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Crosswind correction in holding pattern without wind information

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Instrument Rating, Student Pilot

Hi,  I've heard many articles about crosswind correction in holding pattern. They mainly tell the outbound leg=correction times three things for slow speed aircraft. I notice all of them require wind information in holding pattern. What if somehow under some circumstances the pilot doesn't have the wind information in holding pattern? Is there any method to maintain precise 4 minutes holding pattern in such condition?

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1 Answers



  1. Kris Kortokrax on Sep 30, 2015

    I’m confining this discussion to VOR because if you have GPS, this shouldn’t be a problem and NDB holds are becoming a thing of the past.

    You should have some idea concerning wind direction. You will be holding some wind correction to track to the station unless you have a direct headwind or tailwind.

    If you to a teardrop or parallel entry, you will be tracking a radial outbound and can make a pretty good determination about wind correction.

    The bigger issue is trying to get a 1 minute inbound leg (1 1/2 minute inbound leg > 14,000′). You won’t have a “precise 4 minute holding pattern” unless you are in a no wind environment.

    The easy way to adjust for the 1 minute inbound is to fly a 1 minute outbound leg, do the standard rate turn (1 minute) and then note the time inbound.

    It then becomes a simple proportion problem. With an E6B or CR-1, just set the inner 60 second mark underneath the inbound time (in seconds). Then underneath the 60 second mark on the outside ring, you can read the time (in seconds) to fly outbound that will yield a 60 second inbound leg.

    If you don’t do old school, then use the calculator app on your phone. Divide 60 by the number of seconds you noted on your first inbound leg. The multiply the result by 60. The answer will be the number of seconds to fly outbound to yield a 60 second inbound leg.

    Example: You fly outbound for 60 seconds, do a standard rate turn and end up with a 50 second inbound time.

    60 / 50 = 1.2
    60 * 1.2 = 72

    If you fly 72 seconds outbound and do a standard rate turn, you should end up with a 60 second inbound leg (the true objective).

    +3 Votes Thumb up 3 Votes Thumb down 0 Votes


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