Dispatches from the Creation Wars

New Explanation for Dinosaur Extinction?

Raw Story has a report on new research that suggests that the event that killed off the dinosaurs may have been more complicated than previously thought.

Until now, it has been accepted generally that the Chicxulub impact off the coast of Mexico 65 million years ago wiped out the dinosaurs. Evidence of the crater left by the giant asteroid or comet has been found under the sea off the coast of Yucatan.

But a group of scientists led by Professor Gerta Keller of Princeton and Professor Wolfgang Stinnesbeck of the University of Karlsruhe begged to differ. They uncovered a series of geological clues which suggests the truth may be far more complicated.


I’ve long thought it likely that the Alvarez’ theory regarding the Chixulub impact was only part of the story. I haven’t read the paper for this new research, so I can’t evaluate its accuracy beyond a very superficial level, but I suspect they’re on to something. Here’s the specific argument they seem to be making:

The previous impact theory was beautifully simple and appealing. Much of its evidence was drawn from a thin layer of rock known as the “KT boundary.” This layer is 65 million years old (which is around the time when the dinosaurs disappeared) and is found around the world exposed in cliffs and mines.

For supporters of the impact theory, the KT boundary layers contained two crucial clues. In 1979, scientists discovered that there were high concentrations of a rare element called iridium, which they thought could only have come from an asteroid. Right underneath the iridium was a layer of spherules, tiny balls of rock which seemed to have been condensed from rock which had been vapourised by a massive impact.

But Keller’s team concentrated on a series of rock formations in Mexico where the iridium layer was separated from the spherule layer by many metres of sandstone. Keller found evidence such as ancient worm burrows that suggested that the deposition of the sandstone had been interrupted many times.

Her team concluded that there was a gap of some 300,000 years between the deposition of the spherules (from the Chicxulub crater) and the iridium (from an asteroid). Therefore, there must have been two impacts.

Again, this seems quite reasonable to me without having seen the data. They speculate that a second, possibly larger impact occured in India. So far, they have not found the actual crater. It’s possible that this event may have triggered the astonishing series of volcanic eruptions that produced the Deccan basalts:

The Chicxulub impact conspired with the Deccan Flood Basalt eruptions in India, a period of prolonged and intense volcanic activity, to nudge species towards the brink, said Dr Keller.

Vast amounts of greenhouse gas were pumped into the atmosphere by the Deccan volcanism over a period of more than a million years. By the time Chicxulub struck, land temperatures were seven to eight degrees Celsius warmer than they had been 20,000 earlier.

Weakened by these events, species were finally killed off by the second impact.

I’d like to read the papers to see if the data supports this hypothesis at this point, but my initial reaction is that it seems a more likely scenario than the Alvarez hypothesis by itself.

Comments

  1. #1 Rocky
    November 16, 2006

    Another current theory being presented is that atmospheric oxygen levels fell dramatically, and indeed, reduced atmospheric oxygen may be tied to many past extenction events. O2 may have fallen to 10%-12%. The reasons for the O2 level drops are being examined, and may be tied to the Deccan traps basalt/Chicxulub impact/oceanic methane ice.

  2. #2 NJ
    November 16, 2006

    Having been a grad student at Virginia Tech in the 1980′s (OT: Geez, I’m turning into an old man!), I had a front row seat to the Walter Alvarez (impact) vs. Dewey McLean (Deccan volcanism) arguments. McLean always loved to point out that the paleo data didn’t support the “Bang! You’re dead.” impact idea, and that basaltic volcanism released mantle iridium to the surface.

  3. #3 Jim Lippard
    November 16, 2006

    Donald Prothero’s presentation at this summer’s Skeptics Society conference included quite a bit of criticism of the impact theory–I think it’s safe to say that his opinion is that it’s almost certainly incorrect.

    One of his key points was that marine reptiles and invertebrates had already started to die off long before the impact. There’s a summary of his Skeptics Society talk here.

  4. #4 kehrsam
    November 16, 2006

    I thought they all died leaving the ark. The steps weren’t up to code.

  5. #5 Ed Brayton
    November 16, 2006

    I’ve long thought that the Chixilub v. Deccan traps debate was a false one. There’s no reason why it has to be one or the other, both might have contributed to the decline and extinction of the dinosaurs, and more besides (as this new research may show). It may well be that the long-term changes brought on by the Deccan eruptions dramatically reduced the numbers of living dinosaurs and then the Chixilub impact (or this new possible impact, or a combination of the two) finished them off. It’s never seemed terribly likely to me that a single event was solely responsible for the extinction.

  6. #6 Lorne Ipsum
    November 16, 2006

    There may be doubt about the impact theory being the be-all-and-end-all of the death of the dinosaurs — and I’d argue this is a good thing, since dogmatism is not science’s friend. Still, the Raw Story report is hardly “new” news. The “mounting evidence” it talks about is just a pile of press releases from Gerta Keller, pushing the same stuff she’s been pushing for years.

    If you want something solid to work from, look at peer reviewed papers, not press release journalism.

    Lorne

  7. #7 Lynn
    November 16, 2006

    It always bothers me to hear people talk about “the” reason for an event like a mass extinction. I’ve always felt that was simplistic. Disastrous things of one kind or another happen all the time–it makes far more sense that it was a sort of orchestra of disaster that ganged up on Earth’s life forms, creating mass extinctions. Just the coincidence of a number of bad things happening close together in time.

    I think the evidence that the Chixilub impact happened, and had a lot impact (pardon the expression) on life is extremely good. But accepting that it happened is not the same as accepting that it was “the” cause of the Cretaceous extinctions.

    BTW, I’ve always had an odd little niggling about possible causal relationships between the Chixilub impact and the Deccan traps eruptions. The two locations are pretty much opposite each other on the planet. I keep thinking about compression damage to a skull on the opposite side from a concussive blow to the head LOL!

    Lynn

  8. #8 Tom Ames
    November 16, 2006

    It always bothers me to hear people talk about “the” reason for an event like a mass extinction.

    I agree, except in the case of the current mass extinction, which most definitely does have a single cause.

  9. #9 Miguelito
    November 16, 2006

    I think this Keller stuff is a bit older (like 2003), but it’s still important.

    A good resource and a scathing critique of the Keller multiple-impact theory by Jan Smit:

    http://www.geolsoc.org.uk/template.cfm?name=NSG2349857238495G999999

    I’m not sure I buy into either of their arguments.

    My opinion (I’m a sedimentologist and my opinion actually carries weight for once): I’ve seen alot of bioturbation in my time and the bioturbation evidence that Keller presents is pretty convincing. The Chondrites burrows they document occur only during prolonged periods of quiet water. Furthermore, the glauconite evidence also suggests this.

    I suspect that there are some slumps that Keller are overlooking, adding older debris into the basin fill. I also suspect that the multiple spherule beds may have come from the same upslope source (the sediments on the basin floor had to come from somewhere) and that multiple impacts are not necessary.

    In other words, I doubt that the basin fill on the basin floor was all that rapid, but I’m not sure if there were multiple impacts like Keller suggests.

  10. #10 RickD
    November 16, 2006


    It always bothers me to hear people talk about “the” reason for an event like a mass extinction.

    I agree, except in the case of the current mass extinction, which most definitely does have a single cause.

    Posted by: Tom Ames | November 16, 2006 12:49 PM

    Fluoridation of the drinking water?

  11. #11 JS
    November 16, 2006

    BTW, I’ve always had an odd little niggling about possible causal relationships between the Chixilub impact and the Deccan traps eruptions. The two locations are pretty much opposite each other on the planet. I keep thinking about compression damage to a skull on the opposite side from a concussive blow to the head LOL!

    As I recall, the idea of impact shockwaves contributing to volcanic eruptions on the exact opposite side of the globe was taken fairly seriously back when. I don’t know what happened to that notion, but the symmetry makes sense.

    Additionally, if the impact was big enough to disturb a nearby fault line, earthquakes and volcanic eruptions would occur all along said fault line until the tectonic plates had resettled, a process that (AFAIK) takes on the order of months (my plate tectonics is rather patchwork, so I’m judging by the series of quakes along the southern part of Asia after the big one in Iran a couple of years back).

    - JS

  12. #12 Miguelito
    November 16, 2006

    The eruptions at the Deccan traps started LONG before the Chicxulub impact happened (several million years before).

  13. #13 DuWayne
    November 16, 2006

    Just the coincidence of a number of bad things happening close together in time.

    Close together in time being a relative term. Some of these, likely occured tens, if not hundreds of thousands of years apart. And some of the procceses being described, began million of years before mass-extinction. Certainly, we are talking about a span of time greater than that of human history.

    That said – in terms of earth scale, the whole of human history is a tiny blip.

  14. #14 Dave S.
    November 17, 2006

    I too have heard of the too-ing and fro-ing vis Deccan Trap vulcanism verses bolide impact for some years now. From the geologists I know, the concensus appears to be that the job was being mainly done by the volcanoes and the Chixilub impactor was the coup de gras. Sort of like a boxer being beaten to a pulp by body blows before the wicked right hook to the head fells him. But that doesn’t mean all the now extinct species fell dead immediately after impact like that unfortunate pugilist. Some lived on for a while afterwards, mortally wounded as it were. I suppose that unless you have an inside track into the nuances, all you get is the short and sweet version of the story, which is a sexy one – one asteroid wiping out the dinosaurs, end of story.

    If you look at the Permian Extinction (which makes the K-T extinction look like a picnic at the beach with Rihanna) there is a similar debate going on, with even more complicating factors. In that case its the Siberian Traps verses an Antarctic impact verses glaciation verses the formation of Pangaea (which reduced shallow aquatic environment and brought into contact diverse species, many of which lost out in the ensuing competition) verses a nearby Supernova explosion verses a combo of the above.

  15. #15 SteveF
    November 17, 2006

    Pertinent to this discussion are a couple of recent papers which demonstrate that there was no significant dino diversity decline prior to the K/T event.

    Wang, S.C. and Dodson, P. (2006) Estimating the diversity of dinosaurs. PNAS, early online edition

    Despite current interest in estimating the diversity of fossil and extant groups, little effort has been devoted to estimating the diversity of dinosaurs. Here we estimate the diversity of nonavian dinosaurs at {approx}1,850 genera, including those that remain to be discovered. With 527 genera currently described, at least 71% of dinosaur genera thus remain unknown. Although known diversity declined in the last stage of the Cretaceous, estimated diversity was steady, suggesting that dinosaurs as a whole were not in decline in the 10 million years before their ultimate extinction. We also show that known diversity is biased by the availability of fossiliferous rock outcrop. Finally, by using a logistic model, we predict that 75% of discoverable genera will be known within 60-100 years and 90% within 100-140 years. Because of nonrandom factors affecting the process of fossil discovery (which preclude the possibility of computing realistic confidence bounds), our estimate of diversity is likely to be a lower bound.

    From this paper:

    Given the temporal resolution currently available, then, we find no evidence for a decline in dinosaur diversity near the end of the Cretaceous, a result consistent with a sudden extinction due to bolide impact.

    In addition, we have:

    Fastovsky et al, (2004) Shape of Mesozoic dinosaur richness. Geology, 32, p.877-880.

    The richness of Mesozoic Dinosauria is examined through the use of a new global database. Mesozoic dinosaurs show a steadily increasing rate of diversification, in part attributable to the development of new innovations driving an increasing variety of behavioral strategies. The data do not suggest that dinosaurs were decreasing in richness leading to extinction during the last ;10 m.y. of the Cretaceous. Refinement of the dating of dinosaur fossils, rather than the collection of more dinosaurs, is the best way to resolve globally the rate of the Cretaceous-Tertiary dinosaur extinction.”

  16. #16 Miguelito
    November 17, 2006

    If you look at the Permian Extinction (which makes the K-T extinction look like a picnic at the beach with Rihanna) there is a similar debate going on, with even more complicating factors. In that case its the Siberian Traps verses an Antarctic impact verses glaciation verses the formation of Pangaea (which reduced shallow aquatic environment and brought into contact diverse species, many of which lost out in the ensuing competition) verses a nearby Supernova explosion verses a combo of the above.

    The age of the suspect crater in Antarctica is highly suspect. It is buried under a mile of ice and therefore impossible to accurately date for the time being. It might be as old as the P-T boundary, but it could just as equally be considerably younger.

    Geochemical evidence from sedimentary rocks shows that there was shallow-water anoxia, suggesting that almost the entire ocean at the time was starved of oxygen, probably from the release of volatiles from the Siberian traps. This would have killed off the marine ecosystem quite easily and synchronous climate change could have easily ruined the terrestrial ecosystem as well.

  17. #17 NJ
    November 17, 2006

    I keep thinking about compression damage to a skull on the opposite side from a concussive blow to the head LOL!

    Actually…

    On Mercury, there is a highly disrupted terrain 180 degrees from a major impact crater. I know it has been suggested that the focusing of shockwaves from an impact could have fractured the lithosphere and allowed a large mantle plume to erupt; hence the idea of looking for craters at the antipodes of large igneous provinces (like the Noril’sk flows – Antarctic crater).

