math gun part 1

site/posts/math-gun.html

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+[=post-]
+[#config.stx]
+<p>
+    Hello once more, gentle readers! I hope everyone's having a decent break and won't be suffering from boredom this summer. Today we're covering
+    a much-awaited topic: railguns! When I first learned about magnetism in physics, the very first thing I thought was - "wait, I can make a math gun".
+    Railguns have always fascinated me, and now that I actually understand the fundamental principles behind them, naturally I want to build one.
+</p>
+<p>
+    I'm gonna do this in three sections. The first one (this one) will cover building the initial mathematical model: an equation relating all the properties
+    of a <i>parametric math gun</i> (is that an official term? no. no it is not), including properties of the circuit (like the capacitor grid).
+    The second section will cover building a computational model to visualize (and verify) the results - and the last section will involve actually
+    building the railgun with the kind folks over at
+    <a href="https://scrappysgarage.org/">Scrappy's Garage</a>, who unlike me have the equipment to manufacture a railgun.
+</p>
+<p>
+    <img style="float: right; height: 300px;" src="[^baseurl]/posts/res/railgun-concept.png" /> The basic design of a railgun relies on a property of circuits and magnetics: `F = IL times B` - the force on a current segment is equal to the
+    length of the segment, times the magnitude of the current, times the direction of the current, cross with the magnetic field around it.
+    Railguns, then, work by running a current in the `hat x` direction through a conductive object, and using permanent magnets to produce a sustained
+    magnetic field across that object in the `hat z` direction. This leads to a net force in the `hat x times hat z = hat y` direction. The easiest way
+    to generate that current is with a pair of powered rails.
+    <br><br>
+    The attractive thing about railguns is that they can produce a powerful shot, even in a fairly low-power system. A small drone battery might produce a 5A max current,
+    and you can pretty easily get magnets that will produce a field in the ballpark of 1T. Because the directions are all perpendicular, we can multiply the magnitudes directly
+    on the cross product; a fairly small (5cm) rail weighing 10 grams in that field under these conditions will experience several gravities of acceleration (25 meters per second squared).
+    Consider the highly idealized case of an object experiencing this acceleration across a 50cm rail: `frac 25 2 t^2 = 0.5 -> t = sqrt(frac 1 25)`, working out to an exit velocity of
+    `v = 25t = frac 25 5 = 5` meters per second. That's pretty fast for a launch from a fairly small, weak assembly! However, this is an oversimplified model. To get anything
+    like accuracy, we need to consider air friction, mechanical friction, and self-induction. Let's do air friction first.
+</p>
+<p>
+    The general formula for drag is `F_d = frac 1 2 v^2 p C_d A`: the drag force is one half of the velocity squared, times the density of the fluid, times the drag coefficient,
+    times the cross sectional area. This obviously isn't a perfect model, but I don't want to build a CFD simulation, so it'll have to work. If we assume the object we're launching
+    is a 1cm diameter ball bearing (weighing ~4 grams - it's shockingly hard to find this size, but McMaster comes in clutch as always), we're looking at a circular cross-section
+    of `pi 0.005^2` regardless of orientation. The drag coefficient of a sphere is `0.47`, and the density of atmosphere at sea level is `1.225` kilograms per meter cubed, so we're
+    looking at a drag force of `F_d = frac {0.47 * 1.225} 2 pi 0.005^2 v^2`, or `~9*10^{-5} v^2`. This means the formal equation for force on our marble (normalized to the y direction only) is now
+    `F = 0.01IB - 9*10^{-5} v^2`. Acceleration is simply that divided by mass. Because velocity is the integral of acceleration, and we're starting from 0, we have this little
+    formula: `v = frac 1 0.004 int_0^t 0.01IB - 9*10^{-5} v^2 dt`. This is a very difficult integral to calculate; let's table it for now. Until we reach fairly high velocities, we don't have
+    to worry much about this.
+</p>
+<p>
+    Let's now worry about induction. As the marble moves through the magnetic field, it induces a current in the rails opposing the one running through it - this is Faraday's and Lenz' law.
+    This induced electromotive force is equal to `- frac {d phi} {dt}`, where `phi` is the <i>magnetic flux</i> through a given loop. Magnetic flux is simply the surface integral of magnetic field
