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  • Charge conserving Capacitive Spring.

    Charge conserving Capacitive Spring

    Or

    The "Time Machine"

    Concept:

    http://i210.photobucket.com/albums/b...3/DSCF0071.jpg

    Here can be seen the basic circuit.

    C1 = Capacitor having movable platens.
    C2 = Static capacitance condenser
    R1 = Resistive load (can be replaced with inductive load)


    Basic Embodiment:

    Initial conditions:
    Free Space Permittivity 8.8541E-12
    Area Of Plates 1129420269
    Initial Voltage (both caps) 10
    Charge (C1) 100
    Charge (C2) 100
    Total Charge 200
    C2 Capacitance 10



    1) Capacitor 1 (C1) will be a parallel plate variable condenser with movable platens.

    2) Connected in series to (C1) will be a capacitor of initially equal value (C2) and resistive load (R).

    3) Initially both capacitors will be charged to an equipotential state, so that both have the same voltage across the terminals, and by virtue of having the same capacitance, they will also have an equal charge between them. (See graph for charge distribution)

    4) Entering into the time domain, the platens of capacitor (C1) will beg separated, causing a decrease in capacitance via the equation:



    As the distance between the plates increases, the capacitance of (C1) moves asymptotically towards zero.



    5)Because of the decrease in capacitance (C1), with charge kept constant across the combined capacitor of C1 and C2(law of conservation of charge: Charge conservation - Wikipedia, the free encyclopedia ) The capacitance across combined capacitors (C1 & C2) will decrease and the voltage across combined capacitors (C1 & C2) will increase.

    6) The rise in potential across C1, will cause a potential difference between C1 and C2, meaning they are out of equilibrium. Because of their tendency towards equilibrium, charge will move from C1 to C2 (the sum of both will always equal the total charge initially in the system). Because there is a conduction path through the load, the capacitors C1 and C2 constantly equalize, and as the plate are separated, a new higher voltage is attained across both capacitor simultaneously. This is shown in the graph below.

    Here the values for the capacitors are 10 farads initially, then C1 decreases as discussed. Here we have capacitance of combined capacitor C1 & 2. It starts at 20 farads (parallel reference frame) and as C1 tends towards zero, the total capacitance moves asymptotically towards 10f.



    Here is the total voltage across both capacitors. Starting out at 10 volts initial, as C1 tends towards 0, the voltage asymptotically moves towards 20V.




    The charge distribution across caps (C1) and (C2) changes with respect to time, meaning charge has moved through the load in the equalization process. With initial conditions of C1= 10 Farads and C2= 10 farads, both have an initial charge of 100 coulombs on each condenser. As the capacitance of C1 (blue line) decreases, so does its charge, while the charge across C2 (pink line) increases. The sum of both at any one moment equals 200, or the initial condition of charge once again satisfying the laws of conservation of charge.




    Finally we have the force between the plates of condenser (C1) as a function of distance or time:



    This graph follows the force between the plates of condenser (C1) as they are moved apart and the voltage across both capacitors increases simultaneously. As you can see, while close, there is great force and while they separate, the force decreases. The behavior is that of an exponential decay, much like an inverse square field.


    So what does this equate to?

    By separating the plates of the capacitor, you have raised the energy of the entire system. In doing so you have caused a finite amount of charge to move across the load for a given time, giving you your power. The force exerted will be given by the integral of the equation for force between the plates, so that we can see the total force required to pull apart distance X.

    What is interesting is that while it requires X force to pull the plates apart, they still have a tendency towards moving back together due to the electric forces. -X energy will be exerted by the system returning the plates to their initial position of 10 farads.

    Energy to pull plates apart = X
    Energy to return to initial state = -X

    Total energy = X+ (-x) = 0

    For each investment of mechanical energy in the direction of plate separation, the system returns the energy in the next stroke returning to initial conditions.

    For each stroke (moving the plates out, or moving the plates in) you move (Y) charge in time (T) where power equals (Y)*(T) = Watt

    How quickly the plates can move appart or come together is a function of the TIME CONSTANT of the circuit.

    This can be figured out through calculus solving the following KCL equations.




    These will solve to give an equation similar to the form of




    The heavier the load...the lower the resistance of the load...The faster the time constant.....The more charge moved in less time, equating to more power. Heavier loading = increase in power output.


    If the load is an inductor instead of a resistor (perhaps a transformer)....heavy loading on the secondary equates to decrease in reactance of primary = very small time constants = charge moved very quickly = large power output.

    If the load is an inductor like a motor. Because charge always stays constant, the motors BEMF is not allowed to cancel any charge via its counter generative properties, hence a standard motors BEMF will no longer affect its operation.

