# Thread: [Alsor on Axis wobbling (or not)]

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Originally Posted by macaw
Stokes Theorem

No, it is NOT equipotential, see more below.

Thank you for posting the calculations but they are incorrect. The error lies in your assumption that is constant on the Earth spheroid. It isn't, see HERE.

P_g + P_c = const, in the calculation of the equatorial bulge rotating mass - an isolated mass that is immersed in a constant potential.

So actually it is like this:
P_g + P_c + P_t = const;

P_t - external potential - the other masses (Moon, Sun, etc.).
Considered the body is not insulated.

When we calculate the equatorial bulge, we can ignore the external potential P_t.

In contrast, when we calculate the tidal deformation, then you can take P_c = 0, and then: P = P_g + P_t = const.

Calculating torque and precession, we must take into account the final - the correct shape, i.e. the potential: P = P_g + P_c + P_t = const;

Originally Posted by macaw
While the sum is indeed constant, varies during the astronomical year, therefore is variable so you can't pull it from under the surface integral the way you tried.
P_t changes, but it still is: P = const (because the Earth's shape does not change in the scale of centimeters per month, or year!)

There are some oscillations ('nutation'), but they add up to zero (or almost) in certain cycles, approximately 1 day, month, year, 18 years, ...

Good day.

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Originally Posted by Alsor

P_g + P_c = const, in the calculation of the equatorial bulge rotating mass - an isolated mass that is immersed in a constant potential.
Yes, I have already pointed that out to you.

So actually it is like this:
P_g + P_c + P_t = const;
No, it isn't since is VARIABLE. I have already showed you proof of that.

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Originally Posted by Alsor

There are some oscillations ('nutation'), but they add up to zero (or almost) in certain cycles, approximately 1 day, month, year, 18 years, ...
You are inadvertently mixing two DIFFERENT effects (none of which is null) since wikipedia tells you clearly:

Originally Posted by WP
1. The combined action of the Sun and the Moon is called the lunisolar precession.
2. In addition to the steady progressive motion (resulting in a full circle in about 25,700 years) the Sun and Moon also cause small periodic variations, due to their changing positions. These oscillations, in both precessional speed and axial tilt, are known as the nutation. The most important term has a period of 18.6 years and an amplitude of less than 20 seconds of arc.

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Originally Posted by macaw
No, it isn't since is VARIABLE. I have already showed you proof of that.
Yes.
P_t is variable (because other masses are moving relative to Earth),
but: P = P_g + P_c + P_t = const on Earth (shape is conserved).

According to the current version is as follows:
The Earth spins and the Moon just stands in the distance.
Calculate torque by integrating the tidal potential in intact oblate spheroid (equipotential surface)!

There is already a precession - the static situation!

Then we average the effect over time - integrating along the orbit of the Moon.

The final result is incorrect by 100% - exactly!
Last edited by Alsor; 2011-Jun-19 at 10:29 PM.

5. Originally Posted by Alsor
Yes.
P_t is variable (because other masses are moving relative to Earth),
but: P = P_g + P_c + P_t = const (shape is conserved).

According to the current version is as follows:
The Earth spins and the Moon just stands in the distance.

Then we average the effect over time - integrating along the orbit of the Moon.
But surely the moon doesn't just stand there, it's distance from the Earth varies as this page shows, surely you can't just assume it's static?

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Originally Posted by macaw
You are inadvertently mixing two DIFFERENT effects (none of which is null) since wikipedia tells you clearly:
Precession is zero, and it is also the sum of these nutation.

Is clearly visible throughout the solar system.
Periods of the Moon I showed already.
Transits of Venus even better reveal the lack of precession.
The periods of all the planets are incorrect ... but that's another topic.

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Originally Posted by Garrison
But surely the moon doesn't just stand there, it's distance from the Earth varies as this page shows, surely you can't just assume it's static?
It's not my problem - in this way is calculated axial precession. Check out these calculations.

According to me the moon is moving rather slowly, so in fact it can be assumed in the calculation that does not move (gravity is very fast.)

The error lies in the fact that the surface of the Earth is equipotential - including the Moon, not without it!

Good day.

