Holcomb Energy Systems - the Holy Grail may have arrived

[Rakarskiy]

I'm sure it will work, it's just that the system needs the load. The battery in this case is also a voltage regulator, low level. If it runs for a while, the voltage on the battery should rise. Unfortunately, I don't have all the data from the experiment, namely the condition of the battery itself, so it's hard for me to say anything. What is captured in the video suggests that the system has enough reserve to slowly charge the battery and provide autonomous operation. I'm already getting questions about how to do this myself, so I'm going to produce a brochure with step-by-step instructions for garage do-it-yourselfers.

[bistander]

I'm sure it will work, it's just that the system needs the load. The battery in this case is also a voltage regulator, low level. If it runs for a while, the voltage on the battery should rise. Unfortunately, I don't have all the data from the experiment, namely the condition of the battery itself, so it's hard for me to say anything. What is captured in the video suggests that the system has enough reserve to slowly charge the battery and provide autonomous operation. I'm already getting questions about how to do this myself, so I'm going to produce a brochure with step-by-step instructions for garage do-it-yourselfers.

Hi Rakarskiy,
You say "it's just that the system needs the load."

Then replace the battery with a capacitor and resistor (in parallel). Gives you the buffer and a load.

Prove it works before conning others into building it.
bi

[Rakarskiy]

I'm not encouraging anyone to do anything, that's the first thing. Secondly, in order to do it you need to understand a lot of things, starting with what an electrical circuit is, and most importantly how the electromagnetic current generator system and its magnetic system works. I always warn everyone that it will take more than once to assemble and disassemble, which is not cheap and requires skills and equipment, including measuring equipment. The guy who did this is not a novice in free energy systems.

[bistander]

I'm not encouraging anyone to do anything, that's the first thing. Secondly, in order to do it you need to understand a lot of things, starting with what an electrical circuit is, and most importantly how the electromagnetic current generator system and its magnetic system works. I always warn everyone that it will take more than once to assemble and disassemble, which is not cheap and requires skills and equipment, including measuring equipment. The guy who did this is not a novice in free energy systems.

Hi Rakarskiy,

You say "I'm not encouraging anyone to do anything".

But the last sentence I'm your previous post "I'm going to produce a brochure with step-by-step instructions for garage do-it-yourselfers".

Appears a contradiction.

And we both know that we have disagreement about understanding lots of things, like electric circuits, electromagnetic current generator systems and its magnetic system workings.

So again, I ask, show proof of your claim that the device displayed in that video provided free energy.

Regards,
bi

[Rakarskiy]

No contradiction! In order not to offend everyone who is interested, I will describe how I see this process, with all the calculations “why exactly this is so”. But to do or not to do, everyone makes the decision themselves, at their own peril and risk.

[bistander]

Regarding Holcomb's claimed free energy devices, I saw this post elsewhere today. I'll edit with the link.
bi

MeteoricNinja68

Re: uber darthium post# 4261

Thursday, 02/08/2024 3:35:04 PM

Uber this is what I have learnt today from Holcomb Insiders on how the testing results are manipulated, and people are fooled, much like the videos we are seeing posted here, you can probably dissect the below better than most if not all here:

The ILPG (in line power generation) system uses a massive amount of capacitors on the output side.

The line voltage from the input remains consistent through input to output, but the amperage on the output side can be much higher than the input side.

This turns out to be all “reactive power”, oscillating between the capacitor bank and the inductive coils of the ILPG.

The end result is a drastically distorted power factor (PF) that cannot be corrected and provides no real power to the load. In fact the ILPG and capacitor bank add to the existing load with no benefit in energy savings whatsoever.

At first, the system looks impressive, like it is doing something, raising the amperage up, but that only works when there is a massive amount of capacitance on the load side, which is how the power factor is distorted and no real power, kilowatts (KW) are generated.

If the power meters are correctly used on both the input and output sides of the system, they will show the real power, kilowatts, are higher on the input side and lower on the output side.

https://investorshub.advfn.com/board...e_id=173805839

[Rakarskiy]

2024-02-09_144421.jpg

All you need to know about the author of this information. Their snot won't help them.
For many who want to do the same, I recommend a simple test on two ammeters and one voltmeter everything will be immediately visible (at the end of the post).

Wise Eye OverUnity: Electric generator with solid state magnetic rotor. (rakatskiy.blogspot.com)

[bistander]

For many who want to do the same, I recommend a simple test on two ammeters and one voltmeter everything will be immediately visible (at the end of the post).

So you have no need for power factor (phase angle)?

bi

[Rakarskiy]

Power factor for a dc circuit? (In other words, in a DC circuit there is no power factor).
This is the most effective way to identify opportunities.

[bistander]

Power factor for a dc circuit? (In other words, in a DC circuit there is no power factor).
This is the most effective way to identify opportunities.

So you are not addressing the Holcomb system. OK.
bi

[Rakarskiy]

Holcomb's system is exactly the same as the one tested in the video, with the difference being that Holcomb controls the rotor solenoids with a solid state keying system.
Personally, I have no doubt about the performance of his system. I doubt that it will be quickly integrated into the power grid (games with a control system that protects the electricity producer are always losing), but for a private house as an autonomous source, or even better, a boat with an electric drive, it is reasonable to use it.

Electromagnetic generator OverUnity (youtube.com)

[bistander]

Hi Rakarskiy,

Your system:

2024-02-02_131440.jpg


Here when current to battery is zero and three switches are closed (conducting), and load bulb is drawing power, remove the battery. If load bulb continues to shine continuously forever we have new world order.
Ain't gonna happen!
bi

[Rakarskiy]

I do not recommend this, at least in this unstable installation. The battery acts as a voltage regulator in the DC circuit. If there is no current in the circuit with the battery, the battery does not charge or discharge, but it does a good job of regulating the lower voltage level. Professors joke with you by requiring you to exclude the battery, the formula for the current from the generator in a DC circuit is as follows:

I = (E_gen - U_bat) / (R + r)

Where: E_gen - Alternator phase EMF at idle speed; U_bat - voltage across the terminals of a battery connected to a DC circuit; (R + r) - load and generator phase impedances, respectively.

Even if the battery is excluded from the current components in a DC circuit, it fulfils an important role while remaining charged.

Failure to have this ballast at phase underpower will cause your system to lose the unreduced DC circuit voltage level, resulting in system zeroing.Be careful not to give in to tricky demands.Maintaining the operating voltage in the DC circuit can be accomplished through a capacitor.The approximate formula calculation is:

С = (3200 * I) / (U*k) k - then the ripple factor ( =1-0,0001%), I - current strength, U - mains voltage (online calculator truth in russian)

For a voltage of 13V and current 10A (130W), the calculated smoothing capacitance of the DC capacitor is as follows: 189125295.51 µF (189.1 Farad)
A battery is obviously cheaper to use for this purpose

That's how you can be divorced, but not me, I shared this example with you so that you also do not get sold a losing solution.

[bistander]

I do not recommend this, at least in this unstable installation. The battery acts as a voltage regulator in the DC circuit. If there is no current in the circuit with the battery, the battery does not charge or discharge, but it does a good job of regulating the lower voltage level. Professors joke with you by requiring you to exclude the battery, the formula for the current from the generator in a DC circuit is as follows:

I = (E_gen - U_bat) / (R + r)

Where: E_gen - Alternator phase EMF at idle speed; U_bat - voltage across the terminals of a battery connected to a DC circuit; (R + r) - load and generator phase impedances, respectively.

Even if the battery is excluded from the current components in a DC circuit, it fulfils an important role while remaining charged.

Failure to have this ballast at phase underpower will cause your system to lose the unreduced DC circuit voltage level, resulting in system zeroing.Be careful not to give in to tricky demands.Maintaining the operating voltage in the DC circuit can be accomplished through a capacitor.The approximate formula calculation is:

С = (3200 * I) / (U*k) k - then the ripple factor ( =1-0,0001%), I - current strength, U - mains voltage (online calculator truth in russian)

For a voltage of 13V and current 10A (130W), the calculated smoothing capacitance of the DC capacitor is as follows: 189125295.51 µF (189.1 Farad)
A battery is obviously cheaper to use for this purpose

That's how you can be divorced, but not me, I shared this example with you so that you also do not get sold a losing solution.

Hi Rakarskiy,

I just happened to have a suitable capacitor.

IMG_20240210_132652391.jpg

So you say to put this in place of the battery in your circuit, it will run continuously forever (many years) powering the load bulb at 130 W without any input?

It can not.

Ultra capacitor modules like I display (rated at 15V, 433F) are not too expensive and available. It is possible to procure, so prove your claim. Change the world order.
bi

[Rakarskiy]

Never listen to the provocations of American, Russian and other "professors" who are at the service of the system of global world capital.

In addition to accumulation, a battery is also a regulator of the network voltage when it is included in a DC circuit. A good example of the work of this device in the electric circuit of any car. A simple voltage regulator is used for its charge level.

It is enough to put an ammeter in the battery circuit and see what role it fulfils at a given moment of time (load charge or source-discharge).
About condensers, the variant suggested above, to the garage master who has made a self-propelled generator will be unavailable and to buy a set of condensers with capacity of at least 10 Farad is a very expensive event.

10 Farad is 10000000 mkF if to take a condenser 35V/10000 mkF you need 1000 pcs one piece costs about 2.5-3 dollars, that is 2500-3000 dolars
Ordinary starting battery 65 A*h has a cost in the range of 72-156 USD. At the same time at start-up there is no need to charge the battery, unlike condenser batteries, which still need to be assembled into a proper assembly.