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Tesla Switch

Discussion in 'Alternative Energy' started by j taylor, Jul 9, 2008.

  1. chadj2

    chadj2 New Member

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    There really isnt a point to trying to prove it. It sounds like your mind is pretty much made up. I am not trying to sell anything. For me it is just a hobby that I have fun doing. Some people go wake boarding or surfing for fun, I like to design and build circuit boards for fun.

    Chad
     
  2. crashsite

    crashsite Banned

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    Don't give up the good fight

    No one is suggesting that you stop trying to come up with a true "over unity" system. Just that, while continuing to try, maintain your credibility by displaying that you are well grounded first in the the world of conventional science and physics and then, when you do achieve your breaktrhough, you can explain it to the rest of us in the language we all (you and us) can understand and appreciate.

    It wont do any good to solve the free energy problem only to present it in a gibberish that only you comprehend. For example, I could easily say that I have solved that problem by the use of the following equation:

    x=delta y+(t/infinity)

    But, without defining what y and t represent or how you put a real value on, infinity, it means nothing. I think we're all just saying, don't let your solution man nothing by cloaking it in gibberish.
     
  3. Torben

    Torben Well-Known Member

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    So your theory is that energy producers, who make their money selling energy, *don't* want a cheaper source of power so that they could have lower initial costs to produce the power they sell? If that's true then their finance people need to be slapped with a wet fish and fired.

    Do you honestly believe that oil companies are happy about the price of exploration and extraction? Do you honestly think they wouldn't want to save that money if they could? Hint: as you noted, they are greedy. They want to save that money.

    Now, if this global and rather airtight conspiracy of yours actually did exist, I'd expect it to be run by the oil exploration and extraction unions--but certainly *not* by the energy companies themselves. At least that makes some kind of sense.

    Almost everybody would rather just plug into the grid than take the time and effort to generate their own power, much less lay out the capital cost to design and install the system to do it--regardless of how "free" the energy that system produces is. You could have a 110% overunity device which anybody could build and install for $15,000 (I'd call that a steal for such a device capable of powering a home) and I bet you that you couldn't get 1 in 20 homeowners to install it if they could just plug into the grid, assuming that the grid was similarly powered. A power company can produce and distribute the energy to a large number of consumers cheaper and less wastefully than if everybody installed their own home energy production units.

    May we see your test rig, review your test methodology, and attempt to duplicate your results ourselves?


    Torben
     
    Last edited: Aug 19, 2008
  4. dave

    Dave New Member

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  5. chadj2

    chadj2 New Member

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    Torben,

    I wont argue with you as to whether energy suppression actually exists or does not. After all it is only a theory anyway. It has never been proven conclusively. There has only been testimonials from individuals who claim that they have been harassed, and stories of ppl being killed. But, lets just look at the tesla switch. Can you tell me why it does not work? What is it about the system that is flawed in your opinion, or if you have tested it what failed to perform? I have experimented with this type of system also. Maybe we can compare notes. If you really are honestly curious. I enjoy talking about and experimenting with this stuff. I am not one of the people that are going to use some strange terms to try to explain something that I am unsure of. I will just say I dont know why it is happening. But it appears to be happening.

    Chad
     
  6. Torben

    Torben Well-Known Member

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    I think the last chart on this page pretty much sums it up: energy is being converted, there are losses involved, and available charge in the batteries goes down. Dead simple. I have seen no test results which indicate otherwise--although I have seen a lot of web pages which state that there are results which indicate otherwise, they are rarely published.

    In fact, getting *anybody* to publish exactly what their setup is and what their results are is very hard to do. For instance, I notice that you haven't answered the question: may we see your test rig and results, so that we may attempt to replicate your apparent results? I have not built the system myself because it seems rather pointless, but if you will outline your test rig I will see if I can reproduce your results or not.


    Torben
     
  7. chadj2

    chadj2 New Member

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    Like I said before the process wont go on forever. There are always some losses. But think about this; while the process is going on and you are recapturing a large percentage of your energy why cant you use that energy to run a motor which is connected to a generator? How much output could you get from your generator before the process actually ran down in the batteries which are powering the system. Obviously it would be much more than if you just discharged the batteries through the motor to ground.

    Chad
     
  8. Torben

    Torben Well-Known Member

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    Batteries don't discharge to ground; the circuit path returns to the battery. Current out of the battery equals current back into the battery. What is lost is the charge transduced into other forms of energy (motion, light, sound, heat, etc) by the circuit and inside the battery itself. This is basic electronic theory: http://www.allaboutcircuits.com/vol_1/chpt_1/6.html

    In order to do what you propose, you'd need to somehow reclaim the energy which had been transformed into another form of energy by collecting that energy and converting it back into electrical charge to be stored again in the battery--which is not what you're describing about. Regenerative braking systems do this, for instance.

