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Grandfather clocks

  1. Apr 5, 2005 #1
    Why do grandfather clocks keep going? The only way I see htis working is if the chamber has no air (vacuum).
     
  2. jcsd
  3. Apr 5, 2005 #2

    FredGarvin

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    The ones I know of are all wound. I am assuming you are referring to, what I know as, "anniversary clocks" because they were wound once a year. The dome is definitely not sealed. The rotating ball mechanism is simply a rotating pendulum that is more efficient than a standard pendulum.
     
  4. Apr 5, 2005 #3

    HallsofIvy

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    Grandfather clocks are not perpetual motion machines! The "true" grandfather clocks operate on weights that have to be raised (once a week for an "eight day clock"). Anniversary clocks (those little things in the glass dome- NOT grandfather clocks) operate on springs that have to be wound regularly). Some "fakes" operate on batteries.

    By the way, a vacuum would not be enough for a "perpetual motion" clock- there would still be friction in the gears and other operating parts. I have a grandfather clock sitting about 10 feet from me right now that is rubbing somewhere and I can't find it! I wonder if a good hard kick would help?
     
  5. Apr 5, 2005 #4
    Slightly deviating: Provided a simple pendulum in vacuum, would it swing forever? Close to?

    From there, is it not possible to create a detective system (probably optic) to sense the completion of a period and link that to a clock?
     
  6. Apr 5, 2005 #5

    chroot

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    In the absence of any kind of friction, it would never lose energy, so yes, in theory. Of course, no real pendulum can ever be totally free of all friction, so no, in practice.

    - Warren
     
  7. Apr 5, 2005 #6
    Im saying putting it in a vacuum, as close a vacuum as possible. There will be an extremely small bit of friction from possibly stray air molecules, and some from the junction of string and ceiling.

    My question is in a practical sense, are those small enough to neglect for a certain time (that is reasonable for a clock of such delicate construction) before recalibration? I'm talking at least a week or so. We can attribute the error in the correct time to the energy lost in friction.

    But even if some energy is lost, is the period of a pendulum not the same regardless of the displacement from equilibrium? If not, perhaps we could adjust our detector to compensate for such effects?
     
  8. Apr 5, 2005 #7

    chroot

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    Precisely.
    Sure, you can spend some engineering effort to design a pendulum support that will allow the pendulum to lose only a specified amount of its energy in a specified time. For example, you could design a pendulum that would only lose 1% of its energy to friction in a week, but that design will be more difficult than one that loses 10% of its energy in a week. There is, however, no practical way to design a pendulum that never loses any energy.
    The period does not stay exactly the same when its displacement changes. That approximate relationship is due to taking the "small angle approximation" of the sine function:

    [tex]\sin \theta \approx \theta \,\, \textrm{for small angles}[/tex]

    - Warren
     
  9. Apr 5, 2005 #8

    brewnog

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    I don't know what this rotating ball thing is which Fred mentioned, but in a grandfather clock the long case is designed specifically to exploit the 'small angle' observation which chroot noted.

    Whozum, air resistance is not really a factor in determining the accuracy of a mechanical clock. It would be pretty pointless to use a string (rather than a bar of some kind) as a pendulum, but even so, mechanical resistances are largely the cause of energy losses, traditionally accounted for by using a falling weight.
     
  10. Apr 5, 2005 #9

    dextercioby

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    For an ideal mathematical pendulum (meaning no friction),it can be shown that the period of oscillations,though depending of amplitude for big angles,is independent of time.That means the oscillations are tautochrone...

    One of favorite formulas is this

    [tex] T_{pendulum} (\alpha,l) =4\sqrt{\frac{l}{g}} \int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1-\kappa^{2} \sin^{2}\varphi}} [/tex]

    ,where

    [tex] \kappa^{2} = \sin^{2}\frac{\alpha}{2} [/tex],where [itex] \alpha [/itex] is the angular amplitude of oscillation.

    Note that the modulus of the complete elliptic integral of the first kind is always less than unity which can be used for a series expansion and a proof of the intimate connection between elliptic integrals and Gauss hypergeometric functions.

    Incidentally

    [tex] T_{pendulum} (\alpha,l) =2\pi \sqrt{\frac{l}{g}} \ _{2}F_{1}\left(\frac{1}{2};\frac{1}{2};1;\kappa^{2}\right) [/tex]

    Daniel.

    -----------------------------------------------------------------
    Bibliography:
    Mocioaca,Ghe.,Functii speciale,Ed.Tehnica,Bucuresti,1983.
     
    Last edited: Apr 5, 2005
  11. Apr 6, 2005 #10

    FredGarvin

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    Here you go Brewnog...The anniversary clock has a rotating pendulum. Those are the ones I was referring to.
     
    Last edited: Nov 7, 2005
  12. Apr 6, 2005 #11

    brewnog

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    Cheers for that Mr G, I like the cut of your jib. But not a grandfather clock...
     
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