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Cold Core Model of Earth's Structure


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#35 Cold-co

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Posted 08 June 2009 - 11:12 AM

Boerseum and Stereologist:
You keep dancing around the problem. I contend, as both of you do, there are horizontal vectors of gravity, but the difference lies in the fact that gravity is an elastic force, hence it cannot be cancelled by another elastic force. It is still present and pulling the orb together. You will not feel its effects because it is of equal value all around you, even as you progress into the earth.
When earth tries to flatten from a rotational induced outward force, she must overcome both the effect of vertical gravity and the packing effect of the horizontal component of gravity. In that situation the horizontal component does not cancel or balance out as you contend. It must also be overcome before earth will flatten.
Now, I have trigonometrically calculated the ever present strength of horizontal gravity. I request that you review my calculations. You can then judge for yourselves whether horizontal gravity poses another check to earth's flattening and whether it too should be considered in the flattening equation.

#36 Pyrotex

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Posted 08 June 2009 - 12:39 PM

Cold-Co, that isn't what Boerseun is saying. He is not stating that gravity works in alll works, but rather that it can if there are objects there to cause attraction....

stereo, please be more careful with your spelling, grammar, sentence structure, etc. The two sentences above are almost indecipherable. I can, with effort, translate what *I* would have meant if *I* had written them, but I am not sure at all what *you* meant by them.

On a different note, let's talk about "horizontal gravity".

Somebody is splitting hairs here, and the hair is missing. There is NO distinction between vertical gravity and horizontal gravity. The force of gravity between any two mass particles is along the straight line connecting the two particles. That line could be horizontal or vertical (to the Earth's surface, say), or anything in between. The magnitude of the force does not change in any way.

Now, are there gravitational forces which are "horizontal" in the sense that they exhibit a force that has a horizontal component? The answer is yes, of course yes. The Eiffel Tower is affecting a gravitational force on me, and that force has a significant horizontal component -- because the Eiffel Tower is to the East of me, not directly beneath me.

BTW, if you don't understand horizontal and vertical components of a force, you better go to Wikipedia and find out fast. :naughty:

Is this horizontal force on me from the Eiffel Tower anything special or different from the vertical force I feel on my butt as I sit in this chair? No. No. A thousand times, No.

Is this horizontal force due to some special nature of the mass tugging on me? No. Or to some special nature of ME? No. Gravity works in all directions equally. It works in all directions. Period.

The force of gravity is a force. I never thought I would have to explain this, but given the way this thread is going, it's necessary. It's a force. It's a "pull" on the mass of one object caused by the mass of another object.

Yes, the force of gravity is "bidirectional"! :eek2: Dammit, ALL forces are "bidirectional". Hello! Newton's Laws of Motion: "For every reaction (force) there is an equal but opposite reaction (counter-force)." If my finger pushes against a can of Coca-Cola with a force of 1 pound, then the can is also pushing against my finger with a force of 1 pound. There are no "naked" ("unidirectional") forces in the Universe.

When two opposite but equal forces cancel each other out at a point, there is no "residue" left over. They cancel out. Zero remains. When lots of opposite but equal forces act over the surface of a non-zero volume, like the Coke can, then the contents do not "feel" a force, they "feel" pressure.

A force caused by gravity is basically indistinguishable from any other force, in terms of HOW it affects matter, and HOW the matter behaves as a result. There are "elastic" objects -- there are no "elastic" forces. A force is a pull or push. Period. And a pull equals a -push. Period.

Given a spherical Earth (not true but close enough), then anywhere on its surface, the sum total of ALL vertical and horizontal forces from gravity will be a single vertical force toward the center of the Earth (your "weight"). All horizontal gravitational components will cancel out.

These canceled-out horizontal gravitational components are truly cancelled-out. They do not create a "negative pressure" or a "residue force" or a "bidirectional stretching factor" or a "mortistat gradient" or an "elastic conundrum" or an "inertial fotticyte" or anything else.

#37 Pyrotex

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Posted 08 June 2009 - 01:01 PM

...If you are interested in reviewing the mathematics behind my claim that all orbs experience horizontal gravitational attraction ....

Cold-co,
correct me if I'm wrong, but didn't you say you were doing these calculations in an Excel spreadsheet? I cannot imagine your spreadsheet being greater than 2 Mbytes, but I guess it could be.

Why don't you PLOT the data? Then you can copy the plot, paste it into a Word document and certainly either post it or attach it to a PM to any of us.

That would kill two birds with one stone.

#38 stereologist

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Posted 08 June 2009 - 01:33 PM

stereo, please be more careful with your spelling, grammar, sentence structure, etc. The two sentences above are almost indecipherable. I can, with effort, translate what *I* would have meant if *I* had written them, but I am not sure at all what *you* meant by them.