8. Originally Posted by Alsor
It's not my problem - in this way is calculated axial precession.
Check out these calculations.
I was responding to this statement from you:

The Earth spins and the Moon just stands in the distance.
Which as the perigee and apogee data, confirmed by laser retroreflector data, shows is not correct. If you were unaware of this it reflects badly on your underlying knowledge upon which your views and calculations rest. The moon is not simply a static body at some fixed distance from Earth.

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Originally Posted by Garrison
Which as the perigee and apogee data, confirmed by laser retroreflector data, shows is not correct. If you were unaware of this it reflects badly on your underlying knowledge upon which your views and calculations rest. The moon is not simply a static body at some fixed distance from Earth.
Yes. I talked about that: gravity is very fast, and:
"The error lies in the fact That the surface of the Earth is equipotential - Including the Moon, not without it"

10. Originally Posted by Alsor
Yes. I talked about that: gravity is very fast, and:
What is that even supposed to mean?

"The error lies in the fact That the surface of the Earth is equipotential - Including the Moon, not without it"
But that is based on this claim(my bold):

According to me the moon is moving rather slowly, so in fact it can be assumed in the calculation that does not move
Which is simply not substantiated and in fact seems to contradict the empirical data available.

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Originally Posted by Garrison
What is that even supposed to mean?
It is irrelevant in the context of tidal forces and the lunisolar precession theory.

Therefore, it is all classical mechanics - Newton.

Originally Posted by Garrison
But that is based on this claim(my bold):
"According to me the moon is moving rather slowly, so in fact it can be assumed in the calculation that does not move"

Which is simply not substantiated and in fact seems to contradict the empirical data available.
No. I use only field theory.

Torque and axial precession is calculated from the tidal acceleration, and not from changes of tidal acceleration!

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Originally Posted by Alsor
Yes.
P_t is variable (because other masses are moving relative to Earth),
but: P = P_g + P_c + P_t = const (shape is conserved).

You seem to be able to do vector calculus, so how can you stumble on such a simple algebraic exercise?

According to the current version is as follows:
The Earth spins and the Moon just stands in the distance.
Calculate torque by integrating the tidal potential in intact oblate spheroid (equipotential surface)!
This is not things work in this forum: you put forward a fringe theory and we get to question it.

There is already a precession - the static situation!

Then we average the effect over time - integrating along the orbit of the Moon.

The final result is incorrect by 100% - exactly!
I have no idea what you are talking about, your writing is totally incoherent.

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Originally Posted by Alsor
Precession is zero, and it is also the sum of these nutation.
Twice I have pointed out that experimental data disproves your claim, why do you persist in posting falsities?

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Originally Posted by macaw
Yes, it is a mistake. I meant that P on the surface is the same.

Originally Posted by macaw
I have no idea what you are talking about, your writing is totally incoherent.
The formula for torque in Wikipedia is calculated from a static configuration, ie P = const. So it is obvious that the torque (and averaged over orbits - time), and the final precession are wrong. Check it out.

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Originally Posted by Alsor
Yes, it is a mistake. I meant that P on the surface is the same.
MQ2: On the surface is the same thing, is variable so CANNOT be constant. What algebra rules do you use?

The formula for torque in Wikipedia is calculated from a static configuration, ie P = const.
I am pointing out the error in YOUR formula, so leave the wiki formula alone. I made the effort to follow your calculations and I pointed out the error in your calculations so I would like you to answer the specific questions I am asking you.

So it is obvious that the torque (and averaged over orbits - time), and the final precession are wrong. Check it out.

MQ3: How do you explain the FACT that experimental observation CONTRADICTS your calculations? Your calculations claim (incorrectly) zero precession, experimental observation shows that the precession is not zero. This means that your calculation is wrong. I pointed out where you made the error.

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Originally Posted by macaw
Twice I have pointed out that experimental data disproves your claim, why do you persist in posting falsities?

There are no experimental data.
Dots in the sky is definitely not enough, especially since Sirius is not subject to the mythological precession.

http://en.wikipedia.org/wiki/Sothic_cycle
This gives Sirius the unusual characteristic of appearing to stay stable relative to the equinox and solstices, and for the same reason, the helical rising (or zenith) of Sirius does not slip through the calendar (at the precession rate of about one day per 71.6 years), as other stars do.

