Youtube video of gravity device principle.. ?

[spinner]

Spinner and Tinsel,

  The top h * 1/n of the receiving container will fall a distance of h * 1/n into the first partition.  And the next h * 1/n will fall a distance of h * 1/n into the second partition.  Once all of the partitions are full, THE TOP OF THE TOP PARTITION IS ONLY h * 1/n BELOW THE RECEIVING CONTAINER'S TOP.  The partitions only need to be lifted 2 * h * 1/n.  Please look at the partitioning video and ask questions.

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

I am very good at math.

Understand the partitioning scheme before you try to make a claim to the minimum work convergence.  You assume min. distance is h, when in fact it is 2 * h * 1/n.

Gavin, your "partitioning principle" or the accompanying equations were never a real issue.... Your YT video explains it well, and most people actually understands it...

How F....d up you think we really are? 

But you're obviously missing the point....

"We" never claimed that a min. distance is "h"....

And "we" agreed that a minimal distance is "2*h*1/n" with your part. principle.....

Now, try to evaluate the energy exchanges of your "partitioning scheme" in practice....
Do it with a real values, e.g. m(water)=1000kg, h=1m, g=10, n=10... (easy calcs)....
You can neglect (dismiss) all the possible losses (focusing for an ideal mechanical conversion)

So, a complete cycle energy exchange should be ZERO(0). No doubt about that. A conservation of energy never failed when observing the mass transfers in a gravity field (so far...)

Btw, the only equation you need (evaluating the mass transfers in a gravity field, or an ideal mechanical energy transactions) is a "m*g*h".... A gravitational potential energy says that you can get out of this (closed loop) cycle only what your input was..... Easy. Primary school stuff. Which never failed (to date...)

The complete (idealized) energy eq. for a partitioning scheme principle is: "Ep = n((m/n)*(h/n)*g)" = (mgh)/n
which equals to a known formula: "Ep = mgh" (without partitions ( n=1))....

Now, Wp is the maximum (theoretical) limit of any possible mechanical energy transaction (neglecting any and all the "real life" losses (we're - of course - not considering stuff like the atomic/molecular, zpe, "magic", etc... potentials )).

I do hope you'll see the flaw in your reasoning ("h" wrt. "h/n").
Or (much better), you'll show us something real which will shows us a mistake in "OUR" understanding....

Anyway, good luck!

[wattsup]

Each container has sides and a base. If you split the water volume in 8 containers, you are lifting 8 times more container bases sections then with one. That is additional weight to factor into your idea.  You will also increase the transfer times since each of the 8 containers will have less starting head pressure. You will be transferring at the minimal speed 8 times. Taking the added bases plus the time factor all calculated as an energy expense, this idea is "dead in the water".

[mondrasek]

@ wattsup,

The weight of secondary partitioned containers could be balanced with a counterweight, just like the weight of an elevator is.  The added counterweight would add inertia to the whole system and therefore slow the acceleration of the raising water/container when whatever force is applied to do so.  But it would not raise the amount of force or energy required (minus additional friction... blah, blah, blah).

I don't understand your issue with how much time it takes to transfer the water.  Sure that fact would mean less "output energy' in a given amount of time (longer cycle time), but it would not decrease the "output energy" expected from the "output unit" in each cycle, right?

M.

[GavinPalmer1984]

And "we" agreed that a minimal distance is "2*h*1/n" with your part. principle.....

Now, try to evaluate the energy exchanges of your "partitioning scheme" in practice....
Do it with a real values, e.g. m(water)=1000kg, h=1m, g=10, n=10... (easy calcs)....
You can neglect (dismiss) all the possible losses (focusing for an ideal mechanical conversion)

if the volume of the container is 1 m^3 then,
mass(water) = 1000 kg
work = W = m * d * g
d = 2 * h * 1/n = 2 * 1 * 1/10 = 0.2m
W = 1000 * 0.2 * 10 = 2000 J

If you have 100 partitions
W = 200 J

if you have 1000 partitions
W = 20 J

if you have 10000 partitions
W = 2 J

note that the size of the partitions are unrealistic with h = 1m.  But you should get the idea that the work needed to empty and refill a container is reduced by adding partitions.

Btw, the only equation you need (evaluating the mass transfers in a gravity field, or an ideal mechanical energy transactions) is a "m*g*h".... A gravitational potential energy says that you can get out of this (closed loop) cycle only what your input was..... Easy. Primary school stuff. Which never failed (to date...)

Yes for the sub-systems.
The output unit outputs work = m1 * d1 * g.  And if you lift m1 up, and there are no losses, you can never get more output than input.
The partitioning work = m(water) * d * g.  And if you harness the falling of that water from a higher position to a lower position, you are once again correct.

BUT for the cooperation of the two sub-systems, our current understanding is WRONG.
The partitioning scheme resets the output unit by doing the work = m(water) * d * g via emptying and refilling a container;  creating potential, m1 * d1 * g in the output unit.  I have NOT even addressed using the falling m(water) as a source of energy.

The complete (idealized) energy eq. for a partitioning scheme principle is: "Ep = n((m/n)*(h/n)*g)" = (mgh)/n
which equals to a known formula: "Ep = mgh" (without partitions ( n=1))....

Now, Wp is the maximum (theoretical) limit of any possible mechanical energy transaction (neglecting any and all the "real life" losses (we're - of course - not considering stuff like the atomic/molecular, zpe, "magic", etc... potentials )).

I do hope you'll see the flaw in your reasoning ("h" wrt. "h/n").
Or (much better), you'll show us something real which will shows us a mistake in "OUR" understanding....

Anyway, good luck!

Now listen up...  I will repeat:
Your analysis is right if you look at one of the trees (sub-system = partitioning scheme = output unit).  But if you look at the forest (partitioning scheme AND output unit), you can see what I have been stating all along... work needed to reset the output unit is dependent upon the number of partitions (aka variable) and the work produced by the output unit is constant.

Output unit is reset by emptying and refilling the reservoir.

So... you (and I hope I have changed your mind) and the modern ignorant physicists claim that the maximum amount of work that I can get out of this system (the forest = partitions and output unit) is equal to:
mgh/n  (idealized)

BECAUSE you think that I am going to harness the falling m(water)...  but NO!!  I am going to use the water to create a buoyant force on the output unit!

With hope that I have brought about an understanding, Gavin.

[wattsup]

@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.

@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.