Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
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Stoat

United Kingdom
964 Posts

Posted - 24 Jan 2009 :  05:55:06  Show Profile  Reply with Quote
Thinking a bit more about this, if we have someone here who can model it, then it might be good to talk about what we want in it.

Let's have Joe's planet having an orbital period of 2580 years. The Earth and Mercury. it might be an idea to show the sun's equator as well as the celestial equator. The reason for Mercury rather than the more obvious Jupiter is that Mercury's orbit is important in terms of the concept of curved space time.

I think we have to shove into the model a guestimatee of how fast the solar system is moving as well. From other stuff I reckon that we are moving at about 1.8% of light speed. How does the system know where it's supposed to be? If gravity takes some finite time to propagate, then Joe's planet will always be playing catch up.

I wonder if we could use Dayton Miller's data for this?
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Joe Keller

USA
958 Posts

Posted - 24 Jan 2009 :  21:40:47  Show Profile  Reply with Quote
(email sent 10 min. ago)

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

I had found, with the help of computer programs, that pairs of "disappearing dots" (Barbarossa+Frey) on sky surveys and in prospective photos, are at positions consistent with a mass ratio 0.8771::0.1229 and a circular orbit, for the center of mass, at 198.4 AU. Today I finished a new BASIC computer program, which from observed Barbarossa+Frey positions, finds the barycentric (as defined in the Astronomical Almanac, i.e., including only the familiar planets) XYZ coordinates of the Barbarossa/Frey c.o.m. under the above assumptions, using the Law of Cosines to solve the triangle. I extrapolated the Jan. 20, 2009 barycentric c.o.m., from the Dec. 22, 2008 and March 25, 2007 barycentric centers of mass. Then I converted to J2000 geocentric celestial coordinates.

To get positions for Barbarossa & Frey, from the center of mass, I assumed the same binary separation for Jan. 20, 2009, as seen Dec. 22, 2008. I estimate that the error in this assumption is negligible, because it seems that Frey is near the tip of its almost edge-on binary orbit. So, Frey's apparent distance from Barbarossa is near extremum, and its apparent circumferential binary motion is reduced, by a large factor, by perspective (the orbital path is seen almost end-on).

The crude estimates I gave a few days ago, were essentially by this method, but with rough trigonometry. The precise prediction for Jan. 20, 2009, is

Barbarossa RA 11:28:11.0, Decl -9:16:45

Frey RA 11:28:53.6, Decl -9:07:41


The other computer prediction, used to put the red circle on the Jan. 20 photo, should be adjusted very slightly to

Barbarossa RA 11:28:10.4, Decl -9:16:46

to incorporate my best current understanding of the Barbarossa/Frey binary orbit. This prediction returns to the original data, and fits a solar orbital radius varying linearly with time, to make the acceleration of the c.o.m. precisely Newtonian.

There seem to be many "hot pixels" in the Jan. 20 photo. The file name implies that it is stacked, but I would like to be sure, mainly because a stacked photo might be more sensitive.

At the first predicted position above, for Barbarossa, is a cluster of several mildly bright pixels in the Jan. 20 U. of Iowa photo, typical of what is sometimes seen for a faint galaxy. Though dimmer, these pixels resemble the pixels seen, in this same photo, at the position of the star or galaxy USNO-B 0807-0228782, at 11:28:14.1, -9:17:17, listed as R1 [blank] & R2 19.34. One might expect a planet with effective USNO-B red magnitude ~ +19.5, to look like this, in the Jan. 20 photo.

Faintness could be due to a forward-scattering nebular cloud (there is much evidence for this; see my posts on Dr. Van Flandern's website); smaller than theoretical mass, or greater than theoretical density; and/or an albedo even lower than usual for reddish outer solar system bodies. Outer solar system objects often are reddish in color, so it seems appropriate that I determined the magnitude by comparison with Red sky surveys.

Sincerely,
Joe Keller
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Stoat

United Kingdom
964 Posts

Posted - 25 Jan 2009 :  06:43:15  Show Profile  Reply with Quote
I think we need to discuss the failed binary idea in terms of how it alters the meta model of planet formation.

