NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Navigation without Leap Seconds
From: Gary LaPook
Date: 2008 Apr 15, 13:04 -0700
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:
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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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