Transcription
New Energy Science Part One. The Walton voltage multiplier equals massive gains, or not.
Good day, folks. I hope you're all doing well today. It's been a while since I've actually been behind the camera here to do some real hands-on discussion. And I wanted to talk to you about something. And this might be one of the most important videos that I've done. I know I've done a lot of them, and they're all really good, that's for sure. But to actually get right down to the crunch, to try and help you folks understand basically where I come from and how, basically many years ago, something gave me a riddle to say, "Hey, you have to check out something here. There's something more to all of this."
And essentially, it started, folks, a little bit of backstory on all of this here before you go, "Oh, this is all boring toy stuff." There's a perspective I want to share first here. Believe it or not, in high school here in Canada, Canada, Ontario, in grade nine science, we actually had a brief unit on electronics. And in there, we learned basic RC kind of series-parallel kind of circuit, to know what the electronic blocks and the analog lamps and the big, uh, 1-ohm, 5-watt block resistors and that sort of thing, and the analog meter. And we learned how to read and draw basic schematics and how to divide and whatnot the circuit, and basically, um, Ohm's law by heart and the basics of it, anyways. And, um, that got me thinking when we got into, um, series circuits, DC series, basic high school stuff, right? So I had the idea that, okay, so if a series system can put out a lot of power, maybe why not? Remember, this was high school stuff. You know, based on essentially when you have a battery in a DC series, many batteries, the voltage potential goes up, but you're limited at the basically the lowest of them all in current. So, if they're all equal, you'll get the same current output, but at that voltage level, which is really normal stuff, right? But it got me thinking, and here it is, right here.
This thing here, which is the Walton device, the multiplier on its own. And before you all laugh and say, "Joel, this is just a toy. You should know better by now. You're not impressing us. This trades current for voltage. You've got super high voltage but very low current. It's just a toy. It's a proof of concept. You should move on and not spread this information." Right? That's what the traditional engineer, even my high school teacher, essentially would say. Even the suggestion that this thing essentially, folks, hides a riddle. And I'm just wondering if you guys, for a moment, would know what that riddle is, 'cause it did spark, no pun intended, and not only an interest but a question I had. Why can't all these capacitors add up somehow and give you more current on its output?
So essentially, here's the idea. So essentially, back in the day, what I thought originally, the bright idea, without knowing the big picture, essentially, was, what if we have AC? My, um, markers kind of suck here. We have AC 110 volts. Let's say, I don't know, 1 amp. One amp. And, uh, let's say that's 100 watts. So I'm talking here, if you want to, like, do it with the mains live, right? And you want it to... I'm thinking big here, all of a sudden, right? You know, people use this as a toy usually, but I'm thinking, let's go big. Well, you know, from the AC, not voltage-wise, but anyways, we've got watts here. So, essentially, the idea was, let's have the output be a stack. You know, with my reasoning, it would have been like, with the multiplier, let's say we're at two kilovolts out once we pass this into whatever stage multiplier we're going to get. But now, let's say the input swings from 0.5 to, um, say, 1 amp over here. All of a sudden, this would be assuming we have a perfect series, right? This is, this was, this was my logic at first. This is what kind of got me angry. This is basically wrong, but this is what the intuitive idea that led me to it. Okay. So, my reasoning was, we're going to get like massive, like one kilowatt or more out of this setup, right? And to me, it... don't laugh. You have to start somewhere. But I want to make it so everyone can learn from my mistake and where it led to stubbornness into trying to reverse engineer the solution.
So now, of course, you know, this quote-unquote gain that I had in my mind that I had made up would have came from the stacking, right? It would have came from the stacking. So if you've got a series discharge current, same with all caps, right? You know, but obviously, it's not as easy as that. But to me, it was like, this is what came to me, and it was like, now I was trying to figure out why it doesn't actually work that way, which is what I'm going to get into here. So basically, therefore, the output watts, the input watts, even if there'd be such a, um, massive loss that say 30%, and all of your other losses should still make up for it with this logic, right? But this doesn't work. And the reason it doesn't work is we have to look back into the original circuit diagram of the Walton multiplier here.
So, a little bit of background on how this actually works. During one half of the cycle, the capacitors all charge up, and during the other half of the cycle, there's no conductivity, and there is an output. Now, of course, if this is unloaded, it simply no longer conducts during the negative cycle and then just keeps charging, and you ramp it up. But essentially, what happens is your AC input over here and your high voltage out there. But the way this works with the sine wave, only half, and only charges, and one by one, by the way, which is very important, because I was thinking it must be a parallel capacitive system, and it's not. It charges them one by one, and then when it discharges them, it discharges them as a series capacitive system. But the issue with that is the capacitance drops drastically. It divides on itself when there's a series capacitance. So once you essentially account for that, which I'm going to show you in a moment here, why essentially my previous thought is wrong.
So essentially, what goes on here, this would be your positive output at DC, and this would be your negative, right here, at DC, shared with the AC input here. This is your sine wave, but we have to have a negative somewhere, right? So essentially, you see here, at the DC side, it's a series capacitor output. So that's why, you know, it actually ends up stacking the potentials. But because of the way the cycle works, for one half of the cycle, the capacitors are all charging one by one, and for another half, they're essentially non-conducting. The way this is set up, right? So it takes advantage of the half cycle, is what I'm getting at, to charge all these capacitors one by one due to the diode arrangements here during the input stage here. So by charging one by one, you have to essentially input all the current one by one into them. And of course, as you then output as DC as a series capacitance system, the capacitance drastically drops because you divide.
