Mathematical Analysis of an Ideal ZED

[MarkE]

Are you then stating that an equal and opposite force can not be applied to the transfer of potentials that converts the remaining potential during the transfer?

Is this conversion, in your view, into either heat or a change in KE and nothing else?

I am stating exactly as I have that equalizing potential between two or more potential energy stores results in a loss of energy.  That loss of energy can be quantified as the difference between:  N*(X/N)² and X² for any N > 1.  As it applies to your cylinders, in the Starting / Ending state one and only one cylinder is full charged to a given force potential and as such stores a given amount of energy.  When you attempt to transfer the energy from one cylinder to the other using the stored potential energy which is certainly the case when you speak of manipulating forces, then you will inevitably equalize along the way similarly to the two column water transfer problem shown in the attached graphic.  At that point you will have lost half the energy that you had at the start and will have to add that lost energy in order to continue to the end state, or retreat to the starting state.

Here are some helpful electronic examples since you claim familiarity with electronics:

Take three 1000uF 10V capacitors.  Label them: C1, C2, and C3.

Step 1.
Discharge, all three capacitors completely.
Calculate the stored energy.

Step 2.
Charge C1 to 1V.
Calculate and record the stored energy.

Step 3.
Connect C1 to C2 with a pair of clip leads.
Measure the voltage across C1 and C2.
Calculate and record the stored energy.

Step 4.
Using an additional pair of clip leads connect C3 to C1 and C2.
Measure the voltage across C1 / C2 / C3.
Calculate and record the stored energy.

Step 5.
Discharge all three capacitors completely.

Step 6.
Charge C1 to 1V.
Calculate and record the stored energy.

Step 7.
Using three clip leads, connect C1 to C2 through a 10K Ohm resistor between the anodes, and a clip lead between the cathodes.
Measure the voltage across each C1 and C2.
Record the voltage every 10 seconds for two minutes.
Calculate and record the stored energy on each capacitor and the combination for each recorded data point.

Step 8.
Discharge all C1 and C2 completely.

Step 9.
Charge C1 to 1V.
Calculate and record the stored energy.

Step 10.
Using three clip leads, connect C1 to C2 through a 10K Ohm resistor between the anodes, and a clip lead between the cathodes.
Connect C1 to C2 through a 20K Ohm resistor.
Measure the voltage across each C1 and C2.
Record the voltage every 10 seconds for two minutes.
Calculate and record the stored energy on each capacitor and the combination for each recorded data point.

Questions / Discussion:
Compare the stored energy in Step 3 to the stored energy in Step 2.  How many capacitors were charged in Step 2?  How many were charged in Step 3?
Compare the stored energy in Step 4 to the stored energy in Step 2.  How many capacitors were charged in Step 4?
Compare the stored energy at each 10 second measurement point in Step 7 and Step 10.
Discuss the energy effects of connecting capacitors using a wire as in Steps 3, and 4, versus the 10K resistor in Step 7, or the 20K resistor in Step 10.
Discuss the power effects of connecting capacitors using a wire as in Steps 3, and 4, versus the 10K resistor in Step 7, or the 20K resistor in Step 10.

Extra credit:  Try to think of a way without adding any external energy to get from the measured conditions in Step 4 to the measured conditions in Step 3.  Try to do the same for Step 3 to Step 2.

[MarkE]

You have left too many questions unanswered.

Third base!

[MarkE]

Who"s on first?

That long list made me laugh,, I have stated that me and electronics are not good be fellows,, but I understand what you are trying to get at, and I agree with what you are saying except for one small part.

I can not speak to electrical parts, but with mechanical parts I can state that the condition of a full loss of 1\2 the energy does not have to happen.  Using mechanics, the unused force that is applied can be applied into another mechanical device, without detracting from the primary transfer system.

You have so far failed to show an example of what you claim. I suspect that you simply do not understand and so misinterpret what you think overcomes the problem.

This is done with load sharing stuff all the time and is not anything new.

A Torsen center diff for AWD or 4WD cars, just as an example.

The fact that you are citing a differential as an example of a potential energy behavior reinforces my view that you simply do not understand the N*(X/N)² problem or how it applies to your two cylinder lifter.

[Magluvin]

That long list made me laugh,, I have stated that me and electronics are not good be fellows,, but I understand what you are trying to get at, and I agree with what you are saying except for one small part.

I can not speak to electrical parts, but with mechanical parts I can state that the condition of a full loss of 1\2 the energy does not have to happen. 

"a full loss of 1\2 the energy does not have to happen.  "

Absolutely correct.

We can charge a 100uf cap to 100v and if we allow it to be discharged into another 100uf cap(0v), we end up with 50v in each cap. There is a loss of 50%.  And I say we just lost it stupidly by not making use of the current flow from the full cap to the empty.  For me, I dont believe it was lost due to resistance/heat.  I think that reasoning is just used to cover the simplistic ideals that we started with 1 cap at 100v, and ended up with just half that voltage in each. 

