ORIGINAL: BigJ62
ORIGINAL: witpqs
Well, since I'm getting 2 each that 2 + 2 = 4 (what a cliche!), so maybe the routine takes total # of the plane type into account?
It's almost like the routine thinks there is a fighter group onboard(.6) and whatever is left over is divided between the bomber groups .2 per group out of a truncated 10 or something.
This is exactly correct. How foolish of me to not have realized it. The routine does indeed consider the number of squadrons of fighter/non-fighter in it's calculation. Anything else would tend to over-stuff the carrier.
Once two non-fighter squadrons are detected the calculation is indeed: capacity * .4/2. And, since integers are used, all results will truncate.
If there are two groups, 60% of capacity is reserved for fighters (on British carriers). The remaining 40% is split between the two squadrons detected: (12 * .4)/2 = 2.
Returning Hermes to a single Swordfish squadron seems a very good idea.
And, in anticipation of the next question: no.