Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
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Stoat

United Kingdom
964 Posts

Posted - 09 Jan 2009 :  07:09:57  Show Profile  Reply with Quote
Reading back a few posts I got thinking about Larry's post on force. Just for fun I thought I'd see what we would get from an orbital system which had a small m of the electrons e.m. mass and its gravitational e.m. equivalent.

Now as I'm thinking along the lines of an electron as a toroid, I can imagine that we are dealing with a spiralling e.m. entity which has acceleration.

So we will have e = mc^2 and a further energy equation e = mb^2 for b being the speed of gravity. Both of these equations are binomial expansions of the lorentzian and imply a one to one correspondence. That's an assumption but I think a valid one.

e = mb^2 for the gravitational energy.
1.2355897798E 20J = 9.1093897E-31 * 1.16464217444E 25^2
Divide that through by the speed of light squared to get an equivalent e.m mass.
1.37477903776E 03 kgs
F = GMm / r^2
Take r to be the Compton wavelength, 2.42631060001E-12 metres
1.41946220144E-14 Newtons = 6.67259E-11 * 1.37477903776E 03 * / 2.42631060001E-12^2

F = ma, so an acceleration of the electron mass of 1.55824072544E 16 metres per second squared
and for M an acceleration of 1.03250206939E-17 metres per second squared.

It wobbles about, it will be uncertain in e.m space and even more so in gravitational space. The two spaces are in one to one correspondence, the gravitational energy of the electron is hidden inside of the Shwartzchild radius but that radius can be anywhere within the gravitational wavelength of the particle.

Granted, that sounds pretty horrible but we can consider gravitational space to be larger, or informationally smaller than e.m. space. The smaller the mass, the larger the gravitational wavelength. For a photon, which I think has a rest mass of about 1E-64 kgs the Shwartzchild radius can be anywhere within a gravitational radius of light years. Hence, it seems not to have any rest mass at all but it has momentum, due to it having its own space.

Now this is where I think the Reimann conjecture comes into play. We have quantised e.m matter. I also think that gravity takes an exponential form. What if, real number infinities and natural number infinities have one to one correspondence points, which are legislated by prime numbers? My hunch is that non locality informs a particle of its unique locality.


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Joe Keller

USA
957 Posts

Posted - 09 Jan 2009 :  14:18:59  Show Profile  Reply with Quote
"Talking points" for recruiting amateur, professional, academic or government astrophotographers:

One to two years ago, I studied the Aladin plates, using computers in libraries that had Java software. I found many objects, but it was objects on the 1986 and 1954 Red sky survey plates, together with Joan Genebriera's March 2007 photo of Barbarossa (from Tenerife) and Steve Riley's April 2007 photo of Frey (from southern California) that fit a circular orbit to within a very few arcsec, if the mass ratio Barbarossa::Frey is properly chosen. I found objects on most other relevant plates, that did not fit the orbit so well, but maybe the explanation is that I had picked the wrong "moon" as "Frey". The U. of Iowa photo fits the orbit (about 4" off) too.

The "clincher" is that the four pairs (1954 & 1986 sky surveys, spring 2007 and Dec. 2008 photos) fit a binary orbit perfectly (to 1") with only "one adjustable parameter" (the 5th ellipse point) to fit "two equations" (Kepler's 2nd law between 1st & 2nd and between 2nd & 3rd sectors). (All this information was posted to Dr. Van Flandern's messageboard as I discovered it.) Also, I posted much on the messageboard, refuting the usual objections to the existence of such a planet, and connecting the planet to various unexplained solar system relationships.
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Joe Keller

USA
957 Posts

Posted - 09 Jan 2009 :  22:57:07  Show Profile  Reply with Quote
quote:
Originally posted by Maurol

...the object he refers as Barbarossa is absent from sky catalogs. ...Optical and Infrared catalogs, but also Radio, X-ray and Gamma ray... .
...to the director of University of Córdoba Observatory, here in Argentina, to see if they can take a plate of that region...


