Re: Ice Ages on FE
« Reply #20 on: June 04, 2019, 09:21:39 PM »
You don't get to just shift the burden of evidence onto me.

But I do, if you cannot explain the seismic waves anomalies.

I'm still waiting for any evidence for your claim.

The evidence is very clear.

"The area of land in the northern hemisphere of the earth is to the area of land in the southern hemisphere as three is to one.

The mean weight of the land is two and three-quarter times heavier than that of water; assuming the depth of the seas in both hemispheres to be equal, the northern hemisphere up to sea level is heavier than the southern hemisphere, if judged by sea and land distribution; the earth masses above sea level are additional heavy loads - we include here all the mountains/hills.

But this unequal distribution of masses does not affect the position of the earth, as it does not place the northern hemisphere with its face to the sun. A “dead force” like gravitation could not keep the unequally loaded earth in equilibrium. Also, the seasonal distribution of ice and snow, shifting in a distillation process from one hemisphere to the other, should interfere with the equilibrium of the earth, but fails to do so."


The northern hemisphere has a greater mass than its southern counterpart.

The unequally loaded perfect oblate spheroid (first four layers) DEFIES the law of attractive gravity.

It should rotate with the northern hemisphere facing the sun.

At present, the RE has an unequal distribution of mass: the northern hemisphere has more mass than the southern hemisphere.


It is the unequal mass distribution of the hemispheres upon a perfect oblate spheroid which defies the law of attractive gravity.

Since the northern hemisphere has more mass than its southern counterpart, we have a clear and definite DEFIANCE of the law of attractive gravity.

You can say "lol i win" all you like

Of course I win, since you cannot explain the seismic waves anomalies.

if you can't back up your claim with any evidence

My evidence is very clear and also very direct.

You made a request for torque calculations thinking that it might save your day; it did not.

Now you are going to have to explain why the distribution of the continents on a spherical Earth defies the law of universal attraction.


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Offline Tim Alphabeaver

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Re: Ice Ages on FE
« Reply #21 on: June 04, 2019, 09:55:58 PM »
My evidence is very clear and also very direct.
Okay, maybe I'm not being clear enough.

A “dead force” like gravitation could not keep the unequally loaded earth in equilibrium.
Source?

The unequally loaded perfect oblate spheroid (first four layers) DEFIES the law of attractive gravity.
Source?

It should rotate with the northern hemisphere facing the sun.
Source?

It is the unequal mass distribution of the hemispheres upon a perfect oblate spheroid which defies the law of attractive gravity.
Source?

It is the unequal mass distribution of the hemispheres upon a perfect oblate spheroid which defies the law of attractive gravity.
Source?

Of course I win
Source?

You made a request for torque calculations thinking that it might save your day; it did not.
Source?

All of these claims are just you stating a thing to be true.
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Re: Ice Ages on FE
« Reply #22 on: June 05, 2019, 04:51:17 AM »
Source?

As if you didn't know.

Here is your source:



BASIC NEWTONIAN PHYSICS: we have a center of gravity which is located ABOVE THE EQUATOR, given the fact that the northern hemisphere has more mass than the southern hemisphere. Then, the accepted law of universal gravitation tells us that the Earth should revolve facing the Sun with its North Pole.


According to this formula, the position of the centre of gravity varies according to the shape of the object.

"The area of land in the northern hemisphere of the earth is to the area of land in the southern hemisphere as three is to one.

The mean weight of the land is two and three-quarter times heavier than that of water; assuming the depth of the seas in both hemispheres to be equal, the northern hemisphere up to sea level is heavier than the southern hemisphere, if judged by sea and land distribution; the earth masses above sea level are additional heavy loads - we include here all the mountains/hills.

But this unequal distribution of masses does not affect the position of the earth, as it does not place the northern hemisphere with its face to the sun. A “dead force” like gravitation could not keep the unequally loaded earth in equilibrium. Also, the seasonal distribution of ice and snow, shifting in a distillation process from one hemisphere to the other, should interfere with the equilibrium of the earth, but fails to do so."


