The push at the end of a pair of rails

The current that squeezes what carries it found that a current flowing along a column makes a field that wraps round it and pushes it inward, and that the push can crush a copper tube struck by lightning. The force between pieces of a current found that the force one piece of a circuit exerts on another has two correct values, and that only the force between complete circuits has ever been measured. A railgun is a circuit built to make one piece of itself move. It is the simplest electric machine there is — two rails, a bar lying across them, and a current — and it is worth taking apart, because almost every number in it comes out differently from the first guess.

Two rails and a bar #

Current flows out along one rail, across the sliding bar, called the armature, and back along the other rail to the supply at the breech. Between the rails the two currents, flowing in opposite directions, make fields that add, a field pointing straight through the gap. The armature carries current across that field and is pushed along the rails, away from the breech. Nothing touches the projectile but the armature, there is no gas, and in principle the push can be as large as the current makes it.

How large? The quickest answer is energy. The circuit is a loop whose area grows as the armature moves, so its inductance grows: by $L'$ for every metre of travel, a number fixed by the rails’ shape and spacing, about half a microhenry per metre for rails of practical proportions. Hold the current constant and move the armature a distance $dx$; the field’s energy $\tfrac12 LI^2$ grows by $\tfrac12 L' I^2\,dx$, and a careful account of the supply’s work shows that the armature receives exactly the same amount as mechanical work. So the force is

$$F = \tfrac12\,L'\,I^2.$$

For rails fifteen centimetres apart and an inductance gradient of 0.447 microhenries per metre, a current of 3.74 million amperes gives 3.13 meganewtons — enough to take a ten-kilogram projectile to 2.5 kilometres a second in ten metres of barrel, at an acceleration of thirty-two thousand times gravity, in eight milliseconds.

Where the half comes from #

The energy argument gives the half without saying what it means. The field does.

Far behind the armature the rails look endless in both directions, and the field between them is the field of two long wires, the field that wraps a current twice over. At the armature the rails stop. A straight wire that stops contributes, at a point level with its end, exactly half of what an endless wire would, because by symmetry an endless wire is two half-wires meeting at that point and each contributes equally. Both rails stop at the armature, so the field at the armature is half the field behind it. The current in the armature crosses that half-field, and the force is the current times the field it actually sits in. The ½ in $\tfrac12 L' I^2$ is the half-field at the end of a wire — the same geometric half that makes the field at the mouth of a long solenoid half its interior value.

That reading also says where the force acts. The field falls from full to nearly nothing within a spacing or so of the armature, so the armature is pushed from behind by the field it has left in the barrel and is not pulled by anything ahead. In the language of the force read off a surface that touches nothing, the field behind the armature exerts a pressure $B^2/2\mu_0$ on it, as a gas behind a bullet does, and ahead there is no field to push back.

The voltage a moving bar makes #

The same physics read from the supply’s side gives the half a third way, and it shows what the supply has to do. A conducting bar moving across a magnetic field has a voltage induced along it, the field that makes the other in its simplest form: speed times field times length. For the armature that speed voltage is $L'Iv$, and it opposes the current, so the supply must raise its voltage as the armature gathers speed just to keep the current constant — on top of whatever the resistance needs. At 2.5 kilometres a second and 3.74 million amperes it is 4.2 kilovolts.

Voltage times current is power. The supply pushes 3.74 million amperes against 4.2 kilovolts of speed voltage at the muzzle, about sixteen thousand megawatts, while the armature receives the force times its speed, 3.13 meganewtons at 2.5 kilometres a second, eight thousand megawatts. Exactly half the power the supply delivers against the speed voltage goes into motion. The other half goes into the lengthening field, as the circuit that fights its own change found that any inductance stores what is pushed into it against its own induced voltage. Energy, field and induced voltage are three readings of one fact, and each insists on the same factor of two.

The figures also make plain why the currents are so large. The force grows as the current squared and only linearly with the inductance gradient, and the gradient is set by geometry within narrow limits: rails much further apart than they are thick give a larger $L'$ but a field spread over a larger bore, with a heavier armature to span it. In practice every railgun runs at about half a microhenry per metre, and the only way to a larger push is a larger current. Three to five million amperes for a few milliseconds is a pulse of tens of gigawatts, comparable to the output of every power station in a large country, delivered from a bank of capacitors charged over seconds beforehand.

Who else is pushed #

A force on one piece of a circuit has to be balanced somewhere, and in a railgun the balancing forces are larger than the useful one.

The figure computes every force from the field of the rest of the circuit, segment by segment. The armature’s comes out as $\tfrac12 L' I^2$ to within the integration’s accuracy. The breech — the conductor at the back where the current turns from one rail into the other — sits in the field of the rails’ near ends and is pushed backward by the same amount. That is the gun’s recoil, and it is magnetic: the feed connections at the breech, not the rails, take the reaction to the projectile, and they have to be built to.

Which pieces of a circuit take a recoil was the subject of a long argument. Ampère’s own law of force between current elements, which the force between pieces of a current found agrees with the modern one for every closed circuit and disagrees for pieces, predicts forces along the rails themselves, pushing them lengthwise; in the 1980s it was proposed that railgun rails buckling and wires shattering in high-current experiments were evidence for those longitudinal forces. The totals are the same either way, because the circuit is closed, and the measured effects have generally been explained by heating, thermal stress and the ordinary transverse forces. What a railgun cannot do is settle the question by measuring a total.

The largest forces in the figure are neither of these. The two rails carry opposite currents, and opposite currents repel. Each metre of rail is pushed away from the other with nearly nineteen meganewtons — six times the entire forward push on the projectile, on every metre of the barrel.

Both forces go as the square of the current, so their ratio is fixed by geometry alone and no increase of current improves it. The forward force is the field’s pressure on the small area of the armature; the sideways force is the same pressure acting on the whole length of both rails. A railgun barrel is therefore mostly structure: rails clamped between massive insulating and steel containment, preloaded so that the rails do not separate from the armature as the pulse arrives, because a gap of a fraction of a millimetre opens an arc. The squeeze in the current that squeezes what carries it and the spread here are one force seen from inside and outside the current: parallel currents attract and pinch a column; opposite currents repel and burst a barrel.

The energy left behind #

Constant-current operation has a cost that is easy to miss, and it has nothing to do with resistance.

As the armature moves the circuit lengthens, and at constant current the supply must deliver energy for two purposes: the work done on the armature, $\tfrac12L'I^2\,dx$, and the growth of the field’s own energy, another $\tfrac12L'I^2\,dx$. The two are always equal. When the armature reaches the muzzle, the barrel behind it is full of field holding exactly the projectile’s kinetic energy — 31.3 megajoules here. Where the energy of a field is answered the question of where: in the space between the rails, at $B^2/2\mu_0$ per cubic metre. It is the same bookkeeping that makes charging a capacitor from a fixed voltage lose half the energy drawn, and it has no loss in it; the energy is still there.

What happens to it is the problem. When the armature leaves the rails the circuit opens. The current, carried by a huge inductance, refuses to stop, and an arc forms between the rail ends and the departing armature, through which the field’s energy is discharged as light, heat and noise. The fireball at a railgun’s muzzle is largely the barrel’s stored field, released in a millisecond. Practical systems drive the rails from capacitor banks through pulse-forming circuits whose current falls during the shot, so that less field is left behind, or divert the current through a shunt near the muzzle, and the most efficient designs try to steer the stored energy back to the supply. None of that changes the principle: a constant current can never deliver more than half of what it supplies.

The resistive loss is real too. Rail resistance multiplied by the current squared, integrated over the shot, came to 18.7 megajoules in the figure for a stated rail resistance of fifty microhms per metre of barrel — the current crowds into a thin layer of the rails’ facing surfaces during so short a pulse, so the effective resistance is far higher than the copper’s cross-section suggests. In the figure’s ledger, 39 per cent of the energy supplied ends in the projectile.

A gun that likes being fast #

The resistive loss behaves in a way that makes the railgun unlike any other launcher.

To reach a higher speed in the same barrel the current must be larger, and the heat goes as the current squared — but the current flows for less time, since the shot is over sooner, and the heat in rails whose energised length grows as the armature advances comes out proportional to the muzzle speed. The kinetic energy is proportional to its square. So the ratio of heat to kinetic energy falls as $4R'L/3L'v$, and the faster the gun shoots, the smaller the fraction lost. With the field’s energy discarded the efficiency creeps up towards one half; with it recovered, towards one.

