Re-Inventing The Wheel-Part1-Clemente_Figuera

[Ufopolitics]

UFOP;

Total mass of Part G must match total wattage expended by hi and low primary sets with head room......ie. 500 va over. no need to over complicate.
mass does not need to be equal to primaries just total wattage plus headroom.


MM

Hi MM,

So that means you already have all your primaries set built up and they do the expected output at secondaries in order to build part G Toroid right?


Regards


Ufopolitics

[Ufopolitics]

The order I am working on Figuera...

Hello to All,

While I see how you all guys are developing the Figuera device...I wanted to share the order I am working on it.

This device depends MAINLY on the atmospheric or SPATIAL FLUCTUATIONS of the magnetic fields at PRIMARIES, in order to function properly...So, my FIRST focusing is on Primaries construction, and even reduced to just ONE Primary electromagnet design, in order to observe its expansion and retraction. Before I proceed to reproduce more iron cores and windings...

Once we have just one primary iron core size wound with low resistance wire, then, in order to check FULL FIELD EXPANSION, we must add the secondary core (without any windings whatsoever), and see how this secondary iron core expands our primary field without major decay.

1-If we all resume the primaries operation, we realize it bolts down to a HIGH FIELD and a LOW FIELD basically.

2-However, the TRANSITION between Hi-Lo Field must be done very smoothly...meaning NOT RADICAL, NOT ABRUPT, NOT A STEEP DROP OFF.

But first we need number one....achieving a Hi and a Lo Field without collapsing field all the way.

High Field is basically dependent upon the primary core and wire design...and we are not looking to spent hundred of watts on this endeavor, but exactly all we need to expand field along the secondary core length.

Many do not believe in the screening of a CRT to see the magnetic field(s)...well, too bad because I do...and it helps me to visualize it perfectly and with so much accuracy as the signal we read in our scopes. Basically a simple Black and White old small TV, that I have disconnected its vertical coil at Cathode Tube...so it only shows me a horizontal line...is enough to observe Hi and Lo Field.

Low Field...before spending a lot of money on a bunch of resistors...I used a Potentiometer and the wiper set at the extreme opposite end further away from the straight positive connection or High Field....then I test to reach lowest spot before my horizontal line becomes flat line (dead field)...just a bit above will simulate as Figuera writes..."Field moves further away"...So, once I got this perfect Lo Field, then I measure wiper total resistance and divide it by the number of segments at commutator...that gives me the total series resistors I would need to install, making the "FALL" from Hi to Lo in a smooth fashion. And here is understood that the more segments we have at commutator, the smoother the fluctuations would take place.

Under a B&W CRT horizontal line screening We could see the field response and intensity under fluctuations of current changes...just like flashing a powerful bulb on a dark screen.

The way I understand Part G Toroid...is that Primaries design, basically their iron core TOTAL mass (Which means and INCLUDES ALL Primaries involved in our set up), DICTATES the size of our Full Iron Toroid....

So, basically...this is the reason why... I am still working on Primaries...and again, I am just sharing above the way I am proceeding with the development of my build here, just like "thinking out loud"...and by no means I want anyone to do or necessarily believe in what I have written above...


Regards


Ufopolitics

[Ufopolitics]

""and always in contact with two of their contacts""

Seaad,

I already answer to above...don't you believe me?

It means the brush must be wider than one commutator segment, could be one and a half...and this is done in order NOT to allow Field to EVER Collapse, so like MM wrote..."Make BEFORE Brake"...Which means that if brush is the same width as contacting segment, when both are fully aligned/engaged then either previous or next would be off...NOT GOOD!

""without any more complications than the turning of a brush or group of brushes that move circularly around the cylinder “G” "" Pat. 44267

Back in 1908... Semiconductors, specifically Diodes were not discovered yet...so, Figuera as Buforn were depending on a second commutator plus brushes in order to convert the secondaries output from AC to DC...did you know that was the way back then?...when diodes were still not available on the market?


Hope you understand now.


Ufopolitics

[hanon1492]

Really,

Like i have been telling everyone part G controls the currant not the primaries.

MM

I believe that was not the kind of answer Elcheapo was expecting after the time and effort he surely have dedicated to replicate that design. At least tell him what was the total value of your resistance wire , in ohms, in order to regulate the current.

[seaad]

TWO of their contacts

""and always in contact with two of their contacts""

""without any more complications than the turning of a brush or group of brushes that move circularly around the cylinder “G” "" Pat. 44267

[Elcheapo]

demo

MM,

I need your help. Did a spread sheet with your data for the cop3 demo. Copy is attached.
As you can see, if using 50 volts, the total current for both hi side & low side would be 12 amps.
So 12 x 50 gives you 600 watts as the power input. A far cry from 100W.
This is all dependent of course on the coil's reactance. My coils are wound with 340 turns of 19awg
on iron cores of 1.5" dia. & 2.75" long. XL checks out at 6 ohms using 60 hz. Close to yours.
Not knowing the XL of your coils, I used a value of 7 ohms.

Take note of the chart that the total current for both hi & lo is always at 12 amps.
So current never CHANGES in this 3 coil system. Only the B field should vary up & down.(no lenz)

You said your input was 50 volts @ 2 amps. Did you give me some wrong info?
If not then tell me what I have to change to make this thing work like you did.

[bistander]

Comm jumpers

The jumpers on the commutator just cut in half the number of wires needed between the commutator and resistor bank.

[seaad]

Thank you for answers UFO-p!
I have to consider your answers. But first, have you on purpose excluded some commutator jumpers on your img.?
Arne

[Ufopolitics]

--> Hanon, all

Thanks for reply. I Have more weird text, a new part arrived to my mind from patent 44267.

""the resistance makes the function of a distributor of the current because that current not going to excite some electromagnets excites others and so on;""

Hello Seaad,

It means that Current is always going to be utilized on either one primary or the other primary electromagnet.

When one electromagnet (primary speaking) is filling up with current, the other one is emptying out, and so on per each mating pairs (facing each others with Induced Coil inserted in between..

The above together with this; ""and always in contact with two of their contacts"", makes me think of another post from me in Free Energy | searching for free energy and discussing free energy februari.

That part of text refers to Commutator Elements, and it means that brush must be in contact always with Two Commutator Elements-Segments, and in order to achieve this, the brush must be wider than one commutator element, or close to be two elements in contact face width.

Pls see my pic. with fewer steps below (N.b. L+R is Inductance and coil resistanse )

It seems to me that the electromagnets are to be connected gradually!

Hanon can you check translation?

Arne

Negative, Each Set of Mating Electromagnets in Primaries receive the same currents disbursal, and the minus-plus/ plus-minus relation of currents relate per each pair (N-S).

You could also connect resistors and commutator like I am making it below:

[IMG][/IMG]


And of course that is for a 10 elements commutator like I have...Figuera's Patent is 16 elements. the main point here is that the two positive output terminals to primaries must be 180º apart (on img above it is 1 and 6) in order to do a perfect "plus-minus/minus-plus" disbursal of currents.

Regards


Ufopolitics

[seaad]

More wierd translation or text

--> Hanon, all

Thanks for reply. I Have more weird text, a new part arrived to my mind from patent 44267.

""the resistance makes the function of a distributor of the current because that current not going to excite some electromagnets excites others and so on;""

The above together with this; ""and always in contact with two of their contacts"", makes me think of another post from me in Free Energy | searching for free energy and discussing free energy februari.

Pls see my pic. with fewer steps below (N.b. L+R is Inductance and coil resistanse )

It seems to me that the electromagnets are to be connected gradually!

Hanon can you check translation?

Arne

[Netica]

Hi marathonman,

I have been thinking about something and wondering if you can help.

At the point that the brush makes contact with the G core windings, depending on the direction of the winding on the toroid the effect can be either N,N at the area of brush contact or S,S at the area of brush contact. I am wondering if it matters and if it should relate to the primaries where the primaries may be for example N,N.

[hanon1492]

HI MM,

Thanks for the reply. I was looking for the root source of the 14.8 lbs/kW. Where did you get that figure or how did you calculate/measure it?

References would help.

Thanks,

bi

The way to calculate it is with the equation: Power = Force • Velocity

Taking as velocity the movement of the magnetic lines back and forth cutting the induced coil wires. Yo can go into further detail in this old post link. I recalculated that value of 14.8 lb/KW and I think it is adjusted for an induced coil length of 2 inches and a frequency of 60 Hz. If you are good with english units please confirm if I did it well or not. Thanks

Ufo, Nice graph with the two signals generated in the commutator. I think we were talking about the same concept with two different explanations

Seead, I understand your concern about the translated "small line y" but in spanish is also a weird way to call it ("linea y"). It also crashed my mind to see such a poor explanation of such an important part just with that word, or just calling it "induced circuit". Maybe you can also translated as "row y" referencing to the patent drawing where the 7 induced as placed one after another. Anyway, not even a proper wy to call it. We can take help from the latest Buforn patents where he call it as "another electromagnet between both inducer electromagnets".

[Ufopolitics]

Sure Ufo, I considered the whole primary circuit including the supply.

Yes, the negative connections go from the N and S primary coils to points 180º apart on the toroid winding (represented by 8 and 1 respectively)

PartG is used as a voltage (or current) divider to proportion the current from the source to the primary coils N and S. It is desired to have a certain min/max relationship of current through N and S, where N is max when S is min and vice versa.

At 12:00 (point 8) on the toroid is where primary N connects, so when the brush (being at supply positive) is at point 8, maximum current goes to N. It is desired that at this point, minimum current will flow to primary S which is connected to point 1 on the toroid.

Current must flow through the toroid winding between point 8 and point 1 to have continuity to supply positive via the brush. There are two paths in the toroid winding between 8 and 1, the right side and the left side, which are in parallel with each other and of equal lengths and turns. So to figure the impedance of the toroid winding (8 to 1) we have 2 impedances in parallel, each with equal resistance and inductive reactance. Because the inductance of the 2 coils is linked, the mutual inductance must be considered. Because the current direction is opposite in the 2 halves of the toroid winding, that mutual inductance essentially cancels the inductive reactance for the equivalent impedance.

Therefore, the potential difference between point 8 and point 1 will for all practical purposes be zero. This means the current will not be a minimum in primary S as desired, but essentially maximum, the same as in primary N.

I only looked at this one position (brush at point 8) because this case has the 2 toroid winding halves equal and the parallel combination of 2 equal inductance coils with mutual inductance is easily calculated. Obviously the same is true for brush at point 1. All the other points in between I am unsure and not about to attempt those calculations.

I see the same results looking at it another way using the flux to figure inductance. I mentioned this to you before. When you have half of the toroid mmf CW and the other half equal and CCW, resulting flux is zero and therefore the inductance of the coils on the toroid is zero. As the brush rotates around the toroid, the potential changes and mmf (Ampere-turns) are unlikely to be equal and opposite so there may be inductance in the coils and impedance causing unequal currents in primary N and primary S.

My take anyway,

bi

edit: Weird thought. What if you wound the left half of the toroid backwards from the right side half? Mutual inductance would add instead of subtract.

Hi Bistander,

Yes, exactly, and therefore we must consider the resistance-reactance from both primaries N-S to have a full idea of the whole electromagnetic circuit right?.

But the point here is that we could conclude that Part G Toroid is receiving "some" Negative Input there, at 8 and 1 points, and since their resistance-inductance is up to some value we could then calculate the electromagnetic effects being developed at toroid iron core, which you seem to be disregarding or simply not considering them.

Take a look at my take in the two images below...when brush is sweeping left 180º side:

[IMG][/IMG]

[IMG][/IMG]


Regards


Ufopolitics

[Ufopolitics]

Following the patent guidelines the commutator sweeps all contacts in ascending order (1 to 16). But as 9 is jumped with 8, 10 jumped with 7, 11 jumped with 6, 12 jumped with 5, 13 jumped with 4, 14 jumped with 3, 15 jumped with 2, and 16 jumped with 1, as the drawing shows ( Ufo, please see all rhese jumped in the drawing, not just the two jumped pairs you mentioned )then the resistance is sweep during half cycle
in this direction R1------> and the second half cycle in the way back <---------R2 to complete a whole turn of the brush. In half turn the brush(contact) is approaching the side tap of one set of electromagnets and later in the other direction it is approaching the side tap of the other set of electromagnets.

R1------------------------------------------------------------>
1___ 2____3____4____5____6____7____8
16__15 __14___13___12__11___10___9
<-------------------------------------------------------------R2

Hanon,

That is exactly the same as going at:

1_2_3_4_5_6_7_8_9_10_11_12_13_14_15_16

Following THE SAME rotation (Brush sweeping Commutator NEVER changes directions!!):

R >->->->->->->->->->->->->->->->->->R

Commutator numbers are sequential from 1 to 16, as so the Brush sweep...and the fact that Figuera "reuses" the same sequence of resistors every 180º is what is confusing you.

There is HALF cycle (180º) from 1 to 8, then another HALF Cycle (180º) from 9 to 16 where the same resistors alignment start over again.

Hanon you can NOT show just one signal for both Primaries like You are doing above. Remember they are going at UNISON, which means within the same Space/Time.

Like below:

[IMG][/IMG]

Ufo, I do not know how you get to that interpretation of the patent. Please read again the patent and you will see that this is the sequence proposed. This sequence gives the two 180° unphased signals described in the patent


Image done by Kekho77

Now that Signal is fine (Original)...not just a single one!!

Take a look again at Figuera Image below:

[IMG][/IMG]

Vuelve a leer esa Patente Hanon!!...estudiala más...


Saludos


Ufopolitics

[bistander]

14.8

yes for every kilowatt produced 14.8 lbs pressure is needed to produce it. why do you think generators bog down, because of extreme attraction forces, many, many lbs force. in the Figuera device it is opposite, repelling forces.

MM

HI MM,

Thanks for the reply. I was looking for the root source of the 14.8 lbs/kW. Where did you get that figure or how did you calculate/measure it?

References would help.

Thanks,

bi