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Author Topic: Bedini 10-Coil Alternative Discussion  (Read 66440 times)

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It's not as complicated as it may seem...
@milehigh

The fact of the matter is that a battery will raise it's voltage and resistance in response to the action of "charging" and the inductive discharge will raise it's voltage in proportion to the batteries rise in resistance as it must, ...
Regards
AC

AC, I would think that a battery's internal resistance would decrease as it's state of charge increases.  ???

.99


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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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Actually the term "battery impedance" is common in the industry and does decrease with charging. Testing "battery impedance" is widely used to determine the the quality of the cells.

I have some battery impedance test equipment made by Biddle (AVO). It operates by applying a fixed known AC current across the battery terminals and measuring the AC voltage.

Attached is a paper describing "Battery Impedance Testing" and some battery models.


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It's not as complicated as it may seem...
Thanks ION. good paper.

I was looking for a battery model quite some time ago while I was working on the RA circuit. Nice to have that now.

The paper does seem to support the notion that it is the resistive parts (the dominant portion of the total impedance) of the model that increase in value as the SOC decreases. This makes sense anyway.

.99


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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

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It's not as complicated as it may seem...
AC,

In my experience, an inductor does act very much like a constant current source during it's de-energizing cycle (i.e. during it's inductive kickback). And this is well supported by the theory of how inductors function.

.99


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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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@MileHigh and allcanadian,

interesting points both of you. I am a little confused for sure.

Both arguments seems logical to my mind, but clearly you both have a difference of opinion.

Milehigh, would you mind dissecting allcanadian points one by one in where the error is. Forgive me to ask this, in no way I looking for useless arguments. It is interesting to me to know the real logic in this matters.

many thanks,

Fausto.
   
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Fausto:

I will try to respond to AC's points.  I may be very technical also.

Quote
(An Inductor) will ...  raise it's voltage in proportion to the rate of change of the switching mechanism which is a form of resistance.

This is incorrect.  AC is modeling a switching-off transistor as a resistance that increases in value with respect to time.  That is fine.  An inductor will raise it's voltage in proportion to the instantaneous value of the resistance of the switch.  At the same time the inductor is discharging its stored energy so the amount of current flowing through the resistance of the switch is also decreasing.  Therefore there are two effects happening at the same time.  This could be described by a differential equation and you could solve the differential equation to get the plot for the voltage of the discharging inductor with respect to time.

The most important point here is that the inductor outputs a voltage that is proportional to the instantaneous resistance of the load and the instantaneous amount of current flowing through the inductor, and not the rate of change of the resistance of the load with respect to time.

Quote
If the resistance is infinite to an inductive discharge then the voltage is infinite ( E=IR) and as the energy cannot change the current (I) must approach zero.

This is incorrect.  As the resistance tends towards infinity, the time duration of the current spike tends towards zero.  Note that there ALWAYS IS a measurable current flow, but the length of time of the current flow decreases towards zero.

Quote
No, the collapsing field induced a potential difference by cutting the coil conductors and if the coil is in a circuit then this potential difference will lead to an electric current thus the voltage came first, it is a voltage source first and the current is a result of the voltage discharging itself into a circuit.

This is incorrect.  The magnetic field was built up by the input current flow, and so when the coil is open-circuited, the magnetic field now generates an output current flow.  This demonstrates the natural symmetry in the way an inductor stores energy and releases energy.  The voltage across the inductor terminals are a result of the current flow.  A discharging inductor acts like a current source with an output impedance of infinity.

Quote
Well no it cannot thus it would seem obvious that a discharging capacitor has little in common with a discharging inductance.

This is incorrect.  Imagine two test setups.  The first setup has a 1-Henry inductor connected to a 1-ohm resistor and there is 1 amp of current flowing through the inductor.  The second setup has a 1-Farad capacitor connected to a 1-ohm resistor and there is one volt of potential difference across the capacitor.  For both cases above when you start the two tests the voltage and current waveforms will be identical.  In addition, if you increase the resistance for the inductor test and you decrease the resistance for the capacitor test, then the voltage waveform for the inductor test will be identical to the current waveform for the capacitor test.  Also, the current waveform for the inductor test will be identical to the voltage waveform for the capacitor test.  In all cases they will be exponentially decaying waveforms.  Capacitors and inductors are very similar to each other.

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A battery will raise it's voltage in proportion to the voltage of the source charging it and we call this action "charging" where a resistor will not because it is not "charging" it is dissipating energy as heat, they are not the same thing.

