Requiem for Relativity

Kukulkan pyramid lightbeam: if fake, the faker knows trigonometry

The earthfiles.com website has a recent article about a 2PM July 24, 2009 Apple iPhone photo of the Kukulkan pyramid by U. S. tourist Hector Siliezar. The photo shows an obvious pink lightbeam of constant width, rising vertically from the pyramid. The pyramid is photographed at a slight angle. Assuming that the beam rises vertically from a base centered at the center of the temple, the displacement of the beam relative to the center of the near temple wall, is consistent with the displacement of the center of the near temple wall relative to the center of the top of the stairs, and consistent with the rotation of the temple walls in the view.

I used ruler measurements from the screen, from the earthfiles.com article and from the temple diagrams at world-mysteries.com. The apparent length of visible lefthand temple wall is to the near temple wall, as 5::21. The beam was centered at 3.0mm along a length of near temple wall of 8.3mm, i.e. displaced left 1.15mm. The temple is slightly rectangular, 13.42m wide by 16.5m deep. So if the beam rose from the center of the temple, it should be displaced relative to the near temple wall, by

16.5m/2*8.3mm/13.42m*5/21 = 1.21mm left, in good agreement with 1.15 left, observed.

Also the overhead diagram shows that the distance between the near temple wall (one of the short walls with only one door) and the steps, is to the depth of the temple, as 1.0::9.0. So likewise the temple door should be shifted left relative to the stairs, by

1.21mm*1.0/4.5 = 0.27mm, in good agreement with 0.3mm observed.

More 2012 Doomsaying

Reviewing some results from my recent previous posts:

Hapgood gives the last three Ice Age pole positions only to the nearest degree, but if these are converted from geographic to geocentric latitude, then the opening half-angle of the Earth-centered cone containing them, is 21.65deg.

Now let's relate this to the New Moon of Dec. 13, 2012. The tide will be unusually strong, because not only is Luna at perigee at 23:06h UT Dec. 12, Earth is also near perihelion at this season, and furthermore, perigees are closer when Earth is at perihelion. Other tides have been stronger; e.g., the thousand-year perigee of Jan. 1912, only six minutes after Full Moon and one day after Earth perihelion, whose extraordinary tides are blamed for the iceberg calving that indirectly sank the Titanic on April 15 that year (Sky & Telescope, April 2012, p. 37). However the Dec. 13, 2012 New Moon, at 08:21 UT (minimum astrometric, Sun-Earth-Luna angle) has another unique feature:

As I noted in an earlier post, Luna will be near stationarity in Declination. For a small Sun-Earth-Luna angle, the solar and lunar tidal forces can be added approximately vectorially. Using quadratic interpolation of the basic parameters, on a 6h interval, I find that the maximum tidal force at Earth's center, is at 03:53 UT, Dec. 13, based on the astrometric (includes light delay) JPL Horizons positions and masses. The light delay from the Sun is barely significant at this precision, because Luna moves 20", the aberration angle, in less than one minute of time. Also I find that the direction of tidal force is stationary at minimum actual Declination, at 02:39 UT. That Declination is -21.563deg.

For Declination, I used Earth's mean axis of 2012.948AD (which approximates Dec. 13.25 UT) according to the online NASA Lambda utility's celestial coordinate conversion. I corrected this axis for the primary (i.e. 18.6 year) nutation, finding the nutation in longitude from the position of the Lunar ascending node; and the nutation in obliquity from that in longitude, according to the chart at pietro.org. The nutation correction is barely significant at this precision.

Not only will the New Moon of Dec. 13, 2012, be associated with an unusually strong tidal force; that tidal force will be aligned orthogonally to an axis consistent with Earth's force-free precession. Furthermore the maximum tidal force and the alignment will be simultaneously nearly constant for several hours.

Uranus is stationary in ecliptic longitude at approx. 12:03 UT, Dec. 13.

Ptolemy's precession calculations: hints of 2012?

Olaf Pedersen, "Survey of the Almagest" (1974) (which the ISU library has) cites Anton Pannekoek's article "Ptolemy's precession" in "Vistas in Astronomy 1" (1955) (which the ISU library doesn't have). According to Pedersen, on p. 63 of Pannekoek's article, Pannekoek says that where Ptolemy lists observed right ascensions and declinations of 18 stars at different epochs, the six stars Ptolemy uses for his precession calculation, indeed show that Earth's axis precesses 38"/yr; but, the twelve stars he doesn't use, would have shown 52"/yr, near the modern estimate (e.g. Newcomb's formula)(with secular trend) 49.8"/yr for that epoch.

This suggests that someone extensively rewrote (corrupted) Ptolemy's text to change the correct 52"/yr, to 38"/yr (other, direct longitude data in Ptolemy's text also imply about 36"/yr).

