David Bowling's Continuous Charging Device

[SeaMonkey]

Dieter,

You raise some very interesting possibilities.

Your thought processes are intriguing.

There is an answer to the question of mysterious
"free energy."  Few have found it.

It is not where it is thought to be...

[tinman]

The problem here,is that it is going to be very hard to get accurate power measurements with a high amount of inductive spikes-such as the inverter has.
We can do !on the fly! measurements with some degree of accuracy,but it is not going to be 100%.

I have carried out 4 test run's now,and used different measuring methods for each,and in all test,the inverter consumes and dissipates the same amount of energy,and the remainder of that supplied by the 24volt bank,is returned into the 3rd 12 volt battery-->no free lunch found yet.

In saying that,battery 2(the one connected in series with battery 1,but in parallel with battery 3)dose retain a higher voltage than battery 1(the lead battery in the series connection).So there is obviously energy also being returned to battery 2-along with battery 3,and i suspect it is because of the reason i gave some posts back. But even so,this energy is accounted for,when we are using the series voltage of battery 1 and 2 to make our input power calculations.

I have however,found the exact reactions to that of Davids original circuit he took to the patents office,in that i found a battery for the battery 3 position,that resulted in explosions of current being sent through the circuit--as seen in second video i posted here on the subject. I am not sure why this happens,but i suspect some sort of instantaneous chemical reactions taking place within the dead battery-->i will look further into this.


Brad

[MileHigh]

The problem here,is that it is going to be very hard to get accurate power measurements with a high amount of inductive spikes-such as the inverter has.

Brad

Your comment is nonsense and it's a farce because I already told you that your comments were ridiculous and you intentionally ignored what I said.  Like I told you before, I haven't been on a bench in 25 years, I have never used a DSO, but I could still spin circles around you on a bench with my eyes closed.  Let's do some spoon feeding.

Look at your DSO capture from the current sensing resistor attached to this posting.  You think those are "inductive spikes" and the inverter is acting like a reactive load?  That doesn't make the slightest stitch of sense at all.  Why would the inverter draw power from the battery with "inductive spikes" and look like a reactive load?  WHY?

Look at the undershoot on the "inductive spikes."  The reasonable assumption at this point in time is that it's the inductance in the interconnect wiring and the power resistor that you are using to monitor the current that is causing the undershoot, and it has nothing to do with the inverter itself.  After all, if we assume that it is a train of current pulses going into the inverter, when the inverter shuts off the pulse suddenly, there will be inductive ringing associated with that event in the interconnect wiring.

It appears that the inverter is drawing power from the battery in a continuous train of current pulses, that's all.  Each current pulse will be at the applied voltage for a certain amount of time at a certain current level.  If for the sake of argument you assumed that the current pulses were all identical (just as an example), then with your DSO you could measure the energy in a single pulse, and multiply that by the pulse frequency, and voila you have measured the power draw of the inverter with your DSO.  And there is nothing even remotely inductive or reactive associated with that power draw from the battery.

What is the double sine wave pattern in the spikes?  How long have you been using a DSO and watching other people on the forums use DSOs?  That sine wave pattern has the classic signature of an aliasing pattern because you are under-sampling the waveform.  It's a total fake-out, the sine wave pattern isn't even there, it doesn't exist.  All you have to do it turn up the time base and you should see less spikes, and the sine wave pattern should disappear.

The output of your inverter is 240 VAC at 50 Hz, correct?  So the period for a full AC sine wave cycle is 20,000 microseconds.  Look at your DSO capture, it covers 750 microseconds.  I believe that your load is a good old fashioned tungsten filament light bulb.  Right now I am assuming that you have no idea at all where the under-sampled DSO capture lines up relative to the inverter's output sine wave.  For sure, since the light bulb is a resistive type of load, the inverter is outputting a lot of power at the sine wave peaks.  But right now you don't have the slightest idea how the current pulsing input on the inverter responds to the variable power demand by the inverter output over the full cycle of the 50 Hz AC sine wave.

So in summary, it would appear that the inverter draws power from the battery in a train of current pulses.  You haven't properly captured the pulse train or the individual pulses.  You have no idea what happens on the input when the inverter is drawing much more power at the peaks of the sine wave output.  Do the pulses get more frequent, or does the pulse train remain at the same frequency but the pulse width itself is modulated, or does something else happen?  You have no idea right now.

MileHigh

[tinman]

Your comment is nonsense and it's a farce because I already told you that your comments were ridiculous and you intentionally ignored what I said.  Like I told you before, I haven't been on a bench in 25 years, I have never used a DSO, but I could still spin circles around you on a bench with my eyes closed.  Let's do some spoon feeding.

Look at your DSO capture from the current sensing resistor attached to this posting.  You think those are "inductive spikes" and the inverter is acting like a reactive load?  That doesn't make the slightest stitch of sense at all.  Why would the inverter draw power from the battery with "inductive spikes" and look like a reactive load?  WHY?

