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  1.  
    Posted By: spinner"So the question boils down to, "Does it take more power to saturate the core when the rotor magnet is interacting with it than it does when there is no rotor magnet"?"
    The interaction between energized toroid coil and magnet rotor is relatively week. Why don't you try the experiment, what is an amount of power needed to form and maintain the field in the toroid (where most BUT NOT ALL of the flux is trapped inside) while magnet flies by?


    I wish I could, but I'm afraid I don't have the facilities to do so That's part of the reason for my rant in the first place, to try and provoke someone who does have the appropriate resources to test these conditions and find the answers.
    • CommentAuthorspinner
    • CommentTimeDec 22nd 2009
     
    @overconfident
    But you don't need much...
    For instance, an experiment could be made with a rigid pendulum (magnet) passing by a static toroid coil at BDC...
    The coil would be triggered in the right moment by optical switch () releasing a known energy pulse (programmable power supply, capacitor,.)just in the right moment.
    The gained height on the up-side would show exactly the kinetic advantage - but after substracting ALL the electrical energy inputs (energy pulse, all the losses) you would find out what happened really...
  2.  
    @spinner
    Yes, it makes a difference. The moving magnets induce a magnetic current in the ferrit cores. So the magnetic field of the core varies over the time. I you then power on the toroid coil, the current has to do some work: Changing the field in the core again.
  3.  
    Holy crap. This cannot be made into something like the hunt for the Higgs boson. What is the OU claim here anyway? Is it heat or kinetic energy? Does Steorn even know what they're claiming in that respect, or is it a trade secret or what? Somebody just apply the proper tool(s) and get it over with.

    (PS: Mr. Battery will lose.)
  4.  
    Posted By: Postposterous Is it heat or kinetic energy? Does Steorn even know what they're claiming in that respect, or is it a trade secret or what? )


    How could they know? It's not their job to educate people but just to entertain the investors.
  5.  
    By the way, the "time variant" regarding the Orbo means that the arguments vary from time to time.
  6.  
    Next product: Orbometrically polarized antennae to harvest the subspace omicron waves generated by Orbo.
  7.  
    Posted By: spinner@overconfident
    But you don't need much...
    For instance, an experiment could be made with a rigid pendulum (magnet) passing by a static toroid coil at BDC...
    The coil would be triggered in the right moment by optical switch () releasing a known energy pulse (programmable power supply, capacitor,.)just in the right moment.
    The gained height on the up-side would show exactly the kinetic advantage - but after substracting ALL the electrical energy inputs (energy pulse, all the losses) you would find out what happened really...


    No optical switch, no power supply, no capacitor, no breadboard, no soldering iron, no DMM, no scope, and no toroid coil at the moment.

    I do have plenty of magnets and I can set my WhipMag up as a pendulum, as you have seen.
  8.  
    Posted By: PostposterousNext product: Orbometrically polarized antennae to harvest the subspace omicron waves generated by Orbo.


    That's worth a bet. Guess the next revolutionary Steorn product!
    • CommentAuthorDirtfarmer
    • CommentTimeDec 22nd 2009
     
    I think you have to realise that this is almost surely "proof of concept" demonstrations...
    If they were to really have it, they do have IP to protect, and this would be carefully orchestrated to give the minimum information required...
    So to that end, Yes, if all electrical input is dissipated as heat, (i.i: no CEMF or "inductive" losses) then the rotor turning is "free" energy, no matter how slow, no matter if it takes 5 watts perfectly dissipated as heat to produce .5 watt of rotor power...
    It would indeed be free, and exploitable by other means....
    Goodwill,
    -Dirtfarmer
    •  
      CommentAuthorAngus
    • CommentTimeDec 22nd 2009
     
    Posted By: overconfidentIf the interactions of the PM and core do not adversely affect the power requirements to saturate the core, then the rotor is effectively spun without cost.


    I don't see that. You have to saturate the core to make the motor work. It's a cost of operation. Doesn't matter whether the interaction of the magnet causes an extra cost.
    • CommentAuthorspinner
    • CommentTimeDec 22nd 2009
     
    Posted By: bloodymedia@spinner
    Yes, it makes a difference. The moving magnets induce a magnetic current in the ferrit cores. So the magnetic field of the core varies over the time. I you then power on the toroid coil, the current has to do some work: Changing the field in the core again.

    Ferrite toroids are not very keen to accept any EXTERNALLY induced currents. First, because they are ferrites, and because they're of toroid shape. That's the main point of torroids, to leak as less as possible and to be imune to external fields. So even the CEMF/BEMF or whatever you like to call it is minimalised.

