Firing mechanism

Because there are a lot of mechanisms inside a gun that happen simultaneously it is better to look at every mechanism at a time to put them all together at the end. The first mechanism is the firing mechanism.

When we pull the trigger, the trigger pushes a small piece of metal which will push the sear spring. The sear spring will rotate the sear so the hammer can rotate towards the striking pin. The hammer will rotate because in the handle there is a coil that is constantly pushing the hammer. The sear is the part that prevents that from happening until the trigger is pulled.
The hammer will hit the striking pin which on his turn will strike the primer of the bullet.



Technical details of this mechanism are mainly the characteristics of the coil.

Forces of the coils
The main spring has a normal length of 2,156 inch (=5,476 cm), inside the gun she is compressed to 2,697 cm when the gun is not active. When the gun is active, when the coil is released she will expand to a length of 3,33 cm.
From the technical details we know that the k value which is the spring rate of the coil is equal to 27,69 pounds per inch (= 494,7 kg per meter).

When we use the Hook’s Law we can calculate the forces generated by the coil.
F= k*∆L

In inactive state the compression is equal to : ∆L= 5,476 cm – 2,697 cm = 2,779 cm
F =494,7 kg/m * 0,02779 m = 13,7 kg
We have to multiply by 9,81 m/s² to get F in Newton!
So 13,7*9,81 = 134,4 N

When the gun is fired the compression is equal to : ∆L= 5,476 cm – 3,33 cm = 2,143 cm
F =494,7 kg/m * 0,02143 m = 10,6 kg
We have to multiply by 9,81 m/s² to get F in Newton!
So 10,6*9,81 = 104 N

So we can see that the force that will hit the striking pin will be 104 N, because the coil gets bigger it loses energy and force.