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Robert Tulip
2010-Mar-22, 05:52 AM
In mapping precession of the equinox, three circles provide points of reference. The zodiac ecliptic is the basically unchanging path of the sun against the background stars. The zodiac crosses the equally unchanging plane of the Milky Way Galaxy. Plotted on the celestial dome, the zodiac and the galaxy are unmoving great circles, joined near the opposite constellations of Taurus and Scorpio.

From the terrestrial perspective, the factor that moves, the third circle, is the celestial equator, marking the projection into space of the earth's equator. The precession of earth's orbital spin axis is mapped by the moving nodes where the zodiac crosses the equator. Stars shift between northern and southern hemispheres as a result of the wobble of the axis.

The nodes joining the celestial equator and the ecliptic are the two equinox points where the sun shifts hemisphere. These points move around the zodiac in a similar way to the nodes of the moon precessing around the lunar orbital path (http://en.wikipedia.org/wiki/File:Lunar_perturbation.jpg). Just as the moving intersection between the two circles of the lunar orbit and the ecliptic produces the eclipse cycles, the intersection between the zodiac and the celestial equator produces the precession of the equinox.

As the zodiac node points move around the ecliptic over the course of the precession cycle, the celestial equator also sweeps across different stars. With the polar axis stars (http://commons.wikimedia.org/wiki/File:Precession_S.png) moving around a circle with diameter twice the angle of the earth's tilt of 23.44 degrees, the celestial equator similarly moves up and down across nearly 47 degrees of the sky.

As I understand it, this means that stars which are now just closer to the pole than the equator (ie up to 46.88 degrees latitude) were in the opposite hemisphere about 12,880 years ago. This applies (I think) for the stars at the longitude of the celestial equator furthest from the zodiac node points.

I've posted this just to note down some features of the astronomy of precession and check if my description is correct. I'm not sure if the concept of equinoxes as zodiac nodes has been much discussed, and would welcome comment, especially if my explanation here has any errors.

grant hutchison
2010-Mar-22, 01:28 PM
To avoid confusion I'd rephrase

... stars which are now just closer to the pole than the equator (ie up to 46.88 degrees latitude) were in the opposite hemisphere about 12,880 years ago. This applies (I think) for the stars at the longitude of the celestial equator furthest from the zodiac node points.as
... there are stars which are now just closer to the pole than the equator (ie up to 46.88 degrees latitude) which were in the opposite hemisphere about 12,880 years ago. This applies (I think) for the stars at the longitude of the celestial equator furthest from the zodiac node points.As you suggest, only a very small region of the sky meets these criteria.
Relative to ecliptic plane, the Earth's north pole tilts towards Orion (6h RA, 90º ecliptic longitude) and away from Ophiuchus (18h RA, 270º ecliptic longitude). So there are stars around 6h RA +46º declination which were just south of the celestial equator half a precession cycle ago, and stars at 18h RA -46º declination which were just north of the celestial equator at that time. A prime example of the former (and perhaps the brightest star so affected) is π Aurigae.

(For completeness, it's worth noting that proper motion may have time to adjust the star's position so that this cycle doesn't occur as predicted by a naive coordinate calculation.)

Grant Hutchison

Robert Tulip
2010-Mar-23, 02:23 AM
Thanks Grant.

It seems more than half the sky crosses the equator over the precession cycle.

My understanding is that all points more than 44 degrees of arc away from the focal spot in this map (http://commons.wikimedia.org/wiki/File:Precession_N.gif)are found in the Southern Hemisphere at some time.

grant hutchison
2010-Mar-23, 02:47 AM
My understanding is that all points more than 44 degrees of arc away from the focal spot in this map (http://commons.wikimedia.org/wiki/File:Precession_N.gif)are found in the Southern Hemisphere at some time.That's the north ecliptic pole. Objects need to be more than 66.5º from it to enter the southern celestial hemisphere during the precession cycle. That is, they need to be south of ecliptic latitude +23.5º, the northern limit of the celestial equator.

(To be strictly accurate, we should use the maximum value for the obliquity of the ecliptic, which is +24.5º.)

Grant Hutchison

Robert Tulip
2010-Mar-24, 12:44 AM
That's the north ecliptic pole. Objects need to be more than 66.5º from it to enter the southern celestial hemisphere during the precession cycle. That is, they need to be south of ecliptic latitude +23.5º, the northern limit of the celestial equator.

