Hi Joe, I'm not sure about your contention that that these people had a poor communication network is right. Metal working peoples were inheritors of the great stone circles and presumably some of the knowledge of the builders. They used the circles ut the nature of their ceremonies had changed. For one thin, the associated wood henges would have been long gone by their time. The stone henge, the wood henge and the ceremonial path are associated with ancestor worship. We also have the point, tha the circles were created at a time when early European man was making the painful transition from hunter gatherer to agriculture. The stone circles show the people in a very graphic manner, how the Moon is still mor important than the Sun.

Now even before these sites were set up, we have trade in ceremonial stone axe heads that was extensive. This cultural network carried on into the trading of metal axe heads. That suggests to me that the knowledge of the eighteen year eclipse cycle travelled back and forth between tribes. the hats are local, because the Sun, Moon and ancestors are bound up in the "observations." There's also the point that hill profiles were used for alignments of rising stars and planets.

Incidentally, these hats are the classic witches' hat of children's stories.

One other point, that I find massively intriguing, the stone circles are not circles. Get a rope and two pegs, mark out a circle. Then take your rope and pegs and put one on the circumference at about the south point. Use the other peg to mark off the radius round the circle. So you have six points marked off. Go round two segments on either side of the south starting point. Draw lines from those two points to the centre of the circle.

So it looks like we have a cake with a big slice cut out of it. Now take your rope and half it Walk half way down one of those two lines you've drawn and plant your peg, then from the circumference point draw an arc. Do the same for the other line of your cake slice. Then with a longer rope, go to the southern start point and draw an arc that intersects the two arcs drawn.

So you've now got a circle with a sort of crescent shape cut out of the top. Your stones are placed round this shape, to align with the horizon hills. Why? Nobody knows, Stand in the middle of one of these things and you can't detect that it's not a perfect circle. If it represents the crescent moon, why is is turned through about ninety degrees?

Hello again Joe Keller: With regard to my last request for a calender date on the Pioneer 10. Is the full data set for the doppler velocity measurements available anywhere on the web? Thank You.

...calender date on the Pioneer 10. Is the full data set for the doppler velocity measurements available anywhere on the web? ...

I'm not at home, but I recall that the chart I used, of the Pioneer 10 Doppler data, is from an article in Physical Review by JD Anderson et al, in the 1990s; the title of the article contains the word "anomalous". The arXiv.org preliminary version of the article is on the web. The article is referenced in my own article, Joseph C. Keller et al, Aircraft Engineering and Aerospace Technology, 2002. This would be enough for a librarian with access to "Web of Science" (online Science Citation Index) to find it. Maybe you can find it with a search engine, too.

...these hats are the classic witches' hat of children's stories. ...

Good point! Pliny the Elder reported the ruins of great buildings on the Canary Is. Plato's "Critias" reported that Atlantis was beyond the Pillars of Hercules. The two eclipse tracks, according to my rough estimate, do intersect west of Gibraltar.

The earlier eclipse, 2806BC, has an even Saros number, hence is at the descending node; the later eclipse, 2407BC, has an odd Saros number, hence is at the ascending node. My rough estimate is:

The earlier eclipse, given by NASA as a Julian date, occurs near the spring equinox, assuming that the equinox was March 21 at 325AD. So, if the maximum eclipse was at noon, then Luna's track, with a descending node, is 24-5=19deg north of east, not accounting for Earth's rotation, but Earth's rotation is about half Luna's speed, so the track is double that, i.e. about 38 north of east. By the same reasoning, the track for the later eclipse, which happens a month and a half after the equinox, is 2/3*24+5=21 which doubles to 42, to account for Earth's rotation.

These lines then are almost parallel, so my solution is "poorly conditioned" and especially inaccurate. Anyway, for what it's worth, these lines would intersect near Brazil.

This year's Stonehenge crop circles decoded (Joseph C. Keller, May 18, 2010)

So far this season, there has been one crop circle at Stonehenge (May 9) and one at Old Sarum (May 5). See: "Crop Circle Archive" ( www.x-cosmos.it ). Both these encode the time interval between now, and Dec., 2012.