Thanks!
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Joe Keller

USA
957 Posts

Posted - 10 Jan 2009 :  00:25:08  Show Profile  Reply with Quote
More Defense of the Fourier-Cruttenden Theory of En Bloc Solar System Rotation

Let's consider three frames of reference:

Frame #1. The stars (galaxies, etc., would differ only slightly).
Frame #2. The nodes of the orbits of Jupiter, Saturn and Pluto (i.e., the outer planets for which such points can be measured most accurately).
Frame #3. The perihelia of those planets in Frame #2.

As I mentioned earlier, Frame #2 is, according to old Astronomical Almanacs, about the same as Frame #1. However, old Astronomical Almanacs make a strong and obvious case that Frame #3 rotates retrograde with period roughly (guess what!) 26,000 yr.

A rotating frame is supposed to have centrifugal and Coriolis force. Textbooks would say, that Frame #3 should have both centrifugal and Coriolis force, and Frame #1 neither. So, in Frame #3, according to textbooks, the centrifugal force causes the perihelia to precess (so that they are stationary in Frame #1). Also, in Frame #3, according to textbooks, the Coriolis force causes the nodes to precess (so that they also are stationary in Frame #1). According to textbooks, Frame #1 has neither of these forces, so indeed both the perihelia and the nodes are stationary there.

Really, as near as I can tell from the Astronomical Almanacs, Frame #3 has only the Coriolis force, and Frame #1 the centrifugal force. So, in Frame #1 (i.e., Frame #2) the nodes are stationary and the perihelia precess. In Frame #3, the perihelia are stationary and the nodes precess.

This centrifugal force in Frame #1, equals that due to rotation with period ~26,000 yr. If the sun's gravitational strength is determined from Earth's orbit and again from Mars' orbit, any additional central force that is not inverse-square, will cause a discrepancy between the two calculated values of the sun's gravitational strength. Saying it another way: if the sun's pull on Mars is calculated from the sun's pull on Earth, by applying the inverse square law, the discrepant extra force at Mars from this centrifugal force, in Frame #1, is 2.3% as big as the discrepant extra force that would occur from the anomalous Pioneer acceleration (also, it's of opposite sign). JD Anderson et al, Physical Review 1998, found that they could not rule out discrepant extra centripetal force at Mars, equal to about 1% or less, of what would be observed from the anomalous Pioneer acceleration. So, discrepant extra centrifugal force, of about that magnitude, might exist. For the centrifugal force, increasing as r^1, the outer planets might be a more sensitive test, than Mars; though Anderson said Mars was, in 1998, the most accurate test for constant (i.e., r^0) extra centripetal force.

Pioneer10/11 also show this new centrifugal force. At Earth's orbit, it should be ~1% as strong as the anomalous Pioneer acceleration ("APA"). It is proportional to r^1, so should neutralize the APA, at ~100 AU. A recent determination of the APA, when corrected by me for the tidal acceleration due to Barbarossa, indeed shows a large, steady decrease in the APA very roughly consistent with a linear decrease to zero at 100AU.
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Joe Keller

USA
957 Posts

Posted - 10 Jan 2009 :  01:14:35  Show Profile  Reply with Quote
The Boston banker has emailed me six individual Dec. 22, 2008, image FITS files from the U. of Iowa website, that (?) were median-stacked to make the single image I have (and which "Stoat" processed and posted here). "Stoat", "Marsrocks", "Maurol" and other contributors, might think of better ways to stack these (let me know if you want them). I haven't looked at them yet because this hotel computer's firewall blocks the NRAO FITS reader.

I myself found, on the U. of Iowa website, many "zip" files dated 12/22, possibly all individual images made for me, though the banker doubted that. I only found one such "zip" file of later date, namely 12/27.

On Dec. 22, Barbarossa was near stationarity and moving only 2"/day retrograde (the Dec. 22 photos could have been separated at most by a few hr). Frey would have moved even less because its orbital motion almost would have canceled the retrograde geocentric motion.

