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Author Topic: Motor consumes no charge ?  (Read 8472 times)

Gothic

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Motor consumes no charge ?
« on: May 12, 2016, 12:28:25 AM »
Website is gone and this doc is copyright 18 years, so posting shouldn,t be a problem.
 If there is a problem i,ll remove or the admin will.

Simply for discussion

Free Energy | searching for free energy and discussing free energy

Motor consumes no charge ?
« on: May 12, 2016, 12:28:25 AM »

Offline MagnaProp

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Re: Motor consumes no charge ?
« Reply #1 on: May 12, 2016, 10:10:00 AM »
Thanks for posting that.

This video appears to be related to the first part of that document.

https://www.youtube.com/watch?v=fy2GvLxgSHA


Offline SkyWatcher123

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Re: Motor consumes no charge ?
« Reply #2 on: May 13, 2016, 06:28:56 AM »
Hi folks, Hi gothic, thanks for sharing.
Hi magnaprop, thanks for sharing that video.
The man in the video makes some very good, simple observations.
Is it possible, to then take those 2 caps. and put them in parallel and then use a dc-dc converter to fully charge another single, 2.7 volt-350 farad  capacitor and start the whole process over again and at least keep it going a few times.
The 2 caps in parallel would equal 700 farads at 1.26 volts or greater.
This needs to be tested i think.
peace love light

Offline ALVARO_CS

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Re: Motor consumes no charge ?
« Reply #3 on: May 13, 2016, 12:38:14 PM »
I posted this there:
so in the beginning you have 2.52 V and at the end of vid you have 1.40+1.34= 2.74 V. that´s not half voltage in each cap, that is more total voltage than at the start. . . incomprehensible !

Suppose the video goes on until the voltage in caps is same, ¿would it not be 1.37 Volts in each cap ?
I do not see those 1.26 anywhere.  ???


Offline wistiti

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Re: Motor consumes no charge ?
« Reply #4 on: May 14, 2016, 04:36:27 AM »
verry interesting!
:)
Thanks for sharing!

Free Energy | searching for free energy and discussing free energy

Re: Motor consumes no charge ?
« Reply #4 on: May 14, 2016, 04:36:27 AM »
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Offline tinman

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Re: Motor consumes no charge ?
« Reply #5 on: May 14, 2016, 12:00:01 PM »
Well,so he ended up with more than half the energy in the two caps combined to that of what he started with.
So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.
But there is also one other thing to take into account-->the little motor was running the whole time,and there for was also dissipating power the whole time current was flowing through it,by way of resistive heat loss.

Dont tell the EE guy's,they will start doing back flips.

Great job guys.


Brad

Offline Dog-One

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Re: Motor consumes no charge ?
« Reply #6 on: May 14, 2016, 10:04:54 PM »
Well,so he ended up with more than half the energy in the two caps combined to that of what he started with.
So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.

Yes, I get that we would need about 1.782V to be left in each capacitor to equal the same energy (Joules) we started with.

Okay, screw energy for a moment and just focus on charge:

TStart
C1 -- 2.52V -- 882 Coulombs
C2 -- 0.0V -- 0 Coulombs
QTotal:  882 Coulombs

TFinal
C1 -- 1.468V -- 513.8 Coulombs
C2 -- 1.273V -- 445.55 Coulombs
QTotal:  959.35 Coulombs

So how in the heck do we gain charge, but lose energy?  Isn't this an entirely "closed loop" system?  Charge just jumps in there as energy is split into two chunks?  If we had a system that took charge in two clumps and put them into one clump, would we gain energy and lose charge?

I need a better way to interpret these results Brad.  What is the correct way to think about this simple system?  Apparently you cannot think of energy like water and capacitors like buckets.  This is so very fundamental, I'd hate to waste this moment and beat my head against the wall for another ten years.

M@

Free Energy | searching for free energy and discussing free energy

Re: Motor consumes no charge ?
« Reply #6 on: May 14, 2016, 10:04:54 PM »
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Offline tinman

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Re: Motor consumes no charge ?
« Reply #7 on: May 15, 2016, 10:42:47 AM »
Yes, I get that we would need about 1.782V to be left in each capacitor to equal the same energy (Joules) we started with.

Okay, screw energy for a moment and just focus on charge:

TStart
C1 -- 2.52V -- 882 Coulombs
C2 -- 0.0V -- 0 Coulombs
QTotal:  882 Coulombs



TFinal
C1 -- 1.468V -- 513.8 Coulombs
C2 -- 1.273V -- 445.55 Coulombs
QTotal:  959.35 Coulombs

So how in the heck do we gain charge, but lose energy?  Isn't this an entirely "closed loop" system?  Charge just jumps in there as energy is split into two chunks?  If we had a system that took charge in two clumps and put them into one clump, would we gain energy and lose charge?