So, one long story short here is, in traditional electronics, this is a perfectly linear system that works very much like an AC transformer, but with even a little bit more losses. So essentially, what you put in is what you get out, current-wise. It's a linear, proportional thing. Now, to get a little bit more into why. So I guess the title here would be the capacitor math. So really, to support what I'm saying here, what is going on here? So, um, basically, each cap discharges separately in the ladder here. So because it's a series discharge, it means the same current output of every capacitor is here. Now, of course, without getting into the capacitor losses and all of that right now, 'cause the capacitors all have a loss, but quickly calculating this, just based on this system here. So, hopefully, I can plug this in right here. Just like that, here. So, just give me a moment here to try and express this properly. So half the capacitor that's N. I'm sorry, I have such bad writing. And over here, that's right. So NVPK. So, essentially, here's the math, which means essentially that it's a linear system. So essentially, it means in the series system, like I said, the capacitance drops. This is basically normal stuff. So basically, the key insight here that I'm trying to make is that the energy only grows linearly with the number of stages, N. So in this aspect, which is very important, which would shut down most people, we get to the whole conclusion: no free lunch. You know, it was a for effort, but once we actually get into the hardcore math, we see that as is, this system here is actually more lossy than the transformer would be. But it doesn't mean we have to end there. It opens up the door. This is hiding something. And this is what I want to talk about next.
Because of course, I was stubborn, even in high school, and all these years, you know, up until recently, it was still going in my mind, saying, "There has to be something we can do to make this work." And essentially, I will get to that. And this is why I find this video to be one of the, I think, more important ones, because essentially, those who understand after this will be able to build just about anything. And I don't want anyone coming back at me, you know, if they figure out how to build a 10-kilowatt system and, "Oh, it's Joel," and then they're all knocking at my door after. Right? So, um, try to keep it to yourself, but I want to educate the people here as much as I can who are interested in my channel and what I do. So with that said, the door is open for something else, and I will get into that. And it starts with splitting the positive, as I talked about in my earlier videos. And essentially, in a way, it sort of does this already. So we can extend on this system and make something completely different, which I'm going to talk about next here.
Just zooming in on the math. I realized that it missed some of it here. So my apologies for that. So there it is. But again, one long story short, it just means it's a linear system. All right, folks. So let's talk about a way to break this device into something different. So the first thing that came to mind, and by the way, this is not the only thing we can do, but it's to lead you into the direction that I was heading in. But we want to start with something simple first that we could all relate to, that I've spoken about before, that even got a few people in external forums and whatnot all excited over it. And it was the displacement current taps with capacitors, right? And essentially, it's a big philosophical debate about what we pay for versus what we can get out of it when it comes to displacement current. Because displacement current isn't actually magic hoggy-boogy energy. Maxwell actually later made a correction for it and accounted for it. When you deal with capacitors, there is a real effect of displacement current, but we just, it's usually not relevant. We don't calculate it. We don't even act like it's there. We're concerned about the closed-loop portion of it. So we, in other words, are concerned with real current, closed-loop current, and that's all that matters, what we push through the meter and the watt meter and the closed loop. Essentially, if you are to make a displacement current between capacitors, and people have done this before, I'm not making this up, and the experiment has been replicated over and over again, you can do this basically with two capacitors, one charged, one isn't, and you put a little lamp in between, and the other capacitor charges, the light lit up. So there's an efficiency there. There's an apparent gain of something. And some people, you know, they'll estimate anywhere from 30% and more of energy that quote-unquote wasn't accounted for in the closed loop. Well, there you go. That's the effect of the displacement current, but it's not energy that appeared out of nothing. There's an actual correction for it. And when you factor in the displacement current alongside your real current, then all of a sudden, the math adds up. You've obeyed the laws of conservation of energy. It's just our meters that are hooked up to the mains generator and our closed-loop systems don't account for displacement taps, even though they're there, maybe even naturally occurring within the circuit. We just basically ignore that part. But nothing tells us, hey, stops us from making them. And what's nice about this is, with the displacement tap, it's already there whether we use it or not, involving capacitors. So we may as well utilize it, right? And the whole idea, because you're messing around with fields, like I've been saying before, you want to sip off the field. You don't want to completely collapse the field and ruin any effect. So the idea is not to only trigger displacement tap but do it in a way where the main load here, this output, doesn't really feel that we're doing that.
So let's look at that right now. So let's say here, you see every diode, it's a chain reaction, it's a series, right? And we already concluded earlier that it's a linear thing plus all the losses. This is actually worse than the transformer. Oh my goodness. What can we do? We could recover some of that using displacement taps. Great idea at first. So, let's let's cut these diodes here in every stage and put a displacement tap. Since we're, we can do this even air core. And there's your tap one. And then on and on we go. And set up our displacement tabs like Badini was saying, is to try and be clever. Find extra tabs that we don't normally have that we can do. Now, everybody's going to scream, "Well, Joel, Joel, Joel, it's not going to work. You're your voltage multiplier. You're adding too many taps. You're adding resistance. The load is going to feel it. You won't be able to charge all your capacitors on time in the cycle. You're going to nullify your idea. A for effort though, but haha for you. It's not going to work. Stop thinking naughty, naughty, naughty." Badini had a fix for that, folks, in his chromley converter. What do you do? Air cores is one thing. We put three of them in parallel. Aha. That drops the impedance massively to almost near zero. And we do this times two, as long as every stage. So now we've got our taps, but we drastically reduce the impedance. Now, of course, our circuit is kind of, our closed-loop portion is going to feel part of that. But because it's just a few ohms, it's not really going to significantly affect our output here, which is good. This is what we want, sipping off the field.
