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    Re: Navigation without Leap Seconds
    From: Gary LaPook
    Date: 2008 Apr 15, 13:04 -0700
    Gary LaPook wrote:

    How does one arc second get you to 50 cm? One arc minute is 1852 meters divided by 60 makes 30.9 meters: or 6076 feet divided by 60 equals 101.3 feet (approximately 100 feet in practice.) in an arc second.

    gl
    Fred Hebard wrote:
    Lu,
    
    Why billionths of an arcsecond?  One arcsecond gets one to 1/60th of
    100 feet in traditional surveying, or about 50 cm.  One-thousandth of
    an arcsecond would drop one to 5 mm.  I wonder if refraction is a
    problem here.
    
    Fred
    
    On Apr 15, 2008, at 12:33 PM, Lu Abel wrote:
      
    Fred:
    
    In theory, yes; in practice, no.
    
    To position oneself using star-star distances would require require
    measuring angles to billionths of an arc-second.   Maybe something an
    astronomer could do, but not something you or I are going to do
    with our
    sextants!
    
    BTW, I remember a conversation with a radio-astronomer about 20 years
    ago where he said that his team had measured the distance between two
    radiotelescopes on opposite sides of the US to within a cm or so
    using a
    technique called long-baseline interferometry.   But the whole
    experiment took them a year or so...
    
    Lu Abel
    
    Fred Hebard wrote:
        
    Completely unrelated, but stemming from the same article.
    
    The author states that height can only be known to some few cm or
    whatever because of variations in gravity, if I remember correctly.
    It would seem that this is due to our tradition of assuming we are on
    the surface of a spheroid or ellipsoid when doing navigation.
    Confining ourselves to a surface makes the trig easier, but couldn't
    one position oneself with greater accuracy (with feet firmly planted
    on earth, not on a boat) using only stars or stars plus the sun,
    ignoring the earth's horizon, by measuring star-star distances?  Make
    it a true 3-D problem.  Or would uncertainties in the positions of
    stars still hamper ones efforts, especially uncertainty in their
    distance from us?
    
    Fred Hebard
    
    On Apr 14, 2008, at 9:50 PM, frankreed@HistoricalAtlas.net wrote:
    
          
    The fascinating article which Fred Hebard linked:
     http://www.physicstoday.org/vol-59/iss-3/p10.html
    includes a detailed discussion about the problems of gravitational
    time
    dilation and extremely accurate clocks. That's the main topic, and
    it's
    great stuff.
    
    The article also mentions leap seconds and navigation:
    "Celestial navigators --that vanishing breed-- also like leap
    seconds. The
    Global Positioning System, however, cannot tolerate time jumps and
    employs a
    time scale that avoids leap seconds."
    
    So here's my question: what's the best way of doing celestial
    navigation if
    leap seconds are dropped from official time-keeping? I don't think
    it should
    be all that difficult to work around, but I'm not sure what the best
    approach would be. Assume we get to a point where the cumulative
    time
    difference is, let's say, 60 seconds (that shouldn't happen for
    decades, so
    this is just for the sake of argument). Should we treat the
    difference as a
    60 second clock correction before working the sights? Or should it
    be a 15
    minute of arc longitude correction after working the sights? Or
    something
    else entirely??
    
     -FER
    Celestial Navigation Weekend, June 6-8, 2008 at Mystic Seaport
    Museum:
    www.fer3.com/Mystic2008
    
    
    
    
            
    
          
    
    
    
    
    
      


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