Pressure and reactions

Now we are going to look at how the slide of the gun is pushed backwards and returns to his starting position. The explosion inside the barrel causes a enormous pressure buildup which causes the bullet to go forward. Once the bullet has left the barrel the pressure will only be pushing to the back so the slide will be pushed backwards over the frame. Once the pressure has escaped a coil underneath the barrel will expand again causing the slide to move back to his starting position.

Technical details of this coil.

Forces of the coils

The recoilspring has a normal length of 6,55 inch (=16,6 cm), inside the gun she is compressed to 9,4 cm when the gun is not active. When the gun is active and the slide moves backward the coil is compressed to a length of 4,59 cm which will increase the force that the coil generates.
From the technical details we know that the k value which is the spring rate of the coil is equal to 2,88 pounds per inch (= 51,4 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= 16,6 cm – 9,4 cm = 7,2 cm
F = 51,4 kg/m * 0,072 m = 3,7 kg
We have to multiply by 9,81 m/s² to get F in Newton!
So 3,7*9,81 = 36,3 N

When the gun is fired the compression is equal to : ∆L= 16,6 cm – 4,59 cm = 12 cm
F = 51,4 kg/m * 0,12 m = 6,17 kg
We have to multiply by 9,81 m/s² to get F in Newton!
So 6,17*9,81 = 60,5 N