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Inductive Resistor Open source development of highly efficient inductive resistor circuits.

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  #1  
Old 09-15-2010, 11:20 PM
Erik1n4007 Erik1n4007 is offline
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Distributor for pulse?

Hello and I appologize if this post is in the wrong area as I am new to this medium. Here goes my question. I have been working with the Bendini Circuit and have had no success up to the moment. I am working with some large coils and am trying to create a system of high frequency resonance but need to pulse energy in my coils that are each bifilar, about 5" in diameter toroidal iron core with 13 guage magnetic coil wire, and configured into a triangular setting. Since I have had no luck with the bendini circuit as my circuit wires end up vaporizing due to the incredibly high voltage that I am producing, I am wondering if I could use the same system that a car uses. that is Battery to ignition coil, ignition coil to a condensor (Capacitor), distributor to the coil enteries instead of the spark plugs, then a return to the negative terminal of the 12V battery (small 12V 4 amp.). If heat or high resistance is an issue but it would work, can there be a system of cooling that could be implemented? Any help would be greatly appreciated. Thank you.
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  #2  
Old 09-21-2010, 02:53 AM
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Harvey Harvey is offline
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Hi Erik,

A typical ignition is wired this way:


Battery(+) -------> IGN(+)[Primary Winding] IGN(-) -------> [Contact Points and Condensor] ------> Battery(-)

Battery(-) -------->[Secondary Winding]--------->[Distributor Center Post]----->[Distributor Rotor]---------[High Tension Cylinder Wires]------>[Spark Plug]------->Battery(-)

From this diagram two things should be noted:
1. The condensor sits across the points to minimize burning the contacts by arcing

2. The energy for the secondary path is provided solely by the field collapse of the primary winding. This should be obvious because the secondary path starts with the Battery(-) and ends with the Battery(-). Therefore all the energy is supplied magnetically in the ignition coil.

In your case you are melting wires - presumably the high voltage wires from your explanation. Wires do not melt due to voltage, but instead due to internal heating caused by forcing current through them beyond what they can tolerate. The engineering rule of thumb for copper wire is 700 circular mills of wire mass per ampere. If this is exceeded, the wire will heat up.

This is where the voltage applies: The current is the result of Voltage being applied across a resistance - in this case it would appear to be that resistance of your wires. Using #13AWG (note that British wire gauge is different here) you get exactly 2.00 Ohms per thousand feet of wire. The circular mils of that wire (5180) only supports up to 7.4 Amps without heating the wire. This means that you can only safely have 14.8V applied for every 1000 feet of wire. The formula is E = I x R where E is electromotive force (voltage) and I is intensity (current) and R is resistance.

Using a distributor is a great way to switch High Voltage, but it will not prevent the burning of wires. If you intend to use short lengths of wire that have High Voltage on them then you should have a load between them and ground to limit the current. A spark plug is a good load because the spark itself has a reasonably high resistance. But you will still want to measure the current through your secondary circuit to keep it below the rated values.

I hope that helps
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Old 09-21-2010, 08:14 PM
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Harvey Harvey is offline
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Hi Erik,

Thanks for the PM and link to your blog site - I had a quick look at your setup and tried to gauge visually the size of everything for some general calculations.

It looks like your toroid cores are about 4" in diameter and the coils that are wrapped around them are about 1/2" in diameter.

Without knowing which cores you are using or how much inductance they add to the system these are really very general overview calculations.

First, you have two coils on each core counter wound and you have 3 cores, so you have 6 coils all together. I don't know if these are in series, parallel or a combination, but the following is based on them being in parallel. Also, you have another set underneath, but the core there is not clear. So I just generalized for 8 coils.

The formula for an air core inductance is (rČNČ)/(9r + 10l)

r = 0.25
N = 32
l = 4pi = 12.56

This gives us approximately 64/(2.25 + 125.60)

Which comes to 0.50 microhenries for each coil.

We already established that your coil wire can withstand 7.4A and you are using a 12V battery. So we are looking for an impedance of about 1.6 ohms max on each coil.

There is a specific frequency that those coils will exhibit a 1.6 ohm impedance. The formula for that is f = XL / (2 pi L) where f is hertz and L is Henries. So to convert our microhenries to Henries we need to multiply it by (10^-6) or 0.000001. So it looks like f = 1.6 / (6.28 * 0.5 * 0.000001)

So your target frequency would be around 509.296 kH. That is very interesting because that is close to the same frequency that John Hutchison was playing around with when he started levitating various objects in his 500kV E-Field. I think he was using a base frequency of 456kH and then tuned his secondary oscillator to beat against that at various harmonic modes.

So if you are running around that frequency and your coil inductance is around 0.5”h and they are eight in parallel, you can expect about 60A of current flow out of your battery.

A #4AWG wire will carry 59.7 Amps. This is the size wire used in Heavy Duty jumper cables. The light duty cables are typically #6AWG which only carries 37.5A

So that gives you an idea of what you need to be using to and from the battery and what ball park frequency you should be running your rig at.

If we had a wiring diagram of how things are connected and accurate measurements as well as information regarding the core material then better numbers could be dialed in. But for now, get some jumper cables and use those for the battery connections and don't let your coils get hot, if they do you will ruin the laquer coating and short the counter windings out. so you may want to put some 8A slow blow fuses in line with each coil.

Cheers!
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Old 10-25-2011, 04:03 AM
troylee troylee is offline
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s post is in the wrong area as I am new to this medium. Here goes my question. I have been working with the Bendini Circuit and have had no success up to the moment. I am working with some large coils and am trying to create a system of high frequency resonance


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