topografia 2

UNIVERSIDAD NACIONAL DE SAN ANTONIO ABAD DEL CUSCO

TRABAJO DE TOPOGRAFÍA II 1. Ajuste de un polígono Método Topográfico

2. Compensación de Triángulos

Ingeniería Civil

Topografía II

MGT. ESCOBAR MASIAS, JUAN PABLO

1

TRABAJO DE TOPOGRAFÍA II

. Condiciones Geométricas

I1 + D1 + V1 = 200 g I2 + D2 + V2 = 200 g I3 +D3 +V3 = 200 g

200.0090 g ≠ 200 g 200.0019 g ≠ 200 g 199.9933 g ≠ 200 g

Error = 90

3 = 30 Error=

19

3 = 6.3 Error=

67

3 = 22.3

I4 + D4 + V4 = 200 g I5 + D5 +V5 = 200

200.0055 g ≠ 200 g 199.9943 g ≠ 200 g

Error= 55

3 = 18.3 Error=

57

3 = 19

Ángulos corregidos por triángulos

I1=90.5200 g - 30 D1=41.0010 g - 30 V1=68.4880 g - 30

I2=104.7269 g - 6 D2=53.3590 g - 6 V2=41.9160 g - 7 I3=63.5020 g + 22 D3=84.3718 g + 22 V3=52.1195 g + 23 I4=86.2500 g - 18 D4=60.5590 g - 18 V4=53.1965 g - 19 I5=54.9900 g + 19 D5=59.6453 g + 19 V5=85.3590 + 19

Los ángulos de I en grados centesimales (g)

Los ángulos de D en grados centesimales (g)

Los ángulos de V en grados centesimales (g)

I1=90.5200 g D1=41.0010 g V1=68.4880 g

I2=104.7269 g D2=53.3590 g V2=41.9160 g

I3=63.5020 g D3=84.3718 g V3=52.1195 g

I4=86.2500 g D4=60.5590 g V4=53.1965 g

I5=54.9900 g D5=59.6453 g V5=85.3590 g

2


I1=90.5170 g D1=40.9980 g V1=68.4850 g

I2=104.7263 g D2=53.3584 g V2=41.9153 g I3=63.5042 g D3=84.3740 g V3=52.1218 g I4=86.2482 g D4=60.5572 g V4=53.1946 g I5=54.9919 g D5=59.64 72 g V5=85.3609 g

I1 + I2 + I3 + I4 + I5 = 400 g

399.9876 g ≠ 400 g

Error total = 0.0124

Error= 124

3 = 24.8 25

Ángulos corregidos geométricamente

I1=90.5170 g + 25 D1=40.9980 g - 12 V1=68.4850 g - 13

I2=104.7263 g + 25 D2=53.3584 g - 12 V2=41.9153 g - 13 I3=63.5042 g + 25 D3=84.3740 g - 12 V3=52.1218 g - 13 I4=86.2482 g + 25 D4=60.5572 g - 12 V4=53.1946 g - 13 I5=54.9919 g + 24 D5=59.64 72 g - 12 V5=85.3609 g - 12

I1=90.5195 g D1=40.9968 g V1=68.4837 g

I2=104.7288 g D2=53.3572 g V2=41.9140 g I3=63.5067 g D3=84.3728 g V3=52.1205 g I4=86.2507 g D4=60.5560 g V4=53.1933 g I5=54.9943 g D5=59.64 60 g V5=85.3597 g

. Condición Trigonométrica

3


𝐴𝐹

𝑠𝑖𝑛𝑉1=

𝐵𝐹

𝑠𝑖𝑛𝐷1 ;

𝐵𝐹

𝑠𝑖𝑛𝑉2=

𝐶𝐹

𝑠𝑖𝑛𝐷2 ;

𝐶𝐹

𝑠𝑖𝑛𝑉3=

𝐷𝐹

𝑠𝑖𝑛𝐷3 ;

𝐷𝐹

𝑠𝑖𝑛𝑉4=

𝐸𝐹

𝑠𝑖𝑛𝐷4 ;

𝐸𝐹

𝑠𝑖𝑛𝑉5=

𝐴𝐹

𝑠𝑖𝑛𝐷5

Multiplicando miembro a miembro y luego simplificando

1 =𝑠𝑖𝑛𝑉1𝑠𝑖𝑛𝑉2𝑠𝑖𝑛𝑉3𝑠𝑖𝑛𝑉4𝑠𝑖𝑛𝑉5

𝑠𝑖𝑛𝐷1𝑠𝑖𝑛𝐷2𝑠𝑖𝑛𝐷3𝑠𝑖𝑛𝐷4𝑠𝑖𝑛𝐷5= ∆ Ecuación de la condición trigonométrica

∆ = 0.283919446

0.2839940229= 0.9997373995

∑ cot D = 3.930618999 ∑ cot N = 3.906704474

e = ∆−1

∑cot D + ∆ × ∑ cot N × 636620 =− 21

Ángulos finales

I1=90.5195 g D1=40.9968 g - 21 V1=68.4837 g + 21

I2=104.7288 g D2=53.3572 g - 21 V2=41.9140 g + 21

I3=63.5067 g D3=84.3728 g - 21 V3=52.1205 g + 21

I4=86.2507 g D4=60.5560 g - 21 V4=53.1933 g + 21

I5=54.9943 g D5=59.64 60 g - 21 V5=85.3597 g + 21

I1=90.5195 g D1=40.9947 g V1=68.4858 g

I2=104.7288 g D2=53.3551 g V2=41.9161 g

I3=63.5067 g D3=84.3707 g V3=52.1226 g

I4=86.2507 g D4=60.5539 g V4=53.1954 g

I5=54.9943 g D5=59.64 39 g V5=85.3618 g

Ya teniendo los ángulos corregidos geométrica y trigonométricamente calcularemos un lado por dos caminos y los

