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Dr. Walt Browns Hydroplate Theory – Returning Material


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#1 Glenn Williams

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Posted 16 January 2011 - 04:45 PM

Dr. Walt Browns Hydroplate theory – returning material

Hello everyone my name is Glenn, this is my first post and I apologize in advance for this being somewhat lengthy. I run a local church group in Front Royal Virginia and has been active in speaking and teaching for nearly 20 years. Normally the topics that I speak on has to do with basic Christianity and theological arguments. But for the past few years I've wanted to put together something substantial about Noah's flood.

I don't hold a degree in any of the sciences that would surround this topic so as you can imagine this has been slow going. What I have been studying is the Hydro plate theory by Dr. Walt Brown. I think he is a very intelligent man and puts forth a very good argument that is well-founded in the Scriptures; this is the area that I know, so when someone brings forth something that matches up to God's word I take notice.

While I was studying his theory I came across one area that might be problematic with his theory which is pointed out in the quotation which is attached that was established by someone else. This has to do with the assumed returning material which could potentially overheat the atmosphere and the oceans. If you're familiar with Dr. Brown's theory he talks about a large amount of material that was launched into space. http://www.creations...calNotes16.html

For the record I do not have a problem with the water that escaped from inside of the earth as being too hot. I feel that Dr. Brown covers this sufficiently here:
http://www.creations...eOverview6.html

The biggest problem that I see with the argument that was advanced was the assumptions that were made but I'm hoping that some of the people on this forum can see things and raise arguments that I am not capable of.

One last note is that I have been planning to speak to Dr. Brown directly about this but have not yet found the opportunity, but if and when the opportunity arises any answers that I get from him I will definitely add to this post.

Thank you in advance.


I did an analysis of only one part of Brown's model ... the launch of all the meteroids and comets in the solar system and found if his model were 90% efficient at getting the material out of Earth's gravity, the return of the remaining 10% material would contain enough energy to boil every drop of water on the planet. In short, a flood would have been the least of Noah's problems.

Surface area of Earth 5.14e14 square meters
Mass of water on Earth 1.36e21 kg
Mass of material sent to space according to Brown 2.74e21 kg - http://www.creations...calNotes16.html
Specific heat of ice 2030 j/kg
Specific heat of water 4184 j/kg
Specific heat of steam 1860 j/kg
Latent heat of melting/freezing 334,000 j/kg
Latent heat of condensation/evaporation 2.5e6 j/kg
Velocity of material that did not escape Earth's gravity (average) 5.6 km/sec
Boltzmann constant = 5.67e-8
The energy of re-entry will boil all the water on Earth 10 times.....
Amount of energy to raise all the water on Earth from freezing and boil it.....
Multiply the mass of the oceans times the specific heat of water times the number of degrees (C or K) temperature change. That would be 1.36e21 * 4184 *
100 = 5.39e26 joules to bring all the water on Earth from freezing to boiling. Then multiply the mass of the water times the latent heat of evaporation. This is 1.36e21 * 2.5e6 = 3.4e27 joules.
Adding the two gives us 3.97e27 joules of energy needed to bring all the water on Earth from freezing and boil it.
Amount of energy from material that did not reach escape velocity......
Kinetic energy = one half the mass times the velocity squared. Using Brown's model and an efficiency of 50% we multiply the mass of material sent to space times the velocity squared then divide by two. That gives us 2.74e21 * 5600 * 5600 / 2 = 4.29e28 joules of kinetic energy in material that did not reach escape velocity.
The number of times the energy can evaporate all the water on Earth.....
Divide the total amount of energy available by the amount of energy needed to boil all the water on Earth. This gives us 4.29e28 / 3.97e27 = 10.82 times all the water on Earth can be boiled by the energy of re-entry.
The amount of material that will provide enough heating to boil every drop of water on Earth would need to have a kinetic energy of 3.97e27 joules. With a launch velocity of 5.6 km/sec, the mass of material needed is: m = 3.97e27 * 2 / 5600 / 5600 = 2.53e20 kg will have enough kinetic energy to boil every drop of water on the planet.
Taking the total material launched in Brown's model (2.74e21 kg) and adding 2.53e20 kg that will need to return to boil all the water we find that the minimum total mass launched that will still boil all the water is 2.99e21kg. Calculating the efficiency by 2.74e21 / 2.99e21 = 91.54% efficiency will till boil all the water on the planet.
At least 75% of the returning material will be rock......
Brown says his model sends 2.74e21 kg of material out of Earth's gravity. At 50% efficiency, an equal amount will be sent to space only to return to the surface. He also says about half of the present water in the oceans is the result of his flood model. The mass of Earths surface water is 1.36e21 kg. This means his model uses 6.8e20 kg of water. While Brown says his model continues to send water out of underground storage for some time after the launch phase, that would only reduce the amount of water available to return to Earth. If we take all of the water Brown says ended up in the oceans and compare that to the 2.74e21 kg returning to Earth in a 50% efficient model, we get 6.8e20 / 2.74e21 = 24.82% of the material returning to Earth is water. The rest (75.18%) will be rock. As stated, according to Brown's model, a higher percentage would need to be rock since he doesn't send all his water to space.
If the material takes 40 days to fall, atmospheric temperatures will be over 7700F....
Using the Stefan-Boltzmann equation, we can calculate the average temperature needed to radiate the heat of reentry to space. The equation is w = kat^4 where w= the watts; k= the Boltzmann constant (5.67e-8); a= the radiating surface area and t = the temperature in Kelvin. The energy of the returning material is 4.29e28 joules and 40 days equals 3.46e6 seconds. The number of watts equal the number of joules divided by the number of seconds, so we get 1.24e22 watts. Solving for t, we get t= (w/ka)^.25 or t = (1.24e22 / 5.67e-8 / 5.14e14)^.25 = 4544.58K. Converting 4544K to F, we find 7721F is the average temperature of the atmosphere to radiate that heat to space in 40 days.
If the material takes 150 days to fall, atmospheric temperatures will be over 6000F.....
All we need do for this one is change the time period from 40 days to 150 days. Doing this, we find the number of watts changes from 1.24e22 to 3.32e21.
Using the same equation, we get a temperature of 3266K or about 5400F.
If falling material cools to absolute zero, friction with the atmosphere will still heat it to over 10,000F.....
In order to be as favorable as possible to Brown, we will consider water which requires much more energy to heat than granite. At 5.6 km/sec, each kilogram of falling material will have kinetic energy = 0.5 mass times velocity squared. This is then =.5 * 1 * 5600 *5600 = 1.57e7 joules. The energy to raise the temperature to freezing, melt it, raise it to boiling, and boil it can be calculated by multiplying the specific heat of ice by 273, adding the latent heat of melting, add the specific heat of water times 100 and add the latent heat of evaporation.
This is then (273 * 2030) + 334000 + (100 * 4184) + 2.5e6 = 3.81e6 joules to raise a kilogram of water from absolute zero and boil it.
We can now take the remaining 1.19e7 joules and heat the steam. Dividing 1.19e7 joules by the specific heat of steam (1860 j/kg), we find we can heat the water another 5936C. The final temperature of the water will be 6036C which converts to 10898F.



