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Author Topic: Splitting the Negative  (Read 19818 times)
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What magic is this?

 :D
   

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Not taking me seriously?    >:( 

I think a few tests using "splitting the negative" might be very revealing about this whole "splitting the positive" set of claims.   
   

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Not taking me seriously?    >:( 

I think a few tests using "splitting the negative" might be very revealing about this whole "splitting the positive" set of claims.

Been there-done that.

Oddly enough-the current still seems to flow in the same direction :D,but you kill the !radiant! energy intake C.C


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Agreed!  What sort of tests are you proposing?

Great shot of Mr. Bean.  Love his stuff.

Word is the "Magic" in "splitting the positive" is real
and that "splitting the negative" has none.

Unless a "ground" is attached at precisely the
exact spot in the circuit where it will draw energy
from the ground to keep things going without
expiration.

Naturally, I am very, very doubtful but would love
to pursue your suggestions.


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What should be odd there?
What I see is a difference in potential. And what if on both side are bulbs connected?
   

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Well i gave it a try TK,using 55F caps.

So we have two in series,then the motor,and then the 3rd cap.

Now,the problem using caps  :D
We now have a cap to cap to cap transfer loss of 2/3s--or something like that,as we are now transferring power from cap A to cap B,through the motor,to cap C  C.C

Anyway,tried it both way's-->splitting the negative,and splitting the positive,and guess what --no difference in end results  C.C

I also have 4 neo magnets on the flywheel,and a small gen coil,with an FWBR on the output.
I can charge another 55F cap (cap on right of DUT)to 258mV,and when i connect the gen output, it makes no difference at all to the end result of the other 3 caps voltages. O0


Brad


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So with caps you found no difference between splitting positive and splitting negative, surprise!

But but but... caps are not batteries! Capacitors store energy in dielectric stress, while batteries store it in ionization.
   

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So with caps you found no difference between splitting positive and splitting negative, surprise!

But but but... caps are not batteries! Capacitors store energy in dielectric stress, while batteries store it in ionization.

That's a really simple and elegant way to put it, made me think about how it relates to my own work.. a quick google and this:

http://www.madsci.org/posts/archives/1998-03/889679790.Eg.r.html

Re: what is dielectric strength and dielectric stress??

Consider a parallel plate capacitor. A dielectric (nonconductor) is often put between the plates for two reasons -- one electric; the other mechanical. The electric reason is the molecules within the dielectric may increase the capacitance because the dielectric permittivity is higher than air's. The mechanical reason is to keep the plates apart. A charged capacitor's plates attract each other. The dielectric provides a reaction force keeping the plates apart. It thus feels a mechanical compressive stress. Dielectric strength, according to Electronic Engineers' Handbook, is " ... the maximum (electric) field strength that a dielectric can sustain without breakdown." Electric field strength is measured in volts per meter. What is breakdown? In this case, it's the condition in which the dielectric becomes a conductor, albeit unwilling. Electrons are ripped from the dielectric atoms and their motion is a conduction current. Dielectrics ordinarily do not allow this. Dielectric stress is the electrostatic force divided by the area. It is not a mechanical breakdown limit. Mechanisms for mechanical dielectric failure under voltage stress differ over the long and short term and depend on the dielectric. One long-term failure mechanism reported more than 20 years ago was the growth of metallic "whiskers" that penetrated the dielectric and shorted the plates. The whiskers were perfectly formed crystals that proved to have high mechanical strength. Very nice, but they caused the capacitor to fail. Bother the strength — the things conducted current too! Lawrence Skarin Professor, Electrical/Telecommunications Technology Monroe Community College Rochester, NY


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So with caps you found no difference between splitting positive and splitting negative, surprise!

But but but... caps are not batteries! Capacitors store energy in dielectric stress, while batteries store it in ionization.

Ok,so what about 1.2v ni cad's ?

The test will go from minutes to hours,but i could just leave it running on the bench,while i do other things.


Brad


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So with caps you found no difference between splitting positive and splitting negative, surprise!

But but but... caps are not batteries! Capacitors store energy in dielectric stress, while batteries store it in ionization.

I think most people think the charge is stored on the plates,when in fact,it is stored in the dielectric it self-->or is it? ???


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Ok,so what about 1.2v ni cad's ?

