# Extending run time of 24 hour clock

#### mr_byte

##### Registered User
I managed to get it to run fine for the first 24 hours, after that it goes all to hell. I can't remember if it goes faster or slower after 24, sorry. I've more or less decided to take it back to original.

It's a Seth Thomas, not sure of the model, but it's the same as in my own mini-ogee 30 hour Seth...which I will be experimenting on, likely with a thinner spring.

On another note, I need a way to determine the length of a spring. Not what a barrel will take, but what the length of a given spring is given thickness, arbor size and diameter when fully wound. I can only guess what the original length is, for this clock, and I have another clock that is in need of a spring, and while I can guess what the length it will take using the equation, I'm not sure how to check a spring's length given only the spring, no barrel around.

#### harold bain

NAWCC Member
Deceased
You pretty much have to stretch it out and measure it, with a guesstimate for the last few inches on the inner coil.

#### Tinker Dwight

##### Registered User
Hi
It is relatively easy if fully wound. It is a little more difficult
when unwound but not contained. If unwound in a barrel,
it is similar to the fully wound with the exception of the few
turns between the arbor and the barrel.
I'll assume that you've fully wound it on a winding machine.
From the thickness, you can determine the number
of turns without having to count the layers. Still, counting will
be more accurate if you really want to be picky.
Here is the equation given you know the thickness.

Pi*(D*D-d*d)/(4*Th)

D is the outer diameter
d is the arbor or inner diameter
Th is the thickness of the spring
Pi is 3.1415..... to what ever accuracy you'd want.
At least I think I got it right.
Tinker Dwight

#### shutterbug

##### Moderator
Staff member
NAWCC Member
I think we're talking about a loop end spring. As Harold says, attach one end of it to a secure hook (strong nail or screw) and stretch it out, measuring as you go. Guess the last few inches and order whatever is closest. Sounds to me like a 30 hour clock that loses it after one day. If it's an 8 day clock, that's a huge problem. 30 hour clock .... not so much.

#### mr_byte

##### Registered User
The one that's losing/gaining after a day is the 30 hour I was trying to make run longer, so after I re-install the original springs, it'll be fine.

#### mr_byte

##### Registered User
Here is the equation given you know the thickness.

Pi*(D*D-d*d)/(4*Th)

D is the outer diameter
d is the arbor or inner diameter
Th is the thickness of the spring
Pi is 3.1415..... to what ever accuracy you'd want.
At least I think I got it right.
Tinker Dwight
It would be fully wound. The equation above is a good approximation, from what I can tell in my spreadsheet.

I set up a spreadsheet that calculates the length for each layer like this:

Pi*D (where D is the arbor diameter, at first.)

Then, the next row of the sheet calculates D as follows:
D=D+(T*2)

I know, that's not valid algebraically, but it's ok on a computer. It's telling the sheet to add 2 times the thickness to the diameter.

I then calculate the length of the layer on that diameter, then add 2x thickness, etc.

My spreadsheet is attached.

In a hypothetical case of a spring that's coiled tightly, with the following attributes:

Arbor (d) .398"
Outer Diameter (D) 1.5"
Thickness (T) .018

Tinker's equation of Pi((D^2)-(d^2))/4T=
91.263092053
My sheet works it out as 96.1076024586 at a diameter of 1.514

I will need to get out a loop end spring, coil it, measure, etc and run those numbers, then stretch it out an see where we're at, but I think we're close.

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#### shutterbug

##### Moderator
Staff member
NAWCC Member
Your formula works in my head, and I like it's simplicity. When dealing with loop end springs, accuracy in length becomes less important than when dealing with barrels anyway I think I would eliminate the first line of your calculation from the total though, which would give you 1.25" less.

Last edited:

#### Tinker Dwight

##### Registered User
Hi
I think your calculation gives you about 1 turn
diameter of optimistic length. The length of a turn
is not the outside diameter of that spring, it is closer
to the center of the spring( approximate ).
Both of our calculations are similar except I use
the center of the thickness of the spring and you
are using the outside diameter of the spring.
The reason I say approximate is that the length of the
spring is actually towards the inside of the spring
because the spring has very little compression but
does stretch on the outside of the turn when wound
( the way solids work, more stretch than compression ).
This means that even my calculation is a little on the
long side as I use a line along the middle of the spring's
thickness to calculate the length when in reality, with
a real physical spring, that line is a little more towards
the inside of the turn.
So, even my calculation would give you a slightly longer
measurement than a real spring.
I hope this makes some sense.
Dwight

#### shutterbug

##### Moderator
Staff member
NAWCC Member
So, even my calculation would give you a slightly longer
measurement than a real spring.
I hope this makes some sense.
Dwight
I think that's true, Tinker. His calculation does not account for the length from the wound spool to the movement post though ... so that would help some. It will be interesting to see how the actual tests come out.
Jeff - I should explain my comment above. The first line of your spreadsheet calculates the arbor without a spring, and adds that to the total.

#### Tinker Dwight

##### Registered User
Hi
Also, the thickness of the spring plays a significant
change in the total. One should measure the thickness
of a known number of turns rather than the simple
measurement of one thickness of the unwound spring.
Tinker Dwight

#### mr_byte

##### Registered User
Your formula works in my head, and I like it's simplicity. When dealing with loop end springs, accuracy in length becomes less important than when dealing with barrels anyway I think I would eliminate the first line of your calculation from the total though, which would give you 1.25" less.
I figured the first loop around is at arbor diameter. I like Tinker's formula better, more elegant. Also, if the spring turns out to be longer than the formula gives, it's no problem, but if the spring is shorter, that could be an issue later.

I'll get to measuring one out and run the numbers and stuff after the weekend; I do a local flea market weekends and I'm wiped out when I get home.

#### Tinker Dwight

##### Registered User
Hi
The way I derived the equation is
first that I know that if the spring was all the way
to the center, the average length is just the outside
length divided by 2, for a turn.
All I do then is multiply by the number of turns which
would be the diameter divided by 2 and then divided
by the thickness.
That leaves the squared term of the diameter divided
by 4 times the thickness.
I then do the same calculation to determine how
much to remove for the arbor.
I factored the equation to its simplest form then.
Still, like I said, it should show the spring as being slightly
longer that it really is.
One can apply the same equation to a spring that is
unwound. You'll usually find some spans of the spring where
the spacing between turns is almost equal. Just use
the spacing as the thickness in the equation.
One might split the coil into maybe 4 or 5 sections to
get a more accurate estimate. Remember to count
from zero, not one.
Also, when unwinding to measure, there is always the
tough center that is hard to unwind. Again, the same
formula can be applied to get that last part that can't
be easily measured directly.
Do remember that errors in the turn thickness( or spacing )
causes a greater error for the outside turn of that section
of spring.
Tinker Dwight

#### Jay Fortner

##### Registered User
I've always had a hard time getting a handle on this because each coil is a little longer than the previous. One coil of spring wrapped around a 3/8" arbor is 1.1781" long,a coil at 1 1/2" is 4.7124" I always figured there was a average constant to use as the multiplier but don't know what it is. Then there's the thickness factor. I like Tinks equation at least it gets me in the ballpark then I can look up what's available in that region.

#### mr_byte

##### Registered User
Hi
The way I derived the equation is
<snip: all that modesty> ;-)
Do remember that errors in the turn thickness( or spacing )
causes a greater error for the outside turn of that section
of spring.
Tinker Dwight
I love that formula, and I don't care what they say, you is a math genius.

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