The northern hemisphere has a greater mass than its southern counterpart.

The unequally loaded perfect oblate spheroid (first four layers) DEFIES the law of attractive gravity.


Remember, if you want torque calculations, you must explain the seismic waves anomalies. If you cannot explain the seismic wave anomalies, that means that there is no molten outer core.

Very simple: the northern hemisphere is heavier than the southern hemisphere. Thus, according to the formula put forth by Newton, the force of gravity should place the North Pole facing the Sun.


Here are more gravitational anomalies which defy your request for torque calculations as well as Newton's supposed law of universal gravitation.

Mountainous masses do not exert the gravitational pull expected by the theory of gravitation. The influence of the largest mass on the earth, the Himalaya, was carefully investigated with plumb line on the Indian side. The plumb line is not deflected as calculated in advance. The attraction of the mountain-ground thus computed on the theory of gravitation, is considerably greater than is necessary to explain the anomalies observed. This singular conclusion, I confess, at first surprised me very much. (G. B. Airy.) Out of this embarrassment grew the idea of isostasy. This hypothesis explains the lack of gravitational pull by the mountains in the following way. The interior of the globe is supposed to be fluid, and the crust is supposed to float on it. The inner fluid or magma is heavier or denser, the crust is lighter. Where there is a mountainous elevation, there must also be a protuberance beneath the mountains, this immersed protuberance being of lesser mass than the magma of equal volume. The way seismic waves travel, and computations of the elasticity of the interior of the earth, force the conclusion that the earth must be as rigid as steel; but if the earth is solid for only 2000 miles from the surface, the crust must be more rigid than steel. These conclusions are not reconcilable with the principle of isostasy, which presupposes a fluid magma less than 60 miles below the surface of the earth. There remains a contradiction between isostasy and geophysical data.

Over the oceans, the gravitational pull is greater than over the continents, though according to the theory of gravitation the reverse should be true; the hypothesis of isostasy also is unable to explain this phenomenon. The gravitational pull drops at the coast line of the continents. Furthermore, the distribution of gravitation in the sea often has the peculiarity of being stronger where the water is deeper. In the whole Gulf and Caribbean region the generalization seems to hold that the deeper the water, the more strongly positive the anomalies.

As far as observations could establish, the sea tides do not influence the plumb line, which is contrary to what is expected. Observations on reservoirs of water, where the mass of water could be increased and decreased, gave none of the results anticipated on the basis of the theory of gravitation.


In 1981 a paper was published showing that measurements of G in deep mines, boreholes, and under the sea gave values about 1% higher than that currently accepted. Furthermore, the deeper the experiment, the greater the discrepancy. However, no one took much notice of these results until 1986, when E. Fischbach and his colleagues reanalyzed the data from a series of experiments by Eotvos in the 1920s, which were supposed to have shown that gravitational acceleration is independent of the mass or composition of the attracted body. Fischbach et al. found that there was a consistent anomaly hidden in the data that had been dismissed as random error. On the basis of these laboratory results and the observations from mines, they announced that they had found evidence of a short-range, composition-dependent fifth force. Their paper caused a great deal of controversy and generated a flurry of experimental activity in physics laboratories around the world.

The majority of the experiments failed to find any evidence of a composition-dependent force; one or two did, but this is generally attributed to experimental error. Several earlier experimenters have detected anomalies incompatible with newtonian theory, but the results have long since been forgotten. For instance, Charles Brush performed very precise experiments showing that metals of very high atomic weight and density tend to fall very slightly faster than elements of lower atomic weight and density, even though the same mass of each metal is used. He also reported that a constant mass or quantity of certain metals may be appreciably changed in weight by changing its physical condition. His work was not taken seriously by the scientific community, and the very precise spark photography technique he used in his free-fall experiments has never been used by other investigators. Experiments by Victor Cremieu showed that gravitation measured in water at the earth?s surface appears to be one tenth greater than that computed by newtonian theory.


