New version of flotation device

[Floor]

In step 1, when you insert the float laterally, work needs to be done to expel amount
of water equal the weight of its volume, upward.

Incorrect

Because simultaneous to the float's insertion, a volume which is equal to the volume of the float
exits the tank (as the push rod on the opposite side of the tank).

There is no rise in the level of the fluid height nor is there any work done against fluid pressure
as the float is inserted.

             floor

[Floor]

Here are the diagrams and explanations again.

[Floor]

@telecom

I did the first draft of this / posted it, on the fly.  Here below is a revised version.
    floor


Revised calculations 1  (only slightly)

the energy density is, as should be expected very low, because gravity is a weak force.

assume:

float = 3 meters length by 0.33 meters radius.
approximate 10 feet by 1 foot

cylinder of 0.33 meters diameter or
0.165 meters radius  by 3 meters length
=  0.25669285714285694 cubic meters of water.

0.2566 cu meters of water = 256.6 kilograms of weight or
565.7062 pounds of weight.

256.6 kilograms = 2516.3864 newtons.

assume :

float chamber interior is 10.5 meters height, by 10.5 meters width,
by 0.5 meters depth into the page.

this means that the height of the lifting of the float
=  10.5 meters minus 1 float cylinder diameter.
              this is = 10.5 m - 0.33 m = 10.17 m
the float is lifted to a total of twice this distance
    10.17 m x 2  =  20.34 m  total height of the float rise
...
2516.3864 newtons x 20.34 meters = 5214.899376 joules of energy stored
as weight of the float external to the fluid chamber.  OUTPUT
...
assume:

It requires 25 kilograms of force or  245.1663 newtons, over a displacement of
3 meters to INSERT the float.
               245.1663 newtons x 3 meters = 735.4989 joules of work


It requires 25 kilograms of force or  245.1663 newtons, over a displacement of
3 meters to REMOVE the float.
              245.1663 newtons x 3 meters = 735.4989 joules of work
...
735.4989 newtons x 2 = 1470.9978 joules input work / energy.  INPUT
...
assume:

It requires 25 kilograms of force over a distance of 0.33 meters to initiate / accomplish
a 180 degree rotation of the device.

25 kilograms =  245.1663 newtons
245.1663 newtons x 0.33m = 80.9048 joules of work
...
there are two 180 degree rotations during each complete cycle.
80.9048 joules x 2 = 161.8097 joules           INPUT
...

1470.99 joules input + 80.90 joules input = 1551.9026 joules total input
.............................................................................
5214.89 joules of energy stored
as weight of the lifted float, while external to the fluid chamber.

5214.899376 joules of energy stored - 1551.9026 joules total input
= 3663.0951 joules of excess energy.

assume: a slow operation time to store that energy (not to use it) of 2 minutes

3663.0951 joules per every 2 minutes = 1831.5475 joules per minute.

1831.5475 joules per minute = 30.5233 joules per second or in other words 30.5233 watts
                          (1831.5475 joules / 60 seconds =  30.5233 joules per second)

conservatively arrived at, this is 30.5233 continuous watts gained as output power
                     or
30.5233 continous watts if used for 1 hour =  32 watt hours
                    and this is
732.5592 watt hours if continously used for 24 hours
                   and this is
5.128 kilowatt hours if continously used for one week
                   and this is 
20.51 kilowatt hours if continously used for one month
                   and this is
246.14  kilowatt hours if continously used for one year

probably actually about 1.5 times  this amount of energy.

[telecom]

Incorrect

Because simultaneous to the float's insertion, a volume which is equal to the volume of the float
exits the tank (as the push rod on the opposite side of the tank).

There is no rise in the level of the fluid height nor is there any work done against fluid pressure
as the float is inserted.

             floor

This is a game changer.

[Floor]

@ telecom

There was an error in my math. 

But the outcome is not very significantly effected.

I will post a corrected version of those calcs later.

floor