UMC Time, Right Ascension and Longitude

  • Thread starter whatta
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  • #1
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Where can I find a triplet of UTC Time, Right Ascension (of zenith) and Longitude, preferrably with many, many safe decimal places?
 
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  • #2
russ_watters
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The question doesn't make a whole lot of sense, could you explain in more detail what you are looking for and why? If you are looking for an accurate time signal and your position, GPS is probably the best place to get it. Any good astronomy program will tell you the RA of the zenith to several decimal places (what's a "safe decimal place"?).
 
  • #3
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Any good astronomy program will tell you the RA of the zenith to several decimal places (what's a "safe decimal place"?).
Based on what? To do that, the program needs to know position of Earth relative to stars at some moment. The minimum of data to perform such a computation is a triplet I want.

This can be any place on Earth and any time, but both Longitude and RA have to be measured/calculated really carefully.
 
  • #4
russ_watters
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I use Starry Night ( www.starrynight.com ) to drive my telescope. I enter my lat/long and it takes a time signal from the computer (which gets it from the internet). It generally gives me a pointing accuracy of around 5 arcmin, and most of that error is in the telescope. Assuming you enter a good position and your time signal is accurate to the second, the software would give you 15 arcsec accuracy.
 
  • #5
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do you think you could log a timestamp and zenith RA from its output for some longitude (like, 0) for me?
 
  • #6
russ_watters
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Sure - I'll post a screenshot for you tonight, when I get home.
 
  • #7
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I've found http://maia.usno.navy.mil/conv2000/chapter5/tab5.4.txt [Broken] an impressive equation for what I need (Greenwich Sidereal Time), with two unknowns in it, "(UT1 - UTC)" and "classical expression for the equation of the equinoxes". Why it always has to be splitted across all the internet, and not in a single page.
 
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  • #8
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...okay, so they say at wikipedia that
The ratio of UT1 to mean sidereal time is defined to be 0.997269566329084 − 5.8684×10^−11T + 5.9×10^−15T^2, where T is the number of Julian centuries of 36525 days each that have elapsed since JD 2451545.0 (J2000).
I assume english translation of this is that at "JDJD 2451545.0 (J2000)" local sidereal time of longitude 0 was equal to 0.997269566329084 of UT1 time, which was... ? it is said also that UT1 could be found as UTC (January 1, 2000, 11:58:55.816 UTC) + DUT1 but I have no idea where to look this DUT1 up :( somebody please help me with this mess before I kill myself. EDIT: according to this graph it was around +0.35 sec, but exact value would be nice.
 
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  • #9
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going crrrrazy with this, http://maite152.upc.es/~manuel/tdgps/node18.html [Broken], but he sais that "Tu is the time since J2000 (January 1 2000, 12h UT1) in Julian centuries of 365.25 days" which gives over a minute of difference with above UTC value from wikipedia.

One word: confused. Let's see what russ value will align to.
 
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  • #10
russ_watters
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Here's a screen cap from Starry Night. The cursor disappeared when I did the printscreen, but rest assured, it was right on the zenith...

Part of the problem you may be having with the calculation is that there is more than one standard, as you can see in the cap, and the difference is about half an arcmin.
 

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  • #11
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you know, when I started this thread, I was kind of hoping that, given earth rotation period of 23 h 56 m 4.091 s, one can convert normal time to sidereal (or to RA) with simple linear equation (constructed from values I asked for), but our fellow astronomers f**ed it up beyond belief, and now there are lots of time systems, and lots times lots transformations between these systems.

so. I have ended up looking into three software sources. there are two steps, 1st they convert UTC date directly to "Julian Day"; stellarium and celestia use same code (here, around line 50, or here, around line 435). yoursky's code is slightly different,
Code:
    if ((year < 1582) || ((year == 1582) && ((mon < 9) || (mon == 9 && mday < 5)))) {
	b = 0;
    } else {
	a = ((int) (y / 100));
	b = 2 - a + (a / 4);
    }

    return (((long) (365.25 * (y + 4716))) + ((int) (30.6001 * (m + 1))) +
		mday + b - 1524.5) +
	    ((sec + 60L * (min + 60L * hour)) / 86400.0);
but claims to implement same algorithm (Meeus, Astronomical Algorithms, Chapter 7, page 61).

second step is converting to sidereal time. I could not find code for that in celestia, and both codes from stellarium (here around line 35) and yoursky,
Code:
*  GMST  --  Calculate Greenwich Mean Siderial Time for a given
	      instant expressed as a Julian date and fraction.	*/

double gmst(double jd)
{
    double t, theta0;

    /* Time, in Julian centuries of 36525 ephemeris days,
       measured from the epoch 1900 January 0.5 ET. */

    t = ((floor(jd + 0.5) - 0.5) - 2415020.0) / JulianCentury;

    theta0 = 6.6460656 + 2400.051262 * t + 0.00002581 * t * t;

    t = (jd + 0.5) - (floor(jd + 0.5));

    theta0 += (t * 24.0) * 1.002737908;

    theta0 = (theta0 - 24.0 * (floor(theta0 / 24.0)));

    return theta0;
}
are different.

so. later today, I will put all these codes together and see which is closer to values in your screenshot.
 
  • #12
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The cursor disappeared when I did the printscreen, but rest assured, it was right on the zenith.
Your screenshot sais, "Altitude 88°". Zenith is 90°, isn't it.

Also, your time zone would be helpful. As is, my test computation results are:
Code:
Date: February 13, 2007, 12:05:08 pm
Julian date: 2454144.95856
Stellarium output /15 (1): 20.5457510321
Yoursky output (2): 20.5457282109
Taking 20h away and converting to minutes:
(1) 32.7450619233
(2) 32.7436926529
neither matches 9h 35.742m, but minutes are within 4m margin, which tells us this error can be explained by different time zone.

EDIT: I took another timestamp 11 hours away, and my minutes became worse:
Code:
Date: February 13, 2007, 1:05:08 am
Julian date: 2454144.50023
Stellarium output /15 (1): 9.51563402541
Yoursky output (2): 9.51561121915
Taking 9h away and converting to minutes:
(1) 30.9380415245
(2) 30.9366731487
:( okay. that's it. now i'm pissed.
 
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  • #13
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EDIT: I have found error it was incorrectly computing julian day. now it is like:
Code:
Date: February 13, 2007, 2:05:08 am
Julian date: 2454144.50356
Stellarium output /15 (1): 9.59585307131
Yoursky output (2): 9.59583026493
Taking 9h away and converting to minutes:
(1) 35.7511842786
(2) 35.7498158961
In your screenshot it was 9h 35.742m, so it's a nice match with 2nd value. I will continue to test different codes, but this thread can be considered resolved.
 
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