Some refer to this equal pin stack height as controlled pin stacks. True uniform pin stack height assures that each spring receives consistent pressure when in a relaxed state, thus prolonging the total life of the spring and cylinder. To the contrary, uneven pin stacks put uneven wear on springs and pro- mulgate non-uniform effectiveness of that particular core. However, if the tension level is within the designed limits, it’s OK. Deter- mining the specific pressure amount for each product is up to the various manufacturers (for their respective items). The Control Key for the 6300 Series IC The 6300 Series IC control key bitting is easily created by copying the top master key (TMK) of the key system in positions 1, 2, 5 and 6 on the key blade. The control bittings to be used in positions three and four are selected from the key bitting array (KBA) of the master key system. This method significantly reduces the bittings available in the KBA for any system. Increasing the levels in the master keying sys- tem and cross keying also has a significant im- pact on the yield of keys at each selected level. Pin Stacks for Non- Control Chambers Determine the bottom pin and any master pins needed in each chamber to make the change key and TMK work properly (tem- porarily forget about the control key). Cal- culating pin sizes required for the operating keys (TMK and CKs) is based on the master key system bittings used within the system. The steps are as follows: Determine the smallest key bitting number to be used in each chamber based on the change key and TMK being used. This de- termines the bottom pins for that chamber. Determine which additional combination of master pins will be necessary in each chamber to allow the change key and TMK to turn the plug (this will automati- cally include all master keys). SARGENT Bottom Pins No. Length 1 2 3 4 5 6 .170 .190 .210 .230 .250 .270 7 8 9 .290 .310 .330 10 .350 .020” Increment Two Step Progression .115” Diameter Pins 6300 IC Master/Top Pins No. Length 2 3 4 5 6 7 .040 .060 .080 .100 .120 .140 8 9 12 .160 .180 10 .200 11 .220 13 .260 14 .240 .280 Zero (0) Bitting Depth = 10 [Deepest!] ♥ Control Chambers #3 and #4 Pin Stack Height: (Non-Control) = 15 Pin Stack Height: (Control Chambers) = 20 Decoding: CTRL Bitting: 12 – Top Pin (for CTL Chambers ONLY!) (for CTRL Chambers ONLY!) ☼ Build Up Pin: (CTRL Cut + 8) – (BP + MP) There are Forbidden Combinations in Chambers 3 & 4 →MACS = 7 →→ Figure 1. SARGENT factory method for calculating pin stacks — 6300 series IC. The total value of these two previous steps will create the bottom pins and required master pins to operate the IC. Subtract this total (coded number) from the total pin stack height of 15, and you have the top pin for chambers #1, #2, #5 and #6. Because you have calculated pin stacks for chambers #1, #2, #5 and #6, only the two control chambers are leſt to calculate. Pin Stacks for Control Chambers Since you know the bittings for both the control key and operating keys, you can calculate the stacks for the control cham- bers quite easily. With that in mind, there are two rules to consider. First, because the control key can’t be greater than six steps from the master key (or operating keys) the number of possible bittings is somewhat limited. If the difference were greater, you would need a #1 master pin, which theoretically doesn’t exist. To avoid that problem, stay inside this limit. Second, the cuts on the control key should be different from those of all the operating keys (including masters). Here’s an easy formula to help calculate the CTRL pin stacks: Step 1: Add 8 to your control key bitting for chamber four. Subtract the total from 20, and you’ll have your top pin for the stack. Or use this formula: 12 - CTRL cut = top pin Step 2: Add the total of the bottom pin, master pin and result from Step 1. Subtract the total from 20, and you’ll have your con- trol pin. Or use the formula: (CTRL cut + 8) - plug total = BUP Step 3: Repeat the process for chamber #4. Step 1 uses the number eight not because it is a magical number, but because it is the coded distance control differential between the operating and control shear lines. In real dimensions, the operating shear line is about .498 inch in height, and the control shear line is about .658 inch in height. The difference is .160 inch. If you divide the difference by .020 inch (the system increment) you get the coded eight. Also, you subtracted it from 20 because that’s the total of the pin stack for that chamber. The formula provided does the same, but in a different way. Step 2 simply shows that if you know the total and all of its parts but one, you can add and subtract to find the missing part — using simple math you learned in the third grade. Again, the formula provided calculates the same result, but in a different way. Step 3 just repeats the process. The SARGENT factory method for cal- culating all pin stacks for the 6300 Series IC is included in the accompanying chart (see Figure 1). 25
