New version of flotation device

[telecom]

This may work, but ideally would be nice to allocate weight to each element,
to make it possible to use some rudimental math.

[Floor]

@ Telecom

I will follow your recomendation / request.

   thanks
            floor

[Floor]

@ telecom

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
...
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
the 180 degree rotation of the device.

25 kilograms =  245.1663 newtons
245.1663 newtons x 0.33m = 80.904879 joules of work

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 excess energy.

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

3663.0951 / 2 = 1831.5475 joules per minute.

1831.5475 joules per minute = 30.5233 watts

conservatively arrived at this is 30.5233 watts total output
                     or
43,920 watts over a 24 hour period.

[Floor]

OR

0.25669285714285694 cu meters =  67.78 gallons per two minutes

this is 33.89 gallons  lifted to a 20 foot height per minute.

         floor

[telecom]

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.