#2 MamaElephant

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Posted 18 January 2011 - 07:32 AM

I know that the runaway subduction theory says that the extra heat was miraculously taken care of by God.

I am looking forward to seeing someone answer this OP.

#3 ikester7579

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Posted 18 January 2011 - 03:07 PM

Dr. Walt Browns Hydroplate theory – returning material

Hello everyone my name is Glenn, this is my first post and I apologize in advance for this being somewhat lengthy. I run a local church group in Front Royal Virginia and has been active in speaking and teaching for nearly 20 years.  Normally the topics that I speak on has to do with basic Christianity and theological arguments.  But for the past few years I've wanted to put together something substantial about Noah's flood.

I don't hold a degree in any of the sciences that would surround this topic so as you can imagine this has been slow going.  What I have been studying is the Hydro plate theory by Dr. Walt Brown. I think he is a very intelligent man and puts forth a very good argument that is well-founded in the Scriptures; this is the area that I know, so when someone brings forth something that matches up to God's word I take notice.

While I was studying his theory I came across one area that might be problematic with his theory which is pointed out in the quotation which is attached that was established by someone else. This has to do with the assumed returning material which could potentially overheat the atmosphere and the oceans.  If you're familiar with Dr. Brown's theory he talks about a large amount of material that was launched into space. http://www.creations...calNotes16.html

For the record I do not have a problem with the water that escaped from inside of the earth as being too hot.  I feel that Dr. Brown covers this sufficiently here:
http://www.creations...eOverview6.html

The biggest problem that I see with the argument that was advanced was the assumptions that were made but I'm hoping that some of the people on this forum can see things and raise arguments that I am not capable of.

One last note is that I have been planning to speak to Dr. Brown directly about this but have not yet found the opportunity, but if and when the opportunity arises any answers that I get from him I will definitely add to this post.

Thank you in advance.

View Post


I did an analysis of only one part of Brown's model ... the launch of all the meteroids and comets in the solar system and found if his model were 90% efficient at getting the material out of Earth's gravity, the return of the remaining 10% material would contain enough energy to boil every drop of water on the planet. In short, a flood would have been the least of Noah's problems.

Surface area of Earth 5.14e14 square meters
Mass of water on Earth 1.36e21 kg
Mass of material sent to space according to Brown 2.74e21 kg - http://www.creations...calNotes16.html
Specific heat of ice 2030 j/kg
Specific heat of water 4184 j/kg
Specific heat of steam 1860 j/kg
Latent heat of melting/freezing 334,000 j/kg
Latent heat of condensation/evaporation 2.5e6 j/kg
Velocity of material that did not escape Earth's gravity (average) 5.6 km/sec
Boltzmann constant = 5.67e-8
The energy of re-entry will boil all the water on Earth 10 times.....
Amount of energy to raise all the water on Earth from freezing and boil it.....
Multiply the mass of the oceans times the specific heat of water times the number of degrees (C or K) temperature change. That would be 1.36e21 * 4184 *
100 = 5.39e26 joules to bring all the water on Earth from freezing to boiling. Then multiply the mass of the water times the latent heat of evaporation. This is 1.36e21 * 2.5e6 = 3.4e27 joules.
Adding the two gives us 3.97e27 joules of energy needed to bring all the water on Earth from freezing and boil it.
Amount of energy from material that did not reach escape velocity......
Kinetic energy = one half the mass times the velocity squared. Using Brown's model and an efficiency of 50% we multiply the mass of material sent to space times the velocity squared then divide by two. That gives us 2.74e21 * 5600 * 5600 / 2 = 4.29e28 joules of kinetic energy in material that did not reach escape velocity.
The number of times the energy can evaporate all the water on Earth.....
Divide the total amount of energy available by the amount of energy needed to boil all the water on Earth. This gives us 4.29e28 / 3.97e27 = 10.82 times all the water on Earth can be boiled by the energy of re-entry.
The amount of material that will provide enough heating to boil every drop of water on Earth would need to have a kinetic energy of 3.97e27 joules. With a launch velocity of 5.6 km/sec, the mass of material needed is: m = 3.97e27 * 2 / 5600 / 5600 = 2.53e20 kg will have enough kinetic energy to boil every drop of water on the planet.
Taking the total material launched in Brown's model (2.74e21 kg) and adding 2.53e20 kg that will need to return to boil all the water we find that the minimum total mass launched that will still boil all the water is 2.99e21kg. Calculating the efficiency by 2.74e21 / 2.99e21 = 91.54% efficiency will till boil all the water on the planet.
At least 75% of the returning material will be rock......
Brown says his model sends 2.74e21 kg of material out of Earth's gravity. At 50% efficiency, an equal amount will be sent to space only to return to the surface. He also says about half of the present water in the oceans is the result of his flood model. The mass of Earths surface water is 1.36e21 kg. This means his model uses 6.8e20 kg of water. While Brown says his model continues to send water out of underground storage for some time after the launch phase, that would only reduce the amount of water available to return to Earth. If we take all of the water Brown says ended up in the oceans and compare that to the 2.74e21 kg returning to Earth in a 50% efficient model, we get 6.8e20 / 2.74e21 = 24.82% of the material returning to Earth is water. The rest (75.18%) will be rock. As stated, according to Brown's model, a higher percentage would need to be rock since he doesn't send all his water to space.
If the material takes 40 days to fall, atmospheric temperatures will be over 7700F....
Using the Stefan-Boltzmann equation, we can calculate the average temperature needed to radiate the heat of reentry to space. The equation is w = kat^4 where w= the watts; k= the Boltzmann constant (5.67e-8); a= the radiating surface area and t = the temperature in Kelvin. The energy of the returning material is 4.29e28 joules and 40 days equals 3.46e6 seconds. The number of watts equal the number of joules divided by the number of seconds, so we get 1.24e22 watts. Solving for t, we get t= (w/ka)^.25 or t = (1.24e22 / 5.67e-8 / 5.14e14)^.25 = 4544.58K. Converting 4544K to F, we find 7721F is the average temperature of the atmosphere to radiate that heat to space in 40 days.
If the material takes 150 days to fall, atmospheric temperatures will be over 6000F.....
All we need do for this one is change the time period from 40 days to 150 days. Doing this, we find the number of watts changes from 1.24e22 to 3.32e21.
Using the same equation, we get a temperature of 3266K or about 5400F.
If falling material cools to absolute zero, friction with the atmosphere will still heat it to over 10,000F.....
In order to be as favorable as possible to Brown, we will consider water which requires much more energy to heat than granite. At 5.6 km/sec, each kilogram of falling material will have kinetic energy = 0.5 mass times velocity squared. This is then =.5 * 1 * 5600 *5600 = 1.57e7 joules. The energy to raise the temperature to freezing, melt it, raise it to boiling, and boil it can be calculated by multiplying the specific heat of ice by 273, adding the latent heat of melting, add the specific heat of water times 100 and add the latent heat of evaporation.
This is then (273 * 2030) + 334000 + (100 * 4184) + 2.5e6 = 3.81e6 joules to raise a kilogram of water from absolute zero and boil it.
We can now take the remaining 1.19e7 joules and heat the steam. Dividing 1.19e7 joules by the specific heat of steam (1860 j/kg), we find we can heat the water another 5936C. The final temperature of the water will be 6036C which converts to 10898F.


