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formula for astronomical midnight at Greenwich, in UT?
Hi, I mentioned this in another thread but thought I'd give it a
thread of its own. Can someone help with a formula for astronomical midnight at Greenwich, in UT, given the Julian Day Number? I.e. the UT of the first astronomical midnight (lower culmination of the apparent Sun) following JDN = x. Ideally I need an accuracy of a few seconds. Many thanks in advance. Michael |
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formula for astronomical midnight at Greenwich, in UT?
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formula for astronomical midnight at Greenwich, in UT?
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formula for astronomical midnight at Greenwich, in UT?
On Sep 21, 6:00*pm, "Greg Neill" wrote:
wrote: Hi, I mentioned this in another thread but thought I'd give it a thread of its own. Can someone help with a formula for astronomical midnight at Greenwich, in UT, given the Julian Day Number? I.e. the UT of the first astronomical midnight (lower culmination of the apparent Sun) following JDN = x. Ideally I need an accuracy of a few seconds. Many thanks in advance. Michael For that kind of accuracy, perhaps you could approach it via the equation of time. *Meeus (Astronomical Algorithms) gives a method for determining the E of T for a given date (Julian day). Roughly, the procedure would be: 1. Determine the JD corresponding to Julian midnight for * *the day in question. 2. Determine the E of T for that date from the JD and * *Meeus' method(s). 3. Adjust the JD by the amount of the E of T. 4. Convert the JD into a UT. The determination of the E of T involves determining the Sun's mean anomaly, its apparent right ascention, nutation, and the obliquity of the ecliptic. *It's best that you refer to the book for the mehtods involved, since it would be tedious to reproduce the formulae here in ascii. Doesn't work insofar as the determination of natural noon or the 'mean Sun' as you call it is made using the a 1461 day cycle with 3 years of 365 days and 1 year of 366 days so that you are obliged to give an Equation of Time value for Feb 29th.The ancient kernal component which both separates and links the raw astronomical cycles from the timekeeping averages can be ascertained by the averaging of the noon cycles to 24 hours spread across the orbital cycle which in turn gets applied to the orbital cycle itself therefore the Equation of Time correction is really a new kid on the block in astronomical terms. Expressing the Equation of Time in terms of a wandering 'analemma' Sun is pretty much an assault on the eyes but is no better or worse than everything else surrounding what is basically a simple and stable system of time reckoning which links the number of daily cycles of the Earth to its orbital and annual cycle. |
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formula for astronomical midnight at Greenwich, in UT?
On Sep 21, 11:51*pm, oriel36 wrote:
............................ "Son of a ****ing bitch", lol! |
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formula for astronomical midnight at Greenwich, in UT?
On Sep 22, 3:05*am, William Hamblen
wrote: On Tue, 21 Sep 2010 09:00:24 -0700 (PDT), wrote: Hi, I mentioned this in another thread but thought I'd give it a thread of its own. Can someone help with a formula for astronomical midnight at Greenwich, in UT, given the Julian Day Number? I.e. the UT of the first astronomical midnight (lower culmination of the apparent Sun) following JDN = x. Ideally I need an accuracy of a few seconds. You want to get Jean Meeus' book on Astronomical Algorthims. *The answer is from chapter 11. T = (JD - 2452545.0)/36525 mean sidereal time in seconds = 24110.64841 + 8640184.812866*T + 0.093104*T^2 - 0.0000062 * T^3. This works only for 0 h UT. * Hi and many thanks Greg and Bud. I've now got hold of Jean Meeus's book - what an amazing source! Going by my limited but hopefully growing understanding, I think the equation of time may be the way to go. (I'm not sure whether nutation needs to be taken into account, to get an accuracy of a few seconds rather than 0.01 seconds or so). There is also the following formula with accuracy of around half a minute: E (in minutes) = 9.87 sin 2B - 7.53 cos B - 1.5 sinB where B (in degrees) = (360/365) * (N - 81) (in degrees) and N is day number, counted from B=1 at 1 Jan This seems to work to give astronomical midnight in UT when subtracted from 00:00, although the accuracy is an order of magnitude less than I need. Once I plug the formulae from chap.27 of Meeus into an equation, would it be OK simply to subtract from UT like this, or would this lose too much accuracy? On p.173 Meeus also gives an equation (27.3) which seems to be a version of this but with more accuracy, although it's unclear how much more. (BTW I'm doing this only for the Greenwich meridian; longitude correction is simple). Bud - am I right to think that if I used the formula you posted for GMST, I'd need to keep a table of the UT of the vernal equinox each year? Or am I barking up completely the wrong tree? Excuse my newbie ignorance, but I don't understand how to use that formula to get astronomical midnight on any particular day of the year, in UT. Thanks again! Michael |
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formula for astronomical midnight at Greenwich, in UT?
