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    Re: Navigation without Leap Seconds
    From: Fred Hebard
    Date: 2008 Apr 15, 18:00 -0400

    Oops, divided twice by 60, not once.  You're correct, about 100
    feet.  So .01 arcseconds to 1 foot, or 30 cm, and .0001 to 3 mm.
    Only off by a factor of 60!
    On Apr 15, 2008, at 4:04 PM, Gary J. LaPook wrote:
    > 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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