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Requiem for Relativity
- Joe Keller
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17 years 8 months ago #16603
by Joe Keller
Replied by Joe Keller on topic Reply from
The monotonic trend in Red magnitude of the four Freya sightings, has a correlation coefficient of 0.9275. For n=4, tables give p=0.01, two-tailed, for a correlation coefficient of 0.917; by extrapolation the Freya sightings have p=0.008. Freya has uniform albedo.
Barbarossa's semimajor axis, a = 270 A.U., is given precisely by the discrepancies in giant planet resonances (which give Barbarossa's period). To a good approximation for small eccentricity, the angular speed, when r=a, is independent of eccentricity. The distance estimate of 330 A.U., depends, on the accuracy of this approximation, and on the accuracy of the determination of present angular speed. The latter depends on the accuracy of the COBE & WMAP CMB dipole directions, and on the accuracy of my theoretical correction accounting for the gravity of the four known giant planets. Be all this as it may, 330 A.U. implies plausible albedo and mass for Barbarossa.
Computing Freya's distance from Earth, from Freya's magnitude, the best fit line gives an angle arctan(0.36) between Barbarossa's path and the constant circle. Graphical solution of the orbital equation gives eccentricity 0.36, and Barbarossa 60 degrees past aphelion. Aphelion is 420 A.U. and perihelion 200 A.U.
Takeda & Rasio (ArXiv.org, Oct. 10, 2005) state that the median orbital eccentricity for known extrasolar (giant) planets is 0.31, and the third quartile 0.43, but such planets would be much closer to their stars than is Barbarossa. A more appropriate comparison might be to 61CygniB whose eccentricity is 0.40 and period 650 yr.; the pair 61CygniAB has 4/3 the mass, of the pair sun+Barbarossa, but it is almost equally divided.
If my gravitational theory of the CMB is accurate, the 3-yr WMAP dipole temperature should be 33.6 microK warmer than the 4-yr COBE dipole temperature (25.0 microK from Barbarossa's approach and 8.6 microK from the overall improved alignment of J, S, U, and mainly N). It is 5 microK warmer, but the error bar for the difference is 29 microK.
Barbarossa's semimajor axis, a = 270 A.U., is given precisely by the discrepancies in giant planet resonances (which give Barbarossa's period). To a good approximation for small eccentricity, the angular speed, when r=a, is independent of eccentricity. The distance estimate of 330 A.U., depends, on the accuracy of this approximation, and on the accuracy of the determination of present angular speed. The latter depends on the accuracy of the COBE & WMAP CMB dipole directions, and on the accuracy of my theoretical correction accounting for the gravity of the four known giant planets. Be all this as it may, 330 A.U. implies plausible albedo and mass for Barbarossa.
Computing Freya's distance from Earth, from Freya's magnitude, the best fit line gives an angle arctan(0.36) between Barbarossa's path and the constant circle. Graphical solution of the orbital equation gives eccentricity 0.36, and Barbarossa 60 degrees past aphelion. Aphelion is 420 A.U. and perihelion 200 A.U.
Takeda & Rasio (ArXiv.org, Oct. 10, 2005) state that the median orbital eccentricity for known extrasolar (giant) planets is 0.31, and the third quartile 0.43, but such planets would be much closer to their stars than is Barbarossa. A more appropriate comparison might be to 61CygniB whose eccentricity is 0.40 and period 650 yr.; the pair 61CygniAB has 4/3 the mass, of the pair sun+Barbarossa, but it is almost equally divided.
If my gravitational theory of the CMB is accurate, the 3-yr WMAP dipole temperature should be 33.6 microK warmer than the 4-yr COBE dipole temperature (25.0 microK from Barbarossa's approach and 8.6 microK from the overall improved alignment of J, S, U, and mainly N). It is 5 microK warmer, but the error bar for the difference is 29 microK.
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17 years 8 months ago #16412
by nemesis
Replied by nemesis on topic Reply from
Joe, I hate to ask what may seem a naive question, but when you say Barbarossa is "aliased by moons" exactly what does that mean?
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17 years 8 months ago #16413
by Joe Keller
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<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by nemesis</i>
<br />Joe, I hate to ask what may seem a naive question, but when you say Barbarossa is "aliased by moons" exactly what does that mean?
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Last night I spent two hours telling a retired medical school physiology professor about Barbarossa, Frey, Freya, etc. He said, essentially, "So, Freya or Frey 'stand in' for Barbarossa in almost the same position on the next year's plate."
By "alias", I mean "stand in". I took the term "alias" from signal theory, where it means that a sinusoid of one frequency impersonates a sinusoid of another frequency.
<br />Joe, I hate to ask what may seem a naive question, but when you say Barbarossa is "aliased by moons" exactly what does that mean?
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Last night I spent two hours telling a retired medical school physiology professor about Barbarossa, Frey, Freya, etc. He said, essentially, "So, Freya or Frey 'stand in' for Barbarossa in almost the same position on the next year's plate."
By "alias", I mean "stand in". I took the term "alias" from signal theory, where it means that a sinusoid of one frequency impersonates a sinusoid of another frequency.
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17 years 8 months ago #18847
by Joe Keller
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Error analysis:
The estimated distance to Barbarossa, and the estimated Neptune-like 30% albedo of Freya, imply that Freya's diameter is 12,300 miles. Because the upper error bar allows it, I adopt 17,500 miles because Freya's constant albedo implies that Freya is a Neptune-like gas giant with internal heat, not a Pluto or Earth with nonuniform albedo.
1. If the five USNO-B objects span as little as 4/5 of Barbarossa's track on the (as little as?) "50 years of Schmidt plates", then the angular speed is as great as 69/(4/5*50) times what I used. For a given ellipse, "equal areas in equal times" implies that the angular speed is inversely proportional to the square of the radius. This gives a lower bound of 0.58 times calculated, for Freya's diameter.
