@wattsup
I think we have a language barrier.

@mondrasek

I look at this as an energy in / energy out. Your first full container is emptied into one container that is then lifted to re-fill the first container. Compare that with the same full container emptied into 10 containers that is then requiring more energy to lift because each base adds to the weight, so you need more energy to lift it. You will also have to energize 10 latches to hold it there instead of one. Does not make sense.

As mondrasek and I have said, the mass of the containers themselves can be negated by counter-weight / counter-forces.
You should not assume that we will need to energize latches.  And as I have mentioned, there is energy in my process which I have not even accounted for (aka. falling water, rising water in the partitions).

@GavinPalmer1984

"I am going to use the water to create a buoyant force on the output unit!"

Any force of that type would render the highest and lowest container out of the fill level, meaning the top output container would not fill and there will be water left in the big container because the bottom output container will be too high. Either that or your explanation is really not clear of what you mean by buoyant force.

Usually guys do not make thing to do nothing unless you are working this as a perpetual  motion scheme. But here is nothing perpetual about it because you will never get more energy empty in the big container then you will get back by re-filling it due to the losses of lifting the output containers. So what's the big deal.

Go look at the youtube video:
http://www.youtube.com/watch?v=bIA2rZQgO_c

You must have just come in here and started rambling without reading through the previous posts.  If you had read through the previous posts, you would address problems specifically and point to times in the video clips as reference.  You quote my conclusion...  what is your deal?  Please do not waste other people's time.  I have thought this thing out for over 2 years, and I am intelligent.  Please respect my work by putting forth more effort before claiming:

"But here is nothing perpetual about it because you will never get more energy empty in the big container then you will get back by re-filling it due to the losses of lifting the output containers. So what's the big deal. "

A respectful person will come in and ask questions instead of LEAPING to conclusions... you ignored mondrasek's previous reply to your misunderstanding.  Let me know if your native language is not English so that I might understand your lack of communication skills.

@GavinPalmer1984

On your video at the 0:50 mark, which is the container divided by two, you fill both containers and then you say that both containers only rise by x (d=x). That is not so. They both still rise at 2x and no matter how many containers you use, they will "always" have to rise at 2X otherwise you will not be able to fully empty the containers.

Each container will need to rise 2 * H/n where H/n = the height of one of the small containers and H = the height of the large container.  In the video, X = H.

This thread is only 5 pages long so I did read it through but quickly cause I like spending my time on testing stuff.

No matter how many containers you use, they will always have to rise by 2x, not relative to your big container but to themselves. That's where the first 2x came from. 2 times the distance of the filled contained. So you can lift it once, or with 100 containers but you will always have to lift them at 2 x. So again, what's the big deal.

You are right.  They must all rise two times the height of themselves.  So that every container will rise 2 * H/n.

So the work needed to reset the output unit = M * D * g
M = mass of water
D = 2 * H/n
g = gravitational constant
n = number of partitions

You may also say work = n * (M/n * D * g )
This means that each container with mass = M/n must be lifted a distance D = 2 * H/n.  The work required to lift one partition is = M/n * D * g.

Using more partitions decreases the distance which the water must travel in order to empty and refill the original container.

"Using more partitions decreases the distance which the water must travel in order to empty and refill the original container."

Exactly....... but since each container still has to rise at double its height, using one container or 100 containers will still require the same amount of work.

You are wrong.  Just slow down and think through the problem.

If you use one container which is 1 meter high, and 1000kg of water, and assume gravity = 10m/s^2:
Work = 1000kg * 2 meters * gravity = 20000 J

If you have 2 partitions:
Work = 2 * (500kg * 1 meter * gravity) = 10000 J

If you have 4 partitions:
Work = 4 * (250kg * 0.5 meter * gravity) = 5000 J

Notice how the mass remains 1000kg.  And the distance decreases.

@wattsup

The time it takes to empty and refill the container does not matter because I am only proving that perpetual motion is possible.  I have previously stated that there will be more efficient perpetual motion devices to come... especially when people stop believing in our current interpretation of the laws of physics.

So yes, adding more partitions will increase the time it takes to empty and refill the container.

And it will decrease the total work needed to empty and refill the container.

@Gavin
It seems to me we're just having a big misunderstanding....

If it is of any help for you (so you can go forward with explaining your device/concept), I'd like to confirm that I understand:

>>The partitioning scheme (which enables a proportionally the lowest energy input to actually completely  drain and then fully refill a certain amount of water(mass close loop displacement)).

>> In this thought experiment, we're not considering any losses

>>and (as Gavin said) the energy (mass in a gravity field) mechanical equivalents are not the prime concern of this project...

