Planetary Industry: Difference between revisions

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* The CPU cost of a link of distance l is : c(l) = 15+0.2×l

We can then express the links on a planet as ''n'' the number of started links and ''k'' the ratio of total length compared to minimum length, <math>k=\sum_{link} link.length/ml </math>

We can then express the links on a planet as ''n'' the number of started links and ''k'' the ratio of total length compared to minimum length, <math>k=\sum_{link} link.length/ml </math>. Basically, k is the number of minimal size segments we use

* if the infrastructures are only a launchpad and 6 high-tech production plants (htpp), with the launchpad in the middle, we have 6 links and total 6 minimal segments

* if we add 12 htpp around the 6 existing, half of them will require 2 segments (those just after another htpp) and the other half will require <math>\sqrt 3</math> segments. Therefore the total number of links is 18 , the total length is <math>6×1+6×2+6×\sqrt 3 = 28.4</math> minimal segments.

@@ Line 209: / Line 209: @@
 * The CPU cost of a link of distance l is : c(l) = 15+0.2×l
-We can then express the links on a planet as ''n'' the number of started links and ''k'' the ratio of total length compared to minimum length, <math>k=\sum_{link} link.length/ml </math>
+We can then express the links on a planet as ''n'' the number of started links and ''k'' the ratio of total length compared to minimum length, <math>k=\sum_{link} link.length/ml </math>. Basically, k is the number of minimal size segments we use
 * if the infrastructures are only a launchpad and 6 high-tech production plants (htpp), with the launchpad in the middle, we have 6 links and total 6 minimal segments
 * if we add 12 htpp around the 6 existing, half of them will require 2 segments (those just after another htpp) and the other half will require <math>\sqrt 3</math> segments. Therefore the total number of links is 18 , the total length is <math>6×1+6×2+6×\sqrt 3 = 28.4</math> minimal segments.