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formula for astronomical midnight at Greenwich, in UT?
On Sep 22, 12:34*pm, "Greg Neill" wrote:
wrote: 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. It's a simple curve fit to the equation of time for a particular epoch. *Because the perihelion of the Earth's orbit shifts over time, the "shape" of the E of T curve changes over time too, and the accuracy will not be maintained as one strays from the epoch. The variations in the natural noon cycle are due to two dynamics,the daily rotation of the Earth to the central Sun and the slow orbital turning of the Earth to the central Sun hence the daily daylight/ darkness cycle and the separate orbital daylight/darkness cycle experienced at the polar coordinates as 6 months of daylight followed by 6 months of daylight with a brief orbital twilight/dawn experienced at the equinox which happens to be happening right now. Get back to me when you have a clear idea where the variations in natural noon come from in order to equalize it to a 24 hour average.You are all getting an education you never had before so make good use of it. 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. It should be pretty good (probably within a few seconds I'd guess), since it calculates the parameters for the curve fit from determinations of the current orbital parameters. (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. That is one way to pin down the UT versus Dynamical Time relationship for a given year. *Another is to keep a table of Delta-T values, or calculate Delta-T values from a curve fit. *Future values of Delta-T can really only be roughly predicted, since it depends upon many dynamical factors affecting Earth rotation which accumulate over time. *See Meeus chapter 9 (chapter 10 in the newer editions). |
#12
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formula for astronomical midnight at Greenwich, in UT?
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#13
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formula for astronomical midnight at Greenwich, in UT?
On Sep 22, 3:40*pm, William Hamblen
wrote: On Wed, 22 Sep 2010 02:00:07 -0700 (PDT), wrote: 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 If you wanted to figure when the Sun is at its lowest point you would need to find when the hour angle of the Sun is +/- 12 hours.. Bud Figure out indeed !,a system where there are 1461 days and rotations between Mar 1st 2009 and Feb 29th 2012 and none of you can figure out that the 4 orbital circuits comprise of 365 1/4 rotations and if people can live with 366 1/4 rotations for the same period they suffer an illness of the mind and unfortunately that may very well be the case. All I can think of is that there must be some abiding hatred for astronomy,what or ancestors achieved and all the goodness that is in human nature insofar as that extra rotation or leap day almost squares away the fractional difference lost in non-leap years.Why would any sane person trump up a nonsensical 366 1/4 rotations when the daily cycles correspond to daily rotations within the calendar system while keeping in mind what the adjustment on Feb 29th does in terms of the orbital cycle. This is not good enough as the principles which constitute the 1461 day calendar system are even more fundamental than the arguments for a round Earth and everyone is failing here at a level that nobody really wants. |
#14
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formula for astronomical midnight at Greenwich, in UT?
On Sep 22, 8:28*am, oriel36 wrote:
Get back to me when you have a clear idea where the variations in natural noon come from in order to equalize it to a 24 hour average. http://www.quadibloc.com/science/eot.htm John Savard |
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