5
$\begingroup$

In my previous question a guy called Thorne pointed out that I could make a high-velocity coilgun without using a barrel of enormous length by taking a page out of the LHC's book and using a circular coilgun, allowing the projectile to run several times around the track. That particular question wasn't really about coilguns, though, and I had another previous question that was. In my coilgun question, the most reasonable answer suggested that a coilgun with a 100 meter-long barrel might, just might be able to get a fairly small projectile up to about 30 km/s, but almost certainly couldn't do the same with fairly massive one. So, does the answer change if I'm using a circular track to accelerate the shot? Let's assume the track I have to work with is about 300m in diameter, and once again my goal is to get a 1000kg projectile up to about 30 km/s. I suspect that the mass of the projectile will once again be a problem, though; and I suspect that the design of the accelerator will probably lend itself more to a constant stream of ball-bearing sized iron shot.

EDIT: To clarify, the idea behind the circular coilgun is that a projectile can run around the track more than once before it is fired. The barrel length of the gun is thus effectively infinite (or rather, unlimited), it can be however long it needs to be to get the projectile(s) up to speed. I just don't know whether this makes it practical or not.

And on a related note, what about use of an actual ring-shaped particle accelerator as a weapon? Devices like the LHC are kilometers in diameter, and while they can get their particles up to nearly the speed of light (from what I've heard), the actual destructive power of the beam isn't that great. I don't need a particle stream anywhere near that fast, as long as I can dump enough mass into the beam for its effect on the target to be worthwhile. Is a 300m diameter ring anywhere near enough to be useful? Can such an accelerator be built as a weapon?

Maybe that's why the Enterprise has a circular saucer section...

$\endgroup$
  • $\begingroup$ When the phasers are on full power the forces involved threaten to rip the saucer apart. $\endgroup$ – Joe Bloggs Sep 3 at 20:52
  • $\begingroup$ You've asked two questions here... I'd split them if I were you. Particle accelerators make for pretty terrible weapons, FWIW (thermal and electrostatic blooming ruin your range). For your coiled-coilgun (ho ho) you'll find that TANSTAAFL, because you'll need to expend additional power to bend the projectile into a curved path, and you'll need much more substantial bracing to hold the gun together. $\endgroup$ – Starfish Prime Sep 3 at 21:25
  • $\begingroup$ @StarfishPrime I knew that blooming was a problem, but I figured that electrostatic blooming could be compensated for by running two accelerators, one with a positively-charged beam, the other with a negatively-charged beam, and firing them together. As for thermal, that crops up if the beams are superheated; but do magnetically-accelerated beams need to be superheated? $\endgroup$ – FlyingLemmingSoup Sep 3 at 21:46
  • $\begingroup$ @FlyingLemmingSoup good luck defining the difference between velocity and heat for particles. There's a reason you'll find plasma temperature measured in electronvolts! The act of neutralising a beam will disrupt it slightly, causing defocus; there's no practical way around that. There are uses for charged particle beams as weapons, but they're specialist ones. You'd be better off using your accelerator to drive a free-electron laser for shooting things in space. $\endgroup$ – Starfish Prime Sep 3 at 21:51
  • $\begingroup$ @StarfishPrime Fair enough. I figure that lasers would probably only be useful at short range, though. Useful if you're close enough for a pulse to vaporize hull plating. If you're far enough that they turn into a means of overheating the target, then the target can wear aluminum foil to reflect most of the energy, and your own laser is overheating your own ship more than it is the enemy. $\endgroup$ – FlyingLemmingSoup Sep 3 at 21:56
7
$\begingroup$

The answer given by mcRobusta accounts very well for the circular accelerator. The formula is indeed $F=\frac{m v^2}{r}$. But to compare to the linear case, consider a linear gun with barrel length $d=2 r$. At constant acceleration the exit velocity is $\sqrt{2 d a}$. That is, to get a velocity of $v$ you need a force as follows. $$ F = m \frac{v^2}{2d}= m\frac{v^2}{4r} $$

That is, to keep your mass going round the ring you need to accelerate it 4 times as hard as to get it to the same speed in one diameter. Which isn't that strange since you are in fact flinging the mass back and forth at its maximum speed, meaning you must go from full speed "left" to stopped in one radius, then back to full speed "right" in one radius back.

So if you are looking to use smaller force, the linear gun is best by a factor of 4.

Edit: To respond to the comment. The ring needs to accelerate the mass with a force $F=\frac{m v^2}{r}$ all of the time until it is released, just to keep it in the ring at speed. In addition, you need to apply the force that gets it up to that speed. The only advantage of the ring is the getting-up-to-speed acceleration can be smaller. And that is only an advantage if you have some relatively cheap (from a force and energy point of view) way of having the mass keep turning in the ring. And that's a problem I see no solution for. Spinning a mass with electromagnetism, for example, will use just as much energy to make it move in a ring as along a straight line.

Additional edit: The failure modes of a ring of this nature will be spectacular. The energy in the electromagnets will very likely be many times the energy in the mass. Good possibility you would see the ring turn to plasma if you got any kind of failure. And the mass would exit in some random direction.

Should my military superiors place me on duty on such a device, sabotage and desertion would start to look like pretty good career moves.

Even further edit: I just noticed mcRobusta's paragraph about the electromagnetic ring not communicating the force to the ship. Isaac Newton is not fooled! The force you apply to the projectile using electromagnetism is precisely equaled by the force the ship feels. This is Newton's Third Law. If you push the mass by whatever means, the mass pushes back. Electromagnetic propulsion is not a free lunch in this regard.

