X-wing torpedo yield
Posted: 2004-01-18 03:04pm
Cross posted from SW.com
Ok, since I already knew most of this from my work in the Navy, this only took me and google ~ 5 minutes.
I eyeball the proton torpedo at 50 cms tall and 50 cms wide at the base. Volume of a Cone is 1/3 height * area of base IIRC.
Now taking some assumptions from modern missiles: Fuel is 25% - 75% of the thing. Lets go with 75% to get a low end warhead yield. Now according to Star by Star, the things have ion engines on them. That means its going to use some kind of noble gas for fuel. Lets go with liquid argon: its inert, easy to store, etc. It has a density of 1.5x that of water. Lets assume the rest of the warhead has a density of water (~ what real missles have) That gives it a fuel mass of 36.8 kg and a total mass of ~45 kg.
Now according to I, Jedi proton torps have a flight time of 30 seconds, so it will eject ~1.2 kg per second.
According to here : http://www.stardestroyer.net/Empire/Tec ... pedo2.html they have an acceleration of 72,000 Gs. Since everyone knows Momentum = mass * velocity, and everyone knows the laws of motion, from that we can figure the exit velocit of the exhaust and its specific impulse (not relevent to the yield, but at 2,640,000 thats very impressive).
Now if we say that the sensors, power cell, casing, detonator, etc make up 20% of the remaining volume (not out of line, with the required microcircuirty and solid state electronics we see and the high energy density of power packs it is very reasonable) that gives us 5% for the fuel for the warhead itself. If we assume it it is thermonuclear in nature as indicated by the ROTJ novel, and assume only 50% efficiency (though modern warheads are more efficient, but we want low end), we can estimate its yeild.
Hydrogen fusion is typically what is meant, assuem the density of water, 5% of the total volume, and you get a mass of ~1.64 kg. Fusion releases 6.2*10^14 joules per kg. so multiply .5 * 6.2*10^14*1.6 = 5*10^14
or
120 kilotons
That is ~7.7 times greater then Hiroshima. And note that this si a focused explosion (so all the energy goes into the target) and the near miss barely scortched the paint on the Death Star. That is very tough armor. Now like I stressed, that is both low end and approximate and hinges on a number of assumptions. I'd feel better if I knew the volume:yield ratio for modern nukes and coud extrapolate from there, but what the hey.
Ok, since I already knew most of this from my work in the Navy, this only took me and google ~ 5 minutes.
I eyeball the proton torpedo at 50 cms tall and 50 cms wide at the base. Volume of a Cone is 1/3 height * area of base IIRC.
Now taking some assumptions from modern missiles: Fuel is 25% - 75% of the thing. Lets go with 75% to get a low end warhead yield. Now according to Star by Star, the things have ion engines on them. That means its going to use some kind of noble gas for fuel. Lets go with liquid argon: its inert, easy to store, etc. It has a density of 1.5x that of water. Lets assume the rest of the warhead has a density of water (~ what real missles have) That gives it a fuel mass of 36.8 kg and a total mass of ~45 kg.
Now according to I, Jedi proton torps have a flight time of 30 seconds, so it will eject ~1.2 kg per second.
According to here : http://www.stardestroyer.net/Empire/Tec ... pedo2.html they have an acceleration of 72,000 Gs. Since everyone knows Momentum = mass * velocity, and everyone knows the laws of motion, from that we can figure the exit velocit of the exhaust and its specific impulse (not relevent to the yield, but at 2,640,000 thats very impressive).
Now if we say that the sensors, power cell, casing, detonator, etc make up 20% of the remaining volume (not out of line, with the required microcircuirty and solid state electronics we see and the high energy density of power packs it is very reasonable) that gives us 5% for the fuel for the warhead itself. If we assume it it is thermonuclear in nature as indicated by the ROTJ novel, and assume only 50% efficiency (though modern warheads are more efficient, but we want low end), we can estimate its yeild.
Hydrogen fusion is typically what is meant, assuem the density of water, 5% of the total volume, and you get a mass of ~1.64 kg. Fusion releases 6.2*10^14 joules per kg. so multiply .5 * 6.2*10^14*1.6 = 5*10^14
or
120 kilotons
That is ~7.7 times greater then Hiroshima. And note that this si a focused explosion (so all the energy goes into the target) and the near miss barely scortched the paint on the Death Star. That is very tough armor. Now like I stressed, that is both low end and approximate and hinges on a number of assumptions. I'd feel better if I knew the volume:yield ratio for modern nukes and coud extrapolate from there, but what the hey.