    I would tend to agree with Ed (are you sure you’re not really a geologist?) that a “magic bullet” argument – one thing or the other is simplistic. Blind men feeling the elephant and all that.

  18. #18 John L.
    December 20, 2006

    If anyone is willing to think outside the box, and evaluate a theory that is neither the bolide nor the volcanic one, I would recommend viewing:
    http://www.dinoextinct.com

  19. #19 Coin
    December 20, 2006

    The Earth’s continents merged into a single massive continent prior to the Mesozoic Era. This super-continent is known as Pangea (or Pangaea). This continental consolidation and attendant mountain building, resulting from plate tectonic collisions, created an environment where the ambient gravitational force at the lower elevations and flood-planes of the super-continent was much lower than it is today.

    The effect of the consolidation of the Earth’s land masses and the creation of tall mountain ranges and volcanic peaks lowered the strength of the gravitational field in Pangea, especially at the lower elevations. This is comparable to an object being lowered into a hole toward the Earth’s center as described in the previous section “Review of The Earth’s Gravitational Field.” As mentioned the Earth’s mass above the object would partially cancel the attractive force of the Earth within the sphere below the object.

    The lowered gravitational field permitted the flora and fauna of Pangea to grow to a size which would not be possible today.

    …apparently “outside the box” means “in the complete absence of any support whatsoever by mathematics or physics”.

  20. #20 John L.
    December 21, 2006

    Hello Coin,

    Your point is well taken. “Thinking outside the box” in this instance, to me, means the following:

    If there are two mainstream theories, each of which are deficient so that neither one can overcome the issues or objections raised to establish a “slam dunk”, so to speak, then other theories should be evaluated. I believe this to be true if the alternate theory (or theories) can successfully address those deficiencies.

    Mathematicians, physicists, geologists, paleontologists, etc. hopefully will evaluate this theory. As you know, the scientific method demands a thorough evaluation of a new theory and hopefully, that will come soon.

  21. #21 Coin
    December 22, 2006

    Mathematicians, physicists, geologists, paleontologists, etc. hopefully will evaluate this theory. As you know, the scientific method demands a thorough evaluation of a new theory and hopefully, that will come soon.

    Okay. Well, in that case, speaking as a mathematician, I would like to let you know that I have evaluated the theory and come to the conclusion that it contains no math. None at all. I checked twice.

    As I am not a physicist, geologist, or paleontologist, meanwhile, I will refrain from commenting on those other aspects.

  22. #22 Nebogipfel
    December 22, 2006

    You mean it wasn’t because the dinosaurs started eating soybeans…!??!?

  23. #23 W. Kevin Vicklund
    December 22, 2006

    Assume that the argument John L. links to is accurate in its simple-minded explanation. That is, having really high mountains lowers the surface gravity due to the mass of the mountains, and there are no mitigating factors. Let’s now assume that these mountains are on average as high as Mt. Everest (call it H=9,000 m for ease of calculation). Let’s further assume that these mountains are so shear as to be effectively a solid mass.

    From Wikipedia, we dan derive a usable formula for the resulting acceleration of gravity, g. Neglecting latitude, as it is going to be constant for our purposes here, the formula is g=9.780-0.000003086h-0.0000011(H-h), where h is the height in meters above sea level. The first term is the base value of g, which would normally be modified by latitude. The second term is the free air correction, which accounts for the effect of the additional distance from the center of the earth. The final term is the Bouguer correction, which accounts for the extra mass. So Let’s see how much of a difference these hypothetical mountains would make. At sea level, h=0, reducing the equation to g=9.780-0.0000011(9,000). Multiplying out, we get g=9.780-0.0099, meaning that the maximum effect of the high mountains means the gravitational acceleration is reduced by about 0.1%. The effect of latitude on g is up to 0.5%, with g increasing as we approach the poles. And it should be obvious that the free air correction has a much greater effect as well.

    But that is being highly conservative. The mountains are not going to be effectively solid, as in the assumption above; rather, they would be more accurately approximated as cones or pyramids, reducing the Bouguer correction to a third of its original value.

    The hypothesis doesn’t even survive it’s first brush with Wikipedia. I think it’s safe to call it DOA.

  24. #24 kehrsam
    December 22, 2006

    According to this theory, shouldn’t all marine creatures be giants as well? Surely the reduction of gravity via buoyancy is greater than the hypothetical reduction caused by the Pangaea land mass.

  25. #25 John L.
    December 22, 2006

    Hello W. Kevin,

    The land masses involved here are not merely mountains; they are continents. It would take a very good mathematician to calculate the resultant gravitational force. To do this….he would probably need a computer to calcuate the answer.

    Hello Kehrsam,

    Not all marine creatures would become giants for the same reason that not all land creatures became giants. An animal, sea or land based would reach the optimum size, within the limits of gravity, based on survival criteria. A mouse from 100 million years ago might be identical to one today because it has reached its ideal proportions for survival. For example, low food requirements, a size which allows it to escape into small crevices to escape predators. In other words, there is no advantage to getting bigger.

  26. #26 W. Kevin Vicklund
    December 22, 2006

    The land masses involved here are not merely mountains; they are continents. It would take a very good mathematician to calculate the resultant gravitational force. To do this….he would probably need a computer to calcuate the answer.

    John, if you read what I wrote, especially in light of the Wikipedia article I linked, you’d realize that I assumed continents in my description. I, in fact, performed the calculation most advantageous to your hypothesis. I assumed a 9,000-meter-thick slab of continent that stretches to infinity! Any calculation designed to be more accurate will inevitably make the change in gravitational acceleration even less pronounced than the already miniscule amount I calculated.

    The reality is that simply changing altitude or latitude is going to have a much greater effect on the local acceleration of gravity than clustering the continents together and getting some tall mountain ranges.

    The crux of your error is that you fail to take into account the relatively small mass of the combined continents in relation to the mass of the rest of the earth. As big as the continents are, there simply isn’t enough mass to counter the mass of the earth.

  27. #27 W. Kevin Vicklund
    December 22, 2006

    BTW, I just realized a minor error in the equation I derived. I wrote:

    g=9.780-0.000003086h-0.0000011(H-h)

    The correct equation is:

    g=9.780-0.000003086h-0.0000011(H-2h)

    Unfortunately for John, this makes no difference in the result of my calculation, and more importantly, as altitude increases, the impact of the consolidated landmass decreases even faster than I originally calculated. For those curious, if we assumed Pangaea to be covered by conical or pyramidal mountains instead of a 9,000-meter-thick slab, the proper equation (keeping latitude constant, remember!) would be:

    g=9.780-0.000003086h-0.0000004(H-3h)

    (for the pedants, the unit for g is m/s/s)

  28. #28 John L.
    December 22, 2006

    Hello W. Kevin,

    Thank you for spending the time to research that info.

    When you stated that:
    “The crux of your error is that you fail to take into account the relatively small mass of the combined continents in relation to the mass of the rest of the earth. As big as the continents are, there simply isn’t enough mass to counter the mass of the earth.”

    I believe you have overlooked the inverse square effect of gravitation. As the continents coalesced, their influence on the gravitational strength on the surface of the super-continent would have significantly overcome part of the balance of the earth’s effect. The dense core would have been about 6400 km away and the opposite hemisphere even further away.

  29. #29 W. Kevin Vicklund
    December 26, 2006

    I believe you have overlooked the inverse square effect of gravitation. As the continents coalesced, their influence on the gravitational strength on the surface of the super-continent would have significantly overcome part of the balance of the earth’s effect. The dense core would have been about 6400 km away and the opposite hemisphere even further away.

    Actually, I haven’t overlooked it – it’s part of the equation I derived. But you are failing to consider that only a minuscule part of Pangaea would actually pull an object at sea level away from the center of the earth – the rest would actually pull it towards the center of the earth.

    Let’s continue with the utterly absurd Pangaea I initially constructed, where the average height of Pangaea was taller than the height of Mt. Everest, 9,000 meters. The average height of modern continents is about 750 m. If we make the reasonable assumption of constant volume, that means that the surface area of Pangaea would be about 8% that of modern land surface area. That means Pangaea would cover about 2.4% of the earth, or about 12,000,000 km^2. (See why I said it’s absurd?)

    Now, imagine a tangential plane slicing through the earth at sea level in the middle of Pangaea. The mass above the plane will pull an obsrver away from the earth, whereas a mass below the plane will pull the observe towards the earth. To give you a sense of scale, a spot of paint less than 0.2 mm thick and 40 mm in radius on a basketball would be equivalent to Pangaea on the earth. How much of that spot would be sliced off to reveal the tiniest speck of orange?

    When we do the math, it turns out that the surface area of that slice is about 360,000 km^2. That means that only 3% of Pangaea would reduce g at sea-level (and in fact even less, because I am being generous in my calculations). The remaining 97% is increasing g, and since it is no longer evenly distributed over the face of the earth, but is now closer to the observer, g is going to be higher!

    Now, if you assume a more realistic scenario, that Pangaea is covered by Himalaya-style mountains, you discover that the diameter of the slice is much less than that of the Tibetan Plateau, which is on average 4,500 meters above sea level. In other words, we currently have conditions that mimic a realistic Pangaea!

    I understand you will not accept the reality of your pet hypothesis being disproven. Unfortunately, science is rather impersonal – it doesn’t care what you would like to be true. And what you would like to be true can readily be proven to not be true.

  30. #30 John L.
    December 26, 2006

    Hello W. Kevin,

    After re-reading your original post, I realized you were trying to solve a different problem; the expanding earth scenario.

    Your last post makes the assumption that the surface are of Pangea was 8% of the current continental land mass area. I’m not sure how you arrived at that figure. It’s probably closer to 100%.

    When you take the tangential slice, you fail again to apply the inverse square rule relating to “d” (distance) in the gravitation law (/d squared). Mass closer to the point in question (i.e., near the surface of Pangea) has a much greater influence on the gravitational pull than that of mass further away.

    Land mass beyond the tangential slice has less of an impact for two reasons:

    1. Again (/d squared), i.e. distance is greater.

    2. The component of the gravitational force from land masses beyond the slice are not directing their full gravitational pull toward the center of the earth. If you draw the vector diagram of the forces involved, this will become apparent.

    This is not a back-of-the-envelope calculation. It will take a computer model which simulates the movement of the continents and includes:

    1. Simulation of the consolidation and breakup.
    2. Info on continental land mass densities.
    3. Info on Mesozoic terrestrial topologies including mountain ranges and their height and width.
    4. Mesozoic sea levels, which varied considerably.
    5. The actual calculation of “g” using integral calculus to sum the total contributions from all segments of the land masses. That is in effect, a sum of a lot of vector quantities.
    6. And, that calculation would probably have to be done using the altitude on Pangea as another variable because the gravity gradient (i.e., the change in gravitational pull, a body’s weight, based on elevation) might have been very high.

  31. #31 W. Kevin Vicklund
    December 27, 2006

    After re-reading your original post, I realized you were trying to solve a different problem; the expanding earth scenario.

    Your last post makes the assumption that the surface are of Pangea was 8% of the current continental land mass area. I’m not sure how you arrived at that figure. It’s probably closer to 100%.

    Wrong! In fact, I am not assuming an expanding earth scenario, which is proven by the surface area comment. The mass for Pangaea has to come from somewhere. If the earth isn’t expanding, the continents must have roughly constant mass – which means roughly constant volume. Which means that if height increases, surface area decreases. But if we assume no expansion and same area, then we must also assume same height. Which means I am justified in using current conditions to make arguments.

    When you take the tangential slice, you fail again to apply the inverse square rule relating to “d” (distance) in the gravitation law (/d squared). Mass closer to the point in question (i.e., near the surface of Pangea) has a much greater influence on the gravitational pull than that of mass further away.

    BZZZT! The one failing to take into consideration inverse square law is you. Since you are declaring no expansion and roughly constant area, we can look at the effect of the mass of the Himalayas and specifically the Tibetan Plateau as our starting point for a “proof of concept”. The Plateau is high enough and large enough that a tangential slice will not encompass the whole mass. Therefore, it is a proper base point.

    Land mass beyond the tangential slice has less of an impact for two reasons:

    1. Again (/d squared), i.e. distance is greater.

    2. The component of the gravitational force from land masses beyond the slice are not directing their full gravitational pull toward the center of the earth. If you draw the vector diagram of the forces involved, this will become apparent.

    Let us then draw a vector diagram, one for Pangaea and one for the modern Tibetan Plateau. First, the earth, represented by a circle – this is identical for each diagram. Next, the continents, which on Pangaea is a single large arc on the circle, and on the modern scenario, is a number of smaller arcs distributed about the circle. Finally, the tangential slice, which is identical on both.

    It should be immediately obvious that the effect of the main body of the earth and the effect of the tangential slice are identical, and the only difference is the effect of the continents. For Pangaea, the continents are closer, which means that, guess what, they have a stronger pull than continents on the other side of the earth, though the portion directed to the center of the earth is less (however, the inverse square is a larger factor than the angle). So Pangaea would have a higher g, not a lower g, if we only consider landmass.

