Does The Load Consume The Energy?

~Russ

Does The Load Consume The Energy?
«  »Last edited
well here we are again.


https://www.youtube.com/watch?v=6YTRzTniKc8

shoot away boys and gals.

~Russ

d3x0r

Re: Does The Load Consume The Energy?
« Reply #1,  »
I like your air system analogies... I get it; same with hydraulics....

re; a single wire's magnetic field, this is just the momentum of the air/liquid going through the pipe.

Re: a resonant system... a simple resonant systeem can have a lot of energy stored in it, but if you put a load that does consume it (an LED/incandescent bulb) there are loads that consume ... (maybe)... but if you pull off more power than you put into it you'll drain the reserve stored in the resonance... ... another model : a child on a swing,  You push a little, and they swing a bit, each time you push a little, adding more energy to the reonsonance - and they go higher. and faster... If you have a tray /lefter on a ladder connected to some other load so at a certain height each time they kick the lever, they will lose some of their stored energy, and owon't go higher... each push you add into the system, they go back up and kick the lever, but only that much energy is available.  If the leverl consumes more than you put in, then you will have to push a couple times to get them back to the same height/speed... If the lever is at the very bottom, is a treadmill belt instead maybe, right at the bottom, then each time you push, the energy is immediately transferred and they don't get any height/speed....
This is not to be discouraging; I've seen what seem like OU Effects in the toroidal coils(rodin/abha).  To such a point I was working to try and loop it.  I ended up with an output that was ungrounded (floating) with at least as much potential difference stored in it as was the input; but it was -100V from ground and -112V from ground, and when I grounded it to the same side, it arced and the connecting wire disintegrated immediately. and it was only a few microfarads cap that I was trying to attach, I'm not really sure where so much current came from....

I do have a question though; if you carry your videos forward, how do you represent induction in an air system?  A couple air tanks, with motor and flywheel (each 1V, and 1H inductance and 1A current) they should transfer all potential from one side to the other in some amount of time (takes time to spinup the flywheel, and time for it to spin down) but then with no diode/checkvalve, it will oscillate forever emptying from the other side, spinning up the field, and getting pumped to completion in the other side...  (default circuit on http://www.falstad.com/circuit/ )

I also have a theory that the electrons that induce a magnetic field that in turn induces a magnetic field in another coil are the same electrons coming out as went in, and that they hop wires.  But I've seen no other study that can confirm it; other than that you can get a great deal of power output on an open circuit side of an induction based coupling, so really the electrons must jump from one wire to the other to be re-emitted as photons?  I dunno maybe that's not quite right... maybe it's just that wire itself is also a capacitance and the pico-farads in the wire itself is enough to supply the resonanting current
Anyway as always, look forward to more; I saw your live stream start a few days ago, but didn't get to watch it to the end, and was unable to find it on your live stremas list which the first is 2mo ago for us on the outside... https://www.youtube.com/user/rwg42985/videos?view=2&flow=grid



Cloxxki

Re: Does The Load Consume The Energy?
« Reply #2,  »
@d3x0r
The resonance system of 2 tanks and a flywheel you describe implies a potential that's in the system when two tanks are equal but not present when they're combined in one. Because when the tanks equalize, the flywheel reaches its highest rpm (kinetic energy), right? Russ also touches on it in the video as I type it. It would be interesting if someone could calculate how much energy is stored in the 1000psi tank and how fast the flywheel might spin in a perfect system when it reaches its highest velocity. Is it half? If you think of two identical water tanks, the water level will be halved. You can only let the wheel spin faster if you have a second tank with a bigger bottom surface, or let it flow out of the system at the first tank's bottom.

As Russ said, now how do we extract this potential. I wonder, where do this potential occur? It's in system of large Voltage drops, or very big (tanks) batteries/powerlines? Some people get sick aorund powerlines. Could we spin up wheels alone those lines? I bet we could. The real question is, would it affect the power coming out the other end? My gutt tells me it would register as resistance. And whether it's lost in heat we couldn't recover  or as power we didn't extract before the current reached the end station of the power line, doesn't really matter.

