Does The Load Consume The Energy?

sonnet

Re: Does The Load Consume The Energy?
« Reply #25,  »
Quote from onepower on September 29th, 2017, 12:21 AM
I don't think I expressed my last point very well so I will try again.

Battery >>> unloaded motor>>>Cap   
Here the motor acts like an inductance, the current is low and the cap charge time is slow as in the video.

Battery>>> loaded motor>>> Cap
Here the motor acts like less like an inductance, the current is high and cap charge time is fast.

Thus when the motor is loaded we can still charge a capacitor however we would have to cycle the cap charge/discharge more often creating more losses. When you actually do the test you may understand that the cap with a loaded motor does not charge over 44 seconds as in the video it charges in about 1 second. So you would have to cycle the cap charge/discharge 44 times more often resulting in 44 times more losses.

Don't get me wrong in the video it looks awesome however that is not the way it works in reality with a loaded motor. Here's a neat experiment, if you want to see how a massive inductor acts replace the inductor between two caps with an unloaded DC motor. Initially the motor spins up because it acts like a motor then when the voltage in the caps is almost equal the motor starts acting like a generator in series with the first cap charging the second cap. In effect a common DC motor can act just like a massive inductor acting in slow motion at a very low cycle frequency with minimal losses. How do we know when the motor stops acting like a motor and starts acting like a generator?... you will see the voltage polarity reverse across the motor/generator just like an inductor.

Another experiment is the same as above only we add a boost converter (joule thief) after cap 1 and the motor. In this setup the boost converter will completely drain cap 1 and keep charging cap 2 until the motor comes to a stand still. The boost converter basically scavenges all the energy present in cap 1 and the motor down to the mV level in some cases.

There are many different ways to do things however I think it's important to understand the reality of what is actually happening.
one power your right, Imho I did state earlier that this circuit should be low current and a motor under a load will lead to failure because it pulls current which I referred to as our enemy. A inductor, a simple coil will deliver the magnetic field we want to extract from this circuit. the real energy usage is to be found in another circuit i.e from the magnetic field.. Russ touched on it and a pulse motor could work very well, it gives us a spinning rotor which we can affix a interrupter/commutator or a homapolar generator or both. What we want to do is drive the voltage higher to increase the voltage potential so another coil with a interrupter (could be a shared one from the pulsemotor magnets) attached to a condenser/capacitor will give us the ability to store a even higher voltage. A bit like how a Buzzbox coil works. Homapolar generator  I hear you say, that's high current......yes yes but attached to pulsemotor rotor and the same setup (buzzbox)  but independently wired having its own coil and capacitor we can turn that current into a usable voltage. Homapolar generators with the magnets on the inductor rotor have no drag which a pulse motor should be able keep spinning at a good rpm.
A lot for a mind to picture, but for me a project in the making which I intend to share. When I saw Russ's video he was repeating what I knew and what I believe is the basis for my system to work. Regards

onepower

Re: Does The Load Consume The Energy?
« Reply #26,  »Last edited
Russ
Quote
well, when we connect 2 x 1F caps in series we get .5F worth of  capacity (storage space for our energy ( potential difference))
If we connect them in parallel... well we 2F
so just by connecting caps in series we lose a lot of our capacity. 1/2 a madder a fact.
but if we do the same with battery's... guess what... there always ADDING.
Are you sure about that?... think about it for a moment.
If the capacity of each of the two batteries in series remained the same and the series voltage doubled then we could double the Energy simply by adding them in series however we know this is not true. In fact two batteries in series are subject to the exact same phenomena as our capacitors in series it's just harder to see because a battery acts like a voltage source.

Here is the answer why, when we connect two sources in series the inner (-) and (+) potentials neutralize one another to become a neutral condition and this is why a battery or a capacitor connected in series always lose capacity. Think of it this way, a source is a separation of (-) and (+) potentials and if for any reason the potentials merge then they neutralize one another to become a zero potential condition. So when two sources are connected in series the outer (-) and (+) terminals gain potential at the expense of the inner (-) and (+) terminals losing potential to become a neutral condition.

This is natural law, we produce an energy source by separating a neutral condition into two distinct (-) and (+) potentials and whenever the two (-) and (+) potentials merge for any reason they become a neutral condition and give up there energy.


~Russ

Re: Does The Load Consume The Energy?
« Reply #27,  »
one power, i think you missed this part:

"There for if we take battery's in parallel we all we always just add up the capacity ( in AH)  series only the voltage changes... "

what I'm saying is that the rules dont apply the same between caps and battery. this is one good thing about having the battery...

the resistance changes... thats something to think about.

~Russ

Matt Watts

Come on Man...
« Reply #28,  »
Quote from ~Russ on September 29th, 2017, 02:42 PM
any way. your idea Matt with caps most likely wont work, ...
Such a defeatist attitude gang.  Okay, I'll agree a COP of 3 is off the table, but there's still hope here.


Do you see anything in this picture that says the load eats the charge?

I don't.  All I see is a resistance modifies the time constant.  The load DOES NOT consume the energy.  A cap to cap transfer will equalize eventually where the voltage is now one half the starting voltage.  We lost one half the energy we started with.  Big whoop, let's get back the other half by flipping the polarity of the second capacitor and push that through the load too.  Once complete, then yes, all the energy is gone.

Now think about the three step process again.  We pushed electrical energy through the load three times before we lost it.  Not once, not twice, three times, starting with the same chunk of energy.

If we connected the power source directly to load and pulsed it, with minimal losses we would have near unity operation--simple amperage times voltage; the load gives us heat, light, rotation, whatever at some level of efficiency.

Lets break the 3-step process down in more detail.  The first step pushes energy through the load into a capacitor.  We don't get a nice square wave pulse, but instead a curve like the image above as the capacitor charges.  In step two, we get another such curve, half the amplitude in power, but the same shape.  Step three is identical to step two.

The feeling I get from you guys is the total power pushed through the load in the 3-step process is the same or less than you would get connecting the load directly to the source.  You may be right.  Until someone runs the numbers or has the experimental evidence to lock this down, we won't know for sure.  From my perspective the 3-step process seems as though it would have to be more efficient overall.  We only need to be a tad beyond COP > 1.  That implies self-running right there and the concept doesn't have any restrictions that would prevent it from being scaled up.


Russ initiated another Pulse Motor Build-Off this fall.  How about someone take my 3-step process and prove beyond a reasonable doubt this concept won't in any way shape or form lead us to self-running operation.  I'll be doing my best to prove just the opposite.

See ya on the bench.

~Russ

Re: Does The Load Consume The Energy?
« Reply #29,  »
Good stuff Matt. Looks like you got your pulse motor build off self goals set.

By the way I did this on the bench today.

At a glance. It seemed as if the fun time was longer.

Way better power calculations like Jeremy already did is needed.

I was using 470uf caps and a 100 ohm resistor.

And if you used an inductor you could get 20x3 times more out of that.

You got your self's pulse motor. Thst will run x time longer then straight from the battery. 

Or do you. ;)   

Hehe. Looking forward to your findings.

I beat my head on the desk for a long time about this.

~Russ

Matt Watts

Re: Does The Load Consume The Energy?
« Reply #30,  »Last edited
If you drive an inductor (part of a motor) as the load, the back EMF should partially reverse the discharge of the capacitor(s), provided you add something (diode, high value resistor, tiny cap or neon bulb) across the switch.   So even if my 3-step process turns out to be near unity with a resistive load, chances are good it will work better with an inductive load.
  • Charge
  • Transfer
  • Dump
Could it get any easier...?

Ris

Re: Does The Load Consume The Energy?
« Reply #31,  »
The feeling I get from you guys is the total power pushed through the load in the 3-step process is the same or less than you would get connecting the load directly to the source.  You may be right.
Matt one capacitor completely full can give 90% of its stored energy through 90% efficient load ,now that you take two capacitors one full and another empty you can intercept 50% of the stored energy but you can only get 40% of the useful work
hence the remaining 50% of the stored energy
you again can draw 40%,which is totall only 80%
so Each additional step at best conditions brings about 10% of the losses
 
now you see adding any load between is a bad idea









~Russ

Re: Does The Load Consume The Energy?
« Reply #32,  »
Have a look at what Jeremy did so you can get some. Idea what to do Different.

http://open-source-energy.org/?topic=2679.msg45560#msg45560

Personialy I think the resistance playes a big role.

You might have to match the cap discharge caristics.