According the the "NASA Eclipse Website" ( www.eclipse.gsfc.nasa.gov ), the next solar eclipse, total or partial, is the total eclipse July 11, 2010. There follow five partial solar eclipses (the last of the five is annular) and then the next total eclipse, Nov. 13, 2012.

Let's consider the Old Sarum crop circle. The three full concentric disks whose diameters are in the ratio 4::2::1, signify Earth, Luna's penumbra, and Luna. These full disks signify the imminent total eclipse of July this year. The otherwise identical, but half-hidden, disks signify the next total eclipse to come, Nov. 2012. The six lines through (i.e. behind or after) the full disks, signify the next six eclipses, the sixth of which adjoins the half-hidden disks, signifying that the sixth eclipse to come afterwards, is that signified by the half-hidden disks. The seven simple dark disks in a row, signify the seven eclipses between now and Dec. 2012.

Now let's consider the Stonehenge crop circle. The big central disk is the Sun; the two groups of three simple disks represent Mercury, Venus, & Earth. The big circle around the Sun (through the groups of three disks) is Earth's orbit. As at Old Sarum, the groups of three concentric disks, represent the next two total eclipses. The seven arcs emerging from each "eclipse", represent the seven eclipses (total and partial) between now and Dec. 2012.

July 19, 2009 Liddington Castle, Badbury, crop circle encodes 2012 Venus transit

This crop circle signifies two transits of Venus with a transit of Mercury in between. There were/will be transits of Venus June 2004 & June 2012, and a transit of Mercury Nov. 2006. As at Stonehenge & Old Sarum this month, the groups of three solid disks, represent Mercury, Venus, & Earth. The lines from Venus & Earth, to the concentric disks at each end of the configuration, imply that the concentric disks, are transits of Venus as seen from Earth.

August 18, 2008, Hampshire, and Nov. 15, 2007, Wiltshire, crop circles encode Nov. 13, 2012 total eclipse

Let's first consider the Aug. 18, 2008, crop circle. The distal half of the circumference of the righthand circle, is hollowed (in a stylized way); this represents the half of Luna's orbit farthest from the Sun, and implies that Luna's line of nodes is perpendicular to Earth's orbital radius, so that the hollowed part of Luna's orbit is the part which is below (or above) the ecliptic. The same stylized hollowing, occurs in both connections between the right- and lefthand circles. Each of these connections, together with its two dots, represents a pair of successive (total or partial) solar eclipses, occurring about half a year apart at opposite ends of Luna's line of nodes. This often happens, and happens consecutively; there is little unique about the two pairs Nov. 25, 2011 & May 20, 2012; and May 20, 2012 & Nov. 13, 2012.

The uniqueness is given by the one remaining feature, as shown on Andreas Mueller's diagram ( www.cropcirclescience.org ). This is the slightly off-midline dot at left. With a protractor, I measure this dot as 9.3deg above the midline, but correcting for the flattening, on my screen, judging by the large nearby circle (which is 49x56mm on my screen), I find 10.6deg. The first eclipse of the above-mentioned trio, Nov. 25, 2011, is unusual: it occurs (maximum eclipse) with Luna relatively far from the ecliptic, at ecliptic latitude -1.0575 (ascending node) according to the JPL ephemeris. Using the average inclination, 5.14deg, of Luna's orbit, this implies Luna is 11.8deg clockwise of the line of nodes (at maximum eclipse). This is the same as 11.8*27.32/29.53 = 10.9 "degrees" of the synodic month cycle, where I've used the ratio of the sidereal to the synodic month.

The above-mentioned 2007 crop circle is a simpler, more cryptic version of that in 2008. Here, the smaller disk, on the left, represents Earth; the inner circle around it, Luna's orbit; the inner connecting line, the line of nodes at eclipse season; the two small disks, a pair of eclipses half a year apart. The date of the 2007 crop circle, is exactly 365 days * 5, prior to the Nov. 13, 2012 eclipse.

