### How to Poach Noah’s Ark.

Tuesday, November 21st, 2017This is a reprint from https://www.holysmoke.org/cretins/fludmath.html. I wasn’t able to find any contact information so this is only going to stay up temporarily unless I can get permission to keep it up.

## Dr. Marty Leipzig looks at the mathematics of ‘Noah’s Flood.’

Date: 09-02-99 10:11

From: Marty Leipzig

First- the global flood supposedly (Scripturally) covered the planet, (see that, George? If so, why are you still being so stupid?) and Mount Everest is 8,848 meters tall. The diameter of the earth at the equator, on the other hand, is 12,756.8 km. All we have to do is calculate the volume of water to fill a sphere with a radius of the Earth + Mount Everest; then we subtract the volume of a sphere with a radius of the Earth. Now, I know this won’t yield a perfect result, because the Earth isn’t a perfect sphere, but it will serve to give a general idea about the amounts involved.

So, here are the calculations:

First, Everest

V= 4/3 * pi * r cubed = 4/3 * pi * 6387.248 km cubed = 1.09151 x 10 to the 12 cubic kilometres (1.09151x10^{2}km^{3}) Now, the Earth at sea level V = 4/3 * pi * r cubed = 4/3 * pi * 6378.4 km cubed = 1.08698 x 10 to the 12 cubic kilometres (1.08698x10^{12}km^{3})

The difference between these two figures is the amount of water needed to just cover the Earth:

4.525 x 10 to the ninth cubic kilometres (4.525×10^{09} km^{3}) Or, to put into a more sensible number, 4,525,000,000,000 cubic kilometres

This is one helluva lot of water.

For those who think it might come from the polar ice caps, please don’t forget that water is more dense than ice, and thus that the volume of ice present in those ice caps would have to be more than the volume of water necessary.

Some interesting physical effects of all that water, too. How much weight do you think that is? Well, water at STP weighs in at 1 gram/cubic centimetre (by definition)…so,

4.252x10^{09}km^{3}of water, X 10^{6}(= cubic meters), X 10^{6}(= cubic centimetres), X 1 g/cm^{3}(= grams), X 10^{-3}(= kilograms), (turn the crank) equals 4.525E+21 kg.

Ever wonder what the effects of that much weight would be? Well, many times in the near past (i.e., the Pleistocene), continental ice sheets covered many of the northern states and most all of Canada. For the sake of argument, let’s call the area covered by the Wisconsinian advance (the latest and greatest) was 10,000,000,000 (ten million) km^{2}, by an average thickness of 1 km of ice (a good estimate…it was thicker in some areas [the zones of accumulation] and much thinner elsewhere [at the ablating edges]). Now, 1.00×10^{07} km^{2} X 1 km thickness equals 1.00E+07 km^{3} of ice.

Now, remember earlier that we noted that it would take 4.525×10^{09} km^{3} of water for the flood? Well, looking at the Wisconsinian glaciation, all that ice (which is frozen water, remember?) would be precisely 0.222% […do the math](that’s zero decimal two hundred twenty two thousandths) percent of the water needed for the flood.

Well, the Wisconsinian glacial stade ended about 25,000 YBP (years before present), as compared for the approximately supposedly 4,000 YBP flood event.

Due to these late Pleistocene glaciations (some 21,000 years preceding the supposed flood), the mass of the ice has actually depressed the crust of the Earth. That crust, now that the ice is gone, is slowly rising (called glacial rebound); and this rebound can be measured, in places (like northern Wisconsin), in centimetres/year. Sea level was also lowered some 10’s of meters due to the very finite amount of water in the Earth’s hydrosphere being locked up in glacial ice sheets (geologists call this glacioeustacy).

Now, glacial rebound can only be measured, obviously, in glaciated terranes, i.e., the Sahara is not rebounding as it was not glaciated during the Pleistocene. This lack of rebound is noted by laser ranged interferometery and satellite geodesy [so there], as well as by geomorphology. Glacial striae on bedrock, eskers, tills, moraines, rouche moutenees, drumlins, kame and kettle topography, fjords, deranged fluvial drainage and erratic blocks all betray a glacier’s passage. Needless to say, these geomorphological expressions are not found everywhere on Earth (for instance, like the Sahara). Therefore, although extensive, the glaciers were a local (not global) is scale. Yet, at only 0.222% the size of the supposed flood, they have had a PROFOUND and EASILY recognisable and measurable effects on the lands.