So, of course, taking into consideration capacitor losses and everything, of course, at resonance, we'd be able to recover a perfect image of displacement current alongside whatever the closed loop is. But because, let's say, we can't get perfect resonance, in theory, we can, but in reality, we can't. So, let's say our losses are really, really bad, and our coupling here is just, just 30%. So, anyone can scream here and say, "Loss, loss, loss." But look, I'm telling you, let's just sip 30. Never mind going 100%. Just 30% of every tap. As long as your impedance with the three coils drops drastically, then you can focus a larger pickup coil all around the three, like this. This will concentrate and basically get the same flux but much better coupling of it. And then this becomes your tap, this output here. And then what you're going to do is you rectify that, and then you have a second bus, now a second DC bus that you could actually put in series, which mirrors some of the current, 30% essentially, of what you're already. So now, what I'm getting at is, you're getting whatever you put in plus 30% roughly, and that's with really, really bad losses. I mean, because you should at least be able to do 50%. But I'm being very extremely generous and saying, you know, you might get 30% off the displacement tap because some people were arguing in forums and that, "Oh my goodness, there's a loss." Yeah, there is a loss, but in a series DC, now over here, we're not dealing about the capacitor dividing itself on every stage no more. It's not the same system, but we took advantage of the displacement tap per cycle. So there essentially is a way of doing it. I could give you the math, be a little boring, but we can go through the math. Actually, I think we should, just to shut down the trolls. But essentially, it gets back to what we pay for versus what we can get out of it. But again, this, I, I, I get, would be massive if, you know, whatever. If this is like a 10-stage, a 30-stage, you know, and then you have like all these three coils and then the big coil, three coils, it would be a massive thing. So the practical reality of someone actually building something like this is probably next to zero in reality. But just for sake of argument, yes, in theory, if somebody had the stubbornness to build one, they could, in that sense, and recover at the very least 30%. Now, and you're really just sipping off the field. You're not collapsing it. And with such low impedances all along the way, it, the closed-loop portion doesn't even essentially realize this is happening. So you can recover part of the displacement current, and as you add that up in the series stack, it clones part of your output. You can combine them if you want. Do whatever it is.
So let's take a look at the math real quick here. I'm going to have to, um, All right. So some assumptions to start with. Our input could be 110 volts RMS, 155 volts peak from the mains frequency, 60 Hz. Latter N, 10 stages. Capacitance per stage, 120 microfarads. So the total capac, the total cap, the total capture efficiency is 0.3, as I was discussing above, or 30% of displacement flux harvest in the coils. So that's our base figures, and let's see where it takes us. All right. So looking at energy per stage per cycle. So here's the basic formula here, and plugging in the values here, the example values from earlier, we see that we could recover about 1.4444 joules out of the system here at this efficiency, which is again, very, very, I exaggerated how inefficient the taps could be to show you that once you add it up, this is what you actually get. So now to look at the next stage here. All right. So the main ladder out energy, the basic math, what are given values here, essentially P main output is essentially 865 watts. That's the main closed-loop internal regular, what you would expect from this system. So now let's add the displacement and see what we get. All right. So taking into consideration the displacement tap out energy and all the math here, basically our P total equals 1,125 watts, which is not bad. It's around that, close to a 30% increase. So what is all the big revelation, folks? Well, looking at it this way, the paid input, which is around 865 watts, output is around usable, by the way, is around 1,125 watts because we considered the displacement tap and we used it only 30%. But this gives us a COP / 865 equals 1.30. A COP of 1.3. And that's just a simple modification to the Walton multiplier. It's just a point of concept. This is not something I'd actually want to build because it's still not. But the point is, you hardly need to make it efficient to get a gain here. And depending on how many watts you're putting in, if it's very, very commercial power, this could literally offset the bill quite significantly, just putting it out there. But there is actually much better ways of doing it, which is what we are going to get into. This leads to another revelation, folks. Remember when I said splitting the positive? We're going to get back to that, and I will show you in a moment. But I just wanted to point out the math, which I find is a little bit boring, but there's always a troll who says, "Well, you're not paying attention to traditional." We did it. It's just, this is paid. This is not energy out of nothing. Right? There is a correction for displacement currents, and you can, we didn't bother calculating it because what I'm interested in is what we have to pay for, what we have to pay to put in this extra displacement current. It's there, you whether we use it or not. I decided to use it. If you want to plug in those calculations on your own, you will quickly find out that there's no actual break in the conservation of energy because the total output has to be what we've put in. When we account for the displacement in our input, it obviously matches, right? But in relative to the meter, the mains, the meter, they don't build us, they don't measure displacement current, just real current. That's where this actually is able to happen. It's a subcomponent that's usually ignoring the field energy part of it. So with that said, we will move on to the next part.
All right, folks. Now that I've shown you earlier a possible way to exploit, if you want to call it that, the Walton multiplier device. This is, as I said, I don't think most people would do this unless you're in a lab and you have all the room and everything, 'cause, you know, of course, you're going to have to like take everything into consideration here and really figure out your values and your inductances and all of that, right? And make sure that you're really not taking a lot of impedance here. So in theory, yes, it could be built, but in application, I don't. In other words, it's not somebody in the garage that would probably build this unless you're very stubborn to prove a point, right? So that's where I go, "Okay, I think I've proven my point. I see that with some clever thinking, we can do more with this kind of design, but I think at the end of the day, we have to essentially scrap this because apart, and the displacement tap is clever, but it's not really what I had in mind. And I want to take advantage of capacitor joules and take that boom, right? And even though this is a form of recycling the energy, it's not exactly that boom I was looking for. And it's very complex, but it does, you know, for the sake of argument, complete our point. So moving on, essentially, I came to conclude this, that this design here, for what I want to do at the end of the day, even though there are ways to exploit some of it, it's time to go back to the drawing board. But it's not to say that I still had the idea, why can't I just use all these capacitors in a series and combine the joule discharges and voila, there's the apparent gain. We're done. There you go. Could it be that simple?" Right? That's what I want to do in my head. But then, of course, reality kicks in, folks.
