Intercept a track with controlled closure

How long does an intercept take?

The cross-track component of ground velocity controls closure.

Close the offset with ground velocity

For a straight desired track, the perpendicular component of ground velocity determines how fast an offset closes. If 6 NM right of track, an actual ground intercept angle of 30° left at 180 kt GS gives closure 180 sin 30° = 90 kt. Time to the line is 6/90 h = 4 minutes. Along-track advance is 180 cos 30° × 4/60 = 10.39 NM.

The angle must be the actual ground-track difference. A 30° heading difference can produce a different intercept angle in crosswind. Solve the wind triangle or establish the ground track before using the closure formula. An angle pointing away from the line increases the offset instead of capturing it.

Understand the zero-angle limit

Parallel motion has zero cross-track closure, so an existing offset is never removed. As a small intercept angle approaches zero, time and distance to intercept grow rapidly. At 90°, closure equals GS and along-track advance is zero, but that does not make a perpendicular intercept suitable for every route or procedure.

A steeper intercept leaves less time to recognise capture and roll out. The laboratory assumes an instantaneous track change and does not include turn radius, display lag or airspace protection. Actual capture must use the applicable aircraft guidance and clearance constraints.

Solve for the angle from a time requirement

To remove a 6 NM offset in 6 minutes at 180 kt GS, required cross-track closure is 60 kt. Thus sin α = 60/180 and α ≈ 19.47°. If the requested time implied closure greater than GS, no real angle could satisfy it. This inverse task makes the physical domain explicit. Before accepting a numeric answer, check that the proposed track points toward the line and that the available along-track distance reaches a suitable capture point.

Why begin the capture turn before reaching the line?

During a finite-radius turn, the aircraft still moves sideways.

A finite turn continues to move sideways

For a coordinated level turn in still air, R = V²/(g tan bank), with speed in consistent units. At 180 kt TAS and 25° bank, R is about 1.01 NM. A 90° constant-speed turn therefore uses roughly 1 NM of lateral displacement. Wind makes the ground path different from a circular still-air path.

For a turn from intercept angle α onto a parallel track, lateral movement in the ideal circular model is R(1 − cos α). With R = 1 NM and α = 30°, this is 0.134 NM. Starting the capture turn exactly on the desired line therefore produces an overshoot under this idealisation.

Tracking is a separate phase after capture

The aircraft must finish with the wind-corrected heading that maintains the desired ground track. Pointing parallel to the printed line without accounting for wind can restart drift. A CDI trend may support anticipation, but only with valid guidance and an understood angular or linear sensitivity.

The simple model assumes constant TAS, immediate bank establishment and no wind. It excludes roll-in, roll-out, automation behaviour and approved procedure tolerances. Use it to understand why speed and bank affect capture lead, not to generate a flight-director command or a universal operational lead distance.

Compare speed and bank changes

At the same bank, doubling TAS multiplies radius by four. A 30° intercept capture that needed about 0.134 NM in the R = 1 NM example would need about 0.536 NM at R = 4 NM. Increasing bank reduces radius through tan bank, but bank is constrained by aircraft limits, comfort and the phase of flight. The controls make these dependencies visible. Explain why a capture technique that worked at low speed may overshoot when reused at a much higher speed without changed anticipation.

Can you explain and apply the mechanism?

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