    I think you'd get better runtime out of just wiring the four batteries into the system using automatic failovers so that when one runs down the next one takes over.

    However, if you will please present your setup, methodology, and results, I will attempt to build it and if I'm wrong, I will publicly admit it.


    Torben
     
  9. chadj2

    chadj2 New Member

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    You have go to cut me a little slack. I know that if I connect the positive of a battery to the earth current will not flow unless I had the other terminal connect to earth also. I was just kinda thinking of the ground in a circuit. You know on spice how you always have to put in a ground even though you dont really need one for DC. But anyways what I am getting at is this; are you completely sure that your energy is being converted to heat as it passes through a resistor? Or is the heat a by product of the electrons passing through it. Also, are you sure that just because power is stored in an inductor it is being consumed? Are you certain you cant get that power back? What I have noticed in my experiments is that voltage seems to play a very big role in charging lead acid batteries. That is how the Tesla switch recaptures most of its energy.

    Chad
     
  10. Torben

    Torben Well-Known Member

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    That's a side effect of how spice works, not a side effect of how the real world works. Spice simulators work through mathematical models, not by modelling the actual physics of the materials involved.

    Well, given that the amount of heat produced in a resistor is equal to that predicted by current physics theory for a given resistance, voltage, and current, yes, I'm pretty confident.

    Not "or", "and". Yes, it is. More precisely, it's the charge which is moving more than the electrons (which are also moving but much more slowly than the charge).

    No, and that doesn't make any sense anyway, as it's contradictory. I'm quite certain that the charge which is stored in an inductor is stored in the inductor, not consumed by the inductor. Energy which cannot be reclaimed cannot be said to have been stored.

    No, and I didn't say that. The stored energy in an inductor can be gotten back--otherwise we would not say that the energy had been stored; we'd say it had been consumed. Some energy can be consumed (say, into heat by internal resistance, or into magnetic force) but then we don't say that this energy has been stored by the inductor--although we can store some of the converted energy, for instance, as kinetic energy in a flywheel. This energy can of course be at least partially reclaimed.

    Voltage is important in any battery charger, as is current.

    Again, are you planning on publishing any of these test methodologies, setups, and results of yours which you keep referencing? I really would like to try to replicate some of your results.


    Torben
     
  11. chadj2

    chadj2 New Member

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    Torben,

    One thing I am getting at is that voltage seems to play just as important of a role as amperage in charging a battery. For example a 12 volt battery will charge just as fast whether you use 14 volts with 2 amps or 56 volts with 0.5 amps. Also, you can use a resistor in that setup and the power will still end up in the 12 volt battery. From everything that I have tried concerning lead acid batteries the resistor consumes no power it only limits how much amperage is transferred. The point is that these losses that you talk about may not be as great as you think. I will acknowledge that there are losses in the iron in the core or diode drops but that appears to be it as far as actually consuming power in a motor coil.

    Chad
     
  12. Torben

    Torben Well-Known Member

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    The voltage is certainly related to the current because in charging batteries you typically try to control the current; the voltage must change to compensate given the inherent resistance in the circuit. Ohm's Law.

    "The resistor consumes no power?" Of course it does--the very act of resisting current flow produces heat. A low current through a resistance won't generate enough heat to feel but it's still there.

    I think the losses in a system will be very, very close to the losses predicted by current physics theory, as long as the actual characteristics of the circuit components are known to a reasonably tight tolerance. I'm not sure what you think that I think the losses will be. But assuming we have a 12V supply (rated at, say, 1A) in series with a 500Ω resistor (rated at, say, 1%) then what I think is that the current through the resistor will be 24mA and the power dissipated in the resistor will be 0.29 Watts.

    So are you going to post those test setups and results or what?


    Torben
     
  13. chadj2

    chadj2 New Member

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    Torben,

    Do you have instant messenger ie MSN or Yahoo? I am drilling a heat sink and soldering a circuit I am working on but I would like to keep discussing this. However, I have to keep coming back to check the message board. Would you be willing to continue this conversation on IM?

    Chad
     
  14. Torben

    Torben Well-Known Member

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    Hi Chad,

    I'm afraid I'm on and off the computer at the moment with other things going on so actually the forum method works better for me right now. I'll probably be going offline for a few hours before too long anyway.


    Torben
     
  15. chadj2

    chadj2 New Member

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    Torben,

    Well when you get the chance I would like to chat about this more indepth. But in the example that you gave about the power supply going through the resistor I agree with you completely. But here is the setup that we may be in disagreement about. Assume I had a 36 volt power supply rated at 30 amps feeding into a 100 ohm power resistor and 12 volt battery in series. We can assume that 240 Milliamps of current would flow. This would produce 5.76 watts of heat. The power supply would have to produce 8.64 watts of electrical energy. At the end of 1 second how many joules would be taken from the supply and how many joules would end up in the 12 volt battery? I would submit that about 8.5 joules would end up in the battery ready to be used and 8.64 joules would be taken from the supply.