2. The error bars on the COBE dipole determination imply that Barbarossa's travel between COBE & WMAP, might be 0.33 deg of galactic longitude more or less, than my calculated 0.866 (which includes a correction for planetary effects). This gives a range of 0.72 to 1.60 times calculated, for Freya's diameter.
3. Besides angular speed, the distance to Freya also can be found from Barbarossa's apparent magnitude, assuming Barbarossa's albedo is 7%. Using the five bright Red determinations, the standard error of the mean for Barbarossa's magnitude is 0.225. This gives a range of 0.91 to 1.11 times calculated, for Freya's diameter.
4. By analogy with other solar system bodies, Barbarossa's albedo might be as low as 5%. This gives a lower bound of 0.85 times calculated, for Freya's diameter.
(3) & (4) combine to give a greatest lower bound of 0.77 times calculated. The possibility of Barbarossa albedo > 7%, vitiates the upper bound in (3), so the least upper bound is 1.60, from (2). For simplicity, to make Freya more Neptune-like, I double Freya's area, multiplying its diameter by 1.41, to obtain 17,500 miles.
The estimated distance to Barbarossa, and the estimated Neptune-like 30% albedo of Freya, imply that Freya's diameter is 12,300 miles. Because the upper error bar allows it, I adopt 17,500 miles because Freya's constant albedo implies that Freya is a Neptune-like gas giant with internal heat, not a Pluto or Earth with nonuniform albedo.
1. If the five USNO-B objects span as little as 4/5 of Barbarossa's track on the (as little as?) "50 years of Schmidt plates", then the angular speed is as great as 69/(4/5*50) times what I used. For a given ellipse, "equal areas in equal times" implies that the angular speed is inversely proportional to the square of the radius. This gives a lower bound of 0.58 times calculated, for Freya's diameter.
2. The error bars on the COBE dipole determination imply that Barbarossa's travel between COBE & WMAP, might be 0.33 deg of galactic longitude more or less, than my calculated 0.866 (which includes a correction for planetary effects). This gives a range of 0.72 to 1.60 times calculated, for Freya's diameter.
3. Besides angular speed, the distance to Freya also can be found from Barbarossa's apparent magnitude, assuming Barbarossa's albedo is 7%. Using the five bright Red determinations, the standard error of the mean for Barbarossa's magnitude is 0.225. This gives a range of 0.91 to 1.11 times calculated, for Freya's diameter.
4. By analogy with other solar system bodies, Barbarossa's albedo might be as low as 5%. This gives a lower bound of 0.85 times calculated, for Freya's diameter.
(3) & (4) combine to give a greatest lower bound of 0.77 times calculated. The possibility of Barbarossa albedo > 7%, vitiates the upper bound in (3), so the least upper bound is 1.60, from (2). For simplicity, to make Freya more Neptune-like, I double Freya's area, multiplying its diameter by 1.41, to obtain 17,500 miles.
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17 years 8 months ago #18848
by Joe Keller
Replied by Joe Keller on topic Reply from
Search track for 2007:
Assuming that Barbarossa's period is between 4400 and 6800 yr, the search track is a great-circle segment (approximately a line segment for this small distance) with expected endpoints:
RA 11h 6m 02s Decl -6deg 28' 27" and
RA 11h 7m 24s Decl -6deg 38' 50"
Barbarossa should move 0.5"/day retrograde. If the period is 4400 yr, there's a 50% chance Barbarossa will be on the track west of the west endpoint; if the period is 6800 yr, a 50% chance of being east of the east endpoint. The expected number of stars in the USNO-B catalog, within 1' of this segment, having either Red1 or Red2 magnitude < 18.50, is 28.
Assuming that Barbarossa's period is between 4400 and 6800 yr, the search track is a great-circle segment (approximately a line segment for this small distance) with expected endpoints:
RA 11h 6m 02s Decl -6deg 28' 27" and
RA 11h 7m 24s Decl -6deg 38' 50"
Barbarossa should move 0.5"/day retrograde. If the period is 4400 yr, there's a 50% chance Barbarossa will be on the track west of the west endpoint; if the period is 6800 yr, a 50% chance of being east of the east endpoint. The expected number of stars in the USNO-B catalog, within 1' of this segment, having either Red1 or Red2 magnitude < 18.50, is 28.
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17 years 8 months ago #16415
by Joe Keller
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The ratios of convective energy, to total lightning energy, to optical lightning energy, are about the same for Earth and Jupiter. The convective energy of Barbarossa would equal its 378K (est.) Planck emission. Employing the ratios in Borucki's articles below (I used geometric mean averages with twice the weight for Jupiter as for Earth) I found that the lightning emission from Barbarossa is equivalent to about 3% albedo.
Zarka remarks that the lightning activity on Saturn varied 5-fold between Voyager 1 & 2. Such variation on Barbarossa might explain some of its variable magnitude. The broad peaks of the lightning-related radio emissions for J, S & U, all are near 4 MHz.
References.
Borucki et al, Icarus 52:492+, 1982.
Borucki et al, Reviews of Geophysics and Space Physics 22:363+, 1984.
Zarka et al, Planetary and Space Science, 52:1435+, 2004, p. 1442.
Zarka remarks that the lightning activity on Saturn varied 5-fold between Voyager 1 & 2. Such variation on Barbarossa might explain some of its variable magnitude. The broad peaks of the lightning-related radio emissions for J, S & U, all are near 4 MHz.
References.
Borucki et al, Icarus 52:492+, 1982.
Borucki et al, Reviews of Geophysics and Space Physics 22:363+, 1984.
Zarka et al, Planetary and Space Science, 52:1435+, 2004, p. 1442.
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