    All energy used to cause initial separation of plates, and rise in potential of overall system, will be returned in the subsequent cycle as plates return to initial position and energy state of the entire system decreases. Net input theoretically = Zero

    If all input is returned, and each cycle constitutes a movement of charge, in a definite time dictated by the impedance of the load, then the load dictates the time required to complete a cycle, and how much power is developed. Meaning While loading is constant, Time is the only variable consumed

    The "heaviness" of the load dictates the time constants, the less the time constant, the more power developed


    Put another way, Decrease in time constant = decrease in time domain = increase in possible frequency used. Increase in frequency = increase in power.
    Plates can be moved acoustically, through Super Radiance and Phase transition effects, mechanically, etc.
    Last edited by Armagdn03; 10-12-2010, 06:35 PM.

  • #2
    This is one method of accomplishing what Dr. Lindemann mentions here in the upcoming Renaissance workshop:

    I am going to explain EVERYTHING I know about how and why Back EMF functions in electric motors, how it masks the real efficiency of these machines, and how to overcome it, even in CONVENTIONAL MOTORS.
    Because charge is conserved through the motor, BEMF does not affect the process like it does within a traditional circuit.

    Happy B-Day Sucahyo!

    Sorry i cannot make it to the demonstrations, sounds like fun.

    This is also the same circuit I was referring to when I posted in the parametrics thread after Eric Dollard Spoke:

    Imagine a circuit where action and reaction are separated by TIME.

    The charge in the system is always conserved and stays the same, The work itself oscillates, causing a rise in potential and current simultaneously in the load.

    You raise the energy state of the entire system and cause an internal imbalance. The system then seeks balance, and in so doing accomplishes work, over a set time. After equilibrium is reached, the system is at a higher energy state than in began at.

    The amount of time dictates the power at a given moment. Then the system is allowed to return to its original lower energy state (the reaction separated by time), causing another internal imbalance, which takes TIME to equalize, again causing the same amount of work to be done as the initial impetus.

    It would seem from this circuit that it is possible to create circuits in which time is the dominant factor in output to the load. Dare I say, it is possible to harness the action, reaction, and time is our "source".
    Enjoy!

    Andrew M.
    Last edited by Armagdn03; 10-12-2010, 06:25 PM.

    Comment


    • #3
      Great thread Armagedon
      Makes sense and makes you think. What I don't get is why the separated capacitor plates would pull themselves back again? I guess if HV would be used, then the electrostatic forces could do that, but at low voltage I don't know.
      Thanks,
      Jetijs
      It's better to wear off by working than to rust by doing nothing.

      Comment


      • #4
        very thought provoking ARMA

        Comment


        • #5
          Originally posted by Jetijs View Post
          Great thread Armagedon
          Makes sense and makes you think. What I don't get is why the separated capacitor plates would pull themselves back again? I guess if HV would be used, then the electrostatic forces could do that, but at low voltage I don't know.
          Thanks,
          Jetijs
          The voltage itself causes a mechanical action towards closure of the two plates.

          In a simpler version of the demonstration an only one capacitor is used with moving platens. Charged, the plates are moved appart, and due to the conservation of charge the voltage must rise to compensate for the decrease in charge Q.

          YouTube - MIT Physics Demo -- Adjustable Capacitor with Dielectric

          The energy stored in a condenser is given by U = .5C*V^2. Thus, as the capacitance decreases, the voltage increases, which is the squared unit within the energy equation giving rise to energy. The separation of the plates raises energy state, and the force required move the plates represents the energy expended to create this additional voltage / energy. Thus traditionally you have no energy gain. The resistance to moving the plates due to coulomb electric forces is in response to the gain in energy.

          You put mechanical energy in, and get electric rise in potential energy out.

          Were it not for this force, the energy gained would already be "free".

          As you can see by the graph of the force between the plates, when the area, or the voltage is high enough, force can be rather great.



          Here I have a large area, specifically 1129420269m^2 which I chose to give the capacitor a value of 10 farads. Unrealistically large. But one can easily decrease area and increase voltage I used only 10volts. To those "skilled in the art" the proper proportions will be taken into consideration.

          What is interesting about this setup, is that all charged moved...and all mechanical force used to pull the plates apart...and cause charge re balancing...will be returned when you let the plates go, and the coulomb forces cause them to close once again.

          In reality perhaps you are using several KV, and moving the plates only 1mm or so, Or not really moving the plates at all, but using a property such as optical coherence to achieve a reduction in capacity.
          Last edited by Armagdn03; 10-13-2010, 05:24 PM.

          Comment


          • #6
            Hi, look at Dr. Stifflers last two videos, simplified S- Gate,

            Mike

            Comment


            • #7
              Yes I was looking at that yesterday, thanks. While you may perceive similarities, i do not intend any similarity between this embodiment of this device and Dr. Stifflers work, as it does not use SEC theory. However, other embodiments may have similarities.

