On the Nature of Many Penny Systems

The other day I was on my way through ye olde internets to the Quantum Leap episode index at Wikipedia, but stumbled over a quote by Feynman by accident: "There are 10^11 stars in the galaxy. That used to be a huge number. But it's only a hundred billion. It's less than the national deficit! We used to call them astronomical numbers. Now we should call them economical numbers" So this got me thinking. What if we decided to pay the national debt...in pennies? Seems fair enough, we don't really want to pay it and when you don't want to pay something but you have to you pay it in pennies. Hooray! Last I heard this figure was somewhere around $10,000,000,000,000 (ten trillion), that is: $$ \text{Debt} = 10^{13} \text{dollars} = 10^{15} \text{pennies} $$ Given that a penny is about 1.5 mm thick, how far does stacking them end to end take us? Well: $$ \text{Length} = (10^{15} \text{pennies}) \times ( 0.15 \text{cm}) = 1.5 \times 10^{14} \text{cm} $$ As a bit of comparison, the mean distance from the earth to the sun is $$ \text{a} = 1.5 \times 10^{13} \text{cm,} $$ so our penny stack goes out to 10 AU, which just barely passes by Saturn. Impressive! Fine, so we've got all our pennies out now and we want to fit them in a more economical fashion, taking advantage of all three of our dimensions. How big is the warehouse needed to hold all these shiny Lincolns? Well, we know (or can guess) that one penny weighs about a gram (technically 2.5 g). So if we were to know the density of a penny, we would be good to go. A good guess at density for any metal is around 10 g /cm^3 , and this is true for zinc ( FUN FACT: Pennies were almost entirely made of copper up until 1982, when they switched to 97.5% zinc, 2.5% copper). So, $$ \text{Mass} = (10^{15} \text{pennies}) \times ( 1 \frac{ \text{g}}{\text{penny}})= 10^{15} \text{g} $$ $$ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{10^{15}\text{g}}{10 \frac{\text{g}}{\text{cm}^{3}}} = 10^{14} \text{cm}^{3}= 0.1 \text{km}}^{3} $$ That's a warehouse 1 km wide, 1km long, and a 100 m high. Pretty big. So our next logical progression would be to ask: what are its gravitational properties? To do this let's stack our pennies in a sphere. In that case we have $$ \text{Volume} = \frac{4 \pi}{3} \text{R}^{3} $$ so $$ \text{R} = (\frac{3}{4 \pi} \text{Volume})^{1/3} = (\frac{3}{4 \pi} 10^{14} \text{cm}^{3})^{1/3} \approx 3 \times 10^4 \text{cm}=0.3 \text{km.} $$ To find the gravitational binding energy of a sphere of uniform density we use $$ \text{Binding Energy} = \frac{3}{5} \frac{\text{G}\text{M}^{2}}{R}\text{.} $$ So $$ \text{Binding Energy} = \frac{3}{5} \frac{ (7 \times 10^{-8} \frac{cm^3}{gs^2})(10^{15} \text{g})^{2}}{3 \times 10^{4} \text{cm}} \approx 10^{18} \text{ergs} = 10^{11} \text{Joules.} $$ Noting that the useful measure of destruction, the kiloton of TNT, is about 10^12 J, we see that the binding energy of our spherical mass of pennies is about 100 tons of TNT. So what does this mean to us practically? Well, the binding energy of an object essentially tells us how much energy we need to put in to completely break up the body. So to fully disperse our pennies off to our lendors, we need about 100 tons of TNT worth of energy. An alternative financial fermi problem that might be interesting is how many trees would it take to make the money to equal the national debt or GDP?

On the Superposition of Hip and Hop

image For posterity, completeness and another superfluous Scott Bakula reference I present the following. It is an expression, through the majesty of song (rap), of our love for all things quantum and undead ( or perhaps a superposition of dead/ not dead states...? ).

We all know that the great physicist Paul Dirac, was not a vampire. What this rap presupposes is... maybe he was? If so, he would almost certainly be named Diracula. He would be especially good at physics. But he would especially be totally rad at rapping.

