Why did it break?

Discussion in 'Clock Repair' started by Fred Reiss, Apr 15, 2012.

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  1. Fred Reiss

    Fred Reiss Registered User
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    I'm repairing a shelf clock with a Hermle 351-020 movement. The clock is dated 1982. It runs fine, chimes OK, but doesn't strike the hour. I let down the spring on the strike side and removed the barrel to inspect. I discovered a single tooth missing on the #50 barrel and the second wheels arbor bent. The bent arbor caused binding that stopped the train. I'm replacing the barrel and 2nd wheel from a scrap movement. I'm sure that's all it will need.
    My question is...how did this happen? Everything else in the movement appears fine with very little wear. What could have broken the tooth and bent the 2nd wheel arbor? Just curious.
     
  2. R. Croswell

    R. Croswell Registered User
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    Either the click failed to hold or the spring came loose from the center arbor or broke. Either way, the sudden unwinding of the spring caused the damage. Not uncommon on these.

    RC
     
  3. Fred Reiss

    Fred Reiss Registered User
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    Thanks for the response. The click is fine. The spring is not broken. I guess the spring must have slipped off its arbor and I'm lucky it only broke 1 tooth!.
     
  4. doug sinclair

    doug sinclair Registered User
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    Check out the state of the click and the teeth on the ratchet wheel. Chances are someone's fingers slipped during winding, and the click was unable to grab the ratchet wheel in the nano-second of time that it took for the spring to expend itself. Typical damage under such circumstancesi.
     
  5. Willie X

    Willie X Registered User

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    I agree with others, click failure.

    It is very unusual for the barrel teeth to break but very common for the 2nd wheel arbor to bend.

    If you remove the click wheel and examine it closely with magnification, I'm guessing that there will be some damage. Might just be a slight ding on one tooth. You can clean it up with a file but much better to replace it with the newer milled version.

    Willie X
     
  6. Fred Reiss

    Fred Reiss Registered User
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    Both the ratchet wheel and click are fine. I popped the cap off the barrel and it looks like the spring was the culprit. It has wierd bends right around where the tang connects into the hole. This had to have been an unusual way to break a barrel tooth and bend the arbor on the 2nd wheel.
     
  7. Bob Vance

    Bob Vance Registered User

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    I'm kind of surprised it only took out one tooth. Once all that power is released quite often 4or5 will be taken off. Make sure you have enough pressure with the click spring. I see a lot of these fail on these movements.
    Bob
     
  8. Willie X

    Willie X Registered User

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    If the spring did simply 'unhook' at the arbor, which would be very unusual, you can bet that it will do it again ...

    I would suggest that you check the click spring tension as Bob suggested and replace the stamped click wheel with a milled updated one. Also make sure that the click is held close to the plate by its rivet and that there is no burr at the tip of the click, especially on the back side against the plate.

    Willie X
     
  9. shutterbug

    shutterbug Super Moderator
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    Yes. You want a flat surface butting against a flat surface. Otherwise you're just waiting for another bad experience.
     
  10. Fred Reiss

    Fred Reiss Registered User
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    Maybe my eyes aren't the youngest on the block, but I've inspected the ratchet wheel and the click as closely as possible and I don't see anything resembling
    a burr, flat spot, wear, dings, or anything. The spring on the other hand, is not concentric...it's bent close to where it hooks onto the barrel.
    I've had springs that are hard to start winding because they don't catch easily on the tang. I bend the spring slightly to decrease its' diameter so it will catch hold.

    Only one tooth broken on the barrel...but a very obvious bend in the 2nd wheel!
     
  11. shutterbug

    shutterbug Super Moderator
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    If a spring breaks or comes loose from the arbor, the expansion is absorbed by the barrel, and you won't get internal damage to other wheels. If the click fails though, the barrel spins rapidly during the unwinding of the spring and everything in the train is stressed with potential damage resulting. That's one way to determine what has happened even before you open the plates. Wheel damage like that mentioned in this thread is almost certainly caused by a failed click.
     
