Some Rail Gun Derivations and following questions / Electromagnetic Projectile Accelerators / Forums

Forums

4hv.org :: Forums :: Electromagnetic Projectile Accelerators

« Previous topic | Next topic »

Some Rail Gun Derivations and following questions

Move Thread

LAN_403

EvilTesla-RG

Sat May 23 2009, 07:55PM

Registered Member #1523 Joined: Sat Jun 07 2008, 02:05PM
Location:
Posts: 97

One day, I was thinking about rail guns.

And I had a quesion, How big does a permanent magnet, and current in a railgun have to be, to have equal force to a rail gun without a permanent magnet?

so, I did some equation derivations. Please bear with me.

I started with

f=qvXb

Where f is the force. q is the charge. v is the velocity (of the charges). X is cross product. And b is the magnetic field.

Can be rewritten to

F=ilXb

Where i is the current. And l is the distance between the rails.

So, in a railgun that is not augmented.

F=ilb.

(assuming a railgun where everything is at 90 degrees)

Where b=ui/2(pi)r

Where u is the permeability of free space. And r is the distance.

In a rail gun, the distance would be in the middle of the two rails. And the equation should be doubled.

So, once simplified

B=2ui/(pi)l

So, for a non-augmented railgun

F=2ui2/(pi)

Now, for a railgun augmented by a permanent magnet.

F=ilb (of course)

Where i (for the sake of my purpose) will be replaced with w. Since, although it is a current, it shouldnâ€™t be confused with the current in the original equation, which is larger.

Also, b, instead of being dependent of current. It is dependent on the permanent magnet.

So, for a augmented railgun.
F=wlb.

Now, I will set them equal, to figure out how big the magnetic field and current has to be in an augmented railgun, to equal the force of a non-augmented railgun.

So

Wlb=2ui2/(pi)

So

Wb=2ui2/(pi)l

Yea. So we have a nice equation. But what does it mean?!?!?

Lets plug in some numbers and see what happens.

The common Neodymium magnet is around 1 T.

So, b=1.

U=4(pi)x(10^-7)

And, the common Photo Flash capacitor can put out .4 amps.

It isnâ€™t uncommon to have around ten of these in a small rail gun

So, i=4

Lastly, a good width for a small rail gun is 5 mm. Which is 5x(10^-3)

So, assuming these numbers are good. How big of a current does my augmented rail gun need to be on par with a rail gun powered by 10 photo flash capacitors?

W=4x(10^-7)x42/5x(10^-3)

W=1.28x(10^-3)

So, a small rail gun augmented by one small neodymium magnet. Only requires a current of 1.28 mA to compete with a rail gun powered by ten photo flash capacitors.

So, my question is, why do we bother with high amperages at all? Why donâ€™t we just use some big magnets?

Which I would very much like to know.

Howver, for those who wouldn't mind following some more math. I had anouther quesion. (My curiosity is Unsatiable)

My previous calculation strikes me a a contradiction to the law of conservation of matter an energy.

What if I took that logic a little further, and calculate how big of a magnetic field would be needed to break the law?

So, bear with me as I do some more calculations.

To be a contradiction of the law of conservation of matter and energy, the energy in the projectile must be greater than the energy put into it by the electricity.

Let me start by finding the energy of the projectile.

Assuming no friction. All the energy in the projectile, due to the railgun, is kinetic.

So, we can use, e=.5 m(v^2)

Where e is energy, m is mass. And v is velocity.

And

V=at

Where a is acceleration, and t is time.

So we need to find the acceleration and the time.

Well,

F=ma

And f=ilb

So a=ilb/m

Time is a little harder

If we start with x=a(t^2)

Where x is the distance that the acceleration is applied. (length of the railgun).

To solve for t

T=(x/a)^(1/2)

And we know that

A=ilb/m

So,
T=(xm/ilb)^(1/2)

So, to plug this all in

E=.5 m ((ilb/m)^2) (xm/ilb)

To simplify further

E=.5 xilb

That simplified nicely.

Now, for the energy that is put into the rail gun via the current.

Well,

E=pt

Where p is power

And p=(i^2)r

Where r is resistance. (I could use p=iv, but it seems to me that p=(i^2)r better describes what is going on)

So

E=i2rt

And t=(xm/ilb)^(1/2)

So

E=(i^2)r(xm/ilb)(^1/2)

So for the fun part. To decide if these equations break the laws of physics.

First, I will set the energy in the bullet to be greater than the energy in the electricity. Which inherently implies a breakage in the law of conservation of energy

So

.5xilb>(i^2)r(xm/ilb)^(1/2)

now I will solve for b. to see how big of a magnetic field is needed, before the laws of physics no longer apply

Which comes out to be

b>(4m/xl2i)^(1/3)

hmmâ€¦so, if I have a permanent magnet, augmenting a railgun, that is bigger than that strange number, I will break the law of conservation of energy.

Lets plug in some numbers.

Say,

M=5 grams. Which is .05 kg.

X= 1 meter

L= 5x(10^-3) meters

And I= 4 amps

once pluged in, you come up with

b>12.6 Tesla.

Which, although is bigger magnetic field then I want to make with my college budget. It is only around 12 small neodymium magnets. Not unreasonable.

So, I guess I am wondering. I trust the laws of conservation of matter and energy more than I do my own math and understanding.

Where did I screw up? And what donâ€™t I understand?

rp181

Sat May 23 2009, 08:28PM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
Location:
Posts: 1529

I did not follow the math very well, but....
Math can model equations that theoretically can break laws. There is a point where the math continues to model past practical limits, you need to find the limits.

As for your first question, relating to the above, I think there are also limits to that. You need a starting current that can overcome friction, and to maintain the force. Magnets are considered to have neutral energy, so it does not add power. The magnets mearly improve the affect of the current. Also, with larger railgun's, magnet augmentation becomes pointless, as they will operate above the magnet strength. This is where people use electromagnets rather than permanent magnets, in configurations called parallel augmented or series augmented. This essentially increases the number of turns in a railgun.

I may be wrong, someone confirm/disagree please.

EvilTesla-RG

Sat May 23 2009, 08:51PM

Registered Member #1523 Joined: Sat Jun 07 2008, 02:05PM
Location:
Posts: 97

Hmm...interesting.

I wonder what other limits I was never told about...

Anyway, I did spot some math errors in my second equation.

It isn't much differnt, but the result should be

b>(2mi/x(l^2))^(1/3)

and the final result should be

b>25.2

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob