For this you will need;

4 or 5 Disposable flash cameras (I went to a camera store and asked them for some and they took a big box with about a hundred in it, they sold them to me at ten cents a camera, I only got ten)

aluminum foil

plexiglass

wire

small switch

big switch

aluminum or steel rails

6 nuts and bolts

wire hanger

tools;

screwdriver

soldering iron

duct tape

wire strippers

wire cutters

4 or 5 Disposable flash cameras (I went to a camera store and asked them for some and they took a big box with about a hundred in it, they sold them to me at ten cents a camera, I only got ten)

aluminum foil

plexiglass

wire

small switch

big switch

aluminum or steel rails

6 nuts and bolts

wire hanger

tools;

screwdriver

soldering iron

duct tape

wire strippers

wire cutters

My railgun won't qork, the projectile just welds to the rails. Can I have some help?

<p>You need to use something like C02 to make your projectile have an initial velocity.</p><p>When traveling along the rails the projectile heats up and starts to melt. You need to calculate how fast you need your object to go so that it will melt slightly to help lubricate it without melting it completely.</p>

And it wouldn't let me add it to this one

<p>Rail Guns Need An Initial Velocity! Use it to supplement a gas propelled or explosive device.</p>

I have a problem. My railgun (as shown in the pic below) doesn't work. Whenever I turn it on the projectile just welds to the rails. Can I have some help/advice?

I forgot too add the picture sorry

So you don't need anything to jump start you projectile, I know larger ones use compressed air or springs to avoid welding the projectile to the rails. Also does yours throw cool sparks and stuff, maybe even plasma?

uhm... Yeah, for a railgun to have enough force to create plasma, it would need several megajoules of energy (if you saw the railgun from DARPA or the US Navy, that one shot took so much energy that it could've been used to power a city for a year). <br> <br>To release sparks, a railgun doesn't have to be THAT powerful, but it still requires a fairly large amount of energy. In short, THIS railgun can't be used to kill people or do any heavy damage.

the biggest railgun (navy) was 33 megajoules or 30000000 joules, or a little over 12 horsepower, I see what you mean, but using a 12 hp engine to charge batteries for one hour couldn't power my house for a year.

Yeah, thats just wrong. Joules are a unit of energy. Horsepower is a measure of power.

<p>power X time = energy, though</p>

but power over time is energy right. <br>I just found all of this information, <br>Watt x Time = Joules This is 8948 time 3600 secconds = 32,212,800 joules which is a pretty good conversion for the one minute I put in.<br>Power x Time = Work (energy)<br><br>Joules/distance = Newtons<br>Work/distance = Force<br><br>So you can convert power to energy, its just a little more complicated than most conversions.

Look, you cant just assume time. And the first guy was wrong as a city uses about 2 Gigawatts each year. <br><br>But charging and equations are not linear which you are assuming. By your logic you can convert Joules into cubic meters<br>

I did it, I converted joules to liters of gasoline, one cubic meter equals 1000l so yea I will do that too. So because one liter of gasoline contains about 35 MJ I will just say that we have 35 MJ. That is equivalent too 0.001 cubic meters, hooray. I hate ruining peoples day.

<p>just curious, why is this significant, cant we just go to youtube and look up americas navy</p>

Wow i cant believe i fell for a troll. Kudos

BTW that wasn't off-topic, and I am not a troll.

Look, I'm a double major in physics and chemical engineering and I am telling you that while I understand why you are thinking what your thinking, it is incorrect.<br><br>You are not converting. You are doing calculations. A conversion changes things that are equivalent. You can only change Joules into other energy units.<br><br>Further, you are ignoring inefficency, heat transfer, momentum and impulse, how magnetic fields work, charging and discharging of capacitors and other things.<br><br>Also, the electricity that is used is in watt hours when calculated. <br><br>That is what is what is wrong with what you are doing.

