Cold Core Model of Earth's Structure

[Pyrotex]

Originally Posted by Cold-co

...strain is what horizontal gravity produces. For the earth to bulge she must overcome the force of vertical gravity as well as the strain inherent in her surface layer. ...You are still stuck using directional vectors in explaining horizontal gravity's effect. Pyrotex is on the right track with strain, because in an elastic world, strain works in a 360 degree plane.

Strain may indeed be produced by your "horizontal gravity" -- that is, the ordinary horizontal components of gravity, but we still have to explain why your actual values for HG are hundreds and probably thousands of times greater than makes any sense.

For the Earth to bulge, it is enough that the Earth is spinning. The centripetal force is strongest at the equator, because the equator is farther from the axis than all other areas of the planet.

All vectors are, by definition, directional. A vector's "value" is: (1) a magnitude, and (2) a direction.

Strain can work in ALL directions for ALL 3 dimensions. It is not bound to just a "360 degree plane".

Your work goes to great lengths to calculate the outward strain on a tiny mass within the Earth from the horizontal components of gravity. BUT, what about the in inward strain (compression) from the great pressures within the Earth. Compressive strain DOES CANCEL outward strain. And what about the strain from the Earth's spin? Your work is not finished. You have to calculate ALL strains within the Earth's interior and add them up.

Finally, we need to see the equations for how you add up all the "horizontal gravity" on a tiny mass inside the Earth. This would give us a clue why your values are 1000's of times too big.

[Cold-co]

Pyrotex etal:
I have offered to send you a CD of my method of calculations. The offer still stands.

[Turtle]

Originally Posted by Cold-co

Pyrotex etal:
I have offered to send you a CD of my method of calculations. The offer still stands.

Hey Mr. C. Et Al here; but you can call me Roger. So yeah; no. In this day and age of religious whackos going out & hunting down & killing folks that disagree with them on secular grounds, you have no business asking, let alone pushing, anyone to send you their home address. You're on the internet now and it has its own protocols.

Given that you are the one(s) making the claim(s) that virtually turn(s) all physics, geology, cosmology, and a few other -ologies on their heads, the onus is on you to provide your evidence in the appropriate form for the venue. After all, such a proof as you claim is gonna make you famous and the calculations will end up on the web and in textbooks sooner or later anyway, right?

Suck it up & just do it Mr. C. We've got nothin' but time, or to quote a vulgar phrase, we've got 'til hell freezes over.

[Moontanman]

Originally Posted by Cold-co

Moontainman:
In a cold-core model the bulk of earth's mass moves up into the mantle. Since the volume of the mantle is so much greater than that of the core the density of the mantle only needs to be about 7.5 g/cc, which is about equivalent to the speed of S-waves seen in the mantle.

No Coldco, you cannot just make such a claim with out some sort of back up. If the mantle has a density of 7.5 g/cc it would have to be almost pure iron. We know this is not true, if it were, iron would displace the Ices and metallic hydrogen due to being more dense they would migrate to the core displacing the much less dense "hydrogen" There is also the problem of there not being enough pressure inside the Earth to keep the hydrogen as a metallic solid. You are really going to have to do better than this.

[Boerseun]

I rarely quote myself, but:

Originally Posted by Boerseun

We've danced and parlayed around this topic, but the one question asked which remains to be answered, is simply how do you propose your model to explain the Earth's measurable and demonstrable density?

[modest]

Originally Posted by Pyrotex

Strain can work in ALL directions for ALL 3 dimensions. It is not bound to just a "360 degree plane".

Yes.

Cold-co, when stress is equal on all sides then it is no longer a tensor with direction, but a scalar. That scalar is called pressure. As a direct result of Newton's first law: a static fluid undergoes static ("geostatic", "hydrostatic", "isotropic") pressure. A fluid cannot support shear stress. While this is demanded for a fluid, it is also a good approximation for layers of rock strata near earth's surface:

There is a sense, however, in which one may speak of a "geostatic" pressure analogous to a hydrostatic pressure. At sufficiently large depths of burial (or in sufficiently weak rocks, such as ice or salt) under conditions of large compressive normal stress ("confining pressure") and at sufficiently high temperatures, rocks deform easily in response to relatively low shear stresses, provided that the stresses are maintained for long periods of time. Under these conditions, therefore, shear stresses will tend to be reduced to low values and the three principal normal stresses will tend to become almost equal. Even though such rocks still behave as solids in response to stresses imposed over short time periods (e.g., seismic waves) they respond as fluids to stresses of long duration. Thus, averaged over long periods of time (and over large volumes of rock) shear stresses disappear and the magnitude of the three principal stresses becomes equal to the pressure exerted by a column of rock extending to the surface.

Mechanics in the earth and ... - Google Books

Cold-co, can you tell us what units stress, stain, and pressure have? Can you relate that to the units of acceleration in your data? Can you explain how your data is a measure of stress, strain, or pressure?

It is, by the bye, possible to verify that the pressure deep in the earth is isotropic by looking at crystals that formed down there (like diamonds). They are isotropic.

~modest

[Cold-co]

Moontainman:
You asked about pressures. Here is how I have calculated them. See attachments.

[Pyrotex]

Nice drawing on that first page, Cold-co.
So, that is how you calculated pressure inside the Earth?

Hmmmmm...

Did you find that technique somewhere?
Because I've never seen that approach to calculating pressure inside the Earth?
And I've got to tell you that I'm sceptical that that technique gives a correct answer.
You calculate only the gravitational pull between two halves of a shell.
What about the shells of matter "above" that?
Your technique would conclude that if we took equal-thickness shells down to the core,
then the Earth's internal pressure actually decreases as you approach the core.
That can't be right.
Or is there something here that I'm not seeing?

PS: you do neat work. That doesn't mean I agree with you.

[Moontanman]

Ok coldco, can you give me a model of the Earth with a hydrogen core? How big is the hydrogen core and how thick are the ice layers covering it and what temps do these layers have? So far experiments have shown that at pressures near what is expected at the very center of the earth metallic hydrogen can form at 100K to 300K . Temps at the center of the Earth are much higher and so no metallic hydrogen could form or exist at those pressures. Take away the iron core and you get less pressure and even less likely hood of metallic hydrogen. So far you have failed to show how the mass of the earth could be accounted for by your model or how a metallic hydrogen core could form to start with or it could be at the center of the earth to start with or how it could stay in the metallic state once formed. Please show how your metalic hydrogen core could form to begin with before you postulate it is at the center of the earth.

[Cold-co]

Moontanman:
CharlieO posted a site where my cold-core model is explained both verbally and graphically. I recommend you go back to his post and sit through my powerpoint on the subject. Therein you will see how a planetary body condenses from a 9 Kelvin molecular cloud composed of 75% hydrogen, 23% helium and 2% dirty ice coated dust. At 9 Kelvin all other materials are in a dry dust form save for hydrogen and helium.