valores tienen que ser iguales

4


𝐴𝐹

𝑠𝑖𝑛𝑉1=

𝐵𝐹

𝑠𝑖𝑛𝐷1 ;

𝐵𝐹

𝑠𝑖𝑛𝑉2=

𝐶𝐹

𝑠𝑖𝑛𝐷2

1800

𝑠𝑖𝑛68.4858=

𝐵𝐹

𝑠𝑖𝑛40.9947 ⇒ 𝐵𝐹 =

1800𝑠𝑖𝑛40.9947

𝑠𝑖𝑛68.4858 ⇒ BF = 1228.055605

1228.055605

𝑠𝑖𝑛41.9161=

𝐶𝐹

𝑠𝑖𝑛53.3551 ⇒ CF =

1228.055605𝑠𝑖𝑛53.3551

𝑠𝑖𝑛41.9161 ⇒ CF = 1492.002005

𝐴𝐹

𝑠𝑖𝑛𝐷5=

𝐸𝐹

𝑠𝑖𝑛𝑉5 ;

𝐸𝐹

𝑠𝑖𝑛𝐷4=

𝐷𝐹

𝑠𝑖𝑛𝑉4 ;

𝐷𝐹

𝑠𝑖𝑛𝐷3=

𝐶𝐹

𝑠𝑖𝑛𝑉3

1800

𝑠𝑖𝑛59.64 39 =

𝐸𝐹

𝑠𝑖𝑛85.3618 ⇒ EF =

1800𝑠𝑖𝑛85.3618

𝑠𝑖𝑛59.64 39 = EF = 2175.238572

2175.238572

𝑠𝑖𝑛60.5539=

𝐷𝐹

𝑠𝑖𝑛53.1954 ⇒ DF =

2175.238572𝑠𝑖𝑛53.1954

𝑠𝑖𝑛60.5539 = ⇒ DF = 1981.770138

1981.770138

𝑠𝑖𝑛84.3707=

𝐶𝐹

𝑠𝑖𝑛52.1226 ⇒ C F =

1981.770138𝑠𝑖𝑛52.1226

𝑠𝑖𝑛84.3707 ⇒ CF = 1491.995877

⇒CF = 1492.002005 ≈ 1491.995877

Calculando lados del polígono

𝐴𝐹

𝑠𝑖𝑛𝑉1=

𝐴𝐵

𝑠𝑖𝑛𝐼1 ⇒

1800

𝑠𝑖𝑛68.4858=

𝐴𝐵

𝑠𝑖𝑛90.5195 ⇒ AB =

1800𝑠𝑖𝑛90.5195

𝑠𝑖𝑛68.4858 ⇒ AB = 2022.913461

𝐵𝐹

𝑠𝑖𝑛𝑉2=

𝐵𝐶


1228.055605

𝑠𝑖𝑛41.9161=

𝐵𝐶

𝑠𝑖𝑛104.7288 ⇒BC =

1228.055605𝑠𝑖𝑛104.7288

𝑠𝑖𝑛41.9161 ⇒ BC = 2001.534511

𝐶𝐹

𝑠𝑖𝑛𝑉3=

𝐶𝐷


1492.002005

𝑠𝑖𝑛52.1226 =

𝐶𝐷

𝑠𝑖𝑛63.5067⇒ CD =

1492.002005𝑠𝑖𝑛63.5067

𝑠𝑖𝑛52.1226 ⇒ CD= 1716.460631

𝐶𝐷

𝑠𝑖𝑛𝐼3=

𝐷𝐹

𝑠𝑖𝑛𝐷3⇒

1716.460631

𝑠𝑖𝑛63.5067=

𝐷𝐹

𝑠𝑖𝑛84.3707 ⇒ DF =

1716.460631𝑠𝑖𝑛84.3707

𝑠𝑖𝑛63.5067 ⇒ DF = 1981.778277

𝐷𝐹

𝑠𝑖𝑛𝑉4=

𝐷𝐸


1981.778277

𝑠𝑖𝑛53.1954=

𝐷𝐸

𝑠𝑖𝑛86.2507 ⇒ DE =

1981.778277𝑠𝑖𝑛86.2507

𝑠𝑖𝑛53.1954 ⇒ DE =2609.890167

𝐷𝐹

𝑠𝑖𝑛𝑉4=

𝐸𝐹

𝑠𝑖𝑛𝐷4 ⇒

1981.778277

𝑠𝑖𝑛53.1954=

𝐸𝐹

𝑠𝑖𝑛60.5539 ⇒ EF =

1981.778277𝑠𝑖𝑛60.5539

𝑠𝑖𝑛53.1954 ⇒ EF =2175.247506

𝐸𝐹

𝑠𝑖𝑛𝑉5=

𝐴𝐸


2175.247506

𝑠𝑖𝑛85.3618=

𝐴𝐸

𝑠𝑖𝑛54.9943 ⇒ AE =

2175.247506𝑠𝑖𝑛54.9943

𝑠𝑖𝑛85.3618 ⇒ AE =1698.651646

𝐸𝐹

𝑠𝑖𝑛𝑉5=

𝐴𝐹

𝑠𝑖𝑛𝐷5 ⇒

2175.247506

𝑠𝑖𝑛85.3618=

𝐴𝐹

𝑠𝑖𝑛59.64 39 ⇒ AF =

2175.247506𝑠𝑖𝑛59.64 39

𝑠𝑖𝑛85.3618 ⇒ AF =1800.007393

⇒AF=1800 ≈1800.007393

Lados del polígono

AB=2022.913461