Without further ado, the Diracula Rapula:

Yo, its Diracula and I got something to say
Better listen close cause sometimes my teeth get in the way.

Of my rappin', but never my mathin'
Gonna work it strong, gonna make these monopoles happen.

Regardless of how you feel about me
and my strings with length semi-infinity
Respect my proclamation
cause you find just one and you got charge conservation

So let me take you on a Quantum Leap like Scott Bakula
Rockin' you out with my spectacular vernacular
Amazing math wunda of the undead
Blazin' through complex formulae in my head.

Classified by Linnaeus as Mathematica Rex
I'm peacin' out
watch your equations and watch your necks.

Diracula Out

Fishy Calculation

Aaron Santos over at A Diary of Numbers, author of How Many Licks?, has posted a Fermi Contest. For the uninitiated, a Fermi Problem is a seemingly unanswerable problem, which you can actually estimate reasonable by breaking the problem down into smaller parts. They're really fun, and I intend to post more in the future. The question at hand is: How far would the oceans sink if we took all the fish out? I'll answer in two very different ways.


In both cases, I'll answer how much the ocean depth would decrease by first calculating the total volume of fish in the ocean. Why is this helpful? Because for a spherical shell, you can estimate its volume by just taking its surface area and multiplying by the thickness, i.e. $$ V \sim 4 \pi R^2 \Delta R $$ So, if I can estimate the total volume of fish in the ocean, if I take all of those fish out, change in the height of the ocean will just be this volume divided by the surface area of the ocean, which I will take to be the surface area of the earth times 70% or so. Now I just need to estimate the volume of all of the fish in the oceans...

Energy Budget

First I'll estimate the volume of fish in a rather general way. I'll try to do it on energy grounds. I know how much solar energy hits the earth per meter squared on average (340 W/m^2). I'm going to assume that fish get their energy from plankton, and plankton get their energy from the sun, both with about 10% efficiency. I'll also assume that life occupies as much space as possible, probably about half of the ocean surface counts as liveable. From this I get the total energy available to make fish. How many fish does that allow? I'll assume that fish use about as much energy per kilogram as humans do. I know that humans have to take in about 2000 food Calories or 2,000,000 calories a day to survive. From this I get the total weight of all of the fish in the ocean, and for their volume I assume they're the density of water (fish are bouyant). My calculation: $$ \underset{\text{ \tiny mean solar flux} }{\left( 340 \frac{\text{W}}{\text{m}^2} \right)} \cdot \underset{\text{ \tiny earth surface}}{4 \pi \left( 6 \times 10^6 \text{ m} \right)^2} \cdot \underset{\text{ \tiny frac ocean}}{(0.70)} \cdot \underset{\text{ \tiny frac liveable} }{\left( \frac 12 \right)} = 5.4 \times 10^{20} \text{ W} \text{ (ocean life energy budget)} $$ $$ \cdot \underset{ \text{\tiny plankton eff}}{(0.10)} \cdot \underset{\text{\tiny fish eff}}{(0.10)} = 5.4 \times 10^{18} \text{ W} \text{ (fish energy budget)} $$ $$ \cdot \underset{\text{\tiny energy budget of man}}{\left( \frac{ 75 \text{ kg} }{ 2 \times 10^6 \text{ cal/day} } \right)} \cdot \underset{\text{\tiny cal $\leftrightarrow$ Ws}}{\left( \frac{ 1 \text{ cal} }{ 4 \text{ Ws} } \right)} \cdot \underset{\text{\tiny s $\leftrightarrow$ day}}{\left( \frac{ 60 \cdot 60 \cdot 24 \text{ s} }{ 1 \text{ day} } \right)} = 4.4 \times 10^{14} \text{ kg} \text{ (mass of fish)} $$ $$ \cdot \underset{ \rho_{\text{fish}} = \rho_{\text{water}} }{\left( \frac{ 1 \text{ m}^3 }{ 1000 \text{ kg} } \right)} \cdot \underset{ \text{\tiny ocean surface}}{\left( \frac{ 1}{ 4 \pi (6 \times 10^6 \text{ m})^2 (0.7) } \right)} = \text{ change in depth (m)} = 1.38 \times 10^{-3} \text{ m} \approx 1 \text{ mm} $$ So in the end I get about a millimeter change in the ocean height. Honestly, this is a lot larger than I expected. The ocean is a big place. I suppose there are also a lot of fish in the world. If you look, if we assume the average fish is about 10 kg, I've calculated that the population of fish in the ocean is 4.4E13 fish, or 4 million billion fish. That's a lot of fish. I'm not so sure about my estimate, so I'll do it again from a different direction...


