Modeling an F key mechanism

[DMA]

Turns out, using MX stabilizers is quite tricky with buckling spring mechanisms: barrel top is 0.6mm above the top plate and all the dust and bread crumbs will go straight under the flipper.

So one either needs to severely redesign key travel limiter or add dust caps to the stems.
Added dust caps today. Not sure they warp when molded - I want to have a flat side so that LED can be added, and post-molding shrinking of the assymmetric part can be problematic.. Will think about travel limiter redesign.

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horizontal plates are, from the top, MX cross base (5.6mm above plate), plate, PCB.

[DMA]

attached dust sleeves to barrels relatively easily. Supporting wall came about 0.75mm thick - should be enough for people not explicitly trying to break the key, I guess. Increasing barrel diameter should provide more sturdiness - but can cause barrel to keycap contact.

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I guess I'm close to actually trying to find somebody who can do injection molding of these. Any ideas who can do it?

[DMA]

Printed!

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It's moderately scratchy and moderately wobbly (kinda comparable to MX switches I have, may be tad more. Comparable to model F, actually).
It doesn't bind.
It doesn't click either. I mean it clicks outside of the barrel. It clicks when key is all the way down and flipper disturbed. The spring resets when key gets back up. It just doesn't click when fully assembled. Key travel is 3.8mm - but actuation is somehow even lower.

PS: Noticed a weird thing: assembled model F actuates at around 2mm travel, and original parts assembled on the table at around 3. Turns out, it does matter how spring is mounted: mounting spring in a correct position (see Fig. 1) produces initial incline (see Fig. 2) of about 5 degrees, which - together with initial ~2.5mm of compression pre-bends the spring, setting the direction of future buckling, as well as dictating it's future direction.
Interestingly, Brother flavor of buckling spring is mounted strictly vertically. And they buckle at ~3mm! Except Brother pre-compresses their spring 1mm more (and key stem end actually stops ~0.5mm below the flipper's bottom!) - so I'm kinda lucky my "improperly mounted model F" buckes at 3mm.
Also, Brother springs are essentially model M - so it looks like that extra 15 grams of force are coming solely from that extra 1mm of spring compression!

Fig. 1: correct spring placement

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Fig. 2: initial angling of the spring

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[DMA]

And so, I quickly designed some test flippers with varying degree of spring slant: 5 degree (varying spring support height), and 10 degree with default height.

flippers_galore.jpg (927.08 KiB) Viewed 710 times

Results: (flipper - actuation point)
5° - 3.5mm
5°+0.5mm - 2.75mm
5°+1mm - 2mm
5°+1.5mm - 1.5mm (fails to reset)
10° - 2mm

Somehow 5°+1mm feels crispier than 10°, and 5°+0.5mm even crispier - probably because the spring manages to store more energy before buckling, or there is such thing as too much pre-bending of the spring, so it buckles too readily - who knows!

[DMA]

Quoted barrel and stem at protolabs.com

Stem:
up to 1000 copies: $6385 + $2.83/copy. 1000 shots = $9215 = $9.215/shot
20k - 1 cavity: $9580 + $1.35/copy, total $33780 = $1.689/shot
20k - 8 cavities: $49140 + $0.50/copy, total $59140 = $2.957/shot

Barrel:
up to 1000 copies: $7430 + $2.82/copy. 1000 shots = $10,250.00 = $10.25/shot
20k - 1 cavity: $11145 + $1.21/copy, total $35345 = $1.767/shot
20k - 8 cavities: $52305 + $0.51/copy, total $62505 = $3.125/shot

So, standard 87-key keyboard is $222.72 for just the keys, not even springs or flippers - plus $20725 in tooling.
Or, $87.87 + $101445 in tooling.

Looks like it's not really feasible as a group buy without kickstarter. I kinda had an idea to make a limited run of 100 TKLs with nice 2-digit serials - but this puts the price way north of $2k per unit..

[DMA]

..by experimenting with spring support length & angle I discovered a configuration which only clicks on the keypress - return trip is silent (unless you flick the key - this way stem hits the barrel on the way back with a pretty loud "thud" - but you need to be really intentional about it, even on medium-speed typing the rebound is pretty silent).
I also discovered a configuration that's nearly silent - but still tactile! (Also, turns out tactility is greatly hampered by 3D-printed surface roughness. Not only that - the key whistles as it moves, too )
So I'm kinda ready to mold the switches - if there's anybody alive out there for a group buy.

[kbdfr]

Not being an IBM guy, I have no use for all that,
but I do admire all the energy, care and precision you are unfolding here.

[Green Maned Lion]

Not being an IBM guy, I have no use for all that,
but I do admire all the energy, care and precision you are unfolding here.

How could anyone not be an IBM person?

[vvp]

How could anyone not be an IBM person?

IBM keyboards do not have enough keys
viewtopic.php?p=502674#p502674

[DMA]

IBM keyboards do not have enough keys
viewtopic.php?p=502674#p502674

Easily doable.
In fact the more keys the better - larger number of switches drives cost per unit down.

[DMA]

Today I remembered that with 3D printing one doesn't need to drill into things to see what's inside - can just punch a hole in the model itself and print.

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So I now have a see-thru barrel and a see-thru stem.

PXL_20260215_020917354.jpg (1.91 MiB) Viewed 472 times

Also I wanted to see what will happen if I make the spring rest angle sharper. Turns out, it becomes a momentary switch: the spring buckles, but there's not enough space - so it springs back as you press the key further. On the trip up, it produces another keypress.

Also I experimented with flipper size - thinner paddle, paddle so narrow it's a finger, and no paddle at all. Turns out, loudness of the click is kinda proportional to flipper mass - the heavier the flipper, the louder the click - while tactility is not affected at all. This paves the way for an office friendly model F, keeping all of the wonderful tactility (and supposed ergononic benefits with it), but having almost no click - with a single part change vs standard model! Also, EXTRA-THICC FLIPPERS for the extra-loud slaps!
(probably material also plays a role - ABS-like resin light flipper is 190mg, heavy is 220mg, and original model F one is 260mg and kind of disproportionately louder)

[DMA]

Printed GINORMOUS 320mg and 350mg flippers:

PXL_20260215_213435838.jpg (1.52 MiB) Viewed 414 times

Boy, do those SLAP! Checkmate!
I think it's because heavier flipper spends more time in flight, so spring has time to pump more energy into it (a = F/m, higher m -> lower a -> longer travel time).
Seems like 5-degree are slappier - also likely because of greater flipper impact energy: those activate deeper, so spring stores more energy at buckling time, resulting in greater force (which should result in flipper reaching the impact faster - but maybe total energy is still higher somehow - who knows..)

Also, 40-degree spring stand slope works, not worse than 45 - but should help with spring catching on the spring slide in assembly. Spring can still catch at the slope and not seat properly - Ellipse's keys not seating properly when spring is not totally vertical are likely exactly this problem: stem molds not being polished enough in the spring guide area, resulting in not smooth enough spring slide.
I found that pressing all the way down and then removing my finger tothe side has a good chance to reseat the spring properly - at least on my 3D-printed stems.
Tried to print couple solutions to aid re-seating - nothing worked so far, but I'm not out of ideas either.

[vvp]

Energy pumped into flippers depends only on the force at position r (F(r)) and the tiny position change (dr) over which the force is exerted: E = ∫ F(r) . dr
If two flippers differ in their mass only then the force must be different to result in different energy. Force being different is plausible because the amount of buckling (and therefore the force) can be different a given position r ... this being due to different flipper speed ... as you said.
Different flipper mass will also result in different frequency spectrum of the final sound which may contribute to the feeling of having a different loudness.

[DMA]

Energy pumped into flippers depends only on the force at position r (F(r)) and the tiny position change (dr) over which the force is exerted: E = ∫ F(r) . dr
If two flippers differ in their mass only then the force must be different to result in different energy.

I kinda see heavier flippers moving slower - I don't have a high-enough-speed camera to confirm that tho. I also think same force applied over longer time results in more energy transfer. Tell me where I'm wrong?

Force being different is plausible because the amount of buckling (and therefore the force) can be different a given position r

the spring support part is identical - so the spring initial position is identical and trigger moment as well.
I actually made some experiments: varying spring support angle and distance between pivot point and center of spring support. Within "same angle, same flipper weight" batch, there's ideal (loudest) position, corresponding to deepest buckling point - which kind of confirms my hypothesis of "the energy the spring distributes is proportional to it's compression at the time of buckling". Sure, that force decays over time - I'm not sure if it decays quickly enough to take that decay into account.

Different flipper mass will also result in different frequency spectrum of the final sound which may contribute to the feeling of having a different loudness.

it's not only sound - when holding the assembly aloft and clicking it heavier flippers definitely produce more vibration. Could also be attributed to spectrum - but the slap sounds about the same, just louder. I really don't want to build an anechoic chamber only to find out that spectrum is the same, it's just higher amplitude

[Green Maned Lion]

I'd love for you to upload sound files.

[AndyJ]

You don't think of a keyboard as having a whole lot of engineering behind it. Inventing the buckling spring, sure. But the real meat was in all the tiny details you are rediscovering. Unicomp owns the design now, and presumably has the blueprints, but they almost certainly don't know *why* some blueprints show parts with weird little angles.

It was probably a bunch of guys with drafting boards, sending paper prints to a machine shop. And moldmaking has *always* been expensive. No 3D printers back then.

[DMA]

..so I printed a couple of gigantic, 880mg flippers:

PXL_20260217_023550955.jpg (1.56 MiB) Viewed 296 times

Those aren't that much louder if you hold the thing in your hand.
HOWEVER, should you press the assembly against the table and press the stem, there'll be absolute GUNSHOTS

[kbdfr]

[…] Those aren't that much louder if you hold the thing in your hand.
HOWEVER, should you press the assembly against the table and press the stem, there'll be absolute GUNSHOTS

Same for a tuning fork:

[DMA]

I found a solution to a spring sticking a wrong way.
The solution is a small hole at the top of the stem.
To fix the spring stuck in a wrong position, you unbend a paperclip, stick it about 10-12mm into the hole, and nudge the spring a little.
Then you remove the paperclip, and after 2-3 presses it starts clicking again.
If you don't feel like trying - press the key to compress the spring and _then_ insert the paperclip, nudging the spring until you hear a click (that would be flipper flipping down), after which you remove the paperclick and the key works again.

Now, it is possible (definitely not easy - but doable!) to put the spring in an inop position using the paperclip on the normally working key. This is usually self-resolving - just mash the key quickly 3-4 times - but can also be solved by sticking the paperclip into stem again and removing it quickly.

(Also, I found that 50° spring guide slope instead of the default 45° is virtually impossible to jam).

Finally, a capacitive buckling spring module for the XXI century, cured of all the problems plaguing shoddily produced chinese knock-offs, and also original IBM buckling spring modules!
Just need to find enough volunteers to actually produce it at less-than-exorbitant price.

[vvp]

Energy pumped into flippers depends only on the force at position r (F(r)) and the tiny position change (dr) over which the force is exerted: E = ∫ F(r) . dr
If two flippers differ in their mass only then the force must be different to result in different energy.

I kinda see heavier flippers moving slower - I don't have a high-enough-speed camera to confirm that tho. I also think same force applied over longer time results in more energy transfer. Tell me where I'm wrong?

Heavier flippers will move slower! You have correctly written that this is the result of a = F/m equation. But with heavier flippers (moving more slowly) the force is influencing the flipper movement for longer time. Anyway, at the end of the movement, the kinetic energy transferred into the flipper will not depend on the speed difference in friction-less environment. Notice that the kinetic energy is E=mv²/2. The heavier flippers will move more slowly but they are also heavier: v² will be as smaller as m is bigger.

The overall energy transferred into the flipper is computed using E = ∫ F(r) . dr
It does not depend on mass nor time. Only on the force exerted on the object and the distance over which this force was exerted. That is true if we do not consider (aerodynamic) friction.

But we should also consider the air resistance which I did not consider before. The aerodynamic resistance is proportional to the second power of speed. And it will be more significant just before the flipper hits the PCB (something like ground effect of planes will be there as well). If there is a difference in the energy at the time the flipper hits PCB then it is attributed mostly to:
1) aerodynamic loss
2) force exerted: if the flipper speed profile (over the distance traveled) is different then there might be some dynamic consequences on spring buckling and therefore force (this is despite the flipper spring holder having the same angle)

I'm thinking now that the aerodynamic losses (smaller for heavier flippers) is likely the main difference. Notice that droplets of water falling on tough surface do not splash in vacuum. The splash (droplets of water separating and moving out of the hit location) happen because of aerodynamic forces. So it kind of makes sense the aerodynamic loss may be significant for buckling switch flippers as well.

Thanks for pointing out inconsistencies in my previous response.

[DMA]

@vvp what you say is logical.
But I'm still not convinced somehow: isn't same force applied for longer time results in more work done?
Like, when you plug in an iron, it consumes same wattage, but it definitely becomes warmer with every second passing (until thermostat kicks in) - so it's definitely more energy there with longer application of a constant power.

I can also slam my room's door easily and quickly stop it once it got going - but I tried to close the door of a Minuteman silo once, and let me tell you, first that thing doesn't want to move and then it doesn't want to stop: much larger forces required. And the only real difference there is weight - so I have hard time accepting the fact that energy transfer is a function of only force and distance. Time should definitely be involved somehow.

[vvp]

@vvp what you say is logical.
But I'm still not convinced somehow: isn't same force applied for longer time results in more work done?

No. It is time independent BUT it is trajectory (displacement) dependent.
Imagine the limit scenario where you exert constant force on object which is just not moving any more. In such a case you are not adding any energy to the object despite the fact that the force is active for longer time. Force must cause some displacement to add energy and the more displacement it causes the more energy it adds. It does not matter how log does this displacement last.

Like, when you plug in an iron, it consumes same wattage, but it definitely becomes warmer with every second passing (until thermostat kicks in) - so it's definitely more energy there with longer application of a constant power.

There are two mistakes here:

• Wattage is not the same as force. Wattage is "flow" of energy ... or more precisely energy change per time: W = E/t which is the same as E = W*t. Of course, you increase energy when you increase the time.
• Force is just a force. When you are sitting on a stationary chair you exert gravitational force (corresponding to your mass) on the chair. But you are not adding any energy to it because it is not moving relative to you.

I can also slam my room's door easily and quickly stop it once it got going - but I tried to close the door of a Minuteman silo once, and let me tell you, first that thing doesn't want to move and then it doesn't want to stop: much larger forces required. And the only real difference there is weight - so I have hard time accepting the fact that energy transfer is a function of only force and distance. Time should definitely be involved somehow.

If you exerted the same force then it is just your perception that you added more energy to heavier doors. The feeling you added more energy to the heavier doors is likely because:

• You actually pushed harder on the heavier doors. You likely started to push less (as on light doors) and when they did not move as you expected then you pushed more to get them moving faster.
• The heavier doors have more friction at the hinges. You actually must put more energy to it to overcome bigger frictional energy losses.
• Human (animal) muscles consume a lot of energy when they are under tension and not moving. Allowing muscle to also move does not add that much muscle energy consumption. This "static" energy consumption of muscle adds up when you are closing the door the longer time. So you correctly feel that you spend some additional energy with heavier doors. But this additional energy is the "static" losses of your muscles. It is not transferred to the heavier doors.

[DMA]

[*] The heavier doors have more friction at the hinges. You actually must put more energy to it to overcome bigger frictional energy losses.

You have no idea. That door's hinges are very smooth - you can move that heavy door with a single finger, but it will be a very slow movement.

I figured out where you're wrong: yes, it takes same amount of energy to move something a fixed distance - and that amount is zero, because you simply recover all the energy to arrive at the destination with zero stored energy

However, if you get two identical trucks - one empty, another one loaded to 80000lbs - go full throttle on both, and brake hard at quarter mile - you'll find that second truck a) takes longer to the quarter mile and b) takes longer to stop.

[vvp]

However, if you get two identical trucks - one empty, another one loaded to 80000lbs - go full throttle on both, and brake hard at quarter mile - you'll find that second truck a) takes longer to the quarter mile and b) takes longer to stop.

Ignoring the air resistance and otherwise everything else being the same this should not happen. Notice that the second truck will have significantly slower speed when it starts to brake at the quarter mile location. I have never driven anything heavy so I do not have my own experience.

Otherwise, I guess, We just need to agree that we disagree. Maybe the only thing I can still point out is: If the energy delivered to an object under external force depends on time then why time does not time appear in the official energy equation E = ∫ F(r) . dr

https://en.wikipedia.org/wiki/Energy#Cl ... _mechanics
W is just a different name for E
ds is just a different name for dr

[DMA]

https://en.wikipedia.org/wiki/Work_(physics) says:
"The work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance." So we're both right - depending on what Wikipedia article we believe to be true

Why time doesn't appear in E = ∫ F(r) . dr - I have no idea. Probably gets reduced somewhere.

[DMA]

This thing can't just leave me alone..
Believe it or not, I found a SIMULATOR at https://phet.colorado.edu/sims/cheerpj/ ... moving-man and turns out the word I was looking for wasn't "energy", but "momentum" aka "amount of motion".

Distance: 8 meters.
Acceleration:
1m/s2 -> final speed 4 m/s
0.5 - 2.8
0.25 - 2
0.125 - 1.4
So, momentum = mv = 4, 5.6, 8 и 11.2 newton*seconds, kinetic energy mv^2/2 8, 7.84, 8, 7.84 joules.
Indeed, amount of work by the spring is constant. But it creates "more motion" in a heavier flipper, and it looks like the loudness is proportional to that "amount of motion" in the flipper, not it's kinetic energy.
Which I also can't understand, but I guess one man can't just understand everything

[vvp]

Yes, heavier flippers will have higher momentum (even in friction less environment).

This higher momentum should displace PCB more (higher amplitude) but more slowly (lower frequency). That would be true if the flipper speed at the PCB impact corresponds to the resonant frequency of the PCB. Which is unlikely. Otherwise, the loudness likely depends on the energy. I do not really know.

The loudness difference is likely due to the aerodynamic loses I mentioned before.