[In which Martha is a bit of a math geek and tries to put some science in her fiction…]

So we know the war will arrive in forty years and that the war is now one hundred years wide.

First off, it should be safe to assume that the hundred years are not a solid mass of ships, just that ships will keep arriving during that time period. So you’ll have months (or years) between attacks as each wave hits. Which means I need to sit down and map out on what years ships arrive and what team they are on (assuming that since Side A is running away from Side B that the only time they overlap is when B overtakes A). So off to Excel and some randomizing! ^_^

– assume that the following chances of a ship in range of earth is: years 1-25 35% A, 25-50 80%, 50-75 80%, 75-100 35% (war by bell curve, gotta love it! *grin*)
– assume that the following ratio of B to A are true: years 1-25 21/79 A, 25-50 43/57, 50-75 65/35, 75-100 82/18
– assume a grouping of ships is not probable given the exponential nature of the war
– HOWEVER assume that in-progress ships for Side B are all planet-bound and that ships for Side A are dragged along beside the functioning ship.
– assume there is a 67% chance there is one in-progress ship, 13% chance that there are two ships, and a 20% chance there are no ships. [need to figure out actual exponential build requirements]
– assuming a 34 year build cycle with weapons maturity at 10 years, there is a 71% chance that the second ship is combat ready (ish)
– assuming that Side A will avoid planets it has already harvested, only visits from Side B will matter.

So for the first quarter century there is a 15% chance that a ship will arrive during that year, a 21% chance it will be an enemy ship, a 80% chance that the enemy ship will have at least one ship in tow, and a 71% chance that the ship will make a difference in combat.

Using =RANDBETWEEN(1,100) in Excel the following data set occurs:

Year 5: Side B
Year 8: Side B with combat-ready sidekick
Year 25: Side B with sidekick
Year 26: Side B
Year 27: Side B
Year 28: Side B with sidekick
Year 29: Side B with sidekick
Year 31: Side B with combat-ready sidekick
Year 32: Side B
Year 41: Side B with combat-ready sidekick
Year 44: Side B with combat-ready sidekick
Year 47: Side B with combat-ready sidekick
Year 48: Side B with combat-ready sidekick
Year 49: Side B with sidekick
Year 50: Side B
Year 52: Side B with combat-ready sidekick
Year 53: Side B with sidekick
Year 59: Side B with sidekick
Year 61: Side B with combat-ready sidekick
Year 68: Side B with combat-ready sidekick
Year 69: Side B with combat-ready sidekick
Year 71: Side B with combat-ready sidekick
Year 76: Side B with combat-ready sidekick
Year 77: Side B with combat-ready sidekick
Year 79: Side B with combat-ready sidekick
Year 81: Side B with combat-ready sidekick
Year 85: Side B with combat-ready sidekick
Year 86: Side B with combat-ready sidekick
Year 87: Side B
Year 90: Side B with combat-ready sidekick
Year 100: Side B with combat-ready sidekick

Of course this means that thirty-one ships from Side B will pass close enough to Sol to notice it. Which sort of brings up the point that either the ships beat on Earth for a bit and then move on, or that Earth is destroying them. Otherwise you’d end up with thirty-one ships in orbit. If it took them almost forty years to build one ship they certainly aren’t going to be able to hold off thirty at once!

Hmmm, I really need to work out the actual ship volume through that given system. Thirty-one seems a mite high, although depending on velocity of the expansion it may be low. *ponders*

Assuming Moore’s Law applies to the production times of the ships (they are AIs after all ^_~), we can work backwards to a theoretical start date of the war.

So let’s assume the Ship gave them plans for building two ships. The buildtime required at the start is five thousand years (yikes!). This is an achievable goal, working from the assumption that Earth would be able to complete 20% of the remaining work each year due to advances in technology. Thus the millennial ship quickly becomes a forty-year ship. But we, being clever monkeys, complete 30% a year and manage to polish the ships off fourteen years early. Which means at that point, err– *math* that it would take them three years or so to build another one. (Now of course this learning curve is flipped production-wise so that 30% of the ship is completed in the last year, not the first. ^_~)

Now (as I mentioned above) it takes Side B thirty-four years to build a new ship. This is because they build to the rate of growth, not beyond it. [Which means this number may change!] They build just enough ships so that there is one B ship for every A ship, and A no longer builds ships itself (relying on the planets’ one-for-me-one-for-you builds instead). Since the non-planet-bound A ships avoids battle as much as possible, very few of them are destroyed. Thus their growth rate depends on the number of planets found that are adaptable/advanced enough to be harvested.

Now how many planets has B hit? Well, using the Drake equation (and ignoring the Fermi Paradox, because this is Science Fiction goldarnit), and assuming the following:

R* = 10/year (10 stars formed per year, on the average over the life of the galaxy)
fp = 0.5 (half of all stars formed will have planets)
ne = 2 (2 planets per star will be able to develop life)
fl = 1 (100% of the planets will develop life)
fi = 0.01 0.05 (12% of which will be intelligent life) ((what can I say, I’m an optimist! ^_~)
fc = 0.01 0.12 (12% of which will be able to communicate harvestable)
L = 10,000 years (which will last 10,000 years)

N = 10 × 0.5 × 2 × 1 × 0.05 × 0.12 × 10,000 = 600

Which is still low. Hmm… *digs about a bit* Unless I want to bump up the harvestable percentage the only other one to really play with is L. Unless I want to be silly and say fi is something like 50%. Heh. ^_~ I suppose it depends on how much I want to push the issue. *ponders*