Find the point of equal time

Why is the equal-time point not usually halfway?

It is where time to continue equals time to return under the stated conditions.

Set the two times equal

Let total route distance be D and distance from departure to PET bex. With homeward groundspeed H and onward groundspeed O after the stated event, return time isx/H and onward time is(D−x)/O. Equating them givesx = D H/(O+H). Use the speeds appropriate after the event, not automatically the normal outbound cruise speed.

If homeward GS is lower, less homeward distance is needed for equal time, so the PET moves towards departure. If onward GS is lower, the PET moves towards destination. Equal speeds produce the halfway point, which is a useful check on the formula.

Keep the operating condition explicit

A PET can be calculated for normal cruise, an engine failure, depressurisation or another defined scenario. The event can change TAS, altitude, wind, configuration and fuel flow. State the condition before selecting the two ground speeds. A normal-cruise PET may be irrelevant to a low-level degraded profile.

The route need not be a straight geographical line in a more complete plan. The simplified formula assumes the specified distances share one route axis and constant speeds. Real operational methods can require differing routing, terrain constraints or more detailed segment integration.

Explain what equal time does not mean

PET does not prove equal fuel to the two aerodromes when fuel flows differ, and does not by itself establish that either aerodrome is suitable. It is a comparison of time from a point under supplied conditions. Fuel, weather, terrain and capability remain additional decisions.

The laboratory moves the equal-time point as the two speeds change and shows both computed times. Predict its direction before moving the sliders, then check the equality numerically. This is more transferable than memorising whether a “headwind moves PET forward”, because the answer depends on which branch has which groundspeed.

What can an unexpected PET tell you about the speed assumptions?

Rearrange the equality to test or recover one branch speed.

Recover a missing speed

Fromx/H = (D−x)/O, onward GS O = H(D−x)/x. If a 500 NM route has PET at 300 NM and homeward GS150 kt, onward GS must be 100 kt under the constant-speed model. This inverse calculation checks whether a plotted PET and its speed inputs agree.

The ratiox/(D−x) equals H/O. A PET two thirds of the way along the route means homeward GS is twice onward GS, not half. The longer equal-time branch must have the higher speed.

Distinguish reaching the point from leaving it

Time from departure to the PET can use the planned pre-event outbound GS if the scenario specifies that the event occurs there. Time from PET to either aerodrome uses the post-event branch speeds. Mixing these stages can produce a correct PET location but an incorrect decision clock time or fuel state.

If the route wind changes with position or the branches use different levels, constant ground speeds may be inadequate. Apply the supplied segmented method or planning tool. The algebra here teaches one transparent case and provides a check, rather than replacing a more detailed operational calculation.

Compare with the fuel point separately

A point of safe return is constrained by usable task fuel or endurance. A PET is constrained by equality of branch times. They can occur at different locations and can move differently after a changed fuel state. Do not choose whichever point is farther along the route and call it the other.

In the laboratory, hold route distance fixed and increase only homeward GS. The PET moves away from departure because a longer return distance can now take the same time as the onward branch. Explain that cause using the two time fractions, then verify the inverse relationship with the supplied numerical exercise.

Can you explain and apply the mechanism?

Four original questions. Work from the stated facts, then compare your reasoning with the explanation.