Spring length question

[Dave D]

What would happen if you put a mainspring in a barrel that was too long? Would you not have enough space to actually wind it all up and therefore, actually shorten the running time?
Dave D

 

[shutterbug]

How much length depends on how much thickness too. If you reduce thickness, you can increase length. You need only enough length to run the clock for 8 days. Any more length than necessarily will fill the barrel too much and cause problems. Any more thickness than necessary will over power the train and speed up wear. Somewhere on here is a length calculator.

 

[Willie X]

Short answer, yes. But, one turn (maybe 4 or 5 inches) won't make any noticable difference in the clocks performance.
Willie X

 

[R. Croswell]

What would happen if you put a mainspring in a barrel that was too long? Would you not have enough space to actually wind it all up and therefore, actually shorten the running time?
Dave D

For all practical purposes, if the mainspring is too long you won't be able to get it in the can without great difficulty. Don't forget that a wound spring and an unwound spring are the same size and take up the same amount of space. If the original spring is not available and the original length is unknown, use one of the spring calculators to get an approximation for what to use. Easy to cut hair off, very hard to cut hair back on.

RC

 

[Mike Phelan]

If the correct spring is in a barrel, it will run for the correct length of time, usually 4 or 5 turns.
Too long, too thick, too short or too thin, the running time will be less.

 

[Dave D]

Thank you for helping me understand this.
Dave D

 

[TEACLOCKS]

This might help

https://theindex.nawcc.org/CalcMainspringThickness.php

 

[disciple_dan]

OK, I don't do math. I put this into the spring length calculator Barrel ID- 1.817. Arbor OD. 0.3515. Spring thickness. 0.140. My answer was 9.

 

[gmorse]

Hi Dan,

Spring thickness. 0.140.

I get the same, using the calculator on David Boettcher's website, (it's about 2/3 of the way down). What units are you using?

Regards,

Graham

 

[KurtinSA]

I bet the spring thickness should be 0.014. With that, you would get a spring length of almost 90 inches.

Kurt

 

[disciple_dan]

Yeah, 0.014" and that was a little off. The correct measurements are listed below.
Sorry, I'm terrible with math which is not good in this trade. Thank God for the calculators. Thank God for all of you folks too. I would never be this far along without this forum. Thank you all.
The springs I have in this Mauthe movement are Spring 1 is 0.7250" W x 0.0170" T x 60" L Spring 2 is 0.6675" W x 0.0160" T x 60" L. The depth of the barrel to the underside of the cap is 0.8125". Spring 2 is the closest to being the correct width. When I calculate Spring 2 the suggested length is Results: spring length = 73, turns = 13.2, barrel/arbor = 5.2, barrel/spring = 107, arbor/spring = 21. I'm not sure what all of those calculations mean but the spring is 73" long. What should I consider? Thank you for the help, Danny

 

[disciple_dan]

I hope someone can help me with the thickness of this spring. The clock has not been working for 3 years and the owner doesn't recall how long it would run on a full wind. I'm not sure how long it has been running after these springs were installed but there is no excessive wear at the pivot points.
Does anyone have one of these movements on hand to measure just the thickness? Any advice is welcome, Danny

 

[Mike Phelan]

Don't know about this exact movement, but usually these sorts of striking movements have the same springs in both sides.

Look up one of the UK material dealers such as Cousins and they will show the spring dimensions if you show the barrel size.

 

[shutterbug]

Back to math (sorry) but if you can determine the length needed to run 8 days, you can use that along with the other measurements to determine the thickness needed. However, it would be an educated guess.

 

[disciple_dan]

Math? Would that involve counting teeth to find the gear ratio to see how many times the barrel turns in 24 hours? Where would I find the method to do that? Just google it? Now is a great time to learn that. If someone has the answer to the spring thickness in the meantime, please post it. I'll be counting on you.
Thanks, Danny

 

[Mike Phelan]

Math? Would that involve counting teeth to find the gear ratio to see how many times the barrel turns in 24 hours?
Thanks, Danny

No dentistry needed!
Stick the minute hand on and put a Tipp-ex mark on the barrel, then count how many times the barrel rotates when you turn the hand 24 times. Simples

 

[disciple_dan]

I'm posting this here because it's a question about the same clock. As you can see in this picture, this verge has a ditch in it that the escape wheel has made. I checked it with a file and it's hard. How does brass do that to steel? (I'm not really looking for an answer to that.) My question is whether or not to try to remove the wear or move the wheel. That wear is about a 3rd of the thickness of the pallet. I'm leaning toward moving the wheel.
Also, the pivot on the crutch end of the pallet arbor has visible wear. Not terrible but visible. How so you guys polish that one. I can't do it in my lath or my Rollimat.
So, when working on a particular clock, should I name the post (denoting the movement) I'm working on and make questions on that particular clock on the same thread? Or start a thread for the particular repair to the same movement?
Thanks for the great help, Danny

 

[disciple_dan]

No dentistry needed!
Stick the minute hand on and put a Tipp-ex mark on the barrel, then count how many times the barrel rotates when you turn the hand 24 times. Simples

Man, I wish I had thought of that, I wouldn't feel so dumb right now.
So, What do I do with the information? Feeling dumber now!

