033 Flight Planning and Monitoring topic guide
Point of Equal Time (PET)
The point of equal time (PET, sometimes called the critical point) is the position along track from which it takes the same time to continue to the destination as to turn around and fly back to the departure point. It matters operationally for decisions such as which airfield to head for after a failure, and it is one of the most reliably examined calculations in ATPL Flight Planning.
In still air the PET sits exactly halfway. Wind moves it, and always upwind: with a tailwind outbound the PET moves back towards the departure point, and with a headwind outbound it moves forward past the halfway mark. If your computed PET has moved downwind, you have applied the wind the wrong way around.
The formula and why it works
Call the total distance D, the groundspeed onwards from the PET to the destination O, and the groundspeed home from the PET back to departure H. The distance from departure to the PET is X = D multiplied by H, divided by (O plus H).
The logic is a straight time balance. Time to continue is (D minus X) over O, and time to return is X over H. Setting them equal and rearranging gives the formula. Notice that H, the groundspeed home, sits in the numerator: the stronger the groundspeed home, the further along track the PET can sit while still keeping the return time equal.
The wind is the whole game. A component that is a tailwind on the way out becomes a headwind on the way back, so O and H are built from the same TAS but with the wind component applied in opposite senses. Exam questions are engineered so that forgetting this reversal produces one of the wrong options exactly.
Engine failure variants
Many stems ask for the PET assuming an engine fails at that point. The aircraft then continues or returns at the reduced, single-engine TAS, so both O and H must be recalculated from the single-engine TAS with the wind components applied. The distance D and the wind do not change; only the TAS feeding O and H does.
Keep the two speed regimes separate: normal-operations TAS gets you to the PET (that part only affects the time to reach it), while the failure-case TAS defines O and H inside the formula. Mixing the two is the second most common error after the wind reversal.
Worked example
Worked example: PET with a tailwind outbound
Not quite. The correct answer is B.
Correct answer: B
- A. Halfway. This ignores the wind entirely; the PET only sits at the midpoint in still air.
- B. Correct: X = 480 multiplied by 200 over (280 plus 200) = 200 NM. With a tailwind outbound the PET moves back towards A, and 200 NM is upwind of halfway.
- C. This puts the groundspeed onwards (280 kt) in the numerator instead of the groundspeed home. Remember: groundspeed home on top.
- D. This applies the wind outbound but forgets to reverse it for the return leg (using H = 240 instead of 200), giving 480 x 240 / 520, or about 222 NM.
Step by step
- Groundspeed onwards O: 240 kt TAS plus the 40 kt tailwind gives 280 kt.
- Groundspeed home H: the same wind is now a headwind, so 240 minus 40 gives 200 kt.
- Apply the formula: X = D x H / (O + H) = 480 x 200 / 480 = 200 NM from A.
- Sanity check: continuing from the PET covers 280 NM at 280 kt in 1 hour, and returning covers 200 NM at 200 kt in 1 hour. The times match, so the answer is consistent.
Common mistakes
Applying the wind to one leg only
The same wind component must be added to the TAS in one direction and subtracted in the other. Using the plain TAS for the return leg shifts the PET downwind and lands exactly on a distractor.
Swapping O and H in the formula
Groundspeed home belongs in the numerator. A quick check: with a tailwind outbound, your answer must be less than half the distance. If it is more, you swapped them.
Using normal-operations TAS in a failure-case PET
When the stem says an engine fails at the PET, both onward and return groundspeeds come from the single-engine TAS. Using the all-engines TAS makes the answer plausible but wrong.
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Last reviewed July 2026