Drive train tooth count

Discussion in 'Clock Construction' started by kosmos, Feb 11, 2017.

1. kosmos Registered User

Feb 11, 2017
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Hello everyone,
I have stumbled all over the internet looking for the answer to what may turn out to be a simple question.

On an eight day weight driven clock with a one second pendulum I would like to create, the diameter and tooth count, of the great wheel, which carries the chain/cable and the intermediate wheel that would drive the center wheel which carries the minute hand. I can't seem to understand what goal I seek as far as the pitch diameter/tooth ratio is concerned.
The computations for the diameter and tooth ratios from the center wheel to the escape wheel follow a calculation with (in this case) 3600 bph as the goal.
I try to apply the same logic to the the train of gears I mention above but I can't seem to figure out what goal that this ratio would resolve itself to.
Yikes! hope I'm making some sense.

I am still working with paper and pencil on this project so I can easily erase and or change anything!
Thanks in advance to all who considers this question for me, I appreciate your time
Stephen

2. Tinker Dwight Registered User

Oct 11, 2010
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#2
At the lower end, you want pinions with 8 or more teeth.
At the escapement 6 is a good number.
What are you planning to drive it with?
Tinker Dwight

3. kosmos Registered User

Feb 11, 2017
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Hi Tinker,
thanks for your response. I want to drive the clock with a weight.
Right now I have(on paper) a 3 inch diameter spool mounted to a great wheel that will unwind a cable at 1 1/2 revolutions per day or 12 revolutions in 192 hours (dropping the weight 36 inches).

If I drive a 10 leaf pinion on the center wheel with this great wheel I need 160 teeth on the great wheel.
I would like to have an intermediate wheel, one between the great wheel and the center wheel.
This is the question I tried to ask in my original post.

The thing is I may be approaching this question wrongly and confusing all of you! I actually just repaired a french marble clock that has such a gear train, although it is powered with a mainspring. I could have just counted those teeth. But it's the theory I'm after. I want a greater understanding of it so I can apply it myself.

Stephen

4. Allan Wolff Super Moderator NAWCC Member

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Stephen,
You are on the right track. Tooth ratios from the center wheel to the escape wheel are mostly determined by the beat rate of the clock with the goal being to have the minute hand turn clockwise once per hour.
The two goals of the lower part of the train from the main wheel to the center wheel are to provide enough power to run the clock and the duration the clock will run before it needs to be rewound. You will be working with a number of variables which are tooth count of the main wheel and center pinion, diameter of the cable drum and distance available for the weight to fall. They all affect each other. For example, a lower tooth ratio will deliver more power, but the drum will turn faster and the weight will drop farther. So lowering the tooth count will allow you to make the drum smaller or the weight lighter, but you will need a taller case to run the same duration. Making the drum diameter larger will provide more power, but the weight will drop farther. Making the weight heavier will allow the drum to be smaller or the tooth ratio higher. There are many combinations to get where you need to be.

Allan

5. Tinker Dwight Registered User

Oct 11, 2010
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#5
An intermediate wheel that works as an idler is mainly there to change
rotation or offset. The number of teeth used is not particularly important
because an idler doesn't change ratios. One looks at the distance involved
and determines a practical count to fit. 24 teeth is a common size.
More teeth have slightly less friction.
It is best to put them in line as the wear will mostly go along
the wheels. Even that is not a big concern as the load it so little to turn
hands.
Wheels in the train normally want to get smaller in diameter as they
go towards the escapement so the teeth clear the arbor of the previous
wheel. There are ways around this of course.
Tinker Dwight

6. kosmos Registered User

Feb 11, 2017
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Thank you Allan. I had not considered the size of the drum, the weight, and the tooth ratio in that respect. Your info sure does add to the overall considerations!
I see what you are getting at now Tinker, Thanks! I did discover the wheel sizes getting too large and interfering in some of my earlier drawings.
Thanks to all of you who have considered this.
If anything occurs to anyone please write!

7. jhe.1973 Registered User DonorNAWCC Member

Feb 12, 2011
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#7
Hello Kosmos,

Welcome to the forum. I hope you find it worth your time (pun intended).

Here is a sketch I made years ago when I was building a weight driven regulator:

Once you have arrived at the correct gearing between the minute hand and escape wheel, the great wheel/winding drum sizing has no effect on the movement other than calculating the weight drop for the intended case.

