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  1. #16
    Registered User Jim DuBois's Avatar
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    Default Re: Drive train tooth count (By: Phil Burman)

    The power available to a gear train remains the same no matter how the gear train is assembled, straight line, step train, tightly clumped together, or strung out in what ever form one might place them and still engage mating wheels and pinions properly. The same remains for the power consumed by the train. The wheels and pinions care not where they are planted as long as they are properly depth-ed, one to another. They consume the same power and deliver the same power to the tips of the escape wheel teeth, then to the verge, and then to the pendulum.... there is no increase in torque or loss of torque in a train by relative position changes one wheel assembly to another.

  2. #17
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    Default Re: Drive train tooth count (By: Jim DuBois)

    Quote Originally Posted by Jim DuBois View Post
    The power available to a gear train remains the same no matter how the gear train is assembled, straight line, step train, tightly clumped together, or strung out in what ever form one might place them and still engage mating wheels and pinions properly. The same remains for the power consumed by the train. The wheels and pinions care not where they are planted as long as they are properly depth-ed, one to another. They consume the same power and deliver the same power to the tips of the escape wheel teeth, then to the verge, and then to the pendulum.... there is no increase in torque or loss of torque in a train by relative position changes one wheel assembly to another.
    Hi Jim, I agree with what you say but increase or loss of torque is not the subject being discussed. What is being discussed is the modification of forces on a pivot due to changes in train configuration. Tinker rightly pointed out that it is possible to remove this reaction force from the great wheel pivot by judicious arrangement of the great wheel/second wheel geometry. My argument is that it just transfers the load somewhere else and that you should keep the pivot reaction force related to the driving weight at the great wheel, as this pivot is best able to resist the potential for wear due to its size.

    Phil

  3. #18
    Registered User Jim DuBois's Avatar
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    Default Re: Drive train tooth count (By: Phil Burman)

    Quote Originally Posted by Phil Burman View Post
    Hi Jim, I agree with what you say but increase or loss of torque is not the subject being discussed. What is being discussed is the modification of forces on a pivot due to changes in train configuration. Tinker rightly pointed out that it is possible to remove this reaction force from the great wheel pivot by judicious arrangement of the great wheel/second wheel geometry. My argument is that it just transfers the load somewhere else and that you should keep the pivot reaction force related to the driving weight at the great wheel, as this pivot is best able to resist the potential for wear due to its size.
    Phil
    No matter the term you chose to use to describe the forces at play in a clock movement we are not creating more force, power, torque, reaction force, or whatever than we started with. And it all ends up being consumed in one manner or another. Where a 2nd arbor is positioned around a great wheel does not consume more or less force/power/energy in a properly constructed gear train, certainly a 15% improvement claim should cause a bit of interest....

  4. #19
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    Default Re: Drive train tooth count (By: Jim DuBois)

    First, as Jim points out, the torques are all that are transferred from
    gear to gear.
    The idea that one can unload the pivot by aligning the angle of the
    points of the torques should not be considered levitation. Think about
    a stick. Put a fulcrum in the middle.
    Now put a weight on one end and push down on the other end.
    How much force is on the fulcrum? ( no need to calculate it is
    twice the weight, plus the stick ).
    Now go to the end with the weight.
    Lift it up.
    Now, how much weight is on the fulcrum?
    Was there any levitation there? Did it violate any of Newton's laws?
    It was as though the weight wasn't there, to the fulcrum. You are lifting the weight.
    So, one can see that if you go on opposite sides of the lever, the
    fulcrum sees the most weight. If you are on the same side, the fulcrum
    sees the least weight. No magic, it is just where things are.
    Now instead of a lever arm, we have a gear wheel. Instead of a fulcrum
    we have a pivot and bushing.
    The rules are that same. The closer the two forces are around the gear
    wheel, the less load on the pivot. Still no magic. Not forces have magically
    gone away. It is just how much leverage you are using and where the
    fulcrum is relative to the forces.
    As for the 15%, I assumed a gear ratio of 4:1 for the second wheel to pinion
    and put the forces at 90 degrees.
    Do you still need a force drawing?
    Tinker Dwight

  5. #20
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    Default Re: Drive train tooth count (By: Jim DuBois)

    Quote Originally Posted by Jim DuBois View Post
    No matter the term you chose to use to describe the forces at play in a clock movement we are not creating more force, power, torque, reaction force, or whatever than we started with. And it all ends up being consumed in one manner or another. Where a 2nd arbor is positioned around a great wheel does not consume more or less force/power/energy in a properly constructed gear train, certainly a 15% improvement claim should cause a bit of interest....
    You cannot equate force to power or energy. A force is not consumed, it either exists or doesn't exist. A gear train is nothing more than a set of levers. It is the nature of levers that they increase or decrease the applied force and in so doing increase or decrease the reaction force at the pivot point, this is the essence of their usefulness. You can change the reaction forces on the pivot points of those levers by changing where you apply the driving force. I would recommend that you read up on force vector diagrams.