We can extract power from a river most easily when it's rushing down a steep mountain. Or rather, preventing it to so with a bypass through a pipe and turbine. Even by just tossing the turbine tethered off a bridge. Some fish may prefer maximum current, but any time we don't extract this energy  from the water, we are  pretty much wasting it unless we have stopped using dirtier forms of energy generation.
When a river is wide, deep and slow, it gets harder to extract the energy. It takes bigger turbines or other systems to reach any sort of efficiency. Imagine a few waterwheels with scoops the size of the river cross section. That'd work great. Just need to hoist the ship around it. Takes power to hoist, recovers is lowering, so a wash.
If in a build-off for energy extraction you can only use 1kg in materials, will you make a turbine system for a slow big nearly omnipotent river or a tiny wild one?
Russ seems to want to extract or save energy from an electrical system. I really wonder what he's come up with...

Matt Watts

Re: Does The Load Consume The Energy?
« Reply #3,  »
Quote from ~Russ on September 27th, 2017, 03:35 PM
shoot away boys and gals.
"Holy crap !!"

 :rofl:

 :-D

onepower

Re: Does The Load Consume The Energy?
« Reply #4,  »
Russ
You have a good understanding and you are correct that no load has ever consumed any energy in any process. However what James Joule described was the quantity of work which could be performed in proportion to the quantity of energy discharged by the source. So yes it is correct that the only true energy loss in the system always occurs in the source itself when it discharges.

What Joule said was that if we intend to do practical work by discharging a source then the output work which requires a force acting over a distance is generally in proportion to the rate at which the source discharges to provide the force which acts over a distance. Like this...

Source pressure>>>>>>>>>>>> pressure produces a force which acts on something we call a load discharging the source pressure due to a flow.
Source flow       >>>>>>>>>>>>  flow through load discharges the source pressure.

Thus we can see it was always the source pressure which was being discharging by the outward source flow lowering the source energy density and not a load consuming energy which is absurd. The load simply intercepts some of the energy flowing outward from the source however obviously a load is not required to discharge a source.

I will save you some frustration and here's the catch, logically more output work must require either a greater "Force" or a greater "Flow" of energy within the system however nobody said the total pressure or flow must always come from the source...think about that. A good place to start is Tesla's lectures on energy concepts where he describes the need for a transformation of energy. For example, if I discharge one unit of electrical energy into a resistor I can expect to generate one unit of heat. However if I discharge one unit of electrical energy into a heat pump I can expect to move up to 6 units of heat due to a transformation within an open system. Is a heat pump over unity?, well no it simply discharges the source less to produce the same result by using a transformation and an open system. As well we should note many FE devices supposedly experience a drop in temperature not unlike the evaporator section of a heat pump... coincidence?.

In my opinion understanding energy concepts first then applying them as we see fit is the way forward.

sonnet

Re: Does The Load Consume The Energy?
« Reply #5,  »
Hi Russ, Just wanted to say and let others know your bang on with what your trying to say 100%. This video got the point over well. One point you didn't say although to many it would not need stating, the check valve you forgot to add till later would of course be a diode in an electrical view of the diagram. Your second half of your video also reassured me you have the full picture as I understand it, but your going to have to simplify it somewhat to tell others. Current is not what your trying to achieve it is your enemy. Potential is important, you can produce your magnetic field with potential alone. The magnetic field can go on to create more work in a separate circuit, (i.e. your 'use it or not') but don't use the method you drawn to create current in the transfer from 1000V to 500V  i.e. don't try to produce current in the examples you have grasped. If possible you should stop the current returning to original capacitor. If you have the full picture this will make sense. regards.

jbignes5

Re: Does The Load Consume The Energy?
« Reply #6,  »
 Upon reviewing my favorite Physics teacher's lectures, I found out something quite interesting.

 When talking about resonance, it has to be an open circuit. Like parallel wires and a pulse traveling down the one wire. When it reaches the end of that wire it reflects back but not in the opposite sign. A mountain begets a mountain on a scope. This is how you do not destroy or consume the pulse. When it inverts in a closed loop thats when it gets inverted and equals zero. If you leave the end of the parallel wires unshorted it will reflect back to the source with double the voltage.

 And now for a bunch of math but the proof you asked for with examples:
 Professor Walter Lewin
 https://www.youtube.com/watch?v=dYXjlqwdy44&list=WL&index=21

~Russ


gpssonar

Re: Does The Load Consume The Energy?
« Reply #8,  »
Great video's Russ.

~Russ

Re: Does The Load Consume The Energy?
« Reply #9,  »
Quote from gpssonar on September 28th, 2017, 04:49 PM
Great video's Russ.
Thanks Ronnie,

i hope all is well on your end. its nice to see you back.

~Russ

Matt Watts

Re: Does The Load Consume The Energy?
« Reply #10,  »
And 39 seconds of the 44 seconds of runtime is energy we are throwing away...


