What magic is this?
:D
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.
Quote from: TinselKoala on 2017.11.30, 02:27:02
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
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.
What should be odd there?
What I see is a difference in potential. And what if on both side are bulbs connected?
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
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.
Quote from: TinselKoala on 2017.12.04, 17:26:01
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
Quote from: TinselKoala on 2017.12.04, 17:26:01
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
Quote from: TinselKoala on 2017.12.04, 17:26:01
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? ???
Quote from: TinMan on 2017.12.05, 11:54:08
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? :-[
Quote from: TinMan on 2017.12.07, 15:03:01
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?
Quote from: TinselKoala on 2017.12.07, 16:24:26
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
Quote from: TinselKoala on 2017.12.07, 16:24:26
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?.
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?
Quote from: TinMan on 2017.12.09, 13:10:07
Is the energy stored in the plates,or the dielectric ?
Is the charge stored in the plates,or the dielectric?.
Quote from: muDped on 2017.12.09, 18:13:05
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
Quote from: evolvingape on 2017.12.10, 12:24:28
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
Like in this video, especially from the 2 minute mark:
https://www.youtube.com/watch?v=e2EWeOVCO5o
Itsu
I do love these sort of questions. Here's my take.
Quote from: TinMan on 2017.12.11, 04:08:07
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.
QuoteIf not-does that 100 volts exist across the dielectric sheet?
No. The dielectric loses its polarization while it is being removed.
QuoteIf 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.
Quote from: Smudge on 2017.12.11, 09:50:32
Smudge
Edit to correct an initial boob.
QuoteI 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 ?
QuoteNo. The dielectric loses its polarization while it is being removed.
Are you sure about that one as well ?.
QuoteYes, 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
Quote from: Itsu on 2017.12.11, 09:37:39
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
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
Quote from: Smudge on 2017.12.11, 09:50:32
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
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)
Quote from: Itsu on 2017.12.11, 14:43:05
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
I think gaining a good understanding of what is happening here,would unlock so many doors.
I have one of those old multi plate tuning caps,so I'll try the experiment tonight.
Brad
Quote from: Centraflow on 2017.12.11, 18:13:37
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)
No Mike--havnt seen that one yet--but have now. O0
Brad
Ok,so i placed a diode on one side of a multiplate tuning air capacitor from an old radio.
I then pulsed the capacitor with my SG,using a square wave,at 50% duty cycle, at a frequency of just 6 Hz
I then took one scope shot without air being blown between the plates,and another with air being blown between the plates.
The air supply was from the blow gun of my compressor,at 80psi--so a lot of air passing between all the plates.
Below are two scope shot's.
The first shows the discharge rate without air being blown between the plates.
The second shows the discharge rate with air being blown between the plates.
Even at this low frequency,and a very high volume of air being blown between the plate's,i do not see any difference at all between the scope shots.
If the dielectric (in this case-air) is what holds the charge,why do we see no difference at all between the two scope shot's ?.
Should not the charge be blown away,when uncharged air is continually being injected between the plates?.
Brad
Hi Brad.
You must have some pretty low humidity at the moment?
I once tried a simple experiment using an Aluminium flexible sheet, a little bit heavier gauge than foil. Just by waving it around a bit it in the air it became charged. Could the flowing air be charging the plates?
How do you think the reverse would be? Try evacuating the air by using your compressor in reverse?
Cheers Graham.
Quote from: Grumage on 2017.12.12, 13:19:04
Hi Brad.
You must have some pretty low humidity at the moment?
I once tried a simple experiment using an Aluminium flexible sheet, a little bit heavier gauge than foil. Just by waving it around a bit it in the air it became charged. Could the flowing air be charging the plates?
How do you think the reverse would be? Try evacuating the air by using your compressor in reverse?
Cheers Graham.
No.
I tried blowing the air through the plates,without the SG hooked up--just to look for that very thing.
But not 1mV seen on the scope.
Quote from: TinMan on 2017.12.12, 12:44:09
Ok,so i placed a diode on one side of a multiplate tuning air capacitor from an old radio.
I then pulsed the capacitor with my SG,using a square wave,at 50% duty cycle, at a frequency of just 6 Hz
I then took one scope shot without air being blown between the plates,and another with air being blown between the plates.
The air supply was from the blow gun of my compressor,at 80psi--so a lot of air passing between all the plates.
Below are two scope shot's.
The first shows the discharge rate without air being blown between the plates.
The second shows the discharge rate with air being blown between the plates.
Even at this low frequency,and a very high volume of air being blown between the plate's,i do not see any difference at all between the scope shots.
If the dielectric (in this case-air) is what holds the charge,why do we see no difference at all between the two scope shot's ?.
Should not the charge be blown away,when uncharged air is continually being injected between the plates?.