    Phil

  6. #21
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    Default Re: Drive train tooth count (By: Phil Burman)

    Quote Originally Posted by Phil Burman View Post
    You cannot equate force to power or energy. A force is not consumed, it either exists or doesn't exist. A gear train is nothing more than a set of levers. It is the nature of levers that they increase or decrease the applied force and in so doing increase or decrease the reaction force at the pivot point, this is the essence of their usefulness. You can change the reaction forces on the pivot points of those levers by changing where you apply the driving force. I would recommend that you read up on force vector diagrams.

    Phil
    Jim understands force vectors and is not confusing. He is right, a gear can
    only transfer toque. A force at a distance. No additional torque was transferred to
    the second wheel and no levitation.
    Tinker Dwight
    Last edited by Tinker Dwight; 02-21-2017 at 10:01 AM.

  7. #22
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    Default Re: Drive train tooth count (By: Tinker Dwight)

    Quote Originally Posted by Tinker Dwight View Post
    First, as Jim points out, the torques are all that are transferred from
    gear to gear.
    The idea that one can unload the pivot by aligning the angle of the
    points of the torques should not be considered levitation. Think about
    a stick. Put a fulcrum in the middle.
    Now put a weight on one end and push down on the other end.
    How much force is on the fulcrum? ( no need to calculate it is
    twice the weight, plus the stick ).
    Now go to the end with the weight.
    Lift it up.
    Now, how much weight is on the fulcrum?
    Was there any levitation there? Did it violate any of Newton's laws?
    It was as though the weight wasn't there, to the fulcrum. You are lifting the weight.
    So, one can see that if you go on opposite sides of the lever, the
    fulcrum sees the most weight. If you are on the same side, the fulcrum
    sees the least weight. No magic, it is just where things are.
    Now instead of a lever arm, we have a gear wheel. Instead of a fulcrum
    we have a pivot and bushing.
    The rules are that same. The closer the two forces are around the gear
    wheel, the less load on the pivot. Still no magic. Not forces have magically
    gone away. It is just how much leverage you are using and where the
    fulcrum is relative to the forces.
    As for the 15%, I assumed a gear ratio of 4:1 for the second wheel to pinion
    and put the forces at 90 degrees.
    Do you still need a force drawing?
    Tinker Dwight
    Yes as I previously said I agree that the reaction at the first wheel can be unloaded but you have to consider what is the change in reaction at the second pivot. Surely if the ratio is 4:1 then the distribution of the drive weight is 80% on the second arbour and still with 20% on the first arbour. Also if you put the drum diameter at the same as the first wheel diameter you may just as well have put the drum on the second wheel arbour, with a diameter equal to the second wheel pinion diameter. I think a force vector diagram would help prevent misunderstanding and bring this thread to a quicker conclusion.

    Phil

  8. #23
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    Default Re: Drive train tooth count (By: Tinker Dwight)

    Quote Originally Posted by Tinker Dwight View Post
    Jim understands force vectors and is not confusing. He is right, a gear can
    only transfer toque. A force at a distance. No additional torque was transferred to
    the second wheel and no levitation.
    Tinker Dwight
    Yes and I did agree with Jim on this point. Post #17 "Hi Jim, I agree with what you say" I'm not sure where you are trying to go with is.

    If there is no reaction to the force generating the torque then yes the wheel will fly away. The reaction to the force creating the toque is at the pivot and is equal and opposite to the force generating the torque.

    Phil

  9. #24
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    Default Re: Drive train tooth count (By: Tinker Dwight)

    Quote Originally Posted by Tinker Dwight View Post
    a gear can only transfer toque.
    Tinker Dwight
    As you have pointed out a gear is a lever, levers transfer forces and by mechanical advantage can either increase or degrease the force transferred.

    Phil

  10. #25
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    Default Re: Drive train tooth count (By: Phil Burman)

    As with the lever example, the worst load on the fulcrum is when
    both loads are on opposite sides. The least when they are on the same
    side. 90 degrees is someplace in the middle. My calculation may be in
    error but you will have to agree that it will be reduced by having the
    two at 90 degrees instead of aligned on opposite sides of the pivot
    as shown by the simple lever example.
    It will take me about a week to do the drawing as I'm leaving on a trip.
    Tinker Dwight

  11. #26
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    Default Re: Drive train tooth count (By: Tinker Dwight)

    Quote Originally Posted by Tinker Dwight View Post
    As with the lever example, the worst load on the fulcrum is when
    both loads are on opposite sides. The least when they are on the same
    side. 90 degrees is someplace in the middle. My calculation may be in
    error but you will have to agree that it will be reduced by having the
    two at 90 degrees instead of aligned on opposite sides of the pivot
    as shown by the simple lever example.
    It will take me about a week to do the drawing as I'm leaving on a trip.
    Tinker Dwight
    No problem with the drawing, enjoy your trip.

    I can't seem to persuade you to look at what happens to the reaction force at the second pivot so I guess it's best to just leave it where it is.