 

[disciple_dan]

I found an estimator on the NAWCC site. I don't know how to use it I guess. I tried counting the teeth on the gears and pinions. It asks for the count on the center wheel or pinion. the diagram shows the center wheel as separate from the 5 wheel train. This movement is not set up that way. It has 5 wheels and no center wheel so I counted the pinion of wheel 3 or the center wheel and it gave an estimated spring length of 23.7" I'm having a hard time understanding this. Help!

 

[gmorse]

Hi Danny,

The arbor that rotates once per hour, and carries the minute hand, (indirectly if you like, via the cannon pinion), is usually the centre wheel, and its pinion (not the cannon pinion) is driven by the great wheel on the barrel. If there's another wheel between the barrel and the wheel arbor carrying the minute hand, that's another calculation.

For instance, if, say, there's a pinion of 10 leaves on the centre arbor, and the great wheel on the barrel which is driving it has 120 teeth, one revolution of the barrel will result in 12 revolutions of the centre wheel, or half a day. Now, if the spring calculator says that you'll get 14 turns of the barrel for a given spring thickness and length, that means that a full wind will run the clock for 7 days. (This has nothing to do with counting the rest of the train to discover what the beat rate is supposed to be).

Regards,

Graham

 

[disciple_dan]

If there's another wheel between the barrel and the wheel arbor carrying the minute hand, that's another calculation.

Hey, Graham. Thanks for helping me understand this. I'm not sure I follow you. Isn't the center arbor almost always the third wheel?
Did you go look at the estimated spring length and thickness calculator? https://theindex.nawcc.org/CalcMainspringEstSize.php
I don't understand the center wheel or pinion they want.
Thanks for the help, Danny

 

[gmorse]

Hi Danny,

Isn't the center arbor almost always the third wheel?

The centre wheel is usually the second wheel in a four wheel train, which is the one carrying the minute hand, because the great wheel on the barrel is the first wheel. Does your clock have the extra wheel between the great wheel and the centre wheel, making it a five wheel train? If so, then the example, using the naming and numbers in that diagram, would be:

Centre wheel (carrying the minute hand) = pinion of 16 leaves - has to rotate once per hour
Second wheel (driving the centre wheel) = 48 teeth and pinion of 10 leaves - rotates once in 3 hours (48/16)
Main wheel (on barrel and driving the second wheel) = 64 teeth, (no pinion) - rotates once in 19.2 hours (64*3/10)
So if the spring allows 10 turns, that gives a running time of 192 hours, or 8 days.

Now that we've established what running time is required, the final part of the calculation is to find the size of the barrel and its arbor and the dimensions of the spring which will result in 10 turns, fitting in the barrel which in turn has to fit onto the main wheel. Ideally, the spring should occupy one half of the area between the barrel wall and the arbor, which is what the quoted calculator and also the one in David Boettcher's website use in their calculations.

Regards,

Graham

 

[Willie X]

If it's an 8-Day clock the spring will be around .014" to maybe .016" thick. If it's a 14 Day clock the spring will around .018" thick.

When winding an 8-Day German clock the key will turn about 5 or 6 complete turns in a week's run. A 14-Day clock will only turn only about 2 1/2 or 3 complete turns.

The strike spring is often stronger/thicker on the strike side with German clocks.

Willie X

 

[disciple_dan]

Yeah, I'm pretty sure it's 8 days. It's a Freddrick Mauthe wall regulator. I'm not sure how to tell how many days a clock will run. How do you do that?
I'm not new to clocks and I have a pretty good understanding of what makes them tick (pun intended) but I have never had to deal with this part on any of the repairs I've done. I'm transitioning to full-time clock repair now. I need to start studying these things. I have the books just haven't had the time. I really appreciate your guys' help.
Here are the measurements the calculator is asking for. It's a 5 wheel train. The center wheel has a pinion of 10. The second wheel has 70 teeth with a pinion of 10. The barrel has 66 Teeth. The expected run time is 8 days or more. I'm not sure about how many turns one would need. The barrel Dia. is 1.8165" and the arbor is 0.3515". This is the result of the calculator. Spring Thickness, 0.0258". Spring length, 57" 4.2 turns.

 

[disciple_dan]

One of the springs that were in it was, 0.7250" W x 0.0170" Thick x 60" L

 

[gmorse]

Hi Danny,

I'm not sure how to tell how many days a clock will run. How do you do that?