Is this a consensus?


So, I'd like you to continue with your explanations....
Several posts ago, I asked you about the whole picture (not just the part. scheme).

How is yours output unit supposed to operate?

You are mentioning that the real thing is - indeed -  happening in this "output unit".

Buoyancy... Mind you, the buoyancy is just a gravity "in disguise"....



Looking forward to hear more,
Cheers!

>>The partitioning scheme (which enables a proportionally the lowest energy input to actually completely  drain and then fully refill a certain amount of water(mass close loop displacement)).

Exactly... and the partitioning scheme is not fully developed.

>> In this thought experiment, we're not considering any losses

Yes and No.  Consider them so that the design avoids as much loss as possible. ( I am posting to get help from the online community... there are many brilliant minds out there)

>>and (as Gavin said) the energy (mass in a gravity field) mechanical equivalents are not the prime concern of this project...

The mass of the output unit is the source to lift the partitions.  The emptying and refilling (half-way) of the output unit's container is the source to reset the output unit.

So, I'd like you to continue with your explanations....
Several posts ago, I asked you about the whole picture (not just the part. scheme).

How is yours output unit supposed to operate?

The whole design is a work in progress.  I bring to the table many ideas which are possible.  I like to start broad so that others may work out other designs based upon my original concepts.

I think that the best output unit will be one in which the mass pushes air out of its container, which is used in conjunction with the partitioning scheme (with the partitions inside of another container), and air goes into the bottom of the partitioning container.  The air will then be used to provide buoyant force upon the bottom partitions.

I will not go into the "newer" partitioning model just yet... let me get a new video up with the 2.0 idea which will not only decrease the distance which M(water) will travel, but will also decrease the amount of work which the mass(output unit) must contribute to lifting all of the partitions(because only 2-3 of the bottom partitions will require a buoyant force... all other partitions will have received buoyant force from a lower partition).

I am taking baby steps... it is the best way in my opinion.

@GavinPalmer1984
Try this little quiz I made.

Put the following in order of least to most energy required.

1) Lifting 1000kg - 1 meter high - in one second
2) Lifting 1000kg - 1 meter high - in three seconds
3) Lifting 1000kg - 1/2 meter high - in one second
4) Lifting 1000kg - 1/2 meter high - in one half second

Hope this will get the point across.

Energy required for the partition to get up to max velocity:
W1 = (m * d1 * g) + (1/2 * m * v^2)

Energy required for the partition to move with acceleration = 0, aka constant velocity:
W2 = (m * d2 * g)

Energy required for the partition to slow down to v = 0 and reach destination:
W3 = 0

So the overall work will be:
W = (m * d * g) + (1/2 * m * v^2)
d1 + d2 = d ~= 2 * H/n

So:
Work(1) = 1000kg * 1m * 10 + 1/2 * 1000kg * 1
Work(1) = 10000J + 500J = 10500J
Work(2) = 10000J + 56J   = 10056J
Work(3) = 5000J + 125J   =   5125J
Work(4) = 5000J + 500J   =   5500J

Realistic dimensions would mean that the height (y-dimension) of the output unit's holding container would be roughly equal to the depth(z-dimension).  The width(x-dimension) would be larger than the output unit's width.  The height and depth would be roughly twice the height of the output unit.  So each partition would have a width and depth which is equal to the output unit's container's width and depth.

Let's use these dimensions:
Output Unit:
(y-coordinate)height = 22 meters
(x-coordinate)width = 3.7 meters
(z-coordinate)depth = 3.7 meters

Let us assume that the mass of the output unit itself is zero and only the falling mass (piston) is accounted for.  We will also ignore resistive forces for now.  There are efficient geometric shapes which should be used... I may not be using the most efficient shapes.

mass of piston < Volume of output unit  * density of water
m(p) < V(ou) * p(h2o)
this is where you would account for a temperature window so that the output unit is always buoyant no matter the temperature... temperature affects density.
m(p) < 22 * 3.7^2 * 980kg/m^3
m(p) < 295,156.4 kg

The mass will be able to fall roughly 1/2 of the output unit's height.
Output Work = m * d * g
OW = 295,156.4 * 10 * 9.81
OW = 28954842.84 J - (f1 * 10)
f1 = friction on inner walls over 10 meters

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

mass of water exchanged = m(h2o)
remember, water exchanged must be greater than the height of the output unit.
m(h2o) = Volume of h2o * density of h2o
m(h2o) = 4 * 45 * 22 * 980
m(h2o) = 3,880,800

