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Finding the Correct Length for Mainsprings

From American Horologist magazine, April, 1936

Finding the Correct Length for Mainsprings
By Jacob L. Hagelow

IN SPITE of the fact that a good percentage of mainsprings are now available in watch work that are of the correct length, nevertheless, how often is the watchmaker at the bench confronted with the problem of just what the correct length of that mainspring should be. In many instances the old spring that has been in the watch is broken into so many pieces as to make it impossible to measure, or possibly some difficulty has been had in having the watch run the required length of time on one winding. In order to get the maximum number of turns of power out of a barrel it is necessary to give the mainspring as much area to expand therein as the spring itself contains. If this is not watched closely and a spring is used that is too long or too short, the result will be much the same.

Using a spring too long, not enough area is left in the barrel for all the coils to expand, or if the spring is too short, there is not sufficient spring to uncoil, giving us, naturally, a timepiece that does not run as long as it is made to run on one winding.

It Must Be Remembered 

Many watchmakers will fill one-third of the barrel with mainspring, leaving the other two-thirds of the space for the arbor and the spring to expand in. The arbor usually measures one-third the diameter of the barrel. This method works out with fair results. It must be remembered, however, that the outside one-third of the barrel or space which contains the spring contains more area than does the space in closer to the center where the diameter is much smaller. This would give us a mainspring one or two coils longer than necessary.

There is a definite way, however, to figure the length of mainsprings to overcome this obstacle. Let us take a barrel, for instance, with an inside diameter of 17.8 MM, the diameter of the arbor is 5.5 MM and the thickness of the old spring is .22 MM. By dividing the inside diameter of the barrel by 2, gives us the radius, which in this case is 8.9 MM.

The radius squared X 3.1416 gives us area for the barrel of 248.84 MM. The diameter of the arbor divided by 2 gives us the radius of the arbor, 2.75 MM. This radius squared X 3.1416 gives us the area of the arbor, 23.75 MM. By subtracting the area of the arbor from the area of the barrel we get the area for the spring and the area for the spring to uncoil in, which is 225.09 MM. Now, dividing this answer by 2 gives us the area for spring, 112.54 MM. Dividing this area by thickness of spring gives us the length of mainspring in millimeters, 511.5. As many or most springs come in the inch measurements, and there are 25.4 MM to one inch, this length may be transposed to inches by dividing by 25.4. 511.5 / 7 / 25.4 = 20.13 inches length for spring.


Formula 

Inside diameter of barrel divided by two equals radius of barrel.
Radius of barrel squared times 3.1416 equals area of barrel.
Diameter of arbor divided by two equals radius of arbor.
Radius of arbor squared times 3.1416 equals area of arbor.
Area of barrel minus area for arbor equals area for spring and spring to uncoil in.
Area for spring and spring to uncoil in divided by two equals area for spring.
Area for spring divided by its thickness equals length of spring in millimeters.
Length in millimeters divided by 25.4 equals length in inches. 


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