To make fewer gear cutters necessary, I made all the wheels the same 120 tooth size. I only had two pinion sizes that, for practical purposes, could have been cut with one cutter.

You might find the thread that I started about my regulator covers many questions that you are certain to have:

Hope to 'see' you on the forum often!

8. kosmos Registered User

Feb 11, 2017
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Jim,
Thank you so much for pointing me to your regulator thread! Wow, is about all I can say right now, what a project!
And thanks for the clarification on the gearing below the minute wheel, i.e. the drum, great wheel, etc.
When that question crossed my plans I realized it could be a function of the weight drop but it seemed like too simple of an answer, so I reached out to the forum here and have been rewarded with a lot of great information.

It is true that your thread, a well written and detailed article has answered a lot of questions, and it seems it has generated a few too!
For instance, if you don't mind taking the time to explain, The endstones on the great wheel ,, they have apparently (if I read it right) contribute to a significant frictional loss. I don't understand how this is.

I should explain that I come from the watchmaking world where an endstone functions as a bearing point for pivots on arbors standing upright. As you probably know, this contributes to a much more uniform frictional condition as a watch changes position. Sorry to overstate the obvious, but a clock doesn't change positions, hence my question.

You may have addressed this in your article and I probably missed it, so I apologize ahead of time.

Stephen
PS: your sketch that you attached above is not clickable. The problem might be on my end. But can you revisit that and repost it?
thx.
S.

9. jhe.1973 Registered User DonorNAWCC Member

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#9
Last edited: Feb 18, 2017
Hi Stephen,

Besides consistency during position changes another advantage of an endstone is to limit end shake at the center of rotation of the arbor where it has considerably less surface area than a shoulder around the pivot. Plus, the friction all the way around a shoulder, being farther from the center, has a greater 'leverage' further increasing its effect (power loss).

Although a clock movement doesn't change position, certain misalignments or gearing irregularities can cause a side thrust on an arbor. It seems to me that if this happens at the great wheel, the friction loss would be greater considering this is where the largest driving force exists in the gear train. Because of all that I have learned since my first attempt at clock making, I do know that my gearing is way less than ideal. This may have given me a larger problem at the great wheel than better quality gearing would have had.

I'll try this and see if it works:

You certainly have that right, there are a bunch of really great contributors to this forum who are always willing, even eager, to help.

Glad to be able to help out!

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10. kosmos Registered User

Feb 11, 2017
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Yes of course! I have even seen the offset gouge from a pivot shoulder that is being driven into a plate, an endstone is a great answer to that force. I don't have as much experience with clocks as I do with watches, but over the years I have repaired a few and I have never seen or heard of that application before. Learn something new every day!
And I think you are absolutely correct about being concerned with the great wheel having to deal with the toughest friction. The slow revolution plus the pressure, I was always a bit worried about the lubricant there, wondering if it wasn't getting squeezed out or something.
Speaking of good answers, I also like your solution of the winding of the drum, simple and to the point.

Thanks for the image, it came through, it will definitely help me.

So, I am going to continue to try to get a handle on gear ratio/tooth count and, as Allan Wolff also mentioned above in a post, the influence of the weight, the tooth count, size of the drum, and the distance of the drop etc.
Thanks again
Stephen

11. Tinker Dwight Registered User

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#11
If you put the second wheel off at almost 90 degrees, on the same side as the cable,
it is possible to balance the entire thing so that there is almost zero load
on the first wheel arbor. If the diameter of the drum matches the first
wheel it is even less. It is a different arrangement of the wheels.
The second wheel is the support but no more than it normally takes.
Tinker Dwight

12. kosmos Registered User

Feb 11, 2017
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Thanks Tinker, I will put this idea into the mix.

I see what you are getting at, balancing the great wheel between the cable force on one side and the rest of the train resistance on the other, nice!.
I like the way a problem can be solved by something as thoughtful and simple as this.

13. Phil Burman Registered User

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Doesn't this just shift the drive force to the second wheel arbor. Newton's third law!

Phil

14. Tinker Dwight Registered User

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#14
Last edited: Feb 20, 2017
Actually, if you do the vector analysis, it reduces the load on the
second wheel pivot if it takes the load from the rest of the train and
puts it at 90 degrees from the load of the main wheel. It isn't a lot
but still a little less.
Remember, all the load on the second wheel is torque. If the two are
aligned straight up, the force from the rest of the train and the main
wheel are parallel and add linearly. There is no increase in torque
on the second wheel because of where around the main wheel it is
being pulled on. In the case of the second wheel being off to the side,
that vector is a 90 degrees and is the vector sum on the pivot.
I can make some hand drawings if you need it.
I just did the calculation, it reduces the load on the second bushings
by 15%, assuming equal load on both ends.
Tinker Dwight

15. Phil Burman Registered User

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Hi Tinker, a quick sketch would be most helpful toward my understanding. Also I'm not sure what you mean by "15%, assuming equal load on both ends." both ends of what.

