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Author Topic: The Magneformer-lenzless transformer ?  (Read 39755 times)

Offline TinselKoala

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Re: The Magneformer-lenzless transformer ?
« Reply #30 on: November 12, 2013, 02:03:23 AM »
http://www.youtube.com/watch?v=EQJFuKvrUEs

Of course since MH has narrowed the goalposts to talking about the completely linear, unsaturated behaviour of simple iron laminated cores ... he's right as well.

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Re: The Magneformer-lenzless transformer ?
« Reply #30 on: November 12, 2013, 02:03:23 AM »

Offline MileHigh

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Re: The Magneformer-lenzless transformer ?
« Reply #31 on: November 12, 2013, 03:38:56 AM »
TK:

Yes, thank you for putting it so succinctly.  That's the territory Tinman is in and the principle I am trying to operate with is to understand the basics and give yourself a foundation.

If you are operating in the linear unsaturated region of a transformer and you add the influence of a magnet to the mix, then as long as you are still in the linear unsaturated region then the magnet will have no affect whatsoever on the operation of the transformer.

Tinman:

I am going to try to respond to your postings.

For starters, you asked about the power dissipation across the 18-ohm resistor in your RLC circuit.  This is kind of a case of not seeing the forest for the trees.  Your fancy scope is giving you a real-time Vrms so the answer is right there.  If you want to do it right measure the actual resistance of the 18-ohm resistor with your best multimeter.

Quote
My reference to the lenzless effect,was in relation to absolutly no power increase on the input was shown when we loaded one of the outputs-while remembering that the tank coil is already loaded(power yet to be determond).I am well aware that every transformer uses power at idle,but show me one where the power input dosnt go up when a load is placed on the output-while drawing a load on the second output.

You are in Thane Heinz territory here.  You know that the transformer is drawing power, but you don't have a clear picture of where that power is going.  So when you add a small load to the output of the transformer, what's happening is that the way the power is being split up on the output side changes.  The net result is no increase in input power and you still can't account for precisely where the input power is going.  At least you can say when you add a load and make a measurement that you know were some of the input power is going.  That's all fine, but you should not have made your original statement the way you made it.  You can always qualify statements after the fact but my suggestion is to just be straight from the start.  If you said something like, "I made a power input measurement and when I added a load to the output the power input measurement did not change.  However, I haven't accounted for where all the input power is going therefore I can't make any conclusions," that would have been the real deal.

Quote
but show me one where the power input dosnt go up when a load is placed on the output-while drawing a load on the second output.

Your request for someone to do this is not valid.  Someone else could have a transformer that is drawing input power and when they add a load to the output the "balance of output power" changes with no extra current draw on the input.  There is nothing remarkable about that.

Time to see if there will be a next posting....

MileHigh

Offline MileHigh

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Re: The Magneformer-lenzless transformer ?
« Reply #32 on: November 12, 2013, 04:18:42 AM »
Tinman:

Okay I will try to tackle this one:

Quote
Useing the scope was to show the small signal coming from the SG,which is a visual way for me to show the small amount of power being supplied by the SG-and you have to agree,it is very small.
Also ,please feel free to point me out to my mistake-as this is how we learn. Im just as good at missing things as anyone else. If you are refering to how the transistor distributes that signal power,please remember i have  a diode across the base/emitter junction.

I would like to know what you think is being disipated by the 18 ohm resistor,as far as the scope shot go's.
I am also drawing up a quick scetch for you and all to have a look at,and post your opinions on as to what happens throughout 1 cycle.It will be in 2D,so wont be anything fantastic,but it will show what im am asking.

I will just repeat again that if you want to measure the power that the signal generator is possibly pumping into your setup then you have to do it when the setup is live and running.  Shutting the setup off will invalidate any attempts to measure the signal generator power and looking at the small glitches on the scope display while the setup was switched off was meaningless.

One more time, I am being conservative because I don't have a schematic of your setup.  If the signal generator is only switching on an NPN transistor through a base resistor and an inductor, then for sure you can "eyeball" the input power and make an estimate without really having to make a measurement at all.  But I absolutely refuse to make assumptions like this about your circuit.

Looking forward to seeing your schematic.

Moving on, in post #19 I am sorry to say that your diagram doesn't make much sense.  You have magnetic fields from horseshoe magnets travelling radially across an inductor.  So by definition even if you change the strength of those fields nothing will happen to the inductor coil.  If I can offer you some serious advice, hunker down and look at this guy's videos and watch them from the beginning.

http://www.youtube.com/user/lasseviren1/videos

If it's too much work then just focus on the electricity and magnetism related videos.  Don't be fazed by the math and the guy also talks a lot in plain English.  If you get those clips and go learn stuff then you will be in a better position to experiment.

