Bedini Ferris Wheel Regauging Motor

[genessc]

I understand that my use of 1Joule was arbitrary and not tied to any of the circuit componentry. The 1 Joule was just taken as an example of charge in any sized cap at any volt/uF format.

1 Joule is 1 watt delivered for 1 second of time. Regardless of any circuit component, this is the case.

I believe I stated the conditions to have 10 watts in 1 second correctly, didn't I? The only thing that changes is the overall circuit impedance and you end up with more power or less power, from that same 1 Joule of power, right?

I don't understand how the change of circuit impedance to .1 ohm with all other circuit aspects the same as noted in your example doesn't automatically make the 1 watt for 1 second now 10 watts for 1 second.

What factors/aspects am I missing?

Oh nevermind. I see... the manipulation then is in Time. However its not necessarily tied to the DUTY Cycle is it? It is because the full charge of 1 Joule was discharged in .1 second... but then is the .1 second the same as the .1ohm resistance? Is resistance time?

Could not the same thing be achieved by just lowering the circuit impedance to .1ohms while retaining the 1 second ON duty for the discharge? Wouldn't the result be that the 10watts was delivered instantly at turn ON and slowly tapered down sorta like the 5Time Constants idea? Or is the loss evinced because the Duty was on 9/10s of a second more than it needed to be?

Sorry for the questions, just trying to make sure I understand the concept.
Gene

Something isn't adding up for me.

Thanks,
Gene

Hi Gene,

In my answer the example of 1J was not assigned to any impedance or current value. You cannot assume if the impedance is reduced to .1 Ohm then we have 10 times the watts for the same 1 second, please forget it. You can have 10 times the power i.e. 10W if you take it out from the cap in .1 second, ok? Manipulation of Watt times second.

Gyula

[gyula]

Hi Gene,

In my answer the example of 1J was not assigned to any impedance or current value. You cannot assume if the impedance is reduced to .1 Ohm then we have 10 times the watts for the same 1 second, please forget it. You can have 10 times the power i.e. 10W if you take it out from the cap in .1 second, ok? Manipulation of Watt times second.

Gyula

Hi Gyula,

Thanks for the reply. Given your example then, if 1J is 1 watt of power discharged for a period of 1 second based on a cumulative circuit impedance of 1ohm, and with the only change to the circuit being the cumulative impedance being reduced to .1ohm and now we have 10 times the watts for the same 1 second of ON time, is that correct?
...snip...

[genessc]

Hi Gyula,

Thanks for the reply. Given your example then, if 1J is 1 watt of power discharged for a period of 1 second based on a cumulative circuit impedance of 1ohm, and with the only change to the circuit being the cumulative impedance being reduced to .1ohm and now we have 10 times the watts for the same 1 second of ON time, is that correct?

Now if we change the cumulative impedance to be .00023 ohms (or for ease of math, lets just say .001 ohms) and everything else stays the same, we now have 1000watts output for that 1 second of ON time?

I know we don't use such a slow duty to allow for 1 second of ON time and 1 second of OFF time, but this facilitates an explanation of discharging of caps into a cumulative circuit impedance. (tho some of JB's circuits run at about 3hz or so)

Is there a concept of Current "Density" with capacitive discharge? In some cap discharge circuits one notes a small capacity of 3-10uF and this capacity charges rapidly to a couple hundred volts and is then discharged into a 12volt battery. This happens pretty rapidly in terms of frequency because a small cap climbs to higher voltages faster than a large cap. So the benefit to using a small cap is to get high volt pulses at a relatively high rate of repetition. However this small caps discharges have a "lack of current density" as I've termed it. There is not sufficient Current in phase with the voltage discharge to do work.

Then I can switch that 10uF cap out for a 2400uF cap and this one takes longer to charge up, but the discharge pulses have a much more significant current "density" in their capacity to do work. The larger cap reaches the same max charge voltage but is offset in the extended time it takes to reach that peak voltage. More often this cap is regulated to only charge as JB has noted to double the voltage which keeps the frequency of discharge higher than if the larger cap was allowed to fill to the peak possible voltage.