              Comment


              • #8
                ARMA,

                i just watched the MIT capacitor video you posted.

                interesting, the current flow seemed somewhat proportional to the "speed" that the plates were being seperated at.

                also, at the end, where it shows a plexiglass sheet being used as a "block"....thinking of those patents & inventors ideas where they attempt to alternately block & then allow magnetic flux by using a spinning disk with various materials glud to them ( im sure you know what i mean )... with this idea in mind could we not replace the plexiglass sheet with a plexiglass roter... with segments cut out that would give su the same effect as lowering and raising the sheet....in order to generate current each time.......the motor powering the rotor would not need to be powerful at all as almost zero torque would be needed ( perhaps a pulse motor since they usually only have a very small torque and can also capture the kickback spike too ).

                thoughts?

                David. D

                Comment


                • #9
                  @Rave154

                  Please note patent 3,013,201 "Self Excited Variable Capacitance Electrostatic Generator"

                  Or 3,094,653 "Electrostatic Generator"
                  Or 3,175,104 "High Voltage Electric Generator"

                  Or note the Evert Motor, which is a similar electrostatic version of the "inverse transformer" I posted a while back.

                  evert rotor tech

                  These would all be viable choices for a method to vary the capacitance. Personally I would like to tend towards a solid state method of variable capacitance because it would reduce build time, effort, and money. Two strong candidates for doing so exist that I know of.

                  In the solid state embodiment, the "force between the plates" analogy falls apart, because there is no real change in distance between the plates. The mechanical concept of manual plate separation is one embodiment of the concept of shuttling charge between two capacitors one of variable capacitance. I included it here because it is easiest to imagine the forces on the plates in this state. How when you separate, the charge will move to redistribute, how the voltage of the combined cap will rise towards a 2x potential, how the time constants dictate the forces between plates at any point in time etc. Conceptually it is easy.

                  Comment


                  • #10
                    Originally posted by rave154 View Post
                    ARMA,

                    i just watched the MIT capacitor video you posted.

                    interesting, the current flow seemed somewhat proportional to the "speed" that the plates were being seperated at.


                    thoughts?

                    David. D
                    Actually, the plates in the beginning of the video are being kept at constant voltage. Thus as the plates are separated, charge must leave the capacitor to satisfy the constant voltage criteria. As the man turns the knob, the distance between the plates dictates the capacitance, and thus the charge on them, not the time in which the separation took place. However, the quicker the man turns the knob, the more coulombs per second are moved which equates to higher power at any given moment.

                    Comment


                    • #11
                      Hi Andrew,

                      great thread, thanks! I would like to share a thought. Say we make our variable cap in a very crude way by putting two caps in parallel and vary the capacitance by connecting them in series (a bit like the scalar charger). If the series caps total voltage is 20V and they get discharged into a cap sitting at 10V half of the energy is lost. Are you suggesting that this doesn't happen if the capacitance is changed gradually (stepless)?

                      Also, how would you vary the capacitance solid state? I know about varicaps but am not sure how you would use them in this application. I suppose the easiest way would be to have a cap with a capacitance that could be varied with a control voltage that wouldn't consume almost anything...

                      regards,
                      Mario

                      Comment


                      • #12
                        Originally posted by Mario View Post
                        Hi Andrew,

                        great thread, thanks! I would like to share a thought. Say we make our variable cap in a very crude way by putting two caps in parallel and vary the capacitance by connecting them in series (a bit like the scalar charger). If the series caps total voltage is 20V and they get discharged into a cap sitting at 10V half of the energy is lost. Are you suggesting that this doesn't happen if the capacitance is changed gradually (stepless)?

                        Also, how would you vary the capacitance solid state? I know about varicaps but am not sure how you would use them in this application. I suppose the easiest way would be to have a cap with a capacitance that could be varied with a control voltage that wouldn't consume almost anything...

                        regards,
                        Mario
                        When the capacitor in my scheme reduces its capacitance, if there were no path for it to discharge into (Capacitor no 2) it would rise its voltage to a very large height. For example. if Q=VC and your initial voltage is 10, and your initial capacitance is 10, you would have a Q of 100. Now you reduce your capacitance to 1. Because your charge is constant, your voltage must now also increase 10 fold to 1000v so that 100Q = 1C * 100V.

                        But you DO actually have a discharge path for it, into C2, which is also 10 farads. So...we must re calculate.

                        C1 has now 100V at 1farad,

                        C2 has 10V at 10farads, for a total of 200Q between the two capacitors.

                        The combined capacitor (C1+C2) has a capacitance of 11Farads, with a Q of 200...

                        solving for V=Q/C we get its steady state voltage to be 18.18v. This shows us that as C1 tends towards 0, the combined capacitor (C1+C2) tends towards 10 farads, and the equation V=Q/C = 200q/10f = 20v, and as you can see the system tends towards 20volts.