The question is, should it be done here, or in another thread?

In a failed binary the barycentre will accelerate towards the larger body of the proto system. The larger body is collapsing faster than Joe's planet. It will be dumping angular momentum into Joe's planet and into the cloud that will, much later, become Vega.

Joe's planet has to move outwards to mop up this dumped angular momentum. So the barycentre is slowing in its rate of acceleration inwards. Let's have the barycentre reach a point about half an a.u. from the proto sun. The proto sun is just over twice the mass of the current sun. It is spinning very fast and has to shed half of its mass, in a disk along its equator.

This hurtles out at some percentage of light speed. Very quickly that shed mass is outside of the barycentre radius. So the barycentre has to shift outward towards Joe's planet. Not long after Joe's planet can pick up some of this disk of shed material. That moves the barycentre outward as well.

Meanwhile the proto sun is still in the process of collapse. It needs to shed more mass but now its density is sufficient for it to eject a stream of matter of about one tenth of its mass. This is a stream of gobbets of molten matter, that will later form the planets. Now the problem is, that with a moving barycentre, that's way out in the boonies, the sun is climbing into a new orbit. So it ejects this gobbet stream in a curve. That's good as it allows for the stream to break up. Bad in that it makes the maths rather difficult. The mass ejected should be greater than ten percent and also the ejection velocity will be different to what we'd get if we were looking at a single sun. There's also the problem of the earlier ejection of the equatorial disk. It would be still trying to sort itself out in terms of where the barycentre of the system is. The sun should be able to mop some of that back up, moving the barycentre once again back towards it. Yeek!!!
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Maurol

Argentina
37 Posts

Posted - 26 Jan 2009 :  06:27:59  Show Profile  Reply with Quote
Hi Stoat, sorry for the delay, I was unable to reach a computer during the weekend. Comments below.

quote:
Originally posted by Stoat

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



Short answer: No.
Those 20 minutes are there to align for the effects of lunisolar precession. Lunisolar precession is a well established, defined and verified scientific fact. Moreover: it is a logical effect, due to the Earth being an oblate spheroid.
If you consider the Earth to move at 30 km/s, in 20 minutes it will travel the equivalent of 50 arc secs, on the angle it forms with the Sun.
(Side note: those 20 minutes compensate also(or aggravate, indeed) for some fraction of the "apparent precession" phenomena.)

The problem, as I see it, with binary theories, is just that, as these movements are indeed very subtle, long-term, and complicated to visualize, things tend to go astray with people doing the wrong models, and attributing the wrong causes to the right phenomena, so to speak.

To see the effects of the Sun movement(and the Earth following it, BTW), we'll have to look into the sidereal year, not the tropical year. The tropical year accounts for lunisolar precession, and that's perfectly OK. Remember: Lunisolar precession is relatively easy to observe and verify, because it leaves "physical marks" in the Earth itself(i.e. changes in the orientation of its axis of rotation.)
The "apparent precession" I'm talking about, on the contrary, it is not easy to observe, because it is a subtle, long-term effect, and also because it's not a precession, in the physical sense of the term. It's not the Earth axis changing physically of position(like in lunisolar precession) but rather the Sun moving, and (subtly) changing the geometric relations of the solar system.
There is no way by which this second "precession" could have an effect on the orientation of the Earth axis. Indeed, it is just because the Earth axis does NOT move due to this movement, that we start to have inconsistencies in the long term, and that we will be able one day, by taking these inconsistencies seriously, start to solve them "the right way", so to speak.

People don't like to deal with dynamical, moving things. First it was the Earth, that was fixed in the center of the Universe. Now it is the Sun!

quote:
...
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.



A computer model will be a good help to visualize it(but nevertheless, I think the best way is to "see it" in your mind, at least once). A compulter model can also be useful to make some predictions. More on this below. Now I want to summarize:

1) Lunisolar precession is OK. Cruttenden et al are wrong when they try to relate the Sun movement to the (physical) changes of orientation of the Earth axis. And, as far as I know, Plato never said that the Great Year does have to have 26000 or so years. We could very well have a Great Year of 2800, 3600, 2160, 4320, or so, years.

