David Bowling's Continuous Charging Device

[Dbowling]

SeaMonkey,
That is a perfect description of the basic system as shown by Luc, and even of the system with the pulse motor. All we EVER claimed out of that system was that you could get extended run times. I remember the names of all the people who said that wasn't possible, but refused to test it. They will get coal in their stockings instead of answers. But that is only the beginning of what we have learned about how to use this system. We haven't dumped everything out there because nobody would believe us when we said THIS much of it worked, so have had no reason to share the rest. We know how to SIGNIFICANTLY extend those run times. Whether or not something can be built that will run forever has yet to be determined. Forever isn't here yet and we are far from done. But when you understand how LOOOOOOOOOONG those run times can be, and realize you can use the motor to turn a generator, suddenly you have a system that runs on virtually NOTHING and produces power. That is the advantage of this circuit connected to Matt's modified motor.

[SeaMonkey]

I suppose it should be mentioned that a
discharged lead acid battery does have an
almost magical ability to "bounce back."

The Lead Acid Battery is truly a marvelous
device.  Simple in structure but very complex
in its task of storing and then delivering electrical
energy by chemical reaction activity.  FarmHand
and I have had some interesting discussions
regarding the uniqueness of this over 100 year
old proven technology.

[tinman]

author=Dbowling link=topic=4612.msg512697#msg512697 date=1509772546]

Brad,

24 WATTS IS GOING INTO THE MOTOR AND IT IS CONSUMING 12. REALLY? I DON'T THINK SO

Why do you find this hard to believe,when the motor is one part of a two part series load on the 24 volt battery(the two 12 volt batteries hooked in series).

If we take a 1 volt battery,and series connect 2x 1 ohm resistors,how much power dose each resistor dissipate?
Using your analogy,all the power must go !through! the first resistor--which it dose not.

HARDLY ANYTHING IS CONSUMED BY THE MOTOR.

The exact amount of power can be accurately calculated in this system.

First you have to understand that battery 3 is not series connected to batteries 1 and 2.
Battery 3 is 1 part of a two part load placed on batteries 1 and 2

To work out what is being consumed by the motor and battery 3 is very simple.
The current flowing through the system will remain a constant-same for all components.
So to measure the power consumed by the motor,we multiply the voltage drop across the motor,by the current flowing through the system--the same applies for battery 3.

THERE IS ONLY 12 WATTS AVAILABLE TO THE MOTOR. THE OTHER 12 WATTS+ THAT COMES OUT OF THE CHARGE BATTERIES IS NEUTRALIZED BY THE 12 WATTS+ COMING OUT OF BATTERY 3.

There is no power coming out of battery 3.
If the power was coming out of battery 3,then the battery would be drained-not charged.
As battery 3 is being charged,then it is consuming power to do so.

CAN THIS BE PROVEN? SURE!!! TWO DIFFERENT WAYS. TAKE YOUR TWO BATTERIES IN SERIES AND CONNECT YOUR MOTOR ACROSS THEM AND MEASURE THE RPM. NOW RUN YOUR MOTOR ON JUST ONE OF THEM AND MEASURE THE RPM. THEN PUT THE SYSTEM BACK TOGETHER AND MEASURE THE RPM OF THE MOTOR RUNNING BETWEEN THE POSITIVES. WHAT IS ITS RPM NOW? IS IT RUNNING ON 24 WATTS OR 12 WATTS?

The motor is running on 12 watts

A MOTOR RUNS ON WHATEVER GOES INTO IT. IT DOESN'T HAVE SOME MAGIC BYPASS CIRCUIT THAT ALLOW HALF THE WATTS TO GO ON THROUGH WITHOUT AFFECTING THE MOTOR AT ALL. THIS IS ALL STANDARD ELECTRICAL STUFF, NOT VOODOO MAGIC.

And this is why i believe you and Matt need assistance with power calculation.
This is standard stuff,and you have it wrong.
As i stated before,the motor is 1 part of a two part !series! connected load.
If the motor was hooked to the source(the two batteries)in parallel,then yes,it would consume all of what is going into it. But it is not a parallel connection-->it is 1/2 of a series connected load.
So,as i said,the motors consumption is calculated by the value of the voltage drop across it X's the current flowing through it--standard electrical engineering.

