Author Topic: Youtube video of gravity device principle.. ? (Read 30118 times)

GavinPalmer1984 · « **Reply #60 on:** February 09, 2009, 10:35:46 PM »

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 · « **Reply #61 on:** February 11, 2009, 03:33:00 AM »

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 · « **Reply #62 on:** February 11, 2009, 04:30:46 AM »

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 · « **Reply #63 on:** February 11, 2009, 04:29:55 PM »

Quote from: truth on February 11, 2009, 04:30:46 AM

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 · « **Reply #64 on:** February 12, 2009, 12:41:36 PM »

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

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

GavinPalmer1984 · « **Reply #65 on:** February 12, 2009, 03:51:34 PM »

Quote from: spinner on February 12, 2009, 12:41:36 PM

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 · « **Reply #66 on:** February 12, 2009, 04:16:15 PM »

@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 · « **Reply #67 on:** February 13, 2009, 08:09:35 AM »

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 · « **Reply #68 on:** February 13, 2009, 04:41:10 PM »

Quote from: hartiberlin on February 13, 2009, 08:09:35 AM

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 · « **Reply #69 on:** February 24, 2009, 03:14:15 PM »

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.

spinner · « **Reply #70 on:** February 25, 2009, 09:59:36 PM »

So, Gavin are you surprised with the answer?

The "Physics Journal editor" answered you... (that is a success, most of similar, controversial letters never get any replies....)

What did you expect, claiming a "Perpetual motion", on paper, without any proof...

The partitioning principle.... Aha.

And it seems that a "Flat Earth" stuff wasn't very helpfull, either....

If you want to break "The Laws"... Do that with a decent "proof of a concept" device.
Cheers!

GavinPalmer1984 · « **Reply #71 on:** February 25, 2009, 10:17:31 PM »

HartiBerlin has a model that I may try to construct with my additions which I mentioned previously. Tesla mentioned that everything can be done in the opposite way. I had a device that was opposite to Harti's, but water pressure doubles every 33 feet. The pressure of air is not as substantial.

I was using a device that rotated due to buoyancy with a gravity engaged imbalancer, while Harti's rotates due to gravity and has a buoyancy engaged imbalancer. My design didn't account for the water pressure. I do not see any complications with Harti's model, but I will find out if there are any issues. I still think my design would have worked if I had furthered the design, but I was trying for something simple... and Harti's design retains simplicity.

I hope to let you know how it goes.

spinner · « **Reply #72 on:** March 10, 2009, 04:12:51 AM »

Bump...

GavinPalmer1984 · « **Reply #73 on:** February 02, 2010, 09:29:20 PM »

Remember my concept of partitioning:

http://www.youtube.com/watch?v=bIA2rZQgO_c

And this is an example of how partitioning works in general:

http://www.physorg.com/news184268881.html

==================================

I have had other ideas as well that are similar. Have you ever seen somebody wake surf?

http://www.youtube.com/watch?v=Q9Wy98Btfzk&feature=related

You have many devices that can ride a wake and all of them are in parallel acting as batteries, each generating energy.

At a certain number of wake-riding-energy-gathering-devices along a certain length wake, there would be enough energy to fuel the wake-creating device.

=================================

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Author Topic: Youtube video of gravity device principle.. ? (Read 30118 times)

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