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F ( Φ )=
B sin Φ, G ( Φ )= B cos Φ.
(14.59)
Coordinates of the metric tensor:
G l = ( A + B cos Φ ) 2
B 2 .
0
(14.60)
0
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
( A + B )cos Λ
( A + B )sin Λ
0
Φ =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 .Inaddition,werefertoFig. 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)
= ( A + B ) 2
.
0
0
(14.64)
f 2 ( Φ )
f 2 ( Φ )
0
0
F 2 ( Φ )+ G 2 ( Φ ) = f ( Φ )
f ( Φ )
Λ 1 = F (0)
A + B
A + B cos Φ , Λ 2 =
F ( Φ ) =
.
(14.65)
B
The special case normal cylindric mapping , equidistant on the equator and the set of parallel
circles, the special case normal conformal cylindric mapping , equidistant on the equator, and
the special normal equiareal cylindric mapping , equidistant on the equator are summarized in
Box 14.4 .
 
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