Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
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Jim

1805 Posts

Posted - 16 Feb 2012 :  13:08:23  Show Profile  Reply with Quote
Dr Joe, You are the first person to have a favorable comment on my views about the ice age cycle and climate issues. It is quite clear to me the cycle involves a flux of ~10E14 watts just to make(and unmake) the ice. This energy has to come from the Earth itself from the mantle and easily explains everything about the ice cycle and climate. The mantle flux has been ignored and clearly it is vital to everything on the surface of our planet.
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Joe Keller

USA
944 Posts

Posted - 17 Feb 2012 :  15:09:42  Show Profile  Reply with Quote
The cause of the Ice Ages

As described in Goldstein's "Classical Mechanics" sec. 5.6, or the similar chapter in Synge's mechanics text, a solid spheroid might rotate, not about any of its principal axes, but rather about a body axis which slowly revolves around the pole of the spheroid. For Earth, the axis might revolve in a period of 297 days, because Earth is flattened by 1/297.

If the circle described on the spheroid by the revolving axis, is far from the spheroid's pole, the spheroid will be much less flattened. The actual spheroid, will be the average of all the spheroids that there would be if the axis stayed at one point. Elementary integral calculus shows that if the spheroid is flattened by 1/297 when it rotates about a fixed axis, but the circle of axis revolution is 20.6deg from the spheroid's pole (i.e. latitude 69.4deg), then the average spheroid is flattened about 1/365.

With this amount of flattening, the axis revolves with period one year. So the axis is always over, say, Canada in December and over Siberia in June. Canada then has an arctic winter; but though it is at temperate latitude in summer, the insolation is about the same as an arctic summer. (Elementary integral calculus shows that for an axis tilt of 23.45deg, the summer solstice insolation at latitude 90 - 2*20.6 = 48.8deg, is 8% less than at the pole.) Basically Canada gets an arctic winter and a temperate summer, adding up to an arctic amount of insolation. Likewise Siberia gets a temperate winter and an arctic summer, adding up to a temperate amount of insolation. The axis doesn't exactly move to Canada; rather, it moves to Canada when it counts, namely every winter.

My addition to Goldstein's or Synge's text, is to realize that Earth's flattening becomes somewhat less when the axis revolves yearly, too quickly for Earth to adjust. Then the usual spheroidal shape is averaged over the year and is less flattened. This also affords an estimate of the present latitude line along which the axis will revolve.

I don't know what instigates such a change, but here are some possible clues:

1. In December this year (2012) the New Moon occurs very near perigee, an unusually close perigee, also near Earth's perihelion. Luna's Declination then (either at the New Moon or at the perigee or at the extreme Declination, all of which occur within an interval of 9h) is -20.9, differing only 0.3 degree from the 20.6 estimate found above. This might trigger the change somehow.

2. The ratio 297/365 = 0.8137, where the last one or two digits are not significant. The ratio of the Mayan Long Count to the Barbarossa period (discussed by me above on this thread) is 5125.257/6339.36 = 0.8085. These numbers are close to the number (also discussed by me above on this thread) for which the ratios of the magnitudes of the Legendre polynomials tend to be whole powers of 2.

Edited by - Joe Keller on 17 Feb 2012 15:36:48
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Jim

1805 Posts

Posted - 17 Feb 2012 :  20:51:14  Show Profile  Reply with Quote
Dr Joe, I order to understand why the ice cycle exists the energy required needs to be explored. It is grasping at straws to accept the motion of Earth to explain why energy flows in one direction for a while and then reverses the flow direction. On average to Earth always is one AU from the sun and that indicates the solar flux is constant. The real cause is much better understood by accepting as fact or possibility the mantle of Earth is forcing this cycle as well as some climate changes.
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Joe Keller

USA
944 Posts

Posted - 19 Feb 2012 :  19:01:13  Show Profile  Reply with Quote
Hapgood's estimated pole positions

"Hapgood's examples of recent locations for the North Pole include Hudson Bay (60N, 73W) , the Atlantic Ocean between Iceland and Norway (72N, 10E) and Yukon (63N, 135W)."

- Wikipedia article, "Pole shift hypothesis"

The mean latitude of these points is 65 = 90 - 25 +/- SEM 3.6, vs. my 90 - 20.6 = 69.4 predicted. If I fit a circle (neglecting Earth's oblateness) to these three Hapgood pole points, the center of the circle (which I will call the "Hapgood center") is lat 80.9N, long 90.1W. The opening angle (one might say half-angle) of the cone with vertex at Earth's center and through this circle, is 21.5deg, in good agreement with my calculated 20.6deg (see my previous post) for the circle of yearly axis revolution in force-free rotation.