First you would have to match the sun's energy:

Averaged over an entire year and the entire Earth, the Sun deposits 342 Watts of energy into every square meter of the Earth*. This is a very large amount of heat—1.7 x 1017 watts of power that the Sun sends to the Earth/atmosphere system. For comparison, a large electric power plant would produce 100 million watts of power, or 108 watts. It would take 1.7 billion such power plants to equal the energy coming to the Earth from the Sun—roughly one for every three people on the Earth!
http://earthobservat.../EnergyBalance/

Then you would have to go beyond that several times. Because the clouds were totally blocking the sun's rays so the whole earth was in a cool down. So the heat generated would not only have to totally replace the sun, but go well beyond that.

And as far as enough water to flood the earth is concerned:

http://www.scienceda...71217071316.htm
http://www.livescien...ng_anomoly.html
http://www.physorg.c...ws90171847.html
http://news.wustl.ed...Pages/8749.aspx

The mineral that contains this water is called wadsleyite. It is estimated that this mineral hold 3% water by weight. And according to how much of that mineral there is, that works out to be 30 oceans of water.
http://www.ldolphin....deepwaters.html

So how much water is need to flood the whole earth?

The Radius of the Earth = 3963 miles
The Radius of the earth with 5 miles of water = 3968 Miles.
The volume of the earth = 260711882973.3396 cubic miles
The volume of the earth with water = 261699925947.5533 cubic miles.
261699925947.5533 - 260711882973.3396 = 988042974.2136999965
So the volume of the flood water = 988,042,974.2136999965 cubic miles. But lets round it to 988,042,974 Cubic Miles.
If this water was put into a sphere, it would have a radius of 618 Miles.

And how does evidence for the Pangaea fit?

The earth's crust has actually expanded twice.

First expansion:
In Genesis 1. the whole earth was covered with water. Then it had to recede into one place so that the land could appear. This means that the water went underground. And to go underground, the earth's crust had to expand. This made the Pangaea where all the continents were together.

Second expansion:

When the barometric pressure fell, and the atmosphere released the moisture that it held. It released the 30 oceans worth of water. When the flood water ran off the surface of the earth and into the upper mantle (mineral called wadsleyite). The earth's crust had to expand again in order to receive it. This caused the continents to break apart and why they still look like they were together at one time.

So the Bible supports the evidence, and the evidence supports the flood.

#4 AFJ

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Posted 19 January 2011 - 09:04 AM

Dr. Walt Browns Hydroplate theory – returning material

Hello everyone my name is Glenn, this is my first post and I apologize in advance for this being somewhat lengthy. I run a local church group in Front Royal Virginia and has been active in speaking and teaching for nearly 20 years.  Normally the topics that I speak on has to do with basic Christianity and theological arguments.  But for the past few years I've wanted to put together something substantial about Noah's flood.

I don't hold a degree in any of the sciences that would surround this topic so as you can imagine this has been slow going.  What I have been studying is the Hydro plate theory by Dr. Walt Brown. I think he is a very intelligent man and puts forth a very good argument that is well-founded in the Scriptures; this is the area that I know, so when someone brings forth something that matches up to God's word I take notice.

While I was studying his theory I came across one area that might be problematic with his theory which is pointed out in the quotation which is attached that was established by someone else. This has to do with the assumed returning material which could potentially overheat the atmosphere and the oceans.  If you're familiar with Dr. Brown's theory he talks about a large amount of material that was launched into space. http://www.creations...calNotes16.html

For the record I do not have a problem with the water that escaped from inside of the earth as being too hot.  I feel that Dr. Brown covers this sufficiently here:
http://www.creations...eOverview6.html

The biggest problem that I see with the argument that was advanced was the assumptions that were made but I'm hoping that some of the people on this forum can see things and raise arguments that I am not capable of.

One last note is that I have been planning to speak to Dr. Brown directly about this but have not yet found the opportunity, but if and when the opportunity arises any answers that I get from him I will definitely add to this post.

Thank you in advance.

View Post

PART 1

Using the figures he used to "refute" Brown's model, I wanted to calculate how many bullets it would take to boil a gallon of water 10 times. THIS IS ASSUMING THAT ALL THE KINETIC ENERGY IN THE BULLETS WOULD BE TURNED TO HEAT.

This means that none of the water would move. It would remain static, which is impossible, and I am not qualified to calculate that--I don't know for sure if it could be calculated, because there could be a multitude of variables.

According to the following calculations, and the refuters claims, the returning material was relative to shooting 24,800 .45 caliber magnum bullets into a gallon of water simultaneously--in order to boil it 10 times. That's only if ALL the energy was converted to heat in the water, and the water remained unmoved. Where's the disconnect here?

I will do the math using the figures given in the OP and the joules in a .45 caliber ACP (high velocity/ probably magnum) bullet.

1).45 ACP 230 gr. bullet * Velocity 837 feet per second -- KE= 352.6399 foot pounds Link
2)1 foot pound= 1.355 (app) joules

352.64(1.355)= 477.827 joules for 1 -- .45 caliber ACP bullet
________________________________________________

1)Specific heat of water 4184 j/kg
2) Gallon of water at room temp = 8lbs = 3.636 kg

Now we can multiply 4184 (the specific heat of water) by 3.636 to find how many joules it would take to raise a gallon of water 1 degree C. Answer is 15213.024 joules to raise 1 gallon of water 1 degree celcius.

***Kg per gallon at room temperature-- there is 128 fl oz (8lbs) in a gallon, and .454 kg in a lb. So 1 / .454 = 2.2 lbs per kg.
8 lbs (gallon) / 2.2 = 3.636 kgs in a gallon (app) at room temp.
Or you can say 2.2 lbs (1kg) * 3.636 = 8 lbs (a gallon at room temp)

_________________________________________________

So then to find how many .45 caliber bullets it would take to raise a gallon of water one degree C we divide the total number of joules it takes to raise the temp 1 degree--by the number of joules in a .45 caliber slug traveling at 837 fps (noted above).