  12. Tinker Dwight

    Tinker Dwight Registered User

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    Hi Sbug
    I fail to see the difference between a spring that breaks at the
    arbor and a click failure. The only difference is the inertia of the
    arbor and key, that is small compared to the spring.
    It would be different if the spring broke at the end attached to
    the barrel.
    Tinker Dwight
     
  13. shutterbug

    shutterbug Super Moderator
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    I'm willing to be enlightened here, Tinker ... but if the spring breaks at the arbor, the arbor remains stationary. The spring expands into the barrel (also stationary), which is usually damaged, but the other wheels are spared. If, however, the click fails, the spring expansion is passed to the barrel which is still attached to the arbor via the spring. In that case, the only release for the spring is through the barrel to the train. To me, that's a big difference. I don't think I'm wrong .... but I await your explanation :)
     
  14. Fred Reiss

    Fred Reiss Registered User
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    Now I'm really confused. When I read Tinker Dwight's response, I think he's exactly right and that means the spring slipping off the
    arbor did all the damage. Then I read Shutterbug's response and it makes sense and that means the click, click spring or ratchet wheel is the culprit. I'm even drawing pictures trying to figure out who's right.
    Can anyone else weigh in??
     
  15. Tinker Dwight

    Tinker Dwight Registered User

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    #15 Tinker Dwight, Apr 20, 2012
    Last edited: Apr 20, 2012
    Hi
    The damage is done by the inertia of the spring. Two things
    are happening when something at the center of the spring
    lets go ( arbor or click ).
    The spring pushes outwards. This sometimes splits the barrel.
    It is also rotating very fast and comes to a suddent stop
    when it comes up against the barrel.
    Never under estimate rotational inertia. Any engineer that
    has worked with rotating machinery can tell you that
    it doesn't take much to create a lot of force with a suddent stop.
    If the click breaks, the key and the arbor add to the mass but
    they also slow the rotational speed of the spring by their
    intertia. That amount of energy transfered to the barrel is
    the same and slams the gears teeth.
    It might be slightly greater with the key and arbor spinning
    because there is a slowing effect of the spring coils pushing
    the air between turns but I doubt there is enough to make
    much difference.
    If the spring breaks at the outside of the coil, its intertia
    spins the click ratchet and arbor until friction slows it down.
    Sbug's mistake is to think that the spring is stationary. It
    isn't. When released, it spins very fast and comes to a sudden
    stop when against the barrel. The total inertia is essentally
    that of the energy stored in the spring. It is the same amount
    of energy if the arbor and possibly key come along for the ride.
    Conservation of energy says that it has to go someplace.
    Tinker Dwight
     
  16. bangster

    bangster Super Moderator
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    Here's my thoughts.
    The click is what prevents the barrel & gear from spinning uncontrollably. Click fails, barrel buzzsaws, loses teeth, damages other wheels in the train. A while back I had an unpleasant incident. The letdown tool slipped in my hand, and the spring let go all at once. Didn't lose teeth, but did bend the 2nd wheel arbor. I said several dirty words.

    If the spring breaks or disconnects from the arbor, the barrel & gear won't spin. May move a bit, but won't spin, because the spring is no longer attached at both ends. Other stuff will happen, but the barrel won't buzzsaw. The spring will uncoil inside the barrel. It may bulge or split the barrel, but it's unlikely to damage the 2nd wheel.

    I don't think the OP's damaged barrel and 2nd wheel came from a broken or disconnected spring.
     
  17. harold bain

    harold bain Forums Administrator
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    I don't think the barrel can spin uncontrollably until after it breaks a few teeth, as the second wheel pinion locks it in place. When the click breaks, the arbor and the spring get an uncontrolled letdown. The damage is done by the sudden stop at the end of the spring unwinding.
     
  18. Tinker Dwight

    Tinker Dwight Registered User

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    Hi
    Harold has got it.
    It isn't the fall that kills you, it is the sudden
    stop at the bottom.
    Tinker Dwight
     
  19. bangster

    bangster Super Moderator
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    You're right, of course; I misdescribed the situation. Barrel can't buzzsaw unless it goes out of mesh or loses teeth. Not uncontrolled barrel spin; uncontrolled spring-unwind is what the click prevents. Here's what I observed. Bent 2nd arbor. No lost teeth. Happened when I lost control of the letdown tool. Barrel took a helluva jolt when the SPRING released all of its turning power at once. Enuf to bend the adjoining arbor. Not enuf to shear teeth. Fortunately for me.:p

    Doesn't affect my main point: OP's damage couldn't have been caused by a broken spring, which would eliminate turning power. A broken spring cain't turn a barrel. Had to be something similar to a click failure (e.g. my accident).
     