~Russ

Cycle

Re: Does The Load Consume The Energy?
« Reply #33,  »Last edited by Cycle
Quote from ~Russ on September 29th, 2017, 05:49 PM
one power, i think you missed this part:

"There for if we take battery's in parallel we all we always just add up the capacity ( in AH)  series only the voltage changes... "

what I'm saying is that the rules dont apply the same between caps and battery. this is one good thing about having the battery...

the resistance changes... thats something to think about.

~Russ
Hi, Russ. Great thread, really got me thinking.

So for two capacitors in parallel, the total capacitance is the sum of each capacitor's capacitance, it's the sum total of the plate area of all capacitors in parallel.
Ctotal = C1 + C2

For two capacitors in series, the total capacitance is equivalent to a single capacitor with a plate spacing equal to the sum of the plate spacing of all capacitors in series.
Ctotal = 1 / ((1/C1)+(1/C2))

Now, assuming you put the capacitors in parallel and charged them up, then switched the wiring so they were now in series, does this imply that the series capacitors will deliver less work into a load than the same capacitors in parallel?

{EDIT}
Ah, nope... by placing the caps in series, you're doubling the voltage from what it would be if they were in parallel, so the same amount of energy is still there.
{/EDIT}

Makes one wonder what would happen if one could move the plate spacing. By pulling the plates apart (with a fully charged cap), wouldn't voltage in the cap rise? Then one could merely adjust plate spacing of the two caps to keep current flowing.

~Russ

Re: Does The Load Consume The Energy?
« Reply #34,  »
Cycle.

See here

http://open-source-energy.org/?topic=3074.0

I did this in a simulator.  It's slightly different than you are looking for. But I would get more total voltage in the system by switching from parallel to series.

The idea was that by going from series to parallel you could pull a Negtive voltage on the series cap. And you can. Think about it. The big cap in parall through an inductor tries to pull the smaller series cap negtive.

In the end I could get the total voltage in all caps in series to be higher than the total voltage I started with.

However on the bench this did not work. I never could get that total voltage out. And that make sence.

Why?  Well like I said. Current can't flow inbtween the series caps. That charge is stuck there. (Trapped by Deialetric )  And that's a good thing. We just need to use that to our advantage.

Again. Battery's don't have this problem.

~Russ





onepower

Re: Does The Load Consume The Energy?
« Reply #35,  »Last edited
Part of the issue revolves around appearances and perceptions as to what we think is happening. For instance when I charge a capacitor I force X number of electrons onto the (-) plate which repels X number of electrons from the (+) plate due to coulomb forces. We know this because we can measure the potential on each plate as well as the potential difference across the plates. Hence the term "displacement current" as the electrons forced onto the (-) plate repel or displace the electrons from the (+) plate which is why it now has a (+) potential. Extra electrons is a (-) potential and less electrons is a (+) potential and when more (-) combine with less (+) they become zero or neutral. Not unlike when a high pressure combines with a low pressure to produce no pressure and the condition of pressure has simply ceased to exist. Obviously "something" has moved, a "current" otherwise the pressures could not have neutralized each others condition. Otherwise we are left with the odd notion that nothing has moved but something has changed or changed without changing which is a contradiction in terms.

Thus the "displacement current" qualifies as a real current because the only requirement for a current is that some charges like electrons move in some way and they have in fact moved. From this we can deduce that when we connect two caps in series the inner (-) plate potential and inner (+) plate potential combine to produce zero or no potential. In reality we can prove this by measuring the potential on the wire which connects the two series capacitors and the potential is zero or no potential. However the outer (-) plate and the outer (+) plate have increased in potential... how did this happen?.

It's simple, when we connected the two caps in series and the inner (-) and (+) potential moved as a current to neutralize each other and the potential ceased to exist. Thus the effect is the same as taking two plates and moving them further apart which increases the potential difference. In effect we have simply removed the inner potential between the outer plates leaving only the potential on the outer plates. Which is no different that moving the two plates of one capacitor further apart because in both cases the potential on the outer plates ... is now further apart. The result is the same because the effect on the potential on the outer plates is the same...the plates are further apart. Is that so difficult to understand?, it seems very simple and quite reasonable to me.

Measure the potential on the wire between the two series capacitors which is now zero, there is your proof. However I should note the potential on the outer (-) plate, the inner wire between the caps and the outer (+) plate are to be measured with an electrometer or a DSO with the probe ground connected to a real Earth ground.