According to the U.S. government data at

http://www.ngdc.noaa.gov/geomag/data/poles/NP.xy

the geomagnetic N pole (i.e. equivalent dipole)(not quite the same as the "magnetic" a.k.a. "magnetic dip" pole) passed nearest the Hapgood center in late 1999, being 3.0deg "West by South" (i.e. azimuth ~255deg) of the Hapgood center at 2000.0AD. It was also during 1999, that the rate of geomagnetic pole movement suddenly increased to its present high value: from 1990 through 1998, the movement was never more than 0.0065 radian/yr (much faster than earlier in the 19th & 20th centuries or even in the early 1990s) but from 1999 through 2009, never less than 0.0083 rad/yr, though trending slightly downward from the fastest motion, 0.0113 rad/yr, which is the rate that occurred during 1999.

When I randomly choose 5 of the 21 geomagnetic pole points, 1990.0-2010.0, excluding the 2000.0 point which is taken as the origin, and (to get good spacing) requiring that no pair of points be from adjacent years, I find that 80% of the time, the fitting conic is an hyperbola rather than ellipse, using the gnomonic (i.e. azimuthal from Earth's center) projection to map the geomagnetic pole points onto a plane. So, the "Hapgood center" is a plausible repulsive focus (i.e. focus on the convex side of the curve) of the geomagnetic pole path.

*********

Venus' rotation period: another Mayan Long Count correlation

Tne new article by NT Mueller et al, Icarus 217:474, 2012, correlates 1990-1992 Magellan probe data with 2006-2008 Venus Express probe data, to find Venus' sidereal rotation period as 243.023 +/- 0.002 d, averaged over 16 yr. Mueller mentions in his abstract that the Magellan probe value was 243.0185 +/- 0.0001; I read in the recent [London] Daily Mail article, that this actually was based on four years of Venus' rotation.

If the variation is mainly real, then the values should be weighted according to the duration observed:

(243.0185*4 + 243.023*16)/20 = 243.0221 (this is within Mueller's error bars)

and this best estimate shows that

Mayan Long Count / Venus rotation period
= 360d*5200/243.0221d = 7703.003

where the whole number 7703 is a prime number.

Edited by - Joe Keller on 23 Feb 2012 14:34:52
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Joe Keller

USA
944 Posts

Posted - 24 Feb 2012 :  18:51:33  Show Profile  Reply with Quote
Shift to a pole with yearly torque-free precession is energetically possible for Earth

(This post has been extensively rewritten two months after originally posted. I corrected a significant typographical error in my original computer program. Also I have, perhaps, slightly increased the accuracy of my analysis, by considering Earth's core/mantle structure; I had discussed my analysis with Dr. Ken Driessel, an expert in classical mechanics, who warned that the core/mantle structure might be significant.)

Using data from the 1975 Columbia Encyclopedia (Fig. A, article "Earth") I sometimes will approximate Earth as having a liquid core (the small solid core inside this is without effect) of density 11.0 gm/cm^3 and radius 3500km, and a mantle of density 4.425 gm/cm^3, so the entire Earth has radius 6400km and average density 5.5 gm/cm^3. When Earth undergoes torque-free precession, with its spheroid axis, say, 30deg different from its rotation axis, but the amount of spheroidal flattening about the same, then surface points will find themselves as much as 1/2 radian nearer or farther from the pole than is proper for the amount of bulge at that point. At temperate latitudes this amounts to +/- 6000/300/2 = 10 km of bulge. If this were compensated only by ocean movement, then at most 10 / 2.5 = 4 km of increased ocean depth suffices to equalize surface potential, because the density of water (=1) is almost negligible compared to the average density of Earth (=5.5) and according to the textbook formula (e.g., A. S. Ramsey, "Newtonian Attraction", sec. 7-42) elevation away from an unchanging spheroid is 2.5x as effective, at changing a point's potential, as is elevation caused by additional bulging of the spheroid under the point. Generally, stratigraphic evidence of thousands of yearly 4 km = 13,000 ft rises in ocean level, at temperate latitudes, is lacking. So, the rise might mainly have been provided by movement of crustal rock (density ~2.2) or hydrocarbon layers (density < 1). Thus my theory differs from Hapgood's: he required ~3000km horizontal motions of the crust, to cause pole shift; I require only smooth ~10km vertical motions of it, to reduce flooding when torque-free precessional pole shift occurs (atmospheric pressure would change little, because the atmosphere also seeks the equipotential surface).