    This is not a back-of-the-envelope calculation. It will take a computer model which simulates the movement of the continents and includes:

    But a back-of-the-envelope calculation can tell us whether or not the hypothesis is even reasonable – which it is not.

    1. Simulation of the consolidation and breakup.

    Not necessary for proof of concept.

    2. Info on continental land mass densities.

    Already included in previous calculations, unless you think the densities are going to be wildly different than they currently are (and I seriously doubt you have any supporting evidence)

    3. Info on Mesozoic terrestrial topologies including mountain ranges and their height and width.

    This is a valid point, and I have dealt with it above. Unless you believe that there is going to be an expanding earth or a major reduction in land area, it is quite reasonable to assume Himalayas and Tibetan Plateau as representative.

    4. Mesozoic sea levels, which varied considerably.

    Ah, here we come to the real problem with your hypothesis. The continents and the oceans (including the ocean bed) are floating on the dense mantle. Since they are floating, displacing a continental mass means displacing an equal mass of ocean and mantle. You have failed to consider this effect in your hypothesis – and although I identified it immediately, I purposefully ignored it to give your hypothesis a fair trial. Since everything is floating, the conditions at sea level will be essentially constant, then and now (unless you assume expanding earth)! There will be local variations due to stiffness and elasticity of material, but these are quite tiny. In fact, the largest local variation (for landmasses), as measured by satellite, occurs in the highest mountain ranges and such as the Himalayas. This variation is on the order of 60 mGal, or about 0.0006 m/s/s.

    5. The actual calculation of “g” using integral calculus to sum the total contributions from all segments of the land masses. That is in effect, a sum of a lot of vector quantities.

    Great if you want to find exact numbers, but unnecessary for the proof of concept.

    6. And, that calculation would probably have to be done using the altitude on Pangea as another variable because the gravity gradient (i.e., the change in gravitational pull, a body’s weight, based on elevation) might have been very high.

    I agree – the effect of altitude is going to be a bit higher gradient than the effect of the mass of a tiny sliver of mountain range. It’s also a well known calculation that doesn’t change due to mass anomolies, and compared to the magnitude of g, is quite tiny.

    To sum up: the consolidation of the continents to form Pangaea has a negligible effect on local g. While there may have been some high mountains, this effect is minimal and no different than the effect of high mountains on modern earth.

    Go spread you ignorance to other people, John. Or better yet, let go of your flawed hypothesis and actually learn about how gravity really works.

  32. #32 John L.
    December 28, 2006

    ******************************************************************************
    Hello W. Kevin,

    Your posts are becoming so wordy and filled with assumptions, irrelevant and erroneous statements, I don’t know where to begin.

    1. In response to my statement that Mesozoic terrestrial topologies would have to be an important element in the computer simulation model, you write that “I have dealt with it above.”

    You have not dealt with it. When the Indian sub-continent collided with the Asian continent about 26-8 mya, it raised the Himalayan mountains. When the continental land masses collided some 400 mya to form Pangea, would it not be reasonable to assume that the mountain ranges raised at that time were significantly greater than that produced by the much smaller land mass of the Indian subcontinent? The Pangean terrain then would have been far different from what we see today because of erosion over a few hundred million years. I don’t think it’s unreasonable to assume soaring mountain ranges on Pangea.

    2. “The mass of Pangea has to come from somewhere.” ??????

    Of course it had to come from somewhere. It came from 2 places:

    A. The continents that coalesced to form Pangea.
    B. The mountain building (see 1. Above) with mass supplied by the earth’s upper mantle.

    Throughout your response you erroneously posit a zero-sum-game relating to continental mass and volume vis-a-vis the formation of Pangea. Let me give you one example to show the flaw in your logic. When the Indian sub-continent, mentioned above, drifted north toward the Asian continent, it passed over the Reunion hot spot, a deep mantle plume. This is what created the Deccan Traps. The Indian sub-continent increased in both mass and volume due to that volcanic activity.

    The same is true of land masses where mountain building is the result of colliding tectonic plates. The source of the added mass in this case is the upper mantle. This is a major flaw in your logic. You have not accounted for the significant increase in mass when Pangea was formed……. and that mass increase was at high elevations. This also invalidates your statement “I am justified in using current conditions to make arguments.”

    3. “Since you are declaring no expansion and roughly constant area, we can…….” in your “BZZZT” paragraph.

    Wrong again. When I stated that the surface area of Pangea was close to the combined current continental values, I was referring to the footprint areas. Total topological area (including mountains, valleys, etc.) would be quite different. Based on 2. above, it is clear that Pangea’s initial topological area would have to be well over 100% of current continental measurements…….you claimed 8%. But we are going astray here. IT’S NOT ABOUT TOPOLOGICAL AREA, THAT’S IRRELEVANT. IT’S ABOUT MASS AND ITS DISTRIBUTION.

    4. Your “vector diagram” explanation is a little fuzzy. When you attempt to make the point that the Pangean surface beyond the tangential slice is contributing to the weight of an object on Pangea more than if the continents were in their current position, you correctly mention that a small component is involved but conveniently dispense with any mathematical quantification to verify the former statement. That’s why a computer simulation model is necessary.

    5. Your response to my listing Mesozoic sea levels, which I wrote varied considerably, as an important criteria to be used by computer simulation model was:

    “Here we come to the real problem with your hypothesis.”
    And ,
    “Since everything is floating, the conditions at sea level will be essentially constant, then and now.”

    This is a major scientific faux pas. I suggest you study the Mesozoic literature. Sea levels varied considerably during the Mesozoic. As a matter of fact, they rose dramatically toward the end of the Cretaceous (in the late Campanian-early Maastrichtian). If you would like to confirm that info:

    see “Night Comes To The Cretaceous” by James Laurence Powell
    the Vail Curve graph is on page 167

    This is a well written book but, I believe, the author has arrived at the wrong conclusion for the K-T extinctions.

    The reason sea levels are highly relevant should be apparent to you. You suggested taking a tangential slice of Pangea at SEA LEVEL to argue your views. Clearly, lower sea levels would produce a much larger mass within that tangential slice…………mass that would be acting counter to the earth’s gravitational pull. Another plus for this theory. As a matter of fact, when I viewed that sea level chart with very low sea levels during the Triassic-Jurassic Periods and a dramatic rise in the Cretaceous, I began to wonder if a graph of the sea-level gravity on Pangea would mimic this one very closely. Our repartee has strengthened my belief that the Gravity Theory is the correct one.

    I have to conclude once more with the hope that professionals, who are objective and do not have a bias regarding the present subject………….that leaves out you and me, with the expertise and tools (including computer simulation modeling), will come forth to render a verdict on this theory.

  33. #33 W. Kevin Vicklund
    December 28, 2006

    You keep on harping about the gravitational attraction of Pangaea. It simply is not as large a number as you think it is.

    Forget about the earth. Let’s take Pangaea in isolation. In fact, we are going to go to an even greater level of abstraction.

    Assume the universe consisted of an infinite slab of density p and thickness h. Nothing else existed in this universe, exept for an observer stnding on the slab. What is the gravitational attraction g that that observer will experience?

    To answer that, we need to solve Gauss’s Law for an infinite slab. Gauss’s Law is the general equation for mechanical gravity (we’re not in the ranges where relativity of quantum effects play a part). For spheres and point sources, Gauss’s Law simplifies to the familiar inverse-square law. But for infinite slabs, it has a different formula:

    g=2piGph

    Note that there is no distance variable involved, nor is there a mass variable. Merely a constant (2piG), a density variable (p), and a height variable (h).

    We know G, we know the average density of continental material, and we have previously designated the thickness of the slab. But to refresh your memory:

    G=6.6742×10^-11 m^3/s/s/kg
    p=2,670 kg/m^3 (note that 3,300 is the density of the mantle, creating an upper limit)
    h=9,000 m

    Pay attention to something. G is really small, and we’re multiplying. The magnitudes of p and h are 10^3, or combined 10^6, which means that we are looking at something two or three orders of magnitude larger than 10^-5. Indeed, when we do the math, we discover that:

    g=0.01008 m/s/s or 1.008 Gal

    This is about 0.1% the standard value for g.

    For a finite, thin disk, such that h<

    Therefore, the maximum effect of Pangaea's mass is about 0.1% of current gravity. Note that this is without considering the earth at all!

    Let us now consider the effect of sea level. I don’t have access right now to your source. According to a different source, Mesozoic sea levels started out at current levels, and steadily rose to 150-300 meters by mid-Cretaceous, and then declined. But remember – because the crust is floating on the mantle and the earth is constant mass, rising global sea levels is no worse than a change in altitude. Did I mention that there’s an equation for change in gravity due to altitude? Here it is:

    g=-3.086×10^-6 /s/s times h

    So for a change in sea level of 300 meters, that is equal to:

    g=-9.258×10^-4 m/s/s or 92.58 mGal

    This is less than 0.01% of current gravity. Alternatively, we can calculate the change via the inverse square law, which also gives us slightly less than 0.01%. Or as I said above, essentially constant. Of course, g at sea level increases if the sea level drops…

    Likewise, altitude gives a very small number. For 9,000 meters, the change is:

    g=-0.0278 m/s/s or 2.78 Gal

    Which is a bit under 0.3% of current gravity. Of course, being on top of Pangaea means that Pangaea is now pulling down on you…

    To sum up:

    Pangaea simply could not generate enough gravitational attraction to make a significant dent in surface level gravity. To get it to even 1% of standard g, Pangaea would have to average 90,000 meters above sea level!

  34. #34 W. Kevin Vicklund
    January 2, 2007

    I tried to respond earlier, John, but my comment got tagged for moderation (looks like one of my computers got unlucky in IP Russian Roulette). Since Ed hasn’t put it through, I’m typing a new comment.

    Your hypothesis absolutely hinges on the gravitation exerted by Pangaea. So let’s set aside all the arguments of altitude and sea level changes and placement of continents, and see if we can determine what the maximum effect of Pangaea would be.

    Remove the earth from consideration. Since gravitation is a sum of vectors, we can analyze individual parts and then add it all togethr. So let’s just look at Pangaea as if it were floating in space, with no other objects to distort gravity.

    Now, we need to pick a point of observation. Pangaea would obviously curve to fit the surface of the earth, and we are looking for the effect at sea level, which is where Pangaea and the earth meet. So we should pick a point on the inner curve. But that means that a good portion of Pangaea will have a vector component pulling the wrong way. So rather than try to calculate the opposing forces, let’s flatten out Pangaea to a disk of the same thickness and density – the radius of that disk will be much greater than the thickness. It should be obvious that the gravitational attraction should be now greater, as all points of Pangaea are pulling “up” to some degree. Next, compare an infinite slab of the same thickness and density as Pangaea. It should be obvious that an infinite slab should have a greater gravitational pull than Pangaea, because it obviously has more mass – Pangaea is merely a section of the infinite slab.

    The next question is, can we calculate the gravitational attraction of an infinite slab of known thickness and density? If we can, that puts an upper limit on the gravitational attraction of Pangaea, as logically shown above.

    The answer is: Yes! Gauss’s Law, which for spheres and popint sources simplifies to the familiar inverse square law, has a solution for infinite slabs of thickness h and density p (I’m using p instead of the Greek rho due to the similarity in appearance):

    g=2piGph

    Now, 2pi G is a constant, so we only have two variables that we have to worry about, p and h. We have already defined h as our average thickness of Pangaea; when I suggested 9,000 m, John didn’t have any objection. So for now, let’s take h = 9,000 m. Average continental density is 2.7 g/cm^3, or 2,700 kg/m^3. Since it is floating, it must have a lower density than the mantle, which is 3.3 g/cm^3 – this gives us an upper limit on the density of Pangaea. For now, let’s use the modern number of 2,700 kg/m^3.

    Now, if we plug in these numbers, we get

    g=0.0102 m/s/s or about 0.1% of normal earth gravity at sea level

    Even at the highest possible density, the gravitational attraction of Pangaea is less than 0.13% of normal g. And since g is directly proportional to h and p, we can say that for every 9,000 m of average height at normal continental density, Pangaea reduces gravity at sea level by 0.1%; at the maximum density, every 7,000 m of height reduces gravity by 0.1%. And it should be noted that these are maximum values.

    Again, Pangaea simply is not massive enough to make any significant impact on gravity.

    Now, let’s start looking at sea level effects. John makes a huge deal about the Vail curve, particularly how low it was in Triassic-Jurassic periods. Unfortunately, John has overlooked a vital data point – present day sea levels. In fact, the Vail curve clearly shows that the “very low sea levels” he finds so intriguing are nearly equal to current sea levels. The Mesozoic sea level change ranges from a maximum of about +300 m (late Cretaceous – though more recent estimates put it at only +150 m) to a minimum of -50 m (early Jurassic) of modern levels. That means that sea level changes had an imperceptible effect of much less than 0.01% of modern gravity (not including the effect of altitude).

