this post was submitted on 08 May 2025
118 points (96.8% liked)

Ask Science

10499 readers
96 users here now

Ask a science question, get a science answer.


Community Rules


Rule 1: Be respectful and inclusive.Treat others with respect, and maintain a positive atmosphere.


Rule 2: No harassment, hate speech, bigotry, or trolling.Avoid any form of harassment, hate speech, bigotry, or offensive behavior.


Rule 3: Engage in constructive discussions.Contribute to meaningful and constructive discussions that enhance scientific understanding.


Rule 4: No AI-generated answers.Strictly prohibit the use of AI-generated answers. Providing answers generated by AI systems is not allowed and may result in a ban.


Rule 5: Follow guidelines and moderators' instructions.Adhere to community guidelines and comply with instructions given by moderators.


Rule 6: Use appropriate language and tone.Communicate using suitable language and maintain a professional and respectful tone.


Rule 7: Report violations.Report any violations of the community rules to the moderators for appropriate action.


Rule 8: Foster a continuous learning environment.Encourage a continuous learning environment where members can share knowledge and engage in scientific discussions.


Rule 9: Source required for answers.Provide credible sources for answers. Failure to include a source may result in the removal of the answer to ensure information reliability.


By adhering to these rules, we create a welcoming and informative environment where science-related questions receive accurate and credible answers. Thank you for your cooperation in making the Ask Science community a valuable resource for scientific knowledge.

We retain the discretion to modify the rules as we deem necessary.


founded 2 years ago
MODERATORS
 

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.)

you are viewing a single comment's thread
view the rest of the comments
[–] finkle@lemm.ee 28 points 1 day ago* (last edited 1 day ago) (7 children)

The way the spring is designed is very akin to taking two separate springs, and attaching them in series. On a highly technical level, this creates a quite different response to motion than a single spring would have. On a lower tech level, it's approximately the same as regular spring with a different stiffness value.

I'd like to think that some engineers went down the rabbit hole, doing differential equations or modal analysis to solve some complex design problem. More likely it was just a cheap or easy way to solve a manufacturing or supply problem.

Edit: Upon further reflection, spring stiffness does not care about length. The primary concerns are thumb travel (deflection) and force to click (deflection times stiffness). I suspect that deflection is pretty well set by ergonomic factors. This design would reduce the effective spring stiffness by half (and therefore the force). The most obvious other way to reduce stiffness would be reducing wire thickness, which could reduce reliability.

Noise, on the other hand, is an expected result of clicky-pens. People don't like products behaving unexpectedly. If the design made the pen quieter, you can bet they would advertise it on the packaging.

I kind of hate how much thought I've put into this.

[–] Successful_Try543@feddit.org 14 points 1 day ago* (last edited 1 day ago) (1 children)

Upon further reflection, spring stiffness does not care about length.

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.

This design would reduce the effective spring stiffness by half (and therefore the force)

In fact it's the opposite. The "dead" coils at the end and in the centre increase the overall (compressive) striffness of the spring.

The most obvious other way to reduce stiffness would be reducing wire thickness, which could reduce reliability.

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.

[–] finkle@lemm.ee 2 points 23 hours ago* (last edited 23 hours ago) (1 children)

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.

[–] Successful_Try543@feddit.org 1 points 10 hours ago* (last edited 10 hours ago) (1 children)

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?

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.

[–] untorquer@lemmy.world 1 points 5 hours ago (1 children)

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.

[–] Successful_Try543@feddit.org 1 points 2 hours ago* (last edited 2 hours ago) (1 children)

I was talking about the length of the wire. You are talking about the length of the spring.

By reducing the free length the "dead coils" slightly increase stiffness. They have an impact on the total force at compression.

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.

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.

You're probably right.

[–] untorquer@lemmy.world 2 points 1 hour ago

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...

load more comments (5 replies)