The “hydroplate theory” has far more flaws than current techtonic theory. “Hydroplate Theory” is an attempt to justify a biblical passage, not to reach conclusions form known or observed facts.
The very physics of “hydroplate theory” don’t pass the smell test.
Here’s a pretty good lesson in physics that anyone clinging to hydroplate theory should read and understand, and the most compelling thing you should know before you read this, is it was actually written by a creationist, albeit and old earth creationist not one of these young earth folks:
By the hydproplate theory you have:
The earth constructed as follows: a solid center, a water layer, then above that, a layer of granite or basalt which ‘floats’ on the water. This is as shown below:
^ a mountain ps = 2.1 g/cc height = h
Earth’s crust (granite 2.65 g/cc, or basalt 3.3 g/cc) Thickness = T
Water (density 1.0 g/cc) Thickness = tw
Earth center (solid density > 3.3)
There are several things to notice about this situation. First, the crust must be absolutely impermeable to the water. There must be no earthquakes before the flood since the first crack in this sphere would allow the water to escape. This means that there must be no meteorites before the flood. And heaven help mankind if he ever were to have drilled into the crust for curiosity’s sake.
There must absolutely not have been any elevation differences. The effects of a load on the top of the crust can be seen from using an elastic sheet solution to the load. The 4th order differential equation is:
4
d z
D—— + (pm-pw)zg = P(x)
4
dx
where:
P(x) is the load as a function of horizontal distance
z is the distance the load will sink
g is the acceleration of gravity
pm is the density of the crust
pw is the density of water
D = ET3 /(12(1-sig)^2
where:
E is Young’s modulus, 1011 dynes/cm^2
sig is Poison’s ratio, .25
T is the thickness of the crustal layer into which the load sinks
For a two dimensional load with a ½ width, A, the bending of the crust is:
z max = h(ps-pw)(1-e-^LA cos(LA)) / (pm-ps)
where:
L = 4th root ((pm-pw) g) / (4D))
With a crust thickness of 5 km (5 x 10^5 cm) sig = .25, E = 10^11, we have:
D = 1.1 x 1028
L = 4.37 x 10-7
Now, for a mountain 10 km (1.10^6 cm) in radius and 5 km (5 x 10^5 cm) in height (h), the minimum thickness of the crust must be:
= 4.1km
A crust thinner than this value will be completely broken by the weight of the mountain.
The bending of the crust by 4.1 km will occur by fracture. This would immediately release the water. Thus, there are no mountains. Even a hill one kilometre high would require that the crust bend by 830 meters.
Therefore, the crust must be perfectly smooth. Thus, you must violate the Biblical record where it says that all the high mountains were covered. In your conception of the flood, there could be no mountains or hills.
Secondly, in your model, you must have pillars to retain the physical connection with the core. If you do not do this, you will have the certainty that the crust will eventually crash into the core. Friction between the crust and the water and the water and the core will cause the outer crust to begin to move in a fashion different to that of the earth’s interior. This would cause turbulence and would lead to a crash. The crust is free to move in relation to the core in response to tidal forces. The theoretical height h of the equilibrium tide in a rigid earth is:
h = .5 (M/E)(a/R)^3 a(3cos^2 (theta)-1)
where:
E is the mass of the earth
M is the mass of the moon, 1 and .123 respectively
a is the radius of the earth 6378 km
R is the distance from the earth’s center to the moon’s center, 384,405 km
theta is the angle between the moon and the zenith
Plugging these values into the equation we have h = .00358 km, or h = 3.58 meters. This means that your crust will heave every day by this value. Due to the fact that neither granite nor basalt are single crystalline materials, small fractures will develop in between the individual crystals.
Suppose you placed the water under 5 km of crust, the pressure of the water would be:
5 x 10^5 * 980 * 2.65 = 1.29 x 10^9 dynes = 1281 atmospheres of pressure
The temperature gradient is 1º C for every 30 m so there is a 166º C increase in temperature as we go deeper.
166 + 25º C (the surface temperature) = 191º C
A layer of cave water 2 km thick all around the earth would contain 1 x 10^24 cubic centimeters of water. At 191ºC, the high temperature water would contain 1.7 x 10^26 calories. (1 calorie per degree rise (166 degree rise)). The minute the pressure is released the water will turn to steam and you will cook the earth. Dividing the calories by the surface area of the earth shows that:
heat /cm^2 = 1.7 x 10^26 Calories/5 x 10^14 square cms = 3.3 x 10^7 Cal/cm^2
I don’t think Noah could survive this. This is a poor mechanism for a flood.
I have seen the IPOD* seismic line, 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 22 years.
Water Velocity
Brown has a 10 km thick granite crust with a 1 km thick layer of water. The pressure is enough to raise a tube of water to 17 km (see Brown, pg. 37, Fountains of the Deep). Water squirting up out of the hole will rise to that level. What is the velocity of the water coming out of the crack? Ignoring friction, this can be found by equating the potential energy of the drops at 17 km to the kinetic energy at the surface needed to propel the water that high. Thus:
gh = .5v^2
where:
h is the height of the water, 17 km
v is the velocity
g is the acceleration due to gravity, 9.8
Solving for v, we have v = 577 meters /sec. According to the steam table cited below, there is a 814 times increase of volume in the phase change. The vapor occupies 814 times more volume.
Now, According to Steam Tables (Combustion Engineering Inc., 1940), the pressure needed to keep water a liquid at 250º F, which is the temperature of Brown’s water, is 2.02 atmospheres.
Consider a 1 square meter tube with 577 cubic m/s emanating from it. Due to the fact that 2.02 atmosphere is the weight of 20 meters of water, water coming up the crack will not change to steam until the final 20 meters. With the velocity of 577 meters per second coming out of the crack, this means that 577 cubic meters each second will occupy 814 times the volume that it used to. As a water surface passes the point at which it turns into vapor, it will, within one second, be pushed 577 x 814 = 469,779 m. This is a velocity of 469 kilometers per second. There would be no flood since none of the vapor would remain on the earth. The earth’s escape velocity is about 11 kilometers per second. Any object that exceeds 11 km per second leaves the earth and never returns. How could this theory cause a flood?
In reality these numbers would be somewhat smaller due to frictional effects, but even if they are off by 99%, the steam escaping is still above escape velocity for the earth. The steam would be sent to Alpha Centauri!!”
Very good - now if you just read the rest at his website you’ll see explanations for nearly every problem you have just described. Or do only the critics of YEC theory get to bend your ears?
Bwaaaaahahahahaha.....how DARE you use actual calculations!!
I figured the release of that amount of super-heated water would simply flash-steam everything.