Can Air-Soft Guns Be Transformed into Real Guns Using Physics?

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The discussion revolves around the feasibility of using magnetic rails to accelerate air-soft pellets, with participants debating the implications for recoil. It is established that, despite the idea of zero recoil, the conservation of momentum dictates that some recoil will still be felt by the shooter, similar to traditional firearms. The physics of recoil is analyzed, with emphasis on how the forces involved would still produce a backward momentum equal to the forward momentum of the projectile. Additionally, the challenges of creating a man-portable railgun, including material strength and power source requirements, are acknowledged. Overall, the conversation highlights the complexities of translating theoretical concepts into practical applications in weapon design.
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Let me start with the story first.

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).

:-)
 
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Sounds like you're talking about a Railgun

The rails and projectiles must be built from strong conductive materials; the rails need to survive the violence of an accelerating projectile, and heating due to the large currents and friction involved. The recoil force exerted on the rails is equal and opposite to the force propelling the projectile. The seat of the recoil force is still debated. The traditional equations predict that the recoil force acts on the breech of the railgun. Another school of thought invokes the ampere's force law to assert that it acts along the length of the rails (which is their strongest axis)[2]. The rails also repel themselves via a sideways force caused by the rails being pushed by the magnetic field, just as the projectile is. The rails need to survive this without bending, and must be very securely mounted.
 
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?
 
I vote 'no' on that. The reaction will be in the opposite direction to the action, so it will indeed be transferred 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.
 
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.
 
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.
 
Automatic guns utilize the recoil to load the next round into the chamber, thus minimizing the force experienced by the operator. Look into that.
 
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