~Russ

Re: Does The Load Consume The Energy?
« Reply #36,  »
I agree one power.  Real current is flowing. But it's traped. So if we put in Inductor or resistance inbtweent the caps that forms the netrual zone. We can do "work" but we can't lose any energy. We just need to change the potential on the other side of those caps.  Only move it around. This would be usfull to prove some points.

~Russ
Re: Does The Load Consume The Energy?
« Reply #37,  »
From my understanding. What ever charge you put on one plate. The other "magicaly" tries to match it.   So this is why it appears that current flows. But not through. Rather from the other side.

~Russ

Matt Watts

Re: Does The Load Consume The Energy?
« Reply #39,  »
What I need is those voltage and current formulas multiplied together for power, then integrated so I can determine if my three step process is flawed mathematically or not.  If the math shows what I'm hoping it will, then the bench will tell the story for sure.


What it's looking like to me fits pretty close to Tom Bearden's "Don't kill the dipole."  As long as you don't destroy the dipole, the energy stays alive and can be reused as much as want.  My third step purposely kills the dipole to neutralize it so the cycle can restart.  Have this funny feeling it's not necessary to do that, but I can't see any good way at the moment to avoid it.

~Russ


Mystweaver

Re: Does The Load Consume The Energy?
« Reply #41,  »Last edited
Hi guys, noob here trying to get my head into all this amazing testing you guys are doing.
Watching this video of Russ's I can't help but think back on this one

https://www.youtube.com/watch?v=V7p49pDrxeI&t=652s

Using these 3 caps, it looks to me like the pulse motor is using the collapsing magnetic field to increase the charge.
From my little understanding the back emf charge is negative and the way these caps are setup is asymmetrical(??).

Therefore, if there was a way to swap these three caps about from run to charge and back again whilst the motor is in motion, would there not be a point where there is a constant negative pressure/flow cycle that keeps building and could then be extracted when a greater load is put on the motor?

Please bare in mind that I have very little experience... I'm just saying what I'm seeing and it seems to me to hold true to Russ's vid :)

I hope what is going on here is relevant by adding a 3rd cap and connecting it backwards? (At least I think that's what I am seeing).

itzon

Re: Does The Load Consume The Energy?
« Reply #42,  »
Russ,  I have been thinking the same thing for several years about the load not "consuming" the energy and I am wondering if you or anyone else reading this has done any input versus output current measurements across a heating element?  Everyone has heard that any "missing" energy in a system was converted to heat.  If that is true then wouldn't there be less current flow on the output than on the input due to conversion to heat? 
Just some food for thought.

~Russ

Re: Does The Load Consume The Energy?
« Reply #43,  »
Good stuff guys.

Yes itzon

I have done some basic ones. But I need to do some. More extensive ones for electricity. More in due time for the Vids.

But let's look at this.

Take a glass of water. Now poor it in to another. Did you lose any water? 

Seems to me you only moved it.

There for if the hight of the glass was your voltage And the the amount of water was your energy. Then your only limited to the hight of the water and the amount of water.

If the hight was 10 foot then you could only power 1 turbine. But if it was 50 foot you would power Manny turbines. Same amount of water (energy) needed. Just a diffrent hight (voltage)

In the end you'll end up. With 1 full gass of water at the bottom??? Well that is if your system is closed...

;)   

So proving it with electricity should be the same.

I have 2 massage cap bainks. I may try a heating element in that...  Not sure of the ESR though. This will play a roll in the calculation.

~Russ

Matt Watts

Re: Does The Load Consume The Energy?
« Reply #44,  »Last edited
We should also take another look at this now that we have a better understanding of what a load is and is not.



If our goal is to produce a magnetic field we can harness to perform useful work, while at the same time not consuming energy, resonance clearly comes back into the equation.  In this case to give us a two-fold increase in magnetic field strength for the same input current.

~Russ

Re: Does The Load Consume The Energy?
« Reply #45,  »Last edited
Matt. Can you link us back to where that schematic came from?

And I agree resonance makes alot more sence now that we got our head more around energy. And power.  (Load)

And to note. What the hairpin circuit dose. I would really like to set up a capacitor isolated ossilator.

Then connect a ground on the isolated side. The ground alows us to get more potential in the system. But it's still ossilating on the input side. What ever that input is should not be effected by the output as it's current is completely disconnected.

Put. Our "load" on the ground side and we might just have somthing to talk about. Let it build up then stop it. For a sec. The repeat. Current flow in and back out. As it ramps up and back down. The question becomes dose it bring in and back out more than it takes to ossilate?