Earth's dense core causes Earth's "actual oblateness" k0=1/297, to differ from Earth's "equivalent oblateness" k=1/304, i.e. the oblateness of a homogeneous spheroid having Earth's mass, radius and moments of inertia:

k = ( 6400^5 * 4.425*k0 + 3500^5 * (11.0-4.425)*1/462 ) /
(6400^5 * 4.425 + 3500^5 * (11.0-4.425))

and the converse formula giving k0 from k, is also linear. The core's bulge dr/r = 1/462 is found by equating omega^2 * r^2 / 2 to G*11.0*4/3*pi * r^3/r^2 * dr. (Likewise Earth's bulge would be 1/231 if Earth were homogeneous with density 11.0/2 = 5.5; the actual bulge 1/297 is less, because as discussed above, movement away from a dense core, is more efficient at equalizing potential, than is distortion of a homogeneous spheroid.)

From Earth's "equivalent oblateness" and the online NASA Fact Sheet figure, 8.034*10^44 in cgs units, for Earth's polar-axis moment of inertia, I find the moment of inertia for a sphericized Earth. Then by subtracting the moment of inertia of the core, I find the moment of inertia, i0, for the sphericized mantle. Without correcting oblateness, only subtracting the core's part, I find Earth's mantle's kinetic energy of rotation.

Conservation of energy and of angular momentum, give two equations relating the spheroidal mantle's large & small moments of inertia, i3 & i1, resp., and its spin components parallel & perpendicular to the spheroidal axis, wz & A*wz, resp. (see Goldstein, "Classical Mechanics", sec. 5-6). Because the oblateness is small, constant volume implies approximately i3 + 2*i1 = 3*i0 (I used an exact formula for i1 as a function of i0 & i3, for a homogeneous spheroid, but the exact result hardly differs from that found from the simple formula). The fourth needed equation is (i3 - i1) / i1 = (frequency of tropical year) / wz (see Goldstein).

The conservation of energy equation which I adopt from Goldstein, requires knowledge of the energy liberated as Earth assumes a more spherical shape; this must be added to the kinetic term. I use Earth's gravitational self-energy 2.4884*10^39 erg (Marchenko & Zayats, Stud. Geophys. Geod. 55:35-54, 2011, "Conclusions", p. 52, based on the "piecewise continuous PREM density model"). If Earth were a homogeneous spheroid, the self-energy due to oblateness would equal the total self energy, times 4/5*(k/3)^2, seen by expanding

1 / sqrt((x*(1+k0/3))^2+(y*(1+k0/3))^2+(z*(1-2*k0/3))^2)

in k0, and averaging on the sphere x^2 + y^2 +z^2 = 1. Estimating that 17.9% of Earth's self-energy is core-core self energy, and assuming that the oblateness of the core does not change, I proceeded by lessening the amount given by the above formula, by 17.9%.

The self-energy, above, released by decreasing oblateness, becomes doubled because Earth isn't really incompressible, but instead resembles an adiabatic monatomic gas P * V^gamma = const, where gamma=5/3 and V is proportional to r^3; P = P0/r^4, because both the gravitational force and the reciprocal of the area to which it is applied, are proportional to r^2. Thus P0 * r = const. When decreased oblateness decreases interparticle distances by, say, 1%, then forces and pressure and P0 increase 2%, so r must decrease by 2%, giving in our situation another 2*4/5*(k/3)^2 term of liberated energy. According to the virial theorem, half this becomes increased thermal motion of the particles, and half is available mechanically, giving a grand total (1 + 2/2) * 4/5*(k/3)^2 (times (1 - 0.179), to account for the univolvement of Earth's core in the torque-free precession).

By time-averaging the centrifugal potential at pole and equator, or else by time-averaging the shapes of the equilibrium spheroids, I find that for torque-free precession at an angle alpha, Earth's oblateness becomes less by a factor 1 - 1.5*sin^2(alpha). If I multiply the liberated self-energy by a factor of 1.022, i.e. only 2% different from my estimate, I find that:

1. Torque-free precession with period one year, occurs for alpha = 18.90deg; i.e. the pole moves along 71deg latitude. Hapgood's determinations of the last three Ice Age poles, average latitude 65.0 +/- SEM 3.6, range 60-72deg. So the agreement with Hapgood's poles is acceptable.

2. Earth's actual oblateness agrees with the averaging formula 1 - 1.5*sin^2(alpha), discussed above.

In other words, for torque-free precession to occur at latitudes other than Hapgood's determination for Ice Age poles, Earth's oblateness must assume an implausible value. Only when torque-free precession occurs along the 71st parallel, is Earth's oblateness what it should be according to the law of centrifugal force. Furthermore the energy required for this, is just what I estimate from the two equal terms, due to Earth's loss of oblateness and to Earth's resulting slight shrinkage as a monatomic adiabatic gas.