The banker's message:

"I went to the U of Iowa website and found your original images, but I had to do a lot of searching and downloading to find them. There are six, your name and Barbarossa are in the header of each. They were in the images/external/images/njk/2008/357 folder (x3) and images/external/images/njk/2009/357 folder (x3) ["2009" suggests a new photo, but "357" is the code for the 12/22 photo - JK]. ...In opening many, many other pictures, I noticed the dates indicated next to these files on that website and the dates within the headers almost never matched. Also, many, many pictures have the 12/22 date. So, I'm wondering if somebody is just neglecting to properly enter the date in the header."
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Joe Keller

USA
957 Posts

Posted - 17 Jan 2009 :  12:50:34  Show Profile  Reply with Quote
Note on Social Method

I've been asked, quite reasonably, to gather in one post, all the coordinates and plate numbers for the relevant 1954, 1986, 2007 and 2008 objects, that conform to the solar and binary orbits. I expect to do this soon, but I'm waiting until someone with official academic credentials asks me.

This way, I know that none with any official credentials - none of the journal editors, none of the conference organizers, none of the managers of government-funded telescopes, none of the professors - who ignore my findings or dismiss them with mere pejoratives - know what they are talking about. None of them have bothered to check to see if these points satisfy, against all odds, an accurate orbit, or not. I know they haven't checked, because though all the necessary information is posted, they would at least want to double check with me, whether the points they read about in my old posts, really are the relevant ones. I'm willing to assist the bureaucracy, but they have to talk to me.

Here is one line of thought that some of the more communicative bureaucrats have (I paraphrase):

"We have to keep beating the drum for funding. The undesirable side effect of this, is that the public, instead of merely writing their Congressmen and asking for more astronomy funding, actually thinks about astronomy and burdens us terribly with, maybe, one email a day about a theory that is so unlikely, often little better than 'the moon is made of green cheese', that we really don't have time to check it out. Our solution to this, is to make a rule that we ignore everyone's ideas unless they have a Ph.D. in astronomy. Really, we go further, because there are so many Ph.D.s in astronomy now, that we have to ignore everyone but the journal editors. The National Science Foundation goes by what the journal editors say, therefore so must I.

"I feel some sympathy for you, Dr. Keller, because your father spent his life savings to put you through Harvard, I've verified that you really did graduate cumlaude in Mathematics there, you have some recent job experience indicating that you are able very successfully to apply what you learned as an undergraduate, and this is all undergraduate mathematics. So, I have no reason to doubt what you tell me about your calculations.

"However, if I were to break the rule against paying attention to anyone but a journal editor, where would it end? Certainly not in NSF funding, unless you are correct. Yet there is already an approved list of activities for which I can get NSF funding, correct or not. The central committee (NSF, journal editors, etc.) owns everyone's time; there's no time for you. To do your calculations correctly involves considerable work, which I do not want to duplicate unless I am sure I will be paid. - Sincerely, Prof. ABC"
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Maurol

Argentina
37 Posts

Posted - 17 Jan 2009 :  19:39:39  Show Profile  Reply with Quote
quote:
Originally posted by Joe Keller

Note on Social Method

"However if I were to break the rule against paying attention to anyone but a journal editor where would it end? Certainly not in NSF funding unless you are correct. Yet there is already an approved list of activities for which I can get NSF funding correct or not. The central committee (NSF journal editors etc.) owns everyone's time; there's no time for you. To do your calculations correctly involves considerable work which I do not want to duplicate unless I am sure I will be paid. - Sincerely Prof. ABC"



Which is to say: "As long as we scientists are busy earning a living doing conventional science in our fields, we cannot afford nor have time to do unconventional science in these very own fields, to advance them or produce new discoveries in them."

Which is ludicrous.

As long as I(fortunately it seems) am not a scientist, I can afford to do whatever I want with my time (without getting paid, horror of horrors!) Even spending it in issues and calculations that are clearly not my field. So here we go:

What's Barbarossa period around the Sun?
According to my calculations, based on Joe's data, it's roughly 5142.5 years.

Joe, it could be interesting to know where's Barbarossa now, related to Jupiter and maybe also Saturn (are they now aligned? are they in opposition?)

I don't know if you heard about it, but the Sun is acting "weird" lately. That is, solar cycle 24 is arriving late to the party, so to speak. And nobody seems to know why.
Solar cycles seem to be related to the solar system barycenter being closer and farther of the Sun's surface. Maybe Barbarossa is in opposition or conjunction with Jupiter and/or Saturn right now, and that's changing the Solar system barycenter enough to delay the beginning of solar cycle 24.

