Map Projections of Alternative Structures: Torus, Hyperboloid, Paraboloid, Onion Shape and Others - Map Projections: Cartographic Information Systems - page 636

uniform convergence we can apply termwise integration. A fill in of these integrals of order one,
two, three, four, five and six into ( x − x 0 ,y − y 0 ) ordered according to the power of Δs := s − s 0

le ads directly to ( 5.17 )and( 5.18 ) .
Univariate series inversion

In many applications the clothoid is uncomfortably parameterized in terms of the arc length
s - s 0 with respect to an initial point ( x ( s 0 ) ,y ( s 0 )). We are going beforehand to derive the param-
eterization y

x 0 .
The technique we apply is the inversion of a univariate homogeneous polynomial of degree n , e.g.
according to Grafarend et al. ( 1996 ), being outlined in Table 23.4 .Assoonaswehaveinverted
Δx ( Δs )towards Δs ( Δx ), we replace Δs ( Δx ) within the power series Δy ( Δs ). The result is
presented in

− y 0 = f ( x − x 0 )or Δy ( Δy ) of the clothoid in terms of the abscissa Δx := x −

Lemma 23.2 ( y ( x ) local representation of the clothoid C ).

The series inversion of Δy ( Δs )oftype( 5.17 )byTable 23.4 leads to Δs ( Δx ), namely

Δs = secα 0 Δx + 2 κ 0 sec 2 α 0 tan α 0 Δx 2 + 6 sec 3 α 0 ( κ 0 +3 tan 2 α 0 + κ 0 tan α 0 ) Δx 3 +
1

24 κ 0 sec 4 α 0 (9 κ 0 tan α 0 +15 κ 0 tan 2 α 0 +3 κ 0 +10 κ 0 tan 2 α 0 ) Δx 4 +
1

120 sec 5 α 0 9 κ 0 +90 κ 0 tan 2 α 0 + 105 κ 0 tan 4 α 0 +59 κ 0 κ 0 tanα 0 + 105 κ 0 κ 0 3 α 0 +3 κ 0

+10 κ 0 tan 2 α 0
Δx 5 +

720 κ 0 sec 6 α 0 225 κ 0 tanα 0 + 1050 κ 0 3 α 0 + 1260 κ 0 κ 0 tan 4 α 0 + 153 κ 0 tanα 0
+280 κ 0 tan 3 α 0
Δx 6 + O ( Δx 7 )

1

(23.144)

which is respected in Δy ( Δs )oftype( 5.18 ).

Δy = y − y 0 = Δx tan α 0 + 2 Δx 2 κ 0 sec 3 α 0 + 6 Δx 3 sec 4 α 0 (3 κ 0 tan α 0 + κ 0 )+
1

24 Δx 4 κ 0 sec 5 α 0 (3 κ 0 (1 + 5 tan 2 α 0 )+10 κ 0 tan α 0 )+

1

120 Δx 5 sec 6 α 0 (15 κ 0 tanα 0 (3 + 7 tan 2 α 0 )+ κ 0 κ 0 (19 + 105 tan 2 α 0 )

+10 κ 0 tan α 0 +
1

720 Δx 6 κ 0 sec 7 α 0 (45 κ 0 (1 + 14 tan 2 α 0 +21 tan 4 α 0 ) + 252 κ 0 κ 0 tanα 0 (2 + 5 tan 2 α 0 )+8 κ

0

(6 + 35 tan 2 α 0 ))
+

1

5040 Δx 7 sec 8 α 0 [45 κ 0 tan α 0 (35 + 210 tan 2 α 0 + 231 tan 4 α 0 )+9 κ 0 κ 0
(81 + 1218 tan 2 α 0 + 1925 tan 4 α 0 )+
26 κ 0 κ 0 tan α 0 (86 + 225 tan 2 α 0 )+8 κ 0 (6 + 35 tan 2 α 0 )]+
O ( Δx 8 )

(23.145)

End of Lemma.