We could have a 1,000,000 ohm resistor(pressure limiting valve) inserted between the 2 caps, and it would take a long time to equal out to 50v each, but no extra losses due to the 'added' very large resistance.

Capaitor - air tank

100v - 100psi

50v - 50psi

resistor  -  valve limiting pressure to be released


Now, if we have 2 caps, 100uf/100v  and 100uf/0v  and we add an inductor between the leads of the caps, so that the 100v cap dumps into the empty cap, if we have a switch and time it just right, we can turn the switch on, the 100v cap begins to discharge into the empty cap through the inductor. The inductor starts to build a magnetic field(stored energy), eventually nearly all the energy from the full cap will be in the previously empty cap if we turn the switch off right at that moment.

The inductor acts as a flywheel.  So if we have 2 equal air tanks, one with 100psi, and the other at 0psi, then we run a pipe from the full tank to a valve(switch), and then an efficient turbine or better yet an air driven motor(less leakage) with a flywheel on its shaft, when we open the valve, the pressure in the full tank decreases and the empty tank begins filling. With the air motor/flywheel, as the air is being transferred, the flywheel gets going. When the 2 tanks are at equal pressure(50psi), the flywheel is at its peak rpm, and continues pumping from the originally full tank into the previously empty tank, till we have nearly all the initial pressure of the source tank in the destination tank, and we shut off the valve. Little loss. Not 50%.

Electrical, mechanical, very similar beasts in many ways. When the inductors field is at peak(flywheel at peak rpm) the field begins to collapse, pulling charge from the source cap to the destination cap, turn off the switch.

So if we just let the source tank(100psi) just dump into the destination tank, and we end up with 50% of the total initial energy of the first tank, in both tanks total(50psi each), then we just lost that energy stupidly by just releasing the pressure and not taking advantage of the transfer. Releasing the pressure into a totally larger container(2xcu.in of source container), without doing anything with that release in order to make use of that energy being transferred.   

We could do other work with the air motor, providing power out of the system, and still end up with 50psi in each tank, where the originally lost 50% would have been used mechanically by way of the air motor.     Hope that helps.

Mags

[MarkE]

"a full loss of 1\2 the energy does not have to happen.  "

Absolutely correct.

We can charge a 100uf cap to 100v and if we allow it to be discharged into another 100uf cap(0v), we end up with 50v in each cap. There is a loss of 50%.  And I say we just lost it stupidly by not making use of the current flow from the full cap to the empty.  For me, I dont believe it was lost due to resistance/heat.

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 [/b]power[/b] changes, making the equalization happen faster or slower.

I think that reasoning is just used to cover the simplistic ideals that we started with 1 cap at 100v, and ended up with just half that voltage in each. 

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.

We could have a 1,000,000 ohm resistor(pressure limiting valve) inserted between the 2 caps, and it would take a long time to equal out to 50v each, but no extra losses due to the 'added' very large resistance.

That's right:  The resistor value affects the power not the energy.

Capaitor - air tank

100v - 100psi

50v - 50psi

resistor  -  valve limiting pressure to be released


Now, if we have 2 caps, 100uf/100v  and 100uf/0v  and we add an inductor between the leads of the caps, so that the 100v cap dumps into the empty cap, if we have a switch and time it just right, we can turn the switch on, the 100v cap begins to discharge into the empty cap through the inductor. The inductor starts to build a magnetic field(stored energy), eventually nearly all the energy from the full cap will be in the previously empty cap if we turn the switch off right at that moment.

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.

The inductor acts as a flywheel.  So if we have 2 equal air tanks, one with 100psi, and the other at 0psi, then we run a pipe from the full tank to a valve(switch), and then an efficient turbine or better yet an air driven motor(less leakage) with a flywheel on its shaft, when we open the valve, the pressure in the full tank decreases and the empty tank begins filling. With the air motor/flywheel, as the air is being transferred, the flywheel gets going. When the 2 tanks are at equal pressure(50psi), the flywheel is at its peak rpm, and continues pumping from the originally full tank into the previously empty tank, till we have nearly all the initial pressure of the source tank in the destination tank, and we shut off the valve. Little loss. Not 50%.

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.

Electrical, mechanical, very similar beasts in many ways. When the inductors field is at peak(flywheel at peak rpm) the field begins to collapse, pulling charge from the source cap to the destination cap, turn off the switch.

So if we just let the source tank(100psi) just dump into the destination tank, and we end up with 50% of the total initial energy of the first tank, in both tanks total(50psi each), then we just lost that energy stupidly by just releasing the pressure and not taking advantage of the transfer. Releasing the pressure into a totally larger container(2xcu.in of source container), without doing anything with that release in order to make use of that energy being transferred.   

That is the tyranny of the N*(X/N)² problem.

We could do other work with the air motor, providing power out of the system, and still end up with 50psi in each tank, where the originally lost 50% would have been used mechanically by way of the air motor.     Hope that helps.

Mags

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.