On the basis of newtonian gravity, it might be expected that gravitational attraction over continents, and especially mountains, would be higher than over oceans. In reality, the gravity on top of large mountains is less than expected on the basis of their visible mass while over ocean surfaces it is unexpectedly high. To explain this, the concept of isostasy was developed: it was postulated that low-density rock exists 30 to 100 km beneath mountains, which buoys them up, while denser rock exists 30 to 100 km beneath the ocean bottom. However, this hypothesis is far from proven. Physicist Maurice Allais commented: There is an excess of gravity over the ocean and a deficiency above the continents. The theory of isostasis provided only a pseudoexplanation of this.

The standard, simplistic theory of isostasy is contradicted by the fact that in regions of tectonic activity vertical movements often intensify gravity anomalies rather than acting to restore isostatic equilibrium. For example, the Greater Caucasus shows a positive gravity anomaly (usually interpreted to mean it is overloaded with excess mass), yet it is rising rather than subsiding.

A superb study of the seminal paper published by Roland Eotvos on gravitational anomalies almost 100 years ago:

http://mek.oszk.hu/02000/02054/html/onehund.html

His discoveries remain completely unexplained by modern science.

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Offline Tim Alphabeaver

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Re: Ice Ages on FE
« Reply #23 on: June 05, 2019, 07:08:31 AM »
Source?

As if you didn't know.

Here is your source:


All you've done is written down an equation. If you can't use that equation to back up your argument, it's a non-argument. Just because you state that the Earth's North Pole should face the Sun doesn't mean it's true.
All you need to do is plug your numbers into this equation and show that I'm wrong. Easy, right?

The rest of your post is just you copy-pasting things that I've already read. Your only source is talking about gravitational anomalies - if you were to go up and read my list of things that I asked for a source for, you'd notice that gravitational anomalies aren't on the list. So, could you provide sources for the things I asked about in my previous post?

I'd really much prefer if you stuck to the point we were discussing instead of pasting 50 lines of unrelated information. It means that I have to wade through a bunch of stuff just to find out it has nothing to do with what I'm trying to discuss.

Example: You have a whole paragraph about how gravity in mines was higher than predicted in 1981. Great, I'm sure that's interesting! Not what we're discussing though, is it?
« Last Edit: June 05, 2019, 07:12:21 AM by Tim Alphabeaver »
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Re: Ice Ages on FE
« Reply #24 on: June 05, 2019, 08:17:28 AM »
Are you telling your readers that you cannot use the law of universal gravitation in a very simple context?

Why then would you ask for torque calculations?


BASIC NEWTONIAN PHYSICS: we have a center of gravity which is located ABOVE THE EQUATOR, given the fact that the northern hemisphere has more mass than the southern hemisphere. Then, the accepted law of universal gravitation tells us that the Earth should revolve facing the Sun with its North Pole.

According to this formula, the position of the centre of gravity varies according to the shape of the object.




Here is the proof that the northern hemisphere is heavier than the southern hemisphere, thus will be subjected to a larger gravitational force from the Sun:

"The area of land in the northern hemisphere of the earth is to the area of land in the southern hemisphere as three is to one.

The mean weight of the land is two and three-quarter times heavier than that of water; assuming the depth of the seas in both hemispheres to be equal, the northern hemisphere up to sea level is heavier than the southern hemisphere, if judged by sea and land distribution; the earth masses above sea level are additional heavy loads - we include here all the mountains/hills.

But this unequal distribution of masses does not affect the position of the earth, as it does not place the northern hemisphere with its face to the sun. A “dead force” like gravitation could not keep the unequally loaded earth in equilibrium. Also, the seasonal distribution of ice and snow, shifting in a distillation process from one hemisphere to the other, should interfere with the equilibrium of the earth, but fails to do so."

The northern hemisphere has a greater mass than its southern counterpart.

The unequally loaded perfect oblate spheroid (first four layers) DEFIES the law of attractive gravity.


