# Re: Physics of a Mesopotamian Flood

Glenn Morton (GRMorton@gnn.com)
Sun, 05 May 1996 18:14:54

The Physics of a Mesopotamian Flood.

ABSTRACT:

the ark from Shuruppak to Qardu is calculated. The power exceeds that
reasonably owned by human beings. Dick Fischer's view requires a
sustained human output of 279 watts every day, 24 hours per day to lift
the 1,600,000 kilogram ark to Qardu, 1000 km distant from Shuruppak.
According to Mark Drela and John S. Langford, "Human Powered Flight,"
Scientific American Nov. 1985, p. 144-151 esp. 150) the maximum
sustainable output for a human is about 75 watts over an 8 hour period(40
watts for a 12 hour shift). To output 279 watts over an 8 hour shift
requires 4 people. Since there are only 8 people two 12 hr shifts are
required and the average power output per person drops to 40 watts
for a total of 160 watts available. After a couple of months of this work,
they would be further south than where they started.

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 2000 feet elevation. (Illustrated World Atlas Set, Atlas of the
World, 1993, p. 22) Thus at a minimum the ark must be raised by 1500 feet
(457 m)in elevation for this scenario to be correct.

WEIGHT OF THE ARK

Is this possible. Dick has not stated what he believes the size of the
ark is, but I will assume that the ark is half the widely accepted
dimentions 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.

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

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 (610 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 (610 M (2000 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. Dick seems to
prefer 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 elevation distance, 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 2 mph (.9 m/sec) gives a relative velocity of .93
m/s (3.05 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 that the ark is in a gentle (2 mph) stream
and the ark is half submerged. The friction then is (Giles op cit, p.
56)

F= C rho A(V^2)/2

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 3.05 feet/sec X 75 ft/ (1.217 x 10^-5)= 20
million.

c = .0026

With rho =1000 kg/m^3, V=.93 m/s

F= .0026 X 62.4 lbs/ft^3 X 37.5 X 11.25 .5 X (3.05)^2=316 lbs.

Converting to Newtons,

F = 316 lb x 4.448 Newtons/lb= 1405 Newtons

Thus the total energy needed is

1,656,924 kg X 9.8 m/s^2 X 457 m + 1405 X 1,000,000 = 7,420,699,826 +
1,405,000,000 = 8,825,699,826 joules.

This means that 8 people must output this many joules in one year. This
represents a continuous output of 279 watts output for the entire year.

According to Mark Drela and John S. Langford, "Human Powered Flight,"
Scientific American Nov. 1985, p. 144-151 esp. 150) the maximum
sustainable output for a human is about 75 watts over an 8 hour period.

To output 279 watts over an 8 hour shift requires 4 people. Since there
are only 8 people two 12 hr shifts are required and the average power
output per person drops to 40 watts.

This is clearly an impossibility. After a couple of months of this work,
I bet the ark occupants thought it would have been better to die in the
flood. They would also be further south than where they started.

glenn

Foundation,Fall and Flood
http://members.gnn.com/GRMorton/dmd.htm