Figuera Device, Part G Continuum.( Serious Builders Only)

[marathonman]

Part G

Part G is the sum of all smaller part adding up to part G plus head room including all power used for electronics and or motor so yes it will be ok. you have to take into consideration of how many winding's your part G will use to get your proper currant window ....ie highs and lows. just remember it will not take a lot of winding's to get currant variation. i myself would go for the smaller toroid and stack them to get proper va rating.
unfortunately i haven't the math prowess to calculate that at this time. am working on that as we speak. both will work, you just need to decide on amperage used, wire size and number of turns.

i have fliers out for a geek math wiz but two have failed so far. so for now i am guessing but i know i am close. when i get my timing board assembled i will test it and go from there.

that is the best advice i can give for now.


MM

[Shadow119g]

Thank you MM and UFO!
I will try out what I have because I have it, but I am really looking for more guidance.
I also just ordered the brushes and commutator like Cornboy and Cadman as I made an eight count commutator before.
Thanks again,
SHADOW

[Cornboy 555]

My part G wire is .4 rectangle wire which might be a little to wide as we will see as i have only 16 winds but my core is 5 plus inches deep = a lot of core exposure. my primaries are 18 awg, two separate winding's.( parallel ) this is what i have and all i have. if i had the money and or wire i would rather wind my primaries with 14 awg but all i have is a 10 lb spool of 18 awg wire.

the reason my wire is so wide was i planed on a mechanical part G but changed my mind after i bought wire to switch to electronic timing to mimic brush rotation. also the commutator bars in figueras patent are actual thick winding"s on a core. my suspicion was confermed that the wire count was low from hanon's video and my own test i conducted thus confermed this to be true. 1/4 turn of the variac using DC was all it took to vary the currant.

the whole philosophy behind the madness is why shouldn't the primaries be the least ohmage as possible as part G controls the currant. so this is why the primaries should be lowest as possible to attain the highest magnetization as possible.

the primaries are specifically there to be "ELECTROMAGNETS", NOT GOVERN CURRANT so why not design them to do there job the greatest they can be. so basically if you can get away with it....zero ohms. imagine that. if i had three inch length copper coductor to wrap it with i would do it.

when the primaries are at peak currant, they are there for only 2.08 miliseconds (USA 60 hz) 2.5 miliseconds ( European 50 hz) at 16 figuera intervals. at this length of time almost no amount of currant can do damage to wires (WITHIN REASON) so wind your primaries to take advantage of this peak with no adverse affects. ie. biggest bang for your buck as to say.

part G was made to control the currant NOT THE PRIMARIES so thus we need to configure the primaries with as little resistance and self inductance as possible allowing PART G TO DO IT'S JOB as it so wonderfully does.

do you get my drift.

I love part G.

MM

Thanks MM, so what i gather from your explanation is that we are dealing with an almost dead short circuit for a very short time ?.

If that is the case we should ALL, have a heavy switch to engage the external power source, ONLY after the commutator is brought up to speed.

Also please correct me if i am wrong, the feedback from the primaries into part G creates a resistance to incoming external current?.

Just trying to get my head around the theory before i start my build.

Warm Regards Cornboy.

[marathonman]

Part G

CORNBOY;

QUOTE; "so what i gather from your explanation is that we are dealing with an almost dead short circuit for a very short time ?."

Yes in a sense when at N or S are high for 2 plus milliseconds for 50 or 60 Hz. this will be the bang part prom the primaries, peaking for max output but to short of a time for currant to do damage through heat.

QUOTE; "If that is the case we should ALL, have a heavy switch to engage the external power source, ONLY after the commutator is brought up to speed."

Yes, if using high voltage, if low voltage i would not worry about it.
The delay is only for seconds at most. also connection made from the output back to the field magnets through part G has to be separately operated off the output with it's own switch so you can turn it off or leave it on with or with out an external working load.

QUOTE; "Also please correct me if i am wrong, the feedback from the primaries into part G creates a resistance to incoming external current?."

yes, if referring to original starting supply and the second secondaries incoming pressure as it is only used to replace losses occurred from core, heat and wire losses which are little. it will self regulate if pressure drops. of course the starting supply can be removed.


SHADOW;

Very wise to follow these excellent builders if using mechanical switching keeping in mind all that i have revealed.


