A-08

Appendix

A Preliminary Investigation into
An Evolutionary Boundary

Finding the population of 3-A at the end of the first "year"

In "year one", 2-A mutants come from 1-A.
In each
saturation cycle, 1-A has 10e29 divisions.
One in 1.82*10e12 of these divisions will produce 2-A.
5.49450549*10e16  2-A will be produced per
saturation cycle.
One in 1.82*10e12 is also the number of 3-A generated by 2-A organisms

rem finds populations of 2-A and 3-A at the end of "year one"
countit = 0
original = 5.49450549*10^16
rem          5.49450549*10^16 per saturation cycle (2-A)
A3 = 0                   
population =  0
[start]
population = population + original
population = population * 1.01
A3 = A3 + (population/(1.82 * 10^12))   rem A3 means 3-A
A3 = A3 * 1.02         
countit = countit + 1                           
IF countit = 1000 then [end]                 
GOTO [start]     
[end]
print population     rem  meaning 2-A
print A3                 rem  meaning 3-A
end

answer:
1.16306248*10e23      2-A at end of "year one"
6.31648463*10e16     
3-A at end of "year one"  or about 6.3*10e16

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