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DR position after a forecast wind that was wrong

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DR position after a forecast wind that was wrong

You departed A at 10:00 UTC with a planned TAS of 120 kt on a magnetic track of 085° (variation 12°E, deviation +2° on the current heading). Forecast wind was 340/20. At 10:36 you pass a prominent lake inlet that you identify with high confidence; it lies 66 NM along the planned track. Your heading has been constant, and you have made no heading changes. The next planned turning point B is 48 NM further along track, over featureless terrain, with a 12 NM wide band of low cloud reported ahead.

1. Compute the actual wind velocity and the heading you should now steer to reach B.

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Planned figures. Planned magnetic heading = 085° − 12° (E variation) − 2° (deviation) = 071°M... more precisely, true heading = magnetic heading + variation, so with track 085°M and wind 340/20, first find the planned heading.

Planned: track 085°M. Wind from 340°M is 105° off track to the left. Crosswind = 20 × sin105° ≈ 19.3 kt; headwind = 20 × cos105° ≈ −5.2 kt (i.e. a slight tailwind). Required heading offset = arcsin(19.3/120) ≈ 9.3° into wind, so heading ≈ 085° − 9.3° ≈ 076°M. GS ≈ 120 × cos9.3° + 5.2 ≈ 118.4 + 5.2 ≈ 123.6 kt. Planned time to the inlet = 66/123.6 ≈ 32 min, so you should have arrived at 10:32.

Actual. You arrived at 10:36, 4 min late, having covered 66 NM in 0.60 h: actual GS = 110 kt. You have stayed on track (the inlet is on track), so the wind has no crosswind component relative to your heading - it is a pure headwind. Actual headwind = 123.6 − 110 ≈ 13.6 kt more than planned, i.e. actual wind ≈ 340/34 (approximately, with the same direction).

To B. Remaining 48 NM. With the same wind, GS ≈ 120 − 13.6 − 5.2 ≈ 101 kt (recompute: TAS 120, headwind component ≈ 19 kt → GS ≈ 101 kt). Time ≈ 48/101 ≈ 28.5 min. Keep heading 076°M; no drift correction is needed because you are on track.

2. Cloud is reported ahead and the terrain is featureless. What is your decision logic for continuing versus turning back, and what navigation method do you use if you continue?

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Decision logic:

  • VFR cloud clearance is the hard limit. If the band ahead would force you below safe terrain clearance or into IMC, turn back now while you still have the inlet as a positive fix and 36 minutes of known ground behind you.
  • Compute the escape: returning to A is 66 NM with a tailwind of about 19 kt, GS ≈ 139 kt, about 28 minutes. Continuing to B is 28 minutes with no known feature to fix on. Turning back is the lower-risk option if the cloud is not clearly passable.
  • Do not press on hoping for a gap over featureless terrain: without a fix you cannot tell whether you have drifted, and the cloud removes the horizon and the ground references you would need.

If you continue:

  • Fly a constant heading and use time-and-distance dead reckoning from the inlet fix, with a written log of heading, time and expected position every 10 NM.
  • Use any available navaid or GNSS as a cross-check, but keep DR as primary so a failure does not leave you lost.
  • Set a decision point: if B is not positively identified within 5 minutes of the revised ETA (about 10:36 + 28.5 min ≈ 11:05), turn back or divert to the nearest known VFR airfield rather than searching.

3. What does the 4-minute late arrival tell you about the forecast, and how should you use it for the return leg?

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The forecast wind was 14 kt too weak in the headwind component, a 12% error in groundspeed. That is within normal forecast error but large enough to matter on a 340 NM leg: uncorrected, it would put you 20 NM short at the destination after 3 hours.

For the return leg, do not simply reverse the forecast. The actual wind you measured (about 340/34) is a better estimate than the original forecast for the near term, but it will change with time and altitude. Use it as the initial assumption, then update it at every positive fix using the same method: compare actual time and distance with plan, derive actual GS, and adjust the wind. Also note that the error was a pure headwind on this track; on the reciprocal track it becomes a tailwind, so the return should be faster than the outbound, but the reserve calculation must still assume the forecast, not the optimistic measurement.