One can always hope that people pointing out a fraud in action creates some resistance to that fraud's continuation.  Fraud is often profitable because it is seldom prosecuted.  As John Rohner has found out:  There are exceptions.  I think he has a court date next week for a hearing on charges of contempt in the civil case.  The corporate defense attorneys revealed in public filings that there is a criminal case under seal.  Some of these fraudsters are so full of themselves that they never see the authorities coming.

MarkE,

Thank you so much for providing all your time and hard work!  Your Analysis is a masterpiece and I admit that I came nowhere close to your level of work and presentation.  I especially appreciate where you found the error in my belief that State 1 would have a net buoyant Force of zero.  You are correct that the system would have a net positive buoyancy due to the water being displaced by the riser walls that were assumed to have an SG=0.  So State 1x is correct.  Nice catch.

How much Energy needs to be supplied to the system to go from State 3 back to State 1x?  There is no reason to go from State 3 back to State 2 and then back to State 1x, AFAICS.  If we "pull the plug" from the bottom of the pod chamber I think the system should transition from State 3 back to State 1x.  Do you agree?  Would we need to supply Energy to do that?

M.

Quote from: mondrasek on March 15, 2014, 09:03:33 AM
MarkE,

Thank you so much for providing all your time and hard work!  Your Analysis is a masterpiece and I admit that I came nowhere close to your level of work and presentation.  I especially appreciate where you found the error in my belief that State 1 would have a net buoyant Force of zero.  You are correct that the system would have a net positive buoyancy due to the water being displaced by the riser walls that were assumed to have an SG=0.  So State 1x is correct.  Nice catch.

How much Energy needs to be supplied to the system to go from State 3 back to State 1x?  There is no reason to go from State 3 back to State 2 and then back to State 1x, AFAICS.  If we "pull the plug" from the bottom of the pod chamber I think the system should transition from State 3 back to State 1x.  Do you agree?  Would we need to supply Energy to do that?

M.

Thanks.  Physics always prevails.  The soda bottle / water bottle experiment was proof of the State 1 fallacy.

To go from State 3 to State 1X, let the water out of the pod chamber.  The internal energy decreases from 3.6054mJ of State 3 to 3.4029mJ of State 1X. losing 0.202454mJ. 

Quote from: MarkE on March 15, 2014, 09:59:04 AM
To go from State 3 to State 1X, let the water out of the pod chamber.  The internal energy decreases from 3.6054mJ of State 3 to 3.4029mJ of State 1X. losing 0.202454mJ.

And from what you have taught us all, theoretically, exactly half of that Energy lost (0.5*0.202454mJ) could be transferred to a second Ideal ZED that starts at State 1 (while the former starts at State 3) if, instead of simply letting the water run out of the pod chamber to nowhere, it was coupled to the pod chamber of that second Ideal ZED?  In a similar fashion as how two ZEDs are coupled in LarryC's diagrams in his spreadsheets?

Quote from: mondrasek on March 15, 2014, 10:14:29 AM
And from what you have taught us all, theoretically, exactly half of that Energy lost (0.5*.202454mJ) could be transferred to a second Ideal ZED that starts at State 1 (while the former starts at State 3) if, instead of simply letting the water run out of the pod chamber to nowhere, it was coupled to the pod chamber of that second Ideal ZED?  In a similar fashion as how two ZEDs are coupled in LarryC's diagrams in his spreadsheets?

Why would you want to do that?  Adding more complications or instances of under unity devices just increases percentage loss. Because the risers are unrestrained, going from ST1X to ST3 or back is effectively just the same as pumping water into a column of some dimension or draining it from same, respectively.  The pair reduces to a glorified "U" tube with some water in it.  The system will naturally seek an equilibrium state between ST1X and ST3 on both sides that in total holds less internal energy than when one is at ST3 and the other is at ST1X.