Best regards,
Mauro
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Larry Burford

USA
2207 Posts

Posted - 17 Jan 2009 :  23:30:47  Show Profile  Reply with Quote
Joe, Maurol,

(Welcome to Meta Research, Maurol)

Contrary to popular opinion, comming up with ideas, even good ones, is not even a little bit hard.

The real trick is turning an idea, even a good one, into a successful business. Or into a successful experiment.

If you don't have the money to fund the attempt on your own, you have to try to convince someone else to put their money at risk. I hate it that mainstream scientists, and their funding sources, are not willing to spend a little of their money on odd ball, long shot ideas. But that is the way the world operates. He who pays the piper calls the tune.

===

If you really want to become famous in the world of science, figure out a way to fund some of the more promissing kook theories (like yours, or ours).

Regards,
LB


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Maurol

Argentina
37 Posts

Posted - 18 Jan 2009 :  09:34:49  Show Profile  Reply with Quote
quote:
Originally posted by Larry Burford

Joe, Maurol,

(Welcome to Meta Research, Maurol)

Contrary to popular opinion, comming up with ideas, even good ones, is not even a little bit hard.

The real trick is turning an idea, even a good one, into a successful business. Or into a successful experiment.

If you don't have the money to fund the attempt on your own, you have to try to convince someone else to put their money at risk. I hate it that mainstream scientists, and their funding sources, are not willing to spend a little of their money on odd ball, long shot ideas. But that is the way the world operates. He who pays the piper calls the tune.

===

If you really want to become famous in the world of science, figure out a way to fund some of the more promissing kook theories (like yours, or ours).

Regards,
LB




Hi Larry, thanks.

I'm not sure if I want to become famous, in any world. What I'm sure is that I want to help in the discovery of a new solar system planet or star.

Regarding ideas: ideas are the fuel of scientific progress! And not everything should or need to be profitable, IMO.

Regarding experiments, calculation, verification, etc. In this particular case, is it really that hard? experimentally, the first thing that must be done is to point a good telescope and take a plate of a particular region of the sky, to see if a little dot shows up there, at a very specific point.

Regarding calculations, I calculated Barbarossa's orbital period in less than 30 minutes, using the internet, a units program, and a command line calculator. I plan to do the same for the solar system barycenter now.

Of course a complete verification of Joe's calculations must be done at a certain point in the future, and that could be very hard to do, but now the first thing that is needed is a good picture of that region of the sky. And no one is willing to do that? Because it is so difficult/expensive?

Btw, I'm not the author of the theory of sunspot-barycenter correlation. A quick google search for 'sunspot barycenter' gives a number of interesting results, but it's not clear, at least to me, who the author is. Seems to be a very recent theory.
As you can see, this "simple" idea of sunspot barycenter correlation, seems to have remained undiscovered by more than 500 hundred years(since sunspots are counted).
The only needed thing was to look for other near 11 year cycles in the cosmos(i.e. Jupiter orbital period) and draw two time related charts. Simple, isn't? But, as far as I know, nobody had come up with that idea and done that by more than 500 hundred years.

My idea is that Barbarossa could be playing a part in the sunspot-barycenter cycle.

Regards,
Mauro
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Stoat

United Kingdom
964 Posts

Posted - 18 Jan 2009 :  13:22:57  Show Profile  Reply with Quote
Very roughly, the sun, saturn and barbarossa are in line, Jupiter, uranus and neptune are on the other side. So the barycentre will shift in a little towards the sun's surface. Tricky saying where the sun's surface is but there could well be something in the theory for sunspots.
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Maurol

Argentina
37 Posts

Posted - 18 Jan 2009 :  16:19:35  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

Very roughly, the sun, saturn and barbarossa are in line, Jupiter, uranus and neptune are on the other side. So the barycentre will shift in a little towards the sun's surface. Tricky saying where the sun's surface is but there could well be something in the theory for sunspots.



According to this
http://www.grandunification.com/hypertext/SunspotPredictions.html
(see Radial Distance between Barycenter and Sun graph)
Solar Minima is produced when the solar system barycenter is farther from the Sun center, and Solar Maxima when it is closer. In fact, the solar system barycenter is inside the sun at times. And this very fact seems to give birth to sunspot activity!
So, first hand, Barbarossa actual position would be contrary to the delay in the beginning of sunspot cycle 24, because logic indicates that Barbarossa should be in conjunction with Jupiter to effect the observed delay, if any.

