🚀 go-pugleaf

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
Received: from kontu.utu.fi by tut.fi; id AA24973; Mon, 9 Apr 90 14:58:08 +0300 
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Date: Mon, 9 Apr 90 14:57 EET
From: AHAHMA@kontu.utu.fi
To: kp74615@tut.fi
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Status: RO

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.

3P