17. Originally Posted by Alsor
It is irrelevant in the context of tidal forces and the lunisolar precession theory.

Therefore, it is all classical mechanics - Newton.
You made this claim that 'gravity is very fast' as if it somehow related to your theory so please explain what it means.

No. I use only field theory.

Torque and axial precession is calculated from the tidal acceleration, and not from changes of tidal acceleration!
But if the acceleration changes then the values you calculate would change, a failure to include such changes might explain why your results conflict with observational data.

18. Originally Posted by Alsor
There are no experimental data.
Dots in the sky is definitely not enough, especially since Sirius is not subject to the mythological precession.

http://en.wikipedia.org/wiki/Sothic_cycle

Modern astronomers now measure the rate of precession via radio telescopes fixed on distant quasars and a process known as Very Long Baseline Interferometry (VLBI) confirms the earth changes orientation to the stars at about 50.3 arc seconds p/y, equating to one complete precession of the equinox in about 25,700 years. Nonetheless, Sirius, due to its unusual characteristics, remains practically stationary making it the ideal marker for ancient Egyptian planning purposes.
That is experimental/observational data, if you choose to disagree with that is one thing(though you would need to explain why), to deny it exists is quite another.

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Originally Posted by Alsor
There are no experimental data.
Dots in the sky is definitely not enough, especially since Sirius is not subject to the mythological precession.

http://en.wikipedia.org/wiki/Sothic_cycle

The lunisolar theory of precession requires that the earth wobble enough to lose one complete rotation on its axis and one revolution around the sun (relative to the fixed stars) per precession cycle. Modern astronomers now measure the rate of precession via radio telescopes fixed on distant quasars and a process known as Very Long Baseline Interferometry (VLBI) confirms the earth changes orientation to the stars at about 50.3 arc seconds p/y, equating to one complete precession of the equinox in about 25,700 years.
This is the fourth time I am pointing out to you how observational data contradicts your claims. I am using your own link to prove you wrong. Dang! Garrison beat me to it!

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Originally Posted by macaw
MQ2: On the surface is the same thing, is variable so CANNOT be constant. What algebra rules do you use?
Equal on surface, not constant in time.

Originally Posted by macaw
I am pointing out the error in YOUR formula, so leave the wiki formula alone. I made the effort to follow your calculations and I pointed out the error in your calculations so I would like you to answer the specific questions I am asking you.
No errors, because I calculate in field theory - instantenous potentials.
In practice, slight variations are, of course, and hence the 'nutations', which we observe. Precession is not observed!

Originally Posted by macaw

MQ3: How do you explain the FACT that experimental observation CONTRADICTS your calculations?
The observations confirm my calculations, and mainly because I decided to improve the standard calculations (mindlessly replicated since Newton).

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Originally Posted by Alsor
Equal on surface, not constant in time.
No, it is VARIABLE on the surface, the values on the side opposite to the SUN are DIFFERENT from the values on the side facing the Sun. This is what creates non-zero torque. In ADDITION, also varies with time, so the non-zero torque is also variable in time.

Precession is not observed!
Repeating the same false claim that is contradicted by observational data doesn't make your claim true. It is just as false as the first time you made it.

The observations confirm my calculations, and mainly because I decided to improve the standard calculations (mindlessly replicated since Newton).

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Originally Posted by macaw;1903208"
Very Long Baseline Interferometry (VLBI) confirms the earth changes orientation to the stars at about 50.3 arc seconds p/y, equating to one complete precession of the equinox in about 25,700 years."
VLBI finds no changes in the Earth's orientation respect to the stars, but at best a motion of images of few distant stars (probably quasars).

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Originally Posted by Alsor
VLBI finds no changes in the Earth's orientation respect the stars, but at best a motion of images of few distant stars (probably quasars).
MQ4: Prove it.

While you do that , please answer MQ1-3 since you haven't answered them (evasion attempts don't count).

24. Originally Posted by Alsor

Very Long Baseline Interferometry (VLBI) confirms the earth changes orientation to the stars at about 50.3 arc seconds p/y, equating to one complete precession of the equinox in about 25,700 years."

