Re-Inventing The Wheel-Part1-Clemente_Figuera-THE INFINITE ENERGY MACHINE

[gyulasun]

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Here is an online coil calculator that allows input of insulation thickness and type, coil pitch etc.

https://www.rac.ca/tca/RF_Coil_Design.html

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Hi Cadman,

Thanks for the link and the excel spreadsheet. The online coil calculator is good, the only 'problem' is that it considers single layer coils only.

Unfortunately, I am not good at spreadheets, never had to use it so deeply. Will try to get acquianted with it.

In the meantime I have also found an online calculator which is able to calculate multilayer coils and considers isolation thickness. However, the dielectric constant of the insulating material is not included in it but perhaps this is not a real drawback at 50Hz. Here is the link to it: http://coil32.narod.ru/calc/multi_layer-en.html  from the home site: http://coil32.narod.ru/index-en.html  A good feature is that with selectable plug-in options, it allows to calculate solenoid coil inductance when you insert a ferrite rod into it (hopefully this works for multilayer solenoids too).   I will study it in the next couple of days.

Gyula

[marathonman]

Monster Ferrite Rods http://www.stormwise.com/page26.htm
Paper towel  cardboard centers and toilet paper centers covered in resin are good bobbins to.

[gyulasun]

Hi NRamaswami,

I included my answers in bold between your text lines below.


Since we do not know when the electromagnet would hold the circuit was set up like this..

Mains live wire -- Primary coil input -- Primary coil output - resistive load - Mains Neutral

Secondary coil - Load.

Now when we wind the trifilar or quadfilar coils only down and up or forward and backward, the voltage loss in the primary load is negligible.

When we have more than that the voltage loss in the primary goes up dramatically.

Yes, This happens due to dividing the 220V input voltage between the primary coil and the resistive load which is in series with the primary coil. And in case the coils have got only the down and up turns i.e. relatively short overall wire length, the resulting AC impedance is also relatively small, this means that more voltage is divided across the primary resistive load than across the primary coil. And when the coils have more turns (i.e relatively longer wire length) than in the previous case, meaning higher L inductance, then the coil AC impedance gets also higher than in the previous case, hence two things happen: 1) input current draw decreases from the 220V mains,  2) less voltage is divided across the primary resistive load because the previous relatively low impedance of the primary coil has now increased to a higher impedance which does not let as much voltage to the resistive load than previously.

Amperage at input and load remains the same.  Yes, in a series circuit which the primary coil(s) and the primary resistive load represent in your setup, the current is always the same, regardless from its amplitude.
Amperage does not decrease with increasing wires.  I do not get what you mean on 'increasing wires': increase the wire diameter or increase gauge in your sense?    I suspect this is due to the fact that we are giving same load of 10x200 watts lamps. Possibly due to AC impedance increase and the same load requires the same current or amps, amps do not diminish but amps remain the same but the voltage in the primary load decreases.

However when we increase the layers beyond a certain number the coil loses the ability to transmit current to the load. Probably at this level it is safe for us to make the coil as an electromagnet. Since we do not know how to calculate the whether the coil will hold as an electromagnet or not we did this test. It is a kind of blindmans approach to evaluate whether it is safe to connect the electromagnet to the mains. When we do so after this kind of levels the electromagnet holds. Power draw is low. It is certainly not in the less than 1 amp that you mentioned, possibly it applies to enamal coated windings and not for insulated windings.  No,  current does not depend on the insulation type of the wire in this setup, it depends entirely on AC impedance of the given coil.    Insulation probably takes some amperage on its own. I really do not know.  Insulation takes amperage only if it already got burnt and the gutted parts are able to conduct current towards unwanted directions.  Maybe you wish to check the insulation of such wires.

By low guage wires I meant indeed thin wires. In India wires used are 0.75 sq mm, 1 sq mm and 1.5 sq mm, 2.5 sq mm, 4 sq mm, 6 sq mm, 10 sq mm, 15 sq mm, 25 sq mm, etc.  How they are classed in other countries I do not know. I meant thin wires.  Okay, I have already figured it out but I had to ask it and this may help reduce confusion in other members reading here.

