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    Re: Estimated Position
    From: Steven Tripp
    Date: 1995 Dec 13, 08:57 +0900

    This formula for calculating leeway was in Ocean Navigator #45.
    58.5*heel/boatspeed^3=leeway angle
    Date: Wed, 13 Dec 1995 01:12:20 -0500 (EST)
    From: "J. Shields" 
    Subject: Re: Estimated Position/leeway
    Sender: owner-navigation@ronin.com
    I copied this table into my notebook (from a library book - I think it was
    Anapolis Book Of Seamanship, or something like this).  Would you trust
    these estimates for leeway? or did I just waste my time copying them down
    (also typing them into this posting).  Seriously though - I'm a newbie -
    are they kosher?
                   LEEWAY Angles (degrees) for Wind Strength / point of sail
    BOAT TYPE     FORCE 1-3    FORCE 4-5    FORCE 6-7    FORCE 8-10
                  (3-10 knots) (11-21 knts) (22-33 knts) (34-55 knts)
                  beat  reach  beat  reach  beat  reach  beat  reach
                  ----  -----  ----  -----  ----  -----  ----  -----
    shallow-keel   10     5      8      4    12     10    20     12
    deep keel       6     4      4      2     6      4    12      6
    large cruiser
    -under power    4     2      6     4     10      8    20     12
    -under sail    10     4      8     4     12     12    20     15
    There was no explanation as to how these Leeway angles were determined.
    I just figured the author knew a lot more than me, and it has been the
    only answer I have seen to a pretty important question (in my opinion)
    when you are sailing to weather.
    Date: Thu, 14 Dec 1995 00:35:33 +0100
    To: navigation@ronin.com
    From: J.van.Puffelen@uni4nn.iaf.nl (Jan van Puffelen)
    Subject: Re: Estimated Position
    Sender: owner-navigation@ronin.com
    I have always used the formula 8*A/(v2) for leeway whereby A is the angle of
    heel in degrees and v the speed in knots. This formula seems to work well
    for sailing ships up to 48' or so and for speeds between 3 and 12 knots and
    for angles up to 30 degrees. I cannot vouch for anything outside these

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