Brad
OK, this is interesting. Not being that familiar with the Rigol scopes, what does the "D 85.14ms" indicate and, I assume the "H 500us" is the horizontal sweep rate/division? Just trying to evaluate the velocity of the moving air verses the discharge time period.
Regards,
Pm
Quote from: partzman on 2017.12.12, 14:36:53
OK, this is interesting. Not being that familiar with the Rigol scopes, what does the "D 85.14ms" indicate and, I assume the "H 500us" is the horizontal sweep rate/division? Just trying to evaluate the velocity of the moving air verses the discharge time period.
Regards,
Pm
The "D 85.14 ms" indicates that the trigger point is 85.14 ms from screen center, off to the left. Also indicated by the little orange arrow symbol "T" at top left pointing to the left. The "H 500 us" is the horizontal time per division setting.
Quote from: TinselKoala on 2017.12.12, 15:05:52
The "D 85.14 ms" indicates that the trigger point is 85.14 ms from screen center, off to the left. Also indicated by the little orange arrow symbol "T" at top left pointing to the left. The "H 500 us" is the horizontal time per division setting.
Thanks TK. My knowledge of fluid dynamics is sorely lacking so I'm unable to go any farther with this!
Regards,
Pm
Edit:
Quote from: partzman on 2017.12.13, 14:12:26
Thanks TK. My knowledge of fluid dynamics is sorely lacking so I'm unable to go any farther with this!
Regards,
Pm
Edit:
If the charge is stored in the dielectric,and in this case,the dielectric is the air,i would have thought we should have seen a reduction in time the cap took to discharge,and a decrease in charge, when the air was being blown between the plates,as the charge would have been removed along with the air being removed during the charge and discharge cycles.
But as you can see,there is no change at all--and we are talking about a very small value in capacitance-a low frequency cycle rate,and a high volume of air exchange.
So,in this case,it dose not seem like the charge is stored in the dielectric ???__> have we talked our selves into believing that it is,due to what we !think! we have seen in the video's posted so far?.
Maybe it is like Itsu said--there maybe more than meets the eye here?.
Quote from: TinMan on 2017.12.12, 12:44:09
Ok,so i placed a diode on one side of a multiplate tuning air capacitor from an old radio.
I then pulsed the capacitor with my SG,using a square wave,at 50% duty cycle, at a frequency of just 6 Hz
What is the number of the diode you used and what is the capacitance of the variable cap when closed? It is possible the capacitance of the reverse biased diode is the major contributor to the overall capacitance being discharged which would definitely affect the results. Is a 10Meg scope probe used to discharge the cap?
Quote
I then took one scope shot without air being blown between the plates,and another with air being blown between the plates.
The air supply was from the blow gun of my compressor,at 80psi--so a lot of air passing between all the plates.
Below are two scope shot's.
The first shows the discharge rate without air being blown between the plates.
The second shows the discharge rate with air being blown between the plates.
Another factor that could be affecting these results is that the time constant appears to be ~750us which would mean 5T ~ 3.75ms. At 6Hz, the "on" time would be 83.3ms which is 83.3/3.75 ~22x the discharge time. IOW, you are possibly creating a large number of electrostatic domains around the test cap during the long charge period that could affect the results.
Quote
Even at this low frequency,and a very high volume of air being blown between the plate's,i do not see any difference at all between the scope shots.
If the dielectric (in this case-air) is what holds the charge,why do we see no difference at all between the two scope shot's ?.
Should not the charge be blown away,when uncharged air is continually being injected between the plates?.
Yes, one would assume so.
Perhaps a more conclusive test setup would be to test the variable cap with just a touch from a positive source, then setup the scope to trigger at approximately half that value on a rising edge and discharge the cap with the scope probe. Try charging and discharging this setup until you have consistent results. Then, charge the cap, put a short burst of air to the cap and then discharge with the scope probe and compare the results.
Quote
Brad
Regards,
Pm
Quote from: partzman on 2017.12.15, 14:38:24
What is the number of the diode you used and what is the capacitance of the variable cap when closed? It is possible the capacitance of the reverse biased diode is the major contributor to the overall capacitance being discharged which would definitely affect the results. Is a 10Meg scope probe used to discharge the cap?
Another factor that could be affecting these results is that the time constant appears to be ~750us which would mean 5T ~ 3.75ms. At 6Hz, the "on" time would be 83.3ms which is 83.3/3.75 ~22x the discharge time. IOW, you are possibly creating a large number of electrostatic domains around the test cap during the long charge period that could affect the results.
Yes, one would assume so.
Perhaps a more conclusive test setup would be to test the variable cap with just a touch from a positive source, then setup the scope to trigger at approximately half that value on a rising edge and discharge the cap with the scope probe. Try charging and discharging this setup until you have consistent results. Then, charge the cap, put a short burst of air to the cap and then discharge with the scope probe and compare the results.
