The Perfect Lock Can It Be Picked?

[3-in-1]

Before you go all bumpin, this is purely a mental exercise. Single pin picking, sometimes known as the tentative method or Hobbs method. Relying on the slight and sometimes not so slight differences in machining to pick a lock one tumbler at a time. It really doesn't matter if it pin, wafer, lever or wheel. So can the perfectly machined lock be picked by this method? The keyway is wide open and there are not limitations due to zig zag keyways. The fit of the pins in the pin chambers is is only large enough to ensure a slip fit of the pins and they are all in perfect alignment. Just thinking about this scenario might improve your picking. So what do you think?

[femurat]

No, if the perfect lock exist, it couldn't be SPPed. Unfortunately such a lock can't exist.

Cheers

[3-in-1]

This wasn't meant to be so much of a simple yes or no question. More of a study of what would happen. SPP requires that the pins bind up to successfully pick the lock. In SPP, the side loading torque (turning force) has to be greater than the downward loading of the springs to get the top pins that are binding to stay put as they are pushed up. So there is always a minimum torque required which would vary from lock to lock, and it may vary even within one picking session. In the case of a perfect lock, lets say a 5 pin, at some point of torque, will binding of the top pins occur? Will the torque have to be greater than usual? And will the binding be equal on all pins?

[femurat]

In theory, if all the pins bind at the same time, it's not possible to set one: without a binding order there isn't a binding pin. Since I've never had such a lock in my hands, I'm just speculating, there's not much else I can add.

Cheers

[3-in-1]

Yes just speculating. When a constant torque is applied to the core, the top pins are forced against the walls of the cylinder and core. Assuming all 5 pins are now binding, regardless of whether the binding is equal on all 5, there is a slight deformation of the parts. You can't see or feel this deformation but it is occurring. Once one pin is lifted to the shear line, the constant torque and resulting deformation is now spread among the remaining 4 binding pins, instead of 5. This would increase the load seen by each remaining binding pin, resulting an even greater deformation. Greater deformation would allow for a minute core rotation, resulting in the picked pin to feasibly remain set. Whether you could readily distinguish just when you hit the shearline is yet another question. Now as I said, this is just speculation and hopefully someone can give some more insight here.

[gloves]

Yes just speculating. When a constant torque is applied to the core, the top pins are forced against the walls of the cylinder and core. Assuming all 5 pins are now binding, regardless of whether the binding is equal on all 5, there is a slight deformation of the parts. You can't see or feel this deformation but it is occurring. Once one pin is lifted to the shear line, the constant torque and resulting deformation is now spread among the remaining 4 binding pins, instead of 5. This would increase the load seen by each remaining binding pin, resulting an even greater deformation. Greater deformation would allow for a minute core rotation, resulting in the picked pin to feasibly remain set. Whether you could readily distinguish just when you hit the shearline is yet another question. Now as I said, this is just speculation and hopefully someone can give some more insight here.

Nice question. I'll give my speculations a try, hope you don't mind

I think that this very first assumpion (all 5 pins binding at the same time) is just as plausible as a perfect lock, because they'd need to be on the very same axis for this to occur, just as for the perfect lock. Assuming that slight deformation (which is true in all materials, may it even be a feather on a steel plate) I won't see why it'd be different from pin to pin since they'd all be on the same line (otherwise they won't bind all at the same time). Another 'flaw' I see in this hypothetical physic system is blending perfect, ideal situations with real, practical data.
Even if we assume that the pins are on the same perfectly straight line, yet they're not that perfect to deform in the same way, we're not taking into account that locks also work by means of tolerances.
If the chambers are perfectly fitting, this would make a lock work at 20°C and not at 0°C or 40°C for example, due to materials expansion or not let pins slide because most materials aren't self-lubricant. Or we should arbitrarily choose a x-amount of tolerance to put in our perfect system, and this would detach from theoretical perfect systems.

So while we could discuss this past the end of the world, I feel like saying that a "perfect" lock won't be a lock, or at least not a working one. Tolerances are needed for things to work freely and erasing them won't only impede picking attacks but also intended-key functionality, for which a certain amount of looseness is necessary Some real-world physics laws are different when speaking of negligible measures.

That's just my opinion though, so don't take it as a personal attack. I'm open to discussion.

Cheers

[Legion303]

It is physically impossible to create a perfect physical lock.

Ignoring that for your thought experiment, you then run into the fact that for there to be 0 tolerance error (i.e., "perfect lock"), the pins would have to be exactly the same width as the chambers down to the atom, and thus wouldn't be free to move at all.

-steve

[FarmerFreak]

Aside from the aforementioned fact that a perfect lock wouldn't work.

Normally a lock would have a specific repeatable binding order in which the pins can be set consistently and reliably over and over again. Because, obviously, one pin would be binding at a time. Sometimes you can find a lock where two or more pins will bind at a time, and either pin can be set first. Which, to me at least, would explain what would happen in a so called perfect lock. All the pins would bind equally, and just as with the not so perfect locks. Either pin could be set first, and either pin could be set last.

[FarmerFreak]

Since I didn't mention this in my previous post. It may not be possible to tell if a pin is set correctly.

Normally the way to check if a pin is set correctly is to check and see if it is springy and not binding. But checking if the pin is springy when it is set correctly requires some space between the plug and the housing. And obviously there wouldn't be that space in a perfect lock. But without that space, the lock wouldn't function correctly at all.

