Quote from: fletcher on October 05, 2012, 12:56:58 AM
Does or Is the Work Done capability [Output] of the Pod acting as a piston [rather than slowly floating up & down via water transfer in this simplified approach] greater in one scenario over the other ?

If I understand what you are asking correctly, then I believe this is the case. This also is something that puzzled me while developing the spreadsheet with the PE difference checks you suggested. I've reasoned it in the following way:

If I place a marble on top of the pod then fill the pod tank with the pod unlocked, once the pod hits the top stop, the marble will only have been pushed as high as the pod floats.

However, if I lock the pod down, then I release it once the pod tank is full, the pod has a much grater force behind it to start with, and once it hits the stop, the marble would be thrown further into the air. How far I guess depends on where the stop actually is, too soon and too late will diminish the hight to which the marble would reach. Also I would think that a lighter pod would generate a higher "throw".

I think this can be seen from the fact that the reduced PE difference in the water level is greater than the gained PE difference in the risen pod. In the spreadsheet, the water drop is 71J, while the pod rise is 57J.

The trouble, I think, as mandaresk pointed out, is that when the work gained using this greater value is added to the work value in draining the tank you find it equals that which was required to fill it in the first place.

If I have understood you correctly, this would probably mean that your second option about the layers, or maybe even some garbled combination of the two, is the supposed secret to the whole thing.

Amo

Amolago ..

A simple thought experiment for you to consider.

Take a deep tank with a known water volume in it - have a Pod [with a marble on top] suspended by a TK string so that its bottom surface just touches the water line in the tank - cut the string & let the Pod & marble fall - it will sink into the water & descend past its floatation interface [lets say half the height of the Pod] due to inertia & momentum - at the same time water will rise up the sides of the tank & Pod as the Pod displaces water - the direction of the Pod will reverse & it will then accelerate upwards due to upthrust force & have momentum - viscosity drag force will impede its descent & rise velocity's somewhat.

1. Will the Pod rise all the way out of the water [start conditions re-established] which recedes down the tank sides as the Pod rises ? N.B. equilibrium of forces & water height will be established when the Pod with marble floats.

Clearly the system will show decayed oscillation until equilibrium is achieved where the system CoM is at its lowest height & system PE is least.

2. IF, on the descent phase, the Pod pushes by a one-way pawl [like a one-way clutch or bearing system] & on the way upwards is stopped by the pawl would you expect the marble [not constrained by the pawl] to rise above its starting height ?

Would you expect, at any time or in any configuration you choose, the system CoM to be higher than the starting conditions CoM height & PE ?

N.B. the system starting conditions CoM height & PE is gained by the Work you physically put into the system to run the thought experiment.

Now, view the same thought experiment from the perspective of locking down a Pod & marble & lifting water to the tank & filling the sides with water etc, then releasing the Pod with marble - do you expect a different result in this mode as opposed to the earlier where you still have to do Work to take the system from its preferred gravity induced equilibrium state of lowest PE ?

Quote from: fletcher on October 05, 2012, 05:37:06 AM
Amolago ..
A simple thought for you to consider.

Gents,

With layers or without layers, with air or without air. In the end it all doesn’t matter, if the device remains symmetrical it will also remain symmetrical in its energy process,  in = out. That is what Seamus has been saying all along and he is absolutely correct. That is why we have a law.  Now I am getting into a repeat of a earlier posting which I do not want to do.

AmoLago, I am sure you treat your one layer float as symmetrical, if you think it is, then the “out” will equal the “in”.  The normal standard physics processes are symmetrical ?
Can a multi layer float be symmetrical?, absolutely yes,  although it offers many more possibilities to tamper and/or interfere with the buoyancy processes to change the symmetry and then it would depend on how you use it. 

For example, to go back to my statement in my previous post,
“The reason for short stroke is a requirement due to the physical properties of the multilayer buoyancy device.  This is the only way to minimize water input cost.  The more layers, the more this short stroke becomes an advantage. This is one of the main Travis effect pillars towards OU.” The reason for this is simple and obvious, enough “the only way to minimize water input cost” 

But Wayne didn’t stop there, knowing he had a fantastic lift device, he explored how can other internal ratio’s be promoted and exploited further, by their own natural properties or as needed, were artificially modified to create a non-symmetrical device.