BC=2001.534511

CD=1716.460631

5


DE=2609.890167

EA=1698.651646

Calculando coordenadas

VÉRTICES DISTANCIA ÁNGULOS AZIMUT A 126.3565 g B 2022.913461 30

B 121.8409 g C 2001.534511 108.1591

C 126.2868 g D 1716.460631 181.8723

D 112.6765 g E 2609.890167 269.1958

E 112.8393 g A 1698.651646 356.3565

Vértices D x cos(Azimut) D x sen(Azimut) Coordenadas +Y -Y +X -X Y X

A 52000 18000 B 1802.429092 918.3834931 53802.4291 18918.3835

B C 255.820667 1985.118683 53546.6084 20903.5022

C D 1647.34243 482.1826334 51899.266 21385.6848

D E 1214.14565 2310.2764 50685.1204 19075.4084

E A 1314.876378 1075.415 51999.9968 17999.9934

6


α1 = 45°25ˡ14ˡˡ α2 = 44°10ˡ59ˡˡ α3 = 50°14ˡ35ˡˡ

α4 = 40°09ˡ12ˡˡ α5 = 40°55ˡ59ˡˡ α6 = 48°40ˡ22ˡˡ

α7 = 43°18ˡ47ˡˡ α8 = 47°05ˡ08ˡˡ

I1 = 57°53ˡ15ˡˡ D1 = 71°41ˡ03ˡˡ V1 =50°26ˡ08ˡˡ

I2 = 40°30ˡ34ˡˡ D2 = 76°13ˡ15ˡˡ V2 =63°16ˡ07ˡˡ

I3 = 46°26ˡ15ˡˡ D3 = 60°04ˡ36ˡˡ V3 =73°29ˡ14ˡˡ

I4 = 78°20ˡ34ˡˡ D4 = 34°50ˡ00ˡˡ V4 =66°49ˡ31ˡˡ

I5 = 80°56ˡ42ˡˡ D5 = 57°06ˡ04ˡˡ V5 =41°57ˡ17ˡˡ

I6 = 55°53ˡ15ˡˡ D6 = 65°10ˡ23ˡˡ V6 =58°56ˡ55ˡˡ

β1 = 83°23ˡ18ˡˡ β2 = 70°31ˡ20ˡˡ β3 = 26°05ˡ31ˡˡ

β4 = 29°57ˡ43ˡˡ β5 = 68°49ˡ37ˡˡ β6= 81°12ˡ35ˡˡ

. Corrección del cuadrilátero

α1 = 45°25ˡ14ˡˡ α2 = 44°10ˡ59ˡˡ α3 = 50°14ˡ35ˡˡ

α4 = 40°09ˡ12ˡˡ α5 = 40°55ˡ59ˡˡ α6 = 48°40ˡ22ˡˡ

α7 = 43°18ˡ47ˡˡ α8 = 47°05ˡ08ˡˡ

Condición Geométrica

α1 + α2 = α5 + α6 α3 + α4 = α7 + α8

89°36ˡ13ˡˡ ≠ 89°36ˡ21ˡˡ 90°23ˡ47ˡˡ ≠ 90°23ˡ55ˡˡ

Error = 8ˡˡ Error = 8ˡˡ

8

4 = 2ˡˡ

8

4= 2ˡˡ

Ángulos corregidos

α1 = 45°25ˡ14ˡˡ + 2ˡˡ α2 = 44°10ˡ59ˡˡ + 2ˡˡ α3 = 50°14ˡ35ˡˡ + 2ˡˡ

α4 = 40°09ˡ12ˡˡ + 2ˡˡ α5 = 40°55ˡ59ˡˡ - 2ˡˡ α6 = 48°40ˡ22ˡˡ - 2ˡˡ

α7 = 43°18ˡ47ˡˡ - 2ˡˡ α8 = 47°05ˡ08ˡˡ - 2ˡˡ

α1 = 45°25ˡ16ˡˡ α2 = 44°11ˡ01ˡˡ α3 = 50°14ˡ37ˡˡ