Figuera


MM

[marathonman]

Eddy currants

According to my research the whole reason Figuera chose DC is that it is vastly superior to AC to power the electromagnets in this device. since he is not reversing the flow, the electromagnets it will not produce eddy currants like AC would that rob power from reverse currants and heat. thus by using DC he was left with extremely efficient electromagnets that not only do not reverse polarity (no eddy currant) but also no hysteresis to deal with. the graph below is an example of pure iron hysteresis curve. not only is the curve unbelievably thin but there is no reversal of currant so the electromagnet operates in the top quadrant remaining extremely efficient at all times, unreachable by AC. so we have no hysteresis or eddy currants in the primaries, that is why Figuera chose pure iron for his primaries, biggest bang for the buck. but unfortunately pure iron cost are completely outrageous and we are forced to use laminated cores

unfortunately we have both in the secondary core so it would be advisable to use laminated as eddies and hysteresis are live and kicking. this is why the primaries and secondaries are separate.

last pic is for calculating induction in a toroid. i am thinking Figuera calculated his inductance statically then varied it dynamically as the brush moves. this would not change the currant reduction calculations just changing contact points dynamically on the fly thus constant currant variatiing in complete unison.
i have been confused on how to calculate induction in part G but no more. think about it, why would the calculation change just because you changed contact points, they wouldn't. as long as your static calculations in part G are correct the dynamic currant increase/decrease remain the same. all you are doing is changing contact points over time.

just to tell you how much money i save on wire. at Temco, they want $133.00 for 10 lbs of 18 awg Gp 200 wire. at EIS i paid $68.00 for 9.8 lbs of the same. that is crazy savings if you ask me. plus EIS warehouse is only a few miles from my house. guess who i shop at. be fore warned EIS website is horse and buggy slow as in Amish country.

https://shop.eis-inc.com/sap(bD1lbiZ...?prod_area=442



[IMG][/IMG]



MM

[marathonman]

Opinion not needed

Post was deleted for integity of thread.
[IMG][/IMG]

thank you and have a nice day
MM

[marathonman]

Thread

Post deleted for integrity of thread.

MM

[marathonman]

Pics

UPDATE TO ALL;

Postimage dot org is going down soon be prepared.


MM

[Cadman]

Meanwhile ...

Since we all seem to be occupied with building a switching device or waiting on parts I would like to present some research on the inducer or primary field winding itself.

To summarize the relevant parameters given by Marathonman:
1. primaries need to have low self induction as fast response time in essential.
2. lowest ohms as possible.

Additionally I would like to add:
A. Have the least inductive discharge as practical
B. Have the lowest resistance.
C. Consume the least wattage.

For the example let's assume a requirement of 6000 amp turns, 18 GA wire, 576 turns, and a 4” long coil.

The usual method of winding a generator field coil is to wire it as a single coil. A coil constructed like this would be 3.671 ohms and require a current of 10.4167 amps. Of course this is not practical as that amount of current will destroy the wire. Just for example say it could take the current. The voltage required to drive that current through the resistance would be 38.2151 volts and the watts would be 397.8 watts. Coil heating would be problematic from the wattage alone. This coil would also have a high self inductance and produce a large inductive discharge.

What I would like to present is a way to wind this particular coil in such a manner that it will meet all the above material and performance parameters without difficulty. It will use the same amount and size of wire, provide the requisite amp turns, use a fraction of the voltage and consume only a little over 44 watts instead of nearly 400 watts. The inductive discharge will also be reduced.

Shall I proceed?

CM

[Ufopolitics]

Meanwhile ...

What I would like to present is a way to wind this particular coil in such a manner that it will meet all the above material and performance parameters without difficulty. It will use the same amount and size of wire, provide the requisite amp turns, use a fraction of the voltage and consume only a little over 44 watts instead of nearly 400 watts. The inductive discharge will also be reduced.

Shall I proceed?

CM

Absolutely!!...I am all eyes...

Go for it CM!


Regards


Ufopolitics

[Ufopolitics]

Winding Energy Method...

@Cadman,

I am anxious to see that Method to achieve such energy savings...however, always consider IF it will generate the same stronger magnetic field as the "expensive method" will do...


Regards


Ufopolitics

[Cadman]

To give proper credit I must say that Doug1 at OU opened my eyes to this a couple of years ago. To date I haven't seen it presented in coherent detail.

To summarize so far:
18 AWG wire (6.384 ohms / 1000 ft), 0.006384 ohm / ft
576 turns @ 10.4167A = 6000.02 amp turns (use 10.41A)
576 turns = 575 ft
575 ft = 3.671 ohms

For a single coil of 3.671 ohm resistance
V = I * R
V = 10.41 * 3.671 = 38.2151 V
38.2151 * 10.41 = 397.819 watts
-----

A tested method to reduce sparking was presented in the book The Electromagnet and Electromagnetic Mechanism in the 19th century. Of the different methods presented this one was ranked second for effectiveness. The method was to subdivide the coil into individual layers. This is NOT a twisted multi-filar coil. Each layer(s) is wound as a single coil with it's ends connected to opposite polarities. The coil layers are connected in parallel to each other. This results in each coil segment being slightly different lengths with the shortest length in the first layer and the longest in the last layer, each layer increasing in resistance from inner to outer. This slightly staggers the time involved with the inductive discharge and reduces the sparking.

