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Re-Inventing The Wheel-Part1-Clemente_Figuera

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  • bistander
    replied
    Originally posted by jettis View Post
    Hey bi, Good talking with you,

    To be clearer, sorry, I am talking about increasing magnetic flux, for no increase in amperage and or voltage cost.

    As an example if we increased the coil core and rotor length to 48” (effectively increasing the length and increasing the cross sectional size of our conductor) to match a new 48” rotor with magnets to match. Note we still will keep the coil at 5 ohm’s, it will be of much bigger dimensions than our last comparison, but we are more than ok with that.

    1) We made our rotor coils longer and the amperage draw will be the same at 5 ohms.
    2) The magnetic flux of our 48” coil will be increased over the 24” coil.
    3) Rotor rotational speed should be the same (not taking into account any friction, windage losses etc.) as the 24” version.
    4) Using the 48” version rotor will have an available increase torque production over the 24” version.
    5) Flyback energy will increase as well over the 24” coil.

    That is were I am going, you can increase your mechanical shaft output for no more current used, coupled with more flyback energy that can be recovered.

    As a side note a pulse motor can and often does generate the magnetics (which is additional to the flyback energy) can send a certain amount of energy back to the supply (power source) and also the secondary battery if you have one.

    Are we in agreement?


    Dave Wing
    No agreement by me.
    Sorry, but I never did see value in what I think you're talking about. You'll have to carry on without me. Good luck.
    bi

    BTW, for a given magnetic circuit and given excitation power, one can increase the excitation (mmf) by increasing the mass of copper in the coil.

    ​​​​​
    ​​​​​​

    Leave a comment:


  • jettis
    replied
    Hey bi, Good talking with you,

    To be clearer, sorry, I am talking about increasing magnetic flux, for no increase in amperage and or voltage cost.

    As an example if we increased the coil core and rotor length to 48” (effectively increasing the length and increasing the cross sectional size of our conductor) to match a new 48” rotor with magnets to match. Note we still will keep the coil at 5 ohm’s, it will be of much bigger dimensions than our last comparison, but we are more than ok with that.

    1) We made our rotor coils longer and the amperage draw will be the same at 5 ohms.
    2) The magnetic flux of our 48” coil will be increased over the 24” coil.
    3) Rotor rotational speed should be the same (not taking into account any friction, windage losses etc.) as the 24” version.
    4) Using the 48” version rotor will have an available increase torque production over the 24” version.
    5) Flyback energy will increase as well over the 24” coil.

    That is were I am going, you can increase your mechanical shaft output for no more current used, coupled with more flyback energy that can be recovered.

    As a side note a pulse motor will generate from the passing magnets on the rotor (which is additional and additive to the flyback energy) and the machine can send a certain amount of energy back to the supply (power source) and also the secondary battery if you have one and desire to do so.

    Are we in agreement?


    Dave Wing
    Last edited by jettis; 12-14-2023, 03:18 PM.

    Leave a comment:


  • bistander
    replied
    Originally posted by jettis View Post
    Hey bi, I do not know much at all about Holcomb, however what do you think of this statement? Is there any issues as far as your concerned?

    If you take two iron core coils one made from 40 awg and the other made from 10 awg, each has a 3/4” x 24” laminated steel core acting on a 6” diameter x 24” length rotor with two north magnets measuring 2” wide, one inch thick and 24” long, equally spaced around the rotor.

    When we compare the two coils in pulse motor operation, the coil with 40 awg wire at 5 ohms and then another 10awg wire coil at 5 ohms. Which one has more magnetism to convert into rotational energy and which one will yield more flyback energy? Take note the 10 awg coil could be 30 times the size of the 40 awg coil.

    One could easily see a coil 30 times bigger will produce a lot more rotor torque and flyback energy than the smaller coil. It should also be again noted that each coil (10 awg vs 40 awg) will have the same current draw, from the primary battery, as ohm’s law states.

    Dave Wing
    Hi jettis,
    trick question?
    Couple of issues here.

    Edit. Fat thumb hit wrong button. Dinner ready also. I'll finish later.
    bi

    Edit 2
    ​​​​​ Hi jettis,
    I'll start my reply over. I do see a few issues with what you put out.
    You say "more magnetism to convert into rotational energy".
    I'd say "magnetism" doesn't convert into rotational energy. So in answer to your question of "which one?", both are the same, namely zero.
    Also, " magnetism" is vague. You make it sound as if it is a quantity. What are magnetism's units?

    I did some rough quick calculations. Two coils at 5 ohms each. Inside dimensions to fit on .75 24" core would give 1250 turns for #10 AWG, or 1 turn of #40 AWG Cu wire at room temp.

    As for energy stored in the coil's magnetic field, 1/2Li2, isn't it? And inductance is proportional to turns squared. So energy comparison of the coils is over 6 orders of magnitude.

    Does it make sense to go further? What is your objective here?
    bi

    Last edited by bistander; 12-14-2023, 02:15 AM.