  20. Tinker Dwight

    Tinker Dwight Registered User

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    #20 Tinker Dwight, Apr 20, 2012
    Last edited: Apr 20, 2012
    Hi
    Yes it can! If the spring breaks at the arbor, all of the springs
    energy is transfered to the barrel when it comes to a sudden
    stop against the barrel.
    Since it is making two types of motion at the time, there are
    two effects.
    One is outward. That is the one that can split the barrel.
    The second is rotational. That is the one that can shear
    teeth.
    The effect of a broken click and a spring break at the
    arbor are virtually the same. The rotational inertia of the
    spring is stopped suddenly and it can shear teeth or as
    in your case, bend the arbor of the second wheel.
    The action of the spring stopping against the barrel
    is like hitting the barrel with a hammer. There is both
    radial and rotational hammering.
    All the turns you put on the spring have to go somewhere.
    Where do you suggest that rotational energy goes?
    This is not the same as a break at the outside end
    of the spring. There the energy is released by friction
    of the spinning arbor.
    Tinker Dwight
     
  21. bangster

    bangster Super Moderator
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    I gonna need that explained a little better. I'm no mechanical engineer and don't play one on teevee, so I'm always willing to learn. Here's what I think or thought. If the spring breaks at the arbor, it has no fulcrum to pull against. So it simply unwinds inside itself, rather than trying to turn the barrel. It'll exert outward pressure on the barrel, bulging or splitting the barrel, like an explosion. It won't rotate the barrel. The energy released in the unwinding is expended as impact energy against the walls of the barrel.


    No?
     
  22. shutterbug

    shutterbug Super Moderator
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    Hmmm. I agree with everything you say Tink, except I don't see anything spinning but the spring when it's released from the arbor. I guess we need some more input to clear up this matter. I think it's pretty important to have a clear understanding (even if it means I'm wrong :) )
     
  23. David S

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    I am not sure I am with the rotational inertia part if the spring comes loose at the arbor. We know for sure that the outer coil sure doesn't rotate much since it is hooked to the barrel..expand yes, and there is certainly a sliding / rotation of inner coils as they expand. I would think that most damage would be done if a fully wound spring was being let down and the key slipped while the click was held back..and part way through the uncontrolled let down the click was released, causing a sudden stop. Suddenly the barrel teeth would impact the 2nd pinon. I am not going to test my theory, unless one day it just happens :)
     
  24. Tinker Dwight

    Tinker Dwight Registered User

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    Hi Bang
    Rethink what you just said. You just wound the spring. That is rotational.
    When the spring winds down, it rotates the barrel. In order to fill the space
    that an unwound spring has, it has to not only move outward as you've
    stated but the freed end has to rotate around as well.
    Say it had 14 turns on it when wound. The end has to make up 14 turns.
    It does that really fast ( in less than a tiny fraction of a second ).
    The mass of the moving spring has to be stopped in two directions.
    One as you noted, outward or radially. The other is rotationally. It
    is the rotational energy that is the primary force that drives the clock.
    The average part of the spring travels quite some distance before
    making the abrupt stop. About as much radially as rotationally. When it
    stops, it can break a tooth.
    Conservation of energy says it has to be there.
    Tinker Dwight
     
  25. bangster

    bangster Super Moderator
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    Still don't get it. I think you're saying that when the spring breaks at the arbor, more energy is released in rotation than in expansion. Explain that some more. Maybe I'm dense.

    Suppose when you wind the spring you turn the key clockwise...you wind the spring clockwise. The reason it captures energy is that its outer end is held by the barrel hook, and its inner end is held by the arbor hook, so it can't move. When you wind the spring, its mass isn't rotated. It's squeezed inward toward the center. When it releases its energy, it's not trying to rotate. It's trying to expand outward. When the spring breaks at the arbor, the inner end will rotate in a counterclockwise direction. But except for any friction between the leaves there's no counterclockwise rotational force on the mass of the spring. And if there were, the end result would be that the outer end is pushed off the barrel hook and slides around inside the barrel, rather than trying to rotate the barrel.

    No?
     
  26. David S

    David S Registered User
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    Bang I am sort of in your camp. However as the spring gets to its final diameter it certainly will have a lot of friction agains the barrel, and would try and give it a jerk. I just don't think the mass of the spring and the few turns that it has to uncoil would impart that much damage.
     
