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Requiem for Relativity
14 years 9 months ago #23219
by shando
Replied by shando on topic Reply from Jim Shand
Have a look at "The Lost Book of Enki" ISBN: 1-59143-037-2
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14 years 9 months ago #23866
by nemesis
Replied by nemesis on topic Reply from
On page 162 of contactee Stan Romanek's book "Messages", there is
a chart showing a planetary alignment which is supposed to
signify a date. This chart shows a tenth planet. An astronomer
interpreted this as the plutino Eris (aka Xenia). Using a
computer program called TheSky he found the alignment corresponds
to the date Sept. 21, 2012. But this assumes that "planet 10" is
in fact Eris and not an unknown body like Barbarossa. I am going
to try to contact the publisher and see if the heliocentric
longitudes of the planets in the chart at that date are known. It
might be interesting to compare them to the estimated position of
Barbarossa in 2012.
a chart showing a planetary alignment which is supposed to
signify a date. This chart shows a tenth planet. An astronomer
interpreted this as the plutino Eris (aka Xenia). Using a
computer program called TheSky he found the alignment corresponds
to the date Sept. 21, 2012. But this assumes that "planet 10" is
in fact Eris and not an unknown body like Barbarossa. I am going
to try to contact the publisher and see if the heliocentric
longitudes of the planets in the chart at that date are known. It
might be interesting to compare them to the estimated position of
Barbarossa in 2012.
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14 years 9 months ago #23220
by Joe Keller
Replied by Joe Keller on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by nemesis</i>
<br />On page 162 of contactee Stan Romanek's book "Messages", ...the plutino Eris (aka Xenia). ...Sept. 21, 2012. ...I am going
to try to contact the publisher and see if the heliocentric
longitudes of the planets in the chart at that date are known. It
might be interesting to compare them to the estimated position of
Barbarossa in 2012.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Great! Please post the results here!
<br />On page 162 of contactee Stan Romanek's book "Messages", ...the plutino Eris (aka Xenia). ...Sept. 21, 2012. ...I am going
to try to contact the publisher and see if the heliocentric
longitudes of the planets in the chart at that date are known. It
might be interesting to compare them to the estimated position of
Barbarossa in 2012.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Great! Please post the results here!
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14 years 9 months ago #23867
by Joe Keller
Replied by Joe Keller on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by shando</i>
<br />Have a look at "The Lost Book of Enki" ISBN: 1-59143-037-2
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Thanks for the reference!
<br />Have a look at "The Lost Book of Enki" ISBN: 1-59143-037-2
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Thanks for the reference!
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14 years 9 months ago #23981
by Joe Keller
Replied by Joe Keller on topic Reply from
Update on asteroid axis computer program
I modified the program, to fit with the ten spherical harmonics:
n=0,m=0
n=2,m=0
n=2,m=1 (sin & cos)
n=3,m=1 ( " )
n=2,m=2 (sin & cos)
n=3,m=2 ( " )
This amounts to replacing m=3 in the previous program, with m=1, but avoiding the forbidden n=1 harmonics.
The previous program found, basically, the round solutions; that is, those with the smallest sum of squares, of coefficients divided by the coefficient of the constant term. I made the modified program find, basically, the principal axis rotation solutions; that is, the smallest positive integral over the Gauss sphere, of "K inverse" times sin^2(latitude). This amounts to finding the solutions with the sharpest poles; i.e., the solutions most resembling a prolate spheroid rotating about its (stable) axis of least moment of inertia.
This modified program gave the same answer: the rotation axis of 129 Antigone is within a few degrees of the ecliptic pole. Again, there is prograde/retrograde ambiguity (i.e., N or S ecliptic poles fit about equally well). This answer is different from those in the literature, which cluster at two other axes. So, from this test with Antigone, I can't yet say that I have a program that will find the pole reliably from only two lightcurves.
I modified the program, to fit with the ten spherical harmonics:
n=0,m=0
n=2,m=0
n=2,m=1 (sin & cos)
n=3,m=1 ( " )
n=2,m=2 (sin & cos)
n=3,m=2 ( " )
This amounts to replacing m=3 in the previous program, with m=1, but avoiding the forbidden n=1 harmonics.
The previous program found, basically, the round solutions; that is, those with the smallest sum of squares, of coefficients divided by the coefficient of the constant term. I made the modified program find, basically, the principal axis rotation solutions; that is, the smallest positive integral over the Gauss sphere, of "K inverse" times sin^2(latitude). This amounts to finding the solutions with the sharpest poles; i.e., the solutions most resembling a prolate spheroid rotating about its (stable) axis of least moment of inertia.
This modified program gave the same answer: the rotation axis of 129 Antigone is within a few degrees of the ecliptic pole. Again, there is prograde/retrograde ambiguity (i.e., N or S ecliptic poles fit about equally well). This answer is different from those in the literature, which cluster at two other axes. So, from this test with Antigone, I can't yet say that I have a program that will find the pole reliably from only two lightcurves.
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14 years 9 months ago #23224
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Another little note on something that might help us narrow down the speed of gravity. I think this relates to that lump of copper at absolute zero, which has electrons moving about in it! This is called the Unruh effect and a little u tube lecture on it can be found here. [url][/url]
What I think it should be is,
mc^2/k_b * ((1 - sqrt(1 - v^2 / c^2)) = barh a / 2pi c k_b
Where v is the Fermi velocity of a material
k_b is the Boltzmann constant
a = acceleration
What I think it should be is,
mc^2/k_b * ((1 - sqrt(1 - v^2 / c^2)) = barh a / 2pi c k_b
Where v is the Fermi velocity of a material
k_b is the Boltzmann constant
a = acceleration
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