SFIC Best style mystery

[Raymond]

Today I had to remove an SFIC from a broken lever and replace it in a replacement lever. The core remove key was supposed to be available but the two supplied would not work. Anyway I picked it. I took apart the core to make a control key and am still confused by what I found.

The one mystery column had pins: bottom=5, wafers 2, 14, & 2(top). Yes they total 23. But what would the control cut be and how did they get it into the lockset? I am 100% certain of the order that the pins came out.

I changed the pinning to make a control key and get it back in but still cannot figure out what was going on.

[Evan]

@Raymond:

There is a reason why you were confused by what you found when you decoded the core...

The control key cut for the chamber pinned in the manner you described would have been: 11

As you know the deepest depth possible in an BEST A2 keying system is 9...

So this was either a homebrewed prank by someone who cut an out of spec control key to install the core to begin with OR it was combinated and manipulated into the control lug retracted position and installed with no possible way to remove it other than picking or drilling or knowing which position to cut two increments past a valid bitting depth...

If you do work on BEST A2 SFIC cores on a regular basis you know the math:

Control + ( 10 - Plug Total ) + Build Up = 23

Apply this to your given numbers substituting "X" for the unknown control cut:

X + ( 10 - 7 ) + 2 = 23

How do we solve for "X" ? There are two ways:

Method #1:

In relation to the A2 top (build up) pin --

13 - Build Up = Control Key Cut
Solving for your "X": ( 13 - 2 = 11 )

* This method requires that the core in question was pinned according to the rules, meaning that the locksmith who pinned the core didn't make the stack height 22 or 24 as some sort of decoding frustration security method -- this is the most commonly used decoding technique which can result in an incorrect answer if the core was not pinned exactly to BEST SFIC pinning specifications...

Method #2:

In relation to the A2 Control pin --

Control Pin - ( 10 - Plug Total ) = Control Cut
Solving for your "X": ( 14 - ( 10 - 7 ) = 11 )

* This is the decoding method used by people who know what they are doing because it relies upon the relationship between the keys which operate the core and make the control function possible rather than hoping that the rules were followed by whomever pinned the core...

Do you know who installed the core in question ?

Was it a disgruntled employee or vendor ?

Now you know why your control keys wouldn't work...

~~ Evan

P.S. I am not a mathematician by any means... The ONLY math I am good at is the "A2 SFIC math" and the "mathematics of masterkeying" as it pertains to system sizes... I just thought I should mention that as a suffix to this very technical posting before people here start thinking I am some sort of math or programming genius or something...

[Evan]

If you do work on BEST A2 SFIC cores on a regular basis you know the math:

Control + ( 10 - Plug Total ) + Build Up = 23

I am going to amend this formula and claim a mulligan based on being sick, medicated (NyQuil) and also suffering from insomnia tonight --

Plug total ( Bottom pin + any required master pins ) + [ Control cut + ( 10 - plug total ) ] + Build Up = 23

Anyone with magical moderator powers who wishes to make that change can do so and also eliminate this reply with my thanks...

~~ Evan

[keysman]

I am not sure where this came from , but I DID NOT write it
Keysman
You need a dial caliper, a pinning kit, a capping block, and patience.

1. Take and knock out the pins and keep them in order of the way they came out.
2. Measure the pins and write them down. It's usally the 3rd set of pins. The 4th set is the satck pins.
3. Take a cylinder and cut the top chamber in half and keep it to use for pinning.
4. When you take the measurements and you put the pins in the cut cyl. you will be able to tell and actually see the control shear line and will be able to cut a control key.

Here is some instructions that Terran sent me:
(note: this document refers to A2 system. For other systems,
substitute appropriate numbers.)

It is the theoretically possible to get the control using
only the driver pins. However, doing so assumes that the lock
was correctly pinned to specifications, and that the total
stack height is 23. My experience has been that this is *not*
always the case. I've seen A2 systems which had a stack
height of 24, and some locks which didn't even have a
consistent height for all the chambers. Cutting a control key
using the height of the top pin in one of these cases will
probably result in a key that doesn't work.

For this reason, I prefer to take all my measurements from
the *bottom* of the stack; these are guaranteed to work
because you're measuring the same dimensions that actually
cause the lock to operate.