    I have previously discussed the effect of altitude on gravitation. To put it into the same terms used here, every 3,000 m of altitude reduces g by about 0.1%. That means that altitude is more important than the thickness of Pangaea by a factor of at least 3-1.

    As far as the rest of the continents shifting around, it should be immediately obvious that the impact of the shifting of masses smaler than Pangeaea at distances of thousands of kilometers should certainly be less than the impact of Pangaea’s mass on a nearby observer – regardless of whether the effect is positive or negative.

    Finally, even if we do assume the same footprint (by the way, that is what I was talking about in regards to surface area), the mass of Pangaea is less than 0.1% of the mass of Earth. Therefore, the reduction in surface gravity due to removing the mass of Pangeae from consideration is less than 0.1%.

    Tally all the various effects together, and even at best case for John’s hypothesis, you get much less than 1% variation in gravity. John’s hypothesis simply can’t stand up to any level of scrutiny, no matter how much he tries to hide behind his lack of a computer model.

  35. #35 John L.
    January 3, 2007

    ********************************************************************************************************************
    Hello W. Kevin, Happy New Year.

    I figuratively ran into a brick wall when you suggested “let’s set aside all the arguments of altitude and sea level changes and placement of continents…”

    We can’t do that because they are extremely important elements here. If I refer back to the Vail Curve, referenced earlier, there are certain relevant paleontological facts that makes sea level changes important to the present discussion. The sauropods, our main focus of dinosaur gigantism, began their relatively rapid growth in size at the same point in time that the sea levels dropped to their lowest during the Mesozoic Era. This was at the Triassic-Jurassic boundary (about 208mya). Sauropods thrived until the mid-Cretaceous when the sea levels reached their highest levels. They all disappeared by then except for the titanosaurs. Could these be two strange coincidences? Maybe, but I’d like to treat them as possible clues to the K-T extinctions.

    Is it worthwhile to think about why the sea levels dropped to such low levels at that time? It might be. Not having researched this point and not having a lot of knowledge about geology, I can only speculate. I don’t believe there was anything comparable to today’s polar ice caps, so my guess is as follows. During the formation of Pangea, the continental collisions produced enormous episodes of mountain building, as well as volcanic activity. In order to supply the additional Pangean mass, oceanic plate subduction would have increased proportionately, thereby deepening the ocean floor and lowering global ocean levels.

    As I write this, I’m wondering if I haven’t omitted other factors. If world sea levels dropped over 1000 ft to fill in ocean voids left by plate subduction, it would imply a mass increase on Pangea far beyond what I would have imagined. I’ll have to do further research on this aspect of sea level change.

    Attempting to flatten Pangea into a disc that tangentially contacts the earth at a single point has to be rejected. It’s ok to simplify a complex problem but not to completely disregard the basic structure. By doing that, for example, you eliminate the important effect of a substantial drop in sea level. If you take a spherical object, say a grapefruit for example, and take a tangential slice so that the mid-point of the slice is, say 1/4 ” thick, and then repeat the same at ½”, 3/4″,.., to simulate a deepening sea level, it’s apparent that with each small increase in depth, a geometric (although “geometric” might not be the precise mathematical term) increase in volume is obtained. And, all of this volume contributes to the anti-earth gravity vector. Gauss’s law doesn’t apply here; only the basic Newtonian Universal Law of Gravitation applies.

    To assume that sea levels with the continents in their current position has the same effect as when the continents were combined, with much greater mass, at much higher altitude, is faulty. As I pointed out, the reason why sea levels were low at the Triassic-Jurassic boundary, even lower than today, were for different reasons. Stating that “sea level changes had an imperceptible effect of much less than .01% is unsupported. You must use the “sliced grapefruit” model to calculate the impact, not the “floating disc” model.

    I have not looked at your calculation in detail because, as I pointed out, the “floating disc” model is not valid. Also, the assumed value of 900m for Pangea’s thickness is an under-estimate. That figure is probably accurate for current continental thickness at sea level. Pangea’s average thickness, based on the scenario mentioned above (i.e., extensive mountain building, etc.) would be much greater. And, depending on whether my guess as to why sea levels fell is correct, the thickness would be even much greater than that. After all, if the Indian sub-continent collision could create the Himalayas, collisions by much larger continents could have produced commensurately more mass on Pangea.

    I hate to repeat the statements in the last paragraph of my prior post at the risk of sounding like a broken record, and I won’t. I believe they still apply.
    ///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

  36. #36 Torbjörn Larsson
    January 3, 2007

    Attempting to flatten Pangea into a disc that tangentially contacts the earth at a single point has to be rejected. It’s ok to simplify a complex problem but not to completely disregard the basic structure. By doing that, for example, you eliminate the important effect of a substantial drop in sea level.

    I think you miss the point here. Kevin has consistently calculated the upper limits of what your hypothesis is capable of. If you insist to insert the real geometry, the effect will decrease.

    Discussing sea level will not help much, since the average sea level and density is much less than the block of crust you are discussing. You will still not reach the 1 % figure that Kevin *very* generously compared with.

    Kevin has first identified the reason why the theory is a dud, and then put in all this effort to explain why effects of real geometry are marginal, effectively letting you go on and consider new and possibly correct explanations for the extinction. Isn’t that great?

  37. #37 Torbjörn Larsson
    January 3, 2007

    “the average sea level” – the average sea depth

  38. #38 W. Kevin Vicklund
    January 4, 2007

    I want to congratulate John for proving he is incapable of understanding both the written word and basic laws of geometry and gravity, and also for presenting false data.

    We can’t do that because they are extremely important elements here. If I refer back to the Vail Curve, referenced earlier, there are certain relevant paleontological facts that makes sea level changes important to the present discussion. The sauropods, our main focus of dinosaur gigantism, began their relatively rapid growth in size at the same point in time that the sea levels dropped to their lowest during the Mesozoic Era. This was at the Triassic-Jurassic boundary (about 208mya). Sauropods thrived until the mid-Cretaceous when the sea levels reached their highest levels. They all disappeared by then except for the titanosaurs. Could these be two strange coincidences? Maybe, but I’d like to treat them as possible clues to the K-T extinctions.

    The changes in sea level may very well have something to do with various extinctions. Certainly, the record would suggest there might be a connection. The idea that gravity was the factor is what is in dispute here, and the numbers simply do not add up to a significant change.

    Is it worthwhile to think about why the sea levels dropped to such low levels at that time? It might be. Not having researched this point and not having a lot of knowledge about geology, I can only speculate. I don’t believe there was anything comparable to today’s polar ice caps, so my guess is as follows. During the formation of Pangea, the continental collisions produced enormous episodes of mountain building, as well as volcanic activity. In order to supply the additional Pangean mass, oceanic plate subduction would have increased proportionately, thereby deepening the ocean floor and lowering global ocean levels.

    And my analysis did manage to take that into account, remember? Every 9,000 m of Pangaea thickness reduces sea level gravity by a whopping 0.1% of modern value – if sea level also dropped, add that to the proposed thickness of Pangaea. But you also have to remember inverse square law – when sea levels drop, the distance to the center of the earth also drops, meaning gravity rises – and I demonstrated that the effect of altitude changes is three times greater than the effect of thickness changes.

    As I write this, I’m wondering if I haven’t omitted other factors. If world sea levels dropped over 1000 ft to fill in ocean voids left by plate subduction, it would imply a mass increase on Pangea far beyond what I would have imagined. I’ll have to do further research on this aspect of sea level change.

    Where did you get the 1,000 foot figure from? The Vail curve shows that the world sea levels only dropped by 50 m of current levels, or about 165 feet (by my calculation, about 1/6 of what you claim). (This is about the same as the minimum level during the last Ice Age, btw). Next time, try to offer accepted numbers.

    Attempting to flatten Pangea into a disc that tangentially contacts the earth at a single point has to be rejected. It’s ok to simplify a complex problem but not to completely disregard the basic structure. By doing that, for example, you eliminate the important effect of a substantial drop in sea level. If you take a spherical object, say a grapefruit for example, and take a tangential slice so that the mid-point of the slice is, say 1/4 ” thick, and then repeat the same at ½”, 3/4″,.., to simulate a deepening sea level, it’s apparent that with each small increase in depth, a geometric (although “geometric” might not be the precise mathematical term) increase in volume is obtained. And, all of this volume contributes to the anti-earth gravity vector. Gauss’s law doesn’t apply here; only the basic Newtonian Universal Law of Gravitation applies.

    I have not, in fact, neglected drops in sea level. I specifically stated the problem solution in such a way that any such drop could be accounted for – change in height is directly proportional to change in gravity. So if sea level drops by 90 m, the change in gravity at sea level due to Pangaea’s mass will be 0.01% of current value lower. Of course, when you factor in the fact that you are now closer to the center of the earth, gravity increases by 0.03% for a net increase of 0.02%.

    But let’s look at your grapefruit. You clearly don’t understand the model. Take a grapefruit with part of the peel missing, about 70%. The remaining peel represents the continents, the surface of the grapefruit represents sea level. If the remaining peel is one contiguous blob, that represents Pangaea. Now, take a small slice off the peel such that the slice is tangential to the surface of the grapefruit. The part of the peel you just sliced off is the only part that will reduce gravity – the rest of Pangaea will pull the sea level observer towards the earth. But in my approximation, I flattened the peel so that all of Pangaea was serving to reduce gravity. The tangential slice will always be less massive than the whole of Pangaea, no matter what the sea level is. The relative change in mass of the tangential slice will increase quicker than that of the disk, true, but the mass of the disk will always be greater. Therefore, we can use the thin disk model as an upper limit, knowing that the actual effect will be necessarily less. And again, I reiterate: the thin disk model can account for the lowering in sea level by increasing h by the amount sea levels dropped.

    Gauss’s Law applies to all solid objects (usual caveats for quantum and relativity) – it is the general expression of Newton’s Law for a solid object. Newton’s Universal Law of Gravitation of course applies (same caveats), but it only takes on the simple inverse-square form for spheres and point sources. For non-spheres, you must use Gauss’s Law. Granted, I simplified the problem, but I did so in a conservative manner, in that the actual effect must by the geometry be less than the calculated effect.

    To assume that sea levels with the continents in their current position has the same effect as when the continents were combined, with much greater mass, at much higher altitude, is faulty. As I pointed out, the reason why sea levels were low at the Triassic-Jurassic boundary, even lower than today, were for different reasons. Stating that “sea level changes had an imperceptible effect of much less than .01% is unsupported. You must use the “sliced grapefruit” model to calculate the impact, not the “floating disc” model.

    Since, as I demonstrated above, the “sliced grapefruit” model always has a smaller effect on gravity than the “thin disk” model, I can apply change in sea level to the “thin disk” model and know that the result will be greater than the “sliced grapefruit” model. (I should point out that at the Triassic-Jurassic boundary, sea levels were approximately the same as current, it wasn’t until the early Jurassic that it dropped a small amount below modern levels) But if we take the drop in sea level to be the conversion of sea floor plate into continental mass, that gives us about 2 1/3 times the drop in sea level as additional height in Pangaea, for a total of 3 1/3 times the drop. Using the Vail curve, that’s 50 meters times 3 1/3 or about 165 m (is that where you got the 1,000 feet value?). This gives us a change in gravity of under 0.002% using the “thin disk” model – the “sliced grapefruit” model must be even less than this.

    I have not looked at your calculation in detail because, as I pointed out, the “floating disc” model is not valid. Also, the assumed value of 900m for Pangea’s thickness is an under-estimate. That figure is probably accurate for current continental thickness at sea level. Pangea’s average thickness, based on the scenario mentioned above (i.e., extensive mountain building, etc.) would be much greater. And, depending on whether my guess as to why sea levels fell is correct, the thickness would be even much greater than that. After all, if the Indian sub-continent collision could create the Himalayas, collisions by much larger continents could have produced commensurately more mass on Pangea.

    You obviously didn’t look at my model much at all, because you not only didn’t understand the model, you aren’t even close to the numbers used. I used 9,000 m, not 900 m, which is higher than Mt. Everest! To get even 1% change in gravity, Pangaea would have had to be 90,000 m high! You are also overstating the impact of the collision. Yes, the larger continents would have more mass – but the force of the collision would have been spread out over a larger boundary. You are making a great deal of assumptions here and stating them as fact.

    I hate to repeat the statements in the last paragraph of my prior post at the risk of sounding like a broken record, and I won’t. I believe they still apply.

    I seriously doubt you would accept a professional computer model. You would claim that the model was incorrect, that they didn’t take some factor into account (especially if they used realistic figures, not the fantastically high numbers we’ve been dealing with).

  39. #39 W. Kevin Vicklund
    January 4, 2007

    Another point – the titanosaurs were the largest of the sauropods. Your gravity explanation for extinction of the sauropods ought to have the largest going extinct first, not last. And of course, this doesn’t account for the extinction of the smaller dinosaurs.

    Also, it doesn’t make sense for sauropods to live in mountainous terrain. Think on that.