HERE IS THE SECOND WAY OF PROVING THIS AND IT IS WHAT MATT IS TALKING ABOUT. (I WATCH AND LISTEN MATT, AND I HAVE LEARNED A THING OR TWO FROM YOU) IF YOU HAVE A 12 VOLT BATTERY AND YOU WANT TO MEASURE THE VOLTAGE IN IT, YOU DO NOT TAKE THE TOP OFF AND MEASURE ACROSS FIVE OF THE 6 TWO VOLT SECTIONS INSIDE. YOU MEASURE ACROSS ALL OF THEM. THAT TELLS YOU THE VOLTAGE AVAILABLE IN THE SYSTEM. WHEN YOU PUT TWO 12 VOLT BATTERIES IN SERIES, HOW DO YOU MEASURE THEM? BECAUSE THE TWO BATTERIES ARE CONNECTED IN THE MIDDLE. YOU MEASURE FROM THE POSITIVE OF ONE TO THE NEGATIVE OF THE OTHER. BECAUSE THEY ARE CONNECTED, THEY HAVE BECOME ESSENTIALLY ONE BATTERY. HOW WOULD YOU MEASURE THREE BATTERIES IN SERIES?

And this is where you are making your mistake.
Battery 3 is not hooked in series with batteries 1&2.
Battery 3 is hooked in parallel with batteries 1&2,with a second series connected load between the negatives.
You can measure the voltage drop across each battery individually by placing a voltage measuring  device across the two posts of the battery.

THE SAME WAY, RIGHT? IT IS THE SAME WITH THE THREE BATTERIES THAT ARE IN SERIES HERE, EVEN IF ONE IS IN SERIES BACKWARDS.

There is no such thing as a series connection backwards.

SO YOU WANT TO KNOW THE WATTS THE MOTOR IS RECEIVING? ?? TAKE THE MOTOR OUT AND PUT YOUR METER BETWEEN THE TWO POSITIVES. CONNECT IT EITHER WAY. ONE WAY IT READS POSITIVE 12 AND THE OTHER WAY IT READS NEGATIVE 12. SO WHAT DOES THIS TELL YOU IS AVAILABLE TO THE MOTOR??? IF YOU REPLACE THE METER WITH A MOTOR DO YOU THINK THAT NUMBER IS MAGICALLY GOING TO JUMP TO 24 ON ONE SIDE OF THE MOTOR BUT STAY AT 12 ON THE OTHER SIDE. I DON'T THINK SO TIM.

No,there will be a 12 volt drop across the motor.

WE SAY 12 WATTS. YOU SAID 24 WATTS. WHO IS CORRECT?

No,i said the supply batteries(1&2) were delivering 11.2 watts to the system,and said that the motor was consuming 5.6 watts--which is correct.
Please go back and read my post again.

 

WE DIDN'T DISAGREE WITH ANY OF THE NUMBERS THAT WERE SHOWING ON LUC'S METER, AND I SAID THAT AT THE TIME. WHAT WE DISAGREED WITH WAS WHAT LUC SAID THOSE NUMBERS WERE SHOWING AND THE WAY HE WAS MEASURING THE SYSTEM.[/font]

What Luc said those numbers were showing,is correct.

ONLY 12 WATTS IS AVAILABLE TO THE MOTOR. PERIOD.

If we are still referring to Lucs first video,then only 5.6 watts was being consumed by the motor,as battery 3 was consuming the other 5.6 watts.

To say only 12 watts is available to the motor is also incorrect,as what is available to the motor is what batteries 1&2 can deliver,minus what battery 3 will consume when the motor is placed under load.

THAT IS WHY WE SAID THE MEASUREMENT IS WRONG AND THAT LUC DIDN'T KNOW HOW TO MEASURE THIS SYSTEM CORRECTLY.  AND IF WE ARE CORRECT, AND NOT LUC, THEN THE NUMBERS USED IN ALL THE CALCULATIONS BELOW ARE AUTOMATICALLY INCORRECT. THATS WHY I TOOK EXCEPTION TO ALL OF THE CALCULATIONS BELOW.

Luc's numbers and measurements were correct on his video.

If battery 1 & 2 are connected in series and are 12 volts each = 24 volts and the current is measured at 1 amp = 24 watts entering the motor and if battery 3 is at 12 volts and the current entering it is measured at 1 amp = 12 watts entering battery 3.[/font]

From these numbers we can conclude that the motor is consuming 12 watts,and battery 3 also 12 watts.

However, the motor has converted part of those 12 watts to mechanical power and at best 80% if it is available at the motor shaft which means we have about 9.6 watts in mechanical power at the motor shaft which we can recover back to electrical if we attach a generator to it and can recover at best 80% if it = 7.7 watts and add it to the charge battery which gives a potential total of 19.7 watts recovered from the 24 watts put into the system. HOW MANY WATTS WERE PUT INTO THE MOTOR?

