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.)
It does. If you have two springs with the same diameter made from the same wire, but one spring has the double amount of coils, i.e. the double length, the longer spring will have half the stiffness (or the double compliance) of the shorter spring.
In fact it’s the opposite. The “dead” coils at the end and in the centre increase the overall (compressive) striffness of the spring.
Slightly increasing (or decreasing) spring stiffness by changing the wire diameter is much harder to do, as the diameter of the wire enters the stiffness factor by the power of four
orders of magnitude, i.e. increasing the diameter from 0.4 mm to 0.5 mm increases the stiffness by factor 2.44. Thus, it’s much simpler to introduce ‘dead’ or fewer coils in the spring to increase its stiffness when the outer geometric design properties (spring diameter and length) are given by the design of the pen.When referring to length, a given spring does not care about total length (ignoring situations where you’ll introduce buckling or other modes of loading). Changing the number of coils means you’d be using a different spring design. For a given spring design, it’s the same stiffness whether it’s 5mm long or 10mm. That makes the math of splitting it in series simple.
I am curious about the full impact of the dead coils (I like that term). I was treating then simply as a rigid connections, effectively splitting the spring into series, reducing the effective stiffness. Can you elaborate on how they would work to increase the stiffness?
Your explanation on the effect of diameter makes a lot of sense, especially given how tight tolerances must be at this scale. I just assumed you wouldn’t want to get the spring too thin for strength reasons.
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…