[A-(B,C,D)]
k(S-3)
, [0-(A,B,C,D)]
k(S-3)
。四四组合为三个元素集合(连乘):∑[∏{A
i
k
·B
i
k
·C
i
k
}]
k(S-3)
=∑
{ [(A-1
B-1
C-1
)-1
(B-1
C-1
M-1
)-1
… (A-1
B-1
N-1
)-1
]-1
/
R0
k(S-3)
}
·
R0
k(S-3)
= C4
{ [(A-1
B-1
C-1
)-1
]-1
/
R0
k(S-3)
}
·
R0
k(S-3)
= C4
(1-η2
)
k(S-3)
R0
k(S-3)
(4.1)