Talk:Axial precession

/Archive 2005–2008

Does a Precession Formula Exist?

The astronomical formulas I have seen appear to be based on working back from known observational data (e.g., the earth's precession around the Ecliptic Pole, the Anomalous Precession period, the Lunisolar Precession period, Apsidal (Perihelion) Precession and the "Mercury Question", etc.).

Is there an actual standard formula that exists to calculate precession? Specifically, one that allows the user to enter applicable values (mass of the bodies, distance, orbital period, rotation period, obliquity, etc.), and come up with the precession value? Such a formula should work with any given value (including values given for any hypotheical planetary system).

Unless such a formula exists in the scientific community, I tend to agree with those who hint that determining Precession remains an open-ended mystery (at least regarding both its cause and its affect outside calculations based on direct observation).

Granted, any standard formula would only be accurate to so-many decimal places (due to not being able to account for ALL the multitude of minor gravitational influences from lesser bodies throughout any given system and beyond). Nevertheless, if science has determined the true cause and affect of precession, then a standarized formula (with known-body input values)should exist. Otherwise, any questioning mind may be justified in questioning the existing theories.

The many with greater knowledge may know of such a formula. If so, may it be included on the Wiki page? Oh ... and it would be super if it was written in a manner that is can be interpreted by mathematically limited individuals (who, me?!). --Tesseract501 06:10, 12 October 2008 (UTC)

My dear Tesseract501, thank you for understanding so well the problem of precession and I wish you a lot of success in your future researches. The fact that in front of this problem I am not alone anymore gives me a soul peace that will help me in my own researches. --Abel Cavași (talk) 08:03, 25 October 2008 (UTC)

I am adding a set of equations. — Joe Kress (talk) 17:18, 24 December 2008 (UTC)

Yes, there are many precession nutation models, the most comprehensive is the 2000A model with almost 1400 terms. The problem is the models are "not consistent with dynamical theory" according to the International Astronomical Union (IAU 2006 P03). Also, while the models find a value close to the observed rate they have been unable to predict "changes" in the precession rate (such as the constant increase in the rate over the last few hundred years). Consequently, the models are revised or modified every few years to better match precession observations. This has been going on since d'Alembert substantially modified Newton's formula which did not work. Of course the changes keep getting smaller but one has to wonder if many of the unknown dynamic assumptions (concerning the shape and viscosity of the earth's core, etc.) are included to make the model fit the observable. Interesting area of study. —Preceding unsigned comment added by 69.234.46.179 (talk) 21:33, 16 November 2009 (UTC)

Precession of the equator

The section referring to precession of the equator should be deleted ASAp unless a reference supporting the assertion that `precession of the equinoxes' has been replaced by 'precession of the equator' Terry MacKinnell (talk) 22:52, 27 February 2009 (UTC)

I'll add the official IAU 2006 resolution recommending that lunisolar precession be replaced by precession of the equator and that planetary precession be replaced by precession of the ecliptic. The sum of lunisolar precession and planetary precession was called general precession, which has not changed. I checked the IAU 1938 resolutions, and they refer to lunisolar precession, planetary precession and general precession, so those terms are at least that old. These terms also appear in Simon Newcomb's A Compendium of Spherical Astronomy (1906) who does mention precession of the equinoxes, but only while discussing ancient astronomers. This is becoming ridiculous! How long do new terms have to be in use before the old terms are retired?

A much more serious problem is that the opening sentence stating that precession is the motion of the rotational axis with respect to inerital space is not quite right. That is only its major component formerly called lunisolar precession and now called precession of the equator. This ignores the motion of the reference plane itself, the ecliptic, formerly called planetary precession and now called precession of the ecliptic. — Joe Kress (talk) 01:39, 28 February 2009 (UTC)

I have rewritten the lead due to my concerns. The first paragraph is a bit bloated by the need to include all the various precession terms, which I have included in chronological order. Improvements are welcome. — Joe Kress (talk) 07:40, 28 February 2009 (UTC)

I just reordered this because it didn't flow properly in my opinion. There was an abrupt discontinuity between the two paragraphs in the lead, as if one had been written without any regard for the other. I think we need to begin with a "precession is"-type sentence, and proceed from the general to the specific, rather than launch into detail from the first sentence only to then backtrack to a layman's explanation. Personally I'm happy with the opening statement that precession is the slow change in orientation of the axis, because that gives the ordinary reader a "good enough" (at this point) picture of roughly what the concept is. Then the detail can be expanded (as it is). 86.138.104.133 (talk) 02:58, 2 March 2009 (UTC)

... but bearing in mind your concerns, I just tweaked it again to try to give a more general opening definition that covers all the bases. Please make any necessary changes, but I really do think we need to start with "precession is... <explanation that average layman can understand>". That's my view anyway. 86.138.104.133 (talk) 03:11, 2 March 2009 (UTC).

I have no objection to your reorganization, although I still have more tweaking to do as a result of more research. I've traced the earliest use of the trio lunisolar precession, planetary precession, and general precession back to 1863. Earlier, planetary precession had other names like planetary perturbation, although both lunisolar precession and general precession were in use. I suspect that equivalent German terms were in use as early as 1830. — Joe Kress (talk) 14:38, 2 March 2009 (UTC)

What pre-modern astronomers called it

It's not very clear who these 'pre-modern' astronomers are supposed to have been, who called it the 'precession of the equinoxes'. Plenty of sources show that up to about the end of the sixteenth century AD, which is pre-modern if you like, it was often called 'the motion of the eighth sphere' (e.g. 'De motu octavae sphaerae', I.Werner, Nuremberg, 1522). 'Precession of the equinoxes' is a _relatively_ modern term.

It's also not very clear that the term 'precession of the equinoxes is out of date, either. It's not equal to either the 'precession of the equator' or the 'precession of the ecliptic' -- it's the combined effect of the two of them together. Is there a replacement terminology for that? —Preceding unsigned comment added by 86.14.225.140 (talk) 18:15, 1 March 2009 (UTC)

You are correct that in non-technical discussions (no detailed mathematics), precession of the equinoxes is still in use, which is one of the tweaks I am making. All of the terms now in the lead are in current use, but the motion of the eighth sphere, trepidation, and similar terms are not, so I don't think they belong in the lead. Of course, they do belong in the historical section. The technical replacement for precession of the equinoxes is general precession which is the combination of precession of the equator and precession of the ecliptic. It was not changed in 2006, so it was also the combination of lunisolar precession and planetary precession. — Joe Kress (talk) 14:38, 2 March 2009 (UTC)

Precession direction

Isn't the direction of the precession wrong in the picture? The angular momenta is supposed to be conserved and thus the precession must be in the same direction as the body is spining. Besides I actually found another copy of the same image on the internet with the correct direction of the precession.

