The primary purpose of the rails in a rail gun is to conduct electricity to the projectile, build a magnetic field, and guide the projectile out of the device. The actual shape of the rail is important in two instances. 1) The contact patch between rail and projectile. 2) The structural rigidity of the rail in the horizontal plane (in plane with but perpendicular to projectile motion).

Obviously, the rail surface in contact with the projectile should have the same contour as the projectile itself. The best shape for this interface is flat. This shape is optimal because it is cheaper to make, simplifies the production of high-precision surfaces, facilitates rail compositing as mentioned above, and minimizes match-up errors between projectile and rail. Use of a grooved rail and projectile could be performed, with the benefit of increased contact area and thus a reduction in current density, but in practice, producing these parts would be too costly. Considering the lifespan of rails (around 100 shots, give or take 100), the simplicity of the rail is imperitive.

The shape of the rail not in contact with the projectile must be designed for the utmost in structural rigidity. Fortunately the vast bulk of force experienced by the rails is in one direction, a result of the same equation that dictates the forward motion of the projectile. Thus the rail must be designed to withstand an outward force equal to that exerted on the projectile. As you can imagine this force is huge, as it tends to send the projectile out of the gun at a few km/s.

So, borrowing a few lessons from structural engineering, an ideal rail support (structural backing for contact/conducting surface) would resemble a beam or truss designed for maximum resistance to bending. The cross section of this beam can take on many shapes based on the resources at hand, I or H beam, U channelled, closed rectaungular, closed triangular, and so forth. If you don't already have a bunch of steel girders laying around, I'd get our the old mechanics textbook and calculate the bending moments for some different cross sections and determine which you'll use. The complexity, the strength to weight of the cross-section, availability, and ultimately cost will determine your choice of rail.

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How stiff do the rails need to be?

The rails need to be stiff enough that at maximum force, they do not deflect enough to break the circuit of rails and projectile. The tolerances of your design, sabot geometry, and even pulse shape will determine the structural requirements. You may want to design in some interferance to account for the spreading of the rails, build in some toe-in, design a suspension system to maintain solid contact, or design the sabot to ride on top of the rails rather than between them. This will allow you to account for rail movement at the expense of increased projectile mass, difficulty in maintaining optimal current density, and so forth. As you may have noticed, rail guns are a pile of compromises. It is definitely no easy task to trick the forces of nature into throwing a 2 kg piece of tungsten around at 4 km/s.

Working from the example given earlier (2 kg projectile, 4 km/s muzzle velocity, 6 m rail length) the forces present are as follows.

Assuming constant accelleration, the time the projectile spends in the rails is:

(1) t=d/va

• t= time spent in contact with rails, (seconds)
• d=distance travelled, length of rails, .006 km (kilometers)
• va=average velocity, (vm-vo)/2, muzzle velocity minus initial, 2 km/s (kilometers/second)

Thus t=0.003 seconds, a very short time. The accelleration needed to produce the 4 km/s muzzle velocity is:

(2) a=dv/dt

• a=projectile accelleration (kilometers/second^2)
• dv=change in velocity, 4 km/s (kilometers.sec)
• dt=time spent under accelleration, 0.003 s (seconds)

Thus the accelleration is 1333 km/s^2. Talk about whiplash. The force on the projectile and thus on the rails as well is determined by the accelleration and mass of projectile:

(3) F=ma

• F=force on projectile and rails (Newtons)
• m=mass of projectiile, 2 kg (kilograms)
• a=projectile accelleration, 1333000 m/s^2 (meters/second^2)

Drum roll please...the force is 2666000 N, or 599316 lbf. To get an idea of how much force this is, lets assume we wanted to keep the stress in the rail beam under 50 ksi (345 MPa):

(4) A=F/tau

• A=area of beam in tension (square inches)
• F=force on projectile and rails, 599316 lbf (pound force)
• tau=stress on beam in tension, 50 ksi (thousand pound force per square inch)

The cross-sectional area of beam required to meet this spec would be 11 in^2. That is a lot of metal. The above was for a purely tensile load, but the goemetry of the actual situation is more complicated. The portion of the rail beam nearest the projectile will be in compression, with the farthest portion of the beam cross section in tension. The ratio of these stresses will depend on the cross sectional geometry of the rail and thus the position of the neutral axis. The neutral axis is the line parallel to the rail and the direction of projectile motion where there is no stress, compression and tension are in balance.

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Page last updated 4/24/02