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      CommentAuthorAngus
    • CommentTimeJul 20th 2012
     
    Because we've taken a closed loop through a magnetic field, our trajectory has yielded zero energy - what we gained on the way in we paid out to exit.


    This is wrong. Since you are being so snarky I'll leave it to you to find out why.
    • CommentAuthorVibrator
    • CommentTimeJul 20th 2012
     
    ..just to finish fleshing it out, if we also separate them faster than B can adapt, then we'll have to overcome an even greater input force because B hasn't had time to relax back down its curve - so not only did we forfeit some output energy on the way in, but must also pay extra input energy to leave.

    So a closed loop cycle going fast in, pause, then fast out, has a significant negative energy. It costs input mechanical work, but doesn't pay out any heat at all, let alone a proportionate amount.

    So you can appreciate my curiosity as to what Dr MacDonald thinks about this, since then the argument is constrained to the direction of the asymmetry rather than its thermodynamic plausibility - Orbo would merely be 'creating' energy in exactly the same sense that Sv loss 'destroys' it... indeed, an Orbo run backwards is by definition an Sv loss...
    • CommentAuthorVibrator
    • CommentTimeJul 20th 2012
     
    Posted By: Angus
    Because we've taken a closed loop through a magnetic field, our trajectory has yielded zero energy - what we gained on the way in we paid out to exit.


    This is wrong. Since you are being so snarky I'll leave it to you to find out why.
    lol you're disagreeing with classical mechanics now? A given force yields the same energy, positive as negative, in either direction.

    Just your typical nonsense isn't it - ask me to repeat the question then completely ignore it again. For supposed 'skeptics', you lot are about as scary as the Pathetic Sharks, if slightly more inane...
  1.  
    chirp> An Orbo run backwards is by definition.
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      CommentAuthorAngus
    • CommentTimeJul 20th 2012
     
    Posted By: Vibrator
    Posted By: Angus
    Because we've taken a closed loop through a magnetic field, our trajectory has yielded zero energy - what we gained on the way in we paid out to exit.


    This is wrong. Since you are being so snarky I'll leave it to you to find out why.
    lol you're disagreeing with classical mechanics now? A given force yields the same energy, positive as negative, in either d irection.

    Just your typical nonsense isn't it - ask me to repeat the question then completely ignore it again. For supposed 'skeptics', you lot are about as scary as the Pathetic Sharks, if slightly more inane...


    I see you are going to need help. That's a time varying magnetic field, so it is not necessarily conservative. Go to the back of the class.
    • CommentAuthorjoshs
    • CommentTimeJul 20th 2012
     
    Posted By: Vibrator
    No, to get gain, B would have to change in the same direction as H, before H changes. And that my shaky friend does not happen.
    There's other variables too that could be unrealistically tweaked to provide gain. You work through the possibilities till you find one that IS realistic. But i'm not and haven't talked about gain - only loss, as it pertains to Sv.
    Maybe you do that. Personally, I do not sift each grain of sand on the beach hoping to find a magic leprechaun under one. Should a leprechaun appear, then I will concern myself with the question of from where it might have come.
    • CommentAuthorjoshs
    • CommentTimeJul 20th 2012
     
    Posted By: VibratorWell pay attention then, for the nth time:


    Forget entropic losses for a moment so we have a clean example - suppose we have two low-lag magnets like neos, we let them slowly attract together from some initial distance and measure that energy as F X D. Then we prise them apart back to the same starting distance, again measuring the work.
    Energy is not F X D. If you think it is then that is your first problem. Mechanical work is the integral of F dot ds. I do believe that we have covered that misconception of yours before.


    Because we've taken a closed loop through a magnetic field, our trajectory has yielded zero energy - what we gained on the way in we paid out to exit.

    You need to be precise: We have not gained MPE. We have lost mechanical work.


    If however one or both magnets have appreciable Sv, then their magnetisation will continue to rise for some time after they've attracted together. Then, when we come to separate them, we'll have to overcome this greater force to return them to their starting positions.


    So in both scenarios the magnets travel the same total distance around their curves, but in the latter example the delay has reduced the amount of output work, while keeping the input work maximised. Thus the interaction (the mechanical / MPE conversion) is lopsided, and the loss (of output energy) is due to a reduced force (a delayed increase) caused by the domain lag during the output workload.
    Not quite. Or you are not expressing yourself well. Sv results in an effective velocity dependent counter force to motion in both directions. Hence the name magnetic viscosity. And just like ordinary viscosity, it robs us of energy that we might convert in one direction, and forces us to input more energy than without it in the other. The faster that we go, the more work we lose. The work is not stored, it is lost. Nature's balance for lost work is waste heat.


    Not a spontaneous creation of heat on the input workload. Which of course would indeed be overunity...
    There you go...Third Base! No, we have only well observed losses here. Again, hence the term magnetic viscosity.


    eta: and in case you missed it, this would be interesting because it's thermodynamically under-unity in exactly the same sense that the reversed asymmetry would be OU - it's the same class of interaction, inverted.
    And it no doubt can be correlated to the curl of leprechaun underwear under the right conditions. It is a loss mechanism, no more, and no less.


