Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: North Pole



Jim followed up on this interesting thread with a challenge to
respond with simple yes/no answers.
Here they are, followed by a partial transcript of the Navigation list
thread dealing with confirming the time of the equinoxes
which may be of interest to you.

Brian W

At 01:39 AM 10/1/02, Jim, you wrote:
>The Earth's spin axis precesses such that we get a moving polar star.
>Does the orbital axis precess?

*** Maybe.

>Does the angle between the orbital plane and the spin axis change?
>Jim Green

*** The locus of the Earth orbit is not in a plane.

[transcript of thread from Nav list follows.]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Subject: September Equinox computation
To: NAVIGATION-L@LISTSERV.WEBKAHUNA.COM

Which method would you use to PRECISELY compute
(hh-mm-ss) the September equinox?

Pierre Boucher
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From: Robert Eno <enoid@NUNANET.COM>
Subject: Re: September Equinox computation
To: NAVIGATION-L@LISTSERV.WEBKAHUNA.COM

There are several methods that can be used. Most recently, I used my
nautical almanac and was able to pin it down to the nearest minute. I
don't know if that is precise enough for your purposes.

Essentially you go to the nearest hour to when the sun's declination is
close to zero, note the "v" increment (in this case it is 1.0) then go to
the back of the almanac and scroll through tables until the correction for
the "V" increment takes the declination down to precisely zero. If my
memory serves me right, I believe equinox occurred on 23 September at
04:52 GMT.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From: Herbert Prinz <hprinz@ATTGLOBAL.NET>
Subject: Re: September Equinox computation
To: NAVIGATION-L@LISTSERV.WEBKAHUNA.COM

Pierre,

If this were an astronomy list, I would say that according to the definition
of equinox, you compute the ecliptic longitude of the Sun from a sufficiently
accurate ephemeris for two reasonable guesses t1 and t2 and then solve for t
such that L(t) = 180deg, either by interpolation or by iteration, dependent on
whether you do it manually or with a computer.

But I assume that you are asking how to do do it with the means that the
modern average celestial navigator has at his disposal. The answer is that you
can't do it to the required precision.

For starters, modern nautical almanacs don't tabulate ecliptic longitude
anymore. (Thanks God!). The next best thing is to solve for SHA = 180deg (or
RA = 12 hours) and the worst thing you can do is to solve for Dec = 0deg.
Neither is strictly correct, but using the SHA will get you THEORETICALLY
within a few seconds of the correct time whereas using Dec will get you there
within a minute, or so.

In practice, however, you must compute SHA from the difference of GHA Sun and
GHA Aries from your Nautical Almanac, which means that you have to interpolate
a value that changes only 2.5' per hour from two values that are burdened by
two rounding errors each of up to 0.05'. On top of this, the entries for GHA
Sun in the Nautical Almanac are shifted on purpose by as much as 0.1' from
their correct value. (This has nothing to do with "Selective Availability"; it
facilitates the use of the interpolation table without a need for
v-correction.) In late September, the entries are too high by 0.1' on average.

In short: You can't even rely on getting within a minute of the correct time
of equinox with the Nautical Almanac.

Herbert Prinz
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From: Jay Borseth <jaybo@nomadelectronics.com>

There is a C version of Novas which can use the DE200 ephemeris
available at http://aa.usno.navy.mil/AA. Pocket Stars
(www.nomadelectronics.com) uses the C version of Novas and DE405. You
can look on the bibliography section on the download page for more info
on merging JPL and Novas.

The results I'm getting put the equinox at:
4:51:54 AM GMT on 23 Sep 2002, crossing at E 105 degrees 9' 56".

- Jay
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From: Pierre Boucher <pboucher@LAVOILE.COM>

>Essentially you go to the nearest hour to when the sun's declination is
>close to zero, note the "v" increment (in this case it is 1.0) then go to
>the back of the almanac and scroll through tables until the correction for
>the "V" increment takes the declination down to precisely zero.

From 51 to 56 minutes included gives a v corr. of 0.9
... and from and within those (51 to 56 minutes), there are many pairs of
GHA Sun and Aries that would give a 180d SH.
Example 04-52-04 by Dan Allen works
Thanks to all who contributed.

Pierre Boucher
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


From: Herbert Prinz <hprinz@ATTGLOBAL.NET>
... I plug the numbers into my carefully re-written Pascal version of the
Novas routines using DE200 and obtain (by doing what I described in my
earlier message) for the time of this year's autemnal equinox 4:55:24 UT1.
This value has the advantage that it agrees with what Meeus publishes in
"Tables of the Sun, Moon and Planets", as well as with what the USNO thinks
it should be. (see Earth's Seasons, etc. 1992-2005 ). This is probably not
a coincidence.