Regards,
Pm
PM,
using 2 big variable air capacitors (345pF each) parallel gives 680pF.
Plates on max. capacitance and loaded with 24V.
Attaching the groundlead of the scope to the fixed plates, and touching the probe tip to the variable plates shows this scope signal, see picture.
It that the kind of signal you would expect?
Itsu
Quote from: Itsu on 2017.12.16, 16:05:04
PM,
using 2 big variable air capacitors (345pF each) parallel gives 680pF.
Plates on max. capacitance and loaded with 24V.
Attaching the groundlead of the scope to the fixed plates, and touching the probe tip to the variable plates shows this scope signal, see picture.
It that the kind of signal you would expect?
Itsu
Itsu,
Thanks for doing this test! Yes that is what I would expect only I used a positive polarity but the test will work fine with a negative source and pulse. IOW, if you have the stator plates grounded and apply a quick positive voltage to the rotor plates , then touch your probe to the rotor plate, you should see a fast rising positive positive pulse that slowly decays from the probe resistance. So, the polarity really doesn't matter and it appears that you are ready to test with a short burst of air thru the plates to compare with the static conditions.
If you have consistent static tests, you could save the discharge profile in one of your Ref channels so it could be compared to the discharge profile after air has been applied to the variable cap.
Regards,
Pm
Why do vacuum capacitors work?
https://en.wikipedia.org/wiki/Vacuum_variable_capacitor
Quote from: TinselKoala on 2017.12.16, 17:58:26
Why do vacuum capacitors work?
https://en.wikipedia.org/wiki/Vacuum_variable_capacitor
As I understand it, the dielectric constant of air is 1.0006 compared to a vacuum of 1.0000. Hardly much difference, so electric dipoles would be possible in a vacuum resulting in a capacitance slightly less than than that in air with equal plate dimensions.
Regards,
Pm
Hmmm,
unexpected results.
OK, same setup as above (2x 345pF air variable caps parallel at max. capacity (960pF measured))
Loading with 24V for 1 second the moveable plates with the return lead connected to the static plates.
After 5 second touching with the probe tip (ground lead / return lead still on the static plates) the moveable plates.
Done it twice, see screenshot the first 2 peaks Ref1 and Ref2. (23.xV)
Then in those 5 second waiting time blowing air (with my mouth) inbetween the plates, see 3th peak Ref 3 (out of scale, so clipping).
Then after adjusting the vertical setting again blowing air inbetween the plates, see yellow CH1 peak (110.4V) :o
So there surely is a huge difference.
Itsu
Quote from: Itsu on 2017.12.16, 20:50:56
Hmmm,
unexpected results.
OK, same setup as above (2x 345pF air variable caps parallel at max. capacity (960pF measured))
Loading with 24V for 1 second the moveable plates with the return lead connected to the static plates.
After 5 second touching with the probe tip (ground lead / return lead still on the static plates) the moveable plates.
Done it twice, see screenshot the first 2 peaks Ref1 and Ref2. (23.xV)
Then in those 5 second waiting time blowing air (with my mouth) inbetween the plates, see 3th peak Ref 3 (out of scale, so clipping).
Then after adjusting the vertical setting again blowing air inbetween the plates, see yellow CH1 peak (110.4V) :o
So there surely is a huge difference.
Itsu
Itsu,
OK, if I understand correctly, after charging the capacitor, you blow thru the plates, discharge with your scope probe and then measure a larger peak pulse than without blowing thru the cap. If correct, my interpretation is that since Q=CV or V=Q/C and with the voltage being greater after blowing, you increased the charge between the capacitor plates. The only other option is that your breath has a lower dielectric constant than air and has lowered the capacitance ???. What in the world have you been drinking :). Seriously, the charge has to have increased but I don't know how at this point!
Regards,
Pm
Quote from: partzman on 2017.12.16, 19:04:55
As I understand it, the dielectric constant of air is 1.0006 compared to a vacuum of 1.0000. Hardly much difference, so electric dipoles would be possible in a vacuum resulting in a capacitance slightly less than than that in air with equal plate dimensions.
Regards,
Pm
Dipoles of what?
https://en.wikipedia.org/wiki/Vacuum_polarization
Quote from: partzman on 2017.12.16, 22:11:42
Itsu,
OK, if I understand correctly, after charging the capacitor, you blow thru the plates, discharge with your scope probe and then measure a larger peak pulse than without blowing thru the cap. If correct, my interpretation is that since Q=CV or V=Q/C and with the voltage being greater after blowing, you increased the charge between the capacitor plates. The only other option is that your breath has a lower dielectric constant than air and has lowered the capacitance ???. What in the world have you been drinking :). Seriously, the charge has to have increased but I don't know how at this point!
Regards,
Pm
What's the difference between the relative permittivity of dry air (room air) and moisture-saturated, CO
2 enhanced, oxygen depleted air (Itsu's breath)?