I need a better way to interpret these results Brad.  What is the correct way to think about this simple system?  Apparently you cannot think of energy like water and capacitors like buckets.  This is so very fundamental, I'd hate to waste this moment and beat my head against the wall for another ten years.

M@

Im not even going to pretend i know how that works.
C is I x time,where as Joules is watts per second--> 1 joule is 1 watt  second.

Looks odd when we look at energy stored in cap's
Example
1 volt over 1 Farad= 1 coulomb of charge,but only 500mJ of energy
2 volts over 1 Farad =2 coulombs of charge,and 2 joules of energy
3 volts over 1 Farad =3 coulombs of charge,but now the energy is higher at 4.5 joules.

As we go on,the charge always is the same as the voltage across the cap. If you double the size of the cap,you double the value of charge and energy.
So if we had 3 volts across a 2 Farad cap,we have 6 coulombs of charge,and 9 joules of energy.

I am not sure why at a low voltage,the charge is a higher value than the joules value,and then they switch places as you raise the voltage,where the joules of energy becomes more than the charge value.

I have never had the need to understand or study this situation-Joules V charge,so i can t offer any help there.


Brad

Offline Dog-One

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Re: Motor consumes no charge ?
« Reply #8 on: May 15, 2016, 04:42:30 PM »
Im not even going to pretend i know how that works.

I have never had the need to understand or study this situation-Joules V charge,so i can t offer any help there.

Thank you for giving it a go Brad.

I thought about this all night--charge being the derivative of energy, mathematically.  It kind of seems like energy is stored in a capacitor under pressure, with charge being the medium being compressed.  I suppose there is a way to associate this with an inductor too, but I haven't made any attempt yet to try it.  I did some plots of energy (and charge) transfer between two capacitors, focusing on the break-even point and the curves have some interesting characteristics--much like transferring air between two scuba tanks.

What seems so strange is the actual term "energy".  It's almost as if this is a completely fabricated term, having a very ambiguous meaning, at least in respect to electricity.  We don't actually transfer energy at all, we transfer some medium (charge, voltage, whatever) and in doing so, we can say it took energy to accomplish.  Then of course if it took some amount of time from start to finish, we can say "power" was involved.

Anyway, to quote ol' Ken Wheeler, it's all a bit of a "mind screw".  Hands on the bench is the only way to come to grips with it.

Offline TinselKoala

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Re: Motor consumes no charge ?
« Reply #9 on: May 16, 2016, 01:05:04 AM »
Charge is conserved, as is energy. What may  _not_  be conserved is the capacitance value of large capacitors, which may vary somewhat according to all kinds of things, like temperature, pressure, voltage, charge-discharge currents, number of C-D cycles, etc.

Think about it.



As far as mentioning Kenny Wheeler goes.... even a broken clock could be right twice a day... unless it's a digital clock.

Free Energy | searching for free energy and discussing free energy

Re: Motor consumes no charge ?
« Reply #9 on: May 16, 2016, 01:05:04 AM »
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Gothic

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Re: Motor consumes no charge ?
« Reply #10 on: May 16, 2016, 04:17:40 AM »
Charge is conserved, as is energy. What may  _not_  be conserved is the capacitance value of large capacitors, which may vary somewhat according to all kinds of things, like temperature, pressure, voltage, charge-discharge currents, number of C-D cycles, etc.

Think about it.

I have...

displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of electric displacement field. Displacement current has the units of electric current density, and it has an associated magnetic field just as actual currents do. However it is not an electric current of moving charges, but a time-varying electric field.

and as for the term "energy"  to quote Tom Beardon "I didn,t axe you what it does i axed you what it was"

Gothic

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Re: Motor consumes no charge ?
« Reply #11 on: May 16, 2016, 04:23:55 AM »
Awesome replies, keep it up

This is a work in progress...


Offline Dog-One

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Re: Motor consumes no charge ?
« Reply #12 on: May 16, 2016, 06:42:00 AM »
Charge is conserved, as is energy. What may  _not_  be conserved is the capacitance value of large capacitors, which may vary somewhat according to all kinds of things, like temperature, pressure, voltage, charge-discharge currents, number of C-D cycles, etc.

Think about it.

I sure looks like charge is conserved from what I can see in my experiments.  Energy on the other hand, looks to be consumed in the process of moving charge.

