Requiem for Relativity

Teotihuacan & Cholula: Platonic solids, Arcturus, spring 2013

I recalculated the apparent Declinations of Algieba and Arcturus as precisely as possible. I found that throughout early 2013, the difference in their Declinations is only about 1 arcmin more than the difference in latitude of the respective pyramids, of the Moon at Teotihuacan, and Cholula. For details see "Appendix. Method" at bottom.

Again I considered the hypothetical effect of a likely pole shift. In light of my result for the Serpent Mounds (see my next post) I assume that originally the Pyramid of the Moon was at a vertex of a regular tetrahedron inscribed in Earth's spheroid with one corner at the South Pole. For Earth's spheroid, I adopt flattening 1/298.0 for the reasons given in the Appendix. Analytic geometry gives a quadratic equation for sin(parametric, a.k.a. reduced, latitude); this results in geographic latitude 19.713773, 51.02" greater than the 19.6996 given by Wikipedia for the Pyramid of the Moon today. Now suppose that originally the Pyramid of the Moon was not at this latitude, but rather at the "reduced tetrahedral corner", i.e. the geographic latitude whose corresponding reduced latitude equals the geographic latitude of the inscribed tetrahedron: this greater geographic latitude is 19.775010, 271.48" north of the present geographic latitude of the Pyramid of the Moon.

Because Teotihuacan and Cholula are less than a degree apart, I can use plane trigonometry approximations to find that a pole direction shift, theta, to the west of north, increases the latitude difference between the pyramids of the Moon & Cholula, by a factor cos(38.53490-theta)/cos(38.53490). The larger the pole shift, the farther south Cholula was, assuming we know the latitude of the Pyramid of the Moon as discussed above. During early 2013, Arcturus moves a few arcseconds south and then north again, as its aberration changes. There is a minimum allowable geographic latitude for Cholula, which just equals Arcturus' minimum Declination (of date) achieved about March 25.0. At this time, Algieba (comparison of the epoch 2000.0 position computed in the online Hipparcos catalog, to the Bright Star Catalog, confirms that it is gamma1 Leonis, the brightest component) has apparent Declination only 2.80" S of the assumed original geographic latitude of the Pyramid of the Moon, the tetrahedral latitude for the spheroid. The needed latitude for the pyramid of Cholula, is consistent with a pole shift 2.1012deg W of N.

Months ago, my measurements on the large paper copy of Millon's map at the Iowa State Univ. library, found that from the centers of the top platforms of those pyramids, the alignment is 2.1064 deg W of N. My measurements were, I think, often off by 0.1mm but rarely by 0.3mm; thus the measurement errors roughly are distributed equally over the interval [-0.2mm,+0.2mm]. I had to add several measurements, including the measurements by which I determined the centers; I calculate a sigma error, from the ruler measurements alone, of 0.0336deg. So, the difference between the interpyramid azimuth I measured on Millon's map with a ruler, and what is needed to achieve perfect equality of Arcturus' apparent minimum Declination March 25 with the original geographic latitude of Cholula, is only 0.15 sigma according to my estimated ruler measurement error.

Using the NASA Lambda utility to convert J2000 to J2026 coordinates for Arcturus & Algieba (gamma1 Leonis) I find that at present, the Declinations of these stars are decreasing 18.60" & 18.33"/yr, resp., due to precession and proper motion. So on a timescale of months or a few years, their Declination difference mainly changes due to aberration or nutation. One year's change in the Declination of Arcturus, is about 1/4 of the latitude difference discrepancy which the pole shift west of north is correcting; that is, it corresponds to about 0.5deg, far larger than my map measurement sigma error of 0.03 deg.

Summary. Let's assume that true geographic north then were indicated by the alignment of the pyramids of Moon and Sun at Teotihuacan (2.1deg W of N today). Let's assume also that the original reduced latitude of the Pyramid of the Moon, equaled the geographic latitude of the corners of a regular tetrahedron inscribed in the Earth spheroid (assuming flattening 1/298.0) with one corner at the South Pole. Then Arcturus would have been at the zenith over Cholula at its minimum apparent Declination this spring, and Algieba (gamma1 Leonis, the brighter companion, is the star recorded by the Hipparcos satellite) then would have been only 2.8" south of the zenith over the Pyramid of the Moon.

Sociological comment. Such elaborate warning as these calendars and pyramids provide, implies that the event will be not only catastrophic, but will occur without warning otherwise. Expect a sudden hammerblow, like a giant lightening strike from space, that comes with little warning (only strange unexplained sounds?) because it comes at the speed of light.

The date, Dec 21, 2012, was determined by the need to end the Mayan calendar on a recognized important day so that the calendar could not be dismissed as arbitrary. The pyramids of Teotihuacan and Cholula, playing on the Declination of Arcturus, indicate the spring of 2013 and definitely exclude the springs of 2012 or 2014. The crop circles (see my above posts) seem to be indicating a bell curve for the predicted time, Feb 17 +/- half maximum 64 days = sigma 54.3 d. Algieba's opposition is Feb 18.