I see that it is possible to move the load, due to the driving weight, around within the train. But I don't see how you can eliminate it, it surely must turn up somewhere else on the pivots within the train. Every force and equal and opposite reaction and all that Newton stuff. Otherwise you have levitation.

Phil

16. Jim DuBois Registered User DonorNAWCC Member

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#16
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.

17. Phil Burman Registered User

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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

18. Jim DuBois Registered User DonorNAWCC Member

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#18
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....

19. Tinker Dwight Registered User

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#19
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

20. Phil Burman Registered User

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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

21. Tinker Dwight Registered User

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#21
Last edited: Feb 21, 2017
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

22. Phil Burman Registered User

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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

23. Phil Burman Registered User

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#23
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

24. Phil Burman Registered User

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#24
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

25. Tinker Dwight Registered User

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#25
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

26. Phil Burman Registered User

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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

27. Phil Burman Registered User

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#27
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 â€œwrongâ€ 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

28. kosmos Registered User

Feb 11, 2017
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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

29. Phil Burman Registered User

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#29
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

30. Tinker Dwight Registered User

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#30
Hi Phil
While one may not be able to match the diameters, it still significantly
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

31. John MacArthur Registered User NAWCC Member

#31
What Tinker is driving at is known in chronometer circles as "reverse fusee". The chain from the barrel crosses over to the side of the fusee nearest the next pinion, instead of to the side opposite. As the chain unwinds to the larger end of the fusee, more and more of the side load on the pivots gets shared between the two arbors. This was known early on - something like 200 years ago.

Johnny

32. Phil Burman Registered User

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Hi Tinker, how is the first day of your holiday.

Thanks for the post, I did do the vector diagram and came to your above conclusion in post #27. I'm grateful for your effort and the addition to my understanding. It's another factor to keep in mind when juggling all the parameters that influence train layout.:bang:

Phil

33. Tinker Dwight Registered User

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#33
Hi Phil
It is always good to confirm my math. Even with a calculator and
knowing what buttons to push, I find that solving the problem in two
different ways helps to get the right solution.
I did it once with the sum of squares and then did it again the inverse tangent and
then cosine.
I got the same answer twice. Often a good chance I got it right.
I don't know where I got the 15% number and was not able to recreate it.
We are having a good trip so far.
I did back up, hit a tree and bent the rear bumper.
Tinker Dwight

34. jhe.1973 Registered User DonorNAWCC Member

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#34
You just reminded me of when I did the same thing but might have gone one step farther.

About 45 years ago I had a friend with me and he needed to make a phone call. Soooo I pulled into a large shopping center parking lot & dropped him off at a phone booth (remember them?). I decided to back the car around while he was making his call and the car was stopped with a CRASH.

There were no cars in that entire area of the lot and I managed to hit the concrete footing for a huge, tall parking lot light. I destroyed the rear bumper on the ONLY thing that I could have hit other than the phone booth. At least I remembered where that was!

I think we do need instructions!

35. jhe.1973 Registered User DonorNAWCC Member

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#35
This just jumped out at me while reviewing this thread and I believe that this should be expanded on a bit.

When clock movements become worn, it is the pinion end of the arbor that has the most wear because the wheel it meshes with is trying to push it sideways. For this reason many designers of higher quality clocks try to keep the pinions and wheels close to each other and on one side of the movement. Having the pinion farther away from the pivot allows the arbor to have greater leverage to resist this sideways force.

Before I learned this, I often wondered why regulators often have plates spaced so far apart.

36. Tinker Dwight Registered User

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#36
Yes, the pinion pivot end has the most load. Usually the second wheel is not only a high load
but has the most unbalanced load in the clock. This means if any location should have
a more robust pivot, it is likely the pinion end of the second wheel. Even on a cable with
the cable and second wheel pivot on opposite sides, the second wheel pivot end has almost
the same load as the main wheel that is ( on average ) more evenly distributed.
The second wheel being way off center sees the almost the entire load.
Tinker Dwight

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