ATPL(A) 033 Flight Planning & MonitoringRevision notes and practice questions

033 Flight Planning & MonitoringFree sample, 10 questions

1. 1A performance class B aeroplane is to fly a Commercial Air Transport flight to an isolated destination aerodrome. Using the data below, calculate the minimum fuel (to the nearest 10 kg) that must remain on arrival overhead the isolated aerodrome.
   Flight time: 2 hours 45 minutes
   Top of climb to top of descent: 2 hours
   Normal cruise fuel consumption: 540 kg per hour

   1. 630 kg
   2. 570 kg
   3. 1080 kg
   4. 410 kg
   Show the answer

   Answer B. For a CAT flight to an isolated aerodrome with a reciprocating-engine class B aeroplane, the fuel on arrival overhead must cover 45 minutes plus 15% of the planned cruise time, or 2 hours, whichever is less.
   45 min + 15% x 120 min = 63 min, well under 2 hours.
   Fuel = 63/60 x 540 = 567 kg, about 570 kg. Using the full 2 hours gives the 1080 kg trap, and basing the 15% on the whole 2 h 45 min flight time gives the 630 kg figure.

2. 2The minimum climb gradient required for obstacle clearance is 4%. The TAS for best rate of climb is 230 kt. With a tailwind of 15 kt, what is the minimum rate of climb that must be achieved?

   1. 595 ft/min
   2. 995 ft/min
   3. 875 ft/min
   4. 935 ft/min
   Show the answer

   Answer B. The gradient is measured over the ground, so the ground speed applies: 230 + 15 = 245 kt. At 4% the aircraft must gain 4 ft for every 100 ft travelled over the ground, and 245 kt is 245 × 101.3 ≈ 24 820 ft/min along the ground, so the rate of climb needed is 24 820 × 0.04 ≈ 993 ft/min, about 995 ft/min. Using the TAS of 230 kt instead gives about 935 ft/min, which would not guarantee the gradient with a tailwind.

3. 3For a class B multi-engine aeroplane, what is the range calculation based on?

   1. It is applicable for all temperatures
   2. It is based on ground distance including the wind
   3. It is a function of the pressure altitude and power setting
   4. It is valid for all weights
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   Answer C. Class B range charts are entered with the cruise pressure altitude and the chosen power setting, which together fix the TAS and the fuel flow. They are drawn for a stated weight and give still-air range, with a separate allowance for the ISA deviation. They are therefore not valid for every weight or temperature, and they give an air range, not a ground range including wind.

4. 4Given:
   Distance from departure to destination: 1345 NM
   GS out: 480 kt
   GS home: 360 kt
   What is the time to the PET from the departure point?

   1. 96 min
   2. 50 min
   3. 128 min
   4. 72 min
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   Answer D. The distance to the point of equal time is D × H / (O + H) = 1345 × 360 / (480 + 360) ≈ 576 NM from departure. It is flown at the outbound groundspeed, so the time is 576 / 480 = 1.2 h, or 72 minutes. Dividing by the homebound groundspeed instead gives 96 minutes.

5. 5In Item 15 of an ATS flight plan, how should you indicate that you will overfly waypoint TANGO at 350 kt and flight level 280?

   1. TANGO/K0350FL280
   2. TANGO/F280N0350
   3. TANGO/N0350F280
   4. TANGO/T350F280
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   Answer C. A change of speed or level at a point is written as the point, an oblique stroke, then the speed followed by the level with no space. Speed in knots is N plus four figures (N0350) and a flight level is F plus three figures (F280), giving TANGO/N0350F280. K would mean km/h, FL is not a valid prefix, and the level can never come before the speed.

6. 6Using the data below, what is the ground speed?
   TAS: 150 kt
   Track: 090°
   Wind: 050°/46 kt

   1. 112 kt
   2. 160 kt
   3. 188 kt
   4. 128 kt
   Show the answer

   Answer A. The wind from 050° is 40° off the 090° track, from ahead and to the left: headwind = 46 × cos 40° = 35 kt and crosswind = 46 × sin 40° = 30 kt. The drift correction is asin(30 / 150) ≈ 11°, so GS = 150 × cos 11° − 35 = 147 − 35 = 112 kt. Answers above 150 kt treat the headwind as a tailwind.

7. 7At 14:58 ATC instructs you to fly direct to OAL VOR/DME and reduce speed to M 0.76 for traffic separation. Using the data below, what is the revised ETA at OAL?
   Distance to OAL via the flight-planned route: 120 NM
   Direct distance to OAL: 90 NM
   Magnetic track to OAL from present position: 053°
   Variation: 2°E
   Pre-flight planned Mach: 0.80
   Forecast wind: 350°/40 kt
   Actual wind: 045°/65 kt
   OAT: -35°C
   ETA at OAL: 15:08

   1. 15:12
   2. 15:16
   3. 15:14
   4. 15:10
   Show the answer

   Answer A. At −35°C the speed of sound is about 601 kt, so M 0.76 is a TAS of 457 kt. The track to OAL is 053°M, i.e. 055°T, and the actual wind from 045° at 65 kt is almost on the nose: about 64 kt of headwind, giving a GS of about 393 kt. The 90 NM direct distance takes 90 / 393 × 60 ≈ 14 min, so the revised ETA is 14:58 + 0:14 = 15:12. The planned Mach, the forecast wind and the flight-planned distance are no longer relevant.

8. 8Can a road junction or a railway line be entered as a waypoint in a Global Navigation Satellite System (GNSS) flight plan?

   1. Yes, by inserting the coordinates or referencing to an existing GNSS point.
   2. No, GNSS waypoints cannot be created by the pilot.
   3. Yes, but only with prior approval from the national authority.
   4. Yes, by inserting the coordinates of the airport reference point, then bearing and distance from it.
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   Answer A. GNSS navigators allow user-defined waypoints, so a road junction or railway feature can be entered by typing its latitude and longitude, or by defining it relative to an existing waypoint in the database (place/bearing/distance). No authority permission is needed, and there is no requirement to build it from an aerodrome reference point.

9. 9Given:
   True course (TC): 017°
   Wind: 160°/25 kt
   True airspeed (TAS): 180 kt
   Find the wind correction angle (WCA) and the ground speed (GS).

   1. WCA +5°, GS 199 kt.
   2. WCA +5°, GS 161 kt.
   3. WCA -5°, GS 199 kt.
   4. WCA -5°, GS 161 kt.
   Show the answer

   Answer A. The wind from 160° is 143° off the true course of 017°, coming from behind and to the right. The crosswind is 25 × sin 143° = 15 kt from the right, so the wind correction angle is asin(15/180) ≈ 5° to the right, +5°. The tailwind is 25 × cos 37° = 20 kt, so the ground speed is 180 × cos 5° + 20 ≈ 199 kt. 161 kt treats the tailwind as a headwind, and -5° corrects the wrong way.

10. 10What is the difference between an RNAV STAR and an RNAV STAR overlay?

    1. The RNAV STAR overlay follows a longer route than the RNAV STAR.
    2. The RNAV STAR overlay permits pilots to fly a conventional STAR procedure using GNSS equipment.
    3. The RNAV STAR overlay is a shorter route than the RNAV STAR.
    4. The RNAV STAR overlay will follow a different route from the conventional STAR procedure.
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    Answer B. An RNAV STAR overlay is a conventional arrival, defined by ground-based aids, that has been coded so that it can be flown with area navigation (GNSS or the FMS) along exactly the same path. It is therefore neither longer, shorter nor different from the conventional procedure. A pure RNAV STAR, in contrast, is designed for area navigation from the start and need not match any conventional route.