The cocking phase is where the Colt 1911 differs from single action weapons. It means that this weapon automatically arms the pistol for the next shot, by pulling back the hammer. When the slide is moving backwards it pushes against the hammer which causes the hammer to rotate to its starting position. The mainspring will be pushed back and because the is no more pressure on the trigger, the sear will stop the hammer from rotating and firing again. The gun is now ready to fire the next bullet.
26-11-2010
Ejection and reloading
Now let’s see how the empty bullet case gets removed from the barrel and a new bullet takes place.
Because of the backwards movement of the slide, the bullet case also moves backward. Once the slide is completely at the end, there is an opening in the slide that corresponds with an opening in the frame just large enough for the case to be ejected. Because of the kinetic energy from the backwards movement of the case a little tap of a pin at the end of the barrel is enough to shoot the case out of the barrel. Now the case is gone it is no longer holding down the new bullets inside the magazine from going up. The magazine spring underneath the bullets constantly pushes the bullets upwards. It is the slide that returns to its starting position that pushes the new bullet inside the barrel.
Because of the backwards movement of the slide, the bullet case also moves backward. Once the slide is completely at the end, there is an opening in the slide that corresponds with an opening in the frame just large enough for the case to be ejected. Because of the kinetic energy from the backwards movement of the case a little tap of a pin at the end of the barrel is enough to shoot the case out of the barrel. Now the case is gone it is no longer holding down the new bullets inside the magazine from going up. The magazine spring underneath the bullets constantly pushes the bullets upwards. It is the slide that returns to its starting position that pushes the new bullet inside the barrel.
Barrel linkdown
Barrel linkdown is essential for a number of reasons. First of all it causes the slide to come loose of the frame which makes it able to slide backwards. It also makes it possible to eject empty bullet cartridges and provides an angle with the magazine so a new bullet can slide into the barrel.
Under pressure of the shot the part underneath the barrel called the barrel link will rotate 45 degrees which causes the entire barrel to rotate a few degrees. Because of the recoil spring, the barrel link will rotate back up again after the shot.
Under pressure of the shot the part underneath the barrel called the barrel link will rotate 45 degrees which causes the entire barrel to rotate a few degrees. Because of the recoil spring, the barrel link will rotate back up again after the shot.
16-11-2010
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
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
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.
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.
Different kinds of guns
There are a lot of different kinds of guns, also a because there are a lot of aspects where a gun can differ from another. For example, there are single action guns as there are double action guns. Single action weapons don't automatically recock. Which means that every time a bullet is shot the gun has to be manually recocked to fire the next bullet. Unlike double action weapons who recock automatically when the slide moves backwards.
Another major difference is the type of gun. Revolvers are guns which have a cylindrical bullet magazine, where pistols have a bullet magazine that stacks the bullets on top of each other.
The size of the bullet that the gun uses is called the difference in caliber. The caliber that the Colt 1911 uses is a 9mm. The 9mm stands for the diameter of the bullet.
And maybe the biggest differences in guns is found on the inside. There are a lot of different kinds of mechanisms inside a gun. Every mechanism can vary in guns.
Another major difference is the type of gun. Revolvers are guns which have a cylindrical bullet magazine, where pistols have a bullet magazine that stacks the bullets on top of each other.
The size of the bullet that the gun uses is called the difference in caliber. The caliber that the Colt 1911 uses is a 9mm. The 9mm stands for the diameter of the bullet.
And maybe the biggest differences in guns is found on the inside. There are a lot of different kinds of mechanisms inside a gun. Every mechanism can vary in guns.
How bullets work!
To understand how handguns work we first need to know how a bullet works. A bullet consists out of 5 important parts. The bullet (1), the bullet case (2), the propellant (3), the rim (4) and the primer (5). When the striking pin of the gun hits the primer a small chemical reaction will occur that gives a spark to the proppelant. The proppelant will ignite wich will lead to a massive pressure build-up behind the bullet. The pressure will look for a way out, the least resistant captivator is the bullet, so the bullet will be pushed forward through the barrel by the pressure of the explosion behind it.
The pressure behind the bullet in the Colt 1911 is aproximatelly 82,7 MPa!
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