3 ms·

I can't imagine the amount of electricity (even just for a split second) it would take to pull this off.

zafka 13y ago
In the ideal case, the math is pretty simple. The kinetic energy in 1/2MV^2 . (integral from 0 to mach 7) so if you know the weight of the missile...
- ojii 13y ago
  First sentence of the article: "The U.S. Navy is planning sea trials for a weapon that can fire a low-cost, 23-pound (10-kg) projectile"
- lifeisstillgood 13y ago
  weapon that can fire a low-cost, 23-pound (10-kg) projectile at seven times the speed of sound 340m/s ~ speed of sound so thats 10kg at 2380 m/s (wow!) 523802380 Joules = 28322000 J or 28MJ or 28MW if it needs a whole second of power or well the output of ~1/10 second of a mid sized nuclear power station Yes it needs a lot of power
- jwmerrill 13y ago
  Just the difference between 0 and Mach 7, not the integral.
HCIdivision17 13y ago
If I recall, one of the huge challenges was handling just that. Specifically, I remember something written on it a few years ago that the device was functional, but due to the overwhelming amperage, each shot fused the system up. And now I'll believe it, given that the video shows the casing partially vaporizing as the projectile is exiting. There's something incredibly shocking about something exploding without traditional explosive chemistry. Just a monumental arc flash (or something - I don't know if it's the same mechanism).
jzwinck 13y ago
You don't need to imagine, we can compute it! Mach 7 is 2382 m/s. 23 pounds is 10.4 kg. Since k=mv^2: (1/2) * (10.4 kg) * (2382 m/s)^2 = 29504404 Joules. That's 29.5 MJ, which is 29.5 megawatt-seconds. So it could mean 29.5 megawatts for one second, except of course the rail time is much less than one second. But who cares, let's convert to household units! 29500 kilowatt-seconds is 8.2 kWh. At an average price of US$0.13/kWh, that comes to just $1 worth of electricity to fire a $25000 projectile. Oh, yeah, the losses. Those are probably 90% or so, so let's say $10. And mark it up a factor of ten because the electricity is on a boat, so $100, and a factor of 50 because it's a weapons system. So ballpark $5000 in electricity TCO to fire the $25000 projectile.
bluedino 13y ago
The Popular Mechanics article from a few weeks back said 6 million amps per shot http://www.popularmechanics.com/technology/military/4231461 http://www.popularmechanics.com/technology/military/4231461

zafka 13y ago

In the ideal case, the math is pretty simple. The kinetic energy in 1/2MV^2 . (integral from 0 to mach 7) so if you know the weight of the missile...

ojii 13y ago
First sentence of the article: "The U.S. Navy is planning sea trials for a weapon that can fire a low-cost, 23-pound (10-kg) projectile"
lifeisstillgood 13y ago
weapon that can fire a low-cost, 23-pound (10-kg) projectile at seven times the speed of sound 340m/s ~ speed of sound so thats 10kg at 2380 m/s (wow!) 523802380 Joules = 28322000 J or 28MJ or 28MW if it needs a whole second of power or well the output of ~1/10 second of a mid sized nuclear power station Yes it needs a lot of power
jwmerrill 13y ago
Just the difference between 0 and Mach 7, not the integral.

HCIdivision17 13y ago

If I recall, one of the huge challenges was handling just that. Specifically, I remember something written on it a few years ago that the device was functional, but due to the overwhelming amperage, each shot fused the system up. And now I'll believe it, given that the video shows the casing partially vaporizing as the projectile is exiting. There's something incredibly shocking about something exploding without traditional explosive chemistry. Just a monumental arc flash (or something - I don't know if it's the same mechanism).

jzwinck 13y ago

You don't need to imagine, we can compute it! Mach 7 is 2382 m/s. 23 pounds is 10.4 kg. Since k=mv^2: (1/2) * (10.4 kg) * (2382 m/s)^2 = 29504404 Joules. That's 29.5 MJ, which is 29.5 megawatt-seconds. So it could mean 29.5 megawatts for one second, except of course the rail time is much less than one second. But who cares, let's convert to household units! 29500 kilowatt-seconds is 8.2 kWh. At an average price of US$0.13/kWh, that comes to just $1 worth of electricity to fire a $25000 projectile. Oh, yeah, the losses. Those are probably 90% or so, so let's say $10. And mark it up a factor of ten because the electricity is on a boat, so $100, and a factor of 50 because it's a weapons system. So ballpark $5000 in electricity TCO to fire the $25000 projectile.

bluedino 13y ago

The Popular Mechanics article from a few weeks back said 6 million amps per shot http://www.popularmechanics.com/technology/military/4231461 http://www.popularmechanics.com/technology/military/4231461