Sorry. Bad typing is my norm. Should have proofread.

#39 maddog

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Posted 08 June 2009 - 01:37 PM

To all,

I am in complete agreement with contributions of a sphere above me are "canceled out".
Believe me, I had to work this out one of the *star'd Homework problems for my freshman
physics class and it was a bear. I was very thorough in my proof to work out for a sphere
that any horizontal component of gravity is canceled out due to symmetry. :eek2:

The only thing I can add is the Earth is "not quite" spherical. The deviation in any horizontal
component has got to be less than 3 orders of magnitude lower down, so is negligible. :naughty:

maddog

#40 Cold-co

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Posted 08 June 2009 - 03:16 PM

For All:
I'm in complete agreement with the concept of vertical gravity deminishing as you progress into the earth, that's not the problem. Where the problem lies is in your concept that all horizontal forces cancel out. Due to the nature of matter, those forces are still present. They are balanced, but they are still there and actively pulling. Only when you try to disturb that balance must the strength of their pulls be considered. Since current thinking believes these pulls cancel out, your textbooks do not address these pulls. If you cannot conceive of a latent force being present in matter, a force that hold matter together in the manner put forth by Newton in his universal gravitational equation, then all the arguments I have put forth are mute.

Bulldog:
Go to the site that Charlie O gave you the routing for; step throught the whole PowerPoint presentation, there you will find the schematic and resultant forces displayed.

#41 maddog

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Posted 08 June 2009 - 03:43 PM

For All: I'm in complete agreement with the concept of vertical gravity deminishing as you progress into the earth, that's not the problem.

Then where prey tell is the problem ?

Where the problem lies is in your concept that all horizontal forces cancel out. Due to the nature of matter, those forces are still present. They are balanced, but they are still there and actively pulling. Only when you try to disturb that balance must the strength of their pulls be considered. Since current thinking believes these pulls cancel out, your textbooks do not address these pulls. If you cannot conceive of a latent force being present in matter, a force that hold matter together in the manner put forth by Newton in his universal gravitational equation, then all the arguments I have put forth are mute.

Maybe you do not understand Vectors have both magnitude And Direction. When two
Vectors oppose each other in direction And their respective magnitudes are Exactly Equal, this directly implies that the Resultant will equivicate to Zero === 0, Nada, Nil,
Nothing, no More. Get it. I proved it in a homework in Freshman Physics. We are talking
of the resultant force, not that your pretty little component which didn't vanish per se. It
just met it's match by another force of equal (magnitude) and opposite (direction).
I can conceive of a lot. Though Latent implies some residual amount beyond the Resultancy.

Bulldog: Go to the site that Charlie O gave you the routing for; step throught the whole PowerPoint presentation, there you will find the schematic and resultant forces displayed.

For your edification it is maddog, not Bulldog. I went there already, have the whole presentation,
read it. Been there already. I do not have time now to comment. They have me here
doing three separate projects now in the span of a 9 hour day (9-80). So maybe later
this week. I will then go over what I have read and show that the presentation is making
some huge leaps without working out all the steps. These then become jumps to a
conclusion without founding. Sorry. :eek2::shrug::naughty:

maddog

#42 Cold-co

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Posted 08 June 2009 - 05:06 PM

Maddog:
I understand vectors very well. Maybe you will get the picture if I introduce chemical bonding. The lithosphere, which includes the crust, is held together by chemical bonding of the elements located therein, but no where in the flattening equation is that bonding addressed. The same is true of the horizontal gravity that holds the earth together. No one has bothered to question the horizontal forces in the outer layer of the earth. Top scientists doggedly hold to the conclusion that horizontal forces cancel out, but chemical bonding does not cancel. It is there whether you recognize it or not. I introduce chemical bonding only as an example of a force that has not been considered. It has nothing to do with horizontal gravitational forces although they may play a part in the chemical bonding reaction.

#43 Boerseun

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Posted 08 June 2009 - 11:30 PM

I live in South Africa.

We've got the deepest operational mine shafts in the world. 4½kms down is no joke. The temperature rise to ungodly levels, making permanent air-conditioning a must. The air pressure is something out of this world. But working these shafts have become routine, and we've become quite adept at it.

Yet, not a single case is known of a miner falling towards a shaft wall. They all fall downwards, somehow.

And there is no "horizontal" gravity component necessary to keep a spherical aggregate of dust, rock, stone, water and metals together. That is simply the shape it takes when every individual component falls towards the collective center of mass.