There is something that's very interesting in our digital age, in our semiconductors. Semiconductors are, unless specialty, usually very linear. They're very predictable, and for the sake of our survival, for some reason, very lossy. The IR losses usually are through the roof with this stuff. And we just compensate by cramming in more energy, brute force, into it to compensate and calling it a day, where we put more thermal paste underneath the heat sink and say that's how we deal with that. Okay, but you know, somebody still had to pay for all of that. Why do we need to shove all that extra if we don't really need it? So this is where I get frustrated because I realize that, oh, whether it's diodes, whether it's transistors, whether it's capacitors, whether it's any semiconductor, you're stuck there. There's so much loss internally that even though you'd somehow be able to create an open-loop system and you get a little bit of gain from the environment, by the time you close it back into a circuit like this, no matter what you try to do, it eats up your little bit of gain, the little chance you had. It's like it's designed for us to lose. And not because science says it has to be, but more like society drawn us in that direction for its convenience. Right? The meter loves to bill watts, right? Closed-loop watts. So that's what it's all about. There's a whole market around this, a whole infrastructure, a whole industry. Big tech doesn't want to be shaken. They're quite happy with this technology, actually. But at a certain point, it's going to have to backfire because, you know, you don't want to fry the earth either. And with the rise of AI and all those things, you need more power. How do you do without blowing us up with nuclear, for example? Right? So they're all into the cold fusion is the answer phase. Okay. But there's still a polluting aspect to it. What if we don't have to do cold fusion? Not, and I'm totally against it. Just if there's a better solution that gives us the equivalent or more power, why not explore that as an alternative, at the very least, right? Have all options on the table. As I wanted to do in my mind, I said, let's scrap this circuit and let's make it simpler, a series. But again, as I said, everything has losses. And specifically, and again, it's really funny because people in the outside world and forums and stuff I've read, they try and make fun of me 'cause they think I don't know that something like capacitor losses is a real thing. It's actually the equivalent series resistance, and that's very real, and we have to account for that. Even though you don't see it in the circuits here, every capacitor in between every plate here, there's a little invisible resistor. So in series, what happens to this little invisible resistor is the resistance obviously adds up. Not only does it add up, but what happens is every series, you lose capacitance because it divides on itself. So you stack all of that into the equation. Wherever you may think you've had a gain, it's gone before it even gets to your output. So, uh, that is essentially for capacitors, we have to deal with the equivalent series resistance loss, and those are very real. And it's not to say that capacitive discharges aren't. That's not what I'm saying. It's very real. And if you need an impulse-driven system, and let's say you need those impulses at one kilowatt, in joules a second, so amps a second, let's say we need 1,000 amps a second pulses to run our pulse motor or whatever. Yeah, the series capacitance will give you that at the right calculated values. But the point is, there's no free lunch. If you need 1,000 amps a second, you're going to somehow have to deliver it at the input as well. And not only that, but make up for all your series, your equivalent series resistance losses, and maybe almost have to put double at the input just so you actually get this at the output. But as far as your load is concerned here that requires this, it will run just as normal 'cause it doesn't care that you have to pay more. It just wants to have what it needs. So in that aspect, capacitors are great. You get the joules, you get the amps a second discharges, but it doesn't come for free in normal circumstances. You have to pay for that by the restructuring of the circuit and how the power gets redistributed at the input. So no free lunch, equivalent series resistance. And of course, there's one way to minimize it. If you just use one capacitor, it, you might have a loss of like, not 0.1, but maybe 1% at the discharge. So this way, it's very tolerable. You may not even even want to consider it. But when you're dealing with series systems and that the capacitance drops drastically, this really adds up. And if you had like, you know, a 15% gain factor somewhere, that's gone. You're back there. You have to put more in. Even though you had a gain somewhere, these losses, and that's what really sucks, right? Because our semiconductors, it's almost like I was saying, by design, they, whatever you, it's like you're stuck in this linear trap, no matter what you try and do, right? Almost like they set us up that way. They probably did.
So I have to, I guess, come clean in a certain thing here. This is going to lead me to the next device, which is essentially, you're going to understand how the Moray device works. I actually, in part, built it. And a little bit of a disclosure here. Most people, maybe it's even for my safety, most people know I've been involved in a company for about six months now trying to develop this thing. I am chief engineer in the background and developing all these various concepts. I talk openly about what's already out there in the open source, but we're doing our own thing, our own research in the background. And with that said, I was working on, we wanted to explore, you know, what's already opened, what's known about the Moray effects, right? And from what I understand here, what I was saying is again, all the semiconductors have this major loss, right? But if you take a step back in time, in Moray's days, and Tesla's days, they had basically all vacuum tubes. They didn't really have semiconductors. That wasn't a thing yet. Now, what's really, really great about the vacuum tubes is their efficiency is a lot, losses are a lot less. And no, no, I'm not talking about, "Joel, Joel, what are you talking about? The heater, the heater, the heater, watts, massive watts. Oh my goodness, you're all wrong." Okay, I get that. But taking that part separate for the actual function of the vacuum tubes, nothing beats the efficiency that they had versus semiconductors today. For example, they can handle high voltage, high current, like nothing. Most transistors will pop across the wall and make a crack. That's just what I mean. They are so much more efficient at at high voltage, field energy, low loss, because you have the vacuum dynamics of the tube and everything, right? So that's kind of like an edge that Tesla and Moray had back in their day, 'cause that's happens to be the only thing they had to work with. But they were blessed in a way that they weren't confined to this lossy trap like we are today.
I was experimenting with building a Moray tube device, and he is partially right about using special tubes. And there was this estate, and I got these old tubes back, and they were really, really big, and they had Edison sockets, and they weren't rigged. And I'll get more into that. But one long story short, I was platforming it, and I don't know what happened. But sometimes, you know, being in a small town, I don't think this could happen in Concord, Ontario. I go to the corner store for five minutes, come back. Somebody actually came in through the back door. Looks like they really knew what they were looking for. And it's really weird 'cause this is not like stuff I talked about before. I never showed these tubes 'cause this was something I was doing on the side. I wanted to keep aside for now. Obviously, if you understand the Moray story, you know why I wouldn't want to go flat out public with this right away, right? But I still wanted to experiment and understand what was going on. But it seems like, you know, I kind of laughed it off when people saying, "Oh, people are watching you." Various sentences. I guess they're right to a certain point, because I notice strange vehicles following me around town, going in and out of my parking lot all day. Don't know why. And this would support it. I guess I don't know who they are, what their goal is, but essentially, I guess they knew what they're doing, what they're looking for, and assume that. And what's weird is they didn't take anything else, folks. I was going to eventually make a video, but I couldn't get to that. I wanted to show you the tube as an example here, and I can't 'cause I don't have them. Um, they're like nowhere to be found, unfortunately. So this is like kind of my way of coming clean, kind of like, you know, I don't want to put this name out there, but Bob Lazar, right? When he was all paranoid, just put it out there so if something happens, people know why. So this is my way of putting it out there that I know there's something up. I don't know why. I don't know. Just to let everybody know that it's actually going on. And I guess it's to be expected. That happened with Badini. It happened with Floyd. It happened. It seems to be like a, the commonality once you start getting to a certain level with this stuff. So again, I guess I'll take it as a compliment. It means that I'm in the right direction, and they're really concerned, whoever it is. But at the same time, there's my safety as well, right? By putting it out there, then it becomes more difficult to do nasty things 'cause then everybody's going to know if Joel disappears, ah, he was talking just about this. So it's easier to, I guess, find the criminals at the end of the day. But the point is, you know, more liability, people get discouraged to do things, is what I'm getting at. So this is what we're doing here. So it's not something I wanted to discuss right away, but it's something I feel that just for transparency and safety that I have to get into. And the logic of that, because I'm going to tell you why here.