VLBI finds no changes in the Earth's orientation respect to the stars, but at best a motion of images of few distant stars (probably quasars).
Your claim is the opposite of the quote says, and I remind you you chose the page it comes from as a reference. There is observational data that shows precession, equations that define it, please explain why all of this should be ignored to embrace your model? it isn't enough to create a model that rules out precession, you need to show it matches with observed reality.

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Originally Posted by Garrison
Your claim is the opposite of the quote says, and I remind you you chose the page it comes from as a reference.
I am only saying what in fact we register - the movement of the image.

Cosmology knows the various apparent phenomena such as superluminal motion, multiple images of the same object, distorted images of entire galaxies, various lenses, and many others, including perhaps some as yet unknown.
http://en.wikipedia.org/wiki/Caustic_(optics)

Originally Posted by Garrison
There is observational data that shows precession, equations that define it, please explain why all of this should be ignored to embrace your model? it isn't enough to create a model that rules out precession, you need to show it matches with observed reality.
Very good question.
Solar System after eliminating the Earth's axis precession (the tropical year = period of Earth's orbit) will function correctly - no unnecessary adjustments, and anomalies.

I made some preliminary computer simulations, and there is no chance for any expansion of the period of Earth's orbit about the 20 minutes - phases drift drastically. You can not jump over the elementary laws of geometry.

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Originally Posted by macaw
No, it is VARIABLE on the surface, the values on the side opposite to the SUN are DIFFERENT from the values on the side facing the Sun. This is what creates non-zero torque. In ADDITION, also varies with time, so the non-zero torque is also variable in time.
Prove formally that the mass (gravitationally bound) immersed in the field of inhomogeneous potential has no equipotential surface, and simultaneously it is in equilibrium (the shape does not change).

Also show the calculation of tidal deformations (prolate spheroid).

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Originally Posted by Alsor
Prove formally that the mass (gravitationally bound) immersed in the field of inhomogeneous potential has no equipotential surface, and simultaneously it is in equilibrium (the shape does not change).
This is already shown on the mainstream website pointed out to you several times.
It is not for us to prove to you mainstream science, it is for you to back up your ATM claims.

Also show the calculation of tidal deformations (prolate spheroid).
This has also been pointed out to you several times, though, under the rules I am not obligated to prove to you mainstream science. Having said that, please answer MQ1-4. Thank you.

28. Originally Posted by Alsor
Prove formally that the mass (gravitationally bound) immersed in the field of inhomogeneous potential has no equipotential surface, and simultaneously it is in equilibrium (the shape does not change).

Also show the calculation of tidal deformations (prolate spheroid).
Now would be a good for you to review the rules for this forum. They are linked in my signature line below. The burden of proof here is yours. Members are not obliged to defend the mainstream.

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Originally Posted by macaw
This is already shown on the mainstream website pointed out to you several times. It is not for us to prove to you mainstream science, it is for you to back up your ATM claims.
According to the elementary principles of field theory: the body in equilibrium has the equipotential on boundary surface. Deformation of body is the result of equalizing the potential at the surface: external + internal = uniform at surface).

Traditional Lunisolar precession model is based only on Newton's law, therefore, is not compatible with the more general field theory (Newton did not know full field theory).

-----

A. There are no tidal forces and rotation: a sphere - equipotential surface.

B. Tidal forces present: the sphere is no longer equipotential surface, so changes the shape, and thus formed an elongated spheroid - with the equipotential surface.

C. Rotation: after equalization of potentials, there is oblate spheroid - with the equipotential surface.

B + C = ?
Equipotential too!

30. Originally Posted by Alsor
According to the elementary principles of field theory: the body in equilibrium has the equipotential on boundary surface. Deformation of body is the result of equalizing the potential at the surface: external + internal = uniform at surface).
I think I'm starting to see where you're going wrong.

The Earth is exposed to multiple changing influences, which means the equipotential surface that the physical surface is trying to conform to is changing over time but since the Earth is not a fluid and is therefore rather unable to assume this idealized and dynamic shape instantly, your fundamental assumption is wrong.
At NO TIME is the physical surface of the Earth identical to the equipotential surface.

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