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Posted

Just a quick question, replaced a mainspring which i got from cousins, and measured the old spring which was

H 18 x Th 0. 45 x L 17900 x D x 43 the new one is

18 x 0.45 x 1650 x 45

and when i put it in the barrel it was such i tight fit that it would only turn a couple of times 

what have i done wrong any ideas not posted any photos as I can't see how that will help.

Posted

Hi Willow,  looking at the barrel with its lid off does it follow the 1/3rd rule, the spring in the barrel un wound occupies 1/3rd of the barrel leaving the rest of the space to work in  can you post a pic (sorry) of the barrel with spring fitted.  As Nicklesilver said.    cheers

1 hour ago, nickelsilver said:

There must have been a mistake made in measuring, a shorter spring of same thickness can't take up more space in the barrel.

 

  • Like 1
Posted

Thanks for reply Hectorlooi, i did measure the old spring a few times, the spring does not impinge on the barrel cover  and i have put the clock back together and yesterday manage to turn it a few times so its on test at the mo but how long it will it tick for will have to see. Hi watchweasol, thanks for your reply too, when i put it in the barrel it was a tight fit so had my concerns then that it would not wind up enough like i said its on test at the mo so when it stops i will post a photo. Thanks again both of you.

Posted

What type of clock is it. The 1/3rd rule is the easy way of knowing if the spring is too long, which is what it sounds like as you can only wind it a few times which can cause wear if in fact the spring is too strong. 

  • Like 1
Posted
7 hours ago, oldhippy said:

What type of clock is it. The 1/3rd rule is the easy way of knowing if the spring is too long, which is what it sounds like as you can only wind it a few times which can cause wear if in fact the spring is too strong. 

Hi oldhippy, it is a small smiths wall clock at the moment it ticking away once it stops i will send some photos and will check the spring in the barrel for the 1/3 rule i will also measure the old spring again.

Posted
15 hours ago, oldhippy said:

What type of clock is it. The 1/3rd rule is the easy way of knowing if the spring is too long, which is what it sounds like as you can only wind it a few times which can cause wear if in fact the spring is too strong. 

Hi forget that last reply, it's a W & H Sch small round wall clock.

measured the old spring without the string i used last time😂

this is what the old spring measured 18 x 0-40-0-45 x 1640 x 45 mm

this is the new spring 18 x 0-45 x 1650 x 45 mm.

it's been ticking away for 2 days now on the few turns i managed 

 

Posted

Hi Willow ,  I am still of the opinion that rage spring is too long, the spring you removed may not ne the one it started out with. Just to clarify some points, did the clock work when you got it, if so did you wind it up ? And if you did how many turns did it take. We assume you dismantled and cleaned the clock, why did you change the spring ,  broken or set. Did you physically check the spring size with the new on before fitting. As it does not comply with the 1/3 rd rule the more I think the one in the clock was not correct. The rule is that an unwound spring in the barrel occupy one third, the center arbour. The second third and the spring operating space the final third. As mentioned by Old Hippy an over strong spring can cause damage especially to the operating surfaces of the wheels but also on the pivots and holes. It would be prudent to remove the spring and measure the Barrel following the one third rule. As its a Winterhalder / Hofmeier.   (1910 to 1930) there is a possibility of finding the correct spring.

  • Like 1
Posted
5 minutes ago, watchweasol said:

Hi Willow ,  I am still of the opinion that rage spring is too long, the spring you removed may not ne the one it started out with. Just to clarify some points, did the clock work when you got it, if so did you wind it up ? And if you did how many turns did it take. We assume you dismantled and cleaned the clock, why did you change the spring ,  broken or set. Did you physically check the spring size with the new on before fitting. As it does not comply with the 1/3 rd rule the more I think the one in the clock was not correct. The rule is that an unwound spring in the barrel occupy one third, the center arbour. The second third and the spring operating space the final third. As mentioned by Old Hippy an over strong spring can cause damage especially to the operating surfaces of the wheels but also on the pivots and holes. It would be prudent to remove the spring and measure the Barrel following the one third rule. As its a Winterhalder / Hofmeier.   (1910 to 1930) there is a possibility of finding the correct spring.

