Youtube video of gravity device principle.. ?

[GavinPalmer1984]

@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.

[wattsup]

@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.

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.

[GavinPalmer1984]

@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.

[wattsup]

"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.

But the drawback in using so many containers is the time require to empty the large container will increase since each consecutive level will have the minimal head pressure to work with and that same will apply to the many containers emptying into the big container. More time.

[GavinPalmer1984]

"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.