That seems like a ridiculously huge number. What sort of weapon could hope to penetrate it? If your capital ships can carry weapons with that kind of firepower, I imagine your infantry would have fantastically advanced weapons that coud level large buildings in a singe shot.
Enemy ships with equally or more powerful weapons. Uh...no. CGU infantry use powerful weapons, to be sure, but nothing nearly that powerful. Infantry weapons like that would not be effective at all, and downright dangerous to yourself and your comrades, especially in urban combat. And really, ship weapons like that aren't incredibly powerful in the big scale of things. SW has bigger and crazier shit. Also, remember, that number was for ALL of a Dreadnought's weapons being fired at the same time. It's not ridiculous, just big.
Do your capital ships use kinetic weapons like railguns? Just so that we're clear on the amount of energy you're talkinging about: a 5508 gigaton shield would be able to absorb 2.3*10^25 joules. That would require the impact of a 511 million kilogram mass traveling at the speed of light to penetrate. A 511 million kg block of lead would occupy a volume of 48800 cubic meters. Or perhaps you use energy beams - maybe powered by antimatter, or some other exotic technology that can convert mass directly into energy. In that case a 5508 GT shot would "only" burn through 250 million kilograms of matter/antimatter, or whatever other mass fuel you are using for energy.
When you do the conversion, mass is measured in kilograms, not grams. So it turns out that 5508 GT of energy is equivalent to about 256000 kg. Assuming 100% efficiency, you'd need to fuse 52000000 kg of deuterium to get that same amount of energy. Assuming better than triple-point storage densities, that's over 273600 m^3 of deuterium. If you're using deuterium/tritium fusion, you need even more mass, and more volume. (Yes, I have a spreadsheet for these things. Interesting stuff, really. Assuming I've set it up correctly.)
Err, if it's travelling at C then ANY mass will be infinite won't it? Launch a feather, at least the energy required for the initial acceleration (towards, not TO) C will be less...
I believe Hapsburg meant it to be in terms of 'tons of TNT', the same way they describe yields for nuclear weapons. In Einstein's mass-energy equivalence formula (that nifty E-mc^2 thing, which as a layman I don't pretend to understand where it comes from) the mass is assumed to be the object's 'rest mass'. If you were doing the equally nifty thing for kinetic energy (KE=1/2mv^2), then you'd need to take into account the affects of relativistic speeds.
Agreed, but I read the "railgun" bit and just ignored the rest.... Please Register or Log in to view the hidden image!
Proton cannons, powered by a tritium fusion reactors. I've been through this, like four times, man. Besides, antimatter's for trekkies. Fusion's is for real men. Not at all, really. Compared to SW stuff, nah. And that's kinda how we thought of this, "what could kick star wars' ass?"
This does not seem feasible. A singe deuterium/tritium fusion releases 2.8*10^-12 joules. In order to release 5508 GT you would need to fuse over 68 million kilograms of deuterium and tritium, which if stored in liquid form would occupy a volume of something like 200000 cubic meters. You would also have to fire about 500000 kg of protons. If your proton cannons fire 100 GT blasts, you would need to fuse over 1.2 million kg of deuterium and tritium per shot, assuming you can actually use all 100% of the energy released in the fusion to power the shot. This means that you will have to fuse about 4000 cubic meters of fuel per shot and you would need to eject about nine metric tons of protons. Starwars has technology that can magically create an almost infinite amount of energy from nothing. They don't even try to come up with an explanation for where the energy comes from. If you want to actually use real scientific ideas to back your stories, starwars is a terrible yardstick to use.
You might want to think about it this way; a deuterium/tritium fusion fuel has an energy density of about 80 kilotons/kilogram. Maybe you could figure out how many kilos of fuel you would want to burn through per shot, then multiply that by 80 kilotons to get your energies.