Hi watchweasol no the clock wasn't working when i got it and it would not wind up the spring broke from the arbour end when i got home hence the new spring and i didn't check physically with the new one, i will do what you said requarding the 1/3 rd rule, thanks

Posted
1 hour ago, watchweasol said:

As it does not comply with the 1/3 rd rule …

What kind of rule is that? 

A bit reflection will reveal that such 1/3 rule cannot work.

True is, the arbor takes (used to take) 1/3 of radius or diameter.

The rest of barrel area is divided by two. One for the spring and one unoccupied. That will not result in 3x 1/3 of radius or diameter, however!

Frank

Posted

Now i am getting confused with this 1/3 rd rule your saying praezis it cannot work, can it or can't it, that is the question to be or not to be, help !!!!!!!! what if you have not got a spring what do you do then so pretend there wasn't one in 🤪

Posted (edited)

 

58 minutes ago, Willow said:

Now i am getting confused with this 1/3 rd rule your saying praezis it cannot work, can it or can't it, that is the question to be or not to be, help !!!!!!!! what if you have not got a spring what do you do then so pretend there wasn't one in 🤪

It’s simple: The 1/3 rd rule applies to the radius or diameter not to the area 😉

Edited by Kalanag
  • Like 1
Posted
51 minutes ago, Kalanag said:

It’s simple: The 1/3 rd rule applies to the radius or diameter not to the area 

I am sorry Sir, you are wrong.

Frank

Posted

Before we loose track of what is and isn’t right Donald De Carlo in his book which I have all of them states as Kalnag said. As a professional WATCHMAKER not repairer who built watches for a living came up with the theory so who are we Amateur repair men to disprove his theory, you might as well challenge the theory’s of George Daniels or Henry B Freid. The basic fact remains that the working space for the spring is wrong.     So at this point Willow has a clock that works but the spring winds short (not enough space). The spring was broken when he got it and naturally got a new spring which according to the old spring measurements was almost correct and should work, but no not enough on the wind. Therefore my theory is that the wrong spring was in the clock and probably the reason that it broke being constantly being wound to capacity all the time. So no matter proving or disproving the theory there is not enough space for the spring to operate correctly and is if not corrected going to break again or damage the clock.

 

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Posted

Sorry, this cannot be left uncommented.
Even in the picture you can see, that the released spring takes less than 1/3. He also mentiones it in the text. Mr. de Carles statements are not correct, misleading and omit the main reason for dimensioning a spring.

In short, such "1/3 rule" is nonsense.
Why? 
The optimal mainspring will give the maximum possible number of revolutions. And this spring will occupy exactly 1/2 of the remaining barrel area (a real rule).
In the picture: the area of the released spring is the same as the empty area, which will be occupied by the wound spring then. 

Everyone with some geometry knowledge will see that the released spring packet cannot have the same thickness as the wound packet... For all others: 
released packet = 0.25 R (not 0.33 R!)
wound packet = 0.41 R (not 0.33 R!)

Frank

  • Like 1
  • Thanks 1
Posted

That makes perfectly sense!

Nevertheless calculating areas is a bit difficult for a rule of thumb. So the 1/3 rule is not useless as a first approximation imho.

  • Like 1
Posted (edited)
1 hour ago, Kalanag said:

Nevertheless calculating areas is a bit difficult for a rule of thumb

True. But you can use the numbers above:
Thickness of released spring packet = 1/4 barrel radius 🙂

And I want to apologize. Not you were wrong but Mr. de Carle was.

Frank

Edited by praezis
  • Like 1
Posted

It is just basic there is no real math to it. You do not measure the barrel and work calculations out. I learned it just by word of mouth by my master. 

  • Like 1
Posted

Hi Guys  find attached pages from Laurie Pennman (Master clockmaker) and his calculations for the mainspring acommodation as regards to the barrel fitting. This man is spoken of in the same terms as Brian Loomes also a master clock maker.  Not having done Geometery  scince leaving school in the 50s I would not attempt to dispute Frank's figures but as a rule of thumb its a fair guide, not as proven the hard fact. We are unable to determine in what context De Carle used the diagrams without him telling us which is impossible. So best probably to put this one to bed.

 

Scan_20230203 (3).png

Scan_20230203 (2).png

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