    Chad
     
  16. ArrowHead

    ArrowHead New Member

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    Bad language deleted, such isn't tolerated here - moderator!
     
    Last edited by a moderator: Aug 19, 2008
  17. mneary

    mneary New Member

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    240 milliamps (0.24 coulombs/second) of current will flow, resulting in 0.24*24 = 5.76 watts (5.76 joules/sec) dissipated in the resistor. The battery will receive 0.24 * 12 = 2.88 watts (2.88 joules/sec).

    The supply is providing 0.24A at 36V, which is 8.64 watts. The battery is receiving 0.24 coulombs/second at a potential of 12V, which is 2.88 watts.

    --------------
    In practice it's worse than that. The battery is receiving the charge at 14V and will return it at 12V. The difference heats the battery and you don't get it back.
     
  18. chadj2

    chadj2 New Member

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    mneary,

    This was the first strange effect that I stumbled upon while experimenting with this whole overunity thing. I have noticed during experimentation that the battery seems to charge up much faster than it should according to the books when I use higher voltage. As I said I stumbled upon this by accident. The way I came upon this was when I was experimenting with a motor I had built according to some internet plans. As this motor ran it drew power from one set of batteries and charged another set of batteries simultaneously. I was trying to measure if actual overunity was occuring. I was hoping to be getting more energy out then I was putting in. That is why I happened to be measuring joules out vs joules in. The motor was originally operating at 12 volts and 1 amp. I was hoping to see better results by increasing the voltage. So when I increased the voltage to about 36 volts the motor sped up but the current draw went down to about 650ma as expected. It took me a while to realize that the battery was still charging at the same rate as before when the motor was drawing 1 amp. Once I had an idea of what was happening I took the motor out completely and starting doing load tests with a set of batteries that are discharging into another set of batteries through a resistor. I didnt just do this once or twice. I went through 10 to 15 charge and discharge cycles of about 45 minutes per cycle (talk about a wasted saturday LOL). But the thing that seemed to bear out was that when I doubled the voltage and halved the amperage the battery still charged at the same rate. I know the power was really there because I did the load test by running the load through an inverter which was running a 100 watt light bulb and measured the voltage and amperage that was coming out of the battery. I know that the physics books say that a battery is only supposed to be a chemical reaction and exchange of electrons and that voltage really isnt supposed to play a part. I cant really explain why this happens but I have been building upon this concept for about 2 years now.

    Chad
     
  19. crashsite

    crashsite Banned

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    ...Earth to Chadj2....

    That's not true. You're confusing the storing of energy and then returning it to the circuit later with the notion of the energy being "free". I think the scientific definition of "free" is that there's no cost to get it. In your example of an electric motor, all the energy you get out of the motor had to be supplied (and even then you never get it all back).

    Nobody is questioning if feeding the energy back or storing it (or even converting it to another form and then reconverting it back to electricity) can increase efficiency but, to think of it as "free energy" (while it might appeal to you), is just not seen as free by those of us who are a bit more pragmatic about the performance of the overall circuits.

    Maybe it's grossly unfair that your definition of "free energy" is not the one commonly accepted but, the fact remains that it isn't. It's also unfair that I wasn't born rich and handsome but, it's a fact. And, while it's a real travesty, it still doesn't mean that I'll be offered the lead in the next Hollywood chick flick simply because the producer has such a fine sense of fair play.

    I'll reiterate. If you want to maintain your credibility, you need to conform to the conventions. There's no law or ordinance tha forces you to do so but, likewise, there's no law or ordinance that prohibits us from branding yoiu a nut-fringe lunatic for ranting about "free energy" and "energy conspiracies".
     
  20. crashsite

    crashsite Banned

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    It sounds like you basically built a "genemotor". They used to be pretty common in vehicles and aircraft prior to the advent of solid state electronics that operated efficiently at low voltage. The genemotor acted like a transformer except that the input was DC (the output could be anything the output generator portion supplied...usually 120V 400Hz for aircraft and often HV DC, to run things like mobile transmitters, in vehicles).

    But, I don't think you'll find any manufacturers literature on genemotors that suggest that they may operate as an over-unity power generator if you feed the output back ibnto the battery. You can present your circuit (with your values...designed and measured) and, I'm confident that we can tell you why it works as it does. From your description, it sounds like a pretty simple setup.
     
  21. chadj2

    chadj2 New Member

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    To me the motor was pretty much irrelevent. It just pointed me in the direction of investigating how batteries charged and which loads actually consume power when placed in series with a charging battery. You can try the experiment yourself. However, plan on spending a large part of a day doing the load testing also to verify you have stable results. I would say it would take at least 8 charge and discharge cycles to be pretty confident the results are stable.

    Chad
     

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