# Railgun notes #2

[Following previous railgun notes, which has been updated with corrections]

Force:
```F = B·I·l B = 1 tesla```

``` I: Current = Voltage / Resistance l: Length of armature in meters ```

```F = 1 tesla · V/R · l F = m · a ∴ a = (1 tesla · V/R · l) / m```

Using liquid mercury, let cavity be 1cm square, consider section 1cm long:
```∴ l = 0.01 m Resistivity: 961 nΩ·m ∴ Resistance R = ((961 nΩ·m)*0.01m)/(0.01m^2) = 9.6×10^-7 Ω Volume: 1 millilitre ∴ Mass m = ~13.56 gram = 1.356e-2 kg ∴ a = (1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg)```

Let target velocity = Escape velocity = 11200 m/s = 1.12e4 m/s:
```Railgun length s = 1/2 · a · t^2 And v = a · t ∴ t = v / a ∴ s = 1/2 · a · (v / a)^2 ∴ s = 1/2 · a · v^2 / a^2 ∴ s = 1/2 · v^2 / a ∴ s = 1/2 · ((1.12e4 m/s)^2) / ((1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg))```

@250V: `s = 0.3266 m` (matches previous result)

@1V: ```s = 81.65 m I = V/R = 1V / 9.6×10^-7 Ω = 1.042e6 A P = I · V = 1V · 1.042e6 A = 1.042e6 W```

Duration between rails:
```t = v / a ∴ t = (1.12e4 m/s) / a ∴ t = (1.12e4 m/s) / ( (1 tesla · V/(9.6×10^-7 Ω) · (0.01 m)) / (1.356e-2 kg) )```
(Different formula than before, but produces same values)
@1V: `t = 0.01458 seconds`

Electrical energy usage: E = P · t
@1V: `E = 1.042e6 W · 0.01458 seconds = 1.519e4 joules`

Kinetic energy: `E = 1/2 · m · v^2 = 8.505e5 joules`

Kinetic energy out shouldn’t exceed electrical energy used, so something has gone wrong.