Railgun Voltage Problem

[14927397]

I'm building a railgun for a science project, but I am not certain on how much voltage to use. My project involves firing repeated shots with a set of rails on a railgun to measure the effects of erosion and residue on the rails with shot efficiency. I am planning on firing a railgun 10 times without cleaning the rails after each shot, and measuring the exit velocity and remaining voltage in the capacitors to test efficiency. This process will be done three times. As for the voltage problem, I want to try to find the best voltage possible where it can still propel the projectile without being too expensive. I'm thinking about using capacitors to store the energy. The rails will be copper, and it will be 6 to 9 inches long. The projectile itself is a flat arrowhead design that is .375 in. thick, one inch wide, and 1.5 in. long (Couldn't find metric measurements for parts in US). The projectile is aluminium, because it will melt faster than the rails. For a starting voltage, I am thinking about anywhere from 500 volts to 2k volts, but chances are that I will need to buy at least 50% of that for safety.

Is there a simple formula that I can easily calculate the voltage necessary before building it, or should I just buy a safe amount and guess and check?
Thanks.

[JasonS]

Hello 14927397,

My brief research has shown current is the primary variable that you are trying to determine, not voltage. Wikipedia (http://en.wikipedia.org/wiki/Railgun) provided a couple equations to calculate Force. See below. Once you obtain Force and know the mass of your projectile you can determine the expected velocity of that projectile.

F = ((u_o * I^2)/(2 * pi)) * ln (d/r)

where
F = Force
u_o = permeability constant (http://en.wikipedia.org/wiki/Permeability_constant)
I = Current
pi = 3.1416
ln() = natural log function
d = Distance between the center points of the rails
r = The radius of the rails

OR simplified equation:

F = (L' * I^2)/2

where
F = Force
L' = 0.6 uH/m (ideal, your railgun will be lower)
I = Current

Start with a much lower current than you think you need and gently ramp up in a controlled environment. Have fun and be careful.
Hope this helps!
JasonS

[14927397]

Thank you for the advice! It will help me a lot when doing calculations. However, when I started looking for capacitors, they were measured in farads and voltage, but say nothing about current. A farad measures capacitance, which I'm guessing is how quickly it takes to charge up. Going with your information, the discharge must be very sudden in order to produce a large enough current. What sort of capacitor should I be looking for? For example: oil, ceramic, low mF, high mF, voltage, etc. Thanks.

[Craig_Bridge]

Voltage ratings on capacitors exist to describe the "break down" voltage of a capacitor deals with how much voltage (Electro motive Force) it takes to overcome the dielectric strength which may result in permanent damage to the capacitor. The maximum working voltage is the manufacturer's recomendation on a safe voltage the capacitor can handle.

The capacitance property in Faradays is a measurement of how many electrons can be stored by the capacitor.

There is a differential equation that relates the current to change in voltage with respect to time and the capacitance: i(t) = C dV(t)/dt.
This says that the current flowing at time t is proportional to the rate of change of the voltage with respect to time measured across the capacitor at time t and the proportionality constant is the capacitance in Faradays. Most capacitors are measured in micro Faradays and the ones you probably want are what you are calling "high mFd".

You probably want to think about energy. The amount of energy stored in a capacitor will be affected by the voltage. You won't know what the efficiency of your rail gun is until you can actually experiment with it. You probably should start by assuming 5 to 25% efficiency as an initial guess in order to size the amount of energy you want to store in order to size your capacitor bank.