(To be strictly accurate, we should use the maximum value for the obliquity of the ecliptic, which is +24.5º.)

Grant Hutchison


Thanks very much Grant, I had not previously seen the term ecliptic pole, which makes good sense. The South Ecliptic Pole (http://en.wikipedia.org/wiki/File:Precession_S.gif)is extremely close to the Large Magellanic Cloud, a bit like Kurma the turtle (http://en.wikipedia.org/wiki/Kurma) in the old myth of 'turtles all the way down'.

Here is a Sky Map with zodiac nodes in relation to the ecliptic, equator and galaxy. (http://www.bautforum.com/attachment.php?attachmentid=11889&stc=1&d=1269391036) Do you know, would the celestial equator remain parallel to the line shown here over the course of the precession cycle?

I would be interested to plot the north and south limit points of the celestial equator, and to work out what proportion of the sky is on the equator at some time, but am not sure how to do it.

grapes
2010-Mar-24, 02:06 AM
I would be interested to plot the north and south limit points of the celestial equator, and to work out what proportion of the sky is on the equator at some time, but am not sure how to do it.It would be the proportion of the area of the sphere within 23.5º of the equator. It's a not-well appreciated calculation, that it's just sin 23.5º. (I once casually mentioned that to George Backus (http://igpp.ucsd.edu/people/detail.php?name=Backus_George), and was impressed with myself that he was surprised, until he verified it in his head so fast that I was stunned.)

Robert Tulip
2010-Mar-24, 02:57 AM
It would be the proportion of the area of the sphere within 23.5º of the equator. It's a not-well appreciated calculation, that it's just sin 23.5º. (I once casually mentioned that to George Backus (http://igpp.ucsd.edu/people/detail.php?name=Backus_George), and was impressed with myself that he was surprised, until he verified it in his head so fast that I was stunned.)


Okay, this means on this sky map (http://www.bautforum.com/attachments/astronomy/11889d1269391036-zodiac-nodes-sky-map-zodiac-nodes-ecliptic-equator-galaxy.jpg), (i) the celestial equator moves up and down as a straight line every 26,000 years, and (ii) the stars shown here, up to the distance from the equator of the zodiac sine wave stationary points, and after correcting for proper motion, are in the 47/90 = 52% of the sky that crosses the equator.

Celestial Mechanic
2010-Mar-24, 04:02 AM
Just to clarify, the ecliptic (plane of the Earth's orbit) also moves. Its pole precesses about the invariant pole at a rate of about once every 78 thousand years. The ecliptic is inclined about 1.6 degrees with respect to the invariable plane of the Solar System. The invariable plane is the plane perpendicular to the total angular momentum vector of the Solar System.

grapes
2010-Mar-24, 04:06 AM
and (ii) the stars shown here, up to the distance from the equator of the zodiac sine wave stationary points, and after correcting for proper motion, are in the 47/90 = 52% of the sky that crosses the equator.I think that should be sin (47/2)º, which is 39.88%

Robert Tulip
2010-Mar-24, 07:18 AM
Okay, this means on this sky map (http://www.bautforum.com/attachments/astronomy/11889d1269391036-zodiac-nodes-sky-map-zodiac-nodes-ecliptic-equator-galaxy.jpg), (i) the celestial equator moves up and down as a straight line every 26,000 years, and (ii) the stars shown here, up to the distance from the equator of the zodiac sine wave stationary points, and after correcting for proper motion, are in the 47/90 = 52% of the sky that crosses the equator.
Apologies I was too hasty in this comment. In fact the equator point moves across from left to right due to precession, while the sun moves from right to left over the year, so the zodiac node moves up in the upper half of the diagram and down in the lower half. Does the celestial equator simply move in a band 23.5 degrees either side of its current location?

This map shows all the stars within about 30 degrees of the celestial equator. However, at different times, for example when the zodiac nodes were in Taurus and Scorpio 6000 years ago, were stars not on this map on the celestial equator?

Robert Tulip
2010-Mar-25, 11:51 PM
I think that should be sin (47/2)º, which is 39.88%
Thanks Grapes. The question here is how much of the sky does the celestial equator traverse, and so how we can mark the 'tropics' of the celestial equator on a map.

Modelling on the earth, the proportion is close to the area of the globe surface between the Tropic of Cancer and the Tropic of Capricorn, which are both at 23.5º latitude marking the position of the sun at the solstices.

