geometria do elipsóide

FÓRMULAS

N=
N’=N(1-e2)

M =
Ro =

e’2= n= 1)A grande normal = 0,006470381

N=

N = 6376523 . N N = 6.376.683,24 m 0,999974871

2) A pequena normal

N’=N(1-e2)

N’ = 6.376.683,24x(0,993529619) N' = 6.335.423,67 m

3) As coordenadas retilíneas de P (x, z);

x = 6.351.870,716 m

z = - 558.349,110 m

= 6.376.363,801 m

4)A latitude reduzida (usar 2 processos);

-0,088188868

- 05º02’23,32’’

________

= 6.355.860,255 m

n= = 20.662,745 / 12.732.383,26 = n = 0,001622849

E’’ = ( - ) = n . sen2 - n² . sen4 + n³. sen6 - n4. sen8+ n5. sen10ø sen1’’ 2 sen1’’ 3 sen1’’ 4 sen1’’ 5 sen 1”

0,001622849 . sen( - 10º06’44’’) = -58,77196517 sen1’’

0,000002633638 . sen( - 20º13’28’’) = -0,009389636 2 sen1’’

0,0000000042739982 . sen( - 30º20’12’’) = -0,000148422 3 sen1’’

0,000000000006936053 . sen( - 40º26’56’’) = -0,000000232 4 sen1’’
0,000000000000011256167 . sen (- 50° 33’ 40”) = -0,000000035861 5 sen 1”

E’’ = - 6

E’’ = -60,71310377

E’’ = ( - ) = - 5º03’22’’ – ( - 5º02’21,29’’) = - 0º01’0,71’’

= - E’’ = -5º03’22’’ – ( - 0º01’0,71’’) = - 5 º02’21,29’’

5)A latitude geocêntrica (usar 2 processos)

tg = - 0,087903097

= - 5º01’24,83’’

_______

m = a² - b² = 0,003358431341 a² +b²

E’’ = ( - )= m. sen 2 - m². sen 4 + m³. sen 6 - m4. sen 8 + m5. sen 10 Sen1’’ 2Sen1’’ 3Sen1’’ 4Sen1’’ 5Sen1’’

0,003358431341 . sen(-10º06’44’’) = -121,626602 Sen1’’

0,003358431341² . sen(-20º13’28’’) = -0,402129245