A small correction to my previous post: although sunspots were observed and mentioned as early as the sixteen century, it wasn't until then beginning of the nineteen century that a cycle was noted, by Heinrich Schwabe. So we're are talking of almost 200 years, instead of more than 500. Which is a lot of time, anyways, for a well known cyclical phenomenon in search for a theory.

The correlation between the Sun's barycenter and sunspot activity is very meaningful to me, and I cannot avoid mentioning Astrology: Suddenly we have a mechanism by which planetary positions (conjunctions, oppositions, cuadratures, triangulatures, etc.) effect electromagnetic changes in the Sun, and consequently the Sun's solar wind. Which in turn effect changes in Earth's climate and other events.
Suddenly also, we have a mechanism by which a new or unknown object can gravitationally affect the Sun's electromagnetic activity.

Mauro
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Joe Keller

USA
957 Posts

Posted - 19 Jan 2009 :  19:08:00  Show Profile  Reply with Quote
Thanks Mauro, Bob, & Larry, for your posts. The period Mauro gives is "in the ball park", but for accuracy one must, as someone originally suggested to me on this messageboard, very accurately consider Earth parallax (i.e., convert to heliocentric coordinates). Anyway, the period actually is about 2800 yr, I think.

My margin of error probably is big enough to include the period that the philologist, Sitchin, determined for "Nibiru" from ancient texts. Though Barbarossa's orbit is nearly circular, the position of the solar system center of mass (including Barbarossa) is just at the inner edge of the asteroid belt. So, Barbarossa's small eccentricity might have large effects in the inner solar system, if "ether" theories are valid and the solar system center of mass is of absolute physical importance.

I hope to find time soon, to summarize the orbital data and calculations, but I'm also hoping to get two birds with one stone, by waiting for one of the professionals to contact me, so that when I summarize it for him, I can put the same summary here. I'm working on gaining access to a telescope to use myself, and also will be getting new tires soon, so I might make some road trips to sell this to astronomers who might be able to get "telescope time".
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Maurol

Argentina
37 Posts

Posted - 19 Jan 2009 :  22:15:12  Show Profile  Reply with Quote
quote:
Originally posted by Joe Keller
...
Though Barbarossa's orbit is nearly circular, the position of the solar system center of mass (including Barbarossa) is just at the inner edge of the asteroid belt.
...



That's very interesting. I've made the calculations and obtained 1.13 AU for the Sun-Barbarossa center of mass(I've erroneously supposed first that the barycenter was much closer to the Sun.)

That means also that, at least once in a year, the Earth would be very close to Sun-Barbarossa's barycenter. Geocentric astronomy anyone?

Mauro
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nemesis

84 Posts

Posted - 20 Jan 2009 :  14:31:10  Show Profile  Reply with Quote
This discussion of a possible correlation between the barycenter and solar activity is very interesting. Could there be any connection between Barbarossa's position and the Maunder Minimum of ca. 1645-1715 and thus the "Little Ice Age", I wonder?
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Joe Keller

USA
957 Posts

Posted - 20 Jan 2009 :  16:24:55  Show Profile  Reply with Quote
Hi Prof. *********!

This is great! As I've said before, whether the objects exist or not, it advances science to know, one way or the other.

I'll give you the coordinates sometime within the next 24 hrs. (it's 3PM CST, Jan. 20, 2009 now). I'm not admitting that failure to image on one photo, suffices to disprove their existence (I usually can find missing sky survey objects at least a magnitude dimmer than the supposed limit of a CCD photo; manufacturer's claims are one thing and comparison with sky surveys another), but one photo is better than none.

I'll assume that the effective epoch of the photo is 05:00 Mountain Standard Time, Jan. 20, 2009. The retrograde geocentric motion of these planets now is ~10"/day. So, if the stacked photos are within a 2 hr interval, their range is comparable, to a spread of 1 arcsec "atmospheric seeing".