Now, what you tried to accomplish is to bring the torque computations into the discussion. These calculations, though, are based on certain assumptions, such as the existence of a molten outer core, which in turn rests on the correctness of the seismic wave theory.

Since you are unable to explain the seismic waves anomalies, you can no longer ask for torque calculations; also, it means that there is no molten outer core.

What you are left with is the fact that the northern hemisphere is heavier than the southern hemisphere, thus the North Pole should face the Sun.

BASIC NEWTONIAN PHYSICS: we have a center of gravity which is located ABOVE THE EQUATOR, given the fact that the northern hemisphere has more mass than the southern hemisphere. Then, the accepted law of universal gravitation tells us that the Earth should revolve facing the Sun with its North Pole.

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Offline Tim Alphabeaver

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Re: Ice Ages on FE
« Reply #25 on: June 05, 2019, 08:31:33 AM »
Are you telling your readers that you cannot use the law of universal gravitation in a very simple context?
Correct. I have no idea how to use the universal law of gravitation in this context. The fact that you haven't shown me how it should be used, despite the fact that I've given you ample opportunity to demonstrate it, tells me that you don't know either.

So it's really easy for you, just plug your numbers into this equation to show that the north pole should face the sun.

Sidenote: saying something with large, red text doesn't make it true either.
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Re: Ice Ages on FE
« Reply #26 on: June 05, 2019, 08:38:12 AM »
Very easy.

This is the "law" put forth by Newton in the Principia:

F = GmM/r2

M = mass of the Sun, which stays fixed

m1 = mass of the northern hemisphere

m2 = mass of the southern hemisphere

Ratio:

Gm1M/r2/Gm2M/r2 = m1/m2

Since by hypothesis, m1 > m2, the northern hemisphere will be subjected to a greater gravitational force than the southern hemisphere will.


We have a center of gravity which is located ABOVE THE EQUATOR, given the fact that the northern hemisphere has more mass than the southern hemisphere. Then, the accepted law of universal gravitation tells us that the Earth should revolve facing the Sun with its North Pole.


Convince yourself that the northern hemisphere is indeed heavier (has more mass) than its southern counterpart:

"The area of land in the northern hemisphere of the earth is to the area of land in the southern hemisphere as three is to one.

The mean weight of the land is two and three-quarter times heavier than that of water; assuming the depth of the seas in both hemispheres to be equal, the northern hemisphere up to sea level is heavier than the southern hemisphere, if judged by sea and land distribution; the earth masses above sea level are additional heavy loads - we include here all the mountains/hills.

But this unequal distribution of masses does not affect the position of the earth, as it does not place the northern hemisphere with its face to the sun. A “dead force” like gravitation could not keep the unequally loaded earth in equilibrium. Also, the seasonal distribution of ice and snow, shifting in a distillation process from one hemisphere to the other, should interfere with the equilibrium of the earth, but fails to do so."

This is the reason why heliocentrists will immediately resort to torque calculations involving the molten outer core. However, this is no longer an option since they cannot explain the seismic waves anomalies which defy the accepted molten outer core theory.

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Offline Tim Alphabeaver

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Re: Ice Ages on FE
« Reply #27 on: June 05, 2019, 08:42:24 AM »
Very easy.

This is the "law" put forth by Newton in the Principia:

F = GmM/r2

M = mass of the Sun, which stays fixed

m1 = mass of the northern hemisphere

m2 = mass of the southern hemisphere

Okay, now plug some numbers in and get an actual value of F out...
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Macarios

Re: Ice Ages on FE
« Reply #28 on: June 05, 2019, 08:47:46 AM »
According to this formula, the position of the centre of gravity varies according to the shape of the object.