Two more crop circles, and a Venus transit at Saturn

Months ago, I posted here, my analysis of the famous July 15, 2008 crop circle near Avebury, England, that showed the positions of the planets on or near Dec. 21, 2012. For review, my best effort found that using the longitudes I measured on the most often seen internet photo (with correction for oblique view, but making no assumptions about the orientation of the crop circle), the minimum total variance of these measured longitudes (more precisely, of their differences from Earth's), vs. ephemeris ecliptic longitudes, occurs at 02h UT, Dec. 22, 2012.

There is another crop circle, which shows a large disk representing the Sun, and three smaller disks in line with it; the farthest disk has a ring probably signifying Luna. However, there is no alignment like this (which, if perfect, would be a simultaneous transit of Mercury & Venus) of the three inner planets, within decades of 2012. Yet there is something which, in "projective geometry" or "complex analysis", is analogous to a straight line: namely, a circle.

Because of the positions of Mercury's & Venus' ascending nodes, Mercury's conjunction with Venus in Dec. 2012, brings Mercury especially near the circle in space defined by Earth, the Sun, and Venus. Mercury is closest (in the sense of a local minimum of the distance) to this (ever-changing) circle, at about 03h UT, Dec. 24, 2012.

Also on Dec. 21, 2012, is a transit of Venus visible from the center of Saturn (transits of Venus are about three times commoner at a point on Saturn than at a point on Earth, but it's still remarkable that Venus transits should occur in the same year for both Saturn and Earth). According to the JPL (Jet Propulsion Laboratory, or Jack Parsons Laboratory?) ephemeris' retarded positions as seen from the Sun, the centers of Saturn and Venus align with the edge of the Sun, at 11h35m UT Dec. 21, 2012. Using the mean orbital radii, the light times from the Sun to Venus & Saturn, are 6m & 79m, resp. Therefore at 11h29m, Venus was actually at its "retarded" position, and Saturn was actually 73m of travel, ahead of alignment. Venus' period is 1/50 of Saturn's, so approx. 73/50 = 1.5m later, at 11h30.5m, Venus' and Saturn's centers would align with the edge of the Sun (not apparently as seen from Saturn, but geometrically). The trailing edge of Saturn would align with the leading edge of Venus, and the edge of the Sun, at 11h25.5m, only 14m after Earth's winter solstice.

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New message May 26, 2010 (I'm placing it here because I get a "file full" message when I try to post it as an ordinary new message).

Evidence that the May 22, 2010 crop circle near Wilton, Wiltshire (England) codes for last new moon before winter solstice, 2012

This crop circle is a pie with twelve equal slices, naturally, the twelve synodic months = 29.53*12 = 354 1/3 days making about one year. Some suppose that crop circles like this are designed to tell us something any freshman calculus student learns, like Euler's formula or the first ten digits of pi. Would humans spend thousands or millions of dollars, to monumentalize freshman calculus for a few days in a crop circle? Would a nonhuman risk who-knows-what expense, danger, interplanetary combat or other horror, to deliver such a redundant message? We already know, that anyone or anything that can produce such an impressive crop circle, surely must include on their team, individuals who have learned freshman calculus and the like. They don't need to recite freshman calculus, to convince us that they are intelligent.

The circles bring news. The message is something we don't know, or that isn't yet widely known or believed. The circles are, literally, propaganda. They are just cryptic enough, to avoid mass hysteria, and to command the respect given to that which one must work to understand. Or maybe the circlemakers don't want to bother with, or can't intercept or interpret, our language, and figure that if we can't "get it", then "tough luck".

Andreas Mueller has, according to his website, a Diploma in Graphic-Design from the famous Academy for Fine Arts and Design in Saarbruecken, Germany. Surely someone with such rigorous training would orient his diagram of this crop circle, to exact geographic North (for some reason, I never see a "North" arrow on any photo or diagram of a crop circle!). What seems to be the most complete reproduction of Mueller's full diagram available on the internet, shows (like Mueller's usually shown "preliminary" diagram), after correction for its elliptical shape on my browser, that segment A1A2, which I believe represents the last of the twelve New Moons in 2012 (B1B2 represents the first New Moon in 2012) ends, moving clockwise, 10.0 +/- 0.5 deg clockwise of North. The bars in each sector seem to encode, inter alia, information about the ecliptic latitude of the New Moon relative that of Venus and the Sun. Therefore the New Moon likely is at the center of the sector, that is, 10-30/2 = 5deg counterclockwise, i.e. before, North, which implies that North is Dec. 13+5 = 18, 2012. On the other hand, Marie Asmer's diagram, also published on the internet, places the end of this sector at nearly the exact top, which implies that North is Dec. 13+15 = 28.

