Question

[vidbid]

This is a good question.

.

[Aaron]

Looks like Yes

That reminds me of another similar circuit in Energy Science Forum.

If C2 is a non-polarized cap, then we have 40 volts there with the positive on the left side or S4 side and the negative is on the right side of the cap.

If S4 and S6 is closed, you have the positive voltage on the S4 side flowing over the circuit and to the negative while part of that "Heaviside Flow" is diverged into the copper causing the electrons to move from the S6 side towards the S4 side.

So it looks to me like yes, that C2 is isolated from the rest of the circuit with 40 volts and is lighting the bulb in a very conventional manner - unless there is some other trick.

Looks like you will have electrons moving from S6 to to the S4 direction.

[vidbid]

That reminds me of another similar circuit in Energy Science Forum.

If C2 is a non-polarized cap, then we have 40 volts there with the positive on the left side or S4 side and the negative is on the right side of the cap.

If S4 and S6 is closed, you have the positive voltage on the S4 side flowing over the circuit and to the negative while part of that "Heaviside Flow" is diverged into the copper causing the electrons to move from the S6 side towards the S4 side.

So it looks to me like yes, that C2 is isolated from the rest of the circuit with 40 volts and is lighting the bulb in a very conventional manner - unless there is some other trick.

Looks like you will have electrons moving from S6 to to the S4 direction.

Hmm. Very interesting. That's a great way to look at it.

I'm wondering if electrons can exit the positive terminal of a charged capacitor to go through a load in order to reach the positive terminal of a capacitor with an greater charge?

Perhaps, it is (whatever it is), if one were to look at it another way, a more positive potential trying to equalize with a lower positive potential through a load.

I guess it's possible that it depends on how one looks at the question as to how one perceives the answer.

.

[mbrownn]

This is a good way of understanding electricity in its standard theory. We live in a sea of electrical charge with a potential that is unknown, all a battery, cap or generator for that matter is doing is creating two points of differing potential. (a dipole)

The known voltage of a battery is only the difference in potential between the two terminals, the battery itself may be at a potential of a million volts but without a second point of a differing potential we would never know it.

Typically we use a ground as a reference point of potential but the ground itself could also have a charge that we dont know about.

We could consider the positive terminal of C1 as the earth and the positive terminal of C3 as the ionosphere. Our circuit is C2, switches S4 and S6 and the lamp. S3 is easy to make but S4 and S5 are a little more tricky. Tesla,s patents, and the Tate power module come to mind. During lightning storms we could consider S5 being made. Now comes the dangerous part, and I dont suggest you do it. Fly a kite with a wire, I believe it was Benjamin Franklin that did that, now we have made S3. We have now got more free energy than we can handle.

Franklin, Tesla and the others were all aware of this.

Now consider a lightning conductor as our potential S3 and S5 as the resistance of the air, although high in resistance, the atmosphere surrounds the earth so over that area it should be relatively low. If the conductor was insulated from the building we would get a significant current and voltage provided that the building itself was insulated from the ground.

I could be dreaming here but look at Wardenclyffe tower, Do you see a resemblance to your circuit?

Maybe Tesla's experiments oscillating this circuit was to cause this current to flow in a predictable way.

[vidbid]

Simulation

This is a good way of understanding electricity in its standard theory. We live in a sea of electrical charge with a potential that is unknown, all a battery, cap or generator for that matter is doing is creating two points of differing potential. (a dipole)

The known voltage of a battery is only the difference in potential between the two terminals, the battery itself may be at a potential of a million volts but without a second point of a differing potential we would never know it.

Typically we use a ground as a reference point of potential but the ground itself could also have a charge that we dont know about.

We could consider the positive terminal of C1 as the earth and the positive terminal of C3 as the ionosphere. Our circuit is C2, switches S4 and S6 and the lamp. S3 is easy to make but S4 and S5 are a little more tricky. Tesla,s patents, and the Tate power module come to mind. During lightning storms we could consider S5 being made. Now comes the dangerous part, and I dont suggest you do it. Fly a kite with a wire, I believe it was Benjamin Franklin that did that, now we have made S3. We have now got more free energy than we can handle.

Franklin, Tesla and the others were all aware of this.

Now consider a lightning conductor as our potential S3 and S5 as the resistance of the air, although high in resistance, the atmosphere surrounds the earth so over that area it should be relatively low. If the conductor was insulated from the building we would get a significant current and voltage provided that the building itself was insulated from the ground.

I could be dreaming here but look at Wardenclyffe tower, Do you see a resemblance to your circuit?

Maybe Tesla's experiments oscillating this circuit was to cause this current to flow in a predictable way.

It does look like an inverted Wardenclyffe tower, doesn't it?

Actually, the inspiration for this circuit was from watching videos by Theoria Apophasis and specifically about Uncovering the Missing Secrets of Magnetism videos.




In the following video, B1 = 50v and B2 = 10v. The lamp is rated at 12v. C1 is charged to 50v, and C3 is charged to 10v. After C2 is charged by C1 and C3, the voltages across C1 = 36.67v, C2 = 13.33v, and C3 = 23.33. When the lamp circuit is isolated and run, the voltage across the lamp is 13.33v.

Please disregard the resistor values in the circuit as they are negligible and are there only because they are required in order to make the simulation work.

[VIDEO]https://www.youtube.com/watch?v=J7Xnhu8sh38[/VIDEO]

http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+50+5.0+50%0Av+96+144+144+144+0+0+40.0+50.0+0.0+0.0+0.5%0Av+416+144+368+144+0+0+40.0+10.0+0.0+0.0+0.5%0As+416+144+464+144+0+1+false%0As+320+144+368+144+0+1+false%0As+144+144+192+144+0+1+false%0As+32+144+96+144+0+1+false%0As+240+80+240+144+0+0+false%0As+272+144+272+208+0+0+false%0As+272+48+320+48+0+1+false%0As+176+48+224+48+0+1+false%0As+224+288+304+288+0+0+false%0Ac+240+48+272+48+0+1.0+9.981021581343624E-4%0Aw+240+48+224+48+0%0Ac+96+48+176+48+0+1.0+0.0010000000000000007%0Ac+416+48+320+48+0+1.0+0.0010000000000000007%0Aw+96+48+32+48+0%0Aw+32+144+32+288+0%0Aw+32+288+224+288+0%0Aw+464+288+304+288+0%0Aw+464+288+464+144+0%0Aw+464+48+416+48+0%0Aw+192+96+192+144+0%0Aw+176+96+192+96+0%0Aw+320+96+320+144+0%0Aw+240+48+240+80+0%0Aw+272+96+272+48+0%0Aw+272+96+272+144+0%0Aw+272+240+272+208+0%0Aw+272+240+240+240+0%0A181+240+208+240+144+0+300.00001187427813+1.0+12.0+0.4+0.4%0Aw+240+208+240+240+0%0Ar+176+96+176+48+0+0.001%0Ar+320+96+320+48+0+0.001%0Ar+464+144+464+48+0+5.0E-4%0Ar+32+144+32+48+0+5.0E-4%0Ao+13+64+0+35+0.0048828125+9.765625E-5+0+-1%0Ao+11+64+0+35+0.0048828125+7.8125E-4+1+-1%0Ao+14+64+0+35+0.0048828125+9.765625E-5+2+-1%0A

.

[Aaron]

potential differences

I'm wondering if electrons can exit the positive terminal of a charged capacitor to go through a load in order to reach the positive terminal of a capacitor with an greater charge?

Yes.

Bedini has shown this with batteries for almost 20 years on his old website.

Battery 3 (in bottom left image) is getting charged so positive potential from the 2 batteries in series goes from the two batteries so the current moves from the single battery towards the 2 batteries. It is the same principle - splitting the positive. It isn't about polarity, it is about potential difference as mbrown is mentioning.

As a side note, I don't believe caps are filling up with electrons as the conventional belief states - just like batteries are not. Internal charges are separated so that they polarize the aether and the aether moves to the terminals, goes over the wires, and the electrons actually come from the wire itself.

[vidbid]

Question Part 2

Yes.

Bedini has shown this with batteries for almost 20 years on his old website.

Battery 3 (in bottom left image) is getting charged so positive potential from the 2 batteries in series goes from the two batteries so the current moves from the single battery towards the 2 batteries. It is the same principle - splitting the positive. It isn't about polarity, it is about potential difference as mbrown is mentioning.

As a side note, I don't believe caps are filling up with electrons as the conventional belief states - just like batteries are not. Internal charges are separated so that they polarize the aether and the aether moves to the terminals, goes over the wires, and the electrons actually come from the wire itself.

Thanks for a very thoughtful response.

I suppose the question involves differing quantities of charge, charge polarization, and the potentials, and how they affect each other.

I've made a new video.

[VIDEO]https://www.youtube.com/watch?v=fX3PM9DbFZQ[/VIDEO]

C1 = 50v, C2 = 0v, and C3 = 10v.