Next, I'll try and estimate again, this time guessing based on how much we fish fish. I know that overfishing is a problem, which means that for some fish species we fish more than fish make little fishies, so if I can estimate how much fish we eat, I can estimate how much we fish, so I can estimate how many fish there are, and then I can estimate how much the ocean depth changes. This calculation is very rough, I took some very basic order of magnitude guesses at some of the parameters. In particular, I had to guess how many fish species we fish, which I took to be 1/100 for no very good reason. My calculation is below: $$ \underset{\text{ \tiny fish eaten per person}}{ \left( \frac{ 1 \text{ fish}}{ 1 \text{ week}} \right)} \cdot \underset{\text{ \tiny fish weight}}{\left( \frac{1 \text{ kg}}{1 \text{ fish}}\right)} \cdot \underset{ \text{ \tiny week $\leftrightarrow$ year}}{ \left( \frac{ 52 \text{ weeks}}{ 1 \text{ year} }\right)} \cdot \underset{\text{\tiny people who eat fish}}{\left( 10^9 \right)} =5.2 \times 10^{9} \text{ fish/year fished} $$ $$ \cdot \underset{\text{\tiny fish we eat}}{ \left( \frac{100 \text{ tot fish}}{1 \text{ fish eaten}} \right) } \cdot \underset{\text{\tiny fish lifetime}}{\left( 10 \text{ years} \right)} = 5.2 \times 10^{13} \text{ kg (total fish)}$$ $$ \cdot \underset{ \rho_{\text{fish}} = \rho_{\text{water}} }{\left( \frac{ 1 \text{ m}^3 }{ 1000 \text{ kg} } \right)} \cdot \underset{ \text{\tiny ocean surface}}{\left( \frac{ 1}{ 4 \pi (6 \times 10^6 \text{ m})^2 (0.7) } \right)} = \text{ change in depth (m)} = 1.6 \times 10^{-4} \text{ m} \approx 0.2 \text{ mm} $$ This time I got 0.2 mm or so, which is in relatively good agreement with my other number. At least its not several orders of magnitude off.

Geometric Average

Honestly I trust my first number more than the second, but I'm going to average my two results in order to come up with the number that I'll submit to the contest. But, I'm not going to average my two numbers arithmatically: $$ \mu = \frac{1}{2} ( x_1 + x_2 ) $$ as you normally do to average numbers, instead I am going to average them geometrically, i.e. I'm going to take the square root of their product: $$ \mu = \sqrt{ x_1 x_2 } $$

My Answer

This kind of averaging is logarithmic in nature, and my experience has been that it is much more successful average to use when you are doing fermi problems. Doing this on my two numbers I obtain my entry: $$ \Delta x = \sqrt{ 0.164 \text{ mm} * 1.38 \text{ mm} } \approx \frac{1}{2} \text{ mm} $$ About half a millimeter. I'll be sure to let everyone know how I do in the contest. Drop us an email or leave a comment to let me know how good you think my guess is. Happy Fishing.

Would a laser gun recoil?

