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

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

Originally Posted by Githyanki
This leads me to believe axis-wobbling doesn't really occur; am I right?
Yes. 'Axis-wobbling' induced gravitationally of gravitationally formed body is impossible.
It's easy to prove this mathematically.

2. Originally Posted by Alsor
Yes. 'Axis-wobbling' induced gravitationally of gravitationally formed body is impossible.
It's easy to prove this mathematically.
Perhaps you should take this to ATM, since it's quite solidly against the accepted answer.

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"Earth's axial tilt has only varied between about 22 and 24.5 degrees, because our relatively large Moon helps maintain a stable tilt."

It's incorrect - axis doesn't changes.
Orbital plane precesess only.

24.5 - 22 = 2.5 = 1.25 * 2;

Orbits inclination difference - Earth and Jupiter: 1.57 - 0.32 = 1.25

4. Originally Posted by Alsor
"Earth's axial tilt has only varied between about 22 and 24.5 degrees, because our relatively large Moon helps maintain a stable tilt."

It's incorrect - axis doesn't changes.
Orbital plane precesess only.

24.5 - 22 = 2.5 = 1.25 * 2;

Orbits inclination difference - Earth and Jupiter: 1.57 - 0.32 = 1.25

Dear Alsor, welcome to BAUT.
However, if you want to make claims that are different from mainstream science you will have to do it in the ATM (Against The Mainstream) section of the board.
An example would be your claim "axis doesn't change". Maybe you don't understand that the direction of the axis is meant, which is wel known to change. Then whatever calculation you put at the end of your message, you would have to explain way more that just some numerology. And your numbers are incorrect, inclination difference between Jupiter and the Earth is 1.31 degrees.

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I'm not interested working against any stream.

Orbits precession is simple calculation.

Maybe 1.31; there are other planets, and precession isn't simple - in one plane, but more complicated.

6. Posts above moved from this thread: http://www.bautforum.com/showthread....-Axis-Wobbling

Alsor, your claims are now under the ATM (Against the Mainstream) forum rules. Please check out the links in my signature.

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Originally Posted by Alsor
Yes. 'Axis-wobbling' induced gravitationally of gravitationally formed body is impossible.
It's easy to prove this mathematically.
So you will be providing this easy mathematical proof then?

8. Originally Posted by Alsor
Yes. 'Axis-wobbling' induced gravitationally of gravitationally formed body is impossible.
Please define "axis-wobbling" as you use it.

Originally Posted by Alsor
I'm not interested working against any stream.
You appear to either be using the term "axis-wobbling" in an unconventional way or you are arguing against well accepted ("mainstream") science.

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You known how calculate eqatorial bulge?
So, reverse this process and you find, what was lost in lunisolar theory.

http://en.wikipedia.org/wiki/Earth_tide
"The semidiurnal amplitude of terrestrial tides can reach about 55 cm at the equator which is important in GPS calibration and VLBI measurements."

10. Originally Posted by Alsor
You known how calculate eqatorial bulge?
So, reverse this process and you find, what was lost in lunisolar theory.
What was lost?

11. "Lunisolar theory"? The last time I heard that phrase, it was from somebody claiming we were orbiting Sirius, an idea that has a LOT of problems. Is this more of the same? If so, that would be extreme ATM and has been discussed before here.

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These centimeters of tidal deformations was overlooked.

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Yes, I know that speculation.
I resolved this problem without stars.

14. Originally Posted by Alsor
These centimeters of tidal deformations was overlooked.
By who? When? How? Why? In what context? What was the result of this "overlooking" and how is the result better when not "overlooked"?

Please provide full explanations of your claims.

(Also - it'd help if you quoted the post you are replying to, it makes the thread much easier to read.)

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## Heat Death of the Universe

Originally Posted by Alsor
Yes, I know that speculation.
I resolved this problem without stars.
Without stars? Amazing.

Does this mean without observations? Please refer to the signature line below, and provide us with peer reviewed supporting evidence.

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We have better observations, for example Moon phases and periods:
http://en.wikipedia.org/wiki/Moon

Orbital period T = 27.321582 d
Synodic period T' = 29.530589 d

Can we calculate orbital period of Earth?
Of course, we can:

1/1year = 1/T - 1/T' = 1/365.242124 d;

17. Originally Posted by Alsor
1/1year = 1/T - 1/T' = 1/365.242124 d;
And why not use the sidereal year?

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Originally Posted by pzkpfw
What was the result of this "overlooking" and how is the result better when not "overlooked"?
Actual results are well known: general gravitational anomalies and fantastic numerical coincidences.

Originally Posted by pzkpfw
Please provide full explanations of your claims
Lunisolar precession theory is wrong and superfluous.
Earth's axial precession does not exist.

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You did not understand.

This is result.
I don't use any year, but calculate it from data only.

