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Originally posted by RedFred
Now I really regret wiping the game from my PC. It would have been most interesting to continue to play after transcendance and allow him to build 15+ PBs to launch at me as an experiment.
Perhaps your onetime opponent still has a save of the game?
AS I see it, jt, that meant that it took two tries for the ODPs to shoot down the PB. First missed, second succeeded. Then, two ODPs are changed to 'fired' or 'used' status.
Still don't understand why there are two messages. Seems like each PB is getting two tries to penetrate the defenses.
I think I can answer that one jt. The game played out as I described in my last post. The two scenarios Googlie described were reloads.
Yes, conclusive proof that Mongoose's idea of how the PB/ODB wars works is correct.
I am not sure why the two text messages per PB though. In the first retrial the first message is '12 of 14 deployed' but I had 14 and no one else fired anything at me. So 14 should have been armed. I am not clear on why it doesn't move on to PB2 right away.
Well, Mong, it works like this. You launch your ODPs, and they get in orbit. Then you can move them around. Then someone fires a PB and then the nearest ODP fires. The ODP then plots an intercept course and moves off to the designated coordinates for a manual kill. At the same time, other ODPs (even of other factions) will fire also. The ODP net doesn't know if a given ODP will be sucessful in shooting down a buster so you can get all sorts of similtaneous fire.
The longer it takes to kill the buster, the more potential ODP launches. Depends on positioning. The manual kill is not automatic so the ODP net continues to react until the bogey is down.
It is very complicated Mong, you should stick to nerve gas.
Originally posted by Mongoose
Actually, it's good to know that the pbem system didn't gut the ODP's last ditch defense. Seems all that statistical work needs to be redone, eh?
Why? On the contrary, I often precised that the outcome is EITHER PBhit or ODP sacrifice, depending on how the player/game decided to react.
I think my analysis perfectly stands, so far, and as far as YOU are right (a I relied my work on YOUR assumptions).
Originally posted by Mongoose
Also looks like Firaxis chose MariOne's option: "- always sacrifice the ODP" but only after no ODPs remain undeployed. How rational! How remarkable!!!
Well, it HAS to be that way.
If ther are undeplyued ODP, this means that the PB has been shot down, and there's NO NEED for sacrifice.
If an ODP has to be sacrificed, this mean that you have none left to shoot at the PB, THAT IS, they have been all deployed.
You "praised" FurXs for a tautology
___
About the double messages, the only thing I can think to (APART that they have messed the reports, that is. I didn't say bug? Oh, ça va sans dire...) is that FurXs is giving the player a "halfway stat", just like in combat you get to see partial reports grouping a bung of rounds, not every sisngle shot.
But in such case, the first message should not say "shot down"...
It's not the two messages puzzling me indeed, it's that each PB has to be *shot down* TWICE.
Also when all the ODPs had been deployed, you see that for ONE PB (#5) TWO ODPs get sacrificed.
I wonder, could this be linked to PB's Reactor? Fusion R=2 requires maybe two shots to be destroyed?
THIS would definitely force to alter the statistical analysis parameters!
And would make ODP sacrifices much more likely, you'd have to regard them as *expendable* goods.
I think this deserves a self-hotseat scenarioed test.
Googlie, RedFred, you should share that game's passwords, so that you could perform a reload test until you can verify the sacrifice thing peeking in BOTH palyer's turns.
I don't exactly know what I mean by that, but I mean it (Holden Caulfield)
Originally posted by Chowlett
Indeed, it does involve factorials. Using MariOne's notation (which is non-standard, but works on BBs):
( n k ) = n!/[(k!)(n-k)!]
Thanks Chowlett!
Yeah RedFred that should be written something like
(n)
(k)
with single big vertical parentheses, not split in two as I am forced to write here.
We could also call it C(n;k), where C stands for Combinations.
It is C(n;k) = P(n;k)/D(k)
P are Permutations, and D are Dispositions
P(n;k)= n!/(n-k)!
D(k)= k!
The ABC of combinatorial analysys. (very useful for Poker! )
If you plot down the values of C(n;k) you get the Pascal triangle.
Which rows are used for instance to determine the coefficients for the expansion of (x+y)^N
C(3;5) (actually you should rather invert the arguments, or ignore the negative sign of n-k)
actually yields 10, not 6.
In our formula, with 3 PBs and 5 ODPs, we need to get 6p^5
6 is C(4;2), not C(5;3),
Thus, the formula must use (ODP-1) and (PB-1) in the Combinations formula.
C((5-1);(3-1))p^5 = C(4;2)p^5 = 6p^5
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PS: as I said all this has to be revised if each PB needs TWO shouts to be klilled
I don't exactly know what I mean by that, but I mean it (Holden Caulfield)
+50% defense against conventional missiles for any units up to two squares away. Also, if you have no ODPs, the system can shoot down a non-conventional missile 50% of the time when used inside a two-square radius of the base. Both effects are cumulative; i.e, if there are two Flechette Systems within a two-square radius of a missile's target, then a missile has only a 25% chance of success.
I presume that the no-ODP condition is determined at the time of combat resolution, and not at the beginning of the turn. Who knows?
Thanks.
Yeah, who knows?
And...
Can a Flechette in a base fire as many time as needed in one turn, or only once as the ODPs?
Datalinks talk about "defense...for any UNITS..." what about *empty* bases? And aren't tectonic or fungal missiles considered as non-conventional? In such case, can't they be aimed just at terrain (i.e. an empty square)? And how would Flechette trigger in such case?
(grunt, why do I ask questions for TOTAL USELESS [ i.e. that I will NEVER get to use in a a pbem ] added features, I uselessly paid a whole game price for?)
I don't exactly know what I mean by that, but I mean it (Holden Caulfield)
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