Today I'd like to approach a question near and dear to many a geek heart: do laser guns have recoil? image Let's motivate our question a little bit. I've wondered about this question since I saw star wars. Though I'm no firearms expert, the recoil in guns must come from conservation of momentum principles. Momentum is conserved in a system. The gun starts with zero momentum. We fire, give the bullet momentum, and so to keep the system at zero momentum, the gun must gain equal and opposite momentum. That is, the gun will move backwards. All of that was for conventional guns. Light carries momentum, so if we fire a pulse of light, we expect our laser gun to recoil. So yes, they do have recoil. Satisfied, dear readers? Neither am I. The question we really mean to ask is, does a laser gun have noticeable recoil? We need to make a few reasonable assumptions. Let's assume that the laser gun fires a pulse with as much energy as a bullet has kinetic energy, KE. The energy, E, of light is related to its momentum, p, by E=pc, where c is the speed of light. This gives a momentum of $$ E=KE=pc $$ $$ p=\frac{KE}{c}$$ What is the kinetic energy of a bullet? A little searching reveals that a .22 bullet is \~2.5g and fires with a muzzle velocity of \~330m/s. Kinetic energy is given by KE=1/2mv^2, where m is mass and v velocity. So, the momentum of a laser pulse with equal energy would be $$ p=\frac{mv^2}{2c} $$ $$ p=\frac{.0025kg(330m/s)^2}{23\cdot 10^8m/s} $$ $$ p=4.5\cdot10^{-7}kg \cdot m/s $$ For comparison, the momentum (p=mv) of a .22 bullet is .83 kgm/s. The momentum of a laser gun is 2 million times less than the momentum of a .22. But is momentum all we should consider? I suspect the 'kick' we feel on the recoil is directly related to the force that the gun exerts on the holder. This means that instead of momentum we need to consider impulse, momentum per time. We estimate the time it takes to fire a .22 is \~.1s, so the force delivered 8.3 N. Let's estimate the time it takes a laser gun to fire. Unfortunately, not having a laser gun to fire (feel free to send me one, dear readers), we're more or less going to have to guess at the firing time. Most movies with laser guns show pulses of light (which, incidentally would move so fast we wouldn't see them) on the order of a meter or two long. Given the speed of light, this would give a firing time of \~30 nanoseconds. This would give a force delivered of 15 N. This is close to what we estimated or a .22. So, if movies are to be believed (and really, why wouldn't we believe them?), it seems like laser guns may well have recoil. Note: It is worth questioning if we need the same energy for a laser as for a bullet. That could certainly change our estimate. Maybe we'll return to this question again.*

Ask a Physicist

We've added a new feature here at the Virtuosi. Ask a Physicist. To the right in the sidebar you will see a link to our shared email account. Feel free to send us email asking any question you desire. We will do our best as starting PhD Graduate students to answer. Wondering about some physics principle? Want to know what life as a grad student is like? Want to know our favorite music? Ask away.

Lessons in Prohibition: One

"Holy lack of self control, Batman!" Cornell University Department of Physics-A P6510 Auxiliary Report Introduction: I feel terrible. Batman made me feel terrible. Batman also cost me 20 bucks. Theoretical Background: You see, it's pretty natural for humans to place the mind and body within hypothetical constraints. "What would it be like to be deaf/blind? What if you didn't have an arm? What if you were forced to watch reruns of Quantum Leap for 24 hours straight?". I guess this is in some way a thinking-man's survival instinct. It's always prudent to be prepared, even if just emotionally, and Scott Bakula can be kinda pushy. We here at the Virtuosi believe in preparation. That's why we've decided to convert one of our own (me) into a guinea pig. It is our goal to apply some actual constraints to a real mind (mine) and to observe the results, thereby gleaning valuable insights into how humans (me) respond to adverse conditions (psychological torture). Experimental Procedure: This week we started small. My task was the simple omission of a single word from my written and spoken vocabulary. The word should be innocuous enough that I wouldn't usually say it, but common enough not to be obscure. The word chosen was BATMAN. I was to expressly abstain from the use of BATMAN, else risk a monetary penalty (20 bucks), and if I were successful after 1 week, I would gain a monetary reward (20 bucks). This experiment (bet) began two days ago. I have obviously failed. Data Tables: Data in this case, and always, will be presented with the utmost objectivity (stream of consciousness). Day one: Ha This is really easy Everybody keeps mentioning Shatfan though I don't even know what we're testing I do kinda wanna say it though 20 bucks Shatfan feels kinda dirty Whatever this is dumb I have homework maybe I should just stop thinking about Shatfan Nah This is kinda fun I don't wanna lose 20 bucks I liked Val Kilmer as Shatfan, I can't believe Nic disagrees Catban Batmatt HA thought I lost it there NOWP there are so many alternatives Shut the hell up Corky [possessor of said $20] Oh man everybody knows about this bet I might lose What if someone asks me about fratcan in class? Day two: I actually feel kinda gross thinking about Shatfan he's like in my mouth please get him out this is all people talk about when I'm around oh yeah thanks for that this week will at least be comical hey this might offset the 500 bucks I owe corky maybe I could just say it and not tell anybody ah but no that's not science and this is very obviously sciencemanbat tabnam batfarm man of bats shatbat oh hey dad whats up yea we're doing this thing where I cant say BATMAN... Dad: oops! uh.... Me: crap. I have phone calls to make. I'll talk to you later dad. click Post Failure Data:** Batman is dirty. I don't even want to say it now.