α4 = 40°09ˡ14ˡˡ α5 = 40°55ˡ57ˡˡ α6 = 48°40ˡ20ˡˡ

α7 = 43°18ˡ45ˡˡ α8 = 47°05ˡ06ˡˡ

α1 + α2 + α3 + α4 + α5 + α6 + α7 + α8 = 360°

7


36°00016 ≠ 360°

Error = 16ˡˡ

16

8= 2ˡˡ

α1 = 45°25ˡ16ˡˡ -2ˡˡ α2 = 44°11ˡ01ˡ -2ˡˡ α3 = 50°14ˡ37ˡˡ-2ˡˡ

α4 = 40°09ˡ14ˡˡ -2ˡˡ α5 = 40°55ˡ57ˡˡ -2ˡˡ α6 = 48°40ˡ20ˡˡ-2ˡˡ

α7 = 43°18ˡ45ˡˡ-2ˡˡ α8 = 47°05ˡ06ˡˡ -2ˡˡ

Ángulos corregidos

α1 = 45°25ˡ14ˡˡ α2 = 44°10ˡ59ˡˡ α3 = 50°14ˡ35ˡˡ

α4 = 40°09ˡ12ˡˡ α5 = 40°55ˡ55ˡˡ α6 = 48°40ˡ18ˡˡ

α7 = 43°18ˡ43ˡˡ α8 = 47°05ˡ04ˡˡ

Condiciones Trigonométrica

∆ = 𝑠𝑖𝑛𝛼1𝑠𝑖𝑛𝛼3𝑠𝑖𝑛𝛼5𝑠𝑖𝑛𝛼7

𝑠𝑖𝑛𝛼2𝑠𝑖𝑛𝛼4𝑠𝑖𝑛𝛼6𝑠𝑒𝑛8 =

0.246091536

0.2471608792

∆ = 0.9956734028

∑cotD = 4.023389736 ∑cotN = 4.031185769

eˡˡ = ∆−1

∑cot D + ∆ × ∑ cot N ×206265

eˡˡ = 0.9956734028−1

4.023389736 + 0.9956734028 × 4.031185769 ×206265 = -111.037

α1 = 45°25ˡ14ˡˡ -1ˡ51ˡˡ α2 = 44°10ˡ59ˡˡ - 1ˡ51ˡˡ α3 = 50°14ˡ35ˡˡ +1ˡ51ˡˡ

α4 = 40°09ˡ12ˡˡ -1ˡ51ˡˡ α5 = 40°55ˡ55ˡˡ +1ˡ51ˡˡ α6 = 48°40ˡ18ˡˡ - 1ˡ51ˡˡ

α7 = 43°18ˡ43ˡ + 1ˡ51ˡˡ α8 = 47°05ˡ04ˡˡ -1ˡ51ˡˡ

α1 = 45°27ˡ05ˡˡ α2 = 44°09ˡ08ˡˡ α3 = 50°16ˡ26ˡˡ

α4 = 40°07ˡ21ˡˡ α5 = 40°57ˡ46ˡˡ α6 = 48°38ˡ27ˡˡ

α7 = 43°20ˡ34ˡˡ α8 = 47°03ˡ13ˡˡ

Calculo de distancias

8


𝐴𝐵

𝑠𝑖𝑛𝛼5=

𝐵𝐶

𝑠𝑖𝑛𝛼8 ⇒ AB=1329.890m

𝐵𝐶

𝑠𝑖𝑛𝛼3=

𝐶𝐷

𝑠𝑖𝑛𝛼6 ⇒ CD= 1307.615m

𝐴𝐵

𝑠𝑖𝑛𝛼2=

𝐴𝐷

𝑠𝑖𝑛𝛼7 ⇒ AD= 1182.419m

𝐴𝐷

𝑠𝑖𝑛𝛼4=

𝐶𝐷

𝑠𝑖𝑛𝛼1⇒ CD=1307.613m

. Corrección del polígono

I1 = 57°53ˡ15ˡˡ D1 = 71°41ˡ03ˡˡ V1 =50°26ˡ08ˡˡ

I2 = 40°30ˡ34ˡˡ D2 = 76°13ˡ15ˡˡ V2 =63°16ˡ07ˡˡ

I3 = 46°26ˡ15ˡˡ D3 = 60°04ˡ36ˡˡ V3 =73°29ˡ14ˡˡ

I4 = 78°20ˡ34ˡˡ D4 = 34°50ˡ00ˡˡ V4 =66°49ˡ31ˡˡ

I5 = 80°56ˡ42ˡˡ D5 = 57°06ˡ04ˡˡ V5 =41°57ˡ17ˡˡ

I6 = 55°53ˡ15ˡˡ D6 = 65°10ˡ23ˡˡ V6 =58°56ˡ55ˡˡ

Condiciones geométricas

I1 + D1 + V1 = 180° I2 + D2 +V2 = 180° I3 + D3 +V3 = 180°

180°00ˡ26ˡˡ ≠ 180° 179°59ˡ56ˡˡ ≠ 180° 180°00ˡ05ˡˡ ≠ 180°

Error=26

3= 8.6 Error=