Just an idea I was thinking about.

~Russ

Matt Watts

Re: Does The Load Consume The Energy?
« Reply #46,  »
Quote from ~Russ on October 1st, 2017, 09:38 PM
Matt. Can you link us back to where that schematic came from?
For a guy critical of me missing important details...

Look at my post above, search for the word "this" wrapped with a hyperlink.  :)

~Russ

Re: Does The Load Consume The Energy?
« Reply #47,  »Last edited
Quote from Matt Watts on October 2nd, 2017, 06:29 AM
For a guy critical of me missing important details...

Look at my post above, search for the word "this" wrapped with a hyperlink.  :)
Must be the 5 kids. Lol. Ok!!!

Don't forget. If I'm responding on a weekend. I'm. Most likely trying to see this with my phone and a playground (my house) full of kids jumping on me lol

~Russ

Matt Watts

Re: Does The Load Consume The Energy?
« Reply #48,  »
I'll let it slide this time.   :-P

jbignes5

Re: Does The Load Consume The Energy?
« Reply #49,  »Last edited
 I was going to quote Russ on his video but something went wrong and it grabbed a quote from me.. Go figure..

 I totally agree with you here Russ. What consumes the energy is the returning wave that destructively reacts with the next pulse coming in the opposite direction. Using your setup does not truly use negative energy. It uses the potential difference between the positives. This only charges the target battery or even cap minus intrinsic resistance of the wire in the circuit plus the capacitance and inductive effects. Weather your wires are statically shielded including all test equipment leads determines the electric looping and subsequent hysteresis from the loops. All exposed wires should be statically shielded (not ground shielded) or you will get all kind of cross talk between the unshielded wires.

 I'll link the references to the Relevant Tesla patent:

 static shielding:
 https://teslauniverse.com/nikola-tesla/patents/us-patent-514167-electrical-conductor

 This should allow you to get near zero loss and a gain in the power of the cap dump to the load coil. Plus all the loops from the expansive nature of the electric field would be taken out of the picture.

 This is the lossy line reference:

 https://www.youtube.com/watch?v=cELPT2JfSDU

 Some earlier teachings of the oscillator. Check out the first circuit diagram. You are correct about tying the ground to the plane between the battery +'s.

 https://www.youtube.com/watch?v=W31CCN_ZF34


 Ok so this is how I see your idea here. Even electric motors can be used as an air pump like device that use the potential difference between the positives to create a slope to build up inertia. The motor is used in such a way to impede the flow slightly like free wheeling. Some motors that we have are near 98% efficient. Only two percent would not be flowing into the charge bank of batteries except for real losses. But one thing stops me here. when you try to pick off any flow using traditional generators they have a reactant like effect slowing the current even further.
 This is why I suggested to use a properly oriented bifilar coil to flow through as a primary then add a solenoid as the secondary. Why? Because the resultant reflection is now negated by the coil and it's dual path. Cemf also should lower and very little effect would happen if you put a real load on the transformers secondary. This is due to the separation of the paths in the coil itself and the double pulse countering the previous flipped pulse returning. A net positive is a result. It cost one pulse then after that it should just ring and need only very little additional input.
 Tesla used the bifilar in his greatest potential amplifying machines. It negates the cemf so the impulse passes unharmed or deformed. And what happens when you move a magnet faster by a coil? Riddle me that one BatMan... Just a joke..

 I am also reminded how you just used the abha coil in pure capacitive mode. That might be interesting to see as well. Maybe a solenoid over wrap could be used as a pickoff point since there is a very high complex pathway it utilizes. Maybe just wrap in on the toroid itself? Thicker copper wire, solid core as a secondary? How about a thick copper ring that has been cut once and hooked to with thick leads that are statically shielded from inside the coil? I think you need to match the angle but 90 degrees out of turn with the outer finer wraps. Just some ideas.

 I think we also need to pick certain lengths to these primaries and secondaries. Matching those is a must. So that means we must design generators with specific ratios for the standing waves to be present on the end of the coil. which should dictate the output max wattage as one factor. The others would be traditional mass conversion rates used is transformers.
 We might have to figure out the new max's of the bifilar primaries to secondary ratio though.

 Here is a reference for an old video about capacitors or condenser.

 https://www.youtube.com/watch?v=2bBifpRa890