Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
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Joe Keller

USA
944 Posts

Posted - 25 Apr 2008 :  22:03:00  Show Profile  Reply with Quote
In Barbarossa's Cavern There Are No Stars (Part XIV)

The figures below are the Briggs logarithms x 1000 (i.e., divide by 1000 to get the base-10 log) of "q" (see Part XIII) on 30' radius disks centered at, 1st row, (11h40m,-3deg30'), (11h36m,-3deg30'),...,(10h52m,-3deg30'); 2nd row, same but Decl -4deg30'; etc.; last row, same but Decl -11deg30'. This region is 13deg EW by 9deg NS. It is roughly centered on Barbarossa's 1987 position (approx. 11h18m,-8deg), which is near a "four corners" where degree squares circumscribed around four of the disks, marked by (*), touch. At this Declination, 4m is 0.98deg, so I treat 4m as 1 deg.

-778,-929,-1315,-796,-1431,-559,-393,63,-324,201,653,457,166
987,507,-332,462,664,549,542,740,772,789,1239,752,201
750,488,000,607,627,580,441,1097,966,733,751,787,426
487,833,38,637,745,648,554,903,903,1088,644,727,488
-628,-526,346,525,641,602*,1024*,1124,970,1103,1058,625,675
-204,-556,867,833,862,1003*,1146*,1103,869,566,765,766,534
280,211,655,826,1465,1610,1542,765,804,673,548,258,611
-140,-18,556,622,680,773,987,140,311,-204,-275,-643,372
-84,182,463,160,229,753,708,261,79,577,-186,-204,470

These numbers could undergo a suitable arbitrary linear transformation a*x + b, to give a grayscale optical density, to make a 9x13 pixel grayscale picture. A refinement would be to interpolate at corners (average of 4) and at midpoints of edges (average of 2) to make a 17x25 grayscale picture.
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Joe Keller

USA
944 Posts

Posted - 28 Apr 2008 :  21:02:41  Show Profile  Reply with Quote
In Barbarossa's Cavern There Are No Stars (Part XV)

The USNO-B catalog hardly could correct for the difference between the spectral sensitivity curves, "E" (for the plates used to determine R1)(E is approx. the same as "Rc") and R59F (for the plates used to determine R2). This correction depends drastically on spectral type; most USNO-B stars have only Red magnitudes known, hence nothing known about spectral type.

The long-wavelength tail of the Rc sensitivity curve, which, MS Bessell remarked, is important for Type M stars' magnitudes, lies roughly halfway between the R59F and IsubN sensitivity curves (see MS Bessel's Fig. 1). Optical infrared ("I") magnitude could be substituted for R2 magnitude, and the calculations of Parts XIII & XIV repeated to find another matrix. Because the R2 (R59F) & I (IsubN) sensitivity curves might effectively differ (for Type M stars) from the R1 (E or Rc) curve by equal amounts in opposite directions, averaging the two matrices might find the result that would have been found if R1, R2, and I all had the same sensitivity curve.

I did this. The average matrix is below. In the region described by the matrix, the dates of the most important "I" plates, are approx. 1995.2 & 1997.2. The dates of the most important "R2" plates are approx. 1986.2 & 1987.2. So the effective date of the second measurement is approx. 1991.45 (and of the first, approx. 1954.2).

In this region, 20% more stars list I & R2, than list I & R1. Using I & R2 (instead of I & R1) also lessens the chance of misidentification due to proper motion or long-term variability; because, the plate dates are much closer. So, I compared R2 vs. I as I had compared R2 vs. R1. Averaging was effected by subtracting half the new log(q) for R2 & I, from the log(q) found for R2 & R1. For R2 vs. I, I further increased sample sizes (to about triple those of the R2 vs. R1 study) and hence significance, by using +17 instead of +16 as the dividing line, and by using semi-infinite intervals instead of intervals of length 1 mag.

The matrix entry, corresponding to the degree square containing the 1991.45 c.o.m. Barbarossa/Frey heliocentric position, is marked (**). The three next-closest degree squares are marked (*).

-1140,-1265,-1420,-969,-1519,-608,-421,-162,-686,-186,332,149,-34
507,245,-469,483,728,687,640,576,518,403,844,477,-45
296,231,18,709,684,731,514,913,648,393,461,541,192
64,504,-33,746,818,682,644,706,572,697,316,471,368
-996,-724,214,628,741,636*,1027*,905,683,812,695,368,531
-535,-758,587,546,631,736**,881*,754,500,288,491,598,389
-152,-148,365,394,925,1099,1116,517,499,355,369,137,481
-1180,-395,249,246,288,200,579,-110,-68,-71,-16,-637,329
-498,-213,211,-225,-220,186,329,-118,-274,285,-375,-129,421
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marsrocks

USA
95 Posts

Posted - 29 Apr 2008 :  12:36:51  Show Profile  Visit marsrocks's Homepage  Reply with Quote
Joe, below are the visual images derived from the numbers of your post of 25 Apr 2008 : 22:03:00:

The square marks the coordinates where you placed your asterisks.






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marsrocks

USA
95 Posts

Posted - 29 Apr 2008 :  12:57:22  Show Profile  Visit marsrocks's Homepage  Reply with Quote
When the images are pushed to contrast extremes, it appears the high numbers in your chart are concentrated around two main localities:




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marsrocks

USA
95 Posts

Posted - 29 Apr 2008 :  13:15:25  Show Profile  Visit marsrocks's Homepage  Reply with Quote
And here is the same image again enlarged to 17 x 25:



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marsrocks

USA
95 Posts

Posted - 29 Apr 2008 :  13:20:12  Show Profile  Visit marsrocks's Homepage  Reply with Quote
And this one is the original again enlarged to 50 X 34:



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marsrocks

USA
95 Posts

Posted - 29 Apr 2008 :  13:25:26  Show Profile  Visit marsrocks's Homepage  Reply with Quote
And 416 X 288:



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Joe Keller

USA
944 Posts

Posted - 29 Apr 2008 :  19:48:59  Show Profile  Reply with Quote
In Barbarossa's Cavern There Are No Stars (Part XVI)

Thanks, "marsrocks", for the beautiful artwork! From my data, you've produced the first image in history (and in ten versions!) of a nebula within Barbarossa's solar system, only 200 AU (est.) from Earth.

This might be an advantageous time to start using the name, "Barbarossa", as a label. Otherwise, in the future it might be difficult to prove what the image was supposed to be.