The difference in geocentric position, vs. Dec. 22, is ~7 arcminutes. I don't know whether your team converted to heliocentric coordinates or not (there is a man who writes on Dr. Van Flandern's messageboard, who uses the nom de guerre "Nemesis", who first warned me about the extreme importance of Earth parallax). In any case, I'll give the coordinates, and if by oversight, the wrong coordinates were used, another photo can be taken soon.

By the way, I watched all six hours of President Obama's inauguration coverage this morning, beginning at 6AM. I wish Chief Justice Roberts hadn't surprised Pres. Obama with the pacing of the oath, but all's well that ends well.

Sincerely,
Joe Keller

--------------------------------------------------------------------------------

Date: Tue, 20 Jan 2009 11:00:41 -0600

>Joe - I have a challenge for you.

>I took some images of the Barbarossa field last night using the Dec 22 coordinates. Since the telescope's field of view is 1530 x 1530 arcsec and the object's orbital motion can only be a few km/s (Keplerian orbit at ~100 AU), the angular distance from the Dec 22 position must be about 90 arcsec, so it should be well within the FOV of last night's image. Since I averaged 16 x 120s images, the limiting magnitude is almost 22, so the object and its satellite should definitely be on the image.

If you accurately calculate the coordinates of the objects, I will check your hypothesis.

- *********
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Joe Keller

USA
957 Posts

Posted - 21 Jan 2009 :  16:38:44  Show Profile  Reply with Quote
(email sent 5 min. ago)

Hi Prof. *********!

Here's a (better than the one I sent 5 min. ago) but still crude estimate for Barbarossa's position in geocentric coordinates on the Jan. 20 photo:

RA: 12s West of where it was 12/22
Decl: 54" South of where it was 12/22

This is a rough lowest-order correction for Earth's and Barbarossa's orbital motion around the sun, neglecting Barbarossa's orbital motion around Frey (this binary motion can't be neglected, for Frey).

I'm reworking my calculations from the beginning, using the greater efficiency and generality made possible by hindsight. In a few hours I'll have more accurate information.

Sincerely,
Joe Keller
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Maurol

Argentina
37 Posts

Posted - 23 Jan 2009 :  11:31:37  Show Profile  Reply with Quote
Hi,
On the issue of a binary companion, and precession.

This is lengthy, and relatively difficult to follw, son cling with me for a moment. And please correct me if I'm wrong.

Let's assume an orbital period for a binary companion of 3600 years, for simplicity.

Both the binary(let's call it Barbarossa) and the Sun will orbit their barycenter once in 3600 years. That would be equivalent to a rate of precession for the Earth of 360 degrees/3600 years = 6 arc min/year. The actual rate of Earth's precession is ~ 50 arc sec/year. Even assuming that all of that is due to the Sun's curved path around its barycenter, and assuming a retrograde(similar to Earth's precession) orbit, 6 times that amount is missing from actual Earth's precession.

So, either or:
1) Barbarossa's period must be much bigger (around 10 times the actual
estimation.)
2) Some yet unknown mechanism must account for that difference in
precession rates. By example: Earth must precess by their own
gravitational interactions with Sun and Moon(so called lunisolar
precession), an amount 6 times bigger than it's actually stated, in the opposite direction of Sun's orbit, for the observed value to be 50
arc secs/year.
3) Something is very wrong/unaccounted for in actual precession
calculations. And indeed, with star positions, astronomical coordinates, etc.

Have you heard about so called Bessel's corrections/reductions? I've
read that they were introduced in the past because things (like noon) didn't fall where they were expected.

More recently astronomers have introduced so called "leap seconds",
probably to account for the same.
See http://www.siriusresearchgroup.com/articles/second.shtml
for a good critic of leap seconds, and food for thought.

Well, I can be wrong, but I think that I have comprehended the issue:

This is by no means easy to visualize. The first thing is to visualize the solar system movement through space, and particularly, the Sun movement.
If the Sun is in a binary orbit, it will describe an elliptical movement around the barycenter it conforms with its binary.
Let's assume that Barbarossa's data is correct. So we have, approximately:
Barbarossa-Sun mean distance: 298 AU
Barbarossa mass: 4 Jupiter masses
Barbarossa location: near the ecliptic plane (~ -9 degrees DEC.)
So:
Barbarossa's period: ~3600 years (for simplicity.)
Sun-Barycenter distance: 1.13 AU ~= 1 AU

Now, for all practical issues, we can "forget" about Barbarossa, and
consider only that the Sun orbit is circumscribed and concentric with
Barbarossa's orbit, with a radius of around 1 AU, and that the Sun will complete a revolution around its barycenter in roughly 3600 years.

