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    Re: Tides by bearing of the moon
    From: Brad Morris
    Date: 2009 Apr 8, 17:23 -0400

    New London, Connecticut
    8 April 2009 - 9 April 2009
    41.3550� N, 72.0867� W
    2009-04-08  03:17 EDT  -0.22 feet  Low Tide
    2009-04-08  05:27 EDT   Moonset
    2009-04-08  06:19 EDT   Sunrise
    2009-04-08  09:18 EDT   2.73 feet  High Tide
    2009-04-08  15:27 EDT  -0.01 feet  Low Tide
    2009-04-08  18:42 EDT   Moonrise
    2009-04-08  19:21 EDT   Sunset
    2009-04-08  21:33 EDT   3.26 feet  High Tide
    2009-04-09  04:04 EDT  -0.25 feet  Low Tide
    2009-04-09  05:52 EDT   Moonset
    2009-04-09  06:18 EDT   Sunrise
    2009-04-09  10:02 EDT   2.67 feet  High Tide
    2009-04-09  10:55 EDT   Full Moon
    2009-04-09  16:10 EDT   0.02 feet  Low Tide
    2009-04-09  19:22 EDT   Sunset
    2009-04-09  19:51 EDT   Moonrise
    2009-04-09  22:15 EDT   3.31 feet  High Tide
    -----Original Message-----
    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Hewitt Schlereth
    Sent: Wednesday, April 08, 2009 5:18 PM
    To: NavList@fer3.com
    Subject: [NavList 7883] Re: Tides by bearing of the moon
    The method I used to use was to consider that on the day of the full
    moon, the luni-tidal interval listed in Bowditch for a given place
    would be the time of high tide.
    From my 1933 Bowditch, the High Water Interval for New London is
    9h30m. So on the  day of the full moon - tomorrow 4-9-09 - high tide
    at New London would be 0930 EST or 1030 DST.
    I don't have any tide tables here, so would a kind soul on the List
    look it up and see how old Nathanael did?
    Thanks, Hewitt
    On 4/8/09, frankreed@historicalatlas.com  wrote:
    >  Hi Dave.
    >  Next, try some place more tropical. Hawaii perhaps? :-) That should show 
    significantly worse behavior with respect to azimuth.
    >  I wrote some code a few years back to do tidal calculations. It's much 
    shorter than the widely available X-tide code and only marginally less 
    accurate: http://fer3.com/arc/m2.aspx?i=100454.
    >  Here's a question that's been in the back of my head for a long time... 
    which is a better predictor of tide times: the local hour angle of the actual 
    Moon or the local hour angle of the mean Moon (which we can imagine moving 
    along the celestial equator with constant angular velocity)?
    >  -FER
    >  >
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