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DGCA Air Navigation · 99 questions

Descent Planning: DGCA Air Navigation Questions and Answers

99 practice questions on Descent Planning with answers and explanations, plus the key concepts and formulas, from the TrueHeading question bank.

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About this topic

Use this page to revise Descent Planning for the DGCA Air Navigation paper. Below are the key ideas first, then 12 practice questions with answers and explanations, picked from the 99 questions in the TrueHeading bank for this topic.

In short: Use the 3-to-1 rule for a 3° path.

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Key concepts: Descent Planning

4 ideas to know cold.

How far before the destination should you start a descent?

Use the 3-to-1 rule for a 3° path.

D(NM)≈Δh(ft)1000×3D(\text{NM})\approx\frac{\Delta h(\text{ft})}{1000}\times 3
  • From FL 330 down to 3000 ft: 30 000 ft ÷ 1000 × 3 = 90 NM
  • Add distance to slow down, and allow for wind
height to lose3°distance to gostart of descent3° ≈ 318 ft per NM ≈ 5.2 %quick rule: 3 NM per 1000 ft

What rate of descent holds a 3° path?

About 5 times the ground speed (more precisely 5.3 ×).

ROD(ft/min)≈GS(kt)×5ROD(\text{ft/min})\approx GS(\text{kt})\times 5
  • GS 140 kt → ROD ≈ 700 ft/min
  • A tailwind raises GS and needs a higher rate

Descent gradient from rate and speed?

ft per NM = rate of descent ÷ ground speed × 60.

ft/NM=RODGS×60\text{ft/NM}=\frac{ROD}{GS}\times 60
  • 1000 ft/min at 120 kt = 500 ft/NM (too steep for a 3° path)

How do you calculate the top of descent distance?

Distance (NM) = height to lose (ft) / 300 for a 3° path.

Dist (NM)≈Δh (ft)300\text{Dist (NM)}\approx\dfrac{\Delta h\ (\text{ft})}{300}
  • FL350 to 3 000 ft: 32 000 / 300 ≈ 107 NM
  • Common cockpit rule: about 3 NM per 1 000 ft to lose (3 × 32 = 96 NM)
  • Allow extra distance to decelerate

Practice questions: Descent Planning

Tap “Show answer” after you have tried each one.

Q1To fly a 3° descent path at a groundspeed of 440 kt, the rate of descent is approximately:

  1. 1750 ft/min
  2. 2920 ft/min
  3. 1400 ft/min
  4. 2340 ft/min
Show answer

Answer: D. 2340 ft/min

ROD = GS × 101.3 × tan 3° ≈ 440 × 5.3 ≈ 2335 ft/min (rule of thumb: 5 × GS).

Q2To fly a 3° descent path at a groundspeed of 260 kt, the rate of descent is approximately:

  1. 1590 ft/min
  2. 1380 ft/min
  3. 1030 ft/min
  4. 1930 ft/min
Show answer

Answer: B. 1380 ft/min

ROD = GS × 101.3 × tan 3° ≈ 260 × 5.3 ≈ 1380 ft/min (rule of thumb: 5 × GS).

Q3To fly a 3° descent path at a groundspeed of 400 kt, the rate of descent is approximately:

  1. 1590 ft/min
  2. 3400 ft/min
  3. 2650 ft/min
  4. 2120 ft/min
Show answer

Answer: D. 2120 ft/min

ROD = GS × 101.3 × tan 3° ≈ 400 × 5.3 ≈ 2123 ft/min (rule of thumb: 5 × GS).

Q4An aircraft must descend from 37 000 ft to 2 000 ft on a 3° profile. The distance required is approximately:

  1. 126 NM
  2. 99 NM
  3. 82 NM
  4. 110 NM
Show answer

Answer: D. 110 NM

A 3° path loses ≈ 318 ft per NM (≈ 3 NM per 1 000 ft): 35 000 ÷ 318 ≈ 110 NM.

Q5An aircraft must descend from 35 000 ft to 3 000 ft on a 3° profile. The distance required is approximately:

  1. 101 NM
  2. 85 NM
  3. 161 NM
  4. 75 NM
Show answer

Answer: A. 101 NM

A 3° path loses ≈ 318 ft per NM (≈ 3 NM per 1 000 ft): 32 000 ÷ 318 ≈ 101 NM.

Q6An aircraft must descend from 24 000 ft to 4 000 ft on a 3° profile. The distance required is approximately:

  1. 63 NM
  2. 47 NM
  3. 88 NM
  4. 79 NM
Show answer

Answer: A. 63 NM

A 3° path loses ≈ 318 ft per NM (≈ 3 NM per 1 000 ft): 20 000 ÷ 318 ≈ 63 NM.

Q7Using the 3° rule (3 nm per 1 000 ft), the distance needed to descend from FL360 to 0 ft is about:

  1. 216 nm
  2. 36 nm
  3. 108 nm
  4. 162 nm
Show answer

Answer: C. 108 nm

Height to lose = 36000 ft = 36 thousand ft. 3 × 36 = 108 nm.

Q8Using the 3° rule (3 nm per 1 000 ft), the distance needed to descend from FL290 to 0 ft is about:

  1. 174 nm
  2. 29 nm
  3. 130 nm
  4. 87 nm
Show answer

Answer: D. 87 nm

Height to lose = 29000 ft = 29 thousand ft. 3 × 29 = 87 nm.

Q9To fly a 3° descent path at a ground speed of 320 kt, the rate of descent should be approximately:

  1. 1600 ft/min
  2. 3200 ft/min
  3. 64 ft/min
  4. 960 ft/min
Show answer

Answer: A. 1600 ft/min

ROD (ft/min) ≈ GS × 5 for 3° (300 ft per nm). 320 × 5 = 1600 ft/min.

Q10An aircraft descends from FL360 to 3000 ft at 2000 ft/min with an average ground speed of 360 kt. The ground distance covered in the descent is about:

  1. 99 nm
  2. 69 nm
  3. 129 nm
  4. 121 nm
Show answer

Answer: A. 99 nm

Time = 33000/2000 = 16.5 min. Distance = 360 × 16.5/60 = 99 nm.

Q11An aircraft descends from FL320 to 3000 ft at 2500 ft/min with an average ground speed of 440 kt. The ground distance covered in the descent is about:

  1. 111 nm
  2. 107 nm
  3. 60 nm
  4. 85 nm
Show answer

Answer: D. 85 nm

Time = 29000/2500 = 11.6 min. Distance = 440 × 11.6/60 = 85 nm.

Q12How far (in nm) from the threshold is the 5.0° glide path at a height of 2000 ft above threshold elevation?

  1. 3.8 nm
  2. 6.7 nm
  3. 5.6 nm
  4. 1.9 nm
Show answer

Answer: A. 3.8 nm

Distance = height ÷ tan(angle) = 2000 ÷ (6076 × tan 5.0°) ≈ 3.8 nm.

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