However, the sine formula answer you provided (~40%) gives the area of the cylinder with top and base at the tropics, not the area of the spherical surface. My calculus is a bit rusty to do this spherical geometry calculation, but I will look up how to do it.

This spherical section will give the amount of sky traversed by the equator considering the ecliptic pole as fixed, without taking into account the motion with respect to the invariant pole (as mentioned by Celestial Mechanic).

I would be interested to know if the tropics of the celestial equator have been plotted accurately.

grant hutchison
2010-Mar-26, 12:19 AM
However, the sine formula answer you provided (~40%) gives the area of the cylinder with top and base at the tropics, not the area of the spherical surface.No, it's the spherical surface.
There's a neat result (http://mathworld.wolfram.com/Zone.html) that the surface area S of a spherical zone of height h is:

S = 2πrh

no matter where in the sphere the zone is sliced.

If we base our zone at the equator and extend it to latitude θ, then

h = r.sin(θ)

and therefore:

S = 2πr²sin(θ)

Extend both sides of the equator and we have:

S = 4πr²sin(θ)

or just sin(θ) of the total surface of the sphere.

Grant Hutchison

Robert Tulip
2010-Mar-26, 01:57 AM
Thanks Grant

I hope you won't mind explaining the error of my calculation:

For a sphere with radius 1, and angle θ between equator circle and line from centre to a latitude circle, sin θ (opposite over hypotenuse) equals the vertical line from the latitude circle to the equator circle. This distance is shorter than the arc from the latitude to the equator.

In the attached diagram, for r = 1, sinθ = AC/AO = AC. The formula you and grapes provide suggests that sinθ = AB.

grant hutchison
2010-Mar-26, 08:48 AM
The formula you and grapes provide suggests that sinθ = AB.That's your error: it doesn't suggest that. :)
It says that the surface of revolution generated by AB has an area of 2π.sin(θ) (to be compared to the surface area of the unit sphere, 4π).
The full derivation is given in my link, together with the additional calculations in my post.

ETA: For comparison, the surface of revolution generated by AC has area 2π.sin(θ).cos(θ), which is less than the area generated by AB by a factor of cos(θ).

Grant Hutchison

Robert Tulip
2010-Mar-26, 12:16 PM
That's your error: it doesn't suggest that. :)
It says that the surface of revolution generated by AB has an area of 2π.sin(θ) (to be compared to the surface area of the unit sphere, 4π).
The full derivation is given in my link, together with the additional calculations in my post.

ETA: For comparison, the surface of revolution generated by AC has area 2π.sin(θ).cos(θ), which is less than the area generated by AB by a factor of cos(θ).

Grant HutchisonThanks again Grant, you can see I am rusty on my spherical geometry!

What I am trying to do here is to map the path of the zodiac nodes, the points where the zodiac crosses the celestial equator. I started with the idea that the diagram I posted earlier, showing the current star field 30º each side of the equator, shows the whole area traversed by the celestial equator over the course of the precession cycle. However, this is obviously wrong, since when the nodes reaches the points now at the solstices, in about 6450 years, on this map the equator will still form the same angle with the zodiac as it does now, so will go through stars which are now closer to the poles than any shown on this map.

40%, the sine of the axis inclination, is the fraction of earth that ever has the sun at zenith, and also the amount of the sky that ever stands on the celestial equator.

grant hutchison
2010-Mar-26, 12:51 PM
What you need is a map that is an inverse projection of the one you've linked to: one that shows the ecliptic as a horizontal line, and the celestial equator as a sinusoid which crosses it. By mentally moving the sinusoid along the ecliptic line, you can reproduce the position of the celestial equator for past and future epochs.
Such maps are reasonably common: that projection is often used to plot the position of the sun, moon and planets, for instance, since all these objects tend to stay in the vicinity of the ecliptic.

Grant Hutchison

Robert Tulip
2010-Mar-27, 08:18 PM
What you need is a map that is an inverse projection of the one you've linked to: one that shows the ecliptic as a horizontal line, and the celestial equator as a sinusoid which crosses it. By mentally moving the sinusoid along the ecliptic line, you can reproduce the position of the celestial equator for past and future epochs.
Such maps are reasonably common: that projection is often used to plot the position of the sun, moon and planets, for instance, since all these objects tend to stay in the vicinity of the ecliptic.

