[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**.