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WSJ: Why the U.S. Wants To End the Link Between Time and Sun - Astronomers Say Wait a Sec
Wall Street Journal ^ | July 29, 2005 | KEITH J. WINSTEIN

Posted on 07/29/2005 6:03:12 AM PDT by OESY

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To: r9etb
You won't notice the difference -- unless your bank happens to crap out on a leap second

LOL. Bank? Willie doesn't trust banks. He keeps all his spare change in a large mason jar buried under his front porch. That way it's safe from all the transnational corporations.

101 posted on 07/29/2005 12:44:33 PM PDT by been_lurking
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To: Lonesome in Massachussets
For example, starting from Midnight (0 Hr) Jan. 1, 2000 it took the Earth about an extra 0.2 seconds to rotate back to where is should have been with respect to the fixed stars at (0 Hr) Jan. 1, 2001. The astronomical year was 0.2 seconds longer than the atomic year.

Exactly, but according to the chart the Earth took more than an an extra second to complete the year in 1900. So either the Earth is turning more quickly since then or it's going around the sun more slowly.

102 posted on 07/29/2005 2:25:51 PM PDT by wideminded
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To: wideminded
The x-axis is calendar years, which are only roughly equal to solar years. The comparison is Universal Time (UT) to a posited uniform time scale. Nowadays we implement a uniform timescale using ensembles of atomic clocks in various national laboratories around the world. UT is defined by the rotation of the Earth with respect to the fixed stars. The Earth rotates on its axis about 366.2522 times every year. The motion of the Earth around the sun induces an apparent motion of the Sun from west to east with respect to the fixed stars, which adds up to exactly 360 degrees in a (sidereal) year, making the number of solar days per year exactly one less than the number of sidereal days. (Whew!)

Therefore, the Earths rotates about 360.9856465 degrees per day of UT, with respect to distant stars and pretty close to 360 degrees per day (one average) with respect to the nearest star, the Sun. One day of UT equals 86,400 seconds of UT.

Universal Time (UT) is conventionally defined by an agreed to rotational rate of the Earth, measured against the distant stars. When the Earth rotates a given number of degrees, exactly so many seconds or days of UT time has elapsed. If you do all the math right (and I haven’t given you enough information here to do it) it turns out there are 86,164.09074 seconds of UT in a sidereal day, the time it takes the Earth to rotate once on its axis with respect to the fixed stars. It’s approximately equal to 86,400 times (365.2522/[365.2522+1]).

Imagine that you are an astronomer, and measure the transit of a certain bright star, say Sirius, across the cross hairs in a telescope pointed due south, at the correct elevation to catch it. (You happen to be in the Northern Hemisphere.) At exactly that moment you start a timer that counts atomic seconds. The next evening you snap a photograph of the timer just as Sirius transits again. (Fast aperture!) In the interim you know that precisely 86164.09074 seconds of universal time has elapsed, by definition. The timer may read something else, the difference is the difference between atomic time and UT. You're persistent, and besides you have a NSF grant, so you keep this up for 365 days, at which time precisely 31536057.211 seconds of UT will have passed. (The star transits a littler earlier every day because a sidereal day is shorter than solar day, in 365 days you observe 366 transits.)

You compare your timer to the expected 31536057.211 seconds. If it's less you say the Earth is running fast, if it's more you say the Earth is running slow. The claim is that if you had performed this experiment sometime around 1900, the timer would have indicated over a second more than a expected, in 1500 as second less.

103 posted on 07/30/2005 5:01:17 AM PDT by Lonesome in Massachussets (Lonesome's First Law: Whenever anyone says it's not about the money, it's about the money.)
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To: OESY

You know, when C'thulhu awakes from slumber in R'lyeh and takes over the world, none of this crap is gonna matter.

;-) :-D


104 posted on 07/30/2005 11:32:37 PM PDT by BradyLS (DO NOT FEED THE BEARS!)
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To: Lonesome in Massachussets
The Earth rotates on its axis about 366.2522 times every year

Um, it's more like 365.2422 (.2422 * 4 just about equalling the leap year day - less the slowing-down and occasional added seconds).

105 posted on 07/31/2005 10:25:19 PM PDT by solitas (So what if I support an OS that has fewer flaws than yours? 'Mystic' dual 500 G4's, OSX.4.2)
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To: babyface00
Does this mean we're going to take the current rotation of the earth around its axis, and divide it by 24 to get hours, which will change our definitions of minutes, seconds, etc?

