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    • CommentAuthortruster
    • CommentTimeDec 15th 2009
     
    Sorry if this question has been asked already. I looked around for the information but didn't find it. Let's say that there is no outside energy source being "beamed" to the spinning rig. Let's also assume that there are no additional batteries hidden in its opaque base. Given the minimal friction and presumed heat output, how long could such a device spin as shown on the D size rechargeable battery being used? This live video feed is supposed to last until Dec. 23. Could it possibly operate on the one battery for that long?
  1.  
    Easily, seems to be the consensus.
  2.  
    I guess that's why their demo isn't doing any work. If it were pumping water, or something, the battery would run down too fast
  3.  
    Well, 8 days is 192 hours. The ZDnet article says the battery is a 1.2V 10,000 mAh NiMH rechargeable D cell.

    If it's drawing only 52 mA, it should be able to run for 8 days. At 1.2V that's 0.0625 W, if I've done the calculations correctly. It's been a long time since I've done this stuff, so if I'm wrong, someone please pipe up with the correct numbers.

    0.0625 W doesn't seem like much to me, but maybe it's enough to keep it spinning like that. I don't know.
  4.  
    Lets work backwards. D cell battery contains ~40,000J of energy. 8 days is 192 hours; 691200 seconds. 40000/691200 gives 0.058J. So we have 58mW to play with which should be plenty. As Al has noted reed switches tend to have a MTBF of ~10^6 cycles though so (presuming that switch actually does something) it might be the weak link.
  5.  
    Heh, cross posting. Close enough to each other though.
    • CommentAuthortruster
    • CommentTimeDec 15th 2009
     
    @ hairykrishna

    I'm not sure what you mean by "weak link." Are you saying the reed switches might draw enough power to cause the battery to run down sooner?
  6.  
    It's not obvious to me that 60mW would be plenty. Can you find some examples of comparable mechanical work being done with 60 mW? I'm not having much luck with the Google... all I can find are laser pointers and extremely low power radio transmitters.
  7.  
    Posted By: truster@ hairykrishna

    I'm not sure what you mean by "weak link." Are you saying the reed switches might draw enough power to cause the battery to run down sooner?


    No, they just stop switching after a while. Either because the contacts have stuck together or because they've got crud stopping them making contact.
  8.  
    Posted By: Raymond Luxury YachtIt's not obvious to me that 60mW would be plenty. Can you find some examples of comparable mechanical work being done with 60 mW? I'm not having much luck with the Google... all I can find are laser pointers and extremely low power radio transmitters.


    What mechanical work? The only work they're doing is to overcome frictional and air resistance losses which, if they've built it properly, should be tiny.
  9.  
    Just to be precise, it IS doing work. There will be some friction in the bearings, there will be some air resistance to the spinning rotor, and there will be some heat losses in the coils and metal parts, including the magnets, as well as in the electronics. All this represents work. Not much, it's true, but it can be estimated, and even measured. How much power does it take to keep a similar rotor running at a similar speed, in a similar environment?
    I'm guesstimating that the system must dissipate OTO tens of milliWatts, so it could run for a long time -- weeks -- on the energy contained in the D-cell.
  10.  
    No he means the reed switch will fail from mechanical damage.

    Given the suggested power drain rate from the batttery and the symetrical 4 pole drive 2 pole sense layout where is all the energy that could damage the bearings? The only reason for those bearngs is to lower friction losses so the battery lasts longer.
    only 14,000,000 euro?
    at 419 euro /licence only 33413 need to be sold to break even.
  11.  
    Posted By: alsetalokinJust to be precise, it IS doing work. There will be some friction in the bearings, there will be some air resistance to the spinning rotor, and there will be some heat losses in the coils and metal parts, including the magnets, as well as in the electronics. All this represents work. Not much, it's true, but it can be estimated, and even measured. How much power does it take to keep a similar rotor running at a similar speed, in a similar environment?
    I'm guesstimating that the system must dissipate OTO tens of milliWatts, so it could run for a long time -- weeks -- on the energy contained in the D-cell.