I will still try to answer your questions:

So first,what happens when we switch on the electromagnet?  >>>>>>>> The current will increase in the coil and then level off.  If it's "against the grain" of the magnetic flux in the horseshoe magnet it will take a longer amount of time.

Is there a BEMF or lenz force applied to our electromagnet coil?  >>>>>>>>>>> There is standard BEMF just like when you energize a regular coil.

Will the electromagnet still use the same amount of power ,with and without the inductor and PM being there?.   >>>> Yes.

second-what happens when the electromagnet is switched off?-becomes open circuit.  >>>>  You get an inductive spike that ionizes the air at the switch contacts and the coil discharges its stored energy.
 
Once the electromagnet switches off,and the field of the PM becomes the field within the inductor core-where or what is the BEMF or lenz force between?.  >>>>>>>  I don't understand your question.

If we load the inductive kickback from our electromagnet coil,as in charging a battery,or placing a low value resistor across the output,what happens to the magnetic field within the inductor?.   >>>> The magnetic field will collapse.  The horseshoe magnet is "not really there" because it's a magnet with a constant unchanging field and the coil only responds to changing fields - the same old story.

MileHigh

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Re: The Magneformer-lenzless transformer ?
« Reply #32 on: November 12, 2013, 04:18:42 AM »
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Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #33 on: November 12, 2013, 05:46:05 AM »
Hi Brad,

Referring to your drawing shown above I would say the following comments (with assuming your inductor has an  'I' shape core (i.e. a straight open core) and your electromagnet core has indeed a horse shoe like shape like your permanent magnet):

when the electromagnet is off, most flux from the permanent magnet is directed and closed into the I shaped core of the inductor provided the I core is ferromagnetic i.e. conducts flux much better than air

when you switch on the electromagnet with the polarities shown, I assume two explanations may be valid, depending on distances:

a) either the flux of the electromagnet will join to the poles of the permanent magnet via the upper and lower edge parts of the I core (facing unlike magnetic poles tend to join even via a ferromagnetic core piece in-between), hence the earlier flux of the permanent magnet will move out from the I core lengthwise

b) or the flux from the electromagnet will be directed into the I shaped core just like that of the permanent magnet so that as you say the two opposing flux may neutralize each other lengthwise in the core,  the result is again a flux decrease to near zero in the I core

when you switch off the electromagnet, the flux from the permanent magnet can again penetrate through the I core full in lengthwise due to the assumed good flux conducting properties of the I core

Now on your questions

(So first,what happens when we switch on the electromagnet?)

I discussed above what may happen when you switch on the electromagnet, cases  a)  or  b).

(Is there a BEMF or lenz force applied to our electromagnet coil?)

The moment the current is on in the electromagnet coil (with the proper intensity) AND the R load is hooked to the inductor coil (perhaps together with your tank capacitor not shown) I think the Lenz law effect can manifest only in a smaller amount than in case of a normal Faraday induction because the flux which is causing the main induction in the I core comes mainly from the permanent magnet, especially if flux change really happens like in case a) above.

(Will the electromagnet still use the same amount of power, with and without the inductor and PM being there?)

You may surely have found that placing an I core near to the prongs of a C core changes the inductance of the coil wound onto the C core, how much the inductance changes depends on the air gap left between the I core and the prongs, highest inductance is received when you fully close the gap and smallest when you remove the I core and the prongs of the C core become an open magnetic circuit again. Now considering this, your actual setup already has a certain air gap I suppose which already established a certain L inductance value for the electromagnet coil and once you fixed the distances in a real setup the inductances for the coils are set.
Another factor to consider is the flux coming from the permanent magnet via the I core towards the electromagnet C core, it may influence the permeability of the C core, albeit it can be a small flux influence only,  depending on mainly the thickness of the I core.
So the answer to your question I think is yes, the electromagnet would use very nearly the same amount of input power, with or without the inductor and the PM being there and allowing for the above reasonings. Here I assume that the cores of the I and that of the electromagnet are not driven towards saturation in any instant.

(second-what happens when the electromagnet is switched off?-becomes open circuit)


I discussed this above and I add that when you wish to collect the energy coming from the collapsing field of the electromagnet, you have to be careful with choosing the correct 'on time' for the electromagnet because loading the spike from the collapsing field may extend the 'on time' of the electromagnet. The 'on time' can be conveniently adjusted / compensted by the duty cycle in this case I believe.