Now, there's something very interesting. When people see the Moray device and they saw all these tubes and everything, they just assume, just like everybody would scream at me and go, "What are you talking about? The heater, the heater, the heater, watts, massive watts. That's why we don't do that anymore. So much loss in heat. What are you?" Again, they probably accuse me of this information, but believe it or not, less talked of, unless you're in the specialty field, there is such a thing as cold cathode reactions. And there are a whole family array of vacuum tube rectifiers, usually with the assistance of a gas inside. But needless to say, anywhere from a few volts used mostly in the audio to like radar and pulse systems and the high kilovolts and high current capabilities, zero heater, cold cathode reaction. It's a real vacuum tube, folks. So I had those. Those were the ones, very old ones, that were giving me interesting experiments. They literally looked like a big light bulb, and they kind of look like this, and then you had kind of like a grid here, and you literally had the light bulb. So you had this section here, that section there, and that was the tube. So essentially, as a diode, but completely cold. And if you understand the dynamics of the vacuum tube, this is like your diode now, but at much less loss. And, and what happens is, back to my splitting the positive, right? So this will understand how probably Moray was doing it with equipment they had at their time. So let's say this is your AC line that comes in right here, your two AC wires. And then you need to split the positive, like in Moray's time. So what we're going to do is, first of all, what I have to explain is we're trying to fight this here so we can have a gain, right? So how we do that is we have to have capacitors that have very low, and they have flash capacitors and things like that. But even there, as you get into the capacitance you need with your series, 'cause it drops, but you can compensate by adding more, but you have more of this here. So it like linearly nullifies your gains as you go up, right? But let's say, suppose, and this is, you know, this sounds like far-fetched at first, that there were capacitors that had zero or near zero. Sounds like a pipe dream, doesn't it? But guess what, folks? There is such a thing. Vacuum capacitors have been around since the beginning of radio days. Looks exactly like this. And some of them are even variable. You can tune them with a little knob at the top here. The downside is they're usually in the, um, less than the one microfarad range. But the thing is, of course, it's tiny capacitance, and at low voltage, that won't mean much, but you got to think further than your nose, right? These offer near zero equivalent series resistance 'cause it's vacuum dynamics. So essentially, your joule discharge, you can recover most of it. And you could add a bunch of these in series and still get near zero, which would be impossible with regular today's semiconductor equivalent of capacitors. But yet, this is a capacitor. Very real, used still in radio today, is still used in antenna and ham radios.
So what happens now is if you stack a bunch of these, like times 10, even though the capacitance is lower, you can raise it to a few microfarads. It's going to be a big chain, but who cares if you've got near zero? We're finally doing the original idea I had, and it's actually coming into peace. Great. So, how do we pump this chain now? Well, of course, just a regular AC sine wave into a high voltage transformer, regular stuff at this point, kind of like what Moray would do. And this gives you, let's say, your 2 kilovolts AC, real AC sine wave up and down. Now, back to my cold cathode diodes. Two of these right here as an AV plug of the day. So cathode, anode, anode, cathode. So these are your two diodes here. And this one here goes like this. Sorry for the bad drawing there, but cathode, anode. So here's your AV plug. And then we did the same thing on this wire here. So what happens is you get the split. The positive is on this side, and your negative is over here. And then on the other cycle here, your plus becomes here, and this one becomes, so you split the cycle in two separate quasi. It's like pulse DC, but not real pulse DC, because you still get the curve of the half wave right through the diode. So now we can do the same thing here with the cold cathode tube and do this setup here at the high, 'cause this is all high voltage dynamics. Remember, folks, you're doing it at, there's a reason why, and I'm going to get to that. 2 kilovolts. We then split the positive. So now this side here can happily charge the series vacuum capacitors over here at a few microfarads. Okay. Well, let's say 2 microfarads at very high voltage. That at high voltage, even though it's milliamps, equals real massive watts output because of that high voltage dynamic. So by shifting it to the high voltage, we're able to take advantage of the vacuum capacitors, low ESR, sorry, low ESR dynamics. So at the end of the day, you can get a very strong series joule discharge without having to worry much about these losses here. And we split the positive with nothing but very efficient to rectify cold. So there's no heater. See, no heater. Vacuum, vacuum, vacuum all around, like the Moray device. Vacuum capacitors, high efficiency. Vacuum cold cathode diodes, high efficiency. But we're not done yet. How do you control the discharge? Well, that's pretty easy. You got the AC cycle here on the negative part that can do your switching right here. This is this one here we haven't tapped into. So, what did they have back in those days? Vacuum relays. Yeah, I know, right? It all comes together. So you got another vacuum relay tube here, which is a relay device, and the vacuum tube, and this pulses at H here, here triggers. So this here could literally, if we pass the series through the vacuum relay, it turns on during the discharge. So you literally pump the AC sine wave with its own dynamics, all with analog vacuum tubes. And at this point, we still don't have any loss in heaters 'cause we didn't even use a mechanism that uses heating.