Shatfan is the preferred euphemism for BATMAN. Analysis: We therefore conclude it is impossible (for me) not to say BATMAN for longer than approximately two days. It was assumed by observers that the continual substitution of Shatfan and derivatives (highlighted above) for BATMAN was purely cavalier. This is false. While a bit of joviality was included in the experiment, a vast majority of euphemisms used were to assuage psychological tension. This was accompanied by laughter to maximize psychological relief. Obviously experimental procedures must be altered. We propose now an alternate method of success, via locking the subject in his office for a week. Conclusions: In this report we find that I am a weakling, and that the experiment was kind of lame. However, as a test run for future Prohibitive Tests (PTs) we find the concept promising, in that when Scott Bakula comes around, I'll be the only one ready to take him on. Current developments in PTs include subject which are not fit for print, and thus the authors invite suggestions in which a mundane aspect of my life will be removed for the greater good of mankind. Acknowledgements:** The authors would like to acknowledge that Corky is a shatfan. Stay tuned for more. -Jared

Falling water - hot or cold?

Hello everyone! Since this is my first post as one of the virtuosi, I should probably introduce myself a little. I'm a first year graduate student in physics at Cornell university. I did my undergraduate work at Oberlin college (I know, you've never heard of it), and I'm currently just trying to keep my head above water and take in as much physics as I can. Additionally, I'm trying to find work for the summer, I might post more on that later. Today, the question that is on my mind is: How much does a water droplet heat up when it goes over niagara falls? image (image from http://grandcanyon.free.fr/) Let's begin with a little motivation. Why would a water droplet heat up when it falls? Well, the physical mechanism is that as it falls through air, air resistance dissipates energy. This energy is dissipated mostly as heat, so we expect to put some additional heat into our water droplet as it falls. More quantitatively, assume the water is going to fall from some height h to the ground. The gravitational potential energy of our water droplet is given by $$ PE=mgh $$ where m is the mass of the droplet and g is acceleration due to gravity. We assume that the particle starts with no initial velocity. It is very easy to place an upper bound on how much the droplet will warm. The maximum heating will happen if all of the potential energy were converted to thermal energy. The temperature change wound be $$ mc\Delta T = mgh $$ so $$ \Delta T = \frac{gh}{c} $$ Where c is the specific heat of water. Niagara falls is 51m tall, g is 9.8m/ss and c is 4.1kJ/kgK so this gives a maximum temperature change of .12C=.22F, fractions of a degree. However, we can do better than this. A small droplet will experience a linear drag force, and a bouyant force from the air. The combination of these will result in a terminal velocity of $$ v_t=\frac{2}{9}\frac{(\rho_p-\rho_f)}{\mu} g R^2 $$ Where $\rho_p \text{ and } \rho_f$ are the droplet and air density respectively, $\mu$ is the dynamic viscosity and R is the radius. This gives the kinetic energy the particle gains and we assume the rest of the potential energy goes to thermal energy. So, $$mc\Delta T = mgh-\tfrac{1}{2}mv_t^2$$ $$\Delta T = \frac{gh-v_t^2}{c} $$ Now we can plug in some numbers. A little bit of searching will turn up appropriate values for the density of water (\~1000kg/m^3), the density of air (\~1.2kg/m^3) and the dynamic viscosity of air (\~.8mPas). We estimate the radius of a small droplet of water is 1.5mm (\~1/16"), which seems about a medium sized raindrop. This gives a terminal velocity of 6m/s. Using the above equation this gives a temperature change of $$ \Delta T = .0024C/mh-.009C $$ It looks like our calculation only makes sense after the after droplet has fallen more than 4.5m (this is probably about the distance it takes to reach terminal velocity)! Niagara falls is 51m tall, so this gives a change in temperature of .11C=.2F. The water only heats up by a fraction of a degree. This more realistic estimate is still rather close to our upper bound. I leave it as a question to you, the reader, to estimate the temperature change of a typical raindrop.


So me and Jesse got to thinking today about Railguns. Every year in Physics 213 a common homework problem is a rather simple model of a railgun. We tried to think about a more realistic model. We simulated a rail connected to a voltage source with a limiting resistor, moving under its own magnetic field with a back-emf.


Short answer: Doesn't work very well if you limit the current, but works great if you dump a MA or so down the wire.

See the full solution after the jump.


Imagine you have two closely spaced conducting rails which you connect to a voltage source. Laid across these two rails is another piece of conductor. The first thing to notice is that we expect the conductor, if we assume it can slide along the rails without friction, will be compelled to move because of the magnetic force caused by the magnetic field from the rails acting on the current through the segment. Lets be quantitative, our voltage source provides and emf V. The rails are seperated by some distance L. The initial position of the slug (as it will be termed) is $x_0$ We are interested in the motion of the slug. It will feel a force $$ F = I \int B \cdot dl $$ The current being the current that flows through the circuit. The important contributions to the current is the emf of the source, the resistivity of the rails and slug, and the back emf induced by the motion of the slug. I.e. our equation for the circuit takes the form $$ V - I\lambda( L + 2x ) - \frac{d}{dt} \int B dA = 0 $$ We need to approximate the magnetic field acting on the slug. We will treat this magnetic field as originating wholly from the rails, which we will treat as semi-infinite wires. This approximation should be good in the limit that $L \gg x_0$ The magnetic field of a semi-infinite wire is $$ B = \frac{\mu_0}{4\pi} \frac{I}{r} $$ so our magnetic field acting on the slug, if we treat $z=0$ to be at the bottom rail and $z=L$ to be at the top is given by $$ B = \frac{\mu_0}{4\pi}I \left( \frac{1}{z} + \frac{1}{L-z} \right) $$ We can calculate the influence of this field on the slug. Of course if we integrate this field from 0 to L it will blow up, so we need to set some cutoff, $R$ which will stand in for the fact that our wires have some finite thickness. We proceed: $$ \int B\ dl = \frac{\mu_0}{4\pi} I \int_{R}^{1-R} \frac{1}{z} + \frac{1}{L-z} \ dz = \frac{\mu_0}{4\pi} I \int_{\epsilon}^{1-\epsilon} \frac{1}{\xi} + \frac{1}{1-\xi} \ d\xi $$ where I have non dimensionalized the integral with: $$ \epsilon = R/L $$ We obtain $$ \int B\ dl = \frac{\mu_0}{4\pi} \left[ 2 \log \left( 1/\epsilon - 1 \right) \right] I \equiv \kappa I $$ This gives us the force on our slug $$ F = \kappa I^2 $$ as well as an approximation for the magnetic flux if we assume that this field approximation works for the length of the rails (up to the slug) $$ \int B \ dA = \kappa I x $$ So that we can gather our circuit equation $$ V - I\lambda( L + 2x ) - \kappa \frac{d}{dt}(I x) = 0 $$ and the equation of motion for our slug $$ m \ddot x = \kappa I^2 $$ Lets nondimensionalize these equations with $$ x = L \chi $$ $$ x_0 = f L $$ $$ I = \frac{V}{r + \lambda L (1+ 2f)} \Phi \equiv I_0 \Phi $$ marking the initial current that will flow through the slug. $$ t = \frac{ \kappa}{2 \lambda } \tau $$ giving a characteristic time. Putting it all together, our differential equations become much simpler. $$ f - \chi - \frac{d}{d\tau} \left( \Phi \chi \right) = 0 $$ $$ \frac{d^2 \chi}{d\tau^2} = \beta \Phi^2 $$ $$ \beta = \frac{ \kappa^3 I_0^2 }{ 4 m L \lambda^2 } $$ So that our system is specified completely by the dimensionless constants f, β, f having something to say about the initial geometry of our system, and β speaking to the relative mass of the slug.


These coupled nonlinear differential equations are not easily solved, but by taking the time to nondimensionalize them, we can integrate them numerically. Lets suppose our rails are separated by 1 cm, and our f is taken to be 10 or so (remember the larger f, the better our approximation for the field). And for our material properties, lets take gauge 0 copper wire, i.e. I will use its radius, mass density, and conductivity at room temperature. Lets make the voltage supply be a kilovolt (non unreasonable), and lets choose the limiting resistor in order to keep the current in the gauge 0 wire under its specification: 170 amps. Doing all of this, we have $$ \beta \approx 5.4 \times 10^{-7} $$ $$ f = 10 $$ and for interest our characteristic time is 112 microseconds or so, the mass of our slug is about 5 grams. Our average magnetic field is about 0.07 gauss/amp between the rails, so that our starting field is about 12 gauss.







So it looks like our Railgun doesn't do very much at all. In fact, its quite paltry, achieving very puny speeds. What went wrong? Well, it seems the biggest problem comes from limiting the maximum current that can go down our wire. The truth is that if we want to really project some stuff forward, we're either going to need much higher currents, or some external field. You'll notice that most of the action takes place over a very short time scale, so it could be that we could violate the rated currents of our wire. If that's the case, setting our failsafe resistor to zero, I'll just show two more graphs, that of the velocity and position of the slug:



This is starting to look a lot more like a Railgun, unfortunately, it would draw a current of 15 MA to start! So the real question becomes, how much current can you dump into a copper wire in 0.1 ms or so. 15 MA is a lot and would certainly melt the wire, but would it melt the wire in such a short time? If not, then it looks like we could get some real kick out of our model. In fact, we know that railguns with some kick can be made. According to Wikipedia, the US Navy has a railgun that can accelerate a 3.2 kg projectile at 7 times the speed of sound (\~2.3 km/s). Playing with my code, if I put in by hand a projectile mass of 3.2 kg, and reduce f to 1 (meaning we get a little more kick but our approximation of the field starts to really break down), I can get a final speed of about 2 km/s, which seems to be in relatively good agreement, but this only occurs if the current is \~100 MA for about 30 microseconds. Moral of the story, if you want to build a railgun, you need to be willing to melt some copper. Attached is a Mathematica notebook used for the computation: here.

LaTeX Test

Hopefully $$ \LaTeX $$ is now enabled thanks to the tip at: WatchMath. This is a test. $$ E = mc^2 $$ $$ \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{-\infty}^\infty dx\ \exp \left( -\frac{ (x-\mu)^2 }{ 2 \sigma^2 } \right) = 1 $$ Continue about your day. EDIT: Fixed dx

A New Beginning

Being inspired by a Colloquium given today at Cornell by Chad Orzel over at Uncertain Principles, I'm going to try and resurect this old beast and begin anew. The general plan is that I will try and read more scientific papers, translating them into plainer english, as well as solve little problems or puzzles that I dream up. Wish me luck, perhaps I will get some of my friends involved.