The same formula says something less welcome about length. At a fixed muzzle speed, the heat in the rails is proportional to the barrel’s length: a longer barrel needs a smaller current, but it keeps that current flowing for longer through more rail, and the second effect wins. Halving the barrel halves the resistive loss — at the price of a current larger by the square root of two, a peak force twice as large on the projectile and on every metre of rail, and a contact that must carry the larger current. Every railgun design is a compromise among these, struck where the rails and the armature can survive, and none of the compromises touches the field’s half.

A chemical gun runs the other way. Its projectile is pushed by expanding gas, and gas can follow a receding projectile only so fast — the speed at which it can expand into the space behind is limited by its own sound speed and temperature — so as the projectile approaches that speed the gas does less and less work on it and the efficiency falls. Powder guns rarely exceed two kilometres a second; light-gas guns, using hydrogen for its high sound speed, reach about seven at great expense. The electromagnetic gun has no gas to outrun, and that is the whole reason for building one.

Where the rails give out #

What limits a railgun in practice is not any of the forces or energies above but the contact between the armature and the rails. A solid metal armature must slide at kilometres a second while carrying millions of amperes across a contact patch a few centimetres long. The current density at the contact reaches billions of amperes per square metre, and at high speed the current cannot spread into the fresh rail ahead quickly enough: the time for a current to diffuse into a conductor grows with the square of the depth, the armature moves on before it has, and the current crowds into the trailing edge of the armature, melting it. Above roughly two to three kilometres a second solid armatures tend to lose contact and form an arc, and the arc erodes the rails. Guns built with deliberately plasma armatures — an arc pushing a projectile, rather than a sliding bar — have reached speeds above five kilometres a second in laboratories, at the cost of rails eroded on every shot.

The United States Navy’s programme fired a ten-kilogram-class projectile with about thirty-three megajoules of muzzle energy in 2010, the energy scale of the figures here, and was shelved in 2021. The obstacle was the barrel’s life: rails that must be replaced after a few hundred shots, or fewer, cannot compete with guns whose barrels last thousands. The physics of the push works exactly as the figures say. The engineering of the sliding contact has not been made to.

What the filament picture leaves out #

The figures model each rail as a thin filament carrying the whole current along its centre, a conductor 7.4 centimetres thick for the purpose of excluding its own field. Real rails are wide, flat bars whose current crowds onto their facing surfaces and corners within the microseconds of the pulse, and their inductance gradient is computed from that distribution by solving for the field in the bore, typically 0.4 to 0.6 microhenries per metre; the filament model reproduces the value but not the distribution of force across the rail’s face, which is where the rail actually yields. The armature’s force is computed at the instant its current has fully established; in a real shot the current rises over a millisecond and its distribution in the armature evolves with speed. Friction, air ahead of the projectile and the armature’s own resistance are left out. And the energy figures assume a constant current, which is the simplest case and not what any real supply delivers.

None of these touches the three results the figures are for: that the armature is pushed by half the field behind it, that the rails are pushed apart many times harder per metre than the projectile is pushed forward, and that at constant current the field keeps as much energy as the projectile receives. The domain is a circuit whose current changes slowly compared with the time light takes to cross it, so that the magnetostatic field of the moment governs every force — true for a pulse of milliseconds across a barrel of metres by a factor of a hundred thousand.

Still open: whether a sliding contact can survive #

The unsolved problem of the railgun is a materials problem with an electromagnetic cause. A contact that carries a few million amperes, slides at a few kilometres a second, and leaves the rail it touched fit for the next shot has not been built, and what limits it — the velocity skin effect concentrating current at the armature’s trailing edge, melting and wear at the interface, the onset of arcing as the contact pressure falls behind the armature — is understood qualitatively and modelled with large simulations whose predictions of the transition speed still disagree with measurement by tens of per cent. Whether augmented designs, with extra conductors that strengthen the field without raising the armature’s current, or distributed feeds that energise only the section of rail near the armature, can push the transition past the speeds that would make the gun worth its structure, is being tested one barrel at a time.

The machine itself is entirely understood. A railgun’s armature sits at the end of two rails, where each rail’s field has fallen to half, so it is pushed by $\tfrac12 L'I^2$ — 3.13 MN at 3.74 MA for rails 15 cm apart — while the breech takes an equal recoil, the rails are pushed apart by six times that force on every metre, and at constant current the field behind the armature ends up holding exactly the energy the projectile carries. It is a gun with no gas and therefore no gas speed limit, and the only part of it that physics has not yet made work is the part where metal slides on metal.