[Dov]

Some preliminary assumptions:

Density of Sea Water = 1030 kg/m3 (as per Beicher, Robert J. Physics for Scientists and Engineers. Orlando: Saunders College, 2000.)

Width of the Split = The width of the split is open to suggestions, but I am basing it on the account which says that each tribe passed through in its own column. It is also said that 600,000 males between the ages of 13 and 65 (if I remember correctly) left Egypt. I am going to conservatively assign the total number of people as 850,000 and divide them equally into the 12 tribes.

# of people = 850,000

# of people / tribe = 70,834 (Actually 70,833.33 but we don't have fractional people)

Width of a person = 1m
Space between people in a column = 1m
Total Spaces = 23
Columns / Tribe = 25
Total Width of each tribe = (25 columns*1m)+(24 spaces*1m)= 49m / tribe

Space between tribes = 25m
# of spaces = 11
# of tribes = 12

Total width = (49m*12 tribes)+(25m*11 spaces) = 588+275=863m

I'm going to round it up to 875m to give them some room on the edges.

Now for the length of each column.

Since each tribe has 70,834 people and is divided into 25 columns each column is 5,902 people long. Assigning people a length of 2m to account for strides and putting a space of 3m between people we get:

(5902 people * 2m)+ (5901 spaces*3m) = 11,804m + 17,703m = 29,507m long.

I am going to assume a flat ocean floor and a water height of 30m. I think this height is more than sufficient.

So, what we have now, is basically a cube with dimensions of:

L = 29,507m
W = 875m
H = 30m

Volume = LWH so V = (29507m*875m*30m) = 763,936,230 meters cubed of displaced sea water.

Based on the density of seawater at 1035 kg/m3 we get a mass of:

1035 kg/m3 * 763936230 m3 = 790,673,998,050 kg

I will assume for simplicity that this cube is composed entirely of seawater and that it is suspended vertically at the cube's height of 30 meters. (I understand that it would probably not be this way because the displacement was likely sideways. The sideways displacement causes a serious problem for me as the water at the edges won't move as far as the water in the middle, etc. In addition, some of the water would have to be 'stacked' because it will be higher than it was in its resting state. )

We know that the sea collapsed back in behind them as they moved forward, so it trapped the Egyptians and drowned them.

The normal walking speed of a person carrying things is about 3 km/hr. Assuming that they were in a bit of a hurry (probably not too much of a stretch), let's say that the columns advanced at a rate of 5 km/hr, which probably is a stretch, but I'm going to use it anyway.

Since the cube is 29,507m long, it will take 5.9014 hours to cross it at 5 km/hr. Assuming that the cube caves in behind them in 5m(Long) blocks, then a volume of 129,450m3 of water weighing 133,980,750 kg would fall back down every 3.6 seconds. (The cube of the falling water has dimensions of 5m x 30m x 863m)

So, I guess what's left is to figure out how much energy it would take to hold up all that water for that much time.

Time for some criticism, I guess.

Fire away guys.

BTW, I am deliberately ignoring all the 'stuff' they took with them out of Egypt. I think it will balance some of my other assumptions.