15213.024 / 477.827 = 31.837 (app). So it would take between 31 and 32-- .45 magnums being fired into 1 gallon of water to raise it 1 degree C.
__________________________________________________

Grand Totals


Now, if a gallon is at room temperature (68F=20C), we have 80 degrees C to reach boiling (100C). So we multiply 31 bullets times 80 to go from 20C to 100C and arrive at 2480 bullets. To boil it 10 times is 24,800 bullets.
__________________________________________________

So...
Does anyone here believe that if the ocean was a gallon of water, that the returning material had the relative kinetic energy of 2480 .45 magnums being fired into it simultaneously? Or if you want to believe the "refuters" claim of 10 times, it would be 24,800 .45's

It would destroy the earth before it boiled the ocean!!!

You can check my math. I used a calculator, the info provided in the OP, and
http://curezone.com/conversions.asp

For the info on the .45 caliber:
http://www.thehighro...ad.php?t=344620

#5 AFJ

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Posted 19 January 2011 - 01:52 PM

PART 2

The proposition that falling debris that does not reeach escape velocity in the Hydroplate Theory would "boil the oceans 10 times" has several flaws. The first problem has been shown already--that is a misunderstanding that ALL the kinetic energy would be converted to heat. Much of it would be transferred to the ocean waves and in tsuamis.

The Indian Ocean Tsunami of 2004
The catastrophe began on December 26, 2004, with a magnitude 9.0 earthquake in the deep-water Sunda Trench offshore Sumatra. Within 3-4 minutes, a 1200 kilometer-long rupture opened the seafloor, and a region roughly the length and half the width of California was displaced vertically by about two meters. The work involved is a measure of the raw energy imparted to the tsunami. In this case, it was equivalent to about 100 Hiroshima-sized atomic bombs.1

Footnote 1: 1.Tsunami energy of 8 x 10^15 joules is estimated from disturbance map in: Science News, Jan. 8, 2005. Total energy of the earthquake is 2 x 10^18 joules.


Question: With this many joules, why did we not see an evident increase in the ocean temperature? It is because the energy was converted into a wave.

All energy must be accounted for in thermodynamics, this does not require that all of it be converted to heat. Kinetic energy can be converted into heat, or it can be exchanged to another mass. Just play pool.

Some of the falling debris would also be absorbed by the ocean crust, all dependent on how deep the first oceans were. Some of that mass would have been water, and would have been partially evaporated. Some land would also absorb the impact. If the falling rock/ water and sediments were coming out over 40 days the entire continental crust would already be covered with water and soft sediments, and would also absorb the KE of the debris/slurry.

The second problem is that the refuter makes no difference, that I can see, between the velocities of the OUTGOING ESCAPE VELOCITY debris and the INCOMING MAXIMUM FALL VELOCITY debris. According to my research, this is nearly an 80 fold difference.


He says: "Kinetic energy = one half the mass times the velocity squared. Using Brown's model and an efficiency of 50% we multiply the mass of material sent to space times the velocity squared then divide by two. That gives us 2.74e21 * 5600 * 5600 / 2 = 4.29e28 joules of kinetic energy in material that did not reach escape velocity."

"5600" would be the velocity in the equation. I'm not sure if that's mph, mps or what. But the escape velocity for a rocket is roughly 20 times the speed of sound-- 4.9 miles/s. http://www.redshift-...et_speed-1.html

However, the maximum falling velocity is 9.8 meters per second. http://www.newton.de...00/phy00800.htm
So we are comparing 4.9 miles/second (escape vel. for a rocket--17,640 mph), which is 7885 meters per second to 96 meters per second (maximum fall velocity -- 215 mph). 17,640 mph to 215 mph.

I do not what "5600" is, and how he arrives at this. One thing is sure -- 5600 mph is 2503 meters per second. This is app 26 times faster than the maximum falling velocity. It is evident that the material that reached escape velocity was going MUCH faster than the falling material.

I am suprised that a guy with the knoweldge of physics like this could make such an apparent error. It causes his calculations of the the KE of the falling material to be greatly exaggerated.

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Posted 19 January 2011 - 11:01 PM

The second problem is that the refuter makes no difference, that I can see, between the velocities of the OUTGOING ESCAPE VELOCITY debris and the INCOMING MAXIMUM FALL VELOCITY debris.  According to my research, this is nearly an 80 fold difference. 
He says: "Kinetic energy = one half the mass times the velocity squared. Using Brown's model and an efficiency of 50% we multiply the mass of material sent to space times the velocity squared then divide by two. That gives us 2.74e21 * 5600 * 5600 / 2 = 4.29e28 joules of kinetic energy in material that did not reach escape velocity."

"5600" would be the velocity in the equation.  I'm not sure if that's mph, mps or what.  But the escape velocity for a rocket is roughly 20 times the speed of sound-- 4.9 miles/s.  http://www.redshift-...et_speed-1.html

However, the maximum falling velocity is 9.8 meters per second.  http://www.newton.de...00/phy00800.htm 
So we are comparing 4.9 miles/second (escape vel. for a rocket--17,640 mph), which is 7885 meters per second to 96 meters per second (maximum fall velocity -- 215 mph).  17,640 mph to 215 mph.

I do not what "5600" is, and how he arrives at this.  One thing is sure -- 5600 mph is 2503 meters per second.  This is app 26 times faster than the maximum falling velocity.  It is evident that the material that reached escape velocity was going MUCH faster than the falling material.

I am suprised  that a guy with the knoweldge of physics like this  could make such an apparent error.   It causes his calculations of the the KE of the falling material to be greatly exaggerated.

View Post


Just a quick reply to address some VERY large misconceptions.

9.8 m/s is not a maximum falling speed. It's the acceleration due to gravity. 5600 meters per second is a perfectly acceptable and far more realistic velocity in a discussion of launching projectiles into space than 837 feet per second.

Escape velocity for the earth is 11.2 km/s. Any projectile not launched at that speed will fall back to earth (rockets keep accelerating after they are launched so they are not good examples). This means that any infalling material must have initially been traveling between 0 and 11200 m/s. 5600 meters/second is simply the middle part of that range, which is as good a average velocity as any since all that matters is the initial kinetic energy of the infalling material. The initial kinetic energy is how much energy must be absorbed by the atmosphere and oceans since the objects eventually come to rest.

#7 AFJ

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Posted 20 January 2011 - 03:32 AM

Just a quick reply to address some VERY large misconceptions.

9.8 m/s is not a maximum falling speed.  It's the acceleration due to gravity.  5600 meters per second is a perfectly acceptable and far more realistic velocity in a discussion of launching projectiles into space than 837 feet per second.

