kp74615@nokikana.tut.fi (Karri Tapani Palovuori) (04/09/90)
I'm posting this for my friend because of their faulty mailing
system:
From AHAHMA@kontu.utu.fi Mon Apr 9 14:58:10 1990
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Date: Mon, 9 Apr 90 14:57 EET
From: AHAHMA@kontu.utu.fi
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How efficient is a rail gun? I mean in per cent, how much kinetic energy
can be achieved per joule electrical energy consumed?
I was just calculating the impulse, that a rail gun can give to the projectile.
The force is I*l*B, if B (magnetic flux density) is perpendicular to the
current I. l is the width of the projectile, say, 10 mm. On the other hand
the current equals dQ/dt, Q is electrical charge. The impulse is Fdt, F is the
force. The impulse can also be written m*dv, v is velocity. From these you can
get:
m*dv = l*B*dQ <=> dv = l*B*dQ/m
Since the mass, flux density and width of the projectile are constant, the
equation can be integrated to get
delta-v = l*B*delta-Q
The magnetic flux can be 2 tesla if it is generated with separate iron-cored
coils.
The electrical charge is 10 coulombs with three 1670 uF capacitors
loaded up to 2200 volts. If all of the charge can be brought through the
the projectile (quite reasonable with that high a voltage), then the
change in velocity will be 0.01*2*10/0.01 = 20 m/s, if the projectile weighs
10 grams. The 20 m/s corresponds to an energy of 2 joules only. The energy
in the capacitors is over 10 kilojoules! So the efficiency factor is rather
low, if the initial velocity equals zero. If the projectile had a high
velocity initially, the 20 m/s would be more than 2 joules, but how would
you generate that velocity then?
The coil gun seems to be much better compared to the rail gun. The force
in a magnetic field is V times grad(M.B) ( . means a dot product), where
V is the volume of the projectile, M is the magnetisation of it and
B is the magnetic field flux density. If the projectile is made of iron, that
has a high relative permeability, the equation can be written
F = 2*V*B*grad B/(mu), mu = permeability of vacuum, 4*pi*10^-7
>From this equation you can see there is a small constant in the denominator,
so it is easy to achieve great forces to the projectile. The grad B can also
easily have a big value. A coil with 200 turns can easily generate forces
of several hundreds of newtons to a 1 cm^3 projectile with those same
capacitors, especially if it is wound unsymmetrically to generate a large
gradient in the magnetic field. The force will also last for a longer time
than with the rail gun.
3Plogajan@ns.network.com (John Logajan) (04/10/90)
AHAHMA@kontu.utu.fi writes: >especially if it is wound unsymmetrically to generate a large >gradient in the magnetic field. Hmmm, but what could be MORE unsymmetrical than the coil wound as TIGHTLY as possible. The gradient outside the coil would be as steep as anything you could produce INSIDE the coil. -- - John Logajan @ Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428 - logajan@ns.network.com, john@logajan.mn.org, 612-424-4888, Fax 424-2853
ISW@cup.portal.com (Isaac S Wingfield) (04/11/90)
Karri Palovuori writes: >How efficient is a rail gun? I mean in per cent, how much kinetic energy >can be achieved per joule electrical energy consumed? > >I was just calculating the impulse, that a rail gun can give to the projectile . >The force is I*l*B, if B (magnetic flux density) is perpendicular to the >current I. l is the width of the projectile, say, 10 mm. On the other hand >the current equals dQ/dt, Q is electrical charge. The impulse is Fdt, F is the >force. The impulse can also be written m*dv, v is velocity. From these you can >get: > >m*dv = l*B*dQ <=> dv = l*B*dQ/m > >Since the mass, flux density and width of the projectile are constant, the >equation can be integrated to get > >delta-v = l*B*delta-Q > >The magnetic flux can be 2 tesla if it is generated with separate iron-cored >coils. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^ Back in the early '80's I had a conversation with a Scientist who was actually working on rail guns. According to his description of the technology at that time, no magnetic field other than that generated by the current was involved. He said the force was proportional to I^2 / inductance per unit length (I think; at any rate, it was current squared divided by some function of L). In order to get interesting forces you need (as I've said before), interesting currents -say 1/2 to 1 MegAmp. In order to get currents like this to flow, it's necessary to absolutely minimize the inductance. Regular capacitors, no matter the voltage or capacitance, will not discharge fast enough due to internal L. $pecial low inductance capacitor$ are u$ed for thi$ purpo$e. Note also that a large charged capacitor can produce substantial internal *mechanical* forces. They're usually balanced by opposing charges on the next plate over, but under conditions of instantaneous discharge, large imbalances can exist, which can destroy the capacitor (read "EXPLODE"). If the cap is not designed for fast discharge, don't use it for this kind of application. Connections external to the cap also must be low-L. "For best results, these wires should be wider than they are long". He mentioned that if "muzzle" velocity could be increased by between one and two orders of magnitude, two pellets of deuterium fired at each other would undergo fusion due to sheer mechanical impact. Isaac (don't point that thing at me, buster) isw@cup.portal.com