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I've recently read up about time standards and now understand that UTC is a second-corrected version of international atomic time so that it is kept within 0.9s from UT1. And then UT1 is defined as the mean solar time at 0° longitude. However, it is not realized by measurements of the sun's apparent position, but via a sidereal time or earth rotation angle.

Extensive googling hasn't helped me to understand how this observed earth rotation angle is used to compute UT1. This Phys.SE answer hints that the equation is quite complex. I'd like to understand why it needs to be complex, considering that the "mean sun" that mean solar time refers to is defined to follow a uniform movement (with respect to the stars) along the celestial equator. Therefore, shouldn't the difference between sidereal time and mean solar time be simply a linear function of the time of year?

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As explained in: this lecture

Earth's angle of rotation = 2π(0.7790572732640 + 1.00273781191135448(Julian UT1date - 2451545.0)) radians.

So one observes Earth's angle of rotation and calculates the Julian UT1 date as a decimal date from that formula. Then, subtracting out the "day" portion of the date, you get a fraction of a day that you could convert to an hours, minutes, seconds time.

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  • $\begingroup$ Thanks, that seems to be the answer. So it is a linear function as I suspected. $\endgroup$ – A. Donda Apr 7 '14 at 13:31
  • $\begingroup$ That may be the majority of formulas out there but take a quick look at this funtion of the SOFA C library iausofa.org/2012_0301_C/sofa/era00.c Just an FYI because something doesn't seem right here. $\endgroup$ – Douglas G. Allen Jun 7 '15 at 21:16
  • $\begingroup$ Take the first term and multiply it by 360 and you will get a familiar number which is an offset angle used in many of the GMST formulas. This is just a convenient way to keep it time based. $\endgroup$ – Douglas G. Allen Jun 7 '15 at 21:25

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