Dear Calendar List:
According to the following article, astronomers have determined from eclipse records that the earth's rotation has slowed by 6 hours in 2740 years: http://www.sciencemag.org/news/2016/12/ancienteclipsesshowearthsrotationslowing Does this information provide enough data to compute a good estimate for deltaT? Walter Ziobro 
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There is another website on this topic: servus sepp
Am 07.12.2016 um 16:17 schrieb Walter J Ziobro <[hidden email]>: Dear Calendar List: 
Dear Calendar People, I suspect another significant factor is the fact that earth is a rotating magnet passing through a magnetic field. The interactions between the two must surely have a nonzero influence. And it would change as the poles shift and as Earth moves through a changing magnetic field. Think of it as an electric motor. Victor On Wed, Dec 7, 2016 at 9:24 AM, Sepp ROTHWANGL <[hidden email]> wrote:

In reply to this post by Sepp Rothwangl
And another  Three Gorges Dam. Victor On Wed, Dec 7, 2016 at 9:24 AM, Sepp ROTHWANGL <[hidden email]> wrote:

Is the effect of the Three Gorges Dam large enough to affect eclipse predictions? Sent from AOL Mobile Mail On Wednesday, December 7, 2016 Victor Engel <[hidden email]> wrote: And another  Three Gorges Dam. Victor
On Wed, Dec 7, 2016 at 9:24 AM, Sepp ROTHWANGL <[hidden email]> wrote:

Dear Walter and Calendar People, Using the value in the article, the effect is 0.06 milliseconds a day. That's about 17 days for a millisecond or a million days for a minute. A million days is about 2738 years. Victor On Wed, Dec 7, 2016 at 3:19 PM, Walter J Ziobro <[hidden email]> wrote:

So if I understand correctly, then eclipse predictions for 2700 years from now will be one minute too soon if not adjusted for the slowing effect of the Three Gorges Dam Sent from AOL Mobile Mail On Wednesday, December 7, 2016 Victor Engel <[hidden email]> wrote: Dear Walter and Calendar People, Using the value in the article, the effect is 0.06 milliseconds a day. That's about 17 days for a millisecond or a million days for a minute. A million days is about 2738 years. Victor
On Wed, Dec 7, 2016 at 3:19 PM, Walter J Ziobro <[hidden email]> wrote:

In reply to this post by Victor Engel
On 20161207 21:27, Victor Engel wrote:
> Dear Walter and Calendar People, > > Using the value in the article, the effect is 0.06 milliseconds a day. No, they say 0.06 µs = 60 ns. The uncertainty of the measurement of TIA  UTC is currently some µs, so this is way below measurable. Given the mass of the water in the lake, about 60 Pg, this is to be expected. The mass of water in several Mm² of cloud cover is comparable. Glacial rebound involves much larger masses. Michael Deckers. 
Oh, did I read it wrong? Was I off by a few orders of magnitude? On Wed, Dec 7, 2016 at 3:57 PM, Michael H Deckers <[hidden email]> wrote: On 20161207 21:27, Victor Engel wrote: 
In reply to this post by Sepp Rothwangl
From: East Carolina University Calendar discussion List [[hidden email]] on behalf of Sepp ROTHWANGL [[hidden email]]
Sent: Wednesday, December 07, 2016 10:24
There is another website on this topic:
http://rspa.royalsocietypublishing.org/content/472/2196/20160404#sec33
Irv replies: Note that there is a lot of supplementary data (mostly plain text files) downloadable from the URL given at the end of the paper's introduction. (There is an entirely different URL for supplementary data given at the bottom left of the first page, but that seems to point to copies of the same files.) This paper is a comprehensive update of their previous paper, but they still haven't published polynomials that can be used to approximate Delta T  although perhaps those are buried somewhere in the supplementary data. In figure 20 the paper depicts a power law but gives it as (exp^{n} = 1.3). I assume that means e^{n} = 1.3, but that has no solution, so what do the authors mean by that? And why does the plotted line stop at the top end at the gridline corresponding to a 25year period? The authors persistently use the phrase "tidal friction" without explaining what they mean by that. Can "friction" dissipate angular momentum? Even if it could somehow slow down the Earth rotation rate, that couldn't explain the acceleration of Moon progressively further away from Earth, which is well documented by decades of LASER lunar ranging measurements since the Apollo lunar landing missions. Both of these effects are explained by tidal transfer of angular momentum from Earth to Moon, so why is it that the authors never invoke that explanation and instead repeatedly refer to "friction"?  Irv Bromberg, University of Toronto, Canada http://www.sym454.org/lunar/ 
Dear Irv and Calendar People Irv asked
The authors persistently use the phrase "tidal friction" without explaining what they mean by that. Can "friction" dissipate angular momentum? I reply. No. Friction cannot dissipate angular momentum. The angular momentum is transferred to the moon through gravitational forces. It is this that causes
the moon to move further away. From the moon’s perspective, tidal fiction causes the tidal bulge of Erath to lag and this pulls the moon forward slightly. Karl 16(04(17 From: East Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Irv Bromberg From: East Carolina University Calendar discussion List [[hidden email]]
on behalf of Sepp ROTHWANGL [[hidden email]] Sent: Wednesday, December 07, 2016 10:24 There is another website on this topic:
http://rspa.royalsocietypublishing.org/content/472/2196/20160404#sec33