This is incorrect.  To use the phraseology "a battery will raise its voltage" in this context is a very poor choice of words that would never be spoken by an electrical engineer.  I won't go into a detailed technical discussion here.  I will say that AC's sentence doesn't even make sense.  As a bare minimum, you would have to say that the voltage at a battery's two terminals will increase as a function of the current that you are supplying to the battery.  Certainly you can clamp a battery's terminal voltage to an external voltage source that has a low enough output impedance to overcome the battery's output impedance.   Then you end up charging the battery because current ends up flowing into the battery.  The bottom line is that this is an extremely awkward way of expressing this idea and an extremely poor choice of words that would never be used by someone that truly understands electronics.

Ion posted a model for a battery that I have not looked at yet, but I will assume that the explanation for why a battery terminal voltage will increase when the battery is being charged is explained in the document.

Quote
we call this action "charging" where a resistor will not because it is not "charging" it is dissipating energy as heat, they are not the same thing.

Certainly they are not the same thing, but AC is missing the point.  In both cases, for a simple load resistor and for a charging battery, the energy source is pumping power into the "load."  In one case the "load" is a resistor, and in the other case the "load" is a charging battery.  So in both cases it is valid to consider issues related to impedance matching, because impedance matching is all about how well an energy source can transfer power into a load.

Quote
One could also ask why all the new battery chargers use boost converters exclusively where 20 years ago they used transformers, one word --efficiency....

AC's comment above about boost converters, and the Tesla comment and other comments in the paragraph I quoted from above have absolutely nothing to do with the discussion.

Quote
The fact of the matter is that a battery will raise it's voltage and resistance in response to the action of "charging" and the inductive discharge will raise it's voltage in proportion to the batteries rise in resistance as it must, does this sound like impedance matching?

No, it does not sound like impedance matching at all.  Go back and look at my battery example.  Impedance matching has to do with maximizing the power transfer between the energy source and the load.  In a Bedini motor the load can change impedance and the power transfer remains the same, there is nothing to match.

Quote
My battery chargers and the batteries they charge run cool, not hot, not warm

So what, this is purely anecdotal evidence.  Chances are that you have never measured the power output of one of your buck-boost chargers so you don't know if the battery should get perceptibly hot or not.  Even if the energy transfer is more efficient, the power that gets pumped into the charging battery is still going to be split into power that actually charges the battery and power that becomes waste heat.  If that waste heat production is below a certain threshold then you won't be able to perceive it.  For all you know after 20 minutes the surface temperature of your battery could increase by 1.5 degrees C and you would not be able to perceive this slow increase in temperature by putting your fingers on the battery.

I have never seen any scientific attempt to back up the argument that "Bedini chargers keep the charging batteries cool."  On the other hand, I have read some postings where some Bedini experimenters inadvertently overdo it with their setups and end up boiling their charging batteries.

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@Milehigh,

once again thank you for the detailed explanation.

I am noticing a process of thinking that sometimes is lacking and is not that obvious in others (including me). You mention the relation of one variable as a function of another variable. Many times I see we miss that.

Like in the example:
Quote
Quote
If the resistance is infinite to an inductive discharge then the voltage is infinite ( E=IR) and as the energy cannot change the current (I) must approach zero.

This is incorrect.  As the resistance tends towards infinity, the time duration of the current spike tends towards zero.  Note that there ALWAYS IS a measurable current flow, but the length of time of the current flow decreases towards zero.
You are saying that the voltage is not directly related to the simple way of looking at E=IR but also considering which one is playing the "function" and which one is playing the "variable" plus a hidden variable "time". In your description you say that the "time duration of the current spike" is the function in relation to the "variable current". After that we can apply those two now as being the variables to the function E which than would cause voltage to change, again not losing sight of the hidden variable "time".

In other words we have here TWO functions interrelated by time which makes the whole process not that intuitive when looking at only one function.

So E = IR, would be broken into E = function(I) x function(R). Where I = time x Variable and R = time x Variable (whatever Variable here would be in this example). So changing I or R will dramatically change E as a result. Missing time variable would cause one to incorrectly calculate I or R therefore E too.

Did I got it or am I still missing something in this thinking process?


Fausto.
   
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Fausto:

I think that you are understanding the two variables.  Just for fun, let me look at that issue again hopefully with a new perspective.

Let's just do an imaginary experiment that simulates the two-variable example, but on a much slower time scale.  This is a perfectly valid way of understanding what is going on.