As far as mentioning Kenny Wheeler goes.... even a broken clock could be right twice a day... unless it's a digital clock.

So true and I'm glad you mentioned it because surely in your many hours of building and testing have come across a few bread crumbs that would really be helpful to some of us.  If you ever care to toss a few my way where I can find them, I wouldn't mind following a trail.  Getting to a point in my life where working and watching the mountains of BS pile-up just isn't all that much fun anymore.



Awesome replies, keep it up

This is a work in progress...

My, isn't that interesting.  Looks almost identical to what I have on my bench at the moment.


We can't leave out the Russians, so take a peak at this vid and think about converting it to an electric device.
https://www.youtube.com/watch?v=11r3E4eia0U

Hmmm, a charge pump and the faster you spin it, the more charge it moves.  Any chance the capacitors would fill up?

Maybe one of them...
https://www.youtube.com/watch?v=6Ljmd-ygDs8


Gothic

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Re: Motor consumes no charge ?
« Reply #13 on: May 16, 2016, 11:16:18 PM »
Funny how the mind works,  several times I looked over the pdf posted and at the " buzzer"
 experiment I always saw 4700 uf capacitors being used, but only c1 and c2 are, c3 is
 470...   O.K.  not as hopeful as I was before,  damn.

p.s. I did the experiment with all three caps 400v 2500 uf  and did not get the "overvoltage"
         on c3 as is claimed in the pdf


Offline TinselKoala

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Re: Motor consumes no charge ?
« Reply #14 on: May 17, 2016, 12:26:12 AM »
Im not even going to pretend i know how that works.
C is I x time,where as Joules is watts per second--> 1 joule is 1 watt  second.

No, those two statements are not the same thing. "Joules is watts per second" is wrong. 1 joule is one wattsecond is right. "Watt-second" actually means Watt times second.

Watt=Joule/second, so Joule = Watt x Second. 

The word "per" generally indicates a division operation, as in "Kilometers per hour".... you go 300 km PER 5 hours, how do you find your speed in Km per hour... you divide 300/5 = 60 km per hour.

A watt is a Joule per second. One joule of energy passing your measurement point PER second of time. This is a measurement of power. The more joules of energy per second, the more power.
 
A Joule is NOT a "watt per second". The only way that statement can make sense is describing a rate of change of power, like if you had a circuit that draws one Watt in the first second, two Watts in the second second, three Watts in the third second, and so on... this would be a change in power of one watt per second. Think of turning up a Variac slowly with a light bulb on the output.

You have "C= I x Time" while correct, is kind of misleading. Yes, I (current in Amperes) is equal to Coulombs per second, or rearranging as you have it Coulombs = Amperes x Seconds .... but you are using "C" here to mean Coulombs, where in other places in this discussion "C" is commonly used to represent Capacitance in Farads.
For example we say Q=CV when we mean Coulombs of charge = Capacitance times Voltage.
But then we also use "Q" as a symbol for the "quality" factor of a coil or resonant system, where it has nothing to do with Q= charge in Coulombs.


Quote

Looks odd when we look at energy stored in cap's
Example
1 volt over 1 Farad= 1 coulomb of charge,but only 500mJ of energy
2 volts over 1 Farad =2 coulombs of charge,and 2 joules of energy
3 volts over 1 Farad =3 coulombs of charge,but now the energy is higher at 4.5 joules.

As we go on,the charge always is the same as the voltage across the cap. If you double the size of the cap,you double the value of charge and energy.
So if we had 3 volts across a 2 Farad cap,we have 6 coulombs of charge,and 9 joules of energy.

I am not sure why at a low voltage,the charge is a higher value than the joules value,and then they switch places as you raise the voltage,where the joules of energy becomes more than the charge value.

I have never had the need to understand or study this situation-Joules V charge,so i can t offer any help there.


Brad

The energy, in Joules, on a capacitor goes linearly with capacitance but goes as the square of the voltage: E=(CV2)/2
So if you double the voltage the energy goes up by 22 or 4 times, but if you double the capacitance the energy goes up by 2.

The charge Q in Coulombs on a capacitor goes linearly with voltage and capacitance: Q=CV, or rearranging algebra, C=Q/V or V=Q/C. So of course for a 1 F capacitor, V=C numerically.

Taking both sets of relationships together you can get the full set of numerical values relating E (Joules), C (Farads), V (Volts), and Q (Coulombs).

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Re: Motor consumes no charge ?
« Reply #14 on: May 17, 2016, 12:26:12 AM »

 

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