## Energy of a mainspring

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IRLf
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### Energy of a mainspring

So, first off, I'd like to say thank you and I've been a bit of a lurker here for a while, never saying anything because it wouldn't have provided any further insight than had already been shared.

Anyway, today I began thinking about mainsprings, the coiled springs typically used to power wind up clocks, watches, etc. While reading up on them I began to notice a trend where articles comment on how much energy they can contain (usually described as "a lot" rather than anything useful.) This lead me to try to find a source for the potential energy of a mainspring, something that I've found rather difficult to find.

The closest I've come is a webpage which says:
Mr. Dan Henderson, a Senior Manufacturing Technology Engineer at 3M, shared with me a formula for springs. All other factors being equal, the strength of the spring is proportional to its width: in other words, if we double the width, we double the strength. Similarly, the strength is proportional to the cube (to the power of 3) of the thickness: if we double the thickness, the spring is eight times stronger (2x2x2=8).
$f=kbh^3$

where "b" is the width of the spring and "h" is the thickness. This bothers me because it doesn't provide any form of bending (or whatever the mainspring equivalent of elongation is) into the equation.

Thinking about it further I was wondering if it could simply be classified as a cantilever beam that is being bent way, WAY further than is usually covered (and thus out of the realms of my engineering textbooks I believe, however they are packed and unavailable for consultation for at least 3 weeks), since generally the bending is limited to smaller angles (at least probably less than 360 degrees).

So, to wrap it up, can anyone provide some direction for figuring out (or simply tell me) the potential energy of a mainspring.

Thanks

Goemon
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### Re: Energy of a mainspring

Well, not sure if it's exactly what you're looking for, but:

If you measure the length of the spring before and after winding the clock, then the total energy stored in the spring should be something like [imath]\frac{1}{2}kx^2[/imath], where x is the change in length. So for example, if the circumference of the spring is 1cm, and you turn it 12 times to wind the clock, then you get an energy that looks like [imath]\frac{1}{2}k(144cm^2)[/imath]

k is a constant that depends on how the spring is constructed. It's a lot easier to build a spring and then just measure k than it is to try and calculate what k will be before building the spring. It's possible, but requires a lot of calculations of material strengths and stresses.

You can measure k by measuring how much force is required to wind the clock - to actually turn the handle. The more you wind the clock, the more the force needed to turn the handle increases. Ideally, it's linear: F = kx, where k is the same number as above and x is the amount you've already wound the spring. So k = F/x. If you've wound the spring so its length has decreased by 2cm, and at that point it takes 12 Newtons to turn the handle, then k = 12 Newtons / 0.02 meters = 600 Joules / meter ^2 = 0.06 Joules / cm^2.

The formula you give above applies to k: if you make a second spring twice as long as the first, but otherwise identical, then the energy storage of the second spring will be double the first. If you make a third spring whose material is twice as thick as the first, then the third spring's energy storage (after winding it the same amount) will be 8 times the first.

All purely theoretical, of course
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Tass
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### Re: Energy of a mainspring

IRLf wrote:Thinking about it further I was wondering if it could simply be classified as a cantilever beam that is being bent way, WAY further than is usually covered (and thus out of the realms of my engineering textbooks I believe, however they are packed and unavailable for consultation for at least 3 weeks), since generally the bending is limited to smaller angles (at least probably less than 360 degrees).

Yes, the bending of a cantilever is what you want. Usually it breaks down at large angles because the cantilever can't be approximated as part of a circle anymore, but in a spiral spring circle is a good approximation. Of course you have to use the angle deviation from the relaxed angle and not from straight bar. Also there will be some longitudinal strain as well, but I think is can be neglected as a first approximation.

IRLf
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### Re: Energy of a mainspring

Alright, thanks a lot.

The formula you give above applies to k: if you make a second spring twice as long as the first, but otherwise identical, then the energy storage of the second spring will be double the first. If you make a third spring whose material is twice as thick as the first, then the third spring's energy storage (after winding it the same amount) will be 8 times the first.

I didn't really realize that was the case here, now it makes more sense.

Yes, the bending of a cantilever is what you want. Usually it breaks down at large angles because the cantilever can't be approximated as part of a circle anymore, but in a spiral spring circle is a good approximation. Of course you have to use the angle deviation from the relaxed angle and not from straight bar. Also there will be some longitudinal strain as well, but I think is can be neglected as a first approximation.

I was hoping something like this could be applicable, because then I could use the generally accepted material properties of, say, differing spring steels and using the geometric relationships as described earlier with some good ol' mechanics of materials to develop a relationship. Now to get moved and unpack my library.

uncivlengr
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### Re: Energy of a mainspring

There are straighforward analytical solutions to curved beam bending problems, and from that you could determine a rough model for determining a torsion spring constant.

spring.PNG (6.43 KiB) Viewed 3373 times

Consider a single coil of a spring (see above diagram) with freedom in the radial direction (ie, the fixed ends of the spring are sufficiently far away from the coil being considered and the adjacent coils don't get in the way), apply a torque through the centre of the spring to one end which causes the spring to expand and that end to rotate an angle θ.

Then you can use energy methods to determine the complementary energy of the coil due to the torque:$C=\int_{0}^{2\pi}\frac{M^2}{2EI}(Rdθ)$

Now M and EI can be determined to be constant, at least given the assumptions I've made, so that boils down to $C=\frac{2\pi RM^2}{2EI}$

The total energy is $\Omega=\frac{2\pi RM^2}{2EI}-Tθ$

Then you can take the partial derivative with respect to T, set that equal to zero, solve for θ, which you can use to find an expression for your spring constant.

The only thing left is to figure out an expression for the moment, M, in terms of the torque, T. It should be straightforward, and as I said, a constant throughout the spring, but it's late and I'm not going to get it tonight.
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