Recoil question

1. Apr 7, 2008

iamsomeguy

A group of friends and I were playing air-soft. During one of the breaks between matches one of folks there started waxing poetic and talking about how air-soft guns could be turned into real guns. The discussion turned to using magnetic rails in the barrel to accelerate a .12mm pellet. How heavy the gun would be, the velocity of the pellet, force of impact. Then one of the folks there talked about how great it would be having zero recoil.

The discussion then changed to one of physics and how there wouldn't be zero recoil due to conservation of energy. The force to propel would still have opposite force. Some of us threw some hackneyed math around... but still nothing definitive.

I think it's not that simple of a formula as recoil takes into account design and weight of the weapon as well.

Anyone want to add some real work into this speculation? I'd love to get some input, and it would help settle a bet (at stake, one really nice bottle of booze).

:-)

2. Apr 7, 2008

Briggs

Sounds like you're talking about a Railgun

3. Apr 8, 2008

iamsomeguy

Yep, I am. But a man-portable one. Bascially there are two key science-fictiona assumptions here. One is that the magnetic acceleration can be put into a rifle sized container (M16, AK47, etc.) and two is that a power source is suitable enough to make it worth while.

The part that was missing in our poor math was that the reverse force was not directed back to the shooter, but would be instead directed outward, having the "barrel" absorb the recoil force while the shooter would have no recoil (or miniscule).

Would you say that's an accurate assement?

4. Apr 8, 2008

Danger

I vote 'no' on that. The reaction will be in the opposite direction to the action, so it will indeed be transfered back to the shooter. There will probably be some variances due to the nature of the weapon, but in essence it should be the same as firing any other gun.

5. Apr 8, 2008

ehj

It is exactly the same as with a real gun. In physics the phenomenon is called conservation of momentum, denoted p. Classically momentum is defined as $$p = m*v$$ where m is mass and v is velocity. So let's say you have a gun with a bullet in it with a mass $$m_g_b$$ and because you are holding it, and is standing still, the velocity of the gun and bullet inside of it is zero, and therefore the momentum is zero. After the gun is fired, the momentum must me conserved, meaning the momentum must be zero after the gun was fired, that means
$$p_g_b = p_b_a + p_g_a$$

where $$p_g_b$$ is the momentum of the gun + bullet before it's fired and $$p_b_a$$ is the momentum of the bullet, after the gun is fired and $$p_g_a$$ is the momentum of the gun after it is fired. If we in above equation insert what we earlier concluded, that $$p_g_b$$ is zero we get:

$$0 = p_b_a + p_g_a$$

the bullet has momentum since it has a velocity and a mass, and in order for the equation to hold true, the velocity of the gun must be negative, in other words, directed in the opposite direction - because it's mass is positive.

6. Apr 9, 2008

LURCH

IOW; if you build it so that it spits out a projectile with energy equal to a rifle bullet, you will get the same kick as you would from a rifle (or nearly the same). This will be true whether the projectile is the same mass as a rifle bullet and exits with the same muzzle velocity, or much smaller and lighter than a bullet and exiting at much higher muzzle velocity.

I say "nearly the same," because in a firearm, the chemical explosion in parts an acceleration force that peeks very early in the bullet's travel down the barrel. With a real gun, acceleration is constant all the way down the barrel. Since the entire transfer of energy from the butt of the rifle to the shooter's shoulder takes place in a very small fraction of a second, it would probably not be possible to feel this difference. However, it could be measured if one had precise enough equipment.

7. Apr 9, 2008

skywalker09

Automatic guns utilize the recoil to load the next round into the chamber, thus minimizing the force experienced by the operator. Look into that.