2) Cruttenden et al are right when they realize that a change in the Sun position will affect the apparent position of the Sun related to the stars, seen from Earth. But they haven't realized, as I've done, that this apparent change will be easily hidden/masked/compensated(in the short term) into the length of the sidereal year.

3) To the astronomers, and to the IAU: The Sun is moving! And most probably, it is moving in an accelerated way, orbiting something. How do I know? Instead of saying that I "just know"(which is right, plus, I have my sources) I'll say: because it's logical! All the heavenly bodies we're aware of, and can observe with a certain level of detail, are moving into some kind of orbit! The Sun is no exception to this, of course.

4) Good accelerometers on Earth should be able to detect this. The problem is to decompose the reported acceleration into all its components, and study the "residue".
The same with good (millisecond) pulsars, but here we have to be aware of precision, maybe the effect is so subtle(in the short term) that pulsar data cannot reveal it. I think Joe Keller have made an estimation on this before.

5) A thorough study of the history and reason of so called Bessel corrections, and also of the issue of leap seconds, could be very illuminating, and even provide hints on the nature and amount of the solar system's movement through space.

6) Regarding computer models: It could be useful to do the following:
Model a solar system and measure the length of the sidereal year(in seconds that are not astronomically derived, i.e. atomic clock seconds), first with the Sun immobile. Then, add a binary companion to the model, and measure the length of Earth's sidereal year with the Sun now moving into its elliptical orbit.

The measuring of the length of the sidereal year in this model must be done geometrically, that is, to look for perihelion/aphelion passage, and register the time it takes for the Earth to complete one orbit.

Now you can tune the mass and/or mean distance of the companion, to coincide with the actual "official" sidereal year length. From that, you can derive "good pairs" of mass vs. mean distance for the binary, that match the actual sidereal year.

From then on, it could be a matter of a little of luck, and much patience, to find the companion.

I'll now try to think on an experiment/observation that could shed light into the Sun movement(remember: it's not only the Sun movement. It's all of the solar system), and how to detect it and differentiate it.
I'll post it here, I suppose.

Best regards,
Mauro
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Maurol

Argentina
37 Posts

Posted - 26 Jan 2009 :  11:24:50  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

Thinking a bit more about this, if we have someone here who can model it, then it might be good to talk about what we want in it.



Do the following, and you'll be amazed (you'll also see that I happen to be right ;-)
Download the nifty program called Gravity Simulator (http://www.orbitsimulator.com/gravity/articles/download.html)
It's a pity that it is a windows program. But I can run it on Linux using wine, so it's cool again.

Download the solar system barycenter model: http://www.orbitsimulator.com/gravity/articles/ssbarycenter.html

Run it, starting with today's date. The program has a Graphic Options box, where you can pause the plotting, add Labels, and change between absolute (Sun centered) or floating mode(choose absolute mode).
You can also increase or decrease the time step, to be able to accelerate or decelerate the orbits. You can use 'z' to zoom out (to put the Earth on the screen) and 'Z' to zoom in.

So, play with the program a little bit, to get used to it. It's a very well done program, and with that particular simulation, you can see the effect that the different planets have on the Solar system barycenter.

Now, take note of the time of crossing of the positive X axis, which will correspond roughly to the northern hemisphere vernal equinox. (You can go to View to visualize the Orbital elements box, to see when Y equals or is close to zero.)
If you run it for a complete Earth orbit, and take note of the time of the next crossing of the "vernal equinox"(positive X axis, really), and later subtract the dates, you'll obtain 31540065 seconds for 2008/2009 (~ 365.04705 days.)
This number is 18084.54 seconds (~ 5 hours) smaller than the official sidereal year(!) This is very signification in itself: a program used to very precisely simulate gravitational movements, don't report the official sidereal year for the Earth orbital period, but a number 5 hours smaller.
But let's continue. Now do the following:
Reset the time to today's time, open the ssbarycenter file again (to start over from zero), pause the simulation, and go to Objects -> Create Object
Now add Barbarossa, with the following parameters:
Size: 300000 kmts (not important)
Mass: 4 Jupiter masses
Semi-major axis: 298 AU
Eccentricity: 0.001
Inclination: -9
Leave all the other parameters with their default values. Create the object.