4

3= 1.3 Error=

5

3= 1.6

I4 + D4 + V4 = 180° I5 + D5 +V5 = 180° I6 + D6 +V6 = 180°

180°00ˡ05ˡˡ ≠ 180° 180°00ˡ03ˡˡ ≠ 180° 180°00ˡ33ˡˡ ≠ 180°

Error=5

3= 1.6 Error=

3

3= 1 Error=

33

3= 11

Ángulos corregidos por triángulos

I1 = 57°53ˡ15ˡˡ - 8ˡˡ D1 = 71°41ˡ03ˡˡ - 9ˡˡ V1 =50°26ˡ08ˡˡ - 9ˡˡ

I2 = 40°30ˡ34ˡˡ + 2ˡˡ D2 = 76°13ˡ15ˡˡ + 1ˡˡ V2 =63°16ˡ07ˡˡ + 1ˡˡ

I3 = 46°26ˡ15ˡˡ - 1ˡˡ D3 = 60°04ˡ36ˡˡ - 2ˡˡ V3 =73°29ˡ14ˡˡ - 2ˡˡ

I4 = 78°20ˡ34ˡˡ - 1ˡˡ D4 = 34°50ˡ00ˡˡ - 2ˡˡ V4 =66°49ˡ31ˡˡ -2ˡˡ

I5 = 80°56ˡ42ˡˡ - 1ˡˡ D5 = 57°06ˡ04ˡˡ -1ˡˡ V5 =41°57ˡ17ˡˡ - 1ˡˡ

I6 = 55°53ˡ15ˡˡ -11ˡˡ D6 = 65°10ˡ23ˡˡ -11ˡˡ V6 =58°56ˡ55ˡˡ -11ˡˡ

I1 = 57°53ˡ07ˡˡ D1 = 71°40ˡ54ˡˡ V1 =50°25ˡ59ˡˡ

I2 = 40°30ˡ36ˡˡ D2 = 76°13ˡ16ˡˡ V2 =63°16ˡ08ˡˡ

I3 = 46°26ˡ14ˡˡ D3 = 60°04ˡ34ˡˡ V3 =73°29ˡ12ˡˡ

I4 = 78°20ˡ33ˡˡ D4 = 34°49ˡ58ˡˡ V4 =66°49ˡ29ˡˡ

9


I5 = 80°56ˡ41ˡˡ D5 = 57°06ˡ03ˡˡ V5 =41°57ˡ16ˡˡ

I6 = 55°53ˡ04ˡˡ D6 = 65°10ˡ12ˡˡ V6 =58°56ˡ44ˡˡ

I1 + I2 + I3 + I4 + I5 + I6 = 360°

360°00ˡ00ˡˡ ≠ 360°

Error = 15ˡˡ

15

6= 2.5

Ángulos corregidos por vértice

I1 = 57°53ˡ07ˡˡ -2ˡˡ D1 = 71°40ˡ54ˡˡ +1ˡˡ V1 =50°25ˡ59ˡˡ +1ˡˡ



I4 = 78°20ˡ33ˡˡ - 3ˡˡ D4 = 34°49ˡ58ˡˡ +1ˡˡ V4 =66°49ˡ29ˡˡ +2ˡˡ



I1 = 57°53ˡ05ˡˡ D1 = 71°40ˡ55ˡˡ V1 =50°26ˡ00ˡˡ

I2 = 40°30ˡ33ˡˡ D2 = 76°13ˡ18ˡˡ V2 =63°16ˡ09ˡˡ

I3 = 46°26ˡ12ˡˡ D3 = 60°04ˡ35ˡˡ V3 =73°29ˡ13ˡˡ

I4 = 78°20ˡ30ˡˡ D4 = 34°49ˡ59ˡˡ V4 =66°49ˡ31ˡˡ

I5 = 80°56ˡ39ˡˡ D5 = 57°06ˡ04ˡˡ V5 =41°57ˡ17ˡˡ

I6 = 55°53ˡ01ˡˡ D6 = 65°10ˡ14ˡˡ V6 =58°56ˡ45ˡˡ

Condiciones trigonométricas

∆=𝑠𝑖𝑛𝑉1𝑠𝑖𝑛𝑉2𝑠𝑖𝑛𝑉3𝑠𝑖𝑛𝑉4𝑠𝑖𝑛𝑉5𝑠𝑖𝑛𝑉6

𝑠𝑖𝑛𝐷1𝑠𝑖𝑛𝐷2𝑠𝑖𝑛𝐷3𝑠𝑖𝑛𝐷4𝑠𝑖𝑛𝐷5𝑠𝑖𝑛𝐷6=

0.3475526911

0.3478103017

∆= 0.9992593361

∑cotD = 3.698496404 ∑cotN = 3.768981433

eˡˡ = ∆−1

∑cot D + ∆ × ∑ cot N ×206265

eˡˡ = 0.9992593361−1

3.698496404 + 0.9992593361 × 3.768981433 ×206265 = -20.466 ≈ 20