  40. #40 Torbjörn Larsson
    January 5, 2007

    if sea level also dropped, add that to the proposed thickness of Pangaea

    Ah, sea level drop by for example ice. I assumed it was about the geometrical effects of water redistribution in the model of the land mass. Should have looked at the references to get a clue.

  41. #41 John L.
    January 6, 2007

    *************************************************************************************************

    Hello W. Kevin,

    I’ll try to be concise with my response to your latest post.

    “It doesn’t make sense for sauropods to live in mountain terrain. Think on that.”
    You are 100% correct, we both agree on that. That’s precisely the point. The theory posits that a high gravitational gradient existed on Pangea at that time. At the lowest elevations, the flood plains, inland areas that might be below sea level, etc., would have the lowest values. Some would describe it as a gravitational anomaly. As a matter of fact, if it was discovered that they (the largest dinosaurs) thrived on high altitude plateaus, that would be a serious problem for this theory.

    “The titanosaurs were the largest of the sauropods” and therefore should have gone extinct first. This is not quite right. A few of the largest sauropods were titanosaurs but most were medium-sized sauropods. Also, among the titanosaurs, the largest ones did become extinct earlier than the smaller ones…..well before the K-T boundary. They were significantly different from the earlier sauropods. You’ll have to re-read the section “The Titanosaur Dilemma.” All one can do is look at the available info and then try to piece together the pieces of the puzzle. It is almost impossible to reach a conclusion that is 100% verifiable. It would be like me asking you to prove that Quetzalcoatlus could fly in today’s environment. With all the advanced technical tools at your disposal, you could probably come up with a model that could lift off the ground. But without the real live flying reptile, there would never be 100% certainty.

    Concerning the drop in sea levels, you have to forget about current conditions. One of the main questions about the change in sea level is: What was the amount of the transfer of mass to Pangea from the oceanic plate subduction process? Since the formation of Pangea was a gradual one that started in the Silurian Period (440-395mya) as Laurasia and Gondwanaland formed and continued to coalesce well into the Permian Period (280-225mya), it is the total drop in sea level during this interval that must be used. The Vail Curve specifies around 1000 feet. I hear the grumbling in the background, but we have to follow the information currently available. Again, current conditions cannot be used for comparison. Polar ice caps and millions of years of terrestrial planation and a different continental distribution makes it impossible to extrapolate current conditions back to that period.

    Your 9000m estimate of Pangea’s thickness is so far off that your calculation has no meaning. The following is a section from Wikipedia dealing with plate tectonics. In other words, this is current continental plates (sans Mt. Everest).

    “The plates are around 100 km (60 miles) thick and consist of lithospheric mantle overlain by either of two types of crustal material: oceanic crust (in older texts called 1. sima from silicon and magnesium) and continental crust (sial from silicon and aluminium). The two types of crust differ in thickness, with continental crust considerably thicker than oceanic (50 km vs 5 km).”

    Unless I have made errors in my assumptions, the following things have to be take in account in “The Model.”

    1. A large transfer of mass to Pangea commensurate with a 100 ft drop in worldwide ocean levels.

    2. The transferred mass was of mantle origin and therefore denser than crustal material.

    3. And, by virtue of the above, in effect, elimination of a corresponding mass that would have been part of the earth-centric gravity vector.

    Your calculation of a .1% variation in gravity has to be rejected because you have not considered 1.,2., and 3. above, you have used erroneous estimates of tectonic plate thickness and I’m not sure your model is a valid one. Also, for background information purposes, I have to point out that gravity differentials (i.e., the value of “g”) among various cities of the world are up to about 1%.

    ///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

  42. #42 John L.
    January 6, 2007

    Correction to the typo above. That should be 1000 ft, not 100ftin 1. above.

  43. #43 Torbjörn Larsson
    January 8, 2007

    Polar ice caps and millions of years of terrestrial planation and a different continental distribution makes it impossible to extrapolate current conditions back to that period.

    Besides that you can only gain another 2 – 3000 m (average sea depth) to the 9000 already discussed, you will now also add another unrealistic assumption. “but we have to follow the information currently available” and should use historical variation.

    “The two types of crust differ in thickness, with continental crust considerably thicker than oceanic (50 km vs 5 km).”

    Because crust floats – otherwise you would see 45 km deep seas. The 9 km block considered is already well above what could amass from continental block collisions. Look at Mars’ vulcanos or Ganymede’s or Europa’s faults: the reason why the largest are so much larger than Earth’s is not primarily absence of erosion but absence of developed block dynamics.

    And again, you would not reach much more, since we are discussing well under 1 % difference.

    gravity differentials

    Due to local differences in rock density. Note that in real cases some of the difference goes against what you seem to assume: “The Bouguer anomalies usually are negative in the mountains because of isostasy: the rock density of their roots is lower, compared with the surrounding earth’s mantle.” ( http://en.wikipedia.org/wiki/Gravity_anomaly )

    Also, considering the much less than 1 % difference during average density, you should consider this as an argument against your assumed gravity differences as a factor in evolution.

  44. #44 John L.
    January 8, 2007

    ***************************************************************************************************************
    Hello Torbjörn,

    I think we are going astray here. The point that I and W. Kevin were trying to establish was whether the massing of the continents to form Pangea could alter the gravitational field on Pangea.

    I believe the best way to do that is to do the following:

    Assume the masses involved are concentrated at points. The Earth’s mass is at its center and that of Pangea would also be at its center at a point near but just outside where the circumference of the Earth would be.

    Then calculate the mass of Pangea using the current continental land mass area, the 100km depth and density as specified by Wikipedia. Include also the 350m sea level drop to compute the mass transfer. Wikipedia provides the current ocean area, so this is easy to compute.

    Then using the point where the Earth’s circumference would be, between the Earth’s center and Pangea’s center, compute “g” at that point(using Newton’s law) for the earth and for Pangea. Then compare. The “d” for the Earth would be the radius and the “d’ for Pangea would be a small distance and different values could be used (e.g. 10km, 100km, etc.) Then compare the values to see if the results are reasonable.
    //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

  45. #45 W. Kevin Vicklund
    January 8, 2007

    Let’s skip the discussion portion for now and go straight to the calculable claims.

    Your 9000m estimate of Pangea’s thickness is so far off that your calculation has no meaning. The following is a section from Wikipedia dealing with plate tectonics. In other words, this is current continental plates (sans Mt. Everest).

    “The plates are around 100 km (60 miles) thick and consist of lithospheric mantle overlain by either of two types of crustal material: oceanic crust (in older texts called 1. sima from silicon and magnesium) and continental crust (sial from silicon and aluminium). The two types of crust differ in thickness, with continental crust considerably thicker than oceanic (50 km vs 5 km).”

    Congratulations, John, you have managed to stumble upon a completely meaningless statistic. The “thickness of Pangaea” that I refer to is not the total plate thickness. Rather, it is the average height of Pangaea above sea level. The Wikipedia entry for Earth gives the average continental height as somewhat less than 1/5 of 3,794 m (or less than 760 m). So my assumption of 9,000 m is about 12 times greater than current values – and I’ve stated all this before. For the first level approximation, I set aside the effects of changing sea level, but stated the equation for the solution in such a way that those effects could be added. So all you have to do is modify by the appropriate amount, and I’ve given several examples of how to do so. The fact that you refuse to even make the attempt is quite telling. However, since you seem incapable of performing math, I will do your work for you yet again.

    Unless I have made errors in my assumptions, the following things have to be take in account in “The Model.”

    1. A large transfer of mass to Pangea commensurate with a 1000 ft drop in worldwide ocean levels.

    2. The transferred mass was of mantle origin and therefore denser than crustal material.

    3. And, by virtue of the above, in effect, elimination of a corresponding mass that would have been part of the earth-centric gravity vector.

    Your calculation of a .1% variation in gravity has to be rejected because you have not considered 1.,2., and 3. above, you have used erroneous estimates of tectonic plate thickness and I’m not sure your model is a valid one. Also, for background information purposes, I have to point out that gravity differentials (i.e., the value of “g”) among various cities of the world are up to about 1%.

    First of all, it’s 0.1% per 9,000 m. All you have to do is calculate the additional height due to sea level change. I already did it for 50 m, for 165 m (1000 ft) just multiply by 3.3. But that doesn’t take into account the density, so let’s combine items 1 and 2 into a single calculation.

    As I said previously, if sea level dops due to conversion of mass below the ocean surface to continental mass, the resultant height change is the sea level change plus the height due to conversion. Assuming approximately the same continental footprint as current (which John has stated as part of his model), that means that to maintain equal volume, the additional height must be 2 1/3 times the sea level change. However, density must also be accounted for, so we must multiply the volume factor by the density of mantle (2.7) and divide by the density of continental material (3.3). So our equation for the additional height above sea level is:

    h=s(1+7/3*3.3/2.7) or h=3.85s

    Let’s be generous and round up to 4. So for every meter sea level drops, the average height of Pangaea above sea level increases by 4 times. This means that a drop in sea level of 1000 feet (=165 m) results in an increase in Pangaea’s thickness of 660 m. Thus, the decrease in gravity (not accounting for altitude changes, which hurts John’s model) due to sea level changes is at most a whopping 0.007% of current values.

    As for part 3, technically I’ve already eliminated the corresponding mas, but let’s examine that as if I hadn’t. If we assume the same continental footprint as current, we can calculate the amount of mass to eliminate from the earth-centric vector. The equation is:

    m=pAh

    Plugging in numbers,

    m=(2700 kg/m^3)*(150×10^12 m^2)*(9,660 m)
    m=3.9×10^21 kg

    This seems like a lot, but Earth’s mass is 5.97×10^24 kg. This means that the reduction in gravity due to mass reduction in the earth-centric vector is less than 0.07%.

    Add all three factors up, add it to the 0.1% calculated for the base amount, and you get a grand total of 0.177% change in gravity from current standards.

    Now, John has noted that there is a much more significant change in gravity between various modern cities. That is because we have not addressed the largest factor of all – latitude. Gravity at the poles is 0.5% greater than at the equator. This accounts for almost all the variation in gravity between various locations on the earth’s surface. John has claimed “up to 1%” change between cities. However, Wikipedia lists a number of cities’ gravities on it’s Earth’s gravity page – the biggest difference is 0.4%, and each city listed is very close to the value predicted by its latitude. Perusing the online sources, it looks like John got his 1% number from news sources talking about the release of GRACE data. However, the GRACE press release itself merely said less than 1% – it wasn’t very specific, and a value of 0.5% would be entirely within the model.

    Sorry John. Your pet hypothesis simply doesn’t stand up to the numbers.

  46. #46 W. Kevin Vicklund
    January 8, 2007

    John, your latest post assumes a) dinosaurs walked benath the outermost part of the mantle and could therefore survive submersion in magma, and b) Pangaea can be modelled by compacting it to a sphere more dense than uranium. Neither one of those assumptions is even close to being reasonable. If you insist on using the sphere model, model Pangaea as a set of perfectly packed spheres (multiply the results by 1.25 to account for the gaps). It’ll take a lot of math, but it will be a hell of a lot more accurate than what you are proposing.

  47. #47 W. Kevin Vicklund
    January 9, 2007

    Sorry, correction. I got some previously calculated numbers crossed up. I should have used 305 m, not 165 m. Might as well just double the rest of the numbers that relied on that:

    total additional height = 1,320 m
    reduced gravity due to sea level = 0.014%
    reduced gravity due to mass reduction = 0.07%
    combined reduced gravity = 0.184%

    net difference between this comment and the previous = 0.013%

  48. #48 Torbjörn Larsson
    January 9, 2007

    John:

    I think we are going astray here.

    No, but you keep on discussing a model you assume is correct while not considering what is realistic. I tried to point out why you don’t assume the total crust thickness in a difference calculation, because the block floats. Kevin goes over that again – but I won’t.

    It is evident that your ideas doesn’t work, but instead of learning and go on to more fruitful ideas you try to save a hopeless case by throwing in any objections and numbers you can come up with. That is contraproductive.

    Kevin:

    if sea level dops due to conversion of mass below the ocean surface to continental mass,

    Ah, I missed that assumption. Thanks, now the sea level drop makes sense.

    That is because we have not addressed the largest factor of all – latitude.

    Yes, that provokes a vague memory. It is easy to assume Earth surface is equilibrated with gravity, but rotation has its effects. Thanks again!

  49. #49 John L.
    January 9, 2007

    **************************************************************************************************
    Hello W.Kevin,

    You are correct. The dinosaurs couldn’t walk there. The whole purpose of using the simple model was to determine the rough ratio of Earth/Pangea gravitational pull at that point. The calculation finds that the gravitational pull of Pangea would be far greater than the Earth’s at that point. Therefore, the question is: how far above that point (i.e., moving outward at a higher elevation on Pangea) would we get a ratio that would be in line with this theory. As I tried to point out earlier, this is not a back-of-the-envelope calculation.
    //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

  50. #50 W. Kevin Vicklund
    January 10, 2007

    Assume the masses involved are concentrated at points. The Earth’s mass is at its center and that of Pangea would also be at its center at a point near but just outside where the circumference of the Earth would be.

