Desperatezulu
AH enthusiast
- Joined
- Jul 23, 2015
- Messages
- 398
- Reaction score
- 782
- Location
- South Africa
- Hunted
- South Africa, Zimbabwe, UK
Eish guys - the physics are straightforward albeit not intuitive!
I'll try again...
The Law of conservation of momentum states that m1v1 = -m2v2
I.e. (mass of object 1 * vel of object 1) is equal to and opposite (hence the negative) the (mass of object 2 * vel of object 2)
Let's call object 1 the bullet and powder mass and object 2 the gun. We want to solve for the velocity of object 2 (the gun)
Therefore v2 = -(m1/m2)*v1
v2 is the recoil velocity, or the 'free' velocity of the gun.
Still with me?
V2 = -(39.88g/5000g) * 2000 ft/s
(I have assumed a 500gr bullet and 100gr of powder which equates to 38.88g. Gun weighs 5000g. I have made a big simplification on the velocity of the powder and bullet combo because technically the bullet may be going at 2000ft/s but the powder charge is going at something like 5000ft/s. So the value of m1v1 is actually higher and the recoil velocity would be higher than my simplified answer. But the underlying principle stands - the key factor here is that the mass of the bullet& powder doubles with a double discharge)
Thus V2 = -15.55 ft/s for the single discharge
Hope we're all agreed so far.
If we double discharge, the formula looks like this:
v2 = -((2*39.88)/5000) * 2000 = -31.9 ft/s
Note the "2" in bold above - this is because we have twice the mass of bullet and powder.
We see how the recoil velocity has doubled.
Hope I haven't lost the die hards still following
Now we need to calculate the recoil energy:
I think we're all agreed that the formula for calculating kinetic energy is:
KE = 1/2 mass * velocity^2
Or: recoil energy of gun = 0.5 * mass of gun * recoil velocity * recoil velocity
I hope I can cut to the chase and point out that a doubling (i.e. a factor of 2) of the recoil velocity, once squared, becomes a quadrupling (factor of 4) of the kinetic energy value.
I'll try again...
The Law of conservation of momentum states that m1v1 = -m2v2
I.e. (mass of object 1 * vel of object 1) is equal to and opposite (hence the negative) the (mass of object 2 * vel of object 2)
Let's call object 1 the bullet and powder mass and object 2 the gun. We want to solve for the velocity of object 2 (the gun)
Therefore v2 = -(m1/m2)*v1
v2 is the recoil velocity, or the 'free' velocity of the gun.
Still with me?
V2 = -(39.88g/5000g) * 2000 ft/s
(I have assumed a 500gr bullet and 100gr of powder which equates to 38.88g. Gun weighs 5000g. I have made a big simplification on the velocity of the powder and bullet combo because technically the bullet may be going at 2000ft/s but the powder charge is going at something like 5000ft/s. So the value of m1v1 is actually higher and the recoil velocity would be higher than my simplified answer. But the underlying principle stands - the key factor here is that the mass of the bullet& powder doubles with a double discharge)
Thus V2 = -15.55 ft/s for the single discharge
Hope we're all agreed so far.
If we double discharge, the formula looks like this:
v2 = -((2*39.88)/5000) * 2000 = -31.9 ft/s
Note the "2" in bold above - this is because we have twice the mass of bullet and powder.
We see how the recoil velocity has doubled.
Hope I haven't lost the die hards still following
Now we need to calculate the recoil energy:
I think we're all agreed that the formula for calculating kinetic energy is:
KE = 1/2 mass * velocity^2
Or: recoil energy of gun = 0.5 * mass of gun * recoil velocity * recoil velocity
I hope I can cut to the chase and point out that a doubling (i.e. a factor of 2) of the recoil velocity, once squared, becomes a quadrupling (factor of 4) of the kinetic energy value.