By the way, C1 + C3 = 60v. (I'm not making any claims here.)

Then..

C1 = 36.67v, C2 = 13.33v, and C3 = 23.33v.

By the way, C1 + C3 = 60v. (I'm not making any claims here.)

But now, C2 = 13.33v.

How do we account for the 13.33v in C2, but that's from this simulator?

Someone would have to do the experiment to see what really happens in real life.

.

[vidbid]

Charge Redistribution

[VIDEO]https://www.youtube.com/watch?v=Bl0ncO5b0IA[/VIDEO]

Third video in the series but this time to make sure of the voltages across the caps.

Each capacitor has a capacitance of 1 F.

Starting off with C1 = 50v, C2 = 0v, and C3 = 10v.

C1 = Charge Q = 50 Coulombs; and Energy E = 1250 Joules.

C2 = Charge Q = 0 Coulombs; and Energy E = 0 Joules.

C3 = Charge Q = 10 Coulombs; Energy E = 50 Joules.

For what it's worth, C1 [50v] + C3 [10v] = 60v.

After connection, then..

C1 = 36.67v, C2 = 13.33v, and C3 = 23.33v.

By the way, C1 [36.67v] + C3 [23.33v] = 60v.

Also, C1 [36.67v] + C2 [13.33v] = 50v.

And, C3 [23.33] - C2 [13.33v] = 10v.

C1 = Charge Q = 36.67 Coulombs; Energy E = 672.344 Joules.

C2 = Charge Q = 13.33 Coulombs; Energy E = 88.844 Joules.

C3 = Charge Q = 23.33 Coulombs; Energy E = 272.144 Joules.

http://www.electronics2000.co.uk/cal...calculator.php

http://hyperphysics.phy-astr.gsu.edu...ric/capac.html

.

[Aaron]

splittting the negative

It won't be as clean cut as I mentioned - just stating the basic concepts. It's identical in Bedini's 3 battery concept, the Gray motor and the plasma ignition... all higher voltage positive coming in contact with a lower voltage positive and they share a common ground.

The 50v cap can charge the 10 volt cap up a bit during this process so it goes down a bit and the one on the right goes up a bit - in reality, will probably be closer to what your simulator shows.

Yeah, just have to do the experiment.

---------------------------------------------

I'm working on a circuit that works on the differential between two negatives with a common positive so is like splitting the negative - just the opposite of the splitting the positive. Looking at the flow moving over the wires as a gas under pressure, positive voltage is a pressure and negative voltage is a suction, literally. So if you have a difference between two suctions, you have a potential difference even though it is negative with a positive current - meaning what is between those points is a cold current since no positive voltage multiplied by a positive current is negative watts.

I've done the basic tests in the past and had a validation that someone took this quite far. I'm doing a test right now and if I can show that this is what is between these points and can power a load without the load being reflected back to the front, then it just might be the simplest demonstration of real "cold electricity" that I've seen. A load between these points should drop below ambient temperature.

Anyway, fingers crossed - if my basic demo unit coming together demonstrates something interesting, then I'll be presenting this at the conference. Even if it doesn't do as expected, I might present it anyway because the switching method is something that hardly anyone has ever seen and it is very, very simple.

[vidbid]

Good Luck!

It won't be as clean cut as I mentioned - just stating the basic concepts. It's identical in Bedini's 3 battery concept, the Gray motor and the plasma ignition... all higher voltage positive coming in contact with a lower voltage positive and they share a common ground.

The 50v cap can charge the 10 volt cap up a bit during this process so it goes down a bit and the one on the right goes up a bit - in reality, will probably be closer to what your simulator shows.

Yeah, just have to do the experiment.

---------------------------------------------

I'm working on a circuit that works on the differential between two negatives with a common positive so is like splitting the negative - just the opposite of the splitting the positive. Looking at the flow moving over the wires as a gas under pressure, positive voltage is a pressure and negative voltage is a suction, literally. So if you have a difference between two suctions, you have a potential difference even though it is negative with a positive current - meaning what is between those points is a cold current since no positive voltage multiplied by a positive current is negative watts.

I've done the basic tests in the past and had a validation that someone took this quite far. I'm doing a test right now and if I can show that this is what is between these points and can power a load without the load being reflected back to the front, then it just might be the simplest demonstration of real "cold electricity" that I've seen. A load between these points should drop below ambient temperature.

Anyway, fingers crossed - if my basic demo unit coming together demonstrates something interesting, then I'll be presenting this at the conference. Even if it doesn't do as expected, I might present it anyway because the switching method is something that hardly anyone has ever seen and it is very, very simple.

