Cold Core Model of Earth's Structure

[stereologist]

Trimming a volume is an excellent way to handle points. What sometimes happens is that people use spherical coordinates to lay out shells of points. Just wanted to be sure that your huge calculation was good to go.

PS I haven't used MUMPS since I did a job back in 1987.

[Cold-co]

Modest etal:
There is a powerpoint presentation that shows how Newton did his trigonometric proof. It is located at http://members.cox.net/nchristiansonc/part2.ppt If you can spare a few minutes to view it I think you will get an understanding of how I did my calaulations.
Also I'm attaching an update of the pressures in the earth. I'm prepared to be pilloried again.

[modest]

Originally Posted by CraigD

Here’re the accelerations, in gs, for depths of 0, 0.1, 0.2 ... 0.9 radii, and depth-mass data from various sources...

(Modest’s “5,4,3,2” toy model)
1.0000 0.9633 0.9513 0.9108 0.8399 0.8070 0.6518 0.5197 0.3880 0.2060

If a g is 9.8 m/s² then this can't be right. The total mass of the model is 6.87x10²³ kg. At a distance of 4,000,000 m that's F=(6.87x10²³)(6.67x10^-11[)/(4x10⁶)² =~ 2.86 m/s². Do you, perhaps, mean for "1.0000 g" to represent surface gravity at any numerical value? If that's the case then let's see, the percentage acceleration at depth .5 would be... (2.30646 m/s²)/(2.86561 m/s²) = 0.8049 and yours has 0.8070 at a depth of 0.5. Nevermind, that's what you did.

Please ignore this post

~modest

PS Excellent work Craig. The program is so compact and clever as to border on ridiculous. You always set the bar and there's never enough rep to go around.

[modest]

Originally Posted by Cold-co

Modest:
I like your toy model, but to make it work trigonometrically you have to slice each orb...

There are no orbs in the model as it stands now. I cannot do your method from a vague 3 sentence description. If you are unable to take a step forward (perhaps do the next step yourself) then I have no hope in moving forward on this. Sorry.

~modest

[Cold-co]

Modest:
As I understood your model, each circle represented an orb of constant density. By what you are now posting I really have no idea what your model represents. I can only recommend that you at least take a look at the site I recommended 22 hours ago.

[Cold-co]

Modest:
The attached shows how the matrices were derived and manipulated.

[modest]

Originally Posted by Cold-co

Also I'm attaching an update of the pressures in the earth. I'm prepared to be pilloried again.

I don't know about pillorying, but it does look like you've gone wrong somehow. I once again stipulate that I'm not endorsing your method for finding pressure with the following post, just critiquing the math...

Here are the differences between what I get and what you got for the crust layer of the cold core model:

                yours              mine
                ------             ------
R               6.371x108          6.371x108
r               6.366x108          6.366x108
Density         3.8                3.8
Mass            0.0968x1026        0.165x1026  
Area            0.48x1016          0.2001x1016
Distance (CG)   6.37x108           3.184x108
Force           0.0007x1028        0.0003858x1028
Pressure (KBar) 1.5                1.828

And here's a comparison of pressures in KBar:

yours   mine
------  ------
2.33    1.928
45.4    47.395
94.0    103.534
59.7    87.732
35.0    35.270
202     236.445
46.7    66.234
304     417.329
492     407.908
482     390.400
480     367.730
352     377.618
8.77    7.842
7.59    6.440
5.98    5.035
3.56    3.616
3.21    25.139
1.48    10.874

I can't really spot why our numbers differ. Where the only problem I had with the math in your previous rendition was a conversion of units where it was off by a power of 10, the problem now seems to be more chaotic. I'll attach the excel spreadsheet I used in case you got excel and here is an image of the data:

http://i227.photobucket.com/albums/d...ldpressure.png

Notice in Excel "E" means "x 10 to the power of" so that "5.00E12" is shorthand for 5.00x10¹².

I'll also point out one other thing which didn't seem to cause problems, but might in the future. The formula you gave for the center of mass of a half shell is correct except that you wrote it in a way that is missing a necessary set of parentheses. You gave:

which really would give the wrong answer if followed as shown since division comes before subtraction. It should be:

or, you could write it:

If you want to show your calculations on one of the layers, I could show mine and we can see where they differ. How, for example, did you find the mass and the surface area for the crust layer? If you write out the calculation complete with numbers we can compare notes.

~modest

[Pyrotex]

Originally Posted by Cold-co

Modest etal:
There is a powerpoint presentation that shows how Newton did his trigonometric proof. It is located at http://members.cox.net/nchristiansonc/part2.ppt If you can spare a few minutes to view it I think you will get an understanding of how I did my calaulations.
Also I'm attaching an update of the pressures in the earth. I'm prepared to be pilloried again.

I read the document you linked to.

First, that was NOT the way Newton did his calculations.

Newton used real forces, which were simple pushes or pulls on a point of mass.

Replacing the arrow of force in a diagram with two arrows, heads together, makes no sense and the author gives no justification for this at all.

There is no such thing as an "elastic force". If you really know what the words "force" and "elastic" are, then the concept of "elastic force" makes no sense at all.

Elastic applies to material bodies. It means that if a body is deformed by any combination of forces, then when those forces are removed, the deformed body will resume its original shape. That is what "elastic" means.

How can a force be elastic? Can you deform a force with other forces? No. It makes no sense.

Furthermore, even if you accept (or overlook) that idea, it becomes obvious from his diagrams that the author is doing vectors completely wrong. He has forces pulling on an annulus ring outward from all sides. He ignores the forces that would be pulling inward. He essentially is selecting from ALL the possible forces (real ones) only those forces he wants and ignores the others. He confuses force with strain. He confuses force with pressure.

Sorry, cold-co, but it looks bogus all the way down.

[stereologist]

I did not bother to look at the PPT until Pyrotex commented.

This is a bogus collection of whatever.

The title page is the warning. The force formula states that the gravitational force between two objects is a function of the masses of the objects and the distance between them. To state that "A STRANGE PHENOMENA: WE PULL EARTH WITH THE SAME AMOUNT OF FORCE THAT EARTH PULLS US." is a misleading statement to begin a presentation.

Wouldn't it be odd if the forces were not the same? Plug this force into F=ma and we see that a=F/m causes us to appear to fall and the earth to be motionless.

To call this formula strange suggests that the author does not understand the material they are about to present. This is material that has been reviewed and used for hundreds of years and even put spaceships within a few miles of their target positions billions of kilometers away.

Forget how ridiculous the term "elastic force" is. Also, forget how odd the vector diagrams are. Realize that the force is an attractive force along the line between the masses involved. We're dead in the water if we stop at each gross mistake.

On the next page is a diagram showing a force labeled Fg. I assume that this was meant to be a subscript. Call the force F. For unknown reasons, there are no explanatory notes, the force F is resolved into a force h and v, I suppose for horizontal and vertical.

In the lower right corner there is no explanation of the "horizontal forces acting on an annulus section". There is no similar diagram for the vertical forces acting on an annulus ring. It is the combination of the forces that are the force F on the annulus ring.

On slide 4 is a totally wrong formula. First, I do not see s defined. Second, this is not an equality, rather an inequality. Since slide 4 has a completely bogus formula in it, then its use on slide 6 results in useless malarky.

Slide 5 is baloney as well.

[Cold-co]

Modest:
In going over your spreadsheet, I notice you only use the distance from the hemisphere's cg to its base. For gravitational purposes you need to double that figure. I tried to download your work but for some reason the mathematical functions would not transfer to my system. Maybe the two systems are of two different versions.

Pyrotex:
I guess I'll just have to write to John E. Prussing and Bruce A Conway, the authors of, "Orbital Mechanics," 1993 Oxford University Press and tell them they don't know how Newton did his calculations, because Pyrotex and Stereologist say they don't. Even though neither provided an mathematics to back up their claim. I'm beginning to get the feeling here, that if it isn't in the textbook your are currently familiar with, it must be untrue.