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Author Topic: New version of flotation device  (Read 7772 times)

Offline telecom

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Re: New version of flotation device
« Reply #30 on: June 20, 2019, 04:16:52 AM »
This may work, but ideally would be nice to allocate weight to each element,
to make it possible to use some rudimental math.

Free Energy | searching for free energy and discussing free energy

Re: New version of flotation device
« Reply #30 on: June 20, 2019, 04:16:52 AM »

Offline Floor

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Re: New version of flotation device
« Reply #31 on: June 20, 2019, 05:24:51 PM »
@ Telecom

I will follow your recomendation / request.

   thanks
            floor

Offline Floor

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Re: New version of flotation device
« Reply #32 on: June 20, 2019, 07:15:17 PM »
@ 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.

Free Energy | searching for free energy and discussing free energy

Re: New version of flotation device
« Reply #32 on: June 20, 2019, 07:15:17 PM »
Sponsored links:




Offline Floor

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Re: New version of flotation device
« Reply #33 on: June 20, 2019, 07:35:53 PM »
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

Offline telecom

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Re: New version of flotation device
« Reply #34 on: June 21, 2019, 11:54:37 AM »
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.

Free Energy | searching for free energy and discussing free energy

Re: New version of flotation device
« Reply #34 on: June 21, 2019, 11:54:37 AM »
Sponsored links:




Offline Floor

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Re: New version of flotation device
« Reply #35 on: June 21, 2019, 10:33:10 PM »
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

Offline Floor

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Re: New version of flotation device
« Reply #36 on: June 21, 2019, 10:40:12 PM »
Here are the diagrams and explanations again.

Free Energy | searching for free energy and discussing free energy

Re: New version of flotation device
« Reply #36 on: June 21, 2019, 10:40:12 PM »
Sponsored links:




Offline Floor

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Re: New version of flotation device
« Reply #37 on: June 22, 2019, 06:43:21 PM »
@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.

Offline telecom

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Re: New version of flotation device
« Reply #38 on: June 25, 2019, 07:32:39 PM »
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.

Free Energy | searching for free energy and discussing free energy

Re: New version of flotation device
« Reply #38 on: June 25, 2019, 07:32:39 PM »
3D Solar Panels

Offline Floor

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Re: New version of flotation device
« Reply #39 on: June 26, 2019, 10:43:18 PM »
@ 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

Offline Floor

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Re: New version of flotation device
« Reply #40 on: June 28, 2019, 10:56:06 PM »


Revised calcs.

Free Energy | searching for free energy and discussing free energy

Re: New version of flotation device
« Reply #40 on: June 28, 2019, 10:56:06 PM »
3D Solar Panels

Offline Floor

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Re: New version of flotation device
« Reply #41 on: June 28, 2019, 10:57:33 PM »
revised calculations 3  (there were bad calculations in versions 1 and 2)

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 + 161.8097 joules input = 1632.7997 joules
                       (previously in error)
...........................................................................
5214.89 joules of energy stored as weight of the lifted float, while
external to the fluid chamber.

5214.899376 joules of energy stored - 1632.7997 joules total input
= 3582.0996 joules                              ENERGY YIELD
.........................................................................
assume:

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

3582.0996 joules per every 2 minutes =  1791.0498 joules per minute.

1791.0498 joules per minute =  29.85083 joules per second or in other
words 29.85083 watts
         (1791.0498 joules / 60 seconds =  29.85083 joules per second)

conservatively arrived at, this is 29.85083 continuous watts gained as
output power
                     or
29.85083 continuous watts if used for 1 hour =  29.85083 watt hours
                    and this is
716.41992 watt hours if continuously used for 24 hours
                   and this is
5.0149 kilowatt hours if continuously used for one week
                   and this is 
 20.0597 kilowatt hours if continuously used for one month
                   and this is
240.717  kilowatt hours if continuously used for one year

probably actually about     3      times  this amount of energy.

Offline Floor

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Re: New version of flotation device
« Reply #42 on: July 04, 2019, 06:23:19 PM »
@ All readers

The designs and methods I have presented in this topic and which are novel,
are given into the public domain.

  floor

Offline Floor

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Re: New version of flotation device
« Reply #43 on: July 07, 2019, 07:32:55 AM »
Gravity does all of the work of the lifting in this design.
Then the lifted opject falls.

               floor

Offline Floor

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Re: New version of flotation device
« Reply #44 on: July 14, 2019, 03:51:12 AM »
@ All readers

There is second version of a sucessful gravity based design (presented byCadman)

@ https://overunity.com/18243/cadmans-hydrostatic-displacement-engine/
 
floor

 

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