The Iowa State Univ. library has fifteen books on Frederick Barbarossa that I know of, but only two are in English and one of those is missing. Much of what I know about Barbarossa is from Thomas Carson's "Barbarossa in Italy", a translation of a long, mysterious, almost perfect neoclassical Latin poem rediscovered in the Vatican Library in 1877 and apparently written by a Lombard scholar attached to Barbarossa's army as a translator or guide. In one scene, as I interpret it, someone tried to seize the papacy by having a heavily armed militia of his supporters run around Rome robbing and "vandalizing" (line 675), presumably more explicitly burning out and killing, the supporters (including cardinals) of Barbarossa's just-completed crowning as Emperor by Pope Adrian IV (an Englishman named Nicholas Breakspear)(Bk. II, lines 657-715). Barbarossa's troops soon were galloping through the streets of Rome, slaying these rioters, though Barbarossa took many prisoners (released the next day, to the custody of Adrian IV) and even ordered that the fleeing rioters not be ridden down, shot in the back with crossbows, etc. (lines 739-751). In another scene (Bk. V, lines 3000-3033) Barbarossa, at the suggestion of some of the other military leaders, orders some hostages spectacularly stoned to death, but, quickly enough to save some of them, realizes that this is inhumane. With the help of this precedent, stoning did not become the normal punishment in Europe.

Henry Ford said history is "more or less bunk", and Nietzsche said history amounts mainly to paying close attention to old newspapers. However, at least one person who had worked with Barbarossa during his invasion of Italy, and who was intelligent, educated and sober enough to write almost flawless Latin poetry (Carson thinks the content of the poem indicates that the author was familiar not only with almost the entire Bible, but also with almost the entirety of the work of most major classical Latin writers) had an opinion of Barbarossa which he expressed by writing a favorable eyewitness account, after Barbarossa had departed.

I noted in Part XIII that the "E" (approx. equal to "Rc" at moderate visible wavelengths) sensitivity curve, of the plates on which the USNO-B "R1" is based, extends almost 300A more blueward than does the curve on which "R2" is based. Part XV's averaging involving "I" magnitude, can't correct this. A USNO-B search shows that the fraction of relatively bluish stars (B2 < +17.00) among my sample stars (+15.00 < R2 < +17.00) doubles (from 16% to 29%) for an 8deg shift, from my region southward (the galactic equator is SW by S). (The fraction increases almost as much with a westward or northward shift, and only slightly decreases with an eastward shift; the region of Barbarossa is a 10deg-wide island or isthmus of dim B2 magnitude.) So, my region (9x13deg) is big enough to need detrending, to remove the effect of galactic latitude on R1 vs. R2, due to the better blue sensitivity of R1.

I mapped the region with a projection equating 4m RA to cos(7.5deg)* 1deg. The best-fitting linear detrending, was that which removed the positive trend in the direction 103.6deg clockwise from N (i.e., W by SW). This direction differs 40deg from the perpendicular to the galactic equator, but agrees precisely with the gradient of bluish stars. (I found log(fraction) of relatively bluish stars, as above, in a 10deg diam circle centered on my region, and in additional circles displaced 8deg toward azimuths, clockwise from N, 0, 90, 135, 180, & 270. With Simpson's and Cotes' rules I convolved these with cosine(azimuth) for various phases, finding that the gradient of blue stars lies in the direction 102.3deg.) The correlation coefficient was 0.307, indicating that the trend explains 9% of the variance. The significance ( 9*13 - 2 = 115 degrees of freedom) of the trend is p = 0.0008 (2-tailed). The detrended matrix (this is my best representation of Barbarossa's nebula so far) is

-828,-998,-1198,-791,-1386,-520,-377,-163,-732,-277,197,-31,-259
808,501,-258,650,850,764,673,564,461,302,698,286,-280
586,476,218,865,795,797,536,890,580,281,304,339,-54
343,738,157,891,918,738,655,672,494,574,148,258,111
-728,-501,393,762,830,681*,1027*,860,594,678,516,145,263
-278,-545,755,669,709,770**,870*,698,400,143,301,364,110
94,54,522,506,993,1122,1094,451,388,199,169,-108,191
-945,-204,395,347,345,212,546,-187,-190,-238,-227,-893,28
-273,-33,346,-134,-174,187,285,-206,-407,107,-597,-396,109

(**) denotes the degree-square containing the Barbarossa center of mass in 1991; (*) denotes the three next closest squares.
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Stoat

United Kingdom
964 Posts

Posted - 30 Apr 2008 :  03:49:05  Show Profile  Reply with Quote
Seeing two "lobes" on Marsrock's image reminds me of the famous "ears of Saturn." When Saturn was first looked at, it was thought to have ears. The rings seen almost edge on showed up as two crescent shapes. We should expect each lobe of the nebula to have a slight blue/red shift.
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Joe Keller

USA
944 Posts

Posted - 02 May 2008 :  18:08:36  Show Profile  Reply with Quote
In Barbarossa's Cavern There Are No Stars (Part XVII)

Part XVI's image of Barbarossa's nebula, shows the expected negative correlation (always, relative to the global mean for the region), between extinctions at points separated by vectors equal to Barbarossa's travel from 1954 (approx. date of R1) to 1991 (approx. mean date of R2 & I). (In review, Part XVI's image is constructed essentially from the average, "h", of log(#R2>R1/#R1>R2) & log(#I>R1/#R1>I), on 1deg diam disks, linearly detrended to remove the influence of the star color gradient.)

Approximately, the image maps the extinction due to Barbarossa's nebula in 1991, minus the extinction due to it in 1954 (that is, minus the extinction due to Barbarossa's nebula in 1991, evaluated at a second point 4.25deg E and 2.28deg S of the first). Because R1's spectral sensitivity curve isn't the same as R2's or I's (and my averaging and detrending only crudely correct that) what is mapped isn't exactly the finite difference h(x,y) = g(x,y) - g(x+4.25,y-2.28), where g is the extinction of Barbarossa's nebula. However, negative correlation should occur between lattice points separated by ~ (4,-2), whenever the prograde point of the pair, is near enough Barbarossa's ~1991 position, to be within Barbarossa's nebula.

