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Cold Core Model of Earth's Structure


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#18 Cold-co

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Posted 02 June 2009 - 02:41 PM

Moderation Note: The following 18 posts were moved from the thread "Belief in Earth's Iron Core still puzzling" in favor of having their own topic of discussion here—hopefully allowing for more-useful and more-specific responses.

A MATTER OF GRAVITY

Some two hundred years ago, scientists longed to break free from Church control. To break free, they needed to unseat the Church approved, cold-core cross section that had been taught for over 5000 years. The sleight of hand they devised was so well disguised even the Jesuits, who the Pope directed to derail their efforts, could find no fault in their logic. In time, the Jesuits came to teach the scientist’s view. Today, we perpetuate the scientist’s sleight of hand every time we teach gravitational forces at work within Earth.
Gravity is a bidirectional (elastic) force—Earth pulls you with the same amount of force that you pull the Earth. But since Earth is so much larger than any freely moving body, on or above her surface, we treat her gravity as a directional force. This makes the force of gravity relatively simple, so we teach gravity before we teach elasticity. But in reality, elastic cohesion (the drawing together of particles) is the force identified by Newton in his law of universal gravitation, “Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.”
In addition to our less than accurate treatment of gravity, seismic wave data show Earth’s upper shells to be solids down to her core. Yet, we treat her as a large liquid drop to calculate her moment of inertia from her rotationally induced flattening (f). Since she is thought to at least act like a liquid, her flattening is believed to be held on check solely by the equatorial acceleration of gravity (ge); hence, our flattening equation becomes f = 1.5(C-A/Ma2) + 0.5 rate of rotation squared times Earth's radius (a)/ge). In so doing, we conclude she has a low moment of inertia. In turn, she must have a molten interior to allow heavier particles to sink deep into her core to achieve the low moment of inertia demanded by our flattening equation. However, our hot-core model is valid only if we ignore horizontal elastic cohesion.
Elastic cohesion in a solid imparts a constant pull between all its parts. But, no one ever bothered to calculate the strength of that pull in Earth’s shells, because in a schematic of forces diagram the improper use of directional forces (the scientist’s sleight of hand) makes them appear to cancel out; but a gravitational pull cannot cancel out another gravitational pull—only balance. Their pulls are still present.
Now, if we treat Earth’s outer shell as a structurally sound, hollow sphere, subject to horizontal elastic cohesion; then the mass movement, created by her rotation, must also overcome that shell’s elastic cohesion before she will flatten. Trigonometric calculations of Earth’s gravitational forces show horizontal cohesion in her outermost shell to be an acceleration of equal value to the acceleration of vertical gravity on her surface, thus our flattening equation needs another component, +0.5 rate of rotation squared times earth's radius/gh. Or, since this acceleration is of equal value to the acceleration of vertical gravity, that vertical equatorial acceleration can be doubled to obtain Earth’s moment of inertia. When doubled, our flattening equation yields a moment of inertia equal to the summation of moments of inertia mathematically derived for the individual shells of a condensed, cold-core model, whose shell densities are proportional to the speeds of seismic waves passed through them. Serendipity!!!
The squeeze afforded by horizontal elastic cohesion, gives us a unique way to look at Earth’s mechanics—one of a contracting pressure vessel driven by an ever increasing packing pressure provided by horizontal cohesion. A pressure vessel capable of producing a natural heat-pumping cycle—like the heat-pumping cycle employed in diamond anvil devices used to determine the physical characteristics of solid hydrogen. Just as the test sample in a diamond anvil gives up heat to move to a denser phase, so too does the hydrogen crystal in Earth’s pressure vessel give up heat—heat that shows up as geothermal energy in or on her surface.

#19 Pyrotex

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Posted 05 June 2009 - 02:39 PM

...I could have easily looked at my schematic of forces diagram and concluded that horizontal vectors cancel out, as they do when using calculus, because calculus employes directional vectors; but directional vectors is not the nature of gravity, it is a bidirectional force and a bidirectional force cannot be cancelled by another bidirectional force, only matched in strength. ...

The statement that I have bolded -- I do not understand. I have never heard of a "bidirectional vector". I do not understand why a force of any kind cannot be canceled by an equal but opposite force of the same kind.

Please explain this.

#20 Pyrotex

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Posted 05 June 2009 - 02:50 PM

Hi All:
Here are the results of my calculations.
Radius gv cold gh cold gv hot gh hot gv aver gh aver
6371 9.8331 10.0604 9.9307 8.0385 9.8253 9.7883
6370 9.8346 10.0715 9.8327 8.0434 9.8263 9.7998
6365 9.8381 10.1121 9.8399 8.0735 9.8306 9.8574
6359 9.8475 10.1943 9.8525 8.1349 9.8343 9.9264
6291 9.8625 10.9073 9.9326 8.6772 9.8173 10.6533
6116 9.5896 12.3340 9.8759...