Herbert Prinz

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



From: John Kabel <jjkabel@rogers.com>

I have done the computations with my HP 48GX, using the Sparcom NAV48
Celestial Navigation pack authored by Tom Metcalfe. This system is
based on Meeus, with an internally calculated almanac based on the JPL
ephemeris. Now, the results are nowhere near precise enough to
numerically indicate where the shift from N to S declination occurs.
In fact, the results show a 0d00.0' declination over a span of some
five minutes. But the shift in the letter occurs from 4:55:24 to
4:55:25. And the crossing is at 104d16.8'.

Another "estimate" for the crew to consider. I suspect this program is
using a lot of decimal places internally, then rounding up for the user
only to the nearest 0.1 arc-min. This pac has always been very
accurate for me in previous work.

John Kabel
London, Ontario, Canada
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


From: Peter Fogg <ffive@TPG.COM.AU>

John Kabel wrote:

> ... the results show a 0d00.0' declination over a span of some
> five minutes. But the shift in the letter occurs from 4:55:24 to
> 4:55:25. And the crossing is at 104d16.8'.

With a DR of lat 0d and lon E104d 16'.8 I get a UT for the sun's meridian
transit at 04:55:23,
a LHA for Aries of 179d 59'.8, a SHA of 180d 00'0, an azimuth of 087.5d (!), an
altitude for the sun's centre of 90d 02'.7, and a declination (the holy grail?)
of 0d 00'0.

All of which seems to indicate that according to my system we are very very
close but not quite on the right second, or rather in this case; longitude.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From: Jay Borseth <jaybo@nomadelectronics.com>

...Pocket Stars result is: 4:55:26 AM GMT on 23 Sep
2002, crossing at E 104d16.8'.

I don't know for sure, but I'm guessing the 2 second discrepancy could
be due to differences in DeltaT, for which I've seen multiple different
projections for 2005. Values for intermediate dates such as 23 Sep 2002
are linearly interpolated from the 2000 and 2005 data points. Pocket
Stars is using 69 seconds for 2005, but I've seen 66 seconds used
elsewhere.
...
Julian Date Year DeltaT
2451544.5000000, /*2000.0000,*/ 63.86,
2453371.5000000, /*2005.0000,*/ 69.0, // 66. Alternate values from
other sources
2455197.5000000, /*2010.0000,*/ 74.0, // 70.
...

Jay Borseth
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

From: George Huxtable <george@HUXTABLE.U-NET.COM>

Searchers after the exact moment of Autumn equinox appear to be looking for
the moment when the declination of the Sun is exactly zero, passing from
North to South, and also the Right Ascension of the Sun is exactly 12 hours
or 180 degrees. In this, they are almost certain to be disappointed. Those
two events are unlikely to occur at exactly the same moment.

If the Sun was always exactly on the plane of the ecliptic, then they
would: but in general that is not exactly the case. Because the earth is
perturbed slightly in its path around the Sun by the attractions of the
Moon and other planets, the Sun's latitude (its displacement out of the
plane of the ecliptic) is not always exactly zero, but can vary up to 1.2
seconds of arc.

Note that the effect referred to above is an actual physical shift of the
Earth out of the plane of its orbit round the Sun, by up to 5,000-odd
miles, not a shift of the Earth's polar axis such as precession and
nutation cause.

The moment of autumn Equinox is defined by the Sun's apparent geocentic
longitude (and consequently its Right Ascension also) being 180 degrees,
and NOT by its declination passing through zero. A change in Sun ecliptic
latitude of 1 second of arc would, I think, alter the declination of the
Sun by a similar amount. The Sun's declination around the equinox is
changing at very nearly 24 minutes a day. (I like to remember this by
thinking of the maximum rate of travel of the Sun's geographical position,
North or South, as almost exactly 1 knot).

So a shift in the Sun's position from the ecliptic of 1.2 seconds of arc
would change the moment of zero-crossing of declination from the moment of
the equinox by about 72 seconds of time.

I have not tried to estimate what the ecliptic latitude of the Sun would be
at the 2002 autumn equinox, but for anyone that wishes to, Meeus in
chapters 27 and 25 provides all the necessary information.

I have no wish to sail under false colours, and pose as an authority on
such matters. All that I have said here has been taken from Meeus'
excellent work "Astronomical Algorithms", of which I claim only a partial
understanding. So the conclusions above are somewhat tentative, and stand
to be corrected by anyone who knows more than I do.

George Huxtable.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Brian Whatcott
Altus OK Eureka!

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.