A guess, that Barbarossa's nebula has radius 2deg, was confirmed roughly, by correlations of +0.65, +0.4 and ~ +0.05, between "h" values at points within 2deg of Barbarossa's 1991 position, and points, 1.4, 2.2, or 3deg, resp., away from these (in a generally retrograde direction). The lattice point pairs, one of whose members is within 2deg of Barbarossa's 1991 position, anticorrelate strongest, when the other member is 3deg N and 5deg W of Barbarossa's 1991 position; r = -0.805 and Fisher's normalized z = 3.15 ( > 3 sigma significance). The number of such pairs is n=11; I exclude separation vectors for which the number of pairs is n<8. Linear interpolation of the correlation, and of Fisher's z, between the four sets of lattice point pairs nearest (-4.25,+2.28) separation (relative to the point near Barbarossa)(n=14 or 11, total 50), gives r = -0.678 and normalized z = 2.63, p = 0.0086, 2-tailed.

If "h" is the difference in the extinction of Barbarossa's nebula at two points separated by a constant vector, the autocorrelation of "h" should be -0.5 for points such that h1 = g1-g2 and h2 = g2-g3, if g2 is as big as g1 or g3. Indeed, for the sets of lattice point pairs whose separations are the four vectors nearest (-4.25,2.28), with the pair's first member < 2deg from Barbarossa, the autocorrelation ranges from -0.613 to -0.805; when I pretend that Barbarossa orbits in the opposite direction, the autocorrelation for the analogous four sets nearest (+4.25,-2.28) separation, ranges only from -0.1 to +0.1.

Inspection of my B2/B1, R2/R1, and I/R2 plots, confirms the prograde motion of Barbarossa's inner nebula. Because autocorrelation indicates that the ordinary (inner nebula) extinction extends ~ 2deg from Barbarossa, I considered 6x6deg squares centered on RA11:18,Decl-8. For R2 vs. R1, I found the first difference in column averages of log(q) = log(#R2>R1/#R1>R2), then interpolated linearly between these to find the maximum extinction 0.45deg W of center. (Extinction of R1 & B1 is neglected because it centers 4deg W, at Barbarossa's 1954 position.) B2 vs. B1 had "0" cells prohibiting logarithms, so instead I used (#B2>B1 - #B1>B2) / (#B2>B1 + #B1>B2), weighted within each column, by number of observations. Then I followed the same procedure as for R2 vs. R1, to find maximum extinction 1.40deg W of center.

I vs. R2 are theoretically only 1deg apart, so both extinctions must be considered. Also, the I & R2 spectral windows differ much, so the star color gradient might cause a big linear trend. First, I found first differences in column averages of log(q), as for R2 vs. R1 above. Then I linearly detrended to make the last column average equal the first. If the I and R2 extinctions are the same, the midpoint should be an inflection point. The second difference changed sign in two adjacent instances; interpolation gave two nearby inflection points which I averaged to find the midpoint of I & R2 at 0.77deg E of center.

Thus, the peak extinction, estimated from the USNO-B catalog magnitudes, of B2 vs. B1, was 0.95deg W, of that of R2 vs. R1. Barbarossa's travel in that time would predict 0.5deg W. The estimated midpoint of the extinction peaks of I & R2 was 1.22deg E, of the peak of R2. Barbarossa's travel would predict 1.0/2=0.5deg E. Though the extinction travelled about twice as far as predicted, statistical errors and methodological uncertainties are big. The correct sequence of the three, is significant at p=1/6.
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Joe Keller

USA
944 Posts

Posted - 03 May 2008 :  19:30:01  Show Profile  Reply with Quote
In Barbarossa's Cavern There Are No Stars (Part XVIII)


Dedicated to Frederick Barbarossa: "King, Knight, Hero"
(Helmut Hiller)


Barbarossa Nebula Dynamics Match Observations

The simplest dynamical estimate of the radius of Barbarossa's nebula, would be 197.7 AU * sqrt(0.0103), using the distance from Barbarossa to the sun, and the ratio of Barbarossa's mass to the sun's, to get 20.06 AU. What we want to know, is the radius of the largest possible orbit around Barbarossa. As a start, let's consider only gravity, and ignore the sun's gravitational pull on Barbarossa itself. The case of a face-on circular orbit can be solved exactly: the orbital plane is displaced toward the sun so that Barbarossa's outward pull equals the sun's inward. Barbarossa's best efficiency in this, is max(sin(theta)*cos^2(theta)) = sqrt(4/27), so the biggest possible projected radius is 20.06 * sqrt(sqrt(4/27)) = 12.44 AU.

The case of an edge-on orbit can be estimated using the circular approximation: in half a circular orbit with speed v, Barbarossa vectorially accelerates the body by approx. pi*v, but if the body is bound, again ignoring the sun's gravitational pull on Barbarossa itself, the sun accelerates the dust grain prograde by < (sqrt(2)-1)*v. The sun's efficiency in this, over half an orbit, is only 2/pi, so the biggest possible radius is 20.06 * sqrt((sqrt(2)-1)*1/2) = 9.13 AU. Thus the edge-on and face-on circular cases give approx. equal radii. Though Barbarossa's orbital motion continually converts face-on to edge-on and vice versa, this common radius could be taken as the radius of Barbarossa's nebula, if only gravity were considered.

ISU's copy of Hodge's "Interplanetary Dust", has an anonymous penciled marginal note referring to Burns, Lamy & Soter, Icarus 40:1+, 1979. I took the hint.

According to Burns, et al, Fig. 7a, p. 16, the greatest "beta", i.e., ratio of solar radiation pressure to solar gravity, 5.5, occurs for pure graphite dust of 0.08 micron size (this is for spherical particles, but shape dependence is weak according to, inter alia, Kruegel; and furthermore Burns et al give evidence that rather spherical shapes predominate anyway). So, for classical interstellar dust, i.e., carbonaceous particles of ~ 0.1 mu size, the biggest possible circular orbital radius isn't much bigger than 12/sqrt(5.5) AU (actually, the sun's gravity pulls the same on Barbarossa and on the dust grain, so the net push, equals the radiation pressure); i.e., it can subtend not much more than 1.5 deg from Barbarossa. This matches the 2deg radius found empirically in Part XVII from the autocorrelation, near Barbarossa, of the USNO-B R2 vs. R1 extinction measure (from the third, most accurate version of my image data). Or it can be found simply by looking at the images of that extinction measure, assembled by "marsrocks" above (from the first, least accurate version of my image data).

This small patch around Barbarossa, eight times the diameter of the full moon, is where backscatter of sunlight, from Barbarossa's nebula, might be detected. Farther than 1.5 deg from Barbarossa, solar radiation pressure drives away classical interstellar dust. Ultrafine, smaller-than-classical carbonaceous dust grains, which for any given extinction cause orders of magnitude less scattering, would reach 12/sqrt(1.77) AU --> 2.6 deg from Barbarossa, because (Burns et al Fig. 7a) for graphite of size < 0.004 mu (40A), "beta" has an asymptotic value of only 1.77.