I'll include a copy of the A2 heights here for a convenient
reference:

A2 (Arrow, Falcon, Eagle, Best)
MACS: 9

ROOT BOTTOM MASTER CONTROL
DEPTH PINS PINS PINS DRIVERS
----- ------ ------ ------ -------
# 0 .318 .110 --- ---
# 1 .305 .122 --- --- **
# 2 .293 .135 .025 .025
# 3 .280 .147 .037 .037
# 4 .268 .160 .050 .050 .050
# 5 .255 .172 .062 .062 .062
# 6 .243 .185 .075 .075 .075
# 7 .230 .197 .087 .087 .087
# 8 .218 .210 .100 .100 .100
# 9 .205 .222 .112 .112 .112
#10 .125 .125
#11 .137 .137
#12 .150 .150
#13 .162 .162
#14 .175
#15 .187
#16 .200
#17 .212
#18 .225
#19 .237

To start with, let's consider a non-master-keyed lock, which
has only an operating key and control key. This lock will
have three pins in each chamber; the break between the bottom
two pins operates the lock and the break between the top two
pulls the core.

The operating key is the easiest to understand. Best-style IC
locks have a plug, a ring around the plug, and then the rest
of the core. When the lock operates, only the plug turns; the
ring remains fixed. The bottom pin, therefore, simply
corresponds to the cut heights on the operating key; a
operating key of 210563 would therefore have bottom pins of
2, 1, 0, 5, 6, and 3. Ignore the break between the top two
pins and think of the two combined as the driver pin.

To remove the core, you need to rotate the ring which
surrounds the plug; the plug remains fixed with respect to
the ring. The ring is 1/8" wide. Theoretically, therefore, it
would be possible to remove the lock with a key that was
simply a copy of the the operating key with all of the cuts
1/8" (10 cuts) highger. In practice, however, this is
impossible because such a key would be too high to fit into
the lock. This is where the middle pin comes in. To
understand how the control key works, ignore the break
between the bottom two pins - just think of them as one long
driver. Similarly, disregard the break between the plug and
the ring; just think of the plug and ring together as a plug
with a greater diameter. We want to pin this lock so that a
particular key (the "control" key) operates it. We pin this
exactly like we ordinarily would except that all of the pins
need to be 1/8" longer because of the 1/8" larger effective
diameter of the lock. Recall that 1/8" corresponds to 10
integer units in the A2 system. Therefore if we wanted a
control key of 346346, we'd pin the lock with pins 13, 14,
16, 13, 14, and 16.

Now it's time to remember that this theoretical pin is
actually two pins. But we've already figured out what the
bottom of these should be in the last step! So to put in the
second pin, just figure out what height you need so that the
two together add up to the number you just figured out for
the control key. For example, we have a #2 bottom pin in the
first chamber. We need the total height of the bottom two
pins to be 13, so we put in a #11 pin above it.

At this point, we should have:

Operating Key: 210563
Control Key: 346346
Lock:
11 13 16 8 8 13
2 1 0 5 6 3

Now all that's left is the driver pins. The driver pins are
selected so that the total stack height adds up to 23.
There's nothing magical about the number 23; it's just that
some guys at Best did a bunch of tests and proclaimed that
given the height of the column, the nature of the springs,
etc., a height of 23 was the optimum amount to make the lock
operate well.

In this example, this would make the final pin selection:

10 9 7 10 9 7
11 13 16 8 8 13
2 1 0 5 6 3

Suppose, however, that the lock instead looked like this:

10 10 7 10 10 8
11 13 16 8 8 13
2 1 0 5 6 3

Perhaps the person who pinned it had run out of certain pins,
didn't know how he was "supposed" to do it, or just didn't
care. The difference is small enough that the lock would
almost certainly operate fine. But let's see what would
happen if we tried to make a control key by taking the top
pin and subtracting it from 13. We'd get:

336335

This key might not work (remember that the actual control key
is 346346). How, then, do we decode the lock to get a working
key? We ignore the driver pin, because the operation of the
lock doesn't depend on it. We do the same thing as we did to
figure out how to pin the lock, but in reverse. Getting the
operating key is simple - we just look at the heights of the
bottom pins, and that's the cuts for the key. To get the cuts
on the control key, we *add* the heights of the bottom two
pins in each chamber, and subtract 10 from the result
(remember, this is exactly the same procedure we used to
figure out how to pin it to start with). In the first
chamber, for example, 11+2=13. 13-10=3, which is the correct
cut for the control key. This technique gives us a control
key of 346346, which is correct.

Now, what if the lock is master keyed? First of all, it won't
be possible to distinguish the master key and the change key
from just one lock unless you already know one of the two -
you can't do it with normal locks, and there's nothing
special about ICs that would let you do it with them either.
You can, however, get the control key from just one lock. How
do we do this? Let's consider a hypothetical lock. Just to
make it even more interesting, this lock is master keyed
using a rotating constant system, so not all chambers will
have the same number of pins:

This lock will have the same control key as the one we used
before, and the key that was previously our operating key
will now be our master key. The change key will be: 412366.
The lock should look like:

11 10 8 11 10 8
9 13 14 8 8 10
2 2 2 3
2 1 0 3 6 3

(the pins in chambers 2 and 5 are arranged as they are for
clarity; the gap being where it is doesn't really mean
anything).