    Phil

  12. #27
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    Default Re: Drive train tooth count (By: Phil Burman)

    In the unlikely event that anybody is interested I checked the relative forces for the two scenarios, both with the second wheel offset horizontally from the first wheel, a ratio of 4:1 between the first wheel and the second wheel pinion, and the drive drum diameter equal to the pitch circle diameter of the first wheel. Any effects from the third wheel pinion are not considered.
    The results are interesting in that they show that it is possible to significantly modify the reaction load seen by the first wheel pivot by adjusting the configuration. It was something of a surprise to see that the reaction load was reduced to only 20% of the driving weight (W) from 180% of W, without a change in the reaction load at the second wheel pivot.
    The example configuration is possibly somewhat extreme and may present other problems, not least of which - everything may be rotating in the “wrong” direction. However it does demonstrate that you should not ignore the possibility if you wish to minimise pivot friction and wear and even reduce the power required to drive the movement.

    Scenario 1 - Weight cord off the drum on the same side as second wheel pinion.
    This results in:
    A reaction force at the first wheel pivot 20% of the driving weight (W) and acting up.
    And a reaction force at the second wheel pivot 80% of W and also acting up.

    Scenario 2 - Weight cord off the drum on the far side from second wheel pinion.
    This results in:
    A reaction force at the first wheel pivot 180% of the driving weight (W) and acting up.
    And a reaction force at the second wheel pivot 80% of W and acting down.

    Phil

  13. #28

    Default Re: Drive train tooth count (By: Phil Burman)

    Hi Phil, I am following this discussion with much interest!

    My assumption, back in comment 12, that there could be a "balancing", so to speak, of force was entirely intuitive, so it has been quite an exciting trip down the rabbit hole of engineering and clockmaking for me. And I'm going to have to ask you for some help...

    If I read your scenarios correctly you are showing what I understood from Tinker in comment 11.
    Would it be possible for you to make a drawing for me?
    And it will be interesting to see what Tinker is going to come up with as well. (when he gets back from vacation!)
    Thanks for weighing in on this!
    Stephen

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    Default Re: Drive train tooth count (By: kosmos)

    Quote Originally Posted by kosmos View Post
    Hi Phil, I am following this discussion with much interest!

    My assumption, back in comment 12, that there could be a "balancing", so to speak, of force was entirely intuitive, so it has been quite an exciting trip down the rabbit hole of engineering and clockmaking for me. And I'm going to have to ask you for some help...

    If I read your scenarios correctly you are showing what I understood from Tinker in comment 11.
    Would it be possible for you to make a drawing for me?
    And it will be interesting to see what Tinker is going to come up with as well. (when he gets back from vacation!)
    Thanks for weighing in on this!
    Stephen
    Hi Stephen, yes it looks like Tinker was correct regarding the load on the second wheel arbour remaining the same but not correct that the load on the first wheel pinion will be almost zero. Not unless the gear ratio approaches infinity. Also it is probable that you will be able to have a drum diameter equal to the first wheel diameter due to other practicalities.

    It is not clear what you would like a drawing of.

    Phil

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    Default Re: Drive train tooth count (By: Phil Burman)

    Hi Phil
    While one may not be able to match the diameters, it still significantly
    reduce the load.
    Still, the second wheel seems to be the issue.
    First the torque on the second wheel doesn't change based on where
    around the main wheel it is connected.
    I'm using a 4:1 second wheel ( not that uncommon 10 teeth to 40 teeth ).
    Lets look at what happens in a typical fine regulator with the wheels
    lined up vertically. Say the main wheel rotates counter clock wise.
    I will scale the torque to run the clock for calculation convenience.
    Lets say Main wheel pushes with 4 torque forces on the pinion to left.
    This would cause the wheel to want to turn clockwise.
    At the top of the wheel, the third pinion would need to push back with
    1 torque force on the wheel ( 1:4 loss of torque because of the lever arm ).
    Both the wheel third pinion and the main wheel push parallel. The sum of their
    forces needs to be taken up by the second wheel pivots of 4 + 1 = 5.
    Now lets look at what happens when the third wheel is at 90 degrees instead
    of 180 degrees.
    ( my earlier math was wrong but I have it correct as I was more careful ).
    I place the second wheel at 90 degrees to the left.
    The torques are still the same since that is what is needed to run the clock.
    Now, I have 4 torque forces pushing down on the pinion of the second wheel
    and 1 torque force pushing to the left from the top of the second wheel.
    This is a vector force down to the left of a 1 to 4 triangle.
    The vector force is SqRt( 1^2 + 4^2 ) = 4.123
    4.123 / 5 = 0.825 or a reduction of 17.5%.
    A win-win.
    While the main wheel and the drum diameter may not be desired
    to be matched. What ever ratio you might choose, the minimum
    thrust on the main wheels pivot is still at the minimum when both
    the second pinion and the cord pull down from the same side.
    One can easily draw ones own vector diagram from this description.
    Tinker Dwight

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