Ok, OK, you got me there guys, i won't be drinking and benching (as in working on the bench) anymore. :P
But seriously, PM, you are correct how i did it, so that was not such a good idea i guess to breathe instead of using a blower or fan.
I will repeat that test using such a fan.
But the goal was to replace the charged air (dielectric) by some fresh uncharged air to see if the charge was gone.
It really should not matter if i replace the charged air by moistured air, i think.
We will see, itsu
OK, redid that same test again, (24V from a battery stack) now with both breathing and using a fan.
Again the breathing shows peaking (3th peak) while the fan (4th peak) shows normal amplitude.
Screenshot shows first 2 peaks are without breathing/blowing, 3th is breathing, 4th is using a 12V fan blowing air between the plates.
So it seems that the charge is NOT contained in the air (dielectricum) between the plates.
TK, so what is the difference?
Itsu
Quote from: TinselKoala on 2017.12.17, 00:09:26
Dipoles of what?
https://en.wikipedia.org/wiki/Vacuum_polarization
Electric dipoles as in-
https://en.wikipedia.org/wiki/Electric_displacement_field
Quote "In a dielectric material the presence of an electric field E causes the bound charges in the material (atomic nuclei and their electrons) to slightly separate, inducing a local electric dipole moment."
Pm
Quote from: TinselKoala on 2017.12.17, 00:16:36
What's the difference between the relative permittivity of dry air (room air) and moisture-saturated, CO2 enhanced, oxygen depleted air (Itsu's breath)?
Don't know the answer to this so, let's make some assumptions regarding Itsu's breath in his initial test. If his breath had a higher dielectric constant than the ambient air, this would have the effect of increasing the capacitance between the cap plates. This along with the initial charge should have produced a lower peak discharge voltage but this was not the case. The only option then is that his breath has a dielectric constant <1 so is this the answer?
Pm
All,
Ok, I set up a two plate test capacitor that would allow a .010" thick piece of polycarbonate material to be moved and removed if necessary to confirm the previous test results as described in my post #23 which I have run many times in the past. All my previous results indicated that the charge was stored in the dielectric. At this moment in time however, I can not repeat those test results so I'm presently a little confused with the results I am seeing.
Regards,
Pm
Quote from: partzman on 2017.12.17, 15:46:26
Electric dipoles as in-
https://en.wikipedia.org/wiki/Electric_displacement_field
Quote "In a dielectric material the presence of an electric field E causes the bound charges in the material (atomic nuclei and their electrons) to slightly separate, inducing a local electric dipole moment."
Pm
Certainly. But aren't we talking about _vacuum_ capacitors? What dielectric material contains bound charges that can separate and create a dipole moment in a
vacuum capacitor?
Quote from: partzman on 2017.12.17, 16:05:56
Don't know the answer to this so, let's make some assumptions regarding Itsu's breath in his initial test. If his breath had a higher dielectric constant than the ambient air, this would have the effect of increasing the capacitance between the cap plates. This along with the initial charge should have produced a lower peak discharge voltage but this was not the case. The only option then is that his breath has a dielectric constant <1 so is this the answer?
Pm
Certainly there are differences in relative permittivity between dry ambient air and exhaled air that is saturated with moisture. Not much though. Can we hope to detect the difference between, say, the permittivity of the vacuum, and the permittivity of air moist or dry, with our benchtop equipment? A few parts per ten thousand?
I think that moist exhaled air will be more conductive than dry ambient air.
Quote from: partzman on 2017.12.17, 17:23:50
All,
Ok, I set up a two plate test capacitor that would allow a .010" thick piece of polycarbonate material to be moved and removed if necessary to confirm the previous test results as described in my post #23 which I have run many times in the past. All my previous results indicated that the charge was stored in the dielectric. At this moment in time however, I can not repeat those test results so I'm presently a little confused with the results I am seeing.
Regards,
Pm
Well, you can have energy stored as dielectric stress inside the dielectric material, and you can also have induced electrostatic patch (surface) charges on the surfaces of the material. When a capacitor is made, the metal conductive plates are in direct contact with the dielectric material, so you don't get the patch charges on the surfaces, just the dielectric stress within the material. But when the dielectric is then separated from the conductive plates the patch charges are induced (if the separation is done right.) Then when the dielectric with surface charges is reintroduced between uncharged metal plates, it in turn induces charges on the plates. So it looks like charge is "stored" on the dielectric, but really it is only separated.
I think.
Quote from: TinselKoala on 2017.12.17, 17:26:13
Certainly. But aren't we talking about _vacuum_ capacitors? What dielectric material contains bound charges that can separate and create a dipole moment in a vacuum capacitor?