20. Originally Posted by Alsor
You did not understand.

This is result.
I don't use any year, but calculate it from data only.
Which data?

21. Alsor, you will now present a full description of your idea about the "axis wobbling", including which data you used and which calculations you have made and why.
If your next answer is again a one-two liner, then you will receive an infraction.

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Originally Posted by grapes
Which data?
http://en.wikipedia.org/wiki/Moon

Orbital period T = 27.321582 d
Synodic period T' = 29.530589 d

1/1year = 1/T - 1/T' = 1/27.321582 - 1/29.530589 = 1/365.242124;
then: 1 year = 365.242124 d

Tropical year = ~365.24219 d

365.242124 - 365.24219 = -0.000066 d = -5.7024 seconds;

Sdereal year is longer 1200 s.

It's shorter about 6s than tropical year.

http://en.wikipedia.org/wiki/Gauss%E...Bonnet_theorem
Gauss-Benet theorem: flat curves have 2pi radians per cycle (closed path).
Earth's orbit is nearly flat - little precession only.

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Originally Posted by Grashtel
So you will be providing this easy mathematical proof then?
I'll tell you why there is no axial precession of planets (objects gravitationally bonded).

Gravitational tidal forces stretch the body, and then the principal axes change. Thus, the body is not rotating around the main axis, so precessional moment arises, proportional to the elongation and spin.

You can quickly calculate the approximate elongation of Earth, which balance the torque forces on the equatorial bulge and completely reset the precession.

precession: p = 2 f * W; hence: f = 0.5 P / W;
and elongation in meters is: h = f * R;

Data: W = 2pi / 1 day, p = 2pi/26000 years, R = 6370 km;
ie: h = 0.5 * 6370 km * 1 day / 26,000 years = 33.5cm.

As you can see just such a tidal deformations are measured (average), and this is not a random coincidence, but the condition of natural balance.

Moments of inertia of Earth are adjusted to a hypothetical precession with a period of 26,000 years. Real are likely to compatible with Chandler Wobble.

Greet!

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Originally Posted by Alsor
Actual results are well known: general gravitational anomalies and fantastic numerical coincidences.

Lunisolar precession theory is wrong and superfluous.
Earth's axial precession does not exist.
You sure about it? Mainstream science says otherwise.

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Originally Posted by macaw
You sure about it? Mainstream science says otherwise.
There, the calculations are incomplete: only calculated the torque of tidal forces and immediately precession, completely ignoring tidal deformations.

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Originally Posted by Alsor
There, the calculations are incomplete: only calculated the torque of tidal forces and immediately precession, completely ignoring tidal deformations.
MQ1: They are incomplete? Really? Then, how about you redid them including the tidal deformations? Let's see it.

27. Originally Posted by Alsor
There, the calculations are incomplete: only calculated the torque of tidal forces and immediately precession, completely ignoring tidal deformations.

Alsor, you were asked to explain fully what your idea about the "wobbling" is. Not some mysterious little bit of math that may or may not be just numerology.
This one sentence answer on macaw's comment is way inadequate for a real discussion.
I give you an infraction for not answering questions in an appropriate way.
Now, present your model fully.

28. Originally Posted by Alsor
There, the calculations are incomplete: only calculated the torque of tidal forces and immediately precession, completely ignoring tidal deformations.
I would be inclined to ignore something that is so slight in proportion to the size of the equatorial bulge upon which the solar and lunar gravity are acting to produce the precession-inducing torque. The tidal elongation is on the order of a meter or so, as compared with the roughly 20km difference between the equatorial radius and the polar radius. My educated guess is that any change from this component would be masked by the error bars we can reasonably expect with this lumpy, partially fluid planet. If anything I would expect it to increase the precession rate slightly rather than negate it.

Your series of equations in post 23 may look impressive to a novice with no background in physics, but you have done nothing to convince me that it has any basis in gravitational mechanics.

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Originally Posted by Hornblower
I would be inclined to ignore something that is so slight in proportion to the size of the equatorial bulge upon which the solar and lunar gravity are acting to produce the precession-inducing torque. The tidal elongation is on the order of a meter or so, as compared with the roughly 20km difference between the equatorial radius and the polar radius.
centrifugal acceleration: a_c = W ^ 2 * R = 0.034 [m / ss];
gravitational acceleration: g = GM / R ^ 2 = 9.8;
tidal acceleration: g_t = Gm / d ^ 2 * 2R / d = 1.11e-6 (Moon)

Equatorial bulge: dR = a_c / g * R = 22 km; [at the equator +1/3 dR, the poles -2/3 dR]
Tidal elongation: dr = 3 / 2 * g_t / g = 54 cm;

22km / 54cm = ~ 40000;
a_c / g_t = ~ 30000;

40000/30000 = 3 / 4 - This ratio results from the geometry (the difference between the oblate and elongated spheroids).