You could make a key that would operate the lock by just
taking the bottom pins; this would give you 210363.
Unfortunately, this is probably an incidental master that
doesn't do you much good. If you already knew the change key,
you could of course figure out the master key by taking the
cut that *isn't* on the change key in all those positions
where there is masterkeying. I won't go into that in too much
detail now however because that's more of a masterkeying
concern and not specific to ICs.

You could in this case, also try to make a control key by
using the driver pins. Again, however, this lock isn't pinned
to spec, and this would give 235235, which is wrong. The more
reliable way to do it is to add up the bottom pins and
subtract 10 from the result. But how many bottom pins do you
add up? It's simply - you add up the heights of all the pins
except the top one. This means that in chambers 1, 3, 4, and
6, you'll add up the bottom three pins, and in chambers 2 and
6, you'll add up the bottom two. For example, in chamber 1,
9+2+2=13 and 13-10=3 for a correct control cut. This
procedure gives the control key 346346, which is correct.

I think this covers most of the basics of what's involved. If
you have any questions I'll be happy to try to answer them.
Note that nothing here is guaranteed to be correct, and I did
all the arithmatic in my head as I was typing this in, so I
wouldn't be at all surprised if it has some mistakes in it.

[Raymond]

WOW!!! Did I ever push your buttons. Thanks for the long winded replies. I am sure someone that does not do SFIC will appreciate the class. Really, it must have taken a while to work up all the examples and instruction. So, thanks for real.

I have been doing SFIC for a long time and know the formulas frontward and backward. The pinning I gave at the start of this thread was the true measurement of the pins as they came out of the lock. These numbers do not follow the rules. With a bottom pin of (5+2) 7 and the control of 14, no control key would work. The stack would be 2 steps, (.025) higher than the control shear with no key in the lock. This was the tip column and therefore the first to be taken apart (no loose pins cluttering up my workbench). With complete clarity I removed the cap, spring, and was dismayed to see the 2 wafer first out of the hole, followed by the 14, 2, and 5. I guess it will remain a mystery how they got it into the lever in the first place.

This lock was factory pinned and installed by the Corp of Engineers at an Army building. I was called in by the contractor to trouble shoot warranty problems as three levers had failed.

I didn't really expect an accurate answer as that combination doesn't have one. I just wanted to SHOUT about a difficult experience. Thanks again for the detailed replies.

[Evan]

@Raymond:

It didn't take very long to write my reply...

In the case of your mystery core, 7 (plug total) + 14 (Control pin) = 21 or 1 increment above the control shear line...

So to operate the core, a cut of 5 or 7 (.2562" or .2312" key depth) would have the shear line at the plug... In order to function at the control shear line, an invalid cut of 11 (a theoretical .1812" key depth) would be required on the key... As has been stated that is 2 increments past the deepest cut according to specs...

There is no mystery -- if this was in fact a factory pinned core it must have come with a control key cut to an 11 depth in the tip chamber position... There would have been no other way possible for a core to be installed in a housing unless it was picked to the control lug retracted position and inserted...

I believe that you carefully and accurately measured the pins in the correct order as they existed in the core... All this means is that the BEST factory is apparently able to produce cores that will still function normally even though it would require an out-of-spec control key to remove from the housing...

Sort of like how some lock manufacturers have a deeper bitting depth that can be used on factory produced construction master keys...

That you for telling everyone about your mystery... It is always interesting to hear about an encounter with non-standard factory keyed locks captured in the wild...

~~ Evan

[Quabillion]

Execlent info Keysman

Exactly what I have been looking for to complete my BEST SFIC project. Link here
http://www.lockpicking101.com/viewtopic.php?f=9&t=51721

[Raymond]

I know this is an old topic but new pickers can still find it interesting.
I just thought of a way to make it more difficult for someone to decode the control key for your system.
Split one or two top pins per core into two pins. Operating cut 4, master key cut 6, control cut 5.
Example: bottom to top: normal: 4 2 9 8 (total 23).
Bottom to top: enhanced: 4 2 9 2 6 (total 23).
The confusion factor is how to read the control and top pins. Is the control 4+2+9-10=5 or is it 4+2+9+2-10=7. A person would have to decode several cores to determine the correct control key cut. Otherwise they could cut one test key for each choice. That is, cut the first key to the 5 and if it doesnt work several other cores, cut it down to the 7. If you do this to 2 top pins, the guesser would have to make 4 test keys.