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Given:

Distance from departure to destination: 315 NM
True track: 343°
W/V: 015/15
TAS: 100 kt

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

  • A
    176 NM
  • B
    148 NM
  • C
    167 NM
  • D
    139 NM

Refer to figure.
Note: As per the latest EASA syllabus, no questions address this learning objective. However, they are still part of the UK CAA syllabus, as also confirmed by recent feedback.


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

1. Set wind direction to 015º under the "TRUE HEADING" index at the top.
2. Set the center point on the True Airspeed (TAS) of 100 kt.
3. Mark the wind velocity 15 kt down from the centre point.
4. Initially, set the True Track under the "TRUE HEADING" index.

  • GS home: 163º

5. Note that this heading would result in 4ºR drift and a track of 167º.
6. Reduce the heading value under the index until the heading plus the drift gives a track of 163º. This occurs at a heading of 159º with 5ºR drift.
7. The groundspeed for this track is approximately 112 kt.

  • GS out: 343º

5. Note that this heading would result in 5ºL drift and a track of 338º.
6. Reduce the heading value under the index until the heading minus the drift gives a track of 343º. This occurs at a heading of 348º with 5ºL drift.
7. The groundspeed for this track is approximately 88 kt.


We can now apply the formulas:
Distance to PET = (GS home x Distance) / (GS out + GS home)
Distance to PET = (112 x 315) / (88 + 112)
Distance to PET = 176 NM

Time to PET = Distance to PET / GS out

Your Notes (not visible to others)



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

  • United Kingdom
    1