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This is an old revision of this page, as edited by 63.225.17.34 (talk) at 23:26, 23 March 2017 (Undid revision 771699905 by Vsmith (talk)). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.


Useful Concepts Not Explainable By Plate Tectonics

Growing and Expanding Planets ( like Earth ) have certain problems that can not be reasonably explained by any theory where the mass, density, volume, surface area, and surface gravity do not change appreciably. One area is the Excessive size of land animals in the past, sizes that are no longer possible in the current surface gravity conditions of the Earth, but, were no problem for more than 200 million years between 230 Mya and 65 Mya, and then again when mammals grew excessively large up to about 20 Mya. Either a Growing Earth, or and expanding Earth can easily provide combinations of conditions that allow reduced surface gravities for certain time periods. MWC Golden Co. 71.196.151.6 (talk) 03:42, 27 November 2014 (UTC)[reply]

No, I'm afraid that that won't do. There are several physical explanations as to why physical largeness of animals is not generally the result of evolution in modern times: the ability of intelligence in predators to overcome the utility of largeness as a defence, for one. Using increasing gravity is much more far-fetched than that; in addition it has lots of inescapable consequences which are not observed. As far as I know, Expanding Earth Theory does not have any "Useful Concepts Not Explainable By Plate Tectonics" as the result of consideration. And if you think it does, the right thing to do is edit the article with appropriate sources to convince other editors, or even better, address the geological community through peer-reviewed publication, then use your article as a source, not try to convince anyone in a Talk article. SkoreKeep (talk) 19:28, 25 December 2015 (UTC)[reply]
Considering that the original poster can't see that the span of 235-65 MYA is not "more than 200 million years", I have no problem with dismissing him/her completely. --Khajidha (talk) 11:59, 6 January 2016 (UTC)[reply]

Yes I messed up on the time span by combining concepts into a single sentence. The largest dinosaurs ran from 230 Mya to 66 MYA , and the Largest Mammals from around 65 MYA to approximately 130,000 years ago. The duration exceeds 200 Million Years. The general trend was for animals to get very large, then evolve toward smaller species. This is not something new, but was recognized by engineers in the late 1800's that the largest dinosaurs ( sauropods and their cousins ) are structurally not possible in the current surface gravity of the current sized Earth. The trend showed an absolute peak size before 160 MYA that may maximize at approximately 14.5 Male African Elephant Volumes at a surface gravity of 41 % of the current surface gravity. This would mean that the maximum weight of the largest animals would not exceed 0.41 ( 14.5 ) = 5.945 Male African Elephant Weights. The problem is that Male African Elephants are not standardized in size, or volumetric displacement, and, the ones we have left may not represent the maximum allowable size of animals anymore due to excessive hunting in the last 150 plus years. Even with these limitations, an average of several Male African Elephant is about 5.374 m^3 of volumetric displacement. The average volume of largest sauropods would be 5.374 X 14.5 = 77.923, round to 78 cubic meters of water displacement, but they would only weight about 31.98, round to 32 metric tonnes. An interesting sidebar is that an animal was found in Argentina that was first reported at 100 Tonnes, which was lowered to 77 Tonnes, and after further calculations, was lowered to 70 Tonnes. By conversion this would be approximately 70 cubic meters of water displacement, but the surface gravity would be higher than 41 %, but still estimatable but using the concept that the weight is the 2/3 rds power of volumetric displacement. Dividing 70 by 5.374 gives a value of almost exactly 13.0 Male African Elephant Volumes. Taking 1 divided by the cube root of 13.0 gives a surface gravity of 0.42529 g now. So the weight would be around 13.0 x 0.42529 x 5.374 = 29.71 Tonnes. This number is less than 32 tonnes as would be expected. This animal was 89.66 % of the calculated absolute maximum average volume, but almost 93 % of the maximum allowable average weight of big dinosaurs. The really interesting part is not the size, but the age of the animal. The age is in 100 million year old deposits, not in 160 plus million year old deposits. This is a big deal as it hints that the Earth not only undergoes normal planetary growth ( mass gain ) , but also undergoes periods of expansion that excessively increase the Radius, which lowers the average density of the Earth, and lowers the surface gravity far below what one would expect for that time period. This might also support the concept that size is also limited by food supply, and when the Earth was active 100 MYA, the average temperature was higher ( 26 centigrade ), sea levels were still rising ( maximum elevation at 92 MYA ), and the climate ( rainfall ) allowed for plentiful plant food. In short, the whole of conditions was good for very large animals. Comment By: Michael W. Clark, Golden, Colorado, USA. 73.3.187.143 (talk) 20:56, 24 January 2016 (UTC)[reply]

