Energy required to vaporize a space pirateIt’s a fact that space pirates are taller than Samus, so the question is just how much mass does a typical space pirate have? To find out, I decided I would use Hamwi’s method. [1] Devine’s formula made no sense, as a typical space pirate would be a little over a tonne. For males, the Hamwi formula is 106 lbs. (48.08 kg.) and then an additional 6 lbs. per inch (2.72 kg. per 2.54 cm.). This means that if a typical space pirate stands at 7’6″ (2.286 m.), then 30 in. is an additional height over 5′. This means that a typical space pirate should weigh 286 lbs. (129.727 kg.). That looks like a more reasonable weight.

Based on scan images, we can see space pirates have bones, [2] so what we’re working with here is similar to that of finding out just how much energy is required to atomize a human being. Space pirates appear to be unaffected by high temperatures, as we can observe flying pirates in Metroid Prime standing in Magmoor Caverns next to lava. Samus’ powered armor cannot withstand this temperature, so we should at least expect more energy than 2.99 gigajoules to atomize a space pirate. I will be using the calculation required to find out how to vaporize a human, but apply it to a typical space pirate for the sake of simplicity. After all, organisms are carbon-based and require water to live.

The average density of a human is 985 kg/m^3. [3] A typical space pirate has a mass of 129.727 kg., so 129.727 kg. / 985 kg/m^3 is 0.131702538071066 m^3. This is approximately twice the volume of a human. The composition of a human is 65% oxygen and 9.5% hydrogen. [4] Since water contains two hydrogen atoms and one oxygen atom, adding these together gives us a total of 84%. This means the human body is made up of that much water. I will apply this to space pirates as well. 129.727 kg. multiplied by 84% means that a typical space pirate’s body is made up of 108.97068 kg. of water. Carbon makes up 18.5%, so 129.727 kg. multiplied by 18.5% is 23.999495 kg.

The heat of vaporization of water is 2.257 * 10^6 J/kg. This multiplied by 108.97068 kg. of water gives us a total energy of 245,946,824.76 J. The heat of vaporization of carbon is 2.9624 * 10^7 J/kg. This multiplied by 23.999495 kg. gives us 710,961,039.88 J. Adding these values together gives us the total energy to vaporize a typical space pirate, which would be 956,907,864.64 J. Now that we have this information, we can find out how much energy is required to atomize a space pirate. I expect it to be greater than 2.99 GJ, simply because of the size difference.

We can now use the volume of a typical space pirate, which is 0.131702538071066 m^3. This will be multiplied by 84%, which will give us 0.11063013197969544 m^3. This will be multiplied by the atomization value of water, which is 5.138416 * 10^10 J/m^3. This is 5,684,636,402.47 J. Next is carbon. The volume of a typical space pirate will be multiplied by 18.5%, which will give us 0.02436496954314721 m^3. The atomization value of carbon is 1.565 * 10^11 J/m^3. This multiplied by the volume gives us 3,813,117,733.5 J. Adding these two values of atomization gives us a total of 9,497,754,135.97 J. This is equal to 2.27 tons of TNT.

With these all in mind, this should give us an idea of the amount of energy that also comes from the plasma beam in Metroid Prime and Metroid Prime 3: Corruption, as well as the light beam in Metroid Prime 2: Echoes. Based on the color temperature of the light beam, it may very well produce more energy than the plasma beams, since the hotter an object is, the more energy it’ll produce.

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