The first, least accurate version of my image data (the later versions haven't yet been extended this far) shows roughly vertical strips of R2<R1 about 5.5 deg east, and R2>R1 about 12 deg west, of Barbarossa's 1987 position. These are narrow strips ~ 4 deg wide, not part of the global trend. Barbarossa's 1954 & 1987 positions are separated about 3.9 deg in RA, so if Barbarossa's outermost (gas, or low-beta non-carbonaceous dust) nebula has radius 9 deg, then the 1987 outer nebula position would cover, roughly, a vertical strip 5 to 9 deg east of Barbarossa's 1987 position (vs. ~ 5.5 observed) that wasn't covered in 1954; and the 1954 position would cover, roughly, a vertical strip 9 to 13 deg west of Barbarossa's 1987 position (vs. ~ 12 deg observed). For a 9deg radius, beta ~ 0.15 would be needed ( 12/sqr(0.15) = 31 AU --> 9.0deg ). Really, beta isn't the cause; it's the Jacobi limit, 30 AU (see Technical Detail #2, Part XIX below). So, both the Jacobi limit, and the radiation pressure limit are seen, defining an outer and inner nebula, resp.

These strips, indicating the E & W extreme portions of Barbarossa's outer nebula, give negative extinction (i.e., brightening instead of dimming). These are mainly Type M stars, so this agrees with the negative extinction found for Type K & M stars in my Harvard vs. Johnson magnitude study of the same region (statistically controlled, by samples from distant regions of equal Declination and similar galactic latitude). Reviewing all the stellar photometric studies of Barbarossa's outer nebula that I've done, I find that Type K & M stars consistently undergo paradoxical, negative V extinction there, Type F & A consistently ordinary, positive V extinction. Not only is this seen in the smooth linear variation with spectral type, in my Harvard vs. Johnson study; it's seen in my original, 4deg-resolution, study in which B2 - B1 was greatest E of Barbarossa, but R2 - R1 greatest W of Barbarossa.

I made a 1deg-resolution B2 vs. B1 study, similar to the R2 vs. R1 study the results of which are displayed in the above images by "marsrocks". Even summing results using two pairs of magnitude intervals, [16,18] & [18,20] and [17,19] & [19,21], data are sparse, and my study less thorough, but I do find positive extinction, of B2 vs. B1, out to roughly 4deg from Barbarossa's 1983 position, at least in some directions, as in the R2 vs. R1 case. Unlike the R2 vs. R1 case, the B2 vs. B1 case shows increased, not decreased, extinction of B2 vs. B1, about 9deg E of Barbarossa; and perhaps increased, not decreased, extinction of B1 vs. B2, ~ 5deg W of Barbarossa. Thus the outer nebula finding for B2 & B1, resembles the Harvard vs. Johnson finding for Type F & A stars.

My Hipparcos studies show that B-V changes (it's the reverse of ordinary extinction) in the opposite direction, to V, as though B were much less affected than V. (This holds both for Draper/Cannon Type K0, and for Draper/Cannon Types F5-B9.) This suggests that photons from stars are traded slightly up or down in energy, by a kind of stimulated emission as they traverse Barbarossa's outer nebula which has been pumped by sunlight.

Barbarossa isn't luminous like the sun, so the Poynting-Robertson and Yarkovsky effects are unimportant in Barbarossa's nebula close to Barbarossa. Barbarossa's huge angular momentum replenishes the orbital angular momentum, around the sun, of the particles gravitationally bound to Barbarossa; and the sun's radiation effects average zero on a particle's orbit around Barbarossa.
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Joe Keller

USA
944 Posts

Posted - 04 May 2008 :  16:17:26  Show Profile  Reply with Quote
In Barbarossa's Cavern There Are No Stars (Part XIX)

Technical detail #1. I've seen the claim that the red plexiglass filter used in the Palomar Red sky survey, is about the same as a Wratten #29 filter. The Handbook of Chemistry & Physics, 44th ed., gives (with minor interpolation) 6220A as the 50% cuton for Wratten #29. The next listed, Wratten #26, has 50% cuton at 6070A. MS Bessel's Fig. 1 gives 5890A as the 50% cuton for R59F (i.e., SERC-Red). All these cutons are steep. So, if Wratten #29 is at all correct, the Palomar Red survey cuts on, 300A redder, not 300A bluer, than the SERC-Red survey. However, my data support the other claim I've seen, that Palomar Red's "xx103aE emulsion + red plexiglass filter", together, still were consistent with the "E" passsband (which in turn is about the same as "Rc" at moderate visible wavelengths, allegedly according to a Cambridge Ph.D. thesis; Rc's cuton is 285A bluer than R59F's). R2 - R1 increases toward the galactic equator (or roughly equivalently, along the local gradient of blue star fraction); this gradient in R2 - R1 is seen in the farthest corners of my pure R2>/R1> plot, as much as 20deg away from Barbarossa. So, R1 would seem more sensitive to bluer stars, than R2.

Technical detail #2. The Jacobi limit (Icarus 102:298+, 1993; or 107:304+, 1994), which uses the cube root of the mass ratio (rather than the square root as my estimate does) for Barbarossa, considering only gravity, is 29.8 AU. Empirically, however, the "most distant known satellites" (vis a vis the Jacobi limit), Jupiter's "Pasiphae and Sinope", have semimajor axis only 0.45x the Jacobi limit, corresponding to 13.4 AU for Barbarossa (Icarus 102:298+, p. 304). This agrees well with my estimate, considering only gravity, of ~ 9.13, to 12.44 AU. My estimate easily applies to the virtual negative gravity of radiation pressure, but the Jacobi limit is based on the Roche lobe, and not so easy to apply here.

Technical detail #3. Recent Earth-based photographic searches for satellites of Uranus & Neptune (Icarus 102:298, 1993; 107:304, 1994) have used the same emulsion/filter combination, IIIaJ + GG395 (passband 3950-5350A) and exposure, 60min, as the SERC-J (Blue) sky surveys. If this is best for sunlight reflected from unknown solar system objects, it's presumptively best for Barbarossa's nebula. The removal of red light reduces the red dwarf background.
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Joe Keller

USA
944 Posts

Posted - 07 May 2008 :  20:32:52  Show Profile  Reply with Quote
In Barbarossa's Cavern There Are No Stars (Part XX)

IRAS Imaged Barbarossa's Bow Shock


Schlegel, et al, Astrophysical Journal 500:525+, 1998, Fig. 8, p. 542 ("Dust"), shows a streak of far-IR brightness, measured by the IRAS satellite in 1983, that is only 4 deg from Barbarossa's 1983.2 heliocentric position (Barbarossa was at l=265.9, b=+48.4) and extends ~ 12 deg each way. It's prograde of Barbarossa (i.e., > galactic longitude) and crosses Barbarossa's orbit at approx. 37 deg slope ("beta"). Though presumably its seeming (non-greatcircle) extension near galactic coords l=190, b=+40, is accidental (the alleged Martian canals perhaps also demonstrated the human ability to notice accidentally aligned dots) this streak is one of the most prominent curvilinear "cirrus" "dust clouds" in the sky.

