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Given:

Distance from departure to destination: 165 NM
True track: 055°
W/V: 360/20
TAS: 105 kt

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

  • A
    73 NM
  • B
    132 NM
  • C
    83 NM
  • D
    92 NM
Refer to figure.

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

1. Set wind direction to 360º under the "TRUE HEADING" index at the top.
2. Set the center point on the True Airspeed (TAS) of 105 kt.
3. Mark the wind velocity 20 kt down from the centre point.
4. Initially, set the True Track under the "TRUE HEADING" index.

  • GS home: 055º
5. Note that this heading would result in 10ºR drift and a track of 065º.
6. Reduce the heading value under the index until the heading plus the drift gives a track of 055º. This occurs at a heading of 045º with 10ºR drift.
7. The groundspeed for this track is approximately 92 kt.
  • GS out: 235º

5. Note that this heading would result in 8ºL drift and a track of 227º.
6. Reduce the heading value under the index until the heading minus the drift gives a track of 235º. This occurs at a heading of 243º with 9ºL drift.
7. The groundspeed for this track is approximately 116 kt.

We can now apply the formulas:
Distance to PET = (GS home x Distance) / (GS out + GS home)
Distance to PET = (116 x 165) / (92 + 116)
Distance to PET = 92 NM

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