(Once the electromagnet switches off,and the field of the PM becomes the field within the inductor core-where or what is the BEMF or lenz force between?)

As I mentioned, I assume the Lenz action-reaction force takes place mainly between the permanent magnet flux and the flux in the inductor core, the current taken by the R load surely creates a flux against that of coming from the permanent magnet, effectively reducing it, as if the original flux strength from the permanent magnet would have been weaker.  This may set a certain limit on the amount of output power.
So Lenz law is still valid but mainly acts between the permanent magnet-output coil flux and in a much less rate between the input-output coil flux, this is how I see this, I may be wrong.

Greetings,  Gyula
Gyula
It is good to see some of us can understand a simple drawing,and answer with responces that make sence. From the many test i done with a setup like that shown in the picture above,i can say you are right. When i stated a transformer that showd a 0 lenzz effect when the output was loaded,was in regards to no reflection shown on the input. As you stated,there is a lenz force created between the PM and the inductor,but we dont need to worry about that,as it comes at no cost to us,as it's now the PM that is doing the work !fancy that,a PM doing useful work!

This is where experimenting kills text book physics-actual results befor your eyes,insted of blind faith that all has been looked into and accounted for.

My actual test setup(as pictured above in previous post) showed exactly what you said Gyula.
With or without the inductor and PM there,the electromagnets power draw remaind the same.
Without the PM there,the electromagnets power draw went up,and a voltage was produced across the resistor (standard transformer action)
With the PM in place,the electromagnets power draw went down,and a voltage was produced across the resistor of the same value.

The trick is getting the right amount of current at a set voltage going to the electromagnet,so as it only neutralises the PMs field in the inductor core. This can be done useing a resistor in series with the electromagnets coil,and then measureing the voltage across that resistor. Then we just have to set our SG's signal strength to the transistor so as we get the same voltage across that same resistor during our on period. Then that is when i set up the tank circuit,so as to be able to adjust the frequency until i gained maximum amplitude across that tank circuit.

Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #34 on: November 12, 2013, 05:58:27 AM »
The magnetized core transformer is not an AC transformer as MH keeps suggesting. ::)

Neither is Tinmans input to his primary, in case some didnt look at his scope shot. ;)

When pulsing a primary of a magnetically biased core, the field of the coil should be opposing the cores field, not adding to it to make it stronger and saturate/over saturate the core. Building a field from the coil, working in opposition to the core magnet, gives the field of the coil a lot further to go before saturation than a non magnetized core of the same specs. So the amount of energy of the coil/magcore can be substantially more, up around twice the V/A capability compared to a non biased core.

So talking AC as an input is senseless. ::)

Tinmans project is a special case, as he is using a magnetically biased core. ;) ;D

Mags
Hi Mags

Yes,there is no AC in the drive coil,so not sure where the AC bit is coming from. AC is alternating current-not alternating voltage.The current flow remains in 1 direction when the drive coil is switched on and off,AC is current flowing one way,and then switches and flows the other way-this never happens in the drive coil.

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Re: The Magneformer-lenzless transformer ?
« Reply #34 on: November 12, 2013, 05:58:27 AM »
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Offline Magluvin

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Re: The Magneformer-lenzless transformer ?
« Reply #35 on: November 12, 2013, 06:20:24 AM »
Hi Mags

Yes,there is no AC in the drive coil,so not sure where the AC bit is coming from. AC is alternating current-not alternating voltage.The current flow remains in 1 direction when the drive coil is switched on and off,AC is current flowing one way,and then switches and flows the other way-this never happens in the drive coil.

Hey Brad

"so not sure where the AC bit is coming from."     

It comes from "moving the goal posts" like TK said, in order to put the magnetic biased core in the realm of 'junk'.  ;)   lol

Just keep doin what you do.  ;) Your doing fine. ;D

Mags

Offline TinselKoala

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Re: The Magneformer-lenzless transformer ?
« Reply #36 on: November 12, 2013, 06:57:20 AM »
Many people are confused about what "AC" really means. But "transformer effect" doesn't depend on true alternating current in the primary of the transformer. Recall that induced voltage in a secondary is proportional to the time rate of change of... the magnetic field in the primary. And the magnetic field is proportional to the current (amp-turns). So if the current rises and falls, that's all that is needed for transformer induction effect to happen. The current doesn't have to actually reverse, it can keep going in the same direction, as long as the amplitude of the current rises and falls. This of course produces a change in the magnetic field. The secondary output will be true AC though, even if the input to the primary is pulsed DC or "AC" with enough DC offset so that the current doesn't actually reverse.
"dB/dt", the time rate of change of the B field, can be thought of as the instantaneous slope of the curve describing the B field strength over time. So while the B field is increasing, dB/dt is positive, and when the B field peaks dB/dt is zero, and when the B field is decreasing, dB/dt is negative. So the induced voltage in the secondary flips sign at the peaks (and valleys) of the current which is causing the primary B field to wax and wane.

