (240 / 30) = 8 degrees.

Total Correction = 12 degrees.

Direction: Right of course requires a Left turn.

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