+    over the inside of the loop - in this case with uniform `B`, it's just the magnetic field times the area. Taking `L` to be the distance between wires, flux is thus
+    `B L y`, where `y` is the height of the loop. We don't have to know `y` rigorously; we just need to know the change in `y`, which is `frac {dy} {dt} = v`. Because the other
+    terms are constant, this means `- frac {d phi} {dt} = -BLv`. That's the induced voltage! This means our voltage at any given time is `V = Emf - BLv`. That's very easy to plug
+    into our above formula: Ohm's law gives us `I = frac V R`, so the full equation is now `v = frac 1 0.004 int_0^t 0.01B frac {Emf - BLv} {R} - 9*10^{-5} v^2 dt`.
+</p>
+<p>
+    This is an extremely hard integral to find - even Wolfram|Alpha has no idea. We'll have to build a computational model. 
+</p>
+<p>
+    <img style="float: right; height: 200px;" src="[^baseurl]/posts/res/capacitor-grid-charging.png" /> There's one more thing to do before wrapping this up: we need to consider capacitors. For those not in the know, a capacitor is essentially a short-term battery
+    that charges and discharges comparatively quickly. The attractive thing about capacitors in this situation is that they can be used to turn a battery's continuous
+    low voltage into a brief push of very high voltage. 
+</p>
+<p>
+    You can see in the graphic to the right an example of a voltage source charging a bunch of capacitors in series: I won't go into the nitty-gritty details of how this works
+    on a micro level (although it <i>is</i> quite fascinating). This will take a long time: note that it is in parallel, so the voltage across each capacitor
+    will be charged up to `Emf`. When discharging, the capacitors are switched into series: remember the Loop Rule of a circuit - the voltages will add, meaning the
+    total voltage through the bulb is `5 Emf`. The tradeoff is that, of course, we won't get a <i>sustained</i> voltage this way: we'll have to spend 5x as much time
+    charging as discharging. For a system like a railgun, this is quite ideal: we have a capacitor grid that charges from a battery until we fire a shot, then discharges
+    into the rails until the shot finishes (this would have to be managed by a microcontroller switching solid state relays - we'd use a hall effect sensor to determine
+    when no current is running through the rails, meaning the marble has left the track). We can tune the duration of capacitor discharge with a resistor.
+</p>
+<p>
+    Tuning properties of the capacitor grid and resistor(s?) will likely require brute-force sampling with a computational model; to do that we'll need to have some formulas
+    to compute. Given that the base resistance of the railgun is `R`, and the added resistor is in series with the railgun, our total resistance is `R + R_"add"`.
+    Let's also assume we're using `N` capacitors with capacitance `C`, so our final series voltage is `V = N "Emf"`. Capacitors charge by an exponential function `Q = C"Emf"(1 - e^{- frac t {RC}})`,
+    and `V = frac Q C`, so our capacitor voltage in terms of time is `V(t) = "Emf"(1 - e^{- frac t {RC}})`. This will clearly never reach `"Emf"`, but will quickly <i>approach</i> it -
+    if we want to know the time `t` at which we'll have reached a voltage `V_"targ"`, we solve `V_"targ" = "Emf"(1 - e^{- frac t {RC}})` to get
+    `t = -RC ln(1 - frac {V_"targ"} {"Emf"})`. `1 - frac {V_"targ"} {"Emf"}` is very convenient for our purposes: for instance, if we want to get within 20%, it's just
+    `ln(0.2)`. Assuming we're using larger capacitors on the order of millifarads, and the charging circuit has in the ballpark of 0.01 ohms of initial resistance (this is reasonable,
+    although maybe a bit conservative
+    based on copper wire numbers), then `RC` is on the order of `0.00001`. This gives us 16 <i>micro</i>seconds to an 80% charge, 23 micros to 90%, and about 30 microseconds to
+    95%. Note that this is true for <i>any</i> voltages - a dinky 3.3v battery array and a chonky 120v wall socket supply will both charge our lil' 1mf capacitor to 95% in 30 microseconds
+    (although the wall socket is significantly more likely to blow it up - don't try this at home). Because we're charging in parallel, we logically just multiply by `N` to get
+    the total time to charge the entire grid.
+</p>
+<p>
+    There is a lot more complexity with capacitors (and price... jeez, $50 for a single capacitor), but it's not useful to theorize about the capacitor grid much more
+    without some concrete simulation and bench test results. See you next time with a computational model!
+</p>
+[/]
+[=author "Tyler Clarke"]
+[=date "2025-5-2"]
+[=subject "Physics"]
+[=title "Math Gun pt. 1"]
+[#post.html]

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