Why is the spring strengthened in the middle?
It doesn’t seem to affect the spring’s buckling characteristics.
My speculation is that it’s to reduce spring noise. That strengthened region at the middle is where the spring will buckle outwards most, resting against the barely visible side rails on the inside of the case. Instead of just one wobbly contact point it now has three rigid ones as a “skate” to reduce the stick-slip noise when opening and retracting the tip. Is this right?
(The pen is a Mitsubishi Uni-Ball Power Tank, pretty much my favorite model.)
Yes, if you think of the depicted spring with dead coils in the centre, as two springs of half length combined, they decrease the stiffness of the combined springs, as two springs in row have only half the stiffness.
My thought was starting from the opposite: A spring with the same number of coils, but equally spaced. At its ends, the axial force onto the spring is converted into internal torque. By Hooke’s law, torque and shear strain (here change of twist angle with the coordinate running along the wire) are linked. This twist causes the coiled spring to contract.
However, in the spring with the “dead” coils, this motion is limited to the free coils, as the twist of the “dead” coils is inhibited by their contact. They behave like they were rigid. Thus, there only the wire of the free coils contributes to the (compressive) stiffness of the spring, which is less than the total amount of wire, yielding into a stiffer spring.
It reduces the effective free length of the spring.
Let’s just rearrange the equation for a spring at full compression:
F=-kL
k=-(F/L)
Whether you use one full length spring or two half length springs doesn’t matter, the spring constant is unchanged.
By reducing the free length the “dead coils” slightly increase stiffness. They have an impact on the total force at compression.
I think in this image were looking at what, maybe 10% difference in any of those factors? For the life of me i can’t imagine this matters terribly much in a pen.
I was talking about the length of the wire. You are talking about the length of the spring.
Yes, the part of the wire coiled up in the dead coils does not contribute to the stiffness of the spring. The stiffness if the spring is determined only by the two sections in between. This is the length L in your equation and not the total length of the spring.
You’re probably right.
Right! But those are the same thing as number of coils is the spring length divided by a geometric constant. At free length there is no strain. At compression you reach max strain/torsion. Each coil turn, assuming all are equal, adds equally to the sum of restoring force. Looking at spring free length you’re just paying attention to the summed forces of the active coils.
The dead coils contribute negligibly because they would need to impinge the neighboring wire to deform. (Relegated to pure torsion) Which i think is basically what you were saying…