Samuel —Preceding unsigned comment added by 83.250.175.92 (talk) 22:23, 5 March 2009 (UTC)

No! Precession is in the opposite direction to the spin - otherwise it would be called PROcession of the equinoxes! TomNicholson (talk) 15:33, 18 November 2009 (UTC)

Question

If the Earth's axis of rotation is orthogonally projected to the plane of the solar system, is the projected direction parallel to the major axis of the orbit? If so, is there an explanation of such a phenomenon in gravitational terms? 132.70.50.117 (talk) 10:23, 5 May 2009 (UTC)

The two are not parallel because the two precessions discussed in the article have different periods, the precession of Earth's rotational axis in inertial space and the precession of the major axis of Earth's orbit, also known as anomalistic precession (the last section in the article), which is the advance of the line of apsides (perihelion and aphelion). The projection of Earth's rotational axis onto its orbital plane is from/to the solstices (winter/summer), now about December 21/June 21. Conversely, the major axis of the orbit is from/to the apsides (perihelion/aphelion), now about January 3/July 5. The Gregorian calendar is "tuned" to the equinoxes and solstices, so their dates remain relatively fixed, even though their precession is 50.3"/year in inertial space. But the dates of the apsides move forward in the Gregorian calendar because they advance 11.6"/year in inertial space. The sum of 61.9"/year causes a relatively rapid calendrical advance. In terms of years, the average length of the Gregorian year is 365.2425 days which is nearest the vernal equinox year of 365.242374 days. The anomalistic year at 365.259636 days is even longer than the sidereal year. The lines were parallel about the year 1225 when the length of autumn equalled that of winter and spring equalled summer. The lines will be perpendicular about 3825 causing spring to equal autumn and causing the lengths of winter and summer to be maximally different. This is discussed in The Lengths of the Seasons (on Earth) by Irv Bromberg. — Joe Kress (talk) 03:12, 6 May 2009 (UTC)

Thanks, that's very interesting. Which of the two precessions is supposed to be explained by the equatorial bulge? 132.70.50.117 (talk) 11:46, 6 May 2009 (UTC)

Current precession cycle?

When does the current precession cycle finishes? Echofloripa (talk) 23:14, 16 July 2009 (UTC)

Precession never ends. It takes about 26,000 years for the vernal equinox to precess completely around the ecliptic (the apparent path of the Sun on the celestial sphere). At the same time Earth's axis precesses causing the celestial north pole to trace a circular path on the celestial sphere. So we can say that our current north star, Polaris, will cease to be the north star in only a thousand years and will not be the north star again until 26,000 years from now. You may be thinking of the position of the vernal equinox in specific zodiacal constellations. Because the ecliptic is astrologically divided into twelve constellations, the vernal equinox spends roughtly 2,150 years in each constellation. While the vernal equinox was in the constellation of Pisces, we were in the Age of Pisces. When precession causes the vernal equinox to move into the next zodiacal constellation, Aquarius, the Age of Aquarius begins. Astrologers differ on when this occurs, varying from 1447 AD to 3621 AD, although many think we entered the Age of Aquarius during the 20th century. Also see astrological age. — Joe Kress (talk) 02:33, 17 July 2009 (UTC)

Name change

Zbayz changed the name of this article from "Precession (astronomy)" to "Axial precession (astronomy)" without any discussion whatsoever. His stated reason was "This article only discusses precession of the rotational axis. It says nothing about perhelion precession." both of which are wrong. Although the dominant focus of the article is 'axial precession' it does mention "planetary precession" now renamed "precession of the ecliptic" and "anomalistic precession". The proper name for 'axial precession' is either the old name "lunisolar precession" which was used for at least two centuries or its new name, since 2006, "precession of the equator". Furthermore, this article mentions "planetary precession" now renamed "precession of the ecliptic" which is the gradual change in the inclination of Earth's orbit relative to the invariable plane of the solar system and the combination of "precession of the equator" and "precession of the ecliptic" which is called "general precession" and has been so named for at least one and a half centuries. This article also has a section on "anomalistic precession" at the end of the article, which Zbayz calls "perihelion precession", but refers to an article named apsidal precession.

Zbayz also did not follow Wikipedia naming conventions for disambiguating articles. Do not use a parenthetical qualifier unless two or more articles have the same name. — Joe Kress (talk) 22:42, 25 August 2009 (UTC)

I think that the different motions should have their own articles. Then we can call this one "precession of the equator". Zbayz (talk) 12:35, 26 August 2009 (UTC)

Other thoughts: Since "ecliptic" is about the Earth, "precession of the ecliptic" doesn't seem a very good name for variation in the orbital inclination of celestial objects in general. Also the phrase "anomalistic precession" seems to be hardly used at all. Zbayz (talk) 16:30, 26 August 2009 (UTC)

Astronomy/Astrology splitting suggestion

I suggest to split out astronomic scientific material of this article and through out pseudoscience astrological data. There is no reasons to cover both this topics in the article. Astrology - is a cultural phenomenon, not the science. Thank you. —Preceding unsigned comment added by 95.133.136.138 (talk) 14:14, 7 September 2009 (UTC)

Are you suggesting that science is not a cultural phenomenon? TomNicholson (talk) 15:36, 18 November 2009 (UTC)

Astrology was (and for a long time) the reason for astronomy. I notice it has lost cultural colective favour (except in Biblical instances such as the star of Bethlehem, kings choosing when to procreate etc - where it is regarded as a part of gods plan) When you say ' through out' pseudoscience astrological data. I presume you mean - 'throw out'. This vandalism you blithely propose is just that. 188.220.186.57 (talk) 15:29, 30 August 2011 (UTC)

20 minutes per year - is that right?

From the Effects Para:

Thus, the tropical year, measuring the cycle of seasons (for example, the time from solstice to solstice, or equinox to equinox), is about 20 minutes shorter than the sidereal year, which is measured by the Sun's apparent position relative to the stars. Note that 20 minutes per year is approximately equivalent to one year per 25,771.5 years, so after one full cycle of 25,771.5 years the positions of the seasons relative to the orbit are "back where they started".

I calculate that 20 minutes per year gets 360 degrees of precession in 72 years not in 25,771.5 years. What's going on here? Who did this math?Lkoler (talk) 04:53, 16 October 2009 (UTC)

From Values, the constant rate of precession is 5,028.796195"/Julian century or 50.28796195"/Julian year. A Julian year is 365.25 days. The number of arcseconds in 1° is 3600".

72 years is only 1° of precession via ${\tfrac {3,600}{50.28796195}}=71.58770927$

Multiply by 360° to get 25,771.5 years via $360\times 71.58770927=25,771.57534$

Calculate 20 minutes via ${\tfrac {50.28796195}{1,296,000}}\times 365.25\times 24\times 60=20.40853122$

— Joe Kress (talk) 06:21, 16 October 2009 (UTC)

Thanks for responding so soon. I'm sorry but, I calculated the precession to be 3.34 seconds per year and not 20.4 mins. If you look here [1] in the Journal of Theoretics, you will see how I got there. The problem in your calculations is that you are assuming that the tropical year (365.25631 days of 86400 secs. each) and the sidereal year (366.25631 days of 86164.0905 secs. each) are not the same. But, because of the extra rotation for the sidereal year (see Donald Sauter’s brain teaser at [2] to find out why), you will see why you were wrong. Don't be surprised about this: almost all textbooks and astronomical references are wrong on this point, too. Lkoler (talk) 22:42, 16 October 2009 (UTC)

Wikipedia does not allow fringe theories in its articles. It only allows what "almost all textbooks and astronomical references" state. — Joe Kress (talk) 08:58, 17 October 2009 (UTC)

It's not a theory to state a fact. 366.25631 sidereal days is exactly one (1) day longer than most people use in their calculations (note that calculations are also not theories) for the sidereal year. Please get your definitions correct if you are going to quote Wikipedia rules.Lkoler (talk) 15:27, 17 October 2009 (UTC)

Lkoler, the number of days (sidereal or otherwise) per year (sidereal or otherwise) has no bearing on the statement, "20 minutes per year is approximately equivalent to one year per 25,771.5 years" which can be verified using basic high school maths of proportion and ratio, as patiently demonstrated by Joe Kress above. TomNicholson (talk) 16:13, 18 November 2009 (UTC)

I agree with Lkoler. Common sense indicates otherwise. 20 extra minutes per year? There are 24 hours in a day. 20 minutes is 1/72 of 24 hours. Certainly, the stars(celestial sphere) are not adjusting 1/72 every year. That would put precession at super speed and we would be cycling through the constellations every 72 years. Something is wrong in the math. Common sense just doesn't support it.