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Re: The Magneformer-lenzless transformer ?
« Reply #36 on: November 12, 2013, 06:57:20 AM »
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Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #37 on: November 12, 2013, 07:12:43 AM »
Hi Tinman,

agreed that you will have a bigger current in that time-period of 48 µ sec. Now how come that your scope shows 13,2 V when you drive your circuit with 12 Volt ? Anyway. lets assume 12 Volt , but than you can not use the 12 Volt as driving voltage over the full cycle of 304 µ sec. as the 12 volts are only swiched on and driving the unknown high current for 48 µ sec  , you need the average voltage ( ever lower 1.89  Volt ) which is related to one cycle ( since you do not have the surge-current in that 48 µ sec window9  and multiplying it with the average current you have measured because you only have the average current of 11 mA.. see ?

Your scope shows the false rms-voltage at the top, it does calculate the high value-distribution ( 13,2 V  and this value is wrong also ) ,  which is mainly composed out of the mosfet-off-state. It wrongly calculates 13.2 V related to the 304 - 48 µ sec = 256 µ sec.
You always have to relate the short on-time-voltage to the full cycle.

So again I insist: it looks good...at least not bad

Regards

Kator01
Hi Kator
If you look at the scope shot,you will see that the 13.1 volt peak is from the inductive kickback spike.But you will also see that it is quickly brought down by the cap and battery-more so the battery,as its internal resistance would only be around 1-1.5 ohms-provideing its a healthy battery.
You can see in the rest phase ,that the supply voltage is indeed 12 volts.
If we look at the scope shot,we can see that the efficiency is quite good-as far as the inductive kick back charging go's.
But as far as our tank circuit go's,well thats no so exciting. From that part of the circuit,we have only around .033 watts-but every bit helps.

Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #38 on: November 12, 2013, 07:28:54 AM »
Hey Brad

"so not sure where the AC bit is coming from."     

It comes from "moving the goal posts" like TK said, in order to put the magnetic biased core in the realm of 'junk'.  ;)   lol

Just keep doin what you do.  ;) Your doing fine. ;D

Mags
Experimenting is never junk Mag's lol.
Here is a good example,and it seems that his results are the same as what i get with the setup i posted in a previous post.

http://www.youtube.com/watch?v=2kosQIrDCWM

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Re: The Magneformer-lenzless transformer ?
« Reply #38 on: November 12, 2013, 07:28:54 AM »
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Offline MileHigh

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Re: The Magneformer-lenzless transformer ?
« Reply #39 on: November 12, 2013, 07:30:29 AM »
Tinman:

I can assure you that your transformer input is being excited by "AC."  That is a generic term used in electronics.  It means "changing with respect to time."  If there was no AC input to your transformer then the output wound be dead and the RLC circuit would be dead.  Talking about "moving the goalposts" with respect to the standard definitions for electronics terms is a non-starter.  I am asking you to not go there because that would be another "contest."  I won't even argue the point beyond what I just stated because it would be silly.

With respect to the "other" moving of the goal posts, I already covered that earlier.  The moving is done when you make a general statement about something that relates directly to your setup, and then someone else responds stating "not true" because special case XYZ is not true.  That's the real moving of the goal posts.  I discussed your particular setup and what you are doing and my explanation about the futility of the permanent magnet core is because that's where you start - at the beginning - within the context of your actual experiment.  You learn that non-moving magnets have no affect on transformers because there is no changing magnetic flux with respect to time.  Then you go on the bench and view and measure it for real and confirm it for yourself.  These are the basic building blocks of electronics.  You don't believe me?  That's fine, feel free to do some bench experiments and prove it for yourself.  Also, don't make the mistake of alleging that I am "all theory" because I have done all of this stuff on the bench for real.

From my perspective you are playing with a transformer setup where your excitation is a switching transistor.  That bangs the primary coil with a regular stream of pulses (which is AC) and that excitation is at the resonant frequency for your RLC circuit.  The fact that your core is a magnet has zero affect on your setup.  If you don't believe me then why not put in an equivalent non magnetized core and compare waveforms and make some good measurements?  It would be blind faith on your part to believe that the magnetized core of your transformer does something useful, because it doesn't.  Nor are the magnets doing any work, they are as dead as proverbial doornails.  That's more blind faith on your part, "One day someone will prove that magnets are a source of energy," it will never happen.