A good approximation, is simply to solve 1 - 1.5*sin^2(alpha) = 304/365, for alpha, where Earth's equivalent oblateness, 1/304, is used for better accuracy than Earth's actual oblateness, 1/297. The answer, 19.50deg, agrees well with the result of the more detailed calculation, involving Earth's core, above.

A wandering pole at a large precession angle, would give the sun a more dramatic motion: resembling our temperate summers but our arctic nights. This makes some ancient statements about solar motion seem less exaggerated.

Edited by - Joe Keller on 06 May 2012 18:26:00
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Jim

1805 Posts

Posted - 29 Feb 2012 :  22:19:52  Show Profile  Reply with Quote
Dr Joe, I don't pretend to understand your thesis, but I try and now I wonder what you mean by gravity energy of several kinds. Gravity is a force in my world having nothing in common with energy. Can you explain how you get energy from gravity?
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Joe Keller

USA
944 Posts

Posted - 08 Mar 2012 :  15:36:18  Show Profile  Reply with Quote
quote:
Originally posted by Jim

...energy from gravity?



This is my cue to restate my hypothesis in everyday language. When a spinning skater pulls in his arms, that requires energy (the chemical energy he gets from his food). The energy from pulling in the arms, becomes the increased kinetic energy of rotation.

Likewise, if Earth - either with no change in shape, or with a change toward being more perfectly spherical - rotates about an axis that is someplace away from the pole of the oblate spheroid, then the mass, of the Earth, is, on the average, slightly closer to the axis of rotation. Like the skater, the Earth must rotate slightly faster if the angular momentum is to remain constant. It turns out that the kinetic energy of such rotation is slightly greater than before. Where does this kinetic energy come from? It can come from the "falling" of the Earth into a slightly rounder shape.

If one fixed axis were replaced by another fixed axis, Earth eventually would become an oblate spheroid congruent to the old oblate spheroid, but with a different pole, a pole at the new rotation axis. But under my hypothesis, the axis precesses. So, the flattening effect of each axis position is averaged, and the average of all these oblate spheroids, is a spheroid that is less oblate and with the same pole (geometric pole, not rotation pole) as before. As Earth falls into a more spherical shape, the gravitational potential energy released, becomes the increased kinetic energy of rotation that is needed to conserve angular momentum and make the new state of affairs possible.

It turns out that there is enough gravitational potential energy released, to establish the new state as a precession of Earth's axis along the 69th parallel. This is in good agreement with the position of the three last Ice Age poles according to the geological estimates collected by Hapgood; and in good agreement with the precession rate needed to synchronize with Earth's tropical year, thereby having a climatic effect similar to an outright new fixed axis. (Due to conservation of momentum, the axis is always about the same direction in space - except for astronomical precession - but Earth shifts under the axis so that the position of the axis, relative to Earth, changes grossly, under my hypothesis.)

Earth's area exceeds the area of a sphere of equal volume, by 2.02 parts per million; see oblate spheroid area formula, in CRC Math Tables mensuration formulas, b=a*296/297, e=sqr(1-(b/a)^2). If Earth were less oblate - the oblateness I predict for the torque-free axis precession mode with 1/365 flattening - the excess area would be only 1.34 ppm. So Earth would find itself with an excess skin of 0.68 ppm. This corresponds to a dike 68 cm wide along one edge of a square 1000km on a side, for example. It would entail a significant, but believable and not unprecedented amount of seismic activity.

Edited by - Joe Keller on 08 Mar 2012 15:56:15
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Larry Burford

USA
2064 Posts

Posted - 08 Mar 2012 :  16:01:52  Show Profile  Reply with Quote
IOW, this is not enough to account for a Dec 21, 2012 event? But it might be enough to rid us of California.

I realize that you have only calculated area changes above, but if this dike turned out to be 1000 meters tall instead of 68 cm tall (dikes are seldom taller than they are wide, and you did not specify a height so I just made a guess), the implications would likely be a little different. Any idea which it is more likely to be?
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Jim

1805 Posts

Posted - 08 Mar 2012 :  20:49:29  Show Profile  Reply with Quote
Dr Joe, Thank you for this detailed description about how the surface of Earth excess area due to angular momentum. I don't see how this gets energy enough to cause ice to form and remain for 90,000 years or so. I assume that is way you did this--
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Joe Keller

USA
944 Posts

Posted - 12 Mar 2012 :  16:52:57  Show Profile  Reply with Quote
quote:
Originally posted by Larry Burford

...if this dike turned out to be 1000 meters tall...



