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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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