  27. harold bain

    harold bain Forums Administrator
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    When you wind the spring, you are rotating the arbor end. And you are storing a lot of energy. When all that energy releases at once, the spring does rotate at the arbor end, with lightning speed. When it abruptly stops, that rotational force will make the barrel want to move, against the next pinion.
     
  28. Bill Stuntz

    Bill Stuntz Technical Admin
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    #28 Bill Stuntz, Apr 20, 2012
    Last edited: Apr 20, 2012
    As a newbie, I'm hesitant to chime in here. But here goes anyway. I'm going to restate some of the things from recent posts in my own words.

    As Harold just said: When you twist the key, that rotational energy is stored in the spring. The fact that the diameter of the inner coils of the spring decreases could be considered as a side-effect of the energy storage. The whole purpose of the spring is to store ROTATIONAL energy, NOT to shrink the middle. The RADIAL energy stored by the shrinkage is small compared to the ROTATIONAL energy.

    When the spring breaks at the arbor, that rotational energy is released almost instantly. And the sudden stop transfers that rotational energy to the wheels - tearing them up.
    Some of the energy would be released as friction heat as the coils slide against each other, some of it radially as the coils bulge out against the barrel, MOST of it goes straight to the wheels.

    DAMMIT - I keep finding typos and thinking of things I SHOULD have said! Sorry about all the edits.
     
  29. Tinker Dwight

    Tinker Dwight Registered User

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    Hi Bang
    Take any atom of the spring that is near the arbor and follow its
    path as the spring unwinds. It does no go radially out it rotates.
    Not all of the spring rotates. As was noted, the end connected to the
    barrel doesn't rotate but even one turn in, it rotates just a small
    amount. Each turn rotates more until you get to the end that is attached
    to the arbor.
    Between wound and unwound, some part of the springs mass has to
    rotate to get from the wound position and the unwound position.
    Here is an experiment you can try, Bang.
    Take one of your open coil spring clocks. Wind it up.
    Mark the coils with a sharpie pen radially with a line.
    Unwind it ( safely ). Note where the line is now.
    As well as being farther out, it also is now spread in a spiral.
    if you follow it inward, you'll see that the center part made
    as many turns as you unwound it. ( You could do the same
    experiment with a barreled spring with the cover cap off. )
    Convince yourself that most all of the spring rotates. The center
    where connected to the arbor rotates the most.
    Tinker Dwight
     
  30. Bill Stuntz

    Bill Stuntz Technical Admin
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    GREAT demo, tink. Wish I'd thought of it!
     
  31. David S

    David S Registered User
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    All well and good, but what really is the inertia? Has anyone done the calculations? Perhaps an experiment is in order if calculations are too difficult.
     
  32. bangster

    bangster Super Moderator
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    Right-o. As long as it's still connected to the arbor. We're supposing it's not.
     
  33. harold bain

    harold bain Forums Administrator
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    I don't really see a difference between connected to the arbor, or broken at the arbor. Should have the same quick forces at work as it unwinds. If it breaks closer to or at the barrel hook, most of that energy is directed towards the arbor, which has a stronger lock (the clickwheel and click) so less likely to cause damage than having the force directed to the greatwheel and pinion teeth.
     
  34. Bill Stuntz

    Bill Stuntz Technical Admin
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    The FACT that it is SUDDENLY not connected to the arbor is what causes the almost instant release of all the energy you stored in the spring by twisting it several times. I'm going to INVENT some numbers here - NO idea how accurate they will be. 10 twists x 20 ft-lbs per twist. Is your arm strong enough to twist the entire 200 ft-lbs in .2 seconds? It took approximately 10 seconds of effort to put in those 10 twists. And it's all released almost instantly. That's what makes springs so dangerous.
     
  35. David S

    David S Registered User
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    Bill I think that mass has to be introduced into this equation.
     
  36. Bill Stuntz

    Bill Stuntz Technical Admin
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    Yes, it does. I know I was oversimplifying, hoping to achieve clarity. I wasn't trying to be 100% accurate. Those springs are heavy compared to the weight of the wheels/teeth/arbors. That's why the WHEELS get trashed.
     
  37. David S

    David S Registered User
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    IMHO I don't think so. I think if one could simulate the dynamics of what happens, the mass of the spring is relatively small wrt to the barrel. Also the uncontrolled release is not just BOOM it is out. but rather as it expands there is friction among coils to slow things down.
     