On Sep 22, 10:00*am, wrote:
correction is simple). Bud - am I right to think that if I used the formula you posted for GMST, I'd need to keep a table of the UT of the vernal equinox each year? Or am I barking up completely the wrong tree? Excuse my newbie ignorance, but I don't understand how to use that formula to get astronomical midnight on any particular day of the year, in UT. Thanks again! Michael This is a system of consensual bluffing,anyone who uses calendar dates is automatically using the 1641 calendar system or the steady progression of 24 hour days formatted as 365 days for 3 years and 366 days for one year with the present system ending Feb 29th 2012. This screwing around with 366 1/4 rotations for an orbital circuit arising from 'sidereal time' reasoning is the worst for when you can't explain what the extra Feb 29th rotation does in completing 4 orbital circuits made up of 365 1/4 rotations ,science and astronomy become pointless. There is no phone number to call to inform them of the most serious known breach of logical consistency even to descend on humanity for its breeds a hatred of basic astronomical principles that have to be seen to be believed for even though the spinning Earth/celestial sphere convenience would constitute something which could be considered as a novelty for civil timekeeping purposes and related endeavors it cannot,cannot be a platform for rotational and planetary orbital dynamics. In current terms,this does not constitute systemic risk as the actual core astronomical principles get highlighted with modern imaging and technologies while exposing what went wrong and where,basically mathematicians running amok with timekeeping averages while having no interpretative instincts.The problem is not the unthinking majority but the minority who can spot the problem and don't have the courage to deal with something as straightforward as why a leap day rotation is necessary. |
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formula for astronomical midnight at Greenwich, in UT?
On Sep 22, 3:40*am, oriel36 wrote:
This screwing around with 366 1/4 rotations for an orbital circuit arising from 'sidereal time' reasoning is the worst for when you can't explain what the extra Feb 29th rotation does in completing 4 orbital circuits made up of 365 1/4 rotations ,science and astronomy become pointless. The average length of the solar day is 24 hours. We base the length of the hour on this so that our timekeeping system can accurately tell us when to wake up in the morning. As the Earth's orbit around the Sun is a natural phenomenon, its duration is not necessarily an exact multiple of the 24 hour day. As we use the calendar to decide when to plant crops in the spring, we base our calendar on the tropical year of 365 days, 5 hours, 48 minutes, and 5.1875 seconds which references the Earth's orbit to the tilt of the Earth's axis as a reference, instead of the sidereal year of 365 days, 6 hours, 9 minutes, and 9.7676 seconds which uses the fixed stars as a reference. So the Julian calendar of 365 1/4 days loses about 9 minutes and 10 seconds a year against the sidereal year even as it gains about 11 minutes and 55 seconds a year against the tropical year. A year is not an integer number of days. That's it. There is no "deeper meaning" to the need for an extra 1/4 day in each year. The Earth orbits the Sun, so the direction of the line from the Sun to the Earth changes through 360 degrees in a year. This makes the number of days in a year one less than the number of times the Earth rotates in a year. And we regard the "sidereal day" as the true rotational period of the Earth because that rotation is uniform (except for small changes with understandable physical causes), while the ordinary solar day reflects the Equation of Time and thus is a compound of the Earth's orbital motion and its rotation. Not only does the Universe not rotate around the Earth once a day, it does not rotate around the Sun once a year - or once every 24,000 years either, for that matter. But astronomers tolerate the last apparent error in their choice of coordinate systems as a convenience in reducing older observations to current coordinates with less arithmetic. John Savard |
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formula for astronomical midnight at Greenwich, in UT?
oriel36 wrote:
snip Kelleher crap You're full of ****, Kelleher. And it's rather bad form to be trolling where earnest people are seeking information. Get a life. |
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formula for astronomical midnight at Greenwich, in UT?
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