Yep. Which means that watch and clock company stocks will go through the roof. Everyone will have to get a new watch and a new set of clocks.

I think I will make a long term investment in Seiko and Citizen.

106 posted on 07/31/2005 10:31:27 PM PDT by P-Marlowe
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To: solitas
Um, it's more like 365.2422

NO! No, with respect to the distant stars it's more like 366.2422. The motion of the Earth around the Sun induces an apparent west to east motion of the Sun. If the Earth did not rotate with repect to the fixed stars, every year the Sun would rise once in the west and set once in the east, with six months of night and six months of daylight, everywhere. Stars (other than the very nearest one, the Sun) 'transit the meridian' about four minutes earlier every day. In 365 calendar days, most stars will transit 366 times. Ptolemy knew this, the ancient Chaldeans knew this, (without as much precision as we do today) no later than early in the first milleum BC.

The transits of the distant stars are much, much more regular than the transits of the Sun. As you know, the Sun only transits once every 24 hours on 'average', where average is over a solar year. The timing of Solar transits, which unlike risings and settings are the same regardless of laditude, vary plus or minus about 15 minutes over the course of a year, due to the obliquity of the ecliptic ('tilt of the Earth') and ecentricity of the Earth's orbit. The difference between mean solar time (as in "Greenwich Mean Time") and apparent solar time is given by "the equation of time". The ancient Chaldeans had a pretty good handle on the equation of time.


107 posted on 08/01/2005 4:29:31 AM PDT by Lonesome in Massachussets (Lonesome's First Law: Whenever anyone says it's not about the money, it's about the money.)
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To: P-Marlowe; babyface00
The length of the civil and atomic second were different from 1958 until 1972, to account for the difference in 'mean solar time' and atomic time and nobody noticed or went out and bought new watches every year. With the advent of precision time distribution services such as WWV, it seemed like a good idea to have the two "harmonized" in 1972.

The difference between the two is not great, less than 25 parts in a billion or about one second every 500 days and is expected to change by only about that amount every century for the forseeable future. This is far better than the accuracy of even the best quartz timepiece. There would have to be some new time distribution services, something like WWV-UT to distribute new time codes.

I suggest that most computers running NTP would be completely compatible with such a system, while NTP does not handle leapsecond well at all.

108 posted on 08/01/2005 4:50:10 AM PDT by Lonesome in Massachussets (Lonesome's First Law: Whenever anyone says it's not about the money, it's about the money.)
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To: Lonesome in Massachussets

Yah, okay, you're right - I was thinking solar days; not sidereal days.


109 posted on 08/01/2005 4:57:17 AM PDT by solitas (So what if I support an OS that has fewer flaws than yours? 'Mystic' dual 500 G4's, OSX.4.2)
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To: solitas

(Hate to pile on but) the mean "solar year" is approximately 365.24219 twenty-four hour days. The Gregorian calendar approximates this as 365 +1/4 -1/100 + 1/400 = 365.2425. The Gregorian Calendar gains a day every 3200 years or so, the Julian calendar gained about three days every 400 years. The reason that the sidereal day does not equal 24 x 365.24219/(365.24219 +1) hours is because of the difference between the sidereal year and the solar year. It takes the earth about 365.256363 days to return to the same point in its orbit with respect to the fixed stars. The difference between this and the solar year is due to the precession of the equinox. We like our years to reflect the seasons, not the positions of the sun. If we used a sidereal calendar the seasons would drift about a two days every three centuries.


110 posted on 08/01/2005 5:13:55 AM PDT by Lonesome in Massachussets (Lonesome's First Law: Whenever anyone says it's not about the money, it's about the money.)
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To: solitas

I didn't mean to quibble, but the Earth's rotation is measured with respect to fixed stars, actually distant pulsars billions of light years away. Universal time is actually defined in terms of sidereal time.

It's an important distinction because things like tides and equipotential surface (shape of the earth) are determined by motion with respect to an inertial (sidereal) reference frame.


111 posted on 08/01/2005 5:23:24 AM PDT by Lonesome in Massachussets (Lonesome's First Law: Whenever anyone says it's not about the money, it's about the money.)
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