Energy stored in a clothespin spring (non linear spring) I was wondering how one would go about figuring out the energy stored in a spring in a clothespin.  When opened the spring is deformed, but I don't have any intuition for why it gets stronger with more loops, how it works, or how to calculate the energy or force stored.
Some searching suggests it is a torsion spring, I'll read on that now. Any conceptual help would be appreciated, I find it difficult to understand why so many loops are needed.
Here is the spring:

And it acts as the hinge like so:

 A: Yes, it is a torsion spring. It works by twisting the metal rod that makes up the body of the spring.
The reason for coiling the spring is to fit a long length of metal rod into a short space. You need a long length of rod so that the torsion per unit length remains small. With a shorter length of rod you'd exceed the elastic limit and the rod would be permanently twisted.
Don't be deceived that the body looks like a conventional spring. The only reason for coiling the rod into a helix is to fit a long length of rod into a short space and although it may look like a conventional coiled spring that is not how it works.
A: I would use the well known formula of
$$ U = \int \frac{T^2 }{2 G J} + \frac{F^2 }{2 A G}\;{\rm d}l $$
where $T$ is the torque on the coils, $F$ is the axial force, $A$ is cross section area, $G$ is the modulus of rigidity and $J$ is the polar moment of area. The axial length $l$ is only that if the active coils that counts.
If the applied force $F$ is a distance $a$ from the coil center, then the internal torque in the coils is
$$ T = F a + F \frac{D}{2} \sin\theta $$
where $D$ is the spring diameter and $\theta$ the location along the coil, starting with $\theta=0$. If the active coils are $N$ then the total angle along the helix is $\Theta = N 2 \pi$ and the length increment is ${\rm d} l = \frac{l}{2\pi N} {\rm d}\theta$. Also for a round wire $A=\pi\frac{d^2}{4}$ and $J=\pi\frac{d^4}{64}$, with $d$ the wire diameter.
The total elastic energy is
$$ U = \frac{4 F^2 l ( D^2+8 a^2)}{\pi G d^4} + \frac{2 F^2 l}{\pi G d^2} $$
for integer number of coils only, such that $\cos(2\pi N)=1$.
To get the deflection at the point of force $F$ you calculate $\delta = \frac{\partial U}{\partial F}$.
