Youtube video of gravity device principle.. ?

[TinselKoala]

Gavin, you are still missing the point that your partitioning scheme CANNOT reduce the work needed to reset the unit, past the amount of work gotten out of the output unit. It must do this, in order to repeat cycle or produce energy. But it cannot. Your partitioning scheme reduces the work from mg2h (which is unneccessarily high anyway) to mgh. But it cannot reduce it further.
It just occurred to me: Have you studied calculus, particularly differential calculus? Because you are thinking that an infinite series converges on Zero, when in fact your series converges on One.
For your device to function as intended, you must be able to reduce the work needed to reset the system to a value less than the work extracted. What you have described does not fulfill that criterion, because you are wrong about your partitioning system. Look at the top of the water in the receiving container. The height which this level rises, is the height that goes into the work equation mgh, no matter the number of partitions or how thin they are.

[spinner]

Gavin, you are still missing the point that your partitioning scheme CANNOT reduce the work needed to reset the unit, past the amount of work gotten out of the output unit. It must do this, in order to repeat cycle or produce energy. But it cannot. Your partitioning scheme reduces the work from mg2h (which is unneccessarily high anyway) to mgh. But it cannot reduce it further.
It just occurred to me: Have you studied calculus, particularly differential calculus? Because you are thinking that an infinite series converges on Zero, when in fact your series converges on One.
For your device to function as intended, you must be able to reduce the work needed to reset the system to a value less than the work extracted. What you have described does not fulfill that criterion, because you are wrong about your partitioning system. Look at the top of the water in the receiving container. The height which this level rises, is the height that goes into the work equation mgh, no matter the number of partitions or how thin they are.

Hi, TinselKoala!
Of course!
Gavin's problem is that he sees only the benefits of his partitioning principle....
Energy transactions should always be considered in it's "complete cycle", or in a closed loop, if we want to understand what is really happening....

Cheers!

[GavinPalmer1984]

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.

[mondrasek]

All,

I'm with Gavin on this one.  His partition idea appears to allow the water in the container to move sideways, and not just down.  While the water in the main container does become only half as high, it does not do this by dropping a distance h.  It drops only one partition, or h/n.  It shifts *sideways* out of the main container, into a secondary partitioned container(s).  So to get the water back into the main container the secondary partitioned container must be raise only 2 * h/n.

The main container would have to have as many outlets as the number of partitions in the secondary container.  When lowering the main container, all of these outlets could be opened simultaneously.  In order to refill the main container (after raising the secondary partitioned container 2 *h/n) the partitions must be opened only one at a time, starting from the bottom.

Gavin, I also have realized that the partitions only need to exist in the secondary container and therefore would not be in the main container where I thought they would interfere with the output device falling and flipping.  All in all, I can't see why the whole idea is not possible in theory so far.  Very intriguing.

Thanks,

M.

[GavinPalmer1984]

The main container would have to have as many outlets as the number of partitions in the secondary container.  When lowering the main container, all of these outlets could be opened simultaneously.  In order to refill the main container (after raising the secondary partitioned container 2 *h/n) the partitions must be opened only one at a time, starting from the bottom.

Gavin, I also have realized that the partitions only need to exist in the secondary container and therefore would not be in the main container where I thought they would interfere with the output device falling and flipping.  All in all, I can't see why the whole idea is not possible in theory so far.  Very intriguing.

Thanks,

M.

The main container would have as many outlets as there are partitions, and inlets as their are partitions. (unless an outlet for one stage could be an inlet for another stage or vise versa).  And just to clarify:

The top outlet from the main container opens first.  Once the top partition fills, The next lowest outlet from the main container opens.  This repeats downward until all of the partitions are full.  The partitions are lifted 2 * h/n and the bottom-most inlet on the main container allows the bottom partition to empty back into the main.  Once the bottom partition is empty, the next highest inlet is opened which will allow the next highest partition to enter the main.  Repeat upward.  Note that Opening and Closing of inlet/outlet should be autonomous and harness the Potential within the process itself (ie. not from output unit).

I am glad you are with me M.  Go ahead and ponder this one (I will do another youtube for it soon):

The outer partitions can be surrounded by a fluid so that each partition has enough mass to negate the air within the partition.  The air which exists in the partitions can be directed so that the output unit will only need to provide a small amount of air to lift 2 to 3 partitions only.  The air from the lower partitions will provide a buoyant force for an upper partition.  This concept is relatively new for me.  I will show a diagram ASAP.

Reset work = m * d * g
d = 2 * h/n
m = 2 * M/n  (this is with the added system which I will address soon)
M = mass of liquid exchanged in and out of main container
g = gravitational constant

Remember that this device will only serve to prove that perpetual motion is possible.  There will be many more-efficient devices to be discovered because of our work here.  We will lay the grounds for a new study of meta-physics (or interconnections of sub-systems).

After all, I am of the belief that the world is cyclical and on-going.  So that entropy oscillates within a window (whose min and max may oscillate as well?).  My theory and belief is that the world has existed for all time and will exist for all time.  And this device is a demonstration of my understanding of meta-physics ( a.k.a. nature ).  But let's stay focused on the task at hand -`