Symmetrical: Exhibiting equivalence or correspondence among constituents of an entity or between different entities
Non-Symmetrical: Not exhibiting equivalence or correspondence among constituents of an entity or between different entities

What can be understood with term non-symmetrical in our context. The non-linear effect immediately indicates a disparity between in- an output.  Allowing the device to generate its capable output for a lesser input cost. Effectively internally the full bill is paid in full but not by the external input

Wayne described before the non-linear effect when relating to getting more lift per PSI in the higher than in the lower pressure ranges. (example: a 1.6x increase in pressure gives a 3x force advantage,  5 to 8psi increases lift from 2500 to 7500 pounds),
This characteristic is best expressed with the “pounds/psi” value.  Before you throw up your hands and say this is technically not possible because lift is “area x psi”.  Since area is fixed, the force should follow the psi,  how can the lb/psi be non-linear ? 
Do remember that the psi quoted here is the psi at the bottom of the water column because that relates to our input cost. The pressures on each lift area are determined by each layer’s water head. This for sure can change the lift ratio’s quite substantially.
This means that the relationship from input to output can vary,  one input value can have different output values. In the demo Travis device this change can amount to a 33% variation (based on the example figures above)

This advantage conversion factor for the Zed is available in the upstroke but also in the down stroke, a double advantage.
Lets go back to the example : a 1.6x increase in pressure gives a 3x force advantage,  5 to 8psi increases lift from 2500 to 7500 pounds

To better understand the process, you need to understand the relationships
Keep the following type equivalents in mind : Pressure =distance,  Volume=Weight
1. -  During the upstroke, the input is   â€œvolume x pressure” and the output is “weight x distance”
2. -  During the down-stroke, the input is “weight x distance”  and the output is “volume x pressure”.

If you look carefully, you will see that the upper and down value type relationships are crossed. 

1..  Top “weight” relates to  bottom “pressure ”
and
2..  Top “distance” relates to Bottom “volume”

With the result,
If the non-linear ratio in the upstroke gives us “ 1.6x increase in pressure gives a 3x force (weight) advantage”
That means that the down-stroke will give , a 1x force (weight) will give 60% (5/8) of pressure advantage instead of 1/3 ( this effectively nearly doubles the energy return during the down stroke).

AmoLargo,  Any calculation that is based on symmetry will always give a balance in-output. The laws are not wrong.
So the first to do is,  to look for non-symmetry. The solution to the complex is in the simple.

Regards, Michel

Quote from: fletcher on October 05, 2012, 05:37:06 AM
Now, view the same thought experiment from the perspective of locking down a Pod & marble & lifting water to the tank & filling the sides with water etc, then releasing the Pod with marble - do you expect a different result in this mode as opposed to the earlier where you still have to do Work to take the system from its preferred gravity induced equilibrium state of lowest PE ?

Hi Fletcher,

As I've stated before, I really am a newbie at this, so I'm grateful for the patience shown. Yes, I did have to Google "one-way pawl", just so you know what your up against!

I've read, re-read, and read again, tried a few things in the kitchen, and read once more and I think I've finally understood what you're getting at. That said, I'm not sure of what to post in response to this.

I can see your point and fully agree that if we determine a maximum height a pod might be able to float or sink to, that once it is let go in attempting to reach equilibrium it will never hit those points again.

In answer to the actual question posed though... no, I would not expect any difference in outcome from the two methods.

Ah, as I write that sentence, I think I now get it and subsequently deleted most of my post!

So we're going for:

Quote from: fletcher
IF there is NO advantage to be seen here then the Travis Effect is related to number of risers & pressure differential & is not a function of a Pod acting as a piston rather than floating up as water is transferred back & forth etc.

Amo

@Red_Sunset
GREAT! writeup - well done.

Thanks
Dale