Somewhat as others have done, let's transcribe each sector as a sequence of 8 digits, base 4, with: nothing = 0, a halfbar in the counterclockise part = 1, a halfbar in the clockwise part = 2, and a full bar = 3. Starting with Mueller's B1B2 and moving clockwise to end at A1A2, and also starting at the outside of each sector, the base 4 numerical equivalents of the sectors are:

a) 22212102 (Asmer reads first digit 0, but Keller, Mueller say 2) b) 02212112 c) 12212132 d) 32230102 e) 02203112 f) 12212132 (Keller reads last digit 3, but Mueller, Asmer say 2) g) 12230122 h) 12001122 i) 32331122 j) 02023102 k) 30100312 l) 23313032

The matrix is highly nonrandom:

1. c=f 2. h & i are equal in their last 4 digits 3. the last digit is always 2 4. for a through g, the 3rd digit is always 2 5. the 2nd & 3rd digits are equal except for h, i, j, k 6. for the 1st & 5th digits, a string of 7 is shifted by one 7. the 2nd & 6th digits always add to 3

Only for two months, Nov. & Dec. 2012, do the 2nd & 6th digits not equal 2 & 1, resp. These two months are indeed special. November's New Moon is very near the ecliptic and ascending: there is a total solar eclipse. December's New Moon, still ascending in ecliptic latitude, is the only one of the twelve, which is farther North in ecliptic latitude, than Venus at the same time (the Dec. New Moon is just 0.9deg farther North, in ecliptic latitude, than Venus).

Summary: If the center of each sector represents a New Moon in 2012 (i.e. in the year preceding the winter solstice, 2012) then the diagrammatic "top" of the cropcircle corresponds to Dec. 23 +/- 5, 2012. Also, the 2nd, and, redundantly, the 6th, digits of the code, signify that the Nov. & Dec. New Moons are special. Indeed, these are the New Moons that (Nov.) eclipse the Sun at the ascending node, or (Dec.), still ascending, nearly match the latitude of Venus.

A few are known hoaxes, or even open publicity stunts with no hoax component. Some are true extraterrestrial communications, warning us of the date, Dec. 2012; these use astronomical hints, because apparently they are one-way communications from some kind of "space brothers" who know nothing of our language. Many crop circles are, I think, government-issue disinformation intended to dilute and deflect the effect of the true circles; this includes most of the "geometric" and "mathematical" crop circles, as well as the shallow mathematical explanations for some of the true extraterrestrial crop circles.

The famous "pi" circle really has nothing to do with the digits of pi base 10. It shows conjunctions of Mercury near the ecliptic, alternating superior and inferior conjunctions, between the transit of Mercury in 2006, and the last conjunction of Mercury in late 2012. There are several hints in the circle, that this is what is shown. I'm recovering from H1N1 influenza, but hope in a few days to post the quantitative explanation I have.

The "pi" crop circle: June 1, 2008 near Banbury Castle, Oxfordshire

This crop circle is not intended to represent "pi" in decimal form. This crop circle depicts selected conjunctions of Mercury between the Nov. 2006 transit, and the last Mercury conjunction in 2012. It is an astronomical clock, whose hands point to late 2012.

The three disks at the top, of the usual internet photo of this crop circle, represent Earth, Venus, & Mercury. The lefthand side of the Earth disk is, relative to the center of the central Sun disk, no more than about a degree away from geographic North, assuming that the straight double tracks were made NS according to the Global Positioning System now on many farm tractors (and also according to the schematic of the circle, that I found online). I determined angles by measuring with a protractor, on a printout of an online photo, then correcting for perspective: both slant (flattening of circles) and nearness (varying scale, i.e. converging lines).