<p>And one Watt Hour is 3600 Watt Seconds, and one Watt Second is a Joule. All the other characteristics: efficiency, impulse, etc. are mere complications and have Zero relevance to energy to power equivalence, since all such "losses" are also expressed as energy. If a quantity of fuel produces 1 MJ when completely oxidized, it produces 1 MJ. If your generator is 50% efficient, it produces 500 kJ of electricity and 500 kJ of waste energy in heat, vibration (noise), and electrostatic and electromagnetic losses (which eventually also become heat), but the TOTAL energy is still 1 MJ.</p>

1 watt is equal to 1 joule, "jj.inc" is correct. <br> <br>All you need is an Axiom and you can convert anything you like. <br> <br>For example: <br>Under this situation the Axiom is a Joule, and we know that 1 joule = 1 watt. We also know that Wattage is PxA. <br>So I can have a power supply rated at 2V .5A that = 1Watt or a supply with a power rating of .5V and 2A = 1Watt. <br> <br>I think our education system is in trouble when a person with a Physics and Mechanical engineering degree thinks that Power and Energy are different. <br>Maybe "jj.inc" is wrong about other things but he isn't wrong about that. <br> <br>Einsteins theory or relativity: Energy cannot be created nor destroyed only transformed. <br> <br>Don't tell me you beleive in overunity?

Heyo, I was just looking through and I figure I would see if you had my final results yet Mr. Braniac. If you realized I was right you should at least admit it, and if you know what I did wrong please explain so the world can become a better place. After glancing through my very first comment I see what you caught me on, but I subtley place the time into the factor in the last half of my comment. Just ignore this: "he biggest railgun (navy) was 33 megajoules or 30000000 joules, or a little over 12 horsepower" and read this: but using a 12 hp engine to charge batteries for <strong>ONE HOUR</strong><strong> </strong>couldn't power my house for a year.

You could have just told me I was using the wrong word to explain what I was doing. Its really not that big of a deal. Yea, I did ignore all of that intentionally to say that in absolute best conditions with no energy loss, it still only provided a certain amount of energy. I don't see where I messed up this whole watt hours thing either. I provided time for every "calculation" I did, it doesn't have to be watt hours. I had 33 MJ I calculated that out to watts over a period of time. That is all I did, and that is all I needed to do. Do you understand this?

I am sorry, I just don't like it when someone says no your wrong, I just don't have any real numbers or facts to back it up so I will just muble around and insult you.

No, I am saying 33 gigajoules is equivalent to 12hp over 1hour. Which it is. I just converted the hp to watts, which is perfectly fine, then I plugged in the proper time which is in secconds, one hour = 3600 secconds. 12hp is very roughly rounded, and the concluded answer was extremely comparable. If I directly converted 12hp over to electrical watts with no loss of energy I would still only get roughly 9kw which would be enough energy if stored to run my house which consumes 3kwh for roughly 3 hours. Its a bunch of math, but it is well reasoned. I know it is difficult to understand converting between these to things. <br><br>Joules are a measure of energy (energy is defined as the ability to do work)<br>Work is power * time and is measured as an SI unit of joules<br>Power is the rate at which work is performed or energy is converted<br>Time is, well you know what time is<br><br>By looking closely at these properties you can conclude that power over time is work or hp over one hour is joules<br><br>When I use the word over I don't refer to a fraction or division problem I refer to something occurring over a period of time.

The problem with this calculation is that the railgun mentioned discharged its energy over a fraction of a second. Your calculation provides a power of around 9000W: in reality, it would be more like 3300000000W (almost 4.5million horsepower), based on a guesstimation of a 0.01s discharge time.

