Open discussion for projects on this forum.

[citfta]

Feel free to make any posts to this thread you like. It is for open discussion. I do ask that there be no bashing of a personal nature, but technical discussion is great.

If I had a machine that worked I would never sell it until I had at least a couple of more built that also worked. Then I would feel pretty sure I knew what I was doing. A single machine could be one of those times when everything feel into place by accident but can't be duplicated. After building a few more that worked then I would be more comfortable about what I thought I knew. Just seems like common sense to me to not sell the only one I had that worked.

[seaad]

Hello guys. A tip. Try to squeeze MM gently on his sweet spots. Maybe he will tell you how he made his COP-3 machine. That one he sold. You certainly heard of that on the Re...Figuera thread.

Sorry citfta I couldn't resist writing here.

Summer Regards Arne

[citfta]

You people sound like 80 year old *****es with nothing to do. i've been Very, Very busy with life and i don't have to explain an F-in thing to you loosers. try reading the post instead of running your big fat know NOTHING mouths. better yet why don't you mouth runners leave my name or any thing about my thread out of your pathetic mouths and quite spoiling other threads with your bull **** whining gossip and rhetoric, Grow up and get a life.
How pathetic you people are.

MM

Sounds to me like we hit a sore spot. Sorry MM but this thread is for discussing the technical aspects or lack thereof of any project on this forum or anywhere else. If that bothers you then simply don't read this thread. I have been reading your thread and I do recall very clearly you making the statement that it appears the part G is not quite as you thought it was. So my statement still stands. You will never be able to control the current through the primaries with a part G that has such low resistance. And as built in the pictures I have seen it does not have the proper reactance to control the current through the primaries either.

I don't have any problem at all with it taking several months to get something going as I also have been very busy. But there is no need for your constant bad mouthing of those that disagree with your technical analysis of how the Figuera device and in particular the part G actually works.

So far as I have been able to tell from the two threads, tests done by UFO have confirmed what Bistander and myself have said about the part G.

[bistander]

You people sound like 80 year old *****es with nothing to do. i've been Very, Very busy with life and i don't have to explain an F-in thing to you loosers. try reading the post instead of running your big fat know NOTHING mouths. better yet why don't you mouth runners leave my name or any thing about my thread out of your pathetic mouths and quite spoiling other threads with your bull **** whining gossip and rhetoric, Grow up and get a life.
How pathetic you people are.

MM

try reading the post instead of running your big fat know NOTHING mouths.

I read your posts. Which part do you think I misunderstood?

[marathonman]

Get on with it

You people sound like 80 year old *****es with nothing to do. i've been Very, Very busy with life and i don't have to explain an F-in thing to you loosers. try reading the post instead of running your big fat know NOTHING mouths. better yet why don't you mouth runners leave my name or any thing about my thread out of your pathetic mouths and quite spoiling other threads with your bull **** whining gossip and rhetoric, Grow up and get a life.
How pathetic you people are.

MM

[citfta]

What was interesting to me was that a couple of months ago if I recall correctly MM admitted that maybe part G wasn't what he thought it was. This was after several people had working part Gs that didn't work like MM kept saying they would. How can that happen if he is being given information from a working device?

I have said all along that part G could not be a device that passed a lot of current, yet MM kept insisting it had to have these large copper bars to work properly. As Ufo found out you can't control anything with that design.

I still think the Figuera device might be a real working device with a properly designed part G. I just don't have any spare time right now to pursue it. What little time I have for experimenting is spent working on a device of my own design that is not for this forum. If it works it will be shared with a small group of serious builders on another forum. These guys understand how things really work and are well qualified to evaluate and approve or disprove any claims. If I am going to make any claims I want them checked by qualified people, not some people that have only gotten their education from YouTube.

[bistander]

Moved post from Figuera builders thread

the standard mechanical rotating generator is a complete waste of time

the people of the Tesla time are true leaders in their field thus making our so called leaders from our time seam like school girls playing with dolls.

Hey MM,

If you like that 19th century technology over our modern version, you could join a Mennonite or other such group and avoid using those generators which you call a "waste of time".

Please excuse me for posting on your special thread but nobody else but you uses it lately.

Do you have any idea when we will see an actual demonstration or proof? It's been a long time and appears most interested parties have given up the discussion.