OK, do you have any sources to cover this material? It all sounds like original research, which is not deemed useful in Wikipedia (see WP:NOR). Do that, and then we'll argue about the applicability and utility of your argument. SkoreKeep (talk) 16:28, 25 January 2016 (UTC)[reply]

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Other reference

Other reference:

Visite fortuitement prolongée (talk) 21:26, 25 January 2016 (UTC)[reply]

Reduced Gravity seems to Lead to a Lower Soil Bearing Capacity, and larger feet in Sauropods

Current, The Largest Animals are elephants, and when they are standing still, with all four feet on the ground, they exert a pressure on the Soil of approximately 13 lbs per square inch under each foot. This is approximately 1872 lbs per square foot of surface area under their feet, which is about 75 % of the Ultimate Bearing Capacity of Soils of approximately 2500 lbs per square foot, in the current Surface Gravity of 1.0 G. In comparison, the measured, and Calculated soil pressure under a sauropods foot at the Dinosaur Bulges at Dinosaur Ridge near Morrison Colorado, is approximately 9 lbs per square inch under their feet, which is only 1296 lbs per square feet of foot bottom surface area with all four feet on the ground. Assuming this is also 75% of the Soil Bearing capacity of the Dinosaurs time, the Ultimate Soil Bearing Capacity would be 1728 lbs per square foot. This is only about 69 % of the current soil bearing capacity. Additional calculations indicate that the surface gravity was around 50% of the Current Surface Gravity, the time period was less than 150 MYa, it may be as close as 140 MYa. 0.50 g occurs when the product of the Planets Density and Radius is one half of the current Product. That is when it is one half of 6372.4567 km X 5509.649049 = 35,110,000 km - kg/m^3. one half is 17,555,000 km - kg/m^3. This occurs at roughly 4172.2674 km and a Density of 4207.545 kg/m^3, and 140.61 Million years ago.

If the same Sauropod was alive today, and the bottom of its feet were the same surface area, it would exert a pressure of just over 18 psi (2592 PSF ). Just standing there it exceeds the current Ultimate Bearing Capacity of the current soils in 1.0 g. As soon as it raised one foot, the other three feet would be carrying all the load, and the pressure would go up to 4/3 ( 2592 PSF ) = 3456 PSF which is over 138 % of the ultimate soil bearing capacity of today's soils, in 1.0 g.

Additional Information: " IN Theory " at 0.5 g, Sauropods could be as large as ( 1/0.5)^3 = 8.0 Male African Elephant Volumes, but the one at Dinosaur Ridge was only 2.5317 Male African Elephant Volumes. The bottom surface area of four feet is approximately 1651 square inches X 9.0172 PSI = 14,887.4 lbs in 0.50 g. The reference elephant has a bottom of the four feet surface area of 886.137 in^2 X 13.367 psi = 11,845 lbs. The Ratio ( 2 X 14,887.4 ) / 11,845 = 2.5137 Male African Elephant Volumes. The Volume Ratio is 2.5317 to 1, but the weight ratio is half of that, 1.25685 to 1.

A request for a source led to a comparison. If soils have lower Soil Bearing Capacity on smaller planets, then we would need to look at smaller Planets ( Moon and Mars ). There is an article called " Physical and mechanical properties of Lunar Soils." It is only two pages long, but I noticed that the Density of Lunar Soils ( Table 3 ) in that article, vary between 1.5 grams per cubic centimeter to 1.9 grams per cubic centimeter, and soil bearing capacities range between 25 Kpa and 55 Kpa. On Earth soils typically run between 2.6 grams per cubic centimeter, and 2.72 grams per cubic centimeter. In larger volumes 1500 kg/m^3 to 1900 kg/m^3 ( Moon ), versus 2600 kg/m^3 to 2720 kg/m^3 ( Earth now at 1.0 g). The Moons surface material is less compressed in 1/6 g than the Earth in 1.0 g. I have yet to find anything on Mars, but would expect a gradually increasing Density of material on a planets surface, as the applied force of gravity increases with successively larger planets. As the Density of the materials increases, the Soil Bearing Capacity ( applied load per unit area without shear failure ) would also increase. MWC Golden Colorado. Clarkmwc99 (talk) 21:39, 18 September 2016 (UTC) Check, 14,887.4 / 11,845 = 1.25685 OK. 2.5137 / 2 = 1.25685 OK.[reply]

