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My question upon first glance looks quite elementary, however my specialisation in physics is not circuitry and such I am not sure of my calculations.

A little bit of background: I am trying to create a coilgun (Alas like so many others) to help translate my physical knowledge to a practical application. I am relatively familiar with electromagnetism so I managed to create an equation that should be a good approximation between the physical parameters of a solenoid to the ferrous projectile's final velocity (muzzle velocity). This equation is as followed:

$$V^2 = U^2 + N_c\left(\frac{I^2\mu N^2A}{lm}\right)$$ Whereby $I$ is the current through the coil, $N$ is the number of turns of the coil, $A$ is the coils cross-sectional area; $l$ is the length of the coil, $N_c$ is the number of coils (stages) and $m$ is the mass of the projectile. $U$ is some initial velocity due to an injection mechanism that moves the projectile into the effective range of the magnetic field. Dimensionally speaking, this equation is correct. I had been reading some papers on solenoids (The details of which I will make sure to append to the end of this post) and found an empirically-derived formula that related the force on a plunger inside a solenoid due to the solenoids inductance. This is what I based my equation on and is as followed: $$ F = \frac{1}{2}I^2\frac{L}{x}$$ Where $I$ is current, $L$ is the inductance of the coil and $x$ is the displacement measured from the centre of the coil on the axis perpendicular to the plane of the coil.

Using my first equation, I wanted to have a projectile of mass 18.25g have a final velocity of about $40ms^{-1}$. Temporarily setting $U$ to zero, I came up with the following parameters for the coil:

$N_c$ = 10, I = 40A, N = 100, l = 0.05m, r = 0.025m (An area of $\frac{\pi}{500}m^2)$

I say this is an approximation because the acceleration of the projectile will not be completely constant due to each coil turning off once the tip of the projectile reaches $x = 0$ and therefore there being some period of reduced acceleration as the projectile transitions from one coil the the next.

If the radius of the coil is 0.025m and it has 100 turns, that means I would therefor need $(2\pi(0.025)(100)) = 5\pi $ metres of wire per coil. Using AWG 12 copper wire which has a resistance per metre of $5.21*10^{-3}$ would mean the total resistance of this length of wire is $0.081\Omega$.

To my understanding I can then just use Ohms law to calculate the voltage supplied to the circuit (Whereby only one coil is being powered at a given time during the firing of the projectile) which should just be $(40)(0.081) = 3.24V$

My question is this: If I require such a low voltage, what's stopping me from just connecting 2 or 3 AA batteries (each supplying 1.5V) in series to provide the current needed? The projectile would have a muzzle kinetic energy of around 16J and it seems to me 3 AA batteries are capable of providing this energy, though that would mean they would be outputting 130W of power and that seems slightly insane for a couple of batteries. Have I just gone totally wrong somewhere or is that possible?

Apologies for the lengthy build-up, I have tried to be careful with my wording. Any help would be massively appreciated.

The paper that I quoted the second equation from: https://www.jstor.org/stable/44469406 Page 3, Equation (1)

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You will struggle to get much more than a couple of amps out of an AA battery when you require 40, so that's your first issue. Maybe consider using a capacitor bank?

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  • $\begingroup$ so I would be looking for a setup of capacitors that can charge up to around 3 volts of charge and have an instantaneous discharge current of 40A? $\endgroup$
    – Aidan Daly
    May 11, 2020 at 11:14
  • $\begingroup$ @AidanDaly You better use a step-up DC booster to boost the battery emf to near capacitors max voltage so you can get much higher current output from the capacitors when you discharge them. It is not "instantaneous", they take time to discharge depending on how many coulombs they store and also the resistance of the circuit (formula for 63% discharge is τ=RC, for near 0%, it is 5RC). Note that they also create circuit oscillations (like an RLC) thus producing inductive reactance. However, these oscillations don't last. As the energy is lost to heat due to impedance (total ohms) in the circuit. $\endgroup$
    – Beyondo
    Jun 13, 2021 at 7:16
  • $\begingroup$ @AidanDaly These oscillations aren't necessarily a bad thing, the projectile is gonna be sucked to the center of the coil whichever way the current flows. So you might think that you should disconnect each stage right when the center of mass of the projectile is in the center of the uniform coil. Wrong. When you do that, you don't actually "instantaneously" cut off the magnetic field. Cause the terminals of the coils would temporarily act as capacitors thus creating RLC oscillations again producing magnetic fields that suck the projectile back even after disconnecting reducing efficiency. $\endgroup$
    – Beyondo
    Jun 13, 2021 at 7:31
  • $\begingroup$ @AidanDaly The smart way is to disconnect the coils when the center of mass of the projectile is near the center AND right before the capacitors oscillate back to the coil. You could use a microcontroller to test the potential difference across the capacitor or the coil to test and time that perfectly. So you don't actually lose energy in creating "negative acceleration" thus increasing projectile speed (technically not decreasing it), and also storing additional energy for the next shots. $\endgroup$
    – Beyondo
    Jun 13, 2021 at 7:39
  • $\begingroup$ Btw I'm interested to know what μ did you use in your formula and why? $\endgroup$
    – Beyondo
    Jun 13, 2021 at 7:46

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