The arc from the lefthand side of Earth, to the notch at 6:00, is, as I measure, within about a half degree of 180deg. This arc, I think, signifies Mercury gaining 180deg on Earth, i.e. a time interval between conjunctions. It is a clue that the crop circle is about conjunctions of Mercury. Another arc, from the short line segment at Mercury, to the outermost notch (the notch near 7:00) also is within about a half degree, of 180deg.

By switching the beginnings, there arise two more arcs; these are the ones that are explicitly drawn. From the lefthand side of Earth, to the 7:00 notch, is 146deg; Mercury gains this much vs. Venus, in 58.6d on average, about the same time, 57.9d, as needed for Mercury to gain 180deg vs. Earth. The longer arc, from the short line segment at Mercury, to the 6:00 notch, measures 213deg. I don't fully know why that angle is important, but on average, Mercury gains this much on Venus, in 85.5 days. The six conjunctions of Mercury, which the inner notches of the crop circle seem to depict, are separated by five time intervals. Four of these time intervals, are multiples (4.08, 2.05, 6.05, 8.92) of this mysterious 85.5 day period.

The six inner notches are placed to emphasize their grouping and relationship with each other; they are not exactly where Earth is, at the time of the relevant conjunction of Mercury. None are far from Mercury's (ascending node = 48) line of nodes. So, all represent close conjunctions, i.e. conjunctions for which Mercury's ecliptic latitude is small. Between the transit of 2006, and the end of 2012, there are, according to the online JPL ephemeris, twelve Mercury conjunctions for which Mercury's geocentric ecliptic latitude, at the Greenwich midnight closest to the time of smallest Mercury-Earth-Sun angle, is less than or equal to 1.0deg in magnitude:

The number after the date, is the linear distance from Venus, to the Earth-Mercury line (precisely, using data for the Greenwich midnight closest to conjunction, it is the Earth-Venus distance in A.U., times the sine of the difference between the geocentric ecliptic longitudes of Mercury and Venus). Let's select the Mercury conjunctions which not only have Mercury's ecliptic latitude less than or equal to 1.0deg, but also have Venus less than 0.50 AU from the Earth-Mercury line:

At #3, Venus is only three days away from the ecliptic, whereas at #7, Venus is 1.4deg from the ecliptic. If I keep #3 but discard #7, and exchange #11 for #12, I get:

These are the conjunctions of Mercury shown schematically on the crop circle.

I carefully measured the position of the big dot near the last (i.e. innermost) conjunction (each of the six innermost steps in the spiraling line represents a conjunction). With correction for Earth's orbital eccentricity (Earth's variable orbital angular speed) the dot's position corresponds to Earth's, at Nov. 20, 2012 (or a similar date in other years).

2012 AD transit of Venus starts over Hawaii; 4328 BC summer solstice was inferior conjunction of Venus.

According to NASA's website, the "external ingress" at Honolulu, is 22:09:59 UT, and the "internal ingress" there is 22:27:38. (Due to parallax, the times vary several minutes, for the various locations on Earth.) What might be more important physically, is the "central ingress", i.e. the time when Venus' center crosses the Sun's limb. By interpolation, the time of "central ingress" would be, as seen from Honolulu, about 22:19 UT, June 6, with the Sun at altitude 88deg in Honolulu. Also by interpolation in the NASA table, the "central egress" seen from Honolulu, would be June 7, about 04:35:30 UT, with the Sun at altitude 7deg in Honolulu. The midpoint of the transit, as seen from Honolulu, would be June 7, 01:27 UT.

Hawaii's latitude is about the same as that of Olympus Mons on Mars, or the Great Red Spot on Jupiter. Hawaii is Earth's highest volcano, and Olympus Mons is Mars'. This 2012 transit of Venus, starts when Hawaii is near Earth's subsolar point.

