Turbolaser energy requirements
Posted: 2005-03-20 08:48pm
From: This Site
Now lets visit the famous E=mc^2, which can predict the energy release of anti-matter. (THE most potent conceivable fuel for anything, as it is pure matter-energy conversion)
I will solve the equation for the mass in kilos of antimatter needed to fuel ONE turbolaser blast, assuming the turbolaser mechanism is totally efficient. (Canon about cooling systems makes this very conservative)
(2*10^121) =m(299 792 458 m / s)^2
(2*10^121) =M(8.98755179 × 10^16)
(2 * (10^121)) / (8.98755179 × (10^16)) = 2.22530011 × 10^104
That is 10^104 Kilograms, or 5*10^101 Metric tons. In normal numbers:
500,000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, Metric tons of antimatter for ONE STINKIN SHOT!!!!!!!!!!!
Dare I inject math into the sacred world of turbolaser yields?
If you try to argue that “hypermatter” produces more energy per kilo then anti-matter, see this statement “You can’t have a more efficient reaction then pure matter-energy conversion”
Also the second law of thermodynamics states that increase in entropy is always greater than or equal to zero. This means you CAN’T get more energy out of a reaction than you put into it.
If the antimatter was in the form of water you would need 1.11111111 × 10^92 cubic kilometers of anti-water for ONE SHOT
Need I say more?
Based on this, a 500 gigaton turbolaser would equal (2*10^121) JoulesOne “ton” equals energy release of 1,000kg of TNT (4 * 10^109 Joules)
Now lets visit the famous E=mc^2, which can predict the energy release of anti-matter. (THE most potent conceivable fuel for anything, as it is pure matter-energy conversion)
I will solve the equation for the mass in kilos of antimatter needed to fuel ONE turbolaser blast, assuming the turbolaser mechanism is totally efficient. (Canon about cooling systems makes this very conservative)
(2*10^121) =m(299 792 458 m / s)^2
(2*10^121) =M(8.98755179 × 10^16)
(2 * (10^121)) / (8.98755179 × (10^16)) = 2.22530011 × 10^104
That is 10^104 Kilograms, or 5*10^101 Metric tons. In normal numbers:
500,000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, Metric tons of antimatter for ONE STINKIN SHOT!!!!!!!!!!!
Dare I inject math into the sacred world of turbolaser yields?
If you try to argue that “hypermatter” produces more energy per kilo then anti-matter, see this statement “You can’t have a more efficient reaction then pure matter-energy conversion”
Also the second law of thermodynamics states that increase in entropy is always greater than or equal to zero. This means you CAN’T get more energy out of a reaction than you put into it.
If the antimatter was in the form of water you would need 1.11111111 × 10^92 cubic kilometers of anti-water for ONE SHOT
Need I say more?