    Leave a comment:


  • jettis
    replied
    Hey bi, I do not know much at all about Holcomb, however what do you think of this statement? Is there any issues as far as your concerned?

    If you take two iron core coils one made from 40 awg and the other made from 10 awg, each has a 3/4” x 24” laminated steel core acting on a 6” diameter x 24” length rotor with two north magnets measuring 2” wide, one inch thick and 24” long, equally spaced around the rotor.

    When we compare the two coils in pulse motor operation, the coil with 40 awg wire at 5 ohms and then another 10awg wire coil at 5 ohms. Which one has more magnetism to convert into rotational energy and which one will yield more flyback energy? Take note the 10 awg coil could be 30 times the size of the 40 awg coil.

    One could easily see a coil 30 times bigger will produce a lot more rotor torque and flyback energy than the smaller coil. It should also be again noted that each coil (10 awg vs 40 awg) will have the same current draw, from the primary battery, as ohm’s law states.

    Dave Wing
    Last edited by jettis; 12-13-2023, 07:44 PM.

    Leave a comment:


  • bistander
    replied
    Originally posted by Rakarskiy View Post
    I don't see any judgments for failing to fulfill the terms of sale of Holcomb homes!
    I sent you a couple of examples via PM.

    Originally posted by Rakarskiy View Post
    ... , there have always been crooks. ...
    Once a crook, always a ....


    Originally posted by Rakarskiy View Post

    How does any synchronous generator work? How is the magnetic field in the core amplified? And, most importantly, how can mechanical energy be converted to electrical energy if the wire is in a groove?
    Mr. Rakarskiy,
    You keep repeatedly saying this. Just because you do not understand it, does not mean others also do not understand it. Here, in fact, it is well understood and taught in colleges and many textbooks. Engineers who know the technology and who are skilled in the art apply these principles to design and build generators and motors which convert electric to mechanical energy, and vise versa, responsible for handling about half of the total energy mankind uses on this planet, in an extremely efficient manner. I can not teach you this on a web forum. I've attempted to enlighten (help) you with certain misconceptions and math errors and have received resistance and rudeness. You are a bad student. So I give up trying with you.

    Back to Holcomb. He buys and pays for his proofs. There is no independent verification. Everything he issues has flags.

    Where is the source of energy? He can not make it with iron as he claims.
    bi

    Leave a comment:


  • Rakarskiy
    replied
    I don't see any judgments for failing to fulfill the terms of sale of Holcomb homes! Is this your secret circle of owners? Don't talk to me about Bob, there have always been crooks. These homes are not sold like toilet paper. There are many items that are sold under a contract of sale. If you don't fulfill the terms of the contract, it's a lawsuit!
    How does any synchronous generator work? How is the magnetic field in the core amplified? And, most importantly, how can mechanical energy be converted to electrical energy if the wire is in a groove?

    Leave a comment:


  • bistander
    replied
    Originally posted by Rakarskiy View Post
    bistander, I gave an example of a visit, a pick up and go. About the performance evaluation, only in quantitative terms. Electricity bills are a very good argument, especially for the happy owner of this unit. You can believe it or not, but this fact takes place.
    People ask me, how do I check? I always recommend through a meter for an hour at the input and at the input of the load the generator is providing. Kilowaty for an hour, the best argument: to show that - works or to show that - does not work. And drooling over torque measurements is fancy. Here is an example of how Stepanov made an instant evaluation of all indicators.

    2023-08-29_113449.jpg

    His device is also a power amplifier.
    Hi Rakarskiy,
    what is shown does not resemble the Holcomb devices. And who is Stepanov?

    Yes, the cost of the "power bill" is the indicator. On installations in the US by Holcomb in facilities outside Holcomb's circle, cost has increased. No joy there.

    All these are just reports. The real clue is the science behind the claim. And that does not support Holcomb. All your analysis and calculations never show where the magic source of free energy is. Explain the physics behind Holcomb's claim of extracting endless energy from iron. Dream on.
    bi

    Leave a comment:


  • Rakarskiy
    replied
    bistander, I gave an example of a visit, a pick up and go. About the performance evaluation, only in quantitative terms. Electricity bills are a very good argument, especially for the happy owner of this unit. You can believe it or not, but this fact takes place.
    People ask me, how do I check? I always recommend through a meter for an hour at the input and at the input of the load the generator is providing. Kilowaty for an hour, the best argument: to show that - works or to show that - does not work. And drooling over torque measurements is fancy. Here is an example of how Stepanov made an instant evaluation of all indicators.

    2023-08-29_113449.jpg

    His device is also a power amplifier.