Run the simulation again, taking note of the date of the crossing of the positive X axis. Subtracting the dates for 2008/2009 orbit, you'll obtain 31553160 seconds, which is only 4989.540 seconds (~ 1.4 hours) smaller than the sidereal year.

Do you see what I mean?

Of course, many things should be taken into account(the positive X axis is NOT the real vernal equinox, we are not considering Barbarossa's initial position, we are using absolute mode(fixed frame for the Sun) for simplicity, we are choosing an arbitrary eccentricity, etc.
And indeed, many more experiments can be done, with just this simple setup; creating different Barbarossas with different masses and at different mean distances, to observe the effect on the solar system. Of course, not only the length of the sidereal year changes, but also the eccentricity of the orbits, the velocity and position of the planets, etc.

5 hours in a year, with a Earth traveling at 30 km/s, is a lot of distance to cover, to be attributed only to software errors and imprecise measurements.

I'll see if now I can run the program taking notice of the crossing of the real vernal equinox, and also study the effects of the program "floating mode" on the estimations.

quote:

Let's have Joe's planet having an orbital period of 2580 years. The Earth and Mercury. it might be an idea to show the sun's equator as well as the celestial equator. The reason for Mercury rather than the more obvious Jupiter is that Mercury's orbit is important in terms of the concept of curved space time.
[quote]

Mercury can be important, because the same effect of "apparent precession" will manifest strongly in mercury.

[quote]
I think we have to shove into the model a guestimatee of how fast the solar system is moving as well. From other stuff I reckon that we are moving at about 1.8% of light speed. How does the system know where it's supposed to be? If gravity takes some finite time to propagate, then Joe's planet will always be playing catch up.



I think you can prescind of the solar system movement towards Hercules by the moment, because that will be practically a linear non-accelerated movement, for the periods we are considering(1 Earth year). Anyways, it's not very difficult to consider it: Hercules is at 30 degrees DEC, and the solar system is supposedly moving at 200 km/sec or so towards Hercules, so you must calculate the distance covered in a year, and multiply by the cosine of 30 degrees, to obtain the lateral displacement of the ecliptic in a year, in that direction.
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Stoat

United Kingdom
964 Posts

Posted - 26 Jan 2009 :  11:45:59  Show Profile  Reply with Quote
Hi Mauro, you are right that it's hard to visualise. What I've got in my mind's eye, is a thin metal disk, about 190 cms in diameter. I've got Joe's planet at one edge and the sun at the other. I stick a rod through it at about 1.3 cms and rotate it. Then I've got to say that that rod is pointing towards Vega and the whole system is moving up the rod towards that star. But I have to tilt the disk by nine degrees, and I've got to add a little disk for the earth, which is going to built up on one edge by about 7 degrees.

From above the sun's orbit round the barycentre looks like a perfect circle, the earth's looks like a slight ellipse. Really the sun is describing a spiral. It gently accelerates for more than a thousand years, then it gently deccelerates for the same time interval. It's taht which is the main player in the couple on the oblate earth. Ten spirals of the sun in its movement towards Vega will equal one great year of about 25800 years.
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Maurol

Argentina
37 Posts

Posted - 26 Jan 2009 :  13:54:21  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

Hi Mauro, you are right that it's hard to visualise. What I've got in my mind's eye, is a thin metal disk, about 190 cms in diameter. I've got Joe's planet at one edge and the sun at the other. I stick a rod through it at about 1.3 cms and rotate it. Then I've got to say that that rod is pointing towards Vega and the whole system is moving up the rod towards that star. But I have to tilt the disk by nine degrees, and I've got to add a little disk for the earth, which is going to built up on one edge by about 7 degrees.