    This is getting outside my physics comfort zone. Care to do a back-of-the-envelope calculation on the friction losses in the bearings based on your estimate of the mass of the rotor, the surface area of the bearings, and the coefficient of friction for really good bearings? You work with that kind of stuff all the time, don't you?
    • CommentAuthorCrastney
    • CommentTimeDec 15th 2009 edited
     
    .
  12.  
    Posted By: Raymond Luxury YachtIt's not obvious to me that 60mW would be plenty. Can you find some examples of comparable mechanical work being done with 60 mW? I'm not having much luck with the Google... all I can find are laser pointers and extremely low power radio transmitters.


    Try googling "OCMPMM".
  13.  
    Posted By: Raymond Luxury Yacht
    Posted By: alsetalokinJust to be precise, it IS doing work. There will be some friction in the bearings, there will be some air resistance to the spinning rotor, and there will be some heat losses in the coils and metal parts, including the magnets, as well as in the electronics. All this represents work. Not much, it's true, but it can be estimated, and even measured. How much power does it take to keep a similar rotor running at a similar speed, in a similar environment?
    I'm guesstimating that the system must dissipate OTO tens of milliWatts, so it could run for a long time -- weeks -- on the energy contained in the D-cell.


    This is getting outside my physics comfort zone. Care to do a back-of-the-envelope calculation on the friction losses in the bearings based on your estimate of the mass of the rotor, the surface area of the bearings, and the coefficient of friction for really good bearings? You work with that kind of stuff all the time, don't you?

    I mostly do it empirically, using measurements of rundown time on systems with known rotational inertia and known starting energy. Based on these experiences with devices of what looks to me like approximately the dimensions of the device shown, I come up with my ballpark estimate. OTOO (on the order of) is generally taken to mean "within an order of magnitude either way" so I'm talking about a range of between 1 and 100 milliWatts for the continuous power dissipation figure. It's a guess, dammit, based on my experience and some PDF drawings and jpgs I saw on the internet, ffs. I don need to show you no stinkin envelopes.
    The main bearing losses in that levitated magnetic bearing design will come from the tungsten carbide ball impinging and rotating against the (presumably) hardened steel micrometer head. As I have said elsewhere, I would have designed this bit somewhat differently for even lower friction and less radial play.
    • CommentAuthorAtom
    • CommentTimeDec 15th 2009 edited
     
    [quote][cite]Posted By: Raymond Luxury Yacht[/cite]This is getting outside my physics comfort zone. Care to do a back-of-the-envelope calculation on the friction losses in the bearings based on your estimate of the mass of the rotor, the surface area of the bearings, and the coefficient of friction for really good bearings? You work with that kind of stuff all the time, don't you?[/quote]

    My rough calculation suggests about 0.25mW.

    Assume a 1/1000 coeff bearing supporting 1Kg with an effective 1cm path (if it were a ball type bearing) then we have 9.8 x 1/1000 by 1cm x 2.5rpm =0.245mJ/sec or 0.245mW

    think I fixed up the zeros
  14.  
    Posted By: alsetalokin
    It's a guess, dammit, based on my experience and some PDF drawings and jpgs I saw on the internet


    The guess might be a bit better if they'd post some schematics and electronics parts list.
    • CommentAuthorAtom
    • CommentTimeDec 15th 2009 edited
     
    Given that it is a zero F bearing we should have a lower figure.

    Taking a poorer bearing of say 1/100 we still only lose a measly 2.5mA for a battery life of about 4,000hrs.

    Clearly the wind drag is the big loss factor mechanically. Which is why it spins so slowly!
  15.  
    I wonder what the battery life would be if the specified ‘battery’ was only the outer case and contained lithium batteries.

    It would be most unfortunate if the ‘thing’ rotated for longer than the mathematics predicted.

    There must be a ‘Got You’ in demo, ‘irrefutable mathematical proof’.

    They have their backs to the wall and are seeking money.

    It is just a battery powered motor, until they can convince enough people that it isn’t.