Ángulos finales corregidos

I1 = 57°53ˡ05ˡˡ D1 = 71°40ˡ55ˡˡ -20ˡˡ V1 =50°26ˡ00ˡˡ +20ˡˡ


10


I3 = 46°26ˡ12ˡˡ D3 = 60°04ˡ35ˡˡ-20ˡˡ V3 =73°29ˡ13ˡˡ +20ˡˡ


I5 = 80°56ˡ39ˡˡ D5 = 57°06ˡ04ˡˡ-20ˡˡ V5 =41°57ˡ17ˡˡ +20ˡˡ

I6 = 55°53ˡ01ˡˡ D6 = 65°10ˡ14ˡˡ-20ˡˡ V6 =58°56ˡ45ˡˡ+20ˡˡ

I1 = 57°53ˡ05ˡˡ D1 = 71°40ˡ35ˡˡ V1 =50°26ˡ20ˡˡ

I2 = 40°30ˡ33ˡˡ D2 = 76°12ˡ58ˡˡ V2 =63°16ˡ29ˡˡ

I3 = 46°26ˡ12ˡˡ D3 = 60°04ˡ15ˡˡ V3 =73°29ˡ33ˡˡ

I4 = 78°20ˡ30ˡˡ D4 = 34°49ˡ39ˡˡ V4 =66°49ˡ51ˡˡ

I5 = 80°56ˡ39ˡˡ D5 = 57°05ˡ44ˡˡ V5 =41°57ˡ37ˡˡ

I6 = 55°53ˡ01ˡˡ D6 = 65°09ˡ54ˡˡ V6 =58°57ˡ05ˡˡ

Calculo de distancias CD = 1307.614

𝐶𝐷

𝑠𝑖𝑛𝐼1=

𝐶𝐸

𝑠𝑖𝑛𝑉1 ⇒ CE=1190.228m

𝐼𝐸

𝑠𝑖𝑛𝐷3=

𝐸𝐻

𝑠𝑖𝑛𝑉3 ⇒ EH= 1210.978m

𝐸𝐺

𝑠𝑖𝑛𝐷5=

𝐸𝐹

𝑠𝑖𝑛𝑉5 ⇒ EF= 1552.434m

𝐷𝐸

𝑠𝑖𝑛𝐷1=

𝐶𝐸

𝑠𝑖𝑛𝑉1⇒ 𝐶𝐸 = 1190.138m

𝐸𝐼

𝑠𝑖𝑛𝑉2=

𝐶𝐼

𝑠𝑖𝑛𝐼2 ⇒ CI = 796.061m

𝐸𝐺

𝑠𝑖𝑛𝑉4=

𝐻𝐺

𝑠𝑖𝑛𝐼4 ⇒ HG= 2076.658m

𝐸𝐷

𝑠𝑖𝑛𝑉6=

𝐷𝐹

𝑠𝑖𝑛𝐼6 ⇒ DF= 1416.232m

Corrección de los triángulos

β1 = 83°23ˡ18ˡˡ β2 = 70°31ˡ20ˡˡ β3 = 26°05ˡ31ˡˡ

β4 = 29°57ˡ43ˡˡ β5 = 68°49ˡ37ˡˡ β6= 81°12ˡ35ˡˡ

Condición geométrica

Β1 + β2 + β3 = 180° Β1 + β2 + β3 = 180°

180°00ˡ09ˡˡ ≠ 180° 179°59ˡ55ˡˡ ≠ 180°

Error = 9ˡˡ Error = 5ˡˡ

9

3= 3

5

3= 1.76 ≈2

β1 = 83°23ˡ18ˡˡ - 3ˡˡ β2 = 70°31ˡ20ˡˡ - 3ˡˡ β3 = 26°05ˡ31ˡˡ - 3ˡˡ

11


β4 = 29°57ˡ43ˡˡ + 2ˡˡ β5 = 68°49ˡ37ˡˡ + 2ˡˡ β6= 81°12ˡ35ˡˡ + 1ˡˡ

β1 = 83°23ˡ15ˡˡ β2 = 70°31ˡ17ˡˡ β3 = 26°05ˡ28ˡˡ

β4 = 29°57ˡ45ˡˡ β5 = 68°49ˡ39ˡˡ β6= 81°12ˡ36ˡˡ

Calculo de lados HG = 2076.658

𝐻𝐺

𝑠𝑖𝑛𝛽3=

𝐻𝐽

𝑠𝑖𝑛𝛽1 ⇒ HJ=4690.414m

𝐻𝐽

𝑠𝑖𝑛𝛽1=

𝐺𝐽

𝑠𝑖𝑛𝛽2 ⇒ GJ= 4451.576m

𝐺𝐽

𝑠𝑖𝑛𝛽6=

𝐺𝐾

𝑠𝑖𝑛𝛽5 ⇒ GK= 4200.416m

𝐺𝐾

𝑠𝑖𝑛𝛽5=

𝐽𝐾

𝑠𝑖𝑛𝛽4⇒ 𝐽𝐾 = 2249.649m

-------------------------------------------------------------------------------------------------------------------------------------------------------

