The flatland model of locks can explain effects that involvs more than one pin, but a different model is needed to explain the detailed behavior of a single pin. See Figure 5.1. The pin-column model highlights the relationship between the torque applied and the amount of force needed to lift each pin. It is essential that you understand this relationship.
In order to understand the “feel” of lock picking you need to know how the movement of a pin is effect by the torque applied by your torque wrench (tensioner) and the pressure applied by your pick. A good way to represent this understanding is a graph that shows the minimum pressure needed to move a pin as a function of how far the pin has been displaced from its initial position. The remainder of this chapter will derive that force graph from the pin-column model.
Figure 5.2 shows a single pin position after torque has been applied to the plug. The forces acting of the driver pin are the friction from the sides, the spring contact force from above, and the con tact force from the key pin below. The amount of pressure you apply to the pick determines the contact force from below.
The spring force increases as the pins are pushed in to the hull, but the increase is slight, so we will assume that the spring force is constant over the range of displacements we are interested in. The pins will not move unless you apply enough pressure to overcome the spring force. The binding frictionee is proportional to how hard the driver pin is being scissored between the plug and the hull, which in this case is proportional to the torque. The more torque you apply to the plug, the harder it will be to move the pins. To make a pin move, you need to apply a pressure that is greater than the sum of the spring and friction forces.
When the bottom of the driver pin reaches the sheer line, the situation suddenly changes. See Figure 5.3. The friction binding force drops to zero and the plug rotates slightly (until some other pin binds). Now the only resistance to motion is the spring force. After the top of the key pin crosses the gap between the plug and the hull, new contact force arises from the key pin striking the hull. This force can be quite large, and it causes a peak in the amount of pressure needed to move a pin.
If the pins are pushed further into the hull, the key pin acquires a binding fiction like the driver pin had in the initial situation. See Figure 5.4. Thus, the amount of pressure needed to move the pins before and after the sheer line is about the same. Increasing the torque increases the required pressure. At the sheerline, the pressure increases dramatically due to the key pin hitting the hull. This analysis is summarized graphically in figure 5.5.
Figure 5.1: The pin-column model
Figure 5.2: Binding in the pin-column model
Figure 5.3: Pins at the sheer line
Figure 5.4: Key pin enters hull
Figure 5.5: Pressure required to move pins
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[...] Knowing how a lock works when it is opened by a key is only part of what you need to know. You also need to know how a lock responds to picking. Chapters 3 and 5 present models which will help you understand a lock’s response to picking. [...]
[...] 5 The Pin Column Model [...]
[...] The slow step in basic picking (chapter 4) is locating the pin which is binding the most. The force diagram (Figure 5.5) developed in chapter 5 suggests a fast way to select th correct pin to lift. Assume that all the pins could be characterized by the same force diagram. That is, assume that theyall bind at once and that they allencounter the same friction. Now consider the effect of running the pick over all the pins with a pressure that is great enough to overcome the spring and friction forces but not great enough to overcome the collision force of the key pin hitting the hull. Any pressure that is above the flat portion of the force graph and below the top of the peak will work. As the pick passes over a pin, the pin will rise until it hits the hull, but it will not enter the hull. See Figure 5.3. The collision force at the sheer line resists the pressure of the pick, so the pick rides over the pin without pressing it into the hull. If the proper torque is being applied, the plug will rotate slightly. As the pick leaves the pin, the key pin will fall back to its initial position, but the driver pin will catch on the edge of the plug and stay above the sheer line. See Figure 6.1. In theory one stroke of the pick over the pins will cause the lock to open. [...]
[...] This exercise will teach you the range of torque you will need to apply to a lock. It demonstrates the interaction between torque and pressure which was described in chapter 5. [...]
This is a great guide thanks!
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