Requiem for Relativity

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14 years 8 months ago #23772 by Jim
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Dr Joe, If you alter the mass of Barba you have to relocate the baricenter of the solar center. How does that change all your other reference points?

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14 years 8 months ago #23773 by Joe Keller
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(The post below was originally posted by me at about 8PM CST, Feb. 20, 2010. Within 18hr, this messageboard had become disabled by a hacker, and remained totally disabled for two days. I had sent backup copies of the post to multiple witnesses. Sometime between yesterday and today, Feb. 25, this and all subsequent posts by me, both of which related to this one, were deleted from the messageboard. I don't believe in or disbelieve in manned flying saucers; I simply don't know, and withhold judgment. This post is merely factual: there is a book, the very serious-minded 1987 foreword was by a recently retired British Admiral of the Fleet, who ought to be a responsible person, and the biography of the witness makes him credible. It might be, that the flying saucer story is fictitious, designed to protect the witness and his accomplices, possibly including a major astronomer in California, from prosecution for leaking a government secret.)


Sid Padrick's astronomical information

Professional violinist Timothy Good's "Beyond Top Secret" (1996), like Good's 1987 predecessor book "Above Top Secret", has a Foreword by a retired British Admiral of the Fleet, Peter Hill-Norton. On pp. 365-371, Good gives the history of Sid Patrick's alleged benign encounter with humanlike saucernauts. Padrick, age 45, was a TV repairman (married with a wife and three sons at home, according to Brent Raynes' similar version of the history, on ufoevidence.org), and a U.S. Air Force reservist.

Padrick had insomnia and was out walking near his home near San Francisco, California, at 2AM on Jan. 30, 1965. (UFO researchers have noted, that UFO events peak in likelihood at approx. 2 or 3 AM). A reassuring voice from a flying saucer invited him aboard. He was given a friendly two hour tour including a short ride in his neighborhood. He suffered no apparent ill effects.

The saucernaut who spoke with Padrick, said, in Padrick's words:
"He told me they were from a planet in back of a planet which we observe - but we do not observe them. He did not say we couldn't observe them - he merely said we didn't observe them...I think their planet is in our solar system."

- Padrick, quoted in Good, pp. 368-369

We "observe" Uranus and Mars. On Jan. 30, 1965 Mars' RA was 11h59m (I suppose the Astronomical Almanac gives this in 1950.0 coords) and Uranus' RA was 11h03m. Mars was stationary at maximum RA, at noon (Pacific Standard Time), Jan. 29, 1965. So two planets "which we observe" would have been nearly in front of Barbarossa (then at approx. ecliptic longitude 170 and approx. RA 166) for several months (Mars) or several years (Uranus).

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14 years 8 months ago #23831 by Joe Keller
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(This post was originally posted by me at about 9:40AM CST Feb. 23, 2010.)

<i>Originally posted by pelorus</i>
<br />Wassup

Note re computer sabotage:

Sometime after my post above at 8PM CST Feb. 20 and before 2PM CST Feb. 21, 2010, this entire messageboard became inoperational for the first time I recall in over five years. Sometime between last night and this morning (9AM CST Feb. 23) it became operational again, but "pelorus" had started &gt;300 threads in the "Gravity and Relativity" forum, and threads in several other forums. The threads in "Gravity and Relativity" were started as little as 51sec apart. Often, posts from "pelorus" were made 2sec apart.

Almost always, commercial spam limits itself to a moderate number of posts because occasional spam posts scattered among authentic posts are more tolerated by webmasters and legal authorities, and are more effective advertising, than spam that completely chokes and disables the messageboard. This is automated spam, but only thinly disguised as commercial. The purpose of this spam is to disable this messageboard. I've seen this intensity of spam only during fights between political groups where one sought to disable the messageboard of the other.

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14 years 8 months ago #23868 by Joe Keller
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(This post was originally posted by me at about 10AM CST Feb. 23, 2010. I added the last paragraph yesterday, Feb. 24. I did not delete this post, nor the previous two of mine that also were deleted sometime between yesterday and today. If these three posts are not the reason for the hacker attack, then there's no problem. If they are the reason for the hacker attack, then is it wise to avoid theft, by destroying the only property one has, valuable enough to attract the attention of criminals?)


Addendum to Sid Padrick information

(Above, I misspelled "Padrick" once as "Patrick".)

(coordinate conversions below, from NASA "Lambda" online utility)

From my orbit, I find that for 10h GMT (2h PST) Jan. 30, 1965, the heliocentric (by this I mean, relative to the barycenter of the solar system excluding Barbarossa; due to Jupiter alone, this differs by up to ~5" = 0.0014deg from true heliocentric) J2000.0 ecliptic coordinates of Barbarossa were (170.602, -11.219). Barbarossa's heliocentric J2000.0 celestial coords were (166.948, -6.592). Barbarossa's heliocentric, equinox 1965.0, celestial coords were (166.505, -6.402). Barbarossa's *geocentric* equinox 1965.0 celestial coords were (166.7, -6.5).