All of the solar system is supposedly also moving toward Hercules, so in reality the Sun movement is more like a corkscrew, an helicoid or lemniscata.
But we can forget about that movement for the moment, and concentrate
in the "elliptical" Sun movement.

Have you done that? did you include the Earth, and its axial tilt? what about the other planets, and their markedly different axial tilts?

Done?

So, my conclusions:
On the issue of parallax:
We'll have two parallaxes, so to speak. One parallax, the classic one,
is due to the orbit of the Earth around the Sun, and amounts to 2 AU/year.
The other one, the "new" one, is due to the orbit of the Sun around its barycenter, and will amount to 2.26 (~= 2) AU/1800 years. That is: 0.001255 AU/year, or 187828.437 km/year, which must correspond to an arc of 6 arc min/year of Sun's displacement in the sky, related to the fixed stars.
Side note: Both parallaxes must yet be corrected for the presupposed solar system displacement toward Hercules.

On the issue of precession:
We'll have two precessions, too.
The classic(lunisolar) precession, amounting to ~ 50 arc secs/year, which is effectively due to the gravitational pull of Sun and Moon acting on an oblate Earth. You can clearly observe this precession observing the movement or the Earth axis related to Polaris, by example.
A "new", apparent precession, due to the Sun movement around its
barycenter, amounting to roughly 6 arc min/year. You wouldn't be able to observe this "precession" looking at the relation between Earth axis and Polaris!(because the Earth axis wouldn't precede at all due to this.) Except over long periods of time, and at first it will look more like a "displacement" than like a rotation; it will look as a complete “rotation”(of the Earth itself in space around a "center", not of its axis of rotation, and neither of the Earth around its own axis) only after 3600 years. A rotation with a radius of roughly 1 AU, due to the Earth following the Sun's movement in its binary orbit.

Now a very important question arises:
Why we didn't detect this apparent precession(and the associated
parallax)?
My answer is that we did, but decided to ignore it! For all practical
purposes, astronomers consider the Sun fixed, and the planets in orbit
around a fixed point. So the only "correction" astronomers supposed they must introduce, was to account for the very evident(because you can see it in a relatively short time, and also in the change of the Earth axis related to Polaris) lunisolar precession effect. The other
"precession" (in reality, it's not a precession of the Earth; physically speaking, it is the movement of the Sun) "got lost" into the corrections.
That is, it is actually "hidden" in what we consider to be the duration of the sidereal year!
For that very reason, from time to time astronomers must introduce "leap" seconds, to account for the acceleration of the Sun around its barycenter (to correct the correction, so to speak.)
We relate these leap seconds to the slowing down of the Earth, but, I dare to say: that is a mistake.
The main reason they are there is because the Sun is not only moving, but it is also moving in an accelerated way around its Barycenter.

A rough estimate (based on the previous values) is that the "real" sidereal year would be ~140 minutes shorter, and that those extra 140 minutes are there today to "align" everything up, disregarding that way the Sun's own movement, in order to be able to treat it as stationary.

The most important point is: Apparent precession will not be directly observed, because it is not reflected/based in a real movement of the Earth axis of rotation(better said: not reflected as a precessionary kind of movement; it will only slowly "displace" the Earth, without changing its obliquity). So., it will easily be confused (and hidden) into the duration of the sidereal year.
The only ways to detect it are to find Barbarossa, or to measure the length of the sidereal year against the backdrop of fixed stars.
(See http://siriusresearchgroup.com for details on the later.)

Of course, all this stands(if it's correct, as I think it is) for any binary orbit, and will be gradually more difficult to detect/observe, the longer the orbital period.

Best regards!
Mauro
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Joe Keller

USA
957 Posts

Posted - 23 Jan 2009 :  19:16:58  Show Profile  Reply with Quote
(email sent 15 min. ago)

Hi Prof. *********!

