Youtube video of gravity device principle.. ?

[GavinPalmer1984]

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

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

If you assume that the inlets/outlets are operated by potential within the partitions and that you can transfer the potential from the output unit in a manner that can allow you to dissipate the energy over any amount of time, then there is a surplus in the output unit when using the partitioning scheme.  I want you to see this before we go any further.

The potential of the Output Unit will be something like:
m * x * g
m > 1/26 * M
M = mass of the liquid within the output unit's container. (note that you do not have to empty the whole container and the geometries can be made more efficient which would better the ratio of m : M)
x > 1/5 * H
H = height of the output unit's container
g = gravity.
It simply follows that the Output Unit has a constant amount of output and the partitioning has a variable amount of input which can be reduced to a number close to zero if the dimensions are chosen properly.  But we only need the input to be less than the output.

Assuming we empty the whole container:
IE = mass * distance * gravity + 1/2 * mass * velocity^2
IE = M * 2 * H/n * g + 1/2 * M * V^2
Because we can dissipate the OE over any amount of time 1/2 * M * V^2 can go to zero
IE = M * 2 * H/n * g
OE = 1/5 * 1/26 * M * H * g
with above assumptions:
M * 2 * H/n * g < 1/5 * 1/26 * M * H * g
2 * H/n < 1/130 * H
2 * H < 1/130 * H * n
260 < n

So there you have it.  But realistically, the number of containers will be much less because I assumed we empty the whole output container and the output container's geometry is inefficient.  N would increase depending on the friction encountered and velocity chosen.

[GavinPalmer1984]

@hartiberlin

lift 2 containers and n is even:
(2 * M/n) * (2 * (H + H/n)) * g = 2 * M * 1/n * 2 * g * ( H + H/n)
mass = 2 * M/n
distance = 2 * H + 2 * H/n
g = gravity

lifting all the containers:
2 * H/n * M * g

2 * H * 1/n * M * g ~~ 2 * M * 1/n * 2 * g * (H + H/n)
H ~~ 2 * (H + H/n)

Therefore, lifting all of the containers would be more efficient AND less friction would be encountered.  But thank you for the recognition.  And also notice that even using your less efficient method of partitioning will still decrease the reset work when you increase the number of partitions by decreasing the mass.  The distance which the containers must be lifted will always be greater than H.

I should really post my addition which decreases the mass along with decreasing the distance. (the idea of submerging the partitions within another container of liquid and using the rising air from one partition intelligently so that the air in the bottom partition can be used to provide a buoyant force for at least half of the partitions. And the air in the second lowest partition can provide a buoyant force to the other half of partitions.)

The addition would decrease the work to roughly:
2 * M/n * 2 * H/n * g
mass = 2 * M/n
distance = 2 * h/n
g = gravity

[hartiberlin]

All, have a look at this principle,
it is still in German language, but have a look at the embeded FLASH animation.

Maybe you can use this for your output principle ?

http://www.overunity.de/index.php?topic=74.0

Have a look at the second posting there with the
embeded FLASH animation, wait a few seconds until the
arrow comes up and click the arrow.
Then you will see the great animation there.


Regards, Stefan.

[GavinPalmer1984]

All, have a look at this principle,
it is still in German language, but have a look at the embeded FLASH animation.

Maybe you can use this for your output principle ?

http://www.overunity.de/index.php?topic=74.0

Have a look at the second posting there with the
embeded FLASH animation, wait a few seconds until the
arrow comes up and click the arrow.
Then you will see the great animation there.


Regards, Stefan.

I like it.  I have attempted to do something similar with buoyancy only, but I encountered problems with water pressure.  I will give you my interpretation of the device.  I will not assume that you are using hydraulics because it is not necessary, although it is capable of being a hydraulic system.  I will use your middle point as a reference and pretend that it does not actually exist.

It looks like your large interior mass (LM) is lifting the counterweight (cw).  The narrow cylinder is buoyant, therefore its mass (nm)is less than the mass of the liquid it displaces.

LM > 4 * cw
nm < liquid displaced by narrow cylinder... This provides a mechanism to ensure that the flip will not begin until a sufficient amount of offset has been reached... it is vital.
I will assume nm = 0 for now.
liquid displaced by narrow cylinder's mass is  = qm

So I guess you are trying to ensure that part of LM will remain on the upper portion of the output unit, so that the upper mass of the output unit (UM) is larger than the bottom mass of the output unit (BM)

UM > BM will enable a flip.

I will go ahead and tell you that at 90 degrees in the flip, the masses will start moving again... this means you need a sufficient offset to ensure a complete 180 degree flip.  This is what the buoyant cylinder achieves.  The buoyant cylinder also provides kinetic energy which can be harnessed in multiple ways.

When the buoyant cylinder rises to the top, the BM will increase by the amount equal qm.  As LM falls, the masses of BM and UM are affected.

I also suggest that the counterweight only needs to move from the lower half to the upper half.  It does not need to go beyond the mid-point.