Appendix. Method. This time, I started with original stellar coordinates (Arcturus; and gamma1 Leonis, the brighter member of the Algieba double) from the Hipparcos catalog, in J2000 coordinates but of epoch 1991.25AD; using Julian days, I applied the exact Hipparcos proper motions to these coordinates for all times.

I accounted for precession, mean obliquity and the motion of the ecliptic plane, simply by getting the J2000 coordinates of the mean pole from the NASA Lambda online coordinate conversion utility; these are given in increments of 0.01 Julian yr, so I interpolated quadratically over the range Dec 31, 2012 - Apr 30, 2013, from values at 2013.00, 2013.16 & 2013.32AD. I found nutation, based on a formula by Meeus, from the online utility at www.celnav.de , and interpolated this quadratically too, adding it to the mean pole.

I found Earth's heliocentric position and velocity, by quadric (4th degree) interpolation, on 30 day intervals starting Dec 31, of Earth's JPL Horizons J2000 heliocentric apparent celestial coordinates, and converting to xyz coordinates. For aberration, I divided Earth's velocity by the speed of light, and added that vector to the star's position. For parallax, I multiplied Earth's position vector by the Hipparcos parallax in radians, and subtracted that vector from the star's position. Then I renormalized the star's xyz position vector and reconverted to celestial coordiates.

These calculations required a computer program. I wrote a BASIC program with double precision.

The 1987-88 CRC Handbook of Chemistry & Physics notes that the latest determination it gives of Earth's oblateness, is based on space probe data. The second latest determination given, Zhongolovich 1952, isn't: he approximated Earth as a triaxial ellipsoid with flattening 1/298.0 in the meridian 96W and 1/295.2 in the meridian 6W. Because ancient people would have relied, as Zhongolovich did, on ground-based data (Zhongolovich used data along the Struve arc, which ran mostly through Russia) and because space probe data introduce errors from unknown physical effects, such as the Explorer I rocket or Pioneer probe accelerations, I should use the best, last ground-based determination not contaminated by space probe data, but as applied to the meridians of Teotihuacan or Cholula, 98.8 or 98.3 W.

There is a long tradition of thinking about regular polyhedra: Plato, Kepler, and most recently science journalist Richard Hoagland, who has publicized the occurrence of planetary features (Hawaii, Olympus Mons) near arcsin(1/3), the latitude of three of the corners of a regular tetrahedron inscribed in a sphere with a corner at the South Pole.

Dr Joe, It seems to me all your sources use the same flawed model we have been kicking around here for several years. We have a model with two barycenters that are assumed to be fixed points. These barycenters don't even exist in the real system they attempt to predict, but that is irrelevant because for most users the fact the model is very near to an accurate indicator of how the real system behaves is amazing. you are doing very fine detailed work and the model is never nearly finely tuned enough to give true positions of any of the objects you are looking at.

The Serpent Mounds: another Platonic solid, another pair of stars
(revised Jan 20, 2013)

The Ontario & Ohio Serpent Mounds are to a regular octahedron and Deneb and Vega, as Teotihuacan & Cholula are to a regular tetrahedron and Algieba and Arcturus. I use the same computer program as for my previous post, and the same Hipparcos catalog, to find the Declinations for current times including effects of nutation, aberration & parallax.

The Declination of Algieba1, gamma1 Leonis (when Arcturus is at its minimum Declination this spring due to aberration) was found in my previous post to differ only 2.80" from the geographic latitude whose reduced latitude equals the geographic latitude of a corner of a regular tetrahedron inscribed in the Earth spheroid with a corner at the South Pole. Analogously I can attempt to inscribe a regular octahedron concentrically in the spheroid so that the edges of the base are parallel and perpendicular to the equator, but if the inscribed base is square then the peaks are inside the spheroid. If I modify the base to a rectangle, and place the peaks on the equator anyway, I find that the octahedron is most regular, in the sense that the variance of the lengths of its twelve edges is the smallest, if the corners of the base are at reduced (i.e. parametric) latitude 45.096305; the geographic latitude, whose reduced latitude equals the geographic latitude of the former point, is 45.289021 (I use Earth flattening 1/297.8 because it was found in 1952, from purely ground-based measurements, that the flattening was 1/298.0 at 96W & 1/295.2 at 06W, and for other longitudes I can interpolate those minimum and maximum values). That is, this is the latitude of the "reduced" corners of the most nearly regular rectangular octahedron that can be inscribed orthogonally in the spheroid. At March 18.2, 2013 (see below) Deneb's apparent Declination is 133.15" north of this latitude. If I choose instead the octahedron whose base is square, the reduced latitude is negligibly different: arctan(297.8/296. = 45.096360.

Wikipedia and many other online sources agree that the geographic coordinates of the Serpent Mound of Ohio, are 39deg01'33.09"N, 83deg25'49.60"W; other sources give coordinates differing from these by only fractions of an arcsecond. From a snapshot from Google maps someone put online, I find that such coordinates correspond to points near the midpoint of the serpent curve. Due to the Ohio serpent's orientation, no point on the Ohio serpent is much closer than this, to the Ontario serpent.