Escape velocity for the earth is 11.2 km/s.  Any projectile not launched at that speed will fall back to earth (rockets keep accelerating after they are launched so they are not good examples).  This means that any infalling material must have initially been traveling between 0 and 11200 m/s.  5600 meters/second is simply the middle part of that range, which is as good a average velocity as any since all that matters is the initial kinetic energy of the infalling material.  The initial kinetic energy is how much energy must be absorbed by the atmosphere and oceans since the objects eventually come to rest.

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Your misunderstanding the first post. I was scaling down the earth's ocean to a gallon to show a relative model. This is perfectly acceptable in science. 837 m/s has nothing to do with Brown's model. It is a model where I sought to boil a gallon of water 10 times using specific heat of water in joules divided by the joules in a 45 caliber bullet. You are mixing the models.

Thanks for the help though on the outgoing material. Is 5600 m/s the figure that Brown gives or the refuter, then? Then you are saying the refuter was not saying the falling material boiled the oceans 10 times? So he's saying the atmosphere was heated by the initial KE of the explosion, and the atmosphere then heated the oceans to boil them?

#8 numbers

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Posted 20 January 2011 - 07:57 AM

Your misunderstanding the first post.  I was scaling down the earth's ocean to a gallon to show a relative model.  This is perfectly acceptable in science.  837 m/s has nothing to do with Brown's model.  It is a model where I sought to boil a gallon of water 10 times using specific heat of water in joules divided by the joules in a 45 caliber bullet.  You are mixing the models.

Not really, I understood your model, I don't think you understood why it took so many bullets. The infalling material would absolutely have the same kinetic energy as the number of bullets you calculated (I'm assuming no basic arithmetic errors in the calculations since i didn't redo all of them). There's a reason why the energy of a rock falling from a high altitude is often compared to nuclear weapons. 1/2*m*v^2 gets scary when V is large.

What you did is basically calculate how many childrens water pistols it would take to have a water output equivelant to that of a fire hose, then express disbelief that a fire hose could be real since it would take so many water pistols to do the same thing. Do your bullet calculation with the same bullet mass but a 5600 m/s velocity and you'll probably be surprised at the difference in the result. .5 times (bullet mass) times (velocity squared).

Thanks for the help though on the outgoing material.  Is 5600 m/s the figure that Brown gives or the refuter, then?  Then you are saying the refuter was not saying the falling material boiled the oceans 10 times?  So he's saying the atmosphere was heated by the initial KE of the explosion, and the atmosphere then heated the oceans to boil them?

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It appears to be just an average value from the responder. There would be some average launch velocity for all the material that failed to escape, and half of the possible range seems fair to me. I don't know if brown ever addresses material that fails to escape earth.

His calculations look ok to me (i didn't redo them, just looked at the steps he took) so it does look like there's enough energy from browns idea to boil the oceans many times over.

Basically it works like this: a rock is tossed up into the air. The energy the rock possesses is 1/2 mv^2. The rock eventually falls down and comes to rest somewhere, the energy the rock at rest possesses is 0. All that energy had to go somewhere, and it went into whatever slowed the rock down. Portions of it will go into the atmosphere, portions of it will go into the oceans ( i think it's safe to assume that by the time a rock has reached the bottom of the ocean, it's falling very slowly compared to 5600m/s), and the last of it will go into the ocean floor. Temperature is the average kinetic energy of the molecules in a substance. The rock transfers energy as it falls into the molecules of the air and ocean, increasing the average kinetic energy (aka temperature) of the air and water while slowing the rock.

http://scienceblogs....e_why_doesn.php
This link deals with air movement and temp rather than water, but it's a pretty good explanation for how temperature works. As for your tsunami question, based on the initial calculations and the tsunami energy you provided, the tsunami had the energy to increase ocean temp by about .000000001 degrees, the exact increase depends on the size of the ocean you want to discuss. (10^15 is a drop in the bucket compared to 10^26)

If you want the energy to go into the earth's crust you'd have to explain how the rock managed to avoid getting slowed down signficantly by the air and water first as well as why the heat would stay in the crust instead of spreading into the oceans. The energy can't get radiated into space (see the bottom part of the intial calculations) without the air temp being several thousand degrees. So essentially, yes, it's safe to say that nothing, including noah, would have survived.

#9 Glenn Williams

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Posted 20 January 2011 - 09:06 AM

I appreciate all the input given to this topic so far. I have contacted Dr. Brown and was sent some material to be released and some of the excerpts of that are given below. BTW the refuter is Geno Castagnoli.

For the record, the 5.6km/sec given by Geno is half the minimum require escape velocity mentioned by Walt where Walt says, “escape velocity is 11.2 km/sec” in his on-line material. I think Geno believes that this plays into heating the atmosphere, as I guess we could say that overheating is his general argument. Of course once you read Walt’s argument below I have to wonder would there be any affects of heat , not only because Walt says this became kinetic, but that the water propelling the rock would become super cold. (So I’m guessing that the cold water would absorb any heat of the projectiles.)

From a different position I see that you gentlemen mention that 9.8 m/sec is falling speed and would play into the heat energy to heat the oceans which is considerably less speed than the 5.6km. (BTW I found the .45 magnum example amusing, maybe I can try that in a bulletproof bucket :) .)

Perhaps we can get Geno to weigh in on some of this if I can locate him. Additional difficulty of the argument is one problem I created in trying to argue this from his point but Geno’s objections were bothering me.

Here is the info I received from Dr. & Mrs. Brown:
Quote from Peggy Brown:
“Walt Brown is familiar with the argument raised by Geno Castagnoli, a high school science teacher in Oklahoma.  Walt says the argument is bogus, and frankly, that Geno really doesn't understand what he is talking about.”
-------------
Data sent from Dr. Brown (excerpts):
When a fluid (liquid, vapor, or liquid/vapor mixture) flowing in a uniform channel accelerates, the fluid expands. (Its specific volume increases.) Expansion is a powerful cooling process that provides cooling for refrigerators and air conditioners. The greater the acceleration, the greater the expansion and cooling.

During the initial weeks of the flood, the phenomenal acceleration and expansion was horizontal in the flow under the crust and upward and lateral in the fountains of the great deep.

In this explosive expansion,most of the initially hot subterranean water in the fountains dropped to a temperature of almost
absolute zero (-460°F).

The temperature, T, in an expanding supersonic flow is determined by the Mach number, M, stagnation temperature,T0, and the ratio of specific heats, k, which for a perfect gas is about 1.4

T0/T = 1+ (k-1/2) *(M*M)

The stagnation temperature for the situation is the temperature in the subterranean chamber.
T0 was about 3,000°F. Launch velocities of 32 miles per second were required to place near-parabolic comets in retrograde orbits.

M, T0, and the effective sonic velocity can only be approximated, the flow’s temperature after expansion is
so cold, it can be considered to be nearly absolute zero!