I don't understand quantum field theory on how charge is retained in a vacuum and for that reason, I see no avenue for potential OU energy generation by this means. What I wish to know is where the charge is retained in a capacitance other than a vacuum based device as those to me do have OU potential.
If the charge is not retained in the dielectric of the many types of piezoelectric, ferroelectric, and paraelectric devices in use today that are well documented, then exactly where is it retained?
Pm
Charge is separated in those materials by application of some external force (and the separation can be retained in certain cases without continuing application of the initial external force, such as with electrets). This isn't the same as storage of charge though. It is storage of _energy_ in the separation of charges which would ordinarily be balanced. What is retained is the separation. The overall quantity of charge is unchanged.
Quote from: TinselKoala on 2017.12.17, 18:14:43
Charge is separated in those materials by application of some external force (and the separation can be retained in certain cases without continuing application of the initial external force, such as with electrets). This isn't the same as storage of charge though. It is storage of _energy_ in the separation of charges which would ordinarily be balanced. What is retained is the separation. The overall quantity of charge is unchanged.
So, when I charge an un-poled dielectric between two plates or IOW, I charge this capacitor to any given voltage, where is the charge stored in this simple device?
Pm
Quote from: partzman on 2017.12.17, 21:13:36
So, when I charge an un-poled dielectric between two plates or IOW, I charge this capacitor to any given voltage, where is the charge stored in this simple device?
Pm
On the plates. One has an excess of negative charge (electrons) and the other has an excess of positive charge (holes). What causes the excess of charge? Electrostatic attraction, from the polarization of the dielectric material.
Charges, or rather charge carriers like electrons, only move over great distances in conductors. Think of the "electron sea" of conduction electrons in metals. In dielectric materials, which are insulators by definition, charge does not move that way, since the electrons remain bound to the atoms of the dielectric material, and the material as a whole is electrically neutral. The molecules of a good dielectric are highly polar and elastically bound. The application of an external electric field (by putting charge onto the metal plates) causes alignment of the polar molecules, somewhat in the same way as an applied magnetic field aligns domains in a soft magnetic material. This process stores energy but does not transfer or build up charge in the same way that electrons and holes store charge in the plates. When the voltage source which is responsible for putting excess electrons is removed and a conductive load is applied, the stored energy of twisting and distorting the polar molecules of the dielectric, is released, returning to the unstrained, more random state, as the load allows the excess electrons and "holes" to neutralize.
There are lots of "free energy" implications of this, especially when one considers the vacuum dielectric property. Create enough stress in the vacuum and it will shatter, possibly releasing or de-virtualizing more of those virtual particle pairs talked about in the Wiki article above. This is one reason why I'm interested in high voltage phenomena.
.
author=TinselKoala link=topic=3564.msg66101#msg66101 date=1513552780]
QuoteOn the plates. One has an excess of negative charge (electrons) and the other has an excess of positive charge (holes). What causes the excess of charge? Electrostatic attraction, from the polarization of the dielectric material.
If this were true,once the dielectric material is removed from between the two charged plates,there should be a potential difference between/across the two plates--but there is none.
Once the dielectric is removed,there is no sign of any charge on the two plates--as shown in the two video's posted some time back.
But once the dielectric is returned to it's original position between the two plates,the two plates now have a potential difference across them.
What is needed,is a very large value capacitor(200uF or so),where we can either increase or decrease the amount of dielectric material between the plates,or replace the charged dielectric material for uncharged dielectric material :D
I'll see what i can come up with.
Brad
Vacuum capacitors do work though, with only the vacuum as dielectric. Are you ready to say that the vacuum can hold a net charge?
The dielectric material in an ordinary capacitor is an insulator, that allows much greater voltage to be developed across the plates than would normally be the case if a conductive or ionizable medium, like air, were between the plates. This is also the case for the vacuum capacitor. The capacitance of a vacuum capacitor is nearly the same as an air capacitor of the same dimensions, but can be charged to a much higher voltage (hence energy content) because there aren't any ions (practically) that can conduct and limit the voltage.
In fact this phenomenon is used in high-vacuum "diode" pumps, where electrostatic attraction of the residual gas ions is used to sweep out the remaining gas.
All,
Does anyone reading this thread recognize anything significant in Itsu's posts #39 and #44? Speak up!
Regards,
Pm
Well, what significant is is that when you replace the dry air inbetween the plates of a (variable) air (so NOT vacuum) capacitor, when charged, by
moisture-saturated, CO2 enhanced, oxygen depleted air (my breath) you can up to quadruple the voltage.
Now how can we benefite from that?
I did some further tests, like measuring the capacitance of that capacitor (982pF) while breathing again on / between the plates.
No change in capacitance was measured using my Agilent U1733C LCR meter.
Then i was varying the plates when charged like this, see screenhot.