So now, of course, you're going to say, "Well, Joel, great, great, great, great. But still doesn't make any sense because you're sending it joules, a sudden discharge, and most loads want that sine wave back. And this is not the kind of energy you want." So the very simple fix to that is simply to add the right resistor at the output. And what that does, with the right value with your capacitance, you could literally calculate a AC like slope. It'll filter into it'll shape it essentially. So rather than getting this sharp discharge over here and then it's all gone after like 10% of the time, we can literally slope it into a half AC wave. So essentially, it's our way of preserving the high joule discharge that you get from the capacitors without the equivalent series resistance getting in the way and actually allowing the accumulated joules at the high voltage back to an AC. So there it is, right? So now you're going to tell me, "Okay, how does this translate to actual watts at the load and how does the load react to this?" This is what we're going to get into next. And this is what I'm going to explain, and it's probably all going to be clear from that point on, folks.
All right, just thought I'd do a little correction here. I did this really quick here, but it's not really quite like that. It's actually one is plus, and then the minus comes from there. So you split it. So I was just drawing this really fast. So this is a little, but the rest of it is fine. So I just thought I'd show you the real diagram of splitting the positive because I've been talking about it for a while, but I never showed you in simple terms. So I thought it'd be easier to just actually show you what I mean here. And this is essentially, let's say you're live here, and you're neutral over there. So what happens, you split it essentially with two AV plugs on each side. So I hope everyone can see that. So this one here, for one part of the sine wave, is plus, and then your negative is down here. And then for the other part of the sine wave, the plus is here, and your negative is there. So you have two unique, depending on where the sine wave is going. See, you can see the flow there. I thought I'd, hopefully, you can see that. And this is what I was referring to. But here we'd want to use our vacuum diodes instead, cold cathode, to do this arrangement at high voltage to split the positive. So as I was saying, one side feeds the series capacitors, the other side runs the relay vacuum switch. So during the, the other half, you're discharging. So again, as I was saying, what to do with this energy? It's not shaped yet. So I'm going to explain to you exactly what every load expects and what we can do with that. And this is where the quote, the real apparent gains come in.
Now let's talk about something very important. What we were talking about is essentially how to offer, basically, ways to minimize the losses we normally have in regular semiconductors and capacitive series circuits. Of course, it doesn't mean you have to do the device like that. I was just suggesting if you want to understand a little bit how Moray and those originals were doing it. This is what the parts they had, and this is why they were so successful at high voltages, and why what we can do here is a lot more simpler in today's days. We have much larger capacitors and everything. The big point here that I'd like to make is what's very important is we already know that we can do it now, very lossless, various methods like the tube method, to be able to create a capacitive discharge at the output at the cost of very little waste to get there. So it doesn't mean that this joule energy is magical free energy that came out of nothing. It perfectly calculates. Whatever you put on your input has to be that much to have the equivalent of that joule charge. The difference is it restructures and reshapes the charge to give you a more concentrated, because it's a joule discharge to a capacitive. It basically changes the way the structure of the wave gets distributed. But all in all, it's the same power being distributed in a different way. So essentially, now you're going to go, "Well, you wasted all our time."
Why are you doing this to us? No, no. I'm trying to get to something here. Everything I said so far is completely valid. So, we take the same idea and we already know we get a jewel discharge. What can we use the jewel discharge? The whole idea is to take advantage of what the capacitor does as a jewel discharge. It gives us that instantaneous power. But there's no rule that says we have to waste it all in one shot.
So this is what I was getting at in the last video is with the right RL L network at the output and this isn't relevant to the input capacitance and voltages and their ratings and everything down to the frequency here would be probably my example would be 60 hertz. This can reshape our jewel discharge. So we can extend it if we need to over the course of the half cycle. Very easy with the right R L. So R L L before it goes into our output. And this is our input prompt here on one side. And then this one here would be free. That's after our stack here. We can instead of having the jewel discharge in one we can calculate this so that we end up having an AC like sine wave but what's important is let's stretch it out to about 8.3 NS and we can do that with RL networks.
So what I'm getting at is what we're doing is we are emulating the characteristics of a brute force sine wave to a capacitive discharge with the added advantage of that oomph that goes behind the capacitor that you don't normally get. And of course we're dampening part of it with RCL but we're not completely we're not discharging it over an extremely long time. And this is still a tight. So at this level, you still get a oomph out of it that you wouldn't get in relative to being an always the same sine wave equivalence.
Now what that means is we get into something very interesting here. We get into what looks like apparent power gain. And let me explain why a load would see that. So let's do some ballpark figures. So let's do stack energy. So this is essentially just generally speaking. It's not specifically tied no more to the tube device. I was just explaining that as a possibility. Okay. This is in relevance to any jewel discharge. Okay. But we have to start somewhere to see if our idea holds up. So, so stack energy, let's assume 10.9 joules for this example. And let's say it's an 18 stack at 100 UF they're at 110 volts each or desired this charge rate. So 8.3 NS as I said over here and that's literally half of the 60 Hz AC cycle 10.9 joule energy to be dissipated within this time period and we can do that. It's going to be a bit of a loss but all in all it's going to be pretty good.
So the equivalent resistance for the target discharge. So now let's so for a capacitor stack the energy decay follows this. So essentially to release 95% of that energy within the 8.3 NS window what we want is this right here. So t = t = 3 RC. So here it is now so far. So essentially with all this here we see that the effective series capacitance of 18 * 100 UF equals to and I'm running out of space here. Let's try and do it over. So essentially, one long story short here with these values, we end up with 5.56 UF, which is still not bad if we're going to put in like 110 volts into this, right? So the math seems to check out and I'm going somewhere with all this.
Now, of course, we got to figure out the R as well as it's an RL network. So essentially our resistance is 400 is 498 ohms and of course we have to look at the inductor for smoothing. So let's try and plug it in in the little space I have left here. If we just use R, current is essentially exponential. It's not sine wavelike. So we simply add an L. So the RLC circuit resonates. So the RCL circuit resonates at 60 Hz. Mmh. So yeah, I almost ran out of space here, but long story short, the total L is 127 MH. So essentially what it means is we put something in this ballpark value. So let's say something around 120 to 130 MH the system with these values will want to oscillate at 60 Hz. Therefore give you that slope you need. So pretty clever, right? And that's all that's what thanks to the shaping of the capacitor network.