That's the real deal Tinman.  If you are going to make measurements on the input and all of the outputs and compare total input power to total output power that would be great.  You won't find anything special but you will deserve a lot of credit for making good measurements.

The thing that may be throwing you off is that the magnetic core does nothing.  It's a fact.  The best thing that you could do for yourself is do the research and create some good test procedures to prove it for yourself.  That's the challenge, find the real truth.  If you refuse and just "want to believe" that the magnetized core does something beneficial and the "magnets are doing work" then it would be a shame and you won't benefit from the project.  I'll put it another way, if you do the project and intentionally hunt for something to confirm your beliefs and prejudices and expectations, then you are just cheating yourself to your own detriment.

So it's up to you.  You can increase your knowledge and look for the truth or just look for confirmation of your preconceptions in your measurements.

MileHigh

Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #40 on: November 12, 2013, 07:46:14 AM »
Many people are confused about what "AC" really means. But "transformer effect" doesn't depend on true alternating current in the primary of the transformer. Recall that induced voltage in a secondary is proportional to the time rate of change of... the magnetic field in the primary. And the magnetic field is proportional to the current (amp-turns). So if the current rises and falls, that's all that is needed for transformer induction effect to happen. The current doesn't have to actually reverse, it can keep going in the same direction, as long as the amplitude of the current rises and falls. This of course produces a change in the magnetic field. The secondary output will be true AC though, even if the input to the primary is pulsed DC or "AC" with enough DC offset so that the current doesn't actually reverse.
"dB/dt", the time rate of change of the B field, can be thought of as the instantaneous slope of the curve describing the B field strength over time. So while the B field is increasing, dB/dt is positive, and when the B field peaks dB/dt is zero, and when the B field is decreasing, dB/dt is negative. So the induced voltage in the secondary flips sign at the peaks (and valleys) of the current which is causing the primary B field to wax and wane.
Hi TK
Yes,i know what your saying.Just like it only takes a north field passing over an inductor to produce an AC output.This is why we have an AC output on the tank circuit. But the input is DC,and the primary coil only has a DC current flowwing through it at all times.If we put a DC current into an inductor,a magnetic field will be produced around that inductor. When we switch that coil off,the field collapses-but it never reverses polarity.Now if we load that kickback in the way of charging a battery,or place a resistive load across it,then a BEMF is produced from the coil that apposes that collapsing magnetic field. And as you can see in the scope shot,the current always flows in the same direction.

Definition of AC current:-In alternating current (AC, also ac), the flow of electric charge periodically reverses direction.

Well as we see in the scope,there is no change in direction-only a change in amplitude.
The voltage potential is ofcourse reversed,due to the way the circuit is-much the same as the SSG circuit.This is why the positive of the P/in become's the negative for the P/out,and the collector that use to be our negative side,now becomes our positive potential-thus the diode on the collector.

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Re: The Magneformer-lenzless transformer ?
« Reply #40 on: November 12, 2013, 07:46:14 AM »
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Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #41 on: November 12, 2013, 08:19:20 AM »
MH
Once again,you are clutching at straws,and being far to picky.
Lets have a look at a couple of scope shots from the video-below.
Dose this not clearly show an indication,or give you a visual reference as to how little of the total power supplied to the system,is being supplied by the signal generator?

Also there is no AC current going ito the drive coil,nor coming out of it.Once again you are deviating from the true meaning of AC,and getting it mixed up with something else.

Definition of AC: In alternating current, the flow of electric charge periodically reverses direction. In direct current, the flow of electric charge is only in one direction.

The current flow through the driven coil remains in the same direction at all time's-only the amplitude of that current flow changes !! that is not AC current in the true sence of the definition !!

Edit-well now it seems that i cant post attachments-no !ADD ATACHMENT BUTTON!?-only clear attachments

Offline Newton II

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Re: The Magneformer-lenzless transformer ?
« Reply #42 on: November 12, 2013, 10:12:16 AM »


http://www.youtube.com/watch?v=36W4UdMKVwU


Can somebody explain what is he doing?   Magnetic standing columnar waves?? :'(



Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #43 on: November 12, 2013, 11:08:14 AM »

http://www.youtube.com/watch?v=36W4UdMKVwU


Can somebody explain what is he doing?   Magnetic standing columnar waves?? :'(

Yes-he is playing with his balls ::)
 
A standing wave is just a stationary wave,which means it remains in a constant position.
Not to sure where the columnar part come into it, in regards to his video?.

Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #44 on: November 12, 2013, 11:10:38 AM »
@MH
Scope shots regards to post 41

 

OneLink