    This is an unacceptable model as applied. Since we are dealing with finitely large objects in close proximity, the most basic model would be to model the objects as spheres, knowing that this creates an upper bound for each object. Spheres have an identical calculation as a point source so long as you are at or above the surface, but for points internal to the sphere, only the mass closer to the center counts towards the calculation. If you try to use the total mass while inside the sphere, you will get bogus results. This is very important.

    Then calculate the mass of Pangea using the current continental land mass area, the 100km depth and density as specified by Wikipedia. Include also the 350m sea level drop to compute the mass transfer. Wikipedia provides the current ocean area, so this is easy to compute.

    Hopefully by now John has realized his error in using the thickness of the continents below sea level to argue for the proposed thickness above sea level. So let’s go back to the model I have been using. For simplicity, let’s just use 10 km as the thickness – that will pretty much take care of the proposed changes due to sea level change. As John has obliquely hinted, density is important to the calculation we need to make. In fact, it is the determining factor as to whether his model is reasonable. If mass and density remain the same, then so must volume. Therefore, we can equate the two volume equations to find out how big a sphere must be modeled for the model to approximate reality (units in km^#):

    A*h=4/3*pi*r^3
    150×10^6*10=4/3*pi*r^3
    r=710 km

    So to model a 10 km high Pangaea as a point source with full mass, the point must be located at least 710 km from the obeserver. Any closer than that, and the model for Pangaea must start losing mass or be declared inadequate. So what are the real effects of moving closer? Gravity due to distance is inversely proportional to the square of the radius, but gravity due to mass is directly proportional to the cube of the radius (this is for inside a sphere of constant density). That means that gravity is proportional to the radius; here, that means if the point source is only 10 km from the observer, gravity is only 1/71 that calculated for the full mass at the surface.

    For those interested, gravity at the surface is 0.536 m/s/s, or about 5.5% of modern gravity. For a point source 10 km away, gravity is only 0.00755 m/s/s, or 0.08%. In any case, it should be clear by now that John’s basic model is wildly inaccurate and must be rejected.

    You are correct. The dinosaurs couldn’t walk there. The whole purpose of using the simple model was to determine the rough ratio of Earth/Pangea gravitational pull at that point. The calculation finds that the gravitational pull of Pangea would be far greater than the Earth’s at that point. Therefore, the question is: how far above that point (i.e., moving outward at a higher elevation on Pangea) would we get a ratio that would be in line with this theory. As I tried to point out earlier, this is not a back-of-the-envelope calculation.

    I bolded a very important phrase. According to John, his model predicts the gravitational pull of Pangaea should exceed that of the Earth’s, by a great amount, no less! The first two rules of model building are:

    1. Make sure the model does not make invalid assumptions. (already violated as shown above)

    2. Make sure the results make sense.

    John, the red flags should have been raised as soon as you calculated that Pangaea’s gravity exceeded the Earth’s – this simply does not make sense. Think about it – the mass of the Earth is much greater than Pangaea’s, and it’s density is at least as great as Pangaea’s. So the best you can say is that the 10 km of Pangaea above the observer negates the effect of 10 km of Earth below the observer. And if Pangaea gets higher, it is a simple matter of proportion.

    You complain about me using “back-of-the-envelope” calculations. But these are the very same calculations used by geologists for the base geoid, and they are accurate to within 99.98%. Your models, on the other hand, are so innaccurate that they can only be called bogus – they aren’t even “back-of-the-envelope”.

  51. #51 John L.
    January 12, 2007

    ************************************************************************************************************
    Hello W. Kevin,

    We can both agree on one thing, the simpler the model, the less accurate the result.

    I’ve gone through the calculation you did to determine the radius of a (Pangean) sphere tangent to the reference point. I used the Wikipedia values of area, mass, density, etc. and your 10km number to calculate the radius corresponding to a halving of an object’s weight at the reference point.

    I did use a larger value of the Earth’s radius (6400km instead of 6378) because I believe the equatorial bulge was greater when the continental land masses were consolidated. I don’t know if my estimate is close to being accurate or not. It might even be very conservative. This is where the full 100km of continental plate would have had an impact. I also don’t know if there was an effective increase in Pangean altitude at that time that reached maximum value at the equator.

    The result I got was a radius of about 335km, less than half the value you got. Since Pangea was not a sphere, we have to deform the sphere so that it is closer to the Earth and spread out. As the top of the sphere gets closer to the reference point, it has a greater influence because of ( /d squared) and the bottom of the sphere moves away and has a lesser, but non-negative effect for the same reason. Can I calculate the exact net effect when the flattening out process results in what we believe is the way the Pangean super-continent was formed? No, I’ll leave that to the experts.

    Could some of the assumptions either of us used be wrong? Yes. Could we have ignored other conditions we don’t know about? Yes. For example, experts tell us that in the early Devonian period (395mya-345mya), a period long before the dinosaurs appeared but a time during which Pangea was being formed, the following is true. There were 400 days in the year, the moon was much closer to the Earth and the Earth was spinning faster.

    I mention the above only to make the point that we cannot assume all current conditions have always existed.

    I’ll repeat what I stated earlier. It will take experts in the relevant fields, those that have the resources and expertise, to create a computerized model to simulate the Earth’s Mesozoic conditions. That’s the only way a definitive answer can be determined.

    //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

  52. #52 W. Kevin Vicklund
    January 12, 2007

    John, I’ll compose a reply, but could I ask a favor? I’m having trouble duplicating your numbers. Could you show us your math? I’ve tried to do so for you as much as possible, so it’s not like I’m asking you to do something I wouldn’t do. I’m asking sincerely. I know some people use this request as an attack on a position, which is why I am being so cautious in my request.

    Thank you kindly.

  53. #53 John L.
    January 14, 2007

    *****************************************************************************
    Hello W. Kevin,
    Could you show your calculation using the info I provided in my last post? That is, using the Newtonian Gravity law and the calculation for one half reduction of weight at the reference point.
    ////////////////////////////////////////////////////////////////////////////

  54. #54 W. Kevin Vicklund
    January 14, 2007

    I asked you first. I’m trying to figure out what the hell you mean by it. I either get about 6,500 km (using constant density) or 235 km (ignoring density and the effect of going inside the sphere).

  55. #55 W. Kevin Vicklund
    January 15, 2007

    I have a feeling John is not willing to reveal his innumeracy, so I’m going to go ahead and discuss my own results.

    Essentially, John is asking a non-sensical question, on the order of “If I accelerate at 50 km/hr/hr, how long until I’m going twice the speed of light?” The result is useless because the model exceeds the bounds of utility – you can’t go faster than the speed of light; relativity kicks in first and blows up the model’s assumptions of Newtonian mechanics. In John’s case, I can’t perform a “calculation for one half reduction of weight at the reference point” because there is no reference point using that mass and density that gives a result of g=4.9 (what I believe John means by “one half reduction of weight”). A sphere is very close to the maximum gravity you can get at a point on a surface of an object with uniform density. The object that maximizes surface gravity at a point is described by the equation:

    R=k*sqrt(sin(theta)) (graphed in the polar plane, k is a constant to get the correct radii)

    This gives a slightly deformed sphere similar to an oblate spheroid, except that one hemispheroid has a smaller polar radius than the other. An object at the “flatter” hemispheroid’s pole will experience a gravitational attraction about 3% greater than that of a perfect sphere of the same mass and density. The equatorial radius (re), near polar radius (rn), far polar radius (rf), and ratio of re:rn can be described as:

    re=k*pi/4*cos(51.9)=0.618k
    rn=k*pi/4*sin(51.9)=0.484k
    rf=1-rn=0.516k
    re:rn=tan(51.9)=1.276

    The exact angle (given here in degrees) is calculated as arccos ((pi/4)^2).

    So what all does this mean? Basically, we can squeeze the perfect sphere a little bit and increase the gravity to 3% higher than a perfect sphere. Any more deformation, however, and the gravity begins to decrease. The more you squeeze it, the closer you get to a thin disc. We also know that for an oblate spheroid with a polar radius of rp, the maximum gravity that can be experienced at a point on the surface(specifically one of the poles) will be somewhere in the range rn

    So let's start doing the math. First, the mass of Pangaea, calculated from Wikipedia's land surface area (yes, it is footprint area, not topographical area, I checked the math) and density, and our assumption of Pangaea's average height above sea level:

    m=pAh
    m=2700*150x10^12*10,000
    m=4.05x10^21 kg

    Next, we calculate the radius of our perfect sphere:

    m=pV
    m=p*4/3*pi*r^3
    r^3=4.05x10^21/2700/4*3/pi
    r=710 km

    Now, we can calculate gravity on the surface of the sphere:

    g=Gm/r^2
    g=6.6742x10^-12*4.05x10^21/(710x10^3)^2
    g=0.536 m/s/s or about 5.5% of normal gravity

    Now we calculate the maximum g an object of Pangaea's mass and density can exert:

    gmax=g*1.03
    gmax=0.536*1.03
    gmax=0.552 m/s/s or about 5.6% of normal gravity

    It is physically impossible for Pangaea to reduce the weight of an object by half. (By the way, the equation for the maximizing shape and the 3% figure came from the sci.physics.research newsgroup and was derived by both equation and computer modelling - these are from the experts. The equations for the radii are mine.) And as calculated several times before, the reduction in gravity due to loss of mass is less than 0.1%, for a net max of a 5.7% reduction in weight - nowhere near the 50% John is asking me to find.

    Although we've found the absolute max, it may be of interest to see what the smallest possible reference distance for that max is. Let's set rp=rn:

    Vsphere=Voblate
    4/3*pi*r^3=4/3*pi*rp*re^2
    (710x10^3)^3=rn*(1.276rn)^2
    rn=rp=604 km
    re=770 km

    As you can see, we are quite a ways from a target of 10 km for a reference distance. And we therefore know the gravity will be much less than our max calculated. But let's look at the dimensions of our oblate spheroid when we have a reference distance of rp=10 km.

    r^3=rp*re^2
    (710x10^3)^3=10x10^3*re^2
    re=5,984 km

    In fact, re is so much greater than rp (almost 600 times greater) that it is appropriate to approximate it as a thin disc of radius re and height 2*rp. In fact, a thin disc of those dimensions will have a slightly greater gravity than the enclosed oblate spheroid. But why should we approximate Pangaea as a thin disc twice it's average height when we can more accurately approximate it as a thin disc at it’s average height? That calculation we have already done numerous times.

    You see, we keep coming back to the thin disc model as the simplest model that gives accurate results. And note the following assumptions:

    mass of the earth is virtually unchanged (this is in accordance with experts)
    mass of hydrosphere is virtually unchanged (this is in accordance with experts)
    radius of earth at sea level is as per Vail curve (ie, data from experts)
    ratio of water:land footprint approximately the same (your assertion)
    density of various objects unchanged (you appear to agree)
    angular velocity of earth unchanged (a quick calc assuming a steady slowing puts this effect at less than 0.1% reduction in g – we can throw this in in the future if you want)
    average height of Pangaea more than ten times higher than current (assumed by me, tacitly agreed upon by you, laughed at hysterically by the experts)
    Pangaea is concentrated (assumed by you and me, experts disagree)

    Really, the only assumption we’re really going out on a limb on is the average height of Pangaea – and you’re not going to like what the experts model it as.

  56. #56 John L.
    January 16, 2007

    ********************************************************************************************************
    Hello W. Kevin,

    You have proven what I stated in a prior post:
    “The simpler the model, the less accurate the results.”

    You have also proven what you stated in a prior post:
    “Make sure the model does not make invalid assumptions.”

    Pangea wasn’t a bowling ball touching the Earth at a single point nor was it a floating disc. It was closer to that grapefruit slice mentioned earlier but even that model isn’t exact because the land mass wasn’t a simple circular one. I understand your desire to simplify the problem to make it amenable to basic mathematical principles, but what is needed is an accurate computer model. A model where all the known parameters, whether we think them significant or not, are taken into account and can be easily varied to see their net effect.

    All the assumptions that you describe as “This in accordance with the experts” have to be considered their opinions. For some of those opinions it would be counterintuitive to disagree with them. But, you must remember that until about 40 years ago it would have been not only counterintuitive but downright heresy to believe that the Earth’s continents could slide over the globe.

    The issues relevant to what we are addressing are still being debated today. Plate tectonics, ancient marine transgressions and their causes, climate and glaciation, inner Earth dynamics and changes in mantle viscosity, mantle plumes, polar wander and its effect on sea levels and equatorial bulge, etc., are still subjects of many theories. In other words, there are big gaps in our understanding of how a lot of things work.

    The ratio of water:land footprint area being unchanged over 200 million years is something I assumed. But that could be a false assumption. 200my of terrestrial planation would not only lower the average continental elevation but probably increase the footprint areas over time.

    The Earth’s angular velocity might be basically unchanged but polar wander and its effect on the equatorial bulge and sea levels could be significant.