Cold-co,
what exactly do you expect us to do with this table of numbers?
what conclusion do you expect us to draw from it?

why don't you PLOT this data as three lines in a graph, showing which Earth-model goes with each line, and then INTERPET the PLOT, telling us what conclusion you draw from it all.
I can't speak for everybody here, but when anybody drops a huge table of numbers on me, about all they're going to get in return is a blank stare. :confused:

#21 modest

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Posted 05 June 2009 - 03:06 PM

Now, I could have easily looked at my schematic of forces diagram and concluded that horizontal vectors cancel out, as they do when using calculus, because calculus employes directional vectors; but directional vectors is not the nature of gravity, it is a bidirectional force and a bidirectional force cannot be cancelled by another bidirectional force, only matched in strength.


Like Pyrotex, I'm afraid I don't understand this. Acceleration is change in velocity. Velocity is a vector which means it can either change in magnitude or direction. Your table of data provides two different accelerations for one point. I don't know how to interpret that, and I'm very curious to find out how.

To help me interpret the data you could explain what horizontal acceleration is. I understand vertical acceleration due to gravity. On the surface of the earth if I drop something in a vacuum it will change velocity at 9.8 meters per second squared toward the center of the earth. If I give a straightforward interpretation of your data it seems I should be expecting an object to change velocity when dropped at 10.0604 m/s^2 in the horizontal direction. I doubt you mean for it to be taken that way because that would not agree with observation. So, I'm curious to understand what "horizontal acceleration" means.

Charlie O posted a link to my PowerPoint presentation, which explains the reasoning behind my concern.


I do not have power point installed at the moment.

Modest: I know you would like to be rid of Charlie O and me, but we are talking earth science here. so I think this is where we belong.


After a good deal of effort engaging both you and Charlie in a respectful manner it's difficult to imagine how you came to the conclusion that I would like to 'be rid of you'. Your topic has now been moved into its own thread, here. The staff feels that discussion on your model is better-served having its own thread with its own topic. This will hopefully garner more useful responses. And, I assure you, has nothing to do with 'being rid of you'.

~modest

#22 stereologist

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Posted 05 June 2009 - 03:23 PM

I made a PDF and posted it here:

part0.pdf

#23 modest

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Posted 05 June 2009 - 04:56 PM

Thank you Stereologist. I look forward to looking at it :confused:

~modest

#24 Turtle

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Posted 05 June 2009 - 05:29 PM

I made a PDF and posted it here:

part0.pdf


Erhm...well..ahhh, I think we have another example of anti-science here wherein one starts with a preconception and makes up something to fit it, rather than look at the evidence and draw the conclusions afterward.

I'll leave out of it though except to quote from the horse's mouth now that we can get a look in there. :blink:

However, his cold-core model failed to meet the low moment of inertia needed to keep Earth from flattening.
This impediment bothered him, because the workings of a condensed cold-core model matched well events reported by paleontologists, archaeologists, geologists, and historians. They also matched well events reported in the Bible, including future events foretold by the prophets. Further, they brought reason to the Gloabal Warming debate by inytroducing a natural heat pump cycle of Ice Ages and warming periods.
Fortunately he realized the packing effect of gravity had never been calculated. ...


Something is getting packed here, but it ain't gravity. :confused:

#25 Pyrotex

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Posted 05 June 2009 - 07:28 PM

Packing gravity? Turtle, me lad, I think you're on the trail of something. :hihi: I've got you're back! :)

#26 stereologist

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Posted 05 June 2009 - 08:19 PM

I'm sorry but all of the h's cancel out. These are the h's in the diagram. Then on p6 you end up with doubling, but say "while calculating ... an interesting relationship popped up." Could you show us the math so that we can point out the mistake?

Appreciate it

#27 Cold-co

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Posted 06 June 2009 - 12:39 PM

To All:
Modest: Thank you for providing a new thread. I realize the subject is a major departure from the purpose of earth science.
Stereologist atal: If you are interested in reviewing the spreadsheets I used to calculate the forces of gravity at descending depths within the earth, I will gladly mail you a CD, at no charge. Just send your snail mail address to my email address that Charlie O will provide, I don't have the number of posts needed to submit an email address.
When I did the calculations I selected a diameter for use in calculating conditions at descending depths. Once a diameter is selected the rest of the earth becomes nothing more than a bunch of gram masses surrounding that diameter. It is a practice used to solve engineering problems and is the exact same practice employed by Newton.
For those of you who are confused by the term "bidirectional" I use it to indicate the vector has two arrowheads one at each end of the line. It is similar to the pull produced by a rubber strand. My horizontal gravitational pull occurs in a 360 degree plane. It is similar to the restorative pull in the skin of a rubber balloon, which in a schematic pulls in both directions in a lateral plane. That is why you cannot feel its effect. But as one progresses into the earth the horizontal pull quickly becomes the dominant force. I think it is the force that holds cloud fragments and planetary bodies together.