The test will go from minutes to hours,but i could just leave it running on the bench,while i do other things.


Brad

But but but... NiCd and Lead Acid chemistries are very different...      and then when people put alum in their LA batteries.... what then?  :-[
   
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I think most people think the charge is stored on the plates,when in fact,it is stored in the dielectric it self-->or is it? ???

But but but... charge is not energy !  (Although charge and energy are both conserved, they are not the same.)

For example, you can charge the plates of an air variable capacitor to some charge value. Then you can move the plates closer together or further apart. The quantity of charge doesn't change, but what happens to the stored energy?
   
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But but but... charge is not energy !  (Although charge and energy are both conserved, they are not the same.)

For example, you can charge the plates of an air variable capacitor to some charge value. Then you can move the plates closer together or further apart. The quantity of charge doesn't change, but what happens to the stored energy?

Good day All
OK< so all else being equal, increase the distance between plates until the capacitance is C/2; now the original Voltage (V1) now becomes double, V*2 = V2.
Assuming the Charge: Q stays the same during the plate expansion process; then Energy (stored on cap) = 1/2Q*V; and since V1 is now V2, the Energy stored by the cap has increase by expanding inter-plate distance because the starting voltage V1 has doubled in the process.  (it required external energy applied to the system to physically separate the plates).

take care, peace
lost_bro
« Last Edit: 2017-12-07, 17:47:26 by lost_bro »
   

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But but but... charge is not energy !  (Although charge and energy are both conserved, they are not the same.)

For example, you can charge the plates of an air variable capacitor to some charge value. Then you can move the plates closer together or further apart. The quantity of charge doesn't change, but what happens to the stored energy?

Is the energy stored in the plates,or the dielectric ?

Is the charge stored in the plates,or the dielectric?.


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Could we call it "stored potential energy?"

Also, is there a meaningful definition of what "Splitting the
Positive" or "Splitting the Negative" really means?

What is its purpose and how is it achieved?


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Is the energy stored in the plates,or the dielectric ?

Is the charge stored in the plates,or the dielectric?.

Could we call it "stored potential energy?"

Also, is there a meaningful definition of what "Splitting the
Positive" or "Splitting the Negative" really means?

What is its purpose and how is it achieved?

Good points made by you both, definition is the key to understanding. You cannot analyse competently what you cannot define competently.

I have only devoted a small amount of biological supercomputer time so far to this topic but I quickly came across this link that might offer some insight to the discussion:

~ What's all this Trapped Charge and Dielectric Compression stuff anyhow? ~

http://www.keith-snook.info/capacitor-soakage.html


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Good points made by you both, definition is the key to understanding. You cannot analyse competently what you cannot define competently.

I have only devoted a small amount of biological supercomputer time so far to this topic but I quickly came across this link that might offer some insight to the discussion:

~ What's all this Trapped Charge and Dielectric Compression stuff anyhow? ~

http://www.keith-snook.info/capacitor-soakage.html

Too true EA

We take two ali plates,with a dielectric sheet between them,and we then charge that cap to a potetial of say 100 volts.

We then take the top plate,and lay it on the bench next to the other. We remove the dielectric sheet--will there still be a voltage across the two plates of around 100 volts-or any voltage at all?.

If not-dose that 100 volts exist across the dielectric sheet?

If not,can we then put the cap back together,and still have that 100 volts across the cap?

Im not asking due to not knowing the answers-im asking to see how many here do.

Perhaps a good understanding of how the simple capacitor works,may help in our search for that elusive free energy source.

Brad


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Like in this video, especially from the 2 minute mark:

https://www.youtube.com/watch?v=e2EWeOVCO5o


Itsu
   

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I do love these sort of questions.  Here's my take.

We take two ali plates,with a dielectric sheet between them,and we then charge that cap to a potential of say 100 volts.

We then take the top plate,and lay it on the bench next to the other. We remove the dielectric sheet--will there still be a voltage across the two plates of around 100 volts-or any voltage at all?.
Assuming all this is done without any form of electrical connection to the plates they will retain their original charge values, but since now the mutual capacitance between the plates is much reduced then the voltage has to increase so as to keep the C*V product the same.   For widely separated plates the mutual capacitance is the series combination of each plate's self-capacitance and is independent of the separation distance.  Of course you have to do work to pull the top plate off and to remove the dielectric sheet.

