Lattice Modules
225
For two cells of bending θ 1 and θ 2 per half cell, the total matrix is
ˆ
M x =
⎛
⎜
⎝
cos μ
βsin μ
2l [1 + (1/2) sin (μ/2)] θ 2
− (1/β) sin μ cos μ 2 [1 + (1/2) sin (μ/2)] [1 − sin (μ/2)] θ 2
0
0
1
⎞
⎟
⎠
·
⎛
⎜
⎝
cos μ
βsin μ
2l [1 + (1/2) sin (μ/2)] θ 1
− (1/β) sin μ cos μ 2 [1 + (1/2) sin (μ/2)] [1 − sin (μ/2)] θ 1
0
0
1
⎞
⎟
⎠
=
⎛
⎜
⎝
cos (2μ)
β sin (2μ) d
− (1/β) sin (2μ) cos(2μ) d
0
0
1
⎞
⎟
⎠ ,
where
d =2l
1 +
1
2
sin
μ
2
θ 1 cos μ + 2
1 +
1
2
sin
μ
2
1 − sin
μ
2
θ 1 β sin μ
+ 2l
1 +
1
2
sin
μ
2
θ 2
=2l
1 +
1
2
sin
μ
2
[(2 cos μ + 1) θ 1 + θ 2 ] ,
and
d
=2l
1 +
1
2
sin
μ
2
θ 1
−
1
β
sin μ
+ 2
1 +
1
2
sin
μ
2
1 − sin
μ
2
θ 1 cos μ
+ 2
1 +
1
2
sin
μ
2
1 − sin
μ
2
θ 2
= − 4 sin
2
μ
2
1 +
1
2
sin
μ
2
1 − sin
μ
2
θ 1
+ 2
1 +
1
2
sin
μ
2
1 − sin
μ
2
(cos μ) θ 1
+ 2
1 +
1
2
sin
μ
2
1 − sin
μ
2
θ 2
=2
1 +
1
2
sin
μ
2
1 − sin
μ
2
[(2 cos μ − 1) θ 1 + θ 2 ] .
Note that the relation
β =
2l [1 + sin(μ/2)]
sin μ
was used during the derivations above which lead to the final forms of d and
d
.