I haven't opened your more recent email yet. I wanted to finish my best prediction and give that to you first.

My best prediction, reworking the data from the beginning, for Barbarossa's geocentric celestial coordinates at the Jan. 20, 2009 epoch, is

RA 11h 28m 10.6s
Decl -9deg 16' 49"

This includes the small correction for binary motion.

The prediction I sent two days ago (then, I simply gave the change from my Dec. 22, 2008 position) amounted to

RA 11h 28m 9.6s
Decl -9deg 17' 0"

If I add a correction to that, for binary motion, it's

RA 11h 28m 9.4s
Decl -9deg 16' 58"

Sincerely,
Joe Keller
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Joe Keller

USA
957 Posts

Posted - 23 Jan 2009 :  23:33:59  Show Profile  Reply with Quote
Hi Prof. *********!

I noticed that the star to the east of my "best" coordinates, labeled magnitude 19.5 on the jpg version that I have of the Jan. 20, 2009, U. of Iowa photo, is listed in the USNO-B catalog, as Red1 magnitude 18.03 & Red2 magnitude 18.22. Likewise, the star (or galaxy) to the south, labeled magnitude 20.7, is listed as R1 19.18 & R2 19.27. Glancing at only a few other stars, I see that USNO-B 0806-0230182, at 11:28:23.9, -9:18:58, listed as R1 19.81 & R2 19.33, is far less impressive than the star labeled 20.7, and well might be overlooked. Likewise USNO-B 0807-0228782, at 11:28:14.1, -9:17:17, listed as R1 [blank] & R2 19.34.

I've never found Barbarossa or Frey on a Blue sky survey. My magnitude estimates come almost entirely from comparison to the R1 & R2 magnitudes, according to USNO-B, on Red sky surveys. So, I have been looking at objects roughly a magnitude dimmer than what is labeled "19.5" on this photo, roughly equal to what is labeled "20.7", and only a fraction of a magnitude brighter than many objects hardly imaged at all.

If the limiting magnitude is ~ +21, as measured by this instrument, that would translate to ~ +21-(20.7-19.2) = ~ +19.5, as I've been estimating against Red sky surveys. This hardly seems sufficient, since my best estimate of the magnitudes, by comparison with Red sky surveys, has been ~ +19.

The two position predictions I gave, are based on different assumptions. The first prediction (11:28:9.4, -9:16:58) though less precise, might be more accurate. There is a density at about 11:28:10.0, -9:17:09 (determined with a drafting right angle, with first order correction for differing horizontal and vertical scales on my screen). This is only 14" from my rough estimate of the ~ 194" change in position (mainly due to Earth parallax) between Dec. 22 and Jan. 20.

Sincerely,
Joe Keller
--------------------------------------------------------------------------------
Date: Fri, 23 Jan 2009 18:42:14 -0600
Subject: Re: Jan. 20 U. of Iowa photo
From: *********
To: josephkeller ---
CC: *********

Joe,

Sorry, nothing within ~ 60 arcsec at that position to a limiting magnitude ~21 on the Jan 20 image (see attached.)

*********
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Stoat

United Kingdom
964 Posts

Posted - 24 Jan 2009 :  04:17:01  Show Profile  Reply with Quote
Hi Maurol, I suppose that if we want to lose more than six minutes the first place to look would be in the difference between the tropical year and the sidereal year, about twenty minutes. That's explained by the couple on the earth by the sun. Moving the barycentre out so far would mean that the inner planets have to accelerate slightly for half the year , then decelerate for six months relative to the sun. That would mean that our day would vary slightly over the year. That does happen but it's explained by our orbit not being a circular one.

I don't know, I suppose we'd have to build a computer model to look for a best fit but it's an horrendous job to do.

(Edited) I just took a look at "stars in your backyard" where I had put Joe's planet into the program months ago. I've got the planet at 191 a.u. and so i changed the viewing date to 9999 A.D. Joe's planet was still close to the autumn equinox.

(Edited again) I looked at the info on Joe's planet in starry night and its year is 2640 years. About one tenth of the time it takes our earth's pole to wobble round a complete circle, 25,800 years.
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