geodesia

Asignatura: Geodesia

Tema: Calculo de coordenadas geodésicas

Alumno:

Profesor: José Díaz Chumbirizo.

Sección: TA

[ ] Calculo de coordenadas

PROBLEMA

Dada las coordenadas geodésicas de un punto, la distancia geodésica y azimut a un segundo punto, calcular las coordenadas geodésicas del segundo punto.

DATOS:

PUNTO A:

Latitud: ∅ =18° 14’ 00.452 “

Longitud: λ= 70° 16’ 37.07 “

Azimut geodésico: α = 204° 07’ 45.45 “

Distancia geodésica: S = 40,429.31

Semieje mayor (Elipsoide internacional): a = 6378,388.0

Cuadrado de la segunda excentricidad: e” =0.006722670022

RESOLUCION:

HALLANDO LAS CONSTANTES GEODESICAS:

HALLANDO LA NORMAL(N):

N = a¿¿¿

N = 6378,388.0¿¿¿

N = 6378,388.0

[1−(6.581488119×10−4)]12

N = 6378,388.0

0.9996708714

N = 6380,488.001

HALLANDO EL RADIO DE CURVATURA DEL MERIDIANO (R ):


R = a (1−e )} over {{(1- e sinφ2)32 ¿

R = 6378,388.0(1−0.006722670022)¿¿¿

R = 6335,508.202

[1−(6.581488119×10−4)]32

R = 6335,508.2020.9990129392

R = 6341,767.912

HALLANDO A:

A = 1¿¿

A = 1

(6380,488.001)(4.848136811×10−6)

A = 1

30.93347875

A = 0.03232743424

HALLANDO B:

B= 1R sin 1 ´ ´

B¿ 1

6341,767.912×4.848136811×10−6

B¿ 130.74575846

B¿0.03252481155

HALLANDO C:


C¿ tanφ2RN sin 1´ ´

C¿ tan (18 °14 ´ 00.452 ´ ´ )

2×6341,767.912×6380,488.001×4.848136811×10−6

C¿ 0.3294304892392345885.9

C¿8.396430319×10−10

HALLANDO D:

D¿

32e ´ ´ sinφ cos φ sin 1´ ´

(1−e2 sinφ2)

D¿ 32(0.006722670022)sin(18 ° 14 ´ 00.452 ´ ´ )cos ¿¿¿

D¿ 1.45286893×10−8

0.9993418512

D¿1.453825764×10−8

HALLANDO E:

E¿(1+3 tanφ2) (1−e ´ ´ sinφ2)

6 a2

E¿¿¿

E¿ 1.324700917

2.441030009×1014

E¿5.426811273×10−15

HALLANDO F:


F=( 112 ) senΦcos2Φ sen2(0 ° 00'1' ')

F=( 112 )∗sen (18° 14 '00.452' ' )∗cos2 (18 ° 14' 00.452' ' )∗sen2(0 ° 00'1' ')

F=5.528589768 x10−13

HALLANDO K:

K=S2 sen2αC

K=40,429.312∗sen2 (204 ° 07' 45”45 )∗8.396430319 x10−10

K=0,2293522043

HALLANDO h:

h=Scosα B

h=40,429.31m∗cos ¿

h=−1200.061836

CALCULO DE LAS COORDENADAS USANDO LAS FORMULAS

ANALITICAS


Datos: Constantes Geodésicas

N=6380,488.001

R=6341,767.912

A=0.03232743424

B=0.03252481155

C=8.396430319×10−10

D=1.453825764×10−8

E=5.426811273×10−15

F=5.528589768×10−13

K=0.2293522043

h=−1200.061836

−dϕ=−1199.830705

CALCULANDO LA VARIACION DE FI :

−∆ ϕ=SCosαB+S2Sen2αC+(dϕ )2D−hS2Sen2α E−12S2KE+ 3

2S2 cos2α KE+ 1

2S2 cos2α Sen2ϕ A2K Sen21

SCosαB=h=−1200.061836

S2Sen2αC=K=−1200.061836

(dϕ )2D=(1199.830705 )2×1.453825764×10−8=0.02092918441

h S2Sen2α E=−1200.061836×40 429.312×Sen2204 ᵒ 07 ' 45.45 × 5.426811273× {10} ^ {-15} =-1.778920637× {10} ^ {-3

12S2KE=1

2×40 429.312×0.2293522043×5.426811273×10−15=1.017209248×10−6

32S2 cos2α KE=3

2×40 429.312× cos2 204 ᵒ 07 ' 45.45 × 0.2293522043×5.426811273× {10} ^ {-15} =2.541654812× {10} ^ {-6

12S2 cos2α Sen2ϕ A2 K Sen21 = {1} over {2} × {40 429.31} ^ {2} × {Cos} ^ {2} 204ᵒ07'45.45×Sen2 18 ᵒ 14 ' 0.452× {0.03232743424} ^ {2} × 0.2293522043× {left (4.848136811× {10} ^ {-6} right )} ^ {2


12S2 cos2α Sen2ϕ A2 K Sen21 =3.754266594× {10} ^ {-7

REMPLAZANDO:

−∆ ϕ=−1200.061836±1200.061836+0.02092918441− (−1.778920637×10−3 )−1.017209248×10−6+¿

2.541654812×10−6+3.754266594×10−7

−∆ ϕ=−1199.809774→∆ϕ=1199 ᵒ 48' 35.19

Calculando la Latitud:

ϕ '=ϕ+∆ϕ

ϕ '=18ᵒ 14 ' 0.452 + 1199 ᵒ 48 ' 35.19 →ϕ'=1218 ᵒ 2' 35 .64

Calculando la variación de landa:

∆ λ=S Secϕ ' Senα A

∆ λ=40 429.31×1

cos1218 ᵒ 2 ' .64 } ×Sen204ᵒ07'45.45×0.03232743424¿

∆ λ=718.4673761→∆ λ=718ᵒ 28 ' 2.55

Calculando la Longitud:

λ '=λ+∆ λ

λ '=70 ᵒ 16 ' 37.07 + 718ᵒ28'2.55→λ'=788 ᵒ 44 ' 39 .64

Calculando la variación de Alfa:

−∆ α=∆ λ Sen(ϕ+ϕ '2 )Sec ∆ϕ2 + (∆ λ )3 F

Sen(ϕ+ϕ '2 )=Sen( 18 ᵒ 14 ' 0.452 + 1218ᵒ2 '35 .64

2 )=−0.9786609039

Sec∆ϕ2

= 1

cos1199 ᵒ 48' 35.19} over {2}} = {1} over {Cos 599ᵒ {54} ^ {'} .59=−1.994268646¿


(∆ λ )3F= (718.4673761 )3×5.528589768×10−13=2.050385508×10−4

−∆ α=718.4673761×−0.9786609039×−1.994268646+2.050385508×10−4

−∆ α=1402.242147→∆α=−1402ᵒ 14 ' 31.73

Calculando Alfa prima:

α '=α+∆ α−180 ᵒ

α '=204 ᵒ 07 ' 45.45 -1402ᵒ14'31.73−180ᵒ

α '=−1378 ᵒ 6 ' 46.28

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