Given the following information, calculate the distance from the Equal Time Point (ETP) to position B:

Distance from A to B: 360 NM

Wind component from A to B: -15 kt

Wind component from B to A: +15 kt

TAS: 180 kt

The *Point of Equal Time (PET)* allows a pilot to decide which is the quickest landing area to get to and is the position of a point on a track, where it is as quick to go on, as it is to turn back. The PET can be also called *Equal Time Point (ETP)* or *Critical Point (CP)*.

The formula to calculate the position of the PET is:

*Distance to PET = (D x H) / (O + H)*

Where:

*D: Sector Distance**O: Groundspeed Out**H: Groundspeed Home*

1. The Groundspeed Out (O) is: *TAS - 15 kt = 165 kt.*

2. The Groundspeed Home (H) is: *TAS + 15 kt = 195 kt.*

3. Thus, the distance from A to PET is: *(360 NM x 195 kt) / (165 kt + 195 kt) = 195 NM*.

But, the question asks for the **distance from PET to B**, i.e. what is the remaining distance after passing the PET.

**distance from PET to B**is:

*Distance A to B - Distance to PET = 360 NM - 195 NM =*

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**165 NM**Your Notes (not visible to others)

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