The Important thing to do when using foot comparisons is that the animals have similar feet. Both are four footed, both have large oval feet with internal padding. The make similar foot prints in the Mud. Structurally, Sauropods are not possible in today's surface gravity. Their necks, tails, and legs are too long, and their feet are too small, even in our higher gravity with more soil compression, and higher soil bearing capacity. Michael Clark, Golden Colorado, USA. 63.225.17.34 (talk) 04:27, 18 September 2016 (UTC)[reply]

Are there any reliable sources that make this point? Someguy1221 (talk) 06:08, 18 September 2016 (UTC)[reply]

Graphic: expanding or contracting?

The graphic at the top of the article is confusing. "Read" in the natural top-down direction, it demonstrates a contracting Earth, not an expanding one. --Thnidu (talk) 17:20, 14 October 2016 (UTC)[reply]

How To Calculate the Density and Mass of the Earth, and other Rocky Planets, and the Surface Gravity

The Density of the Earth and other Rocky Planets is variable depending on the composition of the planet and its physical dimensions. The Primary dimension that governs Volume, Surface Area, Density and Mass is the planets Radius in Kilometers. In addition, the Surface Gravity can be calculated as a decimal of the current gravitational acceleration by taking the Ratio of the Radius-Density of a prior time period divided by the current Radius-Density. The Equation for the Density has three parts. The first term is the triaxial coefficient of compression so it has a cubic term. The second term is linear and is the uniaxial coefficient of compression caused by the planets self gravity acting to compress the material vertically downward toward the core. The third term is the "constant" that implies the average composition of the crust of the planet. For Venus, that Constant is 2657.05 kg/m^3 which implies a lighter composition such as granite, or Rhyolite. For the Earth and the Moon, the Constant is 2900 kg/m^3 which is a blend of granite and Basalt. For Mars, the Constant is 2941.05 kg/m^3 which implies a predominately Basaltic Crust.

The Rocky Planet Density Equation is: Density = (1+Pi) X 10^-9 X Radius^3 + (1+ Sqrt.2) X 10^-1 X Radius + Constant. For the Earth and Moon, Constant = 2900 kg/m^3. The Current Radius is 6372.4567 Km at the location of the tilt angle of 23 degrees, 19 minutes, and 39.4 seconds north or south of the equator. Plugging the radius into the Density Equation gives three values for the three terms: 1071.736597 + 1538.447139 + 2900 = Current Density of 5510.183736 Kg/m^ for a Radius of 6372.4567 Km. Note here that if the Radius used is greater than 6371.0, then the Density is lower than 5515, but the resulting mass is similar.

THE DENSITY RADIUS: Multiplying the Radius X the Density gives the Current Radius-Density which is the Denominator for the Ratio of all prior Earth Radius-Densities. The Denominator is 6372.4567 X 5510.183736 = 35,113,407.27 Km-kg/m^3. This is the Current Value of the Radius-Density, but all Values in the Past were smaller. Typically you would want to know when a prior Radius-Density was one half of what it is now so an exponential decay curve could be fitted to estimate the growth rate of the Earth. One Half of 35,113,407.27 is about 17,556,703.63 km-kg/m^2. By taking the square root of this value, it gets you into the ball park of the previous Values for both radius and density. We want to look at values around 4190, but the Radius should be Larger, and the Density smaller than 4190. I will spare you the misery of numerous approximations and just give the results. A Radius of 4172.18 km, and a Density of 4208.03857 kg/m^3 give a Radius-Density = 17,556,694.36 which is very close to 17,556,703.63 km-kg/m^3. Notice that I was wrong as the radius was actually less that 4190, while the Density was actually more than 4190.

Now comes the Problem: When was the Radius-Density one half of what it is now, and thus the surface gravity was also one half of what it is now. The best guess is somewhere in the range of 140 to 150 million years ago the surface gravity passed through the point where it is one half of what it is now.63.225.17.34 (talk) 18:33, 22 March 2017 (UTC)[reply]

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