    Why don't you try with a decent solenoid instead and report what kind of induction you get then?
    Maybe that's why conventional motors are using anything but a toroidal shape stators... And they're getting >95 % efficiency. Of course, they'll never get 300% like Orbo does.
  9.  
    They could have used a bent nail with some wire around it as well. But the nail would become magnetic after some time. Ferrit cores are "hard magnetic", so they return to the former state of (not-) magnetism.
    The toroidal shape helps to keep the field inside of the (powered) coil instead of building up an external field.
    Why do you think toroids are not 'keen' to accept externally induced currents? That's the basic principle of a (toroidal) transformer - transferring the field changes from one coil to the other - electric current in the primary coil -> magnetic current in the core -> electric current in the secondary coil.
  10.  
    Posted By: Angus
    I don't see that. You have to saturate the core to make the motor work. It's a cost of operation.


    For the device Steorn has displayed that is true. I'm trying to learn something about the cost. Is there a surcharge if permanent magnets happen to be flyin nearby?

    And just how much does it cost? Is the energy to saturate the core greater than the energy gained by the passing rotor magnet? I haven't seen any experimental data for that yet.

    Doesn't matter whether the interaction of the magnet causes an extra cost.


    Maybe not in this device.
    • CommentAuthorspinner
    • CommentTimeDec 22nd 2009
     
    Posted By: bloodymediaThey could have used a bent nail with some wire around it as well. But the nail would become magnetic after some time. Ferrit cores are "hard magnetic", so they return to the former state of (not-) magnetism.
    The toroidal shape helps to keep the field inside of the (powered) coil instead of building up an external field.
    Why do you think toroids are not 'keen' to accept externally induced currents? That's the basic principle of a (toroidal) transformer - transferring the field changes from one coil to the other - electric current in the primary coil -> magnetic current in the core -> electric current in the secondary coil.

    We saw a magnet approaching the toroid coil from OUSIDE of the core. That's quite a different situation than having a primary and secondary wound on the same core.

    Switch on a toroidal transformer then try to induce something with the biggest NdFeB magnet you have...
  11.  
    Yes, but take the corresponding moments into account. The magnet approaches the toroid core and magnetises it. The coil is dead at that moment. Then the coil is powerd on and has to change the field in the core. Shortly after the magnet has passed, the current through the coil is switched off again and the core is just a plain simple piece of ferromagnetic metal.
  12.  
    Addendum:
    Just see the core as a part of the magnet, at the moment the magnet is only millimeters away from it.
    It costs more energy to change the magnetic field of a magnet than changing the magnetic field of some ferrit.
    • CommentAuthorDirtfarmer
    • CommentTimeDec 23rd 2009
     
    Posted By: DirtfarmerI think you have to realise that this is almost surely "proof of concept" demonstrations...
    If they were to really have it, they do have IP to protect, and this would be carefully orchestrated to give the minimum information required...
    So to that end, Yes, if all electrical input is dissipated as heat, (i.i: no CEMF or "inductive" losses) then the rotor turning is "free" energy, no matter how slow, no matter if it takes 5 watts perfectly dissipated as heat to produce .5 watt of rotor power...
    It would indeed be free, and exploitable by other means....
    Goodwill,
    -Dirtfarmer

    Whooo--- don't think I Am being a bit garden pathed, here, (though I started to question myself...) because I am NOT saying the only work of the rotor would be in air resistance and Bearing resistence (In that instance..it would still show as merely "heat" (entropy))
    Leaving a "no net work" conclusion (angular momentum all that, ) One must still show net "REAL" power... (even be it .5w for 5)
    (Heavily edited in the typing of this..My regrets for the hard read...)
    Goodwill,
    _Dirtfarmer
    • CommentAuthorDirtfarmer
    • CommentTimeDec 23rd 2009 edited
     
    My POINT in the prior post was that , when all is said and done, regardless of cancellation of CEMF and "inductive losses" (I'm still trying to figure out the practical difference there..)
    One may NOT assume that any CALCULATION of the work done by the rotor is valid, if it is only shown as further entropy, one must REALIZE that work over and above input... now in a calorimetric system , we would see the total expected heat generated by all losses, inductive, resistive, and friction... and could easily say that it is producing "3x" more, but I would be careful with accepting calculations which back out CEMF and "inductive" losses, because, even if they SEEM accurate, by Great Minds... The great accountant usually (always?) has tricks to balance the equation,...
    (It may even have a "Schrodingers cat" aspect to it, that problem) ...as entropy, it is only that...
    And don't forget what Sv really is....
    Goodwill,
    -Dirtfarmer

    EDIT: just to capitalize "realize" EDIT: Now heavy additions with the "now in a calorimetric..."
    •  
      CommentAuthorlegendre
    • CommentTimeDec 23rd 2009
     
    Posted By: maryyugoHey Al, I don't know snot about this stuff but couldn't you visualize the field with fine iron particles or some of those fancy photo methods people have developed for the porpoise?


    You could download a free trial copy of Vizimag..

    http://www.vizimag.com/