With a mm ruler, I measured the positions of 8 points of this cirrus (on Schlegel's polar-projection map, Fig. 8) chosen for their definiteness, and their roughly equal spacing with good linear interpolation between. Then I used cubic Lagrange interpolation on Schlegel's map, to find their galactic coordinates, converted to other spherical coordinates (for a nearly conformal flat map) for which the endmost points of the "cirrus" define the equator, plotted them on Keuffel & Esser graph paper using a sinusoidal projection with Barbarossa's 1983.2 position defining the prime meridian, and graphically found the best-fitting hyperbola (taking Barbarossa as a focus, using symmetry to find the slope of the directrix by eye, then successive approximations to find the eccentricity, ~ 5.5, i.e., very flat). I also made the small correction, longitude --> sin(longitude), because I'm allegedly viewing, approximately, an hyperboloid of revolution, from a finite distance; not a hyperbola. (Hausmann & Slack's Physics, Ch. XXXI, sec. 367, Fig. 355, shows a flat-nosed projectile with an hyperboloidal bow shock with e = 1.5.)

The N & S asymptotes slope 27 and 47deg, resp., to Barbarossa's orbital path. This range of slopes lies well within the theoretical range, 16.9 to 61.5deg, for initial Mach number 3.432, final Mach number > 1, and adiabatic constant k=5/3. (See Hughes & Brighton, "Fluid Dynamics", McGraw-Hill, 1967, Sec. 8.2, eqns. 8.9 & 8.10 and figs. 8-1 to 8-3, pp. 158-159; or, 2nd ed., 1991, pp. 210-211.)

The Mach number "M1" is taken to be the ratio, of Barbarossa's orbital speed in a circular orbit at 197.7 AU, to the speed of sound in rarified dissociated (i.e., atomic) hydrogen at the solar radiation Planck equilibrium temperature (this is independent of albedo, by Kirchhoff's law) at 197.7 AU. Using (HL Johnson's favored) 5800K for the sun's surface, the sun's radius, and the Stefan-Boltzmann T^4 law, gives 28.1K as the equilibrium temperature at Barbarossa. (According to Schlegel, interstellar dust is 15-19K.) The proportionality of the speed of sound, to sqrt(T/m), is so accurate that my result varied little, even when I based it on molecular hydrogen at STP (with the sqrt(T/m) correction). However, I based my calculation on the value 8.894 km/s for dissociated hydrogen at 0.1 bar and 5800K (the lowest pressure in the table; the temperature went up to 6000K, but I wanted to use HL Johnson's favored temperature of the sun)(Vargaftik, Handbook of Physical Properties of Liquids & Gases, 2nd ed., p. 34).

Corrected to 28.1K, the speed of sound in rarified dissociated hydrogen is 0.6176 km/s, giving M1 = 3.432, and a minimum possible "beta" = arccsc(M1) = 16.9deg (Hughes & Brighton, eqn. 8.8). The abscissa of the intersection of the curve M2=1, with the M1=3.432 isocline (from simultaneously solving, numerically, Hughes & Brighton eqns. 8.9 & 8.10) is 61.45deg for the monatomic adiabatic exponent k=5/3 (Fig. 8-3 shows 64deg, but this seems to be for air; k = 1.404 & 1.401 for molecular nitrogen & oxygen, resp., at +15C & std. pressure).

On my R2 vs. R1 chart, I had marked degree squares with ratio < 1 and > 6; these corresponded roughly to lower and upper quartiles. Within the 3x3deg square centered on Barbarossa, the inner nebular extinction shows a northern boundary sloped at ~ 45deg.

On the B2 vs. B1 chart, I had marked squares according to whether none, one or both of the ratios used exceeded 1. Here, the average position of the arrows shows a northern boundary sloped at ~ 67deg, but data are sparse.

The average of these two charts, weighted by number of observations, gives 47deg, i.e., 45 + (67-45)/(10+1) - 27 = 20deg to Barbarossa's orbit. Hughes & Brighton's Fig. 8-3 isocline needs to have its peak shifted 2.5deg left if k=5/3; the left zero doesn't move. So, from the interpolated M1=3.43 isocline, shifted an interpolated 1deg left at the relevant position, I can read that for "beta" (the slope of the shock wave) = 37deg, "theta" (the slope of the obstruction) = 22deg, agreeing with the ~20deg observed in the extinction chart.

Schlegel's Fig. 12 maps the difference between Schlegel's infrared map, and the Burstein & Heiles, Astronomical Journal 87:1165-1189, map based on reddening of external galaxies and HI column densities. The Burstein & Heiles map shows nothing unusual in the vicinity of Barbarossa, so the "cirrus" (really, bow shock) appears on Schlegel's Fig. 12 also. (I chose Fig. 8 for my study, to stay closer to the original data.) Schlegel used DIRBE satellite data (1989-1990) to correct the (higher-resolution) IRAS data for temperature, to convert them into presumed interstellar dust column densities; Schlegel's temperature map shows nothing obviously unusual near Barbarossa.

However, even when temperature is considered, IR emission need not correlate with HI (nor presumably with reddening of external galaxies). Regarding a shell in Eridanus, "When the IR emission is compared with the HI there is little direct positional agreement. ...We are not yet convinced that we can account for apparent variations in IR brightness as resulting exclusively from temperature effects,..." (Verschuur et al, in, IAU Symposium #139, 1989; Conclusion, p. 236).
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nemesis

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Posted - 09 May 2008 :  09:59:52  Show Profile  Reply with Quote
Shouldn't Barbarossa cause observable perturbations of the outer planets?
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Joe Keller

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Posted - 09 May 2008 :  14:43:35  Show Profile  Reply with Quote
quote:
Originally posted by nemesis

Shouldn't Barbarossa cause observable perturbations of the outer planets?



Thanks for bringing up this important question again. Scattered through this long thread, is more information about this.

