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[QUOTE]Originally posted by TSN: [QB] And aside form that, your answer will be in joules/second, which is the same as watts. Watts/second would indicate that the rate of energy production is accelerating. And it doesn't matter how much deuterium there is, since there's more deuterium than antideuterium. You can only count the stuff that actually annihilates, which will be an equal amount of matter and antimatter. So, given the numbers we already have... 3000 m<sup>3</sup> of antimatter an equal mass of matter 5 years (157 784 760 seconds) of power generation [i]c[/i]<sup>2</sup> = 89 875 517 873 681 764 ...we find that the necessary equation is... 2 * 3000 * [i]D[/i] * 89 875 517 873 681 764 / 157 784 760 ...which reduces to approximately... 3 417 650 140 876 * [i]D[/i] ...where [i]D[/i] is the density of the stored antimatter (in kg/m<sup>3</sup>). [/QB][/QUOTE]
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