13-2 Special mapping equations 297
cos Φ = ±
4
π
dt
dΦ
cos
2 t ,
(13.20)
π sin Φ = 2t + sin 2t .
(13.21)
Table 13.1. The solution of the special Kepler equation.
Φ
t
Φ
t
Φ
t
Φ
t
0
◦
0
30
◦
0.415 85
60
◦
0.866 98
90
◦
π/2
10
◦
0.137 24
40
◦
0.559 74
70
◦
1.039 00
20
◦
0.275 48
50
◦
0.709 10
80
◦
1.238 77
The solution of the special Kepler equation is shown in Table 13.1. Thus, the final mapping equations
are provided by (13.22). The left principal stretches are best determined by the numerical solution
of the biquadratic characteristic equation based on the left Jacobi matrix (13.23) as well as the left
Cauchy–Green deformation matrix (13.24) (G r = I 2 ).
x =
2
√
2
π
RΛ cos t , y = R
√
2 sin t ,
2t + sin 2t = π sin Φ ,
(13.22)
J l =
=
R
√
2
π
2 cos t −
πΛ tan t cos Φ
2 cos t
0
π
2 cos Φ
4 cos t
,
(13.23)
C l = J
∗
l G r J l =
=
2R
2
π 2
4 cos
2 t
−πΛ tan t cos Φ
−πΛ tan t cos Φ
π
2 cos
2 Φ(π
2 +4Λ
2 tan
2 t)
16 cos 2 t
,
(13.24)
det (C l − Λ
2
S G l ) =
= R
2
8
π 2 cos
2 t − Λ
2
S cos
2 Φ
−
2
π Λ tan t cos Φ
−
2
π Λ tan t cos Φ
cos
2 Φ
2 cos 2 t (Λ
2 tan
2 t +
π
2
4 ) − Λ
2
S
=
= 0 .
(13.25)
cos Φ = ±
4
π
dt
dΦ
cos
2 t ,
(13.20)
π sin Φ = 2t + sin 2t .
(13.21)
Table 13.1. The solution of the special Kepler equation.
Φ
t
Φ
t
Φ
t
Φ
t
0
◦
0
30
◦
0.415 85
60
◦
0.866 98
90
◦
π/2
10
◦
0.137 24
40
◦
0.559 74
70
◦
1.039 00
20
◦
0.275 48
50
◦
0.709 10
80
◦
1.238 77
The solution of the special Kepler equation is shown in Table 13.1. Thus, the final mapping equations
are provided by (13.22). The left principal stretches are best determined by the numerical solution
of the biquadratic characteristic equation based on the left Jacobi matrix (13.23) as well as the left
Cauchy–Green deformation matrix (13.24) (G r = I 2 ).
x =
2
√
2
π
RΛ cos t , y = R
√
2 sin t ,
2t + sin 2t = π sin Φ ,
(13.22)
J l =
=
R
√
2
π
2 cos t −
πΛ tan t cos Φ
2 cos t
0
π
2 cos Φ
4 cos t
,
(13.23)
C l = J
∗
l G r J l =
=
2R
2
π 2
4 cos
2 t
−πΛ tan t cos Φ
−πΛ tan t cos Φ
π
2 cos
2 Φ(π
2 +4Λ
2 tan
2 t)
16 cos 2 t
,
(13.24)
det (C l − Λ
2
S G l ) =
= R
2
8
π 2 cos
2 t − Λ
2
S cos
2 Φ
−
2
π Λ tan t cos Φ
−
2
π Λ tan t cos Φ
cos
2 Φ
2 cos 2 t (Λ
2 tan
2 t +
π
2
4 ) − Λ
2
S
=
= 0 .
(13.25)