    In and of itself it's also noteworthy in that the additional input work required (over the output work left untapped) is a function of magnetic force and distance - in other words it's interesting precisely because it'snotheat loss. In principle you could dump any amount of mechanical energy into it without converting any of it to heat... it's effectively energy sunk into a magnetic field.
    There is no untapped output work. There is energy lost, and that loss shows up as minute amounts of heat.
    • CommentAuthorjoshs
    • CommentTimeJul 20th 2012
     
    Posted By: Vibrator..just to finish fleshing it out, if we also separate them faster than B can adapt, then we'll have to overcome an even greater input force because B hasn't had time to relax back down its curve - so not only did we forfeit some output energy on the way in, but must also pay extra input energy to leave.

    So a closed loop cycle going fast in, pause, then fast out, has a significant negative energy. It costs input mechanical work, but doesn't pay out any heat at all, let alone a proportionate amount.

    So you can appreciate my curiosity as to what Dr MacDonald thinks about this, since then the argument is constrained to the direction of the asymmetry rather than its thermodynamic plausibility - Orbo would merely be 'creating' energy in exactly the same sense that Sv loss 'destroys' it... indeed, an Orbo run backwards is by definition an Sv loss...
    Given that it is your personal fantasy, I doubt that Dr. MacDonald is even aware of your loopy idea. If he were, I am sure he would dismiss it for the utter nonsense that it is.
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      CommentAuthorDerrickA
    • CommentTimeJul 20th 2012
     
    Perhaps Vibrator can cut to the chase, and teach us all how to make permanent magnets time varying. I certainly haven't had much luck with that. As for playing coy with the possibilities, we've already had our fill from Sean. Where ,exactly, did that get us?
    • CommentAuthorVibrator
    • CommentTimeJul 20th 2012
     
    Posted By: Angus
    Posted By: Vibrator
    Posted By: Angus
    Because we've taken a closed loop through a magnetic field, our trajectory has yielded zero energy - what we gained on the way in we paid out to exit.


    This is wrong. Since you are being so snarky I'll leave it to you to find out why.
    lol you're disagreeing with classical mechanics now? A given force yields the same energy, positive as negative, in either d irection.

    Just your typical nonsense isn't it - ask me to repeat the question then completely ignore it again. For supposed 'skeptics', you lot are about as scary as the Pathetic Sharks, if slightly more inane...


    I see you are going to need help. That's a time varying magnetic field, so it is not necessarily conservative. Go to the back of the class.
    Aha! So are you seriously suggesting time-dependent magnetic interactions are non-classical, non-Newtonian systems? Sheesh well now i've heard everything...

    Non-conservative you say...? Shocking. I. Am. Agasp. I mean who WOULD have thunk it, WHO I ASK YOU!

    So anyway, you're saying it's maybe not a heat loss after all, in the traditional sense, type stuff? And how does this fit with your views on mechanical conservation per se - the whole thermodynamical CoE bit... where's the energy go if not to heat?
  2.  
    It can't be our vibrator, there is no cool avatar icon.


    But that doesn't prevent him from being wrong as usual.

    It's too bad that Sv doesn't have "inertia"... that is, if the attraction could continue to increase, due to the sV delay, while the magnets have actually started moving apart again.... that would be something.
    • CommentAuthorVibrator
    • CommentTimeJul 20th 2012 edited
     
    Energy is not F X D. If you think it is then that is your first problem. Mechanical work is the integral of F dot ds. I do believe that we have covered that misconception of yours before.
    You obviously have no working familiarity with the subject - it's a perfectly standard expression, if one you've clearly not encountered before. W=E=FXD, or the integral, or the "line integral" or whatever, you're nitpicking cuz you're sweating it. Chillax, try to follow what Angus has explained...




    You need to be precise: We have not gained MPE. We have lost mechanical work.
    No, in the symmetrical example input = output.


    Not quite. Or you are not expressing yourself well. Sv results in an effective velocity dependent counter force to motion in both directions.
    No, it is a delayed change in B, never a counter force, and decreases force in one direction while raising it in the other. Usually.


    Hence the name magnetic viscosity. And just like ordinary viscosity, it robs us of energy that we might convert in one direction, and forces us to input more energy than without it in the other. The faster that we go, the more work welose. The work is not stored, it is lost. Nature's balance for lost work is waste heat.
    Fluid viscosity is a poor analog for Sv, having more in common with coercivity imo but regardless, the temperature of the fluid increases in step with the input work, not so with magnetic viscosity. Instead of traditional viscosity, think 'lag'...

    There you go...Third Base! No, we have only well observed losses here. Again, hence the term magneticviscosity.
    see above

    And it no doubt can be correlated to the curl of leprechaun underwear under the right conditions. It is a loss mechanism, no more, and no less.
    But of what kind? What are it's thermodynamic properties? That's the question...