Ok, the Earth's center of mass is little above Earth's geometric center.
What orbits Sun is the center of mass, not geometric center.
Earth's orientation around the center of mass will be more influenced by Earth's rotation than by revolution.

~~~~~

Aether was obsolete and it is shown that equations describe Universe simpler and more accurate without it.
Not even ring interferometer needs any form of aether to utilize Sagnac effect.

Re: Ice Ages on FE
« Reply #29 on: June 05, 2019, 08:49:32 AM »
Since you like calculations so much, go ahead and use the density/volume of the core, outer core, mantle, litosphere (using the land/mountain/oceans mass for each hemisphere) to reach a final result.

An intelligent approach is to realize that F1 will be greater than F2, since m1 > m2.

Remember, it does not matter if the difference is measured in the nth decimal place: according to Newton, who only wrote down the RADIAL component of the acceleration equation, the force of gravitation deals ONLY with mass, distance and the constant G.


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Offline Tim Alphabeaver

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Re: Ice Ages on FE
« Reply #30 on: June 05, 2019, 08:55:07 AM »
Since you like calculations so much, go ahead and use the density/volume of the core, outer core, mantle, litosphere (using the land/mountain/oceans mass for each hemisphere) to reach a final result.

An intelligent approach is to realize that F1 will be greater than F2, since m1 > m2.

Remember, it does not matter if the difference is measured in the nth decimal place: according to Newton, who only wrote down the RADIAL component of the acceleration equation, the force of gravitation deals ONLY with mass, distance and the constant G.
So you have literally no idea how strong this force is, since you haven't calculated it, and yet you're claiming it's strong enough to rotate the whole Earth in a relevant time frame? And that it's strong enough to overcome other effects, such as the rotatiuon of the Earth? If you did the calculation, maybe this force would take 100 billion years to rotate the Earth. You don't know, since you haven't done the calculation.
Maybe it's 50 orders of magnitude less than tidal forces from the moon. Maybe this force would cause some precession of the Earth's rotation that's so small that it's almost undetectable. You don't know, since you haven't done the calculation.
Do you have some other source of information that you're not revealing, or are you actually just making stuff up?

I'm honestly baffled. Even more baffling is this is the exact torque calculation that I wanted you to perform in the first place, as force acting away from an object's centre of mass causes a torque by definition.
« Last Edit: June 05, 2019, 09:06:24 AM by Tim Alphabeaver »
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Re: Ice Ages on FE
« Reply #31 on: June 05, 2019, 09:16:56 AM »
And that it's strong enough to overcome other effects, such as the rotatiuon of the Earth?

But you cannot bring into the discussion ANY ROTATIONAL FORCES.

Newton's law of universal gravitation DOES NOT mention any rotational forces at all.

This is the full acceleration equation: Newton ONLY mentioned/used/stated the first radial component:





Once rotational dynamics come into play, the "universal law" of attraction is shown to be utterly false:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg753387#msg753387

One day, one of the greatest experimental physicists of the 20th century was asked a simple question, by one of his students:

If there was any difference in gravitational effect on a rotating object versus a non-rotating object?

After an extensive search in the literature, no evidence could be found that the experiment had been performed before.

This became one of the most celebrated experiments in modern physics: the spinning ball experiment.


Even more baffling is this is the exact torque calculation that I wanted you to perform in the first place, as force acting away from an object's centre of mass causes a torque by definition.

No problem, except the fact that the torque computations rely on the molten outer core hypothesis, which is denied by the seismic waves anomalies. This is what you should be focusing on, those seismic waves anomalies.


Maybe it's 50 orders of magnitude less than tidal forces from the moon.

Could be, but you cannot bring into our discussion lunar tidal forces, since there are none.

Here is the barometer pressure paradox:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1707294#msg1707294


Maybe this force would cause some precession of the Earth's rotation that's so small that it's almost undetectable.

Could be, but you cannot bring the precession into our discussion, since the acceleration of the annual precession DEFIES Newcomb's equation:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1776082#msg1776082 (two consecutive messages)