But tidal friction can transfer angular momentum between layers of Earth, right? On Fri, Dec 16, 2016 at 7:00 AM, Karl Palmen <[hidden email]> wrote:

In reply to this post by Karl Palmen
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Op 16 dec. 2016, om 14:00 heeft Karl Palmen <[hidden email]> het volgende geschreven:
the Moon raises a todal bulge. The Earth rotates faster around its axis than the Moon revolves around her, and this rotation drags the tidal bulge ahead of the Moon. As a consequence: a) the Moon is pulled forward: accelerated. Being in a gravitationally bound orbit following Kepler's 3rd law nˆ2*aˆ3 = Constant, this forces it in an orbit with larger orbital axis (a) and smaller angular velocity (n); b) the tidal bulge is pulled backward, which brakes the rotation of the Earth: this is the friction. The rotation rate slows down and the length of day increases. Momentum is conserved, there is net transfer of angular momentum from Earth spin to Moon orbit. Energy is conserved, the potential energy of the Moon increases but the rotational energy of the Earth decreases more, the difference is dissipated as heat in the oceans (and possibly in the solid Earth). 

In reply to this post by Irv Bromberg
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Op 15 dec. 2016, om 17:00 heeft Irv Bromberg <[hidden email]> het volgende geschreven:
They did the same back in 1995 and published tables with evaluations of the polynomials for every century or so, but never disclosed what the parameters were. Mind that the DeltaT curve does not follow a neat parabola (2nd degree polynomial). People have tried to fit the entire historical DeltaT curve with a single polynomial, up to degree 20, which is ludicrous I think. The underlying curve should be a parabola (constant deceleration of the rotation due to tidal friction  their formula 1.5) plus a decaying exponential (due to glacial rebound). I posted parameters for this on this list on 6 June 2002. On top of this there are semiperidiodic and erratic excursions of unknown origin, which can not be modeled well by a polynomial or even a sine. So they model phenomenologically using the common practice of fitting cubic splines over short intervals, which can follow a locally smooth curve, and which neatly connect both on the curve and in the first derivative (i.e. no sharp corners at the connection points).
They describe the distribution of power in the frequencies in the signal of the Length of Day. So not the value of DeltaT, but the exponent of a logarithmic curve that fits the Fourier transform of the derivative of DeltaT is 1.3.

In reply to this post by Walter J Ziobro
Op 7 dec. 2016, om 22:32 heeft Walter J Ziobro <000000080342b460[hidden email]> het volgende geschreven:
the rate of rotation  the length of day  increases steadily by about 6 ms every century due to tidal friction (and in addition not so steadily due to other causes). DeltaT is the accumulated difference between the time as indicated by the Earth's rotation angle, and the time counted by a steady (atomic) clock  the integral of the rate in the first line. The rate (length of day) was 86400 SI seconds around 1820; the difference DeltaT was 0 around 1900. It is interesting that the 3 Gorges Dam has a measurable influence on the rate of rotation of the Earth. So have earth quakes, melting polar icecaps, growing and falling leaves every season, and many other things. Tidal friction is the largest and most durable factor.

Op 18 dec. 2016, om 12:01 heeft Tom Peters <[hidden email]> het volgende geschreven:
Sorry, I confused numbers. The rate due to tidal friction should be about +2.3 ms/d/cy (their formula's 1.2 and 1.3 and Figure 18). The actually observed longterm rate is +1.78 ms/d/cy. The additional acceleration that reduces the slowing down due to tidal friction is ascribed to the decrease of the moment of inertia of the Earth because mass moves from the equator to the poles due to glacial rebound: during the ice age the polar regions got depressed by the mass of ice, and rocky mass was bulging out at the equator. The ice has mostly gone but the lithosphere is still recovering. 
In reply to this post by Tom Peters6
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Dear Irv, Tom and Calendar People Tom’s explanation is a more detailed version of my explanation. The lagging of the tidal bulge is Eastward in the direction Earth rotates and the moon goes round Earth. Karl 16(04(20 From: East Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Tom Peters Op 16 dec. 2016, om 14:00 heeft Karl Palmen <[hidden email]> het volgende geschreven:
Dear Irv and Calendar People Irv asked The authors persistently use the phrase "tidal friction" without explaining what they mean by that. Can "friction" dissipate angular momentum? I reply. No. Friction cannot dissipate angular momentum. The angular momentum is transferred to the moon through gravitational forces. It is this that causes
the moon to move further away. From the moon’s perspective, tidal fiction causes the tidal bulge of Erath to lag and this pulls the moon forward slightly. No: the Moon raises a todal bulge. The Earth rotates faster around its axis than the Moon revolves around her, and this rotation drags the tidal bulge ahead of the Moon. As a consequence: a) the Moon is pulled forward: accelerated. Being in a gravitationally bound orbit following Kepler's 3rd law nˆ2*aˆ3 = Constant, this forces it in an orbit with larger orbital axis (a) and smaller angular velocity (n); b) the tidal bulge is pulled backward, which brakes the rotation of the Earth: this is the friction. The rotation rate slows down and the length of day increases. Momentum is conserved, there is net transfer of angular momentum from Earth spin to Moon orbit. Energy is conserved, the potential energy of the Moon increases but the rotational energy of the Earth decreases more, the difference is dissipated as heat in the oceans (and possibly in the solid Earth).  
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