    There is no untapped output work. There is energy lost, and that loss shows up as minute amounts of heat.
    No, we forfeit output work while raising input work, by outpacing changes in B, and the workload is by definition magnetic not thermal!!!
    • CommentAuthorVibrator
    • CommentTimeJul 20th 2012
     
    Posted By: alsetalokinIt can't be our vibrator, there is no cool avatar icon.


    But that doesn't prevent him from being wrong as usual.

    It's too bad that Sv doesn't have "inertia"... that is, if the attraction could continue to increase, due to the sV delay, while the magnets have actually started moving apart again.... that would be something.
    I am not and have not offered Sv as a gain mechanism, the issue is one of classification and thermodynamic characterisation..!
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      CommentAuthorAngus
    • CommentTimeJul 20th 2012 edited
     
    Aha! So are you seriously suggesting time-dependent magnetic interactions are non-classical, non-Newtonian systems? Sheesh well now i've heard everything...


    You may have heard everything but you clearly didn't learn much from it.

    If you move a test magnet around a closed path in a CONSTANT magnetic field, then indeed no net work is done. But if the field varies as you move around, then the situation changes. Indeed you use that fact every time you switch on an electric motor. The field varies as the rotor whizzes round, allowing the rotor to extract power from the mains which are driving the field.

    As Josh's points out, you can't just multiply force times distance to get work if the force varies with the distance. You have to do the path integral over F(s) ds. When the function F(s) itself varies over the time taken to traverse the path, then you have to include that. But the force field F(s) is what is conservative. There is no guarantee on F(s,t)

    Since one of your magnets is inducing magnetisation in the other, with a delay to boot, that's the situation in this case.

    Now can I go back to sleep?
    • CommentAuthorjoshs
    • CommentTimeJul 20th 2012
     
    Posted By: Vibrator
    Energy is not F X D. If you think it is then that is your first problem. Mechanical work is the integral of F dot ds. I do believe that we have covered that misconception of yours before.
    You obviously have no working familiarity with the subject - it's a perfectly standard expression, if one you've clearly not encountered before. W=E=FXD, or the integral, or the "line integral" or whatever, you're nitpicking cuz you're sweating it. Chillax, try to follow what Angus has explained...




    You need to be precise: We have not gained MPE. We have lost mechanical work.
    No, in the symmetrical example input = output.


    Not quite. Or you are not expressing yourself well. Sv results in an effective velocity dependent counter force to motion in both directions.
    No, it is a delayed change in B, never a counter force, and decreases force in one direction while raising it in the other. Usually.


    Hence the name magnetic viscosity. And just like ordinary viscosity, it robs us of energy that we might convert in one direction, and forces us to input more energy than without it in the other. The faster that we go, the more work welose. The work is not stored, it is lost. Nature's balance for lost work is waste heat.
    Fluid viscosity is a poor analog for Sv, having more in common with coercivity imo but regardless, the temperature of the fluid increases in step with the input work, not so with magnetic viscosity. Instead of traditional viscosity, think 'lag'...

    There you go...Third Base! No, we have only well observed losses here. Again, hence the term magneticviscosity.
    see above

    And it no doubt can be correlated to the curl of leprechaun underwear under the right conditions. It is a loss mechanism, no more, and no less.
    But of what kind? What are it's thermodynamic properties? That's the question...

    There is no untapped output work. There is energy lost, and that loss shows up as minute amounts of heat.
    No, we forfeit output work while raising input work, by outpacing changes in B, and the workload is by definition magnetic not thermal!!!
    go learn some elementary physics and your trolling might be more successful.
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      CommentAuthorDerrickA
    • CommentTimeJul 20th 2012
     
    Posted By: Vibratorthe issue is one of classification and thermodynamic characterisation..!


    I hope this thread doesn't drift off in the magnetocaloric direction. We went through all that on the old Steorn forum. Crazy, but some of those forum members were even discussing how to bake magnets in their home ovens.
    • CommentAuthorjoshs
    • CommentTimeJul 20th 2012
     
    Posted By: DerrickA
    Posted By: Vibratorthe issue is one of classification and thermodynamic characterisation..!


    I hope this thread doesn't drift off in the magnetocaloric direction. We went through all that on the old Steorn forum. Crazy, but some of those forum members were even discussing how to bake magnets in their home ovens.
    Every girl and boy will be wanting the hit toy of the season for Christmas: The "Suzi Homemaker Neodymium Iron Boron Overunity Forge" by Scam Co.
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      CommentAuthorAngus
    • CommentTimeJul 20th 2012
     
    Just don't carry them in your pockets.
    • CommentAuthortinker
    • CommentTimeJul 20th 2012
     
    The reason for baking Neos? To demagnetise them. Actually cherry-red for a few seconds with a gas torch does the trick just as well.
    • CommentAuthorloreman
    • CommentTimeJul 20th 2012
     
    Mmmm. Baked neo.....