If time passes by, then the celestial sphere moves. Actually, the celestial sphere moving is the cause of the time measurement, not the other way around. So, if the celestial sphere moves 20 extra minutes from what it was the year before, it must move 1/72 of the angle of the sky. 360 degrees multiplied by 1/72 is 5 degrees of the sky every year. And (divide 5(18,000 arcseconds) degrees by 365 days)would have to move about 50 arcseconds every day. This does not happen.

Remember, an extra 20 minutes per year would mean the celestial sphere shifts 5 degrees every year, and 50 arc seconds per day. It's not 50 arc seconds per day, it's 50 arc seconds per year. Hipparchus could tell you that. It's obvious, but a lot of people are getting it wrong for some reason.--Markblohm (talk) 18:10, 8 December 2009 (UTC)

The math is fine. The difference of 20 time minutes between the tropical and sidereal year means an adjustment in the sky of (20 minutes) / (1 year) or ~0.0038% of the sky, or ~0.86 arc-seconds / year. Dragons flight (talk) 18:29, 8 December 2009 (UTC)

An extra 20 minutes causes the earth to point 5 full degrees different in relation to the stars. Just like an extra 12 hours would cause a 180 degree change. Think about it. 1 tropical year is when the sun is back to the same position in relation to the earth. A tropical year plus 12 more hours would have us facing the opposite way. Likewise, 20 extra minutes would have us facing 5 full degrees off from the previous. Say at time zero you are facing 0 degrees angle. 1 day later you are also facing 0 degrees angle. 100 days later, still 0 degrees angle. 1 tropical year later exactly, and you are facing 0 degrees angle. But what is being said is that 20 more minutes is being added to the tropical year. That is 5 full degrees extra of the celestial sphere(because we have a 24 hour day). 1 tropical year later, the sun is in the same position. The shift is 50 arcseconds of the celestial sphere/year(this shift of 50 arcseconds has been observed for a long time, and is much less than 20 minutes.--Markblohm (talk) 21:06, 8 December 2009 (UTC)

I now understand why the time of around 20 minutes is being found. The math is not wrong. The equation is wrong. Precession is independent of the spin of the earth, so to account for the total seconds of the spin of the earth in a year when finding the real time change of precession is incorrect. Even if we say that it is wobbling, the wobbling is occurring independent of the spin. The "wobble" is only about 50 arcseconds, and is independent of the spin of the earth. We add the extra wobble time onto the total spin at the end of the year. After all, 50 arcseconds of angle(or 3.3 seconds of real time) is the change each year, which is clear to observing astronomers in the past and present.--Markblohm (talk) 01:47, 9 December 2009 (UTC)

Everyone agrees that precession is about 50"/yr, but some disagreement exists concerning its equivalent time per year. The value of 3.34 seconds per year which Houmann and Lkoler use was apparently obtained by multiplying 9.12 ms per sidereal day by the number of sidereal days in a tropical year, which disagrees with the value of 20 minutes per year usually quoted. A solution is to obtain both from angular precession.

In 1900, precession was specified as 50.265"/tropical year, which decreased slightly since then. Since 1984, precession has been specified as 50.29" per Julian year because the length of the Julian year (365.25 days of 86,400 SI seconds each) is known with ultimate precision and it never changes. In astronomy, unlike geometry, any 'year' or 'day' is always 360° or 1,296,000", regardless of its length. Thus a sidereal year, a Julian year, a tropical year, a sidereal day and a solar day all have an angular duration of 1,296,000". To shorten the equations, I'll use 'd' for the dimension of a sidereal day of 86,164 seconds and 'D' for the dimension of the slightly longer solar day of 86,400 seconds. Both the numerical values and their dimensions are critical in this analysis.

First determine the time per sidereal day using the new value of precession:

$50.29''{\text{/yr}}{\tfrac {86,164\ {\text{s/d}}}{(365.25\ {\text{D/yr}})(1,296,000''{\text{/D}})}}=9.154\ {\text{ms/d}}$

Then determine the time per Julian year also from that precession:

$50.29''{\text{/yr}}{\tfrac {(365.25\ {\text{D/yr}})(86,400\ {\text{s/D}})}{1,296,000''{\text{/yr}}}}=1225\ {\text{s/yr}}=20.42\ {\text{minutes/Julian year}}$

To convert time per sidereal day into time per Julian year, combine the conversion factors by rearranging the first equation so that it can replace the angular precession in the second equation:

$9.154\ {\text{ms/d}}{\tfrac {(365.25\ {\text{D/yr}})(1,296,000''{\text{/D}})}{86,164\ {\text{s/d}}}}{\tfrac {(365.25\ {\text{D/yr}})(86,400\ {\text{s/D}})}{1,296,000''{\text{/yr}}}}$

$=9.154\ {\text{ms/d}}{\tfrac {(365.25\ {\text{D/yr}})^{2}}{\ }}{\tfrac {(1,296,000''{\text{/D}})(86,400\ {\text{s/D}})}{(1,296,000''{\text{/yr}})(86,164\ {\text{s/d}})}}=1225\ {\text{s/yr}}=20.42\ {\text{minutes/Julian year}}$

Time per sidereal day must basically be multiplied by the square of the number of days per year, not by just the number of days per year (other terms eliminate the square dimensions). So 3.34 s/yr is wrong, whereas 1225 s/yr is correct. — Joe Kress (talk) 21:26, 9 December 2009 (UTC)

No matter what equations you throw up, 50 arcseconds can never be anything other than 3.3 seconds of real time, just as 180 degrees can never be anything but 12 hours of real time. The spinning earth arcseconds do not get factored into the precession equation. You have made the very common error.--Markblohm (talk) 01:58, 10 December 2009 (UTC)

It is much simpler to find the difference between the tropical and sidereal years. According to the Useful constants of the International Earth Rotation and Reference Systems Service (IERS), the sidereal year is 365.256363004 days and the tropical year is 365.242190402 days, where both are in mean solar days of 86,400 seconds each at 2000 January 1 noon Terrestrial Time. Their difference is 0.014172602 days or 20.408547 minutes.

If the Sun was directly overhead (0°) at the beginning of a tropical year, the only way it could be directly overhead (0°) at the end of one tropical year is if that tropical year had a whole number of days. But a tropical year is 365.2422 days. The extra 0.2422 days beyond an integral number of days means the Sun must be near the western horizon near sunset. — Joe Kress (talk) 05:06, 10 December 2009 (UTC)

It appears to me that Lkoler and Markblohm are consistently confusing two kinds of circular motion. 50 arcsec of daily rotation takes 3.3 sec "of real time", but 50 arcsec of annual revolution takes 365 times as long. But that's so obvious that someone ought to have pointed it out by now, so have I missed something? —Tamfang (talk) 06:04, 10 December 2009 (UTC)

Good point. If the 365.25 D/yr term is removed from my second equation immediately above, then the remaining equation is

$50.29''{\text{/yr}}{\tfrac {86,400\ {\text{s/D}}}{1,296,000''{\text{/yr}}}}=3.35\ {\text{s/D}}$

Because this result is 3.35 seconds, albeit per day rather than per year, it implies that Lkoler and Markblohm have overlooked a term. I find the term "of real time" unnecessary and mysterious. — Joe Kress (talk) 08:37, 10 December 2009 (UTC)

Ah, but we're not measuring 50 arcseconds of annual revolution. If we measure 50 arcseconds of annual revolution(earth's orbital circumference) then we do get 20 minutes. It indeed takes 20 minutes for the earth to travel 50 arcseconds of the orbital circumference. But the observed 50 arcseconds of celestial sky movement is 1/365 of that. Because the earth spins 365 times in its orbital revolution. That is what the famous 50 arcesonds is, 50 arcseconds of celestial sky displacement viewed from the earth every year, cumulative for the whole year. An ancient astronomer like Hipparchus only observed 50 arcseconds of celestial sky displacement per year. He could not have observed 50 arcseconds of the orbital circumference.--Markblohm (talk) 15:26, 10 December 2009 (UTC)

Why do you say "the observed 50 arcseconds of celestial sky movement is 1/365 of that"? You seem to be saying that the annual change in the position of the equinoxes affects the length of the day rather than the length of the (seasonal) year.

