The Crankshaft lock design

[prevariikation]

This lock idea is super cool Coming up with a new idea while being aware of manufacturing costs and restrictions is a heck of a challenge..

How are the number of distinct bittings calculated in the provisional patent? I'm guessing total bittings = (8 wafers)^(4 cuts) - (number violating MAC restrictions)? What is the effective MAC of the system? It seems like there's a tradeoff between the keyway size, ideally small to constrain lockpicking tools, versus the MAC, to preserve physical continuity of the key.

And I feel like you've addressed disadvantages of the leaf springs pretty well, but I had a couple thoughts.

1. It seems like the sidebar can be directly accessed underneath the wafers, and I think it's possible to (weakly) reverse-bias the binding order. By pushing the front end of the sidebar down using an L-shaped tool slipped between the front wafer and lock body, tensioning the lock should cause the rear wafers to bind first, as only the rear spring will be compressed. I am not sure if my logic is sound here.
2. Since the cutout in the bottom of the leaf spring needs to be large enough to accomodate the full sidebar (like a true gate,) one might be able to perform keyspace reduction by pushing a wire into the gap and feeling middle-to-back and middle-to-front for gate positions. In particular, this attack gains effectiveness if a manufacturer opts for only one false gate. Depending on false gate layout, one might be able to further eliminate false gates by applying MAC restrictions.

Totally unrelated, but has anyone built a working prototype? The animation is incredibly informative and I think this mechanism would be wonderfully cool to see in person too.

[Yehonatan Knoll]

Thanks prevariikation,
Calculating the number of distinct bittings for key of length n, B(n), is done through recursion, relative to B(n-1). It's a non trivial combinatorial problem I can elaborate on if math is your thing. B(n) grows very rapidly with n but still sub exponentially.

Your method from point (1) won't work; pushing down on the front of the sidebar would resist tensioning. Nevertheless, if you could insert a tool *from the outside*, reaching the rear end of the sidebar and lifting it, this could indeed pose a problem. You are correct on point (2) but note that, the choice of locating the spring in the middle, is just to help with tolerances stack-up. And at the back, you could have a step, mimicking that of the wafers in fig. 11, as part of the cylinder.

I have not produced a working prototype yet. Too busy with other things. But I do think it could make a cool premium lock/cylinder. Should you or or anyone else here want to collaborate - I'll be glad.

[prevariikation]

Calculating the number of distinct bittings for key of length n, B(n), is done through recursion, relative to B(n-1). It's a non trivial combinatorial problem I can elaborate on if math is your thing.

That'd be cool I've messed around with combinatorics a bit.

Good to know that (1) isn't workable. Getting sidebar access from outside would difficult. And as you said for (2), there are easy modifications to make the leaf spring configuration more robust.

I have not produced a working prototype yet. Too busy with other things. But I do think it could make a cool premium lock/cylinder. Should you or or anyone else here want to collaborate - I'll be glad.

I wish I had the machining skills! I'll be on the lookout for anyone who can though.

[GWiens2001]

If I could get STL files, I could 3D print the parts and test it out. I also know some machinists whom I can see if they would help with making a prototype model, if you like.

Because it is your design, I will not try to create my own STL files. I value the work of those who create and engineer things, and do not support just taking their ideas.

Gordon

[Yehonatan Knoll]

Much appreciated, GWiens2001.
But my invitation to collaborate pertains to actual prototypes in terms of dimensions and materials. The tolerances in the cad model assume certain production methods. Simply printing the parts (from metal?) would not work.

[GWiens2001]

I have some friends who own a business that designs and manufactures parts for several industries with 5 axis CNC machines. They can do very exact tolerances, but I do not know their workload at this time.

Gordon

[prevariikation]

So I've been working out the recursion for distinct bittings, but I seem to be undercounting bittings somehow. If someone could set me right, I'd appreciate the insight.

It seems like the MAC in this system is 1. If Bm(n) is the number of keys of length n that end with a cut of m (cuts considered to be {1,2,3,4},) then I think the recursive properties satisfied would be

B1(n) = B1(n - 1) + B2(n - 1)
B2(n) = B1(n - 1) + B2(n - 1) + B3(n - 1)
B3(n) = B2(n - 1) + B3(n - 1) + B4(n - 1)
B4(n) = B3(n - 1) + B4(n - 1)

aka, a key with an ending cut of 3 could only have a previous cut of 2, 3, or 4.

Since there are exactly four ways to start a key, I populated this chart. (Turns out there's an expression of the total number of keys as a function of the Fibonacci sequence, where Fm = mth number in Fibonacci sequence, given F0 = 0 and F1 = 1.)

Code: Select all

# of wafers, | # of keys with a    | Total distinct keys with
     n       | last bitting of     | n wafers (= 2 * F_[2n+1])
             |    1    2    3    4 |
-------------+---------------------+--------------------------
           1 |    1    1    1    1 |     4 = 2 * 2
           2 |    2    3    3    2 |    10 = 2 * 5
           3 |    5    8    8    5 |    26 = 2 * 13
           4 |   13   21   21   13 |    68 = 2 * 34
           5 |   34   55   55   34 |   178 = 2 * 89
           6 |   89  144  144   89 |   466 = 2 * 233
           7 |  233  377  377  233 |  1220 = 2 * 610
           8 |  610  987  987  610 |  3194 = 2 * 1597
           9 | 1597 2584 2584 1597 |  8362 = 2 * 4181
          10 | 4181 6765 6765 4181 | 21892 = 2 * 10946

So I'm undercounting keys somehow, since I end up with 3194 for an 8-wafer system, instead of the 3482 in the provisional patent. I've calculated that the key counts in the patent numerically match the same recursive properties, but there's something I'm overlooking

[Yehonatan Knoll]

Nice job:) And I didn't realize the Fibonacci connection!

I must have made a mistake. Just calculated the recursion (simplified by the symmetries) manually.

[prevariikation]

Oh, I feel that! I did the addition manually at first and had to redo seven rows Cool, at least I wasn't neglecting something obvious. The closed formula using Fibonacci numbers was a nice surprise.

[prevariikation]

Would you be comfortable sharing dimensions for some of the lock parts? I tried to model the bittings using hand-measurements from the PDF, but I'm now trying to use that model to identify bittings which can't be picked with straight tools.

Taking the "Euro-profile" idea seriously, I cross-referenced the PDF housing length with the length of my euro cylinder, and ended up with a wafer thickness of ~0.068". But now I see the probe in fig. 10 was assigned a width of 0.5mm, which would put the wafer thickness closer to ~0.045". I guess I also assumed the distance between cuts is the same as the wafer thickness, I'm guessing that's not right either?

[Yehonatan Knoll]

The current design only matches Eurocylinder in diameter (length is not an issue...).
Wafer width=1.3 mm (key segments=1.13 each).

My intuition is that the percentage of key bittings not preventing a straight tool insertion is too small to be practically exploited as a weakness (one can also simply discard them).

I can send you some solid file if you wish.

[prevariikation]

I would appreciate that my email address is prevarikation@gmail, if that's easiest.

Thank you for the clarifications, I'll try to update the visualizer today. The current model suggests that, for 10 wafers, ~75% of the keyspace is theoretically pickable front-to-back using a thin allen wrench. (The long end would be held horizontally, shaft near the bottom of the keyway, and twisted to turn the tip of the shorter end to manipulate wafers.)

As you said, even just excluding bittings still leaves >2000 with substantial pick resistance, so none of this is meant as an affront to the design! I was just wondered what a "city rake" kit would statistically be for this lock.

[Yehonatan Knoll]

75% is a lot! Yet another lesson why not to trust your intuition when it comes to nontrivial combinatorics...
Would be interesting to see how this percentage changes with the number of wafers. It clearly vanishes asymptotically.

[Yehonatan Knoll]

To be clear - did you define "pickable combination" as one which does not have *both* maximal right and maximal left gates? If so then 75% still looks much too high...

[prevariikation]

I apologize, I made multiple mistakes and inverted the results of the 8-wafer case With the updated sizing data and a thin hex wrench of thickness 1.27mm = 0.050", it's returning:

• 8 wafers: 1696 of 3194 (≈53%) total bittings prevent straight tool manipulation
• 10 wafers: 15834 of 21892 (≈72.3%)

"Pickable" combinations include any bittings without a maximal right and left gate, but I also tried to account for those that would allow a tool to reach at an angle. In the code, we attempt to measure if there's a path for a straight tool from the opening to the back of the rearmost wafer in rest position.

Paths are checked from -45 to 45 degrees, in 5 degree increments. The algorithm is based on a check for a vertical rod: we identify the potential contact points (the interior corners of the wafers,) and see if the rightmost left-hand contact point and leftmost right-hand contact point are separated by at least the width of the rod, meaning that there is a clear path through. Then we simulate rotation of the rod by instead rotating the contact points by the given angles. I'll try to put up a drawing, it's not a complicated idea but I don't know if it's explained/coded well.