# Limits on EM kinetic weaponry

I am looking for kinetic armament for spaceships. With nuclear reactors providing power, there is plenty of energy for kinetic EM weapons. EM weapons come either as coil guns or as rail guns.

What are the limits on how much velocity you can impart on a projectile of given mass with either of the two systems, if you have limited barrel length/tonnage available for the weapon installation?

• Reduce the mass, increase the speed, and shape your barrel like a torus. – NofP Feb 19 '19 at 22:41
• The most advanced railgun ever built imparted a muzzle velocity of 2.5 km/second, which is remarkably high for artillery, yet pitifully slow for space combat. – AlexP Feb 19 '19 at 22:48
• As I understand it, the chief problem with adding more power isn't the size of the rails - it's keeping them from melting. – Cadence Feb 19 '19 at 22:51
• @AlexP Everything is slow compared to the speed of light – Shadowzee Feb 19 '19 at 23:42
• @AlexP: That depends on what you consider an EM weapon. SLAC en.wikipedia.org/wiki/SLAC_National_Accelerator_Laboratory is capable of accelerating particles almost to the speed of light. Though as a weapon, aiming might prove a bit of a problem :-) – jamesqf Feb 20 '19 at 4:55

I'm gonna neglect the following aspects:

• aerodynamic drag - since you are in space
• cooling - I'm gonna assume ideal cooling
• heating/melting of the projectile - I have no idea how to calculate that

Take a nuclear reactor: e.g. Turkey Point U.S. with a nominal output power of 1000MW. The kinetic energy of a moving mass is $$0.5*m*v^2$$

The question is, how much energy can the projectile gather with the given barrel length?

The velocity over time is $$v(t)=a*t$$ (with constant acceleration $$a$$)

If you integrate it over time, the travelled distance is $$s(t) = 0.5*a*t^2$$

You can only accelerate in the barrel with length $$L$$, so $$L = 0.5*a*t^2$$. $$a = 2L/t^2$$

Power is work (energy) over time ($$P=E/t$$), or force times distance ($$P=F*s$$): $$P = 1000MW = F*L = m*a*L$$. So $$a = 1000MW/(m*L)$$

$$1000MW/(m*L) = 2L/t^2$$

$$t= sqrt((2L*mL)/1000MW) = sqrt((2*m*L^3)/10^9W)$$

So the kinetic energy is (btw. I think this is the part, that is most interesting for you): $$E = P*t = 10^9W * sqrt((2*m*L^3)/10^9W)$$

And the velocity: $$v= sqrt(2*E/m) = sqrt(2*(10^9W * sqrt((2*m*L^3)/10^9W))/m)$$

Estimated projectile velocities with a 1000MW reactor.

You have to consider, that the damage done by the projectile is mostly dependent on the kinetic energy.

Note: These are the ideal conditions. You would have thermal losses in the coils/conductors (you could minimize them with superconductors).

You could increase the power with very large (I mean huge) capacitors. They need to store energy, and discharge, while the projectile is accelerating. This would mean higher projectile energy, but also a recharge time for your vessels.

• Any reason to go with these weird numbers instead of orders of magnitude in the graph? – dot_Sp0T Feb 20 '19 at 8:59
• No special reason. I had to write a script to generate the graph, and add some numbers as input. I did not want to go over 1000kg, because it did not seem realistic, and just the three lines 10, 100, 1000 did not seem to be enough. If you have a better idea, I can modify it. – G. B. Feb 20 '19 at 9:06
• I would've probably gone with [kg]: 0.1, 1, 10, 100, 1000, 10'000 and maybe even add comparisons to it: chocolate bar, bottle of water, a bicycle, newborn elephant cub, small car, truck tractor – dot_Sp0T Feb 20 '19 at 9:44
• This would give you more evenly distributed datapoints and allow easy interpolation – dot_Sp0T Feb 20 '19 at 9:46
• ok, done it, thanks! – G. B. Feb 20 '19 at 9:51