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).

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

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.

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