The summer solstice in 2012 is June 20, 23:09 UT. The weight of my evidence is, that the original Giza-based Egyptian calendar, began at the summer solstice, 4328 BC. So, the time interval between the start of the Egyptian calendar, and the midpoint of the 2012 AD Venus transit as seen from Honolulu, is (2012+4328-1)*365.242187d - (13 + 22/24 - 18/1440)d = 6338.716 * 365.25636d = 10303.735 * 224.70069d, that is, 6338.716 Earth sidereal yr, or 10303.735 Venus sidereal yr. The solstice at 4328 BC occurs with Earth at about right angles to its perihelion, so Earth lags about 1/30 radian = 0.005 cycle, and the difference between Venus and Earth, at the summer solstice 4328 BC, really is only 0.735 - 0.716 - 0.005 = 0.014 cycle = 5deg. That is, the summer solstice of 4328 BC was, at least approximately, an inferior conjunction of Venus.

June 7, 2010 crop circle at Stony Littleton Long Barrow near Wellow, Somerset, England: conjunctions of Venus

An intelligent space probe might locate native astronomical monuments such as Stonehenge, and make crop circles there, to aid the natives in interpreting the crop circle. The June 7, 2010 Somerset crop circle signifies conjunctions of Venus, the last on March 29, 2013. (My source for basic information on this crop circle is Linda Howe's website, www.earthfiles.com ).

There was a transit of Venus June 8, 2004, and will be another June 6, 2012. These are represented by the concentric disks lying at opposite ends of the big circle: transits are inferior conjunctions very near the nodes of Venus. The other three concentric disks, with reversed colors, represent superior conjunctions. The center one, represents the occultation of Venus June 9, 2008 (maximal, from Earth's center, at 03:42 UT). June 7, 2010 (the date of the crop circle) is exactly halfway between June 9, 2008 and June 6, 2012.

There are four other inferior conjunctions of Venus, between the two transits. If the concentric disks at the outside ends, outside the big circle, represent the superior conjunctions immediately before the 2004 transit and after the 2012 transit, then each one is the seventh superior conjunction counting from the other, hence the seventh circle along the arcs drawn. The last superior conjunction is thus March 29, 2013.

(the below is posted here June 22, 2010, not separately, because of a server problem)

The Mayan Long Count, 61 & 243: not just another pretty calendar

The Roman calendar contains a millenium analogous to the Mayan Long Count. We all know it: 365.25 days (Julian style) * 10 yrs/decade * 10 decades/cent * 10 cent/millenium. The Gregorian style adopts, on average, 365.2425 days, the closer to match the actual tropical year, 365.24219 days. Presumably Clavius et al, knew the tropical year more accurately than their Gregorian calendar shows, but sacrificed some accuracy, for the sake of a simpler leapyear formula.

The Mayan Long Count is 360 days * 20 * 20 * 13, where there are Mayan names for the intervals analogous to the Roman decade, century and millenium. The 360 day year isn't unusual for ancient people; it's a round-number compromise between 365 days = 1 Earth year, and 354 days = 12 synodic months. Our European trigonometry of 360 degrees, derives from an Egyptian awareness of this same 360 day approximation. Likewise "base 20" (fingers and toes) is as natural as "base 10". But why 13?

The shocking answer, is that the Mayans, or their forerunners, juggled the factors, to find a simple, Roman-like calendar, usable by the common man, but which also would have a "millenium" (Long Count) equal to a fundamental astronomical period. Consider these from Lang's "Astrophysical Data", p. 41 (Springer, 1992):

Mayan Long Count = 5125.366 tropical year (see above) MLC / sidereal orbital period, in tropical yr, of Jupiter = 5125.366/11.86223 = 432.07 MLC/ 29.4577 (Saturn) = 173.99 MLC/ 84.0139 (Uranus) = 61.006

Thus the Mayan Long Count, as I've noted in earlier posts here to Dr. Van Flandern's messageboard, is significantly close (p=0.003%) to a whole multiple of the orbital periods of Jupiter, Saturn, and Uranus. One must use 243 periods of Uranus, to get a closer common multiple with Jupiter & Saturn, than the MLC which uses 61 Uranus periods. I find that integers with no prime factors bigger than 13, are dense enough, that the existence of a way to construct, by multiplication of days, an MLC so near this desired best approximate common multiple, is not surprising.

