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  1. #1
    Registered User MCCLINTOCK-LOOMIS's Avatar
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    Default Jewelers regulator 3 jar mercury pendulum finding center of oscilation?

    Hello, I'm building another jeweler's regulator case using a typical french pinwheel movement but this time I'm using a 3 jar mercury pendulum instead of the large grid iron style. It's apparent I will need to shorten the pendulum rod as it is running about 20 minutes slow over a 24 hour period. I have loaded the vials with 10 lbs. of mercury and have adjusted it to it's fastest or shortest setting. Is there a reasonable way to determine the center of oscillation with this type of bob? I will need to cut mill drill and tap the top of the rod so I will want to get it right the first time.
    Thanks
    George
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  2. #2

    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    Do you have more mercury? Adding to the level would raise the c.g. and speed it up.

    Johnny

  3. #3
    Registered User Jim DuBois's Avatar
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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: John MacArthur)

    I am not thinking you need the center of oscillation to calculate the needed change in pendulum length to gain 20 minutes in 24 hours. We know the effective length should be a meter +/- a very small amount. We know it is slightly long causing it to run slow by 20 min/24 hr. I would use that to calculate the apparent length. Then by reducing the length of the pendulum rod by that amount you should be well in the adjustment range.

    Or use something like the following to determine how much to reduce the length.

    From http://abbeyclock.com/Pendulum.html Ratios using practical length.
    First example: a clock comes in for repair without a pendulum and no clues as to its length. Make up a pendulum as long as the case will allow, or up to 39.15 inches long (a one second pendulum). Set the time, and take a reading after 24 hours. Record the error and the length of the pendulum (for example, from the middle of the suspension spring to the middle of the pendulum bob).


    e.g. 39 inches long
    lost one hour, 23 minutes in 24 hours.

    1:23 = (1x60x60) + (23x60) = 4980 secs.
    so actual reading = (24x60x60) - 4980 = 81420 secs.
    desired reading = (24x60x60) = 86400 secs. (i.e. 24 hrs.)
    then divide by 24 hours to convert to ratios, using a common denominator:
    81420 ÷ 86400 = 0.942361
    and 86400 ÷ 86400 = 1

    Rearranging the formula:



    |


    Square both sides:



    |


    Eliminate constants:



    |



    |


    The pendulum length in this case needs to be 89 % of its actual length:

    39 x 0.88804 = 34.6 inches

    I would then screw the pendulum bob up as high as it will go, and shorten the pendulum to 35 inches. This is because it is better to have the pendulum a little too long than too short. It can be shortened a little more if it still runs too slow.

  4. #4
    Registered User MCCLINTOCK-LOOMIS's Avatar
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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation?

    Thank you both for your responses. I do have more mercury but the pendulum is a bit heavy and I'm now running it with a temporary 10 lb. weight. This is a couple lbs.heavier than it ran with the grid iron pendulum. I will get more accurate readings to do the calculations. Something I learned long ago as a carpenter is "measure twice cut once"
    Thanks
    George
    Last edited by MCCLINTOCK-LOOMIS; 04-11-2017 at 11:06 AM.

  5. #5
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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    Put it in the center of the adjustment and give us an
    exact amount it is slow and the exact period it was measured
    for.
    We can then tell you how much to remove. It is ok if the
    period is not exactly 24 hour, just that both measurement are done
    as close as you can and that the rating was set to the mid range.
    Tinker Dwight

  6. #6

    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation?

    Regardless of where the pendulum came from, if it is still hanging from its original suspension spring, it should be able to handle the additional weight of mercury necessary to bring the c.g. back up to the correct height for the pendulum to be in time. You have already added a little driving weight to compensate for the pendulum jar having more air resistance than the original gridiron, and thus shouldn't need even more. A heavier pendulum of a given cross-section area doesn't require a heavier driving weight, though this has also been argued endlessly. Adding mercury would look more "correct", also. Under your mantra of "measure twice", give it a try before you cut the rod.

    My two cents for now,
    Johnny
    Last edited by John MacArthur; 04-12-2017 at 07:15 AM.

  7. #7
    Registered User MCCLINTOCK-LOOMIS's Avatar
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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    John I will take your advise and add more mercury. I have around 3 more lbs. I will let you know what happens. While we are on the subject of mercury. I would like to seal the caps to the vials so there is never a risk of a spill. The vials are thick pressed glass with ground tops. Would a bead of silicone calking around the top edge be a good choice,or hi vac. grease, or?
    any thoughts.
    thanks
    George

  8. #8
    Registered User MCCLINTOCK-LOOMIS's Avatar
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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    Tinker, I added mercury to about 1/2 inches from the top around another pound or so. set the adjustments at mid point and checked it for 1 hour. It's now losing 105 seconds per hour or 42 min. in 24 hours. I can adjust the pendulum bob assembly around 1/2 in either direction with it's current setting and their is a weight above the bob for micro adjustment. What do you think?

    George

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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    I calculate that the pendulum should be 94.4% of its current length.
    You should measure from about the center of the bob to the inflection point of your
    suspension spring.
    If it is a short spring, measure from about the center of the spring.
    If it is a long spring you'll have to eyeball it. Depending on the thickness,
    it will be about 1 inch down for a 5 inch spring.
    It is not super critical because a small error here will only be multiplied
    by .04 of the actual value. In other words, a 2 inch error will be .08 inches.


    (1440/1482)^2 = 0.944

    Tinker Dwight
    Last edited by Tinker Dwight; 04-12-2017 at 03:09 PM.