The fountains did not transfer much of its kinetic energy to the atmosphere except for a few seconds after the rupture and the relatively thin boundary layer (edges of) the water.

The energy in the SCW primarily became kinetic energy, not heat.

Some water within the boundary layer was slowed enough to limit its maximum altitude and cause that water to fall back to earth as rain or ice.



#10 AFJ

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Posted 22 January 2011 - 05:26 AM

Just some quick thoughts. 9.8 m/s is wrong, and I corrected it. It is 9.8 m/s^2, which is 214 mph--maximum falling velocity. I read on another site 220 mph, so I know its somewhere in that neighborhood. So 5600 can not be used for the falling debris.

As for 5600 m/s being used for the velocity in the joules formula in the intitial explosion. Numbers and myself have both alluded to the fact that the energy will be accounted for in other ways besides just heat. The energy would radiate 360 degrees, which means it would have converted into seismic waves in the earth, and also tsunamis in the ocean.

I'm not understanding the principle that all the energy would be turned to heat. Even the air is not going to convert all the energy into heat. Since it is technically a fluid, it will react to energy by shock waves and wind. The question should be--how much kinetic friction did the debris produce as it moved through the air. The air is a gas and is fluid. It's not going to produce friction like moving a heavy object on steel.

Heating air requires convection if I remember correctly. This requires time. If half of the debris was moing at 11 km/s as Numbers said, most of the mass of the atmosphere just happens to be in the first 11 km (wiki). So that gives you one second of kinetic friction on the denser atmosphere.



We need to remember also that the second law of thermodynamics will be working. Unless there is continuous input into a system, it will diffuse or move toward disorder. That's why the concusion of an explosion doesn't keep on being felt forever and forever.

Glenn, as for Brown's comment on expansion, that is a fact I was unaware of, and I want to look into it more (when I have the time :wacko: :( ). My only doubt is when he talks about absolute zero!! The water would have frozen, I don't understand. I know that water underpressure can be supercooled under freezing temp, but absolute zero? Maybe I missed something there.

#11 Glenn Williams

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Posted 22 January 2011 - 07:32 AM

It sounds reasonable for you to say that the water would have turned to ice. I know that Dr. Brown says that a lot of ice and snow came down during the flood, even rock ice, the type that contains debris. He uses this to explain the burial of subtropical plants that are found in places like Alaska and the Arctic.

If nothing else I could imagine where all this cold in the environment would have been a factor to offset any increases in temperature.

#12 numbers

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Posted 22 January 2011 - 11:56 AM

Just some quick thoughts.  9.8 m/s is wrong, and I corrected it.  It is 9.8 m/s^2, which is 214 mph--maximum falling velocity. I read on another site 220 mph, so I know its somewhere in that neighborhood.  So 5600 can not be used for the falling debris.

no, 9.8 m/s^2 is acceleration not velocity and it is not terminal velocity. Toss a baseball up in the air, it will accelerate downward (i.e. decelerate) at 9.8 m/s every second until it's upward velocity is zero, then it will start to fall downward. As it falls it will get faster by 9.8 m/s every second. The only limit to how fast it can fall is how aerodynamic it is. Eventually the friction from the air will become so great that it will start to rob the baseball of velocity at the same rate as gravity adds velocity to it. That's terminal velocity and it's not a single value, it varies based on the shape of an object.

For these kinds of calculations the downward velocity is largely irrelevant since we already know how much energy needs to be accounted for from the upward velocity. Downward velocity just tells us how much energy is lost to the atmosphere vs the water of the oceans

The equation is basically this:
1/2 *(mass) * (up velocity)^2 = 1/2 *mass*(down velocity in air)^2 + energy added to atmosphere

Then most of the remaining energy gets transferred to the oceans as the falling material gets slowed significantly before coming to rest on the ocean floor.


As for 5600 m/s being used for the velocity in the joules formula in the intitial explosion.  Numbers and myself have both alluded to the fact that the energy will be accounted for in other ways besides just heat.  The energy would radiate 360 degrees, which means it would have converted into seismic waves in the earth, and also tsunamis in the ocean. 

I'm not understanding the principle that all the energy would be turned to heat.    Even the air is not going to convert all the energy into heat.  Since it is technically a fluid, it will react to energy by shock waves and wind.  The question should be--how much kinetic friction did the debris produce as it moved through the air.  The air is a gas and is fluid.  It's not going to produce friction like moving a heavy object on steel.

How long do you think those shock waves would continue to move around the earth? Are they still moving? If not, where did the energy in those shock waves go?
Remember, we are talking about a quantity of energy that would dwarf trillions of nuclear weapons. There'd be a lot of energy in those shock waves that needs to go somewhere in less than a year in order to allow the ark to come to rest on the top of a mountain without being obliterated by ocean waves.

Heating air requires convection if I remember correctly. This requires time. If half of the debris was moing at 11 km/s as Numbers said, most of the mass of the atmosphere just happens to be in the first 11 km (wiki).  So that gives you one second of kinetic friction on the denser atmosphere.
We need to remember also that the second law of thermodynamics will be working.  Unless there is continuous input into a system, it will diffuse or move toward disorder.  That's why the concusion of an explosion doesn't keep on being felt forever and forever. 

See questions above, remember according to the second law, that diffusion you are referring to is the production of waste heat as the random motion and vibration of air and water molecules increases. If you want the shock waves to diffuse, something has to be absorbing the energy as heat or doing something like storing it as potential energy.

But lets say that there's some process that's 90% efficient at storing the kinetic energy with only 10% energy lost as heat. From the previous calculations that there's enough energy to boil the oceans 10 times over, that would still leave enough waste heat to kill noah and vaporize most or all of the oceans.

#13 AFJ

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Posted 22 January 2011 - 06:22 PM

no, 9.8 m/s^2 is acceleration not velocity and it is not terminal velocity.  Toss a baseball up in the air, it will accelerate downward (i.e. decelerate) at 9.8 m/s every second until it's upward velocity is zero, then it will start to fall downward.  As it falls it will get faster by 9.8 m/s every second.  The only limit to how fast it can fall is how aerodynamic it is.  Eventually the friction from the air will become so great that it will start to rob the baseball of velocity at the same rate as gravity adds velocity to it.  That's terminal velocity and it's not a single value, it varies based on the shape of an object.

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Yeah, I read it again. Sorry. I read the terminal velocity thing and understand it better. When the kinetic energy and the friction of the fluid it's falling in reach equilibrium, it hits terminal velocity. It said a skydiver who falls in a horizontal free fall is only going about 125 mph. Suprisingly slow.