This is incorrect.
The motor is consuming 12 watts unloaded.
As soon as a load(E.G-generator) is placed on the motor,the motor will draw more current,and thus more power.
The charge battery will also receive the same current flow increase,and thus charge faster when a load is placed on the motor.
Luc also showed this in his first video.

So if we have 24 watts coming out from input batteries and 12 watts going in the charge battery it means half of the input power is being used by the motor and potentially half recovered by the charge battery.

This statement is correct.

I hope my explanations here made sense.

I could understand what you are saying,but you are incorrect.

You guys have to realize you are dealing with something a little different here and standard measurement methods can be deceiving.

There is nothing about this system that is a little different,or in any way -hard to take accurate power measurements from.

We know what we are talking about.

After reading all your claims in this post,and others--it is clear that you and Matt do not know how to measure or make power correct calculations--and this is where the problem lies.

But until you UNDERSTAND what is going on, you will have a hard time applying these principles to other things, and that's where it really gets exciting.

I do understand very well what is going on,and have no problem at all with taking accurate power calculations from this simple device.

You are in the same boat UFOpolotics was in some years back,where he two screamed OU galore.

After spending around $180.00 on part's,i had to show him how to do a prony brake test correctly.
Once again,it was a case of some one not knowing how to take correct power measurements of there own machine-as is clearly the case here.

If not, you can continue to believe what you want to believe.

I will continue to make !correct! power measurements.

Dave
I know you may not like what i have said,but it is correct,and you will not find an EE here that will say i am wrong.

You just cant go making up your own type of power measurements to suit your need--to make your DUT work as you claim.

As far as Luc's test showing or exceeding the watt hour rating of the batteries--do not make the mistake that thinking this is something special =-as it is nothing out of the ordinary.

The watt hour rating of a battery is a safety margin ratting only,and the battery could very well put out double the watt hour ratting ,when subject to a deep discharge rate.
It will also kill your batteries much faster.

Battery voltage also has nothing at all to do with the health of the battery,or it's state of charge.


As i said,i still have all my gear i bought for this DUT,and will be happy to shoot a video showing the correct power measurements for each component in the DUT.


Brad

[tinman]

 author=SeaMonkey link=topic=4612.msg512698#msg512698 date=1509772896]
The "Three Battery System" is actually quite simple.

The Load will have applied to it the Voltage Difference
between the Source Pair of Batteries and the Opposing
Battery No. 3.

Correct

The "load" is a device such as an electrical motor
which has been "optimized" to produce pulsations
as it draws load current from the source battery
series pair.  These pulsations are augmented by
brief inductive discharge pulses from the optimized
load motor as it rotates;  the pulses are beneficial
to battery No. 3 as it is being charged.

Correct

As the source battery pair discharges they provide
power to both the "load motor" and the "charging
third battery."  Once the source pair has discharged
sufficiently the weaker of the two is replaced by
swapping it with the "charged" third battery and
operation resumed.

As i did in my tests.

Is there actually any "overunity" or any other kind of
"magic" going on with this circuit?  No.

Correct

Is energy that might otherwise be "wasted" being
recovered and applied to the "charging" battery
No. 3?  Yes.

Energy would be wasted where,if battery 3 was removed?.

There is also the well known (actually by design) ability
of the lead acid battery to increase its ampere hour
rating as it is cycled through several charge-discharge
operations.  Most lead acid batteries experience an
approximately 30% increase in capacity as they are used.
This is due to the composition of the plates "forming"
more active material with each discharge-charge cycle.
It has to do with how the plates are manufactured and
the "paste" which fills the plate grids not being fully
activated at the time of manufacture.

Yes,but in order for there to be extra energy stored in the battery,
You do actually have to supply that energy--it dosnt just appear out of no where.

In the meantime is there any waste of time associated with
evaluating the Three Battery System?  Not at all!  As Luc is
demonstrating with his trials there is much to learn and
seeing how things work with our own eyes is an invaluable
learning experience.  Provided of course that we are able
to truly comprehend what is being seen.

Apparently most are lost,and do not fully understand batteries,or how/what there watt hour ratting means.

The thought that Battery No. 3 is "neutralizing"
12 watts by power coming out of it is certainly
an unorthodox view.  I'll leave it to others to
comment on.  I suspect that in time, Dbowling
will come to realize his error in thinking.

One would hope so.



Brad

[minnie]

   Ever heard of "Flogging a dead horse?"  David Bowling.
   Move on with your life,
              John.