Could you please explain more thoroughly. Time plays no role in these calculations. If something is 33 gigajoules, it is 33 gigajoules for a second, and 33 gigajoules for a day. It is just 33 gigajoules. <br> <br>Ha ha, I was just doing some math and you are just a failure. 33 gigajoules is 3,000,000,000 joules. That number is one 0 short of your number (typo probably). Also sure you may get 4.5 million hp (DON'T QUOTE ME I AM USING YOUR DATA AND IT IS TOO LATE AT NIGHT FOR ME TO CHECK IT), but you only get 4.5 million hp for 0.01 seconds. If you calculate that out which I can't think of how to do at this time you will only have a few horsepower over the course of a short time. <br> <br>-to everyone other than 06sandj Please disregard this post as the information is useless and just a poorly compile explanation of the massive fail seen above.

Yes, the energy is the same, no matter the amount of time; however, you were quoting power, which does include time. The 4.5million hp came from google calculator (3300000000 watts in horsepower) - I'm not exactly a units conversion wizard myself :P I get your point, and it's true what you're saying about powering a city for a year - I was just attempting to point out that you were referring to the power of this particular railgun as 12hp. This may have just been unclear writing or me not understanding your comment particularly well. On the whole, we should probably forget all about this conversation, agree that railguns are pretty awesome, and continue with life :)

oh, no, I wasn't saying the railgun had 12 hp, but the idea that it produces 4.5 million for 0.01 seconds is kinda cool

All cleared up then :) And yes, it is :P Imagine the recoil on that thing - slight problem to firing it from a ship!

ok im not very experienced with this subject, but to my understanding recoil is caused by the expanding gasses from the explosion of gunpowder (or whatever other explosive propelent you could use). so as long as you only used magnetic fields to eccelerate your projectile, and made shure to ventilate the rails, or barrel used to house your projectile, the only recoil you should experience should only be (not shure exactly how to say this, but hopefully this will work) the guns movement related to the inertia of the projectile which would keep it resting. if any of that is wrong plaese correct me.

<p>Um, no. Catapults produce recoil. So do crossbows. Recoil is a function of Newton's "Law" of conservation of linear momentum. If you impart a momentum (Velocity * Mass) to an object, an equal and opposite momentum is imparted to you. You're making a mistake similar to a newspaper writer who assumed that rockets couldn't work in space because there was "no air to push against." If an astronaut in free fall were to throw something, he or she would move in the opposite direction at a velocity proportional to the relative masses. If you'd like a more mundane example, skate out onto an ice rink and throw a shot-put. If you don't brake yourself with your skates, you'll recoil.</p>

<p>yes, i understand that newtons law contributes to the recoil of a weapon and in a comment below i did some rough calculations based on the returned energy of a railgun, but in modern firearms the vast majority of the recoil is generated from chemical expansion causing that equall and opposite reaction, not the projectile which weighs a few grams.</p>

Every action has an equal but opposite reaction (one of them newton laws). When the magnetic field pushes the projectile the projectile pushes back just as hard. Now on a small railgun you won't notice it because the projectiles are usually very small and not going very fast, but the navy gun shoots a very heavy projectile at extreme unimaginable speed, this results in a huge amount of opposite force.

yeah i know about newtons third law but (and again im not very experienced with this) wouldent the force of the "opppsite force" be distributed more or less evenly into the gun (kinda like if a baseball hits a barrel of water, the barrel doesent go flying at the same speed the baseball was traveling) so if a railgun which accelerates a 40 lb projectile to 5,500 mph, which i roughly calculated to require about 54.5 million joules (or 1.36 million joules per pound), was mounted in a platform weighing 5,000 pounds it would only have a recoil of around 272 joules (around twice the recoil of a barret .50) which i think a 16 million pound ship could easily take. now im shure i made some mistake somewere in there so please correct me

I'm not to great with the numbers either, but I think you are probably right, it shouldn't do too much to move the ship, but it would still be 5000 pounds pushing back at 272 joules, which I am sure you would feel. Being anywhere near the thing could be really dangerous I imagine. I wounder what the electromagnetic field around it looks like.

yeah the biggest problem i could imagine is the cannon going off and taking anything magnetic that isnt tied down (and possibly part of the ship) with it. lol i can see the headlines now "navy sends railgun superweapon into battle, oblitorates enemy ship...and sinks own ship in the process"

agreed. <br>seeing where they got the energy and how big the power source is would also be funny.

When you are talking work done by electricity being converted into a working medium, like the magnetic field, strength is a product of it being divided by time. So 1 hour of charging with a 12hp motor will not equate into the aforementioned joules of force. This is why it does NOT require an entire nuclear power facility to reach the fire rate that the weaponized version is projected to reach. You have to take a fairly strong power source, like a compulsator or bank of capacitors, and discharge the stored energy at an extremely rapid pace, far faster than a hundredth of a second. This is why it's so hard, pulsed power sources able to reach the required pulse lengths are extremely crude, inefficient, and bulky. YES, you can make it work on paper, but that's not how it turns out in reality.

Wrong. <br>The energy is being stored then released, hence the capacitors. There is always some energy loss but not that much. <br> <br>Think of your car engine, the energy loss is caused from heat, vibration, friction, noise, just to name a few things, all can be measured and combined they total the energy used to run it. <br>Hydrogen has an exothermic reaction of about 286000 joules per mol, that means the energy released when hydrogen combines with Oxygen is 286,000 Watts, gas is probably pretty close to that too, but you don't see that power at your wheels. <br> <br>In this situation we have less energy loss, no vibration or noise until you release the energy stored. <br> <br>Robert Goddard: "For every action there is an opposit and equal reaction"

Oh, dude, I am so sorry. I have no idea where giga came from, it should have been mega the whole way through, which would drastically bring down your power rating. Sorry I failed as bad as you. <br> <br>-Face Palm-

You can, <br> <br>Joules/Newtons = Distance <br>Work/distance = Force <br> <br>All you need is an Axiom, math does not lie, equations are linear; What isn't linear would be how the rules are applied. <br> <br>Newtons third law: For every action there is an equal and opposite reaction <br> <br>Einsteins theory of relativity: energy cannot be destroyed or created only transfomed

<p>That's Einstein's Theory Of Mass-Energy Equivalence, upon which relativity is based, but it's a separate theory from Special and General Relativity. The common form of E = MC² is actually incomplete. It's more properly E² = M² * C^4, or E = √(M²C^4) = ±MC² which implies the possible existence of negative energy, and possibly, imaginary mass. </p><p>The Special and General Relativity Theories deal with other properties.</p>

<p>i thought work for a circut was amps * volts, </p>

The joules are calculated as <br>(voltage^2*capacitance)/time=joules when you are referring to a railgun. Joules of energy is what is important rather than the work for a railgun. And the proper conversion for figuring the force delivered by the projectile is actually mass times acceleration. Astoundingly this works out in the physical world because, like in a rifle, once the projectile leaves the force acting upon it, in this case the magnetic field is the accelerating force, the projectile is immediately in deceleration unless it is shot along a parabolic path, in which case it could possible experience some acceleration once it passes its apex due to acceleration induced by gravity IF it has decelerated below it's terminal velocity. <br> <br>And horsepower is how long it takes a team of horses (later revised to a force) to drag a 200lb object over a given distance, so it literally has no application in a railgun. Google is NOT reliable for conversions.

<p>to make a plasma current like a electric arc, you just need extremely high current, in a tazer it has about 50k volts at almost 0 amps so no one dies, but you need the high voltage to make the jump</p>

not really. <br>this one throws plasma with a kilojoule. <br>https://sites.google.com/site/futureexperimentalsystems/emla-iii-p

<p>Please correct me if i'm wrong in thsi but, all your doing here is taking the circuit from the camera and giving it a much larger capaciter bank, is that correct?</p>

Help heres a close up of the rail<br>

I made but it won't fire any advice please I need help!!

<p>Nice stuff </p>