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Email explanation sent to me by Ah - March 5,
2003
....each empire allocates it's attack strength in
proportion to the strengths of the ships it is fighting against. So, in
a three way of Emp A, B, and C the amount of damage A gives to B is:
(B's Strength / (B's Strength + C's Strength)) *
A's strength
and to C is:
(C's Strength / (B's Strength + C's Strength)) *
A's strength
In DT's example:
Emp A has 5 BR3s for 45 BP.
Emp B has 5 BR3s for 45 BP.
Emp C has 2 BR3s for 18 BP.
Emp A will give45/(45+18) * 45 = 32.14 points (or,
71.43% of 45 as DT states) to Emp B and 18/(48+18) * 45 = 12.857 points
(or 28.57%) to Emp C.
Emp B will give45/(45+18) * 45 = 32.14 points (or,
71.43% of 45 as DT
states) to Emp A and 18/(48+18) * 45 = 12.857 points (or 28.57%) to Emp
C.
Emp C will give 14/(45+45) * 18 = 9 points to both
A and B.
A and B both receive 41.14 points and have ships
left, while C receives 25.714 points and is eliminated.
If, C had three BR3's A and B would both receive
41.625 damage, and still both have ships left.
In either case, if A was the defender and had a minefield,
while B was set to nuke, the system would get nuked while the minefield
survived. Cool, eh?
In your example:
A has 64BP (4 BR4s)
B has 128BP (8 BR4s)
C has 1BP (1 BR1)
A gives B (128/(128+1))*64 = 63.5 BP damage
A gives C (1/(129+1))*64 = .5 BP damage
B gives A (64/(64+1))*128 = 126.03 BP damage
B gives C (1/(64+1))*128 = 1.97 BP damage
C gives A (64/(64+128))*1 = .33 BP damage
C gives B .67 BP damage
A and C are wipes out and B survives, defending the system. B receives 64.17
BP damage. 2 of his ships are destroyed, while the rest are at a BR of
(96-32.17)/96 * 4 = 2.659.
Whew! It gets *really* messy when more players are
involved and some are at truce!
In a related case...wanna make sure your ally gets
a front line base in a hotly contested area? Have him send a BR1 colony!
Two allies (A and B) are over a
planet and set to nuke it. A has 6 BR3s, B has 1 BR1 colony, and C has 6BR3s.
Bs BR1 colony is guaranteed to survive. You do the math.
-Ah
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