This relationship, between the Long Count and the periods of three giant planets, tells us that the Mayans, or their forerunners, knew of Uranus. Furthermore, they knew the orbital periods of Jupiter, Saturn, and Uranus to about four significant figures or better. For Uranus, Newcomb's 1873 investigation was good to almost five figures, but required a century of telescopic data. The Mayan accuracy also can be achieved, however, by comparison of modern unaided observations, with Ptolemy's occultation records for Jupiter or Saturn, if Earth parallax is accurately calculated. The Mayans must at least have had a good Copernican solar system model, and many centuries of careful visual observation records.

The number 61 apparently does not arise by accident. An even closer common multiple, is 243 times Uranus' period:

The "243" relation also holds, using Lang's 1992 values, for Earth, Venus and Mercury:

243*365.256 day sidereal period (Earth) / 224.701(Venus) = 395.001 243*365.256 / 87.969(Mercury) = 1008.96

(addendum June 23, 2010: I also find that the "243" relation holds for Mars, Earth, and Venus: 243*686.98 day orbital period (Mars) / 365.256(Earth) = 457.04 243*686.98 / 224.701(Venus) = 742.93 )

Finally, recall that Venus' sidereal rotation period is (2007 World Almanac) 243.02 days (these are synodic Earth days). This is only slightly faster than the sidereal rotation period, 243.165d, which would result in Venus, on average, showing exactly the same face to Earth at every closest approach.

The accuracy of the common multiples in the Mayan Long Count, proves that unlike the Roman calendar millenium, the Mayan Long Count is not an arbitrary length of time, but rather is an important astronomical cycle, disguised, for convenience or for deception or both, as an arbitrary length of time. The MLC is based on 61 orbits of Uranus, but the common multiples based on 243 orbits of Uranus or 243 orbits of Earth (243 = 61*4 - 1) reveal that some unknown physical effect causes the "243" cycles and probably the Mayan Long Count too.

Hi Joe, it might be an idea to e mail Linda Howe, as she does have the ear of the media.

I thought I'd pick your brains again, see what you think about this idea. In the thread about why someone believes in an end to the universe, I got thinking about what it must be like to be in orbit round the galactic super massive body. For one thing you could be inside of a neutrino ball. Something with a radius of about a month. Now if you hopped into a space ship and orbited very close to the Swartzchild radius you could watch the universe go out, check your watch and say, "well that's it then. Ten forty gmt."

now let's suppose that the Scwartzchild radius for the universe is one gravitational second. Remember that I'm a bit torn between between it being h = c^2 / b^2 and barh = c^2 / b^2 where b is the speed of gravity. I am leaning more to it being the later.

That gives us a Scwartchild radius of about 2.9E 25 metres. 2.9E 25 = 2GM / c^2

Work out the mass and it's aout the current estimate.

g /r_s (r/c)^2 = 0.5

where r_s, r subscript s, is the Scwartzchild radius. We take g to be G, the universal constant. Working out r we get aout 139 billion light years for the radius of the universe.

t_r / t = sqrt(1 - r_s / r)

Where t_r is time for an observer near or inside Scwartzchild radius. The upshot is that if we could see a galaxy 139 billion years in the past, then it would e in a different time zone but it's not huge, a tenth.

Well, is this a big bang? Yes and no, I suppose. if we take from Dirac the idea that we can have an expanding universe in which either G decreases, or h increases, the we can have a static universe in which the both vary. A universe that always has the Planck lenght at the same value.

Pl = sqrt(barh G / c^3) = about 1.62E-35

Say that's a wavelenght and its mass will be 1.62E-35 = h /mc =1.36E-7 kg

Pop that into the equation for the Scwartzchild radius. r = 2Gm /c^2 = 2.0225E-34

Rough calculations but I reckon that's going to work out at twice barh

Hi Joe, I haven't done the sums yet but there's something interesting going on here I think. all park figures; from the equation g / r_s (r / c)^2 = 0.5 we can work out G when the radius was at about 2.4E-12 metres.