First peak was normal full capacitance (982pF) charged with 24V, not changing the plates, no blowing (control)
second peak was full capacitance (982pF) charged with 24V, then rotating the plates to minimum capacitance (75pF), no blowing
thirth peak was full capacitance (982pF) charged with 24V, blowing , then rotating the plates to minimum capacitance (75pF) <== changed sequence, first blowing then rotating!!
fourth peak was full capacitance (982pF) charged with 24V, not changing the plates, blowing (seems somewhat lower then yesterday).
When changing the capacitance while charged to minimum is causing a higher voltage, but narrower peak.
Itsu
Quote from: Itsu on 2017.12.18, 20:18:01
Well, what significant is is that when you replace the dry air inbetween the plates of a (variable) air (so NOT vacumm) capacitor, when charged, by
moisture-saturated, CO2 enhanced, oxygen depleted air (my breath) you can up to quadruple the voltage.
Now how can we benefite from that?
I did some further tests, like measuring the capacitance of that capacitor (982pF) while breathing again on / between the plates.
No change in capacitance was measured using my Agilent U1733C LCR meter.
Then i was varying the plates when charged like this, see screenhot.
First peak was normal full capacitance (982pF) charged with 24V, not changing the plates, no blowing (control)
second peak was full capacitance (982pF) charged with 24V, then rotating the plates to minimum capacitance (75pF), no blowing
thirth peak was full capacitance (982pF) charged with 24V, then rotating the plates to minimum capacitance (75pF), blowing
fourth peak was full capacitance (982pF) charged with 24V, not changing the plates, blowing (seems somewhat lower then yesterday).
When changing the capacitance while charged to minimum is causing a higher voltage, but smaller peak.
Itsu
Itsu,
Congrats and thanks for analyzing your results O0. I totally agree with you and will add that your breath has enough moisture content to raise the dielectric constant above that of the ambient air between the plates when you charge the capacitor. Now, this is a seeming contradiction to Q=CV or V=Q/C because normally the voltage should be less not greater if we raise the capacitance with an increase in dielectric constant while keeping the charge the same. Now it is possible that there may be some poling of your breath (no offense C.C) that is adding to the original charge but I seriously doubt that.
Attached below is a scope pix of tests I ran to replicate your tests with 2 parallel copper plates held apart with spacers so a piece of polycarbonate sheet could be inserted and removed. I was careful to be sure that the poling of any charge on the sheet was at a minimum or non-existent. Basically, the copper plates with ambient air dielectric were charged with 30vdc then, the polycarb dielectric sheet was slid into position and then discharged with the probe and captured the same as your procedure. The ch1 trace is the normal discharge with no added polycarb and the remaining Refs 1,2,and 3 are the voltages reached with the polycarb dielectric inserted. The results are very consistent and repeatable and show an apparent energy gain of ~(64(avg output peak)/28)^2 = 5.22 .
So, I would classify this as an anomaly and would welcome any expert advice from those more knowledgeable in this area.
Now, what can be done with this? Picture a generator made with segmented conductive stators with segmented dielectric rotors placed between each stator and spun to some reasonably high RPM. Use appropriate circuitry to charge the stators when air is the dielectric and discharge when the dielectric rotors are in place to allow the excess energy to be harvested.
Additionally, poled dielectrics could be used which would add to the "ambient air" charge thus increasing the overall energy gain.
One question would be what the rotor drag would consist of in newtons but I can show an old simulation from my past research on a segmented poled dielectric generator that indicated a stronger pulling force than opposing force which would seemingly indicate a self driving generator!
Regards,
Pm
Could you guys be close to figuring out the Testanika machine? Just seems similar to what you are seeing.
PM, nicely done as usual. I wonder if you can figure out some way (other than the voltage rise!) to determine the actual _work_ you do to the system by sliding the dielectric sheet in between the charged plates.
;)
With rotating electrostatic machines like you describe, if they are hand-cranked one can actually feel the work being done as the dielectric "pushes" against the field when it moves between the charged plates. As the voltage on the accumulators (capacitances) rises they become harder and harder to turn.
Quote from: partzman on 2017.12.18, 22:42:17
Itsu,
Congrats and thanks for analyzing your results O0. I totally agree with you and will add that your breath has enough moisture content to raise the dielectric constant above that of the ambient air between the plates when you charge the capacitor. Now, this is a seeming contradiction to Q=CV or V=Q/C because normally the voltage should be less not greater if we raise the capacitance with an increase in dielectric constant while keeping the charge the same. Now it is possible that there may be some poling of your breath (no offense C.C) that is adding to the original charge but I seriously doubt that.