Estimated lunisolar precession rate during Ice Ages

One way to estimate the angle of Earth's torque-free precession during the Ice Ages, is to find the opening angle of the geocentric circular cone containing Hapgood's three Ice Age pole points; as mentioned previously, this is 21.65deg. Because Earth's oblateness becomes less, Earth's lunisolar precession rate also is less by the oblateness factor

1 - 1.5 * sin^2(21.65) = 0.79583

obtained by averaging the shapes of the spheroids for every pole position along the precession circle.

There is another, "directional", factor. Though Earth's angular momentum axis remains fixed, Earth's spheroidal axis not only varies (during torque-free precession) in obliquity, affecting the torque; but also varies in alignment with the vertical (i.e. perpendicular to the ecliptic) plane through the angular momentum axis, decreasing the relevant component of torque. A periodic function can be integrated numerically to a high degree of precision, by averaging the values at the 0, 180, 90 and 270deg phases. Without torque-free precession, there is only one term,

A = sin(2*23.44)

With torque free precession, there are analogous 0 & 180deg terms,

B = sin(2*(23.44-21.65)) & C = sin(2*(23.44+21.65))

The 90 & 270deg terms both equal, using angles found by Napier's rules for spherical trigonometry,

D = sin(2*31.49) * cos(47.36)

where the cosine factor accounts for the suboptimal direction of the torque. The ratio

(B + C + 2*D)/4 / A = 0.77725

gives the directional factor, and the product of the oblateness and directional factors is

0.61856.

This is so near 1/(golden ratio) = 1/"phi" = 1/((1+sqr(5))/2)
= phi-1 = 0.618034, that I will assume 1/phi is the correct value. The current Newcomb precession rate 50.28"/yr / phi = 31.075"/yr, corresponds to a "Platonic year" of 41,706 yr, near the Milankovitch period.

Using Newcomb's quadratic formula for precession rate with secular trend (quoted in Clemence, Astronomical Journal 52:89+, p. 90) I find that the precession rate one-half precession cycle ago (about 13,000 years ago, i.e. at about the true end of the last Ice Age or at about the beginning of the "Younger Dryas") was 47.4"/yr. Assuming that torque-free precession ended at that time, and that astronomers of that time, averaged the two rates, 47.4" and 47.4"/phi, they would have found a mean precession rate 38.3"/yr, which seems to be what Ptolemy's six stars' coordinates were falsified to demonstrate (while the other twelve stars in Ptolemy's table still have their original truthful coordinates that imply 52"/yr, according to Pannekoek - see previous post for citation - this is near the Newcomb precession rate 49.8"/yr for Ptolemy's time).

Dr Joe, These numbers you keep referencing to determine the force in control of the ice cycle seem a bit of a stretch. The ice cycle requires a flow of energy in and out of the Arctic ocean having nothing at all to do with orbital details. Why not look at stuff like the mass of all the ice that comes and goes during the cycle as well as the location of all that ice. Ice is not all that great a mystery and explaining it does not require orbital calculations. It seems the use of astronomy has shrouded the real force in control of the ice cycle.

This is a good time to reread my Feb. 24, 2012 post to this thread. It has been extensively revised and corrected, but the main findings are the same. It is confusing, but I can say it more simply if I omit the explanation of the calculations:

An oblate spheroid like Earth can wobble like a top. No external forces are needed to make this happen, but in order to continue to have the same angular momentum, extra kinetic energy of rotation must be acquired from somewhere.

If Earth were to wobble like that, it would become more spherical, because with different points of Earth coming to lie under the pole, the equator wouldn't be in the same place all the time: Earth's bulge would average out some, and this can be estimated. I also can estimate how much energy is released by this partial "sphericizing" of Earth.

But let's recognize that I don't really know how much energy becomes available. With the help of a computer, let's just try solving the equations for all the different amounts of energy that might become available. When I try it, I find that for any likely amount of energy, there is a unique solution with a particular angle of wobble, a particular time period of wobble, and the bulge appropriate to that angle of wobble. But for one particular amount of energy (namely 1.022 times my estimate of the gravitational potential energy released by Earth's sphericizing and adiabatic shrinkage) two criteria are satisfied simultaneously:

1. The angle of wobble is 18.9 degrees. This corresponds tolerably well to Hapgood's estimate of the effective locations of the poles during the last three Ice Ages (ave. latitude 65; greatest 72).

2. The period of wobble is 365.24 days. This means that the pole would be in the same place every winter solstice, making it in effect about as good as a pole that is there all year.


A priori, the period might become equal to the tropical year, for some other angle of wobble. But really it can happen only for an angle which corresponds to Hapgood's determinations, and to an energy release only 2% different from my estimate. This is evidence, that Hapgood's pole shifts really happened and that they were due to episodes of large-angle torque-free precession of Earth.