Mathematical Analysis of an Ideal ZED

[Magluvin]

Well then perform all ten steps of the experiment and see if you can defend your belief.  Here is a hint:  No matter what resistance value you use between the two capacitors, you always lose the same amount of energy.  Only the power changes, making the equalization happen faster or slower.This is a natural and common situation.  If one wants to minimize its effects one has some design choices.  One of those choices is to still use equalization, but not to restrict the potential variation between the stores.  Such methods have been used in charge pump power converters since the late 1970s.  Another choice is to convert the energy into another form and then back.  That is the method that pendulums have used for centuries.That's right:  The resistor value affects the power not the energy.That almost works.  What happens is that the inductor current build-up slows down as the second capacitor begins to charge.  This will prevent the first capacitor from fully discharging.  But you are on the right track.  The inductor serves the same purpose as kinetic energy in a pendulum.  In the case of the quasi-resonant converter that you just described the potential energy in the electric field converts into "kinetic" energy in the magnetic field of the inductor. 

If you want high energy transfer efficiency, you need to connect the first capacitor to only the inductor until the capacitor voltage reaches zero, at which point all of the energy is in the inductor's magnetic field, and then connect the inductor to the second capacitor.  Then when the inductor current reaches zero, you need to disconnect the inductor from the second capacitor.  This sort of arrangement can reach very high efficiencies.Actually you suffer a lot of loss due to two factors:  1) The turbine power efficiency will be in the 30% range, and 2) Just as with the inductor, if you do not convert all of the energy out of potential form before transferring it to the second potential store, then you still get burned by the N*(X/N)² problem.That is the tyranny of the N*(X/N)² problem.You have the basic idea right.  Now, all you need is to actually solve the problem for Tom's two cylinder lifter, and then jump in a time machine so that you can provide that answer to Tom so he can apply back when he has been saying he had a solution.

Well, no losses if we are talking 'ideal' 'here' 

Now, if in the electronic example, we cut the switch when the input cap was at near 70v, then we use a free wheel diode to let the inductor finish discharging into the destination cap, the second cap can be charged up to around 70v also. Now we have 2 caps, total energy nearly what was in the originally full cap.  Well, ideal would be 100% of the original energy still available., but in a larger reservoir.

So how do we do that with the air tanks?  I dont think it would work well.  Maybe some alt mech could be made to get it done.  Source tank 100psi, fly wheel accel, second tank filling, cut of the source tank at 70psi.  Flywheel still going, we would have to pull the additional air from the outside with the flywheel/air motor to continue pressurizing the second tank.  Big problem getting to 70 psi though. We just switched our source of 70psi to a 0psi source.  No way to get to 70psi. 

That tank example is just a simple way to get Webby to understand the electronic example in comparison.  If we really want to do it right,  each cap would be compared to 2 tanks(plates). 

More complicated, but problem solved.     But Im sure he can understand the comparison better now.



Mags

[Magluvin]

Well, no losses if we are talking 'ideal' 'here' 

Now, if in the electronic example, we cut the switch when the input cap was at near 70v, then we use a free wheel diode to let the inductor finish discharging into the destination cap, the second cap can be charged up to around 70v also. Now we have 2 caps, total energy nearly what was in the originally full cap.  Well, ideal would be 100% of the original energy still available., but in a larger reservoir.

So how do we do that with the air tanks?  I dont think it would work well.  Maybe some alt mech could be made to get it done.  Source tank 100psi, fly wheel accel, second tank filling, cut of the source tank at 70psi.  Flywheel still going, we would have to pull the additional air from the outside with the flywheel/air motor to continue pressurizing the second tank.  Big problem getting to 70 psi though. We just switched our source of 70psi to a 0psi source.  No way to get to 70psi. 

That tank example is just a simple way to get Webby to understand the electronic example in comparison.  If we really want to do it right,  each cap would be compared to 2 tanks(plates). 

More complicated, but problem solved.     But Im sure he can understand the comparison better now.



Mags

Lol, nearly 2 am and I may need to make corrections on this post tomorrow.  I shouldnt post when sleepy.

Mags

[MarkE]

Well, no losses if we are talking 'ideal' 'here' 

Now, if in the electronic example, we cut the switch when the input cap was at near 70v, then we use a free wheel diode to let the inductor finish discharging into the destination cap, the second cap can be charged up to around 70v also. Now we have 2 caps, total energy nearly what was in the originally full cap.  Well, ideal would be 100% of the original energy still available., but in a larger reservoir.

Well you won't succeed in equalizing at 70.7V that way, but your intuition is on the right track.  If you want to approach 100% efficient energy transfer, you need to transfer only from the first capacitor to the coil, and then from the coil to the second capacitor.  As long as the second capacitor and the first capacitor are connected together whether or not there is a coil in between you suffer from the N*(X/N)² problem.  The coil reduces the effective value of N.

So how do we do that with the air tanks?  I dont think it would work well.  Maybe some alt mech could be made to get it done.  Source tank 100psi, fly wheel accel, second tank filling, cut of the source tank at 70psi.  Flywheel still going, we would have to pull the additional air from the outside with the flywheel/air motor to continue pressurizing the second tank.  Big problem getting to 70 psi though. We just switched our source of 70psi to a 0psi source.  No way to get to 70psi. 

It is tricky business, because you need a highly efficient means to both convert the potential energy to kinetic, and then switch to perform the reverse conversion on the second column.  None of Tom's proposals have included the necessary mechanics.

That tank example is just a simple way to get Webby to understand the electronic example in comparison.  If we really want to do it right,  each cap would be compared to 2 tanks(plates). 

More complicated, but problem solved.     But Im sure he can understand the comparison better now.



Mags

If by problem you mean getting Tom to understand what he is up against, then yes.  If you mean a devised solution that will do substantially better than 50% loss, then it is still at the rough concept phase.  Tom's problem is that he has been claiming to have solved the problem almost two years ago, when it seems that he has yet to understand it.

[powercat]

Well,, why don't you show that you are no different.
Posting the same argument over and over, is indeed repetitive.
I agreed to disagree many times and was fine with the impasse.
But the "tyranny of it all" seems to be ever present even tho that is not the only way nature works.

I'm not denying I'm being repetitive, Wayne Travesty is a fraudulent liar, and needs to be challenged and so does anyone supporting his fraud, if you want things to change I suggest you come up with some proper evidence, like a continuous self running device, but you just continue posting day after day month after month supporting Wayne, how many lies do you want to be associated with ?

I could post that list again of Wayne's promises and claims from two years ago, you know the one, where he says he has 600% overunity and simple physics can show how it works, so you should have no problem building a continuous working device, so are you really surprised that people think you are supporting a conman ?

[minnie]

   Webby,
              what on earth are you after? Nearly everyone is spelling out the facts for you
    and you don't seem to take any notice.
         Could you please take time and describe exactly what you're trying to achieve.
     Is your plan to raise a weight, take it off the apparatus you lifted it with and then use
      the fluid you lifted it with to raise another weight?
                           John.