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)*0.01m)/(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)

Advertisements
Standard

One thought on “Railgun notes

  1. Pingback: Railgun notes #2 | Kitsune Software

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s