Have another look at post #22.

Regards,

Graham

 

[disciple_dan]

Yeah, Using those tooth counts you got 10 turns. Why did my counts get 4.2? I've never seen that few turns before. I do have a Hamilton Westminster chime that has only 5 or 6 turns. What am I missing here? Could I have miss counted? I'll check those counts again and let you know.

 

[Willie X]

The thickness/strength cannot be calculated. If you think it's an 8-Day clock just put in a .015" . 016", of the correct length (can be calculated) and see what happens.

Rather than calculate, just find a spring chart that list barrel sizes. This isn't perfect either but usually pretty close, for a certain known thickness.

I have a German 14-Day Vienna style wall clock that has a very strong strike spring. That spring winds only 2.1 turns per week.

Willie X

 

[disciple_dan]

Rather than calculate, just find a spring chart that list barrel sizes. This isn't perfect either but usually pretty close, for a certain known thickness.

Where would a man find such a list of barrel sizes? That sounds like the SWAG method but, easy. Thanks, Willie X.
Don't give up on me Graham, I'm still counting. Thanks to all, Danny

 

[Mike Phelan]

Where would a man find such a list of barrel sizes?

Any of the UK material dealers - Cousins and others, Dan.

 

[gmorse]

Hi Danny,

The number of turns you can achieve depends only on the ID of the barrel, the OD of the arbor and the spring thickness, nothing else.

You can find the number of turns necessary from the calculator, which for your dimensions and a duration of 8 days gives this:



The resultant spring is 0.0224" thick and 55.7" long, which sounds a little on the thick side. However, if you plug 30 days duration into the calculator you get an implausibly thin (0.0058") and long (215") spring, so I think we can discount that option! (The Boettcher calculator does agree with this estimate).

Regards,

Graham

 

[R. Croswell]

I'm posting this here because it's a question about the same clock. As you can see in this picture, this verge has a ditch in it that the escape wheel has made. I checked it with a file and it's hard. How does brass do that to steel? (I'm not really looking for an answer to that.) My question is whether or not to try to remove the wear or move the wheel. That wear is about a 3rd of the thickness of the pallet. I'm leaning toward moving the wheel.
Also, the pivot on the crutch end of the pallet arbor has visible wear. Not terrible but visible. How so you guys polish that one. I can't do it in my lath or my Rollimat.
So, when working on a particular clock, should I name the post (denoting the movement) I'm working on and make questions on that particular clock on the same thread? Or start a thread for the particular repair to the same movement?
Thanks for the great help, Danny
View attachment 693844

I'm posting this here because it's a question about the same clock. As you can see in this picture, this verge has a ditch in it that the escape wheel has made. I checked it with a file and it's hard. How does brass do that to steel? (I'm not really looking for an answer to that.) My question is whether or not to try to remove the wear or move the wheel. That wear is about a 3rd of the thickness of the pallet. I'm leaning toward moving the wheel.
Also, the pivot on the crutch end of the pallet arbor has visible wear. Not terrible but visible. How so you guys polish that one. I can't do it in my lath or my Rollimat.
So, when working on a particular clock, should I name the post (denoting the movement) I'm working on and make questions on that particular clock on the same thread? Or start a thread for the particular repair to the same movement?
Thanks for the great help, Danny
View attachment 693844

This goes back to questions posed in post #17. Generally, I believe it is best to start a new thread when one has a question about a clock that is not the OP's clock, and especially so if the question is different from the OP's question even if the clocks are the same make and model. The OP's question, and the title of this thread, is about determining the spring length. Dan, your question is about pallet wear and whether to dress the pallet or shift the escape wheel and appears to have gotten lost amidst the discussion of spring length. Others, like myself likely assumed that the spring length question had already been answered and stopped following.

To quickly answer your questions, I would probably shift the escape wheel away from the pallet rut while making sure that you limit the end shake of the escape wheel arbor and the verge. This appears to be a recoil escapement so you could grind the impulse face (where the rut is) until the rut is gone without affecting anything beyond what can be compensated for by a slight adjustment. There is likely a rut on the other pallet as well. Don't do this on a strip deadbeat escapement. Even though the metal is hardened it can still be worn down over time by dirt and dried up oil. You will need to anneal the pallets (heat to dull red and cool slowly) before bending anything. You do not need to anneal before grinding out the rut. The verge pivots do not rotate 360 degrees so as long as they are smooth they should be OK. I use a Dremel buffing wheel to smooth these. If you have other questions about your clock, it will likely get better exposure if you start a new thread just for your clock.

RC

 

[disciple_dan]

Hey, Sorry I hijacked another post. I'll start another post. Thanks for the correction and all the great help. Danny