if the lock...

by **kelly** » 24 Apr 2006 15:10

When locks are configured for multiple keys (ex: 12 doors each have different keys but there's one master that opens them all) are the pins [begin stupid question] in triplet? [end stupid question]

I've been trying to mentally take apart a lock with a master key and the only was I can think it would work would be to have three pins in each possition rather than two. Well, not all possitions, but maybe the first or last three of a six pin lock.
Crude Diagram:

@@@@@@ <---springs
@@@[][][]
[][][][][][] <-pins
[][][][][][]

If this is true, would it not make the lock more comperable to raking? If this is not true, could someone tell me in no uncertain terms why I am stupid (reason #1 is I'm sure not googling it first)?

Many thanks,
-K

by **toomush2drink** » 24 Apr 2006 15:33

You are correct about the raking. Also the pins in master systems can vary in height too to allow more configurations, ie sometimes they may only be a very thin disk rather than a pin as such.

by **Eyes_Only** » 24 Apr 2006 15:37

I never really botherd to study masterkeying in depth but with up to 12 doors in a system I think it would have more then 3 pins in most of the locks pin stacks. Masterkeying ends up creating multiple shear lines so this would make the lock more vulnerable to raking methods. Unless the lock is like a bi-axial interlocking pin system with a sidebar or something.

by **Shrub** » 24 Apr 2006 16:10

Kelly youve got a great grasp of it there especially as youve worked it out yourself, just to add master key pins can be very thin discs and as said do lesson the security a lot in a lock against picking.

by **cheesehead** » 24 Apr 2006 18:41

I've master pinned many a lock. In fact some of the first locks I ever picked were mastered. And in fact, the addition of master pins makes them much easier to pick, since the master pins create extra sheer line options. way to figure it out though - I had to take a lock apart before the concept fully "clicked" for me...

by **toomush2drink** » 24 Apr 2006 19:04

Rekeyed a lock for a customer on saturday and discovered it was mastered, he never even knew !! As i left the building i noticed all the cylinders were the same so not got any problems opening any of them really if they are all mastered, a few hundred flats too.

by **Varjeal** » 24 Apr 2006 21:11

kelly:

To answer your questions, here they are:

#1. Are the pins in triplet? Not necessarily, and unless its a pretty large job, not likely. And btw, that's not a stupid question.

#2. Does it make it easier to rake? Yes.

In a six pin system using 10 possible depths and a MACS of 9, a masterkey system can be potentially created using only one masterpin in a single chamber of each lock.

Let me try and explain with an example:

Let's say we have a master key with the bitting 495610.

All 12 locks are keyed differently with their own key, and the master works them all...it could be done like this.

Lock #1: 495612
Lock #2: 495614
Lock #3: 495616
Lock #4: 495618 (only changed the last chamber of each)

Lock #5: 495630
Lock #6: 495650
Lock #7: 495670
Lock #8: 495690 (only changed position 5 of each)

Lock #9: 495810
Lock #10: 495010
Lock #11: 495210
Lock #12: 495410 (only changed position 4 of each)

If you notice what I've done, for each lock I've only changed one position. (using a safe increment of 2)

If I were to write down the information for lock#1 in a chart, it would look something like this:

C 495612
M 495610
--------------
B 495610
M XXXXX2

This information would tell me that when I actually key the lock, I first need to put in the bottom pin increments of 495610. The "M" portion at the bottom tells me that in the last chamber I need to put a #2 increment master pin to make it all work.

Hope that makes sense.

P.S. Yes, to all you smart people I know that locks 10-12 are unsafe for use as they can be cut down to operate as masters, this is just an example.

by **Mutzy** » 25 Apr 2006 1:44

P.S. Yes, to all you smart people I know that locks 10-12 are unsafe for use as they can be cut down to operate as masters, this is just an example.

what he means, for the people who are new to master keying,
the master key has cuts 495610

the individual keys in question (10-12) have the cuts higher than the master, and can be cut down to have the same cuts as the master key.

i.e.
495610 - Master
495010 - Key 10
495210 - Key 11
495410 - Key 12

So basically, the 4th cut of the individual keys can be cut down from their cut (0/2/4) to the master key cut of 6. Which is something you should make sure cannot happen.

And that's just ONE of the things you have to keep in mind about master key systems. Wait untill we get into phantom keys...