Can you explain this formula please..

Z=sqrt(R2+XL2)     more precisely written: Z=sqrt(R²+X_L²)

Z I believe is the AC impedance.  Yes,  R is DC Resistance Yes,  L is inductance What is X? There is no L as a multiplier, L is a subscript to X so X_L represents the inductive reactance of a coil: X_L=2*Pi*f*L   in the latter formula the L is now indeed the inductance of the coil. What is sqrt.. Is that square root of R squre plus X multiplied by L square.  sqrt is the square root of R squared plus X_L squared and there is no L multiplier.
So Z is directly proportional to inductance and dc resistance. Am I right in this simple understanding of the forumla. Yes basically this is correct because the inductive reactance, X_L, is proportional to the inductance of the coil but this L inductance is not included directly in the formula for Z, now you know.   If DC resistance increases, AC impedance increases and so is inductance of the coil.  This is what you taught me earlier. Is my simple understanding correct.  NO, this latter is NOT correct... It is okay that if DC resistance increases, AC impeadance of a coil also increases BUT the inductance of the coil does NOT increase, I did not teach it to you ever, at least I did not intend or mean that ever.


The bulbs were connected in series only to test when the wire is not able to transmit power to the bulb. At that stage it is safe for us to conclude that the electromagnet will remain stable. Crude method of understanding. Okay.

I actually did not know as the number of coils increase like bifilar, trifilar, quadfilar etc AC impedance will increase and so by making more number of coils, we can get the elcctromagnet to remain stable at lower number of turns as the AC impedance would increase manifold for multifilar wire coil than for a single wire and so the multifilar coil would be stable electromagnet.  Okay.

You see in the absence of technical knowledge, we used a common sense approach to determine at 220 volts when the electromagnet would become stable. When it can be safely connected without the fuse blowing up.  Now I assume you mean here to omit the primary resistive load too?

But I need to test whether the Ammeter actually shows so much of less amps as you suggest. We have never ever seen any amp in any electromagnet less than 5 amps.  I have to notice here that surely you can reach a situation in the setup whereby the primary coil(s) will have a certain AC impedance letting say draw 2A current from the mains but it remains to be seen what power is delivered to the secondary load? [assuming that your 110% or so measured efficiency included the power drawn by the primary resistive load which now will not be present simply because its role was to stabilize the electromagnet(s), right?]

Possibly your calculations based on enamal wires may not directly apply to insulated wires.  If you mean those calculations I showed with the 5A and the 220V about 2 days ago as a kind of 'reverse engineering', then they do apply! This cannot depend so much on the kind of wire insulating material at the 50Hz mains frequency.

It is the thick very thick insulated cables that show a reverse to this line of thinking. But the insulated 3 core wire was only 300 meteres 100x3 meteres and so the insulation could have taken the amps. This is my assumption.  Well, my assumption here is that probably the AC impedance (due to the increased L inductance) of the coil(s) made from the 3 core wire with thick insulation already reached an impedance value which 'blocked' the input mains voltage to reach the primary resistive load as intensely as it let it earlier when the up and down wire length was used.  (if my reasonings applies at all to this situation)

Gyula

[gyulasun]

Impedance equation:  Calculated from resistance (R) and reactance (XL - XC) )

Hi Hanon,

Your formulas include a capacitive reactance, X_C,   besides the inductive reactance, X_L  and I say this to ease the understanding for those learning this. 

Thanks,  Gyula         

[gyulasun]

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my question is: in theory will the resulting wave in the two primary be 90 or 180 out of phase ? . . .or
will it be any out of phase at all ?
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Hi Alvaro,

I think Dieter is correct, it gives a 45° shift,  this may be figured also from dividing a full circle (or turn), 360° by the 8 segments, it gives 45°. 

Gyula