Theory and computations about calculating recoil in firearms
Theory and computations about calculating recoil in firearms
Abstract:
Recoil can be expressed mathematically by the physical law of the conservation of momentum, the law of the conservation of momentum states that for two or more bodies in an isolated system action upon each other, their total momentum remains constant unless an external force is applied. In simply it is stated, the law says that for every action there is an equal and opposite action also the energy can neither be created nor destroyed.
Recoiling of a gun:
Recoil is the backward momentum of a gun when it is discharged. The
recoil caused by the gun exactly balances the forward momentum of the projectile
and exhaust gases according to the Newton’s third law. The theory behind the
recoil is when the gun is fired the projectile starts moving inside the gun barrel
by the pressure created by the expanding gases, in this time the projectile moves
forward and the case that holds the chemical gun powder travels the rearward by
the expanding gas pressure, why it travels rearward because after ignition of the
primer the burning take place inside the cartridge case and the expanding pressure
expands inside the chamber in upward, downward direction and forward and
rearward direction. There is no gap or openings provided in the upward and
downward direction, only it can go either rearward or forward direction, so the
expanding gas pushes the projectile through the opening provided in forward
direction also it pushes the spent case through the opening provided in the back
of the barrel. The backward pushing energy is transferred from the fire arm to
shooter. In simple way after the firing the projectile goes forward and the gun
comes the backward by the pressure of the expanding gas, (1) free recoiling of
firearm is equal and opposite in direction to the momentum of the bullet (or shot
charge/slug and wad column) and the propellant gases. (2) Felt recoil (also
referred to as “Perceived” recoil) can also vary by modifying the rate of
application of force by devices such as recoil pads or damping devices and the
influence of action type. (3) Because the propellant gases are extremely difficult
to weigh, for purposes of this application, the propellant gas weight will be
equated to the powder charge weight. Generally, we can use the recoil force to
cycle the weapon automatically until the ammo exist. (4) Recoil-operated
weapons are those weapons that rely on recoil activity to operate the bolt and
2
related parts. (5) The bolt, locked to the barrel during firing, is released during
recoil after the chamber pressure has become safe. The recoil force varies on the
power of the ammunition and the mass of the firearm.
Free Recoil energy is also called as kinetic energy and it can be expressed as -
FRE = ½MV2.
Where:
M = Mass (or the weight of the firearm, including all attachments)
V = Velocity of the recoiling firearm
To find out the velocity we can use the mentioned formula:
Where:
WE = Weight (in grains) of the projectile
VE = Velocity of projectile in feed per second
WPG = Weight of the propellent gases in grains
VPG = Velocity of the propellent gases in feet per second (fps)
7000 = Conversion factor for grain to pounds
WF = Weight of firearm in pounds
If, therefore:
Finally, we got,
The weight of the propellent gases WPG is equal to the propellent charge weight
WPC. The formula given below is used to find out the recoil energy of the firearm.
The following VPG should be used:
VPG = VEf
High powered rifles VPG = 1.75VE
Shotguns (average length) VPG = 1.50VE
Shotguns (long barrel) VPG = 1.25VE
Pistol & revolvers VPG = 1.50VE
If the calculations are made in metric system, the below formula can be used
If the calculations are made in metric system, the below formula can be used
Let us see a sample problem about calculating recoil,
Problem no 1
How much Free Recoil Energy would be developed by an INSAS rifle weighing
4.6 Kilograms firing a 5.56 x 45 mm with bullet weighing 4.16 grams and loaded
with 27 grains of powder and the velocity of the projectile is 890 mps.
Solution,
Given data
Weight of the fire arm in kilograms Wf = 4.6 kilograms
Weight of the projectile in grams WE = 4.16 grams
Weight of the propellent charge in grams WPC = 1.81 grams
Velocity of the projectile in mpsVE = 890 mps
The following velocity of the propellent gas VPG should be used VPG = VEF
where
for
The f for high powered rifles is VPG = 1.75 VE
Apply the given data into the given formula,
why we choose metric formula because the data's are given in kilograms
FRE = (4.6 ÷ 2 x 9.8) x (4.16 x 890 + 1.81x 890 x 1.75 / 4.6)
2
FRE = (4.6 ÷ 1.96) x (4.16 x 890 + 1.81 x 890 x 1.75 / 4.6)2
FRE = (23.46) x (37024 + 28190 / 4.6)
2
FRE = (23.46) x (65214 / 4.6 )2
FRE = (23.46) x (14.17)
2
FRE = (23.46) x (200.7)
FRE = 4.70 Kg mps
The free recoil would be given in kilo-meters.
References:
1.) Gun recoil technical SAAMI standards (sporting arms and ammunition
manufacturing institute)
2.) Gun recoil technical SAAMI standards (sporting arms and ammunition
manufacturing institute)
3.) Gun recoil technical SAAMI standards (sporting arms and ammunition
manufacturing institute)
4.) Engineering design handbook automatic weapons, 1970 US Army
material command.
5.) Engineering design handbook automatic weapons, 1970 US Army
material command.
6.) Gun recoil technical SAAMI standards (sporting arms and ammunition
manufacturing institute)
7.) Bullet specifications are taken from TECHNOLOGY IN FOCUS paper
about small arms from DRDO
Comments
Post a Comment