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The Pin Column Model
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.
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.
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 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 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. 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
friction like the
driver pin had in the initial situation. 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 5.1.
Figure 5.1
p|
r|
e|
s| maximum due to collision with hull
s| |\
u| | \
r| | \
e|_______ | \_______ spring+friction
| \ |
| \_|spring force only
|_______________________
displacement of pin ->