Goldreich & Ward, PASP 84:737, 1972, claimed that even a moderate Planet X, unless almost exactly coplanar with the rest of the solar system, would move Neptune to high inclination, within much less than the age of the solar system. Their article doesn't consider chaos theory. Also, I've counted four questionable simplifying assumptions explicit in their calculation:

1. p. 737, col. 2, par. 2, sentence 2
2. p. 738, col. 2, par. 1, sen. 3
3. p. 739, col. 1, 3rd to last sentence
4. p. 739, col. 2, last par.


Zakamska & Tremaine, AJ 130:1939+, 2005, measured acceleration relative to those millisecond pulsars (MSPs) which are not within globular clusters. In their sample, 35 of 48 MSPs were binary (typically MSPs are close binaries). The contemporary theory of MSPs is that the original supernova sweeps its environs so clean, that except for companion stars which often endure, and rarely planet-size masses (maybe really stellar fragments), there's nothing nearby to produce an acceleration more than about that of Barbarossa on the sun: a distant brown dwarf companion (there is evidence that many or most stars have these; see my posts) survives the supernova, but planets (closer and smaller) don't; stellar companions, if they survive, are detected and the MSP period corrected for their effects. Thus the scatter in Pdot/P for MSPs, corrected for obvious binary effects, should be about what would come from a brown dwarf companion at a few hundred AU, which endured the supernova! Accordingly, Zamaska & Tremaine believe themselves statistically able, based on the scatter of Pdot/P for their sample of 48 MSPs, to rule out as little as 1/10 ~ 1/sqrt(48) the acceleration due to Barbarossa.

Zakamska & Tremaine used nonparametric statistics (rank correlation) to look for the acceleration effect of Barbarossa, then parametric statistics (scatter of Pdot/P) to calculate the significance of their finding. It would have been better to use parametric statistics for both parts, rather than discard statistical significance by using nonparametric statistics, then use parametric statistics to show that statistical significance is lacking.

That articles resorting to such approaches have been published recently, indicates that ephemerides are considered, at least by some peer reviewers, unable, alone, to rule out bodies like Barbarossa. Standish's use of space probe data to remove residuals from Uranus and Neptune, was complicated. Only a partial orbit of Neptune is available.

"It is true that if you put in the new mass of Neptune, some of the key residuals in Uranus do decrease. But in my opinion it is not correct to say that they disappear entirely."

- Robert Harrington of the U.S. Naval Observatory (New Scientist, Jan. 30, 1993, p. 18, cited by Ken Croswell)


Perhaps the simplest approach is to consider the Pioneer acceleration. Barbarossa's tidal force, i.e., the difference between Barbarossa's pull on Pioneer 10 or 11, and the sun's, would have been, in the outer solar system, of the same order of magnitude as the anomalous Pioneer deceleration (and indeed seems to explain much of the apparent variation in the anomalous acceleration; for discussion and references see, Joe Keller's March 29, 2007 post, p. 10 of this thread, and April 24, 2007 post, near the bottom of the post, p. 13 of this thread). I recall from an article by JD Anderson, that with the help of space probes, Mars' orbital period has been measured accurately enough to rule out any steady unexplained sunward force on Mars, more than ~ 1% as big as the Pioneer deceleration.

On the other hand, Barbarossa's tidal force, for any planet, though also radial, would mostly cancel over a period (it's half as much at quadrature as at opposition/conjunction, so averages to 1/4 of the maximum); for outer planets, whole orbital periods are practically unavailable to measurement anyway. (For Earth at this epoch: net toward the sun in December and June, net away from the sun in March and September.) It would be < 1/10 as big in the inner solar system, as in the outer solar system where it becomes comparable to the Pioneer force. That, combined with averaging to zero, would make it undetectable even by the measurement of Mars' period, which Anderson put forth as the most sensitive way to detect such radial forces.
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Joe Keller

USA
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Posted - 09 May 2008 :  17:14:37  Show Profile  Reply with Quote
Some Political Aspects of Science

According to his online New York Times obituary, Dr. Harrington died in the hospital, of cancer, at age 50, January 23, 1993 ("Saturday"; according to the article "published Jan. 27, 1993") a week before the cover date of the New Scientist article quoting him, as doubting that Standish's work had refuted Planet X. Though Harrington, as recent head of the USNO equatorial division, and well-known Planet X seeker, was an appropriate source for the New Scientist to interview, someone not on his deathbed might have been able to rebut Standish in more detail. Nonetheless, deathbed statements get extra legal weight. This was practically a deathbed statement by Harrington. Surely it was his sincere opinion, free of political influence.

Journalist Peter Jennings waited until he had terminal cancer, to present his expose on UFOs. Another journalist, Bill Moyers, now in his 70s and formally retired since 2004, tried to make a blockbuster statement on the Charlie Rose show last night, but, for the only time in the forty years I've been watching Public Television, the signal was lost for so long, about a minute, that that entire thread of conversation failed to be broadcast. Moyers had been close to Pres. Johnson and was trying to reveal what Johnson really thought about the Vietnam War.
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Joe Keller

USA
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Posted - 10 May 2008 :  23:59:16  Show Profile  Reply with Quote
Lowell Knew Barbarossa Fit Uranus Data

From an excerpt of Lowell's writing (see: WG Hoyt, "Planets X & Pluto", 1980, pp. 136-140) I can summarize Lowell's method. Adams had predicted Neptune from discrepancies in Uranus' orbit. Adams, LeVerrier, Galle et al had > 1 orbit of Uranus to guide them since Flamsteed's (accurate, for its era) 1715 position, but Lowell had < 1 orbit of Neptune. Therefore the remaining discrepancies in Uranus' orbit, unexplained by Neptune, were (and are, I gather from statements by Standish and Harrington) a better guide to any Planet X, than are the discrepancies in Neptune.

What follows is my simplified reconstruction of Lowell's method. Lowell assumed zero inclination (the greatest inclination of any outer planet then known was Saturn's, 2.5deg) and, at first, zero eccentricity. He guessed the mass, eventually as 0.00002 solar masses. Then he searched for the radius and phase, that best explained Uranus' discrepancies. This gave the necessary radius as a function of mass. One point, on this mass-radius curve, was 6.5 Earth masses (0.000020 solar masses) and 44 AU. Lowell preferred these values, because they are consistent not only with the geometric sequence ratio 1.5 for the planetary orbital radii (which describes not only the step from Uranus to Neptune, but also the inner planets from Mercury to Mars, though Venus deviates) but also consistent with the geometric sequence ratio 2.64 = (massJupiter/massNeptune)^(1/3) for planetary masses (which describes the outer planets from Jupiter to Neptune, though Uranus deviates).