                        In one case we have a rise to 100v, and in the other because of the shared C2 (which works as a discharge path) it only rises to 20v. It would be the same situation if we were to let C1 become a 1 farad capacitor, then throw a switch and let it discharge into C2, it would discharge from 100 to 20 volts, for a loss of energy....so yes you are correct...but...

                        Lets consider the energy state of the whole system, Combined capacitor C1+C2 went from 20farads at 10volts, to 11farads at 18.18 volts. The energies are then (according to E = .5CV^2) 1000 joules for the initial state, and 1817.81 for the “reduced capacitance in C1 state”. So you have risen the potential of the entire system (by a factor of 1.817), and…..

                        The Q on C1 was 100Q, after it "discharges" into C2, it now has a voltage of 18.18, with a C of 1farad, giving a Q of 18.18. So 81.82coulombs of charge were moved between the capacitors (100q-18.18q).

                        So you have risen the potential of the entire system AND moved 81.82Coulombs of charge.

                        If you move that in one second, you have moved 81.82coulombs per second, or you have 81.82 watts of power average. If you had higher resistance load and it took two seconds to move the charge you would have an average power of 21.82/2 = 40.91 watts of power average….but over 2 seconds. Same work done, only TIME has changed depending on the impedance of the load…..

                        According to conservation of energy, the work you get out electrically, is equal and opposite to what you put in mechanically. So you put in 817joules of mechanical energy to separate plates, and you get 817 joules of energy in rising the potential of the system, and move your charge per unit time. When you let the plates go, the process reverses, and you get a 817 joule drop, and move another charge per unit time.

                        What you put in you get back out, and work is done.
                        To be continued on next post (this one is getting long)
                        Last edited by Armagdn03; 10-15-2010, 04:46 PM.

                        Comment


                        • #13
                          looking forwards to the continuation

                          Comment


                          • #14
                            Now continuing this and considering force, lets say we start out at our conditions of

                            10 volts across a 20 farad (C1+C2) capacitor with a charge of 200Q. The Separation of the plates is .001meter and our enormous area of 1129420269m^2. The force between the plates is now 500,000newtons.

                            Now we have to pull hard to separate our plate to .002meters.

                            This places our capacitance from 10 farads to 5 farads on C1. This makes our combined capacitance (C1+C2) 15 farads. Its voltage goes from 10V to 13.33.

                            The new force on our plates is now 222,222.22Newtons. N


                            Now our distance goes from .002 to .003, Capacitance goes to 13.333 For (C1+C2), Voltage goes to 15v, and the force on the plates will be 125,000, and this process will continue…

                            …..all the way to say .01m, where it has moved 10x further apart than initially. Here we have our voltage of 18.18, and a force between the plates of 16,528Newtons.

                            We have moved 81.82coulombs of charge, our force has gone from 500,000 to 16,528 or .03% of the initial force.

                            Now the force of the 16,528 although a small fraction of what it was, is still in the direction of closure. If let free the plates would of their own forces move together. From .01 to .009…. At this point, the voltage drops from 18.18 to 18.00v to 20,000newtons of force. Then to .008meters, where the voltage drops to 17.77V and the force grows to 24,691 Newtons.

                            As you can see, if let do its own thing, the device will want to return to its initial voltage of 10V across 20 Farads. As the plates move closer and closer, both physically, and to these values, the force grows quickly,
                            16,528.9256 at .01m
                            20,000.0000
                            24,691.3580
                            31,250.0000
                            40,816.3265
                            55,555.5556
                            80,000.0000
                            125,000.0000
                            222,222.2222
                            500,000.0000 at .001m

                            As you can see, as it moves towards its natural tendency, which initially was working against us, when let go, it works for us, growing the force between the plates, closing them faster and faster, moving more and more charge. In fact every step we went through in their separation is now re-lived as they close and we move ANOTHER 81.82 coulombs of charge. How quickly this charge is moved is given by the RC time constants. The heavier the load, the quicker the time, the quicker the time, the more coulombs per second, which equates to wattage.

                            Comment


                            • #15
                              An interesting story from:

                              http://www.i-am-a-i.org/free-energy/index.html

                              One of my electronics college instructors worked for a high power AM radio station when he was younger. He told us the story of what happened when a million volt capacitor connected to a low voltage variable tuning capacitor. (A capacitor stores electrical charge. The variable tuning capacitor is used in this instance to keep the radio station on frequency. This type of variable capacitor was two sets of parallel mounted metal plates. They were arranged such that one set could move freely between the other set without touching.)
                              The story goes, that the charged million volt capacitor got connected to the normally low voltage tuning capacitor accidentally. There was a loud "thunk". The result was the instantaneous forces generated from Coulomb's Law was so great, that afterwards, the formerly moveable plates were pushed together by such a pressure that "It was impossible to take the tuning capacitor apart."

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