  38. bangster

    bangster Super Moderator
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    Almost got me convinced, Tink. Thought-experiment: The spring (disconnected from the arbor) is unwinding counterclockwise. As it does, its various parts are all following a counterclockwise spiral path. Suppose, halfway through the unwinding, the spring suddenly fuses into a solid mass of steel. When it does that, it won't sit still. It will no longer being expending spring-energy, but inertial energy of the formerly moving parts will cause the entire blob to rotate counter-clockwise for a while. But not because it's still releasing energy from the unwinding spring, because that's no longer getting released, and the remaining stored energy is locked up in the blob as internal stresses. It ceases to be an uncoiling spring, and becomes a flywheel that will eventually run down.

    Maybe so; sounds pretty plausible. So the spring is/was rotating after all. But when the transformation happens, the blob isn't going to yank on the barrel hook and cause the barrel to take off disastrously. It's going in the wrong direction to do that. Instead, it will yank itself OFF the barrel hook, and will flywheel around inside the barrel. The only impetus it gives to the barrel will be from friction, and that doesn't sound to me like enough to strip gears.

    No?
     
  39. Bill Stuntz

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    As I see it, the weight of the (barrel + spring) relative to the wheel is the big problem. I'm being dragged off (kicking & screaming) to dinner. Back in a while.
     
  40. David S

    David S Registered User
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    Bill I think it is the mass of the spring relative to the barrel. Enjoy dinner and thank your bride for being so patience.
     
  41. Tinker Dwight

    Tinker Dwight Registered User

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    Hi
    As I said earlier, never underestimate the rotational inertia. Years ago,
    I helped design a control system for an XY table. The table weighed about
    60 Lbs. It had to ramp up in about 1/4 in to 10 inches per second.
    I needed to calculate how large a motor to do this.
    First, I needed to calculate how big the lead screw and bearing needed
    to be to withstand that much trust. ( not saying I could do that today without
    digging out my physics book ).
    Now, I had to add the inertia of the rotating lead screw and the motors
    armature to determine what the final torque of the motor would be.
    To my surprise, the lead screw ( about 3/8 inches in diameter ) was
    4 times the inertia of the table. When I added the motor it was
    12 times total.
    Almost all my energy was going into rotating things and only a tiny
    part into the movement of the table.
    I've since then had a different respect for any rotating mass. Even
    if one says that only 1/3 of the mass of the spring is rotation when
    released, at the speed that it is going I find it no thought at all that it is
    enough to shear teeth off.
    I don't see what the arbor being attached make any difference other
    than slowing the spring down a little but adding to the rotating mass.
    The total energy released in rotational energy is the same, that
    that was stored in the spring.
    Tinker Dwight
     
  42. shutterbug

    shutterbug Super Moderator
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    Suppose I took the barrel out of the clock, and wound it up. Now, I set it on the floor like a wheel, and release the key. I'm betting I'll get a lot of key rotation, and zero forward motion of the wheel. Course of least resistance. If I got even a little forward motion, I'm thinking in a clock situation the pinion of the next wheel would be more than adequate to arrest that motion without breaking. If Tinker's theory is correct, I should see a lot of forward motion (power being exerted). I just don't think it's going to happen. One of these days I might try that little experiment and see :)
     
  43. Tinker Dwight

    Tinker Dwight Registered User

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    Hi Bang
    It would do that if it could. Try to release the spring from
    the hook on the barrel of an unwound spring by rotating
    the spring backwards. I bet you can't do it.
    Remember the spring has been flattened to the barrel
    before the final smack of the inertia of the spring.
    Tinker Dwight
     
  44. Tinker Dwight

    Tinker Dwight Registered User

    Oct 11, 2010
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    If you release both the barrel and the spring at the same time,
    it won't rotate at all ( conservation of momentum ).
    If you hold the barrel with just enough force to keep it
    from rotating in the direction of the winding, and then
    release the key, it will jump in the opposite direction.
    This is opposite to the direction of the power.
    The trust in the normal powered direction will never
    be more that the winding force.
    Tinker Dwight
     
  45. Bill Stuntz

    Bill Stuntz Technical Admin
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    #45 Bill Stuntz, Apr 20, 2012
    Last edited: Apr 20, 2012
    David: Not exactly a blushing bride any more. We've celebrated 6 Anniversaries. And 18 UNniversaries (NONiversaries?) in between them. We were married on leap year day - 1988. She's used to me by now.