#28 CharlieO

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Posted 06 June 2009 - 06:44 PM

xtainson@juno.com

#29 Boerseun

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Posted 07 June 2009 - 12:05 AM

I'm sitting behind my desk.

I'm being gravitationally attracted by the Earth, the moon, the sun, Jupiter, Saturn, and even the fat guy next door.

The gravitational attraction that manifests in me being pulled straight to the centre of the Earth, is merely the sum total of all the gravity sources in the entire universe. Some are more distant, though.

There are no "horizontal" gravitational forces - there is merely "gravity", which works in all directions.

Like Pyro explained, as you approach the center of the Earth, you might as well ignore the sphere of Earth above you at any given depth, because it all cancels out. You will merely experience the gravitational pull of a shrinking sphere below your feet as you descend.

Gravity working in all directions is an obvious given. Stand on the beach and look at the tides, for instance.

So we don't need any mysterious new force to account for cloud fragments and planetary bodies. The "gravity" experienced by any individual particle in any given system, is merely the sum total of the vectors of gravitational pull of all bodies in the system.

#30 Cold-co

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Posted 07 June 2009 - 02:45 PM

Boerseum:
Thank you, you have made my point when you say; "There are no "horizontal" gravitational forces - there is merely "gravity", which works in all directions."
Since horizontally is a direction then there must be horizontal gravitational attraction.

Charlie O has posted my email address. If you are interested in reviewing the mathematics behind my claim that all orbs experience horizontal gravitational attraction just email your snail mail address. I'll mail you a copy of my trigonometric calculations at no cost to you.

#31 stereologist

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Posted 07 June 2009 - 03:01 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.

Are the files you are talking about too big to attach to an email?

#32 Cold-co

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Posted 07 June 2009 - 03:22 PM

Stereologist:
All I am saying is there are objects in earth's crust that must be attracted by horizontal gravity, just as Boerseun claims. That horizontal gravity exists is evident in the Cavendish balance the device used to determine the gravitational constant. The Cavendish balance does not operate in the vertical, but instead operates in a plane perpendicular to the vertical gravitational vector. Hence. a large deposit of heavy ore will cause a pertabation in any gravity vector.

My server will not allow the transmission of the amount of data that a CD can accomodate. So, I'm sorry it isn't possible to do it in the manner you suggest. Don't worry I'm not going to start bombarding you with unwanted ads or blow your cover.

#33 stereologist

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Posted 07 June 2009 - 03:45 PM

The Cavendish balance measures gravity, not a separate entity, i.e. horizontal gravity. The fact that one vector points down towards the center of the earth does not mean that there are not other vectors points towards other nearby objects. The existence one vector does not say anything about the other vectors does it? For example, suppose we put a Cavendish balance in a 0 gravity location. Does this mean that the attraction seen by the Cavendish balance is no longer what you call horizontal gravity? Does it become regular gravity?

#34 Boerseun

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Posted 08 June 2009 - 12:26 AM

Cold-co, I don't think you understand the implications of what I'm saying.

If you descend in an elevator to, say, a hundred kilometers deep, then surely you'll be gravitationally attracted to the hundred-km thick layer of rock and stone in front of you. Sure. But you will also be gravitationally attracted to the 100km thick layer behind you, to the left and the right of you. All these vectors will cancel out. The only gravitational pull you will experience, will be the resultant of all gravity sources having any sort of impact on you. In other words, the mass to the left, right, in front and behind you might as well be ignored, you won't experience it. You will only fall towards the center of the Earth, experiencing the gravitational pull of a globe underneath your feet, shrinking as you go deeper until you reach the core, where you will be essentially weightless. Because even right at the very core, the total mass of the Earth is all around you - which means gravitational pull will be the sum total of the mass of the Earth in all directions, perfectly canceling out.

Counter-intuitively, this is not to say that the pressure at the core is zero - indeed, that will be the zone of highest pressure. Because all the layers from the core to the surface are being attracted to the core, and supporting those layers above it, which are all pressing down, because the sum total of the gravitational pull they all experience, is downwards, towards a globe of a radius determined by the particular layer's depth.

This globe we're discussing isn't a sort of a gravity "Faraday cage", where gravity is magically made to disappear under the surface. All we're saying, is that because of the shape of the Earth, all mass above, to the front, the left and the right of any particular object inside the Earth can be ignored - they all cancel out (if the Earth was a cube, it would've been a totally different story).All that matters, is the gravitational pull beneath your feet - because there is no other gravity source to cancel that out. That will merely be the resultant of all possible gravity sources you experience.