α1 = 45°27ˡ05ˡˡ α2 = 44°09ˡ08ˡˡ α3 = 50°16ˡ26ˡˡ

α4 = 40°07ˡ21ˡˡ α5 = 40°57ˡ46ˡˡ α6 = 48°38ˡ27ˡˡ

α7 = 43°20ˡ34ˡˡ α8 = 47°03ˡ13ˡˡ

I1 = 57°53ˡ05ˡˡ D1 = 71°40ˡ35ˡˡ V1 =50°26ˡ20ˡˡ

I2 = 40°30ˡ33ˡˡ D2 = 76°12ˡ58ˡˡ V2 =63°16ˡ29ˡˡ

I3 = 46°26ˡ12ˡˡ D3 = 60°04ˡ15ˡˡ V3 =73°29ˡ33ˡˡ

I4 = 78°20ˡ30ˡˡ D4 = 34°49ˡ39ˡˡ V4 =66°49ˡ51ˡˡ

I5 = 80°56ˡ39ˡˡ D5 = 57°05ˡ44ˡˡ V5 =41°57ˡ37ˡˡ

I6 = 55°53ˡ01ˡˡ D6 = 65°09ˡ54ˡˡ V6 =58°57ˡ05ˡˡ

12


β1 = 83°23ˡ15ˡˡ β2 = 70°31ˡ17ˡˡ β3 = 26°05ˡ28ˡˡ

β4 = 29°57ˡ45ˡˡ β5 = 68°49ˡ39ˡˡ β6= 81°12ˡ36ˡˡ

LADOS

AB=1329.890m

CD= 1307.615m

AD= 1182.419m

CD=1307.613m

CE=1190.228m

EH= 1210.978m

EF= 1552.434m

𝐶𝐸 = 1190.138m

CI = 796.061m

HG= 2076.658m

DF= 1416.232m

HJ=4690.414m

GJ= 4451.576m

GK= 4200.416m

𝐽𝐾 = 2249.649m

. Cálculos de azimut.- Estos cálculos se hicieron por la regla de la nemónica

𝐴𝑍𝐴𝐵 = 25°00ˡ00ˡˡ 𝐴𝑍𝐵𝐶 = 135°00ˡ59ˡˡ 𝐴𝑍𝐶𝐷 = 221°55ˡ52ˡˡ

𝐴𝑍𝐴𝐷 = 297°30ˡ18ˡˡ

𝐴𝑍𝐶𝐼 = 76°58ˡ48ˡˡ 𝐴𝑍𝐼𝐻 = 107°16ˡ17ˡˡ 𝐴𝑍𝐻𝐺 = 160°22ˡ11ˡˡ

𝐴𝑍𝐺𝐹 = 263°34ˡ55ˡˡ 𝐴𝑍𝐹𝐷 = 327°32ˡ26ˡˡ 𝐴𝑍𝐶𝐸 = 320°15ˡ17ˡˡ

𝐴𝑍𝐻𝐽 = 89°50ˡ54ˡˡ 𝐴𝑍𝐽𝐾 = 182°23ˡ54ˡˡ

13


.Cálculo de coordenadas

𝑋𝐴 = 170000m 𝑌𝐴= 8500700m

𝑌𝐵 = 𝑌𝐴 + ABcos(𝐴𝑍𝐴𝐵) 𝑌𝐵 = 8501787.569 m

𝑋𝐵 = 𝑋𝐴 + ABsen(𝐴𝑍𝐴𝐵) 𝑋𝐵 = 170507.142m

𝑌𝐶 = 𝑋𝐵 + BCsen(𝐴𝑍𝐵𝑐) 𝑌𝑐 = 8501263.679m

𝑌𝐶 = 𝑌𝐵 + BCcos(𝐴𝑍𝐵𝑐) 𝑋𝑐 = 171740.367m

𝑌𝐷 = 𝑌𝐶 + CDcos(𝐴𝑍𝐶𝐷) 𝑌𝐷 = 8500153.928m

𝑋𝐷 = 𝑋𝐶 + CDsen(𝐴𝑍𝐶𝐷) 𝑋𝐷 =171048.771m

𝑌𝐸 = 𝑌𝐶 + CEcos(𝐴𝑍𝐶𝐸) 𝑌𝐸 = 8502178.839m

𝑋𝐸 = 𝑋𝐶 + CEsen(𝐴𝑍𝐶𝐸) 𝑋𝐸 = 170979.510m

𝑌𝐼 = 𝑌𝐶 + CIcos(𝐴𝑍𝐶𝐼) 𝑌𝐼 = 8501443.025m

𝑋𝐼 = 𝑋𝐶 + CIsen(𝐴𝑍𝐶𝐼) 𝑋𝐼 =172515.962m

𝑌𝐻 = 𝑌𝐼 + IHsen(𝐴𝑍𝐼𝐻) 𝑌𝐻 = 8501171.300m

𝑋𝐻 = 𝑋𝐼 + IHcos(𝐴𝑍𝐼𝐻) 𝑋𝐻 =173389.908m

𝑌𝐺 = 𝑌𝐻 + HGsen(𝐴𝑍𝐻𝐺) 𝑌𝐺 = 8499215.337m

𝑋𝐺 = 𝑋𝐻 + HGcos(𝐴𝑍𝐻𝐺) 𝑋𝐺 =174087.560m

𝑌𝐹 = 𝑌𝐺 + FGcos(𝐴𝑍𝐹𝐺) 𝑌𝐹 = 8498959.029 m

𝑋𝐹 = 𝑋𝐺 + FGsen(𝐴𝑍𝐹𝐺) 𝑋𝐹 =171809.007m

