Bácsatyai László: Magyarországi vetületek - NymE GEO portál

137
A (4.1.4-10)-bıl
⎛ dΦ
⎞
⎜ ⎟
⎝ dΨ
⎠
1
=
N
1
⋅ cosΦ1
. (4.1.4.-12)
M
1
Képezzük a második deriváltakat:
⎛ 1 ⎞ ⎛ 1 ⎞
2
d⎜
⎟ d⎜
⎟
d Ψ r r dΦ
1 dr
dΦ
=
⎝ ⎠
=
⎝ ⎠
⋅ = − ⋅ ⋅ , (4.1.4.-13)
2
2
dx
dx
dΦ
dx
r dΦ
dx
ahol
dΦ
=
dx
1
M
a
dx
= M ⋅ dΦ
miatt. Képezzük a
dr
d
=
dΦ
( N ⋅ cosΦ
)
dΦ
dr
dΦ
deriváltat:
⎡
d⎢a
⋅
= −N
⋅sinΦ
+ cosΦ
⋅
⎣
⎡
⎢a
⋅ e
= −N
⋅sinΦ
+ cosΦ
⋅
⎢
⎣
2
⎤
⋅sinΦ
⋅ cosΦ
⎥ =
3
⎥
2
⎦
2 2
( 1−
e ⋅ sin Φ )
2 2
( 1−
e ⋅sin
Φ )
dΦ
1
−
2
⎤
⎥
⎦
=
⎡
2 2
⎤
2 2
a ⋅ e ⋅ cos Φ
⎡ e ⋅ cos Φ ⎤
= −N ⋅ sinΦ ⋅ ⎢1
−
⎥ = −N
⋅ sinΦ
⋅ ⎢1
−
⎥ ,
2 2
2 2
2 2
⎢⎣
N ⋅ ( 1−
e ⋅sin
Φ ) ⋅ 1−
e ⋅ sin Φ ⎥⎦
⎣ 1−
e ⋅ sin Φ ⎦
mert
a a
N = =
(1.2.1.2.-7. képlet). Végül
W
2 2
1−
e ⋅ sin Φ
dr
⎡1−
e
= −N
⋅ sinΦ
⋅ ⎢
dΦ
⎣
2
2 2
⋅sin
Φ − e ⋅ cos
2 2
1−
e ⋅sin
Φ
2
Φ ⎤
⎥ = −
⎦
1−
e
2
a
⋅sin
2
2
⎡ 1−
e
⋅ sinΦ
⋅ ⎢ 2
Φ ⎣1−
e ⋅sin
2
⎤
⎥
Φ ⎦
és
2
a ⋅ ( 1−
e )
mert M =
3
2 2
( 1−
e ⋅ sin Φ) 2
2
a ⋅ ( 1−
e )
2 2
( 1−
e ⋅sin
Φ )
dr
= −
⋅sinΦ
= −M
⋅sinΦ
, (4.1.4.-14)
3
dΦ
(1.2.1.2.-6. képlet).
Visszahelyettesítve a (4.1.4.-13)-ba:
2
d Ψ 1
1 sinΦ
= − ⋅ ( − M ⋅sinΦ
) ⋅ = . (4.1.4.-15)
2 2
2
dx
r
M r
2
A (4.1.4.-15) értéke a Φ
1
helyen: