# Would a gauss rifle based on generated magnetic fields have any kickback?

In the case of currently developing Gauss rifles, in which a slug is pulled down a line of electromagnets, facilitated by a micro-controller to achieve great speed in managing the switching of the magnets, does the weapon firing produce any recoil? If so, how would you go about calculating that recoil?

• The equal and opposite reaction has to go somewhere, but it may be interesting to try to devise a dissipation mechanism. Commented Nov 27, 2012 at 13:15
• Note that momentum is not recoil. Recoil is the effect that firing has on the wielder, especially their ability to maintain aim. So the force distribution over time would seem to be what is important, and a Gauss rifle would have a lot more control over this than a firearm. Not to mention the ability to reduce it simply by increasing the length of the barrel, which gives almost linear increase in velocity with the same force, while rifles have diminishing returns with barrel length. Commented Nov 27, 2012 at 21:10
• @paulhutton has the basics of what I was hoping to ask. It seems to me that the reverse momentum would be transferred to the coils as each on would be active, so you would be generating a lot of small recoils rather one large one. Commented Nov 28, 2012 at 8:05

If the Gauss rifle shoots a projectile with exit speed of $v_1$ and mass $m_1$, then its momentum will be:

$p=m_1v_1$.

Because of momentum conservation law, the rifle will have the same momentum in opposite direction. If the rifles mass is $m_2$, the rifle will start moving in the opposite direction with end speed of:

$v_2 = \frac{m_1 v_1}{m_2}$.

But, as the projectile is accelerated for longer time than in a gun, the force acting from rifle on its holder will be lower because $F=\frac{dp}{dt}$

• Forgive my ignorance, but I can't help but think there has to be a difference in recoil between a Gauss rifle and a conventional rifle firing the same mass at the same speed -- isn't some of the energy of the normal round's explosion lost (via the cycling mechanism, or because the explosion force is omnidirectional)? And therefore, the charge must be more powerful to provide the same exit velocity, which means slightly greater recoil? Also, doesn't the gas exiting the barrel behind the projectile also exert a force (kind of like thrust in a rocket engine)? Physics was never my strong suit. :) Commented Nov 27, 2012 at 15:10
• @Justinᚅᚔᚈᚄᚒᚔ the projectile exerts an equal force back on the coils which provides the recoil. The gas exiting the barrel after the bullet counts as the momentum of the bullet - the recoil is the total of everything going forward, it doesn't matter if it's bullet/propellant. So yes for a given projectile mass/speed you will have a lower recoil in a Gauss rifle because you don't have the extra momentum of the forward gases to recoil from Commented Nov 27, 2012 at 16:19
• As Paul notes in the comments on the question itself there is also the question of the distribution of the momentum transfer in time. For a fixed projectile velocity the Gauss rifle can (presumably) have a lower peak force. Only experiment will demonstrate how this trades off in terms of single shot accuracy and the difficulty of keeping on target in rock-n-roll mode. Commented Nov 27, 2012 at 23:52

You are imparting a force to accelerate the slug, so you're going to get an equal and opposite reaction. In a normal rifle, the explosion accelerates the bullet rapidly and you get recoil.

In a gauss rifle, the acceleration will be a bit lower, but for a slightly longer time (the entire length of the barrel), so for the same muzzle velocity you will be able to calculate the recoil in the exact same way.

• AFAIK unless you're using low power ammo in a rifle (eg .22Long), the bullet will still be accelerating by a significant amount until it exits the barrel. You will have different acceleration curves and the gauss gun's will presumably be flatter than that of the conventional firearms. The latter will have a sharp peak near the start reaching when the propellant is fully burned and then falling off as the bullets forward movement reduces the pressure of the combustion gas. Commented Nov 27, 2012 at 16:41

About the magnitude. This source claims 2056m/s exit speed, this one around 2382m/s (mach 7).

I have no idea what a personal reilgun's projectile weight should be. Let's take a random pistol round: 7.5g.

Gun length? Let's pick 1m. Gun weight? Let's pick 3kg.

v_gun = - p_projectile / m_gun =
(2382*7.5*(10-3)) / 3 =
5.955m/s


acquired over one meter of travel at let's say constant acceleration.

x = v_0 t + (1/2) a t^2
2 = a t^2
t = sqrt(2/a)
where a = v / t
t = sqrt(2t/v)
t = sqrt(t) sqrt(2/v)
sqrt(t) = sqrt(2/v)
t = 2/v
t ~= 8.4ms


Now the (obviously constant) force is going to be

F = m a
a = 5.955m/s / 8.4*10^-3s = 7089m/ss
F = 3kg * 7089m/ss = 21267N


A typical pro boxer's punch is 5000N. So we are going to need lighter bullets. I wonder what would happen if you shoot this thing in a sandstorm. Or indoors.