Consider many concentric spheroids all with the same flattening, which all then become spheres without changing their internal volumes. All of them will have the same proportion of excess area, i.e. wrinkled skin, but these wrinkles needn't be in the same place. The height of the "dike", i.e. the height of the wrinkle in the uppermost spheroid, depends on the mechanical properties of the material and the strength of the connection between the spheroids. It needn't be very high. My method of calculation only tells me the area of the skin overlap, not its depth.
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Joe Keller

USA
944 Posts

Posted - 12 Mar 2012 :  17:52:58  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 31 Mar 2012 :  18:34:21  Show Profile  Reply with Quote
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.

Edited by - Joe Keller on 30 Apr 2012 16:17:35
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Joe Keller

USA
944 Posts

Posted - 20 Apr 2012 :  21:00:50  Show Profile  Reply with Quote
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).

Edited by - Joe Keller on 21 Apr 2012 12:22:41
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Joe Keller

USA
944 Posts

Posted - 21 Apr 2012 :  12:43:52  Show Profile  Reply with Quote
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).

Edited by - Joe Keller on 30 Apr 2012 20:24:53
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Jim

1805 Posts

Posted - 01 May 2012 :  14:10:59  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 02 May 2012 :  23:55:12  Show Profile  Reply with Quote
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.

Edited by - Joe Keller on 05 May 2012 18:46:07
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Jim

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Posted - 03 May 2012 :  15:57:46  Show Profile  Reply with Quote
Dr Joe, This method using words is much more interesting than all the math stuff. The wobble you are studying is relative to what? I guess the sun, but, two different reference systems are in effect here -right? One system is Earth based with equator base units and the other system is ecliptic based. So, your wobble is projected against the ecliptic and there is where the reaction should be-right?
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Joe Keller

USA
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Posted - 07 May 2012 :  16:25:28  Show Profile  Reply with Quote
quote:
Originally posted by Jim

...The wobble you are studying is relative to what? ...two different reference systems are in effect here... One system is Earth based with equator base units and the other system is ecliptic based.


Jim's summary of the coordinate situation is accurate.

In "lunisolar precession" (i.e. the 26,000 yr precession) Earth's spin angular momentum vector moves in a big cone relative to ecliptic latitude and longitude (i.e. coordinates based on the Sun). In "torque-free precession" Earth's spin angular momentum vector is constant and Earth's spin axis changes only very slightly (this slight change is due to Earth's slight spheroidal shape which is askew during torque-free precession). Relative to Earth coordinates (Earth latitude and longitude) during torque-free precession, the pole moves in a big cone; that is, different parts of Earth (along a line of latitude) come to lie under the rotation pole, which is itself almost exactly fixed in space.

Mainstream experts on the Ice Ages have remarked that cool (not necessarily freezing) summers are especially effective for the accumulation of ice. That is, more ice can accumulate, when summers are cooler, even if the total yearly insolation is somewhat greater. In the approximation of a circular Earth orbit, I find that for torque-free precession at 19deg, the "pole" (i.e. the Earth point under the celestial pole at the winter solstice, so that at the winter solstice, it becomes the north pole) gets 28% more insolation during the entire year than would a real pole (because it gets considerable insolation at the equinoxes when a real pole would get none) but 8% less insolation, than a real pole, at the summer solstice.
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Jim

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Posted - 07 May 2012 :  19:26:58  Show Profile  Reply with Quote
Dr Joe, If you rough in the energy our planet gets from the sun you will see the polar zones on average get about 70 watts per square meter and the equator ~250w/m^2. The average temp at the poles is ~260K +/-? At 260K the poles are radiating 260w/m^2. So, the solar input is less than 25% ot the energy used at the poles even when ice is a mile thick. The idea that orbital details can explain this is not very logical.
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Joe Keller

USA
944 Posts

Posted - 08 May 2012 :  17:46:29  Show Profile  Reply with Quote
quote:
Originally posted by Jim

Dr Joe, If you rough in the energy our planet gets from the sun you will see the polar zones on average get about 70 watts per square meter and the equator ~250w/m^2. The average temp at the poles is ~260K +/-? At 260K the poles are radiating 260w/m^2. So, the solar input is less than 25% ot the energy used at the poles even when ice is a mile thick. The idea that orbital details can explain this is not very logical.



Hi Jim,

Thanks for this timely input of information.

- Joe Keller
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