Let's imagine that you have a very large inductor and a 10-ohm potentiometer.  We are going to imagine that you run a test where you discharge the very large inductor through the potentiometer, and it takes one full minute to fully discharge the inductor when the potentiometer is set to 5 ohms.

Test 1:  You get 1 ampere of current running through the inductor and you discharge it through the potentiometer when it is set to 5 ohms.

You observe the voltage across the inductor for one minute.  You notice that the voltage starts at 5 volts, and as the time goes on the voltage slowly decreases to almost zero volts after one minute.  The voltage and the current in this case is a standard exponentially decaying waveform.

Test 2:  You get 1 ampere of current running through the inductor and you discharge it through the potentiometer when it is set to 5 ohms, but you play with the potentiometer and turn the resistance up as you watch the voltage decrease.

For this test you start with the potentiometer set to 5 ohms.  Therefore the voltage will start to decrease starting from 5 volts.

When the voltage decreases to 4 volts, say after 10 seconds, you turn the potentiometer so that the resistance increases to 10 ohms.  At 10 seconds the current was (4/5) = 0.8 amperes.   Therefore when you turn the potentiometer to 10 ohms, the voltage will automatically increase to 8 volts.  Then the voltage will resume its exponential decay, and it will decay more quickly because now you are burning power more quickly.

Test 3:  You get 1 ampere of current running through the inductor and you discharge it through the potentiometer when it is set to 5 ohms, but you play with the potentiometer and turn the resistance down as you watch the voltage decrease.

For this test you start with the potentiometer set to 5 ohms.  Therefore the voltage will start to decrease starting from 5 volts.

When the voltage decreases to 4 volts, say after 10 seconds, you turn the potentiometer so that the resistance decreases to 1 ohm.  At 10 seconds the current was (4/5) = 0.8 amperes.   Therefore when you turn the potentiometer to 1 ohm, the voltage will automatically decrease to 0.8 volts.  Then the voltage will resume its exponential decay, and it will decay more slowly because now you are burning power more slowly.

So in this thought experiment test #2 looks like the case where an inductor is discharging through a transistor that is in the process of switching from ON to OFF.

MileHigh
« Last Edit: 2010-09-28, 20:36:44 by MileHigh »
   
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To go back to a more general description of what is going on when an inductor discharges through a transistor that is transitioning from ON to OFF where the transistor resistance is a function of time, there are three time-dependent variables.

The equation for the resistance of the transistor is exclusively a function of time and not dependent on anything else:

1) Load_Resistance = r(t) where r(t) is some kind of an equation with time t as a variable

The equations for the coil current and the coil voltage are a function of the initial current, the inductance, the variable resistance, and time:

2) Coil_Current = i(t) where i(t) is some function of i(t=0), L, r(t), and t

3) Coil_Voltage = v(t) where v(t) is some function of i(t=0), L, r(t) and t

With calculus and the differential equations that describe how an inductor works you can determine the solution and derive the equations for 2) and 3) above.

In the Wikipedia entry for "RL circuit" this problem is solved for the case when R is constant.  The solution to this problem is an exponentially decaying voltage and current waveform where the time-constant for the decay is L/R.

MileHigh
   
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Fausto:

Just one more comment about what happens when an inductor discharges through a resistance.  You MUST remember that if the initial current is always the same, then the amount of energy stored in the inductor is always the same.

Therefore every time the inductor discharges it will discharge the same amount of energy, no matter what the load resistance is.

If the resistance is low, it discharges that energy with a low-voltage spike over a long period of time.  The initial discharge current is always the same current that was flowing through the inductor the moment before it started to discharge through the resistor.

If the resistance is high, it discharges that energy with a high-voltage spike over a short period of time.  The initial discharge current is always the same current that was flowing through the inductor the moment before it started to discharge through the resistor.

If the resistance goes towards infinity, it discharges that energy with a voltage spike that goes towards infinity over a period of time that approaches zero.  The initial discharge current is always the same current that was flowing through the inductor the moment before it started to discharge through the resistor.  This is where AC was wrong.

If the resistance is zero, then the inductor does not discharge and the current flows forever.  There is no voltage spike at all, zero volts.

When you go back to the reality of a Bedini motor discharging into a charging battery, you should be able to make the connection with the state of the charging battery.  You can easily draw some general conclusions:

If the battery has a high impedance because it is old and sulfated then the current spikes going into the battery will be shorter and higher in voltage.

If the battery has a low impedance because it is new then the current spikes going into the battery will be longer and lower in voltage.