T = T_oblate + T_prolate = 0;
Deformation of 0.5m at 6000km is only apparently small. This is a global change that can not be ignored.
What is the energy of this movement?

You can ignore the mass movement of local rivers, avalanches, ocean waves, etc.
Summed together give zero global effect, as can be seen: the shape of the Earth does not change permanently in the scale of days or a few years.
---------

Formal proof.

Potentials:
P = P_g + P_t + P_c;
P_g - gravitational potential (self)
P_c - centrifugal potential (rotation)
P_t - tidal potential (external masses)

Gravitationally bound body has the same potential on the surface (equipotential surface: the forces are perpendicular to the surface, so the shape of the body is stable, the mass does not move).

This is a essential and necessary condition, used in the calculation of the equatorial bulge and tidal deformations.

Torque: T = r x F;
Force: F =-grad E;

We integrate on the volume of the body, namely:
dF =-grad E dV / V;
dF = m.da = -grad(P) dV * (m / V);

units of measure are arbitrary (shape and proportions are important), so we accept to simplify the notation: m = 1, V = 1, and we get:
dF =-grad(P) dV; (force density...)

Image:
$\vec{T}=\oint_{V}\vec{r}\times\vec{F}\;dV=\oint_{V}\nabla{P}\times\vec{r}\;dV=\\\oint_{V}\nabla\times(P\cdot\vec{r})\;dV=\oint_{S}(P\cdot\vec{r})\times\,d\vec{S}=P_s\cdot\oint_{S}\vec{r}\times\,d\vec{S}=0$

Explanations:
T = int r x F dV = int r x (-grad(P)) dV = int grad(P) x r dV;

but:
curl(f.r) = grad(f) x r + f curl(r) = grad (f) x r, (because: rot r = 0; f - any scalar func)

thus:
T = int curl(P.r) dV;

Then we use the vector identity:
int curl(F) dV = int F x dS; S - boundary surface of V

ie:
T = int (P.r) x dS = int P. r x dS

S is the equipotential surface, the potential P is constant on the whole integral surface - let the value: P_s;

T = P_s * int r x dS = P_s * 0 = 0;
[vector flux vector position through any closed surface is zero]

Check it out.

Precession of the Earth's axis has never really been observed.

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Originally Posted by Alsor
centrifugal acceleration: a_c = W ^ 2 * R = 0.034 [m / ss];
gravitational acceleration: g = GM / R ^ 2 = 9.8;
tidal acceleration: g_t = Gm / d ^ 2 * 2R / d = 1.11e-6 (Moon)

Equatorial bulge: dR = a_c / g * R = 22 km; [at the equator +1/3 dR, the poles -2/3 dR]
Tidal elongation: dr = 3 / 2 * g_t / g = 54 cm;

22km / 54cm = ~ 40000;
a_c / g_t = ~ 30000;

40000/30000 = 3 / 4 - This ratio results from the geometry (the difference between the oblate and elongated spheroids).

T = T_oblate + T_prolate = 0;
Deformation of 0.5m at 6000km is only apparently small. This is a global change that can not be ignored.
What is the energy of this movement?

You can ignore the mass movement of local rivers, avalanches, ocean waves, etc.
Summed together give zero global effect, as can be seen: the shape of the Earth does not change permanently in the scale of days or a few years.
---------

Formal proof.

Potentials:
P = P_g + P_t + P_c;
P_g - gravitational potential (self)
P_c - centrifugal potential (rotation)
P_t - tidal potential (external masses)

Gravitationally bound body has the same potential on the surface (equipotential surface: the forces are perpendicular to the surface, so the shape of the body is stable, the mass does not move).

Image:
$\vec{T}=\oint_{V}\vec{r}\times\vec{F}\;dV=\oint_{V}\nabla{P}\times\vec{r}\;dV=\\\oint_{V}\nabla\times(P\cdot\vec{r})\;dV=\oint_{S}(P\cdot\vec{r})\times\,d\vec{S}=P_s\cdot\oint_{S}\vec{r}\times\,d\vec{S}=0$

Explanations:
T = int r x F dV = int r x (-grad(P)) dV = int grad(P) x r dV;

but:
curl(f.r) = grad(f) x r + f curl(r) = grad (f) x r, (because: rot r = 0; f - any scalar func)

thus:
T = int curl(P.r) dV;

Then we use the vector identity:
int curl(F) dV = int F x dS; S - boundary surface of V
Stokes Theorem

ie:
T = int (P.r) x dS = int P. r x dS

S is the equipotential surface, the potential P is constant on the whole integral surface
No, it is NOT equipotential, see more below.

- let the value: P_s;

T = P_s * int r x dS = P_s * 0 = 0;
[vector flux vector position through any closed surface is zero]

Check it out.

Precession of the Earth's axis has never really been observed.
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.
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.
Sorry, the axis precesses: "E pur si preceso"

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