Macarios

Re: Ice Ages on FE
« Reply #32 on: June 05, 2019, 09:18:15 AM »
Since you like calculations so much, go ahead and use the density/volume of the core, outer core, mantle, litosphere (using the land/mountain/oceans mass for each hemisphere) to reach a final result.

An intelligent approach is to realize that F1 will be greater than F2, since m1 > m2.

Remember, it does not matter if the difference is measured in the nth decimal place: according to Newton, who only wrote down the RADIAL component of the acceleration equation, the force of gravitation deals ONLY with mass, distance and the constant G.

Intelligen approach will show that the mass above center of mass will be equal to the mass below the center of mass.

Geometric center will have no influence on forces.

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Offline Tim Alphabeaver

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Re: Ice Ages on FE
« Reply #33 on: June 05, 2019, 09:28:07 AM »
No problem, except the fact that the torque computations rely on the molten outer core hypothesis...
[...]
Errrr, okay. I don't think you understand: using Newton's law of gravitation as you're suggesting is a torque calculation by definition. No molten core hypothesis needed, according to you. This is quite contradictory.
T=F*r*sin(x). Torque exists any time a force is acting at a distance r from the centre of mass (CoM) of an object.

You seem also to have reverted to hurling word soup at me; please try to stay on-topic. I'm interested in the magnitude of the force on an uneven sphere, not "the barometer pressure paradox", or "seismic wave anomalies".

Could you just copy paste this text below as your next response? That would make me incredibly happy:
I am making claims about a force, and yet I have no idea how strong it is, not even an order of magnitude estimate.
« Last Edit: June 05, 2019, 09:34:46 AM by Tim Alphabeaver »
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Re: Ice Ages on FE
« Reply #34 on: June 05, 2019, 09:37:09 AM »
Intelligen [sic] approach will show that the mass above center of mass will be equal to the mass below the center of mass.

Wonderful.

Then, you must accept the fact the North Pole should be facing the Sun, since the northern hemisphere is heavier (has more mass) then its southern counterpart.

Re: Ice Ages on FE
« Reply #35 on: June 05, 2019, 09:42:04 AM »
No molten core hypothesis needed, according to you.

According to you, not to me.

At this present moment, you are unable to explain the seismic waves anomalies, therefore you cannot invoke torque calculations. Only the molten outer core would save the day for you, since then you'd be able (just like the  author referenced earlier) to claim that the molten outer core displacement will match the effect of the heavier northern hemisphere.

Torque exists any time a force is acting at a distance r from the centre of mass (CoM) of an object.

Sure, but you need the molten outer core hypothesis to counterbalance the effect of the heavier mass of the northern hemisphere. Reread those two papers (which do have a somewhat poor translation from Polish to English) to understand the arguments involved.

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Re: Ice Ages on FE
« Reply #36 on: June 05, 2019, 09:54:01 AM »
No molten core hypothesis needed, according to you.

According to you, not to me.
[...]
But you said that this force is simply due to the law of universal gravitation, right? How does this law involve any "molten core hypothesis"?

You already said all of the rest, and I acknowledged it. Now show me your calculation for the force from newton's law of gravitation on an uneven sphere.
If you can't calculate it, then you're quite literally just making stuff up. Which is it?

Quote
you are unable to explain the seismic waves anomalies
I thought we were talking about the universal law of gravitation, not seismic wave anomalies. In fact, you quoted the law of gravitation as "your source". There's nothing about seismic waves or molten cores in this equation, so why do you keep trying to drag it back into the discussion?

It seems simple enough to me: you can't answer my questions about the universal law of gravitation because you have no idea what you're talking about. You can't perform any meaningful calculations with it because you have no idea what you're talking about.

I've said it before: you could trivially prove your point by calculating the effect of the universal law of gravitation on an uneven sphere, which you earlier implied is a totally trivial calculation. So go on, I'm waiting.

Here's my guess: instead of doing the calculation, you're going to start talking about "seismic wave anomalies" and link me to another of your forum posts on a different website. You'll post another 50-line reply that starts talking about "barometric pressure anomalies" or how "torque calculations require molten outer core hypotheses" or whatever.

