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    Re: Avoiding Wx Problem
    From: Michael Wescott
    Date: 2000 Dec 15, 12:59 PM

    Bob Owens wrote:
    > You are underway on a course of 050T at a speed of 12 knots. The eye of a
    > hurricane bears 120T, 110 miles from your position. The hurricane is moving
    > toward 285T at 25 knots. If you maneuver at 12 knots to avoid the hurricane,
    > what could be the maximum CPA.(Closest Point of Approach)1
    > The listed answer is 77 miles.
    > I don't have a clue how to figure this one out. My Bowditch must be hiding
    > the obvious but I can't find it.
    This is a Maneuvering Board problem. See NIMA Pub 217 rather than Bowditch.
    Available on the web at:
    The trick to this kind of problem is to convert it from two things moving to
    one thing stationary and one thing moving (with the combined motion of both).
    See the attached diagram.
    The hurricane is at H0. 110 nmi 120 deg T from the Boat. If the boat is
    stationary the hurricane will move in 1 hour to H1, a distance of 25 nmi
    on a course of 285 T (along the green line). We can affect that relative
    movement by 12 nmi in any direction (the blue circle). The course that will
    take the hurricane furthest from the boat is tangent to the blue circle and
    gives the hurricane the (relative) course shown in red. The CPA measures out
    to 76.8 nmi.
            -Mike Wescott

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