If you object to this post let me know and I'll move it to citfta's thread.

bi

So MM did object here:

That would be a good idea Bistander as your opinion or response is entirely not needed or wanted. thank you kindly though and have a good day.

He goes on to tell his serious builders this:

To all:
...

I hope you all understand what i have posted in the last few weeks and have a better mental picture of what is taking place. apparently aligning that up with a working model was a little more difficult then i had originally expected but i or rather we will get there. the two are finally aligning up just nicely. ...

MM

I see three serious builders left, GlenWV, Cornboy555 and Shadow119g. And two interested parties, seaad and myself. It's only been 8 months since he started his serious builders only thread. With all his guidance, preaching and theorizing I don't see anyone close to a working model, except maybe for Ufo. But it seems Ufo stopped pursuing MM's part G design. Wonder why?

I'll go back to the forbidden thread and delete my post as I said. But no answer from MM as to when we might see a demonstration or proof. Back to the sidelines.

bi

[gyula]

See my comments in blue within your text.

Gyula

Thanks for your explanation gyula. I do see in my post you quoted that I made a typo. Where I said the capacitive reactance goes down as the frequency goes down I meant to say as the frequency goes up. I was pretty tired when I typed that. Sorry about that.

Ok, so you are saying in a parallel resonant circuit that below the resonant frequency the reactance is inductive and above the resonant frequency the reactance is capacitive because below the resonant frequency the inductor is carrying more of the current and above the resonant frequency the reactance is capacitive because the capacitor is carrying more of the current?
Yes, that is okay.

I would consider the part of the circuit that is carrying the lesser amount of current to be the one that is more reactive. So I guess it all depends on how you want to look at it.
Well, in reactive networks I consider a component carrying the lesser current has less overall effect in phase shift than the one carrying the higher current, hence a component with lesser current is less reactive, this is how I see this.

Since Bob was asking in reference to a series connected bifilar coil the parallel circuit analysis is more the correct one.
To be fair, we need to consider a series resonant circuit too (I focused on a parallel one so far). Starting from DC and from low frequencies upwards, the capacitor surely has a huge reactance in this range so a series circuit must behave as a capacitive reactance which decreases as we increase the frequency. The coil in this scenario is a series short ( quasi a piece of wire) at DC and at the low frequency ranges i.e. has very low reactance, then its inductive reactance gradually increases as we increase frequency till we reach resonance. Beyond resonance the effect of the capacitive reactance start diminishing as we go up in frequency because Xc will be lower and lower.

To sum up: a series LC has capacitive reactance below the resonant frequency (input current leads voltage) and a parallel LC has an inductive reactance (input current lags voltage) below the resonant frequency.

Thanks for your thoughts. You are welcome.
Carroll

[citfta]

Thanks for your explanation gyula. I do see in my post you quoted that I made a typo. Where I said the capacitive reactance goes down as the frequency goes down I meant to say as the frequency goes up. I was pretty tired when I typed that. Sorry about that.

Ok, so you are saying in a parallel resonant circuit that below the resonant frequency the reactance is inductive and above the resonant frequency the reactance is capacitive because below the resonant frequency the inductor is carrying more of the current and above the resonant frequency the reactance is capacitive because the capacitor is carrying more of the current?

I would consider the part of the circuit that is carrying the lesser amount of current to be the one that is more reactive. So I guess it all depends on how you want to look at it.

Since Bob was asking in reference to a series connected bifilar coil the parallel circuit analysis is more the correct one.

Thanks for your thoughts.
Carroll

[gyula]

Hi Carroll,

I included my comments in blue within your post.

Gyula

Hi gyula,

Good to see you posting here. But you have me confused. I admit I didn't take time to look up everything before replying to Bob. Do you agree that inductive reactance goes up as the frequency goes up Yes I do, there is a linear proportion here, the higher the frequency the higher the inductive reactance becomes till the coils self resonance is reached.

and that capacitive reactance goes down as the frequency goes down. No, I do not agree, capacitive reactance goes down as the frequency increases, there is an inverse proportion here to frequency.