In summation it seems to really charge batteries well one needs to use larger 10,000uF or so sized caps as this is what is doing the direct conversion to the current "density". The frequency drops but the power per discharge is increased significantly given the cophased volts/current per discharge from the large 10,000uF cap versus the small 10uF cap.

Is this observation inline with others on here?

Thanks again for the reply.
Gene

The Joule figure means Wattsecond, so 1J energy means 1Watt power under 1 second, OR ANY combination of two numbers that give 1J when multiplied together. So taking out 1J energy from a capacitor (that stores 1J) under 0.1 second means that you can take out 10W (10W*0.1s=1Ws=1J).
And you can take out this with a switch that is controlled with a given duty cycle, what count is how long one ON time of the switch lasts and how many ON times you let occur within the given total time.
Of course, the amplitude of the current you take out with the controlled switch depends on the total impedance the closed circuit lets develop, including the summed DC resistances of the switch and the capacitor's series resistance and any AC impedance also in series like the combined DC - AC impedance part of the battery, wires etc.

[gyula]

Thanks for catching my mistake. I think that I calculated that more than once and still posted flawed math.

I think we're okay now with the energy dumps into the batteries.

Dave

Ok Dave, finally Vissie got the answer this is what counts.

Gyula

[gyula]

Hi Gene,

The Joule figure means Wattsecond, so 1J energy means 1Watt power under 1 second, OR ANY combination of two numbers that give 1J when multiplied together. So taking out 1J energy from a capacitor (that stores 1J) under 0.1 second means that you can take out 10W (10W*0.1s=1Ws=1J).
And you can take out this with a switch that is controlled with a given duty cycle, what count is how long one ON time of the switch lasts and how many ON times you let occur within the given total time.
Of course, the amplitude of the current you take out with the controlled switch depends on the total impedance the closed circuit lets develop, including the summed DC resistances of the switch and the capacitor's series resistance and any AC impedance also in series like the combined DC - AC impedance part of the battery, wires etc.

So your first question below is answered partly above and partly by John B in the meantime. I cannot add much to it.

Your second question below is answered by you: it is correct the 6.5V is able to drive higher current through 0.1 Ohm than through 10 Ohm (not neccessarily in a linear fashion because it depends also on the 6.5V source inner impedance or any other series nonlinearity).

Your recent question: "Whats the typical resistance inside a biased "ON" BJT MJL21194?" depends on the transistor's saturation resistance which comes from the instantaneous saturation voltage (across collector-emitter) devided by the instantaneous collector current. Study Figure 10 in Page 4 for the Vce(sat) voltage how it changes in the function of the collector current (under condition of Ic/Ib=10) and make the divisions. It is non-linear, so one more reason to select transistors for Beta (hFE) when paralleling them.
Here is the data sheet I refer to: http://www.datasheetcatalog.org/data...a/MJL21193.pdf

rgds, Gyula

Can that Joule figure be accounted for in terms of a current pulse? Meaning what is the .338J of energy equivalent to in terms of the pulse of current being delivered into the battery? The volt difference is 6.5volts...

I suppose this is why its good to keep the discharge path at as low of a resistance as possible, preferably sub-1ohm to make the actual cohered current pulse larger than it would otherwise be with a higher resistance discharge path? 6.5volts pulsed over .1ohms "should" deliver more current than the same 6.5volts pulsed over 10ohms, right?

Of course the battery looks like a high impedance load to current impulses.

Trying to make sense of it still...

Gene

[genessc]

Thanks John B. That makes sense then that the instantaneous amps at "switch on" of the discharge pulse wouldn't register on the meter just cuz the meters are to slow.

Does the instantaneous amps still figure into the resistance on the discharge path to the charging battery? What I mean is, does the discharge path to the battery need to be like .001 ohms or lower so that this instantaneous amp figure can be as big as possible?

Whats the typical resistance inside a biased "ON" BJT MJL21194? I suppose as with diodes, you can just parallel more transistors, presuming they beta match, to bring down the overall resistance in the discharge path. (can you do the same thing with an SCR? )

Thanks again man,
Gene

I guess I should direct this at all of you calculating Joules, you must take into account the instionious peak amps it's what the meters do not catch. When you hit the battery with that the Ions are forced to move, the battery will boil very quick (Cold Boil). I have seen the peak amps reach 600 amps. The battery when discharged still has a very low impedance as the charge rises the impedance lowers to around .00023 Ohms, this is just a ballpark figure (Battery Bible). So re-calculate with the correct impedance. Kickoffs law explains the rest where everything is in the charging system.