A battery's input impedance does not look like a resistor.  It looks like a non-linear resistor.  Just the same, you at least have an idea of what should be going on when you think about how an inductor discharges into a resistor and relate that back to a charging battery.

MileHigh
   
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To whom it may concern:

Read the document called Battery Management that I uploaded earlier. Look closely at the language used throughout the document as it is the language of ENGINEERING. You will notice charts and graphs, testing and results.

Now for comparison read some of the Bhoudini razzle-tazzle and hocus-pocus. Very slippery indeed.

By the way Bhoudini did not invent inductive pulse charging to recover sulphated batteries, it has been well known from before his time.

Newer chargers use boost converters for one main reason: COST. With costs of copper and steel rising, using high frequency switchmode with much smaller magnetics makes good economic sense. You get a few additional benefits which we need not go into here.

Even conventional 60 Hz wall warts have given way to switchmode designs.

Still some folks are content spinning their wheels.


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@MileHigh and ION,

thank you both for such valuable information. It all makes sense.

Now I see why Bedini tries to stress the the "magic" is not on the machine but "in" the battery. In one of his DVDs he mentions that you will not find OU on the machine but in the battery.

Now, It sounds like the fundamental way to prove or disprove Bedini's concepts would be by what he proposes concerning the load test where he claims you can run SSG or its variations (I am not sure which one is the correct one) where one battery would charge 4 others and one should later switch them and repeat the process to infinitum.

You MileHigh, also seems to stress on the process of measuring the input and output that I also think will prove beyond a doubt the lack of OU, it will demonstrate losses in the process and you also stress (I think the fundamental now) that one should now do the looping or switching the batteries test and see if indeed will run continually without stopping.

So Am I right to state that showing the results of the switching batteries would either demonstrate OU or not and if shows OU we have a clear misunderstanding of Bedini's explanation and if does not demonstrate OU at least we can say Bedini's is indeed not communicating correct physics?

Fausto.
   
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Hi Fausto:

Bedini says the "magic" happens in the battery like you said.  Certainly if you measure the power consumption of the Bedini motor drawn from the source battery and then measure the power output going into the charging battery you will find that there are losses in the Bedini motor.

Learning how to make these input and output measurements properly is still a valuable exercise.  Understanding how the discharging drive coil works is also a very valuable exercise.

So, let's take a look at the battery measurement issue, because that's where we are looking for the "magic."

Let's take the example of Rick Friedrich's Yahoo Bedini group because they are supposed to be the serious Bedini experimenters.

From what I understand they do the following:

They charge the charging battery and measure the current going into the charging battery and calculate the ampere-hours.
Then they discharge the charging battery and measure the ampere-hours.
Then they compare the discharging ampere-hours to the charging ampere-hours to calculate the COP.

What are the problems with this test procedure:

1.  They ignore the efficiency of the power transfer from the source battery to the charging battery.  They only measure the current going into the charging battery, they ignore the losses from the source battery.  In a typical Bedini motor you might only get 25%-35% of the power provided by the source battery actually becoming power that goes into the charging battery.

2.  The measurement of the average current going into the charging battery is suspect, you are not sure if your analog ammeter will give you an accurate result because of the "extreme pulsing" nature of the current going through the ammeter.

3.  They are ignoring the corresponding changing voltage of the charging battery when they pulse current into it.

4.  They are ignoring the fact that if the impedance of the charging battery changes as you charge it, then we know that the average current going into the battery will change.  The Yahoo Bedini group is not aware of this important fact.

5.  Points #2 , #3 and #4 above mean that their ampere-hours measurement is invalid.  What they should be doing is measuring the average power going into the charging battery as explained in the Bedini measurement thread.  Then they can calculate the total energy that they put into the battery.

6.  When they do load testing when they discharge the charging battery they measure the ampere-hours.  One more time they are not factoring in the decreasing voltage of the charging battery when they discharge it.  They should be measuring the energy that they can draw from the charging battery.

7.  They are not accounting for the amount of energy that may have been in the charging battery before they start charging it.

When you look at all the points above my conclusion is that the Yahoo Bedini group's method of calculating the battery COP is seriously flawed and will give you invalid data.  And also in the real world you have to make your measurements starting at the power supplied by the source battery, and not at the power supplied into the charging battery.  You simply cannot ignore all of the power that is lost in running the Bedini motor itself.

To be serious, all of the battery COP testing would have to be based on ENERGY in and out, and not based on ampere-hours in and ampere-hours out.  Also you have to account for the energy that is in the charging battery BEFORE you start your COP testing.   All of this is discussed in more detail earlier in this thread.