In other words if you connect a capacitor to DC as soon as the capacitor is fully charged it is not going to pass any more current so you could say the capacitive reactance is infinity. Yes. And as the frequency goes up the capacitor begins to pass more and more current. Yes and this means that its capacitive reactance goes down at the same time to a lower and lower value to pass more and more current. And of course everything is just the opposite for an inductor. Yes, till reaching self resonance for the coil, beyond that the coil starts behaving as if it were a capacitive reactance.

So I am confused as to why you said below the resonant frequency the reactance would be inductive and above would be capacitive? Can you please explain the reason for you saying that?
Well, try to think of a parallel LC circuit drawn with two rectangular blocks instead of the usual L and C symbols. The two blocks represent impedances that of course change when frequency is varied. Say the resonant frequency be at 1 MHz. Starting from DC and up in frequency, the 'coil' block is a short circuit i.e. a very low Z impedance at DC and at low frequencies while the 'capacitor' block first cannot start charging as it should when it were alone because the coil shorts it out. And as we increase frequency the shunting effect from the coil block reduces i.e. its impedance starts increasing while the capacitor block "slowly" starts charging (and discharging) i.e. its reducing impedance slowly starts take effect as getting able to "conduct" AC current more and more. Notice how this capacitor would 'pass' current at low frequencies: hardly because of its high capacitive reactance at the low frequencies. Impedance-wise we have two impedances in parallel and one of them (the coil block) always "remains" at a lower value than the other till reaching resonance. You may look at this as two paralleled 'resistors', one of them (the coil) has a lower value, hence it dominates the combined resistance value. I include two drawings (I combined into one) taken from a website AC Inductance and Inductive Reactance in an AC Circuit which shows the reactance and current curves for a coil and a capacitor in the function of the frequency. Try to mix the two drawings overlayed in your mind: the crossing point of their reactance curve defines the resonant frequency and consider the curves below and above this point.

Take care,
Carroll

PS: I do totally agree with your description of a parallel resonant circuit.

[boguslaw]

Could we have high frequency current and its bemf in the same circuit ?

[Bob Smith]

Hmm. I'm a little confused. I made the changes as I said I would earlier today. I'll let it sit till the dust settles, and if some nuance needs to be made, I'll change them again.

My thanks to both of you for taking the time to clarify this for me, as it is an issue I'm trying to get a handle on, particularly with the series wound bifi pancake coil.

Ideally, I want to be working at the coil's self-resonant frequency, and my initial inspiration in this is the work of Doc Stiffler and his SEC. I wonder if the swbifi pancake coil can be used in a similar way.

Thanks again for the gentlemanly exchange.
Bob

[citfta]

Hi gyula,

Good to see you posting here. But you have me confused. I admit I didn't take time to look up everything before replying to Bob. Do you agree that inductive reactance goes up as the frequency goes up and that capacitive reactance goes down as the frequency goes down.

In other words if you connect a capacitor to DC as soon as the capacitor is fully charged it is not going to pass any more current so you could say the capacitive reactance is infinity. And as the frequency goes up the capacitor begins to pass more and more current. And of course everything is just the opposite for an inductor.

So I am confused as to why you said below the resonant frequency the reactance would be inductive and above would be capacitive? Can you please explain the reason for you saying that?

Take care,
Carroll

PS: I do totally agree with your description of a parallel resonant circuit.

[gyula]

Hi Bob Smith and Carroll,

I made my comments in red for both of you within your own text enviroment below. I commented only where I disagreed with you both or where I felt some additions were needed.

Gyula

Originally Posted by Bob Smith:
I am watching this topic with interest here and on a couple of other forums. I have been working with a set of three series-wound bifilar pancake (swbifip) coils for about a week. My focus is on the interaction between the swbifip coil and the local electrostatic environment.
If I am not mistaken, an inductor pulsed below resonant frequency will be dominated by resistive reactance, while pulsed above resonant frequency, capacitive reactance dominates. In simplest terms, below resonance, the inductor behaves like a resistor, and above resonance, begins to behave like a capacitor. At resonance, neither predominates.
However, given the low reactance and high capacitance of the swbifip coil due to its configuration, what actually happens to its capacitance (let alone reactance) at resonance?
I'm going to throw out a thought - if reactance in an inductor drops to zero at resonance, and we transpose this understanding to the swbifip coil, is it possible that at resonance, its extremely low reactance might drop below zero, effectively giving it negentropic characteristics of a negative resistor?