With a sulphated battery the impedance is extremely high and no power is developed in the system just potential. ( SCR will not work here, it will lock on). The battery under this condition takes a long time to become useable as the sulphation dissolves from the plates. This takes constant use of charging and discharging. Pulse Charging even at a low rate allows the chemical in the battery to absorb the charge, it does this with time and a rest period in-between pulses, "Time Charge". Again I will say that the machine must be tuned to produce double the battery voltage or very close to it, do this with the duty cycle. Some of you have big magnets on the machine timing wheel. My wheel is 16 feet high and has 1 inch magnets, the smaller machines only needs 3/8 magnets this allows the correct space for the circuit to shut off correctly.

Gel cell batteries have a much higher impedance built into them if you boil these they just dry out and become useless ( old ones). But don't believe me read the Battery Bible it will explain what is going on. People have posted that under SG groups. Most chargers are only 3 volts over the battery, but then look at what the transformer in the charger must put out at 10 or 15 amperes to do this. It's all about impedance and voltage.
Hope This Helps.
John B

[John_Bedini]

Bedini GT3

I guess I should direct this at all of you calculating Joules, you must take into account the instionious peak amps it's what the meters do not catch. When you hit the battery with that the Ions are forced to move, the battery will boil very quick (Cold Boil). I have seen the peak amps reach 600 amps. The battery when discharged still has a very low impedance as the charge rises the impedance lowers to around .00023 Ohms, this is just a ballpark figure (Battery Bible). So re-calculate with the correct impedance. Kickoffs law explains the rest where everything is in the charging system.

With a sulphated battery the impedance is extremely high and no power is developed in the system just potential. ( SCR will not work here, it will lock on). The battery under this condition takes a long time to become useable as the sulphation dissolves from the plates. This takes constant use of charging and discharging. Pulse Charging even at a low rate allows the chemical in the battery to absorb the charge, it does this with time and a rest period in-between pulses, "Time Charge". Again I will say that the machine must be tuned to produce double the battery voltage or very close to it, do this with the duty cycle. Some of you have big magnets on the machine timing wheel. My wheel is 16 feet high and has 1 inch magnets, the smaller machines only needs 3/8 magnets this allows the correct space for the circuit to shut off correctly.

Gel cell batteries have a much higher impedance built into them if you boil these they just dry out and become useless ( old ones). But don't believe me read the Battery Bible it will explain what is going on. People have posted that under SG groups. Most chargers are only 3 volts over the battery, but then look at what the transformer in the charger must put out at 10 or 15 amperes to do this. It's all about impedance and voltage.
Hope This Helps.
John B

[Web000x]

Hi Dave,

Yes, you are correct, energy differences should be considered in calculating the remaining energy in the capacitor. Thanks for pointing to my mistake.

Applying the formula for the energy differences I highlighted your math: the energy in the cap when it has got the 36V is (.5*.016*36^2)=10.4J and you somehow mixed it up with the difference, 3.7J?

So (.5*.016*42^2) - (.5*.016*36^2) = 14.1J - 10.4J = 3.7J this is the energy taken out from the capacitor and the remaining 36V in the capacitor represents 10.4J. Is this ok now?

Thanks, Gyula

Thanks for catching my mistake. I think that I calculated that more than once and still posted flawed math.

I think we're okay now with the energy dumps into the batteries.

Dave

[genessc]

Can that Joule figure be accounted for in terms of a current pulse? Meaning what is the .338J of energy equivalent to in terms of the pulse of current being delivered into the battery? The volt difference is 6.5volts...

I suppose this is why its good to keep the discharge path at as low of a resistance as possible, preferably sub-1ohm to make the actual cohered current pulse larger than it would otherwise be with a higher resistance discharge path? 6.5volts pulsed over .1ohms "should" deliver more current than the same 6.5volts pulsed over 10ohms, right?