So yes Fausto, to test for any claims of OU for a Bedini motor, it has to be based on swapping batteries back and forth or making precise energy measurements on the batteries themselves or both.

As I suggested to Jeff (Bit's-n-Bytes) a few weeks ago, he could swap batteries back and forth and between each swap drain 25% of the energy out of the charging battery before he starts the swap.  All of his fancy microcontroller battery switching and swapping systems are just "window dressing."  The most important thing that he should do is do a proof-of-concept first with his 10-coiler.  He doesn't need any fancy extra electronics to do this.  If the proof-of-concept test works then great.  If the test shows no OU, then he can save his time and money by not investing in all sorts of fancy electronics and switching hardware and software.   If he decides after looking at his data that the 10-coiler Bedini motor is not a device that can produce free energy and power his house, that may be the end of his project.

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@MH

Thanks for the last posts. I have a question about what AC also said about the increase in resistance as voltage rises in a battery although my way of asking this is more low end.

So let's say a 12 volt battery has 6 volts in it. You connect a battery charger that pulses a 12 volt discharge into the battery that is at 6 volts. Would it not be unreasonable to assume that when the battery is at 6 volts it is very easy for the charger to bring it up to 7 volts, but when the battery is at 11 volts, the charger will take longer to bring it up to 12 volts. Or is this function purely linear. I think maybe that is what AC was referring to when he says the battery resistance increases as voltage increases in the battery.


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Hi Wattsup:

I still haven't read the battery document that Ion posted, you may want to read it for yourself.

From what I know about standard 12-volt lead-acid batteries, a battery that measures 6 volts would most likely be very old and very sick or damaged in some way.  Apparently the Bedini charging method, a.k.a. charging a battery with pulses of current from a discharging inductor, can rejuvenate a battery like this.  On the other hand, there has to be some limit to this and I suspect that a 12-volt lead-acid battery that measures 6 volts may indeed be too far gone.

To add some more information to this picture, you would want to measure the output impedance of the battery also.  I suspect that a 12-volt battery that measures 6-volts open-circuit with a multimeter would have an extremely high output impedance.  Again, this would be another indication that the battery was most likely shot.  A high output impedance is also sometimes called a "fluffy charge."  In my opinion everyone that experiments with free energy should learn how to measure the output impedance of the batteries that they work with.  This data is more important than the open-circuit battery voltage reading and compliments the battery voltage measurement.

Quote
Would it not be unreasonable to assume that when the battery is at 6 volts it is very easy for the charger to bring it up to 7 volts, but when the battery is at 11 volts, the charger will take longer to bring it up to 12 volts.

This sounds reasonable to me.  Certainly you can expect the voltage to jump up quickly and then start to slow down as you approach 12 volts.  Again, all this would be in the context of the fact that you are working with an extremely unhealthy battery.  I think we have all seen this effect when trying to recharge any type of totally dead battery.

If you look at a healthy discharged lead-acid battery, I think the lower range for the open-circuit voltage might be somewhere around 12 volts.  I will gladly be corrected by someone with more experience with this because I have almost none.  Then as you charge it it gets back to normal, about 12.6 volts.  Of course for up to 90% or more of the discharge cycle for a lead-acid battery, the open-circuit voltage and the voltage under a moderate load is the same, close to 12.6 volts.

My feeling for when people charge batteries up to 14 or 15 volts is that they are really pushing current into their batteries when the battery doesn't want it anymore, it is already fully charged.  So for lack of a better term, you are "frying the battery's brains" when you push it up to 14 or 15 volts.  You are just cooking the battery at this point, you are not even charging it.  People get excited about it and they are making a mistake here.  I will assume that there is a good chance that they are damaging their batteries.

So my gut feel when you push a battery too far like this is as follows:  The input charging impedance of the battery actually starts to increase after the battery is fully charged.  The battery is becoming overstressed when it is fully charged and you still are pushing current into it.  In effect the battery has become a big resistor in this condition and all of the energy that you are putting into it is simply being dissipated as heat internal to the battery.  Since the input charging impedance of the battery is increasing due to the extra electrical stress due to the unwanted current flow, this manifests itself as increased voltage across the terminals of the battery.  This voltage reading is a fake voltage and only exists because you are pumping current into the battery.  I believe that once you disconnect the charging, within less than an hour the voltage will go down much closer to its normal resting voltage.   So here I would agree with AC when he says that the battery resistance increases as the voltage increases.  Again, I will stress that I believe that you are outside of the normal operating "envelope" for the battery in this case and you are actually damaging it and shortening its life.