That's pretty darn dumb of us don't you think?
Re: Does The Load Consume The Energy?
« Reply #11,  »Last edited
So does anyone see the immediate opportunity to build a highly efficient push/pull inverter based from this simple principal?

And what would you do if it happened to end up being 150% efficient?



Just saying...

 :hide:

onepower

Re: Does The Load Consume The Energy?
« Reply #12,  »
I see one little problem with the last video because I have done this test many times. The man in the video is using what amounts to an unloaded motor and the moment he actually loads it up the current flow will increase and the run time will decrease proportionally. The difference between the charge/discharge time will also increase as the inductance of the motor starts to act more like a resistance.

This is why testing at the bench is important.

~Russ

Re: Does The Load Consume The Energy?
« Reply #13,  »Last edited
Quote from onepower on September 28th, 2017, 06:26 PM
I see one little problem with the last video because I have done this test many times. The man in the video is using what amounts to an unloaded motor and the moment he actually loads it up the current flow will increase and the run time will decrease proportionally. The difference between the charge/discharge time will also increase as the inductance of the motor starts to act more like a resistance.

This is why testing at the bench is important.
It dose not change the fact that the energy was not used.

A motor with a fly wheel is loaded. At first  more than later.

Energy was conserved. More or less load. Dose not change the outcome of the energy. It only changed the effenciy. And there for the power genarated (with in the motor) 

At the end he states about the " energy used " that is a wrong statement.
~Russ

PeakPositive

Re: Does The Load Consume The Energy?
« Reply #14,  »
What is not shown in the video is the start voltage of the battery and the end voltage after the motor and cap are used.

He only got 44 seconds of run time from the battery whereas without the cap that motor might of ran for ? say 30 minutes.

The key thing about all this experimenting may the value of ( TIME ). One battery may be rated at say 100 amp hours and 2 batteries in parallel would have 200 amp hours of run time.

We should also note the motor began running slower as the caps approached battery equal voltage.

To me some of the energy in the battery was consumed by charging the cap and spinning the motor.

Easy test keep charging and discharging caps and see if the energy in the battery goes down, ( I'm thinking yes. )

While I think this is very interesting and worth pursuing more, at this point I'm thinking what we are dealing with is just a slowing the discharge of energy compared the using it up fast. ( time factor )

Kind of like using an LED instead of an incandescent light bulb.

Hope I'm wrong so keep up the good work.

I like the direction. :)

Peak Positive

~Russ

Re: Does The Load Consume The Energy?
« Reply #15,  »
Peek. Don't for get. When you say" use up the energy"

It's 100% misleading. The energy was never "consumed".

It was only passing by. Make sure you can wrap your head around that.

:)   

~Russ

PeakPositive

Re: Does The Load Consume The Energy?
« Reply #16,  »
Quote from ~Russ on September 28th, 2017, 08:20 PM
Peek. Don't for get. When you say" use up the energy"

It's 100% misleading. The energy was never "consumed".

It was only passing by. Make sure you can wrap your head around that.

:)   

~Russ
Hi Russ

Yes I understand, I can not comment on what will be only what was.   :)

What I'm saying is if the energy that was is in the caps used to run the motor is gone then it went somewhere, right ?

To me the energy in the caps was used up or consumed. Right ?

No getting around the fact there was voltage in the cap then after running motor there was much less or none if we let it keep going. Right ?

So was some of the energy in the battery transferred to the cap and motor ? I think yes.

Now where did that energy go after the cap was drained?

I guess we could not say "used up" but it is gone somewhere, Right ?

At this point at least in that video the energy in the cap was used up, me thinks.   :)

Keep up the good work, loving this kind of experimenting and thinking.

Next step getting the energy used to turn the motor back into the battery.   :)

Peak Positive

onepower

Re: Does The Load Consume The Energy?
« Reply #17,  »
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.

Matt Watts

Three Step Process
« Reply #18,  »Last edited
Okay guys, I see a framework here we can incorporate into our devices...

Three stages:

1.  Apply power from the source, through the load and into a zeroed capacitor.

2.  Once the capacitor is charged to a point where there is not enough potential between it and the source, we now empty this capacitor through the load and into another zeroed capacitor.

3.  This final stage I really want you all to think about.  Here, the two capacitors have nearly equalized and no longer have enough potential between them to power our load.  So we take the second capacitor and reverse its polarity.  We run it this way until both capacitors have nearly zeroed themselves out.  Once zeroed, we restart this three step process.