Attached below is a scope pix of tests I ran to replicate your tests with 2 parallel copper plates held apart with spacers so a piece of polycarbonate sheet could be inserted and removed. I was careful to be sure that the poling of any charge on the sheet was at a minimum or non-existent. Basically, the copper plates with ambient air dielectric were charged with 30vdc then, the polycarb dielectric sheet was slid into position and then discharged with the probe and captured the same as your procedure. The ch1 trace is the normal discharge with no added polycarb and the remaining Refs 1,2,and 3 are the voltages reached with the polycarb dielectric inserted. The results are very consistent and repeatable and show an apparent energy gain of ~(64(avg output peak)/28)^2 = 5.22 .
So, I would classify this as an anomaly and would welcome any expert advice from those more knowledgeable in this area.
Now, what can be done with this? Picture a generator made with segmented conductive stators with segmented dielectric rotors placed between each stator and spun to some reasonably high RPM. Use appropriate circuitry to charge the stators when air is the dielectric and discharge when the dielectric rotors are in place to allow the excess energy to be harvested.
Additionally, poled dielectrics could be used which would add to the "ambient air" charge thus increasing the overall energy gain.
One question would be what the rotor drag would consist of in newtons but I can show an old simulation from my past research on a segmented poled dielectric generator that indicated a stronger pulling force than opposing force which would seemingly indicate a self driving generator!
Regards,
Pm
My guess is that the anomalous voltage is due to the tribo-electric effect. When dissimilar materials come into contact with each other there can be a transfer of charge from one material to the other. Multiple contacts can produce ever increasing charge transfer which is why aircraft (or any form of missile) gain electric charge when flying through rain or even dust particles, necessitating special wick dischargers to keep the potential gained low enough to stop corona discharge off the various antenna that would otherwise corrupt the signals. So maybe your polycarbonate making contact to and possibly sliding across the electrodes produced the anomalous charge. Don't know where polycarbonate sits in the triboelectric table but it is likely to be removed from copper.
Edit. Seems polycarbonate is near copper so little effect expected. So the next question is how was the dielectric inserted, was it by hand? If so there is the possibility of charge transfer from your body. You say you were careful to ensure that the dielectric was not poled, but how sure were you that the dielectrirc was not carrying surface charge?
Smudge
This subject is now beginning to interest me greatly!
After my introduction to electrostatics earlier this year with the Franklin motor and in particular the statements made by Verpies and Cyril about the tremendous forces that can be got from a pair of like charged poles, it made me want to build a machine.
I have literally Kilo's of Polycarbonate off cuts in various sizes and thicknesses just waiting to be used.
In the new year I'd be willing to build a machine to a design that this group has brainstormed. There must be a way of utilising that repulsive force that TK has mentioned above to our advantage?
My only constraint is a maximum diameter rotor of 14 inches.
Can we design a new machine? I think we can!
Cheers Graham.
Quote from: CITFTA on 2017.12.18, 22:52:08
Could you guys be close to figuring out the Testanika machine? Just seems similar to what you are seeing.
AH-HA
O0
As i stated in post 25 Carroll--> I think gaining a good understanding of what is happening here,would unlock so many doors.
For far to long,we have been trying to replicate and understand other peoples work,while there is so much to be discovered right in front of our noses.
Brad
author=partzman link=topic=3564.msg66119#msg66119 date=1513636937]
So, I would classify this as an anomaly and would welcome any expert advice from those more knowledgeable in this area.
Additionally, poled dielectrics could be used which would add to the "ambient air" charge thus increasing the overall energy gain.
Regards,
Pm
[/quote]
QuoteAttached below is a scope pix of tests I ran to replicate your tests with 2 parallel copper plates held apart with spacers so a piece of polycarbonate sheet could be inserted and removed. I was careful to be sure that the poling of any charge on the sheet was at a minimum or non-existent. Basically, the copper plates with ambient air dielectric were charged with 30vdc then, the polycarb dielectric sheet was slid into position and then discharged with the probe and captured the same as your procedure. The ch1 trace is the normal discharge with no added polycarb and the remaining Refs 1,2,and 3 are the voltages reached with the polycarb dielectric inserted. The results are very consistent and repeatable and show an apparent energy gain of ~(64(avg output peak)/28)^2 = 5.22 .
So,we charge the cap while it has an air dielectric ,and discharge while it has a poly carb dielectric,and we see an energy gain O0--or do we?. :-\
If voltage went up,then capacitance value would have had to of dropped-yea?,or are we about to re-write the ruel book?.
Perhaps leave the plates uncharged,and slide the polycarb dielectric between the plates,and see if you get a voltage reading across the plates.
If the result is 0v across the two plates,then this would eliminate any static charge being introduced into the capacitor when you slide the polycarb sheet between the plates.
QuoteNow, what can be done with this? Picture a generator made with segmented conductive stators with segmented dielectric rotors placed between each stator and spun to some reasonably high RPM. Use appropriate circuitry to charge the stators when air is the dielectric and discharge when the dielectric rotors are in place to allow the excess energy to be harvested.