[gloves]

you then run into the fact that for there to be 0 tolerance error (i.e., "perfect lock"), the pins would have to be exactly the same width as the chambers down to the atom, and thus wouldn't be free to move at all.

that's exactly what I was thinking to

[3-in-1]

Of course there is no perfect lock and I am glad to see at least one of you (Farmerfreak) got into what the topic was all about. Binding and picking. Trouble thinking outside the box? In my perfect lock, the pins do move up and down like they should by the key otherwise it wouldn't be, as was pointed out, a lock. Loop yourselves back to my original 2 postings and reread the scenario. In this lock there is no discernible binding order. At least there doesn't seem to be. The only variable in this perfect lock is the different length of the pins. Everything else is the same. The only goal in the design of the lock is to defeat SPP. Makers have long known that the closer the machining the tougher it is to pick. So called "security pins " exist because they are cheaper to make than extremely fine machining.

[gloves]

Of course there is no perfect lock and I am glad to see at least one of you (Farmerfreak) got into what the topic was all about. Binding and picking. Trouble thinking outside the box? In my perfect lock, the pins do move up and down like they should by the key otherwise it wouldn't be, as was pointed out, a lock. Loop yourselves back to my original 2 postings and reread the scenario. In this lock there is no discernible binding order. At least there doesn't seem to be. The only variable in this perfect lock is the different length of the pins. Everything else is the same. The only goal in the design of the lock is to defeat SPP. Makers have long known that the closer the machining the tougher it is to pick. So called "security pins " exist because they are cheaper to make than extremely fine machining.

I'm sorry if my speculation went too far on your original subject, but I felt like you can't write "apple" without knowing the letters a,e,l,p. Again don't take my words in a bad way, I'd just like to explain my thought to you.

As I and Legion303 pointed out, a perfect lock won't have enough looseness to leave pins travel up and down. If you say my perfect lock can do this, you can do nothing but admit that your perfect lock is no longer perfect. It's like an oxymoron, one of those figures of speech which combines contradictory terms, like saying "She's a hot ugly girl". Is she hot or ugly?
In the same way speaking about a lock, best you could do is to say: a lock with the smallest tolerances.

It'd work, be really precise and difficult to feel while picking, but again the presence of tolerances which makes it work leave an open door to exploitation.

So in my honest opinion I think I have dimostrated by logical speculation, along with fellow members which replied in your thread, that:

a) A perfect lock with pins being able to move would no longer be a perfect lock because it would still have some tolerances.
b) A lock works because it has tolerances, as small as they may be.
c) To defeat SPP the tolerances have to be smaller than the locksmith's ability to feel the lock.

Note on point C: because everyone's ability is different, this is subjective. Also I won't leave out the idea that using hearing/sight aids may shift this human limit to the point that lock tolerances fail (to make a working lock) before human ability to feel does.

So best I'd accept from your "lock with closest possible tolerances to make it work" is that it'd beat a certain percentage of human lockpickers without external improvements. Hope I expressed my thought clearly enough and that they flow flawlessly in your minds.

On a side note out of this theoretical discussion: I know most of you know that many things aren't pursued in this way in the real world because tiny tolerances would make the lock really expensive and all this alone to prevent a certain manipulation could be simply bypassed by other means.

Cheers

[dls]

the perfect lock is a single pin lock no binding order when you realy think about it

[3-in-1]

Gloves, i will split my response here into a couple posts since I don't see that well, Can't type worth a crap and think slowly. First off I am more than happy to see responses that indicate some thinking has gone on ( on topic or off and in agreement or not), so it is tough to upset me. So lets start with a. I didn't take logic or philosophy and I am a college drop out. I suspect that a lock that doesn't work due to lack of the necessary tolerances required for it to function properly would not be described as a perfect lock. I would think that a perfect would not only work but work perfectly. But that is beside the point here and we'll go to b.

b. Locks work and don't work due to tolerances. Yesterday I worked on a Diebold safe lock where the dial ring had bowed (some sort of plastic) to the point of restricting dial movement and the wheels in the lock were rubbing against each other. So in that situation you had both increasing and decreasing tolerances, both of which led to a poor working lock.

c. Now we are talking cause this what the original post was all about. But it will have to wait until I sober up some. And your video of you wearing gloves opening a padlock is just sick.

Oh and Gloves, regarding a,p,l and e's, I can just hear some upper level physics prof saying that "if I have to spell it out for you, you may be in the wrong class". Just a joke son.

[FarmerFreak]

I suspect that a lock that doesn't work due to lack of the necessary tolerances required for it to function properly would not be described as a perfect lock.

In that case, you can ignore my second post entirely.

So ignore this

Normally the way to check if a pin is set correctly is to check and see if it is springy and not binding. But checking if the pin is springy when it is set correctly requires some space between the plug and the housing. And obviously there wouldn't be that space in a perfect lock. But without that space, the lock wouldn't function correctly at all.

We need to ignore that because if the lock has enough tolerances that the plug can rotate. Then there will be enough space between the plug and the housing to check if a pin is springy, and thus set correctly.

And I stand by my first post.

Normally a lock would have a specific repeatable binding order in which the pins can be set consistently and reliably over and over again. Because, obviously, one pin would be binding at a time. Sometimes you can find a lock where two or more pins will bind at a time, and either pin can be set first. Which, to me at least, would explain what would happen in a so called perfect lock. All the pins would bind equally, and just as with the not so perfect locks. Either pin could be set first, and either pin could be set last.

If a lock where made to tight enough tolerances to where all the chambers are perfectly lined up, exactly the same size, and all the pins were the perfect diameter. And the plug had enough room to rotate. I believe that it could be SPP'ed.

Actually, this sentence "And the plug had enough room to rotate." creates another perfect tolerance problem. If the plug has enough room to rotate, then it will have enough room for the front of the plug to be pushed, ever so slightly one direction {left or right} and the back of the plug a different direction. That in itself would make it so you could pick the lock either front to back or back to front.