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DGCA Air Navigation · 106 questions

Critical Point & PNR: DGCA Air Navigation Questions and Answers

106 practice questions on Critical Point & PNR 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 Critical Point & PNR for the DGCA Air Navigation paper. Below are the key ideas first, then 12 practice questions with answers and explanations, picked from the 106 questions in the TrueHeading bank for this topic.

In short: The point on a route from which the flight time to continue to the destination equals the time to return to the departure (or alternate) airfield.

Open the full set in the student zone to answer every question, track your accuracy and take timed mock tests.

Key concepts: Critical Point & PNR

5 ideas to know cold.

What is the critical point (CP) or equal time point (ETP)?

The point on a route from which the flight time to continue to the destination equals the time to return to the departure (or alternate) airfield.

Distance to CP=D×GSbackGSout+GSback\text{Distance to CP}=\frac{D\times GS_{back}}{GS_{out}+GS_{back}}
Time to CP=DGSout+GSback\text{Time to CP}=\frac{D}{GS_{out}+GS_{back}}
  • D is the whole distance between the two airfields
  • The CP moves into the wind

What is the point of no return (PNR)?

The last point from which you can still return to the departure aerodrome using the fuel carried (safe endurance E).

Time to PNR=E×GSbackGSout+GSback\text{Time to PNR}=\frac{E\times GS_{back}}{GS_{out}+GS_{back}}
Distance to PNR=GSout×time to PNR\text{Distance to PNR}=GS_{out}\times\text{time to PNR}
  • Based on fuel, whereas CP is based on time
  • A tailwind outbound brings the PNR further out

CP versus PNR: what is the difference?

CP: equal time to two airfields. PNR: last point of return using fuel.

  • CP depends only on speeds and distance
  • PNR depends on safe endurance as well
  • Both are worked with outbound and homebound ground speeds

What is the point of no return (PNR) and its formula?

The last point from which you can return to the departure airfield with the fuel on board.

TPNR=E×HO+HT_{PNR}=\dfrac{E\times H}{O+H}
  • Endurance E = usable fuel / fuel flow
  • Time to PNR T = E × H / (O + H) where O = outbound GS, H = homebound GS
  • Distance to PNR = T × O

Tip: The wind is the key: a tailwind out means a headwind home, so the PNR moves closer.

What is the critical point (CP) or equal time point?

The point from which the time to continue to the destination equals the time to return to the departure.

Dist to CP=D×HO+H\text{Dist to CP}=\dfrac{D\times H}{O+H}
  • Time to CP = D × H / (O + H) where D = total distance
  • Distance to CP = D × H / (O + H)
  • With no wind CP is the midpoint

Practice questions: Critical Point & PNR

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

Q1Distance A to B is 1400 NM. Groundspeed out 420 kt, groundspeed back 470 kt. The distance from A to the point of equal time (PET) is:

  1. 850 NM
  2. 739 NM
  3. 628 NM
  4. 554 NM
Show answer

Answer: B. 739 NM

PET = D × Gh ÷ (Go + Gh) = 1400 × 470 ÷ 890 = 739 NM.

Q2Distance A to B is 1700 NM. Groundspeed out 380 kt, groundspeed back 310 kt. The distance from A to the point of equal time (PET) is:

  1. 764 NM
  2. 955 NM
  3. 840 NM
  4. 878 NM
Show answer

Answer: A. 764 NM

PET = D × Gh ÷ (Go + Gh) = 1700 × 310 ÷ 690 = 764 NM.

Q3Distance A to B is 1100 NM. Groundspeed out 340 kt, groundspeed back 400 kt. The distance from A to the point of equal time (PET) is:

  1. 832 NM
  2. 446 NM
  3. 535 NM
  4. 595 NM
Show answer

Answer: D. 595 NM

PET = D × Gh ÷ (Go + Gh) = 1100 × 400 ÷ 740 = 595 NM.

Q4Safe endurance is 6 hours. Groundspeed out is 330 kt and groundspeed home is 470 kt. The distance to the point of no return (PNR) is:

  1. 1163 NM
  2. 1629 NM
  3. 698 NM
  4. 1280 NM
Show answer

Answer: A. 1163 NM

PNR distance = E × Go × Gh ÷ (Go + Gh) = 6 × 330 × 470 ÷ 800 = 1163 NM.

Q5Safe endurance is 9 hours. Groundspeed out is 420 kt and groundspeed home is 390 kt. The distance to the point of no return (PNR) is:

  1. 1547 NM
  2. 1820 NM
  3. 1638 NM
  4. 2093 NM
Show answer

Answer: B. 1820 NM

PNR distance = E × Go × Gh ÷ (Go + Gh) = 9 × 420 × 390 ÷ 810 = 1820 NM.

Q6Safe endurance is 8 hours. Groundspeed out is 490 kt and groundspeed home is 370 kt. The distance to the point of no return (PNR) is:

  1. 1518 NM
  2. 1687 NM
  3. 1434 NM
  4. 1939 NM
Show answer

Answer: B. 1687 NM

PNR distance = E × Go × Gh ÷ (Go + Gh) = 8 × 490 × 370 ÷ 860 = 1687 NM.

Q7A flight of 1200 nm from A to B is flown at TAS 220 kt with a wind that gives ground speeds of 180 kt on-track and 260 kt back-track. The critical point (equal time point) is how far from A?

  1. 491 nm
  2. 709 nm
  3. 744 nm
  4. 600 nm
Show answer

Answer: B. 709 nm

CP = D × GS(home) ÷ (GS(out) + GS(home)) = 1200 × 260 ÷ (180 + 260) = 709 nm from A.

Q8A flight of 850 nm from A to B is flown at TAS 360 kt with a wind that gives ground speeds of 330 kt on-track and 390 kt back-track. The critical point (equal time point) is how far from A?

  1. 460 nm
  2. 390 nm
  3. 495 nm
  4. 425 nm
Show answer

Answer: A. 460 nm

CP = D × GS(home) ÷ (GS(out) + GS(home)) = 850 × 390 ÷ (330 + 390) = 460 nm from A.

Q9Distance A to B is 1000 nm; ground speed outbound 340 kt and homebound 360 kt. How long does it take to reach the critical point from A?

  1. 91 min
  2. 105 min
  3. 88 min
  4. 176 min
Show answer

Answer: A. 91 min

CP distance = 1000×360/(340+360) = 514 nm. Time = 514/340 h ≈ 91 min.

Q10Distance A to B is 400 nm; ground speed outbound 480 kt and homebound 420 kt. How long does it take to reach the critical point from A?

  1. 50 min
  2. 32 min
  3. 23 min
  4. 37 min
Show answer

Answer: C. 23 min

CP distance = 400×420/(480+420) = 187 nm. Time = 187/480 h ≈ 23 min.

Q11Endurance is 4.5 h. Ground speed outbound 380 kt, return 320 kt. The time to the point of no return is:

  1. 135 min
  2. 139 min
  3. 123 min
  4. 147 min
Show answer

Answer: C. 123 min

T = E × GS(home) ÷ (GS(out)+GS(home)) = 4.5 × 320 ÷ 700 h = 2.06 h = 123 min.

Q12Endurance is 5.5 h. Ground speed outbound 420 kt, return 380 kt. The time to the point of no return is:

  1. 144 min
  2. 173 min
  3. 165 min
  4. 157 min
Show answer

Answer: D. 157 min

T = E × GS(home) ÷ (GS(out)+GS(home)) = 5.5 × 380 ÷ 800 h = 2.61 h = 157 min.

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