As for what an ancient astronomer could have observed — Is it possible now to observe directly when the equinox occurs? —Tamfang (talk) 18:37, 10 December 2009 (UTC)

Hi Joe(Kress.) May I ask, why are you multiplying 50" by 365?--Markblohm (talk) 16:07, 10 December 2009 (UTC)

Because the relevance of the angle is its relation to the year, not the day. As a thought experiment, consider how the length of the seasonal year would be affected if, leaving everything else unchanged (the amount of tilt, the precession rate, the size of the orbit and so on), the rotation period were doubled. —Tamfang (talk) 19:17, 10 December 2009 (UTC)

Dear Tamfang, You have highlighted the problem exactly. I assume by double rotation period you mean two times faster. But even if you mean two times slower it doesn't matter. If the earth were spinning(rotating) at double it's current speed(and precession rate remained the same), the precession rate would be exactly 3.3 seconds. The seasonal year would still be the tropical year length PLUS the 3.3 seconds. The spinning earth doesn't multiply the precession, which is what is being said in these equations above. You are multiplying the precession time by the amount of days. That is incorrect. The precession movement is independent of the spinning, so you must ADD the extra precession time onto the length of the tropical year.

The sun "travels" in the sky 473040000arcseconds(365(roughly) x 360 degrees) in one year. The celestial sky travels 473040000arcseconds PLUS 50 arcseconds. In other words, the sun travels 31536000 time seconds(525600 minutes) in one year. It then takes 3.3 seconds(50 arcseconds) to catch up on the vernal equinox. These arcseconds/time seconds have been accumulating all year. On the half year the sun would take only 3.3/2 seconds(25 arcseconds to catch up). The sun travels through the sky roughly 473040000arcseconds in one year, the celestial background is travelling 473040050arcseconds.--Markblohm (talk) 20:23, 10 December 2009 (UTC)

The daily rotation needs to catch up SIX HOURS at the end of the year, not 3.3 seconds! The axial precession is about seasons, not days, so it's the yearly motion (not the daily motion) that needs to catch up to it. —Tamfang (talk) 23:02, 10 December 2009 (UTC)

The equinox is the moment when the plane of Earth's equator crosses the center of the Sun. That plane is independent of the phase of the daily rotation. —Tamfang (talk) 05:56, 14 December 2009 (UTC)

Markblohm: My equation for the annual precession time is independent of Earth's rate of rotation. If Earth rotated twice as fast, its day would be half as long, 43,200 seconds, but the number of those shorter days in a year would be twice as many, 730.5 days, so there would be no change in the precession time in a Julian year, 20.42 minutes. The angular speed of the Sun in its apparent orbit around the Earth can also be used to determine the time it takes an equinox to precess along the ecliptic in one year. The Sun moves at an average speed of (1,296,000"/yr)/(365.25 days/yr) = 3548"/day. Hence the annual precession time is (50.29"/yr)/(3548"/day) = 0.014174 day/yr = 20.4 minutes/yr. You agree that this is correct via your statement "It indeed takes 20 minutes for the earth to travel 50 arcseconds of the orbital circumference." But you then state "An ancient astronomer like Hipparchus only observed 50 arcseconds of celestial sky displacement per year. He could not have observed 50 arcseconds of the orbital circumference." To Hipparchus these were a single concept. A little history is warranted:

Hipparchus and Ptolemy regarded the Universe as the stars, five planets and the Sun and Moon all revolving around a motionless Earth at their center. The orbit of the Sun around Earth (the mirror image of Earth's orbit around the Sun) marked the ecliptic, which was also the center of the zodiac, a band which contained the orbits of the Moon and the five planets. But the dominant motion (now known to be due to Earth's own rotation) was parallel to the celestial equator, which was inclined to the ecliptic by 23.5°. Hipparchus discovered that the stars had an additional motion parallel to the ecliptic called precession. A ring mounted in the plane of the equator was used to determine the date/time of the equinoxes, which occurred when the shadow of the Sun cast by the upper part of the ring fell on its lower part, meaning that the Sun was crossing the equator from south to north or vice versa. The location of the equinoxes and solstices among the stars was determined using the Moon as an intermediary because it was visible both during the day and night.

Hipparchus determined that the ecliptic longitude of Regulus, a star within 0.4° of the ecliptic, was 295⁄6° east of the summer solstice (119.8° east of the vernal equinox) in 129 BC. Ptolemy determined that the longitude of this same star was 321⁄2° east of the summer solstice (122.5°) in AD 139, so Ptolemy noted that the star had moved 22⁄3° in 265 years (we now say the summer solstice moved, not the star). From this he concluded that precession was 1° in 100 years or 36"/yr, the same precession that Hipparchus had deduced (precession is now known to be 1° in 72 years or 50"/yr). Today (J2000), Regulus has a longitude of 150.2°, so the solstices and equinoxes have precessed 30.4°, about one zodiacal sign, in 2,129 years, a precession of 51.4"/yr (using Hipparchus' imprecise position), about the same as the modern value. Hipparchus and Ptolemy used twelve zodiacal signs of 30° each to measure longitude. Precession measurements never used or depended upon either the solar day or the sidereal day. They also determined the angular distances (called declinations) of those stars north or south of the equator. They noted that stars near the autumnal equinox which were north of the equator later moved south of the equator (south to north near vernal equinox). Both methods resulted in the same precession. This "celestial sky displacement" was along the ecliptic, which was the "orbital circumference" of the Sun's orbit.

By the 18th century, astronomers no longer needed to use ancient observations, but were able to determine precession by comparing the positions of stars determined at times less than a century apart. These modern astronomers also discovered that some stars had large proper motions (Sirius moves about 1.3" every year), which had to be excluded from those stars used to measure precession. The pointing accuracy of the largest telescopes is now about 5", so precession is observeable within one year and must be considered to even locate stars at high magnifications if a star catalogue several years old is used.

Although you state that the spinning Earth does not multiply the precession, that is exactly what you do via "(365(roughly) x 360 degrees)". — Joe Kress (talk) 22:35, 15 December 2009 (UTC)

Image: Earth_precession.svg‎

This image, currently shown at the top of the page, is confusing in that the rotation of the axis is shown to be in the same direction as the spin of the planet. A reading of the description (click on the image to go to the image page and see the summary below) reveals that the axial rotation shown in this image is "relative to the direction to the Sun at perihelion and aphelion", (ie. it relates to orbital precession - even though the orbit is not shown), rather than axial precession, the title of this page. The commonly understood meaning of the unqualified term "precession" is axial precession relative to the fixed stars, where it moves counter to the spin (hence the term PREcession, rather than PROcession).

This image appears here: http://earthobservatory.nasa.gov/IOTD/view.php?id=541 with the words "orbital precession" at the top. Why have these words been removed? Is someone purposefully trying to confuse here? TomNicholson (talk) 16:57, 18 November 2009 (UTC)

The image Earth precession as originally uploaded in 2008 showed the correct CW precession, but Mysid thought it wrong so he uploaded an incorrect CCW image in 2009. I reverted to the original image for the reason I explained at File talk:Earth precession.svg, so the image at the top of this article now shows the correct CW precession. I have also corrected the file's description, which was indeed wrong, because axial precession has nothing whatsoever to do with orbital precession, also called apsidal precession. I don't know why NASA Earth Observatory has an erroneous CCW precession image on its site. Ironically, the associated animation shows a correct CW precession. — Joe Kress (talk) 02:38, 15 January 2010 (UTC)

Errors in the "Cause" section of the article (still!)