Some online sources give coordinates for the Ontario headquarters building of the Hiawatha Tribe, or for (presumably the headquarters of) Serpent Mounds Park in Ontario, but I found only one source giving coordinates for the Serpent Mound of Ontario itself: www.megalithic.co.uk gives 44deg15'N, 78deg10'W; when I checked the 1:50,000 topographical map in the Iowa State Univ. map room the morning of Jan 16, I found that the megalithic.co.uk coordinates are those of the nearest large town, Keene. ISU's 1978 topographical map shows that the park is only 1km wide at its widest. It occupies a peninsula into Rice Lake. The online coordinates for the park, are about right. My ruler measurement to the marked historical site on the map, gives 44deg 12' 30" N, 78deg 09' 16" W, probably accurate to the nearest arcsecond. I discovered Feb 06, on a website called waymarking.com, that someone else has determined the coordinates as 44deg12.406' = 44deg12'24.36" N, 78deg09.295' = 78deg09'17.70" W. This is about 5.30" closer than mine, to the Ohio serpent mound.

On a spheroid flattened 1/297.8, the Lambert approximation which I copied from Wikipedia, gives geodesic distance 6.486895 deg between these points ("degrees" here signifies length the same as a degree measured along the equator of the spheroid) where I have subtracted 13.24" because of the compression that might occur if this part of Earth's crust were rotated to a NS orientation (see Appendix). The geodesic distance between points on the same meridian of the spheroid, and with geographic latitudes equal to the "octahedron corner" and the apparent Declination of Vega, most nearly equals the Ohio-Ontario distance, at Mar 18.2 2013, when it is only 27.17" less than the adjusted Ohio-Ontario geodesic distance.

If originally the Ontario & Ohio serpent mounds had lain on the same line of longitude, then by spherical trigonometry the pole could have been as far north as 63.17deg. If the Ontario serpent mound had been then at a latitude equal to today's current Declination of Deneb, then the pole would have been at about 63N, 4W, not far from one of Hapgood's positions (not the most recent Ice Age position, but an earlier Ice Age), 72N, 10E.

Appendix. Suppose a unit sphere flattens to an oblate spheroid with axis ratio 1-3*f, f = 1/297.8/3, while maintaining constant volume. The "b" axis of the spheroid is now approx. 1-2*f, and the "a" axis is 1+f to maintain volume. The length of a meridian decreases according to sqrt(((1-2*f)^2+(1+f)^2)/2) = 1-f/2. The north-south height of a typical small sector of Earth's surface, decreases according to a stretch factor approx. 1-f/2, i.e. a stretch of -f/2, a negative stretch, which could be called a compression. The width of the sector increases according to a stretch factor 1+f, a stretch of +f. By spherical trigonometry, the azimuth from the Ohio to the Ontario serpent mound, is th=35.52117deg (near the null compression orientation, arctan(1/sqrt(2)) ) E of N, so the stretch along this line is small, f*(sin(th)^2 - cos(th)^2 / 2 ) = 0.00635*f. When the pole shifted, the stretch increased by (0.00635-(-1/2))*f*6.490495deg (distance measured in "equatorial degrees") = 13.24".

Dr Joe, The ice age cycle requires a change in energy flows and at this time no one even has a guess of how much energy is involved making and melting all the ice. Why would a shift in the poles cause energy to shift from place to place on the surface of our planet?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />...Why would a shift in the poles cause energy to shift from place to place on the surface of our planet?
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

I think I'm ready to try to address this. The "heat of fusion" of water ice is about 80cal/gm. The "solar constant", i.e. the power of the Sun at Earth's distance, is 2.0cal/minute/ sq cm. Since the density of ice is only a little less than 1gm/ml, the Sun puts forth enough power to melt through a cm of ice in 40 miutes. This gives an idealized lower bound of 40*10^5 minutes * pi (to account for Earth's rotation) = 24 yrs to melt through a km of ice (Chicago is thought to have been under 2000 ft of ice during the most recent Ice Age). Really, it takes far longer than this, because much of the incoming solar energy is lost back into space, or carried away by winds and currents to the other places where ice is - allegedly - forming.

The same holds true for the reverse process, of latent energy release and ice deposition, but with yet another problem: if the wettest places on Earth have only about 100 inches (2.5m) of rainfall yearly, how could 1km of ice accumulate in less than 400 yrs? Polar regions tend to have relatively little precipitation: 10 inches/yr and 4000 yrs might be more like it.

Jim, is this related to your ideas?

The ice age cycle requires a shift in energy flows somewhere. Ice will form when the temperature is below 273 kelvin but only when water is involved. If the radiation law is correct a body at 273 kelvin radiates about 280 watts per square meter or so. The sun never provides that much energy anywhere near the poles and still the ice melts. They say greenhouse gas causes this melting. You hope maybe polar shifts can be the cause. I say the cause is mantle heat. It is the mantle that is causing the climate to change.