Of course the battery looks like a high impedance load to current impulses.

Trying to make sense of it still...

Gene

I think we have to consider only the voltage difference i.e. 42.5V-36V=6.5V
So the energy taken out from the charged cap is (0.016* 6.5*6.5)/2=0.338J
Much of the energy stays inside the capacitor.

Gyula

[FRC]

I woke up and couldn't sleep wondering about some of this. My thoughts have been about PN junctions especially in LED's relating to radiant. Some of my testing has been interesting and I'll post more on that elsewhere. but I was following the SSR cap pulser circuits and John's efforts in this part of the circuit. I found this SSR
KD20C40AX: KYOTO ELECTRIC: Electromechanical

But when I ran across this quote it answered some of the things that were troubling me about the cap pulser as a battery charger for the very reasons described here by John B. the SG always was superior in my testing. I am just wondering if anyone else has considered placing a monopole master/slave on the output of the 3GT to charge the batteries?
Or Tesla switch/solar tracker? then with Bit's swapper, this thing would run a
very very long time wouldn't it?

Les

Yes, does anyone know if there are gains achieved with this type or other types of multiple SSG's in series like this ? It does appear that there might
be.

FRC

[Rl2003]

Great work R12003, do you have all 3 coils working on that last video? And if so ,are they all wired with the start of each coil together and ends together? Just curios.
@ minoly, I have never done anything but put water in it. It has been sitting on the shelf in the closet for 5 or 6 months, I put about a cup of water in it about 5 minutes before I took the picture. Good luck at the shop


Brian

Thanks, Yes and yes they are. Now I am just learning and watching how
this is all working as one.

I may have to adjust the timing a little still, and a new timing wheel.
There is a small deflection in it, causing it to roll away from the hall on
one of the magnets. I can only see or hear it when I allow the wheel
to spin very slowly.

Mark P.

[gyula]

Would we not have to take the differences in the energy since the energy storage is not linear?
I have noticed that while watching the voltage on my capacitors with a scope or a multimeter that they always charge to the next step in voltage approximately the same. What I mean is 1-5 volts happens very quickly, 5-10 volts happens quick, 10-15 volts is just a little slower, 15-20 volts is even slower that that, and so forth. This can be attributed to the equation CapVoltage=Sourcevoltage*(1-e^-t/RC).

While watching the cap dumps on any of my Bedini variations, the cap dumps and it recharges to dump again. While it recharges, the time that it takes to go from 12v to 20v is the same as it would have been if I had charged it from 0v and started timing as soon as it got to 12v and let it charge to 20v.

This non linear condition suggests to me that we need to take the differences in the result of the capacitor's energy storage equation which is E=.5*C*V^2.

For John Bedini's capacitor, it would be (.5*.016*42^2) - (.5*.016*36^2) = Capacitor Energy. This equates to approx. 14.1 Joules - 3.7 Joules = 10.4 Joules per capacitor dump.

Hope this helps,

Dave

Hi Dave,

Yes, you are correct, energy differences should be considered in calculating the remaining energy in the capacitor. Thanks for pointing to my mistake.

Applying the formula for the energy differences I highlighted your math: the energy in the cap when it has got the 36V is (.5*.016*36^2)=10.4J and you somehow mixed it up with the difference, 3.7J?

So (.5*.016*42^2) - (.5*.016*36^2) = 14.1J - 10.4J = 3.7J this is the energy taken out from the capacitor and the remaining 36V in the capacitor represents 10.4J. Is this ok now?

Thanks, Gyula

[Les_K]

Kcarring, FRC

I thought I would answer some questions,

First Welcome.

The reason for the low current is allow for the highest possible Radiant Energy that is found in all space that surrounds you, some call it, Radiant, negative energy, special charges and so on, but it's all the same thing. This energy is scattered vectors pointing everywhere thought-out space.

To gather this energy you must build a pump for it. The Coils seem to fit the bill here as you can pump them. It does not take allot of power to do this just potential charge. Most batteries developed currents inside them, The first effect of this is the potentializing of the ions this causes the Ions to move the other direction as it does not take current to do this, just potential (Voltage). The Bedini GT3 machine is much different as the energy is stored in a capacitor to be discharged when the voltage is at max potential ( this is real current in amperes) to the second battery bank.