I will use druckmasse as a time reference and assume that the qm is at the BM and I ignore equivalent liquid masses:
At druckmasse = 31: I would call this max BM:
volume of (1/8 * LM) * density of liquid = vM
BM = qm + cm + 3/8 * LM + vM
UM = -qm + nm + 5/8 * LM
and nm = 0;
UM = -qm + 5/8 * LM
BM = UM + cm + qm + vM - 1/4 * LM
BM > UM iff (cm + qm + vM > 1/4 * LM)

At druckmasse = 32: I would call this equilibrium of cm and LM:
BM = qm + 1/2 * cm + 1/2 * LM
UM = -qm + nm + 1/2 * cm + 1/2 * LM
and nm = 0;
UM = -qm + 1/2 * cm + 1/2 * LM
BM = UM + qm

At druckmasse = 33: This should be max UM
BM = qm + 5/8 * LM
UM = -qm + nm + cm + 3/8 * LM + vM
and nm = 0;
UM = -qm + cm + 3/8 * LM + vM
UM = BM + cm - qm + vM - 1/4 * LM
UM > BM iff (cm + vM > 1/4 * LM + qm)

And to be realistic:
(cm + vM > 1/4 * LM + qm) AND
(cm + qm + vM > 1/4 * LM) AND
(qm > 0) AND
cm < 1/4 * LM AND
2 * cm < cm + 1/4 * LM < 1/2 * LM

if qm = 0, then the equations are satisfied.  But qm must be greater than zero to ensure that we do a successful 180 degree turn.  Substitute (qm - nm) = qm for realistic buoyant cylinder. qm will be constant.

vM - (1/4 * LM - cm) > qm
qm = displaced volume of nm * density of liquid - nm  (liquid is at BM and nm is at UM)
vM = displaced volume of 1/8 * LM * density of liquid
p = density of liquid
Volume (1/8 * LM) * p > Volume of nm * p - nm
This will allow us to begin choosing a valid qm.

A low density matter for LM will be ideal for increasing the mass of vM which will allow for a larger qm.  This helps to ensure that the device achieves 180 degree flip.

Good job.  Let me know if I completely altered your idea.  I know how this would work because I have thoroughly investigated such a device.  It is amazing that we are nearly on the same page.  Also note that the buoyant cylinder would be the main source of output energy via a sliding mass within it and/or it being connected to another pulley system.  I was waiting to disclose my idea for a more efficient device like you have outlined, but your idea uses the same concepts in a better way.  I congratulate your work because I know how unique it is.

You might consider investigating hydraulics instead of pulleys.  Pulleys will have more losses than hydraulics.

I started with my initial design because I wanted to get my partitioning scheme out on the internet.  This is definitely a more efficient perpetual motion endeavor.  There are complications in implementing the device.  I will think about your device and we should definitely collaborate to improve your design with some of my ideas.

I would add that there is no reason why you should have the cm and pulleys in air.  I would have the whole thing operating within the reservoir... pulleys, cm, cables, and all.  The buoyant cylinder would be your output unit.  I am no expert on pulleys, so this might be the crux.

[GavinPalmer1984]

I wrote a physics journal and they write back:

Dear Mr Palmer,

I cannot accept your manuscript "Cooperative Systems" [PoF 090-0170-l] for publication as a Letter in the Physics of Fluids. You propose a perpetual motion machine, which is known to be impossible.

You need a free lift of the water. That is available in tides. In fact there is an interesting "pumping trick" of pumping into the tidal lagoon at high tide, an energy cost which is repaid with interest at low tide. You might wish to read Chapter 14 (pages 81-87) with Appendix G (pages 311-321) of David JC MacKay's book "Sustainable Energy - without the hot air", a free book available on the internet at www.withouthotiar.com

Yours sincerely,
John Hinch
Associate Editor for Letters
Physics of Fluids

_______________________________

I responded with:

Mr. Hinch,
I do not say that I need a free lift.

The lift of partitions will require work =
1/2 * m * v^2 + m * g * d + f * d
m = mass of liquid exchanged (assume partition's mass is countered)
v = max velocity of the lift
g = gravity
f = friction encountered (decreases with distance)
d = distance which the partitions must be lifted.
d = 2 * X/n
X = height of the container amount of liquid exchanged.

Please notice that as the number of partitions increase, the distance which the mass is lifted decreases.  You and the majority of people may assume that perpetual motion is impossible, but we once assumed the earth was flat as well.  You can't see that the earth is NOT flat unless you first acknowledge that our current understanding may be wrong.  I know for a fact that I can defend this proposal... I have not had one person disprove my process.  I beg you to investigate this further.  Please point to your areas of confusion.  I suggest that you stay away from the laws of thermodynamics and conservation of energy when addressing this manuscript.  There is no way to disprove my process with a law which I am attempting to prove inaccurate.

please,
  Gavin Palmer

____________________________

@hartiberlin

Did you ever read through my suggestions on your device?  I hope I helped.