To investigate the effect of eccentricity, Lowell assumed the largest eccentricity of any known planet (0.20, as for Mercury) and studied the two extreme cases: Planet X at perihelion in 1914, and Planet X at aphelion in 1914. Any phase between perihelion and aphelion, or any lesser eccentricity, would amount to an intermediate case, so knowing these two extremes, Lowell would know what range of ecliptic longitude to search. The two cases (unsurprisingly, because of Planet X's assumed large mass and distance) gave perihelia almost 180deg apart, i.e., almost the same ecliptic longitude for Planet X in 1914. This ecliptic longitude must have been either near 202 for both, or near 202-180=22 for both; it must have been 202, because Pluto, discovered near there, was thought for awhile to be the predicted Planet X. Both solutions had periods near 292yr, so the 199yr of data covered ~240deg of Planet X's orbit. Instantaneous data would have implied a 50% difference in major axis (1+0.2 vs. 1-0.2), if one solution were at perihelion and the other at aphelion; 180deg of orbit equally weighted, would average that to zero, and 240deg of orbit to < 10%. So the two fitted major axes also were almost the same, averaging 44AU.

Lowell's eccentricity 0.2 calculation, seems to have been only exploratory, to find the extreme range of ecliptic longitudes to search. So, Lowell's actual solution was, approximately, a 6.7 Earth mass body in a circular orbit of 44AU radius, phase 202deg ecliptic longitude in 1914.

Uranus' orbit is like a frictionless harmonic oscillator. A perturbation drives it 180deg out of phase. The second harmonic perturbation (see my May 11 post below) predominates. Therefore the lag is 1/2 * 1/2 = 1/4 period. For Lowell's 6.7 Earthmass alternative, to have at present the same effect as Lowell's implicit, not explicitly stated, 3430 Earthmass alternative (which I name Barbarossa), they must have the same longitude, when retarded in their own orbits by a quarter of Uranus' orbital period. The 6.7 Earthmass alternative's retarded ecliptic longitude in 1914 was 202-84/4*360/292=176; likewise Barbarossa's retarded longitude in 1987 was 173-84/4*360/2780=170; in 1914, 170-73*360/2780=160.5. So, Lowell in 1914 essentially predicted Barbarossa's heliocentric ecliptic longitude with only 15 deg error, assuming Lowell's very most recent data got almost all the weight when calculating Planet X's longitude. If the effective mean epoch of Lowell's data is 1900, then the *effective* longitudes of Lowell's Planet X and Barbarossa, are equal. (David Todd in 1877 essentially predicted Barbarossa's longitude with only 11deg error; see subsequent posts.)

The radial component of Barbarossa's tidal force was shown above to be smaller than any measurements hitherto would have detected. The maximum tangential component can be estimated by drawing a tangent from Barbarossa to Uranus' orbit. The ratio of this, for Lowell's 6.7 Earthmass alternative, vs. Barbarossa, is 1:1.075. This should be multiplied by a sine factor to correct for the lesser effectiveness of quickly alternating forces; the corrected result is 1:1.19. So, with only 19% error, Lowell's predicted Planet X, based on 1715-1914 discrepancies in Uranus' orbit (unexplained by Neptune), gives a perturbation of the same strength as Barbarossa.
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Joe Keller

USA
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Posted - 11 May 2008 :  18:58:53  Show Profile  Reply with Quote
Planet X: Dynamical Evidence in the Optical Observations

E. Myles Standish's article is entitled, "Planet X: No Dynamical Evidence in the Optical Observations" (Astronomical Journal, 105:2000+, 1993). I'm correcting that title. After Standish carefully removed outliers, adjusted planetary masses to more accurate space probe values, and recomputed orbits, some dynamical evidence for Barbarossa could be seen by inspection of his graphs.

"On 30 Sept 1846, one week after the discovery of Neptune, Le Verrier declared that there may be still another unknown planet out there."

- Paul Schlyter, historian of science (internet)


In 1877 David Todd predicted a planet at ecliptic longitude 170 +/-10 for 1878 (source: Schlyter). Barbarossa's longitude then would have been 159.

By drawing a tangent from the outer to inner orbit, the maximum tangential component of tidal force is found to be approx. (sec^2(theta) - cos(theta))*M/R^2, where theta = arcsin(r/R). The mass of Neptune would need to be increased 7%, to give the magnitude of additional tangential force provided by Barbarossa. So the Voyager/Tyler 0.5% *downward* revision of the 1976 IAU value of Neptune's mass (Standish, Table 1) does little, to explain Lowell's prediction (which was consistent either with Barbarossa or with a smaller, nearer planet). (Lowell explicitly said that he knew only the pull's direction and tidal strength with accuracy; the mass and distance were mutually dependent guesses.) Furthermore, Voyager/Tyler revises Newcomb's 1877 Neptunian mass, likely used by Lowell, only 0.17% downward.

*The* unperturbed orbit, is not the same as *an* unperturbed orbit. A large perturbation can cause a perturbed orbit that looks very like some other unperturbed orbit. This does not imply that there is no perturbation. Likewise, a "poorly conditioned", i.e. very acutely angled, 2x2 system of linear equations, isn't solved accurately by projecting a normal from one line to the other; it's solved by moving along one line to the other. In terms of higher mathematics, one must follow the curve of perturbation in Hilbert space, to find the unperturbed orbit; it is not sufficient simply to move to the nearest unperturbed orbit and say, "Look how close this is."

I express the tidal tangential force, as a Fourier series in the longitude of Uranus (assuming a circular orbit), finding the coefficients numerically by convolution. The 1st harmonic is small and its effect on longitude, can be and is, neutralized by assuming a slight alteration in the magnitude and apse of that relatively gigantic orbital parameter, the eccentricity. Relative to the gravitational force of the sun, the 2nd harmonic coefficient is 0.1461 * 0.0103 * (19.18/197.7)^2 = 14.16 * 10^(-6).

To find displacement in RA, the 2nd harmonic amplitude must be divided by 2^2, to account for two integrations; and it's opposite in sign to the input. The effect on orbital radius can be neglected for a second harmonic perturbation, because the radius is a first harmonic response to that input, and cancels. So the 2nd harmonic amplitude of Uranus' longitude displacement is

-0.25 * 14.16*10^(-6) radians = -0.7302"

where the sign signifies that the displacement is opposite in sign to the input tidal tangential perturbation there.