For 0h GMT Jan. 30, 1965 (according to the JPL "Horizons" ephemeris online) the geocentric, equinox 1965.0, celestial coords of Uranus were (165.681, +6.998). So, at the time of Padrick's alleged information, the RA of Uranus differed only 1.0 degree from that of Barbarossa, and the two objects were, within 0.5 degree, symmetrically placed vis a vis the celestial equator.

Furthermore, Uranus was stationary in J2000.0 RA, only 41 days earlier, on Dec. 20, 1964. Then, Uranus' geocentric J2000.0 celestial coordinates were (166.809, +6.501) and Uranus' geocentric equinox 1965.0 coords were (166.356, +6.690). That is, Uranus recently had been stationary at RA only 0.35deg different from Barbarossa's, and at stationarity had been symmetrically placed vis a vis the celestial equator, within 0.2deg.

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14 years 8 months ago #24036 by Joe Keller
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(Feb. 25: A post above which I made Feb. 18 and updated Feb. 19, and again updated - adding the last two paragraphs - about 2:40PM CST Feb. 24, was altered somehow - not by me - reverting it to its Feb. 19 state. Here is the post as I edited it yesterday, and as it should be.)


Barbarossa system mass = 0.010026 +/- 0.000290 solar

The orbit I fit to the sky survey positions, assuming exactly 0.01 solar mass, implies orbital period 6339.93 Julian yr +/- about 0.1% (see recent discussion of error, above). During the last year, I've considered not only the Egyptian/Mayan calendar, but also various solar system resonances. These considerations imply that Barbarossa's period is closer to 6339.362 Julian yr (figure determined by consideration of resonance with hypothetical proto-Jupiter).

At the epoch of observation, r = a/1.6 for Barbarossa. According to the reduced mass formula in mechanics, a 1% decrease in Barbarossa's mass, has the same effect on its heliocentric orbit, as a 1% increase in "k", which would increase the magnitude of the potential energy at the epoch of observation, by 1%, and the difference in magnitude, potential - kinetic = 1.6 - 1.1 = 0.5, by 1.6/0.5 = 3%. So "a" decreases 3-1 = 2%, and P, which varies according to a^1.5/k^0.5, decreases by 3+0.5 = 3.5%.

So, if Barbarossa's true period is 6339.362yr, and my orbit is perfectly accurate, then the Barbarossa system's mass must be 0.01 + 1*(6339.93/6339.362)^(1/3.5) = 0.010026. My round figure guess for Barbarossa's mass, has turned out to be in error by only one part in 400. The uncertainty in orbital details, which is equivalent to 0.1% uncertainty in period, also must be equivalent, to 0.1/3.5 = .029% solar mass, uncertainty in mass (one part in 35).

Recently (see above) I rechecked my calculation that the torque of all the other planets on Neptune (per degree of inclination), is 3x the torque of Barbarossa on Neptune, assuming that Barbarossa has either a circular orbit at 198 AU with 0.0107 solar mass, or else eccentricity 0.6106 & a=343.8 with 0.0280 solar mass. (Substituting plutinos, or classical KBOs, for Neptune, gives 2x, or 1x, resp.) Without knowing anything about Barbarossa, I can say that for a distant object of any mass, in any orbit, if its torque (per degree of inclination) on the classical Kuiper Belt is to the torque of the planets on the classical Kuiper Belt, in the ratio X::1, then its torques (per degree inclination) on the plutinos and on Neptune, will be to the torques of the planets on them, in the ratios X::2 and X::3, resp.

If X=1, then Barbarossa's mass, estimated from this "precession resonance", is almost three times too great. On the other hand, if 1/X = "e" = 1 + 1 + 1/2 + 1/6 + 1/24 +... = 2.718..., then Barbarossa's mass is 0.01030.

If the classical Edgeworth-Kuiper belt is approximated by circular orbit at 43 AU, then this same recheck (by three-point gaussian quadrature) finds that for a 1::1 ratio of the torque, on the classical Kuiper belt, from Barbarossa, to the torque from the known planets, Barbarossa's mass must be 0.02896 solar. As r, the radius of the Kuiper belt, increases, the mass that Barbarossa must have, decreases as r^5, because the torque from Barbarossa increases as r^2, and the torque from the planets decreases as 1/r^3.

In much earlier posts and in a 2002 article I authored, I discussed the "barrier" distance 52.6 AU. In review, this is the distance at which the maximum gravitational field of a quantum proton, equals the gravitational field of the Sun. Also at this distance, abnormal signals were received from the Pioneer 10 space probe, and, approximately, the Kuiper belt mysteriously ends. Today (Feb. 24, 2010) I find that for a circular orbit at 52.6 AU, the required mass of Barbarossa, for a 1::1 ratio of the torque from Barbarossa to the torque from the planets, is 0.02896 * (52.6/43)^5 = 0.01057 solar. This differs only 5% from the mass implied by my fitted orbit, and corresponds to only a 1% error in the "barrier" distance.

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14 years 8 months ago #23774 by Jim
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Dr Joe, How do you determine the gravity of a proton? It seems to me all you have to work with is Newton and I have have tried without any success to figure the gravity potental of a proton other than multiplying the proton mass by the gravity constant(GM). But, then what about the radius? How do you match that with the solar field at 52.6AU? I think the solar field becomes equal to the gravity of the universe at some distance near 50 or 100AU.

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