Science, Technology

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.

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