My answer to the latter question is no. At resonance the reactances cancel and purely a resistive impedance dominates. This means there is no any phase shift between input current and voltage. However, the coil and the capacitor still maintain their original properties when we try to examine their own current vs their own voltage i.e. they maintain their own phase shifts within themselves as if they were alone. What manifests to the outside is that the input current and voltage will be in phase at resonance. IF there is any phase shift it means you are not at the resonant frequency of the L and C components yet.

There are some decent YT videos out there showing a rise in voltage amplitude at resonance in this coil. Is it another animal entirely, for which another set of parameters has to apply?
Bob

Answer to Bob by Carroll:

First thing is you have a couple of things slightly confused. Capacitive reactance goes DOWN as the frequency goes UP. Inductive reactance goes UP as the frequency goes UP.
So below the natural resonant frequency of a coil the reactance will be capacitive
No, it will be inductive.
and above the resonant frequency the reactance will be inductive.
No, it will be capacitive.
And you are correct that at the resonant frequency they both cancel each other out and the result is that only the DC resistance remains.

Yes, okay but here we need to define whether we talk about a series or parallel LC circuit? Because for a series LC circuit at resonance indeed there is the coil's DC resistance which limits the input current (besides the generator and any other series components in the circuit if there is any). And for a parallel LC circuit the input current is limited by a much higher resistive impedance at resonance than the coil's DC resistance, this impedance is the loaded Q times either the inductive or the capacitive reactance (these latter two are equal in absolute magnitude value at resonance).

I have not seen any evidence that the reactance can drop below zero. If you can link a video or other evidence showing that I would like to see it.

The only real difference in the bifilar series connected coil and a normally connected coil is the added capacitance of the bifilar coil. It also makes no difference if the coil is a pancake coil or a normal solenoid type coil the extra capacitance is still there if the coils are bifilar. Several tests were done on the OU.com site that confirm this.

The added capacitance of the bifilar wound coil is that it allows the coil to be resonant at a much lower frequency than a normally wound coil. As was pointed out on the OU site at the time Tesla was working with high voltages and high frequency, capacitors that met those requirements were hard to make and so Tesla was looking for a way to make a coil resonant without the need for capacitors and thus came up with the bifilar wound coil. Now high voltage capacitors are fairly easy to come by so the need for bifilar wound coils is not as great. They are still an interesting device for studying resonance and especially for learning about inductive and capacitive reactance.

I hope this helps some.

Respectfully,
Carroll

[bistander]

Allen's math

I had started using evostar's thread again as I thought Allen had chased him away but evostar is back using that thread again so I will post this here out of respect for the OP.

@bistander,

I made the point that a more tightly wound coil would generate a stronger field than a sloppily wound coil of equal wire gauge and turns for the same input. Lets say the difference in inductance was 5%. We could equal the strength of the sloppy coil with the tight coil by reducing the input by 5%, right?

Lets say we feed 10 watts into both coils: We would gain 1/2 watt in savings, right? 100 watts would result in 5 watts savings and 500 watts would result in 25 watts savings, got it?

My power input calculations result in a savings of that amount with that amount of difference in inductance between the two types of coils. You say I divided when I should have multiplied. Your approach would throw the answer off in the wrong direction by an "Astronomical Parsec".

I was asked for a simple explanation: Greater inductance results greater efficiency, got it? Your math is really stupid! Stop acting strange.

You say I divided when I should have multiplied.

Here is what Allen wrote:

16.8 mH divided by 10,000 equals 168 Gauss

Which is: 0.0168 / 10000 = 168.

I say that is wrong and that 0.0168 / 10000 = 0.00000168. So I came to the logical conclusion that Allen meant multiply instead of divide. Because:

0.0168 * 10000 = 168.

Silly me. Allen insists he is correct and I am wrong.

Then another example of Allen's math skill can be found in this recent post of his:

...

The inference you make that the tiny wire connection between the bifilar windings accounts a huge 1.9 mH difference in inductance is ludicrous. See the High Perm inductance measurements below:

Allen subtracts 15.9 from 16.8 and gets 1.9. Or "16.8 - 15.9 = 1.9"

I wonder if Allen can see his error and actually admit he is wrong. Probably not.

bi