Chapter 5  
 
The Pin Column Model  
 
 The flatland model of locks can explain effects that involve 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 t 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 contact 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 into 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 friction 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, a 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, they key pin acquires a binding
 friction 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
 sheer line, the pressure increases dramatically due to the key pin hitting the
 hill. This analysis is summarized graphically in figure 5.5.






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