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# Thread: Drive train tooth count

1. ## Drive train tooth count

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. ## Re: Drive train tooth count (By: kosmos)

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

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

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. ## Re: Drive train tooth count (By: Allan Wolff)

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

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. ## Re: Drive train tooth count (By: kosmos)

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. ## Re: Drive train tooth count (By: jhe.1973)

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. ## Re: Drive train tooth count (By: kosmos)

Originally Posted by kosmos
...................so I reached out to the forum here and have been rewarded with a lot of great information.......... 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..............

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? .....................
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:

Originally Posted by kosmos
...................so I reached out to the forum here and have been rewarded with a lot of great information.............
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!

10. ## Re: Drive train tooth count (By: jhe.1973)

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. ## Re: Drive train tooth count (By: kosmos)

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

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

Originally Posted by Tinker Dwight
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
Doesn't this just shift the drive force to the second wheel arbor. Newton's third law!

Phil

14. ## Re: Drive train tooth count (By: Phil Burman)

Originally Posted by Phil Burman
Doesn't this just shift the drive force to the second wheel arbor. Newton's third law!

Phil
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. ## Re: Drive train tooth count (By: Tinker Dwight)

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

Originally Posted by Tinker Dwight
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

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