ATPL(A) 061 General NavigationRevision notes and practice questions

061 General NavigationFree sample, 10 questions

1. 1Using the data below, determine the track error angle.
   Planned track: 197°(T)
   Actual heading: 213°(T)
   Actual in-flight drift: 12° left

   1. 12° right
   2. 16° right
   3. 4° right
   4. 4° left
   Show the answer

   Answer C. With 12° of left drift the aircraft actually tracks 213 − 12 = 201°(T). The track error angle is measured from the planned track to the track made good: 201 − 197 = 4°, and since the actual track lies clockwise of the planned one, the error is to the right. Comparing the heading directly with the planned track (16° right) forgets the drift.

2. 2A ground feature is seen on a relative bearing of 315° and, 3 min later, on a relative bearing of 270°. The W/V is calm and the aircraft GS is 180 kts. What is the minimum distance between the aircraft and the ground feature?

   1. 4.5 NM
   2. 9 NM
   3. 12 NM
   4. 6 NM
   Show the answer

   Answer B. Relative bearing 315° puts the feature 45° left of the nose; 270° puts it exactly abeam on the left. When the angle off the nose doubles from 45° to 90°, the distance flown equals the distance to the feature. In 3 min at 180 kt the aircraft covers 180 × 3/60 = 9 NM. The feature is abeam at that moment, so 9 NM is also the closest point of approach.

3. 3A pilot is planning a VFR cross-country flight using visual checkpoints. Which of the following would make the most suitable checkpoint?

   1. A valley among a range of low hills.
   2. A feature that contrasts with its natural surroundings.
   3. A feature that merges into its natural surroundings.
   4. A small lake in a mountainous region.
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   Answer B. A checkpoint is useful only if it can be found quickly and cannot be confused with anything else, and that depends on contrast with what surrounds it. A feature that blends into its surroundings is hard to spot, one valley among many low hills looks like the next, and small lakes in mountains are numerous and easily mixed up.

4. 4Given the following data for a conic projection chart in the northern hemisphere, calculate the change of longitude:
   Great circle track from P to Q, at P: 112°(T)
   Great circle track from Q to P, at Q: 316°(T)
   Constant of the cone: 0.8

   1. 226°
   2. 255°
   3. 24°
   4. 30°
   Show the answer

   Answer D. The track from Q to P at Q is 316°, so the track from P to Q at Q is 316 − 180 = 136°. The straight line changes direction by 136 − 112 = 24°, which is the chart convergency. Chart convergency = change of longitude × constant of the cone, so the change of longitude is 24/0.8 = 30°. 24° is the convergency itself, not the change of longitude.

5. 5What is the Local Mean Time (LMT) at (35°25'S, 110°45'W) when the Coordinated Universal Time (UTC) is 18:34 on 4th September?

   1. 01:34 on September 5th
   2. 11:34 on September 4th
   3. 01:57 on September 5th
   4. 11:11 on September 4th
   Show the answer

   Answer D. 110°45' of longitude is 7 h 23 min of time (110.75 x 4 min = 443 min). West of Greenwich, local mean time is behind UTC, so subtract: 18:34 - 7:23 = 11:11 on 4 September. Adding the difference instead gives 01:57 on the 5th, the classic east/west mix-up.

6. 6Using trigonometry and the data below, calculate the crosswind component (XWC) to the nearest knot.
   Runway: 32
   Wind (magnetic): 020°/25 kt
   Variation: 15°E

   1. 22 kt
   2. 13 kt
   3. 24 kt
   4. 25 kt
   Show the answer

   Answer A. Runway designators and the tower wind are both magnetic, so the variation plays no part. Runway 32 is about 320°(M) and a wind from 020° is 60° off the runway axis. Crosswind = 25 × sin 60° = 21.7, about 22 kt. Using the cosine gives the 13 kt headwind component instead, and applying the variation to the wind only distorts the angle.

7. 7Using the information below, what is the TAS?
   Pressure altitude: 6000 ft
   OAT: +5°C
   IAS/CAS: 115 kt

   1. 105 kt
   2. 115 kt
   3. 123 kt
   4. 126 kt
   Show the answer

   Answer D. ISA at 6 000 ft is 15 − 12 = +3 °C, so +5 °C is ISA +2. Pressure at 6 000 ft is about 812 hPa, giving a density ratio σ = (812/1013) × (288/278) ≈ 0.83. TAS = CAS/√σ = 115/0.911 ≈ 126 kt, which is what the airspeed window of a flight computer gives with 6 000 ft set against +5 °C. Multiplying instead of dividing gives about 105 kt, and 115 kt ignores the density correction.

8. 8A pilot is planning a VFR flight that crosses controlled airspace. Because of the airspace restrictions, part of the flight must be flown at low level. Which of the following is a navigation problem that may be met at low level on a VFR flight?

   1. Navigation features can be noticed more easily, but are more difficult to identify.
   2. Visual range to the horizon is restricted at low level.
   3. Features with great vertical extent are harder to see.
   4. The impression of flying more slowly over the ground distorts the perception of distance.
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   Answer B. The distance to the visible horizon depends on height, so at low level the pilot sees much less of the surrounding country and checkpoints come into view late. Rising ground and trees shorten the view further. Tall features are actually easier to see from low level, and the sensation at low height is one of higher, not lower, speed over the ground.

9. 9Determine the two smallest map scales among the following:
   A) 1:500 000
   B) 10 cm = 10 NM
   C) 5 inch = 10 NM
   D) 1:110 000

   1. B) and C)
   2. A) and B)
   3. A) and D)
   4. C) and D)
   Show the answer

   Answer B. Convert everything to a ratio. B: 10 cm = 10 NM = 1 852 000 cm, i.e. 1:185 200. C: 5 inches = 12.7 cm for 1 852 000 cm, about 1:145 800. The smallest scales are those with the largest denominators: A (1:500 000) and B (1:185 200); D is the largest scale of the four.

10. 10What are the UTC and date when it is 23:00 LMT on 21st August at position A (11°00'N, 067°30'W)?

    1. 18:30 UTC 21st August
    2. 03:30 UTC 21st August
    3. 18:30 UTC 22nd August
    4. 03:30 UTC 22nd August
    Show the answer

    Answer D. Arc to time: 67°30' of longitude is 4 h 30 min. West of Greenwich, local time is behind UTC, so UTC is LMT plus the difference: 23:00 + 4:30 = 03:30 the next day, 22nd August. 18:30 comes from subtracting the longitude, and 03:30 on the 21st forgets that the time has passed midnight.