Cold co, I will give you the benefit of the doubt and assume that you are reading our responses. But I doubt very much that you are digesting what's being presented here.

#44 Cold-co

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Posted 09 June 2009 - 09:49 AM

Pyrotex
You make a point that the Eiffel Tower attracts you with the same amount of force that you attract the Eiffel Tower. Ok, let us consider a situation where you are surrounded by six Eiffel Towers. You are standing in the middle of the hexagon formed by the towers.
You attract all towers with the same amount of force that towers attract you, right. Does the attraction between you and the first tower get cancelled by the attraction between you and the fourth tower? Does the attraction between you and the second tower get cancelled by the attraction betwen you and the fifth tower? Does the attraction by you on the third tower get cancelled by the attraction between you and the sixth tower? The answer obviously is no, the pulls are balanced, but the physical arrangement I have described has not changed, the pulls are still present. It was that pull that the individual gram mass has on its surrounding gram masses that I have trigonometrically calculated. The results you have already seen.
Boerseun:
You mention deep mines. In Australian mines a gravitometer was taken to very deep levels. The pull of gravity dropped off faster than it should have. They resolved that this could be caused by a heavy mass below the mine shaft or the mass of the earth is located at a higher level than is currently predictated by the hot-core model. Very curious another scientific disconnect. I have identified 26 scientific disconnects that prague the hot-core model. All of them are resolved with a cold-core model.

#45 maddog

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Posted 09 June 2009 - 10:17 AM

Pyrotex: You make a point that the Eiffel Tower attracts you with the same amount of force that you attract the Eiffel Tower. Ok, let us consider a situation where you are surrounded by six Eiffel Towers. You are standing in the middle of the hexagon formed by the towers.
You attract all towers with the same amount of force that towers attract you, right. Does the attraction between you and the first tower get cancelled by the attraction between you and the fourth tower? Does the attraction between you and the second tower get cancelled by the attraction betwen you and the fifth tower? Does the attraction by you on the third tower get cancelled by the attraction between you and the sixth tower? The answer obviously is no, the pulls are balanced, but the physical arrangement I have described has not changed, the pulls are still present. It was that pull that the individual gram mass has on its surrounding gram masses that I have trigonometrically calculated. The results you have already seen.

Cold-co,

Point of note: When the use of the phrase: "cancels out", this does Not mean that it goes away. The force is still present. It has met an equivalent match which makes it's
contribution irrelevant. So in the case of six-equidistant Eiffel Towers in an exact
arraingements as to give complete symmetry, you could create an effect that the sum
total would "cancel out". Non-the-less, all six Eiffel Towers and you would feel the effects
of 1g acceleration of the Earth. So all the Eiffels and you would feel the pull of the Earth.

Boerseun: You mention deep mines. In Australian mines a gravitometer was taken to very deep levels. The pull of gravity dropped off faster than it should have. They resolved that this could be caused by a heavy mass below the mine shaft or the mass of the earth is located at a higher level than is currently predictated by the hot-core model. Very curious another scientific disconnect. I have identified 26 scientific disconnects that prague the hot-core model. All of them are resolved with a cold-core model.

You are not going to be able to explain Van Allen Radiation without a Hot-Core Model.
Your Australian measurement was likely near a Mascon (they have been found on the
Earth as well as the Moon) which cause anomalous Gravity measurements.

maddog

#46 Cold-co

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Posted 09 June 2009 - 10:26 AM

I'm sorry but in my calculations thay do not nicely go away as you claim. They get stronger the deeper you go into the earth. Do the math.

#47 Cold-co

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Posted 09 June 2009 - 11:01 AM

Maddog:
The Van Allen belt would exist with a cold-core model for the same reason it exist for the hot-core model. It is created by earth's magnetic field. Below is my colde-core model.

WHAT DOES EARTH’S COLD-CORE CROSS SECTION LOOK LIKE?

