Donald Smith Devices too good to be true

[boguslaw]

I propose that we try to build such delamorto simple setup
Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

Inverter : 500ватт инвертер.wmv - YouTube

Schematic ??? Delamorto help please. I know this may not be OU just a long running blocking generator but anyway interrupting battery connection and replacing bulb with transformer, diode bridge and electronic circuit can turn it into self-running battery charger I believe.
Delamorto are you here ?

[boguslaw]

Above I posted the simplest device using reactive power to self-power inverter and incandescent bulb with some I think ringing down but very slowly.

[Patrick Kelly]

Yahoo mail failure

Some might wonder how this all relates to what smith and others were doing with their systems. Think of the circuit I presented as a building block, one small portion of the whole. A way to learn, progress and discover.

Here is a diagram showing how this all comes together as a whole... Once you have a basic understanding of what to expect you can get creative and move forward in your own way.

I've presented this in the simplest possible way, starting with the basic building block, so someone could actually build a functioning circuit and somewhat understand what it's doing, what their looking for and how to get there.

And No, it's not the infamous 100kw, solve all, Cosmos sucking energy device that everyone claims to have but I believe you'll be quite happy with the results as long as you can successfully complete the basic circuit. It operates on a voltage low enough that the danger is low but high enough that you can see the effects of what it's doing. A foundation to expand on.

I've learned a great deal from some of the serious contributors on energetic over the years and this is simply my way of "giving back" a little something in hopes it will help someone even in a small way.

Have fun!

Diagram explanations: Top left: is the basic circuit I presented. Lower left: there is a second identical coil and primary to make up a single wire power transfer device. Top right: the two identical coils are stacked and driven by the same circuit forming the basics of a don smith device where the output is formed into an isolated buck converter. Lower right: is my own reduced version of the same using a 3 plate cap system and 2 separate resonant tank circuits.

Patrick Kelly, feel free to use these if you wish.

Hi Dragon,

Thank you for your kind permission to publish your circuits. I would be more than happy to do that but I would need to ask quite a number of questions in order to get the facts right and the description accurate, so if you are willing to join in that project, then please e-mail me direct. The Yahoo mailing service is having major problems at this time and my yahoo.co.uk mail account is effectively dead for incoming mails. People on the Yahoo forums are reporting that they repeatedly send posts which are reported as being "sent" but which never arrive, so I have switched to engpjk@free-energy-info.co.uk for mail, in the hope that it being a 'paid-for' service, will prove reliable.

Thanks,

Patrick

[Guruji]

You should be able to adjust the resistors in the inveror for 50hz. right off hand i would have experiment with it. This is nst 9500 volt 30ma power supply. The resistors in the voltage divider would make the 220 volts you need just vary the resitors. hope this helps

Hang in there

Thanks man. I've built an easy inverter by Aaron but don't have caps as is in his schematics. Now I saw an inverter on youtube with just two transistors and two resistors amazing!!! I'll try this. Than I will try my Don setup to it hope I get some juice out

vid - YouTube

I don't know why this works, but I've read it several times.
You can charge a capacitor inductively using magnetic flux.

I'm NOT the author of the video.

Wow amazing I will try this. Very easy.
Thanks for sharing

[RAD-HHO]

the man who played with tesla hv coils will be so afraid i never thought.
Farmhand use neutral wire instead of live wire as rad HHo forum member did.

have fun light led's for free

rgds

zzzz

You misunderstood. I did use the hot wire. I got slow charging with the hot wire and ground. I used a 2uF 1000V cap. It charged about 2V per second and stopped at 24V.

As I expected and stated, using the neutral an ground does not work. It can't. They are the same, at least in the U.S.

[vidbid]

No doubt, very valuable;
but we all (especially you) have a good information, but let her just show the device that works, that all I'm asking; and you? You're not interested what's really working?

If she wants to show a device, she is welcome to—but only if she feels no pressure on her to do so AND only if no one else is applying pressure on her to do so.

..and me, what I'm interested in is truth—which is freely given, not coerced, cajoled or even coaxed out of..