For a lead acid battery, if the open-circuit voltage is too high or two low, it's bad.  The battery chemistry is functioning normally at about 12.6 volts.  The voltage is determined by the nature of the chemical reactions that generate the voltage for each cell in the battery.  These voltages normally fall in a narrow range.  So a battery that measured 15 volts is outside of that normal range and it's a fake voltage reading.  Like I said before, you are frying the batteries brains like the brain on drugs/fried egg in a skillet cliche.

So I just shared some thoughts and inferences about batteries with you.  I don't play with 12-volt lead-acid batteries.  I would assume that the battery document that Ion uploaded (or linked to) has more accurate and valuable information.

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Thanks again for your kind reply. I understand what you are saying. There is one point though and that is that such conditions can exist with perfectly well working batteries also. I have many times brought down voltage on a 12 vdc 7ah battery down to less then 6 volts and then re-charged it back up to work again just fine. It is not only defective batteries.

Charging a 12 volt battery with a 12 volt charger, the charger is usually putting out higher voltage then the 12 vdc to overcome the batteries minimal 12vdc charge requirement.

The complications start when you are charging back to your battery with voltages in the hundreds of volts but very little amperage. Such a reactive power charging, seems like the battery then changes its internal make-up like a normal battery would have internals like liquid cement but with high reactive re-charge, its becomes more like very light whipped cream.

In general, a battery (mainly car batteries) cannot absorb more then 20% of its amperage rating. This means that if you have a 100 amp 12vdc car battery with the recharge coming from a 100 amp alternator only 20 amps will be effectively absorbed by the battery at any given time.

In looking a Bedini designs, I feel this is where some go very wrong in the re-charge side since they are using regular or deep cycles car batteries. The design has to include a method of paralleling the coil outputs in order to capitalize on the amperage issue. If you are charging your battery with anything less then 20% of the battery amperage then you are cheating yourself out of the maximum re-charge potential.

Maybe one last thing as a side note on the Bedini 10 coil design which I find very interesting. I do think however that instead of using one feed car battery and one charge car battery, such a design would work better with each coil running off its own 7ah 12vdc battery but the flyback from one coil would be sent to the next battery in the loop and so on. This way each battery would see the flyback as coming in from another source other then itself. With such a scheme, then all you have to do to prove OU is to charge all the batteries, then run the system without the re-charge and see how long it takes for the device to die down. Then with the re-charge, if the device keeps turning and turning without any evident voltage drop and the time is now 10 or 20 or 100 times longer, or non-ending, then I think this can be a safe bet of OU activity.

Sorry if all this is not so relevant.


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Wattsup:

Your comments about a method for determining OU based on some sort of battery discharge testing are totally relevant, that's the end-game for any 10-coiler or regular Bedini enthusiast.  I focused a lot on how the motor actually works, how to tune it, how to make measurements on it, so that a true keener would be able to understand what was going on "under the hood" and not just blindly say "radiant spikes" all the time.  They have to understand that the "radiant spikes" are actually inductive current spikes, and the energy in those inductive current spikes comes from the source battery and nowhere else.  If you want to you can measure how much energy is in each individual spike.  Instead of just blindly turning a trim-pot for the base resistance and tuning your motor like that, you can put your scope on the motor to tune it and make sure that it is functioning properly.  I realize that this requires some basic electronics knowledge, and it's all laid out on the table for any keeners to follow-up on if they want to.  I will repeat that any notion that a Bedini motor is "outside of conventional understanding" is ridiculous.

Yes, I forgot about the fact that a new battery could be drained to the point were it would read 7 volts.  I think that battery experts would call that an "over-drained" state for a new battery.  I would assume that it's not healthy to do that to a new battery either.

Quote
Maybe one last thing as a side note on the Bedini 10 coil design which I find very interesting. I do think however that instead of using one feed car battery and one charge car battery, such a design would work better with each coil running off its own 7ah 12vdc battery but the flyback from one coil would be sent to the next battery in the loop and so on. This way each battery would see the flyback as coming in from another source other then itself. With such a scheme, then all you have to do to prove OU is to charge all the batteries, then run the system without the re-charge and see how long it takes for the device to die down. Then with the re-charge, if the device keeps turning and turning without any evident voltage drop and the time is now 10 or 20 or 100 times longer, or non-ending, then I think this can be a safe bet of OU activity.