You can use Gravity Simulator for that. You'll see that both the Sun and Barbarossa are concentric and the Sun's orbit is circumscribed to Barbarossa's orbit. That will help you to visualize all the system.

quote:

From above the sun's orbit round the barycentre looks like a perfect circle, the earth's looks like a slight ellipse. Really the sun is describing a spiral. It gently accelerates for more than a thousand years, then it gently deccelerates for the same time interval. It's taht which is the main player in the couple on the oblate earth. Ten spirals of the sun in its movement towards Vega will equal one great year of about 25800 years.



The Sun is describing an helicoid if you consider the movement of the solar system towards Hercules. If you don't, it describes an ellipse, and the model is easier to assimilate.

The difficult part to visualize is how the Sun's orbital movement affects(or not) the Earth axis, and, as far as I know, no program or model will help you with that, due both to scale and complexity issues.

In my opinion: The _movement_ of the Sun plays practically no part in the precession of the Earth axis. The _mass_ of the Sun(and Moon) plays it. Of course, if the Sun is moving, that will change the Sun distance and location in relation to Earth, and consequently the effect of its mass on the oblate Earth. But I'm assuming that that change is minimum. I'm assuming (I may be wrong) that we can consider 1 AU of mean distance for all precession effects, and also(which is logical), that the Earth will promptly and quickly follow the Sun wherever the Sun goes. I'm assuming a gentle movement of the Sun, so to speak. The effects of precession can, I think, manifest in their most part due to a rotating oblate Earth, and don't need, I think, a moving Sun.

I certainly can be wrong, and the effect of the Sun movement _can_ play a part, (even a significant part), in the precession of the Earth axis. But anyways, what's important to notice is what I've said before: The Sun could be moving much more than 50 arc secs/year, and we wouldn't notice, precisely because that movement will NOT produce nor will be reflected into a change in the orientation of the Earth axis. That movement will be incorporated/masked into the physical characteristics of Earth's and other planet's orbits, i.e. their eccentricity, orbital period, etc.

This will happen even more if, from time to time, to avoid inconsistencies starting to appear, you introduce leap seconds to account for the small differences in the Earth orbital and rotational(measured against the Sun) periods, caused precisely by the Sun movement, and not by the Earth slowing down, or anything like that. If the Earth is effectively slowing down, for what reason we transport the leap seconds to the duration of the sidereal year? they would have to be limited to Earth's rotational period only.

What I'm saying is this: if the Sun really were stationary/not accelerated, the sidereal year would be shorter. And this: we are actually disguising the Sun movement into the increased length of the planet's sidereal years.
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Stoat

United Kingdom
964 Posts

Posted - 26 Jan 2009 :  14:13:01  Show Profile  Reply with Quote
Hi Mauro, I think we had crossed posts there, I was writing my message while you posted a new one. I downloaded that program and the simulator thing, does that need to be put into the gravity program folder? As windows said it's not a program.
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cosmicsurfer

USA
507 Posts

Posted - 26 Jan 2009 :  14:27:12  Show Profile  Visit cosmicsurfer's Homepage  Send cosmicsurfer a Yahoo! Message  Reply with Quote
Because it appears that Joe's Barbarrosa has a circular orbit around a common gravitational center with our Sun, I am now thinking that our solar system is not a failed binary star system. In solar system genesis the proto-sun plasma spliting into a binary system with reverse rotations would match greater binary universe dipole motion and the greater gravitational mass rotation of sun would carry with it a string of orbitals that would become our planets as the plasmas spread apart. So, now we are left with answering questions regarding EPH timelines as described on this site by Tom VanFlandern and Earth catastrophes that may also be alligned with EPH events caused by a great intruder [Bodes law resonance not enough to cause both comets and asteroids too large of an explosion]. Okay, so I now think that the solar system has captured a great intruder that orbits around the barycenter of our binary system on an extreme eliptical orbit that is the cause for short period and long period prograde and retrograde comets and EPH distruction of a fifth planet 65 million years ago and a Mars size planetoid destroyed 3.2 million years ago. A short explanation since have to run to meet someone is that comets probably regenerate orbit from the pioneer effect or graviton field boundary zone 'push' and that these left overs from EPH planetary oceans that are now comets are far older then current predictions. We have a periodic intruder with orbiting planetoids that is probably in a retrograde orbit at a steep angle and caused the Hapgood pole shift but in a very short time causing a global catastrophe at beginning of Younger Dryas. Graviton motion is so high that sling shot effect around barycenter is minimized so that captured intruder system cannot escape from the increased gravitational forces near solar system gravitomagnetic field boundaries. So, if an intruder system was captured by our solar system was there an initial orbital deceleration over time causing increased potential for catastrophes during planetary orbital crossings? I would think that the intruders orbit would stabilize over time. Does anyone know if there is any precedence for solar system planetary capture? John
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Maurol