14


𝑌𝐽 = 𝑌𝐻 + HJsen(𝐴𝑍𝐻𝐽) 𝑌𝐽 = 8501183.716m

𝑋𝐽 = 𝑋𝐻 + HJcos(𝐴𝑍𝐻𝐽) 𝑋𝐽 = 178080.306m

𝑌𝐾 = 𝑌𝐽 + HJcos(𝐴𝑍𝐽𝐾) 𝑌𝑘 =8498936.037m

𝑋𝐾 = 𝑋𝐽 + HJsen(𝐴𝑍𝐽𝐾) 𝑋𝑘 = 177986.165m

PUNTO NORTE ESTE A 8500700.000m 170000.000m B 8501787.569 m 170507.142m C 8501263.679m 171740.367m D 8500153.928m 171048.771m E 8502178.839m 170979.510m F 8498959.029 m 171809.007m G 8499215.337m 174087.560m H 8501171.300m 173389.908m I 8501443.025m 172515.962m J 8501183.716m 178080.306m K 8501183.716m 177986.165m A 8500700.000m 170000.000m

α1 = 63° 11ˡ 14ˡˡ + 1” α4 = 69° 52ˡ 46ˡˡ + 3” α7 = 64° 57ˡ 41ˡˡ + 2”

α2 = 60° 47ˡ 43ˡˡ + 1” α5 = 61° 54ˡ 25ˡˡ + 3” α8 = 63° 43ˡ 25ˡˡ + 1”

α3 = 56° 00ˡ 01ˡˡ+1” α6 = 48° 12ˡ 41ˡˡ + 2” α9 = 51° 18ˡ 50ˡˡ + 1”

∑= 179°59ˡ57ˡˡ ∑= 179°59ˡ52ˡˡ ∑= 179°59ˡ56ˡˡ

ERROR: 3 ERROR: 8 ERROR: 4

15


CONDICIÓN TRIGONOMÉTRICA:

CAMINO I

𝐵𝐶

𝑠𝑖𝑛𝛼2=

𝐴𝐶

𝑠𝑖𝑛𝛼1 𝐴𝐸 =

𝐵𝐶 𝑠𝑖𝑛𝛼1 𝑠𝑖𝑛𝛼4

𝑠𝑖𝑛𝛼2 𝑠𝑖𝑛𝛼6 ………………………… I

𝐴𝐶

𝑠𝑖𝑛𝛼6=

𝐴𝐸

𝑠𝑖𝑛𝛼4

CAMINO II

𝐷𝐸

𝑠𝑖𝑛𝛼9=

𝐶𝐸

𝑠𝑖𝑛𝛼8 𝐴𝐸 =

𝐷𝐸 𝑠𝑖𝑛𝛼4 𝑠𝑖𝑛𝛼8

𝑠𝑖𝑛𝛼9 𝑠𝑖𝑛𝛼5 ………………………… II

𝐶𝐸

𝑠𝑖𝑛𝛼5=

𝐴𝐸

𝑠𝑖𝑛𝛼4

De I y II se tiene:

1 = 𝐵𝐶 𝑠𝑖𝑛𝛼1𝑠𝑖𝑛𝛼1𝑠𝑖𝑛𝛼9

𝐷𝐸 𝑠𝑖𝑛𝛼2𝑠𝑖𝑛𝛼6𝑠𝑒𝑛8 = ∆

eˡˡ = Log∆

∑ i FORMULA NUEVA

1 = 10,6415x10−7 1 = 11,7994x10−7 1 = 11,2387x10−7

1 = 18,8176x10−7 1 = 10,3952x10−7 1 = 16,8598x10−7

1 = 79,722x10−7

eˡˡ = 7,1323x10−6

7,9722x10−6 = 0.89ˡˡ

Esta corrección restamos a 𝛼1, 𝛼5 y 𝛼9 y sumamos a 𝛼2.

Ángulos corregidos

α1 = 63° 11ˡ 14ˡˡ α4 = 69° 52ˡ 49ˡˡ α7 = 64° 57ˡ 43ˡˡ

α2 = 60° 47ˡ 45ˡˡ α5 = 61° 54ˡ 27ˡˡ α8 = 63° 43ˡ 27ˡˡ

α3 = 56° 01ˡ 01ˡˡ α6 = 48° 12ˡ 44ˡˡ α9 = 51° 18ˡ 50ˡˡ

Calculo del lado AE:

𝐴𝐸 = 388.918 𝑠𝑖𝑛𝛼1 𝑠𝑖𝑛𝛼4

𝑠𝑖𝑛𝛼2 𝑠𝑖𝑛𝛼6 = 500.771 m

𝐴𝐸 = 409.573 𝑠𝑖𝑛𝛼4 𝑠𝑖𝑛𝛼8

𝑠𝑖𝑛𝛼9 𝑠𝑖𝑛𝛼5 = 500.772 m

topografia 2

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