@Larry: along with what MH said.... Power is NOT a conserved quantity, like work and energy.

(Too bad too..... or my TinselKoils would be so massively overunity that we could all go home and relax. I have input power of less than a kilowatt, and I have peak output currents of 8 amps... at well over 30 kV. Go ahead and calculate the COP of that one... using Power alone.)

Average power is better, as MH said, but "many is the slip between the cup and the lip" even using average power.

Work is the conserved quantity. (Same thing as Energy: force x distance, even the same units: Joules)

Was there something objectionable or wrong about my question set above? Did I misunderstand the nature of a "cycle"?

We start with a motionless precharged Zed, not counting the work necessary to set this up. We apply a known quantity of WORK, force times distance, both easy to measure, by depressing the input piston (or cylinder, same difference I think). We get out a known quantity of WORK at the peak of the cycle... again, force times distance, easy to measure. And we let the system relax back to the beginning of the cycle which is identical to the start of the cycle, all pressures and displacements the same.... otherwise it's not a cycle, right? Maybe we have to put in more WORK to get it back to the start: this counts as INPUT.
So.... what is this ratio in your spreadsheet simulation? Output WORK/Input WORK. That is the question (whether I suffer the slings and arrows of outrageous fortune....)

Just a little side observation...

The skeptics like me and MH and Seamus10n are asking simple questions that require simple answers, from simple measurements made in simple ways. If I had a Zed or a set-up spreadsheet sim... I really think I could answer the questions, simply and clearly.

But we are getting complicated answers that are nearly uninterpretable, answers that don't include simple measurements, with units attached, combined in simple ways.

Maybe this is caused by the difference in training, education and experience of the people in the conversation, and is just a matter of "talking past" each other and we will eventually settle on a language and a methodology that we can all understand.

Or maybe it's not.

Quote from: seamus102 on August 12, 2012, 10:19:23 PM
We can discount any such analysis based on the knowledge of currently accepted physics if it shows that the system would break the laws of conservation of mass, energy  or momentum.

These laws are mathematically provable as being correct and have  been experimentally validated in every single case. (until now if this claim is to be believed).

If you are getting an overunity result by using any mathematical model that violates these laws then one thing is certain, your analysis is WRONG.

You'd better be prepared to bring to the table a theory that demonstates how and why current physics knowledge
is incorrect. It would necessarily show how gravity (perhaps just in Oklahoma) is not a conservative field.

As such, this claim can only be verified experimentally.  There is absolutely no valid mathematical model that can be constructed using known physics that would support it.

Sir, It is quite poor form on your part to cast dispersions on my work with a preemptive strike before you have even seen it.  How do you know what I have simulated?  Please hold your comments until there is something to comment on.  Then, if you are capable of analyzing my work properly according to the laws of physics, I will welcome your considered analysis.

LarryC:

More for you about your start and end pressure calculations.  It looks to me like you are factoring in the water head pressure but I am not sure if you have the back pressure from the weight factored in properly, like I said before.

Let's split the pressure into two parts.

There is the back pressure associated with the weight itself.  We can treat that as a constant.  I am going to assume that you can calculate the back pressure associated with the Archimedes system.  And of course the back pressure associated with the 4-Riser system is four times higher.

So you know the water volumes in cubic inches, and you know the PSIs for the two cases.   We can do something to make life really easy for you.  We can use energy units that are "CI-PSI."   Cubic inches times pounds per square inch.  Those are the units you are working with if I am correct.  So if you multiply that out you get (Cubic inches x pounds / square inch) = inch-pounds.   You notice that is a unit of energy, it's just a variation on foot-pounds.

Let's say for the sake of argument you use inch-pounds for all of your calculations.  It's perfectly valid.

If you are following me, you should be able to calculate the energy required to lift the weight in both cases, Archimedes and the 4-Riser, and they should be the approximately the same.  You just have to work out the back pressure associated with the weight in each case.

Let's call that your "Lift_Energy" - the amount of energy it takes to lift the weight a certain height.

Now, the other amount of work that has to be done is to push the water into the two types of cylinder and overcome the head pressure.  Be very careful and carefully examine the head pressure in the Archimedes vs. 4-Riser case.  Does the "lever factor" come into play here?  I am sure that you can figure it out.

For the work required in inch-pounds to fill the cylinder with water, don't be intimidated by the fact that the head changes as you add water.  You know the head rises linearly as you add cubic inches of water.  You know the start pressure in PSI, and you know the end pressure in PSI.  You have all of the information that you need.  This is a very simple integration of a ramp function.

The energy that you have to add to overcome the head is the ((End_PSI-squared x Total_Cubic_Inches)/2) - ((Start_PSI-squared x Total_Cubic_Inches)/2)

If the Start_PSI is zero then it simplifies to ((End_PSI-squared x Total_Cubic_Inches)/2).

Lets call that your "Head_Energy" - the energy it takes to pump the water into the cylinder as the head rises.

So, the total energy it takes to lift the weight is simply = Lift_Energy + Head_Energy.

This will apply for both the Archimedes and the 4-Riser setups.  You just have to carefully work out the calculations and you are done.  I think you have enough information on your end to boot-strap yourself and solve for the energy in both cases.

MileHigh

OK, so if a single (or dual) Zed has no real "input" cylinder or piston, and instead must be supplied with an injection of liquid to initiate a cycle... that is even easier. You simply arrange a "syringe" to inject your fluid. You monitor fluid pressure in the syringe, and you know the surface area of the syringe's piston. The pressure in the fluid and the piston area give you the force, the piston travel is your distance. Work in is thus again easy to measure. Work out is the lifted weight times the lift distance, again easy.

Please correct me here if I'm wrong.

And once again... for there to be a "cycle" the end state must be the same as the start state... so syringe again full, with its piston out, and the lifted weight back down. If it takes more work to reestablish this initial state, that counts as input.

Will see3d condescend to put his results in this form so that they may be interpreted by those of us who are straitjacketed by our educations?