    Leave a comment:


  • bistander
    replied
    Originally posted by Rakarskiy View Post
    bistander, who prevents you from contacting the Holcomb Center, arrange a visit and make sure that the system does not work as advertised. Your assertions are not substantiated by anything but your conviction. And it is your conviction that is false.

    https://gehtanders.de/besuch-bei-hol...nergy-systems/
    Hi Rakarskiy,

    ​​​​​There are things which inhibit my ability to travel. That would be off topic. But in a nutshell, I need not travel, or speak to Holcomb, or see his stuff in person to know his claims are false. My assertions are substantiated by several centuries of science. The theories and arguments and demonstrations from this outfit and those they hire don't hold to scrutiny.

    Quick example:
    From the link you just supplied.
    "Measurements

    We were then allowed to carry out our own measurements on a 10 kW generator (ILPG) at a test station. The generator was connected to a continuous load with numerous 3-pas electric motors. Our measurement engineer found that the input current per phase was approximately 2 amps, while the output was approximately 8 amps. P = U * I. This results in: 110 V * 8 amp * 3 phases = 2.6 kW power as output."

    Do I have to walk you through what's obviously false with that?

    So where are the replications? Where is the peer review? Where is the free energy promised? Where is truth?

    bi

    Leave a comment:


  • Rakarskiy
    replied
    bistander, who prevents you from contacting the Holcomb Center, arrange a visit and make sure that the system does not work as advertised. Your assertions are not substantiated by anything but your conviction. And it is your conviction that is false.

    https://gehtanders.de/besuch-bei-hol...nergy-systems/

    Leave a comment:


  • bistander
    replied
    Originally posted by Rakarskiy View Post
    ... My material summarizes the results of my research that I shared with.
    Last edited by Rakarskiy; Today, 01:52 AM.
    Then it should be easy for you to answer.

    Does this method provide power output in excess of input power?

    bi

    Edit:
    is this your conclusion?
    "...synchronous generator with a solid-state rotor (without mechanical rotation) has been patented and implemented in the USA. Holcomb Energy System . This device is direct proof ..."

    quote is from your material.

    Holcomb's "proof" is false.
    Last edited by bistander; 12-08-2023, 02:37 AM. Reason: Typo

    Leave a comment:


  • bistander
    replied
    Originally posted by Rakarskiy View Post
    Good afternoon, try doing the calculation. Suppose we have an oscillator phase nested in two magnetic closed circuits in which we vary the vector and saturation of the magnetic intensity by two electromagnets controlled through a resistive regulator according to the scheme proposed in the Figuer patent of 1908. The flux excitation should be calculated as for an electromagnet, and the EMF excitation in the phase (output winding) as for a generator.

    2023-12-07_072605.jpg
    This is very easy to check first mathematically, then practically, just by using two identical matching cores, the only thing you need is to make an appropriate small gap (0,3-0,5 mm).
    This has not been practically tested, only calculated. If you have the resource, you can test this on a real device by pre-designing the control circuit.
    Hi Rakarskiy,
    do you think that is an answer to my question?
    bi

    Leave a comment:


  • Rakarskiy
    replied
    Good afternoon, try doing the calculation. Suppose we have an oscillator phase nested in two magnetic closed circuits in which we vary the vector and saturation of the magnetic intensity by two electromagnets controlled through a resistive regulator according to the scheme proposed in the Figuer patent of 1908. The flux excitation should be calculated as for an electromagnet, and the EMF excitation in the phase (output winding) as for a generator.

    2023-12-07_072605.jpg
    This is very easy to check first mathematically, then practically, just by using two identical matching cores, the only thing you need is to make an appropriate small gap (0,3-0,5 mm).
    This has not been practically tested, only calculated. If you have the resource, you can test this on a real device by pre-designing the control circuit.

    My material summarizes the results of my research that I shared with.
    Last edited by Rakarskiy; 12-07-2023, 06:52 AM.

    Leave a comment:


  • bistander
    replied
    Originally posted by Rakarskiy View Post
    Hi, bistander! The conclusion is based on the principle of electromagnetic induction in electromagnetic generators with cores and the operation of transformers.
    In the figure is a slice of a Dynamo with an armature that rotates for which position in the wire laid in the slot, will be the maximum EMF (A-A' or B-B').
    2023-12-06_145654.jpg
    Your Electricity * Over Unity: Invention of the Electromagnetic Generator (rakatskiy-blogspot-com.translate.goog)
    Hi Rakarskiy,
    thanks for the reply. But you must have overlooked my questions. So here is a replay.

    What is your conclusion?
    Does this method provide power output in excess of input power?

    Please answer.
    Thanks.
    bi

    Leave a comment:


  • Rakarskiy
    replied
    Hi, bistander! The conclusion is based on the principle of electromagnetic induction in electromagnetic generators with cores and the operation of transformers.
    In the figure is a slice of a Dynamo with an armature that rotates for which position in the wire laid in the slot, will be the maximum EMF (A-A' or B-B').
    2023-12-06_145654.jpg
    Your Electricity * Over Unity: Invention of the Electromagnetic Generator (rakatskiy-blogspot-com.translate.goog)
    Last edited by Rakarskiy; 12-06-2023, 08:55 PM.

    Leave a comment:

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