I'm still studying this whatever you want to call it.

With respect to what you're doing, I wish you abundant success.

I have modified the circuit. It's more simple. By the way, I'm not making any claims about anything.



[VIDEO]https://www.youtube.com/watch?v=J80Y5qguoE8[/VIDEO]

.

[Aaron]

similar concept

Check out: something i do not understand about energy
Look at Dave Wing's post - talking about a similar concept.

[vidbid]

Putting the Genie back in the Bottle

Check out: something i do not understand about energy
Look at Dave Wing's post - talking about a similar concept.

Awesome, Aaron. I'll be sure to check it out. Thanks.

A quick post before I get back to my studies.

So how do we divide the voltages evenly, I wondered.

http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+50+5.0+50%0Ac+112+144+240+144+0+1.0+9.999999999999601E-4%0Ac+432+96+512+96+0+0.5+0.0010000000002973568%0Ac+432+192+512+192+0+0.5+0.0010000000002973568%0Ar+384+96+432+96+0+1.0E-9%0Ar+384+192+432+192+0+1.0E-9%0Ar+512+96+560+96+0+1.0E-9%0Ar+512+192+560+192+0+1.0E-9%0Ar+240+144+288+144+0+1.0E-9%0Ar+112+144+64+144+0+1.0E-9%0As+288+144+384+144+0+1+false%0Aw+384+96+384+144+0%0Aw+384+192+384+144+0%0Aw+560+96+560+144+0%0Aw+560+192+560+144+0%0Av+128+80+224+80+0+0+40.0+50.0+0.0+0.0+0.5%0As+224+80+288+80+0+1+false%0As+64+80+128+80+0+1+false%0Aw+64+144+64+80+0%0Aw+288+144+288+80+0%0Aw+64+144+32+144+0%0Aw+32+144+32+224+0%0Aw+32+224+608+224+0%0Aw+608+224+608+144+0%0Aw+608+144+560+144+0%0Ao+0+64+0+35+0.0048828125+9.765625E-5+0+-1%0Ao+1+64+0+35+0.0048828125+9.765625E-5+1+-1%0Ao+2+64+0+35+0.0048828125+9.765625E-5+2+-1%0A

I realize that this Java program won't play on all browsers. Sorry about that.



[VIDEO]https://www.youtube.com/watch?v=Wc7VQVANlFk[/VIDEO]

There it is. It has to do with capacitor arrangements.

Reference:

1. http://hyperphysics.phy-astr.gsu.edu...ric/capac.html

2. http://hyperphysics.phy-astr.gsu.edu.../capac.html#c2

3. http://hyperphysics.phy-astr.gsu.edu...capcon.html#c1

.

[Matthew Jones]

I'm working on a circuit that works on the differential between two negatives with a common positive so is like splitting the negative - just the opposite of the splitting the positive. Looking at the flow moving over the wires as a gas under pressure, positive voltage is a pressure and negative voltage is a suction, literally. So if you have a difference between two suctions, you have a potential difference even though it is negative with a positive current - meaning what is between those points is a cold current since no positive voltage multiplied by a positive current is negative watts.

So not to beat your dreams or anything, negative side loads do a lot of funny things. Often they appear to do some very wonderful things, like handle big loads with very minuet or NO amounts of loss. Until you start to realize whats really happening. That is when your water turns black.
By running on the negatives you are puling the lead from the negative plates and oxidizing it.
I don't know how big your loads are or if your even using LAB's but its something you should watch for. It is not a good thing as you have no option for recovery of the battery. Monopoles, Alum, ect... The battery is just recked.
It does take while though. So back to the point, Who know where the power may be coming from when you are essentially stripping the raw elements out and burning them.

Matt

[Aaron]

between the negatives

So not to beat your dreams or anything, negative side loads do a lot of funny things. Often they appear to do some very wonderful things, like handle big loads with very minuet or NO amounts of loss. Until you start to realize whats really happening. That is when your water turns black.
By running on the negatives you are puling the lead from the negative plates and oxidizing it.
I don't know how big your loads are or if your even using LAB's but its something you should watch for. It is not a good thing as you have no option for recovery of the battery. Monopoles, Alum, ect... The battery is just recked.
It does take while though. So back to the point, Who know where the power may be coming from when you are essentially stripping the raw elements out and burning them.

Matt

That is possible and time will tell - the switching method is something I haven't seen anyone discuss so I'm anxious to see the results.