© Copyright June 1, 2009 by Neil B. Christianson

Seismic waves show Earth's cross section to have a crust that floats on a thin molten base, five distinct shells, a discontinuity, an outer core, an inner core and an inner-inner core—a cross section condensed from a 9 Kelvin molecular cloud containing 75% hydrogen, 23% helium and 2% ice-coated dust.
Earth’s three top layers contain all manner of elements and compounds. Her crust, primarily composed of granite and basalt, varies in depth from 10 km (6 miles) below oceans to 40 km (25 miles) below continents. If driven, it would be a short commute; say from Miami FL to Miami’s outskirts and Fort Lauderdale. Continental and oceanic crusts are anchored in the Lithosphere, a strong stony shell—currently thought to be a form of olivine, characterized by an efficient transmission of seismic waves. Its lower boundary, at the 80 km (50 mile) mark; or about Delray Beach, shows a sharp drop in seismic wave velocity. This drop in velocity introduces the Asthenosphere—currently thought to be a form of spinel, a weak stony shell that shows a definite softening or weakness wherein seismic waves drop their speed more strongly than anywhere else in the upper Earth. This loss of speed is thought to be caused by partial melting, but it could just as easily be caused by its elevated water content. This shell ends at the 220 km (137 mile) mark; or Vero Beach, still a relatively short drive.
Densities encountered as we cruise into the Earth continue to increase in her first, sec¬ond and third bonded shells. These shells are separated by what scientists describe as phase change zones, because their abrupt increases in density can only be accounted for by a structural change caused by the repackaging of the mate¬rial’s molecules. The first bonded shell of dirty Ice VII, which is known to retain its solid form until it reaches 420 Kelvin, extends from the base of the weak stony shell down to about the 420 km (260 mile) mark; about Daytona Beach, to the first phase change zone, which continues on down to the 470 km (292 mile) mark; or to Palm Coast. It contains a mixture of dirty Ice VII and dirty Ice VIII—Ice VIII forms when Ice VII cools to 278 Kelvin. The second bonded shell contains dirty Ice VIII and extends down to the 670 km (416 mile) mark; about Fernandina Beach GA, to the second phase change zone, which continues on down to the 720 km (447 mile) mark; about Brunswick GA. It contains a mixture of dirty Ice VIII and dirty Ice X. Speeds of seismic waves increase across the third bonded shell of dirty Ice X and then abruptly drop at the Gutenberg discontinuity at the 2885 km (1793 mile) mark; or about Moosenee on the fresh water arm of the Abitibi River, Ontario, Canada, which we will canoe down to board a ship to continue our journey on the open waters of James Bay at the 2895 km (1800 mile) mark and on into Hudson Bay. The Gutenberg discontinuity contains superfluid helium—a perfect conductor of heat. If we take time to look around we will notice how the base of the dirty Ice X shell is scalloped and the outer core is covered with sand hills that have accumulated from dense specks of dust dropping from the Ice X shell and from the purification of the core by the constant flow of superfluid helium. We will also notice an occasional patch of magnetic helium-3—cooled to .0027 Kelvin.
The outer core, which is made of face centered cubic hydrogen, is a quasi-solid down to the 5153 km (3202 mile) mark; about Easter Cape on Baffin Island. It fails to propa¬gate S waves, because its molecules remain free to rotate even though they occupy solid lattice positions. It acts like a liquid. Its crystalline lattice channels and grain boundaries are interlaced with superfluid helium that facilitates the flow of heat coming from ongoing phase changes.
The in¬ner core; where our ship now picks its way through ice floes, extends down to the 5781 km (3592 mile) mark; the Elizabeth Islands. It appears to be denser than its outer companion because it propagates S waves. It is another phase change zone, composed of a mixture of face centered cubic hydrogen and solid, base centered cubic hydrogen. Its structure too is interlaced with superfluid helium.
Seismologists now report another change in density, undoubtedly base centered cubic hydrogen that extends to Earth’s center at the 6371 km (3960 mile) mark; our mode of travel changes to dog sleds for a ride across the ice cap to the North Pole. This innermost ball is a tighter pack of molecular hydrogen molecules, so its grain boundaries and lattice channels can only be laced with helium-3, which gives Earth her reversible magnetic field.

#48 Turtle

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Posted 09 June 2009 - 12:54 PM

Seismic waves show Earth's cross section to have a crust that floats on a thin molten base, five distinct shells, a discontinuity, an outer core, an inner core and an inner-inner core—a cross section condensed from a 9 Kelvin molecular cloud containing 75% hydrogen, 23% helium and 2% ice-coated dust. ...


:doh: This is not the case. Pyrotex described the circumstance of Earth's condensation in one of Chuck's earlier threads. A few thoughts on CharlieO's other questions about origin of Earth's iron core. ...

When we gonna stop dissembling and get the explanation of how this all proves Noah's flood? :help: Just askin'. :Alien:

#49 Cold-co

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Posted 09 June 2009 - 01:49 PM

Turtle:
I read your post before when you were berating CharlieO for his failure to recognize the work that theorists have put forth. I still do not buy your explanation. Mine at least does not require a fantastic separation process to get an iron core in an orb. My model works without an iron core.