To simplify calculations I present each coil as being equal in length. I am also using two layers for each coil. If single layers were used the wattage would be even less. I chose two layers simply to bring the voltage requirement into a more practical range for myself.

The key to all of this is, the amperage per turn is only a small fraction of the total amperage required.

This coil will have 96 turns per layer, 6 layers, 576 turns total. A double layer having 192 turns.
-----------
Parallel coil windings
96 turns per layer, 192 turns in 3 double layers

10.41 A / 576 turns = .01808 A per turn

.006384 * 192 = 1.2257 ohms per double layer
.01818 * 192 = 3.4714 A per double layer
V = I * R
3.4714 * 1.2257 = 4.2549 V
Each double layer = 3.4714 amp @ 4.2549 volt, 14.7705 watts
14.7705 watts * 3 double layers = 44.31 watts

(calculate the parallel resistance)
Rtotal = 1/((1/R1)+(1/R2)+(1/R3))...
1 / 1.2257 = 0.81586
0.81586 * 3 = 2.44758
1 / 2.44758 = 0.40856 ohms parallel
I = V / R
4.2549 / 0.40856 = 10.4143 A, 44.31 watts
-----------
Each individual coil segment is now subject to only 3.47 amps at the peak maximums.
Current from the sine wave is well within the bounds of the coil wire.
Current for each turn is still 0.01808 amp and voltage has dropped to 4.25 volts.

This is presented in the hope that it will be useful and confirmed or disproved through independent testing.

Regards,
CM

[marathonman]

Right

CADMAN;

18 awg chart.


QUOTE; "This slightly staggers the time involved with the inductive discharge and reduces the sparking."

reduces sparking, i thought that was make before break was for and this will also reduce inductive kick to part G. ????

MM

[Cadman]

MM,

Yes??

Irms = 3.4714 / 1.4142 = 2.455

Amps @ 2000A per in^2 = 2.55

[Ufopolitics]

To give proper credit I must say that Doug1 at OU opened my eyes to this a couple of years ago. To date I haven't seen it presented in coherent detail.

To summarize so far:
18 AWG wire (6.384 ohms / 1000 ft), 0.006384 ohm / ft
576 turns @ 10.4167A = 6000.02 amp turns (use 10.41A)
576 turns = 575 ft
575 ft = 3.671 ohms

For a single coil of 3.671 ohm resistance
V = I * R
V = 10.41 * 3.671 = 38.2151 V
38.2151 * 10.41 = 397.819 watts
-----

A tested method to reduce sparking was presented in the book The Electromagnet and Electromagnetic Mechanism in the 19th century. Of the different methods presented this one was ranked second for effectiveness. The method was to subdivide the coil into individual layers. This is NOT a twisted multi-filar coil. Each layer(s) is wound as a single coil with it's ends connected to opposite polarities. The coil layers are connected in parallel to each other. This results in each coil segment being slightly different lengths with the shortest length in the first layer and the longest in the last layer, each layer increasing in resistance from inner to outer. This slightly staggers the time involved with the inductive discharge and reduces the sparking.

To simplify calculations I present each coil as being equal in length. I am also using two layers for each coil. If single layers were used the wattage would be even less. I chose two layers simply to bring the voltage requirement into a more practical range for myself.

The key to all of this is, the amperage per turn is only a small fraction of the total amperage required.

This coil will have 96 turns per layer, 6 layers, 576 turns total. A double layer having 192 turns.
-----------
Parallel coil windings
96 turns per layer, 192 turns in 3 double layers

10.41 A / 576 turns = .01808 A per turn

.006384 * 192 = 1.2257 ohms per double layer
.01818 * 192 = 3.4714 A per double layer
V = I * R
3.4714 * 1.2257 = 4.2549 V
Each double layer = 3.4714 amp @ 4.2549 volt, 14.7705 watts
14.7705 watts * 3 double layers = 44.31 watts

(calculate the parallel resistance)
Rtotal = 1/((1/R1)+(1/R2)+(1/R3))...
1 / 1.2257 = 0.81586
0.81586 * 3 = 2.44758
1 / 2.44758 = 0.40856 ohms parallel
I = V / R
4.2549 / 0.40856 = 10.4143 A, 44.31 watts
-----------
Each individual coil segment is now subject to only 3.47 amps at the peak maximums.
Current from the sine wave is well within the bounds of the coil wire.
Current for each turn is still 0.01808 amp and voltage has dropped to 4.25 volts.

This is presented in the hope that it will be useful and confirmed or disproved through independent testing.

Regards,
CM

Very interesting Cadman,

I have some questions as to understand the "geometry" of your method kind of graphically in my mind...

Horizontal Layers: Are the Layers wound longitudinally, comprehending all core length?...then stacking one on top of the other?

Vertical Layers: Or are Layers wound to gain in Height, while being short in length?...then aligning next to each others from start to end of core?

Thanks


Regards


Ufopolitics