Your Pilot InterviewATPL EASA

Track error, closing angle and the 1-in-60 rule

Turn a known position error into a justified correction.

What does the 1-in-60 rule approximate?

A small angle in radians is approximately lateral error divided by distance flown.

Estimate the track error behind you

For a small angular error, angle in degrees is approximately 60 × off-track distance / distance flown. If a reliable fix shows 4 NM right of track after 80 NM, the estimate is 3° right. The exact local angle is atan(4/80) = 2.86°; 60 is a convenient approximation to 57.3 degrees per radian.

Cancelling only that past 3° error stops further divergence if the cause is a steady wind difference. It does not remove the existing 4 NM offset. A closing angle is additionally needed to reach the intended future point. The distinction separates correcting the velocity error from correcting the position error.

Use distance remaining for the closing angle

With 40 NM remaining, closing angle is approximately 60 × 4/40 = 6° left. A total heading change of about 9° left from the old heading cancels the 3° rightward error and adds the 6° return slope. This assumes stable TAS and wind and a small local-angle geometry.

The exact sum of local angles is atan(4/80) + atan(4/40) ≈ 8.57°. Large offsets or short remaining distance make the approximation less satisfactory. A wrong station, bearing reference or heading input should be repaired directly rather than disguised as a wind-induced track error.

Invert the closing problem

Suppose the available closing angle is limited to 5° and the offset is 3 NM. The approximate distance needed to regain the line is 60 × 3/5 = 36 NM. If only 15 NM remain before the intended fix, that selected return angle cannot achieve the geometry. Use a new valid route or a larger permissible intercept as appropriate to the operational constraints. The formula describes a local path, not a permission to leave protected airspace; confirm terrain, clearance and route limits before acting on it.

When does doubling track error regain the route?

A symmetric construction works only with the stated geometry and steady conditions.

The doubled angle creates symmetric slopes

Suppose the original heading produced a 3° right track error over 40 NM. A heading change of approximately 6° left then gives a ground track about 3° left of the intended line under steady conditions. It removes the accumulated offset over about another 40 NM. Once back on the line, reduce to the 3° left correction needed to prevent renewed drift.

The extra half of the doubled correction is a temporary closing component. Keeping all 6° after regaining the line would begin a new error to the left. The method assumes small angles, a local straight route, stable TAS and wind, and enough remaining distance to complete the symmetric return.

Check whether the construction fits the task

If only 10 NM remain after the initial 40 NM error-building distance, a doubled correction is insufficient to reach the intended fix. Calculate the actual closing angle from offset and distance remaining instead. A new route direction can also change the wind components, so an exact fresh triangle may be preferable.

First confirm that the apparent offset is real and that the return path is clear and permitted. Navigation geometry does not authorise an intercept from outside protected airspace. Terrain, airspace, ATC clearance, fuel and the available navigation accuracy can control the practical choice.

Predict the return fix before turning

With 3° track error over 40 NM, the small-angle offset is about 2 NM. After a doubled correction, predict the crossing of the intended line about 40 NM later. At 160 kt GS that is 15 minutes. A useful return plan therefore contains an angle, expected distance and time, plus a reliable feature or position source to confirm capture. If the check does not agree, reassess the assumptions instead of retaining the correction indefinitely. The animation compares the approximate angles with exact local geometry so that the approximation’s practical limits remain visible.

Can you explain and apply the mechanism?

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

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