That, by the way, means the ring is going to apply a torque to the ship as the mass spins up. Probably that is inconvenient to navigation even if the structure of the ship can stand up to the forces.

$\endgroup$
  • $\begingroup$ This would be true if I only ran the projectiles down the circular track once, but the idea behind a circular track is that I can keep the projectiles on the track and just keep accelerating them until I'm ready to fire. The barrel length of the circular gun is thus effectively infinite. I don't know whether that makes it practical or not, though. I'll edit my original question to make it clearer that I intend for the projectiles to stay on the track for more than one revolution. $\endgroup$ – FlyingLemmingSoup Sep 3 at 22:24
  • 2
    $\begingroup$ To understand what puppetsock means about failure modes, draw the ring, the projectile and draw the vectors. The vector line of the projectile crosses the ring. That means that the ring itself must withstand a substantial percentage of the kinetic energy of the travelling projectile as the projectile is redirected from its straight trajectory into the curved one. The ring has got to be very strong. I agree with the premise of keeping a distance from a ring containing such a projectile. $\endgroup$ – Willk Sep 4 at 14:03
6
$\begingroup$

Our friend Isaac Newton can help us here with his equation of circular motion:

$F = \frac{mv^2}{r}$

where F is the force produced, m is your object mass, v is your object's linear velocity, and r is the radius of motion. As you can see, it's the speed you want to reach that is going to have more of an impact than your mass (which, by the way, is still a problem).

Based on your numbers, you'll be subjecting your accelerator to 6 billion Newtons of force. This will likely cause it to break into tiny pieces. By the way, 30km/s is Mach 88 (88 times the speed of sound). This rotary force is enough to fling New York City into space- you must have some interesting needs.

Running your cannonball at Mach 1 (340 metres per second) will still put a big hole into whatever you fire it at. You'll put 771,000 Newtons of force onto your accelerator, which at a big enough scale it should be able to handle.

On a final, more interesting note, the velocity of an object in a circular path acts at a tangent to the circle, but the acceleration (and thus the force) acts towards the circle's centre. As the direction of velocity is constantly changing in a circle, any object in circular motion is accelerating with this force. Something to consider in your designs.

EDIT: Have you considered modelling the projectile as a point charge in a magnetic field, and accelerating it radially?

My maths may need some touching up here, but using the equations for magnetic flux density we can see: $F = BQV = \frac{mv^2}{r} $

where B is the magnetic field strength, Q is the value of the point charge passing through the field and V is this charge's velocity. Using this, we then get:

$ BQ = \frac{mv}{r} $

Plugging your numbers back in, it turns out that the product of our magnetic field strength and charge of the particle needs be 200,000 Tesla Coulombs. Teslas are harder to increase than Coulombs due to their exponential nature, so lets keep those low. We can already produce up to 45 Teslas, so let's say we can produce 50 on your spacecraft. This means your projectile needs a charge of 4,000 Coulombs (which is still a lot), but assuming you keep it in a vacuum and it's perfectly spherical, you can use Gauss' Laws to calculate the current needed to be held in the projectile and the distance it needs to be from its enclosing magnets to reach this exit velocity in a certain time.

Without number crunching myself, it doesn't look so bad. Plus, with magnetic confinement the force of your projectile is meaningless as it's not actually producing a net force on any surface connected to your ship.

EDIT II: Electric Boogaloo

As suggested by @Efialtes in the comments below, this system would require a ridiculous amount of energy. By substituting in

$ E_K = \frac{1}{2}mv^2 $,

we can calculate the projectile's energy is 0.45 TeraJoules. I cannot begin to quantify how huge this is.

Further, the field would have to be able to shift one million kilograms every five seconds to produce this power. Using equations for electrical energy (and assuming the device is 100% efficient), running this device for ten seconds would require a terawatt of juice. If you prefer, we can assume this device has an internal resistance of 10,000 Ohms (which would be very low). I can then tell you this would need a continuous supply of 2.08 MegaAmps. It takes one MegaAmp to fuse together atomic particles in a superheated plasma, so I really don't envy anyone tasked with designing something like this.

$\endgroup$
  • 1
    $\begingroup$ A missing bit of context here is the coilguns use in space combat. The speed isn't merely needed to inflict damage, but to increase the chance of the fired projectile reaching its target before they've got bored and gone home for tea. $\endgroup$ – Starfish Prime Sep 3 at 21:39
  • 2
    $\begingroup$ Wow. Newton was pretty bright but maybe he's the wrong guy to be calling on this one. Still, a win for peace I guess. $\endgroup$ – mcRobusta Sep 3 at 21:44
  • 1
    $\begingroup$ @FlyingLemmingSoup it isn't obvious that it is much better than a linear coilgun though... you're getting a factor of pi more accelerator length without lengthening the carrying object, but the required radial force is of a similar order of magnitude to the accelerating force (possibly more, too lazy to check) so your weight and power requirements may be rising faster than the improvement in coilgun performance. Hard to tell without actually counting numbers, but the benefit isn't clear. $\endgroup$ – Starfish Prime Sep 3 at 22:00
  • 2
    $\begingroup$ I'm sure you know this has to be said, but this is just another reason why Sir Issac Newton is the deadliest son of a gun in space... $\endgroup$ – Tim B II Sep 4 at 0:03
  • 1
    $\begingroup$ "Plus, with magnetic confinement the force of your projectile is meaningless as it's not actually producing a net force on any surface connected to your ship." Oh, yes it is! That force is transferred via the electromagnetic field directly to the magnetic coils. $\endgroup$ – Logan R. Kearsley Sep 4 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.