I would edit this section directly if it were not for the fact that the image has errors, and I don't have a replacement.

The vertical cyan (pale blue) arrows and the yellow arrows (shown at the equator) indicating the torque should not be present at the equinoxes - the symmetry of the earth-sun relationship at the equinoxes (and hence the lack of torque) is more clearly seen if the orbital motion (which is irrelevant) is ignored, or with reference to the first equation in the following section where it is seen that the torque vanishes at the equinoxes when delta (and sine delta) become zero. This is mentioned in the text: "The magnitude of the torque from the sun (or the moon) varies with the gravitational object's alignment with the earth's spin axis and approaches zero when it is orthogonal" - that is, at the equinoxes.

Is anyone else bothered by the statement, "This average torque is perpendicular to the direction in which the rotation axis is tilted away from the ecliptic pole, so that it does not change the axial tilt itself"? It is the axis of the average torque which is perpendicular to the direction in which the rotation axis is tilted away from the ecliptic pole, and I don't believe it is obvious why this does not change the axial tilt itself. Actually, the "force" of the torque tends to "want" to lessen the axial tilt - a tendency which is thwarted due to the (counter-intuitive) properties of spinning objects. If at this point you think I'm completely crazy :) then try the following ...

Take a bicycle wheel (minus the bike) and hold the spindle in your hands so that your arms are like the front forks of the bike. Spin the wheel. Now (carefully!) try and change the axis of rotation. Weird, isn't it?!

TomNicholson (talk) 18:07, 18 November 2009 (UTC)

Yeah, the image is wrong. The cyan arrows are all wrong, too. I think the author of the image had a misconception of how precession works. I'll try to get the image deleted for now, but it would be nice to have a replacement. ErikHaugen (talk) 19:41, 9 June 2010 (UTC)

On the Introduction

The Introduction mentions a period of 26,000 years and a cone of unspecified size and direction.

A cone is not necessarily circular.

I think that it would be well to indicate, in the introduction (but not necessarily in these terms), that the direction of the Earth's axis of rotation moves in a circle of radius 23.5° about the Earth's orbital axis ; and that the current direction of motion is approximately along a line from Polaris towards some other well-known star. The one which it will be near in AD 4000 might suit ; perhaps in Cepheus, probably Gamma, I think - or maybe use the Great Square.

The circularity and size are present, buried lower down.

82.163.24.100 (talk) 14:33, 24 November 2009 (UTC)

Great Year

Could someone perhaps take a look at the related Great Year article? A number of questionable edits have been made recently. The new "Confusion of the Platonic Year with Precession" section may be of particular concern. Pollinosisss (talk) 07:20, 14 December 2009 (UTC)

Date of vernal equinox inquiry

In 12,850 years, will the date of the vernal equinox in the Northern Hemisphere still be March 20th or will it be October 20th? Why has the date decreased from March 22nd a century ago to March 20th now? Keraunos (talk) 16:14, 1 January 2010 (UTC)

See Gregorian calendar for information on how accurately that calender preserves the position of the vernal equinox. The Gregorian calendar corrects the problems with the Julian calendar by omitting a leap year in centurial years, unless the centurial year is evenly divisible by 400. So this correction was done in 1900, but not in 2000. If you only look at the period between 1910 and 2009, the Gregorian calendar behaves like the Julian calendar. --Jc3s5h (talk) 18:54, 1 January 2010 (UTC)

Does this chart help? —Tamfang (talk) 03:58, 2 January 2010 (UTC)

Thank you so much for the clarification. I had always wondered about that problem. Happy New Year 2010! Keraunos (talk) 11:06, 2 January 2010 (UTC)

Irv Bromberg (Sym454.org) has calculated the drift of the equinox for several calendars (many proposed) at Leap cycle drift relative to the northward equinox, where the upper black line is the equinox drift of the Gregorian calendar. Assuming it continues linearly (a dangerous assumption), there will be about eight days of drift, meaning that the vernal equinox will occur eight days earlier than it does now, about March 12, in 13,000 years. The almost vertical dashed red line shows the rapid drift of the equinox in the Julian calendar, which drifted about ten days in only 1,300 years between roughly AD 300 and AD 1600. This shows that any fixed arithmetic calendar, that is, a calendar that adds a day, week or month at regular intervals, is useless beyond a few thousand years, at least for keeping the vernal equinox near some spcific date. — Joe Kress (talk) 04:19, 15 January 2010 (UTC)

Additional relevant astrology links

If the article is going to discuss Mithraism in detail, it might as well give a nod to "tropical astrology" vs. "sidereal astrology" and the so-called "Age of Aquarius"... AnonMoos (talk) 13:03, 15 January 2010 (UTC)

Voting for Simplified Introduction

The Articles INTRODUCTION is overly complex and therefore poorly explained .The diagrams a bewildering to none-astronomers.Is this the object of wikipedia? A celestial sphere, the sun at the centre with a zodiac ring drawn as a band on the sphere, would be a better introduction.This would: (a) Root the observations to a fairly universally known (and therefore familiar) set of coordinates. (b) Remove the visual confusion (c) Allow for a clear intersection planes to be inscribed on the sphere. I would butte in a fix it myself but its really up to those most impassioned in this debate.Anyway were does a layman get the computer program to generate the diagrams?Ha! —Preceding unsigned comment added by Chasludo (talk • contribs) 23:55, 16 January 2010 (UTC)

Modern estimates of precession

I moved Cuvier's 1825 estimate (recently added by Hmschallenger) to the Middle Ages onwards section, at least temporarily. It certainly does not belong in the Hipparchus section just because it uses observations by Hipparchus. It should be in a "modern estimates" section which currently does not exist. A good start would be Evolution of adopted values for precession by Jay H. Lieske. — Joe Kress (talk) 23:54, 24 January 2010 (UTC)

Article used without reference or credit

I cannot see a credit at

 http://www.crystalinks.com/precession.html

to this article.

There is a simple one link on a line by itself that is labeled "wikipedia" but no indication that the article is copied. The link is not labeled "orginal wikipedia article" or the like. I cannot see that the mere label "wikipedia" constitutes either a reference or credit and the owner appears to have a commercial site (psychic this or astrological that)

In the case of

 http://encycl.opentopia.com/term/Precession_of_the_equinoxes

there is a fine print credit to wikipedia without actually naming this article. —Preceding unsigned comment added by Grshiplett (talk • contribs) 00:52, 4 March 2010 (UTC)

Many sites that use material from Wikipedia do not give adequate credit to Wikipedia. Both of the sites you mention give at least some credit to Wikipedia. The Crystalinks site used the images from an old version of this article but the text is mostly their own. The Wikipedia link they give is to the current version of this article. The Opentopia site is basically a Wikipedia mirror. They also give a pointer to the current version of this article. — Joe Kress (talk) 02:14, 4 March 2010 (UTC)

Strange reference

In the section about Mayans it states that professional scholars do not hold the opinion that the Mayans where aware of precession. The reference is written by a Mayan scholar who does hold that opinion and states her opinion uncategorically in the article by saying... "The end of the baktun on the winter solstice is not a coincidence, and this mathematical feat is certainly a sign of a sophisticated link between Maya astronomy and mathematics."Yourliver (talk) 14:46, 11 May 2010 (UTC)

The sentence you quote does not show she thinks that they were aware of the precession of the equinoxes (and is replying to a rather different claim). The cited article goes no further than to say that it is not impossible that the Mayans were aware of this precession and that some of their records could have been useful in calculating it. A better reference would be good (such as to that author's book) but the current claim is reasonably well supported by this citation, I believe.

