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The Physics of a Mesopotamian Flood.

Copyright 1996,1997 G.R. Morton. This may be freely distributed as long as not changes are made and no monetary charges are made. (home.entouch.net/dmd/physmeso.htm)

ABSTRACT:

From an estimate of the size of the ark the power output needed to push the ark from Shuruppak to Qardu is calculated. This is the suggestion of Dick Fischer in his book The Origins Solution. The power exceeds that reasonably owned by human beings. Some views of Noah's Flood requires that the ark start in Southern Mesopotamia and land in Turkey. Since this requires the ark travel uphill against the flow of water, this document examines the amount of energy required to move the ark in this fashion via human energy output. The ark starts at Shuruppak and ends at Quardu 1000 km northwest. The maximum output of a human being on a daily basis is 7000 calories. Given the crew complement, the conclusion is reached that after a couple of months of pushing the ark northward, the current would push them further south than where they started.

INTRO

What I am going to show is that it is impossible within the current laws of physics for an ark to be picked up at the topographically low delta region and to be deposited anywhere near Qardu in Turkey. Nothing more substantial than Freshman level physics will be used.

A look at a topological map of the region shows that Shuruppak is less than 500 feet in elevation. The lowest land in the region around Qardu in Turkey is 6500 feet elevation. (Illustrated World Atlas Set, Atlas of the World, 1993, p. 22) Thus at a minimum the ark must be raised by 6000 feet (1829 m)in elevation for this scenario to be correct.

WEIGHT OF THE ARK

Is this possible. Assuming that the ark is half the widely accepted dimensions of 137 X 14 X 23 meters (assumed dimensions 68.5 X 7 X 11.4. m) Assuming a foot (.3 m) thick layer of wood (specific gravity .651 g/cc (651 kg/m^3) see Ranald V. Giles, "Fluid Mechanics & Hydraulics, Schaum's outline seris, 1962), p. 37)

68.5 X 11.4 X .3 = 234.2 m^3 of wood for each of the floors and the top and the bottom. The ark has a top and bottom and two interior floors so this number must be multiplied by 4.

top,bottom, 2 floors= 937 m^3 of wood.

There are two ends, a front and a back. This is

11.4 X 7 X .3 = 23.9 m^3 of wood for each one. 47.88 m^3 for both.

There are two sides of the ark a left and a right. This totals to

2 x 68.5 X 7 X .3 = 287.7 m^3 of wood.

Adding this all up gives

287.7 + 47.9 + 937 = 1272.6 m^3 of wood for an ark half the normally accepted dimensions.

The weight of the empty ark is 1272.6 m^3 X 651 kg/m^3=828462 kilograms. A reasonable estimate for the loaded ark would be twice this figure, 1,656,924 kilograms.

LIFTING THE ARK TO MT. QUARDU FOOTHILLS (1982 M ELEVATION)

There are two ways for the ark to be lifted the requisite elevation. First, the water can do it. Boats in locks are raised in this fashion. But in order for this to work, the Mesopotamian region must have been covered by (1982 M (6500 feet of water.). In this case the entire Mesopotamian civilization would be destroyed. This did not happen.

The second way is for the humans to perform the work. Some have suggested this. He writes:

>During that period of floating, the ark was given direction by wind, >or punting poles, or both.

Is this physically reasonable?

The total energy needed to transport the ark to that region (1000 kilometers away) is

Energy = mgh + Fd

where m is the mass of the ark, g is the gravitational acceleration, h is the height the ark must be raised (6000 feet or 1829 m), F is the force used to move the ark and d is the distance.

CALCULATION OF F

To move the ark 1000 km in one year means an average speed of .03 m/s. If the water is flowing at 25 feet per sec. gives a relative velocity of 25.03 ft/s against the water.

Now F is a measure of the friction of the water as it flows down hill against the ark. The ark was poorly designed for frictional reduction. F can be calculated by assuming normal river flood velocities along rivers. Since the Flood of Noah was something large, and memorable, flow rates at the upper end of observed values would probably apply. Luna B. Leopold (Leopold Luna, A View of the River, Harvard University Press, 1994, p. 33) says that the fastest water velocities ever observed are 30 feet per second. Apparently, these were observations made by watching objects float down the river. The fastest velocity ever measured with a current meter was 22.4 feet per second in the Potomac River at Chain Bridge on May 14, 1932. There is a difficulty in measuring the fastest velocities with such a device because they must be dropped into the center of the river from above, which requires a bridge. According to a personal communication by Tom McCloud, the American Red Cross CANOEING first edition 1956 page 351 velocities as high as 24 feet per second have been observed on the Caney Fork, Rock Island Tennessee. The Brazos River, near Waco, Texas, (certainly not a mountainous area) recorded velocities of 22 feet per second.

Since the flood was supposed to be very large and memorable, we can assume that the velocity of water was large. We will assume that the ark is in a stream flowing 25 feet per second and the ark (half the generally accepted size) is half submerged. This means that the ark is 1/8 the volume of the generally accepted figure and that the relative velocity of water against the ark is 25 feet/sec. The narrow end of the ark is 37 feet wide and 22.5 feet high. Half of this area is 416 square feet. We will assume that the narrow end of the ark is what is facing upstream. The friction then is (Giles op cit, p. 56)

F= C rho A(V^2)/2 --this equation must be done in British units.

where C is the coefficient of drag, rho the density, A the area of the surface hit by the water, and V is the velocity of the water.

C = .455/(log Re)^2.58 (Giles, p. 100, 264)

Re (Reynolds number) is 25 feet/sec X 37 ft/ (1.217 x 10^-5)= 76 million.

C = .0021

With rho =1000 kg/m^3=62.4 lb/ft^3, V=25 f/s

F= .0021 X 62.4 lbs/ft^3 X 416 x .5 X (25)^2=17,,035 lbs.

Converting to Newtons,

F = 17,035 lb x 4.448 Newtons/lb= 75772 Newtons

Thus the total energy needed is

total energy = mgh+Fd

The distance from Southern Mesopotamia to Turkey is 1000 kilometers.

this is

1,656,924 kg X 9.8 m/s^2 X 1829 m + 75772 X 1,000,000 = 29,699,037,160 + 75,772,000,000 = 105,471,606,760 joules.

This means that 8 people must output this many joules in one year. This represents an output of 25,207,714 kcal in a year. Or a daily output of 8600 kcal per person per day just to push the boat.

According to Kimberly A. Hammond and Jared Diamond, ("Maximal Sustained Energy Budgets in Humans and Animals," Nature, April 3, 1997, pp 457-462), the maximum sustained daily output of a human being is 7000 kilocalories per day. Because of the need for more calories than humans can output in order to move the ark from Mesopotamia to Ararat, we must conclude that the people in the ark would not be able to keep the ark moving north. They would also not be able to feed any animals on the boat. All they could do would be to slow the southward trip.

This calculation does not take into account that some of the 7000 calories per day must be spent on other activities, like feeding the animals. This also assumes that all of the energy output is efficiently converted to forward motion.

If they ever find a 20 mile per hour rapids, they might be unable to cross it.