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formula for astronomical midnight at Greenwich, in UT?



 
 
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  #1  
Old September 21st 10, 05:00 PM posted to sci.astro.amateur,uk.sci.astronomy
[email protected]
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Posts: 6
Default 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
  #4  
Old September 22nd 10, 07:51 AM posted to sci.astro.amateur,uk.sci.astronomy
oriel36[_2_]
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Posts: 8,478
Default 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.



  #5  
Old September 22nd 10, 09:11 AM posted to sci.astro.amateur,uk.sci.astronomy
rob
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Posts: 43
Default formula for astronomical midnight at Greenwich, in UT?

On Sep 21, 11:51*pm, oriel36 wrote:
............................


"Son of a ****ing bitch", lol!
  #6  
Old September 22nd 10, 10:00 AM posted to sci.astro.amateur,uk.sci.astronomy
[email protected]
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Posts: 6
Default 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


  #7  
Old September 22nd 10, 10:40 AM posted to sci.astro.amateur,uk.sci.astronomy
oriel36[_2_]
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Posts: 8,478
Default 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.





  #8  
Old September 22nd 10, 11:19 AM posted to sci.astro.amateur,uk.sci.astronomy
Quadibloc
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Posts: 7,018
Default 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
  #9  
Old September 22nd 10, 12:14 PM posted to sci.astro.amateur,uk.sci.astronomy
Greg Neill[_6_]
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Posts: 605
Default 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.


  #10  
Old September 22nd 10, 12:34 PM posted to sci.astro.amateur,uk.sci.astronomy
Greg Neill[_6_]
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Posts: 605
Default formula for astronomical midnight at Greenwich, in UT?

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.


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).


 




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