So now what's very interesting is this allows us to emulate half of the sine wave without actually producing that total brute force sine wave current that we normally would push in a regular load. So this is where we get into what I call the apparent gain. But you must remember though at the end of the day as is whatever we get here it's still what we've put in the input. We just found a way to cleverly shape it. But let's see what that actually means to the load which is the final judge the one that does the transducing what we want as the kind of work. So we have to see and look deeper than the traditional system and actually figure out what is a specific load really need specifically for a requirement to give you what you need to operate that load. And once you do the research, you find out very quickly that in most of the cases, it's not even a continuous sine wave current that it needs to do that very work. But we it will happily work like that, but it doesn't need all of that much.
So this is where um we get into transduction of energy domains and whatnot, which I'm going to get into next and explain it in in more simple terms here. I just wanted to explain here that what we can do is we need to shape it. We can do that with everyday part at an actually very minimal loss. So before someone screams, "Yeah, well you're adding resistors and coils, you you're nullifying your jewel discharge." No, not necessarily. We're still got the big impact, but spread out over an 8.3 NS range. And when you have at higher voltages this much capacitors that's going to hit it will hit. So how do you make the system behave as it thinks this is a regular energy signature that it requires. So this is what I call it essentially at the end of the day emulating the sine wave because it's a perfect word for it. We there we're emulating what we're trying to do essentially is trick the load into thinking it's getting more than what it actually is at the input. So at the end of the day, it all falls down to greater efficiency at the load. If we actually know what the load wants, we can shape it with these dynamics.
All right, folks. So back to what I was saying earlier about once you start understanding the actual summary of the waveform and its independent features like stripping it into separate features we notice that dependent on the load that there only specific specific requirements are needed to make that load operate. Now obviously in the full linear always on always the same current sine wave it's kind of an easy universal way to meet that requirement in just about anything you put into it which makes it preferable in the industry but as well also very great for the mains company because they want to push watts and currents right but essentially want to start digging into the inner mechanics of all of this you realize that literally depending on the load it can be as low as 10% of the actual waveform and the specific feature requirement. You give it that and it will work as if the same way if you gave the full current waveform because you specifically gave it the feature it needs to perform. So essentially by cramming all this always on current that the mains and the industry gives us it is such a wasteful way of using the energy universally and we all do it probably at the expense of being able to make them profit and they probably somebody has to know this somewhere folks in the industry and we're trying to keep this quiet because the point is it can range anywhere from 10% which is very good to anywhere up to 40% of the total half cycle energy. So even when you do the research and you get into the loads that require the most energy but specifically giving them the specific feature they want, even that 40% you see there's a massive gain in relative to having to give 100% of that energy to get the same apparent work done. But the thing is, you have to understand your loads and what the loads need to see where I'm coming at with this. It's not magic energy out of nothing. And this is where most people 99.9% run into a big mistake and wall and they don't know. They get the idea but they don't know what they're doing.
So let me explain it. So we have to do the summary of waveform features to really understand this. So there is there are a few of them. There is the volt the volt the second area. This could be the total voltage second per half cycle requirement of your load. And this is expressed in V DT. And essentially what you get with this feature of the waveform magnetic loads such as transformer and motors and part. Now there's another category here and we get into the RMS of the waveform your jewel delivery that governs thermal loads heaters filaments and that sort of thing. So, so let's say heaters and filaments such as your lamps, right? Now, of course, there's another one here which is peak voltage. And by the way, these are all waveform specific features. And once you dig into what your load specifically needs, you just need to specifically provide that requirement and you're in business as far as the load is concerned. It is the final transducer. It will provide the work in that domain for you as if everything would be okay. So this could govern discharge ion loads kind of thing. So for example, fluorescent plasma neons. So let's put keep it simple neon plasma and that sort of thing. And of course we then have the average charge per cycle. The average charge per cycle. And that governs electronic capacitor rectifier loads that kind of thing.
So again in all cases what this concludes is for whatever whatever the load actually is on AC what this tells us is we don't need the full brute force sine wave at the gives us at the same current all the time pushing continuous linear energy. We just need to calculate and there are calculations for every one of these depending on what your load is to actually meet. So essentially if you have a 100 watt load at a certain impedance you calculate how much volts a second or whatever it is for the load the specific load you're trying to run and it's equivalent feature that it needs. Once you meet that you're done. That's it.
So essentially which one that gets very interesting here. And you have to understand again as I've said a lot of times the load is your final transducer. Okay. So why I call this apparent gain is not because it's an actual gain out of nothing but again in some cases we only need to use 10% of the normal energy. It's super efficient. So to us it seems like it's a big gain. Example, if you if you're lighting up a light and you're getting the same let's say it's a big light and you're getting the same one kilowatt equivalent lighting in lumens. The thing is, as long as you filled this requirement here specifically for the lamp, the lamp will glow at at at what it would need if you actually had the 1 kilowatt of full sine wave energy at a fraction of that. But the thing is, it's not actual energy that's being made. It's just you're giving specifically what it needs. So, it's much more efficient than just brute forwarding all that extra that it does not look for you. That's the whole point. But obviously what people don't understand is it's always the load that's your final transducer. And for you to be able to actually lock the kin what those apparent gains is, you want to lock it in. This is what Don Smith was saying. You need to flip it. But when he said flip it, he didn't mean transformers. He meant you need to cross energy domains completely. And that's what the load itself does. Depending on what the application is once you've crossed that domain and here we get lumens which is photons being spit out. Once you've got the photon out there, it doesn't matter what you're doing electrically anymore. There's no. So you're if it if it's giving you what 1 kilowatt would give you at sine wave AC and you're getting the same lumens, it's got it's cuz you simply met the required feature that it would have pulled out at a full sine wave at a loss. You just don't include all of that. You go specifically engineer to what it wants.
