Xref: utzoo sci.electronics:11188 sci.physics:12396 Path: utzoo!attcan!uunet!wuarchive!zaphod.mps.ohio-state.edu!rpi!uupsi!sunic!tut!nokikana.tut.fi!kp74615 From: kp74615@nokikana.tut.fi (Karri Tapani Palovuori) Newsgroups: sci.electronics,sci.physics Subject: Effectiviness of rail vs. coil guns Message-ID: <12336@etana.tut.fi> Date: 9 Apr 90 12:42:25 GMT Sender: News@tut.fi Reply-To: kp74615@nokikana.tut.fi (Karri Tapani Palovuori) Organization: Tampere University of Technology, Finland Lines: 59 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 Message-Id: <9004091158.AA24973@tut.fi> Date: Mon, 9 Apr 90 14:57 EET From: AHAHMA@kontu.utu.fi To: kp74615@tut.fi X-Vms-To: IN%"kp74615@tut.fi" 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