RP General Discussion

We meet again, math, my mortal enemy. Put up your dukes, we shall settle this like gentlemen.

Originally posted by S121
@everyone in EG

I was playing Fallout 3 and decided to create Ethan. Well, if you know about Fallout 3, then you know about the G.O.A.T., the aptitude test that tells the vault dwellers what their career will be. Well, I answered the test 3xactly how Ethan would, and I got "marriage councilor". XD

Just imagine this wild, badass merc sitting down, telling couples how to "keep it fresh". XD

LOL! I imagine Hajime and Eve will be needing a good marriage councilor even though they aren't married (they would probably be bickering and acting like a married couple lol.)

@Mima

Math, demons to some, angels to others. :p

I don't really mind math that much, as long as it doesn't become too repetitive.

@Sus

Yep, I've thought it over, and Ethan will actually have that experience from his previous job. That should be hilarious as a conversation topic, or if he suddenly shows those counciling skills off. XD

@Cucoo

Just another question, do the magnetic properties or electrical conductivity of the metal used in the railgun's construction affect the power needed?

@ S1

The resistance of the wire would affect things. I went and calculated things using a bunch of assumptions.

Also...
What is the unit J supposed to be?
Joules?
If so...

Barrel Length (x) = 28m

distance between rails (d) = 9m

Work (w) = 5E11J

Force (F) = w/x = 1.79E10N

F=Force Magnetic Field (F_B)
F_B = idBsin(90)= idB

two unknowns: current (i) and Magnetic Field (B)

B = 2((u_oi)/(2pi(.5d))) For left and right rail

solve for i

sqrt(F_Bpi/(2u_o))=i

i=1.49E8A

V=i*R

Let's assume there is some resistance and let's say it's 1 ohm because I'm lazy

V=i

V=1.49E8V


Muzzle Velocity (Vf) = 1E6 m/s

Force (F) = w/x = 1.79E10N

Barrel Length (x) = 28m

F=m*a

mass (m) = 1 kg

a = F

a = 1.79E10 m/s²

Vf=vo*t+a*t

solve for t

vo=0

Vf/a=t

t = 5.6E-5s

So I calculated:
Voltage needed to fire

Time needed to fire for lightest projectile


Now voltage for other bullets:
For your 10kg
V = 4.72E8V

For 100kg
V= 1.49E9V (From the 1kg, increased by factor 10)

For 1000kg
V= 4.72E9V (from 10kg, factor of 10)

seems to alternate between the two, increasing by factors of 10.

If you wanted Power in watts, assuming resistance as 1 ohm, just square voltage.

@Cucoo

Well, now I know that the railgun will take a long while to charge between shots. XD

Does t increase by a factor of ten each time as well?

No
Because work is increasing by 10 therefore force increases by 10 and mass of bullet increases by 10

The increase in force cancels the increase of mass therefore acceleration does not change and thus time does not change.

Your increases and velocity makes perfect use of the length of the rail gun.

Oh and your concept of vertical and horizontal modes make no difference cause d cancels out in the formulas.

Also if you do change the resistance of the system to something more realistic, V would increase if R increases and power in watts would be P=I*V = I^2*R

EDIT:

And...on the side
@ Anyone
could I ask for everyone or anyone to go through the timeline in EG and see if something raises any sort of question? I feel like I'm missing something important and it's bugging me...

@Cucoo

Sorry if I seem a little air-headed at the moment. Personal stuff. Susan can fill you in with the details.

Anyway, I'll leave the railgun as a non-transforming weapon.

So, what resistance do you think would be more realistic? To be honest, I have no real resistance in mind for the railgun. All I have down is that the weapon makes heavy use of a transforming element that can change its properties via programming.

Don't worry about it.

For resistance, that would depend on what material you use for the wires. tbh there's a formula to figure out power loss and such but for the sake of simplicity, anything greater than 1 ohm
...heck...maybe it is 1 ohm or even less...
...the resistance is what you can vary to change the voltage and thus the power. do with it as you will. Resistance directly affects voltage and thus power. So if you want to make it require more power, increase resistance. If for some reason you want it to use less, lower it. Of course the hazard with resistance is that too much can make things overheat, and too little can short out the system. So I would say just figure out what kinda power source you will be using, how much power in watts per time interval it can pump out and work from there to change the resistance.


...and I have to ask...
...who's Susan?

@Cucoo

*sees Susan*

Goddammit! This phone had an auto correct feature, so when I typed "Sus" it corrected it. XD *facepalm*

Anyways, I already found a candidate for a power source; antimatter. I made a very rough estimation, and with two reactors, the ship can pump out about 600 billion watts per second. Granted, not all the power would be devoted to the railgun, but even then, it would take a looooooooong time for the gun to charge.

I think I should increase the amount of power the ship can generate.

Another alternative was a black hole generator/reactor, and it would generate around 129 pettawatts. (didn't say timeframe, but I assume in seconds), but the black hole itself would be kilometers wide, much larger than the ship.

Ah autocorrect

Something told me...

Anyway...

...uhm...
...a black hole...generator... as a power supply.
I'm trying to wrap my head around that one...
Don't black holes normally consume energy?
Part of me thinks that's a bit of a stretch
...just like...
Antimatter used as a power source?

I'm almost afraid to ask how that even would work in theory...