Now what's very interesting is these will vary depending on what it is. Now obviously you probably know that it would be this one over here being like the most. So if you're looking for the 40% this would be around here which is fine. Of course you're still doing a lot more than using 100% of the full sine wave energy considering this is 40% of the half cycle. You're saving so much here. But the mistake everyone's going to say is, "Oh my goodness, I'm lighting up a one the equivalent in lumens anyways of a if I were to give regular sine wave 1 kilowatt energy at AC. So I'm really really smart and naive. So since I'm getting a major like a x 10 amplification factor, why am I wasting it on a light? I'm just going to put a little rectifier here and of course be really slick and send it back to my input and I've got a continuous self loop system. Problem solved. Energy crisis over. We're done. Awesome. Right. I have to tell you again, no, it doesn't work that way. Cuz if you actually calculate and you put a meter on here, you're just manipulating energy you already have to do specific work. So it will still balance out. So if you actually even Bedini ran into this if you essentially use capacitor wire capacitor wire and then try and loop it back it's the same domain. You didn't actually gain any energy and we then transduce. That apparent gain isn't happening because you don't have the load transducing it and locking in in another domain which is then free to use without stressing the load. In the case of our light here, we're lucky because we want lumens. We want photons. It does that. We're done. We don't need to do any more work. It did it work. But it is at first deceptive cuz it may give you the impression that, oh jeez, I'm generating an actual 1 kilowatt here. No, you're not. It's just in normal circumstances, you don't use most of that energy to actually power that. And that's what in but in traditional electron dynamics of course it it it equals out and it shows that you use it but the point is the load does not specifically need it.
Back to I said it's clean it's effic it's it's efficient in the way that it's universal and of course you can cram a lot of power and it's easy to calculate but very inefficient and very profitable for the utility company. So essentially one would look at this and again the big secret here is you must transduce to lock in your apparent gains or else it's just apparent there's no real energy gain here. So if you get the load to do the same work that load needs to be so in the sense if you've got motors here spinning and you get the same torque energy that's been converted to mechanical. So now nothing stops you, which this is where the fun begins, from putting a mechanical generator that would spin like it's got 1 kilowatt, but you don't have anywhere near that in the meter. It spins the same efficiency because you're giving it the optimal delivery method. You've got another mechanical generator coupled. It's giving you back real watts and all of a sudden, minus the regular hysteresis and everything, you're getting like 900 watts of real sine wave energy through the generator. That's because you've locked in your gains by transducing it to mechanical. Once it gets into the mechanical domain, it no longer interferes with what's going on electrically. But again, mechanical is still pretty lossy and there's better ways of doing it.
Now, when someone sees all of this here and they say, "You know what? I could put a motor here and lose 40%." That's great. But there's a there's a total revelation that it all clicks. Remember here the volts a second area here. This one here triggers magnetic loads transformers. Essentially, this is big because this one here is the one in some in the math where you only need 10% to achieve for as far as magnetic flux to energize it fully and flip the polarity and produce the same AC sine wave energy on the prime on the secondary. You only need to meet this requirement here. And there's no rule in the book that says you need to do it a perfectly sine wave. It doesn't care which way you feed it. Pulses, half sign, triangle sign, it doesn't matter. As long as at the end of the day you meet this requirement, which opens us to all kinds of wonderful ways, we could efficiently feed our system and get the same thing as if we were to feed it with a lossy sine wave that's pushing continuous current in there. we could do the same work at 10% instead of and that's 10% of the half sign which means that the total AC sine wave you'd be using about 5% of what the mains is giving you to do the same work that's crazy isn't it and that's how literally it can work so what this means is in another reality the utility company the mains could literally have a box that comes into every person's house has this capacitive series setup that discharges with their own analog waveform to takes advantage of their half cycle to charge and discharge and a modest size you know coil and resistor to shape that waveform. Everybody has their own closed loop from that box. They transduce it at the local level. At this point here is the magnetic flux energy. That's the transduction. It changed domains. You gave it what it needs. It doesn't care if it had all these watts or not. Calculate it to your load. You realize you're only using 10%. There's your G. But this would literally break the metering. This would totally break the system, right? But they could lit if they really cared about not harming the environment, not going nuclear or anything like that, they would literally be doing this everywhere, especially in commercial areas. So can you imagine the grid is basically just a trigger. Everyone just uses 5%. They tap into it and you restructure.
So this what I'm getting is even though you can run a heater at 40% direct, why do it that way? At first, you can transduce it, lock your apparent gains in the magnetic domain, which drops it down to 10%. And once you're in the secondary, transduce it back to regular electricity. And then run your heater without having to sacrifice that 40% cuz you took a different energy domain path which offered a higher gain benefit. But the point is, we must always flip the energy domain and then lock in those gains because they're not real electrical gains. You can't just take the raw output and put it back without transducing it and transforming it to something in that amplified form and then taking advantage of it if you have an efficient way of taking some of that energy back. So I wouldn't go with trying to take light because we don't have that efficient solar light. So that's why the transformer seems very appealing to me when you can only drive it with depending on the load just 10%. And I'm not joking. Do the math with the but you got to figure out what your load is. And depending on what it is, it'll range from 10 to 40%. Depending on what the domain is in relevant and the full sine wave or just cramming in absolutely everything and just assuming that's what it needs, that's what it is. That's what a linear system is.
So there's the key here and why everybody usually gets it wrong. They try and and take their gains too soon before locking it in. Once it's transduced into another domain, then you're free to do whatever you want with it. The load doesn't care. Photons are photons. If photons were generated, it means you met the specific requirement, the device. Simple as that. It doesn't do it out of magic. Whether or not you fed it the full sine wave or not, the fact that you got it means that you met the specific waveform feature that system wants. That's all it is. And that's where you get your gains. And that's the whole story essentially of quote unquote over unity systems. No matter which way, which device, whatever, it's all about the transduction, the loads, and how you manage that energy that's already there and drastically being wasted.
So, I hope once and for all that I put it in clearer perspective and I'm pretty eager to see. I'm I'm expecting to get really roasted on this one. So, comments, you know, I won't take too seriously, but I'd like to see where it goes. With that said, until next time, folks, and thank you for watching. I realize this was a very packed featured video, but a lot had to be put out there. Thank you.