Railgun Calculation Problems

I'd say you're on the right track.

But your argument falls apart when you state, "F = (μ∗(V/R)^2)/2π*ln(d/r)

where μ = 1.2366 * 10^-6 N/A², V is 300 volts"
Now go back to Newton's second law and study it more closely. You will see the folly of your assumptions.

M= 0.5kg
WTF are you trying to do knock holes in army surplus Sherman Tanks?

I think it's the sort of application where calculation isn't going to yield an accurate answer due to the transients, high peak voltages/currents, unmeasurable resistances and friction etc.

I'd say the prototype and some experimentation will teach you more.
Del

WTF are you trying to do knock holes in army surplus Sherman Tanks?

Not at .02 m/s!

and you need a railgun for what? is there a ballistic missile in the neighborhood thats annoying you?

Leaving aside questions about the physics of your approach, I'm trying to understand what you're doing with your math analysis.

You go from ln(d/r) to ln(dr) which I assume is just a typo in the last 2 equations. You can take the ln of d/r; you can't take the ln of dr because dr is not dimensionless.

It appears you're using the discharge of a capacitor to provide the voltage to this railgun, and you get the discharge time τ (you use s) as 0.00003 seconds by multiplying R*C? Yet it's 0.000036. (?)

You don't define n.

You convert an acceleration, a, directly into meters/sec/sec? In the last equation you've substituted m/s/s for a; a more conventional substitution would be velocity/time, v/t, and then you could insert the value of tau from your R*C time constant that you call s, for the time, t.

If I put your equations into Mathcad and set n=1 (again, ignoring the physics of whether what you're doing is right, or not) I get v = 1,208 meters/sec.

To get a feel for the figures, consider converting all the energy stored in the capacitor to kinetic energy in the projectile.

½CV² = 0.5 * 0.003 * 300 * 300 = 135J

(C = capacitance in Farads, V = voltage in volts).

Plugging this into ½mv², where m is the projectile mass, 0.5kg, and v is the velocity in meters/second, gives v ≈ 23 ms^-1.

<tap, tap, press equals...>....er, the weight of a crow doing 51mph. Not enough to crack a car windscreen.

Prolly would if it was a frozen crow .

The impressive high (supersonic) demonstrations use high capacitors, high voltages and (tens) of grams of bullets.

Hypersonic results stems from the same light bullet accelerated by a multistage ruptured disk pneumatic gun.

Other than that DG is correct.

Pack it up, Del.

I've got a headache now

I agree with Lyn ( by default ). Your mathematics need some polish. The larger question here is are you simply trying to avoid building a firearm? I think the authorities might conclude this too is a firearm. Of course, they might consider an atlatl to be a firearm.

Do you assume constant acceleration? I have my doubts about this in any practical device, since this assumes a constant dynamic impulse through the extent (length) of the rails. Since your accelerations include should include the integral of the impulse function, combined with the integral of the drag equation, changes in electrical characteristics as the transient is propagated, etc., I do not see you or anyone presenting a closed form solution to this any time soon, other than crude approximation. Build it and they will come.

Sounds pretty cool. Might you be building this to enter in the annual Pumpkin-Chunkin contest?

pumpkins weigh more than 0.5 Kg, nes pas?

nes pas n'est-ce pas

I really should have studied my French lessons. Too busy reading Playboy, I suppose.

I dont read that, just look at he pictures

yeah, I was trying to hard to be PC.

Yes, you are right! And the pumpkins used in the national contest have much thicker walls than a standard pumpkin as they are a well protected variety. Specially grown just for the event. Collapse is one of the biggest problems of accelerating a pumpkin to the velocity required to win the contest.

My point is that a big rail gun has not been used as yet. If I had the time and money, I'd love the challenge, but I would start with something like a standard 0.5Kg and move up from there. Understanding the dynamics is essential. When you have to make adjustments because all pumpkins are not the same weight, it throws in variables that have to be adjusted, but which ones? Can it be done electronically or do things have to be moved?

Just wandering outside the box a little bit....

I believe they call a collapse a "pie"

I think the big secret to optimum punkin' chunkin' is a machine that would slowly add force throughout the range of acceleration to release. High impulse = pie. Steady impulse, i.e. - linear increase in force (and acceleration), or something low linear increasing impulse, (parabolic increase in acceleration), should produce a more useful result. I tend to think of a rail gun as something with pretty high inital impulse.

It is complicated:

no input (energy) = no output (energy)

Duh.