And for 10 points, can someone tell me why keys with cuts only 1 cut different are bad? (i.e. 495610 and 495710) :wink:

by **keysman** » 25 Apr 2006 5:16

Mutzy wrote:
And for 10 points, can someone tell me why keys with cuts only 1 cut different are bad? (i.e. 495610 and 495710)

OK for 10 points! ( to use as I wish ?????) :twisted:

The most common reason would be the masterpins would be small ( thin) enough to get jammed .
Also you get keys that are worn down a bit and then become phanthom keys .. working in unintended locks.

just an aside.. Best IC uses a 1 for the depth difference allowing for massive masterkey systems.

ok do I get the 10 points :?:

by **datagram** » 25 Apr 2006 9:53

Kelly, yes, having masterkey systems creates many shear lines. The equation for this is possible shear lines = 2^n, where n is the number of masterkey pins added. So in a 5 pin cylinder with no masterkey system, you have 2^0 = 1 shear line; only one configuration of pins allows the cylinder to rotate. If pin 1 has a masterkey pin added, that number jumps to 2, and so on and so forth. In a 5 pin lock with 5 masterkey pins (totalling 15 pins, 3 per stack), you get 2^5 = 32 (!) shear line possibilities.

This makes it much easier for picking AND raking to succeed, but with some locks picking and raking still may be very hard.

by **kelly** » 25 Apr 2006 11:48

Wow, that was a pretty thurough answer. Thanks especially to Varjeal. I printed out the thread to read over again. My only hesitancy in taking apart my master keyed lock is that it technically isn't mine. It's on my door, but the door belongs to the school, so off limits.
I really like the idea of masterkeying, really interesting. I'm thinking about buying a set of best cores off ebay later to play with.

ASIDE: with our locks, when the master key is inside, the entire core can be removed by turning it counterclockwise and pulling. What property of the key allows that to occur?

by **p1ckf1sh** » 26 Apr 2006 6:00

datagram wrote:Kelly, yes, having masterkey systems creates many shear lines. The equation for this is possible shear lines = 2^n, where n is the number of masterkey pins added. So in a 5 pin cylinder with no masterkey system, you have 2^0 = 1 shear line; only one configuration of pins allows the cylinder to rotate. If pin 1 has a masterkey pin added, that number jumps to 2, and so on and so forth. In a 5 pin lock with 5 masterkey pins (totalling 15 pins, 3 per stack), you get 2^5 = 32 (!) shear line possibilities.

I think that the 2^n equation only relates to configurations where the masterkey pins/discs are spread over multiple pin chambers, right? I was never good at math and rolled this around in my head for a while, so I might be mistaken, but let me explain my thoughts, maybe someone can point out my mistake.

Simple lock, 3 pins. Regular bitting is

2 - 4 - 6 = only the key cut to 246 will open. 2^0 = 1 holds true.

2 - 4+1 - 6 = adding a masterkey disc of 1 bitting height to middle chamber, now two keys will work: 246 and 256, 2^1 = 2 is true

2 - 4+1+1 - 6 = adding two discs of 1 bitting height each to middle chamber, now three keys will work: 246, 256 and 266. The equation 2^2 gives 4 as the result. Where is the lost key and what is it's bitting? Or is the equation not applicable to this example?

Because in a different example...

2 - 4+1 - 6+1 it holds true. This is one disc per chamber, and this config yields 4 keys: 246, 247, 256 and 257 so 2^2=4 applies.

Can anyone enlighten me?

by **Fogmeister** » 26 Apr 2006 6:16

Hmm... that's true.

Give me a couple of hours and I'll get you a proper equation.

by **Fogmeister** » 26 Apr 2006 6:40

OK, I got a formula but it isn't a very tidy one.

Basically the number of keys can be worked out by multiplying the number of pins in each chanber (is that the correct word).

i.e.

2 - 4+1 - 6 = 1*2*1 = 2 keys
2+1 - 4+1+1 - 6+1 = 2*3*2 = 12 keys

by **datagram** » 26 Apr 2006 6:43

Ah...I hadn't thought of that (bad idea, anyways, isn't it

?) But I went over the math and this is correct (and works globally):

For each chamber in the plug, take the # of shearlines for that chamber and multiply it by the number of shearlines in all other chambers.

Example:

In our 3 pin cylinder, we have:

12(+1+1)3, so we have 1*3*1 = 3 shearline possibilities

With 12(+1+1)3(+1+1), we have 1*3*3 = 9 shearline possibilities.

This should work for any number of masterpins and any number of chambers...

Please let me know if I've made another mistake!

datagram

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