#50 Pyrotex

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Posted 09 June 2009 - 01:59 PM

Pyrotex..., let us consider a situation where you are surrounded by six Eiffel Towers. You are standing in the middle of the hexagon formed by the towers.
You attract all towers with the same amount of force that towers attract you, right. Does the attraction between you and the first tower get cancelled by the attraction between you and the fourth tower? Does the attraction between you and the second tower get cancelled by the attraction betwen you and the fifth tower? ...The answer obviously is no, the pulls are balanced, but the physical arrangement I have described has not changed, the pulls are still present. It was that pull that the individual gram mass has on its surrounding gram masses that I have trigonometrically calculated. ...

Excellent problem. And not a trivial one to solve.

Here I am in the middle of six Eiffel Towers ("Masses"). Each Mass is paired up with an identical Mass equally distant on the other side of me. What is their effect on me? I think we can agree that at the "macro scale" (considering me as a single object) the six gravitational forces do indeed "balance out". I am not pulled in any particular direction. Do they "cancel"?

It occurs to me that you might define "cancel" as some sort of process where two opposing forces "annihilate" each other, leaving no forces behind at all. But in classical physics, when we say two opposing forces "cancel", we mean only that they are equal in magnitude and opposite in direction -- the same as "balance out". So I would say, yes the forces balance out and they cancel, same thing. I'm not implying that the forces make each other go away--they're still there.

Let's change the model a bit. I am now a single Point in the middle of six Masses. We replace the force of gravity with an infinitely thin cable--one end attached to me (the Point) and the other end attached to one of the Masses. The cables are inverse-stretchy. The shorter they become, the MORE force they pull with, with an inverse-square relationship--exactly like gravity does.

The Point is attached to six inverse-stretchy cables, pulling it in six directions. What does the Point feel? I would say it feels nothing at all. All the forces balance exactly out. The Point feels like it is floating, not accellerating in any direction. But the cables are still there. It feels no force, but the cables do not "go away" just because their forces "cancel". Can the Point spin? Yes, it is a dimensionless point so it feels no constraint of any kind. If another external force F is applied to the Point, and the Point has mass M, what will happen? Well, the Point will accellerate, (a), in the direction of the Force, with: a = F/M. Just exactly as if there were NO cables!

Okay, back to original model. It's ME in the middle of six Masses. And I am made up of many, many points. Some of those points are nearer to Mass-1 than to Mass-4. Other points are nearer to Mass-5 than to Mass-2. Since gravity is inverse-square, this means my outstretched fingertips will feel a tiny, tiny residual force OUTWARD from the center of my body. My nose is being pulled OUTWARD from the center of my head, and the hair on the back of my neck is being pulled OUTWARD in the other direction.

This is called, Tidal Force. If I were an exact Point, all gravitational forces would indeed cancel, but I am NOT a Point, I am an "extended" object with parts of me (MOST of me) NOT exactly at the center of the six Masses.

The Tidal Force will pull my extremities OUTWARD (very, very, tiny, tiny force) away from the exact Center. The further from the Center my extremities are, the greater the Tidal Force.

Now this is exactly what creates the Tides on Earth. To a first approximation, the force between Earth and Moon is just a simple force pulling their Centers together. But the water and rock on the "Near Side" of the Earth (relative to the Moon) (that is, the point on the Earth directly "under" the Moon) will be pulled with a greater force than the Center of the Earth -- and the water and rock on the "Far Side" of the Earth will be pulled with a lesser force than the Center of the Earth.

Therefore, the oceans (and rock to a lesser extent) will rise higher under the Moon, and on the opposite side of the Earth.

It occurs to me that what you are trying to calculate with your trigonometry ("horizontal gravity") is something like the Tidal Force.

One thing now. In the final version of the model, where I'm in the Center of six Masses, my outstretched fingers (say, of my left hand) are still being gravitiationally attracted by all six Masses. It's just that those fingers are fractionally closer to, say, Mass-1, and fractionally further from Mass-4. That's why. The opposite is true of my fingers on my right hand. And THIS is why my whole body may experience an "outward tug" all over, as if some mysterious "horizontal force" were trying to expand me outward like a balloon being blown up.

What do you think?

#51 Turtle

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Posted 09 June 2009 - 02:05 PM

Turtle:
I read your post before when you were berating CharlieO for his failure to recognize the work that theorists have put forth. I still do not buy your explanation. Mine at least does not require a fantastic separation process to get an iron core in an orb. My model works without an iron core.


Not good enough C-Man. Dismissing the current understanding simply by calling it fantastic belies the experiments and methodology behind it. You have taken a similar tactic with dodging the physics presented here that contradict your "packing gravity". This is not acceptable. :Alien:

Given that the story of Noah is a re-write of the earlier Epic of Gilgamesh, I'd like to hear about the cold hydrogen core association to that scenario. :help: