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Magnet coil help (turns, gauge, for coilgun)

supak111

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Apr 29, 2012
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Hey everyone I'm building an coil for a coilgun and need help figuring out best wire gauge and # of turns for my coil.

I will be using 3, 13.6v lithium car batteries in series so about 40v, and these batteries can put out about 125 cold cranking amps so I potentially have at least 5000w. I want my coilgun to be without capacitors.

Coilgun projectile will be 3/8" or 9.5mm steel ball bearing that is about 3.5 grams.

My goal is to put as many of those 5000watts energy into the projectile without totally saturating it, so what would be best wire gauge and number of turns?

Thanks everyone.
 

Raven Luni

Oct 15, 2011
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I dont know much about windings and saturation (the maths makes me want to run away) but a straight battery setup isnt going to you much good. You need a very fast pulse for a coil gun which can only be obtained from a capacitor discharge. A battery might be able to supply high current but it has a much bigger ESR which means the voltage will drop off when loaded - or something like that. I'm sure the experts can explain it alot better :p
 

supak111

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Why wouldn't 40v @ 125amp work without caps?

My coilgun will have many stages, like 100 stages, but I don't want caps.

I wish someone would be able to answer my math questions from above
 

CDRIVE

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Are we going to see you in one of those "It only hurts when I laugh" episodes or maybe one of those "Inside Jail" shows? :rolleyes:

Chris
 

supak111

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Lol no nothing illegal about a coilgun, they are legal in all 50 states and don't even required a FOID card, and I'm trying to avoid capacitors so I'm not on one of those "death by electrocution shows" lol
 

CocaCola

Apr 7, 2012
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I wish someone would be able to answer my math questions from above

There is an old saying that if you don't understand what you are doing you probably shouldn't be doing it in the first place...

Honestly though what is you end goal? Why can't you use caps? Is it the cost? Because sometimes things cost what they cost, whether you like it or not...
Railguns are 'generally' built with capacitor banks, I'm sure if you could skip them and just use the batteries with good results no one would be bothered with the capacitors, but quite the contrary is true...
 

supak111

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Capacitors are way to big and heavy, my end goal is to make this coilgun portable and to be able to use it just like you would a hunting rifle.

I'm buying super expensive, but powerful and light lithium car batteries for that reason.

If I do it with heavy/bulky capacitors then there isn't going to be any portability in this coilgun so I shouldn't even buy the lithium batteries in that case.

I want to make a better coilgun then what everyone else has already done. I've researched hundreds of other peoples designs, now I want to take what I've learned and make a better one.
 

CocaCola

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Capacitors are way to big and heavy, my end goal is to make this coilgun portable and to be able to use it just like you would a hunting rifle.

I'm buying super expensive, but powerful and light lithium car batteries for that reason.

If I do it with heavy/bulky capacitors then there isn't going to be any portability in this coilgun so I shouldn't even buy the lithium batteries in that case.

I want to make a better coilgun then what everyone else has already done. I've researched hundreds of other peoples designs, now I want to take what I've learned and make a better one.

LOL...
 
Last edited:

BobK

Jan 5, 2010
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Let's do some calculations

Some constants:

length of barrel L = 1 m (I made this one up, but it seems reasonable)
power from battery P = 5000 W
mass of projectile m = 3.5g = 0.0035 kg

Some variables:

acceleration a
time to accelerate t
muzzle velocity v

Some equations:

v = a t
E = 1/2 m v^2 = P t = 5000 t
L = 1/2 a t^2 = 1

Some maths:

1/2 a t^2 = 1
a = 2/t^2

v = a t
v = 2 / t^2 t = 2 / t

1/2 m v^2 = 5000 t
1 / 2 m (2 / t)^2 = 5000 t
2 m = 5000 t^3
0.007 = 5000 t^3
0.007 / 5000 = t^3
0.011 = t

1/2 m v^2 = 5000 t = 0.11
0.0035 v^2 = 0.11
v^2 = 0.11 / 0.0035 = 31.4
v = sqrt(31) = 5.6

So, your muzzle velocity is a whopping 5.6 m / second. And that is under ideal conditions of no friction and no resistance in your coils and 100% efficiency. You might as well just throw the bullet!

Hopefully, my math is right. Anyone care to check it?

Edited: Oh, oh, the math was wrong!

The last section should be:

1/2 m v^2 = 5000 t = 0.11
0.0035 v^2 = 10000 * 0.11 = 1100
v^2 = 1100 / 0.0035 = 314285
v = sqrt(314285) = 560 m /sec.

So, ideally, it could work.


Bob
 
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CDRIVE

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Uh, I'll take your word for it.

Chris
 

BobK

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You shoudn't I was off by a factor of 10 in my first attempt.

Edited. I can't even get that right. I was off by a factor of 100.

Bob
 
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CDRIVE

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Let's do some calculations

Some constants:

length of barrel L = 1 m (I made this one up, but it seems reasonable)
power from battery P = 5000 W
mass of projectile m = 3.5g = 0.0035 kg

Some variables:

acceleration a
time to accelerate t
muzzle velocity v

Bob
Vital data above!
You shoudn't I was off by a factor of 10 in my first attempt.

Edited. I can't even get that right. I was off by a factor of 100.

Bob

Bob, there must be something contagious going around. This is twice, in just the last two days, that I failed to read vital data. :(

Chris
 

supak111

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Let's do some calculations
v = sqrt(314285) = 560 m /sec.

So, ideally, it could work.
Bob

So at 100% efficiency and 1 meter barrel I could get 560 m/sec or about 550J which is as powerful as most 9mm handguns.

Now I will be lucky to get 20% efficiency so I would need 5x more power or a 5x longer barrel it seems.

Can anyone explain to me difference between using my 40v @125amp batteries, or using 40v caps through same coil? Say the coil has 100 turns and is 3ohms.

Wouldn't both make the same amount of electromagnetism since 3ohms resistance is going to be the bottleneck?
 

CDRIVE

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That equates to a mere 13A load.

Chris

Edit: I missed that you said you have 100 turns. What gage wire is this?!!

Chris
 
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supak111

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I just purchased 120ft, high enamel temp 200c, 14 gauge magnet wire.

I've only bought 2, 13.6v 125amp lithium batteries so far, I will try to purchase 6 or even 8 but need to wait a while.
 

supak111

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Yea I already looked at the chart, I was just asking a hypothetical question.

My actual coil will be 100ft of 14gauge wire (not yet sure on how many turns that will be) So my resistance is .251 ohms

What is the difference using 40v @125amp batteries OR using 40v capacitors?

I know the capacitors pulse the amps really fast but when you have resistance in a wire the caps and batteries will both achieve same amps correct?

Another questions I have, I know @ high amps batteries have voltage drop. Does that mean when I try to draw 125amps from my 40v batteries that the voltage will really only be like 35v or even 30v?

Thanks guys
 

CDRIVE

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Have you ever watched your car's voltmeter when you crank the engine?

Chris
 

supak111

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No I never have, but I'm assuming it goes down just not sure how far down. Maybe to 7 or 8v?
 

CDRIVE

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Dash meters are expanded scale but not known for lab accuracy I'll check mine when I start my truck tomorrow.

Chris
 
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