TRANSFORMAÇÃO DE COORDENADAS PLANAS UTM EM COORDENADAS GEODÉSICAS

MEMORIAL DE CÁLCULO: TRANSFORMAÇÃO DE COORDENADAS PLANAS UTM EM COORDENADAS GEODÉSICAS

DENIEZIO GOMES
Graduado em Engenharia Cartográfica e de Agrimensura, UFPI, 2016.

Trabalho acadêmico apresentado ao curso de Engenharia Cartográfica e de Agrimensura da Universidade Federal do Piauí como requisito avaliativo da disciplina de Geodésia II, sob orientação do Msc. José Lincoln de Sousa Meneses.

DADOS:

N = 7646340,188 m
E = 514513,253 m
Fuso = 19
MC = 69° W

SISTEMA GEODÉSICO DE REFERÊNCIA

SAD-69

a = 6378160,000 m
b = 6356774,719 m
α = 1/298,25

e² = 0,006694542
e’² = 0,006739661


MEMÓRIA DE CÁLCULO

- Referência Inicial.

N₀ = 10000000,000 m
Cte = 500000,00 m

- Coeficientes para o SAD-69.

α = 111133,3486
β = 16038,95511
γ = 16,83348972
δ = 0,021986053
ε = 3,114475*10^-5
ζ = 4,153111*10^-8

- Determinação de q, q², q³, q⁴, q⁵, q⁶.

q = 0,000001*E’
E’ = |Cte – E|
E’ = |500000,00 - 514513,253|
E’ = 14513,253 m q = 0,000001*14513,253
q = 0,014513253
q² = 0,000210635
q³ = 0,000003057
q⁴ = 0,000000044
q⁵ = 0,000000001
q⁶ = 9,345200*10^-12

- Cálculo da Latitude

φ = φ’ – VIII*q² + VIII*q⁴ – D’₆*q⁶

- Determinação de φ’

N’ = 10000000 – N
N’ = 10000000 – 7646340,188
N’ = 2353659,812 m

- 1ª Aproximação:

φ’₁ = (N’/0,9996)*(1/α)
φ’₁ = (2353659,812/0,9996)*(1/111133,3486)
φ’₁ = 21,18717449°

- 2ª Aproximação:

φ’₂ = (1/α)*(B’+βsen2φ’₁ - γsen4φ’₁ +δsen6φ’₁ - εsen8φ’₁ + ζsen10φ’₁)

B’ = N’/K₀
B’ = 2353659,812/0,9996
B’ = 2354601,653

+βsen2φ’₁ = +10809,80212
- γsen4φ’₁ = -16,76283701
+δsen6φ’₁ = +0,017530387
- εsen8φ’₁ = -5,677073*10^-6
+ ζsen10φ’₁ = + (-2,192924*10^-8)
∑ = 2365394,709
φ’₂ = (1/111133,3486)* 2365394,709
φ’₂ = 21,28429257°

- 3ª Aproximação:

φ’₃ = (1/α)*(B’+βsen2φ’₂ - γsen4φ’₂ + δsen6φ’₂ - εsen8φ’₂ + ζsen10φ’₂)

B’ = N’/K₀
B’ = 2353659,812/0,9996
B’ = 2354601,653

+βsen2φ’₂ = +10849,90839
- γsen4φ’₂ = -16,77289758
+δsen6φ’₂ = +0,017394532
- εsen8φ’₂ = -5,261309*10^-6
+ ζsen10φ’₂ = + (-2,252389*10^-8)
∑ = 2365434,806
φ’₃ = (1/111133,3486)* 2365434,806
φ’₃ = 21,28465336°

- 4ª Aproximação:

φ’₄ = (1/α)*(B’+βsen2φ’₃ - γsen4φ’₃ + δsen6φ’₃ - εsen8φ’₃ + ζsen10φ’₃)

B’ = N’/K₀
B’ = 2353659,812/0,9996
B’ = 2354601,653

+βsen2φ’₃ = +10850,05715
- γsen4φ’₃ = -16,77293352
+δsen6φ’₃ = +0,017394024
- εsen8φ’₃ = -5,269763*10^-6
+ ζsen10φ’₃ = + (-2,252609*10^-8)
∑ = 2365434,954
φ’₄ = (1/111133,3486)* 2365434,954
φ’₄ = 21,28465470°

- 5ª Aproximação:

φ’₅ = (1/α)*(B’+βsen2φ’₄ - γsen4φ’₄ + δsen6φ’₄ - εsen8φ’₄ + ζsen10φ’₄)