    Back to the subject: The rotational energy is transferred from the spring into (spring+barrel) and then the (spring+barrel+wheels) essentially unchanged except for frictional losses. The energy being transferred into the larger mass decreases the velocity/acceleration, but the amount of energy (LOTS!) is NOT lost. The energy has to ultimately go SOMEWHERE. That somewhere may be frictional heat generated by bending arbors and wheels, flying wheel teeth, and I don't know where else. But it certainly ends up somewhere we DON'T want it, instead of in the spring where it belongs, and being gradually released to drive the movement.

    At one time, I could have shown you the equations, but not any more. Those are some of the details that (as I have mentioned before) I don't even try to memorize. But I do understand WHY, and given enough time could probably reproduce if I HAD to.
     
  46. bangster

    bangster Super Moderator
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    Not convinced yet. You say the spring has a ton of counterclockwise force at the point where it's fully unwound. And you say that's enough force to shear teeth, but not enough force to pull the outer end loose from the barrel hook? Come on.

    At the point where it's fully unwound it has no rigid connection to the barrel. Just the traction of the outer coil of the spring against the inside of the barrel. At that point it can either grip the barrel and shear off teeth, or slip inside the barrel. My money is on slip.

    You say the connection to the arbor doesn't make any difference? The effect is the same? I'm not convinced. The arbor provides the fixed point —the fulcrum— for the spring to hang onto as it tries to move the moving part —the barrel. Archimedes didn't say "Give me a long enough lever and I'll move the earth without needing a fulcrum." When the spring is intact, its inner end is gripping the arbor as it unwinds, so its outer end can push on the barrel. Remove the arbor and it's no longer pushing on the barrel; it's just unwinding inside the barrel.

    Suppose we invent a way to have a retractable arbor hook. We wind the spring up and then retract the hook. According to your theory, the clock should keep running until the spring runs down, because the barrel will keep turning, regulated by the escapement.. I don't think so. Any movement induced by the unwinding spring will be counterclockwise, and the clock will either start running backwards, or stop altogether.

    But I'm still listening.
     
  47. Bill Stuntz

    Bill Stuntz Technical Admin
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    #47 Bill Stuntz, Apr 20, 2012
    Last edited: Apr 20, 2012
    Bang: I don't know if you will have seen my last reply, since we were both typing at the same time and your reply came out on the next page. If not, please look.
    p.s. I don't think my ST124 would behave the same way, since the spring boxes are riveted to the plate instead of the great wheel as part of the barrel. I'll have to think about that.
    p.p.s. I don't think my ST124 would be vulnerable to the the same type of damage, since the only place for the energy to go is into the plate/box. It might rip the box off the plate or explode the box and damage things because of that, but shouldn't kill the wheels. Unless the shrapnel does it. Breaking at the winding arbor completely disconnects the spring from the vulnerable parts of the movement.
     
  48. bangster

    bangster Super Moderator
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    Oh my no. Mainsprings in cans fixed to the plate are totally different from barreled mainsprings. In your clock, the great wheel is attached to the arbor, rather than to the can.
    Sever the attachment to the arbor, and the great wheel will just sit there while the spring uncoils inside the can.
     
  49. Bill Stuntz

    Bill Stuntz Technical Admin
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    #49 Bill Stuntz, Apr 20, 2012
    Last edited: Apr 20, 2012
    Right, that's what I "saw" when i visualized it. But does visualizing the great wheel attached to the can show you what would happen to the delicate things above & linked to the can when the can/wheel jerks as hard as it would when the spring "hits bottom"? It's not the fall that kills you, it's the sudden stop at the bottom. Or I suppose I should say the "sudden start" of the barrel when the spring hits bottom in the can. Momentarily free as the spring breaks loose from the arbor, and BANG when it hits bottom. Since the can is somewhat restrained by the train while the spring is free after it breaks loose, it forces the momentum to remain in the spring until it hits bottom in the can.

    In mine, what happens when it "hits bottom"? It rips the can loose or something, instead of tearing up the train. As a newbie, I've never seen either one happen. But I CAN visualize the result of each.
     
  50. shutterbug

    shutterbug Super Moderator
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    I guess I'll just have to wait until one comes in for repair. I'm not going to be convinced that a broken spring at the arbor is going to damage the clock until I see it happen. I know it will damage the barrel - I've seen that several times. In the mean time, if any of you guys can supply a pictorial demonstration it would be a real help :)
     
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