Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum Mansfield's Earth Formation Hypothesis: Evidence.

And once again laze shows us he didn't read the paper.

You see, papers have data, and from that data you can answer questions other than the one the paper specifically addresses. In this study, these individuals took paleogravity measures of G from hundreds of sites scattered all over the world. This data directly refutes your hypothesis.

It's simple - if you were correct they would have observed variations in their measured G across the globe. Instead, their data clearly shows a consistent 'G' across the globe, throughout geological history.

Ergo, the paper directly refutes your claims.


Nothing except:
Table 1, which provides the mean and SD of 'G' measured at the varying sites across the globe based on magnetic field line angles.

Table 2, which provides the mean and SD of 'G' measured based on groupings of the varying sites across the globe

And on page 320 (page 5 of the article PDF) the formulas used to calculate radius are measured - and, had you read the article, you'd have noticed that R is calculated based on the changes in surface gravity (dG/dT) measured from the paleomagnetic data.

In otherwords, the "strike" is yours - the radius's they calculated were determined from the measured force of gravity at the various sites they analyzed.


Not working those critical thinking skills much, are we?

I'd first direct you to figure 2 - notice the time scale? Notice the data regarding the very time period we are discussing? Its ironic that you claim to have read the paper, while missing that one of the major points of the paper is the validification and complementation of the methods used to produce figure 2...

And , as with the last paper, the specific conclusion this paper is making is not the same as the one we are asking. But none-the-less, the data within the paper also answers our question.

Now I realize critically analyzing data isn't your forte, but first take into consideration the relationship between gravity, tides and momentum. I realize you're looking desperately for excuses to ignore this paper - like your wikipedia fail below, but none-the-less, that physical relationship is well understood.

Now, the momentum of the earth/moon system is essentially conserved (an assumption shown to be correct on page 55 of the journal; page 19 of the pdf). Ergo, any changes in tidal height must assume a constant momentum. Therefore, the only factor that can lead to broad-scale changes in the tidal heights at a set local is the relative gravitational force between the moon and the earths gravity at the site in question.

Here we have a site, situated on what becomes pangea. And there is no evidence of gravitational changes at that site; only evidence of the moon moving outwards. There is, in figure 2, simular observations acquired from numerous other sites, covering other points in history (including the de facto pangea), also showing no evidence for changes in earthly gravity.

What do YOU conclude from that? The correct conclusion is, of course, that strike 2 also belongs to you...


I direct your attention to figure 5, showing the near-shore bathymetry. And I direct you to figure 10, showing the patter of tides in the regional ocean. And lastly, to table 1, providing information as to the sun-moon alignment, orbital periods, day-length, etc.

In other words, we have all the data that wikipedia correctly states we need to calculate tidal heights. So, given that we have that data, what do you think we can do with it?

LOL, that's a pretty big fail on your part - proving in one quote that you neither read/understood the paper, or what wikipedia was saying.

So once again, strike 3 is also yours.


And now Laze is relying on lying to make his point...always the last tactic of the pseudoscientist when faced with the physical impossibility of their claims.

Simply put, I never made the above statement. What I actually said was:
"Keep in mind we use variations in the earths tidal forces on orbiting satellites to map existent gravitational anomalies (used to map the sea floor). Given that you're proposing changes in gravity approx seven orders of magnitude larger than the ones we observe today, the tidal record should have recorded those changes."

Post #36349

Pretty clear what I said. And in post #36363 I clarified what was said, since you couldn't figure it out on your own:
"You are claiming a decrease of 0.46G. The size of the anomalies measured today are as small as 5uGal. Earth standard gravity is 980mGal.

So you claim: .46*980 = 450.8mGal change
We can detect: 0.005mGal chage
Fold difference: 90,160, AKA ~5 magnitudes of order difference."

I'm assuming that catching you in a bold-faced lie counts as a strike. I'd also like to assume that such on obvious demonstration of reading incomprehension is also worth a strike...


If that were truly the case, you'd be able to list those erroneous assumptions. Since you didn't we can only conclude that once again, you are lying through your teeth.


I've done this before, and once again you didn't provide enough data, so I'll assume the core + mantle that has moved is 50% the mass of the earth. For simplicity I'll assume the gravitational contribution of the rest of the earth remains constant (keep in mind this will underestimate the size of shift you will need - backflow of mantle would increase local G relative to what is calc'd here):

Fg(start) = 100lbs (I'm flattered, btw) = 45kg * G = 441N
Fg(end) = 54lbs = 25kg * G = 245N
Me = 5.98e24kg
Mcore = 0.5*5.98e24kg = 2.99e24kg
Fg(unshifted core) = 0.5G

Portion of our starting Fg determined by the core = 441N * 0.5G = 220.5N

What this has to be reduced to, to reduce Fg(end) to 245N:
delta(Fg) = dFg = 220.5N - (441N-245N) = 220.5N - 196N = 24.5N

So the core has to move to a location where it provides 24.5N of force.

24.5N = G*Me*M/r^2
24.5N = [6.674e-11 * 2.99e24kg * 100kg]/r^2
24.5N = 1.995e16/r^2
r = sqrt(1.995e16/24.5)
r = 2.853945354e+7m = 28539km

r(earth) = 6371km

therefore, the core must be shifted (28539-6371)/6371 = 3.48 earth radi.

Bryan

In other words, to get this shift in my weight, the contribution of the gravitational force of the core must be reduced from 220.5N to 24.5N