310 14 “Ellipsoid-of-revolution to cylinder”: polar aspect
The mapping equations are provided by the following formulae. As “equator”, let us define the
coordinate line Φ = 0 in the X, Y plane.
⎡
⎢
⎢
⎣
X
Y
Z
⎤
⎥
⎥
⎦
Φ=0
=
⎡
⎢
⎢
⎣
(A + B) cos Λ
(A + B) sin Λ
0
⎤
⎥
⎥
⎦ ,
(14.61)
(X
2 + Y
2 ) Φ=0 = (A + B)
2 ,
(X 2 + Y 2 ) Φ=0 = A + B , F (0) = A + B ,
(14.62)
x = (A + B)Λ , y = f (Φ) .
(14.63)
Most notable, we could have alternatively chosen the “equator” as Φ = π. From this, we conclude the
special case F (π) = A − B. In addition, we refer to Fig. 14.2 illustrating the geometry of the torus,
namely its vertical section. As a case study, we present the special forms of the deformation tensor for
the torus as well as its left principal stretches.
C l =
F
2 (0)
0
0
f
2 (Φ)
=
(A + B)
2
0
0
f
2 (Φ)
.
(14.64)
Λ 1 =
F (0)
F (Φ)
=
A + B
A + B cos Φ
, Λ 2 =
f
(Φ)
F 2 (Φ) + G 2 (Φ)
=
f
(Φ)
B
.
(14.65)
A
Φ
A + B
B
boundary of cylinder
Z
boundary of cylinder
X, Y plane
Fig. 14.2. Vertical section. The example of a torus.
The mapping equations are provided by the following formulae. As “equator”, let us define the
coordinate line Φ = 0 in the X, Y plane.
⎡
⎢
⎢
⎣
X
Y
Z
⎤
⎥
⎥
⎦
Φ=0
=
⎡
⎢
⎢
⎣
(A + B) cos Λ
(A + B) sin Λ
0
⎤
⎥
⎥
⎦ ,
(14.61)
(X
2 + Y
2 ) Φ=0 = (A + B)
2 ,
(X 2 + Y 2 ) Φ=0 = A + B , F (0) = A + B ,
(14.62)
x = (A + B)Λ , y = f (Φ) .
(14.63)
Most notable, we could have alternatively chosen the “equator” as Φ = π. From this, we conclude the
special case F (π) = A − B. In addition, we refer to Fig. 14.2 illustrating the geometry of the torus,
namely its vertical section. As a case study, we present the special forms of the deformation tensor for
the torus as well as its left principal stretches.
C l =
F
2 (0)
0
0
f
2 (Φ)
=
(A + B)
2
0
0
f
2 (Φ)
.
(14.64)
Λ 1 =
F (0)
F (Φ)
=
A + B
A + B cos Φ
, Λ 2 =
f
(Φ)
F 2 (Φ) + G 2 (Φ)
=
f
(Φ)
B
.
(14.65)
A
Φ
A + B
B
boundary of cylinder
Z
boundary of cylinder
X, Y plane
Fig. 14.2. Vertical section. The example of a torus.