Science, Technology

Railgun notes

Force on the projectile of a railgun:
F = B·I·l
B: Magnetic field
I: Current
l: Length of armature

Current = Voltage / Resistance

Resistivity of seawater:
ρ = 2.00×10^−1 (Ω·m) (because = (Ω/m-length)*(cross-sectional area))

Let cavity be 1cm square, consider section 1cm long:

Volume: 1 millilitre
mass (m): ~1 gram = 1e-3 kg
Cross-section: 1e-4 m^2
Armature length (l): 1e-2 m
Resistance: ((2.00×10^−1 Ω·m)*0.1m)/(0.01m^2) = 20 Ω
∴ current (I) = Voltage (V) / 20 Ω

Rare earth magnets can be 1 tesla without much difficulty. Assume that here.

F = 1 T · (V/20 Ω) · (1e-2 m)

Target velocity: 11.2 km/s = Escape velocity = 11200 m/s
v = at = 11200 m/s
∴ a = (11200 m/s) / t
s = 1/2 · a · t^2
∴ s = 1/2 · ( (11200 m/s) / t ) · t^2
= 1/2 · (11200 m/s) · t
or: t = s / (1/2 · (11200 m/s))
F = ma = (1e-3 kg) · a
∴ a = F / (1e-3 kg)
∴ t = (11200 m/s) / (F / (1e-3 kg))
= (11200 m/s) · (1e-3 kg) / F
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / F
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (V/20 Ω) · (1e-2 m) )

Say V = 250 volts:
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (250V/20 Ω) · (1e-2 m) ) = ~501760 meters

Say V = 25,000 volts:
∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (25000V/20 Ω) · (1e-2 m) ) = ~5017.6 meters

Liquid mercury instead of seawater:
Resistivity: 961 nΩ·m = 0.961e-6 Ω·m
Resistance: 9.6e-6 Ω
Density: 13.56 times water
F = 1 T · (V/9.6e-6 Ω) · (1e-2 m)
s = 1/2 · (11200 m/s) · (11200 m/s) · (13.56e-3 kg) / ( 1 T · (V/9.6e-6 Ω) · (1e-2 m) )
@250 volts: s = 3.266 meters
@25kV: s = 32.66 millimeters

Power (DC): P = IV where I = V/R,
R = 9.6e-6 Ω
@250 volts: I = 250 / R = 250 V / 9.6e-6 Ω = 2.604e7 amperes
∴ P = 6.51 gigawatts
@25kV: I = 25000 / R = 25000 / 9.6e-6 Ω = 2.604e9 amperes
∴ P = 65.1 terawatts

Duration between rails:
From t = s / (1/2 · (11200 m/s))
@250 volts:
t = 3.266 meters / (1/2 · (11200 m/s)) = 5.8321×10^-4 seconds
@25kV:
t = 32.66 millimeters / (1/2 · (11200 m/s)) = 5.8321×10^-6 seconds

Electrical energy usage:
E = P · t
@250 volts:
E = 6.51 gigawatts · 5.8321×10^-4 seconds = 3.797×10^6 joules
@25kV:
E = 65.1 terawatts · 5.8321×10^-6 seconds = 3.797×10^8 joules
(For reference, 1 litre of aviation turbine fuel is around 3.5e7 joules)

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