The machine does not use allot of current to do this as it is triggered out of the coil that drives the rotation of the wheel. The wheel must just rotate to cause the switching to gather this energy. The original SG machine is self triggered and is a fixed timing so that you can see the effect of Negative energy in the spike, that is then directed to the secondary battery to perform charging with impulse spikes, the power is in the spike to move the Ions in the battery.

The spike is self adjusting to the battery being used as the rotor changes speed to adjust for the battery. This is known as time charging. The SG can be looked at as a self rotating Magneto Energizer, high voltage output. Some of the effects of this is desulphation of bad batteries as it will bring them back to life if the battery is not shorted. You could look at this like a window that is covered with dirt, not a good view. The process cleans the window, now Ions can move in charge mode as the battery develops it's own current inside to charge. "Hope This Helps"

John B

I woke up and couldn't sleep wondering about some of this. My thoughts have been about PN junctions especially in LED's relating to radiant. Some of my testing has been interesting and I'll post more on that elsewhere. but I was following the SSR cap pulser circuits and John's efforts in this part of the circuit. I found this SSR
KD20C40AX: KYOTO ELECTRIC: Electromechanical

But when I ran across this quote it answered some of the things that were troubling me about the cap pulser as a battery charger for the very reasons described here by John B. the SG always was superior in my testing. I am just wondering if anyone else has considered placing a monopole master/slave on the output of the 3GT to charge the batteries?
Or Tesla switch/solar tracker? then with Bit's swapper, this thing would run a very very long time wouldn't it?



Les

[Web000x]

Even when I scale down to the same input as before (19v @ 700mA) when I adjust my PIC pot, better charging takes place. I'm using the two slave coils off of one switch and the two center coil power strands off of one switch, I still cannot get the SCR to unlatch for a pulsed cap dump. There is enough "electron current" to keep any capacitor pulsing from happening.

More switches means better charging. Don't hold back.

Dave

[Web000x]

I have been rather reserved with my posting on what is going on with my replication, but I feel that a relevant point has come up to force me out of my shell.

I assumed that all of the windings were paralleled in the literal sense of the word. After John Bedini posted(http://www.energeticforum.com/renewa...tml#post127253) that the windings were run off of different transistors but still paralleled, I hooked up my second half bedini/cole bipolar circuit that I had constructed on a prototype board.

I had my setup running off of one half bipolar circuit and it was charging my 7800 uF capacitor bank rather weakly with 19v @ 700mA (I am running off of my variac from 120VAC with a bridge rectifier and a smoothing cap). The cap dump was about every second @ 20v to a 12v battery.

When I was in the midst of switching my setup to multiple switches, I tried something different. I just used my center coil (4 wires @ 2 x 16 AWG and 2 x 23 AWG) with both of the 16 AWG paralleled to one bipolar circuit without any slave coils, and to my surprise, the LED on the cap dump circuit (using PG 43 out of FEG book) started blinking even faster. The meter and the scope showed that there was more energy transfer taking place towards the cap.

It seems that the radiant energy doesn't like being added in full parallel and that it must be taken off of multiple different switches just like JB has been telling us.

Now that I have the two slave coils running in parallel in the to one half bipolar switch and the two power wires on the main coil in parallel to one half bipolar switch, the charging is off of the scale comparatively.

I am getting to the point where I have to physically slow the rotor down to allow the cap dump circuit to work properly without going into latch mode (I think this might also be JB's problem with the circuit in FEG).

I will more than likely be adding generator coils to slow the rotor down while feeding the excess energy to the charging or primary batteries. Also, tomorrow I think I'll be making two more half bipolar circuits so that each power coil has its own switch.

So the whole idea of this post is to get more people using individual circuits for each power strand to see better charging, cooler transistors, and whatever other goodies surface.

Attached are some photos of my setup.

Hope to be of assistance,

Dave

P.S. I am not using my timing wheel as you see in the photos. I have been using a PIC16F690 Chip to trigger the hall and run pulse signal back through an optocoupler. I have an analog voltage function that is able to control my pulse duration depending on where I set the potentiometer.