For these kinds of calculations the downward velocity is largely irrelevant since we already know how much energy needs to be accounted for from the upward velocity.  Downward velocity just tells us how much energy is lost to the atmosphere vs the water of the oceans  The equation is basically this:
1/2 *(mass) * (up velocity)^2 = 1/2 *mass*(down velocity in air)^2 + energy added to atmosphere

I'm not sure if it wouldn't be better to deal with the up side and the down side differently. You didn't deal with the convection issue I raised. Most of the mass of the atmosphere is in the lowest 11 km according to wiki. You said escape velocity is 11 km/s, so you have the mass in the atmosphere for one second. You also have supersonic water shooting out. If something is moving that fast and short, I'm thinking more shock wave than heat.


Then the point I'm trying to make, even though I got the terminal velocity wrong, is that on the way down you have 1/2 the original mass times perhaps 100 m/s^2 at most for velocity. It won't fall at 5600 m/s. I realize he did this as an average, but I don't know if that's the best way. That is all I'm trying to say. It's not irrelavent in my opinion.

Then most of the remaining energy gets transferred to the oceans as the falling material gets slowed significantly before coming to rest on the ocean floor.


I didn't get back to you on that. I was talking about some of the debris would fall on the continents, and if the water was not deep or not yet covered. For that matter, we don't know if the preflood oceans were as deep as they are now.

How long do you think those shock waves would continue to move around the earth?  Are they still moving?  If not, where did the energy in those shock waves go?
Remember, we are talking about a quantity of energy that would dwarf trillions of nuclear weapons.   There'd be a lot of energy in those shock waves that needs to go somewhere in less than a year in order to allow the ark to come to rest on the top of a mountain without being obliterated by ocean waves.

Numbers, I know you know shock waves will pass through deep water without hurting anything on the surface, unless it comes to shallower water.

I think the general idea of the model. The waves caused by the debris caused some of the flooding. The water coming out of the earth was cooling the atmosphere and causing precipitation.

I'm not necessarily defending the model, I just don't agree with some of the basic concepts by the refuter. I haven't read the model--only an online summary of some of it.

See questions above, remember according to the second law, that diffusion you are referring to is the production of waste heat as the random motion and vibration of air and water molecules increases.  If you want the shock waves to diffuse, something has to be absorbing the energy as heat or doing something like storing it as potential energy.


What stops any energy process? Why are all the tsunamis, explosive concusions, and other energy causing events that happened last year not continuing indefitely? Because things are always trying to reach equilibrium. If you have a small explosion (not big enough to blow up the room) in one side of a closed room, the air pressure will increase on that side. The concussion wave will move until the air pressure is again at equilibrium. This would be a closed system.

The atmosphere has no walls or ceiling to blow out. It's an open system. What is going to stop the shock waves to displace alot of air into the thinner upper atmosphere where it would cool, or even into space? So the diffusion of heat accelerated by the shock wave of air would diffuse and mix into cooler air and space. This would seem a good application of the second law to me.

You would also have moving water to cool things. This might be part of what Brown was talking about, when he was explaining the cooling effects of expansion, in Glenn's last post.

I'm not saying that none of the debris would not cause heat. I'm saying that 1) not all the energy is going to be converted to heat, otherwise you are saying all energy is converted to heat. This is not true in chemistry (e.g. endothermic reactions). And 2) there are means of cooling the atmosphere before it gets too hot on paper.

But lets say that there's some process that's 90% efficient at storing the kinetic energy with only 10% energy lost as heat.  From the previous calculations that there's enough energy to boil the oceans 10 times over, that would still leave enough waste heat to kill noah and vaporize most or all of the oceans.

To be honest, I'm not necessarily arguing for the hydroplate model. But I do know it's a PhD engineer from MIT (Brown) that has taught for many years versus a high school teacher. I only know the basics in physics, so I'm not qualified to argue with either of them.

#14 Geode

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Posted 22 January 2011 - 06:53 PM

For the record I do not have a problem with the water that escaped from inside of the earth as being too hot. I feel that Dr. Brown covers this sufficiently here:
http://www.creations...eOverview6.html


About the problem with the water being too hot, I find Dr. Brown's explanation unsatisfying for I deal with this problem of heat flow almost every day at work and the evidence that such heat has been present for a long, long time in basin analysis through the use of materials that alter with time in the presence of heat.

Temperatures increase with depth inside the earth. Subterranean water about 10 miles deep would have been extremely hot. Wouldn’t all life on earth have been scalded if that water flooded the earth? 

No. Today’s geothermal heat is a result of the flood. To understand why and to see why life was not scalded, one must first understand tidal pumping and supercritical water (SCW)—a very high-energy, explosive form of water that was discovered in 1822.39 One should also understand why the continents and preflood mountains sank as the subterranean water escaped. [See Endnote 67 on page 216.]

40 Tides in the subterranean water lifted and lowered the massive crust twice daily, stretching and compressing the pillars, thereby generating heat and raising temperatures in the subterranean water. As certain minerals dissolved, this hot, high-pressure water, increasingly contained the ingredients for limestone (CaCO3), salt (NaCl), and quartz (SiO2). In a few chapters, you will see why, after the flood, this dissolved quartz petrified some wood and cemented flood sediments into sedimentary rocks.

SCW.  At a pressure of one atmosphere—about 1.01 bar or 14.7 psi (pounds per square inch)—water boils at a temperature slightly above 212°F (100°C). As pressure increases, the boiling point rises. At a pressure of 3,200 psi (220.6 bars) the boiling temperature is 705°F (374°C). Above this pressure-temperature combination, called the critical point, water is supercritical and cannot boil.

The initial pressure in the 10-mile-deep subterranean chamber was about 62,000 psi (4,270 bars)—far above the critical pressure. After about a century40 of tidal pumping, the subterranean water exceeded the critical temperature, 705°F. As the temperature continued to increase, the pressure grew, the crust stretched and weakened, and the energy from tidal pumping increasingly ionized the water.41

SCW can dissolve much more salt (NaCl) per unit volume than normal water—up to about 840°F (450°C).  At higher temperatures, all salt precipitates out.42 (In a few pages, this fact will show why so much salt is concentrated on the earth and how salt domes formed.)

Hot liquids cool primarily by evaporation from their surfaces.43 SCW consists of microscopic liquid droplets dispersed within water vapor. Most hot objects cool at a rate proportional to their total surface area. The smaller a particle, the larger its surface area is relative to its volume, so more of its heat can be quickly transferred to its surroundings. The liquid in SCW has an area-to-volume ratio that is a trillion (1012) times greater than that of the flood water that covered the earth’s surface. Consequently, the liquid in SCW cools almost instantly if its pressure drops. This is because the myriad of shimmering liquid droplets, each surrounded by vapor, can simultaneously evaporate. A typical SCW droplet at 300 bars and 716°F (380°C) consists of 5–10 molecules. These droplets evaporate, break up, and reform continually.44

This explains how the escaping supercritical liquid transferred its energy into supercritical vapor. How did the vapor lose its energy and cool? Rapid expansion. A remarkable characteristic of supercritical fluids is that a small decrease in pressure produces a gigantic increase in volume—and cooling. So, as the SCW flowed toward the base of the rupture, its pressure dropped and the vapor portion expanded and cooled. As it expanded, it pushed on the surrounding fluid (gas and liquid), giving all fluid downstream ever increasing kinetic energy.