Grant HutchisonI've used SkyGazer 4.5 to produce the attached star map with zodiac as axis and equator as sine curve (http://www.bautforum.com/attachment.php?attachmentid=11909&stc=1&d=1269720976). Shifting the sine curve of the celestial equator to the right shows where the equator will be in future. For example, in 6000 years when the zodiac nodes are in Sagittarius and Gemini, the celestial equator will go through Cetus and close to Denebola in Leo.

Millennium
2010-Jun-14, 05:13 AM
haku Robert,

native Chumash blessings from Central California.

question ... in your thread "Canopus-Sol Relation" you posted two illustrations I am interested in studying and using in more detail: 1) Great Cosmic Year South Celestial Pole, and 2) your un-named "Diagram" re: illustrating "the Science of Precession, with the addition of the designation of stellar axes as golden and iron and approximate year dates for each Age. I have also added the big southern stars which mark the Pole, Canopus, Achernar and Beta Carinae, and the Large Magellanic Cloud which is located at the base of earth's spin cone. The diagram shows how the Vedic imagery matches the zodiacal ages as a model of the temporal structure of the earth."

I am interested in receiving these images in higher resolution so that I can study and compare the with my other sources, and share them with the Sirius Sol discussion group at Facebook which is looking at all the local stars and their mythologies and constructing models of their co-orbits with our Sun and each other. Do you have these illustration online anywhere in high resolution, or can you post them directly to me. If I use them, or portions of them, I can acknowledge you as the author/researcher and also tell people to visit any discussions here at the BautForum related to the subject.

http://www.facebook.com/group.php?gid=109669179052745

I am interested in any wisdom you have on astronomical or astrological traditions which seem to claim or infer dates during the last 26,000 years of a close approach of a stellar companion star such as Alpha Centauri, Sirius or other (or a nearby undiscovered Red Dwarf or Brown Giant), etc. etc. ... anything which can give us dates (and thus data points) for beginning to sketch possible co-orbits for our local interstellar companion stars. [Canopus and similar relatively distant stars would also participate in co-orbits with Sol, Alpha Centauri, Barnard's Star (Red Dwarf), Wolf 359 (Red Dwarf), Lalande 21185 (Brown Dwarf), Luyten 726-8 A (Red Dwarf binary), Sirius, Ross 154 (Red Dwarf), Ross 248 (Red Dwarf), and Epsilon Eridani etc. -- but would not have orbital periodicies in the tens or hundreds of thousands of years, would be in millions of years.)

por todas las cosas,


Millennium Twain

Robert Tulip
2010-Jun-15, 01:21 AM
Dear Millennium Twain

Thank you so much for your kind words and your native blessings. Warm greetings from Australia.

I see this is your first post on the Bad Astronomy Universe Today Forum. BAUT has strict rules, to which I seek to adhere religiously, regarding non-discussion of religious topics and limited discussion of ideas that are outside current mainstream scientific knowledge. These rules recognise the cosmic enlightenment derived from objective knowledge of the universe through mainstream science, and the benefit of consistent and coherent standards of proof for new claims. This thread on Zodiac Nodes, located within a mainstream part of the forum, is restricted purely to mainstream astronomy. Any discussion of religious or mythological matters here requires a clear and obvious relation to mainstream empirical knowledge about space and astronomy.

With that caution, I welcome your question and your interest. The documents you request are available here ( http://rtulip.net/yahoo_site_admin/assets/docs/Astrology_in_the_Bible.164171922.ppt). You are welcome to follow up either by private message or in ways that adhere to BAUT Forum rules.

I have taken an interest in the binary star theories of Walter Cruttenden, but have come to the conclusion that his ideas are not correct because the binary star theory is not compatible with observation. Mainstream physics provides a compelling explanation of the Great Year in terms of earth's spin wobble and lunisolar torque, and refutes the binary theory by reference to the mathematics of gravity. A useful Q&A entry point to this discussion was recently provided by Jim Smith Chiapas (link ( http://www.bautforum.com/showthread.php/103153-Question-on-redshifts-in-Cruttenden-s-Binary-Sun-model)).

My view is that discussion of the mythology of the Great Year is best advanced by rigorous effort to align any claims with the observations of mainstream astronomy. So I have to take issue with your premise that a companion star is a helpful hypothesis. That said, the mythology of Canopus is certainly fascinating, but as I argue in the thread you mention, the relation between Canopus and the Sun is solely that Canopus is a location marker for the South Celestial Pole over the cycle of precession of the equinox.

Best regards
Robert Tulip (http://rtulip.net)