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2+ η 0 (21 226 t 0 + 105 t 0 )+ η 0 (19 34 t 0 +35 t 0 )] ,
1+2 t 0 + η 0
2 N 0
, [30] α = t 0 (1 + 2 t 0 )
6 N 0
t 0
N 0 , [11] α =
[10] α =
,
(20.96)
t 0 (5 + 6 t 0 )
6 N 0
, [13] α = 1+20 t 0 +24 t 0
[12] α =
,
24 N 0
5+28 t 0 +24 t 0
24 N 0
[13] α =
.
Box 20.6 (Coecients of the solution of the boundary value problem with respect to the
Hamilton portrait η 0 := E 2 / (1
E 2 )cos 2 B 0 , V 0 := 1 + η 0 ,t 0 := tan B 0 ).
1
V 0 N 0 ,
(10) = N 0 cos B 0 , (10) =
(20.97)
(20) = 3 η 0 t 0
2 V 0 N 0 , (20) = t 0
2 N 0 cos 2 B 0 ,
(20.98)
t 0
V 0 N 0 cos B 0 ,
(11) =
(20.99)
2+2 η 0 +9 η 0 t 0
6 V 0
N 0 cos B 0 , (21) = 1 3 t 0 + η 0
N 0 cos 2 B 0 ,
(12) =
(20.100)
6 V 0
(30) = η 0 (1
t 0 + η 0 +4 η 0 t 0 )
2 V 0
t 0
6 N 0 cos 3 B 0 ,
N 0 , (30) =
(20.101)
η 0 t 0 (7
3 t 0 +7 η 0 +12 η 0 t 0 )
6 V 0
(13) =
N 0 cos B 0 ,
t 0 (1 t 0 + η 0 )
6 V 0
N 0 cos 3 B 0 ,
(31) =
(20.102)
t 0 + η 0 )
24
(40) = t 0 (1
N 0 cos 4 B 0 ,
(20.103)
t 0 (4 + 3 η 0 +9 η 0 t 0
η 0 )
(22) =
N 0 cos 2 B 0 ,
(20.104)
12 V 0
 
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