Indeed O0
QuoteOne question would be what the rotor drag would consist of in newtons but I can show an old simulation from my past research on a segmented poled dielectric generator that indicated a stronger pulling force than opposing force which would seemingly indicate a self driving generator!
To quote TK :With rotating electrostatic machines like you describe, if they are hand-cranked one can actually feel the work being done as the dielectric "pushes" against the field when it moves between the charged plates. As the voltage on the accumulators (capacitances) rises they become harder and harder to turn.
However,i do not think this is the same,and what we know seems to say the opposite.
We know
1-opposite charges attract
2-Both a negative charged object,and a positive charged object,attract an object with a neutral or no charge.
3 An object with a high negative charge,will attract an object that has a low negative charge-and visa versa for positively charged objects--as long as there is a potential difference in charge.
So it would seem to me,that if one plate has a positive charge,and the other plate has a negative charge,then both would have an attraction toward the polycarb sheet(dielectric),which has a neutral charge.
We would need some rather large segmented disk's to build a generator that would work on this principle,that has any sort of !volume! to the output.
In saying that,this has far more potential than any Bidumy energizer.
Brad
" In saying that,this has far more potential than any Bidumy energizer. "
Brad, was the pun intentional? :)
So, do we open a new thread for a " Freedom group " electrostatic motor generator design? OUR own? :)
Cheers Graham.
Perhaps some one could try this out.
We have two plates(our capacitor plates),and in between them,we have our polycarb dielectric.
What if the capacitor was charged while there was an air gap between the two plates,as well as the polycarb dielectric,and then the two plates were brought together after the capacitor was charges--in other words,the air gap was removed ?
What would be the outcome?.
As far as i know,there should be an attraction between the two plates once charged,and then once discharged,the plates would be able to be separated without much force required.
Brad
Quote from: Grumage on 2017.12.19, 13:16:01
" In saying that,this has far more potential than any Bidumy energizer. "
Brad, was the pun intentional? :)
So, do we open a new thread for a " Freedom group " electrostatic motor generator design? OUR own? :)
Cheers Graham.
Yes,the pun was intentional --i was a dummy to think there was any truth to it.
But all is not lost Graham,as i think we could use the base of our energizer build,to make this capgen Pm was talking about.
I am glad i raised the question about !where! charge and energy is stored in a capacitor,as it would seem that no one is quite clear on just what is going on here.
QuoteSo, do we open a new thread for a " Freedom group " electrostatic motor generator design? OUR own?
I think this would be a great idea,as it is !OUR! idea--!OUR! research on stored energy within a capacitor.
Take note of what Carroll said in post 61,and tie that in with PM's capgen idea O0
Do you want to start the thread Graham,or would you like me to?.
Brad
Quote from: Smudge on 2017.12.19, 09:44:57
My guess is that the anomalous voltage is due to the tribo-electric effect. When dissimilar materials come into contact with each other there can be a transfer of charge from one material to the other. Multiple contacts can produce ever increasing charge transfer which is why aircraft (or any form of missile) gain electric charge when flying through rain or even dust particles, necessitating special wick dischargers to keep the potential gained low enough to stop corona discharge off the various antenna that would otherwise corrupt the signals. So maybe your polycarbonate making contact to and possibly sliding across the electrodes produced the anomalous charge. Don't know where polycarbonate sits in the triboelectric table but it is likely to be removed from copper.
Edit. Seems polycarbonate is near copper so little effect expected. So the next question is how was the dielectric inserted, was it by hand? If so there is the possibility of charge transfer from your body. You say you were careful to ensure that the dielectric was not poled, but how sure were you that the dielectrirc was not carrying surface charge?
Smudge
Smudge,
I am not sure that there was no surface charge on the dielectric. I guess that Itsu's breath could also contain some charge as well but as I view things at this point, I'm not sure it would make any difference in the overall outcome. In an attempt to do a simple analysis of what might be happening, I've concluded that we must be increasing Q (or an equivalent in regards to the plates) with the insertion of a dielectric that had no part in the original charge separation. I do not posses an electrometer at this point but I would like to view how the separated charges are distributed between the plates as this may provide some answers.
Regards,
Pm
Edit:.....
Quote from: TinselKoala on 2017.12.19, 01:50:55
PM, nicely done as usual. I wonder if you can figure out some way (other than the voltage rise!) to determine the actual _work_ you do to the system by sliding the dielectric sheet in between the charged plates.
;)
With rotating electrostatic machines like you describe, if they are hand-cranked one can actually feel the work being done as the dielectric "pushes" against the field when it moves between the charged plates. As the voltage on the accumulators (capacitances) rises they become harder and harder to turn.
Thanks Tk,
Not sure at this point on how to determine the work required to insert the dielectric on my simple test device. Yes, the rotor drag of such a device is a critical key to success so I guess building a single rotor POC is in order to help answer these questions.
Regards,
Pm