Look at the undershoot on the "inductive spikes."  The reasonable assumption at this point in time is that it's the inductance in the interconnect wiring and the power resistor that you are using to monitor the current that is causing the undershoot, and it has nothing to do with the inverter itself.  After all, if we assume that it is a train of current pulses going into the inverter, when the inverter shuts off the pulse suddenly, there will be inductive ringing associated with that event in the interconnect wiring.

It appears that the inverter is drawing power from the battery in a continuous train of current pulses, that's all.  Each current pulse will be at the applied voltage for a certain amount of time at a certain current level.  If for the sake of argument you assumed that the current pulses were all identical (just as an example), then with your DSO you could measure the energy in a single pulse, and multiply that by the pulse frequency, and voila you have measured the power draw of the inverter with your DSO.  And there is nothing even remotely inductive or reactive associated with that power draw from the battery.

What is the double sine wave pattern in the spikes?  How long have you been using a DSO and watching other people on the forums use DSOs?  That sine wave pattern has the classic signature of an aliasing pattern because you are under-sampling the waveform.  It's a total fake-out, the sine wave pattern isn't even there, it doesn't exist.  All you have to do it turn up the time base and you should see less spikes, and the sine wave pattern should disappear.

The output of your inverter is 240 VAC at 50 Hz, correct?  So the period for as full AC sine wave cycle is 20,000 microseconds.  Look at your DSO capture, it covers 750 microseconds.  I believe that your load is a good old fashioned tungsten filament light bulb.  Right now I am assuming that you have no idea at all where the under-sampled DSO capture lines up relative to the inverter's output sine wave.  For sure, since the light bulb is a resistive type of load, the inverter is outputting a lot of power at the sine wave peaks.  But right now you don't have the slightest idea how the current pulsing input on the inverter responds to the variable power demand by the inverter output over the full cycle of the 50 Hz AC sine wave.

So in summary, it would appear that the inverter draws power from the battery in a train of current pulses.  You haven't properly captured the pulse train or the individual pulses.  You have no idea what happens on the input when the inverter is drawing much more power at the peaks of the sine wave output.  Do the pulses get more frequent, or does the pulse train remain at the same frequency but the pulse width itself is modulated, or does something else happen?  You have no idea right now.

MileHigh

I see you have decided to follow me to this thread-just to post more dribble.
It would seem that you are once again making idiotic comments based around assumptions.
No,the CVR is not inductive,nor do i think you have taken the time to look at the circuit to see as to how the spikes you see actually form. Perhaps you missed the constant current draw seen on the scope trace as well.
Do you even know how this inverter opperates?-my guess is no.

So please stop posting dribble on something you know nothing about.

You running rings around me on the bench is laughable,as you wont even answer my challenge on a JT build off,because you know i would wipe the floor with you.

If you want to join in,then dust off that bench,and start experimenting.  If your not prepared  to do that,then run along,as your words carry no weight around here anymore,and you are no help in finding the answers David and the rest of us here seek.


Brad.

[MileHigh]

Brad:

These are your initial comments about the current waveform:

Below is a scope shot across a CVR,showing the input current wave form to the inverter.
As you can see,it is the same as a pure sine wave inverters input wave form.
These are the spikes those on the other forum are speaking of on the input,that is one of the reasons that this setup works as claimed-!i believe!?.
As the batteries will only see the input side of the inverter,and not the output side of the inverter,i am at a loss as to why it has to be a pure sine wave inverter?,as battery regulation has nothing to do with the output of the inverter.

You also said this about the waveform:

The inverter is quite reactive--see scope shot below across CVR to inverter input.

It looks to me like you are confusing standard back-EMF spikes from a coil, with the current spikes that represent the way the inverter is drawing current from the battery feed.  That's why you are saying "quite reactive" and "high amount of inductive spikes."  Assuming that's true, you can see how you made a huge mistake.  You led yourself down a garden path, and took your followers on this thread along with you down that path, with the result being that you and your followers were all lost with respect to how the input side of the inverter draws power from the battery. 

The only reason that I came here is because you refused to respond to the initial comment I made on the other thread so as to not disturb this thread.  I was hoping that you would take a second look at the way the inverter was drawing power from the battery and document it properly so that you and your peers and would all be properly informed about how the input side of the inverter actually worked.

"Dribble" and "idiotic comments" my ass.  You are bluffing.  Just the fact that you refuse to address anything technical in my posting says it all.  For the benefit of yourself and your peers on this thread you need to understand how the input side of the inverter draws current from the battery because that directly affects how the third battery in the charging position gets charged.  If you had listened to me and shown initiative and redid your current sensing resistor measurements and posted what looked like good credible data for yourself and your peers I would not even be here.

MileHigh