All the best. –Syncategoremata (talk) 20:30, 11 May 2010 (UTC)

Actually, you are quite correct. I have reread the article several times. Oddly, she holds the opinion that the Maya were aware that the seasons of winter and summer moved through the solar year and calibrated their calendar to reflect this. However she does not believe that spring and fall could be calculated with any precision at all. I withdraw my statement that Miss Milbrath even remotely can be considered a scholar.Yourliver (talk) 00:37, 12 May 2010 (UTC)

Cause

There's some dispute here about what causes precession of the equinoxes. I got rid of the picture, which I think is misleading. My edits have been largely reverted here: [3]. There is more discussion here: [4] if anyone is interested. ErikHaugen (talk) 23:03, 14 June 2010 (UTC)

Here is the summary, as I see it: There are two competing explanations for precession of the equinoxes. At least one is totally bogus. The first considers the "vertical"(north/south) and "horizontal"(sunward) components of the forces on the halves of the bulge by the sun - the vertical components point in opposite directions and cause a torque. The second, (which I assert is right), is that the difference in magnitude, due to the difference in distance to the sun, causes a torque, since the sun pulling on the closer half of the bulge causes a torque in one direction, and the sun pulling on the far half of the bulge causes a torque in the other direction - the close half wins since the distance to the sun is closer. Note that the two explanations disagree on basic things like "At the solstice, in what direction is the torque caused by the sun pulling just on the far half of the bulge?" And also note that the first explanation does not have anything to do with the inverse square law and gravity getting weaker with the difference in distance. Please say if you do not agree with this assessment. To resolve, can we cite sources? ie, can someone suggest a particular page of a particular astronomy book? (I'd sure appreciate one that freely available online.) I'll offer:

• Encyclopedia of planetary sciences By James H. Shirley, Rhodes Whitmore Fairbridge, p.657 [5] - "Except for twice a year at the vernal or the autumnal equinox when the Sun crosses the equator, there is a tendency for the gravitational attractions of both the Sun and Moon to pull the Earth's equatorial bulge into alignment with their respective planes"
• The ever-changing sky: a guide to the celestial sphere By James B. Kaler p.159 [6] - "the Moon and Sun are continuously moving back and forth across the terrestrial rotation bulge. The force acting to produce precession is at a maximum when these two bodies are at the solstices, and are zero when they are at the equinoxes and in line with the Earth's bulge and center." ErikHaugen (talk) 00:42, 15 June 2010 (UTC)

Also consider Axial_precession_(astronomy)#Equations and its references - the torque varies with the angle of declination, etc. ErikHaugen (talk) 00:53, 15 June 2010 (UTC)

I still think you are mistaken, and so are many others, and in fact it is very difficult to find any literature about precession which try to explain the cause. Most quickly pass this part and then directly jump to the results. There are some however:

• 'Note that the net gravitational force on any portion of the earth’s ring is toward the ecliptic plane, producing a torque in the equatorial plane.' (page 12, physnet2.pa.msu.edu/home/modules/pdf_modules/m77.pdf) It is not for nothing they spell out the idea of the ring.

Looking at this picture we see the vectors of the differential gravitation for different angular distances from the disturbing body. It is true that there is no tangential component whatsoever if the sun/moon are on the equator, so no precession takes place. But that does not mean that there is none at the equinoxes. Because the equatorial bulge is not a single point, it is still a whole ring. That is the crucial issue. Even if during the equinoxes the point of the bulge 'under' the sun is on the equator and not doing anything, the extreme (solstice) points are then on the terminator, and they still are subject the tidal force towards the equator. In fact it does not matter where the sun is somewhere, equinox, solstice or somewhere in between, that part of the bulge north of the equator experiences a force to the south (strongest at the northernmost part, going down to zero to the equator) and vice versa for the southern half. The two forces of course are part of a torque, and there the precession begins. It also means that the precession force is always the same, the whole year around (or the whole month around if you talk about the moon). How otherwise could the precession speed of 50" per year be a constant (ignoring secular changes and the several magnitudes smaller irregularities of the nutation) as everybody agrees ? No astronomer has ever observed that precession would come to a standstill at the equinox and reach full speed at the solstice; it just progresses at a constant rate during the year (and the month).

So the torque does not change over time. What does change on the bulge when the sun (or moon) appears to go round the ecliptic plane is the strength of the radial component, the difference being largest at the equinoxes. But a radial component does not contribute to a torque. If there is anything 'bogus' there is that.

Please restore the picture. It is still correct.

What about the quote of Kaler and others above? I am sorry, no doubt dr Kaler is an excellent astronomer, but he is specialised on stars and nebuale, and therefore less at home in celestial mechanics. His statement on variable precession is utterly wrong. --Tauʻolunga (talk) 08:51, 15 June 2010 (UTC)

Thanks for your response. This quote does not conflict with my (the second above) explanation: 'Note that the net gravitational force on any portion of the earth’s ring is toward the ecliptic plane, producing a torque in the equatorial plane.' Like many quotes that one finds when trying to explain this phenomenon, it is pretty vague. You say "it is very difficult to find any literature about precession which try to explain the cause" and "(Kaler's) statement on variable precession is utterly wrong." - Can you find _any_ that are "right?" It is really problematic to have one unsourced explanation for astronomical phenomenon on Wikipedia when it is a different explanation that is carried in books/etc written by astronomers. ErikHaugen (talk) 13:59, 15 June 2010 (UTC)

Here I'll try to answer some of your points, although ultimately this is beside the point; it doesn't really matter if we "convince" each other, we need to find sources. "How otherwise could the precession speed of 50" per year be a constant" - it's essentially constant from year to year, but during the year it varies. The average over any given 6 month period is always about the same :). "No astronomer has ever observed that precession would come to a standstill at the equinox and reach full speed at the solstice" - I have no idea if this sinusoidal variation can be observed or not; precession happens at a very small rate - again, do you have a source that analyzes this? I want to go back to something I mentioned above - your explanation claims that the torque on the far-side bulge is pulling the far side of the bulge down toward the ecliptic. But, think about what would happen if the ring/donut (consider the bulge in isolation) were fixed about a point in the very center (via tiny spokes of insignificant mass). Now, tie a string to the far side and pull. Does this pulling cause a torque on the donut pulling it toward the plane of the ecliptic? No, it pulls it away from that plane, so that the far side of the donut will go away from the ecliptic. My point is that the way you are breaking down the force arrow components is not right. ErikHaugen (talk) 16:13, 15 June 2010 (UTC)

I already quoted the lab exercise of Physnet. Admittedly not too clear, indeed some more literature would be welcome. I shall keep my eyes open.

50" per year, which is more than 0.1" per day, is a huge value for positional astronomy, where an accuracy of 0.01" is possible. An astronomer who does not correct for precession on a daily basis commits a major error. Any supposedly seasonal variation would have been discovered centuries ago, especially since it is accumulating. None has ever been found. (There is a yearly nutation term with an amplitude of 0.1261", but that is another story).