Eventually, the horizontally flowing liquid-gas mixture began to flow upward through the rupture. As the fluid rose, its pressure dropped to almost zero in seconds, so the electrical energy of ionization was released. The 10,000-fold expansion was a weeks-long, focused explosion of indescribable magnitude, accelerating the mixture, including rocks and dirt, into the vacuum of space.45

In summary, as the flood began, SCW jetted up through a globe-encircling rupture in the crust—as from a ruptured pressure cooker. This huge acceleration expanded the spacing between water molecules, allowing flash evaporation, sudden and extreme cooling, followed by even greater expansion, acceleration, and cooling. Therefore, most of the vast thermal, electrical, chemical, and surface energy46 in the subterranean water ended up not as heat at the earth’s surface but as extreme kinetic energy in all the fountains of the great deep. As you will see, these velocities were high enough to launch rocks into outer space—the final dumping ground for most of the energy in the SCW.


Even if one believes this explanation of heat and water just before and during the flood, geothermal gradients as i understand them, with temperatures rising with depth below the surface, are not addressed. I can't see how Dr. Brown could explain the variation we see in various places around the planet. It does not explain the evidence for such heat flow before the reported time of the flood or why such flows are present today.

What changed between pre-flood and post-flood conditions?

#15 Geode

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Posted 23 January 2011 - 02:14 AM

To be honest, I'm not necessarily arguing for the hydroplate model.  But I do know it's a PhD engineer from MIT (Brown) that has taught for many years versus a high school teacher.  I only know the basics in physics, so I'm not qualified to argue with either of them.

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I seem to have missed who supplied the criticism of the Hydroplate Theory as it is not cited in the OP. However, Glen Morton who does possess a degree in physics (not mechanical engineering like Dr. Brown) and has worked for years as a geophysicist (far more relevant to the subject of plate tectonics than mechanical engineering) has published the most that I have seen in opposition to this theory.

#16 AFJ

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Posted 23 January 2011 - 05:23 AM

I seem to have missed who supplied the criticism of the Hydroplate Theory as it is not cited in the OP. However, Glen Morton who does possess a degree in physics (not mechanical engineering like Dr. Brown) and has worked for years as a geophysicist (far more relevant to the subject of plate tectonics than mechanical engineering) has published the most that I have seen in opposition to this theory.

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I only know that Answers in Genesis does not endorse this model. They tend to go with Baumgardner and CPT. Is there a possibility aspects of both models were involved. It's interesting to ponder, but personally, I'm not sure a model will ever be able to fully account for the flood, no more than anyone can account for the ressurection, or the parting of the Red Sea by natural means.

This is a real gulf for those who insist there is nothing supernatural. You have to think that God brought wild animals to Noah, and he didn't get eaten. God shut probably a very large door to the ark. Then He did something to keep the sewage fumes down--I've always thought hibernation. So it's not a problem with me personally to account for the flood by miracles, because I belive in a God of miracles. That is based on what I know He does even today.

#17 Glenn Williams

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Posted 23 January 2011 - 06:10 AM

Here is the person that originally brought up this objection, I mentioned this earlier in this quote:

Quote from Peggy Brown:
“Walt Brown is familiar with the argument raised by Geno Castagnoli, a high school science teacher in Oklahoma.  Walt says the argument is bogus, and frankly, that Geno really doesn't understand what he is talking about.”


I have a way less understanding of physics than either of you but is someone saying that unless energy is stored up as potential energy then it is always released as heat?

If I throw a rock at a table and it makes a dent and then the rock hits the ground did it all end up as dissipated heat?

#18 Scanman

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Posted 23 January 2011 - 11:49 AM

I seem to have missed who supplied the criticism of the Hydroplate Theory as it is not cited in the OP. However, Glen Morton who does possess a degree in physics (not mechanical engineering like Dr. Brown) and has worked for years as a geophysicist (far more relevant to the subject of plate tectonics than mechanical engineering) has published the most that I have seen in opposition to this theory.

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Glen Morton is a great source...I've had the pleasure of conversing with him in the past.

Morton's critique of Brown's Hydroplate Theory

Peace

#19 Geode

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Posted 23 January 2011 - 08:13 PM

I only know that Answers in Genesis does not endorse this model.  They tend to go with Baumgardner and  CPT.  Is there a possibility aspects  of both models were involved.  It's interesting to ponder, but personally, I'm not sure a model will ever be able to fully account for the flood, no more than anyone can account for the ressurection, or the parting of the Red Sea by natural means.

This is a real gulf for those who insist there is nothing supernatural.  You have to think that God brought wild animals to Noah, and he didn't get eaten.  God shut probably a very large door to the ark.  Then He did something to keep the sewage fumes down--I've always thought hibernation.  So it's not a problem with me personally to account for the flood by miracles,  because I belive in a God of miracles.  That is based on what I know He does even today.

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I think the best way to account for the flood if one insists on a literal interpretation with two of every animal, waters covering the tallest mountaintops, etc. is to invoke miracles for the available scientific evidence does not support the flood. Attempting to force fit available evidence into a flood model requires quite a bit of unnatural twisting.

#20 Geode

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Posted 23 January 2011 - 08:17 PM

Glen Morton is a great source...I've had the pleasure of conversing with him in the past.

Morton's critique of Brown's Hydroplate Theory

Peace

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Glenn finishes with what should be a final coffin nail in the Hydroplate Theory:

On final item.  I have seen the IPOD seismic line, from the continent out into the Atlantic mid oceanic ridge, every  inch of it, and there is ABSOLUTELY NO EVIDENCE OF ANY RESIDUAL BURIED WATER OR DEEPLY BURIED CAVE TO HOLD THE WATER.  There are no indications of collapse structures of the size your model would require anywhere on  any seismic data I have ever examined in the past 25 years.

I would also point out that there is a type of seismic wave, called a shear wave which doesn't travel through water.  In Brown's scenario, there would be no way to absolutely eliminate all the subterranean water.  As the crust collapsed, rubble would block the escape route for some of the water.  Given that shear waves don't travel through liquids, those of us in the seismology business should be able to find places where shear waves don't travel to the receivers.  If the water is shallow, it would only block a few seismology stations. 

But, we don't find any places in the shallow earth which won't transmit shear waves.  That is very strong evidence that there is no residual water remaining under the crust.


The geophysics just does not support the idea.




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