Hand Held Coil Gun cannon - Enough to rip a giant robot limbs into pieces

Question

OK, so my main character has created a hand held coil gun cannon, the whole coil gun is 6 feet long with the barrel taking 5.5 feet of the said coil gun.

The coil gun is equipped with recoil nullifying module(Don't ask what is it because even I don't know yet. I'll make a separate question for that) so recoil at best is like firing a grenade launcher.

Now as for the ammunition, they are like the current ammo used in guns except that the bullet is a ferromagnetic projectile, and the casing or in this case super capacitors which stores 1.1TJ of energy for a single shot and once discharged it is ejected out from the coil gun cannon like spent casings in modern day gun once spent in a single shot.

Now the objective of this design is to be able to breach 1 meter of thick carbon/steel composite armor of possible giant robots, monsters or barriers of enemies that may be encountered.

With 1.1 Tera joule powering a coilgun for a single shot, how powerful a coilgun can be with 50mm ferromagnetic projectiles?

What kind of figures will I be needing to pierce that kind of level of protection 1 meter of carbon/steel composite?

Coil gun specifications that I have

1. I have no idea for the wire, I probably need stronger wire to effectively conduct a 1.1 Tera joule power. Since I need to loop it around a single coil 400 times, it needs to be thin. probably thinner than the current wires of today.

2. There are 8 coils 18 centimeter each that is looped 400 times(Tight and no spacing) in triple layer configuration which is evenly distributed over the 5.5 feet length of the barrel.

3. Of course insulators.

4. "Bullet" or projectile length is 9cm.

5. The projectile is fin stabilized using flexible materials.

6. Projectile is spun by an electric motor running at 60,000 rpm.

7. Internal battery for internal computer and electric motor.

Tell me if I am missing something from the design.

Answer 1

One issue which I note you havn't considered in the written question is how, exactly, the device will be powered? Coilguns, or related devices like railguns use electrical energy, so the soldier is going to need either some sort of generator or a pretty gigantic set of batteries or capacitors to power the weapon. So the physical size is going to be pretty impressive, more like a cannon than an anti material rifle.

Perhaps the easiest way to provide some sort of portability and mitigate the recoil forces (which will be quite substantial) will be to drive the weapon using chemical explosives (an compact, high density energy source) through an MHD generator. The explosives are detonated in an open ended chamber, and the flow of supersonic gasses pass through the MHD generator to provide the energy to drive the coils In many respects, this would be somewhat like a recoilless cannon, and also have some of the disadvantages of a recoils cannon (backblast and visible flash)

One other issue is that coilguns need to be braced against the powerful magnetic forces, and a means of dissipating the heat in the coils and switching devices, so the analogy of a cannon is actually quite accurate, especially if you are firing a very small projectile at megasonic speeds (like Mach 400+ in JerryTheC's answer).

The obligatory Atomic Rockets page will help you calculate the ultimate size of the coilgun, projectile mass and how much energy you will actually need, but here is a start:

Here's a quick method to estimate what kind of performance you can get out of a coilgun. Some folks here might find it interesting.

First, decide on the efficiency of your coilgun. Coilguns are linear brushless electric motors, and brushless electric motors have demonstrated efficiencies of 90% to 95%. Superconductive electric motors might have efficiencies of 98% to 99%. Denote this as a decimal, and call it e; that is e = 0.9 to e = 0.95.

Next, decide on the length and radius of your projectile. Decide on what your projectile is made of and find its mass

mass = density * length * radius2 * &pi (and remember to use consistent units).

Also find the projectile cross-sectional area

area = radius2 * π

Decide how fast you want your projectile to be going and find its final kinetic energy

kinetic energy = 0.5 * mass * velocity2 (again remember to use consistent units).

Given the efficiency of your coilgun, you can find out how much your projectile heats up. You might figure that half of the wasted energy goes into the projectile, and thus your projectile will gain a heat energy of

heat energy = 0.5 * (1/e - 1) * (kinetic energy)

Look up the specific heat of the material your projectile is made of, commonly called C. Then your projectile reaches a temperature of

projectile temperature = (heat energy) / (C * mass) (again make sure your units are consistent).

Now assume that the barrel is filled with field, and that the projectile sweeps the field out of the barrel, turning the field energy into kinetic energy (this is not actually how coilguns work, but it gives the physical upper limit based on energy conservation). The energy density is about 400 kJ/m3/T2 times the square of the magnetic field strength (398,098 J/m3/T2 to six significant figures). Call this value K

K = 400 kJ/m3/T2

You now know the volume needed in the barrel based on how much energy the projectile ends up with

volume = kinetic energy / (K * (magnetic field)2)

Since you know the cross-sectional area of the projectile and thus of the barrel, you know how long the barrel needs to be

length = volume / area

Since you have the parameters, you can calculate what the actual performance of your coilgun will be. Alternatively, you can now work backwards from the amount of energy you want to place on the target and design your coilgun.

Good luck!

Answer 2

As WhatRoughBeast pointed out, watts are power (rate of energy supply) and joules are energy (How much energy in total stored by your capacitor).

If you want to know how much energy can be transferred to your target, then that's your 1.1GJ times the efficiency of your coilgun (less any losses due to atmospheric resistance).

If you work out the mass of your projectile, you can calculate the muzzle velocity ignoring air resistance (which, looking at the result below probably isn't something you can safely ignore).:

kinetic energy of projectile = 1.1GJ * efficiency, and since ke = 1/2 mass times velocity squared, you get (for non-relativistic speeds) velocity squared = ke / (0.5 * mass), so velocity = square root of (ke / (0.5 * mass))

as a quick sanity check, with a 100g (0.1Kg) projectile, and 100% efficiency that gives

V = sqrt(1.1x10^9 / 0.05) = 148,324 m/s or about mach 432.

which is fast but nowhere near lightspeed. But it may be fast enough to worry about whether your projectile burns up en route to the target...

Answer 3

The coilgun specifications are still not making a great deal of sense, since the projectile is 50 mm in diameter and only 90 mm long, making a stubby projectile which is not at all well-adapted to penetrating armor. Nonetheless, we'll go with it.

A steel cylinder of these dimensions will weigh about 1.4 kg. Assuming 100% coilgun efficiency this will produce a velocity of about 40 km/sec for 1.1 GJ, since$$\frac{mv^2}{2} = 1.1\times10^9 $$ and $$v=\sqrt{\frac{2.2\times10^9}{m}}=\sqrt{\frac{22}{1.4}}\times 10^4=12.5\times10^4=40,000$$

This site discusses historical formulae for armor penetration of steel armor, and these use Imperial quantities, so

D = 2 inches

W = 3 lbs

V = 132,000 ft/sec

A representative formula for the penetration referenced to projectile diameter (T/D) is $$\frac{T}{D}= (0.00005021) D^{0.07144}[(\frac{W}{D^3})(V/C)^2(2Cos^3(Ob)]^{0.71429}$$ where C is in the vicinity of 1.2, and Ob is the incidence angle. Assuming normal incidence, the cos term is 1 and drops out, leaving $$\frac{T}{D}= (0.00005021) D^{0.07144}[(\frac{W}{D^3})(V/C)^2]^{0.71429}$$ Crunching through this gives$$\frac{T}{D}=432.6$$ so $$T = 865\text{ inches} = 22\text{ meters} $$

Of course, this is an unrealistic number, as discussed in the link, since the projectile speed is much greater than the speed of sound in either the armor or the projectile. Nonetheless, it suggests that penetration of a meter of armor should be no problem as long the impact angle is fairly close to normal.

Another way to look at the problem is to assume that the projectile creates a hemispherical crater in the armor by vaporizing it. Such a crater will have a volume of $$V = \frac{1}{2}\frac{4\pi r^3}{3} = 2 m^3$$ Assuming an armor density of 8000 kg/$m^3$ this is a mass of 16000 kg. The vaporization energy of steel is about 6.8 MJ/kg, so this worst-case measure of required energy is about 108 GJ.

So the question will be, can the projectile maintain its integrity during the penetration process? If yes, it appears likely that it can penetrate a meter of armor. If 1 meter is enough depth to spread out the impact, the answer is no.

I suspect that, given the high velocities involved, penetration will occur, with lots of energy left over to produce damage in the target. Making the diameter smaller and the length greater will improve performance.

EDIT - It is useful to compare your proposed penetrator to a real one. The M829A1 penetrator weighs 4.6 kg and has a muzzle velocity of 5500 fps. At point blank range it will penetrate 670 mm of steel armor, so something with half the mass but 24 times the velocity would seem a reasonable candidate for 50% better penetration. However, the M829A1 is 2 1/2 times as dense as steel, considerably harder than steel, and is configured as a long slender rod. All of these things will work against your coilgun projectile.

Also, just for comparison, the KE of the M829A is about 6.4 MJ.