..or else it is worthless

[AetherScientist]

You can't have reactive power unless you draw real power first. As soon as
you intercept and use or convert reactive power it is real power. If you stop
reactive power from going back to the source it becomes real. As far as I can
tell anyway.

..

Hi
I can tell you more exactly what I think.

Don Smith explains several times the importance of reactive power. I think indeed that Smith's system is a reactive power amplifier. Why? Read my conclusion.

Mr. Smith explains several times some concepts related with power factor.



In the picture you can see what Smith explains:
In the active power (dissipative), all the electrons rotates in all directions at the same time. This kind of energy is not rotating with order because there is a chaotic form of electron rotation.

In the case of reactive power, all the electrons rotate in the same direction at the same time. First they can rotate clockwise, giving electrical energy, and after their back-spin rotates counter-clockwise giving magnetic energy.
As you can see in the case of reactive power, there is an order in the rotation phenomena. If the electrons rotates in order, there is no
collision-chaos and they don't dissipate as heat.

Another advantage is the reactive power doesn't consume power (watts). Why? Because the power factor is zero and the maximum of voltage/amperes coincides with the zero of the opposite value (voltage or amperes).



As you can see more clearly now, you cannot increase nothing in a system of destruction (dissipative). If you want to increase something, then you need to use a constructive system.

[vidbid]

Can someone list the different practical ways we could use this device on a daily basis?

I can, but I don't feel like it.

I mean what good is technology if it doesn't help us do something?

What is the something that you're wanting it to help us do?

[AetherScientist]

The oersted experiment

The Oersted experiment is based to switch on and off a closed circuit. It proves the wire induces a magnetic field around it.
This is easily shown putting a compass near the wire. The compass needle change if the circuit is switched on.

If you perform the same experiment with AV plug (single wire transmission line), this won't happen.
This proves that longitudinal waves are not dissipative.



Uploaded with ImageShack.us

[Farmhand]

You misunderstood. I did use the hot wire. I got slow charging with the hot wire and ground. I used a 2uF 1000V cap. It charged about 2V per second and stopped at 24V.

As I expected and stated, using the neutral an ground does not work. It can't. They are the same, at least in the U.S.

Yeah it's the same here as far as I know, the neutral wire is connected to
ground.

[Farmhand]

Hi
I can tell you more exactly what I think.

Don Smith explains several times the importance of reactive power. I think indeed that Smith's system is a reactive power amplifier. Why? Read my conclusion.

Mr. Smith explains several times some concepts related with power factor.



In the picture you can see what Smith explains:
In the active power (dissipative), all the electrons rotates in all directions at the same time. This kind of energy is not rotating with order because there is a chaotic form of electron rotation.

In the case of reactive power, all the electrons rotate in the same direction at the same time. First they can rotate clockwise, giving electrical energy, and after their back-spin rotates counter-clockwise giving magnetic energy.
As you can see in the case of reactive power, there is an order in the rotation phenomena. If the electrons rotates in order, there is no
collision-chaos and they don't dissipate as heat.

Another advantage is the reactive power doesn't consume power (watts). Why? Because the power factor is zero and the maximum of voltage/amperes coincides with the zero of the opposite value (voltage or amperes).



As you can see more clearly now, you cannot increase nothing in a system of destruction (dissipative). If you want to increase something, then you need to use a constructive system.

Hi AetherScientist,

Well you can have whatever theories regarding it that you want. What I'm
talking about is the situation with the grid power. If you stop reactive power
returning to the source you pay for it. If you use it you pay. I think it is all
about the net energy transferred/consumed.

AC power - Wikipedia, the free encyclopedia

If you read these quotes with note to the parts in italic and in bold.