Argentina
37 Posts

Posted - 26 Jan 2009 :  14:27:17  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

Hi Mauro, I think we had crossed posts there, I was writing my message while you posted a new one. I downloaded that program and the simulator thing, does that need to be put into the gravity program folder? As windows said it's not a program.



Just install the program (run GravitySimulatorSetup.exe), run it (look for "Gravity Simulator" in the Programs menu) and then open ssbarycenter.gsim using File -> Open from the menu bar of the program.


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Stoat

United Kingdom
964 Posts

Posted - 26 Jan 2009 :  15:39:45  Show Profile  Reply with Quote
Hi Mauro, got it, it was that window didn't recognise its extension, and I hadn't actually ran the program to see that the barycentre data was already there. So, I'll have a bit play with that tomorrow.

I still like the Dayton Miller target of Vega, as we get about 36 degrees plus the nine form joe'splanet, we are moving at almost ninety degrees to the galactic plane.
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nemesis

84 Posts

Posted - 26 Jan 2009 :  19:03:07  Show Profile  Reply with Quote
Precession is a confusing topic to me. Can somebody answer this - 13,000 years in the future, and assuming our calendar is still in use, with it's periodic corrections - will December, January, and February be summer months in the northern hemisphere? Or will they still be winter months?
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Maurol

Argentina
37 Posts

Posted - 26 Jan 2009 :  19:42:17  Show Profile  Reply with Quote
quote:
Originally posted by nemesis

Precession is a confusing topic to me. Can somebody answer this - 13,000 years in the future, and assuming our calendar is still in use, with it's periodic corrections - will December, January, and February be summer months in the northern hemisphere? Or will they still be winter months?



Always will be winter in the northern hemisphere in december, because we are using the tropical year to rule our calendar. As its name imply, one of the functions of the tropical year is to align the calendar to the seasons.
See http://en.wikipedia.org/wiki/Tropical_year, it is well explained.
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Joe Keller

USA
958 Posts

Posted - 26 Jan 2009 :  23:17:32  Show Profile  Reply with Quote
I need suggestions for "cutting & pasting" from the "command prompt" window, into Windows documents. So far, I've been able only to cut & paste into other command prompt windows. I'd like to paste my new BASIC computer program here. [Done - thanks, Mauro.]

I think this new program is the most valid physically. It varies the Barbarossa/Frey mass ratio, the orbital radius, and the time derivative of the orbital radius, consistent with observed positions, to minimize the unexplained work performed. Commonly, three parameters can be varied to make three functions (such as the components of the unexplained force on Barbarossa) become zero. However, I found that for most choices of Barbarossa and Frey among my Red sky survey "disappearing dots", the unexplained force remained large for all physically likely parameter values. One choice allowed a much smaller force, no more than 1/30 of the sun's gravity (which might be supplied by a relatively small, moderately distant moon).

Varying the orbit to minimize the unexplained force, usually produces a "gutter ball" solution with large eccentricity. Minimizing not the force itself, but the work done by that force, produces smaller eccentricities. My one very good combination of "disappearing sky survey dots" gave mass the ratio Barbarossa::Frey = 0.8770::0.1230, a midrange "barycentric" (not including Barbarossa's mass in the barycenter definition) radius 207.4 AU, and decreasing radius, -26 AU per radian (therefore e > 0.1). That is about as before, but I found that this very good combination occurs, with reversal of the Barbarossa and Frey identities on both the 1954 and 1986 plates. For the first time, objects on the 1987 plate fit perfectly, but again, with Barbarossa & Frey reversed from what I had supposed. The very first object I discovered, and named Barbarossa, becomes Frey on the 1987 plate.

