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Figuring the first "year's" population of 3-H
5.10204082*10e14 2-C will be produced per saturation cycle.
One in 1.82*10e12 is the number of 3-H generated by 2-C organisms
countit = 0
original = 5.10204082*10^14
H3 = 0
population = 0
[start]
population = population + original
population = population * 1.01
H3 = H3 + (population/(1.82*10^12))
H3 = H3 * 1.01
countit = countit + 1
IF countit = 1000 then [end]
GOTO [start]
[end]
print H3
end
answer: 5.39428494*10e11 or about 5.4*10e11 3-H at end of "year one"
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