    If the experts laughed hysterically at your estimate of the average height of Pangea above sea level, they will have to explain how they know that a mountain range had eroded away over 200my and, if they can do that, how high it was originally. They would also have to explain where the apparent mass transfer to Pangea went based on the fall in sea levels. I’d also remind them that the Siberian Traps deposited 1.5 million cubic kilometers of lava onto Pangea (compared to the one cubic kilometer resulting from Mt. St. Helens). Most of their study is done in sedimentary environments where microfossils are abundant.

    Earth science is still in its infancy.

    Circumstantial evidence cannot, by itself, confirm any theory but that evidence cannot be ignored. Scientists should be able to explain how land animals approaching the size of the largest whales would be able to inhabit today’s world. Animals that would have to consume sufficient food, with a skull far too small for its body size, to support a herd-like lifestyle; able to rise from a prone position; have the agility to build nests, hatch eggs and care for their just-hatched young. I’d like to see an analysis of the blood pressure necessary to supply blood to their head. I’d like to see an analysis of the above by the experts.

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  57. #57 W. Kevin Vicklund
    January 23, 2007

    Pangea wasn’t a bowling ball touching the Earth at a single point nor was it a floating disc. It was closer to that grapefruit slice mentioned earlier but even that model isn’t exact because the land mass wasn’t a simple circular one. I understand your desire to simplify the problem to make it amenable to basic mathematical principles, but what is needed is an accurate computer model. A model where all the known parameters, whether we think them significant or not, are taken into account and can be easily varied to see their net effect.

    And you continue to illustrate my point. I calculated the maximum effect Pangaea’s mass could have. The grapefruit slice must have a smaller gravitational effect than a thin disc, because a grapefruit slice only partially fills a thin disc. Similarly, by spreading the landmasses into a less compact form reduces Pangaea’s effect even more, because it necessarily moves the mass further away. The more detailed the model gets, the smaller the calculated impact is. We could do a more detailed investigation, but there’s no point if the maximum effect isn’t even close to the proposed model. By the way, I’m not just a layman here. Although I am not a geophysicist or the like, I am an elctrical engineer, and have received extensive training in modelling electrostatics. Electrostatics uses the same equations and concepts as gravity, just different constants and units. I dislike arguments from authority, which is why I have not brought it up before, but I do have the necessary knowledge and experience to evaluate your claims. Frankly, I’ve been extremely kind to your model, which is untenable. Also, I have been taking into account all the variables and expressing them in ways that can be easily varied. Simply put, it is physically impossible to even approach the types of gradients you are suggesting.

    All the assumptions that you describe as “This in accordance with the experts” have to be considered their opinions. For some of those opinions it would be counterintuitive to disagree with them. But, you must remember that until about 40 years ago it would have been not only counterintuitive but downright heresy to believe that the Earth’s continents could slide over the globe.

    It is not just their “opinion” – it is also the result of detailed calculations and computer models, which you claim to trust. Additionally, the only reason Wegener’s theories were disregarded 80 years ago was that there was a lack of mechanism to explain how the plates could move. Once evidence for that mechanism arose, the model was accepted rapidly, as it explained very accurately all geophysical phenomena. It certainly wasn’t heretical to believe in continental drift 40 years ago.

    The issues relevant to what we are addressing are still being debated today. Plate tectonics, ancient marine transgressions and their causes, climate and glaciation, inner Earth dynamics and changes in mantle viscosity, mantle plumes, polar wander and its effect on sea levels and equatorial bulge, etc., are still subjects of many theories. In other words, there are big gaps in our understanding of how a lot of things work.

    Ah, the infamous “gap” problem -the recourse of crackpots everywhere. If we do have these huge gaps, computer modelling won’t mean a damn thing. If, instead, these so-called gaps are merely representative of disagreement over the relative impact various processes have, then we can produce a quite accurate model, within a reasonable margin of error. In that case, a computer model is useful – but it also means that back-of-the-envelope calculations are useful for eliminating erroneous hypotheses. So you have the choice of claiming we don’t know enough to attempt a computer model, or we know enough to destroy your model with carefully evaluated approximations.

    The ratio of water:land footprint area being unchanged over 200 million years is something I assumed. But that could be a false assumption. 200my of terrestrial planation would not only lower the average continental elevation but probably increase the footprint areas over time.

    But you forget that in addition to planation, there is uplift. The earth’s surface remained active during that time. In fact, the water:land footprint has changed a bit – up to 37% land at the lowest sea levels – but almost all of that is from changing sea levels. More on sea level change later.

    The Earth’s angular velocity might be basically unchanged but polar wander and its effect on the equatorial bulge and sea levels could be significant.

    Welcome to Bovine Scatology. In fact, the effect of polar wander on equatorial bulge and sea levels is maximized when there is no polar wander at all. Introduce polar wander, and the equatorial bulge decreases as it tries to shift to the new equator. Until it does, the equatorial g is greater than it was, and the polar g is less than it was, thereby making the difference between the two less than without polar wander. Certainly, it doesn’t help your hypothesis. As far as change in angular velocity goes, using a change rate of 17 s/million years (the highest calculated rate I could find, and thus the most favorable to your hypothesis) gets us a 23-hour day at the beginning of the sauropods. This corresponds to an equatorial bulge less than half a kilometer thicker than current, and that combined with the centripetal force means that the equatorial gravity would be reduced by much less than 0.1% from current. Of course, this also means that the polar gravity would be slightly higher than current.

    If the experts laughed hysterically at your estimate of the average height of Pangea above sea level, they will have to explain how they know that a mountain range had eroded away over 200my and, if they can do that, how high it was originally.

    They have and they do. The Appalachians are one such mountain range. Try googling to see how the experts have modelled them. Using computer models, the experts have determined that the average height of continents is nearly the same as it is now (in fact, the physics of floating bodies tells us it must be so). Over 80% of Pangaea was less than 1000 m above sea level, and only a couple of mountain ranges exceeded 2000 m above sea level.

    They would also have to explain where the apparent mass transfer to Pangea went based on the fall in sea levels.

    You are assuming a lateral transfer of mass. This is incorrect. In fact, the mass transfer was down, not sideways. You see, as oceanic crust cools, it becomes dense, and therefore takes up less volume. With the exception of glaciation, almost all change in sea level is due to the rate at which new seafloor is generated (which is measured by the rate at which it spreads). Slow spreading means that less new seafloor is generated, therefore there is more of the dense old seafloor – which means lower sea levels. Subduction converts old oceanic plate into mantle, although a small amount is converted into continental material as it melts. You have an amazing lack of knowledge of the basic processes that govern plate tectonics for someone who is basing a radical hypothesis based on those same processes.

    I’d also remind them that the Siberian Traps deposited 1.5 million cubic kilometers of lava onto Pangea (compared to the one cubic kilometer resulting from Mt. St. Helens).

    Which, spread out over the entirety of Pangaea, means a layer about 10 m thick. Except Pangeaea is floating, which means that would in reality only increase the average height above sea level by about 2 meters. Colour me unimpressed.

    Most of their study is done in sedimentary environments where microfossils are abundant.

    Not true, and an objection that has no bearing on the issue.

    Circumstantial evidence cannot, by itself, confirm any theory but that evidence cannot be ignored.

    I don’t know, you seem to be doing a really good job at ignoring the evidence.

    Scientists should be able to explain how land animals approaching the size of the largest whales would be able to inhabit today’s world.

    Well, then, it’s a good thing they have.

    Animals that would have to consume sufficient food, with a skull far too small for its body size, to support a herd-like lifestyle;

    Metabolism plays a large role in this – an ectotherm doesn’t need to eat nearly as much as an endotherm of equivalent bodymass. Large animals also don’t need to eat as much per body mass as small animals. A skull need only be big enough to hold a brain and pass food through to the throat. If the largest sauropods didn’t chew their food and instead used gizzards to grind their chow, they would be able to pass a greater amount of forage through their mouths. Morphology of the extant skulls agrees with this. Also, climate models (including computer models) indicate that forage may have been more lush and faster growing during that time, allowing for smaller ranges.

    able to rise from a prone position; have the agility to build nests, hatch eggs and care for their just-hatched young.

    They might not need to rise from a prone position upon reaching adulthood, but you should also consider that their long, muscular necks and tails would provide assistance in doing so. As far as child-rearing, simple nests don’t require much agility at all (assuming that they even laid eggs), eggs don’t necessarily require parental care, and newly hatched young often are able to feed themselves. Parental care may be as simple as protecting the eggs and young from predators.

    I’d like to see an analysis of the blood pressure necessary to supply blood to their head.

    Analyses by the experts (including computer modelling) place blood pressure of Brachiosaurus at about 500 mmHg, 4-5 times that of a healthy human. Reliable sources place the highest sauropod blood pressure at around 600-650 mmHg. This requires a very large heart – and the size heart indicated by fossil remains is the same size heart calculated for the required blood pressure. The analyses are out there, if you deign to look. But that’s your responsibility, not mine.

    I’d like to see an analysis of the above by the experts.

    Then look for them yourself. It’s out there. But I’m not surprised that you haven’t bothered to look. You have shown a willful ignorance throughout this exchange, and your statement about opinions is clear enough – no matter what evidence or computer model is generated, you will blindly cling to your absurd model, shouting “LALALA I CAN’T HEAR YOU” at the top of your voice.

    The final nail in the coffin is that Earth has a Roche limit of 0.14. That means that a 14% or greater gradient will rip the earth apart. It is quite impossible to get the magnitude gravity reduction you were talking about without destroying Earth in the process.

  58. #58 John L.
    January 25, 2007

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    Hello W. Kevin

    Your last post covered a lot of subjects. I will probably have to use several posts to respond to all of them.

    My comment about the belated acceptance of Wegener’s continental drift theory until close to 50 years after he proposed it is relevant here. Quite a bit of circumstantial evidence to support his belief that the continents had merged into a supercontinent had been given. Not only did the coastlines fit together fairly well but fossils, flora (Glossopteris) common to India, South America, southern Africa, Australia, and Antarctica and fauna, (Mesosaurus) common to South America and Africa reenforced his theory. Wegener was ridiculed by many of his contemporaries who thought the continents were unmovable extensions of the Earth. Did they call his theory “bovine scatology?” Probably.

    Let’s move on to the specific issues in your post. My request for the analysis of biologically permissible blood pressure values for sauropods got this response from you “Then look for them yourself. It’s out there.” Your response indicated that you do not have confidence in your “reliable sources” or “experts” since you don’t want to identify them.

    The numbers you cite, 600-650mmHg for the largest sauropods are probably close to the common estimates but I’m not sure at what part of the body those are for. For a giraffe, for example, the numbers given are 200mmHg at heart level and 400mmHg at foot level (due to gravity). The question though, is whether a sauropod’s physiology could withstand this high blood pressure. You don’t even address this issue.

    On the blood pressure issue, sauropods are often compared to giraffes. We know that giraffes have a blood pressure around 200-400mmHg. We also know that they have a unique sophisticated design of their circulatory system to prevent extreme changes in their blood pressure. They also have a large heart (26 lb) compared to similar sized mammals. We also know that they have to splay their forward legs in order to drink water. One has to ask why nature would construct an animal with such a long neck that had to squat to accomplish such a vital function and put itself in a vulnerable position. Could it be that this would mitigate any life-threatening changes in blood pressure that would result from a rapid swing of the skull over the 15 to 17 ft vertical movement if it wasn’t designed the way it is?

    IF YOU SEARCH FOR INFO ON THE INTERNET ABOUT THIS SUBJECT YOU ENCOUNTER NUMEROUS SITES THAT EXPRESS SKEPTICISM ABOUT WHETHER SAUROPODS COULD SUSTAIN THIS LEVEL OF BLOOD PRESSURE. IF YOU CAN FIND A SINGLE SITE THAT CONFIDENTLY SUPPORTS THIS VIEW YOU ARE LUCKY. I’M STILL LOOKING FOR ONE.

    The following is a description of the giraffe’s vascular system from a website:

    A Giraffe’s heart has to generate around double the normal blood pressure for a large mammal in order to maintain blood flow to the brain against gravity. In the upper neck, a complex pressure-regulation system called the rete mitrebile prevents excess blood flow to the brain when the Giraffe lowers its head to drink. Conversely, the blood vessels in the lower legs are under great pressure (because of the weight of fluid pressing down on them). In other animals. such pressure would force the blood out through the capillary walls: Giraffes, however, have a very tight sheath of thick skin over their lower limbs which maintains high extravascular pressure in exactly the same way as a pilot’s g-suit.
    Giraffes can go for days without water. In order to drink water, the giraffe has to spread its front legs and bend its long neck to the water. This is a dangerous position for the giraffe since it can’t see its enemies and can’t get a fast start running.