In a simple alternating current (AC) circuit consisting of a source and a linear load, both the current and voltage are sinusoidal. If the load is purely resistive, the two quantities reverse their polarity at the same time. At every instant the product of voltage and current is positive, indicating that the direction of energy flow does not reverse. In this case, only real power is transferred.
If the loads are purely reactive, then the voltage and current are 90 degrees out of phase. For half of each cycle, the product of voltage and current is positive, but on the other half of the cycle, the product is negative, indicating that on average, exactly as much energy flows toward the load as flows back. There is no net energy flow over one cycle. In this case, only reactive energy flows—there is no net transfer of energy to the load.
Practical loads have resistance, inductance, and capacitance, so both real and reactive power will flow to real loads. Power engineers measure apparent power as the magnitude of the vector sum of real and reactive power. Apparent power is the product of the root-mean-square of voltage and current.
Engineers care about apparent power, because even though the current associated with reactive power does no work at the load, it heats the wires, wasting energy. Conductors, transformers and generators must be sized to carry the total current, not just the current that does useful work.
Another consequence is that adding the apparent power for two loads will not accurately give the total apparent power unless they have the same displacement between current and voltage (the same power factor).
Conventionally, capacitors are considered to generate reactive power and inductors to consume it. If a capacitor and an inductor are placed in parallel, then the currents flowing through the inductor and the capacitor tend to cancel rather than add. This is the fundamental mechanism for controlling the power factor in electric power transmission; capacitors (or inductors) are inserted in a circuit to partially cancel reactive power 'consumed' by the load.

The ratio between real power and apparent power in a circuit is called the power factor. It's a practical measure of the efficiency of a power distribution system. For two systems transmitting the same amount of real power, the system with the lower power factor will have higher circulating currents due to energy that returns to the source from energy storage in the load. These higher currents produce higher losses and reduce overall transmission efficiency. A lower power factor circuit will have a higher apparent power and higher losses for the same amount of real power.

If you allow the reactive power to return to the source- you don't pay for it, but you do pay when it comes back if you use it.

If you don't allow it to return to the source and keep it local by correcting the
power factor you still pay but you can pay less because there are less losses.

Only real power can be used. The reactive power is originally part of the real
power drawn. But if it goes back you only pay for the net "real" power consumed.

Cheers
.

[AetherScientist]

In the post I wasn't talking about to pay or not to pay.
I was talking about reactive power is wattless. If you don't have losses you can amplify it. If you use active power you cannot amplify because you've losses.

I'm not talking about my theories. I'm explaining Smith's theories.

[Farmhand]

And so now for an experiment.

Since I have a source of AC I can see the result of using energy from the
from the different lines, I have the neutral terminals grounded and another
separate ground. What happens when I connect the bare fluro tube (hand held)
to the active terminal is there is a current path into the fluro and a capacitive
action into me and the surroundings, real power is consumed. If I connect a
capacitor plate (metal plate) to the fluro's free end there is more power
consumed and the light is more. if I create a connection to the other ground
there is more power consumed and the light is even brighter.

Experiment video.
One wire lighting power draw - YouTube

There is always power consumed even in one wire lighting and wireless lighting
of stuff. With respect to lighting things and seeing the power input go down,
this is just a result of a system that is wasteful to begin with and made less
wasteful with the addition of the load.



P.S. Of course when the fluro is connected to the neutral terminal it won't
light (by one wire method) even though the voltage is exactly the same as the active terminal, this
is because being grounded it has no effective capacitive path for current.
Current wont flow to me or the surroundings because we are at the same
potential as the neutral terminal. Unless the grounding is poor that is.


..

[vidbid]

The Oersted experiment is based to switch on and off a closed circuit. It proves the wire induces a magnetic field around it.
This is easily shown putting a compass near the wire. The compass needle change if the circuit is switched on.

If you perform the same experiment with AV plug (single wire transmission line), this won't happen.
This proves that longitudinal waves are not dissipative.

I find that bit of information very useful.

Thanks for adding to my understanding.

And if I'm not mistaken, longitudinal waves are the same as electrostatic compression waves.

[dragon]

For those still playing with the circuits I presented - A battery in the same configuration will also charge just as a cap will. Keep in mind charging by induction. We don't have to force it to charge - such as using the driving circuit - we only need to create a condition that allows it to charge because of an imbalance.

If you have further questions you can PM me. I don't plan to post anything of great significance on this forum from here out.