Now let's say that, I think it was de Sitter, that G = 1 /6pi rho t^2

Rho is going to be huge, as we're putting the whole mass of the universe into something the size of an electron. So t is going to be very small. In the region of 1E-60 seconds.

But suppose we say that in the past G was very large and h was proportionately much smaller than it is now. Then from sqrt(barh G / c^3 we can work out what barh was at 1E-60 seconds.

As we are talking about the lorentzian as a refractive index, it makes sense to say that the frequency of a particle doesn't change in a different refractive index space. Because h is much smaller we can have a situation where a subatomic particle has the same energy as the whole universe. I suspect that this primordial particle will turn out to be a pion rather than an electron. First particle on the scene at 1E-60 seconds is a pion which doesn't like what it sees and promptly decays. I suppose the question has to be, how promptly is promptly in this case.

Hi Joe, it can be a problem on the battlefield. Brits and Anzacs fighting alongside American troops can sometimes get into "language" difficulties over usage. Though there was the famous "Houston we have a problem" which was a very British way of stating something.

On the idea of the Scwartzchild radius of the observable universe being one gravitational second. Then the idea of a primordial pion that has the mass energy of the universe ounds like fun to look at, if only because its decay should somehow look like the universe we inhabit. the pion's neutrino carries off the bulk of the energy but this neutrino, I think, has to e considered to have mass. i need to read up on this further, as the few links I've looked at are not saying the neutrino has mass, whereas newer research says it does.

Before getting into that though, I though I'd need to look at red shift. If the "universe" has a radius of about 139 billion light years, then most of it is empty of galaxies, we can still see them after all. eh most distant ones are very young, and if G and h vary proportionately to the planck length then Gand h differ from our values but they are in a different "time zone."

With Doppler shift we have (1-v/c)f or (1+v/c)lambda where v is much smaller than c, let's call that e.m doppler. What about gravitational red shift? (1-c/b)f where b is the speed of gravity. We expand it and can ignore c^2/^2 and C^3/b^3 etc.

So let's accept for the moment that there's a phase transition at the speed of light. Then a photon will have a plus and minus value.

So, let's take a look at the grav energy and the e.m energy again.

E = mc^2 and E =mb^2 we'll use the electron as our mass particle.

m = 9.11E-31 c = 2.998E 8

8.188E-14Js

grav energy will be much greater where I've taken b to be 2.91E 25

Divide 8.188E-14 by barh to get a frequency and we find that grav energy and e.m frequency have the same numerical value. This must mean that barh takes the value one in gravitational space. it's effectively borrowing energy form a time in its own far future. Or it borrows from its past, because with negative refractive indices there's nothing to stop that sort of borrowing. Though at the moment I prefer to have one direction to time but allow negatives with respect to positive. A bit like running the universe off a nine volt battery, set a line at 4.5 volts.

Let's choose one of the nine planets, multiply its orbital period by 243, then divide by the orbital period of one of the other eight planets. Randomly, of these 9*8=72 numbers, 1/10 * 72 = 7.2 should be within 0.05 of an integer. So, it isn't surprising that 8 are, but it is surprising that 5 of the 8, are pairs where the second planet is one or two steps *below* the first (there are only 15 such pairs, so only 1.5 of those should be within 0.05 of an integer). The 8 are:

Also, 7.2 of the 72 should be within 0.05 of a half+integer (1/2 + n), but instead, 12 are. It's surprising that 12 are; and the excess is due to 6 pairs where the second planet is one or two steps *above* the first (there are only 15 such pairs, so only 1.5 of those should be within 0.05 of a half+integer). The 12 are:

Summary: Five of 15 possible pairs, where the inner planet, of the nine including Pluto, is one or two steps below the outer, have the ratio, 243*(outer planet sidereal orbital period)/(inner planet period), within 0.05 of an integer. Six of these same 15 possible pairs, have the ratio, 243*(inner planet period)/(outer planet period), within 0.05 of an (integer plus half). The binomial test gives p=1.27% for the former phenomenon and p=0.22% for the latter.