Arcgis

FORMULAS

N=
N’=N(1-e2)

M =
Ro =

e’2= n= a = 6378137,000 m b = 6356752, 314 m α = = = 0,003352810703

= 0,006644380067

1 – Grande Normal

N = 6378302,826 m

2 – A pequena Normal

N’ = N.(1-e²) → N’ = 6335604,042 m

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

X = N . cos = 6353483,999

Y = N’.sen = - 558365,007

4 – A latitude reduzida ( ) : (2 processos)

tg = . tg

tg = -5° 2’ 21,29”

▪Dada a condição

b= a = 6378137,000 .

b = 6356752,314

=

n = 0,001679220406

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

0,001679220406 . sen( - 10º06’44’’) = -60,41921814 Sen1’’

0,000002819781171² . sen( - 20º13’28’’) = -0,099233592 2 Sen1’’

0,000000004735034082³ . sen( - 30º20’12’’) = -0,000161254 3 Sen1’’

0,0000000000079511658514 . sen( - 40º26’56’’) = -0,000000259 4sen1’’

0,000000000000013351759955 . sen (- 50° 33’ 40”) = 0,00000000000001031 5 sen 1”

E’’= -60,81347248 + 0,100532878 – 0,0001644323282 + 0,0000002660028898 – 0,00000000000001031159864

E’’ = -60,71310377

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

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

5 – Latitude geocêntrica () (2 processos)

tg = - 0,087883278 = arctg (- 0,087883278)

= - 5º01’20,77’’

=

m = 0,00335843134

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