Youtube video of gravity device principle.. ?

[GavinPalmer1984]

For output unit:
From some previous calculations which are not shown on this forum yet, the most output is produced when the mass's height is half of the height of the container which it will slide within.  Having said that, the new and accurate measurement for the mass of the output unit will be figured as follows:

Assume that the output unit itself has no mass.
v = volume of output unit
m = sliding mass within the output unit.
p = density of liquid which output unit floats within

m < 1/2 * v * p

therefore, previous calculation's ammendments:

previously Ignoring friction:
28,954,842.84 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2)
n = 100;
v = 0.22 m/s;
height of a partition = 0.22 meters;

28,954,842.84 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2)
28,954,842.84 > (16751085.12) + (93915.36)
28,954,842.84 > 16,845,000.48
Excess work =  12,109,842.36 J

now ignoring friction with the more realistic mass estimation:
1/2 * 28,954,842.84 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2)
n = 200;
v = 0.11 m/s;
height of a partition = 0.11 meters;

14,477,421.42 > 8,375,542.56 + 23,478.84
14,477,421.42 > 8,399,021.4
Excess work = 6,078,400.02 J

Notice how the mass of the output unit's work was halved, I then compensated by increasing the number of partitions and decreasing the velocity.  Please notice that increasing the partitions will decrease the reset work.  And also notice that I have better  calculations for the mass within the output unit (although I ignored the mass of the output unit itself because I am just proving a point)

please reply.

And I will add that with less mass, the top of the output unit will float above the liquid's water level, thus increasing the amount of water which most be emptied in order for the top of the output unit to fall below the uni-directional latches.  But do remember that the partitioning can minimize the work needed to exchange the water no matter how much water needs to be emptied and refilled.

[hartiberlin]

Hi Gavin,

I had a closer look at your last video
http://www.youtube.com/watch?v=bIA2rZQgO_c

and you are totally right with your theory !

Great results...


If you look at this video for only just these 2 frames at:
1:57 min and
1:59 min
You can clearly see, that only 2 containers are lifted !
All the other containers of water just stay at their place !

So just let us calculate, how much energy we need to lift these
2 containers.

The lower left lowest water container  must be lifted excatly 1.10 Meter, if the height of the
main container is 1 Meter.
and the right lowest water container must be lifted 0.9 Meter.

So the energy required to lift the left container is:
Eleft = mass/11 x 10m/s^2 x 1.1 Meters
and the energy rquired to lift the right container is:
Eright = mass/11 x 10m/s^2 x 0.9 Meters

So now assuming we have a mass of 1000 KG in the main container we get:

The main output container can extract a lifting work for e.g.
a ship of about:
Eoutput= 1000 Kg x 10 m/s^2 x 1 meter= 10.000 Joules

Energy input into the 2 left and right containers is :

Eleft = 1000Kg /11 x 10m/s^2 x 1.1 Meters= 1000 Joules
plus
Eright = 1000Kg/11 x 10m/s^2 x 0.9 Meters= 818
so the total input energy is just only 1818 Joules
and the output energy is 10.000 Joules.

So with 11 Containers we have a COP of 5.5 !

Not too bad !

The only thing which must be seen is, that the boat which will
swimm on the output main tank must not sink too much into the water, so a
wooden or otherwise light substance must be used which does not drag the water down
but can lift the full M x G x H weightenergy.by pulling these external weights up with ropes
from the buoyancy of the swimming boat.

Well done Gavin ! Great system.

Regards, Stefan.

[truth]

 

I have a question:
How much energy will be used opening, closing, and controlling flow from all of the valves required to move this water back and forth?

The buoyant part in the main tank must either weigh more than two partitions full of fluid plus the buoyant force, or have a buoyant force of more than the weight of two partitions full of fluid in order to provide the energy for the lift.

Is this lift to be done with the main tank full or empty?

This is a great idea, but I am not sure all of the energy required to operate and reset is currently being calculated correctly.

[GavinPalmer1984]

I have a question:
How much energy will be used opening, closing, and controlling flow from all of the valves required to move this water back and forth?

The buoyant part in the main tank must either weigh more than two partitions full of fluid plus the buoyant force, or have a buoyant force of more than the weight of two partitions full of fluid in order to provide the energy for the lift.

Is this lift to be done with the main tank full or empty?

This is a great idea, but I am not sure all of the energy required to operate and reset is currently being calculated correctly.

The energy used opening and closing one gate will depend on the design.  If the energy needed to empty and close an inlet/outlet is less than the available potential within a partition and other available non-used potential energy, then the current equations will suffice.  I have one design for the inlets and outlets which would work, but would depend upon the partitions moving at separate times.  There are still obstacles, but I see that this type of device can be made perpetual.

dimensions from previous example:
Output Unit:
(y-coordinate)height = 22 meters
(x-coordinate)width = 3.7 meters
(z-coordinate)depth = 3.7 meters

Output Unit's Container:
height = 45 meters
width = 4 meters
depth = 45 meters

I MADE A MISTAKE ON THE OUTPUT UNIT'S CONTAINER.  THE DEPTH ONLY NEEDS TO BE SLIGHTLY LARGER THAN THE HEIGHT OF THE OUTPUT UNIT FOR THE ROTATION TO TAKE PLACE!!!
new depth = 23 meters

Output Unit's Container = 45 * 4 * 23 = 4140 m^3
Output Unit                  = 23 * 3.7 * 3.7 = 314.8 m^3

the Output Unit is roughly 1/13 * 1/2 = 1/26 of the mass of the Output Unit's Container.  This is using still inefficient geometries.  The Output Unit's Container would best be shaped somewhat like a half-cylinder which would decrease the mass of the liquid which must be emptied and refilled.

The Output Unit will usually be a proportion of the Output Unit's container.  It is in no relation to a constant amount of partitions.  Its relationship to partitions would be variable, depending on the number of partitions used.

The lifting of the partitions should be done with the Output Unit Container's liquid at minimum level... which means that all of the partitions are full.

[spinner]

How, exactly, is your "OU" (output unit) supposed to operate???

The overall energy potential limits are known. So far, no surplus....