Your idea is interesting but it gets a bit more difficult to measure for OU.  It's still possible to do though.  You can measure what the power drain on each battery is, and what the recharging power on each battery is.  You end up with a net power drain on each battery that can be related back to the amount of energy stored in each battery to estimate a run time.  You also would have to factor in the decreasing voltage and decreasing RPMs of the motor over time.  Plus for any battery there is no abrupt cut-off point where it stops outputting power.

Personally if I myself was doing the testing I would cycle my batteries and make measurements on them until got to the point where I really 'knew' them.  Supposing that you have a battery that has a 1 mega-joule capacity.  You could charge it and discharge it to get to understand exactly how it cycles.  Then start your testing where you know with a high degree of confidence that the charging battery has exactly 200 kilo-joules worth of energy in it before you start your test.  Then with your Bedini motor pump exactly 500 kilo-joules into it.  Then discharge it completely and see how much energy is in it.  Do this at least 10 times, and then repeat for two other charging batteries.  If the battery consistently demonstrates that it is charged with more than 700 kilo-joules then you might have something interesting.  In all of this testing I am talking about making charging/discharging measurements on the charging battery only, I am ignoring all of the losses associated with the Bedini motor itself.

The reason I would start the test with 200 kilo-joules in the battery is because I want to avoid starting with a fully discharged battery because I am afraid that there could be non-linear behaviour between a fully discharged state and 200 kilo-joules.  By the same token I want to avoid fully charging the battery because I am afraid it might be non-linear between 700 kilo-joules and 1 mega-joule also.  The reason I would do the testing 10 times consecutively would be for the same reason, there might be some latent 'extra energy' in a brand new battery and multiple test cycles would hopefully flush that out.  As anecdotal evidence, I think we have all had the feeling that the first few cycles of usage with a rechargeable battery seem to contain more energy, and then after that the battery settles in and performs nominally.

The "Plan B" I would advocate was the one I mentioned to Jeff.  Charge and swap, and each time drain 25% of the energy out of the just-charged charging battery.  If there is any Bedini-Bearden "magic" going on, this should run indefinitely, even with all of the energy looses associated with the Bedini motor itself.  This would be just as valid a "brute force" method, and the only thing you have to work out ahead of time is how to discharge 25% of the energy from the charging battery, which is a trivial thing to do.

The "Plan B" "brute force" method demands that the 10-coiler setup produces so much "magic" that the COP you get from charging the battery with inductive current spikes will be big enough to overcome all of the losses associated with the 10-coiler Bedini motor itself, and the 25% energy drain that you put on the freshly charged charging battery.  This is a perfectly reasonable proposition because that's what the believers believe, that they can swap battery banks back and forth and use the "extra energy" derived from the COP to do something useful, like run an inverter to create AC mains power.

MileHigh
« Last Edit: 2010-09-29, 16:37:57 by MileHigh »
   

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Good conversation men! I can only add, from my experience designing battery chargers for a company,  that we considered 10 to 10.5 volts the dead point for a lead acid battery.  At that point 90 percent + of the charge was gone, beyond that you could damage the battery discharging it futher.  Over charging the battery to higher voltages also damaged them decreasing life time.  When charging them we designed the chargers to output between 13.8 and 14.2 volts, but what has not been talked about is that the chargers are designed to trickle down when 14 volts is reached, you do not continue to pump full charging current in at the fully charged voltage.
Battery terminal voltage will always settle back to about 12.5 volts on a good battery when disconnected from the charger.       


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As I have mentioned before, the Bhoudini devices could be laid to rest with a very simple circuit that using a DPDT relay, swaps two small lead acid batteries at a predetermined interval. I could sketch the whole circuit on the back of a folded napkin.

It would use a PUT R-C timer to clock a D-Q flip flop which would change the relay state at the user predetermined interval. You could also count the number of pulses and use that to change relay state.

I would use small capacity (1-2Ah) lead acid batteries so we would not have to wait forever for the test to run uphill or downhill. But large batteries could be used if you happen to have them on hand.

This could run unattended and the battery voltage fed to a strip chart recorder or data logger for record keeping purposes.

This is the test the Bedites will not do.

Room3327:
 Looks like you have some battery experience. I also have designed some battery chargers over the years and was lucky enough to inherit some battery test equipment and good textbooks on Lead Acid Batteries from an elderly fellow who worked at C&D Batteries in Plymouth Meeting PA.

I have in my collection a C&D ferro-resonant constant current charger, used to keep large or small banks trickle charged in series connected systems. I also obtained some AVO battery test equipment some time back.