    When comparing the giraffe to the barosaurus, here’s basically the result:

    Giraffe Barosaurus

    Height 17ft 40ft
    Weight 3000 lbs 50,000 lbs
    Heart wt 26 lbs 3200lbs
    Neck Length 6ft 30ft

    With all the special physical design characteristics necessary to enable the giraffe to exist in today’s gravitational environment, would barosaurs be able to do the same? Some have even proposed that the barosaurs had 8 hearts to supply blood to the brain!

    For some insight by an expert, read the following:
    (This was a discussion about whether sauropods held their necks in a near vertical position vs horizontal position).
    “And while most dinosaur paleontologists also believe long-necked dinosaurs- collectively known as sauropods – behaved this way, a leading Adelaide University researcher has just published new data which suggests otherwise.

    Dr. Roger Seymour, from Adelaide’s Department of Environmental Biology, has co-authored a paper in the prestigious Proceedings of the Royal Society in London, which argues it was physically impossible for sauropods to behave that way. The paper argues that due to the sauropod’s possible heart size and metabolic rates, the only way they could have functioned on land was with a horizontal neck.

    Dr. Seymour based his findings on his research of the factors which determine heart size in animals. He has spent the past 24 years collecting data on heart morphology and arterial blood pressure in reptiles, birds and mammals to determine how blood pressure influences the thickness of the heart wall. His research is directed at understanding the evolution of vertebrate cardiovascular systems.
    His research shows that heart size in all animals depends on two factors: the vertical distance of the head above the heart, and whether the animal was cold- or warm-blooded. For example, the giraffe has exceptionally high blood pressure and an enlarged heart because it has to pump blood up its long neck. Birds and mammals also have relatively large hearts because they are warm-blooded, while cold-blooded reptiles have low metabolic rates, low blood pressures and smaller hearts.”

    “We have determined that the left ventricle in a warm-blooded Barosaurus, for instance, would have needed to weigh about 2000kg to pump the blood its brain needed,” he says. “This is impossible for at least three reasons. First, it would be difficult to fit such a heart in the available space; second, the heart would use more energy than the entire remainder of the body, and third; the thick walls would be mechanically so inefficient that they would expend more energy deforming themselves than in actually pumping the blood.”

    There only two viable solutions to the dinosaurs’ blood pressure problem, Dr Seymour says: either they were restricted to holding their necks horizontally, that they were cold-blooded animals with low blood flow rates.
    “We admit that they could have had a vertical neck, with blood supplied from a smaller heart, but only if they had a low metabolic rate typical of a cold-blooded reptile. Even in this case, however, the heart wall still would have been relatively thick and inefficient for pumping.

    “The question whether dinosaurs were warm or cold-blooded has been debated over the last 30 years, but the metabolic rates of sauropods will probably never be known with certainty. In either case, it appears unlikely that these animals lifted their heads high as commonly depicted.”

    On the issue of whether dinosaurs, sauropods in particular, were cold-blooded, most evidence points to them not being cold-blooded (ectothermic). That will be addressed in my next post.
    Also, it is believed that some of the sauropods did hold their necks in a near vertical position when feeding.

    The following abstract of an article, co-authored by Dr. Seymour, can also be found on the internet:

    Hearts, Neck Posture and Metabolic Intensity of Sauropod Dinosaurs
    Roger S. Seymour, Harvey B. Lillywhite
    Proceedings: Biological Sciences, Vol. 267, No. 1455 (Sep. 22, 2000), pp. 1883-1887

    Abstract
    Hypothesized upright neck postures in sauropod dinosaurs require systemic arterial blood pressures reaching 700 mmHg at the heart. Recent data on ventricular wall stress indicate that their left ventricles would have weighed 15 times those of similarly sized whales. Such dimensionally, energetically and mechanically disadvantageous ventricles were highly unlikely in an endothermic sauropod. Accessory hearts or a siphon mechanism, with sub-atmospheric blood pressures in the head, were also not feasible. If the blood flow requirements of sauropods were typical of ectotherms, the left-ventricular blood volume and mass would have been smaller; nevertheless, the heart would have suffered the serious mechanical disadvantage of thick walls. It is doubtful that any large sauropod could have raised its neck vertically and endured high arterial blood pressure, and it certainly could not if it had high metabolic rates characteristic of endotherms.

    There are many other sites that echo those above. It would be redundant to reference those.

    I’ll address the other issues in my next post.

    TO BE CONTINUED

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  59. #59 W. Kevin Vicklund
    January 26, 2007

    Question, John. Did you actually read the article for which you posted the abstract? It’s quite interesting and freely available online. It also has a very good list of references. Here is the conclusion:

    We conclude that large sauropods were unlikely to have carried their necks erect, but it may have been possible if they had metabolic rates considerably lower than expected for endotherms. The metabolic status of sauropods is debatable (McIntosh et al. 1997; Paladino et al. 1997; Paul 1998) but ectothermy is suggested by the uninsulated skin of well-developed sauropod embryos (Chiappe et al. 1998). Whatever the metabolic reality, it appears that high metabolic rate and upright neck posture were mutually exclusive in sauropods. If some genera such as Diplodocus and Apatosaurus were incapable of lifting the head high, because of movement limitation of the cervical vertebrae (Stevens & Parrish 1999), they avoided high blood pressure and could have had high or low metabolic rates; but if the anatomy of genera such as Brachiosaurus required them to hold the head 7.9m above the heart (Gunga et al. 1999), the animals were unlikely to have been endothermic.

    Notice how this fits in nicely with the possibility that I raised earlier that the sauropods were not endotherms? Also, their model sauropod, Barosaurus, is a diplodocid. As hinted above, there is reason to believe that they couldn’t raise their neck erect due to structural constraints. The brachiosaurids (as I understand it) are the only large sauropods that appear to require an erect neck posture, and they generally don’t have quite as long a neck as the diplodocids. Also, we should consider what is meant by erect. A sauropod that only raises it’s head by 15 degrees still can add 25% of the length of it’s neck to the elevation of it’s head. That’s still quite a bit of elevation considering the length of the neck, but requires nowhere near as much pressure to do the job. At 30 degrees, that’s half the length of the neck, and 45 degrees is about 3/4. At 60 degrees, you get 5/6. So although we talk about the neck being erect, that doesn’t mean it has to be held completely vertical, so the proposed angle can make a big difference.

    Not only does Seymour indicate that ectothermy is a distinct possibility, but other authors do as well. One such author even cites Seymour:

    Dinosaur body temperatures: the occurrence of endothermy and ectothermy
    Frank Seebacher
    Paleobiology, Volume 29, Issue 1 (March 2003), pp 105-122

    Despite numerous studies, the thermal physiology of dinosaurs remains unresolved. Thus, perhaps the commonly asked question whether dinosaurs were ectotherms or endotherms is inappropriate, and it is more constructive to ask which dinosaurs were likely to have been endothermic and which ones ectothermic. Field data from crocodiles over a large size range show that body temperature fluctuations decrease with increasing body mass, and that average daily body temperatures increase with increasing mass. A biophysical model, the biological relevance of which was tested against field data, was used to predict body temperatures of dinosaurs. However, rather than predicting thermal relations of a hypothetical dinosaur, the model considered correct paleogeographical distribution and climate to predict the thermal relations of a large number of dinosaurs known from the fossil record (>700). Many dinosaurs could have had “high” (≥30°C) and stable (daily amplitude ≤2°C) body temperatures without metabolic heat production even in winter, so it is unlikely that selection pressure would have favored the evolution of elevated resting metabolic rates in those species. Recent evidence of ontogenetic growth rates indicates that even the juveniles of large species (3000-4000 kg) could have had biologically functional body temperature ranges during early development. Smaller dinosaurs (<100 kg) at mid to high latitudes (>45°) could not have had high and stable body temperatures without metabolic heat production. However, elevated metabolic rates were unlikely to have provided selective advantage in the absence of some form of insulation, so probably insulation was present before endothermy evolved, or else it coevolved with elevated metabolic rates. Superimposing these findings onto a phylogeny of the Dinosauria suggests that endothermy most likely evolved among the Coelurosauria and, to a lesser extent, among the Hypsilophodontidae, but not among the Stegosauridae, Nodosauridae, Ankylosauridae, Hadrosauridae, Ceratopsidae, Prosauropoda, and Sauropoda.

    I have found the arguments for an ectothermic sauropod metabolism to be the best fit for the data I have seen. An endothermic dinosaur of that size simply doesn’t make sense (dumping of body heat and the quantity of food required, besides the blood pressure issue), although for other, smaller dinosaurs it makes a great deal of sense. I don’t claim any expertise, but it is a topic I have followed for many years.

  60. #60 John L.
    January 27, 2007

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    Hello W. Kevin,

    This is a continuation of my last post. But first I’ll comment on your last post.

    The conclusion of the writers of that article is very important. “It appears that high metabolic rate and upright posture were mutually exclusive in sauropods.” If you read the balance of my post, you will conclude that the sauropods were indeed, endomorphic. Most expert paleontologists also believe that some of the largest sauropods did feed with their necks at a near vertical position:

    http://www.amnh.org/exhibitions/expeditions/treasure_fossil/treasures/barosaurus/barosaurus.gif

    And since the authors posit that the upright posture and the high blood pressure are also incompatible, the conclusion one must reach is that in today’s gravitational environment, sauropods could not live.

    ON THE ROCHE LIMIT

    I don’t know anything about the Roche limit but is seems to be relevant only when celestial bodies approach a strong gravitational field. When a comet’s trajectory takes it too close to, for example, the sun, its cohesive tensile strength may be compromised and as a result, breaks apart. I don’t believe it is relevant here.

    ON COLD-BLOODED (ECTOTHERMIC) VS WARM-BLOODED (ENDOTHERMIC) DINOSAURS.

    This has been a much debated subject. The overwhelming evidence is that they were endothermic. When current reptiles are compared to current mammals, the following observations are made:

    1. Reptiles have much smaller hearts than mammals of comparable size.

    2. Reptiles have splayed legs with close-to-ground bodies. This physiology facilitates their ability to absorb heat from solar- heated sand, rocks, etc. It also enhances the wait-and-ambush modus operandi of carnivore ectotherms. Mammals are constructed in just the opposite way.

    3. Many reptiles are, and were, aquatic animals. Few mammals are and no known dinosaurs were.

    4. Most dinosaurs are considered to have been very active physically. This is especially true of the carnivores who actively pursued their prey unlike the wait-and-ambush reptiles.

    5. Many of the herbivores are believed to have been animals that formed herds and migrated, based on trackways. Reptiles have to remain in an area where they can quickly alternate between exposing themselves to solar heat and then shielding themselves from the same. They also must be familiar with a location that provides them with a place to escape predation when there is little or no solar energy (i.e. at night, cool rainy days, etc.) such as crevices, caves, bodies of water, etc.
    In other words, herd behavior and migration are not characteristic of ectotherms,

    6. The concept of “mass homeothermy” used by some to support ectothermic dinosaurs has a weak foundation. This concept basically means “keeping warm by being huge.” For a better understanding of this concept and why it doesn’t work in attempting to justify dinosaur ectothermism, read what an expert has to say:

    THE DINOSAUR HERESIES by Robert T. Bakker, Ph.D.
    Chapter 4 Dinosaurs Score Where Komodo Dragons Fail

    As the author points out, dinosaurs during the Mesozoic come in all sizes, not just the oversized ones that we consider when “mass homeothermy” is discussed. And I might add, sauropods start out in life as very small sauropods and would therefore not have the protection of “mass homeothermy.”

    7. Dinosaurs inhabited almost every area of land including the arctic regions. How could they cope with extended periods of (6 months of) darkness if they were ectothermic?

    ON THE VIABILITY AND DEXTERITY OF SAUROPODS IN TODAY’S GRAVITATIONAL ENVIRONMENT

    The following is in response to your statement “As far as child-rearing, simple nests don’t require much agility at all (assuming that they even laid eggs)….”

    For more opinions of experts,
    See page 50 of DISCOVERING DINOSAURS by Norell, Gaffney and Dingus

    “Most of our direct evidence comes from alleged dinosaur rookeries. Several have been found in eastern Montana, where a large concentration of dinosaur nests was found at a place now called Egg Mountain. Most of these probably belonged to the hadrosaur Maiasaura. Preserved in these nests are the bones of baby dinosaurs. The finds at Egg Mountain and other sites around the world document that dinosaurs laid their eggs in nests.

    The nests at Egg Mountain are reported to be equally spaced, separated by a space corresponding to the length of an adult Maiasaura. From this arrangement, scientists have inferred that the nests were separated in this way to allow incubation in a tightly packed nesting colony.”

    Your statement that sauropods might have dropped their eggs and left them on their own is not supported by evidence. Those eggs were probably more than a foot in length and would have been dropped from a height of about 10 feet. Leaving those eggs unguarded would be a quick path to extinction.

    I conclude, from the above info, including the prior post that the great physical size of the sauropods would MAKE THEIR EXISTENCE TODAY impossible based the overwhelming evidence that they were endothermic, would have excessively high blood pressure, inadequate food intake to support their metabolic requirement, would lack the hatching and nesting dexterity required and, although only lightly addressed here, the physical constraints of their enormous size.///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

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