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Given:

Distance from departure to destination: 2800 NM
True track: 140°
W/V: 140/100
TAS: 500 kt

What is the distance and time of the PET from the departure point?

  • A
    Distance: 1680 NM Time: 252 min
  • B
    Distance: 1120 NM Time: 134 min
  • C
    Distance: 1400 NM Time: 168 min
  • D
    Distance: 1120 NM Time: 112 min

Refer to figure.
Note: Since the terms PET and PSR are not present in EASA Air Ops, these questions are less likely to appear in the exam.

Solving from Heading (HDG) & Ground Speed (GS), knowing WV, TAS and required track.

1. Set wind direction to 140º under the "TRUE HEADING" index at the top.
2. Set the center point on the True Airspeed (TAS) of 500 kt.
3. Mark the wind velocity 100 kt down from the centre point.
4. Initially, set the True Track under the "TRUE HEADING" index.

  • GS home: 320º

5. The groundspeed for this track is approximately 600 kt.

  • GS out: 140º

5. The groundspeed for this track is approximately 400 kt.

We can now apply the formulas:
Distance to PET = (GS home x Distance) / (GS out + GS home)
Distance to PET = (600 x 2800) / (400 + 600)
Distance to PET = 1680 NM

Time to PET = Distance to PET / GS out 
Time to PET = 1680 NM / 400 kt = 4.2 h = 252 min

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This question has appeared on the real examination, you can find the related countries below.

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