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)*0.01m)/(0.01m^2) = 0.2 Ω (got that wrong first time! Along with all that followed, which is now updated…)

∴ current (I) = Voltage (V) / 0.2 Ω

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

F = 1 T · (V/0.2 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/0.2 Ω) · (1e-2 m) )

Say V = 250 volts:

∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (250V/0.2 Ω) · (1e-2 m) ) = 5020m (not ~501760 meters)

Say V = 25,000 volts:

∴ s = 1/2 · (11200 m/s) · (11200 m/s) · (1e-3 kg) / ( 1 T · (25000V/0.2 Ω) · (1e-2 m) ) = 50.2m (not ~5017.6 meters)

Liquid mercury instead of seawater: Resistivity: 961 nΩ·m = 0.961e-6 Ω·m Resistance: 9.6e-7 Ω (got this one wrong the first time, too!) Density: 13.56 times water

F = 1 T · (V/9.6e-7 Ω) · (1e-2 m)

s = 1/2 · (11200 m/s) · (11200 m/s) · (13.56e-3 kg) / ( 1 T · (V/9.6e-7 Ω) · (1e-2 m) )

@250 volts: s = 0.3266 meters (not 3.266m as before correction) @25kV: s = 3.266 millimetres (not 32.66 millimetres as before) Power (DC): P = IV where I = V/R,

R = 9.6e-7 Ω

@250 volts: I = 250 / R = 250 V / 9.6e-7 Ω = 2.604e8 amperes (x10 more than before correction)

∴ P = 65.1 gigawatts (x10 than before)

@25kV: I = 25000 / R = 25000 V / 9.6e-7 Ω = 2.604e10 amperes (x10 more than before)

∴ P = 651 terawatts (x10 than before)

Duration between rails:

From t = s / (1/2 · (11200 m/s))

@250 volts:

t = 0.3266 meters / (1/2 · (11200 m/s)) = 5.8321×10^-5 seconds (x10 less than before correction)

@25kV:

t = 3.266 millimetres / (1/2 · (11200 m/s)) = 5.8321×10^-7 seconds (x10 less than before)

Electrical energy usage:

E = P · t

@250 volts:

E = 65.1 gigawatts · 5.8321×10^-5 seconds = 3.797×10^6 joules (unchanged by correction)

@25kV:

E = 651 terawatts · 5.8321×10^-7 seconds = 3.797×10^8 joules (unchanged by correction)

(For reference, 1 litre of aviation turbine fuel is around 3.5e7 joules)

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Original post: https://kitsunesoftware.wordpress.com/2017/04/10/railgun-notes/

Original post timestamp: Mon, 10 Apr 2017 18:16:20 +0000

Tags: math, physics, railgun

Categories: Science, Technology