This has implications for the Genebriera, Riley, Turner and U. of Iowa photos. The new predictions are close enough to the old ones, that these photos still are usually of the correct patch of sky, but now there might be other "disappearing" (i.e., not on sky survey) objects on these photos closer to the predicted positions; or, the Barbarossa/Frey roles might need to be reversed. So now, I don't know whether Genebriera, Riley, or either of them, were first to photograph Barbarossa prospectively.

My new program prints out predicted center of mass positions for these photos, and also contains the coordinates of the important "disappearing dots". It will be fairly self-explanatory for someone familiar with engineering mathematics and with BASIC or FORTRAN programming.
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Stoat

United Kingdom
964 Posts

Posted - 27 Jan 2009 :  03:37:48  Show Profile  Reply with Quote
Hi Joe, somewhere in my drink befuddled brain there's a vague memory of someone explaining that program code, cut and pasted into a text document, has some problems with line breaks being seen as paragraph breaks.

Then there's the problem of how good the site is at keeping out hackers. I think that it will see the program as a virus. Best to pop it into a folder then zip it and send it to Larry for advice on how to get it to interested parties.

(Edited) Hi Mouro, is that 365. something figure not really, point two four something? which would make up our leap year exra day?
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Maurol

Argentina
37 Posts

Posted - 27 Jan 2009 :  05:58:20  Show Profile  Reply with Quote
quote:
Originally posted by Joe Keller

I need suggestions for "cutting & pasting" from the "command prompt" window, into Windows documents. So far, I've been able only to cut & paste into other command prompt windows. I'd like to paste my new BASIC computer program here.



Hi Joe,
You can do that by right-clicking into the command prompt window title bar, and choosing Edit -> Mark.
Then you select the text you want to copy with the mouse, by pressing the left mouse button while moving the mouse.
An then, go to the title bar and choose Edit -> Copy again by right-clicking.
It's better to zip the file and attach it in some way, as Stoat says.

Mauro
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Maurol

Argentina
37 Posts

Posted - 27 Jan 2009 :  06:01:05  Show Profile  Reply with Quote
quote:
Originally posted by Stoat
(Edited) Hi Muro, is that 365. something figure not really, point two four something? which would make up our leap year exra day?



Indeed. And you can check by yourself. More on this later.
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nemesis

84 Posts

Posted - 27 Jan 2009 :  12:48:01  Show Profile  Reply with Quote
I'll rephrase my precession question a little differently. Currently, at the winter solstice in late December the Earth is near perigee. In roughly 13,000 years, halfway through the precession cycle, at the winter solstice, still late December by the tropical year, will the Earth be near apogee?
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Maurol

Argentina
37 Posts

Posted - 27 Jan 2009 :  13:50:10  Show Profile  Reply with Quote
quote:
Originally posted by nemesis

I'll rephrase my precession question a little differently. Currently, at the winter solstice in late December the Earth is near perigee. In roughly 13,000 years, halfway through the precession cycle, at the winter solstice, still late December by the tropical year, will the Earth be near apogee?



Hi nemesis,
Short answer: No.
What encompasses the points of perihelion and aphelion is the length of the sidereal year. The tropical year is 20 minutes shorter, to align the calendar with the seasons, that is, to maintain the equinocces(marks of the seasons) in relatively the same days, due to the earth axis precessing, and then changing the occurrence of the equinocces. The tropical year is a convenience.
In 13.000 years, as the difference is cumulative, you'll have a (significant) difference of 13000 * 20 = 26000 minutes, around 18 days, between a calendar based in the sidereal year, and one based in the tropical year.
So in late December, the Earth will still be near perihelion, although in a calendar based in the tropical year perihelion will really occur around December 13.
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nemesis

84 Posts

Posted - 27 Jan 2009 :  16:03:30  Show Profile  Reply with Quote
Thanks, Maurol. You describe the tropical year as a convenience - it sounds more like a fudge factor to me. I don't see the problem with the solstices changing calendar dates, especially over such a long time frame.
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