Partition:
width = 4 meters
depth = 45 meters

ignore inner-operations for now (there is potential energy which can be harvested from other sources within the process)
Input Work = work needed to exchange the m(h2o) = IW
IW = (m * d * g) + (1/2 * m * v^2) + (f2 * d)
d = 2 * 22/n
n = number of partitions
m = m(h2o) = 3,880,800kg
f2 = friction encountered while lifting

We need:
OW > IW
OW = 28,954,842.84 J - (f1 * 10)
IW = (m * d * g) + (1/2 * m * v^2) + (f2 * d)
28,954,842.84 - (f1 * 10) > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2) + (f2 * 2 * 22/n)

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

with friction:
28,954,842.84 - (f1 * 10) - staticfriction1 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2) + (f2 * 2 * 22/n) + staticfriction2
28,954,842.84 - (f1 * 10) - staticfriction1 > (1675108512 / n) + (1,940,400 * v^2) + (f2 * 44/n) + staticfriction2
n = 100;
v = 0.22 m/s;
height of a partition = 0.22 meters;

28,954,842.84 - (f1 * 10 meters) - staticfriction1 > (16751085.12) + (93,915.36) + (f2 * 0.44 meters) + staticfriction2
28,954,842.84 > 16,845,000.48 + (f2 * 0.44) + (f1 * 10) + staticfriction1 + staticfriction2

Relatively bad coefficients of kinetic and static friction are 1.0 and higher.
Force of friction = Normal force * coefficient of static friction + normal force * coefficient of kinetic friction.
Unrealistic worst-case friction:
normal force is related to the area of contact between materials and the pressure exerted.
width of seal = 0.05m
the piston is cube shaped = 4 sides that are 1m long
use 2 seals
normal pressure might be 75,000 N/m^2
for staticfriction = kinetic friction (I am using bad coefficients so this is safe)
f1 = Area * Pressure * coefficient of friction= (0.05m * 1m * 4 * 2) * (75,000) * 1.0 = 30,000 N = staticfriction1

28,954,842.84 > 16,845,000.48 + (f2 * 0.44) + (f1 * 10) + f1 + staticfriction2
28,954,842.84 > 16,845,000.48 + (30,000 * 10) + 30,000 + (f2 * 0.44) + staticfriction2
28,954,842.84 > 17,175,000.48 + (f2 * 0.44) + staticfriction2

28,954,842.84 - 17,175,000.48 > f2 * 0.44 + staticfriction2
11,779,842.36 > f2 * 0.44 + staticfriction2


if f2 + staticfriction2 / 0.44 < 26,772,369, then this device can definitely be perpetual.

check this out:
OW = m1 * d1 * g - f1 * d1
IW = m2 * d2 * g + f2 * d2

Doubling the dimension:
OW = 2*(m1 * d1 * g - f1 * d1)
IW = 2*(m2 * d2 * g + f2 * d2)

All of the masses double.
d1 will double.
d2 can stay the same... aka double the partitions.

If you ignore friction, doubling dimensions and partitions will quadruple OW but only double the IW.
The work done by friction for OW will quadruple because of the increased distance for f1 and the area of contact for f1.

I will not make the claim that the work done by friction for the OW will increase with doubling the dimensions... that just limits the innovation of the design.

Also, I know that there are better geometric shapes which should be used... I happened to use bad ones, and still brought forth great evidence that perpetual motion is possible.

I should add that any other losses within the system are a direct result of the design and implementation to achieve the desired (basic) operation.  For example, creating electricity from kinetic energy has inherent losses.  And storing / transferring that electricity has large inefficiencies.  The designers should conclude, that the use of electricity to maintain a desired state of the device should be avoided... Unless the electrical conversion/storage/transfer processes are increased to an efficiency which is at least equivalent to the inherent resistant forces (friction) within a system which avoids the use of electricity (compare direct mechanical work through pulleys and levers to that of an electric motor) for the same purpose.

Gavin,

I believe there is a problem with how you think the output device will work.  Once it has been lowered below the catches and the water level returned to full height, the output unit will not flip.  Even though it is buoyant and will want to rise, the majority of the weight will be at the bottom.  Much like the ballast in a boat, having the CG of the output unit at the bottom will keep the unit from being able to flip back to the surface.  It will want to rise, but only straight up, not flip.  You will not be able to reset the weight in the output unit using this system as shown so far.

Thanks,

M.

I think you are right.  The mass will need to be smaller and will be determined by the amount of displaced water which is lower than and/or level with the mass.

But the point still stands (although a hole has been shot in the master-plan):
output unit is constant.
reset work is variable.

I appreciate the help... maybe there is a better output unit design... the previous detailed example will not work because the mass in the output unit is too large.