Pull? Which direction do you want to pull? You are not doing the pulling, gravity does, and gravity can only pull in the directions given in the picture again. They all, no exception, have a component towards the equatorial plane, and not away from it. Sorry, you are erring, not me. --Tauʻolunga (talk) 03:42, 16 June 2010 (UTC)

As I explained above, the Physnet quote is totally consistent with my explanation; it is too vague to be useful in this particular discussion. wrt consistent or not observations: "None has ever been found" - do you have a citation? wrt my example: "You are not doing the pulling, gravity does" - the pulling with a string was supposed to be analogous to the Sun's gravity, so I mean, you would stand as if you were the Sun and pull on the far side of the donut with a string - because the far side of the donut appears to you to be above the fixed pivot point (center of the earth), the torque you cause will pull the far side of the bulge up away from the ecliptic. I was just trying to demonstrate, through a simple example, the intuition behind why the Sun's gravity pulling on the night side of the bulge will result in a torque pulling the night side of the bulge away from the ecliptic, but I see it was not clear enough. I said ecliptic, not equatorial plane, and that picture appears to be about tides, not the subject at hand. Unless you find this profitable, let's let it go, since really what we need here for the article is sources, not to convince each other. You keep saying things like "you are erring, not me," but you can not find any sources. I'm not sure what else I can do; I have found quotes from books written by astronomers that clearly contradict your explanation and support mine. ErikHaugen (talk) 05:48, 16 June 2010 (UTC)

That picture belongs to differential gravitation which explains the tides and the precession, and therefore it belongs here too. But what do I care? Have it your way. Apparently we cannot convince each other, and at the end, yes, verifyable sources is what really counts for wikipedia. Without access to libraries it can take a long time before I find one. Until then. --Tauʻolunga (talk) 07:40, 17 June 2010 (UTC)

For what it's worth, I would have to say the Erik is correct in the above discussion. The simplest way to determine this is to simply plug the numbers into the torque equation seen in the article:

${\overrightarrow {T}}={\frac {3Gm}{r^{3}}}(C-A)\sin \delta \cos \delta {\begin{pmatrix}\sin \alpha \\-\cos \alpha \\0\end{pmatrix}}$

Notice that one of the terms is declination (δ). At the equinox, the sun's declination is zero so sinδ = 0. Last time I checked, multiplying any vector by a scalar with a value of 0 gives... ${\overrightarrow {0}}$! So at the equinox, there can be no torque caused by the sun's gravitation.

To think of it another way (one that requires neither derivation of a torque formula nor the assumption that the one given is correct — which it is, by the way), at the equinox, the center of the sun lies IN THE EQUATORIAL PLANE. Since gravitational forces are directed toward that center, they MUST lie in that plane. Since no component of that force can be orthogonal to the equator, there is no resultant torque.

The source of the error lies in confounding two different frames of reference: the equatorial and the ecliptic. At the equinox, there is indeed a small component of the gravitational force that is orthogonal to the ECLIPTIC, but that component is entirely an artifact of the incongruence between the equatorial and ecliptic planes, and it DISAPPEARS when you transform these vectors into the EQUATORIAL plane.

I've tried to be both clear and succinct in this explanation, but if there is any misunderstanding (and any contention must be a misunderstanding), I'll try to clarify. Wilford Nusser (talk) 01:52, 4 July 2010 (UTC)

This disagreement is resolved by noting that torque is not directly related to precession. Instead, torque is directly related to the rate of change of precession+nutation, their derivative. Hence torque must be integrated to find precession+nutation. Precession is separated from nutation by decomposing the torque's sine squared term $sin^{2}\,l$ into ${\tfrac {1}{2}}(1-cos\,2l)$, where $l\,$ is the ecliptic longitude of the Sun, its angular position in its apparent orbit around the Earth relative to the vernal equinox. Note that the torque has a constant term and a periodic term with a semi-amplitude equal to the constant term, so the torque is zero at the equinoxes. Replace $l\,$ with $n't\,$, where $n'\,$ is the angular velocity of the Sun around the Earth, about 3,548"/d. Now $\textstyle \int _{t}{\tfrac {1}{2}}(1-cos\,2n't)={\tfrac {1}{2}}(t-{\tfrac {1}{2n'}}sin\,2n't)$. Thus precession increases linearly throughout any and all years, while nutation is smaller than the torque's corresponding term by ${\tfrac {1}{2n'}}$. Solar precession is about 16" per year, while the semi-annual nutation term has a semi-amplitude of about 1.3". See C.H.H. Cheyne, The Earth's motion of rotation including the theory of precession and nutation (1867) page 41. — Joe Kress (talk) 07:38, 8 July 2010 (UTC)

Thank you all so much pointing out the solution. --Tauʻolunga (talk) 07:13, 10 July 2010 (UTC)

Move

Right now, this page redirects from Axial precession. But since there is no other type of axial precession, there doesn't appear to be a need for a qualifier at the end. Serendi^podous 18:07, 10 April 2011 (UTC)

There may be no other article, but a gyroscope on a stand will precess. —Tamfang (talk) 19:05, 10 April 2011 (UTC)

This article was entitled precession of the equinoxes by Ewlyahoocom when he separated it from precession (where it still has a section now entitled Axial precession (precession of the equinoxes)) on 7 April 2006. On 8 May 2007 The way, the truth, and the light moved it to precession (astronomy) before he renamed the section climatic effects to anomalistic precession. (Where is the move log?) I added several related terms to the lead on 28 February 2009 (general precession, lunisolar precession, planetary precession, precession of the ecliptic, precession of the equator). Zbayz then moved this article to axial precession (astronomy) on 23 August 2009 because he said that it said nothing about perihelion precession. After I noted that both precession of the equator and anomalistic precession were discussed, he moved the anomalistic precession section to apsidal precession). I objected to his move of this article at Name change above, but I couldn't think of a better name so I did not attempt to move it again. Even though I prefer one of the other names I added, a Google search shows that precession of the equinoxes is far and away (100 times) more popular than any other term. Because this article is overwhelming about that part of the astronomical subject, I recommend that it be moved back to precession of the equinoxes. — Joe Kress (talk) 20:50, 10 April 2011 (UTC)

Southern Cross reference

By consequence, the constellation is no longer visible from subtropical northern latitudes, as it was in the time of the ancient Greeks.

How are we defining 'subtropical northern latitudes' here? Because I'm at about 25N, and I can see the top half of the Southern Cross from my driveway, 'in the city', when it's a sufficiently clear night. Around March, to be specific. I imagine it's a much better view in the Keys, with the clearer horizons.

I am hesitant to revise that line without either the original justification for it referenced, or what documentation on my part would be appropriate.

Corgi (talk) 14:28, 15 July 2011 (UTC)

Ecliptic reference

What is the tilt of the precession central axis with reference to the ecliptic of the sun, seeing that tidal torque forces from the moon are stronger than the sun's tidal forces, and the moon's orbit is not aligned with the solar ecliptic, defined by earth's orbit about the sun? The precessional central axis should be on a tidal torque coordinate that is a mathematical mix between the stronger lunar orbit tidal torque, and the weaker solar ecliptic tidal torque, and as such would also shift solar and lunar positions on horizon reference points, with the precession period over the approximately 30,000 year period. With an ecliptic-reference tilted central precession axis, these precession effects would misalign the claimed thousands of year old monuments based on the precision of solar and lunar horizon reference points on claimed ancient precise stone monuments like Stonehenge, and other ancient horizon reference stone marker systems. At least hyperbole, in media claiming such monuments still have "exact" alignments, whatever the word "exact" means in media. Only monuments aligned to the earth's own axis, like The Giza Pyramids, would remain constant, of course pointing to a moving pole, as the star charts show with Polaris today, or Thuban, or Vega, at other points on the precession path. Not to mention systems of stellar alignment, like claimed Nazca Lines, pointing to Sirius and Orion, seem to be media hype, having nothing to do with the Ecliptic, and claims of ancient formation toward these alignments, which will definitely not hold over thousands of years. LoneRubberDragon 76.166.233.62 (talk) 06:54, 5 August 2011 (UTC)

I don't understand your question wrt the moon's orbit. Any astronomical use of the Nazca lines or Stonehenge seems to be theoretical; I don't know what "precision" you are referring to exactly. Can you be more precise? ErikHaugen (talk | contribs) 17:00, 5 August 2011 (UTC)

The orbit of the Moon wobbles around a line perpendicular to the plane of the ecliptic in only 18.6 years, so the average lunar plane is identical to the plane of the ecliptic. Similarly, the average lunar torque has the same direction as the average solar torque so they simply add—the "precessional central axis" does not change direction over thousands of years, so there is no additional effect on archaeological sites. The components of both lunar torque and solar torque in other directions are lumped together under the term nutation because they have no effect over thousands of years. The limits of the tilt of the lunar orbit can be observed and is called lunar standstill which does affect archaeological sites. — Joe Kress (talk) 20:45, 7 August 2011 (UTC)

I found the answer on Wikipedia, too, that the axis of precession in the heart of Draco, is also virtually the north ecliptic pole. en.wikipedia.org at wiki at Ecliptic_pole contains the page with that chart, virtually matching your chart on precession. 76.166.237.179 (talk) 09:48, 25 August 2011 (UTC)

And agreed, with the given data pages and charts of reference, with precisely no axis tilt of the precession circle relative to the solar ecliptic, as shown, will produce solar-lunar monuments that will not have precessional effects, but stellar monuments like some Nazca claimed references, Will suffer general stellar precessional misalignments over thousands of years. 76.166.237.179 (talk) 11:38, 25 August 2011 (UTC)

But Furthermore, looking at the forces, relative to masses and distances, relative to earths oblateness (ellipticity), it shows the sun with 0.5:1 times the precessional gyroscopic torque than the moon, oddly resulting in the precession's central axis at virtually 90 degrees on the heliocentric ecliptic coordinate system. Given: sun 333,000 earth mass, 92,800,000 miles distance, moon 0.0123 earth mass, 238,900 miles distance, earth radius differential 4000 miles radius (approximately). With the given, I get 179:1 gravity force ratio on earth from the sun:moon. And for the earth radius gravitational radial differential effect to distance I get 374:1 for the moon:sun on earth. And if the products are appropriate, that produces the 0.5:1 effect of solar precessional torque to lunar precessional torque, similar to the moon having double the tides compared to the sun's effect on earth. 76.166.237.179 (talk) 14:09, 25 August 2011 (UTC)

As the moon's orbit does gyroscopically precess like the earth, an 18.7 year precession period is impressive, given earth's 28,000 year precession, producing a minor angular 18.7 year Nutation in the precession. The article Nutation does include a paragraph on earth nutation from that lunar 18.7 year cycle that you mention, so perhaps a paragraph to emulate in this article, to go from general precession of the equinoxes, to a more complete specific precession of the equinoxes. To show wikipedia weakness, the moon article does not even directly mention that its orbit plane precesses in 18.7 years, only that it precesses in ambiguous reference, lacking the Saros Cycle being mentioned or that 18.7 year value being given. 76.166.237.179 (talk) 11:38, 25 August 2011 (UTC)

It is interesting, given Saturn's own heliocentric tilt, that Saturn's rings have not differentially precessed from solar gravitational differentials at different radial distances, along with all of its satelite orbits, not being a shambles of inclinations outside of Saturn's rotation axis. Saturn's rings are nearly perfectly flat, which "apparently" indicates their extreme youth, to not have precessed in any percievable way, since their formation. Saturn's moons are another story, and strange, being relatively coplanar. Wiki says "Twenty-four of Saturn's moons are regular satellites; they have prograde orbits not greatly inclined to Saturn's equatorial plane.", and Saturn is tilted to its own orbit 27 degrees like Earth. If the entire Saturn system is young, like its rings, that is odd. 76.166.237.179 (talk) 14:09, 25 August 2011 (UTC)

Nutation is negligible relative to precession. The precession angle of 23.5° is 5000 times larger than the largest nutation component of 17". Nutation is so small that it cannot be observed with the naked eye. So the precession of the Moon's orbit cannot have any observable affect on archaeological sites.

True. 76.166.237.179 (talk) 10:44, 26 August 2011 (UTC)

On the other hand, precession has two components. The largest component is lunisolar precession due to the Moon and Sun. The smaller component is planetary precession where the other planets cause the ecliptic itself to precess (actually advance) around the pole of the invariable plane of the Solar System. This distinction is mentioned at the beginning of this article. The ecliptic wobble is about 2° over a period of about 100,000 years, so it's about 9% of lunisolar precession, which is easily observable and will affect archaeological sites. See Invariable plane#Position.

Good point. I should not have said that the moon has no precession effect, but like you say, an averaged precession effect, from the moon's own Saros Cycle Precession on the earth sun plane. 76.166.237.179 (talk) 10:44, 26 August 2011 (UTC)

The orbits of the satellites of a planet precess around the poles of the Laplace plane of that planet. This is a warped plane that coincides with the planet's equatorial plane for satellites close to the planet, while it coincides with the orbital plane of the planet for satellites far from the planet. The large inner moons of the gas giants precess around their planet's equatorial plane whereas the small outer moons of the gas giants as well as Earth's Moon precess around their planet's orbital plane.

Yeah, I thought about that, too, quite true. With extremely near satellites, the satellite's orbit angular momentum increases, reducing the precession rate effect from the sun for a spehrical main planet, and planetary oblateness effects of the planet itself are increased by that slowness of precession. And with reduced satellite orbit angular momentum at further distances, that solar gyroscopic precession effects increase, as the dipole like dropoff of oblateness effects sharply drop off. Though, it matters not whether the satellites are large or small. If you consider that Saturn's Oblateness effect on extremely near satellite precession, is approximately a sine angular force of the satellite's orbit inclination (zero force at Saturn's Equator leaves solar precession still perfect dominant, and approximately sine, for it is also zero effect for a perpendicular orbit to Saturn's Equatorial Plane), and Oblateness is a differential force based on near bulge torque in difference compared to the opposing far bulge anti-torque-effect at 1/R^2 differences, and Saturn's increased density with depth shrinks the profile of Mass Oblateness Distribution from Optical Oblateness, that it is still odd looking for many of the satellites to be less than 0.5 degrees off Saturn's Plane, many within a tenth of a degree, some within a hundredeth of a degree off Saturn's Plane, all with Saturn's own Precession over millions of years. *Sighs* I'd really need to run a numerical simulation to see the balance of how strong Saturn's Oblateness would modulate solar caused precession, with Saturn 10 times farther, and 100 times more massive than earth, and Saturn Itself precessing over millions of years. 76.166.237.179 (talk) 10:44, 26 August 2011 (UTC)

Spending an hour looking at the dynamics without numerical simulation, further, I note that for a satellite initially on Saturn's Equatorial Plane, it would precess about the Saturn Orbit Normal Axis exclusively according to Solar Gravity Differentials on the satellite orbit with respect to the sun's distance. Then, as the satellite's inclination to Saturn's Equatorial Plane increases, that sine of the inclination approximation starts to work of Saturn Oblateness Gravitational Differentials (near and far oblateness differential distance pull), causing another effect of precession of the satellite orbital plane about the Saturn Rotation Axis. And so the two precessions of Solar Gravitational Differentials on the satellite orbit, and the Saturn Oblateness Gravitational Differentials, work in a Superposition, and get into complex dynamics that go beyond basic paper analysis without Numerical Simulation of rate regiemes of dynamics. 76.166.237.179 (talk) 13:01, 26 August 2011 (UTC)

Saturn's rings are quite close to Saturn so Saturn's equatorial bulge would cause any moonlet whose orbit was inclined to Saturn's equatorial plane to precess around it. But it cannot precess because if its orbit was inclined it would collide with other moonlets whenever it attempted to pass through Saturn's equatorial plane, so all moonlets must lie quite close to Saturn's equatorial plane. — Joe Kress (talk) 04:53, 26 August 2011 (UTC)

For satellites within the ring plane, I assume, colliding with ring chunks, conservatively dragging them into the ring's plane. 76.166.237.179 (talk) 11:41, 26 August 2011 (UTC)