Yet, the supposed flood of Noah, supposedly global in extent, supposedly much more recent, and supposedly orders of magnitude larger in scale; has exactly zero measurable effects and zero evidence for it’s occurrence.

Golly, Wally. I wonder why that may be…?

Further, Mount Everest extends through 2/3 of the Earth’s atmosphere. Since two forms of matter can’t occupy the same space, we have an additional problem with the atmosphere. Its current boundary marks the point at which gasses of the atmosphere can escape the Earth’s gravitational field. Even allowing for partial dissolving of the atmosphere into our huge ocean, we’d lose the vast majority of our atmosphere as it is raised some 5.155 km higher by the rising flood waters; and it boils off into space.

Yet, we still have a quite thick and nicely breathable atmosphere. In fact, ice cores from Antarctica (as well as deep-sea sediment cores) which can be geochemically tested for paleoatmospheric constituents and relative gas ratios; and these records extend well back into the Pleistocene, far more than the supposed 4,000 YBP flood event. Strange that this major loss of atmosphere, atmospheric fractionation (lighter gasses (oxygen, nitrogen, fluorine, neon, etc.) would have boiled off first in the flood-water rising scenario, enriching what remained with heavier gasses (argon, krypton, xenon, radon, etc.)), and massive extinctions from such global upheavals are totally unevidenced in these cores.

Even further, let us take a realistic and dispassionate look at the other claims relating to global flooding and other such biblical nonsense.

Particularly, in order to flood the Earth to the Genesis requisite depth of 10 cubits (~15′ or 5 m.) above the summit of Mt. Ararat (16,900′ or 5,151 m AMSL), it would obviously require a water depth of 16,915′ (5,155.7 m), or over three miles above mean sea level. In order to accomplish this little task, it would require the previously noted additional 4.525 x 10^{9} km^{3} of water to flood the Earth to this depth. The Earth’s present hydrosphere (the sum total of all waters in, on and above the Earth) totals only 1.37 x 10^{9} km^{3}. Where would this additional 4.525 x 10^{9} km^{3} of water come from? It cannot come from water vapour (i.e., clouds) because the atmospheric pressure would be 840 times greater than standard pressure of the atmosphere today. Further, the latent heat released when the vapour condenses into liquid water would be enough to raise the temperature of the Earth’s atmosphere to approximately 3,570 C (6,460 F).

Someone, who shall properly remain anonymous, suggested that all the water needed to flood the Earth existed as liquid water surrounding the globe (i.e., a “vapour canopy”). This, of course, it staggeringly stupid. What is keeping that much water from falling to the Earth? There is a little property called gravity that would cause it to fall.

Let’s look into that from a physical standpoint. To flood the Earth, we have already seen that it would require 4.252 x 10^{9} km^{3} of water with a mass of 4.525 x 10^{21} kg. When this amount of water is floating about the Earth’s surface, it stored an enormous amount of potential energy, which is converted to kinetic energy when it falls, which, in turn, is converted to heat upon impact with the Earth. The amount of heat released is immense:

Potential energy: E=M*g*H, where M = mass of water, g = gravitational constant and, H = height of water above surface.

Now, going with the Genesis version of the Noachian Deluge as lasting 40 days and nights, the amount of mass falling to Earth each day is 4.525 x 10^{21} kg/40 24 hr. periods. This equals 1.10675 x 10^{20}kilograms daily. Using H as 10 miles (16,000 meters), the energy released each day is 1.73584 x 10^{25} joules. The amount of energy the Earth would have to radiate per m^{2}/sec is energy divided by surface area of the Earth times number of seconds in one day. That is: e = 1.735384 x 10^{25}/(4*3.14159* ((6386)^{2}*86,400)) = 391,935.0958 j/m^{2/s}.

Currently, the Earth radiates energy at the rate of approximately 215 joules/m^{2/sec} and the average temperature is 280 K. Using the Stefan- Boltzman 4’th power law to calculate the increase in temperature:

E (increase)/E (normal) = T (increase)/T^{4}(normal)

E (normal) = 215 E (increase) = 391,935.0958 T (normal) = 280.

Turn the crank, and T (increase) equals 1800 K.

The temperature would thusly rise 1800 K, or 1,526.84 C (that’s 2,780.33 F…lead melts at 880 F…ed note). It would be highly unlikely that anything short of fused quartz would survive such an onslaught. Also, the water level would have to rise at an average rate of 5.5 inches/min; and in 13 minutes would be in excess of 6′ deep.

Finally, at 1800 K water would not exist as liquid.