B’ = N’/K₀
B’ = 2353659,812/0,9996
B’ = 2354601,653

+βsen2φ’₄ = +10850,05770
- γsen4φ’₄ = -16,77293365
+δsen6φ’₄ = +0,017394022
- εsen8φ’₄ = -5,259757*10^-6
+ ζsen10φ’₄ = + (-2,252609*10^-8)
∑ = 2365434,955
φ’₅ = (1/111133,3486)* 2365434,955
φ’₅ = 21,28465470°

Como φ’₅ = φ’₄ então:

φ’ = 21,28465470°
φ’ = 21°17’4,757” S

- Cálculo dos Coeficientes:

VII*q²

VII = (tgφ’/(2N²sen1”))*(1+e’²cos²φ’)*(1/K₀²)*10¹²

tgφ’ = 0,389575202

N = 6378160,000/(1-0,006694542*0,3630016862)^1/2
N = 6380975,082 m

sen1” = 0,000004848
cosφ’ = 0,931788482

VII = (0,389575202/(2*6380975,0822*0,000004848)*(1/0,99962)*10¹²
VII = 993,3304526
VII*q² = 993,3304526*0,000210635
VII*q² = 0,209229676”

VIII*q⁴

VIII = A * (5 + B + C – D – E – F)*(1/K₀⁴)*10²⁴

A = tgφ’/(24*N⁴*sen1”) = 2,019561*10^-24
+B = + 3tg²φ’ = +0,045306514
+C = + 6e’²cos²φ’ = +0,035109445
-D = - 6e’²sen²φ’ = -0,005328520
-E = - 3e’⁴cos⁴φ’ = -0,000102723
-F = - 9e’⁴cos²φ’sen²φ’ = - 0,000046770
∑ = 5,484937947

1/K₀⁴ = 1,001601601
VIII = 2,019561*10^-24*5,484937947*1,001601601*10²⁴
VIII = 11,094908635
VIII*q⁴ = 0,000000492”

D’₆*q⁶

D’₆ = A * (61 + B + C + D – E – F) * (1/K₀⁶) * 10³⁶

A = tgφ’/(720*N⁶*sen1”) = 1,653338*10^-39
+B = + 90tg²φ’ = +13,659195429
+C = + 45tg⁴φ’ = +1,036520110
+D = - 107e’²cos²φ’ = +0,626118443
-E = - 16e’²sen²φ’ = -0,143870033
-F = - 45e’²tg²φ’sen²φ’ = - 0,006065274
∑ = 76,171898675

1/K₀⁶ = 1,002403364

D’₆ = 1,653338*10^-39*76,171898675*1,002403364*10³⁶
D’₆ = 0,126240570
D’₆*q⁶ = 0,000000000”

φ = φ’ – VIII*q² + VIII*q⁴ – D’₆*q⁶
φ = 21°17’4,757” – 0,209229676 + 0,000000492 – 0,000000000
φ = 21°17’4,548” S

- Cálculo da Longitude

Δλ = IX*q – X*q³ + E’₅*q⁵

- Determinação dos Coeficientes:

IX*q

IX = (secφ’/(Nsen1”)*(1/K₀)*10¹⁶

secφ’/(Nsen1”) = 0,034691314

1/K₀ = 1,000400160

IX = 0,034691314*1,000400160*10¹⁶
IX = 34705,19582
IX*q = 34705,19582*0,014513253
IX*q = 503,6853”

X*q³

X = A*(1+B+C)*(1/K₀³)*10¹⁸

A = (secφ’/(6N³sen1”) = 1420023*10^-16
+B = + 2tg²φ’ = + 0,303537676
+C = e’²cos²φ’ = + 0,005851574
∑ = 1,309389250

(1/K₀³) = 1,001200961

X = 1420023*10^-16*1,309389250*1,001200961*10¹⁸
X = 186,159596
X*q³ = 186,159596*0,000003057
X*q³ = 0,00056909”

E’₅*q⁵

E₅ = A*(5+B+C+D+E)*(1/K₀⁵)*10³⁰

A = (secφ’/(120N⁵sen1”) = 1,743778*10^-31
+B = + 28tg²φ’ = 4,249527467
+C = + 24tg⁴φ’ = 0,552810725
+D = + 6e’²cos²φ’ = 0,035109445
+E = + 8e’²sen²φ’ = 0,007104693
∑ = 9,844552331

(1/K₀⁵) = 1,002002402
E’₅ = 1,743778*10^-31*9,844552331*1,002002402*10³⁰
E’₅ = 1,720109287

E’₅*q⁵ = 1,720109287*0,000000001
E’₅*q⁵ = 0,000000001”

Δλ = IX*q – X*q³ + E’₅*q⁵
Δλ = 503,6853 – 0,00056909 + 0,000000001
Δλ = 00°08’23,685”

Como o ponto está a leste do MC temos:

λ = MC - Δλ
λ = 69° - 00°08’23,685”
λ = 68°51’36,315” W