ImagingGeek,

You wrote:
“The angle of the crystals they use to measure the strength of paleomagnetism is dependent on the local gravitational force. The local gravitational force is dependent (obviously) on the amount and distribution of mass below a test site.”

You’ll have to explain how the “angle of the crystals” has any relationship to local gravity.

You then listed three hypotheses:
“1) If the mass of the earth increased/decreased over time, this would be apparent as universal changes in paleomagnetism, appearing as a larger/smaller radius, or
2) If the earth had periods of time with areas having significantly different levels of surface gravity, than we would see this as local differences in the paleomagitism measured at different points on the earth, or
3) If the earths surface gravity has been continious(sic) over time, this will be apparent in the form of a constant G, and thus constant radius.”

Again, you have to explain what you believe to be the relationship between paleomagnetism and paleogravity. You also seem to be trying to disprove the Expanding Earth Theory with your references to changes in “radius” and “mass.” The theory we are evaluating requires no change in radius or mass.

You wrote:
“As you can see in table 1 of the paper, no significant variations of paleomagnitism were seen in any of the >200 test sites, which are scattered all over the globe. Ergo, there was never any significant deviations in the local gravitational fields at those sample locations.”

At the risk of sounding like a broken record,
Again, you have to explain what you believe to be the relationship between paleomagnetism and paleogravity.

Your second link works this time. However, I don’t believe your conclusion that variations in sedimentation thickness due to tidal variations can measure changes in surface gravity on the Earth.

You wrote:
“Tidal heights are determined solely by the ratio of lunar gravity to local gravity. The higher the earths gravitational pull at the measurement site, the smaller the tide.”

This is your opinion (i.e., what is local gravity?). Remember that the current theory posits a gravitational gradient; lowest “g” at Pangea’s COM and gradually getting higher toward both poles; and highest at the antipode. Since the moon would be facing significant areas of varying values of “g” on the Earth continuously as the Earth spinned, any attempt to come up with meaningful data on tidal variations would be fruitless.

In response to my example of a 54% change in “g” produced by a shift in the Earth’s COM by a distance of 1/6 th of the diameter of the Earth, you repeatedly write:

“And, as I pointed out several times before, your formula only gives the correct answer if you shift the enter(sic) mass of the earth, meaning it is completely useless for measuring the gravitational shift when only the core moves.”

Again, as I have repeated many times, not only both inner and outer core move but also the densest part of the mantle that surrounds the cores also moves further from Pangea. And yes, the “entire mass of the Earth” (i.e., the COM of the Earth) must shift.

Your previous calculation is totally incomprehensible:
Fg = G*m1*,2/r^2
0.56G = 1/r^2
r = 1.33631 earth radi


Finally, you wrote:
“I'd also point out that your formula ignores all of the things you claim I ignore - rotation of the earth, distortion of the core, etc.”

Red herring alert! My formula is very clear....and simple. The formula simply states that ratio of the local “g” at any point on Pangea to today's "g" is equal to r^2/d^2, where “r” is the radius of the Earth and “d” is the distance between any point on Pangea and the shifted COM of the Earth. It doesn’t have to explain how the outer core becomes distorted or anything else.


Laze