Hi Guys,

Been looking through my spreadsheet again with the single, no layers model, ZED that Fletcher and MT were looking at a bit back. I'll post the spreadsheet later again now I've added some checks as Fletcher suggested, but wanted to get some thoughts first.

With the setup the spreadsheet cam with last time, judging by the amount of volume available to pump the water in to in the pod containing tank, the starting PE of the water is 19.8 J (as the tank can be partially filled without the pod floating), and once filled, is PE 123.5 J. This gives us a minimum amount of work required to fill the pod tank as 103.7 J.

Once the pod is let go, the PE of the water drops to 52.3 J (difference of 71.2 J), whilst the pod only gains 56.9 J. As the force on the pod to go up is that of the water being pulled down (I think?), this means our total potential maximum work out could be 71.2 J. Under normal circumstances, I guess the extra would be accounted for and lost from the pod flying out of the water a little way before falling and bobbing about a bit till it found equilibrium.

Then draining the tank back to it's starting position, which cost us nothing as gravity chips in, we could get out the calculated drop in PE of 56.9 J, which we just gained from the pod rising, as the pod falls.

In PE differnce: 103.7 J
Out PE Difference: 71.2 J + 56.9 J = 128.1 J
Net: +24.4 J

Now building something with as few losses/extra inputs as possible to take advantage of that is a whole different kettle of fish, but does it look right?

Amo

Quote from: webby1 on September 27, 2012, 01:49:51 PM
..............................................................When the precharge has returned the percentage needed to drop the internal pressure by 2\3 the mass on top is no longer able to be held up against the force of gravity and starts to fall, leaving the 1\3 mass potential being applied to the rest of the fluid by using both the precharge residual pressure and gravity pulling the mass down, which ends up with the risers back at the start position with the rest pressure of the system intact.
Now as TK has pointed out it is the precharge that lifts the mass, the extra input then is the apple and since the conditions,, the constants needed for the precharge to exist have not changed, ideally, the full potential from the precharge can be recovered and an additional 1\3 from the mass via gravity. ..................................................................

Hi Webby1,
Reconsider your statements below carefully, from the way I understand your writing,

The statement below does not appear true, if needed go back to Wayne's early postings
        "When the precharge has returned the percentage needed to drop the internal pressure by 2\3 the mass on top is no longer able to be held up against the force of gravity and starts to fall,"

What exactly happens in this scenario what your are describing below on macro level ?
       "leaving the 1\3 mass potential being applied to the rest of the fluid by using both the precharge residual pressure and gravity pulling the mass down, which ends up with the risers back at the start position with the rest pressure of the system intact."

I can not conspire with TK' pointing, the statement is imprecise or I am understanding the statement wrong
         "Now as TK has pointed out it is the precharge that lifts the mass, the extra input then is the apple and since the conditions,, the constants needed for the precharge to exist have not changed, ideally, the full potential from the precharge can be recovered and an additional 1\3 from the mass via gravity."

From what I see in the above is that you are creating your own deviations,
Wayne laid out the path, the facts together with measurements in his posts.  In all discussions thus far, you appear keep your eyes on the payload, because you believe that is where the money is. It is not, the money is in the decoy, the 1/3 overhead (it is a troyan)  I can not stress enough to go back to Wayne's early postings. Look at every statement and declaration Wayne made and question it, ask questions like, how can that be ? How can he say that?

What was misleading at the time were the people who were bluntly stating that Wayne was lying, that he did not have, invent, or was misleading,  if he didn't posted and laid open the design, construction and operations manual with an "open source" clause. This is now referred to as vulture period and had a noticeable influence on today, making the release of information more guarded and restrictive. Sorry for this limitation.

I can assure you, Wayne's statements were made guarded to protect certain interests he has (with good reason), but whatever was written was made truthfully and sufficient to discover and understand the invention.

Michel

Quote from: mondrasek on September 27, 2012, 11:42:49 AM

And that does continue to go through my head:  The output weight is NOT needed to reset the system when the input is removed.  SO, if the input is equal to the output on the upstroke (if it were that case), why can I remove the output and the system still gives back that same input?

Wait a minute. That's not how you described your system working before. Are you telling me now that, at the top of the 4mm lift, you can lock the riser at that point, REMOVE the weight, unlock the riser (doesn't it pop up? ) and then recover your 74 ml water without pushing the risers back down to the start position, they just settle on their own, then you replace the weight, pour the water in and it lifts again?

Regardless of the answer to that:
Back to the issue of the bollard and what is lifting what.

Please, webby and mondrasek, set your riser and weight on the scale. Attach a string to the top of it. Lift upwards, and record the scale reading "just" as the riser starts to actually move and rise up. This is the difference between the actual weight and the weight you are lifting with the string. The scale reading should be.... zero. RIGHT? The thing doesn't move until you have taken up all its true weight on the string, so the scale will read zero just as the thing starts to move upwards.

Now do the same thing with the precharged system.  Set the whole thing on the scale. Record the weight. Now pull up on the string and record the weight "just" as the riser starts to move upwards. Subtract this reading from the first reading. This is the weight you are ACTUALLY lifting.

I am asking you for ten minutes work and a couple of simple weight measurements. My conjecture is that you  are NOT lifting the entire weight of the riser and the moving weight with whatever water you are introducing, because the precharge is already offsetting much of the weight of the moving parts. Please, if you think I am wrong, demonstrate it. Prove me wrong. Then, if I am wrong, we can indeed use the change in GPE of the lifted weight to compute our "output" work. But if I am right...... well, it gets a bit more complicated.

Hoping to do some testing on 3U this weekend and have a question on POD weight.
-- Yes, 3U is using a fully separate POD
As pictured this POD weighs 11oz, is 4.5 inches diameter and 10.5 inches tall / long.
It displaces about an inch free floating.
POD retainer wall is 4.75 ID
Seems there have been a few mentions of POD weight / system setup but I haven't been able to find them in my notes or saved posts.

So 2 related questions for anyone that may have those references:
1. During normal cycles should the POD actually sink or would the setup values preclude that condition?
2. What percentage of the POD should be submerged in a free floating state?

Thanks
Dale

Quote from: seamus103 on September 28, 2012, 07:04:24 AM
163 pages and still not one joule of overunity energy... All somewhat tedious.
Just a few thoughts on the latest discussions.
1. Precharge...  It is present at the start and end of the cycle so it can contribute nothing to the net output. It is not relevant or necessary to include it in the analysis.
2. Take long hard look at the graph of displacement vs input head as shown in #2417. A purely linear result. Somewhat different to the many claims of a non- linear process occurring as expoused by mr wayne some time ago.
This machine doesn't work.. get over it and move on.
Should Mr Travis continue to claim this invention does produces usable energy he should be investigated for fraud. He will by now have realised it does not work but does not admit it.

Hello Seamus,

I love your posts because I appreciate an adversary who can give a good counter argument.

Why are you thinking that the graph in post #2417 should show a non-linear trend ? 
And how would that graph fit into the OU theory espoused by Wayne ?
Please let us know how that impacts a Zed process adversely? and the reason why ?

The reason for you believe that Wayne should know what you already know and he doesn't know yet. 

Legally Mr Travis can claim whatever he likes, since you or nobody else came up with a technical counter argument that disproved his claim conclusively.  When it comes to a public test system, he can take whatever time he wants and he stated already what his plans are in that respect that are in progress.

So the ball is still in your court and has never left your court yet,  so get to it...

Please clarify, because I am lost.

Regards, Michel