I also have a GE ELEK-TRAK tractor fully battery operated and maintained by myself for the last 30 years. Runs on six deep discharge golf cart type batteries. Built in ferro-resonant charger. No gasoline, just plug in to the wall for an overnight charge. These are becoming quite rare and there is a collectors club online.

« Last Edit: 2010-09-30, 00:29:06 by ION »


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ION,
Yes I have some battery experience, but like most things with me it was awhile back.  If I could only remember everything I know.
As I recall ferro-resonant transformers were a pain in the rear to design so I can understand there are not many around, and I just threw out a ferro-resonant charger, 75 volts at something like 20 amps, didn't even think about it, maybe I should have hung on to it but too much junk piled up on me.  Apparently you have quite a collection of equipment, pretty nice stuff. Now if we can just figure out what to do with all the stuff we have?



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"Whatever our resources of primary energy may be in the future, we must, to be rational, obtain it without consumption of any material"  Nicola Tesla

"When bad men combine, the good must associate; else they will fall one by one, an unpitied sacrifice in a contemptible struggle."  Edmund Burke
   
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AC,

In my experience, an inductor does act very much like a constant current source during it's de-energizing cycle (i.e. during it's inductive kickback). And this is well supported by the theory of how inductors function.

.99

Poynty?  What do you mean by constant current?  If the voltage is changing surely the current is changing?  Never heard of a theory that supports a 'constant current' frankly.  I think you could probably average the current.  Unless the resistance across the coil sort of compensates?  Is that what's happening?

   
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Rosemary:

There are bench power supplies that can be configured to work as voltage sources or current sources.  One configuration tries to output a constant voltage for any variable current output and has an output impedance that approximates zero.  The other configuration tries to output a constant current for any variable voltage output and has an output impedance that approximates infinity.

In simple terms you can say that an output impedance of zero means that for any load the output voltage will remain the same.  By the same token you can say that an output impedance of infinity means that for any load the output current will remain the same.

Capacitors and inductors are energy storage devices and in a sense they can be looked at as small temporary power supplies when they are discharging their stored energy.  A capacitor acts like a constant voltage source with an instantaneous output impedance of zero and an inductor acts like a constant current source with an instantaneous output impedance of infinity.  However, both devices store a finite amount of energy and therefore cannot maintain a constant voltage or current indefinitely if they are driving a load.

For a capacitor, the voltage tries to remain constant with a variable current output.  For an inductor the current tries to remain constant with a variable voltage output.  If you examine their behaviour over a very short period of time you can say that they do maintain a constant voltage output or a constant current output.

Some experimenters might want to do their own research to learn how these components work in the time domain and also in the frequency domain.  Needless to say, there are vast information resources available online, in books, and at educational institutions.  If someone is serious about experimenting with a Bedini motor it would be extremely helpful to know this stuff.

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Well stated MH!

One might add that power supplies approximate a pure current source or pure voltage source by means of sensing and regulation circuits. This is the "tries" that MH speaks of.

In the real world, the power supply will have a finite current output capability, or will have a limited headroom of voltage available. You will find yourself "up against the stops" if you exceed these practical limits.


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It's not as complicated as it may seem...
Poynty?  What do you mean by constant current?  If the voltage is changing surely the current is changing?  

By "constant" it means that no matter what load is placed on the inductor, it will deliver the same value of amperes at the time just after the switch opens as was present just before it opened.

In a thought experiment, we will energize the inductor to the same value of current each time, just before we open the switch. For each test we load the inductor with a different value resistor. In each case the de-energizing current will be the same, i.e. "constant"; but the output voltage varies, along with the duration of the kickback.

.99


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Poynt:

Thanks for the simple explanation.  I thought that you would reply also so I decided to give a more generic response to the question ahead of time.

All:

There is enough information in this thread and the Bedini measurement thread to "bootstrap" yourself to get to the point where the information all makes sense and everything starts to fit together in your mind.  Many of the important concepts have been discussed several times over.  To bootstrap yourself you might want to take a trip to your local bookstore and buy a few books on electronics or perhaps find an electronics course online or both.  If you get up the learning curve your experimenting should be much more rewarding.

What would really be interesting at this point in time would be if someone shares their data with us after having read both threads.  The idea would be that we could interactively discuss their setups together and see what happens.  I think that's a long-shot for the experimenters in the Yahoo Bedini groups, but they are certainly welcome if they want to drop in here also.

MileHigh
   
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