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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    Quote Originally Posted by MCCLINTOCK-LOOMIS View Post
    Tinker, I added mercury to about 1/2 inches from the top around another pound or so. set the adjustments at mid point and checked it for 1 hour. It's now losing 105 seconds per hour or 42 min. in 24 hours. I can adjust the pendulum bob assembly around 1/2 in either direction with it's current setting and their is a weight above the bob for micro adjustment. What do you think?

    George
    Of course you still have the issue of the correct mercury volume and distribution for temperature compensation. There must be a procedure for determining the volume and distribution of mercury for correct rate and temperature compensation?

    Without doing the calc it would seem to me that the total volume of mercury would set the temperature compensation and the distribution of mercury between jars would provide a means of adjusting rate. Lowering the mercury level in a jar below the centre of oscillation would give a longer rate and above a shorter rate.

    If you want temperature compensation you shouldn't do any cutting until you full understand the correct calibration procedure, anybody?

    Phil

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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: Phil Burman)

    You need to make an enclosure with a constant temperature
    to put the clock in.
    A hall closet works well.
    You need a way to measure the temperature in the close,
    and measure the rate of the clock.
    The heating should have a fan to keep a uniform temperature.
    the fact that the outside is a different temperature will want
    to make the temperature stratified.
    A good thing to help as well as the fan is to use that insulation sheeting,
    made of Styrofoam and foil covered, to line the closet. It is cheap
    and easily cut with a knife ( messy though ).
    You need to have the clock closed when making the time runs but
    it speeds things up to open the door to allow the air to circulate
    in the movement during the soak time, on changing the temperature.
    I would shorten the pendulum some as 4% is quite a bit.
    You want it close to right before doing the compensation.
    You might work out a method to clamp the top of the pendulum,
    to change the length.
    The incorrect weight distribution near the top has little effect on
    the rate.
    Tinker Dwight

  12. #12

    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    George, I would think that an adequate seal around the top of the jars, at least until you are done experimenting, would be a soft o-ring to fit the top of each jar.

    Tinker, I think the flexure point of most suspension springs is *very* near the top, long or short.

    Phil, calculating the compensation is wickedly complicated - not involving just the expansion difference between mercury and steel, but including that of the glass jars, as well. It also needs to include the differences of angular momentum of the various masses (steel, glass, mercury), and this is quite significant. There's an extremely thorough analysis in Horological Science Newsletter (chapter 161, I think) from a couple of years ago that you might be able to find. If not, I can probably dig it out of my files (piles?). In a nutshell, it finds that most high-grade regulators from the past centuries have been under-compensated by quite a bit. I find it hard to believe that this has not been noticed, but can't find fault with the analysis.

    Johnny

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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: John MacArthur)

    Quote Originally Posted by John MacArthur View Post
    George, I would think that an adequate seal around the top of the jars, at least until you are done experimenting, would be a soft o-ring to fit the top of each jar.

    Tinker, I think the flexure point of most suspension springs is *very* near the top, long or short.

    Phil, calculating the compensation is wickedly complicated - not involving just the expansion difference between mercury and steel, but including that of the glass jars, as well. It also needs to include the differences of angular momentum of the various masses (steel, glass, mercury), and this is quite significant. There's an extremely thorough analysis in Horological Science Newsletter (chapter 161, I think) from a couple of years ago that you might be able to find. If not, I can probably dig it out of my files (piles?). In a nutshell, it finds that most high-grade regulators from the past centuries have been under-compensated by quite a bit. I find it hard to believe that this has not been noticed, but can't find fault with the analysis.

    Johnny
    Mercury has such a high coefficient of expansion that
    it has a significant effect on the self moment of inertia.
    The sum of the moments of inertia are part of the equation.
    moving the center of gravity up without change in moment
    of inertia slows the pendulum.
    Reducing the moment of inertia, speeds the pendulum.
    Also missed is the fact the the liquid also sloshes in the
    tubes. The 1 second beat pendulum has plenty of time
    for the mercury to move.
    The better way is to move a solid weight with a high
    coefficient material.
    This is the typical iron bob with the brass nut near the
    center of the bob, used on a low coefficient rod, such
    as invar, fused quartz or Zerodur.

    invar 1.2 ppm/C
    fused quartz 0.55 ppm/C
    Zerodur 0.1 to 0.007 ppm/C

    Tinker Dwight

  14. #14
    Registered User MCCLINTOCK-LOOMIS's Avatar
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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    Guy's Thanks for all your posts. This pendulum in all reality is just for show as a reasonable timekeeper will suffice from this clock. For now I have made up a temporary rod from a section of old 7/16" steel bar that I have threaded the bottom and fitted the top. I can chew on this until I'm satisfied with the results excluding any temp. compensation. Once length is established I will make adjustments to the original rod. I will post some pic's once clock is finished.
    regards
    George

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    Default Re: Jewelers regulator 3 jar mercury pendulum finding center of oscilation? (By: MCCLINTOCK-LOOMIS)

    Quote Originally Posted by MCCLINTOCK-LOOMIS View Post
    Guy's Thanks for all your posts. This pendulum in all reality is just for show as a reasonable timekeeper will suffice from this clock. For now I have made up a temporary rod from a section of old 7/16" steel bar that I have threaded the bottom and fitted the top. I can chew on this until I'm satisfied with the results excluding any temp. compensation. Once length is established I will make adjustments to the original rod. I will post some pic's once clock is finished.
    regards
    George
    Use rod of the same or close to the same weight per foot.
    The rod effects the rate some as well.
    Tinker Dwight

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