Please just respond directly to the points I'm making - I've had enough word soup today.
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Re: Ice Ages on FE
« Reply #37 on: June 05, 2019, 10:17:58 AM »
But you said that this force is simply due to the law of universal gravitation, right? How does this law involve any "molten core hypothesis"?

I didn't, YOU DID.

You are the one who requested torque calculations. These calculations rest on a very important hypothesis: the existence of the molten outer core. In turn, this hypothesis is generated by the seismic waves theory.

If you want torque calculations, you better explain the seismic waves anomalies.

Until then, we are left with this:

This is the "law" put forth by Newton in the Principia:

F = GmM/r2

M = mass of the Sun, which stays fixed

m1 = mass of the northern hemisphere

m2 = mass of the southern hemisphere

Ratio:

Gm1M/r2/Gm2M/r2 = m1/m2

Since by hypothesis, m1 > m2, the northern hemisphere will be subjected to a greater gravitational force than the southern hemisphere will.


We have a center of gravity which is located ABOVE THE EQUATOR, given the fact that the northern hemisphere has more mass than the southern hemisphere. Then, the accepted law of universal gravitation tells us that the Earth should revolve facing the Sun with its North Pole.


Convince yourself that the northern hemisphere is indeed heavier (has more mass) than its southern counterpart:

"The area of land in the northern hemisphere of the earth is to the area of land in the southern hemisphere as three is to one.

The mean weight of the land is two and three-quarter times heavier than that of water; assuming the depth of the seas in both hemispheres to be equal, the northern hemisphere up to sea level is heavier than the southern hemisphere, if judged by sea and land distribution; the earth masses above sea level are additional heavy loads - we include here all the mountains/hills.

But this unequal distribution of masses does not affect the position of the earth, as it does not place the northern hemisphere with its face to the sun. A “dead force” like gravitation could not keep the unequally loaded earth in equilibrium. Also, the seasonal distribution of ice and snow, shifting in a distillation process from one hemisphere to the other, should interfere with the equilibrium of the earth, but fails to do so."

This is the reason why heliocentrists will immediately resort to torque calculations involving the molten outer core. However, this is no longer an option since they cannot explain the seismic waves anomalies which defy the accepted molten outer core theory.

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Offline Tim Alphabeaver

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Re: Ice Ages on FE
« Reply #38 on: June 05, 2019, 10:48:23 AM »
Until then, we are left with this:

This is the "law" put forth by Newton in the Principia:

F = GmM/r2
That's what I just said. I have a feeling that you're not reading my comments. I stated multiple times that I want you to perform a calculation using Newton's law to calculate the magnitude of this force.
Me: Now show me your calculation for the force from newton's law of gravitation on an uneven sphere.
Also me:you could trivially prove your point by calculating the effect of the universal law of gravitation on an uneven sphere

Hopefully I've made myself crystal clear at this point: I want you to perform the calculation using Newton's law of universal gravitation that shows the Earth's north pole should face the sun.
All I want from you is F = x, where x is a value in Newtons.

I'll try to refrain from using the word "torque" as you don't seem to understand what it means, so any time I use it you just derail the discussion.
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Offline Tim Alphabeaver

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Re: Ice Ages on FE
« Reply #39 on: June 05, 2019, 10:55:35 AM »
If you can't do the calculation, sandokhan, then I really don't see what your argument is. According to you:
- Newton's law means that the north pole should face the sun
- Any time I ask you to use Newton's law to show this, you derail the conversation instead of actually doing the calculation
- Despite the fact that you haven't actually calculated anything, you know for a fact that the north pole should face the sun, because [reasons]
- You then repeatedly state that the north pole should face the sun, as if it's a fact, despite the fact that you can't possibly know this given that you haven't done any kind of calculation or experiment

Are any parts of my summary unfair? Please tell me if they are.
If you want to actually discuss a calculation involving Newton's law, then I'm available to talk. If you're just going to derail for the nth time, then don't even bother responding, please.
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