(The 3rd harmonic amplitude of longitude displacement is, analogously, of the order of 0.044".)

The wave seen on, e.g., the rightward portion of Standish's Fig. 5a, is roughly 0.5" peak-to-peak, which might correspond to 0.125" harmonic (semi)amplitude. Uranus' eccentricity strongly affects both the 1st & 2nd harmonic of longitude displacement directly, and also the 2nd harmonic of tidal tangential force (and thereby also the 2nd harmonic of longitude displacement). Optimum choice of eccentricity and apse might partly nullify both 1st and 2nd harmonics. (Likewise, something with both dipole and quadrupole, whose strengths can be multiplied by the same factor, and which can be rotated by the same angle, has two degrees of freedom with which it might partly nullify a dipole and quadrupole simultaneously.) So, the observed amplitude of longitude displacement vis a vis the best fitted orbit, should be < 0.73".

In 1975.0, Uranus' longitude was 211, Barbarossa's 172, & Barbarossa's retarded longitude 172 - 84/4*360/2780 = 169. At this relative phase, 211-169=42deg (~45deg), the second harmonic longitude displacement should peak. Standish's Figs. 5a & 6a show his most highly corrected data for RA. Because of scatter, the harmonic (semi)amplitude might be estimated as 1/4 the raw peak-to-peak amplitude, i.e. 0.125" from c. 1930 to 1990. The peak is c. 1960 (60deg away from the pred. 1975 peak; this amounts to +0.5 correlation with predicted phase) but the seeming period, c. 1940-1980, is close to the 84/2=42yr predicted. The seeming amplitude (considering that due to cancellation through orbital fitting, my 0.73" is only an upper bound) and period do suggest that the effect of Barbarossa is apparent on Standish's graphs.

Standish (sec. 2, p. 2001) estimates 1.2" error for individual time point observations before 1911. Lowell would have needed ~ (1.2/0.73)^2 / mean(sin^2(theta)) * 2^2 = 22 randomly distributed time point observations total, to detect the second-harmonic discrepany in Uranus longitude, due to Barbarossa, to 2-sigma significance. (With the help of the internet browser's magnification feature, I can see that there are > 1000 dots before 1915 on Standish's Fig. 1a.) Lowell would have known that he sought a second harmonic phenomenon. He might have adjusted Uranus' eccentricity and apse by neutralizing the 1st harmonic precisely, without regard to the 2nd harmonic discrepancy in longitude, which then would not be masked.
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Joe Keller

USA
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Posted - 13 May 2008 :  20:45:55  Show Profile  Reply with Quote
The residuals of Uranus' RA since 1950, seem less grossly sinusoidal in Standish's (1993) Figs. 4a and 6a, than in Fig. 5a. I measured the last 39 data points from the 1600% magnified computer screen (corresponding to the interval 1950-1989), for Figs. 4a, 5a & 6a, and made periodograms. (There was a place, where I found one less point in Figs. 4a & 5a, so interpolated what seemed, by comparison with the abscissas of Fig. 6a, likeliest to be the missing datum.)

Before making the periodograms, I detrended each data set, i.e., adjusted slope and mean to zero, using the average of the first three and average of the last three ordinates, to define the trendline. For Fig. 6a (resp. 4a, 5a), the periodogram peaks at 62 (resp. 68, 60) yr. Fig. 6a's periodogram amplitudes are smaller than Fig. 5a, whose in turn are smaller than Fig. 4a. Well away from the peaks, Figs. 5a & 6a conform well to white (1/f) noise. I used the periodogram amplitude of Fig. 6a at 100 yr, which seemed to be well away from the 62 yr peak and consistent with the amplitude at 10yr, to define the 1/f amplitude curve which I subtracted from Figs. 4a, 5a, & 6a.

Thereby filtering out the white noise, I found the filtered periodogram peaks for Fig. 6a (resp. 4a, 5a) to be 53.9 (resp. 61.5, 55.2) yr. For Figs. 5a & 6a, the signal was almost totally removed by this filtration, leaving only smooth, fairly narrow peaks of heights 0.0983" & 0.0265", resp.

The perturbation signal is a second harmonic; 53.9 yr, the detrended, de-noised periodogram peak for Standish's most highly corrected post-1950 data, corresponds to the net half-period of Uranus vis a vis a Planet X in circular orbit at 52.5 AU. Elsewhere on this messageboard I have discoursed at length on the theoretical and empirical importance of the distance, 52.6 AU.

The amplitude of the RA (essentially, ecliptic longitude) displacement found in Standish's Fig. 6a (resp. 5a), 1950-1989, is only 0.0265"(resp. 0.0983")/0.73" = 4% (resp. 13%) that theorized for Barbarossa (or Lowell's Planet X), but correcting for distance^3 = (52.5(resp. 49.7)/197.7)^3, still should have 2.4 (resp. 7.45) Earth masses.

For all the periodograms, the phase is incoherent. In Fig. 5a, for example, near the de-noised periodogram peak at 55yr, the phase steadily varies ~ 135deg, per year of change in the period. If the discrepancy were due to the tidal gravity of a Planet X, a small change in the period of the fitting sinusoid, would cause only a small change in the best-fitting phase. The periodogram result is consistent with a collection of objects or structures orbiting at ~ 52AU, each with its own phase, affecting Uranus likely nongravitationally. The squares of the interaction amplitudes are additive.

The white noise correction tailored to remove the 100yr & 10yr periodogram amplitudes of Fig. 6a, also removes the 200yr & ~ 10yr periodogram amplitudes of Fig. 5a (the 100yr amplitude of Fig. 5a seems not quite distinct from the 60yr peak); thus the amount of white noise in Figs. 5a & 6a is about the same. The difference between Figs. 5a & 6a, is Earth-based vs. space probe planetary mass determinations. With use of Earth-based determinations, the amplitude of the effect on Uranus' orbit, equals the amplitude (though not the phase, or any phase) of the effect that would be caused by another half of a Uranus (7.45*2 = 14.9 ~ Uranus' 14.6 Earthmasses) orbiting at the special distance, 52.6AU. Use of space probe determinations somehow eliminates most (perhaps, ideally, all) of the effect.
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nemesis

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Posted - 14 May 2008 :  09:50:42  Show Profile  Reply with Quote
Joe, if I'm following you correctly you're saying a planet of ~3.7 Earth masses at 197.7 AU would fit the Uranus perturbation data. A planet of that mass at that distance would be cold and very dim and would require no masking nebula. It could be the "green dot". Wouldn't it be more parsimonious to make that assumption?
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