The question I want you guys to answer for yourselves is why this last step goes completely against the first two steps.  In the first two steps we conserve energy while providing useful work.  In the last step we still provide useful work, but we obliterate the energy.  Why?  Where does the energy actually go?  How?


With this three step process, assuming minimal losses, the opportunity for a COP near three is possible based on the understanding Russ is attempting to express.  We take a single chunk of energy and make it work for us three times before we lose it.  Is this not a viable methodology to bring to the bench and verify?

Ris

Re: Does The Load Consume The Energy?
« Reply #19,  »
did you not read the post above lol
already in the first step you have flaw which you can not bypass.

 

~Russ

Re: Does The Load Consume The Energy?
« Reply #20,  »
Quote from Ris on September 29th, 2017, 10:41 AM
did you not read the post above lol
already in the first step you have flaw which you can not bypass.
one must look at capacity.
voltage is a big factor.

what i'm trying to say is that if you can transfer the energy from one cap to the other... Then the energy is all still there.

put a resistor in between the caps...  what do you get... 1/2 the starting potential. (mined you the energy was not consumed nor is it lost)

now do the same thing... but instead use a matched indicator...  ask your self. did you lose that starting potential? or did you just move it to a new place??? Then ask your self... can you do  " work" ( power)  with that inductor and still get it to transfer that energy???

this is what i'm saying.

1 glass of water is still 1 glass of water. even if you move it to a new cup...

~Russ

onepower

Re: Does The Load Consume The Energy?
« Reply #21,  »Last edited
The two capacitor experiment has great value in my opinion for the following reasons.

1) It is the perfect test platform because no matter how many components we place in the circuit the final energy calculation can be determined using a simple equation... the Energy in each capacitor equals 1/2 C V squared. So we calculate the source capacitors initial energy then do our test and when all the smoke clears we calculate the energy remaining on C1 and C2. If the external work performed or output of our test is less than the energy remaining in C1 and C2 then we lose and it's time to rethink our strategy. If the external work performed or output of our test is more than the energy remaining in C1 and C2 then we win.

2) Two caps of equal value are more accurate that $30,000 worth of power supplies, DSO's and power analyzer's because our caps do not lie or mislead and all we need is one equation... E=1/2 CV^2. This is why I always use my two 5 F super cap packs rated at a max voltage of 24v each for testing. I never use batteries for testing circuits because the batteries energy loss/gain is misleading and calculating the actual loss or gain per cycle is damn near impossible.

The fact remains that we can speculate all we want however if at the end of the day there is no tangible energy gain we can prove to ourselves in reality then there is no tangible energy gain, it is that simple. Personally I have done literally thousands of tests and I wanted to believe however believing is not proving. In retrospect I only started to succeed once I confronted my own demons and told myself this is bs and if the energy gain is not there then it is simply not there and I need to rethink my strategy. Reality is the ultimate indicator your doing something right not speculation or beliefs.

Do the test, prove it to yourself then expand on what works and discard what doesn't.




~Russ

Re: Does The Load Consume The Energy?
« Reply #22,  »
sets go back to the basics.

If we use capacitors. or if we use battery's, everything changes.

Why is that?

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.

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...

now lets think about why.

well if Current flow is needed to do "work" (remember we are moving energy to make " power"... we are using the difference of potential energy )
and if we already stated that we cant create or destroy energy (only move it) then we are saying that the stored charge in between the caps is why we "lose" 1/2 our "energy" 

Its not lost its just not accessible...

How do we fix that? one way is to add a "load" over there too...  in fact you would do " work" in between those caps just by moving the the energy back and forth. and in that idea you would Never "lose" it... ( except ESR of the caps)

just something to think about.

ok so back to battery's.

when we "charge" (split the positive)  a parallel bank of battery with a series ones... in theory we should not lose any of our energy. if we match our AH's

This is what rick is doing. ( But he dose not mach AH's)

and if we use the fly back energy to charge the other battery's... well at some point we are reaching maximum efficiency. 

add some spark gaps in there some where and oh man...

any way. your idea Matt with caps most likely wont work, but with battery's theirs something to think about...

~Russ
Re: Does The Load Consume The Energy?
« Reply #23,  »
oh just to add. why dose a battery do this but not a cap? well a battery is chemical... and that makes all the difference...

~Russ
Re: Does The Load Consume The Energy?
« Reply #24,  »
the only way around the capacitor being used as i wish it to be is to make an adjustable capacitor... 

you would be able to make this work... because you have no other plates to deal with except the 2 you have... ( that change size)

~Russ