Quote
If not-does that 100 volts exist across the dielectric sheet?
No.  The dielectric loses its polarization while it is being removed.

Quote
If not,can we then put the cap back together,and still have that 100 volts across the cap?
Yes, the plates still retain their original charge so when put back together the dielectric gets polarized again.  You can use the attractive forces to do useful work while you reassemble the capacitor.  But you lost that amount when you pulled it apart.

Smudge
 Edit to correct an initial boob.
   

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Smudge
 Edit to correct an initial boob.

Quote
I do love these sort of questions.  Here's my take.Assuming all this is done without any form of electrical connection to the plates they will retain their original charge values, but since now the mutual capacitance between the plates is much reduced then the voltage has to increase so as to keep the C*V product the same.   For widely separated plates the mutual capacitance is the series combination of each plate's self-capacitance and is independent of the separation distance.  Of course you have to do work to pull the top plate off and to remove the dielectric sheet.

Are you sure about that smudge ?

Quote
No.  The dielectric loses its polarization while it is being removed.

Are you sure about that one as well ?.

Quote
Yes, the plates still retain their original charge so when put back together the dielectric gets polarized again.  You can use the attractive forces to do useful work while you reassemble the capacitor.  But you lost that amount when you pulled it apart.

Mmm
 
So,if we separate our capacitor,and short the two plates together,we loose our charge ?

Our dielectric looses it's polarization when removed?

Enjoy Smudge  ;)

A video from Russ

https://www.youtube.com/watch?v=eQGd_jYYfqQ&t=770s


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Like in this video, especially from the 2 minute mark:

https://www.youtube.com/watch?v=e2EWeOVCO5o


Itsu

Yes.

But what about an air capacitor--like the multi plate tuning cap in old radio's.

If we charge that air cap,and then blow the air out,so as it is replaced with new fresh air--will the cap still be charged?.

Brad


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Brad,

my experience is that things are far more complicated then it seems at first glance (like charge is stored in the dielectricum).
Its probably not the dielectricum itself, but a combination of things like distance, matter, etc. and way beyond my present level of comprehencion.

But i am sure someone will be able to explain it in great detail.

 
Itsu
   
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I do love these sort of questions.  Here's my take.Assuming all this is done without any form of electrical connection to the plates they will retain their original charge values, but since now the mutual capacitance between the plates is much reduced then the voltage has to increase so as to keep the C*V product the same.   For widely separated plates the mutual capacitance is the series combination of each plate's self-capacitance and is independent of the separation distance.  Of course you have to do work to pull the top plate off and to remove the dielectric sheet.
No.  The dielectric loses its polarization while it is being removed.
Yes, the plates still retain their original charge so when put back together the dielectric gets polarized again.  You can use the attractive forces to do useful work while you reassemble the capacitor.  But you lost that amount when you pulled it apart.

Smudge
 Edit to correct an initial boob.

Smudge,

I have to respectively disagree here.  The charge is retained in any and all dielectric between the plates.  A quick experiment will prove this.  For example, I took two copper plates approximately  2"x3" and placed a .010" polycarbonate (Lexan) between the plates.  The bottom plate is grounded and the upper plate touched to a positive voltage source.  The charge on this cap is confirmed by touching a 10Meg scope probe to the upper plate and capturing the pulse on a storage scope.  The voltage seen will be somewhat less than the applied voltage but the polarity will be seen to be positive.

Now, if one again charges this cap as above (do not discharge with scope probe) and then carefully remove the upper plate, flip the Lexan dielectric over, replace the upper plate and now touch the probe to the upper plate, we will now see a negative pulse voltage.  Strangely enough this negative voltage will now be larger than the previous positive voltage because we have remnants of stored energy left in the surrounding air dielectric which actually has a polarity based on self discharge of the device before we can reassemble.  This voltage adds to the Lexan dielectric thus producing a larger negative output voltage.  Anyway, this proves that the charge is retained in the dielectric at least in this case.

If you recall years ago, I proposed a segmented rotary generator based on poled dielectrics which I still believe would possibly work.

Regards,
Pm

   

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Brad you should know this one O0

http://ultrabattery.com/

From your part of the world, combined capacitor and battery, very easy to make

Regards

Mike 8)


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