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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.

    Fred Hebard wrote:
    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.
    On Apr 15, 2008, at 12:33 PM, Lu Abel wrote:
    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
    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:
    includes a detailed discussion about the problems of gravitational
    dilation and extremely accurate clocks. That's the main topic, and
    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
    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
    else entirely??
    Celestial Navigation Weekend, June 6-8, 2008 at Mystic Seaport

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